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INSTRUCTOR'S SOLUTIONS MANUALTO ACCOMPANYMECHANICS of FLUIDSnullFOURTH EDITIONMERLE C. POTTER Michigan State UniversityDAVID C. WIGGERT Michigan State UniversityBASSEM RAMADAN Kettering UniversityContents Chapter 1 Basic Considerations Chapter 2 Fluid Statics Chapter 3 Introduction to Fluids in Moti...

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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-r  madan/     

INSTRUCTOR'SSOLUTIONSMANUAL  

TOACCOMPANY  

 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 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.

We used kg = N·s /m 1.10

(C )

1.11

(C )

m

pV RT

800 kN/m2 0.1886 kJ/(kg

3

1

Chapter 1 / Basic Considerations 1.12

(B )

5 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

Chapter 1 Problems: Dimensions, Units, and Physical Quantities 1.14

Conservation of mass — Mass — density

1.17

a) L = [C] T2. [C] = L/T2 b) F = [C]M. [C] = F/M = ML/T2 M = L/T2 3 2 2/3 c) L /T = [C] L L . [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

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 m s N m kg m m e) heat flux: J/s = kg m 2 / s 3 s s s2 J N m kg m m m2 / K s2 f) specific heat: kg K kg K s2 kg K m

m

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

1.26

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

F = 10 F = 10

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

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

3

Chapter 1 / Basic Considerations

1.27

a)

m

4.8

or 0.00043 mm

0.184

b)

m

4.8

or 0.077 mm

0.00103

c)

m d

4.8

or 3.9 mm

0.00002

Pressure and Temperature 1.28

Use the values from Table B.3 in the Appendix.

0.512 ( .488) (628 2 785 + 973) = 873 psf 2 0.512 ( .488) ( 48 + 2 30.1 12.3) = 21.4 F T = 12.3 + 0.512 ( 30.1 + 12.3) + 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.

b) p = 973 + 0.512 (785

1.32

T = 48 +

33,000 35,000

973) +

( 65.8 + 48) = 59 F or ( 59

4

32)

5 = 50.6 C 9

Chapter 1/ Basic Considerations

1.33

1.34

p=

26.5 cos 42 Fn = 1296 MN/m2 = 1296 MPa. = A 152

Fn

Fn2

F= N

Ft

= tan

1

2

= 2.400 N.

0.0004 =0.0095 2.4

Density and Specific Weight

1.39

S=

V

. 1.2

10/ V . 1.94

V = 4.30 ft3

5

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

p RT

W V

mg V

g

From Fig. B.1 at 90°C,

Vpiston

, so that the kinematic viscosity is

mpiston g Dcylinder 2

Dpiston

DL

0.350 kg 9.81 m/s2 0.1205 2 where we used N = kg·m/s2.

6

m

Chapter 1/ Basic Considerations 1.42

du dy From the given velocity

The shear stress can be calculated using distribution, u(y)

y

du dy

y

From Table B.1 at 10 C, du dy

so, at the lower plate where y = 0,

y

At the upper plate where y = 0.05 m,

1.46

Use Eq.1.5.8: T =

2

h

power =

3

2

3

= T 550

2000 60 0.01 /12

2.74 209.4 = 1.04 hp 550

7

= 2.74 ft-lb.

Chapter 1 / Basic Considerations

1.47

Fbelt =

du A dy

10 (0.6 0.002

power =

1.48

F V 746

15.7 10 = 0.210 hp 746

Assume a linear velocity so element shown, dT = dF

4) = 15.7 N.

du dy

r h

r = dA

Due to the area du 2 r dr dy

r=

dr r

r.

C

/

1.51 40

= 2.334

0.001

/293

0.000357

/353

10 6 e1776/313 = 6.80

A = 2.334

10 4 N.s/m2

8

10 6, B = 1776....