Mechanics of Fluids, SI Edition, Student Solutions Manual

196 Pages·2011·2.046 MB·English

by  Merle C. Potter, David C. Wiggert, Bassem H. Ramadan

A STUDENT’S SOLUTIONS MANUAL TO ACCOMPANY 

MECHANICS of FLUIDS,  
TH
4  E , SI 
DITION  

MERLE C. POTTER 
DAVID C. WIGGERT 
BASSEM H. RAMADAN
S ' S M  
TUDENT S  OLUTIONS  ANUAL

A  
TO  CCOMPANY

MECHANICS of FLUIDS 

FOURTH EDITION, SI 

MERLE C. POTTER 
Michigan State University 

DAVID C. WIGGERT 
Michigan State University 

BASSEM RAMADAN 
Kettering University
Contents 

Preface                    iv 

Chapter 1   Basic Considerations          1 

Chapter 2   Fluid Statics              15 

Chapter 3   Introduction to Fluids in Motion        29 

Chapter 4   The Integral Forms of the Fundamental Laws    37 

Chapter 5   The Differential Forms of the Fundamental Laws    59 

Chapter 6   Dimensional Analysis and Similitude            77 

Chapter 7   Internal Flows                     87 

Chapter 8   External Flows                      109 

Chapter 9   Compressible Flow                      131 

Chapter 10  Flow in Open Channels                     141 

Chapter 11   Flows in Piping Systems                    161 

Chapter 12  Turbomachinery                      175 

Chapter 13  Measurements in Fluid Mechanics                           189
Preface for the Student 

This manual provides the solutions to the problems whose answers are provided at the end of our
book, MECHANICS OF FLUIDS, SI. In many cases, the solutions are not as detailed as the
examples in the book; they are intended to provide the primary steps in each solution so you, the
student, are able to quickly review how a problem is solved. The discussion of a subtle point,
should one exist in a particular problem, is left as a task for the instructor. In general, some
knowledge of a problem may be needed to fully understand all of the steps presented. This
manual is not intended to be a self-paced workbook; your instructor is critically needed to provide
explanations, discussions, and illustrations of the myriad of phenomena encountered in the study
of fluids, but it should give you considerable help in working through a wide variety of problems.

The degree of difficulty and length of solution for each problem varies considerably. Some are
relatively easy and others quite difficult. Typically, the easier problems are the first problems for
a particular section.

We continue to include a number of multiple-choice problems in the earlier chapters, similar to
those encountered on the Fundamentals of Engineering Exam (the old EIT Exam) and the
GRE/Engineering Exam. These problems will provide a review for the Fluid Mechanics part of
those exams. They are all four-part, multiple-choice problems and are located at the beginning of
the appropriate chapters.

The examples and problems have been carefully solved with the hope that errors have not been
introduced. Even though extreme care is taken and problems are reworked, errors creep in. We
would appreciate knowing about any errors that you may find so they can be eliminated in future
printings. Please send any corrections or comments to [email protected] have class tested
most of the chapters with good response from our students, but we are sure that there are
improvements to be made.

East Lansing, Michigan Merle C. Potter
David C. Wiggert
Flint, Michigan Bassem Ramadan
Chapter 1 / Basic Considerations
CHAPTER 1

Basic Considerations

FE-type Exam Review Problems: Problems 1.1 to 1.13

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 Pa =23.6×10−9 Pa =23.6 nPa
The mass is the same on earth and the moon, so we calculate the mass using the
weight given on earth as:
m = W/g = 250 N/9.81 m/s2 = 25.484 kg
1.4 (C)
Hence, the weight on the moon is:
W = mg = 25.484 × 1.6 = 40.77 N
The shear stress is due to the component of the force acting tangential to the
area:
1.5 (C) F = Fsinθ=4200sin30(cid:68) =2100 N
shear
F 2100 N
τ= shear = =84×103 Pa or 84 kPa
A 250×10−4 m2
1.6 (B) – 53.6°C
Using Eqn. (1.5.3):
1.7 (D) (T −4)2 (80−4)2
ρ =1000− =1000− =968 kg/m3
water
180 180
du
The shear stress is given by:τ=μ
dr
We determine du/dr from the given expression for u as:
1.8 (A)
du d
= ⎡10(1−2500r2)⎤=−50,000r
⎣ ⎦
dr dr
At the wall r = 2 cm = 0.02 m. Substituting r in the above equation we get:
1

© 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.
Chapter 1 / Basic Considerations
du
=50,000r = 1000 1/s
dr
The density of water at 20°C is 10−3 N⋅s/m2
Now substitute in the equation for shear stress to get
du
τ=μ =10−3 N⋅s/m2×1000 1/s=1 N/m2 =1 Pa
dr
Using Eqn. (1.5.16), β=0(for clean glass tube), and σ=0.0736 N/mfor water
(Table B.1 in Appendix B) we write:
1.9 (D) 4σcosβ 4×0.0736 N/m×1
h= = =3 m or 300 cm
ρgD 1000 kg/m3×9.81 m/s2×10×10−6 m
where we used N = kg×m/s2
1.10 (C) Density
Assume propane (C H ) behaves as an ideal gas. First, determine the gas
3 8
R 8.314 kJ/kmol⋅K
constant for propane using R= u = =0.1885 kJ/kg
M 44.1 kg/kmol
1.11 (C)
2 3
pV 800 kN/m ×4 m
then, m= = =59.99 kg ≈ 60 kg
RT 0.1885 kJ/(kg⋅K)×(10+273) K
Consider water and ice as the system. Hence, the change in energy for the
system is zero. That is, the change in energy for water should be equal to the
change in energy for the ice. So, we write ΔE =ΔE
ice water
m ×320 kJ/kg=m ×c ΔT
ice water water
The mass of ice is calculate using
m =ρV =5 cubes×(1000 kg/m3)×(40×10−6 m3/cube)=0.2 kg
ice
Where we assumed the density of ice to be equal to that of water, namely 1000
1.12 (B)
kg/m3. Ice is actually slightly lighter than water, but it is not necessary for
such accuracy in this problem.
Similarly, the mass of water is calculated using
m =ρV =(1000 kg/m3)×2 liters(10−3 m3/liter)=2 kg
water
Solving for the temperature change for water we get
(cid:68)
0.2 kg×320 kJ/kg=2kg×4.18 kJ/kg⋅K×ΔT ⇒ ΔT =7.66 C
2

© 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.
Chapter 1 / Basic Considerations
Since a dog’s whistle produces sound waves at a high frequency, the speed of
sound is c= RT = 287J/kg⋅K×323K =304 m/s
1.13 (D)
where we used J/kg = m2/s2.

Dimensions, Units, and Physical Quantities

M FT2/L
a) density = = = FT2/L4
L3 L3
1.16 c) power = F×velocity = F×L/T = FL/T
M/T FT2/L
e) mass flux = = = FT/L3
A L2T
1.18 b) N = [C] kg ∴[C] = N/kg = (kg⋅ m/s2)/kg = m/s2
m m
kg +c +km= f. Since all terms must have the same dimensions (units)
s2 s
we require:
1.20
[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
1.22 a) 1.25×108 N c) 6.7×108 Pa e) 5.2 ×10−2 m2
cm m hr
a) 20 cm/hr =20 × × =5.556×10−5m/s
hr 100cm 3600s
745.7 W
c) 500 hp=500 hp× =37,285 W
1.24
hp
2
kN ⎛100cm⎞
2 10 2
e) 2000 kN/cm =2000 × =2×10 N/m
⎜ ⎟
cm2 ⎝ m ⎠
The mass is the same on the earth and the moon, so we calculate the mass, then
calculate the weight on the moon:
1.26
m=27 kg ∴ Wmoon = (27 kg) × (1.63) = 44.01 N

3

© 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.
Chapter 1 / Basic Considerations
Pressure and Temperature

Use the values from Table B.3 in the Appendix:
b) At an elevation of 1000 m the atmospheric pressure is 89.85 kPa. Hence, the
1.28 absolute pressure is 52.3 + 89.85 = 142.2 kPa
d) At an elevation of 10,000 m the atmospheric pressure is 26.49 kPa, and the
absolute pressure is 52.3 + 26.49 = 78.8 kPa
p = p e−gz/RT = 101 e−(9.81 × 4000)/[(287) × (15 + 273)] = 62.8 kPa
o
From Table B.3, at 4000 m: p = 61.6 kPa. The percent error is
1.30
62.8 − 61.6
% error = × 100 = 1.95 %
61.6
Using Table B.3 and linear interpolation we write:
10,600 − 10,000
T = 223.48 + (223.6 − 216.7) = 221.32 K
1.32 12,000 − 10,000
5
or (221.32 − 273.15) = −51.83°C
9
The normal force due pressure is: F =(120,000 N/m2)×0.2×10−4 m2 =2.4N
n
The tangential force due to shear stressis: F =20 N/m2×0.2×10−4 m2 =0.0004N
t
1.34 The total force is F = F2 +F2 = 2.40 N
n t
⎛0.0004⎞
The angle with respect to the normal direction is θ = tan−1 = 0.0095°
⎜ ⎟
⎝ 2.4 ⎠

Density and Specific Weight

Using Eq. 1.5.3 we have
ρ = 1000 − (T − 4)2/180 = 1000 − (70 − 4)2/180 = 976 kg/m3
γ = 9800 − (T − 4)2/18 = 9800 − (70 − 4)2/180 = 9560 N/m3
Using Table B.1 the density and specific weight at 70°C are
ρ=977.8 kg/m3
1.36
γ=977.8×9.81=9592.2 N/m3
976−978
% error for ρ = × 100 = −0.20%
978
9560−9592
% error for γ = × 100 = −0.33%
9592
4

© 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.
Chapter 1 / Basic Considerations
3 −6 3
γV 12,400N/m ×500×10 m
1.38 b) m= = = 0.635 kg
g 9.77 m/s2

Viscosity

Assume carbon dioxide is an ideal gas at the given conditions, then
p 200 kN/m3
ρ= = =2.915 kg/m3
RT (0.189 kJ/kg⋅K)(90+273 K)

W mg
γ= = =ρg =2.915 kg/m3×9.81 m/s2 =28.6 kg/m2⋅s2 =28.6 N/m3
V V
1.40
From Fig. B.1 at 90°C, μ≅2×10−5 N⋅s/m2, so that the kinematic viscosity is
μ 2×10−5 N⋅s/m2
ν= = =6.861×10−6 m2/s
ρ 2.915 kg/m3

The kine matic viscosity cannot be read from Fig. B.2 since the pressure is not at
100 kPa.
The shear stress can be calculated using τ=μdu/dy . From the given velocity

du
distribution, u =120(0.05y− y2) we get ⇒ =120(0.05−2y)

dy
From Table B.1 at 10°C for water,μ=1.308×10−3 N⋅s/m2

So, at the lower plate where y = 0, we have
1.42
du
=120(0.05−0)=6 s−1 ⇒ τ=(1.308×10−3)×6=7.848×10−3 N/m2
dy
At the upper plate where y = 0.05 m,
du
= 120(0.05−2×0.05) =6 s−1 ⇒ τ=7.848×10−3 N/m2
dy
y=0.05
5

© 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.
Chapter 1 / Basic Considerations
The shear stress can be calculated using τ=μdu/dr . From the given velocity

du du
u =16(1−r2 r2) ⇒ we get =16(−2r r2). Hence, =32r r2
o dr o dr o

du
At the centerline, r = 0, so =0, and hence τ = 0.
dr

du 0.25100
At r = 0.25 cm, ⇒ =32r r2 =32 =3200 s−1, ⇒
1.44 dr o (0.5100)2
τ=3200×μ=3200(1×10−3)=3.2 N/m2

du 0.5100
At the wall, r = 0.5 cm, ⇒ =32r r2 =32 =6400 s−1, ⇒
dr o (0.5100)2
τ=6400×μ=6400(1×10−3)=6.4 N/m2

2πR3ωLμ
Use Eq.1.5.8 to calculate the torque, T =
h
where h=(2.6−2.54) 2=0.03 cm=0.03×10−2 m
2π
The angular velocity ω= ×2000 rpm =209.4 rad/s
60
1.46
The viscosity of SAE-30 oil at 21°C is μ=0.2884 Ns/m2(Figure B.2)
2π×(1.27×10−2 m)3×209.4 rad/s×1.2m×0.2884 Ns/m2
T= =3.10 N⋅m
−2
(0.03×10 )m
power = Tω=3.1×209.4 = 650 W = 0.65 kW
Assume a linear velocity in the fluid between the rotating disk and solid surface.
The velocity of the fluid at the rotating disk is V = rω, and at the solid surface
du rω
V = 0. So, = , where h is the spacing between the disk and solid surface,
dy h
and ω=2π×400 60=41.9 rad/s. The torque needed to rotate the disk is
1.48
T = shear force × moment arm
τ dr
Due to the area element shown, dT = dF × r = τdA × r r
where τ = shear stress in the fluid at the rotating disk and
du rω
dA=2πrdr ⇒ dT =μ ×2πrdr×r =μ ×2πr2dr
dy h
6

© 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.