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
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.
1.3 (A) 2.36
D
1.4 (C) The Mass Is The Same On Earth And The Moon: [4(8r)] 32 R.
u
D
1.5 (C) Fshear F Sin 4200sin 30 2100 N. r
=F 2100 N
shEar 84 Kpa
A
1.6 (B)
1.7 (D) 3
Wate 100
D
1.8 (A)
u
D [10 5000r] 10 3
10 5000 0.02 1 Pa.
r
4 Cos 4 0.0736 N/m 1
1.9 (D) H 3m 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/m2 4 m3
1.11 (C) M 59.95 kg
RT 0.1886 kJ/(kg K) (10 273) K
1
© 2012 CEngagE LEarning. All Rights REsErvEd. May not bE scannEd, copiEd or duplicatEd, or postEd to a publicly accEssiblE wEbsitE, in wholE or
, ChaptEr 1 / Basic
1.12 (b) eic ewater. 320 mwater 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,
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 ml / t 2 l2 m / lt 2
2
b) pressure = force/area = f / l
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 2 4
1.16 a) density = ft / l
l3 l3
b) pressure = f/l2
c) power = f × velocity = f l/t = fl/t
d) energy = f×l = fl
e) mass flux = Ft 2 / L
ft / l
m 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 l1/3t
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) m /s = [c] m m . [c] = m3/s m2 m2/3 = m1/3/s
3 2 2/3
2
© 2012 CEngagE LEarning. All Rights REsErvEd. May not bE scannEd, copiEd or duplicatEd, or postEd to a publicly accEssiblE wE bsitE, in wholE or
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
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.
1.3 (A) 2.36
D
1.4 (C) The Mass Is The Same On Earth And The Moon: [4(8r)] 32 R.
u
D
1.5 (C) Fshear F Sin 4200sin 30 2100 N. r
=F 2100 N
shEar 84 Kpa
A
1.6 (B)
1.7 (D) 3
Wate 100
D
1.8 (A)
u
D [10 5000r] 10 3
10 5000 0.02 1 Pa.
r
4 Cos 4 0.0736 N/m 1
1.9 (D) H 3m 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/m2 4 m3
1.11 (C) M 59.95 kg
RT 0.1886 kJ/(kg K) (10 273) K
1
© 2012 CEngagE LEarning. All Rights REsErvEd. May not bE scannEd, copiEd or duplicatEd, or postEd to a publicly accEssiblE wEbsitE, in wholE or
, ChaptEr 1 / Basic
1.12 (b) eic ewater. 320 mwater 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,
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 ml / t 2 l2 m / lt 2
2
b) pressure = force/area = f / l
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 2 4
1.16 a) density = ft / l
l3 l3
b) pressure = f/l2
c) power = f × velocity = f l/t = fl/t
d) energy = f×l = fl
e) mass flux = Ft 2 / L
ft / l
m 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 l1/3t
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) m /s = [c] m m . [c] = m3/s m2 m2/3 = m1/3/s
3 2 2/3
2
© 2012 CEngagE LEarning. All Rights REsErvEd. May not bE scannEd, copiEd or duplicatEd, or postEd to a publicly accEssiblE wE bsitE, in wholE or
