MECHANICS of FLUIDS INSTRUCTOR'S SOLUTIONS MANUAL TO ACCOMPANY FOURTH EDITION. MERLE C. POTTER Michigan State University DAVID C.

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1 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

2 Contents Chapter 1 Basic Considerations 1 Chapter Fluid Statics 15 Chapter Introction to Fluids in Motion 4 Chapter 4 The Integral Fors of the Fundaental Laws 61 Chapter 5 The Differential Fors of the Fundaental Laws 107 Chapter 6 Diensional Analysis and Siilitude 15 Chapter 7 Internal Flows 145 Chapter 8 External Flows 19 Chapter 9 Copressible Flow 7 Chapter 10 Flow in Open Channels 59 Chapter 11 Flows in Piping Systes 0 Chapter 1 Turboachinery 45 Chapter 1 Measureents in Fluid Mechanics 69 Chapter 14 Coputational Fluid Dynaics 75

3 CHAPTER 1 Basic Considerations Chapter 1/ Basic Considerations FE-type Exa Review Probles: Probles 1-1 to (C) = F/a or kg = N//s = N. s /. 1. (B) [μ [τ /] = (F/L )/(L/T)/L = F. T/L. 1. (A) npa. 1.4 (C) The ass is the sae on earth and the oon: [4(8 r)] r. dr 1.5 (C) Fshear F sin 400sin N. 1.6 (B) Fshear 100 N = Pa or 84 kpa A (D) water ( T 4) (80 4) kg/ (A) dr [ r] Pa. 1.9 (D) 1.10 (C) h 4 cos N/ 1 gd kg/ 9.81 /s We used kg = N s / or 00 c (C) pv RT 800 kn/ kj/(kg K) (10 7) K kg 1

4 Chapter 1 / Basic Considerations 1.1 (B) Eice Ewater. ice 0 water cwater T. 6 5 (40 10 ) ( 10 ) T. T 7.66 C. We assued the density of ice to be equal to that of water, naely 1000 kg/. Ice is actually slightly lighter than water, but it is not necessary for such accuracy in this proble. 1.1 (D) For this high-frequency wave, c RT /s. Chapter 1 Probles: Diensions, Units, and Physical Quantities 1.14 Conservation of ass Mass density Newton s second law Moentu velocity The first law of theronaics internal energy teperature 1.15 a) density = ass/volue = M / L b) pressure = force/area = F / L ML / T L M / LT c) power = force velocity = F L / T ML / T L / T ML / T d) energy = force distance = ML / T L ML / T e) ass flux = ρav = M/L L L/T = M/T f) flow rate = AV = L L/T = L /T 1.16 a) density = M FT / L L L b) pressure = F/L FT / L 4 c) power = F velocity = F L/T = FL/T d) energy = F L = FL e) ass flux = M FT / L FT / L T T f) flow rate = AV = L L/T = L /T 1.17 a) L = [C] T. [C] = L/T b) F = [C]M. [C] = F/M = ML/T M = L/T c) L /T = [C] L L /. [C] = L / T L / L 1/ L T Note: the slope S 0 has no diensions a) = [C] s. [C] = /s b) N = [C] kg. [C] = N/kg = kg /s kg = /s c) /s = [C] /. [C] = /s / = 1/ /s

5 Chapter 1/ Basic Considerations 1.19 a) pressure: N/ = kg /s / = kg/ s b) energy: N = kg /s = kg /s c) power: N /s = kg /s d) viscosity: N s/ = kg s 1 kg / s s e) heat flux: J/s = N kg kg / s s s s J N kg f) specific heat: kg K kg K s kg K / K s 1.0 kg c k f. Since all ters ust have the sae diensions (units) we require: s s [c] = kg/s, [k] = kg/s = N s / s N /, [f] = kg / s N. Note: we could express the units on c as [c] = kg / s N s / s N s / 1.1 a) 50 kn b) 57 GPa c) 4 npa d) 17.6 c e) 1. c f) a) N b) s c) Pa d) 5.6 e) f) d d where is in slugs, in slug/ft and d in feet. We used the conversions in the front cover. 0/ a) 0 c/hr = /s 600 b) 000 rev/in = 000 /60 = 09.4 rad/s c) 50 Hp = = 7 85 W d) 100 ft /in = /60 = /s e) 000 kn/c = 10 6 N/c 100 c / = N/ f) 4 slug/in = /60 = kg/s g) 500 g/l = kg/ kg/ h) 500 kwh = = J 1.5 a) F = a = = 400 N. b) F W = a. F = = N. c) F W sin 0 = a. F = = 449 N. 1.6 The ass is the sae on the earth and the oon: 60 = W oon = = lb.

6 Chapter 1 / Basic Considerations 1.7 a) b) c) or d (.7 10 ) d (.7 10 ) 6 6 or or.9 10 d (.7 10 ) Pressure and Teperature 1.8 Use the values fro Table B. in the Appendix. a) = 15.6 kpa. b) = 14. kpa. c) = kpa (use a straight-line interpolation). d) = 78.8 kpa. e) = 5.5 kpa. 1.9 a) = 70 kpa abs. b) c) = 10. psia. d) e) 0 0 = 0.8 in. of Hg abs = 57 of Hg abs. 4 =.6 ft of H O abs. 1.0 p = p o e gz/rt = 101 e /87 (15 + 7) = 6.8 kpa Fro Table B., at 4000 : p = 61.6 kpa. The percent error is % error = 100 = 1.95 %. 61.6,560 0, a) p = 97 + (785 97) = 877 psf 5,000 0,000,560 0,000 T = 1. + ( ) = 1.4 F 5,000 0,000 b) p = (785 97) (.488) ( ) = 87 psf T = ( ) (.488) ( ) = 1.4 F Note: The results in (b) are ore accurate than the results in (a). When we use a linear interpolation, we lose significant digits in the result. 1. T = 48 +,000 0,000 5,000 0,000 ( ) = 59 F or ( 59 ) 5 9 = 50.6 C 4

7 Chapter 1/ Basic Considerations F 1. p = n A = 6.5 cos = 196 MN/ = 196 MPa. 1.4 F F n t (10000) N N 4 F = n t F F =.400 N. = tan = Density and Specific Weight 1.5 = V / 178 = 1.9 slug/ft. = g = 1.9. = 61.8 lb/ft. 1.6 = 1000 (T 4) /180 = 1000 (70 4) /180 = 976 kg/ = 9800 (T 4) /18 = 9800 (70 4) /180 = 9560 N/ % error for = =.0% % error for = =.6% 1.7 S = T = = % error = 100 =.88% a) = W V g g b) = c) = = 0.65 kg = 0.61 kg = 0.6 kg 1.9 S = water V / 10/ V water V = 4.0 ft 5

8 Chapter 1 / Basic Considerations Viscosity 1.40 Assue carbon dioxide is an ideal gas at the given conditions, then p RT W g V V 00 kn/ kj/kg K 90 7 K g Fro Fig. B.1 at 90 C,.915 kg/.915 kg/ 9.81 /s 8.6 kg/ s 8.6 N/ 5 10 N s/, so that the kineatic viscosity is 5 10 N s/ /s.915 kg/ The kineatic viscosity cannot be read fro Fig. B.; the pressure is not 100 kpa At equilibriu the weight of the piston is balanced by the resistive force in the oil e to wall shear stress. This is represented by Wpiston DL where D is the diaeter of the piston and L is the piston length. Since the gap between the piston and cylinder is sall, assue a linear velocity distribution in the oil e to the piston otion. That is, the shear stress is V Vpiston 0 r D D cylinder piston / Using Wpiston piston g, we can write Solve V piston : Vpiston pistong DL D D / cylinder piston V piston g D D piston cylinder piston DL 0.50 kg 9.81 /s kg /N s 0.91 /s where we used N = kg /s N s/

9 Chapter 1/ Basic Considerations 1.4 The shear stress can be calculated using /. distribution, u( y) 10(0.05 y y ) 10(0.05 y) Fro the given velocity Fro Table B.1 at 10 C, N s/ so, at the lower plate where y = 0, y 0 10(0.05 0) 6 s N/ 1 At the upper plate where y = 0.05, y ( ) 6 s N/ = dr = 1.9 0( 1/1) (1/1) = lb/ft ( 1/1) (1/1) dr 0 r r0 [ r / r ] /. r = 0 = 0, r = 0.5 = /100 =. Pa, (0.5 /100) r = 0.5 = /100 = 6.4 Pa (0.5 /100) 1.45 T = force oent ar = RL R = = dr R L = T RL R R = N. s/. R L Use Eq.1.5.8: T = R L h = power = T / / = 1.04 hp 550 =.74 ft-lb. 7

10 Chapter 1 / Basic Considerations 1.47 F belt = A power = F V (0.6 4) = 15.7 N = 0.10 hp r 1.48 Assue a linear velocity so. h eleent shown, dt = df r = da r = Due to the area r dr r. r dr T = R 0 h r dr = (/1) R 60 h /1 5 4 = ft-lb. u 1.49 The velocity at a radius r is r. The shear stress is y. The torque is dt = rda on a differential eleent. We have 1.50 If 0.08 r T = rda= rdx, rad/s where x is easured along the rotating surface. Fro the geoetry x r, so that x/ x T= 0.1 dx x dx (0.08 ) = 56.1 N = cons t and = Ae Cy = cons t. Finally, or u(y) = D e C Cy AeB/T = Ae By/K = Ae Cy, then y 0 = De Cy. = E (e Cy 1) where A, B, C, D, E, and K are constants BT / Ae Ae Ae B/9 B/5 A = , B = = e 1776/1 = N. s/ 8