TEST BANK FOR Mechanics of Fluids 4TH Edition By Merle C. Potter, David C. Wiggert, Bassem H. Ramadan

Exam (elaborations) TEST BANK FOR Mechanics of Fluids 4TH Edition By Merle C. Potter, David C. Wiggert, Bassem H. Ramadan INSTRUCTOR'S SOLUTIONS MANUAL TO ACCOMPANY MECHANICS of FLUIDS FOURTH EDITION MERLE C. POTTER Michigan State University DAVID C. WIGGERT Michigan State University BASSEM RAMADAN Kettering University Contents Chapter 1 Basic Considerations 1 Chapter 2 Fluid Statics 15 Chapter 3 Introduction to Fluids in Motion 43 Chapter 4 The Integral Forms of the Fundamental Laws 61 Chapter 5 The Differential Forms of the Fundamental Laws 107 Chapter 6 Dimensional Analysis and Similitude 125 Chapter 7 Internal Flows 145 Chapter 8 External Flows 193 Chapter 9 Compressible Flow 237 Chapter 10 Flow in Open Channels 259 Chapter 11 Flows in Piping Systems 303 Chapter 12 Turbomachinery 345 Chapter 13 Measurements in Fluid Mechanics 369 Chapter 14 Computational Fluid Dynamics 375 Chapter 1/ Basic Considerations © 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. 1 CHAPTER 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 [τ du/dy] = (F/L2)/(L/T)/L = F.T/L2. 1.3 (A) 2.36 10 8 23.6 10 9 23.6 nPa. 1.4 (C) The mass is the same on earth and the moon: du [4(8r)] 32 r. dr 1.5 (C) Fshear F sin 4200sin30 2100 N. shear 3 4 2 = 2100 N 84 10 Pa or 84 kPa 250 10 m F A 1.6 (B) 1.7 (D) 2 2 3 water 1000 ( 4) 1000 (80 4) 968 kg/m 180 180 T 1.8 (A) du [10 5000r] .02 1 Pa. dr 1.9 (D) 3 2 6 4 cos 4 0.0736 N/m 1 3 m or 300 cm. 1000 kg/m 9.81 m/s 10 10 m h gD We used kg = N·s2/m 1.10 (C) 1.11 (C) m pV 800 kN/m2 4 m3 59.95 kg RT 0.1886 kJ/(kg K) (10 273) K Chapter 1 / Basic Considerations © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part. 2 1.12 (B) ice E Ewater . mice 320 mwater cwater T. 5 (40 10 6) 1000 320 (2 10 3) 1000 4.18 T. T 7.66 C. We assumed the density of ice to be equal to that of water, namely 1000 kg/m3. Ice is actually slightly lighter than water, but it is not necessary for such accuracy in this problem. 1.13 (D) For this high-frequency wave, c RT m/s. Chapter 1 Problems: Dimensions, Units, and Physical Quantities 1.14 Conservation of mass — Mass — density Newton’s second law — Momentum — velocity The first law of thermodynamics — internal energy — temperature 1.15 a) density = mass/volume = M / L3 b) pressure = force/area = F / L2 ML / T2L2 M / LT2 c) power = force velocity = F L / T ML / T2 L / T ML2 / T3 d) energy = force distance = ML / T2 L ML2 / T2 e) mass flux = ρAV = M/L3 × L2 × L/T = M/T f) flow rate = AV = L2 × L/T = L3/T 1.16 a) density = M L FT L L FT L 3 2 3 / 2 / 4 b) pressure = F/L2 c) power = F × velocity = F L/T = FL/T d) energy = F×L = FL e) mass flux = M T FT L T FT L 2 / / f) flow rate = AV = L2 L/T = L3/T 1.17 a) L = [C] T2. [C] = L/T2 b) F = [C]M. [C] = F/M = ML/T2 M = L/T2 c) L3/T = [C] L2 L2/3. [C] = L3 / T L2 L2/3 L1/3T Note: the slope S0 has no dimensions. 1.18 a) m = [C] s2. [C] = m/s2 b) N = [C] kg. [C] = N/kg = kg m/s2 kg = m/s2 c) m3/s = [C] m2 m2/3. [C] = m3/s m2 m2/3 = m1/3/s Chapter 1/ Basic Considerations © 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. 3 1.19 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 d) viscosity: N s/m2 = kg m s s 1 m kg / m s 2 2 e) heat flux: J/s = N m s kg m s m s kg m / s 2 2 3 f) specific heat: J kg K N m kg K kg m s m kg K m / K s 2 2 2 1.20 kg m s m s m 2 c k f . Since all terms must have the same dimensions (units) we require: [c] = kg/s, [k] = kg/s2 = N s2 / m s2 N / m, [f] =kg m / s2 N. Note: we could express the units on c as [c] = kg / s N s2 / m s N s / m 1.21 a) 250 kN b) 572 GPa c) 42 nPa d) 17.6 cm3 e) 1.2 cm2 f) 76 mm3 1.22 a) 1.25 108 N b) 3.21 10 5 s c) 6.7 108 Pa d) 5.6 m3 e) 5.2 10 2 m2 f) 7.8 109 m3 1.23 2 2 2 0.225 0.06854 0.738 0.00194 3.281 m m d d where m is in slugs, in slug/ft3 and d in feet. We used the conversions in the front cover. 1.24 a) 20 cm/hr = 20/100 5.555 10 5m/s 3600 b) 2000 rev/min = 2000 2 /60 = 209.4 rad/s c) 50 Hp = 50 745.7 = 37 285 W d) 100 ft3/min = 100 0.02832/60 = 0.0472 m3/s e) 2000 kN/cm2 = 2 106 N/cm2 1002 cm2/m2 = 2 1010 N/m2 f) 4 slug/min = 4 14.59/60 = 0.9727 kg/s g) 500 g/L = 500 10 3 kg/10 m 500 kg/m3 h) 500 kWh = = 1.8 109 J 1.25 a) F = ma = 10 40 = 400 N. b) F W = ma. F = 10 40 + 10 9.81 = 498.1 N. c) F W sin 30 = ma. F = 10 40 + 9.81 0.5 = 449 N. 1.26 The mass is the same on the earth and the moon: m = 60 32 2 1 863 . . . Wmoon = 1.863 5.4 = 10.06 lb Chapter 1 / Basic Considerations © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part. 4 1.27 a) 26 6 2 10 2 0.225 0.225 4.8 10 0.43 10 m 0.184 (3.7 10 ) m d or 0.00043 mm b) 26 5 2 10 2 0.225 0.225 4.8 10 7.7 10 m 0.00103 (3.7 10 ) m d or 0.077 mm c) 26 2 10 2 0.225 0.225 4.8 10 0.0039 m 0.00002 (3.7 10 ) m d or 3.9 mm Pressure and Temperature 1.28 Use the values from Table B.3 in the Appendix. a) 52.3 + 101.3 = 153.6 kPa. b) 52.3 + 89.85 = 142.2 kPa. c) 52.3 + 54.4 = 106.7 kPa (use a straight-line interpolation). d) 52.3 + 26.49 = 78.8 kPa. e) 52.3 + 1.196 = 53.5 kPa. 1.29 a) 101 31 = 70 kPa abs. b) 760 31 101 760 = 527 mm of Hg abs. c) 14.7 31 101 14.7 = 10.2 psia. d) 34 31 101 34 = 23.6 ft of H2O abs. e) 30 31 101 30 = 20.8 in. of Hg abs. 1.30 p = po e gz/RT = 101 e 9.81 4000/287 (15 + 273) = 62.8 kPa From Table B.3, at 4000 m: p = 61.6 kPa. The percent error is % error = 62.8 61.6 61.6 100 = 1.95 %. 1.31 a) p = 973 + 22,560 20,000 25,000 20,000 (785 973) = 877 psf T = 12.3 + 22,560 20,000 25,000 20,000 ( 30.1 + 12.3) = 21.4 F b) p = 973 + 0.512 (785 973) + 0.512 2 ( .488) (628 2 785 + 973) = 873 psf T = 12.3 + 0.512 ( 30.1 + 12.3) + 0.512 2 ( .488) ( 48 + 2 30.1 12.3) = 21.4 F Note: The results in (b) are more accurate than the results in (a). When we use a linear interpolation, we lose significant digits in the result. 1.32 T = 48 + 33,000 30,000 35,000 30,000 ( 65.8 + 48) = 59 F or ( 59 32) 5 9 = 50.6 C Chapter 1/ Basic Considerations © 2012 Cengage Lea