Fox And McDonald Introduction To Fluid Mechanics 8th Edition Solution Manual Robert W. Fox, Alan T. Mc Donald, Philip J. Pritchard Manu
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Problem 1.1 Given: [Difficulty: 3] Common Substances Tar Sand “Silly Putty” Jello Modeling clay Toothpaste Wax Shaving cream Some of these substances exhibit characteristics of solids and fluids under different conditions. Find: Explain and give examples. Solution: Tar, Wax, and Jello behave as solids at room temperature or below at ordinary pressures. At high pressures or over long periods, they exhibit fluid characteristics. At higher temperatures, all three liquefy and become viscous fluids. Modeling clay and silly putty show fluid behavior when sheared slowly. However, they fracture under suddenly applied stress, which is a characteristic of solids. Toothpaste behaves as a solid when at rest in the tube. When the tube is squeezed hard, toothpaste “flows” out the spout, showing fluid behavior. Shaving cream behaves similarly. Sand acts solid when in repose (a sand “pile”). However, it “flows” from a spout or down a steep incline. Problem 1.2 Given: Five basic conservation laws stated in Section 1-4. Write: A word statement of each, as they apply to a system. Solution: Assume that laws are to be written for a system. [Difficulty: 2] a. Conservation of mass — The mass of a system is constant by definition. b. Newton's second law of motion — The net force acting on a system is directly proportional to the product of the system mass times its acceleration. c. First law of thermodynamics — The change in stored energy of a system equals the net energy added to the system as heat and work. d. Second law of thermodynamics — The entropy of any isolated system cannot decrease during any process between equilibrium states. e. Principle of angular momentum — The net torque acting on a system is equal to the rate of change of angular momentum of the system. Problem 1.3 [Difficulty: 3] Open-Ended Problem Statement: The barrel of a bicycle tire pump becomes quite warm during use. Explain the mechanisms responsible for the temperature increase. Discussion: Two phenomena are responsible for the temperature increase: (1) friction between the pump piston and barrel and (2) temperature rise of the air as it is compressed in the pump barrel. Friction between the pump piston and barrel converts mechanical energy (force on the piston moving through a distance) into thermal energy as a result of friction. Lubricating the piston helps to provide a good seal with the pump barrel and reduces friction (and therefore force) between the piston and barrel. Temperature of the trapped air rises as it is compressed. The compression is not adiabatic because it occurs during a finite time interval. Heat is transferred from the warm compressed air in the pump barrel to the cooler surroundings. This raises the temperature of the barrel, making its outside surface warm (or even hot!) to the touch. Problem 1.4 [Difficulty: 3] Open-Ended Problem Statement: Consider the physics of “skipping” a stone across the water surface of a lake. Compare these mechanisms with a stone as it bounces after being thrown along a roadway. Discussion: Observation and experience suggest two behaviors when a stone is thrown along a water surface: 1. If the angle between the path of the stone and the water surface is steep the stone may penetrate the water surface. Some momentum of the stone will be converted to momentum of the water in the resulting splash. After penetrating the water surface, the high drag* of the water will slow the stone quickly. Then, because the stone is heavier than water it will sink. 2. If the angle between the path of the stone and the water surface is shallow the stone may not penetrate the water surface. The splash will be smaller than if the stone penetrated the water surface. This will transfer less momentum to the water, causing less reduction in speed of the stone. The only drag force on the stone will be from friction on the water surface. The drag will be momentary, causing the stone to lose only a portion of its kinetic energy. Instead of sinking, the stone may skip off the surface and become airborne again. When the stone is thrown with speed and angle just right, it may skip several times across the water surface. With each skip the stone loses some forward speed. After several skips the stone loses enough forward speed to penetrate the surface and sink into the water. Observation suggests that the shape of the stone significantly affects skipping. Essentially spherical stones may be made to skip with considerable effort and skill from the thrower. Flatter, more disc-shaped stones are more likely to skip, provided they are thrown with the flat surface(s) essentially parallel to the water surface; spin may be used to stabilize the stone in flight. By contrast, no stone can ever penetrate the pavement of a roadway. Each collision between stone and roadway will be inelastic; friction between the road surface and stone will affect the motion of the stone only slightly. Regardless of the initial angle between the path of the stone and the surface of the roadway, the stone may bounce several times, then finally it will roll to a stop. The shape of the stone is unlikely to affect trajectory of bouncing from a roadway significantly. Problem 1.5 Given: Dimensions of a room Find: Mass of air [Difficulty: 1] Solution: p Basic equation: ρ= Given or available data p = 14.7psi Rair⋅ T T = ( 59 + 460)R V = 10⋅ ft × 10⋅ ft × 8⋅ ft Then ρ = p Rair⋅ T M = ρ⋅ V ρ = 0.076⋅ Rair = 53.33 ⋅ V = 800⋅ ft lbm ft 3 M = 61.2⋅ lbm ft⋅ lbf lbm⋅ R 3 ρ = 0.00238⋅ slug ft M = 1.90⋅ slug 3 ρ = 1.23 kg 3 m M = 27.8kg Problem 1.6 [Difficulty: 1] Given: Data on oxygen tank. Find: Mass of oxygen. Solution: Compute tank volume, and then use oxygen density (Table A.6) to find the mass. The given or available data is: D = 16⋅ ft p = 1000⋅ psi RO2 = 48.29 ⋅ ft⋅ lbf lbm⋅ R Tc = 279 ⋅ R so the reduced temperature and pressure are: TR = Z = 0.948 T Tc = 1.925 Hence π⋅ D V = 6 M = V⋅ ρ = p⋅ V RO2⋅ T M = 11910 ⋅ lbm (data from NIST WebBook) pR = p = ρ⋅ RO2⋅ T 3 V= p c = 725.2 ⋅ psi p pc = 1.379 Since this number is close to 1, we can assume ideal gas behavior. Therefore, the governing equation is the ideal gas equation where V is the tank volume T = 537 ⋅ R (Table A.6) For oxygen the critical temperature and pressure are: Using a compressiblity factor chart: T = ( 77 + 460 ) ⋅ R π 6 × ( 16⋅ ft) M = 1000⋅ ρ= and lbf 2 in 3 V = 2144.7⋅ ft 3 × 2144.7⋅ ft × 1 M V 3 ⋅ lbm⋅ R 48.29 ft⋅ lbf × 1 ⋅ 1 537 R × ⎛ 12⋅ in ⎞ ⎜ ft ⎝ ⎠ 2 Problem 1.7 Given: [Difficulty: 3] Small particle accelerating from rest in a fluid. Net weight is W, resisting force FD = kV, where V is speed. Find: Time required to reach 95 percent of terminal speed, Vt. Solution: Consider the particle to be a system. Apply Newton's second law. Basic equation: ∑Fy = may FD = kV Assumptions: 1. W is net weight 2. Resisting force acts opposite to V Particle y Then W ∑F y = W − kV = ma y = m dV W dV = g dt dt or dV k = g(1 − V) W dt Separating variables, dV = g dt k 1− W V V t dV W k ( ) V = − ln 1 − = W ∫0 gdt = gt k 1 − Wk V Integrating, noting that velocity is zero initially, ∫ or kgt kgt − − ⎞ W ⎛⎜ k W 1 − V = e ; V = ⎜1 − e W ⎟⎟ k ⎝ W ⎠ 0 kgt But V→Vt as t→∞, so Vt = When V Vt = 0.95 , then e − kgt W W k . Therefore = 0.05 and kgt W − V = 1− e W Vt = 3. Thus t = 3W/gk Problem 1.8 Given: [Difficulty: 2] Small particle accelerating from rest in a fluid. Net weight is W, resisting force is FD = kV, where V is speed. Find: FD = kV Distance required to reach 95 percent of terminal speed, Vt. Solution: Particle Consider the particle to be a system. Apply Newton's second law. Basic equation: ∑Fy = may W Assumptions: 1. W is net weight. 2. Resisting force acts opposite to V. Then, ∑F y = W − kV = ma y = m dV dt = At terminal speed, ay = 0 and V = Vt = Separating variables y W k W g V dV dy or . Then 1 − V Vg 1 − Wk V = V dV g dy = g1 V dV dy V dV = g dy 1 − V1t V Integrating, noting that velocity is zero initially gy = ∫ 0.95Vt 0 ⎡ ⎛ V V dV = ⎢ −VVt − Vt 2 ln ⎜1 − 1 ⎝ Vt 1 − V ⎢⎣ Vt gy = −0.95Vt 2 − Vt 2 ln (1 − 0.95) − Vt 2 ln (1) gy = −Vt 2 [ 0.95 + ln 0.05] = 2.05 Vt 2 ∴y = 2.05 2 W2 Vt = 2.05 2 g gt 0.95Vt ⎞⎤ ⎟⎥ ⎠ ⎥⎦ 0 Problem 1.9 [Difficulty: 2] Given: Mass of nitrogen, and design constraints on tank dimensions. Find: External dimensions. Solution: Use given geometric data and nitrogen mass, with data from Table A.6. The given or available data is: M = 5 ⋅ kg T = ( 20 + 273 ) ⋅ K p = ( 200 + 1 ) ⋅ atm p = 20.4⋅ MPa T = 293 ⋅ K RN2 = 296.8 ⋅ p = ρ⋅ RN2⋅ T The governing equation is the ideal gas equation ρ= and J kg⋅ K (Table A.6) M V 2 where V is the tank volume V= π⋅ D 4 ⋅L L = 2⋅ D where Combining these equations: Hence M = V⋅ ρ = p⋅ V RN2⋅ T = p RN2⋅ T 2 ⋅ π⋅ D 4 ⋅L = 3 2 p RN2⋅ T ⋅ π⋅ D 4 ⋅ 2⋅ D = p ⋅ π⋅ D 2 ⋅ RN2⋅ T 1 1 Solving for D ⎛ 2 ⋅ RN2⋅ T⋅ M ⎞ D= ⎜ p⋅ π ⎝ ⎠ D = 0.239 ⋅ m 3 2 ⎛2 N⋅ m 1 m ⎞ D = ⎜ × 296.8 ⋅ × 293 ⋅ K × 5 ⋅ kg × ⋅ kg⋅ K 6 N ⎜π 20.4 × 10 ⎝ ⎠ L = 2⋅ D These are internal dimensions; the external ones are 2 x 0.5 cm larger: L = 0.477 ⋅ m L = 0.249 ⋅ m D = 0.487 ⋅ m 3 Problem 1.10 [Difficulty: 4] NOTE: Drag formula is in error: It should be: FD = 3 ⋅ π⋅ V⋅ d Given: Data on sphere and formula for drag. Find: Diameter of gasoline droplets that take 1 second to fall 10 in. Solution: Use given data and data in Appendices; integrate equation of motion by separating variables. FD = 3πVd a = dV/dt Mg The data provided, or available in the Appendices, are: − 7 lbf ⋅ s μ = 4.48 × 10 ⋅ ft 2 ρw = 1.94⋅ slug ft 3 SG gas = 0.72 ρgas = SGgas⋅ ρw dV g− Integrating twice and using limits slug ft M⋅ Newton's 2nd law for the sphere (mass M) is (ignoring buoyancy effects) so ρgas = 1.40⋅ 3 ⋅ π⋅ μ⋅ d V( t) = M dV dt 3 = M ⋅ g − 3 ⋅ π⋅ μ⋅ V⋅ d = dt ⋅V − 3⋅ π⋅ μ ⋅ d ⎞ ⎛ ⋅t ⎜ M⋅ g M ⋅⎝1 − e ⎠ 3⋅ π⋅ μ ⋅ d Replacing M with an expression involving diameter d π⋅ d M = ρgas ⋅ 6 ⎡ ⎛ − 3⋅ π⋅ μ ⋅ d ⋅ t ⎢ ⎜ M⋅ g M M x( t) = ⋅⎝e ⋅ ⎢t + − 3⋅ π⋅ μ ⋅ d ⎣ 3⋅ π⋅ μ ⋅ d ⎞⎤ ⎥ 1⎠⎥ ⎦ ⎡⎢ ⎛ − 18⋅ μ ⋅ t 2 ⎜ ρgas ⋅ d ⋅ g ⎢ ρgas ⋅ d ⎜ ρgas⋅ d2 x( t) = ⋅⎝e ⋅ ⎢t + − 18⋅ μ 18⋅ μ ⎣ ⎞⎤⎥ ⎥ 1⎠⎥ ⎦ 2 3 This equation must be solved for d so that x ( 1 ⋅ s) = 10⋅ in. The answer can be obtained from manual iteration, or by using Excel's Goal Seek. −3 ⋅ in 1 10 0.75 7.5 x (in) x (in) d = 4.30 × 10 0.5 0.25 5 2.5 0 0.025 0.05 t (s) 0.075 0.1 0 0.25 0.5 0.75 1 t (s) Note That the particle quickly reaches terminal speed, so that a simpler approximate solution would be to solve Mg = 3πµVd for d, with V = 0.25 m/s (allowing for the fact that M is a function of d)! Problem 1.11 [Difficulty: 3] Given: Data on sphere and formula for drag. Find: Maximum speed, time to reach 95% of this speed, and plot speed as a function of time. Solution: Use given data and data in Appendices, and integrate equation of motion by separating variables. The data provided, or available in the Appendices, are: kg ρair = 1.17⋅ 3 m μ = 1.8 × 10 − 5 N⋅ s ⋅ 2 m Then the density of the sphere is kg ρw = 999 ⋅ 3 m SG Sty = 0.016 ρSty = SGSty⋅ ρw ρSty = 16 d = 0.3⋅ mm kg 3 m 3 π⋅ d kg ( 0.0003⋅ m) M = ρSty⋅ = 16⋅ × π× 6 3 6 m The sphere mass is 3 M = 2.26 × 10 − 10 kg M ⋅ g = 3 ⋅ π⋅ V⋅ d Newton's 2nd law for the steady state motion becomes (ignoring buoyancy effects) so M⋅ g − 10 1 Vmax = = × 2.26 × 10 3 ⋅ π⋅ μ⋅ d 3⋅π ⋅ kg × 9.81⋅ m 2 s 2 × m 1.8 × 10 −5 Newton's 2nd law for the general motion is (ignoring buoyancy effects) dV so g− 3 ⋅ π⋅ μ⋅ d M Integrating and using limits × ⋅ N⋅ s 1 0.0003⋅ m m Vmax = 0.0435 s M⋅ dV dt = M ⋅ g − 3 ⋅ π⋅ μ⋅ V⋅ d = dt FD = 3πVd ⋅V V( t) = − 3⋅ π⋅ μ⋅ d ⎞ ⎛ ⋅t ⎜ M⋅ g M ⋅⎝1 − e ⎠ 3 ⋅ π⋅ μ⋅ d a = dV/dt Mg Using the given data 0.05 V (m/s) 0.04 0.03 0.02 0.01 0 0.01 0.02 t (s) The time to reach 95% of maximum speed is obtained from so t=− M 3 ⋅ π⋅ μ⋅ d ⎛ ⋅ ln⎜ 1 − 0.95⋅ Vmax⋅ 3 ⋅ π⋅ μ⋅ d ⎞ ⎝ The plot can also be done in Excel. M⋅ g ⎠ − 3⋅ π⋅ μ⋅ d ⎞ ⎛ ⋅t ⎜ M⋅ g M ⋅⎝1 − e ⎠ = 0.95⋅ Vmax 3 ⋅ π⋅ μ⋅ d Substituting values t = 0.0133 s Problem 1.12 [Difficulty: 3] Given: Data on sphere and terminal speed. Find: Drag constant k, and time to reach 99% of terminal speed. Solution: Use given data; integrate equation of motion by separating variables. kVt − 13 M = 1 × 10 The data provided are: mg ft Vt = 0.2⋅ s ⋅ slug M⋅ Newton's 2nd law for the general motion is (ignoring buoyancy effects) dV dt M ⋅ g = k ⋅ Vt Newton's 2nd law for the steady state motion becomes (ignoring buoyancy effects) − 13 k = 1 × 10 ⋅ slug × 32.2⋅ ft 2 2 s × 0.2⋅ ft s × lbf ⋅ s = M⋅ g − k⋅ V ⋅ M k − 13 ⋅ slug × M⋅ g − 11 × ⋅ lbf ⋅ s Vt lbf ⋅ s slug⋅ ft g− = dt k M ⋅V ⋅ V⎞ ⎠ V = 0.198 ⋅ 2 ft 1.61 × 10 t = 0.0286 s ⎝ k V = 0.99⋅ Vt We must evaluate this when t = −1 × 10 ⋅ ln⎜⎛ 1 − M⋅ g ft dV t=− k = so − 11 lbf ⋅ s k = 1.61 × 10 slug⋅ ft To find the time to reach 99% of Vt, we need V(t). From 1, separating variables Integrating and using limits (1) ⎛ ⋅ ln⎜ 1 − 1.61 × 10 ⎜ ⎝ ft s − 11 lbf ⋅ s ⋅ ft × 2 1 − 13 1 × 10 × ⋅ slug s 32.2⋅ ft × 0.198 ⋅ ft s × slug⋅ ft ⎞ 2 lbf ⋅ s ⎠ Problem 1.13 [Difficulty: 5] Given: Data on sphere and terminal speed from Problem 1.12. Find: Distance traveled to reach 99% of terminal speed; plot of distance versus time. Solution: Use given data; integrate equation of motion by separating variables. − 13 M = 1 × 10 The data provided are: ft Vt = 0.2⋅ s ⋅ slug mg M⋅ Newton's 2nd law for the general motion is (ignoring buoyancy effects) dV dt = M⋅ g − k⋅ V Newton's 2nd law for the steady state motion becomes (ignoring buoyancy effects) k = 1 × 10 − 13 ⋅ slug × 32.2⋅ ft 2 × s 2 s 0.2⋅ ft × lbf ⋅ s V⋅ dV g− k M ( − 13 y = 1 ⋅ 10 ) 2 ⋅ slug ⋅ 2 M⋅ g Vt ft dV dt = M⋅ dy dV dV ⋅ = M ⋅ V⋅ = M⋅ g − k⋅ V dt dy dy M ⋅g 2 ⋅ ln⎜⎛ 1 − ⎝ k M⋅ g ⋅ V⎞ − ⎠ M k ⋅V V = 0.198 ⋅ ft s 2 2⎛ 2 ⎛ ft 1 s 0.198 ⋅ ft slug⋅ ft ⎞ ⎞ ⋅ ⎜ lbf ⋅ s ⎞ ⋅ ln⎜ 1 − 1.61⋅ 10− 11⋅ lbf ⋅ s ⋅ ⋅ ⋅ ⋅ ⋅⎛ ... ⎜ − 11 − 13 32.2⋅ ft s 2 ⎜ slug⋅ ft ⎠ ft ⎝ 1 ⋅ 10 ⋅ slug lbf ⋅ s ⎠ ⎝ 1.61⋅ 10 ⋅ lbf ⋅ s ⎠ ⎝ ft ⋅ slug × − 11 1.61⋅ 10 −3 ⋅ M⋅ V = 0.99⋅ Vt − 13 y = 4.49 × 10 k = 2 32.2⋅ ft s + 1 ⋅ 10 so = dy k We must evaluate this when M ⋅ g = k ⋅ Vt ⋅V 2 y=− Integrating and using limits (1) − 11 lbf ⋅ s k = 1.61 × 10 slug⋅ ft To find the distance to reach 99% of Vt, we need V(y). From 1: Separating variables kVt × ⋅ lbf ⋅ s 2 0.198 ⋅ ft lbf ⋅ s × s slug⋅ ft ⋅ ft Alternatively we could use the approach of Problem 1.12 and first find the time to reach terminal speed, and use this time in y(t) to find the above value of y: dV From 1, separating variables Integrating and using limits g− k M t=− M k = dt ⋅V ⋅ ln⎜⎛ 1 − ⎝ k M⋅ g ⋅ V⎞ ⎠ (2) V = 0.99⋅ Vt We must evaluate this when t = 1 × 10 − 13 2 ft ⋅ slug × 1.61 × 10 − 11 lbf ⋅ s ⋅ slug⋅ ft ⋅ lbf ⋅ s V = 0.198 ⋅ ⎛ ⋅ ln⎜ 1 − 1.61 × 10 ft s − 11 lbf ⋅ s ⋅ ⎜ ⎝ ft × 2 1 − 13 1 × 10 × ⋅ slug s 32.2⋅ ft × 0.198 ⋅ ft s × slug⋅ ft ⎞ 2 lbf ⋅ s ⎠ t = 0.0286 s V= From 2, after rearranging − 13 ⋅ slug × 32.2⋅ ft 2 s −3 y = 4.49 × 10 dt = M⋅ g k ⎛ − ⎜ ⋅⎝1 − e ⎡ ⎛ − M⋅ g ⎢ M ⎜ y= ⋅⎝e ⋅ ⎢t + k ⎣ k Integrating and using limits y = 1 × 10 dy − 11 1.61 × 10 M ⋅ t⎞ M ⎠ ⋅t ⎞⎤ ⎥ − 1⎠⎥ ⎦ 2 ft × k k ⋅ ⋅ lbf ⋅ s lbf ⋅ s slug⋅ ft ⋅ ⎡0.0291⋅ s ... ⎢ ⎛ ⎜ − ⎢ 2 ft lbf ⋅ s ⎜ ⎢+ 10− 13⋅ slug⋅ ⋅ ⋅⎝e ⎢ − 11 slug⋅ ft 1.61 × 10 ⋅ lbf ⋅ s ⎣ 1.61× 10 − 11 − 13 1× 10 ⋅ ft 5 y (0.001 ft) 3.75 2.5 1.25 0 5 10 15 t (ms) This plot can also be presented in Excel. 20 25 ⎤ ⎞ ⎥ ⋅ .0291 ⎥ ⎥ − 1⎠ ⎥ ⎦ Problem 1.14 [Difficulty: 4] 2 Given: M 70 kg Data on sky diver: k 0.25 N s 2 m Find: Maximum speed; speed after 100 m; plot speed as function of time and distance. Solution: Use given data; integrate equation of motion by separating variables. Treat the sky diver as a system; apply Newton's 2nd law: M Newton's 2nd law for the sky diver (mass M) is (ignoring buoyancy effects): 2 (a) For terminal speed Vt, acceleration is zero, so M g k V 0 dV dt (1) M g Vt so 2 M g k V FD = kV2 k 1 2 2 m m N s Vt 70 kg 9.81 2 2 kg m s 0.25 N s (b) For V at y = 100 m we need to find V(y). From (1) M Separating variables and integrating: dV dt V M m Vt 52.4 s dV dy dV 2 M V M g k V dy dt dt y M g 0 so 2 k V 2 k ln 1 y M g M Mg dV g dy 2 0 k V V 1 a = dV/dt 2 or 2 V M g k 1 e 2 k y M 1 Hence V( y ) Vt 1 e 2 k y M 2 1 For y = 100 m: 2 N s 1 kg m 2 0.25 100 m 2 70 kg s 2 N m m V( 100 m) 52.4 1 e s 2 V( 100 m) 37.4 m s V(m/s) 60 40 20 0 100 200 300 400 500 y(m) (c) For V(t) we need to integrate (1) with respect to t: M Separating variables and integrating: dV dt V 2 M g k V t dV 1 dt M g 2 0 V k V 0 1 M t ln 2 k g so 2 e V( t) Vt 2 e Rearranging k g k g M V t M V V k 1 M ln t 2 k g Vt V M g V k M g 1 t or k V( t) Vt tanh Vt t M 1 V(m/s) 60 40 V ( t) 20 0 5 10 t t(s) The two graphs can also be plotted in Excel. 15 20 Problem 1.15 [Difficulty: 5] 2 Given: M = 70⋅ kg Data on sky diver: k vert = 0.25⋅ N⋅ s 2 2 k horiz = 0.05⋅ m Find: Plot of trajectory. Solution: Use given data; integrate equation of motion by separating variables. N⋅ s m U0 = 70⋅ s 2 m Treat the sky diver as a system; apply Newton's 2nd law in horizontal and vertical directions: Vertical: Newton's 2nd law for the sky diver (mass M) is (ignoring buoyancy effects): M⋅ dV 2 = M ⋅ g − k vert⋅ V (1) dt For V(t) we need to integrate (1) with respect to t: ⌠ ⎮ ⎮ ⎮ ⎮ ⌡ Separating variables and integrating: V t ⌠ dV = ⎮ 1 dt ⌡ M⋅ g 2 0 −V k vert V 0 ⎛ ⎜ 1 M t= ⋅ ⋅ ln⎜ 2 k vert⋅ g ⎜ ⎜ ⎝ ⎛ ⎜ 2⋅ ⎜ M⋅ g ⎝ e oV( t) = ⋅ k vert ⎛ r ⎜ 2⋅ ⎜e ⎝ so Rearranging M⋅ g k vert M⋅ g k vert kvert ⋅ g kvert ⋅ g M dt ⌠ y ( t) = ⎮ ⌡ t 0 ⌠ ⎮ V( t) dt = ⎮ ⎮ ⌡ t M⋅ g k vert M⋅ g k vert ⎛ ⎛ kvert⋅ g ⎞ ⎞ ⋅t M ⎝ ⎠⎠ ⋅ ln⎜ cosh⎜ ⎝ ⋅t =V ⎞ ⎟ ⎟ ⎠ ⎞ − 1⎠ ⋅t ⎞ so V( t) = M⋅ g k vert + 1⎠ or ⌠ y = ⎮ V dt ⌡ ⎛ kvert⋅ g ⎞ ⎛ k vert⋅ g ⎞ ⎞ M⋅ g ⎛ ⋅ t dt = ⋅t ⋅ ln⎜ cosh⎜ k vert ⎝ M M ⎝ ⎠ ⎝ ⎠⎠ ⋅ tanh⎜ 0 y ( t) = −V M dy For y(t) we need to integrate again: +V ⎛ k vert⋅ g ⎞ ⋅ tanh⎜ ⎝ M ⋅t ⎠ 30 600 20 400 y(m) y(m) After the first few seconds we reach steady state: y( t) y( t) 10 200 0 1 2 3 4 5 0 20 40 t t t(s) t(s) M⋅ Horizontal: Newton's 2nd law for the sky diver (mass M) is: dU dt 2 = −k horiz⋅ U For U(t) we need to integrate (2) with respect to t: U Separating variables and integrating: ⌠ 1 ⎮ dU = ⎮ 2 U ⎮ ⌡U t ⌠ k horiz ⎮ − dt ⎮ M ⌡ so 0 − k horiz M ⋅t = − 1 U + 1 U0 0 Rearranging U0 U( t) = 1+ For x(t) we need to integrate again: dx dt =U k horiz⋅ U0 M ⋅t ⌠ x = ⎮ U dt ⌡ or t ⌠ t U0 ⌠ ⎛ k horiz⋅ U0 ⎞ M ⎮ dt = x ( t) = ⎮ U( t) dt = ⎮ ⋅t + 1 ⋅ ln⎜ ⌡ k horiz ⎝ M k ⋅U ⎠ 0 ⎮ 1 + horiz 0 ⋅ t ⎮ M ⌡ 0 x ( t) = M k horiz ⎞ ⎛ khoriz⋅ U0 ⋅t + 1 ⎝ M ⎠ ⋅ ln⎜ 60 (2) 2 x(km) 1.5 1 0.5 0 20 40 60 t(s) Plotting the trajectory: y(km) 0 1 2 −1 −2 −3 x(km) These plots can also be done in Excel. 3 Problem 1.16 [Difficulty: 3] Given: Long bow at range, R = 100 m. Maximum height of arrow is h = 10 m. Neglect air resistance. Find: Estimate of (a) speed, and (b) angle, of arrow leaving the bow. Plot: (a) release speed, and (b) angle, as a function of h Solution: Let V0 = u 0 i + v 0 j = V0 (cos θ 0 i + sin θ 0 j) y ΣFy = m dv = − mg , so v = v0 – gt, and tf = 2tv=0 = 2v0/g dt v2 dv = − mg, v dv = −g dy, 0 − 0 = − gh dy 2 Also, mv Thus h = v 20 2g ΣFx = m v 20 = 2gh From Eq. 2: u0 = x θ0 R (1) 2u v du = 0, so u = u 0 = const, and R = u 0 t f = 0 0 g dt From Eq. 1: h V0 (2) (3) gR gR = 2v 0 2 2gh ∴ u 20 = gR 2 8h 1 Then ⎛ gR 2 ⎞ 2 gR 2 ⎟ + 2 gh and V0 = ⎜⎜ 2 gh + V =u +v = 8h 8h ⎟⎠ ⎝ 2 0 2 0 2 0 (4) 1 ⎛ m 1 ⎞2 9.81 m m ⎟⎟ = 37.7 V0 = ⎜⎜ 2 × 9.81 2 × 10 m + × 100 2 m 2 × 2 s 10 m 8 s s ⎝ ⎠ From Eq. 3: v 0 = 2gh = V0 sin θ , θ = sin −1 2gh V0 (5) 1 ⎡ ⎤ 2 s ⎥ m ⎛ ⎞ θ = sin ⎢⎜ 2 × 9.81 2 ×10 m ⎟ × = 21.8° ⎢⎝ s ⎠ 37.7 m ⎥ ⎣ ⎦ −1 Plots of V0 = V0(h) (Eq. 4) and θ0 = θ 0(h) (Eq. 5) are presented below: V 0 (m/s) Initial Speed vs Maximum Height 80 70 60 50 40 30 20 10 0 0 5 10 15 20 25 30 25 30 h (m) Initial Angle vs Maximum Height 60 50 o θ ( ) 40 30 20 10 0 0 5 10 15 h (m) 20 Problem 1.17 [Difficulty: 2] Given: Basic dimensions M, L, t and T. Find: Dimensional representation of quantities below, and typical units in SI and English systems. Solution: (a) Power Power Energy Time Force Distance Time F L t Force Mass Acceleration From Newton's 2nd law F so M L t Power Hence (b) Pressure (c) Modulus of elasticity (d) Angular velocity (e) Energy Pressure Pressure Force Area Force Area F L t F L F AngularVelocity 2 L 2 M L L 2 M L t t M L 2 t L 2 M L 2 t L Radians Time 2 t 2 kg m 3 L t 3 L t (h) Shear stress (i) Strain (j) Angular momentum Strain Area LengthChange Length F L 2 kg slug 2 1 1 1 t s s Momentum Mass Velocity M ft s 2 M L L M L 2 t MomentOfForce Force Length F L Force 2 m s ft s Energy Force Distance F L ShearStress slug m s M L 2 t L 2 L t 2 2 kg m 2 M L L t (g) Momentum 3 2 2 2 slug ft 2 t 2 s 2 kg m 2 slug ft 2 s M L kg m slug ft t s s kg slug L t 2 2 2 m s L 2 2 s M 2 2 s M L 2 s kg 2 M slug ft s M t (f) Moment of a force 2 2 ft s Dimensionless L AngularMomentum Momentum Distance M L t L M L t 2 2 kg m slugs ft s s 2 Problem 1.18 [Difficulty: 2] Given: Basic dimensions F, L, t and T. Find: Dimensional representation of quantities below, and typical units in SI and English systems. Solution: (a) Power (b) Pressure (c) Modulus of elasticity (d) Angular velocity Power = Energy Pressure = Pressure = Time Force Area Force Area Force × Distance = Time L L s N lbf Radians Time = Force Area = ft N lbf ft 1 1 1 t s s N⋅ m lbf ⋅ ft or M= L t L t F = M⋅ = F⋅ t ⋅ L L⋅ t L t 2 = F⋅ t F L 2 m Force = Mass × Acceleration so Momentum = M ⋅ 2 m 2 From Newton's 2nd law SpecificHeat = s 2 Momentum = Mass × Velocity = M ⋅ (h) Specific heat t F = (f) Momentum ShearStress = lbf ⋅ ft 2 Energy = Force × Distance = F⋅ L (g) Shear stress N⋅ m 2 (e) Energy Hence F⋅ L F = AngularVelocity = = 2 2 N⋅ s lbf ⋅ s N lbf 2 m Energy Mass × Temperature = F⋅ L M⋅ T = F⋅ L ⎛ F⋅ t2 ⎞ ⎜ ⋅T ⎝ L ⎠ = L 2 2 t ⋅T 2 ft L 2 2 m ft 2 s ⋅R s ⋅K F⋅ t 2 LengthChange (i) Thermal expansion coefficient ThermalExpansionCoefficient = (j) Angular momentum Length Temperature = 1 1 1 T K R N⋅ m⋅ s lbf ⋅ ft⋅ s AngularMomentum = Momentum × Distance = F⋅ t⋅ L 2 Problem 1.19 [Difficulty: 1] Given: Viscosity, power, and specific energy data in certain units Find: Convert to different units Solution: Using data from tables (e.g. Table G.2) 2 (a) ⎛ 1 ⋅ ft ⎞ 2 2 ⎜ 2 12 m m ft 1⋅ = 10.76 ⋅ = 1⋅ ×⎜ s s s ⎝ 0.0254⋅ m ⎠ (b) 100 ⋅ W = 100 ⋅ W × (c) 1⋅ kJ kg = 1⋅ kJ kg × 1 ⋅ hp 746 ⋅ W 1000⋅ J 1 ⋅ kJ × = 0.134 ⋅ hp 1 ⋅ Btu 1055⋅ J × 0.454 ⋅ kg 1 ⋅ lbm = 0.43⋅ Btu lbm Problem 1.20 [Difficulty: 1] Given: Pressure, volume and density data in certain units Find: Convert to different units Solution: Using data from tables (e.g. Table G.2) 6895⋅ Pa 1⋅ psi = 1⋅ psi × (b) 1⋅ liter = 1⋅ liter × (c) ⎛ 1 ⋅ ft ⎞ lbf ⋅ s lbf ⋅ s 4.448⋅ N ⎜ 12 N⋅s 1⋅ × = 47.9⋅ = 1⋅ ×⎜ ⋅ ⎠ 2 2 1⋅ lbf 2 ⎝ 0.0254m ft ft m 1⋅ psi × 1⋅ kPa (a) 1⋅ quart 0.946⋅ liter 1000⋅ Pa × = 6.89⋅ kPa 1⋅ gal 4⋅ quart = 0.264⋅ gal 2 Problem 1.21 [Difficulty: 1] Given: Specific heat, speed, and volume data in certain units Find: Convert to different units Solution: Using data from tables (e.g. Table G.2) (a) 4.18⋅ (b) 30⋅ kJ kg⋅ K m s = 4.18⋅ = 30⋅ m s × kJ kg⋅ K 3.281 ⋅ ft 1⋅ m 3 (c) 5⋅ L = 5⋅ L × × 1⋅ m 1000⋅ L ⋅ 1 ⋅ Btu 1.055 ⋅ kJ 1 ⋅ mi 5280⋅ ft ⋅ × 1 ⋅ kg 2.2046⋅ lbm 3600⋅ s hr = 67.1⋅ 3 × × 1⋅ K 1.8⋅ R mi hr ⎛ 100 ⋅ cm × 1⋅ in ⎞ = 305 ⋅ in3 ⎜ 1⋅ m 2.54⋅ cm ⎠ ⎝ = 0.998 ⋅ Btu lbm⋅ R Problem 1.22 [Difficulty: 1] Given: Quantities in English Engineering (or customary) units. Find: Quantities in SI units. Solution: Use Table G.2 and other sources (e.g., Machinery's Handbook, Mark's Standard Handbook) (a) 3.7⋅ acre⋅ ft = 3.7⋅ acre × (b) 150 ⋅ 2 3 (c) (d) in s 3 = 150 ⋅ in gal × 3 ⋅ gpm = 3 ⋅ 3⋅ mph s min = 3⋅ 1 ⋅ acre hr⋅ s 231 ⋅ in × 0.3048⋅ m 1 ⋅ ft 3 3 = 4.56 × 10 ⋅ m 3 ⎛ 0.0254⋅ m ⎞ = 0.00246 ⋅ m ⎜ 1 ⋅ in s ⎝ ⎠ 3 mile × 3 × s 4047⋅ m 1 ⋅ gal 1609⋅ m 1 ⋅ mile 3 3 × ⎛ 0.0254⋅ m ⎞ ⋅ 1⋅ min = 0.000189⋅ m ⎜ 1 ⋅ in s ⎝ ⎠ 60⋅ s × 1 ⋅ hr 3600⋅ s = 1.34⋅ m 2 s Problem 1.23 [Difficulty: 1] Given: Quantities in English Engineering (or customary) units. Find: Quantities in SI units. Solution: Use Table G.2 and other sources (e.g., Google) (a) 100 ⋅ ft 3 m = 100 ⋅ ft 3 min 3 (b) 5 ⋅ gal = 5 ⋅ gal × (c) 65⋅ mph = 65⋅ 231 ⋅ in 1 ⋅ gal mile hr × 3 × ⎛ 0.0254⋅ m ⎞ = 0.0189⋅ m3 ⎜ 1⋅ in ⎝ ⎠ 1852⋅ m 1 ⋅ mile × 3 (d) 3 3 ⎛ 0.0254⋅ m × 12⋅ in ⎞ × 1 ⋅ min = 0.0472⋅ m ⎜ 1 ⋅ in s 1 ⋅ ft ⎠ 60⋅ s ⎝ × 5.4⋅ acres = 5.4⋅ acre × 4047⋅ m 1 ⋅ acre 1 ⋅ hr 3600⋅ s m = 29.1⋅ 4 s 2 = 2.19 × 10 ⋅ m Problem 1.24 Given: Quantities in SI (or other) units. Find: Quantities in BG units. Solution: Use Table G.2. (a) 50⋅ m = 50⋅ m × (b) 250⋅ cc = 250⋅ cm × (c) 100⋅ kW = 100⋅ kW × (d) 5⋅ 2 2 2 ⎛ 1⋅ in × 1⋅ ft ⎞ = 538⋅ ft 2 ⎜ 0.0254m ⋅ 12⋅ in ⎠ ⎝ 3 kg 2 m = 5⋅ kg 2 m × 3 ⎛ 1⋅ m × 1⋅ in × 1⋅ ft ⎞ = 8.83 × 10− 3⋅ ft 3 ⎜ 100⋅ cm 0.0254m 12⋅ in ⎠ ⋅ ⎝ 1000⋅ W 1⋅ kW × 1⋅ hp 746⋅ W = 134⋅ hp 2 ⋅ slug 12⋅ in ⎞ 1⋅ slug ⎛ 0.0254m = 0.0318⋅ ⎜ 1⋅ in × 1⋅ ft × 2 14.95⋅ kg ⎝ ⎠ ft [Difficulty: 1] Problem 1.25 Given: Quantities in SI (or other) units. Find: Quantities in BG units. Solution: Use Table G.2. (a) 180⋅ cc = 180⋅ cm × (b) 300⋅ kW = 300⋅ kW × (c) 50⋅ 3 N⋅s 2 m (d) 2 = 50⋅ N⋅s 2 × m 2 40⋅ m ⋅ hr = 40⋅ m × 3 ⎛ 1⋅ m × 1⋅ in × 1⋅ ft ⎞ = 6.36 × 10− 3⋅ ft 3 ⎜ 100⋅ cm 0.0254m 12⋅ in ⎠ ⋅ ⎝ 1000⋅ W 1⋅ kW 1⋅ lbf 4.448⋅ N × × 1⋅ hp 746⋅ W = 402⋅ hp 2 ⋅ 12⋅ in ⎞ lbf ⋅ s ⎛ 0.0254m = 1.044⋅ ⎜ 1⋅ in × 1⋅ ft 2 ⎝ ⎠ ft 2 ⎛ 1⋅ in × 1⋅ ft ⎞ ⋅ hr = 431⋅ ft 2⋅ hr ⎜ 0.0254m ⋅ 12⋅ in ⎠ ⎝ [Difficulty: 1] Problem 1.26 [Difficulty: 2] Given: Geometry of tank, and weight of propane. Find: Volume of propane, and tank volume; explain the discrepancy. Solution: Use Table G.2 and other sources (e.g., Google) as needed. The author's tank is approximately 12 in in diameter, and the cylindrical part is about 8 in. The weight of propane specified is 17 lb. The tank diameter is D = 12⋅ in The tank cylindrical height is L = 8⋅ in The mass of propane is mprop = 17⋅ lbm The specific gravity of propane is SG prop = 0.495 The density of water is ρ = 998⋅ kg 3 m The volume of propane is given by mprop mprop Vprop = = ρprop SGprop⋅ ρ 3 1 m Vprop = 17⋅ lbm × × × 0.495 998 ⋅ kg 0.454 ⋅ kg 1 ⋅ lbm × ⎛ 1⋅ in ⎞ ⎜ 0.0254⋅ m ⎝ ⎠ 3 3 Vprop = 953 ⋅ in The volume of the tank is given by a cylinder diameter D length L, πD2L/4 and a sphere (two halves) given by πD3/6 2 Vtank = Vtank = The ratio of propane to tank volumes is Vprop Vtank π⋅ D 4 3 ⋅L + π⋅ ( 12⋅ in) 4 π⋅ D 6 2 ⋅ 8 ⋅ in + π⋅ ( 12⋅ in) 6 3 3 Vtank = 1810⋅ in = 53⋅ % This seems low, and can be explained by a) tanks are not filled completely, b) the geometry of the tank gave an overestimate of the volume (the ends are not really hemispheres, and we have not allowed for tank wall thickness). Problem 1.27 [Difficulty: 1] Given: Acreage of land, and water needs. Find: Water flow rate (L/min) to water crops. Solution: Use Table G.2 and other sources (e.g., Machinery's Handbook, Mark's Standard Handbook) as needed. The volume flow rate needed is Q Performing unit conversions Q 4 cm week 10 hectare 4 cm 10 hectare Q 397 week L min 0.04 m 10 hectare week 4 2 1 10 m 1 hectare 1000 L 3 m 1 week 7 day 1 day 24 hr 1 hr 60 min Problem 1.28 Given: Data in given units Find: Convert to different units [Difficulty: 1] Solution: 3 (a) 1⋅ 3 in = 1⋅ min 3 (b) (c) (d) 1⋅ 1⋅ m s 3 3 ⋅ mm 1000⋅ mm⎞ 1⋅ min ⎛ 0.0254m × × = 273⋅ ⎜ min ⎝ 1⋅ in s 1⋅ m ⎠ 60⋅ s in × 3 = 1⋅ liter min m s = 1⋅ 1⋅ gal × 3 × 4 × 0.000946⋅ m liter min × 3 1 ⋅ gal 4 × 0.946 ⋅ liter 3 × 60⋅ s 1⋅ min = 15850⋅ gpm 60⋅ s 1 ⋅ min = 0.264 ⋅ gpm 3 0.0254⋅ m ⎞ m 60⋅ min 1 ⋅ SCFM = 1 ⋅ × = 1.70⋅ × ⎛⎜ min 1 hr 1 ⋅ hr ⋅ ft ⎜ 12 ⎝ ⎠ ft Problem 1.29 Given: Density of mercury. Find: Specific gravity, volume and weight. Solution: Use basic definitions SG = v= ρ ρw = 1.94⋅ From Appendix A ρw 1 ρ [Difficulty: 1] slug ft so v = 1 ⋅ ft 3 so 3 SG = 26.3 SG = 13.6 1.94 3 −5m 3 × 26.3 slug ⋅ ⎞ 1⋅ lbm 1⋅ slug ⎛ 0.3048m × × ⎜ 1⋅ ft 0.4536kg ⋅ 32.2 ⋅ lbm ⎝ ⎠ v = 7.37 × 10 γ = ρ⋅ g Hence on earth γE = 26.3⋅ slug ft On the moon γM = 26.3⋅ 3 × 32.2⋅ slug ft 3 ft s × 5.47⋅ × 2 ft 2 s 1⋅ lbf ⋅ s 2 1⋅ slug ⋅ ft γE = 847 lbf ft 2 × 1 ⋅ lbf ⋅ s 1 ⋅ slug⋅ ft Note that mass-based quantities are independent of gravity γM = 144 3 lbf ft 3 kg Problem 1.30 Given: Definition of kgf. Find: Conversion from psig to kgf/cm2. Solution: Use Table G.2. Define kgf kgf = 1 ⋅ kg × 9.81⋅ m 2 kgf = 9.81N s Then 32⋅ lbf 2 in × 4.448⋅ N 1⋅ lbf × 1⋅ kgf 9.81⋅ N × 2 ⎛ 12⋅ in × 1⋅ ft × 1⋅ m ⎞ = 2.25 kgf ⎜ 1⋅ ft 100⋅ cm ⎠ 0.3048m ⋅ 2 ⎝ cm [Difficulty: 1] Problem 1.31 [Difficulty: 3] Given: Information on canal geometry. Find: Flow speed using the Manning equation, correctly and incorrectly! Solution: Use Table G.2 and other sources (e.g., Google) as needed. The Manning equation is The given data is V= 2 1 3 2 Rh ⋅ S0 which assumes Rh in meters and V in m/s. n 1 S0 = 10 Rh = 7.5⋅ m 1 2 7.5 ⋅ ⎛⎜ 3 V= Hence Using the equation incorrectly: ⎞ ⎝ 10 ⎠ 1 2 V = 86.5⋅ 0.014 Rh = 7.5⋅ m × 1 ⋅ in 0.0254⋅ m 24.6 ⋅ ⎛⎜ 3 V = × m (Note that we don't cancel units; we just write m/s next to the answer! Note also this is a very high speed due to the extreme slope S0.) s 1 ⋅ ft Rh = 24.6⋅ ft 12⋅ in 1 2 Hence n = 0.014 ⎞ ⎝ 10 ⎠ 1 2 V = 191 ⋅ 0.014 ft V = 191 ⋅ This incorrect use does not provide the correct answer (Note that we again don't cancel units; we just write ft/s next to the answer!) s ft s × 12⋅ in 1 ⋅ ft × 0.0254⋅ m 1 ⋅ in This demonstrates that for this "engineering" equation we must be careful in its use! To generate a Manning equation valid for Rh in ft and V in ft/s, we need to do the following: 2 1 2 3 Rh ( m) ⋅ S0 m⎞ 1 ⋅ in 1 ⋅ in 1 ⋅ ft ⎞ ft ⎞ 1 ⋅ ft ⎛ ⎛ = × ⎛⎜ V⎜ = V⎜ × × × n ⎝s⎠ ⎝ s ⎠ 0.0254⋅ m 12⋅ in ⎝ 0.0254⋅ m 12⋅ in ⎠ V = 58.2 m s which is wrong! V⎛⎜ ft ⎞ ⎝s⎠ = 2 1 3 2 Rh ( ft) ⋅ S0 n − × ⎛ 1⋅ in × 1 ⋅ ft ⎞ ⎜ 0.0254⋅ m 12⋅ in ⎝ ⎠ 2 2 1 3 3 2 × 1 3 ⎛ 1 ⋅ in × 1⋅ ft ⎞ = Rh ( ft) ⋅ S0 × ⎛ 1⋅ in × 1 ⋅ ft ⎞ ⎜ 0.0254⋅ m 12⋅ in ⎜ 0.0254⋅ m 12⋅ in n ⎝ ⎠ ⎝ ⎠ 1 3 ⎛ 1 ⋅ 1 ⎞ = 1.49 ⎜ .0254 12 ⎝ ⎠ In using this equation, we ignore the units and just evaluate the conversion factor 2 Hence 3 ft ⎞ 1.49⋅ Rh ( ft) ⋅ S0 ⎛ V⎜ = n ⎝s⎠ 1 2 Handbooks sometimes provide this form of the Manning equation for direct use with BG units. In our case we are asked to instead define a new value for n: 1 2 n BG = n 1.49 n BG = 0.0094 where 24.6 ⋅ ⎛⎜ Using this equation with Rh = 24.6 ft: V = Converting to m/s ⎞ ⎝ 10 ⎠ 1 2 0.0094 V = 284 ⋅ ft s 2 1 2 3 3 ft ⎞ Rh ( ft) ⋅ S0 ⎛ V⎜ = n BG ⎝s⎠ × 12⋅ in 1 ⋅ ft × 0.0254⋅ m 1 ⋅ in V = 284 ft V = 86.6 m s s which is the correct answer! [Difficulty: 2] Problem 1.32 Given: Equation for COPideal and temperature data. Find: COPideal, EER, and compare to a typical Energy Star compliant EER value. Solution: Use the COP equation. Then use conversions from Table G.2 or other sources (e.g., www.energystar.gov) to find the EER. The given data is The COPIdeal is TL ( 20 273) K TL 293 K TH ( 40 273) K TH 313 K 293 COPIdeal 14.65 313 293 The EER is a similar measure to COP except the cooling rate (numerator) is in BTU/hr and the electrical input (denominator) is in W: BTU EER Ideal COPIdeal hr W 2545 EER Ideal 14.65 BTU hr 746 W 50.0 BTU hr W This compares to Energy Star compliant values of about 15 BTU/hr/W! We have some way to go! We can define the isentropic efficiency as ηisen EER Actual EER Ideal Hence the isentropic efficiency of a very good AC is about 30%. Problem 1.33 [Difficulty: 2] Given: Equation for maximum flow rate. Find: Whether it is dimensionally correct. If not, find units of 2.38 coefficient. Write a SI version of the equation Solution: Rearrange equation to check units of 0.04 term. Then use conversions from Table G.2 or other sources (e.g., Google) mmax⋅ T0 2.38 = "Solving" the equation for the constant 2.38: At ⋅ p 0 Substituting the units of the terms on the right, the units of the constant are 1 slug s ×R 2 1 1 1 × ft 2 × 1 psi = slug s ×R 2 1 × ft 2 2 × in lbf × lbf ⋅ s 2 slug ⋅ ft 2 = 2 R ⋅ in ⋅ s ft 3 1 2 c = 2.38⋅ Hence the constant is actually 2 R ⋅ in ⋅ s ft 3 For BG units we could start with the equation and convert each term (e.g., At), and combine the result into a new constant, or simply convert c directly: 1 1 2 c = 2.38⋅ 2 2 R ⋅ in ⋅ s ft 3 1 = 2.38⋅ 2 R ⋅ in ⋅ s ft 3 2 × 2 ⎛ K ⎞ × ⎛ 1⋅ ft ⎞ × 1⋅ ft ⎜ 1.8⋅ R ⎜ 12⋅ in 0.3048m ⋅ ⎝ ⎠ ⎝ ⎠ 1 2 c = 0.04⋅ K ⋅s m so mmax = 0.04⋅ At ⋅ p 0 T0 with At in m2, p 0 in Pa, and T0 in K. Problem 1.34 [Difficulty: 1] Given: Equation for mean free path of a molecule. Find: Dimensions of C for a diemsionally consistent equation. Solution: Use the mean free path equation. Then "solve" for C and use dimensions. The mean free path equation is "Solving" for C, and using dimensions m λ C C ρ d 2 λ ρ d 2 m L C M 3 L M 2 L 0 The constant C is dimensionless. Problem 1.35 [Difficulty: 1] Given: Equation for drag on a body. Find: Dimensions of CD. Solution: Use the drag equation. Then "solve" for CD and use dimensions. The drag equation is 1 2 FD = ⋅ ρ⋅ V ⋅ A ⋅ CD 2 "Solving" for CD, and using dimensions CD = 2⋅ FD 2 ρ⋅ V ⋅ A F CD = M 3 2 × L But, From Newton's 2nd law ⎛ L ⎞ × L2 ⎜t ⎝ ⎠ Force = Mass ⋅ Acceleration F = M⋅ or L t Hence F CD = M 3 L The drag coefficient is dimensionless. × ⎛ L⎞ 2 = 2 ⎜t ×L ⎝ ⎠ M⋅ L t 2 3 × L M × t 2 2 L × 1 2 L =0 2 Problem 1.36 Given: Data on a container and added water. Find: Weight and volume of water added. Solution: Use Appendix A. For the empty container Wc 3.5 lbf For the filled container M total 2.5 slug The weight of water is then Ww M total g Wc [Difficulty: 1] 2 The temperature is ft 1 lbf s Ww 2.5 slug 32.2 3.5 lbf 2 1 slug ft s Ww 77.0 lbf 90°F 32.2°C ρ 1.93 and from Table A.7 slug ft Hence Vw Mw Ww Vw g ρ or ρ 2 3 3 1 s 1 ft 1 slug ft Vw 77.0 lbf 32.2 ft 1.93 slug 2 1 lbf s Vw 1.24ft 3 Problem 1.37 [Difficulty: 1] Given: Equation for vibrations. Find: Dimensions of c, k and f for a dimensionally consistent equation. Also, suitable units in SI and BG systems. Solution: Use the vibration equation to find the diemsions of each quantity 2 m The first term of the equation is d x 2 dt L M The dimensions of this are t 2 Each of the other terms must also have these dimensions. c Hence dx dt M L t t f c: L t t k L so 2 M L and 2 M L t and 2 c M k M t t 2 M L t Suitable units for c, k, and f are 2 M L k x c so 2 kg slug s s k: kg slug 2 kg m f: 2 s 2 s slug ft 2 s s Note that c is a damping (viscous) friction term, k is a spring constant, and f is a forcing function. These are more typically expressed using F (force) rather than M (mass). From Newton's 2nd law: F M L t 2 c Using this in the dimensions and units for c, k, and f we find c: N s lbf s m ft M or k: F t 2 L t F t L N lbf m ft F t 2 L k F t 2 L t 2 F f F L f: N lbf Problem 1.38 Given: Specific speed in customary units Find: Units; Specific speed in SI units [Difficulty: 1] Solution: 1 The units are rpm⋅ gpm 3 2 or ft 3 ft 4 3 4 s 2 Using data from tables (e.g. Table G.2) 1 NScu = 2000⋅ rpm⋅ gpm 2 3 4 ft 3 1 1 ⎛ 1 ⋅ ft ⎞ 2 3 ⎜ 12 rpm⋅ gpm 2 ⋅ π⋅ rad 1 ⋅ min ⎛ 4 × 0.000946⋅ m 1 ⋅ min ⎞ NScu = 2000 × × × ×⎜ ⋅ ×⎜ 1 ⋅ rev 60⋅ s 1 ⋅ gal 3 60⋅ s ⎠ ⎝ ⎝ 0.0254⋅ m ⎠ 2 ft 4 1 ⎛ m3 ⎞ ⋅⎜ s ⎝ s ⎠ NScu = 4.06⋅ 3 rad m 4 2 4 Problem 1.39 Given: "Engineering" equation for a pump Find: SI version [Difficulty: 1] Solution: The dimensions of "1.5" are ft. The dimensions of "4.5 x 10-5" are ft/gpm2. Using data from tables (e.g. Table G.2), the SI versions of these coefficients can be obtained 0.0254m ⋅ 1.5⋅ ft = 1.5⋅ ft × 1 12 −5 4.5 × 10 −5 ft ⋅ = 0.457⋅ m ⋅ ft 2 = 4.5⋅ 10 ft ⋅ gpm 2 gpm × 0.0254m ⋅ 1 12 4.5⋅ 10 −5 ⋅ ft 2 gpm = 3450⋅ m ⎛ m3 ⎞ ⎜ ⎝ s ⎠ 2 The equation is ⎛ ⎛ m3 ⎞ ⎞ H( m) = 0.457 − 3450⋅ ⎜ Q⎜ ⎝ ⎝ s ⎠⎠ 2 ⋅ ft × 60⋅ s ⎞ 1quart ⎛ 1⋅ gal ⋅ ⎜ 4⋅ quart ⋅ 3 1min 0.000946m ⋅ ⎝ ⎠ 2 Problem 1.40 Given: [Difficulty: 2] Air at standard conditions – p = 29.9 in Hg, T = 59°F Uncertainty in p is ± 0.1 in Hg, in T is ± 0.5°F Note that 29.9 in Hg corresponds to 14.7 psia Find: Air density using ideal gas equation of state; Estimate of uncertainty in calculated value. Solution: ρ= in 2 lbf lb ⋅o R 1 p 144 × × = 14.7 2 × ft 2 in 53.3 ft ⋅ lbf 519o R RT The uncertainty in density is given by 1 ⎡⎛ p ∂ρ ⎞ 2 ⎛ T ∂ρ ⎞ 2 ⎤ 2 u ρ = ⎢⎜⎜ u p ⎟⎟ + ⎜⎜ uT ⎟⎟ ⎥ ⎢⎣⎝ ρ ∂p ⎠ ⎝ ρ ∂T ⎠ ⎥⎦ p ∂ρ RT 1 = RT = = 1; ρ ∂p RT RT T ∂ρ T p p = ⋅− =− = −1; 2 ρ ∂T ρ RT ρRT ± 0.1 = ± 0.334% 29.9 ± 0.5 = ± 0.0963% uT = 460 + 59 up = Then [ ] 1 2 2 u ρ = u 2p + (− uT ) [ u ρ = ± 0.348% = ± 2.66 × 10 − 4 ] 1 2 2 = ± 0.334% 2 + (− 0.0963% ) lbm ft 3 Problem 1.41 Given: [Difficulty: 2] Air in hot air balloon p = 759 ± 1 mm Hg T = 60 ± 1°C Find: Solution: (a) Air density using ideal gas equation of state (b) Estimate of uncertainty in calculated value We will apply uncertainty concepts. Governing Equations: ρ= p (Ideal gas equation of state) R⋅ T 1 2 ⎡⎛ x ∂R ⎤2 ⎞ 1 u R = ± ⎢⎜⎜ u x1 ⎟⎟ + L⎥ ⎢⎣⎝ R ∂x1 ⎥⎦ ⎠ We can express density as: 3 ρ = 101⋅ 10 × N 2 × m kg⋅ K 287⋅ N ⋅ m × (Propagation of Uncertainties) 1 333⋅ K kg = 1.06 3 kg ρ = 1.06 m 3 m 1 So the uncertainty in the density is: ⎡⎛ p ∂ρ ⎞ 2 ⎛ T ∂ρ ⎞ 2 ⎤ 2 u ρ = ± ⎢⎜⎜ u p ⎟⎟ + ⎜⎜ uT ⎟⎟ ⎥ ⎢⎣⎝ ρ ∂p ⎠ ⎝ ρ ∂T ⎠ ⎥⎦ Solving each term separately: 1 p ∂ρ = RT =1 ρ ∂p RT T ∂ρ T ⎛ − p = ⎜ ρ ∂T ρ ⎝ RT 2 Therefore: [ p ⎞ = −1 ⎟=− RT ⎠ u ρ = ± (u p ) + (− uT ) 2 up = ] 1 2 2 [ uT = 1 759 1 333 = ± (0.1318% ) + (− 0.3003% ) kg ⎞ ⎛ u ρ = ±0.328% ⎜ ± 3.47 × 10 −3 3 ⎟ m ⎠ ⎝ 2 = 0.1318% ⋅ = 0.3003% ⋅ ] 1 2 2 Problem 1.42 [Difficulty: 2] m = 1.62 ± 0.01oz (20 to 1) D = 1.68 ± 0.01in. (20 to 1) Given: Standard American golf ball: Find: Density and specific gravity; Estimate uncertainties in calculated values. Solution: Density is mass per unit volume, so ρ= m m m 3 6 m =4 3 = = 3 4π (D 2) π D 3 V 3 πR ρ= 6 π ×1.62 oz × 1 0.4536 kg in.3 × × = 1130 kg/m 3 3 3 3 3 16 oz (1.68) in. (0.0254) m SG = and ρ ρH O = 1130 2 kg m3 × = 1.13 m 3 1000 kg 1 2 2 ⎡⎛ m ∂ρ ⎞ ⎛ D ∂ρ ⎞ ⎤2 u ρ = ⎢⎜⎜ u m ⎟⎟ + ⎜⎜ u D ⎟⎟ ⎥ ⎣⎢⎝ ρ ∂m ⎠ ⎝ ρ ∂D ⎠ ⎥⎦ The uncertainty in density is given by m ∂ρ m 1 ∀ = = = 1; ρ ∂m ρ ∀ ∀ um = D ∂ρ D ⎛ 6m ⎞ 6 m = −3; = ⋅ ⎜ − 3 4 ⎟ = −3 ρ ∂D ρ ⎝ πD ⎠ π ρD 4 ± 0.01 = ± 0.617% 1.62 uD = ± 0.1 = ± 0.595% 1.68 Thus [ ] 1 2 2 u ρ = ± u + (− 3u D ) 2 m [ ] 1 2 2 = ± 0.617% + (− 3 × 0.595% ) 2 u ρ = ±1.89% = ± 21.4 u SG = u ρ = ±1.89% = ± 0.0214 Finally, ρ = 1130 ± 21.4 kg/m 3 SG = 1.13 ± 0.0214 (20 to 1) (20 to 1) kg m3 Problem 1.43 Given: [Difficulty: 2] Pet food can H = 102 ± 1 mm (20 to 1) D = 73 ± 1 mm (20 to 1) m = 397 ± 1 g (20 to 1) Find: Magnitude and estimated uncertainty of pet food density. Solution: Density is ρ= m m 4 m or ρ = ρ ( m, D, H ) = = 2 ∀ πR H π D 2 H 1 2 From uncertainty analysis: ⎡⎛ m ∂ρ ⎞ ⎛ D ∂ρ ⎞ ⎛ H ∂ρ ⎞ ⎤ u ρ = ± ⎢⎜⎜ u m ⎟⎟ + ⎜⎜ u D ⎟⎟ + ⎜⎜ u H ⎟⎟ ⎥ ⎢⎣⎝ ρ ∂m ⎠ ⎝ ρ ∂D ⎠ ⎝ ρ ∂H ⎠ ⎥⎦ Evaluating: m ∂ρ m 4 1 ±1 1 4m um = = = = 1; = ±0.252% ρ ∂m ρ π D 2 H ρ πD 2 H 397 D ∂ρ D 4m 1 4m ±1 . = ( −2) = ( −2) = −2; u D = = ±137% 3 2 ρ ∂D ρ ρ 73 πD H πD H 4m 1 4m H ∂ρ H ±1 = ( −1) = ( −1) = −1; u H = = ±0.980% 2 2 2 ρ ∂H ρ ρ πD H 102 πD H 2 Substituting: [ 2 2 2 2 u ρ = ±2.92% ∀= π 4 D2 H = π 4 × (73) 2 mm 2 × 102 mm × m3 9 10 mm 397 g kg m × = 930 kg m 3 ρ= = −4 3 ∀ 4.27 × 10 m 1000 g Thus: ] 1 2 2 u ρ = ± (1 × 0.252 ) + (− 2 × 1.37 ) + (− 1 × 0.980) ρ = 930 ± 27.2 kg m 3 (20 to 1) 3 = 4.27 × 10 −4 m 3 Problem 1.44 Given: [Difficulty: 2] Mass flow rate of water determine by collecting discharge over a timed interval is 0.2 kg/s. Scales can be read to nearest 0.05 kg. Stopwatch can be read to nearest 0.2 s. Find: Estimate precision of flow rate calculation for time intervals of (a) 10 s, and (b) 1 min. Solution: Apply methodology of uncertainty analysis, Appendix F: m = ∆m ∆t 1 Computing equations: 2 2 ⎡⎛ ∆m ∂m ⎞ ⎛ ∆t ∂m ⎞ ⎤2 u ∆t ⎟ ⎥ u ∆m ⎟ + ⎜ u m = ± ⎢⎜ ⎠ ⎝ m ∂∆t ⎠ ⎦⎥ ⎣⎢⎝ m ∂∆m ∆m ∂m 1 = ∆t = 1 and m ∂∆m ∆t Thus ∆t ∂m ∆t 2 ∆m = ⋅ − 2 = −1 m ∂∆t ∆m ∆t The uncertainties are expected to be ± half the least counts of the measuring instruments. Tabulating results: Water Time Interval, ∆t Uncertainty Error in ∆t in ∆t (s) (s) Collected, Error in ∆m ∆m (kg) (%) Uncertainty Uncertainty in ∆m in m (%) (%) (kg) 10 ± 0.10 ± 1.0 2.0 ± 0.025 ± 1.25 ± 1.60 60 ± 0.10 ± 0.167 12.0 ± 0.025 ± 0.208 ± 0.267 A time interval of about 15 seconds should be chosen to reduce the uncertainty in results to ± 1 percent. Problem 1.45 [Difficulty: 3] Given: Nominal mass flow rate of water determined by collecting discharge (in a beaker) over a timed = 100 g s ; Scales have capacity of 1 kg, with least count of 1 g; Timer has least interval is m count of 0.1 s; Beakers with volume of 100, 500, 1000 mL are available – tare mass of 1000 mL beaker is 500 g. Find: Estimate (a) time intervals, and (b) uncertainties, in measuring mass flow rate from using each of the three beakers. Solution: To estimate time intervals assume beaker is filled to maximum volume in case of 100 and 500 mL beakers and to maximum allowable mass of water (500 g) in case of 1000 mL beaker. = m Then Tabulating results ∆m ∆t ∆t = and ∆m ρ∆∀ = m m ∆∀ = 100 mL 500 mL 1000 mL ∆t = 1s 5s 5 s Apply the methodology of uncertainty analysis, Appendix E. Computing equation: 1 2 2 ⎡⎛ ∆m ∂m ⎞ ⎛ ∆t ∂m ⎞ ⎤2 u ∆t ⎟ ⎥ u ∆m ⎟ + ⎜ u m = ± ⎢⎜ ⎠ ⎝ m ∂∆t ⎠ ⎥⎦ ⎣⎢⎝ m ∂∆m The uncertainties are ± half the least counts of the measuring instruments: δ∆m = ±0.5 g 1 ∆m ∂m = ∆t = 1 and m ∂∆m ∆t ∆t ∂m ∆t 2 ∆m = −1 ⋅− = m ∂∆t ∆m ∆t 2 δ∆t = 0.05 s [ 2 ] 1 2 2 ∴ u m = ± u ∆m + (− u ∆t ) Tabulating results: Beaker Volume ∆∀ (mL) 100 500 1000 Water Collected ∆m(g) 100 500 500 Error in ∆m (g) Uncertainty in ∆m (%) ± 0.50 ± 0.50 ± 0.50 ± 0.50 ± 0.10 ± 0.10 Time Interval ∆t (s) 1.0 5.0 5.0 Error in ∆t (s) Uncertainty in ∆t (%) Uncertainty (%) in m ± 0.05 ± 0.05 ± 0.05 ± 5.0 ± 1.0 ± 1.0 ± 5.03 ± 1.0 ± 1.0 Since the scales have a capacity of 1 kg and the tare mass of the 1000 mL beaker is 500 g, there is no advantage in could be reduced to ± 0.50 percent by using the large beaker if a scale using the larger beaker. The uncertainty in m with greater capacity the same least count were available Problem 1.46 Given: [Difficulty: 2] Standard British golf ball: m = 45.9 ± 0.3 g (20 to 1) D = 411 . ± 0.3 mm (20 to 1) Find: Density and specific gravity; Estimate of uncertainties in calculated values. Solution: Density is mass per unit volume, so ρ= m = ∀ ρ= 6 π m 4 πR 3 3 = m 3 6 m = 3 π D3 4π ( D 2) × 0.0459 kg × 1 m 3 = 1260 kg m 3 (0.0411) 3 and ρ SG = ρH 2 O = 1260 kg m3 × m3 = 126 . 1000 kg The uncertainty in density is given by ⎡⎛ m uρ = ± ⎢⎜⎜ ⎢⎣⎝ ρ m ∂ρ m = ρ ∂m ρ 1 2 ∂ρ ⎞ ⎛ D ∂ρ ⎞ um ⎟ + ⎜ uD ⎟ ∂m ⎟⎠ ⎜⎝ ρ ∂D ⎟⎠ 2 ⎤2 ⎥ ⎥⎦ 1 ∀ = = 1; ∀ ∀ um = ± 6 m⎞ D ∂D D ⎛ ⎛ 6m ⎞ = −3⎜ 4 ⎟ = −3; = ⎜− 3 4 ⎟ ρ ∂m ρ ⎝ π D ⎠ ⎝ πD ⎠ Thus [ u ρ = ± u m + (− 3u D ) 2 ] 1 2 2 uD = ± [ 0.3 = ±0.730% 41.1 = ± 0.654 2 + (− 3 × 0.730 ) u ρ = ± 2.29% = ± 28.9 kg m 3 u SG = u ρ = ± 2.29% = ± 0.0289 Summarizing 0.3 = ±0.654% 45.9 ρ = 1260 ± 28.9 kg m 3 (20 to 1) . ± 0.0289 (20 to 1) SG = 126 ] 1 2 2 Problem 1.47 [Difficulty: 3] Given: Soda can with estimated dimensions D = 66.0 ± 0.5 mm, H = 110 ± 0.5 mm. Soda has SG = 1.055 Find: Volume of soda in the can (based on measured mass of full and empty can); Estimate average depth to which the can is filled and the uncertainty in the estimate. Solution: Measurements on a can of coke give m f = 386.5 ± 0.50 g, m e = 17.5 ± 0.50 g ∴ m = m f − m e = 369 ± u m g ⎡⎛ m ∂m f um u m = ⎢⎜ ⎢⎜⎝ m ∂m f f ⎣ 1 2 ⎞ ⎛ me ∂m ⎞ ⎤ ⎟ +⎜ ⎥ ⎟ u m e ⎟ ⎟ ⎜ m ∂m e ⎠ ⎥⎦ ⎠ ⎝ 0.5 g 0.50 =± = ±0.00129, u me = ± = 0.0286 386.5 g 17.5 u mf 2 2 1 2 2 ⎡⎛ 386.5 ⎞ ⎛ 17.5 ⎞ ⎤2 u m = ± ⎢⎜ ×1× 0.00129 ⎟ + ⎜ ×1× 0.0286 ⎟ ⎥ = ±0.0019 ⎠ ⎝ 369 ⎠ ⎦⎥ ⎣⎢⎝ 369 Density is mass per unit volume and SG = ρ/ρΗ2Ο so ∀= m ρ = 1 m m3 kg = 369 g × × × = 350 × 10 −6 m 3 ρH 2 O SG . 1000 kg 1055 1000 g The reference value ρH2O is assumed to be precise. Since SG is specified to three places beyond the decimal point, assume uSG = ± 0.001. Then 1 2 ⎡⎛ m ∂v ⎞ 2 ⎛ SG ∂v ⎞ ⎤2 u SG ⎟ ⎥ um ⎟ + ⎜ u v = ⎢⎜ ⎠ ⎥⎦ ⎢⎣⎝ v ∂m ⎠ ⎝ v ∂SG [ ∀= πD 4 2 2 L or L= ] 1 2 2 u v = ± (1× 0.0019 ) + (− 1× 0.001) = ±0.0021 = ±0.21% 4∀ 4 350 ×10 −6 m 3 10 3 mm = × × = 102 mm πD 2 π 0.066 2 m 2 m 1 ⎡⎛ ∀ ∂L ⎞ 2 ⎛ D ∂L ⎞ 2 ⎤ 2 u L = ⎢⎜ uD ⎟ ⎥ u∀ ⎟ + ⎜ ⎢⎣⎝ L ∂∀ ⎠ ⎝ L ∂D ⎠ ⎥⎦ ∀ ∂L 4 πD 2 = =1 L ∂∀ πD 2 4 4∀ 4∀ 0.5 D ∂L D = ⋅ −2 3 = −2 2 = −2; u D = ± = ±0.0076 πD πD L 66 L ∂D L [ 2 ] 1 2 2 u L = ± (1× 0.0021) + (− 2 × 0.0076 ) = ±0.0153 = ±1.53% Notes: 1. 2. Printing on the can states the content as 355 ml. This suggests that the implied accuracy of the SG value may be over stated. Results suggest that over seven percent of the can height is void of soda. Problem 1.48 Given: Data on water Find: Viscosity; Uncertainty in viscosity [Difficulty: 3] Solution: The data is: 5 Ns A 2.414 10 2 B 247.8 K C 140 K T 303 K m 0.5 K uT The uncertainty in temperature is u T 0.171 % 293 K B Also μ( T) A 10 ( T C) 3 Ns μ ( 293 K ) 1.005 10 Evaluating 2 m A B ln( 10) d μ ( T) dT For the uncertainty B 10 Hence u μ( T) T d μ( T) u T μ( T) dT C T ln( 10) B T u T CT 2 ( C T) 2 Evaluating u μ( T) 1.11 % Problem 1.49 [Difficulty: 4] Given: Dimensions of soda can: D = 66 mm, H = 110 mm Find: Measurement precision needed to allow volume to be estimated with an uncertainty of ± 0.5 percent or less. Solution: H Use the methods of Appendix F: Computing equations: D ∀= πD H 2 4 1 ⎡⎛ H ∂∀ ⎞2 ⎛ D ∂∀ ⎞2 ⎤ 2 u ∀ = ± ⎢⎜ uH ⎟ + ⎜ uD ⎟ ⎥ ⎣⎢⎝ ∀ ∂H ⎠ ⎝ ∀ ∂D ⎠ ⎥⎦ Since ∀ = π D2 H 4 , then ∂∀ ∂H = π D4 and 2 ∂∀ ∂D δx δx = π DH 2 . Letting u D = ± D and u H = ± H , and substituting, 1 1 ⎡⎛ 4H π D 2 δ x ⎞2 ⎛ 4D π DH δ x ⎞ 2 ⎤ 2 ⎡⎛ δ x ⎞ 2 ⎛ 2δ x ⎞ 2 ⎤ 2 u ∀ = ± ⎢⎜ ⎟ +⎜ ⎟ ⎥ = ± ⎢⎜ ⎟ + ⎜ ⎟ ⎥ 2 2 ⎢⎣⎝ π D H 4 H ⎠ ⎝ π D H 2 D ⎠ ⎥⎦ ⎢⎣⎝ H ⎠ ⎝ D ⎠ ⎦⎥ ⎛ δ x ⎞ ⎛ 2δ x ⎞ 2 =⎜ ⎟ +⎜ ⎟ = (δ x) ⎝H⎠ ⎝ D ⎠ 2 Solving, u∀ 2 δx=± 2 u∀ 1 1 2 2 2 2 ⎣⎡( H ) + ( D ) ⎦⎤ =± ⎡ ⎣⎢ ( ⎡⎛ 1 ⎞ 2 ⎛ 2 ⎞ 2 ⎤ ⎢⎜ ⎟ + ⎜ ⎟ ⎥ ⎣⎢⎝ H ⎠ ⎝ D ⎠ ⎦⎥ 0.005 1 110 mm ) +( 2 2 66 mm ) 2 1 ⎤2 ⎦⎥ = ±0.158 mm Check: uH = ± uD = ± δx H δx D =± 0.158 mm = ±1.44 × 10−3 110 mm =± 0.158 mm = ±2.39 × 10−3 66 mm u ∀ = ±[(u H ) 2 + (2u D ) 2 ] 2 = ±[(0.00144) 2 + (0.00478) 2 ] 2 = ±0.00499 1 1 If δx represents half the least count, a minimum resolution of about 2 δx ≈ 0.32 mm is needed. Problem 1.50 [Difficulty: 3] Given: Lateral acceleration, a = 0.70 g, measured on 150-ft diameter skid pad; Uncertainties in Path deviation ±2 ft; vehicle speed ±0.5 mph Find: Estimate uncertainty in lateral acceleration; ow could experimental procedure be improved? Solution: Lateral acceleration is given by a = V2/R. From Appendix F, u a = ±[(2 u v ) 2 + ( u R ) 2 ]1/ 2 From the given data, V 2 = aR; V = aR = 0.70 × 32.2 Then uv = ± and uR = ± δV V δR R = ±0.5 ft ft × 75 ft = 41.1 2 s s mi s ft hr × × 5280 × = ±0.0178 mi 3600 s hr 41.1 ft = ±2 ft × 1 = ±0.0267 75 ft so u a = ± (2 × 0.0178) 2 + (0.0267) 2 1/ 2 = ±0.0445 u a = ±4.45 percent Experimental procedure could be improved by using a larger circle, assuming the absolute errors in measurement are constant. For D = 400 ft; R = 200 ft V 2 = aR; V = aR = 0.70 × 32.2 ft ft × 200 ft = 67.1 = 45.8 mph 2 s s 0.5 2 = ±0.0109; u R = ± = ± 0.0100 45.8 200 2 u a = ± (2 × 0.0109) + 0.0100 2 = ± 0.0240 = ± 2.4% uV = ± [ ] Given data: H= δL = δθ = 57.7 0.5 0.2 ft ft deg For this building height, we are to vary θ (and therefore L ) to minimize the uncertainty u H. Plotting u H vs θ 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 uH 4.02% 2.05% 1.42% 1.13% 1.00% 0.95% 0.96% 1.02% 1.11% 1.25% 1.44% 1.70% 2.07% 2.62% 3.52% 5.32% 10.69% Uncertainty in Height (H = 57.7 ft) vs θ 12% 10% 8% uH θ (deg) 6% 4% 2% 0% 0 10 20 30 40 50 60 70 80 90 θ (o) Optimizing using Solver θ (deg) 31.4 uH 0.947% To find the optimum θ as a function of building height H we need a more complex Solver θ (deg) 50 75 100 125 175 200 250 300 400 500 600 700 800 900 1000 29.9 34.3 37.1 39.0 41.3 42.0 43.0 43.5 44.1 44.4 44.6 44.7 44.8 44.8 44.9 uH 0.992% 0.877% 0.818% 0.784% 0.747% 0.737% 0.724% 0.717% 0.709% 0.705% 0.703% 0.702% 0.701% 0.700% 0.700% Use Solver to vary ALL θ's to minimize the total u H! Total u H's: 11.3% Optimum Angle vs Building Height 50 40 θ (deg) H (ft) 30 20 10 0 0 100 200 300 400 500 H (ft) 600 700 800 900 1000 Problem 1.52 [Difficulty: 4] Given: American golf ball, m = 1.62 ± 0.01 oz, D = 1.68 in. Find: Precision to which D must be measured to estimate density within uncertainty of ± 1percent. Solution: Apply uncertainty concepts Definition: Density, ρ≡ m ∀ ∀ = 34 π R 3 = π D6 3 1 2 ⎡⎛ x ∂R ⎤2 ⎞ u R = ± ⎢⎜ 1 u x1 ⎟ + L⎥ ⎢⎣⎝ R ∂x1 ⎥⎦ ⎠ Computing equation: From the definition, ρ = π Dm = π6Dm = ρ (m, D) 3/6 Thus m ∂ρ ρ ∂m = 1 and D ∂ρ ρ ∂D 3 = 3 , so u ρ = ±[(1 u m ) 2 + (3 u D ) 2 ] 2 1 u 2ρ = u m 2 + 9 u 2D Solving, u D = ± 13 [u ρ 2 − u m2 ] 2 1 From the data given, u ρ = ±0.0100 um = ±0.01 oz = ±0.00617 1.62 oz 1 1 u D = ± [(0.0100) 2 − (0.00617) 2 ] 2 = ±0.00262 or ± 0.262% 3 Since u D = ± δDD , then δ D = ± D u D = ±1.68 in.x 0.00262 = ± 0.00441 in. The ball diameter must be measured to a precision of ± 0.00441 in.( ± 0.112 mm) or better to estimate density within ± 1percent. A micrometer or caliper could be used. Problem 1.53 [Difficulty: 5] δV = 0.001⋅ in δD = 0.0005⋅ in Given: Syringe pump to deliver 100 mL/min Find: (a) Plot uncertainty in flow rate as a function of bore. (b) Find combination of piston speed and bore resulting in minimum uncertainty in flow rate. Solution: min We will apply uncertainty concepts. Q= Governing Equations: π 4 2 ⋅D ⋅V (Flow rate in syringe pump) 1 2 ⎡⎛ x ∂R ⎤2 ⎞ 1 u R = ± ⎢⎜⎜ u x1 ⎟⎟ + L⎥ ⎢⎣⎝ R ∂x1 ⎥⎦ ⎠ Now solving for the piston speed in terms of the bore: V( D) = (Propagation of Uncertainties) 4⋅ Q 2 π⋅ D So the uncertainty in the flow rate is: [ u Q = ± (2u D ) + (uV ) 2 ] 1 2 2 where 1 2 ⎡⎛ D ∂Q ⎞ ⎛ V ∂Q ⎞ ⎤ ⎡⎛ D 2Q ⎞ ⎛ V Q u Q = ± ⎢⎜⎜ u D ⎟⎟ + ⎜⎜ uV ⎟⎟ ⎥ = ± ⎢⎜⎜ u D ⎟⎟ + ⎜⎜ uV Q D Q V ⎢⎣⎝ Q ∂D ⎠ ⎝ Q ∂V ⎥ ⎢ ⎠ ⎦ ⎠ ⎝ ⎣⎝ 2 uD = δD D uv = δV 2 2 The uncertainty is minimized when V ∂u Q ∂D ⎞ ⎟⎟ ⎠ 2 ⎤ ⎥ ⎥⎦ 1 2 =0 1 Substituting expressions in terms of bore we get: ⎡ 32 δD⋅ Q ⎞ 2⎤ ⎥ Dopt = ⎢ ⋅ ⎛⎜ ⎢ π2 ⎝ δV ⎠ ⎥ ⎣ ⎦ Substituting all known values yields Dopt = 1.76⋅ in Plugging this into the expression for the piston speed yields in Vopt = 2.50⋅ min 6 and the uncertainty is u opt = 0.0694⋅ % Graphs of the piston speed and the uncertainty in the flowrate as a function of the bore are shown on the following page. 10 0.3 6 0.2 4 0.1 2 Piston Speed Uncertainty 0 0 1 2 3 Bore (in) 4 0 5 Uncertainty in Flowrate (%) Piston Speed (in/min) 8 Problem 2.1 Given: Velocity fields Find: Whether flows are 1, 2 or 3D, steady or unsteady. [Difficulty: 1] Solution: (1) (2) (3) (4) (5) (6) (7) (8) → → V = V ( x , y) → → V = V ( x , y) → → V = V ( x) → → V = V ( x) → → V = V ( x) → → V = V ( x , y) → → V = V ( x , y) → → V = V ( x , y , z) 2D 2D 1D 1D 1D 2D 2D 3D → → V = V ( t) → → V ≠ V ( t) → → V ≠ V ( t) → → V ≠ V ( t) → → V = V ( t) → → V ≠ V ( t) → → V = V ( t) → → V ≠ V ( t) Unsteady Steady Steady Steady Unsteady Steady Unsteady Steady Problem 2.2 Given: Velocity fields Find: Whether flows are 1, 2 or 3D, steady or unsteady. [Difficulty: 1] Solution: (1) (2) (3) (4) (5) (6) (7) (8) → → V = V ( y) → → V = V ( x) → → V = V ( x , y) → → V = V ( x , y) → → V = V ( x) → → V = V ( x , y , z) → → V = V ( x , y) → → V = V ( x , y , z) 1D 1D 2D 2D 1D 3D 2D 3D → → V = V ( t) → → V ≠ V ( t) → → V = V ( t) → → V = V ( t) → → V = V ( t) → → V ≠ V ( t) → → V = V ( t) → → V ≠ V ( t) Unsteady Steady Unsteady Unsteady Unsteady Steady Unsteady Steady Problem 2.3 Given: Viscous liquid sheared between parallel disks. Upper disk rotates, lower fixed. Velocity field is: rω z V = eˆθ h Find: a. Dimensions of velocity field. b. Satisfy physical boundary conditions. To find dimensions, compare to V = V ( x, y , z ) form. Solution: The given field is V = V (r , z ) . Two space coordinates are included, so the field is 2-D. Flow must satisfy the no-slip condition: 1. At lower disk, V = 0 since stationary. z = 0, so V = eˆθ rω 0 = 0 , so satisfied. h 2. At upper disk, V = eˆθ rω since it rotates as a solid body. z = h, so V = eˆθ rω h = eˆθ rω , so satisfied. h [Difficulty: 2] Problem 2.4 Given: Velocity field Find: Equation for streamlines [Difficulty: 1] Streamline Plots Solution: v u So, separating variables dy dy y dx B x y 2 2 A x y C=1 C=2 C=3 C=4 B y 4 A x B dx A x y (m) For streamlines 5 3 2 Integrating The solution is ln( y ) y B A 1 ln( x ) c ln( x ) c 2 1 C x 0 1 2 3 x (m) The plot can be easily done in Excel. 4 5 Problem 2.5 [Difficulty: 2] Given: Velocity field Find: Equation for streamlines; Plot several in the first quadrant, including one that passes through point (0,0) Solution: Governing equation: For streamlines v u dy dy dx Assumption: 2D flow Hence v u So, separating variables dy dx A y A x y Streamline Plots x 5 4 Integrating ln( y ) ln( x ) c 3 The solution is ln( x y ) c or y C y (m) x y C=1 C=2 C=3 C=4 dx 2 1 x 0 1 The plot can be easily done in Excel. The streamline passing through (0,0) is given by the vertical axis, then the horizontal axis. The value of A is irrelevant to streamline shapes but IS relevant for computing the velocity at each point. 2 3 x (m) 4 5 Problem 2.6 [Difficulty: 1] Given: Velocity field Find: Whether field is 1D, 2D or 3D; Velocity components at (2,1/2); Equation for streamlines; Plot Solution: The velocity field is a function of x and y. It is therefore 2D. u = a⋅ x ⋅ y = 2 ⋅ At point (2,1/2), the velocity components are 1 1 2 v = b ⋅ y = −6 ⋅ v For streamlines = u dy So, separating variables y = dy = dx × 2⋅ m × m⋅ s b⋅ y × m⋅ s 2 a⋅ x ⋅ y 1 2 ⎛ 1 ⋅ m⎞ ⎜2 ⎝ ⎠ ⋅m 2 u = 2⋅ m s 3 m v=− ⋅ 2 s b⋅ y = a⋅ x b dx ⋅ a x b b ln( y ) = Integrating a ⋅ ln( x) + c y = C⋅ x a −3 y = C⋅ x The solution is The streamline passing through point (2,1/2) is given by 1 2 −3 = C⋅ 2 C = 1 3 ⋅2 2 C= 4 y= 4 3 x 20 Streamline for C Streamline for 2C Streamline for 3C Streamline for 4C 16 12 8 4 1 This can be plotted in Excel. 1.3 1.7 2 t=0 x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 c=1 y 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 c=2 y 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 c=3 y 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 t =1 s (### means too large to view) c=1 c=2 c=3 x y y y 0.05 20.00 40.00 60.00 0.10 10.00 20.00 30.00 0.20 5.00 10.00 15.00 0.30 3.33 6.67 10.00 0.40 2.50 5.00 7.50 0.50 2.00 4.00 6.00 0.60 1.67 3.33 5.00 0.70 1.43 2.86 4.29 0.80 1.25 2.50 3.75 0.90 1.11 2.22 3.33 1.00 1.00 2.00 3.00 1.10 0.91 1.82 2.73 1.20 0.83 1.67 2.50 1.30 0.77 1.54 2.31 1.40 0.71 1.43 2.14 1.50 0.67 1.33 2.00 1.60 0.63 1.25 1.88 1.70 0.59 1.18 1.76 1.80 0.56 1.11 1.67 1.90 0.53 1.05 1.58 2.00 0.50 1.00 1.50 t = 20 s x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 c=1 y ###### ###### ###### ###### ###### ###### ###### ###### 86.74 8.23 1.00 0.15 0.03 0.01 0.00 0.00 0.00 0.00 0.00 0.00 0.00 c=2 y ###### ###### ###### ###### ###### ###### ###### ###### 173.47 16.45 2.00 0.30 0.05 0.01 0.00 0.00 0.00 0.00 0.00 0.00 0.00 c=3 y ###### ###### ###### ###### ###### ###### ###### ###### 260.21 24.68 3.00 0.45 0.08 0.02 0.00 0.00 0.00 0.00 0.00 0.00 0.00 Streamline Plot (t = 0) 3.5 c=1 3.0 c=2 c=3 2.5 y 2.0 1.5 1.0 0.5 0.0 0.0 0.5 1.0 1.5 2.0 x Streamline Plot (t = 1 s) 70 c=1 60 c=2 50 c=3 y 40 30 20 10 0 0.0 0.5 1.0 1.5 2.0 x Streamline Plot (t = 20 s) 20 18 c=1 16 c=2 14 c=3 y 12 10 8 6 4 2 0 0.0 0.2 0.4 0.6 x 0.8 1.0 1.2 a= b= C= x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 1 1 0 y 0.16 0.22 0.32 0.39 0.45 0.50 0.55 0.59 0.63 0.67 0.71 0.74 0.77 0.81 0.84 0.87 0.89 0.92 0.95 0.97 1.00 2 y 0.15 0.20 0.27 0.31 0.33 0.35 0.37 0.38 0.39 0.40 0.41 0.41 0.42 0.42 0.43 0.43 0.44 0.44 0.44 0.44 0.45 4 y 0.14 0.19 0.24 0.26 0.28 0.29 0.30 0.30 0.31 0.31 0.32 0.32 0.32 0.32 0.33 0.33 0.33 0.33 0.33 0.33 0.33 6 y 0.14 0.18 0.21 0.23 0.24 0.25 0.26 0.26 0.26 0.27 0.27 0.27 0.27 0.27 0.27 0.27 0.27 0.28 0.28 0.28 0.28 Streamline Plot 1.2 c=0 1.0 c=2 c=4 0.8 c=6 y 0.6 0.4 0.2 0.0 0.0 0.5 1.0 x 1.5 2.0 A = 10 B = 20 C= 1 y 0.50 0.48 0.45 0.43 0.42 0.40 0.38 0.37 0.36 0.34 0.33 0.32 0.31 0.30 0.29 0.29 0.28 0.27 0.26 0.26 0.25 2 y 1.00 0.95 0.91 0.87 0.83 0.80 0.77 0.74 0.71 0.69 0.67 0.65 0.63 0.61 0.59 0.57 0.56 0.54 0.53 0.51 0.50 4 y 2.00 1.90 1.82 1.74 1.67 1.60 1.54 1.48 1.43 1.38 1.33 1.29 1.25 1.21 1.18 1.14 1.11 1.08 1.05 1.03 1.00 6 y 3.00 2.86 2.73 2.61 2.50 2.40 2.31 2.22 2.14 2.07 2.00 1.94 1.88 1.82 1.76 1.71 1.67 1.62 1.58 1.54 1.50 Streamline Plot 3.5 c=1 c=2 3.0 c=4 2.5 c = 6 ((x,y) = (1.2) 2.0 y x 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 1.5 1.0 0.5 0.0 0.0 0.5 1.0 x 1.5 2.0 Problem 2.10 [Difficulty: 2] Given: Velocity field Find: Equation for streamline through (1,3) Solution: For streamlines v u So, separating variables y A dy y dy dx x 2 A y x x dx x Integrating ln( y ) ln( x ) c The solution is y C x which is the equation of a straight line. For the streamline through point (1,3) 3 C 1 C3 and y 3 x For a particle up or x dx A dt x dx dt A x 2 A t c t x 2 2 A c 2 A Hence the time for a particle to go from x = 1 to x = 2 m is 2 ∆t t( x 2 ) t( x 1 ) ∆t ( 2 m) c 2 A 2 2 ( 1 m) c 2 A 2 4 m 1 m 2 2 2 m s ∆t 0.75 s Problem 2.11 [Difficulty: 3] Given: Flow field Find: Plot of velocity magnitude along axes, and y = x; Equation for streamlines Solution: u On the x axis, y = 0, so Plotting M y 2 π 0 v M x 2 π 200 v (m/s) 150 100 50 0 0.2 0.4 0.6 0.8 1 x (km) The velocity is perpendicular to the axis and increases linearly with distance x. This can also be plotted in Excel. u On the y axis, x = 0, so M y v 2 π M x 2 π 0 Plotting 0 0.2 0.4 0.6 u (m/s) 50 100 150 200 y (km) The velocity is perpendicular to the axis and increases linearly with distance y. This can also be plotted in Excel. 0.8 1 u On the y = x axis The flow is perpendicular to line y = x: M y 2 π M x 2 π Slope of line y = x: 2 r x y 2 π u v 1 2 2 2 then along y = r x x x M M 2 x M r 2 2 2 2 u v x x 2 π 2 π 2 π Then the magnitude of the velocity along y = x is V Plotting M x 1 Slope of trajectory of motion: If we define the radial position: v 2 x 200 V(m/s) 150 100 50 0 0.2 0.4 0.6 0.8 1 r (km) This can also be plotted in Excel. For streamlines So, separating variables Integrati ng M x v u dy dx 2 π M y x y 2 π y dy x dx y 2 2 2 2 x 2 2 c The solution x y C is The streamlines form a set of concentric circles. which is the equation of a circle. This flow models a rigid body vortex flow. See Example 5.6 for streamline plots. Streamlines are circular, and the velocity approaches zero as we approach the center. In Problem 2.10, we see that the streamlines are also circular. In a real tornado, at large distances from the center, the velocities behave as in Problem 2.10; close to the center, they behave as in this problem. [Difficulty: 3] Problem 2.12 Given: Flow field Find: Plot of velocity magnitude along axes, and y = x; Equation of streamlines Solution: K y u On the x axis, y = 0, so 2 2 π x y 2 Plotting 0 K x v 2 2 π x y 2 K 2 π x 160 v( m/s) 80 1 0.5 0 0.5 1 80 160 x (km) The velocity is perpendicular to the axis, is very high close to the origin, and falls off to zero. This can also be plotted in Excel. u On the y axis, x = 0, so K y 2 2 2 π x y Plotting K 2 π y v K x 2 2 2 π x y 0 160 v( m/s) 80 1 0.5 0 80 160 y (km) 0.5 1 The velocity is perpendicular to the axis, is very high close to the origin, and falls off to zero. This can also be plotted in Excel. K x u On the y = x axis 2 2 π x x The flow is perpendicular to line y = x: 2 K u v 2 x y Then the magnitude of the velocity along y = x is V 2 2 K 2 4 π Plotting 2 2 K 4 π x 1 r then along y = x u v 2 1 Slope of trajectory of motion: r 2 2 π x x Slope of line y = x: If we define the radial position: K x v 4 π x 1 x 2 1 x 2 K 2 π 2 x x x 2 x K 2 π r 160 v( m/s) 80 1 0.5 0 0.5 1 80 160 x (km) This can also be plotted in Excel. K x For streamlines v u dy dx 2 2 π x y 2 K y 2 2 π x y So, separating variables Integrating 2 x y y dy x dx y 2 2 The solution is 2 2 x 2 2 c x y C which is the equation of a circle. Streamlines form a set of concentric circles. This flow models a vortex flow. See Example 5.6 for streamline plots. Streamlines are circular, and the velocity approaches infinity as we approach the center. In Problem 2.11, we see that the streamlines are also circular. In a real tornado, at large distances from the center, the velocities behave as in this problem; close to the center, they behave as in Problem 2.11. Problem 2.13 [Difficulty: 3] Given: Flow field Find: Plot of velocity magnitude along axes, and y = x; Equations of streamlines Solution: q x u On the x axis, y = 0, so 2 2 π x y 2 Plotting q 2 π x v q y 2 2 π x y 2 0 100 u (m/s) 50 1 0.5 0 0.5 1 50 100 x (km) The velocity is very high close to the origin, and falls off to zero. It is also along the axis. This can be plotted in Excel. q x u On the y axis, x = 0, so 2 2 π x y Plotting 2 0 v q y 2 2 π x y 2 q 2 π y 100 v (m/s) 60 20 1 0.5 20 0 60 100 y (km) The velocity is again very high close to the origin, and falls off to zero. It is also along the axis. This can also be plotted in Excel. 0.5 1 u On the y = x axis q x 2 2 π x x The flow is parallel to line y = x: 2 q v 4 π x v u 2 x y 2 q 2 u v Plotting q 4 π x 4 π 1 x 2 1 r then along y = x 2 V Then the magnitude of the velocity along y = x is 2 1 Slope of trajectory of motion: r 2 2 π x x Slope of line y = x: If we define the radial position: q x 1 x 2 2 2 x x q 2 π 2 x 2 x q 2 π r 100 V(m/s) 60 20 1 0.5 20 0 0.5 60 100 r (km) This can also be plotted in Excel. q y For streamlines v u dy dx 2 2 π x y q x 2 2 π x y So, separating variables dy y 2 y x 2 dx x Integrating ln( y ) ln( x ) c The solution is y C x This flow field corresponds to a sink (discussed in Chapter 6). which is the equation of a straight line. 1 Problem 2.14 [Difficulty: 2] Given: Velocity field Find: Proof that the parametric equations for particle motion are x p = c1 ⋅ e A⋅ t and y p = c2 ⋅ e − A⋅ t ; pathline that was at (2,2) at t = 0; compare to streamline through same point, and explain why they are similar or not. Solution: Governing equations: up = For pathlines dx vp = dt dy v For streamlines dt u = dy dx Assumption: 2D flow Hence for pathlines So, separating variables dx Eliminating t vp = dy y ln( x ) = A⋅ t + C1 x=e The pathlines are = A⋅ x dt = A⋅ dt x Integrating dx up = A⋅ t+ C1 x = c1 ⋅ e t= =e C1 A ⋅ t ⋅e = c1 ⋅ e A⋅ t v u So, separating variables dy y = dy dx =− A⋅ y A⋅ x = =e C2 − A ⋅ t ⋅e A A x ⋅y = c2 ⋅ e = const or x⋅ y = 4 y x x ln( y ) = −ln( x ) + c The solution is ln( x ⋅ y ) = c or x ⋅ y = const or x⋅ y = 4 for given data The streamline passing through (2,2) and the pathline that started at (2,2) coincide because the flow is steady! ( A A) = const ln x ⋅ y dx Integrating − A⋅ t − A⋅ t ⎛ 1 1⎞ ⎜ A A ln⎝ x ⋅ y ⎠ = const or x =− − A⋅ t+ C2 y = c2 ⋅ e so For streamlines = −A⋅ y = −A⋅ dt y=e A⋅ t ⋅ ln⎛⎜ dt ln( y ) = −A⋅ t + C2 ⎞ = − 1 ⋅ ln⎛ y ⎞ A A ⎜ c2 ⎝ c1 ⎠ ⎝ ⎠ 1 dy for given data Problem 2.15 [Difficulty: 2] Given: Velocity field Find: Proof that the parametric equations for particle motion are x p = c1 ⋅ e A⋅ t and y p = c2 ⋅ e 2⋅ A ⋅ t ; pathline that was at (2,2) at t = 0; compare to streamline through same point, and explain why they are similar or not. Solution: Governing equations: up = For pathlines dx vp = dt dy dt v For streamlines u = dy dx Assumption: 2D flow Hence for pathlines So, separating variables up = dx Eliminating t For streamlines x = c1 ⋅ e y = c2 ⋅ e v = u So, separating variables dy y A⋅ t+ C1 dy y = dy dx ln⎛ or y = C⋅ x = 2 ⋅ A⋅ dt C1 A ⋅ t ⋅e = c1 ⋅ e A⋅ t y=e 2⋅ A⋅ t+ C2 y = c2 ⋅ e 2⋅ A ⋅ t = = 2 ⋅ A⋅ y dt ln( y ) = 2 ⋅ A⋅ t + C2 A⋅ t x = c2 ⋅ ⎛⎜ ⎞ ⎝ c1 ⎠ 2 ⋅ A⋅ y 2 ⋅ dx The solution is =e dy vp = ln( x ) = A⋅ t + C1 x=e The pathlines are = A⋅ x dt = A⋅ dt x Integrating dx A⋅ x = 2 so y = c⋅ x =e C2 2⋅ A ⋅ t ⋅e or y= 2⋅ y x ln( y ) = 2 ⋅ ln( x ) + c Integrating y ⎞ ⎜ 2 =c ⎝x ⎠ 2 or y= 1 2 ⋅x 2⋅ A ⋅ t 2⋅ A ⋅ t 2 x = c2 ⋅ e 2 for given data The streamline passing through (2,2) and the pathline that started at (2,2) coincide because the flow is steady! 1 2 ⋅x 2 for given data t=0 x 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 t =1 s C=1 y 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 C=2 y 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 C=3 y 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 x 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.250 0.275 0.300 0.325 0.350 0.375 0.400 0.425 0.450 0.475 0.500 t = 20 s C=1 y 1.00 1.00 0.99 0.99 0.98 0.97 0.95 0.94 0.92 0.89 0.87 0.84 0.80 0.76 0.71 0.66 0.60 0.53 0.44 0.31 0.00 C=2 y 1.41 1.41 1.41 1.41 1.40 1.39 1.38 1.37 1.36 1.34 1.32 1.30 1.28 1.26 1.23 1.20 1.17 1.13 1.09 1.05 1.00 C=3 y 1.73 1.73 1.73 1.73 1.72 1.71 1.71 1.70 1.69 1.67 1.66 1.64 1.62 1.61 1.58 1.56 1.54 1.51 1.48 1.45 1.41 x 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 C=1 y 1.00 1.00 1.00 0.99 0.98 0.97 0.96 0.95 0.93 0.92 0.89 0.87 0.84 0.81 0.78 0.74 0.70 0.65 0.59 0.53 0.45 C=2 y 1.41 1.41 1.41 1.41 1.40 1.40 1.39 1.38 1.37 1.36 1.34 1.33 1.31 1.29 1.27 1.24 1.22 1.19 1.16 1.13 1.10 C=3 y 1.73 1.73 1.73 1.73 1.72 1.72 1.71 1.70 1.69 1.68 1.67 1.66 1.65 1.63 1.61 1.60 1.58 1.56 1.53 1.51 1.48 Streamline Plot (t = 0) 3.5 c=1 c=2 c=3 3.0 2.5 y 2.0 1.5 1.0 0.5 0.0 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 x Streamline Plot (t = 1s) 2.0 c=1 c=2 c=3 1.8 1.6 1.4 y 1.2 1.0 0.8 0.6 0.4 0.2 0.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 x Streamline Plot (t = 20s) 2.0 c=1 c=2 c=3 1.8 1.6 1.4 y 1.2 1.0 0.8 0.6 0.4 0.2 0.0 0.0 0.5 1.0 1.5 x 2.0 2.5 Problem 2.17 [Difficulty: 4] Given: Pathlines of particles Find: Conditions that make them satisfy Problem 2.10 flow field; Also Problem 2.11 flow field; Plot pathlines Solution: The given pathlines are x p a sin ( ω t) The velocity field of Problem 2.12 is u Ky 2 2 π x y y p a cos( ω t) K x v 2 2 2 π x y 2 If the pathlines are correct we should be able to substitute xp and y p into the velocity field to find the velocity as a function of time: Ky u 2 2 π x y v Kx 2 π x 2 y 2 2 K a cos ( ω t) 2 2 2 2 π a sin ( ω t) a cos ( ω t) 2 K ( a sin ( ω t) ) 2 π a sin ( ω t) 2 2 a cos ( ω t) 2 2 K cos ( ω t) 2 π a K sin ( ω t) (1) (2) 2 π a We should also be able to find the velocity field as a function of time from the pathline equations (Eq. 2.9): dxp dt u dxp u dxp dt dt a ω cos( ω t) Comparing Eqs. 1, 2 and 3 u a ω cos( ω t) Hence we see that a ω K 2 π a K cos( ω t) 2 π a or v (2.9) v dyp dt a ω sin( ω t) v a ω sin( ω t) ω K 2 π a (3) K sin( ω t) 2 π a for the pathlines to be correct. 2 The pathlines are a = 300 m a = 400 m a = 500 m 400 To plot this in Excel, compute x p and y p for t ranging from 0 to 60 s, with ω given by the above formula. Plot y p versus xp. Note that outer particles travel much slower! 200 400 200 0 200 This is the free vortex flow discussed in Example 5.6 400 200 400 u The velocity field of Problem 2.11 is M y v 2 π M x 2 π If the pathlines are correct we should be able to substitute xp and y p into the velocity field to find the velocity as a function of time: u v Recall that u M y 2 π M x 2 π dxp dt M ( a cos( ω t) ) 2 π M ( a sin( ω t) ) 2 π u a ω cos( ω t) Hence we see that ω 2 π M a cos( ω t) (4) 2 π M a sin( ω t) (5) 2 π a ω cos( ω t) Comparing Eqs. 1, 4 and 5 M v M a cos( ω t) 2 π for the pathlines to be correct. dyp dt a ω sin( ω t) v a ω sin( ω t) (3) M a sin( ω t) 2 π The pathlines To plot this in Excel, compute x p and y p for t ranging from 0 to 75 s, with ω given by the above formula. Plot y p versus xp. Note that outer particles travel faster! 400 200 400 200 0 200 400 This is the forced vortex flow discussed in Example 5.6 200 400 600 a = 300 m a = 400 m a = 500 m Note that this is rigid body rotation! Problem 2.18 [Difficulty: 2] Given: Time-varying velocity field Find: Streamlines at t = 0 s; Streamline through (3,3); velocity vector; will streamlines change with time Solution: v For streamlines u dy At t = 0 (actually all times!) dx dy So, separating variables y dy dx y dx a y ( 2 cos( ω t) ) a x ( 2 cos( ω t) ) y x x x Integrating ln( y ) ln( x ) c The solution is y C C 3 For the streamline through point (3,3) which is the equation of a hyperbola. x 3 C1 y and 1 x The streamlines will not change with time since dy/dx does not change with time. At t = 0 5 u a x ( 2 cos( ω t) ) 5 1 m u 45 3 v a y ( 2 cos( ω t) ) 5 y 4 2 s v 45 1 3 m 3 s 1 s 3 m 3 m s The velocity vector is tangent to the curve; 0 1 2 3 x 4 5 Tangent of curve at (3,3) is dx Direction of velocity at (3,3) is This curve can be plotted in Excel. dy v u 1 y x 1 Problem 2.19 [Difficulty: 3] Given: Velocity field Find: Plot of pathline traced out by particle that passes through point (1,1) at t = 0; compare to streamlines through same point at the instants t = 0, 1 and 2s Solution: Governing equations: up = For pathlines dx dt vp = dy v For streamlines dt u = dy dx Assumption: 2D flow Hence for pathlines So, separating variables Integrating up = dx dt = A⋅ ( 1 + B⋅ t) A = 1⋅ ⎛ t ⎝ 2 2⎞ ⎠ s B = 1⋅ dy dx = A⋅ ( 1 + B⋅ t) ⋅ dt x = A⋅ ⎜ t + B⋅ m y 1 1 2 y=e The pathlines are Using given data 2 ⎛ t ⎞ x = A⋅ ⎜ t + B⋅ +1 2⎠ ⎝ For streamlines v u So, separating variables Integrating = dy dx 1+ B⋅ t = 2 s 1 y = c2 ⋅ e 1 y=e 1 2 =e C2 ⋅e 2 1 2 ⋅ C⋅ t = c2 ⋅ e dy y = C A ⋅ t⋅ dx C A 2 2 ⋅ C⋅ t 2 2 ⋅ C⋅ t 2 which we can integrate for any given t (t is treated as a constant) ⋅ t⋅ x + c C A ⋅ t⋅ x + const y= ⎛ C ⋅ t⋅ x + const⎞ ⎜ ⎝A ⎠ ( 1+ B⋅ t) 2 ⋅ C⋅ t A⋅ ( 1 + B⋅ t) ( 1 + B⋅ t) ⋅ ln( y ) = y 1 2 ⋅ C⋅ t + C2 2 1 The solution is C = 1⋅ C⋅ y ⋅ t = ( 1 + B⋅ t) ⋅ = C⋅ t ⋅ y ⋅ C⋅ t + C2 1 2 ⎛ t ⎞ x = A⋅ ⎜ t + B⋅ + C1 2⎠ ⎝ dt = C⋅ t⋅ dt ln( y ) = + C1 dy vp = s 1 y=1 For particles at (1,1) at t = 0, 1, and 2s, using A, B, and C data: y=x 2 1 y = (2⋅ x − 1) Streamline and Pathline Plots 5 Streamline (t=0) Streamline (t=1) Streamline (t=2) Pathline 4 y (m) 3 2 1 0 1 2 3 x (m) 4 5 3 Problem 2.20 [Difficulty: 3] Given: Velocity field Find: Plot of pathline traced out by particle that passes through point (1,1) at t = 0; compare to streamlines through same point at the instants t = 0, 1 and 2s Solution: up = dx dt vp = = B⋅ x ⋅ ( 1 + A⋅ t) A = 0.5⋅ For pathlines Governing equations: dy v For streamlines dt u = dy dx Assumption: 2D flow up = Hence for pathlines dx So, separating variables dx dt ⎛ x=e ⎝ 2⎞ t 2⎠ ⎝ 2 ⎠ + C1 B⋅ ⎜t+ A⋅ t C1 = e ⋅e ⎝ x = c1 ⋅ e ⎝ ⎛ B⋅ ⎜t+ A⋅ For streamlines x=e ⎝ v u So, separating variables Integrating = 2⎞ ⎛ ⎛ Using given data 1 vp = s dy dx = ( 1 + A⋅ t) ⋅ y t B⋅ ⎜t+ A ⋅ The pathlines are B = 1⋅ dy ⎛ ln( x ) = B⋅ ⎜ t + A⋅ B⋅ ⎜t+ A⋅ s = B⋅ ( 1 + A⋅ t) ⋅ dt x Integrating 1 2⎠ dt = C⋅ y C = 1⋅ 1 s = C⋅ dt ln( y ) = C⋅ t + C2 + C1 2⎞ dy ⎛ B⋅ ⎜t+ A⋅ = c1 ⋅ e ⎝ 2⎞ t 2⎠ y=e C⋅ t+ C2 =e C2 C⋅ t ⋅e = c2 ⋅ e C⋅ t 2⎞ t 2⎠ y = c2 ⋅ e C⋅ t 2⎞ t 2⎠ y=e C⋅ t C⋅ y B⋅ x ⋅ ( 1 + A⋅ t) dy y = C dx ⋅ B x ( 1 + A⋅ t) ⋅ ln( y ) = C B ⋅ ln( x ) + c which we can integrate for any given t (t is treated as a constant) C The solution is y 1+ A ⋅ t = const ⋅ x B y = const ⋅ x or C y=x For particles at (1,1) at t = 0, 1, and 2s C B y=x C ( 1+ A )B y=x Streamline and Pathline Plots 5 Streamline (t=0) Streamline (t=1) Streamline (t=2) Pathline 4 y (m) 3 2 1 0 1 2 3 x (m) 4 5 ( 1+ 2⋅ A )B Problem 2.21 [Difficulty: 3] Given: Eulerian Velocity field Find: Lagrangian position function that was at point (1,1) at t = 0; expression for pathline; plot pathline and compare to streamlines through same point at the instants t = 0, 1 and 2s Solution: Governing equations: up = For pathlines (Lagrangian description) dx dt vp = dy vp = dy v For streamlines dt u Assumption: 2D flow Hence for pathlines up = dx dt =A So, separating variables dx = A⋅ dt Integrating x = A⋅ t + x 0 A = 2 Using given data The pathlines are given by combining the equations t = For streamlines y ( x ) = y 0 − B⋅ v u So, separating variables = dy dx dy = − = B⋅ t A s = −B⋅ t dt m B = 2 2 s dy = −B⋅ t⋅ dt The Lagrangian description is Hence m ( x − x0) 2 2 2⋅ A t 2 x0 = 1 m y = −B⋅ x ( t) = A⋅ t + x 0 y ( t) = −B⋅ x ( t) = 2 ⋅ t + 1 y ( t) = 1 − t x − x0 A or, using given data y = −B⋅ 2 t + y0 t y0 = 1 m 2 + y0 2 2 2 2 + y 0 = −B⋅ y(x) = 1 − ( x − 1) (x − x0)2 2 2⋅ A + y0 2 4 −B⋅ t A ⋅ dx which we can integrate for any given t (t is treated as a constant) = dy dx y=− The solution is B⋅ t y = − A ⋅x + c B⋅ t A and for the one through (1,1) ⋅ ( x − 1) + 1 1=− B⋅ t A ⋅1 + c c=1+ y = 1 − t⋅ ( x − 1) x = 1 , 1.1 .. 20 Streamline Plots 20 −40 5 10 15 y (m) − 28 − 52 − 76 Streamline (t=0) Streamline (t=1) Streamline (t=2) Pathline − 100 x (m) 20 25 B⋅ t A Problem 2.22 [Difficulty: 3] Given: Velocity field Find: Plot of pathline of particle for t = 0 to 1.5 s that was at point (1,1) at t = 0; compare to streamlines through same point at the instants t = 0, 1 and 1.5 s Solution: Governing equations: up = For pathlines dx dt vp = dy v For streamlines dt u = dy dx Assumption: 2D flow Hence for pathlines So, separating variables up = dx dt = ax ln⎛⎜ ⎞ = a⋅ t x0 ⎝ ⎠ x x ( t) = x 0⋅ e Using given data x ( t) = e v u So, separating variables dy y Hence s = = ln⎛⎜ dy dx x0 = 1 m a⋅ t vp = dy = b ⋅ y ⋅ ( 1 + c⋅ t ) dt 2⋅ t = ln⎛⎜ b = 2 1 2 c = 0.4 s ⎞ = b ⋅ ⎛ t + 1 ⋅ c⋅ t2⎞ ⎜ 2 ⎝ ⎠ ⎝ y0 ⎠ y y ( t) = e dy y = b ⋅ ( 1 + c⋅ t) ⋅ dt y0 = 1 m ⎛ 1 2⎞ b⋅ ⎜t+ ⋅ c⋅ t ⎝ 2 ⎠ y ( t) = e 2 2⋅ t+ 0.4⋅ t b ⋅ y ⋅ ( 1 + c⋅ t ) a⋅ x b ⋅ ( 1 + c⋅ t ) a⋅ x ⋅ dx which we can integrate for any given t (t is treated as a constant) ⎞ = b ⋅ ( 1 + c⋅ t) ⋅ ln⎛ x ⎞ ⎜x ⎝ y0 ⎠ a ⎝ 0⎠ y b The solution is 1 dy = b ⋅ y ⋅ ( 1 + c⋅ t) ⋅ dt Hence For streamlines a = 2 = a⋅ dt x Integrating dx x y = y 0 ⋅ ⎛⎜ ⎞ ⎝ x0 ⎠ a ⋅ ( 1+ c⋅ t) 1 s b t = 0 x y = y 0 ⋅ ⎛⎜ ⎞ ⎝ x0 ⎠ a = x x t = 1 y = y 0 ⋅ ⎛⎜ ⎞ ⎝ x0 ⎠ b ⋅ ( 1+ c⋅ t) a = x 1.4 t = 1.5 x y = y 0 ⋅ ⎛⎜ ⎞ ⎝ x0 ⎠ Streamline and Pathline Plots 10 Streamline (t=0) Streamline (t=1) Streamline (t=1.5) Pathline 8 6 y (m) For b ⋅ ( 1+ c⋅ t) 4 2 0 2 4 6 x (m) 8 10 ⋅ ( 1+ c⋅ t) a = x 1.6 Problem 2.23 [Difficulty: 3] Given: Velocity field Find: Plot of pathline of particle for t = 0 to 1.5 s that was at point (1,1) at t = 0; compare to streamlines through same point at the instants t = 0, 1 and 1.5 s Solution: Governing equations: For pathlines up = dx a = 1 1 vp = dt dy v For streamlines dt u Assumption: 2D flow Hence for pathlines So, separating variables up = dx = a⋅ x dt vp = 5 s = a⋅ dt x Integrating dx ln⎛⎜ dy dt = b⋅ y⋅ t ⎞ = a⋅ t x0 ⎝ ⎠ x ln⎛⎜ x ( t) = x 0⋅ e For streamlines x ( t) = e 5 v = u So, separating variables dy y Hence = = ln⎛⎜ dy dx b⋅ t a⋅ x a⋅ t y y ( t) = y 0⋅ e 1 25 1 2 s = b ⋅ t⋅ dt y0 = 1 m 2 ⋅ b⋅ t 2 2 t y ( t) = e 50 b⋅ y⋅ t a⋅ x ⋅ dx which we can integrate for any given t (t is treated as a constant) ⎞ = b ⋅ t⋅ ln⎛ x ⎞ ⎜x ⎝ y0 ⎠ a ⎝ 0⎠ y b The solution is y ⎞ = b ⋅ 1 ⋅ t2 2 ⎝ y0 ⎠ x0 = 1 m t Using given data dy dy = b ⋅ y ⋅ t⋅ dt 1 Hence b = x y = y 0 ⋅ ⎛⎜ ⎞ ⎝ x0 ⎠ a ⋅t b a = 0.2 x0 = 1 y0 = 1 = dy dx b x y = y 0 ⋅ ⎛⎜ ⎞ ⎝ x0 ⎠ t = 0 = 1 b x y = y 0 ⋅ ⎛⎜ ⎞ x0 ⎝ ⎠ t = 5 x y = y 0 ⋅ ⎛⎜ ⎞ x0 ⎝ ⎠ b = x b t = 10 ⋅t a a ⋅t = 1 ⋅t a = x 2 b a ⋅t = 2 Streamline and Pathline Plots 10 8 6 y (m) For ⋅t a 4 2 Streamline (t=0) Streamline (t=1) Streamline (t=1.5) Pathline 0 2 4 6 x (m) 8 10 Pathline t 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00 3.25 3.50 3.75 4.00 4.25 4.50 4.75 5.00 x 1.00 1.00 1.01 1.03 1.05 1.08 1.12 1.17 1.22 1.29 1.37 1.46 1.57 1.70 1.85 2.02 2.23 2.47 2.75 3.09 3.49 Streamlines t=0 x y 1.00 1.00 1.00 0.78 1.00 0.61 1.00 0.47 1.00 0.37 1.00 0.29 1.00 0.22 1.00 0.17 1.00 0.14 1.00 0.11 1.00 0.08 1.00 0.06 1.00 0.05 1.00 0.04 1.00 0.03 1.00 0.02 1.00 0.02 1.00 0.01 1.00 0.01 1.00 0.01 1.00 0.01 y 1.00 0.78 0.61 0.47 0.37 0.29 0.22 0.17 0.14 0.11 0.08 0.06 0.05 0.04 0.03 0.02 0.02 0.01 0.01 0.01 0.01 t=1s x 1.00 1.00 1.01 1.03 1.05 1.08 1.12 1.17 1.22 1.29 1.37 1.46 1.57 1.70 1.85 2.02 2.23 2.47 2.75 3.09 3.49 t=2s x 1.00 1.00 1.01 1.03 1.05 1.08 1.12 1.17 1.22 1.29 1.37 1.46 1.57 1.70 1.85 2.02 2.23 2.47 2.75 3.09 3.49 y 1.00 0.97 0.88 0.75 0.61 0.46 0.32 0.22 0.14 0.08 0.04 0.02 0.01 0.01 0.00 0.00 0.00 0.00 0.00 0.00 0.00 y 1.00 0.98 0.94 0.87 0.78 0.68 0.57 0.47 0.37 0.28 0.21 0.15 0.11 0.07 0.05 0.03 0.02 0.01 0.01 0.00 0.00 Pathline and Streamline Plots 1.0 Pathline Streamline (t = 0) Streamline (t = 1 s) Streamline (t = 2 s) 0.8 y 0.6 0.4 0.2 0.0 0.0 0.5 1.0 1.5 2.0 x 2.5 3.0 3.5 Problem 2.25 [Difficulty: 3] Given: Flow field Find: Pathline for particle starting at (3,1); Streamlines through same point at t = 1, 2, and 3 s Solution: dx For particle paths dt Separating variables and integrating dx x dy = u = a⋅ x ⋅ t an d = a⋅ t⋅ dt or ln( x ) = or y = b ⋅ t + c2 dy = b ⋅ dt dt =v=b 1 2 ⋅ a⋅ t + c1 2 Using initial condition (x,y) = (3,1) and the given values for a and b c1 = ln( 3 ⋅ m) x = 3⋅ e The pathline is then For streamlines (at any time t) v u dx and y = 4⋅ t + 1 b = a⋅ x ⋅ t So, separating variables dy = b dx ⋅ a⋅ t x Integrating y= b a⋅ t c2 = 1 ⋅ m 2 0.05⋅ t dy = an d ⋅ ln( x ) + c We are interested in instantaneous streamlines at various times that always pass through point (3,1). Using a and b values: c=y− y= 1+ The streamline equation is b a ⋅t 40 t ⋅ ln( x) = 1 − ⋅ ln⎛⎜ 4 0.1⋅ t ⋅ ln( 3) x⎞ ⎝ 3⎠ 30 Pathline Streamline (t=1) Streamline (t=2) Streamline (t=3) 20 y 10 0 1 2 3 4 5 − 10 − 20 These curves can be plotted in Excel. x Problem 2.26 [Difficulty: 4] Given: Velocity field Find: Plot streamlines that are at origin at various times and pathlines that left origin at these times Solution: v For streamlines u = dy dx v 0 ⋅ sin⎡⎢ω⋅ ⎜⎛ t − ⎣ ⎝ = u0 v 0 ⋅ sin⎡⎢ω⋅ ⎛⎜ t − So, separating variables (t=const) x ⎣ ⎝ dy = u0 u0 v 0 ⋅ cos⎡⎢ω⋅ ⎛⎜ t − ⎞⎤ u0 ⎥ ⎠⎦ + c ω v 0 ⋅ ⎡⎢cos⎡⎢ω⋅ ⎜⎛ t − Using condition y = 0 when x = 0 For particle paths, first find x(t) y= dx dt ⎞⎤ ⎥ ⎠⎦ ⋅ dx x ⎣ ⎝ y= Integrating ⎞⎤ u0 ⎥ ⎠⎦ x ⎞⎤ − cos( ω⋅ t)⎤ ⎥ u0 ⎥ ⎠⎦ ⎦ x ⎣ ⎣ ⎝ = u = u0 Separating variables and integrating dx = u 0 ⋅ dt Using initial condition x = 0 at t = τ c1 = −u 0 ⋅ τ x = u 0 ⋅ t + c1 o r x = u 0 ⋅ ( t − τ) x ⎞⎤ = v = v 0 ⋅ sin⎡⎢ω⋅ ⎜⎛ t − ⎥ dt ⎣ ⎝ u 0 ⎠⎦ dy For y(t) we have and dy dt This gives streamlines y(x) at each time t ω so dy ⎡ ⎡ = v = v 0 ⋅ sin⎢ω⋅ ⎢t − dt ⎣ ⎣ u 0 ⋅ ( t − τ) ⎤⎤ u0 ⎥⎥ ⎦⎦ = v = v 0 ⋅ sin( ω⋅ τ) Separating variables and integrating dy = v 0 ⋅ sin( ω⋅ τ) ⋅ dt y = v 0 ⋅ sin( ω⋅ τ) ⋅ t + c2 Using initial condition y = 0 at t = τ c2 = −v 0 ⋅ sin( ω⋅ τ) ⋅ τ y = v 0 ⋅ sin( ω⋅ τ) ⋅ ( t − τ) The pathline is then x ( t , τ) = u 0 ⋅ ( t − τ) y ( t , τ) = v 0 ⋅ sin( ω⋅ τ) ⋅ ( t − τ) These terms give the path of a particle (x(t),y(t)) that started at t = τ. 0.5 0.25 0 1 2 − 0.25 − 0.5 Streamline t = 0s Streamline t = 0.05s Streamline t = 0.1s Streamline t = 0.15s Pathline starting t = 0s Pathline starting t = 0.05s Pathline starting t = 0.1s Pathline starting t = 0.15s The streamlines are sinusoids; the pathlines are straight (once a water particle is fired it travels in a straight line). These curves can be plotted in Excel. 3 Problem 2.27 Given: Velocity field Find: Plot streakline for first second of flow [Difficulty: 5] Solution: Following the discussion leading up to Eq. 2.10, we first find equations for the pathlines in form ( x p( t) = x t , x 0 , y 0 , t0 ) and ( y p( t) = y t , x 0 , y 0 , t0 ) where x 0, y 0 is the position of the particle at t = t0, and re-interprete the results as streaklines ( ) ( x st t0 = x t , x 0 , y 0 , t0 ) and ( ) ( y st t0 = y t , x 0 , y 0 , t0 ) which gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t) For particle paths, first find x(t) dx dt Separating variables and integrating = u = u0 dx = u 0 ⋅ dt ( x = x0 + u0 ⋅ t − t0 o r x ⎞⎤ so = v = v 0 ⋅ sin⎡⎢ω⋅ ⎜⎛ t − u0 ⎥ dt ⎣ ⎝ ⎠⎦ x ⎞⎤ ⎡ ⎛ 0 dy = v = v 0 ⋅ sin⎢ω⋅ ⎜ t0 − ⎥ u0 dt ⎣ ⎝ ⎠⎦ x 0 ⎞⎤ ⎡ ⎛ dy = v 0 ⋅ sin⎢ω⋅ ⎜ t0 − ⎥ ⋅ dt u0 ⎣ ⎝ ⎠⎦ dy For y(t) we have and Separating variables and integrating ( ) ( The streakline is then x st t0 = x 0 + u 0 t − t0 With x0 = y0 = 0 ( ) ( x st t0 = u 0 ⋅ t − t0 ) dy ) ( ⎡ ⎡ x 0 + u 0⋅ t − t0 ⎣ ⎣ u0 = v = v 0 ⋅ sin⎢ω⋅ ⎢t − dt ( ( ) ( ) ( y st t0 = v 0 ⋅ sin⎡ω⋅ t0 ⎤ ⋅ t − t0 ⎣ ⎦ Streakline for First Second y (m) 1 2 4 6 −1 −2 x (m) This curve can be plotted in Excel. For t = 1, t0 ranges from 0 to t. ) ( 2 0 ⎥⎥ ⎦⎦ x 0 ⎞⎤ ⎡ ⎛ y = y 0 + v 0 ⋅ sin⎢ω⋅ ⎜ t0 − ⎥ ⋅ t − t0 u0 ⎣ ⎝ ⎠⎦ x 0 ⎞⎤ ⎡ ⎛ y st t0 = y 0 + v 0 ⋅ sin⎢ω⋅ ⎜ t0 − ⎥ ⋅ t − t0 u0 ⎣ ⎝ ⎠⎦ ( ) ) )⎤⎤ 8 10 ) ) Problem 2.28 [Difficulty: 4] Given: Velocity field Find: Plot of streakline for t = 0 to 3 s at point (1,1); compare to streamlines through same point at the instants t = 0, 1 and 2 s Solution: Governing equations: For pathlines up = dx vp = dt dy v For streamlines dt u = dy dx Following the discussion leading up to Eq. 2.10, we first find equations for the pathlines in form ( x p( t) = x t , x 0 , y 0 , t0 ( ) ( ) x st t0 = x t , x 0 , y 0 , t0 ) ( ) and y p( t) = y t , x 0 , y 0 , t0 and y st t0 = y t , x 0 , y 0 , t0 ( ) ( ) which gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t) Assumption: 2D flow For pathlines So, separating variables up = dx x Integrating dx dt = B⋅ x ⋅ ( 1 + A⋅ t) A = 0.5 1 s B = 1 1 s dy = B⋅ ( 1 + A⋅ t) ⋅ dt y 2 2 t − t0 ⎞ ⎜⎛ x ⎞ ⎛ ln⎜ = B⋅ ⎜ t − t0 + A⋅ 2 x0 ⎝ ⎠ ⎝ ⎠ ln⎛⎜ ⎝ 2 2 t − t0 ⎞ ⎜⎛ B⋅ ⎜t− t0+ A⋅ 2 ⎠ x = x0⋅ e ⎝ The pathlines are vp = dy dt = C⋅ y C = 1 = C⋅ dt y y0 ⎞ = C⋅ t − t ( 0) ⎠ y = y0⋅ e 2 2 ⎛⎜ t − t0 ⎞ B⋅ ⎜t− t0+ A ⋅ 2 ⎠ x p( t) = x 0⋅ e ⎝ ( ) C⋅ t− t0 y p( t) = y 0⋅ e ( ) C⋅ t− t0 where x 0, y 0 is the position of the particle at t = t0. Re-interpreting the results as streaklines: The streaklines are then 2 2 t − t0 ⎞ ⎜⎛ B⋅ ⎜t− t0+ A⋅ 2 ⎠ x st( t0 ) = x 0 ⋅ e ⎝ ( ) y st t0 = y 0 ⋅ e ( ) C⋅ t− t0 where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t) 1 s v For streamlines u So, separating variables = dy dx = ( 1 + A⋅ t) ⋅ C⋅ y B⋅ x ⋅ ( 1 + A⋅ t) dy y = C dx ⋅ B x C ( 1 + A⋅ t) ⋅ ln( y ) = Integrating B which we can integrate for any given t (t is treated as a constant) ⋅ ln( x ) + const C The solution is y 1+ A ⋅ t = const ⋅ x B 2 For particles at (1,1) at t = 0, 1, and 2s y=x y=x 1 3 y=x 2 Streamline and Pathline Plots 10 Streamline (t=0) Streamline (t=1) Streamline (t=2) Streakline 8 y (m) 6 4 2 0 2 4 6 x (m) 8 10 Problem 2.29 [Difficulty: 4] Given: Velocity field Find: Plot of streakline for t = 0 to 3 s at point (1,1); compare to streamlines through same point at the instants t = 0, 1 and 2 s Solution: Governing equations: For pathlines up = dx vp = dt dy v For streamlines dt u = dy dx Following the discussion leading up to Eq. 2.10, we first find equations for the pathlines in form ( x p( t) = x t , x 0 , y 0 , t0 ( ) ( ) x st t0 = x t , x 0 , y 0 , t0 ) ( ) and y p( t) = y t , x 0 , y 0 , t0 and y st t0 = y t , x 0 , y 0 , t0 ( ) ( ) which gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t) Assumption: 2D flow For pathlines So, separating variables up = dx x Integrating dx dt = a⋅ x ⋅ ( 1 + b ⋅ t ) a = 1 = a⋅ ( 1 + b ⋅ t) ⋅ dt 2 2 ⎛⎜ t − t0 ⎞ x ⎞ ⎛ ln⎜ = a⋅ ⎜ t − t 0 + b ⋅ 2 ⎝ ⎠ ⎝ x0 ⎠ 2 2 ⎛⎜ t − t0 ⎞ a⋅ ⎜t− t0+ b⋅ 2 ⎠ x = x0⋅ e ⎝ 1 s b = 1 1 5 s vp = dy y ln⎛⎜ ⎝ dy dt = c⋅ y c = 1 = c⋅ dt y y0 ⎞ = c⋅ t − t ( 0) ⎠ y = y0⋅ e ( ) c⋅ t− t0 1 s 2 2 ⎛⎜ t − t0 ⎞ a⋅ ⎜t− t0+ b⋅ 2 ⎠ x p( t) = x 0⋅ e ⎝ The pathlines are y p( t) = y 0⋅ e ( ) c⋅ t− t0 where x 0, y 0 is the position of the particle at t = t0. Re-interpreting the results as streaklines: The streaklines are then 2 2 ⎛⎜ t − t0 ⎞ a⋅ ⎜t− t0+ b⋅ 2 ⎠ x st( t0 ) = x 0 ⋅ e ⎝ ( ) y st t0 = y 0 ⋅ e ( ) c⋅ t− t0 where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t) v For streamlines u So, separating variables = dy dx = ( 1 + b ⋅ t) ⋅ c⋅ y a⋅ x ⋅ ( 1 + b ⋅ t ) dy y = c dx ⋅ a x ( 1 + b ⋅ t) ⋅ ln( y ) = Integrating c a which we can integrate for any given t (t is treated as a constant) ⋅ ln( x ) + const c The solution is y 1+ b⋅ t = const ⋅ x a 2 y=x For particles at (1,1) at t = 0, 1, and 2s y=x 3 1 y=x 2 Streamline and Pathline Plots 5 Streamline (t=0) Streamline (t=1) Streamline (t=2) Streakline 4 y (m) 3 2 1 0 1 2 3 x (m) 4 5 Problem 2.30 [Difficulty: 4] Given: Velocity field Find: Plot of pathline for t = 0 to 3 s for particle that started at point (1,2) at t = 0; compare to streakline through same point at the instant t = 3 Solution: Governing equations: up = For pathlines dx vp = dt dy dt Following the discussion leading up to Eq. 2.10, we first find equations for the pathlines in form ( x p( t) = x t , x 0 , y 0 , t0 ( ) ( ) x st t0 = x t , x 0 , y 0 , t0 ) ( ) and y p( t) = y t , x 0 , y 0 , t0 and y st t0 = y t , x 0 , y 0 , t0 ( ) ( ) which gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t) Assumption: 2D flow For pathlines So, separating variables up = dx dt = a⋅ x ⋅ t ln⎛⎜ ⎞ = a ⋅ ⎛ t2 − t 2⎞ 0 ⎠ ⎝ ⎝ x0 ⎠ 2 x = x0⋅ e ⋅ ⎛t − t0 2 ⎝ 2 a x p( t) = x 0⋅ e 1 4 1 2 s b = 1 m 3 s vp = dy dt =b dy = b ⋅ dt x a The pathlines are a = = a⋅ t⋅ dt x Integrating dx 2 2⎞ ⎠ ⋅ ⎛t − t0 ⎝ 2 ( ) ( ) y − y0 = b⋅ t − t0 y = y0 + b⋅ t − t0 2⎞ ⎠ ( y p( t) = y 0 + b ⋅ t − t0 ) where x 0, y 0 is the position of the particle at t = t0. Re-interpreting the results as streaklines: a The pathlines are then ( ) x st t0 = x 0 ⋅ e 2 ⋅ ⎛t − t0 ⎝ 2 2⎞ ⎠ ( ) ( y st t0 = y 0 + b ⋅ t − t0 where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t) ) Streakline and Pathline Plots 2 Streakline Pathline y (m) 1.5 1 0.5 0 1 2 x (m) 3 4 Problem 2.31 [Difficulty: 4] Given: 2D velocity field Find: Streamlines passing through (6,6); Coordinates of particle starting at (1,4); that pathlines, streamlines and streaklines coincide Solution: v For streamlines u = a⋅ y Integrating 3 dy dx b = a⋅ y ⌠ ⌠ ⎮ 2 ⎮ a ⋅ y dy = ⎮ b dx ⌡ ⌡ or 2 3 = b⋅ x + c For the streamline through point (6,6) c = 60 and For particle that passed through (1,4) at t = 0 u= dx v= dy dt dt = a⋅ y 3 y = 6 ⋅ x + 180 ⌠ ⌠ ⎮ 2 ⎮ 1 dx = x − x 0 = ⎮ a ⋅ y dt ⌡ ⌡ 2 ⌠ ⌠ ⎮ 1 dy = ⎮ b dt ⌡ ⌡ =b t ⎛ ⌠ 2 x − x 0 = ⎮ a ⋅ y 0 + b ⋅ t dt ⌡ Then ( ) x0 = 1 y0 = 4 2 x = 1 + 16⋅ t + 8 ⋅ t + y = y0 + b⋅ t = y0 + 2⋅ t x = x 0 + a⋅ ⎜ y 0 ⋅ t + b ⋅ y 0 ⋅ t + 2 2 4 3 ⋅t 3 t ⌠ 2 x − x 0 = ⎮ a ⋅ y 0 + b ⋅ t dt ⎮ ⌡t ( ) y = 6⋅ m ⌠ ⌠ ⎮ 1 dy = ⎮ b dt ⌡ ⌡ ⎡ ( y = y0 + b⋅ t − t0 x = x 0 + a⋅ ⎢y 0 ⋅ t − t0 + b ⋅ y 0 ⋅ ⎛ t − t0 ⎝ ⎣ 2 ( ) 2 2⎞ ⎠ + 2 ) ⋅ ⎛ t − t0 3 ⎝ b 3 3 ( ) 3 4 x = −3 + Evaluating at t = 3 x = 31.7⋅ m ⋅ t −1 = ( 3 1 3 ⋅ 4 ⋅ t − 13 This is a steady flow, so pathlines, streamlines and streaklines always coincide ) y = 2⋅ ( t − 1) y = 4⋅ m ⎤ ⎥ ⎠⎦ 3⎞ 0 Hence, with x 0 = -3, y 0 = 0 at t0 = 1 3 x = 26.3⋅ m At t = 1 s y = 4 + 2⋅ t For particle that passed through (-3,0) at t = 1 2 3⎞ b ⋅t ⎝ 0 Hence, with We need y(t) ⎠ Problem 2.32 Solution Pathlines: t 0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60 1.80 2.00 2.20 2.40 2.60 2.80 3.00 3.20 3.40 3.60 3.80 4.00 [Difficulty: 3] The particle starting at t = 3 s follows the particle starting at t = 2 s; The particle starting at t = 4 s doesn't move! Starting at t = 0 x 0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60 1.80 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 Starting at t = 1 s y 0.00 0.40 0.80 1.20 1.60 2.00 2.40 2.80 3.20 3.60 4.00 3.80 3.60 3.40 3.20 3.00 2.80 2.60 2.40 2.20 2.00 Starting at t = 2 s x y x y 0.00 0.20 0.40 0.60 0.80 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 0.00 0.40 0.80 1.20 1.60 2.00 1.80 1.60 1.40 1.20 1.00 0.80 0.60 0.40 0.20 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 -0.20 -0.40 -0.60 -0.80 -1.00 -1.20 -1.40 -1.60 -1.80 -2.00 Streakline at t = 4 s x 2.00 1.80 1.60 1.40 1.20 1.00 0.80 0.60 0.40 0.20 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 Pathline and Streakline Plots 4 3 2 1 y 0 -0.5 0.0 0.5 1.0 1.5 Pathline starting at t = 0 -1 Pathline starting at t = 1 s Pathline starting at t = 2 s -2 Streakline at t = 4 s -3 x 2.0 2.5 y 2.00 1.60 1.20 0.80 0.40 0.00 -0.40 -0.80 -1.20 -1.60 -2.00 -1.80 -1.60 -1.40 -1.20 -1.00 -0.80 -0.60 -0.40 -0.20 0.00 Problem 2.33 [Difficulty: 3] Given: Velocity field Find: Equation for streamline through point (1.1); coordinates of particle at t = 5 s and t = 10 s that was at (1,1) at t = 0; compare pathline, streamline, streakline Solution: Governing equations: v For streamlines u = dy For pathlines dx up = dx vp = dt dy dt Assumption: 2D flow Given data For streamlines a = 1 1 v dy So, separating variables a b Integrating The solution is then 5 s = u b = 1 ⋅ dy = Hence t0 = 0 dx x ( ) b x y = y 0 + ⋅ ln⎛⎜ ⎞ = 5 ⋅ ln( x ) + 1 a x0 up = dx ln⎛⎜ ⎝ The pathlines are y0 = 1 x ⋅ y − y 0 = ln⎛⎜ ⎞ b ⎝ x0 ⎠ a x Integrating x0 = 1 a⋅ x ⎝ Hence for pathlines s b = dx m dx dt ⎠ = a⋅ x = a⋅ dt x x0 dy dt =b dy = b ⋅ dt ⎞ = a⋅ t − t ( 0) ⎠ x = x0⋅ e vp = ( ) a⋅ t− t0 ( ) ( ) y − y0 = b⋅ t − t0 y = y0 + b⋅ t − t0 or b x y = y 0 + ⋅ ln⎛⎜ ⎞ a x0 ⎝ ⎠ For a particle that was at x 0 = 1 m, y 0 = 1 m at t0 = 0 s, at time t = 1 s we find the position is x = x0⋅ e ( ) a⋅ t− t0 1 = e 5 ( For a particle that was at x 0 = 1 m, y 0 = 1 x = x0⋅ e ( ) a⋅ t− t0 = e ( ) a⋅ t− t0 = e 2 m m at t0 = 0 s, at time t = 5 s we find the position is ( ) y = y 0 + b ⋅ t − t0 = 6 m For a particle that was at x 0 = 1 m, y 0 = 1 x = x0⋅ e ) y = y 0 + b ⋅ t − t0 = 2 m m at t0 = 0 s, at time t = 10 s we find the position is ( ) y = y 0 + b ⋅ t − t0 = 11 m m For this steady flow streamlines, streaklines and pathlines coincide Streamline and Position Plots 15 Streamline Position at t = 1 s Position at t = 5 s Position at t = 10 s 12 y (m) 9 6 3 0 2 4 6 x (m) 8 10 Problem 2.34 [Difficulty: 3] Given: Velocity field Find: Equation for streamline through point (2.5); coordinates of particle at t = 2 s that was at (0,4) at t = 0; coordinates of particle at t = 3 s that was at (1,4.25) at t = 1 s; compare pathline, streamline, streakline Solution: Governing equations: v For streamlines u = dy dx up = For pathlines dx dt vp = dy dt Assumption: 2D flow Given data For streamlines a = 2 v u So, separating variables a b Integrating = m b = 1 s dy dx 1 s x0 = 2 y0 = 5 x = 1 x = x b⋅ x = a ⋅ dy = x ⋅ dx 1 2 2 ⋅ y − y0 = ⋅ ⎛ x − x0 ⎞ ⎝ ⎠ 2 b a ( ) 2 The solution is then x 2 2 y = y0 + ⋅ ⎛ x − x0 ⎞ = +4 ⎝ ⎠ 2⋅ a 4 Hence for pathlines up = b dx dt =a Hence dx = a⋅ dt Integrating x − x 0 = a⋅ t − t 0 vp = dy dt = b⋅ x dy = b ⋅ x ⋅ dt ( ) ( ) dy = b ⋅ ⎡x 0 + a⋅ t − t0 ⎤ ⋅ dt ⎣ ⎦ a 2 2 y − y 0 = b ⋅ ⎡⎢x 0 ⋅ t − t0 + ⋅ ⎛ ⎛ t − t0 ⎞ ⎞ − a⋅ t0 ⋅ t − t0 ⎤⎥ ⎝ ⎝ ⎠⎠ 2 ⎣ ⎦ ( The pathlines are ( x = x 0 + a⋅ t − t 0 ) ) ( ) a 2 2 y = y 0 + b ⋅ ⎡⎢x 0 ⋅ t − t0 + ⋅ ⎛ ⎛ t − t0 ⎞ ⎞ − a⋅ t0 ⋅ t − t0 ⎤⎥ ⎝ ⎝ ⎠⎠ 2 ⎣ ⎦ ( ) ( ) For a particle that was at x 0 = 0 m, y 0 = 4 m at t0 = 0s, at time t = 2 s we find the position is ( ) x = x 0 + a⋅ t − t 0 = 4 m a 2 2 y = y 0 + b ⋅ ⎡⎢x 0 ⋅ t − t0 + ⋅ ⎛ ⎛ t − t0 ⎞ ⎞ − a⋅ t0 ⋅ t − t0 ⎥⎤ = 8m ⎝ ⎝ ⎠⎠ 2 ⎣ ⎦ ( ) ( ) For a particle that was at x 0 = 1 m, y 0 = 4.25 m at t0 = 1 s, at time t = 3 s we find the position is ( ) x = x 0 + a⋅ t − t 0 = 5 m a 2 2 y = y 0 + b ⋅ ⎡⎢x 0 ⋅ t − t0 + ⋅ ⎛ ⎛ t − t0 ⎞ ⎞ − a⋅ t0 ⋅ t − t0 ⎥⎤ = 10.25 m ⎝ ⎝ ⎠⎠ 2 ⎣ ⎦ ( ) ( ) For this steady flow streamlines, streaklines and pathlines coincide; the particles refered to are the same particle! Streamline and Position Plots 15 Streamline Position at t = 1 s Position at t = 5 s Position at t = 10 s 12 y (m) 9 6 3 0 1.2 2.4 3.6 x (m) 4.8 6 Problem 2.35 [Difficulty: 4] Given: Velocity field Find: Coordinates of particle at t = 2 s that was at (1,2) at t = 0; coordinates of particle at t = 3 s that was at (1,2) at t = 2 s; plot pathline and streakline through point (1,2) and compare with streamlines through same point at t = 0, 1 and 2 s Solution : Governing equations: For pathlines up = dx dy vp = dt For streamlines dt v u = dy dx Following the discussion leading up to Eq. 2.10, we first find equations for the pathlines in form ( x p( t) = x t , x 0 , y 0 , t0 ( ) ) ( x st t0 = x t , x 0 , y 0 , t0 ) ( ) and y p( t) = y t , x 0 , y 0 , t0 and y st t0 = y t , x 0 , y 0 , t0 ( ) ( ) which gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t) Assumption: 2D flow Given data Hence for pathlines a = 0.2 up = dx dt 1 s b = 0.4 m 2 s = a⋅ y vp = dy dt = b⋅ t Hence dx = a⋅ y ⋅ dt dy = b ⋅ t⋅ dt For x b 2 2 dx = ⎡⎢a⋅ y 0 + a⋅ ⋅ ⎛ t − t0 ⎞⎤⎥ ⋅ dt ⎝ ⎠⎦ 2 ⎣ Integrating ⎡⎢ 3 t 3 ⎥⎤ t 0 2 x − x 0 = a⋅ y 0 ⋅ ( t − t 0 ) + a⋅ ⋅ ⎢ − − t0 ⋅ ( t − t0 )⎥ 3 2 ⎣3 ⎦ The pathlines are ⎡⎢ 3 t 3 ⎥⎤ t 0 2 x ( t ) = x 0 + a⋅ y 0 ⋅ ( t − t 0 ) + a⋅ ⋅ ⎢ − − t0 ⋅ ( t − t0 )⎥ 3 2 ⎣3 ⎦ b 2 2 y − y0 = ⋅ ⎛ t − t0 ⎞ ⎠ 2 ⎝ b b These give the position (x,y) at any time t of a particle that was at (x 0,y 0) at time t0 Note that streaklines are obtained using the logic of the Governing equations, above b 2 2 y ( t) = y0 + ⋅ ⎛ t − t0 ⎞ ⎠ 2 ⎝ The streaklines are ⎡⎢ 3 t 3 ⎥⎤ t 0 2 x ( t 0 ) = x 0 + a⋅ y 0 ⋅ ( t − t 0 ) + a⋅ ⋅ ⎢ − − t0 ⋅ ( t − t0 )⎥ 3 2 ⎣3 ⎦ b 2 2 y t0 = y 0 + ⋅ ⎛ t − t0 ⎞ ⎝ ⎠ 2 ( ) b These gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t) For a particle that was at x 0 = 1 m, y 0 = 2 m at t0 = 0s, at time t = 2 s we find the position is (from pathline equations) ⎡⎢ 3 t 3 ⎥⎤ t 0 2 x = x 0 + a⋅ y 0 ⋅ ( t − t 0 ) + a⋅ ⋅ ⎢ − − t0 ⋅ ( t − t0 )⎥ = 1.91m 3 2 ⎣3 ⎦ b 2 2 y = y 0 + ⋅ ⎛ t − t0 ⎞ = 2.8 m ⎝ ⎠ 2 b For a particle that was at x 0 = 1 m, y 0 = 2 m at t0 = 2 s, at time t = 3 s we find the position is ⎡⎢ 3 t 3 ⎥⎤ t 0 2 x = x 0 + a⋅ y 0 ⋅ ( t − t 0 ) + a⋅ ⋅ ⎢ − − t0 ⋅ ( t − t0 )⎥ = 1.49m 3 2 ⎣3 ⎦ b 2 2 y = y 0 + ⋅ ⎛ t − t0 ⎞ = 3.0 ⎠ 2 ⎝ b For streamlines v u So, separating variables = dy y ⋅ dy = 2 Integrating = dx b a y − y0 ⋅ t⋅ dx 2 = 2 The streamlines are then y = b⋅ t a⋅ y 2 y0 + where we treat t as a constant b⋅ t a ( ⋅ x − x0 2⋅ b⋅ t a ( ) and we have ) ⋅ x − x0 = x0 = 1 m 4 ⋅ t⋅ ( x − 1) + 4 y0 = 2 m m Streamline Plots Pathline Plots 5 15 Pathline (t0=0) Pathline (t0=2) Streakline 12 3 y (m) y (m) 4 9 2 6 1 3 0 0.6 1.2 x (m) Streamline (t=0) Streamline (t=1) Streamline (t=2) Streamline (t=3) 1.8 2.4 3 0 2 4 6 x (m) 8 10 Problem 2.36 [Difficulty: 4] Given: Velocity field Find: Coordinates of particle at t = 2 s that was at (2,1) at t = 0; coordinates of particle at t = 3 s that was at (2,1) at t = 2 s; plot pathline and streakline through point (2,1) and compare with streamlines through same point at t = 0, 1 and 2 s Solution: Governing equations: For pathlines up = dx vp = dt dy v For streamlines dt u = dy dx Following the discussion leading up to Eq. 2.10, we first find equations for the pathlines in form ( x p( t) = x t , x 0 , y 0 , t0 ( ) ) ( x st t0 = x t , x 0 , y 0 , t0 ) ( ) and y p( t) = y t , x 0 , y 0 , t0 and y st t0 = y t , x 0 , y 0 , t0 ( ) ( ) which gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t) Assumption: 2D flow Given data m a = 0.4 2 b = 2 s Hence for pathlines up = dx dt m 2 s = a⋅ t vp = dy dt =b Hence dx = a⋅ t⋅ dt dy = b ⋅ dt Integrating a 2 2 x − x0 = ⋅ ⎛ t − t0 ⎞ ⎝ ⎠ 2 y − y0 = b⋅ t − t0 The pathlines are a 2 2 x ( t) = x0 + ⋅ ⎛ t − t0 ⎞ ⎠ 2 ⎝ y ( t) = y0 + b⋅ t − t0 ( ) ( ) ( ) These give the position (x,y) at any time t of a particle that was at (x 0,y 0) at time t0 Note that streaklines are obtained using the logic of the Governing equations, above The streaklines are a 2 2 x t0 = x 0 + ⋅ ⎛ t − t0 ⎞ ⎠ 2 ⎝ ( ) ( ) y t0 = y 0 + b ⋅ t − t0 These gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t) For a particle that was at x 0 = 2 m, y 0 = 1 m at t0 = 0s, at time t = 2 s we find the position is (from pathline equations) a 2 2 x = x 0 + ⋅ ⎛ t − t0 ⎞ = 2.8 m ⎠ 2 ⎝ ( ) y = y 0 + b ⋅ t − t0 = 5 m For a particle that was at x 0 = 2 m, y 0 = 1 m at t0 = 2 s, at time t = 3 s we find the position is a 2 2 x = x 0 + ⋅ ⎛ t − t0 ⎞ = 3 m ⎝ ⎠ 2 v For streamlines u = dy dx b = ( b a⋅ t So, separating variables dy = Integrating b y − y0 = ⋅ x − x0 a⋅ t The streamlines are then b 5⋅ ( x − 2) y = y0 + ⋅ x − x0 = +1 a⋅ t t a⋅ t ) y = y 0 + b ⋅ t − t0 = 3 m ⋅ dx where we treat t as a constant ( ( ) and we have x0 = 2 m m ) Pathline Plots Streamline Plots 8 8 Pathline (t0=0) Pathline (t0=2) Streakline Streamline (t=0) Streamline (t=1) Streamline (t=2) 6 y (m) 6 y (m) y0 = 1 4 2 4 2 0 1 2 3 x (m) 4 5 0 1 2 3 x (m) 4 5 Problem 2.37 [Difficulty: 2] Given: Sutherland equation Find: Corresponding equation for kinematic viscosity 1 Solution: Governing equation: μ b T 2 1 S p ρ R T Sutherland equation Ideal gas equation T Assumptions: Sutherland equation is valid; air is an ideal gas The given data is 6 b 1.458 10 kg m s K The kinematic viscosity is where ν b' μ ρ μ R T p S 110.4 K 1 R 286.9 1 3 3 2 2 2 R T b T R b T b' T p p S S S 1 1 1 T T T b' 4.129 10 p 2 9 m 1.5 K N m kg K 1.458 10 6 s 2 kg 1 m s K m 3 4.129 10 ν b' T 2 1 S with T b' 4.129 10 9 2 m 3 101.3 10 N 2 s K 3 Hence p 101.3 kPa 2 R b b' 286.9 J kg K 9 2 m 3 s K 2 S 110.4 K 2 Check with Appendix A, Table A.10. At T 0 °C we find 2 5 m T 273.1 K ν 1.33 10 s 3 2 9 m 4.129 10 3 s K ν 1 ( 273.1 K) 2 2 2 5 m ν 1.33 10 110.4 Check! s 273.1 At T 100 °C we find 2 5 m T 373.1 K ν 2.29 10 s 3 2 9 m 4.129 10 3 s K ν 1 ( 373.1 K) 2 2 2 5 m ν 2.30 10 110.4 Check! s 373.1 Viscosity as a Function of Temperature 5 2.5 10 Kinematic Viscosity (m2/s) Calculated Table A.10 2 10 5 5 1.5 10 0 20 40 60 Temperature (C) 80 100 Problem 2.38 Given: Sutherland equation with SI units Find: Corresponding equation in BG units 1 Solution: Governing equation: [Difficulty: 2] μ= b⋅ T 2 1+ S Sutherland equation T Assumption: Sutherland equation is valid The given data is −6 b = 1.458 × 10 kg ⋅ S = 110.4 ⋅ K 1 m⋅ s⋅ K 2 1 Converting constants −6 b = 1.458 × 10 kg ⋅ 1 m⋅ s⋅ K Alternatively b = 2.27 × 10 −8 Also S = 110.4 ⋅ K × lbm 0.454 ⋅ kg × slug 32.2⋅ lbm μ= b⋅ T 2 1+ S ft × ⎛ 5⋅ K ⎞ ⎜ 9⋅ R ⎝ ⎠ 2 −8 b = 2.27 × 10 1 ⋅ × slug 1 ft⋅ s⋅ R lbf ⋅ s −8 b = 2.27 × 10 slug⋅ ft 2 ⋅ 1 2 S = 198.7 ⋅ R 5⋅ K lbf ⋅ s ft 2 2 lbf ⋅ s ft ⋅ R 9⋅ R with T in Rankine, µ in T 0.3048⋅ m 2 1 and × 2 slug ft⋅ s⋅ R × 2 Check with Appendix A, Table A.9. At T = 68 °F we find T = 527.7 ⋅ R μ = 3.79 × 10 − 7 lbf ⋅ s ⋅ ft 1 −8 2.27 × 10 lbf ⋅ s 1 2 ft ⋅ R μ = 1+ × ( 527.7 ⋅ R) 2 2 2 μ = 3.79 × 10 198.7 − 7 lbf ⋅ s ⋅ ft Check! 2 527.7 At T = 200 °F we find T = 659.7 ⋅ R μ = 4.48 × 10 − 7 lbf ⋅ s ⋅ ft 2 1 −8 2.27 × 10 lbf ⋅ s 1 2 μ = ft ⋅ R 1+ × ( 659.7 ⋅ R) 2 2 198.7 659.7 μ = 4.48 × 10 − 7 lbf ⋅ s ⋅ ft 2 Check! Data: Using procedure of Appendix A.3: T (oC) 0 100 200 300 400 µ(x10 ) 1.86E-05 2.31E-05 2.72E-05 3.11E-05 3.46E-05 5 T (K) 273 373 473 573 673 T (K) 273 373 473 573 673 3/2 T /µ 2.43E+08 3.12E+08 3.78E+08 4.41E+08 5.05E+08 The equation to solve for coefficients S and b is 3 2 T µ S ⎛1 ⎞ = ⎜ ⎟T + b b ⎝ ⎠ From the built-in Excel Linear Regression functions: Hence: b = 1.531E-06 S = 101.9 Slope = 6.534E+05 Intercept = 6.660E+07 . . 1/2 kg/m s K K 2 R = 0.9996 Plot of Basic Data and Trend Line 6.E+08 Data Plot 5.E+08 Least Squares Fit 4.E+08 T3/2/µ 3.E+08 2.E+08 1.E+08 0.E+00 0 100 200 300 400 T 500 600 700 800 Problem 2.40 [Difficulty: 2] Given: Velocity distribution between flat plates Find: Shear stress on upper plate; Sketch stress distribution Solution: Basic equation du τyx = μ⋅ dy τyx = − du dy Hence y= dy ⎡ u max⋅ ⎢1 − ⎣ 2 ⎛ 2 ⋅ y ⎞ ⎥⎤ = u ⋅ ⎛ − 4 ⎞ ⋅ 2⋅ y = − 8 ⋅ umax⋅ y ⎜ h max ⎜ 2 2 ⎝ ⎠⎦ h ⎝ h ⎠ 8 ⋅ μ⋅ u max⋅ y h At the upper surface d = h 2 and 2 τyx = −8 × 1.14 × 10 − 3 N⋅ s ⋅ 2 h = 0.1⋅ mm × 0.1⋅ m m s × 0.1 2 u max = 0.1⋅ ⋅ mm × 1⋅ m 1000⋅ mm × m s − 3 N⋅ s μ = 1.14 × 10 ⋅ (Table A.8) 2 m 2 ⎛ 1 × 1000⋅ mm ⎞ ⎜ 0.1⋅ mm 1⋅ m ⎠ ⎝ N τyx = −4.56⋅ 2 m The upper plate is a minus y surface. Since τyx < 0, the shear stress on the upper plate must act in the plus x direction. ⎛ 8 ⋅ μ⋅ umax ⎞ ⋅y ⎜ h2 ⎝ ⎠ τyx( y ) = −⎜ The shear stress varies linearly with y 0.05 0.04 0.03 y (mm) 0.02 0.01 −5 −4 −3 −2 −1 0 1 − 0.01 − 0.02 − 0.03 − 0.04 − 0.05 Shear Stress (Pa) 2 3 4 5 Problem 2.41 [Difficulty: 2] Given: Velocity distribution between parallel plates Find: Force on lower plate Solution: Basic equations du dy so du τyx = μ⋅ dy F = τyx⋅ A = ⎡ d dy τyx = − u max⋅ ⎢1 − ⎣ 8 ⋅ μ⋅ u max⋅ y h At the lower surface y=− h F=− and 2 2 m 2 A = 1⋅ m − 3 N⋅ s μ = 1.14 × 10 − 3 N⋅ s ⋅ 2 m F = 2.28⋅ N ⋅ 2 m F = −8 × 1 ⋅ m × 1.14 × 10 (to the right) 2 2 h = 0.1⋅ mm s 8 ⋅ A⋅ μ⋅ u max⋅ y h and u max = 0.05⋅ Hence 2 ⎛ 2 ⋅ y ⎞ ⎥⎤ = u ⋅ ⎛ − 4 ⎞ ⋅ 2⋅ y = − 8 ⋅ umax⋅ y ⎜ h max ⎜ 2 2 ⎝ ⎠⎦ h ⎝ h ⎠ × 0.05⋅ (Table A.8) m s × −0.1 2 ⋅ mm × 1⋅ m 1000⋅ mm × 2 ⎛ 1 ⋅ 1 × 1000⋅ mm ⎞ ⎜ 0.1 mm 1⋅ m ⎠ ⎝ Problem 2.42 [Difficulty: 2] Open-Ended Problem Statement: Explain how an ice skate interacts with the ice surface. What mechanism acts to reduce sliding friction between skate and ice? Discussion: The normal freezing and melting temperature of ice is 0°C (32°F) at atmospheric pressure. The melting temperature of ice decreases as pressure is increased. Therefore ice can be caused to melt at a temperature below the normal melting temperature when the ice is subjected to increased pressure. A skater is supported by relatively narrow blades with a short contact against the ice. The blade of a typical skate is less than 3 mm wide. The length of blade in contact with the ice may be just ten or so millimeters. With a 3 mm by 10 mm contact patch, a 75 kg skater is supported by a pressure between skate blade and ice on the order of tens of megaPascals (hundreds of atmospheres). Such a pressure is enough to cause ice to melt rapidly. When pressure is applied to the ice surface by the skater, a thin surface layer of ice melts to become liquid water and the skate glides on this thin liquid film. Viscous friction is quite small, so the effective friction coefficient is much smaller than for sliding friction. The magnitude of the viscous drag force acting on each skate blade depends on the speed of the skater, the area of contact, and the thickness of the water layer on top of the ice. The phenomenon of static friction giving way to viscous friction is similar to the hydroplaning of a pneumatic tire caused by a layer of water on the road surface. Problem 2.43 [Difficulty: 2] Given: Velocity profile Find: Plot of velocity profile; shear stress on surface Solution: 2 ⎛ y ⎞ ⎜ ⋅ h⋅ y − ⋅ sin( θ) u= μ ⎝ 2 ⎠ 2 ρ⋅ g The velocity profile is u Hence we can plot u max ⎡y = 2⋅ ⎢ ⎣h − 1 2 ⋅ ⎛⎜ y⎞ u max = so the maximum velocity is at y = h ρ⋅ g h ⋅ ⋅ sin( θ) μ 2 2⎤ ⎥ ⎝h⎠ ⎦ 1 y/h 0.75 0.5 0.25 0 0.25 0.5 0.75 1 u/umax This graph can be plotted in Excel The given data is − 3 lbf ⋅ s h = 0.1⋅ in μ = 2.15 × 10 ⋅ ft τyx = μ⋅ dy At the surface y = 0 τyx = ρ⋅ g ⋅ h ⋅ sin( θ) Hence τyx = 0.85 × 1.94⋅ θ = 45⋅ deg ⎛ y ⎞ τyx = μ⋅ = μ⋅ ⋅ ⎜ h⋅ y − ⋅ sin( θ) = ρ⋅ g ⋅ ( h − y ) ⋅ sin( θ) dy dy μ ⎝ 2 ⎠ du Basic equation 2 du slug ft 3 × 32.2⋅ ft 2 s d ρ⋅ g × 0.1⋅ in × 1 ⋅ ft 12⋅ in 2 2 × sin( 45⋅ deg) × lbf ⋅ s slug⋅ ft The surface is a positive y surface. Since τyx > 0, the shear stress on the surface must act in the plus x direction. lbf τyx = 0.313 ⋅ 2 ft Problem 2.44 Given: Ice skater and skate geometry Find: Deceleration of skater [Difficulty: 2] τ yx = µ y Solution: Governing equation: du τyx = μ⋅ dy ΣFx = M ⋅ ax du dy V = 20 ft/s h x L Assumptions: Laminar flow The given data is W = 100 ⋅ lbf V = 20⋅ − 5 lbf ⋅ s μ = 3.68 × 10 ⋅ ft Then ft L = 11.5⋅ in s w = 0.125 ⋅ in Table A.7 @32oF 2 du V ft 1 12⋅ in − 5 lbf ⋅ s τyx = μ⋅ = μ⋅ = 3.68 × 10 ⋅ × 20⋅ × × dy h 2 s 0.0000575 ⋅ in ft ft lbf τyx = 154 ⋅ 2 ft Equation of motion ΣFx = M ⋅ ax ax = − τyx⋅ A⋅ g W ax = −154 lbf ft ax = −0.495 ⋅ −W τyx⋅ A = ⋅a g x or 2 ft 2 s =− τyx⋅ L⋅ w⋅ g W × 11.5⋅ in × 0.125 ⋅ in × 32.2⋅ ft 2 s × 1 100 ⋅ lbf × ft 2 ( 12⋅ in) 2 h = 0.0000575 ⋅ in Problem 2.45 Given: Block pulled up incline on oil layer Find: Force required to pull the block [Difficulty: 2] Solution: Governing equations: U du τyx = μ⋅ dy y x x f N W d ΣFx = M ⋅ ax θ Assumptions: Laminar flow The given data is W = 10⋅ lbf U = 2⋅ μ = 3.7 × 10 − 2 N⋅ s ⋅ ft w = 10⋅ in s d = 0.001 ⋅ in θ = 25⋅ deg Fig. A.2 @100 oF (38oC) 2 m Equation of motion ΣFx = M ⋅ ax = 0 The friction force is du U 2 f = τyx⋅ A = μ⋅ ⋅ A = μ⋅ ⋅ w dy d Hence F = f + W⋅ sin( θ) = μ⋅ − 2 N⋅ s F = 3.7 × 10 F = 17.1⋅ lbf F − f − W⋅ sin( θ) = 0 s o U d 2 ⋅ w + W⋅ sin( θ) 2 lbf ⋅ s m ft 1 ft 2 ⋅ × 0.0209⋅ ⋅ × 2⋅ × × ( 10⋅ in) × + 10⋅ lbf ⋅ sin( 25⋅ deg) 2 2 N⋅ s s 0.001 ⋅ in 12⋅ in m ft Problem 2.46 Given: Block moving on incline on oil layer Find: Speed of block when free, pulled, and pushed [Difficulty: 2] Solution: y U Governing equations: x x du τyx = μ⋅ dy f N W ΣFx = M ⋅ ax d θ Assumptions: Laminar flow The given data is M = 10⋅ kg W = M⋅ g W = 98.066 N d = 0.025 ⋅ mm θ = 30⋅ deg F = 75⋅ N − 1 N⋅s μ = 10 ⋅ w = 250 ⋅ mm Fig. A.2 SAE 10-39 @30oC 2 m Equation of motion ΣFx = M ⋅ ax = 0 The friction force is du U 2 f = τyx⋅ A = μ ⋅ ⋅ A = μ ⋅ ⋅ w dy d Hence for uphill motion F = f + W ⋅ sin ( θ) = μ ⋅ For no force: U = d ⋅ W⋅ sin( θ) 2 F − f − W ⋅ sin ( θ) = 0 so U d U = d ⋅ ( F − W⋅ sin( θ) ) 2 μ⋅ w U= d ⋅ ( F − W⋅ sin( θ) ) (For downpush change sign of W) 2 μ⋅ w U = 0.196 m U = 0.104 m μ⋅ w Pushing up: 2 ⋅ w + W ⋅ sin ( θ) s s Pushing down: U = d ⋅ ( F + W ⋅ sin ( θ) ) 2 μ⋅w U = 0.496 m s Problem 2.47 [Difficulty: 2] Given: Data on tape mechanism Find: Maximum gap region that can be pulled without breaking tape Solution: Basic equation du τyx μ dy F τyx A and FT 2 F 2 τyx A Here F is the force on each side of the tape; the total force is then The velocity gradient is linear as shown du dy V 0 c V c A w L Combining these results V FT 2 μ w L c L Solving for L c y The area of contact is t F,V x c FT c 2 μ V w L The given data is Hence FT 25 lbf c 0.012 in L 25 lbf 0.012 in 1 ft 12 in μ 0.02 1 2 1 ft s 0.02 slug slug ft s V 3 ft s w 1 in 1 s 1 1 12 in slug ft 3 ft 1 in 2 1 ft s lbf L 2.5 ft Problem 2.48 Given: Flow data on apparatus Find: The terminal velocity of mass m [Difficulty: 2] Solution: Given data: Dpiston = 73⋅ mm Dtube = 75⋅ mm Mass = 2 ⋅ kg Reference data: kg ρwater = 1000⋅ 3 m (maximum density of water) L = 100 ⋅ mm μ = 0.13⋅ From Fig. A.2:, the dynamic viscosity of SAE 10W-30 oil at 25oC is: SG Al = 2.64 N⋅ s 2 m The terminal velocity of the mass m is equivalent to the terminal velocity of the piston. At that terminal speed, the acceleration of the piston is zero. Therefore, all forces acting on the piston must be balanced. This means that the force driving the motion (i.e. the weight of mass m and the piston) balances the viscous forces acting on the surface of the piston. Thus, at r = Rpiston: 2 ⎞⎤ ⎡⎢ ⎛⎜ π⋅ D piston ⋅ L ⎥ ⎢Mass + SGAl⋅ ρwater⋅ ⎜ ⎥ ⋅ g = τrz⋅ A = 4 ⎣ ⎝ ⎠⎦ ⎛ μ⋅ d V ⎞ ⋅ π⋅ D ⎜ piston⋅ L) z ( ⎝ dr ⎠ The velocity profile within the oil film is linear ... d Vz = dr Therefore V ⎛ Dtube − Dpiston ⎞ ⎜ ⎝ ⎠ 2 Thus, the terminal velocity of the piston, V, is: g ⋅ ⎛ SG Al⋅ ρwater⋅ π⋅ Dpiston ⋅ L + 4 ⋅ Mass⎞ ⋅ Dtube − Dpiston ⎝ ⎠ 2 V = or V = 10.2 8 ⋅ μ⋅ π⋅ Dpiston⋅ L m s ( ) Problem 2.49 [Difficulty: 3] Given: Flow data on apparatus Find: Sketch of piston speed vs time; the time needed for the piston to reach 99% of its new terminal speed. Solution: Given data: Dpiston 73 mm Dtube 75 mm L 100 mm Reference data: kg ρwater 1000 3 m (maximum density of water) (From Problem 2.48) μ 0.13 From Fig. A.2, the dynamic viscosity of SAE 10W-30 oil at 25oC is: m V0 10.2 s SG Al 2.64 N s 2 m The free body diagram of the piston after the cord is cut is: Piston weight: 2 π D piston Wpiston SGAl ρwater g L 4 Viscous force: Fviscous( V) τrz A or Fviscous( V) μ 1 π Dpiston L Dtube Dpiston 2 dV mpiston Wpiston Fviscous( V) dt Applying Newton's second law: Therefore dV dt If g a V V g a V V where then The differential equation becomes a dX dt dX dt The solution to this differential equation is: 8 μ SGAl ρwater Dpiston Dtube Dpiston a dV dt a X X( t) X0 e a t where X( 0 ) g a V0 or g a V( t) g a V0 e a t Therefore g ( a t) g V( t) V0 e a a Plotting piston speed vs. time (which can be done in Excel) Piston speed vs. time 12 10 8 V ( t) 6 4 2 0 1 2 t The terminal speed of the piston, Vt, is evaluated as t approaches infinity g Vt a or m Vt 3.63 s The time needed for the piston to slow down to within 1% of its terminal velocity is: V g 0 a t ln a g 1.01 Vt a 1 or t 1.93 s 3 Problem 2.50 [Difficulty: 3] Given: Block on oil layer pulled by hanging weight Find: Expression for viscous force at speed V; differential equation for motion; block speed as function of time; oil viscosity Mg Solution: Governing equations: x y Ft du τyx = μ⋅ dy ΣFx = M ⋅ ax Ft Fv mg N Assumptions: Laminar flow; linear velocity profile in oil layer M = 5 ⋅ kg Equation of motion (block) ΣFx = M ⋅ ax so dV Ft − Fv = M ⋅ dt ( 1) Equation of motion (block) ΣFy = m⋅ ay so dV m⋅ g − Ft = m⋅ dt ( 2) Adding Eqs. (1) and (2) dV m⋅ g − Fv = ( M + m) ⋅ dt The friction force is du V Fv = τyx⋅ A = μ⋅ ⋅ A = μ⋅ ⋅ A dy h Hence m⋅ g − To solve separate variables W = m⋅ g = 9.81⋅ N μ⋅ A h M+m dt = m⋅ g − t=− Hence taking antilogarithms 1− ⋅ V = ( M + m) ⋅ μ⋅ A h ( M + m) ⋅ h μ⋅ A μ⋅ A m⋅ g ⋅ h A = 25⋅ cm 2 The given data is h = 0.05⋅ mm dV dt ⋅ dV ⋅V ⋅ ⎛⎜ ln⎜⎛ m⋅ g − ⎝ ⎝ − ⋅V = e μ⋅ A μ⋅ A ( M+ m) ⋅ h h ⋅t ⋅ V⎞ − ln( m⋅ g ) ⎞ = − ⎠ ⎠ ( M + m) ⋅ h μ⋅ A ⋅ ln⎛⎜ 1 − ⎝ μ⋅ A m⋅ g ⋅ h ⋅ V⎞ ⎠ ⎡ − m⋅ g ⋅ h ⎢ V= ⋅ ⎣1 − e Finally μ⋅ A ( M + m) ⋅ h μ⋅ A ⎤ ⎥ ⎦ ⋅t The maximum velocity is V = m⋅ g ⋅ h μ⋅ A In Excel: The data is M= m= 5.00 1.00 kg kg To find the viscosity for which the speed is 1 m/s after 1 s use Goal Seek with the velocity targeted to be 1 m/s by varying g= 0= 9.81 the viscosity in the set of cell below: 1.30 m/s2 N.s/m2 A= h= 25 0.5 cm 2 mm t (s) 1.00 V (m/s) 1.000 Speed V of Block vs Time t t (s) 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.10 2.20 2.30 2.40 2.50 2.60 2.70 2.80 2.90 3.00 V (m/s) 0.000 0.155 0.294 0.419 0.531 0.632 0.722 0.803 0.876 0.941 1.00 1.05 1.10 1.14 1.18 1.21 1.25 1.27 1.30 1.32 1.34 1.36 1.37 1.39 1.40 1.41 1.42 1.43 1.44 1.45 1.46 1.6 1.4 1.2 1.0 V (m/s) 0.8 0.6 0.4 0.2 0.0 0.0 0.5 1.0 1.5 t (s) 2.0 2.5 3.0 Problem 2.51 [Difficulty: 4] Ff = τ⋅ A x, V, a M⋅ g Given: Data on the block and incline Find: Initial acceleration; formula for speed of block; plot; find speed after 0.1 s. Find oil viscosity if speed is 0.3 m/s after 0.1 s Solution: Given data M = 5 ⋅ kg From Fig. A.2 μ = 0.4⋅ A = ( 0.1⋅ m) 2 d = 0.2⋅ mm θ = 30⋅ deg N⋅ s 2 m Applying Newton's 2nd law to initial instant (no friction) so M ⋅ a = M ⋅ g ⋅ sin( θ) − Ff = M ⋅ g ⋅ sin( θ) ainit = g ⋅ sin( θ) = 9.81⋅ m 2 × sin( 30⋅ deg) s M ⋅ a = M ⋅ g ⋅ sin( θ) − Ff Applying Newton's 2nd law at any instant so M⋅ a = M⋅ dV g ⋅ sin( θ) − − Integrating and using limits or V = 5 ⋅ kg × 9.81⋅ m 2 s V( 0.1⋅ s) = 0.404 ⋅ m s M⋅ d μ⋅ A μ⋅ A M⋅ d ⋅ ln⎛⎜ 1 − ⎝ m 2 s du V Ff = τ⋅ A = μ⋅ ⋅ A = μ⋅ ⋅ A dy d and μ⋅ A = M ⋅ g ⋅ sin( θ) − dV Separating variables At t = 0.1 s dt ainit = 4.9 d ⋅V = dt ⋅V μ⋅ A M ⋅ g ⋅ d ⋅ sin( θ) ⋅ V⎞ = t ⎠ − μ⋅ A ⎞ ⎛ ⋅t ⎜ M ⋅ g ⋅ d ⋅ sin( θ) M⋅ d V( t) = ⋅⎝1 − e ⎠ μ⋅ A 2 × 0.0002⋅ m⋅ sin( 30⋅ deg) × m 0.4⋅ N⋅ s⋅ ( 0.1⋅ m) ⎛ 0.4⋅ 0.01 ⋅ 0.1⎞⎤ ⎡ −⎜ ⎢ N⋅ s 5⋅ 0.0002 ⎠⎥ × × ⎣1 − e ⎝ ⎦ 2 2 kg⋅ m The plot looks like V (m/s) 1.5 1 0.5 0 0.2 0.4 0.6 0.8 t (s) To find the viscosity for which V(0.1 s) = 0.3 m/s, we must solve − μ⋅ A ⎤ ⎡ ⋅ ( t= 0.1⋅ s ) ⎢ ⎥ M ⋅ g ⋅ d ⋅ sin( θ) M⋅ d V( t = 0.1⋅ s) = ⋅ ⎣1 − e ⎦ μ⋅ A The viscosity µ is implicit in this equation, so solution must be found by manual iteration, or by any of a number of classic root-finding numerical methods, or by using Excel's Goal Seek Using Excel: μ = 1.08⋅ N⋅ s 2 m 1 Problem 2.52 [Difficulty: 3] Given: Block sliding on oil layer Find: Direction of friction on bottom of block and on plate; expression for speed U versus time; time required to lose 95% of initial speed Solution: U Governing equations: du τyx = μ⋅ dy ΣFx = M ⋅ ax Fv y h Assumptions: Laminar flow; linear velocity profile in oil layer x The bottom of the block is a -y surface, so τyx acts to the left; The plate is a +y surface, so τyx acts to the right Equation of motion ΣFx = M ⋅ ax The friction force is du U 2 Fv = τyx⋅ A = μ⋅ ⋅ A = μ⋅ ⋅ a dy h Hence − 2 1 U dU ⋅ U = M⋅ h ⋅ dU = − μ⋅ a dt 2 ⋅ dt M⋅ h 2 ⎞ = − μ⋅ a ⋅ t M⋅ h ⎝ U0 ⎠ ln⎛⎜ dU Fv = M ⋅ dt U U To solve separate variables μ⋅ a so 2 − Hence taking antilogarithms U = U0 ⋅ e μ⋅ a M⋅ h ⋅t t t=− Solving for t M⋅ h μ⋅ a Hence for U U0 = 0.05 t = 3.0⋅ 2 ⋅ ln⎛⎜ ⎞ ⎝ U0 ⎠ M⋅ h μ⋅ a U 2 Problem 2.53 Given: Varnish-coated wire drawn through die Find: Force required to pull wire [Difficulty: 2] r F x D d Solution: Governing equations: du τyx = μ⋅ dy ΣFx = M ⋅ ax L Assumptions: Laminar flow; linear velocity profile in varnish layer The given data is D = 1 ⋅ mm d = 0.9⋅ mm L = 50⋅ mm Equation of motion ΣFx = M ⋅ ax F − Fv = 0 The friction force is du V Fv = τyx⋅ A = μ⋅ ⋅ A = μ⋅ ⋅ π⋅ d ⋅ L dr ⎛D − d⎞ so V = 50⋅ −2 m μ = 20 × 10 s poise for steady speed ⎜ ⎝ 2 ⎠ Hence F − Fv = 0 so F = 2 ⋅ π⋅ μ⋅ V⋅ d ⋅ L D− d −2 F = 2 ⋅ π × 20 × 10 F = 2.83 N poise × 0.1⋅ kg m⋅ s⋅ poise × 50⋅ m s × 0.9⋅ mm × 50⋅ mm × 1 ( 1 − 0.9) ⋅ mm × m 1000⋅ mm Problem 2.54 [Difficulty: 3] Given: Data on annular tube Find: Whether no-slip is satisfied; location of zeroshear stress; viscous forces Solution: The velocity profile is Check the no-slip condition. When 2 2 Ro Ri r 2 2 u z( r) ln Ri r 4 μ L Ri Ri ln Ro ∆p 1 2 2 Ro Ri Ro 2 2 u z R o ln Ri Ro 4 μ L Ri Ri ln Ro 1 r Ro ∆p 1 ∆p 2 2 2 2 u z Ro R Ro Ro Ri 0 4 μ L i When r Ri 2 2 Ro Ri Ri 2 2 u z R i ln Ri Ri 0 Ri 4 μ L Ri ln Ro 1 ∆p The no-slip condition is satisfied. The given data is The viscosity of the honey is Ri 5 mm Ro 25 mm μ 5 N s 2 m ∆p 125 kPa L 2 m The plot looks like Radial Position (mm) 25 20 15 10 5 0 0.25 0.5 0.75 Velocity (m/s) For each, shear stress is given by du τrx μ dr τrx μ duz( r) dr 2 2 Ro Ri 1 ∆p 2 r 2 ln μ Ri r dr 4 μ L Ri Ri ln Ro d Ro Ri 1 ∆p τrx 2 r 4 L Ri ln r Ro 2 Hence 2 For zero stress 2 r Ro Ri 2 2 Ri ln r Ro 2 0 r or 2 2 Ro Ri 2 Fo ∆p π Ro Ri 2 ln Ro 2 2 Ri 2 ln Ro Ro Ri 1 ∆p Fo τrx A 2 Ro 2 π R o L 4 L Ri ln Ro Ro 2 On the outer surface Ri Ro r 13.7 mm 2 2 1 m 2 ( 25 mm) ( 5 mm) 1 m 3 N 1000 mm Fo 125 10 π 25 mm 2 1000 mm 5 m 2 ln 25 Fo 172 N Ro Ri 1 ∆p Fi τrx A 2 Ri 2 π R i L 4 L Ri ln Ri Ro 2 On the inner surface 2 2 2 Ro Ri 2 Fi ∆p π Ri Ri 2 ln Ro Hence 2 2 2 1 m ( 25 mm ) ( 5 mm ) 2 1 m 3 N 1000 mm Fi 125 10 π 5 mm 2 1000 mm 5 m 2 ln 25 Fi 63.4 N Note that Fo Fi 236 N and ∆p π Ro Ri 2 2 236 N The net pressure force just balances the net viscous force! Problem 2.55 [Difficulty: 3] Given: Data on flow through a tube with a filament Find: Whether no-slip is satisfied; location of zero stress;stress on tube and filament Solution: V( r) The velocity profile is Check the no-slip condition. When r r d 2 ∆p 2 ln V D 2 1 ∆p 16 μ L ∆p d ln 2 2 2 2 D d d d 2 2 2 r d 16 μ L 1 d D D d 16 μ L 1 2 D 2 V( d ) 2 D d 2 d 4 r D V( D) When 16 μ L 1 ∆p 2 2 d D 2 D d ln 2 d D ln D d 0 2 2 d ln D ln d d 0 The no-slip condition is satisfied. The given data is d 1 μm The viscosity of SAE 10-30 oil at 100 oC is (Fig. A.2) D 20 mm ∆p 5 kPa 2 N s μ 1 10 2 m L 10 m The plot looks like Radial Position (mm) 10 8 6 4 2 0 0.25 0.5 0.75 1 Velocity (m/s) du τrx μ dr For each, shear stress is given by dV( r) d τrx μ μ dr dr 2 2 1 ∆p 2 D d 2 r 2 ln d 4 r 16 μ L d Di ln D 1 ∆p D d τrx( r) 8 r 16 L d ln r D 2 2 8 r For the zero-stress point D d 2 2 d ln r D 2 0 or r 2 d D d 8 ln D r 2.25 mm Radial Position (mm) 10 7.5 5 2.5 3 2 1 0 1 2 3 4 Stress (Pa) Using the stress formula D τrx 2.374 Pa 2 d τrx 2.524 kPa 2 Problem 2.56 Given: Flow between two plates Find: Force to move upper plate; Interface velocity [Difficulty: 2] Solution: The shear stress is the same throughout (the velocity gradients are linear, and the stresses in the fluid at the interface must be equal and opposite). Hence du1 du2 τ μ1 μ2 dy dy Solving for the interface velocity V i Then the force required is Vi Vi V Vi μ1 μ2 h1 h2 or V μ1 h 2 1 μ2 h 1 1 where V i is the interface velocity m s 0.1 0.3 1 0.15 0.5 m Vi 0.714 s Vi N s m 1 1000 mm 2 F τ A μ1 A 0.1 0.714 1 m h1 2 s 0.5 mm 1 m m F 143 N Problem 2.57 [Difficulty: 2] Given: Flow of three fluids between two plates Find: Upper plate velocity; Interface velocities; plot velocity distribution Solution: The shear stress is the same throughout (the velocity gradients are linear, and the stresses in the fluids at the interfaces must be equal and opposite). F 100 N Given data h 1 0.5 mm 2 A 1 m μ1 0.15 h 2 0.25 mm N s 2 μ2 0.5 m τ The (constant) stress is τ μ For each fluid ∆V or ∆y τ h 1 V12 μ1 τ h 2 V23 V12 μ2 τ h 3 V V23 μ3 Hence Hence Hence F 2 μ3 0.2 m N s 2 m τ 100 Pa A ∆V N s h 3 0.2 mm τ ∆y μ where ΔV is the overall change in velocity over distance Δy m V12 0.333 s where V 12 is the velocity at the 1 - 2 interface m V23 0.383 s where V 23 is the velocity at the 2 - 3 interface V 0.483 m where V is the velocity at the upper plate s 1 Position (mm) 0.75 0.5 0.25 0 0.1 0.2 0.3 Velocity (m/s) 0.4 0.5 Problem 2.58 [Difficulty: 2] Problem 2.59 [Difficulty: 2] Problem 2.60 [Difficulty: 2] Problem 2.61 Given: Data on the viscometer Find: Time for viscometer to lose 99% of speed [Difficulty: 3] Solution: The given data is R = 50⋅ mm H = 80⋅ mm 2 a = 0.20⋅ mm I = 0.0273⋅ kg⋅ m μ = 0.1⋅ N⋅ s 2 m I⋅ α = Torque = −τ⋅ A⋅ R The equation of motion for the slowing viscometer is where α is the angular acceleration and τ is the viscous stress, and A is the surface area of the viscometer The stress is given by τ = μ⋅ du dy = μ⋅ V− 0 μ⋅ V = a a = μ⋅ R⋅ ω a where V and ω are the instantaneous linear and angular velocities. Hence Separating variables μ⋅ R⋅ ω I⋅ α = I⋅ dω dω μ⋅ R ⋅ A ω dt =− a 2 μ⋅ R ⋅ A ⋅ A⋅ R = a ⋅ω 2 =− a⋅ I ⋅ dt 2 − Integrating and using IC ω = ω0 μ⋅ R ⋅ A ω( t) = ω0 ⋅ e a⋅ I ⋅t 2 − A = 2 ⋅ π⋅ R ⋅ H a⋅ I 0.01⋅ ω0 = ω0 ⋅ e The time to slow down by 99% is obtained from solving Note that μ⋅ R ⋅ A t=− so a⋅ I 2 ⋅ ln( 0.01) μ⋅ R ⋅ A a⋅ I t=− so ⋅t ⋅ ln( 0.01) 3 2 ⋅ π⋅ μ⋅ R ⋅ H 2 t = − 0.0002⋅ m⋅ 0.0273⋅ kg⋅ m 2⋅ π 2 ⋅ m 0.1⋅ N⋅ s ⋅ 1 ( 0.05⋅ m) ⋅ 1 2 ⋅ N⋅ s 3 0.08⋅ m kg⋅ m ⋅ ln( 0.01) t = 4.00 s Problem 2.62 Difficulty: [2] Problem 2.63 [Difficulty: 4] Problem 2.64 [Difficulty: 3] Given: Shock-free coupling assembly Find: Required viscosity Solution: du τrθ = μ⋅ dr Basic equation Shear force F = τ⋅ A V2 = ω2(R + δ) δ τrθ = μ⋅ V1 = ω1R P = T⋅ ω2 = F⋅ R⋅ ω2 = τ⋅ A2 ⋅ R⋅ ω2 = P= Hence ( P = T⋅ ω Power ⎡⎣ω1⋅ R − ω2 ⋅ ( R + δ)⎤⎦ du ∆V τrθ = μ⋅ = μ⋅ = μ⋅ dr ∆r δ Assumptions: Newtonian fluid, linear velocity profile Then Torque T = F⋅ R ) ( ) μ⋅ ω1 − ω2 ⋅ R δ (ω1 − ω2)⋅ R Because δ << R δ ⋅ 2 ⋅ π⋅ R⋅ L⋅ R⋅ ω2 3 2 ⋅ π⋅ μ⋅ ω2 ⋅ ω1 − ω2 ⋅ R ⋅ L δ P⋅ δ μ= ( ) 3 2 ⋅ π⋅ ω2 ⋅ ω1 − ω2 ⋅ R ⋅ L μ = 10⋅ W × 2.5 × 10 μ = 0.202 ⋅ 2⋅ π N⋅ s 2 m −4 ⋅m × 1 ⋅ min 9000 rev μ = 2.02⋅ poise × 1 ⋅ min 1000 rev × 1 ( .01⋅ m) 3 × 1 0.02⋅ m × N⋅ m s⋅ W 2 × ⎛ rev ⎞ × ⎛ 60⋅ s ⎞ ⎜ 2 ⋅ π⋅ rad ⎜ ⎝ ⎠ ⎝ min ⎠ which corresponds to SAE 30 oil at 30oC. 2 Problem 2.65 [Difficulty: 4] Part 1/2 Problem 2.65 [Difficulty: 4] Part 2/2 Problem 2.66 [Difficulty: 4] Problem 2.67 [Difficulty: 4] The data is N (rpm) µ (N·s/m ) 10 0.121 20 0.139 30 0.153 40 0.159 50 0.172 60 0.172 70 0.183 80 0.185 2 The computed data is ω (rad/s) ω/θ (1/s) η (N·s/m x10 ) 1.047 120 121 2.094 240 139 3.142 360 153 4.189 480 159 5.236 600 172 6.283 720 172 7.330 840 183 8.378 960 185 2 3 From the Trendline analysis k = 0.0449 n - 1 = 0.2068 n = 1.21 The fluid is dilatant The apparent viscosities at 90 and 100 rpm can now be computed N (rpm) ω (rad/s) 90 9.42 100 10.47 ω/θ (1/s) 1080 1200 η (N·s/m x10 ) 191 195 2 3 Viscosity vs Shear Rate 2 3 η (N.s/m x10 ) 1000 Data Power Trendline 100 η = 44.94(ω/θ) 2 R = 0.9925 0.2068 10 100 1000 Shear Rate ω/θ (1/s) Problem 2.69 [Difficulty: 4] Given: Data on insulation material Find: Type of material; replacement material Solution: The velocity gradient is du/dy = U/ δ Data and computations where δ = τ (Pa) 50 100 150 163 171 170 202 246 349 444 U (m/s) 0.000 0.000 0.000 0.005 0.01 0.03 0.05 0.1 0.2 0.3 0.001 m du/dy (s-1) 0 0 0 5 10 25 50 100 200 300 Hence we have a Bingham plastic, with τy = 154 µp = 0.963 Pa 2 N·s/m At τ = 450 Pa, based on the linear fit du/dy = 307 s For a fluid with τy = 250 Pa -1 we can use the Bingham plastic formula to solve for µ p given τ , τ y and du/dy from above µp = N·s/m 0.652 2 Shear Stress vs Shear Strain 500 450 τ (Pa) 400 350 300 250 Linear data fit: τ = 0.9632(du/dy ) + 154.34 2 R = 0.9977 200 150 100 50 0 0 50 100 150 200 du/dy (1/s) 250 300 350 Problem 2.70 [Difficulty: 3] Given: Viscometer data Find: Value of k and n in Eq. 2.17 Solution: τ (Pa) du/dy (s-1) 0.0457 0.119 0.241 0.375 0.634 1.06 1.46 1.78 5 10 25 50 100 200 300 400 Shear Stress vs Shear Strain 10 Data Power Trendline τ (Pa) The data is 1 1 10 100 τ = 0.0162(du/dy)0.7934 2 R = 0.9902 0.1 0.01 du/dy (1/s) k = 0.0162 n = 0.7934 Hence we have The apparent viscosity from Blood is pseudoplastic (shear thinning) η = -1 du/dy (s ) η (N·s/m ) 2 5 10 25 50 100 200 300 400 0.0116 0.0101 0.0083 0.0072 0.0063 0.0054 0.0050 0.0047 k (du/dy )n -1 2 o µ water = 0.001 N·s/m at 20 C Hence, blood is "thicker" than water! 1000 Problem 2.71 [Difficulty: 5] Problem 2.72 [Difficulty: 5] Problem 2.73 [Difficulty: 4] Given: Conical bearing geometry Find: Expression for shear stress; Viscous torque on shaft Solution: Basic equation ds τ = μ⋅ du dT = r⋅ τ⋅ dA dy dz z Infinitesimal shear torque AA r Assumptions: Newtonian fluid, linear velocity profile (in narrow clearance gap), no slip condition Section AA tan( θ) = r r = z⋅ tan( θ) so z U = ωr a Then τ = μ⋅ du dy = μ⋅ ∆u ∆y = μ⋅ ( ω⋅ r − 0 ) (a − 0) = μ⋅ ω⋅ z⋅ tan( θ) a As we move up the device, shear stress increases linearly (because rate of shear strain does) But from the sketch dz = ds ⋅ cos( θ) The viscous torque on the element of area is dA = 2 ⋅ π⋅ r⋅ ds = 2 ⋅ π⋅ r⋅ dT = r⋅ τ⋅ dA = r⋅ T= Solving for µ μ= Using given data cos( θ) μ⋅ ω⋅ z⋅ tan( θ) a 3 Integrating and using limits z = H and z = 0 dz ⋅ 2 ⋅ π⋅ r⋅ 3 dz dT = cos( θ) 2 ⋅ π⋅ μ⋅ ω⋅ z ⋅ tan( θ) a⋅ cos( θ) 4 π⋅ μ⋅ ω⋅ tan( θ) ⋅ H 2 ⋅ a⋅ cos( θ) 2 ⋅ a⋅ cos( θ) ⋅ T 3 4 π⋅ ω⋅ tan( θ) ⋅ H H = 25⋅ mm θ = 30⋅ deg μ = a = 0.2⋅ mm 2 ⋅ a⋅ cos( θ) ⋅ T 3 4 π⋅ ω⋅ tan( θ) ⋅ H From Fig. A.2, at 20oC, CASTOR OIL has this viscosity! ω = 75⋅ rev T = 0.325 ⋅ N⋅ m s μ = 1.012 ⋅ N⋅ s 2 m 3 ⋅ dz Problem 2.74 [Difficulty: 5] Problem 2.75 [Difficulty: 5] Problem 2.76 Given: [Difficulty: 5] Geometry of rotating bearing Find: Expression for shear stress; Maximum shear stress; Expression for total torque; Total torque Solution: τ = μ⋅ Basic equation du dT = r⋅ τ⋅ dA dy Assumptions: Newtonian fluid, narrow clearance gap, laminar motion From the figure h = a + R⋅ ( 1 − cos( θ) ) dA = 2 ⋅ π⋅ r⋅ dr = 2 ⋅ π R⋅ sin( θ) ⋅ R⋅ cos( θ) ⋅ dθ To find the maximum τ set du dy = dy = u−0 u = ω⋅ r = ω⋅ R⋅ sin( θ) τ = μ⋅ Then du r = R⋅ sin( θ) h u = h μ⋅ ω⋅ R⋅ sin( θ) a + R⋅ ( 1 − cos( θ) ) d ⎡ μ⋅ ω⋅ R⋅ sin( θ) ⎤ ⎢ ⎥=0 dθ ⎣ a + R⋅ ( 1 − cos( θ) ) ⎦ R⋅ cos( θ) − R + a⋅ cos( θ) = 0 R⋅ μ⋅ ω⋅ ( R⋅ cos( θ) − R + a⋅ cos( θ) ) so ( R + a − R⋅ cos( θ) ) θ = acos⎛⎜ 2 ⎞ = acos⎛ 75 ⎞ ⎜ 75 + 0.5 ⎝ R + a⎠ ⎝ ⎠ R =0 θ = 6.6⋅ deg kg τ = 12.5⋅ poise × 0.1⋅ τ = 79.2⋅ m⋅ s poise 2 × 2 ⋅ π⋅ 70 rad 1 N⋅ s ⋅ × 0.075 ⋅ m × sin( 6.6⋅ deg) × × 60 s [ 0.0005 + 0.075 ⋅ ( 1 − cos( 6.6⋅ deg) ) ] ⋅ m m⋅ kg N 2 m The torque is ⌠ T = ⎮ r⋅ τ⋅ A dθ = ⌡ θ ⌠ max 4 2 μ⋅ ω⋅ R ⋅ sin( θ) ⋅ cos( θ) ⎮ dθ ⎮ a + R⋅ ( 1 − cos( θ) ) ⌡ wher e ⎛ R0 ⎞ θmax = asin⎜ ⎝R⎠ 0 This integral is best evaluated numerically using Excel, Mathcad, or a good calculator T = 1.02 × 10 −3 ⋅ N⋅ m θmax = 15.5⋅ deg Problem 2.77 [Difficulty: 2] Problem 2.78 Given: Data on size of various needles Find: Which needles, if any, will float [Difficulty: 2] Solution: For a steel needle of length L, diameter D, density ρs, to float in water with surface tension σ and contact angle θ, the vertical force due to surface tension must equal or exceed the weight 2 2 ⋅ L⋅ σ⋅ cos( θ) ≥ W = m⋅ g = 4 ⋅ ρs⋅ L⋅ g π⋅ SG ⋅ ρ⋅ g θ = 0 ⋅ deg m 8 ⋅ σ⋅ cos( θ) π⋅ ρs⋅ g and for water ρ = 1000⋅ kg 3 m SG = 7.83 From Table A.1, for steel 8 ⋅ σ⋅ cos( θ) ⋅ D≤ or −3 N σ = 72.8 × 10 From Table A.4 Hence π⋅ D = 8 π⋅ 7.83 × 72.8 × 10 −3 N ⋅ m 3 × m 999 ⋅ kg 2 × s 9.81⋅ m Hence D < 1.55 mm. Only the 1 mm needles float (needle length is irrelevant) × kg⋅ m 2 N⋅ s −3 = 1.55 × 10 ⋅ m = 1.55⋅ mm Problem 2.79 Given: Caplillary rise data Find: Values of A and b [Difficulty: 3] Solution: D (in.) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 h (in.) 0.232 0.183 0.090 0.059 0.052 0.033 0.017 0.010 0.006 0.004 0.003 A = 0.403 b = 4.530 2 The fit is a good one (R = 0.9919) Capillary Rise vs. Tube Diameter 0.3 h (in.) h = 0.403e-4.5296D R2 = 0.9919 0.2 0.1 0.0 0.0 0.2 0.4 0.6 D (in.) 0.8 1.0 1.2 Problem 2.80 [Difficulty: 2] Open-Ended Problem Statement: Slowly fill a glass with water to the maximum possible level before it overflows. Observe the water level closely. Explain how it can be higher than the rim of the glass. Discussion: Surface tension can cause the maximum water level in a glass to be higher than the rim of the glass. The same phenomenon causes an isolated drop of water to “bead up” on a smooth surface. Surface tension between the water/air interface and the glass acts as an invisible membrane that allows trapped water to rise above the level of the rim of the glass. The mechanism can be envisioned as forces that act in the surface of the liquid above the rim of the glass. Thus the water appears to defy gravity by attaining a level higher than the rim of the glass. To experimentally demonstrate that this phenomenon is the result of surface tension, set the liquid level nearly as far above the glass rim as you can get it, using plain water. Add a drop of liquid detergent (the detergent contains additives that reduce the surface tension of water). Watch as the excess water runs over the side of the glass. Problem 2.81 [Difficulty: 5] Open-Ended Problem Statement: Plan an experiment to measure the surface tension of a liquid similar to water. If necessary, review the NCFMF video Surface Tension for ideas. Which method would be most suitable for use in an undergraduate laboratory? What experimental precision could be expected? Discussion: Two basic kinds of experiment are possible for an undergraduate laboratory: 1. Using a clear small-diameter tube, compare the capillary rise of the unknown liquid with that of a known liquid (compare with water, because it is similar to the unknown liquid). This method would be simple to set up and should give fairly accurate results. A vertical traversing optical microscope could be used to increase the precision of measuring the liquid height in each tube. A drawback to this method is that the specific gravity and co ntact angle of the two liquids must be the same to allow the capillary rises to be compared. The capillary rise would be largest and therefore easiest to measure accurately in a tube with the smallest practical diameter. Tubes of several diameters could be used if desired. 2. Dip an object into a pool of test liquid and measure the vertical force required to pull the object from the liquid surface. The object might be made rectangular (e.g., a sheet of plastic material) or circular (e.g., a metal ring). The net force needed to pull the same object from each liquid should be proportional to the surface tension of each liquid. This method would be simple to set up. However, the force magnitudes to be measured would be quite small. A drawback to this method is that the contact angles of the two liquids must be the same. The first method is probably best for undergraduate laboratory use. A quantitative estimate of experimental measurement uncertainty is impossible without knowing details of the test setup. It might be reasonable to expect results accurate to within ± 10% of the true surface tension. *Net force is the total vertical force minus the weight of the object. A buoyancy correction would be necessary if part of the object were submerged in the test liquid. Problem 2.82 [Difficulty: 2] Problem 2.83 [Difficulty: 2] Given: Boundary layer velocity profile in terms of constants a, b and c Find: Constants a, b and c Solution: u = a + b ⋅ ⎛⎜ Basic equation y⎞ ⎝δ⎠ + c⋅ ⎛⎜ y⎞ 2 ⎝δ⎠ Assumptions: No slip, at outer edge u = U and τ = 0 At y = 0 0=a a=0 At y = δ U= a+ b+ c b+c=U (1) At y = δ τ = μ⋅ b + 2⋅ c = 0 (2) 0= du dy d dy a + b ⋅ ⎛⎜ y⎞ ⎝δ⎠ From 1 and 2 c = −U Hence u = 2 ⋅ U⋅ ⎛⎜ Dimensionless Height =0 + c⋅ ⎛⎜ y⎞ ⎝δ⎠ 2 = b δ + 2 ⋅ c⋅ y 2 = δ b δ + 2⋅ c δ b = 2⋅ U y⎞ ⎝δ⎠ − U⋅ ⎛⎜ y⎞ ⎝δ⎠ 2 u U = 2 ⋅ ⎛⎜ y⎞ ⎝δ⎠ − ⎛y⎞ ⎜δ ⎝ ⎠ 2 1 0.75 0.5 0.25 0 0.25 0.5 Dimensionless Velocity 0.75 1 Problem 2.84 [Difficulty: 2] Given: Boundary layer velocity profile in terms of constants a, b and c Find: Constants a, b and c Solution: Basic equation u = a + b ⋅ ⎛⎜ + c⋅ ⎛⎜ y⎞ ⎝δ⎠ y⎞ 3 ⎝δ⎠ Assumptions: No slip, at outer edge u = U and τ = 0 At y = 0 0=a a=0 At y = δ U= a+ b+ c b+c=U (1) At y = δ τ = μ⋅ b + 3⋅ c = 0 (2) 0= From 1 and 2 Dimensionless Height dy d dy c=− u= Hence du =0 a + b ⋅ ⎛⎜ y⎞ ⎝δ⎠ U 2 ⋅ ⎛⎜ y⎞ ⎝δ⎠ − U 2 y⎞ ⎝δ⎠ b= 2 3⋅ U + c⋅ ⎛⎜ 3 2 ⋅ ⎛⎜ 3 = b δ + 3 ⋅ c⋅ y 2 3 = b ⋅ ⎛⎜ y⎞ δ δ + 3⋅ c δ ⋅U y⎞ ⎝δ⎠ 3 u U = 3 2 ⎝δ⎠ − 1 2 ⋅ ⎛⎜ y⎞ 3 ⎝δ⎠ 1 0.75 0.5 0.25 0 0.25 0.5 Dimensionless Velocity 0.75 1 Problem 2.85 Given: Local temperature Find: Minimum speed for compressibility effects [Difficulty: 1] Solution: Basic equation V = M⋅ c c= Hence and k⋅ R⋅ T M = 0.3 for compressibility effects For air at STP, k = 1.40 and R = 286.9J/kg.K (53.33 ft.lbf/lbmoR). V = M ⋅ c = M ⋅ k ⋅ R⋅ T 1 V = 0.3 × ⎡1.4 × 53.33 ⋅ ⎢ ⎣ ft⋅ lbf lbm⋅ R × 32.2⋅ lbm⋅ ft 2 lbf ⋅ s 2 × ( 60 + 460 ) ⋅ R⎤ ⋅ ⎥ ⎦ 60⋅ mph 88⋅ ft s V = 229 ⋅ mph Problem 2.86 [Difficulty: 3] Given: Geometry of and flow rate through tapered nozzle Find: At which point becomes turbulent Solution: Basic equation Re = For pipe flow (Section 2-6) ρ⋅ V⋅ D μ = 2300 for transition to turbulence 2 π⋅ D Q= Also flow rate Q is given by 4 ⋅V We can combine these equations and eliminate V to obtain an expression for Re in terms of D and Q Re = ρ⋅ V⋅ D μ = ρ⋅ D 4 ⋅ Q 4 ⋅ Q⋅ ρ ⋅ = μ π⋅ μ⋅ D 2 π⋅ D Re = 4 ⋅ Q⋅ ρ π⋅ μ⋅ D For a given flow rate Q, as the diameter is reduced the Reynolds number increases (due to the velocity increasing with A -1 or D -2). Hence for turbulence (Re = 2300), solving for D The nozzle is tapered: Carbon tetrachloride: Din = 50⋅ mm μCT = 10 D= 4 ⋅ Q⋅ ρ 2300⋅ π⋅ μ Dout = − 3 N⋅ s ⋅ Din Dout = 22.4⋅ mm 5 (Fig A.2) For water 2 ρ = 1000⋅ 3 m m SG = 1.595 kg ρCT = SG⋅ ρ (Table A.2) ρCT = 1595 kg 3 m For the given flow rate Q = 2⋅ L 4 ⋅ Q⋅ ρCT min π⋅ μCT⋅ Din For the diameter at which we reach turbulence But L = 250 ⋅ mm D = = 1354 4 ⋅ Q⋅ ρCT 2300⋅ π⋅ μCT LAMINAR 4 ⋅ Q⋅ ρCT π⋅ μCT⋅ Dout D − Din Dout − Din Lturb = 186 ⋅ mm TURBULENT D = 29.4⋅ mm and linear ratios leads to the distance from D in at which D = 29.4⋅ mm Lturb = L⋅ = 3027 Lturb L = D − Din Dout − Din Problem 2.87 [Difficulty: 2] Given: Data on water tube Find: Reynolds number of flow; Temperature at which flow becomes turbulent Solution: Basic equation At 20oC, from Fig. A.3 ν 9 10 For the heated pipe Hence Re For pipe flow (Section 2-6) Re ν V D ν V D 2300 2 7 m and so s 2300 1 2300 ρ V D μ Re 0.25 m s V D ν 0.005 m 9 10 for transition to turbulence 0.25 m s 0.005 m From Fig. A.3, the temperature of water at this viscosity is approximately ν 5.435 10 T 52 C 1 2 7m s 7 s 2 m Re 1389 Problem 2.88 [Difficulty: 3] Given: Data on supersonic aircraft Find: Mach number; Point at which boundary layer becomes turbulent Solution: Basic equation V = M⋅ c Hence M= V c c= and k⋅ R⋅ T For air at STP, k = 1.40 and R = 286.9J/kg.K (53.33 ft.lbf/lbmoR). V = k ⋅ R⋅ T At 27 km the temperature is approximately (from Table A.3) T = 223.5 ⋅ K 1 2 2 ⎛ ⎞ ⋅ ⎜ 1 × 1 ⋅ kg⋅ K × 1⋅ N⋅ s × 1 ⋅ 1 ⎞ M = 2.5 M = ⎜⎛ 2700 × 10 ⋅ × hr 3600⋅ s ⎠ ⎝ 1.4 286.9 N⋅ m kg⋅ m 223.5 K ⎠ ⎝ 3 m For boundary layer transition, from Section 2-6 Then Retrans = ρ⋅ V⋅ x trans 1 ⋅ hr Retrans = 500000 μ ⋅ Retrans x trans = so μ ρ⋅ V We need to find the viscosity and density at this altitude and pressure. The viscosity depends on temperature only, but at 223.5 K = - 50oC, it is off scale of Fig. A.3. Instead we need to use formulas as in Appendix A μ= b ⋅T 2 1+ S where 3 −6 b = 1.458 × 10 1 m⋅ s ⋅ K 3 S = 110.4 ⋅ K 2 − 5 N⋅s − 5 kg μ = 1.459 × 10 m⋅ s ⋅ 2 m − 5 kg x trans = 1.459 × 10 kg m kg ⋅ T μ = 1.459 × 10 Hence ρ = 0.0297 m 1 For µ kg ρ = 0.02422× 1.225⋅ At this altitude the density is (Table A.3) ⋅ m⋅ s × 500000× 3 m 1 1 hr 3600⋅ s ⋅ × × ⋅ × 0.0297 kg 2700 3 m 1⋅ hr 10 1 x trans = 0.327m Problem 2.89 Given: Type of oil, flow rate, and tube geometry Find: Whether flow is laminar or turbulent [Difficulty: 2] Solution: ν= Data on SAE 30 oil SG or density is limited in the Appendix. We can Google it or use the following At 100 oC, from Figs. A.2 and A.3 − 3 N⋅ s μ = 9 × 10 ⋅ ν = 1 × 10 2 − 3 N⋅ s ⋅ 2 1 × 10 m Hence The specific weight is SG = 1 × ρ −5 ⋅ s 2 × ⋅ s kg⋅ m ρ = 900 2 s ⋅N m γ = ρ⋅ g γ = 900 ⋅ kg 3 2 Q= π⋅ D 4 ⋅V −6 Q = 100 ⋅ mL × Then Hence V = Re = 4 π V= so 10 m 2 2 N⋅ s × 3 N γ = 8.829 × 10 ⋅ kg⋅ m s 4⋅ Q 2 ⋅m 1 1 × ⋅ 9 s 3 −5 m ⋅ s × Q = 1.111 × 10 2 ⎛ 1 ⋅ 1 × 1000⋅ mm ⎞ ⎜ 12 mm 1⋅ m ⎠ ⎝ V = 0.0981 ρ⋅ V⋅ D μ Re = 900 ⋅ kg 3 m Flow is laminar 3 m π⋅ D 3 1 ⋅ mL × 1.11 × 10 3 SG = 0.9 × 9.81⋅ m For pipe flow (Section 2-6) kg m kg ρwater = 1000⋅ 3 m ρwater ρ= so ρ 2 −5 m m ρ = 9 × 10 μ × 0.0981⋅ m s × 0.012 ⋅ m × 1 9 × 10 2 ⋅ m − 3 N⋅ s 2 × N⋅ s kg⋅ m Re = 118 m s 3 −5m s μ ν Problem 2.90 Given: Data on seaplane Find: Transition point of boundary layer [Difficulty: 2] Solution: For boundary layer transition, from Section 2-6 Retrans 500000 Then Retrans At 45oF = 7.2 oC (Fig A.3) ρ V x trans μ 2 5 m ν 0.8 10 s V x trans ν 10.8 5 ft ft V 5 ft s ν 8.64 10 m 2 s s 2 500000 s ν Retrans 2 2 1 x trans 8.64 10 x trans so 1 100 mph 60 mph 88 x trans 0.295 ft ft s As the seaplane touches down: At 45oF = 7.2 oC (Fig A.3) 2 5 m ν 1.5 10 s 10.8 4 ft 2 4 ft s ν 1.62 10 2 1 x trans 1.62 10 ft m 2 s s 2 s 500000 1 100 mph 60 mph 88 ft s x trans 0.552 ft Problem 2.91 [Difficulty: 3] Given: Data on airliner Find: Sketch of speed versus altitude (M = const) Solution: Data on temperature versus height can be obtained from Table A.3 At 5.5 km the temperature is approximately 252 c The speed of sound is obtained from where k = 1.4 R = 286.9 J/kg·K c = 318 m/s V = 700 km/hr V = 194 m/s K k R T (Table A.6) We also have or Hence M = V/c or M = 0.611 V = M · c = 0.611·c To compute V for constant M , we use V = 677 At a height of 8 km: km/hr NOTE: Realistically, the aiplane will fly to a maximum height of about 10 km! T (K) 4 262 5 259 5 256 6 249 7 243 8 236 9 230 10 223 11 217 12 217 13 217 14 217 15 217 16 217 17 217 18 217 19 217 20 217 22 219 24 221 26 223 28 225 30 227 40 250 50 271 60 256 70 220 80 181 90 181 c (m/s) V (km/hr) 325 322 320 316 312 308 304 299 295 295 295 295 295 295 295 295 295 295 296 298 299 300 302 317 330 321 297 269 269 Speed vs. Altitude 713 709 750 704 695 686 677 668 658 700 649 649 649 649 649 649 649 649 Speed V (km/hr) z (km) 650 649 649 651 654 600 657 660 663 697 725 705 653 592 592 550 0 20 40 60 Altitude z (km) 80 100 Problem 2.92 [Difficulty: 4] Open-Ended Problem Statement: How does an airplane wing develop lift? Discussion: The sketch shows the cross-section of a typical airplane wing. The airfoil section is rounded at the front, curved across the top, reaches maximum thickness about a third of the way back, and then tapers slowly to a fine trailing edge. The bottom of the airfoil section is relatively flat. (The discussion below also applies to a symmetric airfoil at an angle of incidence that produces lift.) NACA 2412 Wing Section It is both a popular expectation and an experimental fact that air flows more rapidly over the curved top surface of the airfoil section than along the relatively flat bottom. In the NCFMF video Flow Visualization, timelines placed in front of the airfoil indicate that fluid flows more rapidly along the top of the section than along the bottom. In the absence of viscous effects (this is a valid assumption outside the boundary layers on the airfoil) pressure falls when flow speed increases. Thus the pressures on the top surface of the airfoil where flow speed is higher are lower than the pressures on the bottom surface where flow speed does not increase. (Actual pressure profiles measured for a lifting section are shown in the NCFMF video Boundary Layer Control.) The unbalanced pressures on the top and bottom surfaces of the airfoil section create a net force that tends to develop lift on the profile. Problem 3.1 [Difficulty: 2] Given: Data on nitrogen tank Find: Pressure of nitrogen; minimum required wall thickness Assumption: Ideal gas behavior Solution: Ideal gas equation of state: p ⋅V = M⋅R⋅T where, from Table A.6, for nitrogen R = 55.16⋅ Then the pressure of nitrogen is p = ft⋅ lbf lbm⋅ R ⎛ 6 ⎞ 3⎟ ⎝ π⋅ D ⎠ M⋅ R⋅ T = M⋅ R⋅ T⋅ ⎜ V p = 140⋅ lbm × 55.16⋅ p = 3520⋅ ft⋅ lbf lbm⋅ R ⎡ ⎤ × ⎛ ft ⎞ ⎜ ⎟ 3⎥ ⎝ 12⋅ in ⎠ ⎣ π × ( 2.5⋅ ft) ⎦ × ( 77 + 460) ⋅ R × ⎢ 6 lbf 2 in To determine wall thickness, consider a free body diagram for one hemisphere: π⋅ D ΣF = 0 = p ⋅ 2 − σc ⋅ π ⋅ D ⋅ t pπD2/4 p⋅ D σcπDt 4 where σc is the circumferential stress in the container Then t= p⋅ π⋅ D 2 4 ⋅ π ⋅ D ⋅ σc t = 3520 ⋅ lbf 2 in t = 0.0733⋅ ft × = 4 ⋅ σc 2.5 ⋅ ft 4 2 × in 3 30 × 10 ⋅ lbf t = 0.880⋅ in 2 Problem 3.2 [Difficulty: 2] Given: Pure water on a standard day Find: Boiling temperature at (a) 1000 m and (b) 2000 m, and compare with sea level value. Solution: We can determine the atmospheric pressure at the given altitudes from table A.3, Appendix A The data are Elevation (m) 0 1000 2000 p/p o p (kPa) 1.000 0.887 0.785 101.3 89.9 79.5 We can also consult steam tables for the variation of saturation temperature with pressure: p (kPa) 70 80 90 101.3 T sat (°C) 90.0 93.5 96.7 100.0 We can interpolate the data from the steam tables to correlate saturation temperature with altitude: Elevation (m) 0 1000 2000 p/p o p (kPa) T sat (°C) 1.000 0.887 0.785 101.3 89.9 79.5 The data are plotted here. They show that the saturation temperature drops approximately 3.4°C/1000 m. 100.0 96.7 93.3 Saturation Temperature (°C) Variation of Saturation Temperature with Pressure Sea Level 100 1000 m 98 96 2000 m 94 92 90 88 70 75 80 85 90 95 Absolute Pressure (kPa) 100 105 Problem 3.3 [Difficulty: 2] Given: Data on flight of airplane Find: Pressure change in mm Hg for ears to "pop"; descent distance from 8000 m to cause ears to "pop." Solution: Assume the air density is approximately constant constant from 3000 m to 2900 m. From table A.3 ρSL 1.225 kg ρair 0.7423 ρSL 3 m ρair 0.909 kg 3 m We also have from the manometer equation, Eq. 3.7 ∆p ρair g ∆z Combining ∆hHg ρair ρHg ∆hHg ∆z 0.909 13.55 999 ∆p ρHg g ∆hHg and also ρair SGHg ρH2O SGHg 13.55 from Table A.2 ∆z 100 m ∆hHg 6.72 mm For the ear popping descending from 8000 m, again assume the air density is approximately constant constant, this time at 8000 m. From table A.3 ρair 0.4292 ρSL ρair 0.526 kg 3 m We also have from the manometer equation ρair8000 g ∆z8000 ρair3000 g ∆z3000 where the numerical subscripts refer to conditions at 3000m and 8000m. Hence ∆z8000 ρair3000 g ρair8000 g ∆z3000 ρair3000 ρair8000 ∆z3000 ∆z8000 0.909 0.526 100 m ∆z8000 173 m Problem 3.4 [Difficulty: 3] Given: Boiling points of water at different elevations Find: Change in elevation Solution: From the steam tables, we have the following data for the boiling point (saturation temperature) of water Tsat (oF) p (psia) 10.39 8.39 195 185 The sea level pressure, from Table A.3, is pSL = 14.696 psia Hence Altitude vs Atmospheric Pressure o p/pSL 195 185 0.707 0.571 From Table A.3 p/pSL 0.7372 0.6920 0.6492 0.6085 0.5700 15000 12500 Altitude (ft) Tsat ( F) Altitude (m) 2500 3000 3500 4000 4500 Altitude (ft) 8203 9843 11484 13124 14765 Data 10000 Linear Trendline 7500 z = -39217(p/pSL) + 37029 5000 R2 = 0.999 2500 0.55 0.60 0.65 0.70 p/pSL Then, any one of a number of Excel functions can be used to interpolate (Here we use Excel 's Trendline analysis) p/pSL 0.707 0.571 Altitude (ft) 9303 14640 Current altitude is approximately The change in altitude is then 5337 ft Alternatively, we can interpolate for each altitude by using a linear regression between adjacent data points p/pSL p/pSL For 0.7372 0.6920 Altitude (m) 2500 3000 Altitude (ft) 8203 9843 0.6085 0.5700 Altitude (m) 4000 4500 Altitude (ft) 13124 14765 Then 0.7070 2834 9299 0.5730 4461 14637 The change in altitude is then 5338 ft 9303 ft 0.75 Problem 3.5 Given: Data on system Find: Force on bottom of cube; tension in tether [Difficulty: 2] Solution: dp Basic equation dy = − ρ⋅ g or, for constant ρ ∆p = ρ⋅ g⋅ h where h is measured downwards The absolute pressure at the interface is pinterface = patm + SGoil⋅ ρ⋅ g⋅ hoil Then the pressure on the lower surface is pL = pinterface + ρ⋅ g⋅ hL = patm + ρ⋅ g⋅ SGoil⋅ hoil + hL For the cube ( V = 125⋅ mL 1 3 V = 1.25 × 10 Then the size of the cube is d = V d = 0.05 m Hence hL = hU + d hL = 0.35 m The force on the lower surface is F L = pL ⋅ A where ( −4 ) 3 ⋅m and the depth in water to the upper surface is hU = 0.3⋅ m where hL is the depth in water to the lower surface A = d 2 2 A = 0.0025 m ) FL = ⎡patm + ρ⋅ g⋅ SGoil⋅ hoil + hL ⎤ ⋅ A ⎣ ⎦ ⎡ kg m N⋅ s ⎥⎤ 3 N 2 FL = ⎢101 × 10 ⋅ + 1000⋅ × 9.81⋅ × ( 0.8 × 0.5⋅ m + 0.35⋅ m) × × 0.0025⋅ m 2 3 2 ⎢ kg⋅ m⎥ m m s ⎣ ⎦ 2 FL = 270.894 N For the tension in the tether, an FBD gives Note: Extra decimals needed for computing T later! ΣFy = 0 FL − FU − W − T = 0 ( ) where FU = ⎡patm + ρ⋅ g⋅ SGoil⋅ hoil + hU ⎤ ⋅ A ⎣ ⎦ or T = FL − FU − W ( ) Note that we could instead compute ∆F = FL − FU = ρ⋅ g⋅ SGoil⋅ hL − hU ⋅ A Using FU ⎡ kg m N⋅ s ⎥⎤ 3 N 2 FU = ⎢101 × 10 ⋅ + 1000⋅ × 9.81⋅ × ( 0.8 × 0.5⋅ m + 0.3⋅ m) × × 0.0025⋅ m 2 3 2 ⎢ kg⋅ m⎥ m m s ⎣ ⎦ T = ∆F − W 2 FU = 269.668 N For the oak block (Table A.1) and Note: Extra decimals needed for computing T later! SGoak = 0.77 W = 0.77 × 1000⋅ W = SGoak⋅ ρ⋅ g⋅ V so kg 3 m T = FL − FU − W × 9.81⋅ m 2 × 1.25 × 10 s T = 0.282 N −4 3 ⋅m × 2 N⋅ s kg⋅ m W = 0.944 N Problem 3.6 Given: Data on system before and after applied force Find: Applied force [Difficulty: 2] Solution: Basic equation dp dy For initial state p1 patm ρ g h For the initial FBD ΣFy 0 For final state p2 patm ρ g H For the final FBD ΣFy 0 F1 p1 A ρ g h A and F2 p2 A ρ g H A F2 W F 0 4 p y0 patm with (Gage; F1 is hydrostatic upwards force) W F 1 ρ g h A and π D p patm ρ g y y0 F1 W 0 F ρH2O SG g From Fig. A.1 or, for constant ρ ρ g (Gage; F2 is hydrostatic upwards force) F F2 W ρ g H A ρ g h A ρ g A ( H h ) 2 ( H h) SG 13.54 F 1000 kg 3 m F 45.6 N 13.54 9.81 m 2 s π 4 2 ( 0.05 m) ( 0.2 0.025) m 2 N s kg m Problem 3.7 [Difficulty: 1] Given: Pressure and temperature data from balloon Find: Plot density change as a function of elevation Assumption: Ideal gas behavior Solution: Density Distribution 0.078 p (psia) 14.71 14.62 14.53 14.45 14.36 14.27 14.18 14.10 14.01 13.92 13.84 T (oF) 53.6 52.0 50.9 50.4 50.2 50.0 50.5 51.4 52.9 54.0 53.8 (lbm/ft3) 0.07736 0.07715 0.07685 0.07647 0.07604 0.07560 0.07506 0.07447 0.07380 0.07319 0.07276 Density (lbm/ft3) Using the ideal gas equation, = p/RT 0.077 0.076 0.075 0.074 0.073 0.072 0 1 2 3 4 5 6 Elevation Point 7 8 9 10 Problem 3.8 [Difficulty: 2] Given: Properties of a cube floating at an interface Find: The pressures difference between the upper and lower surfaces; average cube density Solution: The pressure difference is obtained from two applications of Eq. 3.7 pU p0 ρSAE10 g ( H 0.1 d) pL p0 ρSAE10 g H ρH2O g 0.9 d where pU and pL are the upper and lower pressures, p0 is the oil free surface pressure, H is the depth of the interface, and d is the cube size Hence the pressure difference is ∆p pL pU ρH2O g 0.9 d ρSAE10 g 0.1 d From Table A.2 ∆p ρH2O g d 0.9 SGSAE10 0.1 SGSAE10 0.92 kg ∆p 999 3 m 9.81 2 m 2 0.1 m ( 0.9 0.92 0.1) s Ns kg m ∆p 972 Pa For the cube density, set up a free body force balance for the cube ΣF 0 ∆p A W Hence 2 W ∆p A ∆p d ρcube m d 3 ρcube 972 2 W 3 d g N 2 m ∆p d 3 d g 1 0.1 m ∆p d g 2 s 9.81 m kg m 2 Ns ρcube 991 kg 3 m Problem 3.9 Given: Data on tire at 3500 m and at sea level Find: Absolute pressure at 3500 m; pressure at sea level [Difficulty: 2] Solution: At an elevation of 3500 m, from Table A.3: pSL 101 kPa patm 0.6492 pSL patm 65.6 kPa and we have pg 0.25 MPa pg 250 kPa p pg patm At sea level patm 101 kPa p 316 kPa Meanwhile, the tire has warmed up, from the ambient temperature at 3500 m, to 25oC. At an elevation of 3500 m, from Table A.3 Tcold 265.4 K and Thot ( 25 273) K Thot 298 K Hence, assuming ideal gas behavior, pV = mRT, and that the tire is approximately a rigid container, the absolute pressure of the hot tire is phot Thot Tcold p phot 354 kPa Then the gage pressure is pg phot patm pg 253 kPa Problem 3.10 Given: Data on air bubble Find: Bubble diameter as it reaches surface [Difficulty: 2] Solution: dp Basic equation dy ρsea g and the ideal gas equation M p ρ R T V R T We assume the temperature is constant, and the density of sea water is constant For constant sea water density p patm SGsea ρ g h Then the pressure at the initial depth is p1 patm SGsea ρ g h1 The pressure as it reaches the surface is p2 patm M R T V For the bubble p Hence p1 V 1 p2 V 2 Then the size of the bubble at the surface is From Table A.2 but M and T are constant p1 D 2 D 1 p2 SGsea 1.025 1 3 M R T const p V P1 V2 V1 p2 or or 1 3 3 3 p1 D2 D1 p2 patm ρsea g h1 ρsea g h1 D1 1 patm patm 1 3 D 1 (This is at 68oF) slug ft D2 0.3 in 1 1.025 1.94 D2 0.477 in where p is the pressure at any depth h 3 32.2 ft 2 s 2 2 1 ft lbf s 14.7 lbf 12 in slugft 2 100 ft in 1 3 Problem 3.11 [Difficulty: 2] Given: Properties of a cube suspended by a wire in a fluid Find: The fluid specific gravity; the gage pressures on the upper and lower surfaces Solution: From a free body analysis of the cube: ( ) 2 ΣF = 0 = T + pL − pU ⋅ d − M⋅ g where M and d are the cube mass and size and pL and pU are the pressures on the lower and upper surfaces For each pressure we can use Eq. 3.7 p = p 0 + ρ ⋅ g⋅ h Hence pL − pU = ⎡p0 + ρ⋅ g⋅ ( H + d)⎤ − p0 + ρ⋅ g⋅ H = ρ⋅ g⋅ d = SG⋅ ρH2O⋅ d ⎣ ⎦ ( ) where H is the depth of the upper surface 2 ⋅ slug× 32.2⋅ M⋅ g − T SG = Hence the force balance gives ρH2O ⋅ g ⋅ d 2 × s SG = 3 2 ft 1.94 ⋅ slug ft 3 × 32.2⋅ lbf ⋅ s − 50.7 ⋅ lbf slug⋅ ft SG = 1.75 2 ft 2 × s lbf⋅ s slug⋅ ft × ( 0.5 ⋅ ft) 3 From Table A.1, the fluid is Meriam blue. The individual pressures are computed from Eq 3.7 p = p 0 + ρ ⋅ g⋅ h pg = 1.754 × 1.94⋅ For the upper surface pg = ρ⋅ g⋅ h = SG⋅ ρH2O⋅ h or slug ft pg = 1.754 × 1.94⋅ For the lower surface 3 slug ft 3 × 32.2⋅ ft 2 s × 32.2⋅ ft 2 s × 2 3 2 ⋅ ft × ⎛ 1⋅ ft ⎞ ⎟ slug⋅ ft ⎝ 12⋅ in ⎠ lbf ⋅ s 2 ×⎜ 2 ⎛ 2 + 1 ⎞ ⋅ ft × lbf ⋅ s × ⎛ 1⋅ ft ⎞ ⎟ ⎜ ⎟ slug⋅ ft ⎝ 12⋅ in ⎠ ⎝ 3 2⎠ ×⎜ pg = 0.507⋅ psi 2 pg = 0.888⋅ psi Note that the SG calculation can also be performed using a buoyancy approach (discussed later in the chapter): Consider a free body diagram of the cube: ΣF = 0 = T + FB − M ⋅ g where M is the cube mass and FB is the buoyancy force FB = SG ⋅ ρH2O ⋅ L ⋅ g Hence 3 T + SG ⋅ ρH2O ⋅ L ⋅ g − M ⋅ g = 0 3 or SG = M⋅ g − T ρH2O ⋅ g ⋅ L 3 as before SG = 1.75 Problem 3.12 [Difficulty: 4] Given: Model behavior of seawater by assuming constant bulk modulus Find: (a) Expression for density as a function of depth h. (b) Show that result may be written as ρ = ρo + bh (c) Evaluate the constant b (d) Use results of (b) to obtain equation for p(h) (e) Determine depth at which error in predicted pressure is 0.01% Solution: 5 SGo = 1.025 From Table A.2, App. A: dp Governing Equations: Ev = 2.42⋅ GPa = 3.51 × 10 ⋅ psi = ρ⋅ g dh Ev = (Hydrostatic Pressure - h is positive downwards) dp (Definition of Bulk Modulus) dρ ρ ρ Then dρ dp = ρ⋅ g⋅ dh = Ev⋅ ρ or dρ ρ 2 = g Ev dh h ⌠ ⌠ g 1 ⎮ dρ = ⎮ dh ⎮ 2 ⎮ Ev ⎮ ρ ⌡0 ⌡ρ Now if we integrate: o After integrating: Now for ρ o⋅ g⋅ h Ev ρ − ρo ρ⋅ ρo = g⋅ h Ev Therefore: ρ = E v⋅ ρ o E v − g⋅ h ⋅ ρ o ρ and ρo 1 = 1− <<1, the binomial expansion may be used to approximate the density: ρ o⋅ g⋅ h Ev ρ ρo 2 In other words, ρ = ρo + b⋅ h where b = Since ρo ⋅ g = 1+ ρ o⋅ g⋅ h Ev (Binomial expansion may be found in a host of sources, e.g. CRC Handbook of Mathematics) Ev dp = ρ⋅ g⋅ dh then an approximate expression for the pressure as a function of depth is: h ⌠ papprox − patm = ⎮ ⌡0 ( ρo + b ⋅ h )⋅ g dh → papprox − patm = ( g⋅ h ⋅ 2⋅ ρo + b ⋅ h 2 ) Solving for papprox we get: papprox = patm + ( g⋅ h ⋅ 2⋅ ρ o + b⋅ h 2 ) = patm + ρo⋅ g⋅ h + b⋅ g ⋅ h 2 2 ⎛ 2 b⋅ h ⎞ ⎝ 2 ⎠ = patm + ⎜ ρo⋅ h + ⎟⋅g Now if we subsitiute in the expression for b and simplify, we get: 2 ⎛ ρo ⋅ g h2 ⎞⎟ ⎛ ρ o⋅ g⋅ h ⎞ ⎜ ⋅ g = patm + ρo⋅ g⋅ h⋅ ⎜ 1 + papprox = patm + ρo⋅ h + ⋅ ⎟ ⎜ Ev 2 ⎟ 2⋅ E v ⎝ ⎠ ⎝ ⎠ ⎛ ρ o⋅ g⋅ h ⎞ papprox = patm + ρo⋅ g⋅ h⋅ ⎜ 1 + ⎟ 2Ev ⎝ ⎠ The exact soution for p(h) is obtained by utilizing the exact solution for ρ(h). Thus: ρ ⌠ E v ⎛ρ⎞ pexact − patm = ⎮ dρ = Ev⋅ ln ⎜ ⎟ ⎮ ρ ⎝ ρo ⎠ ⌡ρ Subsitiuting for ρ ρo we get: o If we let x = ρ o⋅ g⋅ h Ev For the error to be 0.01%: ∆pexact − ∆papprox ∆pexact = 1− This equation requires an iterative solution, e.g. Excel's Goal Seek. The result is: h= x⋅ E v ρ o⋅ g ⎛ ρ o⋅ g⋅ h ⎞ pexact = patm + Ev⋅ ln ⎜ 1 − ⎟ Ev ⎝ ⎠ ⎛ x⎞ ρo⋅ g⋅ h⋅ ⎜ 1 + ⎟ ⎝ 2⎠ Ev⋅ ln ⎡⎣( 1 − x) − 1⎤ = 1− ⎦ ln ⎡⎣( 1 − x) − 1⎤ = 0.0001 ⎦ x = 0.01728 Solving x for h: 3 2 2 ft s 5 lbf ⎛ 12⋅ in ⎞ × slug⋅ ft × × ×⎜ ⎟ 2 1.025 × 1.94⋅ slug 32.2⋅ ft ⎝ ft ⎠ 2 h = 0.01728 × 3.51 × 10 ⋅ in ⎛ x⎞ x⋅ ⎜ 1 + ⎟ ⎝ 2⎠ −1 4 h = 1.364 × 10 ⋅ ft lbf ⋅ s This depth is over 2.5 miles, so the incompressible fluid approximation is a reasonable one at all but the lowest depths of the ocean. Problem 3.13 [Difficulty: 3] Given: Model behavior of seawater by assuming constant bulk modulus Find: The percent deviations in (a) density and (b) pressure at depth h = 6.5 mi, as compared to values assuming constant density. Plot results over the range of 0 mi - 7 mi. Solution: 5 SGo 1.025 From Table A.2, App. A: dp Governing Equations: dh Ev 2.42 GPa 3.51 10 psi ρ g (Hydrostatic Pressure - h is positive downwards) dp Ev (Definition of Bulk Modulus) dρ ρ ρ dρ dp ρ g dh Ev ρ Then dρ or ρ 2 g Ev h 6.5 mi Now if we integrate: dh h g 1 dρ dh 2 Ev ρ 0 ρ o ρ ρo After integrating: ρ ρo g h Therefore: ρ Ev ∆ρ ρo ρ ρo ρo 1 1 ρ ρo 1 1 1 ρ o g h 1 1 Ev E v ρ o and Ev ρo g h ρ o g h Ev ρ o g h To determine an expression for the percent deviation in pressure, we find ρ For variable density and constant bulk modulus: For constant density: 1 ρo g h Ev pconstρ patm ρo g dh ρo g h 0 ρ o g h Ev ∆ρ ρo ρo g h Ev 1 p patm for variable ρ, and then for constant ρ. E v ρ p patm dρ Ev ln ρ ρo ρ o h ρo 1 ρo g h Ev 1 Ev ρ ρ o g h Ev ∆p pconstρ p pconstρ If we let x pconstρ Ev ρ ρ g h o ρo Ev ln ρ o g h x 3.51 10 ρ o g Ev ρ o g h 1 ρo g h ln 1 1 pconstρ ρ o g h Ev ∆p ρ 1 ρo ln Ev 2 3 2 1 1 ft 1 s 5 lbf 12 in slug ft mi 2 5280 ft 2 1.025 1.94 slug 32.2 ft ft x 149.5 mi lbf s in Substituting into the expressions for the deviations we get: ∆ρ h x h h devρ ρo h xh 149.5 mi h 1 x ∆p x devp ln 1 pconstρ h h 1 1 x 149.5 mi h ln 1 1 1 149.5 mi h devρ 4.55 % For h = 6.5 mi we get: The plot below shows the deviations in density and pressure as a function of depth from 0 mi to 7 mi: Errors in Density and Pressure Assuming Constant Density 8 Errors in Density and Pressure (%) Density Pressure 6 4 2 0 2 4 Depth (mi) 6 devp 2.24 % Problem 3.14 Given: [Difficulty: 3] Cylindrical cup lowered slowly beneath pool surface Air Find: H Expression for y in terms of h and H. Plot y/H vs. h/H. D y Air H–y Solution: y Governing Equations: dp dh ρ g (Hydrostatic Pressure - h is positive downwards) p V M R T Assumptions: (Ideal Gas Equation) (1) Constant temperature compression of air inside cup (2) Static liquid (3) Incompressible liquid First we apply the ideal gas equation (at constant temperature) for the pressure of the air in the cup: Therefore: π 2 π 2 p V pa D H p D ( H y) 4 4 and upon simplification: p V constant pa H p ( H y) Now we look at the hydrostatic pressure equation for the pressure exerted by the water. Since ρ is constant, we integrate: p pa ρ g ( h y) at the water-air interface in the cup. Since the cup is submerged to a depth of h, these pressures must be equal: pa H pa ρ g ( h y) ( H y) pa H pa y ρ g ( h y) ( H y) Explanding out the right hand side of this expression: 2 0 pa y ρ g ( h y) ( H y) ρ g h H ρ g h y ρ g H y ρ g y pa y 2 2 pa y ρ g y pa ρ g ( h H) y ρ g h H 0 ρ g ( h H) y h H 0 2 We now use the quadratic equation: pa pa ( h H) ( h H) 4 h H ρ g ρ g y 2 we only use the minus sign because y can never be larger than H. Now if we divide both sides by H, we get an expression for y/H: 2 pa pa h h h 1 1 4 y H ρ g H H ρ g H H H 2 The exact shape of this curve will depend upon the height of the cup. The plot below was generated assuming: pa 101.3 kPa H 1 m Height Ratio, y/H 0.8 0.6 0.4 0.2 0 20 40 60 Depth Ratio, h/H 80 100 Problem 3.15 Given: Geometry of straw Find: Pressure just below the thumb Assumptions: [Difficulty: 1] (1) Coke is incompressible (2) Pressure variation within the air column is negligible (3) Coke has density of water Solution: Basic equation dp dy ρ g or, for constant ρ ∆p ρ g h where h is measured downwards This equation only applies in the 15 cm of coke in the straw - in the other 30 cm of air the pressure is essentially constant. The gage pressure at the coke surface is Hence, with hcoke 15 cm pcoke ρ g hcoke because h is measured downwards pcoke 1000 kg 3 m pcoke 1.471 kPa pcoke 99.9 kPa 9.81 m 2 s gage 2 15 cm 2 m N s kPa m 100 cm kg m 1000 N Problem 3.16 [Difficulty: 2] Given: Data on water tank and inspection cover Find: If the support bracket is strong enough; at what water depth would it fail pbaseA Assumptions: Water is incompressible and static Cover Solution: Basic equation patmA dp dy ρ g or, for constant ρ ∆p ρ g h where h is measured downwards The absolute pressure at the base is pbase patm ρ g h h 16 ft The gage pressure at the base is pbase ρ g h This is the pressure to use as we have patm on the outside of the cover. The force on the inspection cover is F pbase A where where 2 A 1 in 1 in A 1 in F ρ g h A F 1.94 slug ft 3 32.2 ft 2 2 2 2 ft lbf s 12 in slug ft 16 ft 1 in s F 6.94 lbf The bracket is strong enough (it can take 9 lbf). To find the maximum depth we start with F 9.00 lbf h F ρ g A h 9 lbf h 20.7 ft 1 ft 3 1.94 slug 1 2 s 32.2 ft 1 2 in 2 12 in slug ft ft lbf s2 Problem 3.17 Given: [Difficulty: 4] Container of mercury with vertical tubes of known diameter, brass cylinder of known dimensions introduced into larger tube, where it floats. d1 39.5 mm d2 12.7 mm D 37.5 mm H 76.2 mm SGHg 13.55 SGb 8.55 Find: (a) Pressureon the bottom of the cylinder (b) New equlibrium level, h, of the mercury Solution: We will analyze a free body diagram of the cylinder, and apply the hydrostatics equation. ΣFz 0 Governing equations: dp dz (Vertical Equilibrium) ρ g (Hydrostatic Pressure - z is positive upwards) ρ SG ρwater Assumptions: (Definition of Specific Gravity) (1) Static liquid (2) Incompressible liquid If we take a free body diagram of the cylinder: ΣFz p π 4 π 2 2 D ρ b g D H 0 4 p 8.55 1000 kg 3 m 9.81 m 2 thus: p ρb g H SGb ρwater g H m 76.2 mm p 6.39 kPa (gage) 3 10 mm s pA This pressure must be generated by a column of mercury h+x in height. Thus: p ρHg g ( h x) SGHg ρwater g ( h x) SGb ρwater g H Thus: hx SGb SGHg H The value of x can be found by realizing that the volume of mercury in the system remains constant. Therefore: π π 2 2 2 D x d1 D h d2 h 4 4 4 π 2 These expressions now allow us to solve for h: h 8.55 13.55 ( 37.5 mm) 2 z mg 2 2 d2 d1 x 1 h D D Now if we solve for x: h SGb D 2 2 2 SGHg d1 d2 H Substituting in values: 2 ( 39.5 mm) ( 12.7 mm) 2 76.2 mm h 39.3 mm Problem 3.18 [Difficulty: 2] Given: Data on partitioned tank Find: Gage pressure of trapped air; pressure to make water and mercury levels equal Solution: The pressure difference is obtained from repeated application of Eq. 3.7, or in other words, from Eq. 3.8. Starting from the right air chamber pgage SGHg ρH2O g ( 3 m 2.9 m) ρH2O g 1 m pgage ρH2O g SGHg 0.1 m 1.0 m 2 kg m Ns pgage 999 9.81 ( 13.55 0.1 m 1.0 m) 3 2 kg m m s pgage 3.48 kPa If the left air pressure is now increased until the water and mercury levels are now equal, Eq. 3.8 leads to pgage SGHg ρH2O g 1.0 m ρH2O g 1.0 m pgage ρH2O g SGHg 1 m 1.0 m 2 kg m Ns pgage 999 9.81 ( 13.55 1 m 1.0 m) 3 2 kg m m s pgage 123 kPa Problem 3.19 [Difficulty: 2] Given: Data on partitioned tank Find: Pressure of trapped air required to bring water and mercury levels equal if right air opening is sealed Solution: First we need to determine how far each free surface moves. In the tank of Problem 3.18, the ratio of cross section areas of the partitions is 0.75/3.75 or 1:5. Suppose the water surface (and therefore the mercury on the left) must move down distance x to bring the water and mercury levels equal. Then by mercury volume conservation, the mercury free surface (on the right) moves up (0.75/3.75)x = x/5. These two changes in level must cancel the original discrepancy in free surface levels, of (1m + 2.9m) - 3 m = 0.9 m. Hence x + x/5 = 0.9 m, or x = 0.75 m. The mercury level thus moves up x/5 = 0.15 m. Assuming the air (an ideal gas, pV=RT) in the right behaves isothermally, the new pressure there will be pright Aright Lrightold Lrightold patm patm p Vrightnew Aright Lrightnew Lrightnew atm Vrightold where V, A and L represent volume, cross-section area, and vertical length Hence pright 3 3 0.15 101 kPa pright 106 kPa When the water and mercury levels are equal application of Eq. 3.8 gives: pleft pright SGHg ρH2O g 1.0 m ρH2O g 1.0 m pleft pright ρH2O g SGHg 1.0 m 1.0 m pleft 106 kPa 999 kg m pgage pleft patm 3 9.81 m 2 2 ( 13.55 1.0 m 1.0 m ) s pgage 229 kPa 101 kPa N s kg m pleft 229 kPa pgage 128 kPa Problem 3.20 Given: [Difficulty: 2] Two-fluid manometer as shown l 10.2 mm SGct 1.595 (From Table A.1, App. A) Find: Pressure difference Solution: We will apply the hydrostatics equation. Governing equations: dp dh ρ g (Hydrostatic Pressure - h is positive downwards) ρ SG ρwater Assumptions: (Definition of Specific Gravity) (1) Static liquid (2) Incompressible liquid z Starting at point 1 and progressing to point 2 we have: d p1 ρwater g ( d l) ρct g l ρwater g d p2 Simplifying and solving for p2 p1 we have: ∆p p2 p1 ρct g l ρwater g l SGct 1 ρwater g l Substituting the known data: ∆p ( 1.591 1) 1000 kg 3 m 9.81 m 2 s 10.2 mm m 3 10 mm ∆p 59.1 Pa Problem 3.21 Given: [Difficulty: 2] U-tube manometer, partiall filled with water, then a given volume of Meriam red oil is added to the left side 3 D 6.35 mm Voil 3.25 cm SGoil 0.827 (From Table A.1, App. A) Find: Equilibrium height, H, when both legs are open to atmosphere. Solution: We will apply the basic pressure-height relation. Governing Equations: dp dh ρ g (Hydrostatic Pressure - h is positive downwards) ρ SG ρwater Assumptions: (Definition of Specific Gravity) (1) Static liquid (2) Incompressible liquid Integration of the pressure equation gives: Thus: pB pA ρoil g L and p2 p1 ρ g h2 h1 A pD pC ρwater g ( L H) B Since pA pC patm and pB pD since they are at the same height: ρoil g L ρwater g ( L H) or Solving for H: H L 1 SGoil SGoil L L H The value of L comes from the volume of the oil: Solving for L: L 4 Voil π D We can now calculate H: 2 Voil 3 L 4 3.25 cm π ( 6.35 mm) 2 H 102.62 mm ( 1 0.827) π 4 2 D L 10 mm cm 3 L 102.62 mm H 17.75 mm C L L–H D Problem 3.22 Given: [Difficulty: 2] Two fluid manometer contains water and kerosene. With both tubes open to atmosphere, the difference in free surface elevations is known Ho = 20⋅ mm SGk = 0.82 (From Table A.1, App. A) Find: The elevation difference, H, between the free surfaces of the fluids when a gage pressure of 98.0 Pa is applied to the right tube. Solution: We will apply the hydrostatics equation. Governing Equations: dp dh = ρ⋅ g (Hydrostatic Pressure - h is positive downwards) ρ = SG⋅ ρwater Assumptions: (Definition of Specific Gravity) (1) Static liquid (2) Incompressible liquid When the gage pressure Δp is applied to the right tube, the water in the right tube is displaced downward by a distance, l. The kerosene in the left tube is displaced upward by the same distance, l. Δp l Under the applied gage pressure Δp, the elevation difference, H, is: H0 H = H o + 2⋅ l l H1 Since points A and B are at the same elevation in the same fluid, their pressures are the same. Initially: ( pA = ρ k⋅ g⋅ H o + H 1 ) A pB = ρwater⋅ g⋅ H1 B Setting these pressures equal: ( ) ρk⋅ g⋅ Ho + H1 = ρwater⋅ g⋅ H1 Solving for H1 H1 = ρ k⋅ H o ρwater − ρk = SGk⋅ Ho 1 − SGk H1 = 0.82 × 20⋅ mm 1 − 0.82 Now under the applied gage pressure: ( ) pA = ρk⋅ g⋅ Ho + H1 + ρwater⋅ g⋅ l H h ( ) pB = ∆p + ρwater⋅ g⋅ H1 − l H1 = 91.11⋅ mm A B Setting these pressures equal: ( ) ( ) ∆p SGk⋅ Ho + H1 + l = + H1 − l ρwater⋅ g l= ∆p ⎤ + H1 − SGk⋅ (Ho + H1)⎥ ⎢ 2 ρwater⋅ g ⎣ ⎦ 1⎡ Substituting in known values we get: l = 1 2 ⎡ N ⎢ ⎣ m × ⎢98.0⋅ 2 3 × 1 m 999 kg × 1 2 ⋅ s 9.81 m × kg⋅ m 2 N⋅ s + [ 91.11⋅ mm − 0.82 × ( 20⋅ mm + 91.11⋅ mm) ] × ⎤ ⎥ 3 ⎥ 10 ⋅ mm⎦ m l = 5.000⋅ mm Now we solve for H: H = 20⋅ mm + 2 × 5.000⋅ mm H = 30.0⋅ mm Problem 3.23 Given: Data on manometer Find: Deflection due to pressure difference [Difficulty: 2] Solution: Basic equation dp dy = − ρ⋅ g or, for constant ρ ∆p = ρ⋅ g⋅ ∆h where h is measured downwards Starting at p1 pA = p1 + SGA⋅ ρ⋅ g⋅ ( h + l ) Next, from A to B pB = pA − SGB⋅ ρ⋅ g⋅ h Finally, from A to the location of p2 p2 = pB − SGA⋅ ρ⋅ g⋅ l Combining the three equations p2 = pA − SGB⋅ ρ⋅ g⋅ h − SGA⋅ ρ⋅ g⋅ l = ⎡p1 + SGA⋅ ρ⋅ g⋅ ( h + l) − SGB⋅ ρ⋅ g⋅ h⎤ − SGA⋅ ρ⋅ g⋅ l ⎣ ⎦ ( where l is the (unknown) distance from the level of the right interface ) ( ) p2 − p1 = SGA − SGB ⋅ ρ⋅ g⋅ h h= p1 − p2 (SGB − SGA)⋅ ρ⋅ g h = 18⋅ lbf ft 2 h = 0.139⋅ ft × 1 ( 2.95 − 0.88) × 1 ⋅ ft 3 1.94 slug h = 1.67⋅ in × 1 2 ⋅ s 32.2 ft × slug⋅ ft 2 s ⋅ lbf Problem 3.24 Given: Data on manometer Find: Gage pressure at point a Assumption: [Difficulty: 2] e c Water, liquids A and B are static and incompressible d Solution: Basic equation dp dy or, for constant ρ ρ g ∆p ρ g ∆h where ∆h is height difference Starting at point a p1 pa ρH2O g h1 where h1 0.125 m 0.25 m Next, in liquid A p2 p1 SGA ρH2O g h2 where h2 0.25 m Finally, in liquid B patm p2 SGB ρH2O g h3 where h3 0.9 m 0.4 m h1 0.375 m h3 0.5 m Combining the three equations patm p1 SGA ρH2O g h2 SGB ρH2O g h3 pa ρH2O g h1 SGA ρH2O g h2 SGB ρH2O g h3 pa patm ρH2O g h1 SGA h2 SGB h3 or in gage pressures pa ρH2O g h1 SGA h2 SGB h3 pa 1000 kg 3 9.81 m 3 pa 4.41 10 Pa m 2 2 [ 0.375 ( 1.20 0.25) ( 0.75 0.5) ] m s pa 4.41 kPa (gage) N s kg m Problem 3.25 [Difficulty: 2] SGoil 0.827 from Table A.1 Given: Two fluid manometer, Meriam red oil is the second fluid Find: The amplification factor which will be seen in this demonstrator Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g (Hydrostatic Pressure - h is positive downwards) ρ SG ρwater Assumptions: (Definition of Specific Gravity) (1) Static liquid (2) Incompressible liquid h Integrating the hydrostatic pressure equation we get: p p o ρ g h For the left leg of the manometer: b hA hB pa patm ρwater g hA pb pa ρwater g l patm ρwater g hA l l a pa patm ρwater g hB For the right leg: pb pa ρoil g l patm ρwater g hB SGoil l Combining the right hand sides of these two equations: Upon simplification: hA l hB SGoil l AF l ∆h 1 1 SGoil patm ρwater g hA l patm ρwater g hB SGoil l ∆h hA hB l 1 SGoil For Meriam red AF 1 1 0.827 so the amplification factor would be: 5.78 AF 5.78 Problem 3.26 [Difficulty: 2] Given: Water flow in an inclined pipe as shown. The pressure difference is measured with a two-fluid manometer L 5 ft h 6 in SGHg 13.55 (From Table A.1, App. A) Find: Pressure difference between A and B Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g (Hydrostatic Pressure - h is positive downwards) ρ SG ρwater Assumptions: (Definition of Specific Gravity) (1) Static liquid (2) Incompressible liquid (3) Gravity is constant Integrating the hydrostatic pressure equation we get: ∆p ρ g ∆h Progressing through the manometer from A to B: pA ρwater g L sin( 30 deg) ρwater g a ρwater g h ρHg g h ρwater g a pB Simplifying terms and solving for the pressure difference: ∆p pA pB ρwater g h SGHg 1 L sin( 30 deg) Substituting in values: ∆p 1.94 slug ft 3 32.2 ft 2 s 6 in ft 12 in ( 13.55 1) 5 ft sin( 30 deg) 2 ft slugft 12 in lbf s 2 ∆p 1.638 psi Problem 3.27 Given: Data on fluid levels in a tank Find: Air pressure; new equilibrium level if opening appears [Difficulty: 2] Solution: Using Eq. 3.8, starting from the open side and working in gage pressure pair = ρH2O × g × ⎡SGHg × ( 0.3 − 0.1) ⋅ m − 0.1 ⋅ m − SGBenzene× 0.1 ⋅ m⎤ ⎣ ⎦ Using data from Table A.2 pair = 999⋅ kg m 3 × 9.81⋅ m 2 2 × ( 13.55 × 0.2 ⋅ m − 0.1 ⋅ m − 0.879 × 0.1 ⋅ m ) × s N ⋅s kg ⋅ m pair = 24.7⋅ kPa To compute the new level of mercury in the manometer, assume the change in level from 0.3 m is an increase of x. Then, because the volume of mercury is constant, the tank mercury level will fall by distance (0.025/0.25)2x. Hence, the gage pressure at the bottom of the tan can be computed from the left and the right, providing a formula for x ⎡ ⎛ 0.025 ⎞ ⎤⎥ ⋅ m ... SGHg × ρH2O × g × ( 0.3⋅ m + x) = SGHg × ρH2O × g × ⎢0.1⋅ m − x⋅ ⎜ ⎟ ⎣ ⎝ 0.25 ⎠ ⎦ + ρH2O × g × 0.1 ⋅ m + SGBenzene × ρH2O × g × 0.1 ⋅ m 2 Hence The new manometer height is x = [ 0.1⋅ m + 0.879 × 0.1⋅ m + 13.55 × ( 0.1 − 0.3) ⋅ m] 2 ⎡ 0.025 ⎞ ⎤ ⎢1 + ⎛⎜ ⎟ ⎥ × 13.55 ⎣ ⎝ 0.25 ⎠ ⎦ h = 0.3⋅ m + x x = −0.184 m (The negative sign indicates the manometer level actually fell) h = 0.116 m Problem 3.28 Given: [Difficulty: 2] Reservoir manometer with vertical tubes of knowm diameter. Gage liquid is Meriam red oil D = 18⋅ mm d = 6⋅ mm SGoil = 0.827 (From Table A.1, App. A) Find: The manometer deflection, L when a gage pressure equal to 25 mm of water is applied to the reservoir. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh = ρ⋅ g (Hydrostatic Pressure - h is positive downwards) ρ = SG⋅ ρwater Assumptions: (Definition of Specific Gravity) (1) Static liquid (2) Incompressible liquid Integrating the hydrostatic pressure equation we get: ∆p = ρ⋅ g⋅ ∆h Beginning at the free surface of the reservoir, and accounting for the changes in pressure with elevation: patm + ∆p + ρoil⋅ g⋅ ( x + L) = patm ∆p x+L = Upon simplification: The gage pressure is defined as: ρoil⋅ g Combining these two expressions: x+L = ρwater⋅ g⋅ h ρoil⋅ g x and L are related through the manometer dimensions: L= Therefore: ∆h 2⎤ ⎡ d SGoil⋅ ⎢1 + ⎛⎜ ⎞⎟ ⎥ D ⎣ (Note: s = L ∆h which yields = π 4 ∆h SGoil 2 ⋅D ⋅x = π 2 ⋅d ⋅L 4 2 ⎛d⎞ L ⎟ ⎝D⎠ x=⎜ Substituting values into the expression: L = 25⋅ mm ⎡ 6⋅ mm ⎞ ⎤ ⎟⎥ ⎝ 18⋅ mm ⎠ ⎦ 0.827⋅ ⎢1 + ⎛⎜ ⎝ ⎠⎦ s = 1.088 ∆p = ρwater⋅ g⋅ ∆h where ∆h = 25⋅ mm ⎣ for this manometer.) 2 L = 27.2⋅ mm Problem 3.29 [Difficulty: 2] Given: A U-tube manometer is connected to the open tank filled with water as shown (manometer fluid is Meriam blue) D1 2.5 m D2 0.7 m d 0.2 m SGoil 1.75 (From Table A.1, App. A) Find: The manometer deflection, l Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g (Hydrostatic Pressure - h is positive downwards) ρ SG ρwater Assumptions: (Definition of Specific Gravity) (1) Static liquid (2) Incompressible liquid Integrating the hydrostatic pressure equation we get: ∆p ρ g ∆h When the tank is filled with water, the oil in the left leg of the manometer is displaced downward by l/2. The oil in the right leg is displaced upward by the same distance, l/2. D1 Beginning at the free surface of the tank, and accounting for the changes in pressure with elevation: d D2 l patm ρwater g D1 D2 d ρoil g l patm 2 Upon simplification: ρwater g D1 D2 d l ρoil g l 2 D1 D2 d l l SGoil l 2 2.5 m 0.7 m 0.2 m 1.75 1 2 l D1 D2 d 1 SGoil 2 l 1.600 m Problem 3.30 Given: [Difficulty: 2] Reservoir manometer with dimensions shown. The manometer fluid specific gravity is given. D 5 8 in d 3 16 in SGoil 0.827 Find: The required distance between vertical marks on the scale corresponding to Δp of 1 in water. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dz ρ g (Hydrostatic Pressure - z is positive upwards) ρ SG ρwater Assumptions: (Definition of Specific Gravity) (1) Static liquid (2) Incompressible liquid Integrating the hydrostatic pressure equation we get: ∆p ρ g ∆z h Beginning at the free surface of the tank, and accounting for the changes in pressure with elevation: patm ∆p ρoil g ( x h) patm Upon simplification: ∆p ρoil g ( x h) Therefore: ρwater g l ρoil g ( x h) The applied pressure is defined as: x and h are related through the manometer dimensions: Solving for h: h l 2 d SGoil 1 D x xh π 4 2 D x ∆p ρwater g l where l SGoil π 2 d h 4 Substituting values into the expression: 2 d h D x h 1 in 2 0.1875 in 0.625 in 0.827 1 h 1.109 in l 1 in Problem 3.31 Given: [Difficulty: 2] A U-tube manometer is connected to the open tank filled with water as shown (manometer fluid is mercury). The tank is sealed and pressurized. D1 2.5 m D2 0.7 m d 0.2 m po 0.5 atm Find: The manometer deflection, l Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g (Hydrostatic Pressure - h is positive downwards) ρ SG ρwater Assumptions: SGHg 13.55 (From Table A.1, App. A) (Definition of Specific Gravity) (1) Static liquid (2) Incompressible liquid Integrating the hydrostatic pressure equation we get: ∆p ρ g ∆h When the tank is filled with water and pressurized, the mercury in the left leg of the manometer is displaced downward by l/2. The mercury in the right leg is displaced upward by the same distance, l/2. D1 Beginning at the free surface of the tank, and accounting for the changes in pressure with elevation: d D2 l patm po ρwater g D1 D2 d ρHg g l patm 2 Upon simplification: po ρwater g D1 D2 d po l ρHg g l 2 l ρwater g D1 D2 d 1 SGHg 2 Substituting values into the expression: 5 3 2 0.5 atm 1.013 10 N 1 m 1 s 2.5 m 0.7 m 0.2 m 2 1000 kg 9.8 m m atm l 13.55 1 2 l 0.549 m Problem 3.32 [Difficulty: 3] Given: Inclined manometer as shown. D 96 mm d 8 mm Angle θ is such that the liquid deflection L is five times that of a regular U-tube manometer. Find: Angle θ and manometer sensitivity. Solution: We will apply the hydrostatics equations to this system. Governing Equation: dp dz Assumptions: ρ g (Hydrostatic Pressure - z is positive upwards) (1) Static liquid (2) Incompressible liquid Integrating the hydrostatic pressure equation we get: ∆p ρ g ∆z Applying this equation from point 1 to point 2: p1 ρ g ( x L sin ( θ) ) p2 Upon simplification: x p1 p2 ρ g ( x L sin ( θ) ) Since the volume of the fluid must remain constant: π 4 2 D x π 2 d L 4 2 d L D x d Therefore: p1 p2 ρ g L sin ( θ) D 2 Now for a U-tube manometer: p1 p2 ρ g h For equal applied pressures: L Hence: d 2 sin ( θ) h D p1incl p2incl p1U p2U Since L/h = 5: d 2 sin ( θ) D ρ g L sin ( θ) ρ g h h L 2 d 1 8 mm 5 96 mm D 2 θ 11.13 deg The sensitivity of the manometer: s L ∆he L SG h s 5 SG Problem 3.33 Given: Data on inclined manometer Find: Angle θ for given data; find sensitivity [Difficulty: 3] Solution: Basic equation dp dy ρ g or, for constant ρ where Δh is height difference ∆p ρ g ∆h Under applied pressure ∆p SGMer ρ g ( L sin( θ) x) From Table A.1 SGMer 0.827 and Δp = 1 in. of water, or ∆p ρ g h ∆p 1000 kg 3 9.81 h 0.025 m 2 m 2 0.025 m s The volume of liquid must remain constant, so x Ares L Atube Solving for θ h 25 mm where m Combining Eqs 1 and 2 (1) x L Atube Ares N s ∆p 245 Pa kg m d D 2 L (2) 2 d ∆p SGMer ρ g L sin ( θ) L D sin ( θ) ∆p SGMer ρ g L sin ( θ) 245 N 2 m d D 2 1 0.827 1 The sensitivity is the ratio of manometer deflection to a vertical water manometer L h 0.15 m 0.025 m m 1000 kg θ 11 deg s 3 s6 1 2 s 9.81 m 1 1 0.15 m kg m 2 s N 2 8 0.186 76 Problem 3.34 Given: [Difficulty: 4] Barometer with water on top of the mercury column, Temperature is known: h2 6.5 in h1 28.35 in SGHg 13.55 T 70 °F (From Table A.2, App. A) pv 0.363 psi (From Table A.7, App. A) Find: (a) Barometric pressure in psia (b) Effect of increase in ambient temperature on length of mercury column for the same barometric pressure: Tf 85 °F Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g (Hydrostatic Pressure - h is positive downwards) ρ SG ρwater Assumptions: (Definition of Specific Gravity) (1) Static liquid (2) Incompressible liquid Water vapor Water Integrating the hydrostatic pressure equation we get: h2 ∆p ρ g ∆h Mercury Start at the free surface of the mercury and progress through the barometer to the vapor pressure of the water: h1 patm ρHg g h1 ρwater g h2 pv patm pv ρwater g SGHg h1 h2 patm 0.363 lbf 2 in 1.93 slug ft 3 32.2 ft 2 s 2 lbf s slug ft ft 12 in ( 13.55 28.35 in 6.5 in) 3 patm 14.41 At the higher temperature, the vapor pressure of water increases to 0.60 psi. Therefore, if the atmospheric pressure were to remain constant, the length of the mercury column would have to decrease - the increased water vapor would push the mercury out of the tube! lbf 2 in Problem 3.35 Given: [Difficulty: 3] U-tube manometer with tubes of different diameter and two liquids, as shown. d1 = 10⋅ mm d2 = 15⋅ mm SGoil = 0.85 Find: Solution: ∆p = 250⋅ (a) the deflection, h, corresponding to (b) the sensitivity of the manometer N 2 m We will apply the hydrostatics equations to this system. Governing Equations: dp dz = − ρ⋅ g (Hydrostatic Pressure - z is positive upwards) ρ = SG⋅ ρwater Assumptions: (Definition of Specific Gravity) (1) Static liquid (2) Incompressible liquid patm patm + Δp patm patm Integrating the hydrostatic pressure equation we get: ( ) ( ) p − po = −ρ⋅ g⋅ z − zo = ρ⋅ g⋅ zo − z h From the left diagram: l2 pA − patm = ρwater⋅ g⋅ l1 = ρoil⋅ g⋅ l2 l3 lw l1 ( 1) l4 A B From the right diagram: pB − patm + ∆p = ρwater⋅ g⋅ l3 ( ) ( 2) pB − patm = ρwater⋅ g⋅ l4 + ρoil⋅ g⋅ l2 ( 3) Combining these three equations: From the diagram we can see ( ) ( ) ∆p = ρwater⋅ g⋅ l4 − l3 + ρoil⋅ g⋅ l2 = ρwater⋅ g⋅ l4 + l1 − l3 lw = l1 − l3 ( ∆p = ρwater⋅ g⋅ h + lw ) and h = l4 Therefore: ( 4) 2 We can relate lw to h since the volume of water in the manometer is constant: π 4 2 π 2 ⋅ d1 ⋅ lw = ⋅ d2 ⋅ h 4 ⎛ d2 ⎞ lw = ⎜ ⎟ ⋅ h ⎝ d1 ⎠ 2⎤ ⎡ ⎢ ⎛ d2 ⎞ ⎥ ∆p = ρwater⋅ g⋅ h⋅ 1 + ⎜ ⎟ ⎢ ⎥ ⎣ ⎝ d1 ⎠ ⎦ Substituting this into (4) yields: Substituting values into the equation: N h = 250⋅ 2 × m The sensitivity for the manometer is defined as: Therefore: s= 1 ⎛ d2 ⎞ 1+⎜ ⎟ ⎝ d1 ⎠ 2 s= s = 1 3 ⋅ m 999 kg h × Solving for h: 2 1 s 9.81 m × where ∆he 1 ⎛ 15⋅ mm ⎞ 1+⎜ ⎟ ⎝ 10⋅ mm ⎠ ∆p h= 1 ⎛ 15⋅ mm ⎞ 1+⎜ ⎟ ⎝ 10⋅ mm ⎠ 2 2⎤ ⎡ ⎢ ⎛ d2 ⎞ ⎥ ρwater⋅ g⋅ 1 + ⎜ ⎟ ⎢ ⎥ ⎣ ⎝ d1 ⎠ ⎦ × kg⋅ m 2 N⋅ s 3 × 10 ⋅ mm h = 7.85⋅ mm m ∆p = ρwater⋅ g⋅ ∆he s = 0.308 2 The design is a poor one. The sensitivity could be improved by interchanging d2 and d1 , i.e., having d2 smaller than d1 A plot of the manometer sensitivity is shown below: Sensitivity 1 0.5 0 1 2 Diameter Ratio, d2/d1 3 4 5 Problem 3.36 Given: [Difficulty: 3] Water column standin in glass tube ∆h = 50⋅ mm D = 2.5⋅ mm σ = 72.8 × 10 −3N m (From Table A.4, App. A) Find: (a) Column height if surface tension were zero. (b) Column height in 1 mm diameter tube Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh = ρ⋅ g (Hydrostatic Pressure - h is positive downwards) ΣFz = 0 Assumptions: (Static Equilibrium) (1) Static, incompressible liquid (2) Neglect volume under meniscus (3) Applied pressure remains constant (4) Column height is sum of capillary rise and pressure difference Δhc Δh Δhp ∆h = ∆hc + ∆hp Assumption #4 can be written as: Choose a free-body diagram of the capillary rise portion of the column for analysis: ΣFz = π⋅ D⋅ σ⋅ cos ( θ) − 4⋅ σ Therefore: ∆hc = ⋅ cos ( θ) ρ⋅ g⋅ D π 2 ⋅ D ⋅ ρ⋅ g⋅ ∆hc = 0 4 θ Substituting values: ⎛ 103⋅ mm ⎞ ⎟ ×⎜ ∆hc = 4 × 72.8 × 10 ⋅ × ⋅ × ⋅ × ⋅ × 2 ⎝ m 999 kg 9.81 m 2.5 mm m ⎠ N⋅ s −3 N 1 3 m 1 2 s 1 kg⋅ m 1 2 Δhc ∆hc = 11.89⋅ mm Therefore: ∆hp = ∆h − ∆hc ∆hp = 50⋅ mm − 11.89⋅ mm π Dδ ∆hp = 38.1⋅ mm Mg = ρgV (result for σ = 0) For the 1 mm diameter tube: ⎛ 103⋅ mm ⎞ ⎟ ∆hc = 4 × 72.8 × 10 ⋅ × ⋅ × ⋅ × ⋅ × ×⎜ m 999 kg 9.81 m 1 mm N⋅ s2 ⎝ m ⎠ −3 N ∆h = 29.7⋅ mm + 38.1⋅ mm 1 3 m 1 2 s 1 1 kg⋅ m 2 ∆hc = 29.71⋅ mm ∆h = 67.8⋅ mm Problem 3.37 Given: [Difficulty: 4] Sealed tank is partially filled with water. Water drains slowly from the tank until the system attains equilibrium. U-tube manometer is connected to the tank as shown. (Meriam blue in manometer) L 3 m D1 2.5 m D2 0.7 m d 0.2 m Find: The manometer deflection, l, under equilibrium conditions Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh Assumptions: ρ g SGoil 1.75 (From Table A.2, App. A) (Hydrostatic Pressure - h is positive downwards) p V M R T (Ideal gas equation of state) ρ SG ρwater (Definition of Specific Gravity) (1) Static liquid (2) Incompressible liquid (3) Air in tank behaves ideally patm p0 Integrating the hydrostatic pressure equation we get: L ∆p ρ g ∆h D1 d H D2 To determine the surface pressure po under equilibrium conditions po V o M R Ta Thus, M R To po L D1 Simplifying: po p ( L H) a L D 1 p ( L H) a ρwater g H pa Va Vo pa L D 1 A p ( L H) A a po ρwater g H pa Now under equilibrium conditions: Upon rearranging: Now we apply the quadratic formula to solve for H: l we assume that the air expands at constant temperature: pa Va 2 Combining these expressions: ρwater g H pa ρwater g L H D1 pa 0 a ρwater g a 999 kg 3 9.81 m b pa ρwater g L m 3 Pa a 9.8 10 2 s kg m 5 N b 1.013 10 999 9.81 3 m 2 3 2 m m s 5 N 2 c D 1 pa m 5 b 1.307 10 Pa 5 c 2.5 m 1.013 10 c 2.532 10 Pa m m Hupper b 2 b 4 a c 2 a 1.307 105 Pa2 4 9.8 103 Pa 2.532 105 Pa m 5 1.307 10 Pa m Hupper 3 Pa 2 9.8 10 m Hupper 10.985 m Hlower b b 4 a c 2 a 1.307 105 Pa2 4 9.8 103 Pa 2.532 105 Pa m 5 1.307 10 Pa 2 m Hlower 3 Pa 2 9.8 10 m Hlower 2.352 m H 2.352 m Since H can not be greater than 3 m (otherwise the tank would overflow!), we must select the lower value for H: Solving for the pressure inside the tank: po ( 3 m 2.5 m) 5 1.013 10 Pa ( 3 m 2.352 m) l po ρwater g H D2 d ρoil g l pa 2 Applying the hydrostatic pressure equation to the manometer: Solving for the manometer deflection: pa po l ρwater g H D2 d 1 SGoil 1 2 1 m 1 s kg m 1 5 N 4 N l 1.013 10 7.816 10 2.352 m 0.7 m 0.2 m 2 2 2 999 kg 9.81 m 1 m m N s 1.75 2 3 4 po 7.816 10 Pa 2 l 0.407 m Problem 3.38 [Difficulty :2] Fluid 1 Fluid 2 Given: Two fluids inside and outside a tube Find: (a) An expression for height Δh (b) Height difference when D =0.040 in for water/mercury Assumptions: ρ1gΔhπD2/4 (1) Static, incompressible fluids (2) Neglect meniscus curvature for column height and volume calculations Solution: A free-body vertical force analysis for the section of fluid 1 height Δh in the tube below the "free surface" of fluid 2 leads to ∑ 2 F = 0 = ∆p⋅ π⋅ D 2 − ρ1⋅ g⋅ ∆h⋅ 4 π⋅ D 4 + π⋅ D⋅ σ⋅ cos ( θ) where Δp is the pressure difference generated by fluid 2 over height Δh, 2 π⋅ D Hence ∆p⋅ Solving for Δh ∆h = − 4 2 − ρ1⋅ g⋅ ∆h⋅ 4⋅ σ⋅ cos ( θ) ( σπDcosθ π⋅ D 4 ∆p = ρ2⋅ g⋅ ∆h 2 2 π⋅ D π⋅ D = ρ2⋅ g⋅ ∆h⋅ − ρ1⋅ g⋅ ∆h⋅ = −π⋅ D⋅ σ⋅ cos ( θ) 4 4 ) g⋅ D ⋅ ρ 2 − ρ 1 For fluids 1 and 2 being water and mercury (for mercury σ = 375 mN/m and θ = 140o, from Table A.4), solving for Δh when D = 0.040 in ∆h = −4 × 0.375⋅ ∆h = 0.360⋅ in N m × lbf 4.448⋅ N × 0.0254m in 2 × cos ( 140⋅ deg) × s 32.2⋅ ft × 1 0.040⋅ in × 3 3 1 slugft ⋅ ⎛ 12⋅ in ⎞ × × ⎟ 2 1.94⋅ slug ⎝ ft ⎠ ( 13.6 − 1) lbf⋅ s ft ×⎜ Problem 3.39 [Difficulty: 2] h1 Oil Air h4 h2 Hg Given: Data on manometer before and after an "accident" Find: Change in mercury level Assumptions: h3 x (1) Liquids are incompressible and static (2) Pressure change across air in bubble is negligible (3) Any curvature of air bubble surface can be neglected in volume calculations Solution: Basic equation dp dy ρ g or, for constant ρ For the initial state, working from right to left where ∆h is height difference ∆p ρ g ∆h patm patm SGHg ρ g h3 SGoil ρ g h1 h2 SGHg ρ g h3 SGoil ρ g h1 h2 (1) Note that the air pocket has no effect! For the final state, working from right to left patm patm SGHg ρ g h3 x SGoil ρ g h4 SGHg ρ g h3 x SGoil ρ g h4 (2) The two unknowns here are the mercury levels before and after (i.e., h3 and x) Combining Eqs. 1 and 2 SGHg ρ g x SGoil ρ g h1 h2 h4 From Table A.1 SGHg 13.55 The term h1 h2 h4 h1 h2 h4 Then from Eq. 3 x 1.4 13.55 x SGoil SGHg h1 h2 h4 (3) is the difference between the total height of oil before and after the accident ∆V π d2 4 1.019 in 2 1 0.2 in3 1.019 in π 0.5 in 4 x 0.1053 in Problem 3.40 [Difficulty: 2] Water Given: Water in a tube or between parallel plates Find: Height Δh for each system Solution: a) Tube: A free-body vertical force analysis for the section of water height Δh above the "free surface" in the tube, as shown in the figure, leads to ∑ 2 F = 0 = π⋅ D⋅ σ⋅ cos ( θ) − ρ⋅ g⋅ ∆h⋅ π⋅ D 4 Assumption: Neglect meniscus curvature for column height and volume calculations Solving for Δh ∆h = 4⋅ σ⋅ cos ( θ) ρ ⋅ g⋅ D b) Parallel Plates: A free-body vertical force analysis for the section of water height Δh above the "free surface" between plates arbitrary width w (similar to the figure above), leads to ∑ F = 0 = 2⋅ w⋅ σ⋅ cos(θ) − ρ⋅ g⋅ ∆h⋅ w⋅ a Solving for Δh ∆h = 2⋅ σ⋅ cos ( θ) ρ ⋅ g⋅ a For water σ = 72.8 mN/m and θ = 0o (Table A.4), so 4 × 0.0728⋅ a) Tube ∆h = 999⋅ kg 3 × 9.81⋅ m m 2 ∆h = 999⋅ kg 3 m m × kg⋅ m 2 −3 ∆h = 5.94 × 10 m ∆h = 5.94⋅ mm m ∆h = 2.97⋅ mm N⋅ s × 0.005⋅ m s 2 × 0.0728⋅ b) Parallel Plates N × 9.81⋅ m 2 s N m × 0.005⋅ m × kg⋅ m 2 N⋅ s −3 ∆h = 2.97 × 10 σ= ρ= 0.005 1.94 lbf/ft 3 slug/ft Using the formula above Δh (in) 0.0400 0.0200 0.0133 0.0100 0.0080 0.0067 0.0057 0.0050 0.0044 0.0040 0.0036 0.0033 0.0031 0.0029 0.0027 0.0025 0.0024 0.0022 0.0020 Capillary Height Between Vertical Plates 0.045 Height Δh (in) a (in) 0.004 0.008 0.012 0.016 0.020 0.024 0.028 0.032 0.036 0.040 0.044 0.048 0.052 0.056 0.060 0.064 0.068 0.072 0.080 0.040 0.035 0.030 0.025 0.020 0.015 0.010 0.005 0.000 0.00 0.01 0.02 0.03 0.04 Gap a (in) 0.05 0.06 0.07 0.08 p SL = R = = 101 286.9 999 kPa J/kg.K kg/m3 The temperature can be computed from the data in the figure. The pressures are then computed from the appropriate equation. From Table A.3 Atmospheric Pressure vs Elevation 1.00000 0 10 20 30 40 50 60 70 80 90 0.10000 Pressure Ratio p /p SL 0.01000 0.00100 Computed 0.00010 Table A.3 0.00001 0.00000 Elevation (km) Agreement between calculated and tabulated data is very good (as it should be, considering the table data are also computed!) 100 z (km) T (oC) T (K) 0.0 2.0 4.0 6.0 8.0 11.0 12.0 14.0 16.0 18.0 20.1 22.0 24.0 26.0 28.0 30.0 32.2 34.0 36.0 38.0 40.0 42.0 44.0 46.0 47.3 50.0 52.4 54.0 56.0 58.0 60.0 61.6 64.0 66.0 68.0 70.0 72.0 74.0 76.0 78.0 80.0 82.0 84.0 86.0 88.0 90.0 15.0 2.0 -11.0 -24.0 -37.0 -56.5 -56.5 -56.5 -56.5 -56.5 -56.5 -54.6 -52.6 -50.6 -48.7 -46.7 -44.5 -39.5 -33.9 -28.4 -22.8 -17.2 -11.7 -6.1 -2.5 -2.5 -2.5 -5.6 -9.5 -13.5 -17.4 -20.5 -29.9 -37.7 -45.5 -53.4 -61.2 -69.0 -76.8 -84.7 -92.5 -92.5 -92.5 -92.5 -92.5 -92.5 288.0 275.00 262.0 249.0 236.0 216.5 216.5 216.5 216.5 216.5 216.5 218.4 220.4 222.4 224.3 226.3 228.5 233.5 239.1 244.6 250.2 255.8 261.3 266.9 270.5 270.5 270.5 267.4 263.5 259.5 255.6 252.5 243.1 235.3 227.5 219.6 211.8 204.0 196.2 188.3 180.5 180.5 180.5 180.5 180.5 180.5 m = 0.0065 (K/m) T = const m = -0.000991736 (K/m) m = -0.002781457 (K/m) T = const m = 0.001956522 (K/m) m = 0.003913043 (K/m) T = const p /p SL z (km) p /p SL 1.000 0.784 0.608 0.465 0.351 0.223 0.190 0.139 0.101 0.0738 0.0530 0.0393 0.0288 0.0211 0.0155 0.0115 0.00824 0.00632 0.00473 0.00356 0.00270 0.00206 0.00158 0.00122 0.00104 0.000736 0.000544 0.000444 0.000343 0.000264 0.000202 0.000163 0.000117 0.0000880 0.0000655 0.0000482 0.0000351 0.0000253 0.0000180 0.0000126 0.00000861 0.00000590 0.00000404 0.00000276 0.00000189 0.00000130 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 13.0 14.0 15.0 16.0 17.0 18.0 19.0 20.0 22.0 24.0 26.0 28.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 1.000 0.942 0.887 0.835 0.785 0.737 0.692 0.649 0.609 0.570 0.533 0.466 0.406 0.352 0.304 0.262 0.224 0.192 0.164 0.140 0.120 0.102 0.0873 0.0747 0.0638 0.0546 0.0400 0.0293 0.0216 0.0160 0.0118 0.00283 0.000787 0.000222 0.0000545 0.0000102 0.00000162 Problem 3.43 [Difficulty: 3] Given: Data on isothermal atmosphere Find: Elevation changes for 3% pressure change and 5% density change; plot of pressure and density versus elevation Solution: Assumptions: Static, isothermal fluid,; g = constant; ideal gas behavior dp Basic equations dz dp Then dz For an ideal gas with T constant p2 p1 g ρ1 Rair T Rair 287 Evaluating C For a 3% reduction in pressure For a 5% reduction in density To plot p2 p1 and ρ2 ρ1 we rearrange Eq. 1 p2 p1 ρ2 ρ1 ρ2 ρ1 p2 p1 ln ρ2 Rair T From Table A.6 and Rair T Rair T0 p ρ R T and p g ρ g ∆z Integrating ρ g dp p where T T0 so ∆z ρ2 ρ1 g Rair T dz Rair T0 g ρ2 ρ2 C ln ρ1 ρ1 ln (1) N m kg K Rair T0 287 g N m kg K ( 30 273) K 1 2 s 9.81 m kg m 2 C 8865 m N s 0.97 so from Eq. 1 ∆z 8865 m ln ( 0.97) ∆z 270 m 0.95 so from Eq. 1 ∆z 8865 m ln ( 0.95) ∆z 455 m p2 p1 e ∆z C 5000 Elevation (m) 4000 3000 2000 1000 0.5 0.6 0.7 0.8 Pressure or Density Ratio This plot can be plotted in Excel 0.9 1 Problem 3.44 [Difficulty: 3] Given: Atmospheric conditions at ground level (z = 0) in Denver, Colorado are p0 = 83.2 kPa, T0 = 25°C. Pike's peak is at elevation z = 2690 m. Find: p/p0 vs z for both cases. Solution: dp Governing Equations: Assumptions: ρ g dz p ρ R T (1) Static fluid (2) Ideal gas behavior (a) For an incompressible atmosphere: dp dz At z ρ g p p 0 ρ g dz 0 becomes p (b) For an adiabatic atmosphere: ρ dp dz m s p 83.2 kPa 1 9.81 z 2690 m k g z p p0 ρ0 g z p0 1 R T0 or 2 2690 m p ρ ρ 0 p0 const p dp ρ0 g dz p0 becomes 287 N m or 1 p p But k1 1 k k dp p p0 1 k1 k p p 1 298 K 2 N s p 57.5 kPa kg m 1 k 1 k ρ g kg K (1) 1 k dp ρ 0 g p0 1 k dz k1 k1 ρ 0 g k k k p p0 g z 1 k1 hence p0 k 0 Solving for the pressure ratio k 1 ρ0 p 1 g z p0 k p0 At z 2690 m k k1 or 1.4 1 1.4 p 83.2 kPa 1 9.81 p k 1 g z 1 p0 k R T0 m 2 s 2690 m kg K 287 N m k k1 (2) 1 298 K 2 N s kg m 1.4 1.41 p 60.2 kPa Elevation above Denver (m) Equations 1 and 2 can be plotted: 510 3 410 3 310 3 210 3 110 3 Temperature Variation with Elevation Incompressible Adiabatic 0 0.4 0.6 0.8 Pressure Ratio (-) 1 Problem 3.45 Given: [Difficulty: 3] Martian atmosphere behaves as an idel gas, constant temperature m Mm 32.0 T 200 K g 3.92 2 s ρo 0.015 kg 3 m Find: Density at z = 20 km Plot the ratio of density to sea level density versus altitude, compare to that of earth. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp ρ g dz (Hydrostatic Pressure - z is positive upwards) p ρ R T Ru R Mm Assumptions: (Ideal Gas Equation of State) (Definition of Gas Constant) (1) Static fluid (2) Constant gravitational acceleration (3) Ideal gas behavior dp R T dρ Taking the differential of the equation of state (constant temperature): R T Substituting into the hydrostatic pressure equation: ρ Integrating this expression: dρ z 1 g dρ dz R T ρ 0 ρ N m kg mol K 1 32.0 3.92 ρ 0.015 kg m 3 e For the Martian atmosphere, let m s 2 ρ g z R T ρo kg mol R 259.822 kg 3 20 10 m dρ Therefore: ln o Evaluating: R 8314.3 dz ρ g ρ ρ or ρo g R T e dz g z RT ( 1) N m kg K 1 kg K 1 1 N s 259.822 N m 200 K kg m 2 3 kg 3 ρ 3.32 10 m x g R T x 3.92 m 2 s 1 kg K 259.822 N m 1 1 200 K 2 N s kg m x 0.07544 1 km ρ Therefore: ρo x z e These data are plotted along with the data for Earth's atmosphere from Table A.3. Density Ratios of Earth and Mars versus Elevation 20 Elevation z (km) 15 10 5 Earth Mars 0 0 0.2 0.4 0.6 Density Ratio (-) 0.8 1 Problem 3.46 Given: [Difficulty: 3] Door located in plane vertical wall of water tank as shown a = 1.5⋅ m b = 1⋅ m c = 1⋅ m ps Atmospheric pressure acts on outer surface of door. Find: c Resultant force and line of action: (a) for (b) for y ps = patm y’ a psg = 0.3⋅ atm Plot F/Fo and y'/yc over range of ps/patm (Fo is force determined in (a), yc is y-ccordinate of door centroid). Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dy b = ρ⋅ g (Hydrostatic Pressure - y is positive downwards) ⌠ ⎮ p dA ⎮ ⌡ ⌠ ⎮ y'⋅ FR = y⋅ p dA ⎮ ⌡ FR = Assumptions: (Hydrostatic Force on door) (First moment of force) (1) Static fluid (2) Incompressible fluid We will obtain a general expression for the force and line of action, and then simplify for parts (a) and (b). Since dp = ρ⋅ g⋅ dh Now because patm it follows that p = ps + ρ⋅ g⋅ y acts on the outside of the door, psg is the surface gage pressure: c+ a c+ a ⌠ ⌠ ⌠ ρ⋅ g 2 ⎮ ⎡ ⎤ FR = p dA = ⎮ p⋅ b dy = ⎮ psg + ρ⋅ g⋅ y ⋅ b dy = b⋅ ⎢psg⋅ a + ⋅ a + 2⋅ a⋅ c ⎥ ⎮ ⌡ 2 ⌡c ⎣ ⎦ ⌡ c ( ⌠ ⎮ y'⋅ FR = y⋅ p dA ⎮ ⌡ Therefore: Evaluating the integral: y' = ( ) ) c+ a 1 ⌠ 1 ⌠ ⎮ y' = y⋅ p dA = ⋅⎮ y⋅ psg + ρ⋅ g⋅ y ⋅ b dy FR ⎮ FR ⌡ ⌡ c ( b ⎡ psg ⎡ ρ⋅ g ⎡ 2 2 3 3⎤ ⎢ ⎣( c + a) − c ⎤⎦ + ⋅ ⎣( c + a) − c ⎤⎦⎥ FR ⎣ 2 3 ⎦ ) p = psg + ρ⋅ g⋅ y ( 1) ( psg = 0 For part (a) we know Fo = ) ⎤ b ⎡ psg 2 ρ⋅ g ⎡ 3 ⋅⎢ a + 2⋅ a⋅ c + ⋅ ⎣a + 3⋅ a⋅ c⋅ ( a + c)⎤⎦⎥ FR ⎣ 2 3 ⎦ Simplifying: y' = 1 2 kg × 999⋅ 3 so substituting into (1) we get: × 9.81⋅ m m 1 3 × 999⋅ kg 3 Fo = 2 2 s m × 9.81⋅ 2 m ρ ⋅ g⋅ b 2 (2 y' = 1 × 1⋅ m⋅ 2 N⋅ s Fo = 25.7⋅ kN kg⋅ m ρ ⋅ g⋅ b ⎡ 3 ⋅ ⎣a + 3⋅ a⋅ c⋅ ( a + c)⎤⎦ 3⋅ F o ⋅ 25.7 × 10 s ) ⋅ a + 2⋅ a⋅ c × 1⋅ m × ⎡⎣( 1.5⋅ m) + 2 × 1.5⋅ m × 1⋅ m⎤⎦ × Substituting into (2) for the line of action we get: y' = ( 2) 1 × ⎡⎣( 1.5⋅ m) + 3 × 1.5⋅ m × 1⋅ m × ( 1.5⋅ m + 1⋅ m)⎤⎦ × 3 3 N 2 N⋅ s kg⋅ m y' = 1.9 m psg = 0.3⋅ atm . Substituting into (1) we get: For part (b) we know ⎡ 1.013 × 10 ⋅ N ⎢ ⎣ m ⋅ atm 5 FR = 1⋅ m × ⎢0.3⋅ atm × 2 × 1.5⋅ m + 1 2 × 999⋅ kg 3 × 9.81⋅ m m 2 × ⎡⎣( 1.5⋅ m) + 2 × 1.5⋅ m × 1⋅ m⎤⎦ × 2 s 2 N⋅ s ⎤⎥ kg⋅ m⎥ ⎦ FR = 71.3⋅ kN kg m ⎤ 999⋅ × 9.81⋅ ⎢ 5 3 2⎥ 2 m s 3 3 N⋅ s ⎥ ⎢ 0.3⋅ atm × 1.013 × 10 ⋅ N × ⎡( 1.5) 2 + 2⋅ 1.5⋅ 1⎤ ⋅ m2 + 1⋅ m × × ⎡⎣( 1.5) + 3⋅ 1.5⋅ 1⋅ ( 1.5 + 1)⎤⎦ ⋅ m × ⎣ ⎦ ⎢ 2 2 kg⋅ m⎥ 3 m ⋅ atm ⎣ ⎦ y' = Substituting ⎡into (2) for the line of action we get: 3 71.3 × 10 ⋅ N y' = 1.789 m The value of F/Fo is obtained from Eq. (1) and our result from part (a): F Fo = ⎡ ⎣ b⋅ ⎢psg⋅ a + ρ ⋅ g⋅ b 2 For the gate yc = c + a 2 ρ⋅ g 2 (2 (2 )⎤ ⋅ a + 2⋅ a⋅ c ⎥ ) ⋅ a + 2⋅ a⋅ c ⎦ = 1+ 2⋅ psg ρ⋅ g⋅ ( a + 2⋅ c) Therefore, the value of y'/yc is obtained from Eqs. (1) and (2): ⎡ psg 2 ρ⋅ g ⎡ 3 ⎤⎦⎥⎤ ( ) ⎢ a + 2 ⋅ a ⋅ c + ⋅ a + 3 ⋅ a ⋅ c ⋅ ( a + c ) ⎣ ⎡ psg 2 y' 2⋅ b 3 ⎦ (a + 2⋅ a⋅ c) + ρ⋅ g ⋅ ⎡⎣a3 + 3⋅ a⋅ c⋅ (a + c)⎤⎦⎤⎥ = 2⋅ b ⋅ ⎣ 2 = ⋅⎢ yc FR⋅ ( 2⋅ c + a) ⎣ 2 3 ρ ⋅ g ( 2 ⋅ c + a ) 2 ⎦ ⎡ ⎡ ⎤⎤ ⋅ (a + 2⋅ a⋅ c)⎥⎥ ⎢b⋅ ⎢psg⋅ a + 2 ⎣ ⎣ ⎦⎦ Simplifying this expression we get: y' yc = 2 ( 2⋅ c + a) ( ⋅ ) psg 2 ρ⋅ g ⎡ 3 a + 2⋅ a⋅ c + ⋅ ⎣a + 3⋅ a⋅ c⋅ ( a + c)⎤⎦ 2 3 psg⋅ a + ρ⋅ g 2 (2 ) ⋅ a + 2⋅ a⋅ c Based on these expressions we see that the force on the gate varies linearly with the increase in surface pressure, and that the line of action of the resultant is always below the centroid of the gate. As the pressure increases, however, the line of action moves closer to the centroid. Plots of both ratios are shown below: Force Ratio vs. Surface Pressure 40 Force Ratio F/Fo 30 20 10 0 0 1 2 3 4 5 4 5 Surface Pressure (atm) Line of Action Ratio vs. Surface Pressure 1.05 Line of Action Ratio y'/yc 1.04 1.03 1.02 1.01 1 0 1 2 Surface Pressure (atm) 3 Problem 3.47 Given: [Difficulty: 2] Door of constant width, located in plane vertical wall of water tank is hinged along upper edge. b 1 m D 1 m ps L 1.5 m D x y (a) Force F, if ps patm Find: Hinge h Atmospheric pressure acts on outer surface of door; force F is applied at lower edge to keep door closed. L (b) Force F, if p 0.5 atm sg Plot F/Fo over tange of ps/patm (Fo is force determined in (a)). pdA F Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp ρ g dh FR (Hydrostatic Pressure - h is positive downwards) p dA (Hydrostatic Force on door) ΣMz 0 Assumptions: (Rotational Equilibrium) (1) Static fluid (2) Constant density (3) Door is in equilibrium Taking moments about the hinge: F L y p dA 0 dA b dy L 1 Solving for the force: F b y p dy L 0 Since dp ρ g dh and hence it follows that p p s ρ g ( D y ) L ( 1) We will obtain a general expression for F and then simplify for parts (a) and (b). p p s ρ g h Now because patm 1 From Equation (1): F b y psg ρ g ( D y) dy L 0 where h Dy acts on the outside of the door, L psg is the surface gage pressure. b 2 F psg ρ g D y ρ g y dy L 0 After integrating: F 2 3 b L L psg ρ g D ρ g L 2 3 (a) For ps patm it follows that psg 0 Fo 999 kg 3 9.81 m m 2 F b psg or L 2 D L 2 3 F o ρ g b L Therefore: D L 2 3 ρ g L ( 2) ( 3) 2 1 m 1.5 m N s 3 kg m 2 1 m 1.5 m s Fo 14.7 kN (b) For psg 0.5 atm we substitute variables: 101 kPa atm F 1 m 0.5 atm 1.5 m 2 kg 3 9.81 m F From Equations (2) and (3) we have: 999 Fo m 2 2 1 m 1.5 m N s 3 kg m 2 1.5 m s F 52.6 kN D L psg 2 2 3 1 D L D L ρ g b L 2 ρ g 2 3 2 3 b psg L ρ g L Here is a plot of the force ratio as a function of the surface pressure: Force Ratio, F/Fo 30 20 10 0 0 1 2 3 Surface Pressure Ratio, ps/patm 4 5 Problem 3.48 [Difficulty: 5] Discussion: The design requirements are specified except that a typical floor height is about 12 ft, making the total required lift about 36 ft. A spreadsheet was used to calculate the system properties for various pressures. Results are presented on the next page, followed by a sample calculation. Total cost dropped quickly as system pressure was increased. A shallow minimum was reached in the 100-110 psig range. The lowest-cost solution was obtained at a system pressure of about 100 psig. At this pressure, the reservoir of 140 gal required a 3.30 ft diameter pressure sphere with a 0.250 in wall thickness. The welding cost was $155 and the material cost $433, for a total cost of $588. Accumulator wall thickness was constrained at 0.250 in for pressures below 100 psi; it increased for higher pressures (this caused the discontinuity in slope of the curve at 100 psig). The mass of steel became constant above 110 psig. No allowance was made for the extra volume needed to pressurize the accumulator. Fail-safe design is essential for an elevator to be used by the public. The control circuitry should be redundant. Failures must be easy to spot. For this reason, hydraulic actuation is good: leaks will be readily apparent. The final design must be reviewed, approved, and stamped by a professional engineer since the design involves public safety. The terminology used in the solution is defined in the following table: Symbol Definition Units p System pressure psig Ap Area of lift piston in2 Voil Volume of oil gal Ds Diameter of spherical accumulator ft t Wall thickness of accumulator in Aw Area of weld in2 Cw Cost of weld $ Ms Mass of steel accumulator lbm Cs Cost of steel $ Ct Total Cost $ A sample calculation and the results of the system simulation in Excel are presented below. p πD S2 4 πD S tσ Results of system simulation: Problem 3.49 Given: Geometry of chamber system Find: Pressure at various locations Assumptions: [Difficulty: 2] (1) Water and Meriam Blue are static and incompressible (2) Pressure gradients across air cavities are negligible Solution: Basic equation dp or, for constant ρ ρ g dy where ∆h is height difference ∆p ρ g ∆h For point A pA patm ρH2O g h1 or in gage pressure pA ρH2O g h1 Here we have h1 8 in h1 0.667 ft pA 1.94 slug ft 3 32.2 ft 2 2 0.667 ft s lbf s ft slugft 12 in For the first air cavity pair1 pA SGMB ρH2O g h2 where From Table A.1 SGMB 1.75 lbf pair1 0.289 2 1.75 1.94 slug in ft 3 32.2 ft 2 2 pA 0.289 psi h2 4 in h2 0.333 ft 2 0.333 ft ft slug ft 12 in where h3 6 in s lbf s (gage) 2 pair1 0.036 psi (gage) Note that p = constant throughout the air pocket For point B pB pair1 SGHg ρH2O g h3 pB 0.036 lbf 2 1.75 1.94 in For point C slug ft 3 32.2 ft 2 2 0.5 ft s pC pair2 SGHg ρH2O g h4 pC 0.416 lbf 2 1.75 1.94 in slug ft 3 32.2 ft 2 lbf 2 in 1.75 1.94 ft 3 32.2 pB 0.416 psi h4 10 in 0.833 ft ft slug ft 12 in 2 s slug 2 where For the second air cavity pair2 pC SGHg ρH2O h5 pair2 1.048 ft slug ft 12 in lbf s h3 0.5 ft ft 2 s lbf s h5 6 in 0.5 ft ft slug ft 12 in 2 lbf s h4 0.833 ft 2 where (gage) pC 1.048 psi (gage) h5 0.5 ft 2 pair2 0.668 psi (gage) Problem 3.50 Given: Geometry of gate Find: Force FA for equilibrium [Difficulty: 3] h H = 25 ft FA A y R = 10 ft y B x z Solution: ⌠ ⎮ F R = ⎮ p dA ⌡ Basic equation or, use computing equations dp = ρ⋅ g dh ΣMz = 0 FR = pc⋅ A Ixx y' = yc + A ⋅ yc where y would be measured from the free surface Assumptions: static fluid; ρ = constant; patm on other side; door is in equilibrium Instead of using either of these approaches, we note the following, using y as in the sketch ΣMz = 0 FA = FA ⋅ R = 1 ⌠ ⎮ ⋅ y⋅ ρ ⋅ g⋅ h dA R ⎮ ⌡ ⌠ ⎮ y⋅ p d A ⎮ ⌡ with with dA = r⋅ dr⋅ dθ p = ρ ⋅ g⋅ h and (Gage pressure, since p = patm on other side) y = r⋅ sin ( θ) h = H−y π Hence ⌠ π R 3 4 1 ⌠ ⌠ ρ⋅ g ⎮ ⎛ H ⋅ R R 2⎞ FA = ⋅ ⎮ ⎮ ρ⋅ g⋅ r⋅ sin ( θ) ⋅ ( H − r⋅ sin ( θ) ) ⋅ r dr dθ = ⋅⎮ ⎜ ⋅ sin ( θ) − ⋅ sin ( θ) ⎟ dθ R ⌡0 ⌡0 R 3 4 ⎠ ⌡0 ⎝ FR = Using given data 3 4 ⎛ 2⋅ H ⋅ R 2 π ⋅ R 3 ⎞ ρ ⋅ g ⎛ 2⋅ H ⋅ R π⋅ R ⎞ ⎟ = ρ ⋅ g⋅ ⎜ ⎟ ⋅⎜ − − R ⎝ 3 8 ⎠ 8 ⎠ ⎝ 3 FR = 1.94⋅ slug ft 3 × 32.2⋅ 2 2 π 3⎤ lbf ⋅ s ⎡2 × ⎢ × 25⋅ ft × ( 10⋅ ft) − × ( 10⋅ ft) ⎥ × 2 ⎣3 8 ⎦ slug⋅ ft s ft 4 FR = 7.96 × 10 ⋅ lbf Problem 3.51 Given: Geometry of access port Find: Resultant force and location [Difficulty: 2] y’ y a = 1.25 ft dy w FR SG = 2.5 b = 1 ft Solution: Basic equation FR = ⌠ ⎮ p dA ⎮ ⌡ dp dy ⌠ ⌠ ⎮ ⎮ ΣMs = y'⋅ FR = y dF R = y⋅ p dA ⎮ ⎮ ⌡ ⌡ = ρ⋅ g Ixx y' = yc + A ⋅ yc FR = pc⋅ A or, use computing equations We will show both methods Assumptions: Static fluid; ρ = constant; patm on other side FR = ⌠ ⌠ ⎮ ⎮ p dA = SG⋅ ρ⋅ g⋅ y dA ⎮ ⎮ ⌡ ⌡ a w dA = w⋅ dy and but b y = w = a b a ⋅y a Hence 2 ⌠ ⌠ b b 2 SG⋅ ρ⋅ g⋅ b⋅ a FR = ⎮ SG⋅ ρ⋅ g⋅ y⋅ ⋅ y dy = ⎮ SG⋅ ρ⋅ g⋅ ⋅ y dy = ⎮ ⎮ a a 3 ⌡0 ⌡0 Alternatively FR = pc⋅ A Hence FR = For y' 3 ⌠ ⌠ b 3 SG⋅ ρ⋅ g⋅ b⋅ a ⎮ y'⋅ FR = y⋅ p dA = ⎮ SG⋅ ρ⋅ g⋅ ⋅ y dy = ⎮ ⎮ a 4 ⌡ ⌡0 Alternatively Ixx y' = yc + A ⋅ yc 2 pc = SG⋅ ρ⋅ g⋅ yc = SG⋅ ρ⋅ g⋅ ⋅ a 3 and A = with 1 2 ⋅ a⋅ b 2 SG⋅ ρ⋅ g⋅ b⋅ a 3 a Using given data, and SG = 2.5 (Table A.1) and 3 y' = 3 b⋅ a Ixx = 36 and 2.5 FR = y' = 3 3 4 ⋅a ⋅ 1.94⋅ slug ft 3 (Google it!) × 32.2⋅ ft 2 y' = SG⋅ ρ⋅ g⋅ b⋅ a 4⋅ F R 2 3 2 y' = 0.938⋅ ft 3 4 ⋅a 3 ⋅a + × 1⋅ ft × ( 1.25⋅ ft) × s = b⋅ a 2 3 3 ⋅ ⋅ = ⋅a 36 a⋅ b 2⋅ a 4 2 lbf ⋅ s slug⋅ ft FR = 81.3⋅ lbf Problem 3.52 Given: Geometry of plane gate Find: Minimum weight to keep it closed [Difficulty: 3] L=3m h y L/2 dF W w=2m Solution: FR = Basic equation ⌠ ⎮ p dA ⎮ ⌡ dp dh = ρ⋅ g ΣMO = 0 Ixx y' = yc + A ⋅ yc F R = pc ⋅ A or, use computing equations Assumptions: static fluid; ρ = constant; patm on other side; door is in equilibrium Instead of using either of these approaches, we note the following, using y as in the sketch ⌠ L ⎮ W⋅ ⋅ cos ( θ) = y dF ⎮ 2 ⌡ ΣMO = 0 We also have dF = p⋅ dA Hence W= with p = ρ⋅ g⋅ h = ρ⋅ g⋅ y⋅ sin ( θ) (Gage pressure, since p = patm on other side) ⌠ ⌠ 2 ⎮ ⎮ y⋅ p dA = ⋅ y⋅ ρ⋅ g⋅ y⋅ sin ( θ) ⋅ w dy ⎮ L⋅ cos ( θ) ⌡ L⋅ cos ( θ) ⎮ ⌡ 2 ⋅ L ⌠ 2⋅ ρ⋅ g⋅ w⋅ tan( θ) ⌠ 2 2 2 ⎮ W= ⋅ y⋅ p dA = ⋅ ⎮ y dy = ⋅ ρ⋅ g⋅ w⋅ L ⋅ tan( θ) ⎮ ⌡ L⋅ cos ( θ) ⌡ L 3 0 2 Using given data W = 2 3 ⋅ 1000⋅ kg 3 m × 9.81⋅ m 2 s 2 × 2⋅ m × ( 3⋅ m) × tan( 30⋅ deg) × 2 N⋅ s kg⋅ m W = 68⋅ kN Problem 3.53 [Difficulty: 4] Given: Semicylindrical trough, partly filled with water to depth d. Find: (a) General expressions for FR and y' on end of trough, if open to the atmosphere. (b) Plots of results vs. d/R between 0 and 1. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dy = ρ⋅ g FR = ⌠ ⎮ p dA ⎮ ⌡ y'⋅ FR = Assumptions: (Hydrostatic Force on door) ⌠ ⎮ y⋅ p dA ⎮ ⌡ (First moment of force) R–d (1) Static fluid (2) Incompressible fluid Integrating the pressure equation: Therefore: (Hydrostatic Pressure - y is positive downwards) p = ρ ⋅ g⋅ h where h = y − ( R − d) ⎡ y − ⎛1 − ⎜ ⎣R ⎝ p = ρ⋅ g⋅ [ y − ( R − d) ] = ρ⋅ g⋅ R⋅ ⎢ Expressing this in terms of θ and α in the figure: For the walls at the end of the trough: d ⎞⎤ ⎟⎥ R ⎠⎦ θ y h d dy α p = ρ⋅ g⋅ R⋅ ( cos ( θ) − cos ( α) ) dA = w⋅ dy = 2⋅ R⋅ sin ( θ) ⋅ dy Now since y = R⋅ cos ( θ) it follows that dy = −R⋅ sin ( θ) ⋅ dθ Substituting this into the hydrostatic force equation: R 0 ⌠ ⌠ FR = ⎮ p⋅ w dy = ⎮ ρ⋅ g⋅ R⋅ ( cos ( θ) − cos ( α) ) ⋅ 2⋅ R⋅ sin ( θ) ⋅ ( −R⋅ sin ( θ) ) dθ ⌡R−d ⌡α Upon simplification: 3⌠ 2 2 3 ⎡ ( sin ( α) ) ⎛ α sin ( α) ⋅ cos ( α) ⎞⎥⎤ FR = 2⋅ ρ⋅ g⋅ R ⎮ ⎡⎣sin ( θ) ⋅ cos ( θ) − ( sin ( θ) ) ⋅ cos ( α)⎤⎦ dθ = 2⋅ ρ⋅ g⋅ R ⋅ ⎢ − cos ( α) ⋅ ⎜ − ⎟ ⌡0 2 ⎣ 3 ⎝2 ⎠⎦ α 3 3 ⎡ ( sin ( α) ) FR = 2⋅ ρ⋅ g⋅ R ⋅ ⎢ ⎣ Non-dimensionalizing the force: ⎡ ( sin ( α) ) 3 FR ρ⋅ g⋅ R 3 3 = 2⋅ ⎢ ⎣ 3 ⎛ α − sin ( α) ⋅ cos ( α) ⎞⎥⎤ ⎟ 2 ⎝2 ⎠⎦ 3 − cos ( α) ⋅ ⎜ ⎛ α − sin ( α) ⋅ cos ( α) ⎞⎥⎤ ⎟ 2 ⎝2 ⎠⎦ − cos ( α) ⋅ ⎜ To find the line of action of the force: R 0 ⌠ ⌠ y'⋅ FR = ⎮ y⋅ p⋅ w dy = ⎮ R⋅ cos ( θ) ⋅ ρ⋅ g⋅ R⋅ ( cos ( θ) − cos ( α) ) ⋅ 2⋅ R⋅ sin ( θ) ⋅ ( −R⋅ sin ( θ) ) dθ ⌡R−d ⌡α Upon simplification: 3 ( sin ( α) ) ⎤ sin ( 4⋅ α) ⎞ 4 ⎡1 ⎛ 2 2 2 ⎥ y'⋅ FR = 2⋅ ρ⋅ g⋅ R ⋅ ⎮ ⎡⎣( sin ( θ) ) ⋅ ( cos ( θ) ) − cos ( α) ⋅ ( sin ( θ) ) ⋅ cos ( θ)⎤⎦ dθ = 2⋅ ρ⋅ g⋅ R ⋅ ⎢ ⋅ ⎜ α − − cos ( α) ⋅ ⎟ ⌡0 4 3 ⎣8 ⎝ ⎠ ⎦ 4⌠ α 3 ( sin ( α) ) ⎤ sin ( 4⋅ α) ⎞ 4 ⎡1 ⎛ ⎥ y'⋅ FR = 2⋅ ρ⋅ g⋅ R ⋅ ⎢ ⋅ ⎜ α − ⎟ − cos ( α) ⋅ 4 3 ⎣8 ⎝ ⎠ ⎦ and therefore y' = Simplifying the expression: y' R = y'⋅ FR FR or y' R = y'⋅ FR R ⋅ FR sin ( 4⋅ α) ⎞ ( sin ( α) ) 1 ⎛ ⋅ ⎜α − ⎟ − cos ( α) ⋅ 4 8 ⎝ 3 ⎠ ( sin ( α) ) 3 3 ⎛ α − sin ( α) ⋅ cos ( α) ⎞ ⎟ 2 ⎝2 ⎠ − cos ( α) ⋅ ⎜ Plots of the non-dimensionalized force and the line of action of the force are shown in the plots below: Non-dimensional Force 0.8 0.6 0.4 0.2 0 0 0.5 1 d/R 1 0.8 y'/R 0.6 0.4 0.2 0 0 0.5 d/R 3 1 Problem 3.54 Given: Gate geometry Find: Depth H at which gate tips [Difficulty: 3] Solution: This is a problem with atmospheric pressure on both sides of the plate, so we can first determine the location of the center of pressure with respect to the free surface, using Eq.3.11c (assuming depth H) 3 Ixx y' = yc + A ⋅ yc w⋅ L Ixx = and with 12 yc = H − L 2 where L = 1 m is the plate height and w is the plate width Hence ⎛ ⎝ 3 L⎞ y' = ⎜ H − w⋅ L ⎟+ 2⎠ ⎛ 12⋅ w⋅ L⋅ ⎜ H − ⎝ L⎞ ⎟ ⎛ ⎝ = ⎜H − 2⎠ 2 L⎞ L ⎟+ 2⎠ ⎛ 12⋅ ⎜ H − ⎝ L⎞ ⎟ 2⎠ But for equilibrium, the center of force must always be at or below the level of the hinge so that the stop can hold the gate in place. Hence we must have y' > H − 0.45⋅ m L Combining the two equations ⎛⎜ H − ⎞⎟ + 2⎠ ⎝ Solving for H H ≤ L 2 + L 2 ⎛ ⎝ 12⋅ ⎜ H − L L⎞ ≥ H − 0.45⋅ m ⎟ 2⎠ 2 ⎛L ⎞ 12⋅ ⎜ − 0.45⋅ m⎟ ⎝2 ⎠ H ≤ 1⋅ m 2 + ( 1⋅ m) 2 ⎛ 1⋅ m − 0.45⋅ m⎞ 12 × ⎜ ⎟ ⎝ 2 ⎠ H ≤ 2.17⋅ m Problem 3.55 Given: Geometry of cup Find: Force on each half of cup Assumptions: [Difficulty: 1] (1) Tea is static and incompressible (2) Atmospheric pressure on outside of cup Solution: ⌠ ⎮ p dA ⎮ ⌡ Basic equation FR = or, use computing equation FR = pc⋅ A dp dh = ρ⋅ g The force on the half-cup is the same as that on a rectangle of size FR = ⌠ ⌠ ⎮ ⎮ p dA = ρ⋅ g⋅ y dA ⎮ ⎮ ⌡ ⌡ h Hence ⌠ ρ⋅ g⋅ w⋅ h FR = ⎮ ρ⋅ g⋅ y⋅ w dy = ⌡0 2 Alternatively FR = pc⋅ A Using given data FR = 1 2 × 999⋅ 3 m × 9.81⋅ w = 6.5⋅ cm and dA = w⋅ dy but 2 h ρ⋅ g⋅ w⋅ h FR = pc ⋅ A = ρ⋅ g⋅ yc⋅ A = ρ⋅ g⋅ ⋅ h ⋅ w = 2 2 and kg h = 8⋅ cm m 2 s 2 3 2 ⎛ m ⎞ × N⋅ s ⎟ ⎝ 100⋅ cm ⎠ kg⋅ m × 6.5⋅ cm × ( 8⋅ cm) × ⎜ Hence a teacup is being forced apart by about 2 N: not much of a force, so a paper cup works! 2 FR = 2.04⋅ N Problem 3.56 Given: Geometry of lock system Find: Force on gate; reactions at hinge [Difficulty: 3] Ry Rx Solution: Basic equation or, use computing equation FR = ⌠ ⎮ p dA ⎮ ⌡ dp = ρ⋅ g dh FR FR = pc⋅ A Assumptions: static fluid; ρ = constant; patm on other side The force on each gate is the same as that on a rectangle of size h = D = 10⋅ m w = and W 2⋅ cos ( 15⋅ deg) ⌠ ⌠ ⎮ ⎮ FR = ⎮ p dA = ⎮ ρ⋅ g⋅ y dA ⌡ ⌡ h but Hence ⌠ ρ ⋅ g⋅ w ⋅ h F R = ⎮ ρ ⋅ g⋅ y⋅ w dy = ⌡0 2 Alternatively FR = pc⋅ A Using given data FR = 1 2 ⋅ 1000⋅ 3 × 9.81⋅ m dA = w⋅ dy 2 h ρ ⋅ g⋅ w ⋅ h FR = pc⋅ A = ρ⋅ g⋅ yc⋅ A = ρ⋅ g⋅ ⋅ h⋅ w = 2 2 and kg Fn m 2 s × 34⋅ m 2⋅ cos ( 15⋅ deg) 2 2 2 × ( 10⋅ m) × N⋅ s kg⋅ m FR = 8.63⋅ MN For the force components Rx and Ry we do the following FR w ΣMhinge = 0 = FR⋅ − Fn⋅ w⋅ sin ( 15⋅ deg) 2 Fn = ΣFx = 0 = FR⋅ cos ( 15⋅ deg) − Rx = 0 Rx = FR⋅ cos ( 15⋅ deg) Rx = 8.34⋅ MN ΣFy = 0 = −Ry − FR⋅ sin ( 15⋅ deg) + Fn = 0 Ry = Fn − FR⋅ sin ( 15⋅ deg) Ry = 14.4⋅ MN R = ( 8.34⋅ MN , 14.4⋅ MN) R = 16.7⋅ MN 2⋅ sin ( 15⋅ deg) Fn = 16.7⋅ MN Problem 3.57 [Difficulty: 2] Given: Liquid concrete poured between vertical forms as shown t = 0.25⋅ m H = 3⋅ m W = 5⋅ m SGc = 2.5 (From Table A.1, App. A) Find: (a) Resultant force on form (b) Line of application Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dy = ρ⋅ g (Hydrostatic Pressure - y is positive downwards) FR = pc⋅ A (Hydrostatic Force) Ixx y' = yc + A ⋅ yc (Location of line of action) Ixy x' = xc + A ⋅ yc Assumptions: For a rectangular plate: Liquid Concrete (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface and on the outside of the form. W⋅ H Ixx = t = 0.25 m 3 12 H=3m y’ x’ Ixy = 0 xc = 2.5⋅ m yc = 1.5⋅ m Integrating the hydrostatic pressure equation: The density of concrete is: ρ = 2.5 × 1000⋅ kg p = ρ ⋅ g⋅ y ρ = 2.5 × 10 3 m W=5m FR 3 kg 3 m Therefore, the force is: FR = ρ⋅ g⋅ yc⋅ H⋅ W Substituting in values gives us: m 3 kg × 9.81⋅ × 1.5⋅ m × 3⋅ m × 5⋅ m 3 2 FR = 2.5 × 10 ⋅ m FR = 552⋅ kN s To find the line of action of the resultant force: 3 2 W⋅ H H y' = yc + = yc + 12⋅ W⋅ H⋅ yc 12⋅ yc Since Ixy = 0 it follows that x' = xc y' = 1.5⋅ m + ( 3⋅ m) 2 12⋅ 1.5⋅ m y' = 2.00 m x' = 2.50⋅ m Problem 3.58 Given: [Difficulty: 4] Window, in shape of isosceles triangle and hinged at the top is located in the vertical wall of a form that contains concrete. a 0.4 m b 0.3 m c 0.25 m SGc 2.5 (From Table A.1, App. A) Find: The minimum force applied at D needed to keep the window closed. Plot the results over the range of concrete depth between 0 and a. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g FR Assumptions: p dA (Hydrostatic Force on door) y' FR y p dA (First moment of force) ΣM 0 (Rotational equilibrium) b (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface and on the outside of the window. Integrating the pressure equation yields: p ρ g ( h d) p0 Summing moments around the hinge: h d a w for h > d for h < d d ac where FD (Hydrostatic Pressure - h is positive downwards) FD a d 0.15 m dA D h p dA 0 a a 1 1 ρ g h p dA h ρ g ( h d ) w dh h ( h d) w dh a a d a d From the law of similar triangles: w b ah a h Therefore: w b a a ( a h) dF = pdA FD a a ρ g b ρ g b 3 2 h ( a d) h a d h dh FD h ( h d) ( a h) dh 2 a a d a d Into the expression for the force at D: Evaluating this integral we get: FD 4 4 3 3 2 2 ρ g b a d ( a d) a d a d a d 2 4 3 2 a 4 2 1 d 1 FD ρ g b a 1 1 4 a 3 The density of the concrete is: and after collecting terms: 3 2 d d 1 d d 1 1 a a 2 a a ρ 2.5 1000 kg 3 ρ 2.5 10 m 3 kg 3 m ( 1) d a 0.15 0.4 0.375 Substituting in values for the force at D: 2 m 1 0.375 3 kg 2 1 4 3 2 N s 9.81 0.3 m ( 0.4 m) 1 ( 0.375) ( 1 0.375) 1 ( 0.375) 1 ( 0.375) 3 2 3 2 4 kg m FD 2.5 10 m s To plot the results for different values of c/a, we use Eq. (1) and remember that Therefore, it follows that d a 1 c d ac FD 32.9 N In addition, we can maximize the force by the maximum force a (when c = a or d = 0): 2 2 1 1 ρ g b a 12 4 3 Fmax ρ g b a and so 1 d 4 1 d d 3 1 d d 2 12 1 1 1 1 Fmax 4 a 3 a a 2 a a FD 1.0 Force Ratio (FD/Fmax) 0.8 0.6 0.4 0.2 0.0 0.0 0.5 Concrete Depth Ratio (c/a) 1.0 Problem 3.59 [Difficulty: 2] Given: Door as shown; Data from Example 3.6. Find: Force to keep door shut using the two seperate pressures method. Solution: We will apply the computing equations to this system. F R pc A Governing Equations: p0 3 Ixx y' yc yc A bL Ixx 12 p0 h1 ’ h2 ’ F1 F2 F 1 p0 A F1 100 lbf ft 2 3 ft 2 ft F2 pc A ρ g hc L b γ hc L b F1 600 lbf F2 100 lbf ft For the rectangular door 1 3 Ixx b L 12 2 Ixx 1 L h'2 hc hc b L hc 12 hc x' 1 ft z' 1.5 ft 1.5 ft 3 ft 2 ft 3 F2 900 lbf 2 h'2 1.5 m 1 ( 3 m) 12 1.5 m h'2 2 m The free body diagram of the door is then h1’ h2’ F1 ΣMAx 0 L Ft F1 L h'1 F2 L h'2 L h'1 h'2 Ft F1 1 F2 1 L L F2 Az Ft Ay Ft 600 lbf 1 1.5 2 900 lbf 1 3 3 Ft 600 lbf Problem 3.60 Given: [Difficulty: 2] γ = 62.4⋅ Plug is used to seal a conduit. lbf ft 3 Find: Magnitude, direction and location of the force of water on the plug. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh Assumptions: =γ (Hydrostatic Pressure - y is positive downwards) F R = pc ⋅ A (Hydrostatic Force) Ixx y' = yc + A ⋅ yc (Location of line of action) (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts on the outside of the plug. Integrating the hydrostatic pressure equation: π 2 FR = pc⋅ A = γ⋅ hc⋅ ⋅ D 4 p = γ⋅ h FR = 62.4⋅ lbf ft π For a circular area: ⋅D 4 3 × 12⋅ ft × π 4 × ( 6⋅ ft) 2 2 π 4 64 D Ixx = ⋅ D Therefore: y' = yc + = yc + 64 π 2 16⋅ yc ⋅ D ⋅ yc 4 y' = 12⋅ ft + 4 FR = 2.12 × 10 ⋅ lbf ( 6⋅ ft) 2 16 × 12⋅ ft y' = 12.19⋅ ft The force of water is to the right and perpendicular to the plug. Problem 3.61 Given: Description of car tire Find: Explanation of lift effect [Difficulty: 1] Solution: The explanation is as follows: It is true that the pressure in the entire tire is the same everywhere. However, the tire at the top of the hub will be essentially circular in cross-section, but at the bottom, where the tire meets the ground, the cross section will be approximately a flattened circle, or elliptical. Hence we can explain that the lower cross section has greater upward force than the upper cross section has downward force (providing enough lift to keep the car up) two ways. First, the horizontal projected area of the lower ellipse is larger than that of the upper circular cross section, so that net pressure times area is upwards. Second, any time you have an elliptical cross section that's at high pressure, that pressure will always try to force the ellipse to be circular (thing of a round inflated balloon - if you squeeze it it will resist!). This analysis ignores the stiffness of the tire rubber, which also provides a little lift. Problem 3.62 Given: [Difficulty: 2] Circular access port of known diameter in side of water standpipe of known diameter. Port is held in place by eight bolts evenly spaced around the circumference of the port. Center of the port is located at a know distance below the free surface of the water. d 0.6 m D 7 m L 12 m Find: (a) Total force on the port (b) Appropriate bolt diameter Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp ρ g dh (Hydrostatic Pressure - y is positive downwards) FR pc A F σ A Assumptions: (Hydrostatic Force) h (Normal Stress in bolt) L (1) Static fluid (2) Incompressible fluid (3) Force is distributed evenly over all bolts (4) Appropriate working stress in bolts is 100 MPa (5) Atmospheric pressure acts at free surface of water and on outside of port. D p ρ g h Integrating the hydrostatic pressure equation: The resultant force on the port is: d π 2 FR pc A ρ g L d 4 FR 999 kg 3 9.81 m m 2 12 m π s 4 2 ( 0.6 m) 2 N s kg m FR 33.3 kN To find the bolt diameter we consider: 2 Therefore: 2 π db FR σ σ FR A where A is the area of all of the bolts: Solving for the bolt diameter we get: 1 3 db 33.3 10 N 2 π FR 2 π σ A 8 π 4 2 db 2 π db 2 1 2 db 1 100 10 2 m 6 N 1 2 3 10 mm m db 7.28 mm Problem 3.63 Given: Geometry of rectangular gate Find: Depth for gate to open [Difficulty: 3] Solution: Basic equation Computing equations L dp dh ρ g FR pc A y’ D ΣMz 0 Ixx y' yc A yc Ixx b D 3 F1 12 F2 Assumptions: Static fluid; ρ = constant; patm on other side; no friction in hinge For incompressible fluid p ρ g h where p is gage pressure and h is measured downwards The force on the vertical gate (gate 1) is the same as that on a rectangle of size h = D and width w Hence D ρ g w D F1 pc A ρ g yc A ρ g D w 2 2 The location of this force is 3 Ixx D w D 1 2 2 y' yc D 12 w D D A yc 2 3 2 The force on the horizontal gate (gate 2) is due to constant pressure, and is at the centroid F 2 p ( y D ) A ρ g D w L Summing moments about the hinge L 2 L ΣMhinge 0 F1 ( D y') F2 F1 D D F2 2 3 2 F 1 D 3 ρ g w D 2 2 D L L F2 ρ g D w L 3 2 2 3 ρ g w D ρ g D w L 6 2 D 3 L D 8.66 ft 3 5ft 2 Problem 3.64 Given: [Difficulty: 3] Gate AOC, hinged along O, has known width; Weight of gate may be neglected. Gate is sealed at C. b = 6⋅ ft Find: Force in bar AB Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh Assumptions: = ρ⋅ g (Hydrostatic Pressure - h is positive downwards) FR = pc⋅ A (Hydrostatic Force) Ixx y' = yc + A ⋅ yc (Location of line of action) ΣMz = 0 (Rotational equilibrium) (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface of water and on outside of gate (4) No resisting moment in hinge at O (5) No vertical resisting force at C FAB L1 h1 ’ L1 p = ρ ⋅ g⋅ h Integrating the hydrostatic pressure equation: F1 L2 The free body diagram of the gate is shown here: F1 is the resultant of the distributed force on AO F2 is the resultant of the distributed force on OC x2’ FAB is the force of the bar Cx is the sealing force at C First find the force on AO: F1 = 1.94⋅ slug ft 3 × 32.2⋅ ft 2 s F1 = pc ⋅ A1 = ρ⋅ g⋅ hc1⋅ b ⋅ L1 2 × 6⋅ ft × 6⋅ ft × 12⋅ ft × lbf⋅ s slugft ⋅ F1 = 27.0⋅ kip F2 b⋅ L 1 Ixx 3 L1 2 h'1 = hc1 + = hc1 + = hc1 + A⋅ hc1 12⋅ b⋅ L1⋅ hc1 12⋅ hc1 Next find the force on OC: F2 = 1.94⋅ slug ft 3 × 32.2⋅ ft 2 h'1 = 6⋅ ft + ( 12⋅ ft) 2 12 × 6⋅ ft h'1 = 8⋅ ft 2 × 12⋅ ft × 6⋅ ft × 6⋅ ft × s lbf ⋅ s slug⋅ ft Since the pressure is uniform over OC, the force acts at the centroid of OC, i.e., ( ) ( F2 = 27.0⋅ kip ) FAB⋅ L1 + L3 − F1⋅ L1 − h'1 + F2⋅ x'2 = 0 Summing moments about the hinge gives: FAB x'2 = 3⋅ ft L1 h1 ’ L1 Solving for the force in the bar: Substituting in values: FAB = FAB = 1 12⋅ ft + 3⋅ ft FAB = 1800⋅ lbf ( ) F1⋅ L1 − h'1 − F2⋅ x'2 F1 L2 L1 + L3 3 3 ⋅ ⎡⎣27.0 × 10 ⋅ lbf × ( 12⋅ ft − 8⋅ ft) − 27.0 × 10 ⋅ lbf × 3⋅ ft⎤⎦ Thus bar AB is in compression x2’ F2 Problem 3.65 Given: [Difficulty: 3] Gate shown with fixed width, bass of gate is negligible. Gate is in equilibrium. b = 3⋅ m Find: Water depth, d Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh Assumptions: = ρ⋅ g (Hydrostatic Pressure - h is positive downwards) FR = pc⋅ A (Hydrostatic Force) Ixx y' = yc + A ⋅ yc (Location of line of action) ΣMz = 0 (Rotational equilibrium) (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface of water and on outside of gate M h p = ρ ⋅ g⋅ h Integrating the hydrostatic pressure equation: FR = pc⋅ A = ρ⋅ g⋅ hc⋅ A hc = d A = b⋅ 2 d y d l sin ( θ) L 2 ρ ⋅ g⋅ b⋅ d Therefore, FR = 2⋅ sin ( θ) To find the line of application of this force: θ 3 Ixx y' = yc + A ⋅ yc Since 3 b⋅ l and Ixx = 12 A = b⋅ l it follows that 2 b⋅ l l y' = yc + = yc + 12⋅ b⋅ l⋅ yc 12⋅ yc where l is the length of the gate in contact with the water (as seen in diagram) l= l and d are related through: d l d Therefore, yc = and = sin ( θ) 2 2⋅ sin ( θ) y' = d 2⋅ sin ( θ) d + 2 ( sin ( θ) ) 2 ⋅ 2⋅ sin ( θ) 12⋅ d The free body diagram of the gate is shown here: = 2⋅ d 3⋅ sin ( θ) T y d FR y’ Summing moments about the hinge gives: T⋅ L − ( l − y') ⋅ FR = 0 Solving for l: l= d sin ( θ) = M⋅ g⋅ L FR where T = M⋅ g + y' ⎛ 2⋅ M⋅ g⋅ L ⋅ sin ( θ) + 2⋅ d ⎞ ⋅ sin ( θ) or d=⎜ 2 3⋅ sin ( θ) ⎟ ⎝ ρ ⋅ g⋅ b⋅ d ⎠ Solving for d: ⎡ 6⋅ M⋅ L ⋅ ( sin ( θ) ) 2⎤ ⎥ ⎣ ρ⋅ b ⎦ d=⎢ θ So upon further substitution we get: d 3 = 2⋅ M⋅ L⋅ ( sin ( θ) ) ρ ⋅ b⋅ d Ahoriz A vertical 2 2 1 3 Substituting in values: ⎡ d = ⎢6 × 2500⋅ kg × 5⋅ m × ⎣ d = 2.66 m 1 3 ⋅ m 999 kg × 1 3m 2⎤ × ( sin ( 60⋅ deg) ) ⎥ ⎦ 1 3 Problem 3.66 Given: Geometry of gate Find: Force at A to hold gate closed [Difficulty: 3] y h Solution: Basic equation Computing equations D y’ dp dh ρ g ΣMz 0 FR pc A FR Ixx y' yc A yc Ixx w L FA 3 12 Assumptions: Static fluid; ρ = constant; patm on other side; no friction in hinge For incompressible fluid p ρ g h where p is gage pressure and h is measured downwards The hydrostatic force on the gate is that on a rectangle of size L and width w. Hence L FR pc A ρ g hc A ρ g D sin ( 30 deg) L w 2 FR 1000 kg 3 9.81 m m 2 s 1.5 3 2 2 sin ( 30 deg) m 3 m 3 m N s kg m FR 199 kN Ixx where y' and y are measured along the plane of the gate to the free surface The location of this force is given by y' yc A yc c yc D sin ( 30 deg) L 2 yc 1.5 m sin ( 30 deg) 3 m 2 yc 4.5 m 3 2 2 Ixx w L 1 1 L ( 3 m) y' yc yc yc 4.5 m A yc 12 w L yc 12 yc 12 4.5 m Taking moments about the hinge y' 4.67 m D F L ΣMH 0 FR y' A sin ( 30 deg ) D y' sin ( 30 deg) FA FR L 1.5 4.67 sin ( 30 deg) FA 199 kN 3 FA 111 kN Problem 3.67 Given: Block hinged and floating Find: SG of the wood [Difficulty: 3] Solution: dp Basic equation dh ρ g ΣMz 0 Ixx y' yc A yc FR pc A Computing equations Assumptions: Static fluid; ρ = constant; patm on other side; no friction in hinge p ρ g h For incompressible fluid where p is gage pressure and h is measured downwards The force on the vertical section is the same as that on a rectangle of height d and width L Hence d ρ g L d F1 pc A ρ g yc A ρ g d L 2 2 2 Mg y’ y The location of this force is F1 3 Ixx d L d 1 2 2 y' yc d A yc 2 12 L d d 3 x F2 The force on the horizontal section is due to constant pressure, and is at the centroid F2 p ( y d) A ρ g d L L Summing moments about the hinge L L ΣMhinge 0 F1 ( d y') F2 M g 2 2 Hence F 1 d SG ρ g L 2 4 ρ g L d 2 2 d 3 2 L ρ g d L 2 2 L 3 L d F2 SG ρ L g 3 2 2 3 SG 1 d d 3 L L 3 SG 1 0.5 0.5 3 1 1 SG 0.542 Problem 3.68 [Difficulty: 4] Given: Various dam cross-sections Find: Which requires the least concrete; plot cross-section area A as a function of α Solution: For each case, the dam width b has to be large enough so that the weight of the dam exerts enough moment to balance the moment due to fluid hydrostatic force(s). By doing a moment balance this value of b can be found a) Rectangular dam Straightforward application of the computing equations of Section 3-5 yields D 1 2 FH = p c ⋅ A = ρ⋅ g ⋅ ⋅ w⋅ D = ⋅ ρ⋅ g ⋅ D ⋅ w 2 2 D 3 Ixx D 2 w⋅ D y' = y c + = + = ⋅D A⋅ y c 2 3 D 12⋅ w⋅ D⋅ 2 so Also y = D − y' = FH y mg O D b 3 m = ρcement⋅ g ⋅ b ⋅ D⋅ w = SG ⋅ ρ⋅ g ⋅ b ⋅ D⋅ w Taking moments about O ∑ M0. = 0 = −FH⋅y + 2 ⋅m⋅g so ⎛ 1 ⋅ ρ⋅ g⋅ D2⋅ w⎞ ⋅ D = b ⋅ ( SG⋅ ρ⋅ g ⋅ b⋅ D⋅ w) ⎜2 ⎝ ⎠ 3 2 Solving for b b= The minimum rectangular cross-section area is A = b⋅ D = For concrete, from Table A.1, SG = 2.4, so A= b D 3 ⋅ SG 2 D 3 ⋅ SG 2 D 3 ⋅ SG 2 = D 3 × 2.4 2 A = 0.373 ⋅ D b) Triangular dams FV D Instead of analysing right-triangles, a general analysis is made, at the end of which right triangles are analysed as special cases by setting α = 0 or 1. x FH Straightforward application of the computing equations of Section 3-5 yields y m 1g m 2g O D 1 2 FH = p c⋅ A = ρ⋅ g ⋅ ⋅ w⋅ D = ⋅ ρ⋅ g ⋅ D ⋅ w 2 2 αb b 3 Ixx w⋅ D D 2 y' = y c + = + = ⋅D A⋅ y c D 2 3 12⋅ w⋅ D⋅ 2 so Also y = D − y' = D 3 FV = ρ⋅ V⋅ g = ρ⋅ g ⋅ α⋅ b ⋅ D 2 ⋅w = 1 2 ⋅ ρ⋅ g ⋅ α⋅ b ⋅ D⋅ w x = ( b − α⋅ b ) + 2 3 ⋅ α⋅ b = b ⋅ ⎛⎜ 1 − α⎞ ⎝ 3⎠ For the two triangular masses 1 m1 = ⋅ SG ⋅ ρ⋅ g ⋅ α⋅ b ⋅ D⋅ w 2 x 1 = ( b − α⋅ b ) + 1 m2 = ⋅ SG ⋅ ρ⋅ g ⋅ ( 1 − α) ⋅ b ⋅ D⋅ w 2 x2 = 2 3 1 3 ⋅ α⋅ b = b ⋅ ⎛⎜ 1 − ⎝ 2⋅ α ⎞ ⋅ b ( 1 − α) Taking moments about O ∑ M0. = 0 = −FH⋅y + FV⋅x + m1⋅g⋅x1 + m2⋅g⋅x2 so Solving for b 1 1 D α 2 −⎛⎜ ⋅ ρ⋅ g ⋅ D ⋅ w⎞ ⋅ + ⎛⎜ ⋅ ρ⋅ g ⋅ α⋅ b ⋅ D⋅ w⎞ ⋅ b ⋅ ⎛⎜ 1 − ⎞ ... 3⎠ ⎝2 ⎠ 3 ⎝2 ⎠ ⎝ 2⋅ α ⎞ ⎡ 1 2 1 ⎞ ⎛ ⎛ + ⎜ ⋅ SG ⋅ ρ⋅ g ⋅ α⋅ b ⋅ D⋅ w ⋅ b ⋅ ⎜ 1 − + ⎢ ⋅ SG ⋅ ρ⋅ g ⋅ ( 1 − α) ⋅ b ⋅ D⋅ w⎥⎤ ⋅ ⋅ b ( 1 − α) 3 2 2 ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ 3 b= D (3⋅α − α2) + SG⋅(2 − α) For a right triangle with the hypotenuse in contact with the water, α = 1 , b= The cross-section area is =0 D 3 − 1 + SG = D b = 0.477 ⋅ D 3 − 1 + 2.4 A= b⋅ D 2 and 2 = 0.238 ⋅ D For a right triangle with the vertical in contact with the water, α = 0, and 2 A = 0.238 ⋅ D 3 ⎠ b= A= The cross-section area is A= For a general triangle D 2 ⋅ SG b⋅ D 2 b⋅ D 2 D = b = 0.456 ⋅ D 2 ⋅ 2.4 2 2 = 0.228 ⋅ D A = 0.228 ⋅ D 2 2 D = (3⋅α − α2) + SG⋅(2 − α) 2⋅ D A= 2⋅ (3⋅α − α2) + 2.4⋅(2 − α) 2 D A= The final result is 2 2 ⋅ 4.8 + 0.6⋅ α − α The dimensionless area, A /D 2, is plotted A /D 2 0.2282 0.2270 0.2263 0.2261 0.2263 0.2270 0.2282 0.2299 0.2321 0.2349 0.2384 Dam Cross Section vs Coefficient Dimensionless Area A /D 2 Alpha 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0.240 0.238 0.236 0.234 0.232 0.230 0.228 0.226 Solver can be used to find the minimum area Alpha 0.300 0.224 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Coefficient A /D 2 0.2261 From the Excel workbook, the minimum area occurs at α = 0.3 2 Amin = D 2 A = 0.226 ⋅ D 2 2 ⋅ 4.8 + 0.6 × 0.3 − 0.3 The final results are that a triangular cross-section with α = 0.3 uses the least concrete; the next best is a right triangle with the vertical in contact with the water; next is the right triangle with the hypotenuse in contact with the water; and the cross-section requiring the most concrete is the rectangular cross-section. 1.0 Problem 3.69 Given: Geometry of dam Find: Vertical force on dam Assumption: [Difficulty: 2] Water is static and incompressible Solution: Basic equation: For incompressible fluid dp dh = ρ⋅ g p = patm + ρ⋅ g⋅ h where h is measured downwards from the free surface The force on each horizontal section (depth d = 0.5 m and width w = 3 m) is ( ) F = p⋅ A = patm + ρ⋅ g⋅ h ⋅ d⋅ w Hence the total force is ( ) ( ) ( ) ( ) FT = ⎡patm + patm + ρ⋅ g⋅ h + patm + ρ⋅ g⋅ 2⋅ h + patm + ρ⋅ 3⋅ g⋅ h + patm + ρ⋅ g⋅ 4⋅ h ⎤ ⋅ d⋅ w ⎣ ⎦ where we have used h as the height of the steps ( FT = d⋅ w⋅ 5⋅ patm + 10⋅ ρ⋅ g⋅ h ⎛ ) 2 kg m N⋅ s ⎟⎞ 3 N + 10 × 999⋅ × 9.81⋅ × 0.5⋅ m × 2 3 2 kg⋅ m ⎟ FT = 0.5⋅ m × 3⋅ m × ⎜ 5 × 101 × 10 ⋅ ⎜ ⎝ FT = 831⋅ kN m m s ⎠ Problem 3.70 Given: Geometry of dam Find: Vertical force on dam Assumptions: [Difficulty: 2] (1) water is static and incompressible (2) since we are asked for the force of the water, all pressures will be written as gage Solution: dp Basic equation: dh ρ g p ρ g h For incompressible fluid where p is gage pressure and h is measured downwards from the free surface The force on each horizontal section (depth d and width w) is F p A ρ g h d w (Note that d and w will change in terms of x and y for each section of the dam!) Hence the total force is (allowing for the fact that some faces experience an upwards (negative) force) F T p A Σ ρ g h d w ρ g d Σ h w Starting with the top and working downwards FT 1.94 slug ft 3 32.2 ft 2 s 4 FT 2.70 10 lbf 2 3 ft [ ( 3 ft 12 ft) ( 3 ft 6 ft) ( 9 ft 6 ft) ( 12 ft 12 ft) ] lbf s slug ft The negative sign indicates a net upwards force (it's actually a buoyancy effect on the three middle sections) Problem 3.71 Given: [Difficulty: 3] Parabolic gate, hinged at O has a constant width. 2 b 1.5 m a 1.0 m D 1.2 m H 1.4 m Find: (a) Magnitude and moment of the vertical force on the gate due to water (b) Horizontal force applied at A required to maintain equilibrium Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g (Hydrostatic Pressure - h is positive downwards) ΣMz 0 Fv Assumptions: (Rotational equilibrium) p dA y (Vertical Hydrostatic Force) x' Fv x dF v (Moment of vertical force) y' FH y dF H (Moment of Horizontal Hydrostatic Force) y (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface of water and on outside of gate Integrating the hydrostatic pressure equation: FA x’ FV p ρ g h FH (a) The magnitude and moment of the vertical component of hydrostatic force: Ox y’ x Oy 3 2 Fv p dA y ρ g h b dx where h D y x a y dx 3 a y dy D Substituting back into the relation for the force: D 2 Fv ρ g ( D y) b 3 a y dy 3 ρ g b a 0 0 4 D4 D4 D ρ g b a Evaluating the integral: Fv 3 ρ g b a 4 4 3 H D y2 y3 dy y Substituting values we calculate the force: Fv 999 kg 9.81 3 m m 1.5 m 1.0 2 s 1 2 ( 1.2 m) 4 2 4 m FA x’ N s FH Fv 7.62 kN x' Fv x dFv x p dAy D D 3 2 2 x' Fv a y ρ g ( D y) b 3 a y dy 3 ρ g a b 0 0 D7 2 x' Fv 3 ρ g a b 6 kg x'Fv 999 3 9.81 m 7 D 2 m 2 s D y5 y6 dy D 3 7 x Using the derivation for the force: Evaluating the integral: 7 ρ g a2 b D7 ρ g a2 b 7 42 14 y’ Oy Ox To find the associated moment: H FV kg m Now substituting values into this equation: 2 1.0 1.5 m ( 1.20 m) N s 2 14 kg m m x'Fv 3.76 kN m (positive indicates counterclockwise) (b) Horizontal force at A to maintain equilibrium: we take moments at O: x' FV y' FH H FA 0 Solving for the force at A: FA 1 H x' Fv y' FH To get the moment of the horizontal hydrostatic force: y' FH D D 2 y dF H y p dA x y ρ g h b dy ρ g b y ( D y) dy ρ g b D y y dy 0 0 D D D ρ g b Evaluating the integral: y' FH ρ g b 3 6 2 3 y'FH 999 kg 3 9.81 m Therefore: FA m 2 1.5 m s 1 1 1.4 m ( 1.20 m) 6 3 3 3 Now substituting values into this equation: 2 N s kg m ( 3.76 kN m 4.23 kN m) y'FH 4.23 kN m (counterclockwise) FA 5.71 kN Problem 3.72 Given: [Difficulty: 3] Parabolic gate, hinged at O has a constant width. 1 b 2 m c 0.25 m D 2 m H 3 m Find: (a) Magnitude and line of action of the vertical force on the gate due to water (b) Horizontal force applied at A required to maintain equilibrium (c) Vertical force applied at A required to maintain equilibrium Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g (Hydrostatic Pressure - h is positive downwards) ΣMz 0 Fv Assumptions: (Rotational equilibrium) p dA y (Vertical Hydrostatic Force) x' Fv x dFv (Location of line of action) F H pc A (Horizontal Hydrostatic Force) Ixx h' hc A hc (Location of line of action) (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface of water and on outside of gate Integrating the hydrostatic pressure equation: Fv x’ h’ FH p ρ g h Ox (a) The magnitude and line of action of the vertical component of hydrostatic force: D y D D D ( 1) B x Oy c c c c 2 p dA y ρ g h b dx ρ g ( D y) b dx ρ g D c x b dx ρ g b 0 0 0 0 3 3 3 2 2 2 D 1 D 2 ρ g b D Evaluating the integral: Fv ρ g b 1 1 3 1 3 2 2 2 c c c FV D c x2 dx 2 Fv Substituting values: 3 kg 999 3 9.81 m 2 1 2 2 1 m N s kg m 0.25 2 m ( 2 m) s x' Fv To find the line of action of this force: m 3 2 1 1 x dFv Therefore, x' x dF v x p dA y Fv Fv D Using the derivation for the force: x' 999 kg 3 m 9.81 m 2 D c c 1 ρ g b 2 3 x' x ρ g D c x b dx D x c x dx Fv 0 Fv 0 2 2 ρ g b D ρ g b D D c D Fv 2 c 4 c F v 4 c Evaluating the integral: x' 2 m 1 73.9 10 s Fv 73.9 kN 1 3 N 1 4 Now substituting values into this equation: 2 1 2 ( 2 m) m 0.25 N s x' 1.061 m kg m To find the required force at A for equilibrium, we need to find the horizontal force of the water on the gate and its line of action as well. Once this force is known we take moments about the hinge (point O). 2 D D FH pc A ρ g hc b D ρ g b D ρ g b 2 2 FH 999 kg 3 9.81 m m 2 2 m s hc since ( 2 m) 2 2 D 2 Therefore the horizontal force is: 2 N s FH 39.2 kN kg m To calculate the line of action of this force: 3 Ixx D b D D D 2 1 2 h' hc D 2 2 3 12 b D D 6 A hc h' 2 2 m 3 h' 1.333 m y Now we have information to solve parts (b) and (c): (b) Horizontal force applied at A for equilibrium: take moments about O: FA H Fv x' FH ( D h') 0 Solving for FA FA Fv x' FH ( D h') D h’ FH H FV H 1 1 [ 73.9 kN 1.061 m 39.2 kN ( 2 m 1.333 m) ] 3 m x Oy Ox FA FA x’ FA 34.9 kN (c) Vertical force applied at A for equilibrium: take moments about O: FA L Fv x' FH ( D h') 0 Solving for FA FA y Fv x' FH ( D h') L D L is the value of x at y = H. Therefore: L FA H c L 3 m 1 0.25 L x’ h’ FH m L 3.464 m 1 1 [ 73.9 kN 1.061 m 39.2 kN ( 2 m 1.333 m) ] 3.464 m Ox FA 30.2 kN Oy FA FV x Problem 3.73 [Difficulty: 2] Given: Liquid concrete is poured into the form shown R = 2⋅ ft w = 15⋅ ft SGc = 2.5 (Table A.1, App. A) Find: Magnitude and line of action of the vertical force on the form Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh = ρ⋅ g Fv = ⌠ ⎮ p dA y ⎮ ⌡ x'⋅ Fv = Assumptions: (Vertical Hydrostatic Force) ⌠ ⎮ x dF v ⎮ ⌡ (Moment of vertical force) (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface of concrete and on outside of gate p = ρ ⋅ g⋅ h Integrating the hydrostatic pressure equation: Fv = (Hydrostatic Pressure - h is positive downwards) ⌠ ⌠ ⎮ ⎮ p dA y = ρ⋅ g⋅ h⋅ sin ( θ) dA ⎮ ⎮ ⌡ ⌡ where dA = w⋅ R⋅ dθ and π ⌠2 h = R − y = R − R⋅ sin ( θ) π ⌠2 ⎮ 2⎮ 2 Therefore, Fv = ⎮ ρ⋅ g⋅ ( R − R⋅ sin ( θ) ) ⋅ w⋅ R⋅ sin ( θ) dθ = ρ⋅ g⋅ w⋅ R ⋅ ⎮ ⎡⎣sin ( θ) − ( sin ( θ) ) ⎤⎦ dθ ⌡0 ⌡0 Evaluating the integral: The density of concrete is: 2 ⎡ ⎣ ⎛ π − 0⎞ + ( 0 − 0)⎤ = ρ⋅ g⋅ w⋅ R2⋅ ⎛ 1 − π ⎞ ⎟ ⎥ ⎜ ⎟ 4⎠ ⎝4 ⎠ ⎦ ⎝ Fv = ρ⋅ g⋅ w⋅ R ⋅ ⎢−( 0 − 1) − ⎜ ρ = 2.5 × 1.94⋅ slug ft Substituting values we calculate the force: ρ = 4.85⋅ 3 Fv = 4.85⋅ ft slug ft To find the line of action: x'⋅ Fv = slug 3 ⌠ ⌠ ⎮ ⎮ x dF v = x⋅ p dAy ⎮ ⎮ ⌡ ⌡ × 32.2⋅ ft 2 s 3 2 ⎛ ⎝ × 15⋅ ft × ( 2⋅ ft) × ⎜ 1 − π⎞ 2 lbf⋅ s ⎟× 4 ⎠ slugft ⋅ Using the derivation for the force: Fv = 2011⋅ lbf π ⌠2 x'⋅ Fv = ⌠ 3⎮ 2 ⎮ R⋅ cos ( θ) ⋅ ρ⋅ g⋅ ( R − R⋅ sin ( θ) ) ⋅ w⋅ R⋅ sin ( θ) dθ = ρ⋅ g⋅ w⋅ R ⋅ ⎮ ⎡⎣sin ( θ) ⋅ cos ( θ) − ( sin ( θ) ) ⋅ cos ( θ)⎤⎦ dθ ⎮ ⌡ ⌡ 0 Evaluating the integral: x' = x'⋅ Fv Fv ρ ⋅ g⋅ w ⋅ = R Therefore the line of action of the force is: 3 6 ⎛ ρ ⋅ g⋅ w ⋅ R ⋅ ⎜ 1 − ⎟ 4⎠ ⎝ 2 3 R 1⎞ − ⎟ = ρ ⋅ g⋅ w ⋅ 2 3 6 ⎝ ⎠ 3 ⎛1 x'⋅ Fv = ρ⋅ g⋅ w⋅ R ⋅ ⎜ π⎞ = R π⎞ ⎛ 6⋅ ⎜ 1 − ⎟ 4⎠ ⎝ Substituting values: x' = 2⋅ ft ⎛ 6⋅ ⎜ 1 − ⎝ π⎞ ⎟ 4⎠ x' = 1.553⋅ ft Problem 3.74 [Difficulty: 2] Given: Open tank as shown. Width of curved surface b = 10⋅ ft Find: (a) Magnitude of the vertical force component on the curved surface (b) Line of action of the vertical component of the force Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh =γ Fv = − ⌠ ⎮ p dA y ⎮ ⌡ x’ (Vertical Hydrostatic Force) ⌠ ⎮ x dF v ⎮ ⌡ x'⋅ Fv = Assumptions: (Hydrostatic Pressure - h is positive downwards) FRy y (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface of water and on outside of wall p = γ⋅ h Integrating the hydrostatic pressure equation: L (Moment of vertical force) x ( dAy = b ⋅ dx We also define the incremental area on the curved surface as: 2 2 h = L− R −x We can define along the surface ) 1 2 Substituting these into the force equation we get: R Fv = − ⌠ ⎮ ⎮ ⌡ ⌠ 1⎤ ⎡⎢ ⎮ ⎥ R 2 ⎮ ⌠ 2 2 ⎥ ⎢ L− p dAy = −⎮ γ⋅ ⎣L − R − x ⎦ ⋅ b dx = −γ⋅ b ⋅ ⎮ ⌡0 ⌡0 ( ( ) lbf π ⎞⎤ ⎡ ⎛ Fv = −⎢62.4⋅ × 10⋅ ft × 4⋅ ft × ⎜ 10⋅ ft − 4⋅ ft × ⎟⎥ 3 4⎠ ⎝ ft ⎣ ⎦ To find the line of action of the force: Therefore: x' = x'⋅ Fv Fv Evaluating the integral: Substituting known values: x'⋅ Fv = ⌠ ⎮ x dF v ⎮ ⌡ ⎛ γ ⋅ b ⋅ R⋅ ⎜ L − R⋅ ⎟ 4⎠ ⎝ x' = ) ⎛ ⎝ 3 R = 2 Fv = −17.12 × 10 ⋅ lbf ( dFv = −γ⋅ b ⋅ L − where ( ⌠ ⋅ ⎮ x⋅ γ⋅ b ⋅ L − π ⎞ ⌡0 1 2 R − x dx = − γ ⋅ b ⋅ R ⋅ ⎜ L − R ⋅ R 2 − x ) dx = ⎟ 4⎠ (negative indicates downward) 2 2 ) R − x ⋅ dx R 1 2 π⎞ ⌠ ⋅⎮ π ⎞ ⌡0 ⎛ R⋅ ⎜ L − R⋅ ⎟ 4⎠ ⎝ ( L⋅ x − x⋅ 2 2 R −x ) dx 2 4⋅ R 4⋅ R 2 1 3⎞ ⎛1 ⎛L R⎞ ⎛L R⎞ ⋅ ⎜ ⋅ L⋅ R − ⋅ R ⎟ = ⋅⎜ − ⎟ = ⋅⎜ − ⎟ R ⋅ ( 4⋅ L − π ⋅ R ) ⎝ 2 3 ⎠ R ⋅ ( 4 ⋅ L − π ⋅ R ) ⎝ 2 3 ⎠ 4⋅ L − π ⋅ R ⎝ 2 3 ⎠ 4 x' = 4⋅ 4⋅ ft ⎛ 10⋅ ft − 4⋅ ft ⎞ ⋅⎜ ⎟ 4⋅ 10⋅ ft − π⋅ 4⋅ ft ⎝ 2 3 ⎠ x' = 2.14⋅ ft Problem 3.75 [Difficulty: 2] Given: Gate formed in the shape of a circular arc has width w. Liquid is water; depth h = R Find: (a) Magnitude of the net vertical force component due to fluids acting on the gate (b) Line of action of the vertical component of the force Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dy = ρ⋅ g Fv = − ⌠ ⎮ p dA y ⎮ ⌡ x'⋅ Fv = Assumptions: (Hydrostatic Pressure - y is positive downwards) (Vertical Hydrostatic Force) ⌠ ⎮ x dF v ⎮ ⌡ (Moment of vertical force) (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface of water and on outside of gate Integrating the hydrostatic pressure equation: p = ρ ⋅ g⋅ y Instead of y, we use θ as our variable of integration: Therefore, dy = R⋅ cos ( θ) ⋅ dθ In addition, y = R⋅ sin ( θ) dAy = w⋅ R⋅ sin ( θ) ⋅ dθ π π ⌠2 ⌠2 ⎮ π 2 ⎮ 2 2 Therefore, Fv = −⎮ ρ⋅ g⋅ R⋅ sin ( θ) ⋅ w⋅ R⋅ sin ( θ) dθ = −ρ⋅ g⋅ R ⋅ w⋅ ⎮ ( sin ( θ) ) dθ = −ρ⋅ g⋅ R ⋅ w⋅ ⌡0 ⌡0 4 2 Fv = − π ⋅ ρ ⋅ g⋅ R ⋅ w 4 (negative indicates downward) To find the line of action of the vertical component of the force: 2 2 dFv = −ρ⋅ g⋅ R ⋅ w⋅ ( sin ( θ) ) ⋅ dθ x' = x'⋅ Fv Fv =− π ⌠2 4 2 π⋅ ρ⋅ g⋅ R ⋅ w x'⋅ Fv = ⌠ ⎮ x dFv where x = R⋅ cos ( θ) and the elemental force is ⎮ ⌡ Substituting into the above integral yields: π ⌠2 ⎮ 4⋅ R ⎮ 4⋅ R 1 2 2 2 ⋅ ⎮ −( R⋅ cos ( θ) ) ⋅ ⎡⎣ρ⋅ g⋅ R ⋅ w⋅ ( sin ( θ) ) ⎤⎦ dθ = ⋅ ⎮ ( sin ( θ) ) ⋅ cos ( θ) dθ = ⋅ ⌡0 π ⌡0 π 3 x' = 4⋅ R 3⋅ π Problem 3.76 [Difficulty: 3] Given: Dam with cross-section shown. Width of dam b 160 ft Find: (a) Magnitude and line of action of the vertical force component on the dam (b) If it is possible for the water to overturn dam Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g Fv (Hydrostatic Pressure - h is positive downwards from free surface) p dA y (Vertical Hydrostatic Force) FH pc A (Horizontal Hydrostatic Force) x dF v Ixx h' hc hc A x' Fv (Moment of vertical force) (Line of action of vertical force) ΣMz 0 Assumptions: y (Rotational Equilibrium) A x’ (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface of water and on outside of dam FV B Integrating the hydrostatic pressure equation: y’ p ρ g h x x Into the vertical force equation: h’ FH x B B Fv p dAy ρ g h b dx ρ g b ( H y) dx x x A A From the definition of the dam contour: x y A y B Therefore: y B xA and xA 10 ft 9 ft 2 1 ft xA 2.11 ft xB F v ρ g b x Into the force equation: A Fv 1.94 slug ft 3 ft 32.2 2 s x' Fv 1 Fv xB x' Fv x dFv where dFv ρ g b H xB xB A H xB xA B ln xA xA A 1 5 x' xB A Fv 2.71 10 lbf B dx x A Therefore: H x B x dx x A xB A H 2 2 xB xA B xB xA B A ln 2 xA A Evaluating the integral: 2 7.0 1 lbf s 2.11 1 slug ft 2 B x ρ g b H dx x A xA Substituting known values: 160 ft 9 ft ( 7.0 ft 2.11 ft) 10 ft ln To find the line of action of the force: x' Fv H B dx ρ g b H x x B ln xB A B A x A xA A Substituting known values we get: H xB xA B ln x 9 ft x' 2 2 2 2 2 A A 71 2.11 1 2 7 2.11 ft 10 ft ( 7 2.11) ft 10 ft 1 ft ln x' 4.96 ft 71 2.11 1 2 9 ft ( 7 2.11) ft 10 ft ln To determine whether or not the water can overturn the dam, we need the horizontal force and its line of action: H ρ g b H FH pc A ρ g H b 2 2 FH Substituting values: For the line of action: Therefore: h' H 2 1 2 2 1.94 ft Ixx h' hc hc A b H 3 12 slug 3 32.2 ft 2 2 160 ft ( 9 ft) s where hc 2 1 H H 2 H H b H 2 6 3 H 2 3 5 FH 4.05 10 lbf slug ft A H b 2 h' 2 lbf s Ixx 9 ft b H 3 12 h' 6.00 ft Taking moments of the hydrostatic forces about the origin: Mw FH ( H h') Fv x' 5 5 Mw 4.05 10 lbf ( 9 6) ft 2.71 10 lbf 4.96 ft 5 Mw 1.292 10 lbf ft The negative sign indicates that this is a clockwise moment about the origin. Since the weight of the dam will also contribute a clockwise moment about the origin, these two moments should not cause the dam to tip to the left. Therefore, the water can not overturn the dam. Problem 3.77 [Difficulty: 3] w 35 m Given: Tainter gate as shown Find: Force of the water acting on the gate Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g (Hydrostatic Pressure - h is positive downwards from free surface) dF p dA Assumptions: (Hydrostatic Force) (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface of water and on outside of gate Integrating the hydrostatic pressure equation: p ρ g h ρ g R sin( θ) Resolving the hydrostatic force into horizontal and vertical components: dFH dF cos ( θ) p dA cos ( θ) ρ g R sin ( θ) w R dθ cos ( θ) since θ1 2 FH ρ g R w sin ( θ) cos ( θ) dθ 0 Integrating this expression: 30 deg F H ρ g R w 0 2 FH 1 8 999 kg 3 2 sin ( θ) cos ( θ) dθ ρ g R w 9.81 m m 2 ( sin ( 30 deg) ) 2 10 m where θ1 asin 30 deg 20 m 2 2 30 deg s 2 8 Substituting known values: N s 7 FH 1.715 10 N kg m 2 π 3 12 8 ( sin ( θ) ) dθ ρ g R w 2 2 dFv dF sin ( θ) p dA sin ( θ) ρ g R w ( sin ( θ) ) dθ Substituting known values: 2 π 3 kg m N s 2 999 3 9.81 2 ( 20 m) 35 m 12 8 kg m m s Fv ρ g R w 2 2 ( 20 m) 35 m Similarly, we can calculate the vertical component of the hydrostatic force: 2 F v ρ g R w 0 dA w R dθ 6 Fv 6.21 10 N Now since the gate surface in contact with the water is a circular arc, all elements dF of the force, and hence the line of action of the resulta must pass through the pivot. Thus: Magnitude of the resultant force: FR 2 FH Fv 2 FR 1.715 107 N2 6.21 106 N2 7 FR 1.824 10 N The line of action of the force: Fv FH α atan 6.21 106 N 1.715 107 N α atan α 19.9 deg The force passes through the pivot at an angle α to the horizontal. Problem 3.78 Given: Gate geometry Find: Force on stop B [Difficulty: 4] x y’ Solution: Basic equations 4R/3π R/2 D FV dp dh = ρ⋅ g W1 A R FB ΣMA = 0 WGate FH y W2 x Weights for computing FV F1 Assumptions: static fluid; ρ = constant; patm on other side p = ρ⋅ g⋅ h For incompressible fluid where p is gage pressure and h is measured downwards We need to compute force (including location) due to water on curved surface and underneath. For curved surface we could integrate pressure, but here we use the concepts that FV (see sketch) is equivalent to the weight of fluid above, and FH is equivalent to the force on a vertical flat plate. Note that the sketch only shows forces that will be used to compute the moment at A For FV FV = W1 − W2 with kg m N⋅ s W1 = ρ⋅ g⋅ w⋅ D⋅ R = 1000⋅ × 9.81⋅ × 3⋅ m × 4.5⋅ m × 3⋅ m × 3 2 kg⋅ m m s 2 W2 = ρ⋅ g⋅ w⋅ π⋅ R 2 4 = 1000⋅ 397 189 × 3⋅ m 2 m 2 × 3⋅ m × π s 4 2 × ( 3⋅ m) × 2 N⋅ s kg⋅ m FV = 189⋅ kN R 4⋅ R FV⋅ x = W1⋅ − W2⋅ 2 3⋅ π x = For FH 3 × 9.81⋅ m FV = W1 − W2 with x given by kg − 208 189 Computing equations × or 4 3⋅ π × 3⋅ m FH = pc⋅ A x= W1 R W2 4⋅ R ⋅ − ⋅ Fv 2 F v 3⋅ π x = 1.75 m Ixx y' = yc + A ⋅ yc W1 = 397⋅ kN W2 = 208⋅ kN Hence R⎞ ⎛ FH = pc⋅ A = ρ⋅ g⋅ ⎜ D − ⎟ ⋅ w⋅ R 2⎠ ⎝ kg FH = 1000⋅ × 9.81⋅ 3 m ⎛ ⎝ m × ⎜ 4.5⋅ m − 2 s 2 3⋅ m ⎞ N⋅ s 2 ⎠ kg⋅ m ⎟ × 3⋅ m × 3⋅ m × FH = 265⋅ kN The location of this force is 3 2 Ixx R ⎞ w⋅ R 1 R R ⎛ y' = yc + = ⎜D − ⎟ + × = D− + A ⋅ yc ⎝ 2⎠ 12 R⎞ 2 R⎞ ⎛ ⎛ w ⋅ R⋅ ⎜ D − ⎟ 12⋅ ⎜ D − ⎟ 2 2⎠ ⎝ ⎠ ⎝ y' = 4.5⋅ m − 3⋅ m 2 ( 3⋅ m) + ⎛ ⎝ 2 12 × ⎜ 4.5⋅ m − y' = 3.25 m 3⋅ m ⎞ ⎟ 2 ⎠ The force F1 on the bottom of the gate is F1 = p⋅ A = ρ⋅ g⋅ D⋅ w⋅ R F1 = 1000⋅ kg 3 × 9.81⋅ m 2 m × 4.5⋅ m × 3⋅ m × 3⋅ m × 2 s N⋅ s F1 = 397⋅ kN kg⋅ m For the concrete gate (SG = 2.4 from Table A.2) WGate = SG⋅ ρ⋅ g⋅ w⋅ FB = 4 3⋅ π 2 = 2.4⋅ 1000⋅ 4 ⋅ WGate + 4 3⋅ π x R × 499⋅ kN + FB = 278⋅ kN kg 3 × 9.81⋅ m F B ⋅ R + F 1⋅ Hence, taking moments about A FB = π⋅ R R − WGate⋅ 2 ⋅ FV + 1.75 3 4⋅ R 3⋅ π [ y' − ( D − R) ] R × 189⋅ kN + m 2 × 3⋅ m × s π 4 2 2 × ( 3⋅ m) × N⋅ s kg⋅ m − FV⋅ x − FH⋅ [ y' − ( D − R) ] = 0 ⋅ FH − 1 2 ⋅ F1 [ 3.25 − ( 4.5 − 3) ] 3 × 265⋅ kN − 1 2 × 397⋅ kN WGate = 499⋅ kN Problem 3.79 Given: Sphere with different fluids on each side Find: Resultant force and direction [Difficulty: 4] Solution: The horizontal and vertical forces due to each fluid are treated separately. For each, the horizontal force is equivalent to that on a vertical flat plate; the vertical force is equivalent to the weight of fluid "above". For horizontal forces, the computing equation of Section 3-5 is FH pc A where A is the area of the equivalent vertical plate. For vertical forces, the computing equation of Section 3-5 is FV ρ g V where V is the volume of fluid above the curved surface. The data is ρ 999 For water kg 3 m For the fluids SG1 1.6 SG2 0.8 For the weir D 3 m L 6 m (a) Horizontal Forces For fluid 1 (on the left) D 1 2 FH1 pc A ρ1 g D L SG1 ρ g D L 2 2 FH1 For fluid 2 (on the right) 2 1.6 999 kg 3 9.81 m m 2 2 ( 3 m) 6 m s 2 N s kg m FH1 423 kN D D 1 2 FH2 pc A ρ2 g L SG2 ρ g D L 4 2 8 FH2 The resultant horizontal force is 1 1 8 0.8 999 kg 3 m 9.81 m 2 s 2 ( 3 m) 6 m 2 N s kg m FH FH1 FH2 (b) Vertical forces For the left geometry, a "thought experiment" is needed to obtain surfaces with fluid "above" FH2 52.9 kN FH 370 kN Hence π D 4 FV1 SG1 ρ g 2 L 2 kg FV1 1.6 999 3 9.81 m m 2 s π ( 3 m) 2 2 6 m 8 N s kg m FV1 333 kN (Note: Use of buoyancy leads to the same result!) For the right side, using a similar logic π D 4 FV2 SG2 ρ g 2 L 4 FV2 0.8 999 kg 3 9.81 m The resultant vertical force is FV FV1 FV2 m 2 s π ( 3 m) 16 2 2 6 m N s kg m FV2 83.1 kN FV 416 kN Finally the resultant force and direction can be computed F 2 FH FV FV FH α atan 2 F 557 kN α 48.3 deg Problem 3.80 [Difficulty: 3] Given: Cylindrical weir as shown; liquid is water Find: Magnitude and direction of the resultant force of the water on the weir Solution: We will apply the hydrostatics equations to this system. Governing Equations: Assumptions: dp = ρ⋅ g dh ⎯→ ⎯ → dFR = −p ⋅ dA (Hydrostatic Pressure - h is positive downwards from free surface) (Hydrostatic Force) (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts on free surfaces and on the first quadrant of the cylinder Using the coordinate system shown in the diagram at the right: h1 3⋅ π ⌠ 2 ⎮ FRx = ⎮ ⌡0 y x D1 ⎯ →→ →→ ⌠ ⌠ ⌠ ⎮ ⎮ ⎮ FRx = FR⋅ i = − p dA ⋅ i = − p ⋅ cos ( θ + 90⋅ deg) dA = ⎮ p ⋅ sin( θ) dA ⎮ ⎮ ⌡ ⌡ ⌡ ⎯ →→ →→ ⌠ ⌠ ⎮ ⎮ FRy = FR⋅ j = − p dA⋅ j = − p⋅ cos ( θ) dA ⎮ ⎮ ⌡ ⌡ θ h2 Now since dA = L⋅ R⋅ dθ it follows that 3⋅ π ⌠ 2 p⋅ L⋅ R⋅ sin ( θ) dθ and ⎮ FRy = −⎮ ⌡0 Next, we integrate the hydrostatic pressure equation: p⋅ L⋅ R⋅ cos ( θ) dθ p = ρ ⋅ g⋅ h Now over the range Over the range 0≤θ≤π π≤θ≤ 3⋅ π 2 h1 = R ( 1 − cos ( θ) ) h2 = −R⋅ cos ( θ) Therefore we can express the pressure in terms of θ and substitute into the force equations: 3⋅ π 3⋅ π ⌠ 2 ⌠ 2 π ⎮ ⎮ ⌠ ⎮ ρ⋅ g⋅ R⋅ ( 1 − cos ( θ) ) ⋅ L⋅ R⋅ sin ( θ) dθ − ⎮ FRx = ⎮ p⋅ L⋅ R⋅ sin ( θ) dθ = ρ⋅ g⋅ R⋅ cos ( θ) ⋅ L⋅ R⋅ sin ( θ) dθ ⌡0 ⌡0 ⌡π π 3⋅ π ⌠ 2 2 ⌠ 2 ⎮ FRx = ρ⋅ g⋅ R ⋅ L⋅ ⎮ ( 1 − cos ( θ) ) ⋅ sin ( θ) dθ − ρ⋅ g⋅ R ⋅ L⋅ ⎮ ⌡0 ⌡π cos ( θ) ⋅ sin ( θ) dθ D2 3⋅ π ⎤ ⎡ ⎢ π ⎥ ⌠ 2 ⎮ 3 1⎞ 2 ⎛ 2 ⎢⌠ ⎥ 2 FRx = ρ⋅ g⋅ R ⋅ L⋅ ⎮ ( 1 − cos ( θ) ) ⋅ sin ( θ) dθ − ⎮ cos ( θ) ⋅ sin ( θ) dθ = ρ⋅ g⋅ R ⋅ L⋅ ⎜ 2 − ⎟ = ⋅ ρ⋅ g⋅ R ⋅ L ⎢⌡ ⎥ ⌡ ⎝ 2⎠ 2 π ⎣ 0 ⎦ Substituting known values: FRx = 3 2 × 999⋅ kg 3 × 9.81⋅ m m 2 2 2 × ( 1.5⋅ m) × 6⋅ m × s N⋅ s FRx = 198.5⋅ kN kg⋅ m Similarly we can calculate the vertical force component: 3⋅ π 3⋅ π ⎡ ⎤ ⎢ π ⎥ ⌠ 2 ⌠ 2 ⎮ ⎮ ⎢⌠ ⎥ FRy = −⎮ p⋅ L⋅ R⋅ cos ( θ) dθ = − ⎮ ρ⋅ g⋅ R⋅ ( 1 − cos ( θ) ) ⋅ L⋅ R⋅ cos ( θ) dθ − ⎮ ρ⋅ g⋅ R⋅ cos ( θ) ⋅ L⋅ R⋅ cos ( θ) dθ ⎢⌡ ⎥ ⌡0 ⌡ π ⎣ 0 ⎦ 3⋅ π ⎡ ⎤ ⎢ π ⎥ ⌠ 2 ⎮ 3⋅ π 2 ⎢⌠ 2 ⎥ 2 ⎛ π 3⋅ π π ⎞ 2 FRy = −ρ⋅ g⋅ R ⋅ L⋅ ⎮ ( 1 − cos ( θ) ) ⋅ cos ( θ) dθ − ⎮ ( cos ( θ) ) dθ = ρ⋅ g⋅ R ⋅ L⋅ ⎜ + − ⎟ = ⋅ ρ ⋅ g⋅ R ⋅ L ⎢⌡ ⎥ ⌡ 2 4 2 4 ⎝ ⎠ π ⎣ 0 ⎦ Substituting known values: FRy = 3⋅ π 4 × 999⋅ kg 3 m × 9.81⋅ m 2 2 × ( 1.5⋅ m) × 6⋅ m × s 2 N⋅ s kg⋅ m FRy = 312⋅ kN Now since the weir surface in contact with the water is a circular arc, all elements dF of the force, and hence the line of action of the resultant force, must pass through the pivot. Thus: 2 Magnitude of the resultant force: FR = ( 198.5⋅ kN) + ( 312⋅ kN) The line of action of the force: α = atan ⎜ ⎛ 312⋅ kN ⎞ ⎟ ⎝ 198.5⋅ kN ⎠ 2 FR = 370⋅ kN α = 57.5⋅ deg Problem 3.81 [Difficulty: 3] Given: Cylindrical log floating against dam Find: (a) Mass per unit length of the log (b) Contact force per unit length between log and dam Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh = ρ⋅ g (Hydrostatic Pressure - h is positive downwards from free surface) ⎯ ⎯ → → dF = p ⋅ dA Assumptions: (Hydrostatic Force) (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts on free surfaces and on the first quadrant of the log Integrating the hydrostatic pressure equation: dFH R = D/2 dFV dF h p = ρ⋅ g⋅ h = ρ⋅ g⋅ R⋅ ( 1 − cos ( θ) ) θ Resolving the incremental force into horizontal and vertical components: 2 dF = p ⋅ dA = p ⋅ w⋅ R⋅ dθ = ρ⋅ g⋅ R⋅ ( 1 − cos ( θ) ) ⋅ w⋅ R⋅ dθ = ρ⋅ g⋅ R ⋅ w⋅ ( 1 − cos ( θ) ) 2 dFH = dF⋅ sin( θ) = ρ⋅ g⋅ R ⋅ w⋅ ( 1 − cos ( θ) ) ⋅ dθ⋅ sin( θ) 2 dFv = dF⋅ cos ( θ) = ρ⋅ g⋅ R ⋅ w⋅ ( 1 − cos ( θ) ) ⋅ dθ⋅ cos ( θ) Integrating the expression for the horizontal force will provide us with the contact force per unit length: 3⋅ π ⌠ 2 ⎮ FH = ⎮ ⌡0 3⋅ π ⌠ 2 2 2 ⎮ ρ⋅ g⋅ R ⋅ w⋅ ( 1 − cos ( θ) ) ⋅ sin( θ) dθ = ρ⋅ g⋅ R ⋅ w⋅ ⎮ ⌡0 2 ⎛ 1 + 1 ⎞ = ρ⋅ g⋅ R ⋅ w ⎟ 2 ⎝ 2 ⎠ 2 ( sin( θ) − sin( θ) ⋅ cos ( θ) ) dθ = ρ⋅ g⋅ R ⋅ w⋅ ⎜ − Therefore: FH w = ρ⋅ g⋅ R 2 2 Integrating the expression for the vertical force will provide us with the mass per unit length of the log: 3⋅ π 3⋅ π ⌠ 2 ⌠ 2 ⎮ 3⋅ π ⎞ 2 2 ⎮ 2 ⎛ Fv = ⎮ ρ⋅ g⋅ R ⋅ w⋅ ( 1 − cos ( θ) ) ⋅ cos ( θ) dθ = ρ⋅ g⋅ R ⋅ w⋅ ⎮ ( 1 − cos ( θ) ) ⋅ cos ( θ) dθ = ρ⋅ g⋅ R ⋅ w⋅ ⎜ −1 − ⎟ ⌡0 ⌡0 4 ⎠ ⎝ Therefore: Fv w 2 ⎛ ⎝ = − ρ ⋅ g⋅ R ⋅ ⎜ 1 + 3⋅ π ⎞ ⎟ 4 ⎠ From a free-body diagram for the log: ΣFy = 0 − m w ⋅g − Solving for the mass of the log: Fv w m =0 m w w 2 ⎛ ⎝ =− = ρ⋅ R ⋅ ⎜ 1 + Fv w⋅ g 3⋅ π ⎞ ⎟ 4 ⎠ Problem 3.82 Given: [Difficulty: 3] Curved surface, in shape of quarter cylinder, with given radius R and width w; water stands to depth H. R = 0.750⋅ m w = 3.55⋅ m H = 0.650⋅ m Find: Magnitude and line of action of (a) vertical force and (b) horizontal force on the curved surface Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh = ρ⋅ g (Hydrostatic Pressure - h is positive downwards from free surface) ⌠ ⎮ p dA y ⎮ ⌡ Fv = (Vertical Hydrostatic Force) FH = pc⋅ A (Horizontal Hydrostatic Force) ⌠ ⎮ x dF v ⎮ ⌡ Ixx h' = hc + hc⋅ A x'⋅ Fv = Assumptions: (Moment of vertical force) (Line of action of horizontal force) dF (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts on free surface of the water and on the left side of the curved surface Integrating the hydrostatic pressure equation: From the geometry: h = H − R⋅ sin ( θ) ⎛H⎞ θ1 = asin ⎜ ⎟ ⎝R⎠ R θ h H p = ρ ⋅ g⋅ h y = R⋅ sin ( θ) ⎛ 0.650 ⎞ θ1 = asin ⎜ ⎟ ⎝ 0.750 ⎠ x = R⋅ cos ( θ) dA = w⋅ R⋅ dθ x’ dF FV θ1 = 1.048⋅ rad h’ R Therefore the vertical component of the hydrostatic force is: θ 1 ⌠ ⌠ ⌠ ⎮ ⎮ ⎮ Fv = ⎮ p dAy = ⎮ ρ⋅ g⋅ h⋅ sin ( θ) dA = ρ⋅ g⋅ ( H − R⋅ sin ( θ) ) ⋅ sin ( θ) ⋅ w⋅ R dθ ⌡0 ⌡ ⌡ θ ( ) 1 ⌠ ⎡ ⎛ θ1 sin 2⋅ θ1 ⎞⎤ 2 Fv = ρ⋅ g⋅ w⋅ R⋅ ⎮ ⎡⎣H⋅ sin ( θ) − R⋅ ( sin ( θ) ) ⎤⎦ dθ = ρ⋅ g⋅ w⋅ R⋅ ⎢H⋅ 1 − cos θ1 − R⋅ ⎜ − ⎟⎥ ⌡0 4 ⎣ ⎝2 ⎠⎦ ( ( )) θ FH y’ H kg Fv = 999⋅ 3 × 9.81⋅ m m 2 2 ⎡ ⎣ ⎛ 1.048 − sin ( 2 × 1.048⋅ rad) ⎞⎤ × N⋅ s ⎟⎥ 4 ⎝ 2 ⎠⎦ kg⋅ m × 3.55⋅ m × 0.750⋅ m × ⎢0.650⋅ m × ( 1 − cos ( 1.048⋅ rad) ) − 0.750⋅ m × ⎜ s Fv = 2.47⋅ kN To calculate the line of action of this force: θ 1 ⌠ 2⌠ ⎡ 2 ⎮ ⎮ x'⋅ Fv = R⋅ cos ( θ) ⋅ ρ⋅ g⋅ h⋅ sin ( θ) dA = ρ⋅ g⋅ w⋅ R ⋅ ⎣H⋅ sin ( θ) ⋅ cos ( θ) − R⋅ ( sin ( θ) ) ⋅ cos ( θ)⎤⎦ dθ ⎮ ⌡ ⌡ 0 2 H 2 R 3 Evaluating the integral: x'⋅ Fv = ρ⋅ g⋅ w⋅ R ⋅ ⎡⎢ ⋅ sin θ1 − ⋅ sin θ1 ⎥⎤ 2 3 ⎣ ⎦ ( ( )) x' = x'⋅ Fv Fv x' = 999⋅ 2 = kg 3 ( ( )) ρ ⋅ g⋅ w ⋅ R ⎡ H 2 R 3⎤ ⋅ ⎢ ⋅ sin θ1 − ⋅ sin θ1 ⎥ Fv 2 3 ⎣ ⎦ × 9.81⋅ m m 2 ( ( )) ( ( )) 2 × 3.55⋅ m × ( 0.750⋅ m) × s Therefore we may find the line of action: Substituting in known values: ( ) 0.650 sin θ1 = 0.750 ⎡ 0.650⋅ m ⎛ 0.650 ⎞ 2 0.750⋅ m ⎛ 0.650 ⎞ 3⎤ N⋅ s2 ×⎜ ×⎜ ⎟ − ⎟ ⎥× 3 N ⎣ 2 3 ⎝ 0.750 ⎠ ⎝ 0.750 ⎠ ⎦ kg⋅ m 2.47 × 10 1 ⋅ 1 ×⎢ x' = 0.645 m 2 For the horizontal force: FH = 1 2 × 999⋅ kg 3 m H ρ ⋅ g⋅ H ⋅ w FH = pc⋅ A = ρ⋅ g⋅ hc⋅ H⋅ w = ρ⋅ g⋅ ⋅ H⋅ w = 2 2 × 9.81⋅ m 2 2 2 × ( 0.650⋅ m) × 3.55⋅ m × s For the line of action of the horizontal force: N⋅ s FH = 7.35⋅ kN kg⋅ m Ixx h' = hc + hc⋅ A 3 Ixx H w⋅ H 2 1 H H 2 h' = hc + = + ⋅ ⋅ = + = ⋅H 12 H w⋅ H 6 hc⋅ A 2 2 3 where h' = 2 3 Ixx = w⋅ H 12 × 0.650⋅ m 3 A = w⋅ H Therefore: h' = 0.433 m Problem 3.83 Given: Canoe floating in a pond Find: What happens when an anchor with too short of a line is thrown from canoe [Difficulty: 2] Solution: Governing equation: FB w gVdisp W Before the anchor is thrown from the canoe the buoyant force on the canoe balances out the weight of the canoe and anchor: FB1 Wcanoe Wanchor w gVcanoe1 The anchor weight can be expressed as Wanchor a gVa so the initial volume displaced by the canoe can be written as Vcanoe1 Wcanoe a Va w g w After throwing the anchor out of the canoe there will be buoyant forces acting on the canoe and the anchor. Combined, these buoyant forces balance the canoe weight and anchor weight: FB2 Wcanoe Wanchor w gVcanoe2 w gVa Vcanoe 2 Wcanoe Wa Va w g w g Vcanoe 2 Wcanoe a Va Va w g w Using the anchor weight, Hence the volume displaced by the canoe after throwing the anchor in is less than when the anchor was in the canoe, meaning that the canoe is floating higher. Problem 3.84 Given: [Difficulty: 3] Curved surface, in shape of quarter cylinder, with given radius R and width w; liquid concrete stands to depth H. R = 1⋅ ft w = 4⋅ ft Fvmax = 350⋅ lbf SG = 2.50 From Table A.1, App A Find: (a) Maximum depth of concrete to avoid cracking (b) Line of action on the form. (c) Plot the vertical force and line of action over H ranging from 0 to R. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh = ρ⋅ g Fv = (Hydrostatic Pressure - h is positive downwards from free surface) ⌠ ⎮ p dA y ⎮ ⌡ (Vertical Hydrostatic Force) x ⌠ ⎮ x'⋅ Fv = x dF v ⎮ ⌡ Assumptions: d (Moment of vertical force) (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts on free surface of the concrete θ1 θ x’ h FV p = ρ ⋅ g⋅ h Integrating the hydrostatic pressure equation: From the geometry: y = R⋅ sin ( θ) y x = R⋅ cos ( θ) h = y−d d = R−H dA = w⋅ R⋅ dθ Therefore the vertical component of the hydrostatic force is: π Fv = ⌠ ⌠ ⎮ ⎮ p dA y = ⎮ ⎮ ⌡ ⌡ ⌠2 ⎮ ρ⋅ g⋅ h⋅ sin ( θ) dA = ⎮ ρ⋅ g⋅ ( R⋅ sin ( θ) − d) ⋅ sin ( θ) ⋅ w⋅ R dθ ⌡θ where 1 π ⌠2 ( ) ⎛d⎞ θ1 = asin ⎜ ⎟ ⎝ R⎠ ⎮ ⎡ ⎛ π θ1 sin 2⋅ θ1 ⎞ ⎤ 2 Fv = ρ⋅ g⋅ w⋅ R⋅ ⎮ ⎡⎣R⋅ ( sin ( θ) ) − d⋅ ( sin ( θ) )⎤⎦ dθ = ρ⋅ g⋅ w⋅ R⋅ ⎢R⋅ ⎜ − + ⎟ − d⋅ cos θ1 ⎥ 4 ⌡θ ⎣ ⎝4 2 ⎠ ⎦ ( ) In terms of H: 1 ( ) sin θ1 = R−H R ( ) cos θ1 = 2 R − ( R − H) R 2 = 2⋅ R ⋅ H − H R 2 ( ) ( ) ( ) sin 2⋅ θ1 = 2⋅ sin θ1 ⋅ cos θ1 = 2⋅ ( R − H ) ⋅ 2⋅ R ⋅ H − H R 2 2 ⎡⎢ ⎡⎢ ⎛ H⎞ ⎤ ⎤ asin ⎜ 1 − ⎟ 2⎥ 2⎥ R π ( R − H ) ⋅ 2 ⋅ R ⋅ H − H 2 ⋅ R ⋅ H − H ⎝ ⎠ ⎥ − ( R − H) ⋅ ⎥ Fv = ρ⋅ g⋅ w⋅ R⋅ ⎢R⋅ ⎢ − + 2 2 R ⎢ ⎢4 ⎥ ⎥ 2R ⎣ ⎣ ⎦ ⎦ This equation can be solved iterative for H: H = 0.773⋅ ft To calculate the line of action of this force: π x'⋅ Fv = ⌠ ⎮ ⎮ ⌡ ⌠2 2 ⎮ ⎡ 2 x⋅ ρ⋅ g⋅ h⋅ sin ( θ) dA = ρ⋅ g⋅ R ⋅ w⋅ ⎮ ⎣R⋅ ( sin ( θ) ) ⋅ cos ( θ) − d⋅ sin ( θ) ⋅ cos ( θ)⎤⎦ dθ ⌡θ 1 Evaluating the integral: R d 2 3 2⎤ x'⋅ Fv = ρ⋅ g⋅ R ⋅ w⋅ ⎡⎢ ⋅ ⎡1 − sin θ1 ⎤ − ⋅ cos θ1 ⎥ ⎣ ⎦ 2 ⎣3 ⎦ Therefore we may find the line of action: x' = ( ( )) ( ) sin θ1 = Substituting in known values: ⎛ slug ⎞ ⎝ ft ⎠ x' = ⎜ 2.5 × 1.94⋅ 3 ⎟ × 32.2⋅ ft 2 x'⋅ Fv Fv 1 − 0.773 1 2 × ( 1⋅ ft) × 4⋅ ft × s 2 = ρ ⋅ g⋅ R ⋅ w ⎡ R ⎡ d 3 2⎤ ⋅ ⎢ ⋅ 1 − sin θ1 ⎤ − ⋅ cos θ1 ⎥ ⎣ ⎦ Fv 2 ⎣3 ⎦ ⋅ 1 350 lbf ( ( )) ( ) = 0.227 1 ( ( )) cos θ1 = ( ( )) 2 1 − 0.227 = 0.9739 2 ⎡ 1⋅ ft × ⎡1 − ( 0.227) 3⎤ − 0.227⋅ ft × ( 0.9739) 2⎤ × lbf ⋅ s ⎣ ⎦ ⎥ 2 ⎣ 3 ⎦ slug⋅ ft ×⎢ x' = 0.396⋅ ft We may use the equations we developed above to plot the vertical force and line of action as a function of the height of the concrete in the Vertical Force vs. Depth Ratio Line of Action vs. Depth Ratio 500.0 0.4 Line of Action (ft) Vertical Force (lbf) 400.0 300.0 200.0 0.2 100.0 0.0 0.0 0.5 Depth Ratio H/R 1.0 0.0 0.0 0.5 Depth Ratio H/R 1.0 Problem 3.85 Given: [Difficulty: 3] Model cross section of canoe as a parabola. Assume constant width W over entire length L 2 y a x a 1.2 ft 1 W 2 ft L 18 ft Find: Expression relating the total mass of canoe and contents to distance d. Determine maximum allowable total mass without swamping the canoe. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g Fv Assumptions: (Hydrostatic Pressure - h is positive downwards from free surface) p dA y (Vertical Hydrostatic Force) (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts on free surface of the water and inner surface of the canoe. At any value of d the weight of the canoe and its contents is balanced by the net vertical force of the water on the canoe. Integrating the hydrostatic pressure equation: Fv p ρ g h p dA y ρ g h L dx where h ( H d) y y a x To determine the upper limit of integreation we remember that y Hd F v 2 0 Therefore, x Hd a Hd a At the surface and so the vertical force is: 2 ρ g ( H d) a x L dx 2 ρ g L 0 Upon simplification: Fv 2 ρ g L 2 ( H d) a 3 2 Hd a 3 3 2 2 ( H d ) a ( H d ) 2 ( H d) a x dx 2 ρ g L 3 a a 3 1 4 ρ g L ( H d) 2 M g 1 3 3 a or M 4 ρ L ( H d) 3 a 3 2 where M is the mass of the canoe. 3 ft 2 32.174 lb ( 2.4 ft) 1.2 slug 4 slug The limit for no swamping is d=0, and so: M 1.94 18 ft 3 3 ft This leaves us no margin, so if we set d=0.2 ft we get M 4 3 1.94 slug ft 3 18 ft ft 1.2 3 M 5.08 10 lb 3 2 32.174 lb ( 2.2 ft) slug 3 M 4.46 10 lb Clearly the answer is highly dependent upon the allowed risk of swamping! Problem 3.86 [Difficulty: 4] Given: Cylinder of mass M, length L, and radius R is hinged along its length and immersed in an incompressilble liquid to depth Find: General expression for the cylinder specific gravity as a function of α=H/R needed to hold the cylinder in equilibrium for α ranging from 0 to 1. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh = ρ⋅ g Fv = (Hydrostatic Pressure - h is positive downwards from free surface) ⌠ ⎮ p dA y ⎮ ⌡ (Vertical Hydrostatic Force) ΣM = 0 Assumptions: H = αR (Rotational Equilibrium) h (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts on free surface of the liquid. dFV θ dF dFH The moments caused by the hydrostatic force and the weight of the cylinder about the hinge need to balance each other. Integrating the hydrostatic pressure equation: p = ρ ⋅ g⋅ h dFv = dF⋅ cos ( θ) = p⋅ dA⋅ cos ( θ) = ρ⋅ g⋅ h⋅ w⋅ R⋅ dθ⋅ cos ( θ) Now the depth to which the cylinder is submerged is H = h + R⋅ ( 1 − cos ( θ) ) h = H − R⋅ ( 1 − cos ( θ) ) and into the vertical force equation: Therefore 2 ⎡H dFv = ρ⋅ g⋅ [ H − R⋅ ( 1 − cos ( θ) ) ] ⋅ w⋅ R⋅ cos ( θ) ⋅ dθ = ρ⋅ g⋅ w⋅ R ⋅ ⎢ ⎣R ⎤ ⎦ − ( 1 − cos ( θ) )⎥ ⋅ cos ( θ) ⋅ dθ 1 + cos ( 2⋅ θ)⎤ 2 2 2⎡ dFv = ρ⋅ g⋅ w⋅ R ⋅ ⎡⎣( α − 1) ⋅ cos ( θ) + ( cos ( θ) ) ⎤⎦ ⋅ dθ = ρ⋅ g⋅ w⋅ R ⋅ ⎢( α − 1) ⋅ cos ( θ) + ⎥ ⋅ dθ 2 ⎣ ⎦ Now as long as α is not greater than 1, the net horizontal hydrostatic force will be zero due to symmetry, and the vertical force is: θmax ⌠ Fv = ⎮ ⌡− θ max θmax ⌠ 1 dF v = ⎮ ⌡0 2 dF v where ( ) cos θmax = R−H R = 1−α or θmax = acos ( 1 − α) 2⌠ ⎮ Fv = 2ρ⋅ g⋅ w⋅ R ⋅ θmax ⎮ ⌡0 ⎡( α − 1) ⋅ cos ( θ) + 1 + 1 ⋅ cos ( 2⋅ θ)⎤ dθ ⎢ ⎥ 2 2 ⎣ ⎦ Now upon integration of this expression we have: 2 Fv = ρ⋅ g⋅ w⋅ R ⋅ ⎡⎣acos ( 1 − α) − ( 1 − α) ⋅ α⋅ ( 2 − α)⎤⎦ The line of action of the vertical force due to the liquid is through the centroid of the displaced liquid, i.e., through the center of the cylinde 2 The weight of the cylinder is given by: W = M⋅ g = ρc⋅ V⋅ g = SG⋅ ρ⋅ π⋅ R ⋅ w⋅ g where ρ is the density of the fluid and SG = ρc ρ The line of action of the weight is also throught the center of the cylinder. Taking moment about the hinge we get: ΣMo = W⋅ R − Fv⋅ R = 0 2 or in other words W = Fv and therefore: 2 SG⋅ ρ⋅ π⋅ R ⋅ w⋅ g = ρ⋅ g⋅ w⋅ R ⋅ ⎡⎣acos ( 1 − α) − ( 1 − α) ⋅ α⋅ ( 2 − α)⎤⎦ SG = 1 π ⋅ ⎡⎣acos ( 1 − α) − ( 1 − α) ⋅ α⋅ ( 2 − α)⎤⎦ Specific Gravity, SG 0.6 0.4 0.2 0 0 0.5 alpha (H/R) 1 Problem 3.87 Given: [Difficulty: 4] Canoe, modeled as a right semicircular cylindrical shell, floats in water of depth d. The shell has outer radius R and leng R = 1.2⋅ ft L = 17⋅ ft d = 1⋅ ft Find: (a) General expression for the maximum total mass that can be floated, as a function of depth, (b) evaluate for the given conditions (c) plot for range of water depth between 0 and R. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dy = ρ⋅ g Fv = Assumptions: (Hydrostatic Pressure - y is positive downwards from free surface) ⌠ ⎮ p dA y ⎮ ⌡ (Vertical Hydrostatic Force) (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts on free surface of the liquid. θmax y is a function of θ for a given depth d: y = d − ( R − R⋅ cos ( θ) ) = d − R + R⋅ cos ( θ) θ y The maximum value of θ: ⎡ ( R − d) ⎤ ⎥ ⎣ R ⎦ d dF θmax = acos ⎢ A free-body diagram of the canoe gives: ΣFy = 0 = M⋅ g − Fv where is the vertical force of the water on the canoe. Fv θ θ max max ⌠ ⌠ ⌠ ⌠ ⎮ ⎮ Fv = p dA y = p⋅ cos ( θ) dA = ⎮ ρ⋅ g⋅ y⋅ L⋅ R⋅ cos ( θ) dθ = 2⋅ ρ⋅ g⋅ L⋅ R⋅ ⎮ ( d − R + R⋅ cos ( θ) ) ⋅ cos ( θ) dθ ⎮ ⎮ ⌡0 ⌡− θ ⌡ ⌡ max θmax ⌠ F v = 2⋅ ρ ⋅ g⋅ L ⋅ R ⋅ ⎮ ⌡0 Since For M= Fv g ⎛ θmax sin (2⋅ θmax) ⎞⎤ ⎡⎣( d − R) ⋅ cos ( θ) + R⋅ ( cos ( θ) ) 2⎤⎦ dθ = 2⋅ ρ⋅ g⋅ L⋅ R⋅ ⎡⎢( d − R) ⋅ sin θ + R ⋅ ⎜ + ⎟⎥ ( ) max 4 ⎣ ⎝ 2 ⎠⎦ it follows that R = 1.2⋅ ft L = 17⋅ ft and d = 1⋅ ft ⎡ ( ) ⎛ θmax M = 2⋅ ρ⋅ L⋅ R⋅ ⎢( d − R) ⋅ sin θmax + R⋅ ⎜ ⎣ we can determine the mass: ⎝ 2 + ( ) ⎟⎥ sin 2⋅ θmax ⎞⎤ 4 ⎠⎦ ⎡ ( 1.2 − 1)⎤ ⎥ ⎣ 1.2 ⎦ θmax = acos ⎢ θmax = 1.403⋅ rad M = 2 × 1.94⋅ slug ft 3 ⎡ ⎣ ⎛ 1.403⋅ rad + sin ( 2 × 1.403⋅ rad) ⎞⎤ × 32.2⋅ lbm ⎟⎥ 2 4 slug ⎝ ⎠⎦ × 17⋅ ft × 1.2⋅ ft × ⎢( 1⋅ ft − 1.2⋅ ft) × sin ( 1.403⋅ rad) + 1.2⋅ ft × ⎜ M = 1895⋅ lbm When we enter the values of d/R into the expressions for θmax and M, we get the following graph: Mass versus Submersion Depth Ratio Mass, M (kg) 1000 500 0 0 0.5 Submersion Depth Ratio (d/R) 1 Problem 3.88 Given: Geometry of glass observation room Find: Resultant force and direction Assumptions: [Difficulty: 4] Water in aquarium is static and incompressible Solution: The x, y and z components of force due to the fluid are treated separately. For the x, y components, the horizontal force is equivalent to that on a vertical flat plate; for the z component, (vertical force) the force is equivalent to the weight of fluid above. For horizontal forces, the computing equation of Section 3-5 is FH pc A where A is the area of the equivalent vertical plate. For the vertical force, the computing equation of Section 3-5 is FV ρ g V where V is the volume of fluid above the curved surface. The data are ρ 1.94 For water slug ft For the fluid (Table A.2) SG 1.025 For the aquarium R 5 ft 3 H 35 ft (a) Horizontal Forces Consider the x component yc H The center of pressure of the glass is Hence FHx pc A SG ρ g yc FHx 1.025 1.94 slug ft 3 π R 4 R 3 π yc 32.88 ft 2 4 32.2 ft 2 32.88 ft s π ( 5 ft) 4 2 2 lbf s slug ft 4 FHx 4.13 10 lbf The y component is of the same magnitude as the x component 4 FHy FHx FHy 4.13 10 lbf The resultant horizontal force (at 45o to the x and y axes) is FH 2 FHx FHy 2 4 FH 5.85 10 lbf (b) Vertical forces The vertical force is equal to the weight of fluid above (a volume defined by a rectangular column minus a segment of a sphere) 3 π R 2 The volume is V H Then FV SG ρ g V 4 4 π R 3 8 V 621.8 ft 3 FV 1.025 1.94 slug ft 3 32.2 ft 2 s 3 621.8 ft 2 lbf s slug ft 4 FV 3.98 10 lbf Finally the resultant force and direction can be computed F 2 FH FV FV FH α atan 2 4 F 7.07 10 lbf α 34.3 deg Note that α is the angle the resultant force makes with the horizontal Problem *3.89 [Difficulty: 2] Given: Hydrometer as shown, submerged in nitric acid. When submerged in water, h = 0 and the immersed volume is 15 cubic cm. SG 1.5 d 6 mm Find: The distance h when immersed in nitric acid. Solution: We will apply the hydrostatics equations to this system. Fbuoy ρ g Vd Governing Equations: Assumptions: (1) Static fluid (2) Incompressible fluid ΣFz 0 M g Fbuoy 0 Taking a free body diagram of the hydrometer: Solving for the mass of the hydrometer: When immersed in water: M ρw V w M Fbuoy g ρ V d When immersed in nitric acid: Since the mass of the hydrometer is the same in both cases: π 2 When the hydrometer is in the nitric acid: Vn Vw d h 4 π 2 Therefore: ρw Vw SG ρw Vw d h 4 Vw SG Vw π 2 d h 4 SG 1 4 h Vw SG π d2 (Buoyant force is equal to weight of displaced fluid) ρ w V w ρ n V n ρn SG ρw Solving for the height h: Vw ( 1 SG) SG 3 M ρ n V n π 2 d h 4 4 1.5 1 10 mm 2 1.5 π ( 6 mm) cm h 15 cm 3 h 177 mm Problem *3.90 [Difficulty: 3] Given: Data on sphere and weight Find: SG of sphere; equilibrium position when freely floating T Solution: Basic equation FB F B ρ g V where Hence T M g M g ρ g V 2 γ Weight Volume ΣFz 0 T FB W M 10 kg SG ρ g V 0 SG 10 kg The specific weight is ΣFz 0 and m 3 1000 kg SG 1 0.025 m SG ρ g V V FB ρ g 3 M ρ V V W 1 2 1 SG 0.9 2 SG ρ g W SG ρ g V 2 γ 0.9 1000 kg 3 9.81 m 2 m 2 s N s kg m γ 8829 W FB with FB ρ g Vsubmerged Vsubmerged From references (trying Googling "partial sphere volume") 3 V 4 π R where h is submerged depth and R is the sphere radius π h 2 3 1 3 3 0.025 m3 4 π 2 Hence π h W SG ρ g V FB ρ g ( 3 R h) 3 2 h ( 3 0.181 m h) 3 3 0.9 .025 m π ( 3 R h) 1 3 R 2 h ( 3 R h) R 0.181 m 3 SG V π 2 h ( 0.544 h) 0.0215 This is a cubic equation for h. We can keep guessing h values, manually iterate, or use Excel's Goal Seek to find 3 m For the equilibriul position when floating, we repeat the force balance with T = 0 FB W 0 N h 0.292 m Problem *3.91 [Difficulty: 2] Given: Specific gravity of a person is to be determined from measurements of weight in air and the met weight when totally immersed in water. Find: Expression for the specific gravity of a person from the measurements. Solution: We will apply the hydrostatics equations to this system. Fbuoy ρ g Vd Governing Equation: Assumptions: (Buoyant force is equal to weight of displaced fluid) (1) Static fluid (2) Incompressible fluid Fnet Taking a free body diagram of the body: Fnet ΣFy 0 Fnet M g Fbuoy 0 Fbuoy is the weight measurement for the immersed body. Fnet M g Fbuoy M g ρw g Vd Therefore the weight measured in water is: However in air: Fair M g Fnet Fair ρw g Vd and Vd Fair Fnet Mg ρw g Now in order to find the specific gravity of the person, we need his/her density: Fair M g ρ g Vd ρ g Fair Fnet ρw g Now if we call the density of water at 4 deg C ρ Simplifying this expression we get: Fair F Fnet ρw air ρw4C then: ρ ρ SG w4C F F Fair air net SG Fair Fnet ρ w w ρw4C Solving this expression for the specific gravity of the person SG, we get: SG SGw F Fair air Fnet Problem *3.92 [Difficulty: 2] Given: Iceberg floating in seawater Find: Quantify the statement, "Only the tip of an iceberg shows (in seawater)." Solution: We will apply the hydrostatics equations to this system. Fbuoy ρ g Vd Governing Equations: Assumptions: (Buoyant force is equal to weight of displaced fluid) (1) Static fluid (2) Incompressible fluid Taking a free body diagram of the iceberg: M g Fbuoy ρsw g Vd Combining these expressions: ΣFz 0 M g Fbuoy 0 M ρice Vtot But the mass of the iceberg is also: ρice Vtot g ρsw g Vd The volume of the iceberg above the water is: Therefore we may define a volume fraction: Mg ρice SGice Fbuoy Vd Vtot Vtot ρsw SGsw SGice Vshow Vtot Vd Vtot 1 SGsw VF Vshow Vtot 1 Substituting in data from Tables A.1 and A.2 we get: VF 1 SGice SGsw 0.917 1.025 VF 0.1054 Only 10% of the iceberg is above water Problem *3.93 [Difficulty: 2] Given: Geometry of steel cylinder Find: Volume of water displaced; number of 1 kg wts to make it sink Solution: The data is For water ρ 999 kg 3 m For steel (Table A.1) SG 7.83 For the cylinder D 100 mm The volume of the cylinder is Vsteel δ The weight of the cylinder is W SG ρ g Vsteel H 1 m π D 2 π D H 4 Vsteel 3.22 10 kg W 7.83 999 3 9.81 m δ 1 mm m 2 3.22 10 4 3 m s 4 3 m 2 N s W 24.7 N kg m At equilibium, the weight of fluid displaced is equal to the weight of the cylinder Wdisplaced ρ g Vdisplaced W Vdisplaced W ρ g 3 24.7 N m 999 kg 2 s 9.81 m kg m Vdisplaced 2.52 L 2 N s To determine how many 1 kg wts will make it sink, we first need to find the extra volume that will need to be dsiplaced Distance cylinder sank x1 Vdisplaced x1 0.321 m π D 2 4 x2 H x1 Hence, the cylinder must be made to sink an additional distance We deed to add n weights so that 1 kg n g ρ g π D 4 x2 0.679 m 2 x2 2 n ρ π D x2 4 1 kg Hence we need n 6 weights to sink the cylinder 999 kg 3 m π 4 2 ( 0.1 m) 0.679 m 1 1 kg 2 N s kg m n 5.33 Problem *3.94 [Difficulty: 2] Given: Experiment performed by Archimedes to identify the material conent of King Hiero's crown. The crown was weighed in air and in water. Find: Expression for the specific gravity of the crown as a function of the weights in water and air. Solution: We will apply the hydrostatics equations to this system. F b ρ g V d (Buoyant force is equal to weight of displaced fluid) (1) Static fluid (2) Incompressible fluid Ww Governing Equations: Assumptions: ΣFz 0 Taking a free body diagram of the body: Ww M g Fb 0 Ww is the weight of the crown in water. Mg Ww M g Fbuoy M g ρw g Vd However in air: Therefore the weight measured in water is: so the volume is: Vd Wa Ww ρw g Wa M g Fb Ww Wa ρw g Vd M ρw g Wa M Now the density of the crown is: ρc ρ Vd Wa Ww Wa Ww w Therefore, the specific gravity of the crown is: SG ρc ρw Wa Wa Ww SG Wa Wa Ww Note: by definition specific gravity is the density of an object divided by the density of water at 4 degrees Celsius, so the measured temperature of the water in the experiment and the data from tables A.7 or A.8 may be used to correct for the variation in density of the water with temperature. Problem *3.95 [Difficulty: 2] Open-Ended Problem Statement: Gas bubbles are released from the regulator of a submerged Scuba diver. What happens to the bubbles as they rise through the seawater? Discussion: Air bubbles released by a submerged diver should be close to ambient pressure at the depth where the diver is swimming. The bubbles are small compared to the depth of submersion, so each bubble is exposed to essentially constant pressure. Therefore the released bubbles are nearly spherical in shape. The air bubbles are buoyant in water, so they begin to rise toward the surface. The bubbles are quite light, so they reach terminal speed quickly. At low speeds the spherical shape should be maintained. At higher speeds the bubble shape may be distorted. As the bubbles rise through the water toward the surface, the hydrostatic pressure decreases. Therefore the bubbles expand as they rise. As the bubbles grow larger, one would expect the tendency for distorted bubble shape to be exaggerated. Problem *3.96 [Difficulty: 2] Given: Balloons with hot air, helium and hydrogen. Claim lift per cubic foot of 0.018, 0.066, and 0.071 pounds force per cubic f for respective gases, with the air heated to 150 deg. F over ambient. Find: (a) evaluate the claims of lift per unit volume (b) determine change in lift when air is heated to 250 deg. F over ambient. Solution: We will apply the hydrostatics equations to this system. Governing Equations: Assumptions: L ρa g V ρg g V (Net lift force is equal to difference in weights of air and gas) p ρ R T (Ideal gas equation of state) (1) Static fluid (2) Incompressible fluid (3) Ideal gas behavior The lift per unit volume may be written as: LV ρg g ρa ρg ρa g 1 V ρa L we take into account that the pressure inside and outside the balloon are equal: lbf At standard conditions the specific weight of air is: γa 0.0765 3 ft Rg 386.1 For helium: For hydrogen: ft lbf lbm R Rg 766.5 Tg Ta ft lbf Tg Ta lbm R and therefore: now if we take the ideal gas equation and Ra Ta Ra Ta L ρa g 1 γa 1 V R g T g R g T g the gas constant is: Ra 53.33 ft lbf lbm R lbf 53.33 LVHe 0.0765 1 3 386.1 ft and therefore: and Ta 519 R lbf LVHe 0.0659 3 ft lbf 53.33 LVH2 0.0765 1 3 766.5 ft lbf LVH2 0.0712 3 ft For hot air at 150 degrees above ambient: Rg Ra Tg Ta 150 R and therefore: lbf 519 lbf LVair150 0.0765 1 LVair150 0.0172 3 3 519 150 ft ft The agreement with the claims stated above is good. For hot air at 250 degrees above ambient: Rg Ra LVair250 LVair150 Tg Ta 250 R and therefore: 1.450 lbf 519 LVair250 0.0765 1 3 519 250 ft lbf LVair250 0.0249 3 ft Air at ΔT of 250 deg. F gives 45% more lift than air at ΔT of 150 deg.F! Problem *3.97 [Difficulty: 2] V FB y FD W = Mg Given: Data on hydrogen bubbles Find: Buoyancy force on bubble; terminal speed in water Solution: Basic equation F B ρ g V ρ g FB 1.94 slug ft 3 π 3 d 6 32.2 FB FD W 0 Hence V 2 π 6 2 1 ft lbf s 12 in slug ft 11 FB 1.89 10 5 lbf s 2 from Table A.7 at 68oF ft V 1.89 10 11 lbf 3 ft s 1 3 π 1 2.10 10 V 0.825 ft 2 5 lbf s for terminal speed lbf where we have ignored W, the weight of the bubble (at STP most gases are about 1/1000 the density of water) μ 2.10 10 with 3 π μ d ΣFy 0 FB FD W 3 0.001 in F D 3 π μ V d F B FB V 1.15 10 ft s For terminal speed ΣFy M ay and 1 0.001 in in min As noted by Professor Kline in the film "Flow Visualization", bubbles rise slowly! 12 in 1 ft Problem *3.98 [Difficulty: 3] Given: Data on hot air balloon Find: Maximum mass of balloon for neutral buoyancy; mass for initial acceleration of 2.5 ft/s2. Assumptions: Fbuoyancy Whot air Air is treated as static and incompressible, and an ideal gas Solution: y Basic equation Hence FB ρatm g V ΣFy M ay and Wload ΣFy 0 FB Whotair Wload ρatm g V ρhotair g V M g M V ρatm ρhotair 3 M 320000 ft 14.7 lbf 2 in V patm R 1 Tatm 1 for neutral buoyancy Thotair 2 1 1 12 in lbm R 53.33 ft lbf ( 160 460 ) R ft ( 48 460 ) R M 4517 lbm Initial acceleration ΣFy FB Whotair Wload ρatm ρhotair g V Mnew g Maccel a Mnew 2 ρhotair V a Solving for Mnew ρatm ρhotair g V Mnew g Mnew 2 ρhotair V a Mnew V ρatm ρhotair g 2 ρhotair a V patm g 1 1 2 a ag Tatm Thotair Thotair ag 2 2 lbf 12 in lbm R s 1 1 1 3 ft Mnew 320000 ft 14.7 32.2 2 2.5 2 2 ft 53.33 ft lbf ( 2.5 32.2) ft ( 160 460) ( 48 460) ( 160 460) in s R Mnew 1239 lbm To make the balloon move up or down during flight, the air needs to be heated to a higher temperature, or let cool (or let in ambient air). Problem *3.99 [Difficulty: 4] Given: Spherical balloon filled with helium lifted a payload of mass M=230 kg. At altitude, helium and air were in thermal equilibrium. Balloon diameter is 120 m and specific gravity of the skin material is 1.28. Find: The altitude to which the balloon rose. Solution: We will apply the hydrostatics equations to this system. Governing Equations: Assumptions: Fbuoy ρ g Vd (Buoyant force is equal to mass of displaced fluid) p ρ R T (Ideal gas equation of state) (1) Static, incompressible fluid (2) Static equilibrium at 49 km altitude (3) Ideal gas behavior D t Taking a free body diagram of the balloon and payload: ΣFz Fbuoy MHe g Ms g M g 0 z M Substituting for the buoyant force and knowing that mass is density times volume: ρair g Vb ρHe g Vb ρs g Vs M g 0 The volume of the balloon: ρair ρHe p T 6 π D p 6 T π 3 6 π D 3 π Vb 6 2 π ρs t D M 2 1 ( 120 m) 3 1 kg 3 m 2 V s π D t Substituting these into the force equation: From the ideal gas equation of state and remembering that pressure and temperature of the air and helium are equal: Substituting known values and consulting Appendix A for gas constants: 1 1 R R He air π 1280 3 D The volume of the skin: π ρs t D M ρair Vb ρHe Vb ρs Vs M 0 0.013 10 m ( 120 m) 230 kg 3 2 1 2 1 287 1 N m Pa m 4 kPa 3.616 10 kg K N K 2080 To determine the altitude, we need to check this ratio against data from Table A.3. We find that the ratio of pressure to temperature matches the result above at: h 48.3 km Problem *3.100 [Difficulty: 3] Given: A pressurized balloon is to be designed to lift a payload of mass M to an altitude of 40 km, where p = 3.0 mbar and T = -25 deg C. The balloon skin has a specific gravity of 1.28 and thickness 0.015 mm. The gage pressure of the helium is 0.45 mbar. The allowable tensile stress in the balloon is 62 MN/m2 Find: (a) The maximum balloon diameter (b) The maximum payload mass Solution: We will apply the hydrostatics equations to this system. Governing Equations: Assumptions: t D Fbuoy = ρ⋅ g ⋅ Vd (Buoyant force is equal to mass of displaced fluid) p = ρ⋅ R⋅ T (Ideal gas equation of state) M (1) Static, incompressible fluid (2) Static equilibrium at 40 km altitude (3) Ideal gas behavior πD tσ The diameter of the balloon is limited by the allowable tensile stress in the skin: ΣF = π 4 2 ⋅ D ⋅ ∆p − π⋅ D⋅ t⋅ σ = 0 −3 Dmax = 4 × 0.015 × 10 4⋅ t⋅ σ Dmax = ∆p Solving this expression for the diameter: 6 N ⋅ m × 62 × 10 ⋅ 2 m 2 1 × −3 0.45⋅ 10 πD 2∆p/4 × ⋅ bar bar ⋅ m Fbuoyant Dmax = 82.7m 5 10 ⋅ N z To find the maximum allowable payload we perform a force balance on the system: ΣFz = Fbuoy − M He⋅ g − M b ⋅ g − M ⋅ g = 0 Solving for M: ( The air density: ρa ⋅ g ⋅ Vb − ρHe⋅ g ⋅ Vb − ρs ⋅ g ⋅ Vs − M ⋅ g = 0 Mg ) M = ρa − ρHe ⋅ Vb − ρs ⋅ Vs The volume of the skin is: 2 Vs = π⋅ D ⋅ t pa ρa = Ra⋅ T Repeating for helium: The payload mass is: M = 6 ⋅ bar × 287⋅ N ⋅ m × π 6 1 ( 273 − 25) ⋅ K ( ) 3 2 ⋅ ρa − ρHe ⋅ D − π⋅ ρs⋅ D ⋅ t 5 × 10 ⋅ N 2 bar ⋅ m − 3 kg ρa = 4.215 × 10 3 m − 4 kg ρHe = 6.688 × 10 3 m − 3 kg × ( 4.215 − 0.6688) × 10 M = 638 kg kg⋅ K 3 Vb = ⋅ D 6 M= Therefore, the mass is: −3 π π The volume of the balloon is: ρa = 3.0 × 10 p ρHe = R⋅ T M bg ⋅ 3 m 3 3 kg × ( 82.7⋅ m) − π × 1.28 × 10 ⋅ 3 m 2 −3 × ( 82.7⋅ m) × 0.015 × 10 ⋅m Problem *3.101 [Difficulty: 3] Given: Geometry of block and rod F hinge,y (L + c)/2 Find: Angle for equilibrium Fhinge,x L/2 c θ Assumptions: Water is static and incompressible a F BR Solution: F BB ΣM Hinge = 0 Basic equations FB = ρ⋅ g ⋅ V WR (Buoyancy) L WB The free body diagram is as shown. FBB and F BR are the buoyancy of the block and rod, respectively; c is the (unknown) exposed length of the rod Taking moments about the hinge ( L + c) (WB − FBB)⋅ L⋅ cos( θ) − FBR⋅ with WB = M B⋅ g FBB = ρ⋅ g ⋅ VB ( MB = ( 2⋅ L 1 2 2 ⋅ L −c × 1000⋅ M B = 29.1 kg 2 ) + ρ⋅VB − 12 ⋅MR kg 3 m 2 ρ⋅ A⋅ L − c We can solve for M B ρ⋅ A FBR = ρ⋅ g ⋅ ( L − c) ⋅ A (MB − ρ⋅ VB)⋅ L − ρ⋅ A⋅ ( L − c)⋅ Combining equations MB = 2 L ⋅ cos( θ) + WR⋅ ⋅ cos( θ) = 0 2 2 × 20⋅ cm × 2 ( L + c) 2 WR = M R⋅ g L + MR⋅ = 0 2 ) = 2⋅⎛⎜ MB − ρ⋅VB + 12 ⋅MR⎞ ⋅L ⎝ and since ⎠ c= 2 ⎛ m ⎞ × 1 ⋅ ⎡⎢( 5⋅ m) 2 − ⎜ 5⋅ m ⎣ ⎝ 100 ⋅ cm ⎠ a sin( θ) MB = ρ⋅ A 2⋅ L ⎡ ⋅ ⎢L − ⎣ 2 2 ⎛ a ⎞ ⎤⎥ + ρ⋅ V − 1 ⋅ M ⎜ sin( θ) B 2 R ⎝ ⎠⎦ 2 ⎛ 0.25⋅ m ⎞ ⎥⎤ + 1000⋅ kg × 0.025 ⋅ m3 − 1 × 1.25⋅ kg ⎜ 2 3 ⎝ sin( 12⋅ deg) ⎠ ⎦ m Problem *3.102 [Difficulty: 3] Given: Glass hydrometer used to measure SG of liquids. Stem has diameter D=5 mm, distance between marks on stem is d=2 mm per 0.1 SG. Hydrometer floats in kerosene (Assume zero contact angle between glass and kerosene). Find: Magnitude of error introduced by surface tension. Solution: We will apply the hydrostatics equations to this system. Fbuoy = ρ⋅ g ⋅ Vd Governing Equations: Assumptions: (Buoyant force is equal to weight of displaced fluid) D = 5 mm (1) Static fluid (2) Incompressible fluid (3) Zero contact angle between ethyl alcohol and glass The surface tension will cause the hydrometer to sink ∆h lower into the liquid. Thus for this change: ΣFz = ∆Fbuoy − Fσ = 0 ∆Fbuoy = ρ⋅ g ⋅ ∆V = ρ⋅ g ⋅ The change in buoyant force is: ρ⋅ g ⋅ π 4 Solving for ∆h: 2 ⋅ D ⋅ ∆h = π⋅ D⋅ σ ∆h = 4 y Fσ Kerosene 2 ⋅ D ⋅ ∆h ∆F B Fσ = π⋅ D⋅ σ⋅ cos( θ) = π⋅ D⋅ σ The force due to surface tension is: Thus, π d= 2 mm/0.1 SG ρ⋅ g ⋅ D⋅ ∆h Upon simplification: 4 4⋅ σ =σ From Table A.2, SG = 1.43 and from Table A.4, σ = 26.8 mN/m ρ⋅ g ⋅ D −3 N Therefore, ∆h = 4 × 26.8 × 10 ⋅ m 3 × m 1430⋅ kg 2 × s 9.81⋅ m So the change in specific gravity will be: ∆SG = 1.53 × 10 1 × −3 −3 5 × 10 ⋅m × × ⋅m kg⋅ m 2 −3 m s ⋅N 0.1 ∆SG = 0.0765 −3 2 × 10 ∆h = 1.53 × 10 ⋅m From the diagram, surface tension acts to cause the hydrometer to float lower in the liquid. Therefore, surface tension results in an indicated specific gravity smaller than the actual specific gravity. Problem *3.103 [Difficulty:4] Given: Sphere partially immersed in a liquid of specific gravity SG. Find: (a) Formula for buoyancy force as a function of the submersion depth d (b) Plot of results over range of liquid depth Solution: We will apply the hydrostatics equations to this system. Fbuoy = ρ⋅ g ⋅ Vd Governing Equations: Assumptions: (Buoyant force is equal to weight of displaced fluid) (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts everywhere d We need an expression for the displaced volume of fluid at an arbitrary depth d. From the diagram we see that: ( ( )) d = R 1 − cos θmax at an arbitrary depth h: h = d − R⋅ ( 1 − cos( θ) ) R dθ r = R⋅ sin( θ) Rsin θ So if we want to find the volume of the submerged portion of the sphere we calculate: θmax θ θ ⌠ max 2 ⌠ max 2 3 2 3⌠ ( sin( θ) ) dθ R ⋅ ( sin( θ) ) ⋅ R⋅ sin( θ) dθ = π⋅ R ⋅ ⎮ Vd = ⎮ π r dh = π⋅ ⎮ ⌡ ⌡ ⌡ Evaluating the integral we get: 0 0 0 h θmax ⎡⎢ cos θ ( ( max))3 Vd = π⋅ R ⋅ ⎢ − cos( θmax) + 3 ⎣ 3 ⎤ 3 ⎡ d d⎞ 2⎤ d 3 1⎛ we get: Now since: ⎢ cos θ = 1 − V π ⋅ R 1 − = ⋅ − ⎛⎜ 1 − ⎞ + ⎥ ( ) ⎜ ⎥ max d 3⎦ R R⎠ R ⎠ 3⎦ ⎣3 ⎝ ⎝ 2⎥ ⎡ d 2⎤ d 3 1 Fbuoy = ρw⋅ SG⋅ g ⋅ π⋅ R ⋅ ⎢ ⋅ ⎛⎜ 1 − ⎞ − ⎜⎛ 1 − ⎞ + ⎥ R⎠ R ⎠ 3⎦ ⎣3 ⎝ ⎝ 3 Thus the buoyant force is: If we non-dimensionalize by the force on a fully submerged sphere: Fd = Fbuoy 4 3 ρw⋅ SG⋅ g ⋅ ⋅ π⋅ R 3 = 3 ⎡1 3 ⎡1 d 2⎤ d Fd = ⎢ ⋅ ⎛⎜ 1 − ⎞ − ⎜⎛ 1 − ⎞ + ⎥ 4 ⎣3 ⎝ R⎠ R ⎠ 3⎦ ⎝ 3 ⎤ d ⎢ ⋅ ⎛⎜ 1 − ⎞ − ⎜⎛ 1 − d ⎞ + 2⎥ 4 ⎣3 ⎝ R⎠ R ⎠ 3⎦ ⎝ 3 Force Ratio Fd 1.0 0.5 0.0 0.0 0.5 1.0 Submergence Ratio d/R 1.5 2.0 Problem 3.104 [Difficulty: 2] Given: Geometry of rod Find: (L + c)/2 How much of rod is submerged; force to lift rod out of water L/2 c Solution: Basic equations θ ΣM Hinge = 0 FB = ρ⋅ g ⋅ V (Buoyancy) FBR WR The free body diagram is as shown. FBR is the buoyancy of the rod; c is the (unknown) exposed length of the rod L Taking moments about the hinge −FBR⋅ ( L + c) 2 L ⋅ cos( θ) + WR⋅ ⋅ cos( θ) = 0 2 with FBR = ρ⋅ g ⋅ ( L − c) ⋅ A Hence −ρ⋅ A⋅ ( L − c) ⋅ We can solve for c ( 2 2 2 L⋅ M R L − ρ⋅ A 3 2 c = L + M R⋅ = 0 2 ) = MR⋅ L 2 ρ⋅ A⋅ L − c c= ( L + c) WR = M R⋅ g ( 5 ⋅ m) − 5 ⋅ m × m 1000⋅ kg × 1 ⋅ 20 2 1 cm 2 ⎛ 100 ⋅ cm ⎞ × 1.25⋅ kg ⎜ 1⋅ m ⎝ ⎠ × c = 4.68 m Then the submerged length is L − c = 0.323 m To lift the rod out of the water requires a force equal to half the rod weight (the reaction also takes half the weight) F= 1 2 ⋅ MR⋅ g = 1 2 × 1.25⋅ kg × 9.81⋅ m 2 s 2 × N⋅ s kg⋅ m F = 6.1 N a Problem *3.105 [Difficulty: 2] FB y x H = 60 cm W θ h = 5 cm Given: Data on river Find: Largest diameter of log that will be transported Solution: Basic equation FB = ρ⋅ g ⋅ Vsub ΣFy = 0 where FB = ρ⋅ g ⋅ Vsub = ρ⋅ g ⋅ Asub ⋅ L and ΣFy = 0 = FB − W W = SG⋅ ρ⋅ g ⋅ V = SG⋅ ρ⋅ g ⋅ A ⋅ L 2 R Asub = ⋅ ( θ − sin ( θ) ) 2 From references (e.g. CRC Mathematics Handbook) Hence ρ⋅ g ⋅ R 2 2 where R is the radius and θ is the included angle 2 ⋅ ( θ − sin ( θ) ) ⋅ L = SG⋅ ρ⋅ g ⋅ π⋅ R ⋅ L θ − sin( θ) = 2 ⋅ SG ⋅ π = 2 × 0.8 × π This equation can be solved by manually iterating, or by using a good calculator, or by using Excel's Goal Seek θ = 239 ⋅ deg From geometry the submerged amount of a log is H − h Hence H − h = R + R⋅ cos⎛⎜ π − Solving for R R= ⎝ R + R⋅ cos⎛⎜ π − ⎝ θ⎞ 2⎠ θ⎞ 2⎠ H− h θ⎞ 1 + cos⎛⎜ 180deg − 2⎠ ⎝ D = 2⋅ R and also R = ( 0.6 − 0.05) ⋅ m 239 ⎞ ⋅ deg⎥⎤ 1 + cos⎡⎢⎜⎛ 180 − 2 ⎠ ⎣⎝ ⎦ D = 0.737 m R = 0.369 m Problem *3.106 [Difficulty: 4] Given: Data on sphere and tank bottom Find: Expression for SG of sphere at which it will float to surface; minimum SG to remain in position y FU FB x Assumptions: (1) Water is static and incompressible (2) Sphere is much larger than the hole at the bottom of the tank Solution: FL FB = ρ⋅ g ⋅ V Basic equations and FL = p atm⋅ π⋅ a where ΣFy = FL − FU + FB − W 2 FU = ⎡p atm + ρ⋅ g ⋅ ( H − 2 ⋅ R)⎤ ⋅ π⋅ a ⎣ ⎦ 2 4 3 2 Vnet = ⋅ π⋅ R − π⋅ a ⋅ 2 ⋅ R 3 FB = ρ⋅ g ⋅ Vnet W = SG ⋅ ρ⋅ g ⋅ V W V= with 4 3 ⋅ π⋅ R 3 Now if the sum of the vertical forces is positive, the sphere will float away, while if the sum is zero or negative the sphere will stay at the bottom of the tank (its weight and the hydrostatic force are greater than the buoyant force). Hence 4 4 2 2 3 2 3 ΣFy = p atm⋅ π⋅ a − ⎡p atm + ρ⋅ g ⋅ ( H − 2 ⋅ R)⎤ ⋅ π⋅ a + ρ⋅ g ⋅ ⎜⎛ ⋅ π⋅ R − 2 ⋅ π⋅ R⋅ a ⎞ − SG⋅ ρ⋅ g ⋅ ⋅ π⋅ R ⎣ ⎦ 3 3 ⎝ 4 3 2 ΣFy = π⋅ ρ⋅ g ⋅ ⎢⎡( 1 − SG ) ⋅ ⋅ R − H⋅ a ⎥⎤ 3 ⎣ ⎦ This expression simplifies to ΣFy = π × 1.94⋅ slug ft ΣFy = −0.012 ⋅ lbf ⎠ 3 × 32.2⋅ ft 2 s × 2 3 2⎤ ⎡4 ⎢ × ( 1 − 0.95) × ⎛⎜ 1⋅ in × ft ⎞ − 2.5⋅ ft × ⎛⎜ 0.075 ⋅ in × ft ⎞ ⎥ × lbf ⋅ s 12⋅ in ⎠ 12⋅ in ⎠ ⎦ slug⋅ ft ⎣3 ⎝ ⎝ Therefore, the sphere stays at the bottom of the tank. Problem *3.107 [Difficulty: 4] Given: Cylindrical timber, D = 1 ft and L = 15 ft, is weighted on the lower end so that is floats vertically with 10 ft submerged in sea water. When displaced vertically from equilibrium, the timber oscillates in a vertical direction upon release. Find: Estimate the frequency of the oscillation. Neglect viscous forces or water motion. Solution: We will apply the hydrostatics equations to this system. Governing Equations: Assumptions: Fbuoy ρ g Vd (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts everywhere (4) Viscous effects and water motion are negligible. ΣFy Fbuoy M g 0 At equilibrium: (Buoyant force is equal to weight of displaced fluid) D = 1 ft M ρ Vd ρ A d 2 ΣFy Fbuoy M g M Once the timber is displaced: d y dt 2 2 ρ g A ( d y ) M g M d y 2 2 ρ g A d ρ g A y ρ A d g M dt d y dt 2 Thus we have the equation: M 2 2 d y dt ρ g A y 0 2 or: d y 2 dt ρ g A ρ A d ω g d To express this as a frequency: ω 32.2 ft s f ω 2 π 2 2 y 0 d y 2 dt 2 This ODE describes simple harmonic motion with the natural frequency ω described by: Solving for ω: d =10 ft (Equilibrium Depth) L ω g d y 0 g d 1 ω 1.7944 10 ft 1.7944 f 2 π rad 1 s f 0.286 Hz s Problem *3.108 [Difficulty: 3] Given: Data on boat Find: Effective density of water/air bubble mix if boat sinks Floating H = 8 ft Solution: Basic equations Sinking h = 7 ft FB = ρ⋅ g ⋅ V ΣFy = 0 and θ = 60 o We can apply the sum of forces for the "floating" free body ΣFy = 0 = FB − W FB = SGsea⋅ ρ⋅ g ⋅ Vsubfloat where 2 1 2⋅ h ⎞ L⋅ h Vsubfloat = ⋅ h ⋅ ⎛⎜ ⋅L = 2 ⎝ tan⋅ θ ⎠ tan( θ) Hence W= SGsea⋅ ρ⋅ g ⋅ L⋅ h SGsea = 1.024 (Table A.2) 2 (1) tan( θ) We can apply the sum of forces for the "sinking" free body 2 ΣFy = 0 = FB − W 1 2⋅ H ⎞ L⋅ H Vsubsink = ⋅ H⋅ ⎛⎜ ⋅L = 2 ⎝ tan⋅ θ ⎠ tan( θ) FB = SGmix⋅ ρ⋅ g ⋅ Vsub where 2 Hence W= Comparing Eqs. 1 and 2 SGmix⋅ ρ⋅ g ⋅ L⋅ H (2) tan( θ) W= SGsea⋅ ρ⋅ g ⋅ L⋅ h 2 tan( θ) h SGmix = SGsea ⋅ ⎛⎜ ⎞ H ⎝ ⎠ The density is ρmix = SGmix⋅ ρ 2 = SGmix⋅ ρ⋅ g ⋅ L⋅ H tan( θ) 2 SGmix = 1.024 × ⎛7⎞ ⎜8 ⎝ ⎠ ρmix = 0.784 × 1.94⋅ 2 SGmix = 0.784 slug ft 3 ρmix = 1.52⋅ slug ft 3 Problem *3.109 [Difficulty: 2] FB F 20 cm D = 10 cm 8 cm 2 cm W y x Given: Data on inverted bowl and dense fluid Find: Force to hold in place Assumption: Fluid is static and incompressible Solution: Basic equations FB = ρ⋅ g ⋅ V Hence F = FB − W For the buoyancy force FB = SG fluid ⋅ ρH2O⋅ g ⋅ Vsub For the weight W = SG bowl⋅ ρH2O⋅ g ⋅ Vbowl Hence ( and ΣFy = 0 ΣFy = 0 = FB − F − W with Vsub = Vbowl + Vair ) F = SGfluid⋅ ρH2O⋅ g ⋅ Vbowl + Vair − SG bowl⋅ ρH2O⋅ g ⋅ Vbowl ( ) F = ρH2O⋅ g ⋅ ⎡SG fluid⋅ Vbowl + Vair − SGbowl⋅ Vbowl⎤ ⎣ ⎦ 3 2 3 ⎞⎤ 2 ⎡ ⎛ ⎡ m π⋅ ( 0.1⋅ m) ⎤ ⎥ − 5.7 × ⎜ 0.9⋅ L × m ⎥ × N ⋅ s F = 999⋅ × 9.81⋅ + ( 0.08 − 0.02) ⋅ m⋅ × ⎢15.6 × ⎢0.9⋅ L × 3 2 ⎣ 1000⋅ L 4 1000⋅ L ⎠⎦ kg⋅ m ⎣ ⎦ ⎝ m s kg F = 159.4 N m Problem *3.110 [Difficulty: 4] Open-Ended Problem Statement: In the “Cartesian diver” child's toy, a miniature “diver” is immersed in a column of liquid. When a diaphragm at the top of the column is pushed down, the diver sinks to the bottom. When the diaphragm is released, the diver again rises. Explain how the toy might work. Discussion: A possible scenario is for the toy to have a flexible bladder that contains air. Pushing down on the diaphragm at the top of the liquid column would increase the pressure at any point in the liquid. The air in the bladder would be compressed slightly as a result. The volume of the bladder, and therefore its buoyancy, would decrease, causing the diver to sink to the bottom of the liquid column. Releasing the diaphragm would reduce the pressure in the water column. This would allow the bladder to expand again, increasing its volume and therefore the buoyancy of the diver. The increased buoyancy would permit the diver to rise to the top of the liquid column and float in a stable, partially submerged position, on the surface of the liquid. Problem *3.111 [Difficulty: 4] Open-Ended Problem Statement: Consider a conical funnel held upside down and submerged slowly in a container of water. Discuss the force needed to submerge the funnel if the spout is open to the atmosphere. Compare with the force needed to submerge the funnel when the spout opening is blocked by a rubber stopper. Discussion: Let the weight of the funnel in air be Wa. Assume the funnel is held with its spout vertical and the conical section down. Then Wa will also be vertical. Two possible cases are with the funnel spout open to atmosphere or with the funnel spout sealed. With the funnel spout open to atmosphere, the pressures inside and outside the funnel are equal, so no net pressure force acts on the funnel. The force needed to support the funnel will remain constant until it first contacts the water. Then a buoyancy force will act vertically upward on every element of volume located beneath the water surface. The first contact of the funnel with the water will be at the widest part of the conical section. The buoyancy force will be caused by the volume formed by the funnel thickness and diameter as it begins to enter the water. The buoyancy force will reduce the force needed to support the funnel. The buoyancy force will increase as the depth of submergence of the funnel increases until the funnel is fully submerged. At that point the buoyancy force will be constant and equal to the weight of water displaced by the volume of the material from which the funnel is made. If the funnel material is less dense than water, it would tend to float partially submerged in the water. The force needed to support the funnel would decrease to zero and then become negative (i.e., down) to fully submerge the funnel. If the funnel material were denser than water it would not tend to float even when fully submerged. The force needed to support the funnel would decrease to a minimum when the funnel became fully submerged, and then would remain constant at deeper submersion depths. With the funnel spout sealed, air will be trapped inside the funnel. As the funnel is submerged gradually below the water surface, it will displace a volume equal to the volume of the funnel material plus the volume of trapped air. Thus its buoyancy force will be much larger than when the spout is open to atmosphere. Neglecting any change in air volume (pressures caused by submersion should be small compared to atmospheric pressure) the buoyancy force would be from the entire volume encompassed by the outside of the funnel. Finally, when fully submerged, the volume of the rubber stopper (although small) will also contribute to the total buoyancy force acting on the funnel. Problem *3.112 [Difficulty: 2] Given: Steel balls resting in floating plastic shell in a bucket of water Find: What happens to water level when balls are dropped in water Solution: Basic equation FB ρ Vdisp g W for a floating body weight W When the balls are in the plastic shell, the shell and balls displace a volume of water equal to their own weight - a large volume because the balls are dense. When the balls are removed from the shell and dropped in the water, the shell now displaces only a small volume of water, and the balls sink, displacing only their own volume. Hence the difference in displaced water before and after moving the balls is the difference between the volume of water that is equal to the weight of the balls, and the volume of the balls themselves. The amount of water displaced is significantly reduced, so the water level in the bucket drops. Volume displaced before moving balls: V1 Wplastic Wballs ρ g Wplastic Volume displaced after moving balls: V2 Change in volume displaced ∆V V2 V1 Vballs ρ g Vballs ∆V Vballs 1 SG balls Wballs ρ g Vballs SGballs ρ g Vballs ρ g Hence initially a large volume is displaced; finally a small volume is displaced (ΔV < 0 because SGballs > 1) Problem *3.113 [Difficulty: 4] Open-Ended Problem Statement: A proposed ocean salvage scheme involves pumping air into “bags” placed within and around a wrecked vessel on the sea bottom. Comment on the practicality of this plan, supporting your conclusions with analyses. Discussion: This plan has several problems that render it impractical. First, pressures at the sea bottom are very high. For example, Titanic was found in about 12,000 ft of seawater. The corresponding pressure is nearly 6,000 psi. Compressing air to this pressure is possible, but would require a multi-stage compressor and very high power. Second, it would be necessary to manage the buoyancy force after the bag and object are broken loose from the sea bed and begin to rise toward the surface. Ambient pressure would decrease as the bag and artifact rise toward the surface. The air would tend to expand as the pressure decreases, thereby tending to increase the volume of the bag. The buoyancy force acting on the bag is directly proportional to the bag volume, so it would increase as the assembly rises. The bag and artifact thus would tend to accelerate as they approach the sea surface. The assembly could broach the water surface with the possibility of damaging the artifact or the assembly. If the bag were of constant volume, the pressure inside the bag would remain essentially constant at the pressure of the sea floor, e.g., 6,000 psi for Titanic. As the ambient pressure decreases, the pressure differential from inside the bag to the surroundings would increase. Eventually the difference would equal sea floor pressure. This probably would cause the bag to rupture. If the bag permitted some expansion, a control scheme would be needed to vent air from the bag during the trip to the surface to maintain a constant buoyancy force just slightly larger than the weight of the artifact in water. Then the trip to the surface could be completed at low speed without danger of broaching the surface or damaging the artifact. Problem *3.114 Given: [Difficulty: 3] Cylindrical container rotating as in Example 3.10 R = 0.25⋅ m h o = 0.3⋅ m f = 2 ⋅ Hz Find: (a) height of free surface at the entrance (b) if solution depends on ρ Solution: We will apply the hydrostatics equations to this system. Governing Equations: Assumptions: (Hydrostatic equation) (1) Incompressible fluid (2) Atmospheric pressure acts everywhere In order to obtain the solution we need an expression for the shape of the free surface in terms of ω, r, and h o. The required expression was derived in Example 3.10. The equation is: z = ho − ( ω⋅ R) 2 2⋅ g ⎡1 ⋅⎢ ⎣2 − 2 ⎛ r ⎞ ⎥⎤ ⎜R ⎝ ⎠⎦ The angular velocity ω is related to the frequency of rotation through: h1 = ho − Now since h1 is the z value which corresponds to r = 0: Substituting known values: h 1 = 0.3⋅ m − 1 4 × ⎛⎜ 12.57 ⋅ ⎝ ω = 2 ⋅ π⋅ f rad s ( ω⋅ R) rad s = 12.57 ⋅ 2 4⋅ g 2 × 0.25⋅ m⎞ × ⎠ ω = 2⋅ π × 2⋅ 2 s 9.81⋅ m The solution is independent of ρ because the equation of the free surface is independent of ρ as well. h 1 = 0.05 m rad s Problem *3.115 [Difficulty: 2] Given: U-tube accelerometer Find: Acceleration in terms of h and L Solution: We will apply the hydrostatics equations to this system. Governing Equations: (Hydrostatic equation in x-direction) (Hydrostatic equation in y-direction) Assumptions: (1) Incompressible fluid (2) Neglect sloshing (3) Ignore corners (4) Both ends of U-tube are open to atmosphere In the coordinate system we are using, we can see that: Thus, ∂p = − ρa ∂x ∂p = − ρg ∂y ax = a ay = 0 g x = 0 g y = −g dp = Now if we evaluate Δp from left to right in the U-tube: ∂p ∂p ∆x + ∆y ∂x ∂y We may also write this expression as: ∆p = Simplifying this expression: ∆p = ρ⋅ a⋅ L − ρ⋅ g ⋅ h = 0 ∂p ∂p dy dx + ∂x ∂y ∆p = ( −ρ⋅ g ) ⋅ ( −b ) + ( −ρ⋅ a) ⋅ ( −L) + ( −ρ⋅ g ) ⋅ ( b + h ) = 0 Solving for h: h= a⋅ L g Problem *3.116 [Difficulty: 2] Given: Rectangular container with constant acceleration Find: Slope of free surface Solution: Basic equation In components p ρ g x ρ ax x p ρ g y ρ ay y p ρ g z ρ az z We have ay az 0 g x g sin( θ) g y g cos( θ) Hence p ρ g sin( θ) ρ ax x p ρ g cos( θ) 0 y (1) From Eq. 3 we can simplify from p p ( x y z) Hence a change in pressure is given by dp At the free surface p = const., so x p dx dp 0 x y p dy p dx y p dy dy or dx x y Hence at the free surface, using Eqs 1 and 2 dy dx x y p ρ g sin( θ) ρ ax dy dx ρ g cos( θ) p 9.81 ( 0.5) m 2 3 s 9.81 ( 0.866 ) dy dx 0.224 m 2 s m 2 s At the free surface, the slope is (3) p p ( x y ) to p 0 z (2) gz 0 g sin( θ) ax g cos( θ) p at the free surface p Problem *3.117 Given: Spinning U-tube sealed at one end Find: Maximum angular speed for no cavitation [Difficulty: 2] Assumptions: (1) water is incompressible (2) constant angular velocity Solution: Basic equation 2 ⎛∂ ⎞ V 2 = −ρ⋅ ω ⋅ r −⎜ p = ρ⋅ ar = −ρ⋅ r ⎝∂r ⎠ In components ∂ Between D and C, r = constant, so ∂z ∂ Between B and A, r = constant, so ∂z ∂ ∂z p = −ρ⋅ g p = −ρ⋅ g and so p D − p C = −ρ⋅ g ⋅ H (1) p = −ρ⋅ g and so p A − p B = −ρ⋅ g ⋅ H (2) and so ⌠ C ⌠ 2 ⎮ 1 dp = ⎮ ρ⋅ ω ⋅ r dr ⌡ ⌡p 0 p ∂ Between B and C, z = constant, so ∂r 2 p = ρ⋅ ω ⋅ r B 2 Integrating 2 L p C − p B = ρ⋅ ω ⋅ 2 Since p D = p atm, then from Eq 1 p C = p atm + ρ⋅ g ⋅ H From Eq. 3 From Eq. 2 2 L L (3) 2 2 L 2 p B = p C − ρ⋅ ω ⋅ 2 so p B = p atm + ρ⋅ g ⋅ H − ρ⋅ ω ⋅ 2 p A = p B − ρ⋅ g ⋅ H so 2 L p A = p atm − ρ⋅ ω ⋅ 2 2 Thus the minimum pressure occurs at point A (not B). Substituting known data to find the pressure at A: p A = 14.7⋅ lbf 2 in − 1.94⋅ slug ft 3 × ⎛⎜ 1600⋅ ⎝ rev min × 2 ⋅ π⋅ rad rev At 68oF from steam tables, the vapor pressure of water is × min ⎞ 60⋅ s ⎠ 2 × 1 2 × ⎛⎜ 3 ⋅ in × ⎝ p v = 0.339 ⋅ psi 2 2 2 ⎞ × lbf ⋅ s × ⎛ ft ⎞ = 2.881 ⋅ lbf ⎜ 12⋅ in ⎠ slug⋅ ft ⎝ 12⋅ in ⎠ 2 in ft which is less than the pressure at A. Therefore, cavitation does not occur.: Problem *3.118 [Difficulty: 2] Given: Spinning U-tube sealed at one end Find: Pressure at A; water loss due to leak Assumption: Water is incompressible; centripetal acceleration is constant Solution: Basic equation From the analysis of Example Problem 3.10, solving the basic equation, the pressure p at any point (r,z) in a continuous rotating fluid is given by 2 p = p0 + ρ⋅ ω ⋅ ⎛ r − r0 2 ⎝ 2 2⎞ ⎠ − ρ⋅ g⋅ ( z − z0) (1) In this case p = pA The speed of rotation is ω = 300 ⋅ rpm ω = 31.4⋅ p D = 0 ⋅ kPa (gage) The pressure at D is 2 Hence pA = ρ⋅ ω 2 p0 = pD where p 0 is a reference pressure at point (r0,z0) z = zA = zD = z0 = H ( 2) − ρ⋅g⋅(0) = − ρ⋅ω2 ⋅L 2 2 ⋅ −L p A = −0.42⋅ psi =− r= 0 r0 = rD = L rad 1 2 s × 1.94⋅ slug ft 3 × ⎛⎜ 31.4⋅ ⎝ rad ⎞ s 2 ⎠ 2 × ( 3 ⋅ in) × 4 2 ⎛ 1 ⋅ ft ⎞ × lbf ⋅ s ⎜ slug⋅ ft ⎝ 12⋅ in ⎠ (gage) When the leak appears,the water level at A will fall, forcing water out at point D. Once again, from the analysis of Example Problem 3.10, we can use Eq 1 In this case p = pA = 0 Hence 0= 2 ρ⋅ ω 2 p0 = pD = 0 The amount of water lost is z0 = zD = H r= 0 r0 = rD = L ( 2) − ρ⋅g⋅(zA − H) ⋅ −L 2 zA = H − z = zA ω ⋅L 2⋅ g 2 = 12in − 1 2 × ⎛⎜ 31.4⋅ ⎝ rad ⎞ s ⎠ ∆h = H − zA = 12⋅ in − 0.52⋅ in 2 2 × ( 3 ⋅ in) × 2 s 32.2⋅ ft × 1 ⋅ ft 12⋅ in ∆h = 11.48 ⋅ in zA = 0.52⋅ in Problem *3.119 [Difficulty: 2] ω R Given: Centrifugal manometer consists of pair of parallel disks that rotate to develop a radial pressure difference. There is no flow between the disks. Find: (a) an expression for the pressure difference, ∆p, as a function of ω, R, and ρ. (b) find ω if ∆p = 8 µm H2O and R = 50 mm Solution: We will apply the hydrostatics equations to this system. G G − ∇ p + ρg = ρa Governing Equations: ∂p + ρ g r = ρa r ∂r − Assumptions: (1) Incompressible fluid (2) Standard air between disks (3) Rigid body motion (4) Radial direction is horizontal For rigid body motion: ar = − 2 V r (Hydrostatic equation) =− ( r⋅ ω) 2 r 2 = −r⋅ ω (Hydrostatic equation in radial direction) In addition, since r is horizontal: gr = 0 ∂p = ρ rω 2 ∂r Thus, the hydrostatic equation becomes: We can solve this expression by separating variables and integrating: 2⌠ R ∆p = ρ⋅ ω ⋅ ⎮ ⌡ 2 Evaluating the integral on the right hand side: r dr 0 ω= Solving for the rotational frequency: 2 ⋅ ∆p ρ⋅ R Therefore: Substituting in values: ω= ω = 2⋅ 2× The pressure differential can be expressed as: 2 ∆p = ρ⋅ ω ⋅ R 2 2 ∆p = ρ⋅ g ⋅ ∆h ρw g ⋅ ∆h ⋅ ρair 2 R 999 1.225 × 9.81⋅ m 2 s −6 × 8 × 10 ⋅m × 1 (50 × 10− 3⋅m) 2 ω = 7.16⋅ rad s Problem *3.120 [Difficulty: 2] ω = 1000 s-1 ρ r1 r2 Given: Test tube with water Find: (a) Radial acceleration (b) Radial pressure gradient (c) Rotational speed needed to generate 250 MPa pressure at the bottom of the tube Solution: We will apply the hydrostatics equations to this system. G G − ∇ p + ρg = ρa Governing Equations: − ∂p + ρ g r = ρa r ∂r (1) Incompressible fluid (2) Rigid body motion (3) Radial direction is horizontal For rigid body motion: ar = − 2 In addition, since r is horizontal: r =− ( r⋅ ω) 2 r gr = 0 r2 = 130 mm (Hydrostatic equation) Assumptions: V r1 = 50 mm (Hydrostatic equation in radial direction) 2 2 = −r⋅ ω ar = −r⋅ ω ∂p = ρ rω 2 ∂r Thus, the hydrostatic equation becomes: We can solve this expression by separating variables and integrating: r2 2⌠ ∆p = ρ⋅ ω ⋅ ⎮ r dr ⌡r 1 2 ∆p = Evaluating the integral on the right hand side: Substituting in values: ω = 6 N 2 × 250 × 10 ⋅ 2 m ω = 938 ⋅ Hz ρ⋅ ω ⋅ ⎛ r − r1 2 ⎝2 2 2⎞ ⎠ Solving for ω: 3 × m 999 ⋅ kg × ω= 1 (130 × 10− 3⋅m) − (50 × 10− 3⋅m) 2 2 × 2 ⋅ ∆p ρ⋅ ⎛ r2 − r1 ⎝ 2 kg⋅ m 2 N⋅ s × 2⎞ ⎠ rev 2 ⋅ π⋅ rad Problem *3.121 [Difficulty: 3] Given: Rectangular container of base dimensions 0.4 m x 0.2 m and a height of 0.4 m is filled with water to a depth of d = 0.2 m. Mass of empty container is M c = 10 kg. The container slides down an incline of θ = 30 deg with respect to the horizontal. The coefficient of sliding friction is 0.30. Find: The angle of the water surface relative to the horizontal. y x Solution: α We will apply the hydrostatics equations to this system. Governing Equations: Assumptions: θ G G − ∇p + ρg = ρa (Hydrostatic equation) G G (Newton's Second Law) F = Ma (1) Incompressible fluid (2) Rigid body motion Writing the component relations: − ∂p = ρa x ∂x We write the total differential of pressure as: ∂p = − ρa x ∂x dp = ∂p ∂p dy dx + ∂x ∂y dy a x and ∂p ∂x dy α = atan⎛⎜ − ⎞ =− =− ⎝ dx ⎠ dx ∂p ∂y g + ay M = M c + M w = M c + ρw⋅ Vw − ∂p − ρg = ρa y ∂y ∂p = − ρ (g + a y ) ∂y Now along the free surface of the water dp = 0. Thus: To determine the acceleration components we analyze a free-body diagram: M = 10⋅ kg + 999 ⋅ kg 3 × 0.4⋅ m × 0.2⋅ m × 0.2⋅ m M = 25.98 kg m ΣFy' = 0 = N − M ⋅ g ⋅ cos( θ) N = M ⋅ g ⋅ cos( θ) N = 25.98 ⋅ kg × 9.81⋅ m 2 2 N⋅ s × cos( 30⋅ deg) × kg⋅ m s ΣFx' = M ⋅ ax' = M ⋅ g ⋅ sin( θ) − Ff = M ⋅ g ⋅ sin( θ) − μ⋅ N ax' = 9.81⋅ m 2 × sin( 30⋅ deg) − 0.30 × 220.7 ⋅ N × s ax' = g ⋅ sin( θ) − μ⋅ 1 25.98 ⋅ kg × kg⋅ m 2 N⋅ s y N x M ax' = 2.357 m 2 × cos( 30⋅ deg) ax = 2.041 s Thus, m 2 s α = atan⎛⎜ 2.041 ⎞ ⎝ 9.81 − 1.178 ⎠ α = 13.30 ⋅ deg ay = −2.357 ⋅ 2 s m 2 s F f = µN y’ m Now that we have the acceleration in the x'-y' system, we transform it to the x-y system: ax = 2.357 ⋅ N = 220.7 N x’ ax = ax'⋅ cos( θ) × sin( 30⋅ deg) θ N Mg ay = −ax'⋅ sin( θ) ay = −1.178 m 2 s Problem *3.122 [Difficulty: 3] Given: Rectangular container of base dimensions 0.4 m x 0.2 m and a height of 0.4 m is filled with water to a depth of d = 0.2 m. Mass of empty container is M c = 10 kg. The container slides down an incline of θ = 30 deg with respect to the horizontal without friction. Find: (a) The angle of the water surface relative to the horizontal. (b) The slope of the free surface for the same acceleration up the plane. Solution: We will apply the hydrostatics equations to this system. G G − ∇ p + ρg = ρa G G F = Ma Governing Equations: Assumptions: y x α (Hydrostatic equation) (Newton's Second Law) θ (1) Incompressible fluid (2) Rigid body motion − Writing the component relations: ∂p = ρa x ∂x We write the total differential of pressure as: ax ∂p ∂x dy =− =− dx ∂p ∂y g + ay M = M c + M w = M c + ρw⋅ Vw dp = α = atan⎛⎜ − and ∂p = − ρa x ∂x − ∂p ∂p dy dx + ∂x ∂y dy ⎞ Now along the free surface of the water dp = 0. Thus: To determine the acceleration components we analyze a free-body diagram: ⎝ dx ⎠ M = 10⋅ kg + 999 ⋅ ∂p = − ρ (g + a y ) ∂y ∂p − ρg = ρa y ∂y kg 3 × 0.4⋅ m × 0.2⋅ m × 0.2⋅ m M = 25.98 kg m ΣFx' = M ⋅ ax' = M ⋅ g ⋅ sin( θ) ax' = g ⋅ sin( θ) ax = ax'⋅ cos( θ) = g ⋅ sin( θ) ⋅ cos( θ) ay = −ax'⋅ sin( θ) = g ⋅ ( sin( θ) ) Thus, dy dx =− g ⋅ sin( θ) ⋅ cos( θ) 2 g ⎡⎣1 − ( sin( θ) ) ⎤⎦ =− sin( θ) ⋅ cos( θ) ( cos( θ) ) 2 slope = g ⋅ sin( θ) ⋅ cos( θ) 2⎤ g ⎡⎣1 + ( sin( θ) ) ⎦ =− sin( θ) ⋅ cos( θ) 1 + ( sin( θ) ) 2 x 2 Ff = µN y’ =− sin( θ) cos( θ) For the acceleration up the incline: ax = −g ⋅ sin( θ) ⋅ cos( θ) Thus, y = −tan( θ) α = 30⋅ deg x’ ay = g ⋅ ( sin( θ) ) slope = θ N Mg 2 sin( 30⋅ deg) ⋅ cos( 30⋅ deg) 1 + ( sin( 30⋅ deg) ) 2 slope = 0.346 Problem *3.123 [Difficulty: 3] Given: Cubical box with constant acceleration Find: Slope of free surface; pressure along bottom of box Solution: Basic equation In components ∂ − p + ρ⋅ g x = ρ⋅ ax ∂x ∂ − p + ρ⋅ g y = ρ⋅ ay ∂y ∂ − p + ρ⋅ g z = ρ⋅ az ∂z We have ax = ax ay = 0 az = 0 gz = 0 Hence ∂ ∂ (3) ∂x gx = 0 ∂ p = −SG⋅ ρ⋅ ax (1) ∂y p = −SG⋅ ρ⋅ g From Eq. 3 we can simplify from p = p ( x , y , z) Hence a change in pressure is given by dp = dy Hence at the free surface dx L = 80⋅ cm at the midpoint x = L y= 2 On the bottom y = 0 so ∂x p ⋅ dy p ⋅ dx + x=0 y= 5 8 ⋅L ∂ ∂ ∂y p ⋅ dy dy or =− dx ∂x ∂ p =− p ax =− g 0.25⋅ g g = −0.25 x 4 +C and through volume conservation the fluid rise in the rear balances the fluid fall in the front, so at the midpoint the free surface has not moved from the rest position (box is half filled) 2 so p =0 (4) 1 L =− ⋅ +C 2 4 2 L dp = −SG ⋅ ρ⋅ ax ⋅ dx − SG ⋅ ρ⋅ g ⋅ dy Combining Eqs 1, 2, and 4 p = p atm when L ∂ ∂y ∂z p = p( x , y) to ∂ (2) ∂y y=− The equation of the free surface is then We have ∂x p ⋅ dx + dp = 0 = At the free surface p = const., so For size ∂ g y = −g or 5 p atm = −SG⋅ ρ⋅ g ⋅ ⋅ L + c 8 C= 5 8 ⋅L y= 5 8 ⋅L − x 4 p = −SG ⋅ ρ⋅ ax ⋅ x − SG ⋅ ρ⋅ g ⋅ y + c 5 c = p atm + SG ⋅ ρ⋅ g ⋅ ⋅ L 8 5 5 x p ( x , y ) = p atm + SG⋅ ρ⋅ ⎜⎛ ⋅ g ⋅ L − ax ⋅ x − g ⋅ y⎞ = p atm + SG⋅ ρ⋅ g ⋅ ⎛⎜ ⋅ L − −y 4 ⎝8 ⎠ ⎝8 2 kg 5 x N⋅ s m p ( x , 0 ) = p atm + SG⋅ ρ⋅ g ⋅ ⎛⎜ ⋅ L − ⎞ = 101 + 0.8 × 1000⋅ × × 9.81⋅ × 3 4⎠ kg⋅ m 2 ⎝8 m s p ( x , 0 ) = 105 − 1.96⋅ x kPa x⎞ ⎛5 × ⎜ × 0.8⋅ m − 3 4⎠ ⎝8 10 ⋅ Pa (p in kPa, x in m) Problem *3.124 [Difficulty: 3] Given: Gas centrifuge, with maximum peripheral speed Vmax = 950 ft/s contains uranium hexafluoride gas (M = 352 lb/lbmol) at 620 deg F. Find: (a) Ratio of maximum pressure to pressure at the centrifuge axis (b) Evaluate pressure ratio at 620 deg F. Solution: We will apply the hydrostatics equations to this system. G G − ∇ p + ρg = ρa Governing Equations: − Assumptions: Vmax = ωr 2 r2 (Hydrostatic equation) ∂p + ρg r = ρa r (Hydrostatic equation radial component) ∂r (1) Incompressible fluid (2) Rigid body motion (3) Ideal gas behavior, constant temperature 2 2 V ( r⋅ ω) 2 For rigid body motion: ar = − =− = −r⋅ ω r r Thus: ∂p p = − ρa r = rω 2 ∂r Rg T Separating variables and integrating: p r 2 2 2 r ⎛ p2 ⎞ ω 2 ln⎜ = ⋅ ⎝ p1 ⎠ Rg⋅ T 2 2 2 ⌠ 2 1 ω ⌠ ⎮ ⋅ ⎮ r dr dp = Rg ⋅ T ⌡0 ⎮ p ⌡p where we define: Vmax = ω⋅ r2 therefore: ⎛ p2 ⎞ Vmax ln⎜ = p1 ⎝ ⎠ 2 ⋅ Rg ⋅ T 1 ⎛ V 2⎞ ⎜ max ⎜ 2⋅ R ⋅ T p2 g ⎠ p rat = = e⎝ p Solving for the pressure ratio: 1 The gas constant: Rg = 1545 ft⋅ lbf ⋅ 352 lbm⋅ R Substituting in all known values: Rg = 4.39⋅ ft⋅ lbf lbm⋅ R 2 2 ⎤ ⎡⎛ ft 1 lbm ⋅ R 1 lbf ⋅ s ⎢⎜950⋅ ⎞ × × × × ⎥ s ⎠ 2 4.39⋅ ft⋅ lbf ( 620+ 460) ⋅ R 32.2⋅ lbm⋅ ft⎦ p rat = e⎣⎝ p rat = 19.2 Problem *3.125 Given: [Difficulty: 3] Pail is swung in a vertical circle. Water moves as a rigid body. Point of interest is the top of the trajectory. Find: d V = 5 m/s d (a) Tension in the string (b) Pressure on pail bottom from the water. Solution: R= 1 m T We will apply the hydrostatics equations to this system. G G − ∇ p + ρg = ρa Governing Equations: − Assumptions: (Hydrostatic equation radial component) (1) Incompressible fluid (2) Rigid body motion (3) Center of mass of bucket and water are located at a radius of 1 m where V = rω = 5 m/s ( ) ( ( ) −T − M b + M w ⋅ g = M b + M w ar ⎛V ⎞ T = Mb + Mw ⋅ ⎜ −g ⎝ r ⎠ π 2 and: M w = ρ⋅ V = ρ⋅ ⋅ d ⋅ h 4 Now we find T: (Hydrostatic equation) ∂p + ρ g r = ρa r ∂r Summing the forces in the radial direction: Thus the tension is: r 2 2 ) M w = 999 ⋅ 2 kg 3 m × π 4 T = ( 1.529 + 25.11 ) ⋅ kg × ar = − where M b = 15⋅ N × where: 2 × ( 0.4⋅ m) × 0.2⋅ m s 9.81⋅ m × V r kg⋅ m 2 N⋅ s M w = 25.11 ⋅ kg 2 ⎡⎛ m 2 1 m ⎤ N⋅ s ⎢⎜ 5⋅ ⎞ × − 9.81⋅ ⎥ × 2⎥ ⎢⎝ s ⎠ 1⋅ m kg⋅ m s ⎦ ⎣ If we apply this information to the radial hydrostatic equation we get: − M b = 1.529 ⋅ kg ∂p V2 − ρg = − ρ ∂r r T = 405 ⋅ N Thus: ⎞ ⎛V 2 ∂p = ρ ⎜⎜ − g ⎟⎟ ∂r ⎝ r ⎠ If we assume that the radial pressure gradient is constant throughout the water, then the pressure gradient is equal to: p r = 999 ⋅ kg 3 m × 2 2 ⎡ ⎤ ⎢⎜⎛ 5⋅ m ⎞ × 1 − 9.81⋅ m ⎥ × N⋅ s ⎢⎝ s ⎠ 2⎥ 1⋅ m kg⋅ m s ⎦ ⎣ and we may calculate the pressure at the bottom of the bucket: p r = 15.17 ⋅ kPa m ∆p = p r⋅ ∆r ∆p = 15.17 ⋅ kPa m × 0.2⋅ m ∆p = 3.03⋅ kPa Problem *3.126 [Difficulty: 3] D = 2.5 in. z H/2 r R = 5 ft Given: rev s Find: (a) Slope of free surface (b) Spin rate necessary for spillage (c) Likelihood of spilling versus slipping Solution: We will apply the hydrostatics equations to this system. G G − ∇ p + ρg = ρa Governing Equations: Assumptions: 2 ar = − r SG = 1.05 Half-filled soft drink can at the outer edge of a merry-go-round ω = 0.3⋅ V H = 5 in. =− ( r⋅ ω) r (Hydrostatic equation) − ∂p + ρ g r = ρa r ∂r − ∂p + ρg z = ρa z ∂z (Hydrostatic equation radial component) (Hydrostatic equation z component) (1) Incompressible fluid (2) Rigid body motion (3) Merry-go-round is horizontal 2 2 = −r⋅ ω ∂p ∂p dr + dp = dz ∂r ∂z az = 0 gr = 0 g z = −g Thus: ∂p = ρ rω 2 ∂r ∂p ∂r ρrω 2 rω 2 dz =− = =− ∂p ∂z − ρg dr g For the free surface the pressure is constant. Therefore: slope = 5 ⋅ ft × ⎛⎜ 20⋅ So the slope at the free surface is ⎝ H To spill, the slope must be slope sp = rev min H × min 60⋅ s × slope sp = D 2 ⋅ π⋅ rad ⎞ rev 5 2.5 ⎠ ∂p = − ρg So p = p(r,z) ∂z 2 2 × s 32.2⋅ ft slope = 0.681 slope sp = 2.000 D Thus, ωsp = g dz ⋅ r dr ωsp = 32.2⋅ ft 2 s × 1 5 ⋅ ft ×2 rad ωsp = 3.59⋅ s This is nearly double the original speed (2.09 rad/s). Now the coefficient of static friction between the can and the surface of the merry-go-round is probably less than 0.5.Thus the can would not likely spill or tip; it would slide off! Problem *3.127 Discussion: [Difficulty: 2] Separate the problem into two parts: (1) the motion of the ball in water below the pool surface, and (2) the motion of the ball in air above the pool surface. Below the pool water surface the motion of each ball is controlled by buoyancy force and inertia. For small depths of submersion ball speed upon reaching the surface will be small. As depth is increased, ball speed will increase until terminal speed in water is approached. For large depths, the actual depth will be irrelevant because the ball will reach terminal speed before reaching the pool water surface. All three balls are relatively light for their diameters, so terminal speed in water should be reached quickly. The depth of submersion needed to reach terminal speed should be fairly small, perhaps 1 meter or less (The initial water depth required to reach terminal speed may be calculated using the methods of Chapter 9). Buoyancy is proportional to volume and inertia is proportional to mass. The ball with the largest volume per unit mass should accelerate most quickly to terminal speed. This will probably be the beach ball, followed by the table-tennis ball and the water polo ball. The ball with the largest diameter has the smallest frontal area per unit volume; the terminal speed should be highest for this ball. Therefore, the beach ball should have the highest terminal speed, followed by the water polo ball and the table-tennis ball. Above the pool water surface the motion of each ball is controlled by aerodynamic drag force, gravity force, and inertia (see the equation below). Without aerodynamic drag, the height above the pool water surface reached by each ball will depend only on its initial speed (The maximum height reached by a ball in air with aerodynamic drag may be calculated using the methods of Chapter 9). Aerodynamic drag reduces the height reached by the ball. Aerodynamid drag is proportional to frontal area. The heaviest ball per unit frontal area (probably the water polo ball) should reach the maximum height and the lightest ball per unit area (probably the beach ball) should reach the minimum height. dV 1 2 ΣFy = −FD − m⋅ g = m⋅ ay = m⋅ = −CD⋅ A⋅ ⋅ ρ⋅ V − m⋅ g since dt 2 V0 y Thus, − 1 2 CD⋅ A⋅ ⋅ ρ⋅ V 2 m −g= dV dt = V⋅ dV (1) 1 2 FD = CD⋅ A⋅ ⋅ ρ⋅ V 2 We solve this by separating variables: dy FD V⋅ dV W = mg Solving for the maximum height: CD⋅ A⋅ ρ V2 ⋅ 1+ m⋅ g 2 = −g ⋅ dy Integrating this expression over the flight of the ball yields: 2 ⎛⎜ ρ⋅ CD⋅ A Vo ⎞ − ⋅ ⋅ ln⎜ 1 + = −g ⋅ y max m⋅ g 2 ⎠ ρ⋅ CD⋅ A ⎝ m⋅ g 2 ⎛⎜ ρ⋅ CD⋅ A Vo ⎞ FDo ⎞ ⎛ m Simplifying: y max = − (2) ⋅ ln⎜ 1 + ⋅ ln⎜ 1 + y max = − ⋅ 2 ⎠ m⋅ g ρ⋅ CD⋅ A ⎝ ρ⋅ CD⋅ A ⎝ m⋅ g ⎠ m If we neglect drag, equation (1) becomes: −g ⋅ dy = V⋅ dV Checking the limiting value predicted by Eq (2) as lim y max C D →0 Integrating and solving for the maximum height: CD → 0 y max = − Vo 2⋅ g : we remember that for small x that ln(1+x) = -x. Thus: ⎛ m ρC D A Vo2 ⎞ Vo2 ⎟⎟ = = lim ⎜⎜ C D →0 ρC A mg 2 ⎝ D ⎠ 2g 2 which is the result in Equation (3). (3) Problem *3.128 Given: [Difficulty: 4] A steel liner is to be formed in a spinning horizontal mold. To insure uniform thickness the minimum angular velocity should be at least 300 rpm. For steel, SG = 7.8 θ y x ri Find: (a) The resulting radial acceleration on the inside surface of the liner (b) the maximum and minimum pressures on the surface of the mold Solution: We will apply the hydrostatics equations to this system. G G − ∇ p + ρg = ρa Governing Equations: Assumptions: 2 ar = − Hence: Thus: dp = V r =− ro (gravity is downward in this diagram) ( r⋅ ω) r − ∂p + ρ g r = ρa r ∂r − 1 ∂p + ρg θ = ρaθ r ∂θ − ∂p + ρg z = ρa z ∂z (Hydrostatic equation) (Hydrostatic equation radial component) (Hydrostatic equation transeverse component) (Hydrostatic equation z component) (1) Incompressible fluid (2) Rigid body motion 2 2 = −r⋅ ω ar = 4 ⋅ in × ⎜⎛ 300 × ⎝ aθ = 0 rev min × 2 ⋅ π⋅ rad rev az = 0 × min ⎞ 2 60⋅ s ⎠ ∂p = ρg r − ρa r = ρrω 2 − ρg cos θ ∂r × g r = −g ⋅ cos( θ) g θ = g ⋅ sin( θ) ft ar = 329 ⋅ 12⋅ in 2 ar = 10.23 ⋅ g s ∂p = ρrg θ − ρraθ = ρrg sin θ ∂θ ∂p ∂p dr + dθ = ρrω 2 − ρg cos θ dr + (ρrg sin θ )dθ ∂r ∂θ ( ft gz = 0 ∂p = ρg z − ρa z = 0 ∂z ) ( ) We can integrate to find pressure as a function of r and θ. r ( p ri , θ = p atm ) ⌠ 2 Therefore, we integrate: p − p atm = ⎮ ρ⋅ r⋅ ω − ρ⋅ g ⋅ cos( θ) dr + f ( θ) ⌡r i ∂p ⎞ 2 ⎟ = ρrω − ρg cos θ ∂r ⎠θ ⎛ r2 − r 2⎞ i ⎠ ⎝ p = p atm + ρ⋅ ω ⋅ − ρ⋅ g ⋅ cos( θ) ⋅ ( r − ri) + f ( θ) 2 2 df ∂p ⎞ = ρrg sin θ ⎟ = ρ (r − ri )g sin θ + dθ ∂θ ⎠ r Taking the derivative of pressure with respect to θ: Thus, the integration function f(θ) is: ⎛ r2 − r 2⎞ i ⎠ ⎝ p = p atm + ρ⋅ ω ⋅ − ρ⋅ g ⋅ ( r − ri) ⋅ cos( θ) − ρ⋅ g ⋅ ri⋅ cos( θ) + C 2 2 Therefore, the pressure is: The integration constant is determined from the boundary condition: ( ) p ri , θ = p atm ⎛ r 2 − r 2⎞ i ⎠ ⎝i p atm = p atm + ρ⋅ ω ⋅ − ρ⋅ g ⋅ ( ri − ri) ⋅ cos( θ) − ρ⋅ g ⋅ ri⋅ cos( θ) + C 2 2 Therefore, the pressure is: f ( θ) = −ρ⋅ g ⋅ ri⋅ cos( θ) + C −ρ⋅ g ⋅ ri⋅ cos( θ) + C = 0 C = ρ⋅ g ⋅ ri⋅ cos( θ) ⎛ r2 − r 2⎞ i ⎠ ⎝ p = p atm + ρ⋅ ω ⋅ − ρ⋅ g ⋅ ( r − ri) ⋅ cos( θ) 2 2 The maximum pressure should occur on the mold surface at θ = π: p maxgage = ⎡⎛ ⎢ 7.8⋅ 1.94⋅ slug ⎞ × 3 ⎢⎜ ft ⎠ ⎣⎝ 2 2 2 ⎛ 31.42 ⋅ rad ⎞ × 1 ⋅ ⎛⎜ 6 − 4 ⎞ ⋅ ft2 − ⎜ 2 s ⎠ 2 ⎜ ⎝ ⎝ 12 ⎠ 2 ⎛ 7.8⋅ 1.94⋅ slug ⎞ × 32.2⋅ ft × ⎛ 6 − 4 ⎞ ⋅ ft ⋅ cos( π)⎥⎤ ⋅ lbf ⋅ s ⎜ ⎜ ⎥ slug⋅ ft 3 2 ⎝ 12 ⎠ ft ⎠ s ⎝ ⎦ p maxgage = 1119⋅ psf p maxgage = 7.77⋅ psi The minimum pressure should occur on the mold surface at θ = 0: 2 2 2 ⎡⎛ slug ⎞ ⎛ rad ⎞ 1 ⎛⎜ 6 − 4 ⎞ 2 ⎢ p mingage = 7.8⋅ 1.94⋅ × ⋅ × ⎜ 31.42 ⋅ ⋅ ft − 3 2 ⎢⎜ s ⎠ 2 ⎜ ⎝ ft 12 ⎣⎝ ⎠ ⎝ ⎠ 2 ⎛ 7.8⋅ 1.94⋅ slug ⎞ × 32.2⋅ ft × ⎛ 6 − 4 ⎞ ⋅ ft ⋅ cos( 0 )⎥⎤ ⋅ lbf ⋅ s ⎜ ⎜ ⎥ slug⋅ ft 3 2 ⎝ 12 ⎠ ft ⎠ s ⎝ ⎦ p mingage = 956 ⋅ psf (In both results we divided by 144 to convert from psf to psi.) p mingage = 6.64⋅ psi Problem *3.129 Discussion: [Difficulty: 4] A certain minimum angle of inclination would be needed to overcome static friction and start the container into motion down the incline. Once the container is in motion, the retarding force would be provided by sliding (dynamic) friction. the coefficient of dynamic friction usually is smaller than the static friction coefficient. Thus the container would continue to accelerate as it moved down the incline. This acceleration would procide a non-zero slope to the free surface of the liquid in the container. In principle the slope could be measured and the coefficent of dynamic friction calculated. In practice several problems would arise. To calculate dynamic friction coefficient one must assume the liquid moves as a solid body, i.e., that there is no sloshing. This condition could only be achieved if there were nminimum initial disturbance and the sliding distance were long. It would be difficult to measure the slope of the free surface of liquid in the moving container. Images made with a video camera or a digital still camera might be processed to obtain the required slope information. α Ff = µN y x θ N mg ΣFy = 0 = N − M ⋅ g ⋅ cos( θ) N = M ⋅ g ⋅ cos( θ) ΣFx = M ⋅ ax = M ⋅ g ⋅ sin( θ) − Ff Ff = μk ⋅ N = μk ⋅ M ⋅ g ⋅ cos⋅ ( θ) Thus the acceleration is: ax = g ⋅ sin( θ) − μk ⋅ g ⋅ cos( θ) G G − ∇ p + ρg = ρa − ∂p + ρg sin θ = ρa x = ρ ( g sin θ − µ k g cos θ ) ∂x ∂p = ρgµ k cos θ ∂x − ∂p − ρg cos θ = ρa x = 0 ∂y ∂p = − ρg cos θ ∂y dp = ∂p ∂p dy dx + ∂x ∂y ρgµ k cosθ ∂p ∂x dy =− = µk =− ∂p ∂y − ρg cos θ dx For the free surface the pressure is constant. Therefore: So the free surface angle is: Now for a static liquid: ( ) α = atan μk Now since it is necessary to make the container slip along the surface, ( ) ( ) θ > atan μs > atan μk = α Thus, α < θ, as shown in the sketch above. Problem 4.1 Given: Data on mass and spring Find: Maximum spring compression [Difficulty: 2] Solution: The given data is M = 5 ⋅ lb h = 5 ⋅ ft k = 25⋅ lbf ft Apply the First Law of Thermodynamics: for the system consisting of the mass and the spring (the spring has gravitional potential energy and the spring elastic potential energy) E1 = 0 Total mechanical energy at initial state Note: The datum for zero potential is the top of the uncompressed spring E2 = M ⋅ g ⋅ ( −x ) + Total mechanical energy at instant of maximum compression x But E1 = E2 so 0 = M ⋅ g ⋅ ( −x ) + Solving for x x = 1 2 ⋅ k⋅ x x = 2 × 5 ⋅ lb × 32.2⋅ k ft 2 s x = When just resting on the spring M⋅ g × ft 25⋅ lbf × 32.2⋅ lb⋅ ft 2 s ⋅ lbf x = 0.200 ft k When dropped from height h: E1 = M ⋅ g ⋅ h Total mechanical energy at initial state Total mechanical energy at instant of maximum compression x E2 = M ⋅ g ⋅ ( −x ) + Note: The datum for zero potential is the top of the uncompressed spring E1 = E2 so M ⋅ g ⋅ h = M ⋅ g ⋅ ( −x ) + 2 ⋅ k⋅ x 2 2 2⋅ M⋅ g But 1 1 2 ⋅ k⋅ x 2 1 2 ⋅ k⋅ x 2 x = 0.401 ⋅ ft 2 x − Solving for x x = x = 5 ⋅ lb × 32.2⋅ ft 2 s × ft 25⋅ lbf 2⋅ M⋅ g k M⋅ g k × + ⋅x − 2⋅ M⋅ g⋅ h k =0 2 ⎛ M⋅ g ⎞ + 2⋅ M⋅ g⋅ h ⎜ k k ⎝ ⎠ 32.2⋅ lb⋅ ft 2 s ⋅ lbf 2 + ⎛ 5 ⋅ lb × 32.2⋅ ft × ft × 32.2⋅ lb⋅ ft ⎞ + 2 × 5⋅ lb × 32.2⋅ ft × ft × 32.2⋅ lb⋅ ft × 5 ⋅ f ⎜ 2 25⋅ lbf 2 25⋅ lbf 2 2 s s ⋅ lbf ⎠ s ⋅ lbf s ⎝ x = 1.63⋅ ft Note that ignoring the loss of potential of the mass due to spring compression x gives x = 2⋅ M⋅ g⋅ h k x = 1.41⋅ ft Problem 4.2 Given: An ice-cube tray with water at 15oC is frozen at –5oC. Find: Change in internal energy and entropy Solution: Apply the Tds and internal energy equations Governing equations: Assumption: Tds = du + pdv du = cv dT Neglect volume change Liquid properties similar to water The given or available data is: T1 = (15 + 273) K = 288 K cv = 1 kcal kg ⋅ K T2 = (− 5 + 273) K = 268 K ρ = 999 Then with the assumption: Tds = du + pdv = du = c v dT or ds = cv Integrating ⎛T s 2 − s1 = cv ln⎜⎜ 2 ⎝ T1 kg m3 dT T ∆S = 999 ⎞ ⎟⎟ ⎠ or ⎛T ⎞ ∆S = m(s 2 − s1 ) = ρVcv ln⎜⎜ 2 ⎟⎟ ⎝ T1 ⎠ J kcal 10 −6 m 3 kg ⎛ 268 ⎞ × × ×1 × ln⎜ 250 mL ⎟ × 4190 3 kcal kg ⋅ K mL m ⎝ 288 ⎠ ∆S = −0.0753 Also [Difficulty: 2] kJ K u 2 − u1 = cv (T2 − T1 ) ∆U = 999 or ∆U = mcv (T2 − T1 ) = ρVcv ∆T kcal J 10 −6 m 3 kg 250 mL × × ×1 × (− 268 − 288)K × 4190 3 mL kg ⋅ K kcal m ∆U = −20.9 kJ Problem 4.3 [Difficulty: 2] Given: Data on ball and pipe Find: Speed and location at which contact is lost θ V Fn Solution: The given data is r = 1 ⋅ mm M R = 50⋅ mm ∑ Fn = Fn − m⋅g⋅cos(θ) = m⋅an 2 an = − V R+r 2 Contact is lost when Fn = 0 or 2 V = g ⋅ ( R + r) ⋅ cos( θ) −m⋅ g ⋅ cos ( θ) = −m⋅ V R+r (1) 2 For no resistance energy is conserved, so E = m⋅ g ⋅ z + m⋅ V 2 2 = m⋅ g ⋅ ( R + r) ⋅ cos( θ) + m⋅ 2 V = 2 ⋅ g ⋅ ( R + r) ⋅ ( 1 − cos( θ) ) Hence from energy considerations 2 Combining 1 and 2, V = 2 ⋅ g ⋅ ( R + r) ⋅ ( 1 − cos( θ) ) = g ⋅ ( R + r) ⋅ cos( θ) Hence θ = acos⎛⎜ Then from 1 V = 2⎞ ⎝3⎠ ( R + r) ⋅ g ⋅ cos( θ) θ = 48.2⋅ deg V = 0.577 m s V 2 = E0 = m⋅ g ⋅ ( R + r) (2) or 2 ⋅ ( 1 − cos( θ) ) = cos( θ) Problem 4.4 Given: Data on Boeing 777-200 jet Find: Minimum runway length for takeoff [Difficulty: 2] Solution: Basic equation dV dV ΣFx = M ⋅ = M ⋅ V⋅ = Ft = constant dt dx Separating variables M ⋅ V⋅ dV = Ft⋅ dx Integrating x= Note that the "weight" is already in mass units! 2 x = For time calculation Integrating M⋅ t= M⋅ V 2 ⋅ Ft 1 2 × 325 × 10 kg × ⎛⎜ 225 km 3 ⎝ dV = Ft dt dV = hr Ft M × 1 ⋅ km 1000⋅ m × 2 2 1 N⋅ s 1 ⎞ × ⋅ × 3600⋅ s ⎠ kg⋅ m 3 N 2 × 425 × 10 1 ⋅ hr x = 747 m ⋅ dt M⋅ V Ft 3 t = 325 × 10 kg × 225 km hr × 1 ⋅ km 1000⋅ m × 1 ⋅ hr 3600⋅ s 1 × 2 × 425 × 10 Aerodynamic and rolling resistances would significantly increase both these results ⋅ 1 3 N 2 × N⋅ s kg⋅ m t = 23.9 s Problem 4.5 Given: Car sliding to a stop Find: Initial speed; skid time [Difficulty: 2] Solution: Governing equations: ΣFx = M ⋅ ax Ff = μ ⋅ W Assumptions: Dry friction; neglect air resistance Given data L = 200 ⋅ ft μ = 0.7 W W d2 ΣFx = −Ff = −μ ⋅ W = M ⋅ ax = ⋅ ax = ⋅ x g g dt 2 or d 2 dt 2 x = −μ ⋅ g Integrating, and using I.C. V = V0 at t = 0 Hence dx dt Integrating again (1) 1 1 2 2 x = − ⋅ g ⋅ t + V0 ⋅ t + c2 = − ⋅ g ⋅ t + V0 ⋅ t 2 2 We have the final state, at which From Eq. 1 = −μ⋅ g ⋅ t + c1 = −μ⋅ g ⋅ t + V0 dx x f = L and dt =0 dx = 0 = −μ⋅ g ⋅ tf + V0 dt since x = 0 at t = 0 at t = tf so tf = (2) V0 μ⋅ g 2 2 Substituting into Eq. 2 V0 V0 ⎛ V0 ⎞ x = x f = L = − ⋅ g ⋅ t + V0 ⋅ t = − ⋅ g ⋅ tf + V0 ⋅ tf = − ⋅ g ⋅ ⎜ + V0 ⋅ = μ⋅ g 2 ⋅ μ⋅ g 2 2 2 ⎝ μ⋅ g ⎠ Solving L= Using the data V0 = 64.7⋅ mph 1 V0 2 ⋅ μ⋅ g 2 1 1 2 or The skid time is V0 = tf = 2 ⋅ μ⋅ g ⋅ L V0 μ⋅ g tf = 4.21 s Problem 4.6 [Difficulty: 2] Given: Block sliding to a stop Find: Distance and time traveled; new coeeficient of friction Solution: Governing equations: ΣFx = M ⋅ ax Ff = μ⋅ W Assumptions: Dry friction; neglect air resistance m V0 = 5 ⋅ s μ = 0.6 Given data M = 2 ⋅ kg W W d2 ΣFx = −Ff = −μ⋅ W = M ⋅ ax = ⋅ ax = ⋅ x g g dt 2 L = 2⋅ m d or 2 dt 2 x = −μ⋅ g Integrating, and using I.C. V = V0 at t = 0 Hence dx dt Integrating again (1) 1 1 2 2 x = − ⋅ g ⋅ t + V0 ⋅ t + c2 = − ⋅ g ⋅ t + V0 ⋅ t 2 2 We have the final state, at which From Eq. 1 = −μ⋅ g ⋅ t + c1 = −μ⋅ g ⋅ t + V0 dx x f = L and dt dx = 0 = −μ⋅ g ⋅ tf + V0 dt =0 since x = 0 at t = 0 (2) t = tf at tf = so V0 μ⋅ g Using given data 2 Substituting into Eq. 2 Solving tf = 0.850 s 2 V0 V0 ⎛ V0 ⎞ x = x f = L = − ⋅ g ⋅ t + V0 ⋅ t = − ⋅ g ⋅ tf + V0 ⋅ tf = − ⋅ g ⋅ ⎜ + V0 ⋅ = μ⋅ g 2 ⋅ μ⋅ g 2 2 2 ⎝ μ⋅ g ⎠ 1 x = V0 1 2 1 2 2 2 ⋅ μ⋅ g For rough surface, using Eq. 3 with x = L (3) μ = Using give data V0 2 2⋅ g⋅ L μ = 0.637 tf = V0 μ⋅ g x = 2.12 m tf = 0.800 s Problem 4.7 Given: Car entering a curve Find: Maximum speed [Difficulty: 2] Solution: 2 Governing equations: ΣFr = M ⋅ ar V Ff = μ⋅ W ar = μwet = 0.3 r = 100 ⋅ ft r Assumptions: Dry friction; neglect air resistance Given data μdry = 0.7 2 V ΣFr = −Ff = −μ⋅ W = −μ⋅ M ⋅ g = M ⋅ ax = M ⋅ r or V= μ⋅ r⋅ g Hence, using given data V = μdry⋅ r⋅ g V = 32.4⋅ mph V = μwet⋅ r⋅ g V = 21.2⋅ mph Problem 4.8 Given: Data on air compression process Find: Work done [Difficulty: 2] Solution: Basic equation δw = p ⋅ dv Assumptions: 1) Adiabatic 2) Frictionless process δw = pdv Given data p 1 = 1 ⋅ atm p 2 = 4 ⋅ atm From Table A.6 R = 286.9 ⋅ J kg⋅ K T1 = 20 °C T1 = 293 K and k = 1.4 Before integrating we need to relate p and v. An adiabatic frictionless (reversible) process is isentropic, which for an ideal gas gives k p⋅ v = C δw = p ⋅ dv = C⋅ v Integrating w= w= C k−1 k= where ⋅ ⎛ v2 −k 1 −k ⎝ cp cv ⋅ dv 1 −k ⎞ − v2 1 k 1 −k k 1 −k ⎞ ⎛ − p1⋅ v1 ⋅ v2 ⎠ = ( k − 1) ⋅ ⎝ p 2⋅ v 2 v 2 ⎠ R ⋅ T1 ⎛ T2 ⎞ ⋅ T2 − T1 = ⋅⎜ −1 ( k − 1) T1 ( k − 1) R ( ) ⎝ (1) ⎠ k But k p⋅ v = C means k p1⋅ v1 = p2⋅ v2 k k− 1 Rearranging ⎛ p2 ⎞ =⎜ T1 ⎝ p1 ⎠ T2 or ⎛ R ⋅ T1 ⎞ ⎛ R ⋅ T2 ⎞ p1⋅ ⎜ = p2⋅ ⎜ ⎝ p1 ⎠ ⎝ p2 ⎠ k k− 1 ⎤ ⎡ ⎢ ⎥ k R⋅ T1 ⎢⎛ p 2 ⎞ ⎥ Combining with Eq. 1 w = ⋅ ⎢⎜ − 1⎥ k−1 ⎣⎝ p 1 ⎠ ⎦ 1.4− 1 ⎡⎢ ⎤⎥ 1.4 ⎢ 4 ⎥ 1 J w = × 286.9 ⋅ × ( 20 + 273 ) K × ⎢⎛⎜ ⎞ − 1⎥ 0.4 kg⋅ K ⎣⎝ 1 ⎠ ⎦ w = 102 kJ kg k Problem 4.9 [Difficulty: 2] Given: Data on cooling of a can of soda in a refrigerator Find: How long it takes to warm up in a room Solution: The First Law of Thermodynamics for the can (either warming or cooling) is M ⋅ c⋅ dT dt ( ) = −k ⋅ T − Tamb dT or dt ( ) = −k ⋅ T − Tamb where M is the can mass, c is the average specific heat of the can and its contents, T is the temperature, and Tamb is the ambient temperature Separating variables Integrating dT T − Tamb = −A⋅ dt ( ) T( t) = Tamb + Tinit − Tamb ⋅ e − At where Tinit is the initial temperature. The available data from the coolling can now be used to obtain a value for constant A Given data for cooling Hence Tinit = 80 °F Tinit = 540 ⋅ R Tamb = 35 °F Tamb = 495 ⋅ R T = 45 °F T = 505 ⋅ R when τ = 2 ⋅ hr k= 1 τ ⎛ Tinit − Tamb ⎞ ⋅ ln⎜ ⎝ T − Tamb ⎠ = 1 2 ⋅ hr × 1 ⋅ hr 3600⋅ s × ln⎛⎜ 540 − 495 ⎞ ⎝ 505 − 495 ⎠ −4 −1 k = 2.09 × 10 s Then, for the warming up process Tinit = 45 °F Tinit = 505 ⋅ R Tend = 60 °F Tend = 520 ⋅ R ( ) with Tend = Tamb + Tinit − Tamb ⋅ e Hence the time τ is τ= 1 k ⎛ Tinit − Tamb ⎞ ⋅ ln⎜ ⎝ Tend − Tamb ⎠ = Tamb = 72 °F Tamb = 532 ⋅ R − kτ s 2.09⋅ 10 −4 ⋅ ln⎛⎜ 505 − 532 ⎞ ⎝ 520 − 532 ⎠ 3 τ = 3.88 × 10 s τ = 1.08⋅ hr Problem 4.10 Given: Data on heating and cooling a copper block Find: Final system temperature [Difficulty: 2] Solution: Basic equation Q W ∆E Assumptions: 1) Stationary system ΔE = ΔU 2) No work W = 0 3) Adiabatic Q = 0 Then for the system (water and copper) ∆U 0 or M copper ccopper Tcopper M w cw Tw M copper ccopper M w cw Tf where Tf is the final temperature of the water (w) and copper (copp er) The given data is M copper 5 kg ccopper 385 J kg K Tcopper ( 90 273 ) K Also, for the water ρ 999 kg 3 so Tf M w ρ V M copper ccopper Tcopper M w cw Tw Mcopper ccopper Mw cw Tf 291 K J kg K Tw ( 10 273 ) K m Solving Eq. 1 for Tf cw 4186 Tf 18.1 °C M w 4.00 kg V 4 L (1) Problem 4.11 [Difficulty: 2] Given: Data on heat loss from persons, and people-filled auditorium Find: Internal energy change of air and of system; air temperature rise Solution: Basic equation Q W ∆E Assumptions: 1) Stationary system ΔE =ΔU 2) No work W = 0 Then for the air For the air and people ∆U Q 85 W person 6000 people 15 min 60 s ∆U 459 MJ min ∆U Qsurroundings 0 The increase in air energy is equal and opposite to the loss in people energy For the air Hence From Table A.6 ∆U Q ∆T but for air (an ideal gas) Q Rair 286.9 ∆T 286.9 717.4 M ρ V with p V Rair T Rair Q T M cv ∆U M cv ∆T cv p V J kg K and cv 717.4 459 10 J ( 20 273 ) K This is the temperature change in 15 min. The rate of change is then kg K 2 1 6 J m 3 N 101 10 ∆T 15 min 6.09 1 3.5 10 K hr 5 1 3 m ∆T 1.521 K Problem 4.12 [Difficulty: 3] Given: Data on velocity field and control volume geometry Find: Several surface integrals Solution: dA1 = wdzˆj − wdykˆ dA1 = dzˆj − dykˆ dA2 = − wdykˆ dA2 = − dykˆ ( V = aˆj + bykˆ (a) (b) ) ( V = 10 ˆj + 5 ykˆ ( )( ) ) V ⋅ dA1 = 10 ˆj + 5 ykˆ ⋅ dzˆj − dykˆ = 10dz − 5 ydy 1 1 1 5 1 V ⋅ dA1 = 10dz − 5 ydy = 10 z 0 − y 2 = 7.5 A1 2 0 ∫ ∫ ∫ 0 0 ( )( ) (c) V ⋅ dA2 = 10 ˆj + 5 ykˆ ⋅ − dykˆ = −5 ydy (d) V V ⋅ dA2 = − 10 ˆj + 5 ykˆ 5 ydy (e) z ( ) ( ) ) ∫( ) 1 1 1 25 3 ˆ V V ⋅ dA2 = − 10 ˆj + 5 ykˆ 5 ydy = − 25 y 2 ˆj − y k = −25 ˆj − 8.33kˆ 0 A2 3 0 ∫ ( 0 Control volume y Problem 4.13 [Difficulty: 3] Given: Data on velocity field and control volume geometry Find: Volume flow rate and momentum flux z Solution: 3m First we define the area and velocity vectors y dA = dydziˆ + dydyxkˆ V = axiˆ + byˆj 4m 5m V = xiˆ + yˆj or x We will need the equation of the surface: z = 3 − 4 3 x or x = 4 − z 4 3 Then a) Volume flow rate )( ) 3 3 5 3 2 ⎞ 4 ⎞ ⎛ ⎛ Q = ∫ V ⋅ dA = ∫ xiˆ + yˆj ⋅ dydziˆ + dxdykˆ = ∫∫ xdydz = 5∫ ⎜ 4 − z ⎟dz = 5⎜ 4 z − z 2 ⎟ A A 3 ⎠ 3 ⎠0 ⎝ 0 0 0⎝ ( m3 m3 = 30 s s Q = (60 − 30 ) b) Momentum flux ( ) ( ) ρ ∫ V V ⋅ dA = ρ ∫ (xiˆ + yˆj )(xiˆ + yˆj )⋅ dydziˆ + dxdykˆ = ρ ∫ (xiˆ + yˆj )( xdydz ) A A A 2 4 ⎞ y2 ⎛ = ρ ∫ ∫ x dydziˆ + ρ ∫ ∫ xydydzˆj = 5∫ ⎜ 4 − z ⎟ dziˆ + 3 ⎠ 2 0 0 0 0 0⎝ 3 5 3 5 3 2 3 5 3 ⎛ 4 ⎞ ∫ ⎜⎝ 4 − 3 z ⎟⎠dzˆj 00 3 32 16 ⎞ 25 ⎛ 2 ⎞ 16 16 3 ⎞ ˆ 25 ⎛ ⎛ = 5∫ ⎜16 − z + z 2 ⎟dziˆ + ⎜ 4 z − z 2 ⎟ ˆj == 5⎜16 z − z 2 + z ⎟ i + (12 − 6) ˆj 3 9 2 3 3 27 2 ⎝ ⎠0 ⎝ ⎠0 ⎠ 0⎝ = 5(48 − 48 + 16)iˆ + 75 ˆj 3 Momentum flux = 80iˆ + 75 ˆj N Problem 4.14 [Difficulty: 3] Given: Data on velocity field and control volume geometry Find: Surface integrals z 4m 3m Solution: 5m First we define the area and velocity vectors dA = dydziˆ + dxdzˆj V = axiˆ + byˆj + ckˆ or V = 2 xiˆ + 2 yˆj + kˆ We will need the equation of the surface: y = y 3 2 x or x = y 3 2 x Then ∫V ⋅ dA = ∫ (− axiˆ + byˆj + ckˆ )⋅ (dydziˆ − dxdzˆj ) A A 2 3 2 2 2 3 2 3 2 2 3 1 3 = ∫∫ − axdydz − ∫∫ bydxdz = −a ∫ dz ∫ ydy − b ∫ dz ∫ xdx = − 2a y 2 − 2b x 2 3 2 3 0 4 0 0 0 0 0 0 0 0 Q = (− 6a − 6b ) = −24 m s ∫ ( A 3 2 3 x or x = y , and also dy = dx and a = b 2 3 2 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ − axi + byj + ck − axi + byj + ck ⋅ dydzi − dxdzj ) ( )( )( = ∫ (− axiˆ + byˆj + ckˆ )(− axdydz − bydxdz ) A 0 3 We will again need the equation of the surface: y = V V ⋅ dA = ∫ 2 ) A 3 3 3 ⎞ ⎛ ⎞⎛ = ∫ ⎜ − axiˆ + axˆj + ckˆ ⎟⎜ − ax dxdz − a xdxdz ⎟ A 2 2 2 ⎠ ⎝ ⎠⎝ 3 ⎞ ⎛ = ∫ ⎜ − axiˆ + axˆj + ckˆ ⎟(− 3axdxdz ) A 2 ⎠ ⎝ 2 2 2 2 2 2 9 = 3∫ ∫ a x dxdziˆ − ∫ ∫ a 2 x 2 dxdzˆj − 3∫ ∫ acxdxdzkˆ 200 0 0 0 0 2 2 3 ⎛ ⎞ ⎟iˆ − (9)⎜ a 2 x ⎜ ⎟ 3 0⎠ ⎝ m4 = 64iˆ − 96 ˆj − 60kˆ s2 ⎛ x3 = (6)⎜ a 2 ⎜ 3 ⎝ 2 2 ⎛ ⎞ ⎟ ˆj − (6)⎜ ac x ⎜ ⎟ 2 0⎠ ⎝ 2 ⎞ ⎟ = 16a 2 iˆ − 24a 2 ˆj − 12ackˆ ⎟ 0⎠ 2 Problem 4.15 [Difficulty: 2] Given: Control Volume with linear velocity distribution Find: Volume flow rate and momentum flux Solution: Apply the expressions for volume and momentum flux Governing equations: Assumption: ∫ ( Q = V ⋅ dA mf = ρ V V ⋅ dA A A ∫ ) (1) Incompressible flow For a linear velocity profile V V = yiˆ h dA = − w dyiˆ and also For the volume flow rate: h Q= ∫ ( ) Vw V ˆ i ⋅ − w dyiˆ = − h h y =0 h ∫ Vw y 2 y dy = − h 2 y =0 h 0 1 Q = − Vhw 2 The momentum flux is h mf = h V ˆ ⎛ V 2w y3 Vw V 2w ⎞ i ⋅⎜− ρ ydy ⎟ = − ρ 2 iˆ y 2 dy = − ρ 2 iˆ 3 h ⎝ h h h ⎠ y =0 y =0 ∫ 1 mf = − ρV 2 whiˆ 3 ∫ h 0 Problem 4.16 Given: Control Volume with linear velocity distribution Find: Kinetic energy flux Solution: Apply the expression for kinetic energy flux V2 ρV ⋅ dA A 2 ∫ Governing equation: kef = Assumption: (1) Incompressible flow For a linear velocity profile [Difficulty: 2] V V = yiˆ h V (y) = V y h dA = − w dyiˆ and also The kinetic energy flux is h 2 V 3w Vw 1 ⎛V ⎞ ⎛ ⎞ ydy ⎟ = − ρ kef = ⎜ y⎟ ⎜− ρ h 2⎝h ⎠ ⎝ 2h 3 ⎠ y =0 ∫ 1 kef = − ρV 3 wh 8 h ∫ V 3w y 4 y dy = − ρ 2h 3 4 y =0 h 3 0 Problem 4.17 [Difficulty: 2] Given: Control Volume with parabolic velocity distribution Find: Volume flow rate and momentum flux Solution: Apply the expressions for volume and momentum flux Governing equations: Assumption: ∫ ( Q = V ⋅ dA mf = ρ V V ⋅ dA A A ∫ ) (1) Incompressible flow For a linear velocity profile ⎡ ⎛ r ⎞2 ⎤ V = uiˆ = umax ⎢1 − ⎜ ⎟ ⎥iˆ and also ⎢⎣ ⎝ R ⎠ ⎥⎦ dA = 2πrdriˆ For the volume flow rate: h R ⎡ ⎛ r ⎞2 ⎤ ⎡ ⎡ r2 ⎡ R2 R2 ⎤ r3 ⎤ r4 ⎤ ˆ ˆ 2 u Q = umax ⎢1 − ⎜ ⎟ ⎥i ⋅ 2πrdri = 2πumax = − π ⎢ r − 2 ⎥ dy = 2πumax ⎢ − ⎥ ⎢ ⎥ max 2 4 ⎦⎥ R ⎦⎥ ⎢ ⎢⎣ ⎝ R ⎠ ⎥⎦ ⎣⎢ 2 4 R ⎦⎥ 0 ⎣⎢ 2 r =0 y =0 ⎣ R ( ∫ Q= ) ∫ 1 πumax R 2 2 The momentum flux is R ⎫ ⎫⎪ ⎡ ⎛ r ⎞ 2 ⎤ ⎧⎪ ⎡ ⎛ r ⎞2 ⎤ ⎡ ⎛ r ⎞ 2 ⎤ ⎧⎪ ⎡ r3 ⎤ ⎪ ˆ ˆ ˆ ˆ mf = umax ⎢1 − ⎜ ⎟ ⎥i ⎨umax ⎢1 − ⎜ ⎟ ⎥i ⋅ 2πrdri ⎬ = umax ⎢1 − ⎜ ⎟ ⎥i ⎨2πumax ⎢ r − 2 ⎥ dr ⎬ R ⎥⎦ ⎪ ⎢ ⎢⎣ ⎝ R ⎠ ⎥⎦ ⎪⎩ ⎢⎣ ⎝ R ⎠ ⎥⎦ ⎢⎣ ⎝ R ⎠ ⎥⎦ ⎪ ⎪⎭ r =0 y =0 ⎣ ⎭ ⎩ R ( ∫ R ⎡ 2r 3 r 5 ⎤ ⎢ r − 2 + 4 ⎥ dr R R ⎥⎦ ⎢ y =0 ⎣ ∫ = 2 ˆ 2πumax i = 2 ˆ⎡ r 2πumax i⎢ h r6 ⎤ − + ⎥ 2 6 R 4 ⎥⎦ 0 ⎢⎣ 2 4 R 2 r4 2 R2 R2 ⎤ 2 ˆ⎡ R = 2πumax − + i⎢ ⎥ 4 6 ⎦⎥ ⎣⎢ 2 1 2 mf = πumax R 2iˆ 3 ) ∫ Problem 4.18 Given: Control Volume with parabolic velocity distribution Find: Kinetic energy flux Solution: Apply the expressions for kinetic energy flux V2 ρV ⋅ dA A 2 ∫ Governing equation: kef = Assumption: (1) Incompressible flow For a linear velocity profile [Difficulty: 2] ⎡ ⎛ r ⎞2 ⎤ V = uiˆ = umax ⎢1 − ⎜ ⎟ ⎥iˆ ⎣⎢ ⎝ R ⎠ ⎦⎥ ⎡ ⎛ r ⎞2 ⎤ V = u = umax ⎢1 − ⎜ ⎟ ⎥ ⎣⎢ ⎝ R ⎠ ⎦⎥ For the volume flow rate: 2 ⎫⎪ ⎡ ⎛ r ⎞ 2 ⎤ ⎫⎪ ⎧⎪ ⎡ ⎛ r ⎞2 ⎤ 1 ⎧⎪ kef = ⎨umax ⎢1 − ⎜ ⎟ ⎥ ⎬ ρ ⎨umax ⎢1 − ⎜ ⎟ ⎥iˆ ⋅ 2πrdriˆ ⎬ 2⎪ ⎢⎣ ⎝ R ⎠ ⎥⎦ ⎪⎭ ⎩⎪ ⎢⎣ ⎝ R ⎠ ⎥⎦ ⎪⎭ r =0 ⎩ R ( ∫ 2 4 ⎡ ⎛ r ⎞ 2 ⎤ ⎫⎪ 1 2 ⎡ ⎛ r ⎞ ⎛ r ⎞ ⎤ ⎧⎪ umax ⎢1 − 2⎜ ⎟ + ⎜ ⎟ ⎥ ρ ⎨2πumax ⎢1 − ⎜ ⎟ ⎥ rdr ⎬ 2 ⎝ R ⎠ ⎝ R ⎠ ⎥⎦ ⎩⎪ ⎢⎣ ⎢⎣ ⎝ R ⎠ ⎥⎦ ⎪⎭ r =0 R = ∫ R = ⎢⎣ ⎛r⎞ ⎝R⎠ ⎡ r3 ⎢⎣ R2 2 ⎛r⎞ ⎝R⎠ 4 6 ⎛r⎞ ⎤ ⎝ R ⎠ ⎥⎦ ∫ ∫ 3 πρumax ⎢r − 3 r =0 R = ⎡ 3 πρumax ⎢1 − 3⎜ ⎟ + 3⎜ ⎟ − ⎜ ⎟ ⎥ rdr r =0 +3 r5 R4 − r7 ⎤ ⎥ dr R 6 ⎥⎦ h 2 r6 r8 ⎤ 3r 4 3 ⎡r = πρumax + − 6⎥ ⎢ − 2 4 2R 8R ⎦⎥ 0 ⎣⎢ 2 4 R 1 3 kef = πρumax R2 8 ) and also dA = 2πrdriˆ Problem 4.19 Given: Data on flow through nozzles Find: Exit velocity in each jet; velocity in pipe [Difficulty: 1] Solution: Basic equation →→ ( ∑ V⋅A) = 0 CS Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow The given data is Area of each nozzle Q = 2.2⋅ gpm A = π 4 ⋅d 2 d = 1 32 ⋅ in n = 50 −4 A = 7.67 × 10 2 in Area of the pipe π 2 Apipe = ⋅ D 4 Apipe = 0.442in Total area of nozzles Atotal = n ⋅ A Atotal = 0.0383in Then for the pipe flow 2 2 V = The jet speeds are then Q Atotal V = 18.4 ft →→ ( ∑ V⋅A) = −Vpipe⋅Apipe + n⋅V⋅A = 0 n⋅ A d Vpipe = V⋅ = V⋅ n ⋅ ⎛⎜ ⎞ Apipe ⎝ D⎠ 2 ⎛ ⎜ ft Vpipe = 18.4⋅ × 50 × ⎜ s ⎜ ⎝ 1 ⎞ 32 ⎟ 3 4 ⎠ 3 (Note that gal = 231 in ) s CS Hence D = (Number of nozzles) 2 ft Vpipe = 1.60⋅ s 3 4 ⋅ in Problem 4.20 Given: Data on flow through nozzles Find: Average velocity in head feeder; flow rate [Difficulty: 1] Solution: Basic equation →→ ( ∑ V⋅A) = 0 CS Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow Then for the nozzle flow →→ ( ∑ V⋅A) = −Vfeeder⋅Afeeder + 10⋅Vnozzle⋅Anozzle = 0 CS Hence ⎛ Dnozzle ⎞ Vfeeder = Vnozzle⋅ = Vnozzle⋅ 10⋅ ⎜ Afeeder ⎝ Dfeeder ⎠ 10⋅ Anozzle ⎛ 1⎞ ⎜ 8 ft Vfeeder = 10⋅ × 10 × ⎜ s ⎝1⎠ 2 2 ft Vfeeder = 1.56⋅ s 2 The flow rate is Q = Vfeeder⋅ Afeeder = Vfeeder⋅ Q = 1.56⋅ ft s × π 4 × ⎛⎜ 1 ⋅ in × ⎝ π⋅ Dfeeder 1 ⋅ ft 4 2 ⎞ × 7.48⋅ gal × 60⋅ s 12⋅ in ⎠ 1 ⋅ min 3 1 ⋅ ft Q = 3.82⋅ gpm Problem 4.21 [Difficulty: 3] Given: Data on flow into and out of tank Find: Time at which exit pump is switched on; time at which drain is opened; flow rate into drain Solution: ∂ Basic equation ∂t M CV + →→ ( ρ⋅ V⋅ A ) = 0 ∑ CS Assumptions: 1) Uniform flow 2) Incompressible flow ∂ After inlet pump is on ∂t M CV + →→ ∂ ( ρ⋅ V⋅ A ) = M tank − ρ⋅ Vin⋅ Ain = 0 ∑ ∂t CS ⎛ Din ⎞ = Vin⋅ = Vin⋅ ⎜ Atank dt ⎝ Dtank ⎠ Ain dh Hence the time to reach hexit = 0.7 m is texit = h exit dh ∂ dh M tank = ρ⋅ Atank ⋅ = ρ⋅ Vin⋅ Ain dt ∂t 2 where h is the level of water in the tank ⎛ Dtank ⎞ = ⋅⎜ Vin ⎝ Din ⎠ h exit 2 texit = 0.7⋅ m × 1 s ⋅ × 5 m ⎛ 3⋅ m ⎞ ⎜ 0.1⋅ m ⎝ ⎠ 2 texit = 126 s dt ∂ After exit pump is on ∂t →→ dh ∂ ( = Vin⋅ Ain − Vexit ⋅ Aexit ρ⋅ V⋅ A) = M tank − ρ⋅ Vin⋅ Ain + ρ⋅ Vexit ⋅ Aexit = 0 Atank⋅ ∑ dt ∂t M CV + CS 2 ⎛ Din ⎞ ⎛ Dexit ⎞ = Vin⋅ − Vexit ⋅ = Vin⋅ ⎜ − Vexit ⋅ ⎜ Atank Atank dt ⎝ Dtank ⎠ ⎝ Dtank ⎠ dh Ain Hence the time to reach hdrain = 2 m is Aexit tdrain = texit + (hdrain − hexit) dh = (hdrain − hexit) 2 ⎛ Din ⎞ ⎛ Dexit ⎞ Vin⋅ ⎜ − Vexit ⋅ ⎜ ⎝ Dtank ⎠ ⎝ Dtank ⎠ dt tdrain = 126 ⋅ s + ( 2 − 0.7) ⋅ m × 2 2 1 2 0.1⋅ m ⎞ 0.08⋅ m ⎞ m 5 ⋅ × ⎛⎜ − 3 ⋅ × ⎛⎜ s s ⎝ 3⋅ m ⎠ ⎝ 3⋅ m ⎠ m 2 tdrain = 506 s The flow rate into the drain is equal to the net inflow (the level in the tank is now constant) 2 Qdrain = Vin⋅ π⋅ Din 4 − Vexit ⋅ π⋅ Dexit 4 2 m π m π 2 2 Qdrain = 5 ⋅ × × ( 0.1⋅ m) − 3 ⋅ × × ( 0.08⋅ m) s 4 s 4 3 m Qdrain = 0.0242 s Problem 4.22 [Difficulty: 1] Given: Data on wind tunnel geometry Find: Average speeds in wind tunnel; diameter of section 3 Solution: Basic equation Q V A Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow 3 Given data: Q 15 m D1 1.5 m s 2 Between sections 1 and 2 Hence Q V1 A1 V1 V1 π 4 For section 3 we can use V1 Q D1 π D1 4 2 2 V3 π D1 2 V2 A2 V2 4 m V1 8.49 s π D2 4 V2 π 4 π D3 4 m V3 75 s D2 1 m 2 or Q D2 2 V1 D3 D1 V3 m V2 19.1 s D3 0.505 m Problem 4.23 Difficulty: 4] Given: Data on flow into and out of cooling tower Find: Volume and mass flow rate of cool water; mass flow rate of moist and dry air Solution: →→ ( ρ⋅ V⋅ A) = 0 ∑ Basic equation Q = V⋅ A and at each inlet/exit CS Assumptions: 1) Uniform flow 2) Incompressible flow 5 lb mwarm = 2.5⋅ 10 ⋅ hr Given data: At the cool water exit The mass flow rate is Qcool = V⋅ A mcool = ρ⋅ Qcool D = 6 ⋅ in V = 5⋅ mcool = 1.94⋅ ft 3 × 0.982 ⋅ ft lb ρmoist = 0.065 ⋅ 3 ft s 3 ft π 2 Qcool = 5 ⋅ × × ( 0.5⋅ ft) s 4 slug ft 3 s ft Qcool = 0.982 ⋅ s Qcool = 441 ⋅ gpm slug mcool = 1.91⋅ s 5 lb mcool = 2.21 × 10 ⋅ hr NOTE: Software does not allow dots over terms, so m represents mass flow rate, not mass! For the water flow we need →→ ( ρ⋅ V⋅ A) = 0 ∑ to balance the water flow CS We have −mwarm + mcool + mv = 0 mv = mwarm − mcool This is the mass flow rate of water vapor. To obtain air flow rates, from psychrometrics where x is the relative humidity. It is also known (try Googling "density of moist air") that lb mv = 29341 ⋅ hr x= mv mair ρmoist ρdry We are given lb ρmoist = 0.065 ⋅ 3 ft = 1+x 1 + x⋅ RH2O Rair For dry air we could use the ideal gas equation ρdry = 0.002377⋅ slug ft 3 p ρdry = R⋅ T but here we use atmospheric air density (Table A.3) ρdry = 0.002377⋅ slug ft 3 × 32.2⋅ lb slug lb ρdry = 0.0765⋅ 3 ft Note that moist air is less dense than dry air! Hence 0.065 0.0765 x = 1+x = 1 + x⋅ 0.065 ⋅ Hence mv mair 53.33 0.0765 − 0.065 =x 85.78 53.33 using data from Table A.6 85.78 x = 0.410 − .0765 leads to Finally, the mass flow rate of moist air is mv mair = x lb 1 mair = 29341 ⋅ × hr 0.41 mmoist = mv + mair 5 lb mmoist = 1.01 × 10 ⋅ hr lb mair = 71563 ⋅ hr Problem 4.24 Given: Data on flow through box Find: Velocity at station 3 [Difficulty: 1] Solution: Basic equation →→ ( ∑ V⋅A) = 0 CS Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow Then for the box ∑( ) →→ V⋅ A = −V1⋅ A1 + V2⋅ A2 + V3⋅ A3 = 0 CS Note that the vectors indicate that flow is in at location 1 and out at location 2; we assume outflow at location 3 Hence A1 A2 V3 = V1 ⋅ − V2 ⋅ A3 A3 ft 0.5 ft 0.1 V3 = 10⋅ × − 20⋅ × s 0.6 s 0.6 Based on geometry Vx = V3 ⋅ sin( 60⋅ deg) ft Vx = 4.33⋅ s Vy = −V3 ⋅ cos( 60⋅ deg) ft Vy = −2.5⋅ s → ⎯ ft ft V3 = ⎛⎜ 4.33⋅ , −2.5⋅ ⎞ s s⎠ ⎝ ft V3 = 5 ⋅ s Problem 4.25 Given: Data on flow through device Find: Volume flow rate at port 3 [Difficulty: 1] Solution: Basic equation →→ ( ∑ V⋅A) = 0 CS Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow Then for the box ∑( ) →→ V⋅ A = −V1⋅ A1 + V2⋅ A2 + V3⋅ A3 = −V1⋅ A1 + V2⋅ A2 + Q3 CS Note we assume outflow at port 3 Hence Q3 = V1 ⋅ A1 − V2 ⋅ A2 The negative sign indicates the flow at port 3 is inwards. m m 2 2 Q3 = 3 ⋅ × 0.1⋅ m − 10⋅ × 0.05⋅ m s s Flow rate at port 3 is 0.2 m3/s inwards 3 m Q3 = −0.2⋅ s Problem 4.26 Given: Water needs of farmer Find: Number of supply pipes needed [Difficulty: 1] Solution: Basic equation Q V A Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow The given data is: Then A 150 m 400 m Q If n is the number of pipes The farmer needs 5 pipes. A h t 4 2 A 6 10 m h 7.5 cm t 1 hr or n D 37.5 cm 3 Q 1.25 Q V π 4 m s 2 D n 4 Q 2 π V D n 4.527 V 2.5 m s Problem 4.27 Given: Data on filling of glass carboy Find: Time to fill [Difficulty: 1] Solution: We can treat this as an unsteady problem if we choose the CS as the entire carboy Basic equation ∂ ∂t →→ ( ρ⋅ V⋅ A) = 0 ∑ M CV + CS Assumptions: 1) Incompressible flow 2) Uniform flow Given data: Q = 3 ⋅ gpm Hence ∂ ∂t M CV = ρ⋅ A⋅ D = 15⋅ in dh dt = ρ⋅ A⋅ h τ h = 2 ⋅ ft →→ ( ρ⋅ V⋅ A) = ρ⋅ Q ∑ =− CS where Q is the fill rate, A is the carboy cross-section area, dh/dt is the rate of rise in the carboy, and τ is the fill time π Hence τ = 4 2 ⋅D ⋅h Q τ = 6.12⋅ min Problem 4.28 Given: Data on filling of a sink Find: Time to half fill; rate at which level drops [Difficulty: 1] Solution: This is an unsteady problem if we choose the CS as the entire sink Basic equation ∂ ∂t →→ ( ρ⋅ V⋅ A) = 0 ∑ M CV + CS Assumptions: 1) Incompressible flow Given data: mrate = 4 ⋅ gpm Hence ∂ To half fill: Then, using Eq 1 ∂t w = 18⋅ in d = 12⋅ in →→ ( ρ⋅ V⋅ A) = Inflow − Outflow ∑ M CV = − V Q = 4 ⋅ gpm Qdrain = 1 ⋅ gpm (1) CS V = τ L = 2 ⋅ ft 1 2 ⋅ L⋅ w⋅ d =Q After the drain opens, Eq. 1 becomes V = 1.5 ft τ = dV dt Qdrain Vlevel = − L⋅ w V Q 3 V = 11.2 gal τ = 168 s τ = 2.81 min = L⋅ w⋅ Vlevel = −Qdrain where V level is the speed of water level drop − 4 ft Vlevel = −7.43 × 10 s in Vlevel = −0.535 min Problem 4.29 [Difficulty: 1] Given: Air flow system Find: Flow rate and velocity into each room; narrowest supply duct Solution: Basic equation Q V A Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow The given data is: L Qperson 8 s n rooms 6 n students 20 h 200 mm w 500 mm m Vmax 1.75 s Then for each room Qroom n students Qperson L Qroom 160 s and Vroom Qroom m Vroom 1.6 s For the supply duct Q n rooms Qroom and w h Q 960 L s 3 m Qroom 0.16 s 3 Q 0.96 m s Q Vmax A Vmax w h where w and h are now the supply duct dimensions w Q Vmax h w 1.097 m h 500 mm Problem 4.30 Given: Data on filling of a basement during a storm Find: Flow rate of storm into basement [Difficulty: 1] Solution: This is an unsteady problem if we choose the CS as the entire basement Basic equation ∂ ∂t →→ ( ρ⋅ V⋅ A) = 0 ∑ M CV + CS Assumptions: 1) Incompressible flow Given data: Hence or dh Qpump = 27.5⋅ gpm ∂ ∂t M CV = ρ⋅ A⋅ dh dt dt = 4⋅ in hr A = 30⋅ ft⋅ 20⋅ ft →→ ( ρ⋅ V⋅ A) = ρ⋅ Qstorm − ρ⋅ Qpump ∑ =− CS dh Qstorm = Qpump − A⋅ dt gal Qstorm = 27.5⋅ − 30⋅ ft × 20⋅ ft × min Qstorm = 2.57⋅ gpm ⎛ 4 ⋅ ft ⎞ × 7.48⋅ gal × 1 ⋅ hr ⎜ 12 hr 60⋅ min 3 ⎝ ⎠ ft where A is the basement area and dh/dt is the rate at which the height of water in the basement changes. Data on gals from Table G.2 Problem 4.31 Given: Data on compressible flow Find: Downstream density [Difficulty: 1] Solution: Basic equation →→ ( ρ⋅ V⋅ A ) = 0 ∑ CS Assumptions: 1) Steady flow 2) Uniform flow Then for the box →→ ( ρ⋅ V⋅ A ) = −ρu ⋅ Vu⋅ Au + ρd ⋅ Vd ⋅ Ad = 0 ∑ CS Hence Vd ⋅ Ad ρu = ρd ⋅ Vu⋅ Au 1000⋅ m 2 kg s 0.1⋅ m ρu = 1⋅ ⋅ ⋅ 3 m 2 m 1500⋅ 0.25⋅ m s kg ρu = 0.267 3 m Problem 4.32 [Difficulty: 2] Given: Data on flow through device Find: Velocity V3; plot V3 against time; find when V3 is zero; total mean flow Solution: Governing equation: − V3 = V1⋅ A1 + V2⋅ A2 A3 − V3 = 6.67⋅ e →→ V⋅ A = 0 ∑ −V1⋅ A1 − V2⋅ A2 + V3⋅ A3 = 0 Applying to the device (assuming V3 is out) The velocity at A3 is ⌠ →→ ⎮ ⎮ V dA = ⌡ For incompressible flow (Eq. 4.13) and uniform flow 10⋅ e t 2 m ⋅ = s 2 × 0.1⋅ m + 2⋅ cos ( 2⋅ π⋅ t) ⋅ m s 2 × 0.2⋅ m 2 0.15⋅ m t 2 + 2.67⋅ cos( 2 ⋅ π⋅ t) The total mean volumetric flow at A3 is ∞ ⌠ ⎮ ∞ ⎮ ⌠ Q = ⎮ V3 ⋅ A3 dt = ⎮ ⌡ ⌡ 0 ⎛ − ⎜ ⎝ 6.67⋅ e ⎛ − ⎜ Q = lim ⎜ −2 ⋅ e t→∞⎝ 1 ⎞ t 2 + 2.67⋅ cos( 2 ⋅ π⋅ t) ⎠ ⋅ 0.15 dt⋅ ⎜⎛ ⎝s 0 2⎞ ⋅m ⎠ ⎞ t 2 m + 5⋅ π ⋅ sin( 2 ⋅ π⋅ t) 3 ⎠ − ( −2 ) = 2 ⋅ m The time at which V3 first is zero, and the plot of V3 is shown in the corresponding Excel workbook 3 Q = 2⋅ m t = 2.39⋅ s t (s) V 3 (m/s) 9.33 8.50 6.86 4.91 3.30 2.53 2.78 3.87 5.29 6.41 6.71 6.00 4.48 2.66 1.15 0.48 0.84 2.03 3.53 4.74 5.12 4.49 3.04 1.29 -0.15 -0.76 Exit Velocity vs Time 10 8 V 3 (m/s) 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.10 2.20 2.30 2.40 2.50 6 4 2 0 0.0 0.5 1.0 1.5 2.0 -2 t (s) The time at which V 3 first becomes zero can be found using Goal Seek t (s) V 3 (m/s) 2.39 0.00 2.5 Problem 4.33 Given: Data on flow down an inclined plane Find: Find u max [Difficulty: 2] Solution: Basic equation ⌠ mflow = ⎮ ρu dA ⌡ Assumptions: 1) Steady flow 2) Incompressible flow Evaluating at 1 and 2 h h 0 0 ⌠ ⌠ 2 2 ⎮ ρ ⋅ g ⋅ sin ( θ) ⋅ w ⎮ y ⎞ ρ⋅ g ⋅ sin ( θ) ⎛ mflow = ⎮ ρ⋅ ⋅⎜h ⋅y − ⋅ w dy = ⋅⎮ 2 ⎠ μ μ ⎮ ⎝ ⎮ ⌡ ⌡ 2 mflow = ρ ⋅ g ⋅ sin ( θ) ⋅ w μ 2 Hence mflow = ⎛ h3 ⋅⎜ ⎝ 2 ρ ⋅ g ⋅ sin( θ) ⋅ w⋅ h 3⋅ μ 3 − h 3⎞ 6 ⎠ 2 ⎛ y ⎞ ⎜h ⋅y − dy 2 ⎠ ⎝ Problem 4.34 [Difficulty: 2] y 2h Given: Data on flow at inlet and outlet of channel Find: Find u max x CS Solution: Basic equation ∫ ρ V ⋅ dA = 0 CS Assumptions: 1) Steady flow 2) Incompressible flow Evaluating at 1 and 2 ⌠ −ρ⋅ U⋅ 2 ⋅ h ⋅ w + ⎮ ⌡ h ρ⋅ u ( y ) dy = 0 −h ⎡ u max⋅ ⎢[ h − ( −h ) ] − ⎢ ⎣ Hence u max = 3 2 ⋅U = 3 2 ⌠ ⎮ ⎮ ⎮ ⌡ h −h ⎡ 3 ⎛ 3 ⎞⎤⎤ ⎢ h − ⎜ − h ⎥⎥ = 2⋅ h ⋅ U ⎢ 3 ⋅ h2 ⎜ 3 ⋅ h2 ⎥⎥ ⎣ ⎝ ⎠⎦⎦ × 2.5⋅ m s ⎡ u max⋅ ⎢1 − ⎣ 2 ⎛ y ⎞ ⎥⎤ dy = 2 ⋅ h⋅ U ⎜h ⎝ ⎠⎦ 4 u max⋅ ⋅ h = 2 ⋅ h ⋅ U 3 u max = 3.75⋅ m s Problem 4.35 Given: Data on flow at inlet and outlet of pipe Find: Find U [Difficulty: 2] Solution: Basic equation ∫ ρ V ⋅ dA = 0 CS Assumptions: 1) Steady flow 2) Incompressible flow Evaluating at inlet and exit 2 ⌠ −ρ⋅ U⋅ π⋅ R + ⎮ ⌡ R ρ⋅ u ( r) ⋅ 2 ⋅ π⋅ r dr = 0 0 u max⋅ ⎜⎛ R − 2 ⎝ Hence U= 1 2 × 3⋅ 1 2 m s ⋅R ⌠ ⎮ ⎮ ⎮ ⌡ R ⎡ u max⋅ ⎢1 − ⎣ 0 2⎞ ⎠ 2 = R ⋅U U= 1 2 ⋅ u max U = 1.5⋅ m s 2 ⎛ r ⎞ ⎤⎥ ⋅ 2⋅ r dr = R2⋅ U ⎜R ⎝ ⎠⎦ Problem 4.36 Given: Data on flow at inlet and outlet of channel Find: Find u max [Difficulty: 2] Solution: ∫ ρ V ⋅ dA = 0 Basic equation CS Assumptions: 1) Steady flow 2) Incompressible flow Evaluating at 1 and 2 ⌠ −ρ⋅ V1 ⋅ H⋅ w + ⎮ ⌡ H ρ⋅ V2 ( y ) ⋅ w dy = 0 −H or ⌠ V1 ⋅ H = ⎮ ⎮ ⌡ H −H Hence π Vm = ⋅ V1 4 π⋅ y ⎞ dy = Vm⋅ cos⎛⎜ ⎝ 2⋅ H ⎠ ⌠ 2⋅ ⎮ ⎮ ⌡ H 0 4 ⋅ H⋅ Vm 2⋅ H ⎛ ⎛ π ⎞ π⋅ y ⎞ dy = 2 ⋅ Vm⋅ ⋅ ⎜ sin⎜ − sin( 0 ) ⎞ = Vm⋅ cos⎛⎜ π ⎝ ⎝2⎠ π ⎝ 2⋅ H ⎠ ⎠ Problem 4.37 [Difficulty: 3] Given: Velocity distribution in annulus Find: Volume flow rate; average velocity; maximum velocity; plot velocity distribution Solution: Governing equation For the flow rate (Eq. 4.14a) and average velocity (Eq. 4.14b) The given data is Ro = 5 ⋅ mm ∆p Ri = 1 ⋅ mm L = −10⋅ ⌠ →→ ⎮ Q = ⎮ V dA ⌡ kPa μ = 0.1⋅ m Q Vav = A N⋅ s (From Fig. A.2) 2 m 2 2 ⎛ Ro − Ri ⎛ Ro ⎞ ⎞ −∆p ⎜ 2 2 u ( r) = ⋅ ln⎜ ⋅ R −r + 4 ⋅ μ⋅ L ⎜ o ⎝ r ⎠⎟ ⎛ Ri ⎞ ln⎜ ⎜ ⎝ ⎝ Ro ⎠ ⎠ R The flow rate is ⌠ o Q = ⎮ u ( r) ⋅ 2 ⋅ π⋅ r dr ⌡R ⎡ ⎛ R 2 − R 2⎞ ⎤ i ⎠ ⎛ 2 2 2⎞ ⎢ ⎝ o 2⎞⎥ ⎛ Q = ⋅ R − Ri ⋅ ⎠ ⎢ ⎛ R ⎞ − ⎝ Ri + Ro ⎠⎥ 8 ⋅ μ⋅ L ⎝ o o ⎢ ln⎜ ⎥ Ri ⎣ ⎝ ⎠ ⎦ i ∆p⋅ π Considerable mathematical manipulation leads to Substituting values Q = 3 ⋅ ( −10⋅ 10 ) ⋅ 8 π Q = 1.045 × 10 The average velocity is Q Vav = = A N 2 2 ( 0.1⋅ N⋅ s 2 ⋅ m ⋅m m 3 −5m Q = 10.45 ⋅ s 2 2 2⎡ 2 2 2 ⎞ ⋅ ⎢ 5 − 1 − 5 2 + 12 ⎥⎤ ⋅ ⎛ m ⎞ ⎜ ⎥ ⎝ 1000 ⎠ ⎝ 1000 ⎠ ⎢ ln⎛ 5 ⎞ ⎢⎣ ⎜⎝ 1 ⎠ ⎥⎦ 2⎞ ⎠ ) mL s 1 −5 m Vav = × 1.045 × 10 ⋅ × π s 1 2 5 −1 2 ⋅ ⎛⎜ 1000 ⎞ 2 ⎝ m ⎠ ⎛ ⎡ ⎡ Ro − Ri ⎛ Ro ⎞ ⎞⎥⎤ ∆p ⎢ 2 d ⎢ −∆p ⎜ 2 ⋅ ln⎜ =0= ⋅ Ro − r + =− ⋅ −2 ⋅ r − dr 4 ⋅ μ⋅ L ⎢ dx ⎢ 4 ⋅ μ⋅ L ⎜ ⎝ r ⎠ ⎟⎥ ⎛ Ri ⎞ ln⎜ ⎢ ⎜ ⎥ ⎢ ⎣ ⎝ ⎝ Ro ⎠ ⎠⎦ ⎣ 2 The maximum velocity occurs when ( m 3 Q π⋅ ⎛ R o − R i ⎝ ) ⋅ 5 − 1 ⋅ ⎛⎜ du 2 m Vav = 0.139 s ⎛ R 2 − R 2⎞ ⎤ i ⎠⎥ ⎝ o ⎛ Ri ⎞ ⎥ ln⎜ ⋅r ⎥ Ro ⎝ ⎠ ⎦ 2 r = Then Ri − Ro 2 ⎛ Ri ⎞ 2 ⋅ ln⎜ ⎝ Ro ⎠ r = 2.73⋅ mm Substituting in u(r) u max = u ( 2.73⋅ mm) = 0.213 ⋅ The maximum velocity using Solver instead, and the plot, are also shown in an Excel workbook Ro = Ri= ¬p /L = 5 1 -10 mm mm kPa/m ¬◊ϕ 0.1 N.s/m 2 r (mm) u (m/s) 0.000 0.069 0.120 0.157 0.183 0.201 0.210 0.213 0.210 0.200 0.186 0.166 0.142 0.113 0.079 0.042 0.000 Annular Velocity Distribution 6 5 r (mm) 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00 3.25 3.50 3.75 4.00 4.25 4.50 4.75 5.00 4 3 2 1 0 0.00 The maximum velocity can be found using Solver r (mm) u (m/s) 2.73 0.213 0.05 0.10 0.15 u (m/s) 0.20 0.25 m s Problem 4.38 [Difficulty: 2] Given: Data on flow at inlet and outlet of a reducing elbow Find: Find the maximum velcoity at section 1 Solution: ∫ ρ V ⋅ dA = 0 Basic equation CS Assumptions: 1) Steady flow 2) Incompressible flow Evaluating at 1, 2 and 3 h ⌠ 1 −⎮ V1 ( y ) ⋅ w dy + V2 ⋅ w⋅ h 2 + V3 ⋅ w⋅ h 3 = 0 ⌡ 0 or Hence h 2 V1max ⌠ 1 V1max h 1 ⋅ = V2 ⋅ h 2 + V3 ⋅ h 3 ⋅ ⎮ y dy = 2 h 1 ⌡0 h1 ( 2 V1max = ⋅ V3 ⋅ h 3 + V2 ⋅ h 2 h1 ) 2 ⎛ m m V1max = ⋅ ⎜ 5 ⋅ × 0.15⋅ m + 1 ⋅ × 0.2⋅ m⎞ 0.5⋅ m ⎝ s s ⎠ m V1max = 3.80 s Problem 4.39 Given: Data on flow at inlet and outlet of channel Find: Find u max Solution: Basic equation [Difficulty: 2] ∫ ρ V ⋅ dA = 0 CS Assumptions: 1) Steady flow 2) Incompressible flow h Evaluating at inlet and exit ⌠ −U⋅ w⋅ h + ⎮ Vexit ( x ) ⋅ w dx = 0 ⌡ 0 ( ) Here we have x Vexit = Vmax − Vmax − Vmin ⋅ h Hence x Vexit = 2 ⋅ Vmin − Vmin⋅ h h ⌠ ⌠ ⎮ Vexit ( x ) ⋅ w dx = ⎮ ⎮ ⌡ ⌡ 0 h But we also have 2 ⎛ 2 ⋅ V − V ⋅ x ⎞ ⋅ w dx = ⎛⎜ 2⋅ V ⋅ h − V ⋅ h ⎞ ⋅ w = 3 ⋅ V ⋅ h⋅ w ⎜ min min min 2 ⋅ h min h 2 min ⎝ ⎠ ⎝ ⎠ 0 3 Hence 2 ⋅ Vmin⋅ h ⋅ w = U⋅ w⋅ h 2 m Vmin = × 7.5⋅ 3 s Vmax = 2 ⋅ Vmin 2 Vmin = ⋅ U 3 m Vmin = 5.00⋅ s Problem 4.30 Problem 4.40 [Difficulty: 2] Problem 4.27 Problem 4.41 [Difficulty: 2] Problem 4.31 Problem 4.42 [Difficulty: 2] Problem 4.28 Problem 4.43 [Difficulty: 2] Problem 4.33 Problem 4.44 [Difficulty: 2] Problem 4.45 [Difficulty: 2] CS Outflow Given: Data on airflow out of tank Find: Find rate of change of density of air in tank Solution: Basic equation ∂ ρ + ρ d V V ∫ ∫ ⋅ dA = 0 ∂t CV CS Assumptions: 1) Density in tank is uniform 2) Uniform flow 3) Air is an ideal gas Hence Vtank⋅ dρtank dρtank dt dt + ρexit ⋅ V⋅ A = 0 3 N = −300 × 10 ⋅ 2 m × 250 ⋅ dρtank dt m s =− ρexit ⋅ V⋅ A Vtank 2 × 100 ⋅ mm × =− p exit ⋅ V⋅ A Rair⋅ Texit ⋅ Vtank 2 1 1 ⎛ 1⋅ m ⎞ × 1 ⋅ kg⋅ K × × ⎜ 1000⋅ mm ( −20 + 273 ) ⋅ K 3 286.9 N⋅ m ⎝ ⎠ 0.4⋅ m kg dρtank Hence dt 3 = −0.258 ⋅ m s The mass in the tank is decreasing, as expected Problem 4.32 Problem 4.46 [Difficulty: 2] Problem 4.35 Problem 4.47 [Difficulty: 2] Problem 4.48 [Difficulty: 3] Given: Data on draining of a tank Find: Depth at various times; Plot of depth versus time Solution: Basic equation ∂ ρ + ρ d V V ∫ ∫ ⋅ dA = 0 ∂t CV CS Assumptions: 1) Uniform flow 2) Incompressible flow 3) Neglect air density Treating the tank as the CV the basic equation becomes ∂ ⌠ ⎮ ∂t ⌡ y ρ⋅ Atank dy + ρ⋅ V⋅ Aopening = 0 or ρ⋅ π 0 V= Using Separating variables dy 1 y Solving for y Using the given data 2 dy ⋅D ⋅ dt + ρ⋅ π 4 2 ⋅d ⋅V = 0 1 2⋅ g⋅ y and simplifying dy dt 2 = 4 ⎛ d ⎞ ⋅ 2⋅ g ⋅ dt ⎜D ⎝ ⎠ and integrating = −⎛⎜ 2 ⎞ ⋅ 2⋅ g⋅ y 2 ⎝ D⎠ d 1⎞ ⎛⎜ 1 2 d 2 2 2⋅ ⎜ y − y0 = −⎛⎜ ⎞ ⋅ 2 ⋅ g t ⎝ ⎠ ⎝ D⎠ 2 2 ⎡⎢ g ⎛ d ⎞ ⎥⎤ y ( t) = y 0⋅ 1 − ⋅⎜ ⋅t ⎢ 2⋅ y0 ⎝ D ⎠ ⎥ ⎣ ⎦ y ( 1 ⋅ min) = 1.73⋅ ft 2 y ( 2 ⋅ min) = 0.804 ⋅ ft y ( 3 ⋅ min) = 0.229 ⋅ ft 3 Depth (ft) 2.5 2 1.5 1 0.5 0 0.5 1 1.5 t (min) 2 2.5 3 Problem 4.49 [Difficulty: 3] Given: Data on draining of a tank Find: Times to a depth of 1 foot; Plot of drain timeversus opening size Solution: Basic equation ∂ ρ + ρ d V V ∫ ∫ ⋅ dA = 0 ∂t CV CS Assumptions: 1) Uniform flow 2) Incompressible flow 3) Neglect air density Treating the tank as the CV the basic equation becomes y ∂⌠ ⎮ ρ⋅ Atank dy + ρ⋅ V⋅ Aopening = 0 ∂t ⌡ or ρ⋅ π 0 Separating variables dy 1 y Solving for t 2 dy ⋅D ⋅ dt + ρ⋅ π 4 2 ⋅d ⋅V = 0 1 V= Using 4 2⋅ g⋅ y and simplifying dt 2 = dy ⎛ d ⎞ ⋅ 2⋅ g ⋅ dt ⎜D ⎝ ⎠ and integrating = −⎛⎜ d⎞ ⎝ D⎠ 2 ⋅ 2⋅ g⋅ y 2 1⎞ ⎛⎜ 1 2 d 2 2 2⋅ ⎜ y − y0 = −⎛⎜ ⎞ ⋅ 2 ⋅ g t ⎝ ⎠ ⎝ D⎠ 2 t= 2⋅ y0 g ⋅ ⎛⎜ D⎞ 2 ⎛ ⋅⎜1 − ⎝d⎠ ⎝ y ⎞ y0 ⎠ Using the given data t( 2 ⋅ ft) = 45.6 s Hence for the first drop of 1 foot ∆t = t( 2 ⋅ ft) ∆t = 45.6 s For the second drop of 1 foot ∆t = t( 1 ⋅ ft) − t( 2 ⋅ ft) ∆t = 59.5 s t( 1 ⋅ ft) = 105 s This is because as the level drops the exit speed, hence drain rate, decreases. Drain Time (min) 15 10 5 0.1 0.2 0.3 d (in) 0.4 0.5 Problem 4.38 Problem 4.50 [Difficulty: 3] Problem 4.39 Problem 4.51 [Difficulty: 3] Problem 4.40 Problem 4.52 [Difficulty: 3] Problem 4.41 Problem 4.53 P4.48. [Difficulty: 3] Problem 4.42 Problem 4.54 [Difficulty: 4] Problem 4.55 [Difficulty: 4] Given: Data on draining of a funnel Find: Formula for drain time; time to drain from 12 in to 6 in; plot drain time versus hole diameter Solution: Basic equation ∂ ρ + ρ d V V ∫ ∫ ⋅ dA = 0 ∂t CV CS Assumptions: 1) Uniform flow 2) Incompressible flow 3) Neglect air density Treating the funnel as the CV the basic equation becomes y ∂⌠ ⎮ ρ⋅ Afunnel dy + ρ⋅ V⋅ Aopening = 0 ∂t ⌡ 0 2 2 For the funnel Afunnel = π⋅ r = π⋅ ( y ⋅ tan( θ) ) Hence y π 2 2∂⌠ 2 ρ⋅ π⋅ ( tan( θ) ) ⋅ ⎮ y dy + ρ⋅ V⋅ ⋅ d = 0 ⌡ 4 ∂t 0 Then 2 2 dy ( tan( θ) ) ⋅ y ⋅ dt = − 2⋅ g⋅ y⋅ d or 2 ⎛ y3 ⎞ d ⎜ = − 2⋅ g⋅ y⋅ 4 dt ⎝ 3 ⎠ 2d ( tan( θ) ) ⋅ 2 4 3 Separating variables 2 y ⋅ dy = − 2⋅ g⋅ d 2 4 ⋅ tan( θ) 2 ⋅ dt 0 Hence ⌠ 3 ⎮ ⎮ 2⋅ g⋅ d 2 ⋅t ⎮ y dy = − 2 ⌡y 4 ⋅ tan( θ) 0 5 or 2 5 ⋅ y0 2 2⋅ g⋅ d = 4 ⋅ tan( θ) 5 2 Solving for t 8 tan( θ) ⋅ y 0 t= ⋅ 5 2⋅ g⋅ d 2 and using the given data t = 2.55⋅ min 2 ⋅t To find the time to drain from 12 in to 6 in., we use the time equation with the two depths; this finds the time to drain from 12 in and 6 in, so the difference is the time we want 5 2 8 tan( θ) ⋅ y 0 ∆t1 = ⋅ 5 2 2⋅ g⋅ d y 1 = 6 ⋅ in 2 5 2 8 tan( θ) ⋅ y 1 − ⋅ 5 2 2⋅ g⋅ d 2 ∆t1 = 2.1⋅ min 5 2 8 tan( θ) ⋅ y 1 ∆t2 = ⋅ 2 5 2⋅ g⋅ d 2 ∆t2 = 0.451 ⋅ min ∆t1 + ∆t2 = 2.55⋅ min Note that The second time is a bit longer because although the flow rate decreases, the area of the funnel does too. Drain Time (min) 3 2 1 0.25 0.3 0.35 0.4 d (in) 0.45 0.5 Problem 4.56 [Difficulty: 4] Given: Data on draining of a funnel Find: Diameter that will drain in 1 min.; plot diamter versus depth y 0 Solution: ∂ ρdV + ∫ ρV ⋅ dA = 0 ∫ ∂t CV CS Basic equation Assumptions: 1) Uniform flow 2) Incompressible flow 3) Neglect air density Treating the funnel as the CV the basic equation becomes y ∂⌠ ⎮ ρ⋅ Afunnel dy + ρ⋅ V⋅ Aopening = 0 ∂t ⌡ 0 2 2 For the funnel Afunnel = π⋅ r = π⋅ ( y ⋅ tan( θ) ) Hence y π 2 2∂⌠ 2 ρ⋅ π⋅ ( tan( θ) ) ⋅ ⎮ y dy + ρ⋅ V⋅ ⋅ d = 0 4 ∂t ⌡0 2 ⎛ y3 ⎞ d ⎜ = − 2⋅ g⋅ y⋅ 4 dt ⎝ 3 ⎠ 2d ( tan( θ) ) ⋅ or 3 2 2 dy ( tan( θ) ) ⋅ y ⋅ Then dt = − 2⋅ g⋅ y⋅ d 2 2 y ⋅ dy = − Separating variables 4 2⋅ g⋅ d 2 4 ⋅ tan( θ) 2 ⋅ dt 0 ⌠ 3 ⎮ ⎮ 2⋅ g⋅ d 2 ⋅t ⎮ y dy = − 2 ⌡y 4 ⋅ tan ( θ ) 0 Hence 5 2 or 5 ⋅ y0 2 = 2⋅ g⋅ d 4 ⋅ tan( θ) 2 ⋅t 5 2 d = Solving for d 8 tan( θ) ⋅ y 0 ⋅ 5 2⋅ g⋅ t 2 t = 1 min and using the given data, for d = 0.399 in 1 d (in) 0.8 0.6 0.4 0.2 0 2 4 6 8 10 12 y0 (in) 14 16 18 20 22 24 Problem 4.57 [Difficulty: 4] Part 1/2 Problem 4.57 For p = 500 kPa, solving Eq. 2 for t we find t = 42.2 days [Difficulty: 4] Part 2/2 Problem 4.58 Given: Data on flow through a control surface Find: Net rate of momentum flux Solution: Basic equation: We need to evaluate ∫ CS [Difficulty: 3] VρV ⋅ dA Assumptions: 1) Uniform flow at each section From Problem 4.24 ft V1 = 10⋅ s Then for the control surface A1 = 0.5⋅ ft ft V2 = 20⋅ s 2 A2 = 0.1⋅ ft 2 A3 = 0.6⋅ ft 2 ft V3 = 5 ⋅ s It is an outlet VρV ⋅ dA = V1ρV1 ⋅ A1 + V2 ρV2 ⋅ A2 + V3 ρV3 ⋅ A3 CS = V1iˆρ V1 ⋅ A1 + V2 ˆjρ V2 ⋅ A2 + V3 sin(60)iˆ − V3 cos(60) ˆj ρ V3 ⋅ A3 = −V1iˆρV1 A1 + V2 ˆjρV2 A2 + V3 sin(60)iˆ − V3 cos(60 ) ˆj ρV3 A3 ∫ ( ) ( [ [ ] ) [ [ ]( ] ) ] = ρ − V12 A1 + V32 A3 sin (60) iˆ + ρ V22 A2 − V32 A3 cos(60 ) ˆj Hence the x component is ρ [− V12 A1 + V32 A3 sin (60 )] = 65⋅ lbm ft and the y component is 3 ( 2 2 ⋅s ) ft2 × lbf = −2406⋅ lbf lbm⋅ ft 4 2 × −10 × 0.5 + 5 × 0.6 × sin( 60⋅ deg) ⋅ s ρ [V22 A2 − V32 A3 cos(60 )] = 65⋅ lbm ft 3 ft lbf ⋅ s 2 2 × ( 20 × 0.1 − 5 × 0.6 × cos( 60⋅ deg) ) ⋅ × = 2113⋅ lbf 2 lbm⋅ ft 4 s 2 Problem 4.59 [Difficulty: 3] y 2h Given: Data on flow at inlet and outlet of channel Find: Ratio of outlet to inlet momentum flux Solution: x CS mf x = ∫ uρV ⋅ dA Basic equation: Momentum flux in x direction at a section A Assumptions: 1) Steady flow 2) Incompressible flow 2 Evaluating at 1 and 2 mfx1 = U⋅ ρ⋅ ( −U⋅ 2 ⋅ h ) ⋅ w mfx1 = 2 ⋅ ρ⋅ w⋅ U ⋅ h Hence ⌠ ⎮ ⌠ 2 2 mfx2 = ⎮ ρ⋅ u ⋅ w dy = ρ⋅ w⋅ u max ⋅ ⎮ ⎮ ⌡ −h ⌡ h h −h ⎡ ⎢1 − ⎣ 2⎤ ⎛y⎞ ⎥ ⎜h ⎝ ⎠⎦ 2 ⌠ ⎮ 2 dy = ρ⋅ w⋅ u max ⋅ ⎮ ⎮ ⌡ h −h ⎡ ⎢1 − ⎣ 2 ⋅ ⎛⎜ y⎞ ⎝h⎠ 2 + 4 ⎛ y ⎞ ⎤⎥ dy ⎜ ⎝h⎠ ⎦ 2 4 2 2 16 mfx2 = ρ⋅ w⋅ u max ⋅ ⎛⎜ 2 ⋅ h − ⋅ h + ⋅ h⎞ = ρ⋅ w⋅ u max ⋅ ⋅ h 5 3 15 ⎝ ⎠ Then the ratio of momentum fluxes is 16 mfx2 mfx1 But, from Problem 4.34 u max = = 15 2 ⋅ ρ⋅ w⋅ u max ⋅ h 2 2 ⋅ ρ⋅ w⋅ U ⋅ h 3 2 ⋅U ⎛ u max ⎞ = ⋅⎜ 15 ⎝ U ⎠ 2 8 ⎛ mfx2 8 ⎜ = ⋅⎜ mfx1 15 ⎝ 3 2 ⋅U ⎞ U ⎠ 2 = 6 5 = 1.2 Hence the momentum increases as it flows in the entrance region of the channel. This appears to contradict common sense, as friction should reduce flow momentum. What happens is the pressure drops significantly along the channel so the net force on the CV is to the right. Problem 4.60 Given: Data on flow at inlet and outlet of pipe Find: Ratio of outlet to inlet momentum flux Solution: [Difficulty: 3] mf x = ∫ uρV ⋅ dA Basic equation: Momentum flux in x direction at a section A Assumptions: 1) Steady flow 2) Incompressible flow ( ) 2 2 Evaluating at 1 and 2 mfx1 = U⋅ ρ⋅ −U⋅ π⋅ R mfx1 = ρ⋅ π⋅ U ⋅ R Hence ⌠ ⎮ ⌠ 2⎮ 2 mfx2 = ⎮ ρ⋅ u ⋅ 2⋅ π⋅ r dr = 2⋅ ρ⋅ π⋅ u max ⋅ ⎮ ⌡ 0 ⌡ 2 R ⎡ r ⋅ ⎢1 − ⎣ R 2⎤ ⎛r ⎞⎥ ⎜R ⎝ ⎠⎦ 0 2 ⌠ 2⎮ dr = 2⋅ ρ⋅ π⋅ u max ⋅ ⎮ ⎮ ⌡ R 5⎞ 3 ⎛ ⎜ r − 2⋅ r + r dy 4 ⎜ 2 R ⎠ R ⎝ 0 ⎛ R R 2 R mfx2 = 2⋅ ρ⋅ π⋅ u max ⋅ ⎜ − + 2 6 2 ⎝ 2 2 2⎞ ⎠ 2 R = ρ⋅ π⋅ u max ⋅ 2 3 Then the ratio of momentum fluxes is 1 mfx2 mfx1 But, from Problem 4.35 = 3 u max = 2 ⋅ U 2 ⋅ ρ⋅ π⋅ u max ⋅ R 2 ρ⋅ π⋅ U ⋅ R 2 2 ⎛ u max ⎞ = ⋅⎜ 3 ⎝ U ⎠ 2 1 mfx2 mfx1 = 1 3 ⋅ ⎛⎜ ⎝ 2⋅ U ⎞ U ⎠ 2 = 4 3 = 1.33 Hence the momentum increases as it flows in the entrance region of the pipe This appears to contradict common sense, as friction should reduce flow momentum. What happens is the pressure drops significantly along the pipe so the net force on the CV is to the right. Problem 4.61 Given: Data on flow through a bend Find: Find net momentum flux Solution: Basic equations [Difficulty: 3] ∫ ρ V ⋅ dA = 0 Momentum fluxes: mfx = mfy = CS Assumptions: 1) Steady flow 2) Incompressible flow h ⌠ 1 −⎮ V1 ( y ) ⋅ w dy + V2 ⋅ w⋅ h 2 + V3 ⋅ w⋅ h 3 = 0 ⌡ Evaluating mass flux at 1, 2 and 3 0 or h h 2 V1max h 1 ⌠ 1 ⌠ 1 y dy − V2 ⋅ h 2 = V3 ⋅ h 3 = ⎮ V1 ( y ) dy − V2 ⋅ h 2 = ⎮ V1max⋅ ⋅ − V2 ⋅ h 2 h1 h1 2 ⌡ ⎮ 0 ⌡ 0 Hence ( 2 V1max = ⋅ V3 ⋅ h 3 + V2 ⋅ h 2 h1 ) Using given data m V1max = 3.8 s For the x momentum, evaluating at 1, 2 and 3 h ⌠ 1 mfx = −⎮ V1 ( y ) ⋅ ρ⋅ V1 ( y ) ⋅ w dy + V3 ⋅ cos( θ) ⋅ ρ⋅ V3 ⋅ h 3 ⋅ w ⌡ 0 h 2 3 ⌠ 1 2 V1max h 1 ⎮ ⎛ y ⎞ 2 2 mfx = −⎮ ⎜ V1max⋅ ⋅ ρ⋅ w dy + V3 ⋅ ρ⋅ h 3 ⋅ cos( θ) ⋅ w = − ⋅ ⋅ ρ⋅ w + V3 ⋅ ρ⋅ h 3 ⋅ w⋅ cos( θ) 2 3 h 1⎠ ⎮ ⎝ h1 ⌡ 0 h ⎛ ⎞ 2 1 2 mfx = ρ⋅ w⋅ ⎜ −V1max ⋅ + V3 ⋅ cos( θ) ⋅ h 3 3 ⎝ ⎠ Using given data mfx = 841 N Using given data mfy = −2075 N For the y momentum, evaluating at 1, 2 and 3 mfy = −V2 ⋅ ρ⋅ V2 ⋅ h 2 ⋅ w + V3 ⋅ sin( θ) ⋅ ρ⋅ V3 ⋅ h 3 ⋅ w mfy = ρ⋅ w⋅ ⎛ −V2 ⋅ h 2 − V3 ⋅ sin( θ) ⋅ h 3⎞ ⎝ ⎠ 2 2 Problem 4.49 Problem 4.62 [Difficulty: 2] Problem 4.63 Given: Water jet hitting object Find: Jet speed; Force generated [Difficulty: 2] CS y Solution: x U Basic equations: Continuity and Momentum flux in x direction Rx Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow Given data Q = 1 ⋅ gpm d = 0.01⋅ in ρ = 1.94⋅ slug ft Using continuity Q = V⋅ A = U⋅ π d ⋅d 2 U = Using data Q π 4 Using momentum ⋅d 3 U = 4085 2 ft s 2 2 2 π⋅ D Rx = u 1 ⋅ ρ⋅ −u 1 ⋅ A1 = −ρ⋅ U ⋅ A = −ρ⋅ U ⋅ 4 ( ) 2 π⋅ d Rx = −ρ⋅ U ⋅ Hence 2 ft Rx = −1.94⋅ × ⎛⎜ 4085⋅ ⎞ × 3 s⎠ ⎝ ft slug π⋅ ⎛⎜ 2 4 .01 ⋅ ft⎞ 2 2 ⎝ 12 ⎠ × lbf ⋅ s 4 slug⋅ ft Rx = −17.7⋅ lbf U = 2785⋅ mph FAST! Problem 4.64 Given: Fully developed flow in pipe Find: Why pressure drops if momentum is constant [Difficulty: 1] Solution: Basic equation: Momentum flux in x direction Assumptions: 1) Steady flow 2) Fully developed flow Hence ∆p Fx = − τw⋅ As = 0 L ∆p = L⋅ τw⋅ As where ∆p is the pressure drop over length L, τw is the wall friction and As is the pipe surface area The sum of forces in the x direction is zero. The friction force on the fluid is in the negative x direction, so the net pressure force must be in the positive direction. Hence pressure drops in the x direction so that pressure and friction forces balance Problem 4.65 Given: Data on flow and system geometry Find: Force required to hold plug [Difficulty: 2] Solution: Basic equation: 3 The given data is Then D1 = 0.25⋅ m A1 = π⋅ D1 D2 = 0.2⋅ m Q = 1.5⋅ m p 1 = 3500⋅ kPa s ρ = 999 ⋅ kg 3 m 2 2 Q V1 = A1 m V1 = 30.6 s 2 Q V2 = A2 m V2 = 84.9 s A1 = 0.0491 m 4 π 2 2 A2 = ⋅ ⎛ D1 − D2 ⎞ ⎠ 4 ⎝ A2 = 0.0177 m Applying the basic equation ( ) ( −F + p 1 ⋅ A2 − p 2 ⋅ A2 = 0 + V1 ⋅ −ρ⋅ V1 ⋅ A1 + V2 ⋅ ρ⋅ V2 ⋅ A2 Hence ) and p2 = 0 (gage) F = p 1 ⋅ A1 + ρ⋅ ⎛ V1 ⋅ A1 − V2 ⋅ A2⎞ ⎝ ⎠ 2 F = 3500 × kN 2 m 2 2 ⋅ 0.0491⋅ m + 999 ⋅ kg 3 m × 2 2 ⎡⎛ ⎤ m ⎢⎜ 30.6⋅ ⎞ ⋅ 0.0491⋅ m2 − ⎛⎜ 84.9⋅ m ⎞ ⋅ 0.0177⋅ m2⎥ s⎠ s⎠ ⎣⎝ ⎝ ⎦ F = 90.4⋅ kN Problem 4.66 Given: Nozzle hitting stationary cart Find: Value of M to hold stationary; plot M versu θ [Difficulty: 2] Solution: Basic equation: Momentum flux in x direction for the tank Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow 5) Exit velocity is V 2 Rx = −M ⋅ g = V⋅ ρ⋅ ( −V⋅ A) + V⋅ cos( θ) ⋅ ( V⋅ A) = ρ⋅ V ⋅ A⋅ ( cos( θ) − 1 ) Hence 2 When θ = 40o M = s 9.81⋅ m × 1000⋅ kg 3 m × ⎛⎜ 10⋅ ⎝ m⎞ s ⎠ 2 2 × 0.1⋅ m × ( 1 − cos( 40⋅ deg) ) 2 M= ρ⋅ V ⋅ A g ⋅ ( 1 − cos( θ) ) M = 238 kg M (kg) 3000 2000 1000 0 45 90 Angle (deg) This graph can be plotted in Excel 135 180 Problem 4.67 [Difficulty: 2] Given: Large tank with nozzle and wire Find: Tension in wire; plot for range of water depths Solution: Basic equation: Momentum flux in x direction for the tank Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow Hence When y = 0.9 m 2 Rx T V ρ ( V A) ρ V A ρ ( 2 g y ) T π 2 1000 kg 3 m 9.81 m 2 π d 2 T 4 2 0.9 m ( 0.015 m) s 1 2 ρ g y π d 2 T is linear with y! 2 N s kg m T 3.12 N 4 T (N) 3 2 1 0 0.3 0.6 y (m) This graph can be plotted in Excel 0.9 Problem 4.68 Given: Water flowing past cylinder Find: Horizontal force on cylinder [Difficulty: 2] y V x Solution: CS Rx Basic equation: Momentum flux in x direction V Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow ( ) ( ) Hence Rx = u 1 ⋅ ρ⋅ −u 1 ⋅ A1 + u 2 ⋅ ρ⋅ u 2 ⋅ A2 = 0 + ρ⋅ ( −V⋅ sin( θ) ) ⋅ ( V⋅ a⋅ b ) For given data Rx = −1000⋅ kg 3 m × ⎛⎜ 3 ⋅ m⎞ ⎝ s⎠ 2 θ 2 Rx = −ρ⋅ V ⋅ a⋅ b ⋅ sin( θ) 2 × 0.0125⋅ m × 0.0025⋅ m × sin( 20⋅ deg) × This is the force on the fluid (it is to the left). Hence the force on the cylinder is N⋅ s kg⋅ m Rx = −Rx Rx = −0.0962 N Rx = 0.0962 N Problem 4.69 [Difficulty: 2] Given: Water jet hitting plate with opening Find: Force generated on plate; plot force versus diameter d CS y Solution: x V V Basic equation: Momentum flux in x direction Rx Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow 2 2 π⋅ D 2 π⋅ d Rx = u 1 ⋅ ρ⋅ −u 1 ⋅ A1 + u 2 ⋅ ρ⋅ u 2 ⋅ A2 = −ρ⋅ V ⋅ + ρ⋅ V ⋅ 4 4 ( Hence ) ( ) 2 For given data π slug ⎛ ft Rx = − ⋅ 1.94⋅ × ⎜ 15⋅ ⎞ × 4 3 ⎝ s⎠ ft 2 ⎛ 1 ⋅ ft⎞ × ⎢⎡1 − ⎜ ⎝3 ⎠ ⎣ 2 2 2 ⎛ 1 ⎞ ⎤⎥ × lbf ⋅ s ⎜ ⎝ 4 ⎠ ⎦ slug⋅ ft 2 Rx = − 2 π⋅ ρ⋅ V ⋅ D 4 ⎡ ⋅ ⎢1 − ⎣ 2 ⎛ d ⎞ ⎤⎥ ⎜D ⎝ ⎠⎦ Rx = −35.7⋅ lbf From Eq 1 (using the absolute value of Rx) Force (lbf) 40 30 20 10 0 0.2 0.4 0.6 Diameter Ratio (d/D) This graph can be plotted in Excel 0.8 1 (1) Problem 4.70 [Difficulty: 4] y V x CS W Rx Given: Water flowing into tank Find: Mass flow rates estimated by students. Explain discrepancy Solution: Basic equation: Momentum flux in y direction Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow For the first student m1 = ρ⋅ V where m1 represents mass flow rate (software cannot render a dot above it!) t kg 1 3 m1 = 1000⋅ × 3⋅ m × 3 60⋅ s m For the second student M m2 = t kg m1 = 50.0 s where m2 represents mass flow rate 1 m2 = 3150⋅ kg × 60⋅ s kg m2 = 52.5 s There is a discrepancy because the second student is measuring instantaneous weight PLUS the force generated as the pipe flow momentum is "killed". There is a discrepancy because the second student is measuring instantaneous weight PLUS the force generated as the pipe flow momentum is "killed". To analyse this we first need to find the speed at which the water stream enters the tank, 10 m below the pipe exit. This would be a good place to use the Bernoulli equation, but this problem is in the set before Bernoulli is covered. Instead we use the simple concept that the fluid is falling under gravity (a conclusion supported by the Bernoulli equation). From the equations for falling under gravity: 2 2 Vtank = Vpipe + 2 ⋅ g ⋅ h where V tank is the speed entering the tank, Vpipe is the speed at the pipe, and h = 10 m is the distance traveled. Vpipe is obtained from m1 Vpipe = 2 ρ⋅ π⋅ d pipe 4 = 4 ⋅ m1 2 π⋅ ρ⋅ d pipe 3 4 kg m Vpipe = × 50⋅ × × π s 1000⋅ kg Then Vtank = ⎛ 1 ⎞ ⎜ 0.05⋅ m ⎝ ⎠ 2 m Vpipe = 25.5 s 2 2 Vpipe + 2 ⋅ g ⋅ h ⎛ 25.5⋅ m ⎞ + 2 × 9.81⋅ m × 10m ⎜ s⎠ 2 ⎝ s Vtank = m Vtank = 29.1 s We can now use the y momentum equation for the CS shown above ( ) Ry − W = −Vtank⋅ ρ⋅ −Vtank⋅ Atank Vtank⋅ Atank = Vpipe⋅ Apipe where A tank is the area of the water flow as it enters the tank. But for the water flow 2 Hence ∆W = Ry − W = ρ⋅ Vtank⋅ Vpipe⋅ π⋅ d pipe 4 This equation indicate the instantaneous difference ΔW between the scale reading (Ry) and the actual weight of water (W) in the tank ∆W = 1000⋅ kg 3 × 29.1⋅ m m s × 25.5⋅ m s ∆m = Inducated as a mass, this is × π 4 × ( 0.05⋅ m) 2 ∆W g ∆W = 1457 N ∆m = 149 kg Hence the scale overestimates the weight of water by 1457 N, or a mass of 149 kg For the second student M = 3150⋅ kg − 149 ⋅ kg Hence M m2 = t M = 3001 kg where m2 represents mass flow rate 1 kg m2 = 3001⋅ kg × m2 = 50.0 60⋅ s s Comparing with the answer obtained from student 1, we see the students now agree! The discrepancy was entirely caused by the fact that the second student was measuring the weight of tank water PLUS the momentum lost by the water as it entered the tank! Problem 4.71 [Difficulty: 3] Given: Water tank attached to mass Find: Whether tank starts moving; Mass to just hold in place Solution: Basic equation: Momentum flux in x direction for the tank Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure at exit 4) Uniform flow Hence 2 2 π⋅ D Rx = V⋅ cos( θ) ⋅ ρ⋅ ( V⋅ A) = ρ⋅ V ⋅ 4 ⋅ cos( θ) V= We need to find V. We could use the Bernoulli equation, but here it is known that V = 2 × 9.81⋅ m 2 × 2⋅ m V = 6.26 s Hence Rx = 1000⋅ kg 3 m × ⎛⎜ 6.26⋅ ⎝ m⎞ s ⎠ This force is equal to the tension T in the wire 2 × π 4 2⋅ g⋅ h where h = 2 m is the height of fluid in the tank m s 2 × ( 0.05⋅ m) × cos( 60⋅ deg) Rx = 38.5 N T = Rx T = 38.5 N Fmax = M ⋅ g ⋅ μ For the block, the maximum friction force a mass of M = 10 kg can generate is m where µ is static friction 2 N⋅ s Fmax = 10⋅ kg × 9.81⋅ × 0.55 × 2 kg⋅ m s Fmax = 54.0 N Hence the tension T created by the water jet is less than the maximum friction F max; the tank is at rest The mass that is just sufficient is given by M= Rx g⋅ μ M ⋅ g ⋅ μ = Rx M = 38.5⋅ N × 1 2 ⋅ s 9.81 m × 1 0.55 × kg⋅ m 2 N⋅ s M = 7.14 kg Problem 4.72 Given: Gate held in place by water jet Find: Required jet speed for various water depths [Difficulty: 4] Solution: Basic equation: Momentum flux in x direction for the wall Note: We use this equation ONLY for the jet impacting the wall. For the hydrostatic force and location we use computing equations Ixx FR = p c⋅ A y' = y c + A⋅ y c Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow Hence ( 2 2 π⋅ D ) Rx = V⋅ ρ⋅ −V⋅ Ajet = −ρ⋅ V ⋅ 4 This force is the force generated by the wall on the jet; the force of the jet hitting the wall is then 2 2 π⋅ D Fjet = −Rx = ρ⋅ V ⋅ 4 For the hydrostatic force where D is the jet diameter w⋅ h h 1 2 FR = p c⋅ A = ρ⋅ g ⋅ ⋅ h ⋅ w = ⋅ ρ⋅ g ⋅ w⋅ h 2 2 3 Ixx h 2 12 y' = y c + = + = ⋅h A⋅ y c 2 h 3 w⋅ h ⋅ 2 where h is the water depth and w is the gate width For the gate, we can take moments about the hinge to obtain h −Fjet⋅ h jet + FR⋅ ( h − y') = −Fjet⋅ h jet + FR⋅ = 0 3 where h jet is the height of the jet from the ground 2 Hence For the first case (h = 1 m) For the second case (h = 0.5 m) For the first case (h = 0.25 m) h 1 2 π⋅ D 2 h Fjet = ρ⋅ V ⋅ ⋅ h jet = FR⋅ = ⋅ ρ⋅ g ⋅ w⋅ h ⋅ 4 3 2 3 V = V = V = 2 3⋅ π 2 3⋅ π 2 3⋅ π × 9.81⋅ m 2 m 2 m 2 s V = 28.9 m V = 10.2 m V = 3.61 m 2 3 × 1 ⋅ m × ( 0.5⋅ m) × 3 ⎛ 1 ⎞ × 1 ⎜ 0.05⋅ m 1⋅ m ⎝ ⎠ × 1 ⋅ m × ( 0.25⋅ m) × 2 3 3 ⋅ π⋅ D ⋅ h j ⎛ 1 ⎞ × 1 ⎜ 1⋅ m ⎝ 0.05⋅ m ⎠ s × 9.81⋅ 2 ⋅ g ⋅ w⋅ h 2 3 × 1 ⋅ m × ( 1 ⋅ m) × s × 9.81⋅ V= 2 ⎛ 1 ⎞ × 1 ⎜ 1⋅ m ⎝ 0.05⋅ m ⎠ s s s Problem 4.55 Problem 4.73 [Difficulty: 2] Problem 4.56 Problem 4.74 [Difficulty: 2] Problem 4.75 [Difficulty: 2] Problem 4.76 [Difficulty: 3] Given: Flow into and out of CV Find: Expressions for rate of change of mass, and force Solution: Basic equations: Mass and momentum flux Assumptions: 1) Incompressible flow 2) Uniform flow dMCV For the mass equation dt dMCV →→ ( + ρ⋅ ( −V1 ⋅ A1 − V2 ⋅ A2 + V3 ⋅ A3 + V4 ⋅ A4 ) = 0 ρ⋅ V⋅ A) = ∑ dt + CS dMCV dt Fx + For the x momentum ( = ρ⋅ V1 ⋅ A1 + V2 ⋅ A2 − V3 ⋅ A3 − V4 ⋅ A4 p 1 ⋅ A1 + 2 5 13 ⋅ p 2 ⋅ A2 − 4 5 ⋅ p 3 ⋅ A3 − 5 13 ) ⋅ p 4 ⋅ A4 = 0 + V1 ( 2 p 1 ⋅ A1 − 2 5 13 ⋅ p 2 ⋅ A2 + Fy + For the y momentum 4 5 p 1 ⋅ A1 2 ⋅ p 3 ⋅ A3 + − 12 13 5 13 ⋅ p 4 ⋅ A4 + ρ⋅ ⎛⎜ − ⎝ ⋅ p 2 ⋅ A2 − 3 5 ⋅ p 3 ⋅ A3 + p 1 ⋅ A1 2 + 12 13 ⋅ p 2 ⋅ A2 + 3 5 ⋅ p 3 ⋅ A3 − 12 13 ⋅ p 4 ⋅ A4 + ρ⋅ ⎛⎜ − ⎝ ) ) ( ) 5 4 5 2 2 2 2 ⋅ V1 ⋅ A1 − ⋅ V2 ⋅ A2 + ⋅ V3 ⋅ A3 + ⋅ V3 ⋅ A3⎞ 13 5 13 2 ⎠ 1 12 13 ⋅ p 4 ⋅ A4 = 0 + V1 ( 2 ) ( 12 ) ⋅ V ⋅ −ρ⋅ V2 ⋅ A2 ... 13 2 12 3 + ⋅ V3 ⋅ ρ⋅ V3 ⋅ A3 − ⋅ V ⋅ ρ⋅ V3 ⋅ A3 13 3 5 ⋅ −ρ⋅ V1 ⋅ A1 − ( Fy = − ( 5 ⋅ V ⋅ −ρ⋅ V2 ⋅ A2 ... 13 2 4 5 + ⋅ V3 ⋅ ρ⋅ V3 ⋅ A3 + ⋅ V ⋅ ρ⋅ V3 ⋅ A3 5 13 3 ( Fx = − ) ⋅ −ρ⋅ V1 ⋅ A1 + ) ( ) 12 3 12 2 2 2 2 ⋅ V1 ⋅ A1 − ⋅ V2 ⋅ A2 + ⋅ V3 ⋅ A3 − ⋅ V3 ⋅ A3⎞ 13 5 13 2 ⎠ 1 Problem 4.77 [Difficulty: 2] y x CS Rx Given: Water flow through elbow Find: Force to hold elbow Solution: Basic equation: Momentum flux in x direction for the elbow Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow Hence ( From continuity V2 ⋅ A2 = V1 ⋅ A1 Hence ) ( Rx + p 1g ⋅ A1 + p 2g ⋅ A2 = V1⋅ −ρ⋅ V1⋅ A1 − V2⋅ ρ⋅ V2⋅ A2 ⎛ D1 ⎞ V2 = V1⋅ = V1⋅ ⎜ A2 ⎝ D2 ⎠ A1 so 3 N Rx = −350 × 10 ⋅ 2 m × π⋅ ( 0.2⋅ m) 4 2 3 N − 75 × 10 ⋅ 2 m × Rx = −p 1g ⋅ A1 − p 2g ⋅ A2 − ρ⋅ ⎛ V1 ⋅ A1 + V2 ⋅ A2⎞ ⎝ ⎠ ) 2 2 m 0.2 ⎞ V2 = 0.8⋅ ⋅ ⎛⎜ s ⎝ 0.04⎠ π⋅ ( 0.04⋅ m) 4 2 2 m V2 = 20⋅ s 2 ... 2 2 2 2 2 kg ⎡⎛ π⋅ ( 0.2⋅ m) π⋅ ( .04⋅ m) ⎤ N⋅ s m m ⎥× × ⎢⎜ 0.8⋅ ⎞ × + −1000⋅ + ⎛⎜ 20⋅ ⎞ × 3 ⎣⎝ 4 4 s⎠ ⎝ s⎠ ⎦ kg⋅ m m Rx = −11.6⋅ kN The force is to the left: It is needed to hold the elbow on against the high pressures, plus it generates the large change in x momentum Problem 4.78 [Difficulty: 2] y CS x Rx Given: Water flow through elbow Find: Force to hold elbow Solution: Basic equation: Momentum flux in x direction for the elbow Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure at exit 4) Uniform flow Hence ( ) ( Rx + p 1g ⋅ A1 = V1⋅ −ρ⋅ V1⋅ A1 − V2⋅ ρ⋅ V2⋅ A2 From continuity V2 ⋅ A2 = V1 ⋅ A1 Hence Rx = −15⋅ lbf 2 in so 2 × 4⋅ in − 1.94⋅ slug ft 3 × ) Rx = −p 1g ⋅ A1 − ρ⋅ ⎛ V1 ⋅ A1 + V2 ⋅ A2⎞ ⎝ ⎠ 2 A1 V2 = V1⋅ A2 2 ft 4 V2 = 10⋅ ⋅ s 1 2 2 ⎡⎛ ft ⎞ 2 2 ⎛ ft ⎞ 2 ⎤ 1⋅ ft ⎞ lbf ⋅ s 2 ⎢⎜ 10⋅ × ⋅ 4⋅ in + ⎜ 40⋅ ⋅ 1⋅ in ⎥ × ⎛⎜ slug ⋅ ft ⎣⎝ s ⎠ ⎝ s⎠ ⎦ ⎝ 12⋅ in ⎠ ft V2 = 40⋅ s Rx = −86.9⋅ lbf The force is to the left: It is needed to hold the elbow on against the high pressure, plus it generates the large change in x momentum Problem 4.79 Given: Water flow through nozzle Find: Force to hold nozzle [Difficulty: 2] Solution: Basic equation: Momentum flux in x direction for the elbow Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow Hence ( From continuity V2 ⋅ A2 = V1 ⋅ A1 Hence ) ( Rx + p 1g⋅ A1 + p 2g⋅ A2 = V1 ⋅ −ρ⋅ V1 ⋅ A1 + V2 ⋅ cos( θ) ⋅ ρ⋅ V2 ⋅ A2 3 N Rx = −15 × 10 ⋅ 2 m Rx = −668 ⋅ N × ⎛ D1 ⎞ = V1 ⋅ ⎜ V2 = V1 ⋅ A2 ⎝ D2 ⎠ A1 s o π⋅ ( 0.3⋅ m) 4 2 + 1000⋅ kg 3 m × 2 ) Rx = −p 1g⋅ A1 + ρ⋅ ⎛ V2 ⋅ A2 ⋅ cos( θ) − V1 ⋅ A1⎞ ⎝ ⎠ 2 m 30 V2 = 1.5⋅ ⋅ ⎛⎜ ⎞ s ⎝ 15 ⎠ 2 2 m V2 = 6 ⋅ s 2 2 2 ⎡⎛ m 2 π⋅ ( 0.15⋅ m) 2 m π⋅ ( .3⋅ m) ⎤ N⋅ s ⎢⎜ 6 ⋅ ⎞ × ⎥× ⋅ cos( 30⋅ deg) − ⎛⎜ 1.5⋅ ⎞ × s⎠ 4 4 ⎣⎝ s ⎠ ⎝ ⎦ kg⋅ m The joint is in tension: It is needed to hold the elbow on against the high pressure, plus it generates the large change in x momentum Problem 4.61 Problem 4.80 [Difficulty: 2] Problem 4.63 Problem 4.81 [Difficulty: 2] Problem 4.82 [Difficulty: 2] CS y x Given: Water flow through orifice plate Find: Force to hold plate Rx Solution: Basic equation: Momentum flux in x direction for the elbow Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow ( ) ( Hence Rx + p 1g⋅ A1 − p 2g⋅ A2 = V1 ⋅ −ρ⋅ V1 ⋅ A1 + V2 ⋅ ρ⋅ V2 ⋅ A2 From continuity Q = V1 ⋅ A1 = V2 ⋅ A2 so Q ft V1 = = 20⋅ × A1 s 3 4 1 π⋅ ⎛⎜ ⋅ ft⎞ ⎝3 ⎠ 2 = 229 ⋅ ft s ) Rx = −p 1g⋅ A1 + ρ⋅ ⎛ V2 ⋅ A2 − V1 ⋅ A1⎞ ⎝ ⎠ 2 2 and 2 A1 D ft V2 = V1 ⋅ = V1 ⋅ ⎛⎜ ⎞ = 229 ⋅ × A2 d s ⎝ ⎠ 2 ⎛ 4 ⎞ = 1628⋅ ft ⎜ 1.5 s ⎝ ⎠ NOTE: problem has an error: Flow rate should be 2 ft3/s not 20 ft3/s! We will provide answers to both Hence Rx = −200 ⋅ lbf 2 × π⋅ ( 4 ⋅ in) 2 + 1.94⋅ 4 in slug ft 3 × 2 2 2 2⎤ ⎡⎛ ft ft ⎢⎜ 1628⋅ ⎞ × π⋅ ( 1.5⋅ in) − ⎛⎜ 229 ⋅ ⎞ × π⋅ ( 4⋅ in) ⎥ × s⎠ s⎠ 4 4 ⎣⎝ ⎝ ⎦ × 2 2 2 2⎤ 2 2 ⎡⎛ ft ft ⎢⎜ 163 ⋅ ⎞ × π⋅ ( 1.5⋅ in) − ⎛⎜ 22.9⋅ ⎞ × π⋅ ( 4 ⋅ in) ⎥ × ⎛⎜ 1⋅ ft ⎞ × lbf ⋅ s s⎠ s⎠ 4 4 slug⋅ ft ⎣⎝ ⎝ ⎦ ⎝ 12⋅ in ⎠ 2 2 ⎛ 1 ⋅ ft ⎞ × lbf ⋅ s ⎜ 12⋅ in slug⋅ ft ⎝ ⎠ Rx = 51707 ⋅ lbf With more realistic velocities Hence Rx = −200 ⋅ lbf 2 in Rx = −1970⋅ lbf × π⋅ ( 4 ⋅ in) 4 2 + 1.94⋅ slug ft 3 Problem 4.64 Problem 4.83 [Difficulty: 2] Problem 4.84 [Difficulty: 2] CS Ve y x Rx Given: Data on rocket motor Find: Thrust produced Solution: Basic equation: Momentum flux in x direction for the elbow Assumptions: 1) Steady flow 2) Neglect change of momentum within CV 3) Uniform flow Hence Rx p eg Ae Ve ρe Ve Ae me Ve Rx p eg Ae me Ve where p eg is the exit pressure (gage), me is the mass flow rate at the exit (software cannot render dot over m!) and V e is the exit velocity For the mass flow rate kg kg me mnitricacid maniline 80 32 s s Hence Rx ( 110 101 ) 10 3 N 2 m π ( 0.6 m) 4 2 112 kg me 112 s kg s 180 m s 2 N s kg m Rx 22.7 kN Problem 4.65 Problem 4.85 [Difficulty: 2] Problem 4.86 [Difficulty: 3] Problem 4.87 [Difficulty: 2] Given: Data on flow and system geometry Find: Deflection angle as a function of speed; jet speed for 10o deflection Solution: The given data is kg ρ = 999⋅ 2 A = 0.01⋅ m 3 L = 2⋅ m k = 500⋅ m N x 0 = 1⋅ m m Basic equation (y momentum): Applying this to the current system in the vertical direction Fspring = V⋅ sin( θ) ⋅ ( ρ⋅ V⋅ A) ( ) But ) 2 Hence k ⋅ x 0 − L⋅ sin( θ) = ρ⋅ V ⋅ A⋅ sin( θ) Solving for θ θ = asin⎜ For the speed at which θ = ( Fspring = k ⋅ x = k ⋅ x 0 − L⋅ sin( θ) k⋅ x0 ⎛ ⎞ ⎜ k⋅ L + ρ⋅ A⋅ V2 ⎝ ⎠ 10o, solve V= ( k ⋅ x 0 − L⋅ sin( θ) ) ρ⋅ A⋅ sin( θ) 500 ⋅ V = 999 ⋅ N m ⋅ ( 1 − 2 ⋅ sin( 5 ⋅ deg) ) ⋅ m kg 3 ⋅ 2 kg⋅ m 2 V = 21.8 ⋅ 0.01⋅ m ⋅ sin( 5 ⋅ deg) N⋅ s m Angle (deg.) 35 30 25 20 15 10 5 0 5 10 15 V (m/s) 20 25 m s Problem 4.69 Problem 4.88 [Difficulty: 3] Problem 4.71 Problem 4.89 [Difficulty: 3] Problem 4.90 [Difficulty: 2] y x Ry Rx CS Given: Data on nozzle assembly Find: Reaction force Solution: Basic equation: Momentum flux in x and y directions Assumptions: 1) Steady flow 2) Incompressible flow CV 3) Uniform flow For x momentum ( ) 2 Rx = V2 ⋅ cos( θ) ⋅ ρ⋅ V2 ⋅ A2 = ρ⋅ V2 ⋅ π⋅ D2 4 2 ⋅ cos( θ) ⎛ D1 ⎞ V2 = V1 ⋅ = V1 ⋅ ⎜ A2 ⎝ D2 ⎠ A1 2 m V2 = 2 ⋅ × s From continuity A1 ⋅ V1 = A2 ⋅ V2 Hence 2 2 m⎞ N⋅ s π 2 ⎛ Rx = 1000⋅ × × ( 0.025 ⋅ m) × cos( 30⋅ deg) × × ⎜ 18⋅ 3 ⎝ s⎠ kg⋅ m 4 m For y momentum Ry − p 1 ⋅ A1 − W − ρ⋅ Vol ⋅ g = −V1 ⋅ −ρ⋅ V1 ⋅ A1 − V2 ⋅ sin( θ) ⋅ ρ⋅ V2 ⋅ A2 ( π⋅ D1 2 + W + ρ⋅ Vol ⋅ g + 4 W = 4.5⋅ kg × 9.81⋅ m m V2 = 18 s 2 3 N Ry = 125 × 10 ⋅ 2 × × kg 3 m Ry = 554 ⋅ N × π 4 ρ⋅ π ( ) ⋅ ⎛ V ⋅ D1 − V2 ⋅ D2 ⋅ sin( θ) ⎞ ⎠ 4 ⎝ 1 N⋅ s 2 2 2 2 kg⋅ m 4 × 3 W = 44.1 N π⋅ ( 0.075 ⋅ m) m + 1000⋅ ) Rx = 138 ⋅ N 2 s Hence 2 kg Ry = p 1⋅ where ⎛ 7.5 ⎞ ⎜ 2.5 ⎝ ⎠ Vol = 0.002 ⋅ m 2 + 44.1⋅ N + 1000⋅ ⎡⎛ m ⎞ 2 2 ⎢⎜ 2 ⋅ × ( 0.075 ⋅ m) − s ⎣⎝ ⎠ kg 3 m 3 × 0.002 ⋅ m × 9.81⋅ m 2 s 2 × N⋅ s kg⋅ m ... 2 2 ⎛ 18⋅ m ⎞ × ( 0.025 ⋅ m) 2 × sin( 30⋅ deg)⎤⎥ × N⋅ s ⎜ s ⎝ ⎠ ⎦ kg⋅ m Problem 4.91 Given: Data on water jet pump Find: Speed at pump exit; pressure rise [Difficulty: 3] Solution: Basic equation: Continuity, and momentum flux in x direction Assumptions: 1) Steady flow 2) Incompressible flow CV 3) Uniform flow From continuity ⎛ 0.75 − 0.1 ⎞ + 100 ⋅ ft × 0.1 ⎜ 0.75 0.75 s ⎝ ⎠ ft V2 = 10⋅ × s For x momentum As Aj Aj ⎛ A2 − Aj ⎞ V2 = Vs⋅ + Vj⋅ = Vs⋅ ⎜ + Vj⋅ A2 A2 A2 ⎝ A2 ⎠ −ρ⋅ Vs⋅ As − ρ⋅ Vj⋅ Aj + ρ⋅ V2 ⋅ A2 = 0 ( ) ( ft V2 = 22⋅ s ) ( p 1 ⋅ A2 − p 2 ⋅ A2 = Vj⋅ −ρ⋅ Vj⋅ Aj + Vs⋅ −ρ⋅ Vs⋅ As + V2 ⋅ ρ⋅ V2 ⋅ A2 ) A ⎛ 2 Aj 2 s 2⎞ ∆p = p 2 − p 1 = ρ⋅ ⎜ Vj ⋅ + Vs ⋅ − V2 A2 A2 ⎝ ⎠ ∆p = 1.94⋅ slug ft Hence ∆p = 1816⋅ 3 lbf ft 2 × 2 2 2 2 ⎡⎛ ft ft ft ⎤ ⎢⎜ 100 ⋅ ⎞ × 0.1 + ⎛⎜ 10⋅ ⎞ × ( 0.75 − 0.1) − ⎛⎜ 22⋅ ⎞ ⎥ × lbf ⋅ s s⎠ s⎠ 0.75 ⎝ 0.75 ⎣⎝ ⎝ s ⎠ ⎦ slug⋅ ft ∆p = 12.6⋅ psi Problem 4.73 Problem 4.92 [Difficulty: 3] Problem 4.93 [Difficulty: 3] V1 V2 CS p1 p2 Rx y x Given: Data on adiabatic flow of air Find: Force of air on pipe Solution: Basic equation: Continuity, and momentum flux in x direction, plus ideal gas equation p ρ R T Assumptions: 1) Steady flow 2) Ideal gas CV 3) Uniform flow From continuity ρ1 V1 A1 ρ2 V2 A2 0 ρ1 V1 A ρ2 V2 A For x momentum Rx p 1 A p 2 A V1 ρ1 V1 A V2 ρ2 V2 A ρ1 V1 A V2 V1 Rx p 2 p 1 A ρ1 V1 A V2 V1 For the air P1 ρ1 Rair T1 kg K 1 3 N ρ1 ( 200 101 ) 10 2 286.9 N m ( 60 273 ) K m 3 N Rx ( 80 200 ) 10 2 m Hence ρ1 V1 ρ2 V2 2 0.05 m 3.15 kg 3 m 150 m s 2 0.05 m ( 300 150 ) ρ1 3.15 kg 3 m m s 2 N s kg m Rx 2456 N This is the force of the pipe on the air; the pipe is opposing flow. Hence the force of the air on the pipe is Fpipe 2456 N The air is dragging the pipe to the right Fpipe Rx Problem 4.74 Problem 4.94 [Difficulty: 3] Problem 4.95 [Difficulty: 3] V1 V2 CS p1 p2 ρ1 Rx y V3 ρ2 x Given: Data on heated flow of gas Find: Force of gas on pipe Solution: Basic equation: Continuity, and momentum flux in x direction p = ρ⋅ R⋅ T Assumptions: 1) Steady flow 2) Uniform flow From continuity m3 ρ1 V2 = V1 ⋅ − ρ2 ρ2 ⋅ A −ρ1 ⋅ V1 ⋅ A1 + ρ2 ⋅ V2 ⋅ A2 + m3 = 0 3 m 6 kg m V2 = 170 ⋅ × − 20⋅ × × s 2.75 s 2.75⋅ kg For x momentum ( m V2 = 322 s 2 0.15⋅ m ( ) Rx + p 1 ⋅ A − p 2 ⋅ A = V1 ⋅ −ρ1 ⋅ V1 ⋅ A + V2 ⋅ ρ2 ⋅ V2 ⋅ A Rx = ⎡( p − p ) + ρ ⋅ V 2 − ρ ⋅ V 2⎤ ⋅ A 2 2 1 1⎦ ⎣ 2 1 ⎡ 3 N ⎢ ⎣ m Rx = ⎢( 300 − 400 ) × 10 ⋅ Hence ) 1 where m3 = 20 kg/s is the mass leaving through the walls (the software does not allow a dot) Rx = 1760 N 2 ⎡ kg ⎢ ⎣ m + ⎢2.75⋅ 3 × ⎜⎛ 322 ⋅ ⎝ m⎞ s ⎠ 2 − 6⋅ kg 3 m × ⎜⎛ 170 ⋅ ⎝ 2⎤ ⎥ × N⋅ s ⎥ × 0.15⋅ m2 s ⎠ ⎥ kg⋅ m⎥ ⎦ ⎦ m⎞ 2⎤ Problem 4.96 Given: Data on flow out of pipe device Find: Velocities at 1 and 2; force on coupling [Difficulty: 3] Solution: Basic equations (continuity and x and y mom.): The given data is ρ = 999⋅ 3 kg D = 20⋅ cm 3 L = 1⋅ m t = 20⋅ mm p 3g = 50⋅ kPa Q = 0.3⋅ m From continuity Q = A⋅ Vave Note that at the exit V( x ) = V1 + Hence Q= 1 ( Applying y momentum V3 = π Q 2 ⋅D ) ⋅x ( ) m V1 = 10 s ⌠ Ry = −⎮ ⌡ L V2 = 2 ⋅ V1 m V2 = 20 s m V3 = 9.549 s π 2 Rx + p 3g⋅ ⋅ D = −V3 ⋅ ρ⋅ Q 4 0 Expanding and integrating L ) 4 Applying x momentum (V2 − V1) ( 1 Vave = ⋅ V1 + V2 2 1 ⋅ V1 + V2 ⋅ L⋅ t = ⋅ V1 + 2 ⋅ V1 ⋅ L⋅ t 2 2 2⋅ Q V1 = 3 ⋅ L⋅ t At the inlet (location 3) due to linear velocity distribution π 2 Rx = −p 3g⋅ ⋅ D − V3 ⋅ ρ⋅ Q 4 ⌠ ⎮ V( x ) ⋅ ρ⋅ V( x ) ⋅ t dx = −ρ⋅ t⋅ ⎮ ⎮ ⌡ L Rx = −4.43⋅ kN (V2 − V1) ⎤ 2 ⎡ ⎢V1 + ⋅ x⎥ dx L ⎣ ⎦ 0 2 ⎡⎢ ⎛ V2 − V1 ⎞ L2 ⎛ V2 − V1 ⎞ L3⎥⎤ 2 Ry = −ρ⋅ t⋅ ⎢V1 ⋅ L + 2 ⋅ V1 ⋅ ⎜ ⋅ +⎜ ⋅ ⎥ ⎣ ⎝ L ⎠ 2 ⎝ L ⎠ 3⎦ Ry = −4.66⋅ kN m s Problem 4.78 Problem 4.97 [Difficulty: 3] Problem 4.79 Problem 4.98 [Difficulty: 3] Problem 4.99 [Difficulty: 4] Given: Data on flow in wind tunnel Find: Mass flow rate in tunnel; Maximum velocity at section 2; Drag on object Solution: Basic equations: Continuity, and momentum flux in x direction; ideal gas equation p = ρ⋅ R⋅ T Assumptions: 1) Steady flow 2) Uniform density at each section From continuity mflow = ρ1 ⋅ V1 ⋅ A1 = ρ1 ⋅ V1 ⋅ π⋅ D1 2 where mflow is the mass flow rate 4 p atm ρair = Rair⋅ Tatm We take ambient conditions for the air density kg m π⋅ ( 0.75⋅ m) mflow = 1.2⋅ × 12.5⋅ × 3 s 4 m ⌠ ⌠ mflow = ⎮ ρ2 ⋅ u 2 dA2 = ρair⋅ ⎮ ⎮ ⎮ ⌡ ⌡ Also R N kg⋅ K 1 kg ρair = 101000⋅ × × ρ = 1.2 2 286.9 ⋅ N⋅ m 293 ⋅ K air 3 m m 2 kg mflow = 6.63 s 2 ⋅ π⋅ ρair⋅ Vmax ⌠ R 2 2 ⋅ π⋅ ρair⋅ Vmax⋅ R Vmax⋅ ⋅ 2 ⋅ π⋅ r dr = ⋅ ⎮ r dr = ⌡ R 3 R 0 0 Vmax = 3 ⋅ mflow 2 ⋅ π⋅ ρair⋅ R 3 3 kg m Vmax = × 6.63⋅ × × 2⋅ π s 1.2⋅ kg 2 2 r ⎛ 1 ⎞ ⎜ 0.375 ⋅ m ⎝ ⎠ 2 m Vmax = 18.8 s ⌠ Rx + p 1 ⋅ A − p 2 ⋅ A = V1 ⋅ −ρ1 ⋅ V1 ⋅ A + ⎮ ρ2 ⋅ u 2 ⋅ u 2 dA2 ⎮ ⌡ R 2 ⌠ 2 ⎮ 2 ⋅ π⋅ ρair⋅ Vmax ⌠ R 3 r⎞ ⎛ Rx = p 2 − p 1 ⋅ A − V1 ⋅ mflow + ⎮ ρair⋅ ⎜ Vmax⋅ ⋅ 2 ⋅ π⋅ r dr = p 2 − p 1 ⋅ A − V1 ⋅ mflow + ⋅ ⎮ r dr ⌡ 2 R⎠ ⎝ ⎮ 0 R ⌡ ( For x momentum ( ) ) ( ) 0 ( ) π 2 2 Rx = p 2 − p 1 ⋅ A − V1 ⋅ mflow + ⋅ ρair⋅ Vmax ⋅ R 2 We also have p 1 = ρ⋅ g ⋅ h 1 p 1 = 1000⋅ kg 3 × 9.81⋅ m Hence Rx = ( 147 − 294 ) ⋅ N 2 m Rx = −54 N × m 2 × 0.03⋅ m p 1 = 294 Pa p 2 = ρ⋅ g ⋅ h 2 p 2 = 147 ⋅ Pa s π⋅ ( 0.75⋅ m) 4 2 ⎡ + ⎢−6.63⋅ ⎢ ⎣ kg s × 12.5⋅ m s The drag on the object is equal and opposite + π 2 × 1.2⋅ kg 3 m × ⎛⎜ 18.8⋅ ⎝ Fdrag = −Rx m⎞ s ⎠ 2 × ( 0.375 ⋅ m) ⎤ 2⎥ ⎥ ⎦ × Fdrag = 54.1 N N k Problem 4.100 Given: Data on wake behind object Find: An expression for the drag [Difficulty: 2] Solution: Basic equation: Momentum Applying this to the horizontal motion 2 F U ρ π 1 U 1 u ( r) ρ 2 π r u ( r) dr 0 Integrating and using the limits 1 2 2 F π ρ U 2 r u ( r) dr 0 2 F π ρ U 1 2 2 π r 2 r 1 cos dr 2 2 F π ρ U 1 2 4 π r r cos π r dr 2 r 2 r cos 2 2 F π ρ U 1 3 2 8 2 π 2 1 0 1 0 F 2 5 π 2 8 π ρ U Problem 4.101 Given: Data on flow in 2D channel Find: Maximum velocity; Pressure drop [Difficulty: 3] y 2h x Solution: Basic equations: Continuity, and momentum flux in x direction CS Assumptions: 1) Steady flow 2) Neglect friction 3 Given data w = 25⋅ mm h = 50⋅ mm Q U1 = 2⋅ w⋅ h From continuity Q = U1 ⋅ 2 ⋅ h ⋅ w Also ⌠ −ρ⋅ U1⋅ A1 + ⎮ ρ⋅ u 2 dA = 0 ⎮ ⌡ ⌠ ⎮ U1 ⋅ 2 ⋅ h ⋅ w = w⋅ ⎮ ⎮ ⌡ h ⎛ u max⋅ ⎜ 1 − ⎜ ⎝ −h Hence For x momentum u max = 3 2 Q = 0.025⋅ y h ⋅ U1 ρ = 750⋅ s 3 m ⎠ h 4 h dy = w⋅ u max⋅ ⎡⎢[ h − ( −h ) ] − ⎡⎢ − ⎛⎜ − ⎞⎥⎤⎥⎤ = w⋅ u max⋅ ⋅ h 3 3 3 ⎣ ⎣ ⎝ ⎠⎦⎦ u max = 15 m s ⌠ Note that there is no Rx (no friction) p 1 ⋅ A − p 2 ⋅ A = V1 ⋅ −ρ1 ⋅ V1 ⋅ A + ⎮ ρ2 ⋅ u 2 ⋅ u 2 dA2 ⎮ ⌡ h ⌠ 2 2 ⎮ 2⎞ ρ⋅ u max 1 w ⎮ y 2 2 2 2⎛ ⎜ dy = −ρ⋅ U1 + ⋅ ⎡⎢2 ⋅ h − 2 ⋅ ⎛⎜ ⋅ h⎞ + 2 ⋅ ⎛⎜ ⋅ h⎞⎤⎥ p 1 − p 2 = −ρ⋅ U1 + ⋅ ⎮ ρ⋅ u max ⋅ 1 − h ⎜ A 2 ⎣ ⎝3 ⎠ ⎝ 5 ⎠⎦ h ⎠ ⎮ ⎝ ⌡ ( ) −h ⎡8 3 ⎤ 8 2 2 ∆p = p 1 − p 2 = −ρ⋅ U1 + ⋅ ρ⋅ u max = ρ⋅ U1 ⋅ ⎢ ⋅ ⎛⎜ ⎞ − 1⎥ 15 ⎣ 15 ⎝ 2 ⎠ ⎦ 2 Hence kg m U1 = 10.0 s 2⎞ 2 m ∆p = 1 5 ⋅ ρ⋅ U1 2 ∆p = 15.0⋅ kPa Problem 4.102 Given: Data on flow in 2D channel Find: Maximum velocity; Pressure drop [Difficulty: 3] y 2h x Solution: Basic equations: Continuity, and momentum flux in x direction CS Assumptions: 1) Steady flow 2) Neglect friction 3 R = 75⋅ mm Given data From continuity Q = 0.1⋅ Q = U1 ⋅ π⋅ R 2 m ρ = 850⋅ s π⋅ R 3 m Q U1 = kg m U1 = 5.66 s 2 ⌠ −ρ⋅ U1⋅ A1 + ⎮ ρ⋅ u 2 dA = 0 ⎮ ⌡ Also ⌠ 2 ⎮ U1 ⋅ π⋅ R = ⎮ ⎮ ⌡ R 2 2 4 2 ⎛ ⎛ R2 r ⎞ R R ⎞ R ⎜ ⎜ u max⋅ 1 − − = 2 ⋅ π⋅ u max⋅ = π⋅ u max⋅ ⋅ 2 ⋅ π⋅ r dr = 2 ⋅ π⋅ u max⋅ 2 2 ⎜ ⎜ 2 2 4 R 4 ⋅ R ⎝ ⎠ ⎝ ⎠ 0 u max = 2 ⋅ U1 Hence For x momentum ( u max = 11.3 m s ⌠ p 1 ⋅ A − p 2 ⋅ A = V1 ⋅ −ρ1 ⋅ V1 ⋅ A + ⎮ ρ2 ⋅ u 2 ⋅ u 2 dA2 ⎮ ⌡ ( ⌠ ⎮ 2 2 2 ⎮ p 1 − p 2 ⋅ π⋅ R = −ρ⋅ π⋅ R ⋅ U1 + ⎮ ⎮ ⌡ ) Note that there is no Rx (no friction) R 2 2 2 4 6 ⎛ r ⎞ R R ⎞ 2 2 2 ⎛R ⋅ 2 ⋅ π⋅ r dr = −ρ⋅ π⋅ R ⋅ U1 + 2 ⋅ π⋅ ρ⋅ u max ⋅ ⎜ ρ⋅ u max ⋅ ⎜ 1 − − 2⋅ + ⎜ ⎜ 2 2 2 4 R ⎠ 4⋅ R 6⋅ R ⎠ ⎝ ⎝ ) 2 0 1 1 1 1 2 2 2 2 2 2 ∆p = p 1 − p 2 = −ρ⋅ U1 + ⋅ ρ⋅ u max = −ρ⋅ U1 + ⋅ ρ⋅ 2 ⋅ U1 = ρ⋅ U1 ⋅ ⎢⎡ ⋅ ( 2 ) − 1⎥⎤ = ⋅ ρ⋅ U1 3 3 3 3 ⎣ ⎦ ( Hence ∆p = 1 3 × 850 ⋅ kg 3 m × ⎛⎜ 5.66⋅ ⎝ m⎞ s ⎠ 2 ) 2 × N⋅ s kg⋅ m ∆p = 9.08⋅ kPa Problem 4.84 Problem 4.103 [Difficulty: 3] Problem 4.86 Problem 4.104 [Difficulty: 3] Problem 4.105 [Difficulty: 4] b CS c y x a d Ff Given: Data on flow of boundary layer Find: Plot of velocity profile; force to hold plate Solution: Basic equations: Continuity, and momentum flux in x direction Assumptions: 1) Steady flow 2) Incompressible 3) No net pressure force ρ = 750 ⋅ Given data kg m U0 = 10⋅ s w = 1⋅ m 3 m L = 1⋅ m δ = 5 ⋅ mm 1 0.8 y δ 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 u( y ) U0 δ From continuity ⌠ −ρ⋅ U0 ⋅ w⋅ δ + mbc + ⎮ ρ⋅ u ⋅ w dy = 0 ⌡ 0 δ Hence where mbc is the mass flow rate across bc (Note: sotware cannot render a dot!) ⌠ mbc = ⎮ ρ⋅ U0 − u ⋅ w dy ⌡ ( ) 0 For x momentum δ ⌠ ⌠ −Ff = U0 ⋅ −ρ⋅ U0 ⋅ w⋅ δ + U0 ⋅ mbc + ⎮ u ⋅ ρ⋅ u ⋅ w dy = ⎮ ⌡ ⌡ 0 ( ) δ ⎡−U 2 + u2 + U ⋅ ( U − u)⎤ ⋅ w dy 0 0 ⎣ 0 ⎦ 0 Then the drag force is δ δ ⌠ ⌠ u ⎞ 2 u ⎛ Ff = ⎮ ρ⋅ u ⋅ U0 − u ⋅ w dy = ⎮ ρ⋅ U0 ⋅ ⋅⎜1 − dy U U ⎮ ⌡ 0 0 ⎝ ⎠ 0 ⌡ ( ) 0 1 But we have u U0 = 3 2 ⌠ =⎮ w ⎮ ⌡ Ff ⋅η − η= 1 0 1 2 ⋅η 3 where we have used substitution 1 u ⎞ 2 ⌠ ⋅ ⎛⎜ 1 − dη = ρ⋅ U0 ⋅ δ⋅ ⎮ ρ⋅ U0 ⋅ δ⋅ ⎮ U0 U0 ⎝ ⎠ ⌡ 0 u 2 Ff ⎛ 3 ⋅ η − 9 ⋅ η2 − 1 ⋅ η3 + 3 ⋅ η4 − 1 ⋅ η6⎞ dη ⎜2 4 2 2 4 ⎝ ⎠ 3 1 3 1 ⎞ 3 2 2 − + − = ρ⋅ U0 ⋅ δ⋅ ⎛⎜ − = 0.139 ⋅ ρ⋅ U0 ⋅ δ w ⎝ 4 4 8 10 28 ⎠ Hence Ff w Ff w = 0.139 × 750 ⋅ kg 3 m = 52.1 N m × ⎛⎜ 10⋅ ⎝ m⎞ s ⎠ 2 2 × 0.05⋅ m × N⋅ s kg⋅ m y = δ⋅ η Problem 4.106 [Difficulty: 4] b CS c y x a d Ff Given: Data on flow of boundary layer Find: Force on plate per unit width Solution: Basic equations: Continuity, and momentum flux in x direction Assumptions: 1) Steady flow 2) Incompressible 3) No net pressure force δ From continuity ⌠ −ρ⋅ U0 ⋅ w⋅ δ + mbc + ⎮ ρ⋅ u ⋅ w dy = 0 ⌡ where mbc is the mass flow rate across bc (Note: sotware cannot render a dot!) 0 δ Hence ⌠ mbc = ⎮ ρ⋅ U0 − u ⋅ w dy ⌡ ( ) 0 For x momentum δ ⌠ ⌠ −Ff = U0 ⋅ −ρ⋅ U0 ⋅ w⋅ δ + U0 ⋅ mbc + ⎮ u ⋅ ρ⋅ u ⋅ w dy = ⎮ ⌡ ⌡ 0 ( ) δ ⎡−U 2 + u2 + U ⋅ ( U − u)⎤ ⋅ w dy 0 0 ⎣ 0 ⎦ 0 Then the drag force is δ δ ⌠ ⌠ u ⎞ 2 u ⎛ Ff = ⎮ ρ⋅ u ⋅ U0 − u ⋅ w dy = ⎮ ρ⋅ U0 ⋅ ⋅⎜1 − dy U0 U0 ⎮ ⌡ ⎝ ⎠ 0 ⌡ ( ) 0 But we have u U0 = y y = δ⋅ η where we have used substitution δ ⌠ =⎮ w ⎮ ⌡ Ff η= 1 1 u ⎞ 2 ⌠ ⋅ ⎛⎜ 1 − dη = ρ⋅ U0 ⋅ δ⋅ ⎮ η⋅ ( 1 − η) dη ρ⋅ U0 ⋅ δ⋅ ⌡ U0 U0 0 ⎝ ⎠ u 2 0 Ff w Hence Ff w Ff w 1 1 1 2 2 = ρ⋅ U0 ⋅ δ⋅ ⎛⎜ − ⎞ = ⋅ ρ⋅ U0 ⋅ δ 3 6 2 ⎝ ⎠ = 1 6 × 1.225 ⋅ = 0.163 ⋅ kg 3 m N m × ⎛⎜ 20⋅ ⎝ m⎞ s ⎠ 2 × 2 1000 2 ⋅m × N⋅ s kg⋅ m (using standard atmosphere density) Problem 4.107 Difficulty: 4] Part 1/2 Problem 4.107 [Difficulty: 4] Part 2/2 Problem 4.108 [Difficulty: 4] Problem 4.109 [Difficulty: 4] Problem *4.91 Problem *4.110 [Difficulty: 4] Problem *4.111 [Difficulty: 4] CS Given: Air jet striking disk Find: Manometer deflection; Force to hold disk Solution: Basic equations: Hydrostatic pressure, Bernoulli, and momentum flux in x direction 2 p ρ + V + g ⋅ z = constant 2 Assumptions: 1) Steady flow 2) Incompressible 3) No friction 4) Flow along streamline 5) Uniform flow 6) Horizontal flow (g x = 0) Applying Bernoulli between jet exit and stagnation point p ρair 2 + V 2 = p0 ρair +0 p0 − p = 1 But from hydrostatics p 0 − p = SG⋅ ρ⋅ g ⋅ ∆h ∆h = so 2 1 2 2 ⋅ ρair⋅ V 2 ⋅ ρair⋅ V SG ⋅ ρ⋅ g 3 2 2 = ρair⋅ V 2 ⋅ SG ⋅ ρ⋅ g 2 ft ft s 1 ∆h = 0.002377⋅ × × × ⎜⎛ 225 ⋅ ⎞ × 3 s⎠ 2 ⋅ 1.75 1.94⋅ slug 32.2⋅ ft ⎝ ft slug For x momentum ( ) ∆h = 0.55⋅ ft 2 2 π⋅ D Rx = V⋅ −ρair⋅ A⋅ V = −ρair⋅ V ⋅ 4 2 ft Rx = −0.002377⋅ × ⎛⎜ 225 ⋅ ⎞ × 3 s⎠ ⎝ ft slug The force of the jet on the plate is then F = −Rx π⋅ ⎛⎜ 0.5 ⋅ ft⎞ 2 2 ⎝ 12 ⎠ × lbf ⋅ s 4 slug⋅ ft Rx = −0.164 ⋅ lbf F = 0.164 ⋅ lbf ∆h = 6.6⋅ in Problem *4.112 [Difficulty: 3] CS y x V, A Rx Given: Water jet shooting upwards; striking surface Find: Flow rate; maximum pressure; Force on hand Solution: Basic equations: Bernoulli and momentum flux in x direction p ρ 2 V 2 g z constant Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow Given data h 10 m ρ 1000 kg D 1 cm 3 m p atm Using Bernoulli between the jet exit and its maximum height h ρ or Then V Q 2 g h π 4 V 14.0 2 D V Q 66.0 2 V V 2 p atm ρ g h m s L min For Dr. Pritchard the maximum pressure is obtained from Bernoulli p atm ρ 2 2 p max ρ p 1 2 2 ρ V p 98.1 kPa (gage) 2 For Dr. Pritchard blocking the jet, from x momentum applied to the CV Rx u 1 ρ u 1 A1 ρ V A Hence Repeating for Dr. Fox 2 π F ρ V 4 2 D h 15 m p 1 2 V 2 ρ V 2 π F ρ V F 15.4 N 4 2 g h p 147.1 kPa 2 D F 23.1 N V 17.2 (gage) m s Q π 4 2 D V Q 80.8 L min Problem *4.113 [Difficulty: 3] Problem *4.114 [Difficulty: 3] CS Given: Water jet striking disk Find: Expression for speed of jet as function of height; Height for stationary disk Solution: Basic equations: Bernoulli; Momentum flux in z direction p ρ 2 + V 2 + g ⋅ z = constant Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow The Bernoulli equation becomes V0 2 2 2 V + g⋅ 0 = 2 + g⋅ h 2 ( 2 V = V0 − 2 ⋅ g ⋅ h ) V= 2 V0 − 2 ⋅ g ⋅ h 2 Hence −M ⋅ g = w1 ⋅ −ρ⋅ w1 ⋅ A1 = −ρ⋅ V ⋅ A But from continuity ρ⋅ V0 ⋅ A0 = ρ⋅ V⋅ A Hence we get M ⋅ g = ρ⋅ V⋅ V⋅ A = ρ⋅ V0 ⋅ A0 ⋅ V0 − 2 ⋅ g ⋅ h Solving for h h= V⋅ A = V0 ⋅ A0 so 2 1 ⎢⎡ 2 ⋅ V − 2⋅ g ⎢ 0 ⎣ ⎛ M⋅ g ⎞ ⎜ ρ⋅ V ⋅ A ⎝ 0 0⎠ 2⎤ ⎥ ⎥ ⎦ ⎡⎢ 2 m h = × × ⎢⎛⎜ 10⋅ ⎞ − 2 9.81⋅ m ⎢⎝ s⎠ ⎢⎣ 1 h = 4.28 m 2 s 3 ⎡ ⎤ s 4 ⎢2⋅ kg × 9.81⋅ m × m ⎥ × × 1000⋅ kg 10⋅ m 2⎥ 2 ⎢ 25 s ⋅ m⎞ ⎥ π⋅ ⎜⎛ ⎢ ⎣ ⎝ 1000 ⎠ ⎦ 2⎤ ⎥ ⎥ ⎥ ⎥⎦ Problem *4.96 Problem *4.115 [Difficulty: 4] Part 1/2 Problem *4.96 cont'd Problem *4.115 [Difficulty: 4] Part 2/2 Problem *4.116 Given: Stream of water striking a vane Find: Water speed; horizontal force on vane [Difficulty: 3] Solution: Basic equations: Bernoulli; Momentum flux in x direction p ρ 2 + V + g ⋅ z = constant 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow Given or available data From Bernoulli Combining D = 50⋅ mm p0 = p + 1 2 2 ⋅ ρwater⋅ V 1 kg ρwater = 1000⋅ 3 m ρHg = 13.6⋅ ρwater and for the manometer p 0 − p = ρHg⋅ g ⋅ ∆h 2 ⋅ρ ⋅ V = ρHg⋅ g ⋅ ∆h 2 water Applying x momentum to the vane V = or 2 ⋅ ρHg⋅ g ⋅ ∆h ρwater θ = 30⋅ deg V = 14.1 ∆h = 0.75⋅ m m s π 2 π 2 Rx = ρwater⋅ V⋅ ⎛⎜ −V⋅ ⋅ D ⎞ + ρwater⋅ ( −V⋅ cos( θ) ) ⋅ ⎛⎜ V⋅ ⋅ D ⎞ 4 4 ⎝ ⎠ 2 π 2 Rx = −ρwater⋅ V ⋅ ⋅ D ⋅ ( 1 + cos( θ) ) 4 ⎝ ⎠ Rx = −733 N Assuming frictionless, incompressible flow with no net pressure force is realistic, except along the vane where friction will reduce flow momentum at the exit. Problem *4.117 [Difficulty: 2] Given: Data on flow and venturi geometry Find: Force on convergent section; water pressure Solution: Basic equations: 2 p Bernoulli equation and x momentum + ρ ρ = 999 ⋅ The given data is kg V 2 + g ⋅ z = const D = 100 ⋅ mm 3 d = 50⋅ mm p 1 = 200 ⋅ kPa Q = 1000⋅ A1 = 0.00785 m π 2 A2 = ⋅ d 4 A2 = 0.00196 m m V1 = 2.12 s Q V2 = A2 m V2 = 8.49 s m L min For pressure we first need the velocities 2 A1 = π⋅ D V1 = π Then 2 4 4 Q 2 ⋅D p1 Applying Bernoulli between inlet and throat + ρ V1 2 = 2 2 p2 ρ + V2 2 2 ρ 2 2 p 2 = p 1 + ⋅ ⎛ V1 − V2 ⎞ ⎠ 2 ⎝ Solving for p 2 p 2 = 200 ⋅ kPa + 1 2 ⋅ 999 ⋅ kg 3 ( ) 2 2 m 2 × 2.12 − 8.49 ⋅ m 2 s 2 × N⋅ s kg⋅ m × kN 1000⋅ N p 2 = 166 ⋅ kPa Applying the horizontal component of momentum ( ) ( −F + p 1 ⋅ A2 − p 2 ⋅ A2 = V1 ⋅ −ρ⋅ V1 ⋅ A1 + V2 ⋅ ρ⋅ V2 ⋅ A2 ) F = p 1 ⋅ A1 − p 2 ⋅ A2 + ρ⋅ ⎛ V1 ⋅ A1 − V2 ⋅ A2⎞ ⎝ ⎠ 2 or F = 200 ⋅ kN 2 m F = 1.14 kN 2 × 0.00785 ⋅ m − 166 ⋅ kN 2 m 2 2 × 0.00196 ⋅ m + 999 ⋅ kg 3 m × 2 ⎡⎛ ⎢⎜ 2.12⋅ m ⎞ ⋅ 0.00785 ⋅ m2 − s⎠ ⎣⎝ 2 2 ⎛ 8.49⋅ m ⎞ ⋅ 0.00196 ⋅ m2⎤⎥ ⋅ N⋅ s ⎜ s⎠ ⎝ ⎦ kg × m Problem *4.118 Given: Nozzle flow striking inclined plate Find: Mimimum gage pressure [Difficulty: 3] Solution: Basic equations: Bernoulli and y momentum p ρ The given data is 2 V g z const 2 ρ 999 kg 3 L q 1200 s m m q V2 W For the exit velocity and nozzle velocity Then from Bernoulli p1 ρ ρ 2 2 V1 p atm V2 2 2 W 80 mm h 0.25 m w 20 mm m V2 15.0 s w V1 V2 W or p1 θ 30 deg m V1 3.75 s V V1 2 2 ρ H 7.5 m 2 2 ρ g h p 1 103 kPa (gage) Applying Bernoulli between 2 and the plate (state 3) p atm ρ 2 ρ 2 2 V2 p atm V3 ρ g H 2 V3 2 V2 2 g H m V3 19.3 s For the plate there is no force along the plate (x momentum) as there is no friction. For the force normal to the plate (y momentum) we have Ry V3 cos( θ) ρ V3 A3 V3 cos( θ) ( ρ q ) Ry V3 cos( θ) ρ q Ry 20.0 kN m Problem *4.119 Given: Water faucet flow Find: Expressions for stream speed and diameter; plot [Difficulty: 3] Solution: p Basic equation: Bernoulli ρ 2 + V + g ⋅ z = const 2 Assumptions: Laminar, frictionless, uniform flow D0 = 5 ⋅ mm The given data is V0 = π The initial velocity is 4 h = 50⋅ mm Q ⋅ D0 Q = p atm ρ V( z) = Evaluating at h 2 V0 + 2 ⋅ g ⋅ z V( h ) = 1.03 Q = 0.333⋅ L min m V0 = 0.283 s 2 Then applying Bernoulli between the exit and any other location Then 1⋅ L 3⋅ min Also m π π 2 2 V0⋅ ⋅ D0 = V⋅ ⋅ D 4 4 2 + V0 2 = ρ 2 + V 2 D( z) = so − g⋅ z (z downwards) D0 1 2⋅ g ⋅ z ⎞ ⎜⎛ 1 + 2 ⎜ V0 ⎝ ⎠ D( h ) = 2.62⋅ mm s p atm 4 1.25 10 Height (mm) V (m/s) 1 0.75 0.5 20 30 40 0.25 0 10 20 30 z (mm) 40 50 − 2.5 − 1.5 − 0.5 0.5 Diameter (mm) 1.5 2.5 Problem *4.99 Problem *4.120 [Difficulty: 4] Problem *4.100 Problem *4.121 [Difficulty: 4] Problem *4.102 Problem *4.122 [Difficulty: 4] Problem *4.123 [Difficulty: 4] Part 1/2 Problem *4.123 [Difficulty: 4] Part 2/2 Problem *4.124 [Difficulty: 5] Given: Plates coming together Find: Expression for velcoity field; exit velocity; plot Solution: Apply continuity using deformable CV as shown Basic equation: =0 Assumptions: Incompressible, uniform flow m V0 0.01 s Given data: Continuity becomes or 2 dh π r R 100 mm or t 2 V 2 π r h π r V0 V 2 π r h 0 dt If V0 is constant h h 0 V0 t Evaluating V( R 0 ) 0.250 Exit Velocity (m/s) h 0 2 mm so m s V( r t) V0 r 2 h 0 V0 t V( R 0.1 s) 0.500 πr2h V2πrh 0 Hence r V( r) V0 2 h Note that tmax h0 V0 tmax 0.200 s m s 6 4 2 0 0.05 0.1 0.15 t (s) The velocity greatly increases as the constant flow rate exits through a gap that becomes narrower with time. 0.2 Problem *4.104 Problem *4.125 [Difficulty: 5] Part 1/2 Problem *4.104 cont'd Problem *4.125 [Difficulty: 5] Part 2/2 Problem *4.126 [Difficulty: 4] Part 1/2 Problem *4.126 [Difficulty: 4] Part 2/2 Problem *4.127 [Difficulty: 3] CS (moves at speed U) y x Rx Ry Given: Water jet striking moving vane Find: Force needed to hold vane to speed U = 5 m/s Solution: Basic equations: Momentum flux in x and y directions Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Jet relative velocity is constant Then ( ) ( ) Rx = u 1 ⋅ −ρ⋅ V1 ⋅ A1 + u 2 ⋅ ρ⋅ V2 ⋅ A2 = −( V − U) ⋅ [ ρ⋅ ( V − U) ⋅ A] + ( V − U) ⋅ cos( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A] 2 Rx = ρ( V − U) ⋅ A⋅ ( cos( θ) − 1 ) A = π 4 ⋅ ⎜⎛ 40 ⎝ 1000 ⋅ m⎞ 2 −3 A = 1.26 × 10 ⎠ 2 m Using given data Rx = 1000⋅ kg 3 × ⎡⎢( 25 − 5 ) ⋅ ⎣ m Then ( ) ( m⎤ 2 2 N⋅ s −3 2 × 1.26 × 10 ⋅ m × ( cos( 150 ⋅ deg) − 1 ) × ⎥ s⎦ kg⋅ m Rx = −940 N ) Ry = v 1 ⋅ −ρ⋅ V1 ⋅ A1 + v 2 ⋅ ρ⋅ V2 ⋅ A2 = −0 + ( V − U) ⋅ sin( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A] 2 Ry = ρ( V − U) ⋅ A⋅ sin( θ) Ry = 1000⋅ kg 3 m × ⎡⎢( 25 − 5 ) ⋅ ⎣ m⎤ 2 2 N⋅ s −3 2 R = 252 N × 1.26 × 10 ⋅ m × sin( 150 ⋅ deg) × ⎥ s⎦ kg⋅ m y Hence the force required is 940 N to the left and 252 N upwards to maintain motion at 5 m/s Problem 4.128 [Difficulty: 3] CS (moves at speed U) y Rx Ry Given: Water jet striking moving vane Find: Force needed to hold vane to speed U = 10 m/s x Solution: Basic equations: Momentum flux in x and y directions Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Jet relative velocity is constant Then ( ) ( ) Rx = u 1 ⋅ −ρ⋅ V1 ⋅ A1 + u 2 ⋅ ρ⋅ V2 ⋅ A2 = −( V − U) ⋅ [ ρ⋅ ( V − U) ⋅ A] + ( V − U) ⋅ cos( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A] 2 Rx = ρ( V − U) ⋅ A⋅ ( cos( θ) − 1 ) Using given data Rx = 1000⋅ kg 3 × ⎡⎢( 30 − 10) ⋅ ⎣ m Then ( ) ( m⎤ 2 2 N⋅ s 2 × 0.004 ⋅ m × ( cos( 120 ⋅ deg) − 1 ) × ⎥ s⎦ kg⋅ m Rx = −2400 N ) Ry = v 1 ⋅ −ρ⋅ V1 ⋅ A1 + v 2 ⋅ ρ⋅ V2 ⋅ A2 = −0 + ( V − U) ⋅ sin( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A] 2 Ry = ρ( V − U) ⋅ A⋅ sin( θ) Ry = 1000⋅ kg 3 m × ⎡⎢( 30 − 10) ⋅ ⎣ m⎤ 2 2 N⋅ s 2 × 0.004 ⋅ m × sin( 120 ⋅ deg) × ⎥ s⎦ kg⋅ m Hence the force required is 2400 N to the left and 1390 N upwards to maintain motion at 10 m/s Ry = 1386 N Problem 4.129 [Difficulty: 2] Problem 4.130 [Difficulty: 3] Given: Data on jet boat Find: Formula for boat speed; flow rate; value of k; new speed and flow rate Solution: CV in boat coordinates Basic equation: Momentum Given data m D = 75⋅ mm Vj = 15⋅ s V = 10⋅ m kg ρ = 1000⋅ s 3 m Applying the horizontal component of momentum Fdrag = V⋅ ( −ρ⋅ Q) + Vj⋅ ( ρ⋅ Q) 2 2 Fdrag = k ⋅ V or, with k ⋅ V = ρ⋅ Q⋅ Vj − ρ⋅ Q⋅ V 2 k ⋅ V + ρ⋅ Q⋅ V − ρ⋅ Q⋅ Vj = 0 Solving for V For the flow rate To find k from Eq 1, let V= − ρ⋅ Q 2⋅ k 2 ⎛ ρ⋅ Q ⎞ + ρ⋅ Q⋅ Vj ⎜ 2⋅ k k ⎝ ⎠ + π 2 Q = Vj⋅ ⋅ D 4 α= ρ⋅ Q 3 Q = 0.0663 m s 2 2 2 For k = ρ⋅ Q 2⋅ α m Vj = 25⋅ s k = 3.31 α + 2 ⋅ α⋅ Vj 2 ( V + α) = V + 2 ⋅ α⋅ V + α = α + 2 ⋅ α⋅ Vj Hence 2 V = −α + then 2⋅ k 2 (1) α = or ( V ) 2 ⋅ Vj − V α = 10 m s N ⎛ m⎞ ⎜ ⎝s⎠ π 2 Q = Vj⋅ ⋅ D 4 2 3 Q = 0.11 m s 2 ⎡ ρ⋅ Q ρ⋅ Q⋅ Vj⎤ ρ⋅ Q ⎞ ⎥ V = 16.7 m + ⎛⎜ + k ⎦ s ⎣ 2⋅ k ⎝ 2⋅ k ⎠ V = ⎢− Problem 4.110 Problem 4.131 [Difficulty: 2] Problem 4.112 Problem 4.132 [Difficulty: 2] Problem 4.133 [Difficulty: 3] CS (moves at speed U) y Ry Rx Given: Water jet striking moving vane Find: Expressions for force and power; Show that maximum power is when U = V/3 Solution: Basic equation: Momentum flux for inertial CV Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Jet relative velocity is constant Then Rx u 1 ρ V1 A1 u 2 ρ V2 A2 ( V U) [ ρ ( V U) A ] ( V U) cos ( θ) [ ρ ( V U) A ] 2 Rx ρ( V U) A ( cos ( θ) 1) This is force on vane; Force exerted by vane is equal and opposite 2 Fx ρ ( V U) A ( 1 cos( θ) ) The power produced is then 2 P U Fx ρ U ( V U) A ( 1 cos( θ) ) To maximize power wrt to U dP dU Hence 2 ρ ( V U) A ( 1 cos( θ) ) ρ ( 2 ) ( 1 ) ( V U) U A ( 1 cos( θ) ) 0 V U 2 U V 3 U 0 Note that there is a vertical force, but it generates no power U V 3 for maximum power x Problem 4.134 [Difficulty: 3] CS (moves to left at speed Vc) Vj + Vc Vj + Vc y R Rx x t Given: Water jet striking moving cone Find: Thickness of jet sheet; Force needed to move cone Solution: Basic equations: Mass conservation; Momentum flux in x direction Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Jet relative velocity is constant Then Hence ( −ρ⋅ V1 ⋅ A1 + ρ⋅ V2 ⋅ A2 = 0 t= Dj ) −ρ⋅ Vj + Vc ⋅ π⋅ Dj 4 2 ( ) + ρ⋅ Vj + Vc ⋅ 2 ⋅ π⋅ R⋅ t = 0 (Refer to sketch) 2 t = 8⋅ R 1 8 2 × ( 4 ⋅ in) × 1 t = 0.222 ⋅ in 9 ⋅ in Using relative velocities, x momentum is ( ) ( ) ( ) ( ) ( ) ( ) Rx = u 1 ⋅ −ρ⋅ V1 ⋅ A1 + u 2 ⋅ ρ⋅ V2 ⋅ A2 = − Vj + Vc ⋅ ⎡ρ⋅ Vj + Vc ⋅ Aj⎤ + Vj + Vc ⋅ cos( θ) ⋅ ⎡ρ⋅ Vj + Vc ⋅ Aj⎤ ⎣ ⎦ ⎣ ⎦ ( )2 Rx = ρ Vj + Vc ⋅ Aj⋅ ( cos( θ) − 1 ) Using given data 2 ft Rx = 1.94⋅ × ⎡⎢( 100 + 45) ⋅ ⎥⎤ × 3 s⎦ ⎣ ft slug π⋅ ⎛⎜ 4 ⋅ ft⎞ 2 2 ⎝ 12 ⎠ × ( cos( 60⋅ deg) − 1 ) × lbf ⋅ s 4 Hence the force is 1780 lbf to the left; the upwards equals the weight slug⋅ ft Rx = −1780⋅ lbf Problem 4.114 Problem 4.135 [Difficulty: 3] Problem 4.116 Problem 4.136 [Difficulty: 3] Problem 4.117 Problem 4.137 [Difficulty: 3] Problem 4.138 [Difficulty: 2] Problem 4.139 Given: Jet impacting a splitter vane Find: Mass flow rate ratio; new speed U Solution: Apply momentum equation to inertial CV [Difficulty: 4] Assumptions: No pressure force; neglect water mass on vane; steady flow wrt vane; uniform flow; no change of speed wrt the vane Basic equation Given data V = 25⋅ −5 m A = 7.85⋅ 10 s 2 ⋅m U = 10⋅ ) Hence 0 = 0 + ( V − U) ⋅ m2 − ( V − U) ⋅ sin( θ) ⋅ m3 Note that m1 = ρ⋅ A⋅ ( V − U) m1 = m2 + m3 so s ρ = 999⋅ kg 3 m 0 = v 1⋅ −m1 + v 2⋅ m2 + v 3⋅ m3 For no vertical force, y momentum becomes and θ = 30⋅ deg V− U For constant speed wrt the vane, the jet velocity at each location is ( m where v i and mi are the vertical components of velocity and mass flow rates, respectively, at the inlet and exits, wrt the vane coordinates m2 1 m2 = m3 ⋅ sin( θ) = sin( θ) = m3 2 or kg m1 = 1.18 s m3 m1 = m3 ⋅ sin( θ) + m3 ( m1 = 1 m3 1 + sin( θ) m1 ) ( = kg m3 = 0.784 s 2 3 ) and using x momentum Rx = u 1 ⋅ −m1 + u 2 ⋅ m2 + u 3 ⋅ m3 = ( V − U) ⋅ −m1 + 0 + ( V − U) ⋅ cos( θ) ⋅ m3 Writing in terms of m1 Rx = ( V − U) ⋅ m1 ⋅ ⎛⎜ Instead, the force is now Rx = −16⋅ N Hence Rx = ( V − U) ⋅ ρ⋅ A⋅ ⎛⎜ Solving for U U = V− cos( θ) ⎝ 1 + sin( θ) Rx = −7.46 N ⎠ Rx = ( V − U) ⋅ m1 ⋅ ⎛⎜ but 2 − 1⎞ cos( θ) ⎝ 1 + sin( θ) cos( θ) ⎝ 1 + sin( θ) ⎡ρ⋅ A⋅ ⎛ − 1⎞⎥⎤ ⎢ ⎜ ⎣ ⎝ 1 + sin( θ) ⎠⎦ cos( θ) − 1⎞ ⎠ − 1⎞ Rx kg m2 = 0.392 s ⎠ U = 3.03 m s and m1 = ρ⋅ A⋅ ( V − U) Problem 4.120 Problem 4.133 Problem 4.140 [Difficulty: 3] Problem 4.141 [Difficulty: 2] Problem 4.142 [Difficulty: 3] Problem 4.143 [Difficulty: 4] Given: Data on vane/slider Find: Formula for acceleration and speed; plot Solution: The given data is ρ = 999 ⋅ kg 2 M = 30⋅ kg 3 A = 0.005 ⋅ m m V = 20⋅ dU The equation of motion, from Problem 4.141, is dt 2 ρ⋅ ( V − U) ⋅ A = M − g ⋅ μk 2 The acceleration is thus a= ρ⋅ ( V − U) ⋅ A − g ⋅ μk M μk = 0.3 s m dU Separating variables ρ⋅ ( V − U) ⋅ A M Substitute u= V− U du dU = −du ρ⋅ A⋅ u and u = V - U so Using initial conditions ⌠ ⎮ ⎮ ⎮ ⎮ ⌡ − − 1 ⎛ ρ⋅ A⋅ u2 ⎞ ⎜ − g ⋅ μk ⎝ M ⎠ M g ⋅ μk ⋅ ρ⋅ A M g ⋅ μk ⋅ ρ⋅ A V− U= U= V− Note that ⎛ ρ⋅ A ⎝ g ⋅ μk ⋅ M ⎡ ρ⋅ A ⋅ atanh⎜ ⋅ atanh⎢ ⎣ g ⋅ μk⋅ M g ⋅ μk ⋅ M ρ⋅ A g ⋅ μk ⋅ M ρ⋅ A ⎛ ρ⋅ A ⎝ g ⋅ μk ⋅ M atanh⎜ du = − − g ⋅ μk ⎛ M g ⋅ μk ⋅ ρ⋅ A ρ⋅ A ⋅ atanh⎜ M ⎠ g ⋅ μk ⋅ ρ⋅ A ⎞ ⎡ ρ⋅ A ⎣ g ⋅ μk ⋅ M ⋅ atanh⎢ ⎤ M ⎦ g ⋅ μk ⋅ ρ⋅ A ⋅ ( V − U)⎥ + ⋅u ⎝ g⋅ μk ⋅ M ⎠ ⎞ ⋅u = − − g ⋅ μk = −dt 2 M But = dt 2 ⎛ ⋅ atanh⎜ ⎤ ⋅ ( V − U)⎥ ⎦ ρ⋅ A ⎞ ⋅ V = −t ⎝ g ⋅ μk⋅ M ⎠ ⎛ g⋅ μk ⋅ ρ⋅ A ⎛ ρ⋅ A ⋅ V⎞ ⎞ ⋅ t + atanh⎜ ⎜⎝ M ⎝ g⋅ μk ⋅ M ⎠ ⎠ ⋅ tanh⎜ ⎛ g⋅ μk ⋅ ρ⋅ A ⎛ ρ⋅ A ⋅ V⎞ ⎞ ⋅ t + atanh⎜ ⎜⎝ M ⎝ g⋅ μk ⋅ M ⎠ ⎠ ⋅ tanh⎜ ⎞ π ⎠ 2 ⋅ V = 0.213 − ⋅i which is complex and difficult to handle in Excel, so we use the identity atanh( x ) = atanh⎛⎜ 1⎞ ⎝x⎠ − π 2 ⋅i for x > 1 so U= V− and finally the identity tanh⎛⎜ x − ⎝ g ⋅ μk ⋅ M ρ⋅ A π 2 ⋅ i⎞ = ⎠ ⎛ g⋅ μk ⋅ ρ⋅ A 1 ⎞ − π ⋅ i⎞ ⋅ t + atanh⎛ ⎜ M 2 ⎟ ρ⋅ A ⎜ ⋅V ⎜ ⎜ ⎝ ⎝ g⋅ μk ⋅ M ⎠ ⎠ ⋅ tanh⎜ 1 tanh( x ) g ⋅ μk ⋅ M to obtain ρ⋅ A U( t) = V − ⎛ g⋅ μk ⋅ M 1 ⎞ ⎞ ⎛ g⋅ μk ⋅ ρ⋅ A ⋅ t + atanh⎜ ⋅ M ⎝ ⎝ ρ⋅ A V ⎠ ⎠ tanh⎜ g ⋅ μk ⋅ M 2 a= Note that ρ⋅ ( V − U) ⋅ A M − g ⋅ μk ⎛ g ⋅ μk⋅ ρ⋅ A ⎛ g ⋅ μk⋅ M 1 ⎞ ⎞ ⋅ t + atanh⎜ ⋅ M ⎝ ⎝ ρ⋅ A V ⎠ ⎠ tanh⎜ g ⋅ μk a( t ) = Hence ρ⋅ A V− U= and ⎛ g⋅ μk ⋅ ρ⋅ A ⎛ g⋅ μk ⋅ M 1 ⎞ ⎞ ⋅ t + atanh⎜ ⋅ M ⎝ ⎝ ρ⋅ A V ⎠ ⎠ 2 − g ⋅ μk tanh⎜ The plots are presented below 20 U (m/s) 15 10 5 0 0.5 1 1.5 2 2.5 3 2 2.5 3 t (s) a (m/s2) 60 40 20 0 0.5 1 1.5 t (s) Problem 1.24 Problem 4.133 Problem 4.144 [Difficulty: 3] Problem 4.145 [Difficulty: 4] CS (moves at speed instantaneous speed U) y x Given: Water jet striking moving vane/cart assembly Find: Angle θ at t = 5 s; Plot θ(t) Solution: Basic equation: Momentum flux in x direction for accelerating CV Assumptions: 1) No changes in CV 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Constant jet relative velocity Then ( ) ( ) −M ⋅ arfx = u 1 ⋅ −ρ⋅ V1 ⋅ A1 + u 2 ⋅ ρ⋅ V2 ⋅ A2 = −( V − U) ⋅ [ ρ⋅ ( V − U) ⋅ A] + ( V − U) ⋅ cos( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A] 2 Since −M ⋅ arfx = ρ( V − U) ⋅ A⋅ ( cos( θ) − 1 ) or arfx = constant U = arfx⋅ t ⎡ θ = acos⎢1 − ⎢ ⎣ then M ⋅ arfx cos( θ) = 1 − cos( θ) = 1 − 2 ρ⋅ ( V − U) ⋅ A M ⋅ arfx ( ) 2 ρ⋅ V − arfx⋅ t ⋅ A ⎤ ⎥ 2 ρ⋅ ( V − arfx⋅ t) ⋅ A⎥ ⎦ M ⋅ arfx Using given data ⎡⎢ ⎢ ⎢ ⎢⎣ θ = acos 1 − 55⋅ kg × 1.5⋅ m 2 3 × s m 1000⋅ kg 1 × ⎛ 15⋅ m − 1.5⋅ m × 5 ⋅ s⎞ ⎜ s 2 s ⎝ ⎠ 2 × ⎥⎤ 2⎥ 0.025 ⋅ m ⎥ ⎥⎦ 1 20 Angle Speed 135 15 90 10 45 5 0 at t = 5 s 0 2.5 5 7.5 Time t (s) The solution is only valid for θ up to 180 o (when t = 9.14 s). This graph can be plotted in Excel 0 10 Speed U (m/s) Angle (deg) 180 θ = 19.7⋅ deg Problem 4.146 [Difficulty: 3] Given: Vaned cart with negligible resistance Find: Initial jet speed; jet and cart speeds at 2.5 s and 5 s; what happens to V - U? Solution: Apply x momentum Assumptions: 1) All changes wrt CV 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Constant jet area Given data ρ = 999 ⋅ kg 2 M = 5 ⋅ kg 3 A = 50⋅ mm a = 2.5⋅ m Then 2 Hence a⋅ M = ρ⋅ ( V − U) ⋅ ( 1 − cos( θ) ) ⋅ A Solving for V V( t) = a⋅ t + Also, for constant acceleration θ = 120 ⋅ deg 2 s −a⋅ M = u 1 ⋅ [ −ρ⋅ ( V − U) ⋅ A] + u 1 ⋅ [ ρ⋅ ( V − U) ⋅ A] Hence, evaluating m where u1 = V − U and u 2 = ( V − U) ⋅ cos( θ) From this equation we can see that for constant acceleration V and U must increase at the same rate! M⋅ a ρ⋅ ( 1 − cos( θ) ) ⋅ A V( 0 ) = 12.9 U( t) = a⋅ t m s V( 2.5⋅ s) = 19.2 so m s V( 5 ⋅ s) = 25.4 V− U= m s M⋅ a ρ⋅ ( 1 − cos( θ) ) ⋅ A = const! Problem 4.147 [Difficulty: 3] Part 1/2 Problem 4.147 [Difficulty: 3] Part 2/2 Problem 4.148 [Difficulty: 3] Problem 4.149 [Difficulty: 3] Part 1/2 Problem 4.149 [Difficulty: 3] Part 2/2 Problem 4.130 Problem 4.150 [Difficulty: 3] Problem 4.151 [Difficulty: 4] Given: Vaned cart being hit by jet Find: Jet speed to stop cart in 1s and 2 s; distance traveled Solution: Apply x momentum Assumptions: 1) All changes wrt CV 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Constant jet area Given data ρ = 999 ⋅ kg M = 5 ⋅ kg 3 D = 35⋅ mm θ = 60⋅ deg m π A = 4 2 2 ⋅D A = 962 ⋅ mm Then −arf ⋅ M = u 1 ⋅ [ −ρ⋅ ( V + U) ⋅ A] + u 2 ⋅ [ ρ⋅ ( V + U) ⋅ A] where arf = Hence or − dU − dU dt dt dU u 1 = −( V + U) dt 2 2 2 ⋅ M = ρ⋅ ( V + U) ⋅ A − ρ⋅ ( V + U) ⋅ A⋅ cos( θ) = ρ⋅ ( V + U) ⋅ A⋅ ( 1 − cos( θ) ) 2 ⋅ M = ρ⋅ ( V + U) ⋅ A⋅ ( 1 − cos( θ) ) (1) − Hence U0 V⋅ V + U0 V= − U0 2 ) + = 1 V ρ⋅ ( 1 − cos( θ) ) ⋅ A⋅ t M U0 4 2 + d ( V + U) ( V + U) Integrating from U0 at t = 0 to U = 0 at t ( u 2 = −( V + U) ⋅ cos( θ) and Note that V is constant, so dU = d(V+U), separating variables Solving for V m U0 = 5 ⋅ s U0 ⋅ M ρ⋅ A⋅ ( 1 − cos( θ) ) ⋅ t or − 2 = 1 V + U0 ρ⋅ ( 1 − cos( θ) ) ⋅ A M = ⋅ dt ρ⋅ ( 1 − cos( θ) ) ⋅ A M M ⋅ U0 2 V + V⋅ U0 − ρ⋅ ( 1 − cos( θ) ) ⋅ A⋅ t ⋅t To find distances note that dU dt so Eq. 1 can be rewritten as Separating variables −U⋅ = dU dU dx ⋅ = U⋅ dx dx dt dU ⋅ M = ρ⋅ ( V + U) ⋅ A⋅ ( 1 − cos( θ) ) dx 2 U⋅ dU ( V + U) 2 =− ρ⋅ A⋅ ( 1 − cos( θ) ) M ⋅ dx 0 It can be shown that ⌠ V ⎞ V V U ⎮ dU = ln⎛⎜ + − ⎮ 2 ⎝ V + U0 ⎠ V V + U0 ⎮ ( V + U) ⌡U (Remember that V is constant) 0 ln⎛⎜ V ⎞+1− ⎝ V + U0 ⎠ x=− Solving for x V V + U0 M ρ⋅ A⋅ ( 1 − cos( θ) ) ⋅ ⎜⎛ ln⎛⎜ ⎝ =− ρ⋅ A⋅ ( 1 − cos( θ) ) V M ⎞+1− ⎝ V + U0 ⎠ ⋅x V ⎞ V + U0 ⎠ Substituting values: To stop in To stop in t = 1⋅ s V = − and x = − U0 2 + U0 2 + 4 U0 ⋅ M ρ⋅ A⋅ ( 1 − cos( θ) ) ⋅ t ⋅ ⎛⎜ ln⎛⎜ ⎞+1− V ⎞ ρ⋅ A⋅ ( 1 − cos( θ) ) V + U0 ⎝ ⎝ V + U0 ⎠ ⎠ M t = 2⋅ s V = − and x = − U0 2 + U0 4 2 + V U0 ⋅ M ρ⋅ A⋅ ( 1 − cos( θ) ) ⋅ t ⎞+1− V ⎞ ρ⋅ A⋅ ( 1 − cos( θ) ) V + U0 ⎝ ⎝ V + U0 ⎠ ⎠ M ⋅ ⎛⎜ ln⎛⎜ V V = 5.13 m s x = 1.94 m V = 3.18 m s x = 3.47 m Problem 4.132 Problem 4.152 [Difficulty: 3] Problem 4.153 [Difficulty: 4] Given: Data on vane/slider Find: Formula for acceleration, speed, and position; plot Solution: Apply x momentum Assumptions: 1) All changes wrt CV 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Constant jet area The given data is ρ 999 kg 2 M 30 kg 3 A 0.005 m m Then k U M arf u 1 [ ρ ( V U) A ] u 2 m2 u 3 m3 where arf Hence or dU u1 V U dt k U M dU dt dU dt ρ ( V U) A M a s k 7.5 Ns m u3 0 2 k U M 2 The acceleration is thus m ρ ( V U) A 2 u2 0 V 20 ρ ( V U) A M k U M The differential equation for U can be solved analytically, but is quite messy. Instead we use a simple numerical method - Euler's method ρ ( V U( n) ) 2 A k U( n ) ∆t U( n 1 ) U( n ) M M For the position x dx dt so U x ( n 1 ) x ( n ) U( n ) ∆t The final set of equations is U( n 1 ) U( n ) ρ ( V U( n) ) 2 A k U( n ) ∆t M M 2 a( n ) ρ ( V U( n ) ) A M x ( n 1 ) x ( n ) U( n ) ∆t k U( n ) M where Δt is the time step The results can be plotted in Excel Position x vs Time 45 x (m) 40 35 30 25 20 a (m/s 2) 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 0.0 0.0 0.7 1.6 2.7 3.9 5.2 6.6 7.9 9.3 10.8 12.2 13.7 15.2 16.6 18.1 19.6 21.1 22.6 24.1 25.7 27.2 28.7 30.2 31.7 33.2 34.8 36.3 37.8 39.3 40.8 0.0 6.7 9.5 11.1 12.1 12.9 13.4 13.8 14.1 14.3 14.5 14.6 14.7 14.8 14.9 15.0 15.0 15.1 15.1 15.1 15.1 15.1 15.2 15.2 15.2 15.2 15.2 15.2 15.2 15.2 15.2 66.6 28.0 16.1 10.5 7.30 5.29 3.95 3.01 2.32 1.82 1.43 1.14 0.907 0.727 0.585 0.472 0.381 0.309 0.250 0.203 0.165 0.134 0.109 0.0889 0.0724 0.0590 0.0481 0.0392 0.0319 0.0260 0.0212 5 0 -5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 2.5 3.0 t (s) Velocity U vs Time 16 14 U (m/s) U (m/s) 12 10 8 6 4 2 0 0.0 0.5 1.0 1.5 2.0 t (s) 70 Acceleration a vs Time 60 2 x (m) a (m/s ) t (s) 15 10 50 40 30 20 10 0 0 1 1 2 t (s) 2 3 3 Problem 4.134 Problem 4.154 [Difficulty: 3] Problem 4.136 Problem 4.155 [Difficulty: 3] Problem 4.156 [Difficulty: 3] Given: Data on system Find: Jet speed to stop cart after 1 s; plot speed & position; maximum x; time to return to origin Solution: Apply x momentum Assumptions: 1) All changes wrt CV 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Constant jet area The given data is kg ρ = 999 ⋅ 2 M = 100⋅ kg 3 A = 0.01⋅ m m Then −arf ⋅ M = u 1⋅ [ −ρ⋅ ( V + U) ⋅ A ] + u 2⋅ m2 + u 3⋅ m3 where arf = Hence − dU dt dU u 1 = −( V + U) dt 2 ⋅ M = ρ⋅ ( V + U) ⋅ A or dU dt u2 = u3 = 0 and 2 =− ρ⋅ ( V + U) ⋅ A d ( V + U) which leads to M ( V + U) V + U0 U = −V + Integrating and using the IC U = U0 at t = 0 m U0 = 5⋅ s 1+ ( ρ⋅ A⋅ V + U0 2 = −⎛⎜ ρ⋅ A ⎝ M ⋅ dt⎞ ⎠ ) ⋅t M To find the jet speed V to stop the cart after 1 s, solve the above equation for V, with U = 0 and t = 1 s. (The equation becomes a quadratic in V). Instead we use Excel's Goal Seek in the associated workbook From Excel V = 5⋅ m s dx For the position x we need to integrate dt The result is x = −V⋅ t + ⎡ ρ⋅ A ⎣ M ⋅ ln⎢1 + V + U0 = U = −V + 1+ ( ρ⋅ A⋅ V + U0 M ) ( ρ⋅ A⋅ V + U0 M ) ⋅t ⎤ ⎦ ⋅ t⎥ This equation (or the one for U with U = 0) can be easily used to find the maximum value of x by differentiating, as well as the time for x to be zero again. Instead we use Excel's Goal Seek and Solver in the associated workbook From Excel x max = 1.93⋅ m The complete set of equations is t( x = 0 ) = 2.51⋅ s V + U0 U = −V + ρ⋅ A⋅ V + U0 1+ ⋅t M ( ) x = −V⋅ t + ⎡ ρ⋅ A ⎣ M ⋅ ln⎢1 + ( ρ⋅ A⋅ V + U0 M ) ⎤ ⎦ ⋅ t⎥ The plots are presented in the Excel workbook: t (s) x (m) U (m/s) 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 0.00 0.82 1.36 1.70 1.88 1.93 1.88 1.75 1.56 1.30 0.99 0.63 0.24 -0.19 -0.65 -1.14 5.00 3.33 2.14 1.25 0.56 0.00 -0.45 -0.83 -1.15 -1.43 -1.67 -1.88 -2.06 -2.22 -2.37 -2.50 To find V for U = 0 in 1 s, use Goal Seek t (s) U (m/s) V (m/s) 1.0 0.00 5.00 To find the maximum x , use Solver t (s) x (m) 1.0 1.93 To find the time at which x = 0 use Goal Seek t (s) x (m) 2.51 0.00 Cart Position x vs Time 2.5 2.0 x (m) 1.5 1.0 0.5 0.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 2.5 3.0 -1.0 -1.5 t (s) Cart Speed U vs Time 6 5 U (m/s) 4 3 2 1 0 -1 0.0 0.5 1.0 1.5 -2 -3 t (s) 2.0 Problem 4.157 [Difficulty: 2] Given: Mass moving betweem two jets Find: Time st slow to 2.5 m/s; plot position; rest position; explain Solution: Apply x momentum Assumptions: 1) All changes wrt CV 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Constant jet area The given data is kg ρ = 999 ⋅ M = 5⋅ kg 3 V = 20⋅ m m m U0 = 10⋅ s s Then −arf ⋅ M = u 1⋅ [ −ρ⋅ ( V − U) ⋅ A ] + u 2⋅ [ −ρ⋅ ( V + U) ⋅ A ] + u 3⋅ m3 where arf = − Hence dU dt dU u1 = V − U dt ⋅ M = ρ⋅ A ⋅ ⎡⎣−( V − U) + ( V + U) 2 dU Separating and integrating U Solving for t For position x t = − =− dx ⋅ ln⎛⎜ 4 ⋅ ρ⋅ A⋅ V M ⎞ 4 ⋅ ρ⋅ V⋅ A U0 ⎝ ⎠ M − U m 2 A = 100⋅ mm s u3 = 0 2⎤ ⎦ = 4⋅ ρ⋅ A ⋅ V⋅ U or ( ) ln( U) − ln U0 = − 4 ⋅ ρ⋅ A⋅ V M − ⋅t 4⋅ ρ ⋅ A ⋅ V t = 0.867 s for ⋅t M U = U0 ⋅ e U = 2.5 (1) m s ⋅t M = U = U0 ⋅ e dt x final = ⋅ dt and and using given data 4⋅ ρ ⋅ A ⋅ V and a straightforward integration leads to For large time u 2 = −( V + U) U = 2.5⋅ M ⋅ U0 4 ⋅ ρ⋅ V⋅ A x ( t) = M ⋅ U0 4 ⋅ ρ⋅ V⋅ A ⎛ − ⎜ ⋅⎝1 − e 4⋅ ρ ⋅ V ⋅ A M ⎞ ⋅t ⎠ For t = 0.867 s x ( t) = 4.69 m x final = 6.26 m 8 x (m) 6 4 2 0 1 2 t (s) 3 4 Problem *4.158 [Difficulty: 3] Given: Water jet striking moving disk Find: Acceleration of disk when at a height of 3 m CS moving at speed U Solution: Basic equations: Bernoulli; Momentum flux in z direction (treated as upwards) for linear accelerating CV p ρ 2 + V 2 + g ⋅ z = constant Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow The Bernoulli equation becomes V0 2 2 + g⋅ 0 = V1 2 2 ( + g ⋅ z − z0 ) V1 = 2 ⎛ 15⋅ m ⎞ + 2 × 9.81⋅ m ⋅ ( 0 − 3) ⋅ m ⎜ 2 ⎝ s⎠ s V1 = (All in jet) ( 2 V0 + 2 ⋅ g ⋅ z0 − z ) m V1 = 12.9 s The momentum equation becomes ( ) ( ) ( ) ( ) −W − M ⋅ arfz = w1 ⋅ −ρ⋅ V1 ⋅ A1 + w2 ⋅ ρ⋅ V2 ⋅ A2 = V1 − U ⋅ ⎡−ρ⋅ V1 − U ⋅ A1⎤ + 0 ⎣ ⎦ Hence arfz = ( )2 ρ⋅ V1 − U ⋅ A1 − W arfz = 1000⋅ M kg 3 m = × ⎡⎢( 12.9 − 5 ) ⋅ ⎣ ( )2 ρ⋅ V1 − U ⋅ A1 M m⎤ 2 V0 2 ρ⋅ V1 − U ⋅ A0 ⋅ V1 ( −g= 15 ) M 1 m 2 × − 9.81⋅ × 0.005 ⋅ m × ⎥ s⎦ 2 12.9 30⋅ kg s −g arfz = 2.28 using m 2 s V1 ⋅ A1 = V0 ⋅ A0 Problem 4.159 [Difficulty: 4] M = 35 kg CS moving at speed U D = 75 mm Given: Water jet striking disk Find: Plot mass versus flow rate to find flow rate for a steady height of 3 m Solution: Basic equations: Bernoulli; Momentum flux in z direction (treated as upwards) p ρ 2 V 2 g z constant Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow V0 The Bernoulli equation becomes 2 2 g 0 V1 (All in jet) 2 2 g h V1 2 V0 2 g h The momentum equation becomes M g w1 ρ V1 A1 w2 ρ V2 A2 V1 ρ V1 A1 0 2 M Hence M ρ V1 A1 V1 A1 V0 A0 but from continuity g ρ V1 V0 A0 g 2 π ρ V0 D0 2 V0 2 g h g 4 and also Q V0 A0 This equation is difficult to solve for V 0 for a given M. Instead we plot first: M (kg) 150 100 50 0.02 0.04 0.06 0.08 Q (cubic meter/s) 3 Goal Seek or Solver in Excel feature can be used to find Q when M = 35 kg Q 0.0469 m s Problem 4.160 [Difficulty: 3] Problem 4.161 [Difficulty: 3] Problem 4.142 Problem 4.162 [Difficulty: 3] Part 1/2 Problem 4.142 cont'd Problem 4.162 Difficulty: [3] Part 2/2 Problem 4.163 [Difficulty: 4] Given: Rocket sled on track Find: Plot speed versus time; maximum speed; effect of reducing k Solution: Basic equation: Momentum flux in x direction Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow Given data M 0 = 5000⋅ kg k = 50⋅ N⋅ s m Ve = 1750⋅ s m −FR − arf ⋅ M = u e⋅ mrate = −Ve⋅ mrate The momentum equation becomes From continuity M = M 0 − mrate⋅ t Hence, combining dU −k ⋅ U − M 0 − mrate⋅ t ⋅ = −Ve⋅ mrate dt Separating variables Integrating Simplifying Solving for U Using given data ( ) dU Ve⋅ mrate − k ⋅ U 1 = ( dt M 0 − mrate⋅ t (( ( ) k ⎛ Ve⋅ mrate − k ⋅ U ⎞ ⋅ ln⎜ ⎝ Ve⋅ mrate ⎠ ⎡ ⎢ Ve⋅ mrate ⎢ U( t) = ⋅ ⎢1 − k ⎣ U( 10⋅ s) = 175 m s 1 k and FR = k ⋅ U dU dt ))) = ⋅ ln⎛⎜ 1 − ⎝ kg mrate = 50⋅ s = Ve⋅ mrate − k ⋅ U M 0 − mrate⋅ t ) ( = M fuel = 1000⋅ kg or ⋅ ln Ve⋅ mrate − k ⋅ U − ln Ve⋅ mrate k 1 (All in jet) 1 mrate ( ( ) ( )) ⋅ ln M 0 − mrate⋅ t − ln M 0 ⎞ = 1 ⋅ ln⎛ M0 − mrate⋅ t ⎞ = 1 ⋅ ln⎛ 1 − mrate⋅ t ⎞ ⎜ ⎜ M0 M0 Ve⋅ mrate ⎠ mrate ⎝ ⎠ mrate ⎝ ⎠ k⋅ U k mrate⋅ t ⎞ ⎛ ⎜1 − M 0 ⎠ ⎝ ⎤ mrate⎥ ⎥ ⎥ ⎦ and fuel is used up when tfuel = M fuel mrate tfuel = 20 s This is when the speed is maximum With 10% reduction in k The percent improvement is ( ) Umax = U tfuel k 2 = 0.9⋅ k Umax2 − Umax Umax m Umax = 350 s k2 ⎤ ⎡ ⎢ ⎥ mrate⎥ ⎢ mrate⋅ tfuel ⎞ Ve⋅ mrate ⎢ ⎛ ⎥ Umax2 = ⋅ 1 − ⎜1 − ⎢ ⎥ M0 k2 ⎣ ⎝ ⎠ ⎦ = 1.08⋅ % When the fuel runs out the momentum equation simplifies from ( ) dU −k ⋅ U − M 0 − mrate⋅ t ⋅ = −Ve⋅ mrate dt − The solution to this (with U = Umax when t = tfuel) m Umax2 = 354 s Uempty( t) = Umax⋅ e ( k⋅ t− tfuel to −k ⋅ U − dU dt ) M0− Mfuel 400 U (m/s) 300 200 100 0 0 20 40 t (s) 60 =0 Problem 4.164 [Difficulty: 3] CS at speed U y x Ve Y X Given: Data on rocket sled Find: Minimum fuel to get to 265 m/s Solution: Basic equation: Momentum flux in x direction Assumptions: 1) No resistance 2) p e = p atm 3) Uniform flow 4) Use relative velocities From continuity dM dt Hence from momentum = mrate = constant −arfx⋅ M = − M = M 0 − mrate⋅ t so dU dt ( ) ( (Note: Software cannot render a dot!) ) ⋅ M 0 − mrate⋅ t = u e⋅ ρe⋅ Ve⋅ Ae = −Ve⋅ mrate Ve⋅ mrate Separating variables dU = Integrating M0 mrate⋅ t ⎞ ⎞ ⎛ ⎛ U = Ve⋅ ln⎜ = −Ve⋅ ln⎜ 1 − or M0 ⎝ M0 − mrate⋅ t ⎠ ⎝ ⎠ M 0 − mrate⋅ t ⋅ dt ⎛⎜ − ⎜ The mass of fuel consumed is mf = mrate⋅ t = M 0 ⋅ ⎝ 1 − e Hence ⎛ − ⎜ mf = 900 ⋅ kg × ⎝ 1 − e U Ve 265 ⎞ 2750 ⎠ ⎛⎜ − M0 ⎜ t= ⋅⎝1 − e mrate ⎞ ⎠ mf = 82.7 kg U Ve ⎞ ⎠ Problem 4.165 [Difficulty: 3] CS at speed U y x Ve Y X Given: Data on rocket weapon Find: Expression for speed of weapon; minimum fraction of mass that must be fuel Solution: Basic equation: Momentum flux in x direction Assumptions: 1) No resistance 2) p e = p atm 3) Uniform flow 4) Use relative velocities 5) Constant mass flow rate From continuity dM dt = mrate = constant M = M 0 − mrate⋅ t so ( ) ( (Note: Software cannot render a dot!) ) dU Hence from momentum −arfx⋅ M = − ⋅ M 0 − mrate⋅ t = u e⋅ ρe⋅ Ve⋅ Ae = −Ve⋅ mrate dt Separating variables dU = Ve⋅ mrate M 0 − mrate⋅ t ⋅ dt Integrating from U = U0 at t = 0 to U = U at t = t ( ( ) ⎛ ( )) = −Ve⋅ ln⎜ 1 − U − U0 = −Ve⋅ ln M 0 − mrate⋅ t − ln M 0 ⎛ U = U0 − Ve⋅ ln⎜ 1 − ⎝ Rearranging mrate⋅ t ⎞ ⎝ M0 ⎠ mrate⋅ t ⎞ M0 MassFractionConsumed = ⎠ mrate⋅ t M0 − =1−e ( U−U0) Ve − =1−e ( 3500− 600) 6000 = 0.383 Hence 38.3% of the mass must be fuel to accomplish the task. In reality, a much higher percentage would be needed due to drag effects Problem 4.166 [Difficulty: 3] Part 1/2 Problem 4.166 [Difficulty: 3] Part 2/2 Problem 4.147 Problem 4.167 [Difficulty: 3] Problem 4.168 [Difficulty: 3] Part 1/2 Problem 4.168 [Difficulty: 3] Part 2/2 Problem 4.148 Problem 4.169 [Difficulty: 3] Problem 4.170 [Difficulty: 3] CS at speed V y x Y Ve X Given: Data on rocket Find: Speed after 5 s; Maximum velocity; Plot of speed versus time Solution: Basic equation: Momentum flux in y direction Assumptions: 1) No resistance 2) p e = p atm 3) Uniform flow 4) Use relative velocities 5) Constant mass flow rate From continuity dM dt Hence from momentum Separating variables = mrate = constant M = M 0 − mrate⋅ t so ( ) −M ⋅ g − arfy⋅ M = u e⋅ ρe⋅ Ve⋅ Ae = −Ve⋅ mrate dV = or (Note: Software cannot render a dot!) arfy = dV dt = Ve⋅ mrate M −g= Ve⋅ mrate M 0 − mrate⋅ t ⎞ ⎛ Ve⋅ mrate ⎜ M − m ⋅ t − g ⋅ dt rate ⎝ 0 ⎠ Integrating from V = at t = 0 to V = V at t = t ( ( ) ⎛ ( )) − g⋅ t = −Ve⋅ ln⎜ 1 − V = −Ve⋅ ln M 0 − mrate⋅ t − ln M 0 At t = 5 s ⎝ mrate⋅ t ⎞ M0 ⎠ ⎛ V = −Ve⋅ ln⎜ 1 − − g⋅ t ⎝ m Vmax = −2500⋅ ⋅ ln⎛⎜ 1 − 10⋅ × × 5 ⋅ s⎞ − 9.81⋅ × 5 ⋅ s s ⎝ s 350 ⋅ kg 2 ⎠ s m kg mrate⋅ t ⎞ 1 For the motion after 5 s, assuming the fuel is used up, the equation of motion becomes M0 ⎠ − g⋅ t m Vmax = 336 s a = −M ⋅ g 500 V (m/s) 300 100 − 100 0 20 40 − 300 − 500 Time (s) 60 −g Problem 4.151 Problem 4.171 [Difficulty: 3] Problem 4.172 [Difficulty: 4] y x CS (moves at speed U) Ry Ff Given: Water jet striking moving vane Find: Plot of terminal speed versus turning angle; angle to overcome static friction Solution: Basic equations: Momentum flux in x and y directions Assumptions: 1) Incompressible flow 2) Atmospheric pressure in jet 3) Uniform flow 4) Jet relative velocity is constant ( ) ( ) −Ff − M ⋅ arfx = u 1 ⋅ −ρ⋅ V1 ⋅ A1 + u 2 ⋅ ρ⋅ V2 ⋅ A2 = −( V − U) ⋅ [ ρ⋅ ( V − U) ⋅ A] + ( V − U) ⋅ cos( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A] Then 2 arfx = ρ( V − U) ⋅ A⋅ ( 1 − cos( θ) ) − Ff (1) M ( ) Ry − M ⋅ g = v 1 ⋅ −ρ⋅ V1 ⋅ A1 + v 2 ⋅ ρ⋅ V2 ⋅ A2 = 0 + ( V − U) ⋅ sin( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A] Also 2 Ry = M ⋅ g + ρ( V − U) ⋅ A⋅ sin( θ) At terminal speed arfx = 0 and Ff = µkRy. Hence in Eq 1 0= or ρ⋅ V − Ut ⋅ A⋅ ( 1 − cos( θ) ) − μk ⋅ ⎡M ⋅ g + ρ⋅ V − Ut ⋅ A⋅ sin( θ)⎤ ⎣ ⎦ ( )2 ( )2 M V − Ut = ( μk ⋅ M ⋅ g ρ⋅ A⋅ 1 − cos( θ) − μk ⋅ sin( θ) ) = ( )2 ( ρ⋅ V − Ut ⋅ A⋅ 1 − cos( θ) − μk ⋅ sin( θ) Ut = V − M ( μk ⋅ M ⋅ g ρ⋅ A⋅ 1 − cos( θ) − μk ⋅ sin( θ) The terminal speed as a function of angle is plotted below; it can be generated in Excel ) ) − μk ⋅ g Terminal Speed (m/s) 20 15 10 5 0 10 20 30 40 50 60 70 80 Angle (deg) For the static case Ff = μs⋅ Ry and arfx = 0 (the cart is about to move, but hasn't) Substituting in Eq 1, with U = 0 2 0= or ( ρ⋅ V ⋅ A⋅ ⎡1 − cos( θ) − μs⋅ ρ⋅ V ⋅ A⋅ sin( θ) + M ⋅ g ⎣ cos( θ) + μs⋅ sin( θ) = 1 − 2 ) M μs⋅ M ⋅ g 2 ρ⋅ V ⋅ A We need to solve this for θ! This can be done by hand or by using Excel's Goal Seek or Solver Note that we need θ = 19o, but once started we can throttle back to about θ = 12.5 o and still keep moving! θ = 19.0⋅ deg 90 Problem 4.173 [Difficulty: 3] Problem 4.174 [Difficulty: 3] Problem 4.175 [Difficulty: 3] Problem 4.176 [Difficulty: 4] CS at speed V y x Y Ve X Given: Data on rocket Find: Maximum speed and height; Plot of speed and distance versus time Solution: Basic equation: Momentum flux in y direction Assumptions: 1) No resistance 2) p e = p atm 3) Uniform flow 4) Use relative velocities 5) Constant mass flow rate From continuity dM dt = mrate = constant M = M 0 − mrate⋅ t so ( ) Hence from momentum −M ⋅ g − arfy⋅ M = u e⋅ ρe⋅ Ve⋅ Ae = −Ve⋅ mrate Hence arfy = Separating variables dV = dV dt = Ve⋅ mrate M (Note: Software cannot render a dot!) −g= Ve⋅ mrate M 0 − mrate⋅ t −g ⎞ ⎛ Ve⋅ mrate ⎜ M − m ⋅ t − g ⋅ dt rate ⎝ 0 ⎠ Integrating from V = at t = 0 to V = V at t = t ( ( ) ⎛ ( )) − g⋅ t = −Ve⋅ ln⎜ 1 − V = −Ve⋅ ln M 0 − mrate⋅ t − ln M 0 mrate⋅ t ⎞ ⎛ V = −Ve⋅ ln⎜ 1 − − g⋅ t M0 ⎝ ⎠ ⎝ for t ≤ tb To evaluate at tb = 1.7 s, we need V e and mrate mf mrate = tb mrate = Also note that the thrust Ft is due to momentum flux from the rocket Ft = mrate⋅ Ve Ft Ve = mrate Hence ⎛ Vmax = −Ve⋅ ln⎜ 1 − ⎝ mrate⋅ tb ⎞ M0 ⎠ mrate⋅ t ⎞ M0 ⎠ − g⋅ t (burn time) 12.5⋅ gm 1.7⋅ s (1) − 3 kg mrate = 7.35 × 10 Ve = 5.75⋅ N 7.35 × 10 − 3 kg ⋅ s × kg⋅ m 2 s ⋅N m Ve = 782 s s − g ⋅ tb m m 1 − 3 kg Vmax = −782 ⋅ ⋅ ln⎜⎛ 1 − 7.35 × 10 ⋅ × × 1.7⋅ s⎞ − 9.81⋅ × 1.7⋅ s 2 s ⎝ s 0.0696⋅ kg ⎠ s m Vmax = 138 s To obtain Y(t) we set V = dY/dt in Eq 1, and integrate to find Y= Ve⋅ M 0 mrate mrate⋅ t ⎞ ⎡⎛ ⋅ ⎢⎜ 1 − ⎣⎝ M0 ⎠⎝ ⎝ m Yb = 782 ⋅ × 0.0696⋅ kg × s At t = tb +− 1 2 × 9.81⋅ m 2 mrate⋅ t ⎞ ⎛ ⎛ ⋅ ⎜ ln⎜ 1 − M0 s 7.35 × 10 × ( 1.7⋅ s) −3 ⋅ kg ⎠ ⎞ ⎤ 1 ⎠ ⎦ 2 − 1 + 1⎥ − ⋅ ⎡⎢⎛⎜ 1 − ⎣⎝ ⋅ g⋅ t 2 t ≤ tb tb = 1.7⋅ s (2) 0.00735 ⋅ 1.7 ⎞ ⎛ .00735⋅ 1.7 ⎞ ⎛ − 1⎞ + 1⎥⎤ ... ⎜ ln⎜ 1 − .0696 ⎠⎝ ⎝ ⎠ ⎠ ⎦ 0.0696 2 s Yb = 113 m After burnout the rocket is in free assent. Ignoring drag ( V( t) = Vmax − g ⋅ t − tb ) ( (3) ) ( ) 1 2 Y( t) = Yb + Vmax⋅ t − tb − ⋅ g ⋅ t − tb 2 t > tb The speed and position as functions of time are plotted below. These are obtained from Eqs 1 through 4, and can be plotted in Excel 150 V (m/s) 100 50 0 5 10 15 20 − 50 Time (s) Y (m) 1500 1000 500 0 5 10 15 20 Time (s) Using Solver, or by differentiating y(t) and setting to zero, or by setting V(t) = 0, we find for the maximum y t = 15.8 s y max = 1085 m (4) Problem 4.177 [Difficulty: 3] Given: Data on "jet pack" rocket Find: Initial exhaust mass flow rate; mass flow rate at end; maximum time of flight Solution: Basic equation: Momentum flux in y direction Assumptions: 1) Jet pack just hovers 2) Steady flow 3) Uniform flow 4) Use relative velocities Given data m Ve = 3000⋅ s M 0 = 200⋅ kg mrateinit = M 0 ⋅ g moon Ve −M ⋅ g moon = −v 1 ⋅ mrate = −Ve⋅ mrate M f ⋅ g moon mratefinal = dM dt = Ve M ⋅ g moon so Ve Integrating Solving for t t=− g moon M f = M 0 − M fuel dM = mrate dt dM M ⎛ M0 ⎞ gmoon ln⎜ = ⋅t Ve ⎝M⎠ Ve m ⋅ ln⎛⎜ ⎞ M0 ⎝ ⎠ M mrate = or 2 M ⋅ g moon Ve M f = 100 kg kg mratefinal = 0.0556 s Flight ends as fuel is used up. To find this, from continuity Hence g moon = 1.67 kg mrateinit = 0.111 s Finally, when all the fuel is just used up, the mass is Then g moon = 0.17⋅ g s At all instants, the momentum becomes Hence, initially M fuel = 100⋅ kg = g moon Ve − or M = M0⋅ e so when M = Mf mrate = but M ⋅ g moon Ve ⋅ dt gmoon ⋅t Ve tfinal = − Ve g moon ⎛ Mf ⎞ ⋅ ln⎜ ⎝ M0 ⎠ tfinal = 20.8 min Problem 4.178 [Difficulty: 5] Part 1/3 Problem 4.178 [Difficulty: 5] Part 2/3 Problem 4.178 [Difficulty: 5] Part 3/3 Problem 4.179 [Difficulty: 5] Part 1/2 Problem 4.179 [Difficulty: 5] Part 2/2 Problem 4.180 [Difficulty: 5] Given: Water jet striking moving disk Find: Motion of disk; steady state height CS moving at speed U Solution: Basic equations: Bernoulli; Momentum flux in z direction (treated as upwards) for linear accelerating CV p ρ 2 + V 2 + g ⋅ z = constant Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure 4) Uniform flow 5) velocities wrt CV V0 The Bernoulli equation becomes 2 + g⋅ 0 = 2 V1 2 + g⋅ h 2 2 ( 2 arfz = d h U= and 2 dt (1) m V1 = 12.9 s ( ) ( ) 2 dh −M ⋅ g − M ⋅ we get dt d h ( 2 = −ρ⋅ ⎛⎜ V1 − ⎝ dt 2 d h V1 ⋅ A1 = V0 ⋅ A0 Using Eq 1, and from continuity ) V0 − 2 ⋅ g ⋅ h ) −M ⋅ g − M ⋅ arfz = w1 ⋅ −ρ⋅ V1 ⋅ A1 + w2 ⋅ ρ⋅ V2 ⋅ A2 = V1 − U ⋅ ⎡−ρ⋅ V1 − U ⋅ A1⎤ + 0 ⎣ ⎦ The momentum equation becomes With 2 V1 = ⎛ 15⋅ m ⎞ + 2 × 9.81⋅ m ⋅ ( 0 − 3) ⋅ m ⎜ s 2 ⎝ ⎠ s V1 = (All in jet) 2 dt = dh ⎞ dt ⎠ 2 ⋅ A1 2 ρ⋅ A0 ⋅ V0 ⎛ V 2 − 2 ⋅ g⋅ h − dh ⎞ ⋅ −g ⎜ 0 dt ⎠ ⎝ 2 M ⋅ V0 − 2 ⋅ g ⋅ h (2) This must be solved numerically! One approach is to use Euler's method (see the Excel solution) dh At equilibrium h = h 0 dt 2 d h =0 2 =0 so dt ⎛ V 2 − 2 ⋅ g⋅ h ⎞ ⋅ ρ⋅ A ⋅ V − M⋅ g = 0 0⎠ 0 0 ⎝ 0 Hence ⎡ ⎢ ⎛ h0 = × ⎜ 15⋅ × × 1− 2 ⎝ s⎠ 9.81⋅ m ⎢ ⎣ 1 m⎞ 2 2 s and h0 = V0 2 2⋅ g ⎡ ⋅ ⎢1 − ⎢ ⎣ ⎛⎜ M⋅ g ⎞ ⎜ ρ⋅ V02⋅ A0 ⎝ ⎠ 2 3 ⎡ s ⎞ 1 ⎥⎤ ⎢30⋅ kg × 9.81⋅ m × m × ⎛⎜ × ⎢ 2 1000⋅ kg ⎝ 15⋅ m ⎠ 2⎥ s .005⋅ m ⎦ ⎣ 2⎤ 2⎤ ⎥ ⎥ ⎦ ⎥ ⎥ h 0 = 10.7 m ⎦ In Excel: Ξt = 0.05 s 2 A 0 = 0.005 m 2 9.81 m/s 15 m/s 30 kg Ξ = 1000 kg/m3 2 2 h (m) 2.000 2.000 2.061 2.167 2.310 2.481 2.673 2.883 3.107 3.340 3.582 3.829 4.080 4.333 4.587 4.840 dh/dt (m/s) 0.000 1.213 2.137 2.852 3.412 3.853 4.199 4.468 4.675 4.830 4.942 5.016 5.059 5.074 5.066 5.038 d h/dt (m/s ) 24.263 18.468 14.311 11.206 8.811 6.917 5.391 4.140 3.100 2.227 1.486 0.854 0.309 -0.161 -0.570 -0.926 0.800 0.850 0.900 0.950 1.000 1.050 1.100 1.150 1.200 1.250 1.300 1.350 1.400 1.450 1.500 1.550 1.600 1.650 1.700 1.750 1.800 1.850 1.900 1.950 2.000 2.050 2.100 2.150 2.200 2.250 2.300 2.350 2.400 2.450 5.092 5.341 5.588 5.830 6.069 6.302 6.530 6.753 6.969 7.179 7.383 7.579 7.769 7.952 8.127 8.296 8.457 8.611 8.757 8.896 9.029 9.154 9.272 9.384 9.488 9.587 9.679 9.765 9.845 9.919 9.989 10.052 10.111 10.166 4.991 4.930 4.854 4.767 4.669 4.563 4.449 4.328 4.201 4.069 3.934 3.795 3.654 3.510 3.366 3.221 3.076 2.931 2.787 2.645 2.504 2.365 2.230 2.097 1.967 1.842 1.720 1.602 1.489 1.381 1.278 1.179 1.085 0.997 -1.236 -1.507 -1.744 -1.951 -2.130 -2.286 -2.420 -2.535 -2.631 -2.711 -2.776 -2.826 -2.864 -2.889 -2.902 -2.904 -2.896 -2.878 -2.850 -2.814 -2.769 -2.716 -2.655 -2.588 -2.514 -2.435 -2.350 -2.261 -2.167 -2.071 -1.972 -1.871 -1.769 -1.666 12 6 10 5 8 4 Position 6 3 Speed 4 2 2 1 0 0 0 1 2 3 4 Time t (s) t (s) 2.950 3.000 3.050 3.100 3.150 3.200 3.250 3.300 3.350 3.400 3.450 3.500 3.550 3.600 3.650 3.700 3.750 3.800 3.850 3.900 3.950 4.000 4.050 4.100 4.150 4.200 4.250 4.300 4.350 4.400 4.450 4.500 4.550 h (m) dh/dt (m/s) 10.506 0.380 10.525 0.341 10.542 0.307 10.558 0.275 10.571 0.246 10.584 0.220 10.595 0.197 10.604 0.176 10.613 0.157 10.621 0.140 10.628 0.124 10.634 0.111 10.640 0.098 10.645 0.087 10.649 0.078 10.653 0.069 10.656 0.061 10.659 0.054 10.662 0.048 10.665 0.043 10.667 0.038 10.669 0.033 10.670 0.030 10.672 0.026 10.673 0.023 10.674 0.021 10.675 0.018 10.676 0.016 10.677 0.014 10.678 0.013 10.678 0.011 10.679 0.010 10.679 0.009 2 2 2 d h/dt (m/s ) -0.766 -0.698 -0.634 -0.574 -0.519 -0.469 -0.422 -0.380 -0.341 -0.306 -0.274 -0.245 -0.219 -0.195 -0.174 -0.155 -0.138 -0.123 -0.109 -0.097 -0.086 -0.077 -0.068 -0.060 -0.053 -0.047 -0.042 -0.037 -0.033 -0.029 -0.026 -0.023 -0.020 5 Speed (m/s) 2 t (s) 0.000 0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.400 0.450 0.500 0.550 0.600 0.650 0.700 0.750 Position (m) g= V= M= Problem 4.181 [Difficulty: 5] Part 1/2 Problem 4.181 [Difficulty: 5] Part 2/2 Problem 4.182 [Difficulty: 5] Part 1/3 Problem 4.133 Problem 4.182 [Difficulty: 5] Part 2/3 Problem 4.182 [Difficulty: 5] Part 3/3 Problem 4.183 [Difficulty: 5] Part 1/2 4.184 4.184 4.184 Problem 4.183 [Difficulty: 5] Part 2/2 Problem 4.184 4.137 [Difficulty: 5] Part 1/4 Problem 4.184 [Difficulty: 5] Part 2/4 Problem 4.184 [Difficulty: 5] Part 3/4 Problem 4.184 [Difficulty: 5] Part 4/4 Problem 4.185 [Difficulty: 3] Part 1/2 Problem 4.185 [Difficulty: 3] Part 2/2 Problem *4.165 Problem 4.186 [Difficulty: 2] Example 4.6 4.6 Problem 4.187 [Difficulty: 3] Problem 4.188 [Difficulty: 3] Given: Data on rotating spray system Find: Torque required to hold stationary; steady-state speed Solution: Basic equation: Rotating CV Assumptions: 1) No surface force; 2) Body torques cancel; 3) Sprinkler stationary; 4) Steady flow; 5) Uniform flow; 6) L<50 mm Wave Speed (m/s) 0.4 0.3 0.2 0.1 0 0 0.05 Wavelength (m) 0.1 N m Problem 8.1 Given: Air entering duct Find: Flow rate for turbulence; Entrance length Solution: The basic equations are The given data is Re V D π 2 Recrit 2300 Q D 125 mm From Table A.10 ν 2.29 10 Llaminar 0.06 Recrit D or, for turbulent, Lturb = 25D to 40D ν Q π Hence [Difficulty: 1] Recrit 4 4 D V 2 5 m s D 2 D or ν Q Recrit π ν D 4 For laminar flow Llaminar 0.06 Recrit D Llaminar 17.3 m For turbulent flow Lmin 25 D Lmin 3.13 m 3 3m Q 5.171 10 Lmax 40 D s Lmax 5.00 m Problem 8.2 [Difficulty: 2] Problem 8.3 [Difficulty: 3] Given: Air entering pipe system Find: Flow rate for turbulence in each section; Which become fully developed Solution: From Table A.10 The given data is ν = 1.69 × 10 2 −5 m ⋅ L = 2⋅ m s D1 = 25⋅ mm D2 = 15⋅ mm D3 = 10⋅ mm or Q= Recrit = 2300 The critical Reynolds number is Writing the Reynolds number as a function of flow rate Re = V⋅ D ν = Q π 4 ⋅ D 2 ν ⋅D Re⋅ π⋅ ν⋅ D 4 Then the flow rates for turbulence to begin in each section of pipe are Q1 = Q2 = Q3 = Recrit⋅ π⋅ ν⋅ D1 Q1 = 7.63 × 10 4 Recrit⋅ π⋅ ν⋅ D2 Q2 = 4.58 × 10 4 Recrit⋅ π⋅ ν⋅ D3 Q3 = 3.05 × 10 4 3 −4m s 3 −4m s 3 −4m s Hence, smallest pipe becomes turbulent first, then second, then the largest. For the smallest pipe transitioning to turbulence (Q3) For pipe 3 Re3 = 2300 Llaminar = 0.06⋅ Re3 ⋅ D3 or, for turbulent, Lmin = 25⋅ D3 Lmin = 0.25 m For pipes 1 and 2 Llaminar = 0.06⋅ ⎜ ⎛ 4⋅ Q3 ⎞ ⎝ π⋅ ν⋅ D1 ⎠ ⋅ D1 Llaminar = 1.38 m Lmax = 40⋅ D3 Lmax = 0.4 m Llaminar = 1.38 m Llaminar < L: Fully developed Lmax/min < L: Fully developed Llaminar < L: Fully developed ⎛ 4⋅ Q3 ⎞ Llaminar = 0.06⋅ ⎜ ⎝ π⋅ ν⋅ D2 ⎠ ⋅ D2 Llaminar = 1.38 m Llaminar < L: Fully developed For the middle pipe transitioning to turbulence (Q2) For pipe 2 Re2 = 2300 Llaminar = 0.06⋅ Re2 ⋅ D2 Llaminar = 2.07 m or, for turbulent, Lmin = 25⋅ D2 Lmin = 1.23⋅ ft Lmax = 40⋅ D2 Llaminar > L: Not fully developed Lmax = 0.6 m Lmax/min < L: Fully developed For pipes 1 and 3 ⎛ 4 ⋅ Q2 ⎞ L1 = 0.06⋅ ⎜ ⎝ π⋅ ν⋅ D1 L3min = 25⋅ D3 ⎠ ⋅ D1 L3min = 0.25⋅ m L1 = 2.07⋅ m L3max = 40⋅ D3 Llaminar > L: Not fully developed L3max = 0.4 m Lmax/min < L: Fully developed For the large pipe transitioning to turbulence (Q1) For pipe 1 Re1 = 2300 Llaminar = 0.06⋅ Re1 ⋅ D1 Llaminar = 3.45 m or, for turbulent, Lmin = 25⋅ D1 Lmin = 2.05⋅ ft Lmax = 40⋅ D1 Llaminar > L: Not fully developed Lmax = 1.00 m Lmax/min < L: Fully developed For pipes 2 and 3 L2min = 25⋅ D2 L2min = 1.23⋅ ft L2max = 40⋅ D2 L2max = 0.6 m Lmax/min < L: Fully developed L3min = 25⋅ D3 L3min = 0.82⋅ ft L3max = 40⋅ D3 L3max = 0.4 m Lmax/min < L: Fully developed Problem 8.4 [Difficulty: 2] Given: That transition to turbulence occurs at about Re = 2300 Find: Plots of average velocity and volume and mass flow rates for turbulence for air and water Solution: The basic equations are From Tables A.8 and A.10 For the average velocity V⋅ D Re = Recrit = 2300 ν kg ρair = 1.23⋅ 3 m V= νair = 1.45 × 10 ⋅ s Vair = 2 −5 m ⋅ 2 s Vair = D Hence for air Vw = ⋅ π 4 2 ⋅D ⋅V = π π 4 2 Vw = Recrit⋅ ν D 2 −5 m Qair = × 2300 × 1.45⋅ 10 4 ⋅ s = π⋅ Recrit⋅ ν 4 m s D ⋅D 2 ⋅D 2 For water D 2 s ⋅D ⋅ s 0.00262 ⋅ D Q = A⋅ V = m 0.0334⋅ 2 −6 m For the volume flow rates νw = 1.14 × 10 D 2300 × 1.14 × 10 For water 2 −6 m kg ρw = 999 ⋅ 3 m Recrit⋅ ν 2300 × 1.45 × 10 Hence for air 2 −5 m π −6 m Qw = × 2300 × 1.14⋅ 10 ⋅ ⋅D 4 s m Qair = 0.0262⋅ ×D s 2 m Qw = 0.00206 ⋅ ×D s Finally, the mass flow rates are obtained from volume flow rates mair = ρair⋅ Qair kg mair = 0.0322⋅ ×D m⋅ s mw = ρw⋅ Qw kg mw = 2.06⋅ ×D m⋅ s These results can be plotted in Excel as shown below in the next two pages ⋅ s From Tables A.8 and A.10 the data required is ◊ air = 1.23 kg/m 3 2 ◊ air = 1.45E-05 m /s ◊w = 999 kg/m 3 ◊ w = 1.14E-06 m /s 2 0.0001 0.001 0.01 0.05 V air (m/s) 333.500 33.350 3.335 0.667 2.62 0.262 D (m) V w (m/s) 26.2 1.0 2.5 5.0 7.5 10.0 3.34E-02 1.33E-02 6.67E-03 4.45E-03 3.34E-03 5.24E-02 2.62E-03 1.05E-03 5.24E-04 3.50E-04 2.62E-04 3 Q air (m /s) 2.62E-06 2.62E-05 2.62E-04 1.31E-03 2.62E-02 6.55E-02 1.31E-01 1.96E-01 2.62E-01 Q w (m 3/s) 2.06E-07 2.06E-06 2.06E-05 1.03E-04 2.06E-03 5.15E-03 1.03E-02 1.54E-02 2.06E-02 m air (kg/s) 3.22E-06 3.22E-05 3.22E-04 1.61E-03 3.22E-02 8.05E-02 1.61E-01 2.42E-01 3.22E-01 m w (kg/s) 2.06E-04 2.06E-03 2.06E-02 1.03E-01 2.06E+00 5.14E+00 1.03E+01 1.54E+01 2.06E+01 Average Velocity for Turbulence in a Pipe 1.E+04 V (m/s) 1.E+02 Velocity (Air) Velocity (Water) 1.E+00 1.E-02 1.E-04 1.E-04 1.E-03 1.E-02 1.E-01 D (m ) 1.E+00 1.E+01 Flow Rate for Turbulence in a Pipe Q (m3/s) 1.E+01 1.E-01 Flow Rate (Air) Flow Rate (Water) 1.E-03 1.E-05 1.E-07 1 .E-04 1.E-03 1.E-02 1.E-01 1.E+00 1.E+01 D (m) Mass Flow Rate for Turbulence in a Pipe m flow (kg/s) 1.E+02 1.E+00 Mas s Flow Rate (Air) Mas s Flow Rate (Water) 1.E-02 1.E-04 1.E-06 1.E-04 1.E-03 1.E-02 1.E-01 D (m) 1.E+00 1.E+01 Problem 8.5 [Difficulty: 4] Part 1/2 Problem 8.5 [Difficulty: 4] Part 2/2 Problem 8.6 [Difficulty: 2] Problem 8.7 [Difficulty: 2] Problem 8.8 [Difficulty: 3] Problem 8.9 [Difficulty: 3] D p1 F a L Given: Piston cylinder assembly Find: Rate of oil leak Solution: Q Basic equation l 3 3 = a ⋅ ∆p Q= 12⋅ μ⋅ L π⋅ D⋅ a ⋅ ∆p (from Eq. 8.6c; we assume laminar flow and verify this is correct after solving) 12⋅ μ⋅ L F 4⋅ F ∆p = p 1 − p atm = = A 2 π⋅ D For the system 4 ∆p = × 4500⋅ lbf × π ⎛ 1 × 12⋅ in ⎞ ⎜ 4⋅ in 1⋅ ft ⎝ ⎠ 2 μ = 0.06 × 0.0209⋅ At 120 oF (about 50oC), from Fig. A.2 ∆p = 358 ⋅ psi lbf ⋅ s ft Q = π 12 × 4 ⋅ in × ⎜⎛ 0.001 ⋅ in × Check Re: ⎝ V= Re = Q A μ = 1.25 × 10 2 − 3 lbf ⋅ s ⋅ ft 2 2 2 3 3 ft 1 − 5 ft ⎞ × 358 ⋅ lbf × 144 ⋅ in × × Q = 1.25 × 10 ⋅ 12⋅ in ⎠ 2 −3 2 ⋅ in s 2 1 ⋅ ft 1.25 × 10 lbf ⋅ s in 1 ⋅ ft = Q a ⋅ π⋅ D V⋅ a 1 π × 1.25 × 10 ν = 6 × 10 ν Re = 0.143 ⋅ V = ft s × 0.001 ⋅ in × 1 ⋅ ft 12⋅ in −5 × − 5 ft 3 s × 10.8 ft × 1 .001⋅ in × 1 4 ⋅ in 2 × ⎛ 12⋅ in ⎞ ⎜ 1⋅ ft ⎝ ⎠ − 4 ft ν = 6.48 × 10 s s −4 6.48 × 10 ft 2 ⋅ Re = 0.0184 3 Q = 0.0216⋅ 2 V = 0.143 ⋅ Q ⎛ π⋅ D2 ⎞ ⎜ ⎝ 4 ⎠ 3 4 − 5 ft × Vp = × 1.25 × 10 π s 12⋅ in ⎞ ⎛ 1 ⎜ 4 ⋅ in × 1 ⋅ ft ⎝ ⎠ s ft s 2 (at 120 oF, from Fig. A.3) s so flow is very much laminar The speed of the piston is approximately Vp = in 2 The piston motion is negligible so our assumption of flow between parallel plates is reasonable − 4 ft Vp = 1.432 × 10 ⋅ s Problem 8.10 [Difficulty: 2] Problem 8.11 [Difficulty: 2] y 2h Given: Laminar flow between flat plates Find: Shear stress on upper plate; Volume flow rate per width Solution: du τyx = μ⋅ dy Then τyx = At the upper surface y=h The volume flow rate is ⌠ h 2 ⌠ ⌠ h ⋅ b dp ⎮ Q = ⎮ u dA = ⎮ u ⋅ b dy = − ⋅ ⋅⎮ ⌡ 2⋅ μ dx ⎮ ⌡ −h ⌡ −h 2 2 ⋅ u(y) = − dp dx h 2 Basic equation ⋅ ⎡ dp 2⋅ μ dx ⋅ ⎢1 − ⎣ b =− 2 ⎛ y ⎞ ⎥⎤ ⎜h ⎝ ⎠⎦ (from Eq. 8.7) 2⋅ y ⎞ dp ⎜ 2 = −y⋅ dx ⎝ h ⎠ ⋅⎛− 1⋅ m 3 N τyx = −1.5⋅ mm × × 1.25 × 10 ⋅ 1000⋅ mm 2 m ⋅m h −h Q x 2 3 × ⎛⎜ 1.5⋅ mm × ⎝ 1⋅ m 3 ⎡ ⎢1 − ⎣ 2 ⎛ y ⎞ ⎥⎤ dy ⎜h ⎝ ⎠⎦ τyx = −1.88Pa 3 Q= − 2 ⎞ × 1.25 × 103⋅ N × m 1000⋅ mm ⎠ 0.5⋅ N⋅ s 2 m ⋅m Q b 2⋅ h ⋅ b dp ⋅ 3⋅ μ dx = −5.63 × 10 2 −6 m s Problem 8.12 Given: Piston-cylinder assembly Find: Mass supported by piston [Difficulty: 3] Solution: Basic equation Q l Available data 3 a ∆p This is the equation for pressure-driven flow between parallel plates; for a small gap a, the flow between the piston and cylinder can be modeled this way, with l = πD 12 μ L L 4 inD 4 in a 0.001 in From Fig. A.2, SAE10 oil at 20oF is Q 0.1 gpm μ 0.1 N s 68 °F 20 °C 3 lbf s μ 2.089 10 or 2 m Hence, solving for ∆p ∆p 12 μ L Q π D a 2 Note the following M π D 4 2 4 ∆p 2.133 10 psi 3 F ∆p A ∆p A force balance for the piston involves the net pressure force Hence ft ∆p π 4 Q Vave a π D ft Vave 2.55 s Hence an estimate of the Reynolds number in the gap is W M g and the weight 5 M 8331 slug g 2 D M 2.68 10 lb ν 10 Re 2 4 m s a Vave ν ν 1.076 10 3 ft 2 s Re 0.198 This is a highly viscous flow; it can be shown that the force on the piston due to this motion is much less than that due to ∆p! Note also that the piston speed is Vpiston Vpiston Vave 4 Q 2 π D 0.1 % ft Vpiston 0.00255 s so the approximation of stationary walls is valid Problem 8.13 [Difficulty: 3] Problem 8.14 [Difficulty: 3] Problem 8.15 [Difficulty: 3] Given: Hydrostatic bearing Find: Required pad width; Pressure gradient; Gap height Solution: Basic equation Q l Available data h 3 12 μ dp dx F 1000 lbf l 1 ft p i 35 psi 212 °F 100 °C At 100 oC from Fig. A.2, for SAE 10-30 (F is the load on width l) μ 0.01 Q 2.5 N s gal per ft hr 4 lbf s μ 2.089 10 2 m ft 2 For a laminar flow (we will verify this assumption later), the pressure gradient is constant 2 x p ( x ) p i 1 W where p i = 35 psi is the inlet pressure (gage), and x = 0 to W/2 F l p dx Hence the total force in the y direction due to pressure is where b is the pad width into the paper W 2 2 x dx F 2 l p i 1 W F 1 2 p i l W 0 W This must be equal to the applied load F. Hence dp The pressure gradient is then dx ∆p W 2 ∆p W 2 2 F pi l 35 lbf 2 W 0.397 ft in 1 0.397 ft 176 psia ft 2 Q From the basic equation l Check Re: From Fig. A.3 Re V D ν ν 1.2 10 h 3 12 μ 1 dp dx Q 12 μ l h dp dx we can solve for 3 h 2.51 10 3 in D Q h Q ν A ν l h 2 5 m s 4 ft ν 1.29 10 2 s Re Q ν l Re 0.72 so flow is very laminar Problem 8.16 [Difficulty: 3] Problem 8.17 Given: Navier-Stokes Equations Find: Derivation of Eq. 8.5 [Difficulty: 2] Solution: The Navier-Stokes equations are 4 3 ∂u ∂v ∂w + + =0 ∂x ∂y ∂z 1 4 5 3 (5.1c) 6 4 3 ⎛∂ u ∂ u ∂ u⎞ ⎛ ∂u ∂u ∂u ∂u ⎞ ∂p + µ ⎜⎜ 2 + 2 + 2 ⎟⎟ +u +v + w ⎟⎟ = ρg x − ∂z ⎠ ∂y ∂x ∂x ∂y ∂z ⎠ ⎝ ∂t ⎝ ∂x 2 ρ ⎜⎜ 1 4 5 3 2 4 2 5 (5.27a) 3 ⎛ ∂ 2v ∂ 2v ∂ 2v ⎞ ⎛ ∂v ∂p ∂v ⎞ ∂v ∂v ρ ⎜⎜ + u + v + w ⎟⎟ = ρg y − + µ ⎜⎜ 2 + 2 + 2 ⎟⎟ ∂z ⎠ ∂y ∂y ∂z ⎠ ∂y ∂x ⎝ ∂t ⎝ ∂x 1 3 3 3 3 3 3 (5.27b) 3 3 ⎛∂ w ∂ w ∂ w⎞ ⎛ ∂w ∂p ∂w ⎞ ∂w ∂w + µ ⎜⎜ 2 + 2 + 2 ⎟⎟ ⎟⎟ = ρg z − +w +v +u ∂z ⎠ ∂y ∂z ∂z ⎠ ∂y ∂x ⎝ ∂t ⎝ ∂x 2 ρ ⎜⎜ 2 2 (5.27c) The following assumptions have been applied: (1) Steady flow (given). (2) Incompressible flow; ρ = constant. (3) No flow or variation of properties in the z direction; w= 0 and ∂/∂z = 0. (4) Fully developed flow, so no properties except pressure p vary in the x direction; ∂/∂x = 0. (5) See analysis below. (6) No body force in the x direction; gx = 0 Assumption (1) eliminates time variations in any fluid property. Assumption (2) eliminates space variations in density. Assumption (3) states that there is no z component of velocity and no property variations in the z direction. All terms in the z component of the Navier–Stokes equation cancel. After assumption (4) is applied, the continuity equation reduces to ∂v/∂y = 0. Assumptions (3) and (4) also indicate that ∂v/∂z = 0 and ∂v/∂x = 0. Therefore v must be constant. Since v is zero at the solid surface, then v must be zero everywhere. The fact that v = 0 reduces the Navier–Stokes equations further, as indicated by (5). Hence for the y direction ∂p = ρg ∂y which indicates a hydrostatic variation of pressure. In the x direction, after assumption (6) we obtain ∂ 2u ∂p µ 2 − =0 ∂y ∂x Integrating twice u= 1 ∂p 2 c1 y + y + c2 µ 2 µ ∂x To evaluate the constants, c1 and c2, we must apply the boundary conditions. At y = 0, u = 0. Consequently, c2 = 0. At y = a, u = 0. Hence 0= 1 ∂p 2 c1 a + a µ 2 µ ∂x which gives c1 = − 1 ∂p a 2 µ ∂x and finally u= 2 a 2 ∂p ⎡⎛ y ⎞ ⎛ y ⎞⎤ ⎢⎜ ⎟ − ⎜ ⎟ ⎥ 2 µ ∂x ⎢⎣⎝ a ⎠ ⎝ a ⎠⎥⎦ Problem 8.18 [Difficulty: 4] Problem 8.19 [Difficulty: 5] Problem 8.20 [Difficulty: 2] Problem 8.21 [Difficulty: 3] Given: Laminar velocity profile of power-law fluid flow between parallel plates Find: Expression for flow rate; from data determine the type of fluid Solution: ⎡⎢ n h ∆p ⎞ n⋅ h ⎢ u = ⎛⎜ ⋅ ⋅ ⋅ 1− ⎝ k L ⎠ n + 1 ⎢⎣ n+ 1⎤ 1 The velocity profile is ⌠ Q = w⋅ ⎮ ⌡ The flow rate is then ⎛y⎞ ⎜h ⎝ ⎠ n ⎥ ⎥ ⎥⎦ h h or, because the flow is symmetric u dy −h ⌠ ⎮ ⎮ ⎮ 1− ⎮ ⌡ The integral is computed as 0 n+ 1 ⎛y⎞ ⎜h ⎝ ⎠ ⌠ Q = 2 ⋅ w⋅ ⎮ u dy ⌡ n 2⋅ n+ 1⎤ ⎡⎢ ⎥ n ⎢ ⎥ n y dy = y ⋅ ⎢1 − ⋅ ⎛⎜ ⎞ ⎥⎦ ⎣ 2⋅ n + 1 ⎝ h ⎠ 1 2⋅ n+ 1⎤ ⎡ n ⎢ ⎥ h ∆p ⎞ n⋅ h n n Q = 2 ⋅ w⋅ ⎛⎜ ⋅ ⋅ ⋅ h ⋅ ⎢1 − ⋅ ( 1) ⎥ ⎝ k L ⎠ n + 1 ⎣ 2⋅ n + 1 ⎦ Using this with the limits An Excel spreadsheet can be used for computation of n. The data is dp (kPa) Q (L/min) 10 0.451 20 0.759 30 1.01 40 1.15 50 1.41 60 1.57 70 1.66 80 1.85 90 2.05 1 n This must be fitted to Q= 1 2 ⎛ h ⋅ ∆p ⎞ ⋅ 2 ⋅ n⋅ w⋅ h or ⎜ L ⎝k ⎠ 2⋅ n + 1 Q = k ⋅ ∆p n 100 2.25 1 n Q= 2 ⎛ h ⋅ ∆p ⎞ ⋅ 2 ⋅ n⋅ w⋅ h ⎜k L ⎝ ⎠ 2⋅ n + 1 We can fit a power curve to the data Flow Rate vs Applied Pressure for a Non-Newtonian Fluid 10.0 Q (L/min) Data Power Curve Fit 1.0 Q = 0.0974dp0.677 2 R = 0.997 0.1 10 Hence dp (kPa) 1/n = It's a dilatant fluid 0.677 n = 1.48 100 Problem 8.22 [Difficulty: 2] Given: Laminar flow between moving plates Find: Expression for velocity; Volume flow rate per depth Solution: Given data ft U1 = 5⋅ s d = 0.2⋅ in ft U2 = 2⋅ s Using the analysis of Section 8.2, the sum of forces in the x direction is ⎡ ∂ dy ⎛ ⎛ dy ⎞⎤ dx dx ⎞ ⎢τ + τ ⋅ − ⎜ τ − ∂ τ ⋅ ⎥ ⋅ b ⋅ dx + ⎜ p − ∂ p ⋅ − p + ∂ p ⋅ ⋅ b ⋅ dy = 0 2 ⎠⎦ ∂y ∂x 2 ∂x 2 ⎠ ⎣ ∂y 2 ⎝ ⎝ Simplifying dτ dy = dp dx 2 =0 μ⋅ or d u dy Integrating twice u = c1 ⋅ y + c2 Boundary conditions: u ( 0 ) = −U1 c2 = −U1 ( ) 2 =0 ft c2 = −5 s u ( y = d ) = U2 u in ft/s, y in ft Hence y u ( y ) = U1 + U2 ⋅ − U1 d u ( y ) = 420 ⋅ y − 5 The volume flow rate is ⌠ ⌠ Q = ⎮ u dA = b ⋅ ⎮ u dy ⌡ ⌡ ⌠ Q = b⋅ ⎮ ⎮ ⌡ d Q = b⋅ d⋅ (U2 − U1) Q 2 b = d⋅ (U2 − U1) 2 3 ft Q b = 0.2⋅ in × 1 ⋅ ft 12⋅ in × 1 2 × ( 2 − 5) × ft Q s b = −.025⋅ s ft 3 ft Q b = −.025⋅ s ft × 7.48⋅ gal 1 ⋅ ft 3 × 60⋅ s Q 1 ⋅ min b U1 + U2 d = −11.2⋅ gpm ft −1 c1 = 420 s U + U2 d 2 ⎞ ⎡ U + U ⋅ y − U ⎤ dx = b⋅ ⎛⎜ 1 ⋅ − U ⋅ d ⎢( 1 ) ⎥ 2 d 1 1 2 ⎣ ⎦ ⎝ d ⎠ 0 Hence c1 = Problem 8.23 [Difficulty: 2] Problem 8.24 [Difficulty: 3] Given: Properties of two fluids flowing between parallel plates; applied pressure gradient Find: Velocity at the interface; maximum velocity; plot velocity distribution Solution: dp Given data =k dx k = −50⋅ kPa m h = 5 ⋅ mm μ1 = 0.1⋅ N⋅ s μ2 = 4 ⋅ μ1 2 μ2 = 0.4⋅ m N⋅ s 2 m (Lower fluid is fluid 1; upper is fluid 2) Following the analysis of Section 8.2, analyse the forces on a differential CV of either fluid The net force is zero for steady flow, so dτ dy ⎞⎤ dp dx ⎞⎤ ⎡ dτ dy ⎛ ⎡ dp dx ⎛ ⎢τ + dy ⋅ 2 − ⎜ τ − dy ⋅ 2 ⎥ ⋅ dx⋅ dz + ⎢p − dx ⋅ 2 − ⎜ p + dx ⋅ 2 ⎥ ⋅ dy⋅ dz = 0 ⎣ ⎝ ⎠⎦ ⎣ ⎝ ⎠⎦ dτ Simplifying dy = dp dx =k μ⋅ so for each fluid d 2 dy 2 u =k Applying this to fluid 1 (lower fluid) and fluid 2 (upper fluid), integrating twice yields u1 = k 2 ⋅ y + c1 ⋅ y + c2 2 ⋅ μ1 u2 = k 2 2 ⋅ μ2 ⋅ y + c3 ⋅ y + c4 For convenience the origin of coordinates is placed at the centerline y = −h We need four BCs. Three are obvious u1 = 0 y=0 (1) u 1 = u 2 (2) y=h u2 = 0 The fourth BC comes from the fact that the stress at the interface generated by each fluid is the same du1 du2 μ1 ⋅ = μ2 ⋅ dy dy y=0 Using these four BCs 0= k 2 ⋅ μ1 2 ⋅ h − c1 ⋅ h + c2 c2 = c4 0= k 2 ⋅ μ2 (4) 2 ⋅ h + c3 ⋅ h + c4 μ1 ⋅ c1 = μ2 ⋅ c3 Hence, after some algebra c1 = k⋅ h 2 ⋅ μ1 ⋅ (μ2 − μ1) (μ2 + μ1) c4 = − k⋅ h 2 μ2 + μ1 c2 = c4 c3 = k⋅ h 2 ⋅ μ2 ⋅ (μ2 − μ1) (μ2 + μ1) (3) c1 = −750 1 c2 = 2.5 s m c3 = −187.5 s 1 c4 = 2.5 s m s The velocity distributions are then u1( y) = k 2 ⋅ μ1 ⎡ 2 ⋅ ⎢y + y ⋅ h ⋅ ⎣ (μ2 − μ1)⎤ ⎥− (μ2 + μ1)⎦ k⋅ h 2 u2( y) = μ2 + μ1 k 2 ⋅ μ2 ⎡ 2 ⋅ ⎢y + y ⋅ h ⋅ ⎣ (μ2 − μ1)⎤ ⎥− (μ2 + μ1)⎦ k⋅ h μ2 + μ1 Evaluating either velocity at y = 0, gives the velocity at the interface u interface = − k⋅ h 2 u interface = 2.5 μ2 + μ1 m s The plots of these velocity distributions can be done in Excel. Typical curves are shown below 5 y (mm) 2.5 0 0.5 1 1.5 2 2.5 3 3.5 − 2.5 −5 u (m/s) Clearly, u 1 has the maximum velocity, when ( ( h μ2 − μ1 y max = − ⋅ 2 μ2 + μ1 ) ) du1 dy =0 y max = −1.5 mm (We could also have used Excel's Solver for this.) 2 ⋅ y max + h ⋅ or ( ) u max = u 1 y max u max = 3.06 (μ2 − μ1) (μ2 + μ1) m s =0 2 Problem 8.25 Given: Laminar flow of two fluids between plates Find: Velocity at the interface [Difficulty: 3] Solution: Using the analysis of Section 8.2, the sum of forces in the x direction is ⎛ ∂ dx dx ⎞ ⎡ ∂ dy ⎛ ∂ dy ⎞⎤ ∂ ⋅ b ⋅ dy = 0 ⎢τ + τ ⋅ − ⎜ τ − τ ⋅ ⎥ ⋅ b ⋅ dx + ⎜ p − p ⋅ − p + p ⋅ ∂x 2 ⎠ ⎣ ∂y 2 ⎝ ∂y 2 ⎠⎦ ⎝ ∂x 2 Simplifying dτ dy = dp dx 2 =0 μ⋅ or dy y=0 u1 = 0 2 =0 u 1 = c1 ⋅ y + c2 Applying this to fluid 1 (lower fluid) and fluid 2 (upper fluid), integrating twice yields We need four BCs. Three are obvious d u y = h u1 = u2 y = 2⋅ h u 2 = c3 ⋅ y + c4 u2 = U The fourth BC comes from the fact that the stress at the interface generated by each fluid is the same y=h du1 du2 μ1⋅ = μ2⋅ dy dy Using these four BCs 0 = c2 c1⋅ h + c2 = c3⋅ h + c4 Hence c2 = 0 From the 2nd and 3rd equations c1⋅ h − U = −c3⋅ h Hence μ1 c1⋅ h − U = −c3⋅ h = − ⋅ h ⋅ c1 μ2 and Hence for fluid 1 (we do not need to complete the analysis for fluid 2) 20⋅ Evaluating this at y = h, where u 1 = u interface u interface = ft s 1⎞ ⎛1 + ⎜ 3⎠ ⎝ U = c3⋅ 2⋅ h + c4 μ1 ⋅ c1 = μ2 ⋅ c3 c1 = U ⎛ μ1 ⎞ ⎝ μ2 ⎠ h⋅ ⎜ 1 + u1 = U ⎛ h ⋅⎜1 + ⎝ μ1 ⎞ μ2 ⎠ u interface = 15⋅ ft s ⋅y μ1⋅ c1 = μ2⋅ c3 Problem 8.26 [Difficulty: 2] Given: Computer disk drive Find: Flow Reynolds number; Shear stress; Power required Solution: For a distance R from the center of a disk spinning at speed ω V R ω The gap Reynolds number is Re V 25 mm ρ V a Re 22.3 m s 1000 mm 8500 rpm 2 π rad rev 1 min ν 1.45 10 ν 6 0.25 10 V 22.3 60 s 2 5 m V a μ 1 m 5 1.45 10 s from Table A.10 at 15oC s s m m Re 0.384 2 m The flow is definitely laminar The shear stress is then τ μ du dy μ τ 1.79 10 5 N s V μ 1.79 10 a 5 N s 2 m The power required is P T ω T τ A R P τ A R ω P 1600 N 2 m from Table A.10 at 15oC 2 m 22.3 m s 1 0.25 10 6 τ 1.60 kPa m where torque T is given by A ( 5 mm) with 2.5 10 5 2 m 25 mm 2 A 2.5 10 1 m 1000 mm 5 8500 rpm 2 m 2 π rad rev 1 min 60 s P 0.890 W Problem 8.27 [Difficulty: 2] Given: Velocity profile between parallel plates Find: Pressure gradients for zero stress at upper/lower plates; plot Solution: From Eq. 8.8, the velocity distribution is The shear stress is u= U⋅ y a + ⎛ ∂ ⎞ ⎡⎢⎛ y ⎞ 2 − p ⋅ ⎜ 2 ⋅ μ ⎝ ∂x ⎠ ⎣⎝ a ⎠ a 2 ⋅⎜ y⎤ ⎥ a⎦ 2 1 ⎛∂ ⎞ y τyx = μ⋅ − ⎞ = μ⋅ + ⋅ ⎜ p ⋅ ⎛ 2⋅ ⎜ dy a 2 ⎝ ∂x ⎠ a 2 a du U a ⎝ (a) For τyx = 0 at y = a The velocity distribution is then (b) For τyx = 0 at y = 0 The velocity distribution is then 0 = μ⋅ u= The velocity distributions can be plotted in Excel. a U⋅ y a 0 = μ⋅ u= U U a U⋅ y a + − − + a ∂ ⋅ p 2 ∂x a 2 2⋅ μ ⋅ ∂ ∂x 2 ⋅ U⋅ μ a 2 ⎡ y 2 ⎞ − a ⎣⎝ ⎠ ⋅ ⎢⎛⎜ y⎤ ⎥ a⎦ a ∂ ⋅ p 2 ∂x a 2 2⋅ μ ⋅ ⎠ u U ∂ ∂x 2 ⋅ U⋅ μ a 2 ⎡ y ⎞2 − ⎣⎝ a ⎠ ⋅ ⎢⎛⎜ y⎤ ⎥ a⎦ u U p =− 2 ⋅ U⋅ μ a = 2⋅ p = y − a ⎛y⎞ ⎜a ⎝ ⎠ 2 ⋅ U⋅ μ a = 2 ⎛y⎞ ⎜a ⎝ ⎠ 2 2 2 y /a 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 (a) u /U 0.000 0.190 0.360 0.510 0.640 0.750 0.840 0.910 0.960 0.990 1.00 (b) u /U 0.000 0.010 0.040 0.090 0.160 0.250 0.360 0.490 0.640 0.810 1.000 Zero-Stress Velocity Distributions 1.00 Zero Stress Upper Plate Zero Stress Lower Plate y /a 0.75 0.50 0.25 0.00 0.00 0.25 0.50 u /U 0.75 1.00 Problem 8.28 [Difficulty: 2] Problem 8.29 [Difficulty: 2] Problem 8.30 [Difficulty: 3] Given: Data on flow of liquids down an incline Find: Velocity at interface; velocity at free surface; plot Solution: 2 Given data h = 10⋅ mm θ = 60⋅ deg ν1 ν2 = 5 m ν1 = 0.01⋅ s ν2 = 2 × 10 2 −3m s (The lower fluid is designated fluid 1, the upper fluid 2) From Example 5.9 (or Exanple 8.3 with g replaced with gsinθ), a free body analysis leads to (for either fluid) d 2 dy 2 u =− ρ⋅ g ⋅ sin( θ) μ Applying this to fluid 1 (lower fluid) and fluid 2 (upper fluid), integrating twice yields u1 = − ρ⋅ g ⋅ sin( θ) 2 ⋅ μ1 2 ⋅ y + c1 ⋅ y + c2 We need four BCs. Two are obvious u2 = − ρ⋅ g ⋅ sin( θ) 2 ⋅ μ2 2 ⋅ y + c3 ⋅ y + c4 y=0 u1 = 0 (1) y=h u1 = u2 (2) The third BC comes from the fact that there is no shear stress at the free surface y = 2⋅ h du2 μ2 ⋅ =0 dy (3) The fourth BC comes from the fact that the stress at the interface generated by each fluid is the same du1 du2 μ1 ⋅ = μ2 ⋅ dy dy y=h Using these four BCs c2 = 0 − ρ⋅ g ⋅ sin( θ) 2 ⋅ μ1 2 ⋅ h + c1 ⋅ h + c2 = − −ρ⋅ g ⋅ sin( θ) ⋅ 2 ⋅ h + μ2 ⋅ c3 = 0 Hence, after some algebra c1 = 2 ⋅ ρ⋅ g ⋅ sin( θ) ⋅ h μ1 c2 = 0 ρ⋅ g ⋅ sin( θ) 2 ⋅ μ2 (4) 2 ⋅ h + c3 ⋅ h + c4 −ρ⋅ g ⋅ sin( θ) ⋅ h + μ1 ⋅ c1 = −ρ⋅ g ⋅ sin( θ) ⋅ h + μ2 ⋅ c3 c3 = 2 ⋅ ρ⋅ g ⋅ sin( θ) ⋅ h μ2 2 c4 = 3 ⋅ ρ⋅ g ⋅ sin( θ) ⋅ h ⋅ (μ2 − μ1) 2 ⋅ μ1 ⋅ μ2 The velocity distributions are then u1 = ρ⋅ g ⋅ sin( θ) 2 ⋅ μ1 ( ⋅ 4⋅ y⋅ h − y 2 ) u2 = ⎡ ρ⋅ g ⋅ sin( θ) 2 ⋅ ⎢3 ⋅ h ⋅ 2 ⋅ μ2 (μ2 − μ1) μ1 ⎣ + 4⋅ y⋅ h − y 2⎤ ⎥ ⎦ Rewriting in terms of ν1 and ν2 (ρ is constant and equal for both fluids) u1 = g ⋅ sin( θ) 2 ⋅ ν1 ( ⋅ 4⋅ y⋅ h − y ) 2 u2 = g ⋅ sin( θ) 2 ⋅ ν2 ⎡ 2 ⋅ ⎢3 ⋅ h ⋅ (ν2 − ν1) ⎣ ν1 + 4⋅ y⋅ h − y 2⎤ ⎥ ⎦ (Note that these result in the same expression if ν1 = ν2, i.e., if we have one fluid) 2 u interface = Evaluating either velocity at y = h, gives the velocity at the interface Evaluating u 2 at y = 2h gives the velocity at the free surface 2 3 ⋅ g ⋅ h ⋅ sin( θ) u freesurface = g ⋅ h ⋅ sin( θ) ⋅ u freesurface⋅ h Note that a Reynolds number based on the free surface velocity is ν2 2 ⋅ ν1 (3⋅ ν2 + ν1) = 1.70 2 ⋅ ν1 ⋅ ν2 u interface = 0.127 0.000 0.0166 0.0323 0.0472 0.061 0.074 0.087 0.098 0.109 0.119 0.127 u 2 (m/s) indicating laminar flow Velocity Distributions down an Incline 24 Lower Velocity 20 0.127 0.168 0.204 0.236 0.263 0.287 0.306 0.321 0.331 0.338 0.340 y (mm) 0.000 1.000 2.000 3.000 4.000 5.000 6.000 7.000 8.000 9.000 10.000 11.000 12.000 13.000 14.000 15.000 16.000 17.000 18.000 19.000 20.000 Upper Velocity 16 12 8 4 0 0.0 0.1 0.2 u (m/s) 0.3 s u freesurface = 0.340 The velocity distributions can be plotted in Excel. y (mm) u 1 (m/s) m 0.4 m s Problem 8.31 [Difficulty: 2] Given: Velocity distribution on incline Find: Expression for shear stress; Maximum shear; volume flow rate/mm width; Reynolds number Solution: u(y) = From Example 5.9 τ = μ⋅ For the shear stress τ is a maximum at y = 0 ρ⋅ g ⋅ sin( θ) μ du dy ⎛ y ⎝ 2 ⋅ ⎜ h⋅ y − 2⎞ ⎠ = ρ⋅ g ⋅ sin( θ) ⋅ ( h − y ) τmax = ρ⋅ g ⋅ sin( θ) ⋅ h = SG ⋅ ρH2O⋅ g ⋅ sin( θ) ⋅ h τmax = 1.2 × 1000 kg 3 × 9.81⋅ m m 2 2 × sin( 15⋅ deg) × 0.007 ⋅ m × s N⋅ s τmax = 21.3 Pa kg⋅ m This stress is in the x direction on the wall ⌠ h ⎮ ⌠ ⌠ Q = ⎮ u dA = w⋅ ⎮ u ( y ) dy = w⋅ ⎮ ⌡ ⌡ ⎮ 0 ⌡ The flow rate is h ρ⋅ g ⋅ sin( θ) μ 2 ⎛ y ⎞ dy ⋅ ⎜ h⋅ y − 2 ⎠ ⎝ Q= ρ⋅ g ⋅ sin( θ) ⋅ w⋅ h 3⋅ μ 0 3 m Q w = 1 3 × 1.2 × 1000 kg 3 × 9.81⋅ m m 2 2 m 3 × sin( 15⋅ deg) × ( 0.007 ⋅ m) × s The gap Reynolds number is V= Re = Q A = Q V = 217 ⋅ w⋅ h −4 = 2.18 × 10 s Q m w = 217 mm 3 s mm × 1 V = 31.0⋅ 7 ⋅ mm mm μ kg 3 m × 31⋅ mm s 2 × 7 ⋅ mm × m 1.60⋅ N⋅ s × ⎛ 1⋅ m ⎞ ⎜ ⎝ 1000⋅ mm ⎠ 3 s ρ⋅ V⋅ h Re = 1.2 × 1000 The flow is definitely laminar N⋅ s 1.60⋅ N⋅ s kg⋅ m mm The average velocity is 2 ⋅ mm 2 Re = 0.163 s 3 Problem 8.32 [Difficulty: 3] Given: Flow between parallel plates Find: Shear stress on lower plate; Plot shear stress; Flow rate for pressure gradient; Pressure gradient for zero shear; Plot Solution: u(y) = From Section 8-2 U⋅ y a + ⎡ y 2 ⎞ − 2 ⋅ μ dx ⎣⎝ a ⎠ a 2 ⋅ dp ⋅ ⎢⎛⎜ y⎤ ⎥ a⎦ 3 ft a u = U⋅ For dp/dx = 0 a ⌠ ⌠ U⋅ a y = ⎮ u ( y ) dy = w⋅ ⎮ U⋅ dy = ⎮ ⌡ 2 l a 0 ⌡ y Q a 1 Q = 2 × 5⋅ ft × s 0.1 12 ⋅ ft Q = 0.0208⋅ s ft 0 τ = μ⋅ For the shear stress du dy = μ⋅ U − 7 lbf ⋅ s μ = 3.79 × 10 when dp/dx = 0 a ⋅ ft (Table A.9) 2 The shear stress is constant - no need to plot! τ = 3.79 × 10 − 7 lbf ⋅ s ⋅ ft 2 × 5⋅ ft s × 12 0.1⋅ ft × ⎛ 1⋅ ft ⎞ ⎜ 12⋅ in ⎝ ⎠ 2 −6 τ = 1.58 × 10 Q will decrease if dp/dx > 0; it will increase if dp/dx < 0. τ = μ⋅ For non- zero dp/dx: du dy = μ⋅ U a + a⋅ τ( y = 0.25⋅ a) = μ⋅ At y = 0.25a, we get U a dp dx ⋅ ⎛⎜ + a⋅ y − ⎝a dp dx ⋅ ⎛⎜ 1⎞ 2⎠ 1 ⎝4 − 1⎞ 2⎠ = μ⋅ U a − a dp ⋅ 4 dx lbf Hence this stress is zero when dp dx = 4 ⋅ μ⋅ U a 2 − 7 lbf ⋅ s = 4 × 3.79 × 10 ⋅ ft 2 × 5⋅ ft s × 2 2 ⎛ 12 ⎞ = 0.109 ⋅ ft = 7.58 × 10− 4 psi ⎜ 0.1⋅ ft ft ft ⎝ ⎠ 0.1 y (in) 0.075 0.05 0.025 − 1× 10 −4 0 1× 10 −4 Shear Stress (lbf/ft3) 2× 10 −4 −4 3× 10 ⋅ psi Problem 8.33 [Difficulty: 3] Given: Flow between parallel plates Find: Location and magnitude of maximum velocity; Volume flow in 10 s; Plot velocity and shear stress Solution: From Section 8.2 For u max set du/dx = 0 Hence From Fig. A.2 at U⋅ y u(y) = du b =0= dy ⎡ y 2 ⎞ − 2 ⋅ μ dx ⎣⎝ b ⎠ b + U u = u max dp ⋅ y⎤ ⋅ ⎢⎛⎜ ⎥ b⎦ 2 dp 2 ⋅ y 1 1 dp U ⋅ ⋅⎛ − ⎞= + ⋅ ⋅ (2⋅ y − b) 2 ⋅ μ dx ⎜ 2 a 2 ⋅ μ dx b ⎝b ⎠ b + b 2 b y= at 59 °F = 15⋅ °C 2 μ = 4⋅ μ⋅ U − b⋅ dp dx N⋅ s μ = 0.0835⋅ 2 m y = Hence 0.1⋅ in 2 + 0.0835⋅ ft U⋅ y u max = lbf ⋅ s b u max = 2 ⋅ ft s 2 × 2⋅ lbf ⋅ s ft ft s 2 1 × 2 0.1⋅ in × in ⋅ ft y = 0.0834⋅ in 50⋅ lbf + ⎡ y 2 ⎞ − 2 ⋅ μ dx ⎣⎝ b ⎠ × 2 2 2 50⋅ psi ⎛ 12⋅ in ⎞ ft ⎛ .0834 ⎞ + 1 × ⎛ 0.1 ⋅ ft⎞ × × − × ⎜ 0.1 ⎜ ⎜ 1 ⋅ ft × ft .0835 ⋅ lbf ⋅ s ⎝ ⎠ 2 ⎝ 12 ⎠ ⎝ ⎠ b 2 ⋅ dp ⋅ ⎢⎛⎜ y⎤ ⎥ y = 0.0834⋅ in with b⎦ 2 ⎡ ⎤ ⎢⎛⎜ .0834 ⎞ − ⎛⎜ .0834 ⎞⎥ ⎣⎝ 0.1 ⎠ ⎝ 0.1 ⎠⎦ u max = 2.083 ⋅ b ⌠ =⎮ w ⌡0 Q Q w = 1 2 ⌠ ⎮ u ( y ) dy = w⋅ ⎮ ⎮ ⌡ b ⎡ U⋅ y b 2 dp ⎡⎛ y ⎞ 2 ⎢ + ⋅ ⋅ ⎢⎜ − 2 ⋅ μ dx ⎣⎝ b ⎠ ⎣ b × 2⋅ ft s × 0.1 12 ⋅ ft − 1 12 3 × = 0.0125 s Q ft w 2 50⋅ psi ⎞ ⎛ 12⋅ in ⎞ ft ⎛ 0.1 ⎞ × ⎛⎜ − ×⎜ ⎜ 12 ⋅ ft × ft ⎠ ⎝ 1 ⋅ ft ⎠ .0835 ⋅ lbf ⋅ s ⎝ ⎝ ⎠ ft w 3 ⎥⎥ dy = U⋅ b − b ⋅ dp b⎦⎦ 12⋅ μ dx 2 0 3 Q y⎤⎤ = 5.61⋅ gpm ft 2 ft s Flow = Q w ⋅ ∆t = 5.61⋅ The velocity profile is gpm ft 1 ⋅ min × 60⋅ s u U = y b × 10⋅ s + Flow = 0.935 ⋅ ⎡ y 2 ⎞ − 2 ⋅ μ⋅ U dx ⎣⎝ b ⎠ b 2 ⋅ dp ⋅ ⎢⎛⎜ y⎤ ⎥ b⎦ gal ft τ = μ⋅ For the shear stress du dy = μ⋅ U b + b dp ⎡ ⎛ y ⎞ ⋅ ⋅ ⎢2 ⋅ ⎜ − 1⎤⎥ 2 dx ⎣ ⎝ b ⎠ ⎦ The graphs below can be plotted in Excel 1 0.8 y/b 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 1 1.2 u/U 1 0.8 y/b 0.6 0.4 0.2 − 500 0 500 1× 10 3 Shear Stress (Pa) 3 1.5× 10 3 2× 10 2.5× 10 3 Problem 8.34 [Difficulty: 3] Given: Flow between parallel plates Find: Pressure gradient for no flow; plot velocity and stress distributions; also plot for u = U at y = a/2 Solution: Basic equations Available data From Eq 2 for Q = 0 u(y) = U⋅ y a U = 1.5⋅ dp dx = + m ⋅ dp ⋅ ⎢⎛⎜ y⎤ Q ⎥ (1) a⎦ l = U⋅ a 2 − a 3 dp ⋅ 12⋅ μ dx τ = μ⋅ (2) U a μ = 1⋅ From Fig. A.2 for castor oil at 20oC + a⋅ dp dx N⋅ s 2 m 6 ⋅ μ⋅ U a 2 a = 5 ⋅ mm s 2 ⎡ y 2 ⎞ − 2 ⋅ μ dx ⎣⎝ a ⎠ a = 6 × 1⋅ N⋅ s 2 × 1.5⋅ m m s × 1 dp ( 0.005 ⋅ m) dx 2 The graphs below, using Eqs. 1 and 3, can be plotted in Excel 1 y/a 0.75 0.5 0.25 − 0.5 0 0.5 1 1.5 u (m/s) 1 y/a 0.75 0.5 0.25 −1 − 0.5 0 0.5 1 Shear Stress (kPa) The pressure gradient is adverse, to counteract the flow generated by the upper plate motion 1.5 = 360 ⋅ kPa m ⋅ ⎜⎛ y ⎝a − 1⎞ 2⎠ (3) u(y) = For u = U at y = a/2 we need to adjust the pressure gradient. From Eq. 1 2 a⎤ ⎢⎡⎛ a ⎞ ⎥ 2 2 a dp ⎢⎜ 2 2⎥ U= ⋅ ⋅ ⎜ − + a 2 ⋅ μ dx ⎢⎣⎝ a ⎠ a ⎥⎦ U⋅ Hence U⋅ y a ⎡ y 2 ⎞ − 2 ⋅ μ dx ⎣⎝ a ⎠ a + 2 ⋅ dp ⋅ ⎢⎛⎜ y⎤ ⎥ a⎦ a dp or dx dp dx =− 4 ⋅ U⋅ μ a 2 = −240 ⋅ = −4 × 1 ⋅ N⋅ s 2 × 1.5⋅ m m s kPa m 1 y/a 0.75 0.5 0.25 0 0.5 1 1.5 2 u (m/s) 1 y/a 0.75 0.5 0.25 −1 − 0.5 0 0.5 1 Shear Stress (kPa) The pressure gradient is positive to provide the "bulge" needed to satisfy the velocity requirement 1.5 × 1 ( 0.005 ⋅ m) 2 Problem 8.35 [Difficulty: 3] Given: Flow between parallel plates Find: Shear stress on lower plate; pressure gradient for zero shear stress at y/a = 0.25; plot velocity and shear stress Solution: u(y) = Basic equations U⋅ y a q = 1.5⋅ Available data + ⎡ y 2 ⎞ − 2 ⋅ μ dx ⎣⎝ a ⎠ a 2 ⋅ dp gpm ⋅ ⎢⎛⎜ y⎤ ⎥ (1) a⎦ a = 0.05⋅ in ft From Fig. A.2, Carbon tetrachloride at 20oC μ = 0.001 ⋅ Q = l U⋅ a 2 a − 3 ⋅ dp 12⋅ μ dx τ = μ⋅ (2) U a + a⋅ U= From Eq. 3, when y = 0, with U = 1.60 ft N⋅ s s y ⎝a − 1⎞ 2⎠ (3) − 5 lbf ⋅ s μ = 2.089 × 10 2 2⋅ Q or a⋅ l τyx = dx ⋅ ⎛⎜ 68°F = 20°C ⋅ m From Eq. 2, for zero pressure gradient dp ft U = 2⋅ q U = 1.60⋅ a μ⋅ U ft s −5 τyx = 5.58 × 10 a 2 ⋅ psi A mild adverse pressure gradient would reduce the flow rate. For zero shear stress at y/a = 0.25, from Eq. 3 0 = μ⋅ U a + a⋅ dp dx ⋅ ⎛⎜ 1 ⎝4 − 1⎞ 2⎠ dp or dx dp 2 dx a = 0.0536⋅ 0.75 y/a 0.75 y/a 4 ⋅ μ⋅ U 1 1 0.5 0.5 0.25 0.25 − 0.5 = 0 0.5 1 1.5 2 u (ft/s) Note that the location of zero shear is also where u is maximum! − 1× 10 −4 0 −4 1× 10 Shear Stress (psi) −4 2× 10 psi ft Problem 8.36 Using the result for average velocity from Example 8.3 [Difficulty: 3] Problem 8.37 [Difficulty: 3] Problem 8.38 [Difficulty: 5] Problem 8.39 [Difficulty: 5] Problem 8.40 [Difficulty: 2] Given: Expression for efficiency Find: Plot; find flow rate for maximum efficiency; explain curve Solution: η 0.0% 7.30% 14.1% 20.3% 25.7% 30.0% 32.7% 33.2% 30.0% 20.8% 0.0% Efficiency of a Viscous Pump 35% 30% 25% η q 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 20% 15% 10% 5% 0% 0.00 0.10 0.20 0.30 0.40 q For the maximum efficiency point we can use Solver (or alternatively differentiate) q 0.333 η 33.3% The efficiency is zero at zero flow rate because there is no output at all The efficiency is zero at maximum flow rate ∆p = 0 so there is no output The efficiency must therefore peak somewhere between these extremes 0.50 Problem 8.41 Problem 2.66 [Difficulty: 5] Problem 8.42 [Difficulty: 5] Part 1/2 Problem 8.42 [Difficulty: 5] Part 2/2 Problem 8.43 [Difficulty: 3] Given: Data on a journal bearing Find: Time for the bearing to slow to 100 rpm; visocity of new fluid Solution: The given data is D = 35⋅ mm L = 50⋅ mm δ = 1 ⋅ mm ωi = 500 ⋅ rpm ωf = 100 ⋅ rpm μ = 0.1⋅ 2 I = 0.125 ⋅ kg⋅ m N⋅ s 2 m I⋅ α = Torque = −τ⋅ A⋅ The equation of motion for the slowing bearing is D 2 where α is the angular acceleration and τ is the viscous stress, and A = π⋅ D⋅ L is the surface area of the bearing τ = μ⋅ As in Example 8.2 the stress is given by U δ = μ⋅ D⋅ ω 2⋅ δ where U and ω are the instantaneous linear and angular velocities. Hence Separating variables μ⋅ D⋅ ω I⋅ α = I⋅ dω dω μ⋅ π⋅ D ⋅ L ω =− dt 2⋅ δ ⋅ π⋅ D⋅ L⋅ D 2 3 =− μ⋅ π⋅ D ⋅ L 4⋅ δ ⋅ω 3 =− 4 ⋅ δ⋅ I ⋅ dt 3 − Integrating and using IC ω = ω0 μ⋅ π⋅ D ⋅ L ω( t) = ωi⋅ e 4⋅ δ ⋅ I ⋅t 3 − The time to slow down to ω f 4 ⋅ δ⋅ I t = − 3 μ⋅ π⋅ D ⋅ L For the new fluid, the time to slow down is t = 10⋅ min Rearranging the equation μ = − 4 ⋅ δ⋅ I 3 π ⋅ D ⋅ L⋅ t 4⋅ δ ⋅ I ωf = ωi⋅ e = 10 rpm is obtained from solving so μ⋅ π⋅ D ⋅ L ⎛ ωf ⎞ ⋅ ln⎜ ⎝ ωi ⎠ Hence 3 t = 1.19 × 10 s ⎛ ωf ⎞ ⋅ ln⎜ ⎝ ωi ⎠ μ = 0.199 kg m⋅ s ⋅t t = 19.9⋅ min It is more viscous as it slows down the rotation in a shorter time Problem 8.44 Given: Navier-Stokes Equations Find: Derivation of Example 8.3 result [Difficulty: 2] Solution: The Navier-Stokes equations are (using the coordinates of Example 8.3, so that x is vertical, y is horizontal) 4 3 ∂u ∂v ∂w + + =0 ∂x ∂y ∂z 1 4 5 (5.1c) 3 4 3 ⎛ ∂ 2u ∂ 2u ∂ 2u ⎞ ⎛ ∂u ∂u ∂u ∂u ⎞ ∂p ρ ⎜⎜ + u + v + w ⎟⎟ = ρg x − + µ ⎜⎜ 2 + 2 + 2 ⎟⎟ ∂x ∂y ∂z ⎠ ∂x ∂y ∂z ⎠ ⎝ ∂t ⎝ ∂x 1 4 5 3 4 6 5 (5.27a) 3 ⎛∂ v ∂ v ∂ v⎞ ⎛ ∂v ∂v ∂v ∂v ⎞ ∂p + u + v + w ⎟⎟ = ρg y − + µ ⎜⎜ 2 + 2 + 2 ⎟⎟ ∂x ∂y ∂z ⎠ ∂y ∂z ⎠ ∂y ⎝ ∂t ⎝ ∂x 2 ρ ⎜⎜ 1 3 3 3 3 2 2 3 3 (5.27b) 3 3 ⎛∂ w ∂ w ∂ w⎞ ⎛ ∂w ∂p ∂w ⎞ ∂w ∂w + µ ⎜⎜ 2 + 2 + 2 ⎟⎟ ⎟⎟ = ρg z − +w +v +u ∂z ⎠ ∂y ∂z ∂z ⎠ ∂y ∂x ⎝ ∂t ⎝ ∂x 2 ρ ⎜⎜ 2 2 (5.27c) The following assumptions have been applied: (1) Steady flow (given). (2) Incompressible flow; ρ = constant. (3) No flow or variation of properties in the z direction; w= 0 and ∂/∂z = 0. (4) Fully developed flow, so no properties except possibly pressure p vary in the x direction; ∂/∂x = 0. (5) See analysis below. (6) No body force in the y direction; gy = 0 Assumption (1) eliminates time variations in any fluid property. Assumption (2) eliminates space variations in density. Assumption (3) states that there is no z component of velocity and no property variations in the z direction. All terms in the z component of the Navier–Stokes equation cancel. After assumption (4) is applied, the continuity equation reduces to ∂v/∂y = 0. Assumptions (3) and (4) also indicate that ∂v/∂z = 0 and ∂v/∂x = 0. Therefore v must be constant. Since v is zero at the solid surface, then v must be zero everywhere. The fact that v = 0 reduces the Navier–Stokes equations further, as indicated by (5). Hence for the y direction ∂p =0 ∂y which indicates the pressure is a constant across the layer. However, at the free surface p = patm = constant. Hence we conclude that p = constant throughout the fluid, and so ∂p =0 ∂x In the x direction, we obtain µ ∂ 2u + ρg = 0 ∂y 2 Integrating twice u=− c 1 ρgy 2 + 1 y + c2 2µ µ To evaluate the constants, c1 and c2, we must apply the boundary conditions. At y = 0, u = 0. Consequently, c2 = 0. At y = a, du/dy = 0 (we assume air friction is negligible). Hence τ (y = δ ) = µ which gives du dy =− y =δ 1 µ ρgδ + c1 µ =0 c1 = ρgδ and finally 2 1 ρg ρg 2 ⎡ ⎛ y ⎞ 1 ⎛ y ⎞ ⎤ 2 δ ⎢⎜ ⎟ − ⎜ ⎟ ⎥ ρgy + u=− y= µ µ ⎢⎣⎝ δ ⎠ 2 ⎝ δ ⎠ ⎥⎦ 2µ Problem 8.45 [Difficulty: 3] Problem 8.46 Given: Paint flow (Bingham fluid) Find: Maximum thickness of paint film before flow occur [Difficulty: 3] Solution: Basic equations: du τyx = τy + μp ⋅ dy Bingham fluid: Use the analysis of Example 8.3, where we obtain a force balance between gravity and shear stresses: dτyx dy The given data is τy = 40⋅ Pa ρ = 1000⋅ = −ρ⋅ g kg 3 m From the force balance equation, itegrating Hence Motion occurs when τyx = −ρ⋅ g ⋅ ( δ − y ) τmax ≥ τy Hence the maximum thickness is or τyx = −ρ⋅ g ⋅ y + c and we have boundary condition τyx( y = δ) = 0 τmax = ρ⋅ g ⋅ δ and this is a maximum at the wall ρ⋅ g ⋅ δ ≥ τy δ = τy ρ⋅ g δ = 4.08 × 10 −3 m δ = 4.08 mm Problem 8.47 [Difficulty: 4] Given: Equation for fluid motion in the x-direction. Find: Expression for peak pressure Solution: Begin with the steady-state Navier-Stokes equation – x-direction Governing equation: The Navier-Stokes equations are 4 3 ∂u ∂v ∂w + + =0 ∂x ∂y ∂z 1 4 5 (5.1c) 3 4 3 ⎛ ∂ 2u ∂ 2u ∂ 2u ⎞ ⎛ ∂u ∂u ∂p ∂u ⎞ ∂u ⎟ ⎜ + u = − v + + g w + ρ⎜ ρ x µ ⎜⎜ 2 + 2 + 2 ⎟⎟ ∂y ∂x ∂x ∂z ⎟⎠ ∂y ∂z ⎠ ⎝ ∂t ⎝ ∂x 1 4 5 3 4 6 5 (5.27a) 3 ⎛ ∂ 2v ∂ 2v ∂ 2v ⎞ ⎛ ∂v ∂v ⎞ ∂v ∂v ∂p ρ ⎜⎜ + u + v + w ⎟⎟ = ρg y − + µ ⎜⎜ 2 + 2 + 2 ⎟⎟ ∂z ⎠ ∂z ⎠ ∂y ∂y ∂x ∂y ⎝ ∂t ⎝ ∂x 1 3 3 3 3 3 3 (5.27b) 3 3 ⎛∂ w ∂ w ∂ w⎞ ⎛ ∂w ∂p ∂w ⎞ ∂w ∂w +w +v +u + µ ⎜⎜ 2 + 2 + 2 ⎟⎟ ⎟⎟ = ρg z − ∂z ⎠ ∂y ∂z ∂z ⎠ ∂y ∂x ⎝ ∂t ⎝ ∂x ρ ⎜⎜ 2 2 2 (5.27c) The following assumptions have been applied: (1) Steady flow (given). (2) Incompressible flow; ρ = constant. (3) No flow or variation of properties in the z direction; w= 0 and ∂/∂z = 0. (4) Fully developed flow, so no properties except possibly pressure p vary in the x direction; ∂/∂x = 0. (5) See analysis below. (6) No body force in the y direction; gy = 0 Assumption (1) eliminates time variations in any fluid property. Assumption (2) eliminates space variations in density. Assumption (3) states that there is no z component of velocity and no property variations in the z direction. All terms in the z component of the Navier–Stokes equation cancel. After assumption (4) is applied, the continuity equation reduces to ∂v/∂y = 0. Assumptions (3) and (4) also indicate that ∂v/∂z = 0 and ∂v/∂x = 0. Therefore v must be constant (except of course in a more realistic model v ≠ 0 near the transition. Since v is zero at the solid surface, then v must be zero everywhere. The fact that v = 0 reduces the Navier–Stokes equations further, as indicated by (5). Hence for the y direction ∂p =0 ∂y which indicates the pressure is a constant across the flow. Hence we conclude that p is a function at most of x. In the x direction, we obtain 0=− ∂p ∂ 2u +µ 2 ∂x ∂y (1) Integrating this twice for the first region u1 = where 1 dp 2 c1 y + y + c2 µ 2 µ dx 1 dp denotes the pressure gradient in region 1. Note that we change to regular derivative as p is a function of x only. Note that dx 1 ⎛ ∂p ⎞ ⎟ and a function of y only ⎝ ∂x ⎠ Eq 1 implies that we have a function of x only ⎜ ⎛ ∂ 2u ⎞ ⎜⎜ 2 ⎟⎟ that must add up to be a constant (0); hence ⎝ ∂y ⎠ EACH is a constant! This means that p dp = const = s L1 dx 1 using the notation of the figure. To evaluate the constants, c1 and c2, we must apply the boundary conditions. We do this separately for each region. In the first region, at y = 0, u = U. Consequently, c2 = U. At y = h1, u = 0. Hence 0= 1 dp 2 c1 h1 + h1 + U 2 µ dx 1 µ so c1 = − 1 dp µU h1 − 2 dx 1 h1 Hence, combining results u1 = ⎛ y ⎞ 1 dp y 2 − h1 y + U ⎜⎜ − 1⎟⎟ 2 µ dx 1 ⎝ h1 ⎠ ( ) Exactly the same reasoning applies to the second region, so u2 = where 1 dp 2 µ dx (y 2 2 ⎞ ⎛ y − h2 y + U ⎜⎜ − 1⎟⎟ ⎝ h2 ⎠ ) p dp = const = − s L2 dx 2 What connects these flow is the flow rate Q. h1 h2 0 0 q = ∫ u1dy = ∫ u 2 dy = − 1 dp 3 Uh1 1 dp 3 Uh2 h2 − h1 − =− 12 µ dx 1 2 12µ dx 2 2 Hence 1 p s 3 Uh1 1 p s 3 Uh2 h1 + h2 + =− 12µ L1 2 12µ L2 2 Solving for ps, p s ⎛ h13 h23 ⎞ Uh2 Uh1 ⎜ + ⎟= − 12 µ ⎜⎝ L1 L2 ⎟⎠ 2 2 or ps = 6 µU (h2 − h1 ) ⎛ h13 h23 ⎞ ⎜⎜ + ⎟⎟ ⎝ L1 L2 ⎠ Problem 8.48 [Difficulty: 2] Problem 8.49 [Difficulty: 2] Problem 8.50 Given: Data on water temperature and tube Find: Maximum laminar flow; plot [Difficulty: 3] B Solution: 5 Ns A 2.414 10 From Appendix A B 247.8 K 2 C 140 K μ ( T) A 10 in T C m D 7.5 mm ρ 1000 kg Recrit 2300 3 m T1 20 °C 3 Ns T1 253 K μ T1 3.74 10 T2 120 °C 2 T2 393 K 4 Ns μ T2 2.3 10 m 2 m The plot of viscosity is 0.01 μ N s 2 m 1 10 3 1 10 4 20 0 20 40 60 80 100 120 T (C) For the flow rate ρ Vcrit D Recrit Qmax T1 5.07 10 μ 3.00 5m s ρ Qmax D μ π 4 4 ρ Qmax 2 D μ π D Qmax( T) π μ( T) D Recrit 4 ρ L Qmax T1 182 hr Qmax T2 3.12 10 3 6m s L Qmax T2 11.2 hr 200 Q (L/hr) 150 100 50 20 0 20 40 60 T (C) 80 100 120 Problem 8.51 [Difficulty: 2] d p1 D F L Given: Hyperdermic needle Find: Volume flow rate of saline Solution: π⋅ ∆p⋅ d 4 Basic equation Q= For the system F 4⋅ F ∆p = p 1 − p atm = = A 2 π⋅ D 4 ∆p = At 68oF, from Table A.7 (Eq. 8.13c; we assume laminar flow and verify this is correct after solving) 128⋅ μ ⋅ L π × 7.5⋅ lbf × 12⋅ in ⎞ ⎛ 1 ⎜ 0.375⋅ in × 1⋅ ft ⎝ ⎠ ∆p = 67.9⋅ psi − 5 lbf ⋅ s μH2O = 2.1 × 10 ⋅ ft Q = π 128 × 67.9⋅ lbf 2 Q = 8.27 × 10 V= Q A ⋅ Q = 144 ⋅ in π⋅ d 2 ⋅ 1 ⋅ ft 2 4 2 1 12⋅ in ft ⎞ × × × 12⋅ in ⎠ 1 ⋅ in 1 ⋅ ft −4 1.05 × 10 lbf ⋅ s 1 ⋅ ft × ⎛⎜ 0.005 ⋅ in × ⎝ 3 3 − 3 in Q = 1.43 × 10 s V = μ = 1.05 × 10 ft 2 × − 4 lbf ⋅ s μ = 5⋅ μH2O 2 in − 7 ft Check Re: 2 4 π × 8.27 × 10 − 7 ft 3 s ⋅ Q = 0.0857⋅ s 2 × 3 ⎛ 1 ⎞ × ⎛ 12⋅ in ⎞ ⎜ .005⋅ in ⎜ ⎝ ⎠ ⎝ 1⋅ ft ⎠ 2 V = 6.07⋅ in min ft s 4 Re = ρ⋅ V⋅ d ρ = 1.94⋅ μ Re = 1.94⋅ slug ft slug ft 3 × 6.07⋅ ft s (assuming saline is close to water) 3 × 0.005 ⋅ in × 1 ⋅ ft 12⋅ in × ft 2 1.05 × 10 −4 × ⋅ lbf ⋅ s slug⋅ ft 2 s ⋅ lbf Re = 46.7 Flow is laminar 2 Problem 8.52 [Difficulty: 3] Given: Data on a tube Find: "Resistance" of tube; maximum flow rate and pressure difference for which electrical analogy holds for (a) kerosine and (b) castor oil Solution: L = 250 ⋅ mm The given data is D = 7.5⋅ mm From Fig. A.2 and Table A.2 Kerosene: μ = 1.1 × 10 − 3 N⋅ s ⋅ ρ = 0.82 × 990 ⋅ 2 m Castor oil: μ = 0.25⋅ 3 = 812 ⋅ m N⋅ s ρ = 2.11 × 990 ⋅ 2 m For an electrical resistor kg kg 3 3 m = 2090⋅ m V = R⋅ I kg kg 3 m (1) The governing equation for the flow rate for laminar flow in a tube is Eq. 8.13c 4 Q= or π⋅ ∆p⋅ D 128 ⋅ μ⋅ L 128 ⋅ μ⋅ L ∆p = 4 ⋅Q (2) π⋅ D By analogy, current I is represented by flow rate Q, and voltage V by pressure drop Δp. Comparing Eqs. (1) and (2), the "resistance" of the tube is R= 128 ⋅ μ⋅ L 4 π⋅ D The "resistance" of a tube is directly proportional to fluid viscosity and pipe length, and strongly dependent on the inverse of diameter The analogy is only valid for Re < 2300 ρ⋅ Writing this constraint in terms of flow rate Q π 4 or ρ⋅ V⋅ D μ < 2300 ⋅D 2 ⋅D μ < 2300 or Qmax = 2300⋅ μ⋅ π⋅ D 4⋅ ρ The corresponding maximum pressure gradient is then obtained from Eq. (2) ∆pmax = 128 ⋅ μ⋅ L 4 π⋅ D 2 ⋅ Qmax = 32⋅ 2300⋅ μ ⋅ L 3 ρ⋅ D Substituting values (a) For kerosine (b) For castor oil 3 −5m Qmax = 1.84 × 10 s 3 −3m Qmax = 1.62 × 10 s l Qmax = 1.10⋅ min ∆pmax = 65.0⋅ Pa l Qmax = 97.3⋅ min ∆pmax = 1.30⋅ MPa The analogy fails when Re > 2300 because the flow becomes turbulent, and "resistance" to flow is then no longer linear with flow rate Problem 8.53 [Difficulty: 3] Problem 8.54 [Difficulty: 3] Problem 8.56 [Difficulty: 4] Part 1/2 Problem 8.56 [Difficulty: 4] Part 2/2 Problem 8.57 [Difficulty: 4] Problem 8.52 8.56 Problem 8.58 Given: Tube dimensions and volumetric flow rate Find: Pressure difference and hydraulic resistance [Difficulty: 2] Solution: The flow rate of a fully developed pressure-driven flow in a pipe is Q = flow rate π∆pR 4 8µL . Rearranging it, one obtains ∆p = 8µLQ . For a πR 4 Q = 10µl / min , L=1 cm, µ = 1.0 × 10 −3 Pa.s , and R = 1 mm, ∆p = m 8 10 ×10 −9 m 3 0.01 × × × × 4 ×1.0 ×10 −3 Pa.s = 0.00424 Pa −12 s 1×10 m 60 π Similarly, the required pressure drop for other values of R can be obtained. The hydraulic resistance Rhyd = ∆p 8µL = . Substituting the values of the viscosity, length and radius of the tube, one obtains the Q πR 4 value of the hydraulic resistance. R (mm) 1 10-1 10-2 10-3 10-4 ∆p 0.00424 Pa 42.4 Pa 424 kPa 4.24 GPa 4.24 x 104 GPa Rhyd (Pa.s/m3) 2.55 x 107 2.55 x 1011 2.55 x 1015 2.55 x 1019 2.55 x 1023 (3) To achieve a reasonable flow rate in microscale or nanoscale channel, a very high pressure difference is required since ∆p is proportional to R−4. Therefore, the widely used pressure-driven flow in large scale systems is not appropriate in microscale or nanoscale channel applications. Other means to manipulate fluids in microscale or nanoscale channel applications are required. Problem 8.59 . Given: Definition of hydraulic resistance Find: Hydraulic resistance in a diffuser Solution: Basic equation: Rhyd = ∆p 8µ z2 1 8µ z2 1 dz dz = = 4 ∫ ∫ z z Q π 1 r π 1 (ri + αz ) 4 Rhyd = = 8µ π =− ∆p Q ∫ z 0 1 d (ri + αz ) (ri + αz ) 4 8µ 1 8µ −3 (ri + αz ) −3 0z = − [(ri + αz ) −3 − ri ] πα 3 3πα Rhyd = − 8µ ⎡ 1 1⎤ − ⎢ ⎥ 3πα ⎣ (ri + αz ) 3 ri 3 ⎦ [Difficulty: 2] Problem 8.60 [Difficulty: 4] Given: Relationship between shear stress and deformation rate; fully developed flow in a cylindrical blood vessel Find: Velocity profile; flow rate Solution: Similar to the Example Problem described in Section 8.3, based on the force balance, one obtains τ rx = r dp 2 dx (1) This result is valid for all types of fluids, since it is based on a simple force balance without any assumptions about fluid rheology. Since the axial pressure gradient in a steady fully developed flow is a constant, Equation (1) shows that τ = 0 < τc at r = 0. Therefore, there must be a small region near the center line of the blood vessel for which τ < τc. If we call Rc the radial location at which τ = τc, the flow can then be divided into two regions: r > Rc: The shear stress vs. shear rate is governed by τ = τc + µ du dr (2) r < Rc: τ = 0 < τc. We first determine the velocity profile in the region r > Rc. Substituting (1) into (2), one obtains: r dp du = τc + µ 2 dx dr (3) Using equation (3) and the fact that du/dr at r = Rc is zero, the critical shear stress can be written as Rc dp = τc . 2 dx (4) Rearranging eq. (4), Rc is Rc = 2τ c / dp . dx Inserting (4) into (3), rearranging, and squaring both sides, one obtains (5) µ du 1 dp = [r − 2 rRc + Rc ] dr 2 dx (6) Integrating the above first-order differential equation using the non-slip boundary condition, u = 0 at r = R: u=− 1 dp ⎡ 2 8 ⎤ 3/ 2 (R − r 2 ) − Rc ( Rc − r 3 / 2 ) + 2 Rc ( R − r )⎥ for Rc ≤ r ≤ R ⎢ 4µ dx ⎣ 3 ⎦ (7) In the region r < Rc, since the shear stress is zero, fluid travels as a plug with a plug velocity. Since the plug velocity must match the velocity at r = Rc, we set r = Rc in equation (7) to obtain the plug velocity: u=− [ ] 1 dp 2 2 ( R − Rc ) + 2 Rc ( R − Rc ) for r ≤ Rc 4µ dx (8) The flow rate is obtained by integrating u(r) across the vessel cross section: R Rc R Q = ∫ u (r )2πrdr = ∫ u (r )2πrdr + ∫ u (r )2πrdr 0 Rc 0 4 πR 4 dp ⎡ 16 Rc 4 Rc 1 ⎛ Rc ⎞ ⎤ =− + − ⎜ ⎟ ⎥ ⎢1 − 8µ dx ⎣⎢ 7 R 3 R 21 ⎝ R ⎠ ⎦⎥ Given R = 1mm = 10-3 m, µ = 3.5 cP = 3.5×10-3 Pa⋅s, and τc = 0.05 dynes/cm2 = 0.05×10-1 Pa, and From eq. (5), Rc = 2τ c / (9) dp = −100 Pa / m . dx dp dx 2 × 0.05 10 ×10 −6 N / m 2 Rc = = 0.1mm Pa / m 100 Substituting the values of R, µ, Rc, and Q=− π 1× 10−12 m 4 8 3.5 × 10 −3 Pa.s dp into eq. (9), dx × (−100) Pa / m × [1 − 16 0.1 mm 4 0.1 mm 1 0.1 mm 4 + − ( ) ] = 3.226 × 10 −9 m3 / s 7 1 mm 3 1 mm 21 1 mm Problem 8.61 [Difficulty: 4] Given: Fully developed flow, Navier-Stokes equations; Non-Newtonian fluid Find: Velocity profile, flow rate and average velocity Solution: According to equation (8.10), we can write the governing equation for Non-Newtonian fluid velocity in a circular tube n r ∂p c1 ⎛ du ⎞ τ rx = k ⎜ ⎟ = + 2 ∂x r ⎝ dr ⎠ (1) However, as for the Newtonian fluid case, we must set c1 = 0 as otherwise we’d have infinite stress at r = 0. Hence, equation (1) becomes n r ∂p ⎛ du ⎞ k⎜ ⎟ = 2 ∂x ⎝ dr ⎠ (2) The general solution for equation (2), obtained by integrating, is given by 1 1+ 1 ⎛ 1 ∂p ⎞ n r n + c2 u =⎜ ⎟ ⎝ 2k ∂x ⎠ ⎛1 + 1 ⎞ ⎜ ⎟ ⎝ n⎠ 1 (3) Apply the no slip boundary condition at r = R into equation (3), we get 1 1+ 1 ⎛ 1 ∂p ⎞ n R n c 2 = −⎜ ⎟ ⎝ 2k ∂x ⎠ ⎛1 + 1 ⎞ ⎜ ⎟ ⎝ n⎠ 1 (4) The fluid velocity then is given as 1 n +1 n +1 ⎞ n ⎛ 1 ∂p ⎞ n ⎛⎜ n n ⎟ u (r ) = r R − ⎜ ⎟ ⎜ ⎟ (n + 1) ⎝ 2k ∂x ⎠ ⎝ ⎠ (5) The volume flow rate is 1 Q = ∫ udA == ∫ A Hence R 0 n +1 n +1 ⎞ n ⎛ 1 ∂p ⎞ n ⎛ n 2πr ⎟ ⎜⎜ r − R n ⎟⎟dr ⎜ (n + 1) ⎝ 2k ∂x ⎠ ⎝ ⎠ (6) 1 1 R 3 n +1 n +1 3 n +1 r2 n ⎤ 2nπ ⎛ 1 ∂p ⎞ n ⎡ n 2nπ ⎛ 1 ∂p ⎞ n n ⎛ n 1⎞ n Q= = − r R − ⎟ R ⎜ ⎟ ⎜ ⎟ ⎢ ⎜ ⎥ (n + 1) ⎝ 2k ∂x ⎠ ⎣ 3n + 1 2 ⎝ 3n + 1 2 ⎠ ⎦ 0 (n + 1) ⎝ 2k ∂x ⎠ (7) Simplifying 1 nπ ⎛ 1 ∂p ⎞ n Q=− ⎜ ⎟ R (3n + 1) ⎝ 2k ∂x ⎠ When n = 1, then k = µ, and 3 n +1 n (8) πa 4 ∂p Q=− 8µ ∂x , just like equation (8.13b) in the textbook. The average velocity is given by 1 Q Q n ⎛ 1 ∂p ⎞ n V = = 2 =− ⎟ R ⎜ (3n + 1) ⎝ 2k ∂x ⎠ A πR n +1 n (9) Based on Eq. (7), the pressure gradient is ⎤ ⎡ ∂p Q(3n + 1) ⎥ = −2k ⎢ 3 n +1 ⎢ ⎥ ∂x ⎣ nπR n ⎦ n (10) Substituting Q = 1µL/min= 1 × 10-9/60 m3/s, R = 1mm = 10-3 m, and n = 0.5 into eq.(10): ⎡ 10 −9 ⎤ ( 3(0.5) + 1) ⎥ ⎢ ∂p = −2k ⎢ 60 3(0.5 )+1 ⎥ ∂x ⎢ 0.5π 0.5 ⎥ ⎢⎣ ⎥⎦ 0.5 = −325k Pa/m for n = 0.5 (11) Similarly, the required pressure gradients for n = 1 and n = 1.5 can be obtained: ∂p = −42.4k Pa/m, for n = 1 ∂x (12) ∂p ≈ −5.42k Pa/m, for n = 1.5 ∂x (12) Obviously, the magnitude of the required pressure gradient increases as n decreases. Among the three types of fluids (pseudoplastic for n = 0.5, Newtonian for n = 1, and dilatant for n = 1.5), the dilatant fluid requires the lowest pressure pump for the same pipe length. Problem 8.62 [Difficulty: 2] Given: Fully developed flow in a pipe; slip boundary condition on the wall Find: Velocity profile and flow rate Solution: Similar to the example described in Section 8.3, one obtained u= r 2 ∂p + c2 4 µ ∂x (1) The constant c2 will be determined by the slip velocity boundary condition at r = R: u =l ∂u ∂r (2) and one obtains c2 = R 2 ∂p ⎛ l ⎞ ⎜ 2 − 1⎟ 4 µ ∂x ⎝ R ⎠ (3) Substituting c2 into Eq.(1), one obtains u=− 1 ∂p 2 R − r 2 + 2lR 4 µ ∂x ( ) (4) The volume flow rate is R Q = ∫ u 2πrdr = − 0 l⎤ πR 4 ∂p ⎡ 1+ 4 ⎥ ⎢ R⎦ 8µ ∂x ⎣ Substituting R = 10 µm, µ = 1.84 x 10-5 N⋅s/m2, mean free path l = 68 nm, and − Q=− π (10 ×10 −6 ) 4 m 4 8 1.84 × 10 −5 Pa.s × (−1.0 × 10 6 )Pa/m × [1 + 4 (5) ∂p = 1.0×106 Pa/m into eq. (5), ∂x 68 ×10 −9 m ] = 2.19 × 10 −10 m 3 /s . 10 × 10 −6 m Problem 8.63 [Difficulty: 3] Given: Fully developed flow, velocity profile, and expression to calculate the flow rate Find: Velocity and flow rate Solution: For the fully developed flow, the N-S equations can be simplified to Substituting the trial solution in equation (1), one obtains ⎛ ∂ 2 u ∂ 2 u ⎞ ∂p = constant µ ⎜⎜ 2 + 2 ⎟⎟ = ∂z ⎠ ∂x ⎝ ∂y 1 ⎞ ∂p ⎛ 1 − 2µu 0 ⎜ 2 + 2 ⎟ = b ⎠ ∂x ⎝a u0 = − Rearrange it and one obtains a 2b 2 ∂p 2 2 2 µ (a + b ) ∂x Q = ∫ u ( y, z )dydz = ab ∫ The flow rate is 2π 0 Substituting u ( ρ ,φ ) = u 0 (1 − ρ ) into Eq. (4) and integrating twice: 2 Substituting u0 into (5), one obtains Q = ab ∫ 2π 0 0 1 0 πa 3b 3 1 ∂p Q = πabu0 = − 2 2 2 4 µ (a + b ) ∂x For a pipe radius R, a = b = R, from equation (6), 1 ⎛ πR 4 ∂p ⎞ ⎟ Q pipe = ⎜⎜ − 8 ⎝ µ ∂x ⎟⎠ which is the same as equation (8.13b) in the book. For a channel with an elliptic cross-section with a = R and b = 1.5R, from equation (6), one has Q pipe = 29 ⎛ πR 4 ∂p ⎞ ⎜− ⎟. 104 ⎜⎝ µ ∂x ⎟⎠ (2) (3) 1 ∫ ρu ( ρ , φ )dρdφ ∫ ρu (1 − ρ 0 (1) 2 (4) 1 )dρdφ = πabu0 2 (6) (5) Problem 8.64 Given: The expression of hydraulic resistance of straight channels with different cross sectional shapes Find: Hydraulic resistance [Difficulty: 2] Solution: Based on the expressions of hydraulic resistance listed in the table, one obtains Using the circle as the example, Rhyd 1 8 1×10 −3 × 10 × 10 −3 Pa ⋅ s × m = µL 4 = = 0.254 ×1012 Pa ⋅ s/m 3 −4 4 4 π π (1×10 ) m a 8 The results are Shape Rhyd (1012 Pa·s/m3) Circle 0.25 Ellipse 3.93 Triangle 18.48 Two plates 0.40 Rectangle 0.51 Square 3.24 Comparing the values of the hydraulic resistances, a straight channel with a circular cross section is the most energy efficient to pump fluid with a fixed volumetric flow rate; the triangle is the worst. Problem 8.65 Given: Two-fluid flow in tube Find: Velocity distribution; Plot [Difficulty: 3] Solution: D = 5 ⋅ mm Given data L = 5⋅ m ∆p = −5 ⋅ MPa μ1 = 0.5⋅ N⋅ s μ2 = 5 ⋅ 2 m N⋅ s 2 m From Section 8-3 for flow in a pipe, Eq. 8.11 can be applied to either fluid 2 u= ⎛ ∂ ⎞ c1 ⋅ ln( r) + c2 p + 4 ⋅ μ ⎝ ∂x ⎠ μ r ⋅⎜ Applying this to fluid 1 (inner fluid) and fluid 2 (outer fluid) 2 u1 = r 4 ⋅ μ1 ⋅ ∆p L + 2 c1 ⋅ ln( r) + c2 μ1 r= We need four BCs. Two are obvious u2 = D r 4 ⋅ μ2 u2 = 0 2 ⋅ ∆p L + (1) c3 μ2 ⋅ ln( r) + c4 r= D 4 u1 = u2 (2) The third BC comes from the fact that the axis is a line of symmetry du1 r= 0 dr =0 (3) The fourth BC comes from the fact that the stress at the interface generated by each fluid is the same r= du1 du2 μ1 ⋅ = μ2 ⋅ dr dr D 4 2 Using these four BCs ⎛ D⎞ ⎜2 ⎝ ⎠ ⋅ ∆p + c3 ⋅ ln⎛ D ⎞ + c = 0 ⎜ 4 μ2 ⎝ 2 ⎠ 4 ⋅ μ2 L lim c1 r → 0 μ1 ⋅ r (4) 2 ⎛ D⎞ ⎜4 ⎝ ⎠ ⋅ ∆p + c1 ⋅ ln⎛ D ⎞ + c = ⎜ 2 μ1 ⎝ 4 ⎠ 4 ⋅ μ1 L 2 ⎛ D⎞ ⎜4 ⎝ ⎠ ⋅ ∆p + c3 ⋅ ln⎛ D ⎞ + c ⎜ 4 μ2 ⎝ 4 ⎠ 4 ⋅ μ2 L 4 ⋅ c1 4 ⋅ c3 D ∆p D ∆p ⋅ + = ⋅ + 8 L D D 8 L =0 Hence, after some algebra c1 = 0 (To avoid singularity) c2 = − ( 2 D ⋅ ∆p μ2 + 3 ⋅ μ1 64⋅ L μ1 ⋅ μ2 ) 2 c3 = 0 c4 = − D ⋅ ∆p 16⋅ L⋅ μ2 u 1 ( r) = The velocity distributions are then ⎡⎢ 2 4 ⋅ μ1 ⋅ L ⎢ ⎣ ∆p ⋅ r − 2 ⎛ D ⎞ ⋅ ( μ2 + 3⋅ μ1 )⎤⎥ ⎜2 ⎥ 4 ⋅ μ2 ⎝ ⎠ ⎦ u 2 ( r) = ∆p 4 ⋅ μ2 ⋅ L ⎡2 ⋅ ⎢r − ⎣ 2 ⎛ D ⎞ ⎤⎥ ⎜2 ⎝ ⎠⎦ (Note that these result in the same expression if µ 1 = µ 2, i.e., if we have one fluid) Evaluating either velocity at r = D/4 gives the velocity at the interface 2 u interface = − 3 ⋅ D ⋅ ∆p u interface = − 64⋅ μ2 ⋅ L 3 64 × ( 0.005 ⋅ m) × ⎛ −5 × 10 ⋅ 2 2 ⎞× m × 1 5 ⋅ N⋅ s 5 ⋅ m 2 m ⎠ 6 N ⎜ ⎝ u interface = 0.234 Evaluating u 1 at r = 0 gives the maximum velocity 2 u max = − ( D ⋅ ∆p⋅ μ2 + 3 ⋅ μ1 ) 64⋅ μ1 ⋅ μ2 ⋅ L u max = − 1 64 × ( 0.005 ⋅ m) × ⎛ −5 × 10 ⋅ 2 2 m ⎞ × 5 + 3 × 0.5 ⋅ m × 1 u max = 1.02 s 2 N⋅ s 5 ⋅ m 5 × .5 m ⎠ 6 N ⎜ ⎝ 2.5 Inner fluid Outer fluid r (mm) 2 1.5 1 0.5 0 0.2 0.4 0.6 Velocity (m/s) The velocity distributions can be plotted in Excel 0.8 1 1.2 m s Problem 8.66 Given: Turbulent pipe flow Find: Wall shear stress [Difficulty: 2] Solution: Basic equation (Eq. 4.18a) Assumptions 1) Horizontal pipe 2) Steady flow 3) Fully developed flow With these assumptions the x momentum equation becomes 2 p1 π D 4 2 π D τw π D L p 2 0 4 or τw p2 p1 D 4 L 3 12 1 τw 750 psi 4 15 Since τw is negative it acts to the left on the fluid, to the right on the pipe wall τw 3.13 psi ∆p D 4 L Problem 8.67 Given: Pipe glued to tank Find: Force glue must hold when cap is on and off [Difficulty: 2] Solution: Basic equation (Eq. 4.18a) First solve when the cap is on. In this static case 2 π D Fglue 4 p1 where p 1 is the tank pressure Second, solve for when flow is occuring: Assumptions 1) Horizontal pipe 2) Steady flow 3) Fully developed flow With these assumptions the x momentum equation becomes 2 p1 π D 4 2 π D τw π D L p 2 0 4 Here p1 is again the tank pressure and p 2 is the pressure at the pipe exit; the pipe exit pressure is p atm = 0 kPa gage. Hence 2 Fpipe Fglue τw π D L π D 4 p1 We conclude that in each case the force on the glue is the same! When the cap is on the glue has to withstand the tank pressure; when the cap is off, the glue has to hold the pipe in place against the friction of the fluid on the pipe, which is equal in magnitude to the pressure drop. π lbf 2 Fglue ( 3 in) 30 4 2 in Fglue 212 lbf Problem 8.68 [Difficulty: 2] Given: Data on pressure drops in flow in a tube Find: Which pressure drop is laminar flow, which turbulent Solution: Given data ∂ ∂x p 1 = −4.5⋅ kPa ∂ m ∂x p 2 = −11⋅ kPa m D = 30⋅ mm From Section 8-4, a force balance on a section of fluid leads to R ∂ D ∂ τw = − ⋅ p = − ⋅ p 2 ∂x 4 ∂x Hence for the two cases D ∂ τw1 = − ⋅ p 1 4 ∂x τw1 = 33.8 Pa D ∂ τw2 = − ⋅ p 2 4 ∂x τw2 = 82.5 Pa Because both flows are at the same nominal flow rate, the higher pressure drop must correspond to the turbulent flow, because, as indicated in Section 8-4, turbulent flows experience additional stresses. Also indicated in Section 8-4 is that for both flows the shear stress varies from zero at the centerline to the maximums computed above at the walls. The stress distributions are linear in both cases: Maximum at the walls and zero at the centerline. Problem 8.69 Given: Flow through channel Find: Average wall stress [Difficulty: 2] Solution: Basic equation (Eq. 4.18a) Assumptions 1) Horizontal pipe 2) Steady flow 3) Fully developed flow With these assumptions the x momentum equation becomes H p 1 W H τw 2 L ( W H) p 2 W H 0 or W H τw p 2 p 1 2 ( W H) L τw ∆p L 2 1 1 lbf τw 1 2 2 in 1 in 2 144 in ft 2 H W 1 ft 12 in 30 ft 1 1 ft 9.5 in 12 in 1 30 ft Since τw < 0, it acts to the left on the fluid, to the right on the channel wall lbf τw 0.195 2 ft 3 τw 1.35 10 psi Problem 8.70 [Difficulty: 3] Problem 8.71 Given: Data from a funnel viscometer filled with pitch. Find: Viscosity of pitch. [Difficulty: 1] Solution: V πD 4 ρg ⎛ h ⎞ (Volume flow rate) = ⎜1 + ⎟ t 128µ ⎝ L ⎠ where Q is the volumetric flow rate, V flow volume, t is the time of flow, D is the diameter of the funnel stem, ρ is the density of the Basic equation: Q = pitch, µ is the viscosity of the pitch, h is the depth of the pitch in the funnel body, and L is the length of the funnel stem. Assumption: Viscous effects above the stem are negligible and the stem has a uniform diameter. The given or available data is: Calculate the flow rate: V = 4.7 ×10 −5 m 3 t = 17,708days D = 9.4mm h = 75mm L = 29mm ρ = 1.1×103 Q= V = t kg m3 4.7 × 10 −5 m 3 m3 = 3.702 × 10 −14 24hour 3600s s 17708day × × day hour Solve the governing equation for viscosity: µ= µ= πD 4 ρg ⎛ h⎞ ⎜1 + ⎟ 128Q ⎝ L ⎠ µ = 2.41×108 N ⋅s m2 4 m m ⎛ ⎞ 3 kg ⎟ ×1.1× 10 3 × 9.81 2 m s ⎛ 75mm ⎞ N × s 2 ⎝ 1000mm ⎠ ⎟× ⎜1 + 3 ⎝ 29mm ⎠ kg × m −14 m 128 × 3.702 ×10 s π × (9.4mm )4 × ⎜ Compare this to the viscosity of water, which is 10-3 N·s/m2! Relate this equation to 8.13c for flow driven by a pressure gradient: Q= π∆pD 4 πD 4 ∆p . = × 128µL 128µ L For this problem, the pressure (Δp) is replaced by the hydrostatic force of the pitch. Consider the pressure variation in a static fluid. ∆p ∆p dp = − ρg = − ρg = = . ∆z L + h dz Replacing the term in 8.13c Q= Hence which is the same as the given equation. πD 4 ∆p πD 4 ρg × (L + h ) πD 4 ⎛ h⎞ × = × = × ρg × ⎜1 + ⎟ L 128µ L 128µ 128µ ⎝ L⎠ V πD 4 ρg ⎛ h ⎞ Q= = ⎜1 + ⎟ t 128µ ⎝ L ⎠ Problem 8.72 [Difficulty: 3] Problem 8.73 [Difficulty: 3] Given: Data on mean velocity in fully developed turbulent flow Find: Trendlines for each set; values of n for each set; plot Solution: y/R 0.898 0.794 0.691 0.588 0.486 0.383 0.280 0.216 0.154 0.093 0.062 0.041 0.024 u/U 0.996 0.981 0.963 0.937 0.907 0.866 0.831 0.792 0.742 0.700 0.650 0.619 0.551 y/R 0.898 0.794 0.691 0.588 0.486 0.383 0.280 0.216 0.154 0.093 0.062 0.037 u/U 0.997 0.998 0.975 0.959 0.934 0.908 0.874 0.847 0.818 0.771 0.736 0.690 Equation 8.22 is Mean Velocity Distributions in a Pipe u/U 1.0 0.1 0.01 0.10 1.00 y/R Re = 50,000 Re = 500,000 Power (Re = 500,000) Power (Re = 50,000) Applying the Trendline analysis to each set of data: At Re = 50,000 At Re = 500,000 u/U = 1.017(y/R )0.161 u/U = 1.017(y/R )0.117 2 with R = 0.998 (high confidence) Hence 1/n = 0.161 n = 6.21 with R 2 = 0.999 (high confidence) Hence Both sets of data tend to confirm the validity of Eq. 8.22 1/n = 0.117 n = 8.55 Problem 8.74 [Difficulty: 3] Problem 8.75 [Difficulty: 3] Part 1/2 Problem 8.75 [Difficulty: 3] Part 2/2 Problem 8.76 Given: Laminar flow between parallel plates Find: Kinetic energy coefficient, α [Difficulty: 3] Solution: Basic Equation: The kinetic energy coefficient, α is given by ∫ α= A ρ V 3dA (8.26b) m V 2 From Section 8-2, for flow between parallel plates 2 2 ⎡ ⎛ ⎞ ⎤ 3 ⎡ ⎛ y ⎞ ⎤ y ⎟ ⎥ = V ⎢1 − ⎜ ⎟ ⎥ u = umax ⎢1 − ⎜ ⎢ ⎜a ⎟ ⎥ 2 ⎢ ⎜a ⎟ ⎥ ⎢⎣ ⎝ 2 ⎠ ⎥⎦ ⎢⎣ ⎝ 2 ⎠ ⎥⎦ since umax = 3 V . 2 Substituting α= ∫ A ρV 3dA m V 2 = ∫ A ρu 3dA ρV A V 2 3 = 1 ⎛u⎞ 1 dA = ⎟ ⎜ ∫ A A⎝V ⎠ wa a 2 a 2 3 3 2 ⎛u⎞ ⎛u⎞ ∫a ⎜⎝ V ⎟⎠ wdy = a ∫0 ⎜⎝ V ⎟⎠ dy − 2 Then 3 3 31 1 3 2 a ⎛ u ⎞ ⎛ umax ⎞ ⎛⎜ y ⎞⎟ ⎛ 3 ⎞ ⎜⎜ ⎟⎟ ⎜ α= = ⎜ ⎟ ∫ (1 − η 2 ) dη ⎟ d⎜ ∫ a ⎟ ⎝2⎠ 0 a 2 0 ⎝ umax ⎠ ⎝ V ⎠ ⎝ 2⎠ where η = y a 2 Evaluating, (1 − η ) 2 3 = 1 − 3η 2 + 3η 4 − η 6 The integral is then ⎛ 3⎞ α =⎜ ⎟ ⎝2⎠ 31 ⎛ 3⎞ ∫0 (1 − 3η + 3η − η )dη = ⎜⎝ 2 ⎟⎠ 2 4 6 3 1 3 5 1 7 ⎤ 27 16 ⎡ 3 ⎢⎣η − η + 5 η − 7 η ⎥⎦ = 8 35 = 1.54 0 Problem 8.77 [Difficulty: 3] Problem 8.78 [Difficulty: 3] Given: Definition of kinetic energy correction coefficient α Find: α for the power-law velocity profile; plot Solution: Equation 8.26b is α= ⌠ ⎮ 3 ⎮ ρ⋅ V dA ⌡ 2 mrate⋅ Vav where V is the velocity, mrate is the mass flow rate and Vav is the average velocity 1 n For the power-law profile (Eq. 8.22) V = U⋅ ⎛⎜ 1 − For the mass flow rate mrate = ρ⋅ π⋅ R ⋅ Vav Hence the denominator of Eq. 8.26b is mrate⋅ Vav = ρ⋅ π⋅ R ⋅ Vav We next must evaluate the numerator of Eq. 8.26b ⎝ ⎞ R⎠ r 2. 2 3 ⌠ 3 ⎮ ⎮ n r 3 3 ρ⋅ V dA = ⎮ ρ⋅ 2 ⋅ π⋅ r⋅ U ⋅ ⎛⎜ 1 − ⎞ dr ⎮ R⎠ ⎝ ⌡ ⌠ ⎮ ⎮ ⌡ ⌠ ⎮ ⎮ ⎮ ⎮ ⌡ 2 R 3 2 2 3 n 2 ⋅ π⋅ ρ⋅ R ⋅ n ⋅ U r⎞ ⎛ dr = ρ⋅ 2 ⋅ π⋅ r⋅ U ⋅ ⎜ 1 − ( 3 + n) ⋅ ( 3 + 2⋅ n) R⎠ ⎝ 3 0 To integrate substitute Then m=1− r R r = R⋅ ( 1 − m) ⌠ ⎮ ⎮ ⎮ ⎮ ⌡ dm = − R dr = −R⋅ dm R 0 dr 0 ⌠ 3 ⎮ n ⎮ r n 3 ρ⋅ 2 ⋅ π⋅ r⋅ U ⋅ ⎛⎜ 1 − ⎞ dr = −⎮ ρ⋅ 2 ⋅ π⋅ R⋅ ( 1 − m) ⋅ m ⋅ R dm ⌡ R⎠ ⎝ 1 3 1 ⌠ 3 ⎞ ⎮ ⎛ 3 +1 ⎜ ⎮ n 3 n ρ⋅ V dA = ⎮ ρ⋅ 2 ⋅ π⋅ R⋅ ⎝ m − m ⎠ ⋅ R dm ⌡ ⌠ ⎮ ⎮ ⌡ Hence 0 2 2 3 ⌠ 2 ⋅ R ⋅ n ⋅ ρ⋅ π⋅ U ⎮ 3 d ρ ⋅ V A = ⎮ ( 3 + n) ⋅ ( 3 + 2⋅ n) ⌡ α= Putting all these results together ⌠ ⎮ 3 ⎮ ρ⋅ V dA ⌡ 2 2 = 3 ( 3+ n) ⋅ ( 3+ 2⋅ n) 2 3 ρ⋅ π⋅ R ⋅ Vav mrate⋅ Vav α= 2 2⋅ R ⋅ n ⋅ ρ⋅ π⋅ U 3 2 2⋅ n ⎛ U ⎞ ⋅ ⎜V ⎝ av ⎠ ( 3 + n) ⋅ ( 3 + 2⋅ n ) To plot α versus ReVav we use the following parametric relations ( ) n = −1.7 + 1.8⋅ log Reu Vav U = 2⋅ n (Eq. 8.23) 2 (Eq. 8.24) ( n + 1) ⋅ ( 2⋅ n + 1) Vav ReVav = ⋅ ReU U α= 3 2 2⋅ n ⎛ U ⎞ ⋅ ⎜V ⎝ av ⎠ ( 3 + n) ⋅ ( 3 + 2⋅ n ) (Eq. 8.27) A value of ReU leads to a value for n; this leads to a value for Vav/U; these lead to a value for ReVav and α The plots of α, and the error in assuming α = 1, versus ReVav can be done in Excel. Re U 1.00E+04 2.50E+04 5.00E+04 7.50E+04 1.00E+05 2.50E+05 5.00E+05 7.50E+05 1.00E+06 2.50E+06 5.00E+06 7.50E+06 1.00E+07 n 5.50 6.22 6.76 7.08 7.30 8.02 8.56 8.88 9.10 9.82 10.4 10.7 10.9 V av/U 0.776 0.797 0.811 0.818 0.823 0.837 0.846 0.851 0.854 0.864 0.870 0.873 0.876 Re Vav Alpha 7.76E+03 1.09 1.99E+04 1.07 4.06E+04 1.06 6.14E+04 1.06 8.23E+04 1.05 2.09E+05 1.05 4.23E+05 1.04 6.38E+05 1.04 8.54E+05 1.04 2.16E+06 1.03 4.35E+06 1.03 6.55E+06 1.03 8.76E+06 1.03 Error 8.2% 6.7% 5.9% 5.4% 5.1% 4.4% 3.9% 3.7% 3.5% 3.1% 2.8% 2.6% 2.5% Kinetic Energy Coefficient vs Reynolds Number Alpha 1.10 1.08 1.05 1.03 1.00 1E+03 1E+04 1E+05 1E+06 1E+07 1E+06 1E+07 Re Vav Error in assuming Alpha = 1 vs Reynolds Number 10.0% Error 7.5% 5.0% 2.5% 0.0% 1E+03 1E+04 1E+05 Re Vav Problem 8.79 Given: Data on flow through elbow Find: Head loss Solution: Basic equation [Difficulty: 2] 2 2 ⎛⎜ p ⎞ ⎜⎛ p ⎞ h V1 V2 1 lT 2 + α ⋅ + z + α ⋅ + z − ⎜ ρ⋅ g ⎜ ρ⋅ g 1 2 = g = HlT 2 ⋅ g 2 ⋅ g ⎝ ⎠ ⎝ ⎠ Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 Then HlT = p1 − p2 ρ⋅ g 2 + V1 − V2 3 2⋅ g 2 + z1 − z2 2 ( ) 2 2 m kg⋅ m s 1 m s 3 N 2 2 HlT = ( 70 − 45) × 10 ⋅ × × × + × 1.75 − 3.5 ⋅ ⎛⎜ ⎞ × + ( 2.25 − 3 ) ⋅ mHlT = 1.33 m 2 1000⋅ kg 2 9.81⋅ m 2 s 9.81 ⋅m ⎝ ⎠ m s ⋅N In terms of energy/mass h lT = g ⋅ HlT h lT = 9.81⋅ m 2 s 2 × 1.33⋅ m × N⋅ s kg⋅ m h lT = 13.0⋅ N⋅ m kg Problem 8.80 [Difficulty: 2] Given: Data on flow in a pipe Find: Head loss for horizontal pipe; inlet pressure for different alignments; slope for gravity feed Solution: 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 1 2 2 2 = h lT ⎝ ⎠ ⎝ρ ⎠ The basic equation between inlet (1) and exit (2) is Given or available data D = 75⋅ mm Horizontal pipe data p 1 = 275 ⋅ kPa Equation 8.29 becomes h lT = V = 5⋅ m s p 2 = 0 ⋅ kPa p1 − p2 ρ = 999 ⋅ kg m (Gage pressures) h lT = 275 ⋅ ρ 3 μ = 0.001 ⋅ (8.29) N⋅ s 2 m z1 = z2 V1 = V2 J kg For an inclined pipe with the same flow rate, the head loss will be the same as above; in addition we have the following new data z1 = 0 ⋅ m Equation 8.29 becomes z2 = 15⋅ m ( ) p 1 = p 2 + ρ⋅ g ⋅ z2 − z1 + ρ⋅ h lT p 1 = 422 ⋅ kPa For a declining pipe with the same flow rate, the head loss will be the same as above; in addition we have the following new data z1 = 0 ⋅ m Equation 8.29 becomes z2 = −15⋅ m ( ) p 1 = p 2 + ρ⋅ g ⋅ z2 − z1 + ρ⋅ h lT p 1 = 128 ⋅ kPa For a gravity feed with the same flow rate, the head loss will be the same as above; in addition we have the following new data p 1 = 0 ⋅ kPa Equation 8.29 becomes h lT z2 = z1 − g (Gage) z2 = −28.1 m Problem 8.81 Given: Data on flow through elbow Find: Inlet velocity [Difficulty: 2] Solution: Basic equation 2 2 ⎛⎜ p ⎞ ⎜⎛ p ⎞ h V1 V2 1 lT 2 ⎜ ρ⋅ g + α⋅ 2⋅ g + z1 − ⎜ ρ⋅ g + α⋅ 2 ⋅ g + z2 = g = HlT ⎝ ⎠ ⎝ ⎠ Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 Then 2 2 ( V2 − V1 = 2 ⋅ V1 )2 − V12 = 3⋅ V12 = ( 2⋅ p1 − p2 ) ( ) + 2 ⋅ g ⋅ z1 − z2 − 2 ⋅ g ⋅ HlT ρ ⎡ (p1 − p2) ⎤ + g ⋅ ( z1 − z2 ) − g ⋅ HlT⎥ 3 ⎣ ρ ⎦ V1 = 2 V1 = 2 ⋅⎢ 3 ⎡ × ⎢50 × 10 ⋅ ⎢ ⎣ 3 N 2 m 3 × m 1000⋅ kg × kg⋅ m 2 s ⋅N + 9.81⋅ m 2 s × ( −2 ) ⋅ m − 9.81⋅ m 2 s ⎤ × 1 ⋅ m⎥ ⎥ ⎦ m V1 = 3.70 s Problem 8.82 Given: A given piping system and volume flow rate with two liquid choices. Find: Which liquid has greater pressure loss [Difficulty: 2] Solution: Governing equation: ⎞ ⎛ P1 ⎞ ⎛P V2 V2 ⎜⎜ + α1 1 + gz1 ⎟⎟ − ⎜⎜ 2 + α 2 2 + gz 2 ⎟⎟ = hlT 2 2 ⎠ ⎠ ⎝ρ ⎝ρ 2 2 LV V hlT = hl + hlm = f +K D 2 2 Assumption: 1) Steady flow 2) Incompressible 3) Neglect elevation effects 4)Neglect velocity effects LV2 V2 ∆P = ρf + ρK 2 D 2 From Table A.8 it is seen that hot water has a lower density and lower kinematic viscosity than cold water. The lower density means that for a constant minor loss coefficient (K) and velocity the pressure loss due to minor losses will be less for hot water. The lower kinematic viscosity means that for a constant diameter and velocity the Reynolds number will increase. From Figure 8.13 it is seen that increasing the Reynolds number will either result in a decreased friction factor (f) or no change in the friction factor. This potential decrease in friction factor combined with a lower density for hot water means that the pressure loss due to major losses will be less for hot water as well. Cold water has a greater pressure drop Problem 8.83 Given: Increased friction factor for water tower flow Find: How much flow is decreased [Difficulty: 2] Solution: Basic equation from Example 8.7 V2 = ( 2 ⋅ g ⋅ z1 − z2 f ⋅ ⎛⎜ L ⎝D ) + 8⎞ + 1 ⎠ where L = 680 ⋅ ft D = 4 ⋅ in With f = 0.0308, we obtain ft V2 = 8.97⋅ s and Q = 351 gpm We need to recompute with f = 0.035 V2 = 2 × 32.2⋅ ft 2 × 80⋅ ft × s z1 − z2 = 80⋅ ft 1 0.035 ⋅ ⎜⎛ ⎜ ⎝ 680 4 12 + 8⎞ + 1 ft V2 = 8.42⋅ s ⎠ 2 Hence π⋅ D Q = V2 ⋅ A = V2 ⋅ 4 Q = 8.42⋅ ft s × π 4 2 × ⎛ 4 ⋅ ft⎞ × 7.48⋅ gal × 60⋅ s ⎜ 12 1 ⋅ min 3 ⎝ ⎠ 1 ⋅ ft Q = 330 ⋅ gpm (From Table G.2 1 ft3 = 7.48 gal) Hence the flow is decreased by ( 330 − 309 ) ⋅ gpm = 21⋅ gpm Problem 8.84 [Difficulty: 2] Given: Increased friction factor for water tower flow, and reduced length Find: How much flow is decreased Solution: Basic equation from Example 8.7 V2 = ( 2 ⋅ g ⋅ z1 − z2 f ⋅ ⎛⎜ L ⎝D where now we have L = 530 ⋅ ft We need to recompute with f = 0.04 V2 = ) + 8⎞ + 1 ⎠ D = 4 ⋅ in 2 × 32.2⋅ ft 2 × 80⋅ ft × s z1 − z2 = 80⋅ ft 1 0.035 ⋅ ⎜⎛ ⎜ ⎝ 530 4 12 + 8⎞ + 1 ft V2 = 9.51⋅ s ⎠ 2 Hence π⋅ D Q = V2 ⋅ A = V2 ⋅ 4 Q = 9.51⋅ ft s × π 4 2 × ⎛ 4 ⋅ ft⎞ × 7.48⋅ gal × 60⋅ s ⎜ 12 1 ⋅ min 3 ⎝ ⎠ 1 ⋅ ft Q = 372 ⋅ gpm (From Table G.2 1 ft3 = 7.48 gal) Problem 8.85 [Difficulty: 2] Problem 8.86 Given: Data on flow through Alaskan pipeline Find: Head loss Solution: Basic equation [Difficulty: 2] 2 2 ⎛ p ⎞ ⎛ p ⎞ h V1 V2 lT ⎜ 1 ⎜ 2 + α⋅ + z1 − + α⋅ + z2 = = HlT ⎜ ρ ⋅g ⎜ ρ ⋅g 2 ⋅ g 2 ⋅ g g oil oil ⎝ ⎠ ⎝ ⎠ Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) SG = 0.9 (Table A.2) Then p1 − p2 HlT = + z1 − z2 SGoil⋅ ρH2O⋅ g 3 2 1 m kg⋅ m s 3 N HlT = ( 8250 − 350 ) × 10 ⋅ × × × × + ( 45 − 115 ) ⋅ m 2 0.9 1000⋅ kg 2 9.81⋅ m m s ⋅N In terms of energy/mass h lT = g ⋅ HlT h lT = 9.81⋅ m 2 s HlT = 825 m 2 × 825 ⋅ m × N⋅ s kg⋅ m h lT = 8.09⋅ kN⋅ m kg Problem 8.87 [Difficulty: 2] Problem 8.88 Given: Data on flow through a tube Find: Head loss Solution: Basic equation [Difficulty: 2] 2 2 ⎞ ⎜⎛ p ⎞ h ⎛⎜ p V1 V2 1 lT 2 + α ⋅ + z + α ⋅ + z − ⎜ ρ⋅ g ⎜ ρ⋅ g 1 2 = g = HlT 2 ⋅ g 2 ⋅ g ⎝ ⎠ ⎝ ⎠ Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 Given or available data The basic equation reduces to Q = 10⋅ h lT = L min ∆p ρ D = 15⋅ mm ∆p = 85⋅ kPa ρ = 999 ⋅ kg 3 m 2 h lT = 85.1 m 2 s HlT = h lT g HlT = 8.68 m Problem 8.89 [Difficulty: 2] Problem 8.90 Given: Data on flow from reservoir Find: Head from pump; head loss Solution: Basic equations [Difficulty: 3] 2 2 ⎛⎜ p ⎞ ⎜⎛ p ⎞ h V3 V4 3 lT 4 + α ⋅ + z + α ⋅ + z − ⎜ ρ⋅ g ⎜ ρ⋅ g 3 4 = g = HlT 2 ⋅ g 2 ⋅ g ⎝ ⎠ ⎝ ⎠ for flow from 3 to 4 2 2 ⎛⎜ p ⎞ ⎜⎛ p ⎞ ∆h V3 V2 3 pump 2 = Hpump for flow from 2 to 3 ⎜ ρ⋅ g + α⋅ 2⋅ g + z3 − ⎜ ρ⋅ g + α⋅ 2 ⋅ g + z2 = g ⎝ ⎠ ⎝ ⎠ Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) V2 = V3 = V4 (constant area pipe) Then for the pump Hpump = p3 − p2 ρ⋅ g 3 2 m kg⋅ m s 3 N Hpump = ( 450 − 150 ) × 10 ⋅ × × × 2 1000⋅ kg 2 9.81⋅ m m s ⋅N In terms of energy/mass h pump = g ⋅ Hpump h pump = 9.81⋅ m 2 2 × 30.6⋅ m × s For the head loss from 3 to 4 HlT = p3 − p4 ρ⋅ g Hpump = 30.6 m N⋅ s kg⋅ m 3 h lT = g ⋅ HlT N⋅ m kg + z3 − z4 2 m kg⋅ m s 3 N HlT = ( 450 − 0 ) × 10 ⋅ × × × + ( 0 − 35) ⋅ m 2 1000⋅ kg 2 9.81⋅ m m s ⋅N In terms of energy/mass h pump = 300 ⋅ h lT = 9.81⋅ m 2 s HlT = 10.9 m 2 × 10.9⋅ m × N⋅ s kg⋅ m h lT = 107 ⋅ N⋅ m kg Problem 8.91 [Difficulty: 2] Given: Data on flow in a pipe Find: Friction factor; Reynolds number; if flow is laminar or turbulent Solution: Given data From Appendix A ∆p D = 75⋅ mm ρ = 1000⋅ L kg = 0.075 ⋅ μ = 4 ⋅ 10 3 Pa m kg mrate = 0.075 ⋅ s − 4 N⋅ s ⋅ m 2 m The governing equations between inlet (1) and exit (2) are 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 ⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h l (8.29) ⎝ ⎠ ⎝ ⎠ 2 hl = f ⋅ L V ⋅ D 2 (8.34) For a constant area pipe V1 = V2 = V Hence Eqs. 8.29 and 8.34 become f = 2⋅ D 2 ⋅ (p1 − p2) ρ L⋅ V For the velocity mrate V = ρ⋅ π 4 f = The Reynolds number is Re = 2 ⋅ D ∆p ⋅ 2 L ρ⋅ V V = 0.017 2 ⋅D 2 ⋅ D ∆p ⋅ 2 L ρ⋅ V Hence = ρ⋅ V⋅ D μ m s f = 0.0390 Re = 3183 This Reynolds number indicates the flow is turbulent. (From Eq. 8.37, at this Reynolds number the friction factor for a smooth pipe is f = 0.043; the friction factor computed above thus indicates that, within experimental error, the flow corresponds to turbulent flow in a smooth pipe) Problem 8.92 [Difficulty: 2] Problem 8.93 [Difficulty: 3] Using the above formula for f 0, and Eq. 8.37 for f 1 e/D = 0 0.0001 0.0002 0.0005 0.001 0.002 0.005 0.01 0.02 0.05 Re 1.00E+04 2.50E+04 5.00E+04 7.50E+04 1.00E+05 2.50E+05 5.00E+05 7.50E+05 1.00E+06 5.00E+06 1.00E+07 5.00E+07 1.00E+08 0.0310 0.0244 0.0208 0.0190 0.0179 0.0149 0.0131 0.0122 0.0116 0.0090 0.0081 0.0066 0.0060 0.0311 0.0247 0.0212 0.0195 0.0185 0.0158 0.0145 0.0139 0.0135 0.0124 0.0122 0.0120 0.0120 0.0313 0.0250 0.0216 0.0200 0.0190 0.0167 0.0155 0.0150 0.0148 0.0140 0.0139 0.0138 0.0137 0.0318 0.0258 0.0226 0.0212 0.0204 0.0186 0.0178 0.0175 0.0173 0.0168 0.0168 0.0167 0.0167 0.0327 0.0270 0.0242 0.0230 0.0223 0.0209 0.0204 0.0201 0.0200 0.0197 0.0197 0.0196 0.0196 0.0342 0.0291 0.0268 0.0258 0.0253 0.0243 0.0239 0.0238 0.0237 0.0235 0.0235 0.0234 0.0234 0.0383 0.0342 0.0325 0.0319 0.0316 0.0309 0.0307 0.0306 0.0305 0.0304 0.0304 0.0304 0.0304 0.0440 0.0407 0.0395 0.0390 0.0388 0.0383 0.0381 0.0380 0.0380 0.0379 0.0379 0.0379 0.0379 0.0534 0.0508 0.0498 0.0494 0.0493 0.0489 0.0488 0.0487 0.0487 0.0487 0.0486 0.0486 0.0486 0.0750 0.0731 0.0724 0.0721 0.0720 0.0717 0.0717 0.0716 0.0716 0.0716 0.0716 0.0716 0.0716 0.001 0.002 0.005 0.01 0.02 0.05 0.0338 0.0288 0.0265 0.0256 0.0251 0.0241 0.0238 0.0237 0.0236 0.0235 0.0234 0.0234 0.0234 0.0376 0.0337 0.0322 0.0316 0.0313 0.0308 0.0306 0.0305 0.0305 0.0304 0.0304 0.0304 0.0304 0.0431 0.0402 0.0391 0.0387 0.0385 0.0381 0.0380 0.0380 0.0380 0.0379 0.0379 0.0379 0.0379 0.0523 0.0502 0.0494 0.0492 0.0490 0.0488 0.0487 0.0487 0.0487 0.0486 0.0486 0.0486 0.0486 0.0738 0.0725 0.0720 0.0719 0.0718 0.0716 0.0716 0.0716 0.0716 0.0716 0.0716 0.0716 0.0716 f0 Using the add-in function Friction factor from the Web e/D = Re 1.00E+04 2.50E+04 5.00E+04 7.50E+04 1.00E+05 2.50E+05 5.00E+05 7.50E+05 1.00E+06 5.00E+06 1.00E+07 5.00E+07 1.00E+08 0 0.0001 0.0002 0.0005 f 0.0309 0.0245 0.0209 0.0191 0.0180 0.0150 0.0132 0.0122 0.0116 0.0090 0.0081 0.0065 0.0059 0.0310 0.0248 0.0212 0.0196 0.0185 0.0158 0.0144 0.0138 0.0134 0.0123 0.0122 0.0120 0.0120 The error can now be computed 0.0312 0.0250 0.0216 0.0200 0.0190 0.0166 0.0154 0.0150 0.0147 0.0139 0.0138 0.0138 0.0137 0.0316 0.0257 0.0226 0.0212 0.0203 0.0185 0.0177 0.0174 0.0172 0.0168 0.0168 0.0167 0.0167 0.0324 0.0268 0.0240 0.0228 0.0222 0.0208 0.0202 0.0200 0.0199 0.0197 0.0197 0.0196 0.0196 e/D = 0 0.0001 0.0002 0.0005 0.001 0.002 0.005 0.01 0.02 0.05 Re 1.00E+04 2.50E+04 5.00E+04 7.50E+04 1.00E+05 2.50E+05 5.00E+05 7.50E+05 1.00E+06 5.00E+06 1.00E+07 5.00E+07 1.00E+08 0.29% 0.39% 0.63% 0.69% 0.71% 0.65% 0.52% 0.41% 0.33% 0.22% 0.49% 1.15% 1.44% 0.36% 0.24% 0.39% 0.38% 0.33% 0.04% 0.26% 0.41% 0.49% 0.51% 0.39% 0.15% 0.09% 0.43% 0.11% 0.19% 0.13% 0.06% 0.28% 0.51% 0.58% 0.60% 0.39% 0.27% 0.09% 0.06% 0.61% 0.21% 0.25% 0.35% 0.43% 0.64% 0.64% 0.59% 0.54% 0.24% 0.15% 0.05% 0.03% Error (%) 0.88% 1.27% 0.60% 1.04% 0.67% 1.00% 0.73% 0.95% 0.76% 0.90% 0.72% 0.66% 0.59% 0.47% 0.50% 0.37% 0.43% 0.31% 0.16% 0.10% 0.10% 0.06% 0.03% 0.02% 0.02% 0.01% 1.86% 1.42% 1.11% 0.93% 0.81% 0.48% 0.31% 0.23% 0.19% 0.06% 0.03% 0.01% 0.00% 2.12% 1.41% 0.98% 0.77% 0.64% 0.35% 0.21% 0.15% 0.12% 0.03% 0.02% 0.01% 0.00% 2.08% 1.21% 0.77% 0.58% 0.47% 0.24% 0.14% 0.10% 0.08% 0.02% 0.01% 0.00% 0.00% 1.68% 0.87% 0.52% 0.38% 0.30% 0.14% 0.08% 0.06% 0.05% 0.01% 0.01% 0.00% 0.00% The maximum discrepancy is 2.12% at Re = 10,000 and e/D = 0.01 0.100 f0 0.010 0.001 1E+04 e/D = 0 e/D = 0.0001 e/D = 0.0002 e/D = 0.0005 e/D = 0.001 e/D = 0.002 e/D = 0.005 e/D = 0.01 e/D = 0.02 e/D = 0.05 1E+05 1E+06 Re 1E+07 1E+08 Problem 8.94 [Difficulty: 3] Solution: Using the add-in function Friction factor from the web site e/D = Re 500 1.00E+03 1.50E+03 2.30E+03 1.00E+04 1.50E+04 1.00E+05 1.50E+05 1.00E+06 1.50E+06 1.00E+07 1.50E+07 1.00E+08 0 0.0001 0.0002 0.0005 0.001 0.002 0.005 0.01 0.02 0.04 0.1280 0.0640 0.0427 0.0489 0.0338 0.0313 0.0251 0.0246 0.0236 0.0235 0.0234 0.0234 0.0234 0.1280 0.0640 0.0427 0.0512 0.0376 0.0356 0.0313 0.0310 0.0305 0.0304 0.0304 0.0304 0.0304 0.1280 0.0640 0.0427 0.0549 0.0431 0.0415 0.0385 0.0383 0.0380 0.0379 0.0379 0.0379 0.0379 0.1280 0.0640 0.0427 0.0619 0.0523 0.0511 0.0490 0.0489 0.0487 0.0487 0.0486 0.0486 0.0486 0.1280 0.0640 0.0427 0.0747 0.0672 0.0664 0.0649 0.0648 0.0647 0.0647 0.0647 0.0647 0.0647 f 0.1280 0.0640 0.0427 0.0473 0.0309 0.0278 0.0180 0.0166 0.0116 0.0109 0.0081 0.0076 0.0059 0.1280 0.0640 0.0427 0.0474 0.0310 0.0280 0.0185 0.0172 0.0134 0.0130 0.0122 0.0121 0.0120 0.1280 0.0640 0.0427 0.0474 0.0312 0.0282 0.0190 0.0178 0.0147 0.0144 0.0138 0.0138 0.0137 0.1280 0.0640 0.0427 0.0477 0.0316 0.0287 0.0203 0.0194 0.0172 0.0170 0.0168 0.0167 0.0167 0.1280 0.0640 0.0427 0.0481 0.0324 0.0296 0.0222 0.0214 0.0199 0.0198 0.0197 0.0197 0.0196 Friction Factor vs Reynolds Number 1.000 0.100 f e/D = 0.010 0.001 1.0E+02 0 0.0001 0.0002 0.0005 0.001 0.002 0.005 0.01 0.02 0.04 1.0E+03 1.0E+04 Re 1.0E+05 1.0E+06 1.0E+07 1.0E+08 Problem 8.95 [Difficulty: 2] Problem 8.96 [Difficulty: 3] Using the above formula for f 0, and Eq. 8.37 for f 1 e/D = 0 0.0001 0.0002 0.0005 0.001 0.002 0.005 0.01 0.02 0.05 Re 1.00E+04 2.50E+04 5.00E+04 7.50E+04 1.00E+05 2.50E+05 5.00E+05 7.50E+05 1.00E+06 5.00E+06 1.00E+07 5.00E+07 1.00E+08 0.0309 0.0244 0.0207 0.0189 0.0178 0.0148 0.0131 0.0122 0.0116 0.0090 0.0081 0.0066 0.0060 0.0310 0.0245 0.0210 0.0193 0.0183 0.0156 0.0143 0.0137 0.0133 0.0123 0.0122 0.0120 0.0120 0.0311 0.0248 0.0213 0.0197 0.0187 0.0164 0.0153 0.0148 0.0146 0.0139 0.0139 0.0138 0.0138 0.0315 0.0254 0.0223 0.0209 0.0201 0.0183 0.0176 0.0173 0.0172 0.0168 0.0168 0.0167 0.0167 0.0322 0.0265 0.0237 0.0226 0.0220 0.0207 0.0202 0.0200 0.0199 0.0197 0.0197 0.0197 0.0197 0.0335 0.0285 0.0263 0.0254 0.0250 0.0241 0.0238 0.0237 0.0236 0.0235 0.0235 0.0235 0.0235 0.0374 0.0336 0.0321 0.0316 0.0313 0.0308 0.0306 0.0305 0.0305 0.0304 0.0304 0.0304 0.0304 0.0430 0.0401 0.0391 0.0387 0.0385 0.0382 0.0381 0.0381 0.0380 0.0380 0.0380 0.0380 0.0380 0.0524 0.0502 0.0495 0.0492 0.0491 0.0489 0.0488 0.0488 0.0488 0.0487 0.0487 0.0487 0.0487 0.0741 0.0727 0.0722 0.0720 0.0719 0.0718 0.0717 0.0717 0.0717 0.0717 0.0717 0.0717 0.0717 0.001 0.002 0.005 0.01 0.02 0.05 0.0338 0.0288 0.0265 0.0256 0.0251 0.0241 0.0238 0.0237 0.0236 0.0235 0.0234 0.0234 0.0234 0.0376 0.0337 0.0322 0.0316 0.0313 0.0308 0.0306 0.0305 0.0305 0.0304 0.0304 0.0304 0.0304 0.0431 0.0402 0.0391 0.0387 0.0385 0.0381 0.0380 0.0380 0.0380 0.0379 0.0379 0.0379 0.0379 0.0523 0.0502 0.0494 0.0492 0.0490 0.0488 0.0487 0.0487 0.0487 0.0486 0.0486 0.0486 0.0486 0.0738 0.0725 0.0720 0.0719 0.0718 0.0716 0.0716 0.0716 0.0716 0.0716 0.0716 0.0716 0.0716 f0 Using the add-in function Friction factor from the Web e/D = Re 1.00E+04 2.50E+04 5.00E+04 7.50E+04 1.00E+05 2.50E+05 5.00E+05 7.50E+05 1.00E+06 5.00E+06 1.00E+07 5.00E+07 1.00E+08 0 0.0001 0.0002 0.0005 f 0.0309 0.0245 0.0209 0.0191 0.0180 0.0150 0.0132 0.0122 0.0116 0.0090 0.0081 0.0065 0.0059 0.0310 0.0248 0.0212 0.0196 0.0185 0.0158 0.0144 0.0138 0.0134 0.0123 0.0122 0.0120 0.0120 0.0312 0.0250 0.0216 0.0200 0.0190 0.0166 0.0154 0.0150 0.0147 0.0139 0.0138 0.0138 0.0137 0.0316 0.0257 0.0226 0.0212 0.0203 0.0185 0.0177 0.0174 0.0172 0.0168 0.0168 0.0167 0.0167 0.0324 0.0268 0.0240 0.0228 0.0222 0.0208 0.0202 0.0200 0.0199 0.0197 0.0197 0.0196 0.0196 The error can now be computed e/D = Re 1.00E+04 2.50E+04 5.00E+04 7.50E+04 1.00E+05 2.50E+05 5.00E+05 7.50E+05 1.00E+06 5.00E+06 1.00E+07 5.00E+07 1.00E+08 0 0.01% 0.63% 0.85% 0.90% 0.92% 0.84% 0.70% 0.59% 0.50% 0.07% 0.35% 1.02% 1.31% 0.0001 0.15% 0.88% 1.19% 1.30% 1.34% 1.33% 1.16% 0.99% 0.86% 0.17% 0.00% 0.16% 0.18% 0.0002 0.26% 1.02% 1.32% 1.40% 1.42% 1.25% 0.93% 0.72% 0.57% 0.01% 0.09% 0.18% 0.19% 0.0005 0.001 0.002 0.005 0.01 0.02 0.05 0.46% 1.20% 1.38% 1.35% 1.28% 0.85% 0.48% 0.30% 0.20% 0.11% 0.15% 0.19% 0.20% Error (%) 0.64% 0.73% 1.22% 1.03% 1.21% 0.84% 1.07% 0.65% 0.94% 0.52% 0.47% 0.16% 0.19% 0.00% 0.07% 0.07% 0.01% 0.10% 0.15% 0.18% 0.18% 0.19% 0.20% 0.20% 0.20% 0.20% 0.55% 0.51% 0.28% 0.16% 0.09% 0.07% 0.13% 0.16% 0.17% 0.19% 0.20% 0.20% 0.20% 0.19% 0.11% 0.00% 0.06% 0.09% 0.15% 0.18% 0.18% 0.19% 0.20% 0.20% 0.20% 0.20% 0.17% 0.14% 0.16% 0.17% 0.18% 0.19% 0.20% 0.20% 0.20% 0.20% 0.20% 0.20% 0.20% 0.43% 0.29% 0.24% 0.23% 0.22% 0.21% 0.20% 0.20% 0.20% 0.20% 0.20% 0.20% 0.20% The maximum discrepancy is 1.42% at Re = 100,000 and e/D = 0.0002 0.100 f 0.010 0.001 1E+04 e/D = 0 e/D = 0.0001 e/D = 0.0002 e/D = 0.0005 e/D = 0.001 e/D = 0.002 e/D = 0.005 e/D = 0.01 e/D = 0.02 e/D = 0.05 1E+05 1E+06 Re 1E+07 1E+08 Problem 8.97 Given: Flow through gradual contraction Find: Pressure after contraction; compare to sudden contraction Solution: [Difficulty: 3] 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 = h lm 2 2 ⎝ ⎠ ⎝ρ ⎠ Basic equations h lm = K⋅ V2 2 2 Q = V⋅ A Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Horizontal Available data Q = 25⋅ 3 L Q = 0.025 s m D1 = 75⋅ mm D2 = 37.5⋅ mm s p 1 = 500⋅ kPa 3 2 ⎛ D2 ⎞ ⎛ 37.5 ⎞ 2 For an included angle of 150 and an area ratio =⎜ = 0.25 we find from Table 8.3 =⎜ A1 ⎝ D1 ⎠ ⎝ 75 ⎠ Hence the energy equation becomes kg m A2 o ρ = 999⋅ 2 2 2 ⎛⎜ p V1 ⎞ ⎛⎜ p 2 V2 ⎞ V2 1 ⎜ ρ + 2 − ⎜ ρ + 2 = K⋅ 2 with ⎝ ⎠ ⎝ ⎠ V1 = K = 0.35 4⋅ Q π⋅ D1 2 V2 = 4⋅ Q π⋅ D2 2 2 ρ 8 ⋅ ρ⋅ Q ⎡ ( 1 + K) 1 ⎤ 2 2 p 2 = p 1 − ⋅ ⎡( 1 + K) ⋅ V2 − V1 ⎤ = p 2 − ⋅⎢ − ⎥ ⎦ 2 ⎣ 2 4 4 ⎢ π D1 ⎥ ⎣ D2 ⎦ 2 3 2 ⎛ m ⎞ 1 1 ⎤ × N⋅ s p = 170 ⋅ kPa p 2 = 500 × 10 ⋅ − × 999 ⋅ − × ⎜ 0.025 ⋅ × ⎡( 1 + 0.35) × ⎢ 2 2 3 ⎝ s ⎠ 4⎥ 4 kg⋅ m 2 m π m ( 0.075 ⋅ m) ⎦ ( 0.0375⋅ m) ⎣ 3 N 8 kg Repeating the above analysis for an included angle of 180 o (sudden contraction) 2 K = 0.41 2 3 ⎛ 1 m ⎞ 1 ⎤ × N⋅ s p = 155 ⋅ kPa × ⎜ 0.025 ⋅ × ⎡( 1 + 0.41) × p 2 = 500 × 10 ⋅ − × 999 ⋅ − ⎢ 4 kg⋅ m 2 2 2 3 ⎝ s ⎠ 4⎥ ( 0.0375⋅ m) m π m ( 0.075 ⋅ m) ⎦ ⎣ 3 N 8 kg Problem 8.98 [Difficulty: 3] Problem 8.99 Given: Flow through sudden contraction Find: Volume flow rate [Difficulty: 3] Solution: Basic equations 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 = h lm 2 2 ⎝ ⎠ ⎝ρ ⎠ h lm = K⋅ V2 2 2 Q = V⋅ A Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Horizontal Hence the energy equation becomes 2 2 2 ⎛⎜ p V1 ⎞ ⎛⎜ p 2 V2 ⎞ V2 1 ⎜ ρ + 2 − ⎜ ρ + 2 = K⋅ 2 ⎝ ⎠ ⎝ ⎠ From continuity A2 V1 = V2 ⋅ = V2 ⋅ AR A1 Hence 2 2 2 2 ⎛⎜ p V2 ⋅ AR ⎞ ⎛⎜ p 2 V2 ⎞ V2 1 + = K⋅ −⎜ ⎜ρ + 2 2 ⎠ 2 ⎝ ⎠ ⎝ρ Solving for V 2 Hence ( 2⋅ p1 − p2 V2 = ( 2 2 ) ρ⋅ 1 − AR + K V2 = 2 × 0.5⋅ lbf 2 Q = V2 ⋅ A2 = π 4 π⋅ D2 4 2 × ⎛ D2 ⎞ ⎛ 1 ⎞ 2 AR = ⎜ =⎜ = 0.25 ⎝ D1 ⎠ ⎝ 2 ⎠ 2 × in Q = ) so from Fig. 8.14 3 1 slug⋅ ft ⎛ 12⋅ in ⎞ × ft × × ⎜ 1 ⋅ ft 2 2 1.94⋅ slug ⎝ ⎠ 1 − 0.25 + 0.4 lbf ⋅ s ( ) ft V2 = 7.45⋅ s 2 ⋅ V2 3 ⎛ 1 ⋅ ft⎞ × 7.45⋅ ft Q = 0.0406⋅ ft ⎜ 12 s s ⎝ ⎠ Q = 2.44⋅ ft 3 min Q = 18.2⋅ gpm K = 0.4 Problem 8.100 Given: Flow through sudden expansion Find: Inlet speed; Volume flow rate [Difficulty: 3] Solution: Basic equations 2 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 V1 1 2 ⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h lm h lm = K⋅ 2 ⎝ ⎠ ⎝ ⎠ Q = V⋅ A ∆p = ρH2O⋅ g ⋅ ∆h Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Horizontal Hence the energy equation becomes 2 2 2 ⎛⎜ p V1 ⎞ ⎛⎜ p 2 V2 ⎞ V1 1 ⎜ ρ + 2 − ⎜ ρ + 2 = K⋅ 2 ⎝ ⎠ ⎝ ⎠ From continuity A1 V2 = V1 ⋅ = V1 ⋅ AR A2 Hence 2 2 2 2 ⎛⎜ p V1 ⎞ ⎛⎜ p 2 V1 ⋅ AR ⎞ V1 1 = K⋅ ⎜ρ + 2 −⎜ρ + 2 2 ⎝ ⎠ ⎝ ⎠ Solving for V 1 Also Hence V1 = ( 2⋅ p2 − p1 ( ) 2 2 ⎛ D1 ⎞ ⎛ 75 ⎞ 2 AR = ⎜ =⎜ = 0.111 ⎝ D2 ⎠ ⎝ 225 ⎠ ) ρ⋅ 1 − AR − K kg m × 1 so from Fig. 8.14 K = 0.8 2 N⋅ s 5 p 2 − p 1 = ρH2O⋅ g ⋅ ∆h = 1000⋅ × 9.81⋅ × ⋅m × = 49.1⋅ Pa 3 2 1000 kg⋅ m m s V1 = 2 × 49.1⋅ 3 N 2 m Q = V1 ⋅ A1 = π⋅ D1 4 × m 1.23⋅ kg 2 ⋅ V1 (1 − 0.1112 − 0.8) × kg⋅ m 2 N⋅ s 3 2 75 m Q = ⋅ m⎞ × 20.6⋅ × ⎛⎜ 4 ⎝ 1000 ⎠ s π m V1 = 20.6 s Q = 0.0910⋅ m s 3 Q = 5.46⋅ m min Problem 8.101 Given: Data on a pipe sudden contraction Find: Theoretical calibration constant; plot [Difficulty: 4] Solution: Given data D1 = 45⋅ mm D2 = 22.5⋅ mm The governing equations between inlet (1) and exit (2) are 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 ⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h l ⎝ ⎠ ⎝ ⎠ where h l = K⋅ V2 (8.29) 2 (8.40a) 2 Hence the pressure drop is (assuming α = 1) 2 ⎛⎜ V 2 V 2 V2 ⎞ 2 1 ∆p = p 1 − p 2 = ρ⋅ ⎜ − + K⋅ 2 2 ⎠ ⎝ 2 For the sudden contraction so π π 2 2 V1 ⋅ ⋅ D1 = V2 ⋅ ⋅ D2 = Q 4 4 ∆p = 4 ⎡ ⎢⎛ D1 ⎞ ⋅ ⎜ ( 1 + K) − 2 ⎢ D2 ⎣⎝ ⎠ ρ⋅ V1 2 ⎛ D1 ⎞ V2 = V1 ⋅ ⎜ ⎝ D2 ⎠ or ⎤ ⎥ 1 ⎥ ⎦ For the pressure drop we can use the manometer equation ∆p = ρ⋅ g ⋅ ∆h Hence In terms of flow rate Q ρ⋅ g ⋅ ∆h = 4 ⎡ ⎢⎛ D1 ⎞ ⋅ ⎜ ( 1 + K) − 2 ⎢ D2 ⎣⎝ ⎠ ρ⋅ V1 2 ⎤ ⎥ 1 ⎥ ⎦ ⎡⎛ D ⎞ 4 ⎢ 1 ρ⋅ g ⋅ ∆h = ⋅ ⋅ ⎜ ( 1 + K) − 2 ⎢ D2 2 ⎣ ⎝ ⎠ π 2 ⎛ ⋅D ⎞ ⎜4 1 ⎝ ⎠ ρ 2 Q ⎤ ⎥ 1 ⎥ ⎦ 2 or ⎡ D ⎞4 ⎢⎛ 1 g ⋅ ∆h = ⋅ ⎜ ( 1 + K) − 2 4 ⎢ D2 ⎠ π ⋅ D1 ⎣⎝ Hence for flow rate Q we find Q = k ⋅ ∆h 2 8⋅ Q 2 k= where g ⋅ π ⋅ D1 4 4 ⎡ ⎢⎛ D1 ⎞ 8⋅ ⎜ ⎢ D ( 1 + K) − ⎣⎝ 2 ⎠ For K, we need the aspect ratio AR ⎛ D2 ⎞ AR = ⎜ ⎝ D1 ⎠ From Fig. 8.15 K = 0.4 ⎤ ⎥ ⎥ ⎦ 1 ⎤ ⎥ ⎥ ⎦ 1 2 AR = 0.25 5 2 Using this in the expression for k, with the other given values k = g ⋅ π ⋅ D1 4 ⎡⎛ D ⎞ 4 ⎢ 1 8⋅ ⎜ ⎢ D ( 1 + K) − ⎣⎝ 2 ⎠ −3 m k = 1.52 × 10 ⎤ ⎥ 1 ⎥ ⎦ L For Δh in mm and Q in L/min k = 2.89⋅ min 1 mm 2 The plot of theoretical Q versus flow rate Δh can be done in Excel. Calibration Curve for a Sudden Contraction Flow Meter 60 Q (L/mm) 50 40 30 20 10 0 0 50 100 150 200 Dh (mm) It is a practical device, but is a) Nonlinear and b) has a large energy loss 250 300 350 ⋅ 2 s Problem 8.102 Given: Flow through a reentrant device Find: Head loss [Difficulty: 3] Solution: Basic equations 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 = h lT 2 2 ⎝ ⎠ ⎝ρ ⎠ 2 V2 L V2 h lT = h l + h lm = f ⋅ ⋅ + K⋅ 2 D 2 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) L << so ignore hl 5) Reentrant 3 Available data D1 = 100 ⋅ mm D2 = 50⋅ mm Q = 0.01⋅ π 2 A1 = ⋅ D1 4 A1 = 7.85 × 10 mm 3 m π 2 A2 = ⋅ D2 4 2 K = 0.78 and from Table 8.2 s 3 2 A2 = 1.96 × 10 mm Hence between the free surface (Point 1) and the exit (2) the energy equation becomes p1 ρ From continuity Hence Solving for h + V1 2 2 − V2 2 2 − Q = V1 ⋅ A1 = V2 ⋅ A2 2 p2 ρ = K⋅ V2 2 2 ρ 2 1 Q ⎞ 1 Q ⎞ Q ⎞ g ⋅ h + ⋅ ⎛⎜ − ⋅ ⎛⎜ = K⋅ ⋅ ⎛⎜ 2 A1 2 A2 2 A2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 1 ⎛Q⎞ ⎜A ⎝ 2⎠ h = p1 − p2 and also = 2 2 2⎤ ⎡ ⎛ A2 ⎞ ⎥ ⎢ ⋅ 1+ K− ⎜ ⎥ 2⋅ g ⎢ ⎣ ⎝ A1 ⎠ ⎦ h = 2.27 m ρ⋅ g ⋅ h ρ = g⋅ h where h is the head loss Q = V⋅ A Problem 8.103 [Difficulty: 4] Given: Contraction coefficient for sudden contraction Find: Expression for minor head loss; compare with Fig. 8.15; plot Solution: We analyse the loss at the "sudden expansion" at the vena contracta The governing CV equations (mass, momentum, and energy) are Assume: 1) Steady flow 2) Incompressible flow 3) Uniform flow at each section 4) Horizontal: no body force 5) No shaft work 6) Neglect viscous friction 7) Neglect gravity The mass equation becomes Vc⋅ Ac = V2 ⋅ A2 The momentum equation becomes p c⋅ A2 − p 2 ⋅ A2 = Vc⋅ −ρ⋅ Vc⋅ Ac + V2 ⋅ ρ⋅ V2 ⋅ A2 or (using Eq. 1) Ac p c − p 2 = ρ⋅ Vc⋅ ⋅ V2 − Vc A2 The energy equation becomes pc p2 ⎛ ⎛ 2⎞ 2⎞ Qrate = ⎜ u c + + Vc ⋅ −ρ⋅ Vc⋅ Ac + ⎜ u 2 + + V2 ⋅ ρ⋅ V2 ⋅ A2 ρ ρ ⎝ ⎠ ⎝ ⎠ or (using Eq. 1) (1) ( ) ( h lm = u 2 − u c − = mrate ) ) (2) ( Qrate ( ) 2 Vc − V2 2 ( 2 + pc − p2 ρ (3) ) 2 h lm = Combining Eqs. 2 and 3 Vc − V2 2 ⎡ ⎢ h lm = ⋅ 1− 2 ⎢ ⎣ Vc Ac Cc = From Eq. 1 h lm = h lm = Vc Ac + Vc⋅ ⋅ V2 − Vc A2 ( 2⎤ A V ⎤ ⎛ V2 ⎞ ⎥ 2 c ⎡⎛ 2 ⎞ ⎜ V ⎥ + Vc ⋅ A ⋅ ⎢⎜ V − 1⎥ 2 ⎣⎝ c ⎠ ⎝ c⎠ ⎦ ⎦ Vc 2 ⋅ ⎛ 1 − Cc 2 ⎝ Vc 2 Vc 2⎞ 2 ⎠ + Vc ⋅ Cc⋅ ( Cc − 1 ) ⋅ ⎛ 1 − C c + 2 ⋅ C c − 2 ⋅ C c⎞ ⎝ ⎠ 2 2 ) V2 = A2 h lm = Hence 2 2 2 2 ( ⋅ 1 − Cc 2 2 )2 (4) 2 2 2 Vc ⎛ V2 ⎞ 2 h lm = K⋅ = K⋅ ⋅⎜ = K⋅ ⋅ Cc 2 2 2 Vc ⎝ ⎠ V2 But we have K= Hence, comparing Eqs. 4 and 5 Vc (5) ( 1 − Cc) 2 Cc 2 ⎛ 1 − 1⎞ ⎜C ⎝ c ⎠ 2 So, finally K= where ⎛ A2 ⎞ Cc = 0.62 + 0.38⋅ ⎜ ⎝ A1 ⎠ 3 This result,can be plotted in Excel. The agreement with Fig. 8.15 is reasonable. 0.5 0.4 K 0.3 0.2 0.1 0 0.2 0.4 0.6 AR 0.8 1 Problem 8.104 Given: Flow through short pipe Find: Volume flow rate; How to improve flow rate [Difficulty: 3] Solution: Basic equations 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 ⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h lT ⎝ ⎠ ⎝ ⎠ 2 V2 L V2 h lT = h l + h lm = f ⋅ ⋅ + K⋅ 2 D 2 2 Q = V⋅ A Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) L << so ignore hl 5) Reentrant Hence between the free surface (Point 1) and the exit (2) the energy equation becomes V1 2 + g ⋅ z1 − 2 From continuity Hence V2 2 = K⋅ 2 V2 2 2 A2 V1 = V2 ⋅ A1 2 2 2 V2 V2 ⎛ A2 ⎞ ⋅⎜ + g⋅ h − = K⋅ 2 2 2 ⎝ A1 ⎠ V2 2 Solving for V 2 V2 = Hence V2 = 2⋅ g⋅ h 2 × 9.81⋅ m 2 × 1⋅ m × s Q = V2 ⋅ A2 Q = 3.33⋅ K = 0.78 and from Table 8.2 2 ⎡ ⎛ A2 ⎞ ⎤⎥ ⎢ ⎢1 + K − ⎜ A ⎥ ⎣ ⎝ 1⎠ ⎦ m s 1 m V2 = 3.33 s 2 ⎡ 350 ⎞ ⎤ ⎢1 + 0.78 − ⎛⎜ ⎥ ⎣ ⎝ 3500 ⎠ ⎦ 2 × 350 ⋅ mm × ⎛ 1⋅ m ⎞ ⎜ ⎝ 1000⋅ mm ⎠ 2 3 −3m Q = 1.17 × 10 s The flow rate could be increased by (1) rounding the entrance and/or (2) adding a diffuser (both somewhat expensive) 3 Q = 0.070 ⋅ m min Problem 8.105 [Difficulty: 3] Problem 8.106 [Difficulty: 3] Problem 8.107 [Difficulty: 3] Given: Flow out of water tank Find: Volume flow rate using hole; Using short pipe section; Using rounded edge Solution: Basic equations 2 2 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 V2 1 L V2 2 ⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h lT h lT = hl + h lm = f ⋅ D ⋅ 2 + K⋅ 2 Q = V⋅ A ⎝ ⎠ ⎝ ⎠ Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Vl << 5) L << so hl = 0 Available data D = 25⋅ mm r = 5 ⋅ mm h = 5⋅ m Hence for all three cases, between the free surface (Point 1) and the exit (2) the energy equation becomes 2 g ⋅ z1 − V2 2 2 = K⋅ V2 and solving for V 2 2 From Table 8.2 Khole = 0.5 for a hole (assumed to be square-edged) Also, for a rounded edge r D Hence for the hole = V2 = 5⋅ mm 25⋅ mm = 0.2 > 0.15 2 × 9.81⋅ m s Q = V2 ⋅ A2 Hence for the pipe V2 = 2 × 9.81⋅ Q = 8.09⋅ m s Q = V2 ⋅ A2 V2 = 1 s × π 4 × ( 0.025⋅ m) 2 × 9.81⋅ m 2 × 5⋅ m × 2 1 s × π 4 × ( 0.025⋅ m) L ( 1 + 0.04) L s 3 −3m Q = 3.97 × 10 s Q = 3.97⋅ L Q = 3.64⋅ L s 2 3 −3m Q = 3.64 × 10 s s The pipe leads to a LOWER flow rate s 1 4.77 − 3.97 = 0.8⋅ Kround = 0.04 m V2 = 7.42 s ( 1 + 0.78) m (1 + K) m V2 = 8.09 s ( 1 + 0.5) m 2⋅ g ⋅ h Kpipe = 0.78 for a short pipe (rentrant) so from Table 8.2 3.64 − 3.97 = −0.33⋅ s Hence the change in flow rate is 2 × 5⋅ m × Q = 7.42⋅ Hence the change in flow rate is Hence for the rounded 2 × 5⋅ m × V2 = m V2 = 9.71 s Q = V2⋅ A2 Q = 4.77⋅ L s The rounded edge leads to a HIGHER flow rate Problem 8.108 Given: Data on inlet and exit diameters of diffuser Find: Minimum lengths to satisfy requirements [Difficulty: 2] Solution: Given data D1 = 100 ⋅ mm D2 = 150 ⋅ mm The governing equations for the diffuser are h lm = K⋅ V1 ( ) V1 = Cpi − Cp ⋅ 2 2 1 Cpi = 1 − and 2 2 (8.44) (8.42) 2 AR Combining these we obtain an expression for the loss coefficient K 1 K= 1− − Cp 2 (1) AR The area ratio AR is ⎛ D2 ⎞ AR = ⎜ ⎝ D1 ⎠ 2 AR = 2.25 The pressure recovery coefficient Cp is obtained from Eq. 1 above once we select K; then, with Cp and AR specified, the minimum value of N/R1 (where N is the length and R1 is the inlet radius) can be read from Fig. 8.15 (a) K = 0.2 1 Cp = 1 − 2 −K Cp = 0.602 AR From Fig. 8.15 N R1 = 5.5 R1 = N = 5.5⋅ R1 (b) K = 0.35 Cp = 1 − D1 2 R1 = 50⋅ mm N = 275 ⋅ mm 1 2 −K Cp = 0.452 AR From Fig. 8.15 N R1 =3 N = 3 ⋅ R1 N = 150 ⋅ mm Problem 8.109 Given: Data on geometry of conical diffuser; flow rate Find: Static pressure rise; loss coefficient Solution: Basic equations Cp = p2 − p1 1 2 ⋅ ρ⋅ V1 V1 2 [Difficulty: 3] 2 ( (8.41) h lm = K⋅ 2 Given data D1 = 2 ⋅ in From Eq. 8.41 1 2 ∆p = p 2 − p 1 = ⋅ ρ⋅ V1 ⋅ Cp (1) 2 ) V1 2 = Cpi − Cp ⋅ 2 D2 = 3.5⋅ in 1 Cpi = 1 − (8.44) (8.42) 2 AR N = 6 ⋅ in Q = 750 ⋅ gpm (N = length) K= 1− Combining Eqs. 8.44 and 8.42 we obtain an expression for the loss coefficient K 1 2 − Cp (2) AR The pressure recovery coefficient Cp for use in Eqs. 1 and 2 above is obtained from Fig. 8.15 once compute AR and the dimensionless length N/R1 (where R1 is the inlet radius) The aspect ratio AR is ⎛ D2 ⎞ AR = ⎜ ⎝ D1 ⎠ R1 = 2 D1 R1 = 1 ⋅ in 2 2 ⎛ 3.5 ⎞ ⎜ 2 ⎝ ⎠ AR = AR = 3.06 N Hence R1 =6 From Fig. 8.15, with AR = 3.06 and the dimensionless length N/R1 = 6, we find Cp = 0.6 V1 = π To complete the calculations we need V1 4 3 4 gal 1 ⋅ ft 1 ⋅ min V1 = × 750 ⋅ × × × π min 7.48⋅ gal 60⋅ s Q ⋅ D1 2 ∆p = We can now compute the pressure rise and loss coefficient from Eqs. 1 and 2 ∆p = 1 2 K= 1− × 1.94⋅ slug ft 1 2 AR 3 − Cp × ⎛⎜ 76.6⋅ ⎝ ft ⎞ s⎠ K = 1− 2 2 × 0.6 × 1 3.06 2 lbf ⋅ s slug⋅ ft − 0.6 × 1 2 ⎛ 1 ⎞ V = 76.6⋅ ft 1 ⎜ 2 s ⎜ ⋅ ft ⎝ 12 ⎠ 2 ⋅ ρ⋅ V1 ⋅ Cp 2 ⎛ 1 ⋅ ft ⎞ ⎜ 12⋅ in ⎝ ⎠ 2 ∆p = 23.7⋅ psi K = 0.293 Problem 8.110 [Difficulty: 4] Problem 8.111 [Difficulty: 4] Given: Sudden expansion Find: Expression for minor head loss; compare with Fig. 8.15; plot Solution: The governing CV equations (mass, momentum, and energy) are Assume: 1) Steady flow 2) Incompressible flow 3) Uniform flow at each section 4) Horizontal: no body force 5) No shaft work 6) Neglect viscous friction 7) Neglect gravity The mass equation becomes V1 ⋅ A1 = V2 ⋅ A2 The momentum equation becomes p 1 ⋅ A2 − p 2 ⋅ A2 = V1 ⋅ −ρ⋅ V1 ⋅ A1 + V2 ⋅ ρ⋅ V2 ⋅ A2 or (using Eq. 1) A1 p 1 − p 2 = ρ⋅ V1 ⋅ ⋅ V2 − V1 A2 The energy equation becomes p1 p2 ⎛ ⎛ 2⎞ 2⎞ Qrate = ⎜ u 1 + + V1 ⋅ −ρ⋅ V1 ⋅ A1 + ⎜ u 2 + + V2 ⋅ ρ⋅ V2 ⋅ A2 ρ ρ ⎝ ⎠ ⎝ ⎠ or (using Eq. 1) ( ) ( h lm = u 2 − u 1 − = mrate h lm = V1 − V2 2 ⎡ ⎢ h lm = ⋅ 1− 2 ⎢ ⎣ V1 2 2 ) V1 − V2 2 ( 2⎤ ( 2 A1 + V1 ⋅ ⋅ V2 − V1 A2 ⎛ V2 ⎞ ⎜ ⎝ V1 ⎠ ) (2) 2 Qrate ( ) ( 2 Combining Eqs. 2 and 3 (1) + p1 − p2 ρ ) A V ⎤ ⎥ 2 1 ⎡⎛ 2 ⎞ + V1 ⋅ ⋅ ⎢⎜ − 1⎥ ⎥ A2 ⎦ ⎣⎝ V1 ⎠ ⎦ (3) ) AR = From Eq. 1 h lm = Hence h lm = A1 A2 V1 2 V1 2 V1 ( 2 ( 2 ⋅ 1 − AR 2 h lm = K⋅ V2 = 2 ) 2 ⋅ 1 − AR + 2 ⋅ AR − 2 ⋅ AR V1 2 2 K = ( 1 − AR) Finally ) + V12⋅AR⋅(AR − 1) 2 = ( 1 − AR) ⋅ V1 2 2 2 This result, and the curve of Fig. 8.15, are shown below as computed in Excel. The agreement is excellent. AR 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 K CV K Fig. 8.15 1.00 0.81 0.64 0.49 0.36 0.25 0.16 0.09 0.04 0.01 0.00 1.00 0.60 0.38 0.25 0.10 0.01 0.00 (Data from Fig. 8.15 is "eyeballed") Loss Coefficient for a Sudden Expansion 1.0 Theoretical Curve 0.8 Fig. 8.15 K 0.5 0.3 0.0 0.00 0.25 0.50 Area Ratio AR 0.75 1.00 Problem 8.112 [Difficulty: 3] Problem 8.113 Given: Sudden expansion Find: Expression for upstream average velocity [Difficulty: 2] Solution: 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 ⎜ ρ + α1⋅ 2 + g⋅ z1 − ⎜ ρ + α2⋅ 2 + g⋅ z2 = h lT ⎝ ⎠ ⎝ ⎠ The basic equation is (8.29) 2 V h lT = h l + K ⋅ 2 Assume: 1) Steady flow 2) Incompressible flow 3) h l = 0 4) α1 = α2 = 1 5) Neglect gravity The mass equation is V1⋅ A1 = V2⋅ A2 so V2 = AR⋅ V1 Equation 8.29 becomes p1 ρ or (using Eq. 1) ∆p ρ Solving for V1 If the flow were frictionless, K = 0, so + = V1 = V1 2 = 2 (1) p1 + ρ p2 − p1 V1 2 + K⋅ 2 V1 = ρ 2 2 2 ⋅ ∆p ( compared to ( ( 2 ⋅ ∆p 2 < V1 ) 2 ρ⋅ 1 − AR ∆pinvscid = ∆p = 2 ) 2 V1 2 2 ( V1 2 2 ( ) 2 ⋅ 1 − AR 2 ) ⋅ 1 − AR − K Hence a given flow rate would generate a larger Δp for inviscid flow 2 ) ⋅ 1 − AR − K Hence the flow rate indicated by a given Δp would be lower If the flow were frictionless, K = 0, so V1 ρ⋅ 1 − AR − K Vinviscid = A1 V2 = V1⋅ A2 Problem 8.114 [Difficulty: 4] e d Flow Nozzle Short pipe Given: Flow out of water tank through a nozzle Find: Change in flow rate when short pipe section is added; Minimum pressure; Effect of frictionless flow Solution: Basic equations 2 2 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 V2 1 L V2 2 ⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h lT h lT = hl + h lm = f ⋅ D ⋅ 2 + K⋅ 2 Q = V⋅ A ⎝ ⎠ ⎝ ⎠ Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Vl << 5) L << so hl = 0 Available data D2 = 25⋅ mm r = 0.02⋅ D2 D3 = 50⋅ mm r = 0.5⋅ mm z1 = 2.5⋅ m ρ = 999 ⋅ kg 3 m Knozzle = 0.28 For a rounded edge, we choose the first value from Table 8.2 Hence for the nozzle case, between the free surface (Point 1) and the exit (2) the energy equation becomes g ⋅ z1 − V2 2 V2 2 = Knozzle⋅ 2 2 2 ⋅ g ⋅ z1 Solving for V 2 V2 = (1 + Knozzle) Hence V2 = 2 × 9.81⋅ m 2 × 2.5⋅ m × s Q = V2 ⋅ A2 Q = 6.19⋅ m s 1 m V2 = 6.19 s ( 1 + 0.28) × π 4 × ( 0.025 ⋅ m) 2 3 −3m Q = 3.04 × 10 s Q = 3.04 L When a small piece of pipe is added the energy equation between the free surface (Point 1) and the exit (3) becomes g ⋅ z1 − From continuity V3 2 2 V2 = Knozzle⋅ 2 A2 V3 = V2 ⋅ = V2 ⋅ AR A3 2 V2 + Ke⋅ 2 2 s V2 = Solving for V 2 2 ⋅ g ⋅ z1 ⎛ AR2 + K ⎞ nozzle + Ke⎠ ⎝ 2 ⎛ D2 ⎞ ⎛ 25 ⎞ 2 AR = =⎜ =⎜ = 0.25 A3 ⎝ D3 ⎠ ⎝ 50 ⎠ A2 We need the AR for the sudden expansion AR = 0.25 Ke = 0.6 From Fig. 8.15 for AR = 0.25 V2 = V2 = Hence 2 ⋅ g ⋅ z1 ⎛ AR2 + K ⎞ nozzle + Ke⎠ ⎝ 2 × 9.81⋅ m 2 × 2.5⋅ m × s Q = V2 ⋅ A2 Q = 7.21⋅ m s × π 4 × ( 0.025 ⋅ m) 1 m V2 = 7.21 s (0.252 + 0.28 + 0.6) 3 −3m 2 Q = 3.54 × 10 ∆Q Comparing results we see the flow increases from 3.04 L/s to 3.54 L/s Q = s 3.54 − 3.04 3.04 Q = 3.54 = 16.4⋅ % The flow increases because the effect of the pipe is to allow an exit pressure at the nozzle LESS than atmospheric! The minimum pressure point will now be at Point 2 (it was atmospheric before adding the small pipe). The energy equation between 1 and 2 is 2 2 ⎛⎜ p V2 ⎞ V2 2 g ⋅ z1 − ⎜ + = Knozzle⋅ 2 ⎠ 2 ⎝ρ Solving for p 2 Hence 2 ⎡⎢ V2 p 2 = ρ⋅ ⎢g ⋅ z1 − ⋅ ( Knozzle + 2 ⎣ p 2 = 999 ⋅ kg 3 m ⎡ m ⎢ ⎣ s × ⎢9.81⋅ 2 ⎥⎤ )⎦ 1⎥ × 2.5⋅ m − 1 2 × ⎛⎜ 7.21⋅ ⎝ m⎞ s ⎠ 2 ⎤ lbf ⋅ s ⎥ ⎦ slug⋅ ft × ( 0.28 + 1 )⎥ × 2 p 2 = −8.736 ⋅ kPa If the flow were frictionless the the two loss coeffcients would be zero. Instead of Instead of V2 = 2 ⋅ g ⋅ z1 ⎛ AR2 + K ⎞ nozzle + Ke⎠ ⎝ we'd have If V2 is larger, then p2, through Bernoulli, would be lower (more negative) V2 = 2 ⋅ g ⋅ z1 2 AR which is larger L s Problem 8.115 [Difficulty: 2] Problem 8.116 [Difficulty: 4] Problem 8.117 Given: Data on water flow from a tank/tubing system Find: Minimum tank level for turbulent flow [Difficulty: 3] Solution: Basic equations: 2 2 ⎞ ⎛⎜ p ⎞ ⎛⎜ p V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 1 2 2 2 = h lT = ⎝ ⎠ ⎝ρ ⎠ Re = f = ρ⋅ V⋅ D μ 64 2 hl = f ⋅ (8.36) L V ⋅ D 2 hl + major ∑ h lm (8.29) minor 2 h lm = K⋅ (8.34) (Laminar) 2 g ⋅ d − α⋅ This can be solved expicitly for height d, or solved using Solver V (8.40a) 2 ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ Re The energy equation (Eq. 8.29) becomes ∑ V 2 2 = f⋅ 2 L V V ⋅ + K⋅ D 2 2 h lm = f ⋅ (8.37) Le V2 (8.40b) ⋅ D 2 (Turbulent) Problem 8.118 [Difficulty: 2] Problem 8.119 Given: Data on water flow from a tank/tubing system Find: Minimum tank level for turbulent flow [Difficulty: 3] Solution: Basic equations: 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 ⎜ ρ + α1⋅ 2 + g⋅ z1 − ⎜ ρ + α2⋅ 2 + g⋅ z2 = h lT = ⎝ ⎠ ⎝ ⎠ Re = 2 ρ⋅ V⋅ D hl = f ⋅ μ ∑ hl + major ∑ h lm (8.29) minor 2 L V ⋅ D 2 h lm = K ⋅ (8.34) ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ (8.37) V h lm = f ⋅ (8.40a) 2 Le V2 (8.40b) ⋅ D 2 (Turbulent) Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Velocity at free surface is << The available data is D = 7.5⋅ mm L = 500 ⋅ mm From Table A.8 at 10oC ρ = 1000 − 3 N⋅ s kg μ = 1.3⋅ 10 3 m Re = 10000 From Re = Kent = 0.5 ρ⋅ V⋅ D Re = μ π 4 Hence V = π⋅ μ⋅ D⋅ Re 2 4⋅ ρ V = 1.73 ⎛ π⋅ D2 ⎞ ⎜ ⎝ 4 ⎠ f Q = ⋅D Q 1 Assuming a smooth tube or = −2 ⋅ log⎛⎜ ⎞ ⎝ Re⋅ f ⎠ 2.51 so m s 2 2 V ⋅ ⎛⎜ f ⋅ L + Kent + Kexit⎞ Solving for d d = FOR r > 0.15D) Kent = 0.04 (Table 8.2) 2⋅ g ⎝ D ⎠ 3 −5m Q = 7.66 × 10 f = 0.0309 g⋅ d = f ⋅ The energy equation (Eq. 8.29) becomes 2 m Kexit = 1 (Table 8.2) ρ⋅ Q⋅ D ⋅ 2 2 L V V V ⋅ + Kent⋅ + Kexit ⋅ D 2 2 2 d = 545 ⋅ mm d = 475 ⋅ mm s Q = 0.0766⋅ l s Problem 8.120 Given: Data on a tube Find: "Resistance" of tube for flow of kerosine; plot [Difficulty: 4] Solution: The basic equations for turbulent flow are 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 1 2 2 2 = hl ⎝ ⎠ ⎝ρ ⎠ ⎛ e ⎞ 2 ⎜ D L V 1 2.51 (8.34) hl = f ⋅ ⋅ + = −2.0⋅ log ⎜ D 2 f ⎝ 3.7 Re⋅ f ⎠ The given data is L = 250 ⋅ mm From Fig. A.2 and Table A.2 μ = 1.1 × 10 (8.37) D = 7.5⋅ mm − 3 N⋅ s ⋅ ρ = 0.82 × 990 ⋅ 2 m For an electrical resistor (8.29) kg 3 = 812 ⋅ m kg (Kerosene) 3 m V = R⋅ I (1) Simplifying Eqs. 8.29 and 8.34 for a horizontal, constant-area pipe ⎛ ⎜ ⎜ 2 p1 − p2 L V L ⎝ = f⋅ ⋅ = f⋅ ⋅ ρ D 2 D ⎞ Q π 4 2 2 ⋅D ⎠ ∆p = or 2 8 ⋅ ρ⋅ f ⋅ L 2 5 2 ⋅Q (2) π ⋅D By analogy, current I is represented by flow rate Q, and voltage V by pressure drop Δp . Comparing Eqs. (1) and (2), the "resistance" of the tube is R= ∆p Q = 8 ⋅ ρ⋅ f ⋅ L⋅ Q 2 5 π ⋅D The "resistance" of a tube is not constant, but is proportional to the "current" Q! Actually, the dependence is not quite linear, because f decreases slightly (and nonlinearly) with Q. The analogy fails! The analogy is hence invalid for Re > 2300 or ρ⋅ V⋅ D μ > 2300 ρ⋅ Writing this constraint in terms of flow rate Q π 4 ⋅D 2 ⋅D μ > 2300 or Q> 2300⋅ μ⋅ π⋅ D 4⋅ ρ 3 −5m Q = 1.84 × 10 Flow rate above which analogy fails s The plot of "resistance" versus flow rate cab be done in Excel. "Resistance" of a Tube versus Flow Rate 9 "R" 3 (10 Pa/m /s) 1.E+01 1.0E-05 1.0E-04 1.0E-03 1.E-01 1.E-03 3 Q (m /s) 1.0E-02 Problem 8.121 Given: Data on tube geometry Find: Plot of reservoir depth as a function of flow rate [Difficulty: 3] Solution: Basic equations: 2 2 ⎞ ⎛⎜ p ⎞ ⎛⎜ p V1 V2 1 2 ⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h lT = ⎝ ⎠ ⎝ ⎠ Re = f = ρ⋅ V⋅ D μ 64 2 hl = f ⋅ (8.36) L V ⋅ D 2 (Laminar) g ⋅ d − α⋅ This can be solved expicitly for height d, or solved using Solver 2 In Excel: V 2⋅ g ⋅ ⎜⎛ α + f ⋅ ⎝ L D + K⎞ ⎠ major ∑ h lm (8.29) minor V (8.40a) 2 ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ 2 d= hl + 2 h lm = K⋅ (8.34) Re The energy equation (Eq. 8.29) becomes ∑ V 2 2 = f⋅ 2 L V V ⋅ + K⋅ D 2 2 h lm = f ⋅ (8.37) Le V2 (8.40b) ⋅ D 2 (Turbulent) Required Reservoir Head versus Flow Rate 75 50 d (m) 25 0 0 2 4 6 Q (L/min) 8 10 12 Problem 8.122 Given: Flow of oil in a pipe Find: Percentage change in loss if diameter is reduced Solution: Basic equations Available data 2 hl = f ⋅ ν = 7.5⋅ 10 V= Here Re = Then L V ⋅ D 2 Q A f = − 4 ft ⋅ s 2 V = π⋅ D V⋅ D 4 π D = 1 ⋅ in × 0.1⋅ Re = 18.3⋅ ν hl = f ⋅ Laminar Re L = 100 ⋅ ft 2 The flow is LAMINAR ⎛ e ⎞ ⎜ 1 2.51 D + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ Turbulent 2 4⋅ Q = 64 [Difficulty: 3] ft s ft 3 × s 1 × 12 ⎛ 12 ⋅ 1 ⎞ ⎜ 1 ft ⎝ ⎠ hl = 2 V = 18.3⋅ 7.5 × 10 −4 ⋅ ft ft 3 s s Re = 2033 2 2 ⎛ 18.3 ft ⎞ ⎜ s⎠ 64 100 ⎝ hl = × × 2033 1 2 64 L V ⋅ ⋅ Re D 2 Q = 0.100 ft s ⋅ ft × 2 L V ⋅ D 2 Q = 45⋅ gpm h l = 6326⋅ ft 2 2 s 12 D = 0.75⋅ in When the diameter is reduced to V= Re = Q A 4⋅ Q = 2 π⋅ D V⋅ D V = 4 π × 0.1⋅ Re = 32.6⋅ ν The flow is TURBULENT For drawn tubing, from Table 8.1 Given ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ s 3 s × × ⎛ 12 ⋅ 1 ⎞ ⎜ 0.75 ft ⎝ ⎠ 0.75 12 2 V = 32.6⋅ ft s s ⋅ ft × 7.5 × 10 −4 ⋅ ft 2 L V ⋅ D 2 Re = 2717 e = 0.000005⋅ ft f = 0.0449 2 ⎛ 32.6 ft ⎞ ⎜ s⎠ 100 ⎝ h l = .0449 × × 2 0.75 2 hl = f ⋅ ft ft 12 4 The increase in loss is 3.82 × 10 − 6326 6326 = 504 ⋅ % 4 ft h l = 3.82 × 10 ⋅ This is a HUGH increase! The main increase is because the diameter reduction causes the velocity to increase; the loss goes as V2, and 1/D, so it increases very rapidly 2 2 s Problem 8.123 Given: Data on water system Find: Minimum tank height; equivalent pressure [Difficulty: 4] Solution: 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 1 2 2 2 = h lT = ⎝ ⎠ ⎝ρ ⎠ Basic equations: 2 ρ⋅ V⋅ D Re = hl = f ⋅ μ ∑ hl + major ∑ (8.29) h lm minor 2 L V ⋅ D 2 h lm = K⋅ (8.34) ⎛ e ⎞ ⎜ D 1 2.51 (8.37) + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ V (8.40a) 2 h lm = f ⋅ Le V2 (8.40b) ⋅ D 2 (Turbulent) Available data D = 7.5⋅ mm L = 1⋅ m Re = 100000 From Section 8.7 Kent = 0.5 Lelbow45 = 16⋅ D Lelbow90 = 30⋅ D LGV = 8 ⋅ D Lelbow45 = 0.12 m Lelbow90 = 0.225 m LGV = 0.06 m From Table A.8 at 10oC ρ = 1000 Q = π⋅ μ⋅ D⋅ Re 4⋅ ρ μ = 1.3⋅ 10 3 Q = 7.66 × 10 d = ⎛ 2⋅ g ⎝ V V 2⋅ g ⋅⎜1 + f ⋅ IF INSTEAD the reservoir was pressurized Q = 0.766 s 2 d− 2 Hence ⋅ 2 m 3 −4m The energy equation becomes f = 0.0180 − 3 N⋅ s kg m Then and so L D V = s Q V = 17.3 ⎛ π⋅ D2 ⎞ ⎜ ⎝ 4 ⎠ m s Lelbow90 Lelbow45 LGV ⎞ ⎛ L + 2⋅ f ⋅ + 2⋅ f ⋅ + f⋅ 2⋅ g ⎝ D D D D ⎠ 2 = l V + 2⋅ f ⋅ ⋅⎜f ⋅ Lelbow90 D + 2⋅ f ⋅ ∆p = ρ⋅ g ⋅ d Lelbow45 D + f⋅ LGV ⎞ D ∆p = 781 ⋅ kPa ⎠ d = 79.6⋅ m Unrealistic! which is feasible at this Re Problem 8.124 Given: Flow from pump to reservoir Find: Pressure at pump discharge Solution: [Difficulty: 2] 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 = h lT 2 2 ⎝ ⎠ ⎝ρ ⎠ Basic equations 2 V1 L V1 h lT = h l + h lm = f ⋅ ⋅ + Kexit ⋅ D 2 2 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) V2 << Hence the energy equation between Point 1 and the free surface (Point 2) becomes 2 2 ⎛ p1 V2 ⎞ L V V ⎜ + − ( g ⋅ z2 ) = f ⋅ ⋅ + Kexit ⋅ D 2 2 2 ⎠ ⎝ρ ⎛ V ⎝ 2 Solving for p 1 p 1 = ρ⋅ ⎜ g ⋅ z2 − From Table A.7 (68oF) ρ = 1.94⋅ slug ft Re = 2 3 V⋅ D ν For commercial steel pipe e = 0.00015 ⋅ ft 2⎞ 2 + f⋅ L V V ⋅ + Kexit ⋅ D 2 2 ν = 1.08 × 10 Re = 10⋅ ft Kexit = 1.0 so we find ft 4 ⋅ mile ft 3 ⎡ ⎢ ⎣ 2 s × 50⋅ ft + .0150 × 12 s s ⋅ ft × −5 1.08 × 10 ⋅ ft Re = 6.94 × 10 2 e so ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ For the exit × ⎢32.2⋅ 9 2 D Given slug ⋅ (Table 8.1) Flow is turbulent: p 1 = 1.94⋅ s × − 5 ft ⎠ 0.75⋅ ft p 1 = ρ⋅ ⎜ g ⋅ z2 + f ⋅ ⎝ 5280⋅ ft 1mile × 1 2 × ⎛⎜ 10⋅ ⎝ = 0.000200 2⎤ 2⎞ L V ⋅ D 2 ⎠ 2 ⎥ × lbf ⋅ s s ⎠ ⎥ slug⋅ ft ⎦ ft ⎞ Turbulent f = 0.0150 ⎛ × 5 4 lbf p 1 = 4.41 × 10 ⋅ ft 2 p 1 = 306 ⋅ psi Problem 8.125 [Difficulty: 3] Given: Data on reservoir/pipe system Find: Plot elevation as a function of flow rate; fraction due to minor losses Solution: L = D = e/D = K ent = K exit = 250 50 0.003 0.5 1.0 Required Head versus Flow Rate m mm 200 150 z (m) = 1.01E-06 m2/s 3 Q (m /s) V (m/s) 0.0000 0.0005 0.0010 0.0015 0.0020 0.0025 0.0030 0.0035 0.0040 0.0045 0.0050 0.0055 0.0060 0.0065 0.0070 0.0075 0.0080 0.0085 0.0090 0.0095 0.0100 0.000 0.255 0.509 0.764 1.02 1.27 1.53 1.78 2.04 2.29 2.55 2.80 3.06 3.31 3.57 3.82 4.07 4.33 4.58 4.84 5.09 Re 0.00E+00 1.26E+04 2.52E+04 3.78E+04 5.04E+04 6.30E+04 7.56E+04 8.82E+04 1.01E+05 1.13E+05 1.26E+05 1.39E+05 1.51E+05 1.64E+05 1.76E+05 1.89E+05 2.02E+05 2.14E+05 2.27E+05 2.40E+05 2.52E+05 100 f z (m) h lm /h lT 0.000 0.0337 0.562 0.0306 2.04 0.0293 4.40 0.0286 7.64 0.0282 11.8 0.0279 16.7 0.0276 22.6 0.0275 29.4 0.0273 37.0 0.0272 45.5 0.0271 54.8 0.0270 65.1 0.0270 76.2 0.0269 88.2 0.0269 101 0.0268 115 0.0268 129 0.0268 145 0.0267 161 0.0267 179 50 0.882% 0.972% 1.01% 1.04% 1.05% 1.07% 1.07% 1.08% 1.09% 1.09% 1.09% 1.10% 1.10% 1.10% 1.10% 1.11% 1.11% 1.11% 1.11% 1.11% 0 0.0000 0.0025 0.0050 3 Q (m /s) 0.0075 0.0100 Minor Loss Percentage versus Flow Rate 1.2% 1.1% h lm /h lT 1.0% 0.9% 0.8% 0.0000 0.0025 0.0050 3 Q (m /s) 0.0075 0.0100 Problem 8.126 Given: Flow through three different layouts Find: Which has minimum loss Solution: Basic equations [Difficulty: 3] 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ 2 V1 V2 1 L V 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z h = h + h f ⋅ ⋅ − = h = + ⎜ρ ⎜ 1 2 l lm lT lT 2 2 D 2 ⎝ ⎠ ⎝ρ ⎠ ∑ Minor ⎛ Le V2 ⎞ ⎜f ⋅ ⋅ ⎝ D 2⎠ Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) Ignore additional length of elbows For a flow rate of For water at 20oC Q = 350 ⋅ L V= min ν = 1.01 × 10 For Case (a) = A ⋅ 4⋅ Q 2 Re = s V⋅ D ν e e = 0.15⋅ mm 2 D 2 p1 s 0.001 ⋅ m 1⋅ L × 0.05⋅ m × 2 1 ⋅ min × × 60⋅ s ⎛ 1 ⎞ V = 2.97 m ⎜ 0.05⋅ m s ⎝ ⎠ s −6 1.01 × 10 2 Re = 1.47 × 10 ⋅m −4 = 6.56 × 10 L = 5.81 m 2 p2 m min 3 × f = 0.0201 5.25 + 2.5 ⋅ m Hence the energy equation is π L × 350 ⋅ Re = 2.97⋅ ⎛ e ⎞ ⎜ 1 2.51 D + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ L = 4 V = π⋅ D 2 −6 m Flow is turbulent. From Table 8.1 Given Q Two 45o miter bends (Fig. 8.16), for each Le D = 13 2 Le V L V − = f⋅ ⋅ + 2⋅ f ⋅ ⋅ ρ ρ D 2 D 2 Le ⎞ ⎛L ∆p = p 1 − p 2 = ρ⋅ f ⋅ ⋅ ⎜ + 2⋅ 2 ⎝D D⎠ 2 Solving for Δp V ∆p = 1000⋅ kg 3 × .0201 × ⎛⎜ 2.97⋅ ⎝ m For Case (b) L = ( 5.25 + 2.5) ⋅ m Hence the energy equation is p1 ρ − p2 ρ = f⋅ m⎞ s ⎠ 2 × 2 ⎛ 5.81 + 2⋅ 13⎞ × N⋅ s ⎜ ⎝ 0.05 ⎠ kg⋅ m L = 7.75 m 2 Le V2 L V ⋅ + f⋅ ⋅ D 2 D 2 One standard 90o elbow (Table 8.4) ∆p = 25.2⋅ kPa Le D = 30 5 Solving for Δp 2 Le ⎞ V ⎛L ∆p = p 1 − p 2 = ρ⋅ f ⋅ ⋅⎜ + 2 ⎝D D⎠ ∆p = 1000⋅ kg 3 × .0201 × ⎛⎜ 2.97⋅ ⎝ m For Case (c) Hence the energy equation is L = ( 5.25 + 2.5) ⋅ m p1 ρ Solving for Δp − p2 ρ m⎞ s 2 × ⎠ L = 7.75 m 2 2 ⎛ 7.75 + 30⎞ × N⋅ s ⎜ ⎝ 0.05 ⎠ kg⋅ m Three standard 90o elbows, for each ∆p = 32.8⋅ kPa Le D = 30 2 Le V L V ⋅ + 3⋅ f ⋅ ⋅ D 2 D 2 = f⋅ 2 Le ⎞ V ⎛L ∆p = p 1 − p 2 = ρ⋅ f ⋅ ⋅ ⎜ + 3⋅ 2 ⎝D D⎠ ∆p = 1000⋅ kg 3 m × .0201 × ⎛⎜ 2.97⋅ ⎝ Hence we conclude Case (a) is the best and Case (c) is the worst m⎞ s ⎠ 2 × 2 ⎛ 7.75 + 3 × 30⎞ × N⋅ s ⎜ ⎝ 0.05 ⎠ kg⋅ m ∆p = 43.4⋅ kPa Problem 8.127 Given: Flow through rectangular duct Find: Pressure drop Solution: Basic equations [Difficulty: 2] 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ 2 V1 V2 1 L V 2 ⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h lT h lT = hl + h lm = f ⋅ D ⋅ 2 + ⎝ ⎠ ⎝ ⎠ ∑ Minor ⎛ Le V2 ⎞ ⎜f ⋅ ⋅ ⎝ D 2⎠ 4 ⋅ a⋅ b Dh = 2⋅ (a + b) Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 Available data Q = 1750⋅ cfm At 50oF, from Table A.9 ρ = 0.00242 ⋅ L = 1000⋅ ft slug ft Hence 1 f Hence or, in in water V = 15.6⋅ ρ⋅ V⋅ Dh 2.51 ⎞ ⎝ Re⋅ f ⎠ L Dh ∆p ρw⋅ g ⋅ ft 5 so V 2 ∆p = 0.031 ⋅ psi h = 0.848 ⋅ in a = 0.75⋅ ft ρw = 1.94⋅ slug ft and 2 ⋅ ρ⋅ 2 s Re = 1.18 × 10 μ = −2 ⋅ log⎛⎜ − 7 lbf ⋅ s ft a⋅ b ∆p = f ⋅ h = 3 Q V = Re = For a smooth duct μ = 3.69⋅ 10 b = 2.5⋅ ft f = 0.017 3 4 ⋅ a⋅ b Dh = 2⋅ ( a + b) Dh = 1.15⋅ ft Problem 8.128 Given: Data on circuit Find: Plot pressure difference for a range of flow rates [Difficulty: 3] Solution: 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 ⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h lT = ⎝ ⎠ ⎝ ⎠ Basic equations: Re = f = ρ⋅ V⋅ D 64 μ 2 hl = f ⋅ (8.36) L V ⋅ D 2 ∑ hl + major ∑ h lm (8.29) minor 2 h lm = K⋅ (8.34) (Laminar) Re V h lm = f ⋅ (8.40a) 2 ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ Le V2 (8.40b) ⋅ D 2 (8.37) (Turbulent) The energy equation (Eq. 8.29) becomes for the circuit ( 1 = pump inlet, 2 = pump outlet) p1 − p2 ρ In Excel: 2 = f⋅ 2 2 L V V V ⋅ + 4 ⋅ f ⋅ Lelbow⋅ + f ⋅ Lvalve⋅ D 2 2 2 2 or ∆p = ρ⋅ f ⋅ V 2 ⎛L ⋅⎜ ⎝D + 4⋅ Lelbow D + Lvalve ⎞ D ⎠ Required Pressure Head for a Circuit 1200 Dp (kPa) 1000 800 600 400 200 0 0.00 0.01 0.02 0.03 Q (m3/s) 0.04 0.05 0.06 0.07 Problem 8.129 [Difficulty: 3] c h LA d e LB Given: Pipe friction experiment Find: Required average speed; Estimate feasibility of constant head tank; Pressure drop over 5 m Solution: Basic equations 2 2 p p V1 V2 1 2 ρ α 2 g z1 ρ α 2 g z2 h lT 2 LA VA LB VB h lT h A h B fA fB DA 2 DB 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) Ignore minor losses We wish to have ReB 10 Hence, from ReB 5 VB DB VB ν 2 6 m 5 VB 10 1.01 10 We will also need DB VA VB DA ReA VA DA ν s ReB ν and for water at 20oC DB 1 m VA 4.04 s 2.5 5 2 m ReA 1.01 0.05 m s m VA 1.01 s s 1.01 10 4 6 2 m ReA 5 10 Both tubes have turbulent flow For PVC pipe (from Googling!) e 0.0015 mm For tube A For tube B Given Given 2 6 m m VB 4.04 s 0.025 m 2 ν 1.01 10 e D 1 2.51 A 2.0 log 3.7 fA ReA fA e D 1 2.51 B 2.0 log 3.7 fB ReB fB fA 0.0210 fB 0.0183 s 2 Applying the energy equation between Points 1 and 3 VB g LA h LA 2 2 g 1 2 LA 2 2 LA VA LB VB fA fB DA 2 DB 2 LB 1 fB g h 2 DB VB Solving for LA 2 2 fA VA DA 2 4.04 m s 2 9.81 1 0.0183 9.81 m 0.5 m 2 0.025 s 20 m 2 0.0210 2 s 1 0.05 m 1.01 m s 2 LA 12.8 m Most ceilings are about 3.5 m or 4 m, so this height is IMPRACTICAL Applying the energy equation between Points 2 and 3 2 2 2 p VB p 3 VB 2 L VB ρ 2 ρ 2 fB D 2 B ∆p 1000 kg 3 m 0.0183 2 5 m 0.025 m 4.04 L m s 2 2 VB ∆p ρ fB DB 2 or N s kg m ∆p 29.9 kPa 2 Problem 8.130 [Difficulty: 3] Part 1/2 Problem 8.130 [Difficulty: 3] Part 2/2 Problem 8.131 Given: Same flow rate in various ducts Find: Pressure drops of each compared to round duct Solution: Basic equations [Difficulty; 3] 2 2 ⎞ ⎛⎜ p ⎞ ⎛⎜ p V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 = hl 2 2 ⎝ ⎠ ⎝ρ ⎠ 4⋅ A Dh = Pw e = 0 (Smooth) Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) Ignore minor losses The energy equation simplifies to 2 L V ∆p = p 1 − p 2 = ρ⋅ f ⋅ ⋅ Dh 2 But we have From Table A.9 Hence For a round duct For a rectangular duct But Q V= V = 1250⋅ A ν = 1.62 × 10 Re = V⋅ Dh − 4 ft ⋅ 3 min × 1 ⋅ min 60⋅ s b ft s × ar 2 h = b⋅ h ar 2 slug = ρ⋅ f 2 ⋅ V Dh 2 V = 20.8⋅ ft s at 68oF 3 s 5 × Dh = 1.284 × 10 ⋅ Dh −4 2 1.62 × 10 ⋅ ft 4 Dh = π so 1 ⋅ ft ft π 4⋅ A 4⋅ b⋅ h 2 ⋅ h ⋅ ar Dh = = = Pw 2⋅ ( b + h) 1 + ar h= 1 × ρ = 0.00234 ⋅ s 4⋅ A Dh = D = L 2 Re = 20.8⋅ ν ft ∆p or = A × 1 ⋅ ft 2 where or ar (Dh in ft) Dh = 1.13⋅ ft ar = h= b h A and ar 2 ⋅ ar Dh = ⋅ A 1 + ar The results are: Round Given ar = 1 Dh = 1.13⋅ ft 5 1 Re = 1.284 × 10 ⋅ ⋅ Dh ft ⎛ e ⎞ ⎜ D h 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ 2 ⋅ ar Dh = ⋅ A 1 + ar f = 0.0167 Dh = 1 ⋅ ft Re = 1.45 × 10 ∆p L = ρ⋅ f 5 2 ⋅ V Dh 2 ∆p L = 7.51 × 10 5 1 5 Re = 1.284 × 10 ⋅ ⋅ Dh Re = 1.28 × 10 ft − 3 lbf ⋅ ft 3 Given ⎛ e ⎞ ⎜ D h 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ f = 0.0171 ∆p L = ρ⋅ f 2 ⋅ ∆p V Dh 2 L 8.68 × 10 Hence the square duct experiences a percentage increase in pressure drop of = 8.68 × 10 −3 − 3 lbf ⋅ ft 3 −3 − 7.51 × 10 = 15.6⋅ % −3 7.51 × 10 ar = 2 Given 2 ⋅ ar Dh = ⋅ A 1 + ar Dh = 0.943 ⋅ ft ⎛ e ⎞ ⎜ Dh 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ 5 1 Re = 1.284 × 10 ⋅ ⋅ Dh ft f = 0.0173 ∆p L Re = 1.21 × 10 = ρ⋅ f 2 ⋅ ∆p V Dh 2 9.32 × 10 Hence the 2 x 1 duct experiences a percentage increase in pressure drop of 5 = 9.32 × 10 L −3 − 3 lbf ft Given 2 ⋅ ar Dh = ⋅ A 1 + ar Dh = 0.866 ⋅ ft = 24.1⋅ % −3 ⎛ e ⎞ ⎜ Dh 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ 5 1 Re = 1.284 × 10 ⋅ ⋅ Dh ft f = 0.0176 ∆p L Re = 1.11 × 10 = ρ⋅ f 2 ⋅ 5 ∆p V Dh 2 L = 0.01⋅ lbf ft 3 −3 Hence the 3 x 1 duct experiences a percentage increase in pressure drop of 0.01 − 7.51 × 10 −3 7.51 × 10 Note that f varies only about 7%; the large change in Δp/L is primarily due to the 1/Dh factor 3 −3 − 7.51 × 10 7.51 × 10 ar = 3 ⋅ = 33.2⋅ % Problem 8.132 Given: Flow down corroded iron pipe Find: Pipe roughness; Power savings with new pipe [Difficulty: 4] Solution: 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 = hl 2 2 ⎝ ⎠ ⎝ρ ⎠ Basic equations 2 hl = f ⋅ L V ⋅ D 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) No minor losses 3 Available data D = 50⋅ mm ∆z = 40⋅ m L = ∆z p 1 = 750⋅ kPa p 2 = 250⋅ kPa Q = 0.015⋅ m s ρ = 999⋅ kg 3 m Hence the energy equation becomes 2 ⎛ p1 ⎞ ⎛ p2 ⎞ L V ⎜ + g ⋅ z1 − ⎜ + g ⋅ z2 = f ⋅ ⋅ D 2 ⎝ρ ⎠ ⎝ρ ⎠ Here V= Q = A 4⋅ Q V= 2 π⋅ D 4 π 3 × 0.015⋅ m s × 1 ( 0.05⋅ m) V = 7.64 2 m s In this problem we can compute directly f and Re, and hence obtain e/D Solving for f f = ⎛ p1 − p2 2⋅ D 2 L⋅ V f = 2× ⋅⎜ ⎝ 0.05 40 From Table A.8 (20oF) ν = 1.01 × 10 Flow is turbulent: ρ × ( + g z1 − z2 ⎞ )⎠ 3 2 ⎤ kg⋅ m m ⎛ s ⎞ × ⎡⎢( 750 − 250 ) × 103⋅ N × m × + 9.81⋅ × 40⋅ m⎥ f = 0.0382 ⎜ 7.64⋅ m 1000⋅ kg 2 2 ⎥ 2 ⎝ ⎠ ⎢⎣ s ⋅N s m ⎦ 2 −6 m ⋅ s Re = ⎛ e ⎞ ⎜ 1 D 2.51 = −2.0⋅ log ⎜ + f ⎝ 3.7 Re⋅ f ⎠ V⋅ D ν Re = 7.64⋅ m s × 0.05⋅ m × s −6 1.01 × 10 2 ⋅m Re = 3.78 × 10 5 Solving for e New pipe (Table 8.1) ⎛ − ⎜ e = 3.7⋅ D⋅ ⎜ 10 ⎝ ⎞ 1 2⋅ f − e e = 0.15⋅ mm 2.51 e e = 0.507 mm Re⋅ f ⎠ D = 0.0101 = 0.003 D ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ Given f = 0.0326 In this problem Hence ⎡ L V ∆p = p 1 − p 2 = ρ⋅ ⎢g ⋅ z2 − z1 + f ⋅ ⋅ D 2 ⎣ ( ∆pnew = 1000⋅ kg 3 m ∆pold = p 1 − p 2 Compared to ∆pold = 500 ⋅ kPa we find ⎡ ) × ⎢9.81⋅ ⎢ ⎣ 2⎤ m 2 ⎥ ⎦ × ( −40⋅ m) + s 0.0326 2 × 40 0.05 × ⎛⎜ 7.64⋅ ⎝ ∆pold = 500 kPa ∆pold − ∆pnew ∆pold = 26.3⋅ % As power is ΔpQ and Q is constant, the power reduction is the same as the above percentage! 2⎤ 2 ⎥ × N⋅ s s ⎠ ⎥ kg⋅ m ⎦ m⎞ ∆pnew = 369 ⋅ kPa Problem 8.133 Given: Flow through fire hose and nozzle Find: Supply pressure Solution: Basic equations [Difficulty: 3] 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 = h lT 2 2 ⎝ ⎠ ⎝ρ ⎠ 2 L V h lT = h l + h lm = f ⋅ ⋅ + D 2 ∑ Minor ⎛ V2 ⎞ ⎜ K⋅ ⎝ 2⎠ Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) p 2 = p atm so p 2 = 0 gage Hence the energy equation between Point 1 at the supply and the nozzle exit (Point n); let the velocity in the hose be V p1 ρ 2 Vn − 2 2 ( ) and V= 2 From continuity Vn = ⎛ D ⎞ ⋅V ⎜D ⎝ 2⎠ Solving for p 1 ⎡⎢ L p1 = ⋅ f ⋅ + Ke + 4⋅ Kc + 2 ⎢ D ⎣ From Table A.7 (68oF) ρ = 1.94⋅ 2 ρ⋅ V slug ft Re = For the hose e D Flow is turbulent: 2 2 Vn L V V = f⋅ ⋅ + Ke + 4⋅ Kc ⋅ + Kn ⋅ D 2 2 2 V⋅ D Re = 15.3⋅ ν A = 4⋅ Q V= 2 π⋅ D 4 π × 0.75⋅ 4 ⎛ D ⎞ ⋅ 1 + K ⎥⎤ ( n)⎥ ⎜D ⎝ 2⎠ ⎦ ν = 1.08 × 10 3 Q ft s − 5 ft × ⋅ 3 12 ft 3 s 1 × ⎛ 1 ⋅ ft ⎞ ⎜4 ⎝ ⎠ 2 V = 15.3⋅ ft s 2 s s ⋅ ft × −5 1.08 × 10 ⋅ ft 2 Re = 3.54 × 10 5 Turbulent = 0.004 ⎞ ⎛ e ⎜ 1 2.51 D = −2.0⋅ log ⎜ + 3.7 f Re⋅ f ⎠ ⎝ Given p1 = 1 2 × 1.94⋅ slug ft 3 4 lbf p 1 = 2.58 × 10 ⋅ ft 2 × ⎛⎜ 15.3⋅ ⎝ ft ⎞ s⎠ 2 ⎡⎢ ⎢ ⎢⎣ × 0.0287 × p 1 = 179 ⋅ psi 250 1 4 f = 0.0287 + 0.5 + 4 × 0.5 + 4 2 ⎛ 3 ⎞ × ( 1 + 0.02)⎤⎥ × lbf ⋅ s ⎜ ⎥ slug⋅ ft ⎝1⎠ ⎥⎦ Problem 8.134 Given: Proposal for bench top experiment Find: Design it; Plot tank depth versus Re [Difficulty: 4] Solution: Basic equations: 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 1 2 2 2 = h lT = ⎝ ⎠ ⎝ρ ⎠ Re = f = ρ⋅ V⋅ D μ 64 2 hl = f ⋅ (8.34) (8.36) (Laminar) The energy equation (Eq. 8.29) becomes 2 2 2 L V V g ⋅ H − α⋅ = f⋅ ⋅ + K⋅ 2 D 2 2 This can be solved explicity for reservoir height H 2 H= In Excel: V 2⋅ g ⋅ ⎜⎛ α + f ⋅ ⎝ L D + K⎞ ⎠ major hl + ∑ h lm (8.29) minor 2 L V ⋅ D 2 Re V ∑ h lm = K ⋅ V (8.40a) 2 ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ h lm = f ⋅ (8.37) Le V2 (8.40b) ⋅ D 2 (Turbulent) Computed results: Q (L/min) V (m/s) 0.200 0.225 0.250 0.275 0.300 0.325 0.350 0.375 0.400 0.425 0.450 0.472 0.531 0.589 0.648 0.707 0.766 0.825 0.884 0.943 1.002 1.061 Re Regime f H (m) 1413 1590 1767 1943 2120 2297 2473 2650 2827 3003 3180 Laminar Laminar Laminar Laminar Laminar Laminar Turbulent Turbulent Turbulent Turbulent Turbulent 0.0453 0.0403 0.0362 0.0329 0.0302 0.0279 0.0462 0.0452 0.0443 0.0435 0.0428 0.199 0.228 0.258 0.289 0.320 0.353 0.587 0.660 0.738 0.819 0.904 The flow rates are realistic, and could easily be measured using a tank/timer system The head required is also realistic for a small-scale laboratory experiment Around Re = 2300 the flow may oscillate between laminar and turbulent: Once turbulence is triggered (when H > 0.353 m), the resistanc e to flow increases requiring H >0.587 m to maintain; hence the flow reverts to la minar, only to trip over again to turbulent! This behavior will be visible: the exit flow will switch back and forth between smooth (laminar) and chaotic (turbulent) Required Reservoir Head versus Reynolds Number 1.00 0.75 H (m) 0.50 Laminar 0.25 Turbulent 0.00 1000 1500 2000 Re 2500 3000 3500 Problem 8.135 [Difficulty: 3] Problem 8.136 [Difficulty: 3] Given: Drinking of a beverage Find: Fraction of effort of drinking of friction and gravity Solution: Basic equations 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 ⎜ ρ + α ⋅ 2 + g ⋅ z1 − ⎜ ρ + α ⋅ 2 + g ⋅ z2 = h l ⎝ ⎠ ⎝ ⎠ 2 hl = f ⋅ L V ⋅ D 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) No minor losses Hence the energy equation becomes, between the bottom of the straw (Point 1) and top (Point 2) g ⋅ z1 − 2 ⎛ p2 ⎞ L V ⎜ + g ⋅ z2 = f ⋅ ⋅ D 2 ⎝ρ ⎠ where p 2 is the gage pressure in the mouth The negative gage pressure the mouth must create is therefore due to two parts ( 2 ) p grav = −ρ⋅ g ⋅ z2 − z1 p fric = −ρ⋅ f ⋅ 12 Q = Assuming a person can drink 12 fluid ounces in 5 s Assuming a straw is 6 in long diameter 0.2 in, with roughness V= 128 4⋅ Q Re = Given Then − 5 ft ⋅ −5 4 π Hence the fraction due to friction is ⋅ 3 s in (from Googling!) − 3 ft 3 s 2 × ⎛ 1 × 12⋅ in ⎞ V = 11.5⋅ ft ⎜ 0.2⋅ in 1 ⋅ ft s ⎝ ⎠ (for water, but close enough) ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ slug 3 slug ft − 3 ft Q = 2.51 × 10 2 ν p fric = −1.94⋅ 3 7.48⋅ gal × 2.51 × 10 ft Re = 11.5⋅ p grav = −1.94⋅ 1⋅ ft s V⋅ D ft and × e = 5 × 10 π⋅ D From Table A.7 (68oF) ν = 1.08 × 10 ⋅ gal 5⋅ s V = 2 L V ⋅ D 2 3 ft × 32.2⋅ 2 p fric p fric + p grav 0.2 12 ⋅ ft × s 1.08 × 10 4 −5 ft Re = 1.775 × 10 2 f = 0.0272 × s × 0.0272 × s × 1 2 6 0.2 2 ⋅ ft × × = 77⋅ % 1 2 lbf ⋅ s p grav = −31.2⋅ slug⋅ ft × ⎛⎜ 11.5⋅ ⎝ lbf ft ft ⎞ s⎠ 2 2 × lbf ⋅ s slug⋅ ft and gravity is These results will vary depending on assumptions, but it seems friction is significant! p fric = −105 ⋅ lbf ft p grav p fric + p grav 2 2 p grav = −0.217 ⋅ psi p fric = −0.727 ⋅ psi = 23⋅ % Problem 8.137 Given: Draining of swimming pool with larger hose Find: Flow rate and average velocity Solution: Basic equations [Difficulty: 4] 2 2 ⎞ ⎛⎜ p ⎞ ⎛⎜ p V1 V2 1 2 ⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h l ⎝ ⎠ ⎝ ⎠ 2 2 L V ⋅ D 2 hl = f ⋅ V h lm = Kent⋅ 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) No minor losses Available data D = 25⋅ mm L = 30⋅ m e = 0.2⋅ mm h = 3⋅ m 2 g ⋅ ( ∆z + h ) = f ⋅ Hence the energy equation becomes Solving for V V= 2 ⋅ g ⋅ ( ∆z + h ) f⋅ We also have Re = L D ∆z = 1.5⋅ m 2 2 −6 m Kent = 0.5 ν = 1 ⋅ 10 ⋅ s 2 L V V V ⋅ + Kent⋅ + D 2 2 2 (1) + Kent + 1 V⋅ D (2) ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ In addition ν (3) Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f, which we can solve iteratively Make a guess for f f = 0.1 then V = 2 ⋅ g ⋅ ( ∆z + h ) f⋅ using Eq 3, at this Re V = Re = V⋅ D ν Re = 2.13 × 10 4 L D V = 1.37 m V = 1.38 m + Kent + 1 s Re = V⋅ D ν Re = 3.41 × 10 4 Re = 3.46 × 10 4 f = 0.0371 V = Then, repeating 2 ⋅ g ⋅ ( ∆z + h ) f⋅ f = 0.0371 Q = V⋅ π⋅ D 4 L D + Kent + 1 s which is the same as before, so we have convergence 2 The flow rate is then s + Kent + 1 2 ⋅ g ⋅ ( ∆z + h ) f⋅ Using Eq 3, at this Re D m f = 0.0382 Then, repeating using Eq 3, at this Re L V = 0.852 3 −4m Q = 6.79 × 10 Note that we could use Excel's Solver for this problem s Q = 0.679 l s Re = V⋅ D ν Problem 8.138 Given: Flow in horizontal pipe Find: Flow rate Solution: Basic equations [Difficulty: 4] 2 2 ⎞ ⎛⎜ p ⎞ ⎛⎜ p V1 V2 1 2 ⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h l ⎝ ⎠ ⎝ ⎠ 2 hl = f ⋅ L V ⋅ D 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) No minor losses Available data L = 200 ⋅ m D = 75⋅ mm e = 2.5⋅ mm ∆p = 425 ⋅ kPa ρ = 1000⋅ kg 3 μ = 1.76⋅ 10 − 3 N⋅ s ⋅ m 2 m Hence the energy equation becomes p1 ρ Solving for V We also have p2 − ρ ∆p ρ 2 = f⋅ L V ⋅ D 2 2 ⋅ D⋅ ∆p V= Re = = V= L⋅ ρ⋅ f ρ⋅ V⋅ D k 2 ⋅ D⋅ ∆p f Re = c⋅ V or μ ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ In addition k = (1) (2) k = 0.565 L⋅ ρ c = where ρ⋅ D μ m s c = 4.26 × 10 4 s m (3) Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f Make a guess for f Given Given f = 0.1 V = then ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ ⎞ ⎛ e ⎜ 1 2.51 D = −2.0⋅ log ⎜ + 3.7 f Re⋅ f ⎠ ⎝ Q = V⋅ π⋅ D 4 V = 5.86⋅ ft V = 7.74⋅ ft f f = 0.0573 V = k f f = 0.0573 V = k V= ⋅ f 2 The flow rate is then k 3 Q = 0.0104 Note that we could use Excel's Solver for this problem m s Q = 10.42 l s s s ft s Q = 165 ⋅ gpm Re = c⋅ V Re = 7.61 × 10 4 Re = c⋅ V Re = 1.01 × 10 5 Re = c⋅ V Re = 1.01 × 10 5 Problem 8.139 [Difficulty: 2] Problem 8.140 [Difficulty: 4] . Given: Two potential solutions to improve flowrate. Find: Which solution provides higher flowrate Solution: Basic equations: ⎛ p1 ⎞ ⎛p ⎞ V2 V2 ⎜⎜ + α 1 1 + gz1 ⎟⎟ − ⎜⎜ 2 + α 2 2 + gz 2 ⎟⎟ = hlT 2 2 ⎝ρ ⎠ ⎝ ρ ⎠ ⎛e/ D 1 2.51 LV2 V2 = −2.0 log⎜ + hlT = hl + hlm = f +K ⎜ 3.7 Re f D 2 2 f ⎝ Assumptions: 1) Steady flow 2) Incompressible 3) Neglect minor losses 4) Option 1: let z1 = 0 Given data Q = VA V1 2 V2 = α2 2 2 2 p 2 = p atm = 0 kPa gage p1 = 200 kPa gage The energy equation becomes: Solving for V: α1 ⎞ ⎟ ⎟ ⎠ V = D = 0.019 m p1 ρ − gz 2 = f ⎛p ⎞ 2 ⋅ D ⋅ ⎜⎜ 1 − g ⋅ z 2 ⎟⎟ ⎝ρ ⎠ f ⋅L e =0 D z 2 = 15 m LV2 D 2 V = k f (1) L = 23 m k= ⎞ ⎛p 2 ⋅ D ⋅ ⎜⎜ 1 − g ⋅ z 2 ⎟⎟ 3 ⎠ = 2 × 0.019 m × ⎛⎜ 200,000 N × m × kg ⋅ m − 9.81 m × 15 m ⎞⎟ × 1 ⎝ρ ⎜ ⎟ 23 m L m 2 999 kg N ⋅ s 2 s2 ⎝ ⎠ k = 0.296 m s Re = We also have ρ ⋅V ⋅ D µ Re = c ⋅ V (2) or where c = 999 Assuming water at 20oC (ρ = 999 kg/m3, µ = 1 x 10-3 kg/(m·s)): ⎛ 2.51 = −2.0 log⎜ ⎜ Re f f ⎝ 1 In addition: Given 1 f Given 1 f Given 1 f f ⎛ 2.51 = −2.0 log⎜ ⎜ Re f ⎝ ⎛ 2.51 = −2.0 log⎜ ⎜ Re f ⎝ ⎛ 2.51 = −2.0 log⎜ ⎜ Re f ⎝ The flowrate is then: Option 2: let Given data f = 0.015 z1 = 0 ⎞ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎠ π s m⋅s kg = 18981 × 0.019 m × 3 -3 m 1 × 10 kg m V , Re and f V = k f = 0.0213 V = k f = 0.0222 V = k f = 0.0223 V = k then m Q1 = × (0.019) m × 1.98 = s 4 2 ρ⋅D µ ⎞ ⎟ (3) ⎟ ⎠ Equations 1, 2 and 3 form a set of simultaneous equations for Make a guess for c= 2 f f f f = 2.42 m s Re = c ⋅ V = 4.59 × 10 4 = 2.03 m s Re = c ⋅ V = 3.85 × 10 4 = 1.99 m s Re = c ⋅ V = 3.77 × 10 4 = 1.98 m s Re = c ⋅ V = 3.76 × 10 4 5.61 × 10 −4 m3 s p 2 = p atm = 0 kPa gage p1 = 300 kPa gage D = 0.0127 m The analysis for Option 2 is identical to Option 1: The energy equation becomes: p1 ρ − gz 2 = f LV2 D 2 z 2 = 15 m e = 0.05 D L = 16 m V = Solving for V: k= ⎞ ⎛p 2 ⋅ D ⋅ ⎜⎜ 1 − g ⋅ z 2 ⎟⎟ ⎝ρ ⎠ f ⋅L V = k (4) f ⎛p ⎞ 2 ⋅ D ⋅ ⎜⎜ 1 − g ⋅ z 2 ⎟⎟ 3 ⎝ρ ⎠ = 2 × 0.0127 m × ⎛⎜ 300,000 N × m × kg ⋅ m − 9.81 m × 15 m ⎞⎟ × 1 ⎟ 16 m ⎜ L m 2 999 kg N ⋅ s 2 s2 ⎝ ⎠ k = 0.493 We also have m s Re = ρ ⋅V ⋅ D µ c = 999 Re = c ⋅ V (5) or ρ⋅D µ kg m⋅s s = 12687.3 × 0.0127 m × 3 -3 m m 1 × 10 kg ⎛e/ D 2.51 = −2.0 log⎜ + ⎜ f ⎝ 3.7 Re f 1 In addition: c= where ⎞ ⎛ ⎟ = −2.0 log⎜ 0.05 + 2.51 ⎟ ⎜ 3.7 Re f ⎠ ⎝ ⎞ ⎟ ⎟ ⎠ (6) Equations 4, 5 and 6 form a set of simultaneous equations for V , Re and f Make a guess for Given Given f f = 0.07 ⎛ 0.05 2.51 = −2.0 log⎜ + ⎜ 3.7 Re f f ⎝ ⎛ 0.05 1 2.51 = −2.0 log⎜ + ⎜ f ⎝ 3.7 Re f 1 The flowrate is then: V = k f = 0.0725 V = k f = 0.0725 V = k then ⎞ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎠ π f f m s Re = c ⋅ V = 2.36 × 10 4 = 1.83 m s Re = c ⋅ V = 2.32 × 10 4 = 1.83 m s Re = c ⋅ V = 2.32 × 10 4 3 m −4 m Q2 = × (0.0127 ) m × 1.83 = 2.32 × 10 s s 4 2 2 This problem can also be solved explicitly: LV2 − gz 2 = f ρ D 2 p1 The energy equation becomes: or: f = 1.86 f = Plugging this into the Colebrook equation: 1 V ⎛p ⎞ 2 D⎜⎜ 1 − gz 2 ⎟⎟ ⎝ρ ⎠ L Option 1 is 2.42 times more effective! 1 f Noting that the =V L ⎛p ⎞ 2 D⎜⎜ 1 − gz 2 ⎟⎟ ⎝ρ ⎠ ⎛ ⎛ ⎜ ⎜ ⎜ e / D 2.51µ ⎜ = −2.0 log⎜ + V ρV D ⎜⎜ ⎜ 3.7 ⎜ ⎜ ⎝ ⎝ V s on the right hand side cancel provides: ⎛ ⎜ ⎜ e / D 2.51µ V = −2.0 log⎜ + 3 . 7 ρD ⎜ ⎜ ⎝ ⎛ 2 D⎜⎜ ⎝ ⎛ 2 D⎜⎜ ⎝ ⎞⎞ ⎟⎟ ⎟⎟ L ⎟⎟ p1 ⎞ − gz 2 ⎟⎟ ⎟ ⎟ ρ ⎠ ⎟⎠ ⎟⎠ ⎞ ⎞⎛ ⎟⎜ 2 D⎛⎜ p1 − gz 2 ⎞⎟ ⎟ ⎜ρ ⎟⎟ ⎟⎜ L ⎝ ⎠ ⎟ ⎟⎜ L p1 ⎞ ⎟⎜ ⎟ − gz 2 ⎟⎟ ⎟ ⎜ ⎟ ρ ⎠ ⎠⎝ ⎠ Assuming water at 20oC (ρ = 999 kg/m3, µ = 1 x 10-3 kg/(m·s)) gives the remaining information needed to perform the calculation. For Option 1: ⎛ ⎞ kg m3 1 ⎜ 2.51 × 1 × 10 −3 ⎟ × × m ⋅ s 999 kg 0.019 m ⎜ ⎟ ⎜ ⎟ V = −2.0 log⎜ 23 m 1 ⎟ × × ⎜ 2 × 0.019 m ⎛ ⎞⎟ N m3 kg ⋅ m m ⎜⎜ 200,000 2 × − 9.81 2 × 15 m ⎟⎟ ⎟ × ⎜⎜ 2 999 kg ⋅ m N s s ⎝ ⎠ ⎟⎠ ⎝ ⎛ 2 × 0.019 m ⎛ ⎞ ⎞⎟ N m3 kg ⋅ m m ⎟⎟ 9 . 81 15 m ×⎜ × ⎜⎜ 200,000 2 × × − × ⎜ 999 kg N ⋅ s 2 23 m m s2 ⎝ ⎠ ⎟⎠ ⎝ m V = 1.98 s and: Q1 = Option 2: let Given data z1 = 0 π 4 × (0.019) m 2 × 1.98 2 m3 m = 5.61 × 10 − 4 s s p 2 = p atm = 0 kPa gage p1 = 300 kPa gage z 2 = 15 m D = 0.0127 m e = 0.05 D The analysis for Option 2 results in the same equations as used in Option 1 once again giving: ⎛ ⎜ ⎜ e / D 2.51µ V = −2.0 log⎜ + ρD ⎜ 3.7 ⎜ ⎝ ⎛ 2 D⎜⎜ ⎝ ⎞ ⎞⎛ ⎟⎜ 2 D⎛⎜ p1 − gz 2 ⎞⎟ ⎟ ⎟⎟ ⎜ρ ⎟⎜ L ⎠ ⎝ ⎟ ⎟⎜ L p1 ⎞ ⎟⎜ ⎟ − gz 2 ⎟⎟ ⎟ ⎜ ⎟ ρ ⎠ ⎠⎝ ⎠ L = 16 m ⎞ ⎛ 0.05 kg m3 1 ⎟ ⎜ + 2.51 × 1 × 10 −3 × × m ⋅ s 999 kg 0.0127 m ⎟ ⎜ 3.7 ⎟ ⎜ V = −2.0 log⎜ 16 m 1 ⎟ × × 3 ⎜ 2 × 0.0127 m ⎛ ⎞⎟ kg ⋅ m N m m ⎜⎜ 300,000 2 × × − 9.81 2 × 15 m ⎟⎟ ⎟ ⎜⎜ 2 999 kg N ⋅ s s m ⎠ ⎟⎠ ⎝ ⎝ ⎛ 2 × 0.0127 m ⎛ ⎞ ⎞⎟ N m3 kg ⋅ m m ⎟⎟ − × ×⎜ × ⎜⎜ 300,000 2 × × 9 . 81 15 m ⎜ 999 kg N ⋅ s 2 16 m m s2 ⎠ ⎟⎠ ⎝ ⎝ V = 1.83 The flowrate is then: Q2 = π 4 × (0.0127 ) m 2 × 1.83 2 m = s m s 2.32 × 10 − 4 m3 s Problem 8.141 Given: Kiddy pool on a porch. Find: Time to fill pool with a hose. [Difficulty: 3] Solution: ⎛ p1 ⎞ ⎛ p2 ⎞ V1 2 V 22 ⎜ ⎟ ⎜ Basic equations: ⎜ ρ + α 1 2 + gz1 ⎟ − ⎜ ρ + α 2 2 + gz 2 ⎟⎟ = hlT ⎝ ⎠ ⎝ ⎠ 2 2 ⎛e/ D 1 2.51 LV V = −2.0 log⎜ + hlT = hl + hlm = f +K ⎜ D 2 2 f ⎝ 3.7 Re f Assumptions: 1) Steady flow 2) Incompressible 3) Neglect minor losses 4) Given data p1 = 60 z1 = 0 ft D = 0.625 in z 2 = 20.5 ft e =0 D L = 50 ft k= V = Q = VA V1 2 V2 = α2 2 2 2 lbf gage in 2 The energy equation becomes: Solving for V: α1 ⎞ ⎟ ⎟ ⎠ p2 = 0 lbf gage in 2 LV2 − gz 2 = f D 2 ρ p1 ⎛p ⎞ 2 ⋅ D ⋅ ⎜⎜ 1 − g ⋅ z 2 ⎟⎟ ⎝ρ ⎠ f ⋅L V = k (1) f ⎛p ⎞ 2 ⋅ D ⋅ ⎜⎜ 1 − g ⋅ z 2 ⎟⎟ 2 3 ⎞ slug ⋅ ft ft 1 ⎝ρ ⎠ = 2 × 0.625 in × ft × ⎛⎜ 60 lbf × 144 in × ft × − 32.2 2 × 20.5 ft ⎟⎟ × 2 2 2 ⎜ L 12 in ⎝ in 1.94 slug lbf ⋅ s ft s ⎠ 50 ft k = 2.81 ft s We also have Re = ρ ⋅V ⋅ D µ Re = c ⋅ V (2) or where c= ρ⋅D µ Assuming water at 68oF (ρ = 1.94 slug/ft3, µ = 2.1 x 10-5 lbf·s/ft2): c = 1.94 s slug ft ft 2 lbf ⋅ s 2 = 4811.5 × × × × 0 . 625 in 3 -5 ft 12 in 2.1 × 10 lbf ⋅ s slug ⋅ ft ft ⎛ 2.51 = −2.0 log⎜ ⎜ Re f f ⎝ 1 In addition: ⎞ ⎟ (3) ⎟ ⎠ Equations 1, 2 and 3 form a set of simultaneous equations for V , Re and f Given Given V = k f = 0.0177 V = k f = 0.0179 V = k f = 0.015 Make a guess for f ⎛ 2.51 = −2.0 log⎜ ⎜ Re f f ⎝ ⎛ 2.51 1 = −2.0 log⎜ ⎜ Re f f ⎝ 1 then ⎞ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎠ f f f = 22.94 ft s Re = c ⋅ V = 1.1 × 10 5 = 21.12 ft s Re = c ⋅ V = 1.02 × 10 5 = 21.0 ft s Re = c ⋅ V = 1.01 × 10 5 The flowrate is then: Q = π × (0.625) in 2 × 2 4 ft 2 ft ft 3 × = 0 . 0447 21 . 0 s s 144 in 2 Volume pool = 2.5 ft × time = Volume pool Q = π 4 × (5) ft 2 = 49.1 ft 3 2 49.1 ft 3 = 1097 s = ft 3 0.0447 s This problem can also be solved explicitly in the following manner: p1 The energy equation becomes: ρ f = or: 1 V − gz 2 = f LV2 D 2 ⎛p ⎞ 2 D⎜⎜ 1 − gz 2 ⎟⎟ ⎝ρ ⎠ L Plugging this into the Colebrook equation: 1 =V f L ⎞ ⎛p 2 D⎜⎜ 1 − gz 2 ⎟⎟ ⎝ρ ⎠ ⎛ ⎛ ⎜ ⎜ ⎜ 2.51µ ⎜ = −2.0 log⎜ ⎜V ⎜ ρV D ⎜ ⎜ ⎜ ⎝ ⎝ ⎛ 2 D⎜⎜ ⎝ ⎞⎞ ⎟⎟ ⎟⎟ L ⎟⎟ p1 ⎞ − gz 2 ⎟⎟ ⎟ ⎟ ρ ⎠ ⎟⎠ ⎟⎠ 18.3 min. Noting that the V s on the right hand side cancel provides: ⎞ ⎞⎛ ⎛ ⎟⎜ 2 D⎛⎜ p1 − gz 2 ⎞⎟ ⎟ ⎜ ⎜ ⎟ ⎟⎜ ⎜ 2.51µ L ⎝ρ ⎠⎟ V = −2.0 log⎜ ⎜ ⎟ ⎟ ρ D L p ⎛ ⎞ 1 ⎟ ⎜ ⎜ 2 D⎜⎜ − gz 2 ⎟⎟ ⎟ ⎜ ⎟ ⎟ ⎜ ⎝ρ ⎠ ⎠⎝ ⎝ ⎠ Assuming water at 68oF (ρ = 1.94 slug/ft3, µ = 2 x 10-5 lbf·s/ft2) and g = 32.2 ft/s2 gives the remaining information needed to perform the calculation finding: ⎞ ⎛ slug ⋅ ft 1 12 in lbf ⋅ s ft 3 ⎟ ⎜ 2.51 × 2.1 × 10 −5 × × × × 2 2 0.625 in ft 1.94 slug lbf ⋅ s ft ⎟ ⎜ ⎟ ⎜ V = −2.0 log⎜ 50 ft 12 in 1 ⎟ × × × 3 ⎟ ⎜ 2 × 0.625 in ft ⎞ ⎛ lbf 144 in 2 slug ⋅ ft ft ft ⎜⎜ 60 2 × × × − 32.2 2 × 20.5 ft ⎟⎟ ⎟ ⎜⎜ 2 2 1.94 slug lbf ⋅ s ft s ⎠ ⎟⎠ ⎝ in ⎝ ⎛ 2 × 0.625 in ⎞ ⎞⎟ ⎛ ft lbf 144 in 2 ft 3 slug ⋅ ft ft ⎟⎟ ×⎜ × × ⎜⎜ 60 2 × × × − × 32 . 2 20 . 5 ft ⎜ 50 ft 12 in ⎝ 1.94 slug lbf ⋅ s 2 in ft 2 s2 ⎠ ⎟⎠ ⎝ V = 21.0 ft s and: Q = π 4 × (0.052) ft 2 × 21.0 2 Volume pool = 2.5 ft × time = Volume pool Q = π 4 ft ft 3 = 0.0447 s s × (5) ft 2 = 49.1 ft 3 2 49.1 ft 3 = 1097 s = ft 3 0.0447 s 18.3 min. Problem 8.142 [Difficulty: 3] Problem 8.143 [Difficulty: 3] Problem 8.144 [Difficulty: 3] Problem 8.145 [Difficulty: 3] Problem 8.146 Given: Flow from large reservoir Find: Flow rates in two pipes Solution: Basic equations [Difficulty: 4] 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 ⎜ ρ + α ⋅ 2 + g ⋅ z1 − ⎜ ρ + α ⋅ 2 + g ⋅ z2 = h l ⎝ ⎠ ⎝ ⎠ 2 hl = f ⋅ 2 L V ⋅ D 2 V h lm = Kent ⋅ 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 Available data D = 50⋅ mm H = 10⋅ m L = 10⋅ m e = 0.15⋅ mm (Table 8.1) ν = 1⋅ 10 2 −6 m ⋅ Kent = 0.5 (Table A.8) s The energy equation becomes 2 V2 L V2 g ⋅ z1 − z2 − ⋅ V2 = f ⋅ ⋅ + Kent⋅ 2 D 2 2 ( Solving for V ) 2 2 and V2 = V z1 − z2 = H 2⋅ g⋅ H V= f⋅ We also have 1 L + Kent + 1 D ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ (1) Re = (2) V⋅ D (3) ν We must solve Eqs. 1, 2 and 3 iteratively. Make a guess for V and Then V = 1⋅ m Then s ⎞ ⎛ e ⎜ 1 2.51 D = −2.0⋅ log ⎜ + f ⎝ 3.7 Re⋅ f ⎠ 2⋅ g⋅ H V = f⋅ L + Kent + 1 D Re = V⋅ D ν f = 0.0286 V = 5.21 m s Re = 5.00 × 10 4 Repeating and Then Re = V⋅ D ⎞ ⎛ e ⎜ 1 2.51 D = −2.0⋅ log ⎜ + f ⎝ 3.7 Re⋅ f ⎠ 2⋅ g⋅ H V = f⋅ Repeating and Then Re = L ⎞ ⎛ e ⎜ 1 2.51 D = −2.0⋅ log ⎜ + f ⎝ 3.7 Re⋅ f ⎠ 2⋅ g⋅ H V = Q = π 4 V = 5.36 m s Re = 2.68 × 10 ν L 5 f = 0.0267 + Kent + 1 D V⋅ D f⋅ Hence Re = 2.61 × 10 ν 5 f = 0.0267 V = 5.36 + Kent + 1 D m s 3 2 ⋅D ⋅V Q = 0.0105 m Q = 10.5⋅ s l s We repeat the analysis for the second pipe, using 2L instead of L: Make a guess for V and Then V = 1⋅ m ⎞ ⎛ e ⎜ 1 2.51 D = −2.0⋅ log ⎜ + 3.7 f Re⋅ f ⎠ ⎝ 2⋅ g⋅ H V = f⋅ Repeating and Then Then s Re = 2⋅ L ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ 2⋅ g⋅ H 2⋅ L + Kent + 1 D Re = 5.00 × 10 ν f = 0.0286 V = 3.89 m s Re = 1.95 × 10 ν f⋅ V⋅ D + Kent + 1 D V⋅ D V = Re = f = 0.0269 V = 4.00 m s 5 4 Repeating and Then Re = V⋅ D ⎞ ⎛ e ⎜ 1 2.51 D = −2.0⋅ log ⎜ + f ⎝ 3.7 Re⋅ f ⎠ 2⋅ g⋅ H V = f⋅ Hence Re = 2.00 × 10 ν Q = π 4 2⋅ L + Kent + 1 D 2 ⋅D ⋅V As expected, the flow is considerably less in the longer pipe. 5 f = 0.0268 V = 4.00 m s Q = 7.861 × 10 3 −3m s Q = 7.86⋅ l s Problem 8.147 Given: Galvanized drainpipe Find: Maximum downpour it can handle Solution: [Difficulty: 3] 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 = hl 2 2 ⎝ ⎠ ⎝ρ ⎠ Basic equations 2 hl = f ⋅ L V ⋅ D 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) No minor losses Available data D = 50⋅ mm e = 0.15⋅ mm h=L (Table 8.1) ( From Table A.7 (20oC) 2 ) The energy equation becomes L V g ⋅ z1 − g ⋅ z2 = g ⋅ z1 − z2 = g ⋅ h = f ⋅ ⋅ D 2 Solving for V V= k= 2 ⋅ D⋅ g ⋅ h L⋅ f 2 ⋅ D⋅ g = 2 ⋅ D⋅ g V= f k = k (1) f 2 × 0.05⋅ m × 9.81⋅ m k = 0.99 2 s Re = We also have V⋅ D Re = c⋅ V or ν s c = 0.05⋅ m × (2) m s c= where D ν 4 s −6 1.01 × 10 c = 4.95 × 10 ⋅ 2 ⋅m ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ In addition ν = 1.01 × 10 m (3) Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f Make a guess for f f = 0.01 V = then k V = 9.90 m f Given s Re = c⋅ V 5 Re = 4.9 × 10 ⎛ e ⎞ ⎜ 1 D 2.51 = −2.0⋅ log ⎜ + 3.7 f Re⋅ f ⎠ ⎝ f = 0.0264 V = k f V = 6.09 m s Re = c⋅ V Re = 3.01 × 10 5 2 −6 m ⋅ s Given ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ f = 0.0266 k V = V = 6.07 m V = 6.07 m s f Given Re = c⋅ V Re = 3.00 × 10 5 Re = c⋅ V Re = 3.00 × 10 5 ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ f = 0.0266 k V = s f 2 The flow rate is then Q = V⋅ 3 π⋅ D Q = 0.0119⋅ 4 m s 3 The downpour rate is then Q Aroof 0.0119⋅ = m s 2 × 500 ⋅ m Note that we could use Excel's Solver for this problem 100 ⋅ cm 1⋅ m × 60⋅ s 1 ⋅ min = 0.143 ⋅ cm min The drain can handle 0.143 cm/min Problem 8.148 [Difficulty: 3] Part 1/2 Problem 8.148 [Difficulty: 3] Part 2/2 Problem 8.149 Given: Flow in a tube Find: Effect of tube roughness on flow rate; Plot [Difficulty: 3] Solution: Governing equations: 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 ⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h lT = ⎝ ⎠ ⎝ ⎠ Re = f = ρ⋅ V⋅ D 64 μ 2 hl = f ⋅ L V ⋅ D 2 (8.36) hl + major ∑ h lm (8.29) minor 2 (8.34) (Laminar) Re ∑ h lm = K⋅ V (8.40a) 2 ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ h lm = f ⋅ (8.37) Le V2 (8.40b) ⋅ D 2 (Turbulent) The energy equation (Eq. 8.29) becomes for flow in a tube 2 L V p 1 − p 2 = ∆p = ρ⋅ f ⋅ ⋅ D 2 This cannot be solved explicitly for velocity V, (and hence flow rate Q) because f depends on V; solution for a given relative roughness e/D requires iteration (or use of Solver) Flow Rate versus Tube Relative Roughness for fixed Dp 8 6 3 Q (m /s) x 10 4 4 2 0 0.00 0.01 0.02 e/D 0.03 0.04 0.05 It is not possible to roughen the tube sufficiently to slow the flow down to a laminar flow for this Δp. Even a relative roughness of 0.5 (a physical impossibility!) would not work. Problem 8.150 Given: Flow in a tube Find: Effect of tube length on flow rate; Plot [Difficulty: 3] Solution: Governing equations: 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 1 2 2 2 = h lT = ⎝ ⎠ ⎝ρ ⎠ Re = f = ρ⋅ V⋅ D 64 μ 2 hl = f ⋅ L V ⋅ D 2 (8.36) hl + major ∑ h lm (8.29) minor 2 (8.34) (Laminar) Re ∑ h lm = K⋅ V (8.40a) 2 ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ h lm = f ⋅ (8.37) Le V2 (8.40b) ⋅ D 2 (Turbulent) The energy equation (Eq. 8.29) becomes for flow in a tube 2 L V p 1 − p 2 = ∆p = ρ⋅ f ⋅ ⋅ D 2 This cannot be solved explicitly for velocity V, (and hence flow rate Q) because f depends on V; solution for a given L requires iteration (or use of Solver) Flow Rate vs Tube Length for Fixed Dp 10.0 Laminar Q (m3/s) Turbulent 4 x 10 1.0 0.1 0 5 10 15 L (km) 20 25 30 35 The "critical" length of tube is between 15 and 20 km. For this range, the fluid is making a transition between laminar and turbulent flow, and is quite unstable. In this range the flow oscillates between laminar and turbulent; no consistent solution is found (i.e., an Re corresponding to turbulent flow needs an f assuming laminar to produce the Δp required, and vice versa!) More realistic numbers (e.g., tube length) are obtained for a fluid such as SAE 10W oil (The graph will remain the same except for scale) Problem 8.151 [Difficulty: 5] Given: Flow from a reservoir Find: Effect of pipe roughness and pipe length on flow rate; Plot Solution: Governing equations: 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 ⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h lT = ⎝ ⎠ ⎝ ⎠ Re = f = ρ⋅ V⋅ D 64 μ 2 hl = f ⋅ hl + major ∑ h lm (8.29) minor 2 L V ⋅ D 2 (8.36) ∑ h lm = K⋅ (8.34) V (8.40a) 2 ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ (Laminar) Re h lm = f ⋅ (8.37) Le V2 (8.40b) ⋅ D 2 (Turbulent) The energy equation (Eq. 8.29) becomes for this flow (see Example 8.5) ⎛ p pump = ∆p = ρ⋅ ⎜ g ⋅ d + f ⋅ ⎝ 2⎞ L V ⋅ D 2 ⎠ We need to solve this for velocity V, (and hence flow rate Q) as a function of roughness e, then length L. This cannot be solved explicitly for velocity V, (and hence flow rate Q) because f depends on V; solution for a given relative roughness e/D or length L requires iteration (or use of Solver) It is not possible to roughen the tube sufficiently to slow the flow down to a laminar flow for this Δp. Flow Rate versus Tube Relative Roughness for fixed Dp 0.020 0.015 3 Q (m /s) 0.010 0.005 0.000 0.00 0.01 0.01 0.02 0.02 0.03 0.03 0.04 0.04 0.05 0.05 e/D Flow Rate versus Tube Length for fixed Dp 0.010 Q (m3/s) 0.005 0.000 0 200 400 600 L (m) 800 1000 1200 Problem 8.152 [Difficulty: 4] Given: System for fire protection Find: Height of water tower; Maximum flow rate; Pressure gage reading Solution: Governing equations: 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 ⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h lT = ⎝ ⎠ ⎝ ⎠ Re = f = ∑ hl + major ∑ h lm(8.29) minor 2 ρ⋅ V⋅ D hl = f ⋅ μ 64 (8.36) L V ⋅ D 2 (8.34) h lm = 0.1⋅ h l h lm = f ⋅ ⎛ e ⎞ ⎜ D 1 2.51 (8.37) + = −2.0⋅ log ⎜ 3.7 f Re ⋅ f ⎝ ⎠ (Laminar) Re Le V2 (8.40b) ⋅ D 2 (Turbulent) For no flow the energy equation (Eq. 8.29) applied between the water tower free surface (state 1; height H) and pressure gage is g⋅ H = p2 or ρ H= p2 ρ⋅ g (1) The energy equation (Eq. 8.29) becomes, for maximum flow (and α = 1) 2 g⋅ H − 2 V = h lT = ( 1 + 0.1) ⋅ h l 2 or g⋅ H = V 2 ⋅ ⎛⎜ 1 + 1.1⋅ f ⋅ ⎝ L⎞ (2) D⎠ This can be solved for V (and hence Q) by iterating, or by using Solver The energy equation (Eq. 8.29) becomes, for restricted flow g⋅ H − p2 ρ 2 + The results in Excel are shown below: V = h lT = ( 1 + 0.1) ⋅ h l 2 2 p 2 = ρ⋅ g ⋅ H − ρ⋅ V 2 ⋅ ⎜⎛ 1 + 1.1⋅ ρ⋅ f ⋅ ⎝ L⎞ D⎠ (3) Problem 8.153 Given: Syphon system Find: Flow rate; Minimum pressure Solution: [Difficulty: 4] 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 = h lT 2 2 ⎝ ⎠ ⎝ρ ⎠ Basic equations 2 h lT = f ⋅ L V ⋅ + h lm D 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 Hence the energy equation applied between the tank free surface (Point 1) and the tube exit (Point 2, z = 0) becomes g ⋅ z1 − V2 2 2 2 2 2 Le V2 V L V V = g ⋅ z1 − = f⋅ ⋅ + Kent⋅ + f⋅ ⋅ 2 D 2 2 D 2 Kent = 0.78 From Table 8.2 for reentrant entrance For the bend R D =9 The two lengths are Le = 56⋅ D We also have Re = D (1) ⎡ ⎛ L Le ⎞⎤ ⎢1 + Kent + f ⋅ ⎜ + ⎥ ⎣ ⎝ D D ⎠⎦ Le = 2.8 m V⋅ D or ν ν = 1.14 × 10 2 −6 m ⋅ s then h = 2.5⋅ m and L = 4.51 m Re = c⋅ V (2) where c = 0.05⋅ m × ⎞ ⎛ e ⎜ 1 2.51 D = −2.0⋅ log ⎜ + 3.7 f Re⋅ f ⎠ ⎝ f = 0.01 = 56 L = ( 0.6 + π⋅ 0.45 + 2.5) ⋅ m Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f Make a guess for f Le for a 90o bend so for a 180 o bend 2⋅ g⋅ h V= In addition = 28 D Solving for V From Table A.7 (15oC) Le so from Fig. 8.16 s D ν 4 s −6 1.14 × 10 (3) c= 2 c = 4.39 × 10 ⋅ ⋅m e = 0.0015⋅ mm (Table 8.1) m V = 2⋅ g⋅ h ⎡ ⎛ L Le ⎞⎤ ⎢1 + Kent + f ⋅ ⎜ + ⎥ ⎣ ⎝ D D ⎠⎦ ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ Given 2⋅ g⋅ h V = ⎛ L Le ⎞⎤ ⎡ ⎢1 + Kent + f ⋅ ⎜ + ⎥ ⎣ ⎝ D D ⎠⎦ ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ Given 2⋅ g⋅ h V = ⎡ ⎛L ⎢1 + Kent + f ⋅ ⎜ + ⎥ ⎣ ⎝ D D ⎠⎦ ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ Given V = Le ⎞⎤ 2⋅ g⋅ h Le ⎞⎤ ⎡ ⎛L ⎢1 + Kent + f ⋅ ⎜ + ⎥ ⎣ ⎝ D D ⎠⎦ V = 3.89 m Re = c⋅ V s Re = 1.71 × 10 5 f = 0.0164 V = 3.43 m s Re = c⋅ V Re = 1.50 × 10 5 Re = c⋅ V Re = 1.49 × 10 5 Re = c⋅ V Re = 1.49 × 10 5 f = 0.0168 V = 3.40 m s f = 0.0168 V = 3.40 m s Note that we could use Excel's Solver for this problem. 2 Q = The flow rate is then π⋅ D 4 3 −3m ⋅V Q = 6.68 × 10 s The minimum pressure occurs at the top of the curve (Point 3). Applying the energy equation between Points 1 and 3 2 ⎛⎜ p ⎞ 2 2 V3 ⎛ p3 V2 Le V2 ⎞ 3 V L V g ⋅ z1 − ⎜ + + g ⋅ z3 = g ⋅ z1 − ⎜ + + g ⋅ z3 = f ⋅ ⋅ + Kent⋅ + f⋅ ⋅ 2 2 D 2 2 D 2 ⎝ρ ⎠ ⎝ρ ⎠ where we have Le D = 28 for the first 90o of the bend, and L = ⎜⎛ 0.6 + ⎝ π × 0.45 ⎞ 2 ⎠ ⋅m L = 1.31 m 2 ⎡ ⎛ L Le ⎞⎤⎤ V ⎡ ⋅ ⎢1 + Kent + f ⋅ ⎜ + p 3 = ρ⋅ ⎢g ⋅ z1 − z3 − ⎥⎥ 2 ⎣ ⎣ ⎝ D D ⎠⎦⎦ ( ) ⎡ ⎢ kg ⎢ m p 3 = 1000⋅ × 9.81⋅ × ( −0.45⋅ m) − 3 ⎢ 2 m s ⎣ 2 ⎤ ⎛ 3.4⋅ m ⎞ ⎥ ⎜ 2 s⎠ ⎡ 1.31 ⎝ ⎥ N⋅ s p = −20.0⋅ kPa ⋅ ⎢1 + 0.78 + 0.0168⋅ ⎛⎜ + 28⎞⎥⎤ × 2 ⎣ ⎝ 0.05 ⎠⎦⎥⎦ kg⋅ m 3 Problem 8.154 [Difficulty: 4] Given: Tank with drainpipe Find: Flow rate for rentrant, square-edged, and rounded entrances Solution: 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 = h lT 2 2 ⎝ ⎠ ⎝ρ ⎠ Basic equations 2 h lT = f ⋅ 2 L V V ⋅ + Kent⋅ D 2 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 Available data D = 1 ⋅ in L = 2 ⋅ ft e = 0.00085 ⋅ ft h = 3 ⋅ ft (Table 8.1) r = 0.2⋅ in Hence the energy equation applied between the tank free surface (Point 1) and the pipe exit (Point 2, z = 0) becomes V2 g ⋅ z1 − Solving for V V= We also have Re = From Table A.7 (20oC) In addition 2 2 2 2 2 V L V V = g ⋅ z1 − = f⋅ ⋅ + Kent⋅ 2 D 2 2 2⋅ g⋅ h (1) ⎛1 + K + f ⋅ L ⎞ ⎜ ent D⎠ ⎝ V⋅ D Re = c⋅ V (2) or ν ν = 1.01 × 10 2 −6 m ⋅ − 5 ft ν = 1.09 × 10 s ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ c= where 2 c = s D ν Make a guess for f f = 0.01 (3) Re = c⋅ V Given Kent = 0.78 V = then Re = 7.49 × 10 2⋅ g⋅ h ⎛ ⎜ 1 + Kent + f ⋅ D⎠ ⎝ 4 ⎛ e ⎞ ⎜ 1 D 2.51 = −2.0⋅ log ⎜ + f ⎝ 3.7 Re⋅ f ⎠ L⎞ f = 0.0389 ν c = 7665⋅ Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f For a reentrant entrance, from Table 8.2 D V = 2.98 m s s ft 2⋅ g⋅ h V = Given L⎞ ⎛ ⎜ 1 + Kent + f ⋅ D ⎝ ⎠ e ⎛ ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ V = 2.57 L⎞ ⎛ ⎜ 1 + Kent + f ⋅ D ⎝ ⎠ e ⎛ ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ V = m s Re = c⋅ V Re = 6.46 × 10 4 Re = c⋅ V Re = 6.46 × 10 4 Re = c⋅ V Re = 6.46 × 10 4 f = 0.0391 2⋅ g⋅ h V = Given V = 2.57 m s f = 0.0391 2⋅ g⋅ h V = 2.57 L⎞ ⎛ ⎜ 1 + Kent + f ⋅ D ⎝ ⎠ m s Note that we could use Excel's Solver for this problem 2 The flow rate is then Q = V⋅ π⋅ D Q = 0.0460 4 Given then ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ V = Given f = 0.01 V = 2⋅ g⋅ h ⎛ ⎜ 1 + Kent + f ⋅ D⎠ ⎝ L⎞ 2⋅ g⋅ h V = 2.71 L⎞ m s 2⋅ g⋅ h V = 2.71 L⎞ ⎛ ⎜ 1 + Kent + f ⋅ D⎠ ⎝ Q = V⋅ π⋅ D m Q = 0.0485 4 For a rounded entrance, from Table 8.2 r D m Re = 8.07 × 10 4 Re = c⋅ V Re = 6.83 × 10 4 Re = c⋅ V Re = 6.82 × 10 4 V = 3.21 s Re = c⋅ V f = 0.0390 2 The flow rate is then Q = 20.6⋅ gpm s f = 0.0389 ⎛ ⎜ 1 + Kent + f ⋅ D ⎝ ⎠ ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ V = 3 Kent = 0.5 For a square-edged entrance, from Table 8.2 Make a guess for f ft = 0.2 s ft 3 s Q = 21.8⋅ gpm Kent = 0.04 Make a guess for f f = 0.01 V = then V = 3.74 m V = V = 2⋅ g⋅ h V = m V = 3.02 L⎞ 4 s Re = c⋅ V Re = 7.59 × 10 4 Re = c⋅ V Re = 7.58 × 10 4 Re = c⋅ V Re = 7.58 × 10 4 f = 0.0389 2⋅ g⋅ h m V = 3.01 ⎛ ⎜ 1 + Kent + f ⋅ D ⎝ ⎠ ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ Given Re = 9.41 × 10 f = 0.0388 ⎛ ⎜ 1 + Kent + f ⋅ D ⎝ ⎠ ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ Given ⎛1 + K + f ⋅ L ⎞ ⎜ ent D⎠ ⎝ Re = c⋅ V s ⎞ ⎛ e ⎜ 1 2.51 D = −2.0⋅ log ⎜ + f ⎝ 3.7 Re⋅ f ⎠ Given 2⋅ g⋅ h L⎞ s f = 0.0389 2⋅ g⋅ h m V = 3.01 ⎛ ⎜ 1 + Kent + f ⋅ D ⎝ ⎠ L⎞ s Note that we could use Excel's Solver for this problem 2 The flow rate is then In summary: Q = V⋅ π⋅ D 4 Renentrant: Q = 20.6⋅ gpm Q = 0.0539 ft 3 s Square-edged: Q = 24.2⋅ gpm Q = 21.8⋅ gpm Rounded: Q = 24.2⋅ gpm Problem 8.155 [Difficulty: 4] Given: Tank with drainpipe Find: Flow rate for rentrant, square-edged, and rounded entrances Solution: 2 2 ⎞ ⎛⎜ p ⎞ ⎛⎜ p V1 V2 1 2 ⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h lT ⎝ ⎠ ⎝ ⎠ Basic equations 2 h lT = f ⋅ 2 L V V ⋅ + Kent⋅ D 2 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 Available data D = 1 ⋅ in L = 2 ⋅ ft e = 0.00085 ⋅ ft h = 3 ⋅ ft (Table 8.1) r = 0.2⋅ in Hence the energy equation applied between the tank free surface (Point 1) and the pipe exit (Point 2, z = 0) becomes V2 g ⋅ z1 − Solving for V V= 2 2 2 2 2⋅ g⋅ H (1) ⎛1 + K + f ⋅ L ⎞ ⎜ ent D⎠ ⎝ H = h+L where now we have We also have From Table A.7 (20oC) In addition Re = 2 V L V V = g ⋅ z1 − = f⋅ ⋅ + Kent⋅ 2 D 2 2 V⋅ D Re = c⋅ V (2) or ν ν = 1.01 × 10 H = 5 ft 2 −6 m ⋅ − 5 ft ν = 1.09 × 10 s ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ c= where 2 s c = D ν Make a guess for f f = 0.01 (3) Re = c⋅ V Given Kent = 0.78 V = then Re = 9.67 × 10 2⋅ g⋅ H L⎞ ⎛ ⎜ 1 + Kent + f ⋅ D⎠ ⎝ 4 ⎛ e ⎞ ⎜ 1 D 2.51 = −2.0⋅ log ⎜ + 3.7 f Re⋅ f ⎠ ⎝ f = 0.0388 ν c = 7665⋅ Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f For a reentrant entrance, from Table 8.2 D V = 3.85 m s s ft 2⋅ g⋅ H V = Given V = 3.32 L⎞ ⎛ ⎜ 1 + Kent + f ⋅ D ⎝ ⎠ e ⎛ ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ V = m s Re = c⋅ V Re = 8.35 × 10 4 Re = c⋅ V Re = 8.35 × 10 4 f = 0.0389 2⋅ g⋅ H V = 3.32 L⎞ ⎛ ⎜ 1 + Kent + f ⋅ D ⎝ ⎠ m s Note that we could use Excel's Solver for this problem 2 The flow rate is then Q = V⋅ π⋅ D Q = 0.0594 4 f = 0.01 Given Given then ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ V = V = 2⋅ g⋅ H ⎛ ⎜ 1 + Kent + f ⋅ D ⎝ ⎠ L⎞ V = 3.51 L⎞ V = 3.51 ⎛ ⎜ 1 + Kent + f ⋅ D ⎝ ⎠ L⎞ π⋅ D m s m Q = 0.0627 4 For a rounded entrance, from Table 8.2 r D Make a guess for f f = 0.01 V = 4.83 s s ft 5 Re = c⋅ V Re = 8.82 × 10 4 Re = c⋅ V Re = 8.82 × 10 4 Re = c⋅ V s 3 Q = 28.2⋅ gpm s = 0.2 then m m Re = 1.04 × 10 V = 4.14 f = 0.0388 2⋅ g⋅ H Q = V⋅ Q = 26.7⋅ gpm 2⋅ g⋅ H 2 The flow rate is then s f = 0.0387 ⎛ ⎜ 1 + Kent + f ⋅ D ⎝ ⎠ ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ V = 3 Kent = 0.5 For a square-edged entrance, from Table 8.2 Make a guess for f ft Kent = 0.04 V = 2⋅ g⋅ H ⎛1 + K + f ⋅ L ⎞ ⎜ ent D⎠ ⎝ Re = c⋅ V Re = 1.22 × 10 5 ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ Given V = f = 0.0386 2⋅ g⋅ H L⎞ ⎛ ⎜ 1 + Kent + f ⋅ D ⎝ ⎠ e ⎞ ⎛ ⎜ D 1 2.51 = −2.0⋅ log ⎜ + f ⎝ 3.7 Re⋅ f ⎠ Given V = V = s Re = c⋅ V Re = 9.80 × 10 4 Re = c⋅ V Re = 9.80 × 10 4 Re = c⋅ V Re = 9.80 × 10 4 f = 0.0388 2⋅ g⋅ H m V = 3.89 L⎞ ⎛ ⎜ 1 + Kent + f ⋅ D ⎝ ⎠ e ⎛ ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ Given m V = 3.90 s f = 0.0388 2⋅ g⋅ H m V = 3.89 L⎞ ⎛ ⎜ 1 + Kent + f ⋅ D ⎝ ⎠ s Note that we could use Excel's Solver for this problem 2 The flow rate is then In summary: Q = V⋅ π⋅ D 4 Renentrant: Q = 26.7⋅ gpm Q = 0.0697 ft 3 s Square-edged: Q = 31.3⋅ gpm Q = 28.2⋅ gpm Rounded: Q = 31.3⋅ gpm Problem 8.156 [Difficulty: 5] Given: Tank with drain hose Find: Flow rate at different instants; Estimate of drain time Solution: 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 ⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h l ⎝ ⎠ ⎝ ⎠ Basic equations 2 hl = f ⋅ L V ⋅ D 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) Ignore minor loss at entrance (L >>; verify later) Available data L = 1⋅ m D = 15⋅ mm e = 0.2⋅ mm 3 Vol = 30⋅ m Hence the energy equation applied between the tank free surface (Point 1) and the hose exit (Point 2, z = 0) becomes g ⋅ z1 − Solving for V V= We also have Re = From Fig. A.2 (20oC) In addition V2 2 2 2 2 V L V = g ⋅ z1 − = f⋅ ⋅ 2 D 2 2⋅ g⋅ h (1) ⎛1 + f ⋅ L ⎞ ⎜ D⎠ ⎝ V⋅ D Re = c⋅ V (2) or ν 2 −6 m ν = 1.8 × 10 ⋅ c = s D c = 8333 ν ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ h = 10⋅ m initially where c= and D ν s m (3) Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f Make a guess for f f = 0.01 V = then 2⋅ g⋅ h ⎛1 + f ⋅ L ⎞ ⎜ D⎠ ⎝ V = 10.8 m s Re = c⋅ V Re = 9.04 × 10 4 Given Given ⎛ e ⎞ ⎜ D 1 2.51 + f = 0.0427 = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ V = ⎛ e ⎞ ⎜ D 1 2.51 + f = 0.0427 = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ V = 2⋅ g⋅ h ⎛1 + f ⋅ ⎜ D⎠ ⎝ L⎞ 2⋅ g⋅ h ⎛1 + f ⋅ ⎜ D⎠ ⎝ L⎞ Note that we could use Excel's Solver for this problem π⋅ D 4 Given 3 −3m Q = 1.26 × 10 ⎛ e ⎞ ⎜ D 1 2.51 + f = 0.0430 = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ V = 2 Q = V⋅ π⋅ D 4 Given Q = 8.9 × 10 ⎛1 + f ⋅ L ⎞ ⎜ D⎠ ⎝ 2⋅ g⋅ h ⎛1 + f ⋅ ⎜ D⎠ ⎝ L⎞ L D Re = 5.95 × 10 4 Re = c⋅ V Re = 5.95 × 10 4 = 2.8 Ke = 0.5 h lm < h l l s V = 5.04 m V = 5.04 m Q = 0.890 ⋅ s V = ⎛ e ⎞ ⎜ D 1 2.51 + f = 0.0444 = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ 2 Q = V⋅ 2⋅ g⋅ h 3 −4m ⎛ e ⎞ ⎜ D 1 2.51 + f = 0.0444 = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ The flow rate is then f⋅ s Re = c⋅ V s s Re = c⋅ V Re = 4.20 × 10 4 Re = c⋅ V Re = 4.20 × 10 4 Re = c⋅ V Re = 1.85 × 10 4 Re = c⋅ V Re = 1.85 × 10 4 l s h = 1⋅ m Next we recompute everything for Given m Q = 1.26⋅ s V = ⎛ e ⎞ ⎜ D 1 2.51 + f = 0.0430 = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ The flow rate is then V = 7.14 s h = 5⋅ m Next we recompute everything for Given m Note: 2 Q = V⋅ The flow rate is then V = 7.14 π⋅ D 4 V = 3 −4m Q = 3.93 × 10 s 2⋅ g⋅ h ⎛1 + f ⋅ ⎜ D⎠ ⎝ L⎞ 2⋅ g⋅ h ⎛1 + f ⋅ L ⎞ ⎜ D⎠ ⎝ V = 2.23 m V = 2.23 m Q = 0.393 ⋅ s s l s Initially we have dQ/dt = -1.26L/s, then -.890 L/s, then -0.393 L/s. These occur at h = 10 m, 5 m and 1 m. The corresponding volumes in the tank are then Q = 30,000 L, 15,000 L, and 3,000 L3. Using Excel we can fit a power trendline to the dQ/dt versus Q data to find, approximately 1 dQ dt = −0.00683 ⋅ Q t = 293 ⋅ ( 30 − 2 Q) where dQ/dt is in L/s and t is s. Solving this with initial condition Q = -1.26 L/s when t = 0 gives Hence, when Q = 3000 L (h = 1 m) t = 293 ⋅ ( 30000 − 3000) ⋅ s 4 t = 3.47 × 10 s t = 9.64⋅ hr Problem 8.157 [Difficulty: 4] Problem 8.158 [Difficulty: 4] Part 1/2 Problem 8.158 [Difficulty: 4] Part 2/2 Problem 8.159 [Difficulty: 4] Part 1/2 Problem 8.159 [Difficulty: 4] Part 2/2 Problem 8.160 [Difficulty: 5] Applying the energy equation between inlet and exit: ∆p ρ = f L V 2 D 2 "Old school": or ∆p ρf V 2 = D 2 L ∆p ⎛ ∆p ⎞ ⎛ Q0 ⎞ =⎜ ⎟ ⎜ ⎟ L ⎝ L ⎠0 ⎜⎝ Q ⎟⎠ Q (gpm) 20 18 16 Flow (gpm) 14 12 10 8 6 4 2 0 0.00 0.01 Your boss was wrong! 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00 3.25 3.50 3.75 4.00 4.25 4.50 4.75 5.00 5.25 5.50 5.75 6.00 6.25 6.50 6.75 7.00 7.25 7.50 7.75 0.02 8.00 8.25 8.50 8.75 9.00 Q (ft3/s) D= e= 2 ν = ρ = V (ft/s) Re f 1 in 0.00015 ft 2 1.08E-05 ft /s 3 1.94 slug/ft ∆p (old ∆p (psi/ft) school) (psi) 0.00279 0.511 3940 0.0401 0.00085 0.00085 0.00334 0.613 4728 0.0380 0.00122 0.00115 0.00390 0.715 5516 0.0364 0.00166 0.00150 0.00446 0.817 6304 0.0350 0.00216 0.00189 0.00501 0.919 7092 0.0339 0.00274 0.00232 1.021Pressure 7881 Drop 0.0329 0.00338 0.00278 Flow0.00557 Rate versus 0.00613 1.123 8669 0.0321 0.00409 0.00328 0.00668 1.226 9457 0.0314 0.00487 0.00381 0.00724 1.328 10245 0.0307 0.00571 0.00438 0.00780 1.430 11033 0.0301 0.00663 0.00498 0.00836 1.532 11821 0.0296 0.00761 0.00561 0.00891 1.634 12609 0.0291 0.00865 0.00628 0.00947 1.736 13397 0.0286 0.00977 0.00698 0.01003 1.838 14185 0.0282 0.01095 0.00771 0.01058 1.940 14973 0.0278 0.01220 0.00847 0.01114 2.043 15761 0.0275 0.01352 0.00927 0.01170 2.145 16549 0.0272 0.01491 0.01010 0.01225 2.247 17337 0.0268 0.01636 0.01095 0.01281 2.349 18125 0.0265 0.01788 0.01184 0.01337 2.451 18913 0.0263 0.01947 0.01276 0.01393 2.553 19701 0.0260 0.02113 0.01370 0.01448 2.655 20489 0.0258 0.02285 0.01468 School" 0.01504 2.758 21277 0.0255 "Old 0.02465 0.01569 0.01560 2.860 22065 0.0253 Exact0.02651 0.01672 0.01615 2.962 22854 0.0251 0.02843 0.01779 0.01671 3.064 23642 0.0249 0.03043 0.01888 0.01727 3.166 24430 0.0247 0.03249 0.02000 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.01783 3.268 25218 0.0245 0.03462 0.02115 Pressure Drop (psi/ft) 0.01838 3.370 26006 0.0243 0.03682 0.02233 0.01894 3.472 26794 0.0242 0.03908 0.02354 0.01950 3.575 27582 0.0240 0.04142 0.02477 0.02005 3.677 28370 0.0238 0.04382 0.02604 Problem 8.161 Given: Flow from large reservoir Find: Diameter for flow rates in two pipes to be same Solution: Basic equations [Difficulty: 5] 2 2 ⎞ ⎛⎜ p ⎞ ⎛⎜ p V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 = hl 2 2 ⎝ ⎠ ⎝ρ ⎠ 2 hl = f ⋅ 2 L V ⋅ D 2 V h lm = Kent⋅ 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 Available data D = 50⋅ mm H = 10⋅ m L = 10⋅ m e = 0.15⋅ mm (Table 8.1) ν = 1 ⋅ 10 2 −6 m ⋅ Kent = 0.5 (Table A.8) s For the pipe of length L the energy equation becomes 2 V2 L V2 g ⋅ z1 − z2 − ⋅ V2 = f ⋅ ⋅ + Kent⋅ 2 D 2 2 ( Solving for V ) 2 2 and V2 = V z1 − z2 = H 2⋅ g⋅ H V= f⋅ We also have 1 L + Kent + 1 D ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ (1) Re = (2) V⋅ D (3) ν We must solve Eqs. 1, 2 and 3 iteratively. Make a guess for V and Then V = 1⋅ m s ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ 2⋅ g⋅ H V = f⋅ Repeating Then Re = L + Kent + 1 D V⋅ D ν Re = V⋅ D Re = 5.00 × 10 ν f = 0.0286 V = 5.21 m s Re = 2.61 × 10 5 4 and Then ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ 2⋅ g⋅ H V = f⋅ Repeating and Then Re = V = 5.36 L + Kent + 1 D V⋅ D ⎛ e ⎞ ⎜ 1 2.51 D + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ π Q = 4 s 5 f = 0.0267 2⋅ g⋅ H V = m Re = 2.68 × 10 ν f⋅ Hence f = 0.0267 V = 5.36 L + Kent + 1 D m s 3 2 ⋅D ⋅V Q = 0.0105 m Q = 10.5⋅ s l s This is the flow rate we require in the second pipe (of length 2L) For the pipe of length 2L the energy equation becomes 2 V2 2 ⋅ L V2 g ⋅ z1 − z2 − ⋅ V2 = f ⋅ ⋅ + Kent⋅ 2 2 D 2 ( ) 2 Hence H= V 2⋅ g ⋅ ⎜⎛ f ⋅ ⎝ 1 2⋅ L D + Kent + 1⎞ and Re = V⋅ D Re = 2.23 × 10 ν 3 and V2 = V z1 − z2 = H Q = 0.0105 m Q V = 3.72 m D = 0.06⋅ m Then we have π 4 5 ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ V = 2 ⋅D f = 0.0256 2 Using Eq 4 to find H V ⎛ 2⋅ L ⋅⎜f ⋅ Hiterate = + Kent + 1⎞ 2⋅ g ⎝ D ⎠ Hence the diameter is too large: Only a head of s (4) ⎠ We must make a guess for D (larger than the other pipe) Then 2 2 Hiterate = 7.07 m Hiterate = 7.07 m But H = 10 m would be needed to generate the flow. We make D smaller s Try Then and D = 0.055 ⋅ m Re = V⋅ D Re = 2.43 × 10 ν Q V = Then we have π 4 5 ⎞ ⎛ e ⎜ D 2.51 = −2.0⋅ log ⎜ + f ⎝ 3.7 Re⋅ f ⎠ 1 V = 4.43 2 ⋅D m s f = 0.0261 2 Using Eq 4 to find H 2⋅ L Hiterate = + Kent + 1⎞ ⋅ ⎛⎜ f ⋅ 2⋅ g ⎝ D ⎠ V Hence the diameter is too small: A head of Try Then and D = 0.056 ⋅ m Re = Q V = Re = 2.39 × 10 ν But H = 10 m Hiterate = 10.97 m would be needed. We make D slightly larger Then we have V⋅ D Hiterate = 10.97 m π 4 5 ⎞ ⎛ e ⎜ 1 2.51 D = −2.0⋅ log ⎜ + f ⎝ 3.7 Re⋅ f ⎠ V = 4.27 2 ⋅D m s f = 0.0260 2 Using Eq 4 to find H V ⎛ 2⋅ L ⋅⎜f ⋅ Hiterate = + Kent + 1⎞ 2⋅ g ⎝ D ⎠ Hence the diameter is too large A head of Try Then and Hiterate = 10.02 m D = 0.05602 ⋅ m Then we have Re = V⋅ D Q π 4 5 ⎞ ⎛ e ⎜ 1 2.51 D = −2.0⋅ log ⎜ + f ⎝ 3.7 Re⋅ f ⎠ But H = 10 m would be needed. We can make D smaller V = Re = 2.39 × 10 ν Hiterate = 10.02 m V = 4.27 2 ⋅D m s f = 0.0260 2 Using Eq 4 to find H V ⎛ 2⋅ L Hiterate = + Kent + 1⎞ ⋅⎜f ⋅ 2⋅ g ⎝ D ⎠ Hiterate = 10 m Hence we have D = 0.05602 m V = 4.27 3 Check Q = 0.0105 D = 56.02 ⋅ mm m π s 4 2 ⋅ D ⋅ V = 0.0105 3 m s m s But H = 10 m Problem 8.162 Given: Hydraulic press system Find: Minimum required diameter of tubing Solution: Basic equations [Difficulty: 3] 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 = hl 2 2 ⎝ ⎠ ⎝ρ ⎠ L V2 hl = f ⋅ ⋅ D 2 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Ignore minor losses The flow rate is low and it's oil, so try assuming laminar flow. Then, from Eq. 8.13c 1 ∆p = 128⋅ μ ⋅ Q⋅ L D= or 4 π⋅ D − 2 N⋅s μ = 3.5 × 10 ⋅ ft × 2 1⋅ m 4 lbf ⋅ s 0.0209⋅ For SAE 10W oil at 100 oF (Fig. A.2, 38oC) ⎛ 128⋅ μ ⋅ Q⋅ L ⎞ ⎜ π⋅ ∆p ⎝ ⎠ 2 − 4 lbf ⋅ s μ = 7.32 × 10 N⋅s ⋅ ft 2 2 m 1 Hence 3 2 2 ⎡ 128 ft in 1⋅ ft ⎞ ⎤⎥ − 4 lbf ⋅ s ⎢ D = × 7.32 × 10 × 0.02⋅ × 165⋅ ft × × ⎛⎜ 2 s ( 3000 − 2750) ⋅ lbf ⎝ 12⋅ in ⎠ ⎥ ⎢π ft ⎣ ⎦ Check Re to assure flow is laminar From Table A.2 SG oil = 0.92 Re = 0.92 × 1.94⋅ V= Q A 4⋅ Q = 2 π⋅ D Re = so slug ft 3 V= × 15.4⋅ ft s × 0.488 ⋅ ft × 12 4 π × 0.02⋅ ft 3 s × ⎛ 12 ⋅ 1 ⎞ ⎜ 0.488 ft ⎝ ⎠ 4 D = 0.0407⋅ ft D = 0.488⋅ in 2 V = 15.4⋅ ft s SG oil⋅ ρH2O⋅ V⋅ D μ ft 2 −4 7.32 × 10 × lbf ⋅ s lbf ⋅ s 2 slug ⋅ ft Hence the flow is laminar, Re < 2300. The minimum diameter is 0.488 in, so 0.5 in ID tube should be chosen Re = 1527 Problem 8.163 Given: Flow out of reservoir by pump Find: Smallest pipe needed Solution: [Difficulty: 4] 2 2 2 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 Le V2 V2 1 L V2 2 ⎜ ρ + α ⋅ 2 + g ⋅ z1 − ⎜ ρ + α ⋅ 2 + g⋅ z2 = h lT h lT = h l + hlm = f ⋅ D ⋅ 2 + Kent ⋅ 2 + f ⋅ D ⋅ 2 ⎝ ⎠ ⎝ ⎠ Basic equations Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Vl << Hence for flow between the free surface (Point 1) and the pump inlet (2) the energy equation becomes − Solving for h 2 = p 2/ρg 2 p2 − g ⋅ z2 − ρ 2 =− 2 2 2 Le V2 V L V V − g ⋅ z2 − = f⋅ ⋅ + Kent ⋅ + f ⋅ ⋅ and ρ 2 D 2 2 D 2 p2 2 Le ⎞ ⎤ V ⎡ ⎛L h 2 = −z2 − ⋅ ⎢f ⋅ ⎜ + + Kent⎥ 2⋅ g ⎣ ⎝ D D⎠ ⎦ Kent = 0.78 From Table 8.2 V2 and we are given Q = 6⋅ (1) for rentrant, and from Table 8.4 two standard elbows lead to Le D e = 0.046 ⋅ mm (Table 8.1) ν = 1.51 × 10 We also have p = ρ⋅ g ⋅ h 2 −6 m ⋅ Q = 6 × 10 s (Table A.8) s 3 −3m L = 2 × 30 = 60 z2 = 3.5⋅ m L = ( 3.5 + 4.5) ⋅ m L = 8 m s h 2 = −6 ⋅ m Equation 1 is tricky because D is unknown, so V is unknown (even though Q is known), L/D and Le/D are unknown, and Re and hence f are unknown! We COULD set up Excel to solve Eq 1, the Reynolds number, and f, simultaneously by varying D, but here we try guesses: D = 2.5⋅ cm V = 4⋅ Q 2 π⋅ D Given ⎞ ⎛ e ⎜ 1 2.51 D = −2.0⋅ log ⎜ + f ⎝ 3.7 Re⋅ f ⎠ V = 12.2 m s Re = V⋅ D ν f = 0.0238 2 Le ⎞ ⎤ V ⎡ ⎛L ⋅ ⎢f ⋅ ⎜ + + Kent⎥ h 2 = −78.45 m h 2 = −z2 − 2⋅ g ⎣ ⎝ D D⎠ ⎦ but we need -6 m! Re = 2.02 × 10 5 D = 5 ⋅ cm V = 4⋅ Q V = 3.06 2 π⋅ D Given ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ m s V = 4⋅ Q 2 π⋅ D Given ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ V⋅ D ν Re = 1.01 × 10 5 Re = 9.92 × 10 4 f = 0.0219 2 Le ⎞ ⎤ V ⎡ ⎛L h 2 = −z2 − ⋅ ⎢f ⋅ ⎜ + + Kent⎥ h 2 = −6.16 m 2⋅ g ⎣ ⎝ D D⎠ ⎦ D = 5.1⋅ cm Re = V = 2.94 m s but we need -6 m! Re = f = 0.0219 2 Le ⎞ ⎤ V ⎡ ⎛L h 2 = −z2 − ⋅ ⎢f ⋅ ⎜ + + Kent⎥ h 2 = −5.93 m D⎠ 2⋅ g ⎣ ⎝ D ⎦ To within 1%, we can use 5-5.1 cm tubing; this corresponds to standard 2 in pipe. V⋅ D ν Problem 8.164 Given: Flow of air in rectangular duct Find: Minimum required size Solution: [Difficulty: 4] 2 2 ⎞ ⎛⎜ p ⎞ ⎛⎜ p V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 = hl 2 2 ⎝ ⎠ ⎝ρ ⎠ Basic equations 2 hl = f ⋅ 4⋅ A Dh = Pw L V ⋅ Dh 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Ignore minor losses 3 Q = 1⋅ Available data m L = 100 ⋅ m s kg ρH2O = 999 ⋅ 3 m ρ = 1.25⋅ ∆h = 25⋅ mm ar = 3 e = 0⋅ m 2 −5 m kg ν = 1.41⋅ 10 3 ⋅ (Table A.10) s m Hence for flow between the inlet (Point 1) and the exit (2) the energy equation becomes p1 ρ − p2 ρ = 2 ∆p = f⋅ ρ ∆p = ρH2O⋅ g ⋅ ∆h and For a rectangular duct ∆p = 245 Pa 2 4⋅ b⋅ h 2 ⋅ h ⋅ ar 2 ⋅ h ⋅ ar Dh = = = 2⋅ ( b + h) h ⋅ ( 1 + ar) 1 + ar 2 Hence L V ⋅ Dh 2 ∆p = ρ⋅ f ⋅ L⋅ V 2 ⋅ ( 1 + ar) 2 ⋅ h ⋅ ar 2 = ρ⋅ f ⋅ L⋅ 1 Solving for h ⎡ ρ⋅ f ⋅ L⋅ Q2 ( 1 + ar) ⎤ ⎢ ⎥ h= ⋅ ⎢ 4⋅ ∆p 3 ⎥ ar ⎣ ⎦ Q 2 2 b A = b⋅ h = h ⋅ and also ⋅ ( 1 + ar) 2⋅ A 2 ⋅ h ⋅ ar 2 = ρ⋅ f ⋅ L⋅ Q 4 ⋅ h 2 = h ⋅ ar ( 1 + ar) 1 ⋅ 5 3 h ar 5 (1) Equation 1 is tricky because h is unknown, so Dh is unknown, hence V is unknown (even though Q is known), and Re and hence f are unknown! We COULD set up Excel to solve Eq 1, the Reynolds number, and f, simmultaneously by varying h, but here we try guesses: 1 f = 0.01 ⎡ ρ⋅ f ⋅ L⋅ Q2 ( 1 + ar)⎤ ⎥ h = ⎢ ⋅ ⎢ 4 ⋅ ∆p 3 ⎥ ar ⎣ ⎦ 5 h = 0.180 m V = Q 2 h ⋅ ar V = 10.3 m s 2 ⋅ h ⋅ ar Dh = 1 + ar Given Dh = 0.270 m Re = ⎛ e ⎞ ⎜ D h 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ V⋅ Dh ν Re = 1.97 × 10 5 f = 0.0157 1 ⎡ ρ⋅ f ⋅ L⋅ Q2 ( 1 + ar)⎤ ⎥ h = ⎢ ⋅ ⎢ 4 ⋅ ∆p 3 ⎥ ar ⎣ ⎦ 2 ⋅ h ⋅ ar Dh = 1 + ar Given 5 h = 0.197 m V = Q 2 V = 8.59 m V = 8.53 m V = 8.53 m h ⋅ ar Dh = 0.295 m Re = ⎛ e ⎞ ⎜ D h 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ V⋅ Dh ν s 5 Re = 1.8 × 10 f = 0.0160 1 ⎡ ρ⋅ f ⋅ L⋅ Q2 ( 1 + ar)⎤ ⎥ h = ⎢ ⋅ ⎢ 4 ⋅ ∆p 3 ⎥ ar ⎣ ⎦ 2 ⋅ h ⋅ ar Dh = 1 + ar Given Hence 5 h = 0.198 m Dh = 0.297 m h = 198 mm Q 2 h ⋅ ar Re = ⎛ e ⎞ ⎜ D h 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ h = 0.198 m V = V⋅ Dh ν Re = 1.79 × 10 Dh = 0.297 m 5 f = 0.0160 b = 2⋅ h b = 395 ⋅ mm V = Q 2 h ⋅ ar 2 ⋅ h ⋅ ar Dh = 1 + ar s Re = V⋅ Dh In this process h and f have converged to a solution. The minimum dimensions are 198 mm by 395 mm ν s Re = 1.79 × 10 5 Problem 8.165 [Difficulty: 4] Problem 8.166 Given: Flow of air in square duct Find: Minimum required size Solution: Basic equations [Difficulty: 4] 2 2 ⎞ ⎛⎜ p ⎞ ⎛⎜ p 2 V1 V2 1 L V 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z h = f ⋅ ⋅ − = h ⎜ρ ⎜ 1 2 l l 2 2 Dh 2 ⎝ ⎠ ⎝ρ ⎠ 4⋅ A Dh = Pw Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Ignore minor losses Available data Q = 1500⋅ cfm ρH2O = 1.94⋅ L = 1000⋅ ft − 4 ft slug ft e = 0.00015 ⋅ ft ν = 1.47⋅ 10 3 ⋅ 2 s ρ = 0.00247 ⋅ (Table 8.1) slug ft ∆h = 0.75⋅ in (Table A.9) 3 Hence for flow between the inlet (Point 1) and the exit (2) the energy equation becomes p1 ρ − p2 ρ ∆p = ρ 2 = f⋅ For a square duct 4⋅ h⋅ h Dh = =h 2⋅ ( h + h) Hence ∆p = ρ⋅ f ⋅ L⋅ L V ⋅ Dh 2 2 V 2⋅ h ∆p = 3.90 lbf ft A = h⋅ h = h and also 2 = ρ⋅ f ⋅ L⋅ ∆p = ρH2O⋅ g ⋅ ∆h and 2 ∆p = 0.0271⋅ psi 2 2 Q 2 2⋅ h⋅ A = ρ⋅ f ⋅ L⋅ Q 2⋅ h 5 1 Solving for h ⎛ ρ⋅ f ⋅ L⋅ Q2 ⎞ h=⎜ ⎝ 2⋅ ∆p ⎠ 5 (1) Equation 1 is tricky because h is unknown, so Dh is unknown, hence V is unknown (even though Q is known), and Re and hence f are unknown! We COULD set up Excel to solve Eq 1, the Reynolds number, and f, simmultaneously by varying h, but here we try guesses: 1 f = 0.01 Dh = h ⎛ ρ⋅ f ⋅ L⋅ Q2 ⎞ h = ⎜ ⎝ 2 ⋅ ∆p ⎠ Dh = 1.15⋅ ft 5 h = 1.15⋅ ft V = Q h Re = V⋅ Dh ν V = 19.0⋅ 2 Re = 1.48 × 10 5 ft s ⎛ e ⎞ ⎜ Dh 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ Given f = 0.0174 1 ⎛ ρ⋅ f ⋅ L⋅ Q2 ⎞ h = ⎜ ⎝ 2 ⋅ ∆p ⎠ 5 h = 1.28⋅ ft V = Q h Dh = h Dh = 1.28⋅ ft Re = ⎛ e ⎞ ⎜ Dh 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ Given V = 15.2⋅ 2 V⋅ Dh ν ft s Re = 1.33 × 10 5 f = 0.0177 1 ⎛ ρ⋅ f ⋅ L⋅ Q2 ⎞ h = ⎜ ⎝ 2 ⋅ ∆p ⎠ Dh = h h = 1.28⋅ ft V = Re = ⎛ e ⎞ ⎜ D h 1 2.51 + = −2.0⋅ log ⎜ f ⎝ 3.7 Re⋅ f ⎠ h = 1.28⋅ ft Q h Dh = 1.28⋅ ft Given Hence 5 Dh = h Dh = 1.28⋅ ft V⋅ Dh ν Re = 1.32 × 10 2 V⋅ Dh ν ft s Re = 1.32 × 10 f = 0.0177 V = Q h Re = V = 15.1⋅ 2 V = 15.1⋅ ft s 5 In this process h and f have converged to a solution. The minimum dimensions are 1.28 ft square (15.4 in square) 5 Problem 8.167 Given: Flow in a tube Find: Effect of diameter; Plot flow rate versus diameter [Difficulty: 3] Solution: Basic equations: 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 1 2 2 2 = hl ⎝ ⎠ ⎝ρ ⎠ Re = f = ρ⋅ V⋅ D 64 μ (8.29) 2 hl = f ⋅ L V ⋅ D 2 (8.36) (8.34) (Laminar) Re ⎛ e ⎞ ⎜ D 1 2.51 (8.37) + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ (Turbulent) The energy equation (Eq. 8.29) becomes for flow in a tube 2 L V p 1 − p 2 = ∆p = ρ⋅ f ⋅ ⋅ D 2 This cannot be solved explicitly for velocity V (and hence flow rate Q), because f depends on V; solution for a given diameter D requires iteration (or use of Solver) Flow Rate versus Tube Diameter for Fixed Dp 0.8 0.6 3 Q (m /s) x 10 4 Laminar Turbulent 0.4 0.2 0.0 0.0 2.5 5.0 D (mm) 7.5 10.0 Problem 8.169 [Difficulty: 3] Part 1/2 Problem 8.169 [Difficulty: 3] Part 2/2 Problem 8.170 [Difficulty: 3] Problem 8.171 [Difficulty: 4] Problem 8.172 Problem 8.151 [Difficulty: 5] Part 1/2 Problem 8.172 [Difficulty: 5] Part 2/2 Problem 8.173 Given: Flow through water pump Find: Power required Solution: Basic equations [Difficulty: 2] 2 2 ⎞ ⎜⎛ p ⎞ Vs ⎜⎛ p d Vd s h pump = ⎜ + + g ⋅ zd − ⎜ + + g ⋅ zs 2 2 ⎝ρ ⎠ ⎝ρ ⎠ V= Q A = 4⋅ Q 2 π⋅ D Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow 3 Hence for the inlet 4 lbm 1 ⋅ slug ft Vs = × 25⋅ × × × π s 32.2⋅ lbm 1.94⋅ slug For the outlet 4 lbm 1 ⋅ slug ft Vd = × 25⋅ × × × π s 32.2⋅ lbm 1.94⋅ slug 3 Then h pump = pd − ps ρ 2 + ⎛ 12 ⋅ 1 ⎞ ⎜ 3 ft ⎝ ⎠ 2 ⎛ 12 ⋅ 1 ⎞ ⎜ 2 ft ⎝ ⎠ 2 ft Vs = 8.15⋅ s p s = −2.5⋅ psi ft Vd = 18.3⋅ s p d = 50⋅ psi 2 Vd − Vs Wpump = mpump⋅ h pump and 2 2 2 ⎛⎜ p − p Vd − Vs ⎞ d s Wpump = mpump⋅ ⎜ + 2 ⎝ ρ ⎠ Note that the software cannot render a dot, so the power is Wpump and mass flow rate is mpump! ⎡ lbm 1 ⋅ slug lbf Wpump = 25⋅ × × ⎢( 50 − −2.5) ⋅ × s 32.2⋅ lbm ⎢ 2 in ⎣ Wpump = 5.69⋅ hp For an efficiency of η = 70⋅ % 2 2 3 2 1 lbf ⋅ s ⎤⎥ 1 ⋅ hp 2 2 ⎛ ft ⎞ ⎛ 12⋅ in ⎞ × ft + × 18.3 − 8.15 × × ⋅ ⎜ 1 ⋅ ft ⎜s 2 ⎥ 1.94 ⋅ slug slug ⋅ ft ft⋅ lbf ⎝ ⎠ ⎝ ⎠ ⎦ 550 ⋅ ( ) s Wrequired = Wpump η Wrequired = 8.13⋅ hp Problem 8.174 Given: Flow through water pump Find: Power required Solution: Basic equations [Difficulty: 1] 2 2 ⎞ ⎜⎛ p ⎞ Vs ⎜⎛ p d Vd s h pump = ⎜ + + g ⋅ zd − ⎜ + + g ⋅ zs 2 2 ⎝ρ ⎠ ⎝ρ ⎠ Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow In this case we assume Ds = Dd The available data is ∆p = 35⋅ psi Then h pump = Vs = Vd so Q = 500 ⋅ gpm pd − ps ρ = ∆p η = 80⋅ % Wpump = mpump⋅ h pump and ρ ∆p ∆p Wpump = mpump⋅ = ρ⋅ Q⋅ ρ ρ Wpump = Q⋅ ∆p Wpump = 5615 ft⋅ lbf s Wpump = 10.2⋅ hp Note that the software cannot render a dot, so the power is Wpump and mass flow rate is mpump! For an efficiency of η = 80 % Wrequired = Wpump η Wrequired = 12.8⋅ hp Problem 8.175 Given: Flow in pipeline with pump Find: Pump pressure Δp Solution: Basic equations [Difficulty: 3] 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ ∆p V1 V2 1 pump 2 = h lT ⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 + ρ ⎝ ⎠ ⎝ ⎠ 2 hl = f ⋅ L V ⋅ D 2 h lm = f ⋅ Le V2 ⋅ D 2 2 h lm = K⋅ V 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 Available data From Section 8.8 l L = 50⋅ m D = 125 ⋅ mm Q = 50⋅ e = 0.15⋅ mm p 1 = 150 ⋅ kPa p 2 = 0 ⋅ kPa z1 = 15⋅ m z2 = 30⋅ m Kent = 0.5 Lelbow90 = 30⋅ D Lelbow90 = 3.75 m LGV = 8 ⋅ D LAV = 150 ⋅ D LAV = 18.75 m ρ = 1000 s − 3 N⋅ s kg μ = 1.3⋅ 10 3 m Hence and V = Q V = 4.07 ⎛ π⋅ D ⎜ ⎝ 4 ⎠ 2⎞ m Re = s ⋅ (Table A.8) 2 m ρ⋅ V⋅ D 5 Re = 3.918 × 10 μ ⎛ e ⎞ ⎜ D 1 2.51 + = −2 ⋅ log⎜ 3.7 f Re⋅ f ⎠ ⎝ Given LGV = 1 m f = 0.0212 The loss is then Lelbow90 LGV LAV ⎞ ⎛ L + 7⋅ f ⋅ + 5⋅ f ⋅ + f⋅ + Kent 2 ⎝ D D D D ⎠ 2 h lT = V The energy equation becomes ⋅⎜f ⋅ p1 − p2 ρ 2 h lT = 145 m 2 s 2 ∆ppump V + g ⋅ z1 − z2 − + = h lT 2 ρ ( ( ) ) 2 ( V ∆ppump = ρ⋅ h lT + ρ⋅ g ⋅ z2 − z1 + ρ⋅ + p2 − p1 2 ) ∆ppump = 150 ⋅ kPa Problem 8.176 [Difficulty: 3] Problem 8.177 [Difficulty: 3] Problem 8.178 Given: Flow in air conditioning system Find: Pressure drop; cost [Difficulty: 3] Solution: Basic equations 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 = hl 2 2 ⎝ ⎠ ⎝ρ ⎠ 2 hl = f ⋅ L V ⋅ D 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 3 Available data L = 5 ⋅ km ρ = 1000 D = 0.75⋅ m kg 3 − 3 N⋅ s μ = 1.3⋅ 10 m V = Then so Q ⎛ π⋅ D ⎜ ⎝ 4 ⎠ Given The energy equation becomes 2⎞ e = 0.046 ⋅ mm ⋅ Q = 0.65⋅ m s (Table A.8) 2 m V = 1.47 m Re = s ⎛ e ⎞ ⎜ D 1 2.51 + = −2 ⋅ log⎜ f ⎝ 3.7 Re⋅ f ⎠ ∆p = f ⋅ L D ρ⋅ V⋅ D μ f = 0.0131 2 ⋅ ρ⋅ V 2 and Wpump = Q⋅ ∆p The required power is Power = The daily cost is then C = cost⋅ Power⋅ day Wpump ηp ⋅ ηm ∆p = 94.4⋅ kPa Wpump = 61.3⋅ kW Power = 84.9⋅ kW C = 285 dollars ηp = 85⋅ % cost = ηm = 85⋅ % 0.14 (dollars) kW⋅ hr Re = 8.49 × 10 5 Problem 8.179 [Difficulty: 4] Given: Fire nozzle/pump system Find: Design flow rate; nozzle exit velocity; pump power needed Solution: Basic equations 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V2 V3 2 3 ⎜ ρ + α⋅ 2 + g⋅ z2 − ⎜ ρ + α⋅ 2 + g⋅ z3 = h l ⎝ ⎠ ⎝ ⎠ L V2 hl = f ⋅ ⋅ D 2 2 for the hose Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 2 and 3 is approximately 1 4) No minor loss p3 ρ + V3 2 p4 + g ⋅ z3 = + ρ 2 V4 2 2 + g ⋅ z4 for the nozzle (assuming Bernoulli applies) 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V2 V1 2 1 ⎜ ρ + α⋅ 2 + g⋅ z2 − ⎜ ρ + α⋅ 2 + g⋅ z1 = h pump ⎝ ⎠ ⎝ ⎠ for the pump Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) No minor loss Hence for the hose ∆p ρ = p2 − p3 ρ 2 L V = f⋅ ⋅ D 2 or V= 2 ⋅ ∆p⋅ D ρ⋅ f ⋅ L We need to iterate to solve this for V because f is unknown until Re is known. This can be done using Excel's Solver, but here: ∆p = 750 ⋅ kPa L = 100 ⋅ m e = 0 D = 3.5⋅ cm ρ = 1000⋅ kg ν = 1.01 × 10 3 2 −6 m ⋅ m Make a guess for f Given f = 0.01 2 ⋅ ∆p⋅ D ρ⋅ f ⋅ L ⎞ ⎛ e ⎜ 1 2.51 D = −2.0⋅ log ⎜ + 3.7 f Re⋅ f ⎠ ⎝ V = Given V = 2 ⋅ ∆p⋅ D ρ⋅ f ⋅ L V = 5.92 m s ⎛ e ⎞ ⎜ 1 D 2.51 = −2.0⋅ log ⎜ + 3.7 f Re⋅ f ⎠ ⎝ V = 7.25 m s Re = V⋅ D Re = 2.51 × 10 ν f = 0.0150 Re = V⋅ D ν f = 0.0156 Re = 2.05 × 10 5 5 s 2 ⋅ ∆p⋅ D V = V = 5.81 ρ⋅ f ⋅ L ⎛ e ⎞ ⎜ D 1 2.51 + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ Given V = 5.80 ρ⋅ f ⋅ L 2 Q = V⋅ A = p3 For the nozzle ρ + V3 V4 = 2 π⋅ D 4 ⋅V p4 + g ⋅ z3 = + ρ 3 N 2 × 700 × 10 ⋅ 2 m V4 2 m Re = s 2 × ( 0.035 ⋅ m) × 5.80⋅ 4 p 1 = 350 ⋅ kPa Ppump = ρ⋅ Q⋅ (p2 − p1) ρ Ppump η ( = Q⋅ p 2 − p 1 ) ν Re = 2.01 × 10 5 V⋅ D Re = 2.01 × 10 5 ν 3 −3m m Q = 5.58 × 10 s + g ⋅ z4 m 1000⋅ kg × so kg⋅ m 2 s ⋅N + ⎛⎜ 5.80⋅ ⎝ m⎞ p 2 = 700 ⋅ kPa + 750 ⋅ kPa The pump power is Ppump = mpump⋅ h pump V⋅ D s 2 3 Q = 0.335 ⋅ s ( 2 3 × s 2⋅ p3 − p4 V4 = ρ ) m min + V3 2 m V4 = 37.9 s ⎠ 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V2 V1 2 1 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 2 1 = h pump 2 2 ⎝ ⎠ ⎝ρ ⎠ For the pump Prequired = π Q = 2 Re = f = 0.0156 2 ⋅ ∆p⋅ D V = m so h pump = p2 − p1 ρ p 2 = 1450⋅ kPa P pump and mpump are pump power and mass flow rate (software can't do a dot!) Ppump = 5.58 × 10 Prequired = 3 −3 m 6.14⋅ kW 70⋅ % ⋅ s 3 N × ( 1450 − 350 ) × 10 ⋅ 2 Ppump = 6.14⋅ kW m Prequired = 8.77⋅ kW Problem 8.180 [Difficulty: 4] Part 1/2 Problem 8.180 [Difficulty: 4] Part 2/2 Problem 8.181 Given: Flow in water fountain Find: Daily cost [Difficulty: 2] Solution: Basic equations Wpump Q ∆p ∆p ρ g H Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 3 Available data m Q 0.075 ρ 999 s kg 3 m Hence H 10 m Cost ηp 85 % 0.14 kW hr (dollars) ∆p ρ g H ∆p 98 kPa Wpump Q ∆p Wpump 7.35 kW Power Wpump ηp ηm C Cost Power day Power 10.2 kW C 34.17 (dollars) ηm 85 % Problem 8.182 [Difficulty: 3] Problem 8.183 Given: Flow in a pump testing system Find: Flow rate; Pressure difference; Power [Difficulty: 4] Solution: Governing equations: 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 1 2 2 2 = h lT = ⎝ ⎠ ⎝ρ ⎠ Re = f = 2 ρ⋅ V⋅ D hl = f ⋅ μ 64 L V ⋅ D 2 (8.36) h lm = f ⋅ (8.34) ∑ major Le V2 ⋅ D 2 hl + ∑ h lm (8.29) minor (8.40b) ⎛ e ⎞ ⎜ D 1 2.51 (8.37) + = −2.0⋅ log ⎜ 3.7 f Re ⋅ f ⎝ ⎠ (Laminar) Re (Turbulent) The energy equation (Eq. 8.29) becomes for the circuit (1 = pump outlet, 2 = pump inlet) p1 − p2 ρ 2 = f⋅ 2 or ∆p = ρ⋅ f ⋅ V 2 2 2 L V V V ⋅ + 4 ⋅ f ⋅ Lelbow⋅ + f ⋅ Lvalve⋅ D 2 2 2 ⎛L ⋅⎜ ⎝D + 4⋅ Lelbow D + Lvalve ⎞ D ⎠ (1) This must be matched to the pump characteristic equation; at steady state, the pressure generated by the pump just equals that lost to friction in the circuit 4 2 ∆p = 750 − 15 × 10 ⋅ Q (2) Finally, the power supplied to the pump, efficiency η, is Power = In Excel: Q⋅ ∆p η (3) Circuit and Pump Pressure Heads 1200 Dp (kPa) 1000 800 600 Circuit 400 Pump 200 0 0.00 0.01 0.02 0.03 0.04 3 Q (m /s) 0.05 0.06 0.07 Problem 8.184 Equations Given: Pump/pipe system Find: Flow rate, pressure drop, and power supplied; Effect of roughness [Difficulty: 4] Solution: Re = f = 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 1 2 2 2 = h lT − ∆hpump ⎝ ⎠ ⎝ρ ⎠ ⎛ e ⎞ ⎜ D 1 2.51 (Laminar) (Turbulent) + = −2.0⋅ log ⎜ 3.7 f Re ⋅ f ⎝ ⎠ ρ⋅ V⋅ D μ 64 Re 2 h lT = f ⋅ L V ⋅ D 2 The energy equation becomes for the system (1 = pipe inlet, 2 = pipe outlet) 2 ∆hpump = f ⋅ 2 L V ⋅ D 2 ∆ppump = ρ⋅ f ⋅ or L V ⋅ D 2 (1) This must be matched to the pump characteristic equation; at steady state, the pressure generated by the pump just equals that lost to friction in the circuit 2 ∆ppump = 145 − 0.1⋅ Q (2) Finally, the power supplied to the pump, efficiency η, Power = In Excel: Q⋅ ∆p η (3) is Pum p and Pipe Pressure Heads Pipe (e = 0.5 in) 160 Pipe (e = 0.25 in) Pump Dp (psi) 120 80 40 0 10 15 20 3 Q (ft /s) 25 30 Problem 8.185 Given: Fan/duct system Find: Flow rate [Difficulty: 3] Solution: 2 2 ⎛⎜ p ⎞ ⎜⎛ p ⎞ V1 V2 1 2 ⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h lT − ∆hfan ⎝ ⎠ ⎝ ⎠ 2 h lT = f ⋅ 2 L V V ⋅ + K⋅ Dh 2 2 The energy equation becomes for the system (1 = duct inlet, 2 = duct outlet) 2 ∆hfan = f ⋅ 2 or ∆ppump = 2 L V V ⋅ + K⋅ Dh 2 2 ρ⋅ V 2 ⋅ ⎛⎜ f ⋅ L ⎝ Dh 2 + K⎞ (1) ⎠ where 4⋅ A 4⋅ h Dh = = =h Pw 4⋅ h This must be matched to the fan characteristic equation; at steady state, the pressure generated by the fan just equals that lost to friction in the circuit 2 ∆pfan = 1020 − 25⋅ Q − 30⋅ Q In Excel: (2) Fan and Duct Pressure Heads 2500 Dp (Pa) 2000 1500 Duct 1000 Fan 500 0 0.0 0.5 1.0 1.5 3 Q (m /s) 2.0 2.5 3.0 Problem 8.186 [Difficulty: 4] Part 1/2 Problem 8.186 [Difficulty: 4] Part 2/2 Problem 8.187 Given: Pipe system Find: Flow in each branch Solution: Governing equations: [Difficulty: 5] 2 2 p p V1 V2 1 2 ρ α1 2 g z1 ρ α2 2 g z2 h l f 64 (Laminar) Re (8.36) 2 (8.29) e D 1 2.51 2.0 log 0.5 3.7 0.5 f Re f h lT f L V D 2 (Turbulent) (8.37) 2 The energy equation (Eq. 8.29) can be simplified to ∆p ρ f L V D 2 This can be written for each pipe section In addition we have the following contraints Q0 Q1 Q4 (1) ∆p ∆p0 ∆p1 (3) ∆p2 ∆p3 (5) Q4 Q2 Q3 ∆p ∆p0 ∆p4 ∆p2 We have 5 unknown flow rates (or, equivalently, velocities) and five equations In Excel: (8.34) (2) (4) Problem 8.188 Given: Pipe system Find: Flow in each branch if pipe 3 is blocked Solution: Governing equations: [Difficulty: 5] 2 2 p p V1 V2 1 2 ρ α1 2 g z1 ρ α2 2 g z2 h l f 64 (Laminar) Re (8.36) 2 (8.29) e D 1 2.51 2.0 log 0.5 3.7 0.5 f Re f h lT f L V D 2 (Turbulent) (8.37) 2 The energy equation (Eq. 8.29) can be simplified to ∆p ρ f L V D 2 This can be written for each pipe section In addition we have the following contraints Q0 Q1 Q4 (1) ∆p ∆p0 ∆p1 (3) Q4 Q2 ∆p ∆p0 ∆p4 ∆p2 We have 4 unknown flow rates (or, equivalently, velocities) and four equations In Excel: (8.34) (2) (4) Problem 8.189 Given: Water pipe system Find: Flow rates [Difficulty: 5] Solution: Basic equations 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 + α ⋅ + g ⋅ z + α ⋅ + g ⋅ z − ⎜ρ ⎜ 1 2 1 2 2 2 = hl ⎝ ⎠ ⎝ρ ⎠ f = 64 2 h lT = f ⋅ L V ⋅ D 2 ⎛ e ⎞ ⎜ D 1 2.51 (Turbulent) + = −2.0⋅ log ⎜ 3.7 f Re⋅ f ⎠ ⎝ (Laminar) Re 2 The energy equation can be simplified to ∆p = ρ⋅ f ⋅ L V ⋅ D 2 This can be written for each pipe section 2 Pipe A (first section) LA VA ∆pA = ρ⋅ fA ⋅ ⋅ DA 2 Pipe B (1.5 in branch) LB VB ∆pB = ρ⋅ fB⋅ ⋅ DB 2 Pipe C (1 in branch) LC VC ∆pC = ρ⋅ fC⋅ ⋅ DC 2 Pipe D (last section) LD VD ∆pD = ρ⋅ fD⋅ ⋅ DD 2 (4) QA = QD (5) QA = QB + QC (6) ∆p = ∆pA + ∆pB + ∆pD (7) ∆pB = ∆pC (8) (1) 2 (2) 2 (3) 2 In addition we have the following contraints We have 4 unknown flow rates (or velocities) and four equations (5 - 8); Eqs 1 - 4 relate pressure drops to flow rates (velocities) In Excel: Problem 8.190 [Difficulty: 4] Problem 8.191 [Difficulty: 5] Part 1/2 Problem 8.191 [Difficulty: 5] Part 2/2 Problem 8.192 [Difficulty: 2] Problem 8.193 Given: Flow through an orifice Find: Pressure drop [Difficulty: 2] Solution: ( ) Basic equation mactual = K⋅ At⋅ 2 ⋅ ρ⋅ p 1 − p 2 = K⋅ At⋅ 2 ⋅ ρ⋅ ∆p For the flow coefficient K = K⎜ ReD1 , At 65oC,(Table A.8) ⎞ ⎛ Dt ⎝ D1 ⎠ ρ = 980 ⋅ kg ν = 4.40 × 10 3 2 −7 m ⋅ m V= Q V = A ReD1 = β= Note that mactual is mass flow rate (the software cannot render a dot!) V⋅ D 4 π × s 1 ( 0.15⋅ m) 2 × 20⋅ L s m ReD1 = 1.13⋅ × 0.15⋅ m × s ν Dt β = D1 3 × 0.001 ⋅ m V = 1.13 1⋅ L s −7 4.40 × 10 2 ⋅m 75 K = 0.624 Then 2 2 ⎛ mactual ⎞ 1 ρ⋅ Q ⎞ 1 ρ Q ⎞ ∆p = ⎜ ⋅ = ⎛⎜ ⋅ = ⋅ ⎛⎜ ⎝ K⋅ At ⎠ 2⋅ ρ ⎝ K⋅ At ⎠ 2⋅ ρ 2 ⎝ K⋅ At ⎠ s ReD1 = 3.85 × 10 β = 0.5 150 From Fig. 8.20 m 2 ⎡ L 0.001 ⋅ m3 ⎤ 1 4 1 ⎥ ∆p = × 980 ⋅ × × × × ⎢20⋅ × 2 3 ⎢ s 1⋅ L 0.624 π 2⎥ m ( 0.075 ⋅ m) ⎦ ⎣ 1 kg 2 ∆p = 25.8⋅ kPa 5 Problem 8.194 Given: Reservoir-pipe system Find: Orifice plate pressure difference; Flow rate [Difficulty: 3] Solution: Basic equations: 2 2 ⎛⎜ p ⎞ ⎛⎜ p ⎞ V1 V2 1 2 ⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h lT = hl + Σh lm(8.29) ⎝ ⎠ ⎝ ⎠ 2 hl = f ⋅ f = 64 L V ⋅ D 2 2 (8.34) (Laminar) h lm = K⋅ (8.36) Re V (8.40a) 2 e ⎞ ⎜⎛ D 1 2.51 ⎟ + = −2.0⋅ log ⎜ 0.5 3.7 0.5 ⎜ f Re⋅ f ⎠ ⎝ (Turbulent) (8.37) 2 There are three minor losses: at the entrance; at the orifice plate; at the exit. For each 2 The energy equation (Eq. 8.29) becomes (α = 1) g ⋅ ∆H = V 2 ⋅ ⎛⎜ f ⋅ L ⎝ D h lm = K⋅ + Kent + Korifice + Kexit⎞ ⎠ V 2 (1) (ΔH is the difference in reservoir heights) This cannot be solved for V (and hence Q) because f depends on V; we can solve by manually iterating, or by using Solver The tricky part to this problem is that the orifice loss coefficient Korifice is given in Fig. 8.23 as a percentage of pressure differential ∆p across the orifice, which is unknown until V is known! The mass flow rate is given by mrate = K⋅ At⋅ 2 ⋅ ρ⋅ ∆p (2) where K is the orifice flow coefficient, At is the orifice area, and Δp is the pressure drop across the orifice Equations 1 and 2 form a set for solving for TWO unknowns: the pressure drop Δp across the orifice (leading to a value for Korifice) and the velocity V. The easiest way to do this is by using Solver In Excel: Problem 8.195 [Difficulty: 2] Given: Flow through a venturi meter (NOTE: Throat is obviously 3 in not 30 in!) Find: Flow rate Solution: Basic equation C⋅ At mactual = 4 ( C⋅ At ) ⋅ 2 ⋅ ρ⋅ p 1 − p 2 = ⋅ 2 ⋅ ρ⋅ ∆p Note that mactual is mass flow rate (the software cannot render a dot!) 4 1−β 1−β For ReD1 > 2 x 105, 0.980 < C < 0.995. Assume C = 0.99, then check Re β= Also Then Dt β = D1 β = 0.5 6 ∆p = ρHg⋅ g ⋅ ∆h = SG Hg⋅ ρ⋅ g ⋅ ∆h Q= mactual ρ V= C⋅ At = Q 4⋅ Q ReD1 = ⋅ 2 ⋅ ρ⋅ SGHg⋅ ρ⋅ g ⋅ ∆h = 4 ⋅ ρ⋅ 1 − β π⋅ C⋅ Dt π⋅ D1 ⎛ 1 ⋅ ft⎞ × 2 × 13.6 × 32.2⋅ ft × 1⋅ ft ⎜4 2 ⎝ ⎠ s V = 2 − 6 ft ⋅ 4 π 1 × ⎛ 1 ⋅ ft⎞ ⎜2 ⎝ ⎠ 2 2 4 ⋅ 2 ⋅ SGHg⋅ g ⋅ ∆h 4⋅ 1 − β 2 4 At 75oF,(Table A.7) ν = 9.96 × 10 2 4 × 0.99 × 1 − 0.5 A ⋅ 2 ⋅ ρ⋅ ∆p = ρ⋅ 1 − β 4× = π⋅ C⋅ Dt 4 π Q = Hence 3 × 1.49⋅ ft Q = 1.49⋅ ft V = 7.59⋅ ft 3 s 3 s s 2 s V⋅ D1 ν Thus ReD1 > 2 x 105. The volume flow rate is ft 1 ReD1 = 7.59⋅ × ⋅ ft × s 2 Q = 1.49⋅ ft 3 s s −6 9.96 × 10 ⋅ ft 2 ReD1 = 3.81 × 10 5 Problem 8.196 Given: Flow through an venturi meter Find: Flow rate [Difficulty: 2] Solution: Basic equation C At mactual 4 2 ρ p 1 p 2 C At 2 ρ ∆p 4 1 β 1 β Note that mactual is mass flow rate (the software cannot render a dot!) For ReD1 > 2 x 105, 0.980 < C < 0.995. Assume C = 0.99, then check Re Available data D1 2 in Dt 1 in ∆p 25 psi ρ 1.94 slug ft β Then Q Q Dt β 0.5 D1 mactual ρ C At At 68oF(Table A.7) V C 0.99 and assume 2 ρ ∆p 4 ρ 1 β π C Dt 2 2 ∆p 4 1 β ρ Q V 4 Hence 3 A Q 0.340 4 Q π D1 5 ft ν 1.08 10 2 V 15.6 2 s Thus ReD1 > 2 x 105. The volume flow rate is ReD1 ft 3 s Q 152 gpm ft s V D1 ν Q 152 gpm 5 ReD1 2.403 10 Problem 8.197 [Difficulty: 2] Problem 8.198 [Difficulty: 3] Given: Flow through a venturi meter Find: Maximum flow rate for incompressible flow; Pressure reading Solution: Basic equation C⋅ At mactual = 4 ( C⋅ At ) ⋅ 2 ⋅ ρ⋅ p 1 − p 2 = ⋅ 2 ⋅ ρ⋅ ∆p Note that mactual is mass flow rate (the software cannot render a dot!) 4 1−β 1−β Assumptions: 1) Neglect density change 2) Use ideal gas equation for density ρ= Then p ρ = 60⋅ Rair⋅ T lbf 2 2 × in 1 − 3 slug ⎛ 12⋅ in ⎞ × lbm⋅ R × 1 ⋅ slug ⋅ ρ = 9.53 × 10 ⋅ ⎜ 1⋅ ft 32.2⋅ lbm ( 68 + 460 ) ⋅ R 3 53.33 ⋅ ft⋅ lbf ⎝ ⎠ ft For incompressible flow V must be less than about 100 m/s or 330 ft/s at the throat. Hence mactual = ρ⋅ V2 ⋅ A2 mactual = 9.53 × 10 − 3 slug ft β= Dt β = D1 3 3 ∆p = ρHg⋅ g ⋅ ∆h ∆h = and in addition ⎛ mactual ⎞ 4 ∆p = ⋅⎜ ⋅ 1−β 2⋅ ρ ⎝ C⋅ At ⎠ 2 ( × s π 4 × ⎛ 1 ⋅ ft⎞ ⎜4 ⎝ ⎠ 2 slug mactual = 0.154 ⋅ s β = 0.5 6 Also 1 ft × 330 ⋅ ) ∆h = so ∆p ρHg⋅ g (1 − β4) ⋅⎛ mactual ⎞ 2 2 ⋅ ρ⋅ ρHg⋅ g ⎜ C⋅ A t ⎠ ⎝ For ReD1 > 2 x 105, 0.980 < C < 0.995. Assume C = 0.99, then check Re ∆h = (1 − 0.54) × 2 Hence At 68oF,(Table A.7) 2 ⎡ slug 4 ⎛ 4 ⎞⎤ 1 ⎥ × × × × × ⎢0.154 ×⎜ 13.6⋅ 1.94⋅ slug 32.2⋅ ft ⎣ s −3 0.99 π ⎝ 1 ⋅ ft ⎠ ⎦ 9.53 × 10 slug ft V= 3 Q A ft = 4 ⋅ mactual π⋅ ρ⋅ D1 ν = 1.08 × 10 ReD1 = 2 − 5 ft ⋅ 3 2 2 s V = 4 π ft × 3 9.53 × 10 −3 1 × ⎛ 1 ⋅ ft⎞ ⎜2 ⎝ ⎠ slug 2 × 0.154 slug s ∆h = 6.98⋅ in V = 82.3⋅ 2 s V⋅ D1 ν Thus ReD1 > 2 x 105. The mass flow rate is ft 1 ReD1 = 82.3⋅ × ⋅ ft × s 2 slug mactual = 0.154 ⋅ s s −5 1.08 × 10 ⋅ ft 2 and pressure ReD1 = 3.81 × 10 ∆h = 6.98⋅ in 6 Hg ft s Problem 8.199 [Difficulty: 3] Problem 8.200 Given: Flow through venturi Find: Maximum flow rate before cavitation [Difficulty: 3] Solution: Basic equation C At mactual 4 C At 2 ρ p 1 p 2 2 ρ ∆p 4 1β 1β Note that mactual is mass flow rate (the software cannot render a dot!) For ReD1 > 2 x 105, 0.980 < C < 0.995. Assume C = 0.99, then check Re Available data D1 100 mm Dt 50 mm p 1g 200 kPa C 0.99 p atm 101 kPa p v 1.23 kPa Steam tables - saturation pressure at 10oC ρ 1000 kg ν 1.3 10 3 2 6 m At β π Dt 2 2 At 1963 mm 4 Dt A1 Hence the largest Δp is ∆p p 1 p t Then mrate 4 pt pv 2 A1 7854 mm p t 1.23 kPa ∆p 300 kPa C At kg mrate 49.2 s 2 ρ ∆p 4 1β mrate ρ Q V1 A1 Re1 2 p 1 301 kPa The smallest allowable throat pressure is the saturation pressure Check the Re π D1 β 0.5 D1 p 1 p atm p 1g Q (Table A.8) s m Then 3 Q 0.0492 m s m V1 6.26 s V1 D1 ν 5 Re1 4.81 10 3 Thus ReD1 > 2 x 105. The volume flow rate is Q 0.0492 m s (Asumption - verify later) Q 49.2 L s Problem 8.201 [Difficulty: 1] V 1, A 1 Given: Flow through a diffuser Find: Derivation of Eq. 8.42 Solution: Basic equations Cp = p2 − p1 p1 1 ρ 2 ⋅ ρ⋅ V1 2 + V1 2 2 p2 + g ⋅ z1 = + ρ V2 2 V 2, A 2 2 + g ⋅ z2 Q = V⋅ A Assumptions: 1) All the assumptions of the Bernoulli equation 2) Horizontal flow 3) No flow separation From Bernoulli p2 − p1 ρ 2 2 2 2 ⎛ A1 ⎞ = − ⋅⎜ = − 2 2 2 2 ⎝ A2 ⎠ V1 V2 V1 V1 2 using continuity Hence ⎡ 2 V 2 A ⎞ 2⎤ 1 ⎛ 1 ⎥ ⎢ V1 Cp = = ⋅ − ⋅⎜ ⎢ ⎥=1− 2 2 1 1 2 2 ⎝ A2 ⎠ ⎦ ⋅ ρ⋅ V1 ⋅ V1 ⎣ 2 2 Finally Cp = 1 − p2 − p1 1 1 ⎛ A1 ⎞ ⎜A ⎝ 2⎠ which is Eq. 8.42. 2 AR This result is not realistic as a real diffuser is very likely to have flow separation 2 Problem 8.202 [Difficulty: 4] Part 1/2 Problem 8.202 [Difficulty: 4] Part 2/2 Problem 8.203 [Difficulty: 3] Problem 8.204 [Difficulty: 5] Part 1/2 Problem 8.204 [Difficulty: 5] Part 2/2 Problem 8.205 [Difficulty: 5] Part 1/2 Problem 8.205 [Difficulty: 5] Part 2/2 Problem 9.1 Given: Minivan traveling at various speeds Find: Plot of boundary layer length as function of speed [Difficulty: 2] Solution: Governing equations: The critical Reynolds number for transition to turbulence is VL crit/ =500000 Re crit = The critical length is then L crit = 500000/V Tabulated or graphical data: = 3.79E-07 = lbf.s/ft 2 3 slug/ft 0.00234 o (Table A.9, 68 F) Computed results: V (mph) L crit (ft) 5.52 4.42 3.68 3.16 2.76 1.84 1.38 1.10 0.920 0.789 0.690 0.614 10 13 15 18 20 30 40 50 60 70 80 90 Length of Laminar Boundary Layer on the Roof of a Minivan 6 5 4 L crit (ft) 3 2 1 0 0 10 20 30 40 50 V (mph) 60 70 80 90 100 Problem 9.2 Given: Model of riverboat Find: Distance at which transition occurs [Difficulty: 2] Solution: Basic equation For water at 10oC Hence For the model Rex ρ U x μ ν 1.30 10 xp xm ν Rex U xp 18 U x 2 6 m 5 and transition occurs at about Rex 5 10 (Table A.8) and we are given ν s x p 0.186 m x p 18.6 cm x m 0.0103 m x m 10.3 mm U 3.5 m s Problem 9.3 [Difficulty: 3] Given: Boeing 757 Find: Point at which BL transition occurs during takeoff and at cruise Solution: Basic equation For air at 68oF Hence At 33,000 ft ρ U x Rex μ ν 1.62 10 xp U x 4 ft ν Rex U T 401.9 R 5 and transition occurs at about Rex 5 10 (Table A.9) and we are given ν 2 U 160 s x p 0.345 ft x p 4.14 in (Intepolating from Table A.3) T 57.8 °F mi hr 234.7 ft s We need to estimate Ȟ or ȝ at this temperature. From Appendix A-3 b T μ S 1 Hence μ 6 b 1.458 10 kg 1 T b T 1 S m s K 5 N s μ 1.458 10 2 S 110.4 K 2 7 lbf s μ 3.045 10 m ft 2 T For air at 10,000 m (Table A.3) ρ ρSL ν Hence xp 0.3376 ρSL 0.002377 slug ft 4 ft μ ν 3.79 10 ρ ν Rex U x p 0.244 ft 3 ρ 0.3376 ρSL ρ 8.025 10 4 slug ft 2 and we are given s x p 2.93 in U 530 mi hr 3 Problem 9.4 [Difficulty: 2] Given: Experiment with 1 cm diameter sphere in SAE 10 oil Find: Reasonableness of two flow extremes Solution: Basic equation For SAE 10 For For ReD ρ U D μ ReD 2.5 10 (Fig. A.3 at 20 oC) and s ReD 1 we find 5 ReD 2.5 10 For water ν 1.01 10 5 ν ReD D ν ReD D D 1 cm U 0.011 U 2750 m s m U 1.10 cm which is reasonable s which is much too high! s we need to increase the sphere diameter D by a factor of about 1000, or reduce the viscosity ν by the same factor, or some combination of these. One possible solution is 2 6 m ReD 2.5 10 U U 5 ReD 2.5 10 5 and transition occurs at about ν 2 4 m ν 1.1 10 Note that for For U D (Table A.8 at 20 oC) s we find U D 10 cm and ν ReD D Hence one solution is to use a 10 cm diameter sphere in a water tank. U 2.52 m s which is reasonable Problem 9.5 [Difficulty: 2] Given: Flow around American and British golf balls, and soccer ball Find: Speed at which boundary layer becomes turbulent Solution: Basic equation For air ReD ρ U D μ ν 1.62 10 For the American golf ball D 1.68 in For the British golf ball For soccer ball U D 5 ReD 2.5 10 and transition occurs at about ν 4 ft 2 (Table A.9) s Hence D 41.1 mm Hence D 8.75 in Hence U U U ν ReD D ν ReD D ν ReD D U 289 ft U 300 ft s U 55.5 s ft s U 197 mph U 88.2 m U 205 mph U 91.5 m U 37.9 mph U 16.9 m s s s Problem 9.6 [Difficulty: 2] Given: Sheet of plywood attached to the roof of a car Find: Speed at which boundary layer becomes turbulent; Speed at which 90% is turbulent Solution: Rex Basic equation ρ U x μ ν 1.50 10 For air U x Rex 5 10 and transition occurs at about ν 2 5 m 5 (Table A.10) s Now if we assume that we orient the plywood such that the longer dimension is parallel to the motion of the car, we can say: Hence U ν Rex x U 3.8 m s When 90% of the boundary layer is turbulent x 0.1 2 m Hence U ν Rex x U 37.5 m s x 2 m U 13.50 km U 135.0 km hr hr Problem 9.7 [Difficulty: 2] Given: Laminar boundary layer (air & water) Find: Plot of boundary layer length as function of speed (at various altitudes for air) Solution: Governing equations: The critical Reynolds number for transition to turbulence is Re crit = UL crit/ = 500000 The critical length is then L crit = 500000/U For air at sea level and 10 km, we can use tabulated data for density from Table A.3. For the viscosity , use the Sutherland correlation (Eq. A.1) = bT 1/2/(1+S /T ) b = 1.46E-06 kg/m.s.K1/2 S = 110.4 K Air (sea level, T = 288.2 K): = 1.225 kg/m3 (Table A.3) = 1.79E-05 N.s/m2 (Sutherland) Water (20 oC): Air (10 km, T = 223.3 K): = 0.414 kg/m3 = (Table A.3) = 1.46E-05 N.s/m2 (Sutherland) 998 slug/ft3 = 1.01E-03 N.s/m2 (Table A.8) Computed results: Water Air (Sea level) Air (10 km) U (m/s) L crit (m) L crit (m) L crit (m) 0.05 0.10 0.5 1.0 5.0 15 20 25 30 50 100 200 10.12 5.06 1.01 0.506 0.101 0.0337 0.0253 0.0202 0.0169 0.0101 0.00506 0.00253 146.09 73.05 14.61 7.30 1.46 0.487 0.365 0.292 0.243 0.146 0.0730 0.0365 352.53 176.26 35.25 17.63 3.53 1.18 0.881 0.705 0.588 0.353 0.176 0.0881 1000 0.00051 0.0073 0.0176 Length of Laminar Boundary Layer for Water and Air 100.0 1.0 L crit (m) 0.0 0.0 1.E-02 Water Air (Sea level) Air (10 km) 1.E+00 1.E+02 U (m/s) 1.E+04 Problem 9.8 Given: Aircraft or missile at various altitudes Find: Plot of boundary layer length as function of altitude Solution: Governing equations: The critical Reynolds number for transition to turbulence is Re crit = UL crit/ = 500000 The critical length is then L crit = 500000/U Let L 0 be the length at sea level (density 0 and viscosity 0). Then L crit/L 0 = (/0)/(/0) The viscosity of air increases with temperature so generally decreases with elevation; the density also decreases with elevation, but much more rapidly. Hence we expect that the length ratio increases with elevation For the density , we use data from Table A.3. For the viscosity , we use the Sutherland correlation (Eq. A.1) = bT 1/2/(1+S /T ) b = S = 1.46E-06 110.4 kg/m.s.K1/2 K [Difficulty: 2] Computed results: z (km) T (K) /0 /0 L crit/L 0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 6.0 288.2 284.9 281.7 278.4 275.2 271.9 268.7 265.4 262.2 258.9 255.7 249.2 1.0000 0.9529 0.9075 0.8638 0.8217 0.7812 0.7423 0.7048 0.6689 0.6343 0.6012 0.5389 1.000 0.991 0.982 0.973 0.965 0.955 0.947 0.937 0.928 0.919 0.910 0.891 1.000 1.04 1.08 1.13 1.17 1.22 1.28 1.33 1.39 1.45 1.51 1.65 7.0 8.0 9.0 10.0 11.0 12.0 13.0 14.0 15.0 16.0 17.0 18.0 19.0 20.0 22.0 24.0 26.0 28.0 30.0 242.7 236.2 229.7 223.3 216.8 216.7 216.7 216.7 216.7 216.7 216.7 216.7 216.7 216.7 218.6 220.6 222.5 224.5 226.5 0.4817 0.4292 0.3813 0.3376 0.2978 0.2546 0.2176 0.1860 0.1590 0.1359 0.1162 0.0993 0.0849 0.0726 0.0527 0.0383 0.0280 0.0205 0.0150 0.872 0.853 0.834 0.815 0.795 0.795 0.795 0.795 0.795 0.795 0.795 0.795 0.795 0.795 0.800 0.806 0.812 0.818 0.824 1.81 1.99 2.19 2.41 2.67 3.12 3.65 4.27 5.00 5.85 6.84 8.00 9.36 10.9 15.2 21.0 29.0 40.0 54.8 Length of Laminar Boundary Layer versus Elevation 60 50 40 L/L 0 30 20 10 0 0 10 20 z (m) 30 Problem 9.9 Given: [Difficulty: 2] Sinusoidal velocity profile for laminar boundary layer: u A sin( B y ) C Find: (a) Three boundary conditions applicable to this profile (b) Constants A, B, and C. Solution: For the boundary layer, the following conditions apply: u 0 at y 0 (no slip condition) u U at y δ (continuity with freestream) y u 0 at y δ (no shear stress at freestream) Applying these boundary conditions: ( 1 ) u ( 0 ) A sin( 0 ) C 0 C0 ( 2 ) u ( δ) A sin( B δ) U ( 3) y u A B cos( B y ) Back into (2): A sin Thus: u ( δ) A B cos( B δ) 0 y Therefore: B δ δ U Therefore: A U 2δ π So the expression for the velocity profile is: u U sin π y 2 δ π 2 or B π 2 δ Problem 9.10 Linear, sinusoidal, and parabolic velocity profiles Plots of y/δ vs u/U for all three profiles Here are the profiles: Laminar Boundary Layer Velocity Profiles Linear Sinusoidal Parabolic 0.8 Dimensionless Distance y/δ Given: Find: Solution: [Difficulty: 2] 0.6 0.4 0.2 0 0 0.2 0.4 0.6 Dimensionless Velocity u/U 0.8 Problem 9.11 Given: Laminar boundary layer profile Find: If it satisfies BC’s; Evaluate */ and / [Difficulty: 2] Solution: 3 The boundary layer equation is u 3 y 1 y for which u = U at y = U 2 2 The BC’s are u 0 0 At y = 0 At y = du dy 0 y u 3 1 3 0 0 0 U 2 2 3 1 3 y2 3 1 32 du 0 U U 3 3 dy 2 2 y 2 2 For *: u u * 1 dy 1 dy U U 0 0 Then 1 1 * 1 u u u y 1 dy 1 d 1 d 0 U U 0 U 0 with Hence u 3 1 3 U 2 2 1 1 1 u 1 3 3 2 1 4 3 * 3 1 d 1 d 0.375 2 2 4 8 0 8 0 U 0 u u u u 1 dy 1 dy U U U U 0 0 For : Then 1 1 1 u u u u u u y 1 dy 1 d 1 d 0U U U U 0 U U 0 Hence 1 1 1 9 1 3 1 1 3 1 u u 3 3 1 d 3 1 3 d 2 3 4 6 d 2 2 4 2 2 4 2 2 2 0U U 0 0 39 3 2 3 3 1 4 3 5 1 7 0.139 4 4 8 10 28 0 280 1 Problem 9.12 Given: Laminar boundary layer profile Find: If it satisfies BC’s; Evaluate */ and / [Difficulty: 2] Solution: 3 4 The boundary layer equation is u y y y 2 2 for which u = U at y = U The BC’s are u 0 0 0 y u 3 4 20 20 0 0 U 1 1 2 3 du y2 y3 U 2 6 3 4 4 0 U 2 6 3 4 4 y dy At y = 0 At y = du dy 0 u u dy 1 dy U U 0 For *: * 1 Then 1 1 * 1 u u u y 1 dy 1 d 1 d 0 U U 0 U 0 with u 2 2 3 4 U u 1 1 3 * 0 .3 1 d 1 2 2 3 4 d 2 4 5 U 2 5 10 0 0 0 1 1 1 Hence For : u u u u 1 dy 1 dy U U U U 0 0 Then 1 1 1 u u u u y u u 1 dy 1 d 1 d U U 0 U U 0U U 0 Hence u u 1 d 2 3 4 1 2 3 4 d 2 4 2 2 3 9 4 4 5 4 6 4 7 8 d 0U U 0 0 1 1 1 2 4 3 1 4 9 5 4 7 1 8 1 9 37 0.117 3 2 5 7 2 9 0 315 1 Problem 9.13 Given: Laminar boundary layer profile Find: If it satisfies BC’s; Evaluate */ and / [Difficulty: 3] Solution: The boundary layer equation is u y 2 U u y 2 2 U u 0 0 The BC’s are At y = 0 At y = du dy 0 y 2 1 2 2 y for which u = U at y = 0 y u 2 0 0 U 1 du U 2 2 0 so it fails the outer BC. y dy This simplistic distribution is a piecewise linear profile: The first half of the layer has velocity gradient second half has velocity gradient U 1.414 U , and the 2 2 U 0.586 U . At y = , we make another transition to zero velocity gradient. For *: u u * 1 dy 1 dy U U 0 0 Then 1 1 * 1 u u u y 1 dy 1 d 1 d 0 U U 0 U 0 u 1 2 0 U 2 u 2 2 2 1 U with 2 1 1 2 Hence u * 1 d U 0 1 1 12 0 2 d 1 2 2 1 12 1 2 1 d 2 2 12 1 2 2 1 1 2 0 2 1 2 2 1 2 2 1 2 3 2 * 1 0.396 2 8 4 8 4 4 For : u u u u 1 dy 1 dy U U U U 0 0 Then 1 u 0 U 1 u u 1 dy U U 0 1 u u y 1 d U 0 U u 1 d U Hence, after a LOT of work u u 1 d 0U U 1 12 2 1 2 d 0 12 2 1 1 2 2 2 3 3 0 2 2 1 2 1 1 2 2 2 1 d 12 1 1 2 1 2 2 1 2 2 2 1 2 2 1 0.152 2 8 12 24 6 12 1 2 Problem 9.14 Given: Find: Solution: [Difficulty: 2] Power law velocity profiles Plots of y/δ vs u/U for this profile and the parabolic profile of Problem 9.10 Here are the profiles: Boundary Layer Velocity Profiles Power Parabolic Dimensionless Distance y/δ 0.8 0.6 0.4 0.2 0 0 0.2 0.4 0.6 0.8 Dimensionless Velocity u/U Note that the power law profile gives and infinite value of du/dy as y approaches zero: du U d u U U y dy d y 7 6 7 as y0 Problem 9.15 Given: Find: Solution: [Difficulty: 2] Linear, sinusoidal, and parabolic velocity profiles The momentum thickness expressed as /δ for each profile We will apply the definition of the momentum thickness to each profile. θ Governing Equation: δ u U 1 u dy (Definition of momentum thickness) U 0 1 δ δ θ If we divide both sides of the equation by δ, we get: δ u U 1 u dy However, we can change U 0 the variable of integration to u For the linear profile: U = y/δ, resulting in: dη 1 δ dy Therefore: 1 1 2 η ( 1 η) dη η η dη Evaluating this integral: δ 0 0 u For the sinusoidal profile: U sin π 2 η u U 1 u dη U θ δ 1 2 1 3 1 θ 6 δ 0.1667 Into the momentum thickness: 1 π θ π sin η 1 sin η dη δ 2 2 0 1 0 η Into the momentum thickness: θ 1 π π 2 sin η sin η dη 2 2 0 Evaluating this integral: θ δ For the parabolic profile: u U 1 δ θ 2 π 2 π π 4 2 η η 2 θ δ 0.1366 Into the momentum thickness: 1 2 2 2 3 4 2 η η 1 2 η η dη 2 η 5 η 4 η η dη δ 0 0 θ Evaluating this integral: θ δ 1 5 3 1 1 5 2 θ 15 δ 0.1333 Problem 9.16 Given: Find: Solution: [Difficulty: 2] Linear, sinusoidal, and parabolic velocity profiles The displacement thickness expressed as δ*/δ for each profile We will apply the definition of the displacement thickness to each profile. δdisp Governing Equation: infinity 1 u dy U 0 δ 1 u dy U (Definition of displacement thickness) 0 If we divide both sides of the equation by δ, we get: δdisp δ 1 δ δ 1 u dy U However, we can change 0 the variable of integration to η = y/δ, resulting in: u For the linear profile: U δdisp δ 1 ( 1 η ) dη Evaluating this integral: u U δ 1 δdisp δ 0 δ dy Therefore: δdisp δ sin π 2 η 1 1 1 u dη U 0 η Into the displacement thickness: For the sinusoidal profile: δdisp dη 1 1 2 1 δdisp 2 δ 0.5000 Into the displacement thickness: δ 1 sin π η dη Evaluating this integral: disp 1 2 2 δ π δdisp δ 0.3634 0 For the parabolic profile: u U δdisp δ 1 2 η η 2 Into the displacement thickness: 1 2 1 2 η η2 dη 1 2 η η dη Evaluating this integral: 0 0 δdisp δ 11 1 3 1 δdisp 3 δ 0.3333 Problem 9.17 Given: Find: Solution: [Difficulty: 2] Power law and parabolic velocity profiles The displacement and momentum thicknesses expressed as δ*/δ and /δ for each profile We will apply the definition of the displacement and momentum thickness to each profile. δdisp Governing Equations: infinity 1 u dy U 0 θ δ u U δ 1 u dy U (Definition of displacement thickness) 0 1 u dy (Definition of momentum thickness) U 0 If we divide both sides of the equations by δ, we get: δdisp δ 1 δ δ 1 δ δ 1 u dy U θ 0 However, we can change the variable of integration to δdisp δ 1 1 u dη U 0 δ 1 u U 1 dη 1 δ u η 7 dy U dy Therefore: dη 0 Into the displacement thickness: δdisp δ 0 Evaluating this integral: δdisp δ 1 Into the momentum thickness: u U 1 U U 1 u 1 For the power law profile: u 0 = y/δ, resulting in: θ δ 1 1 1 θ 7 7 η 1 η dη δ 0 0 1 7 1 η dη 1 2 1 7 7 η η dη 7 8 1 δdisp 8 δ θ Evaluating this integral: δ 7 θ 0.0972 δ For the parabolic profile: u U 2 η η 2 Into the displacement thickness: δdisp δ 1 0 1 2 1 2 η η2 dη 1 2 η η dη 0 0.1250 8 7 9 7 72 Evaluating this integral: δdisp δ Into the momentum thickness: 11 1 1 3 1 δdisp 3 δ 1 0.3333 2 2 3 4 2 2 η η 1 2 η η dη 2 η 5 η 4 η η dη δ 0 0 θ Evaluating this integral: θ δ 1 5 3 1 1 5 2 θ 15 δ Power Law δdisp 0.1250 δ 0.0972 δ Parabolic 0.3333 δ 0.1333 δ Profile θ 0.1333 Problem 9.18 Given: Data on fluid and boundary layer geometry Find: Mass flow rate across ab; Drag [Difficulty: 3] CV Solution: The given data is ρ 1.5 slug ft Governing equations: U 10 3 ft s d L 10 ft δ 1 in b 3 ft c Rx Mass Momentum Assumptions: (1) Steady flow (2) No pressure force (3) No body force in x direction (4) Uniform flow at a Applying these to the CV abcd δ Mass ( ρ U b δ) ρ u b dy mab 0 0 For the boundary layer u U y δ dy η δ dη 1 Hence 1 mab ρ U b δ ρ U η δ dy ρ U b δ ρ U b δ 2 0 1 mab ρ U b δ 2 slug mab 1.875 s δ Momentum Rx U ( ρ U δ) mab u ab u ρ u b dy 0 u ab U Note that and 1 δ 2 2 u ρ u b dy ρ U b δ η dη 0 2 Rx ρ U b δ 0 1 2 2 ρ U b δ U ρ U b δ η dy 2 1 0 2 Rx ρ U b δ 1 2 2 ρ U δ 1 3 2 ρ U δ 1 2 Rx ρ U b δ 6 Rx 6.25 lbf We are able to compute the boundary layer drag even though we do not know the viscosity because it is the viscosity that creates the boundary layer in the first place Problem 9.19 [Difficulty: 3] Given: Data on fluid and boundary layer geometry Find: Mass flow rate across ab; Drag; Compare to Problem 9.18 Solution: The given data is Governing equations: ρ 1.5 slug ft U 10 3 ft L 3 ft s δ 0.6 in b 10 ft Mass Momentum Assumptions: (1) Steady flow (2) No pressure force (3) No body force in x direction (4) Uniform flow at a Applying these to the CV abcd δ Mass ( ρ U b δ) ρ u b dy mab 0 0 For the boundary layer u U y δ dy η δ dη 1 Hence 1 mab ρ U b δ ρ U η δ dy ρ U b δ ρ U b δ 2 0 1 mab ρ U b δ 2 slug mab 3.75 s δ Momentum Rx U ( ρ U δ) mab u ab u ρ u b dy 0 u ab U Note that and 1 δ 2 2 u ρ u b dy ρ U b δ η dη 0 0 1 1 2 2 2 Rx ρ U b δ ρ U b δ U ρ U b δ η dy 2 0 2 Rx ρ U b δ 1 2 1 2 Rx ρ U b δ 6 2 ρ U δ 1 3 2 ρ U δ Rx 12.50 lbf We should expect the drag to be larger than for Problem 9.18 because the viscous friction is mostly concentrated near the leading edge (which is only 3 ft wide in Problem 9.18 but 10 ft here). The reason viscous stress is highest at the front region is that the boundary layer is very small (δ <<) so τ = μdu/dy ~ μU/δ >> Problem 9.20 [Difficulty: 3] δ = 1 in Given: Find: Flow over a flat plate with parabolic laminar boundary layer profile Solution: We will apply the continuity and x-momentum equations to this system. (a) Mass flow rate across ab (b) x component (and direction) of force needed to hold the plate in place Governing Equations: d V V CS dA 0 t CV udV uV dA Fsx Fbx CS t CV Assumptions: (Continuity) (x- Momentum) (1) Steady flow (2) No net pressure forces (3) No body forces in the x-direction (4) Uniform flow at da CV d ρ U b δ ρ u b dy mab 0 From the assumptions, the continuity equation becomes: c Rx δ The integral can be written as: 0 δ δ ρ u b dy ρ b u dy ρ U b δ 0 0 1 2η η2 dη where η 0 y δ This integral is equal to: ρ U b δ 1 1 2 ρ U b δ 3 3 2 1 mab ρ U b δ ρ U b δ ρ U b δ Substituting known values: 3 3 Solving continuity for the mass flux through ab we get: 1 slug ft ft mab 1.5 10 3.0 ft 1 in 3 3 s 12 in ft slug mab 1.250 s δ From the assumptions, the momentum equation becomes: Rx u da ( ρ U b δ) u ab mab u ρ u b dy where u da u ab U 0 1 2 2 Thus: Rx ρ U b δ ρ U b δ 3 δ 2 2 u ρ u b dy ρ U b δ 3 0 δ δ 2 2 u ρ u b dy ρ b u dy ρ U b δ 0 0 ρ U b δ 2 4 3 1 1 2 η η 2 2 δ u ρ u b dy The integral can be written as: 0 dη ρ U b δ 1 2 4η2 4η3 η4 dη This integral is equal to: 0 0 1 8 2 2 2 8 2 15 ρ U b δ Therefore the force on the plate is: Rx 3 ρ U b δ 15 ρ U b δ 5 15 Substituting known values: Rx 2 15 1.5 slug ft 3 10 This force must be applied to the control volume by the plate. ft s 2 3.0 ft 1 in ft 12 in 2 lbf s slug ft Rx 5.00 lbf (to the left) Problem 9.21 [Difficulty: 2] Given: Data on wind tunnel and boundary layers Find: Displacement thickness at exit; Percent change in uniform velocity through test section Solution :The solution involves using mass conservation in the inviscid core, allowing for the fact that as the boundary layer grows it reduces the size of the core. One approach would be to integrate the 1/7 law velocity profile to compute the mass flow in the boundary layer; an easier approach is to simply use the displacement thickness! Basic equations δdisp (4.12) δ 1 u dy U 0 Assumptions: 1) Steady flow 2) Incompressible 3) No friction outside boundary layer 4) Flow along streamline 5) Horizontal 1 For this flow The design data is The volume flow rate is We also have Hence u ρ U A const and ft Udesign 160 s w 1 ft U y δ 7 h 1 ft Adesign w h ft 3 Q Udesign Adesign Q 160 δin 0.4 in δexit 1 in δdisp δ 0 1 u dy U δ 1 1 7 y dy δ Adesign 1 ft s δ 0 1 1 7 1 η dη where 0 η y δ Hence at the inlet and exit δin δdispin 8 δdispin 0.05 in δdispexit δexit 8 δdispexit 0.125 in δ δdisp 8 2 Hence the areas are Ain w 2 δdispin h 2 δdispin Ain 0.9834 ft Aexit w 2 δdispexit h 2 δdispexit 2 Aexit 0.9588 ft 2 Applying mass conservation between "design" conditions and the inlet ρ Udesign Adesign ρ Uin Ain 0 or Also Uin Udesign Adesign Ain Adesign Uexit Udesign Aexit ft Uin 162.7 s ft Uexit 166.9 s The percent change in uniform velocity is then Uexit Uin Uin 2.57 % The exit displacement thickness is δdispexit 0.125 in Problem 9.22 [Difficulty: 2] Given: Data on boundary layer in a cylindrical duct Find: Velocity U2 in the inviscid core at location 2; Pressure drop Solution: The solution involves using mass conservation in the inviscid core, allowing for the fact that as the boundary layer grows it reduces the size of the core. One approach would be to integrate the 1/7 law velocity profile to compute the mass flow in the boundary layer; an easier approach is to simply use the displacement thickness! The given or available data (from Appendix A) is ρ 1.23 kg m U1 12.5 s 3 m D 100 mm δ1 5.25 mm δ2 24 mm Governing equations: Mass p Bernoulli ρ 2 V 2 g z constant (4.24) The displacement thicknesses can be computed from boundary layer thicknesses using Eq. 9.1 1 u δdisp 1 dy δ U 0 1 δ 7 1 η dη δ 8 0 Hence at locations 1 and 2 δ1 δdisp1 8 Applying mass conservation at locations 1 and 2 δdisp1 0.656 mm δ2 δdisp2 8 δdisp2 3 mm ρ U1 A1 ρ U2 A2 0 A1 U2 U1 A2 or The two areas are given by the duct cross section area minus the displacement boundary layer π 2 A1 D 2 δdisp1 4 Hence A1 7.65 10 3 2 m A1 U2 U1 A2 For the pressure drop we can apply Bernoulli to locations 1 and 2 to find π 2 A2 D 2 δdisp2 4 A2 6.94 10 m U2 13.8 s ρ 2 2 p 1 p 2 ∆p U2 U1 ∆p 20.6 Pa 2 3 2 m Problem 9.23 [Difficulty: 2] Given: Data on wind tunnel and boundary layers Find: Uniform velocity at exit; Change in static pressure through the test section Solution: Basic equations δdisp (4.12) δ 1 u dy U p ρ 2 V 2 g z const 0 Assumptions: 1) Steady flow 2) Incompressible 3) No friction outside boundary layer 4) Flow along streamline 5) Horizontal 1 For this flow The given data is We also have Hence δdisp δ 0 u an d U m U1 25 s δ1 20 mm h 25 cm A h 1 u dy U δ y δ ρ U A const 7 2 A 625 cm 2 δ2 30 mm 1 1 7 y dy δ δ 1 1 7 1 η dη 0 0 η wher e y δ δ δdisp 8 Hence at the inlet and exit δ1 δdisp1 8 Hence the areas are δdisp1 2.5 mm 2 2 A2 h 2 δdisp2 δ2 δdisp2 8 δdisp2 3.75 mm A1 h 2 δdisp1 A1 600 cm 2 A2 588 cm 2 Applying mass conservation between Points 1 and 2 ρ U1 A1 ρ U2 A2 0 p1 The pressure change is found from Bernoulli ρ Hence ∆p U U2 2 1 ρ 2 2 A1 U2 U1 A2 or U1 2 2 p2 ∆p 15.8 Pa ρ U2 2 m U2 25.52 s 2 with ρ 1.21 kg 3 m The pressure drops slightly through the test section Problem 9.24 [Difficulty: 2] Given: Data on wind tunnel and boundary layers Find: Uniform velocity at Point 2; Change in static pressure through the test section Solution: Basic equations (4.12) δdisp δ 1 u dy U p ρ 2 V 2 g z const 0 Assumptions: 1) Steady flow 2) Incompressible 3) No friction outside boundary layer 4) Flow along streamline 5) Horizontal 1 u ρ U A const and The given data is m U1 20 s W 40 cm We also have δ1 1 cm δ2 1.3 cm For this flow Hence δdisp δ 0 U 1 u dy U δ 1 1 7 y dy δ 1 y δ 7 2 2 A W 1 7 1 η dη A 0.1600 m η where 0 0 Hence at the inlet and exit δ1 δdisp1 δdisp1 0.125 cm 8 Hence the areas are δ δ2 δdisp2 8 y δ δ δdisp 8 δdisp2 0.1625 cm 2 2 A2 W 2 δdisp2 2 A1 W 2 δdisp1 A1 0.1580 m 2 A2 0.1574 m Applying mass conservation between Points 1 and 2 ρ U1 A1 ρ U2 A2 0 The pressure change is found from Bernoulli p1 ρ Hence ∆p ρ 2 U1 U2 2 2 or U1 2 2 p2 ρ U2 2 A1 U2 U1 A2 m U2 20.1 s with ρ 1.21 2 kg 3 m 4 ∆p 2.66 10 psi ∆p 1.835 Pa Problem 9.25 Given: Data on wind tunnel and boundary layers Find: Pressure change between points 1 and 2 [Difficulty: 2] Solution: Basic equations (4.12) p ρ 2 V 2 g z const Assumptions: 1) Steady flow 2) Incompressible 3) No friction outside boundary layer 4) Flow along streamline 5) Horizontal For this flow ρ U A const The given data is ft U0 100 s We also have δdisp2 0.035 in Hence at the Point 2 A2 h 2 δdisp2 U1 U0 h 3 in 2 p1 ρ Hence ∆p U U2 2 1 ρ 2 2 A1 9 in 2 The pressure change is found from Bernoulli 2 A2 8.58 in Applying mass conservation between Points 1 and 2 ρ U1 A1 ρ U2 A2 0 A1 h o r U1 2 2 2 The pressure drops by a small amount as the air accelerates p2 ρ U2 2 A1 U2 U1 A2 ft U2 105 s wit h ρ 0.00234 2 slug ft 3 ∆p 8.05 10 psi ∆p 1.16 lbf ft 2 3 Problem 9.26 [Difficulty: 3] Given: Find: Developing flow of air in flat horizontal duct. Assume 1/7-power law velocity profile in boundary layer. Solution: We will apply the continuity and x-momentum equations to this problem. (a) Displacement thickness is 1/8 times boundary layer thickness (b) Static gage pressure at section 1. (c) Average wall shear stress between entrance and section 2. Governing Equations: δdisp infinity δ 1 u dy U 0 0 dV V dA 0 1 u dy U (Definition of displacement thickness) CS t CV ud V u V CS dA Fsx Fbx t CV Assumptions: (Continuity) (x- Momentum) (1) Steady, incompressible flow (2) No body forces in the x-direction (3) No viscous forces outside boundary layer (4) Boundary layers only grow on horizontal walls L = 20 ft H = 1 ft V1 = 40 ft/s δ 2 = 4 in δdisp If we divide both sides of the displacement thickness definition by δ, we get: δ 1 δ δ 1 u dy U 0 However, we can change the variable of integration to η = y/δ, resulting in: dη 1 δ dy Therefore: δdisp δ 1 1 u dη U 0 1 1 For the power law profile: u U η 7 Into the displacement thickness: δdisp δ 0 Evaluating this integral: δdisp δ 1 7 1 η dη 1 7 8 1 δdisp 8 δ 1 8 V1 A1 V2 A2 or After applying the assumptions from above, continuity reduces to: Solving for the velocity at 2: H V2 V1 V1 H 2 δdisp2 H H p0 ρ 1 2 p 1g p 1 p 0 ρ V1 2 p 1g 1 2 p 2g p 2 p 0 ρ V2 2 p 2g 1 2 1 2 Substituting known values: 4 2 p V ρ 0.00234 1 1 1 4 ft 1 4 12 ft V2 43.6 s along a streamline. Therefore: 2 slug ft 0.00234 δ2 ft V2 40 1 ft s From Bernoulli equation, since z = constant: V1 w H V2 w H 2 δdisp2 3 slug ft 3 40 43.6 2 2 ft lbf s s slug ft 12 in ft 2 2 2 ft lbf s s slug ft 12 in ft p 1g 0.01300 psi 2 p 2g 0.01545 psi Now if we apply the momentum equation to the control volume (considering the assumptions shown): Fsx uV dA CS p1 p2 w 2 H H τ w L V1 ρ V1 w 2 δ2 0 H u ρ u w dy V2 ρ V2 δ2 w 2 1 2 7 2 2 2 7 The integral is equal to: ρ w u dy ρ V2 δ2 w η dη ρ V2 δ2 w 9 0 0 δ2 p1 p2 w 2 H τ τ 2 H τ w L ρ V1 2 Therefore the momentum equation becomes: 2 2 H w ρ V2 δ2 w Simplifying and solving for the shear stress we get: 2 9 H 2 2 H 2 H p 1 p 2 ρ V1 V2 δ2 Substituting in known values we get: L 2 2 2 9 1 1 20 ft [ ( 0.01328 ) ( 0.01578 ) ] lbf s ft lbf 1 ft slug ft ft 1 ft 1 2 4 40 43.6 ft 0.00234 2 2 3 s 2 s 2 9 12 slug ft 12 in in ft 2 2 2 5 τ 5.46 10 psi 2 Problem 9.27 [Difficulty: 3] Given: Find: Air flow in laboratory wind tunnel test section. Solution: We will apply the continuity and Bernoulli equations to this problem. (a) Freestream speed at exit (b) Pressure at exit δdisp Governing Equations: infinity 1 u dy U 0 δ 1 u dy U (Definition of displacement thickness) 0 dV V dA 0 CS t CV 2 p V gz const 2 Assumptions: (Continuity) (Bernoulli) (1) Steady, incompressible flow (2) No body forces in the x-direction (3) No viscous forces outside boundary layer (4) Streamline exists between stations 1 and 2 (5) Uniform flow outside boundary layer (6) Boundary layer is the same on all walls (7) Neglect corner effects (8) Constant elevation between 1 and 2 If we divide both sides of the displacement thickness definition by δ, we get: δdisp δ 1 δ δ U1 = 80 ft/s 1 dy U u 0 L = 2 ft However, we can change the variable of integration to η = y/δ, resulting in: dη δdisp Therefore: δ 1 0 1 δdisp δ 0 1 7 1 η dη 1 δdisp1 0.8 in 8 W = 1 ft 1 δ dy 1 1 u dη u 7 For the power law profile: η U U Evaluating this integral: δdisp δ δdisp1 0.100 in 1 7 8 1 Into the displacement thickness: So the displacement thicknesses are: 8 1 δdisp2 1 in 8 δdisp2 0.125 in W 2 δdisp1 U2 U1 W 2 δdisp2 Solving for the speed at 2: 2 U2 W 2 δdisp22 U1 A1 U2 A2 or U1 W 2 δdisp1 After applying the assumptions from above, continuity reduces to: 2 ft U2 80 s Substituting known values: 2 1 2 0.100 1 2 0.125 ft U2 91.0 s From Bernoulli equation, since z = constant: p1 ρ ∆p12 U U2 2 1 ρ 2 2 ∆p12 1 2 U1 2 0.00239 2 slug ft 3 p2 U2 ρ 2 along a streamline. Therefore: 2 2 80 91 2 ft 2 2 s 2 lbf s slug ft ft 12 in 2 ∆p12 0.01561 psi From ambient to station 1 we see a loss at the tunnel entrance: 2 2 p0 U0 p1 U1 ρ 2 ρ 2 h lT Since U0 0 and p 0 p atm 0 we can solve for the pressure at 1: p 1 ρ h lT 1 2 ρ U1 2 where ρhlT 0.3 12 ft 1.94 slug ft 3 32.2 ft 2 2 s 2 2 lbf 1 slug ft lbf s Therefore: p 1 0.01085 0.00239 80 2 2 3 slug ft s in ft p 2 p 1 ∆p12 it follows that: lbf s slug ft ft 12 in 2 ρhlT 0.01085 psi 2 ft 0.0640 psi So the pressure at 2 is: 12 in p 2 0.0640 psi 0.01561 psi 0.0796 psi Since the pressure drop can be expressed as h2 p2 ρ g lbf So in terms of water height: h 2 0.0796 2 in 2 3 p 2 ρ g h 2 2 s slug ft 12 in 12 in ft ft 2 ft 1.94 slug 32.2 ft lbf s p 2 0.0796 psi h 2 2.20 in Problem 9.28 Given: Data on fluid and boundary layer geometry Find: Gage pressure at location 2; average wall stress [Difficulty: 3] Solution: The solution involves using mass conservation in the inviscid core, allowing for the fact that as the boundary layer grows it reduces the size of the core. One approach would be to integrate the 1/7 law velocity profile to compute the mass flow in the boundary layer; an easier approach is to simply use the displacement thickness! The average wall stress can be estimated using the momentum equation for a CV The given and available (from Appendix A) data is ρ 0.00234 slug ft 3 ft U1 50 s L 20 ft D 15 in δ2 4 in Governing equations: Mass Momentum Bernoulli p ρ 2 V g z constant 2 (4.24) Assumptions: (1) Steady flow (2) No pressure force (3) No body force in x direction The displacement thickness at location 2 can be computed from boundary layer thickness using Eq. 9.1 1 δdisp2 δ2 0 Hence u 1 dy δ 2 U 0 δ2 δdisp2 8 Applying mass conservation at locations 1 and 2 1 δ2 7 1 η dη 8 δdisp2 0.500 in ρ U1 A1 ρ U2 A2 0 π 2 A1 D 4 A1 U2 U1 A2 or A1 1.227 ft 2 The area at location 2 is given by the duct cross section area minus the displacement boundary layer π 2 A2 D 2 δdisp2 4 A2 1.069 ft 2 Hence A1 U2 U1 A2 ft U2 57.4 s For the pressure change we can apply Bernoulli to locations 1 and 2 to find Hence ρ 2 2 p 1 p 2 ∆p U2 U1 2 ∆p 6.46 10 p 2 ( gage ) p 1 ( gage ) ∆p p 2 6.46 10 3 psi 3 p 2 ∆p psi For the average wall shear stress we use the momentum equation, simplified for this problem D 2 2 2 π 2 ∆p A1 τ π D L ρ U1 A1 ρ U2 D 2 δ2 ρ 4 D 2 2 2 π r u dr δ2 1 where y u ( r) U2 δ2 7 D 2 y dr dy 0 2 2 2 ρ 2 π r u dr 2 π ρ U2 D δ δ 2 D The integral is r and 2 2 7 D y y dy 2 δ 2 2 D 2 ρ D 2 Hence τ δ2 D δ2 2 2 2 π r u dr 7 π ρ U2 δ2 8 9 D δ2 2 2 π 2 2 ∆p A1 ρ U1 A1 ρ U2 D 2 δ2 7 π ρ U2 δ2 4 8 9 τ 6.767 10 π D L 5 psi Problem 9.29 [Difficulty: 5] Given: Air flow in laboratory wind tunnel test section. Find: (a) Displacement thickness at station 2 (b) Pressure drop between 1 and 2 (c) Total drag force caused by friction on each wall We will apply the continuity, x-momentum, and Bernoulli equations to this problem. Solution: Governing Equations: δdisp infinity 1 u dy U 0 δ 1 u dy U (Definition of displacement thickness) 0 dV V dA 0 CS t CV 2 p V gz const 2 (Continuity) (Bernoulli) udV uV dA Fsx Fbx CS t CV Assumptions: (1) Steady, incompressible flow (2) No body forces in the x-direction (3) No viscous forces outside boundary layer (4) Streamline exists between stations 1 and 2 If we divide both sides of the displacement thickness definition by δ, we get: (x- Momentum) (5) Uniform flow outside boundary layer (6) Boundary layer is the same on all walls (7) Neglect corner effects (8) Constant elevation between 1 and 2 δdisp δ 1 δ δ 1 u dy U 0 However, we can change the variable of integration to η = y/δ, resulting in: dη 1 δ dy Therefore: δdisp δ 1 0 u 1 dη U 1 1 u If we assume the power law profile (turbulent BL): η U Evaluating this integral: δdisp δ 1 7 8 1 7 Into the displacement thickness: 1 δdisp2 20.3 mm 8 So the displacement thickness is: 8 2 H U2 U1 H 2 δdisp2 1 7 1 η dη δdisp2 2.54 mm 2 U1 A1 U2 A2 or U1 H U2 H 2 δdisp2 After applying the assumptions from above, continuity reduces to: Solving for the speed at 2: δdisp δ 0 2 m U2 50.2 s Substituting known values: 305 305 2 2.54 2 m U2 51.9 s From Bernoulli equation, since z = constant: p1 ρ ∆p12 U U2 2 1 ρ 2 2 ∆p12 1 2 1.23 U1 2 2 kg 3 p2 ρ U2 2 along a streamline. Therefore: 2 2 50.2 51.9 2 2 m m 2 2 s N s ∆p12 106.7 Pa kg m To determine the drag on the walls, we choose the control volume shown above and apply the x-momentum equation. From the assumptions, the equation reduces to: CS uV dA Fsx Applying this to the control volume: δ 2 p 1 H δ2 FD p 2 H δ2 U1 ρ U1 H δ2 Uavg mtop u ρ u H dy The mass flow rate through the top of the CV δ 2 mtop m1 m2 ρ U1 H δ2 ρ u H dy 0 can be determined using the continuity equation across the control volume: 0 1 1 δ 2 7 7 ρ u H dy ρ U2 H δ2 η dη ρ U2 H δ2 Therefore: 8 0 0 This integral can be evaluated using the power law profile: 7 mtop ρ H δ2 U1 U2 8 The average speed can be approximated as the mean of the speeds at 1 and 2: Uavg U1 U2 Finally the integral in the momentum equation may also be evaluated using the power law profile: 1 2 7 2 2 7 u ρ u H dy ρ U2 H δ2 η dη ρ U2 H δ2 9 0 0 δ2 p 1 H δ2 FD p 2 H δ2 U1 ρ U1 H δ2 U1 U2 2 Thus, the momentum equation may be rewritten as: ρ H δ2 U1 7 8 U2 7 9 2 ρ U2 H δ2Solving for the drag force: 2 FD p 1 p 2 ρ U1 2 U1 U2 7 7 U1 U2 U2 2 H δ2 Substituting in all known values yields: 8 2 9 N kg FD 106.7 1.23 2 3 m m 2 2 2 2 50.2 m ( 50.2 51.9) 50.2 7 51.9 m 7 51.9 m N s 0.305 m 0.020 s 8 s kg m 2 s2 9 FD 2.04 N The viscous drag force acts on the CV in the direction shown. The viscous drag force on the wall of the test section is equal and opposite: Problem 9.30 [Difficulty: 2] Given: Find: Blasius exact solution for laminar boundary layer flow Solution: The Blasius solution is given in Table 9.1; it is plotted below. u Plot and compare to parabolic velocity profile: U 2 y δ y δ 2 Parabolic Blasius Dimensionless Height y/δ 0.8 0.6 0.4 0.2 0 0 0.2 0.4 0.6 Dimensionless Velocity u/U 0.8 Problem 9.31 [Difficulty: 3] Given: Find: Blasius exact solution for laminar boundary layer flow Solution: We will apply the shear stress definition to both velocity profiles. (a) Evaluate shear stress distribution (b) Plot τ/τw versus y/δ u π y (c) Compare with results from sinusoidal velocity profile: sin U 2 δ Governing Equation: τ μ y For Blasius: u U f'( η) and η y Therefore: τ 2 ρ U μ ρ U From the above equation: U f''( η) τ τw For the sinusoidal profile: τ ν x (Shear stress in Newtonian fluid) u U The shear stress is: τ μ ( U f'( η ) ) U μ ( f''( η ) ) η U μ f''( η ) ν x ν x y y U f''( η) f''( 0 ) f''( η) τ is proportional to f''(η) Rex f''( η) 0.33206 Since y δ at η 5 it follows that μ U d u μ U π cos π y 2 δ δ y U δ 2 d δ τw μ U π δ 2 y δ η 5 Thus: τ τw cos π y 2 δ Both profiles are plotted here: Dimensionless Height y/δ 0.8 0.6 0.4 0.2 Sinusoidal Blasius 0 0 0.2 0.4 0.6 Dimensionless Shear Stress τ/τw 0.8 Problem 9.32 Given: Find: Blasius exact solution for laminar boundary layer flow (a) Evaluate shear stress distribution (b) Plot τ/τw versus y/δ (c) Compare with results from sinusoidal velocity profile: u 2 y U δ Solution: y δ 2 We will apply the shear stress definition to both velocity profiles. Governing Equation: τ μ y τ 2 ρ U μ ρ U From the above equation: f''( η) τ τw For the parabolic profile: τ (Shear stress in Newtonian fluid) u U For Blasius: u U f'( η) and η y Therefore: [Difficulty: 3] ν x U ν x f''( η) f''( 0 ) U The shear stress is: τ μ ( U f'( η) ) U μ ( f''( η) ) η U μ f''( η) ν x y y f''( η) τ is proportional to f''(η) Rex f''( η) 0.33206 Since y δ at η 5 it follows that μ U d u μ U 2 2 y δ y U δ δ d δ τw μ U δ 2 y δ Thus: η 5 τ τw 1 y δ Both profiles are plotted here: Dimensionless Height y/δ 0.8 0.6 0.4 0.2 Parabolic Blasius 0 0 0.2 0.4 0.6 Dimensionless Shear Stress τ/τw 0.8 Problem 9.33 Given: Find: Solution: Blasius exact solution for laminar boundary layer flow 5 Plot v/U versus y/δ for Rex 10 We will apply the stream function definition to the Blasius solution. For Blasius: u U f'( η) and η y U ν x The stream function is: 1 ν U From the stream function: v ψ f ( η) x 2 x Thus [Difficulty: 3] 1 ν U f ( η) v 2 x ψ U ν x f ( η) d f η But η 1 y U 1 η 2 x 2 x ν x x dη x ν U x d f 1 η 1 ν U ( η f'( η) f ( η) ) and dη 2 x 2 x ν U x v U 1 2 ν U x ( η f'( η) f ( η) ) v U Since y δ at η 5 it follows that y δ η η f'( η) f ( η) 2 Rex Plotting v/U as a function of y/δ: 5 Dimensionless Height y/δ 0.8 0.6 0.4 0.2 0 0 1 10 3 3 2 10 Dimensionless flow Velocity v/U 3 3 10 Problem 9.34 [Difficulty: 3] Given: Find: Blasius exact solution for laminar boundary layer flow Solution: We will apply the stream function definition to the Blasius solution. (a) Prove that the y component of velocity in the solution is given by Eq. 9.10. (b) Algebraic expression for the x component of a fluid particle in the BL (c) Plot ax vs η to determine the maximum x component of acceleration at a given x For Blasius: u U f'( η) and η y U 1 ν U From the stream function: v ψ f ( η) x 2 x Thus 1 ν U f ( η) v 2 x ψ The stream function is: ν x U ν x f ( η) d f η But η 1 y U 1 η 2 x 2 x ν x x dη x ν U x d f 1 η 1 ν U ( η f'( η) f ( η) ) dη 2 x 2 x ν U x which is Eq. 9.10. v The acceleration in the x-direction is given by: x u U ax u x u v η 1 η U f''( η) d d f'( η) η U f''( η) x 2 dη 2 x dx y u y where u U f'( η) u U 1 2 ν U x ( η f'( η) f ( η) ) Evaluating the partial derivatives: U d d f'( η) η U f''( η) ν x dη dy 2 Therefore: 2 1 η U f'' 1 ν U 1 U U 1 U ax U f'( η) ( η f' f ) U f'' η f' f'' ( η f' f'' f f'') 2 x ν x 2 x 2 x 2 x Simplifying yields: 2 ax U 2x f ( η) f''( η) If we plot f(η)f''(η) as a function of η: f(η)f''(η) The maximum value of this function is 0.23 at η of approximately 3. 0.2 2 axmax 0.115 0.1 0 0 1 2 3 η 4 5 U x Problem 9.35 [Difficulty: 4] Given: Blasius solution for laminar boundary layer Find: Point at which u = 0.95U; Slope of streamline; expression for skin friction coefficient and total drag; Momentum thickness Solution: Basic equation: Use results of Blasius solution (Table 9.1 on the web), and f' f' u U u U at η 3.5 0.9555 at η 4.0 f' 0.95 From Table A.10 at 20oC ν 1.50 10 The streamline slope is given by dx dy dx We have v 1 Rex η 3.5 2 5 m U 5 and s ν x y η dy U 0.9130 Hence by linear interpolation, when Hence ν x η y x 20 cm s u U f' where u 2 m ( 0.95 0.9130)η 3.94 y 0.305 cm U ν U ( 4 3.5) ( 0.9555 0.9130) x ( η f' f ) 1 U f' 1 2 U x ν U x v and ( η f' f ) f' 1 2 Rex 1 2 ν U x ( η f' f ) ( η f' f ) f' 4 Rex 6.67 10 ν From the Blasius solution (Table 9.1 on the web) Hence by linear interpolation f 1.8377 at η 3.5 f 2.3057 at η 4.0 f 1.8377 dy dx The shear stress is 1 2 Rex ( 2.3057 1.8377) ( 4.0 3.5) ( η f' f ) f' ( 3.94 3.5) f 2.25 0.00326 u v μ u at y = 0 (v = 0 at the wall for all x, so the derivative is zero there) x y y τw μ 2 2 U d f τw μ U ν x 2 dη and at η = 0 d f dη 2 0.3321 (from Table 9.1) τw 0.3321 U The friction drag is 2 ρ U μ μ ρ U 2 τw 0.3321 ρ U 0.3321 ρ U x Rex x FD τw dA L τw b dx where b is the plate width 0 L L 2 1 ρ U 2 ν dx FD 0.3321 b dx 0.3321 ρ U 1 U Rex 2 0 x 1 2 ν 2 FD 0.3321 ρ U b 2 L U For the momentum integral τw 2 ρ U θL We have dθ or dx 2 FD ρ U b L dθ τw 2 0.6642 ReL dx ρ U L 1 FD 0.6642 L τw dx 2 2 b ReL ρ U 0 ρ U 1 L 1 m θL 0 ReL 0.6642 L ReL U L ν ReL 3.33 10 θL 0.115 cm 5 Problem 9.36 Given: Data on flow over flat plate Find: Plot of laminar thickness at various speeds Solution: Given or available data: Governing Equations: δ x 5.48 2 5 m ν 1.5 10 (9.21) and Rex The critical Reynolds number is [Difficulty: 2] (from Table A.10 at 20oC) s U x Rex ν δ 5.48 so ν x U Recrit 500000 Hence, for velocity U the critical length xcrit is x crit 500000 ν U The calculations and plot were generated in Excel and are shown below: U (m/s) x c rit (m) 1 7.5 2 3.8 3 2.5 4 1.9 5 1.5 10 0.75 x (m) δ (mm) δ (mm) δ (mm) δ (mm) δ (mm) δ (mm) 0.000 0.025 0.050 0.075 0.100 0.2 0.5 1.5 1.9 2.5 3.8 5.0 0.00 3.36 4.75 5.81 6.71 9.49 15.01 25.99 29.26 33.56 41.37 47.46 0.00 2.37 3.36 4.11 4.75 6.71 10.61 18.38 20.69 23.73 29.26 0.00 1.94 2.74 3.36 3.87 5.48 8.66 15.01 16.89 19.37 0.00 1.68 2.37 2.91 3.36 4.75 7.50 13.00 14.63 0.00 1.50 2.12 2.60 3.00 4.24 6.71 11.62 0.00 1.06 1.50 1.84 6.0 7.5 51.99 58.12 Laminar Boundary Layer Profiles δ (mm) 70 60 U = 1 m/s 50 U = 2 m/s U = 3 m/s 40 U = 4 m/s 30 U = 5 m/s 20 U = 10 m/s 10 0 0 2 4 x (m) 6 8 Problem 9.37 Given: Blasius nonlinear equation Find: Blasius solution using Excel [Difficulty: 5] Solution: The equation to be solved is 2 d3 f d 3 f d2 f d 2 0 (9.11) The boundary conditions are f 0 and df 0 at 0 d df 1 at d Recall that these somewhat abstract variables are related to physically meaningful variables: f (9.12) u f U and y U x y Using Euler’s numerical method f n1 f n f n (1) f n1 f n f n (2) f n1 f n f n h In these equations, the subscripts refer to the nth discrete value of the variables, and = 10/N is the step size for (N is the total number of steps). But from Eq. 9.11 f 1 f f 2 so the last of the three equations is 1 f n1 f n f n f n 2 (3) Equations 1 through 3 form a complete set for computing f , f , f . All we need is the starting condition for each. From Eqs. 9.12 f 0 0 and f 0 0 We do NOT have a starting condition for f ! Instead we must choose (using Solver) f 0 so that the last condition of Eqs. 9.12 is met: f N 1 Computations (only the first few lines of 1000 are shown): = 0.01 Make a guess for the first f ''; use Solver to vary it until f 'N = 1 Count 0 1 2 3 4 5 6 7 10 8 9 10 8 11 12 6 13 14 4 15 16 2 17 18 0 19 0.0 20 21 22 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.2 0.20 0.21 0.22 f f' f'' 0.0000 0.0000 0.3303 0.0000 0.0033 0.3303 0.0000 0.0066 0.3303 0.0001 0.0099 0.3303 0.0002 0.0132 0.3303 0.0003 0.0165 0.3303 0.0005 0.0198 0.3303 Blasius 0.0007 Velocity 0.0231 Profile 0.3303 0.0009 0.0264 0.3303 0.0012 0.0297 0.3303 0.0015 0.0330 0.3303 0.0018 0.0363 0.3303 0.0022 0.0396 0.3303 0.0026 0.0429 0.3303 0.0030 0.0462 0.3303 0.0035 0.0495 0.3303 0.0040 0.0528 0.3303 0.0045 0.0562 0.3303 0.0051 0.0595 0.3303 0.0056 0.0628 0.3303 0.6 0.0063 0.4 0.0661 0.3302 0.0069 0.0694 0.3302 u/U = f ' 0.0076 0.0727 0.3302 0.8 1.0 Problem 9.38 Given: Parabolic solution for laminar boundary layer Find: Plot of δ, δ*, and τ w versus x/L [Difficulty: 2] Solution: Given or available data: Basic equations: u U 2 ν 1.08 10 y δ δ y 2 5 ft δ x 2 s (From Table A.8 at 68 oF) L 9 in 5.48 cf Rex τw 1 2 2 ρ U U 5 ft s 0.730 Rex 1 1 1 3 u y u 2 2 Hence: * 1 dy 1 d 1 2 d U U 3 0 3 0 0 0 1 The computed results are from Excel, shown below: Laminar Boundary Layer Profiles δ (in) 0.000 0.019 0.026 0.032 0.037 0.042 0.046 0.050 0.053 0.056 0.059 0.062 0.065 0.067 0.070 0.072 0.075 0.077 0.079 0.082 0.084 δ * (in) τ w (psi) 0.000 0.006 0.1344 0.009 0.0950 0.011 0.0776 0.012 0.0672 0.014 0.0601 0.015 0.0548 0.017 0.0508 0.018 0.0475 0.019 0.0448 0.020 0.0425 0.021 0.0405 0.022 0.0388 0.022 0.0373 0.023 0.0359 0.024 0.0347 0.025 0.0336 0.026 0.0326 0.026 0.0317 0.027 0.0308 0.028 0.0300 0.09 0.16 0.08 0.14 0.07 0.12 δ 0.06 0.10 0.05 τw (psi) 0.00 0.45 0.90 1.35 1.80 2.25 2.70 3.15 3.60 4.05 4.50 4.95 5.40 5.85 6.30 6.75 7.20 7.65 8.10 8.55 9.00 Re x 0.00.E+00 1.74.E+04 3.47.E+04 5.21.E+04 6.94.E+04 8.68.E+04 1.04.E+05 1.22.E+05 1.39.E+05 1.56.E+05 1.74.E+05 1.91.E+05 2.08.E+05 2.26.E+05 2.43.E+05 2.60.E+05 2.78.E+05 2.95.E+05 3.13.E+05 3.30.E+05 3.47.E+05 δ and δ * (in) x (in) 0.08 0.04 τw 0.03 0.06 0.04 δ* 0.02 0.02 0.01 0.00 0.00 0 3 6 x (in) 9 Problem 9.39 Given: Parabolic solution for laminar boundary layer Find: Derivation of FD; Evaluate FD and θ L [Difficulty: 2] Solution: Basic equations: u U 2 y δ y δ L 9 in Assumptions: 2 δ x b 3 ft 5.48 Rex U 5 ft ρ 1.94 s slug ft p 0, and U = const x 2) δ is a function of x only 3) Incompressible 4) Steady flow 3 1) Flat plate so The momentum integral equation then simplifies to τw ρ d dx U θ 2 θ where δ u U 1 u dy U 0 2 dθ For U = const τw ρ U dx The drag force is then FD τw dA For the given profile θ δ 1 0 θ From Table A.7 at 68 oF 2 15 θL L L 2 dθ 2 ρ U b dx ρ U b 1 dθ τw b dx dx 0 0 1 0 1 2 FD ρ U b θL u u 2 2 2 3 4 2 1 dη 2 η η 1 2 η η dη 2 η 5 η 4 η η dη U U 15 0 0 δ ν 1.08 10 δL L 5 ft 5.48 ReL 2 θL δ 15 L 2 FD ρ U b θL 2 s ReL U L ν δL 0.0837 in θL 0.01116 in FD 0.1353 lbf ReL 3.47 10 5 Problem 9.40 Given: [Difficulty: 3] Thin flat plate installed in a water tunnel. Laminar BL's with parabolic profiles form on both sides of the plate. L 0.3 m Find: Solution: b 1 m U 1.6 m s ν 1 10 2 6 m u U s 2 y δ δ y 2 Total viscous drag force acting on the plate. We will determine the drag force from the shear stress at the wall U L First we will check the Reynolds number of the flow: ReL ν 5 4.8 10 Therefore the flow is laminar throughout. L The viscous drag for the two sides of the plate is: FD 2 τw b dx The wall shear stress τw is: 0 2 at y = 0, which for the parabolic profile yields: 2 0 2 μ U u τw μ U δ 2 δ y δ τw μ The BL thickness δ is: L 1 δ 5.48 2 x Therefore: FD 2 b U ν L 1 2 μ U 4 U 2 dx b μ U x dx 1 5.48 ν 0 ν 2 x 5.48 U 0 Evaluating this integral: FD 8 b μ U 5.48 U L ν FD 1.617 N Problem 9.41 [Difficulty: 2] Given: Data on fluid and plate geometry Find: Drag at both orientations using boundary layer equation Solution: The given data is ρ 1.5 slug ft μ 0.0004 3 lbf s ft ReL First determine the nature of the boundary layer 2 ρ U L μ U 10 ft L 10 ft s ReL 3.75 10 b 3 ft 5 The maximum Reynolds number is less than the critical value of 5 x 105 Hence: Governing equations: cf τw 1 2 cf (9.22) 2 ρ U 0.730 (9.23) Rex L The drag (one side) is FD τw b dx 0 Using Eqs. 9.22 and 9.23 FD ρ U b 2 1 L 2 0.73 ρ U x dx μ 0 3 Repeating for FD 0.73 b μ L ρ U FD 5.36 lbf L 3 ft b 10 ft 3 FD 0.73 b μ L ρ U FD 9.79 lbf (Compare to 6.25 lbf for Problem 9.18) (Compare to 12.5 lbf for Problem 9.19) Problem 9.42 Given: Triangular plate Find: Drag [Difficulty: 3] Solution: Basic equations: cf τw 1 2 cf 2 ρ U L 2 ft 3 0.730 Rex L 1.732 ft 2 W 2 ft U 15 ft s Assumptions: (1) Parabolic boundary layer profile (2) Boundary layer thickness is based on distance from leading edge (the "point" of the triangle). From Table A.9 at 70 oF ν 1.63 10 4 ft 2 ρ 0.00233 s ft ReL First determine the nature of the boundary layer The drag (one side) is We also have slug U L ν 3 ReL 2 10 5 so definitely laminar L FD τw dA FD τw w( x ) dx w( x ) W 0 x L 1 1 2 2 0.730 τw cf ρ U ρ U 2 2 Rex L Hence 1 2 W FD ρ U 2 L L 1 0.730 0.730 x W 2 2 ρ U ν x dx dx 2 L U x 0 3 ν 0 L The integral is 3 1 2 2 2 x dx L 3 0 so 3 3 FD 0.243 ρ W ν L U FD 1.11 10 Note: For two-sided solution 2 FD 2.21 10 lbf 3 lbf Problem 9.43 [Difficulty: 3] Plate is reversed from this! Given: Triangular plate Find: Drag Solution: Basic equations: cf τw 1 2 2 ρ U L 2 ft From Table A.9 at 70 oF cf 3 4 ft 2 ρ 0.00233 s W 2 ft ReL U 15 U L ν ReL 2 10 5 so definitely laminar FD τw w( x ) dx w( x ) W 1 0 We also have 1 1 2 2 0.730 τw cf ρ U ρ U 2 2 Rex Hence 1 2 FD ρ U W 2 L 0.730 1 U x x L s 3 L FD τw dA ft slug ft First determine the nature of the boundary layer The drag (one side) is Rex L 1.732 ft 2 ν 1.63 10 0.730 3 0.730 2 dx ρ U W ν 2 L x 1 2 x L 2 x dx L 1 0 ν 0 The integral is L x 1 2 3 1 2 2 2 L 4 x 2 d 2 L x L 3 L 3 L 1 0 3 FD 0.487 ρ W ν L U FD 2.22 10 Note: For two-sided solution The drag is much higher (twice as much) compared to Problem 9.42. This is because τ w is largest near the leading edge and falls off rapidly; in this problem the widest area is also at the front 3 lbf 3 2 FD 4.43 10 lbf Problem 9.44 Given: Parabolic plate Find: Drag [Difficulty: 3] Solution: Basic equations: cf τw 1 2 0.730 cf 2 ρ U Rex W 2 L W 1 ft 2 L 0.25 ft 1 ft U 15 ft s Note: "y" is the equation of the upper and lower surfaces, so y = W/2 at x = L From Table A.9 at 70 oF ν 1.63 10 4 ft 2 ρ 0.00233 s ft ReL First determine the nature of the boundary layer The drag (one side) is slug 3 U L 4 ReL 2.3 10 ν so just laminar L FD τw dA FD τw w( x ) dx w( x ) W 0 We also have 1 1 2 2 0.730 τw cf ρ U ρ U 2 2 Rex Hence 1 2 FD ρ U W 2 x L L 0.730 U x x L 3 dx 0.730 2 L ν ρ U W 1 dx L 2 0 ν 0 3 FD 0.365 ρ W ν L U FD 3.15 10 Note: For two-sided solution 4 lbf 4 2 FD 6.31 10 lbf Problem 9.45 [Difficulty: 4] Note: Plate is now reversed! Given: Parabolic plate Find: Drag Solution: Basic equations: cf τw 1 2 0.730 cf 2 ρ U Rex W 2 L W 1 ft 2 L 0.25 ft 1 ft U 15 ft s Note: "y" is the equation of the upper and lower surfaces, so y = W/2 at x = 0 From Table A.10 at 70oF ν 1.63 10 4 ft 2 ρ 0.00234 s ft U L ReL First determine the nature of the boundary layer The drag (one side) is slug ν 3 4 ReL 2.3 10 so just laminar L FD τw dA FD τw w( x ) dx w( x ) W 1 0 We also have 1 1 2 2 0.730 τw cf ρ U ρ U 2 2 Rex Hence 1 2 FD ρ U W 2 x L L x 0.730 1 L U x 3 dx 0.730 2 ρ U W ν L 2 1 x 1 L dx 0 ν 0 The tricky integral is (this might be easier to do numerically!) 0.730 2 x 1 3 FD 2 L x x so dx x L ln x L L 2 L x x 1 2 ρ U W ν i L 1 x 1 L dx 0.434 m 0 L 1 x 1 L dx 0 Note: For two-sided solution The drag is much higher compared to Problem 9.44. This is because τ w is largest near the leading edge and falls off rapidly; in this problem the widest area is also at the front FD 4.98 10 4 lbf 4 2 FD 9.95 10 lbf Problem 9.46 Given: Pattern of flat plates Find: Drag on separate and composite plates Solution: Basic equations: cf 1 0.730 cf 2 Rex ρ U 2 Parabolic boundary layer profile Assumption: For separate plates τw [Difficulty: 3] L 3 in W 3 in U 3 We also have From Table A.7 at 70 oF ν 1.06 10 s ReL First determine the nature of the boundary layer The drag (one side) is ft U L ReL 7.08 10 ν 4 5 ft 2 ρ 1.93 slug s so definitely laminar L FD τw dA FD τw W dx 0 1 1 2 2 0.730 τw cf ρ U ρ U 2 2 Rex L Hence 1 2 FD ρ U W 2 L 1 0.730 0.730 2 2 ρ U W ν x dx dx 2 U x 0 3 ν 0 L The integral is 1 1 2 2 dx 2 L x so 3 FD 0.730 ρ W ν L U FD 0.0030 lbf FTotal 4 FD FTotal 0.0119 lbf 0 This is the drag on one plate. The total drag is then For both sides: For the composite plate L 4 3 in L 1.00 ft ReL U L ν 2.83 10 5 2 FTotal 0.0238 lbf so still laminar 3 FComposite 0.730 ρ W ν L U FComposite 0.0060 lbf For both sides: 2 FComposite 0.0119 lbf The drag is much lower on the composite compared to the separate plates. This is because τ w is largest near the leading edges and falls off rapidly; in this problem the separate plates experience leading edges four times! ft 3 Problem 9.47 [Difficulty: 3] u y Given: Laminar boundary layer flow with linear velocity profile: Find: Expressions for δ/x and Cf using the momentum integral equation Solution: We will apply the momentum integral equation Governing Equations: U τw ρ Cf 1 η (Momentum integral equation) τw 2 Assumptions: 2 d d U θ δdisp U U dx dx δ (Skin friction coefficient) 2 ρ U (1) Zero pressure gradient, so U is constant and dp/dx = 0 (2) δ is a function of x only, and δ = 0 at x = 0 (3) Incompressible flow 1 u u 2 d Applying the assumptions to the momentum integral equation yields: τw ρ U θ ρ U δ 1 dη U U dx dx 0 1 2 2 2 1 Substituting for the velocity profile: τw ρ U d δ η η dη ρ U d δ dx 6 dx 0 2 d at y = 0 u y τw μ Now the wall shear stress is also: 2 δ 2 6 μ ρ U Solving for the boundary layer thickness: Substituting the velocity profile: μ U 2 1 d ρ U δ 6 dx δ Setting both expressions for the wall shear stress equal: Integrating this expression: τw μ U δ 6 μ Separating variables: δ dδ ρ U dx 2 x C However, we know that C = 0 since δ = 0 when x = 0. Therefore: δ From the definition for skin friction coefficient: 12 μ ρ U Cf x δ or x τw 1 2 2 ρ U μ U δ δ 2 12 μ δ ρ U x x 2 2 ρ U 2 μ ρ U δ 2 μ x ρ U x δ 2 6 μ ρ U x 3.46 Rex Rex Rex 3.46 Upon simplification: Cf 0.577 Rex Problem 9.48 Given: [Difficulty: 2] Horizontal surface immersed in a stream of standard air. Laminar BL with sinusoidal profile forms. L 1.8 m b 0.9 m Find: Solution: U 3.2 m s ν 1.46 10 2 5 m u U s π y 2 δ sin Plot δ, δ*, and τw versus x/L for the plate We will determine the drag force from the shear stress at the wall Governing τw ρ U2 d θ μ u at y = 0 dx y Equations: δdisp δ 1 1 u dη U (Wall shear stress) (Displacement thickness) 0 δ θ 1 u U 1 u dη (Momentum thickness) U 0 1 π θ π sin η 1 sin η dη δ 2 2 For the sinusoidal velocity profile: 0 Evaluating this integral: θ δ 4π π μ U 2 π 2 4 π dx δ dδ or δ dδ 2 ρ U Solving this expression for δ/x: Also, δdisp δ 1 4 π d d θ θ δ δ 2 π x dx dδ x π 0 π π μ U 2 4 π τw μ U cos ρ U δ 2 δ 2 π x 2 δ 2 δ To determine the wall shear stress: Separating variables yields: π π 2 sin η sin η dη 2 2 0 0.1366 Therefore it follows that 2 π 1 δ x π 2 4π 1 sin π η dη 2 π 2 μ 4 π ρ U 2 dx Integrating yields: 2 π 2 μ x 4 π ρ U μ δ ρ U x x Evaluating this integral: δdisp δ 1 2 π 0.363 0 The Reynolds number is related to x through: δ 5 Rex 2.19 10 x Plots of δ, δdisp and τ w as functions of x are shown on the next page. where x is measured in meters. δdisp δ 4.80 Rex 0.363 BL Thickness Disp. Thickness Wall Shear 0.04 10 0.03 0.02 5 0.01 0 0 0.5 1 x (m) 1.5 0 Wall Shear Stress (Pa) Boundary Layer and Displacement Thicknesses (mm) 15 Problem 9.49 Given: Water flow over flat plate Find: Drag on plate for linear boundary layer [Difficulty: 3] Solution: Basic equations: FD 2 τw dA du τw μ dy W 1 m 2 6 m ρ 1000 s ReL First determine the nature of the boundary layer y The velocity profile is u U U η δ du U Hence τw μ μ dy δ We also have 2 dδ τw ρ U The integral is (1) U 0.8 u dη U ν m s 3 m U L ReL 2.15 10 5 so laminar but we need δ(x) 1 1 u η η2 dx 16 so U 1 2 dδ τw μ ρ U δ 6 dx Separating variables δ dδ δ 6 μ 12 μ ρ U FD 2 2 dx or x or ρ U 1 2 dδ 2 dδ τw ρ U ρ U dx 6 dx δ 2 δ x L U τw dA 2 W μ dx 2 W δ 0 L 0 6 μ ρ U 12 Rex 1 2 dx 2 L x so 0 FD 2 3 3 ρ W ν L U (2) but δ(0) = 0 so c = 0 x c 3.46 Rex L 1 μWU U ρ U 2 2 x dx x dx μ U ν 0 12 μ 3 1 L The integral is 1 kg 1 Comparing Eqs 1 and 2 Then u U u 2 dδ 1 dη ρ U η ( 1 η) dη dx U U dx 0 0 0 Hence τw ρ U dx at y = 0, and also 1 0 L 0.35 m From Table A.8 at 10 oC ν 1.30 10 2 dδ FD 2 μ W U 3 FD 0.557N U L ν Problem 9.50 Given: [Difficulty: 2] Horizontal surface immersed in a stream of standard air. Laminar BL with linear profile forms. L 0.8 m b 1.9 m Find: Solution: Plot δ, δ*, U 5.3 m s ν 1.46 10 2 5 m u U s y δ and τw versus x/L for the plate We will determine the drag force from the shear stress at the wall Governing τw ρ U2 d θ μ u at y = 0 dx y Equations: δdisp δ 1 1 u dη U (Wall shear stress) (Displacement thickness) 0 δ θ 1 u U 1 u dη (Momentum thickness) U 0 For the linear velocity profile: 1 1 2 η ( 1 η ) dη η η dη δ 0 0 θ δ 1 Therefore it follows that d θ d θ δ δ To determine the wall shear stress: dx dδ x 6 x Separating variables yields: Also, δdisp δ 6 μ ρ U 2 dx δ dδ Evaluating this integral: 0 The Reynolds number is related to x through: 2 1 ( 1 η ) dη δ Integrating yields: δdisp δ 5 Rex 3.63 10 x Plots of δ, δdisp and τ w as functions of x are shown on the next page. 6 μ ρ U θ Evaluating this integral: x τw μ U δ 1 6 2 ρ U 6 0.1667 δ x Solving this expression for δ/x: δ x 1 δdisp 2 δ where x is measured in meters. 3.46 Rex 1 2 BL Thickness Disp. Thickness Wall Shear 0.04 4 0.03 2 0.02 0 0 0.2 0.4 x (m) 0.6 0.01 0.8 Wall Shear Stress (Pa) Boundary Layer and Displacement Thicknesses (mm) 6 Problem 9.51 Given: Horizontal surface immersed in a stream of standard air. Laminar BL with linear profile forms. L 0.8 m b 1.9 m Find: Solution: [Difficulty: 2] U 5.3 m s ν 1.46 10 2 5 m u U s y δ Algebraic expressions for wall shear stress and drag; evaluate at given conditions We will determine the drag force from the shear stress at the wall Governing τw ρ U2 d θ μ u at y = 0 dx y Equations: δ θ 1 u U 1 u dη (Wall shear stress) (Momentum thickness) U 0 1 1 2 η ( 1 η ) dη η η dη δ 0 0 θ For the linear velocity profile: δ 1 Therefore it follows that d θ d θ δ δ To determine the wall shear stress: dx dδ x 6 x Separating variables yields: 6 μ ρ U 2 dx δ dδ 2 Substituting this back into the expression for wall shear stress: The drag force is given by: For the given conditions: FD τw dA ReL U L ν δ Integrating yields: τw μ U δ 6 μ ρ U x x μ U δ 1 6 2 ρ U 6 12 1 12 μ U Rex x 0.1667 δ x Solving this expression for δ/x: μ U τw δ x 12 Rex μ U τw 0.289 Rex x Rex θ L L L 2 2 dθ ρ U b dx b ρ U dθ τw b dx dx 0 0 2.90 10 θ Evaluating this integral: 2 FD ρ U b θL 0 5 12 δL L 5.14 mm ReL δL θL 0.857 mm 6 2 FD ρ U b θL 0.0563 N FD 0.0563 N Problem 9.52 Given: Data on flow in a channel Find: Static pressures; plot of stagnation pressure [Difficulty: 3] Solution: The given data is h 1.2 in Appendix A ρ 0.00239 δ2 0.4 in w 6 in slug ft Governing equations: ft U2 75 s 3 Mass Before entering the duct, and in the the inviscid core, the Bernoulli equation holds 2 p V g z constant ρ 2 (4.24) Assumptions: (1) Steady flow (2) No body force in x direction For a linear velocity profile, from Table 9.2 the displacement thickness at location 2 is δ2 δdisp2 2 δdisp2 0.2 in From the definition of the displacement thickness, to compute the flow rate, the uniform flow at location 2 is assumed to take place in the entire duct, minus the displacement thicknesses at top and bottom Then 2 A2 w h 2 δdisp2 A2 4.80 in Q A2 U2 Q 2.50 ft 3 s Mass conservation (Eq. 4.12) leads to U2 U1 A1 U2 A2 A2 U1 U A1 2 where A1 w h 2 A1 7.2 in ft U1 50 s The Bernoull equation applied between atmosphere and location 1 is p atm ρ p1 ρ U1 2 2 or, working in gage pressures 1 2 p 1 ρ U1 2 p 1 0.0207 psi (Static pressure) Similarly, between atmosphere and location 2 (gage pressures) 1 2 p 2 ρ U2 2 p 2 0.0467 psi (Static pressure) The static pressure falls continuously in the entrance region as the fluid in the central core accelerates into a decreasing core. The stagnation pressure at location 2 (measured, e.g., with a Pitot tube as in Eq. 6.12), is indicated by an application of the Bernoulli equation at a point pt ρ p ρ u 2 2 where p t is the total or stagnation pressure, p = p 2 is the static pressure, and u is the local velocity, given by u U2 y y δ2 δ2 h δ2 y 2 u U2 (Flow and pressure distibutions are symmetric about centerline) Hence y (in) 0.00 0.04 0.08 0.12 0.16 0.20 0.24 0.28 0.32 0.36 0.40 0.44 0.48 0.52 0.56 0.60 1 2 p t p 2 ρ u 2 The plot of stagnation pressure is shown in the Excel sheet below Stagnation Pressure Distibution in a Duct u (ft/s) p t (psi) 0.00 7.50 15.00 22.50 30.00 37.50 45.00 52.50 60.00 67.50 75.00 75.00 75.00 75.00 75.00 75.00 0.000 0.000 0.002 0.004 0.007 0.012 0.017 0.023 0.030 0.038 0.047 0.047 0.047 0.047 0.047 0.047 0.6 0.5 0.4 y (in) 0.3 0.2 0.1 0.0 0.00 0.01 0.02 0.03 p t (psi gage) The stagnation pressure indicates total mechanical energy - the curve indicates significant loss close to the walls and no loss of energy in the central core. 0.04 0.05 Problem 9.53 Given: [Difficulty: 3] Turbulent boundary layer flow of water, 1/7-power profile The given or available data (Table A.9) is U 1 m s Find: L 1 m ν 1.00 10 2 6 m ρ 999 s kg 3 m (a) Expression for wall shear stress (b) Integrate to obtain expression for skin friction drag (c) Evaluate for conditions shown Solution: Basic Equation: Cf τw 1 2 0.0594 (Skin friction factor) 1 2 ρ U Rex 5 Assumptions: 1) Steady flow 2) No pressure force 3) No body force in x direction 4) Uniform flow at ab 1 τw 0.0594 ρ U Rex 2 Solving the above expression for the wall shear stress: 2 1 5 1 U 2 τw 0.0594 ρ U 2 ν 1 5 x 1 5 L L 1 1 1 L 5 1 U 2 5 5 FD τw b dx 0.0594 ρ U dx where c is defined: x b dx c b x 2 ν 0 0 Integrating to find the drag: 0 1 U 2 c 0.0594 ρ U 2 ν 1 4 5 Therefore the drag is: 5 5 1 U L 5 2 FD c b L 0.0594 ρ U b L 4 4 2 ν Upon simplification: 1 5 1 0.0721 2 FD ρ U b L 1 2 ReL Evaluating, with b 1 m ReL U L ν 1 10 6 1 0.0721 2 FD ρ U b L 1 2 ReL 5 FD 2.27 N 5 Problem 9.54 [Difficulty: 3] Note: Figure data applies to problem 9.18 only Given: Data on fluid and turbulent boundary layer Find: Mass flow rate across ab; Momentum flux across bc; Distance at which turbulence occurs Solution: CV Mass Basic equations: d Momentum c Rx Assumptions: 1) Steady flow 2) No pressure force 3) No body force in x direction 4) Uniform flow at ab The given or available data (Table A.10) is U 50 m s δ 19 mm Consider CV abcd b 3 m ρ 1.23 kg ν 1.50 10 3 2 5 m m kg mad 3.51 s mad ρ U b δ (Note: Software cannot render a dot) 1 δ mad ρ u b dy mab 0 Mass s and in the boundary layer u U 0 7 1 y η7 δ dy dη δ 1 Hence 1 7 7 mab ρ U b δ ρ U η δ dη ρ U b δ ρ U b δ 8 0 1 mab ρ U b δ 8 kg mab 0.438 s 1 δ The momentum flux across bc is δ mfbc u ρ V dA 0 0 2 7 2 2 7 u ρ u b dy ρ U b δ η dη ρ U b δ 9 0 7 2 mfbc ρ U b δ 9 mfbc 136.3 2 s From momentum Rx U ( ρ U δ) mab u ab mfbc Transition occurs at Rex 5 10 5 kg m and 2 Rx ρ U b δ mab U mfbc U x Rex ν x trans Rx 17.04 N Rex ν U x trans 0.1500 m Problem 9.55 U 10 [Difficulty: 3] m L 5 m ν 1.45 10 2 5 m Given: Data on flow over a flat plate Find: Plot of laminar and turbulent boundary layer; Speeds for transition at trailing edge s (from Table A.10) s Solution: Governing For laminar flow δ 5.48 Equations: x Rex The critical Reynolds number is (9.21) and U x Rex ν δ 5.48 so Recrit 500000 Hence, for velocity U the critical length xcrit is δ x 0.382 (9.26) Rex δ 0.382 so 1 For (a) completely laminar flow Eq. 1 holds; for (b) completely turbulent flow Eq. 3 holds; for (c) transitional flow Eq.1 or 3 holds depending on xcrit in Eq. 2. Results are shown below from Excel. Re x 0.00 0.125 0.250 0.375 0.500 0.700 0.75 1.00 1.50 2.00 3.00 4.00 0.00E+00 8.62E+04 1.72E+05 2.59E+05 3.45E+05 4.83E+05 5.17E+05 6.90E+05 1.03E+06 1.38E+06 2.07E+06 2.76E+06 5.00 3.45E+06 (a) Laminar (b) Turbulent (c) Transition δ (mm) δ (mm) δ (mm) 0.00 0.00 0.00 2.33 4.92 2.33 3.30 8.56 3.30 4.04 11.8 4.04 4.67 14.9 4.67 5.52 19.5 5.5 5.71 20.6 20.6 6.60 26.0 26.0 8.08 35.9 35.9 9.3 45.2 45.2 11.4 62.5 62.5 13.2 78.7 78.7 14.8 94.1 94.1 4 5 x U 5 x (m) ν 5 (1) U x crit 500000 1 For turbulent flow ν x (3) ν U (2) Boundary Layer Profiles on a Flat Plate 100 75 δ (mm) Laminar Turbulent Transitional 50 25 0 0 0.5 1 1.5 2 2.5 x (m) 3 The speeds U at which transition occurs at specific points are shown below x trans (m) 5 4 3 2 1 U (m/s) 1.45 1.81 2.42 3.63 7.25 3.5 4 4.5 5 Problem 9.56 Turbulent boundary layer flow of water L 1 m Find: Solution: Governing Equations: Plot δ, δ*, U 1 1 2 6 m m ν 1.00 10 s u U s y δ 7 and τw versus x/L for the plate We will determine the drag force from the shear stress at the wall δ x 0.382 (Boundary layer thickness) 1 Rex δdisp δ Cf 5 1 (Displacement thickness) 8 τw 1 2 0.0594 (Skin friction factor) 1 2 ρ U Rex 5 Assumption: Boundary layer is turbulent from x = 0 For the conditions given: ReL U L ν 6 1.0 10 1 q 2 2 ρ U 500 Pa τw 0.0594 1 Rex 30 Boundary Layer and Displacement Thicknesses (mm) Here is the plot of boundary layer thickness and wall shear stress: q 29.7 Pa Rex 1 5 5 3 BL Thickness Disp. Thickness Wall Shear 20 2 10 1 0 0 0.5 x (m) 0 1 Wall Shear Stress (Pa) Given: [Difficulty: 2] Problem 9.57 Given: Triangular plate Find: Drag [Difficulty: 3] Solution: Basic equations: cf τw 1 2 2 ρ U L 2 ft From Table A.10 at 70oF cf We also have 2 4 ft ρ 0.00234 s ReL U 80 ft s slug ft 3 U L ReL 9 10 ν 5 so definitely still laminar over a significant portion of the plate, but we are told to assume turbulent! L FD τw dA FD τw w( x ) dx w( x ) W 0 x L 1 1 2 2 0.0594 τw cf ρ U ρ U 1 2 2 1 2 W FD ρ U 2 L L The integral is W 2 ft 2 Rex Hence 5 L 1.732 ft First determine the nature of the boundary layer The drag (one side) is 1 Rex 3 ν 1.62 10 0.0594 9 4 5 5 5 x dx L 9 0 L 5 1 9 0.0594 x 1 U x ν dx 0.0594 2 L 4 5 5 ν x dx L 0 5 W ρ U 5 0 1 so 4 9 FD 0.0165 ρ W L ν U Note: For two-sided solution 5 FD 0.0557 lbf 2 FD 0.1114 lbf Problem 9.58 Given: Parabolic plate Find: Drag [Difficulty: 3] Solution: Basic equations: cf τw 1 2 cf 2 ρ U 0.0594 1 Rex 5 W 2 L W 1 ft 2 L 3 in 1 ft U 80 ft s Note: "y" is the equation of the upper and lower surfaces, so y = W/2 at x = L From Table A.9 at 70 oF ν 1.63 10 4 ft 2 ρ 0.00233 s ft ReL First determine the nature of the boundary layer The drag (one side) is We also have 3 U L ReL 1.23 10 ν 5 so still laminar, but we are told to assume turbulent! L FD τw dA FD τw w( x ) dx w( x ) W 0 x L 1 1 2 2 0.0594 τw cf ρ U ρ U 1 2 2 Rex Hence slug 1 2 FD ρ U W 2 5 L x 0.0594 L 1 U x ν 9 dx 0.0594 2 1 1 L 3 5 2 5 10 ρ U W L ν x dx 0 5 0 1 4 9 FD 0.0228 ρ W ν L U 5 FD 0.00816 lbf Note: For two-sided solution 2 FD 0.01632 lbf Problem 9.59 Given: Pattern of flat plates Find: Drag on separate and composite plates [Difficulty: 3] Solution: Basic equations: cf τw 1 2 For separate plates From Table A.7 at 70 oF cf 2 ρ U We also have L 3 in 5 W 3 in ν 1.06 10 5 ft 2 ρ 1.93 s ft s slug 3 U L ReL ReL 1.89 10 ν 6 so turbulent L FD τw dA 1 U 80 ft FD τw W dx 0 1 2 τw cf ρ U 2 2 2 0.0594 ρ U 1 Rex Hence 1 Rex First determine the nature of the boundary layer The drag (one side) is 0.0594 2 FD ρ U W 2 5 1 9 L 1 0.0594 1 U x ν dx 0.0594 2 L 1 5 5 5 ρ U W ν x dx 0 5 0 L The integral is 4 1 5 5 5 dx L x 4 0 This is the drag on one plate. The total drag is then 1 so 4 9 FD 0.0371 ρ W ν L U 5 FTotal 4 FD FD 1.59 lbf FTotal 6.37 lbf For both sides: 2 FTotal 12.73 lbf For the composite plate L 4 3 in L 12.00 in and since the Reynolds number for the single plate was turbulent, we know that the flow around the composite plate will be turbulent as well. 1 4 9 FComposite 0.0371 ρ W ν L U 5 FComposite 4.82 lbf For both sides: 2 FComposite 9.65 lbf The drag is much lower on the composite compared to the separate plates. This is because τ w is largest near the leading edges and falls off rapidly; in this problem the separate plates experience leading edges four times! Problem 9.60 [Difficulty: 3] 1 u 1 6 y η 6 δ Given: Turbulent boundary layer flow with 1/6 power velocity profile: Find: Expressions for δ/x and Cf using the momentum integral equation; compare to 1/7-power rule results. Solution: We will apply the momentum integral equation U τw Governing Equations: ρ τw Cf 1 2 Assumptions: 2 d d U θ δdisp U U dx dx (Momentum integral equation) (Skin friction coefficient) 2 ρ U (1) Zero pressure gradient, so U is constant and dp/dx = 0 (2) δ is a function of x only, and δ = 0 at x = 0 0.25 (3) Incompressible flow 2 ν τ 0.0233 ρ U U δ (4) Wall shear stress is: w 1 u u 2 d τw ρ U θ ρ U δ 1 dη U U dx dx 0 1 2 1 2 6 6 2 6 Setting our two τ w's equal: Substituting for the velocity profile: τw ρ U d δ η η dη ρ U d δ dx 56 x d 0 Applying the assumptions to the momentum integral equation yields: 2 d 1 1 0.0233 ρ U 2 U δ ν 0.25 d δ 56 dx 2 6 ρ U Simplifying and separating variables: 4 δ dδ 0.0233 56 6 ν 4 dx U 4 1 4 4 4 4 56 ν 5 56 ν δ 0.0233 x C but C = 0 since δ = 0 at x = 0. Therefore: δ 0.0233 x 5 6 U 6 U 4 5 Integrating both sides: 1 5 In terms of the Reynolds number: δ x 0.353 1 Rex 5 For the skin friction factor: 1 1 0.0233 ρ U 2 Cf τw 1 2 2 ρ U 1 2 U δ ν 2 ρ U 4 4 1 1 Re 5 4 4 x ν x 4 0.0466 0.0466 Rex 0.353 Upon simplification: U x δ 1 1 Cf 0.0605 1 Rex These results compare to δ x 0.353 1 Rex 5 and Cf 0.0605 1 Rex 5 for the 1/7-power profile. 5 Problem 9.61 [Difficulty: 3] 1 Given: u Turbulent boundary layer flow with 1/6 power velocity profile: U The given or available data (Table A.9) is U 1 m s L 1 m 1 6 y η 6 δ 2 6 m ν 1.00 10 s ρ 999 Expressions for δ/x and Cf using the momentum integral equation; evaluate drag for the conditions given Solution: We will apply the momentum integral equation τw ρ τw Cf 1 2 Assumptions: 2 d d U θ δdisp U U dx dx 3 m Find: Governing Equations: kg (Momentum integral equation) (Skin friction coefficient) 2 ρ U (1) Zero pressure gradient, so U is constant and dp/dx = 0 (2) δ is a function of x only, and δ = 0 at x = 0 0.25 (3) Incompressible flow ν 2 τw 0.0233 ρ U (4) Wall shear stress is: U δ 1 u u 2 d Applying the assumptions to the momentum integral equation yields: τw ρ U θ ρ U δ 1 dη U U dx dx 0 1 2 1 2 6 6 2 6 Setting our two τ w's equal: Substituting for the velocity profile: τw ρ U d δ η η dη ρ U d δ dx 56 dx 0 2 d 1 1 0.0233 ρ U 2 U δ ν 0.25 d δ 56 dx 2 6 ρ U Simplifying and separating variables: 4 δ dδ 0.0233 56 ν 4 dx 6 U 4 1 4 4 56 ν 5 56 ν 4 4 δ 0.0233 x C but C = 0 since δ = 0 at x = 0. Therefore: δ 0.0233 x 6 U 6 U 5 4 5 Integrating both sides: 1 5 δ In terms of the Reynolds number: x 0.353 1 Rex 5 For the skin friction factor: 1 1 0.0233 ρ U 2 Cf τw 1 2 1 2 ρ U 2 U δ ν 4 2 ρ U 4 1 1 Re 5 4 4 x ν x 4 0.0466 0.0466 Rex 0.353 Upon simplification: U x δ 1 1 Cf 0.0605 1 Rex 5 L The drag force is: 1 1 L 1 1 L 5 5 0.0605 1 2 ν 5 2 ν 5 FD τw b dx 0.0605 ρ U x ρ U b x dx b dx 2 2 U U 0 0 0 1 Evaluating the integral: FD 0.0605 2 ρ U 2 ν U 5 4 5 5 b L 4 2 In terms of the Reynolds number: FD 0.0378 ρ U b L 1 ReL For the given conditions and assuming that b = 1 m: 6 ReL 1.0 10 5 and therefore: FD 2.38 N Problem 9.62 [Difficulty: 3] 1 u 1 8 y η8 δ Given: Turbulent boundary layer flow with 1/8 power velocity profile: Find: Expressions for δ/x and Cf using the momentum integral equation; compare to 1/7-power rule results. Solution: We will apply the momentum integral equation U τw Governing Equations: ρ τw Cf 1 2 Assumptions: 2 d d U θ δdisp U U dx dx (Momentum integral equation) (Skin friction coefficient) 2 ρ U (1) Zero pressure gradient, so U is constant and dp/dx = 0 (2) δ is a function of x only, and δ = 0 at x = 0 0.25 (3) Incompressible flow 2 ν τ 0.0233 ρ U U δ (4) Wall shear stress is: w 1 u u 2 d τw ρ U θ ρ U δ 1 dη U U dx dx 0 1 2 1 2 8 8 2 8 Setting our two τ w's equal: Substituting for the velocity profile: τw ρ U d δ η η dη ρ U d δ dx 90 x d 0 Applying the assumptions to the momentum integral equation yields: 2 d 1 1 0.0233 ρ U 2 U δ ν 0.25 d δ 56 dx 2 6 ρ U Simplifying and separating variables: δ dδ 0.262 4 ν 4 dx U 4 1 5 Integrating both sides: 4 5 δ 4 0.262 ν 4 x C U 1 4 5 ν but C = 0 since δ = 0 at x = 0. Therefore: δ 0.262 x 4 U In terms of the Reynolds number: δ x 5 0.410 1 Rex 5 For the skin friction factor: 1 1 0.0233 ρ U 2 Cf τw 1 2 2 ρ U 1 2 U δ ν 2 ρ U 4 4 1 1 Re 5 4 4 ν x x 4 0.0466 0.0466 Rex Upon simplification: U x δ 0.410 1 1 Cf 0.0582 1 Rex These results compare to δ x 0.353 1 Rex 5 and Cf 0.0605 1 Rex 5 for the 1/7-power profile. 5 Problem 9.63 Given: [Difficulty: 3] Turbulent boundary layer flow of water, 1/7-power profile The given or available data (Table A.9) is U 20 m s Find: L 1.5 m b 0.8 m ν 1.46 10 2 5 m s ρ 1.23 kg 3 x 1 0.5 m m (a) δ at x = L (b) τw at x = L (c) Drag force on the portion 0.5 m < x < L Solution: Basic equations: δ x 0.382 Rex Cf (Boundary Layer Thickness) 1 5 τw 1 2 0.0594 (Skin friction factor) 1 2 ρ U Rex 5 Assumptions: 1) Steady flow 2) No pressure force 3) No body force in x direction At the trailing edge of the plate: ReL U L ν 2.05 10 6 Therefore 0.382 δL 31.3 mm 1 ReL 1 2 0.0594 Similarly, the wall shear stress is: τwL ρ U 1 2 ReL δL L 5 τwL 0.798 Pa 5 L To find the drag: L 1 1 1 L 5 1 U 5 2 5 x b dx c b x FD τw b dx 0.0594 ρ U dx where c is defined: 2 ν x 0 x 1 1 U 1 2 c 0.0594 ρ U 2 ν 1 4 5 Therefore the drag is: 5 5 1 5 2 FD c b L ρ U b L CfL x 1 Cfx1 4 4 2 At x = x1: Rex1 U x 1 ν 6.849 10 5 Cfx1 0.0594 1 Rex1 3 4.043 10 and at x = L CfL 5 0.0594 1 ReL 5 Therefore the drag is: Alternately, we could solve for the drag using the momentum thickness: At x = L δL 31.304 mm 7 θL δ 3.043 mm At x = x1: 72 L 2 FD ρ U b θL θx1 δx1 x 1 0.382 1 Rex1 3 3.245 10 FD 0.700 N where θ 12.999 mm θx1 7 72 δ 7 δ 1.264 mm 72 x1 5 Therefore the drag is: FD 0.700 N Problem 9.64 Given: [Difficulty: 3] Air at standard conditions flowing over a flat plate The given or available data (Table A.10) is U 30 ft s Find: x 3 ft ν 1.57 10 4 ft 2 ρ 0.00238 s slug ft 3 δ and τw at x assuming: (a) completely laminar flow (parabolic velocity profile) (b) completely turbulent flow (1/7-power velocity profile) Solution: (Laminar Flow) Basic equations: δ x (Turbulent Flow) 5.48 δ Rex x 0.382 Rex Cf τw 1 2 2 ρ U 0.730 U x ν δlam x For laminar flow: 1 2 Rex The Reynolds number is: 5.73 10 δturb x 5.48 0.382 Comparing results: δlam (Skin friction factor) 1 2 ρ U Rex 5 τwlam 7.17 10 5 The turbulent boundary layer has a much larger skin friction, which causes it to grow more rapidly than the laminar boundary layer. 4.34 5 τwturb 3.12 10 3.72 τwlam psi 5 Rex τwturb 6 δturb 0.970 in 1 1 2 0.0594 τwturb ρ U 1 2 δturb 0.0594 δlam 0.261 in Rex Rex 5 1 2 0.730 τwlam ρ U 2 Rex For turbulent flow: 5 τw Cf Rex (Boundary Layer Thickness) 1 psi Problem 9.65 Given: [Difficulty: 3] Air at standard conditions flowing through a plane-wall diffuser with negligible BL thickness. Walls diverge slightly to accomodate BL growth, so p = constant. The given or available data (Table A.9) is U 60 m s Find: L 1.2 m W1 75 mm 2 5 m ν 1.46 10 s ρ 1.23 kg 3 m (a) why Bernoulli is applicable to this flow. (b) diffuser width W2 at x = L Solution: p1 Basic equations: ρ V1 2 2 p2 g z1 ρ V2 2 g z2 2 ρ dV ρ V dA 0 t Assumptions: (Bernoulli Equation) (Continuity) (1) Steady flow (2) Turbulent, 1/7-power velocity profile in boundary layer (3) z = constant (4) p = constant The Bernoulli equation may be applied along a streamline in any steady, incompressible flow in the absence of friction. The given flow is steady and incompressible. Frictional effects are confined to the thin wall boundary layers. Therefore, the Bernoulli equation may be applied along any streamline in the core flow outside the boundary layers. (In addition, since there is no streamline curvature, the pressure is uniform across sections 1 and 2. From the assumptions, Bernoulli reduces to: V1 V2 and from continuity: ρ V1 A1 ρ V2 A2eff 0 or A2eff W2 2 δdisp2 b W1 b The Reynolds number is: ReL U L ν Therefore: W2 W1 2 δdisp2 6 4.932 10 From turbulent BL theory: δ2 L 0.382 1 ReL 5 21.02 mm The displacement thickness is determined from: δdisp2 δ2 1 1 1 u dη U where u U η 7 η y δ 0 Substituting the velocity profile and valuating the integral: δdisp2 δ2 1 1 δ2 7 1 η dη 0 Therefore: W2 W1 2 δdisp2 8 δdisp2 2.628 mm W2 80.3 mm Problem 9.66 Given: [Difficulty: 3] Laboratory wind tunnel has flexible wall to accomodate BL growth. BL's are well represented by 1/7-power profile. Information at two stations are known: The given or available data (Table A.9) is U 90 ft s Find: H1 1 ft W1 1 ft δ1 0.5 in δ6 0.65 in ν 1.57 10 4 ft 2 ρ 0.00238 s slug ft 3 (a) Height of tunnel walls at section 6. (b) Equivalent length of flat plate that would produce the inlet BL (c) Estimate length of tunnel between stations 1 and 6. Solution: Basic equations: Assumptions: ρ dV ρ V dA 0 t (Continuity) (1) Steady flow (2) Turbulent, 1/7-power velocity profile in boundary layer (3) z = constant (4) p = constant Applying continuity between 1 and 6: A1 U1 A6 U6 where A is the effective flow area. The velocities at 1 and 6 must be equal since pressure is constant. In terms of the duct dimensions: W1 2 δdisp1H1 2 δdisp1 W1 2 δdisp6 H6 2 δdisp6 solving for the height at 6: H6 W1 2 δdisp1H1 2 δdisp1 2 δdisp6 W1 2 δdisp6 The displacement thickness is determined from: δdisp δ 1 1 1 u dη U u where U η 7 η y δ 0 Substituting the velocity profile and valuating the integral: δdisp δ 1 0 1 δ 7 1 η dη δdisp1 0.0625 in Therefore: 8 We may now determine the height at 6: δdisp6 0.0813 in H6 1.006 ft 1 For a flat plate turbulent boundary layer with 1/7-power law profile: δ1 L1 5 1 4 δ1 U 4 L1 0.382 ν 0.382 1 Re1 0.382 ν 5 4 5 L1 Solving for L1: U 5 L1 1.725 ft To estimate the length between 1 and 6, we determine length necessary to build the BL at section 6: 5 1 4 L6 δ6 U 4 2.394 ft 0.382 ν Therefore, the distance between 1 and 6 is: L L6 L1 L 0.669 ft Problem 9.67 Given: [Difficulty: 3] Laboratory wind tunnel has fixed walls. BL's are well represented by 1/7-power profile. Information at two stations are known: The given or available data (Table A.9) is m U1 26.1 s Find: H 305 mm W 305 mm δ1 12.2 mm δ2 16.6 mm ν 1.46 10 2 5 m ρ 1.23 s kg 3 m (a) Change in static pressure between 1 and 2 (b) Estimate length of tunnel between stations 1 and 2. Solution: Basic equations: ρ dV ρ V dA 0 t Assumptions: (Continuity) (1) Steady flow (2) Turbulent, 1/7-power velocity profile in boundary layer (3) z = constant A1 U1 A2 U2 Applying continuity between 1 and 6: where A is the effective flow area. In terms of the duct dimensions: W 2 δdisp1H 2 δdisp1 U1 W 2 δdisp2 H 2 δdisp2 U2 solving for the speed at 2: W 2 δdisp1 H 2 δdisp1 U2 U1 W 2 δdisp2 H 2 δdisp2 δdisp δ The displacement thickness is determined from: 1 1 1 dη U u u where U η 7 η y δ 0 Substituting the velocity profile and valuating the integral: δdisp δ 1 0 1 δ 7 1 η dη δdisp1 1.525 mm Therefore: 8 We may now determine the speed at 2: Applying Bernoulli between 1 and 2: p1 ρ U1 2 2 p2 ρ U2 2 2 Solving for the pressure change: ∆p Substituting given values: δdisp2 2.075 mm m U2 26.3 s 1 2 ρ U1 U2 2 ∆p 6.16Pa 2 1 For a flat plate turbulent boundary layer with 1/7-power law profile: δ x 5 1 4 4 δ1 U1 x1 0.494 m 0.382 ν 0.382 1 Rex 0.382 ν 5 4 5 Solving for location at 1: x U 5 To estimate the length between 1 and 6, we determine length necessary to build the BL at section 2: 5 1 4 4 δ2 U2 x2 0.727 m 0.382 ν Therefore, the distance between 1 and 2 is: L x2 x1 L 0.233 m Problem 9.68 [Difficulty: 3] Given: Data on flow in a duct Find: Velocity at location 2; pressure drop; length of duct; position at which boundary layer is 20 mm Solution: The given data is D 6 in δ1 0.4 in Table A.9 ρ 0.00234 slug ft Governing equations Mass In the boundary layer δ x ν 1.56 10 3 ft U1 80 s δ2 1.2 in 0.382 4 ft 2 s (9.26) 1 5 Rex In the the inviscid core, the Bernoulli equation holds p ρ 2 V 2 g z constant (4.24) Assumptions: (1) Steady flow (2) No body force (gravity) in x direction For a 1/7-power law profile, from Example 9.4 the displacement thickness is Hence δ δdisp 8 δ1 δdisp1 8 δdisp1 0.0500 in δ2 δdisp2 8 δdisp2 0.1500 in From the definition of the displacement thickness, to compute the flow rate, the uniform flow at locations 1 and 2 is assumed to take place in the entire duct, minus the displacement thicknesses π 2 A1 D 2 δdisp1 4 A1 0.1899 ft 2 π 2 A2 D 2 δdisp2 4 A2 0.1772 ft 2 Mass conservation (Eq. 4.12) leads to U2 ρ U1 A1 ρ U2 A2 0 or A1 U2 U1 A2 ft U2 85.7 s The Bernoulli equation applied between locations 1 and 2 is p1 ρ or the pressure drop is U1 2 2 p2 ρ U2 2 2 ρ 2 2 p 1 p 2 ∆p U2 U1 2 ∆p 7.69 10 3 psi (Depends on ρ value selected) The static pressure falls continuously in the entrance region as the fluid in the central core accelerates into a decreasing core. If we assume the stagnation pressure is atmospheric, a change in pressure of about 0.008 psi is not significant; in addition, the velocity changes by about 5%, again not a large change to within engineering accuracy To compute distances corresponding to boundary layer thicknesses, rearrange Eq.9.26 1 δ x 0.382 1 Rex 0.382 U x ν 5 5 so x δ 0.382 4 1 U ν 5 Applying this equation to locations 1 and 2 (using U = U1 or U2 as approximations) 5 1 4 4 δ1 U1 x1 0.382 ν For location 3 x 1 1.269 ft 5 1 δ2 U2 x2 0.382 ν 4 4 x 2 x 1 3.83 ft (Depends on ν value selected) δ3 0.6 in δ3 δdisp3 8 x 2 5.098 ft δdisp3 0.075 in π 2 A3 D 2 δdisp3 4 A3 0.187 ft A1 U3 U1 A3 ft U3 81.4 s 5 1 δ3 U2 x3 0.382 ν 4 4 x 3 x 1 0.874 ft (Depends on ν value selected) x 3 2.143 ft 2 4 Problem 9.69 [Difficulty: 3] Given: Linear, sinusoidal and parabolic velocity profiles Find: Momentum fluxes Solution: δ The momentum flux is given by 2 mf ρ u w dy 0 where w is the width of the boundary layer For a linear velocity profile u U For a sinusoidal velocity profile u U For a parabolic velocity profile u U y δ η (1) sin π y π sin η 2 δ 2 2 y δ (2) 2 y 2 η ( η) 2 δ For each of these u U f ( η) Using these in the momentum flux equation 2 mf ρ U δ w f ( η) dη (3) y δ η 1 2 (4) 0 1 For the linear profile Eqs. 1 and 4 give 2 mf ρ U δ w η dη 2 mf 1 mf 1 mf 8 0 3 2 ρ U δ w 1 For the sinusoidal profile Eqs. 2 and 4 give 2 π 2 mf ρ U δ w sin η dη 2 2 2 ρ U δ w 0 For the parabolic profile Eqs. 3 and 4 give 2 mf ρ U δ w 1 2 2 η ( η) 2 dη 0 The linear profile has the smallest momentum, so would be most likely to separate 15 2 ρ U δ w Problem 9.70 [Difficulty: 3] Given: Data on a large tanker Find: Cost effectiveness of tanker; compare to Alaska pipeline Solution: The given data is L 360 m B 70 m D 25 m kg ρ 1020 U 6.69 3 m s m 4 P 1.30 10 hp (Power consumed by drag) P 9.7 MW The power to the propeller is P Pprop 70 % Pprop 1.86 10 hp The shaft power is Ps 120% Pprop Ps 2.23 10 hp The efficiency of the engines is η 40 % Hence the heat supplied to the engines is Q t The journey time is Ps 4 4 8 BTU Q 1.42 10 η x hr t 134 hr U 10 Qtotal Q t The total energy consumed is x 2000 mi Qtotal 1.9 10 BTU From buoyancy the total ship weight equals the displaced seawater volume M ship g ρ g L B D 9 M ship ρ L B D M ship 1.42 10 lb Hence the mass of oil is M oil 75% M ship M oil 1.06 10 lb The chemical energy stored in the petroleum is q 20000 E q M oil The total chemical energy is The equivalent percentage of petroleum cargo used is then 9 BTU lb 13 E 2.13 10 BTU Qtotal E The Alaska pipeline uses epipeline 120 BTU but for the ton mi ship The ship uses only about 15% of the energy of the pipeline! 0.089 % eship Qtotal M oil x eship 17.8 BTU ton mi Problem 9.71 Given: Plane-wall diffuser Find: (a) For inviscid flow, describe flow pattern and pressure distribution as φ is increased from zero (b) Redo part (a) for a viscous fluid (c) Which fluid will have the higher exit pressure? [Difficulty: 2] Solution: For the inviscid fluid: With φ = 0 (straight channel) there will be no change in the velocity, and hence no pressure gradient. With φ > 0 (diverging channel) the velocity will decrease, and hence the pressure will increase. For the viscous fluid: With φ = 0 (straight channel) the boundary layer will grow, decreasing the effective flow area. As a result, velocity will increase, and the pressure will drop. With φ > 0 (diverging channel) the pressure increase due to the flow divergence will cause in increase in the rate of boundary layer growth. If φ is too large, the flow will separate from one or both walls. The inviscid fluid will have the higher exit pressure. (The pressure gradient with the real fluid is reduced by the boundary layer development for all values of φ.) Problem 9.72 [Difficulty: 4] Given: Laminar (Blasius) and turbulent (1/7-power) velocity distributions Find: Plot of distributions; momentum fluxes Solution: δ The momentum flux is given by 2 mf ρ u dy per unit width of the boundary layer 0 Using the substitutions u U y f ( η) δ η 1 the momentum flux becomes 2 mf ρ U δ f ( η ) dη 2 0 For the Blasius solution a numerical evaluation (a Simpson's rule) of the integral is needed 2 ∆η 2 2 2 2 mflam ρ U δ f η0 4 f η1 2 f η2 f ηN 3 where Δη is the step size and N the number of steps The result for the Blasius profile is 2 mflam 0.525 ρ U δ 1 For a 1/7 power velocity profile 2 2 7 mfturb ρ U δ η dη 0 7 2 mfturb ρ U δ 9 The laminar boundary has less momentum, so will separate first when encountering an adverse pressure gradient. The computed results were generated in Excel and are shown below: (Table 9.1) (Simpsons Rule) η 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Laminar Weight Weight x u/U 0.000 0.166 0.330 0.487 0.630 0.751 0.846 0.913 0.956 0.980 0.992 w 1 4 2 4 2 4 2 4 2 4 1 Simpsons': y /δ = η 2 (u/U ) 0.00 0.11 0.22 0.95 0.79 2.26 1.43 3.33 1.83 3.84 0.98 0.525 0.0 0.0125 0.025 0.050 0.10 0.15 0.2 0.4 0.6 0.8 1.0 t u/U 0.00 0.53 0.59 0.65 0.72 0.76 0.79 0.88 0.93 0.97 1.00 Laminar and Turbulent Boundary Layer Velocity Profiles 1.00 0.75 y /δ 0.50 Laminar Turbulent 0.25 0.00 0.00 0.25 0.50 0.75 u/U 1.00 Problem 9.73 [Difficulty: 5] Given: Channel flow with laminar boundary layers Find: Maximum inlet speed for laminar exit; Pressure drop for parabolic velocity in boundary layers Solution: Basic equations: Retrans 5 10 δ 5 x 5.48 p Rex ρ 2 V g z const 2 Assumptions: 1) Steady flow 2) Incompressible 3) z = constant From Table A.10 at 20oC Then For For a parabolic profile ν 1.50 10 2 5 m ρ 1.21 s Umax L Retrans Umax ν 5 Retrans 5 10 δdisp δ 1 0 δ2 L 3 m Retrans ν L 5.48 Retrans L 3 m h 15 cm m Umax 2.50 s U1 Umax m U1 2.50 s δ2 0.0232 m 1 1 2 1 u dλ 1 2 λ λ dλ U 3 0 1 δdisp2 δ2 3 From continuity kg where δtrans is the displacement thickness δdisp2 0.00775 m U1 w h U2 w h 2 δdisp2 h U2 U1 h 2 δdisp2 m U2 2.79 s Since the boundary layers do not meet Bernoulli applies in the core p1 ρ ∆p From hydrostatics U1 2 ρ 2 2 p2 ρ U2 U1 2 ∆p ρH2O g ∆h ∆h U2 ∆p ρH2O g 2 2 2 ρ 2 2 ∆p p 1 p 2 U2 U1 2 ∆p 0.922 Pa with kg ρH2O 1000 3 m ∆h 0.0940 mm ∆h 0.00370 in Problem 9.74 Given: [Difficulty: 3] u Laminar boundary layer with velocity profile U 2 a b λ c λ d λ 3 λ y δ Separation occurs when shear stress at the surface becomes zero. Find: (a) Boundary conditions on the velocity profile at separation (b) Appropriate constants a, b, c, d for the profile (c) Shape factor H at separation (d) Plot the profile and compare with the parabolic approximate profile Solution: Basic equations: u U 2 y δ δ y 2 (Parabolic profile) The boundary conditions for the separation profile are: The velocity gradient is defined as: du dy Applying the boundary conditions: at y 0 u0 τ μ du at y δ uU τ μ du U d u U 2 b 2 c λ 3 d λ δ δ dλ U y0 λ0 u U du dy 2 3 a b 0 c 0 d 0 0 The velocity profile and gradient may now be written as: δ U b 2 c 0 3 d 0 u c λ d λ U 2 2 0 du 3 dy U δ dy dy 0 Four boundary conditions for four coefficients a, b, c, d 0 Therefore: a0 Therefore: b0 2 c λ 3 d λ 2 Applying the other boundary conditions: yδ λ1 u U du dy The velocity profile is: u U δdisp δ 1 0 1 2 3 λ 2 λ 3 2 3 c 1 d 1 1 U δ 2 c 1 3 d 1 2 0 1 δdisp θ c3 d 2 δdisp δ δ θ 2 3 2 3 3 λ 2 λ 1 3 λ 2 λ dλ Expanding out the δ 0 integrand yields: θ 1 9 4 9 2 3 4 5 6 3 λ 2 λ 9 λ 12 λ 4 λ dλ 1 2 δ 0 2 5 7 70 θ 2 c 3 d 0 H The shape parameter is defined as: 1 3λ2 2λ3 dλ 1 1 12 12 Solving this system of equations yields: cd1 Thus H 1 2 70 9 H 3.89 The two velocity profiles are plotted here: Height y/δ 1 0.5 Separated Parabolic 0 0 0.5 Velocity Distribution u/U 1 Problem 9.75 [Difficulty: 4] Discussion: Shear stress decreases along the plate because the freestream flow speed remains constant while the boundary-layer thickness increases. The momentum flux decreases as the flow proceeds along the plate. Momentum thickness θ (actually proportional to the defect in momentum within the boundary layer) increases, showing that momentum flux decreases. The forct that must be applied to hold the plate stationary reduces the momentum flux of the stream and boundary layer. The laminar boundary layer has less shear stress than the turbulent boundary layer. Therefore laminar boundary layer flow from the leading edge produces a thinner boundary layer and less shear stress everywhere along the plate than a turbulent boundary layer from the leading edge. Since both boundary layers continue to grow with increasing distance from the leading edge, and the turbulent boundary layer continues to grow more rapidly because of its higher shear stress, this comparison will be the same no matter the distance from the leading edge. Problem 9.76 Given: [Difficulty: 5] Laboratory wind tunnel has fixed walls. BL's are well represented by 1/7-power profile. Information at two stations are known: The given or available data (Table A.9) is ft U1 80 s dp dx 0.035 Find: H1 1 ft in H2 O W1 1 ft L 10 in in δ1 0.4 in ν 1.62 10 4 ft 2 ρ 0.00234 s slug ft 3 (a) Reduction in effective flow area at section 1 (b) dθ/dx at section 1 (c) θ at section 2 Solution: Basic equations: ρ dV ρ V dA 0 t τw ρ Assumptions: (Continuity) 2 d d U θ δdisp U U dx dx (Momentum integral equation) (1) Steady flow (2) Turbulent, 1/7-power velocity profile in boundary layer (3) z = constant The percent reduction in flow area at 1 is given as: Aeff A A The displacement thickness is determined from: δdisp δ 1 W1 2 δdisp H1 2 δdisp W1 H1 W1 H1 1 1 u dη U u where U η 7 η y δ 0 Substituting the velocity profile and valuating the integral: δdisp δ 1 0 1 δ 7 1 η dη Therefore: 8 Thus: δdisp1 0.0500 in Aeff A A 1.66 % Solving the momentum integral equation for the momentum thickness gradient: dθ dx 1 At station 1: τw1 ρ U1 2 0.0233 U1 δ1 L 1 θ u u 1 dη δ U U 0 ν 1 4 0.0233 8 9 0 Solving for the velocity gradient: p 1 2 τw 2 ( H 2) ρ U θ dU U dx 4 2.057 10 3 U1 δ1 ν 2 1 7 7 7 7 7 η η dη Now outside the boundary layer 2 ρ U constant 72 δdisp1 7 Thus: θ1 δ1 0.0389 in H 1.286 72 θ1 from the Bernoulli equation. Then: dp dx ρ U dU dx 1 dp 1 1 dU 0.1458 Substituting all of this information into the above expression: 2 dx ft U dx ρ U dθ dx We approximate the momentum thickness at 2 from: dθ θ2 θ1 L dx 4.89 10 4 0.00587 in ft θ2 0.0438 in Problem 9.77 Given: [Difficulty: 5] Laboratory wind tunnel of Problem 9.76 with a movable top wall: The given or available data (Table A.9) is ft U1 80 s Find: H1 1 ft W 1 ft δ 0.4 in L 10 in ν 1.57 10 4 ft 2 ρ 0.00234 s slug ft 3 (a) Velocity distribution needed for constant boundary layer thickness (b) Tunnel height distribution h(x) from 0 to L Solution: Basic equations: ρ dV ρ V dA 0 t τw ρ Assumptions: (Continuity) 2 d d U θ δdisp U U dx dx (Momentum integral equation) (1) Steady flow (2) Incompressible flow (3) Turbulent, 1/7-power velocity profile in boundary layer (4) δ = constant δ From the 1/7-power profile: δdisp 8 θ 7 72 δ H 72 56 τw After applying assumptions, the momentum integral equation is: 2 ( H 2) ρ U To integrate, we need to make an assumption about the wall shear stress: τw Integrating: Case 1: assume constant τw: U dU dx ρ θ ( H 2 ) U U1 1 2 τw ρ U1 x 2 θ ( H 2 ) 2 d U U dx θ U U1 2 2 τw ρ θ ( H 2 ) which may be rewritten as: x U U1 1 Cf θ ( H 2 ) 1 ν 2 Case 2: assume τ w has the form: τw 0.0233 ρ U U δ 4 Substituting and rearranging yields the following expression: x 1 τw 2 ρ U 4 ( H 2) θ d U U dx U δ 0.0233 ν dU or 0.75 0.0233 ν 0.25 δ U dx Integrating this yields: ( H 2) θ or: 0.25 ν x 1 0.00583 U1 ( H 2 ) θ U1 δ From continuity: U1 A1 U A which may be rewritten as: U1 W 2 δdisp H1 2 δdisp U W 2.δdisp h 2 δdisp 0.25 0.25 ν 4 U U1 0.0233 δ Thus: A A1 U U 1 0.25 1 Evaluating using the given data: Cf 0.0466 Reδ1 0.25 x ( H 2) θ h 2 δdisp and H1 2 δdisp U1 δ δdisp 0.0500 in 8 W U θ 7 72 δ 0.0389 in h solving for h: 1 2 Reδ1 U1 δ ν δdisp U1 δdisp 2 H1 H1 U 4 1.699 10 3 4.082 10 The results for both wall profiles are shown in the plot here: Top Surface Height (in) 14 13.5 13 12.5 Case 1 Case 2 12 0 2 4 U 4 6 Distance along tunnel (in) 8 10 Problem 9.78 Given: [Difficulty: 3] Barge pushed upriver L 80 ft B 35 ft Find: 5 ft D 5 ft From Table A.7: ν 1.321 10 2 s ρ 1.94 slug ft 3 Power required to overcome friction; Plot power versus speed Solution: CD Basic equations: FD 1 2 CD (9.32) 2 ρ U A From Eq. 9.32 1 2 FD CD A ρ U 2 The power consumed is P FD U 0.455 logReL 2.58 A L ( B 2 D) and 1 3 P CD A ρ U 2 1610 (9.37b) ReL A 3600 ft Re L CD P (hp) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 9.70E+06 1.94E+07 2.91E+07 3.88E+07 4.85E+07 5.82E+07 6.79E+07 7.76E+07 8.73E+07 9.70E+07 1.07E+08 1.16E+08 1.26E+08 1.36E+08 1.45E+08 0.00285 0.00262 0.00249 0.00240 0.00233 0.00227 0.00222 0.00219 0.00215 0.00212 0.00209 0.00207 0.00205 0.00203 0.00201 0.0571 0.421 1.35 3.1 5.8 9.8 15 22 31 42 56 72 90 111 136 150 120 P (hp) 90 60 30 0 6 ν The calculated results and the plot were generated in Excel: U (mph) 3 U L 2 Power Consumed by Friction on a Barge 0 ReL 9 U (mph) 12 15 Problem 9.79 Given: Pattern of flat plates Find: Drag on separate and composite plates [Difficulty: 3] Solution: Basic equations: CD FD 1 2 For separate plates From Table A.8 at 70 oF 2 ρ V A L 3 in W 3 in ν 1.06 10 5 ft First determine the Reynolds number ReL CD 2 s V L ν 0.0742 1 ReL ρ 1.93 A W L ft 3 ReL 7.08 10 5 5 This is the drag on one plate. The total drag is then FTotal 4 FD L 4 3 in L 12.000 in ReL First determine the Reylolds number CD so use Eq. 9.34 CD 0.00502 FD 0.272 lbf For the composite plate V 30 slug 1 2 FD CD ρ V A 2 The drag (one side) is then 2 A 9.000 in 0.0742 1 V L ν FTotal 1.09 lbf For both sides: 2 FTotal 2.18 lbf A W L A 36 in ReL 2.83 10 2 6 so use Eq. 9.34 CD 0.00380 5 The drag (one side) is then ReL 1 2 FD CD ρ V A 2 FD 0.826 lbf For both sides: 2 FD 1.651 lbf The drag is much lower on the composite compared to the separate plates. This is because τ w is largest near the leading edges and falls off rapidly; in this problem the separate plates experience leading edges four times! ft s Problem 9.80 Given: [Difficulty: 3] Towboat model at 1:13.5 scale to be tested in towing tank. Lm 3.5 m Find: Bm 1 m d m 0.2 m m Up 7 knot 3.601 s Disp m 5500 N (a) Estimate average length of wetted surface on the hull (b) Calculate skin friction drag force on the prototype Solution: Basic equations: 1 2 FD CD ρ U A 2 CD (Drag) 0.455 logReL 2.58 1610 (Drag Coefficient) ReL We will represent the towboat as a rectangular solid of length L av, with the displacement of the boat. From buoyancy: W ρ g V ρ g Lav Bm d m thus: For the prototype: Lav W ρ g Bm d m Lp 13.5 Lav Lp 37.9 m ReL The Reynolds number is: Lav 2.80 m Up Lp ReL 1.36 10 ν 8 This flow is predominantly turbulent, so we will use a turbulent analysis. The drag coefficient is: CD The area is: 2 0.455 logReL 2.58 1610 ReL 0.00203 2 A 13.5 Lav Bm 2 d m 716 m The drag force would then be: 1 2 FD CD ρ Up A 2 FD 9.41 kN This is skin friction only. Problem 9.81 Given: Aircraft cruising at 12 km Find: Skin friction drag force; Power required [Difficulty: 3] Solution: Basic equations: CD FD 1 2 ρ V A 2 We "unwrap" the cylinder to obtain an equivalent flat plate L 38 m From Table A.3, with D 4 m ρ z 12000 m ρSL ρ 0.2546 ρSL 2 A L π D kg ρSL 1.225 3 m 0.2546 kg ρ 0.3119 A 478 m and also T 216.7 K kg S 110.4 K 3 m 1 From Appendix A-3 μ b T 2 1 S with 6 b 1.458 10 T 1 m s K 2 1 Hence b T 2 1 S μ μ 1.42 10 5 N s ReL Next we need the Reynolds number 0.455 log ReL 2 m T CD 2.58 ρ V L μ CD 0.00196 The drag is then 1 2 FD CD ρ V A 2 FD 7189 N The power consumed is P FD V P 1.598 MW ReL 1.85 10 8 so use Eq. 9.35 V 800 km hr Problem 9.82 Given: [Difficulty: 3] Towboat model at 1:13.5 scale to be tested in towing tank. Lm 7.00 m Find: Bm 1.4 m d m 0.2 m Vp 10 knot (a) Model speed in order to exhibit similar wave drag behavior (b) Type of boundary layer on the prototype (c) Where to place BL trips on the model (d) Estimate skin friction drag on prototype Solution: Basic equations: 1 2 FD CD ρ V A 2 (Drag) Vm The test should be conducted to match Froude numbers: Rep The Reynolds number is: g Lm A L ( B 2 d ) 0.0594 ReL Therefore CDm 0.0743 0.2 0.2 0.2 2.97 10 3 Rem For the prototype: CDp g Lp Vm 2.72 knot 8 Rep 4.85 10 ν 5 Ret 5 10 so xt L Ret Rep 0.00155 x t 0.0109 m We calculate the drag coefficient from turbulent BL theory: 0.0743 ReL Lm Vm Vp Lp x t 0.00155 Lm Thus the location of transition would be: CD 1.25 Cf 1.25 Vp Vp Lp Therefore the boundary layer is turbulent. Transition occurs at The wetted area is: 0.455 logRep 2.56 For the model: Lm 7 m Rem Vm Lm ν 9.77 10 6 2 Am 12.6 m 1 2 and the drag force is: FDm CDm ρ Vm Am 2 1610 Rep CDp 1.7944 10 3 FDm 36.70 N 3 2 Ap 2.30 10 m 1 2 FDp CDp ρ Vp Ap 2 FDp 54.5 kN Problem 9.83 Given: Stabilizing fin on Bonneville land speed record auto z 1340 m Find: [Difficulty: 2] V 560 km hr H 0.785 m L 1.65 m (a) Evaluate Reynolds number of fin (b) Estimate of location for transition in the boundary layer (c) Power required to overcome skin friction drag Solution: Basic equations: Assumptions: At this elevation: 1 2 FD CD ρ V A 2 (Drag) (1) Standard atmosphere (use table A.3) T 279 K ρ 0.877 1.23 kg 1.079 3 m The Reynolds number on the fin is: Assume transition occurs at: ReL μ 1.79 10 3 m m s ρ V L 5 Ret 5 10 7 ReL 1.547 10 μ The location for transition would then be: From Figure 9.8, the drag coefficient is: CD 0.0029 The drag force would then be: 5 kg kg The area is: xt x t 53.3 mm 2 FD 98.0 N P FD V If we check the drag coefficient using Eq. 9.37b: ρ V A 2 L H 2.591 m (both sides of the fin) 1 2 FD CD ρ V A 2 The power required would then be: Ret μ P 15.3 kW CD 0.455 log ReL 2.58 1610 ReL 0.0027 This is slightly less than the graph, but still reasonable agreement. Problem 9.84 Given: Nuclear submarine cruising submerged. Hull approximated by circular cylinder L 107 m Find: [Difficulty: 4] D 11.0 m V 27 knot (a) Percentage of hull length for which BL is laminar (b) Skin friction drag on hull (c) Power consumed Solution: Basic equations: 1 2 FD CD ρ V A 2 5 Transition occurs at Ret 5 10 (Drag) so the location of transition would be: xt L Ret ν xt V L L 0.0353% We will therefore assume that the BL is completely turbulent. The Reynolds number at x = L is: The wetted area of the hull is: ReL V L ν 1.42 10 9 For this Reynolds number: CD 3 0.455 logReL 2.58 1.50 10 2 A π D L 3698 m So the drag force is: The power consumed is: 1 2 FD CD ρ V A 2 P FD V 5 FD 5.36 10 N P 7.45 MW Problem 9.85 Given: Racing shell for crew approximated by half-cylinder: L 7.32 m Find: [Difficulty: 3] D 457 mm V 6.71 m s (a) Location of transition on hull (b) Thickness of turbulent BL at the rear of the hull (c) Skin friction drag on hull Solution: Basic equations: 1 2 FD CD ρ V A 2 5 Transition occurs at Ret 5 10 so the location of transition would be: δ For the turbulent boundary layer x The Reynolds number at x = L is: The wetted area of the hull is: A (Drag) 0.382 Rex ReL π D 2 0.2 V L ν Therefore δ x t 0.0745 m V 0.2 so the BL thickness is: δ 0.382 L ReL 2 L 5.2547 m So the drag force is: 7 Ret ν 0.382 L ReL 4.91 10 xt For this Reynolds number: 1 2 FD CD ρ V A 2 Note that the rowers must produce an average power of CD δ 0.0810 m 0.2 3 0.455 logReL 2.58 2.36 10 FD 278 N P FD V 1.868 kW to move the shell at this speed. Problem 9.86 Given: Plastic sheet falling in water Find: Terminal speed both ways [Difficulty: 3] Solution: Basic equations: ΣFy 0 FD CD for terminal speed 1 2 h 0.5 in W 4 ft L 2 ft SG 1.7 2 ρ V A CD 0.0742 (9.34) (assuming 5 x 105 < ReL < 107) 1 ReL 5 From Table A.8 at 70 oF ν 1.06 10 5 ft A W L 2 ρ 1.94 s slug ft 3 A free body diagram of the sheet is shown here. Summing the forces in the vertical (y) direction: FD Fb Wsheet 0 FD Wsheet Fb ρ g h A ( SG 1 ) Fb Also, we can generate an expression for the drag coefficient in terms of the geometry of the sheet and the water properties: FD y V x 4 1 9 W sheet 1 0.0742 1 0.0742 2 2 2 5 5 5 FD 2 CD A ρ V 2 A ρ V W L ρ V 0.0742 W L ν ρ V 1 2 2 1 ReL 5 V L ν (Note that we double FD because drag acts on both sides of the sheet.) 5 5 9 Hence ρH2O g h W ( SG 1 ) 0.0742 W L Check the Reynolds number Repeating for ReL ReL 5 1 9 5 5 ν ρ V Solving for V V L ν L 4 ft Check the Reynolds number 1 V L ν 5 1 g h ( SG 1) L 5 V 0.0742 ν 1 g h ( SG 1) L 5 ft V V 15.79 s 0.0742 ν ReL 2.98 10 6 Hence Eq. 9.34 is reasonable 6 Eq. 9.34 is still reasonable 9 V 17.06 ft s ReL 6.44 10 The short side vertical orientation falls more slowly because the largest friction is at the region of the leading edge (τ tails off as the boundary layer progresses); its leading edge area is larger. Note that neither orientation is likely - the plate will flip around in a chaotic manner. Problem 9.87 Given: [Difficulty: 4] 600-seat jet transport to operate 14 hr/day, 6 day/wk L 240 ft D 25 ft Find: V 575 mph z 12 km TSFC 0.6 lbm hr lbf (a) Skin friction drag on fuselage at cruise (b) Annual fuel savings if drag is reduced by 1% Solution: Basic equations: 1 2 FD CD ρ V A 2 (Drag) T 216.7 K 390.1 R ρ 0.2546 0.002377 From the atmosphere model: slug ft From the Sutherland model for viscosity: μ b T 1 ReL ρ V L μ 4.1247 10 8 S 5 kg 1.422 10 ft 3 So the Reynolds number is CD 4 A π D L 1.885 10 ft 3 0.455 logReL 2.58 1.76 10 2 So the drag force is: If there were a 1% savings in drag, the drop in drag force would be: The savings in fuel would be: 4 slug 6.0518 10 T For this Reynolds number: The wetted area of the fuselage is: m s 3 hr Δmfuel TSFC ΔFD 14 day 1 2 3 FD CD ρ V A FD 7.13 10 lbf 2 ΔFD 1 % FD 71.31 lbf 6 52 day 7 yr 4 lbm Δmfuel 2.670 10 yr If jet fuel costs $1 per gallon, this would mean a savings of over $4,400 per aircraft per year. Problem 9.88 Given: [Difficulty: 4] Supertanker in seawater at 40oF L 1000 ft B 270 ft D 80 ft ν 1.05 1.65 10 Find: 5 ft V 15 knot 25.32 2 s 5 ft 1.73 10 2 ρ 1.9888 s (a) Thickness of the boundary layer at the stern of the ship (b) Skin friction drag on the ship (b) Power required to overcome the drag force ft SG 1.025 s slug ft 3 Solution: Basic equations: 1 2 FD CD ρ V A 2 The Reynolds number is δ x 0.382 Rex ReL V L At the stern of the ship: 0.20 ν (Drag) 9 1.4613 10 So the BL is turbulent. The BL thickness is calculated using: δ L 0.382 ReL The wetted area of the hull is: δ 5.61 ft 0.20 5 2 A L ( B 2 D) 4.30 10 ft For this Reynolds number: CD So the drag force is: 1 2 FD CD ρ V A 2 The power consumed to overcome the skin friction drag is: 3 0.455 logReL 2.58 1.50 10 5 FD 4.11 10 lbf P FD V 4 P 1.891 10 hp 7 ft lbf P 1.040 10 s Problem 9.89 Given: "Resistance" data on a ship Lp 130 m Lm Find: [Difficulty: 4] Lp 80 1.625 m ρ 1023 2 Ap 1800 m Am Ap 80 2 kg 3 3 N s μ 1.08 10 m 2 m 2 0.281 m Plot of wave, viscous and total drag (prototype and model); power required by prototype Solution: Basic equations: CD FD 1 2 From Eq. 9.32 (9.32) 2 Fr U gL ρ U A 1 2 FD CD A ρ U 2 This applies to each component of the drag (wave and viscous) as well as to the total The power consumed is P FD U From the Froude number U Fr gL 1 3 P CD A ρ U 2 The solution technique is: For each speed Fr value from the graph, compute U; compute the drag from the corresponding "resistance" value from the graph. The results were generated in Excel and are shown below: Model Fr Wave "Resistance" Viscous "Resistance" 0.10 0.20 0.30 0.35 0.40 0.45 0.50 0.60 0.00050 0.00075 0.00120 0.00150 0.00200 0.00300 0.00350 0.00320 0.0052 0.0045 0.0040 0.0038 0.0038 0.0036 0.0035 0.0035 Total Wave U (m/s) Drag (N) "Resistance" 0.0057 0.0053 0.0052 0.0053 0.0058 0.0066 0.0070 0.0067 0.40 0.80 1.20 1.40 1.60 1.80 2.00 2.40 0.0057 0.0344 0.1238 0.2107 0.3669 0.6966 1.0033 1.3209 Viscous Drag (N) 0.0596 0.2064 0.4128 0.5337 0.6971 0.8359 1.0033 1.4447 Total Power (W) Drag (N) 0.0654 0.2408 0.5366 0.7444 1.0640 1.5324 2.0065 2.7656 Drag on a Model Ship 3.0 2.5 Total Wave Viscous 2.0 F (N) 1.5 1.0 0.5 0.0 0.0 0.5 1.0 1.5 2.0 U (m/s) 2.5 3.0 2.5 3.0 Power Requirements for a Model Ship 7.0 6.0 5.0 P (W) 4.0 3.0 2.0 1.0 0.0 0.0 0.5 1.0 1.5 2.0 U (m/s) 0.0261 0.1923 0.6427 1.0403 1.6993 2.7533 4.0057 6.6252 Prototype Fr Wave "Resistance" Viscous "Resistance" 0.10 0.20 0.30 0.35 0.40 0.45 0.50 0.60 0.00050 0.00075 0.00120 0.00150 0.00200 0.00300 0.00350 0.00320 0.0017 0.0016 0.0015 0.0015 0.0013 0.0013 0.0013 0.0013 Total U (m/s) "Resistance" 0.0022 0.0024 0.0027 0.0030 0.0033 0.0043 0.0048 0.0045 3.6 7.1 10.7 12.5 14.3 16.1 17.9 21.4 Wave Drag (MN) 0.0029 0.0176 0.0634 0.1079 0.1879 0.3566 0.5137 0.6763 Viscous Drag (MN) 0.0100 0.0376 0.0793 0.1079 0.1221 0.1545 0.1908 0.2747 Total Drag (MN) 0.0129 0.0552 0.1427 0.2157 0.3100 0.5112 0.7045 0.9510 Drag on a Prototype Ship 1.0 F (MN) 0.8 Total 0.6 Wave Viscous 0.4 0.2 0.0 0 5 10 15 U (m/s) 20 25 Power Required by a Prototype Ship 25000 20000 P (kW) 15000 10000 5000 0 0 5 10 15 U (m/s) 20 For the prototype wave resistance is a much more significant factor at high speeds! However, note that for both scales, the primary source of drag changes with speed. At low speeds, viscous effects dominate, and so the primary source of drag is viscous drag. At higher speeds, inertial effects dominate, and so the wave drag is the primary source of drag. 25 Power (kW) Power (hp) 46.1 394.1 1528.3 2696.6 4427.7 8214.7 12578.7 20377.5 61.8 528.5 2049.5 3616.1 5937.6 11015.9 16868.1 27326.3 Problem 9.90 Given: Flag mounted vertically H 194 ft W 367 ft Find: [Difficulty: 2] V 10 mph 14.67 ft s ρ 0.00234 slug ft 3 ν 1.62 10 4 ft 2 s Force acting on the flag. Was failure a surprise? Solution: Basic equations: 1 2 FD CD ρ V A 2 (Drag) We should check the Reynolds number to be sure that the data of Fig. 9.10 are applicable: ReW V W ν 3.32 10 7 (We used W as our length scale here since it is the lesser of the two dimensions of the flag.) Since the Reynolds number is less than 1000, we may use Figure 9.10 to find the drag coefficient. 4 The area of the flag is: A H W 7.12 10 ft So the drag force is: 2 AR W H 1.89 1 2 FD CD ρ V A 2 From Fig. 9.10: CD 1.15 4 FD 2.06 10 lbf This is a large force. Failure should have been expected. Problem 9.91 Given: Fishing net Find: Drag; Power to maintain motion [Difficulty: 3] 3 8 in 9.525 mm Solution: Basic equations: CD FD 1 2 ρ V A 2 We convert the net into an equivalent cylinder (we assume each segment does not interfere with its neighbors) L 12 m W 2 m d 0.75 mm Spacing: D 1 cm Total number of threads of length L is Total number of threads of length W is Total length of thread n1 W n2 L LT L1 L2 ρ 999 D 3 Red V d ν s L1 n 1 L L1 2400 m n 2 1200 Total length L2 n 2 W L2 2400 m LT 4800 m LT 2.98 mile A lot! 2 Note that L W 24.00 m ν 1.01 10 2 6 m m The Reynolds number is m Total length A 3.60 m kg V 3.09 n 1 200 2 The frontal area is then A LT d From Table A.8 D V 6 knot s Red 2292 For a cylinder in a crossflow at this Reynolds number, from Fig. 9.13, approximately Hence 1 2 FD CD ρ V A 2 FD 13.71 kN The power required is P FD V P 42.3 kW CD 0.8 Problem 9.92 Given: [Difficulty: 2] Rotary mixer rotated in a brine solution R 0.6 m ω 60 rpm d 100 mm SG 1.1 ρ ρw SG ρ 1100 kg 3 m ν 1.05 1.55 10 Find: 2 6 m s 2 6 m 1.63 10 s (a) Torque on mixer (b) Horsepower required to drive mixer Solution: Basic equations: 1 2 FD CD ρ V A 2 (Drag) T 2 R FD (Torque) P T ω (Power) Assumptions: Drag on rods and motion induced in the brine can be neglected. The speed of the disks through the brine is: The area of one disk is: A π 4 2 V R ω 3.77 m s From Table 9.2: CD 1.17 for a disk. 2 d 0.00785 m So the drag force is: 1 2 FD CD ρ V A 71.8 N 2 and the torque is: T 2 R FD The power consumed to run the mixer is: P T ω 542 W T 86.2 N m P 0.726 hp Problem 9.93 Given: Data on a rotary mixer Find: New design dimensions [Difficulty: 3] Solution: The given data or available data is R 0.6 m P 350 W ω 60 rpm ρ 1099 kg 3 m For a ring, from Table 9.3 CD 1.2 The torque at the specified power and speed is T P T 55.7 N m ω The drag on each ring is then 1 T FD 2 R FD 46.4 N The linear velocity of each ring is V R ω V 3.77 m s The drag and velocity of each ring are related using the definition of drag coefficient FD CD 1 2 Solving for the ring area A FD 1 2 But A 2 ρ A V π 4 3 A 4.95 10 2 ρ V CD do di 2 The outer diameter is d o 125 mm Hence the inner diameter is di 2 do 2 4 A π d i 96.5 mm 2 m Problem 9.94 Given: Man with parachute W = 250 ⋅ lbf V = 20⋅ Find: [Difficulty: 2] ft ρ = 0.00234 ⋅ s slug ft 3 Minimum diameter of parachute FD Solution: Basic equations: 1 2 FD = CD⋅ ⋅ ρ⋅ V ⋅ A 2 (Drag) Assumptions: (1) Standard air x (2) Parachute behaves as open hemisphere (3) Vertical speed is constant For constant speed: ΣFy = M ⋅ g − FD = 0 In terms of the drag coefficient: Solving for the area: A= 1 2⋅ W 2 π 4 FD = W W 2 CD⋅ ⋅ ρ⋅ V ⋅ A = W 2 CD⋅ ρ⋅ V Setting both areas equal: Therefore: V y 2 ⋅D = From Table 9.2: CD = 1.42 for an open hemisphere. 2⋅ W Solving for the diameter of the parachute: 2 CD⋅ ρ⋅ V The area is: D = 8 π ⋅ A= π 4 2 ⋅D W 2 CD⋅ ρ⋅ V Therefore the diameter is: D = 21.9⋅ ft Problem 9.95 Given: Data on airplane landing M 9500 kg Find: [Difficulty: 3] km Vi 350 hr km Vf 100 hr x f 1200 m CD 1.43 (Table 9.3) Solution: 1 2 FD CD ρ V A 2 (Drag) Assumptions: (1) Standard air (2) Parachute behaves as open hemisphere (3) Vertical speed is constant Newton's second law for the aircraft is M dV 1 2 CD ρ A V 2 dt where A and CD are the single parachute area and drag coefficient Separating variables dV 2 V Integrating, with IC V = Vi CD ρ A 2 M 1 Integrating again with respect to t x ( t) Eliminating t from Eqs. 1 and 2 x dt Vi V( t) CD ρ A 2 M 2 M CD ρ A (1) Vi t CD ρ A 2 M ln 1 Vi t Vi CD ρ A V 2 M ln (2) (3) To find the minimum parachute area we must solve Eq 3 for A with x = xf when V = Vf A 2 M CD ρ x f Vi Vf ln (4) For three parachutes, the analysis is the same except A is replaced with 3A, leading to A 2 M 3 CD ρ x f Vi Vf ln kg 3 m Single and three-parachute sizes; plot speed against distance and time; maximum "g''s Basic equations: ρ 1.23 (5) dV The "g"'s are given by 2 dt CD ρ A V g which has a maximum at the initial instant (V = Vi) 2 M g The results generated in Excel are shown below: Single: A = D = Triple: 11.4 m 3.80 m 2 A = 3.8 m2 D = 2.20 m "g "'s = -1.01 Max t (s) x (m) V (km/hr) 0.00 2.50 5.00 7.50 10.0 12.5 15.0 17.5 20.0 22.5 24.6 0.0 216.6 393.2 542.2 671.1 784.7 886.3 978.1 1061.9 1138.9 1200.0 350 279 232 199 174 154 139 126 116 107 100 Aircraft Velocity versus Time 350 300 250 V (km/hr) 200 150 100 50 0 0 5 10 15 t (s) 20 25 Aircraft Velocity versus Distance 350 300 250 V (km/hr) 200 150 100 50 0 0 200 400 600 800 x (m) 1000 1200 Problem 9.96 [Difficulty: 3] Given: Data on airplane and parachute Find: Time and distance to slow down; plot speed against distance and time; maximum "g"'s Solution: The given data or available data is M 8500 kg km Vi 400 hr km Vf 100 hr π 2 2 Asingle Dsingle 28.274 m 4 Newton's second law for the aircraft is CD 1.42 ρ 1.23 kg 3 Dsingle 6 m Dtriple 3.75 m m π 2 2 Atriple Dtriple 11.045 m 4 M dV 1 2 CD ρ A V dt 2 where A and C D are the single parachute area and drag coefficient Separating variables dV 2 V Integrating, with IC V = Vi CD ρ A 2 M Vi V( t) 1 Integrating again with respect to t x ( t) Eliminating t from Eqs. 1 and 2 x dt CD ρ A 2 M 2 M CD ρ A (1) Vi t CD ρ A 2 M ln 1 Vi t Vi CD ρ A V 2 M ln (2) (3) To find the time and distance to slow down to 100 km/hr, Eqs. 1 and 3 are solved with V = 100 km/hr (or use Goal Seek) dV The "g"'s are given by dt g 2 CD ρ A V 2 M g which has a maximum at the initial instant (V = Vi) For three parachutes, the analysis is the same except A is replaced with 3A. leading to Vi V( t) 1 x ( t) 3 CD ρ A 2 M 2 M 3 CD ρ A Vi t 3 CD ρ A 2 M ln 1 Vi t The results generated in Excel are shown here: t (s) x (m) V (km/hr) t (s) x (m) V (km/hr) 0.0 0.0 1.0 96.3 2.0 171 3.0 233 4.0 285 5.0 331 6.0 371 7.0 407 8.0 439 9.0 469 9.29 477 0.0 0.0 1.0 94.2 2.0 165 3.0 223 4.0 271 5.0 312 6.0 348 7.0 380 7.93 407 9.0 436 9.3 443 400 302 243 203 175 153 136 123 112 102 100 400 290 228 187 159 138 122 110 100 91 89 "g "'s = -3.66 Max Aircraft Velocity versus Time 400 350 One Parachute Three Parachutes 300 V (km/hr) 250 200 150 100 50 0 0 1 2 3 4 5 6 t (s) 7 8 9 10 450 500 Aircraft Velocity versus Distance 400 350 V (km/hr) 300 One Parachute 250 Three Parachutes 200 150 100 50 0 0 50 100 150 200 250 300 350 x (m) 400 Problem 9.97 Given: Windmills are to be made from surplus 55 gallon oil drums D 24 in Find: [Difficulty: 2] H 29 in Which configuration would be better, why, and by how much Solution: Basic equations: 1 2 FD CD ρ V A 2 (Drag) Assumptions: (1) Standard air (2) Neglect friction in the pivot (3) Neglect interference between the flow over the two halves For the first configuration: ΣM D ΣM D 2 Fu D 2 Fd D 2 Fu Fd Where Fu is the force on the half "catching" the wind and F d is the force on the half "spilling" the wind. 1 1 1 D 2 2 2 CDu ρ V A CDd ρ V A CDu CDd ρ V A 2 2 2 2 2 For the second configuration: ΣM H ΣM H 2 Fu H 2 Fd H 2 Fu Fd 1 1 1 H 2 2 2 CDu ρ V A CDd ρ V A CDu CDd ρ V A 2 2 2 2 2 Since H > D, the second configuration will be superior. The improvement will be: H D D 20.8 % Problem 9.98 Given: Bike and rider at terminal speed on hill with 8% grade. W 210 lbf A 5 ft Find: [Difficulty: 2] ft Vt 50 s 2 CD 1.25 (a) Verify drag coefficient (b) Estimate distance needed for bike and rider to decelerate to 10 m/s after reaching level road Solution: Basic equations: 1 2 FD CD ρ V A 2 (Drag) Assumptions: (1) Standard air (2) Neglect all losses other than aerodynamic drag θ atan( 9 %) 5.143 deg Summing forces in the x-direction: ΣFx FG FD 0 1 2 Expanding out both force terms: M g sin( θ) CD ρ Vt A Solving this expression for the drag coefficient: 2 The angle of incline is: CD 2 W sin( θ) CD 1.26 2 ρ Vt A The original estimate for the drag coefficient was good. W d W d Once on the flat surface: ΣFx FD V V V Therefore: g dt g ds Separating variables: ds 2 W dV CD ρ g A V W g Integrating both sides yields: d V C 1 ρ V2 A D 2 ds V ∆s V2 CD ρ g A V1 2 W ln ∆s 447 ft Problem 9.99 Given: Data on cyclist performance on a calm day Find: Performance hindered and aided by wind [Difficulty: 2] Solution: The given data or available data is FR 7.5 N M 65 kg CD 1.2 ρ 1.23 2 A 0.25 m kg V 30 3 m The governing equation is 1 2 FD ρ A V CD 2 km hr FD 12.8 N The power steady power generated by the cyclist is Now, with a headwind we have P FD FR V P 169 W km Vw 10 hr V 24 km hr The aerodynamic drag is greater because of the greater effective wind speed 1 2 FD ρ A V Vw CD 2 FD 16.5 N The power required is that needed to overcome the total force FD + FR, moving at the cyclist's speed P V FD FR P 160 W This is less than the power she can generate She wins the bet! With the wind supporting her the effective wind speed is substantially lower km VW 10 hr V 40 1 2 FD ρ A V VW CD 2 km hr FD 12.8 N The power required is that needed to overcome the total force FD + FR, moving at the cyclist's speed P V FD FR This is more than the power she can generate P 226 W She loses the bet P 0.227 hp Problem 9.100 Given: [Difficulty: 2] Ballistic data for .44 magnum revolver bullet m m Vi 250 Vf 210 ∆x 150 m M 15.6 gm D 11.2 mm s s Average drag coefficient Find: Solution: Basic 1 2 FD CD ρ V A equations: 2 (Drag) Assumptions: (1) Standard air (2) Neglect all losses other than aerodynamic drag Newton's 2nd law: Separating variables: 1 2 d d d ΣFx FD M V M V V Therefore: M V V CD ρ V A 2 t s x d d d dx 2 M dV CD ρ A V Solving this expression for the drag coefficient: Integrating both sides yields: CD Vf ∆x ρ A Vi 2 M ln ∆x Vf CD ρ A Vi 2 M The area is: ln A π 4 2 2 D 98.52 mm Therefore the drag coefficent is: CD 0.299 Problem 9.101 Given: Data on cyclist performance on a calm day Find: Performance on a hill with and without wind [Difficulty: 3] Solution: The given data or available data is FR 7.5 N CD 1.2 2 M 65 kg A 0.25 m kg ρ 1.23 V 30 3 m The governing equation is 1 2 FD ρ A V CD 2 Power steady power generated by the cyclist is P FD FR V Riding up the hill (no wind) θ 5 deg km hr FD 12.8 N P 169 W P 0.227 hp For steady speed the cyclist's power is consumed by working against the net force (rolling resistance, drag, and gravity) Cycling up the hill: 1 2 P FR ρ A V CD M g sin( θ) V 2 This is a cubic equation for the speed which can be solved analytically, or by iteration, or using Excel's Goal Seek or Solver. The solution is obtained from the associated Excel workbook V 9.47 From Solver Now, with a headwind we have km hr km Vw 10 hr The aerodynamic drag is greater because of the greater effective wind speed 1 2 FD ρ A V Vw CD 2 The power required is that needed to overcome the total force (rolling resistance, drag, and gravity) moving at the cyclist's speed is Uphill against the wind: 1 2 P FR ρ A V Vw CD M g sin( θ) V 2 This is again a cubic equation for V From Solver V 8.94 km hr Pedalling downhill (no wind) gravity helps increase the speed; the maximum speed is obtained from Cycling down the hill: 1 2 P FR ρ A V CD M g sin( θ) V 2 This cubic equation for V is solved in the associated Excel workbook V 63.6 From Solver km hr Pedalling downhill (wind assisted) gravity helps increase the speed; the maximum speed is obtained from Wind-assisted downhill: 1 2 P FR ρ A V Vw CD M g sin( θ) V 2 This cubic equation for V is solved in the associated Excel workbook V 73.0 From Solver km hr Freewheeling downhill, the maximum speed is obtained from the fact that the net force is zero Freewheeling downhill: 1 2 FR ρ A V CD M g sin( θ) 0 2 V M g sin( θ) FR 1 2 Wind assisted: V 58.1 km V 68.1 km ρ A CD hr 1 2 FR ρ A V Vw CD M g sin( θ) 0 2 V Vw M g sin( θ) FR 1 2 ρ A CD hr Problem 9.102 [Difficulty: 3] Given: Data on cyclist performance on a calm day Find: Performance hindered and aided by wind; repeat with high-tech tires; with fairing Solution: The given data or available data is FR 7.5 N M 65 kg CD 1.2 ρ 1.23 2 A 0.25 m kg V 30 3 m The governing equation is 1 2 FD ρ A V CD 2 Power steady power generated by the cyclist is P FD FR V Now, with a headwind we have km Vw 10 hr km hr FD 12.8 N P 169 W The aerodynamic drag is greater because of the greater effective wind speed 1 2 FD ρ A V Vw CD 2 (1) The power required is that needed to overcome the total force FD + FR, moving at the cyclist's speed is P V FD FR (2) Combining Eqs 1 and 2 we obtain an expression for the cyclist's maximum speed V cycling into a headwind (where P = 169 W is the cyclist's power) Cycling into the wind: 1 2 P FR ρ A V Vw CD V 2 (3) This is a cubic equation for V; it can be solved analytically, or by iterating. It is convenient to use Excel's Goal Seek (or Solver). From the associated Excel workbook V 24.7 From Solver km hr By a similar reasoning: Cycling with the wind: 1 2 P FR ρ A V Vw CD V 2 (4) P 0.227 hp V 35.8 From Solver km hr With improved tires FR 3.5 N Maximum speed on a calm day is obtained from 1 2 P FR ρ A V CD V 2 This is a again a cubic equation for V; it can be solved analytically, or by iterating. It is convenient to use Excel's Goal Seek (or Solver). From the associated Excel workbook V 32.6 From Solver km hr Equations 3 and 4 are repeated for the case of improved tires From Solver Against the wind V 26.8 km V 29.8 km With the wind hr V 39.1 km V 42.1 km hr For improved tires and fairing, from Solver V 35.7 km hr Against the wind hr With the wind hr Problem 9.103 [Difficulty: 3] FBnet V FD y x T Given: Series of party balloons Find: Wind velocity profile; Plot Wlatex Solution: Basic equations: CD FD 1 FB ρair g Vol 2 ρ V A 2 The above figure applies to each balloon For the horizontal forces FD T sin( θ) 0 T cos( θ) FBnet Wlatex 0 Here π D FBnet FB W ρair ρHe g 6 (2) 3 D 20 cm M latex 3 gm RHe 2077 Rair 287 N m p He 111 kPa kg K N m p air 101 kPa kg K Applying Eqs 1 and 2 to the top balloon, for which Wlatex 0.02942 N p He kg THe 293 K ρHe ρHe 0.1824 RHe THe 3 m p air kg Tair 293 K ρair ρair 1.201 Rair Tair 3 m FBnet 0.0418 N θ 65 deg FBnet Wlatex cos( θ) Wlatex M latex g 3 π D FBnet ρair ρHe g 6 FD T sin( θ) This problem is ideal for computing and plotting in Excel, but we will go through the details here. (1) For the vertical forces We have (Table A.6) ΣF 0 sin( θ) Hence FD FBnet Wlatex tan( θ) FD 0.0266 N But we have 1 1 2 2 π D FD CD ρair V A CD ρair V 2 2 4 2 V 8 FD 2 CD ρair π D From Table A.9 ν 1.50 10 2 5 m s V 1.88 CD 0.4 with from Fig. 9.11 (we will check Re later) m s The Reynolds number is Red V D ν 4 Red 2.51 10 We are okay! For the next balloon θ 60 deg 8 FD V 2 V 1.69 CD ρair π D The Reynolds number is Red For the next balloon V D 2 V 1.40 CD ρair π D The Reynolds number is Red For the next balloon V D 2 V 1.28 CD ρair π D The Reynolds number is Red For the next balloon V D 2 CD ρair π D The Reynolds number is Red For the next balloon V D 2 CD ρair π D The Reynolds number is Red For the next balloon V D 2 CD ρair π D The Reynolds number is Red For the next balloon V D V 0.77 CD ρair π D The Reynolds number is Red For the next balloon V D ν θ 10 deg CD 0.4 We are okay! FD 0.00452 N with CD 0.4 with CD 0.4 m s 4 Red 1.03 10 with s 2 FD 0.00717 N m FD FBnet Wlatex tan( θ) 8 FD V 4 θ 20 deg CD 0.4 We are okay! Red 1.30 10 ν with s V 0.97 FD 0.00870 N m FD FBnet Wlatex tan( θ) 8 FD V 4 θ 30 deg CD 0.4 We are okay! Red 1.43 10 ν with s V 1.07 FD 0.01043 N m FD FBnet Wlatex tan( θ) 8 FD V 4 θ 35 deg CD 0.4 We are okay! Red 1.57 10 ν with s V 1.18 FD 0.01243 N m FD FBnet Wlatex tan( θ) 8 FD V 4 θ 40 deg CD 0.4 We are okay! Red 1.71 10 ν with s FD FBnet Wlatex tan( θ) 8 FD FD 0.01481 N m 4 θ 45 deg V Red 1.87 10 ν CD 0.4 We are okay! FD FBnet Wlatex tan( θ) 8 FD with s 4 θ 50 deg FD 0.0215 N m Red 2.25 10 ν V FD FBnet Wlatex tan( θ) We are okay! FD FBnet Wlatex tan( θ) FD 0.002191 N 8 FD V 2 V 0.54 CD ρair π D The Reynolds number is Red V D m s Red 7184.21 ν We are okay! V ( 0.54 0.77 0.97 1.07 1.18 1.28 1.40 1.69 1.88 ) In summary we have m s h ( 1 2 3 4 5 6 7 8 9 )m 10 h (m) 8 6 4 2 0 0.5 1 1.5 2 V (m/s) This does not seem like an unreasonable profile for the lowest portion of an atmospheric boundary layer - over cities or rough terrain the atmospheric boundary layer is typically 300-400 meters, so a near-linear profile over a small fraction of that distance is not out of the question. Problem 9.104 [Difficulty: 2] FB V FD y T W x Given: Sphere dragged through river Find: Relative velocity of sphere Solution: CD Basic equations: FD 1 FB ρ g Vol 2 ρ V A ΣF 0 2 The above figure applies to the sphere For the horizontal forces FD T sin( θ) 0 For the vertical forces Here V 5 m s (1) T cos( θ) FB W 0 D 0.5 m The Reynolds number is Red (2) SG 0.30 V D 6 Red 1.92 10 ν and from Table A.8 ν 1.30 10 2 6 m s ρ 1000 kg 3 m Therefore we estimate the drag coefficient: CD 0.15 (Fig 9.11) FB W FD T sin( θ) sin( θ) ρ g Vol ( 1 SG ) tan( θ) cos( θ) 3 Hence Therefore π D FD ρ g 6 1 ( 1 SG ) tan( θ) 2 2 π D CD ρ V 2 4 But we have 1 2 1 2 2 π D FD CD ρ V A CD ρ V 2 2 4 3 ρ g π D 6 ( 1 SG) tan( θ) 2 Solving for θ: tan( θ) 3 CD V 4 g D ( 1 SG ) 2 CD V 3 θ atan 4 g D ( 1 SG) The angle with the horizontal is: α 90 deg θ α 50.7 deg Problem 9.105 [Difficulty: 2] Problem 9.106 Given: Data on dimensions of anemometer Find: Calibration constant; compare to actual with friction [Difficulty: 3] Solution: The given data or available data is D 2 in R 3 in ρ 0.00234 slug ft 3 The drag coefficients for a cup with open end facing the airflow and a cup with open end facing downstream are, respectively, from Table 9.3 CDopen 1.42 CDnotopen 0.38 1 2 The equation for computing drag is FD ρ A V CD 2 (1) 2 A where π D A 0.0218 ft 4 2 Assuming steady speed ω at steady wind speed V the sum of moments will be zero. The two cups that are momentarily parallel to the flow will exert no moment; the two cups with open end facing and not facing the flow will exert a moment beacuse of their drag forces. For each, the drag is based on Eq. 1 (with the relative velocity used!). In addition, friction of the anemometer is neglected 1 1 2 2 ΣM 0 ρ A ( V R ω) CDopen R ρ A ( V R ω) CDnotopen R 2 2 or 2 2 ( V R ω) CDopen ( V R ω) CDnotopen This indicates that the anemometer reaches a steady speed even in the abscence of friction because it is the relative velocity on each cup that matters: the cup that has a higher drag coefficient has a lower relative velocity Rearranging for k V ω 2 V R C ω Dopen 2 V R C Dnotopen ω Hence CDnotopen 1 CDopen R k CDnotopen 1 CDopen k 9.43 in k 0.0561 mph rpm For the actual anemometer (with friction), we first need to determine the torque produced when the anemometer is stationary but about to rotate Minimum wind for rotation is Vmin 0.5 mph The torque produced at this wind speed is Tf 2 1 2 ρ A Vmin CDopen R Tf 3.57 10 1 ρ A V 2 C 2 min Dnotopen R 6 ft lbf A moment balance at wind speed V, including this friction, is ΣM 0 or 1 ρ A ( V R ω) 2 C 2 Dopen R 1 ρ A ( V R ω) 2 C 2 Dnotopen R Tf 2 Tf 2 2 ( V R ω) CDopen ( V R ω) CDnotopen R ρ A This quadratic equation is to be solved for ω when V 20 mph After considerable calculations ω 356.20 rpm This must be compared to the rotation for a frictionless model, given by V ωfrictionless k The error in neglecting friction is ωfrictionless 356.44 rpm ω ωfrictionless ω 0.07 % Problem 9.107 Given: Circular disk in wind Find: Mass of disk; Plot α versus V [Difficulty: 2] Solution: CD Basic equations: ΣM 0 FD 1 2 2 ρ V A Summing moments at the pivotW L sin( α) Fn L 0 1 M g sin( α) Hence ρ 1.225 The data is 2 2 2 π D ρ ( V cos( α) ) kg 2 V Rearranging 4 V 15 3 m M 1 2 Fn ρ Vn A CD 2 and 2 CD m D 25 mm s M 0.0451 kg 8 g sin( α) 2 π ρ D CD CD 1.17 2 π ρ V cos( α) D CD 8 M g α 10 deg tan( α) V 35.5 cos( α) m s tan( α) cos( α) We can plot this by choosing α and computing V 80 V (m/s) 60 40 20 0 10 20 30 40 Angle (deg) This graph can be easily plotted in Excel 50 60 70 (Table 9.3) Problem 9.108 [Difficulty: 2] Given: Mass, maximum and minimum drag areas for a skydiver Find: (a) Terminal speeds for skydiver in each position (b) Time and distance needed to reach percentage of terminal speed from given altitude Solution: CD Basic equation FD 1 2 2 ρ U A W 170 lbf ACDmin 1.2 ft The given or available data are: ρ From Table A.3 we can find the density: ρSL 0.7433 Ut 2 W ρ A CD ACDmax 9.1 ft ρ 0.002377 slug ft To find terminal speed, we take FBD of the skydiver: ΣFy 0 Solving for the speed: 2 3 2 H 9800 ft 2987 m 3 slug 0.7433 1.767 10 ft M g FD 0 For the minimum drag area: Utmax For the maximum drag area: Utmin 3 1 2 FD ρ U A CD M g W 2 2 W ρ ACDmin 2 W ρ ACDmax ft Utmax 400 s ft Utmin 145.4 s To find the time needed to reach a fraction of the terminal velocity, we re-write the force balance: ΣFy M ay M g FD M ay W In terms of the weight: W Ut W dU d U g dt Ut g dt 1 2 and M g 2 ρ U A CD W g U dU dy 1 2 2 ρ U A CD M W dU W dU U g dt g dy W Ut g 2 U d U Ut dy Ut dU dt 2 Simplifying this expression: 2 dU dy To normalize the derivatives by the terminal speed: We may now re-write the above equation as: 2 2 W Ut U d U U 2 W W Ut d U W ρ A CD g g dt Ut Ut dy Ut 2 Ut ρ A CD 1 M U where we have substituted for the terminal speed. Ut Ut U d U U d U 1 Now we can integrate with respect to time and g g dt Ut Ut Ut dy Ut distance: U Un Ut If we let Ut dUn 1 Un g dt 2 we can rewrite the equations: t g Separating variables: dt Ut 0 0.90 1 1 Un 2 g t dUn Integrating we get: Ut atanh( 0.9) atanh( 0 ) 0 Evaluating the inverse hyperbolic tangents: Now to find the distance: 2 Ut 2 t 1.472 Ut dUn 1 Un Un g dy g so: tmin 1.472 Utmin g Separating variables: y 0 Integrating we get: 1.472 Utmax 6.65 s tmax dy 2 Ut g 18.32 s 0.9 Un g 1 Un 2 dUn 0 2 1 0.92 0.8304 Ut 0.8304 Solving for the distance: y ln g 2 2 10 Ut g y 1 so: y min 0.8304 Utmin g 2 166.4 m y max 0.8304 Utmax g 2 1262 m Problem 9.109 [Difficulty: 3] Problem 9.110 [Difficulty: 3] Problem 9.111 [Difficulty: 3] Problem 9.112 [Difficulty: 3] Given: Data on a bus Find: Power to overcome drag; Maximum speed; Recompute with new fairing; Time for fairing to pay for itself Solution: Basic equation: 1 2 FD ρ A V CD 2 The given data or available data is 1 2 FD ρ A V CD 2 P FD V V 80 km V 22.2 hr FD 2096 N P FD V m s 2 A 7.5 m CD 0.92 ρ 1.23 kg 3 m P 46.57 kW The power available is Pmax 465 hp 346.75 kW The maximum speed corresponding to this maximum power is obtained from 1 1 2 Pmax ρ A Vmax CD Vmax 2 or Pmax Vmax 1 ρ A CD 2 We repeat these calculations with the new fairing, for which 1 2 FD ρ A V CD 2 FD 1959 N 3 m km Vmax 43.4 Vmax 156.2 s hr CD 0.86 Pnew FD V Pnew 43.53 kW 1 The maximum speed is now The initial cost of the fairing is Pmax Vmax 1 ρ A CD 2 3 Cost 4500 dollars The cost per day is reduced by improvement in the bus performance at 80 km/h The new cost per day is then Hence the savings per day is The initial cost will be paid for in m km Vmax 44.4 Vmax 159.8 s hr Costday 300 The fuel cost is Gain Pnew P Gain 93.5 % Costdaynew Gain Costday Costdaynew 280 Saving Costday Costdaynew Saving 19.6 τ Cost Saving dollars day dollars τ 7.56 month day dollars day Problem 9.113 [Difficulty: 2] Given: Data on 1970's and current sedans Find: Plot of power versus speed; Speeds at which aerodynamic drag exceeds rolling drag Solution: CD Basic equation: FD 1 2 2 ρ V A The aerodynamic drag is 1 2 FD CD ρ V A 2 Total resistance FT FD FR ρ = The rolling resistance is FR 0.015 W The results generated in Excel are shown below: 0.00234 3 slug/ft (Table A.9) Computed results: V (mph) F D (lbf) 1970's Sedan F T (lbf) P (hp) F D (lbf) 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 12.1 18.9 27.2 37.0 48.3 61.2 75.5 91.4 109 128 148 170 193 218 245 273 302 79.6 86.4 94.7 104 116 129 143 159 176 195 215 237 261 286 312 340 370 4.24 5.76 7.57 9.75 12.4 15.4 19.1 23.3 28.2 33.8 40.2 47.5 55.6 64.8 74.9 86.2 98.5 6.04 9.44 13.6 18.5 24.2 30.6 37.8 45.7 54.4 63.8 74.0 84.9 96.6 109 122 136 151 58.5 61.9 66.1 71.0 76.7 83.1 90.3 98.2 107 116 126 137 149 162 175 189 204 3.12 4.13 5.29 6.63 8.18 10.0 12.0 14.4 17.1 20.2 23.6 27.5 31.8 36.6 42.0 47.8 54.3 V (mph) F D (lbf) 67.5 F R (lbf) 67.5 V (mph) F D (lbf) 52.5 F R (lbf) 52.5 47.3 The two speeds above were obtained using Solver 59.0 Current Sedan F T (lbf) P (hp) Power Consumed by Old and New Sedans 120 1970's Sedan Current Sedan 100 80 P (hp) 60 40 20 0 20 30 40 50 60 V (mph) 70 80 90 100 Problem 9.114 [Difficulty: 4] Given: Data on a sports car Find: Speed for aerodynamic drag to exceed rolling resistance; maximum speed & acceleration at 100 km/h; Redesign change that has greatest effect Solution: 1 2 Basic equation: FD ρ A V CD 2 P FD V The given data or available data is M 1250 kg 2 A 1.72 m CD 0.31 Pengine 180 hp 134.23 kW FR 0.012 M g To find the speed at which aerodynamic drag first equals rolling resistance, set the two forces equal 2 FR ρ A CD V 21.2 m V 76.2 s V 100 1 2 FD ρ V A CD 2 η Pused 1 2 ρ V A CD FR 2 hr The power consumed by drag and rolling resistance at this speed is Hence the drive train efficiency is km V 27.8 hr m s Pengine 17 hp 12.677 kW FD 253 N Pused FD FR V Pused 11.1 kW η 87.7 % Pengine The acceleration is obtained from Newton's second law where T is the thrust produced by the engine, given by M a ΣF T FR FD P T V The maximum acceleration at 100 km/h is when full engine power is used. Pengine 180 hp 134.2 kW Because of drive train inefficiencies the maximum power at the wheels is Pmax η Pengine Hence the maximum thrust is Tmax The maximum acceleration is then Pmax V 3 km To find the drive train efficiency we use the data at a speed of The aerodynamic drag at this speed is kg m FR 147.1 N The rolling resistance is then Hence V ρ 1.23 Pmax 118 kW Tmax 4237 N amax Tmax FD FR M amax 3.07 m 2 s The maximum speed is obtained when the maximum engine power is just balanced by power consumed by drag and rolling resistance Pmax For maximum speed: 1 ρ V 2 A C F V 2 max D R max This is a cubic equation that can be solved by iteration or by using Excel's Goal Seek or Solver km Vmax 248 hr We are to evaluate several possible improvements: For improved drive train η η 6 % η 93.7 % Pmax Pmax η Pengine Pmax 126 kW 1 ρ V 2 A C F V 2 max D R max km Vmax 254 hr Solving the cubic (using Solver) Improved drag coefficient: CDnew 0.29 Pmax Pmax 118 kW 2 1 2 ρ Vmax A CDnew FR Vmax km This is the Vmax 254 hr best option! Solving the cubic (using Solver) Reduced rolling resistance: FRnew 0.91 % M g FRnew 111.6 N 1 2 Pmax ρ Vmax A CD FRnew Vmax 2 Solving the cubic (using Solver) km Vmax 250 hr The improved drag coefficient is the best option. Problem 9.115 [Difficulty: 3] Given: zero net force acting on the particle; drag force and electrostatic force Find: Solution: (a) Under steady-state, the net force acting on the particle is zero. The forces acting on the particle contain the electrostatic force FE and the drag force FD (Page 418, the first equation right after Fig.8.11). FE Fd 0 Qs E 6Ua 0 (1) where U is the particle velocity relative to the stationary liquid. Qs E (2) Then one obtains U 6a V FE y Qs FD x (b) From the solution, we can know that the particle velocity depends on its size. Smaller particles run faster than larger ones, and thus they can be separated. (c) Substituting the values of a, Qs, E, and into equation (2), we obtains the velocity for a=1 m U C 1000 V/m 10 12 N 0.053 0.053m/s 3 6 m Pa s m 6 10 Pa s 1 10 and U = 0.0053 m/s for a = 10 m. The negatively charged particle moves in the direction opposite to that of the electric field applied. Problem 9.116 Given: Data on dimensions of anemometer Find: Calibration constant [Difficulty: 5] Solution: The given data or available data is D 2 in R 3 in ρ 0.00234 slug ft 3 The drag coefficients for a cup with open end facing the airflow and a cup with open end facing downstream are, respectively, from Table 9.3 CDopen 1.42 CDnotopen 0.38 Assume the anemometer achieves steady speed ω due to steady wind speed V k The goal is to find the calibration constant k, defined by V ω We will analyze each cup separately, with the following assumptions 1) Drag is based on the instantaneous normal component of velocity (we ignore possible effects on drag coefficient of velocity component parallel to the cup) 2) Each cup is assumed unaffected by the others - as if it were the only object present 3) Swirl is neglected 4) Effects of struts is neglected R Relative velocity = Vcos - R Vcos V In this more sophisticated analysis we need to compute the instantaneous normal relative velocity. From the sketch, when a cup is at angle θ, the normal component of relative velocity is Vn V cos( θ) ω R (1) The relative velocity is sometimes positive, and sometimes negatiive. From Eq. 1, this is determined by ω R θc acos V For V n( θ ) 0 θ θc Vn 0 θc θ 2 π θc Vn 0 θc θ 2 π Vn 0 0 90 (2) 180 270 360 θ The equation for computing drag is 1 2 FD ρ A Vn CD 2 (3) 2 A where π D 2 A 3.14 in 4 In Eq. 3, the drag coefficient, and whether the drag is postive or negative, depend on the sign of the relative velocity For 0 θ θc CD CDopen FD 0 θc θ 2 π θc CD CDnotopen FD 0 θc θ 2 π CD CDopen FD 0 The torque is 1 2 T FD R ρ A Vn CD R 2 The average torque is Tav 2 π 1 2 π 0 1 T dθ π π T dθ 0 where we have taken advantage of symmetry Evaluating this, allowing for changes when θ = θ c θc π 1 1 1 1 2 2 Tav ρ A Vn CDopen R dθ ρ A Vn CDnotopen R dθ π 2 π 2 θ 0 c Using Eq. 1 and note that θ π c 2 2 ( V cos( θ) ω R) dθ CDnotopen ( V cos( θ) ω R) dθ Tav C 2 π Dopen 0 θ c π θ c 2 2 2 V ρ A R ω V Tav cos( θ) R dθ CDnotopen cos( θ) R dθ CDopen ω 2 π ω θ 0 c ρ A R V ω k The integral is 1 2 2 1 2 ( k cos( θ) R) dθ k 2 cos( θ) sin( θ) 2 θ 2 k R sin( θ) R θ For convenience define 1 2 1 2 f ( θ) k cos( θ) sin( θ) θ 2 k R sin( θ) R θ 2 2 Hence Tav ρ A R 2 π CDopen f θc CDnotopen f ( π) f θc For steady state conditions the torque (of each cup, and of all the cups) is zero. Hence 0 CDopen f θc CDnotopen f ( π) f θc or CDnotopen f θc f ( π) CDopen CDnotopen Hence CDnotopen 1 2 1 2 2 π 2 k cos θc sin θc θc 2 k R sin θc R θc k R π 2 CDopen CDnotopen 2 2 Recall from Eq 2 that ω R θc acos V Hence or R θc acos k CDnotopen R R 1 R R 2 1 R 2 π 2 2 k sin acos acos 2 k R sin acos R acos k R π k k 2 2 k 2 k k C C Dopen Dnotopen This equation is to be solved for the coefficient k. The equation is highly nonlinear; it can be solved by iteration or using Excel's Goal Seek or Solver From the associated Excel workbook k 0.990 ft The result from Problem 9.106 was k 0.0561 k 0.0707 mph mph rpm This represents a difference of 20.6%. The difference can be attributed rpm to the fact that we had originally averaged the flow velocity, rather than integrated over a complete revolution. Problem 9.117 [Difficulty: 4] Problem 9.118 [Difficulty: 4] Problem 9.119 [Difficulty: 4] Problem 9.120 [Difficulty: 2] Given: Data on advertising banner Find: Power to tow banner; Compare to flat plate; Explain discrepancy Solution: Basic equation: 1 2 FD ρ A V CD 2 P FD V V 55 mph The given data or available data is 1 2 FD ρ A V CD 2 FD 771 lbf P FD V 7 0.455 2.58 log ReL 1 2 FD ρ A V CD 2 s A 180 ft 4 ft V L ReL ν ft A L h ν 1.62 10 For a flate plate, check Re CD V 80.7 2 L 45 ft CD 0.05 h 4 ft slug ft L CD 0.563 h 4 ft lbf P 6.22 10 s P 113 hp 2 (Table A.9, 69oF) s ReL 2.241 10 so flow is fully turbulent. Hence use Eq 9.37b 1610 CD 0.00258 ReL ρ 0.00234 FD 3.53 lbf This is the drag on one side. The total drag is then 2 FD 7.06 lbf . This is VERY much less than the banner drag. The banner drag allows for banner flutter and other secondary motion which induces significant form drag. 3 Problem 9.121 [Difficulty: 5] Problem 9.122 Given: Data on car antenna Find: Bending moment [Difficulty: 1] FD Solution: V Basic equation: 1 2 FD ρ A V CD 2 The given or available data is V 120 km 33.333 hr s L 1.8 m D 10 mm x A 0.018 m kg 3 M 2 ν 1.50 10 5 m m For a cylinder, check Re Re V D ν y 2 A L D ρ 1.225 m Re 2.22 10 (Table A.10, 20 oC) s 4 From Fig. 9.13 CD 1.0 1 2 FD ρ A V CD 2 The bending moment is then L M FD 2 M 11.0 N m FD 12.3 N Problem 9.123 Given: Data on wind turbine blade Find: Bending moment [Difficulty: 1] FD V Solution: Basic equation: 1 2 FD ρ A V CD 2 The given or available data is V 85 knot 143.464 A L W slug ft ReL CD The bending moment is then s L 1.5 ft W 115 ft 3 2 4 ft ReL 1.32 10 ν 0.0742 1 1740 ReL x M ν 1.63 10 V L ReL ft A 172.5 ft ρ 0.00233 For a flat plate, check Re y CD 0.00311 5 1 2 FD 2 ρ A V CD 2 FD 25.7 lbf W M FD 2 M 1480 ft lbf 6 2 (Table A.9, 70oF) s so use Eq. 9.37a Problem 9.124 Given: Data on wind turbine blade Find: Power required to maintain operating speed [Difficulty: 4] Solution: Basic equation: 1 2 FD ρ A V CD 2 The given or available data is ω 25 rpm ρ 0.00233 L 1.5 ft slug ft W 115 ft ν 1.63 10 3 4 ft 2 (Table A.9, 70oF) s The velocity is a function of radial position, V( r) r ω, so Re varies from 0 to Remax V( W) L Remax 2.77 10 ν 6 The transition Reynolds number is 500,000 which therefore occurs at about 1/4 of the maximum radial distance; the boundary layer is laminar for the first quarter of the blade. We approximate the entire blade as turbulent - the first 1/4 of the blade will not exert much moment in any event Re( r) Hence L ν V( r) L ω ν r 1 CD Using Eq. 9.37a 0.0742 1 ReL The drag on a differential area is Hence M 1 dM dFD 2 W ReL 0.0742 1 L ω r ν 5 1740 L ω ν 1 2 2 ρ dA V CD ρ L V CD dr 2 2 W 1 1740 1 2 ρ L V CD r dr 0 1 2 M ρ L ω 2 5 M W 0.0742 r L ω r ν 1 5 1740 The bending moment is then 1 5 1 ν 2 3 ρ L ω r 0.0742 r 2 L ω 1 5 r 1 L ω ν dM dFD r ν 1 1740 r dr L ω 0 1 1 14 19 5 5 1 1740 ν 3 5 2 5 0.0742 ν 5 ν ν 2 0.0742 L ω r 1740 L ω r dr M 2 ρ L ω 19 L ω W 3 L ω W 0 M 1666 ft lbf 5 Hence the power is P M ω P 7.93 hp Problem 9.125 Given: A runner running during different wind conditions. Find: Calories burned for the two different cases Solution: Governing equation: CD FD FD 1 V 2 A 2 1 C D V 2 A 2 Assumption: 1) CDA = 9 ft2 2) Runner maintains speed of 7.5 mph regardless of wind conditions No wind: 0.00238 slug/ft 3 V 7.5 mph 11 ft/s The drag force on the runner is: FD 2 1 slug 2 ft 9 ft 2 0.00238 3 11 2 1.296 lbf 2 ft s Energy burned: E Power time FD Vrunner time Where time 4 mi Hence E 1.296 lbf hr 3600 s 1920 s 7.5 mi hr 0.0003238 kcal 11f 1920 s s ft lbf E 8.86 kcal With 5 mph wind: Going upwind: Vrel 12.5 mph 18.33 The drag force on the runner is: time 2 mi FD ft s 2 slug 1 2 ft 9 ft 2 0.00238 3 18.33 2 3.598 lbf 2 ft s hr 3600 s 960 s 7.5 mi hr [Difficulty: 2] Eupwind 3.598 lbf 0.0003238 kcal 11f 960 s 12.30 kcal s ft lbf Vrel 2.5 mph 3.67 Going downwind: The drag force on the runner is: FD ft s 2 slug 1 2 ft 9 ft 2 0.00238 3 3.67 2 0.144 lbf 2 ft s hr 3600 s 960 s 7.5 mi hr 0.0003238 kcal 11f 0.144 lbf 960 s 0.49 kcal s ft lbf time 2 mi E downwind Hence the total energy burned to overcome drag when the wind is 5 mph is: E 12.30 kcal 0.49 kcal 12.79 kcal this is 44% higher Problem 9.126 [Difficulty: 2] Problem 9.127 [Difficulty: 2] Problem 9.128 Given: Find: Solution: [Difficulty: 2] Data on helium-filled balloon, angle balloon string makes when subjected to wind FB V Drag coefficient for the balloon FD 1 2 Basic equations: FD ρ A V CD 2 The given or available data is D 20 in FB 0.3 lbf ρ 0.00233 slug ft Based on a free body diagram of the balloon, The reference area for the balloon is: A π 4 3 y ΣFy 0 V 10 ν 1.63 10 ft s 4 ft x θ 55 deg 2 T (Table A.9, 70oF) s FD FB tan( 90 deg θ) 0.2101 lbf 2 D 2.182 ft 2 so the drag coefficient is: CD FD 1 2 2 ρ V A 0.826 Problem 9.129 [Difficulty: [ 2] Problem 9.130 Given: 3 mm raindrop Find: Terminal speed [Difficulty: 2] Solution: Basic equation: 1 2 FD ρ A V CD 2 Given or available data is D 3 mm Summing vertical forces ΣF 0 kg ρH2O 1000 3 m kg ρair 1.225 3 m 1 2 M g FD M g ρair A V CD 0 2 M 1.41 10 Assume the drag coefficient is in the flat region of Fig. 9.11 and verify Re later V Re 2 M g CD ρair A V D ν V 8.95 2 5 m (Table A.10, 20 oC) s Buoyancy is negligible 3 π D M ρH2O 6 Check Re ν 1.50 10 5 2 kg A π D 4 CD 0.4 m s 3 Re 1.79 10 which does place us in the flat region of the curve Actual raindrops are not quite spherical, so their speed will only be approximated by this result 6 A 7.07 10 2 m Problem 9.131 [Difficulty: 3] Problem 9.132 [Difficulty: 3] Problem 9.133 [Difficulty: 3] F n2 Fn1 W Given: Circular disk in wind Find: Mass of disk; Plot α versus V Solution: Basic equations: CD ΣM 0 FD 1 2 2 ρ V A 1 D Summing moments at the pivot W L sin( α) Fn1 L L Fn2 0 (1) and for each normal drag 2 2 1 2 Fn ρ Vn A CD 2 Assume 1) No pivot friction 2) CD is valid for Vn = Vcos(α) The data is ρ 1.225 kg μ 1.8 10 3 5 N s m 2 m D 25 mm d 3 mm CD1 1.17 (Table 9.3) Red ρ V d μ V 15 m s L 40 mm α 10 deg Red 3063 so from Fig. 9.13 CD2 0.9 2 Hence 1 2 π D Fn1 ρ ( V cos( α) ) CD1 2 4 Fn1 0.077 N 1 D 2 Fn2 ρ ( V cos( α) ) L d CD2 2 2 Fn2 0.00992 N The drag on the support is much less than on the disk (and moment even less), so results will not be much different from those of Problem 9.105 2 Hence Eq. 1 becomes 1 1 D D 1 2 π D 2 M L g sin( α) L ρ ( V cos( α) ) CD1 L ρ ( V cos( α) ) L d CD2 2 4 2 2 2 2 2 M ρ V cos( α) 2 1 D D 2 π D CD1 1 L d CD2 4 g sin( α) 2 2 L 2 M 0.0471 kg V Rearranging 4 M g ρ tan( α) cos( α) We can plot this by choosing α and computing V 1 1 π D2 C 2 D1 1 D L D d C D2 2 L 2 V 35.5 80 V (m/s) 60 40 20 0 10 20 30 40 Angle (deg) This graph can be easily plotted in Excel 50 60 70 m s tan( α) cos( α) Problem 9.134 [Difficulty: 3] Given: Data on a tennis ball Find: Terminal speed time and distance to reach 95% of terminal speed Solution: The given data or available data is M 57 gm 2 A Then π D 4 2 5 m D 64 mm ν 1.45 10 3 A 3.22 10 At terminal speed drag equals weight FD M g The drag at speed V is given by 1 2 FD ρ A V CD 2 Hence the terminal speed is Vt Check the Reynolds number Re s kg 3 m 2 CD 0.5 (from Fig. 9.11) M g 2 ρ 1.23 m Assuming high Reynolds number 1 ρ A CD Vt D m Vt 23.8 s Re 1.05 10 ν 5 Check! For motion before terminal speed Newton's second law applies M a M dV 1 2 M g ρ V A CD dt 2 Separating variables 2 d V g k V dt or V 1 2 dV t g k V 0 Hence Evaluating at V = 0.95Vt For distance x versus time, integrate g V( t) k 0.95 Vt dx dt g k k where 1 2 ρ A CD k 0.0174 2 M dV g k V 1 g k k V g atanh tanh g k t g k tanh g k t tanh g k t 1 t g k x atanh 0.95 Vt t 0 g k tanh g k t dt k g t 4.44 s 1 m Note that 1 tanh( a t) dt ln( cosh( a t) ) a Hence x ( t) Evaluating at V = 0.95Vt 1 k t 4.44 s ln cosh g k t so x ( t) 67.1 m Problem 9.135 [Difficulty: 4] Problem 9.136 [Difficulty: 3] Problem 9.137 Given: Data on model airfoil Find: Lift and drag coefficients [Difficulty: 3] Solution: Basic equation: CD FD 1 2 ρ A V 2 Given or available data is D 2 cm V 30 m CL FL where A is plan area for airfoil, frontal area for rod 1 2 ρ A V 2 (Rod) L 25 cm FL 50 N s b 60 cm c 15 cm (Airfoil) FH 6 N Note that the horizontal force F H is due to drag on the airfoil AND on the rod ρ 1.225 kg ν 1.50 10 3 2 5 m m For the rod Rerod V D 4 Rerod 4 10 ν Arod L D Arod 5 10 1 2 FDrod CDrod ρ Arod V 2 Hence for the airfoil A b c CD 2 so from Fig. 9.13 3 2 ρ A V CD 0.0654 CDrod 1.0 2 m FDrod 2.76 N FD FH FDrod FD 1 (Table A.10, 20 oC) s CL FD 3.24 N FL 1 2 2 ρ A V CL 1.01 CL CD 15.4 Problem 9.138 [Difficulty: 3] Problem 9.139 [Difficulty: 4] Problem 9.140 [Difficulty: 4] Problem 9.141 [Difficulty: 4] Given: Data on a tennis ball Find: Terminal speed time and distance to reach 95% of terminal speed Solution: The given data or available data is M 57 gm 2 Then From Problem 9.132 A CD π D ν 1.45 10 3 2 A 3.22 10 4 24 m Re 1 Re 24 CD 2 5 m D 64 mm 1 Re 400 0.646 Re CD 0.5 400 Re 3 10 0.4275 5 5 6 CD 0.000366 Re 3 10 Re 2 10 CD 0.18 Re 2 10 6 At terminal speed drag equals weight FD M g The drag at speed V is given by 1 2 FD ρ A V CD 2 Assume CD 0.5 Hence the terminal speed is Vt Check the Reynolds number Re 2 M g ρ A CD Vt D ν This is consistent with the tabulated CD values! m Vt 23.8 s Re 1.05 10 5 s ρ 1.23 kg 3 m For motion before terminal speed, Newton's second law is M a M dV 1 2 M g ρ V A CD dt 2 Hence the time to reach 95% of terminal speed is obtained by separating variables and integrating t 0.95 Vt 1 g ρ A CD 2 M dV 2 V 0 For the distance to reach terminal speed Newton's second law is written in the form M a M V dV 1 2 M g ρ V A CD dx 2 Hence the distance to reach 95% of terminal speed is obtained by separating variables and integrating x 0.95 V t V g ρ A CD 2 M dV 2 V 0 These integrals are quite difficult because the drag coefficient varies with Reynolds number, which varies with speed. They are best evaluated numerically. A form of Simpson's Rule is ∆V f ( V) dV f V0 4 f V1 2 f V2 4 f V3 f VN 3 where ΔV is the step size, and V0, V1 etc., are the velocities at points 0, 1, ... N. Here From the associated Excel calculations (shown below): V0 0 0.95 Vt VN 0.95 Vt ∆V t 4.69 s x 70.9 m N These results compare to 4.44 s and 67.1 m from Problem 9.132, which assumed the drag coefficient was constant and analytically integrated. Note that the drag coefficient IS essentially constant, so numerical integration was not really necessary! For the time: V (m/s) Re 0 1.13 2.26 3.39 4.52 5.65 6.78 7.91 9.03 10.2 11.3 12.4 0 4985 9969 14954 19938 24923 29908 34892 39877 44861 49846 54831 5438 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 W f (V ) W xf (V ) 1 0.102 0.102 4 0.102 0.409 2 0.103 0.206 4 0.104 0.416 2 0.106 0.212 4 0.108 0.432 2 0.111 0.222 4 0.115 0.458 2 0.119 0.238 4 0.125 0.499 2 0.132 0.263 4 0.140 0.561 13.6 14.7 15.8 16.9 18.1 19.2 20.3 21.5 22.6 59815 64800 69784 74769 79754 84738 89723 94707 99692 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 2 4 2 4 2 4 2 4 1 CD 0.151 0.165 0.183 0.207 0.241 0.293 0.379 0.550 1.05 0.302 0.659 0.366 0.828 0.483 1.17 0.758 2.20 1.05 Total time: 4.69 For the distance: f (V ) W xf (V ) s 0.00 0.115 0.232 0.353 0.478 0.610 0.752 0.906 1.08 1.27 1.49 1.74 0.000 0.462 0.465 1.41 0.955 2.44 1.50 3.62 2.15 5.07 2.97 6.97 2.05 2.42 2.89 3.51 4.36 5.62 7.70 11.8 23.6 4.09 9.68 5.78 14.03 8.72 22.5 15.4 47.2 23.6 Total distance: 70.9 m Problem 9.142 Given: Data on an air bubble Find: Time to reach surface [Difficulty: 4] Solution: The given data or available data is h 100 ft 30.48 m ρ 1025 kg CD 0.5 (Fig. 9.11) (Table A.2) 3 p atm 101 kPa m 1 dx V dt where To find the location we have to integrate the velocity over time: V patm ρ g h 3 CD p atm ρ g ( h 4 g d 0 The results (generated in Excel) for each bubble diameter are shown below: d 0 = 0.3 in d 0 = 7.62 mm d0= t (s) x (m) V (m/s) t (s) x (m) V (m/s) 0 5 10 15 20 25 30 35 40 45 50 63.4 0 2.23 4.49 6.76 9.1 11.4 13.8 16.1 18.6 21.0 23.6 30.5 0.446 0.451 0.455 0.460 0.466 0.472 0.478 0.486 0.494 0.504 0.516 0.563 5 mm 0 5 10 15 20 25 30 35 40 45 50 55 0 1.81 3.63 5.47 7.32 9.19 11.1 13.0 14.9 16.9 18.8 20.8 0.362 0.364 0.367 0.371 0.374 0.377 0.381 0.386 0.390 0.396 0.401 0.408 60 65 70 75 77.8 22.9 25.0 27.1 29.3 30.5 0.415 0.424 0.435 0.448 0.456 d0 = 15 mm t (s) x (m) V (m/s) 0.0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0 45.1 0 3.13 6.31 9.53 12.8 16.1 19.5 23.0 26.6 30.5 0.626 0.635 0.644 0.655 0.667 0.682 0.699 0.721 0.749 0.790 Use Goal Seek for the last time step to make x = h ! Depth of Air Bubbles versus Time 30 25 20 x (m) 15 10 Initial Diameter = 5 mm Initial Diameter = 0.3 in 5 Initial Diameter = 15 mm 0 0 10 20 30 40 50 t (s) 60 70 80 x) 6 Problem 9.143 Given: Data on a tennis ball Find: Maximum height [Difficulty: 4] Solution: The given data or available data is M 57 gm 2 5 m m Vi 50 s D 64 mm ν 1.45 10 2 Then From Problem 9.132 A CD CD π D s 3 2 A 3.22 10 4 24 m Re 1 Re 24 1 Re 400 0.646 Re CD 0.5 400 Re 3 10 0.4275 The drag at speed V is given by 5 6 CD 0.000366 Re 3 10 Re 2 10 CD 0.18 Re 2 10 6 1 2 FD ρ A V CD 2 dV 1 2 ρ V A CD M g dt 2 For motion before terminal speed, Newton's second law (x upwards) is M a M For the maximum height Newton's second law is written in the form M a M V 0 Hence the maximum height is 5 V dV x max ρ A CD 2 V g 2 M V dV 1 2 ρ V A CD M g dx 2 V i V dV ρ A CD 2 V g 2 M i 0 This integral is quite difficult because the drag coefficient varies with Reynolds number, which varies with speed. It is best evaluated numerically. A form of Simpson's Rule is ∆V f ( V) dV f V0 4 f V1 2 f V2 4 f V3 f VN 3 where ΔV is the step size, and V0, V1 etc., are the velocities at points 0, 1, ... N. ρ 1.23 kg 3 m Here V0 0 From the associated Excel workbook (shown here) If we assume the integral VN Vi x max 48.7 m ∆V V (m/s) Re 0 11034 22069 33103 44138 55172 66207 77241 88276 99310 110345 121379 132414 143448 154483 165517 176552 187586 198621 209655 220690 0.0 2.5 5.0 7.5 10.0 12.5 15.0 17.5 20.0 22.5 25.0 27.5 30.0 32.5 35.0 37.5 40.0 42.5 45.0 47.5 50.0 CD 0.000 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 W 1 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 1 Vi N f (V ) W xf (V ) 0.000 0.252 0.488 0.695 0.866 1.00 1.09 1.16 1.19 1.21 1.21 1.20 1.18 1.15 1.13 1.10 1.06 1.03 1.00 0.970 0.940 0.000 1.01 0.976 2.78 1.73 3.99 2.19 4.63 2.39 4.84 2.42 4.80 2.36 4.62 2.25 4.38 2.13 4.13 2.00 3.88 0.940 CD 0.5 V i V dV x max ρ A CD 2 V g 2 M 0 becomes x max M ρ A CD ρ A CD ln 2 M g 2 Vi 1 x max 48.7 m The two results agree very closely! This is because the integrand does not vary much after the first few steps so the numerical integral is accurate, and the analytic solution assumes CD = 0.5, which it essentially does! Problem 9.144 [Difficulty: 4] Problem 9.145 [Difficulty: 3] Given: Data on rooftop carrier Find: Drag on carrier; Additional fuel used; Effect on economy; Effect of "cheaper" carrier Solution: Basic equation: Given or available data is CD FD 1 2 ρ A V 2 w 1 m V 100 h 50 cm km V 27.8 hr kg ρH2O 1000 3 m ρ 1.225 kg r 10 cm m FE 12.75 s r h Additional power is Additional fuel is 0.2 ∆P FD V s o FE 30.0 L A 0.5 m mi gal BSFC 0.3 2 5 m m From the diagram km 2 A w h ν 1.50 10 3 ηd 85 % kg kW hr (Table A.10, 20oF) s 1 2 FD CD ρ A V 2 CD 0.25 FD 59.1 N ∆P 1.93 kW ηd ∆FC BSFC ∆P 4 kg ∆FC 1.61 10 ∆FC 0.00965 s kg min Fuel consumption of the car only is (with SGgas 0.72 from Table A.2) FC FE Fuel economy with the carrier is r h Additional power is 0 ∆P FE SG gas ρH2O FCT FC ∆FC The total fuel consumption is then For the square-edged: V s o FD V ηd V FCT SG gas ρH2O CD 0.9 ∆P 6.95 kW FC 1.57 10 3 kg FCT 1.73 10 FE 11.6 s 3 kg km L 1 2 FD CD ρ A V 2 s FC 0.0941 kg min FCT 0.104 FE 27.2 mi gal FD 213 N kg min Additional fuel is ∆FC BSFC ∆P The total fuel consumption is then Fuel economy withy the carrier is now 4 kg ∆FC 5.79 10 s 3 kg FCT FC ∆FC FE V FCT FCT 2.148 10 SG gas ρH2O The cost of the trip of distance d 750 km for fuel costing p $ 3.50 gal ∆FC 0.0348 FE 9.3 km L s FCT 0.129 FE 21.9 The cost of the trip of with the rounded carrier ( FE 11.6 Cost d FE p discount Cost 69.47 plus the rental fee Cost 59.78 plus the rental fee km ) is then L d FE p min kg min mi gal with a rental discount $ 5 less than the rounded carrier is then Cost kg Hence the "cheaper" carrier is more expensive (AND the environment is significantly more damaged!) Problem 9.146 Given: Data on barge and river current Find: Speed and direction of barge Solution: Basic equation: CD FD 1 2 Given or available data is 2 ρ A V W 8820 kN w 10 m kg CDa 1.3 ρw 998 3 m [Difficulty: 4] 2 6 m νw 1.01 10 s L 30 m h 7 m kg ρa 1.21 3 m m Vriver 1 s νa 1.50 10 h sub W ρw g w L 3.00 m Vsub w L h sub CDw 1.3 2 5 m (Water data from Table A.8, air s data from Table A.10, 20 oC) First we need to calculate the amount of the barge submerged in the water. From Archimedes' Principle: The submerged volume can be expressed as: m Vwind 10 s W ρw g Vsub Combining these expressions and solving for the depth: h air h h sub 4.00 m Therefore the height of barge exposed to the wind is: Assuming the barge is floating downstream, the velocities of the water and air relative to the barge is: Vw Vriver Vbarge Assuming that the barge is rectangular, the areas exposed to the air and water are: Va Vwind Vbarge 2 Aa L w 2 ( L w) h air 620 m 2 Aw L w 2 ( L w) h sub 540 m In order for the barge to be traveling at a constant speed, the drag forces due to the air and water must match: 1 2 2 CDw ρw Vw Aw CDa ρa Va Aa 2 2 1 Solving for the speed relative to the water: 2 ρ Aa 2 2 a Vw Va ρw Aw ρa Aa In terms of the barge speed: Vw Va ρw Aw So solving for the barge speed: 2 Since the drag coefficients are equal, we can simplify: ρw Vw Aw ρa Va Aa Since the speeds must be in opposite directions: ρa Aa Vriver Vbarge Vwind Vbarge ρw Aw ρa Aa Vriver Vwind ρw Aw m Vbarge Vbarge 1.426 s ρa Aa 1 ρw Aw downstream Problem 9.147 [Difficulty: 4] Problem 9.148 [Difficulty: 4] Given: Data on sonar transducer Find: Drag force at required towing speed; minimum depth necessary to avoid cavitation Solution: CD Basic equation: FD 1 2 Given or available data is 2 ρ A V D 15 in A π 4 2 D 1.227 ft 2 p p inf 1 2 V 55 The Reynolds number of the flow is: Re The area is: CP V D ν ft s p min 5 psi ρ 1.93 6.486 10 h 6 ρ g 3 ν 1.06 10 5 ft 2 (Table A.7, 70oF) s From Fig. 9.11, we estimate the drag coefficient: Therefore the drag force is: p inf p atm slug ft From Fig. 9.12 the minimum pressure occurs where CP 1.2 Solving for the required depth: p p atm ρ g h 2 ρ V 1 2 FD CD ρ V A 2 Therefore: CD 0.18 FD 645 lbf 1 2 p inf p min CP ρ V 29.326 psi 2 h 33.9 ft Problem 9.149 [Difficulty: 4] Problem 9.150 [Difficulty: 4] Given: Data on a rocket Find: Plot of rocket speed with and without drag Solution: From Example 4.12, with the addition of drag the momentum equation becomes FB y FS y CV a rf y dV t CV v xyz dV CV v xyz V xyz dA where the surface force is FS y 1 AV 2 C D 2 Following the analysis of the example problem, we end up with 2 dVCV Ve m e 12 AVCV C D g dt M 0 m e t This can be written (dropping the subscript for convenience) dV f V , t dt (1) where f V , t Ve m e 12 AV 2 C D M 0 m e t g (2) Equation 1 is a differential equation for speed V. It can be solved using Euler’s numerical method Vn 1 Vn t f n where Vn+1 and Vn are the n + 1th and nth values of V, fn is the function given by Eq. 2 evaluated at the nth step, and t is the time step. The initial condition is V0 0 at t 0 Given or available data: M 0 = 400 kg m e = 5 kg/s V e = 3500 m/s = 1.23 kg/m D = 700 mm C D = 0.3 3 Computed results: 2 A = 0.38 m N = 20 t = 0.50 s With drag: n t n (s) V n (m/s) 0 0.0 0.0 1 0.5 17.0 2 1.0 34.1 3 1.5 51.2 4 2.0 68.3 5 2.5 85.5 6 3.0 102 7 3.5 119 8 4.0 136 9 4.5 152 10 5.0 168 11 5.5 184 33.9 34.2 34.3 34.3 34.2 34.0 33.7 33.3 32.8 32.2 31.5 30.7 17.0 34.1 51.2 68.3 85.5 102 119 136 152 168 184 200 0.0 17.0 34.1 51.3 68.7 86.2 104 122 140 158 176 195 33.9 34.2 34.5 34.8 35.1 35.4 35.6 35.9 36.2 36.5 36.9 37.2 17.0 34.1 51.3 68.7 86.2 104 122 140 158 176 195 213 12 6.0 13 6.5 14 7.0 15 7.5 16 8.0 17 8.5 18 9.0 19 9.5 20 10.0 29.8 28.9 27.9 26.9 25.8 24.7 23.6 22.5 21.4 214 229 243 256 269 282 293 305 315 213 232 251 270 289 308 328 348 368 37.5 37.8 38.1 38.5 38.8 39.1 39.5 39.8 40.2 232 251 270 289 308 328 348 368 388 200 214 229 243 256 269 282 293 305 f n V n+1 (m/s) Without drag: V n (m/s) f n V n+1 (m/s) Trajectory of a Rocket 400 300 V (m/s) 200 Without Drag 100 With Drag 0 0 2 4 6 8 t (s) 10 12 Problem 9.151 Given: Find: Solution: [Difficulty: 4] Baseball popped up, drag estimates based on Reynolds number Time of flight and maximum height CD Basic equation: FD 1 2 Given or available data is 2 ρ A V M 0.143 kg ΣFy M ay Here are the calculations performed in Excel: ρ = 1.21 dVy dt m V0y 25 D 0.073 m s We solve this problem by discretizing the flight of the ball: Given or available data: M = 0.143 V 0y = 25 D = 0.073 ay ΣFy ∆Vy ay ∆t ∆t M ∆y Vy ∆t kg m/s m kg/m 3 ν = 1.50E-05 m /s 2 Computed results: A = 0.00419 m Δt = 0.25 s 2 CD a y (m/s2 ) V ynew (m/s) 25.0 22.3 19.6 17.0 14.4 11.9 9.3 6.8 4.4 1.9 0.0 -2.5 Re 1.22E+05 1.08E+05 9.54E+04 8.26E+04 7.01E+04 5.77E+04 4.54E+04 3.33E+04 2.13E+04 9.30E+03 0.00E+00 1.19E+04 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.47 0.47 0.10 -10.917 -10.688 -10.490 -10.320 -10.177 -10.059 -9.964 -9.893 -9.844 -9.840 -9.810 -9.799 22.3 19.6 17.0 14.4 11.9 9.3 6.8 4.4 1.9 0.0 -2.5 -4.9 -4.9 -7.3 -9.8 -12.2 -14.6 -16.9 -19.3 -21.5 -23.7 2.39E+04 3.57E+04 4.76E+04 5.93E+04 7.09E+04 8.24E+04 9.37E+04 1.05E+05 1.15E+05 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 -9.767 -9.714 -9.641 -9.547 -9.434 -9.303 -9.154 -8.988 -8.816 -7.3 -9.8 -12.2 -14.6 -16.9 -19.3 -21.5 -23.7 t n (s) y (m) V y (m/s) 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.44 2.69 0.0 5.9 11.1 15.7 19.6 22.9 25.6 27.6 29.0 29.8 30.0 29.7 2.94 3.19 3.44 3.69 3.94 4.19 4.44 4.69 4.93 28.7 27.2 25.1 22.3 19.0 15.0 10.5 5.4 0.0 The results are plotted below. The answers are: height = 30.0 m time = 4.93 s Trajectory of Baseball 35 30 25 y (m) 20 15 10 5 0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 t (s) 3.5 4.0 4.5 5.0 Problem 9.152 [Difficulty: 5] Problem 9.153 [Difficulty: 5] Problem 9.154 [Difficulty: 5] Part 1/2 Problem 9.154 [Difficulty: 5] Part 2/2 Problem 9.155 [Difficulty: 5] Part 1/2 Problem 9.155 [Difficulty: 5] Part 2/2 Problem 9.156 [Difficulty: 3] Given: Data on airfoil and support in wind tunnel, lift and drag measurements Find: Lift and drag coefficients of airfoil FL Solution: Basic equations: CD 1 2 The given or available data is FD 2 ρ A V L 6 in CL V FL 1 2 2 ρ A V W 30 in FL 10 lbf FD V 100 y ft s Dcyl 1 in FD 1.5 lbf ρ 0.00233 slug ft Re V Dcyl 4 Re 5.112 10 ν 3 ν 1.63 10 4 ft 2 s FD FDcyl FDairfoil We need to determine the cylindrical support's contribution to the total drag force: Compute the Reynolds number x Lcyl 10 in Therefore: CDcyl 1 1 2 So the drag force on the support is: FDcyl CDcyl ρ V Lcyl Dcyl 0.809 lbf 2 So the airfoil drag is: FDairfoil FD FDcyl 0.691 lbf The reference area for the airfoil is: A L W 1.25 ft The lift and drag coefficients are: CL FL 1 2 CD 2 ρ V A 2 CL 0.687 FDairfoil 1 2 2 ρ V A CD 0.0474 Problem 9.157 [Difficulty: 2] Given: Antique airplane guy wires Find: Maximum power saving using optimum streamlining Solution: Basic equation: Given or available data is CD FD 1 2 ρ A V 2 L 50 m The Reynolds number is Hence Re V 175 km hr V 48.6 m s A 0.25 m kg 3 m V D ν D 5 mm 2 A L D ρ 1.21 P FD V ν 1.50 10 2 5 m Re 1.62 10 (Table A.10, 20 oC) s 4 1 2 P CD ρ A V V 2 so from Fig. 9.13 CD 1.0 P 17.4 kW with standard wires Figure 9.19 suggests we could reduce the drag coefficient to CD 0.06 Hence 1 2 Pfaired CD ρ A V V 2 Pfaired 1.04 kW The maximum power saving is then ∆P P Pfaired ∆P 16.3 kW Thus ∆P P 94 % which is a HUGE savings! It's amazing the antique planes flew! Problem 9.158 [Difficulty: 4] Problem 9.159 [Difficulty: 5] Problem 9.160 [Difficulty: 1] Problem 9.161 Given: Aircraft in level flight Find: Effective lift area; Engine thrust and power [Difficulty: 1] Solution: Basic equation: CD FD 1 2 ρ A V For level, constant speed 2 FD T Given or available data is V 225 km ρ 1.21 kg hr CL FL 1 2 FL W P T V 2 ρ A V V 62.5 m s CL 0.45 CD 0.065 M 900 kg (Table A.10, 20 oC) 3 m Hence Also 1 2 FL CL ρ A V M g 2 FL FD The power required is then CL CD FL M g T FD T 1275 N P T V P 79.7 kW A 2 M g 2 2 CL ρ V FL 8826 N A 8.30 m CD FD FL CL FD 1275 N Problem 9.162 Given: Data on a hydrofoil Find: Minimum speed, power required, top speed [Difficulty: 2] FL V Solution: Assumption: y FD W x The drag on the hydrofoil is much greater than any other drag force on the craft once the foil supports the craft. The given data or available data is ρ 1.94 slug ft A 7.5 ft 3 To support the hydrofoil, the lift force must equal the weight: Based on the required lift force, the speed must be: Vmin 2 CL 1.5 CD 0.63 W 4000 lbf Pmax 150 hp FL W 4000 lbf 2 FL ft Vmin 19.1 s ρ A CL The drag force at this speed is 1 2 FD ρ A Vmin CD 2 FD 1680 lbf Engine thrust required T FD T 1680 lbf The power required is P T Vmin P 58.5 hp As the speed increases, the lift will increase such that the lift and weight are still balanced. Therefore: CD Pmax W Vmax CL Solving for the maximum speed: Vmax Pmax CL W CD ft Vmax 49.1 s Problem 9.163 Given: Data on an airfoil Find: Maximum payload; power required [Difficulty: 2] Solution: The given data or available data is ρ 0.00234 slug ft 3 L 5 ft w 7 ft V 40 Then A w L A 35 ft The governing equations for steady flight are W FL and ft s CL 0.75 CD 0.19 2 T FD where W is the model total weight and T is the thrust The lift is given by 1 2 FL ρ A V CL 2 W M g FL The payload is then given by or FL 49.1 lbf M FL g M 49.1 lb The drag is given by 1 2 FD ρ A V CD 2 FD 12.4 lbf Engine thrust required T FD T 12.4 lbf The power required is P T V P 498 ft lbf s P 0.905 hp The ultralight model is just feasible: it is possible to find an engine that can produce about 1 hp that weighs less than about 50 lb. Problem 9.164 [Difficulty: 3] Given: Data on F-16 fighter Find: Minimum speed at which pilot can produce 5g acceleration; flight radius, effect of altitude on results Solution: The given data or available data is ρ 0.00234 slug ft 3 A 300 ft 2 CL 1.6 W 26000 lbf At 5g acceleration, the corresponding force is: FL 5 W 130000 lbf The minimum velocity corresponds to the maximum lift coefficient: Vmin 2 FL ρ A CL 481 ft ft Vmin 481 s s To find the flight radius, we perform a vertical force balance: β 90 deg asin FL sin ( 90 deg β) W 0 Now set the horizontal force equal to the centripetal acceleration: W 78.5 deg FL W FL cos ( 90 deg β) a g c ac g FL W cos ( 90 deg β) ac 157.6 ft s 2 The flight radius corresponding to this acceleration is: R As altitude increases, the density decreases, and both the velocity and radius will increase. Vmin ac R 1469 ft 2 Problem 9.165 Given: Data on an airfoil Find: Maximum payload; power required [Difficulty: 3] Solution: V 40 The given data or available data is Then the area is A b c and the aspect ratio is ar ft s ρ 0.00234 slug ft A 35.00 ft b 3 c 5 ft b 7 ft 2 ar 1.4 c The governing equations for steady flight are W FL and T FD where W is the model total weight and T is the thrust CL 1.2 At a 10o angle of attack, from Fig. 9.17 CDi 0.010 where CDi is the section drag coefficient The wing drag coefficient is given by Eq. 9.42 The lift is given by 2 CD CDi π ar 1 2 FL ρ A V CL 2 CD 0.337 FL 78.6 lbf W M g FL The payload is then given by M or CL FL M 78.6 lb g The drag is given by 1 2 FD ρ A V CD 2 Engine thrust required T FD The power required is P T V FD 22.1 lbf T 22.1 lbf P 1.61 hp NOTE: Strictly speaking we have TWO extremely stubby wings, so a recalculation of drag effects (lift is unaffected) gives b 3.5 ft and A b c 2.00 A 1.63 m ar b c c 5.00 ft ar 0.70 CL 2 CD CDi π ar so the wing drag coefficient is The drag is 1 2 FD 2 ρ A V CD 2 Engine thrust is T FD The power required is P T V CD 0.665 FD 43.6 lbf T 43.6 lbf P 3.17 hp In this particular case it would seem that the ultralight model makes more sense - we need a smaller engine and smaller lift requirements. However, on a per unit weight basis, the motor required for this aircraft is actually smaller. In other words, it should probably be easier to find a 3.5 hp engine that weighs less than 80 lb (22.9 lb/hp) than a 1 hp engine that weighs less than 50 lb (50 lb/hp). Problem 9.166 [Difficulty: 3] Given: Data on a light airplane Find: Angle of attack of wing; power required; maximum "g" force Solution: The given data or available data is ρ = 1.23⋅ V = 63⋅ m s The governing equations for steady flight are kg 3 2 M = 1000⋅ kg A = 10⋅ m CL = 0.72 CD = 0.17 W = M ⋅ g = FL T = FD m where W is the weight T is the engine thrust The lift coeffcient is given by 1 2 FL = ⋅ ρ⋅ A⋅ V ⋅ Cd 2 Hence the required lift coefficient is CL = M⋅ g 1 2 2 ⋅ ρ⋅ A⋅ V From Fig 9.17, for at this lift coefficient α = 3 ⋅ deg and the drag coefficient at this angle of attack is CD = 0.0065 CL = 0.402 (Note that this does NOT allow for aspect ratio effects on lift and drag!) Hence the drag is 1 2 FD = ⋅ ρ⋅ A⋅ V ⋅ CD 2 FD = 159 N and T = FD T = 159 N The power required is then P = T⋅ V P = 10⋅ kW The maximum "g"'s occur when the angle of attack is suddenly increased to produce the maximum lift From Fig. 9.17 CL.max = 1.72 1 2 FLmax = ⋅ ρ⋅ A⋅ V ⋅ CL.max 2 The maximum "g"s are given by application of Newton's second law M ⋅ aperp = FLmax where a perp is the acceleration perpendicular to the flight direction FLmax = 42⋅ kN Hence In terms of "g"s aperp = aperp g FLmax aperp = 42 M m 2 s = 4.28 Note that this result occurs when the airplane is banking at 90 o, i.e, when the airplane is flying momentarily in a circular flight path in the horizontal plane. For a straight horizontal flight path Newton's second law is M ⋅ aperp = FLmax − M ⋅ g Hence In terms of "g"s aperp = aperp g FLmax M = 3.28 −g aperp = 32.2 m 2 s Problem 9.167 [Difficulty: 3] Given: Data on a light airplane Find: Cruising speed achieved using a new airfoil design Solution: V 150 mph 220.00 The given data or available data is Then the area is A b c and the aspect ratio is ar ft ρ 0.00234 s ft A 192.50 ft b slug 3 c 5.5 ft b 35 ft 2 ar 6.36 c The governing equations for steady flight are W FL and T FD where W is the total weight and T is the thrust CL 0.3 For the NACA 23015 airfoil: CDi 0.0062 where CDi is the section drag coefficient The wing drag coefficient is given by Eq. 9.42 The drag is given by 1 2 FD ρ A V CD 2 Engine thrust required T FD The power required is P T V The wing drag coefficient is given by Eq. 9.42 P 1 3 ρ A V CD 2 2 CD CDi π ar T 116.7 lbf P 46.66 hp CDi 0.0031 CL 2 CD CDi π ar 3 so the new speed is: CD 0.011 FD 116.7 lbf CL 0.2 For the NACA 66 2-215 airfoil: The power is: CL Vnew 2 P ρ A CD 3 CD 5.101 10 ft Vnew 282 s Vnew 192.0 mph Problem 9.168 Given: Data on an airfoil Find: Solution: Maximum payload; power required The given data or available data is Vold 150 mph ρ 0.00234 [Difficulty: 3] slug ft 3 A 192.5 ft 2 35 arold 5.5 Assuming the old airfoil operates at close to design lift, from Fig. 9.19 CL 0.3 CDi 0.0062 CL 2 Then CDold CDi π arold The new wing aspect ratio is arnew 8 Hence The power required is CL CDold 0.0107 2 CDnew CDi π arnew CDnew 0.00978 1 2 P T V FD V ρ A V CD V 2 If the old and new designs have the same available power, then 1 2 2 ρ A Vnew CDnew Vnew ρ A Vold CDold Vold 2 2 1 3 or CDold Vnew Vold CDnew ft Vnew 227 s arold 6.36 (CDi is the old airfoil's section drag coefficient) Problem 9.169 [Difficulty: 3] Problem 9.170 Given: Aircraft in circular flight Find: Drag and power [Difficulty: 3] Solution: Basic equations: CD FD 1 CL 2 ρ A V 2 The given data or available data are ρ 0.002377 FL 1 2 slug ft 2 ρ A V R 3250 ft 3 V 150 mph V 220 ft P FD V Σ F M a M 10000 lbm M 311 slug A 225 ft s 2 ar 7 Assuming the aircraft is flying banked at angle β, the vertical force balance is FL cos( β) M g 0 or 1 2 2 ρ A V CL cos( β) M g (1) The horizontal force balance is 2 M V FL sin( β) M ar R or 1 2 2 ρ A V CL sin( β) 2 Then from Eq 1 M g FL cos( β) Hence CL 2 V R g β 24.8 deg 4 FL 1.10 10 lbf FL 1 tan( β) (2) R V2 β atan R g 2 Equations 1 and 2 enable the bank angle β to be found M V 2 ρ A V CL 0.851 CL For the section, CDinf 0.0075 at CL 0.851 (from Fig. 9.19), so 2 CD CDinf π ar Hence CD FD FL CL FD 524 lbf The power is P FD V P 1.15 10 5 ft lbf s P 209 hp CD 0.040 Problem 9.171 [Difficulty: 4] Given: Aircraft in circular flight Find: Maximum and minimum speeds; Drag and power at these extremes Solution: Basic equations: FD CD 1 CL 2 ρ A V 2 The given data or available data are ρ 0.002377 1 2 slug ft A 225 ft FL 2 ρ A V R 3250 ft 3 2 P FD V Σ F M a M 10000 lbm M 311 slug ar 7 The minimum velocity will be when the wing is at its maximum lift condition. From Fig . 9. 17 or Fig. 9.19 CL 1.72 CDinf 0.02 where CDinf is the section drag coefficient CL 2 CD CDinf π ar The wing drag coefficient is then CD 0.155 Assuming the aircraft is flying banked at angle β, the vertical force balance is FL cos( β) M g 0 or 1 2 1 2 ρ A V CL cos( β) M g 2 (1) The horizontal force balance is 2 M V FL sin( β) M ar R or ρ A V CL sin( β) 2 2 M V (2) R Equations 1 and 2 enable the bank angle β and the velocity V to be determined 2 2 M V 2 R M g 2 2 1 sin( β) cos( β) 1 1 2 2 ρ A V C ρ A V C 2 L L 2 2 or R 2 4 M V 2 M g 4 2 4 ρ A V CL 2 2 2 4 2 2 M g V 2 2 ρ A CL 4 2 tan( β) V R g V 149 2 M R 2 ft s V 102 mph 2 V2 R g β atan β 12.0 deg The drag is then 1 2 FD ρ A V CD 2 FD 918 lbf 5 ft lbf P FD V The power required to overcome drag is P 1.37 10 P 249 hp s The analysis is repeated for the maximum speed case, when the lift/drag coefficient is at its minimum value. From Fig. 9.19, reasonable values are CL 0.3 corresponding to α = 2 o (Fig. 9.17) 47.6 CL 2 CD CDinf π ar The wing drag coefficient is then 4 From Eqs. 1 and 2 CL CDinf 2 2 M g V 2 2 ρ A CL V ( 309.9 309.9i) 2 4 M 2 CD 0.0104 ft s Obviously unrealistic (lift is just too low, and angle of attack is too low to generate sufficient lift) 2 R We try instead a larger, more reasonable, angle of attack CL 0.55 CL The wing drag coefficient is then 4 From Eqs. 1 and 2 2 2 CD 0.0203 m V 204 mph 4 V 91.2 2 2 tan( β) V R g 1 2 FD ρ A V CD 2 The power required to overcome drag is M R 2 2 CD CDinf π ar 2 2 M g V ρ A CL The drag is then corresponding to α = 4 o (Fig. 9.17) CDinf 0.0065 s 2 V2 R g β atan β 40.6 deg FD 485 lbf P FD V 5 ft lbf P 1.45 10 s P 264 hp Problem 9.172 [Difficulty: 3] Problem 9.173 [Difficulty: 4] Problem 9.174 [Difficulty: 5] Part 1/2 Problem 9.174 [Difficulty: 5] Part 2/2 Problem 9.175 Given: Car spoiler Find: Whether they are effective [Difficulty: 4] Solution: To perform the investigation, consider some typical data For the spoiler, assume b 4 ft c 6 in ρ 1.23 kg 3 A b c m From Fig. 9.17 a reasonable lift coefficient for a conventional airfoil section is Assume the car speed is V 55 mph Hence the "negative lift" is 1 2 FL ρ A V CL 2 CL 1.4 FL 21.7 lbf This is a relatively minor negative lift force (about four bags of sugar); it is not likely to produce a noticeable difference in car traction The picture gets worse at 30 mph: FL 6.5 lbf For a race car, such as that shown on the cover of the text, typical data might be b 5 ft In this case: c 18 in A b c FL 1078 lbf Hence, for a race car, a spoiler can generate very significant negative lift! A 7.5 ft 2 V 200 mph A 2 ft 2 Problem 9.176 [Difficulty: 5] Problem 9.177 [Difficulty: 5] Problem 9.178 [Difficulty: 2] Problem 9.179 [Difficulty: 5] Problem 9.180 Given: Data on rotating cylinder Find: Lift force on cylinder [Difficulty: 2] Solution: CL = Basic equations: FL 1 2 The given or available data is 2 ⋅ ρ⋅ A⋅ V ρ = 1.21⋅ kg 3 2 −5 m ν = 1.50⋅ 10 ⋅ m The spin ratio is: The area is ω⋅ D 2⋅ V = 0.419 s L = 30⋅ cm D = 5 ⋅ cm ω = 240 ⋅ rpm V = 1.5⋅ m s From Fig. 9.29, we can estimate the maximum lift coefficient: CL = 1.0 2 A = D⋅ L = 0.015 m Therefore, the lift force is: 1 2 FL = ⋅ CL⋅ ρ⋅ A⋅ V 2 FL = 0.0204 N Problem 9.181 [Difficulty: 2] Problem 9.182 Given: Find: [Difficulty: 2] Data on original Flettner rotor ship Maximum lift and drag forces, optimal force at same wind speed, power requirement Solution: CL Basic equations: FL 1 2 The given or available data is 2 ρ A V ρ 0.00234 slug ft ν 1.62 10 The spin ratio is: The area is ω D 2 V 9.52 A D L 500 ft L 50 ft 3 4 ft D 10 ft ω 800 rpm V 30 mph 44 ft s 2 s From Fig. 9.29, we can estimate the lift and drag coefficients: CL 9.5 CD 3.5 2 Therefore, the lift force is: 1 2 FL CL ρ A V 2 FL 1.076 10 lbf The drag force is: 1 2 FD CD ρ A V 2 FD 3.964 10 lbf This appears to be close to the optimum L/D ratio. The total force is: F 2 4 3 2 4 FL FD To determine the power requirement, we need to estimate the torque on the cylinder. F 1.147 10 lbf T τ A R τ π L D D 2 2 π τ D L 2 In this expression τ is the average wall shear stress. We can estimate this stress using the flat plate approximation: D V ω 2 D 2.857 107 Re ν τ FD A τ 1 2 For a cylinder at this Reynolds number: CD 0.003 Therefore, the shear stress is: ρ V CD 6.795 10 2 3 lbf ft 2 2 So the torque is: T π τ D L 2 The power is: P T ω 4471 53.371 ft lbf ft lbf s P 8.13 hp Problem 9.183 [Difficulty: 4] x R L Given: Baseball pitch Find: Spin on the ball Solution: Basic equations: 1 2 The given or available data is Σ F M a FL CL 2 ρ A V ρ 0.00234 slug ft M 5 oz Compute the Reynolds number C 9 in ν 1.62 10 3 D 2 L 60 ft s 2 C D 2.86 in π V D Re 4 ft Re 1.73 10 ν A π D 2 A 6.45 in 4 V 80 mph 5 This Reynolds number is slightly beyond the range of Fig. 9.27; we use Fig. 9.27 as a rough estimate The ball follows a trajectory defined by Newton's second law. In the horizontal plane ( x coordinate) 2 V FL M aR M ax M R 1 2 FL ρ A V CL 2 and where R is the instantaneous radius of curvature of the trajectory From Eq 1 we see the ball trajectory has the smallest radius (i.e. it curves the most) when C L is as large as possible. From Fig. 9.27 we see this is when CL 0.4 Solving for R Also, from Fig. 9.27 Hence From the trajectory geometry R ω D 2 V 2 M (1) CL A ρ 1.5 ω 1.5 to 2 V D x R cos( θ) R Solving for x x R R 1 L R 2 V 1.8 ω 1.8 where sin( θ) 2 x R 1 ω D ω 14080 rpm L R R Hence R 463.6 ft 2 x 3.90 ft 2 V D L R defines the best range ω 16896 rpm Problem 9.184 [Difficulty: 3] Problem 9.185 [Difficulty: 4] x R L Given: Soccer free kick Find: Spin on the ball Solution: 1 2 2 ρ A V ρ 1.21 The given or available data is Σ F M a FL CL Basic equations: 2 5 m kg ν 1.50 10 3 m M 420 gm C 70 cm Compute the Reynolds number D 2 C D 22.3 cm π V D Re L 10 m s A Re 4.46 10 ν π D 4 5 The ball follows a trajectory defined by Newton's second law. In the horizontal plane ( x coordinate) 2 and 1 2 FL ρ A V CL 2 where R is the instantaneous radius of curvature of the trajectory 2 M Hence, solving for R R From the trajectory geometry x R cos( θ) R Hence x R 1 Solving for R R Hence, from Eq 1 CL For this lift coefficient, from Fig. 9.27 Hence ω D 2 V L where sin( θ) 2 R R L 2 x 2 2 x 2 M R A ρ R 50.5 m CL 0.353 1.2 ω 1.2 (And of course, Beckham still kind of rules!) (1) CL A ρ 2 V D ω 3086 rpm L R 2 A 0.0390 m This Reynolds number is beyond the range of Fig. 9.27; however, we use Fig. 9.27 as a rough estimate V FL M aR M ax M R x 1 m V 30 m s Problem 10.1 [Difficulty: 2] Problem 10.2 Given: Geometry of centrifugal pump Find: Estimate discharge for axial entry; Head [Difficulty: 2] Solution: Basic equations: (Eq. 10.2b) (Eq. 10.2c) The given or available data is ρ 999 kg 3 r1 10 cm r2 20 cm b 1 4 cm β1 30 deg β2 15 deg b 2 4 cm m ω 1600 rpm From continuity Q Vn w sin( β) 2 π r b w Vn sin( β) From geometry Vn Q Vt U w cos( β) U cos( β) U cot( β) sin( β) 2 π r b For an axial entry Vt1 0 so Using given data U1 ω r1 m U1 16.755 s Hence Q 2 π r1 b 1 U1 tan β1 Q U1 cot β1 0 2 π r1 b 1 3 Q 0.2431 m s To find the power we need U 2, Vt2, and m rate The mass flow rate is kg mrate 242.9 s mrate ρ Q m U2 33.5 s U2 ω r2 Q Vt2 U2 cot β2 2 π r2 b 2 Hence Wm U2 Vt2 U1 Vt1 mrate The head is H Wm mrate g m Vt2 15.5 s 5 J Wm 1.258 10 s Wm 126 kW H 52.8 m Problem 10.3 Given: Data on centrifugal pump Find: Estimate basic dimensions [Difficulty: 2] Solution: Basic equations: (Eq. 10.2b, directly derived from the Euler turbomachine equation) The given or available data is ρ 999 kg 3 3 Q 0.6 m ω 3000 rpm 3 m Q 0.0100 min m w2 5.4 s Vt1 0 From the outlet geometry Vt2 U2 Vrb2 cos β2 U2 Hence, in Eq. 10.2b Wm U2 mrate r2 ω mrate with Wm η Win and mrate ρ Q Hence r2 From continuity Hence s Win 5 kW η 72 % and U2 r2 ω β2 90 deg For an axial inlet Also m 2 2 2 Wm 3.6 kW kg mrate 9.99 s Wm 2 r2 0.06043 m m Vn2 5.40 s mrate ω Vn2 w2 sin β2 Q Vn2 2 π r2 b 2 Q b2 2 π r2 Vn2 r2 6.04 cm 3 b 2 4.8776 10 m b 2 0.488 cm Problem 10.4 [Difficulty: 2] Problem 10.5 [Difficulty: 2] Given: Geometry of centrifugal pump Find: Theoretical head; Power input for given flow rate Solution: Basic equations: (Eq. 10.2b) (Eq. 10.2c) The given or available data is ρ 1.94 slug ft 3 ω 575 rpm r1 15 in r2 45 in b 1 4.75 in b 2 3.25 in β1 40 deg β2 60 deg Q 80000 gpm Q 178 Q Vn Vrb sin( β) 2 π r b From geometry Vn Q Vt U Vrb cos( β) U cos( β) U cot( β) sin( β) 2 π r b Using given data U1 ω r1 U2 ω r2 ft Vt1 6.94 s Q Vt2 U2 cot β2 2 π r2 b 2 ft Vt2 210 s The mass flow rate is mrate ρ Q slug mrate 346 s Hence Wm U2 Vt2 U1 Vt1 mrate The head is H Wm mrate g s Vrb sin( β) ft U1 75.3 s 3 Vn From continuity Q Vt1 U1 cot β1 2 π r1 b 1 ft ft U2 226 s 7 ft lbf Wm 1.62 10 s 4 Wm 2.94 10 hp H 1455 ft Problem 10.6 [Difficulty: 2] Given: Geometry of centrifugal pump Find: Theoretical head; Power input for given flow rate Solution: Basic equations: (Eq. 10.2b) (Eq. 10.2c) The given or available data is ρ 1.94 slug ft 3 ω 1250 rpm r1 3 in r2 9.75 in b 1 1.5 in b 2 1.125 in β1 60 deg β2 70 deg Q 1500 gpm Q 3.34 Q Vn Vrb sin( β) 2 π r b From geometry Vn Q Vt U Vrb cos( β) U cos( β) U cot( β) sin( β) 2 π r b Using given data U1 ω r1 U2 ω r2 ft Vt1 22.9 s Q Vt2 U2 cot β2 2 π r2 b 2 ft Vt2 104 s The mass flow rate is mrate ρ Q slug mrate 6.48 s Hence Wm U2 Vt2 U1 Vt1 mrate The head is H Wm mrate g s Vrb sin( β) ft U1 32.7 s 3 Vn From continuity Q Vt1 U1 cot β1 2 π r1 b 1 ft Wm 66728 ft U2 106.4 s ft lbf s Wm 121 hp H 320 ft Problem 10.7 [Difficulty: 2] Given: Geometry of centrifugal pump Find: Rotational speed for zero inlet velocity; Theoretical head; Power input Solution: Basic equations: (Eq. 10.2b) (Eq. 10.2c) The given or available data is ρ 1.94 slug ft 3 r1 3 in r2 9.75 in b 1 1.5 in b 2 1.125 in β1 60 deg β2 70 deg Q 4000 gpm Q 8.91 Q Vn Vrb sin( β) 2 π r b From geometry Vn Q Vt U Vrb cos( β) U cos( β) U cot( β) 2 π r b sin( β) For Vt1 0 we get Q U1 cot β1 0 2 π r1 b 1 Hence, solving for ω ω Q 2 2 π r1 b 1 We can now find U2 or ω r1 ω 105 Q 2 π r1 b 1 ω 1001 rpm s ft U2 85.2 s ft Vt2 78.4 s The mass flow rate is mrate ρ Q slug mrate 17.3 s Hence Eq 10.2b becomes Wm U2 Vt2 mrate Wm 1.15 10 H Wm mrate g cot β1 0 rad Q Vt2 U2 cot β2 2 π r2 b 2 The head is s Vrb sin ( β) cot β1 U2 ω r2 3 Vn From continuity ft 5 ft lbf s Wm 210 hp H 208 ft Problem 10.8 [Difficulty: 2] Given: Geometry of centrifugal pump Find: Theoretical head; Power input for given flow rate Solution: Basic equations: (Eq. 10.2b) (Eq. 10.2c) The given or available data is ρ 1000 kg 3 r2 7.5 cm m ω 1750 rpm b 2 2 cm β2 65 deg 3 Q 225 3 m Q 0.0625 hr m s From continuity Q Vn2 2 π r2 b 2 m Vn2 6.63 s From geometry Vn2 Vt2 U2 Vrb2 cos β2 U2 cos β2 sin β2 Using given data U2 ω r2 Hence Q Vt2 U2 cot β2 2 π r2 b 2 m Vt2 10.7 s The mass flow rate is mrate ρ Q kg mrate 62.5 s Hence Wm U2 Vt2 mrate The head is H m U2 13.7 s Wm mrate g Vt1 0 (axial inlet) Wm 9.15 kW H 14.9 m Problem 10.9 [Difficulty: 2] Problem 10.10 [Difficulty: 2] Given: Geometry of centrifugal pump Find: Draw inlet and exit velocity diagrams; Inlet blade angle; Power Solution: Q Vn 2 π r b Basic equations: The given or available data is R1 1 in R2 7.5 in b 2 0.375 in Q 800 gpm Q 1.8 U1 ω R1 ft U1 17.5 s U2 ω R2 ft U2 131 s Q Vn2 2 π R 2 b 2 ft Vn2 14.5 s R2 Vn1 V R1 n2 ft Vn1 109 s ρ 1.94 slug ft Velocity diagrams: 3 ft ω 2000 rpm 3 β2 75 deg s Vt2 Vrb1 V n1 = V1 (Vt1 = 0) Vrb2 2 1 V2 2 Vn2 U2 U1 Then Vn1 β1 atan U1 From geometry Vt1 U1 Vn1 cos β1 Then Wm U2 Vt2 U1 Vt1 ρ Q β1 80.9 deg (Essentially radial entry) ft Vt1 0.2198 s Vt2 U2 Vn2 cos β2 4 ft lbf Wm 5.75 10 s ft Vt2 127.1 s Wm 105 hp Problem 10.11 [Difficulty: 3] Given: Geometry of centrifugal pump Find: Shutoff head; Absolute and relative exit velocitiesTheoretical head; Power input Solution: Basic equations: (Eq. 10.2b) (Eq. 10.2c) The given or available data is kg ρ 999 3 m ω 1800 rpm R1 2.5 cm R2 18 cm b 2 1 cm 3 β2 75 deg Q 30 m min 3 Q 0.500 m s m At the exit U2 ω R2 U2 33.9 s At shutoff Vt2 U2 m Vt2 33.9 s At design. from continuity Q Vn2 2 π R 2 b 2 m Vn2 44.2 s From the velocity diagram Vn2 w2 sin β2 Vn2 w2 sin β2 V2 with Vt2 α2 atan Vn2 For Vt1 0 we get Wm U2 Vt2 ρ Q 374 kW H0 117 m m Vt2 22.1 s m V2 49.4 s 2 Hence we obtain m w2 45.8 s Vt2 U2 Vn2 cot β2 2 1 H0 U2 Vt2 g Vn2 Vt2 α2 26.5 deg H Wm ρ Q g 76.4 m Problem 10.12 [Difficulty: 3] Problem 10.13 [Difficulty: 2] Given: Geometry of centrifugal pump Find: Inlet blade angle for no tangential inlet velocity at 125,000 gpm; Head; Power Solution: Basic equations: (Eq. 10.2b) (Eq. 10.2c) The given or available data is ρ 1.94 slug ft 3 ω 575 rpm r1 15 in r2 45 in b 1 4.75 in β2 60 deg Q 125000 gpm Q 279 ft 3 s Vn Vrb sin( β) From continuity Q Vn Vrb sin( β) 2 π r b From geometry Vn Q Vt U Vrb cos( β) U cos( β) U cot( β) sin( β) 2 π r b For Vt1 0 we obtain Q U1 cot β1 0 2 π r1 b 1 Using given data U1 ω r1 Hence β1 acot Also U2 ω r2 b 2 3.25 in or cot β1 2 π r1 b 1 U1 Q ft U1 75.3 s 2 π r1 b 1 U1 Q β1 50 deg ft U2 226 s Q Vt2 U2 cot β2 2 π r2 b 2 ft Vt2 201 s The mass flow rate is mrate ρ Q slug mrate 540 s Hence Wm U2 Vt2 U1 Vt1 mrate The head is H Wm mrate g 7 ft lbf Wm 2.45 10 s Wm 44497 hp H 1408 ft Problem 10.14 [Difficulty: 3] Problem 10.15 Given: Data on a centrifugal pump Find: Estimate exit angle of impeller blades [Difficulty: 3] Solution: The given or available data is ρ 999 kg 3 Q 50 m ω 1750 rpm L Win 45 kW s b 2 10 mm η 75 % D 300 mm The governing equation (derived directly from the Euler turbomachine equation) is Wm Vt2 U2 ρ Q For an axial inlet Vt1 0 hence We have D U2 ω 2 m U2 27.5 s Hence Wm Vt2 U2 ρ Q m Vt2 24.6 s From continuity Q Vn2 π D b 2 m Vn2 5.31 s an d Wm η Win Wm 33.8 kW With the exit velocities determined, β can be determined from exit geometry tan( β) Vn2 U2 Vt2 or U2 Vt2 β atan Vn2 β 61.3 deg Problem 10.16 [Difficulty: 3] Given: Data on a centrifugal pump Find: Flow rate for zero inlet tangential velocity; outlet flow angle; power; head developed Solution: The given or available data is ρ 999 kg 3 ω 1200 rpm η 70 % β1 25 deg r2 150 mm m r1 90 mm b 1 10 mm b 2 7.5 mm β2 45 deg The governing equations (derived directly from the Euler turbomachine equation) are We also have from geometry Vt2 α2 atan Vn2 From geometry Vn1 Vt1 0 U1 Vrb1 cos β1 r1 ω cos β1 sin β1 and from continuity Q Vn1 2 π r1 b 1 Hence (1) Q r1 ω 0 2 π r1 b 1 tan β1 2 Q 2 π r1 b 1 ω tan β1 Q 29.8 L s 3 Q 0.0298 The power, head and absolute angle α at the exit are obtained from direct computation using Eqs. 10.2b, 10.2c, and 1 above U1 r1 ω m U1 11.3 s m U2 18.8 s U2 r2 ω From geometry Vn2 Vt2 U2 Vrb2 cos β2 r2 ω cos β2 sin β2 and from continuity Q Vn2 2 π r2 b 2 m Vn2 4.22 s m Vt1 0 s m s Hence Vn2 Vt2 r2 ω tan β2 m Vt2 14.6 s Using these results in Eq. 1 Vt2 α2 atan Vn2 α2 73.9 deg Using them in Eq. 10.2b Wm U2 Vt2 U1 Vt1 ρ Q Using them in Eq. 10.2c H 1 g Wm 8.22 kW H 28.1 m U2 Vt2 U1 Vt1 This is the power and head assuming no inefficiency; with η = 70%, we have (from Eq. 10.4c) Wh η Wm Wh 5.75 kW Hp η H Hp 19.7 m (This last result can also be obtained from Eq. 10.4a Wh ρ Q g Hp) Problem 10.17 Given: Impulse turbibe Find: Optimum speed using the Euler turbomachine equation [Difficulty: 1] Solution: The governing equation is the Euler turbomachine equation In terms of the notation of Example 10.13, for a stationary CV r1 = r2 = R U1 = U2 = U Vt1 = V − U Vt2 = ( V − U) ⋅ cos( θ) and mflow = ρ⋅ Q Hence Tshaft = [ R⋅ ( V − U) ⋅ cos( θ) − R⋅ ( V − U) ] ⋅ ρ⋅ Q Tout = Tshaft = ρ⋅ Q⋅ R⋅ ( V − U) ⋅ ( 1 − cos( θ) ) The power is Wout = ω⋅ Tout = ρ⋅ Q⋅ R⋅ ω⋅ ( V − U) ⋅ ( 1 − cos( θ) ) Wout = ρ⋅ Q⋅ U⋅ ( V − U) ⋅ ( 1 − cos( θ) ) These results are identical to those of Example 10.13. The proof that maximum power is when U = V/2 is hence also the same and will not be repeated here. Problem 10.18 Given: Data on centrifugal pump Find: Pressure rise; Express as ft of water and kerosene [Difficulty: 1] Solution: Basic equations: The given or available data is η ρ Q g H Wm ρw 1.94 slug ft Wm 18 hp 3 Q 350 gpm H For kerosene, from Table A.2 SG 0.82 ft 3 s η 82 % η Wm Solving for H Q 0.780 H 166.8 ft ρw Q g η Wm Hk SG ρw Q g Hk 203 ft Problem 10.19 [Difficulty: 3] Given: Geometry of centrifugal pump Find: Draw inlet velocity diagram; Design speed for no inlet tangential velocity; Outlet angle; Head; Power Solution: Basic equations: (Eq. 10.2b) (Eq. 10.2c) The given or available data is r1 4 in ρ 1.94 r2 7 in slug ft 3 Velocity diagrams: b 1 0.4 in Q 70 cfm Q 1.167 b 2 0.3 in ft s V t2 w1 Vn1 = V 1 (Vt1 = 0) w2 2 V2 2 Vn2 U2 U1 Q Vn w sin( β) 2 π r b w Vn Vn1 sin( β) Vn2 From geometry Vn Q Vt U Vrb cos( β) U cos( β) U cot( β) sin( β) 2 π r b For Vt1 0 we obtain Q U1 cot β1 0 2 π r1 b 1 Solving for ω ω Q 2 2 π r1 b 1 Hence β2 45 deg 3 1 From continuity β1 20 deg U1 ω r1 or ω r1 Q 2 π r1 b 1 ω 138 ft U1 45.9 s U2 ω r2 cot β1 rad s A2 A1 cot β1 0 ω 1315 rpm ft U2 80.3 s r2 b 2 r1 b 1 Q Vn2 2 π r2 b 2 From the sketch Hence The head is ft Vn2 12.73 s Vt2 α2 atan Vn2 Q Vt2 U2 cot β2 2 π r2 b 2 ft Vt2 67.6 s α2 79.3 deg Wm U2 Vt2 ρ Q H Wm ρ Q g 4 ft lbf Wm 1.230 10 s Wm 22.4 hp H 169 ft Problem 10.20 [Difficulty: 4] Given: Geometry of centrifugal pump with diffuser casing Find: Flow rate; Theoretical head; Power; Pump efficiency at maximum efficiency point Solution: Basic equations: (Eq. 10.2b) (Eq. 10.2c) The given or available data is ρ 1000 kg 3 r2 7.5 cm b 2 2 cm β2 65 deg m ω 1750 rpm ω 183 rad s Using given data U2 ω r2 Illustrate the procedure with Q 0.065 m U2 13.7 s 3 m s From continuity Q Vn2 2 π r2 b 2 m Vn2 6.9 s From geometry Vn2 Vt2 U2 Vrb2 cos β2 U2 cos β2 sin β2 Hence Q Vt2 U2 cot β2 2 π r2 b 2 V2 2 m Vt2 10.5 s m V2 12.6 s 2 Vn2 Vt2 Hideal U2 Vt2 Hideal 14.8 m g Tfriction 10 % Vt1 0 Wmideal Tfriction 10 % ω 10 % Q ρ g Hideal ω ρ Q Hideal ω Tfriction 5.13 N m (axial inlet) V2 2 2 Vn2 Hactual 60 % 0.75 2 g 2 g η Q ρ g Hactual Q ρ g Hideal ω Tfriction Hactual 3.03 m η 18.7 % 25 Efficiency (%) 20 15 10 5 0 0.02 0.04 0.06 0.08 0.1 Q (cubic meter/s) The above graph can be plotted in Excel. In addition, Solver can be used to vary Q to maximize η. The results are 3 Q 0.0282 m s Wm Q ρ g Hideal ω Tfriction η 22.2 % Hideal 17.3 m Wm 5.72 kW Hactual 4.60 m Problem 10.21 [Difficulty: 4] Problem 10.22 [Difficulty: 2] Problem 10.23 [Difficulty: 2] Problem 10.24 [Difficulty: 3] Given: Data on suction pump Find: Plot of performance curves; Best effiiciency point Solution: ηp Basic equations: ρ = 1.94 slug/ft Ph Ph ρ Q g H Pm 3 (Note: Software cannot render a dot!) Fitting a 2nd order polynomial to each set of data we find 2 H =-0.00759Q + 0.390Q + 189.1 Q (cfm) H (ft) P m (hp) P h (hp) η (%) 36 50 74 88 125 190 195 176 162 120 25 30 35 40 46 12.9 18.4 24.6 27.0 28.4 η =-6.31x10 Q + 0.01113Q + 0.207 -5 51.7% 61.5% 70.4% 67.4% 61.7% 2 Finally, we use Solver to maximize η by varying Q : Q (cfm) H (ft) η (%) 88.2 164.5 69.8% Pump Performance Curve 250 100% H BEP 200 η 75% 50% 100 25% 50 0 0% 0 20 40 60 80 Q (cfm) 100 120 140 η (%) H (ft) 150 Problem 10.25 [Difficulty: 3] Given: Data on suction pump Find: Plot of performance curves; Best effiiciency point Solution: ηp Basic equations: ρ = 1.94 slug/ft Ph Ph ρ Q g H Pm 3 Ns N Q ( g H) (Note: Software cannot render a dot!) 0.75 Fitting a 2nd order polynomial to each set of data we find -5 2 -4 H =-1.062x10 Q + 6.39x10 Q + 22.8 Q (cfm) H (ft) P m (hp) P h (hp) η (%) 0 200 400 600 800 1000 23.0 22.3 21.0 19.5 17.0 12.5 15.2 17.2 24.4 27.0 32.2 36.4 0.0 8.4 15.9 22.1 25.7 23.6 η =-1.752x10 Q + 0.00237Q + 0.0246 -6 0.0% 49.0% 65.1% 82.0% 79.9% 65.0% 2 Finally, we use Solver to maximize η by varying Q : Q (cfm) H (ft) η (%) 676 18.4 82.6% Pump Performance Curve 100% 25 H BEP η 20 75% 50% 10 25% 5 0 0% 0 200 400 600 800 1000 Q (cfm) The Specific Speed for this pump is: 2.639 1200 η (%) H (ft) 15 Problem 10.26 [Difficulty: 3] Given: Data on axial flow fan Find: Volumetric flow rate, horsepower, flow exit angle Solution: Basic equations: (Eq. 10.2b) (Eq. 10.2c) The given or available data is ρ 0.002377 slug ft 3 ω 1350 rpm d tip 3 ft The mean radius would be half the mean diameter: Therefore, the blade speed is: U r ω d root 2.5 ft r U 194.39 1 d tip d root 2 2 U Vn1 V1 cos α1 So the entrance velocity components are: The volumetric flow rate would then be: Since axial velocity does not change: The exit speed relative to the blade is: Vt2 U w2 cos β2 The flow exit angle is: β2 60 deg r 1.375 ft s V1 sin α1 w1 cos β1 U V1 cos α1 w1 sin β1 ft V1 107.241 s cos α1 sin α1 tan β1 β1 30 deg ft From velocity triangles we can generate the following two equations: Combining the two equations: V1 α1 55 deg cos α1 w1 V1 sin β1 ft Vn1 61.511 s (axial component) (tangential component) ft w1 123.021 s Vt1 V1 sin α1 2 2 Q Vn1 d tip d root 4 π ft Vt1 87.846 s Q 132.9 ft 3 s Vn2 Vn1 Vn2 w2 sin β2 ft Vt2 158.873 s Vt2 α2 atan Vn2 ft so the tangential component of absolute velocity is: w2 71.026 s Into the expression for power: Wm U Vt2 Vt1 ρ Q Wm 7.93 hp α2 68.8 deg Problem 10.27 [Difficulty: 2] Problem 10.28 Given: Data on centrifugal pump Find: Electric power required; gage pressure at exit [Difficulty: 3] Solution: Basic equations: (Eq. 10.8a) (Eq. 10.8b) (Eq. 10.8c) The given or available data is ρ 1.94 slug ft 3 T 4.75 lbf ft ηp 75 % ηe 85 % Q 65 gpm Q 0.145 p 1 12.5 psi z1 6.5 ft ft V1 6.5 s z2 32.5 ft ft V2 15 s From Eq. 10.8c ω T ηp Hp ρ Q g Hence, from Eq. 10.8b ρ 2 2 p 2 p 1 V1 V2 ρ g z1 z2 ρ g Hp 2 p 2 53.7 psi Also Wh ρ g Q Hp Wh 1119 The shaft work is then Hence, electrical input is ft ω 3000 rpm s Hp 124 ft Wh Wm ηp Wm We ηe 3 ft lbf s Wm 1492 We 1756 ft lbf s ft lbf s Wh 2.03 hp Wm 2.71 hp We 2.38 kW Problem 10.29 [Difficulty: 2] Problem 10.30 [Difficulty: 2] Problem 10.31 [Difficulty: 2] Given: Data on small centrifugal pump Find: Specific speed; Sketch impeller shape; Required power input Solution: Basic equation: (Eq. 7.22a) (Eq. 10.3c) The given or available data is ρ 1000 kg 3 3 ω 2875 rpm ηp 70 % m Q 0.016 m s 2 Hence h g H h 392 1 Then NS ω Q (H is energy/weight. h is energy/mass) 2 s 2 3 h m NS 0.432 4 From the figure we see the impeller will be centrifugal The power input is (from Eq. 10.3c) Wh Wm ηp Wm ρ Q g H ηp Wm 8.97 kW H 40 m Problem 10.32 [Difficulty: 2] Problem 10.33 [Difficulty: 3] Given: Data on a pump Find: Shutoff head; best efficiency; type of pump; flow rate, head, shutoff head and power at 900 rpm Solution: The given or available data is ρ 999 3 kg Ns 1.74 3 D 500 mm Q 0.725 m H 10 m s m Wm 90 kW ω' 900 rpm 1 Wh ρ Q g H The governing equations are ω Q Ns Q1 ω1 D1 3 Q2 h1 ω2 D2 3 2 ω1 D1 2 H0 C1 3 h Similarity rules: 2 2 P1 2 ω2 D2 g 4 h2 U2 2 3 ρ1 ω1 D1 5 P2 3 ρ2 ω2 D2 5 3 h g H 98.1 J Ns h ω Hence kg ω 63.7 1 Q H0 The shutoff head is given by 4 rad Wh ρ Q g H 71.0 kW s Wh ηp Wm 78.9 % 2 U2 2 D m U2 ω 2 g U2 15.9 s H0 Hence U2 g 2 H0 25.8 m with D1 = D2: Q1 ω1 Q2 Q or ω2 ω Q' Q' Q ω' H0 Also 2 ω P1 ρ ω1 3 P2 ρ ω2 ω 3 1.073 m s H'0 h1 ω1 2 h2 ω2 2 Wm 3 ω W' m 3 ω' or 2 ω' H'0 H0 ω ω' or 3 ω' H 2 ω H' 2 ω' H' H H'0 56.6 m 2 21.9 m ω 2 ω' W' m Wm ω ω' 3 W' m 292 kW Problem 10.34 [Difficulty: 3] Given: Data on a pump at BEP Find: (a) Specific Speed (b) Required power input (c) Curve fit parameters for the pump performance curve. (d) Performance of pump at 820 rpm Solution: The given or available data is ρ 1.94 slug ft 3 η 87% D 16 in The governing equations are Ns Q 2500 cfm H 140 ft ω Q ( g H) Wh ρ Q g H 0.75 ω 1350 rpm W Wh η ω' 820 rpm 2 H0 U2 g Ns 1.66 The specific speed is: W 761 hp The power is: At shutoff Since D U2 ω 2 2 H H0 A Q ft U2 94.248 s A it follows that H0 Therefore: U2 2 H0 276.1 ft g H0 H 2 5 min A 2.18 10 2 Q Another way to write this is: At BEP: Q' Q ω' ω At 5 H( ft) 276.1 2.18 10 ω' H'0 H0 ω ω' 820 rpm ft Q( cfm) 2 2 and Q' 1519 cfm A' A 5 H'0 101.9 ft Thus: A' 2.18 10 2 5 min ft H' H ω' ω 5 2 H' 51.7 ft η' η 87 % ω' Wm W ω 3 Wm 170.5 hp Problem 10.35 Given: Data on pumping system Find: Number of pumps needed; Operating speed [Difficulty: 3] Solution: Wh ρ Q g H Basic equations: ηp Wh Wm The given or available data is kg 3 6 L Qtotal 110 10 day m Qtotal 1.273 s Then for the system Wh ρ Qtotal g H Wh 125 kW The required total power is Wh Wm η Wm 192 kW ρ 1000 3 m Hence the total number of pumps must be The flow rate per pump will then be Q 192 37.5 5.12 , or at least six pumps Qtotal 6 From Fig. 10.15 the peak effiiciency is at a specific speed of about NScu 2000 We also need H 32.8 ft Q 3363 gpm 3 Hence N NScu H 4 1 Q H 10 m N 473 2 The nearest standard speed to N 473 rpm should be used 3 Q 0.212 m s Q 212 L s η 65 % Problem 10.36 Given: Data on centrifugal pump Find: Head at 1150 rpm [Difficulty: 2] Solution: Basic equation: (Eq. 10.2c) The given or available data is ρ 1000 kg 3 3 Q 0.025 m ω 1750 rpm m Q Vn2 2 π r2 b 2 Hence Q r2 2 π b 2 Vn2 Then V'n2 From the outlet geometry Finally ω' ω r2 0.0909 m Vn2 V'n2 2.30 U'2 ω' r2 V't2 U'2 V'n2 cos β2 H' U'2 V't2 g b 2 1.25 cm m Vn2 3.5 s ω' 1150 rpm From continuity Also β2 60 deg s m s U'2 11.0 m V't2 9.80 m H' 10.9 m s s r2 9.09 cm Problem 10.37 Given: Data on pumping system Find: Total delivery; Operating speed [Difficulty: 3] Solution: Basic equations: Wh Wh ρ Q g H ηp Wm 30 kW H 30 m Wm The given or available data is ρ 1000 kg 3 H 98.425 ft η 65 % m Then for the system WmTotal 8 Wm WmTotal 240 kW The hydraulic total power is WhTotal WmTotal η The total flow rate will then be QTotal The flow rate per pump is Q WhTotal 156 kW 3 WhTotal m QTotal 0.53 s ρ g H QTotal 8 3 Q 0.066 From Fig. 10.15 the peak effiiciency is at a specific speed of about NScu 2500 3 Hence N NScu H 4 1 Q 7 L QTotal 4.58 10 day N 2410 2 The nearest standard speed to N 2410 rpm should be used m s Q 1051 gpm Problem 10.38 [Difficulty: 2] Problem 10.39 [Difficulty: 3] Given: Data on Peerless Type 10AE12 pump at 1720 rpm Find: Data at speeds of 1000, 1200, 1400, and 1600 rpm Solution: Q1 The governing equations are the similarity rules: ω1 D1 For scaling from speed ω1 to speed ω2: Speed (rpm) = 1760 Q (gal/min) 0 500 1000 1500 2000 2500 3000 3500 4000 3 Q2 Q1 ω1 Q (gal/min) 0 284 568 852 1136 1420 1705 1989 2273 ω2 D2 h1 3 2 ω1 D1 ω2 H2 H1 ω1 ω2 Speed (rpm) = 1000 2 Q H (ft) H (fit) 0 170 161 250000 160 160 1000000 155 157 2250000 148 152 4000000 140 144 6250000 135 135 9000000 123 123 12250000 110 109 16000000 95 93 Q2 Q (gal/min) 0 341 682 1023 1364 1705 2045 2386 2727 where 2 ω2 D2 2 h g H 2 Here are the results generated in Excel: Speed (rpm) = 1200 H (ft) 52.0 51.7 50.7 49.0 46.6 43.5 39.7 35.3 30.2 2 h2 H (ft) 74.9 74.5 73.0 70.5 67.1 62.6 57.2 50.8 43.5 Speed (rpm) = 1400 Q (gal/min) 0 398 795 1193 1591 1989 2386 2784 3182 H (ft) 102.0 101.3 99.3 96.0 91.3 85.3 77.9 69.2 59.1 Speed (rpm) = 1600 Q (gal/min) 0 455 909 1364 1818 2273 2727 3182 3636 H (ft) 133.2 132.4 129.7 125.4 119.2 111.4 101.7 90.4 77.2 Data from Fig. D.8 is "eyeballed" The fit to data is obtained from a least squares fit to H = H 0 - AQ 2 161 ft H0= A = 4.23E-06 ft/(gal/min) Performance Curves for Pump at various Speeds Fig. D.8 Data 180 1000 rpm 160 1200 rpm 1400 rpm 140 H (ft) 1600 rpm 120 100 80 60 40 20 0 0 500 1000 1500 2000 2500 Q (gal/min) 3000 3500 4000 4500 Problem 10.40 [Difficulty: 3] Problem 10.41 [Difficulty: 3] Problem 10.42 [Difficulty: 3] 10.6 Problem 10.43 [Difficulty: 3] 10.6: Problem 10.44 [Difficulty: 4] Problem 10.45 10.20 10 0.2 20 10-4 1 10 0-4 10.20 [Difficulty: 3] Part 1/2 Problem 10.45 [Difficulty: 3] Part 2/2 Problem 10.46 [Difficulty: 5] Problem 10.47 [Difficulty: 3] Given: Data on a model fan, smaller scale similar fan Find: Scale factor and volumetric flow rate of similar fan Solution: Basic equations: Q1 ω1 D1 3 The given or available data is Q2 ω2 D2 H1 3 2 ω1 D1 2 H2 2 ω2 D2 ω1 1440 rpm 2 ω2 1800 rpm 3 m Q1 6.3 s Solving the head equation for the scale D 2/D1: We can use this to find the new flowrate: D2 D1 H1 0.15 m ω1 ω2 H2 H1 0.8 D2 Q2 Q1 ω1 D1 ω2 3 H2 H1 0.15 m D2 D1 0.8 3 m Q2 4.03 s Problem 10.48 Given: Data on a model pump Find: Prototype flow rate, head, and power at 125 rpm [Difficulty: 3] Solution: Wh ρ Q g H Basic equation: Q1 ω1 D1 3 Q2 ω2 D2 and similarity rules h1 (10.19a) 3 The given or available data is 2 ω1 D1 2 h2 2 ω2 D2 Nm 100 rpm P1 (10.19b) 3 2 ρ1 ω1 D1 Np 125 rpm ρ 1000 P2 5 3 (10.19a) ρ2 ω2 D2 5 kg 3 m 3 From Eq. 10.8a From Eq. 10.19a (with Dm/Dp = 1/3) m Qm 1 s Hm 4.5 m Whm ρ Qm g Hm Whm 44.1 kW Qp ωp Dp Qm 3 3 ωm Dm 3 Np Qp 27 Qm Nm From Eq. 10.19b (with Dm/Dp = 1/3) hp 2 ωp Dp 2 m Qp 33.8 s hm g Hp or 2 2 2 ωp Dp ωm Dm 2 2 ωp Dp ωp 2 Hp Hm 3 Hm ωm Dm ωm From Eq. 10.19c (with Dm/Dp = 1/3) Pp 3 ρ ωp Dp 5 3 ωp Dp 3 Qp Qm 3 Qm ωm Dm ωm ωp or or 5 ρ ωm Dm Np Whp 243 Whm Nm 2 2 ωm Dpm Np Hp 9 Hm Nm 3 Pm 3 2 2 g Hm 5 2 Hp 63.3 m ωp Dp ωp 5 Whp Whm 3 Whm ωm Dm ωm 3 Whp 20.9 MW 3 Problem 10.49 [Difficulty: 2] Given: Data on a model pump Find: Temperature for dynamically similar operation at 1800 rpm; Flow rate and head; Comment on NPSH Solution: Basic equation: Re1 Re2 Q1 and similarity rules ω1 D1 3 Q2 ω2 D2 H1 3 2 ω1 D1 2 H2 2 ω2 D2 2 3 The given or available data is ω1 3600 rpm From Table A.8 at 15 oC ν1 1.14 10 For D = constant V1 D Re1 ν1 Q1 ω1 D Re2 2 ω1 D or ν2 ω2 ν2 ν1 ω1 2 7m ν2 5.7 10 ( 5.52 6.02) Q2 ( 5.70 6.02) H2 T2 48 or ω2 Q2 Q1 ω1 or ω2 H2 H1 ω1 3 2 ω2 D degrees C 3 m Q2 0.0500 s 2 H2 6.75 m The water at 48 oC is closer to boiling. The inlet pressure would have to be changed to avoid cavitation. The increase between runs 1 and 2 would have to be ∆p p v2 p v1 where p v2 and pv1 are the vapor pressures at T 2 and T1. From the steam tables: p v1 1.71 kPa s , we find, by linear interpolation ( 50 45) 2 s ω2 D D ω2 D H1 2 ν1 s 3 ω1 D D 2 7m T2 45 and also H1 27 m 2 6 m From Table A.8, at ν2 5.7 10 From similar operation m Q1 0.1 s ω2 1800 rpm p v2 11.276 kPa ∆p p v2 p v1 ∆p 9.57 kPa Problem 10.50 [Difficulty: 4] Problem 10.51 Given: Data on a NPSHR for a pump Find: Curve fit; Maximum allowable flow rate Solution: [Difficulty: 2] The results were generated in Excel: 2 Q (cfm) 20 40 60 80 100 120 140 Q 4.00E+02 1.60E+03 3.60E+03 6.40E+03 1.00E+04 1.44E+04 1.96E+04 NPSHR (ft) 7.1 8.0 8.9 10.3 11.8 14.3 16.9 NPSHR (fit) 7.2 7.8 8.8 10.2 12.0 14.2 16.9 The fit to data is obtained from a least squares fit to NPSHR = a + bQ a = b = 7.04 ft 2 5.01E-04 ft/(cfm) 2 Q (cfm) NPSHR (ft) 160.9 20.00 Use Goal Seek to find Q ! NPSHR Curve for a Pump NPSHR (m) 18 16 Data at 1450 rpm 14 12 10 Curve Fit 8 6 4 2 0 0 20 40 60 80 3 100 3 Q (m /s x 10 ) 120 140 160 Problem 10.52 [Difficulty: 3] Given: Data on a boiler feed pump Find: NPSHA at inlet for field temperature water; Suction head to duplicate field conditions Solution: Basic equation: 1 2 NPSHA p t p v p g p atm ρ V p v 2 Given or available data is Ds 10 cm Dd 7.5 cm H 125 m Q 0.025 p inlet 150 kPa p atm 101 kPa zinlet 50 cm ρ 1000 3 m s kg 3 ω 3500 rpm m For field conditions p g p inlet ρ g zinlet From continuity Vs 4 Q 2 π Ds From steam tables (try Googling!) at 115oC Hence p g 145 kPa m Vs 3.18 s p v 169 kPa 1 2 NPSHA p g p atm ρ Vs p v 2 Expressed in meters or feet of water NPSHA 82.2 kPa NPSHA ρ g 8.38m In the laboratory we must have the same NPSHA. From Table A.8 (or steam tables - try Googling!) at 27 oC Hence 1 2 p g NPSHA p atm ρ Vs p v 2 The absolute pressure is p g p atm 80.7 kPa p g 20.3 kPa NPSHA ρ g 27.5 ft p v 3.57 kPa Problem 10.53 Given: Pump and supply pipe system Find: Maximum operational flow rate [Difficulty: 3] Solution: Basic equations: H 2 2 p V1 V2 p2 1 α g z α g z ρ 1 2 1 ρ 2 2 2 h lT h lT f NPSHA 2 2 Le V2 L V V f K D 2 D 2 2 Le for the elbow, and K for the square entrance pt pv 2 Hr H0 A Q ρ g Assumptions: 1) p 1 = 0 2) V1 = 0 3) α 2 = 1 4) z 2 = 0 We must match the NPSHR (=Hr) and NPSHA From the energy equation 2 2 p2 V2 Le V2 V L V f f K g H D 2 2 2 D 2 ρ NPSHA pt pv ρ g p2 ρ g p atm ρ g V2 2 2 g pv ρ g 2 g L K ρ g D D 2 Le patm pv V L NPSHA H K f 2 g D D ρ g p2 2 H V 1 f Le Calculated results and plot were generated using Excel: Given data: Computed results: L = 20 ft e = 0.00085 ft D = 6.065 in K e nt = 0.5 L e /D = 30 H0= 10 ft Q (cfs) V (ft/s) 0.2 1.00 0.4 1.99 0.6 2.99 0.8 3.99 1.0 4.98 Re 4.75E+04 9.51E+04 1.43E+05 1.90E+05 2.38E+05 f NPSHA (ft) NPSHR (ft) 0.0259 55.21 10.32 0.0243 55.11 11.26 0.0237 54.95 12.84 0.0234 54.72 15.06 0.0232 54.43 17.90 5.98 6.98 7.98 8.97 2.85E+05 3.33E+05 3.80E+05 4.28E+05 0.0231 0.0230 0.0229 0.0229 54.08 53.66 53.18 52.63 21.38 25.48 30.22 35.60 A= H = p atm = pv= 7.9 22 14.7 0.363 ft/(cfs)2 ft psia psia 1.2 1.4 1.6 1.8 = 1.93 slug/ft 3 2.0 9.97 4.75E+05 0.0228 52.02 41.60 2.2 2.4 2.6 10.97 11.96 12.96 5.23E+05 0.0228 5.70E+05 0.0227 6.18E+05 0.0227 51.35 50.62 49.82 48.24 55.50 63.40 2.28 11.36 5.42E+05 0.0228 51.07 51.07 = 1.06E-05 ft /s 2 Crossover point: Error 0.00 NPSHA and NPSHR 70 60 Head (ft) 50 40 NPSHA NPSHR 30 20 10 0 0.0 0.5 1.0 1.5 Q (cfs) 2.0 2.5 3.0 Problem 10.54 [Difficulty: 2] Problem 10.55 [Difficulty: 5] H Given: Pump and supply pipe system Find: Maximum operational flow rate as a function of temperature Solution: Basic equations: 2 2 p 2 2 Le V2 V1 V2 p 2 1 L V V ρ α1 2 g z1 ρ α2 2 g z2 h lTh lT f D 2 f D 2 K 2 NPSHA Le for the elbow, and K for the square entrance 2 Hr H0 A Q pt pv ρ g Assumptions: 1) p 1 = 0 2) V1 = 0 3) α 2 = 0 4) z 2 = 0 We must match the NPSHR (=Hr) and NPSHA From the energy equation NPSHA g H pt pv ρ g 2 2 2 p2 V2 L f L V f e V K V D 2 2 D 2 2 ρ p2 ρ g p atm ρ g V2 2 2 g The results generated using Excel are shown on the next page. Given data: Computed results: pv ρ g p2 ρ g V L Le K D D 1 f L Le patm pv K 2 g D D ρ g 2 NPSHA H 2 g 2 H V f Given data: Computed results: o T ( C) p v (kPa) ρ 0 0.661 5 0.872 10 1.23 15 1.71 20 2.34 L= e = D= K ent = L e /D = H0 = 6 0.26 15 0.5 30 3 m A= H= p atm = 3000 6 101 m/(m /s) m kPa = 1000 kg/m m mm cm 3 2 3 2 3 (kg/m 1000 1000 1000 999 998 3 3 ) ν (m /s) Q (m /s) V (m/s) Re 1.76E-06 0.06290 3.56 3.03E+05 1.51E-06 0.06286 3.56 3.53E+05 1.30E-06 0.06278 3.55 4.10E+05 1.14E-06 0.06269 3.55 4.67E+05 1.01E-06 0.06257 3.54 5.26E+05 f NPSHA (m)NPSHR (m) Error 0.0232 0.0231 0.0230 0.0230 0.0229 14.87 14.85 14.82 14.79 14.75 14.87 14.85 14.82 14.79 14.75 0.00 0.00 0.00 0.00 0.00 25 30 35 3.17 4.25 5.63 997 996 994 8.96E-07 0.06240 8.03E-07 0.06216 7.25E-07 0.06187 3.53 3.52 3.50 5.91E+05 0.0229 6.57E+05 0.0229 7.24E+05 0.0228 14.68 14.59 14.48 14.68 14.59 14.48 0.00 0.00 0.00 40 7.38 992 6.59E-07 0.06148 3.48 7.92E+05 0.0228 14.34 14.34 0.00 = 1.01E-06 m /s 45 9.59 990 6.02E-07 0.06097 3.45 8.60E+05 0.0228 14.15 14.15 50 12.4 988 5.52E-07 0.06031 3.41 9.27E+05 0.0228 13.91 13.91 55 15.8 986 5.09E-07 0.05948 3.37 9.92E+05 0.0228 13.61 13.61 60 19.9 983 4.72E-07 0.05846 3.31 1.05E+06 0.0228 13.25 13.25 65 25.0 980 4.40E-07 0.05716 3.23 1.10E+06 0.0227 12.80 12.80 70 31.2 978 4.10E-07 0.05548 3.14 1.15E+06 0.0227 12.24 12.24 75 38.6 975 3.85E-07 0.05342 3.02 1.18E+06 0.0227 11.56 11.56 80 47.4 972 3.62E-07 0.05082 2.88 1.19E+06 0.0227 10.75 10.75 85 57.8 969 3.41E-07 0.04754 2.69 1.18E+06 0.0227 9.78 9.78 90 70.1 965 3.23E-07 0.04332 2.45 1.14E+06 0.0227 8.63 8.63 95 84.6 962 3.06E-07 0.03767 2.13 1.05E+06 0.0228 7.26 7.26 100 101 958 2.92E-07 0.02998 1.70 8.71E+05 0.0228 5.70 5.70 Use Solver to make the sum of absolute errors between NPSHA and NPSHR zero by varying the Q 's 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 NPSHR increases with temperature because the p v increases; NPHSA decreases because ρ decreases and p v increases Maximum Flow Rate Versus Water temperature 0.07 0.06 0.04 3 Q (m /s) 0.05 0.03 0.02 0.01 0.00 0 10 20 30 40 50 o T ( C) 60 70 80 90 100 Problem 10.56 [Difficulty: 3] Given: Pump and reservoir system Find: System head curve; Flow rate when pump off; Loss, Power required and cost for 1 m 3/s flow rate Solution: Basic equations: 2 2 p 2 2 V1 V2 p 2 1 L V V ρ α1 2 g z1 ρ α2 2 g z2 h lT hp h lT f D 2 Σ K 2 (K for the exit) where points 1 and 2 are the reservoir free surfaces, and h p is the pump head H Note also h ηp Pump efficiency: g Wh Wm Assumptions: 1) p 1 = p2 = patm 2) V1 = V2 = 0 3) α 2 = 0 4) z1 0, z2 15 m 4) K Kent Kent 1.5 2 From the energy equation g z2 f 2 L V V h p K D 2 2 Given or available data L 300 m ρ 1000 2 D 40 cm kg ν 1.01 10 3 2 2 L V V h p g z2 f K D 2 2 e 0.26 mm 2 6 m m (Table 8.1) (Table A.8) s The set of equations to solve for each flow rate Q are 4 Q V 2 Re V D π D ν e D 1 2.51 2.0 log 3.7 f Re f 3 For example, for Q 1 m s V 7.96 m Re 3.15 10 s 2 2 L V V Hp z2 f K D 2 g 2 g 6 f 0.0179 Hp 33.1 m 40 Head (m) 30 20 10 10 0 0.2 0.4 0.6 20 Q (cubic meter/s) 0.8 2 L V V Hp z2 f K D 2 g 2 g 1 3 The above graph can be plotted in Excel. In Excel, Solver can be used to find Q for H p = 0 Q 0.557 3 At Q 1 m we saw that s Hp 33.1 m 4 Assuming optimum efficiency at Q 1.59 10 gpm from Fig. 10.15 ηp 92 % Then the hydraulic power is Wh ρ g Hp Q Wh 325 kW The pump power is then Wh Wm ηp Wm 2 706 kW If electricity is 10 cents per kW-hr then the hourly cost is about $35 If electricity is 15 cents per kW-hr then the hourly cost is about $53 If electricity is 20 cents per kW-hr then the hourly cost is about $71 m s (Zero power rate) Problem 10.57 [Difficulty: 2] Problem 10.58 [Difficulty: 3] Part 1/2 Problem 10.58 [Difficulty: 3] Part 2/2 Problem 10.59 Given: Data on pump and pipe system Find: Delivery through system [Difficulty: 3] Solution: Governing Equations: For the pump and system where the total head loss is comprised of major and minor losses and the pump head (in energy/mass) is given by (from Example 10.6) Hpump( ft) 55.9 3.44 10 5 Q( gpm) 2 Hence, applied between the two reservoir free surfaces (p 1 = p 2 = 0, V1 = V2 = 0, z1 = z2 ) we have 0 h lT ∆hpump h lT g Hsystem ∆hpump g Hpump HlT Hpump or (1) 2 where L1 L2 V1 V2 HlT f1 Kent f2 Kexit D1 2 g D2 2 Results generated in Excel are shown on the next page. 2 Given or available data: L1 = D1 = L2 = D2 = e = 3000 9 1000 6 0.00085 ft ν= in K ent = ft K exp = in Q loss = ft (Table 8.1) 2 1.23E-05 ft /s (Table A.7) 0.5 (Fig. 8.14) 1 75 gpm The system and pump heads are computed and plotted below. To find the operating condition, Goal Seek is used to vary Q 1 so that the error between the two heads is zero. Q 1 (gpm) Q 2 (gpm) V 1 (ft/s) V 2 (ft/s) Re 1 Re 2 f1 f2 H lT (ft) H pump (ft) 100 200 300 400 500 600 700 25 125 225 325 425 525 625 0.504 1.01 1.51 2.02 2.52 3.03 3.53 0.284 1.42 2.55 3.69 4.82 5.96 7.09 30753 61506 92260 123013 153766 184519 215273 11532 57662 103792 149922 196052 242182 288312 0.0262 0.0238 0.0228 0.0222 0.0219 0.0216 0.0215 0.0324 0.0254 0.0242 0.0237 0.0234 0.0233 0.0231 0.498 3.13 8.27 15.9 26.0 38.6 53.6 55.6 54.5 52.8 50.4 47.3 43.5 39.0 Q 1 (gpm) Q 2 (gpm) V 1 (ft/s) V 2 (ft/s) Re 1 Re 2 f1 f2 H lT (ft) H pump (ft) 627 552 3.162 6.263 192785 254580 0.0216 0.0232 42.4 42.4 Error) 0% 700 800 Pump and System Heads 60 50 H (ft) 40 30 Pump System 20 10 0 0 100 200 300 400 Q (gal/min) 500 600 Problem 10.60 Given: Pump and reservoir/pipe system Find: Flow rate using different pipe sizes [Difficulty: 3] Solution: 2 2 p V1 V2 p 2 1 α g z α g z ρ 1 2 1 ρ 2 2 2 h lT h p Basic equations: 2 h lT f 2 2 Le V L V V Σ f Σ K D 2 D 2 2 H and also Le for the elbows, and K for the square entrance and exit h g Le 2) V = V = 0 3) α = 1 4) z1 0, z2 24 ft 4) K Kent Kexp 5) is for two elbows atm 1 2 D Assumptions: 1) p = p = p 1 Hence 2 h lT f 2 2 Le V2 L V V f K D 2 D 2 2 z2 HlT Hp or and also We want to find a flow that satisfies these equations, rewritten as energy/weight rather than energy/mass L Le V2 HlT f K D D 2 g H1T z2 Hp Here are the results calculated in Excel: Given or available data (Note: final results will vary depending on fluid data selected): L = 1750 ft e = 0.00015 ft (Table 8.1) D = 7.981 in K ent = K exp = L e/D elbow = 0.5 1 60 (Fig. 8.14) ν = 1.06E-05 ft /s (Table A.8) z2 = 24 ft L e/D va lve = 8 (Table 8.4) 2 (Two) H1T Hp z2 2 The pump data is curve-fitted to H pump = H 0 - AQ . The system and pump heads are computed and plotted below. To find the operating condition, Solver is used to vary Q so that the error between the two heads is minimized. A plot of the pump and system heads is shown for the 8 in case - the others will look similar. Q (cfm) Q2 V (ft/s) H p (fit) H lT + z 2 (ft) H p (ft) Re f 0.000 0 90.0 0.00 0 0.0000 89 24.0 50.000 2500 87.0 2.40 150504 0.0180 87 28.5 100.000 10000 81.0 4.80 301007 0.0164 81 40.4 150.000 22500 70.0 7.20 451511 0.0158 72 59.5 200.000 40000 59.0 9.59 602014 0.0154 59 85.8 250.000 62500 43.0 11.99 752518 0.0152 42.3 119.1 300.000 90000 22.0 14.39 903022 0.0150 21.9 159.5 H0= 89 ft A = 7.41E-04 ft/(cfm) Q (cfm) V (ft/s) 167.5 8.03 Repeating for: 2 Re H p (fit) H lT + z 2 (ft) Error) f 504063 0.0157 67.9 67.9 0.00% D = 10.02 in Q (cfm) V (ft/s) 179.8 8.63 H p (fit) H lT + z 2 (ft) Error) Re f 541345 0.0156 64.7 64.7 0.00% D = Repeating for: Q (cfm) V (ft/s) 189.4 9.09 12 in H p (fit) H lT + z 2 (ft) Error) Re f 570077 0.0155 62.1 62.1 0.00% Pump and System Heads (8 in pipe) 180 160 140 120 H (ft) 100 80 60 40 20 0 Pump Curve Fit Pump Data Total Head Loss 0 50 100 150 Q (cfm) 200 250 300 350 Problem 10.61 Given: Data on pump and pipe system Find: Delivery through system, valve position to reduce delivery by half [Difficulty: 3] Solution: Governing Equations: For the pump and system where the total head loss is comprised of major and minor losses Hence, applied between the two reservoir free surfaces (p 1 = p 2 = 0, V1 = V2 = 0, z1 - z2 = z) we have g ∆z h lT ∆hpump h lT g ∆z g Hsystem g ∆z ∆hpump g Hpump or HlT ∆z Hpump 2 where Le Le V L HlT f 2 Kent Kexit 2 g Delbow Dvalve D The calculations performed using Excel are shown on the next page: Given or available data (Note: final results will vary depending on fluid data selected): L = 1200 ft D = 12 in e = 0.00015 ft (Table 8.1) K e nt = K e xp = L e/D elbow = 0.5 1 30 (Fig. 8.14) ν = 1.23E-05 ft /s (Table A.7) z = -50 ft L e/D valve = 8 (Table 8.4) 2 2 The pump data is curve-fitted to H pump = H 0 - AQ . The system and pump heads are computed and plotted below. To find the operating condition, Solver is used to vary Q so that the error between the two heads is minimized. Q (gpm) Q 2 (gpm) H pump (ft) 0 500 1000 1500 2000 2500 3000 0 250000 1000000 2250000 4000000 6250000 9000000 H0= 180 179 176 165 145 119 84 43 0.00 1.42 2.84 4.26 5.67 7.09 8.51 Re 0 115325 230649 345974 461299 576623 691948 H pum p (fit) H lT + z (ft) f 0.0000 0.0183 0.0164 0.0156 0.0151 0.0147 0.0145 180 176 164 145 119 84.5 42.7 50.0 50.8 52.8 56.0 60.3 65.8 72.4 ft A = 1.52E-05 ft/(gpm) Q (gpm) V (ft/s) 2705 V (ft/s) 7.67 2 Re f H pum p (fit) H lT + z (ft) Error) 623829 0.0146 68.3 68.3 0% Pump and System Heads 200 180 160 140 H (ft) 120 100 80 60 40 20 0 Pump Curve Fit Pump Data Total Head Loss 0 500 1000 1500 2000 Q (gal/min) 2500 3000 3500 For the valve setting to reduce the flow by half, use Solver to vary the value below to minimize the error. L e/D valve = Q (gpm) V (ft/s) 1352 3.84 26858 H pum p (fit) H lT + z (ft) Error) Re f 311914 0.0158 151.7 151.7 0% Problem 10.62 [Difficulty: 3] Given: Data on pump and pipe system Find: Delivery through series pump system; valve position to reduce delivery by half Solution: Governing Equations: For the pumps and system where the total head loss is comprised of major and minor losses Hence, applied between the two reservoir free surfaces (p 1 = p 2 = 0, V1 = V2 = 0, z1 - z2 = z) we have g ∆z h lT ∆hpump h lT g ∆z g Hsystem g ∆z ∆hpump g Hpump or HlT ∆z Hpump 2 where Le Le V L HlT f 2 Kent Kexit Delbow Dvalve D 2 g 2 For pumps in series Hpump 2 H0 2 A Q where for a single pump Hpump H0 A Q The calculations in Excel are shown on the next page. 2 Given or available data (Note: final results will vary depending on fluid data selected): L = D = e = 1200 12 0.00015 ft in ft (Table 8.1) ν = Δz = 1.23E-05 -50 ft /s (Table A.7) ft 2 K ent = K exp = L e /D elbow = 0.5 1 30 (Fig. 8.14) L e/D valve = 8 (Table 8.4) 2 The pump data is curve-fitted to H pump = H 0 - AQ . The system and pump heads are computed and plotted below. To find the operating condition, Solver is used to vary Q so that the error between the two heads is minimized. Q (gpm) Q 2 (gpm) H pum p (ft) H pump (fit) V (ft/s) 0 500 1000 1500 2000 2500 3000 3250 0 250000 1000000 2250000 4000000 6250000 9000000 179 176 165 145 119 84 43 180 176 164 145 119 85 43 0.00 1.42 2.84 4.26 5.67 7.09 8.51 9.22 Re 0 115325 230649 345974 461299 576623 691948 749610 f 0.0000 0.0183 0.0164 0.0156 0.0151 0.0147 0.0145 0.0144 f 0.0145 H pumps (par) H lT + Δz (ft) 73.3 73.3 Error) 0% H0= 180 A= 1.52E-05 Q (gpm) V (ft/s) 3066 8.70 H pum ps (par) H lT + Δz (ft) 359 351 329 291 237 169 85 38 50.0 50.8 52.8 56.0 60.3 65.8 72.4 76.1 ft 2 ft/(gpm) Re 707124 Pump and System Heads Pump Curve Fit Pump Data Total Head Loss Pumps in Series 400 350 300 H (ft) 250 200 150 100 50 0 0 1000 2000 Q (gal/min) 3000 For the valve setting to reduce the flow by half, use Solver to vary the value below to minimize the error. L e /D valve = 50723 Q (gpm) V (ft/s) 1533 4.35 Re 353562 f 0.0155 H pumps (par) 287.7 H lT + Δz (ft) 287.7 Error) 0% 4000 Problem 10.63 [Difficulty: 4] Given: Data on pump and pipe system, and their aging Find: Reduction in delivery through system after 20 and 40 years (aging and non-aging pumps) Solution: Given or available data (Note: final results will vary depending on fluid data selected) : L = D = e = 1200 12 0.00015 ft in ft (Table 8.1) K ent = K exp = L e/D elbow = 0.5 1 30 (Fig. 8.14) = z = 1.23E-05 -50 ft2/s (Table A.7) ft L e/D valve = 8 (Table 8.4) The pump data is curve-fitted to H pump = H 0 - AQ 2. The system and pump heads are computed and plotted below. To find the operating condition, Solver is used to vary Q so that the error between the two heads is minimized. New System: Q (gpm) Q 2 (gpm) H pump (ft) V (ft/s) 0 500 1000 1500 2000 2500 3000 0 250000 1000000 2250000 4000000 6250000 9000000 179 176 165 145 119 84 43 0.00 1.42 2.84 4.26 5.67 7.09 8.51 Re 0 115325 230649 345974 461299 576623 691948 f 0.0000 0.0183 0.0164 0.0156 0.0151 0.0147 0.0145 H pump (fit) H lT + z (ft) 68.3 68.3 Error) 0% H0 = 180 A = 1.52E-05 H pump (fit) H lT + z (ft) 180 176 164 145 119 84.5 42.7 50.0 50.8 52.8 56.0 60.3 65.8 72.4 ft Q (gpm) V (ft/s) 2705 7.67 ft/(gpm)2 Re 623829 f 0.0146 Pump and System Heads -When New 200 180 160 140 H (ft) 120 100 80 60 40 20 0 Pump Curve Fit Pump Data Total Head Loss 0 500 1000 1500 Q (gal/min) 2000 2500 3000 20-Year Old System: f = 2.00 f new Q (gpm) V (ft/s) 2541 7.21 Re 586192 f 0.0295 H pump (fit) H lT + z (ft) 81.4 81.4 Re 572843 f 0.0354 H pump (fit) H lT + z (ft) 85.8 85.8 H pump (fit) H lT + z (ft) 79.3 79.3 H pump (fit) H lT + z (ft) 78.8 78.8 Error) 0% Flow reduction: 163 gpm 6.0% Loss Error) 0% Flow reduction: 221 gpm 8.2% Loss Error) 0% Flow reduction: 252 gpm 9.3% Loss Error) 0% Flow reduction: 490 gpm 18.1% Loss 40-Year Old System: f = 2.40 f new Q (gpm) V (ft/s) 2484 7.05 20-Year Old System and Pump: H pump = 0.90 H new f = 2.00 f new Q (gpm) V (ft/s) 2453 6.96 Re 565685 f 0.0296 40-Year Old System and Pump: H pump = 0.75 H new f = 2.40 f new Q (gpm) V (ft/s) 2214 6.28 Re 510754 f 0.0358 3500 Problem 10.64 [Difficulty: 3] Given: Data on pump and pipe system Find: Delivery through parallel pump system; valve position to reduce delivery by half Solution: Governing Equations: For the pumps and system where the total head loss is comprised of major and minor losses Hence, applied between the two reservoir free surfaces (p 1 = p 2 = 0, V1 = V2 = 0, z1 - z2 = z) we have g ∆z h lT ∆hpump h lT g ∆z g Hsystem g ∆z ∆hpump g Hpump or HlT ∆z Hpump 2 where Le Le V L HlT f 2 Kent Kexit Delbow Dvalve D 2 g For pumps in parallel 1 2 Hpump H0 A Q 4 where for a single pump Hpump H0 A Q 2 The calculations performed using Excel are shown on the next page. Given or available data (Note: final results will vary depending on fluid data selected): L = D = e = 1200 12 0.00015 ft in ft (Table 8.1) ν = Δz = 1.23E-05 -50 ft /s (Table A.7) ft 2 K ent = K exp = L e /D elbow = 0.5 1 30 (Fig. 8.14) L e/D valve = 8 (Table 8.4) 2 The pump data is curve-fitted to H pump = H 0 - AQ . The system and pump heads are computed and plotted below. To find the operating condition, Solver is used to vary Q so that the error between the two heads is minimized. Q (gpm) Q 2 (gpm) 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 0 250000 1000000 2250000 4000000 6250000 9000000 H0= 180 A= 1.52E-05 Q (gpm) V (ft/s) 4565 12.95 H pum p (ft) 179 176 165 145 119 84 43 H pump (fit) 180 176 164 145 119 85 43 V (ft/s) 0.00 1.42 2.84 4.26 5.67 7.09 8.51 9.93 11.35 12.77 14.18 Re 0 115325 230649 345974 461299 576623 691948 807273 922597 1037922 1153247 f 0.0000 0.0183 0.0164 0.0156 0.0151 0.0147 0.0145 0.0143 0.0142 0.0141 0.0140 f 0.0141 H pumps (par) 100.3 H lT +Δz (ft) 100.3 Error) 0% H pum ps (par) 180 179 176 171 164 156 145 133 119 103 85 H lT + Δz (ft) 50.0 50.8 52.8 56.0 60.3 65.8 72.4 80.1 89.0 98.9 110.1 ft 2 ft/(gpm) Re 1053006 Pump and System Heads Pump Curve Fit Pump Data Total Head Loss Pumps in Parallel 200 150 H (ft) 100 50 0 0 1000 2000 Q (gal/min) 3000 For the valve setting to reduce the flow by half, use Solver to vary the value below to minimize the error. L e /D valve = Q (gpm) V (ft/s) 2283 6.48 9965 Re 526503 f 0.0149 H pumps (par) 159.7 H lT + z (ft) 159.7 Error) 0% 4000 5000 Problem 10.65 [Difficulty: 4] Given: Data on pump and pipe system Find: Delivery through parallel pump system; reduction in delivery after 20 and 40 years Solution: Given or available data (Note: final results will vary depending on fluid data selected) : L = D = e = 1200 12 0.00015 ft in ft (Table 8.1) K ent = K exp = L e/D elbow = 0.5 1 30 (Fig. 8.14) = z = 1.23E-05 -50 ft2/s (Table A.7) ft L e/D valve = 8 (Table 8.4) 2 The pump data is curve-fitted to H pump = H 0 - AQ . The system and pump heads are computed and plotted below. To find the operating condition, Solver is used to vary Q so that the error between the two heads is minimized. Q (gpm) Q 2 (gpm) H pump (ft) H pump (fit) V (ft/s) 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 0 250000 1000000 2250000 4000000 6250000 9000000 179 176 165 145 119 84 43 180 176 164 145 119 85 43 0.00 1.42 2.84 4.26 5.67 7.09 8.51 9.93 11.35 12.77 14.18 Re 0 115325 230649 345974 461299 576623 691948 807273 922597 1037922 1153247 f 0.0000 0.0183 0.0164 0.0156 0.0151 0.0147 0.0145 0.0143 0.0142 0.0141 0.0140 f 0.0141 H pumps (par) H lT + z (ft) 100.3 100.3 Error) 0% H0 = 180 A = 1.52E-05 Q (gpm) V (ft/s) 4565 12.95 H pumps (par) H lT + z (ft) 180 179 176 171 164 156 145 133 119 103 85 50.0 50.8 52.8 56.0 60.3 65.8 72.4 80.1 89.0 98.9 110.1 ft ft/(gpm)2 Re 1053006 Pump and System Heads Pump Curve Fit Pump Data Total Head Loss Pumps in Parallel 200 150 H (ft) 100 50 0 0 1000 2000 Q (gal/min) 3000 4000 20-Year Old System: f = 2.00 f new Q (gpm) V (ft/s) 3906 11.08 Re 900891 f 0.0284 H pumps (par) H lT + z (ft) 121.6 121.6 Re 855662 f 0.0342 H pump (fit) H lT + z (ft) 127.2 127.2 f 0.0285 H pump (fit) H lT + z (ft) 114.6 114.6 f 0.0347 H pump (fit) H lT + z (ft) 106.4 106.4 Error) 0% Flow reduction: 660 gpm 14.4% Loss Error) 0% Flow reduction: 856 18.7% Error) 0% Flow reduction: 860 gpm 18.8% Loss Error) 0% Flow reduction: 1416 31.0% 40-Year Old System: f = 2.40 f new Q (gpm) V (ft/s) 3710 10.52 20-Year Old System and Pumps: H pump = 0.90 H new f = 2.00 f new Q (gpm) V (ft/s) 3705 10.51 Re 854566 40-Year Old System and Pumps: f = 2.40 f new H pump = 0.75 H new Q (gpm) V (ft/s) 3150 8.94 Re 726482 5000 Problem 10.66 Given: Data on pump and pipe system Find: Delivery through series pump system; reduction after 20 and 40 years [Difficulty: 4] Solution: Given or available data (Note: final results will vary depending on fluid data selected) : L = D = e = 1200 12 0.00015 ft in ft (Table 8.1) K ent = K exp = L e/D elbow = 0.5 1 30 (Fig. 8.14) = z = 1.23E-05 -50 ft2/s (Table A.7) ft L e/D valve = 8 (Table 8.4) The pump data is curve-fitted to H pump = H 0 - AQ 2. The system and pump heads are computed and plotted below. To find the operating condition, Solver is used to vary Q so that the error between the two heads is minimized. Q (gpm) Q 2 (gpm) H pump (ft) H pump (fit) V (ft/s) 0 500 1000 1500 2000 2500 3000 3250 0 250000 1000000 2250000 4000000 6250000 9000000 179 176 165 145 119 84 43 180 176 164 145 119 85 43 0.00 1.42 2.84 4.26 5.67 7.09 8.51 9.22 Re 0 115325 230649 345974 461299 576623 691948 749610 f 0.0000 0.0183 0.0164 0.0156 0.0151 0.0147 0.0145 0.0144 H pumps (par) H lT + z (ft) 359 351 329 291 237 169 85 38 50.0 50.8 52.8 56.0 60.3 65.8 72.4 76.1 H0 = 180 A = 1.52E-05 Q (gpm) V (ft/s) 3066 8.70 ft ft/(gpm)2 Re 707124 f 0.0145 H pumps (par) H lT + z (ft) 73.3 73.3 Error) 0% Pump and System Heads Pump Curve Fit Pump Data Total Head Loss Pumps in Series 400 350 300 H (ft) 250 200 150 100 50 0 0 1000 2000 Q (gal/min) 3000 20-Year Old System: f = 2.00 f new Q (gpm) V (ft/s) 2964 8.41 Re 683540 f 0.0291 H pumps (par) H lT + z (ft) 92.1 92.1 Re 674713 f 0.0349 H pump (fit) H lT + z (ft) 98.9 98.9 f 0.0291 H pump (fit) H lT + z (ft) 90.8 90.8 f 0.0351 H pump (fit) H lT + z (ft) 94.1 94.1 Error) 0% Flow reduction: 102 gpm 3.3% Loss Error) 0% Flow reduction: 141 gpm 4.6% Loss Error) 0% Flow reduction: 151 gpm 4.9% Loss Error) 0% Flow reduction: 294 gpm 9.6% Loss 40-Year Old System: f = 2.40 f new Q (gpm) V (ft/s) 2925 8.30 20-Year Old System and Pumps: f = 2.00 f new H pump = 0.90 H new Q (gpm) V (ft/s) 2915 8.27 Re 672235 40-Year Old System and Pumps: H pump = 0.75 H new f = 2.40 f new Q (gpm) V (ft/s) 2772 7.86 Re 639318 4000 Problem 10.67 [Difficulty: 3] Given: Water supply for Englewood, CO Find: (a) system resistance curve (b) specify appropriate pumping system (c) estimate power required for steady-state operation at two specified flow rates Solution: 2 2 p V1 V2 p 2 1 α g z α g z ρ 1 2 1 ρ 2 2 2 h lT h p Basic equations: 2 h lT f Assumptions: 1) p 1 = p2 = patm Hence 2 2) V 1 = V2 = 0 3) Kent = 0 2 L V g z1 z2 f 1 h p or D 2 2 Le V L V V Σ f Σ K D 2 D 2 2 H 4) Kexit = 1 h Wp g 5) L e/D = 0 2 L V Hp z2 z1 f 1 D 2 g The results calculated using Excel are shown below: Given or available data (Note: final results will vary depending on fluid data selected): L = e = 1770 0.046 m mm (Table 8.1) z 1 = 1610 m z 2 = 1620 m D = 68.5 cm ρ = ν = 1.01E-06 m /s (Table A.8) 2 The required pump head is computed and plotted below. 3 Q (m /hr) V (m/s) Re f H p (m) 0 500 1000 1500 2000 2500 3000 3200 3500 3900 4000 0.00 0.38 0.75 1.13 1.51 1.88 2.26 2.41 2.64 2.94 3.01 0.00E+00 2.56E+05 5.11E+05 7.67E+05 1.02E+06 1.28E+06 1.53E+06 1.64E+06 1.79E+06 1.99E+06 2.04E+06 0.0000 0.0155 0.0140 0.0133 0.0129 0.0126 0.0124 0.0124 0.0123 0.0122 0.0122 10.0 10.3 11.1 12.3 14.0 16.1 18.6 19.8 21.6 24.3 25.0 998 kg/m 3 ρ Q g Hp ηp Required Pump Head 30 H (m) 25 Pump Head 20 Flow Rates of Interest 15 10 5 0 0 500 1000 1500 2000 2500 3000 3500 3 Q (m /hr) The maximum flow rate is: 17172 gpm The associated head is: 80 ft Based on these data and the data of Figures D.1 and D.2, we could choose two 16A 18B pumps in parallel, or three 10AE14 (G) pumps in parallel. The efficiency will be approximately 90% Therefore, the required power would be: 3 191.21 kW at Q = 3200 m /hr 286.47 kW at Q = 3900 m /hr 3 4000 Problem 10.68 [Difficulty: 3] Given: System shown, design flow rate Find: Head losses for suction and discharge lines, NPSHA, select a suitable pump Solution: We will apply the energy equation for steady, incompressible pipe flow. Basic equations: 2 2 p p V1 V2 1 2 ρ α1 2 g z1 ρ α2 2 g z2 h lT hp 2 h lT f 2 2 Le V L V V Σ f Σ K D 2 D 2 2 H h g Assumptions: 1) p ent = pexit = patm 2) V ent = Vexit = 0 The given or available data is Q 800 From Table A.8 at 20 oC ν 1.01 10 L min 2 6 m s D 10 cm p v 2.34 kPa D 4.6 10 At the inlet: 4 ρ 998 V Q A 4 Q 3 2 V 1.698 p 2tabs p v ρ g Re s V D ν NPSHA 1.681 10 5 2 0.019 In this case: Le 75 D K 0.78 L 2 m z2 8.7 m z1 7.2 m 2 L Le V2 V p 2t ρ g z2 z1 f K 2 D D 2 NPSHA m e 1.11 6.9 f 1.8 log Re 3.7 D 2 p 2 V2 L Le V2 2 V g z1 α2 g z2 f Σ K 2 2 ρ D D 2 The NPSHA can be calculated: kg π D Therefore we can calculate the friction factor: Solving for total pressure at 2: p atm 101.3 kPa m At the specified flow rate, the speed of the water is: e e 0.046 mm p 2t 18.362 kPa (gage) p 2t p atm p v ρ g NPSHA 8.24 m For the entire system: In this case: 2 2 Le V2 L V V g z1 z2 f f K hp D 2 D 2 2 z1 7.2 m z2 88 m L 2 m 400 m 402 m Solving for the required head at the pump: In U.S. Customary units: Q 211 gpm Hp z2 z1 Le ( 75 55 8 2 30) D L 2 f L e K V D D 2 g K 0.78 1 Hp 92.7 m Hp 304 ft A pump would be selected by finding one for which the NPSHR is less than the NPSHA. Based on these data and the information in Appendix D, a 2AE11 or a 4AE12 pump would be capable of supplying the required head at the given flow rate. The pump should be operated at a speed between 1750 and 3500 rpm, but the efficiency may not be acceptable. One should consult a complete catalog to make a better selection. Problem 10.69 8. 8 8.155 .15 155 [Difficulty: 3] Problem 10.70 [Difficulty: 3] Given: Flow system and data of Problem 10.68; data for pipe aging from Problem 10.63 Find: Pumps to maintain system flow rates; compare delivery to that with pump sized for new pipes only Solution: We will apply the energy equation for steady, incompressible pipe flow. Basic equations: 2 2 p V1 V2 p1 2 α g z α g z ρ 1 2 1 ρ 2 2 2 h lT h p 2 h lT f 2 2 Le V L V V Σ f Σ K D 2 D 2 2 H h g Assumptions: 1) p ent = pexit = patm 2) V ent = Vexit = 0 The given or available data is Q 800 L min 2 6 m From Table A.8 at 20 oC ν 1.01 10 s At the specified flow rate, the speed of the water is: e D 4 4.6 10 e 0.046 mm p v 2.34 kPa ρ 998 p atm 101.3 kPa kg 3 m V Q A 4 Q V 1.698 2 π D m s Re V D ν e 1.11 6.9 f 1.8 log Re 3.7 D Therefore we can calculate the friction factor: For the entire system: In this case: D 10 cm 5 1.681 10 2 0.019 2 2 Le V2 L V V g z1 z2 f f K hp D 2 D 2 2 z1 7.2 m z2 87 m L 2 m 400 m 402 m Solving for the required head at the pump: Hp z2 z1 Le ( 75 55 8 2 30) D L 2 f L e K V D D 2 g For old pipes, we apply the multipliers from Problem 10.63: f20 5.00 fnew The results of the analysis, computed in Excel, are shown on the next page. f40 8.75 fnew K 0.78 1 The required pump head is computed and plotted below. New Q (L/min) V (m/s) H p (m) Q (gpm) Re f 0 200 400 600 800 922 1000 1136 1200 1400 1600 1800 2000 0.000 0.424 0.849 1.273 1.698 1.957 2.122 2.410 2.546 2.971 3.395 3.820 4.244 0.00E+00 4.20E+04 8.40E+04 1.26E+05 1.68E+05 1.94E+05 2.10E+05 2.39E+05 2.52E+05 2.94E+05 3.36E+05 3.78E+05 4.20E+05 0.0000 0.0231 0.0207 0.0196 0.0189 0.0187 0.0185 0.0183 0.0182 0.0180 0.0178 0.0177 0.0176 79.80 80.71 83.07 86.77 91.80 95.52 98.14 103.20 105.80 114.77 125.04 136.62 149.51 0.00 52.84 105.68 158.52 211.36 243.64 264.20 300.08 317.04 369.88 422.72 475.56 528.40 New H p (ft) 20 yo H p (ft) 40 yo H p (ft) Pump (ft) 261.81 264.81 272.52 284.67 301.17 313.38 321.99 338.59 347.12 376.54 410.25 448.24 490.51 261.81 276.57 314.52 374.19 455.19 515.10 557.37 638.8 680.64 824.95 990.27 1176.6 1383.9 261.81 287.59 353.89 458.10 599.58 704.21 778.03 920.2 993.31 1245.3 1534.0 1859.4 2221.4 856.54 840.48 792.30 712.00 599.58 515.10 455.04 338.59 278.38 69.59 -171.31 -444.33 -749.47 If we assume that the head at 800 L/min for 40 year old pipe is 70% of the maximum head for the pump, 2 and that the pump curve has the form H = H 0 - AQ : H 800 = 599.58 ft We plot the pump curve along with the head loss on the graph below: 856.54 ft H0= A = 0.005752 ft/gpm2 Required Pump Head 800 New Pipe 20 Years Old 40 Years Old Pump Curve 700 H (ft) 600 500 400 300 200 100 150 200 Q (gpm) 250 300 Sizing the pump for 800 L/min for at 40 years would (assuming no change in the pump characteristics) produce 922 L/min at 20 years and 1136 L/min for new pipe. Since the head increases by a factor of two, the extra head could be obtained by placing a second identical pump in series with the pump of Problem 10.68. 350 Problem 10.71 [Difficulty: 3] Problem 10.72 Given: Flow from pump to reservoir Find: Select a pump to satisfy NPSHR Solution: Basic equations [Difficulty: 3] 2 2 p V1 V2 p2 1 α g z α g z ρ 1 ρ 2 h lT h p 2 2 2 V1 L V1 h lT h l h lm f Kexit D 2 2 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 is approximately 1 4) V 2 << V 1 2 2 p1 V2 L V V z2 f Kexit Hp D 2 g 2 g ρ g 2 g Note that we compute head per unit weight, H, not head per unit mass, h, so the energy equation between Point 1 and the free surface (Point 2) becomes Solving for H p 2 2 2 p1 V L V V Hp z2 f Kexit ρ g 2 g D 2 g 2 g From Table A.7 (68oF) ρ 1.94 slug ft For commercial steel pipe 3 e 0.00015 ft ν 1.08 10 (Table 8.1) Given For the exit Kexit 1.0 Note that for an NPSHR of 15 ft this means 2 Re s e so D V D Re 6.94 10 ν p1 ρ g π D 4 V 0.0002 f 0.0150 2 p1 L V Hp z2 f ρ g D 2 g 15 ft Q 4.42 5 2 p1 L V Hp z2 f ρ g D 2 g so we find 2 Q e 1 2.51 D 2.0 log 3.7 f Re f Flow is turbulent: Note that 5 ft ft Hp 691 ft 3 Q 1983 gpm s For this combination of Q and Hp, from Fig. D.11 the best pump appears to be a Peerless two-stage 10TU22C operating at 1750 rpm After 10 years, from Problem 10.63, the friction factor will have increased by a factor of 2.2 f 2.2 0.150 We now need to solve 2 p1 L V Hp z2 f ρ g D 2 g V for the new velocity V p1 Hp z2 f L ρ g 2 D g 2 Q π D 4 V f 0.330 Q 0.94 V 2.13 ft ft s and f will still be 2.2 0.150 3 s Q 423 gpm Much less! Problem 10.73 Given: Water pipe system Find: Pump suitable for 300 gpm [Difficulty: 3] Solution: 2 2 p p V1 V2 1 2 α g z α g z ρ 1 2 1 ρ 2 2 2 h l f 64 (Laminar) Re 2 h lT f L V D 2 e D 1 2.51 2.0 log (Turbulent) 3.7 f Re f 2 The energy equation can be simplified to ∆p ρ f L V D 2 This can be written for each pipe section 2 Pipe A (first section) LA VA ∆pA ρ fA DA 2 Pipe B (1.5 in branch) LB VB ∆pB ρ fB DB 2 Pipe C (1 in branch) LC VC ∆pC ρ fC DC 2 (1) 2 (2) 2 (3) 2 LD VD ∆pD ρ fD DD 2 In addition we have the following contraints Pipe D (last section) (4) QA QD Q Q QB QC (5) ∆p ∆pA ∆pB ∆pD ∆pB ∆pC (7) (6) (8) We have 2 unknown flow rates (or, equivalently, velocities); We solve the above eight equations simultaneously Once we compute the flow rates and pressure drops, we can compute data for the pump ∆ppump ∆p and The calculations, performed in Excel, are shown on the next page. Qpump QA Wpump ∆ppump Qpump Pipe Data: Pipe A B C D L (ft) D (in) e (ft) 150 150 150 150 1.5 1.5 1 1.5 0.00085 0.00085 0.00085 0.00085 Fluid Properties: 3 ρ= 1.94 slug/ft μ = 2.10E-05 lbf-s/ft Q= 300 gpm 0.668 ft /s 2 Flow Rate: = Flows: Heads: Constraints: 3 3 3 3 3 Q A (ft /s) 0.668 Q B (ft /s) 0.499 Q C (ft /s) 0.169 Q D (ft /s) 0.668 V A (ft/s) 54.47 V B (ft/s) 40.67 V C (ft/s) 31.04 V D (ft/s) 54.47 Re A 6.29E+05 Re B 4.70E+05 Re C 2.39E+05 Re D 6.29E+05 fA 0.0335 fB 0.0336 fC 0.0384 fD 0.0335 Δp A (psi) 804.0 Δp B (psi) 448.8 Δp C (psi) 448.8 Δp D (psi) 804.0 (8) Δp B = Δp C 0.00% (6) Q = Q B + Q C 0.00% Error: 0.00% Vary Q B and Q C using Solver to minimize total error Q (gpm) Δp (psi) P (hp) 2057 300 360 This is a very high pressure; a sequence of pumps would be needed For the pump: Problem 10.74 [Difficulty: 3] Problem 10.75 Given: Pump and supply pipe system Find: Head versus flow curve; Flow for a head of 85 ft [Difficulty: 4] Solution: 2 2 p p V1 V2 1 2 Basic equations: α1 g z1 α2 g z2 h lT h pump 2 2 ρ ρ 2 Applying to the 70 ft branch (branch a) 2 h lT f 2 2 Le V2 L V V f K D 2 D 2 2 2 Le Va Va L Va g Ha f f K g Hpump 2 D 2 D 2 Lea where Ha 70 ft and is due to a standard T branch (= 60) and a standard elbow (= 30) from Table 8.4, and D K Kent Kexit 1.5 from Fig. 8.14 L Lea Va (1) Hpump Ha f K D D 2 g Applying to the 50 ft branch (branch b) L Leb Vb Hpump Hb f K D D 2 g (2) Leb where Hb 50 ft and is due to a standard T run (= 20) and two standard elbows (= 60), and K Kent Kexit 1.5 D Here are the calculations, performed in Excel: Given data: L e D K L e a /D L e b /D Ha Hb = 1000 ft = 0.00085 ft = 6.065 in = 1.5 = 90 = 80 = 70 ft = 50 ft ρ = 3 1.94 slug/ft ν = 1.06E-05 ft2/s Computed results: Set up Solver so that it varies all flow rates to make the total head error zero H p ump (ft) Q (ft3 /s) Q a (ft3 /s) V a (ft/s) 72.0 1.389 0.313 1.561 74.0 1.574 0.449 2.237 76.0 1.724 0.553 2.756 78.0 1.857 0.641 3.195 80.0 1.978 0.718 3.581 82.0 2.090 0.789 3.931 84.0 2.195 0.853 4.252 Re a 7.44E+04 1.07E+05 1.31E+05 1.52E+05 1.71E+05 1.87E+05 2.03E+05 fa H pump (Eq. 1) Q b (ft3 /s) V b (ft/s) 0.0248 72.0 1.076 5.364 0.0241 74.0 1.125 5.607 0.0238 76.0 1.171 5.839 0.0237 78.0 1.216 6.063 0.0235 80.0 1.260 6.279 0.0234 82.0 1.302 6.487 0.0234 84.0 1.342 6.690 Re b 2.56E+05 2.67E+05 2.78E+05 2.89E+05 2.99E+05 3.09E+05 3.19E+05 fb H pu mp (Eq. 2) H (Errors) 0.0232 72.0 0.00 0.0231 74.0 0.00 0.0231 76.0 0.00 0.0231 78.0 0.00 0.0231 80.0 0.00 0.0230 82.0 0.00 0.0230 84.0 0.00 85.0 2.246 0.884 4.404 2.10E+05 0.0233 85.0 1.362 6.789 3.24E+05 0.0230 85.0 0.00 86.0 88.0 90.0 92.0 94.0 2.295 2.389 2.480 2.567 2.651 0.913 0.970 1.023 1.074 1.122 4.551 4.833 5.099 5.352 5.593 2.17E+05 2.30E+05 2.43E+05 2.55E+05 2.67E+05 86.0 88.0 90.0 92.0 94.0 1.382 1.420 1.457 1.494 1.529 6.886 7.077 7.263 7.445 7.622 3.28E+05 3.37E+05 3.46E+05 3.55E+05 3.63E+05 86.0 88.0 90.0 92.0 94.0 0.00 0.00 0.00 0.00 0.00 0.0233 0.0233 0.0232 0.0232 0.0231 For the pump head less than the upper reservoir head flow will be out of the reservoir (into the lower one) 0.0230 0.0230 0.0230 0.0230 0.0229 Total Error: 0.00 Head Versus Flow Rate 100 Head (ft) 95 90 85 80 75 70 1.0 1.5 2.0 3 Q (ft /s) 2.5 3.0 Problem 10.76 Given: Data on flow from reservoir/pump Find: Appropriate pump; Reduction in flow after 10 years Solution: [Difficulty: 4] 2 2 p V1 V4 p 4 1 ρ g α 2 g z1 ρ g α 2 g z4 HlT Hp Basic equation: for flow from 1 to 4 2 2 Le V2 L V V HlT f f K D 2 g D 2 g 2 g Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) V 2 = V3 = V4 (constant area pipe) ρ 1000 Given or available data kg ν 1.01 10 3 2 6 m m s p v 2.34 kPa (Table A.8) Q 0.075 3 p 2 150 kPa For minor losses we have p 3 450 kPa D 15 cm e 0.046 mm z1 20 m z4 35 m V Le Four elbows: D At the pump inlet NPSHA The head rise through the pump is Hp p2 p3 p2 ρ g 1 2 4 Q V 4.24 2 π D 4 12 48 (Fig. 8.16) Square inlet: m s m s Kent 0.5 2 ρ V p v NPSHA 16.0 m ρ g Hp 30.6 m 3 Hence for a flow rate of Q 0.075 m or Q 1189 gpm and Hp 30.6 m or Hp 100 ft, from s Appendix D. Fig. D3 a Peerless4AE11 would suffice 2 2 Le V2 L V V We do not know the pipe length L! Solving the energy equation for it:z1 z4 HlT Hp f f Kent Hp D 2 g D 2 g 2 g For f Given Re V D ν 5 Re 6.303 10 e 1 D 2.51 2.0 log f 3.7 Re f and e D f 0.0161 4 3.07 10 L Hence, substituting values Le Kent D z1 z4 Hp D 2 f D f V 2 g D From Problem 10.63, for a pipe D 0.15 m or D 5.91 in, the aging over 10 years leads to L 146 m fworn 2.2 f We need to solve the energy equation for a new V Vworn 2 Hence Qworn π D 4 2 g z1 z4 Hp Le L fworn D D Kent 3 Vworn m Qworn 0.0510 s 3 ∆Q Qworn Q Check f Reworn m Vworn 2.88 s Vworn D ν ∆Q 0.0240 Given Hence using 2.2 x 0.0161 is close enough to using 2.2 x 0.0165 m ∆Q s Q e D 1 2.51 2.0 log 3.7 f Reworn f 32.0 % f 0.0165 Problem 10.77 8.124 [Difficulty: 4] Part 1/2 Problem 10.77 [Difficulty: 4] Part 2/2 Problem 10.78 [Difficulty: 4] 8.158 Problem 10.79 [Difficulty: 4] Given: Sprinkler system for lakeside home Find: (a) Head loss on suction side of pump (b) Gage pressure at pump inlet (c) Hydraulic power requirement for the pump (d) Change in power requirement if pipe diameter is changed (e) Change in power requirement if the pump were moved Solution: L34 = 45 m 30 m L12 = 20 m 3m We will apply the energy equation for steady, incompressible pipe flow. Basic equations: 2 2 p p V1 V2 1 2 α g z α g z ρ 1 2 1 ρ 2 2 2 h lT h p 2 h lT f 2 2 Le V L V V Σ f Σ K D 2 D 2 2 H h g Assumptions: 1) p 1 = patm 2) V 1 = 0 The given or available data is Q 40 L min D 2 cm e 0.15 mm z1 0 m z2 3 m z3 z2 From Table A.8 at 20 oC ν 1.01 10 2 6 m s At the specified flow rate, the speed of the water is: e D 7.5 10 3 Between 1 and 2: p2 z4 33 m p v 2.34 kPa ρ 998 p 4 300 kPa (gage) L12 20 m L34 45 m kg 3 m V Q A 4 Q 2 π D Therefore we can calculate the friction factor: p atm 101.3 kPa V 2.122 m s Re V D ν e 1.11 6.9 f 1.8 log Re 3.7 D 2 2 L12 Le V2 V V α2 g z2 f K ρ 2 D 2 2 D 4.202 10 4 2 0.036 In this case: Le ( 30 16) D K 0.78 2 L12 Le V2 V HlT12 f K D 2 g 2 g D The head loss before the pump is: HlT12 8.844 m V2 p 2 ρ Solving for pressure at 2: 2 L12 Le V2 V g z2 f K 2 D 2 2 D p 2 54.946 kPa (gage) To find the pump power, we need to analyze between 3 and 4: 2 p4 p3 L34 Le V2 V g z3 g z4 f K D 2 2 ρ ρ D L34 Le V2 p 3 p 4 ρ g z4 z3 f D 2 D Now we can calculate the power: In this case: Le ( 16 16) D p 3 778.617 kPa Hp Thus the pump head is: V A 4 Q 2 V 0.531 π D m s V D Re ν 2.101 10 4 e D 3 3.75 10 2 L12 Le V2 V g z2 f K 2 2 D 2 D p 2 ρ Le ( 16 16) D K 0 L34 Le V2 p 3 p 4 ρ g z4 z3 f D 2 D Hp p3 p2 ρ g 58.44 m D 4 cm e 1.11 6.9 f 1.8 log 3.7 D Re V2 K 0.78 85.17 m Wp 556 W Le ( 30 16) D ρ g Wp ρ g Q Hp Changing to 4 centimeter pipe would reduce the mean velocity and hence the head loss and minor loss: Q p3 p2 K 0 2 0.032 p 2 26.922 kPa (gage) p 3 778.617 kPa (gage) Wpnew ρ g Q Hp 381.283 W ∆Wp Wpnew Wp Wp 31 % 2 The pump should not be moved up the hill. The NPSHA is: NPSHA V p 2 p atm ρ pv 2 If anything, the pump should be moved down the hill to increase the NPSHA. ρ g 4.512 m for 2-cm pipe. Problem 10.80 [Difficulty: 4] Given: Fire nozzle/pump system Find: Appropriate pump; Impeller diameter; Pump power input needed Solution: Basic equations 2 2 2 p V2 V3 p3 2 L V2 ρ α 2 g z2 ρ α 2 g z3 h l h l f D 2 for the hose Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 2 and 3 is approximately 1 4) No minor loss 2 2 p V2 V1 p1 2 ρ α 2 g z2 ρ α 2 g z1 h pump for the pump Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) No minor loss The first thing we need is the flow rate. Below we repeat Problem 8.179 calculations Hence for the hose ∆p ρ p2 p3 ρ 2 f L V D 2 or 2 ∆p D V ρ f L We need to iterate to solve this for V because f is unknown until Re is known. This can be done using Excel's Solver, but here: ∆p 750 kPa L 100 m Make a guess for f f 0.01 Given 2 ∆p D V 7.25 ρ f L 2 ∆p D ρ f L V 5.92 2 ∆p D ρ f L V 5.81 m Re s 2 ∆p D ρ f L s Re V D kg ν 1.01 10 3 2 6 m m ν Re 2.51 10 5 V D ν Re 2.05 10 5 Re 2.01 10 5 Re 2.01 10 5 f 0.0156 m Re s e 1 2.51 D 2.0 log 3.7 f Re f V m ρ 1000 f 0.0150 e D 1 2.51 2.0 log f 3.7 Re f V Given V D 3.5 cm e D 1 2.51 2.0 log 3.7 f Re f V Given e 0 V 5.80 V D ν f 0.0156 m s Re V D ν s 2 Q π D V 4 Q 5.578 10 3 3m 3 Q 0.335 s We have p 1 350 kPa For the pump 2 2 p V2 V1 p1 2 ρ α 2 g z2 ρ α 2 g z1 h pump so h pump p 2 700 kPa 750 kPa p2 p1 Hpump or ρ m min p 2 1450 kPa p2 p1 Hpump 112 m ρ g 3 We need a pump that can provide a flow of Q 0.335 m min or Q 88.4 gpm, with a head of Hpump 112 m or Hpump 368 ft From Appendix D, Fig. D.1 we see that a Peerless 2AE11 can provide this kind of flow/head combination; it could also handle four such hoses (the flow rate would be 4 Q 354 gpm). An impeller diameter could be chosen from proprietary curves. The required power input is Wh Wm ηp Wm Prequired Ppump η where we choose ηp 75 % from Fig. 10.15 ρ Q g Hpump Prequired ηp 6.14 kW 70 % Wm 8.18 kW for one hose or Prequired 8.77 kW or 4 Wm 32.7 kW for four 4 Prequired 35.1 kW for four Problem 10.81 [Difficulty: 4] Given: Manufacturer data for a pump Find: (a) Plot performance and develop curve-fit equation. (b) Calculate pump delivery vs discharge height for length of garden hose Solution: h lT f Basic equations: 2 2 Le V2 L V V f K D 2 D 2 2 H h 2 Hp H0 A Q g 2 h lT f For this case, Le = K = 0, therefore: Given data: L = e = D = ρ = Here are the results calculated in Excel: Here are the data for the head generated by the pump, as well as the head losses for the hose and the pipe: 15 0 20 m ft mm 998 kg/m3 D = e = ν = 1.01E-06 m /s 2 H 0= 7.48727 m A= L V D 2 0.0012 m/(L/min) 2 2 20 0 mm mm D = e = 25 0.15 Head (m) fa H L (m) Q (L/min) Q z fit (m) V (m/s) Re a fa H L (m) V (m/s) 0.3 0.7 1.5 77.2 75.0 71.0 5959.840 5625.000 5041.000 0.320 0.722 1.425 4.096 3.979 3.767 8.11E+04 7.88E+04 7.46E+04 0.0188 0.0189 0.0191 12.1 11.4 10.4 2.621 2.546 2.411 6.49E+04 0.0334 6.30E+04 0.0334 5.97E+04 0.0335 7.0 6.6 6.0 3.0 4.5 6.0 8.0 61.0 51.0 26.0 0.0 3721.000 2601.000 676.000 0.000 3.012 4.359 6.674 7.487 3.236 2.706 1.379 0.000 6.41E+04 5.36E+04 2.73E+04 0.00E+00 0.0198 0.0206 0.0240 0.0000 7.9 5.8 1.7 0.0 2.071 1.732 0.883 0.000 5.13E+04 4.29E+04 2.19E+04 0.00E+00 4.4 3.1 0.8 0.0 10 9 8 7 6 5 4 3 2 1 0 Data Fit Hose Pipe 10 Re a z (m) Head Versus Flow Rate for Pump 0 mm mm 20 30 40 Q (L/min) 50 60 70 80 0.0337 0.0340 0.0356 0.0000 To determine the discharge heights for the hose and the pipe, For the hose: Re a Q (L/min) V (m/s) 0.0 0.000 0.00E+00 10.0 0.531 1.05E+04 20.0 1.061 2.10E+04 30.0 1.592 3.15E+04 40.0 2.122 4.20E+04 50.0 2.653 5.25E+04 60.0 3.183 6.30E+04 we subtract the head loss from the head generated by the pump. For the pipe: Re a fa fa H L (m) Disch (m) V (m/s) 0.0000 0.000 7.487 0.000 0.00E+00 0.0000 0.0305 0.328 7.039 0.340 8.40E+03 0.0398 0.0256 1.101 5.906 0.679 1.68E+04 0.0364 0.0232 2.248 4.157 1.019 2.52E+04 0.0351 0.0217 3.740 1.823 1.358 3.36E+04 0.0345 0.0207 5.558 -1.077 1.698 4.20E+04 0.0340 0.0199 7.689 -4.531 2.037 5.04E+04 0.0337 H L (m) Disch (m) % Diff 0.000 7.487 0% 0.140 7.227 -3% 0.514 6.492 -9% 1.115 5.290 -21% 1.943 3.620 -50% 2.998 1.483 4.279 -1.122 Flow Rate Versus Discharge Height Flow Rate (L/min) 60 50 Hose Pipe 40 30 20 10 0 0 1 2 3 4 5 6 7 Discharge Height (m) The results show that the 15% performance loss is an okay "ball park" guess at the lower flow rates, but not very good at flow rates above 30 L/min. 8 Problem 10.82 [Difficulty: 4] Swimming pool filtration system, filter pressure drop is Δp=0.6Q2, with Δp in psi and Q in gpm Given: Find: Speed and impeller diameter of suitable pump; estimate efficiency Solution: We will apply the energy equation for steady, incompressible pipe flow. Basic equations: 2 2 p p 2 2 Le V2 V1 V2 1 L V V 2 ρ α1 2 g z1 ρ α2 2 g z2 h lT hp h lT f D 2 Σ f D 2 Σ K 2 Q 30 gpm Q 1.893 10 The given or available data are: ρ 1.93 slug ft 3 ρ 995 3 3m s kg ν 1.06 10 D 20 mm 3 5 ft 2 s ν 9.848 10 H h g 2 7m s e 0 mm m Setting state 1 at the pump discharge, state 2 at the tee, state 3a downstream of the filter, and state 3b after the 40 ft pipe, we can look at the pressure drop between 1 and 2: V1 V2 e D e D 0 0 V Q A 4 Q 2 V 6.025 π D m Re s V D ν 1.224 10 5 e 1.11 6.9 f 1.8 log Re 3.7 D Therefore we can calculate the friction factor: 2 0.017 2 Le 0 K 0 and therefore the pressure drop is: Since this is a straight run of pipe: ∆p12 ρ f V L 2 D ∆p12 47.04 kPa Since both legs exhaust to the same pressure, the pressure drops between the two must be equal, and the flow rates must equal the total flow rate of the system. This requires an iterative solution, using Solver in Excel. The result is: 3 3 m Qa 1.094 10 s 3 4 m Qb 7.99 10 The resulting pressure drop is s ∆p23 42.96 kPa Neglecting any pressure at the pump inlet, the pump must supply: ∆ppump ∆p12 ∆p23 90.0 kPa ∆ppump The resulting head is: Hpump 9.226 m in U.S. units: Hpump 30.269 ft ρ g This head is too low for any of the pumps in Fig. D.1. Therefore, assuming a speed of 3500 rpm: In customary units: The pump power is: Ncu 2733 N 1485 So from Figure 10.9 we can estimate the efficiency: Wp ρ Q g Hpump η 262.056 W N ω Q g Hpump 0.75 0.544 η 65 % Wp 262.1 W Problem 10.83 [Difficulty: 4] Part 1/2 Problem 10.83 [Difficulty: 4] Part 2/2 Problem 10.84 Given: Data on centrifugal fan Find: Plot of performance curves; Best effiiciency point [Difficulty: 3] Solution: ηp Basic equations: Wh Wh Q ∆p Wm ∆p ρw g ∆h (Note: Software cannot render a dot!) Here are the results, calculated using Excel: ρw = slug/ft3 1.94 Fitting a 2nd order polynomial to each set of data we find Δp =-1.51x10 Q + 2.37x10 Q + 0.0680 -6 Q (ft /s) Δp (psi) Pm (hp) P h (hp) η (%) 106 141 176 0.075 0.073 0.064 2.75 3.18 3.50 2.08 2.69 2.95 75.7% 84.7% 84.3% 211 246 282 0.050 0.033 0.016 3.51 3.50 3.22 2.76 2.13 1.18 78.7% 60.7% 36.7% 2 -4 η =-3.37x10 Q + 0.0109Q -0.0151 -5 3 2 Finally, we use Solver to maximize η by varying Q : Q (ft /s) Δp (psi) η (%) 161.72 0.0668 86.6% 3 Fan Performance Curve BEP 0.08 100% 0.07 0.06 75% η Δp 0.04 50% 0.03 25% 0.02 0.01 0.00 100 120 140 160 180 200 3 Q (ft /s) 220 240 260 280 0% 300 η (%) Δp (psi) 0.05 Problem 10.85 [Difficulty: 3] Given: Data on centrifugal fan and square metal duct Find: Minimum duct geometry for flow required; Increase if fan speed is increased Solution: Wh Basic equations: ηp Wh Q ∆p and for the duct L V ∆p ρair f Dh 2 and fan scaling Q 200 Wm 2 ft ∆p ρw g ∆h (Note: Software cannot render a dot!) 2 4 A 4 H Dh H P 4 H 3 s ω 750 rpm ω' 1000 rpm Q' ω' ω Q Q' 266.67 Here are the results, calculated using Excel: ρw = 1.94 slug/ft3 ρ air = 0.00237 slug/ft Fitting a 2nd order polynomial to each set of data we find Δp =-1.51x10 Q + 2.37x10 Q + 0.0680 3 -6 2 ν air = 1.58E-04 ft /s L= 50 ft Assume smooth ducting Note: Efficiency curve not needed for this problem. We use the data to get a relationship for pressure increase. Q (ft3 /s) Δp (psi) Pm (hp) P h (hp) η (%) 106 141 176 211 246 282 0.075 0.073 0.064 0.050 0.033 0.016 2.75 3.18 3.50 3.51 3.50 3.22 2.08 2.69 2.95 2.76 2.13 1.18 75.7% 84.7% 84.3% 78.7% 60.7% 36.7% -4 Now we need to match the pressure loss in the duct with the pressure rise across the fan. To do this, we use Solver to vary H so the error in p is zero Fan Q (ft3 /s) Δp (psi) 266.67 H (ft) V (ft/s) 1.703 91.94 Answers: Q (ft3/s) 200.00 A plot of the performance curve is shown on the next page. 2 Re 9.91.E+05 f 0.0117 0.0238 Duct Δp (psi) 0.0238 Error in Δp 0.00% H (ft) Q (ft3 /s) H (ft) 1.284 266.67 1.703 ft 3 s Fan Performance Curve 0.08 100% 0.07 0.06 75% 0.04 50% 0.03 0.02 25% 0.01 0.00 100 120 140 160 180 200 3 Q (m /s) 220 240 260 280 0% 300 η (%) Δp (mm) 0.05 Problem 10.86 Given: Data on centrifugal fan and various sizes Find: Suitable fan; Fan speed and input power [Difficulty: 3] Solution: Q' Basic equations: Q ω' D' ω D 3 h' h ω' ω 2 D' D 2 P' P ω' ω 3 5 D' D We choose data from the middle of the table above as being in the region of the best efficiency Q 176 ft 3 s ∆p 0.064 psi P 3.50 hp and ω 750 rpm D 3 ft ρw 1.94 slug ft The flow and head are Q' 600 ft 3 h' 1 in At best efficiency point: h s ∆p ρw g 3 1.772 in These equations are the scaling laws for scaling from the table data to the new fan. Solving for scaled fan speed, and diameter using the first two equations 1 ω' ω Q 2 3 h' 1 4 Q' h ω' 265 rpm D' D Q' 2 1 h 4 Q h' D' 76.69 in This size is too large; choose (by trial and error) Q 246 ω' ω ft 3 h s 1 3 2 4 h' Q' h Q 0.033 psi ρw g ω' 514 rpm 0.914 in P 3.50 hp D' D 1 1 2 4 h Q h' Q' D' 54.967 in Hence it looks like the 54-inch fan will work; it must run at about 500 rpm. Note that it will NOT be running at best efficiency. The power will be P' P 3 D' ω D ω' 5 P' 9.34 hp Problem 10.87 [Difficulty: 3] Given: Data on centrifugal fan Find: Fan outlet area; Plot total pressure rise and power; Best effiiciency point Solution: ηp Basic equations: Wh p dyn At Q 200 ft Wh Q ∆pt Wm 1 2 ∆p ρw g ∆ht (Note: Software cannot render a dot!) 2 ρair V 3 we have h dyn 0.25in s ρw V Hence ρair Q V A 2 g h dyn and h dyn and A ∆ht ∆h h dyn ρw g ρair V2 ρw 2 Q V The velocity V is directly proportional to Q, so the dynamic pressure at any flow rate Q is The total pressure Δh t will then be p dyn h dyn 0.25 in 3 ft 200 s Q 2 Δh is the tabulated static pressure rise Here are the results, generated in Excel: At Q = h dyn = 200 0.25 3 ft /s in Hence V = 33.13 ft/s A = 6.03749 ft2 ρw = 1.94 slug/ft3 ρ air = 0.00237 slug/ft Fitting a 2nd order polynomial to each set of data we find 3 -5 2 -3 h t =-3.56x10 Q + 6.57x10 Q + 1.883 3 Q (ft /s) Δp (psi) Pm (hp) h dyn (in) h t (in) Ph (hp) η (%) 106 141 176 211 0.075 0.073 0.064 0.050 2.75 3.18 3.50 3.51 0.07 0.12 0.19 0.28 2.15 2.15 1.97 1.66 2.15 2.86 3.27 3.32 246 282 0.033 0.016 3.50 3.22 0.38 0.50 1.29 0.94 3.01 2.51 -4 2 P h = -1.285x10 Q + 0.0517Q - 1.871 78.2% η =-3.37x10 Q + 0.0109Q -0.0151 90.0% 93.5% Finally, we use Solver to maximize η by varying Q : 94.5% 3 Q (ft /s) η (%) P h (hp) h t (in) 85.9% -5 77.9% A plot of the performance curves is shown on the next page. 161.72 2 2.01 3.13 86.6% Fan Performance Curve 3.5 100% 3.0 h t (cm), Ph (kW) 2.0 50% 1.5 1.0 25% 0.5 0.0 100 120 140 160 180 200 3 Q (ft /s) 220 240 260 280 0% 300 η (%) 75% 2.5 Problem 10.88 [Difficulty: 4] Problem 10.89 [Difficulty: 4] 10.88 10.88 Problem 10.90 [Difficulty: 3] Problem 10.91 Given: Data on turbine system Find: Model test speed; Scale; Volume flow rate [Difficulty: 3] 1 Solution: Wh ρ Q g H Basic equations: η Wmech NS Wh slug ft 3 2 1 5 2 4 ρ h The given or available data is ρ 1.94 ω P Wp 36000 hp Hp 50 ft ωp 95 rpm Hm 15 ft Wm 50 hp where sub p stands for prototype and sub m stands for model Note that we need h (energy/mass), not H (energy/weight) h p Hp g h p 1609 Hence for the prototype NS 5 NS 2 4 ωm Wm For dynamically similar conditions 5 2 4 Hp 2 ωp Dp Also Qp ωp Dp 3 2 1 ρ hm 2 ωm NS 1 5 2 4 ρ hm Hm 2 so Dp 2 Qm Dm so 3 ωm Dm rad ωm 59.3 s 2 ωp ωm Hm Hp 0.092 Dm Qm Qp ωp Dp ωm 3 To find Q p we need efficiency. At Wp 36000 hp and Hp 50 ft from F ig. 10.17 we find (see below), for 1 NScu ft 2 2 s 1 Wm ωm Dm 2 h m 482.6 NS 3.12 1 Then for the model h m Hm g 2 1 ρ hp 2 s 1 ωp Wp ft N( rpm) P( hp) 5 H( ft) 4 2 135.57 η 93 % ωm 566 rpm Hence from and also η Wmech Wh Wmech ρ Q g H Wm Qm ρ g Hm η Wp Qp ρ g Hp η 6 Qp 3.06 10 gpm 4 Qm 1.418 10 gpm Problem 10.92 [Difficulty: 2] Problem 10.93 [Difficulty: 2] V1 U = R Vj D Given: Pelton turbine Find: 1) Power 2) Operating speed 3) Runaway speed 4) Torque 5) Torque at zero speed Solution: 2 2 p V1 Vj pj h lT 1 α z α z ρ g 1 ρ g j g 2 g 2 g Basic equations 2 V h lT h l h lm K 2 and from Example Tideal ρ Q R Vj U ( 1 cos( θ) ) θ 165 deg 10.5 Assumptions: 1) p j = pamt 2) Incompressible flow 3) α at 1 and j is approximately 1 4) Only minor loss at nozzle 5) z 1 = z j Given data Then and Hence p 1g 700 psi V1 15 mph d 7.5 in D 8 ft p 1g V1 2 2 K Vj 2 g 2 g ρ g g 2 π d Vj 2 Vj o r ft 3 Q Vj 4 Q 97.2 P η ρ Q g H P 15392 hp s Urun Vj T η Tideal Stall occurs when U 0 K 0.04 p 1g ρ g U 149 ρ 1.94 slug ft 1K ωrun From Example 10.5 Tideal ρ Q R Vj U ( 1 cos( θ) ) Hence H η 86 % 2 p V1 1g 2 2 ρ 2 From Fig. 10.10, normal operating speed is around U 0.47 Vj At runaway ft V1 22 s D R 2 V1 H 1622 ft ω s ft Vj 317 s 2 2 g ft 3 U R Urun ω 37.2 rad s rad ωrun 79.2 s D 2 ω 356 rpm ωrun 756 rpm 5 Tideal 2.49 10 ft lbf 5 T 2.14 10 ft lbf Tstall η ρ Q R Vj ( 1 cos ( θ) ) 5 Tstall 4.04 10 ft lbf Problem 10.94 Given: Find: Solution: Basic equations: [Difficulty: 2] Data on Francis turbines at Niagra Falls Specific speed, volume flow rate to each turbine, penstock size 1 Wh ρ Q g H η Wmech NS Wh ω P 2 1 5 2 4 ρ h 2 h g H h lT f L V D 2 The given or available data is ρ 998 kg 3 Wmech 54 MW ω 107 rpm η 93.8% H 65 m Lpenstock 400 m Hnet H 83% m 1 2 h g H 637.4 The specific energy of the turbine is: m NS The specific speed is: 2 s ω Wmech 1 5 2 4 ρ h Solving for the flow rate of the turbine: Wmech Q ρ h η NS 0.814 3 90.495 3 m Q 90.5 s 2 m h lT g H Hnet 108.363 2 s Based on the head loss: 2 Since V Q A 4 Q m s into the head loss equation: 2 π D 1 2 2 L 1 4 Q 8 f L Q h lT f D 2 2 5 2 π D D π Assuming concrete-lined penstocks: D (m) 2.000 3.510 3.414 3.418 V (m/s) 28.807 9.354 9.888 9.862 8 f L Q2 Solving for the diameter: D π 2 h lT e 3 mm Re 5.70E+07 3.25E+07 3.34E+07 3.34E+07 5 This will require an iterative solution. If we assume a diameter of 2 m, we can iterate to find the actual diameter: e /D 0.001500 0.000855 0.000879 0.000878 f 0.02173 0.01892 0.01904 0.01904 D (m) 3.510 3.414 3.418 3.418 D 3.42 m Problem 10.95 [Difficulty: 2] Problem 10.96 [Difficulty: 3] 10.39 10.39 Problem 10.97 Given: Find: Solution: [Difficulty: 3] Data on Pelton wheel Rotor radius, jet diameter, water flow rate. The given or available data is ρ 999 kg 3 2 6 m Wmech 26.8 MW ω 225 rpm H 360 m ν 1.14 10 m From Bernoulli, the jet velocity is: m Vj Cv 2 g H 82.35 s Vi 2 g H Assuming a velocity coefficient of s Cv 0.98 From Fig. 10.36, at maximum efficiency: U R ω 0.47 Vj (4% loss in the nozzle): So the radius can be calculated: R 0.47 From Fig. 10.37 the efficiency at full load is η 86% Thus: η Wmech Q ρ g H ω 1.643m Solving for the flow rate: Q π 2 Q Therefore, We can now calculate the jet velocity: Aj Dj 4 Vj Vj Q Dj 2 0.37 m π Vj 3 Wmech η ρ g H 8.836 m s Dj 37.0 cm mrate ρ Q 8.83 10 3 kg s Problem 10.98 [Difficulty: 3] Given: Impulse turbine requirements Find: 1) Operating speed 2) Wheel diameter 4) Jet diameter 5) Compare to multiple-jet and double-overhung 1 Solution: Vj Basic equations: 2 g H NS ω P 2 1 5 2 4 ρ h Model as optimum. This means. from Fig. 10.10 Given or available data H 350 m η P U 0.47 Vj and from Fig. 10.17 NScu 5 P 15 MW ρ 1.94 Vj m Vj 82.9 s 2 g H U 0.47 Vj U 38.9 D Q 4.91 η ρ g H 1 5 2 4 ρ ( g H) 2 U s NScu NS 43.46 NS 0.115 m s (1) 1 P The wheel radius is m 3 P ω NS For a single jet η 89 % 3 We need to convert from N Scu (from Fig. 10.17) to NS (see discussion after Eq. 10.18b). The water consumption is Q with slug ft Then Q Vj Aj ρ Q g H ω 236 rpm Dj 4 Q (2) π Vj Dj 0.275 m 2 D 3.16 m (3) ω For multiple (n) jets, we use the power and flow per jet ωn ω n From Eq 1 Results: n From Eq. 2 ωn ( n ) rpm Djn Dj an d n Djn( n ) 0.275 Dn n Dn ( n ) m 1 236 2 333 0.194 2.23 3 408 0.159 1.82 4 471 0.137 1.58 5 527 0.123 1.41 A double-hung wheel is equivalent to having a single wheel with two jets D 3.16 m from Eq. 3 Problem 10.99 Given: Data on impulse turbine Find: Plot of power and efficiency curves [Difficulty: 2] Solution: Basic equations: T F R H = 33 ρ = R = 1.94 0.50 P ω T η P Here are the results calculated in Excel: ρ Q g H ω (rpm) Q (cfm) F (lbf) T (ft-lbf) P (hp) ft slug/ft3 ft 0 1000 1500 1900 2200 2350 2600 2700 7.74 7.74 7.74 7.44 7.02 5.64 4.62 4.08 2.63 2.40 2.22 1.91 1.45 0.87 0.34 0.09 1.32 1.20 1.11 0.96 0.73 0.44 0.17 0.05 0.000 0.228 0.317 0.345 0.304 0.195 0.084 0.023 η (%) 0.0% 47.3% 65.6% 74.4% 69.3% 55.3% 29.2% 9.1% Turbine Performance Curves 100% 0.40 90% 0.35 80% 0.30 70% 60% 50% 0.20 40% 0.15 30% 0.10 20% 0.05 10% 0.00 0% 0 500 1000 1500 ω (rpm) 2000 2500 3000 η (%) P (hp) 0.25 Problem 10.100 [Difficulty: 4] Problem 10.101 Given: Find: [Difficulty: 3] Published data for the Tiger Creek Power Plant (a) Estimate net head at the site, turbine specific speed, and turbine efficiency (b) Comment on consistency of the published data Solution: NScu Basic Equations: N P 5 H N P NS 5 4 ρ ( g H) η P ρ Q g Hnet 4 The given or available data is kg ρ 999 3 3 P 58 MW Q 21 m m Hgross 373 m s ν 1.14 10 2 6 m s Using data from Fig. 10.37, we will assume η 87% We can take this to estimate the net head: Therefore: Hnet Hgross 86.875 % P Hnet 324 m ρ Q g η This is close to 87%, so the assumption for the efficiency was a good one. From the same figure, we will assume NScu 5 Therefore the dimensionless specific speed is NScu NS 0.115 43.46 5 We may then calculate the rotational speed for the turbine: N NS ρ g Hnet 4 108.8 rpm P The power output seems low for a turbine used for electricity generation; several turbines are probably used in this one plant. To check the claims: 58 MW 58 MW 24 hr 1 day s 3 21 m 365 day yr hr 3600 s 8 kW hr 5.081 10 0.767 kW hr 2 m m This number is 50% higher than the claim. yr This is in excellent agreement with the claim. Problem 10.102 Given: Hydraulic turbine site Find: Minimum pipe size; Fow rate; Discuss [Difficulty: 4] Solution: 2 hl L V Hl f g D 2 g Basic equations: ∆z and also, from Example 10.15 the optimum is when Hl 3 As in Fig. 10.41 we assume L 2 ∆z and Then, for a given pipe diameter D V 2 Q V Also f = f 0.02 2 g D Hl f L g D 3 f 2 π D V Ph ρ Q 2 4 Pm η Ph Here are the results in Excel: 0.02 ρ = 998.00 kg/m3 η = 83% 3 D (cm) V (m/s) Q (m /s) P h (kW) 25 30 35 40 45 50 6.39 7.00 7.56 8.09 8.58 9.04 0.314 0.495 0.728 1.016 1.364 1.775 6.40 12.12 20.78 33.16 50.09 72.42 Pm (kW) 5.31 10.06 17.25 27.53 41.57 60.11 41.0 8.19 1.081 36.14 30.00 Turbine efficiency varies with specific speed Pipe roughness appears to the 1/2 power, so has a secondary effect. A 20% error in f leads to a 10% change in water speed and 30% change in power. A Pelton wheel is an impulse turbine that does not flow full of water; it directs the stream with open buckets. A diffuser could not be used with this system. Use Goal Seek or Solver to vary D to make Pm 30 kW! Power Versus Pipe Diameter 70 60 Pm (kW) 50 40 30 20 10 0 20 25 30 35 40 D (cm) 45 50 55 Problem 10.103 [Difficulty: 4] Problem 10.104 [Difficulty: 3] V2 = V3 = V y 2h x V1 Given: Data on boat and propeller Find: Propeller diameter; Thrust at rest; Thrust at 15 m/s Solution: CS V4 Basic equation: (4.26) Assumption: 1) Atmospheric pressure on CS 2) Horizontal 3) Steady w.r.t. the CV 4) Use velocities relative to CV The x-momentum is then V It can be shown (see Example 10.13) that For the static case T u 1 mrate u 4 mrate V4 V1 mrate m V1 0 s 1 2 V4 V1 kg where mrate 50 is the mass flow rate s m V4 45 s so V 1 2 V4 V1 2 From continuity π D mrate ρ V A ρ V 4 4 mrate Hence D For V1 = 0 T mrate V4 V1 When in motion m V1 15 s Hence for V1 = 15 m/s T mrate V4 V1 V 22.5 m s kg 3 m D 1.52 m ρ π V ρ 1.23 with T 2250 N and V 1 V4 V1 2 T 750 N so V4 2 V V1 m V4 30 s Problem 10.105 Given: Data on fanboat and propeller Find: Thrust at rest; Thrust at 12.5 m/s [Difficulty: 3] Solution: Assume the aircraft propeller coefficients in Fi.g 10.40 are applicable to this propeller. At V = 0, J = 0. Extrapolating from Fig. 10.40b We also have D 1.5 m n 1800 rpm n 30 The thrust at standstill (J = 0) is found from At a speed V 12.5 m s CF 0.16 J V n D The thrust and power at this speed can be found rev and s kg 3 m 2 4 FT CF ρ n D J 0.278 ρ 1.225 and so from Fig. 10.40b 2 4 FT CF ρ n D FT 893 N (Note: n is in rev/s) FT 809 N CP 0.44 3 and CF 0.145 5 P 111 kW P CP ρ n D Problem 10.106 Given: Data on jet-propelled aircraft Find: Propulsive efficiency [Difficulty: 3] y x U V FD CS Y Solution: X Basic equation: (4.26) (4.56) Assumption: 1) Atmospheric pressure on CS 2) Horizontal 3) Steady w.r.t. the CV 4) Use velocities relative to CV The x-momentum is then kg is the mass flow rate where mrate 50 s FD U mrate ( V U) U The useful work is then The energy equation simplifies to W η Hence U2 mrate 2 mrate ( V U) U mrate 2 U 225 FD mrate ( V U) or With FD u 1 mrate u 2 mrate ( U) mrate ( V) mrate m s and 2 2 V U η 45% V2 mrate 2 2 mrate V U 2 2 2 ( V U) U V2 U2 2 1 V U V U 2 η 1 V 775 m s Problem 10.107 9.89 [Difficulty: 4] Problem 10.108 [Difficulty: 4] V2 = V3 = V y 2h x V1 CS V4 Given: Definition of propulsion efficiency η Find: η for moving and stationary boat Solution: Assumption: 1) Atmospheric pressure on CS 2) Horizontal 3) Steady w.r.t. the CV 4) Use velocities relative to CV The x-momentum (Example 10.3): T u 1 mrate u 4 mrate mrate V4 V1 Applying the energy equation to steady, incompressible, uniform flow through the moving CV gives the minimum power input requirement V 2 V 2 4 1 Pmin mrate 2 2 On the other hand, useful work is done at the rate of Puseful V1 T V1 mrate V4 V1 Combining these expressions or η η V1 mrate V4 V1 2 2 V1 V4 mrate 2 2 V1 V4 V1 1 2 V4 V1 V4 V1 2 V1 V1 V4 When in motion V1 30 mph For the stationary case V1 0 mph and V4 90 mph η η 2 V1 V1 V4 2 V1 V1 V4 η 50 % η 0 % Problem 10.109 [Difficulty: 4] 9.174 Problem 10. ['LIILFXOW\ 5] Problem 10.111 [Difficulty: 2] Given: NASA-DOE wind turbine generator Find: Estimate rotor tip speed and power coefficient at maximum power condition Solution: CP Basic equations: Pm 1 2 and we have ρ 0.00237 slug ft 3 3 ρ V π R X 2 ω 45 rpm 4.712 rad s ω R V U ω R η Pm Pideal R 63 ft V 16 knot 27.005 ft s P 135 hp U ω R 297 The blade tip speed is: The tip speed ratio is: X ω R V 10.994 η 74% ft s (X will decrease at the wind speed increases.) P The mechanical work out is: Pm 182.4 hp η From this we can calculate the power coefficient: CP Pm 1 2 3 ρ V π R 0.345 2 Problem 10.112 [Difficulty: 3] Problem 10.113 Given: Model of farm windmill Find: Angular speed for optimum power; Power output [Difficulty: 2] Solution: Basic equations: CP P 1 2 From Fig. 10.45 Hence, for Also 3 ρ V π R X 2 CPmax 0.3 V 10 m s 1 3 2 P CPmax ρ V π R 2 ω R ρ 1.225 kg 3 m X 0.8 at ω and we have V X V R P 144 W and D 1 m ω 16 R rad s D 2 R 0.5 m ω 153 rpm Problem 10.114 [Difficulty: 2] Given: NASA-DOE wind turbine generator Find: Estimate rotor tip speed and power coefficient at maximum power condition Solution: Pm CP Basic equations: 1 2 and we have ρ 1.23 kg X 3 ρ V π R ω 70 rpm 3 2 R 5 m ω R V U ω R H 18 m η Pm Pideal 2 A 110 m U ω R 36.652 m 2 m s From Fig. 10.45: CP 0.34 when X 5.3 (maximum power condition) If we replace the π R term in the power coefficient 1 3 with the swept area we will get: P CP ρ V A 2 Here are the results, calculated using Excel: A = 110.00 m ρ = U= 2 Power coefficient data were taken from Fig. 10.45 1.23 kg/m 36.65 m/s V (kt) V (m/s) 10.0 12.5 15.0 17.5 20.0 22.5 25.0 30.0 5.14 6.43 7.72 9.00 10.29 11.57 12.86 15.43 3 X 7.125 5.700 4.750 4.071 3.562 3.167 2.850 2.375 CP 0.00 0.30 0.32 0.20 0.10 0.05 0.02 0.00 P (kW) 0.00 5.40 9.95 9.87 7.37 4.72 2.88 0.00 Power Versus Wind Speed 12 10 P (kW) 8 6 4 2 0 5 10 15 20 V (knots) 25 30 35 Problem 10.115 [Difficulty: 5] Part 1/3 Problem 10.115 [Difficulty: 5] Part 2/3 Problem 10.115 [Difficulty: 5] Part 3/3 Problem 10.116 [Difficulty: 5] Part 1/3 Problem 10.116 [Difficulty: 5] Part 2/3 Problem 10.116 [Difficulty: 5] Part 3/3 Problem 10.117 [Difficulty: 2] Given: Prototype air compressor, 1/5 scale model to be built Find: Mass flow rate and power requirements for operation at equivalent efficiency Solution: M R T01 ω D p D2 c01 01 M p 8.9 Given data: M R T01 ω D f2 2 3 5 c01 ρ01 ω D p 01 D Wc η f1 Basic equations: kg s ωp 600 rpm Dm Dp 1 5 Wcp 5.6 MW Since the efficiencies are the same for the prototype and the model, it follows that: M m Rm T01m 2 M p Rp T01p p 01m Dm p 01p Dp 2 ωm Dm c01m ωp Dp Wcm c01p ρ01m ωm Dm 3 5 Wcp 3 ρ01p ωp Dp 5 Given identical entrance conditions for model and prototype and since the working fluid for both is air: Mm 2 Dm Mp Dp Solving for the mass flow rate of the model: 2 ωm Dm ωp Dp Wcm 3 5 ωm Dm Solving for the speed of the model: 3 2 M m 0.356 Solving for the power requirement for the model: 5 kg s Dp ωm ωp 3000 rpm Dm 3 Wcp ωp Dp Dm Mm Mp Dp ωm Dm Wcm Wcp ωp Dp 5 Wcm 0.224 MW Problem 10.118 Given: Find: Solution: Prototype air compressor equipped with throttle to control entry pressure Speed and mass flow rate of compressor at off-design entrance conditions Basic equations: M T01 η f1 Given data: [Difficulty: 3] p 01 p 01d 14.7 psi T01 ∆T01 ω T01 T01d 70 °F M T01 f2 p 01 T01 ω ωd 3200 rpm T01 58 °F Since the normalized speed is equal to that of the design point, it follows that: ω T01 Solving for the required speed: At similar conditions: M T01 p 01 M d 125 lbm s p 01 8.0 psi ωd T01d T01 ω ωd T01d M d T01d p 01d Solving for the actual mass flow rate: ω 3164 rpm M Md T01d p 01 T01 p 01d M 68.8 lbm s Problem 10.119 Given: Find: Solution: Design conditions for jet turbine, off-design actual conditions New operating speed, mass flow rate, and exit conditions for similar operation Basic equations: M T01 η f1 p 01 Given data: [Difficulty: 3] p 01d 160 psi ωd M T01 T01 M d T01d p 02 T01 p 01 ω lbm s p 02 M T01 f3 p 02d 80 psi p 01 T01 ω T02d 1350 °F ∆T0d Solving for the actual mass flow rate: Solving for the temperature drop: T01d p 01d p 02d T01 ω ωd T01d p 01d M d 500 Solving for the required speed: T01 T01 T02 T01d T02d T01d p 01 p 01 T01d T01 ∆T0 T01 M T01 f2 T01 1600 °F ω p 01 ∆T0 T01d 1700 °F ωd 500 rpm p 01 140 psi At similar conditions: T01 ω M Md T01 ∆T0 ∆T0d T01d T01 T02 T01 T01d T02d T01d Solving for the exit pressure: p 02d p 02 p 01 p 01d ω 488 rpm T01d p 01 T01 p 01d M 448 lbm s Substituting in temperatures: T02 1266 °F p 02 70 psi Problem 10.120 [Difficulty: 4] Discussion: When we change the working fluid, we need to be sure that we use the correct similitude relationships. Specifically, we would need to keep fluid-specific parameters (gas constant and specific heat ratio) in the relationships. The functional relationships are: h0 s ND 2 , , m 01 ND 2 ND P , , ,k f 1 3 c01 01 N 3 D 5 01 ND So these dimensionless groups need to be considered. When we replace air with helium, both the gas constant R and the specific heat ratio k will increase. Given a fixed inflow pressure and temperature and a fixed geometry, the effect would be to decrease density and increase sound speed. Therefore, replacing air with helium should result in decreased mass flow rate and power, and an increased operating speed. When considering dimensional parameters, the important thing to remember is that the operability maps for compressors and/or turbines were constructed for a single working fluid. Therefore, to be safe, an engineer should reconstruct an operability map for a new working fluid. Problem 11.1 [Difficulty: 2] Given: Trapezoidal channel Find: Derive expression for hydraulic radius; Plot R/y versus y for two different side slopes Solution: b 2 m Available data α1 30 deg α2 60 deg The area is (from simple geometry of a rectangle and triangles) 1 A b y 2 y y cot( α) y ( b y cot( α) ) 2 The wetted perimeter is (from simple geometry) P b 2 Hence the hydraulic radius is R R We are to plot y A P y ( b y cot( α) ) b 2 which is the same as that listed in Table 11.1 y sin( α) b y cot( α) b 2 y sin( α) with y for α = 30o and 60o, and 0.5 < y < 3 m. b 2 m sin( α) The graph is shown below; it can be plotted in Excel. 0.75 30 Degrees 60 Degrees R/y 0.5 0.25 0 0.5 1 1.5 2 2.5 3 y (m) As the depth increases, the hydraulic radius becomes smaller relative to depth y - wetted perimeter becomes dominant over area Problem 11.2 [Difficulty: 2] Given: Circular channel Find: Derive expression for hydraulic radius; Plot R/D versus D for a range of depths Solution: The area is (from simple geometry - a segment of a circle plus two triangular sections) 2 2 2 1 D α D α D α D α ⋅ α + 2 ⋅ ⋅ ⋅ sin⎛⎜ π − ⎞ ⋅ ⋅ cos⎛⎜ π − ⎞ = ⋅α + ⋅ sin⎛⎜ π − ⎞ ⋅ cos⎛⎜ π − ⎞ 8 2 2 2⎠ 2 2⎠ 4 2⎠ 8 2⎠ ⎝ ⎝ ⎝ ⎝ A= D A= D 2 8 2 ⋅α + D 2 ⋅ sin( 2 ⋅ π − α) = 8 D 8 2 ⋅α − D 8 2 D ⋅ sin( α) = P= The wetted perimeter is (from simple geometry) D 2 8 ⋅ ( α − sin( α) ) ⋅α 2 D Hence the hydraulic radius is R= A P = 8 ⋅ ( α − sin( α) ) D 2 R We are to plot D = We will need y as a function of α: 1 4 ⋅ ⎛⎜ 1 − ⎝ y= = ⋅α 1 4 ⋅ ⎛⎜ 1 − ⎝ sin( α) ⎞ α ⎠ ⋅D which is the same as that listed in Table 11.1 sin( α) ⎞ ⎠ α D 2 + D 2 ⋅ cos⎛⎜ π − ⎝ α⎞ 2⎠ = D 2 ⋅ ⎛⎜ 1 − cos⎛⎜ ⎝ α ⎞⎞ ⎝ 2 ⎠⎠ or y D = 1 2 ⋅ ⎛⎜ 1 − cos⎛⎜ ⎝ ⎝ 2 ⎠⎠ The graph can be plotted in Excel. 0.4 R/D 0.3 0.2 0.1 0 0.2 0.4 0.6 y/D 0.8 α ⎞⎞ 1 Problem 11.3 Given: Wave from a passing boat Find: Estimate of water depth [Difficulty: 1] Solution: Basic equation c g y Available data c 10 mph or c 14.7 ft s We assume a shallow water wave (long wave compared to water depth) c g y so y c 2 g y 6.69 ft Problem 11.4 Given: Pebble dropped into flowing stream Find: Estimate of water speed [Difficulty: 1] Solution: Basic equation c g y Available data y 2 m and relative speeds will be and 7 m Vwave 1 s We assume a shallow water wave (long wave compared to water depth) c Hence g y so Vstream Vwave c c 4.43 m s m Vstream 2.57 s Vwave Vstream c m Vwave 7 s Problem 11.5 Given: Pebble dropped into flowing stream Find: Estimate of water depth and speed [Difficulty: 2] Solution: Basic equation c g y Available data 5 ft Vwaveupstream 1 s ft Vwaveupstream 5 s 13 ft Vwavedownstream 1 s ft Vwavedownstream 13 s and relative speeds will be But we have Vwavedownstream Vstream c Adding Vstream Subtracting c and Vwavedownstream Vwaveupstream 2 Vwavedownstream Vwaveupstream 2 Vwave Vstream c Vwaveupstream Vstream c ft Vstream 4 s c9 ft s We assume a shallow water wave (long wave compared to water depth) Hence c g y so y c 2 g y 2.52 ft Problem 11.6 [Difficulty: 3] Given: Speed of surface waves with no surface tension Find: Speed when λ/y approaches zero or infinity; Value of λ/y for which speed is 99% of this latter value Solution: g⋅ λ Basic equation c= For λ/y << 1 tanh⎛⎜ (1) 2 ⋅ π⋅ y ⎞ 2 ⋅ π⋅ tanh⎛⎜ ⎝ λ ⎠ 2 ⋅ π⋅ y ⎞ approaches 1 ⎝ λ ⎠ Hence c is proportional to so as λ/y approaches ∞ λ We wish to find λ/y when c = 0.99⋅ g ⋅ y Combining this with Eq 1 0.99⋅ g ⋅ y = g⋅ λ 2 ⋅ π⋅ tanh⎛⎜ ⎝ 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜ 2 Hence 2 ⋅ π⋅ y ⎞ ⎝ λ ⎠ = λ y c= c= so g⋅ λ 2⋅ π g⋅ y g⋅ λ 2 0.99 ⋅ g ⋅ y = or 2 ⋅ π⋅ y ⎞ λ tanh( ∞) → 1 ⎠ 2 ⋅ π⋅ tanh⎛⎜ 2 ⋅ π⋅ y ⎞ ⎝ λ ⎠ Letting λ/y = x we find 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜ 2 2⋅ π ⎞ ⎝ x ⎠ =x This is a nonlinear equation in x that can be solved by iteration or using Excel's Goal Seek or Solver Hence x = 1 x = 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜ 2⋅ π ⎞ x = 6.16 x = 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜ 2⋅ π ⎞ x = 4.74 x = 4.74 x = 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜ 2⋅ π ⎞ x = 5.35 x = 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜ 2⋅ π ⎞ x = 5.09 x = 5.09 x = 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜ 2⋅ π ⎞ x = 5.2 x = 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜ 2⋅ π ⎞ x = 5.15 x = 5.15 x = 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜ 2⋅ π ⎞ x = 5.17 x = 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜ 2⋅ π ⎞ x = 5.16 x = 5.16 x = 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜ 2⋅ π ⎞ x = 5.17 x = 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜ 2⋅ π ⎞ x = 5.16 λ y = 5.16 2 ⎝ x ⎠ 2 ⎝ x ⎠ 2 ⎝ 2 ⎝ 2 ⎝ x x x ⎠ ⎠ ⎠ 2 ⎝ x ⎠ 2 ⎝ x ⎠ 2 ⎝ x ⎠ 2 ⎝ x ⎠ 2 ⎝ x ⎠ Problem 11.7 Given: Expression for capillary wave length Find: Length of water and mercury waves [Difficulty: 1] Solution: σ Basic equation λ 2 π Available data Table A.2 (20oC) ρ g SG Hg 13.55 SG w 0.998 ρ 1000 kg 3 m σHg 484 10 Table A.4 (20oC) Hence λHg 2 π λ w 2 π σHg SG Hg ρ g σw SGw ρ g 3 N m σw 72.8 10 λHg 12 mm λHg 0.472 in λw 17.1 mm λw 0.675 in 3 N m Problem 11.8 [Difficulty: 2] Given: Expression for surface wave speed Find: Plot speed versus wavelength for water and mercury waves Solution: ⎛ g ⋅ λ 2 ⋅ π⋅ σ ⎞ ⎛ 2 ⋅ π⋅ y ⎞ ⎜ 2 ⋅ π + ρ⋅ λ ⋅ tanh⎜ λ ⎝ ⎠ ⎝ ⎠ Basic equation c= Available data Table A.2 (20oC) SG Hg = 13.55 SG w = 0.998 ρ = 1000⋅ kg 3 m Table A.4 (20oC) Hence cw( λ) = σHg = 484 × 10 −3 N ⋅ m 2 ⋅ π⋅ σw ⎞ ⎛ g⋅ λ ⎛ 2 ⋅ π⋅ y ⎞ ⎜ 2 ⋅ π + SG ⋅ ρ⋅ λ ⋅ tanh⎜ λ ⎝ ⎠ w ⎝ ⎠ σw = 72.8 × 10 cHg( λ) = −3 N ⋅ m y = 7 ⋅ mm 2 ⋅ π⋅ σHg ⎞ ⎛ g⋅ λ ⎛ 2 ⋅ π⋅ y ⎞ ⎜ 2⋅ π + SG ⋅ ρ⋅ λ ⋅ tanh⎜ λ ⎝ ⎠ Hg ⎝ ⎠ 0.7 Water Mercury Wave speed (m/s) 0.6 0.5 0.4 0.3 0.2 0.1 20 40 60 Wavelength (mm) 80 100 Problem 11.9 Given: Sharp object causing waves Find: Flwo speed and Froude number [Difficulty: 1] Solution: Basic equation c g y Available data y 150 mm θ 30 deg We assume a shallow water wave (long wave compared to water depth) c g y so c 1.21 m s From geometry Hence Also sin( θ) Fr c V c so V Fr 2 V c sin( θ) or V 2.43 Fr m s 1 sin( θ) Fr 2 Problem 11.10 Given: Shallow water waves Find: Speed versus depth [Difficulty: 2] Solution: c( y ) Basic equation g y We assume a shallow water wave (long wave compared to water depth) 10 Wave Speed (m/s) Rapid Flow: Fr > 1 1 Tranquil Flow: Fr < 1 0.1 3 1 10 0.01 0.1 Depth (m) 1 10 Problem 11.11 Given: Motion of sumerged body Find: Speed versus ship length [Difficulty: 2] Solution: c Basic equation g y We assume a shallow water wave (long wave compared to water depth) In this case we want the Froude number to be 0.5, with Fr 0.5 V and c c g x where x is the ship length V 0.5 c 0.5 g x Hence Ship Speed (m/s) 100 10 1 1 10 100 Ship Length (m) 3 1 10 Problem 11.12 Given: Flow in a rectangular channel Find: Froude numbers [Difficulty: 1] Solution: V Basic equation Fr Available data y 750 mm Hence Fr 1 Fr 2 g y V1 g y V2 g y m V1 1 s m V2 4 s Fr 1 0.369 Subcritical flow Fr 2 1.47 Supercritical flow Problem 11.12 Given: Flow in a rectangular channel with wavy surface Find: Froude numbers [Difficulty: 2] Solution: V Basic equation Fr Available data b 10 ft g y y 6 ft A "wavy" surface indicates an unstable flow, which suggests critical flow Hence Then V Fr g y V 13.9 ft Q V b y Q 834 ft Fr 1 s 3 s 5 Q 3.74 10 gpm Problem 11.14 Given: Data on sluice gate Find: Downstream depth; Froude number [Difficulty: 2] Solution: Basic equation: p1 ρ⋅ g + V1 2 2⋅ g p2 V2 2 + y1 = + + y2 + h 2⋅ g ρ⋅ g The Bernoulli equation applies because we have steady, incompressible, frictionless flow. Noting that p 1 = p 2 = p atm, (1 = upstream, 2 = downstream) the Bernoulli equation becomes V1 2 2⋅ g V2 2 + y1 = + y2 2⋅ g 3 The given data is For mass flow m b = 5⋅ m y 1 = 2.5⋅ m Q = 10⋅ Q = V⋅ A so Q and V1 = b⋅ y1 2 Using these in the Bernoulli equation ⎛ Q ⎞ ⎜ b⋅ y ⎝ 1⎠ + y = 1 2⋅ g s Q V2 = b⋅ y2 2 ⎛ Q ⎞ ⎜ b⋅ y ⎝ 2⎠ + y 2 2⋅ g (1) 2 The only unknown on the right is y2. The left side evaluates to ⎛ Q ⎞ ⎜ b⋅ y ⎝ 1 ⎠ + y = 2.53 m 1 2⋅ g To find y 2 we need to solve the non-linear equation. We must do this numerically; we may use the Newton method or similar, or Excel's Solver or Goal Seek. Here we interate manually, starting with an arbitrary value less than y 1. 2 ⎛ Q ⎞ ⎜ b⋅ y ⎝ 2 ⎠ + y = 2.57 m 2 2⋅ g 2 ⎛ Q ⎞ ⎜ b⋅ y ⎝ 2 ⎠ + y = 2.54 m 2 2⋅ g y 2 = 0.25⋅ m ⎛ Q ⎞ ⎜ b⋅ y ⎝ 2 ⎠ + y = 3.51 m For y = 0.3⋅ m 2 2 2⋅ g For y 2 = 0.305 ⋅ m ⎛ Q ⎞ ⎜ b⋅ y ⎝ 2 ⎠ + y = 2.50 m For y = 0.302 ⋅ m 2 2 2⋅ g Hence y 2 = 0.302 m is the closest to three figs. Then Q V2 = b⋅ y2 m V2 = 6.62 s For Fr 2 = V2 g⋅ y2 2 2 Fr 2 = 3.85 Problem 11.15 [Difficulty: 3] Given: Rectangular channel Find: Plot of specific energy curves; Critical depths; Critical specific energy Solution: Given data: b= 20 ft Specific energy: ⎛ Q2 E = y + ⎜⎜ 2 ⎝ 2 gb ⎞ 1 ⎟⎟ 2 ⎠ y Critical depth: yc Specific Energy, E (ft·lb/lb) y (ft) 0.5 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.5 4.0 4.5 5.0 ⎛ Q = ⎜⎜ ⎝ gb 2 2 ⎞ ⎟⎟ ⎠ 1 3 5 Q = 0 0.50 0.60 0.80 1.00 1.20 1.40 1.60 1.80 2.00 2.20 2.40 2.60 2.80 3.00 3.50 4.00 4.50 5.00 Q = 25 0.60 0.67 0.84 1.02 1.22 1.41 1.61 1.81 2.01 2.21 2.40 2.60 2.80 3.00 3.50 4.00 4.50 5.00 Q = 75 1.37 1.21 1.14 1.22 1.35 1.51 1.69 1.87 2.05 2.25 2.44 2.63 2.83 3.02 3.52 4.01 4.51 5.01 Q = 125 2.93 2.28 1.75 1.61 1.62 1.71 1.84 1.99 2.15 2.33 2.51 2.69 2.88 3.07 3.55 4.04 4.53 5.02 Q = 200 6.71 4.91 3.23 2.55 2.28 2.19 2.21 2.28 2.39 2.52 2.67 2.83 3.00 3.17 3.63 4.10 4.58 5.06 y c (ft) E c (ft) 0.365 0.547 0.759 1.14 1.067 1.60 1.46 2.19 4 3 y (ft) Q=0 2 Q = 25 cfs Q = 75 cfs Q = 125 cfs Q = 200 cfs 1 0 0 2 4 E (ft) 6 Problem 11.16 Given: Rectangular channel flow Find: Critical depth [Difficulty: 1] 1 Solution: Basic equations: ⎛ Q2 ⎞ yc = ⎜ ⎜ g⋅ b2 ⎝ ⎠ Given data: b = 2.5⋅ m 3 3 Q = 3⋅ m s 1 Hence ⎛ Q2 ⎞ yc = ⎜ ⎜ g⋅ b2 ⎝ ⎠ 3 y c = 0.528 m Problem 11.17 Given: Data on trapezoidal channel Find: Critical depth and velocity [Difficulty: 3] Solution: 2 V Basic equation: Ey The given data is: b 20 ft 2 g α atan ( 2) S0 0.0016 α 63.4deg n 0.025 ft Q 400 3 s 2 In terms of flow rate Ey Q A y ( b y cot ( α ) ) where (Table 11.1) 2 2 A g 2 Hence in terms of y Q Ey 2 2 2 ( b y cot ( α ) ) y g For critical conditions dE dy 2 2 Q 01 3 g y ( b y cot( α) ) 3 3 2 2 g y ( b y cot( α) ) 1 3 Q ( b 2 y cot( α) ) 3 g y ( b y cot( α) ) 3 2 Hence g y ( b y cot( α) ) Q ( b 2 y cot( α) ) 0 Let f ( y ) g y ( b y cot( α) ) Q ( b 2 y cot( α) ) 3 2 Q cot( α) 3 2 We can iterate or use Excel's Goal Seek or Solver to find y when f(y) = 0 Guess y 2 ft f ( y ) 1.14 10 6 ft 7 2 y 2.25 ft f ( y ) 1.05 10 5 ft 7 2 s y 2.35 ft 5 ft f ( y ) 3.88 10 s y 2.3 ft 5 ft f ( y ) 1.36 10 7 2 y 2.275 ft s Hence critical depth is y 2.27 ft and critical speed is V Q A and 4 ft f ( y ) 1.38 10 7 2 y 2.272 ft s A y ( b y cot( α) ) V 8.34 ft s 2 s The solution is somewhere between y = 2.25 ft and y = 2.35 ft, as the sign of f(y) changes here. f ( y ) 657 ft 2 s A 48.0 ft 2 7 7 Problem 11.18 Given: Data on rectangular channel Find: Minimum specific energy; Flow depth; Speed [Difficulty: 2] Solution: 2 Basic equation: E=y+ V 2⋅ g In Section 11-2 we prove that the minimum specific energy is when we have critical flow; here we rederive the minimum energy point 3 ft For a rectangular channel Q = V⋅ b ⋅ y or 2 Hence, using this in the basic equation E is a minimum when The speed is then given by E=y+ ⎛ Q ⎞ ⋅ 1 =y+ ⎜ b⋅ y 2⋅ g ⎝ ⎠ ⎛ Q2 ⎞ 1 =1−⎜ ⋅ =0 dy ⎜ b 2⋅ g y 3 ⎝ ⎠ dE V = V= Q Q Q b = 10⋅ s ft ⎛ Q2 ⎞ 1 ⎜ ⋅ ⎜ 2 ⋅ b 2⋅ g y 2 ⎝ ⎠ = constant 1 ⎛ Q2 ⎞ y = ⎜ ⎜ b 2⋅ g ⎝ ⎠ or V = 6.85⋅ b⋅ y with b⋅ y 3 ft s 1 Note that from Eq. 11.22 we also have Vc = ⎛ g⋅ Q ⎞ ⎜ b ⎝ ⎠ 3 ft Vc = 6.85⋅ s 2 The minimum energy is then Emin = y + V 2⋅ g Emin = 2.19⋅ ft which agrees with the above y = 1.46⋅ ft Problem 11.19 Given: Data on rectangular channel Find: Depths for twice the minimum energy [Difficulty: 3] Solution: 2 E=y+ Basic equation: V 2⋅ g 3 ft Q = V⋅ b ⋅ y For a rectangular channel or 2 Hence, using this in the basic eqn. E=y+ We have a nonlinear implicit equation for y y+ V= ⎛ Q ⎞ ⋅ 1 =y+ ⎜ b⋅ y 2⋅ g ⎝ ⎠ Q b⋅ y Q with ⎛ Q2 ⎞ 1 ⎜ ⋅ ⎜ 2 ⋅ b 2⋅ g y 2 ⎝ ⎠ b and = 10⋅ s ft = constant E = 2 × 2.19⋅ ft E = 4.38⋅ ft ⎛ Q2 ⎞ 1 ⎜ ⋅ =E ⎜ 2 ⋅ b2⋅ g y 2 ⎝ ⎠ This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We start with a y larger than the critical, and evaluate the left side of the equation so that it is equal to E = 4.38⋅ ft For For y = 2 ⋅ ft ⎛ Q2 ⎞ 1 y+⎜ ⋅ = 2.39⋅ ft ⎜ 2 ⋅ b2⋅ g y 2 ⎝ ⎠ For y = 4 ⋅ ft ⎛ Q2 ⎞ 1 y+⎜ ⋅ = 4.10⋅ ft ⎜ 2 ⋅ b2⋅ g y 2 ⎝ ⎠ y = 4.5⋅ ft y+ ⎛ Q2 ⎞ 1 ⎜ ⋅ = 4.58⋅ ft ⎜ 2 ⋅ b2⋅ g y 2 ⎝ ⎠ For y = 4.30⋅ ft y+ ⎛ Q2 ⎞ 1 ⎜ ⋅ = 4.38⋅ ft ⎜ 2 ⋅ b2⋅ g y 2 ⎝ ⎠ Hence y = 4.30⋅ ft y = 0.5⋅ ft y+ ⎛ Q2 ⎞ 1 ⎜ ⋅ = 6.72⋅ ft ⎜ 2 2 ⎝ 2⋅ b ⋅ g ⎠ y ⎛ Q2 ⎞ 1 y+⎜ ⋅ = 4.33⋅ ft ⎜ 2 ⋅ b2⋅ g y 2 ⎝ ⎠ For the shallow depth y = 1 ⋅ ft y+ ⎛ Q2 ⎞ 1 ⎜ ⋅ = 2.55⋅ ft ⎜ 2 2 ⎝ 2⋅ b ⋅ g ⎠ y For For y = 0.6⋅ ft ⎛ Q2 ⎞ 1 y+⎜ ⋅ = 4.92⋅ ft ⎜ 2 ⋅ b2⋅ g y 2 ⎝ ⎠ For y = 0.65⋅ ft For y = 0.645 ⋅ ft y+ Hence y = 0.645 ⋅ ft For ⎛ Q2 ⎞ 1 ⎜ ⋅ = 4.38⋅ ft ⎜ 2 2 y 2 ⋅ b ⋅ g ⎝ ⎠ Problem 11.20 Given: Trapezoidal channel Find: Critcal depth [Difficulty: 2] Solution: 2 Basic equation: Ey V 2 g The critical depth occurs when the specific energy is minimized For a trapezoidal channel (Table 11.1) A y ( b cot(α) y ) Hence for V Using this in Eq. 11.14 E is a minimum when Q V A Ey dE dy Q y ( b cot(α) y ) 2 Q 1 y ( b cot(α) y) 2 g 2 2 Q cot(α) 1 2 g y ( b y cot(α)) 2 2 g y ( b y cot(α)) This can be simplified to 3 Q g y ( b y cot(α)) This expression is the simplest one for y; it is implicit 3 3 g y ( b y cot(α)) 2 Q ( b 2 y cot(α)) 3 3 g y ( b y cot(α)) 2 Q cot(α) Hence we obtain for y 3 Q 1 2 1 2 0 Problem 11.21 Given: Data on trapezoidal channel Find: Critical depth [Difficulty: 3] Solution: 2 E=y+ Basic equation: V 2⋅ g In Section 11-2 we prove that the minimum specific energy is when we have critical flow; here we rederive the minimum energy point For a trapezoidal channel (Table 11.1) A = ( b + cot(α)⋅ y ) ⋅ y Q V= Hence for V A = E=y+ Using this in the basic equation ( b + cot(α)⋅ y ) ⋅ y b = 10⋅ ft and Q = 400 ⋅ =1− dy 2 2 ⎝1⎠ ft α = 71.6 deg 3 s g ⋅ y ⋅ ( b + y ⋅ cot(α)) 3 Q − 3 g ⋅ y ⋅ ( b + y ⋅ cot(α)) 2 =0 2 Q ⋅ cot(α) 2 3⎞ 2 Q ⋅ cot(α) g ⋅ y ⋅ ( b + y ⋅ cot(α)) Hence we obtain for y α = atan⎛⎜ 2 Q ⎡ ⎤ ⋅ 1 ⎢ ( b + cot(α)⋅ y ) ⋅ y⎥ 2 ⋅ g ⎣ ⎦ 2 dE E is a minimum when Q and 3 + 2 Q 3 g ⋅ y ⋅ ( b + y ⋅ cot(α)) 2 =1 Q ⋅ ( b + 2 ⋅ y ⋅ cot(α)) or 3 g ⋅ y ⋅ ( b + y ⋅ cot(α)) 3 This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below, to make the left side equal unity 2 y = 5 ⋅ ft Q ⋅ ( b + 2 ⋅ y ⋅ cot( α) ) 3 g ⋅ y ⋅ ( b + y ⋅ cot( α) ) 3 2 = 0.3 y = 4 ⋅ ft Q ⋅ ( b + 2 ⋅ y ⋅ cot( α) ) 3 g ⋅ y ⋅ ( b + y ⋅ cot( α) ) 3 3 g ⋅ y ⋅ ( b + y ⋅ cot( α) ) 2 y = 3.5⋅ ft Q ⋅ ( b + 2 ⋅ y ⋅ cot( α) ) Q ⋅ ( b + 2 ⋅ y ⋅ cot( α) ) 3 g ⋅ y ⋅ ( b + y ⋅ cot( α) ) 3 = 0.7 2 = 1.03 y = 3.55⋅ ft Q ⋅ ( b + 2 ⋅ y ⋅ cot( α) ) 3 g ⋅ y ⋅ ( b + y ⋅ cot( α) ) 2 y = 3.53⋅ ft 3 = 1.00 The critical depth is y = 3.53⋅ ft 3 = 0.98 =1 Problem 11.22 Given: Data on venturi flume Find: Flow rate [2] Solution: Basic equation: p1 ρ⋅ g + V1 2 p2 V2 2 + y1 = + + y2 2⋅ g ρ⋅ g 2⋅ g At each section Q = V⋅ A = V⋅ b ⋅ y The given data is b 1 = 2 ⋅ ft The Bernoulli equation applies because we have steady, incompressible, frictionless flow V= or y 1 = 1 ⋅ ft Q b⋅ y b 2 = 1 ⋅ ft y 2 = 0.75⋅ ft 2 Hence the Bernoulli equation becomes (with p 1 = p 2 = p atm) Solving for Q Q = ( 2⋅ g⋅ y1 − y2 2 ) ⎛ ⎞ −⎛ ⎞ ⎜ b ⋅y ⎜ b ⋅y ⎝ 2 2⎠ ⎝ 1 1⎠ 1 1 2 ⎛ Q ⎞ ⎜ b ⋅y ⎝ 1 1⎠ + y = 1 2⋅ g Q = 3.24⋅ ft 3 s 2 ⎛ Q ⎞ ⎜ b ⋅y ⎝ 2 2⎠ + y 2 2⋅ g Problem 11.23 Given: Data on rectangular channel and a bump Find: Elevation of free surface above the bump [Difficulty: 3] Solution: p1 Basic equation: ρ g V1 2 2 g p2 V2 2 y1 y 2 h The Bernoulli equation applies because we have steady, 2 g ρ g incompressible, frictionless flow. Note that at location 2 (the bump), the potential is y 2 + h, where h is the bump height 2 E Recalling the specific energy V 2 g y and noting that p1 = p 2 = p atm, the Bernoulli equation becomes Q At each section Q V A V b y or V The given data is b 10 ft y 1 1 ft h 4 in Q V1 b y1 ft V1 10 s Hence we find E1 and V1 b y Q 100 ft 3 s 2 2 g y1 V2 E1 2.554 ft 2 2 E1 E2 h y2 h 2 g Hence E1 E2 h 2 Q 2 2 g b y2 y2 h 2 Q or 2 2 g b y2 2 y 2 E1 h This is a nonlinear implicit equation for y 2 and must be solved numerically. We can use one of a number of numerical root finding techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We select y 2 so the left side of the equation equals E1 h 2.22 ft 2 For y 2 1 ft Q 2 2 g b y2 2 y 2 2.55 ft 2 For y 2 2.19 ft 2 For y 2 1.5 ft 2 For y 2 1.4 ft Q 2 2 g b y2 Q V2 b y2 so we have Fr 1 V1 g y1 2 2 g b y2 2 y 2 2.19 ft 2 y 2 2.22 ft 2 y 2 1.3 ft y 2 1.30 ft Hence Note that Q ft V2 7.69 s Fr 1 1.76 and Fr 2 V2 g y2 Fr 2 1.19 Q 2 2 g b y2 Problem 11.24 Given: Data on rectangular channel and a bump Find: Local change in flow depth caused by the bump [Difficulty: 3] Solution: Basic equation: p1 ρ g V1 2 2 g p2 V2 2 y1 y 2 h The Bernoulli equation applies because we have steady, 2 g ρ g incompressible, frictionless flow. Note that at location 2 (the bump), the potential is y 2 + h, where h is the bump height 2 E Recalling the specific energy V y 2 g Q At each section Q V A V b y or V The given data is b 10 ft y 1 1 ft h 0.25 ft Q 20 Q V1 b y1 ft V1 2 s y2 h 2 or Hence we find and Hence E1 V1 E1 E2 h and noting that p1 = p 2 = p atm, the Bernoulli equation becomes b y ft 3 s 2 2 g y1 E1 1.062 ft V2 2 2 E1 E2 h y2 h 2 g Q 2 2 g b y2 2 Q 2 2 g b y2 y 2 E1 h 2 This is a nonlinear implicit equation for y 2 and must be solved numerically. We can use one of a number of numerical root finding techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We select y 2 so the left side of the equation equals E1 h 0.812 ft 2 For y 2 0.75 ft 2 Q 2 2 g b y2 y 2 0.861 ft 2 For y 2 0.797 ft 2 For y 2 0.7 ft Q 2 2 g b y2 2 For Hence y 2 0.65 ft 2 g b y2 y 2 0.676 ft Note that Q V2 b y2 so we have Fr 1 V1 g y1 y 2 0.827 ft 2 y 2 0.812 ft 2 Q 2 2 and y 2 0.676 ft y2 y1 y1 Q 2 2 g b y2 32.4 % ft V2 2.96 s Fr 1 0.353 and Fr 2 V2 g y2 Fr 2 0.634 Problem 11.25 Given: Data on rectangular channel and a bump Find: Local change in flow depth caused by the bump [Difficulty: 3] Solution: Basic equation: p1 ρ g V1 2 2 g p2 V2 2 y1 y2 h 2 g ρ g The Bernoulli equation applies because we have steady, incompressible, frictionless flow. Note that at location 2 (the bump), the potential is y 2 + h, where h is the bump height 2 Recalling the specific energy E V 2 g y Q At each section Q V A V b y or V The given data is b 10 ft y 1 0.3 ft h 0.1 ft Q V1 b y1 ft V1 6.67 s Hence we find E1 and V1 b y Q 20 ft 3 s 2 2 g y1 E1 0.991 ft V2 2 2 E1 E2 h y2 h 2 g Hence E1 E2 h and noting that p1 = p 2 = p atm, the Bernoulli equation becomes 2 Q 2 2 g b y2 y2 h 2 or Q 2 2 g b y2 2 y 2 E1 h This is a nonlinear implicit equation for y 2 and must be solved numerically. We can use one of a number of numerical root finding techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We select y 2 so the left side of the equation equals E1 h 0.891 ft 2 For y 2 0.3 ft Q 2 2 g b y2 2 y 2 0.991 ft 2 For y 2 0.901 ft 2 For y 2 0.35 ft 2 For Hence y 2 0.33 ft Q 2 2 g b y2 Q V2 b y2 so we have Fr 1 V1 g y1 2 2 g b y2 2 y 2 0.857 ft 2 y 2 0.891 ft 2 y 2 0.334 ft Note that Q y 2 0.334 ft y2 y1 and y1 11.3 % ft V2 5.99 s Fr 1 2.15 and Fr 2 V2 g y2 Fr 2 1.83 Q 2 2 g b y2 Problem 11.26 Given: Data on wide channel Find: Stream depth after rise [Difficulty: 3] Solution: p1 Basic equation: ρ g V1 2 2 g p2 V2 2 y1 y2 h 2 g ρ g The Bernoulli equation applies because we have steady, incompressible, frictionless flow. Note that at location 2 (the bump), the potential is y 2 + h, where h is the bump height 2 Recalling the specific energy E V y 2 g At each section Q V A V1 b y 1 V2 b y 2 The given data is y 1 2 ft Hence Then E1 V1 E1 E2 h and noting that p1 = p 2 = p atm, the Bernoulli equation becomes y1 V2 V1 y2 ft V1 3 s h 0.5 ft 2 2 g y1 V2 E1 2.14 ft 2 2 E1 E2 h y2 h 2 g V1 y 1 2 2 g 2 1 y2 y2 h 2 or V1 y 1 2 2 g 1 y2 2 y 2 E1 h This is a nonlinear implicit equation for y 2 and must be solved numerically. We can use one of a number of numerical root finding techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We select y 2 so the left side of the equation equals E1 h 1.64 ft 2 For y 2 2 ft V1 y 1 y 2 1.3 ft Hence y 2 1.31 ft Note that y1 V2 V1 y2 so we have Fr 1 V1 g y1 V1 y 1 2 g 2 1 2 g 2 For 2 y2 y 2 2.14 ft 2 y 2 1.5 ft For 2 V1 y 1 y2 2 y 2 1.63 ft y 2 1.31 ft For ft V2 4.58 s Fr 1 0.37 and Fr 2 V2 g y2 Fr 2 0.71 2 g 2 1 2 V1 y 1 2 g 1 y2 2 y 2 1.75 ft 2 y 2 1.64 ft 2 1 y2 Problem 11.27 Given: Data on sluice gate Find: Water level upstream; Maximum flow rate [Difficulty: 2] Solution: Basic equation: p1 ρ⋅ g + V1 2 p2 V2 2 + y1 = + + y2 + h 2⋅ g ρ⋅ g 2⋅ g The Bernoulli equation applies because we have steady, incompressible, frictionless flow. Noting that p 1 = p 2 = p atm, and V 1 is approximately zero (1 = upstream, 2 = downstream) the Bernoulli equation becomes y1 = The given data is Q b V2 2 + y2 2⋅ g 2 = 10⋅ m y 2 = 1.25⋅ m s Hence Q = V2 ⋅ A2 = V2 ⋅ b ⋅ y 2 Then upstream 2 ⎞ ⎜⎛ V2 y1 = ⎜ + y2 ⎝ 2⋅ g ⎠ Q V2 = b⋅ y2 or m V2 = 8 s y 1 = 4.51 m The maximum flow rate occurs at critical conditions (see Section 11-2), for constant specific energy In this case V2 = Vc = Hence we find y1 = Hence yc = g⋅ yc 2 g⋅ yc 3 + yc = + yc = ⋅ yc 2⋅ g 2⋅ g 2 Vc 2 3 ⋅ y1 y c = 3.01 m Vc = 3 m Q b = Vc⋅ y c Q b = 16.3⋅ s m (Maximum flow rate) g⋅ yc m Vc = 5.43 s Problem 11.28 Given: Data on sluice gate Find: Flow rate [Difficulty: 2] Solution: Basic equation: p1 ρ⋅ g 2 + V1 p2 2 V2 + y1 = + + y2 2⋅ g ρ⋅ g 2⋅ g The Bernoulli equation applies because we have steady, incompressible, frictionless flow. Noting that p 1 = p 2 = p atm, (1 = upstream, 2 = downstream) the Bernoulli equation becomes 2 V1 2⋅ g 2 V2 + y1 = + y2 2⋅ g The given data is b = 3 ⋅ ft y 1 = 6⋅ ft y 2 = 0.9⋅ ft Also Q = V⋅ A so Q V1 = b ⋅ y1 2 ⎛ Q ⎞ ⎜ b⋅ y ⎝ 1⎠ + y = 1 2⋅ g Using these in the Bernoulli equation 2 Solving for Q Note that Q = 2 2⋅ g⋅ b ⋅ y1 ⋅ y2 y1 + y2 2 Q = 49.5⋅ ft and Q V2 = b ⋅ y2 2 ⎛ Q ⎞ ⎜ b⋅ y ⎝ 2⎠ + y 2 2⋅ g 3 s Q V1 = b⋅ y1 ft V1 = 2.75⋅ s Fr 1 = Q V2 = b⋅ y2 ft V2 = 18.3⋅ s Fr 2 = V1 g⋅ y1 V2 g⋅ y2 Fr 1 = 0.198 Fr 2 = 3.41 Problem 11.29 Given: Data on sluice gate Find: Water depth and velocity after gate [Difficulty: 2] Solution: E1 = Basic equation: y3 y2 = V1 2 2⋅ g 1 2 p2 V2 + y1 = + = E2 2⋅ g ρ⋅ g ⎛ ⎝ ⋅ −1 + 1 + 8 ⋅ Fr 2 For the gate 2⎞ For the jump (state 2 before, state 3 after) ⎠ m V1 = 0.2⋅ s y 1 = 1.5⋅ m The given data is 2 2 q = y 1 ⋅ V1 Hence Then we need to solve V2 q = 0.3 m s E1 = 2 + y 2 = E1 2⋅ g q or V1 2 2⋅ g + y1 E1 = 1.50 m 2 2⋅ g⋅ y2 2 + y 2 = E1 with E1 = 1.50 m We can solve this equation iteratively (or use Excel's Goal Seek or Solver) 2 y 2 = 0.5⋅ m ⎛q⎞ ⎜y ⎝ 2 ⎠ + y = 0.518 m 2 2⋅ g y 2 = 0.055 ⋅ m ⎛q⎞ ⎜y ⎝ 2 ⎠ + y = 1.57 m 2 2⋅ g For y 2 = 0.0563⋅ m ⎛q⎞ ⎜y ⎝ 2 ⎠ + y = 1.50 m 2 2⋅ g Then q V2 = y2 m V2 = 5.33 s For 2 For y 2 = 0.05⋅ m ⎛q⎞ ⎜y ⎝ 2 ⎠ + y = 1.89 m 2 2⋅ g y 2 = 0.057 ⋅ m ⎛q⎞ ⎜y ⎝ 2 ⎠ + y = 1.47 m 2 2⋅ g 2 For 2 For 2 Hence Note that y 2 = 0.056 m Fr 2 = is the closest to three figs. V2 g⋅ y2 Fr 2 = 7.17 Problem 11.30 [Difficulty: 2] Given: Rectangular channel flow with hump and/or side wall restriction Find: Whether critical flow occurs 1 Solution: Basic equations: ⎛ Q2 ⎞ yc = ⎜ ⎜ 2 ⎝ g⋅ b ⎠ 3 2 Q E=y+ 2 3 A = b⋅ y Emin = h = 350 ⋅ mm Q = 2.4⋅ 2⋅ g⋅ A 2 ⋅ yc (From Example 11.4) 3 Given data: b = 2⋅ m y = 1⋅ m h = 35⋅ cm E1 = y + 2 (a) For a hump with Then for the bump Ebump = E1 − h Q 2⋅ g⋅ b 2 ⋅ 1 y m s E1 = 1.07 m 2 Ebump = 0.723 m (1) 1 ⎡⎢ ⎛ Q 2⎥⎤ ⎞ ⎢ ⎜⎝ b ⎠ ⎥ yc = ⎢ ⎣ g ⎥⎦ For the minimum specific energy 3 y c = 0.528 m Emin = 3 2 ⋅ yc Emin = 0.791 m (2) Comparing Eqs. 1 and 2 we see that the bump IS sufficient for critical flow (b) For the sidewall restriction with b const = 1.5⋅ m as in Example 11.4 we have Econst = E1 Econst = 1.073 m (3) 1 With b const: ⎡ ⎛ Q ⎞ 2⎤ ⎢ ⎥ ⎢ ⎜⎝ b const ⎠ ⎥ yc = ⎢ ⎥ g ⎣ ⎦ 3 y c = 0.639 m Eminconst = 3 2 ⋅ yc Eminconst = 0.959 m (4) Comparing Eqs. 3 and 4 we see that the constriction is NOT sufficient for critical flow (c) For both, following Example 11.4 Eboth = E1 − h Eboth = 0.723 m (5) Eminboth = Eminconst Eminboth = 0.959 m (6) Comparing Eqs. 5 and 6 we see that the bump AND constriction ARE sufficient for critical flow (not surprising, as the bump alone is sufficient!) Problem 11.31 Given: Hydaulic jump data Find: Energy consumption; temperature rise [Difficulty: 2] Solution: Basic equations: P ρ g Hl Q (1) Hl is the head loss in m of fluid); multiplying by ρg produces energy/vol; multiplying by Q produces energy/time, or power Urate ρ Q cH2O ∆T (2) Urate is the rate of increase of internal energy of the flow; cH20∆T is the energy increase per unit mass due to a ∆T temperature rise; multiplying by ρQ converts to energy rise of the entire flow/time 3 Given data: From Eq. 1 From Example 11.5 P ρ g Hl Q Equating Eqs. 1 and 2 Q 9.65 m Hl 0.258 m s kg ρ 999 and 3 cH2O 1 m P 24.4 kW kg K a significant energy consumption ρ g Hl Q ρ Q cH2O ∆T or ∆T g Hl ∆T 6.043 10 cH2O The power consumed by friction is quite large, but the flow is very large, so the rise in temperature is insignificant. In English units: P 32.7 hp kcal 5 Q 1.53 10 gpm ∆T 1.088 10 3 ∆°F 4 ∆°C Problem 11.32 Given: Data on rectangular channel and hydraulic jump Find: Flow rate; Critical depth; Head loss [Difficulty: 2] 1 Solution: Basic equations: The given data is 2⎞ ⎛ = ⋅ −1 + 1 + 8⋅ Fr 1 ⎝ ⎠ y1 2 2 2 ⎛⎜ V1 ⎞ ⎛⎜ V2 ⎞ Hl = E1 − E2 = ⎜ y 1 + − ⎜y + 2⋅ g ⎠ ⎝ 2 2⋅ g ⎠ ⎝ b = 4⋅ m y 1 = 0.4⋅ m y2 1 y 2 = 1.7⋅ m 2 1 + 8⋅ Fr 1 = 1 + 2⋅ We can solve for Fr 1 from the basic equation ⎛ Q2 ⎞ yc = ⎜ ⎜ g⋅ b 2 ⎝ ⎠ y2 y1 2 Fr 1 = Hence y2 ⎞ ⎛ −1 ⎜ 1 + 2⋅ y1 ⎝ ⎠ 8 V1 = Fr 1 ⋅ g ⋅ y 1 Fr 1 = 3.34 Fr 1 = and V1 g⋅ y1 m V1 = 6.62 s 3 Then Q = V1 ⋅ b ⋅ y 1 Q = 10.6⋅ m s 1 3 The critical depth is ⎛ Q2 ⎞ yc = ⎜ ⎜ g⋅ b2 ⎝ ⎠ Also Q V2 = b⋅ y2 The energy loss is 2 2 ⎛⎜ V2 ⎞ V1 ⎞ ⎛⎜ − ⎜y + Hl = ⎜ y 1 + 2⋅ g ⎠ 2⋅ g ⎠ ⎝ 2 ⎝ Note that we could used y c = 0.894 m m V2 = 1.56 s Hl = Fr 2 = V2 g⋅ y2 Hl = 0.808 m ( y2 − y1)3 4⋅ y1⋅ y2 Hl = 0.808 m Fr 2 = 0.381 3 Problem 11.33 Given: Data on wide channel and hydraulic jump Find: Jump depth; Head loss [Difficulty: 2] Solution: Basic equations: 2 2 ⎛⎜ V1 ⎞ ⎛⎜ V2 ⎞ Hl = E1 − E2 = ⎜ y 1 + − ⎜y + 2⋅ g ⎠ ⎝ 2 2⋅ g ⎠ ⎝ 2⎞ ⎛ = ⋅ −1 + 1 + 8⋅ Fr 1 ⎝ ⎠ y1 2 y2 1 3 m The given data is Q b = 10⋅ s m Also Q = V⋅ A = V⋅ b ⋅ y Hence Q V1 = b⋅ y1 Then y2 = ⎛ 2 ⎝ y1 ⋅ −1 + y 1 = 1⋅ m m V1 = 10.0 s 1 + 8 ⋅ Fr 1 2⎞ The energy loss is Note that we could use 2 2 ⎛⎜ V1 ⎞ ⎛⎜ V2 ⎞ Hl = ⎜ y 1 + − ⎜y + 2⋅ g ⎠ ⎝ 2 2⋅ g ⎠ ⎝ Hl = ( y2 − y1)3 4⋅ y1⋅ y2 V1 g⋅ y1 Fr 1 = 3.19 y 2 = 4.04 m ⎠ Q V2 = b⋅ y2 Fr 1 = m V2 = 2.47 s Fr 2 = V2 g⋅ y2 Hl = 1.74 m Hl = 1.74 m Fr 2 = 0.393 Problem 11.34 Given: Data on wide channel and hydraulic jump Find: Jump depth [Difficulty: 1] Solution: y2 Basic equations: y1 ⎛ 2 ⎝ 1 = ⋅ −1 + 1 + 8 ⋅ Fr 1 2⎞ ⎠ 3 m The given data is Q b = 2⋅ s y 1 = 500 ⋅ mm m Also Q = V⋅ A = V⋅ b ⋅ y Hence Q V1 = b⋅ y1 Then y2 = Note: Q V2 = b⋅ y2 ⎛ 2 ⎝ y1 ⋅ −1 + m V1 = 4.00 s 1 + 8 ⋅ Fr 1 2⎞ Fr 1 = V1 g⋅ y1 Fr 1 = 1.806 y 2 = 1.05⋅ m ⎠ ft V2 = 6.24⋅ s Fr 2 = V2 g⋅ y2 Fr 2 = 0.592 Problem 11.35 Given: Data on wide channel and hydraulic jump Find: Jump depth; Head loss [Difficulty: 2] Solution: Basic equations: The given data is 1 Q = 200 ⋅ ft 3 s Also Q = V⋅ A = V⋅ b ⋅ y Hence Q V1 = b⋅ y1 Then y2 = ⎛ 2 ⎝ y1 ⋅ −1 + Q V2 = b⋅ y2 The energy loss is Note that we could use 2 2 ⎛⎜ V1 ⎞ ⎛⎜ V2 ⎞ Hl = E1 − E2 = ⎜ y 1 + − ⎜y + 2⋅ g ⎠ ⎝ 2 2⋅ g ⎠ ⎝ 2⎞ ⎛ = ⋅ −1 + 1 + 8⋅ Fr 1 ⎝ ⎠ y1 2 y2 1 + 8 ⋅ Fr 1 b = 10⋅ ft y 1 = 1.2⋅ ft ft V1 = 16.7⋅ s Fr 1 = 2⎞ V1 g⋅ y1 Fr 1 = 2.68 y 2 = 3.99⋅ ft ⎠ ft V2 = 5.01⋅ s 2 2 ⎛⎜ V1 ⎞ ⎛⎜ V2 ⎞ Hl = ⎜ y 1 + − ⎜y + 2⋅ g ⎠ ⎝ 2 2⋅ g ⎠ ⎝ Hl = Fr 2 = V2 g⋅ y2 Hl = 1.14⋅ ft ( y2 − y1)3 4⋅ y1⋅ y2 Hl = 1.14⋅ ft Fr 2 = 0.442 Problem 11.36 Given: Data on wide channel and hydraulic jump Find: Flow rate; Head loss [Difficulty: 2] Solution: Basic equations: The given data is 2⎞ ⎛ = ⋅ −1 + 1 + 8 ⋅ Fr 1 ⎝ ⎠ y1 2 2 2 V1 ⎞ ⎜⎛ V2 ⎞ ⎜⎛ Hl = E1 − E2 = ⎜ y 1 + − ⎜y + 2⋅ g ⎠ ⎝ 2 2⋅ g ⎠ ⎝ b = 5 ⋅ ft y 1 = 0.66⋅ ft y2 1 We can solve for Fr 1 from the basic equation 2 1 + 8 ⋅ Fr 1 = 1 + 2 ⋅ y 2 = 3.0⋅ ft y2 y1 2 Fr 1 = Hence Then Also y2 ⎞ ⎛ −1 ⎜ 1 + 2⋅ y1 ⎝ ⎠ 8 Fr 1 = 3.55 ft g⋅ y1 3 Q = V1 ⋅ b ⋅ y 1 Q = 54.0⋅ Q V2 = b⋅ y2 ft V2 = 3.60⋅ s s 2 ⎛⎜ V2 ⎞ − ⎜y + ⎠ ⎝ 2 2⋅ g ⎠ Fr 2 = V2 g⋅ y2 2⎞ Note that we could use V1 ft V1 = 16.4⋅ s V1 = Fr 1 ⋅ g ⋅ y 1 ⎛⎜ V1 The energy loss is Hl = ⎜ y 1 + 2⋅ g ⎝ Fr 1 = and Hl = 1.62⋅ ft Hl = ( y2 − y1)3 4⋅ y1⋅ y2 Hl = 1.62⋅ ft Fr 2 = 0.366 Problem 11.37 [Difficulty: 2] Given: Data on wide spillway flow Find: Depth after hydraulic jump; Specific energy change Solution: Basic equations: 2⎞ ⎛ = ⋅ −1 + 1 + 8 ⋅ Fr 1 ⎠ y1 2 ⎝ y2 1 m V1 = 25 s The given data is y 1 = 0.9⋅ m Then Fr 1 is Fr 1 = Hence y2 = Then Q = V1 ⋅ b ⋅ y 1 = V2 ⋅ b ⋅ y 2 For the specific energies V1 ⎛ 2 ⎝ ⋅ −1 + V1 Note that we could use 1 + 8 ⋅ Fr 1 2⎞ ⎠ y 2 = 10.3 m y1 V2 = V1 ⋅ y2 m V2 = 2.19 s 2 E1 = y 1 + 2⋅ g V2 The energy loss is Fr 1 = 8.42 g⋅ y1 y1 2 2 V1 ⎞ ⎜⎛ V2 ⎞ ⎜⎛ Hl = E1 − E2 = ⎜ y 1 + − ⎜y + 2⋅ g ⎠ ⎝ 2 2⋅ g ⎠ ⎝ E1 = 32.8 m 2 E2 = y 2 + 2⋅ g E2 = 10.5 m Hl = E1 − E2 Hl = 22.3 m Hl = ( y2 − y1)3 4⋅ y1⋅ y2 E2 E1 = 0.321 Hl = 22.3⋅ m Problem 11.38 [Difficulty: 2] Given: Data on rectangular channel flow Find: Depth after hydraulic jump; Specific energy change Solution: 2⎞ ⎛ = ⋅ −1 + 1 + 8 ⋅ Fr 1 ⎝ ⎠ y1 2 2 2 V1 ⎞ ⎜⎛ V2 ⎞ ⎜⎛ Hl = E1 − E2 = ⎜ y 1 + − ⎜y + 2⋅ g ⎠ ⎝ 2 2⋅ g ⎠ ⎝ y 1 = 0.4⋅ m b = 1⋅ m Then Q = V1 ⋅ b ⋅ y 1 = V2 ⋅ b ⋅ y 2 Q V1 = b⋅ y1 Then Fr 1 is Fr 1 = Hence y2 = and Q V2 = b⋅ y2 Basic equations: y2 1 3 The given data is For the specific energies V1 y1 2 ⎛ ⎝ ⋅ −1 + Note that we could use 1 + 8 ⋅ Fr 1 2⎞ ⎠ y 2 = 4.45 m m V2 = 1.46 s V1 2 E1 = y 1 + 2⋅ g V2 The energy loss is Fr 1 = 8.20 g⋅ y1 E1 = 13.9 m 2 E2 = y 2 + 2⋅ g E2 = 4.55 m Hl = E1 − E2 Hl = 9.31 m Hl = ( y2 − y1)3 4⋅ y1⋅ y2 Hl = 9.31⋅ m Q = 6.5 m s m V1 = 16.3 s Problem 11.39 Given: Data on sluice gate Find: Water depth before and after the jump [Difficulty: 3] Solution: E1 = Basic equation: y3 y2 = V1 2 2⋅ g 1 2 p2 V2 + y1 = + = E2 2⋅ g ρ⋅ g ⎛ ⎝ ⋅ −1 + 1 + 8 ⋅ Fr 2 For the gate 2⎞ For the jump (state 2 before, state 3 after) ⎠ m V1 = 0.2⋅ s y 1 = 1.5⋅ m The given data is 2 2 q = y 1 ⋅ V1 Hence Then we need to solve V2 q = 0.3 m E1 = s 2 + y 2 = E1 2⋅ g q or V1 2 2⋅ g + y1 E1 = 1.50 m 2 2⋅ g⋅ y2 2 + y 2 = E1 with E1 = 1.50 m We can solve this equation iteratively (or use Excel's Goal Seek or Solver) 2 y 2 = 0.5⋅ m ⎛q⎞ ⎜y ⎝ 2 ⎠ + y = 0.518 m 2 2⋅ g y 2 = 0.055 ⋅ m ⎛q⎞ ⎜y ⎝ 2 ⎠ + y = 1.57 m 2 2⋅ g For y 2 = 0.0563⋅ m ⎛q⎞ ⎜y ⎝ 2 ⎠ + y = 1.50 m 2 2⋅ g Then q V2 = y2 m V2 = 5.33 s For 2 For y 2 = 0.05⋅ m ⎛q⎞ ⎜y ⎝ 2 ⎠ + y = 1.89 m 2 2⋅ g y 2 = 0.057 ⋅ m ⎛q⎞ ⎜y ⎝ 2 ⎠ + y = 1.47 m 2 2⋅ g 2 For 2 For 2 For the jump (States 2 to 3) y3 = ⎛ 2 ⎝ y2 ⋅ −1 + Hence Note that 1 + 8 ⋅ Fr 2 2⎞ ⎠ y 2 = 0.056 m Fr 2 = y 3 = 0.544 m is the closest to three figs. V2 g⋅ y2 Fr 2 = 7.17 Problem 11.40 Given: Surge wave Find: Surge speed [Difficulty: 3] V2 At rest y1 Solution: 2 Basic equations: V1 ⋅ y 1 g + y1 2 2 = 2 V2 ⋅ y 2 g + y2 V 2 = VSurge 2 2 (This is the basic momentum equation for the flow) V1 ⋅ y 1 = V2 ⋅ y 2 Then 2 2 y 2 − y 1 = ⋅ ⎛ V1 ⋅ y 1 − V2 ⋅ y 2⎞ = ⎠ g ⎝ 2 2 2 2 2 ⋅ V2 2 y2 y1 2 2 ⎡ ⎡ V 2 ⎤ ⎤ ⎢⎛ 1 ⎞ ⎥ 2 ⋅ V2 ⎢⎛ y2 ⎞ ⎥ ⋅ y1 − y2 = ⋅ y1 − y2 ⋅ ⎜ ⋅ ⎜ ⎢ V ⎥ ⎢ y ⎥ g g ⎣⎝ 2 ⎠ ⎦ ⎣⎝ 1 ⎠ ⎦ 2 ⋅ V2 2 2 2 y2 y2 + y1 = 2⋅ ⋅ g y1 g⋅ y1 = ⎛ 2 ⎞ 2 ⋅ V 2⋅ y y − y 2 2 ( 2 1) ⎜ y2 ⋅ − y2 = ⋅ g ⎜ y1 g y1 ⎝ ⎠ V2 V2 = But V2 2 y2 − y1 = Dividing by (y 2 - y 1) V1 or ⎛ y1 ⎞ ⎝ y2 ⋅⎜1 + V2 = VSurge ( or y2 + y1 g 2 V2 = ⋅ y 1 ⋅ 2 y2 so VSurge = ) ⎠ g⋅ y1 2 ⎛ y1 ⎞ ⎝ y2 ⋅⎜1 + ⎠ y2 Problem 11.41 Given: Tidal bore Find: Speed of undisturbed river [Difficulty: 3] At rest V1 = Vr + Vbore Solution: 2 Basic equations: V2 ⋅ y 2 g + y2 2 2 2 V1 ⋅ y 1 = g + y1 y2 y1 2 2 (This is the basic momentum equation for the flow) V2 V2 ⋅ y 2 = V1 ⋅ y 1 or Given data Vbore = 18⋅ mph ft Vbore = 26.4⋅ s Then 2 2 y 1 − y 2 = ⋅ ⎛ V2 ⋅ y 2 − V1 ⋅ y 1⎞ = ⎝ ⎠ g 2 2 2 2 2 2 ⋅ V1 y1 − y2 = Dividing by (y 2 - y 1) 2 y2 y 1 = 8 ⋅ ft y 2 = y 1 + 12⋅ ft y 2 = 20⋅ ft 2 ⎡ 2 ⎡ V ⎞2 ⎤ ⎤ ⎢⎛ 2 ⎥ 2 ⋅ V1 ⎢⎛ y1 ⎞ ⎥ ⋅ ⎜ ⋅ y2 − y1 = ⋅ ⎜ ⋅ y2 − y1 ⎢ ⎥ ⎢ ⎥ g g ⎣⎝ V1 ⎠ ⎦ ⎣⎝ y2 ⎠ ⎦ 2 ⋅ V1 2 2 2 y1 y1 + y2 = 2⋅ ⋅ g y2 g⋅ y2 y1 ⎛ 2 ⎞ 2 ⋅ V 2⋅ y y − y 1 1 ( 1 2) ⎜ y1 ⋅ − y1 = ⋅ ⎜ g g y2 ⎝ y2 ⎠ V1 V1 = But V1 = or ⎛ y2 ⎞ ⎝ y1 ⋅⎜1 + V1 = Vr + Vbore ⎠ or ( y1 + y2 g 2 V1 = ⋅ y 2 ⋅ 2 y1 ) ft V1 = 33.6⋅ s V1 = 22.9⋅ mph Vr = V1 − Vbore ft Vr = 7.16⋅ s Vr = 4.88⋅ mph Problem 11.42 Given: Rectangular channel flow Find: Discharge [Difficulty: 1] Solution: Basic equation: Q 1 n 2 1 3 2 A Rh Sb Note that this is an "engineering" equation, to be used without units! For a rectangular channel of width b 2 m and depth y 1.5 m we find from Table 11.1 2 A b y A 3.00 m n 0.015 Manning's roughness coefficient is Q 1.49 2 1 3 2 A Rh Sb n Rh and b y b 2 y Sb 0.0005 3 Q 3.18 m s Rh 0.600 m Problem 11.43 Given: Data on rectangular channel Find: Depth of flow [Difficulty: 3] Solution: Basic equation: Q= 1 2 1 3 2 ⋅ A⋅ Rh ⋅ Sb n Note that this is an "engineering" equation, to be used without units! 3 For a rectangular channel of width b = 2.5⋅ m and flow rate Q = 3 ⋅ Manning's roughness coefficient is n = 0.015 Q= 1 n b⋅ y ⋅ b ⋅ y ⋅ ⎛⎜ s we find from Table 11.1 A = b⋅ y R= b⋅ y b + 2⋅ y Sb = 0.0004 and 2 Hence the basic equation becomes m 1 3 ⎞ ⋅S 2 b ⎝ b + 2⋅ y ⎠ 2 3 ⎞ = Q⋅ n 1 ⎝ b + 2⋅ y ⎠ y ⋅ ⎛⎜ Solving for y b⋅ y b ⋅ Sb 2 This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below, to make the Q⋅ n left side evaluate to = 0.900 . 1 b ⋅ Sb 2 2 For y = 1 ( m) y ⋅ ⎛⎜ b⋅ y 2 3 ⎞ = 0.676 ⎝ b + 2⋅ y ⎠ For y = 1.2 ( m) y ⋅ ⎛⎜ b⋅ y 2 For y = 1.23 ( m) The solution to three figures is y ⋅ ⎛⎜ b⋅ y 2 3 ⎞ = 0.894 ⎝ b + 2⋅ y ⎠ 3 ⎞ = 0.865 ⎝ b + 2⋅ y ⎠ For y = 1.24 ( m) y = 1.24 (m) y ⋅ ⎛⎜ b⋅ y 3 ⎞ = 0.904 ⎝ b + 2⋅ y ⎠ Problem 11.44 Given: Data on trapzoidal channel Find: Depth of flow [Difficulty: 3] Solution: Basic equation: Q= 1.49 2 1 3 2 ⋅ A⋅ Rh ⋅ Sb n Note that this is an "engineering" equation, to be used without units! α = atan ⎛⎜ b = 8 ⋅ ft For the trapezoidal channel we have 1⎞ ⎝ 2⎠ α = 26.6deg Q = 100⋅ ft 3 S0 = 0.0004 s n = 0.015 A = y ⋅ ( b + y ⋅ cot ( α ) ) = y ⋅ ( 8 + 2⋅ y ) Hence from Table 11.1 Q= Hence 1.49 2 1 3 2 ⋅ A⋅ Rh ⋅ Sb n Rh = 2 = 1.49 0.015 ⋅ y ⋅ ( 8 + 2 ⋅ y ) ⋅ y ⋅ ⎡⎢ y⋅ ( 8 + 2⋅ y)⎤ y ⋅ ( b + y ⋅ cot(α)) 2⋅ y b + = y ⋅ ( 8 + 2⋅ y ) 8 + 2⋅ y ⋅ 5 sin ( α ) 1 3 2 ⎥ ⋅ 0.0004 = 100(Note that we don't use units!) ⎣ 8 + 2⋅ y⋅ 5 ⎦ 5 Solving for y [ y⋅ ( 8 + 2⋅ y) ] 3 2 = 50.3 ( 8 + 2⋅ y⋅ 5) 3 This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. 5 For For y = 2 y = 2.6 ( ft) ( ft) [ y⋅ ( 8 + 2⋅ y) ] 5 3 2 = 30.27 For y = 3 ( ft) 3 2 ( 8 + 2⋅ y⋅ 5) 3 ( 8 + 2⋅ y⋅ 5) 3 5 5 [ y⋅ ( 8 + 2⋅ y) ] 3 2 = 49.81 For y = 2.61 ( ft) ( 8 + 2⋅ y⋅ 5) 3 The solution to three figures is [ y⋅ ( 8 + 2⋅ y) ] [ y⋅ ( 8 + 2⋅ y) ] 3 2 ( 8 + 2⋅ y⋅ 5) 3 y = 2.61 (ft) = 65.8 = 50.18 Problem 11.45 Given: Data on trapezoidal channel Find: Depth of flow [Difficulty: 3] Solution: Basic equation: Q= 1 2 1 3 2 ⋅ A⋅ Rh ⋅ Sb n Note that this is an "engineering" equation, to be used without units! b = 2.5⋅ m For the trapezoidal channel we have α = atan⎛⎜ 1⎞ ⎝2⎠ 3 α = 26.6 deg Q = 3⋅ m S0 = 0.0004 s n = 0.015 A = y ⋅ ( b + cot( α) ⋅ y ) = y ⋅ ( 8 + 2 ⋅ y ) Hence from Table 11.1 R= y ⋅ ( b + y ⋅ cot( α) ) b+ 2 Q= Hence 1 n 2 1 3 ⋅ A⋅ Rh ⋅ Sb 2 = 1 0.015 ⋅ y ⋅ ( 2.5 + 2 ⋅ y ) ⋅ ⎡⎢ ( 2.5 + 2 ⋅ y ) ⋅ y⎤ 2⋅ y = y ⋅ ( 2.5 + 2 ⋅ y ) 2.5 + 2 ⋅ y ⋅ 5 cot( α) 1 3 2 ⎥ ⋅ 0.0004 = 3 ⎣ 2.5 + 2⋅ y ⋅ 5 ⎦ (Note that we don't use units!) 5 Solving for y [ y ⋅ ( 2.5 + 2 ⋅ y ) ] 3 2 = 2.25 ( 2.5 + 2⋅ y⋅ 5) 3 This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. 5 For For y = 1 y = 0.81 ( m) ( m) [ y ⋅ ( 2.5 + 2 ⋅ y ) ] 5 3 2 = 3.36 For y = 0.8 ( m) 3 2 ( 2.5 + 2⋅ y⋅ 5) 3 ( 2.5 + 2⋅ y⋅ 5) 3 5 5 [ y ⋅ ( 2.5 + 2 ⋅ y ) ] 3 2 = 2.23 For y = 0.815 ( m) ( 2.5 + 2⋅ y⋅ 5) 3 The solution to three figures is [ y ⋅ ( 2.5 + 2 ⋅ y ) ] [ y ⋅ ( 2.5 + 2 ⋅ y ) ] 3 2 ( 2.5 + 2⋅ y⋅ 5) 3 y = 0.815 (m) = 2.17 = 2.25 Problem 11.46 Given: Data on flume Find: Discharge [Difficulty: 1] Solution: Basic equation: Q 1.49 n 2 1 3 2 A Rh Sb Note that this is an "engineering" equation, to be used without units! For a rectangular channel of width b 6 ft and depth y 3 ft we find from Table 11.1 A b y A 18 ft n 0.013 For concrete (Table 11.2) Q 2 1.49 2 1 3 2 A Rh Sb n Rh and b y b 2 y 1 ft Sb 1000 ft Q 85.5 ft 3 s Rh 1.50 ft Sb 0.001 Problem 11.47 Given: Data on flume Find: Slope [Difficulty: 1] Solution: Basic equation: Q 1.49 2 1 3 2 A Rh Sb n Note that this is an "engineering" equation, to be used without units! For a rectangular channel of width b 3 ft and depth y 6 ft we find A b y A 18 ft 2 Rh b y Rh 1.20 ft b 2 y n 0.0145 For wood (not in Table 11.2) a Google search finds n = 0.012 to 0.017; we use Sb n Q 2 1.49 A R 3 h 2 Sb 1.86 10 3 with Q 90 ft 3 s Problem 11.48 Given: Data on square channel Find: Dimensions for concrete and soil cement [Difficulty: 2] Solution: Basic equation: Q 1 n 2 1 3 2 A Rh Sb Note that this is an "engineering" equation, to be used without units! A b For a square channel of width b we find 2 R b y b 2 y b b 2 b 3 2 Hence 1 8 1 2 3 Sb b 3 2 Q b Sb b n 3 2 1 2 n 3 3 or 2 3 3 Q b n 1 Sb 2 3 The given data is Q 20 m s For concrete, from Table 11.2 (assuming large depth) Sb 0.003 n .013 b 2.36 m For soil cement from Table 11.2 (assuming large depth) n .020 b 2.77 m 2 8 b 3 Problem 11.49 Given: Data on trapezoidal channel Find: Bed slope [Difficulty: 1] Solution: Basic equation: Q 1 2 1 3 2 A Rh Sb n Note that this is an "engineering" equation, to be used without units! 3 For the trapezoidal channel we have b 2.4 m α 45 deg For bare soil (Table 11.2) n 0.020 Hence from Table 11.1 A y ( b cot( α) y ) Hence Sb Q n 2 A R 3 h y 1.2 m 2 A 4.32 m 2 Sb 1.60 10 Rh 3 Q 7.1 y ( b y cot( α) ) b 2 y sin( α) m s Rh 0.746 m Problem 11.50 Given: Data on triangular channel Find: Required dimensions [Difficulty: 1] Solution: Basic equation: Q= 1 n 2 1 3 2 ⋅ A⋅ Rh ⋅ Sb Note that this is an "engineering" equation, to be used without units! 3 α = 45⋅ deg Sb = 0.001 For concrete (Table 11.2) n = 0.013 (assuming y > 60 cm: verify later) Hence from Table 11.1 A = y ⋅ cot( α) = y For the triangular channel we have 2 2 Hence Q= 1 3 ⋅ A⋅ Rh ⋅ Sb n 2 Rh = y ⋅ cos( α) 2 2 1 2 = 1 n ⋅ y ⋅ ⎛⎜ 2 3 Q = 10⋅ = m s y 2⋅ 2 8 1 1 8 1 3 ⎞ ⋅S = 1 ⋅y 3 ⋅⎛ 1 ⎞ ⋅S 2 = 1 ⋅y 3 ⋅S 2 ⎜8 b n b b 2⋅ n ⎝ ⎠ ⎝ 2⋅ 2 ⎠ y 3 Solving for y 2⋅ n⋅ Q ⎞ y = ⎛⎜ ⎝ Sb ⎠ 8 y = 2.20 m (The assumption that y > 60 cm is verified) Problem 11.51 Given: Data on semicircular trough Find: Discharge [Difficulty: 2] Solution: Basic equation: Q 1 2 1 3 2 A Rh Sb n Note that this is an "engineering" equation, to be used without units! For the semicircular channel D 1 m y 0.25 m Hence, from geometry y D 2 α 2 asin 180 deg D 2 α 120 deg n 0.022 For corrugated steel, a Google search leads to Hence from Table 11.1 1 A 8 Rh Then the discharge is Q 1 n 2 A 0.154 m D Rh 0.147 m ( α sin( α) ) D 1 4 Sb 0.01 1 sin( α) α 2 1 3 3 2 m A Rh Sb s 2 3 Q 0.194 m s Problem 11.52 Given: Data on semicircular trough Find: Discharge [Difficulty: 1] Solution: Basic equation: Q= 1 2 1 3 2 ⋅ A⋅ Rh ⋅ Sb n Note that this is an "engineering" equation, to be used without units! For the semicircular channel D = 1⋅ m α = 180 ⋅ deg For corrugated steel, a Google search leads to (Table 11.2) Hence from Table 11.1 1 A = 8 Rh = Then the discharge is Q= 1 n 2 A = 0.393 m ⋅D Rh = 0.25 m ⋅ ( α − sin( α) ) ⋅ D 1 4 ⋅ ⎛⎜ 1 − ⎝ sin( α) ⎞ α n = 0.022 ⎠ 2 1 3 3 2 m ⋅ A⋅ Rh ⋅ Sb ⋅ s 2 3 Q = 0.708 m s Sb = 0.01 Problem 11.53 Given: Data on flume with plastic liner Find: Depth of flow [Difficulty: 3] Solution: Basic equation: Q= 1.49 n 2 1 3 2 ⋅ A⋅ Rh ⋅ Sb Note that this is an "engineering" equation, to be used without units! For a rectangular channel of width b = 6 ⋅ ft and depth y we find from Table 11.1 A = b⋅ y = 6⋅ y n = 0.010 and also R= 1 ⋅ ft Sb = 1000⋅ ft and 2 Q= Hence 1.49 n 3 2 2 = 1.49 0.010 6⋅ y ⋅ 6 ⋅ y ⋅ ⎛⎜ y 1 ⎞ ⋅ 0.001 2 = 85.5 (Note that we don't use units!) ⎝ 6 + 2⋅ y ⎠ 5 3 2 ( 6 + 2⋅ y) 6⋅ y 6 + 2⋅ y 3 5 Solving for y = Sb = 0.001 1 ⋅ A⋅ Rh ⋅ Sb b⋅ y b + 2⋅ y 3 85.5⋅ 0.010 = or 1 2 2 3 1.49⋅ .001 ⋅ 6 ⋅ 6 y 3 2 ( 6 + 2⋅ y) = 0.916 3 This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We start with Problem 11.46's depth 5 For y = 3 ( feet) y 5 3 2 ( 6 + 2⋅ y) = 1.191 For y = 2 ( feet) 3 y 2 ( 6 + 2⋅ y) 5 For y = 2.5 ( feet) y 2 = 0.931 For y = 2.45 ( feet) 3 y = 2.47 ( feet) y 3 2 ( 6 + 2⋅ y) 3 y 3 3 2 ( 6 + 2⋅ y) 5 For = 0.684 5 3 ( 6 + 2⋅ y) 3 = 0.916 y = 2.47 (feet) 3 = 0.906 Problem 11.54 Given: Data on trapzoidal channel Find: New depth of flow Solution: Basic equation: Q= 1 n 2 1 3 2 ⋅ A⋅ Rh ⋅ Sb [Difficulty: 3] Note that this is an "engineering" equation, to be used without units! 3 b = 2.4⋅ m For the trapezoidal channel we have α = 45⋅ deg For bare soil (Table 11.2) n = 0.020 Hence from Table 11.1 A = y ⋅ ( b + cot( α) ⋅ y ) = y ⋅ ( 2.4 + y ) Q = 10⋅ R= m y ⋅ ( b + y ⋅ cot( α) ) b+ 2 Q= Hence 1 2 1 3 ⋅ A⋅ Rh ⋅ Sb n 2 = 1 0.020 ⋅ y ⋅ ( 2.4 + y ) ⋅ ⎡⎢ y ⋅ ( 2.4 + y ) Sb = 0.00193 s 2⋅ y = y ⋅ ( 2.4 + y ) 2.4 + 2 ⋅ y ⋅ 2 sin( α) 1 3 2 ⎤ ⎥ ⋅ 0.00193 = 10 ⎣ 2.4 + 2 ⋅ y⋅ 2⎦ (Note that we don't use units!) 5 Solving for y [ y ⋅ ( 2.4 + y ) ] 3 2 = 4.55 ( 2.4 + 2⋅ y⋅ 2) 3 This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We start with a larger depth than Problem 11.49's. 5 For y = 1.5 ( m) [ y ⋅ ( 2.4 + y ) ] 5 3 2 = 5.37 For y = 1.4 ( m) ( 2.4 + 2⋅ y⋅ 2) 3 [ y ⋅ ( 2.4 + y ) ] 2 y = 1.35 ( m) [ y ⋅ ( 2.4 + y ) ] 5 3 2 = 4.41 For y = 1.37 ( m) ( 2.4 + 2⋅ y⋅ 2) 3 The solution to three figures is = 4.72 ( 2.4 + 2⋅ y⋅ 2) 3 5 For 3 [ y ⋅ ( 2.4 + y ) ] 3 2 ( 2.4 + 2⋅ y⋅ 2) 3 y = 1.37 (m) = 4.536 Problem 11.55 Given: Data on trapzoidal channel Find: New depth of flow Solution: Basic equation: Q= 1 2 1 3 2 ⋅ A⋅ Rh ⋅ Sb n [Difficulty: 3] Note that this is an "engineering" equation, to be used without units! 3 b = 2.4⋅ m For the trapezoidal channel we have α = 45⋅ deg For bare soil (Table 11.2) n = 0.010 Hence from Table 11.1 A = y ⋅ ( b + y ⋅ cot( α) ) = y ⋅ ( 2.4 + y ) Q= Hence 1 2 1 3 2 ⋅ A⋅ Rh ⋅ Sb n Q = 7.1⋅ Rh = 2 = 1 0.010 ⋅ y ⋅ ( 2.4 + y ) ⋅ ⎡⎢ y ⋅ ( 2.4 + y ) m Sb = 0.00193 s y ⋅ ( b + y ⋅ cot( α) ) b+ = 2⋅ y y ⋅ ( 2.4 + y ) 2.4 + 2 ⋅ y ⋅ 2 sin( α) 1 3 2 ⎤ ⎥ ⋅ 0.00193 = 7.1 (Note that we don't use units!) ⎣ 2.4 + 2 ⋅ y⋅ 2⎦ 5 Solving for y [ y ⋅ ( 2.4 + y ) ] 3 2 = 1.62 ( 2.4 + 2⋅ y⋅ 2) 3 This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We start with a shallower depth than that of Problem 11.49. 5 For y = 1 ( m) [ y ⋅ ( 2.4 + y ) ] 5 3 2 = 2.55 For y = 0.75 ( m) ( 2.4 + 2⋅ y⋅ 2) 3 [ y ⋅ ( 2.4 + y ) ] 2 y = 0.77 ( m) [ y ⋅ ( 2.4 + y ) ] 5 3 2 = 1.60 For y = 0.775 ( m) ( 2.4 + 2⋅ y⋅ 2) 3 The solution to three figures is = 1.53 ( 2.4 + 2⋅ y⋅ 2) 3 5 For 3 [ y ⋅ ( 2.4 + y ) ] 3 2 ( 2.4 + 2⋅ y⋅ 2) 3 y = 0.775 (m) = 1.62 Problem 11.56 Given: Data on semicircular trough Find: New depth of flow [Difficulty: 4] Solution: Q= Basic equation: 1 2 1 3 2 ⋅ A⋅ Rh ⋅ Sb n Note that this is an "engineering" equation, to be used without units! 3 D = 1⋅ m For the semicircular channel Sb = 0.01 2 Q= Hence − Solving for α 1 A= α 1 1 8 ⋅ ( α − sin( α) ) 3 ⋅ A⋅ Rh ⋅ Sb n 2 Rh = 2 1 2 3 8 2 ⋅ ( α − sin( α) ) ⋅ D = = m s n = 0.022 For corrugated steel, a Google search leads to (Table 11.2) From Table 11.1 Q = 0.5⋅ 1 4 ⋅ ⎛⎜ 1 − ⎝ sin( α) ⎞ α ⎠ ⋅D = 1 4 ⋅ ⎜⎛ 1 − ⎝ sin( α) ⎞ α 1 3 sin( α) ⎞⎤ 1 1 2 ⋅ 0.01 = 0.5 (Note that we don't use units!) ⋅ ⎢⎡ ⋅ ( α − sin( α) )⎤⎥ ⋅ ⎡⎢ ⋅ ⎛⎜ 1 − ⎥ 0.022 ⎣ 8 α ⎠⎦ ⎦ ⎣4 ⎝ 1 5 ⋅ ( α − sin( α) ) 3 = 2.21 This is a nonlinear implicit equation for α and must be solved numerically. We can use one of a number of numerical root finding techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We start with a half-full channel − For α = 180 ⋅ deg α − For α = 159 ⋅ deg α 2 3 5 ⋅ ( α − sin( α) ) 3 − = 3.14 For α = 160 ⋅ deg 5 2 3 ⋅ ( α − sin( α) ) 3 y = D 2 ⋅ ⎜⎛ 1 − cos⎛⎜ ⎝ α − = 2.20 For α = 159.2 ⋅ deg The solution to three figures is α = 159 ⋅ deg From geometry ⎠ α ⎞⎞ ⎝ 2 ⎠⎠ y = 0.410 m α 2 3 5 ⋅ ( α − sin( α) ) = 2.25 5 2 3 3 ⋅ ( α − sin( α) ) 3 = 2.212 Problem 11.57 [Difficulty: 3] Given: Triangular channel Find: Proof that wetted perimeter is minimized when sides meet at right angles Solution: From Table 11.1 2 A = y ⋅ cot( α) P= 2⋅ y sin( α) y= We need to vary z to minimize P while keeping A constant, which means that Hence we eliminate y in the expression for P For optimizing P dP dα or =− 2 ⋅ ( A⋅ cos( α) − A⋅ sin( α) ⋅ tan( α) ) sin( 2 ⋅ α) ⋅ A⋅ tan( α) A⋅ cos( α) − A⋅ sin( α) ⋅ tan( α) = 0 P = 2⋅ A ⋅ A with A = constant cot( α) 1 cot( α) sin( α) =0 1 tan( α) = tan( α) tan( α) = 1 α = 45⋅ deg For α = 45o we find from the figure that we have the case where the sides meet at 90o. Note that we have only proved that this is a minimum OR maximum of P! It makes sense that it's the minimum, as, for constant A, we get a huge P if we set α to a large number (almost vertical walls); hence we can't have a maximum value at α = 45o. Problem 11.58 Given: Data on trapezoidal channel Find: Normal depth and velocity [Difficulty: 3] Solution: Basic equation: Q= 1.49 2 1 3 2 ⋅ A⋅ Rh ⋅ Sb n Note that this is an "engineering" equation, to be used without units! b = 20⋅ ft For the trapezoidal channel we have α = atan( 2 ) α = 63.4 deg A = y ⋅ ( b + y ⋅ cot( α) ) = y ⋅ ⎛⎜ 20 + Hence from Table 11.2 ⎝ 1 2 Q = 400 ⋅ ⋅ y⎞ ⎠ Hence 3 s Rh = 2 Sb = 0.0016 n = 0.025 y ⋅ ( b + y ⋅ cot( α) ) b+ 2⋅ y y ⋅ ⎜⎛ 20 + ⎝ = 1 2 ⋅ y⎞ 20 + y ⋅ 5 sin( α) 3 1 ⎡ y ⋅ ⎛ 20 + 1 ⋅ y⎞ ⎤ ⎜ ⎢ ⎥ 1 1 1 2 ⎠ 2 3 2 ⎝ Q = ⋅ A⋅ Rh ⋅ Sb = ⋅ y ⋅ ⎛⎜ 20 + ⋅ y⎞ ⋅ ⎢ ⎥ ⋅ 0.0016 = 400 n 0.025 ⎝ 2 ⎠ ⎣ 20 + y ⋅ 5 ⎦ 2 ft 1 (Note that we don't use units!) 5 Solving for y 3 ⎡y⋅ ⎛ 20 + 1 ⋅ y⎞⎤ ⎢ ⎜ ⎥ 2 ⎠⎦ ⎣ ⎝ = 250 This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We start with an arbitrary depth 2 ( 20 + y⋅ 5) 3 5 For y = 5 ( ft) ⎡y⋅ ⎛ 20 + 1 ⋅ y⎞⎤ ⎢ ⎜ ⎥ 2 ⎠⎦ ⎣ ⎝ 5 3 = 265 2 For y = 4.9 ( ft) ( 20 + y⋅ 5) 3 3 ⎡y⋅ ⎛ 20 + 1 ⋅ y⎞⎤ ⎢ ⎜ ⎥ 2 ⎠⎦ ⎣ ⎝ = 256 2 ( 20 + y⋅ 5) 3 5 For y = 4.85 ( ft) ⎡y⋅ ⎛ 20 + 1 ⋅ y⎞⎤ ⎢ ⎜ ⎥ 2 ⎠⎦ ⎣ ⎝ 2 5 3 = 252 For y = 4.83 ( ft) ( 20 + y⋅ 5) 3 The solution to three figures is y = 4.83⋅ ft Finally, the normal velocity is V = Q A ⎡y⋅ ⎛ 20 + 1 ⋅ y⎞⎤ ⎢ ⎜ ⎥ 2 ⎠⎦ ⎣ ⎝ 2 ( 20 + y⋅ 5) 3 Then A = ( b + y ⋅ cot( α) ) ⋅ y V = 3.69⋅ ft s A = 108 ⋅ ft 2 3 = 250 ⎠ Problem 11.59 Given: Data on trapezoidal channel Find: Geometry for greatest hydraulic efficiency Solution: Basic equation: Q= 1 n 2 1 3 2 ⋅ A⋅ Rh ⋅ Sb [Difficulty: 5] Note that this is an "engineering" equation, to be used without units! For the trapezoidal channel we have From Table 11.1 α = atan⎛⎜ 3 1⎞ α = 26.6⋅ deg ⎝2⎠ A = y ⋅ ( b + y ⋅ cot( α) ) Q = 250⋅ P=b + m Sb = 0.001 s n = 0.020 2⋅ y sin ( α ) We need to vary b and y to obtain optimum conditions. These are when the area and perimeter are optimized. Instead of two independent variables b and y, we eliminate b by doing the following b= Taking the derivative w.r.t. y But at optimum conditions Hence ∂ ∂y ∂ ∂y A y P= 0=− Hence Then A y − y ⋅ cot ( α ) + 2⋅ y sin ( α ) 2 1 ∂ A − cot ( α ) + ⋅ A − sin ( α ) y ∂y 2 y ∂ and A − cot( α) + 2 2⋅ y sin( α) ∂y 2 or sin( α) A = y ⋅ ( b + y ⋅ cot( α) ) b= P= and so P=0 y Comparing to − y ⋅ cot( α) we find A= 2⋅ y 2 sin( α) 2 − y ⋅ cot( α) A = y ⋅ ( b + y ⋅ cot( α) ) = 2⋅ y 2 sin( α) 2 − y ⋅ cot( α) − 2 ⋅ y ⋅ cot( α) A = y ⋅ ( b + y ⋅ cot( α) ) = y ⋅ ⎜⎛ 2⋅ y ⎝ sin( α) P=b+ A=0 2⋅ y sin( α) = 4⋅ y sin( α) − 2 ⋅ y ⋅ cot( α) + y ⋅ cot( α) ⎞ = y ⋅ ⎜⎛ 2 ⎠ − 2 ⋅ y ⋅ cot( α) = 2 ⋅ y ⋅ ⎜⎛ 2 ⎝ sin( α) 2 ⎝ sin( α) − cot( α) ⎞ ⎠ − cot( α) ⎞ ⎠ y ⋅ ⎜⎛ 2 and Rh = A P = ⎝ sin( α) ⎠ = y 2 2 − cot( α) ⎞ 2 ⋅ y ⋅ ⎛⎜ ⎝ sin( α) ⎠ 2 Hence Q= 1 n − cot( α) ⎞ 2 3 ⋅ A⋅ Rh ⋅ Sb 2 1 2 = 1 n ⋅ ⎡⎢y ⋅ ⎛⎜ 2 2 ⎣ ⎝ sin( α) − cot( α) ⎞⎥⎤ ⋅ ⎛⎜ ⎠⎦ y⎞ ⎝2⎠ 1 3 ⋅ Sb 2 1 8 2 3 y ⋅ Sb 2 ⎛ ⎞ Q= ⎜ − cot( α) ⋅ 2 ⎝ sin( α) ⎠ n⋅ 2 3 3 Solving for y Finally 2 ⎡ ⎤ ⎢ ⎥ 3 2 ⋅ n⋅ Q ⎢ ⎥ y = ⎢ 1⎥ ⎢ 2 ⎥ 2 − cot( α) ⎞ ⋅ Sb ⎥ ⎢ ⎛⎜ ⎣ ⎝ sin( α) ⎠ ⎦ b = 2⋅ y sin( α) − 2 ⋅ y ⋅ cot( α) 8 y = 5.66 (m) b = 2.67 (m) Problem 11.60 Given: Data on trapezoidal channel Find: Normal depth Solution: Q= Basic equation: 1.49 2 1 3 2 ⋅ A⋅ Rh ⋅ Sb n [Difficulty: 3] Note that this is an "engineering" equation, to be used without units! b = 20⋅ ft For the trapezoidal channel we have α = atan⎛⎜ ⎞ ⎝ 1.5 ⎠ 1 α = 33.7 deg Q = 1000⋅ ft 3 s S0 = 0.0002 n = 0.014 A = y ⋅ ( b + y ⋅ cot( α) ) = y ⋅ ( 20 + 1.5⋅ y ) Hence from Table 11.1 Hence Q= 1.49 2 1 3 2 ⋅ A⋅ Rh ⋅ Sb n Rh = 2 = 1.49 0.014 ⋅ y ⋅ ( 20 + 1.5⋅ y ) ⋅ ⎡⎢ y ⋅ ( 20 + 1.5⋅ y ) ⎤ y ⋅ ( b + y ⋅ cot( α) ) b+ 2⋅ y = y ⋅ ( 20 + 1.5⋅ y ) 20 + 2 ⋅ y ⋅ 3.25 sin( α) 1 3 2 ⎥ ⋅ 0.0002 = 1000 (Note that we don't use units!) ⎣ 20 + 2 ⋅ y⋅ 3.25⎦ 5 Solving for y [ ( 20 + 1.5⋅ y ) ⋅ y ] 3 2 ( 20 + 2⋅ y⋅ 3.25) = 664 3 This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. 5 For y = 7.5 ( ft) [ ( 20 + 1.5⋅ y ) ⋅ y ] 3 2 ( 20 + 2⋅ y⋅ 5 3.25) = 684 For y = 7.4 ( ft) [ ( 20 + 1.5⋅ y ) ⋅ y ] 2 ( 20 + 2⋅ y⋅ 3 3.25) 5 For y = 7.35 ( ft) [ ( 20 + 1.5⋅ y ) ⋅ y ] ( 20 + 2⋅ y⋅ The solution to three figures is 3.25) = 667 3 5 3 2 3 = 658 For y = 7.38 ( ft) [ ( 20 + 1.5⋅ y ) ⋅ y ] 2 ( 20 + 2⋅ y⋅ 3 y = 7.38 (ft) 3 3.25) 3 = 663 Problem 11.61 Given: Trapezoidal channel Find: Geometry for greatest hydraulic efficiency [Difficulty: 5] Solution: From Table 11.1 A = y ⋅ ( b + y ⋅ cot ( α ) ) P=b+ 2⋅ y sin( α) We need to vary b and y (and then α!) to obtain optimum conditions. These are when the area and perimeter are optimized. Instead of two independent variables b and y, we eliminate b by doing the following b = Taking the derivative w.r.t. y But at optimum conditions Hence ∂ ∂y ∂ ∂y A P=0 But ∂ ∂α − cot( α) + 2 2 cot( α) + 1 = We can now evaluate A from Eq 1 A= − y ⋅ cot ( α ) + 2⋅ y sin ( α ) 2 sin( α) A =0 2 )=0 2⋅ y 2 2 − y ⋅ cot( α) sin( α) − or cos( α) 2 2 2 sin( α) + cos( α) +1= sin( α) α = acos⎛⎜ 2 1⎞ 2⋅ y − y ⋅ cot( α) = 2 − 1 2 ⋅y = 3 3 = 2 ( 2 1 sin( α) 2 ⎛ 4 − 1 ⎞ ⋅ y2 = 3⋅ y2 ⎜ 3⎠ ⎝ 3 2 But for a trapezoid Comparing the two A expressions A = y ⋅ ( b + y ⋅ cot( α) ) = y ⋅ ⎛⎜ b + ⎝ A = ⎛⎜ b + ⎝ 1 3 ⋅ y⎞ ⋅ y = ⎠ 3⋅ y 2 1 3 ⋅ y⎞ ⎠ we find b= ) + cot( α) + 1 = 0 α = 60 deg ⎝2⎠ 2 2 (1) 2 ⋅ cos( α) sin( α) −2 ⋅ cos( α) = −1 2⋅ y ( 2 − y ⋅ −1 − cot( α) 2 sin( α) Hence y A= or sin( α) 2 ⋅ y ⋅ cos( α) sin( α) ∂y 2 2 A=− ∂ and A 0=− A P= and so 2 1 ∂ A − cot ( α ) + ⋅ A − sin ( α ) y ∂y 2 y P= y Now we optimize A w.r.t. α − y ⋅ cot ( α ) y 1 ⎞ 2 ⎛ ⋅y = ⋅y ⎜ 3− 3⎠ 3 ⎝ But the perimeter is In summary we have and P=b+ 2⋅ y sin( α) = b + 2⋅ y⋅ 2 3 =b+ 4 ⋅ y = b + 2⋅ b = 3⋅ b 3 α = 60 deg b= 1 3 P− ⋅P so each of the symmetric sides is 1 3 2 ⋅P = 1 3 ⋅P We have proved that the optimum shape is equal side and bottom lengths, with 60 angles i.e., half a hexagon! Problem 11.62 Given: Rectangular channel flow Find: Critical depth 1 Solution: Basic equations: [Difficulty: 1] ⎛ Q2 ⎞ yc = ⎜ ⎜ g⋅ b2 ⎝ ⎠ 3 Q= 1.49 2 1 3 2 ⋅ A⋅ Rh ⋅ Sb n For a rectangular channel of width b = 2 ⋅ m and depth y = 1.5⋅ m we find from Table 11.1 2 A = b⋅ y A = 3.00⋅ m n = 0.015 Manning's roughness coefficient is Q= 1.49 2 1 3 2 ⋅ A⋅ Rh ⋅ Sb n Rh = and 1 Hence ⎛ Q2 ⎞ yc = ⎜ ⎜ g⋅ b2 ⎝ ⎠ 3 y c = 0.637 m b⋅ y b + 2⋅ y Sb = 0.0005 3 Q = 3.18⋅ m s Rh = 0.600 ⋅ m Problem 11.63 Given: Data on rectangular channel and weir Find: If a hydraulic jump forms upstream of the weir [Difficulty: 4] 1 Solution: Q= Basic equations: 1 n 2 1 3 2 ⋅ A⋅ Rh ⋅ Sb ⎛ Q2 ⎞ yc = ⎜ ⎜ g⋅ b2 ⎝ ⎠ 3 Note that the Q equation is an "engineering" equation, to be used without units! For a rectangular channel of width b = 2.45⋅ m and depth y we find from Table 11.1 A = b ⋅ y = 2.45⋅ y b⋅ y Rh = b + 2⋅ y 2.45⋅ y = 2.45 + 2 ⋅ y 2 Q= Hence 1 n 3 and also n = 0.015 2 1 3 ⋅ A⋅ Rh ⋅ Sb 2 = 1 0.015 ⋅ 2.45⋅ y ⋅ ⎛⎜ 2.45⋅ y y Q = 5.66⋅ ⎞ ⋅ 0.0004 2 = 5.66 ⎝ 2.45 + 2⋅ y ⎠ 2 ( 2.45 + 2 ⋅ y ) 5.66⋅ 0.015 = 3 2 2 3 .0004 ⋅ 2.54⋅ 2.54 y or 1 (Note that we don't use units!) 3 2 ( 2.54 + 2 ⋅ y ) = 0.898 3 This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding techniques, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We start with the given depth 5 For y = 1.52 ( m) y 5 3 2 ( 2.54 + 2 ⋅ y ) = 0.639 For y = 2 ( m) 3 y y = 1.95 ( m) y ( 2.54 + 2 ⋅ y ) = 0.908 3 5 3 2 ( 2.54 + 2 ⋅ y ) 3 2 5 For = 0.879 For y = 1.98 3 ( m) y 3 2 ( 2.54 + 2 ⋅ y ) = 0.896 3 1 y = 1.98 (m) This is the normal depth. s 1 5 3 m 3 5 Solving for y Sb = 0.0004 and We also have the critical depth: ⎛ Q2 ⎞ yc = ⎜ ⎜ g⋅ b2 ⎝ ⎠ 3 y c = 0.816 m Hence the given depth is 1.52 m > y c, but 1.52 m < y n, the normal depth. This implies the flow is subcritical (far enough upstream it is depth 1.98 m), and that it draws down to 1.52 m as it gets close to the wier. There is no jump. Problem 11.64 Given: Data on rectangular flume Find: Optimum geometry [Difficulty: 2] Solution: Basic equations: Q 1.49 n 2 1 3 2 A Rh Sb and from Table 11.3, for optimum geometry b 2 yn Note that the Q equation is an "engineering" equation, to be used without units! Available data ft Sb 10 mile Sb 0.00189 A b yn 2 yn ft 3 s n 0.013 For wood (unplaned), a Google seach gives Hence Q 40 2 Rh A P 2 y n 2 y n 2 y n y n yn 2 2 2 Then 1 3 1 y 1.49 1.49 2 n 3 2 2 Q A Rh Sb 2 y n Sb n n 2 3 Solving for y n 2 Q n 2 3 yn 1 4 1.49 Sb 2 5 y n 2.00 (ft) b 2y n b 4.01 (ft) Problem 11.65 [Difficulty: 2] Given: Data on rectangular channel Find: Expressions valid for critical depth at optimum geometry Solution: Basic equations: Q= 1 n 2 1 3 2 ⋅ A⋅ Rh ⋅ Sb b = 2⋅ yn and from Table 11.3, for optimum geometry Note that the Q equation is an "engineering" equation, to be used without units! Hence A = b⋅ yn = 2⋅ yn 2 Rh = A P = 2⋅ yn yn + 2⋅ yn + yn 2 2 Then 1 1 1 3 2 y 1 1 2 ⎛ n⎞ 3 2 2 Q = ⋅ A⋅ Rh ⋅ Sb = ⋅ 2 ⋅ y n ⋅ ⎜ ⋅ Sb n n ⎝2⎠ or Q= or Fr = 2 3 = 8 1 3 2 ⋅ y ⋅ Sb n n yn 2 We can write the Froude number in terms of Q V Fr = g⋅ y = Q A⋅ g ⋅ y Q = 1 2 2⋅ yn ⋅ g⋅ yn 2 Q 5 2⋅ g⋅ yn 2 5 1= Hence for critical flow, Fr = 1 and y n = y c, so Q or 5 2⋅ g⋅ yc Q = 2⋅ g⋅ yc 5 2 Q = 6.26⋅ y c 2 24.7⋅ n 2 2 To find Sc, equate the expressions for Q and set Sb = Sc 1 Q= 2 3 8 1 3 2 ⋅ y ⋅ Sc n c 4 5 = 2⋅ g⋅ yc 2 or 3 − 2 Sc = 2 ⋅ g ⋅ n ⋅ y c 1 3 Sc = 1 yc 3 Problem 11.66 Given: Data on trapezoidal canal Find: Critical slope [Difficulty: 3] Solution: Q= Basic equations: 1.49 2 1 3 2 ⋅ A⋅ Rh ⋅ Sb n A = y ⋅ b + y ⋅ cot ( α ) and y ⋅ ( b + y ⋅ cot( α) ) Rh = b+ 2⋅ y sin( α) Note that the Q equation is an "engineering" equation, to be used without units! α = atan⎛⎜ b = 10⋅ ft Available data 2⎞ α = 63.4⋅ deg ⎝1⎠ Q = 600⋅ ft 3 s n = 0.015 For brick, a Google search gives For critical flow y = yc Vc = g⋅ yc so Q = A⋅ Vc = y c⋅ b + y c⋅ cot( α) ⋅ g ⋅ y c ( ) (yc⋅ b + yc⋅ cot( α))⋅ g⋅ yc = Q Q = 600⋅ with ft 3 s This is a nonlinear implicit equation for y c and must be solved numerically. We can use one of a number of numerical root finding techniques, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We start with the given depth For yc = 5 ( ft) (yc⋅ b + yc⋅ cot( α))⋅ g ⋅ y c = 666 For y c = 4.5 ( ft) (yc⋅ b + yc⋅ cot( α))⋅ g ⋅ y c = 569 For y c = 4.7 ( ft) (yc⋅ b + yc⋅ cot( α))⋅ g ⋅ y c = 607 For y c = 4.67 ( ft) (yc⋅ b + yc⋅ cot( α))⋅ g ⋅ y c = 601 Hence y c = 4.67 (ft) and Acrit = 49.0 (ft2) Rhcrit = 2.818 (ft) Acrit = y c⋅ b + y c⋅ cot( α) Rhcrit = ( y c⋅ b + y c⋅ cot( α) b+ Solving the basic equation for Sc Q= 1.49 2⋅ yc ) sin( α) 2 1 3 2 ⋅ A⋅ Rh ⋅ Sb n Sbcrit = n⋅ Q ⎛ ⎞ ⎜ 2 ⎜ ⎟ 3 ⎜ 1.49⋅ A ⋅ R crit hcrit ⎠ ⎝ 2 Sbcrit = 0.00381 Problem 11.67 Given: Data on wide channel Find: Critical slope [Difficulty: 2] Solution: Basic equations: Q 1.49 n 2 1 3 2 A Rh Sb A b y and Rh y Note that the Q equation is an "engineering" equation, to be used without units! 3 ft Available data q 20 s ft From Table 11.2 n 0.015 For critical flow y yc Vc g yc 2 Q A Vc b y c g y c so Hence Q b g 3 yc 3 q g y c 2.316 (ft) Solving the basic equation for Sc Sbcrit Q 1.49 Sbcrit 2 1 3 2 A Rh Sb n n Q 2 1.49 b y y 3 c c n q 5 1.49 y 3 c 1.49 2 1 3 2 b y c y c Sb n 2 Sbcrit n q 5 1.49 y 3 c n 0.013 Note from Table 11.2 that a better roughness is and then yc or 2 2 Sbcrit 0.00185 2 Sbcrit 0.00247 Problem 11.68 Given: Data on optimum rectangular channel Find: Channel width and slope [Difficulty: 2] Solution: Basic equations: Q 1.49 n 2 1 3 2 A Rh Sb b 2 yn and from Table 11.3, for optimum geometry Note that the Q equation is an "engineering" equation, to be used without units! Available data Hence Q 100 ft 3 n 0.015 s A b yn 2 yn 2 Rh A P (Table 11.2) 2 yn 2 yn 2 yn yn yn 2 We can write the Froude number in terms of Q Fr V g y Q A g y Q or 1 2 2 yn g yn Fr 2 Q 5 2 g yn 2 5 1 Hence for critical flow, Fr = 1 and y n = y c, so yc Q 2 g or Q 2 g yc (ft) and 5 2 g yc 2 Hence Q 2 2 5 y c 2.39 b 2 yc 2 2 Then Hence 1 3 Sc Using (from Table 11.2) n Q 8 1 1.49 2 3 y 3 c 1.49 1 1 y 2 c 3 2 2 Q A Rh Sb 2 yc Sc n n 2 1.49 b 4.78 or Q 1.49 2 n 3 8 1 3 2 y c Sc (ft) 2 Sc 0.00615 n 0.013 Sc n Q 8 1 1.49 2 3 y 3 c 2 Sc 0.00462 Problem 11.69 Given: Data on broad-crested wier Find: Maximum flow rate/width Solution: 3 Basic equation: Q = Cw⋅ b ⋅ H Available data H = 1 ⋅ ft 2 P = 8 ⋅ ft − 1 ⋅ ft 3 ft 3 Then [Difficulty: 1] Q b = q = Cw⋅ H 2 = 3.4⋅ s ft P = 7 ⋅ ft Cw = 3.4 Problem 11.70 Given: Data on rectangular, sharp-crested weir Find: Required weir height [Difficulty: 3] Solution: 3 Basic equations: 2 2 Q Cd 2 g b' H 3 where Given data: b 1.6 m Q 0.5 Cd 0.62 and b' b 0.1 n H n 2 with 3 Hence we find m s 3 3 2 2 2 2 Q Cd 2 g b' H Cd 2 g ( b 0.1 n H) H 3 3 3 Rearranging ( b 0.1 n H) H 2 3 Q 2 2 g Cd This is a nonlinear implicit equation for H and must be solved numerically. We can use one of a number of numerical root finding techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. 5 The right side evaluates to 3 Q 2 2 g Cd 0.273 m 2 3 For H 1 m ( b 0.1 n H) H 2 5 1.40 m 2 3 For H 0.3 m ( b 0.1 n H) H 2 H 0.31 m ( b 0.1 n H) H 2 0.253 m H 0.316 m But from the figure ( b 0.1 n H) H H P 2.5 m 2 H 0.5 m ( b 0.1 n H) H 2 0.265 m 2 For H 0.35 m ( b 0.1 n H) H 2 0.530 m 2 For H 0.315 m ( b 0.1 n H) H H 0.316 m P 2.5 m H P 2.18 m 2 2 5 0.317 m 3 5 0.273 m 2 5 3 5 3 For For 5 3 For 3 2 5 0.272 m 2 Problem 11.71 Given: Data on rectangular, sharp-crested weir Find: Discharge [Difficulty: 1] Solution: 3 Basic equation: Q Cw b H 2 where Cw 3.33 and b 8 ft Note that this is an "engineering" equation, to be used without units! 3 Q Cw b H 2 Q 26.6 ft 3 s P 2 ft H 1 ft Problem 11.72 Given: Data on rectangular, sharp-crested weir Find: Required weir height [Difficulty: 3] Solution: 3 Basic equations: 2 2 Q Cd 2 g b' H 3 where Given data: b 1.5 m Q 0.5 Cd 0.62 and b' b 0.1 n H with n 2 3 Hence we find m s 3 3 2 2 2 2 Q Cd 2 g b' H Cd 2 g ( b 0.1 n H) H 3 3 3 Rearranging ( b 0.1 n H) H 2 3 Q 2 2 g Cd This is a nonlinear implicit equation for H and must be solved numerically. We can use one of a number of numerical root finding techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. 5 The right side evaluates to For H 1 m 3 Q 2 2 g Cd 0.273 m ( b 0.1 n H) H 2 3 2 5 1.30 m 2 3 For H 0.3 m ( b 0.1 n H) H 2 H 0.34 m ( b 0.1 n H) H 2 0.237 m H 0.331 m But from the figure ( b 0.1 n H) H H P 2.5 m 2 H 0.5 m ( b 0.1 n H) H 2 0.284 m 2 For H 0.35 m ( b 0.1 n H) H 2 0.495 m 2 For H 0.33 m ( b 0.1 n H) H H 0.331 m P 2.5 m H P 2.17 m 2 2 5 0.296 m 3 5 0.273 m 2 5 3 5 3 For For 5 3 For 3 2 5 0.272 m 2 Problem 11.73 Given: Data on V-notch weir Find: Flow head [Difficulty: 1] Solution: 5 Basic equation: 8 θ 2 Q = Cd ⋅ ⋅ 2⋅ g ⋅ tan ⎛⎜ ⎞ ⋅ H 15 ⎝2⎠ where Cd = 0.58 2 H = 5 Q ⎛ ⎞ ⎜ θ 8 ⎜ Cd ⋅ ⋅ 2⋅ g ⋅ tan ⎛⎜ ⎞ 15 ⎝ ⎝2⎠⎠ H = 0.514m θ = 60⋅ deg Q = 150⋅ L s Problem 11.74 Given: Data on V-notch weir Find: Discharge [Difficulty: 1] Solution: 5 Basic equation: Q Cw H 2 where H 1.5 ft Cw 2.50 Note that this is an "engineering" equation in which we ignore units! 5 Q Cw H 2 Q 6.89 ft 3 s for θ 90 deg Problem 11.75 Given: Data on V-notch weir Find: Weir coefficient [Difficulty: 1] Solution: 5 Basic equation: Q Cw H 2 where H 180 mm Note that this is an "engineering" equation in which we ignore units! Cw Q 5 H 2 Cw 1.45 Q 20 L s Problem 12.1 Given: Air flow through a filter Find: Change in p, T and ρ [Difficulty: 2] Solution: Basic equations: h 2 h 1 c p T2 T1 p ρ R T Assumptions: 1) Ideal gas 2) Throttling process In a throttling process enthalpy is constant. Hence h2 h1 0 s o T2 T1 0 or T constant The filter acts as a resistance through which there is a pressure drop (otherwise there would be no flow. Hence p 2 p 1 From the ideal gas equation p1 p2 ρ1 T1 ρ2 T2 The governing equation for entropy is Hence p2 p1 ∆s R ln so T1 p 2 p2 ρ2 ρ1 ρ1 T2 p 1 p1 T2 p2 ∆s cp ln R ln T1 p1 p2 and 1 p1 Entropy increases because throttling is an irreversible adiabatic process Hence ρ2 ρ1 so ∆s 0 Problem 12.2 [Difficulty: 2] Problem 12.3 Given: Data on an air compressor Find: Whether or not the vendor claim is feasible [Difficulty: 2] Solution: Basic equation: T2 p2 ∆s cp ln R ln T1 p1 The data provided, or available in the Appendices, is: p 1 14.7 psi T1 ( 50 460 ) R p 2 ( 150 14.7) psi T2 ( 200 460 ) R Then BTU lb R T2 p2 Rgas ln ∆s cp ln T1 p1 Rgas 53.33 ∆s 0.1037 ft lbf lb R 0.0685 or for all real processes ∆s 0 lb R lb R We have plotted the actual process in red (1-2) on this temperature-entropy diagram, and the ideal compression (isentropic) in blue (1-2s). The line of constant pressure equal to 150 psig is shown in green. However, can this process actually occur? The second law of thermodynamics states that, for an adiabatic process ∆s 0 BTU BTU 2s Temperature T cp 0.2399 2 1 Hence the process is NOT feasible! Entropy s Problem 12.4 Given: Data on turbine inlet and exhaust Find: Whether or not the vendor claim is feasible [Difficulty: 2] Solution: Basic equation: T2 p2 ∆s cp ln R ln T1 p1 The data provided, or available in the Appendices, is: T1 ( 2200 460 ) R T1 1.478 10 K p 2 1 atm 14.696 psi T2 ( 850 460 ) R T2 727.778 K BTU Rgas 53.33 lb R T2 p2 ∆s cp ln Rgas ln T1 p1 ∆s 0.0121 An example of this type of process is plotted in green on the graph. Also plotted are an isentropic process (blue - 1-2s) and one with an increase in entropy (red: 1-2i). All three processes expand to the same pressure. The constant pressure curve is drawn in purple.The second law of thermodynamics states that, for an adiabatic process ∆s 0 or for all real processes ∆s 0 ft lbf lb R 0.0685 BTU lb R BTU lb R 1 Temperature T cp 0.2399 Then 3 p 1 10 atm 146.959 psi 2i 2 2s Hence the process is NOT feasible! Entropy s Problem 12.5 Given: Air before and after expansion; process Find: Final temperature and change in entropy [Difficulty: 2] Solution: Basic equations: T2 p2 ∆s cp ln R ln T1 p1 p V m R T The data provided, or available in the Appendices, is: p 1 50 psi T1 660 R p 2 1 atm 14.696 psi cp 0.2399 From the process given: p 1 V1 Btu Rgas 53.33 lb R 1.3 p 2 V2 1.3 ft lbf lb R 0.0685 From the ideal gas equation of state: Btu lb R p 2 V2 p 1 V1 T2 V1 T1 V2 p 2 T1 p 1 T2 1 p2 When we combine these two equations we get: p1 1 So the final temperature is: Then p1 T2 T1 p2 V1 V 2 1.3 p 2 T1 p 1 T2 1 1.3 T2 p2 ∆s cp ln Rgas ln T1 p1 T2 497.5 R ∆s 0.0161 Btu lb R 1.3 Solving for temperature ratio: T1 T2 p2 p 1 1.3 1 Problem 12.6 Given: Adiabatic air compressor Find: Lowest delivery temperature; Sketch the process on a Ts diagram [Difficulty: 2] Solution: Basic equation: T2 p2 ∆s cp ln R ln T1 p1 1 k The lowest temperature implies an ideal (reversible) process; it is also adiabatic, so Δs = 0, and The data provided, or available in the Appendices, is: p 1 101 kPa p 2 ( 500 101 ) kPa 1 k Hence p1 T2 T1 p2 Temperature T T2 864 R p2 2 The process is k p1 1 Entropy s p1 T2 T1 p2 T1 288.2 K k k 1.4 Problem 12.7 Given: Data on turbine inlet and exhaust Find: Whether or not the vendor claim is feasible [Difficulty: 2] k 1 Solution: Basic equations: T2 p2 ∆s cp ln R ln T1 p1 η h1 h2 h 1 h 2s p2 T1 p2 T2 ∆h cp ∆T k when s = constant T1 T2 T1 T2s The data provided, or available in the Appendices, is: 3 p 1 10 bar 1 10 kPa T1 1400 K η 80 % P 1 MW p 2 1 bar 100 kPa cp 1004 J kg K Rgas 287 J kg K k 1.4 k 1 p2 If the expansion were isentropic, the exit temperature would be: T2s T1 p1 Since the turbine is not isentropic, the final temperature is higher: Then kJ ∆h cp T1 T2 542.058 kg k 725.126 K T2 T1 η T1 T2s 860.101 K T2 p2 ∆s cp ln Rgas ln T1 p1 The mass flow rate is: ∆s 171.7157 m P ∆h J kg K 1.845 kg s Problem 12.8 Given: Test chamber with two chambers Find: Pressure and temperature after expansion [Difficulty: 2] Solution: Basic equation: p ρ R T ∆u q w (First law - closed system) ∆u cv ∆T ∆u 0 or for an Ideal gas Vol 2 2 Vol 1 so 1 ρ2 ρ1 2 so p2 Assumptions: 1) Ideal gas 2) Adiabatic 3) No work For no work and adiabatic the first law becomes We also have From the ideal gas equation M ρ Vol const p2 p1 Hence Note that and ρ2 T2 1 ρ1 T1 2 T2 20 °F p2 200 kPa 2 T2 p2 1 ∆s cp ln R ln R ln 0.693 R 2 T1 p1 ∆T 0 T2 T1 1 2 p1 p 2 100 kPa so entropy increases (irreversible adiabatic) Problem 12.9 [Difficulty: 2] Given: Supercharger Find: Pressure, temperature and flow rate at exit; power drawn Solution: Basic equation: T2 p2 ∆s cp ln R ln T1 p1 p ρ Rair T ∆h q w ∆h cp ∆T (First law - open system) Assumptions: 1) Ideal gas 2) Adiabatic In an ideal process (reversible and adiabatic) the first law becomes ∆h w or for an ideal gas wideal cp ∆T k 1 For an isentropic process T2 p2 ∆s 0 cp ln R ln T1 p1 The given or available data is T1 ( 70 460 ) R ft p2 T1 p1 T2 or p 1 14.7 psi p 2 ( 200 14.7) psi k 1.4 cp 0.2399 T2 1140 R T2 681 °F p 2 215 psi ρ1 Q2 Q1 ρ2 p 1 T2 Q2 Q1 p 2 T1 ft Q2 0.0737 s 3 Q1 0.5 s k Btu lbm R η 70 % Rair 53.33 ft lbf lbm R k 1 Hence p2 T2 p1 k T1 We also have mrate ρ1 Q1 ρ2 Q2 For the power we use Pideal mrate wideal ρ1 Q1 cp ∆ T p1 From the ideal gas equation ρ1 Rair T1 Hence ρ1 0.00233 ft Pideal ρ1 Q1 cp T2 T1 The actual power needed is Pactual Pideal η slug or 3 3 ρ1 0.0749 lbm ft Pideal 5.78 kW Pactual 8.26 kW A supercharger is a pump that forces air into an engine, but generally refers to a pump that is driven directly by the engine, as opposed to a turbocharger that is driven by the pressure of the exhaust gases. 3 Problem 12.10 [Difficulty: 2] Given: Cooling of air in a tank Find: Change in entropy, internal energy, and enthalpy Solution: Basic equation: p ρ R T T2 p2 ∆s cp ln R ln T1 p1 ∆u cv ∆T ∆h cp ∆T Assumptions: 1) Ideal gas 2) Constant specific heats Given or available data M 5 kg T1 ( 250 273) K cp 1004 J cv 717.4 kg K For a constant volume process the ideal gas equation gives J p2 T2 p2 T1 T2 p2 ∆s cp ln R ln T1 p1 ∆s 346 ∆u cv T2 T1 cp k 1.4 cv T2 p 1 3 MPa p T1 1 R cp cv R 287 p 2 1.85 MPa J kg K ∆u 143 ∆h cp T2 T1 ∆h 201 ∆S M ∆s ∆S 1729 ∆U M ∆u ∆U 717 kJ ∆H M ∆h ∆H 1004 kJ kJ kg kJ kg J K Here is a plot of the T-s diagram: T-s Diagram for Constant Volume Cooling 750 1 T (K) Total amounts are k kg K p1 Then T2 ( 50 273) K 500 250 0 -400 2 -350 -300 -250 -200 Δs (J/kg.K) -150 -100 -50 0 J kg K Problem 12.11 [Difficulty: 3] Given: Air in a piston-cylinder Find: Heat to raise temperature to 1200oC at a) constant pressure and b) constant volume Solution: The data provided, or available in the Appendices, is: T1 ( 100 273 ) K T2 ( 1200 273 ) K a) For a constant pressure process we start with Hence, for p = const. R 287 J kg K J kg K cv cp R cv 717 J kg K T ds dh v dp ds dh T dT cp T But δq T ds Hence δq cp dT b) For a constant volume process we start cp 1004 q c p dT q c p T2 T1 q c v dT q c v T2 T1 q 1104 q 789 kJ kg T ds du p dv du dT cv T T Hence, for v = const. ds But δq T ds Hence δq cv dT Heating to a higher temperature at constant pressure requires more heat than at constant volume: some of the heat is used to do work in expanding the gas; hence for constant pressure less of the heat is available for raising the temperature. Constant volume: q ∆u From the first law: Constant pressure: q ∆u w The two processes can be plotted using Eqs. 11.11b and 11.11a, simplified for the case of constant pressure and constant volume. a) For constant pressure T2 p2 s2 s1 cp ln R ln T1 p1 so T2 ∆s cp ln T1 b) For constant volume T2 v2 s2 s1 cv ln R ln T1 v1 so T2 ∆s cv ln T1 The processes are plotted in Excel and shown on the next page kJ kg T-s Diagram for Constant Pressure and Constant Volume Processes 1500 T (K) 1250 1000 750 500 a) Constant Pressure 250 b) Constant Volume 0 0 250 500 750 Δs (J/kg.K) 1000 1250 1500 Problem 12.12 Given: Data on Otto cycle Find: Plot of pV and Ts diagrams; efficiency [Difficulty: 4] Solution: The data provided, or available in the Appendices, is: cp 1004 J kg K p 1 100 kPa R 287 J kg K T1 ( 20 273 ) K J cv cp R cv 717 T3 ( 2750 273 ) K V1 500 cc k kg K cp cv V1 V2 8.5 k 1.4 V2 58.8 cc V4 V1 Computed results: M For process 1-2 we have isentropic behavior T v Hence V1 T2 T1 V2 R T1 k 1 V1 p2 p1 V2 kg (12.12 a and 12.12b) k p 2 2002 kPa k V2 p 1 V1 p 2 V2 p ( V) dV W12 k 1 V1 T3 p3 p2 T2 p v constant T2 690 K V3 V2 4 k constant V1 p ( V) p 1 V For process 2 - 3 we have constant volume Hence M 5.95 10 k 1 The process from 1 -2 is The work is p 1 V1 and s constant W12 169 J V3 58.8 cc p 3 8770 kPa Q12 0 J (Isentropic) V V2 constant The process from 2 -3 is and T ∆s cv ln T2 W23 0 J (From 12.11a) Q23 M ∆u M cv dT Q23 M cv T3 T2 Q23 995 J For process 3 - 4 we again have isentropic behavior Hence V3 T4 T3 V4 k 1 V3 p4 p3 V4 T4 1284 K The process from 3 - 4 is V3 p ( V) p 3 V The work is W34 k p 4 438 kPa k and p 3 V3 p 4 V4 k1 s constant W34 742 J Q34 0 J T ∆s cv ln T4 W41 0 J For process 4-1 we again have constant volume The process from 4 -1 is V V4 constant and (From 12.11a) Q41 M cv T1 T4 The net work is Wnet W12 W23 W34 W41 The efficiency is η Wnet Q23 This is consistent with the expression for the Otto efficiency Q41 422 J Wnet 572 J η 57.5 % ηOtto 1 1 k 1 r where r is the compression ratio r V1 V2 r 8.5 ηOtto 57.5 % Plots of the cycle in pV and Ts space, generated using Excel, are shown on the next page. p - V Diagram for Otto Cycle 10000 p (kPa) 8000 6000 4000 2000 0 0 100 200 300 400 500 1000 1250 V (cc) T - s Diagram for Otto Cycle 3500 T (K) 3000 2500 2000 1500 1000 500 0 0 250 500 750 s (J/kg.K) Problem 12.13 Given: Data on diesel cycle Find: Plot of pV and Ts diagrams; efficiency [Difficulty: 4] Solution: The data provided, or available in the Appendices, is: cp 1004 J kg K R 287 J cv cp R kg K kg K k cp T1 ( 20 273 ) K T3 ( 3000 273 ) K V1 500 cc V1 V2 12.5 V2 40 cc V4 1.75 V3 V4 70 cc M V3 V2 p 1 V1 M 5.95 10 R T1 T v For process 1-2 we have isentropic behavior V1 T2 T1 V2 k 1 V1 p2 p1 V2 T2 805 K V5 V1 4 k constant (12.12a) k 1 k 1.4 cv p 1 100 kPa Computed results: Hence J cv 717 p v constant kg (12.12c) k p 2 3435 kPa k The process from 1 -2 is V1 p ( V) p 1 V The work is p 1 V1 p 2 V2 2 p ( V) dV W12 k1 V and s constant V 1 For process 2 - 3 we have constant volume Hence T3 p3 p2 T2 V3 V2 V3 40 cc p 3 13963 kPa W12 218 J Q12 0 J (Isentropic) The process from 2 -3 is V V2 constant T ∆s cv ln T2 and W23 0 J (From Eq. 12.11a) Q23 M ∆u M cv dT p4 p3 For process 3 - 4 we have constant pressure The process from 3 - 4 is p p 3 constant Q23 M cv T3 T2 Q23 1052 J p 4 13963 kPa V4 T4 T3 V3 and T ∆s cp ln T3 T4 5728 K (From Eq. 12.11b) W34 p 3 V4 V3 W34 419 J V4 T5 T4 V5 For process 4 - 5 we again have isentropic behavior Q34 M cp T4 T3 Q34 1465 J k 1 T5 2607 K k Hence V4 p5 p4 V5 The process from 4 - 5 is V4 p ( V) p 4 V The work is W45 p 5 890 kPa k and p 4 V4 p 5 V5 s constant W45 1330 J k1 Q45 0 J For process 5-1 we again have constant volume The process from 5 -1 is V V5 constant and T ∆s cv ln T5 (From Eq. 12.11a) Q51 M cv T1 T5 Q51 987 J The net work is Wnet W12 W23 W34 W45 W51 The heat added is Qadded Q23 Q34 The efficiency is η Wnet Qadded Qadded 2517 J η 60.8 % W51 0 J Wnet 1531 J This is consistent with the expression from thermodynamics for the diesel efficiency r k1 c ηdiesel 1 k 1 k rc 1 r 1 where r is the compression ratio r V1 V2 V4 rc V3 and rc is the cutoff ratio r 12.5 rc 1.75 ηdiesel 58.8 % The plots of the cycle in pV and Ts space, generated using Excel, are shown here: p - V Diagram for Diesel Cycle 16000 p (kPa) 14000 12000 10000 8000 6000 4000 2000 0 0 100 200 300 400 500 V (cc) T - s Diagram for Diesel Cycle 7000 T (K) 6000 5000 4000 3000 2000 1000 0 0 500 1000 s (J/kg.K) 1500 2000 Problem 12.14 [Difficulty: 3] Given: Air is compressed from standard conditions to fill a tank Find: (a) Final temperature of air if tank is filled adiabatically and reversibly (b) Heat lost if tank is filled isothermally (c) Which process results in a greater mass of air in the tank Solution: The data provided, or available in the Appendices, is: cp 1004 J kg K 3 V 1 m R 287 J cv cp R kg K p 1 0.1 MPa cv 717 T1 ( 20 273) K J kg K k cp k 1.4 cv p 2 2 MPa k 1 p2 T2s T1 p1 Adiabatic, reversible process is isentropic: For the isothermal process, we look at the first law: The work is equal to: w p dv v2 From Boyle's law: p 1 v 1 p 2 v 2 w 252 kJ kg v1 R T1 v k T2s 689.9 K ∆u q w cv ∆T 1 p1 substituting this into the above equation: p2 kJ R T2s (The negative sign indicates heat loss) p V M R T M p2 V R T1 23.8 kg 3 Q 5.99 10 kJ The mass in the tank after compression isothermally is: p2 V p1 w R T1 ln p2 kg Q M q M qw v The mass of the air can be calculated from the ideal gas equation of state: For the isentropic compression: ∆u 0 and 2 1 v2 dv R T1 dv R T1 ln v v1 v Therefore the heat transfer is q w 252 So the actual heat loss is equal to: but ΔT = 0 so: 10.1 kg M t 23.8 kg Therefore the isothermal compression results in more mass in the tank. Problem 12.15 [Difficulty: 3] Problem 12.16 [Difficulty: 2] Problem 12.17 [Difficulty: 3] Problem 12.18 [Difficulty: 3] Given: Data on flow through compressor Find: Efficiency at which power required is 30 MW; plot required efficiency and exit temperature as functions of efficiency Solution: The data provided, or available in the Appendices, is: R 518.3 J kg K cp 2190 J cv cp R kg K cv 1672 J k kg K T1 ( 13 273 ) K p 1 0.5 MPa 101 kPa m V1 32 s p 2 8 MPa 101 kPa Wcomp 30 MW D 0.6 m cp cv k 1.31 The governing equation is the first law of thermodynamics for the compressor 2 2 V2 V1 M flow h 2 h 2 Wcomp 2 1 2 2 V2 V1 Wcomp M flow cp T2 T1 2 or We need to find the mass flow rate and the temperature and velocity at the exit p1 π 2 M flow ρ1 A1 V1 D V1 R T1 4 The exit velocity is then given by M flow p2 R T2 M flow π p1 R T1 π 2 D V1 4 2 D V2 4 V2 M flow 36.7 kg s 4 M flow R T2 (1) 2 π p 2 D The exit velocity cannot be computed until the exit temperature is determined! Using Eq. 1 in the first law Wcomp M flow cp T2 T1 2 4 Mflow R T2 V12 π p D2 2 2 In this complicated expression the only unknown is T2, the exit temperature. The equation is a quadratic, so is solvable explicitly for T2, but instead we use Excel's Goal Seek to find the solution (the second solution is mathematically correct but physically unrealistic - a very large negative absolute temperature). The exit temperature is T2 660 K 1 k If the compressor was ideal (isentropic), the exit temperature would be given by T p k constant (12.12b) 1 k Hence k p1 T2s T1 p2 T2s 529 K For a compressor efficiency η, we have η h 2s h 1 η or h2 h1 with V2 η 65.1 % T2 T1 T2 T1 To plot the exit temperature and power as a function of efficiency we use 4 M flow R T2 T2s T1 T2s T1 η 2 2 V2 V1 Wcomp M flow cp T2 T1 2 and 2 π p 2 D The dependencies of T2 and Wcomp on efficiency are plotted in Excel and shown here: Required Compressor Power as a Function of Efficiency 140 W comp (MW) 120 100 80 60 40 20 0 0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 80% 90% 100% η Gas Exit Temperature as a Function of Efficiency 2500 T (K) 2000 1500 1000 500 0 0% 10% 20% 30% 40% 50% η 60% 70% 100% Problem 12.19 [Difficulty: 3] Given: Data on performance degradation of turbine Find: Time necessary for power output to drop to 950 kW Solution: The data provided, or available in the Appendices, is: 3 p 1 10 bar 1 10 kPa T1 1400 K ηinitial 80 % p 2 1 bar 100 kPa cp 1004 J kg K Pinitial 1 MW Pfinal 950 kW Rgas 287 J kg K If the turbine expansion were isentropic, the actual output would be: So when the power output drops to 950 kW, the new efficiency is: Since the efficiency drops by 1% per year, the time elapsed is: k 1.4 Pinitial Pideal 1.25 MW ηinitial ηfinal Pfinal Pideal ∆t 4 yr 76 % Problem 12.20 [Difficulty: 4] Problem 12.21 Given: Data on flow rate and balloon properties Find: "Volumetric ratio" over time [Difficulty: 3] Solution: The given or available data are: Rair 53.3 ft lbf Tatm 519 R lbm R p atm 14.7 psi Standard air density p atm lbm ρair 0.0765 Rair Tatm 3 ft Mass flow rate M rate Vrate ρair 1.275 10 From a force balance on each hemisphere p patm π r2 σ 2 π r Hence p p atm 2 σ The instantaneous volume is 4 3 Vball π r 3 The instantaneous mass is M ball Vball ρ The time to fill to radius r from r = 5 in is t s 2 or p p atm 8 π k r Rair Tair M ball ( r) M ball ( r 5in) M rate ∆V Vball ( t ∆t) Vball ( t) The results, calculated using Excel, are shown on the next page: 4 lbm σ k A k 4 π r p ρ 3 where r Density in balloon 3 lbf ft Basic equation: The volume change between time steps t is k 200 ft Vrate 0.1 min r (in) p (psi) ρ (lb/ft3 ) V ball (ft ) M ball (lb) t (s) ΔV/V rate 5.00 5.25 5.50 5.75 6.00 6.25 6.50 6.75 7.00 29.2 30.0 30.7 31.4 32.2 32.9 33.6 34.3 35.1 0.152 0.156 0.160 0.164 0.167 0.171 0.175 0.179 0.183 0.303 0.351 0.403 0.461 0.524 0.592 0.666 0.746 0.831 0.0461 0.0547 0.0645 0.0754 0.0876 0.101 0.116 0.133 0.152 0.00 67.4 144 229 325 433 551 683 828 0.00 42.5% 41.3% 40.2% 39.2% 38.2% 37.3% 36.4% 35.5% 3 Volume Increase of Balloon as Percentage of Supplied Volume 44% ΔV/V flow 42% 40% 38% 36% 34% 0 250 500 t (s) 750 1000 Problem 12.22 [Difficulty: 3] Given: Sound wave Find: Estimate of change in density, temperature, and velocity after sound wave passes Solution: Basic equation: p ρ R T T2 p2 ∆s cp ln R ln T1 p1 du cv dT dh cp dT Assumptions: 1) Ideal gas 2) Constant specific heats 3) Isentropic process 4) infinitesimal changes Given or available data T1 ( 20 273 ) K c k R T1 For small changes, from Section 11-2 p 1 100 kPa c 343 dp 20 Pa The air density is ρ1 R T1 Then dVx 1 ρ1 c 2 dp c dρ ρ1 1.19 so dρ dp dρ 1.70 10 2 4 kg Dividing by the ideal gas equation we find m kg 3 m dVx 0.049 dp p dp dρ dT T1 p 1 ρ1 dρ ρ kg K a very small change! 3 m dp J s This is the velocity of the air after the sound wave! s For the change in temperature we start with the ideal gas equation p ρ R T Hence R 286.9 m c p1 k 1.4 and differentiate dp dρ R T ρ R dT dT T dT 0.017 K dT 0.030 ∆°F a very small change! Problem 12.23 Given: Five different gases at specified temperature Find: Sound speeds for each gas at that temperature Solution: Basic equation: c k R T The data provided, or available in the Appendices, is: k H2 1.41 [Difficulty: 3] RH2 4124 J kg K J k CH4 1.31 RCH4 518.3 kg K T ( 20 273 ) K J k He 1.66 RHe 2077 k N2 1.40 RN2 296.8 kg K J kg K J k CO2 1.29 RCO2 188.9 kg K cH2 k H2 RH2 T cH2 1305 m cHe k He RHe T cHe 1005 m cCH4 cN2 cCO2 k CH4 RCH4 T k N2 RN2 T k CO2 RCO2 T s s cCH4 446 cN2 349 m s m s cCO2 267 m s Problem 12.24 [Difficulty: 3] Given: Sound wave Find: Estimate of change in density, temperature, and velocity after sound wave passes Solution: Basic equations: p ρ R T Ev dp dρ ρ Assumptions: 1) Ideal gas 2) Constant specific heats 3) Infinitesimal changes dp To find the bulk modulus we need in dρ Ev dp dρ ρ dp dρ ρ p For rapid compression (isentropic) k const and so ρ Hence Ev ρ k p ρ dp dρ k p ρ Ev k p For gradual compression (isothermal) we can use the ideal gas equation Hence Ev ρ ( R T) p p ρ R T so dp dρ R T Ev p We conclude that the "stiffness" (Ev) of air is equal to kp when rapidly compressed and p when gradually compressed. To give an idea of values: For water Ev 2.24 GPa For air ( k 1.4) at p 101 kPa Rapid compression Ev k p Gradual compression Ev p Ev 141 kPa Ev 101 kPa Problem 12.25 Given: Device for determining bulk modulus Find: Time delay; Bulk modulus of new material [Difficulty: 2] Solution: Basic equation: Hence for given data c Ev ρ Ev 200 GN 2 L 1 m and for steel SG 7.83 kg ρw 1000 3 m Δt 0.198 ms Δt 198 μs m For the steel c Ev SG ρw Δt Hence the time to travel distance L is For the unknown material M 0.25 kg The density is then ρ M 2 L The speed of sound in it is Hence th bulk modulus is c c 5054 s 4 L Δt 1.98 10 c D 1 cm ρ 3183 π D Δt 0.5 ms kg 3 m 4 L c 2000 Δt Ev ρ c m 2 m s Ev 12.7 GN 2 m s Problem 12.26 Given: Hunting dolphin Find: Time delay before it hears prey at 1/2 mile [Difficulty: 2] Solution: Basic equation: Given (and Table A.2) data c Ev ρ 3 L 0.5 mi 2.64 10 ft SG 1.025 5 Ev 3.20 10 psi ρw 1.94 slug ft For the seawater c Ev SG ρw Hence the time for sound to travel distance L is c 4814 ∆t L c ft s ∆t 0.548 s ∆t 548 ms 3 Problem 12.27 Given: Submarine sonar Find: Separation between submarines [Difficulty: 2] Solution: Basic equation: Given (and Table A.2) data c Ev ρ ∆t 3.25 s SG 1.025 Ev 2.42 GN 2 m For the seawater c Ev SG ρw c 1537 Hence the distance sound travels in time Δt is L c ∆t The distance between submarines is half of this x L 2 m s L 5 km x 2.5 km kg ρw 1000 3 m Problem 12.28 Given: Airplane cruising at two different elevations Find: Mach numbers [Difficulty: 1] Solution: Basic equation: c k R T M J Available data R 286.9 At z 1500 m Hence c kg K k R T Repeating at z 15000 m Hence c The Mach number is k R T c k 1.4 T 278.4 K c 334 M The Mach number is V from Table A.3 m s V c 1204 km and we have hr V 550 km hr M 0.457 c T 216.7 K c 295 M V c m s c 1062 M 1.13 km hr and we have V 1200 km hr Problem 12.29 [Difficulty: 2] Given: Scramjet-powered missile traveling at fixed Mach number and altitude Find: Time necessary to cover specified range Solution: Basic equation: c k R T M J Available data R 286.9 At z 85000 ft kg K V c k 1.4 M 7 6 ∆x 600 nmi 3.65 10 ft z 25908 m interpolating from Table A.3 T 220.6 K ( 222.5 K 220.6 K) 25908 24000 26000 24000 T 222 K Hence c k R T The time needed to cover the range is: c 299 m ∆t ∆x s V c 981 531 s ft and we have s ∆t 8.85 min V M c 6864 ft s This is about ten times as fast as the Tomahawk! Problem 12.30 [Difficulty: 1] Problem 12.31 [Difficulty: 1] Problem 12.32 Given: Airplane cruising at 550 mph Find: Mach number versus altitude [Difficulty: 2] Solution: c k R T M V Here are the results, generated using Excel: c V = 500 mph R = 286.90 J/kg-K k = 1.40 (Table A.6) Data on temperature versus height obtained from Table A.3 z (m) T (K) 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 6000 7000 8000 9000 10000 288.2 284.9 281.7 278.4 275.2 271.9 268.7 265.4 262.2 258.9 255.7 249.2 242.7 236.2 229.7 223.3 c (m/s) c (mph) 340 338 336 334 332 330 329 326 325 322 320 316 312 308 304 299 661 658 654 650 646 642 639 635 631 627 623 615 607 599 590 582 M 0.756 0.760 0.765 0.769 0.774 0.778 0.783 0.788 0.793 0.798 0.803 0.813 0.824 0.835 0.847 0.859 Mach Number versus Elevation 0.90 0.85 M Basic equation: 0.80 0.75 0.70 0 1000 2000 3000 4000 5000 z (m) 6000 7000 8000 9000 10000 Problem 12.33 Given: Fireworks displays! Find: How long after seeing them do you hear them? [Difficulty: 2] Solution: Basic equation: c k R T Assumption: Speed of light is essentially infinite (compared to speed of sound) The given or available data is TJuly ( 75 460 ) R L 1 mi Hence cJuly cJuly 1134 Then the time is ∆tJuly k Rair TJuly L cJuly In January TJan ( 5 460 ) R Hence cJan Then the time is ∆tJan k Rair TJan L cJan k 1.4 ft s ∆tJuly 4.66 s cJan 1057 ft s ∆tJan 5.00 s Rair 53.33 ft lbf lbm R Problem 12.34 Given: X-15 rocket plane speed and altitude Find: Mach number [Difficulty: 2] Solution: Basic equation: Available data At c k R T R 286.9 M J kg K z 58400 m V c k 1.4 V 7270 interpolating from Table A.3 km hr T 270.7 K ( 255.8 K 270.7 K) T 258 K Hence c k R T c 322 m s c 1159 km hr and we have M V c 6.27 58400 50000 60000 50000 Problem 12.35 [Difficulty: 2] Given: Mach number and altitude of hypersonic aircraft Find: Speed assuming stratospheric temperature, actual speed, speed assuming sea level static temperature Solution: Basic equation: c k R T M J Available data Rair 286.9 Assuming T 390 R 217 K Hence c At kg K k Rair T V c k 1.4 c 295 M 7 m and we have s m Vstrat M c 2065 s z 120000 ft z 36576 m interpolating from Table A.3 T 226.5 K ( 250.4 K 226.5 K) 36576 30000 40000 30000 T 242 K Hence c k Rair T c 312 m and we have s m Vactual M c 2183 s The error is: Assuming T 288.2 K Hence c k Rair T Vstrat Vactual Vactual c 340 m s and we have 5.42 % m Vsls M c 2382 s The error is: Vsls Vactual Vactual 9.08 % Problem 12.36 Given: Shuttle launch Find: How long after seeing it do you hear it? [Difficulty: 2] Solution: Basic equation: c k R T Assumption: Speed of light is essentially infinite (compared to speed of sound) The given or available data is T ( 80 460 ) R L 3.5 mi Hence c k Rair T c 1139 ∆t L ∆t 16.23 s Then the time is c ft s In the winter: T ( 50 460 ) R Hence c k Rair T c 1107 ∆t L ∆t 16.7 s Then the time is c ft s k 1.4 Rair 53.33 ft lbf lbm R Problem 12.37 [Difficulty: 2] Given: Echo heard while hammering near mountain lake, time delay of echo is known Find: How far away are the mountains Solution: Basic equation: c k R T Assumption: Speed of light is essentially infinite (compared to speed of sound) The given or available data is T ( 25 273 ) K Hence c The distance covered by the sound is: k 1.4 k Rair T L c ∆t Rair 287 c 346 L 1038 m J kg K ∆t 3 s m s but the distance to the mountains is half that distance: L 2 519 m Problem 12.38 Given: Data on water specific volume Find: Speed of sound over temperature range [Difficulty: 2] Solution: c Basic equation: As an approximation for a liquid c ρ at isentropic conditions p ∆p using available data. ∆ρ We use compressed liquid data at adjacent pressures of 5 MPa and 10 MPa, and estimate the change in density between these pressures from the corresponding specific volume changes ∆p p 2 p 1 1 ∆ρ v2 1 and v1 c ∆p ∆ρ at each temperature Here are the results, calculated using Excel: p2 = p1 = p = 10 5 5 MPa MPa MPa Data on specific volume versus temperature can be obtained fro any good thermodynamics text (try the Web!) p1 Speed of Sound versus Temperature p2 T ( C) v (m /kg) v (m /kg) Δρ (kg/m ) c (m/s) 3 0 20 40 60 80 100 120 140 160 180 200 0.0009977 0.0009996 0.0010057 0.0010149 0.0010267 0.0010410 0.0010576 0.0010769 0.0010988 0.0011240 0.0011531 3 0.0009952 0.0009973 0.0010035 0.0010127 0.0010244 0.0010385 0.0010549 0.0010738 0.0010954 0.0011200 0.0011482 3 2.52 2.31 2.18 2.14 2.19 2.31 2.42 2.68 2.82 3.18 3.70 1409 1472 1514 1528 1512 1470 1437 1366 1330 1254 1162 1600 1500 1400 c (m/s) o 1300 1200 1100 1000 0 50 100 o T ( C) 150 200 Problem 12.39 Section 12-2 [Difficulty: 3] Problem 12.40 [Difficulty: 2] Given: Data on atmospheric temperature variation with altitude Find: Sound of speed at sea level; plot speed as function of altitude Solution The given or available data is: R = k = 286.9 1.4 J/kg.K Computing equation: c kRT Computed results: (Only partial data is shown in table) z (m) T (K) c (m/s) 288.2 284.9 281.7 278.4 275.2 271.9 268.7 265.4 262.2 258.9 255.7 249.2 242.7 236.2 229.7 223.3 340 338 336 334 332 330 329 326 325 322 320 316 312 308 304 299 Speed of Sound Variation with Altitude 350 325 c (m/s) 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 6000 7000 8000 9000 10000 300 275 250 0 10000 20000 30000 40000 50000 z (m) 60000 70000 80000 90000 100000 Problem 12.41 [Difficulty: 3] Given: Data on atmospheric temperature variation with altitude Find: Lapse rate; plot rate of change of sonic speed with altitude Solution: Rair 286.9 dz c dc dz z (km) T (K) -1 dc/dz (s ) 0 1 2 3 4 5 6 7 8 9 10 288.2 281.7 275.2 268.7 262.2 255.8 249.3 242.8 236.3 229.8 223.3 -0.00383 -0.00387 -0.00392 -0.00397 -0.00402 -0.00407 -0.00412 -0.00417 -0.00423 -0.00429 -0.00435 T T0 m m Hence T0 288.2 K T10k 223.3 K z 10000 m T T0 m z dT For an ideal gas k 1.4 kg K T10k T0 z k R T which can be evaluated at z = 10 km z m k R 2 c 3K 6.49 10 k R T0 m z m Here are the results, calculated using Excel: Rate of Change of Sonic Speed with Altitude -0.0038 -0.0039 -1 For a linear temperature variation J dc/dz (s ) The given or available data is: -0.0040 -0.0041 -0.0042 -0.0043 -0.0044 0 2 4 6 z (km) 8 10 Problem 12.42 Given: Air flow at M = 1.9 Find: Air speed; Mach angle [Difficulty: 1] Solution: Basic equations: c k R T M T ( 77 460 ) R M 1.9 Hence c c 1136 Then the air speed is The Mach angle is given by V M c α asin 1 k 1.4 ft V 2158 M 1 M c The given or available data is k Rair T α asin V s ft s α 31.8 deg V 1471 mph Rair 53.33 ft lbf lbm R Problem 12.43 [Difficulty: 3] x h Given: Hypersonic aircraft flying overhead Find: Time at which airplane is heard, how far aircraft travelled Solution: c Basic equations: k R T M 7 Given or available data M α asin V k 1.4 R 286.9 x At h 120000 ft h 36576 m J kg K The time it takes to fly from directly overhead to where you hear it is ∆t If the temperature is constant then 1 M c x V h tan( α) T 226.5 K ( 250.4 K 226.5 K) interpolating from Table A.3 T 242.2 K c Using this temperature Hence α asin M 1 k R T α 8.2 deg c 312 m and s x h tan( α) V M c V 2183 m s x 253.4 km ∆t x V ∆t 116.06 s 36576 30000 40000 30000 Problem 12.44 Given: Projectile fired into a gas, Mach cone formed Find: Speed of projectile [Difficulty: 3] Solution: Basic equations: Given or available data c k R T p 450 kPa ρ 4.5 α asin M V M c kg 3 k 1.625 m Combining ideal gas equation of state and the sonic speed: From the Mach cone angle: M 1 sin( α) M 4.62 c k p ρ α 1 25 2 p ρ R T deg 12.5 deg c 403.1 Therefore the speed is: m s V M c V 1862 m s Problem 12.45 [Difficulty: 1] Problem 12.46 [Difficulty: 2] Problem 12.47 [Difficulty: 2] Problem 12.48 [Difficulty: 2] x h Given: High-speed jet flying overhead Find: Estimate speed and Mach number of jet Solution: Basic equations: c k R T Given or available data T ( 25 273 ) K M α asin V 1 M c h 3000 m k 1.4 R 286.9 J kg K The time it takes to fly from directly overhead to where you hear it is ∆t 7.5 s The distance traveled, moving at speed V, is x V ∆t tan( α) The Mach angle is related to height h and distance x by 1 sin( α) and also we have M c cos( α) h x h (1) V ∆t (2) V c V ∆t c ∆t h h cos( α) Dividing Eq. 2 by Eq 1 sin( α) V Note that we could have written this equation from geometry directly! We have Hence Then the speed is c k R T M 1 sin( α) V M c c 346 m s so α acos c ∆t h M 1.99 V 689 m s Note that we assume the temperature of the air is uniform. In fact the temperature will vary over 3000 m, so the Mach cone will be curved. This speed and Mach number are only rough estimates. α 30.1 deg Problem 12.49 [Difficulty: 2] Problem 12.50 [Difficulty: 3] x h Given: Supersonic aircraft flying overhead Find: Time at which airplane heard Solution: Basic equations: Given or available data c k R T V 1000 M m α asin V h 3 km s k 1.4 The time it takes to fly from directly overhead to where you hear it is ∆t x If the temperature is constant then 1 M c R 286.9 J kg K x V h tan ( α ) The temperature is not constant so the Mach line will not be straight. We can find a range of Δt by considering the temperature range At h 3 km we find from Table A.3 that Using this temperature Hence c T 268.7 K k R T α asin 1 c 329 M Hence c k R T α asin M and s α 19.2 deg x h tan ( α ) V c x 8625m M 3.04 ∆t x V ∆t 8.62s T 288.2 K At sea level we find from Table A.3 that Using this temperature m M 1 c 340 m M and s α 19.9 deg x h tan ( α ) V c x 8291m Thus we conclude that the time is somwhere between 8.62 and 8.29 s. Taking an average M 2.94 ∆t x V ∆t 8.55 s ∆t 8.29s Problem 12.51 [Difficulty: 3] x h x = Vt Given: Supersonic aircraft flying overhead Find: Location at which first sound wave was emitted Solution: Basic equations: Given or available data c k R T V 1000 M m α asin M V c h 3 km s We need to find Δx as shown in the figure 1 k 1.4 R 286.9 J kg K ∆x h tan( α) The temperature is not constant so the Mach line will not be straight (α is not constant). We can find a range of α and Δx by considering the temperature range At h 3 km we find from Table A.3 that Using this temperature Hence c T 268.7 K k R T α asin 1 c 329 M Hence c k R T α asin s α 19.2 deg V an d M ∆x h tan( α) ∆x 1043 m an d M ∆x h tan( α) ∆x 1085 m c M 3.04 T 288.2 K At sea level we find from Table A.3 that Using this temperature m M 1 c 340 m s α 19.9 deg Thus we conclude that the distance is somwhere between 1043 and 1085 m. Taking an average V c ∆x 1064 m M 2.94 Problem 12.52 [Difficulty: 4] Part 1/2 Problem 12.52 [Difficulty: 4] Part 2/2 Problem 12.53 Given: Speed of automobile Find: Whether flow can be considered incompressible [Difficulty: 2] Solution: Consider the automobile at rest with 60 mph air flowing over it. Let state 1 be upstream, and point 2 the stagnation point on the automobile The data provided, or available in the Appendices, is: R 287 J k 1.4 kg K V1 60 mph p 1 101 kPa T1 ( 20 273 ) K 1 The basic equation for the density change is ρ0 ρ 1 ( k 1) 2 M 2 k 1 (12.20c) 1 ( k 1) ρ0 ρ1 1 or For the Mach number we need c m V1 26.8 s ρ0 ρ1 1 k1 2 M1 2 k 1 ρ1 1.201 c1 c1 343 V1 3 m s M 1 0.0782 c1 ρ0 1.205 kg m k R T1 k 1 2 p1 ρ1 R T1 M1 1 2 M1 kg The percentage change in density is ρ0 ρ1 ρ0 3 m 0.305 % This is an insignificant change, so the flow can be considered incompressible. Note that M < 0.3, the usual guideline for incompressibility V1 120 mph For the maximum speed present m V1 53.6 s M1 1 ρ0 ρ1 1 k1 2 M1 2 k 1 ρ0 1.216 kg 3 m The percentage change in density is This is still an insignificant change, so the flow can be considered incompressible. V1 c1 M 1 0.156 ρ0 ρ1 ρ0 1.21 % Problem 12.54 Given: Supersonic transport aircraft Find: Explanation of sound wave refraction [Difficulty: 5] Solution: A sound wave is refracted when the speed of sound varies with altitude in the atmosphere. (The variation in sound speed is caused by temperature variations in the atmosphere, as shown in Fig. 3.3) Imagine a plane wave front that initially is vertical. When the wave encounters a region where the temperature increase with altitude (such as between 20.1 km and 47.3 km altitude in Fig. 3.3), the sound speed increases with elevation. Therefore the upper portion of the wave travels faster than the lower portion. The wave front turns gradually and the sound wave follows a curved path through the atmosphere. Thus a wave that initially is horizontal bends and follows a curved path, tending to reach the ground some distance from the source. The curvature and the path of the sound could be calculated for any specific temperature variation in the atmosphere. However, the required analysis is beyond the scope of this text. Problem 12.55 [Difficulty: 2] Given: Mach number range from 0.05 to 0.95 Find: Plot of percentage density change; Mach number for 1%, 5% and 10% density change Solution: k 1.4 The given or available data is: Basic equation: 1 ρ0 ρ 1 ( k 1) 2 M 2 1 k 1 (12.20c) ∆ρ Hence ρ0 ρ0 ρ ρ0 1 ρ so ρ0 ∆ρ ρ0 1 1 Here are the results, generated using Excel: M 0.05 0.10 0.15 0.20 0.25 0.30 0.35 Δρ /ρ o 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 7.6% 9.4% 11% 14% 16% 18% 21% 23% 26% 29% 31% 34% 0.1% 0.5% 1.1% 2.0% 3.1% 4.4% 5.9% To find M for specific density changes use Goal Seek repeatedly Δρ /ρ o M 0.142 1% 0.322 5% 0.464 10% Note: Based on ρ (not ρ o) the results are: 0.142 0.314 0.441 Density Variation with Mach Number 40% Δρ/ρo 30% 20% 10% 0% 0.0 0.1 0.2 0.3 0.4 0.5 M 0.6 0.7 0.8 0.9 1.0 ( k 1) 2 M 2 1 k Problem 12.56 Given: Scramjet-powered missile traveling at fixed Mach number and altitude Find: Stagnation temperature at the nose of the missile Solution: T0 T 1 k1 Available data R 286.9 J At z 85000 ft Basic equation: 2 kg K M [Difficulty: 2] 2 k 1.4 M 7 z 25908 m interpolating from Table A.3 T 220.6 K ( 222.5 K 220.6 K) T 222 K So the stagnation temperature is T0 T 1 k1 2 M 2 T0 2402 K 25908 24000 26000 24000 Problem 12.57 [Difficulty: 2] Given: X-15 rocket plane traveling at fixed Mach number and altitude Find: Stagnation temperature at the nose of the plane Solution: Basic equation: Available data At T0 T 1 k1 R 286.9 J 2 kg K z 58400 m M 2 c k 1.4 k R T M V 7270 interpolating from Table A.3 V c km hr T 270.7 K ( 255.8 K 270.7 K) 58400 50000 60000 50000 T 258 K Hence c k R T So the stagnation temperature is c 322 m T0 T 1 c 1159 s k1 2 M 2 km hr and we have M V c 6.27 T0 2289 K Problem 12.58 [Difficulty: 1] Given: Car and F-1 race car traveling at sea level Find: Ratio of static to total pressure in each case; are compressiblilty effects experienced? Solution: k Basic equations: Given or available data At sea level, from Table A.3 Hence c k R T M V p0 c p Vcar 55 mph ft Vcar 80.7 s k 1.4 Rair 53.33 T 288.2 K or c k Rair T p p0 1 k1 c 1116 2 M car M F1 VF1 2 p p0 k 1 ft VF1 323 s VF1 220 mph ft lbf lbm R ft M car s ρ 0.002377 slug ft Vcar 3 p 14.696 psi M car 0.0723 c 0.996 2 ρ Vcar p 1 2 p p0 1 0.996 M F1 0.289 c 1 2 k 1 The pressure ratio is 2 M k Note that the Bernoulli equation would give the same result! For the Formula One car: T 519 R The pressure ratio is k1 1 k1 2 M F1 2 k k 1 0.944 Note that the Bernoulli equation would give almost the same result: 2 ρ VF1 p 1 2 p p0 1 0.945 Incompressible flow can be assumed for both cases, but the F1 car gets very close to the Mach 0.3 rule of thumb for compressible vs. incompressible flow. Problem 12.59 [Difficulty: 2] Given: Scramjet-powered missile traveling at fixed Mach number and altitude Find: Stagnation and dynamic pressures k Solution: Basic equation: Available data At c k R T R 286.9 M J kg K z 85000 ft V p0 c p k 1.4 M 7 1 k1 2 M p SL 14.696 psi 2 k 1 p dyn ρSL 0.2377 1 2 2 ρ V slug ft 3 z 25908 m interpolating from Table A.3 T 220.6 K ( 222.5 K 220.6 K) 25908 24000 26000 24000 T 222 K Hence c k R T c 299 m s c 981 ft and we have s V M c 6864 ft s The static pressure and density can be found by interpolation: k 25908 24000 p p SL 0.02933 ( 0.02160 0.02933 ) p 0.323 psi 26000 24000 25908 24000 slug ρ ρSL 0.03832 ( 0.02797 0.03832 ) ρ 0.00676 26000 24000 3 ft p 0 p 1 p dyn 1 2 k1 2 2 ρ V M 2 k 1 p 0 1336 psi p dyn 1106 psi Problem 12.60 [Difficulty: 2] Given: X-15 rocket plane traveling at fixed Mach number and altitude Find: Stagnation and dynamic pressures k Solution: Basic equation: Available data At c k R T R 286.9 M J kg K z 58400 m V p0 c p k 1.4 1 V 7270 interpolating from Table A.3 k1 2 M km 2 k 1 2 2 ρ V kg ρSL 1.225 3 m p SL 101.3 kPa hr 1 p dyn T 270.7 K ( 255.8 K 270.7 K) 58400 50000 60000 50000 T 258 K Hence c k R T c 322 m s c 1159 km M and we have hr V c 6.27 The static pressure and density can be found by interpolation: 58400 50000 p p SL 0.0007874 ( 0.0002217 0.0007874 ) 60000 50000 p 0.0316 kPa k p 0 p 1 58400 50000 ρ ρSL 0.0008383 ( 0.0002497 0.0008383 ) 60000 50000 k1 2 M 2 k 1 p 0 65.6 kPa 4 kg ρ 4.21 10 3 m p dyn 1 2 2 ρ V p dyn 0.86 kPa Problem 12.61 Given: Aircraft flying at 250 m/s Find: Stagnation pressure [Difficulty: 1] k Solution: Basic equations: Given or available data First we need c k R T V 250 c M m V p0 c p T ( 50 273 ) K s k R T c 299 m s then 1 k1 2 p 0 p 1 k1 2 M 2 2 M k 1 p 0 44.2 kPa V c k 1 p 28 kPa k Finally we solve for p0 M k 1.4 M 0.835 R 286.9 J kg K Problem 12.62 [Difficulty: 2] Given: Pressure data on aircraft in flight Find: Change in air density; whether flow can be considered incompressible Solution: The data provided, or available in the Appendices, is: k 1.4 p 0 48 kPa p 27.6 kPa T ( 55 273 ) K Governing equation (assuming isentropic flow): p k constant (12.12c) ρ 1 Hence ρ ρ0 p p 0 k 1 so ∆ρ ρ ρ0 ρ ρ k p0 1 1 ρ p ρ0 ∆ρ ρ 48.5 % NOT an incompressible flow! Problem 12.63 Given: Aircraft flying at 12 km Find: Dynamic and stagnation pressures [Difficulty: 2] k Solution: Basic equations: Given or available data At h 12 km ,from Table A.3 c k R T M V p0 c p M 2 h 12 km kg ρSL 1.225 3 m p SL 101.3 kPa ρ 0.2546 ρSL ρ 0.312 Also Hence p 0 p 1 k1 c 2 k R T p dyn 1 2 M 2 2 2 M 2 k 1 p dyn R 286.9 p 0.1915 p SL p 19.4 kPa p 0 152 kPa m s p dyn 54.3 kPa V M c V 590 m s 1 2 2 ρ V J k 1.4 k 1 c 295 ρ V 3 k1 m k Hence kg 1 kg K T 216.7 K Problem 12.64 [Difficulty: 1] Problem 12.65 [Difficulty: 1] Problem 12.66 [Difficulty: 1] Problem 12.67 [Difficulty: 2] Given: Mach number of aircraft Find: Pressure difference; air speed based on a) compressible b) incompressible assumptions Solution: The data provided, or available in the Appendices, is: R 287 J cp 1004 kg K J kg K T 223.3 K From Table A.3, at 10 km altitude k 1.4 M 0.65 p 0.2615 101 kPa p 26.4 kPa k p0 The governing equation for pressure change is: p 1 k1 2 M 2 k 1 (12.20a) k Hence The pressure difference is p 0 p 1 k1 2 M 2 k 1 p 0 35.1 kPa p 0 p 8.67 kPa a) Assuming compressibility c k R T c 300 m V M c s V 195 m s b) Assuming incompressibility Here the Bernoulli equation applies in the form p ρ For the density Hence ρ p V 2 p0 ρ so V 2 p0 p ρ 0.412 R T V 205 2 ρ kg 3 V 2 p0 p m ρ m s In this case the error at M = 0.65 in computing the speed of the aircraft using Bernoulli equation is 205 195 195 5.13 % Problem 12.68 [Difficulty: 1] Problem 12.69 [Difficulty: 2] Given: Flight altitude of high-speed aircraft Find: Mach number and aircraft speed errors assuming incompressible flow; plot k Solution: The governing equation for pressure change is: p0 p Hence 1 k1 2 M 2 k 1 (12.20a) k k 1 k 1 2 ∆p p 1 M 1 2 p0 ∆p p 0 p p 1 p (1) For each Mach number the actual pressure change can be computed from Eq. 1 p Assuming incompressibility, the Bernoulli equation applies in the form ρ 2 V 2 p0 V so ρ 2 ∆p and the Mach number based on this is Using Eq. 1 M incomp V c ρ k R T 2 ∆p k ρ R T k k 1 2 k 1 2 M incomp M 1 1 k 2 The error in using Bernoulli to estimate the Mach number is ∆M M M incomp M M For errors in speed: Actual speed: V M c Speed assuming incompressible flow: The error in using Bernoulli to estimate the speed from the pressure difference is V M k R T Vinc M incomp k R T ∆V V The computations and plots are shown below, generated using Excel: Vincomp V V 2 p0 p ρ 2 ∆p ρ The given or available data is: R = k = T = 286.9 1.4 216.7 J/kg.K K (At 12 km, Table A.3) Computed results: c = M 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 295 M in comp 0.100 0.201 0.303 0.408 0.516 0.627 0.744 0.865 0.994 m/s ΔM/M V (m/s) 0.13% 0.50% 1.1% 2.0% 3.2% 4.6% 6.2% 8.2% 10.4% 29.5 59.0 88.5 118 148 177 207 236 266 V incomp (m/s) 29.5 59.3 89.5 120 152 185 219 255 293 ΔV/V 0.13% 0.50% 1.1% 2.0% 3.2% 4.6% 6.2% 8.2% 10.4% Error in Mach Number Using Bernoulli 12% 10% ΔM/M 8% 6% 4% 2% 0% 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 M Error in Speed Using Bernoulli 12% 10% ΔV/V 8% 6% 4% 2% 0% 0 50 100 150 V (m/s) 200 250 300 Problem 12.70 Given: Wind tunnel at M = 2.5 Find: Stagnation conditions; mass flow rate [Difficulty: 2] k Solution: Basic equations: Given or available data Then c k R T M V p0 c p M 2.5 T ( 15 273 ) K k 1.4 R 286.9 T0 T 1 k1 2 M 1 k1 2 M 2 k 1 T0 T 1 2 A 0.175 m T0 648 K T0 375 °C J kg K 2 p 0 p 1 k1 2 M 2 k 1 p 0 598 kPa The mass flow rate is given by mrate ρ A V We need c k R T c 340 ρ p ρ 0.424 and also Then R T mrate ρ A V m V M c s kg 3 m kg mrate 63.0 s 2 p 35 kPa k Also k1 V 850 m s M 2 Problem 12.71 [Difficulty: 2] Problem 12.72 Given: Wind tunnel test of supersonic transport Find: Lift and drag coefficients [Difficulty: 3] k Solution: Basic equations: c k R T FL CL 1 2 Given or available data M CD 2 ρ V A M 1.8 Finally 2 T0 k1 1 We also need k1 p p 0 1 T and 2 k Rair T CD 2 ρ V A FD 1 2 1 2 ρ V A 2 k 1 T0 T 1 k1 2 M 2 2 ρ V A p 0 200 psi Rair 53.33 k FL 12000 lbf ft lbf lbm R k 1 p 34.8 psi T 123 °F 2 c 1183 ft c 807 mph s ft V 1452 mph s slug ft FL 2 2 M FD ρ 0.00501 Rair T 1 2 V 2129 p CL M k1 T 583 R M V M c ρ p k 1.4 c c 1 T0 ( 500 460 ) R 2 Then p0 2 A 100 in We need local conditions V CL 1.52 CD 0.203 3 FD 1600 lbf Problem 12.73 [Difficulty: 2] Problem 12.74 [Difficulty: 2] Problem 12.75 [Difficulty: 2] Given: Data on air flow in a duct Find: Stagnation pressures and temperatures; explain velocity increase; isentropic or not? Solution: The data provided, or available in the Appendices, is: Rair 287 At altitude: J kg K cp 1004 J k 1.4 kg K M 9.68 p SL 101.3 kPa kg ρSL 1.225 3 m 33528 30000 z 110000 ft z 33528 m T1 226.5 K ( 250.4 K 226.5 K) T 234.9 K 40000 30000 1 33528 30000 p 1 p SL 0.01181 ( 0.002834 0.01181 ) 40000 30000 The sound speed is: c k Rair T1 307.239 So the stagnation temperature and pressure are: m p 1 0.8756 kPa V M c 2974 so the flight speed is: s T01 T1 1 k1 2 M 2 m V 9757 s T01 4638 K ft s T01 8348 R k p 01 p 1 1 k1 2 As the air passes through the shock wave, stagnation pressure decreases: M 2 k 1 p 01 29.93 MPa p 02 p 01 ( 1 0.996 ) Therefore, the total head probe sees a pressure of Since there is no heat transfer through the shock wave, the stagnation temperature remains the same: p 02 119.7 kPa T02 T01 T02 8348 R Problem 12.76 [Difficulty: 2] Given: Data on air flow in a duct Find: Stagnation pressures and temperatures; explain velocity increase; isentropic or not? Solution: The data provided, or available in the Appendices, is: R 287 M 1 0.1 J cp 1004 kg K T1 ( 20 273 ) K For stagnation temperatures: J k 1.4 kg K p 1 1000 kPa T01 T1 1 k1 T02 T2 1 k1 2 2 M 2 0.7 M1 M2 2 2 T2 ( 5.62 273 ) K p 2 136.5 kPa T01 293.6 K T01 20.6 C T02 293.6 K T02 20.6 C (Because the stagnation temperature is constant, the process is adiabatic) k For stagnation pressures: p 01 p 1 1 k1 2 M1 2 k 1 p 01 1.01 MPa k p 02 p 2 1 The entropy change is: Note that k1 2 M2 2 k 1 p 02 189 kPa T2 p2 ∆s cp ln R ln T1 p1 V1 M 1 k R T1 m V1 34.3 s ∆s 480 V2 M 2 k R T2 Although there is friction, suggesting the flow should decelerate, because the static pressure drops so much, the net effect is flow acceleration! The entropy increases because the process is adiabatic but irreversible (friction). δq From the second law of thermodynamics ds : becomes ds > 0 T J kg K m V2 229 s Problem 12.77 [Difficulty: 2] Given: Data on air flow in a duct Find: Stagnation temperatures; explain; rate of cooling; stagnation pressures; entropy change Solution: The data provided, or available in the Appendices, is: R 287 T1 ( 500 273 ) K p 1 500 kPa M 1 0.5 M 2 0.2 For stagnation temperatures: J kg K M rate 0.05 T01 T1 1 k1 T02 T2 1 k1 2 2 M1 M2 cp 1004 2 2 J kg K k 1.4 T2 ( 18.57 273 ) K p 2 639.2 kPa T01 811.7 K T01 539 C T02 256.5 K T02 16.5 C kg s The fact that the stagnation temperature (a measure of total energy) decreases suggests cooling is taking place. For the heat transfer: Q M rate cp T02 T01 Q 27.9 kW k For stagnation pressures: p 01 p 1 1 k1 2 M1 2 k 1 p 01 593 kPa k p 02 p 2 1 The entropy change is: k1 2 M2 2 k 1 T2 p2 ∆s cp ln R ln T1 p1 p 02 657 kPa ∆s 1186 J kg K The entropy decreases because the process is a cooling process (Q is negative). δq From the second law of thermodynamics: ds becomes ds ve T Hence, if the process is reversible, the entropy must decrease; if it is irreversible, it may increase or decrease Problem 12. ['LIILFXOW\ 3] Problem 12.79 Given: Air flow in duct with heat transfer and friction Find: Heat transfer; Stagnation pressure at location 2 [Difficulty: 3] k Solution: Basic equations: Given or available data c k R T and from V p0 c p V1 h1 p 1 400 kPa T1 325 K p 2 275 kPa T2 450 K J dm h2 V2 k1 2 J m ρ V A const ρ1 V2 V1 ρ2 m V2 302 s q c p T2 T1 c2 V2 V1 ρ2 2.13 kg k1 q 160 2 M2 2 m s s o M2 kJ kg V2 c2 k 1 p 02 385 kPa 3 m 2 c2 425 2 3 2 k kg 2 k R T2 p 02 p 2 1 kg K p2 ρ2 R T2 q h2 h1 k 1 2 R 286.9 V2 V1 2 2 ρ1 4.29 δQ M m V1 150 s k 1.4 kg K 2 Hence δQ 1 p1 ρ1 R T1 dm We also have 2 2 Also 2 ρ V A const cp 1004 Then M M 2 0.711 Problem 12.80 [Difficulty: 2] Given: Data on air flow in a ramjet combustor Find: Stagnation pressures and temperatures; isentropic or not? Solution: The data provided, or available in the Appendices, is: Rair 53.33 M 1 0.2 ft lbf lbm R T1 ( 600 460 ) R For stagnation temperatures: cp 0.2399 BTU k 1.4 lbm R p 1 7 psi T01 T1 1 k1 T02 T2 1 k1 2 2 M rate 0.1 M 2 0.9 M1 M2 2 2 lbm s T2 ( 1890 460 ) R T01 1068.5 R T01 608.8 °F T02 2730.7 R T02 2271 °F p 2 4.1 psi Since we are modeling heat addition, the stagnation temperature should increase. The rate of heat addition is: Q M rate cp T02 T01 Q 39.9 BTU s k For stagnation pressures: The entropy change is: p 01 p 1 1 k1 2 M1 2 k k 1 7.20 psi T2 p2 BTU ∆s cp ln Rair ln 0.228 lbm R T1 p1 p 02 p 2 1 k1 2 M2 2 k 1 6.93 psi T p02 T02 p2 The entropy increases because heat is being added. Here is a Ts diagram of the process: p01 T01 T2 p1 T1 s Problem 12.81 [Difficulty: 2] Given: Data on air flow in a ramjet combustor Find: Stagnation pressures and temperatures; isentropic or not? Solution: The data provided, or available in the Appendices, is: Rair 53.33 M 1 0.2 ft lbf lbm R T1 ( 600 460 ) R For stagnation temperatures: cp 0.2399 k 1.4 lbm R p 1 7 psi T01 T1 1 k1 T02 T2 1 k1 The rate of heat addition is: BTU 2 2 M rate 0.1 M 2 0.9 M1 M2 Q M rate cp T02 T01 2 2 lbm s T2 ( 1660 460 ) R T01 1068.5 R T01 608.8 °F T02 2463.4 R T02 2003.8 °F Q 33.5 p 2 1.6 psi BTU s k For stagnation pressures: p 01 p 1 1 k1 2 M1 2 k 1 p 01 7.20 psi k p 02 p 2 1 The entropy change is: k1 2 M2 2 k 1 T2 p2 ∆s cp ln Rair ln T1 p1 p 02 2.71 psi ∆s 0.267 The friction has increased the entropy increase across the duct, even though the heat addition has decreased. BTU lbm R Problem 12.82 [Difficulty: 2] Problem 12.83 [Difficulty: 2] Given: Air flow through turbine Find: Stagnation conditions at inlet and exit; change in specific entropy; Plot on Ts diagram k Solution: p0 Basic equations: p Given or available data k1 1 2 2 k 1 T0 T 1 k1 2 M 2 M 1 0.4 p 1 625 kPa T1 ( 1250 273) K M 2 0.8 p 2 20 kPa T2 ( 650 273) K k 1.4 R 286.9 cp 1004 Then M J kg K T01 T1 1 k1 2 2 M1 T2 p2 ∆s cp ln R ln T1 p1 J kg K T01 1572K T01 1299 °C k p 01 p 1 1 k1 T02 T2 1 k1 2 2 M1 k 1 2 M2 p 01 698 kPa 2 T02 1041 K k p 02 p 2 1 k1 2 M2 2 k 1 p 02 30 kPa T2 p2 ∆s cp ln R ln T1 p1 p01 T T 01 p1 p 02 T1 T 02 p2 T2 s ∆s 485 J kg K T02 768 °C Problem 12.84 [Difficulty: 3] Problem 12.85 Given: Air flow leak in window of airplane Find: Mass flow rate [Difficulty: 2] 1 Solution: Basic equations: mrate ρ V A Vcrit 2 k ρ0 R T0 k1 ρcrit k 1 k 1 2 The interior conditions are the stagnation conditions for the flow Given or available data kg ρSL 1.225 3 m T0 271.9 K ρ0 0.7812 ρSL ρ0 0.957 kg 3 m (Above data from Table A.3 at an altitude of 2500 m) 2 A 1 mm Then ρcrit cp 1004 ρ0 1 k 1 k 1 J k 1.4 kg K ρcrit 0.607 kg Vcrit 3 m 2 The mass flow rate is mrate ρcrit Vcrit A 4 kg mrate 1.83 10 s R 286.9 2 k k1 R T0 J kg K m Vcrit 302 s Problem 12.86 Given: Air leak in ISS Find: Mass flow rate [Difficulty: 2] 1 Solution: mrate ρ V A Basic equations: 2 k Vcrit k1 ρ0 R T0 ρcrit k 1 k 1 2 The interior conditions are the stagnation conditions for the flow Given or available data T0 ( 65 460 ) R The density of air inside the ISS would be: Then ρ0 ρcrit 1 k 2 The mass flow rate is 1 p 0 14.7 psi Rair 53.33 p0 ρ0 Rair T0 ρcrit 1.49 10 k 1 ft lbf lbm R ρ0 2.35 10 ft 3 2 k Vcrit 2 A 0.001 in 3 slug ft 3 slug k 1.4 k1 3 Rair T0 ft Vcrit 1025 s mrate ρcrit Vcrit A mrate 1.061 10 5 slug s 4 lbm mrate 3.41 10 s Problem 12.87 [Difficulty: 1] Problem 12.88 Given: Data on helium in reservoir Find: Critical conditions [Difficulty: 1] Solution: The data provided, or available in the Appendices, is: RHe 386.1 ft lbf lbm R For critical conditions k 1.66 T0 Tcrit T0 3600 R k1 Tcrit 2 p 0 ( 725 14.7)psi T0 p 0 740 psi Tcrit 2707 R k 1 2 k p0 p crit k 1 k 1 2 p0 p crit k k 2 Vcrit k RHe Tcrit 1 ft Vcrit 7471 s k 1 p crit 361 psi absolute Problem 12.9 ['LIILFXOW\ 1] Problem 12.90 Given: Data on hot gas stream Find: Critical conditions [Difficulty: 1] Solution: The data provided, or available in the Appendices, is: R 287 J kg K For critical conditions k 1.4 T0 Tcrit T0 ( 1500 273) K k1 Tcrit 2 T0 T0 1773K p 0 140 kPa Tcrit 1478K k 1 2 k p0 p crit k 2 1 k 1 p0 p crit k k 2 Vcrit k R Tcrit m Vcrit 770 s 1 k 1 p crit 74.0 kPa absolute Problem 12.91 Given: Data on air flow in a ramjet combustor Find: Critical temperature and pressure at nozzle exit [Difficulty: 1] Solution: The data provided, or available in the Appendices, is: k 1.4 p 0 1.7 MPa T0 1010 K The critical temperature and pressure are: T0 Tcrit k1 Tcrit 2 T0 Tcrit 841.7 K k 1 2 k p0 p crit k 2 1 k 1 p0 p crit k k 2 1 k 1 p crit 0.898 MPa Problem 12.92 Given: Data on air flow in a ramjet combustor Find: Critical temperature and pressure at nozzle exit [Difficulty: 1] Solution: k 1.4 The data provided, or available in the Appendices, is: Stagnation conditions are: k1 T02 T2 1 2 M2 2 M 2 0.9 T2 ( 1660 460) R T02 2463.4R p 2 1.6 psi T02 2003.8°F k p 02 p 2 1 k1 2 2 M2 k 1 p 02 2.71 psi The critical temperature and pressure are: T02 Tcrit2 k1 Tcrit2 2 T02 Tcrit2 2052.9 R k 1 2 k p 02 p crit2 k 2 1 k 1 p crit2 p 02 k k 2 1 k 1 p crit2 1.430 psi Tcrit2 1593.2 °F Problem 13.1 Given: Air extracted from a large tank Find: Mass flow rate [Difficulty: 2] Solution: h1 V1 2 Basic equations: mrate ρ V A h2 Given or available data T0 ( 70 273 ) K p 0 101 kPa D 15 cm cp 1004 mrate ρ A V A 2 V2 ( 1 k) 2 p 2 k const We need the density and velocity at the nozzle. In the tank 1 From the isentropic relation p ρ ρ0 p0 J k 1.4 kg K R 286.9 J kg K 2 A 0.0177 m 4 p0 ρ0 R T0 ρ 0.379 const p 25 kPa π D k k ρ 2 The mass flow rate is given by T p ρ0 1.026 kg 3 m kg 3 m We can apply the energy equation between the tank (stagnation conditions) and the point in the nozzle to find the velocity 2 h0 h V V 2 2 h0 h 2 c p T0 T ( 1 k ) p0 T T0 p Fot T we again use insentropic relations Then The mass flow rate is V 2 c p T0 T mrate ρ A V Note that the flow is supersonic at this point Hence we must have a converging-diverging nozzle V 476 k T 230.167 K T 43.0 °C m s kg mrate 3.18 s c k Rc T 304 m s M V c M 1.57 Problem 13.2 [Difficulty: 2] Problem 13.3 [Difficulty: 2] Given: Steam flow through a nozzle Find: Speed and Mach number; Mass flow rate; Sketch the shape Solution: Basic equations: 2 mrate ρ V A h1 2 V1 h2 2 V2 2 Assumptions: 1) Steady flow 2) Isentropic 3) Uniform flow 4) Superheated steam can be treated as ideal gas Given or available data T0 ( 450 273 ) K p 0 6 MPa p 2 MPa D 2 cm k 1.30 R 461.4 J (Table A.6) kg K From the steam tables (try finding interactive ones on the Web!), at stagnation conditions Hence at the nozzle section J s0 6720 kg K h 0 3.302 10 6 J J an s s0 6720 kg K d p 2 MPa T 289 °C From these values we find from the steam tables that Hence the first law becomes The mass flow rate is given by Hence For the Mach number we need V 2 h0 h mrate ρ A V mrate c A V v k R T v 6 J h 2.997 10 v 0.1225 s 2 A kg 3 m V 781 A V kg π D 4 4 A 3.14 10 2 m kg mrate 2.00 s c 581 The flow is supersonic starting from rest, so must be converging-diverging m s M V c M 1.35 m kg Problem 13.4 [Difficulty: 2] Given: Data on flow in a passage Find: Pressure and Mach number at downstream location Solution: The given or available data is: R = k = p1 = M1 = 296.8 1.4 450 0.7 A1 = 0.15 m A2 = 0.45 m p 01 = 624 kPa J/kg-K kPa 2 2 Equations and Computations: From M 1 and p 1, and Eq. 13.7a (using built-in function Isenp (M ,k )) From M 1, and Eq. 13.7d (using built-in function IsenA (M ,k )) A For isentropic flow (p 01 = p 02, A * * 1 = 0.1371 2 m * 2 = A 1) p 02 = A * A 2/A * 624 2 = 0.1371 2 = 3.2831 kPa 2 m * From A 2/A 2, and Eq. 13.7d (using built-in function IsenMsubfromA (M ,k )) Since there is no throat, the flow stays subsonic M2 = 0.1797 From M 2 and p 02, and Eq. 13.7a (using built-in function Isenp (M ,k )) p2 = 610 kPa Problem 13.5 [Difficulty: 2] Given: Data on flow in a passage Find: Temperature and Mach number at downstream location Solution: R = k = T1 = T1 = M1 = 296.8 1.4 30 303 1.7 °C A1 = 0.15 m2 A2 = 0.45 m2 T 01 = 478 K A *1 = 0.1121 m2 478 K = 0.1121 m2 A 2/A *2 = 4.0128 The given or available data is: J/kg-K K Equations and Computations: From M 1 and T 1, and Eq. 13.7b (using built-in function Isent (M ,k )) From M 1, and Eq. 13.7d (using built-in function IsenA (M ,k )) For isentropic flow (T 01 = T 02, A *2 = A *1) T 02 = A * 2 * From A 2/A 2, and Eq. 13.7d (using built-in function IsenMsupfromA (M ,k )) Since there is no throat, the flow stays supersonic! M2 = 2.94 From M 2 and T 02, and Eq. 13.7b (using built-in function Isent (M ,k )) T2 = T2 = 175 -98 K °C Problem 13.6 Given: Air flow in a passage Find: Mach number; Sketch shape [Difficulty: 2] Solution: k Basic equations: Given or available data The speed of sound at state 1 is Hence p0 p 1 k1 2 M 2 k 1 c T1 ( 10 273 ) K p 1 150 kPa m V1 120 s p 2 50 kPa k 1.4 R 286.9 c1 c1 337 M1 k R T1 V1 Solving for M2 M2 s k k p 0 p 1 1 m M 1 0.356 c1 For isentropic flow stagnation pressure is constant. Hence at state 2 Hence k R T k1 2 M1 2 p0 p2 1 k1 2 k 1 k 1 k 2 p 0 1 k1 p2 p 0 164 kPa M 2 1.42 Hence, as we go from subsonic to supersonic we must have a converging-diverging nozzle M2 2 k 1 J kg K Problem 13.7 [Difficulty: 2] Given: Data on flow in a passage Find: Pressure at downstream location Solution: The given or available data is: ft·lbf/lbm·oR R = k = 53.33 1.4 T1 = p1 = V1 = M2 = 560 30 1750 2.5 c1 = 1160 M1 = 1.51 p 01 = 111 psi p 02 = 111 psi p2 = 6.52 psi o R psi ft/s Equations and Computations: From T 1 and Eq. 12.18 Then ft/s From M 1 and p 1, and Eq. 13.7a (using built-in function Isenp (M ,k )) For isentropic flow (p 01 = p 02) From M 2 and p 02, and Eq. 13.7a (using built-in function Isenp (M ,k )) Problem 13.8 [Difficulty: 2] Given: Data on flow in a passage Find: Stagnation conditions; whether duct is a nozzle or diffuser; exit conditions Solution: The given or available data is: R = k = p1 = T1 = V1 = 259.8 1.4 200 420 200 kPa K m/s A1 = 0.6 m2 A2 = 0.5 m2 c1 = 391 m/s M1 = 0.512 T 01 = 442 K p 01 = 239 kPa J/kg-K Equations and Computations: From T 1 and Eq. 12.18 Then From M 1 and T 1, and Eq. 13.7b (using built-in function Isent (M ,k )) From M 1 and p 1, and Eq. 13.7a (using built-in function Isenp (M ,k )) Since the flow is subsonic and the area is decreasing, this duct is a nozzle. From M 1, and Eq. 13.7d (using built-in function IsenA (M ,k )) A *1 = 0.4552 For isentropic flow (p 01 = p 02, T 01 = T 02, A *2 = A *1) p 02 = 239 T 02 = 442 A *2 = 0.4552 m2 kPa K m2 * A 2/A 2 = 1.0984 From A 2/A * 2, and Eq. 13.7d (using built-in function IsenMsubfromA (M ,k )) Since there is no throat, the flow stays subsonic! M2 = 0.69 From M 2 and stagnation conditions: (using built-in functions) p2 = T2 = 173 403 kPa K Problem 13.9 [Difficulty: 3] Given: Data on flow in a passage Find: Shape of flow passage; exit area provided the flow is reversible Solution: The given or available data is: R = k = m= p1 = T1 = T1 = 53.33 1.4 20 30 1200 1660 A1 = M2 = 8 1.2 ft-lbf/lbm-°R lbm/s psia °F °R in2 Equations and Computations: Using the ideal gas law we calculate the density at station 1: lbm/ft3 ρ1 = 0.04880 Now we can use the area and density to get the velocity from the mass flow rate: V1 = 7377 ft/s From T 1 and Eq. 12.18 Then c1 = 1998 M1 = 3.69 ft/s Since the flow is supersonic and the velocity is decreasing, this duct is converging. From M 1, and Eq. 13.7d (using built-in function IsenA (M ,k )) A *1 = 0.9857 in2 A *2 = 0.9857 in2 A 2/A *2 = 1.0304 A2 = 1.016 For isentropic flow ( A *2 = A *1) Therefore the exit area is: in2 Problem 13.10 [Difficulty: 3] Given: Data on flow in a nozzle Find: Mass flow rate; Throat area; Mach numbers Solution: The given or available data is: R = k = T0 = p1 = A = 286.9 1.4 523 200 1 J/kg·K K kPa p2 = 50 kPa 2 cm Equations and Computations: We don't know the two Mach numbers. We do know for each that Eq. 13.7a applies: Hence we can write two equations, but have three unknowns (M 1, M 2, and p 0)! We also know that states 1 and 2 have the same area. Hence we can write Eq. 13.7d twice: We now have four equations for four unknowns (A *, M 1, M 2, and p 0)! We make guesses (using Solver) for M 1 and M 2, and make the errors in computed A * and p 0 zero. For: M1 = 0.512 M2 = 1.68 from Eq. 13.7a: p0 = 239 kPa p0 = 239 kPa 0.00% and from Eq. 13.7d: A* = 0.759 cm 2 A* = 0.759 cm 2 0.00% Note that the throat area is the critical area Sum The stagnation density is then obtained from the ideal gas equation 0 = 1.59 3 kg/m The density at critical state is obtained from Eq. 13.7a (or 12.22c) * = Errors 1.01 kg/m3 The velocity at critical state can be obtained from Eq. 12.23) V* = 418 m/s m rate = 0.0321 kg/s The mass flow rate is *V *A * 0.00% Problem 13. ['LIILFXOW\ 2] Problem 13.12 Given: Air flow in a passage Find: Speed and area downstream; Sketch flow passage [Difficulty: 3] k 1 Solution: Basic equations: T0 T Given or available data 1 k1 2 M 2 c k R T T1 ( 32 460 ) R p 1 25 psi M 1 1.75 T2 ( 225 460 ) R k 1.4 Rair 53.33 D1 3 ft Hence 2 ( k 1) 1 k 1 M2 2 A 1 k 1 Acrit M 2 A1 T0 T1 1 k1 2 M1 2 π D1 2 4 T0 793 R A1 7.07 ft ft lbf lbm R 2 T0 334 °F For isentropic flow stagnation conditions are constant. Hence We also have 2 T0 M2 k1 c2 k Rair T2 Hence V2 M 2 c2 From state 1 Acrit T2 1 M 2 0.889 c2 1283 ft s ft V2 1141 s A1 M 1 k 1 Acrit 5.10 ft 2 ( k 1) 1 k 1 M 2 1 2 k 1 2 k 1 Hence at state 2 2 ( k 1) 1 k 1 M 2 2 Acrit 2 A2 M2 k 1 2 A2 5.15 ft Hence, as we go from supersonic to subsonic we must have a converging-diverging diffuser 2 2 Problem 13.13 [Difficulty: 2] Problem 13.14 [Difficulty: 3] Given: Data on flow in a passage Find: Mach numbers at entrance and exit; area ratio of duct Solution: The given or available data is: R = k = T1 = p1 = T2 = T 02 = p2 = 286.9 1.4 310 200 294 316 125 J/kg-K K kPa K K kPa Equations and Computations: Since the flow is adiabatic, the stagnation temperature is constant: T 01 = 316 K Solving for the Mach numbers at 1 and 2 using Eq. 13.7b (using built-in function IsenMfromT (Tratio ,k )) Then M1 = 0.311 M2 = 0.612 Using the ideal gas equation of state, we can calculate the densities of the gas: kg/m3 ρ1 = 2.249 ρ2 = 1.482 kg/m3 c1 = c2 = V1 = V2 = 352.9 343.6 109.8 210.2 m/s m/s m/s m/s From static temperatures and Eq. 12.18 Since flow is steady, the mass flow rate must be equal at 1 and 2. So the area ratio may be calculated from the densities and velocities: A 2/A 1 = 0.792 Note that we can not assume isentropic flow in this problem. While the flow is adiabatic, it is not reversible. There is a drop in stagnation pressure from state 1 to 2 which would invalidate the assumption of isentropic flow. Problem 13.15 [Difficulty: 3] Given: Flow in a converging nozzle to a pipe Find: Plot of mass flow rate Solution: The given or available data is R = k = T0 = p0 = Dt = 287 1.4 293 101 1 J/kg·K K kPa cm 2 A t = 0.785 cm Equations and Computations: The critical pressure is given by p * = 53.4 kPa Hence for p = 100 kPa down to this pressure the flow gradually increases; then it is constant c V = M ·c = p /RT (m/s) (m/s) (kg/m3) 343 41 1.19 342 58 1.18 342 71 1.18 341 82 1.17 341 92 1.16 340 101 1.15 337 138 1.11 335 168 1.06 332 195 1.02 329 219 0.971 326 242 0.925 322 264 0.877 318 285 0.828 315 306 0.778 313 313 0.762 313 313 0.762 313 313 0.762 313 313 0.762 313 313 0.762 Flow Rate (kg/s) 0.00383 0.00539 0.00656 0.00753 0.00838 0.0091 0.0120 0.0140 0.0156 0.0167 0.0176 0.0182 0.0186 0.0187 0.0187 0.0187 0.0187 0.0187 0.0187 Flow Rate in a Converging Nozzle 0.020 0.018 0.016 0.014 Flow Rate (kg/s) p M T (K) (kPa) (Eq. 13.7a) (Eq. 13.7b) 100 0.119 292 99 0.169 291 98 0.208 290 97 0.241 290 96 0.270 289 95 0.297 288 90 0.409 284 85 0.503 279 80 0.587 274 75 0.666 269 70 0.743 264 65 0.819 258 60 0.896 252 55 0.974 246 53.4 1.000 244 53 1.000 244 52 1.000 244 51 1.000 244 50 1.000 244 0.012 0.010 0.008 0.006 0.004 0.002 0.000 Using critical conditions, and Eq. 13.9 for mass flow rate: 53.4 1.000 244 313 313 0.762 0.0185 (Note: discrepancy in mass flow rate is due to round-off error) 50 60 70 80 p (kPa) 90 100 Problem 13.16 [Difficulty: 2] Given: Data on flow in a passage Find: Flow rate; area and pressure at downstream location; sketch passage shape Solution: The given or available data is: R = k = 286.9 1.4 J/kg.K A1 = T1 = p1 = V1 = T2 = M2 = 0.25 283 15 590 410 0.75 m2 K kPa m/s Equations and Computations: From T 1 and Eq. 12.18 Then (12.18) c1 = 337 M1 = 1.75 m/s Because the flow decreases isentropically from supersonic to subsonic the passage shape must be convergent-divergent From p 1 and T 1 and the ideal gas equation 1 = 0.185 kg/m3 m rate = 27.2 kg/s The mass flow rate is m rate = 1A 1V 1 From M 1 and A 1, and Eq. 13.7d (using built-in function IsenA (M ,k )) (13.7d) A* = 0.180 m2 A2 = 0.192 m2 From M 2 and A *, and Eq. 13.7d (using built-in function IsenA (M ,k )) From M 1 and p 1, and Eq. 13.7a (using built-in function Isenp (M ,k )) (13.7a) p 01 = 79.9 kPa p 02 = 79.9 kPa p2 = 55.0 kPa For isentropic flow (p 01 = p 02) From M 2 and p 02, and Eq. 13.7a (using built-in function Isenp (M ,k )) Problem 13.17 [Difficulty: 3] Given: Data on tank conditions; isentropic flow Find: Plot cross-section area and pressure distributions Solution: The given or available data is: R = k = 53.33 1.4 T0 = p0 = pe = m rate = 500 45 14.7 2.25 ft·lbf/lbm·oR o R psia psia lbm/s Equations and Computations: From p 0, p e and Eq. 13.7a (using built-in function IsenMfromp (M,k)) (13.7a) Me = 1.37 Because the exit flow is supersonic, the passage must be a CD nozzle We need a scale for the area. From p 0, T 0, m flow, and Eq. 13.10c (13.10c) Then At = A* = 0.0146 ft2 For each M , and A *, and Eq. 13.7d (using built-in function IsenA (M ,k ) (13.7d) we can compute each area A . From each M , and p 0, and Eq. 13.7a (using built-in function Isenp (M ,k ) we can compute each pressure p . L (ft) M 0.069 0.086 0.103 0.120 0.137 0.172 0.206 0.274 0.343 0.412 0.480 0.549 0.618 0.686 0.755 0.823 0.892 0.961 1.000 1.098 1.166 1.235 1.304 1.372 1.00 1.25 1.50 1.75 2.00 2.50 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00 12.00 13.00 14.00 14.6 16.00 17.00 18.00 19.00 20.00 A (ft 2) p (psia) 0.1234 0.0989 0.0826 0.0710 0.0622 0.0501 0.0421 0.0322 0.0264 0.0227 0.0201 0.0183 0.0171 0.0161 0.0155 0.0150 0.0147 0.0146 0.0146 0.0147 0.0149 0.0152 0.0156 0.0161 44.9 44.8 44.7 44.5 44.4 44.1 43.7 42.7 41.5 40.0 38.4 36.7 34.8 32.8 30.8 28.8 26.8 24.9 23.8 21.1 19.4 17.7 16.2 14.7 Area Variation in Passage 0.14 0.12 A (ft2) 0.10 0.08 0.06 0.04 0.02 0.00 0 5 10 15 20 L (ft) p (psia) Pressure Variation in Passage 50 45 40 35 30 25 20 15 10 5 0 0 2 4 6 8 10 L (ft) 12 14 16 18 20 Problem 13.18 [Difficulty: 2] Given: Flow in a converging-diverging nozzle to a pipe Find: Plot of mass flow rate Solution: The given or available data is R = k = T0 = p0 = Dt = 286.9 1.4 293 101 1 J/kg·K K kPa cm 2 At = 0.785 cm p* = 53.4 kPa De = Ae = 2.5 cm 4.909 cm2 Equations and Computations: The critical pressure is given by This is the minimum throat pressure For the CD nozzle, we can compute the pressure at the exit required for this to happen 2 A* = 0.785 cm A e/A * = 6.25 M e = 0.0931 or p e = 100.4 or (= A t) 3.41 67.2 (Eq. 13.7d) kPa (Eq. 13.7a) Hence we conclude flow occurs in regimes iii down to v (Fig. 13.8); the flow is ALWAYS choked! p* M T * (K) c* V * = c * = p /RT (kPa) (Eq. 13.7a) (Eq. 13.7b) (m/s) (m/s) (kg/m3) 53.4 1.000 244 313 313 0.762 (Note: discrepancy in mass flow rate is due to round-off error) Flow Rate (kg/s) 0.0187 0.0185 (Using Eq. 13.9) Problem 13.19 Given: Isentropic air flow in converging nozzle Find: Pressure, speed and Mach number at throat [Difficulty: 2] Solution: Basic equations: k T0 T Given or available data k1 1 2 M p0 2 p 1 k1 2 p 1 350 kPa m V1 150 s k 1.4 R 286.9 M 2 k 1 M 1 0.5 p b 250 kPa J kg K The flow will be choked if p b/p0 < 0.528 k k1 p 0 p 1 1 2 M1 2 k 1 pb p 0 415 kPa p0 0.602 (Not choked) k Hence p0 pt so 1 Mt k1 2 Mt 2 k 1 where k 1 k 2 p 0 1 k1 pt Also V1 M 1 c1 M 1 k R T1 or Then T0 T1 1 Hence Then Finally k1 Tt ct 2 T0 1 k1 2 k R Tt Vt M t ct Mt 2 M1 2 pt pb p t 250 kPa M t 0.883 V1 T1 k R M1 1 2 T1 224 K T0 235 K T0 37.9 °C Tt 204 K Tt 69.6 °C ct 286 m s m Vt 252 s T1 49.1 °C Problem 13.20 Given: Air flow in a converging nozzle Find: Mass flow rate [Difficulty: 2] Solution: k Basic equations: mrate ρ V A Given or available data p b 35 psi pb p0 T p 0 60 psi k 1.4 Since T0 p ρ R T Rair 53.33 ft lbf lbm R 0.583 is greater than 0.528, the nozzle is not choked and Hence Mt and Tt ct k 1 k 2 p 0 1 k1 pt T0 1 k1 2 Mt k Rair Tt 2 1 k1 2 M pt pb ft Vt 1166 s mrate ρt At Vt slug mrate 0.528 s At 0.0873 ft Vt ct 3 slug ft k1 π 2 At Dt 4 Tt 106 °F p 1 Dt 4 in Tt 566 R ρt 5.19 10 p0 T0 ( 200 460 ) R M t 0.912 pt ρt Rair Tt 2 3 lbm mrate 17.0 s 2 2 M 2 k 1 Problem 13.21 [Difficulty: 2] Given: Data on flow in a passage Find: Possible Mach numbers at downstream location Solution: The given or available data is: R = k = M1 = 286.9 1.4 1 A1 = 0.2 m2 A2 = 0.5 m2 A *1 = 0.2 m2 A *2 = 0.2 m2 J/kg-K Equations and Computations: Since the flow is sonic at the entrance: For isentropic flow (A *2 = A *1) A 2/A *2 = 2.5 Now there are two Mach numbers which could result from this area change, one subsonic and one supersonic. From A 2/A * 2, and Eq. 13.7d (using built-in functions) M 2sub = 0.2395 2.4428 M 2sup = Problem 13.22 [Difficulty: 3] Given: Data on three tanks Find: Mass flow rate; Pressure in second tank Solution: The given or available data is: R = k = 286.9 1.4 At = 1 J/kg.K 2 cm We need to establish whether each nozzle is choked. There is a large total pressure drop so this is likely. However, BOTH cannot be choked and have the same flow rate. This is because Eq. 13.9a, below (13.9b) indicates that the choked flow rate depends on stagnation temperature (which is constant) but also stagnation pressure, which drops because of turbulent mixing in the middle chamber. Hence BOTH nozzles cannot be choked. We assume the second one only is choked (why?) and verify later. Temperature and pressure in tank 1: T 01 = 308 p 01 = 650 We make a guess at the pressure at the first nozzle exit: p e1 = 527 NOTE: The value shown is the final answer! It was obtained using Solver ! This will also be tank 2 stagnation pressure: p 02 = 527 65 Pressure in tank 3: p3 = K kPa kPa kPa kPa Equations and Computations: From the p e1 guess and Eq. 13.17a: Then at the first throat (Eq.13.7b): M e1 = T e1 = 0.556 290 K The density at the first throat (Ideal Gas) is: Then c at the first throat (Eq. 12.18) is: Then V at the first throat is: Finally the mass flow rate is: e1 = 6.33 341 190 0.120 kg/m m/s m/s kg/s c e1 = V e1 = m rate = 3 First Nozzle! For the presumed choked flow at the second nozzle we use Eq. 13.9a, with T 01 = T 02 and p 02: m rate = 0.120 kg/s For the guess value for p e1 we compute the error between the two flow rates: m rate = 0.000 Use Solver to vary the guess value for p e1 to make this error zero! Note that this could also be done manually. kg/s Second Nozzle! Problem 13.23 [Difficulty: 2] Problem 13.24 [Difficulty: 2] Problem 13.25 [Difficulty: 2] Given: Data on converging nozzle; isentropic flow Find: Pressure and Mach number; throat area; mass flow rate Solution: The given or available data is: R = k = 286.9 1.4 J/kg.K A1 = T1 = V1 = p atm = 0.05 276.3 200 101 m2 K m/s kPa Equations and Computations: From T 1 and Eq. 12.18 Then (12.18) c1 = 333 M1 = 0.60 m/s To find the pressure, we first need the stagnation pressure. If the flow is just choked pe = p atm = p* = 101 kPa From p e = p * and Eq. 12.22a (12.22a) p0 = 191 kPa From M 1 and p 0, and Eq. 13.7a (using built-in function Isenp (M ,k ) (13.7a) Then p1 = 150 kPa The mass flow rate is m rate = 1A 1V 1 Hence, we need 1 from the ideal gas equation. 1 = 1.89 kg/m3 m rate = 18.9 kg/s The mass flow rate m rate is then The throat area A t = A * because the flow is choked. From M 1 and A 1, and Eq. 13.7d (using built-in function IsenA (M ,k ) (13.7d) Hence A* = 0.0421 m2 At = 0.0421 m2 Problem 13.26 [Difficulty: 2] Problem 13.27 [Difficulty: 2] Problem 13.28 [Difficulty: 3] Given: Data on flow in a passage Find: Exit temperature and mass flow rate of air assuming isentropic flow Solution: The given or available data is: R = k = T1 = p1 = p 01 = 53.33 1.4 450 45 51 ft-lbf/lbm-°R A1 = 4 ft2 A2 = 3 ft2 °R psia psia Equations and Computations: From the static and stagnation pressures we can calculate M 1: M1 = 0.427 T 01 = 466.38 °R A *1 = 2.649 ft2 From the M 1 and T 1 we can get T 01: From M 1, and Eq. 13.7d (using built-in function IsenA (M ,k )) For isentropic flow (p 02 = p 01, T 02 = T 01, A *2 = A *1) p 02 = 51 T 02 = 466.38 A *2 = 2.649 A 2/A *2 = 1.1325 psia °R ft2 Given subsonic flow in the duct, we can find the exit Mach number using Equation 13.7d M2 = 0.653 From the Mach number and stagnation state we can calculate the static pressure and temperature: p2 = 38.28 psia T2 = 430 °R From T 2 and Eq. 12.18 c2 = V2 = 1016.38 664.11 ft/s ft/s Using the ideal gas law we calculate the density at station 2: lbm/ft3 ρ2 = 0.2406 Now we can use the area, density, and velocity to calculate the mass flow rate: m = 479 lbm/s Problem 13.29 [Difficulty: 2] Given: Temperature in and mass flow rate from a tank Find: Tank pressure; pressure, temperature and speed at exit Solution: The given or available data is: R = k = T0 = 286.9 1.4 273 J/kg.K At = m rate = 0.001 2 m2 K kg/s Equations and Computations: Because p b = 0 Hence the flow is choked! pe = p* Hence Te = T* From T 0, and Eq. 12.22b (12.22b) T* = 228 Te = 228 -45.5 K K o C Also Hence Me = Ve = 1 V* = From T e and Eq. 12.18 ce (12.18) Then ce = 302 m/s Ve = 302 m/s To find the exit pressure we use the ideal gas equation after first finding the exit density. The mass flow rate is m rate = eA eV e Hence e = 6.62 kg/m3 pe = 432 kPa From the ideal gas equation p e = eRT e From p e = p * and Eq. 12.22a (12.22a) p0 = 817 kPa We can check our results: From p 0, T 0, A t, and Eq. 13.9a (13.9a) Then m choked = m choked = 2.00 m rate kg/s Correct! Problem 13.30 [Difficulty: 2] Problem 13.31 [Difficulty: 3] Given: Temperature and pressure in a tank; nozzle with specific area Find: Mass flow rate of gas; maximum possible flow rate Solution: The given or available data is: R = k = T0 = p0 = 296.8 1.4 450 150 At = 30 cm2 At = pb = 0.003 100 m2 J/kg.K K kPa kPa Equations and Computations: Assuming that the nozzle exit pressure is the back pressure: pe = 100 kPa Then the nozzle exit Mach number is: Me = 0.7837 This nozzle is not choked. The exit temperature is: Te = 400.78 K From T e and Eq. 12.18 Then (12.18) ce = 408.08 m/s Ve = 319.80 m/s From the ideal gas equation of state, we can calculate the density: kg/m3 e = 0.8407 Therefore the mass flow rate is: m = 0.807 kg/s When the room pressure can be lowered, we can choke the nozzle. p* pe = T* Te = From T 0, and Eq. 12.22b (12.22b) Also Hence T* = p* = 375 79.24 Te = 375 Me = Ve = 1 K V* = From T e and Eq. 12.18 Then K kPa ce (12.18) ce = 395 m/s Ve = 395 m/s To find the mass flow rate we calculate the density from the ideal gas equation of state: Hence e = 0.7120 kg/m3 m max = 0.843 kg/s Therefore the mass flow rate is: We can check our results: From p 0, T 0, A t, and Eq. 13.9a (13.9a) Then m choked = m choked = 0.843 m rate kg/s Correct! Problem 13.32 [Difficulty: 2] Given: Isentropic air flow into a tank Find: Initial mass flow rate; Ts process; explain nonlinear mass flow rate Solution: Basic equations: k T0 T Given or available data Then k1 1 2 M p0 2 p p 0 101 kPa p b p 0 10 kPa k 1.4 R 286.9 A π 4 2 D J kg K Avena 65 % A pb The flow will be choked if p b/p0 < 0.528 p0 0.901 1 k1 M 2 2 p b 91 kPa k 1 mrate ρ A V T0 ( 20 273 ) K D 5 mm 2 Avena 12.8 mm (Not choked) k Hence p0 p vena 1 so M vena Then Tvena Then cvena k1 2 M 2 k 1 wher e k 1 k 2 p 0 1 k1 pvena T0 1 k1 2 M vena 2 k R Tvena p vena p b p vena 91 kPa M vena 0.389 Tvena 284 K cvena 338 Tvena 11.3 °C m s and Vvena M vena cvena m Vvena 131 s Also p vena ρvena R Tvena ρvena 1.12 mrate ρvena Avena Vvena mrate 1.87 10 Finally kg 3 m 3 kg s The Ts diagram will be a vertical line (T decreases and s = const). After entering the tank there will be turbulent mixing (s increases) and the flow comes to rest (T increases). The mass flow rate versus time will look like the curved part of Fig. 13.6b; it is nonlinear because V AND ρ vary Problem 13.33 Given: Spherical cavity with valve Find: Time to reach desired pressure; Entropy change [Difficulty: 3] k Solution: Basic equations: T0 T 1 k1 2 M p0 2 p 1 k1 2 M 2 k 1 T2 p2 ∆s cp ln R ln T1 p1 k 1 Given or available data Then the inlet area is p ρ R T c k R T mrate ρ A V p 0 101 kPa Tatm ( 20 273 ) K p f 45 kPa Tf Tatm π 2 At d 4 At 0.785 mm T0 Tatm k 1.4 2 k 2 mchoked At p 0 R T0 k 1 R 286.9 pf ρf R Tf ρf 0.535 kg k 1 k 2 We have choked flow so mrate At p 0 R T0 k 1 ∆t J cp 1004 π 3 pb so p0 3 J kg K 3 D V 0.131 m 0.446 (Choked) and final mass is M ρf V M 0.0701 kg m Since the mass flow rate is constant (flow is always choked) Hence 3 D 50 cm kg K and tank volume is V The flow will be choked if p b/p0 < 0.528; the MAXIMUM back pressure is p b p f The final density is d 1 mm 2 ( k 1 ) M 2 ( k 1) ∆t 374 s mrate M mrate ∆t mrate 1.873 10 ∆t or M mrate 4 kg s ∆t 6.23 min The air in the tank will be cold when the valve is closed. Because ρ =M/V is constant, p = ρRT = const x T, so as the temperature rises to ambient, the pressure will rise too. T2 p2 For the entropy change during the charging process is given by ∆s cp ln R ln p where T1 1 and p1 p0 p2 pf Hence T2 p2 ∆s cp ln R ln T1 p1 T1 Tatm T2 Tatm ∆s 232 J kg K Problem 13.34 [Difficulty: 3] Problem 13.35 [Difficulty: 3] Problem 13.36 [Difficulty: 3] Problem 13.37 [Difficulty: 3] Given: Air-driven rocket in space Find: Tank pressure; pressure, temperature and speed at exit; initial acceleration Solution: R = k = T0 = 286.9 1.4 398 J/kg.K At = M = m rate = 25 25 0.05 mm2 kg kg/s Because p b = 0 Hence the flow is choked! pe = p* Hence Te = T* The given or available data is: K Equations and Computations: From T 0, and Eq. 12.22b (12.22b) Also Hence T* = 332 Te = 332 K 58.7 o Me = Ve = C 1 V* = From T e and Eq. 12.18 Then K ce (12.18) ce = 365 m/s Ve = 365 m/s To find the exit pressure we use the ideal gas equation after first finding the exit density. The mass flow rate is m rate = eA eV e Hence e = 0.0548 pe = 5.21 kg/m3 From the ideal gas equation p e = eRT e kPa From p e = p * and Eq. 12.22a (12.22a) p0 = 9.87 kPa We can check our results: From p 0, T 0, A t, and Eq. 13.9a (13.9a) Then m choked = m choked = 0.050 m rate kg/s Correct! The initial acceleration is given by: (4.33) which simplifies to: pe At Max mrateV ax = or: 1.25 ax m/s2 m rate V p e At M Problem 13.38 [Difficulty: 3] Given: Air flow through a converging-diverging nozzle Find: Nozzle exit area and mass flow rate Solution: The given or available data is: R = k = p0 = T0 = pe = 286.9 1.4 2 313 200 At = 20 J/kg-K MPa K kPa cm2 Equations and Computations: Using the stagnation to exit static pressure ratio, we can find the exit Mach number: (using built-in function Isenp (M ,k )) Me = 2.1572 A e/A * = 1.9307 From M e, and Eq. 13.7d (using built-in function IsenA (M ,k )) At the throat the flow is sonic, so At = A*. Therefore: Ae = 38.6 cm2 To find the mass flow rate at the exit, we will use the choked flow equation: From p 0, T 0, A t, and Eq. 13.9a (13.9a) m = 17.646 kg/s Problem 13.39 [Difficulty: 1] Given: Hydrogen flow through a converging-diverging nozzle Find: Nozzle exit Mach number Solution: The given or available data is: R = k = p0 = T0 = T0 = pe = 766.5 1.41 100 540 1000 20 ft-lbf/lbm-°R psia °F °R psia Equations and Computations: Using the stagnation to exit static pressure ratio, we can find the exit Mach number: (using built-in function Isenp (M ,k )) Me = 1.706 Problem 13.40 Given: Gas cylinder with broken valve Find: Mass flow rate; acceleration of cylinder [Difficulty: 3] k Solution: T0 Basic equations: T 1 k1 2 M 2 p0 p 1 k1 2 M 2 k 1 p ρ R T c k R T mrate ρ A V (4.33) Given or available data p atm 101 kPa k 1.66 R 2077 p 0 20 MPa p atm 20.101 MPa J kg K The exit temperature is Te p b p atm T0 Ve ce The exit pressure is pe p0 k 1 3 5.025 10 M CV 65 kg (Choked: Critical conditions) Te 52.8 °C ce p e 9.8 MPa and exit density is pe ρe R Te k 1 mrate ρe Ae Ve ax p0 The momentum equation (Eq. 4.33) simplifies to Hence pb 2 Ae 78.5 mm k R Te m Ve 872 s 1 k 2 Then so Te 220 K 1 k 1 2 The exit speed is π 2 Ae d 4 d 10 mm so the nozzle area is The flow will be choked if p b/p0 < 0.528: T0 ( 20 273 ) K kg mrate 1.468 s pe patm Ae MCV ax Ve mrate pe patm Ae Ve mrate M CV ax 31.4 m 2 s The process is isentropic, followed by nonisentropic expansion to atmospheric pressure ρe 21 kg 3 m Problem 13.41 [Difficulty: 3] Problem 13.42 Given: Spherical air tank Find: Air temperature after 30s; estimate throat area [Difficulty: 4] Solution: Basic equations: T0 T 1 k1 2 M 2 p k ρ dVCV t const ρ ρ V dACS 0 (4.12) Assumptions: 1) Large tank (stagnation conditions) 2) isentropic 3) uniform flow Given or available data p atm 101 kPa p 1 2.75 MPa T1 450 K D 2 m V ∆M 30 kg ∆t 30 s k 1.4 R 286.9 J p b p atm The flow will be choked if p b/p1 < 0.528: so pb p1 0.037 π 6 3 D 3 V 4.19 m kg K (Initially choked: Critical conditions) We need to see if the flow is still choked after 30s The initial (State 1) density and mass are The final (State 2) mass and density are then For an isentropic process p k const ρ The final temperature is T2 To estimate the throat area we use p2 ρ2 R so p1 ρ1 R T1 ρ1 21.3 M 2 M 1 ∆M M 2 59.2 kg kg 3 M 1 ρ1 V M 1 89.2 kg M2 ρ2 V ρ2 14.1 m ρ2 p2 p1 ρ1 k T2 382 K p 2 1.55 MPa pb p2 0.0652 mtave ρtave At Vtave ∆t or The average stagnation temperature is The average stagnation pressure is T0ave p 0ave (Still choked) ∆M At ∆t ρtave Vtave where we use average values of density and speed at the throat. T1 T2 2 p1 p2 2 T0ave 416 K p 0ave 2.15 MPa 3 m T2 109 °C ∆M kg Hence the average temperature and pressure (critical) at the throat are Ttave Hence Finally T0ave 1 k 1 2 Vtave k R Ttave ∆M At ∆t ρtave Vtave Ttave 347 K and p 0ave p tave k 1 k 2 p tave ρtave R Ttave m Vtave 373 s 4 At 2.35 10 2 m 2 At 235 mm This corresponds to a diameter Dt 4 At π Dt 0.0173 m Dt 17.3 mm The process is isentropic, followed by nonisentropic expansion to atmospheric pressure 1 p tave 1.14 MPa k 1 ρtave 11.4 kg 3 m Problem 13.43 Given: Ideal gas flow in a converging nozzle Find: Exit area and speed [Difficulty: 4] k 1 k Solution: T0 Basic equations: T k1 1 2 p 1 35 psi Given or available data M p0 2 p ρ1 0.1 lbm ft c1 Check for choking: Hence M1 V1 k1 M2 2 M1 2 A1 1 ft c1 2 k p 2 25 psi p1 c1 1424 ρ1 k 1.25 ft s p 0 37.8 psi p crit 21.0 psi k 1 k 1 k 1 k 2 p 0 1 k1 p2 k 1 2 ( k 1) 1 k 1 M 2 1 2 k 1 2 k p ρ const ρ A V const Hence p2 > pcrit, so NOT choked k 1 M 1 A1 Acrit Finally from continuity 2 ft V1 500 s 3 p0 k 2 For isentropic flow M k The critical pressure is then p crit From M1 we find 2 k R T1 or, replacing R using the ideal gas equation p 0 p 1 1 Then we have k1 M 1 0.351 c1 Then 1 1 k 1 M2 2 A 1 k 1 Acrit M 2 k 1 2 ( k 1) so so M 2 0.830 Acrit 0.557 ft 2 k 1 1 k 1 M 2 2 2 A2 M2 k 1 2 Acrit 1 p1 ρ2 ρ1 p2 k A1 ρ1 V2 V1 A2 ρ2 ρ2 0.131 lbm ft ft V2 667 s 3 2 ( k 1) A2 0.573 ft 2 Problem 13.44 [Difficulty: 4] Part 1/3 Problem 13.44 [Difficulty: 4] Part 2/3 Problem 13.44 [Difficulty: 4] Part 3/3 Problem 13.45 [Difficulty: 3] Given: Air flow through a converging-diverging nozzle Find: Nozzle mass flow rate Solution: The given or available data is: R = k = V1 = p1 = T1 = T1 = 53.33 1.4 50 15 70 530 ft-lbf/lbm-°R At = 1 ft2 c1 = 1128.80 ft/s M1 = 0.0443 ft/s psia °F °R Equations and Computations: At station 1 the local sound speed is: So the upstream Mach number is: So now we can calculate the stagnation temperature and pressure: p0 = T0 = 15.021 530.21 psia °R To find the mass flow rate, we will use the choked flow equation: From p 0, T 0, A t, and Eq. 13.10a (13.10a) m = 50.0 lbm/s Problem 13.46 Given: CD nozzle attached to large tank Find: Flow rate [Difficulty: 2] k Solution: Basic equations: T0 T Given or available data 1 2 M 2 p0 p 1 k1 2 M p 0 150 kPa T0 ( 35 273 ) K k 1.4 R 286.9 For isentropic flow Me Then Te Also ce Finally k1 J kg K k 1 k 2 p 0 1 k1 p e 1 k 1 M 2 e 2 ce 332 k 1 mrate ρ V A p e 101 kPa D 2.75 cm π 2 Ae D 4 Ae 5.94 cm M e 0.773 T0 k R Te 2 m s pe ρe R Te ρe 1.28 mrate ρe Ve Ae kg mrate 0.195 s kg 3 m Te 275 K Te 1.94 °C Ve M e ce m Ve 257 s 2 Problem 13.47 [Difficulty: 4] Problem 13.48 [Difficulty: 4] Part 1/2 Problem 13.48 [Difficulty: 4] Part 2/2 Problem 13.49 [Difficulty: 2] Given: Design condition in a converging-diverging nozzle Find: Tank pressure; flow rate; throat area Solution: The given or available data is: R = k = 53.33 1.4 T0 = 560 Ae = pb = Me = 1 14.7 2 pe = pb pe = 14.7 o ft.lbf/lbm. R o R 2 in psia Equations and Computations: At design condition psia From M e and p e, and Eq. 13.7a (using built-in function Isenp (M ,k ) (13.7a) p0 = 115 psia From M e and A e, and Eq. 13.7d (using built-in function IsenA (M ,k ) (13.7d) Hence 2 A* = 0.593 in At = 0.593 in 2 From p 0, T 0, A t, and Eq. 13.10a (13.10a) m choked = 1.53 lb/s Problem 13.50 [Difficulty: 3] Given: Wind tunnel test section with blockage Find: Maximum blockage that can be tolerated; air speed given a fixed blockage Solution: The given or available data is: R = k = M1 = T1 = T1 = 53.33 1.4 1.2 70 530 ft-lbf/lbm-°R °F °R At = 1 ft2 Equations and Computations: * The test section will choke if the blockage decreases the area to A . In the test section: * A 1/A = 1.0304 So the minimum area would be * 2 A = ft 0.9705 And the blockage would be the difference between this and the test section area: 2 * ft A1 - A = 0.0295 A1 - A = 4.25 in2 A1 - A = 3.0000 in A actual = The resulting isentropic area ratio is: 0.9792 ft2 A actual/A * = 1.0090 * If we have a blockage of: 2 Then the actual area would be: and the actual Mach number is: 1.1066 M actual = (remember that since we're already supersonic, we should use the supersonic solution) The stagnation temperature for the wind tunnel is (based on test section conditions) T0 = 682.64 °R So the actual static temperature in the tunnel is: T actual = 548.35 °R The sound speed would then be: c actual = 1148.17 ft/s And so the speed in the test section is: V actual = 1270.5 ft/s Problem 13.51 [Difficulty: 3] Given: Air flow through a converging-diverging nozzle equipped with pitot-static probe Find: Nozzle velocity and mass flow rate Solution: The given or available data is: R = k = p1 = p 01 = T1 = T1 = 286.9 1.4 75 100 20 293 A1 = 10 in2 A1 = 0.006452 m2 J/kg-K kPa kPa °C K Equations and Computations: At station 1 the local sound speed is: c1 = 343.05 m/s Based on the static and pitot pressures, the Mach number is: M1 = 0.6545 Therefore the velocity is: V1 = 225 m/s The local density can be calculated using the ideal gas equation of state: kg/m3 ρ1 = 0.8922 So the mass flow rate is: m = 1.292 kg/s Problem 13.52 [Difficulty: 2] Problem 13.53 [Difficulty: 2] Problem 13.54 [Difficulty: 3] Given: Methane discharging from one tank to another via a converging nozzle Find: Mass flow rate at two different back pressures Solution: The given or available data is: R = k = p0 = T0 = T0 = 96.32 1.31 75 80 540 Ae = 1 ft-lbf/lbm-°R psia °F °R in2 Equations and Computations: If the nozzle were choked, the exit Mach number is 1 and the pressure would be: p* = 40.79 psia Therefore, in part a, when pe = 15 psia The nozzle is choked, and we can use the choked flow equation: From p 0, T 0, A t, and Eq. 13.9a (13.9a) m = 1.249 lbm/s In part b, when pe = 60 psia The nozzle is not choked. The exit Mach number is: Me = 0.5915 The exit temperature can be found from the Mach number: Te = 512.2 °R The sound speed at the exit is: ce = 1442.6 ft/s And so the exit flow speed is: Ve = 853.3 ft/s The density can be calculated using the ideal gas equation of state: lbm/ft3 ρe = 0.1751 The mass flow rate can then be calculated directly from continuity: m= 1.038 lbm/s Problem 13.55 [Difficulty: 2] Problem 13.56 [Difficulty: 2] Problem 13.57 [Difficulty: 3] Part 1/2 Problem 13.57 [Difficulty: 3] Part 2/2 Problem 13.58 Given: Rocket motor on test stand Find: Mass flow rate; thrust force [Difficulty: 3] k Solution: Basic equations: T0 T 1 k1 2 M 2 patm pe Ae Rx mrate Ve Given or available data p e 75 kPa k1 2 p 0 4 MPa so the nozzle exit area is T0 1 k 1 M 2 e 2 Then mrate ρe Ae Ve kg mrate 19.3 s p ρ R T c k R T mrate ρ A V T0 3250 K k 1.25 π 2 Ae d 4 Ae 491 cm k R Te pe patm Ae MCV ax Ve mrate Rx p e p atm Ae Ve mrate R 300 J kg K 2 M e 3.12 and m Ve 2313 s k 1 ce Ve M e ce Hence 2 Te 1467 K The exit speed is The momentum equation (Eq. 4.33) simplifies to M Momentum for pressure pe and velocity Ve at exit; Rx is the reaction force k 1 k 2 p 0 1 k1 p e Me The exit temperature is Te p 1 p atm 101 kPa d 25 cm From the pressures p0 Rx 43.5 kN ce 742 pe ρe R Te m s kg ρe 0.170 3 m Problem 13.59 [Difficulty: 3] Problem 13.60 [Difficulty: 3] Problem 13.61 [Difficulty: 3] Problem 13.62 [Difficulty: 4] Given: Compressed CO 2 in a cartridge expanding through a nozzle Find: Throat pressure; Mass flow rate; Thrust; Thrust increase with diverging section; Exit area Solution: Basic equations: Assumptions: 1) Isentropic flow 2) Stagnation in cartridge 3) Ideal gas 4) Uniform flow Given or available data: J k 1.29 R 188.9 p 0 35 MPa T0 ( 20 273 ) K p0 From isentropic relations p crit k 1 k 2 Since p b << pcrit, then p t p crit Throat is critical so mrate ρt Vt At Tt Vt At k1 d t 0.5 mm p crit 19.2 MPa k 1 Tt 256 K 2 k R Tt π d t p atm 101 kPa p t 19.2 MPa T0 1 1 kg K m Vt 250 s 2 4 At 1.963 10 7 pt ρt R Tt ρt 396 mrate ρt Vt At kg mrate 0.0194 s kg 3 m 2 m Rx p tgage At mrate Vt For 1D flow with no body force the momentum equation reduces to Rx mrate Vt p tgage At p tgage p t p atm Rx 8.60 N When a diverging section is added the nozzle can exit to atmospheric pressure p e p atm 1 k 1 k 2 p 0 Me 1 k 1 p e Hence the Mach number at exit is Te ce T0 1 k1 2 Me 2 k R Te 2 M e 4.334 Te 78.7 K ce 138 m s m Ve 600 s Ve M e ce The mass flow rate is unchanged (choked flow) Rx mrate Ve From the momentum equation The percentage increase in thrust is 11.67 N 8.60 N 8.60 N mrate ρe Ve Ae The exit area is obtained from mrate Ae ρe Ve T 35.7 % and pe ρe R Te ρe 6.79 6 T0 pt Tt Conv. Nozzle CD Nozzle Te s kg 3 m Ae 4.77 10 p0 pb Rx 11.67 N 2 m 2 Ae 4.77 mm Problem 13.63 [Difficulty: 3] Given: Rocket motor Find: Nozzle exit area, velocity, and thrust generated Solution: The given or available data is: R = k = p0 = T0 = 70.6 1.25 175 5400 At = pe = 1 14.7 ft-lbf/lbm-°R psia °R 2 in psia Equations and Computations: The exit Mach number can be calculated based on the pressure ratio: Me = 2.2647 The isentropic area ratio at this Mach number is: * 2.4151 Ae = 2.42 A e/A = So the nozzle exit area is: 2 in The exit temperature can be found from the Mach number: Te = 3290.4 °R The sound speed at the exit is: ce = 3057.8 ft/s And so the exit flow speed is: Ve = 6925.2 ft/s The density can be calculated using the ideal gas equation of state: 3 ρ e = 0.009112 lbm/ft The nozzle is choked, and we can use the choked flow equation: From p 0, T 0, A t, and Eq. 13.10a (13.10a) m = 1.058 lbm/s Based on the momentum equation, we can calculate the thrust generated: Rx = 228 lbf Note that since the flow expanded perfectly (the nozzle exit pressure is equal to the ambient pressure), the pressure terms drop out of the thrust calculation. Problem 13.64 [Difficulty: 4] Given: Rocket motor with converging-only nozzle Find: Nozzle exit pressure and thrust Solution: The given or available data is: R = k = p0 = T0 = 70.6 1.25 175 5400 At = pb = 1 14.7 ft-lbf/lbm-°R psia °R in2 psia Equations and Computations: If the diverging portion of the nozzle is removed, the exit Mach number is 1: The exit Mach number can be calculated based on the pressure ratio: Me = 1.0000 The isentropic area ratio at this Mach number is: A e/A * = 1.0000 So the nozzle exit area is: At = 1.00 in2 The exit temperature and pressure can be found from the Mach number: Te = 4800.0 °R pe = 97.1 psia The sound speed at the exit is: ce = 3693.2 ft/s And so the exit flow speed is: Ve = 3693.2 ft/s The nozzle is choked, and we can use the choked flow equation: From p 0, T 0, A t, and Eq. 13.9a (13.9a) m = 1.058 lbm/s Based on the momentum equation, we can calculate the thrust generated: F= 204 lbf Problem 13.65 [Difficulty: 3] Given: CO2 cartridge and convergent nozzle Find: Tank pressure to develop thrust of 15 N Solution: The given or available data is: R = k = T0 = pb = Dt = 188.9 1.29 293 101 0.5 J/kg·K K kPa mm At = 0.196 mm2 Equations and Computations: The momentum equation gives R x = m flowV e Hence, we need m flow and V e For isentropic flow pe = pe = pb 101 kPa If we knew p 0 we could use it and p e, and Eq. 13.7a, to find M e. Once M e is known, the other exit conditions can be found. Make a guess for p 0, and eventually use Goal Seek (see below). p0 = 44.6 MPa From p 0 and p e, and Eq. 13.7a (using built-in function IsenMfromp (M ,k ) (13.7a) Me = 4.5 From M e and T 0 and Eq. 13.7b (using built-in function IsenT (M ,k ) (13.7b) Te = 74.5 From T e and Eq. 12.18 K (12.18) Then ce = 134.8 m/s Ve = 606 m/s The mass flow rate is obtained from p 0, T 0, A t, and Eq. 13.10a (13.10a) m choked = 0.0248 kg/s Finally, the momentum equation gives R x = m flowV e = 15.0 We need to set R x to 15 N. To do this use Goal Seek to vary p 0 to obtain the result! N Problem 13.66 Given: Normal shock due to explosion Find: Shock speed; temperature and speed after shock [Difficulty: 3] V Shock speed Vs Shift coordinates: (Vs – V) (Vs) Solution: Basic equations: 2 M2 p2 p1 k1 V M c M k R T 2 k 2 k 1 M1 1 2 k 2 k1 Given or available data k 1.4 From the pressure ratio M1 Then we have Shock at rest 2 2 M1 M1 k1 k 1 2 2 1 2 M1 k M1 2 T1 2 k 1 M 2 2 1 T2 k1 k1 R 286.9 J kg K k 1 p2 k 2 k p1 k 1 1 k1 k 2 2 1 2 M1 k M1 2 T2 T1 2 k 1 M 2 2 1 2 M2 M1 p 2 30 MPa T2 14790 K T2 14517 °C 1 2 k1 M 2 0.382 2 k M 2 1 k 1 1 V1 M 1 k R T1 m V1 5475 s After the shock (V2) the speed is V2 M 2 k R T2 m V2 930 s V Vs V2 V 4545 V2 Vs V T1 ( 20 273 ) K M 1 16.0 Then the speed of the shock (Vs = V1) is But we have p 1 101 kPa Vs V1 m s These results are unrealistic because at the very high post-shock temperatures experienced, the specific heat ratio will NOT be constant! The extremely high initial air velocity and temperature will rapidly decrease as the shock wave expands in a spherical manner and thus weakens. m Vs 5475 s Problem 13.67 [Difficulty: 2] Given: Standing normal shock Find: Pressure and temperature ratios; entropy increase Solution: R = cp = k = M1 = 286.9 1004 1.4 1.75 p 2/p 1 = 3.41 T 2/T 1 = The entropy increase across the shock is: Δs = 1.495 The given or available data is: J/kg-K J/kg-K Equations and Computations: The pressure ratio is: The tempeature ratio is: 51.8 J/kg-K Problem 13.68 [Difficulty: 3] Given: Air flowing into converging duct, normal shock standing at duct exit Find: Mach number at duct entrance, duct area ratio Solution: The given or available data is: R = cp = k = M3 = p 2/p 1 = 286.9 1004 1.4 0.54 2 J/kg-K J/kg-K Equations and Computations: For the given post-shock Mach number, there can be only one Mach number upstream of the shock wave: M2 = 2.254 M3 = 0.5400 (We used Solver to match the post-shock Mach number by varying M 2.) The stagnation pressure is constant in the duct: p 0/p 2 = 11.643 p 0/p 1 = 23.285 So the duct entrance Mach number is: M1 = 2.70 The isentropic area ratios at stations 1 and 2 are: A 1/A * = * 3.1832 A 2/A = 2.1047 A 1/A 2 = 1.512 So the duct area ratio is: Problem 13.69 [Difficulty: 2] Given: Normal shock near pitot tube Find: Air speed Solution: Basic equations: k p 1 p 2 ρ1 V1 V2 V1 Given or available data T1 285 R p p 1 1.75 psi k 1.4 Rair 53.33 k 1 k p 2 02 1 k1 p 2 At state 2 M2 From momentum p 1 p 2 ρ2 V2 ρ1 V1 2 2 M1 Also c1 Then 2 p2 2 1 k M 2 1 k p1 1 k Rair T1 k1 2 M 2 p 02 10 psi k 1 p 2 8 psi ft lbf lbm R 2 2 2 but ρ V ρ c M or p1 1 k M1 p 2 R T k R T M k p M 2 2 2 p2 1 k M2 M 1 2.01 c1 827 ft s ft V1 1666 s V1 M 1 c1 ft V1 1822 s Note: With p1 = 1.5 psi we obtain (Using normal shock functions, for 1 M 2 0.574 2 p1 p2 k p2 M2 k p1 M1 Hence p0 (Momentum) p2 p1 4.571 we find M 1 2.02 M 2 0.573 Check!) Problem 13.70 [Difficulty: 3] Given: C-D nozzle with normal shock Find: Mach numbers at the shock and at exit; Stagnation and static pressures before and after the shock k 1 Solution: 1 k 1 M2 2 A 1 Basic equations: Isentropic flow k 1 Acrit M 2 2 M2 Normal shock Given or available data p2 k1 2 k M 2 1 k 1 1 k 1.4 Rair 53.33 2 At 1.5 in k p0 p1 k1 1 p 2 M 2 k 1 k k 1 M 2 1 2 k1 2 1 M1 2 2 2 M1 2 ( k 1) 2 k k1 2 M1 k1 p 02 k1 p 01 1 2 k M 2 k k 1 1 k ft lbf lbm R 2 As 2.5 in k 1 p 01 125 psi T0 ( 175 460 ) R (Shock area) Ae 3.5 in 1 k 1 1 2 Because we have a normal shock the CD must be accelerating the flow to supersonic so the throat is at critical state. Acrit At At the shock we have As Acrit k 1 1.667 At this area ratio we can find the Mach number before the shock from the isentropic relation 1 k 1 M 2 1 2 1 k 1 Acrit M1 2 2 ( k 1) As Solving iteratively (or using Excel's Solver, or even better the function isenMsupfromA from the Web site!) M 1 1.985 The stagnation pressure before the shock was given: p 01 125 psi The static pressure is then p1 p 01 k k 1 1 k 1 M 2 1 2 p 1 16.4 psi 2 After the shock we have M2 M1 2 k1 M 2 0.580 2 k 2 k 1 M1 1 k Also k 1 M 2 1 2 k1 2 1 M1 2 p 02 p 01 2 k M 2 k k 1 1 k and k 1 1 p 02 91.0 psi 1 k 1 1 2 k k 1 2 p 2 p 1 M1 k 1 k 1 p 2 72.4 psi Finally, for the Mach number at the exit, we could find the critical area change across the shock; instead we find the new critical area from isentropic conditions at state 2. 1 k 1 M 2 2 2 Acrit2 As M 2 k 1 2 At the exit we have Ae Acrit2 k 1 2 ( k 1) 2 Acrit2 2.06 in 1.698 At this area ratio we can find the Mach number before the shock from the isentropic relation k 1 1 k 1 M 2 e 2 1 k 1 Acrit2 Me 2 2 ( k 1) Ae Solving iteratively (or using Excel's Solver, or even better the function isenMsubfromA from the Web site!) These calculations are obviously a LOT easier using the Excel functions available on the Web site! M e 0.369 Problem 13.71 [Difficulty: 2] Problem 13.72 [Difficulty: 2] Problem 13.73 [Difficulty: 3] Given: Pitot probe used in supersonic wind tunnel nozzle Find: Pressure measured by pitot probe; nozzle exit velocity Solution: The given or available data is: R = k = M1 = p1 = T0 = 286.9 1.4 5 10 1450 J/kg-K kPa K Equations and Computations: Downstream of the normal shock wave, the Mach number is: M2 = 0.4152 The static and stagnation pressure ratios are: p 2/p 1 = 29.000 p 02/p 01 = 0.06172 So the static pressure after the shock is: p2 = 290 kPa The pitot pressure, however, is the stagnation pressure: p 02/p 2 = 1.12598 p 02 = 327 kPa The static temperature at the nozzle exit can be calculated: T 01/T 1 = 6.000 T1 = 241.67 K At the nozzle exit the sound speed is: c2 = 311.56 m/s Therefore the flow velocity at the nozzle exit is: V2 = 1558 m/s Problem 13.74 [Difficulty: 3] Given: Air approaching a normal shock Find: Pressure and velocity after the shock; pressure and velocity if flow were decelerated isentropically Solution: The given or available data is: R = k = V1 = p1 = T1 = 286.9 1.4 900 50 220 J/kg-K c1 = 297.26 m/s m/s kPa K Equations and Computations: The sonic velocity at station 1 is: So the Mach number at 1 is: M1 = 3.028 Downstream of the normal shock wave, the Mach number is: M2 = 0.4736 The static pressure and temperature ratios are: p 2/p 1 = 10.528 T 2/T 1 = 2.712 So the exit temperature and pressure are: p2 = 526 kPa T2 = 596.6 K At station 2 the sound speed is: c2 = 489.51 m/s Therefore the flow velocity is: V2 = 232 m/s If we decelerate the flow isentropically to M 2s = 0.4736 The isentropic pressure ratios at station 1 and 2s are: p 0/p 1 = 38.285 p 0/p 2s = 1.166 p 2s/p 1 = 32.834 So the final pressure is: p 2s = 1642 The temperature ratios are: T 0/T 1 = 2.833 T 0/T 2s = 1.045 T 2s/T 1 = 2.712 So the final temperature is: 596.6 T 2s = The sonic velocity at station 2s is: c 2s = 489.51 Therefore the flow velocity is: V 2s = 232 kPa K m/s m/s Problem 13.75 [Difficulty: 3] Given: Air accelerating through a converging-diverging nozzle, passes through a normal shock Find: Mach number before and after shock; entropy generation Solution: The given or available data is: R = k = p 01 = T 01 = T 01 = 53.33 1.4 150 400 860 ft-lbf/lbm-°R psia °F °R At = 3 in2 A1 = A2 = 6 in2 Equations and Computations: The isentropic area ratio at the station of interest is: A 1/A 1* = 2.00 So the Mach number at 1 is: M1 = 2.20 Downstream of the normal shock wave, the Mach number is: M2 = 0.547 The total pressure ratio across the normal shock is: p 02/p 01 = 0.6294 Since stagnation temperature does not change across a normal shock, the increase in entropy is related to the stagnation pressure loss only: ft-lbf/lbm-°R Δs 1-2 = 24.7 Btu/lbm-°R Δs 1-2 = 0.0317 Problem 13.76 [Difficulty: 2] Given: Normal shock Find: Speed and temperature after shock; Entropy change Solution: R = k = cp = The given or available data is: 53.33 1.4 0.2399 T 01 = p1 = M1 = 1250 20 2.5 1 = 0.0432 V1 = 4334 T 01 /T 1 = 2.25 ft·lbf/lbm·R 0.0685 Btu/lbm·R Btu/lbm·R o R psi Equations and Computations: From p1 1 RT1 slug/ft3 ft/s Using built-in function IsenT (M,k): T1 = o R o F o R 728 o F 143 psi 556 96 Using built-in function NormM2fromM (M,k): M2 = 0.513 Using built-in function NormTfromM (M,k): T 2 /T 1 = Using built-in function NormpfromM (M,k): p 2 /p 1 = From V 2 M 2 kRT 2 From T s c p ln 2 T1 2.14 T2 = 7.13 V2 = 867 s = 0.0476 37.1 p2 = ft/s p R ln 2 p1 Btu/lbm·R ft·lbf/lbm·R 1188 Problem 13.77 [Difficulty: 2] Problem 13.78 [Difficulty: 2] Given: Normal shock Find: Pressure after shock; Compare to isentropic deceleration Solution: R = k = T 01 = p 01 = M1 = 286.9 1.4 550 650 2.5 Using built-in function Isenp (M,k): p 01 /p 1 = 17.09 Using built-in function NormM2fromM (M,k): M2 = 0.513 Using built-in function NormpfromM (M,k): p 2 /p 1 = 7.13 Using built-in function Isenp (M,k) at M 2: p 02 /p 2 = 1.20 The given or available data is: J/kg·K K kPa Equations and Computations: But for the isentropic case: Hence for isentropic deceleration: p1 = 38 kPa p2 = 271 kPa p2 = 543 kPa p 02 = p 01 Problem 13.79 [Difficulty: 2] Given: Normal shock Find: Speed and Mach number after shock; Change in stagnation pressure Solution: The given or available data is: R = k = 53.33 1.4 T1 = p1 = V1 = 445 5 2000 c1 = M1 = 1034 2.84 ft·lbf/lbm·R o 0.0685 Btu/lbm·R R psi mph 2933 ft/s 793 ft/s Equations and Computations: From Then c1 kRT1 Using built-in function NormM2fromM (M,k): M2 = 0.486 Using built-in function NormdfromM (M,k): 2 / 1 = 3.70 Using built-in function Normp0fromM (M,k): p 02 /p 01 = 0.378 Then V2 1 V 2 1 V2 = 541 Using built-in function Isenp (M,k) at M 1: p 01 /p 1 = 28.7 ft/s mph From the above ratios and given p 1: p 01 = p 02 = p 01 – p 02 = 143 54.2 89.2 psi psi psi Problem 13.80 [Difficulty: 2] Given: Normal shock Find: Speed; Change in pressure; Compare to shockless deceleration Solution: The given or available data is: R = k = 53.33 1.4 T1 = p1 = V1 = 452.5 14.7 1750 c1 = M1 = 1043 2.46 ft·lbf/lbm·R o 0.0685 Btu/lbm·R R psi mph 2567 ft/s p2 = 101 psi p2 – p1 = 86.7 psi 781 ft/s p2 = 197 psi p2 – p1 = 182 psi Equations and Computations: From Then c1 kRT1 Using built-in function NormM2fromM (M,k): M2 = 0.517 Using built-in function NormdfromM (M,k): 2 / 1 = 3.29 Using built-in function NormpfromM (M,k): p 2 /p 1 = 6.90 Then V2 1 V 2 1 V2 = 532 Using built-in function Isenp (M,k) at M 1: p 01 /p 1 = 16.1 Using built-in function Isenp (M,k) at M 2: p 02 /p 2 = 1.20 From above ratios and p 1, for isentropic flow (p 0 = const): ft/s mph Problem 13.81 [Difficulty: 2] Problem 13.82 [Difficulty: 2] Problem 13.83 [Difficulty: 2] Problem 13.84 [Difficulty: 3] Given: Stagnation pressure and temperature probes on the nose of the Hyper-X Find: Pressure and temperature read by those probes Solution: The given or available data is: R = k = M1 = z= z= p SL = T SL = 53.33 1.4 9.68 110000 33528 14.696 518.76 ft-lbf/lbm-°R ft m psia °R Equations and Computations: At this altitude the local pressure and temperature are: p 1/p SL = 0.008643 p1 = 0.12702 psia °R T1 = 422.88 The stagnation pressure and temperature at these conditions are: p 01/p 1 = 34178.42 p 01 = 4341.36 psia T 01/T 1 = 19.74 °R T 01 = 8347.81 Downstream of the normal shock wave, the Mach number is: M2 = 0.3882 The total pressure ratio across the normal shock is: p 02/p 01 = 0.003543 So the pressure read by the probe is: p 02 = 15.38 psia Since stagnation temperature is constant across the shock, the probe reads: °R T 02 = 8348 Problem 13.85 Given: Normal shock Find: Rankine-Hugoniot relation [Difficulty: 4] Solution: Basic equations: 2 Momentum: p 1 ρ1 V1 p 2 ρ2 V2 Energy: h1 2 1 1 2 2 V1 h 2 V2 2 2 Mass: ρ1 V1 ρ2 V2 Ideal Gas: p ρ R T 2 2 From the energy equation 2 h 2 h 1 2 cp T2 T1 V1 V2 V1 V1 V1 V2 From the momentum equation p 2 p 1 ρ1 V1 ρ2 V2 ρ1 V1 V1 V2 Hence Using this in Eq 1 2 2 (1) where we have used the mass equation p2 p1 V1 V2 ρ1 V1 p2 p1 p2 p1 V2 p 2 p 1 ρ1 1 1 2 c p T2 T1 V1 V2 1 1 p 2 p 1 ρ1 V1 ρ1 ρ1 V1 ρ2 ρ1 ρ2 where we again used the mass equation Using the ideal gas equation p2 2 cp ρ2 R 1 1 p 2 p 1 ρ1 R ρ1 ρ2 p1 Dividing by p 1 and multiplying by ρ2, and using R = c p - cv, k = cp/cv 2 Collecting terms cp R p2 p1 p2 p1 2 k k 1 2 k p2 p1 For an infinite pressure ratio ρ2 ρ2 p 2 ρ2 k p2 2 1 1 ρ1 k 1 p1 ρ1 p1 ρ1 ρ2 ρ2 k 1 ρ1 ρ2 ρ1 2 k k 1 1 ( k 1) ( k 1) 2 k ρ2 ρ2 k 1ρ ρ 1 ρ1 1 1 1 ρ2 ρ1 1 ρ2 ρ1 0 ( k 1 ) ρ2 1 ( k 1 ) ρ1 ( k 1) ( k 1) or ρ2 or p2 p1 ( k 1) ρ1 ρ1 ( k 1) ( k 1) ( k 1) ρ1 ρ2 ρ2 k1 k1 ρ2 ρ1 (= 6 for air) Problem 13.86 [Difficulty: 3] Problem 13.87 [Difficulty: 3] Problem 13.88 [Difficulty: 3] Problem 13.89 [Difficulty: 4] Problem 13.90 [Difficulty: 3] Problem 13.91 [Difficulty: 4] Given: Air flowing through a wind tunnel, stagnation and test section conditions known Find: Throat area, mass flow rate, static conditions in test section, minumum diffuser area Solution: The given or available data is: R = k = p 01 = T 01 = T 01 = 53.33 1.4 14.7 75 535 A1 = M1 = 1 2.3 ft-lbf/lbm-°R psia °F °R ft2 A schematic of this wind tunnel is shown here: Equations and Computations: For the Mach number in the test section, the corresponding area ratio is: A 1/A 1* = 2.193 So the throat area is: ft2 0.456 At = The mass flow rate can be calculated using the choked flow equation: m= 22.2 lbm/s The static conditions in the test section are: p 01/p 1 = 12.5043 2.0580 T 01/T 1 = 1.176 psia p1 = °R T1 = 260 The strongest possible shock that can occur downstream of the first throat is when the shock wave is in the test section. The post-shock Mach number is then M2 = 0.5344 The area ratio corresponding to this Mach number is: A 2/A 2* = Therefore, the minimum diffuser throat area is A 2* = 1.2792 0.782 ft2 Problem 13.92 [Difficulty: 2] Problem 13.93 [Difficulty: 3] Problem 13.94 [Difficulty: 2] Problem 13.95 [Difficulty: 2] Problem 13.96 [Difficulty: 2] Problem 13.97 [Difficulty: 3] Given: Air accelerating through a converging-diverging nozzle Find: Pressure ratios needed to operate with isentropic flow throughout, supersonic flow at exit (third critical); isentropic flow throughout, subsonic flow at exit (first critical point); and isentropic flow throughout, supersonic flow in the diverging portion, and a normal shock at the exit (second critical point). Solution: The given or available data is: k = Md = 1.4 2.5 Equations and Computations: The pressure ratio for the third critical can be found from the design point Mach number: p 0inlet/p b,3rd = 17.0859 p b,3rd/p 0inlet = 0.0585 The area ratio for this nozzle is: A /A * = 2.637 So to operate at first critical the exit Mach number would be: M 1st = 0.226 Since at first critical the flow is isentropic, the pressure ratio is: p 0inlet/p b,1st = 1.0363 0.9650 p b,1st/p 0inlet = At second critical, the flow is isentropic to the exit, followed by a normal shock. At the design Mach number, the pressure ratio is: p b,2nd/p b,3rd = 7.125 Therefore, the back pressure ratio at the second critical is: 0.4170 p b,2nd/p 0inlet = p b,1st/p 0inlet = p b,2nd/p 0inlet = p b,3rd/p 0inlet = 0.9650 0.4170 0.0585 Problem 13.98 [Difficulty: 3] Given: Oxygen accelerating through a converging-diverging nozzle Find: Pressure ratios for critical points, show that a shock forms in the nozzle, pre- and postshock Mach numbers, exit Mach number Solution: The given or available data is: R = k = p 0inlet = pb = A e/A t = 48.29 1.4 120 50 3 ft-lbf/lbm-°R psia psia Equations and Computations: Based on the area ratio, the design Mach number is: Md = 2.637 The pressure ratio for the third critical can be found from the design point Mach number: 21.1422 p 0inlet/p b,3rd = p b,3rd/p 0inlet = 0.04730 If a normal shock exists in the nozzle, the pressure ratio should be between the first and second critical points. At the first critical point the exit Mach number is 0.197 M 1st = Since at first critical the flow is isentropic, the pressure ratio is: 1.0276 p 0inlet/p b,1st = p b,1st/p 0inlet = 0.9732 At second critical, the flow is isentropic to the exit, followed by a normal shock. At the design Mach number, the pressure ratio is: p b,2nd/p b,3rd = 7.949 Therefore, the back pressure ratio at the second critical is: p b,2nd/p 0inlet = 0.3760 The actual back pressure ratio is 0.4167 p b/p 0inlet = This pressure ratio is between those for the first and second critical points, so a shock exists in the nozzle. We need to use an iterative solution to find the exact location of the shock wave. Specifically, we iterate on the pre-shock Mach number until we match the exit pressure to the given back pressure: M1 = 2.55 A 1/A t = 2.759 p 0inlet/p 1 = 18.4233 p1 = 6.513 psia M2 = 0.508 p 2/p 1 = 7.4107 p2 = 48.269 psia A e/A 2 = 1.0873 A 2/A 2* = * A e/A 2 = Me = p 02/p 2 = p 02/p e = pe = (We used Goal Seek in Excel for this solution.) 1.324 1.440 0.454 1.193 1.152 50.000 psia Problem 13.99 [Difficulty: 2] Problem 13.100 [Difficulty: 3] Given: Normal shock in CD nozzle Find: Exit pressure; Throat area; Mass flow rate Solution: The given or available data is: R = k = T 01 = p 01 = M1 = 286.9 1.4 550 700 2.75 A1 = 25 cm2 Ae = 40 cm2 J/kg·K K kPa Equations and Computations (assuming State 1 and 2 before and after the shock): Using built-in function Isenp (M,k): p 01 /p 1 = 25.14 p1 = 28 kPa Using built-in function IsenT (M,k): T 01 /T 1 = 2.51 T1 = 219 K 3.34 A 1* = A t = 7.49 cm2 284 kPa Using built-in function IsenA (M,k): A 1 /A 1* = Then from the Ideal Gas equation: Also: So: Then the mass flow rate is: 1 = 0.4433 c1 = V1 = 297 815 m rate = m rate = kg/m3 m/s m/s 1 V 1A 1 0.904 kg/s For the normal shock: Using built-in function NormM2fromM (M,k): M2 = 0.492 Using built-in function Normp0fromM (M,k) at M 1: p 02 /p 01 = 0.41 p 02 = For isentropic flow after the shock: Using built-in function IsenA (M,k): But: Hence: A 2 /A 2* = A2 = 1.356 A1 A 2* = 18.44 cm2 Using built-in function IsenAMsubfromA (Aratio,k): A e /A 2* = 2.17 Me = 0.279 Using built-in function Isenp (M,k): p 02 /p e = 1.06 pe = 269 For: kPa Problem 13.101 [Difficulty: 2] Problem 13.102 [Difficulty: 3] Problem 13.103 [Difficulty: 3] Problem 13.104 [Difficulty: 3] Problem 13.105 [Difficulty: 3] Problem 13.106 [Difficulty: 3] Problem 13.107 [Difficulty: 3] Problem 13.108 [Difficulty: 4] Problem 13.109 [Difficulty: 3] Given: Air flowing through a converging-diverging nozzle with standing normal shock Find: Exit Mach number and static pressure; design point pressure Solution: The given or available data is: R = k = p 0inlet = T 01 = T 01 = A e/A t = A 1/A t = 53.33 1.4 150 200 660 1.76 1.2 ft-lbf/lbm-°R psia °F °R Equations and Computations: The pre-shock Mach number can be found based on the area ratio: 1.5341 M1 = The static pressure before the shock wave is: 3.8580 p 0inlet/p 1 = p1 = 38.881 psia The Mach number and static pressure after the shock wave are: M2 = 0.689 p 2/p 1 = 2.5792 p2 = 100.282 psia The area ratio for the remainder of the nozzle is: A e/A 2 = 1.4667 Based on this and the post-shock Mach number, we can determine the exit Mach number: A 2/A 2* = * A e/A 2 = Me = 1.102 1.617 0.392 Therefore the exit pressure is: p 02/p 2 = 1.374 p 02/p e = 1.112 pe = 123.9 psia Based on the area ratio, the design Mach number is: Md = 2.050 The pressure ratio for the third critical can be found from the design point Mach number: 8.4583 p 0inlet/p b,3rd = p b,3rd/p 0inlet = 0.1182 So the design pressure is: pd = 17.73 psia Problem 13.110 [Difficulty: 4] Problem 13.111 [Difficulty: 3] Problem 13.112 [Difficulty: 5] Given: Air flowing through a converging-diverging nozzle followed by duct with friction Find: Back pressure needed for (a) normal shock at nozzle exit, (b) normal shock at duct exit, (c) back pressure for shock-free flow Solution: The given or available data is: R = k = p 0inlet = T 0inlet = A e/A t = L/D = f= 286.9 1.4 1 320 2.5 10 0.03 J/kg-K MPa K Equations and Computations: (a) For a shock wave at the nozzle exit: The pre-shock Mach number can be found based on the area ratio: M1 = 2.4428 The static pressure before the shock wave is: 15.6288 p 0inlet/p 1 = p1 = 63.984 kPa The Mach number and static pressure after the shock wave are: M2 = 0.5187 p 2/p 1 = 6.7950 p2 = 434.770 kPa The friction length and critical pressure ratio after the shock wave are: fL/D 2 = 0.9269 p 2/p 2* = 2.0575 The friction length for the duct is: fL/D 2-3 = 0.3000 Therefore, the friction length at the duct exit is: fL/D 3 = 0.6269 Iterating on Mach number with Solver to match this friction length yields: M3 = 0.5692 fL/D 3 = 0.6269 The critical pressure ratio for this condition is: 1.8652 p 3/p 3* = Since the critical pressure at 2 and 3 are equal, the back pressure is: 394 kPa pb = p3 = (b) For a shock wave at the duct exit: We use the same nozzle exit Mach number and pressure: M1 = 2.443 p1 = 63.984 kPa The friction length and critical pressure ratio at this condition are: fL/D 1 = 0.4195 p 1/p 1* = 0.3028 The friction length for the duct is: fL/D 1-2 = 0.3000 Therefore, the friction length at the duct exit is: fL/D 2 = 0.1195 Iterating on Mach number with Solver to match this friction length yields: M2 = 1.4547 fL/D 2 = 0.1195 The critical pressure ratio for this condition is: p 2/p 2* = 0.6312 Since the critical pressure at 1 and 2 are equal, the pressure is: p2 = 133.388 kPa The Mach number and static pressure after the shock wave are: M3 = 0.7178 p 3/p 2 = 2.3021 pb = p3 = 307 kPa (c) For shock-free flow, we use the conditions from part b before the shock wave: 133.4 kPa pb = p3 = Problem 13.113 [Difficulty: 4] Given: Air flowing through a converging-diverging nozzle followed by diabatic duct Find: Mach number at duct exit and heat addition in duct Solution: The given or available data is: R = cp = k = p 0inlet = T 0inlet = A 1/A t = Te = 286.9 1004 1.4 1 320 2.5 350 J/kg-K J/kg-K MPa K K Equations and Computations: The Mach number at the nozzle exit can be found based on the area ratio: 2.4428 M1 = The static temperature is: T 0inlet/T 1 = 2.1934 T1 = 145.891 K The Rayliegh flow critical ratios at this condition are: T 1/T 1* = 0.39282 * T 01/T 01 = 0.71802 Since all we know is the static temperature at the exit, we need to iterate on a solution. We can guess at a pre-shock Mach number at the duct exit, and iterate on that value until we match the exit temperature: M2 = 1.753 T 2/T 2* = * 0.62964 T 02/T 02 = 0.84716 T2 = 233.844 K T 02 = 377.553 K M3 = 0.6274 T 3/T 2 = 1.4967 T3 = 350.000 K In this case we used Solver to match the exit temperature. Therefore, the exit Mach number is: M3 = 0.627 The rate of heat addition is calculated from the rise in stagnation temperature: q 1-2 = 57.78 kJ/kg Problem 13.114 [Difficulty: 2] Problem 13.115 [Difficulty: 5] Problem 13.116 [Difficulty: 4] Problem 13.117 [Difficulty: 2] Given: Nitrogen traveling through duct Find: Inlet pressure and mass flow rate Solution: The given or available data is: R = k = D = M2 = T2 = p2 = T1 = 296.8 1.4 30 0.85 300 200 330 J/kg-K cm K kPa K Equations and Computations: We can find the critical temperature and pressure for choking at station 2: T 2/T * = 1.0485 * T = 286.1 * 1.2047 p 2/p = K * p = 166.0 kPa Knowing the critical state, the Mach number at station 1 can be found: (we will use Goal Seek to match the Mach number) T 1/T * = M1 = 1.1533 0.4497 T 1/T * = 1.1533 The static to critical pressure ratio is a function of Mach number. Therefore: p1 = 396 kPa The sound speed at station 1 is: c1 = 370.30 m/s So the velocity at 1 is: V1 = 166.54 m/s The density at 1 can be calculated from the ideal gas equation of state: kg/m3 ρ1 = 4.0476 The area of the duct is: A = 0.0707 m2 m = 47.6 kg/s So the mass flow rate is: Problem 13.118 [Difficulty: 2] Given: Air flow in an insulated duct Find: Mass flow rate; Range of choked exit pressures k 1 Solution: T0 Basic equations: T Given or available data 1 k1 2 M 2 c 2 ( k 1) 1 k 1 M2 2 A 1 k 1 Acrit M 2 k R T T0 ( 80 460 ) R p 0 14.7 psi k 1.4 Rair 53.33 p 1 13 psi ft lbf lbm R D 1 in 2 A π D 2 A 0.785 in 4 Assuming isentropic flow, stagnation conditions are constant. Hence M1 k 1 k p 2 0 1 k1 p1 c1 Also M 1 0.423 k Rair T1 c1 341 p1 ρ1 Rair T1 m s ρ1 0.0673 mrate ρ1 V1 A When flow is choked M 2 1 hence We also have c2 From continuity ρ1 V1 ρ2 V2 k Rair T2 1 k1 2 M1 2 T1 521 R V1 M 1 c1 m V1 144 s T2 450 R T2 9.7 °F V2 c2 ft V2 1040 s T1 61.7 °F 3 lbm mrate 0.174 s T0 T2 k1 1 2 ft c2 1040 s V1 ρ2 ρ1 V2 p 2 ρ2 Rair T2 T0 lbm ft Hence Hence T1 ρ2 0.0306 lbm ft 3 p 2 5.11 psi The flow will therefore choke for any back pressure (pressure at the exit) less than or equal to this pressure (From Fanno line function p1 p crit 2.545 at M 1 0.423 so p crit p1 2.545 p crit 5.11 psi Check!) Problem 13.119 [Difficulty: 4] Given: Air flow from converging nozzle into pipe Find: Plot Ts diagram and pressure and speed curves Solution: The given or available data is: R = k = 53.33 1.4 ft·lbf/lbm·oR cp = 0.2399 Btu/lbm· R 187 T0 = p0 = pe = 710 25 24 Me= 0.242 Using built-in function IsenT (M ,k ) Te = 702 Using p e, M e, and function Fannop (M ,k ) p* = 5.34 Using T e, M e, and function FannoT (M ,k ) T* = 592 Equations and Computations: From p 0 and p e, and Eq. 13.7a (using built-in function IsenMfromp (M ,k )) o o ft·lbf/lbm· R R psi psi o o R psi o R We can now use Fanno-line relations to compute values for a range of Mach numbers: M T /T * 0.242 0.25 0.26 0.27 0.28 0.29 0.3 0.31 0.32 0.33 0.34 0.35 0.36 0.37 0.38 0.39 0.4 0.41 0.42 0.43 0.44 0.45 1.186 1.185 1.184 1.183 1.181 1.180 1.179 1.177 1.176 1.174 1.173 1.171 1.170 1.168 1.166 1.165 1.163 1.161 1.159 1.157 1.155 1.153 o c (ft/s) 702 701 701 700 699 698720 697 697700 696 695680 694 660 o 693 T ( R) 692640 691 690620 689 600 688 687580 686 0 685 684 682 1299 1298 1298 1297 1296 1296 1295 1294 1293 1292 1292 1291 1290 1289 1288 1287 1286 1285 1284 1283 1282 1281 T ( R) V (ft/s) p /p * p (psi) 315 4.50 24.0 325 4.35 23.2 337 4.19 22.3 350 Ts Curve 4.03 (Fanno) 21.5 363 3.88 20.7 376 3.75 20.0 388 3.62 19.3 401 3.50 18.7 414 3.39 18.1 427 3.28 17.5 439 3.19 17.0 452 3.09 16.5 464 3.00 16.0 477 2.92 15.6 489 2.84 15.2 502 2.77 14.8 514 2.70 14.4 527 2.63 14.0 539 2.56 13.7 10 20 30 552 2.50 13.4 . o s (ft lbf/lbm 564 2.44 13.0R) 576 2.39 12.7 s o (ft·lbf/lbm· R ) Eq. (12.11b) 0.00 1.57 3.50 5.35 7.11 8.80 10.43 11.98 13.48 14.92 16.30 17.63 18.91 20.14 21.33 22.48 23.58 24.65 25.68 26.67 27.63 28.55 40 50 0.46 0.47 0.48 0.49 0.5 0.51 0.52 0.53 0.54 0.55 0.56 0.57 0.58 0.59 0.6 0.61 0.62 0.63 0.64 0.65 0.66 0.67 0.68 0.69 0.7 0.71 0.72 0.73 0.74 0.75 0.76 0.77 0.78 0.79 0.8 0.81 0.82 0.83 0.84 0.85 0.86 0.87 0.88 0.89 0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1 1.151 1.149 1.147 1.145 1.143 1.141 1.138 1.136 1.134 1.132 1.129 1.127 1.124 1.122 1.119 1.117 1.114 1.112 1.109 1.107 1.104 1.101 1.098 1.096 1.093 1.090 1.087 1.084 1.082 1.079 1.076 1.073 1.070 1.067 1.064 1.061 1.058 1.055 1.052 1.048 1.045 1.042 1.039 1.036 1.033 1.029 1.026 1.023 1.020 1.017 1.013 1.010 1.007 1.003 1.000 681 1280 589 2.33 12.4 29.44 680 1279 601 2.28 12.2 30.31 679 1277 613 2.23 11.9 31.14 Velocity V 2.18 Versus M11.7 (Fanno) 31.94 677 1276 625 676 1275 638 2.14 11.4 32.72 675 1400 1274 650 2.09 11.2 33.46 674 1273 662 2.05 11.0 34.19 672 1200 1271 674 2.01 10.7 34.88 671 1270 686 1.97 10.5 35.56 669 1000 1269 698 1.93 10.3 36.21 668 1267 710 1.90 10.1 36.83 800 667 1266 722 1.86 9.9 37.44 V (ft/s) 665 600 1265 733 1.83 9.8 38.02 664 1263 745 1.80 9.6 38.58 662 400 1262 757 1.76 9.4 39.12 661 1260 769 1.73 9.2 39.64 659 200 1259 781 1.70 9.1 40.14 658 1258 792 1.67 8.9 40.62 0 656 1256 804 1.65 8.8 41.09 0.2 0.3 0.4 0.5 0.6 0.7 0.8 655 1255 815 1.62 8.6 41.53 M 8.5 653 1253 827 1.59 41.96 652 1252 839 1.57 8.4 42.37 650 1250 850 1.54 8.2 42.77 648 1248 861 1.52 8.1 43.15 647 1247 873 1.49 8.0 43.51 645 1245 884 Pressure p 1.47 Versus M7.8 (Fanno) 43.85 643 1244 895 1.45 7.7 44.18 642 30 1242 907 1.43 7.6 44.50 640 1240 918 1.41 7.5 44.80 638 25 1239 929 1.38 7.4 45.09 636 1237 940 1.36 7.3 45.36 635 1235 951 1.35 7.2 45.62 20 633 1234 962 1.33 7.1 45.86 631 1232 973 1.31 7.0 46.10 p (psi) 15 629 1230 984 1.29 6.9 46.31 628 1228 995 1.27 6.8 46.52 10 626 1227 1006 1.25 6.7 46.71 624 1225 1017 1.24 6.6 46.90 5 622 1223 1027 1.22 6.5 47.07 620 1221 1038 1.20 6.4 47.22 0 619 1219 1049 1.19 6.3 47.37 0.2 0.3 0.4 0.5 0.6 0.7 0.8 617 1218 1059 1.17 6.3 47.50 M 6.2 615 1216 1070 1.16 47.63 613 1214 1080 1.14 6.1 47.74 611 1212 1091 1.13 6.0 47.84 609 1210 1101 1.11 6.0 47.94 607 1208 1112 1.10 5.9 48.02 605 1206 1122 1.09 5.8 48.09 603 1204 1132 1.07 5.7 48.15 601 1202 1142 1.06 5.7 48.20 600 1201 1153 1.05 5.6 48.24 598 1199 1163 1.04 5.5 48.27 596 1197 1173 1.02 5.5 48.30 594 1195 1183 1.01 5.4 48.31 592 1193 1193 1.00 5.3 48.31 0.9 1.0 0.9 1.0 Problem 13.120 [Difficulty: 4] Given: Air flow from converging-diverging nozzle into pipe Find: Plot Ts diagram and pressure and speed curves Solution: The given or available data is: R = k = 53.33 1.4 ft·lbf/lbm·oR cp = 0.2399 Btu/lbm·oR 187 T0 = p0 = pe = 710 25 2.5 Me = 2.16 Using built-in function IsenT (M ,k ) Te = 368 Using p e, M e, and function Fannop (M ,k ) p* = 6.84 Using T e, M e, and function FannoT (M ,k ) T* = 592 Equations and Computations: From p 0 and p e, and Eq. 13.7a (using built-in function IsenMfromp (M ,k )) ft·lbf/lbm·oR R o psi psi o R psi o R We can now use Fanno-line relations to compute values for a range of Mach numbers: M T /T * 2.157 2 1.99 1.98 1.97 1.96 1.95 1.94 1.93 1.92 1.91 1.9 1.89 1.88 1.87 1.86 1.85 1.84 1.83 1.82 1.81 1.8 1.79 1.78 1.77 1.76 1.75 1.74 1.73 1.72 1.71 0.622 0.667 0.670 0.673 0.676 0.679 0.682 0.685 0.688 0.691 0.694 0.697 0.700 0.703 0.706 0.709 0.712 0.716 0.719 0.722 0.725 0.728 0.731 0.735 0.738 0.741 0.744 0.747 0.751 0.754 0.757 T (oR) 368 394 396 398 400 402 650 403 405 600 407 409 550 410 500 412 T (oR) 414 450 416 418 400 420 350 421 423 300 425 0 427 429 431 433 435 436 438 440 442 444 446 448 c (ft/s) 940 974 976 978 980 982 985 987 989 991 993 996 998 1000 1002 1004 1007 1009 1011 5 1013 1015 1018 1020 1022 1024 1027 1029 1031 1033 1036 1038 V (ft/s) p /p * p (psi) s (ft·lbf/lbm·oR) Eq. (12.11b) 2028 0.37 2.5 1948 0.41 2.8 1942 0.41 2.8 1937 Ts Curve 0.41 (Fanno) 2.8 1931 0.42 2.9 1926 0.42 2.9 1920 0.42 2.9 1914 0.43 2.9 1909 0.43 2.9 1903 0.43 3.0 1897 0.44 3.0 1892 0.44 3.0 1886 0.44 3.0 1880 0.45 3.1 1874 0.45 3.1 1868 0.45 3.1 1862 0.46 3.1 1856 0.46 3.1 1850 0.46 10 15 20 3.2 25 1844 0.47 3.2 . o s (ft lbf/lbm R) 1838 0.47 3.2 1832 0.47 3.2 1826 0.48 3.3 1819 0.48 3.3 1813 0.49 3.3 1807 0.49 3.3 1801 0.49 3.4 1794 0.50 3.4 1788 0.50 3.4 1781 0.50 3.5 1775 0.51 3.5 0.00 7.18 7.63 8.07 8.51 8.95 9.38 9.82 10.25 10.68 11.11 11.54 11.96 12.38 12.80 13.22 13.64 14.05 14.46 30 14.87 15.28 15.68 16.08 16.48 16.88 17.27 17.66 18.05 18.44 18.82 19.20 35 40 1.7 1.69 1.68 1.67 1.66 1.65 1.64 1.63 1.62 1.61 1.6 1.59 1.58 1.57 1.56 1.55 1.54 1.53 1.52 1.51 1.5 1.49 1.48 1.47 1.46 1.45 1.44 1.43 1.42 1.41 1.4 1.39 1.38 1.37 1.36 1.35 1.34 1.33 1.32 1.31 1.3 1.29 1.28 1.27 1.26 1.25 1.24 1.23 1.22 1.21 1.2 1.19 1.18 1.17 1.16 1.15 1.14 1.13 1.12 1.11 1.1 1.09 1.08 1.07 1.06 1.05 1.04 1.03 1.02 1.01 1 0.760 0.764 0.767 0.770 0.774 0.777 0.780 0.784 0.787 0.790 0.794 0.797 0.800 0.804 0.807 0.811 0.814 0.817 0.821 0.824 0.828 0.831 0.834 0.838 0.841 0.845 0.848 0.852 0.855 0.859 0.862 0.866 0.869 0.872 0.876 0.879 0.883 0.886 0.890 0.893 0.897 0.900 0.904 0.907 0.911 0.914 0.918 0.921 0.925 0.928 0.932 0.935 0.939 0.942 0.946 0.949 0.952 0.956 0.959 0.963 0.966 0.970 0.973 0.976 0.980 0.983 0.987 0.990 0.993 0.997 1.000 450 1040 452 1042 454 1045 456 1047 458 1049 460 1051 462 1054 464 1056 466 1058 468 1060 470 1063 472 1065 2500 474 1067 476 1069 478 2000 1072 480 1074 482 1500 1076 484 1078 V (ft/s) 486 1080 488 1000 1083 490 1085 492 500 1087 494 1089 496 1092 0 498 1094 2.0 500 1096 502 1098 504 1101 506 1103 508 1105 510 1107 512 1110 514 1112 516 1114 8 518 1116 520 7 1118 522 1121 6 524 1123 527 5 1125 529 1127 p (psi) 4 531 1129 533 3 1132 535 1134 2 537 1136 539 1 1138 541 1140 0 543 1143 2.0 545 1145 547 1147 549 1149 551 1151 553 1153 555 1155 557 1158 559 1160 561 1162 564 1164 566 1166 568 1168 570 1170 572 1172 574 1174 576 1176 578 1179 580 1181 582 1183 584 1185 586 1187 588 1189 590 1191 592 1193 1768 0.51 3.5 19.58 1761 0.52 3.5 19.95 1755 0.52 3.6 20.32 1748 0.53 3.6 20.69 1741 0.53 3.6 21.06 1735 0.53 3.7 21.42 1728 0.54 3.7 21.78 1721 0.54 3.7 22.14 1714 0.55 3.7 22.49 1707 Velocity V 0.55 Versus M 3.8 (Fanno) 22.84 1700 0.56 3.8 23.18 1693 0.56 3.8 23.52 1686 0.57 3.9 23.86 1679 0.57 3.9 24.20 1672 0.58 3.9 24.53 1664 0.58 4.0 24.86 1657 0.59 4.0 25.18 1650 0.59 4.0 25.50 1642 0.60 4.1 25.82 1635 0.60 4.1 26.13 1627 0.61 4.1 26.44 1620 0.61 4.2 26.75 1612 0.62 4.2 27.05 1605 0.62 4.3 27.34 1597 0.63 4.3 27.63 1.8 1.6 1.4 1589 0.63 4.3 27.92 M 4.4 1582 0.64 28.21 1574 0.65 4.4 28.48 1566 0.65 4.5 28.76 1558 0.66 4.5 29.03 1550 0.66 4.5 29.29 1542 Pressure p 0.67 Versus M 4.6 (Fanno) 29.55 1534 0.68 4.6 29.81 1526 0.68 4.7 30.06 1518 0.69 4.7 30.31 1510 0.69 4.8 30.55 1502 0.70 4.8 30.78 1493 0.71 4.8 31.01 1485 0.71 4.9 31.24 1477 0.72 4.9 31.46 1468 0.73 5.0 31.67 1460 0.74 5.0 31.88 1451 0.74 5.1 32.09 1443 0.75 5.1 32.28 1434 0.76 5.2 32.48 1426 0.76 5.2 32.66 1417 0.77 5.3 32.84 1.8 1.6 1.4 1408 0.78 5.3 33.01 M 5.4 1399 0.79 33.18 1390 0.80 5.4 33.34 1381 0.80 5.5 33.50 1372 0.81 5.6 33.65 1363 0.82 5.6 33.79 1354 0.83 5.7 33.93 1345 0.84 5.7 34.05 1336 0.85 5.8 34.18 1327 0.86 5.9 34.29 1318 0.87 5.9 34.40 1308 0.87 6.0 34.50 1299 0.88 6.0 34.59 1290 0.89 6.1 34.68 1280 0.90 6.2 34.76 1271 0.91 6.2 34.83 1261 0.92 6.3 34.89 1251 0.93 6.4 34.95 1242 0.94 6.5 34.99 1232 0.96 6.5 35.03 1222 0.97 6.6 35.06 1212 0.98 6.7 35.08 1203 0.99 6.8 35.10 1193 1.00 6.8 35.10 1.2 1.2 1.0 1.0 Problem 13.121 [Difficulty: 3] Given: Oxygen traveling through duct Find: Inlet and exit Mach numbers, exit stagnation conditions, friction factor and absolute roughness Solution: The given or available data is: R = k = D = L = m = p1 = T1 = p2 = 259.8 1.4 35 5 40.0 200 450 160 J/kg-K A = 0.0962 m2 cm m kg/s kPa K kPa Equations and Computations: The area of the duct is: The sound speed at station 1 is: c1 = 404.57 m/s The density at 1 can be calculated from the ideal gas equation of state: kg/m3 ρ1 = 1.7107 So the velocity at 1 is: V1 = 243.03 m/s and the Mach number at 1 is: M1 = 0.601 The critical temperature and pressure may then be calculated: p 1/p * = 1.7611 * p = 113.6 * 1.1192 T 1/T = kPa * T = 402.1 K Since the critical pressure is equal at 1 and 2, we can find the pressure ratio at 2: p 2/p * = 1.4089 The static to critical pressure ratio is a function of Mach number. Therefore: M2 = 0.738 p 2/p * = 1.4089 (we used Solver to find the correct Mach number to match the pressure ratio) The exit temperature is: T 2/T * = T2 = 1.0820 435.0 K Based on the exit Mach number, pressure, and temperature, stagnation conditions are: p 02 = 230 kPa T 02 = 482 K The maximum friction lengths at stations 1 and 2 are: fL 1/D = 0.48802 0.14357 fL 2/D = So the friction length for this duct is: fL /D = 0.34445 and the friction factor is: f = 0.02411 Now to find the roughness of the pipe, we need the Reynolds number. From the LMNO Engineering website, we can find the viscosities of oxygen: 2 μ 1 = 2.688E-05 N-s/m 2 μ 2 = 2.802E-05 N-s/m Therefore the Reynolds number at station 1 is: Re1 = 5.413E+06 At station 2, we will need to find density and velocity first. From ideal gas equation: kg/m3 ρ2 = 1.4156 The sound speed at 2 is: c2 = 397.79 m/s So the velocity at 2 is: V2 = 293.69 m/s and the Reynolds number is: Re2 = 5.193E+06 So the Reynolds number does not change significantly over the length of duct. We will use an average of the two to find the relative roughness: Re = 5.303E+06 The relative roughness for this pipe is: e/D = 0.00222 f = 0.02411 (we used Solver to find the correct roughness to match the friction factor.) Therefore, the roughness of the duct material is: e = 0.0776 cm Problem 13.122 [Difficulty: 3] Given: Air flow in a converging nozzle and insulated duct Find: Pressure at end of duct; Entropy increase k Solution: T0 Basic equations: T Given or available data 1 k1 2 M 2 p0 p k1 1 2 T0 ( 250 460 ) R p 0 145 psi k 1.4 cp 0.2399 M 2 k 1 T2 p2 ∆s cp ln Rair ln T1 p1 p 1 125 psi BTU Rair 53.33 lbm R c k R T T2 ( 150 460 ) R ft lbf lbm R Assuming isentropic flow in the nozzle k 1 k p 2 0 1 k1 p1 M1 M 1 0.465 In the duct T0 (a measure of total energy) is constant, soM 2 At each location Then k Rair T1 c1 1279 ft c2 k Rair T2 c2 1211 ft p 2 ρ2 Rair T2 p 2 60.8 psi (Note: Using Fanno line relations, at M 1 0.465 Finall y Tcrit p crit T2 Tcrit 1.031 so M 2 0.907 p2 p crit 2 M1 2 T1 681 R T1 221 °F M 2 0.905 V1 M 1 c1 ft V1 595 s V2 M 2 c2 ft V2 1096 s V1 ρ2 ρ1 V2 ρ2 0.269 lbm so p1 Then s ft T1 1 k1 s ρ1 0.4960 mrate ρ1 V1 A ρ2 V2 A Hence T0 T0 1 k1 T2 2 c1 p1 ρ1 Rair T1 Also T1 1.150 3 lbm ft 3 T2 p2 BTU ∆s cp ln Rair ln ∆s 0.0231 lbm R T1 p1 T1 Tcrit Tcrit 329 K 1.150 p1 2.306 p crit 1.119 p 2 1.119 p crit 2.3060 p crit 54.2 psi p 2 60.7 psi Check!) Problem 13.123 [Difficulty: 2] Problem 13.124 [Difficulty: 3] Problem 13.125 [Difficulty: 3] Problem 13.126 [Difficulty: 3] Given: Nitrogen traveling through C-D nozzle and constant-area duct with friction Find: Exit temperature and pressure Solution: The given or available data is: R = k = p 01 = T 01 = T 01 = A e/A t = fL /D = 55.16 1.4 105 100 560 4 0.355 ft-lbf/lbm-°R psia °F °R Equations and Computations: Based on the area ratio of the nozzle, we can find the nozzle exit Mach number: M1 = 2.940 The pressure and temperature at station 1 are therefore: p1 = 3.128 psia °R T1 = 205.2 The critical temperature, pressure, and maximum friction length at 1 are: p 1/p * = 0.2255 * p = 13.867 * 0.4397 T 1/T = psia * T = °R 466.7 fL 1/D = 0.51293 Based on the maximum and actual friction lengths, the maximum friction length at station 2 is: fL 2/D = 0.15793 So the exit Mach number is: M2 = 1.560 fL 2/D = 0.15793 (we used Solver to find the correct Mach number to match the friction length) The critical pressure and temperature ratios at station 2 are: p 2/p * = * T 2/T = So the exit temperature and pressure are: p2 = T2 = 0.5759 0.8071 7.99 377 psia °R Problem 13.127 [Difficulty: 3] Given: Air flow in a CD nozzle and insulated duct Find: Temperature at end of duct; Force on duct; Entropy increase Solution: Basic equations: Given or available data Fs p 1 A p 2 A Rx mrate V2 V1 T0 T1 ( 100 460 ) R p 1 18.5 psi k 1.4 cp 0.2399 T k1 1 2 M T2 p2 ∆s cp ln Rair ln T1 p1 2 M1 2 2 M2 1 BTU Rair 53.33 lbm R A 1 in ft lbf lbm R Assuming isentropic flow in the nozzle k1 2 1 M1 T0 T2 2 so T1 T0 k1 2 1 M2 2 Also c1 mrate ρ1 V1 A ρ1 0.0892 lbm ft 1 3 2 k1 2 M1 2 M2 2 c2 T2 840 R k Rair T2 V2 M 2 c2 V1 ρ2 ρ1 V2 so p 2 ρ2 Rair T2 Hence Rx p 2 p 1 A mrate V2 V1 Finally T2 p2 ∆s cp ln Rair ln T1 p1 (Note: Using Fanno line relations, at M 1 2 k1 mrate ρ1 V1 A ρ2 V2 A2 lbm mrate 1.44 s T2 T1 ft V1 2320 s k Rair T1 V1 M 1 c1 p1 ρ1 Rair T1 1 T1 Tcrit p1 p crit T1 T2 p1 Rx 13.3 lbf p2 0.6667 0.4083 ft V2 1421 s ρ2 0.146 (Force is to the right) BTU lbm R T2 p2 p1 0.4083 T1 0.667 p 2 45.3 psi lbm ft p 2 45.3 psi ∆s 0.0359 T2 380 °F T2 840 R Check!) 3 Problem 13.128 [Difficulty: 3] Problem 13.129 [Difficulty: 4] Problem 13.130 [Difficulty: 4] Given: Air traveling through converging nozzle and constant-area duct with friction; Find: choked flow at duct exit. Pressure at end of duct; exit conditions if 80% of duct were removed Solution: The given or available data is: R = k = p1 = T1 = 286.9 1.4 600 550 J/kg-K kPa K Equations and Computations: Station 1 is a stagnation state, station 2 is between the nozzle and friction duct, and station 3 is at the duct exit. For part (a) we know: fL 2-3/D = 5.3 M3 = 1 Therefore, we can make the following statements: fL 3/D = 0 fL 2/D = 5.300 So the Mach number at the duct entrance is: M2 = 0.300 fL 2/D = 5.300 (we used Solver to find the correct Mach number to match the friction length) The pressure at station 2 can be found from the Mach number and stagnation state: p 1/p 2 = 1.0644 p2 = 563.69 kPa Since state 3 is the critical state, we can find the pressure at state 3: p 2/p * = * 3.6193 p = 155.75 kPa p3 = 155.7 kPa For part (a) we know that if we remove 80% of the duct: fL 2-3/D = 1.06 M2 = 0.300 fL 2/D = 5.300 p2 = 563.69 kPa Since we know state 2 and the friction length of the duct, we can find state 3: fL 3/D = 4.240 So the Mach number at the duct exit is: M3 = 0.326 fL 2/D = 4.240 (we used Solver to find the correct Mach number to match the friction length) To find the exit pressure: p 2/p * = * 3.6193 p = 155.75 p 3/p * = 3.3299 kPa At state 3 the pressure ratio is: So the pressure is: p3 = 519 kPa These processes are plotted in the Ts diagram below: T p1 T1 p2 p 3short p* * s Problem 13.131 [Difficulty: 4] Problem 13.132 [Difficulty: 3] Problem 13.133 [Difficulty: 2] Given: Air flow in a converging nozzle and insulated duct Find: Length of pipe Solution: Basic equations: Fanno-line flow equations, and friction factor Given or available data T0 ( 250 460 ) R p 0 145 psi p 1 125 psi D 2 in k 1.4 cp 0.2399 T2 ( 150 460 ) R BTU Rair 53.33 lbm R ft lbf lbm R 1 From isentropic relations k 1 k 2 p0 M1 1 k 1 p1 T0 T1 Then for Fanno-line flow 1 k1 fave Lmax1 Dh 2 M 1 so 2 1 M1 k M1 2 2 1 2 M 1 0.465 T1 T0 T1 681 R 1 k 1 M 2 1 2 ( k 1) M 2 1 ln 1.3923 k1 2 k 2 M1 2 1 2 k1 2 k 1 p1 p1 1 2 2.3044 p crit p2 M1 k1 2 1 M1 2 p crit p1 2.3044 k 1 T1 Tcrit p crit 54.2 psi 2 1 k1 Also, for Tcrit Then 1.031 T2 Tcrit 2 1 fave Lmax2 Dh k1 2 2 1.150 M1 2 Tcrit 592 R k 1 T2 M2 1 M2 k M2 2 leads to 2 2 T1 221 °F M2 2 k1 T1 1.150 Tcrit 132 °F k 1 Tcrit 2 ( k 1) M 2 2 ln 0.01271 2 k k1 2 M2 2 1 2 k1 Tcrit T2 1 M 2 0.906 Also p1 ρ1 Rair T1 ρ1 0.496 For air at T1 221 °F, from Table A.9 (approximately) lbm ft 7 lbf s μ 4.48 10 ft For commercial steel pipe (Table 8.1) e 0.00015 ft Hence at this Reynolds number and roughness (Eq. 8.37) Combining results e D 4 9 10 so 2 and Re1 ρ1 V1 D μ 6 Re1 3.41 10 f 0.01924 2 ft f L f L 12 D ave max2 ave max1 L12 .01924 ( 1.3923 0.01271 ) Dh Dh f These calculations are a LOT easier using the Excel Add-ins! ft V1 595 s V1 M 1 k Rair T1 3 L12 12.0 ft Problem 13.134 [Difficulty: 2] Problem 13.135 [Difficulty: 3] Given: Air traveling through a square duct Find: Entrance static and stagnation conditions; friction factor Solution: The given or available data is: R = k = s= L= M1 = M2 = T2 = p2 = 53.33 1.4 2 40 3 1.7 500 110 ft-lbf/lbm-°R ft ft °R psia Equations and Computations: From the entrance Mach number we can calculate: p 01/p 1 = 36.7327 T 01/T 1 = 2.8000 p 1/p * = * T 1/T = fL 1/D = From the exit Mach number we can calculate: p 2/p * = 0.2182 0.4286 0.52216 0.5130 * 0.7605 T 2/T = 0.20780 fL 2/D = Since we know static conditions at 2, we can find the critical pressure and temperature: p* = 214.4 psia * T = °R 657.5 Therefore, the static conditions at the duct entrance are: p1 = 46.8 psia °R 282 T1 = and from the isentropic relations we can find stagnation conditions: p 01 = 1719 psia °R T 01 = 789 To find the friction factor of the duct, first we need to friction length: fL 1-2/D = 0.31436 The area and perimeter of the duct are: ft2 A= 4.0 P= 8.0 ft Therefore the hydraulic diameter of the duct is: 2.0 ft DH = From the hydraulic diameter, length, and friction length, the friction factor is: f= 0.01572 Problem 13.136 [Difficulty: 3] Given: Air traveling through a cast iron pipe Find: Friction factor needed for sonic flow at exit; inlet pressure Solution: The given or available data is: R = k = D= L= M1 = T1 = T1 = M2 = p2 = 53.33 1.4 3.068 10 0.5 70 530 1 14.7 ft-lbf/lbm-°R in ft °F °R psia Equations and Computations: From the entrance Mach number we can calculate: p 1/p * = fL 1/D = From the exit Mach number we can calculate: 2.1381 1.06906 p 2/p * = 1.0000 fL 2/D = 0.00000 To find the friction factor of the duct, first we need to friction length: fL 1-2/D = 1.06906 Based on this, and the pipe length and diameter, the friction factor is: f= 0.0273 We can calculate the critical pressure from the exit pressure: p* = 14.7 Therefore, the static pressure at the duct entrance is: p1 = 31.4 psia psia Problem 13.137 [Difficulty: 3] Part 1/2 Problem 13.137 [Difficulty: 3] Part 2/2 Problem 13.138 [Difficulty: 3] Problem 13.139 Example 13.8 [Difficulty: 3] Problem 13.140 [Difficulty: 3] Problem 13.141 [Difficulty: 4] Part 1/2 Problem 13.141 [Difficulty: 4] Part 2/2 Problem 13.142 Given: Air flow through a CD nozzle and tube. Find: Average friction factor; Pressure drop in tube [Difficulty: 2] Solution: Assumptions: 1) Isentropic flow in nozzle 2) Adiabatic flow in tube 3) Ideal gas 4) Uniform flow Given or available data: J k 1.40 R 286.9 p 0 1.35 MPa T0 550 K kg K p 1 15 kPa where State 1 is the nozzle exit D 2.5 cm L 1.5 m 1 k 1 k 2 p0 From isentropic relations M 1 1 k 1 p1 2 M 1 3.617 Then for Fanno-line flow (for choking at the exit) ( k 1) M 2 1 ln 0.599 Dh 2 k1 2 k 2 k M1 M1 2 1 2 2 ( k 1) M 2 D 1 M1 1 k1 fave fave 0.0100 ln 2 L k1 2 k 2 k M 2 1 M 1 1 2 fave Lmax Hence 1 M1 2 k1 1 2 k 1 p1 p1 1 2 0.159 p crit p2 M1 k1 2 1 M1 2 p2 p1 1 2 k 1 1 2 M k1 2 M1 1 1 2 ∆p p 1 p 2 p 2 94.2 kPa ∆p 79.2 kPa These calculations are a LOT easier using the Excel Add-ins! Problem 13.143 [Difficulty: 3] Given: Air flow in a CD nozzle and insulated duct Find: Duct length; Plot of M and p Solution: Basic equations: Fanno-line flow equations, and friction factor Given or available data T1 ( 100 460 ) R p 1 18.5 psi k 1.4 cp 0.2399 Then for Fanno-line flow at M 1 2 M1 2 BTU Rair 53.33 lbm R 2 and at M 2 1 Also p crit p1 Dh 1 M2 k M2 2 2 lbm R fave Lmax1 Dh 1 M1 k M1 2 2 ( k 1) M 2 1 ln 0.305 k1 2 k 2 M1 2 1 2 k1 ( k 1) M 2 2 ln 0 k1 2 k 2 M2 2 1 2 k1 p1 lbm ρ1 ρ1 0.089 Rair T1 3 ft V1 M 1 k Rair T1 7 lbf s For air at T1 100 °F, from Table A.9 μ 3.96 10 ft For commercial steel pipe (Table 8.1) ft lbf p crit 45.3 psi 0.4082 fave Lmax2 A 1 in 1 k 1 p1 p1 1 2 0.4082 p crit p2 M1 k1 2 1 M1 2 so 2 M2 1 e 0.00015 ft Hence at this Reynolds number and roughness (Eq. 8.37) e D 1.595 10 3 ft V1 2320 s so 2 and 4 A D Re1 π D 1.13 in ρ1 V1 D μ 6 Re1 1.53 10 f .02222 1.13 Combining results ft fave Lmax1 D fave Lmax2 12 L12 .02222 ( 0.3050 0 ) Dh Dh f L12 1.29 ft L12 15.5 in These calculations are a LOT easier using the Excel Add-ins! The M and p plots are shown in the Excel spreadsheet on the next page. The given or available data is: M 2.00 1.95 1.90 1.85 1.80 1.75 1.70 1.65 1.60 1.55 1.50 1.45 1.40 1.35 1.30 1.25 1.20 1.15 1.10 1.05 1.00 fL m ax/D ΔfL ma x/D 0.305 0.000 0.290 0.015 0.274 0.031 0.258 0.047 0.242 0.063 0.225 0.080 0.208 0.097 0.190 0.115 0.172 0.133 0.154 0.151 0.136 0.169 0.118 0.187 0.100 0.205 0.082 0.223 0.065 0.240 0.049 0.256 0.034 0.271 0.021 0.284 0.010 0.295 0.003 0.302 0.000 0.305 f = 0.0222 p * = 45.3 kPa D = 1.13 in Fanno Line Flow Curves(M and p ) x (in) p /p * p (psi) 0 0.8 1.6 2.4 3.2 4.1 4.9 5.8 6.7 7.7 8.6 9.5 10.4 11.3 12.2 13.0 13.8 14.5 15.0 15.4 15.5 0.408 0.423 0.439 0.456 0.474 0.493 0.513 0.534 0.557 0.581 0.606 0.634 0.663 0.695 0.728 0.765 0.804 0.847 0.894 0.944 1.000 18.49 19.18 19.90 20.67 21.48 22.33 23.24 24.20 25.22 26.31 27.47 28.71 30.04 31.47 33.00 34.65 36.44 38.37 40.48 42.78 45.30 2.0 45 1.9 40 1.8 1.7 35 1.6 M 1.5 30 p (psi) 1.4 25 1.3 M 1.2 20 Pressure 1.1 1.0 15 0 4 8 x (in) 12 16 Problem 13.144 [Difficulty: 3] Problem 13.145 [Difficulty: 4] Given: Natural gas pumped through a pipe Find: Required entrance pressure and power needed to pump gas through the pipe Solution: The given or available data is: R = cp = k = D= L= f= T1 = T1 = T2 = m= p2 = 96.32 0.5231 1.31 30 60 0.025 140 600 600 40 150 Equations and Computations: At the exit of the pipe we can calculate the density: p2 = 21.756 ft-lbf/lbm-°R Btu/lbm-°R in mi °F °R °R lbm/s kPa psia lbm/ft3 ρ2 = 0.05421 A= 4.909 ft2 V2 = 150.32 ft/s c2 = 1561.3 ft/s The pipe area is: Therefore, the flow velocity is: The local sound speed is: So the Mach number is: M2 = 0.09628 From the exit Mach number we can calculate: T 02/T 2 = 1.0014 fL 2/D = 76.94219 Given the length, diameter, and friction factor, we know: fL 1-2/D = 3168.0 Therefore: fL 1/D = 3244.9 So from this information we can calculate the entrance Mach number: M1 = 0.01532 fL 1/D = 3244.9 (We use Solver to calculate the Mach number based on the friction length) The entrance sound speed is the same as that at the exit: c1 = 1561.3 ft/s So the flow velocity is: V1 = 23.91 ft/s We can calculate the pressure ratio from the velocity ratio: p1 = 136.8 psi From the entrance Mach number we can calculate: T 01/T 1 = 1.0000 So the entrance and exit stagnation temperatures are: °R T 01 = 600.02 °R T 02 = 600.86 The work needed to pump the gas through the pipeline would be: W = 17.5810 Btu/s W = 24.9 hp Problem 13.146 [Difficulty: 5] Given: Air flowing through a tube Find: Mass flow rate assuming incompressible, adiabatic, and isothermal flow Solution: R = k = 53.33 1.4 ν = D= L= f= p1 = T1 = p2 = 0.000163 1 10 0.03 15 530 14.7 ft2/s in ft A= 0.005454 ft2 ρ1 = 0.07642 lbm/ft3 V1 = 100.56 ft/s m incomp = 0.0419 lbm/s The given or available data is: ft-lbf/lbm-°R psia °R psia Equations and Computations: The tube flow area is: For incompressible flow, the density is: The velocity of the flow is: The mass flow rate is: For Fanno flow, the duct friction length is: 3.600 fL 1-2/D = and the pressure ratio across the duct is: p 1/p 2 = 1.0204 To solve this problem, we have to guess M 1. Based on this and the friction length, we can determine a corresponding M 2. The pressure ratios for M 1 and M 2 will be used to check the validity of our guess. M1 M2 fL 1/D fL 2/D fL 1-2/D p 1/p 2 0.0800 0.0813 106.72 103.12 3.600 1.0167 0.0900 0.0919 83.50 79.90 3.600 1.0213 0.1000 0.1027 66.92 63.32 3.600 1.0266 0.1100 0.1136 54.69 51.09 3.600 1.0326 Here we used Solver to match the friction length. When both the friction length and the pressure ratios match the constraints set above, we have our solution. Therefore our entrance and exit Mach numbers are: M1 = 0.0900 M2 = 0.0919 The density at 1 was already determined. The sound speed at 1 is: c1 = 1128.8 ft/s so the velocity at 1 is: V1 = 101.59 ft/s and the mass flow rate is: m Fanno = 0.0423 lbm/s To solve this problem for isothermal flow, we perform a calculation similar to that done above for the Fanno flow. The only difference is that we use the friction length relation and pressure ratio relation for isothermal flow: M1 M2 fL 1/D fL 2/D fL 1-2/D p 1/p 2 0.0800 0.0813 105.89216 102.29216 3.600 1.0167 0.0900 0.0919 82.70400 79.10400 3.600 1.0213 0.1000 0.1027 66.15987 62.55987 3.600 1.0266 0.1100 0.1136 53.95380 50.35380 3.600 1.0326 Here we used Solver to match the friction length. When both the friction length and the pressure ratios match the constraints set above, we have our solution. Therefore our entrance and exit Mach numbers are: M1 = 0.0900 M2 = 0.0919 The density and sound speed at 1 were already determined. The velocity at 1 is: V1 = 101.59 ft/s and the mass flow rate is: m Isothermal = 0.0423 lbm/s Note that in this situation, since the Mach number was low, the assumption of incompressible flow was a good one. Also, since the Fanno flow solution shows a very small change in Mach number, the temperature does not change much, and so the isothermal solution gives almost identical results. Problem 13.147 [Difficulty: 4] Given: Oxygen supplied to astronaut via umbilical Find: Required entrance pressure and power needed to pump gas through the tube Solution: The given or available data is: R = cp = k = Q= D= L= f= T1 = T1 = T2 = p2 = 259.8 909.4 1.4 10 1 15 0.01 20 293 293 30 J/kg-K J/kg-K L/min cm m °C K K kPa Equations and Computations: At the exit of the pipe we can calculate the density: kg/m3 ρ2 = 0.39411 m= 6.568E-05 kg/s A= 7.854E-05 m2 V2 = 2.12 m/s c2 = 326.5 m/s so the mass flow rate is: The pipe area is: Therefore, the flow velocity is: The local sound speed is: So the Mach number is: M 2 = 0.006500 From the exit Mach number we can calculate: T 02/T 2 = 1.0000 fL 2/D = 16893.2 Given the length, diameter, and friction factor, we know: fL 1-2/D = 15.0 Therefore: fL 1/D = 16908.2 So from this information we can calculate the entrance Mach number: M 1 = 0.006498 fL 1/D = 16908.2 (We use Solver to calculate the Mach number based on the friction length) The entrance sound speed is the same as that at the exit: c1 = 326.5 m/s So the flow velocity is: V1 = 2.12 m/s We can calculate the pressure ratio from the velocity ratio: p1 = 30.0 kPa From the entrance Mach number we can calculate: T 01/T 1 = 1.0000 So the entrance and exit stagnation temperatures are: T 01 = 293.00 K T 02 = 293.00 K The work needed to pump the gas through the pipeline would be: W = 1.3073E-07 W W = 0.1307 microwatts Problem 13.148 Given: Isothermal air flow in a pipe Find: Mach number and location at which pressure is 500 kPa [Difficulty: 5] Solution: Basic equations: Given or available data From continuity Since Then At M 1 0.176 At M 2 0.529 Hence f Lmax 1 k M mrate ρ V A p ρ R T T1 ( 15 273 ) K p 1 1.5 MPa m V1 60 s D 15 cm k 1.4 R 286.9 ρ1 V1 ρ2 V2 or T1 T2 and c1 c1 340 p1 M2 M1 p2 D D D 1 k M1 f Lmax2 L12 18.2 D D f 2 2 m s ln k M 1 2 ln k M 2 2 f Lmax1 D 2 2 M1 18.819 0.614 18.819 0.614 18.2 L12 210 m V1 c1 ln k M 2 f 0.013 p 2 500 kPa J kg K p2 T2 V2 V M c M k R T 2 1 k M2 k M2 2 k M V1 M 2 0.529 k M1 f Lmax2 f L12 p1 T1 k R T1 f Lmax1 D p1 M2 M1 p2 M 1 0.176 Problem 13.149 [Difficulty: 2] Given: Isothermal air flow in a duct Find: Downstream Mach number; Direction of heat transfer; Plot of Ts diagram Solution: Basic equations: h1 V1 2 2 δQ h2 dm V2 2 2 T0 T 1 k1 2 M 2 mrate ρ V A Given or available data T1 ( 20 273 ) K p 1 350 kPa M 1 0.1 From continuity mrate ρ1 V1 A ρ2 V2 A so ρ1 V1 ρ2 V2 Also p ρ R T M Hence continuity becomes p1 R T1 T1 T2 Hence M2 But at each state p2 M 1 c1 Since From energy and p1 p2 R T2 V p 2 150 kPa V M c or c M 2 c2 c1 c2 p1 M1 p2 M2 so M1 M 2 0.233 2 2 V2 V1 h h 2 h 02 h 01 cp T02 T01 2 1 dm 2 δQ T0 T 1 k1 2 M 2 or T0 T 1 k1 2 M 2 p02 Since T = const, but M 2 > M 1, then T02 > T01, and δQ dm 0 T p01 T02 T 01 so energy is ADDED to the system p2 p1 s Problem 13.150 [Difficulty: 4] Part 1/2 Problem 13.150 [Difficulty: 4] Part 2/2 Problem 13.151 [Difficulty: 2] Problem 13.152 [Difficulty: 5] Part 1/2 Problem 13.152 [Difficulty: 5] Part 2/2 Problem 13.153 [Difficulty: 4] Given: Air flow from converging nozzle into heated pipe Find: Plot Ts diagram and pressure and speed curves Solution: The given or available data is: R = k = 53.33 1.4 ft·lbf/lbm·oR cp = 0.2399 Btu/lbm·oR 187 T0 = p0 = pe= 710 25 24 Me = 0.242 Using built-in function IsenT (M ,k ) Te = 702 Using p e, M e, and function Rayp (M ,k ) p* = 10.82 Using T e, M e, and function RayT (M ,k ) T* = 2432 Equations and Computations: From p 0 and p e, and Eq. 13.7a (using built-in function IsenMfromp (M ,k )) ft·lbf/lbm·oR R psi psi o o R psi o R We can now use Rayleigh-line relations to compute values for a range of Mach numbers: M T /T * 0.242 0.25 0.26 0.27 0.28 0.29 0.3 0.31 0.32 0.33 0.34 0.35 0.36 0.37 0.38 0.39 0.4 0.41 0.42 0.43 0.44 0.45 0.46 0.289 0.304 0.325 0.346 0.367 0.388 0.409 0.430 0.451 0.472 0.493 0.514 0.535 0.555 0.576 0.595 0.615 0.634 0.653 0.672 0.690 0.708 0.725 T (oR) 702 740 790 841 892 9433000 994 1046 2500 1097 1149 2000 1200 1250 T (oR) 1500 1301 1351 1000 1400 1448500 1496 1543 0 1589 0 1635 1679 1722 1764 c (ft/s) 1299 1334 1378 1422 1464 1506 1546 1586 1624 1662 1698 1734 1768 1802 1834 1866 1897 1926 1955 1982 2009 2035 2059 V (ft/s) 50 p /p * p (psi) 315 2.22 24.0 334 2.21 23.9 358 2.19 23.7 384 Ts Curve 2.18(Rayleigh) 23.6 410 2.16 23.4 437 2.15 23.2 464 2.13 23.1 492 2.12 22.9 520 2.10 22.7 548 2.08 22.5 577 2.07 22.4 607 2.05 22.2 637 2.03 22.0 667 2.01 21.8 697 2.00 21.6 728 1.98 21.4 759 1.96 21.2 790 1.94 21.0 821 1.92 100 150 20.8 852 1.91 20.6 o s (ft.lbf/lbm 884 1.89 20.4R) 916 1.87 20.2 947 1.85 20.0 Δs (ft·lbf/lbm·oR) Eq. (12.11b) 0.00 10.26 22.81 34.73 46.09 56.89 67.20 77.02 86.40 95.35 103.90 112.07 119.89 127.36 134.51 141.35 147.90 154.17 160.17 200 165.92 171.42 176.69 181.73 250 300 0.47 0.48 0.49 0.5 0.51 0.52 0.53 0.54 0.55 0.56 0.57 0.58 0.59 0.6 0.61 0.62 0.63 0.64 0.65 0.66 0.67 0.68 0.69 0.7 0.71 0.72 0.73 0.74 0.75 0.76 0.77 0.78 0.79 0.8 0.81 0.82 0.83 0.84 0.85 0.86 0.87 0.88 0.89 0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1 0.742 0.759 0.775 0.790 0.805 0.820 0.834 0.847 0.860 0.872 0.884 0.896 0.906 0.917 0.927 0.936 0.945 0.953 0.961 0.968 0.975 0.981 0.987 0.993 0.998 1.003 1.007 1.011 1.014 1.017 1.020 1.022 1.024 1.025 1.027 1.028 1.028 1.029 1.029 1.028 1.028 1.027 1.026 1.025 1.023 1.021 1.019 1.017 1.015 1.012 1.009 1.006 1.003 1.000 1805 1845 1884 1922 1958 3000 1993 2027 2500 2060 2091 2000 2122 2150 V (ft/s) 1500 2178 2204 1000 2230 2253 500 2276 2298 0 2318 0.2 2337 2355 2371 2387 2401 2415 2427 2438 2449 30 2458 2466 25 2474 2480 20 2486 p 2490 (psi) 15 2494 2497 10 2499 2501 5 2502 2502 0 2501 0.2 2500 2498 2495 2492 2488 2484 2479 2474 2468 2461 2455 2448 2440 2432 2083 979 1.83 19.8 186.57 2106 1011 1.81 19.6 191.19 2128 1043 1.80 19.4 195.62 Velocity V Versus M (Rayleigh) 2149 1075 1.78 19.2 199.86 2170 1107 1.76 19.0 203.92 2189 1138 1.74 18.8 207.80 2208 1170 1.72 18.6 211.52 2225 1202 1.70 18.4 215.08 2242 1233 1.69 18.2 218.48 2258 1265 1.67 18.0 221.73 2274 1296 1.65 17.9 224.84 2288 1327 1.63 17.7 227.81 2302 1358 1.61 17.5 230.65 2315 1389 1.60 17.3 233.36 2328 1420 1.58 17.1 235.95 2339 1450 1.56 16.9 238.42 2350 1481 1.54 16.7 240.77 2361 1511 1.53 16.5 243.01 0.3 0.4 0.5 0.6 0.7 0.8 2370 1541 1.51 16.3 245.15 M 16.1 2379 1570 1.49 247.18 2388 1600 1.47 15.9 249.12 2396 1629 1.46 15.8 250.96 2403 1658 1.44 15.6 252.70 2409 1687 1.42 15.4 254.36 2416 1715 1.41 15.2 255.93 Pressure p Versus M (Rayleigh) 2421 1743 1.39 15.0 257.42 2426 1771 1.37 14.9 258.83 2431 1799 1.36 14.7 260.16 2435 1826 1.34 14.5 261.41 2439 1853 1.33 14.4 262.59 2442 1880 1.31 14.2 263.71 2445 1907 1.30 14.0 264.75 2447 1933 1.28 13.9 265.73 2449 1959 1.27 13.7 266.65 2450 1985 1.25 13.5 267.50 2451 2010 1.24 13.4 268.30 2452 2035 1.22 13.2 269.04 2452 2060 1.21 13.1 269.73 2452 2085 1.19 12.9 270.36 2452 2109 1.18 12.8 270.94 0.3 0.4 0.5 0.6 0.7 0.8 2451 2133 1.17 12.6 271.47 M 2450 2156 1.15 12.5 271.95 2449 2180 1.14 12.3 272.39 2448 2203 1.12 12.2 272.78 2446 2226 1.11 12.0 273.13 2444 2248 1.10 11.9 273.43 2441 2270 1.09 11.7 273.70 2439 2292 1.07 11.6 273.92 2436 2314 1.06 11.5 274.11 2433 2335 1.05 11.3 274.26 2429 2356 1.04 11.2 274.38 2426 2377 1.02 11.1 274.46 2422 2398 1.01 10.9 274.51 2418 2418 1.00 10.8 274.52 0.9 1.0 0.9 1.0 Problem 13.154 [Difficulty: 2] Given: Air flow through a duct with heat transfer Find: Exit static and stagnation temperatures; magnitude and direction of heat transfer Solution: The given or available data is: R = cp = k = M1 = T1 = M2 = 286.9 1004 1.4 3 250 1.6 J/kg-K J/kg-K K Equations and Computations: We can determine the stagnation temperature at the entrance: T 01/T 1 = 2.8000 So the entrance stagnation temperature is: T 01 = 700.00 K The reference stagnation temperature for Rayliegh flow can be calculated: T 01/T 0* = 0.6540 * 1070.4 K T0 = Since the reference state is the same at stations 1 and 2, state 2 is: T 02/T 0* = 0.8842 T 02 = 946 K T 02/T 2 = 1.5120 T2 = 626 K The heat transfer is related to the change in stagnation temperature: q 1-2 = 247 kJ/kg Problem 13.155 [Difficulty: 4] Given: Air flow from converging-diverging nozzle into heated pipe Find: Plot Ts diagram and pressure and speed curves Solution: The given or available data is: R = k = 53.33 1.4 ft·lbf/lbm·oR cp = 0.2399 Btu/lbm·oR 187 T0 = p0 = pe = 710 25 2.5 Me = 2.16 Using built-in function IsenT (M ,k ) Te = 368 Using p e, M e, and function Rayp (M ,k ) p* = 7.83 Using T e, M e, and function RayT (M ,k ) T* = 775 Equations and Computations: From p 0 and p e, and Eq. 13.7a (using built-in function IsenMfromp (M ,k )) ft·lbf/lbm·oR R o psi psi o R psi o R We can now use Rayleigh-line relations to compute values for a range of Mach numbers: M T /T * 2.157 2 1.99 1.98 1.97 1.96 1.95 1.94 1.93 1.92 1.91 1.9 1.89 1.88 1.87 1.86 1.85 1.84 1.83 1.82 1.81 1.8 1.79 1.78 1.77 1.76 1.75 1.74 1.73 1.72 1.71 0.475 0.529 0.533 0.536 0.540 0.544 0.548 0.552 0.555 0.559 0.563 0.567 0.571 0.575 0.579 0.584 0.588 0.592 0.596 0.600 0.605 0.609 0.613 0.618 0.622 0.626 0.631 0.635 0.640 0.645 0.649 T (oR) 368 410 413 416 418 421 800 424 750 427 430 700 433 650 436 600 440 o T ( R) 550 443 500 446 449 450 452 400 455 350 459 300 462 0 465 468 472 475 479 482 485 489 492 496 499 503 c (ft/s) 940 993 996 1000 1003 1007 1010 1014 1017 1021 1024 1028 1032 1035 1039 1043 1046 1050 105410 1057 1061 1065 1069 1073 1076 1080 1084 1088 1092 1096 1100 V (ft/s) p /p * p (psi) Δs (ft·lbf/lbm·oR) Eq. (12.11b) 2028 0.32 2.5 1985 0.36 2.8 1982 0.37 2.9 1979 Ts Curve 0.37 (Rayleigh) 2.9 1976 0.37 2.9 1973 0.38 2.9 1970 0.38 3.0 1966 0.38 3.0 1963 0.39 3.0 1960 0.39 3.0 1957 0.39 3.1 1953 0.40 3.1 1950 0.40 3.1 1946 0.40 3.2 1943 0.41 3.2 1939 0.41 3.2 1936 0.41 3.2 1932 0.42 3.3 1928 0.42 20 30 40 3.3 50 1925 0.43 3.3 . o s (ft lbf/lbm R) 1921 0.43 3.4 1917 0.43 3.4 1913 0.44 3.4 1909 0.44 3.5 1905 0.45 3.5 1901 0.45 3.5 1897 0.45 3.6 1893 0.46 3.6 1889 0.46 3.6 1885 0.47 3.7 1880 0.47 3.7 0.00 13.30 14.15 14.99 15.84 16.69 17.54 18.39 19.24 20.09 20.93 21.78 22.63 23.48 24.32 25.17 26.01 26.86 27.70 60 28.54 29.38 30.22 31.06 31.90 32.73 33.57 34.40 35.23 36.06 36.89 37.72 70 80 1.7 1.69 1.68 1.67 1.66 1.65 1.64 1.63 1.62 1.61 1.6 1.59 1.58 1.57 1.56 1.55 1.54 1.53 1.52 1.51 1.5 1.49 1.48 1.47 1.46 1.45 1.44 1.43 1.42 1.41 1.4 1.39 1.38 1.37 1.36 1.35 1.34 1.33 1.32 1.31 1.3 1.29 1.28 1.27 1.26 1.25 1.24 1.23 1.22 1.21 1.2 1.19 1.18 1.17 1.16 1.15 1.14 1.13 1.12 1.11 1.1 1.09 1.08 1.07 1.06 1.05 1.04 1.03 1.02 1.01 1 0.654 0.658 0.663 0.668 0.673 0.677 0.682 0.687 0.692 0.697 0.702 0.707 0.712 0.717 0.722 0.727 0.732 0.737 0.742 0.747 0.753 0.758 0.763 0.768 0.773 0.779 0.784 0.789 0.795 0.800 0.805 0.811 0.816 0.822 0.827 0.832 0.838 0.843 0.848 0.854 0.859 0.865 0.870 0.875 0.881 0.886 0.891 0.896 0.902 0.907 0.912 0.917 0.922 0.927 0.932 0.937 0.942 0.946 0.951 0.956 0.960 0.965 0.969 0.973 0.978 0.982 0.986 0.989 0.993 0.997 1.000 507 1104 510 1107 514 1111 517 1115 521 1119 525 1123 529 1127 532 1131 536 1135 540 1139 544 1143 548 1147 2500 551 1151 555 1155 559 2000 1159 563 1164 567 1500 1168 571 1172 V (ft/s) 575 1176 579 1000 1180 583 1184 587 500 1188 591 1192 595 1196 0 599 1200 2.0 603 1204 607 1208 612 1213 616 1217 620 1221 624 1225 628 1229 632 1233 636 1237 9 641 1241 645 8 1245 649 7 1249 653 1253 6 657 1257 662 5 1261 p (psi) 666 4 1265 670 1269 3 674 1273 678 2 1277 682 1 1281 686 1285 0 690 1288 2.0 694 1292 699 1296 703 1300 706 1303 710 1307 714 1310 718 1314 722 1318 726 1321 730 1324 733 1328 737 1331 741 1334 744 1337 747 1341 751 1344 754 1347 757 1349 761 1352 764 1355 767 1358 769 1360 772 1362 775 1365 1876 0.48 3.7 38.54 1872 0.48 3.8 39.36 1867 0.48 3.8 40.18 1863 0.49 3.8 41.00 1858 0.49 3.9 41.81 1853 0.50 3.9 42.62 1849 0.50 3.9 43.43 1844 0.51 4.0 44.24 1839 0.51 4.0 45.04 1834 0.52 4.1 45.84 Velocity V Versus M (Rayleigh) 1829 0.52 4.1 46.64 1824 0.53 4.1 47.43 1819 0.53 4.2 48.22 1814 0.54 4.2 49.00 1809 0.54 4.3 49.78 1803 0.55 4.3 50.56 1798 0.56 4.3 51.33 1793 0.56 4.4 52.10 1787 0.57 4.4 52.86 1782 0.57 4.5 53.62 1776 0.58 4.5 54.37 1770 0.58 4.6 55.12 1764 0.59 4.6 55.86 1758 0.60 4.7 56.60 1752 0.60 4.7 57.33 1.8 1.6 1.4 1746 0.61 4.8 58.05 M 4.8 1740 0.61 58.77 1734 0.62 4.9 59.48 1728 0.63 4.9 60.18 1721 0.63 5.0 60.88 1715 0.64 5.0 61.56 1708 0.65 5.1 62.24 Pressure p Versus M (Rayleigh) 1701 0.65 5.1 62.91 1695 0.66 5.2 63.58 1688 0.67 5.2 64.23 1681 0.68 5.3 64.88 1674 0.68 5.3 65.51 1667 0.69 5.4 66.14 1659 0.70 5.5 66.76 1652 0.71 5.5 67.36 1645 0.71 5.6 67.96 1637 0.72 5.6 68.54 1629 0.73 5.7 69.11 1622 0.74 5.8 69.67 1614 0.74 5.8 70.22 1606 0.75 5.9 70.75 1598 0.76 6.0 71.27 1.8 1.6 1.4 1589 0.77 6.0 71.78 M 6.1 1581 0.78 72.27 1573 0.79 6.2 72.75 1564 0.80 6.2 73.21 1555 0.80 6.3 73.65 1546 0.81 6.4 74.08 1537 0.82 6.4 74.50 1528 0.83 6.5 74.89 1519 0.84 6.6 75.27 1510 0.85 6.7 75.63 1500 0.86 6.7 75.96 1491 0.87 6.8 76.28 1481 0.88 6.9 76.58 1471 0.89 7.0 76.86 1461 0.90 7.1 77.11 1451 0.91 7.1 77.34 1441 0.92 7.2 77.55 1430 0.93 7.3 77.73 1420 0.94 7.4 77.88 1409 0.95 7.5 78.01 1398 0.97 7.6 78.12 1387 0.98 7.6 78.19 1376 0.99 7.7 78.24 1365 1.00 7.8 78.25 1.2 1.2 1.0 1.0 Problem 13.156 [Difficulty: 2] Problem 13.157 [Difficulty: 2] Problem 13.158 [Difficulty: 3] Given: Air flow through a duct with heat transfer Find: Heat addition needed to yield maximum static temperature and choked flow Solution: The given or available data is: R = cp = k = D= V1 = p1 = T1 = T1 = 53.33 0.2399 1.4 6 300 14.7 200 660 ft-lbf/lbm-°R Btu/lbm-°R in ft/s psia °F °R c1 = 1259.65 ft/s Equations and Computations: The sound speed at station 1 is: So the Mach number is: M1 = 0.2382 We can determine the stagnation temperature at the entrance: T 01/T 1 = 1.0113 So the entrance stagnation temperature is: °R T 01 = 667.49 The reference stagnation temperature for Rayliegh flow can be calculated: T 01/T 0* = 0.2363 * °R T0 = 2824.4 For the maximum static temperature, the corresponding Mach number is: M2 = 0.8452 Since the reference state is the same at stations 1 and 2, state 2 is: T 02/T 0* = 0.9796 °R T 02 = 2767 The heat transfer is related to the change in stagnation temperature: q 1-2 = 504 Btu/lb For acceleration to sonic flow the exit state is the * state: q 1-* = 517 Btu/lb Problem 13.159 Given: Frictionless flow of Freon in a tube Find: Heat transfer; Pressure drop [Difficulty: 2] NOTE: ρ2 is NOT as stated; see below Solution: Basic equations: mrate ρ V A p ρ R T BTU Given or available data h 1 25 lbm ft A mrate The pressure drop is π 4 3 2 V1 8.03 s mrate ft V2 944 s ∆p ρ1 V1 V2 V1 h 2 65 BTU ρ2 0.850 lbm h 02 h 2 p 1 p 2 ρ1 V1 V2 V1 2 h 01 h 1 Q 107 V 2 2 V2 s ∆p 162 psi 3 lbm mrate 1.85 s V1 BTU lbm ft A 0.332 in V1 ρ1 A Q mrate h 02 h 01 2 h0 h 2 D ft V2 ρ2 A The heat transfer is lbm ρ1 100 D 0.65 in Then Q mrate h 02 h 01 2 h 01 25.0 BTU h 02 82.8 BTU lbm 2 lbm (74 Btu/s with the wrong ρ2!) (-1 psi with the wrong ρ 2!) Problem 13.160 Given: Frictionless air flow in a pipe Find: Heat exchange per lb (or kg) at exit, where 500 kPa [Difficulty: 2] Solution: Basic equations: mrate ρ V A δQ p ρ R T Given or available data T1 ( 15 273 ) K dm cp T02 T01 p 1 1 MPa M 1 0.35 D 5 cm k 1.4 cp 1004 p1 ρ1 R T1 ρ1 12.1 V1 M 1 c1 m V1 119 s From momentum p1 p2 V2 V1 ρ1 V1 m V2 466 s From continuity ρ1 V1 ρ2 V2 V1 ρ2 ρ1 V2 ρ2 3.09 T2 564 K T2 291 °C At section 1 p2 T2 and T02 T2 1 k1 T01 T1 1 k1 with Then ρ2 R δQ 3 c1 p 2 500 kPa J kg K k R T1 m Hence kg 2 2 M2 M1 2 2 p 1 p 2 ρ1 V1 V2 V1 (Momentum) (Energy) R 286.9 c1 340 M2 1 J kg K m s kg 3 m T02 677 K T02 403 °C T01 295 K T01 21.9 °C Btu kJ cp T02 T01 164 383 dm lbm kg T0 (Note: Using Rayleigh line functions, for M 1 0.35 0.4389 T0crit so T0crit T01 0.4389 T0crit 672K close to T2 ... Check!) Problem 13.161 [Difficulty: 3] Problem 13.162 [Difficulty: 3] Problem 13.163 [Difficulty: 3] Given: Nitrogen flow through a duct with heat transfer Find: Heat transfer Solution: The given or available data is: R = cp = k = M1 = T 01 = p1 = p2 = 55.16 0.2481 1.4 0.75 500 24 40 ft-lbf/lbm-°R Btu/lbm-°R °R psia psia Equations and Computations: We can find the pressure and stagnation temperature at the reference state: p 1/p * = 1.3427 * 0.9401 T 01/T 0 = So the reference pressure and stagnation temperature are: p* = 17.875 psia °R 531.9 T0 = We can now find the exit Mach number through the reference pressure: * p 2/p * = M2 = 2.2378 0.2276 p 2/p * = 2.2378 (We used Solver to match the reference pressure ratio by varying M 2.) Since the reference state is the same at stations 1 and 2, state 2 is: T 02/T 0* = 0.2183 °R T 02 = 116 The heat transfer is related to the change in stagnation temperature: q 1-2 = -95.2 Btu/lb (The negative number indicates heat loss from the nitrogen) Problem 13.164 [Difficulty: 3] Problem 13.165 [Difficulty: 3] Given: Frictionless flow of air in a duct Find: Heat transfer without choking flow; change in stagnation pressure k Solution: Basic equations: T0 k1 1 T M 2 mrate p1 p2 A p0 2 A π 4 2 D At state 1 From continuity From momentum A 78.54 cm 2 k 1.4 kg mrate 0.5 s D 10 cm M2 1 cp 1004 J R 286.9 kg K 2 2 From continuity p1 p1 ρ1 V1 M 1 c1 M k R T1 R T1 R T1 1 p1 M1 T2 T02 T2 1 k1 or 1 k M 2 1 p2 p1 1 k M 2 2 k p1 M1 ρ2 V2 R T1 p2 M2 T2 T1 p1 M1 p2 M2 T1 2 M2 2 T02 1394 K p 2 31.1 kPa k p2 M2 R T2 2 T2 1161 K T01 T1 1 T2 888 °C k1 2 M1 2 k p 02 p 2 1 k1 Finally J kg K p1 kg m ρ1 ρ1 0.894 c 1 k R T1 c1 331 R T1 3 s m mrate V1 m then V1 V1 71.2 M1 M 1 0.215 ρ1 A c1 s mrate p 2 2 2 2 2 2 2 p1 p2 V2 V1 ρ2 V2 ρ1 V1 but ρ V ρ c M k R T M k p M A R T p1 p2 k p2 M2 k p1 M1 Then 2 cp T02 T01 Hence Hence k 1 p 1 70 kPa dm M 2 mrate ρ A V δQ Given or available data T1 ( 0 273 ) K k1 p ρ R T p V2 V1 1 δQ M2 2 2 p 02 58.8 kPa MJ cp T02 T01 1.12 kg dm (Using Rayleigh functions, at M 1 0.215 T01 T0crit k k 1 T01 T01 276 K p 01 p 1 1 ∆p0 p 02 p 01 T01 0.1975 T02 0.1975 T02 k1 2 M1 2 k 1 p 01 72.3 kPa ∆p0 13.5 kPa T02 1395 K and ditto for p02 ...Check!) Problem 13.166 [Difficulty: 3] Problem 13.167 [Difficulty: 2] Problem 13.168 [Difficulty: 3] Given: Air flow through a duct with heat transfer Find: Exit conditions Solution: The given or available data is: R = cp = k = m= 286.9 1004 1.4 20 kg/s A= p1 = T1 = q 1-2 = 0.06 320 350 650 m2 kPa K kJ/kg ρ1 = 3.1868 kg/m3 V1 = 104.5990 m/s c1 = 374.9413 m/s J/kg-K J/kg-K Equations and Computations: The density at the entrance is: So the entrance velocity is: The sonic velocity is: So the Mach number is: M1 = 0.2790 We can determine the stagnation temperature at the entrance: T 01/T 1 = 1.0156 So the entrance stagnation temperature is: T 01 = 355.45 K The reference conditions for Rayliegh flow can be calculated: T 01/T 0* = 0.3085 * T0 = 1152.2 * T 1/T = 0.3645 T* = 960.2 * 2.1642 p 1/p = K K p* = 147.9 kPa The heat transfer is related to the change in stagnation temperature: T 02 = 1002.86 K The stagnation temperature ratio at state 2 is: T 02/T 0* = We can now find the exit Mach number: M2 = 0.8704 0.652 * T 02/T 0 = 0.8704 (We used Solver to match the reference pressure ratio by varying M 2.) We can now calculate the exit temperature and pressure: T 2/T * = T2 = * p 2/p = T2 = 0.9625 924 K 1.5040 222 kPa Problem 13.169 [Difficulty: 3] Given: Air flow through a duct with heat transfer Find: Heat transfer needed to choke the flow Solution: The given or available data is: R = cp = k = p1 = T1 = V1 = 286.9 1004 1.4 135 500 540 c1 = 448.1406 J/kg-K J/kg-K kPa K m/s Equations and Computations: The sonic velocity at state 1 is: m/s So the Mach number is: M1 = 1.2050 We can determine the stagnation temperature at the entrance: T 01/T 1 = 1.2904 So the entrance stagnation temperature is: T 01 = 645.20 K The reference conditions for Rayliegh flow can be calculated: T 01/T 0* = * T0 = 0.9778 659.9 K Since the flow is choked, state 2 is: M2 = 1.000 T 02 = 659.85 K The heat transfer is related to the change in stagnation temperature: q 1-2 = 14.71 kJ/kg To choke a flow, heat must always be added . Problem 13.170 [Difficulty: 2] Problem 13.171 [Difficulty: 2] Problem 13.172 [Difficulty: 2] Problem 13.173 [Difficulty: 4] Given: Air flow through a duct with heat transfer followed by converging duct, sonic at exit Find: Magnitude and direction of heat transfer Solution: The given or available data is: R = cp = k = M1 = T1 = p1 = A 2/A 3 = M3 = 53.33 0.2399 1.4 2 300 70 1.5 1 ft-lbf/lbm-°R Btu/lbm-°R °R psia Equations and Computations: We can determine the stagnation temperature at the entrance: T 01/T 1 = 1.8000 So the entrance stagnation temperature is: °R T 01 = 540.00 The reference stagnation temperature ratio at state 1 is: 0.7934 T 01/T 0* = The reference conditions for Rayliegh flow can be calculated: °R T 0* = 680.6 Since the flow is sonic at state 3, we can find the Mach number at state 2: M2 = 1.8541 We know that the flow must be supersonic at 2 since the flow at M 1 > 1. The reference stagnation temperature ratio at state 2 is: 0.8241 T 02/T 0* = Since the reference stagnation temperature at 1 and 2 are the same: °R T 02 = 560.92 The heat transfer is related to the change in stagnation temperature: q 1-2 = 5.02 Btu/lbm The heat is being added to the flow. Problem 13.174 [Difficulty: 3] Problem 13.175 [Difficulty: 3] Given: Data on flow through gas turbine combustor Find: Maximum heat addition; Outlet conditions; Reduction in stagnation pressure; Plot of process Solution: R = k = cp = T1 = p1 = M1 = The given or available data is: 286.9 1.4 1004 773 1.5 0.5 p02 J/kg·K T02 J/kg·K K MPa p2 T2 p01 T T01 T1 Equations and Computations: p1 From p1 1 RT1 1= 6.76 kg/m From V1 M 1 kRT1 V1 = 279 m/s 3 s Using built-in function IsenT (M,k): T 01 /T 1 = 1.05 T 01 = 812 K Using built-in function Isenp (M,k): p 01 /p 1 = 1.19 p 01 = 1.78 MPa For maximum heat transfer: M2 = 1 Using built-in function rayT0 (M,k), rayp0 (M,k), rayT (M,k), rayp (M,k), rayV (M,k): * T 01 /T 0 = * p 01 /p 0 = * T /T = * p /p = / = * 0.691 T0 = * 1174 K ( = T 02) 1.114 * 1.60 MPa ( = p 02) 978 K ( = T 02) p0 = * T = 0.790 * 1.778 p = 0.444 = * 0.844 MPa 3.01 kg/m -182 kPa Note that at state 2 we have critical conditions! Hence: From the energy equation: p 012 – p 01 = Q dm -0.182 MPa c p T02 T01 Q /dm = 364 kJ/kg ( = p 2) 3 ( = 2) Problem 13.176 [Difficulty: 3] Problem 13.177 [Difficulty: 3] Problem 13.178 [Difficulty: 4] Part 1/2 Problem 13.178 [Difficulty: 4] Part 2/2 Problem 13.179 [Difficulty: 4] Part 1/2 Problem 13.179 [Difficulty: 4] Part 2/2 Problem 13.180 [Difficulty: 3] Given: Normal shock Find: Approximation for downstream Mach number as upstream one approaches infinity Solution: 2 Basic equations: 2 M 2n M 1n 2 k1 (13.48a) 2 k 2 k 1 M1n 1 2 M 1n Combining the two equations M2 M 2n sin( β θ) 1 M2 M 2n M 2 sin( β θ) (13.47b) 2 k1 2 k M 2 1 k 1 1n sin( β θ) 2 M 1n 2 k 2 k1 2 k 1 M1n 1 sin( β θ) 2 2 2 ( k 1 ) M 1n 2 k 1 sin( β θ) 2 k 1 M1n2 As M1 goes to infinity, so does M1n, so M2 1 2 k sin( β θ) 2 k 1 M2 k1 2 k sin( β θ) 2 Problem 13.181 [Difficulty: 3] Given: Air deflected at an angle, causing an oblique shock Find: Possible shock angles; pressure and temperature corresponding to those angles Solution: The given or available data is: R = k = M1 = T1 = p1 = θ = 286.9 1.4 1.8 400 100 14 J/kg-K K kPa ° Equations and Computations: There are two possible shock angles for a given deflection, corresponding to the weak and strong shock solutions. To find the shock angle, we have to iterate on the shock angle until we match the deflection angle, which is a function of Mach number, specific heat ratio, and shock angle. The weak shock solution is: β weak = 49.7 ° θ = 14.0000 ° The strong shock solution is: β strong = 78.0 ° θ = 14.0000 ° We used Solver in Excel to iterate on the shock angles. For the weak shock, the pre-shock Mach number normal to the wave is: 1.3720 M 1nweak = The pressure and temperature ratios across the shock wave are: 2.0295 p 2/p 1weak = 1.2367 T 2/T 1weak = Therefore, the post-shock temperature and pressure are: p 2weak = 203 kPa 495 K T 2weak = For the weak shock, the pre-shock Mach number normal to the wave is: 1.7608 M 1nstrong = The pressure and temperature ratios across the shock wave are: p 2/p 1strong = 3.4505 1.5025 T 2/T 1strong = Therefore, the post-shock temperature and pressure are: 345 kPa p 2strong = T 2strong = 601 K Problem 13.182 [Difficulty: 3] Given: Oblique shock in flow at M = 3 Find: Minimum and maximum , plot of pressure rise across shock Solution: The given or available data is: R = k = M1 = 286.9 1.4 3 J/kg.K Equations and Computations: The smallest value of is when the shock is a Mach wave (no deflection) = sin-1(1/M 1) The largest value is = 19.5 o = 90.0 o The normal component of Mach number is M 1n = M 1sin() (13.47a) For each , p2/p1 is obtained from M1n, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) Computed results: (o) M 1n p 2/p 1 19.5 20 30 40 50 60 70 75 80 85 90 1.00 1.03 1.50 1.93 2.30 2.60 2.82 2.90 2.95 2.99 3.00 1.00 1.06 2.46 4.17 5.99 7.71 9.11 9.63 10.0 10.3 10.3 Pressure Change across an Oblique Shock 12.5 10.0 7.5 p 2/p 1 5.0 2.5 0.0 0 30 60 ( ) o 90 Problem 13.183 [Difficulty: 3] Given: Data on an oblique shock Find: Mach number and pressure downstream; compare to normal shock Solution: R = k = p1 = M1 = The given or available data is: = 286.9 1.4 80 2.5 35 J/kg.K kPa o Equations and Computations: From M 1 and M 1n = M 1t = 1.43 2.05 From M1n and p1, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) p2 = 178.6 V t1 = V t2 The tangential velocity is unchanged Hence c t1 M t1 = (T 1) 1/2 c t2 M t2 M t1 = (T 2)1/2 M t2 M 2t = (T 1/T 2)1/2 M t1 From M1n, and Eq. 13.48c (using built-in function NormTfromM (M ,k )) Hence T 2/T 1 = 1.28 M 2t = 1.81 kPa Also, from M1n, and Eq. 13.48a (using built-in function NormM2fromM (M ,k )) (13.48a) M 2n = 0.726 The downstream Mach number is then M 2 = (M 2t2 + M 2n2)1/2 M2 = 1.95 Finally, from geometry V 2n = V 2sin( - ) Hence = - sin-1(V 2n/V 2) or = - sin-1(M 2n/M 2) = 13.2 o 570 kPa For the normal shock: From M1 and p1, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) p2 = Also, from M1, and Eq. 13.48a (using built-in function NormM2fromM (M ,k )) M2 = 0.513 For the minimum : The smallest value of is when the shock is a Mach wave (no deflection) = sin-1(1/M 1) = 23.6 o Problem 13.184 [Difficulty: 3] Given: Data on an oblique shock Find: Deflection angle ; shock angle ; Mach number after shock Solution: The given or available data is: R = k = M1 = T1 = p 2 /p 1 = 286.9 1.4 3.25 283 5 J/kg.K K Equations and Computations: From p 2/p 1, and Eq. 13.48d (using built-in function NormpfromM (M ,k ) and Goal Seek or Solver ) (13.48d) For p 2 /p 1 = 5.00 M 1n = 2.10 From M 1 and M 1n, and Eq 13.47a M 1n = M 1sin() = 40.4 (13.47a) o From M 1 and , and Eq. 13.49 (using built-in function Theta (M ,, k ) (13.49) = 23.6 o To find M 2 we need M 2n. From M 1n, and Eq. 13.48a (using built-in function NormM2fromM (M ,k )) (13.48a) M 2n = 0.561 The downstream Mach number is then obtained from from M 2n, and , and Eq. 13.47b M 2n = M 2sin( - ) Hence M2 = 1.94 (13.47b) Problem 13.185 [Difficulty: 3] Given: Velocities and deflection angle of an oblique shock Find: Shock angle ; pressure ratio across shock Solution: The given or available data is: R = k = V1 = V2 = = 286.9 1.4 1250 650 35 J/kg.K m/s m/s o Equations and Computations: From geometry we can write two equations for tangential velocity: For V 1t V 1t = V 1cos() (1) For V 2t V 2t = V 2cos( - ) (2) For an oblique shock V 2t = V 1t, so Eqs. 1 and 2 give V 1cos() = V 2cos( - ) Solving for (3) = tan-1((V 1 - V 2cos())/(V 2sin())) = (Alternatively, solve Eq. 3 using Goal Seek !) 62.5 o For p 2/p 1, we need M 1n for use in Eq. 13.48d (13.48d) We can compute M 1 from and , and Eq. 13.49 (using built-in function Theta (M ,, k )) (13.49) For = 35.0 o = 62.5 o M1 = 3.19 This value of M 1 was obtained by using Goal Seek : Vary M 1 so that becomes the required value. (Alternatively, find M 1 from Eq. 13.49 by explicitly solving for it!) We can now find M 1n from M 1. From M 1 and Eq. 13.47a M 1n = M 1sin() Hence M 1n = 2.83 Finally, for p 2/p 1, we use M 1n in Eq. 13.48d (using built-in function NormpfromM (M ,k ) p 2 /p 1 = 9.15 (13.47a) Problem 13.186 [Difficulty: 4] Given: Airfoil with included angle of 60o Find: Plot of temperature and pressure as functions of angle of attack Solution: R = k = T1 = p1 = V1 = The given or available data is: 286.9 1.4 276.5 75 1200 = 60 c1 = 333 M1 = 3.60 J/kg.K K kPa m/s o Equations and Computations: From T 1 Then m/s Computed results: (o) (o) (o) Needed (o) 0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 16.00 18.00 20.00 22.00 24.00 26.00 28.00 30.00 47.1 44.2 41.5 38.9 36.4 34.1 31.9 29.7 27.7 25.7 23.9 22.1 20.5 18.9 17.5 16.1 30.0 28.0 26.0 24.0 22.0 20.0 18.0 16.0 14.0 12.0 10.0 8.0 6.0 4.0 2.0 0.0 30.0 28.0 26.0 24.0 22.0 20.0 18.0 16.0 14.0 12.0 10.0 8.0 6.0 4.0 2.0 Sum: Error 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% o M 1n p 2 (kPa) T 2 ( C) 2.64 2.51 2.38 2.26 2.14 2.02 1.90 1.79 1.67 1.56 1.46 1.36 1.26 1.17 1.08 1.00 597 539 485 435 388 344 304 267 233 202 174 149 126 107 90 75 357 321 287 255 226 198 172 148 125 104 84 66 49 33 18 3 597 357 Max: To compute this table: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) Type the range of Type in guess values for Compute Needed from = /2 - Compute from Eq. 13.49 (using built-in function Theta (M ,, k ) Compute the absolute error between each and Needed Compute the sum of the errors Use Solver to minimize the sum by varying the values (Note: You may need to interactively type in new values if Solver generates values that lead to no ) For each , M 1n is obtained from M 1, and Eq. 13.47a For each , p 2 is obtained from p 1, M 1n, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) For each , T 2 is obtained from T 1, M 1n, and Eq. 13.48c (using built-in function NormTfromM (M ,k )) Pressure on an Airfoil Surface as a Function of Angle of Attack 700 600 p 2 (kPa) 500 400 300 200 100 0 0 5 10 15 20 25 30 25 30 ( ) o Temperature on an Airfoil Surface as a Function of Angle of Attack 400 350 T 2 (oC) 300 250 200 150 100 50 0 0 5 10 15 ( ) o 20 Problem 13.187 [Difficulty: 4] Given: Airfoil with included angle of 20o Find: Mach number and speed at which oblique shock forms Solution: The given or available data is: R = k = T1 = = 286.9 1.4 288 10 J/kg.K K o Equations and Computations: From Fig. 13.29 the smallest Mach number for which an oblique shock exists at a deflection = 10o is approximately M 1 = 1.4. By trial and error, a more precise answer is (using built-in function Theta (M ,, k ) M1 = 1.42 = 67.4 o = 10.00 o c1 = V1 = 340 483 A suggested procedure is: 1) Type in a guess value for M 1 2) Type in a guess value for m/s m/s 3) Compute from Eq. 13.49 (using built-in function Theta (M ,, k )) (13.49) 4) Use Solver to maximize by varying 5) If is not 10 o, make a new guess for M 1 o 6) Repeat steps 1 - 5 until = 10 Computed results: M1 ( o) ( o) 1.42 1.50 1.75 2.00 2.25 2.50 3.00 4.00 5.00 6.00 7.00 67.4 56.7 45.5 39.3 35.0 31.9 27.4 22.2 19.4 17.6 16.4 10.0 10.0 10.0 10.0 10.0 10.0 10.0 10.0 10.0 10.0 10.0 Error 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% Sum: 0.0% To compute this table: 1) Type the range of M 1 2) Type in guess values for 3) Compute from Eq. 13.49 (using built-in function Theta (M ,, k ) o 4) Compute the absolute error between each and = 10 5) Compute the sum of the errors 6) Use Solver to minimize the sum by varying the values (Note: You may need to interactively type in new values if Solver generates values that lead to no , or to values that correspond to a strong rather than weak shock) Oblique Shock Angle as a Function of Aircraft Mach Number 90 75 60 (o) 45 30 15 0 1 2 3 4 M 5 6 7 Problem 13.188 [Difficulty: 3] Given: Data on airfoil flight Find: Lift per unit span Solution: The given or available data is: R = k = p1 = M1 = = c = 286.9 1.4 70 2.75 7 1.5 J/kg.K kPa o m Equations and Computations: The lift per unit span is L = (p L - p U)c (1) (Note that p L acts on area c /cos(), but its normal component is multiplied by cos()) For the upper surface: pU = p1 pU = 70.0 kPa For the lower surface: We need to find M 1n = The deflection angle is = 7 o From M 1 and , and Eq. 13.49 (using built-in function Theta (M , ,k )) (13.49) = 7.0 o = 26.7 o M 1n = 1.24 For (Use Goal Seek to vary so that = ) From M 1 and From M 1n and p 1, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) From Eq 1 p2 = 113 kPa pL = p2 pL = 113 kPa L = 64.7 kN/m Problem 13.189 [Difficulty: 4] Given: Airfoil with included angle of 60o Find: Angle of attack at which oblique shock becomes detached Solution: The given or available data is: R = k = T1 = p1 = V1 = 286.9 1.4 276.5 75 1200 = 60 c1 = 333 M1 = 3.60 J/kg.K K kPa m/s o Equations and Computations: From T 1 Then m/s From Fig. 13.29, at this Mach number the smallest deflection angle for which an oblique shock exists is approximately = 35o. By using Solver , a more precise answer is (using built-in function Theta (M ,, k ) M1 = 3.60 = 65.8 o = 37.3 o A suggested procedure is: 1) Type in a guess value for 2) Compute from Eq. 13.49 (using built-in function Theta (M ,, k )) (13.49) 3) Use Solver to maximize by varying For a deflection angle the angle of attack is = - /2 = 7.31 o Computed results: (o) (o) (o) Needed (o) 0.00 1.00 2.00 3.00 4.00 5.50 5.75 6.00 6.25 6.50 6.75 7.00 7.25 7.31 47.1 48.7 50.4 52.1 54.1 57.4 58.1 58.8 59.5 60.4 61.3 62.5 64.4 65.8 30.0 31.0 32.0 33.0 34.0 35.5 35.8 36.0 36.3 36.5 36.8 37.0 37.3 37.3 30.0 31.0 32.0 33.0 34.0 35.5 35.7 36.0 36.2 36.5 36.7 37.0 37.2 37.3 Sum: Error 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% M 1n p 2 (kPa) T 2 (oC) 2.64 2.71 2.77 2.84 2.92 3.03 3.06 3.08 3.10 3.13 3.16 3.19 3.25 3.28 597 628 660 695 731 793 805 817 831 845 861 881 910 931 357 377 397 418 441 479 486 494 502 511 521 533 551 564 931 564 0.0% Max: To compute this table: Type the range of Type in guess values for Compute Needed from = + /2 Compute from Eq. 13.49 (using built-in function Theta (M ,, k ) Compute the absolute error between each and Needed Compute the sum of the errors Use Solver to minimize the sum by varying the values (Note: You may need to interactively type in new values if Solver generates values that lead to no ) For each , M 1n is obtained from M 1, and Eq. 13.47a For each , p 2 is obtained from p 1, M 1n, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) For each , T 2 is obtained from T 1, M 1n, and Eq. 13.48c (using built-in function NormTfromM (M ,k )) 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) Pressure on an Airfoil Surface as a Function of Angle of Attack 1000 p 2 (kPa) 900 800 700 600 500 0 2 4 6 8 6 8 o ( ) Temperature on an Airfoil Surface as a Function of Angle of Attack 600 T 2 (oC) 550 500 450 400 350 300 0 2 4 (o) Problem 13.190 [Difficulty: 3] Given: Oblique shock Mach numbers Find: Deflection angle; Pressure after shock Solution: The given or available data is: k = p1 = M1 = 1.4 75 4 M2 = 2.5 = 33.6 kPa Equations and Computations: We make a guess for : o From M 1 and , and Eq. 13.49 (using built-in function Theta (M , ,k )) (13.49) From M 1 and From M 2, , and = 21.0 M 1n = M 2n = 2.211 0.546 o (1) We can also obtain M 2n from Eq. 13.48a (using built-in function normM2fromM (M ,k )) (13.48a) M 2n = 0.546 (2) We need to manually change so that Eqs. 1 and 2 give the same answer. Alternatively, we can compute the difference between 1 and 2, and use Solver to vary to make the difference zero Error in M 2n = 0.00% Then p 2 is obtained from Eq. 13.48d (using built-in function normpfromm (M ,k )) (13.48d) p2 = 415 kPa Problem 13.191 [Difficulty: 3] Given: Data on airfoil flight Find: Lift per unit span Solution: The given or available data is: R = k = p1 = M1 = 286.9 1.4 75 2.75 U = 5 o L = c = 15 2 o J/kg.K kPa m Equations and Computations: The lift per unit span is L = (p L - p U)c (1) (Note that each p acts on area c /cos(), but its normal component is multiplied by cos()) For the upper surface: We need to find M 1n(U) The deflection angle is U = U U = 5 o From M 1 and U, and Eq. 13.49 (using built-in function Theta (M , ,k )) (13.49) For U = 5.00 o U = 25.1 o (Use Goal Seek to vary U so that U = U) From M 1 and U M 1n(U) = 1.16 From M 1n(U) and p 1, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) p2 = 106 kPa pU = p2 pU = 106 L = L L = 15 o L = 15.00 o L = 34.3 o kPa For the lower surface: We need to find M 1n(L) The deflection angle is From M 1 and L, and Eq. 13.49 (using built-in function Theta (M , ,k )) For (Use Goal Seek to vary L so that L = L) From M 1 and L M 1n(L) = 1.55 From M 1n(L) and p 1, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) From Eq 1 p2 = 198 kPa pL = p2 pL = 198 kPa L = 183 kN/m Problem 13.192 [Difficulty: 3] Given: Air deflected at an angle, causing an oblique shock Find: Post shock pressure, temperature, and Mach number, deflection angle, strong or weak Solution: The given or available data is: R = k = M1 = T1 = T1 = p1 = β = 53.33 1.4 3.3 100 560 20 45 ft-lbf/lbm-°R °F °R psia ° Equations and Computations: The pre-shock Mach numbers normal and parallel to the wave are: M 1n = 2.3335 M 1t = 2.3335 The sound speed upstream of the shock is: c1 = 1160.30 ft/s Therefore, the speed of the flow parallel to the wave is: V 1t = 2707.51 ft/s The post-shock Mach number normal to the wave is: M 2n = 0.5305 The pressure and temperature ratios across the shock wave are: p 2/p 1 = 6.1858 T 2/T 1 = 1.9777 Therefore, the post-shock temperature and pressure are: p2 = 124 psia 1108 °R T2 = 648 °F T2 = The sound speed downstream of the shock is: c2 = 1631.74 ft/s So the speed of the flow normal to wave is: V 2n = 865.63 ft/s The speed of the flow parallel to the wave is preserved through the shock: 2707.51 ft/s V 2t = Therefore the flow speed after the shock is: V2 = 2842.52 ft/s and the Mach number is: M2 = 1.742 Based on the Mach number and shock angle, the deflection angle is: θ = 27.3 ° Since the Mach number at 2 is supersonic, this is a weak wave. This can be confirmed by inspecting Fig. 13.29 in the text. Problem 13.193 [Difficulty: 3] Given: Air passing through jet inlet Find: Pressure after one oblique shock; pressure after two shocks totaling same overall turn Solution: The given or available data is: R = k = M1 = p1 = θ = 53.33 1.4 4 8 8 ft-lbf/lbm-°R psia ° Equations and Computations: To find the shock angle, we have to iterate on the shock angle until we match the deflection angle, which is a function of Mach number, specific heat ratio, and shock angle. β = 20.472 ° θ = 8.0000 ° We used Solver in Excel to iterate on the shock angle. The pre-shock Mach number normal to the wave is: 1.3990 M 1n = The pressure ratio across the shock wave is: p 2/p 1 = 2.1167 Therefore, the post-shock pressure is: p2 = 16.93 psia Now if we use two 4-degree turns, we perform two oblique-shock calculations. For the first turn: β 1-2a = 17.258 ° θ = 4.0000 ° We used Solver in Excel to iterate on the shock angle. The pre-shock Mach number normal to the wave is: 1.1867 M 1n = The post-shock Mach number normal to the wave is: M 2an = 0.8506 The pressure ratio across the shock wave is: p 2a/p 1 = 1.4763 Therefore, the post-shock pressure is: p 2a = 11.8100 psia So the Mach number after the first shock wave is: M 2a = 3.7089 For the second turn: β 2a-2b = 18.438 ° θ = 4.0000 ° We used Solver in Excel to iterate on the shock angle. The pre-shock Mach number normal to the wave is: 1.1731 M 2an = The post-shock Mach number normal to the wave is: M 2bn = 0.8594 The pressure ratio across the shock wave is: p 2b/p 2a = 1.4388 Therefore, the post-shock pressure is: p 2b = 16.99 psia The pressure recovery is slightly better for two weaker shocks than a single stronger one! Problem 13.194 [Difficulty: 4] Given: Air turning through an incident and reflected shock wave Find: Pressure, temperature, and Mach number after each wave Solution: The given or available data is: R = k = M1 = p1 = T1 = T1 = θ = 53.33 1.4 2.3 14.7 80 540 10 ft-lbf/lbm-°R psia °F °R ° Equations and Computations: To find the shock angle, we have to iterate on the shock angle until we match the deflection angle, which is a function of Mach number, specific heat ratio, and shock angle. For the first turn: β 1-2 = 34.326 ° θ = 10.0000 ° We used Solver in Excel to iterate on the shock angle. The pre-shock Mach numbers normal and parallel to the wave are: M 1n = 1.2970 1.8994 M 1t = The post-shock Mach number normal to the wave is: M 2n = 0.7875 The pressure and temperature ratios across the shock wave are: p 2/p 1 = 1.7959 1.1890 T 2/T 1 = Therefore, the post-shock pressure and temperature are: p2 = 26.4 psia T2 = 642 °R Since the parallel component of velocity is preserved across the shock and the Mach number is related to the square root of temperature, the new parallel component of Mach number is: M 2t = 1.7420 So the Mach number after the first shock wave is: M2 = 1.912 For the second turn: β 2-3 = 41.218 ° θ = 10.0000 ° We used Solver in Excel to iterate on the shock angle. The pre-shock Mach numbers normal and parallel to the wave are: M 1n = 1.2597 M 1t = 1.4380 The post-shock Mach number normal to the wave is: M 2an = 0.8073 The pressure and temperature ratios across the shock wave are: p 3/p 2 = 1.6845 T 2/T 1 = 1.1654 Therefore, the post-shock pressure is: p3 = 44.5 psia T3 = 748 °R Since the parallel component of velocity is preserved across the shock and the Mach number is related to the square root of temperature, the new parallel component of Mach number is: M 2t = 1.3320 So the Mach number after the second shock wave is: M2 = 1.558 Problem 13.195 [Difficulty: 3] Given: Wedge-shaped projectile Find: Speed at which projectile is traveling through the air Solution: The given or available data is: R = k = p1 = T1 = T1 = θ = p2 = 53.33 1.4 1 10 470 10 3 ft-lbf/lbm-°R psia °F °R ° psia Equations and Computations: The pressure ratio across the shock wave is: p 2/p 1 = 3.0000 For this pressure ratio, we can iterate to find the Mach number of the flow normal to the shock wave: M 1n = 1.6475 p 2/p 1 = 3.0000 We used Solver in Excel to iterate on the Mach number. With the normal Mach number, we can iterate on the incident Mach number to find the right combination of Mach number and shock angle to match the turning angle of the flow and normal Mach number: M1 = 4.9243 β 1-2 = 19.546 ° θ = 10.0000 ° The pre-shock Mach numbers normal and parallel to the wave are: M 1n = 1.6475 M 1t = 4.6406 We used Solver in Excel to iterate on the Mach number and shock angle. Now that we have the upstream Mach number, we can find the speed. The sound speed upstream of the shock wave is: c 1 = 1062.9839 ft/s Therefore, the speed of the flow relative to the wedge is: V1 = 5234 ft/s Problem 13.196 [Difficulty: 4] Given: Flow turned through an expansion followed by a oblique shock wave Find: Mach number and pressure downstream of the shock wave Solution: The given or available data is: R = k = M1 = p1 = θ = 53.33 1.4 2 1 16 ft-lbf/lbm-°R atm ° Equations and Computations: The Prandtl-Meyer function of the flow before the expansion is: ω1 = 26.380 ° Since we know the turning angle of the flow, we know the Prandtl-Meyer function after the expansion: ω2 = 42.380 ° We can iterate to find the Mach number after the expansion: M2 = 2.6433 ω2 = 42.380 ° The pressure ratio across the expansion wave is: p 2/p 1 = 0.3668 Therefore the pressure after the expansion is: p2 = 0.3668 atm We can iterate on the shock angle to find the conditions after the oblique shock: β 2-3 = 36.438 ° θ = 16.0000 ° We used Solver in Excel to iterate on the shock angle. The pre-shock Mach numbers normal and parallel to the wave are: M 2n = 1.5700 M 2t = 2.1265 The post-shock Mach number normal to the wave is: M 3n = 0.6777 The pressure and tempreature ratios across the shock are: p 3/p 2 = 2.7090 T 3/T 2 = 1.3674 The pressure after the shock wave is: p3 = 0.994 atm We can get the post-shock Mach number parallel to the shock from the temperature ratio: M 3t = 1.8185 So the post-shock Mach number is: M3 = 1.941 Problem 13.197 [Difficulty: 4] Given: Air passing through jet inlet Find: Pressure after one oblique shock; after two shocks totaling same overall turn, after isentropic compression Solution: The given or available data is: R = k = M1 = p1 = θ = 53.33 1.4 2 5 20 ft-lbf/lbm-°R psia ° Equations and Computations: To find the shock angle, we have to iterate on the shock angle until we match the deflection angle, which is a function of Mach number, specific heat ratio, and shock angle. β = 53.423 ° θ = 20.0000 ° We used Solver in Excel to iterate on the shock angle. The pre-shock Mach number normal to the wave is: 1.6061 M 1n = The pressure ratio across the shock wave is: p 2/p 1 = 2.8429 Therefore, the post-shock pressure is: p2 = 14.21 psia Now if we use two 10-degree turns, we perform two oblique-shock calculations. For the first turn: β 1-2a = 39.314 ° θ = 10.0000 ° We used Solver in Excel to iterate on the shock angle. The pre-shock Mach number normal to the wave is: M 1n = 1.2671 The post-shock Mach number normal to the wave is: M 2an = 0.8032 The pressure ratio across the shock wave is: p 2a/p 1 = 1.7066 Therefore, the post-shock pressure is: p 2a = 8.5329 psia So the Mach number after the first shock wave is: M 2a = 1.6405 For the second turn: β 2a-2b = 49.384 ° θ = 10.0000 ° We used Solver in Excel to iterate on the shock angle. The pre-shock Mach number normal to the wave is: M 2an = 1.2453 The post-shock Mach number normal to the wave is: M 2bn = 0.8153 The pressure ratio across the shock wave is: p 2b/p 2a = 1.6426 Therefore, the post-shock pressure is: p 2b = 14.02 psia For the isentropic compression, we need to calculate the Prandtl-Meyer function for the incident flow: ω1 = 26.3798 ° The flow out of the compression will have a Prandtl-Meyer function of: ω 2i = 6.3798 ° To find the exit Mach number, we need to iterate on the Mach number to match the Prandtl-Meyer function: M 2i = 1.3076 ω 2i = 6.3798 ° The pressure ratio across the compression wave is: p 2i/p 1 = 2.7947 Therefore, the exit pressure is: p 2i = 13.97 psia Problem 13.198 [Difficulty: 3] Given: Air flow in a duct Find: Mach number and pressure at contraction and downstream; Solution: The given or available data is: k = M1 = 1.4 2.5 = p1 = 7.5 50 o kPa Equations and Computations: For the first oblique shock (1 to 2) we need to find from Eq. 13.49 (13.49) We choose by iterating or by using Goal Seek to target (below) to equal the given Using built-in function theta (M, ,k ) = 7.50 o = 29.6 o Then M 1n can be found from geometry (Eq. 13.47a) M 1n = 1.233 Then M 2n can be found from Eq. 13.48a) Using built-in function NormM2fromM (M,k ) (13.48a) M 2n = 0.822 Then, from M 2n and geometry (Eq. 13.47b) M2 = 2.19 From M 1n and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) p 2/p 1 = p2 = 1.61 80.40 Pressure ratio We repeat the analysis of states 1 to 2 for states 2 to 3, to analyze the second oblique shock We choose for M 2 by iterating or by using Goal Seek to target (below) to equal the given Using built-in function theta (M, ,k ) = 7.50 o = 33.5 o Then M 2n (normal to second shock!) can be found from geometry (Eq. 13.47a) M 2n = 1.209 Then M 3n can be found from Eq. 13.48a) Using built-in function NormM2fromM (M,k ) M 3n = 0.837 Then, from M 3n and geometry (Eq. 13.47b) M3 = 1.91 From M 2n and Eq. 13.48d (using built-in function NormpfromM (M ,k )) p 3/p 2 = p3 = 1.54 124 Pressure ratio Problem 13.199 [Difficulty: 4] Given: Air flow into engine Find: Pressure of air in engine; Compare to normal shock Solution: The given or available data is: k = p1 = M1 = 1.4 50 3 = 7.5 kPa o Equations and Computations: Assuming isentropic flow deflection p 0 = constant p 02 = p 01 For p 01 we use Eq. 13.7a (using built-in function Isenp (M , k )) (13.7a) p 01 = p 02 = For the deflection = 1837 1837 7.5 kPa kPa o From M 1 and Eq. 13.55 (using built-in function Omega (M , k )) (13.55) 1 = Deflection = Applying Eq. 1 49.8 2 - 1 = (M 2) - (M 1) o (1) 2 = 1 - 2 = 42.3 (Compression!) o From 2, and Eq. 13.55 (using built-in function Omega (M , k )) For 2 = M2 = 42.3 2.64 o (Use Goal Seek to vary M 2 so that 2 is correct) Hence for p 2 we use Eq. 13.7a (using built-in function Isenp (M , k )) p 2 = p 02/(p 02/p 2) For the normal shock (2 to 3) p2 = 86.8 M2 = 2.64 kPa From M 2 and p 2, and Eq. 13.41d (using built-in function NormpfromM (M ,k )) (13.41d) p3 = 690 kPa For slowing the flow down from M 1 with only a normal shock, using Eq. 13.41d p = 517 kPa Problem 13.200 [Difficulty: 3] Given: Deflection of air flow Find: Pressure changes Solution: R k p M The given or available data is: = = = = 286.9 1.4 95 1.5 J/kg.K kPa 1 = 15 o 2 = 15 o Equations and Computations: We use Eq. 13.55 (13.55) and Deflection = a - b = (M a) - (M b) From M and Eq. 13.55 (using built-in function Omega (M , k )) = 11.9 1 = 1 - 1 = 1 + 1 = 26.9 o For the first deflection: Applying Eq. 1 o From 1, and Eq. 13.55 (using built-in function Omega (M , k )) For 1 = 26.9 o (1) M1 = 2.02 (Use Goal Seek to vary M 1 so that 1 is correct) Hence for p 1 we use Eq. 13.7a (13.7a) The approach is to apply Eq. 13.7a twice, so that (using built-in function Isenp (M , k )) p 1 = p (p 0/p )/(p 0/p 1) p1 = 43.3 kPa For the second deflection: We repeat the analysis of the first deflection Applying Eq. 1 2 + 1 = 2 - 2 = 2 + 1 + 2 = 41.9 o (Note that instead of working from the initial state to state 2 we could have worked from state 1 to state 2 because the entire flow is isentropic) From 2, and Eq. 13.55 (using built-in function Omega (M , k )) For 2 = 41.9 M2 = 2.62 o (Use Goal Seek to vary M 2 so that 2 is correct) Hence for p 2 we use Eq. 13.7a (using built-in function Isenp (M , k )) p 2 = p (p 0/p )/(p 0/p 2) p2 = 16.9 kPa Problem 13.201 [Difficulty: 3] Given: Air flow in a duct Find: Mach number and pressure at contraction and downstream; Solution: The given or available data is: k = M1 = 1.4 2.5 = p1 = 30 50 o kPa Equations and Computations: For the first oblique shock (1 to 2) we find from Eq. 13.49 (13.49) Using built-in function theta (M, ,k ) = 7.99 o Also, M 1n can be found from geometry (Eq. 13.47a) M 1n = 1.250 Then M 2n can be found from Eq. 13.48a) Using built-in function NormM2fromM (M,k ) (13.48a) M 2n = 0.813 Then, from M 2n and geometry (Eq. 13.47b) M2 = 2.17 From M 1n and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) p 2/p 1 = p2 = 1.66 82.8 Pressure ratio We repeat the analysis for states 1 to 2 for 2 to 3, for the second oblique shock We choose for M 2 by iterating or by using Goal Seek to target (below) to equal the previous , using built-in function theta (M, ,k ) = 7.99 o = 34.3 o Then M 2n (normal to second shock!) can be found from geometry (Eq. 13.47a) M 2n = 1.22 Then M 3n can be found from Eq. 13.48a) Using built-in function NormM2fromM (M,k ) M 3n = 0.829 Then, from M 3n and geometry (Eq. 13.47b) M3 = 1.87 From M 2n and Eq. 13.48d (using built-in function NormpfromM (M ,k )) p 3/p 2 = p3 = 1.58 130 Pressure ratio Problem 13.202 [Difficulty: 4] Given: Mach number and deflection angle Find: Static and stagnation pressures due to: oblique shock; compression wave Solution: The given or available data is: R = k = p1 = M1 = 286.9 1.4 50 3.5 J/kg.K kPa = 35 o = 35 o Equations and Computations: For the oblique shock: We need to find M 1n The deflection angle is From M 1 and , and Eq. 13.49 (using built-in function Theta (M , ,k )) (13.49) For = 35.0 o = 57.2 o (Use Goal Seek to vary so that = 35o) From M 1 and M 1n = 2.94 From M 1n and p 1, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) p2 = 496 kPa To find M 2 we need M 2n. From M 1n, and Eq. 13.48a (using built-in function NormM2fromM (M ,k )) (13.48a) M 2n = 0.479 The downstream Mach number is then obtained from from M 2n, and , and Eq. 13.47b M 2n = M 2sin( - ) Hence M2 = (13.47b) 1.27 For p 02 we use Eq. 12.7a (using built-in function Isenp (M , k )) (13.7a) p 02 = p 2/(p 02/p 2) p 02 = 1316 kPa For the isentropic compression wave: For isentropic flow p 0 = constant p 02 = p 01 p 01 = 3814 kPa p 02 = 3814 kPa For p 01 we use Eq. 13.7a (using built-in function Isenp (M , k )) (Note that for the oblique shock, as required by Eq. 13.48b (13.48b) p 02/p 01 = 0.345 (using built-in function Normp0fromM (M ,k ) p 02/p 01 = 0.345 (using p 02 from the shock and p 01) For the deflection = = -35.0 (Compression ) o We use Eq. 13.55 (13.55) and Deflection = 2 - 1 = (M 2) - (M 1) From M 1 and Eq. 13.55 (using built-in function Omega (M , k )) Applying Eq. 1 1 = 58.5 2 = 1 + 2 = 23.5 o 2 = M2 = 23.5 1.90 o o From 2, and Eq. 13.55 (using built-in function Omega (M , k )) For (Use Goal Seek to vary M 2 so that 2 = 23.5o) Hence for p 2 we use Eq. 13.7a (using built-in function Isenp (M , k )) p 2 = p 02/(p 02/p 2) p2 = 572 kPa (1) Problem 13.203 [Difficulty: 3] Given: Deflection of air flow Find: Mach numbers and pressures Solution The given or available data is: R = k = p2 = M2 = 286.9 1.4 10 4 1 = 15 o 2 = 15 o J/kg.K kPa Equations and Computations: We use Eq. 13.55 (13.55) and Deflection = a - b = (M a) - (M b) From M and Eq. 13.55 (using built-in function Omega (M , k )) 2 = 65.8 o For the second deflection: Applying Eq. 1 1 = 2 - 2 1 = 50.8 o From 1, and Eq. 13.55 (using built-in function Omega (M , k )) For 1 = 50.8 M1 = 3.05 o (Use Goal Seek to vary M 1 so that 1 is correct) (1) Hence for p 1 we use Eq. 13.7a (13.7a) The approach is to apply Eq. 13.7a twice, so that (using built-in function Isenp (M , k )) p 1 = p 2(p 0/p 2)/(p 0/p 1) p1 = 38.1 kPa For the first deflection: We repeat the analysis of the second deflection Applying Eq. 1 2 + 1 = 2 - = 2 - (2 + 1) = 35.8 o (Note that instead of working from state 2 to the initial state we could have worked from state 1 to the initial state because the entire flow is isentropic) From , and Eq. 13.55 (using built-in function Omega (M , k )) For = 35.8 M = 2.36 o (Use Goal Seek to vary M so that is correct) Hence for p we use Eq. 13.7a (using built-in function Isenp (M , k )) p = p 2(p 0/p 2)/(p 0/p ) p = 110 kPa Problem 13.204 [Difficulty: 4] Given: Mach number and airfoil geometry Find: Lift and drag per unit span Solution: The given or available data is: R = k = p1 = M1 = = c = 286.9 1.4 50 1.75 18 1 J/kg.K kPa o m Equations and Computations: F = (p L - p U)c The net force per unit span is Hence, the lift force per unit span is L = (p L - p U)c cos() (1) D = (p L - p U)c sin() (2) The drag force per unit span is For the lower surface (oblique shock): We need to find M 1n The deflection angle is = = 18 o From M 1 and , and Eq. 13.49 (using built-in function Theta (M , ,k )) (13.49) For = 18.0 o = 62.9 o (Use Goal Seek to vary so that is correct) From M 1 and M 1n = 1.56 From M 1n and p 1, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) p2 = 133.2 pL = p2 pL = 133.2 kPa kPa For the upper surface (isentropic expansion wave): For isentropic flow p 0 = constant p 02 = p 01 For p 01 we use Eq. 13.7a (using built-in function Isenp (M , k )) (13.7a) For the deflection p 01 = 266 kPa p 02 = 266 kPa = = 18.0 (Compression ) o We use Eq. 13.55 (13.55) and Deflection = 2 - 1 = (M 2) - (M 1) From M 1 and Eq. 13.55 (using built-in function Omega (M , k )) Applying Eq. 3 1 = 19.3 2 = 1 + 2 = 37.3 o o (3) From 2, and Eq. 13.55 (using built-in function Omega (M , k )) For 2 = M2 = 37.3 2.42 o (Use Goal Seek to vary M 2 so that 2 is correct) Hence for p 2 we use Eq. 13.7a (using built-in function Isenp (M , k )) p 2 = p 02/(p 02/p 2) p2 = 17.6 kPa pU = p2 pU = 17.6 kPa From Eq. 1 L = 110.0 kN/m From Eq. 2 D = 35.7 kN/m Problem 13.205 [Difficulty: 3] Given: Wedge-shaped airfoil Find: Lift per unit span assuming isentropic flow Solution: The given or available data is: R k p M = = = = = c = 286.9 1.4 70 2.75 7 1.5 J/kg.K kPa o m Equations and Computations: The lift per unit span is L = (p L - p U)c (1) (Note that p L acts on area c /cos(), but its normal component is multiplied by cos()) For the upper surface: pU = p pU = 70 kPa For the lower surface: = = -7.0 o We use Eq. 13.55 (13.55) and Deflection = L - = (M L) - (M ) (2) From M and Eq. 13.55 (using built-in function Omega (M , k )) = 44.7 = L - L = + L = 37.7 o L = ML = 37.7 2.44 o o Applying Eq. 2 From L, and Eq. 13.55 (using built-in function Omega (M , k )) For (Use Goal Seek to vary M L so that L is correct) Hence for p L we use Eq. 13.7a (13.7a) The approach is to apply Eq. 13.7a twice, so that (using built-in function Isenp (M , k )) p L = p (p 0/p )/(p 0/p L) From Eq 1 pL = 113 kPa L = 64.7 kN/m Problem 13.206 [Difficulty: 4] Given: Mach number and airfoil geometry Find: Drag coefficient Solution: The given or available data is: R = k = p1 = M1 = 286.9 1.4 95 2 J/kg.K kPa = 0 o = 10 o Equations and Computations: The drag force is D = (p F - p R)cs tan(/2) (1) (s and c are the span and chord) This is obtained from the following analysis Airfoil thickness (frontal area) = 2s (c /2tan(/2)) Pressure difference acting on frontal area = (p F - p R) (p F and p R are the pressures on the front and rear surfaces) The drag coefficient is 2 C D = D /(1/2V A ) But it can easily be shown that V 2 = pkM 2 (2) Hence, from Eqs. 1 and 2 C D = (p F - p R)tan(/2)/(1/2pkM 2) (3) For the frontal surfaces (oblique shocks): We need to find M 1n The deflection angle is = /2 = 5 o From M 1 and , and Eq. 13.49 (using built-in function Theta (M , ,k )) (13.49) = 5.0 o = 34.3 o M 1n = 1.13 For (Use Goal Seek to vary so that = 5o) From M 1 and From M 1n and p 1, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) p2 = 125.0 pF = p2 pF = 125.0 kPa kPa To find M 2 we need M 2n. From M 1n, and Eq. 13.48a (using built-in function NormM2fromM (M ,k )) (13.48a) M 2n = 0.891 The downstream Mach number is then obtained from from M 2n, and , and Eq. 13.47b M 2n = M 2sin( - ) Hence M2 = (13.47b) 1.82 For p 02 we use Eq. 13.7a (using built-in function Isenp (M , k )) (13.7a) p 02 = 742 kPa For the rear surfaces (isentropic expansion waves): Treating as a new problem Here: M 1 is the Mach number after the shock and M 2 is the Mach number after the expansion wave p 01 is the stagnation pressure after the shock and p 02 is the stagnation pressure after the expansion wave M 1 = M 2 (shock) M1 = 1.82 p 01 = p 02 (shock) p 01 = For isentropic flow For the deflection 742 kPa p 0 = constant p 02 = p 01 p 02 = 742 = = 10.0 kPa o We use Eq. 13.55 (13.55) and Deflection = 2 - 1 = (M 2) - (M 1) From M 1 and Eq. 13.55 (using built-in function Omega (M , k )) Applying Eq. 3 1 = 21.3 2 = 1 + 2 = 31.3 o o From 2, and Eq. 13.55 (using built-in function Omega(M, k)) For 2 = M2 = (Use Goal Seek to vary M 2 so that 2 = 31.3o) 31.3 2.18 o (3) Hence for p 2 we use Eq. 13.7a (using built-in function Isenp (M , k )) p 2 = p 02/(p 02/p 2) Finally, from Eq. 1 p2 = 71.2 pR = p2 pR = 71.2 CD = 0.0177 kPa kPa Problem 13.207 Given: Mach number and airfoil geometry Find: Plot of lift and drag and lift/drag versus angle of attack Solution: The given or available data is: k = p1 = M1 = 1.4 50 1.75 = c = 12 1 kPa o m Equations and Computations: The net force per unit span is F = (p L - p U)c Hence, the lift force per unit span is L = (p L - p U)c cos() (1) The drag force per unit span is D = (p L - p U)c sin() (2) For each angle of attack the following needs to be computed: [Difficulty: 4] For the lower surface (oblique shock): We need to find M 1n Deflection = From M 1 and , and Eq. 13.49 (using built-in function Theta (M , ,k )) (13.49) find (Use Goal Seek to vary so that is the correct value) From M 1 and find M 1n From M 1n and p 1, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) find p2 and pL = p2 For the upper surface (isentropic expansion wave): For isentropic flow p 0 = constant p 02 = p 01 For p 01 we use Eq. 13.7a (using built-in function Isenp (M , k )) (13.7a) find p 02 = = Deflection 266 kPa we use Eq. 13.55 (13.55) and Deflection = 2 - 1 = (M 2) - (M 1) (3) From M 1 and Eq. 13.55 (using built-in function Omega (M , k )) find Applying Eq. 3 1 = 19.3 2 = 1 + o From 2, and Eq. 12.55 (using built-in function Omega (M , k )) From 2 find M2 (Use Goal Seek to vary M 2 so that 2 is the correct value) (4) Hence for p 2 we use Eq. 13.7a (using built-in function Isenp (M , k )) p 2 = p 02/(p 02/p 2) pU = p2 Finally, from Eqs. 1 and 2, compute L and D Computed results: (o) (o) (o) 0.50 1.00 1.50 2.00 4.00 5.00 10.00 15.00 16.00 16.50 17.00 17.50 18.00 35.3 35.8 36.2 36.7 38.7 39.7 45.5 53.4 55.6 56.8 58.3 60.1 62.9 0.50 1.00 1.50 2.00 4.00 5.00 10.0 15.0 16.0 16.5 17.0 17.5 18.0 Error 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% Sum: 0.0% M 1n p L (kPa) 2 (o) 2 from M 2 (o) 1.01 1.02 1.03 1.05 1.09 1.12 1.25 1.41 1.44 1.47 1.49 1.52 1.56 51.3 52.7 54.0 55.4 61.4 64.5 82.6 106.9 113.3 116.9 121.0 125.9 133.4 19.8 20.3 20.8 21.3 23.3 24.3 29.3 34.3 35.3 35.8 36.3 36.8 37.3 19.8 20.3 20.8 21.3 23.3 24.3 29.3 34.3 35.3 35.8 36.3 36.8 37.3 Error 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% Sum: 0.0% M2 p U (kPa) L (kN/m) D (kN/m) 1.77 1.78 1.80 1.82 1.89 1.92 2.11 2.30 2.34 2.36 2.38 2.40 2.42 48.7 47.4 46.2 45.0 40.4 38.3 28.8 21.3 20.0 19.4 18.8 18.2 17.6 2.61 5.21 7.82 10.4 20.9 26.1 53.0 82.7 89.6 93.5 97.7 102.7 110 0.0227 0.091 0.205 0.364 1.46 2.29 9.35 22.1 25.7 27.7 29.9 32.4 35.8 L/D 115 57.3 38.2 28.6 14.3 11.4 5.67 3.73 3.49 3.38 3.27 3.17 3.08 To compute this table: 1) Type the range of 2) Type in guess values for 3) Compute from Eq. 13.49 (using built-in function Theta (M ,, k ) 4) Compute the absolute error between each and 5) Compute the sum of the errors 6) Use Solver to minimize the sum by varying the values (Note: You may need to interactively type in new values if Solver generates values that lead to no ) 7) For each , M 1n is obtained from M 1, and Eq. 13.47a 8) For each , p L is obtained from p 1, M 1n, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) 9) For each , compute 2 from Eq. 4 10) For each , compute 2 from M 2, and Eq. 13.55 (using built-in function Omega (M ,k )) 11) Compute the absolute error between the two values of 2 12) Compute the sum of the errors 13) Use Solver to minimize the sum by varying the M 2 values (Note: You may need to interactively type in new M 2 values) if Solver generates values that lead to no ) 14) For each , p U is obtained from p 02, M 2, and Eq. 13.47a (using built-in function Isenp (M , k )) 15) Compute L and D from Eqs. 1 and 2 Lift and Drag of an Airfoil as a Function of Angle of Attack L and D (kN/m) 120 100 80 Lift 60 Drag 40 20 0 0 2 4 6 8 10 12 14 16 18 20 () o Lift/Drag of an Airfoil as a Function of Angle of Attack 140 120 L/D 100 80 60 40 20 0 0 2 4 6 8 10 (o) 12 14 16 18 20 Problem 13.208 [Difficulty: 4] Given: Mach number and airfoil geometry FU 1 Find: Lift and Drag coefficients FL RU RL Solution: R = k = p1 = M1 = The given or available data is: 286.9 1.4 95 2 J/kg.K kPa = 12 o = 10 o Equations and Computations: Following the analysis of Example 13.14 the force component perpendicular to the major axis, per area, is F V/sc = 1/2{(p FL + p RL) - (p FU + p RU)} (1) and the force component parallel to the major axis, per area, is F H/sc = 1/2tan(/2){(p FU + p FL) - (p RU + p RL)} (2) using the notation of the figure above. (s and c are the span and chord) The lift force per area is F L/sc = (F Vcos() - F Hsin())/sc (3) The drag force per area is F D/sc = (F Vsin() + F Hcos())/sc C L = F L/(1/2V 2A ) The lift coefficient is (4) (5) But it can be shown that V 2 = pkM 2 (6) Hence, combining Eqs. 3, 4, 5 and 6 C L = (F V/sc cos() - F H/sc sin())/(1/2pkM 2) (7) Similarly, for the drag coefficient C D = (F V/sc sin() + F H/sc cos())/(1/2pkM 2) (8) For surface FL (oblique shock): We need to find M 1n The deflection angle is = + /2 = 17 o From M 1 and , and Eq. 13.49 (using built-in function Theta (M , ,k )) (13.49) For = 17.0 o = 48.2 o (Use Goal Seek to vary so that = 17o) From M 1 and M 1n = 1.49 From M 1n and p 1, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) p2 = 230.6 p FL = p2 p FL = 230.6 kPa kPa To find M 2 we need M 2n. From M 1n, and Eq. 13.48a (using built-in function NormM2fromM (M ,k )) (13.48a) M 2n = 0.704 The downstream Mach number is then obtained from from M 2n, and , and Eq. 13.47b M 2n = M 2sin( - ) Hence M2 = (13.47b) 1.36 For p 02 we use Eq. 13.7a (using built-in function Isenp (M , k )) (13.7a) p 02 = 693 kPa For surface RL (isentropic expansion wave): Treating as a new problem Here: M 1 is the Mach number after the shock and M 2 is the Mach number after the expansion wave p 01 is the stagnation pressure after the shock and p 02 is the stagnation pressure after the expansion wave M 1 = M 2 (shock) M1 = 1.36 p 01 = p 02 (shock) p 01 = For isentropic flow For the deflection 693 kPa p 0 = constant p 02 = p 01 p 02 = 693 = = 10.0 kPa o We use Eq. 13.55 (13.55) and Deflection = 2 - 1 = (M 2) - (M 1) (3) From M 1 and Eq. 13.55 (using built-in function Omega (M , k )) Applying Eq. 3 1 = 7.8 2 = 1 + 2 = 17.8 o o From 2, and Eq. 13.55 (using built-in function Omega (M , k )) 2 = M2 = For 17.8 1.70 o (Use Goal Seek to vary M 2 so that 2 = 17.8o) Hence for p 2 we use Eq. 13.7a (using built-in function Isenp (M , k )) p 2 = p 02/(p 02/p 2) p2 = 141 p RL = p2 p RL = 141 kPa kPa For surface FU (isentropic expansion wave): M1 = For isentropic flow 2.0 p 0 = constant p 02 = p 01 p 01 = p 02 = 743 743 For p 01 we use Eq. 13.7a (using built-in function Isenp (M , k )) For the deflection = - /2 = 7.0 kPa o We use Eq. 13.55 and Deflection = 2 - 1 = (M 2) - (M 1) (3) From M 1 and Eq. 13.55 (using built-in function Omega (M , k )) Applying Eq. 3 1 = 26.4 2 = 1 + 2 = 33.4 o o From 2, and Eq. 13.55 (using built-in function Omega(M, k)) 2 = M2 = For 33.4 2.27 o (Use Goal Seek to vary M 2 so that 2 = 33.4o) Hence for p 2 we use Eq. 13.7a (using built-in function Isenp (M , k )) p 2 = p 02/(p 02/p 2) p2 = 62.8 p FU = p2 p FU = 62.8 kPa kPa For surface RU (isentropic expansion wave): Treat as a new problem. Flow is isentropic so we could analyse from region FU to RU but instead analyse from region 1 to region RU. M1 = For isentropic flow TOTAL deflection 2.0 p 0 = constant p 02 = p 01 p 01 = p 02 = 743 743 = + /2 = 17.0 kPa kPa o We use Eq. 13.55 and Deflection = 2 - 1 = (M 2) - (M 1) (3) From M 1 and Eq. 13.55 (using built-in function Omega (M , k )) Applying Eq. 3 1 = 26.4 2 = 1 + 2 = 43.4 o o From 2, and Eq. 13.55 (using built-in function Omega(M, k)) 2 = M2 = For 43.4 2.69 o (Use Goal Seek to vary M 2 so that 2 = 43.4o) Hence for p 2 we use Eq. 13.7a (using built-in function Isenp (M , k )) p 2 = p 02/(p 02/p 2) p2 = 32.4 kPa p RU = p2 p RU = 32.4 kPa p FL = p RL = p FU = p RU = 230.6 140.5 62.8 32.4 kPa kPa kPa kPa The four pressures are: From Eq 1 F V/sc = 138 kPa From Eq 2 F H/sc = 5.3 kPa From Eq 7 CL = 0.503 From Eq 8 CD = 0.127 Problem 13.209 [Difficulty: 3] Given: The gas dynamic relations for compressible flow Find: The shock values and angles in each region Solution: Begin with the 1-D gas dynamic relations for compressible flow Governing equations: k 1 2 V1 M 1 kRT1 ; T01 T1 1 M1 2 Assumption: The flow is compressible and supersonic V1 M 1 kRTa 5 1.4 287 216.7 1475.4 V f V1 1475.4 m s m s k 1 2 T01 T1 1 M 1 1300 K 2 From (1) to (2) there is an oblique shock with M 1 =5 and 100 From the oblique shock figure (or tables) 1 19.38 M1n M1 sin( ) M1n 1.659 1 ( ) M 1n M1 M2 M 2n From Normal Shock Tables M 1n 1.659 M 2 n 0.65119 10 T2 1.429 T1 M2 M 2n 4.0 sin( ) M2 M1 Problem 13.210 [Difficulty: 3] Given: The gas dynamic relations for compressible flow Find: The shock values and angles in each region Solution: Begin with the 1-D gas dynamic relations for compressible flow Governing equations: k 1 2 V1 M 1 kRT1 ; T01 T1 1 M1 2 Assumption: The flow is compressible and supersonic V1 M 1 kRTa 5 1.4 287 216.7 1475.4 V f V1 1475.4 m s m s k 1 2 T01 T1 1 M 1 1300 K 2 From (2) to (3) A second oblique shock with M 2 4.0 and 100 From the oblique shock tables 2 22.230 and M 2 n M 2 sin 1.513 From normal shock tables M 3n 0.698 M 3n 0.698 sin( ) sin12.23 M 3 3.295 M3 V1 10° V1 10° Problem 13.211 [Difficulty: 4] Given: The gas dynamic relations for compressible flow Find: Exit Mach number and velocity Solution: Begin with the 1-D gas dynamic relations for compressible flow Governing equations: k 1 2 M1 V1 M 1 kRT1 ; T01 T1 1 2 Assumption: The flow is compressible and supersonic V1 M 1 kRTa 5 1.4 287 216.7 1475.4 m s Assuming M2 = 4.0, M3 = 3.295, and M4 = 1.26 A 4* 1.05 A T04 and 1.317 T4 A5 A5 A4 5 1.05 5.25 A* A4 A* M 5 3.23 With k 1 2 M 5 3.11 T5 2 To find the temperature at state 5, we need to express the temperature in terms of the entrance temperature and known temperature ratios: T T T T0 T0 T T5 T1 2 3 4 4 5 5 T1 T2 T3 T4 T04 T05 T05 1 Now since the stagnation temperatures at 4 and 5 are equal (isentropic flow through the nozzle): 1 T5 216.7 K 1.429 1.333 3.744 1.317 1 3.11 T5 654.5 K Therefore, the exhaust velocity is: m V5 M 5 kRT5 3.23 1.4 287 654.5 1656 s
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File Type : PDF File Type Extension : pdf MIME Type : application/pdf PDF Version : 1.5 Linearized : No Page Count : 2184 XMP Toolkit : XMP toolkit 2.9.1-13, framework 1.6 About : uuid:4d7635d5-a7d3-11f1-0000-422aa637022d Producer : GPL Ghostscript 9.19 Modify Date : 2016:08:31 17:08:59-03:00 Create Date : 2016:08:31 17:08:59-03:00 Creator Tool : PScript5.dll Version 5.2.2 Document ID : uuid:4d7635d5-a7d3-11f1-0000-422aa637022d Format : application/pdf Title : Fox and McDonald Introduction to Fluid Mechanics 8th edition Solution Manual Creator : John Leylegian Description : Solutions Manual Author : John Leylegian Subject : Solutions ManualEXIF Metadata provided by EXIF.tools