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Instant download and all chapters Solutions Manual Mechanics of Fluids 4th Edition Potter, Wiggert, Ramadan https://testbankdata.com/download/3460            

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

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

1

59.95 kg

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 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

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 1.17

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. b) N = [C] kg. c) m3/s = [C] m2 m2/3.

[C] = m/s2 [C] = N/kg = kg m/s2 kg = m/s2 [C] = m3/s m2 m2/3 = m1/3/s

2

ML2 / T 3

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 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 1.20

kg

m s2

m km f . Since all terms must have the same dimensions (units) we require: s [c] = kg/s, [k] = kg/s2 = N s 2 / m s 2 N / m, [f] = kg m / s 2 N. c

Note: we could express the units on c as [c] = kg / s 1.21

a) 250 kN e) 1.2 cm2

1.22

a) 1.25 108 N m3 d) 5.6

1.23

1.24

0.225

b) 572 GPa f) 76 mm3

2 2

0.738

N s/m

d) 17.6 cm3

c) 42 nPa

b) 3.21 10 5 s e) 5.2 10 2 m2

0.06854m

N s2 / m s

c) 6.7 108 Pa f) 7.8 109 m3

m

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 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 a) 20 cm/hr =

1.25

a) F = ma = 10 40 = 400 N. b) F W = ma. c) F W sin 30 = ma.

1.26

The mass is the same on the earth and the moon: 60 1.863. m= Wmoon = 1.863 5.4 = 10.06 lb 32.2

F = 10 F = 10

40 + 10 9.81 = 498.1 N. 40 + 9.81 0.5 = 449 N.

3

Chapter 1 / Basic Considerations

1.27

a)

0.225

b)

0.225

c)

0.225

m d

2

m d

2

m d2

0.225

0.225 0.225

4.8 10

26

0.184 (3.7 10

4.8 10

)

26

0.00103 (3.7 10 4.8 10

0.43 10 6 m or 0.00043 mm

10 2

10 2

7.7 10 5 m or 0.077 mm

10 2

0.0039 m or 3.9 mm

)

26

0.00002 (3.7 10

)

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 c) 14.7 e) 30

31 = 70 kPa abs.

b) 760

31 14.7 = 10.2 psia. d) 34 101 31 30 = 20.8 in. of Hg abs. 101

31 760 = 527 mm of Hg abs. 101 31 34 = 23.6 ft of H2O 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

1.31

a) p = 973 +

1.32

T = 48 +

22,560 20,000 (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 35,000

30,000 ( 65.8 + 48) = 59 F or ( 59 30,000

4

32)

5 = 50.6 C 9

Chapter 1/ Basic Considerations

1.33

1.34

p= Fn Ft

26.5 cos 42 Fn = = 1296 MN/m2 = 1296 MPa. 4 A 152 10 (120 000) 0.2 10 20 0.2 10

4

4

2.4 N

0.0004 N

Fn2

F=

= tan

1

Ft2 = 2.400 N.

0.0004 =0.0095 2.4

Density and Specific Weight

m V

0.2 = 1.92 slug/ft3. 180 / 1728

1.35

=

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

1.38

W a) m = g

0.0024T = 13.6 0.0024 50 = 13.48. 13.48 13.6 % error = 100 = .88% 13.6 V g

12 400 500 10 9.81

12 400 500 10 b) m = 9.77 12 400 500 10 c) m = 9.83 1.39

= g = 1.92

S=

m/ V water

water

6

6

. 1.2

6

= 0.632 kg

= 0.635 kg = 0.631 kg

10/ V . 1.94

V = 4.30 ft3

5

32.2 = 61.8 lb/ft3.

Chapter 1 / Basic Considerations Viscosity 1.40

Assume carbon dioxide is an ideal gas at the given conditions, then 200 kN/m3 0.189 kJ/kg K 90 273 K

p RT

W V

mg V

2.915 kg/m3

2.915 kg/m3 9.81 m/s2

g

From Fig. B.1 at 90°C,

28.6 kg/m2 s2

28.6 N/m3

2 10 5 N s/m2 , so that the kinematic viscosity is

2 10 5 N s/m2 2.915 kg/m3

6.861 10 6 m2 /s

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 Vpiston 0

V r

Using Wpiston

Dcylinder

Dpiston / 2

mpiston g , we can write Vpiston

mpiston g

Dcylinder

Dpiston / 2

DL

Solve Vpiston :

Vpiston

mpiston g Dcylinder 2

Dpiston

DL

0.350 kg 9.81 m/s2 0.1205 0.120 m 2

2 0.025 N s/m

2

0.12 0.10 m

where we used N = kg·m/s2.

6

0.91 kg m2 /N s3

0.91 m/s

Chapter 1/ Basic Considerations 1.42

du /dy . From the given velocity

The shear stress can be calculated using distribution,

120(0.05 2 y )

1.308 10 3 N s/m2 so, at the lower plate where y = 0,

From Table B.1 at 10 C, du dy

du dy

y2 )

u ( y ) 120(0.05 y

120(0.05 0) 6 s 1

1.308 10 3

6 7.848 10 3 N/m2

y 0

At the upper plate where y = 0.05 m, du dy

1.43

1.44

120(0.05 2 0.05)

2

r = 0.25

r = 0.5

1.45

T = force =

30(2 1/12) (1/12)

du dr

30(2 1/12) (1/12)

7.848 10 3 N/m2

y 0.05

du = 1.92 dr

=

1

6s

= 32

= 32

[32r / r02 ] 32 r / r02 .

1 1

= 0.014 lb/ft2

2

10

0.25 /100

3

10 3

(0.5 /100) 2 0.5 /100 (0.5 /100) 2

moment arm = 2 RL

T

2 R3 L 1.46 Use Eq.1.5.8: T = h power =

= 6.4 Pa

du 2 R2L = dr

R=

0.4 1000 2 12

= T 550

2

0.5/12

.012 0.2

3

2000 2 60 0.01/12

2.74 209.4 = 1.04 hp 550

7

= 0,

= 3.2 Pa,

0.0026

0.4 1000 2 R 2 L 2 R

r=0

0.4 R

2

1000 2 R2L.

= 0.414 N.s/m2.

4 0.006

= 2.74 ft-lb.

Chapter 1 / Basic Considerations 1.47

Fbelt =

du A 1.31 10 dy power =

1.48

F V 746

3

10 (0.6 0.002

15.7 10 = 0.210 hp 746

r . Due to the area h du r = dA r = 2 r dr r. dy

Assume a linear velocity so element shown, dT = dF

T=

R

2 2 3 r dr = h h

0

1.49

4) = 15.7 N.

du dy

400 2 60 2 0.08/12

2.36 10

R4 4

5

dr r

(3/12) 4

= 91

10 5 ft-lb.

u . y

The velocity at a radius r is r . The shear stress is The torque is dT = rdA on a differential element. We have T=

rdA=

0.08 0

2000 2 60

r 2 rdx , 0.0002

209.4 rad/s

where x is measured along the rotating surface. From the geometry x T=

0.08

0.1

0

1.50

209.4 x / 2 2 0.0002

x dx 329 000 2

0.08

x 2dx

0

2 r, so that

329 000 (0.083 ) = 56.1 N . m 3

du = 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 Cy 1) where A, B, C, D, E, and K are constants. e 0 C

If

1.51

Ae 40

B/T

= 2.334

0.001 Ae B/293 0.000357

Ae

A = 2.334

B /353

10 6 e1776/313 = 6.80

10 4 N.s/m2

8

10 6, B = 1776.