engineering fluid mechanics, 9th edition

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Engineering
Fluid
Mechanics
Ninth Edition
Clayton T. Crowe
W
ASHINGTON STATE UNIVERSITY, PULLMAN
Donald F. Elger
UNIVERSITY OF IDAHO, MOSCOW
Barbara C. Williams

UNIVERSITY OF IDAHO, MOSCOW
John A. Roberson
WASHINGTON STATE UNIVERSITY, PULLMAN
John Wiley & Sons, Inc.
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Crowe, C. T. (Clayton T.)
Engineering fluid mechanics/Clayton T. Crowe, Donald F. Elger, Barbara C. Williams, John A. Roberson. —9th ed.

ISBN-13: 978-0470-25977-1
Printed in the United States of America
10 9 8 7 6 5 4 3 2 1
∞
To our spouses, Jeannette and Linda
and in memory of Roy
and to our students past, present, and future
and to those who share our love of learning
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Contents
PREFACE vii
CHAPTER 1 Introduction 1
1.1 Liquids and Gases 1
1.2 The Continuum Assumption 2
1.3 Dimensions, Units, and Resources 4
1.4 Topics in Dimensional Analysis 5
1.5 Engineering Analysis 11
1.6 Applications and Connections 12
CHAPTER 2 Fluid Properties 15
2.1 Properties Involving Mass and Weight 15
2.2 Ideal Gas Law 16
2.3 Properties Involving Thermal Energy 17
2.4 Viscosity 18
2.5 Bulk Modulus of Elasticity 24
2.6 Surface Tension 25
2.7 Vapor Pressure 27
2.8 Summary 28
CHAPTER 3 Fluid Statics 33
3.1 Pressure 33
3.2 Pressure Variation with Elevation 37

3.3 Pressure Measurements 43
3.4 Forces on Plane Surfaces (Panels) 48
3.5 Forces on Curved Surfaces 52
3.6 Buoyancy 55
3.7 Stability of Immersed and Floating Bodies 56
3.8 Summary 61
CHAPTER 4 Flowing Fluids
and Pressure Variation 77
4.1 Descriptions of Fluid Motion 78
4.2 Acceleration 82
4.3 Euler’s Equation 86
4.4 Pressure Distribution in Rotating Flows 89
4.5 The Bernoulli Equation Along a Streamline 92
4.6 Rotation and Vorticity 100
4.7 The Bernoulli Equation in Irrotational Flow 105
4.8 Separation 111
4.9 Summary 113
CHAPTER 5 Control Volume
Approach and Continuity Equation 127
5.1 Rate of Flow 129
5.2 Control Volume Approach 133
5.3 Continuity Equation 138
5.4 Cavitation 144
5.5 Differential Form of the Continuity Equation 147
5.6 Summary 149
CHAPTER 6 Momentum Equation 163
6.1 Momentum Equation: Derivation 163
6.2 Momentum Equation: Interpretation 165
6.3 Common Applications 168
6.4 Additional Applications 179

6.5 Moment-of-Momentum Equation 192
6.6 Navier-Stokes Equation 196
6.7 Summary 201
CHAPTER 7 The Energy Equation 217
7.1 Energy, Work, and Power 217
7.2 Energy Equation: General Form 219
7.3 Energy Equation: Pipe Flow 222
7.4 Power Equation 227
7.5 Contrasting the Bernoulli Equation and
the Energy Equation 229
7.6 Transitions 230
7.7 Hydraulic and Energy Grade Lines 233
7.8 Summary 236
CHAPTER 8 Dimensional Analysis
and Similitude 249
8.1 Need for Dimensional Analysis 249
8.2 Buckingham
⌸
Theorem 251
8.3 Dimensional Analysis 252
8.4 Common
␲-Groups 256
8.5 Similitude 259
8.6 Model Studies for Flows Without
Free-Surface Effects 263
8.7 Model-Prototype Performance 265
8.8 Approximate Similitude at High
Reynolds Numbers 267
8.9 Free-Surface Model Studies 270
8.10 Summary 273

vi
CONTENTS
CHAPTER 9 Surface Resistance 281
9.1 Surface Resistance with Uniform
Laminar Flow 281
9.2 Qualitative Description of the
Boundary Layer 286
9.3 Laminar Boundary Layer 288
9.4 Boundary Layer Transition 292
9.5 Turbulent Boundary Layer 292
9.6 Pressure Gradient Effects on
Boundary Layers 304
9.7 Summary 306
CHAPTER 10 Flow in Conduits 315
10.1 Classifying Flow 316
10.2 Specifying Pipe Sizes 319
10.3 Pipe Head Loss 320
10.4 Stress Distributions in Pipe Flow 322
10.5 Laminar Flow in a Round Tube 324
10.6 Turbulent Flow and the Moody Diagram 327
10.7 Solving Turbulent Flow Problems 332
10.8 Combined Head Loss 336
10.9 Nonround Conduits 341
10.10 Pumps and Systems of Pipes 342
10.11 Summary 349
CHAPTER 11 Drag and Lift 363
11.1 Relating Lift and Drag to
Stress Distributions 364
11.2 Calculating Drag Force 365
11.3 Drag of Axisymmetric and 3D Bodies 370

11.4 Terminal Velocity 374
11.5 Vortex Shedding 376
11.6 Reducing Drag by Streamlining 377
11.7 Drag in Compressible Flow 378
11.8 Theory of Lift 379
11.9 Lift and Drag on Airfoils 383
11.10 Lift and Drag on Road Vehicles 389
11.11 Summary 392
CHAPTER 12 Compressible Flow 401
12.1 Wave Propagation in Compressible Fluids 401
12.2 Mach Number Relationships 407
12.3 Normal Shock Waves 412
12.4 Isentropic Compressible Flow
Through a Duct with Varying Area 416
12.5 Summary 428
CHAPTER 13 Flow Measurements 435
13.1 Measuring Velocity and Pressure 435
13.2 Measuring Flow Rate (Discharge) 443
13.3 Measurement in Compressible Flow 458
13.4 Accuracy of Measurements 463
13.5 Summary 464
CHAPTER 14 Turbomachinery 475
14.1 Propellers 476
14.2 Axial-Flow Pumps 481
14.3 Radial-Flow Machines 485
14.4 Specific Speed 488
14.5 Suction Limitations of Pumps 490
14.6 Viscous Effects 492
14.7 Centrifugal Compressors 493
14.8 Turbines 496

14.9 Summary 505
CHAPTER 15 Flow in Open
Channels 511
15.1 Description of Open-Channel Flow 511
15.2 Energy Equation for Steady
Open-Channel Flow 514
15.3 Steady Uniform Flow 514
15.4 Steady Nonuniform Flow 522
15.5 Rapidly Varied Flow 523
15.6 Hydraulic Jump 533
15.7 Gradually Varied Flow 538
15.8 Summary 545
Appendix A-1
Answers A-11
Index I-1
Preface
Audience
This book is written for engineering students of all majors who are taking a first or second
course in fluid mechanics. Students should have background knowledge in statics and calculus.
This text is designed to help students develop meaningful and connected knowledge of
main concepts and equations as well as develop the skills and approaches that work effectively
in professional practice.
Approach
Through innovative ideas and professional skills, engineers can make the world a better place.
In particular, fluid mechanics plays a very important role in the design, development, and anal-
ysis of systems from microscale applications to giant hydroelectric power generation. For this
reason, the study of fluid mechanics is essential to the background of an engineer. The approach
in this text is to emphasize both professional practice and technical knowledge.
This text is organized to support the acquisition of deep and connnected knowledge. Each
chapter begins by informing students what they should be learning (i.e., learning outcomes) and

why this learning is relevant. Topics are linked to previous topics so that students can see how
knowledge is building and connecting. Seminal equations, defined as those that are commonly
used, are carefully derived and explained. In addition, Table F.2 in the front of the book orga-
nizes the main equations.
This text is organized to support the development of skills for problem solving. Example
problems are presented with a structured approach that was developed by studying the research
literature that describes how experts solve technical problems. This structured approach, la-
beled as “Engineering Analysis,” is presented in Chapter 1. Homework problems are organized
by topic, and a variety of types of problems are included.
Organization of Knowledge
Chapters 1 to 11 and 13 are devoted to foundational concepts of fluid mechanics. Relevant con-
tent includes fluid properties; forces and pressure variations in static fluids; qualitative descrip-
tions of flow and pressure variations; the Bernoulli equation; the control volume concept;
control volume equations for mass, momentum, and energy; dimensional analysis; head loss
viii
PREFACE
in conduits; measurements; drag force; and lift force. Nearly all professors cover the material
in Chapters 1 to 8 and 10. Chapters 9, 11, and 13 are covered based on instructor preference.
Chapters 12, 14, and 15 are devoted to special topics that are optional for a first course in fluid
mechanics.
In this 9th edition, there is some reorganization of sections and some additions of new
technical material. Chapter 1 provides new material on the nature of fluids, unit practices, and
problem solving. Sections in Chapter 4 were reordered to provide a more logical development
of the Bernoulli equation. Also, the material on the Eulerian and Lagrangian approaches was
moved from Chapter 4 to Chapter 5. Chapter 7 provides new material on energy and power and
a new section to describe calculation of power. In Chapter 10, new material on standard pipe
sizes was added and new sections were added to describe flow classification and to describe
how to solve turbulent flow problems. The open channels flow topics that were in Chapter 10
were moved to Chapter 15. Chapters 11 and 15 were modified by introducing new sections to
better organize the material. Also, a list of main equations and a detailed list of unit conversions

were added to the front of the book.
Features of this Book*
*Learning Outcomes. Each chapter begins with learn-
ing outcomes so students can identify what knowledge they
should be gaining by studying the chapter.
*Rationale. Each section describes what content is pre-
sented and why this content is relevant so students can con-
nect their learning to what is important to them.
*Visual Approach. The text uses sketches and photo-
graphs to help students learn more effectively by connect-
ing images to words and equations.
*Foundational Concepts. This text presents major con-
cepts in a clear and concise format. These concepts form
building blocks for higher levels of learning. Concepts are
identified by a blue tint.
*Seminal Equations. This text emphasizes technical
derivations so that students can learn to do the derivations
on their own, increasing their levels of knowledge. Features
include
• Derivations of each main equation are presented in a
step-by-step fashion.
• The assumptions associated with each main equation are
stated during the derivation and after the equation is
developed.
• The holistic meaning of main equations is explained
using words.
• Main equations are named and listed in Table F.2.
Chapter Summaries. Each chapter concludes with a
summary so that students can review the key knowledge in
the chapter.

*Engineering Analysis. Example problems and solu-
tions to homework problems are structured with a step-by-
step approach. As shown in Fig. 1 (next page), the solution
process begins with problem formulation, which involves
interpreting the problem before attempting to solve the
problem. The plan step involves creating a step-by-step
plan prior to jumping into a detailed solution. This struc-
tured approach provides students with an approach that can
generalize to many types of engineering problems.
*Grid Method. This text presents a systematic process,
called the grid method, for carrying and canceling units.
Unit practice is emphasized because it helps students spot
and fix mistakes and because it helps student put meaning
on concepts and equations.
Traditonal and SI Units. Examples and homework
problems are presented using both SI and traditional unit
systems. This presentation helps students develop unit
practices and gain familiarity with units that are used on
professional practice.
Example Problems. Each chapter has approximately
10 example problems, each worked out in detail. The pur-
pose of these examples is to show how the knowledge is
used in context and to present essential details needed for
application.
* Asterisk denotes a new or major modification to this edition.
PREFACE
ix
Homework Problems. The text includes several types
of end-of-chapter problems, including
• Preview (or prepare) questions

ސޑ᭣
. A preview (or pre-
pare) question is designed to be assigned prior to in-class
coverage of the topic. The purpose is so that students
come to class with some familiarity with the topics.
• Analysis problems. These problems are traditional prob-
lems that require a systematic, or step-by-step, approach.
They may involve multiple concepts and generally can-
not be solved by using a memorization approach. These
problems help students learn engineering analysis.
• Computer problems. These problems involve use of a
computer program. Regarding the choice of software, we
have left this open so that instructors may select a pro-
gram or may allow their students to select a program.
• Design problems. These problems have multiple possi-
ble solutions and require assumptions and decision
making. These problems help students learn to manage
the messy, ill-structured problems that typify profes-
sional practice.
Solutions Manual. The formatting of the instructor’s so-
lutions manual parallels the engineering-analysis approach
presented in the text. Each solution includes a description of
the situation, statement of the problem goals, statements of
main assumptions, summary of the solution approach, a de-
tailed solution, and review comments. In addition, each
problem analysis is organized using text labels, such as
“momentum equation (x-direction),” so that the labels
themselves provide a summary of the solution approach.
Image Gallery. The figures from the text are available
in PowerPoint format, for easy inclusion in lecture pre-

sentations. This resource is available only to instructors.
To request access to this password-protected resource,
visit the Instructor Companion Site portion of the web site
located at
www.wiley.com/college/crowe
, and register
for a password.
Interdisciplinary Approach. Historically, this text
was written for the civil engineer. We are retaining this
approach while adding material so that the text is also ap-
propriate for other engineering disciplines. For example,
the text presents the Bernoulli equation using both head
terms (civil engineering approach) and terms with units of
EXAMPLE 3.1 LOAD LIFTED BY A HYDRAULIC
JACK
A hydraulic jack has the dimensions shown. If one exerts a
force F of 100 N on the handle of the jack, what load, F
2
, can
the jack support? Neglect lifter weight.
Problem Definition
Situation: A force of is applied to the handle of a
jack.
Find: Load F
2
in kN that the jack can lift.
Assumptions: Weight of the lifter component (see sketch) is
negligible.
Sketch:
Plan

1.
Calculate force F
1
acting on the small piston by applying
moment equilibrium.
2.
Calculate pressure p
1
in the hydraulic fluid by applying
force equilibrium.
3.
Calculate the load F
2
by applying force equilibrium.
Solution
1.
Moment equilibrium
2.
Force equilibrium (small piston)
Thus
3.
Force equilibrium (lifter)
• Note that because they are at the same eleva-
tion (this fact will be established in the next section).
• Apply force equilibrium:
Review
The jack in this example, which combines a lever and a
hydraulic machine, provides an output force of 12,200 N
from an input force of 100 N. Thus, this jack provides a
mechanical advantage of 122 to 1!

F 100 Nϭ
30 cm
F
BC
3.0 cm
1.5 cm diameter
A
1
A
2
Check valve
5 cm diameter
Lifter
F
2
M
C
Α
0ϭ
0.33 m() ϫ 100 N()0.03 m()F
1
– 0ϭ
F
1
0.33 m ϫ 100 N()
0.03 m

1100 Nϭϭ
F
small piston

Α
p
1
A
1
F
1
– 0ϭϭ
p
1
A
1
F
1
1100 Nϭϭ
p
1
F
1
A
1

1100 N
␲d
2
4⁄
6.22 ϫ10
6
() Nm⁄
2

ϭϭ ϭ
p
1
p
2
ϭ
F
lifter
Α
F
2
p
1
A
2
– 0ϭϭ
F
2
p
1
A
2
6.22 ϫ10
6
N
m
2

©¹
¨¸

§·
␲
4

ϫ 0.05 m()
2
©¹
§·
12.2 kNϭϭ ϭ
The first steps model
how engineers
change a problem
statement into a
meaningful and clear
problem definition.
The next steps model
how engineers find a
solution path by
using main concepts.
Figure 1
Structured problem solving is used throughout the text.
x
PREFACE
pressure (the approach used by chemical and mechanical
engineers). We include problems that are relevant to prod-
uct development as practiced by mechanical and electrical
engineers. Some problems feature other disciplines, such
as exercise physiology. The reason for this interdiscipli-
nary approach is that the world of today’s engineer is be-
coming more and more interdisciplinary.

Also Available from the Publisher. Practice Prob-
lems with Solutions: A Guide for Learning Engineering
Fluid Mechanics, 9th Edition, by Crow, Elger, and Rober-
son. ISBN 978 0470 420867. This is a companion manual
to the textbook, and presents additional example prob-
lems with complete solutions for students seeking addi-
tional practice problems and solution guidance.
Visit
www.wiley.com/college/crowe
for ordering in-
formation. It is also available in a set with the textbook at
a discounted price.
Author Team
Most of the book was originally written by Profes-
sor John Roberson, with Clayton Crowe adding the
material on compressible flow. Professor Roberson
retired from active authorship after the 6th edition,
Professor Donald Elger joined on the 7th edition,
and Professor Barbara Williams joined on the 9th
Edition.
Acknowledgments
We wish to thank the following instructors for their help in reviewing and offering feedback
on the manuscript for the 9th edition: David Benson, Kettering University; John D. Dietz,
University of Central Florida; Howard M. Liljestrand, University of Texas; Nancy Ma, North
Carolina State University; Saad Merayyan, California State University, Sacramento; Sergey
A. Smirnov, Texas Tech University; and Yaron Sternberg, University of Maryland.
Special recognition is given to our colleagues and mentors. Clayton Crowe acknowledges
the many years of professional interaction with his colleagues at Washington State University,
and the loving support of his family. Ronald Adams, Engineering Dean at Oregon State Uni-
versity, mentored Donald Elger during his Ph.D. research and introduced him to a new way of

thinking about fluid mechanics. Ralph Budwig, a fluid mechanics researcher and educator at
The University of Idaho, has provided many hours of delightful discussion. Wilfried Brutsuert
at Cornell University and George Bloomsburg at the University of Idaho inspired Barbara Wil-
liams’s passion for fluid mechanics. Roy Williams, husband and colleague, always pushed for
excellence. Charles L. Barker introduced John Roberson to the field of fluid mechanics and mo-
tivated him to write a textbook.
Finally, we owe a debt of gratitude to Cody Newbill and John Crepeau, who helped
with both the text and the solutions manual. We are also grateful to our families for their en-
couragement and understanding during the writing and editing of the text.
We welcome feedback and ideas for interesting end-of-chapter problems. Please con-
tact us at the e-mail addresses given below.
Clayton T. Crowe ()
Donald F. Elger ()
Barbara C. Williams ()
John A. Roberson (emeritus)
The author team. Left to right:
Donald Elger, Barbara Williams,
and Clayton Crowe.
CHAPTER
Introduction
Prior to fluid mechanics, students take courses such as physics, statics, and dynamics, which
involve solid mechanics. Mechanics is the field of science focused on the motion of material
bodies. Mechanics involves force, energy, motion, deformation, and material properties.
When mechanics applies to material bodies in the solid phase, the discipline is called solid
mechanics. When the material body is in the gas or liquid phase, the discipline is called fluid
mechanics. In contrast to a solid, a fluid is a substance whose molecules move freely past
each other. More specifically, a fluid is a substance that will continuously deform—that is,
flow under the action of a shear stress. Alternatively, a solid will deform under the action of a
shear stress but will not flow like a fluid. Both liquids and gases are classified as fluids.
This chapter introduces fluid mechanics by describing gases, liquids, and the contin-

uum assumption. This chapter also presents (a) a description of resources available in the ap-
pendices of this text, (b) an approach for using units and primary dimensions in fluid
mechanics calculations, and (c) a systematic approach for problem solving.
Liquids and Gases
This section describes liquids and gases, emphasizing behavior of the molecules. This
knowledge is useful for understanding the observable characteristics of fluids.
Liquids and gases differ because of forces between the molecules. As shown in the first
row of Table 1.1, a liquid will take the shape of a container whereas a gas will expand to fill
a closed container. The behavior of the liquid is produced by strong attractive force between
the molecules. This strong attractive force also explains why the density of a liquid is much
Fluid mechanics applies concepts
related to force and energy to
practical problems such as the
design of gliders. (Photo courtesy
of DG Flugzeugbau GmbH.)
SIGNIFICANT LEARNING OUTCOMES
Conceptual Knowledge
• Describe fluid mechanics.
• Contrast gases and liquids by describing similarities and differences.
• Explain the continuum assumption.
Procedural Knowledge
• Use primary dimensions to check equations for dimensional
homogeneity.
• Apply the grid method to carry and cancel units in calculations.
• Explain the steps in the “Structured Approach for Engineering
Analysis” (see Table 1.4).
1.1
2
INTRODUCTION
higher than the density of gas (see the fourth row). The attributes in Table 1.1 can be gen-

eralized by defining a gas and liquid based on the differences in the attractive forces between
molecules. A gas is a phase of material in which molecules are widely spaced, molecules
move about freely, and forces between molecules are minuscule, except during collisions. Al-
ternatively, a liquid is a phase of material in which molecules are closely spaced, molecules
move about freely, and there are strong attractive forces between molecules.
The Continuum Assumption
This section describes how fluids are conceptualized as a continuous medium. This topic is
important for applying the derivative concept to characterize properties of fluids.
Table 1.1 COMPARISON OF SOLIDS, LIQUIDS, AND GASES
Attribute
Solid Liquid Gas
Typical Visualization
Macroscopic
Description
Solids hold their shape; no need
for a container
Liquids take the shape of the
container and will stay in open
container
Gases expand to fill a closed
container
Mobility of Molecules Molecules have low mobility
because they are bound in a
structure by strong
intermolecular forces
Liquids typically flow easily even
though there are strong intermolecular
forces between molecules
Molecules move around freely
with little interaction except

during collisions; this is why
gases expand to fill their container
Typical Density Often high; e.g., density of steel
is 7700 kg m
3
Medium; e.g., density of water is
1000 kg m
3
Small; e.g., density of air at sea
level is 1.2 kg m
3
Molecular Spacing Small—molecules are close
together
Small—molecules are held close
together by intermolecular forces
Large—on average, molecules are
far apart
Effect of Shear Stress Produces deformation Produces flow Produces flow
Effect of Normal Stress Produces deformation that may
associate with volume change;
can cause failure
Produces deformation associated with
volume change
Produces deformation associated
with volume change
Viscosity NA High; decreases as temperature
increases
Low; increases as temperature
increases
Compressibility Difficult to compress; bulk

modulus of steel is 160 10
9
Pa
Difficult to compress; bulk modulus
of liquid water is 2.2 10
9
Pa
Easy to compress; bulk modulus
of a gas at room conditions is
about 1.0 10
5
Pa
⁄⁄ ⁄
××
×
1.2
1.2 THE CONTINUUM ASSUMPTION
3
While a body of fluid is comprised of molecules, most characteristics of fluids are due
to average molecular behavior. That is, a fluid often behaves as if it were comprised of con-
tinuous matter that is infinitely divisible into smaller and smaller parts. This idea is called the
continuum assumption. When the continuum assumption is valid, engineers can apply limit
concepts from differential calculus. Recall that a limit concept, for example, involves letting
a length, an area, or a volume approach zero. Because of the continuum assumption, fluid pa-
rameters such as density and velocity can be considered continuous functions of position
with a value at each point in space.
To gain insight into the validity of the continuum assumption, consider a hypothetical
experiment to find density. Fig. 1.1a shows a container of gas in which a volume has
been identified. The idea is to find the mass of the molecules inside the volume and then
to calculate density by

The calculated density is plotted in Fig. 1.1b. When the measuring volume is very small
(approaching zero), the number of molecules in the volume will vary with time because of
the random nature of molecular motion. Thus, the density will vary as shown by the wiggles
in the blue line. As volume increases, the variations in calculated density will decrease until
the calculated density is independent of the measuring volume. This condition corresponds to
the vertical line at If the volume is too large, as shown by then the value of density
may change due to spatial variations.
In most applications, the continuum assumption is valid. For example, consider the
volume needed to contain at least a million molecules. Using Avogadro’s number of
molecules mole, the limiting volume for water is , which corresponds
to a cube less than mm on a side. Since this dimension is much smaller than the flow di-
mensions of a typical problem, the continuum assumption is justified. For an ideal gas (1 atm
and 20
o
C) one mole occupies 24.7 liters. The size of a volume with more than molecules
would be , which corresponds to a cube with sides less than mm (or one
micrometer). Once again this size is much smaller than typical flow dimensions. Thus, the
continuum assumption is usually valid in gas flows.
The continuum assumption is invalid for some specific applications. When air is in
motion at a very low density, such as when a spacecraft enters the earth’s atmosphere,
then the spacing between molecules is significant in comparison to the size of the
spacecraft. Similarly, when a fluid flows through the tiny passages in nanotechnology
devices, then the spacing between molecules is significant compared to the size of these
passageways.
Figure 1.1
When a measuring
volume is large
enough for random
molecular effects to
average out, the

continuum assumption is
valid
ΔV

Gas
(a) (b)
Selected
volume = ΔV
Continuum assumption
is valid.
ΔV
2
Δm
ΔV
Volume ΔV
ΔV
1
Gas molecules
ΔV

ΔM
␳
ΔM
ΔV

ϭ
ΔV

ΔV

1
. ΔV

2
,
10
6
()
610
23
×
⁄
10
13–
mm
3
10
4–
10
6
10
10–
mm
3
10
3–
4
INTRODUCTION
Dimensions, Units, and Resources
This section describes the dimensions and units that are used in fluid mechanics. This

information is essential for understanding most aspects of fluid mechanics. In addition, this
section describes useful resources that are presented in the front and back of this text.
Dimensions
A dimension is a category that represents a physical quantity such as mass, length, time, mo-
mentum, force, acceleration, and energy. To simplify matters, engineers express dimensions
using a limited set that are called primary dimensions. Table 1.2 lists one common set of pri-
mary dimensions.
Secondary dimensions such as momentum and energy can be related to primary dimen-
sions by using equations. For example, the secondary dimension “force” is expressed in pri-
mary dimensions by using Newton’s second law of motion, F ma. The primary
dimensions of acceleration are so
(1.1)
In Eq. (1.1), the square brackets mean “dimensions of.” This equation reads “the primary di-
mensions of force are mass times length divided by time squared.” Note that primary dimen-
sions are not enclosed in brackets.
Units
While a dimension expresses a specific type of physical quantity, a unit assigns a number so
that the dimension can be measured. For example, measurement of volume (a dimension) can
be expressed using units of liters. Similarly, measurement of energy (a dimension) can be ex-
pressed using units of joules. Most dimensions have multiple units that are used for measure-
ment. For example, the dimension of “force” can be expressed using units of newtons,
pounds-force, or dynes.
Unit Systems
In practice, there are several unit systems in use. The International System of Units (abbrevi-
ated SI from the French “Le Système International d'Unités”) is based on the meter,
Table 1.2 PRIMARY DIMENSIONS
Dimension Symbol Unit (SI)
Length L meter (m)
Mass M kilogram (kg)
Time T second (s)

Temperature
␪
kelvin (K)
Electric current i ampere (A)
Amount of light C candela (cd)
Amount of matter N mole (mol)
1.3
ϭ
LT
2
⁄ ,
F[] ma[]M
L
T
2

ML
T
2

ϭϭϭ
1.4 TOPICS IN DIMENSIONAL ANALYSIS
5
kilogram, and second. Although the SI system is intended to serve as an international stan-
dard, there are other systems in common use in the United States. The U.S. Customary Sys-
tem (USCS), sometimes called English Engineering System, uses the pound-mass (lbm) for
mass, the foot (ft) for length, the pound-force (lbf) for force, and the second (s) for time. The
British Gravitational (BG) System is similar to the USCS system that the unit of mass is the
slug. To convert between pounds-mass and kg or slugs, the relationships are
Thus, a gallon of milk, which has mass of approximately 8 lbm, will have a mass of about

0.25 slugs, which is about 3.6 kg.
For simplicity, this text uses two categories for units. The first category is the familiar
SI unit system. The second category contains units from both the USCS and the BG systems
of units and is called the “Traditional Unit System.”
Resources Available in This Text
To support calculations and design tasks, formulas and data are presented in the front and
back of this text.
Table F.1 (the notation “F.x” means a table in the front of the text) presents data for con-
verting units. For example, this table presents the factor for converting meters to feet (1 m
3.281 ft) and the factor for converting horsepower to kilowatts (1 hp 745.7 W). Notice that
a given parameter such as viscosity will have one set of primary dimensions (M LT) and
several possible units, including pascal-second ( ), poise, and Table F.1 lists
unit conversion formulas, where each formula is a relationship between units expressed using
the equal sign. Examples of unit conversion formulas are 1.0 m 3.281 ft and 3.281
ft km 1000. Notice that each row of Table F.1 provides multiple conversion formulas. For
example, the row for length conversions,
(1.2)
has the usual conversion formulas such as 1 m 39.37 in, and the less common formulas
such as 1.094 yd
Table F.2 presents equations that are commonly used in fluid mechanics. To make them
easier to remember, equations are given descriptive names such as the “hydrostatic equa-
tion.” Also, notice that each equation is given an equation number and page number corre-
sponding to where it is introduced in this text.
Tables F.3, F.4, and F.5 present commonly used constants and fluid properties. Other
fluid properties are presented in the appendix. For example, Table A.3 (the notation “A.x”
means a table in the appendix) gives properties of air.
Table A.6 lists the variables that are used in this text. Notice that this table gives the
symbol, the primary dimensions, and the name of the variable.
Topics in Dimensional Analysis
This section introduces dimensionless groups, the concept of dimensional homogeneity of an

equation, and a process for carrying and canceling units in a calculation. This knowledge is
1.0 lbm
1
2.2

kg
1
32.2

slugϭϭ
ϭ
ϭ
⁄
Pa sؒ lbf sؒ ft⁄
2
.
ϭ
ϭ ⁄
1 m 3.281 ft 1.094 yd 39.37 in
km
1000
10
6
␮mϭϭ ϭ ϭϭ
ϭ
ϭ 10
6
␮m.
1.4
6

INTRODUCTION
useful in all aspects of engineering, especially for finding and correcting mistakes during cal-
culations and during derivations of equations.
All of topics in this section are part of dimensional analysis, which is the process for
applying knowledge and rules involving dimensions and units. Other aspects of dimensional
analysis are presented in Chapter 8 of this text.
Dimensionless Groups
Engineers often arrange variables so that primary dimensions cancel out. For example,
consider a pipe with an inside diameter D and length L. These variables can be grouped to
form a new variable which is an example of a dimensionless group. A dimensionless
group is any arrangement of variables in which the primary dimensions cancel. Another
example of a dimensionless group is the Mach number M, which relates fluid speed V to
the speed of sound c:
Another common dimensionless group is named the Reynolds number and given the symbol
Re. The Reynolds number involves density, velocity, length, and viscosity :
(1.3)
The convention in this text is to use the symbol [-] to indicate that the primary dimensions of
a dimensionless group cancel out. For example,
(1.4)
Dimensional Homogeneity
When the primary dimensions on each term of an equation are the same, the equation is
dimensionally homogeneous. For example, consider the equation for vertical position s of an
object moving in a gravitational field:
In the equation, g is the gravitational constant, t is time, v
o
is the magnitude of the vertical
component of the initial velocity, and s
o
is the vertical component of the initial position. This
equation is dimensionally homogeneous because the primary dimension of each term is

length L. Example 1.1 shows how to find the primary dimension for a group of variables us-
ing a step-by-step approach. Example 1.2 shows how to check an equation for dimensional
homogeneity by comparing the dimensions on each term.
Since fluid mechanics involves many differential and integral equations, it is useful to
know how to find primary dimensions on integral and derivative terms.
To find primary dimensions on a derivative, recall from calculus that a derivative is de-
fined as a ratio:
LD,⁄
M
V
c

ϭ
␮
Re
␳VL
␮
ϭ
Re[]
␳VL
␮

-[]ϭϭ
s
gt
2
2

v
o

ts
o
++ϭ
yd
df
Δf
Δy

Δy 0
→
limϭ
1.4 TOPICS IN DIMENSIONAL ANALYSIS
7
Thus, the primary dimensions of a derivative can be found by using a ratio:
EXAMPLE 1.1 PRIMARY DIMENSIONS OF THE
REYNOLDS NUMBER
Show that the Reynolds number, given in Eq. 1.4, is a
dimensionless group.
Problem Definition
Situation: The Reynolds number is given by
Find: Show that Re is a dimensionless group.
Plan
1.
Identify the variables by using Table A.6.
2.
Find the primary dimensions by using Table A.6.
3.
Show that Re is dimensionless by canceling primary
dimensions.
Solution

1.
Var iab les
• mass density,
• velocity, V
• Length, L
• viscosity,
2.
Primary dimensions
3.
Cancel primary dimensions:
Since the primary dimensions cancel, the Reynolds number
is a dimensionless group.
EXAMPLE 1.2 DIMENSIONAL HOMOGENEITY
OF THE IDEAL GAS LAW
Show that the ideal gas law is dimensionally homogeneous.
Problem Definition
Situation: The ideal gas law is given by
Find: Show that the ideal gas law is dimensionally
homogeneous.
Plan
1.
Find the primary dimensions of the first term.
2.
Find the primary dimensions of the second term.
3.
Show dimensional homogeneity by comparing the terms.
Solution
1.
Primary dimensions (first term)
• From Table A.6, the primary dimensions are:

2.
Primary dimensions (second term).
• From Table A.6, the primary dimensions are
• Thus
3.
Conclusion: The ideal gas law is dimensionally
homogeneous because the primary dimensions of each
term are the same.
yd
df
f
y

f
[]
y[]
ϭϭ
Re ␳VL()␮.⁄ϭ
␳
␮
␳[] M L
3
⁄ϭ
V[] LT⁄ϭ
L[] Lϭ
␮[] M LT⁄ϭ
␳VL
␮

M

L
3

L
T

L[]
LT
M

[-]ϭϭ
␳VL()␮⁄
p
␳RT.ϭ
p[]
M
LT
2

ϭ
␳[] M L
3
⁄ϭ
R[] L
2
␪T
2
⁄ϭ
T[] ␪ϭ
␳RT[]

M
L
3

©¹
¨¸
§·
L
2
␪T
2

©¹
¨¸
§·
␪()
M
LT
2
ϭϭ
8
INTRODUCTION
The primary dimensions for a higher-order derivative can also be found by using the basic
definition of the derivative. The resulting formula for a second-order derivative is
(1.5)
Applying Eq. (1.5) to acceleration shows that
To find primary dimensions of an integral, recall from calculus that an integral is defined as a
sum:
Thus
(1.6)

For example, position is given by the integral of velocity with respect to time. Checking pri-
mary dimensions for this integral gives
In summary, one can easily find primary dimensions on derivative and integral terms by
applying fundamental definitions from calculus. This process is illustrated by Example 1.3
EXAMPLE 1.3 PRIMARY DIMENSIONS OF
A DERIVATIVE AND INTEGRAL
Find the primary dimensions of where is viscosity,
u is fluid velocity, and y is distance. Repeat for
where t is time, is volume, and
␳ is density.
Problem Definition
Situation: A derivative and integral term are specified above.
Find: Primary dimensions on the derivative and the integral.
Plan
1.
Find the primary dimensions of the first term by applying
Eq. (1.5).
2.
Find the primary dimensions of the second term by
applying Eqs. (1.5) and (1.6).
Solution
1.
Primary dimensions of
• From Table A.6:
• Apply Eq. (1.5):
• Combine the previous two steps:
d
2
f
dy

2

Δ df dy⁄()
Δy

Δy 0
→
lim
f
y
2

f
[]
y
2
[]

ϭϭϭ
d
2
y
dt
2

y
t
2

L

T
2

ϭϭ
fyd
Ύ
f Δy
i
i 1ϭ
N
Α
N ∞
→
limϭ
fyd
Ύ
f[]y[]ϭ
Vtd
Ύ
V[]t[]
L
T

Tؒ Lϭϭϭ
␮
d
2
u
dy
2

, ␮
td
d
␳ V

d
V

Ύ
V

␮
d
2
u
dy
2

␮[] M LT⁄ϭ
u[] LT⁄ϭ
x[] Lϭ
d
2
u
dy
2

u
y

2

LT⁄
L
2

ϭϭ
␮
d
2
u
dy
2

␮[]
d
2
u
dy
2

M
LT

©¹
§·
LT⁄
L
2

©¹
§·
M
L
2
T
2

ϭϭ ϭ
1.4 TOPICS IN DIMENSIONAL ANALYSIS
9
Sometimes constants have primary dimensions. For example, the hydrostatic equation
relates pressure p, density , the gravitational constant g, and elevation z:
For dimensional homogeneity, the constant C needs to have the same primary dimensions as
either or Thus the dimensions of C are Another example involves ex-
pressing fluid velocity V as a function of distance y using two constants a and b:
For dimensional homogeneity both sides of this equation need to have primary dimensions of
Thus, [b] L and .
The Grid Method
Because fluid mechanics involves complex equations and traditional units, this section pre-
sents the grid method, which is a systematic way to carry and cancel units. For example, Fig.
1.2 shows an estimate of the power P required to ride a bicycle at a speed of V 20 mph.
The engineer estimated that the required force to move against wind drag is F 4.0 lbf and
applied the equation As shown, the calculation reveals that the power is 159 watts.
As shown in Fig. 1.2, the grid method involves writing an equation, drawing a grid, and
carrying and canceling units. Regarding unit cancellations, the key idea is the use of unity
conversion ratios, in which unity (1.0) appears on one side of the equation. Examples of
unity conversion ratios are
Figure 1.2 shows three conversion ratios. Each of these ratios is obtained by using informa-
tion given in Table F.1. For example, the row in Table F.1 for power shows that

Dividing both sides of this equation by gives
Table 1.3 shows how to apply the grid method. Notice how the same process steps can apply
to different situations.
2.
Primary dimensions of
• Find primary dimensions from Table A.6:
• Apply Eqs. (1.5) and (1.6) together:
Figure 1.2
Grid method
td
d
␳ V

d
V

Ύ
t[] Tϭ
␳[] M L
3
⁄ϭ
V

[] L
3
ϭ
td
d
␳ V

d
V

Ύ
␳V

t

M
L
3

©¹
¨¸
§·
L
3
T

©¹
¨¸
§·
M
T

ϭϭ ϭ
␳
p
␳gz+ constant Cϭϭ
p

␳gz. C[] M LT
2
⁄ .ϭ
Vy() ay b y–()ϭ
LT⁄[]. ϭ a[] L
1–
T
1–
ϭ
ϭ
ϭ
P FV.ϭ
1.0
1 m/s
2.237 mph

ϭ 1.0
1.0 N
0.2249 lbf

ϭ
1 W N mؒ s⁄().ϭ Nmؒ s⁄
1.0
Wsؒ
Nmؒ

ϭ
4 lbf 20 mph 1.0 m/s
2.237 mph
1.0 N

0.2248 lbf
W*s
N*m
P = FV =
P = 159 W
10
INTRODUCTION
Since fluid mechanics problems often involve mass units of slugs or pounds-mass
(lbm), it is easy to make a mistake when converting units. Thus, it is useful to have a system-
atic approach. The idea is to relate mass units to force units through F ma. For example, a
force of 1.0 N is the magnitude of force that will accelerate a mass of 1.0 kg at a rate of 1.0 m s
2
.
Thus,
Rewriting this expression gives a conversion ratio
(1.7)
When mass is given using slugs, the corresponding conversion ratio is
(1.8)
A force of 1.0 lbf is defined as the magnitude of force that will accelerate a mass of 1.0 lbm
at a rate of 32.2 ft s
2
. Thus,
Thus, the conversion ratio relating force and mass units becomes
(1.9)
Example 1.4 shows how to apply the grid method. Notice that calculations involving
traditional units follow the same process as calculations involving SI units.
Table 1.3 APPLYING THE GRID METHOD (TWO EXAMPLES)
Process Step Example 1 Example 2
Situation: Convert a pressure of 2.00
psi to pascals.

Situation: Find the force in newtons that is needed to
accelerate a mass of 10 g at a rate of 15 ft s
2
.
Step 1. Problem: Write the term or
equation. Include numbers and units.
Step 2. Conversion Ratios: Look up
unit conversion formula(s) in Table F.1
and represent these as unity conversion
ratios.
Step 3. Algebra: Multiply variables and
cancel units. Fix any errors.
Step 4. Calculations: Perform the
indicated calculations. Round the
answer to the correct number of
significant figures.
⁄
p
2.00 psiϭ
F maϭ
F (N) 0.01 kg()15 ft/s
2
()ϭ
1.0
1 Pa
1.45 10
4–
psi×
ϭ
1.0

1.0 m
3.281 ft

ϭ 1.0
Ns
2
ؒ
kg mؒ

ϭ
p
2.00 psi[]
1 Pa
1.45 10
4–
psi×

ϭ
F 0.01 kg[]
15 ft
s
2

1.0 m
3.281 ft

Ns
2
ؒ
kg mؒ

ϭ
p
13.8 kPaϭ F 0.0457 Nϭ
ϭ
⁄
1.0 N()1.0 kg()1.0 m/s
2
()ϵ
1.0
kg mؒ
Ns
2
ؒ
ϭ
1.0
slug ftؒ
lbf s
2
ؒ

ϭ
⁄
1.0 lbf()1.0 lbm()32.2 ft/s
2
()ϵ
1.0
32.2 lbm ftؒ
lbf s
2

ؒ

ϭ
1.5 ENGINEERING ANALYSIS
11
Engineering Analysis
In fluid mechanics, many problems are messy and open-ended. Thus, this section presents a
structured approach to problem solving.
Engineering analysis is a process for idealizing or representing real-world situations using
mathematics and scientific principles and then using calculations to extract useful informa-
tion. For example, engineering analysis is used to find the power required by a pump, wind
force acting on a building, and pipe diameter for a given application. Engineering analysis
EXAMPLE 1.4 GRID METHOD APPLIED TO
A ROCKET
A water rocket is fabricated by attaching fins to a 1-liter plastic
bottle. The rocket is partially filled with water and the air space
above the water is pressurized, causing water to jet out of the
rocket and propel the rocket upward. The thrust force T from
the water jet is given by where is the rate at which
the water flows out of the rocket in units of mass per time and
V is the speed of the water jet. (a) Estimate the thrust force in
newtons for a jet velocity of where the
mass flow rate is (b) Estimate the
thrust force in units of pounds-force (lbf). Apply the grid
method during your calculations.
Problem Definition
Situation:
1.
A rocket is propelled by a water jet.
2.

The thrust force is given by
Find: Thrust force supplied by the water jet. Present the
answer in N and lbf.
Sketch:
Plan
Find the thrust force by using the process given in
Table 1.3. When traditional units are used, apply Eq. (1.9).
Solution
1.
Thrust force (SI units)
• Insert numbers and units:
• Insert conversion ratios and cancel units:
2.
Thrust force (traditional units)
• Insert numbers and units:
• Insert conversion ratios and cancel units:
Review
1.
Validation. Since 1.0 lbf 4.45 N, answer (a) is the same
as answer (b).
2.
Tip! To validate calculations in traditional units, repeat the
calculation in SI units.
Tm
.
V,ϭ m
.
V 30 m/s 98.4 ft/s()ϭ
m
.

9 kg/s 19.8 lbm/s().ϭ
Tm
.
V.ϭ
Pressure = p
Water jet
Velocity = V = 30 m/s = 98.4 ft/s
Mass flow rate = m = 9 kg/s = 19.8 lbm/s
Air
Water
Tm
.
Vϭ
T N() m
.
V 9 kg/s()30 m/s()ϭϭ
T N()
9 kg
s

30 m
s

Ns
2
ؒ
kg mؒ

ϭ
T 270 Nϭ

Tm
.
Vϭ
T lbf()m
.
V 19.8 lbm/s()98.4 ft/s()ϭϭ
T lbf()
19.8 lbm
s

98.4 ft
s

lbf s
2
ؒ
32.2 lbm ftؒ

ϭ
T 60.5 lbfϭ
ϭ
1.5