, 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
, 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) [ [ du/dy)] = (F/L2)/(L/T)/L = F.T/L2
8 9
1.3 (A) 2.36 10 Pa 23.6 10 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:
1.4 (C) m = W/g = 250 N/9.81 m/s2 = 25.484 kg
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) Fshear F sin 4200 sin 30 2100 N
Fshear 2100 N
= 84 103 Pa or 84 kPa
A 4 2
250 10 m
1.6 (B) – 53.6 C
Using Eqn. (1.5.3):
1.7 (D) (T 4)2 (80 4)2 3
water 1000 1000 968 kg/m
180 180
du
The shear stress is given by:
dr
1.8 (A) We determine du/dr from the given expression for u as:
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.
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
, 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) [ [ du/dy)] = (F/L2)/(L/T)/L = F.T/L2
8 9
1.3 (A) 2.36 10 Pa 23.6 10 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:
1.4 (C) m = W/g = 250 N/9.81 m/s2 = 25.484 kg
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) Fshear F sin 4200 sin 30 2100 N
Fshear 2100 N
= 84 103 Pa or 84 kPa
A 4 2
250 10 m
1.6 (B) – 53.6 C
Using Eqn. (1.5.3):
1.7 (D) (T 4)2 (80 4)2 3
water 1000 1000 968 kg/m
180 180
du
The shear stress is given by:
dr
1.8 (A) We determine du/dr from the given expression for u as:
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.
