Solution-Manual-for-Mechanics-of-Fluids-4th-Edition-by-Potter.pdf

Chapter 1/ Basic ConsiderationsCHAPTER 1 Basic Considerations FE-type Exam Review Problems: Problems 1-1 to 1-14. 1.1 (C) m = F/a or kg = N/m/s2 = N.s2/m. 1.2 (B) [μ 1.3 (A) 2.36 10 1.4 (C) The mass is the same on earth and the moon: 1.5 (C) Fshear 1.6 (B) 1.7 (D) 1.8 (A) 1.9 (D) [τ du/dy] = (F/L2)/(L/T)/L = F.T/L2. 8 F sin F = shear A 1000 water du dr h 23.6 10 4 cos gD 9 23.6 nPa. 4200sin 30 2100 N 250 10 (T 4)2 180 4 m du dr [4(8r )] 32 r. 2100 N. 84 103 Pa or 84 kPa 2 (80 4)2 180 1000 [10 5000r ] 10 3 968 kg/m3 10 5000 0.02 1 Pa. 4 0.0736 N/m 1 1000 kg/m3 9.81 m/s2 10 10 6 m 3 m or 300 cm. We used kg = N·s2/m 1.10 1.11 (C) (C) m pV RT 800 kN/m 2 4 m3 0.1886 kJ/(kg K) (10 273) K 59.95 kg 1 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 15 a) density = mass/volume = M / L3 b) pressure = force/area = F / L2 ML / T 2 L2 M / LT 2 c) power = force velocity = F L / T ML / T 2 L / T d) energy = force distance = ML / T 2 L ML2 / T 2 e) mass flux = ρAV = M/L3 × L2 × L/T = M/T f) flow rate = AV = L2 × L/T = L3/T 1.16 M FT 2 / L a) density = 3 L L3 b) pressure = F/L2 ML2 / T 3 FT 2 / L4 c) power = F × velocity = F L/T = FL/T d) energy = F×L = FL M FT 2 / L e) mass flux = FT / L T T f) flow rate = AV = L2 L/T = L3/T 1. in whole or in part. 1. 1. c RT 287 323 304 m/s. c) m3/s = [C] m2 m2/3. Units. copied or duplicated. May not be scanned. . All Rights Reserved. [C] = F/M = ML/T2 M = L/T2 c) L3/T = [C] L2 L2/3.13 (D) For this high-frequency wave. Chapter 1 Problems: Dimensions. Ice is actually slightly lighter than water. and Physical Quantities 1. b) N = [C] kg. mice 320 mwater cwater T .18 a) m = [C] s2.66 C.17 a) L = [C] T2. We assumed the density of ice to be equal to that of water. [C] = m/s2 [C] = N/kg = kg m/s2 kg = m/s2 [C] = m3/s m2 m2/3 = m1/3/s 2 © 2012 Cengage Learning. namely 1000 kg/m3. but it is not necessary for such accuracy in this problem. T 7.14 Conservation of mass — Mass — density Newton’s second law — Momentum — velocity The first law of thermodynamics — internal energy — temperature 1. [C] = L3 / T L2 L2 / 3 L1/ 3 T Note: the slope S0 has no dimensions.Chapter 1 / Basic Considerations 1. 5 (40 10 6 ) 1000 320 (2 10 3 ) 1000 4. or posted to a publicly accessible we bsite.12 (B) Eice Ewater .18 T . [C] = L/T2 b) F = [C]M. c) F W sin 30 = ma. 1. [k] = kg/s2 = N s 2 / m s 2 N / m.21 10 5 s e) 5.1 N.281 d d2 where m is in slugs.25 108 N d) 5. m= Wmoon = 1.555 10 5 m/s 3600 b) 2000 rev/min = 2000 2 /60 = 209.59/60 = 0. c Note: we could express the units on c as [c] = kg / s 1.06 lb 32. b) F W = ma. Since all terms must have the same dimensions (units) we require: s [c] = kg/s.9727 kg/s g) 500 g/L = 500 10 3 kg/10 m 500 kg/m3 h) 500 kWh = 500 1000 3600 = 1.22 a) 1. in whole or in part.Chapter 1/ Basic Considerations 1.81 = 498.06854m N s2 / m s F = 10 F = 10 40 + 10 9.8 109 J 1.4 = 10.5 = 449 N. 40 + 9.225 b) 572 GPa f) 76 mm3 2 2 c) 6.26 0.19 a) pressure: N/m2 = kg m/s2/m2 = kg/m s2 b) energy: N m = kg m/s2 m = kg m2/s2 c) power: N m/s = kg m2/s3 kg m 1 s 2 kg / m s d) viscosity: N s/m2 = 2 s m N m kg m m kg m 2 / s 3 e) heat flux: J/s = 2 s s s J N m kg m m m 2 / K s2 f) specific heat: 2 kg K kg K s kg K 1. .21 a) 250 kN e) 1. The mass is the same on the earth and the moon: 60 1.20 kg m s2 m km f .0472 m3/s e) 2000 kN/cm2 = 2 106 N/cm2 1002 cm2/m2 = 2 1010 N/m2 f) 4 slug/min = 4 14.738 N s/m d) 17.23 0.863.2 cm2 1.2 3 © 2012 Cengage Learning.02832/60 = 0. All Rights Reserved. We used the conversions in the front cover.8 109 m3 m 0.4 rad/s c) 50 Hp = 50 745.24 a) 20 cm/hr = 1.7 = 37 285 W d) 100 ft3/min = 100 0.00194 3.863 5.81 0. or posted to a publicly accessible website.6 cm3 c) 42 nPa b) 3. [f] = kg m / s 2 N.2 10 2 m2 0.7 108 Pa f) 7. 20/100 5.25 a) F = ma = 10 40 = 400 N. May not be scanned.6 m3 1. in slug/ft3 and d in feet. copied or duplicated. 488) ( 48 + 2 30. 101 31 760 = 527 mm of Hg abs.32 T = 48 + 22.6 1.00103 (3.8 + 48) = 59 F or ( 59 30.225 4.077 mm 26 0.560 20.000 ( 65.30 p = po e gz/RT = 101 e 9. 1.8 10 0.8 10 ) 26 0.85 = 142.000 35.0039 m or 3.184 (3.7 10 5 m or 0. b) 760 31 14. d) 34 101 31 30 = 20.8 kPa From Table B. The percent error is 62.512 T = 12.3 + 1.6 kPa.3) + ( .7 10 10 2 ) 0. All Rights Reserved.6 % error = 100 = 1.3 = 153.4 = 106. copied or duplicated.512 b) p = 973 + 0.4 F 2 Note: The results in (b) are more accurate than the results in (a).225 m d 2 m d 2 m d2 0.81 4000/287 (15 + 273) = 62.2 kPa.000 20. a) 52. b) 52. d) 52.1 + 12.000 0.225 0.1 12.3) = 21. in whole or in part.5 kPa. e) 52.8 10 26 0.00002 (3.7 kPa (use a straight-line interpolation).225 0.560 20.3 + 26. c) 52. 101 31 34 = 23.196 = 53. 101 1.3 + 101. . or posted to a publicly accessible we bsite.29 a) 101 c) 14.7 = 10.8 61. When we use a linear interpolation.00043 mm 10 2 10 2 ) 7. at 4000 m: p = 61.31 a) p = 973 + 1.43 10 6 m or 0.8 in.488) (628 2 785 + 973) = 873 psf 2 0.6 ft of H2O abs.3) = 21.6 kPa.9 mm Pressure and Temperature 1.2 psia.225 0.7 10 4. we lose significant digits in the result. May not be scanned.95 %.49 = 78.27 a) 0. 33.225 b) c) 0.1 + 12. of Hg abs.000 (785 973) = 877 psf 25.3 + 89.6 C 9 4 © 2012 Cengage Learning.000 32) 5 = 50.512 (785 973) + ( .3 in the Appendix.7 e) 30 31 = 70 kPa abs.3 + 0.000 30.000 T = 12.512 ( 30.000 22.4 F 25.3.28 Use the values from Table B.3 + 54.Chapter 1 / Basic Considerations 1.3 + ( 30.8 kPa.000 20. 61.7 10 4. 400 N.0004 =0.83 1.20% 978 9560 978 9.77 12 400 500 10 c) m = 9.92 m/ V S= water water 6 = 0.48. copied or duplicated.631 kg 10/ V .81 12 400 500 10 b) m = 9.2 = 61.6 0.632 kg 6 = 0. All Rights Reserved. .8 lb/ft3.2 10 4 4 2.2 10 20 0. 180 / 1728 1.Chapter 1/ Basic Considerations 1. in whole or in part.94 V = 4.30 ft3 5 © 2012 Cengage Learning.0095 2. 0.2 = 0.38 W a) m = g 32. 0.37 S = 13. 1.0004 N = tan 1 Ft2 = 2.6 % error = 100 = .0024 50 = 13.39 = g = 1.33 1.48 13.635 kg 6 .36 = 1000 (T 4)2/180 = 1000 (70 4)2/180 = 976 kg/m3 = 9800 (T 4)2/18 = 9800 (70 4)2/180 = 9560 N/m3 976 978 % error for = 100 = .6 V g 12 400 500 10 9.92 slug/ft3. May not be scanned.34 p= Fn Ft 26.81 1. or posted to a publicly accessible website.81 % error for = 100 = .6 1.35 = 1.2 = 1.5 cos 42 Fn = = 1296 MN/m2 = 1296 MPa.4 N Fn2 F= 0. 1.88% 13. 13.4 Density and Specific Weight m V 0.0024T = 13. 4 A 152 10 (120 000) 0.36% 978 9. 915 kg/m3 9. then 200 kN/m3 0. B.10 m 0.189 kJ/kg K 90 273 K p RT W V mg V 2.1 at 90°C.120 m 2 2 0.91 kg m2 /N s3 0.025 N s/m 2 0. 6 © 2012 Cengage Learning.915 kg/m3 6. All Rights Reserved.Chapter 1 / Basic Considerations Viscosity 1. the shear stress is Vpiston 0 V r Using Wpiston Dcylinder Dpiston / 2 mpiston g .40 Assume carbon dioxide is an ideal gas at the given conditions.1205 0. we can write Vpiston mpiston g Dcylinder Dpiston / 2 DL Solve Vpiston : Vpiston mpiston g Dcylinder 2 Dpiston DL 0.12 0. .915 kg/m3 2. copied or duplicated. 28.6 kg/m2 s2 28. in whole or in part.2. the pressure is not 100 kPa.41 At equilibrium the weight of the piston is balanced by the resistive force in the oil due to wall shear stress.6 N/m3 2 10 5 N s/m2 .81 m/s2 0. May not be scanned.91 m/s where we used N = kg·m/s2. This is represented by Wpiston DL where D is the diameter of the piston and L is the piston length. Since the gap between the piston and cylinder is small. or posted to a publicly accessible we bsite.81 m/s2 g From Fig. so that the kinematic viscosity is 2 10 5 N s/m2 2. That is.861 10 6 m2 /s The kinematic viscosity cannot be read from Fig. assume a linear velocity distribution in the oil due to the piston motion. B.350 kg 9. 1. 43 1.74 ft-lb.5 /100) 2 moment arm = 2 RL T 2 R3 L 1.4 R 2 1000 2 R2L. du 2 R2L = dr R= r=0 0. du dy du dy y2 ) u ( y ) 120(0. in whole or in part.006 = 2.4 1000 2 12 2 0.45 T = force = 30(2 1/12) (1/12) du dr 30(2 1/12) (1/12) 7. du dy 1.5 /100 3 (0. 120(0. = 0.308 10 3 N s/m2 so. copied or duplicated. = 3.5 1. All Rights Reserved.05 y 120(0.Chapter 1/ Basic Considerations 1.848 10 3 N/m2 y 0 At the upper plate where y = 0.05) 2 r = 0.04 hp 550 7 © 2012 Cengage Learning. From Table B.25 /100 3 10 (0.4 1000 2 R 2 L 2 R 0.44 120(0.308 10 3 6 7.05 2 y ) 1. 1 1 = 0.012 0.1.42 du /dy .014 lb/ft2 2 10 0. .05 du = 1.0026 0.1 at 10 C.2 Pa.74 209.5.4 = 1. May not be scanned.5/12 = T 550 = 0.414 N.05 0) 6 s 1 1.2 3 2000 2 60 0.01/12 4 0.25 r = 0. .05 2 0.s/m2. From the given velocity The shear stress can be calculated using distribution.8: T = h power = = 6.05 m. at the lower plate where y = 0. or posted to a publicly accessible website.4 Pa 0.848 10 3 N/m2 y 0.92 dr = 1 6s = 32 = 32 [32r / r02 ] 32 r / r02 .46 Use Eq.5 /100) 2 0. 2. 001 Ae B/293 0.0002 2000 2 60 209. From the geometry x 0.08/12 2. in whole or in part.1 N . Due to the area h du r = dA r = 2 r dr r. or posted to a publicly accessible we bsite.210 hp 746 r .47 Fbelt = du A 1.Chapter 1 / Basic Considerations 1.002 15. dT = dF R 2 2 3 r dr = h h T= 0 1. May not be scanned. du dy 400 2 60 2 0.49 4) = 15.7 N. m 3 du = cons’t and = AeB/T = AeBy/K = AeCy. y The velocity at a radius r is r .000357 40 = 2.31 10 dy power = 1. B. D.1 0 1. copied or duplicated. . dy Assume a linear velocity so element shown.334 B /353 10 6 e1776/313 = 6.51 B/T 0. 10 4 N.08 x 2dx 0 2 r.6 0.4 rad/s where x is measured along the rotating surface.0002 x dx 329 000 2 0. 0.08 T= rdA= 0 r 2 rdx . C.36 10 R4 4 5 dr r (3/12) 4 = 91 10 5 ft-lb. so that 329 000 (0.80 10 6.083 ) = 56. e 0 C If Ae 1. The shear stress is The torque is dT = rdA on a differential element.334 Ae A = 2.48 F V 746 3 10 (0.08 T= 0. B = 1776. u . and K are constants. = De Cy. All Rights Reserved. then dy du du AeCy = cons’t. or u(y) = = E (e Cy 1) where A. We have 0.4 x / 2 2 0.7 10 = 0.s/m2 8 © 2012 Cengage Learning. E.50 209. dy dy D Cy y Finally. The pressure inside the droplet is greater than the outside pressure of 8000 kPa. or V 1. 9 © 2012 Cengage Learning.54 Use c = 1450 m/s.52 m= V . L = c t = 1450 V p V 2200 MPa.59 Use Table B. diesel fuel is usually pumped to a pressure of about 20 000 kPa before it is injected into the engine. 1. 1.0002 0.0736 0.866 1000 9.00504 = 7.000 144 /1. All Rights Reserved. Assume mass to be constant in a volume subjected dV d . in whole or in part.55 1.000 144 /1.5 MPa 20 p= 1. copied or duplicated.25 MPa 1 Surface Tension 2 R 2 0.8 10 4 20 2 10 = 0.00504 lb/ft. May not be scanned. 2 r pinside p= poutside 4 R Bubbles: p = 4 /R = 59.Chapter 1/ Basic Considerations Compressibility 1. to a pressure increase.1: 1.3 kPa 4 0.81 0.62 = 899 m 1.025 N/m 10 kPa 5 10 6 m Hence.96 104 Pa or 29.57 2100 V b) c 1 = 0.5.61 See Example 1.0538 psi 1/(32 12) 2 0.74 psf or 0. 1.00909 m3 or 9090 cm3 2200 0.6 kPa. d V = V d .0076 m3.0076 = 17. The difference is given by Eq. p= B 327.4: h= 4 cos gD 4 0. or posted to a publicly accessible website.000 144 /1.87 = 4870 fps V =3. B V = V V p B 1.3 = 136. . pinside poutside 10 kPa 8000 10 8010 kPa In order to achieve this high pressure in the droplet.13: 5 10 p 6 = 2.60 The droplet is assumed to be spherical.93 = 4940 fps V V 2270 0.0741 p= 1.93 = 4670 fps c) c 308. then dm = 0.58 = 0.130 m.56 a) c 327. Then dm = d V + V d .53 B= 1. May not be scanned. 10 © 2012 Cengage Learning.94 13. 1.968 < 0.2 0.1482 No Each surface tension force = D.63 force up = L 1.65 1. so T = 50 C is a maximum temperature.0174 in 2 cos = force down = ghtL. gt L needle 2 L. or posted to a publicly accessible we bsite. The water would “boil” above this temperature. There is a force on the outside and one on the inside of the ring. 1. All Rights Reserved. dl h h(x) F dW Vapor Pressure 1.4: h = 4 cos gD 4 0.6 32. copied or duplicated.032cos130 1. .81 2 0.62 See Example 1.64 Draw a free-body diagram: The force must balance: d2 W = 2 L or L g 4 d 1. At 50 C water has a vapor pressure of 12. F=2 D neglecting the weight of the ring. a force balance yields: d2 (0. .0741 4 4 0.00145 ft or 1.66 0. h= L 2 cos .8/12 = 0.67 D From the infinitesimal free-body shown: dx d cos gh x dx.68 The absolute pressure is p = 80 + 92 = 12 kPa. W 8 g From the free-body diagram in No.2 kPa.Chapter 1 / Basic Considerations 1.004)2 Is g< 2 ? 7850 9. cos = d d dx/d h g xdx g x We assumed small so that the element thickness is x. in whole or in part.47. the temperature from Table B.71 The absolute pressure is 14. Assuming atmospheric pressure to be 100 kPa. m V 0. to be 141 F.81 9333 N.5 11.74 in 101.287 293 1.226 101. p RT 1. 0.3 0. 1. the elevation that has a pressure of 50.72 The inlet pressure to a pump cannot be less than 0 kPa absolute. At 82 C the vapor pressure from Table B. x = 16. Ideal Gas p RT 1.76 W 1. 1. If bubbles were observed to form at 3.1 is interpolated. 0.226 kg/m3.1 is 50. copied or duplicated.8 (by interpolation). This would be the minimum pressure that could be obtained since the water would vaporize below this pressure.1 is 7. A circulation is set up and the air moves from the outside in and the inside out: infiltration.287 248 Yes. 1. p2 (35 14.03 N/m3 85 1.7 psi gage. 0.3 1.01 slug. in whole or in part.287 (273 15) p RT 1. or posted to a publicly accessible website.69 The engineer knew that water boils near the vapor pressure.77 Assume that the steel belts and tire rigidity result in a constant volume so that m1 = m2: V1 p2 V 2 p1 or T2 T1 m1RT1 p1 m2 RT2 .0 psia.70 At 40 C the vapor pressure from Table B. All Rights Reserved. = 1.5 = 3. The heavier air outside enters at the bottom and the lighter air inside exits at the top.19 kg/m3. May not be scanned.4 kPa. From Table B. using vapor pressure.287 (15 273) out 9. This is the “chimney” effect.3.226 kg/m3 .4 psia or 52.83 km.1339 15 2.8 kPa is interpolated to be 5500 m. 100 (10 20 4) 9.81 = 12.73 1. 11 © 2012 Cengage Learning. . we have 10 000 + 100 = 600 x.7) 150 460 10 460 67.0 psia (this is boiling).1339 slug/ft 3.Chapter 1/ Basic Considerations 1.75 750 44 1716 470 p Vg RT 0. or posted to a publicly accessible we bsite. Hence. 1 10 402 0. 1 15(V f2 102 ).9 fps.) 12 © 2012 Cengage Learning.15 m/s.6 m or d PE 1 mV 2 mg ( 10).4 fps.79 0 KE 1 mV 2 mg ( 20).80 W1-2 KE. May not be scanned. 2 0 1. 2 20 32. Hence. (See Problem 1. 2 1 15(V f2 102 ).2.78 The pressure holding up the mass is 100 kPa.81 40 32.2u1 0 u2 .8 C where cv comes from Table B. we have 100 000 1 m 9. u2 u1 40 000.82 E2 E1.27 m/s. 2 20sds b) V2 V2 1 15(V f2 102 ). copied or duplicated. 2 40 000 u cv T.4. in whole or in part. T 55. 35. C from Table B.81.2. 717 The following shows that the units check: E1 E2 . 0.42 m/s. The First Law 1. T 69. 1. All Rights Reserved.287 288 r 12. 100 1000 3600 2 1000 2000 10 6 4180 T . 1 5(V f2 102 ). We used c = 4180 J/kg. 2 Vf V V 25.2 C.Chapter 1 / Basic Considerations 1. 1 1500 2 1 mV 2 2 C mH2Oc T . 2 s ds 20 200sin Vf 15.2 m.m/s2 from Newton’s 2nd law. a) 200 0 10 0 102 20 2 10 200 cos c) 0 20 1. 25.5. m pV RT 100 4 r 3 / 3 10 200. 2 2 Vf 16.75 for a units check. 1 15(V f2 102 ). mcar V 2 mair c kg m2 / s 2 m2 kg C m2 kg C kg J/(kg C) N m s2 (kg m/s2 ) m s 2 where we used N = kg. using pA = W. 19. . m 10 200 kg. Note: We used kJ on the left and kJ on the right.32 2970 m.83 1.004 0.1 C. copied or duplicated.287 c) W 2 0. The specific heat c was found in Table B.287 293 200 2 b) Q mcv T (1. or posted to a publicly accessible website. Also. a measured pressure is a gage pressure.88 T1 2 V 1 ). May not be scanned.287 373 200 2 c) Q mcv T (1.Chapter 1/ Basic Considerations 1.310 ft-lb.287 473 1. 1.287)(2 373 373) 999 kJ 0.4 /1. mRT T1 T2 since V the temperature doubles if the pressure doubles. 32.84 m f h f mwater c T . All Rights Reserved.2 2 W pd V The 1st law states: Q W u mcv T 0.287)(2 293 293) 999 kJ 0.4 T 423 p2 p1 2 (150 100) 904 kPa abs or 804 kPa gage.4 / 0. T1 293 Note: We assumed patm = 100 kPa since it was not given.310 ft-lb or 101 Btu.2 C We assume an isentropic process for the maximum pressure: k /k 1 1.7. L c t 357 8.4 151. If p = const. 0. 200 2 a) Q mcv T (1. 2V 1 and W a) W 2 0.2 40 000 100 4.287 b) W 2 0. Q W 78. .287 2 c= kRT T2 p( V p(2 V 1 V 1) 333 191 kJ 423 243 kJ 473 272 kJ T1 V1 pV 1 T2 so if T2 V2 mRT1.85 If the volume is fixed the reversible work is zero since the boundary does not move.5.87 1. 4 1 W1-2 1716 530ln 78.8 K or 121. using .287)(2 473 473) 999 kJ 0.17. Finally. 1. in whole or in part.18 T . Hence.004 0. p1 V 1 p2 V 2 . p p1 p2 Table B. T 19.004 0. p mRT dV V2 d V mRT mRT ln mRT ln 2 V V V1 p1 since. for the T = const process.4 287 318 357 m/s. 2T1.86 W pd V then V 1. p2 p1 k 1/ k (20 273) 500 5000 0.4 and Eq. Also. 13 © 2012 Cengage Learning. 1. c 340 299 % decrease 100 12.1 m/s b) c kRT 1.4 287 293 343.32 2970 m. or posted to a publicly accessible we bsite.4 188.9 m/s d) c kRT 1.92 a) c= kRT 1.K in the above equations.4 287 223 At 10 000 m the speed of sound c kRT 1. At sea level. in whole or in part.4 296.5 293 424.7.004 0. b) c= kRT 1.89 p2 w p1 T2 / T1 u k /k 1 100 473 / 293 cv (T2 T1) 1.4 534 kPa abs.4 287 253 319 m/s.9 293 266.90 a) c kRT 1. 340 1. We used Eq.32 2854 m. (1. 1. L c t 319 8. 299 m/s. All Rights Reserved. L c t 357 8.287)(473 293) 129 kJ/kg.4 461.4 287 318 357 m/s.4/0.91 kRT 1.32 2654 m. .4 4124 293 1301 m/s e) c kRT 1. Speed of Sound 1.4 287 293 343 m/s. May not be scanned. 1.17 for cv. copied or duplicated.9 m/s c) c kRT 1.06 %.4 287 288 340 m/s.8 293 348. L c t 343 8. 14 © 2012 Cengage Learning.1 m/s Note: We must use the units on R to be J/kg.Chapter 1 / Basic Considerations 1. c) c= kRT 1.