Instant download and all chapters Solutions Manual Mechanics of Fluids 4th Edition Potter, Wiggert, Ramadan https://testbankdata.com/download/solutions-manual-mechanics-fluids-4th-edition-potter-wiggert-ra madan/ 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 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. 8 9 1.3 (A) 2.36 10 23.6 10 23.6 nPa. du 1.4 (C) The mass is the same on earth and the moon: [4(8r )] 32 r. dr 1.5 (C) Fshear F sin 4200sin 30 2100 N. F 2100 N = shear 4 2 84 103 Pa or 84 kPa A 250 10 m 1.6 (B) (T 4)2 (80 4)2 1.7 (D) water 1000 1000 968 kg/m3 180 180 du 3 1.8 (A) [10 5000r ] 10 10 5000 0.02 1 Pa. dr 4 cos 4 0.0736 N/m 1 1.9 (D) h 3 m or 300 cm. gD 1000 kg/m3 9.81 m/s2 10 10 6 m We used kg = N·s2/m 1.10 (C) pV 800 kN/m 2 4 m3 1.11 (C) m 59.95 kg RT 0.1886 kJ/(kg K) (10 273) K 1 Chapter 1 / Basic Considerations 1.12 (B) Eice 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 287 323 304 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 / T 2 L2 M / LT 2 c) power = force velocity = F L / T ML / T 2 L / T ML2 / T 3 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 M FT 2 / L 1.16 a) density = 3 FT 2 / L4 L L3 b) pressure = F/L2 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 1.17 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/ 3 T 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 2 Chapter 1/ Basic Considerations 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 kg m 1 d) viscosity: N s/m2 = 2 s 2 kg / m s s m N m kg m m e) heat flux: J/s = 2 kg m 2 / s 3 s s s J N m kg m m f) specific heat: 2 m 2 / K s2 kg K kg K s kg K m m 1.20 kg c km f . Since all terms must have the same dimensions (units) we require: s2 s [c] = kg/s, [k] = kg/s2 = N s 2 / m s 2 N / m, [f] = kg m / s 2 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 0.06854m m 1.23 0.225 2 2 0.738 0.00194 3.281 d d2 where m is in slugs, in slug/ft3 and d in feet. We used the conversions in the front cover. 20/100 1.24 a) 20 cm/hr = 5.555 10 5 m/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 = 500 1000 3600 = 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: 60 m= 1.863. Wmoon = 1.863 5.4 = 10.06 lb 32.2 3 Chapter 1 / Basic Considerations 26 m 4.8 10 1.27 a) 0.225 2 0.225 10 2 0.43 10 6 m or 0.00043 mm d 0.184 (3.7 10 ) 26 m 4.8 10 b) 0.225 2 0.225 10 2 7.7 10 5 m or 0.077 mm d 0.00103 (3.7 10 ) 26 m 4.8 10 c) 0.225 0.225 0.0039 m or 3.9 mm d2 0.00002 (3.7 10 10 2 ) 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. 31 1.29 a) 101 31 = 70 kPa abs. b) 760 760 = 527 mm of Hg abs. 101 31 31 c) 14.7 14.7 = 10.2 psia. d) 34 34 = 23.6 ft of H2O abs. 101 101 31 e) 30 30 = 20.8 in. of Hg abs. 101 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 62.8 61.6 % error = 100 = 1.95 %. 61.6 22,560 20,000 1.31 a) p = 973 + (785 973) = 877 psf 25,000 20,000 22,560 20,000 T = 12.3 + ( 30.1 + 12.3) = 21.4 F 25,000 20,000 0.512 b) p = 973 + 0.512 (785 973) + ( .488) (628 2 785 + 973) = 873 psf 2 0.512 T = 12.3 + 0.512 ( 30.1 + 12.3) + ( .488) ( 48 + 2 30.1 12.3) = 21.4 F 2 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. 33,000 30,000 5 1.32 T = 48 + ( 65.8 + 48) = 59 F or ( 59 32) = 50.6 C 35,000 30,000 9 4 Chapter 1/ Basic Considerations Fn 26.5 cos 42 1.33 p= = 4 = 1296 MN/m2 = 1296 MPa. A 152 10 4 Fn (120 000) 0.2 10 2.4 N 1.34 F= Fn2 Ft2 = 2.400 N. 4 Ft 20 0.2 10 0.0004 N 1 0.0004 = tan =0.0095 2.4 Density and Specific Weight m 0.2 1.35 = = 1.92 slug/ft3. = g = 1.92 32.2 = 61.8 lb/ft3. V 180 / 1728 1.36 = 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 = .20% 978 9560 978 9.81 % error for = 100 = .36% 978 9.81 1.37 S = 13.6 0.0024T = 13.6 0.0024 50 = 13.48. 13.48 13.6 % error = 100 = .88% 13.6 6 W V 12 400 500 10 1.38 a) m = = 0.632 kg g g 9.81 6 12 400 500 10 b) m = = 0.635 kg 9.77 6 12 400 500 10 c) m = = 0.631 kg 9.83 m/ V 10/ V 1.39 S= . 1.2 . V = 4.30 ft3 water water 1.94 5 Chapter 1 / Basic Considerations Viscosity 1.40 Assume carbon dioxide is an ideal gas at the given conditions, then p 200 kN/m3 2.915 kg/m3 RT 0.189 kJ/kg K 90 273 K W mg g 2.915 kg/m3 9.81 m/s2 28.6 kg/m2 s2 28.6 N/m3 V V From Fig. B.1 at 90°C, 2 10 5 N s/m2 , so that the kinematic viscosity is 2 10 5 N s/m2 6.861 10 6 m2 /s 2.915 kg/m3 The kinematic viscosity cannot be read from Fig. B.2; the pressure is not 100 kPa. 1.41 At equilibrium the weight of the piston is balanced by the resistive force in the oil due to wall shear stress. 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, assume a linear velocity distribution in the oil due to the piston motion. That is, the shear stress is V Vpiston 0 r Dcylinder Dpiston / 2 Using Wpiston mpiston g , we can write Vpiston mpiston g DL Dcylinder Dpiston / 2 Solve Vpiston : mpiston g Dcylinder Dpiston Vpiston 2 DL 0.350 kg 9.81 m/s2 0.1205 0.120 m 2 2 0.91 kg m2 /N s3 0.91 m/s 2 0.025 N s/m 0.12 0.10 m where we used N = kg·m/s2. 6 Chapter 1/ Basic Considerations 1.42 The shear stress can be calculated using du /dy . From the given velocity distribution, du u ( y ) 120(0.05 y y2 ) 120(0.05 2 y ) dy From Table B.1 at 10 C, 1.308 10 3 N s/m2 so, at the lower plate where y = 0, du 120(0.05 0) 6 s 1 1.308 10 3 6 7.848 10 3 N/m2 dy y 0 At the upper plate where y = 0.05 m, du 1 120(0.05 2 0.05) 6s 7.848 10 3 N/m2 dy y 0.05 du 30(2 1/12) 1.43 = = 1.92 2 = 0.014 lb/ft2 dr (1/12) 30(2 1/12) du 1.44 2 [32r / r02 ] 32 r / r02 . r=0 = 0, (1/12) dr 3 0.25 /100 r = 0.25 = 32 1 10 = 3.2 Pa, (0.5 /100) 2 3 0.5 /100 r = 0.5 = 32 1 10 = 6.4 Pa (0.5 /100) 2 du 0.4 1.45 T = force moment arm = 2 RL R= 2 R2L = 2 1000 2 R2L. dr R T 0.0026 = = 0.414 N.s/m2. 0.4 0.4 2 1000 2 R 2 L 1000 2 .012 0.2 R 12 3 2000 2 2 0.5/12 4 0.006 2 R3 L 60 1.46 Use Eq.1.5.8: T = = = 2.74 ft-lb. h 0.01/12 T 2.74 209.4 power = = 1.04 hp 550 550 7 Chapter 1 / Basic Considerations du 3 10 1.47 Fbelt = A 1.31 10 (0.6 4) = 15.7 N. dy 0.002 F V 15.7 10 power = = 0.210 hp 746 746 du r 1.48 Assume a linear velocity so . Due to the area dy h dr du r element shown, dT = dF r = dA r = 2 r dr r. dy 400 2 5 R 2.36 10 (3/12) 4 2 3 2 R4 60 T= r dr = = 91 10 5 ft-lb. 0 h h 4 2 0.08/12 u 1.49 The velocity at a radius r is r . The shear stress is . y The torque is dT = rdA on a differential element. We have 0.08 r 2000 2 T= rdA= 2 rdx , 209.4 rad/s 0 0.0002 60 where x is measured along the rotating surface. From the geometry x 2 r, so that 0.08 0.08 209.4 x / 2 x 329 000 T= 0.1 2 dx 329 000 x 2dx (0.083 ) = 56.1 N . m 0 0.0002 2 0 3 du 1.50 If = cons’t and = AeB/T = AeBy/K = AeCy, then dy du du AeCy = cons’t. = De Cy. dy dy D Cy y Finally, or u(y) = e = E (e Cy 1) where A, B, C, D, E, and K are constants. C 0 B/T 0.001 Ae B/293 1.51 Ae A = 2.334 10 6, B = 1776. B /353 0.000357 Ae 40 = 2.334 10 6 e1776/313 = 6.80 10 4 N.s/m2 8