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ELEMENTARY
FLUID MECHANICS
BY
JOHN K. VENNARD
Assistant Professor of Fluid Mechanics
New York University
NEW YORK
JOHN WILEY & SONS, INC.
LONDON: CHAPMAN & HALL, LIMITED
1940
COPYRIGHT, 1940, BY
JOHN K. VENNARD
All Rights Reserved
This book or any part thereof must not
be reproduced in any form without
the written permission of the publisher.
PRINTED IN U. S. A.
PRESS OP
BRAONWORTH A CO.. INC.
BRIDGEPORT. CONN.
ELEMENTARY
FLUID MECHANICS
PREFA^R
Fluid mechanics is the study of all ttuYds under all possible condi
tions of rest and motion. Its approaches analytical, rational, and
mathematical rather than empirical ; it concerns itself with those basic
principles which lead to the solution of numerous diversified problems,
and it seeks results which are widely applicable to similar fluid sit
uations and not limited to isolated special cases. Fluid mechanics
recognizes no arbitrary boundaries between fields of engineering knowl
edge but attempts to solve all fluid problems, irrespective of their
occurrence or of the characteristics of the fluids involved.
This textbook is intended primarily for the beginner who knows
the principles of mathematics and mechanics but has had no previous
experience with fluid phenomena. The abilities of the average
beginner and the tremendous scope of fluid mechanics appear to be in
conflict, and the former obviously determine limits beyond which it is
not feasible to go; these practical limits represent the boundaries of
the subject which I have chosen to call elementary fluid mechanics.
The apparent conflict between scope of subject and beginner f s ability
is only along mathematical lines, however, and the physical ideas of
fluid mechanics are well within the reach of the beginner in the field.
Holding to the belief that physical concepts are the sine qua non of
mechanics, I have sacrificed mathematical rigor and detail in develop
ing physical pictures and in many cases have stated general laws only
(without numerous exceptions and limitations) in order to convey basic
ideas; such oversimplification is necessary in introducing a new subject
to the beginner.
Like other courses in mechanics, fluid mechanics must include
disciplinary features as well as factual information the beginner must
follow theoretical developments, develop imagination in visualizing
physical phenomena, and be forced to think his way through problems
of theory and application. The text attempts to attain these objec
tives in the following ways: omission of subsidiary conclusions is
designed to encourage the student to come to some conclusions by
himself; application of bare principles to specific problems should
develop ingenuity; illustrative problems are included to assist in
overcoming numerical difficulties; and many numerical problems for
vi PREFACE
the student to solve are intended not only to develop ingenuity but to
show practical applications as well.
Presentation of the subject begins with a discussion of funda
mentals, physical properties and fluid statics. Frictionless flow is
then discussed to bring out the applications of the principles of con
servation of mass and energy, and of impulsemomentum law, to fluid
motion. The principles of similarity and dimensional analysis are
next taken up so that these principles may be used as tools in later
developments. Frictional processes are discussed in a semiquanti
tative fashion, and the text proceeds to pipe and openchannel flow.
A chapter is devoted to the principles and apparatus for fluid measure
ments, and the text ends with an elementary treatment of flow about
immersed objects. Throughout the text, the footpoundsecond sys
tem of dimensions has been used, and problems of conversion from the
metric system, which so frequently divert the beginner's attention
from the physical ideas, have been avoided; justifications for experi
mental results and empirical formulas have been presented except
at points where the student should discover them for himself;
bibliographies have been included to guide the inquiring reader to
more exhaustive treatments of the subject.
For criticism of my Notes on Elementary Fluid Mechanics which
have been expanded into the present text, I wish to extend my appre
ciation to many of my colleagues at New York University, Professor
Boris A. Bakhmeteff of Columbia University, and Professor William
Allan of the College of the City of New York.
I am deeply indebted to Mr. William H. Peters of the Curtiss
Wright Corporation for carefully reviewing the first eight chapters of
the manuscript, and to Professor Frederick K. Teichmann of New
York University for critical comments on the last chapter. I also
wish to thank Mr. J. Charles Morgan for general comments and
assistance in reading proof and Miss Katherine Williams for her care
and patience in typing the manuscript.
JOHN K. VENNARD
NEW YORK, N.Y.
May, 1040
CONTENTS
CHAPTER FACE
I . FUNDAMENTALS 1
Art. 1 Development of Fluid Mechanics. 2. Physical Character
istics of the Fluid State^. 3. Density, Specific Weight, Specific
Volume, ,and Specific Gravity. 4. Compressibility, Elasticity.
5. Viscosity. 6. Surface Tension, Capillarity. 7. Vapor Pressure.
II. FLUID STATICS 22
Art.^PressureDensityHeight Relationships. 9. Absolute and
Gage Pressures. 10. Manometry. 11. Forces on Submerged Plane
Surfaces. 12. Forces on Submerged Curved Surfaces. 13. Buoy
ancy and Flotation. 14. Stresses in Circular Pipes and Tanks.
15. Fluid Masses Subjected to Acceleration.
III. THE FLOW OF AN IDEAL FLUID 55
Art. 16 Definitions. 17. Equation of Continuity. 18. Euler's
Equation. 19. Bernoulli's Equation. 20. Energy Relationships.
21. Flow of an Incompressible Fluid. 22. Flow of a Compressible
Fluid. 23. ImpulseMomentum Relationships. 24. Flow Curva
tures, Types of Vortices, Circulation.
IV. THE FLOW OF A REAL FLUID 106
Art. 25 Laminar and Turbulent Flow. 26. Fluid Flow Past Solid
Boundaries. 27. Fluid Flow between Parallel Plates. 28. Flow
about Immersed Objects. 29. Stability Secondary Flows.
V. SIMILARITY AND DIMENSIONAL ANALYSIS 125
Art. 30 Similarity and Models. 31. Dimensional Analysis.
VI. FLUID FLOW IN PIPES 139
Art. 32 Energy Relationships. 33. General Mechanics of Fluid
Flow in Pipe Lines. 34. Laminar Flow. 35. Dimensional Analysis
of the PipeFriction Problem. 36. Results of PipeFriction Experi
ments. 37. Velocity Distribution in Circular Pipes The Pipe
Coefficient. 38. Approximate Thickness of the Laminar Film.
39. Pipe Friction for Compressible Fluids. 40. Pipe Friction in
NonCircular Pipes. 41. PipeFriction Calculations by the Hazen
Williarns Method. 42. Minor Losses in Pipe Lines. 43. The
Pressure Grade Line and Its Use. 44. Branching Pipes.
vii
viii CONTENTS
CHAPTER PAGE
VII. FLUID FLOW IN OPEN CHANNELS 196
Art. 45 Fundamentals. 46. Uniform Flow The Chezy Equation.
47. The Chezy Coefficient. 48. Best Hydraulic Cross Section.
49. Variation of Velocity and Rate of Flow with Depth in Closed
Conduits. 50. Specific Energy. 51. Critical Depth Relationships.
52. Occurrence of Critical Depth. 53. Varied Flow. 54. The
Hydraulic Jump.
VIII. FLUID MEASUREMENTS 229
Art. 55 Measurement of Fluid Properties. 56. Measurement of
Static Pressure. 57. Measurement of Surface Elevation. 5&. Meas
urement of Stagnation Pressure. 59. The Pitot (PitotStatic)
Tube. y '60. The Venturi Tube. 6t^The PitotVenturi, 62. Ane
mometers and Current Meters. 63. Total Quantity Methods.
&4. Venturi Meters. 65. Nozzles. &fl Orifices. 67. Flow Bends.
S&r 1 Pitot Tube Methods. 69. Dilution anj Thermal Methods.
70. SaltVelocity Method. 71. Weirs. ^^ Current Meter Meas
urements. v ?3. Float Measurements.
IX. FLOW ABOUT IMMERSED OBJECTS 294
Art. 74 Fundamentals and Definitions. 75. Dimensional Analysis
of the Drag Problem. 76. Frictional Drag. 77. Profile Drag.
78. Drag at High Velocities. 79. Lift by Change of Momentum.
80. Circulation Theory of Lift. 81. Origin of Circulation. 82. Foils
of Finite Length. 83. Lift and Drag Diagrams.
APPENDICES 335
I. Description and Dimensions of Symbols. II. Specific Weight and
Density of Water. III. Velocity of a Pressure Wave through a
Fluid. IV. Viscosities of Liquids and Gases. V. Cavitation.
INDEX 347
ELEMENTARY FLUID MECHANICS
CHAPTER I
FUNDAMENTALS
1. Development of Fluid Mechanics. Man's desire for knowl
edge of fluid phenomena began with his problems of water supply
and disposal and the use of water for obtaining power. With only a
rudimentary appreciation for the physics of fluid flow he dug wells,
operated crude water wheels and pumping devices, and, as his cities
increased in size, constructed ever larger aqueducts, which reached their
greatest size and grandeur in those of the City of Rome. However,
with the exception of the thoughts of Archimedes (250 B.C.) on the
principles of buoyancy little of the scant knowledge of the ancients
appears in modern fluid mechanics. After the fall of the Roman
Empire (A.D. 476) no progress was made in fluid mechanics until the
time of Leonardo Da Vinci (14521519). This great genius designed
and built the first chambered canal lock near Milan and ushered in a
new era in hydraulic engineering; he also studied the flight of birds and
developed some ideas on the origin of the forces which support them.
After the time of Da Vinci, the accumulation of hydraulic knowledge
rapidly gained momentum, the contributions of Galileo, Torricelli,
Newton, Pitot, D. Bernoulli, and D'Alembert to the fundamentals of
the science being outstanding. Although the theories proposed by
these scientists were in general confirmed by crude experiments, diver
gences between theory and fact led D'Alembert to observe in 1744 that,
"The theory of fluids must necessarily be based upon experiment."
D'Alembert showed that there is no resistance to motion when a body
moves through an ideal (nonviscous) fluid, yet obviously this con
clusion is not valid for bodies moving through real fluid. This dis
crepancy between theory and practice is called the "D'Alembert
paradox" and serves to demonstrate the limitations of theory alone in
solving fluid problems.
Because of the conflict between theory and practice, two schools
of thought arose in the treatment of fluid problems, one dealing with
2 FUNDAMENTALS
the theoretical and the other with the practical aspects of fluid flow,
and in a sense these two schools of thought have persisted down to the
present day, resulting in the theoretical field of " hydrodynamics" and
the practical one of "hydraulics." Notable contributions to theo
retical hydrodynamics have been made by Euler, La Grange, Helm
hoi tz, Kirchhoff, Lord Rayleigh, Rankine, Lord Kelvin, and Lamb.
In a broad sense, experimental hydraulics became a study of the laws
of fluid resistance, mainly in pipes and open channels. Among the
many scientists who devoted their energies to this field were Brahms,
Bossut, Chezy, Dubuat, Fabre, Coulomb, Eytelwein, Belanger, Dupuit,
d'Aubisson, Hagen, and Poisseuille.
Toward the middle of the last century, Navier and Stokes succeeded
in modifying the general equations for ideal fluid motion to fit that
of a viscous fluid and in so doing showed the possibilities of adjusting
the differences between hydraulics and hydrodynamics. At about the
same time, theoretical and experimental work on vortex motion by
Helmholtz was aiding in explaining away many of the divergent results
of theory and practice.
Meanwhile, hydraulic research went on apace, and large quantities
of excellent data were collected or formulas proposed for fluid re
sistance, notably by Darcy, Bazin, Weisbach, Fanning, Ganguillet,
Kutter, and Manning; among researchers on other hydraulic prob
lems were Thomson, Fteley, Stearns, and H. Smith. Unfortunately,
researches led frequently to empirical formulas obtained by fitting
curves to experimental data or merely presenting the results in tabular
form, and in many cases the relationship between the physical facts
and the resulting formula was not apparent.
Toward the end of the last century, new industries arose which
demanded data on the flow of fluids other than water; this fact and
many significant advances in knowledge tended to arrest the increasing
empiricism of hydraulics. These advances were: (1) the theoretical
and experimental researches of Reynolds; (2) the development of di
mensional analysis by Lord Rayleigh ; (3) the use of models by Froude,
Reynolds, Fargue, and Engels in the solution of fluid problems; and
(4) the rapid progress of theoretical and experimental aeronautics in
the work of Lanchester, Lilienthal, Kutta, Joukowski, and Prandtl.
These advances allowed new tools to be applied to the solution of
fluid problems and gave birth to modern fluid mechanics.
Since the beginning of the present century, empiricism has waned
and fluid problems have been solved by increasingly rational methods;
these methods have produced so many fruitful results and have aided
PHYSICAL CHARACTERISTICS OF THE FLUID STATE 3
so materially in increasing our knowledge of the details of fluid phenom
ena that the trend appears likely to continue into the future. Among
the foremost contributors to modern fluid mechanics are Prandtl,
Blasius, Karman, Stanton, Nikuradse, Bakhmeteff, Koch, Bucking
ham, Gibson, Rehbock, Durand, and Taylor.
2. Physical Characteristics of the Fluid State. Matter exists in
two states the solid and the fluid, the fluid state being commonly
divided into the liquid and gaseous states.
Solids differ from liquids and liquids from gases in the spacing and
latitude of motion of their molecules, these variables being large in a
gas, smaller in a liquid, and extremely small in a solid. It follows
that inter molecular cohesive forces are large in a solid, smaller in a
liquid, and extremely small in a gas^ These fundamental facts account
for the familiar compactness and rigidity of form possessed by solids,
the ability of liquid molecules to move freely within a liquid mass, and
the capacity of gases to fill com
pletely the containers in which
they are placed.
A more fruitful and rigorous
mechanical definition of the solid
and fluid states may be made on
the basis of their actions under the
various types of stress. Applica FIG. 1.
tion of tension7~c6rnpression , or
shear stresseVto a solid results first in elastic deformation, and later,
if these~stresses exceed the elastic limits, in permanent distortion
of the material. Fluids, however, possess elastic properties under
compression stress, but application of infinitesimal shear stress results
in continual and permanent distortion. This inability to resist shear
stress gives fluids their characteristic ability to " flow". Fluids
will support tension stress to the extent of the cohesive forces be
tween their molecules. Since such forces are extremely small, it is
customary in engineering problems to assume that fluid can support
no tension stress.
Since shear stress applied to fluids always results in distortion or
"flow," it is evident that in fluids at rest no shear stresses can. exist
and compression stress, or "pressure/' becomes the only stress to be
considered.
Fluids being continuous media, it follows that pressures occurring
or imposed at a point in a fluid will ^transmitted undiminished to all
other points in the fluid (neglecting the weight of the fluid).
U UU H I i V
132 SIMILARITY AND DIMENSIONAL ANALYSIS
Before examining the methods of dimensional analysis, recall that
there are two different systems by which the dimensions of physical
quantities may be expressed. These systems are the forcelengthtime
system and the masslengthtime system. The former system, gener
ally preferred by engineers, becomes the familiar "footpoundsecond"
system when expressed in English dimensions; the latter system in
English dimensions becomes the "footslugsecond" system. The
latter system is generally preferred in dimensional analysis, and, since
the student is familiar with the former system, the use of the latter
will serve to develop versatility in the use of dimensions.
A summary of the fundamental quantities of fluid mechanics and
their dimensions in the various systems is given in Table VI, the con
ventional system of capital letters being followed to indicate the dimen
sions of quantities. The basic relation between the forcelengthtime
and masslengthtime systems of dimensions is given by the Newtonian
law, force or weight = (Mass) X (Acceleration) and, therefore,
dimensionally,
p = yr .
r m ^2
from which the dimensions of any quantity may be converted from
one system to the other.
To illustrate the mathematical steps in a simple dimensional prob
lem, consider the familiar equation of fluid statics
p wh
but assume that the dimensions of w and h are known and those of p
unknown. The dimensions of p can be only some combination of
M, L, and T, and this combination may be discovered by writing the
equation dimensionally as
Unknown dimensions of p (Dimensions of w) X (Dimensions of h)
or
M*L b T* = (jj^ X (L)
in which a, b, and c are unknowns. The principle of dimensional
homogeneity being applied, the exponents of each of the quantities is
the same on each side of the equation, giving
a = 1, b =~ 2 + 1 = 1, c = 2
DIMENSIONAL ANALYSIS
133
TABLE VI
DIMENSIONS OF FUNDAMENTAL QUANTITIES USED IN FLUID MECHANICS
Quantity
Symbol
English Engi
neering
Dimensions
ForceLength
Time
Dimensions
MassLength
Time
Dimensions
Acceleration
d
ft/sec 2
L/T 2
L/T 2
Acceleration due to gravity
Area
g
A
ft/sec 2
ft 2
L/T 2
L 2
L/T 2
L 2
Density . . . ...
p
Ib sec 2 /ft 4
FT 2 /L 4
lf/L 3
Force
F
Ib
F
ML/T 2
Kinematic viscosity . . .
p
ft 2 /sec
i: 2 /r 2
I 2 /T
Length
I
ft
L
L
Mass
M
Ib sec 2 /ft
FT 2 /L
M
Power
P
ft Ib/sec
PL /T
ML 2 /T*
Pressure ....
b
lb/ft 2
F/L 2
MILT"
Rate of flow
Q
ft 3 /scc
L*/T
L*/T
Specific weight
IV
lb/ft 3
F/L*
M/L 2 T 2
Time
t
sec
T
T
Velocity
V
ft /sec
L/T
L/T
Viscosity
fj,
Ib sec/ft 2
FT/L 2
M/LT
Weight
ir
Ib
F
ML/T 2
Weight rate of flow
G
Ib/soc
F/T
ML/T*
whence
Dimensions of p ML 1 T 2
M
It is obvious, of course, that this result might have been obtained more
directly by cancellation of L on the righthand side of the equation,
for this has been, and will continue to be, the usual method of obtain
ing the unknown dimensions of a quantity, It is of utmost impor
tance, however, to note the mathematical steps which lie unrevealed
in this hasty cancellation, if the basis of dimensional analysis is to be
understood.
The above methods may now be used in quite another and more
important way. To illustrate by another familiar example, suppose
that it is known that the power P, which can be derived from a hy
draulic motor, is dependent upon the rate of flow through the motor
Q, the specific weight of the fluid flowing w, and the unit energy E
which is given up by every pound of fluid as it passes through the
4 FUNDAMENTALS
Since pressure ^cornyBr^sioruatress, the equilibrium of a mass of
fluid at rest occurs from pressure acting inward upon its boundary
surface as shown in Fig. 1. If this mass of fluid is reduced to infini
tesimal size, it becomes evident that at a point in a fluid the pressure
is the same in all directions. Since fluids are unable to support tan
gential (shear) stresses, no component of force can exist along the solid
boundary of Fig. 1, and thus pressure must be transmitted from a
fluid to a solid boundary normal to the boundary at every point.
PHYSICAL PROPERTIES OF FLUIDS
3. Density, Specific Weight, Specific Volume, and Specific Gravity.
Density * is the mass of fluid contained in a unit of volume ; specific
weight, 1 the weight of fluid contained in a unit of volume. Both these
terms are fundamentally measures of the number of molecules per
unit of volume. Since molecular activity and spacing increase with
temperature fewer molecules will exist in a given unit volume as tem
perature rises, thus causing density and specific weight to decrease
with increasing temperature. 2 Since a larger number of molecules can
be forced into a given volume by application of pressure, it will be
found that density and specific weight will increase with increasing
pressure.
Density, p (rho), will be expressed in the masslengthtime system
of dimensions and will have the dimensions of mass units (slugs) per
cubic foot (slugs/ft 3 ).
Specific weight, w, will be expressed in the forcelengthtime system
of dimensions and will have the dimensions of pounds per cubic
foot (lb/ft 8 ).
Since a mass, M, is related to its weight, W, by the equation
M W
M =
g
in which g is the acceleration due to gravity, density and specific
weight (the mass and weight of a unit volume of fluid) will be related
by a similar equation
p = or w = pg
g
1 In American engineering practice, specific weight is frequently termed " density r '
and density " mass density/'
2 A variation in temperature from 32Fto212F will decrease the specific weight
of water 4 per cent (Appendix II) and will decrease the density of gases 37 per cent
(assuming no pressure variation).
PHYSICAL PROPERTIES OF FLUIDS 5
Using the fact that physical equations are dimensionally homogeneous,
the footpoundsecond dimensions of p (which are equivalent to slugs
per cubic foot) may be calculated as follows :
Ib
_ . . f Dimensions of w ft 3 Ib sec 2
Dimensions of p = : =  = 4
Dimensions of g ft ft
sec 2
This algebraic use of the dimensions of quantities in the equation ex
pressing physical relationship will be employed extensively and will
prove to be an invaluable check on engineering calculations. 3
The specific volume, v, defined as volume per unit of weight, will
have dimensions of cubic feet per pound (ft 3 /lb). This definition
identifies specific volume as the reciprocal of specific weight and intro
duces the equations
1 1
i) = or w = 
W V
Specific gravity, 5, is the ratio of specific weight or density of a sub
stance to the specific weight or density of pure water. Since all these
items vary with temperature, temperatures must be quoted when spe
cific gravity is used in precise calculations of specific weight or density.
Specific gravities of a few common liquids at 68 F. (except as noted),
are presented in Table I, from which the specific weights of liquids
TABLE I *
SPECIFIC GRAVITIES, S, OF VARIOUS LIQUIDS AT 68 F f
(Referred to water at 39.2 F)
Ethyl alcohol 0. 789
Turpentine (dpinene) . 862
Benzene 0.888
Linseed oil 0.934 (59.9 F)
Castor oil 0.960
Water 0.998
Glycerine 1 .264 (57 F)
Carbon tetrachloride 1 .594
Mercury 13 . 546
* Smithsonian Physical Tables, Eighth Ed., 1933, Smithsonian Institution.
t Except as noted.
3 A summary of quantities and their dimensions is given in Appendix I.
6 FUNDAMENTALS
may be readily calculated by
w = 5 X 62.45 lb/ft 3
The specific weight of gases may be calculated by means of Boyle's
law and Charles' law. Using the specific volume of a gas, Boyle's
law may be stated as 4
pv = Constant
which expresses the law of compression or expansion of a gas at con
stant temperature. Charles' law, expressing the variation of pressure
with temperature in a constant volume of gas, is 4
= Constant
Obviously the only combination of variables which will satisfy both
Boyle's and Charles' laws simultaneously is
which is called the "equation of state" of the gas in which the constant,
JR, is called the "gas constant" and has dimensions of feet/degree
Fahrenheit absolute. Since w = l/v 9 the above equation may be
transformed into
, = 
W RT
from which specific weights of gases may be readily calculated.
Application of Avogadro's law, that "all gases at the same pressures
and temperatures have the same number of molecules per unit of
volume," allows the calculation of a "universal gas constant." Con
sider two gases having constants R\ and R%, specific weights w\ and w^
and existing at the same pressure and temperature, p and T. Dividing
their equations of state
P _
w\T
4 p is the absolute pressure in pounds per square foot, T is the temperature in
degrees F absolute (degrees F } 459.6), and the above "constants" are constant if the
gas is " perfect." Common gases in the ordinary engineering range of pressures and
temperatures may be considered to be "perfect" for most engineering calculations.
PHYSICAL PROPERTIES OF FLUIDS
results in
' but, according to Avogadro's principle, the specific weight of a gas must
be proportional to its molecular weight, giving w<2/Wi = w 2 /%i, in
which mi and ra 2 are the respective molecular weights of the gases.
Combining this equation with the preceding one gives m^/mi
or
In other words, the product of molecular weight and gas constant is
the same 5 for all gases. This product mR is called the "universal
gas constant" and is preferred for general use by many engineers.
Values of these gas constants are given in Table II.
TABLE II
GAS CONSTANTS FOR COMMON GASES *
R, ft/F abs mR
Sulphur dioxide ..................... 23.6 1512
Carbon dioxide ........... .......... 34.9 1536
Oxygen ............................ 48.3 1546
Air ................................ 53.3 1545
Nitrogen ........................... 55.1 1543
Ammonia .......................... 89.5 1516
Hydrogen .......................... 767.0 1546
* O. W. Eshbach, Handbook of Engineering Fundamentals, p. 716, John Wiley & Sons, 1936.
ILLUSTRATIVE PROBLEM
Calculate the density of carbon dioxide at a temperature of 80 F and absolute
pressure of 100 lb/in. 2
, x 1536
m = 12 + 2(16) =44, R =  = 34.9
44
p 100 X 144 Ib
w __ _ __  __ n 703 
RT 34.9 X (80 + 460) ' ft 3
0.793 / ,
P   0.0246 slugs/ft 3
6 The constancy of mR is particularly true for the monatomic and diatomic gases.
Gases having more than two atoms per molecule tend to deviate from the law
mR Constant. See Table II.
8 FUNDAMENTALS
4. Compressibility, Elasticity. All fluids may be compressed with
consequent increase in density, the process of compression taking
place at the expense of the space between molecules. Therefore, as
fluids are compressed the molecular spacing is diminished and the
fluids become increasingly difficult to compress further. Fluids also
become more difficult to compress as temperature increases because
of increased molecular activity reducing the molecular spacing available
for compression. Owing to these facts, it is obviously an approxima
tion to express elastic compression of fluids by Hooke's law
Stress
*' == 7^ ~.
Strain
because E is not a constant but increases with increased temperature
or pressure. Such an approximation, however, is justified for ordinary
engineering calculations since the range of pressure encountered in
engineering is comparatively small and the change in E over this pres
sure range is usually negligible. 6 The above equation then becomes
V
the strain ( AF/F) being the decrease in volume (AF) per original
volume (F) obtained by an increment of pressure (A/>).
Compression of gases may take place according to various laws of
thermodynamics. The isothermal compression of a volume of perfect
gas, FI, existing at an absolute pressure, pi, to a volume Fg at a
pressure p2 will be accomplished according to Boyle's law.
P\V\ = p2^2 or pV = Constant
Using the specific volume, v, in this equation there results
P
pv = Constant or = Constant
w
Frequently expansion of gases occurs so rapidly that there is
no opportunity for flow of heat during the process. Such an expansion
follows the adiabatic law
t
PiVf = p2V 2 k or pV k = Constant
8 E for water is commonly taken as a constant, 300,000 lb/in. 2
COMPRESSIBILITY, ELASTICITY 9
which, written in terms of specific volume and specific weight, becomes
t P
pv K = Constant or ^ = Constant
in which k, called the "adiabatic constant/' is the ratio of the two
specific heats of the gas, that at constant pressure, c p , to that at con
stant volume, c v . Values of k for common gases are given in Table III.
TABLE III
ADIABATIC CONSTANTS, k t FOR COMMON GASES *
Sulphur dioxide 1.26
Carbon dioxide 1 . 30
Oxygen 1 . 40
Air 1.40
Nitrogen 1 . 40
Ammonia 1.32
Hydrogen 1 . 40
* O. W. Eshbach, Handbook of Engineering Fundamentals, p. 717, John Wiley & Sons, 1936.
Values of modulus of elasticity of gases, E, may be derived for
isothermal and adiabatic processes for use in subsequent developments.
Writing the law of elastic compression in differential form
V
For isothermal compression
pV = Constant
Differentiating this equation in respect to V results in
whence
dp_ __ p_
dV~ V
and substituting this in the first expression
Ep
for an isothermal process.
10 FUNDAMENTALS
By a similar analysis E for an adiabatic process may be shown to be
given by
E = kp
Imposed pressures or pressure disturbances are not transmitted
instantaneously from point to point in a fluid, but move in waves at
finite velocity. The velocity or celerity of propagation of such waves is
dependent upon the elastic properties of the fluid ; fluids which are more
easily compressible (having low values of E) transmit pressures with
smaller velocity than those which are difficult to compress. A pressure
disturbance is transmitted in a fluid with a celerity, c, which is given by
the equation 7
in which c is frequently termed the "sonic" or "acoustic" velocity since
it is the velocity with which sound, a pressure disturbance, travels. In
a gas, sound moves by a series of adiabatic compressions and rarefac
tions. Thus the sonic velocity in a gas may be calculated from
P
an equation which is accurately confirmed by experiment.
ILLUSTRATIVE PROBLEMS
Ten cubic feet of water exist at atmospheric pressure. When a pressure of
2000 lb/in. 2 is applied, what reduction in volume results?
A 2000
E =  77P ^ 300,000 =  
A V A V
~v To""
 AF = 0.0667 ft 8
Calculate the velocity with which sound travels in water.
300,000 X 144
457  4720 ft/sec
32.2
7 For derivation see Appendix III.
VISCOSITY
11
T
^
_^
~*~
*j.
 T
~^~
3J Viscosity. The property of viscosity, which is exhibited by all
fltFWns, is due fundamentally to the existence of cohesion and inter
action between fluid molecules. As fluids flow these cohesions and
interactions result in tangential or
shear stresses between the moving
fluid layers. Consider the thin viscous
fluid layers shown in relative motion
in Fig. 2. Let them have a thickness,
dy, and areas of contact, A, the lower
layer moving with velocity, v, the
upper one with velocity v + dv. To maintain this velocity difference
a force F must be continually exerted on the upper layer as indicated,
which results in a shear or friction stress, r (tau), between the layers,
given by
F
f TTTf T
p
T
FIG. 2.
For viscous fluid motion the shear stress, r, has been found to be por
portional to the rate of change of velocity along y or
r oc
dv
~dy
and the "coefficient of viscosity," M (mu), 8 is defined as the constant
of proportionality in the above equation, or
dv
dy
n
Indicating
r, and thus the coemaenj^ofjdscosit^ independent 9 of pressure.
Viscosity varies ^widely with temperature, but temperature varia
tion has an opposite effect upon the viscosities of liquids and gases due
to their fundamentally different intermolecular characteristics. In
gases, where intermolecular cohesion is negligible, the shear stress, r,
8 The dimensions of /* may be obtained by writing
Dimensions of r lb/ft 2
Dimensions of
Dimensions of
dy
dv ft/se
= Ib sec/ft 2
9 Viscosity actually increases slightly with pressure, but this variation is negligible
in most engineering problems. Oils manifest the greatest increase of viscosity with
pressure.
12
FUNDAMENTALS
between moving layers of fluid results from an exchange of momentum
between these layers brought about by molecular agitation normal to
.060
.050
.040
,030
.020
100
200 300 400
Temperature in Degrees T.
FIG. 3. Viscosities of Gases. 10
500
600
the general direction of motion. Since this molecular activity is known
to increase with temperature, the shear stress, and thus the viscosity
10 Data on viscosities from Smithsonian Physical Tables, Eighth Edition, 1933,
Smithsonian Institution. (For original data see Appendix IV.) Data on viscos
ity of steam from Fluid Meters, Their Theory and Application, Fourth Edition,
1937, A.S.M.E.
VISCOSITY
13
10,000
5,000
1,000
500
100
50
10
1.0
0.5
0.1
Gycer
\
I enzene
.Ca
stor oil
x0 ive oil
Mercury
Carbon tetr chloride
Linsee oil
50 100 150
Temperature in Degrees F.
FIG. 4. Viscosities of Liquids. 10
200
Water
Turpentin
250
10 Data on viscosities from Smithsonian Physical Tables, Eighth Edition, 1933,
Smithsonian Institution. (For original data, see Appendix IV.) Data on viscos
ity of steam from Fluid Meters, Their Theory and Application, Fourth Edition,
1937, A.S.M.E.
14 FUNDAMENTALS
of gases, will increase with temperature (Fig. 3). This reasoning is
borne out by tests and by considerations of the kinetic theory of gases
which indicate that gas viscosities vary directly with the square root
of
In a viscous liquid, momentum exchange due to molecular agitation
is small compared to the cohesive forces between the molecules, and
thus shear stress, r, and viscosity, jo,, are primarily dependent on the
magnitude of these cohesive forces. Since these forces decrease rapidly
with increases of temperature, liquid viscosities decrease as tempera
ture increases (Fig. 4).
jj,
Owing to the continual appearance of the ratio  in subsequent
P
developments, this term has been defined by
z/(nu) = 
P
in which v is called the " kinematic viscosity. " The kinematic viscosity
embraces both the viscosity and density properties of a fluid. Dimen
sional consideration of the above equation shows the dimensions of v
to be square feet per second, a combination of kinematic terms, which
explains the name "kinematic" viscosity.
ILLUSTRATIVE PROBLEM
Calculate the kinematic viscosity of glycerine at 80 F.
From Fig. 4, ju = 0.0103 Ib sec/ft. 2
From Table I, S = 1.26
JLI 0.0103
v =  =    = 0.00423 ft 2 /sec
p 62.4 X 1.26
32.2
6. Surface Tension, Capillarity. The apparent tension effects,
which occur on the free surfaces of liquids, where the surfaces are in
contact with another fluid or a solid, depend fundamentally upon the
relative sizes of intermolecular cohesive and adhesive forces. On a
free liquid surface in contact with the atmosphere, surface tension
manifests itself as an apparent "skin" over the surface which will
support small loads. 11 The magnitude of surface tension, T, is the
force in the surface and normal to a line of unit length drawn in the
11 A small needle placed gently upon a water surface will not sink but will be
supported by the tension in the liquid surface.
SURFACE TENSION, CAPILLARITY
15
liquid surface ; thus it will have dimensions of pounds per foot. Since
surface tension is directly dependent upon intermolecular cohesive
forces, its magnitude will decrease as temperature increases. 12 Surface
tension is also dependent upon the gas in contact with the liquid surface,
thus surface tensions are usually quoted "in contact with air" as indi
cated in Table IV.
TABLE IV
SURFACE TENSION, T, OF COMMON LIQUIDS *
(At 68 F in contact with air)
Ib/ft
Ethyl alcohol 0.001527
Carbon tetrachloride 001832
Turpentine 0.001857
Benzene .001980
Olive oil 002295
Water 0.004985
Mercury 0.03562
* International Critical Tables, First Edition, 19261933, McGrawHill Book Company.
Surface tension in the surface of a droplet of liquid causes the pres
sure inside of the droplet to be greater than that outside. The rela
tion of this excess pressure to the surface tension can be found by a
simple mechanical analysis as follows. Consider the droplet of diam
Tension in surface
V= T Ib. per ft.
FIG. 5.
eter d, indicated in Fig. 5. If the droplet is halved the forces on one
half are seen to be (1) the force due to surface tension, T, existing
around the circumference of the droplet and acting to the right, and
12 For example, the surface tension of water decreases from 0.00498 Ib/ft at 68 F
to 0,00421 Ib/ft at 212 F,
16 FUNDAMENTALS
(2) the force due to the excess pressure, p, acting to the left. These
forces are in equilibrium resulting in
Surface tension force Pressure force
or
vdT  p =
4
giving
thus relating excess pressure to surface tension and indicating that
these pressures increase as the size of droplet decreases.
The angle of contact made by a liquid on a horizontal surface illus
trates another surfacetension phenomenon of more complex nature.
Consider the mercury and water on a glass surface illustrated by Fig. 6.
Glass T a t * A
surta c,v * yrxX"
/S x g /\ } C _s?S y
Water Mercury
FIG. 6.
The familiar large angle of contact assumed by the mercury indicates
a comparatively large affinity of mercury molecules for each other
(cohesion) and small affinity of these molecules for those of the glass
(adhesion). The opposite effect is exhibited by the water. The
water is said to "wet" the glass since its angle of contact is less than
90 degrees. The stability of these liquids on a solid surface may be
characterized by the equilibrium of assumed surface tensions 18 at
their points of contact. Thus the following equation may be written
Tis + T a i cos a = T a8
Frequently equilibrium^does not exist between these tensions, and the
following inequality results
T a8 > Ti 8 + T a i cos a
18 T a i = surface tension between liquid and air.
Tis = surface tension between liquid arid solid.
T a8 = surface tension between air and solid.
SURFACE TENSION, CAPILLARITY
17
causing the liquid to spread over the surface on which it is placed.
Such a condition exists when certain types of oil are placed on a water
surface.
Surfacetension effects like the above, existing when surfaces of
liquids come in contact with vertical solid surfaces, result in the
phenomenon known as "capillarity." Water and mercury in contact
with a vertical clean glass plate are illustrated in Fig. 7. Here again
are demonstrated the results
of attractions and repulsions,
cohesions and adhesions, be
tween the molecules of liquid
and solid.
When a vertical tube
is
Glass
surface 
Water
Glass
surface 
Mercury
by*
FIG. 7.
FIG. 8.
placed. in a liquid as in Fig. 8, these surface phenomena form a "menis
cus/' or curved surface, in the tube, and the liquid in the tube
stands above or below that outside, depending upon the size of the
angle of contact. The "capillary rise," h, in such a tube may be cal
culated approximately by considering the equilibrium of the vertical
forces on the mass of fluid A BCD. Neglecting the fluid above the low
point of the meniscus the weight of A BCD is given by
which acts downward. The vertical component, FT, of the force due
to surface tension is given by
FT = irdT cos ft
which acts upward and is in equilibrium with the downward force, thus
,^ 2 JT
wh = irdl cos ft
4
18 FUNDAMENTALS
giving
42" cos ft
h = ~
wd
allowing the capillary rise to be calculated approximately and confirm
ing the familiar fact that capillary rise becomes greater as tube diam
eter is decreased.
Similarly it may be shown that the capillary rise between vertical
parallel plates is given by
_ 2Tcos/3
wd
where d is the distance between the plates.
ILLUSTRATIVE PROBLEM
Of what diameter must a droplet of water be to have the pressure within it
0.1 lb/in. 2 greater than that outside?
From Table IV, T = 0.004985 Ib/ft.
4T 4 X 0.004985
p t d = = 0.001385 ft = 0.0166 in.
a 0.1 X 144
7. Vapor Pressure. All liquids possess a tendency to vaporize,
i.e., to change from the liquid state into the gaseous state. Such
vaporization occurs because molecules are continually projected
through the free liquid surface and lost from the body of liquid. Such
molecules, being gaseous, are capable of exerting a partial pressure, the
" vapor pressure" of the liquid, and since this pressure is dependent
primarily upon molecular activity it will increase with increasing tem
perature. The variation of the vapor pressure of water with tem
perature is indicated in Fig. 9.
For boiling to occur a liquid's temperature must be raised sufficiently
for the vapor pressure to become equal to the pressure imposed on the
liquid. This means that the boiling point of a liquid is depehdent
upon its pressure as well as its temperature. 14
Table V offers a comparison of the vapor pressures of a few common
liquids at the same temperature. The low vapor pressure of mercury
along with its high density makes this liquid well suited for use in
barometers and other pressuremeasuring devices.
14 For instance, water boils at 212 F when exposed to an atmospheric pressure of
14.7 Ib/sq in., but will boil at 200 F if the imposed pressure is reduced to 1 1.4 Ib/sq in.
VAPOR PRESSURE
19
Vapor Pressure in Pounds per Square Inch
t^co ; rolo'^cn ** ro u> *> in o u
}
/
f^
jf
f
j
r
/
/
^
/
j
f
f
/
y
^
,
/
/
^
/
^
/
/
y
yf
r
y
/
/
/
'
>
'
/
'
50 100 150 200
Temperature in Degrees F.
FIG. 9. Vapor Pressure of Water.
TABLE V*
VAPOR PRESSURE, p v , OF COMMON LIQUIDS AT 68 F
lb/ft 2 lb/in. 2
Ether 1231. 8.55
Carbon tetrachloride 250 . 1 . 738
Benzene 208 . 1 .448
Ethyl alcohol 122 .4 .850
Water 48.9 0.339
Turpentine 1 . 115 .00773
Mercury .00362 .0000251
* Smithsonian Physical Tables, Eighth Edition, 1933, Smithsonian Institution.
20 FUNDAMENTALS
BIBLIOGRAPHY
HISTORICAL
W. F. DURAND, "The Development of Our Knowledge of the Laws of Fluid Me
chanics," Science, Vol. 78, No. 2025, p. 343, October 20, 1933.
R. GIACOMELLI and E. PISTOLESI. Historical Sketch, Aerodynamic Theory, Vol. I,
p. 305, Julius Springer, Berlin, 1934.
C. E. BARDSLEY, "Historical Resume of the Development of the Science of Hydrau
lics," Pub. 39, Engineering Experiment Station of Oklahoma Agricultural and
Mechanical College, April, 1939.
VISCOSITY
E. C. BINGHAM, Fluidity and Plasticity, McGrawHill Book Co., 1922.
E. HATSCHEK, The Viscosity of Liquids, Van Nostrand, 1928.
PROBLEMS
1. The two pistons A and B have respectively crosssectional areas of 2 in. 2 and
50 in. 2 What force, F, must be applied to piston A to support a weight of 100 Ib
on B1
2. If 186 ft 3 of a certain oil weigh 9860 Ib, calculate the
specific weight, density, and specific gravity of this oil.
3. Calculate the specific weight and density of mercury
at 68 F.
4. Calculate the specific weight and density of glycerine
at57F.
5. The density of alcohol is 1.53 slugs/ft 3 . Calculate its specific weight, specific
gravity, and specific volume.
6. A cubic foot of air at 14,7 lb/in. 2 and 59 F weighs 0.0765 lb/ft 3 . What is
its specific volume?
7. Calculate the specific weight, specific volume, and density of air at 40 F and
50 lb/in. 2 absolute.
8. Calculate the density, specific weight, and specific volume of carbon dioxide
at 100 lb/in. 2 absolute and 200 F.
9. Calculate the density, specific weight, and specific volume of chlorine gas at
50 lb/in. 2 absolute and 100 F.
10. Calculate the density of carbon monoxide at 20 lb/in. 2 absolute and 50 F.
11. The specific volume of a certain perfect gas at 30 lb/in. 2 absolute and 100 F
is 10 ft, 3 Calculate its gas constant and molecular weight.
12. If h = . what are the dimensions of h?
w
13. If V v2gh, calculate the dimensions of V,
14. If F = QwV/g, what are the dimensions of Fl
15. Twelve cubic feet of water are placed under a pressure of 1000 lb/in. 2 . Calcu
late the volume at this pressure.
16. If the volume of a liquid is reduced 0.035 per cent by application of a pressure
of 100 lb/in. 2 , what is its modulus of elasticity?
17. What pressure must be applied to water to reduce its volume 1 per cent?
18. Ten cubic feet of air at 100 F and 50 lb/in. 2 absolute are compressed isother
PROBLEMS 21
mally to 2 cu ft. What is the pressure when the air is reduced to this volume? What
is the modulus of elasticity at the beginning and end of the compression?
19. If the air in the preceding problem is compressed adiabatically to 2 cu ft,
calculate the final pressure and temperature and the modulus of elasticity at beginning
and end of the compression.
20. Calculate the velocity of sound in air of standard conditions (32 F and
14.7 lb/in. 2 absolute).
21. Calculate the velocity of sound in fresh water.
22. Calculate the kinematic viscosity of turpentine at 68 F.
23. Calculate the kinematic viscosity of castor oil at 68 F.
24. Calculate the kinematic viscosity of nitrogen at 100 F and 80 lb/in. 2 absolute.
25. What is the ratio between the viscosities of air and water at 50 F? What is
the ratio between their kinematic viscosities at this temperature and standard baro
metric pressure?
26. A space of 1in. width between two large plane surfaces is filled with glycerine
at 68.5 F. What force is required to drag a very thin plate of 5 ft 2 area between the
surfaces at a speed of 0.5 ft/sec if this plate remains equidistant from the two sur
faces? If it is at a distance of 0.25 in. from one of the surfaces?
27. Castor oil at 68 F fills the space between two concentric cylinders of 10in.
height and 6in. and 6.25in. diameters. What torque is required to rotate the inner
cylinder at 12 rpm, the outer cylinder remaining stationary?
28. What force is necessary to overcome viscous action when removing the above
inner cylinder from the outer one at a speed of 1 ft/sec?
29. A circular disk of diameter d is rotated in a liquid of viscosity /x at a small
distance A/& from a fixed surface. Derive an expression for the torque T, necessary to
maintain an angular velocity co. Neglect centrifugal effects.
30. Calculate the excess pressure within a droplet of water at 68 F if the droplet
has a diameter of 0.01 in.
31. What excess pressure may be caused within a 0.20in. diameter cylindrical
jet of water by surface tension?
32. Calculate the capillary rise of water in a glass tube of 1mm diameter at 68 F.
(ft  0.)
33. Calculate the capillary rise of water (68 F) between two vertical, clean glass
plates spaced 1 mm apart. (3 = 0.)
34. Develop the equation for theoretical capillary rise between parallel plates.
35. Calculate the capillary depression of mercury in a glass tube of 1mm diameter
at 68 F. (ft = 140.)
36. A soap bubble 2 in. in diameter contains a pressure (in excess of atmospheric)
of 0.003 lb/in. 2 Calculate the surface tension of the soap film.
37. What force is necessary to lift a thin wire ring of 1in. diameter from a water
surface at 68 F? Neglect weight of ring.
38. What is the minimum absolute pressure which may be maintained in the space
above the liquid, in a can of ether at 68 F?
39. To what value must the absolute pressure over carbon tetrachloride be re
duced to make it boil at 68 F?
40. What reduction below standard atmospheric pressure must occur to cause
water to boil at 150 F?
41. A 6in.diameter cylinder containing a tightfitting piston is completely filled
with water at 150 F. What force is necessary to withdraw the piston if atmospheric
pressure is 14.70 lb/in. 2 ?
CHAPTER II
FLUID STATICS
The subject of fluid statics involves fluid problems in which there
is no relative motion between fluid particles. If no relative motion
exists between particles of a fluid, viscosity can have no effect, and the
fluids involved may be treated as if they were completely devoid of
viscosity. With the effects of viscosity excluded from fluid statics
exact solutions of problems may be obtained by analytical methods
without the aid of experiment.
8. PressureDensityHeight Relationships. The fundamental
equation of fluid statics is that relating pressure, density, and vertical
distance in a fluid. This equation may be derived readily by consider
ing the vertical equilibrium of an element of fluid such as the small cube
of Fig. 10. Let this cube be differentially small and have dimensions
dx, dy, and dz, and assume that the density of the fluid in the cube is
uniform. If the pressure upward on the bottom face of this cube is p,
the force due to this pressure will be given by p dx dy. Assuming an
increase of pressure in the positive direction of z, the pressure down
dp
ward on the top face of the cube will be p + dz, and the force due
dz
to this pressure will be ( p \  dz) dx dy. The other vertical force
\ dz /
involved is the weight, dW, of the cube, given by
dW = w dxdy dz
The vertical equilibrium of the cube will be expressed by
/ dp \
[p + dzj
dxdy + wdxdydz pdxdy =
gvng
dp
=
dz
22
PRESSUREDENSITYHEIGHT RELATIONSHIPS
23
the fundamental equation of fluid statics, which must be integrated for
the solution of engineering problems. Such integration may be
accomplished by transposing the terms w and dz, resulting in
dp
w
FIG. 10.
which may be integrated as follows :
r pi d P r'
/ =  / dz = z 2 
J w J*i
giving
24
FLUID STATICS
in which pi is the greater pressure existing at the lower point 1, p2
the lesser pressure existing at the upper point 2, and h the vertical
distance between these points. The integration of the lefthand side
of the equation cannot be carried out until w = f(p) is known. For
gases this relationship may be obtained from certain laws of thermo
dynamics. For liquids the specific weight, w, is sensibly constant
allowing integration of the equation to
Pi 
w
= h or pi
permitting ready calculation of the increase in pressure in a liquid
as depth is gained. It should be noted that equation 1 embodies
certain basic and familiar facts concerning fluids at rest. It shows
that, if h = 0, the pressure difference is zero and thus pressure is con
stant over horizontal planes
a fluid. Conversely, if
Piezometer m
columns
T~
20.35 in.
of
mercury
Manometer
Mercury
Water
in
the pressure is constant over
a horizontal plane the height
23.ift of fluid above that plane is
of i i
water constant, resulting in the
tendency of liquids to "seek
their own levels. "
Equation 1 also indicates
the fact that pressure at a
point in a liquid of given
density is dependent solely
upon the height of the liquid
above the point, allowing
this vertical height, or " head," of liquid to be used as an indication
of pressure. Thus pressures maybe quoted in "inches of mercury,"
" feet of water/' etc. The relation of pressure and head r is illus
trated numerically by the " manometer " and "piezometer columns"
of Fig. 11.
1 For use in problem solutions it is advisable to keep in mind certain pressure and
head equivalents for common liquids. The use of " conversion factors," whose physi
cal significance is rapidly lost, may be avoided by remembering that standard atmos
pheric pressure is 14.70 lb/in. 2 , 29.92 in. of mercury (32 F), or 33.9 ft of water (60 F).
10 Ib. per sq. in.
FlG. 11.
ABSOLUTE AND GAGE PRESSURES
25
ILLUSTRATIVE PROBLEMS
A closed tank is partially filled with carbon tetrachloride. The pressure on the
surface of the liquid is 10 lb/in. 2 . Calculate the pressure 15 ft below the surface.
w = 1.59 X 62.4 = 99.1 lb/ft 3
p l  pz wk, pi  10 X 144  99.1 X 15
pi = 2927 lb/ft 2 = 20.3 lb/in. 2
If the atmospheric pressure at the earth is 14.70 lb/in. 2 and the air tempera
ture there 60 F, calculate the pressure 1000 ft above the earth, assuming that the
air temperature does not vary between the earth's surface and this elevation.
Isothermal condition, therefore, pi/wi = pz/wz = p/w.
i
w
14.70 X 144
53.3(60 + 460)
pi 14.70 X 144
0.0763
. r 1 ^ =
Jpz W
1000
1000
27,750
0.0361
i pl
In
 0.0763 lb/ft 8
= 27,750ft
27,750
14.70
In
14.70
log = 0.0361 X 2.303 = 0.083
Therefore
P2
14.70
= 1.21,
12.15 lb/in. 2
9. Absolute and Gage Pressures. Although
pressures are measured and quoted by two
different systems, one relative, the other
absolute, no confusion will result if the rela
tion between these systems is completely FlG 12 .
understood.
" Gage pressures " may be best understood by examination of the
common Bourdon pressure gage, a diagrammatic sketch of which is
shown in Fig. 12. A bent tube (^4) is held rigidly at B and its free
end connected to a pointer (C) by means of the link (D). When pres
sure is admitted to the tube at B, the tube tends to straighten, thus
actuating the lever system which moves the pointer over a graduated
26
FLUID STATICS
scale. When the gage is in proper adjustment the pointer rests at
zero on the scale when the gage is disconnected, and in this condi
tion the pressure inside and outside of the tube will be the same, thus
giving the tube no tendency to deform. Since atmospheric pressure
usually exists outside of the tube, it is apparent that pressure gages
are actuated by the difference between the pressure inside and that
outside of the tube. Thus, in the gage, or relative, system^of pressure
measurement, the atmospheric pressure becomes the zero of pressure.
For pressure greater than atmospheric the pointer will move to the
right ; for pressure less than atmospheric the tube will tend to contract,
moving the pointer to the left.
The reading in the first case is
positive and is called "gage pres
sure" or simply "pressure," 2 and
is usually measured in pounds
per square inch ; the reading in
Gageo the second case is negative, is
designated as " vacuum," and is
usually measured in inches of
mercury.
The absolute zero of pressure
will exist only in a completely
Abs. o evacuated space (perfect vacu
um). Atmospheric pressure as
measured by a barometer will
be seen to be pressure in excess of this absolute zero and is, there
fore, an "absolute pressure." 2 The magnitudes of the atmospheric
pressure in both the absolute and gage system, being known, the
following equation may be written
A
Gage
pressure A
\ Movable
t
Vacuum B
datum 
^ ! .
Atmospheric
Absolute B
pressure
pressure A
Absolute
(vanes with
weather and
pressure B
altitude)
, 1
Fixed 1
datum
FIG. 13.
Absolute pressure = Atmospheric pressure
Vacuum
+ Gage pressure
which allows easy conversion from one system to the other. Possibly
a better picture of these relationships can be gained from a diagram
such as that of Fig. 13 in which are shown two typical pressures, A and
5, one above, the other below, atmospheric pressure, with all the rela
tionships indicated graphically.
2 Throughout the remainder of the book "pressure" should be understood to mean
"gage pressure"; when "absolute pressure" is meant, it will be designated as such.
MANOMETRY 27
ILLUSTRATIVE PROBLEM
A Bourdon gage registers a vacuum of 12.5 in. of mercury when the barometric
pressure is 14.50 lb/in. 2 Calculate the corresponding absolute pressure.
14.70
Vacuum = 12.5 X = 6.15 lb/in. 2
Absolute pressure = 14.50  6.15 = 8.35 lb/in. 2
10. Manometry. Bourdon pressure gages, owing to their inevi
table mechanical limitations, are not in general satisfactory for precise
measurements of pressure; when greater precision is required, measure
ments of the height of liquid columns of known density are commonly
used. Such measurements may be accomplished by means of "manom
eters" like those of Fig. 14, pressures being obtained by the application
of the pressuredensityheight relationships.
Consider the elementary manometer of Fig. 14a, consisting of a
glass tube connected to a reservoir of liquid. With both the reservoir
and the tube open to the atmosphere the liquid surfaces will stand on
the horizontal line 00. If the reservoir is now connected to a volume
of gas having an unknown pressure, p x , the surface of the reservoir
liquid will drop to the line 11 and that in the tube will rise to the
point 2. Since pressures over horizontal planes in a fluid are constant,
the pressure p x existing on the reservoir surface will also exist as in
dicated in the tube. Applying equation 1
Px  p2 = *
and with the tube open to the atmosphere, p 2 = and
p x = wh
thus p x may be obtained by measurement of the distance h.
Figure 146 illustrates the measurement of a pressure less than
atmospheric with the same type of manometer. Here the reservoiris
open to the atmosphere and the unknown pressure, p xt admitted to the
tube. Applying the same principles to the liquid column of height h,
pi  Px = wh
in which
Pi =0
giving
p x = wh
28
FLUID STATICS
the negative sign indicating the pressure to be less than the atmos
pheric pressure.
The familiar mercury barometer is shown in Fig. 14c. Such a
barometer is constructed by filling a glass tube, closed at one end, with
r
lr
ff
ft
M_
h
^^P
FIG. 14. Manometers.
mercury and inverting the tube, keeping its open end below the surface
of the mercury in a reservoir. The mercury column, hj will be sup
ported by the atmospheric pressure, leaving an evacuated space, con
taining only mercury vapor, in the top of the tube. Using the absolute
MANOMETRY 29
system of pressures, equilibrium of the mercury column, h, will be
expressed by
P*tm.  Pv = Wh
in which p v for mercury has been seen to be negligible at ordinary
temperatures (Table V) resulting in
thus allowing atmospheric pressure to be easily obtained by measure
ment of the height of the mercury column.
Calculation of the pressure, p x , measured by the Utube manometer
of Fig. 14J may be obtained easily by noting that
Pi  p2
and that
Pi = Px +
giving
p x + wl =
resulting in
p x = wih wl
allowing p x to be calculated. 3
Utube manometers are frequently used to measure the difference
between two unknown pressures p x and p y , as in Fig. 14e. Here,
as before,
P* = Pd
and
p4 = Px +
p5 = py +
giving
Px + v>il = Py +
and
Px Py = ^2^2 +
thus allowing direct calculation of the pressure difference, p x p y .
" Differential manometers*' of the above type are frequently made with
the Utube inverted, a liquid of small density existing in the top of the
inverted U; the pressure difference measured by manometers of this
3 The use of formulas for manometer solutions is not recommended until experi
ence has been gained in their limitations.
30 FLUID STATICS
type may be readily calculated by application of the foregoing prin
ciples.
When large pressures or pressure differences are to be measured a
number of Utube manometers may be connected in series. Several
applications of the above principles will allow solution for the unknown
pressure or pressure difference.
There are many forms of precise manometers, two of the most
cpmmon of which are shown in Figs. 14/ and 14g. The former is the
"ordinary " draft gage" used in measuring the comparatively small
pressures in drafts of all types. Its equilibrium position is shown at
A, and when it is submitted to a pressure, p x , a vertical deflection, h,
is obtained in which p x = wh. In this case, however, the liquid is
forced down a gently inclined tube so that the manometer " deflect ion,"
/, is much greater than h and, therefore, more accurately read. The
draft gage is usually calibrated to read directly in inches of water.
The principle of the sloping tube is also employed in the alcohol
'micromanometer of Fig. 14g, used in aeronautical research work. Here
the gently sloping glass tube is mounted on a carriage, C, which is
moved vertically by turning the dial, D, which actuates the screw, S.
When p x is zero the carriage is adjusted so that the liquid in the tube
is brought to the hair line, X, and the reading on the dial recorded.
When the unknown pressure, p x , is admitted to the reservoir the alcohol
runs upward in the tube toward B and the carriage is then raised until
the liquid surface in the tube rests again at the hair line, X. The
difference between the dial reading at this point and the original one
gives the vertical travel of the carriage, h, which is the head of alcohol
equivalent to the pressure p x .
Along with the above principles of manometry the following prac
tical considerations should be appreciated: (1) manometer liquids, in
changing their specific gravities with temperature, will induce errors
in pressure measurements if this factor is overlooked ; (2) errors due to
capillarity may usually be canceled by selecting manometer tubes of
uniform size; (3) although some liquids appear excellent (from density
considerations) for use in manometers, their surfacetension effects
may give poor menisci and thus inaccurate readings ; (4) fluctuations of
the manometer liquids will reduce accuracy of pressure measurement,
but these fluctuations may be reduced by a throttling device in the
manometer line, a short length of small tube proving excellent for this
purpose; (5) when fluctuations are negligible refined optical devices
and verniers may be used for extremely precise readings of the liquid
surfaces.
FORCES ON SUBMERGED PLANE SURFACES
31
ILLUSTRATIVE PROBLEM
This vertical pipe line with attached gage and manom
eter contains oil and mercury as shown. The manometer
is open to the atmosphere. What will be the gage reading,
Since
Pi
Pr
Pi
Px
Px + (0.90 X 62.4)10
(13.55 X 62.4) yf
Oil (.90)
Mercury
(13.55)"
10'
505 lb/ft 2 =3.51 lb/in. 2
11. Forces on Submerged Plane Surfaces. The calculation of
the magnitude, direction, and location of the total forces on surfaces
submerged in a liquid is essential in the design of dams, bulkheads,
gates, tanks, etc.
For a submerged, plane, horizontal area the calculation of these
force properties is simple, because the pressure does not vary over the
FIG. 15.
area; for nonhorizontal planes the problem is complicated by pressure
variation. Pressure in liquids, however, has been shown to vary
linearly with depth (equation 1), resulting in the typical pressure
diagrams and resultant forces of Fig. 15.
Now consider the general case 4 of a plane submerged area A B, such
as that of Fig. 16, located in any inclined plane XX. Let the center of
4 A general solution for the magnitude, direction, and location of the resultant
force on this area will allow easy calculation of the forces on areas of more regular
shape.
32
FLUID STATICS
gravity of this area be located as shown, at a depth h g and at a distance
l g from the line of intersection, 00, of plane XX and the liquid surface.
Calculating the force, dF y on the small area, dA,
dF = pdA = whdA
but h = / sin a, and substituting this value for h
dF = wl sin a <L4
FIG. 16.
and the total force on the area AB will result from the integration of
this expression, giving
r A
F = w sin a / IdA (2)
C A
but I IdA is recognized as the statical moment of the area AB,
about the line 00 which is also given by the product of the area, A,
and the perpendicular distance, / g , from 00 to the center of gravity of
the area. Thus
IdA = l g A
FORCES ON SUBMERGED PLANE SURFACES 33
and substituting this in equation 2
F = wAlg sin a
but kg = l g sin a, giving
F = whgA (3)
indicating that the magnitude of the resultant force on a submerged
plane area may be calculated by multiplying the area, A, by the pres
sure at its center of gravity, wh g .
The magnitude of the resultant force having been calculated, its
direction and location must be considered. Its direction, because of
the inability of liquids to support shear stress, is necessarily normal to
the plane, and its point of application may be found if the moment of
the force can be calculated and divided by the magnitude of the force.
Referring again to Fig. 16, the moment, dM, of the force, dF, about
the line 00 is given by
dM = IdF
in which
dF = wl sin a dA
Therefore, by substitution,
dM = wl 2 sin a dA
and integrating to obtain the total moment, M,
/ A
l 2 dA
C
in which / l 2 dA is the moment of inertia of the area A , about the
^ A
/ PdA
line 00, thus
Ioo
M = wl
Designating the point of intersection of the resultant force and
the plane as the "center of pressure" and its distance from 00 as
lp, lp will be given by
i ^
h ~ F
in which
M = w/o_osin a
and
F = wl g A sin a
34
FLUID STATICS
Substituting these values above gives
o~o
wLA sin
thus locating the resultant force in respect to the line 00, and com
pleting the solution of the general problem.
The above equation may be made more usable by .placing it in
terms of the moment of inertia, / g , about an axis parallel to 00
through the center of gravity of the area. Using the equation for
transferring moment of inertia of an area from one axis to another,
/oo = I* + fa
and substituting in the equation for l p
, 
~
which may be written as
or
k =
FIG. 17.
*g
LA
(4)
allowing direct calculation of the dis
tance along the XX plane between
center of gravity and center of pres
sure. This equation also indicates
that center of pressure is always below
center of gravity except for a hori
zontal area, but that the distance
between center of pressure and center
of gravity diminishes as the depth of
submergence of the area is increased. 6
The lateral location of the center
of pressure for regular plane areas,
such as that of Fig. 17, is readily cal
culated by considering the area to be
composed of a large number of rec
6 This fact allows the approximation made for small areas under great submer
gence, or pressure, that the resultant force acts at their centers of gravity,
FORCES ON SUBMERGED CURVED SURFACES 35
tangles of differentially small height, dh. The center of gravity and
center of pressure of each of these small rectangles will be coincident
and at the center of the rectangle, and, therefore, all the forces on
these rectangles will act on the median line AB. The resultant of
these forces must also act on the median line. The vertical and
lateral location of the center of pressure of areas of more irregular
form may be obtained by dividing the area into regular areas, locat
ing the forces on these, and finding the location of the resultant of
these forces by taking moments about any convenient axis. The point
where the line of action of the resultant force pierces the area is the
center of pressure of the whole area.
ILLUSTRATIVE PROBLEM
A circular gate 8 ft in diameter lies in a plane sloping 60 with the horizontal.
If water stands above the center of the gate to a depth of 10 ft, calculate the mag
nitude, direction, and location of the total force exerted by water on gate.
Direction: normal to gate
Magnitude:
F = whgA a
F = 62.4 X 10 X  (8) 2 = 31,400 Ib
4
Location:
7 *  71 W Mw ft "
64
l g = = 11.55ft
0.866
A =  (8) 2 = 50.3 ft 2
4
I* 647T
IP Ig = = = 0.346 ft
l g A 11.55 X 50.3
Therefore force passes through a point (c.p.) located 0.346 ft below the center of
gravity measured down the plane.
12. Forces on Submerged Curved Surfaces. The total forces on
submerged curved areas cannot be calculated by the foregoing methods.
These forces may be readily obtained, however, by calculating the
horizontal and vertical components of the forces as indicated below.
The curved area, AB, of Fig. 18a is exposed to liquid pressure on
its upper and lower surfaces. Obviously, the vertical component of
36
FLUID STATICS
the pressure force is downward on the upper surface, upward on the
lower surface, and these two force components have the same line of
action and the same magnitude. If the liquid vertically above the
area AB is isolated by drawing the lines BC and AD, it becomes appa
rent that no vertical force can be transmitted across these lines because
of the inability of the liquid to support shear stress. Hence, the
vertical component of force on the area AB is simply the weight of
liquid, A BCD, thus
FV = WABCD
and the line of action of this force will pass through the center of gravity
of ABCD.
(a)
F v ~ W ABCD
(c)
The horizontal component of force may be readily established by
considering the horizontal equilibrium of the mass of liquid ABE, EB
being the projection of AB on a vertical plane. If FH is the horizontal
component of force exerted by the area on the liquid ABE horizontal
equilibrium is expressed by
F'H = FED
in which the magnitude, direction, and location of FEE may be calcu
lated by the methods of Art. 11. The horizontal component of force
exerted by the liquid on the area AB will have the direction, magnitude,
and line of action of FEE* and the resultant force, F, may be obtained
by composition of the horizontal and vertical components as indicated
in the two typical cases of Figs. 186 and 18c.
ILLUSTRATIVE PROBLEM
37
ILLUSTRATIVE PROBLEM
Calculate magnitude, direction, and location of the total force exerted by the
water on the area AB which is a quarter of a circular cylinder and is 8 ft long
(normal to plane of paper).
Horizontal component, FH
Direction: Horizontal to right
Magnitude:
FH =
FH = 8 X 5 X 62.4 X 12.5 = 31,200 Ib
?
c
1
1
'
h
i
A> i
^
W
~ W AODC
A
T ^tp
If 1 1 .424r2.12'fi
'4
kr.5'
WAOB
.166
t 1 ! _
1F H
Location:
AA*B* = 8 X5 = 40ft 2
8 X5 3
12
83.2
12.5 X 40
83.2 ft 4
= 0.166ft
Vertical component , Fv
Direction: Vertically downward
Magnitude:
Fy = WAODC + WAOB
F v = 10 X 5 X 8 X 62.4
X 8 X 62.4  34,750 Ib
Location: e = distance, between FV and line DB, and taking moments about
point B
34,750 X e  24,950 X 2.5 f 9800 X 2.12
e  2.40 ft.
38
FLUID STATICS
31,200 Total force F:
Direction: Downward to right
48 with horizontal
34,750
Magnitude: F = 1000 A/31.2 2 + 34.7S 2 = 46,600 Ib
Location: Through a point located 2.334 ft above B
and 2.40 ft to the right of B
13. Buoyancy and Flotation. The familiar principles of buoyancy
(Archimedes' principle) and flotation are usually stated Respectively:
(1) A body immersed in a fluid is buoyed up by a force equal to the
weight of fluid displaced by the body;
and (2) a floating body displaces its
own weight of the fluid in which it
floats. These principles may be easily
proved by the methods of Art. 12.
A body, A BCD, suspended in a
liquid of specific weight, w, is illustrated
on Fig. 19. After the vertical lines AE
and CF are drawn, it is obvious that
the force F\, acting vertically downward
on the upper surface A DC, is given by
F l = ^(Volume ADCFE)
and F 2 , the force upward on the lower
surface ABC, by
F 2 = ^(Volume ABCFE)
The net vertical force, F&, exerted by the liquid on the body is upward
and given by
FB = 7*2 FI
or
F B = ^(Volume ABCFE  Volume ADCFE)
Performing the indicated subtraction results in
FB = ^(Volume A BCD) = w( Volume of body)
thus "the buoyant force is equal to the weight of fluid displaced by the
body." The vertical equilibrium of the body is expressed by
F + F B  W =
For a floating body (Fig. 20) vertical equilibrium is expressed by
F B  W =
BUOYANCY AND FLOTATION
39
and the vertical component of force on the immersed area, A BCD,
will (Art. 12) be given by
thus
F B = ^(Volume ABCD)
W = ^(Volume ABCD)
1W of body
e.g. of body
and the body "displaces its own weight of the fluid in which it floats."
The above principles find many applications in engineering, such as
calculations: of the draft of sur
face vessels; of the weight of a
ship's cargo from the increment
in depth of flotation; of the lift
of balloons; etc. \ F
The stability of submerged or P 20
floating bodies is dependent upon
the relative location of the buoyant force and the weight of the body.
The buoyant force acts upward through the center of gravity of the
displaced volume; the weight acts downward at the center of gravity
FIG. 21.
of the body. Stability or instability will be determined by whether
a righting or overturning moment is developed when the center of
gravity and center of buoyancy move out of vertical alignment. Obvi
ously, for the submerged bodies, such as the balloon and submarine
of Fig. 21, stability requires the center of buoyancy to be above the
40 FLUID STATICS
center of gravity. In surface vessels, however, the center of gravity
is usually above the center of buoyancy, and stability exists because
of movement of the center of buoyancy to a position outboard of the
center of gravity as the ship "heels over," thus producing a righting
moment. An overturning moment, resulting in capsizing, occurs if
the center of gravity moves outboard of the center of buoyancy.
ILLUSTRATIVE PROBLEM
A ship has a crosssectional area of 4000 ft 2 at the water line when the draft
is 10 ft. How many pounds of cargo will increase the draft 2 in.? Assume salt
water.
Since the ship floats, the weight of water displaced by the cargo equals the
weight of the cargo. Therefore
Weight of cargo = 4000 X A X 64.0 = 42,700 Ib
14. Stresses in Circular Pipes and Tanks. The circumferential
tension stresses in pipes and tanks under pressure may be readily
calculated if pressure variation is
neglected and if the thickness of
the pipe or tank is small com
pared to the diameter.
A section of pipe of length /,
having an internal diameter d, is
shown in Fig. 22. This pipe con
tains a fluid whose pressure is p;
the circumferential tension stress
in the walls is s t . Pass a vertical
plane through the center of the pipe and consider the horizontal
equilibrium of the forces acting on the section of pipe to the right of
this plane. This equilibrium is expressed by
2T  pdl =
in which T is the total tension force in the wall of length / and thick
ness t due to the stress s^ and therefore is given by
T = s t tl
Substituting this in the equation above
2s t tl = pdl
\pd
FLUID MASSES SUBJECTED TO ACCELERATION
41
thus allowing the wall stress to be calculated when the internal pres
sure and the dimensions of the pipe or tank are known. The final
equation also indicates that the stress caused by a given pressure may
be reduced by decreasing the diameter d, or increasing the wall thick
ness t. Since increasing the wall thickness increases the cost it be
comes evident why smallbore tubing is in general use in highpressure
work.
A mechanical analysis similar to the above may be applied in the
design of wooden tanks or pipes where the tension is carried by external
circumferential hoops, and in the design of concrete structures of this
type where the tension stress is carried by circumferential reinforcing
rods. t
15. Fluid Masses Subjected
to Acceleration. Fluid masses
may be subjected to various
types of acceleration without
relative motion occurring be
tween fluid particles or between
fluid particles and boundaries.
Such fluid masses will be found
to conform to the laws of fluid statics, modified to account for the
inevitable inertia forces which exist when acceleration occurs.
An open container of liquid subjected to a vertical upward accelera
tion, a (Fig. 23), will contain greater pressures than if the liquid is
at rest, owing to the forces of inertia which act in the opposite direction
to that of acceleration. The general relationships for this type of
accelerated motion may be obtained by considering the vertical
equilibrium of a cylinder of fluid, of height A, crosssectional area A,
and having its upper base in the liquid surface.
The force F, upward on the bottom of the cylinder due to pressure,
will balance the forces of inertia Fj and weight W, acting downward.
Therefore, __
in which ,, . ,
r = pA
W = Mg
FI = M a
in which M, the mass of the liquid cylinder, is expressed by
^Static pressure pwk
FlG. 23.
M = phA =
g
42
FLUID STATICS
and substituting these values in the first equation
and solving for p
w w
pA =  hAg +  hAa
g g
p =
wh
indicating that pressure variation with depth is linear and that the
pressure at any point will be given by the product of (  ) and the
static pressure, wh, at the point. These facts are indicated graphically
in Fig. 23.
Similarly it may be shown that, for a fluid mass undergoing a ver
tical downward acceleration a, pressure p, at a depth h, will be given by
p =
If a = g in this equation, the pressure becomes zero, showing that a
freely falling unconfined fluid mass exerts no pressure, a fact which
will have many applications in subsequent problems.
Horizontal acceleration of a liquid mass in an open container is
indicated in Fig. 24, acceler
ation of this kind causing the
liquid surface to drop at the
front of the tank and to
rise at the rear. The forces
which a liquid particle at
the surface will exert on its
neighboring particles will be
its weight, W, acting down
ward, and its inertia force,
Fj, acting horizontally and
in a direction opposite to
that of acceleration. How
ever, for a liquid surface to be stable the resultant of these forces
must act normal to the liquid surface; thus, referring to Fig. 24, the
liquid surface will stand at an angle, 0, with the horizontal, and
FIG. 24.
Ma a
FLUID MASSES SUBJECTED TO ACCELERATION 43
proving that the liquid surface and other lines of constant pressure
are straight lines having a slope a/g.
A fruitful means of examining this problem further is to imagine
the magnitude and direction of the acceleration due to gravity, g, to
be changed to those of g f . When this is done the pressurevariation
problem may be analyzed as one of simple fluid statics in which the
relation of pressure, p, to depth, /, along the direction of g f will be
given by
p = w'l
in which w' is the apparent specific weight of the liquid in a system
where acceleration due to gravity is g' instead of g. The density of the
liquid, p, is the same in both systems and, therefore,
w w f
from which
/
Substituting above
p = (^)wl (5)
but from similar triangles
and substitution of this relation in equation 5 gives the familiar
equation
p = wh
The fact that this equation applies to fluid masses while they are being
accelerated horizontally means that the total forces on vertical areas,
such as the ends of the container of Fig. 24, may be calculated by the
principles of Art. 11. These forces are indicated as F and F%, and if
the mass of fluid in the container is designated by M, it will be found
that
FI  F 2 = Ma
the unbalanced force, FI F 2j being equal to the product of mass and
acceleration, thus checking the Newtonian relationship, F = Ma.
44
FLUID STATICS
ILLUSTRATIVE PROBLEM
A rectangular tank 15 ft long, 5 ft high, and 8 ft wide is filled with water and
accelerated along the direction of its length, at 6 ft/sec 2 . Calculate the volume of
water spilled, and check the equal
y ity of force to accelerate the final
I mass and the force exerted by
liquid on the ends of the tank.
15'
a =6 Ft. per sec. 2
32.2
X 15 = 2.80 ft
Volume spilled = J X 2.80 X 15 X 8 = 168 ft 3
Force, F, for accelerat ion = Ma
s 62.4
F = (5 X 15 X 8  168) X 6
o L ,L
5020 Ib
F l = 5 X 8 X 2.5 X 62.4 = 6230 Ib
2.2
F 2 = X 2.2 X 8 X 62.4 = 1210 Ib
p l _ F 2 = 6230  1210 = 5020 Ib (Check)
Fluid masses subjected to rotation at constant angular velocity
will contain pressure variations unlike those in fluids at rest, because
of the centrifugal forces exerted by fluid particles. The centrifugal
force exerted by a mass M, rotating about an axis with circumferential
velocity V, at a radius r, is given by
(6)
^ '
in which V 2 /r is termed the "centrifugal acceleration."
The simple case of fluid rotation about a vertical axis is indicated
FIG. 25.
in Fig. 25 where the forces on a small element of fluid are considered.
Let this element have dimensions dx, dy, and dz and rotate with the
circumferential velocity, v, at a distance x from the axis of rotation Z.
FLUID MASSES SUBJECTED TO ACCELERATION 45
Since the mass of fluid does not move in the radial direction an equi
librium of forces exists in this direction which may be stated as
Centrifugal force == Centripetal force
in which the centripetal force can result only from pressure variation
along x. Taking the pressure on the inner vertical face of the element
to be p and assuming the pressure to increase with x, the pressure on
the outer face will be
dp
p + dp or p + ~dx
dx
thus
Centripetal force = ( p H dx ) dy dz p dy dz
\ dx /
= dx dy dz
dx
The centrifugal force may be calculated from equation 6 in which,
for the fluid element,
w
M = p dx dy dz = dx dy dz
g
V = v cox
r = x
co being the angular velocity of rotation. Therefore
Centrifugal force dx dy dz
g x
Equating centrifugal and centripetal forces,
dp . , , w (ux) 2
dx dy dz dx dy dz
dx g x
results in
dp w 2
 = o> x
dx g
a differential equation expressing the variation of pressure in the radial
direction for fluid masses subjected to constant angular velocity. Since
dp/dx is a positive quantity it may be concluded directly that pressure
will increase as radius increases. This equation may be integrated for
practical use between the eixis of rotation where x = and p = p c
46
FLUID STATICS
and any point, x, where a pressure, p, exists. Separating the variables
and integrating,
{***= f X "
J<P C w Jo
g
gives
w w
(7)
the physical meaning of which is given on Fig. 26.
Figure 26a shows an open cylindrical container being rotated
about a central vertical axis. The terms p/w and p c /w are recognized
*L
w \
^
(a)
FIG. 26.
as the "heads" of liquid corresponding respectively to the pressures p
and p c . The equation
y = w ~ w
is obvious from the figure and hence equation 7 becomes
_ 2
y Y g x
indicating the liquid surface to be of paraboloidal form. Since no
acceleration exists in a vertical direction the variation of pressure with
depth (as for horizontal linear acceleration) will follow the static law
p ^ wh
which is indicated on the figure.
ILLUSTRATIVE PROBLEM
47
Figure 26b illustrates a closed container filled with liquid having
an initial pressure before rotation begins. This initial pressure, p$, at
some point in the liquid is represented by the corresponding head of
liquid PQ/W. Since there is no velocity on the axis of rotation, no
increase in pressure will exist here and the parabolic curve of pressure
variation becomes as indicated on the figure.
The foregoing analysis indicates that pressure
may be created by the rotation of a fluid mass.
This principle is utilized in centrifugal pumps
and blowers to create pressure in order to cause
fluids to flow.
ILLUSTRATIVE PROBLEM
A vertical cylindrical tank 5 ft high and 3 ft in
diameter is filled with water to a depth of 4 ft. The
tank is then closed and the pressure in the space above
the water surface raised to 10 lb/in. 2 What pressure
will exist at the intersection of wall and tank bottom
when the tank is rotated about a central vertical axis
at 150 rpm?
Since no liquid escapes, space above liquid remains
same.
Volume of this space =  X (3) 2 X 1
7.08 ft 3
7.08
but
Therefore
in which
Solving,
7.08
2?r
co = 150 X = 15.7 rad./sec.
ou
xi = 1.044ft, yi = 4.16ft
15/7 2 X D?
y 2 = = 8.60 ft
p = (8.60 + 0.84) 62.4 + 10 X 144 = 2029 lb/ft 2
2029
P = 14.11b/in ?
144
48 FLUID STATICS
PROBLEMS
42. Calculate the pressure in an open tank of benzene at a point 8 ft below the
surface.
43. If the pressure 10 ft below the free surface of a liquid is 20 lb/in. 2 , calculate its
specific weight and specific gravity.
44. If the pressure at a point in the ocean is 200 lb/in. 2 , what is the pressure 100 ft
below this point? Specific weight of salt water is 64.0 lb/ft 3 .
45. An open vessel contains carbon tetrachloride to a depth of 6 ft and water on
the CCU to a depth of 5 ft. What is the pressure at the bottom of the vessel?
46. How many inches of mercury are equivalent to a pressure of 20 lb/in. 2 ?
How many feet of water?
47. One foot of air at 60 F and 14.7 lb/in. 2 is equivalent to how many pounds per
square inch? inches of mercury? feet of water?
48. The barometric pressure at sea level is 30.00 in. of mercury when that on a
mountain top is 29.00 in. If the air temperature is constant at 60 F and the specific
weight of air is assumed constant at 0.075 lb/ft 3 , calculate the elevation of the moun
tain top.
49. If atmospheric pressure at the ground is 14.7 lb/in. 2 and temperature is 60 F,
calculate the pressure 10,000 ft above the ground, assuming (a) no density variation;
(b) an isothermal variation of density with pressure; (c) an adiabatic variation of
density with pressure.
50. Prove that the depth of an assumed isothermal atmosphere is infinitely great.
51. Calculate the depth of an adiabatic atmosphere if temperature and pressure
at the ground are respectively 60 F and 14.7 lb/in. 2
52. With atmospheric pressure at 14.5 Ib/in. 2 , what absolute pressure corresponds
to a gage pressure of 20 lb/in. 2 ?
53. When the barometer reads 30 in. of mercury, what absolute pressure corre
sponds to a vacuum of 12 in. of mercury?
54. If a certain absolute pressure is 12.35 lb/in. 2 , what is the corresponding
vacuum if atmospheric pressure is 29.92 in. of mercury?
55. A Bourdon pressure gage attached to a closed tank of air reads 20.47 lb/in. 2
with the barometer at 30.50 in. of mercury. If barometric pressure drops to 29.18 in.
of mercury, what will the gage read?
56. The compartments of these tanks are closed
and filled with air. Gage A reads 30 lb/in. 2 Gage B
registers a vacuum of 10 in. of mercury. What will
gage C read if it is connected to compartment 1 but
inside of compartment 2? Barometric pressure is 14.6
lb/in. 2
57. Assuming the liquid in Fig. 14a to be mercury and h to be 16 in., calculate
the pressure p x .
58. Calculate the pressure p x in Fig. 146 if the liquid is carbon tetrachloride and
h is 8.0 feet.
59. If the barometer of Fig. Uc is filled with ether (S = 0.94) at 68 F, calculate
h if the barometric pressure is 14.7 lb/in. 2
60. Calculate the height of the column of a water barometer for an atmospheric
pressure of 14.5P lb/in. 2 if the water is at 50 F (w = 62.42 lb/ft 8 ); at 150 F (w
61.15 lb/ft 3 ; j at 212 F (w = 59.83 lb/ft 8 ).
61. Barometric pressure is 29.43 in. of mercury. Calculate h.
PROBLEMS
49
62. Calculate the pressure p x in Fig. 14d if / =30 in., h = 20 in.; liquid w is water
and w\ mercury.
63. With the manometer reading as shown, calculate p x .
Oi
9.52" Mercury
vacuum
PROBLEM 61.
PROBLEM 63.
Mercury."*
PROBLEM 64.
64. The specific gravity of the liquid in the left side of this open Utube is un
known. Calculate it.
65. In Fig. 140, h = 50 in., h = 20 in., lz = 30 in., liquid w\ is water, wz benzene,
W3 mercury. Calculate p x p y .
r
65"
Oil (S=.90)
10"
.Water 
Wate
PROBLEM 66.
^Mercury'
PROBLEM 67.
66. Calculate p x p y for this inverted Utube manometer.
67. Two manometers as shown are connected in series. Calculate p x
Mercury
Mercury
PROBLEM 69.
PROBLEM 70.
68. An inclined gage having a tube of 1/8in. bore laid on a slope of 1 : 20, and
reservoir of 1in. diameter, contains linseed oil. What distance will the oil move
along the tube when a pressure of 1 in. of water is connected to the gage?
69. Calculate the gage reading.
70. Calculate the gage reading. Specific gravity of oil is 0.85. Barometric pres
sure is 29.75 in. mercury.
so
FLUID STATICS
71. A rectangular gate 6 ft long and 4 ft high lies in a vertical plane with its center
7 ft below a water surface. Calculate magnitude, direction, and location of the total
force on the gate.
72. A circular gate 10 ft in diameter has its center 8 ft below a water surface and
lies in a plane sloping at 60. Calculate magnitude, direction, and location of total
force on this gate.
73. A triangular area of 6ft base and 5ft altitude has its base horizontal and lies
in a 45 plane with its apex 9 ft below a water surface. Calculate magnitude, direc
tion, and location of total force on this area.
74. A square 9 ft by 9 ft lies in a vertical plane. Calculate the distance between
Water
_i
Water
^4'^
5'
1
4'
V"
3'
\
PROBLEM 75.
PROBLEM 76.
8'
PROBLEM 77.
the center of pressure and the center of gravity, and the total force on the square,
when its upper edge is (a) in the water surface and (b) 50 ft below the water surface.
75. Calculate the vertical and lateral location of the center of pressure of this
triangle, which is located in a vertical plane.
76. Calculate magnitude and location of the total force on this vertical plane area.
77. Calculate magnitude and location of the total force on this vertical plane area.
78. An 8 ft square gate lies in a vertical plane. If one diagonal of this gate is
vertical and its center is 10 ft below a water surface, calculate magnitude, direction,
and location of the total force on the gate.
Hinge
Sill
PROBLEM 80.
_20' J
PROBLEM 82.
79. A vertical rectangular gate 10 ft high and 6 ft wide has a depth of water on
its upper edge of 15 ft. What is the location of a horizontal line which divides this
area (a) so that the forces on the upper and lower portions are the same; (&) so that
the moments of these forces about the line are the same?
80. This rectangular gate is hinged at the upper edge and is 4 ft wide. Calculate
the total force on the sill, neglecting weight of the gate.
81. The center of pressure of an isosceles triangle of 9ft altitude and 6ft base,
lying in a vertical plane, is at a depth of 12 ft. Calculate the depth of water over the
apex of the triangle.
PROBLEMS
51
82. Is this concrete dam (w = 150 lb/ft 8 ) safe against overturning? Neglect
uplift.
83. Water will rise behind a 10ft concrete wall of rectangular cross section to a
depth of 6 ft. How thick must the wall be to prevent overturning? w for concrete
= 150 lb/ft 3 .
84. A sliding gate 10 ft wide and 5 ft high situated in a vertical plane has a coeffi
cient of friction between itself and guides of 0.20. If the gate weighs 2 tons and if
its upper edge is at a depth of 30 ft, what vertical force is required to raise it? Neglect
buoyancy of the gate.
85. A butterfly valve, consisting essentially of a circular area pivoted on a hori
zontal axis through its center, is 7 ft in diameter and lies in a 60 plane with its center
10 ft below a water surface. What torque must be exerted on the valve's axis to
just open it?
86. A rectangular tank 5 ft wide, 6 ft high, and 10 ft long contains water to a depth
of 3 ft and oil (S 0.85) on the water to a depth of 2 ft. Calculate magnitude and
location of the force on one end of this tank.
PROBLEM 87.
PROBLEM
PROBLEM 89.
87. This 6 ft by 6 ft square gate is hinged at the upper edge. Calculate the total
force on the sill.
88. This wicket dam is 15 ft high and 4 ft wide and is pivoted at its center.
Assume a hydrostatic pressure distribution, and calculate the vertical and horizontal
reactions at the two sills. Neglect weight of the dam and consider all joints to be
pin connected.
PROBLEM 91.
PROBLEM 92.
PROBLEM 93.
89. The flashboards on a spillway crest are 4 ft high and supported on steel posts
spaced 2 ft on centers. The posts are designed to fail under a bending moment of
5000 ftlb. What depth over the flashboards will cause the posts to fail? Assume
hydrostatic pressure distribution.
90. Calculate the horizontal and vertical components of the force on a gate 4 ft
square, located in a 60 plane, and having its upper edge 10 ft below a water surface.
52
FLUID STATICS
91. Using force components, calculate the load in the strut AB, if these struts
have 5ft spacing along the small dam A C. Consider all joints to be pin connected.
92. Calculate the magnitude, direction, and location of the total foice on the up
stream face of a section of this dam 1 ft wide. What is the moment of this force
about 0?
93. This tainter gate is pivoted at and is 30 ft long. Calculate the horizontal
and vertical components of force on the face of the gate.
94. A concrete pedestal having the shape of the frustum of a right pyramid of
lower base 4 ft square, upper base 2 ft square, and height 3 ft, is to be poured. Taking
the specific weight of concrete as 150 lb/ft 3 , calculate the vertical force of uplift on the
forms.
95. A hemispherical shell 4 ft in diameter is connected to the vertical wall of a
tank containing water. If the center of the shell is 6 ft below the water surface,
what are the vertical and horizontal force components on the shell?
96. This half conical buttress is used to support a half
cylindrical tower on the upstream face of a dam. Calculate
the vertical and horizontal components of force exerted by the
water on the buttress.
Water
5'R.
50'
PROBLEM 96.
PROBLEM 97.
40' \
PROBLEM 103.
97. This rectangular tank 10 ft wide has a quarter cylinder AB joining its end and
bottom. Calculate the magnitude, direction, and location of the total force on AB.
98. A 12in. diameter hole in a vertical wall between two water tanks is closed
by an 18in. diameter sphere in the tank of higher water surface elevation. The dif
ference in the water surface elevations in the two tanks is 5 ft. Calculate the force
exerted by the water on the sphere. Neglect buoyancy.
99. A stone weighs 60 Ib in air and 40 Ib in water. Calculate its volume and
specific gravity.
100. If the specific gravity of ice is 0.90, what percentage of the volume of an
iceberg will remain above sea water (S = 1.025)?
101. Six cubic inches of lead (S 11.4) arc attached to the apex of a conical can,
having an altitude of 12 in. and a base of 6in. diameter, and weighing 1 Ib. When
placed in water, to what depth will the can be immersed?
102. A cylindrical can 3 in. in diameter and 7 in. high weighing 4 oz contains
water to a depth of 3 in. When this can is placed in water, how deep will it sink?
103. The barge shown weighs 40 tons and carries a cargo of 40 tons. Calculate its
draft in fresh water.
104. A barge having water line area of 2000 ft 2 sinks 2 in. when a certain load is
added. Calculate the load.
105. A balloon having a total weight of 800 Ib contains 15,000 ft 3 of hydrogen.
How many pounds of ballast are necessary to keep the balloon on the ground? Baro
metric pressure = 14.7 lb/in. 2 Temperature of air and hydrogen, 60 F. Pressure
of the hydrogen is the same as that of the atmosphere.
PROBLEMS 53
106. A concrete (w  150 lb/ft 8 ) slab 2 ft thick and 10 ft square is dragged on
rollers up a 30 incline under water, by a force exerted parallel to the incline. Calcu
late this force if the coefficient of friction between slab and incline may be taken as
0.05.
107. Calculate the tension stress in the \ in. wall of a 24in. steel pipe containing
fluid under a pressure of 100 lb/in. 2
108. What is the minimum thickness allowable for a 5ftdiameter steel pipe line
to carry a fluid at a pressure of 150 lb/in. 2 ? Take allowable tension stress for steel to
be 16,000 lb/in. 2
109. This wood stave pipe line is to withstand a pressure of
50 lb/in. 2 Calculate the stress developed in the steel hoops.
110. A 5ftdiameter wood stave pipe line is to carry water
under a 60ft head. If allowable steel stress is taken as 16,000
lb/in. 2 , what spacing of f in. diameter steel hoops will be
necessary?
111. A concrete water tank 60 ft high and 20 ft in diam
eter is reinforced with 1 in. diameter steel hoops spaced 3 in.
on centers. When the tank contains water 50 ft deep, what ^% * Dia. steel hoops
is the stress in the hoops, assuming that the concrete takes centers
no tension?
112. An open cylindrical container containing 1.0 cu ft of water at a depth of
2 ft is accelerated vertically upward at 20 ft/sec. 2 Calculate the pressure and total
force on the bottom of the container. Calculate this total force by application of
Newton's law.
113. The container of the preceding problem is accelerated vertically downward
Y ,, , / at 20 ft/sec. 2 Calculate pressure and total force on the
] / bottom of the container.
4' Water / 60 o 114. Calculate the total forces on the ends and
1 p * bottom of this container when at rest and when being
^ p ... accelerated upward at 10 ft /sec. 2 Container is 5 ft
wide.
115. A closed tank 10 ft high is filled with water, and the pressure at the top of
the tank raised to 30 lb/in. 2 Calculate the pressure at top and bottom of this tank
when accelerated vertically downward at 15 ft /sec. 2
116. An open conical container 6 ft high is filled with water and moves vertically
downward with a deceleration of 10 ft/sec. 2 Calculate the pressure at the bottom of
the container.
117. A rectangular tank 5 ft wide, 10 ft long, and 6 ft deep contains water to a
depth of 4 ft. When it is accelerated horizontally at 10 ft/sec. 2 in the direction of its
length, calculate the depth of water at each end of the tank and the total force on
each end of the tank. Check the difference between these forces by calculating the
inertia force of the accelerated mass.
118. When the tank of the preceding problem is accelerated at 15 ft /sec 2 how
much water is spilled? Calculate the forces on the ends of the tank, and check as
indicated above.
119. The above tank is accelerated at 20 ft /sec 2 . Calculate the water spilled and
the forces on the ends, and check as indicated above.
120. The tank of problem 117 contains water to a depth of 2.0 ft and is accelerated
horizontally to the right at 15 ft /sec 2 . Calculate the depth at the left end of the tank.
54 FLUID STATICS
121. A closed rectangular tank 4 ft high, 8 ft long, and 5 ft wide is filled with
water, and the pressure at the top raised to 20 lb/in. 2 Calculate the pressures in the
corners of this tank when it is accelerated horizontally along the direction of its length
at 15 ft /sec 2 . Calculate the forces on the ends of the tank, and check their difference
by Newton's law.
122. The tank of the preceding problem is f full of water, and the pressure in the
air space above the water is 20 lb/in. 2 Calculate the pressures and forces, and check
as indicated above for the same acceleration.
123. An open container of liquid accelerates down a 30 inclined plane at 15
ft /sec 2 . What is the slope of its free surface?
124. An open container of liquid accelerates up a 30 inclined plane at 10 ft/sec 2 .
What is the slope of its free surface?
125. When this Utube containing water is accelerated hori
zontally to the right at 10 ft /sec 2 , what are the pressures at A, B,
and C?
126. An open cylindrical tank 3 ft in diameter and 5 ft deep is
filled with water and rotated about its axis at 100 rpm. How
much liquid is spilled? What are the pressures at the center of
the bottom of the tank and at a point on the bottom 1 ft from the
center?
127. The above tank contains water to a depth of 3 ft. What
will be the depth at the wall of the tank when rotated at 60 rpm?
128. The above tank contains water to a depth of 4 ft and is rotated at 100 rpm.
How much water is spilled?
129. The above tank contains water to a depth of 1 ft. At what speed must it
be rotated to uncover a bottom area 1 ft in diameter?
130. The above tank is filled with water and closed, the pressure at its top is
raised to 20 lb/in. 2 , and the tank is rotated at 200 rpm. Calculate the pressure on the
axis and at the wall on the top and bottom of the tank. If the tank is of steel (allow
able tension stress 16,000 lb/in. 2 ) how thick must its walls be?
131. A tube 2 in. in diameter and 4 ft long is filled with water and closed. It is
then rotated at 150 rpm in a horizontal plane about one end as a pivot. Calculate
the pressure on the outer end of the tube using the equation in the text, and check by
calculating the centrifugal force of the rotating mass.
132. The pressure at a point 12 in. from the axis of rotation in a closed filled vessel
of mercury is 100 lb/in. 2 before rotation starts. What will this pressure become
when the vessel is rotated at 500 rpm?
133. The impeller of a closed filled centrifugal water pump is rotated at 1750 rpm.
If the impeller is 2 ft in diameter, what pressure is developed by rotation?
134. When the Utube of problem 125 is rotated at 200 rpm about its central axis,
what are the pressures at A, B, and C?
CHAPTER III
THE FLOW OF AN IDEAL FLUID
An insight into the basic laws of fluid flow can best be obtained
from a study of the flow of a hypothetical ''ideal fluid." An ideal or
"perfect" fluid is a fluid assumed to have no viscosity. In such a fluid,
therefore, there can be no frictional effects between moving fluid layers
or between these layers and boundary walls, and, thus, no cause for
eddy formation or energy loss due to friction and turbulence. The
assumption of an ideal fluid allows a fluid to be treated as an aggrega
tion of small spheres which will support pressure forces normal to their
surfaces, but will slide over one another without resistance. Thus,
the motion of these ideal fluid particles becomes analogous to the
motion of a solid body on a resistanceless plane, and leads to the
FIG. 27.
conclusion that unbalanced forces existing on particles of an ideal
fluid will result in the acceleration of these particles according to
Newton's law.
By assuming an ideal fluid, simple equations can be derived based
on familiar physical concepts. Later these equations must be modified
to suit the flow of real fluids after an understanding of the mechanics
of fluid friction is obtained.
16. Definitions. Fluid flow may be steady or unsteady. Steady
flow occurs in a system when none of the variables involved changes
with time; if any variable changes with time, the condition of un
steady flow exists. In the pipe of Fig. 27, leading from a large
reservoir of fixed surface elevation, unsteady flow exists while the valve
55
56
THE FLOW OF AN IDEAL FLUID
A is being adjusted; with the valve opening fixed, steady flow occurs.
The problems of steady flow are more elementary than those of un
steady flow and have more general engineering application ; therefore,
the latter will be omitted from subsequent treatment, with the excep
tion of certain simple principles and examples.
If the paths of ideal fluid particles in steady flow are traced , the
result is a series of smooth curves, called "streamlines," and the sketch
ing of such streamlines results in a "streamline picture." In an un
steady flow the picture changes from instant to instant, but smooth
curves may be drawn in the flow indicating the instantaneous directions
(a) Instantaneous Streamlines
(Unsteady Flow),
Streamlines in a Passage
(Steady Flow).
(c) Streamlines About
an Object.
(d) "Absolute" Streamlines
About a Moving Object
FIG. 28.
of fluid particles; such curves are called "instantaneous streamlines"
(Fig; 28a).
Streamlines may be relative or absolute, depending upon the
motion of the observer. Figures 2Sb and c illustrate relative stream
lines in a steady flow through a passage and about an object as seen
by an observer fixed to the passage or object. Such a streamline pic
ture will remain the same to the observer (1) if the fluid flows through
the passage (or about the object) or (2) the passage or object moves in
the opposite direction through fluid at rest. Fixed at a point in the
fluid as the object passes, the observer sees an unsteady flow picture
on which he may note the paths of individual fluid particles. How
ever, an instantaneous observation of fluid motion as the object passes
DEFINITIONS
57
Streamlines
results in the streamline picture of Fig. 2Sd. The lines occurring
thereon are sometimes termed ''absolute" streamlines, and move with
the moving object.
Generally, streamline pictures are of more qualitative than quan
titative value to the engineer ; they allow him to
visualize fluid flows and (as will be seen directly)
to locate regions of high and low velocity, and of
high and low pressure.
When streamlines are drawn through a closed
curve in a steady flow, these streamlines form
boundaries across which the fluid particles cannot
pass. Thus, the space between streamlines be
comes a tube, called a "streamtube," and such
a tube may be treated as if isolated from the
remaining fluid (Fig. 29).
The velocity of a particle moving along a streamline in a fluid
flow may be expressed by
dl
FIG. 29.
in which (Fig. 30) dl is the distance covered by the particle in time dt.
If the velocity changes as the particle covers the distance dl, an
acceleration, a, exists which is expressed by
dv dl dv
dv
for steady motion. In unsteady flow,
however, a change of velocity with re
spect to time occurs, not only because
FIG. 30. the particle moves from one point to
another, but also because the whole flow
picture is changing and at any point a change of velocity occurs with
time. If this latter change of velocity with respect to time is desig
nated by dv/dt, the total acceleration is given by
(8)
in which the first term is called "conyective" acceleration, and the
second "lorql" arr^W^tinn Obviously, local acceleration is a term
peculiar to unsteady flow and vanishes from the above equation when
it is applied to steady flow.
58
THE FLOW OF AN IDEAL FLUID
17. Equation of Continuity. The application of the principle of
conservation of mass to fluid flow in a streamtube results in the
"equation of continuity/' expressing the continuity of the flow from
point to point along the streamtube.
If the crosssectional areas and
average velocities at sections 1 and
2 in the streamtube of Fig. 31 are
designated by AI, A 2 , Vi and F 2 ,
respectively, the .quantity of fluid
passing section 1 per unit of time
will be expressed by AI FI, and the
mass of fluid passing section 1 per
unit of time will be AI Vip\. Simi
larly, the mass of fluid passing sec
tion 2 will be A 2 V 2 p 2 , Obviously,
no fluid mass is being created or destroyed between sections 1 and 2,
and therefore
= A 2 V 2 ^ 2
FIG. 31.
Thus the mass of fluid passing any point in a streamtube per unit of
time is the same.
If this equation is multiplied by g, the acceleration due to gravity,
there results
= A 2 V 2 w 2 = G
giving the equation of continuity in terms of weight. The product,
G, will be found to have dimensions of pounds per second and is termed
the "weight rate of flow" or "weight flow." Its calculation will be
necessary to express concisely the rate of flow of gases, whose densities
may vary during the flow process.
For liquids, and for gases when pressure and temperature changes
are negligible, w\ w 2 , resulting in
A l V l = A 2 V 2 = Q
indicating that for fluids of constant density the product of cross
sectional area and velocity at any point in a streamtube will be the
same. This product Q, is designated as the "rate of flow" or "flow"
and will have dimensions of cubic feet per second.
The fact that the product A V remains constant along a streamtube
allows partial interpretation of the streamline picture. As the distance
EULER'S EQUATION
59
between streamlines (^4) increases, the velocity must decrease, hence
the conclusion: Streamlines widely spaced indicate regions of low
velocity; streamlines closely spaced indicate regions of high velocity.
ILLUSTRATIVE PROBLEM
Twelve pounds per minute of air flow through a 6in. diameter pipe line. If
the gage pressure in the line is 30 lb/in. 2 and the temperature 100 F, calculate the
average velocity in the line if the barometric pressure is 15.0 lb/in. 2
G = if = 0.20 Ib/sec
(30 4 15) 144
= 0.2171b/ft 3
53.3(100 + 460)
A ^YO^ft*
4\12/
G = 0.20 = AwV = 0.196 X 0.217 X V
V =4.70 ft/sec
18. Euler's Equation. By applying Newton's law to the motion
of fluid masses, Leon hard Euler (1750) laid the groundwork for the
study of the dynamics of ideal fluids. Although Kuler's original
equations were entirely general and
written in terms of the components of
force and acceleration along the three
axes, the mathematics may be sim
plified considerably by writing the
equation in the direction of motion,
that is, along a streamtube.
Consider a differentially small sec
tion of streamtube having the dimen
sions shown in Fig. 32. The forces
tending to accelerate the fluid mass
contained therein are: (1) the component of weight in the direction
of motion, and (2) the forces on the ends of the element in the
direction of motion due to pressure. Assuming that motion is in
an upward direction and that pressure and velocity increase in this
direction, the force dF w in the direction of motion due to the weight of
the element is given by
dz
dF w = pg dl dA cos a = pgdldA
dl
60 THE FLOW OF AN IDEAL FLUID
The force dFp, in the direction of motion, due to the pressure on the
ends of the element, is
dF P = pdA 
\ al J dl
The mass dM of fluid being accelerated is
dM = pdldA
and the total acceleration a (equation 8) is
dv dv
ft as y  j 
dl dt
Substituting the above values in the Newtonian equation,
dF w + dF P = (dM)a
there results
~pg dl dA ~  dl dA = r . ,
a/ dl \ dl dt
dl
Dividing by p~dA gives
dl
dp dv
+ vdv + gdz + dl = (9)
p dt
which is Euler's differential equation for unsteady fluid motion in a
streamtube. For steady motion dv/dt = and the Eulerian equation
simplifies to
dp
+ vdv + gdz = (10)
P
the fundamental equation of steady fluid motion. By dividing this
equation by g an alternate form of the equation is obtained
dp vdv
+ + dz = (10)
w g
19. Bernoulli's Equation. Euler's equation may be integrated
along the streamtube with the following result
h / vdv + I gdz = Constant
P J . J
and if the fluid is a liquid, or a gas flowing with negligible change of
density, the integrations may be carried out giving
p V 2
 + + gz = Constant
P 2
or, multiplying by p,
or, dividing by w,
BERNOULLI'S EQUATION
F 2
p + P + pgz = Constant
p V 2
r ~ h z = Constant
w 2g
61
(ID
(12)
The existence of these equations was first recognized by Bernoulli,
and they were first presented in
his Ilydrodynamica (1738), a few
years before the development of
Euler's equations.
The Bernoulli equations
impose another mathematical
condition upon flow in a stream
tube. It has already been shown
that (for an incompressible fluid)
the product of area and velocity
is everywhere constant along
a streamtube. Now from the
Bernoulli equation it becomes
evident that the sum of three
terms involving pressure, veloc
ity, and vertical elevation will
also be a constant at every point
along the streamtube.
Writing equation 12 between two points on the typical streamtube
of Fig. 33
FIG. 33.
w
2g
w
2g
(12a)
it becomes evident that the terms involved are linear distances, allowing
a simple graphical interpretation of the equation. The terms pi/w and
p2/w are the familiar " pressure heads" of fluid statics and may be rep
resented by the piezometer columns; the terms z\ and z^ the " potential
heads," are the vertical heights of sections 1 and 2 above a horizon
tal datum plane; the terms V\/2g and Fl/2g, the "velocity heads,"
represent the head due to motion of the fluid. The sum of these terms
is the same at all points in the streamtube and is equal to the vertical
distance between the upper and lower parallel lines. The Bernoulli
terms in equation 12 thus are seen to have the dimensions of feet, or,
62
THE FLOW OF AN IDEAL .FLUID
more rigorously, "feet of the fluid flowing," since w, the specific weight
of the flowing fluid, appears in one of the terms. The Bernoulli terms
in equation 11 will have the dimensions of pressure (pounds per square
foot) and are designated respectively as " pressure" or " static pres
sure," " velocity pressure," and " potential pressure."
Bernoulli's equation gives further aid in the interpretation of
streamline pictures, equations 11 and 12 indicating that, when velocity
increases, the sum of pressure and potential head must decrease. In
the usual streamline picture, the potential head varies little, allowing
the general statement : where velocity is high pressure is low. Regions
of closely spaced streamlines have been shown to indicate regions of
high velocity, and now from the Bernoulli equation these are seen
also to be regions of low pressure.
ILLUSTRATIVE PROBLEM
Ten cubic feet of water flow per second downward through this pipe line.
When the upper pressure gage reads 10 lb/in. 2 , calculate the reading of the lower
gage and the height to which water will rise in the open piezometer tubes.
10.0
= 12.7 ft/sec, F 2 
 37.5 ft/sec
Taking datum plane at section 2 and using
Bernoulli equation,
72 t T ^2
i
Pi
 + 4
w 2g
,.
' 1 r 22
w 2g
62.4
2g
23.1 + 2.5 + 14
14 .
w
^ = 17.8ft,
w
to +
w
17.8 X 62.4
144
2g
21.8 40
= 7.70 lb/in. 2
Height of column 1 = 23.1 ft
Height of column 2 = 17.8 ft
At this point it should be noted that with increase of velocity or
potential head the pressure within a flowing fluid can drop no farther
than to the absolute zero of pressure, thus placing a restriction upon
the Bernoulli equation. Such a condition is not possible in gases,
however, owing to their expansion with reduction in pressure, but
BERNOULLI'S EQUATION
63
frequently assumes great importance in the flow of liquids. In liquids
the absolute pressure can drop only to the vapor pressure of the liquid,
whereupon vaporization takes place and "cavitation" 1 may occur,
with accompanying vibration, destructive action, and other deleterious
effects.
Before the time of Bernoulli, Torricelli (1643) discovered that the
velocity of efflux, F, of a fluid from an orifice under a head h was given
theoretically by
V = V2gh
the velocity being equal to that attained by a solid body falling from
rest through a height h. Torri
celli 's theorem is now recognized
as a special case of Bernoulli's
theorem involving certain assump
tions and conditions which con
tinually appear in engineering
problems. The above equation
may be derived from Bernoulli's
equation by considering steady flow
through the reservoir and orifice of
Fig. 34. Taking section 1 at the FIG. 34.
free reservoir surface, section 2 in
the jet immedieitely outside of the orifice, and the datum plane at the
center of the orifice, Bernoulli's equation may be written as
Horizontal
datum plane
Pi
W
F?
2g
P2
w
vl
2g
But, since the tank is very large compared to the orifice, V\ will be
very small and when squared usually becomes negligible. The pressure
on the reservoir surface, pi, is atmospheric and may be taken as zero.
Atmospheric pressure surrounds the free jet, and thus the pressure in
the jet at section 2 will be zero. Obviously, zi = IIandz 2 = 0; there
fore, the Bernoulli equation becomes
Vl
giving
F 2 = V2gh
as demonstrated by Torricelli.
1 For a description of the cavitation phenomenon see Appendix V,
64
THE FLOW OF AN IDEAL FLUID
Another useful special application of the Bernoulli principle is to
the streamtube which approaches and remains adjacent to a solid
body placed in a flowing fluid (Fig. 35). Let this streamtube have an
infinitesimal cross section and be represented by the streamline AB.
Because of the interference of the body, the fluid particles moving on
the streamline AB will decelerate as they approach the body and will
FIG. 35.
temporarily come to rest at the point S, called the stagnation point;
they then will move around the contour of the body with a variation
in velocity approximately as shown on the figure. From Bernoulli's
equation 11 the pressure variation with these velocity changes will be
about as shown, and the pressure at the stagnation point, the " stag
nation pressure," p s , may be calculated from
po and V being respectively the pressure and velocity in the undis
turbed fluid ahead of the solid body. In this equation V 8 = 0;
therefore
P.P. + IP Vt (13)
ENERGY RELATIONSHIPS
65
allowing the pressure developed on the front of objects in a flowing
fluid to be readily calculated.
ILLUSTRATIVE PROBLEM
A submarine moves through salt water at a depth of 50 ft and at a speed of
15 mph. Calculate the pressure on the nose of the submarine.
pa ~ po + %pV
rt 164.0/15 X 5280\ 2
p a = 50 X 64.0 + 1 )
2 32.2 \ 3600 /
p a 3200 + 480 = 3680 lb/ft 2 = 25.6 lb/in. 2
2
Energy 1
Internal
Pressure
Velocity
Potential
Heat
energy
added
Mechanical
energy
added
Datum plane
FIG. 36.
20. Energy Relationships. The principles of dynamics and of
conservation of mass having been applied to fluid flow in a strearntube,
application of the principle of energy conservation is in order.
Figure 36 illustrates, in a qualitative manner, the various types of
energy involved in fluid flow and allows the following general energy
equation to be written
f Energy in )
j fluid at
section 1
f Energy added to j
f  fluid between sec [
I tions 1 and 2
f Energy in 
= j fluid at
[ section 2 j
The energy which a flowing fluid possesses is composed of four types :
internal energy, due to molecular agitation and manifested by tem
66
THE FLOW OF AN IDEAL FLUID
perature; and the energies due to the pressure, velocity, and height
of the fluid above datum. Heat energy may be added to or subtracted
from a flowing fluid through the walls of the tube, or mechanical
energy may be added to or subtracted from the fluid by a pump or
turbine. Thus the above equation may be written
Internal energy
+ Pressure energy
+ Velocity energy
+ Potential energy
Heat energy
and/or
Mechanical
energy added
Internal energy
j Pressure energy
f Velocity energy
+ Potential energy
L_
!
* p
ft ,
1
r~
!i
FIG. 37
H
all the energies of which must now be obtained in terms of the variables
involved in the flow process. It is most advantageous to write this
equation, not in terms of the total energies supplied to or existing in
the flowing fluid, but rather in terms of the energies existing in or
supplied to a single typical pound of the fluid. Thus, the energies of
the equation will have the dimensions
of footpounds per pound (ftlb/lb) of
the fluid flowing.
Let the internal energy contained in
a pound of fluid be I ftlb. The kinetic
energy of a pound of fluid moving with
velocity V may be calculated from
M V 2 /2, in which M = 1/g, indicating the kinetic energy to be
V 2 /2g ftlb/lb. The potential energy of a weight, W, at a height, 2,
above a plane is given by Wz\ therefore, the potential energy of 1 Ib
of fluid, z ft above the datum, will be z ftlb.
The pressure energy contained in a pound of fluid may be calculated
from the work that may be done by this pressure. If a quantity of
fluid at a pressure p is admitted to the cylinder of Fig. 37, the force
exerted on the piston is pA ; the work done as the piston moves a
distance / is pAl\ and the weight of fluid which does this work is wAL
Therefore, the pressure energy of, or work done by, a pound of fluid is
w
wAl
If the heat energy, in British thermal units, added to a pound of
fluid is designated by //, its equivalent in footpounds will be 778 E#,
since 778 ftlb are equivalent to 1 Btu. Finally, the mechanical energy
added to a pound of fluid being designated by EM ftlb/lb, the general
energy equation becomes
^
E M = / 2 +  + ^ + *2  (14)
2g
2g
ENERGY RELATIONSHIPS 67
Since this general energy equation is the basis for the solution of the
majority of engineering problems on fluid flow, its application to certain
important special cases must be examined.
If the temperature of the fluid is nearly equal to that of the sur
roundings, the heat added to or given off by the fluid is usually negli
gible. With negligible changes in temperature and density there is
negligible change in internal energy. If the fluid passes through no
pump or motor, E M will vanish from the equation. For these special
conditions, then, TI = 7 2 , EH = 0, EM = 0, Wi = w 2 = w, the general
energy equation reduces to the familiar Bernoulli equation,
fc + * f!
w 2g w 2g
Thus, this form of the Bernoulli equation is seen to be an equation of
energy as well as an equation of linear distances, a fact which is further
confirmed by the equivalence of the dimensions of the unit energy,
footpounds per pound, and of length, feet.
When a gas or vappr is the fluid flowing, the general energy equation
is written
F? 7?
7787/1 + ~ + zi + 778 H + E M = 778// 2 + ^  + z 2
in which
778/A = 7i +
Wi
778// 2 = 7 2 +
1V2
the term 77 being called the "total heat content" or the "enthalpy" of
the gas or vapor. In problems of gas or vapor flow the difference
zi z 2 is usually negligible compared to the other terms of the general
energy equation. Frequently the flow process is an adiabatic one,
occurring so rapidly that no heat energy, EH, is lost or gained. With
no pump or motor involved in the problem, the equation for the adia
batic flow of gases and vapors becomes
a familiar flow equation of thermodynamics. In vaporflow problems
7/1 and 7/ 2 are obtained in Btu/lb from tables or diagrams, and since
68 THE FLOW OF AN IDEAL FLUID
the equation applies to an adiabatic process conditions 1 and 2 must
have the same entropy.
For a perfect gas, thermodynamics shows that
i
778(7*!  H a ) =
k ~ i ki~i
,"  p 2 k J
K '
thus for the adiabatic flow of a perfect gas equation 15 Becomes
i
This equation may also be obtained directly from Euler's equation
dp
+ vdv + dZ = (10)
w
by the assumption of an adiabatic process in Fig. 32. Neglecting the
third term of the equation and integrating
w
gives
vl v\
and the righthand side of the equation may be evaluated by the
insertion of the adiabatic relation of p and w
A = A
w k wi k
The result is i
as before.
The calculation of the power P of a machine which supplies a unit
energy EM to a flowing fluid is an important engineering problem and
may be accomplished readily as follows: The number of pounds of
fluid flowing per second will be given by G or Qw, and the energy in
footpounds given to every pound of the flowing fluid is EM Obvi
ously then
P = GE M = QwE M (17)
FLOW OF AN INCOMPRESSIBLE FLUID
69
therefore
Horsepower of machine
GE
M
550 550
550
(18)
ILLUSTRATIVE PROBLEM
How many horsepower must theoretically
be supplied to this pump to maintain a flow
of 2.0 cfs under the given pressure conditions?
Consider flow from point 1 to point 2, taking
datum plane through point 1.
PI = 
X 14.7 X 144 =  425 lb/ft 2
29.92 (6" Vacuum
2.0
7T
4
0,
5.72 ft/sec
43 ft, p 2 = 0, F 2 =
_ Pi , V{ 425 (5.72) 2
1_  t zi  1 
w 2g 62.4 2g
+ = 63 ft
40
2 =   + + 22 = + + 43 = 43.0 ft
w 2g
E P = E 2  Ei 
Pump horsepower
43  (6.3) = 49.3 ft (ftlb/lb)
QwEp 2.0 X 62.4 X 49.3
~~
550
550
= 11.2 hp
21. Flow of an Incompressible Fluid. A constriction in a stream
tube or pipe line is frequently used
as a device for metering fluid flow.
Simultaneous application of the con
tinuity and Bernoulli principles to
such a constriction will allow direct
calculation of the rate of flow when
certain variables are measured. To
develop these relationships for an
incompressible fluid, consider the
general type of constriction illus
trated in Fig. 38 and write the
continuity and Bernoulli equations
between sections 1 and 2. For an incompressible fluid, temperature
FIG. 38.
70
THE FLOW OF AN IDEAL FLUID
and density will not vary appreciably as the pressure changes from
Pi to p2 Thus the simultaneous equations become
Q = A 1 V 1 = A 2 V 2
i \ rl fio V O
*   * ^ I f 
r "I r 1 == r ~ ~r #2
which may be solved for Q by substituting
Vi = and F 2 = 7
^AI ^2
in the second equation, resulting in
^2 ~^
Q
w
Thus is becomes evident that the quantity of flow through a constric
tion in a streamtube (1) is dependent upon the
20 Lb er difference of the sums of pressure and potential
w sq. in. heads at points 1 and 2 and (2) varies with the
/ t \ square root of this difference.
12"
ILLUSTRATIVE PROBLEM
= /7\40Lb. per Calculate rate of flow of water through this pipe
sq ' ln> line when the gages read as shown.
IT /6V
 X ( I
4 \12/
40 X 14_4 20 X 144
62.4
62.4
Q = 10.6 cfs
22. Flow of a Compressible
Fluid. When a compressible fluid,
such as a vapor or gas, flows
through a constriction (Fig. 39)
in a streamtube, large changes in
density and temperature may WlTl
occur as flow takes place. When
the constriction is used as a metering device, it is usually short and
FLOW OF A COMPRESSIBLE FLUID 71
thus the expansion from pressure p\ to pressure p 2 occurs rapidly
enough to prevent loss or gain of heat by the fluid, and, therefore,
the adiabatic flow equations 15 and 16 may be applied. The simul
taneous equations (in which the pressures are absolute) become for
a perfect gas
G = AiViWi = A 2 V 2 w 2
fcii
2 J
and when Vi = G/ ' A\w\ and V 2 = G/A 2 w 2 , from the first equation, are
substituted in the second there results
This equation may be improved for practical use, and for further
analysis by the elimination of w% by substituting
1 1
k
and
./>i>
derived from the adiabatic equation, giving
an equation first derived by St. Venant (1839). The equation indicates
that rate of flow of a compressible fluid is dependent upon pressure
ratios and not upon pressure differences, as was the case for the incom
pressible fluid. It shows also that the rate of flow cannot be calcu
lated from pressure measurements only, but that the temperature of
the fluid at section 1 must be measured to obtain w\. In most engineer
ing applications of this equation A 2 is small compared to A\ and thus
the term
'v*
72
THE FLOW OF AN IDEAL FLUID
may be taken to be unity and the simplified equation becomes
A 2
k! P ^L\ Pl
(21)
Further investigation of this equation leads to an interesting and
significant paradox: If p% = 0, it seems reasonable to expect flow to
occur from section 1 to section 2,
yet if p2 is placed in the equa
tion the weight flow, G, becomes
zero also! This inconsistency means
that some new factor has entered
the problem ; this new factor must
now be examined.
If equation 21 is plotted the
curve of Fig. 40 results, indicating
G be zero where the pressure ratio
p2/Pi is either zero or unity, and
having a maximum value, G max ., at
a certain "critical pressure ratio,"
(p2/Pi)c The properties of this
maximum point on the curve may be obtained by differentiating
equation 21 in respect to p2/pi and equating the result to zero. How
ever, differentiation of the whole equation is unnecessary, since the
only variables involved are
*+i
i
LW \Pi'
and the properties of the maximum point may be obtained from
i.o
FIG. 40.
which, when differentiated and solved for p2/Pi, now (p%/pi) c > gives
fe).
\pJc
\pl
showing that the critical pressure ratio is dependent only upon the
adiabatic constant, k.
FLOW OF A COMPRESSIBLE FLUID 73
An equation for G ma x, may now be obtained by substituting this
expression in equation 21 and multiplying by Wz/^/w&Uz, giving
2gk p l w l \( 2 yi ( 2 V" 1
Gmax, = W 2 A 2 \ 7 7 I TT~; / ~ V Z~T~1 J
* & 1 W 2 W2 L\& + I/ \fe + I/ J
but from the adiabatic relation between pressure, p, and specific weight,
w, at critical pressure conditions
__2_
m, = / 2_y^i
which may be substituted in the above equation, reducing it to
(22)
From the equation of continuity
and by comparison of these two equations
V 
V2 max.
P2
or, in other words, when maximum flow takes place the velocity of
flow at the constricted section equals the acoustic velocity (Art. 4), the
velocity with which a pressure disturbance, such as sound, will travel
in the fluid.
The significance of the fluid attaining the acoustic velocity at the
constricted section and an explanation of the paradox mentioned
above can best be obtained from a study of the flow under various
pressure conditions through a constriction formed by a smooth nozzle
(Fig. 41) installed in a pipe line. With p% = pi(A),p2/pi = 1.0,
p2 = Pz> and no flow will occur. With p 2 less than p\(B), but with
1.0 > p2/Pi > (p2/Pi)c, p2 = p2> and flow will take place according
to equation 21. As the maximum flow condition is reached, (C)
p2/Pi = (p2/Pi)c and p 2 p2 and the velocity of flow from the nozzle
equals the acoustic velocity as derived above. If now the pressure p^
74
THE FLOW OF AN IDEAL FLUID
surrounding the jet of fluid at the nozzle exit, is further lowered, (D) no
further reduction in p 2 , the pressure in the jet, occurs. A discon
tinuity of pressure then exists between the surrounding fluid and the
interior of the jet a discontinuity which is established and main
tained by the acoustic velocity; the outside pressure tends to enter the
jet with the velocity of sound, but since the fluid itself is moving with
this velocity, the surrounding pressure cannot penetrate the jet. The
1 A
P'
. A
/
"5"
/
u
D
r\
uation
^C
23
/
x
N
B
G
i
I
CM
1
Equatior
1.0
FIG. 41.
pressure in the jet is, therefore, fixed by conditions upstream from the
constriction and remains equal to the critical pressure regardless of the
magnitude of the surrounding pressure, p 2 . With the pressure p' 2
remaining constant no change in the flow will occur and thus the
maximum flow will exist for all pressure ratios below the critical. These
facts are shown graphically on Fig. 41, and may be summarized as
follows. With
< ( ) ,
pi \pi/c '
< Pcyp2 =Pc,~ = . ^
^1 \l
In engineering practice it is customary to simplify calculations by
designing constrictions so that flow will occur with pressure ratios
below the critical. Equation 22 is not convenient for the calculation
FLOW OF A COMPRESSIBLE FLUID 75
of Gmax. since the temperature at section 2 is not in general measured.
This equation may be transformed into
fe+1
x*^)*"" 1 (23)
by the substitution of
2
Since Wi = p\/RTi, equation 23 becomes
Cmax. ^2
a simple equation for calculating flow under maximum conditions after
pressure and temperature, pi and 7\, are measured. The equation
may be further simplified, by collecting the constants, to
^ \ 7, I 1/ ^ /T (24)
or
x~i rr * ~* 6 r 1.
^max. ^ /~~T
in which
and is obviously a characteristic constant of the gas, dependent only
upon its physical constants, R and &, and upon acceleration due to
gravity, g.
When a vapor flows through a constriction in a stream tube its
behavior is similar to that of a perfect gas, the expansion being adia
batic if rapid and each vapor having a critical pressure ratio below
which the conditions of maximum flow will be maintained. Unlike
that of perfect gases the critical pressure ratio of a vapor will depend
upon its thermodynamic conditions, steam, for example, having a
critical pressure ratio of 0.58 when saturated and 0.55 when highly
superheated.
76
THE FLOW OF AN IDEAL FLUID
The flow of a vapor under adiabatic conditions has been character
ized by equation 15 and by the equation of continuity, thus
Vl V\
_.J. 778(ffl _^
which may be solved simultaneously for G, giving
which simplifies to
G = 223.84 2 4
if AZ
!  H 2
(25)
AI is small, which it usually is in engineering practice.
In applying equation 25 to the flow of a vapor through a nozzle
installed in a pipe line (Fig. 42) a temperatureentropy or total heat
entropy diagram is used to find the values of II \ arid H' 2 . With the
Super heat Region
Lines of
constant
pressure
Super heat
Region
Dry saturation line
8
FIG. 42.
thermodynamic properties (pressure, temperature, quality) of the
vapor known for section 1 of the streamtube, point 1 may be located
on either diagram. Since the process is adiabatic a line of con
stant entropy (vertical) must be followed to a pressure, p' 2 , on the
diagram, corresponding to section 2 of the streamtube. If the pressure
IMPULSEMOMENTUM RELATIONSHIPS 77
ratio is above the critical, p' 2 = pz\ but if below the critical, p'< 2
(P'2/Pi)cpi With point 2' established on either diagram the enthalpy,
11*2, and the specific volume, v> 2 , may be obtained, allowing calculation
of G, the weight flow.
ILLUSTRATIVE PROBLEM
Calculate the rate of flow of carbon dioxide through this 1 in. diameter oiifice
installed in a 6in. pipe line when the downstream pressure gage reads (<z) 80 lb/in. 2 ,
(b) 30 lb/in. 2 Barometric pressure is 15.0 lb/in. 2
1 30
. \ / VLM^l 100 Lb per (a) 80 Lb. per sq. in.
El  = [ ] _ C4f SQ >n ^ 30 Lb< P er sq> in>
>iA~\i.3o + i/ ~ ' 15 ; F 
Wl =
u.oz,o i
pi 100 + 15 V
V U_t IV
\
1" 6 \
115X144 073lb'ft II
r i
34 V (4AO f 1 SO) W
e *Jft>~[&
G = " X  \ (115 X 144)0.783 [(0.825) 1 ' 64  (0.825) 1
4 \12/ \ 1.30  1
= 1.89 Ib/sec
(&) ^ = J*L_15_ = 0.391
pi 100 f 15
  115 X 144
V, ^ \Tfii7 "~ ' I I 1JLO A A ^^
32. 2 X 1.3 / 2 V' 67 4 \12/ n ^ .
( ) \ . 2.36 Ib/sec
34.9 \1.3 + I/ V460 + 150
23. ImpulseMomentum Relationships. The impulsemomentum
law provides another useful tool for the solution of fluid flow problems.
Its application to complex fluid flow processes sometimes allows circum
vention of the complexities and gives a simple answer to problems
which cannot be solved by the use of foregoing principles or more
advanced energy considerations.
78
THE FLOW OF AN IDEAL FLUID
The impulsemomentum law applied to a particle (Fig. 43) is
usually stated.
FAt = A(MF)
in which the product F&t is called the
"impulse," the righthand side of the
equation being recognized as the change
of momentum. Fora particle whose mass,
M , remains constant, the above equation
may be written
FM = MAF
FIG. 43.
and the impulsemomentum law may be stated: An external force, F,
acting on a moving particle of mass M, for a time A/, will change its
velocity by an amount AF. Examination of the equation discloses the
fact that A2 and M are scalar quantities, and F and A V are vector quan
v i Force and components
exerted by the fluid
Force and components
acting on the fluid
FIG. 44.
tities. Since these vector quantities are related by a simple algebraic
equation their directions must be the same. Thus if the direction of
AFcan be obtained the direction of the force becomes established also.
Applying the above principles to the stream of moving fluid of
IMPULSEMOMENTUM RELATIONSHIPS 79
Fig. 44, the force, F, necessary to change the velocity of the mass M
from FI to 2 is given by
F&t = M&V
in which
AF = (F 2  FO
the > indicating a vectorial subtraction carried out as indicated in the
vector diagram. The distance A/ is covered by the mass M in time
A/, giving
A/ = FiA/
and
M = pA A/ = p<4 Ft A/ =
substitution of these values in the fundamental equation above results
in
F = Qp(V 2 > F x ) = ^ (F 2 > Fi) (26)
an equation which may be used for the calculation of force components
F x and F y as well as the total force, F. Since the direction of F is the
same as that of (F 2 FI)
from the similar force and velocity triangles, and, therefore,
Ow
>V l ) x (27)
and
F y = ^(V 2 ^V 1 ) y (28)
from which the component of force in a given direction necessary to
accomplish a component of momentum change in this direction may
be calculated.
In the above equations 26, 27, and 28, the forces involved are
external forces exerted on the fluid in order to accomplish a certain
change of velocity. In certain engineering problems, however, the
forces exerted by the fluid on its surroundings (equal and opposite to
80
THE FLOW OF AN IDEAL FLUID
the above forces) are of more immediate practical value. Designating
these forces by ( F), ( F x ), and ( F v ), the above equations become
(F)
o
The great advantage of the impulsemomentum law is evident
from the fundamental equa
tions: the forces are seen to
be dependent only upon initial
and final velocity conditions
and are entirely independent
of the flow complexities occur
ring between these conditions.
The physical meaning of
the impulsemomentum law
can best be illustrated by its
application to a number of
specific engineering problems.
The forces on a fixed
blade, or deflector, as a fluid
stream passes over it may
be readily calculated. From
Fig. 45,
(Vi+ V 2 ) x = Vi  V 2 cosa
= F 2 sin a
Therefore
Qw
_/y = ^(F! F 2 cosa)
V(Vi  V 2 cos a) 2 + (F 2 sin a) 2
g
IMPULSEMOMENTUM RELATIONSHIPS
81
ILLUSTRATIVE PROBLEM
A jet of water having 2in. diameter strikes a fixed blade and is deflected 120
from its original direction. Calculate magnitude and direction of the total force
on the blade if velocity of jet is 100 ft/sec.
100  X X 62.4
32.2
X 173.2 = 733 Ib
F,100Ft.\
per sec v x
per sec.
i ^^ ^r,_
V l  100
The computation of the force exerted by a fluid stream on a moving
blade forms the basis for elementary calculations on impulse turbines.
Let the single moving blade of Fig. 46 have a velocity, v, in the same
direction as the fluid stream.
Since the work done on the
blade will be derived from the
component of force ( F x ) in
the direction of motion, this
component only will be calcu
lated. The velocity of fluid
relative to the blade, u\, at its
entrance will be given by
u\ = Vi v
and if the blade is assumed
frictionless the velocity relative
to the blade at exit, u 2 , will be
the same
v
FIG. 46.
The absolute velocity of the fluid stream as it leaves the blade, V%,
will be a composition of its velocity along the blade and the velocity
of the blade itself and may be calculated as indicated on the vector
diagrams; therefore
> V 2 ) x =V l (v+ (Vi  v) cos a)
82
THE FLOW OF AN IDEAL FLUID
or
cos a)
For a single blade, however, the quantity of flow being deflected
per unit of time is not equal to the quantity of flow in the jet since the
blade is moving away from the jet. The deflected rate of flow, Q f ,
will be given by
Q' =
allowing the force component, ( F x ), exerted on the single blade to
be calculated as
<=.
cos
In a closely spaced series of moving blades of the above type such
as occurs in an impulse water turbine (Fig. 47) all the fluid flowing is
Vj
2
v l
FIG. 47.
deflected by the blades and Q = (/. Thus the force exerted on the
blades in the direction of motion may be calculated from
(F x ) =
g
tO(l 
and since the power transferred from jet to blade is given by
P = (~FJv
p =(F!
g
allowing the theoretical power developed by the turbine to be calcu
lated. For a given quantity of flow and size of jet, the theoretical
IMPULSEMOMENTUM RELATIONSHIPS 83
power developed will depend only upon the peripheral speed of the
wheel, v, and will be zero when v is zero and when v is equal to the jet
velocity V\. Confirmation of these facts is given by the plot of the
above equation in Fig. 47, which also exhibits a point of maximum
power at which the turbine should be operated for best efficiency.
The properties of the maximum point on this curve may be obtained
by the usual differentiation
~ (Vi w)(l  cosa)J =
dvL g J
resulting in v V\/2 for maximum theoretical power from which the
maximum power, P ma x. may be calculated by substitution in the
general expression, giving
= (1  cos a)
g 4
which becomes
yl
W 2g
if a = 180. The power available in the jet according to equation 17 is
the same expression as given above. Therefore, all the jet power may
be theoretically transferred to the turbine (1) if the speed of operation
is correct and (2) if the blades are designed with an angle a = ISO .
The force on a reducer, enlargement, or bend in a pipe line may be
determined from the Bernoulli equation by tedious calculations and
graphical integrations, but this force may be obtained directly and
easily by use of the impulsemomentum law.
The pipe reducer shown in Fig. 48 represents a typical problem
of this type. As flow takes place certain forces act on the fluid within
the reducer and continually change the momentum of the fluid con
tained therein. The forces acting on the fluid within the reducer are
due to pressure exerted at the ends by the adjacent fluid and at the
84
THE FLOW OF AN IDEAL FLUID
sides by the boundary walls. The forces at the ends Fi and F% are
given by
F l = p l A l
On the side walls a pressure variation about as shown will exist accord
ing to the Bernoulli principle. This results in a force indicated sche
matically by T 7 , whose horizontal component is F x
impulsemomentum law.
Ow
Fi F 2 F X = *
o
Applying the
allowing F x to be calculated. 2 The force exerted by the fluid on the
reducer will, of course, be equal to F x but opposite in direction.
The force on a reducing pipe bend may be calculated (Fig. 49) by
applying the same principles. With notation similar to that above,
(F 2
Ow
F v  F 2 sin = ^ (F 2  Vi) v
o
2 Note that a net force exerted in a given direction accomplishes a change of
momentum in this direction. In this case, the direction of the net force (Fi F%
F x ) is obviously to the right and vectorial considerations indicate the vectorial
difference of velocities ( V% > FI) to be to the right also.
IMPULSEMOMENTUM RELATIONSHIPS
85
from which F x and F y may be calculated, the total force, F, exerted
by the bend on the fluid being given by
F = V Fl + Fl
and the force exerted by the fluid on the bend is, as before, equal to
F but opposite in direction.
FIG. 49.
ILLUSTRATIVE PROBLEM
When 10 cfs of water flow upward through this conical enlargement, the pressure
gage reads 15 lb/in. 2 Calculate magnitude and direction of the total force exerted
by liquid on enlargement.
Vi = 12.72 ft per sec, F 2 = 3,18 ft per sec, p 2 = 14.72 lb/in. 2
Isolate liquid between sections 1 and 2. F
The velocity of the liquid is being decreased I
by a net force acting on the liquid opposed
to motion. The net force is (Fi f W
F FI) in which F is the force exerted on
^ sq. in Pef the liquid by the enlargement. The de
crease of momentum per unit time is
(Qw/g)(Vz > FI).
Equating these,
F 2 + W  F  FI = Q^ (F 2 > Fi)
g
F 2 = 14.72 X  X (24) 2 = 6660 Ib,
4
86
THE FLOW OF AN IDEAL FLUID
F 1 = 15.0 X X (12) 2 = 1700 Ib
4
6600 + 343  F  1700 =
X 62 ' 4
(3.18  12.72) = 185
32.2
F = 5118 Ib
Thus the force exerted by liquid on enlargement has a magnitude of 5118 Ib and
acts vertically downward.
When fluid discharges from an orifice in a large container, the jet
of fluid will cause a reactive force on the container which may be calcu
lated easily by the impulsemomentum law. From Fig. 50 it is obvious
that the force exerted on the container will be due to the reduction in
pressure on the side of the container which contains the orifice, but
this pressure reduction cannot in general be calculated without assump
tions or experimental data. Isolating a mass of fluid A BCD from
which the jet issues, the horizontal force, F\, exerted on AD by the
Q
FIG. 50.
vertical wall, may be calculated assuming the fluid to be at rest and
using the laws of fluid statics. On the wall BC the pressure distribu
tion will not be static because a pressure reduction will occur, accord
ing to the Bernoulli principle, as flow takes place toward the orifice.
The force F% exerted on BC by the wall will occur from a pressure
distribution about as shown and obviously will be less than F\. Now,
applying the impulsemomentum law, F, the net force exerted by the
container on the fluid, is given by
F = F l  F 2 = ^
g
v,)
in which V\, the velocity within the tank, is negligible. Thus
Qw
F = F 1  F 2 =
g
IMPULSEMOMENTUM RELATIONSHIPS
87
allowing the net force, F, to he calculated and the reactive force
exerted by the fluid on the container will be equal to F in magnitude
but opposite in direction.
A different expression for F may be obtained by applying Torri
celli's theorem and the continuity principle,
V 2 = 2gh and Q = A 2
and substituting these values in the above equation for F,
F  A
or, in other words, the reactive force exerted on the container by the
moving fluid is just twice the force exerted on an area the size of the
orifice submerged at a depth, h, below the liquid surface.
FIG. 51.
Another application of the impulsemomentum law may be made
to ship or airplane propellers which obtain their thrusts by changing
the momentum of a mass of fluid. A screw propeller with its slip
stream is shown in Fig. 51 and may be considered to be (1) moving to
88 THE FLOW OF AN IDEAL FLUID
the left with a velocity Vi through still fluid, or (2) stationary in a
flow of fluid from left to right with velocity Vi ; the relative motion in
both cases is the same. For such a propeller operating in an uncon
fined fluid, the pressures pi and p, at some distance ahead of and
behind the propeller, are obviously the same.
The thrust, F, of the propeller on the fluid results in a change in
the velocity of the fluid from Vi to F 4 and therefore
g
The useful power output P and thrust developed by a propeller
moving at a velocity Vi is P = FV\. To obtain this output it is
necessary to supply energy enough to create an increase of velocity
from FI to F 4 . From equation 17 the power input, P;, to create this
velocity difference is given by
which may be written
Thus
and the propeller efficiency, ?, may be calculated from
Po
**
(Vi
~2
denominator; then
indicating that the efficiency of a propeller operating in an ideal fluid
cannot be 100 per cent 8 since there must always be a sizable velocity
3 Ship and airplane propellers may have efficiencies of about 80 per cent.
IMPULSEMOMENTUM RELATIONSHIPS
89
difference (V FI), making the denominator of the above expression
always greater than the numerator.
Further information on the screw propeller may be obtained by
means of the Bernoulli principle. Bernoulli's equation written
between sections 1 and 2 is
Pi
and between sections 3 and 4
=p s
in which V is the velocity through the plane of the propeller. Since
Pi = Pi, subtraction of the first equation from the second results in
~ F?)
but another equation for the thrust Fcan be obtained from the pressure
difference (p% p%) acting on the area A, which is
F = (Pa  Pz)A
and substituting the equivalent of (p% p%) from the above equation
F = A\ P (V\  V\)
This expression may be equated to the impulsemomentum expression
for thrust, giving
Qw ___ _
g
T (j) ) '* /Jf
Since Q = A V and p = w/g, this equa
tion reduces to
vl)
showing that the velocity through the
plane of a propeller operating in an ideal
fluid is the numerical average of the
velocities at some distance ahead of and FIG. 52.
behind the propeller.
The impulsemomentum law, when applied to the rotation of a
particle about an axis, is stated as
in which (Fig. 52) T is the torque which must act on the particle for
90 THE FLOW OF AN IDEAL FLUID
time A/ in order to accomplish the change in angular momentum
A(Af Vtr), M being the mass of the particle, V t , its tangential compo
nent of velocity, and r the distance between the particle and axis of
rotation. Since the mass of the particle is constant, the equation may
be written
which, for application in fluid problems (as in equation 26), becomes
or
r = ^(F,/ 2  F^n) (29)
g
an equation which forms the basis of reaction turbine and centrifugal
pump design.
To illustrate the use of this equation consider its application to a
simplified hydraulic reaction turbine, sections through which are shown
in Fig. 53. Water flows inward through the fixed guide vanes 01,
acquiring a "whirl," and thus possesses a tangential component of
velocity, V^ as it leaves them. It then passes through the blades 12
of the rotating element or "runner," and discharges downward as
shown.
As the water leaves the guide vanes at point 1 , it will have an abso
lute velocity FI, having tangential and radial components, V^ and
F ri , given by
V tl FI cos ai
V ri = FI sin ai
The radial component, F n , is dependent on the rate of flow through
the turbine, Q, and may be calculated by applying the continuity
principle, giving
Q = 7 ri 2*Ti/
For best operating conditions the water in leaving the guide vanes
must pass smoothly into the moving runner at point 1. To accom
plish this the tangential velocity of the runner, co^, at this point must
be such that the component of relative velocity, u, is tangential to
IMPULSEMOMENTUM RELATIONSHIPS
91
the blade as is indicated on the vector diagram. From this vector
diagram the following equation for V tl may be written:
V tl = wr ! + V Tl cot ft
FIG. 53.
At the exit of the runner, the relative velocity, u%, will be tan
gential to the blade. The peripheral velocity of the blade, co^, is
fixed by the speed of rotation and by the radius at exit, and the
component of velocity in a radial direction is determined by the
continuity equation
Q = V r2 27rr 2 l
The vector diagram at the runner exit then becomes as shown and
Vti = wr 2 + V rz COt ftj
92
THE FLOW OF AN IDEAL FLUID
The torque equation 29 allows calculation of the torque exerted by
the runner on the fluid. The torque exerted by the fluid on the runner
will be equal and opposite to this and given by
~
o
(30)
To obtain a more detailed expression for the theoretical torque
exerted on a hydraulic turbine runner when operating under ideal
conditions the quantities derived above may be substituted in equation
30; the horse power developed may be calculated from
hp =
550
24. Flow Curvatures, Types of Vortices, Circulation. Many fore
going examples have demonstrated the fact that pressure distributions
across fluid flows may be calculated from
the laws of fluid statics when fluid par
ticles move in straight parallel paths.
Since curvilinear fluid motion occurs so
frequently in practice it is important to
understand the properties of pressure
distribution in this type of flow.
In the streamline picture of Fig. 54, a
differentially small fluid mass moves along
a curved streamline of radius r with a
velocity V. If a pressure p exists on the
inner face, that on the outer face will be
p + dp, assuming an increase in pressure
with increase in radius. Since the mass
of fluid is in equilibrium in a radial direction, the centrifugal force
must be balanced by the centripetal force due to pressure. These
forces are
V 2 V 2
Centrifugal force = dM = pdrdldn
r r
Centripetal force = (p + dp)dldn pdldn = dpdldn
which, when equated, give
FIG. 54.
dpdldn = pdrdldn
V 2
FLOW CURVATURES, TYPES OF VORTICES, CIRCULATION 93
or
V 2
dp = pdr
r
(31)
indicating that pressure will increase with radius in curved flow, or,
more generally, pressure at the outside of a curvilinear flow will be
greater than that toward the center of curvature.
These facts will explain some of the details of flow through a sharp
edged orifice, a problem which will be
treated more exhaustively in a later
chapter. Owing to inertia, particles
of fluid issuing from the vertical
sharpedged orifice of Fig. 55 will
not move horizontally, but rather in
smooth curves as shown, the curves
resulting in the contraction of the jet
to a diameter less than that of
the orifice. At the point where the
streamlines become parallel at the
contracted section of the jet, the pres
sure in the jet is zero and the velocity
is given from Torricelli's theorem
(Art. 19) by
V 2
FIG. 55.
Upstream from point 2, centers of curvature of the jet streamlines are
outside of the jet. This means that in moving away from the center
of curvature the jet is penetrated, and from equation 31 it may be con
cluded that the pressure within the curved region of the jet is greater
than that outside. This pressure variation is indicated in the figure.
To appreciate more fully the significance of equation 31, let it be
applied to two specific types of rotational motion, one of which has
been discussed previously (Art. 15).
Certain facts and equations have been brought out concerning
the fluid motion obtained by rotating a container of fluid about a
vertical axis (Art. 15). After equilibrium sets in, the motion of the
fluid at any point in the container is like that of a solid body, possessing
a circumferential velocity, V, at a radius, r, given by
V
car
94
THE FLOW OF AN IDEAL FLUID
in which co is the angular velocity of rotation. If this relation between
velocity and radius is substituted in equation 31 the result is
coV
dp = pdr
or
dp
w
c 2 rdr
g
FIG. 56.
the same equation as developed in Art. 15. The
streamline picture in this type of rotational mo
tion is shown in Fig. 56, and a particle moving
on a streamline is indicated. Inspection of the
motion of this particle shows that it will occupy
successive positions 1234, and as the container
makes one revolution the particle rotates once on
its own axis. Hence it may be concluded that
in this type of fluid motion all the fluid particles
rotate about their own axes. Such fluid motion
is designated as "rotational," and the specific
motion indicated in Fig. 56 is sometimes called a "rotational vortex"
or "forced vortex."
Introducing now a new term, F (gamma), the "circulation," which
will prove useful later, F is defined by the equation,
F = (Dvds
in which <p should be read, "the line integral around a closed curve
of  ." The meaning of circulation may be
obtained from the closed curve drawn in the
flow picture of Fig. 57. A differential amount
of circulation, dT, is defined as the compon
ent, v, of velocity along ds, multiplied by ds,
thus
dT = vds
and the circulation along the entire closed
curve is given by the integration of this expression along the curve.
Therefore, /
F = (p vds
FLOW CURVATURES, TYPES OF VORTICES, CIRCULATION 95
Now the circulation around the forced vortex of Fig. 56 may be
calculated by taking for simplicity a circle of radius r as the closed
curve. The velocity along such a curve is V = ur, the length of curve
is 2wr, and since V does not vary along the curve no integration is
necessary. Circulation, F, therefore, becomes
T = (o;r)27rr = 2irur 2
and the circulation is seen to vary with the size and location of the
closed curve, another charac
teristic of "rotational" fluid
motion.
Another type of fluid mo
tion is obtained when the sum
of the Bernoulli terms for every
streamline is the same. Such
motion may exist when all the
streamlines originate in the
same field of energy, such as
in the reservoir and pipe bend
of Fig. 58. The characteristics
of the curved flow in this bend
may be obtained from the fact
that
P+ ipl
Elevation
Plan
FIG. 58.
= Constant
at any point on a horizontal plane in the flow. Therefore, by differ
entiation
dp + pVdV =
giving a relation between p and V which may be placed in equation 31.
The result is
V 2
pVdV = pdr
or
^r_
V r
Integrating,
InV + In r = C
and eliminating logarithms
Vr = Ci
96
THE FLOW OF AN IDEAL FLUID
v=rConstant
showing that in this case of curved fluid motion the velocities will be
least at great distances from the centers of curvature and greatest
near these centers.
If flow of this type occurs about a vertical axis a socalled free
vortex is formed, such as develops when a container is drained through
an opening in the bottom. The streamline picture and hyperbolic
velocity variation with radius for such a vortex are shown in, the plan
view of Fig. 59. The profile of the liquid surface in the vortex may be
obtained by neglecting any
radial velocity and applying
the Bernoulli equation. Over
the surface the pressure is
zero, and, therefore, this
equation becomes
V 2
 h z = Constant
2g
the constant being the dis
tance between datum plane
and liquid surface at a great
distance from the axis of
rotation. Thus the surface
profile within the vortex
may be readily calculated.
To investigate the rota
tional properties of particles
in this fluid flow, let a typical
fluid particle be designated
as before (Fig. 59). As flow
occurs it will be noticed that the particle exhibits no rotation about
its own axis as it occupies the successive positions 1234. Since
none of the particles rotate about their own axes the motion is described
as "irrotational" and the free vortex motion described above is termed
an "irrotational vortex/'
The circulation, F, about a free vortex may now be obtained and
some useful conclusions drawn from the result. For simplicity select
a circular closed curve of radius r, whose center is the center of the
vortex. The velocity along such a curve is constant and given by
FIG. 59.
FLOW CURVATURES, TYPES OF VORTICES, CIRCULATION 97
or
Therefore, the circulation F becomes
F = 2vr = 2wCi
showing that the circulation F is independent of the size of curve
selected for the calculation of the circulation and that the circulation
around the center of a free vortex is constant.
By a more rigorous and generalized treat
ment it may be shown that this constant
circulation is independent of the shape of the
closed curve, provided that the vortex center
is included within its boundaries.
In order to obtain more general conclu
sions concerning the properties of circulation
in irrotational motion, the circulation around
a closed curve that excludes the center of the
vortex may be calculated. A simple curve, A BCD, of this type is
shown in Fig. 60 ;
= TAB + TBC + TCD + TDA
FIG. 60.
and may be evaluated from the above principles with the following
results
TABCD =
6
i i
+  2*r + =
It may be concluded that there is no circulation in irrotational motion
around a closed curve which excludes
the vortex center. Such a vortex
center is called in mathematics a
"singular point," and is denned as a
point where velocity becomes in
finite a physical impossibility.
The circulation about a vortex
center is a constant of the vortex
F IG . 6i. and a measure of its strength. A
vortex may, therefore, be designated
by F **> to indicate its strength and direction of rotation. Spine other
98
THE FLOW OF AN IDEAL FLUID
I
useful conclusions which may be obtained from further investigation of
vortex properties are summarized graphically in Fig. 61.
A combination of the "forced" and "free" vortices, called the "com
pound" vortex, frequently occurs on the surface of a liquid when the
surface is disturbed by a blunt
object, such as an oar or paddle,
moving through it. The familiar
shape of the compound vortex is
shown in Fig. 62. In the regions
at a distance from the axis of
rotation the surface profile of the
free vortex appears, but in the
center, where the free vortex would require a great drop in the liquid
surface, a core of liquid is rotated by the motion of the free vortex.
On this core of liquid forms the characteristic paraboloidal surface
curve of the forced vortex.
forced vortex 
 Free vortex
Rotating core
of liquid
FIG. 62.
BIBLIOGRAPHY
PROPELLERS AND HYDRAULIC TURBINES
W. SPANNHAKE, Centrifugal Pumps, Turbines and Propellers, Technology Press, 1934.
PROPELLERS
F. E. WEICK, Aircraft Propeller Design, McGrawHill Book Co., 1930.
HYDRAULIC TURBINES
R. L. DAUGHERTY, Hydraulic Turbines, Third Edition, McGrawHill Book Co., 1920.
G. E. RUSSELL, Textbook on Hydraulics, fourth edition, Henry Holt & Co., 1934.
A. H. GIBSON, Hydraulics and Its Applications, fourth edition, D. Van Nostrand,
1930.
PROBLEMS
135. The flow in this pipe line is reduced linearly from 10 ft 8 /sec
to zero in 30 sec by a valve at the end of the line. Calculate the
local, convective, and total accelerations at points A, B, and C,
after 10 sec of valve closure.
136. The velocity of water in a 4in. pipe line is 7 ft/sec. Cal
culate the rate of flow in cubic feet per second, gallons per minute,
pounds per second, and slugs per second.
137. One hundred pounds of water per minute flow through a
6in. pipe line. Calculate the velocity.
138. One hundred gallons per minute of glycerine flow in a 3in.
pipe line. Calculate the velocity.
139. Air flows in a 6in. pipe at a pressure of 20 lb/in. 2 and a temperature of
100 F. If barometric pressure is 14.7 lb/in 2 . and velocity of flow is 12 ft/sec,
calculate the weight flow in pounds per second.
PROBLEMS
99
140. Water flows in a pipe line composed of 3in. and 6in. pipe. Calculate the
velocity in the 3in. pipe when that in the 6in. pipe is 8 ft/sec. What is its ratio to
the velocity in the 6in. pipe?
141. A smooth nozzle with tip diameter 2 in. terminates a 6in. water line. Calcu
late the velocity of efflux from the nozzle when the velocity in the line is 10 ft/sec.
142. Air discharges from a 12in. duct through a 4in. nozzle into the atmosphere.
The pressure in the duct is 10 lb/in. 2 , and that in the nozzle stream is atmospheric.
The temperature in the duct is 100 F, and that in the nozzle stream 23 F. The
barometric pressure is 14.7 lb/in. 2 , and the velocity in the duct is 73.5 ft /sec. Calcu
late the velocity of the nozzle stream.
143. Air flows with a velocity of 15 ft/sec in a 3in. pipe line at a point where the
pressure is 30 lb/in. 2 and temperature 60 F. At a point downstream the pressure
is 20 lb/in. 2 and temperature 80 F. Calculate the velocity at this point. Barometric
pressure is 14.7 lb/in. 2
144. At a point in a twodimensional fluid flow, two streamlines are parallel and
3 in. apart. At another point these streamlines are parallel but only 1 in. apart.
If the velocity at the first point is 10 ft/sec, calculate the velocity at the second.
145. Water flows in a pipe line. At a point in the line where the diameter is 7 in.,
the velocity is 12 ft/sec and the pressure is 50 lb/in. 2 At a point 40 ft away, the
diameter reduces to 3 in. Calculate the pressure here when the pipe is (a) horizontal,
(6) vertical with flow downward.
146. A horizontal 6in. pipe in which 1000 gpm of carbon tetrachloricle is flowing
contains a pressure of 30 lb/in. 2 If this pipe reduces to 4in. diameter, calculate the
pressure in the 4in. pipe.
147. In a pipe 1 ft in diameter, 10 cfs of water are pumped up a hill. On the hill
top (elevation 160) the line reduces to 8in. diameter. If the pump maintains a
pressure of 100 lb/in. 2 at elevation 70, calculate the pressure in the pipe on the hilltop.
148. In a 3in. horizontal pipe line containing a pressure of 8 lb/in. 2 100 gpm of
water flow. If the pipe line reduces to 1in. diameter, calculate the pressure in the
1in. section.
149. If benzene flows through this pipe
line and its velocity at A is 8 ft/sec, where is
the benzene level in the open tube C?
150. Water flows through a 1in. con
striction in a horizontal 3in. pipe line. If
the water temperature is 150 F (w = 61.2
Elev.
500 g"
12'Dia.
Elev.
400
PROBLEM 149.
PROBLEM 151.
lb/ft 3 ) and the pressure in the line is maintained at 40 lb/in. 2 , what is the maxi
mum rate of flow which may occur? Barometric pressure is 14.7 lb/in. 2
151. If 8 cfs of water flow through this pipe line, calculate the pressure at point A.
152. When the head of water on a 2in. diameter smooth orifice is 10 ft, calculate
the rate of flow therefrom.
100
THE FLOW OF AN IDEAL FLUID
153. Five gallons of water flow out of a vertical 1in. pipe per minute. Calculate
the diameter of the stream 2 ft below the end of the pipe.
154. A 3in. horizontal pipe is connected to a water tank 15 ft below the surface.
The pipe terminates in a 1in. diameter smooth nozzle. Calculate the pressure in
the line.
155. Calculate the rate of flow through this pipe
line and the pressures at A , B, C, and D.
156. A smooth nozzle 2 in. in diameter is con
nected to a water tank. Connected to the tank at
the same elevation is an open Utube manometer
containing mercury and registering a deflection of
25 in. The lower mercury surface is 20 in. below
the tank connection. What flow will be obtained
from the nozzle?
157. A 3in. horizontal pipe is connected to a
tank of water 5 ft below the water surface. The
pipe is gradually enlarged to 3.5in. diameter and
discharges freely into the atmosphere. Calculate rate of flow and pressure in the
3in. pipe.
158. A siphon consisting of a 1in. hose is used to drain water from a tank. The
outlet end of the hose is 8 ft below the water surface, and the bend in the hose 3 ft
above the water surface. Calculate the rate of flow and pressure in the bend.
159. A 1in. nozzle on a horizontal 3in. pipe discharges a stream of water with
a velocity of 60 ft/sec. Calculate the pressure in the pipe and the velocity of the
nozzle stream at a point 20 ft below the nozzle. (Neglect air friction.)
160. A smooth 2in. nozzle terminates a 6in. pipe and discharges water vertically
upward. If a pressure gage in the pipe 4 ft below the nozzle tip reads 50 lb/in. 2 ,
calculate the velocity at the nozzle tip and rate of flow. What is the velocity of the
stream at a point 30 ft above the nozzle tip?
161. Water flows through a 3in. constriction in a horizontal 6in. pipe. If the
pressure in the 6in. section is 40 lb/in. 2 and that in the constriction 20 lb/in. 2 , calcu
late the velocity in the constriction and the rate of flow.
162. Water discharges from a tank of water through a 2in. nozzle into a tank of
gasoline (sp. gr. 0.72). The nozzle is 10 ft below the water surface and 11 ft below
the surface of the gasoline. Calculate the rate of flow.
163. Water discharges through a 1in. nozzle under a 20ft head, into a tank of air
in which a vacuum of 10 in. of mercury is maintained. Calculate the rate of flow.
164. A closed tank contains water with air above it. The air is maintained at
a pressure of 15 lb/in. 2 , and 10 ft below the water surface an orifice discharges into
the atmosphere. At what velocity will water emerge from the orifice?
165. A pump draws water from a reservoir through a 12in. pipe. When 12 cfs
are being pumped, what is the pressure in the pipe at a point 8 ft above the reservoir
surface: (a) in pounds per square inch; (b) in feet of water?
166. The pressure in the testing section of a wind tunnel is 1.07 in. of water
when the velocity is 60 mph. Calculate the pressure on the nose of an object when
placed in the testing section of this tunnel. Assume w for air = 0.0763 lb/ft 3 .
167. The pressure in a 4in. pipe line carrying 1000 gpm of perfect fluid weighing
70 lb/ft 8 is 20 lb/in. 2 Calculate the pressure on the upstream end of a small object
placed in this pipe line.
PROBLEMS 101
168. An airship flies through still air at 50 mph. What is the pressure on the nose
of the ship if the air temperature is 40 F and pressure 13.0 lb/in. 2 ?
169. A submarine moves at 10 knots/hr through salt water (S 1.025) at a depth
of 50 ft. Calculate the pressure on the bow of the submarine. (1 knot = 6080 ft.)
170. Benzene discharges from an orifice in a tank under a 10ft head. What pres
sure will exist on the nose of a small object placed in the jet close to the orifice, in
pounds per square inch, in feet of benzene?
171. A circular cylinder 6 in. in diameter is placed in a wind tunnel where the
velocity is 80 mph, the air density 0.0763 lb/ft 3 . The cylinder is placed with its axis
normal to the flow. Calculate the pressure on the front of the cylinder and the pres
sure at a point on the cylinder's surface 90 from the front where the velocity is
160 mph.
172. The pressure on the front of an object in a stream of water is 8 in. of water
above the static pressure in the stream. Calculate the velocity of the stream.
173. A horizontal 6in. water line contains a flow of 5 cfs and a pressure of 25
lb/in. 2 Taking a datum 5 ft below the pipe's centerline, calculate the energy avail
able in the flow.
174. A 2in. nozzle discharges 1 cfs of water vertically upward. Calculate the
energy in the jet (a) at the tip of the nozzle; (b) 20 ft above the nozzle tip, taking
a datum plane at the nozzle tip; (c) the energy of flow in the pipe (4in.) 3 ft below
the nozzle tip, taking datum plane at the nozzle tip.
175. In a perfectly insulated section of horizontal 12in. pipe, 8 cfs of water flow
at a pressure of 40 lb/in. 2 and temperature of 50 F. The pipe bends vertically
upward, is reduced to 6in. diameter, and becomes horizontal again 10 ft above the
12in section. A heating coil in the vertical section delivers 1000 Btu/sec to the
flow. Calculate temperature and pressure in the 6in. horizontal section assuming
that no heat is lost through the pipe walls.
176. Six pounds per second of superheated steam flow upward in a 12in. pipe
line. At elevation 100 the pressure and temperature are 150 lb/in. 2 and 396 F,
and at elevation 200, 138 lb/in. 2 and 390 F. The barometer reads 15.0 lb/in. 2
Calculate the heat lost through the pipe walls between the above two points, using
the following data from the steam tables:
w m = 0.345 lb/ft 3
w 2 ()0 = 0.322 lb/ft 3
H m = 1213.6 Btu/lb
7/200 = 1212.0 Btu/lb
177. A tank containing superheated steam at 10 lb/in. 2 and 314.5 F discharges
adiabatically through a small orifice into the atmosphere. Calculate the velocity of
the steam jet if steam tables indicate that H in the tank is 1196.1 Btu/lb and in the
jet 1155.4 Btu/lb. The barometer reads 15 lb/in. 2
178. A tank containing air at 10 lb/in. 2 and 314. 5 F. discharges adiabatically
through a small orifice into the atmosphere. Calculate the velocity and temperature
of the air jet. The barometer reads 15.0 lb/in. 2
179. A pump takes 1000 gpm of benzene from an open tank through an 8in. pipe.
It discharges this flow through a 6in. pipe, and at a point on this pipe 10 ft above the
liquid surface (in the tank) a pressure gage reads 35 lb/in. 2 What horsepower is
being supplied by the pump?
102 THE FLOW OF AN IDEAL FLUID
180. A pump having 4in. suction pipe and 3in. discharge pipe pumps 500 gpm
of water. At a point on the suction pipe a vacuum gage reads 6 in. of mercury;
on the discharge pipe 12 ft above this point, a pressure gage reads 48 lb/in. 2 Calcu
late the horsepower suplied by the pump.
181. What horsepower pump is theoretically required to raise 200 gpm of water
from a reservoir of surface elevation 100 to one of surface elevation 250?
182. Through a 4in. pipe, 1.0 cfs of water enters a small hydraulic motor and
discharges through a 6in. pipe. The inlet pipe is lower than the discharge pipe, and
at a point on the former a pressure gage reads 60 lb/in. 2 ; 14 ft above this on the dis
charge pipe a pressure gage reads 30 lb/in. 2 What horsepower is developed by the
motor?
183. If 12 cfs of water are pumped over a hill through an 18in. pipe line, and the
hilltop is 200 ft above the surface of the reservoir from which the water is being taken,
calculate the pump horsepower required to maintain a pressure of 25 lb/in. 2 on the
hilltop.
184. A hydraulic turbine in a power plant takes 100 cfs of water from a reservoir
of surface elevation 235 and discharges it into a river of surface elevation 70. What
theoretical horsepower is available in this flow?
185. A pump takes water from a tank and discharges it into the atmosphere
through a horizontal 2in. nozzle. The nozzle is 15 ft above the water surface and is
connected to the pump's 4in. discharge pipe. What horsepower pump is required
to maintain a pressure of 40 lb/in. 2 just behind the nozzle?
186. Water flows through a 4in. constriction in a horizontal 6in. pipe. The
pressure in the pipe is 40 lb/in. 2 and in the constriction 25 lb/in. 2 Calculate the flow.
187. Water flows upward through a 6in. constriction in a vertical 12 in. pipe.
In the constriction there is a vacuum of 8 in. of mercury, and at a point on the pipe
5 ft below the constriction a pressure gage reads 30 lb/in. 2 Calculate the flow.
188. Carbon tetrachloride flows downward through a 2in. constriction in a 3in.
vertical pipe line. If a differential manometer conteiining mercury is connected to the
constriction and to a point in the pipe 4 ft above the constriction and this manometer
reads 14 in., calculate the rate of flow. (CCU fills manometer tubes to mercury
surfaces.)
189. A lin. smooth nozzle is connected to the end of a 6in. water line. The
pressure in the pipe behind the nozzle is 40 lb/in. 2 Calculate the rate of flow.
190. A smooth 2in. nozzle terminates a 4in. pipe line and discharges water
vertically upward. If a pressure gage on the pipe 6 ft below the nozzle tip reads
50 lb/in. 2 , calculate the discharge.
191. Air flows through a 3in. constriction in a 6~in. pipe. Pressure gages con
nected to pipe and constriction read respectively 50 lb/in. 2 and 35 lb/in. 2 The tem
perature in the pipe is 200 F, and the barometric pressure 14.7 lb/in. 2 Calculate the
(weight) rate of flow.
192. Solve the preceding problem when the pressure gage reading at the con
.sttiction is maintained at (a) 15 lb/in. 2 , (b) 5 lb/in. 2
193. Carbon dioxide discharges from a 6in. pipe through a 2in. nozzle into the
atmosphere. If the gage pressure in the pipe is 10 lb/in. 2 , the temperature 100 F,
and the barometric pressure 30.5 in. mercury, calculate the weight of CO2 discharged
per second. Calculate the pressure and temperature within the jet.
194. Solve the preceding problem when the pressure gage reads 20 lb/in. 2
195. Superheated steam at 5 lb/in. 2 and 328 F. (H = 1203.5 Btu/lb) dis
PROBLEMS
103
charges from a 6in. pipe through a 2in. nozzle into the atmosphere. Calculate the
rate of flow. The barometer reads 15.0 lb/in. 2 In the steam jet H = 1179.7 Btu/lb,
w = 0.0347 lb/ft 3 .
196. If in the preceding problem discharge takes place into a tank where the
pressure is 8 lb/in. 2 abs, calculate the rate of flow and pressure in the steam jet.
Critical pressure ratio is 0.55 (in jet, H = 1155.8 Btu/lb, w = 0.0274 lb/ft 3 ).
197. A 250gpm horizontal jet of water 1in. in diameter strikes a stationary blade
which deflects it 60 from its original direction. Calculate the vertical and horizontal
components of force exerted by the liquid on the blade. Find the magnitude and
direction of the total force on the blade.
198. Solve the preceding problem with a deflection of 150.
199. A 2in. jet of water moving at 120 ft/sec has its direction reversed by a
smooth stationary deflector. Calculate the magnitude
and direction of the force on the deflector.
200. The jet of the preceding problem strikes a
stationary flat plate whose surface is normal to the jet.
Calculate the magnitude and direction of the force on the
plate.
201. This 2in. diameter jet moving at 100 ft /sec is
45
divided in half by a "splitter" on the stationary flat plate. Calculate the magnitude
and direction of the force on the plate.
202. Calculate the magnitude and direction of the vertical and horizontal com
ponents and the total force exerted on this
stationary blade by a 2in. jet of water mov
ing at 50 ft /sec.
203. Calculate the magnitude and direc
tion of the force required to move this single
blade horizontally against the direction of the
jet at a velocity of 50 ft/sec. What horse
power is required to accomplish this motion?
204. What force is exerted on a single flat plate moving at 20 ft/sec by a jet of
1in. diameter having a velocity of 50 ft/sec, if they both move in the same direction
and the surface of the plate is normal to the jet?
205. A 6in. pipe line equipped with a 2in. nozzle supplies water to an impulse
turbine 6 ft in diameter having blade angles of 165. Plot a curve of theoretical
horsepower vs. rpm when the pressure behind the nozzle is 100 lb/in. 2 What is the
force on the blades when the maximum horsepower is being developed?
206. A horizontal 6in. pipe in which 2.2 cfs of water are flowing contracts to
3in. diameter. If the pressure in the 6in. pipe is 40 lb/in. 2 , calculate the magnitude
and direction of the horizontal force exerted on the contraction.
104
THE FLOW OF AN IDEAL FLUID
207. A horizontal 2in. pipe in which 400 gpm of water are flowing enlarges to a
4in. diameter. If the pressure in the smaller pipe is 20 lb/in. 2 , calculate magnitude
and direction of the horizontal force on the enlargement.
208. A conical enlargement in a vertical pipe line is 5 ft long and enlarges the pipe
from 12in. to 24in. in diameter. Calculate the magnitude and direction of the verti
cal force on this enlargement when 10 cfs of water flow upward through the line and
the pressure at the smaller end of the enlargement is 30 lb/in. 2
209. A 2in. nozzle terminates a 6in. horizontal water line. The pressure behind
the nozzle is 60 lb/in. 2 Calculate the magnitude and direction of the force on the
nozzle.
210. A 90 bend occurs in a 12in. horizontal pipe in which the pressure is 40 lb/in. 2
Calculate the magnitude and direction of the force on the bend when 10 cfs of water
flow therein.
211. A 6in. horizontal pipe line bends through 90 and while bending changes its
diameter to 3 in. The pressure in the 6in. pipe is 30 lb/in. 2 , and the direction of flow
is from larger to smaller. Calculate the magnitude and direction of the total force
on the bend when 2.0 cfs of water flow therein.
212. Solve the preceding problem if the bend is 120.
213. A 2in. smooth nozzle discharges horizontally from a tank under a 30ft head
of water. Calculate the force exerted on the tank.
214. A 3in. vertical pipe line discharges 50 gpm of water into a tank of water
whose free surface is 10 ft below the end of the pipe. What force is exerted on the
tank?
215. An airplane flies at 120 mph through still air (w = 0.0763 lb/ft 8 ). The pro
peller is 6 ft in diameter, and its slipstream has a velocity of 200 mph relative to the
fuselage. Calculate (a) the propeller efficiency, (6) the velocity through the plane of
the propeller, (c) the horsepower input, (d) the horsepower output, (e) the thrust of
the propeller, (f) the pressure difference across the propeller plane.
216. A ship moves up a river at 20 mph (relative to shore). The river current
has a velocity of 5 mph. The velocity of water a short distance behind the propellers
is 40 mph relative to the ship. If the velocity of 100 cfs of water is changed by the
propeller, calculate the thrust.
217. This stationary blade is pivoted
at point 0. Calculate the torque exerted
thereon when a 2in. water jet moving at
100 ft /sec passes over it as shown.
218. A radial reaction turbine has
r i = 3 ft, f2 = 2 ft, and its flow cross section is 1 ft high. The guide vanes are set
so that ai = 30. When 100 cfs of water flow through this turbine the angle exg is
found to be 60. Calculate the torque exerted on the turbine runner. If the angle
02 is 150, calculate the speed of rotation of the turbine runner and the angle ft
PROBLEMS 105
necessary for smooth flow into the runner. Calculate the horsepower developed by
the turbine at the above speed.
219. A centrifugal pump impeller having dimensions and angles as shown rotates
at 500 rpm. Assuming a radial direction of velocity at the blade entrance, calculate
the rate of flow, the pressure difference between inlet and outlet of blades, and the
torque and horsepower required to meet these conditions.
220. An open cylindrical tank 5 ft in diameter and 10 ft high containing water
to a depth of 6 ft is rotated about its vertical axis at 75 rpm. Calculate the Bernoulli
constants taking the datum at the bottom of the tank for the two streamlines lying in
a horizontal plane 1 ft above the bottom of the tank and having radii of 1 ft and 2 ft.
Calculate also the circulation along these streamlines.
221. In a free vortex the velocity 0.5 ft from the center of rotation is 10 ft /sec.
Calculate the Bernoulli constants for the two streamlines lying in a horizontal plane
5 ft below the water surface and having radii of 1 ft and 2 ft. Calculate also the
cirrulation,along these streamlines.
CHAPTER IV
THE FLOW OF A REAL FLUID
The flow of a real fluid is vastly more complex than that of an ideal
one owing to phenomena caused by the existence of viscosity. Vis
cosity introduces a resistance to motion by causing shear or friction
forces to exist between fluid particles. For flow to take place work
must be done to overcome these resistance forces, and in the process
energy is lost in heat. The study of real fluid flow is essentially one
of the relation of viscosity and other variables to these resistance forces
and lost energies.
25. Laminar and Turbulent Flow. The effects of viscosity cause
the flow of a real fluid to occur under two very different conditions,
or regimes: that of "laminar" flow and that of "turbulent" flow. The
characteristics of these regimes were first demonstrated by Reynolds, 1
with an apparatus similar to that of Fig. 63. Water flows from a glass
Dye
^c
,: '. I ' , '.=&=?.
H
Water
~
B
^
A
L= x
__j u
FIG. 63.
tank through a bellmouthed glass pipe, the flow being controlled by
the valve A. A thin tube, B, leading from a reservoir of dye, C,
has its opening within the entrance of the glass pipe. Reynolds dis
covered that, for low velocities of flow in the glass pipe, a thin filament
of dye issuing from the tube did not diffuse, but was maintained intact
1 O. Reynolds, " An experimental investigation of the circumstances which deter
mine whether the motion of water shall be direct or sinuous and of the law of resistance
in parallel channels," Philosophical Transactions of the Royal Society, Vol. 174,
Part III, p. 935, 1883.
106
LAMINAR AND TURBULENT FLOW 107
throughout the pipe, forming a thin straight line parallel to the axis
of the pipe. As the valve was opened, however, and greater velocities
obtained, the dye filament wavered and broke, diffusing through the
flowing water in the pipe. Reynolds found that the velocity at which
the filament of dye diffused was dependent upon the degree of quiet
ness of the water in the tank, higher velocities being obtainable with
increased quiescence. He also discovered that if the dye filament had
once diffused it became necessary to decrease the velocity in order to
restore it, but the restoration always occurred at approximately the
same velocity in the pipe.
Since intermingling of fluid particles during flow would cause dif
fusion of the dye filament, Reynolds deduced from his experiments
that at low velocities this intermingling was entirely absent and that
the fluid particles moved in parallel layers, or laminae, sliding over
particles in adjacent laminae, but not mixing with them; this is the
regime of ''laminar flow." Since at higher velocities the dye filament
diffused through the pipe, it was obvious that some intermingling of
fluid particles was occurring, or, in other words, the flow was " tur
bulent." Laminar flow broke down into turbulent flow at some
critical velocity above that at which turbulent flow was restored to
the laminar condition; the former velocity is called the upper critical
velocity, and the latter, the lower critical velocity.
Reynolds was able to generalize his conclusions from these and
other experiments by the introduction of a dimensionless term, NR,
called the Reynolds number, and defined by
in which V is the average velocity in the pipe, d the diameter of the
pipe, and p and ju the density and viscosity of the fluid flowing therein.
Reynolds found that certain critical values of the Reynolds number,
(NR)CI defined the upper and lower critical velocities for all fluids
flowing in all sizes of pipes, thus deducing the fact that a single number
defines the limits of laminar and turbulent pipe flow for all fluids.
The upper limit of laminar flow was found by Reynolds to corre
spond to a Reynolds number of 12,000 to 14,000, but unfortunately this
upper critical Reynolds number is indefinite, being dependent upon
several incidental conditions such as: (1) initial quietness of the fluid, 2
2 Ekman, working in 1910 with Reynolds' original apparatus, was able to obtain
laminar flow up to a Reynolds number of 50,000 by quieting the water for days before
running his tests.
Fluid (dyer
Fluid layer
v+dv

108 THE FLOW OF A REAL FLUID
(2) shape of pipe entrance, and (3) roughness of pipe. However,
these high values of the upper critical Reynolds number are of only
academic interest and the engineer may take the upper limit of laminar
flow to be defined by (N R ) C = 2700 to 4000.
The lower limit of turbulent flow, defined by the lower critical
Reynolds number, is of greater engineering importance; it defines a
condition below which all turbulence entering the flow from any
source will eventually be damped out by viscosity. This lower critical
Reynolds number thus sets a limit below which laminar flow will
always occur; many experiments have indicated the lower critical
Reynolds number to have a value of 2100.
Between Reynolds numbers of 2100 and 4000 a region of uncer
tainty exists and the engineer must make conservative selection in
this region of the variables which are dependent upon the Reynolds
number.
In laminar flow, agitation of
fluid particles is of a molecular
nature only, and these particles
are restrained to motion in
FlG 64 parallel paths by the action of
viscosity. The shearing stress
between adjacent moving layers is determined in laminar flow by the
viscosity and is completely defined by the differential equation
(Art. 5) ,
dv
the stress being the product of viscosity and velocity gradient (Fig. 64).
In turbulent flow, fluid particles are not retained in layers by viscous
action, but move in heterogeneous fashion through the flow, sliding
past some fluid particles, and colliding with others in an entirely hap
hazard manner, causing a complete mixing of the fluid as flow takes
place. These fluid particles, moving in all directions through the flow,
cause at any point a rapid and irregular pulsation in velocity along the
general direction of motion and across this direction as well. If the
components in the direction of motion of the instantaneous velocities
. at a point could be measured, a variation of high frequency and small
magnitude would be obtained and the time average of all these velocity
components would be the velocity which is taken to exist at the point.
Across the general direction of motion, rapid variation would be found
to exist also, and the time average of these variations would obviously
be zero.
LAMINAR AND TURBULENT FLOW 109
Since turbulence is an entirely chaotic motion of individual fluid
particles through short distances in every direction as flow takes place,
the motion of individual fluid particles is impossible to trace and
characterize mathematically, but mathematical relationships may be
obtained by considering the average motion of aggregations of fluid
particles.
The shear stress between two fluid layers in highly turbulent flow
is not due to viscosity, but rather to the momentum exchanges occur
ring as fluid particles move from one of these layers to the other as the
result of the turbulent mixing process. Taking the velocities of fluid
particles due to turbulence as v y and v x , respectively normal to and
along the direction of general motion, it is evident that if homogeneous
turbulence is assumed
V X = Vy
Using this assumption, Reynolds 3 showed that the shear stress
between moving fluid layers in turbulent flow is given by
r = p v x v y
in which v x v y is the time average of the product of v x and v y . Prandtl 4
succeeded in relating the above velocities of turbulence to the general
flow characteristics by proposing that fluid particles are transported by
turbulence a certain average distance, /, from regions of one velocity to
A ,/
Fluid layer
Fluid layer

y
dy
FIG. 65.
regions of another and in so doing suffer changes in their general
velocities of motion. Prandtl termed the distance / the "mixing
length" and suggested that the change in velocity, Av, incurred by a
fluid particle moving through the distance / was proportional to v x and
to v y . From Fig. 65,
At, = /
8 O. Reynolds, " On the dynamical theory of incompressible viscous fluids and the
determination of the criterion," Philosophical Transactions of the Royal Society, Al,
Vol. 186, p. 123, 1895.
4 L. Prandtl, " Ueber die ausgebildete Turbulenz," Proceedings Second Inter
national Congress Applied Mechanics, Zurich, 1926, p. 62.
110
THE FLOW OF A REAL FLUID
and, from the foregoing statement and substitution of this equation
in the Reynolds expression for shear stress, there results
thus indicating the variables which determine the shear stress in tur
bulent flow.
ILLUSTRATIVE PROBLEM
Water flows in a 6in. pipe line at a velocity of 12 ft/sec and at a temperature
of 50 F. Is the flow laminar or turbulent?
V = 0.0000273 Ib sec/ft 2
NR _
*
0.0000273
Since 355,000 > 2100, flow is turbulent.
= 355,000
26. Fluid Flow Past Solid Boundaries. A knowledge of flow
phenomena near a solid boundary is of great value in engineering prob
lems because in actuality flow is never encountered which is not affected
by the solid boundaries over which it passes.
The classic aeronautical problem is the flow
of fluid over the surfaces of an object such
as a wing or fuselage, and in other branches
of engineering the problem of flow between
solid boundaries, such as in pipes and
channels, is of paramount importance.
One experimentally determined fact
concerning fluid flow over smooth solid
boundaries is that no motion of fluid par
ticles, relative to the boundary, exists
adjacent to the boundary. This means that
a diagram of velocity distribution must
always indicate a velocity of zero at the
boundary. The physical explanation of
this phenomenon is that a very thin layer of fluid, possibly having a
thickness of but a few molecules, adheres to the solid boundary and
motion takes place relative to this layer.
Laminar flow occurring over smooth or rough boundaries possesses
the same properties (Fig. 66), the velocity being zero at the bound 
(&) Rough Boundary.
FIG. 66.
FLUID FLOW PAST SOLID BOUNDARIES
111
ary, and the shear stress between moving layers at any distance,
y, being given by

r
'/////////y.
(a) Smooth Boundary.
dy/
Thus, in laminar flow, surface roughness has no effect on the flow
properties.
In turbulent flow, however, the rough
ness of the boundary surface will affect the
physical properties of fluid motion. When
turbulent flow occurs over smooth solid
boundaries it is always separated from the
boundary by a film of laminar flow (Fig.
67). This laminar film has been observed
experimentally, and its existence may be
justified theoretically by the following sim
ple reasoning : The presence of a boundary
in a turbulent flow will curtail the freedom
of the turbulent mixing process by reduc
ing the available mixing length, and, in a
region very close to the boundary, the
available mixing length is reduced to zero, resulting in a film of
laminar flow over the boundary.
In the laminar film the shear stress, r, is given by
(b) Rough Boundary.
FlG. 67.
r =
dv
l Ty
and at a distance from the boundary where turbulence is completely
developed
r oc
Y*\
w
Between the latter region and the laminar film lies a transition region
in which shear stress results from a complex combination of turbulent
and viscous action, turbulent mixing being restricted by the viscous
effects due to the proximity of the wall. The fact that there is a
transition from fully developed turbulence to no turbulence at the
boundary surface shows that the laminar film, although given an
arbitrary thickness, <5, in Fig. 67, possesses no real line of demarcation
Between the laminar and turbulent regions.
112 THE FLOW OF A REAL FLUID
The roughness of boundary surfaces will affect the physical proper
ties of turbulent flow, and the effect of this roughness is dependent
upon the thickness of the laminar film. A boundary surface is said
to be "smooth" if its projections or protuberances are so completely
submerged in the laminar film (Fig. 67a) that they have no effect on
the turbulent mixing process. However, when the height of the
roughness projections is equal to or greater than the thickness of the
laminar film (shown schematically in Fig. 676), the projections serve
to augment the turbulence and, if roughness is excessive, even to pre
vent the existence of such a film. Since the thickness of the laminar
film, which is a variable quantity, is a criterion of effective roughness,
it is possible for the same boundary surface to behave as a smooth one
or a rough one, depending upon the thickness of the laminar film which
covers it.
Since surface roughness serves to increase the turbulence in a
flowing fluid and thus decrease the effect of viscous action some pre
diction may be made as to the effect of roughness on energy losses.
In turbulent flow over rough surfaces energy is consumed by the work
done in the continual generation of turbulence by the roughness
protuberances. The energy involved in this turbulence is composed
of the kinetic energy of fluid masses, which is known to be proportional
to the squares of their velocities. Since these velocities are in turn
proportional to the velocities of general motion it may be concluded
that energy losses caused by rough surfaces vary with the squares of
velocities.
As turbulent flow takes place over smooth surfaces, work is done
at the expense of available fluid energy against the shear stress due to
viscous action in the laminar film. No predictions will be made as
to the relation of energy losses and velocities in this case where a com
bination of turbulent and viscous action exists, but many experiments
have indicated that for turbulent flow over smooth surfaces energy
losses will vary with the 1.75 to 1.85 power of velocities.
In laminar flow over boundary surfaces, energy losses will sub
sequently be shown theoretically and experimentally to be directly
proportional to velocities.
The velocity profiles of Fig. 67 indicate a small velocity gradient
in the turbulent region at a comparatively small distance from the
boundary. In other words, flow with a uniform velocity, undisturbed
by the presence of the boundary, is taking place in close proximity
to it, and this flow, although turbulent, possesses the velocity and
pressure characteristics of an ideal fluid flow. The suggestions
FLUID FLOW BETWEEN PARALLEL PLATES
113
(1) that the viscous action of a real fluid on a solid surface is confined
to a thin region close to the boundary, and (2) that outside of this
region fluid can be treated as an ideal one were made by Prandtl 5
and served to revolutionize the treatment of the subject. The sug
gestions justified the use of the ideal fluid in determining the velocities,
pressures, and shapes of the streamlines about an object in a real fluid
flow.
27. Fluid Flow between Parallel Plates. To develop some of the
fundamental mechanical relationships of fluid flow in passages, con
sider an incompressible fluid flowing in a section between two vertical
parallel plates of infinite extent.
FIG. 68.
Figure 68 represents two such vertical plates with spacing, 2y ,
between which flow through a section of height b is to be analyzed.
From the continuity principle the average velocities of flow, Vi and F 2 ,
through any two sections, 1 and 2, are the same, and thus any mass of
fluid, ABCD, of width 2y, which is isolated in the flow, moves with
constant velocity. Constant velocity motion requires that the net
force acting on this fluid mass be zero. Since the force F T , due to the
shear stress r, opposes motion, the pressure force Fp, which balances
this force, must act in the direction of motion, requiring that pi > p%,
and showing that a pressure drop occurs in the direction of flow. If
sections 1 and 2 are separated by a distance /, and the pressures over
6 Proceedings Third International Mathematics Congress, Heidelberg, 1904.
114 THE FLOW OF A REAL FLUID
these sections are uniform, the shear force, F T , and pressure force,
may be calculated directly as
F r = 26/r
and
F P = 2by(pi  p2)
which may be equated
2blr = 2by(pi  p 2 )
giving the fundamental result
and proving that r varies linearly with y, being zero midway between
the two plates and possessing a maximum value, T O , at the walls of the
passage.
The lost power, PL, consumed by the total flow through the section
of height b in overcoming the shear force F To on its boundary surface
may be readily calculated from
PL = F TO V
in which
F TO = r 2bl = (p l  p 2 )2y b
giving
PL = (Pi  p2)2y bV
but lost power may also be calculated from
P L = QwE L
in which EL is the lost energy per pound of fluid flowing. Equating
these two expressions for lost power in which
Q = b2y V
(Pi  p2)lyobV = 2y bwE L
whence
Pl P2
E L =
w
or the energy consumed per pound of fluid between two sections is
simply the loss in pressure head between the two sections. The same
FLUID FLOW BETWEEN PARALLEL PLATES 115
result may be obtained directly from the following energy balance by
the Bernoulli equation
Pi , Vl P2.VI
 r =  r  h EL
w 2g w 2g
and since V\ = V%
w
The foregoing developments are perfectly general, applying to
both laminar and turbulent flow. Now for mathematical simplicity
assume that laminar flow exists between the two parallel plates to
illustrate the application of the shear stress equations. In laminar
flow the shear stress, r, is given by 6
dv
r == ~~ M T~
dy
and between parallel plates by
For laminar flow between parallel plates both these equations must be
satisfied; therefore
~ M ^ = v r~) y
the variables of which may be separated as follows :
The lefthand side of this equation may be integrated between the
center velocity, V c , and the variable velocity, v, at a distance, y, from
the center; the righthand side may be integrated between the cor
responding limits, zero and y. Integrating
v c v
' (32)
6 The minus sign appears since y in the above problem is taken in a direction
opposite to that assumed in previous problems; here v decreases with increasing y,
116
THE FLOW OF A REAL FLUID
showing that for laminar flow between parallel plates the velocity
distribution is a parabolic curve having a maximum value midway
between the plates, as illustrated in Fig. 68.
28. Flow About Immersed Objects. 7 As real fluid flows in turbu
lent condition over the surface of a solid object placed in the flow,
the effects of viscosity will create velocity conditions at the surface
similar to those described above. However, the laminar film which
forms over the surface of the object in general is not of constant thick
ness since it must begin from no thickness at the front of the object
where the fluid first contacts it, and increase along the surface of the
object in the direction of motion ; this type of laminar film is termed a
"laminar boundary layer." Under certain conditions this layer may
change into a "turbulent boundary layer" which possesses a thin
laminar film beneath it and adjacent to the object. These phenomena
are illustrated in Fig. 69 for a simple type of object, a smooth flat
plate, the thicknesses of the various layers and films being greatly ex
aggerated. Although the bound
ary layer occupies an extremely
small and usually invisible part
of the flow picture it is, neverthe
less, of great importance since it is
the essential reason for the exis
tence of a frictional "drag" force
exerted by the fluid on the object.
The boundary layer in adhering to, or separating from, the object
on which it forms brings about different flow phenomena and different
effects upon the drag force. On a streamlined object (Fig. 70a and b)
the boundary layer will adhere to the surface of the object and the
flow picture appears to be identical with that of an ideal fluid. On a
blunt object, however, the boundary layer will cause the flow to
separate from the object, resulting in a flow picture (Fig. 70d) vastly
different from that of an ideal fluid (Fig. 70c). The phenomenon of
separation thus becomes an important factor in determining the
characteristics of fluid flow about objects; its mechanism and proper
ties must now be investigated.
To discover the fundamental properties of separation, compare
ideal fluid flow and real fluid flow about a blunt object such as a circular
cylinder (Fig. 71). Let the cylinder be placed in an ideal flow possess
ing a pressure p and velocity V  A symmetrical streamline picture
layer
layer
FiG. 69.
7 See Chapter IX for a more complete treatment of this subject.
FLOW ABOUT IMMERSED OBJECTS
117
and pressure distribution will result, and, of course, no energy will be
lost as flow takes place about the object. For a real fluid having the
(a) Ideal Fluid.
(c) Ideal Fluid.
(6) Real Fluid.
<d) Real Fluid.
FIG. 70.
same velocity and pressure, energy will be consumed in overcoming
resistance caused by the shear stresses in the boundary layer as flow
Pressure Variation
Ideal fluid
Real fluid
Surface of discontinuity
FIG. 71.
passes over the surface of the object. The result is that real fluid
particles possess less energy when they arrive at point X than ideal
118 THE FLOW OF A REAL FLUID
fluid particles, although these particles started with the same energy
content. The fluid particles, in moving from point X to S f along the
surface of the object, are moving into a region of high pressure and must
possess enough energy to accomplish this motion against the pressure
gradient. The ideal fluid particles can do this because of their satis
factory energy content, but the real fluid particles are unable to do
the same since some of their energy has been dissipated. The result
is that these particles are unable to move beyond a point Y on the
rear surface of the cylinder; here they come to rest, accumulate, and
are given a rotary motion by the surrounding flow. An eddy of in
creasing size is then developed at point Y, and the momentum of this
eddy becomes so great that the eddy cannot be retained by the
cylinder but must break away from it, allowing another one to
form and the process to repeat itself.
The result of separation and eddy formation is the formation behind
solid objects of a turbulent wake, the turbulence of which is of an
"eddying" nature in contrast
to the "normal" turbulence
discussed in Art. 25. In the
^ O O O O creation of the eddies of the
turbulent wake, fluid energy
 has been stored in the eddies
FIG. 72. and, therefore, made unavail
able. As these eddies die out
owing to the influence of viscosity, their energy is converted into heat
and lost from the fluid flow. The turbulent wake thus becomes
another fluid mechanism in which energy may be lost.
The turbulent wake behind a blunt object is separated from the
"live" flow by a "surface of discontinuity" on each side of which
pressures may be the same, but velocities differ greatly. Such a sur
face of discontinuity is indicated schematically on Fig. 71. Actually
this surface does not possess the symmetry, uniformity, and stability
implied by the sketch, but wavers in a transverse direction as the
eddies form; besides this property the discontinuity surface itself
has an inherent tendency to break up into eddies. To illustrate this
tendency, consider a simple type of a discontinuity surface which may
be created between adjacent fluid streams moving with different veloc
ities (Fig. 72). If an observer moves in the direction of the streams
with a velocity equal to the numerical average of those of the streams,
he sees the relative velocity profile shown at the right from which the
tendency for eddy creation is immediately evident.
FLOW ABOUT IMMERSED OBJECTS 119
To return now to the problem of drag forces on objects in a fluid
flow, the effects of boundary layers, separation, and wakes may be
observed. Fundamentally, drag is caused by the components of the
normal and tangential forces
transmitted from the fluid to
the surface elements of the gin
solid object. These normal
forces are those of pressure,
which in general may be
calculated by applying the
Bernoulli principle to the IG *
strearntube adjacent to the object. The tangential forces are those
of shear at the surface of the object arising from viscous effects in the
boundary layer. From Fig. 73, the total drag force, J9, may be ex
pressed mathematically as
r 8 r s
D = I pdA sin a + I TodA cos a (33)
r s
in which / designates ''integral over the surface of the object."
The drag resulting from the pressure variation over the surface of the
object is called the "form" or "profile" drag, D p , since its magnitude
will be found to depend primarily on the "form" or "profile" of the
object. The drag force, D f , incurred by the shear stresses over the
surface of the object due to frictional effects is termed the "frictional"
or "skin friction drag." Hence the relations:
r s
Profile drag, D p = I pdA sin a
(34)
r s
Frictional drag, D/ = / r dA cos a. (35)
and from equation (33)
D = D p + D f (36)
The approximate relative magnitudes of these drag forces occurring
on various objects when placed in a turbulent flow and the effect of
the turbulent wake upon them may be obtained from a study of objects
having different shapes but the same cross section, placed in the same
fluid flow. Three such objects are: the thin circular disk, sphere, and
streamlined form of Fig. 74.
For the disk, the streamline picture indicates a stagnation point
at the center of the upstream side and a greatly reduced pressure at
120
THE FLOW OF A REAL FLUID
the edges. This reduced pressure, being adjacent to the turbulent
wake, is transmitted into it, causing the downstream side of the disk
to be exposed to reduced pressure which will contribute to the profile
drag force. Designating the pressure reduction below that in the
FIG. 74.
undisturbed flow by vectors directed away from the surface of the disk
(and that in excess of the pressure in the undisturbed flow by vectors
toward the surface), a pressure diagram results whose net area is the
profile drag. Thus the profile drag is given by
D p = 2(ABC + BFGH  FCE)
and the frictional drag, Df, will then be zero since none of the shear
stresses on the surface of the disk have components in the direction of
motion.
For the sphere, the turbulent wake is smaller than that of the disk
and, from examination of the streamline picture, will possess a some
what higher pressure. To obtain the profile drag graphically the dia
gram of the pressure components is necessary. Here
D p = 2(ABC + EFG  CDE)
which is seen to be much smaller than that of the disk, a fact which is
confirmed by many experiments indicating the profile drag of the
sphere to be roughly onethird that of the disk. The frictional drag
of the sphere is a finite quantity since the shear stresses at its surface
have components in the direction of motion; however, shearstress
variations are extremely difficult to calculate and their small magni
STABILITYSECONDARY FLOWS 121
tudes will result in a frictional drag which is negligible compared to
the profile drag of the sphere.
For the streamlined form the turbulent wake is extremely small and
the pressure surrounding it and within it is comparatively large since
the gentle contour of the body has allowed deceleration of the flow and
consequent regain of pressure, without incurring separation. The
pressure diagrams lead directly to the conclusion that the profile drag
of the streamlined form is very small, and experimental results indicate
it to be about 1/40 that of the disk. The frictional drag for objects
of this shape is much greater than that for the sphere since streamlining
has brought more surface area into contact with the flow. For well
streamlined objects frictional drag assumes a magnitude comparable to
that of profile drag.
The foregoing examples illustrate the fact that the viscosity prop
erty of a fluid is the root of the drag problem. Viscosity has been seen
to cause drag either by frictional effects on the surface of an object or
through profile drag by causing separation and the creation of a low
pressure region behind the object. By streamlining an object the
size of its lowpressure turbulent wake is decreased and a reduction in
profile drag is accomplished, but in general an increase in frictional
drag is incurred.
For an ideal fluid in which there is no viscosity and thus no cause
for frictional effects or formation of a turbulent wake, regardless of
the shape of the object about which flow is occurring, it is evident that
the drag of the object is zero. Two centuries ago, D'Alembert's dis
covery that all objects in an ideal fluid exhibit no drag was a funda
mental and disturbing paradox ; today this fact is a logical consequence
of the fundamental reasoning presented above.
29. Stability Secondary Flows. The flow phenomena about a
symmetrical body, such as the sphere, suggests the existence of a
general law of fluid motion which will prove to be widely applicable.
On the front half of the sphere the flow picture has been seen to be
practically identical with that of the ideal fluid, but on the rear half,
where a turbulent wake forms, the flow picture bears no resemblance to
that of the ideal fluid. On the front half of the sphere the flow is
being accelerated and pressure head (or energy) is being converted
into velocity head. For ideal flow to occur behind the sphere, decelera
tion equivalent to the above acceleration must occur and velocity head
must be reconverted into pressure head. Owing to boundary layer
phenomena this deceleration of the fluid flow is not accomplished
without the formation of .a turbulent wake, the eddies of which serve
122
THE FLOW OF A REAL FLUID
to consume fluid energy. This single example illustrates the following
general law of fluid motion: Acceleration of a moving fluid, identified
by convergent streamlines and a decrease in pressure in the direction of
motion, is an efficient and stable fluid process, accompanied by no
eddy formation and small energy losses; on the other hand, decelera
tion of fluid flow is an inefficient process, accompanied by instability,
eddy formation, and large energy losses.
These principles may be seen to apply to flow through passages
such as the convergentdivergent tube or the nozzle of Fig. 75, where
smooth, stable flow exists in the convergent passages but separation
and eddying turbulence occur as the flow is decelerated.
Acceleration
Deceleration
Acceleration
Deceleration
> Surface of discontinuity
FIG. 75.
Another engineering application of these principles occurs in a
comparison of the efficiencies of hydraulic turbines and centrifugal
pumps. In the turbine the flow passages are convergent, causing
continual acceleration of the fluid through the machine. In the pump,
which creates pressure head by dynamic means, the passages are diver
gent. Hydraulic turbine efficiencies have been obtained up to 94 per
cent but maximum centrifugal pump efficiencies range around 87
per cent, and the difference between these efficiencies may be attributed
to the inherent efficiency and inefficiency of the acceleration and
deceleration processes.
Another consequence of the boundary layer is the creation of a
flow within a flow, a "secondary flow" superimposed upon the main,
or "primary," flow. A classical and useful example of the creation of
an eddy motion and dissipation of energy by a secondary flow occurs
when fluid flows through a smooth bend in a circular pipe (Fig. 76).
BIBLIOGRAPHY
123
For an ideal fluid flowing under these conditions it has been shown
(Art. 24) that a pressure gradient develops across the bend due to the
centrifugal forces of fluid particles as they move through the bend.
Stability occurs in the ideal fluid when this pressure gradient brings
about a balance between the centrifugal and centripetal forces on the
fluid particles. In a real fluid this stability is disrupted by the velocity
being reduced to zero at the walls owing to the existence of the laminar
film. The reduction of velocity at the outer part (A) of the bend
reduces the centrifugal force of the particles moving near the wall,
causing the pressure at the wall to be below that which would be main
tained in an ideal fluid. However, the velocities of fluid particles
Laminar film
FlG. 76.
toward the center of the bend are about the same as those of the ideal
fluid, and the pressure gradient .developed by their centrifugal forces
is about the same. The " weakening" of the pressure gradient at the
outer wall will cause a flow to be set up from the center of the pipe
toward the wall which will develop into the twin eddy motion shown,
and this secondary motion added to the main flow will cause a double
spiral motion, the energy of which will be dissipated in heat as the
motion is destroyed by viscosity. The energy of the secondary motion
has been derived from the available fluid energy, and as viscosity
causes this energy to be dissipated, fluid energy is lost in much the same
way as it is lost by eddies in a turbulent wake,
BIBLIOGRAPHY
B. A. BAKHMETEFF, The Mechanics of Turbulent Flow, Princeton University Press,
1936.
H. ROUSE, " Modern Conceptions of the Mechanics of Fluid Turbulence," Trans.
A.S.C.E., Vol. 102, 1937.
Fluid Mechanics for Hydraulic Engineers, McGrawHill Book Co., 1938.
L. PRANDTL and O. G. TIETJENS, Applied Hydro and Aeromechanics, McGrawHill
Book Co., 1934.
124 THE FLOW OF A REAL FLUID
L. PRANDTL, The Physics of Solids and Fluids, Part II, Blackie & Son, 1936.
C. V. DRYSDALE, The Mechanical Properties of Fluids, Blackie & Son, 1925.
R. A. DODGE and M. J. THOMPSON, Fluid Mechanics, McGrawHill Book ,Co., 1937.
M. P. O'BRIEN and G. H. HICKOX, Applied Fluid Mechanics, McGrawHill Book Co.,
1937.
PROBLEMS
222. If 30 gpm of water flow in a 3in. pipe line at 70 F, is the flow laminar or
turbulent?
223. Glycerine flows in a 1in. pipe at a velocity of 1 ft/sec and temperature of
80 F. Is the flow laminar or turbulent?
224. Superheated steam at 400 F and absolute pressure 100 lb/in. 2 (w  0.202
lb/ft 3 ) flows in a 1in. pipe at a velocity of 5 ft/sec. Is the flow laminar or turbulent?
225. Linseed oil flows at 80 F in a in. pipe at 2 ft/sec. Is the flow laminar or
turbulent?
226. Carbon dioxide flows in a 2in. pipe at a velocity of 5 ft /sec, temperature of
150 F, and pressure 40 lb/in. 2 The barometer reads 15.0 lb/in. 2 Is the flow
laminar or turbulent?
227. What is the maximum flow of water which may occur in a 6in. pipe at 80 F
at laminar condition?
228. What is the maximum flow of air that may occur at laminar condition in a
4in. pipe at 30 lb/in. 2 abs and 100 F.?
229. What is the largest diameter pipe line that may be used to carry 100 gpm
of Unseed oil at 80 F if the flow is to be laminar?
230. A fluid flows in a 3in. pipe line which discharges into a 6in. line. Calculate
the Reynolds number in the 6in. pipe if that in the 3in. pipe is 20,000.
231. The loss of energy in a certain pipe line flow is 3 ftlb/lb of fluid flowing.
What loss of energy will occur when the flow is doubled, assuming (a) laminar flow,
(b) turbulent flow and smooth pipe, (c) turbulent flow and rough pipe?
232. Water flows horizontally between vertical parallel plates spaced 2 ft apart.
If the pressure drop in the direction of flow is 4 lb/in. 2 per 100 ft, calculate the shear
stress in the flow at the surfaces of the plates, 3 in., 6 in., and 9 in. from them, and
at the midpoint between them.
233. A liquid flows in laminar condition between two parallel plates 2 ft apart.
If the velocity at the midpoint between the plates is 4 ft/sec, calculate the velocities
at the plates and at distances of 3 in., 6 in., and 9 in. from them. If the viscosity of
this liquid is 0.1 Ib sec/ft 2 and its density 1.8 slugs/ft 3 , calculate the loss of pressure
and loss of head in a distance of 100 ft along the flow. What is the rate of flow
through a section having b 3 ft? (See Fig, 68.)
CHAPTER V
SIMILARITY AND DIMENSIONAL ANALYSIS
30. Similarity and Models. Near the latter part of the last cen
tury, models began to be used to study flow phenomena which could
not be solved by mathematical methods or by means of available
experimental results. At the present time the use of models is increas
ing: the aeronautical engineer obtains data and checks his designs by
model tests in wind tunnels; the ship designer tests ship models in
towing basins; the mechanical engineer tests models of turbines and
pumps and predicts the performance of the fullscale machines from
these tests; the civil engineer works with models of hydraulic structures
and rivers to obtain more reliable solutions to his design problems.
The justification for the use of models is an economic one a model,
being small, costs little compared to the "prototype" from which it is
built, and its results may lead to savings of many times its cost; a
model also adds a certainty to design which can never be obtained
from calculations alone.
Similarity of flow phenomena not only occurs between a prototype
and its model but also may exist between various natural phenomena
if certain laws of similarity are satisfied. Similarity thus becomes a
means of correlating the apparently divergent results obtained from
similar fluid phenomena and as such becomes a valuable tool of modern
fluid mechanics ; the application of the laws of similarity will be found
to lead to more comprehensive solutions and thus to a better under
standing of fluid phenomena in general.
There are many types of similarity, all of which must be obtained
if complete similarity is to exist between fluid phenomena. The first
and simplest of these is the familiar geometrical similarity which states
that the flow pictures of model and prototype have the same shape,
and, therefore, that the ratios between corresponding lengths in model
and prototype are the same. In the model and prototype of Fig. 77,
for example,
125
126
SIMILARITY AND DIMENSIONAL ANALYSIS
Corollaries of geometric similarity are that areas vary with the
squares of lengths, thus
= (AY = (L\ 2
\dj \lj
and that volumes vary with the cubes of lengths.
 1
Prototype
Model
FIG. 77.
If the two similar objects of Fig. 77 are placed in the similar fluid
flows of Fig. 78, another type of similarity, kinematic similarity, exists
if motion of the fluid about the objects is the same. Such similarity of
motion is characterized by ratios of corresponding velocities and accel
erations being the same throughout the flow picture; for example,
FIG. 78.
In order to maintain geometric and kinematic similarity between
flow pictures there must be forces acting on corresponding fluid masses
which are related by ratios similar to those above, and this similarity,
governed by the existence of such force ratios, is called "dynamic
similarity/' Without defining the nature of the forces, dynamic
similarity may be indicated schematically (Fig. 78) by
SIMILARITY AND MODELS 127
or
(*) = (Is]
\F!/ model V^l/p
prototype
But these force ratios must be maintained for all the corresponding
fluid masses throughout the flow pictures, and thus it is evident that
they can be governed only by relations between dynamic and kinematic
properties of the flow and the physical properties of the fluids involved.
The forces which may act on fluid masses in a fluid flow are those
of pressure, Fp; inertia, F/; gravity, FQ\ viscosity, Fy\ elasticity, FE\
and surface tension, FT Since these forces are taken to be those on
any fluid mass, they may be generalized by the following fundamental
relationships:
F P = pA = pi 2
Fj = Ma = p/ 3 ^ = P F 2 / 2
/
FQ = M g = pl 3 g
F V = ^A = M ^?
dy I
F E = EA = El 2
F T = Tl
To obtain dynamic similarity between two flow pictures when all
these forces act, all independent force ratios which can be written
must be the same in model and prototype; thus dynamic similarity
between two flow pictures when all possible forces are acting is
expressed by the following five simultaneous equations :
,Fi/P \Fi/m \pV 2 /P \pFVm
FI\ ( FT\ ( F/p\ / F/p\ _ /Reynolds\ _/Reynolds\
J*r/p ~\Fv/ \~M~A "" \^T/ W "" \ number JP\ number ) m
FI\ / Fi\ / F 2 \ / F 2 \ __ / Froude \ f Froude \
WP \7fc/* = \lg)p ~ \ !g/ ~ \ number /P " \ number ) m
JF^s/p \~FJs )m "" \"E"/P \~/m "" \ number / p ""\ number ) m
Fi\ /Fi\ fplV 2 \ (plV*\ t Weber \ f Weber \
JT/P ~\FT/'* \~/p~\T~/~ \ number ) p \ number ) m
128
SIMILARITY AND DIMENSIONAL ANALYSIS
in which the quantities, , V, and / may be any pressure, velocity,
and length, provided that the quantities used are the corresponding
ones in model and prototype ; the force ratios are named for the experi
menters who first derived and used them.
Fortunately, in most engineering problems the above five equa
tions are not necessary since some of the forces stated above (1) may
not act, (2) may be of negligible magnitude, or (3) may be related by
certain known laws and, therefore, are not independent. In each new
problem of similarity a good understanding of fluid phenomena is
necessary to determine how the problem may be satisfactorily simpli
fied by the elimination of irrelevant or negligible forces. The reason
ing involved in such an analysis is best illustrated by citing certain
simple and recurring engineering examples.
In the classical aeronautical problem where a model of a wing is to
be built and tested (Fig. 79), the model must first have the same
shape as its prototype and be placed in the flow at the same "angle of
attack/ 1 a. After these requirements are met, dynamic/ kinematic,
and geometric similarity of the flow pictures will be obtained if all the
relevant force ratios are made equal in model and prototype.
Certain forces, however, may be eliminated immediately: Surface
tension forces are negligible; if the fluids are taken to be incom
pressible, the elasticity forces drop out; the gravity forces, although
acting on all fluid particles, do not affect the flow picture and, there
fore, may be omitted. Thus, in this problem there exist only the
forces of pressure, inertia, and viscosity which must be related by
some physical equation. Since all corresponding forces in model and
prototype have the same ratio only one ratio is necessary to character
ize complete dynamic similarity. This ratio is
and from this equation it is evident that complete dynamic similarity
fixes no theoretical restrictions as to the fluid which may be used in
SIMILARITY AND MODELS 129
the testing of a model ; any fluid may be used provided that the Rey
nolds number of the model can be made equal to the Reynolds number
of the prototype.
The above reasoning may be applied without change to the tur
bulent flow of fluids in circular pipes (Fig. 80). Here geometric simi
larity requires that the roughness pattern of the pipe surfaces be
similar, and complete dynamic similarity is
obtained when f p l J
(AJ , , , * ( ^ *
(N R )i = (N R ) 2
Here, again, it is immaterial which fluids are ^  )
involved; complete dynamic similarity is ob '  ;p   ^
tained when the Reynolds numbers of the two FIG. 80.
flows are the same. 1
It has been shown (Art. 25) that the laminar flow regime is defined
by low values of the Reynolds number and that low Reynolds numbers
indicate that inertia forces are small compared to those of viscosity.
For laminar flow, pressure forces are relatively large, but, owing to
large resistance, fluid motion is slow and inertia forces may be neg
lected. Complete dynamic similarity is defined in laminar flow by
Fv/P \F V /m \fjLVj P
an equation whose shape is confirmed by equation 32, one of the laws
of laminar motion between parallel plates.
The laws of shipmodel testing were first developed and used by
William Froude in England about 1870. In this specialized field of
engineering the problems of similarity are different from those pre
sented above, but the fluid phenomena and reasoning involved will
find many other useful engineering applications. In this problem the
fluids suffer inappreciable compression during flow, and the elasticity
forces may be neglected. If the models for testing are not extremely
small the forces of surface tension are entirely negligible, leaving the
forces of pressure, inertia, viscosity, and gravity to be considered.
As a ship moves through the free surface of a liquid it encounters
resistance due to skin friction and to pressure variation, like a sub
1 The significance of this simple statement should be fully appreciated. Complete
dynamic similarity implies similarity of the complex turbulent flow processes in the
two pipes.
130
SIMILARITY AND DIMENSIONAL ANALYSIS
merged object (Art. 28). For a surface vessel the pressures over the
hull are determined by the Bernoulli equation applied to the stream
tubes adjacent to the hull. This pressure variation, however, is mani
fested by a rise or depression of the liquid surface since the pressure
at a point below the surface is related to its depth by the static law
d m t * m '' /m
FIG. 81.
p = wh. The shape of the water surface adjacent to a surface vessel
becomes that of Fig. 81, and, from the static law, the forces of pres
sure (p) become equivalent to those of gravity (wh). Therefore, pres
sure forces are eliminated from the problem and complete dynamic
similarity is defined by the simultaneous equations
VI P
FA /FA (Vlp\ (
~T) = VvJ = V  ) = V
FV/P \FV/ \ M /P \
Solution of these equations results in
^G") 1
v m ^m'
indicating that a relation between the kinematic viscosities of the
fluids involved is determined when the model scale is selected. This
means (1) that a fluid for the towing basin must be found whose kine
matic viscosity is a certain proportion of that of water, or (2) that if
water is used in the towing basin the model scale must be unity, result
ing in a fullscale model ! This fact proves an unsurmoun table obstacle
to complete dynamic similarity for ship models and necessitates a
DIMENSIONAL ANALYSIS 131
compromise between the Reynolds and Froude laws. This com
promise is effected by obtaining ''incomplete dynamic similarity" by
making the Froude numbers in model and prototype the same and cor
recting the test results .by experimental data dependent upon the
Reynolds law. This method of treatment is justified since the flow
conditions resulting from viscosity are small compared to those result
ing from the wave pattern which is an inertiagravity phenomenon.
ILLUSTRATIVE PROBLEMS
Water flows at 86 F in a 3in. pipe line at a velocity of 5 ft/sec. With what
velocity must linseed oil flow in a 1in. pipe line at the same temperature for the
two flows to be dynamically similar?
The Reynolds numbers for the two flows are the same.
.
12 32 " 2 3 . y =37.2 ft/sec
0.00001667 0.000692
A surface vessel of 500ft length is to be tested by a model 10 ft long. If the
vessel travels at 25 mph, at what speed must the model move in order to have
approximate similarity between model and prototype?
For incomplete similarity the Froude numbers are the same in model and
prototype.
5280\ 2
36007 V 2
500 X 32.2 10 X 32.2
518 ft/sec
31. Dimensional Analysis. Another useful tool of modern fluid
mechanics and a necessary adjunct to the principle of similarity is that
field of mathematics known as ' 'dimensional analysis," the mathe
matics of the dimensions of quantities.
The methods of dimensional analysis are built up on the principle
of dimensional homogeneity which states that an equation expressing a
physical relationship between quantities must be dirnensionally homo
geneous; i.e., the dimensions of each side of the equation must be the
same. This principle has already been utilized in Chapter I in obtain
ing the dimensions of mass density and kinematic viscosity, and it has
been recommended as a valuable means of checking engineering calcu
lations. Now, further investigation of the principle will reveal that it
affords a means of constructing physical equations from a knowledge
of the variables involved and their dimensions.
132 SIMILARITY AND DIMENSIONAL ANALYSIS
Before examining the methods of dimensional analysis, recall that
there are two different systems by which the dimensions of physical
quantities may be expressed. These systems are the forcelengthtime
system and the masslengthtime system. The former system, gener
ally preferred by engineers, becomes the familiar ' 'footpoundsecond"
system when expressed in English dimensions; the latter system in
English dimensions becomes the "foot slugsecond" systern. The
latter system is generally preferred in dimensional analysis, and, since
the student is familiar with the former system, the use of the latter
will serve to develop versatility in the use of dimensions.
A summary of the fundamental quantities of fluid mechanics and
their dimensions in the various systems is given in Table VI, the con
ventional system of capital letters being followed to indicate the dimen
sions of quantities. The basic relation between the forcelengthtime
and masslengthtime systems of dimensions is given by the Newtonian
law, force or weight = (Mass) X (Acceleration) and, therefore,
dimensionally,
A
F ~ M Y 2
from which the dimensions of any quantity may be converted from
one system to the other.
To illustrate the mathematical steps in a simple dimensional prob
lem, consider the familiar equation of fluid statics
p = wh
but assume that the dimensions of w and h are known and those of p
unknown. The dimensions of p can be only some combination of
M, L, and T, and this combination may be discovered by writing the
equation dimensionally as
Unknown dimensions of p = (Dimensions of w) X (Dimensions of K)
or
M L b T c * (jrlf*) X (L)
in which a, 6, and c are unknowns. The principle of dimensional
homogeneity being applied, the exponents of each of the quantities is
the same on each side of the equation, giving
a = 1, jrs2 + 1* 1, =~2
DIMENSIONAL ANALYSIS
133
TABLE VI
DIMENSIONS OF FUNDAMENTAL QUANTITIES USED IN FLUID MECHANICS
Quantity
Symbol
English Engi
neering
Dimensions
ForceLength
Time
Dimensions
MassLength
Time
Dimensions
Acceleration
a
ft/sec 2
L/T 2
L/T 2
Acceleration due to gravity .
Area
g
A
ft/sec 2
ft 2
L/T 2
L 2
L/T 2
L 2
Density
p
Ib sec 2 /ft 4
FT 2 /L 4
M/L*
Force
F
Ib
F
ML/T 2
Kinematic viscosity
v
ft 2 /sec
L 2/ T 2
L 2 /T
Length
I
ft
L
L
Mass
M
Ib sec 2 /ft
FT 2 /L
M
Power
p
ft Ib/sec
FL/T
ML 2 /T Z
Pressure
P
lb/ft 2
F/L 2
M/LT 2
Rate of flow
Q
ft 3 /sec
L*/T
L*IT
Specific weight . .
w
lb/ft 3
F/L Z
M/L 2 T 2
Time
t
sec
T
T
Velocity
V
ft/sec
L/T
L/T
Viscosity
p.
Ib sec/ft 2
FT/L 2
M/LT
Weight
17
Ib
F
ML/T 2
Weight rate of flow
G
Ib/sec
F/T
ML/T*
whence
Dimensions of p = ML
M
LT*
It is obvious, of course, that this result might have been obtained more
directly by cancellation of L on the righthand side of the equation,
for this has been, and will continue to be, the usual method of obtain
ing the unknown dimensions of a quantity. It is of utmost impor
tance, however, to note the mathematical steps which lie unrevealed
in this hasty cancellation, if the basis of dimensional analysis is to be
understood.
The above methods may now be used in quite another and more
important way. To illustrate by another familiar example, suppose
that it is known that the power P, which can be derived from a hy
draulic motor, is dependent upon the rate of flow through the motor
Q, the specific weight of the fluid flowing w, and the unit energy E
which is given up by every pound of fluid as it passes through the
134 SIMILARITY AND DIMENSIONAL ANALYSIS
*
machine. Suppose that the relation between these four variables is
unknown but it is known that these are the only variables involved in
the problem. 2 With this meager knowledge the following mathe
matical statement may be made :
P~f(Q,w,E)
From the principle of dimensional homogeneity it is obvious^ that the
quantities involved cannot be added or subtracted since their dimen
sions are different. This principle limits the equation to a combina
tion of products and quotients of powers and roots of the quantities
involved, which may be expressed in the general form
P = CQ a w b E c
in which Cis a dimensionless constant which may exist in the equation
but cannot, of course, be obtained by dimensional methods. Writing
the equation dimensionally
ML 2 _ /ZA 8 (_M_\ b
T 3 ~\T/ \L 2 T 2 ) ( >
equations are obtained in the exponents of the dimensions as follows :
M : 1 = b
L : 2 = 3a  2b  c
T : 3 = a  2b
whence
a = 1, 6 = 1, c = 1
and resubstitution of these values in the above equation gives
P CQwE
The shape of the equation (confirmed by equation 17) has, therefore,
been derived without physical analysis solely from consideration of the
dimensions of the quantities which were known to enter the problem.
The magnitude of C may be obtained either (1) from a physical analy
sis of the problem, or (2) from experimental measurements of P, <2,
w, and E.
From the above problem it appears that in dimensional analysis
only three equations can be written since there are only three funda
2 Note that experience and analytical ability in determining the relevant variables
are necessary before the methods of dimensional analysis can be successfully applied.
DIMENSIONAL ANALYSIS 135
mental dimensions: M, L, and T. This fact limits the completeness
with which a problem with more than three unknowns may be solved,
but does not limit the utility of dimensional analysis in obtaining the
shape of the equation. This point may be illustrated by considering a
more complex problem of fluid mechanics, that of the calculation of
the drag of a surface vessel, and in this problem it may also be observed
how the similarity principle may be utilized in interpreting the results.
Considering the surface vessel of Fig. 81, having a certain shape and
draft d, the force 3 D y necessary to tow or propel the ship will depend
upon the size of the ship (characterized by its length, /) the viscosity
Vj and density p of the fluid in which the ship moves, the velocity of
motion V, and the acceleration due to gravity g, since it has been
shown in Art. 30 that the wave pattern is a gravity phenomenon.
Thus, with no further knowledge than the variables involved in the
problem, an equation may be written
which for dimensional reasons must have the shape
D = CP P yV d g'
and the equation of the dimensions of the terms is
M = (L]
resulting in the three equations of the exponents of M, L, and T
M : 1 = b + c
L:l = a  3b c + d + e
T : 2 = c  d  2e
whence
b = 1 c, d = 2  c  2e, a = 2 + e c
and substituting these values in the second equation
and by collecting terms
8 The force, D, is the equal and opposite of the total drag of the ship.
136 SIMILARITY AND DIMENSIONAL ANALYSIS
But
F 2 Vlp
N F = and N R =
lg M
allowing the equation to be written in the more general form
D = f(N F ,N R )pl 2 V 2
The drag of objects moving through a fluid is usually expressed by
in which CD is a dimensionless "drag coefficient, " the magnitude and
properties of which are usually determined by experiment. Com
parison of the last two equations reveals that
C D = 2f(N F N R ) =f'(N Pt N B )
showing without experiment, but from dimensional analysis alone,
that the drag coefficient depends upon the Froude and Reynolds
numbers.
The principles of dynamic similarity have demonstrated that the
flow picture about a surface vessel is completely similar to the flow
picture about its model if
(ftp) prototype = (^F) model
and
(NR) prototype = (^Vfl)model
but since
CD = f'(N Pt N R )
it is evident that one of the results of obtaining complete dynamic
similarity between a model and prototye is the equality of the drag
coefficients in the model and prototype. Thus the fundamental reason
for obtaining dynamic similarity between a model and its prototype is
to cause their dimensionless coefficients to be the same, allowing them
to be measured in the model and used for the prototype.
BIBLIOGRAPHY
SIMILARITY
A. H. Gibson, "The Principle of Dynamic Similarity with Special Reference to Model
Experiments," Engineering, Vol. 117, 1924, pp. 325, 357 r 391, 422.
A. C. CHICK, "The Principle of Similitude," Hydraulic Laboratory Practice, p. 796,
A.S.M.E. 1929.
PROBLEMS 137
K. C. REYNOLDS, "Notes on the Laws of Hydraulic Similitude as Applied to Experi
ments with Models," Hydraulic Laboratory Practice, p. 759, A.S.M.E., 1929.
O. G. TIETJENS, "Use of Models in Aerodynamics and Hydrodynamics," Trans.
A.S.M.E., Vol. 54, 1932, p. 225.
DIMENSIONAL ANALYSIS
J. R. FREEMAN, "Introduction," p. 775, Hydraulic Laboratory Practice, A.S.M.E.,
1929.
A. C. CHICK, "Dimensional Analysis," p. 782, Hydraulic Laboratory Practice,
A.S.M.E., 1929.
E. BUCKINGHAM, "Model Experiments and the Forms of Empirical Equations,"
Trans. A.S.M.E., Vol. 37, 1915, p. 263.
P. W. BRIDGMAN, Dimensional Analysis, Yale University Press, 1922.
PROBLEMS
234. An airplane wing of chord length 10 ft moves through still air at 60 F and
14.7 lb/in. 2 at a speed of 200 mph. A 1 : 20 scale model of this wing is placed in a
wind tunnel, and dynamic similarity between model and prototype is desired,
(a) What velocity is necessary in a tunnel where the air has the same pressure and
temperature as that in flight? (b) What velocity is necessary in a variabledensity
wind tunnel where pressure is 200 lb/in. 2 abs and temperature 60 F? (c) At what
speed must the model move through water (60 F) for dynamic similarity?
235. An airship 600 ft long is to be tested by a 1 : 100 scale model. If the ship
moves at 80 mph and the windtunnel velocity is 60 mph, calculate model and flight
Reynolds numbers, assuming that both tunnel and flight air is the same (14.7 lb/in. 2 ,
60 F).
236. A submerged submarine moves at 10 mph. At what theoretical speed must
a 1 : 20 model be towed for dynamic similarity between model and prototype,
assuming sea water and towingtank water the same?
237. It is desired to obtain dynamic similarity between 2 cfs of water at 50 F
flowing in a 6in. pipe and linseed oil flowing at a velocity of 30 ft/sec at 90 F.
What size of pipe is necessary for the linseed oil?
238. The flow of air in a 2in. pipe at 10 lb/in. 2 and 60 F is to be similar to the
water flow of the preceding problem. If barometer is standard, what air velocity is
required?
230. Water flows at 60 F in a 3in. pipe at 5 ft /sec. What velocity must water
of the same temperature have in a 6in. pipe for the two pipe flows to be dynamically
similar?
240. Water (60 F) flows in a 2in. pipe at 3 ft/sec. What velocity must glyc
erine at 80 F have in a 6in. pipe for the two flows to be dynamically similar?
241. When castor oil flows at 60 F in a 2in. horizontal pipe line at 10 ft/sec, a
pressure drop of 320 lb/in. 2 occurs in 200 ft of pipe. Calculate the pressure drop in
the same length of 1in. pipe when linseed oil at 100 F flows therein at a velocity of
1 ft/sec.
242. A ship 200 ft long is to be tested by a 1 : 50 scale model. If the ship is to
travel at 30 mph, at what speed must the model be towed to obtain incomplete
dynamic similarity with its prototype?
138 SIMILARITY AND DIMENSIONAL ANALYSIS
243. A ship model 3 ft in length is tested in a towing basin at a speed of 3 ft/sec.
To what ship velocity does this speed correspond if the ship has a length of (a) 200 ft,
(b) 400 ft?
244. The discharge of a perfect fluid through an orifice under a static head is an
inertiagravity phenomenon and one to which the Froude law may be applied.
Using such an orifice and a geometrically similar model of the same, derive ratios
between the following quantities in model and prototype (in terms of the head):
velocity, rate of flow, horsepower of jet.
245. By dimensional analysis prove that: Kinetic energy = constant X MV^.
246. By dimensional analysis prove that: Centrifugal force = constant X MV 2 /r.
247. By dimensional analysis prove that: G constant X AwV.
248. Prove by dimensional analysis that a body of mass M and radius of gyration
r, rotating at angular velocity co, possesses kinetic energy given by
Kinetic energy ~ constant X Mr 2 co 2
249. Prove by dimensional analysis that: w p/RT.
250. By dimensional analysis, prove that the force F necessary to accomplish a
change A F in the velocity of a mass M in time t is given by
F constant X
t
251. By dimensional analysis, prove that: Pressure drop = constant X
laminar flow. (See equation 32 and Art. 30.)
CHAPTER VI
FLUID FLOW IN PIPES
The problems of fluid flow in pipe lines the prediction of rate of
flow through pipes of given characteristics, the calculation of energy
losses therein, etc. have wide application in engineering practice;
they afford an opportunity of applying many of the foregoing principles
to fluid flows of a comparatively simple and controlled nature.
32. Energy Relationships. The flow of a real fluid differs from
that of a perfect fluid in that energy in the real fluid is continually
converted into heat through the processes of turbulence and friction,
brought about by the existence of viscosity. Therefore the energy
equations of Art. 20 must be modified for application to the real fluid
by the introduction of a lost energy term. This being designated as
EL (footpounds per pound), or as the "lost head," HL (feet), it is ob
vious from previous reasoning that
E L = h L
Introducing the lost energy to the various energy equations, the
general energy equation 14
1 V\ fa Vl
/i+ +^+z l + 778E H +E M = I 2 +^ + ~ *+z 2 (14)
Wi 2g w 2 2g
remains unchanged. This results from the fact that the heat energy
generated by the turbulence and frictional processes is merely an
exchange of energy between the terms of the equation and hence does
not appear therein as a separate term. If we assume that no machine
exists between the sections 1 and 2 of the streamtube EM = 0, and
write the above equation in differential form
= dl + d + d ~
substituting for w its equivalent l/v, the equation becomes
(V 2 \
77SdE H = dl + d (pv) + d () + dz
139
140 FLUID FLOW IN PIPES
or
77SdE ff = dl + pdv + vdp + d[ ) + dz
\2g /
Considering now the thermodynamic aspects of the problem, the
total amount of heat added to every pound of flowing fluid will be the
sum of that added from external sources and that generated by fric
tional processes; thus the total heat added to the fluid becomes"
77SdE H + dh L
which goes into changing internal energy and doing work upon the
flowing fluid. Thus
77SdE H + dh L = dl + pdv
Solving this equation simultaneously with the above general energy
equation yields
>2>
which is the differential form of the Bernoulli equation for the flow
of all fluids.
For compressible fluids, the equation becomes, upon substitution
of l/w for v and integration,
^ + lf + Zl== Zl + 22 + ^ (37)
in which the evaluation of the integral term is dependent upon the
variation of density with pressure.
For fluids which may be treated as incompressible there is no
variation of density with pressure, and the integration may be carried
out, giving
O TT2
n\ V 1 Do V o
I * \ ___ x <a  *>   7 /'JQN
w 2g l w 2g 2
the familiar Bernoulli equation, modified to include the lost energy
term, HL.
33. General Mechanics of Fluid Flow in Pipe Lines. The meaning
of lost head and energy in pipe flow can best be obtained from a study
of the mechanics of fluid motion in circular pipes, similar to that of
Art. 27. Figure 82 represents a section of a long, straight, uniform,
GENERAL MECHANICS OF FLUID FLOW IN PIPE LINES 141
sloping circular pipe in which laminar or turbulent flow of an incom
pressible fluid is fully established. 1 Flow being assumed from section 1
to section 2, the Bernoulli equation (38) modified for lost head may be
described graphically as indicated, and it is observed: (1) that the
"pressure (hydraulic) grade line" must always slope downward in the
direction of flow, and (2) that the vertical drop in this grade line
between any two points must represent the lost head in the pipe
between these two points.
FIG. 82.
Since the pipe is of uniform size, the continuity principle requires
motion in the pipe to take place at constant velocity, and constant
velocity motion in turn requires that the net force acting on any mov
ing fluid mass be zero. The forces acting along the axis of the pipe
on the fluid mass of length / and radius r are those of pressure on the
ends, shear on the sides, and the component of weight in the direction
of motion. The equation involving these forces may be seen from the
figure to be
piirr 2 p 2 irr 2 + W sin a  r2wrl =
or
(Pi Pz)Trr 2 + wr 2 wlsin a r2irrl =
1 That is, the section considered is a good distance from the pipe entrance or other
source of eddying turbulence.
142 FLUID FLOW IN PIPES
<2>1 ~~ 2>o
in which sin a =   Substituting this value and canceling TIT,
and dividing by
2lr
w
But from the figure
, fpi P2 , \
h L = I  h zi  z 2 1
w w /
giving
/w/fciA
r = ()r (39)
which shows that here, as between parallel plates (Art. 27), the distri
bution of shear stress in both flow regimes is linear, shear stress being
maximum at the pipe wall and zero at the pipe center. At the wall the
shear stress is designated by T O , and from the above general equation
2 4
The lost power PL accompanying the lost head fa may be calcu
lated directly from
PL = Qwh L
but more fruitfully from the work done by the moving fluid against
the shear stress, T O . Here
PL = (r irdt)V
but
wdfa
which when substituted above results in
_ 2 \
P L    = ^ VJ wh L = Qwh L
as before.
LAMINAR FLOW
143
ILLUSTRATIVE PROBLEM
Water flows through a section of a 12 in. pipe line 1000 ft long, running from
elevation 300 to elevation 150. A pressure gage at elevation 300 reads 40 lb/in. 2 ,
and one at elevation 150 reads 90 lb/in. 2 Calculate loss of head, direction of flow,
and shear stress at the wall of the pipe.
92.3
150'
M _ 40 >< 144 _ 92 s ft
I I yi,O 1L
\w/ m 62.4
90 X 144
= 208.0 ft
'150 62.4
h L = 300 + 92.3  150  208.0 = 34.3 ft
Direction of flow: downward
wdh L 62.4 X 1 X 34.3
4 X 1000
= 0.535 lb/ft 2
34. Laminar Flow. For laminar flow the relation between shear
stress and velocity gradient,
dv
T = JJ,
dy
allows a simple theoretical derivation of the relationships between the
other variables involved, the results of which are confirmed by
experiment.
The general equation for the shear stress, r, has been shown to be
but in laminar flow r is also given by 5
dv
' dr
2 See footnote 6, page 115.
144
FLUID FLOW IN PIPES
Equating these two expressions for r,
dv
"^dr* 21
the variables of which may be separated
wh L
rdr
and the relation between velocity and radius obtained by integrating
the equation between the center of the pipe, where r = and v = V cy
and the variable radius, where r = r and v = v. Integrating
/ V C 7
, whL
dv= ~w
(40)
showing that for laminar flow in circular pipes the velocity profile is
a parabolic curve (Fig. 83) having a maximum velocity at the center
of the pipe and reducing to zero at the walls.
FIG. 83.
From the fact that, when v = 0, r = R, the fundamental equation of
laminar flow may be obtained from equation 40
_
y c
The relation between the average velocity and center velocity may
LAMINAR FLOW 145
be derived by equating two expressions for the rate of flow, Q.
Obviously,
Q =
in which V is the average velocity. But
/A. s+R
vdA = / v2vrdr
in which the variable velocity, v, may be shown from equation 40 to be
given by
Substituting this value for v and integrating,
and equating the two expressions for Q
(42)
proving that the average velocity is always onehalf of the center
velocity for laminar flow in circular pipes.
Equation 41 may now be obtained in terms of the average velocity,
V, by substituting V c = 2 V, giving
a more practical version of the basic laminar flow equation. Trans
position of equation 43 allows calculation of lost head occurring in
laminar flow from ,, 7Tr
32fj.IV
hL =~^ (44)
and indicates that the head lost in laminar flow varies directly with the
velocity of flow. By multiplying the righthand side of equation 44 by
2 V/2 V and substituting pg for w, equation 44 may be transformed for
use in subsequent developments to
32/1/7 27 64 \ I V 2 64 / V 2
Pgd 2 2V ~[ Vdpd 2g " N R d2g
~
146 FLUID FLOW IN PIPES
Another statement of the equation for laminar flow may be derived
in terms of the rate of flow, Q, by substituting for V in equation 43
the relation
4
which gives
>rrd 4 wh L
Q =  (46)
This equation shows that in laminar flow the rate of flow, Q, which will
occur in a circular pipe varies directly with the lost head and with the
fourth power of the diameter but inversely with the length of pipe and
viscosity of the fluid flowing. These facts of laminar flow were estab
lished experimentally, independently, and almost simultaneously by
Hagen (1839) and Poiseuille (1840), and thus the law of laminar flow
expressed by the above equations is termed the " HagenPoiseuille
law."
The experimental verification, by Hagen, Poiseuille, and later
investigators, of the above theoretical derivations serves to confirm
the assumptions (1) that there is no velocity adjacent to a solid
boundary and (2) that in laminar flow the shear stress is given by
dv
T = M
ay
which were taken for granted in the above derivations.
ILLUSTRATIVE PROBLEM
One hundred gallons of oil (S = 0.90 and /x = 0.0012 Ib sec/ft 2 ) flow per minute
through a 3in. pipe line. Calculate the center velocity, the lost head in 1000 ft.
of this pipe, and the shear stress and velocity at a point 1 in. from the center line.
100 1 =4.53 ft/sec
60 X 7,
Vdp 4.53 X A X (0.90 X 1.935) ,_
iv R =* = '   = 1645
M 0.0012
DIMENSIONAL ANALYSIS OF THE PIPEFRICTION PROBLEM 147
Therefore laminar flow exists and F c = 2 X 4.53 = 9.06 ft/sec
64 / F 2 64 1000 (4.53) 2
X
N R d2g 1645
v = V c [ 1  ) = 9.06  1 
. , M
X   = 49.6 ft of oil
2g
(l 
 5.03 ft/sec
2/
" = 0.116 Ib/ft*
2 X 1000 12
35. Dimensional Analysis of the PipeFriction Problem. Although
Prandtl and von Karman have met with some success in a theoretical
treatment of turbulent flow in pipes, the advanced mathematics
involved places their analysis of the subject beyond the scope of an
elementary textbook.
However, a good understanding of turbulent pipe flow may be
attained from a study of experimental results, interpreted by the
methods of dimensional analysis and similarity. A general investiga
FIG. 84.
tion of pipe flow by these methods will, of course, lead to results which
are equally applicable to laminar and turbulent flow and, therefore,
will allow the inclusion not only of experimental results of tests in
turbulent flow, but also of the relationships of laminar flow which were
derived in Art. 34.
For a dimensional analysis of the lost head due to pipe friction,
consider a rough pipe (Fig. 84), of diameter d, in which a fluid of
viscosity M and density p is flowing with an average velocity V. In a
148 FLUID FLOW IN PIPES
length of this pipe, /, a lost head, HL, is caused by pipe friction, and
the loss of pressure PL equivalent to this head loss is given by
PL W}IL
Assume that the roughness of the pipe has a definite pattern and that
the average height of the roughness protuberances is e. Making a selec
tion of the variables upon which the lost pressure, pi,, depends,
and by the methods of dimensional analysis,
p L = Cl a V b d c x p y e z
Before writing the equations involving the dimensions of these quanti
ties it may be predicted that
a = 1
or, in other words, the loss of head may be expected to vary directly
with the length of pipe. The dimensional equation thus becomes
M
LT 2 ~ w>
from which the equations of exponents are
M: 1 = x + y
L: 1 = 1 +& + #
nr* , _____ f\ ___ _____ "L ____ _,
Solving in terms of x and z
b = 2 x
c=l x z
y = 1 x
which, when substituted above, give
PL = <
or by arrangement of terms and insertion of 2/2
d~7\Vdp) W
DIMENSIONAL ANALYSIS OF THE PIPEFRICTION PROBLEM 149
but since
PL =
wd
But w/p = g, Vdp/fjL = NR, and since x and z are unknown the equa
tion may be written in the general form
d^ (47)
The equation used by Darcy (1857), Weisbach, and others for the
calculation of lost head due to pipe friction was obtained experi
mentally from numerous tests on the flow of water in pipes. This
socalled " Darcy equation" is the basis of pipefriction calculations
today and is
I V 2
/*=/ (48)
in which/ is called the ''friction factor." The early hydraulic experi
menters discovered that the friction factor was apparently dependent
upon the pipe diameter d, the average velocity of flow V, and the
roughness of the pipe, but with the use of dimensional analysis these
facts and more may be safely predicted today without the aid of experi
ment, as may be seen from a comparison of the last two equations,
which gives
Dimensional analysis has thus allowed isolation of all the variables
upon which the friction factor depends and has shown that these
variables exist in two dimensionless combinations, Vdp/n, the Rey
nolds number; and e/d, the "relative roughness" 3 of the pipe.
Dimensional analysis has indicated furthermore a single general solu
tion of the pipefriction problem which is applicable to all fluids.
The physical significance of equation 49 may be stated briefly:
The friction factors of pipes will be the same if their Reynolds numbers,
roughness patterns, and relative roughnesses are the same. This state
ment being interpreted by the principle of similarity, it becomes
8 " Relative " roughness since e/d expresses the size of the roughness protuberances
relative to the diameter of the pipe.
150
FLUID FLOW IN PIPES
evident that its basic meaning is : The friction factors of pipes are the
same if their flow pictures are in every detail geometrically and dynami
cally similar.
Dimensional analysis cannot, of course, give the exact mathematical
relationship between /, NR, and e/d but indicates only that there is a
relationship between these variables which may be found by a theo
retical or experimental analysis. The former has been used in the
case of laminar flow, resulting in equation 45
which shows that
N R d 2g
(50)
and that in laminar flow the friction factor / is independent of the
surface roughness, bearing out the fact (Art. 26) that surface roughness
can have no effect upon laminar flow.
For turbulent flow the following review of experimental results is
necessary to obtain the relationship of/, NR, and e/d.
36. Results of PipeFriction Experiments. The results of recent
tests by Nikuradse 4 demonstrate perfectly the relationship of /, NR,
and e/d for both laminar and turbulent flow. In these tests geometrical
FIG. 85. Relation of friction factor, Reynolds number, and relative roughness for
similar pipes.
similarity of the roughness patterns was obtained artificially by fixing
a coating of uniform sand grains to the pipe wall. The results of the
tests, plotted logarithmically on Fig. 85, illustrate the following
important fundamentals.
4 J. Nikuradse, " Stromungsgesetze in rauhen Rohren," V. D. I. Forschungsheft,
361, 1933.
RESULTS OF PIPEFRICTION EXPERIMENTS 151
1. The physical difference of the laminar and turbulent flow regimes
is indicated by the change in the relationship of / to NR at the critical
Reynolds number of 2100.
2. The laminar regime is characterized by a single curve, given by
the equation / = 64=/Nn for all surface roughness and thus shows that
lost head in laminar flow is independent of surface roughness.
3. In turbulent flow a curve of / vs. NR exists for every relative
roughness, e/d, and from the shape of the curves it may be concluded
that for rough pipes the roughness is more important than the Reynolds
number in determining the magnitude of the friction factor.
4. At high Reynolds numbers, the friction factors of rough pipes
become constant dependent wholly upon the roughness of the pipe and
thus independent of the Reynolds number. Thus, for highly turbu
lent flow over rough surfaces the Darcy equation becomes
/ V 2
}IL = (Constant)  
d 2g
showing that
H L a V 2 (51)
for turbulent flow over rough surfaces.
5. In turbulent flow a single curve expresses the relationship of /
and NR for all pipes which are hydraulically smooth, showing that
surface roughness, when submerged in the laminar film, can have no
effect on the friction factor and thus that lost head in smooth pipes is
caused by viscosity effects alone. Using, for simplicity, the equation
developed by Blasius, 5 expressing the relation between / and NR for
turbulent flow in smooth pipes,
(52)
(AW
and substituting this in the Darcy equation
0.3164 I V 2
it is evident that
h L a V 1  75 (53)
5 H. Blasius, Forschungsarbeiten auf dem Gebiete des Ingenieurwesens, 131,
1913.
152 FLUID FLOW IN PIPES
indicating approximately how head loss varies with velocity for turbu
lent flow over smooth surfaces.
6. The series of curves for the rough pipes diverge from the smooth
pipe curve as the Reynolds number increases. In other words, pipes
which are smooth at low values of NR become rough at high values of
NR. This may be explained by the thickness of the laminar film
decreasing (Art. 38) as the Reynolds number increases, thus exposing
smaller roughness protuberances to the turbulent region and causing
the pipe to exhibit the properties of a rough pipe.
Unfortunately the excellent results of Nikuradse cannot be applied
directly to engineering problems since the roughness patterns of com
mercial pipe are entirely different and much more variable than the
artificial roughnesses used by Nikuradse. At present, because of this
lack of uniformity, the surface roughnesses encountered in engineering
practice cannot be classified practically by anything more than a
descriptive statement. This fact gives an inevitable uncertainty to
the selection of the friction factor in engineering problems, an uncer
tainty which may be overcome only by practical experience.
A practical summary of the friction factors for pipes of commercial
roughness has been developed by Pigott. 6 A portion of this summary
and the Blasius and Nikuradse curves for smooth pipes are presented
in Fig. 86 and may be used in the solution of problems.
The accuracy of pipefriction calculations is necessarily lessened by
the unpredictable change in the roughness and friction factor due to
the accumulation of dirt and rust on the pipe walls. This accumulation
not only increases surface roughness but also reduces the effective pipe
diameter as well, and may lead to an extremely large increase in the
friction factor after the pipe has been given a long period of service.
ILLUSTRATIVE PROBLEM
If 90 gpm of water at 68 F flow through a smooth 3in. pipe line, calculate the
loss of head in 3000 ft of the pipe.
90 1
X r^ = 4.08 ft/sec
60 X 7.48
4.08 XAX 1.935
0.000021
6 R. J. S. Pigott, "The flow of fluids in closed conduits," Mechanical Engineering,
Vol. 55, No. 8, p. 497, August, 1933.
RESULTS OF PIPEFRICTION EXPERIMENTS
153
154
FLUID FLOW IN PIPES
From the Plot of Fig. 86,
/ = 0.0181
kL =/ . 0.0181
d 2g fV
2g
 56.1 ft
37. Velocity Distribution in Circular Pipes The Pipe Coefficient.
For laminar flow in circular pipes it has been shown theoretically
(Art. 34) and may be proved experimentally that the variation of
velocity along a diameter follows a parabolic curve, and one of the
characteristics of a parabolic velocity distribution in a circular pipe
has been seen to be the fixed relationship
V c = 2V
between the center velocity and average velocity. This equation may
be written
_F _ 1
V c " 2
in which V/V C , the ratio of average to center velocity, is frequently
called the " pipe coefficient " or " pipe factor."
FIG. 87. Velocity distributions in circular pipes. 7
In turbulent flow the curve of velocity distribution is not deter
mined by viscous shear between moving layers, but depends upon the
strength and extent of the turbulent mixing process. Turbulent mix
7 H. Rouse, " Modern Conceptions of the Mechanics of Fluid Turbulence,"
Trans. A.S.C.E., Vol. 102, 1937, p. 463.
VELOCITY DISTRIBUTION IN CIRCULAR PIPES
155
ing of fluid particles tends to cause them to move at the same velocity
and thus the velocity distribution curve becomes increasingly flattened
as the Reynolds number increases, as shown by the curves of Fig. 87,
which are drawn for the same average velocity, V. A glance at these
curves indicates that the pipe coefficient, V/V C , increases with the
Reynolds number and suggests a practical method of describing veloc
ity distribution properties by the single curve (1) of Fig. 88. A sup
plementary curve (2) of pipe coefficient vs. Vjlp/ii is useful in obtain
.50
FIG. 88. Relation of pipe coefficient and Reynolds number. 8
ing direct solutions of problems in which the center velocity is known
and the pipe coefficient is to be found.
A relationship between pipe coefficient and friction factor may be
obtained from the equation
V c  V = 4.07 \P
which was proposed by Prandtl 9 from the results of Nikuradse'sexperi
8 Data from T. E. Stanton and J. R. Pannell, "Similarity of Motion in Relation
to the Surface Friction of Fluids," Philosophical Transactions of the Royal Society,
A 214, 1914, p. 199.
9 L. Prandtl, "Neuere Ergebnisse der Turbulenzforschung," V. d. I. Zeit., Vol. 77,
No. 5, Feb. 4, 1933.
156 FLUID FLOW IN PIPES
ments. A more useful expression for T O may be calculated from the
general equation
ivdhi,
since it is now known that HL is given by
, . / v 2
hL = /7T
d 2g
Substituting this in the expression for T O
*3'%i'
or
in which \/^o/j> is termed the " friction velocity." When this expres
sion for \/~r /P is substituted in Prand til's equation above
VcV V c
__ = __ 1=4 . c
giving
& (ss)
"" i + i
thus relating pipe coefficient and friction factor.
In deriving the Bernoulli equation from the principle of energy
conservation for the perfect fluid (Art. 20), it was seen that the kinetic
energy per pound of fluid flowing was given by V 2 /2g. For the perfect
fluid all fluid particles will pass a point in a circular pipe with the same
velocity, causing the distribution of velocity to be uniform, as shown
in Fig. 89. For the real fluid, however, the existence of a velocity dis
tribution curve having a maximum velocity at the center of the pipe
and no velocity at the wall will cause the term V 2 /2g to be an erroneous
expression for the kinetic energy of the flow.
To demonstrate this fact, consider laminar flow in a circular pipe
VELOCITY DISTRIBUTION IN CIRCULAR PIPES
157
(Fig. 89), where the velocity distribution curve is given by the simple
equation
The power due to the kinetic energy of fluid passing with velocity v
through the differential area dA is given by
v 2 v 2
P = dQw = v2irrdrw
2 2
which must be integrated to obtain the total kinetic energy of the flow.
Real Fluid  Laminar flow
Perfect Fluid
FIG. 89.
Substituting the above expression for v and integrating,
results in
P = vR VcW
But
giving
V c = 2V
v 2 v 2
P = (wR 2 V)w~ = Qw
158
FLUID FLOW IN PIPES
Thus for laminar flow the kinetic energy term is not V 2 /2g but V 2 /g,
and, because of the existence of velocity variation across the pipe in
real fluid flow, a correction term, a, must be introduced to the Bernoulli
equation, giving
in which a = 2 for laminar flow. In turbulent flow the flattening of
the velocity distribution curve is an approach to the straightline veloc
ity distribution of the perfect fluid, and hence in turbulent flow the
correction term a. has a value close to unity, a magnitude of 1.05 being
a satisfactory average value for ordinary turbulent flows.
The erfect of the term a in many engineering calculations is entirely
negligible since the magnitude of the V 2 /2g terms are frequently very
small compared to the other terms of the Bernoulli equation and a
slight change in the velocityhead terms will have no effect upon the
final results. The use of the term a is, of course, not justified unless
the other Bernoulli terms are known precisely and unless flow cross
sections are chosen where known velocity distributions will exist.
38. Approximate Thickness of the Laminar Film. The approxi
mate thickness of the laminar film may be established by the following
simple analysis and approxima
te * tions. Assume for simplicity
that the film has a precise
thickness, d (Fig. 90), and that
within the film the flow is
wholly laminar. Let the veloc
ity at the outer boundary of
the film be designated by V w .
At the wall the shear stress is
given by
FIG. 90.
r =
dv
"Ty
But since 6 is very small the velocity distribution within the film may
be assumed linear. Thus
dv V w
and
_
TO M
APPROXIMATE THICKNESS OF THE LAMINAR FILM 159
But r is also given (Art. 37) by
J T72
r. S p7
Equating the two expressions for r and solving for 6
which, by insertion of d/d, may be written
5 = / ~v W P d
and therefore
d = / ~F ~Vd P ( 5
The ratio V W /V may be found from the relationship of von
Karman 10
T7
= = 11.6
p
for smooth pipes. But (Art. 37)
Thus
which may be substituted in equation 56, giving
5 32.8
(57)
allowing the approximate thickness of the laminar film to be calculated
and showing that the thickness of the film relative to the pipe diameter
is dependent upon both friction factor and Reynolds number and
decreases in thickness with increasing Reynolds number.
10 Th. von Karman, "Turbulence and Skin Friction," J. Aero. Sciences, Vol. I,
No. 1, p. 1, January, 1934.
160 FLUID FLOW IN PIPES
39. Pipe Friction for Compressible Fluids. The calculation of
pressure loss due to pipe friction when gases and vapors flow in insu
lated pipe lines is, in general, a rather complex process and therefore
will not be treated exhaustively here. The complications will be
evident upon examination of the flow of a gas in a perfectly insu
lated pipe. Although through the insulation no heat is added to or
abstracted from the fluid flowing between sections 1 and 2, the expan
sion from pressure pi to pressure p2 is not a reversible adiabatic process
since heat is generated by fluid friction and is added to the flowing
fluid. The expansion thus becomes a polytropic process, the nature
of which depends upon the lost pressure, which is unknown. The
solution of such a problem can be obtained only by a tedious trialand
error procedure, applied to short lengths of the pipe.
However, a direct solution for the pipefriction loss may be
obtained for a perfect gas flowing isothermally in a pipe line. Iso
thermal fluid flow can occur only when the transfer of heat through
the pipe walls and the addition of heat to the fluid from the pipe
friction process are adjusted in such a manner that the temperature
of the fluid remains constant. Such an adjustment of heat transfers
is approximated naturally in uninsulated pipes where velocities are
small and where temperatures inside and outside of the pipe are
about the same; frequently the flow of gases in long pipe lines may
be treated isothermally.
Gas flow in a uniform pipe line is characterized by the fact that
the velocity does not remain constant but continually increases in
the direction of flow, the drop in pressure along the line, caused by
pipe friction, bringing about a continuous reduction in density in the
direction of flow. From the continuity equation
G = AwV
it is evident that a decrease in specific weight must cause an increase
in velocity since the weight flow G and the area A are both constant.
The continual change of pressure, velocity, and specific weight as
flow takes place necessitates writing the Bernoulli equation in differen
tial form and subsequently integrating to obtain practical results.
Neglecting the difference in the h terms, equation 37 may be written
in differential form as
dp VdV J7
H h dh L =
w g
PIPE FRICTION FOR COMPRESSIBLE FLUIDS
161
and applied to the differentially small element of fluid shown in Fig. 91.
Substituting the Darcy equation for the lost head term, this equation
becomes
dp VdV f dl V 2
d 2g
Dividing by V 2 /2g
FIG. 91.
But from the continuity principle V = G/Aw, which, when substituted
in the first term, gives
dV f
+ 2+dl0
The specific weight, w, is given by the equation of state of the gas
w = p/RT, in which, for an isothermal process, T is constant. Sub
stituting this expression for w in the first term, the equation becomes
which may be integrated between the indicated limits, giving
2 2 G 2 RT \
Pi p2  772
(58)
and allowing the pressure p% to be calculated when the other variables
are known. Theoretically the solution of the equation must be accom
plished by trial since V% cannot be calculated until p 2 is known, but
162 FLUID FLOW IN PIPES
usually the term 2ln(V<2,/V\) is so small in comparison to fl/d that it
may be neglected, reducing the equation to
thus making possible a direct solution.
The Reynolds number is, of course, necessary to obtain the friction
factor, /. Although the velocity and density of the fluid change con
tinually throughout the pipe, the Reynolds number of the flow remains
constant and may be calculated as follows:
Vdp Vdw
jy R  _ 
M Mg
but V = G/wA which, when substituted above, gives
Gd
N R = 
vgA
in which all the terms are constant for isothermal flow in a uniform
pipe.
Compressible fluids may be frequently treated as incompressible
in pipefriction calculations if the pressure and density changes are
not large; the lost head or pressure may be calculated from the
Darcy equation
which, under the above conditions, may be written
Pi P* f l Vi
~^T =f dT g
The limits of application of equation 60 may be seen by comparing
equation 60 with equation 59. To accomplish this, substitute in
equation 59 the following relations :
Pi ~pl = (Pi p2)(Pi +Pz)
%Vlwl
which give
(Pi 
d 2g
PIPE FRICTION FOR COMPRESSIBLE FLUIDS 163
Rearranging and inserting 2/2 on the righthand side of the equation,
^1
Substituting pi/RT for w\, this equation becomes
Pi ~ P2 _
or
Pi ~ P2
(61)
Comparison of equations 60 and 61 shows that equation 60 can be
multiplied by a correction factor dependent upon the pressure ratio,
p2/Pi> to obtain the result given by equation 59. For a pressure
ratio p2/Pi = 0.96, the magnitude of the correction factor is 1.02,
indicating that an error of 2 per cent is incurred if density change is
neglected and pressure drop calculated from equation 60. An allow
able error of 2 per cent being assumed, it is apparent that equation 59
must be used when p2/Pi < 0.96 but that equation 60 will give satis
factory results when p2/pi > 0.96.
ILLUSTRATIVE PROBLEM
If 40 Ib/min of air flow isothermally through a horizontal, smooth 3in. pipe
line at a temperature of 100 F, and the pressure at a point in this line is 50 lb/in. 2 ,
abs, calculate the pressure in the line 2000 ft downstream from this point.
* 8x
*
0.000000402 X 32.2
From the plot of Fig. 86,
/ = 0.0145
2 _ 2 = G 2 RT I
& A 2 d
,. ft v , A ^ J (m 2 X 53.3 X (100 + 460) 2000
(50 X 144)  p 2 = r /,\,i  0145 ^7
51,900,000  pi = 19,860,000
/> 2 = 5660 lb/ft 2 = 39.3 lb/in. 2 ab
164 FLUID FLOW IN PIPES
40. Pipe Friction in NonCircular Pipes. Although the majority
of pipes used in engineering practice are of circular cross section,
occasions arise when calculations must be carried out on friction loss
in rectangular passages and other conduits of noncircular form. The
foregoing equations for circular pipes may be adapted to these special
problems by the means of a new term, called the "hydraulic radius/'
The hydraulic radius is defined as the area of flow cross section
divided by the wetted perimeter. In a circular pipe of diameter"^,
4 d
Hydraulic radius R =  = 
7T0 4
or
d = 4
for a pipe of circular cross section.
This value may be substituted in the Darcy equation for lost head
and into the expression for the Reynolds number with the following
results :
and
N R = (63)
M
from which the lost head in conduits of any form may be calculated
with the aid of the plot of Fig. 86.
The calculation of lost head in noncircular conduits involves the
calculation of the hydraulic radius, R, of the flow cross section and
using the friction factor obtained on an "equivalent" circular pipe
having a diameter d given by
d = 4R
In view of the complexities of laminar films, turbulence, roughness,
shear stress, etc., it seems surprising at first that a circular pipe
"equivalent" to a noncircular conduit may be obtained so easily,
and it would, therefore, be expected that the method might be subject
to certain limitations. The method gives satisfactory results when the
problem is one of turbulent flow over rough surfaces, but if used for
laminar flow large errors are introduced.
PIPE FRICTION IN NONCIRCULAR PIPES 165
The foregoing facts may be justified theoretically by examining
further the structure of the Darcy equation
h  flv *
4*^
in which obviously
From the definition of the hydraulic radius, its reciprocal is the "wetted
perimeter per unit of flow cross section" and is, therefore, an index of
the extent of the rough surface in contact with the flowing fluid.
The hydraulic radius may be safely used in the above equation when
resistance to flow and head loss are primarily dependent upon the
extent of the rough boundary surface, as for turbulent flow in which
pipe friction phenomena are confined to a thin region adjacent to the
boundary surface and thus vary with the size of this surface. In
laminar flow, however, friction phenomena result from the viscosity
properties of the fluid and are independent of surface roughness.
The magnitude of the boundary surface plays a secondary role in
these phenomena, and so in laminar flow the use of the hydraulic radius
to obtain a circular pipe equivalent to a noncircular one is not possible.
ILLUSTRATIVE PROBLEM
Calculate the loss of head and pressure drop when air at standard conditions
(14.7 lb/in. 2 , 60 F) flows through 200 ft of 18 in. by 12 in. smooth rectangular
duct with an average velocity of 10 ft/sec. Sp. wt. of standard air = 0.0763
lb/ft. 3
18 X 12
p =  = 3.6 in. = 0.30 ft.
2 X 18 + 2 X 12
.0763
10 X 4 X .30 X 
75,900
/* 0.000000375
From the plot of Fig. 86,
/ = 0.019
H L  /  0.019 X 200 X = 49.2 ft of air
R2g 4 X .30 2g
p L = w h L = 0.0763 X 49.2  5.82 lb/ft 2  0.0405 lb/in 2 .
166 FLUID FLOW IN PIPES
41. PipeFriction Calculations by the Hazen Williams Method.
In order to circumvent the difficulties encountered because surface
roughness is a relative quantity, causing friction factors to be different
in pipes having the same roughness but having different diameters,
Hazen and Williams 11 proposed a formula in which the friction factor
was a function of surface roughness only. The formula was originally
proposed for the solution of hydraulic problems but should give correct
results for the flow of any fluid provided that flow occurs at high
Reynolds numbers. Although the formula is empirical and thus does
not possess dimensional homogeneity it gives good results and is in
general use by American engineers.
The shape of the HazenWilliams formula compares favorably
with that of the "Chezy equation," which may be derived from the
Darcy equation as follows. The Darcy equation states
which may be solved for F, giving
in which VSg/f = C, the " Chezy coefficient," and h L /l = 5, the lost
head per foot of pipe; substitution of these values results in
V = CV~RS = CR  ro S 5 (64)
as proposed by Chezy (1775).
Hazen and Williams found that experimental results were best
satisfied by the formula
V = 1.318C* W # 63 S 54 (65)
Values of the coefficient Chw are given in Table VII; it is obvious
that here again experience is necessary in the selection of coefficients
if reliable results are to be obtained.
The advantages and disadvantages of the HazenWilliams method
are evident from the formula and foregoing discussion. Among the
advantages are: (1) the coefficient depends only upon roughness,
(2) the effect of roughness and the other variables upon the velocity of
11 A. Hazen and G. S. Williams, Hydraulic Tables, Third edition, 1920, John Wiley
&Sons.
MINOR LOSSES IN PIPE LINES 167
TABLE VII
HAZEN WILLIAMS COEFFICIENT, Chw
Pipes extremely straight and smooth 140
Pipes very smooth 130
Smooth wood, smooth masonry 120
New riveted steel, vitrified clay 110
Old cast iron, ordinary brick 100
Old riveted steel 95
Old iron in bad condition 6080
flow and capacity of pipe are given directly by the formula. Its*dis
ad vantages are: (1) its lack of dimensional homogeneity, and (2) the
impossibility of applying it to the flow of all fluids under all conditions.
Although the formula appears cumbersome with its fractional expon
ents, this disadvantage is overcome in engineering practice by the use
of tables and diagrams in its solution.
ILLUSTRATIVE PROBLEM
If 90 gpm of water flow through a smooth 3in. pipe line, calculate the loss of
head in 3000 ft of this pipe.
90 1
4.08 ft/sec.
60 X 7.48 TT / 3
R =  ^"' = 0.0625 ft.
From Table VII,
C hw = 140
4.08 = 1.318 X 140 X (0.0625) 68 S' 54
S = 0.0218 =  = , h L = 65.3 ft
/ 3000
Compare results with the illustrative problem of Art. 36.
42. Minor Losses in Pipe Lines. Into the category of minor
losses in pipe lines fall those losses incurred by change of section,
bends, elbows, valves, and fittings of all types. Although in long pipe
lines these are distinctly "minor" losses and can often be neglected
168
FLUID FLOW IN PIPES
without serious error, in shorter pipe lines an accurate knowledge of
their effects must be known for correct engineering calculations.
The general aspects of minor losses in pipe lines may be obtained
from a study of the flow phenomena about an abrupt obstruction
placed in a pipe line (Fig. 92), which creates flow conditions typical
of those which consume energy and cause minor losses. Minor losses
generally result from changes of velocity, velocity increases causing
small losses but decreases of velocity causing large losses because of
the creation of eddying turbulence. In Fig, 92, energy is consumed
FIG. 92.
in the creation of eddies as the fluid decelerates between sections 2 and
3, and this energy is dissipated in heat as the eddies decay between
sections 3 and 4. Minor losses in pipe flow are, therefore, accom
plished in the pipe downstream from the source of the eddies, and the
pipe friction processes in this length of pipe are hopelessly complicated
by the superposition of eddying turbulence upon the normal turbulence
pattern. To make minor loss calculations possible it is necessary to
assume separate action of the normal turbulence and eddying tur
bulence although in reality a complex combination and interaction
of the two processes exists. Assuming the processes separate allows
MINOR LOSSES IN PIPE LINES 169
calculation of the losses due to normal pipe friction AL^ and h^ 4 ,
and also permits the loss, HL, due to the obstruction alone, to be
assumed concentrated at section 2. This is a great convenience for
engineering calculations since the total lost head in a pipe line may be
obtained by a simple addition of pipe friction and minor losses without
considering the above complications.
In order that they may be inserted readily into the Bernoulli
equation, minor losses are expressed by
V 2
in which KL is a coefficient usually determined by experiment. A
dimensional analysis of the flow past an obstruction of a given shape
(Fig. 92) leads to the conclusion that
vdp e
or if the shape is changed and the conclusion made more general,
fvdp \
KL = /I , roughness, shape )
The effect of surface roughness upon minor losses is generally very
small since the obstruction is usually short and has little contact with
the flowing fluid; however the large irregularities of the obstruction
act in similar fashion to the roughness protuberances of a very rough
pipe, and Nikuradse's tests (Fig. 85) show that for turbulent flow the
friction factors of such pipes are practically independent of the Rey
nolds number of the flow. From these two statements it may be
concluded that for turbulent flow the experimental coefficients,
KL, for most minor losses are dependent primarily upon the shape of
the obstruction and are practically independent of roughness and
Reynolds number. This conclusion is borne out by experiment and
is useful in the interpretation of the following experimental results
which were obtained by hydraulic tests; since the hydraulic tests
were all made at high Reynolds numbers it may be concluded that
these results are adaptable to other fluids at high Reynolds numbers
as well.
When a sudden enlargement of section (Fig. 93) occurs in a pipe
line, a rapid deceleration in flow takes place accompanied by the
characteristic eddying turbulence, which may persist in the larger pipe
170
FLUID FLOW IN PIPES
for a distance of 50 pipe diameters before dying out and allowing
restoration of normal turbulence pattern.
In Fig. 92, the distance between the two pressure grade lines at
section 2 proved to be the loss of head due to the obstruction. For the
FIG. 93. Sudden Enlargement.
sudden enlargement, however, this does not hold because a change oi
velocity occurs and brings about a change of pressure according to the
Bernoulli principle. The distance between the two grade lines at
section 2 may be obtained readily by writing the Bernoulli equation
between sections 1 and 3.
r
w 2g
whence
(66)
MINOR LOSSES IN PIPE LINES
171
in which the lefthand side of the equation is obviously the vertical
distance between the grade lines as shown in Fig. 93.
Application of the impulsemomentum law to the sudden enlarge
ment (Fig. 94) allows theoretical calculation of the lost head HL, and
2 3
FIG. 94. Sudden Enlargement.
the theoretical analysis gives results which are confirmed closely by
experiment. The fluid between sections 2 and 3 experiences a reduc
tion in momentum caused by the difference of pressure p% p\.
Neglecting pipe friction
Pi = p2
and the impulsemomentum law gives
which may be written (since A% = A 2 )
Ps Pi = Q Vi  V*
A 3
~ F 3 )
 F 3 )
w
g
g
Neglecting pipe friction, equation 66 becomes
p3 ~ Pi
vl  vl
w
h L
giving another expression for Equating these two expres
w
sions
2V 3 (V 1  7 8 ) V\ 
hi
from which
 F 3 ) 2
(67)
thus allowing calculation of the lost head for any sudden enlargement
where the pipe sizes are known. This loss of head due to sudden
172
FLUID FLOW IN PIPES
enlargement is frequently termed the " BordaCarnot loss" after the
men who made its original development.' It may be expressed
rigorously as
in which the coefficient KL has been found experimentally to be close
to unity and may be assumed
so for most engineering calcu
lations. Empirical formulas are
available if greater precision is
desired.
A special case of a sudden
enlargement exists when a pipe
discharges into a large tank or
reservoir (Fig. 95). Here the
velocity downstream from the
enlargement may be taken to be
zero, and the lost head, called
FIG. 95. Pipe Exit.
the "exit loss," may be calculated from
(V,  O) 2
V\
which simply states that when a pipe discharges into a large volume
of fluid the velocity energy of the flow is lost. This agrees with the
result that would be expected from a nonmathematical analysis as
indicated in Fig. 95.
The loss of head due to gradual enlargement is, of course, dependent
upon the shape of the enlargement. Tests have been carried out by
Gibson on the losses in conical enlargements, or "diffusor tubes,"
and the results expressed as a proportion of the loss occurring in a
sudden enlargement by
2g
in which KL is primarily dependent upon the cone angle, but is also a
function of the area ratio, as shown in Fig. 96. Because of the large
surface of the conical enlargement which contacts the fluid, the coeffi
cient KL embodies the effects of friction as well as those of eddying
turbulence. In an enlargement of small central angle, KL will result
MINOR LOSSES IN PIPE LINES
173
almost wholly from surface friction; but as the angle increases and
the enlargement becomes more abrupt, the surfaces are reduced, and
here the energy consumed in eddies determines the magnitude of KL.
1.3
160
FIG. 96. Loss Coefficients for Conical Enlargements. 12
From the plot it may be observed that: (1) there is an optimum cone
angle of 7 where the combination of the effects of surface friction and
eddying turbulence is a minimum; (2) it is better to use a sudden
enlargement than one of cone angle around 60, since KL is smaller
for the former.
ILLUSTRATIVE PROBLEM
A 12in.diameter horizontal water line enlarges to a 24in. line through a 20
conical enlargement. When 10 cfs flow through this line the pressure in the
smaller pipe is 20 lb/in. 2 Calculate the pressure in the larger pipe, neglecting pipe
friction.
12.7 ft/sec, F 24 = 3.18 ft/sec
From the plot of Fig. 96,
K L = 0.43
Pu .
+ () +0.43
3.18) 2
20 X 144 (12.7) 2
j _j_ Q _
62.4 2g w
= 47.93 ft p 20.7 lb/in. 2
IV
12 A. H. Gibson, Hydraulics and Its Applications, Fourth edition, 1930, p. 93,
D. Van Nostrand Co.
174
FLUID FLOW IN PIPES
The physical properties of flow through a sudden contraction are
shown in Fig. 97. Inertia prevents the fluid from following the solid
boundary, and the " live " stream of fluid contracts at section 3 to a
FIG. 97. Sudden Contraction.
diameter less than d. From section 3 to section 4 eddying turbulence
similar to that of the sudden enlargement accounts for most of the
energy which is consumed by the contraction.
Writing Bernoulli's equation between sections 1 and 4
w
whence
2g
which is shown in the figure to be the distance between the two
theoretical pressure grade lines.
MINOR LOSSES IN PIPE LINES
175
The loss of head, hi, in a sudden contraction is expressed by
h L K L ^
in which 4 is the velocity in the smaller pipe. The coefficient KL
depends primarily upon the diameter ratio d 4 /di, which determines
the shape of the sudden contraction ; the relation of KL to d/di and
velocity is shown on Fig. 98.
0.5
0.4
0.3
0.2
0.1
F 4 40tt. per sec.
1.0
FIG. 98. Loss Coefficients for Sudden Contractions. 13
A sharpedged pipe entrance from a large body of fluid is given by
the condition d/d\ = in Fig. 98, and it should be noted that the loss
coefficient KL for such an entrance is close to 0.5 for ordinary velocities.
The values of KL for this and other pipe entrances are shown in Fig. 99,
and here again it may be observed that the magnitude of KL depends
primarily upon the amount of deceleration and consequent eddying
turbulence caused by the entrance.
Because of smooth acceleration of the fluid in gradual contractions
the losses are usually so small that they may be neglected in most
engineering calculations.
13 Data from H. W. King and C. O. Wisler, Hydraulics, Third edition, 1933,
page 182, John Wiley & Sons.
176 FLUID FLOW IN PIPES
Smooth pipe bends will cause losses of head due to the energy con
sumed by the twin eddy motion set up by secondary flows (Art. 29).
The nature and magnitude of bend losses are shown in Fig. 100;
they are comparable to those of Fig. 92. The loss of head in a bend
is expressed by
V 2
in which V is the average velocity in the pipe. The loss coefficient
KL has been shown experimentally by Hofmann M to be a function of
shape, roughness, and Reynolds
I number. Figure 101 gives a par
'  tial summary of his results for
90 circular bends, the shape of
which may be defined by the
l~ .8 simple ratio r/d. The tests were
carried out on polished brass
bends to obtain the "smooth "
curve ; the bends were then arti
ficially roughened by applying a
mixture of sand and paint, result
ing in the " rough " curve. The
curves thus give the extremes of
roughness variation and illus
trate the dependency of KL upon
roughness. Hofmann's tests
were made at Reynolds numbers
between 60,000 and 225,000, but
the curves of Fig. 101 give an
approximate summary of his
results for Reynolds numbers
r
' M from 100,000 to 225,000 in which
FIG. 99. Pipe Entrances. range the values of KL became
practically constant 15 and thus
independent of the Reynolds number. Because of the constancy
of loss coefficients at high Reynolds numbers, it is probable that
14 Trans, of the Hydraulic Institute of the Munich Technical University, Bulletin 3,
p. 29, 1935, A.S.M.E.
15 This constancy of KL is more true of the rough bends than of the smooth ones.
Compare this with the friction factors for smooth and rough pipe, Figs. 85 and 86.
MINOR LOSSES IN PIPE LINES
177
turbulence
FlG. 100.
0.5
0,4
0.3
0.2
.Rough
0.1
Smooth
10
FIG. 101. Loss Coefficients for 90 Circular Bends.
178 FLUID FLOW IN PIPES
Hofmann's results may be applied beyond the Reynolds numbers
attained in his experiments.
The losses of head caused by commercial pipe fittings occur because
of their rough and irregular shapes which cause excessive turbulence
to be created. The shapes of commercial pipe fittings are determined
more by structural properties, ease in handling, and production
methods than by headloss considerations, and it is, therefore, not
feasible or economically justifiable to build pipe fittings having com
pletely streamlined interiors in order to minimize head loss. The loss
of head in commercial pipe fittings is usually expressed by a loss
coefficient, K^ and the velocity head in the pipe, as
72
h L = K L ~
in which KL is a constant (at high Reynolds numbers), the magnitude
of which depends upon the shape of the fitting. Values of KL for
various common fittings, compiled by the Crane Co., 16 are presented
in Table VIII.
TABLE VIII
Loss COEFFICIENT, KL, FOR COMMERCIAL PIPE FITTINGS
Globe valve, wide open ............................... 10 .0
Angle valve, wide open ............................... 5.0
Gate valve, wide open ......... . ....................... 19
f open .......................................... 1.15
f open .......................................... 5.6
Jopen .......................................... 24.0
Return bend ......................................... 2.2
Standard tee ......................................... 1.8
90 elbow ............................................ 90
45 elbow ............................................ 42
It is generally recognized that when fittings are placed in close
proximity the total loss obtained through them is less than their
numerical sum obtained by the foregoing methods. Systematic tests
have not been made on this subject because a simple numerical sum
of the losses gives a result in excess of the actual losses and thus pro
duces an error on the conservative side when predictions of pressures
and rates of flow are to be made.
18 Engineering Data on Flow of Fluids in Pipes and Heat Transmission, 1935,
THE PRESSURE GRADE LINE AND ITS USE
179
43. The Pressure Grade Line and Its Use. The utility of plotting
above the center line of a pipe the pressure head therein has been
apparent in many of the foregoing examples. The result is the
hydraulic or pressure grade line which may be used to give a graphical
significance to the equations of pipe flow.
Some of the properties of the pressure grade line which have been
noted in the preceding problems are (1) its characteristic downward
slope in the direction of flow due to pipe friction, (2) the increase of
this slope with velocity, and (3) the rather abrupt rises and drops
in the grade line when minor losses are incurred.
Pressure (hydraulic)
grade line (exact)
FIG. 102.
Several illustrations of the meaning of the pressure grade line and
its relation to the Bernoulli equation are given in the following exam
ples. Frequently, in long pipe lines where velocities are small and
minor losses of little significance, an approximate grade line may be
sketched which will allow useful engineering conclusions to be drawn
directly.
In a long pipe line between two reservoirs (Fig. 102) the exact
and approximate hydraulic grade lines are as shown. A drop in the
exact grade line occurs as the fluid enters the pipe, caused by (1) the
increases of velocity at the expense of pressure and (2) the head lost
180 FLUID FLOW IN PIPES
at the pipe entrance. Considering minor losses and velocity heads
and writing Bernoulli's equation between the two reservoir surfaces
V 2 I V 2 V 2
+ + h = + + + . 5 _+/ +
2g d 2g 2g
From the exact pressure grade line of Fig. 102
V 2 V 2 I V 2
the same result as obtained from the Bernoulli equation but of more
significance because all the terms may be visualized graphically.
The minor losses and velocity heads being neglected, the Bernoulli
equation becomes
d 2g
and similarly from the approximate pressure grade line
ILLUSTRATIVE PROBLEM
A 6in. pipe line 2000 ft long connects two reservoirs, one of surface elevation
300, the other of surface elevation 200. Assuming that / = 0.024, calculate the
rate of flow through the line (a) including and (b) neglecting minor losses.
.300
F 2 2000 F 2 F 2
(a) + + 300  + + 200 + 0.5 + 0.024r +
2g A 2g 2g
V  8.13 ft/sec Q  1.595 cfs
2000 F 2
(b) + + 300 = + + 200 + 0.024
T 2g
V  8.20 ft/sec Q = 1.61 cfs
Consider now a pipe line of more complicated nature (Fig. 103)
consisting of three pipes of different sizes, with enlargement, contrac
THE PRESSURE GRADE LINE AND ITS USE
181
FIG. 103.
tion, entrance, and exit losses. The Bernoulli equation may be
written as before
2g di 2g
From the exact pressure grade line
V! VI
~T~ + T"
2g 2g
182 FLUID FLOW IN PIPES
which is identical with the result obtained from the Bernoulli
equation.
Neglecting minor losses and pressure variation incurred by velocity
changes, the Bernoulli equation gives
+ + * o + o + 04^ +/,' +/,
and from the approximate pressure grade line the same result
7 T/ 2 7 T/ 2 7V 2
h = flT^+/2J^+f3J^
di 2g d 2 2g d% 2g
is obtained.
The above examples serve to illustrate the properties of the exact
and approximate pressure grade lines and the use of these lines in
lending a significance to pipe flow problems which cannot be obtained
from equations alone.
FIG. 104.
When a pump or motor is present in a pipe line the pressure grade
line assumes a special shape as a result of the fluid energy which is
given to or taken from the fluid by the machine. The approximate
grade line for a pipe in which a pump is installed is shown in Fig. 104.
Here as the flow takes place from A to B the pump must supply
THE PRESSURE GRADE LINE AND ITS USE
183
sufficient energy (1) to raise the fluid through a height h, and (2) to
overcome the friction loss in the line. This is seen clearly from the
pressure grade line, which gives
E p = h + fi 
~
2g
/2
d 2 2g
The same result may, of course, be obtained by writing the Bernoulli
equation between the reservoir surfaces; here
+ + + E p
2g
d 2 2g
At points in a pipe line where the velocity is high or where the
elevation of the pipe is well above the datum plane, the pressure grade
line may be below the pipe (Fig. 105). According to the method of
plotting the grade line, the pressure in a pipe which lies above the
grade line must have a negative value and thus be less than the
atmospheric pressure. Regions of negative pressure in pipe lines are
frequently a source of trouble in pipeline operation since gases dis
solved in the flowing fluid tend to come out of solution and collect
^Pressure grade line
(approximate)
FIG. 105.
in the line, thus reducing the capacity of the line; negative pressures
within the line also give an opportunity for air to leak into the line,
increase the pressure, and reduce the flow. Regions of negative pres
sure are avoided as much as possible in the design of pipe lines, but
where they exist a source of vacuum is required to draw off collections
of gas. Gases which collect at high points in a pipe line when the
pipe is below the hydraulic grade line may be vented by opening a
cock at the highest point in the pipe and allowing the pressure of the
fluid to force the gases out.
184
FLUID FLOW IN PIPES
If the curves in the elevational view of a pipe line are abrupt,
they are frequently loosely termed "siphons," although the only
resemblance to siphonic action occurs in a pipe which is convex upward
and runs above the hydraulic grade line. Pipeline curves which are
convex upward are called " siphons" whether they are above or below
the hydraulic grade line; curves which are convex downwards are
termed "inverted siphons," an obvious misnomer. This confused
situation may be somewhat clarified by the study of the action of the
"true siphon" of Fig. 106 and by comparison of true siphonic action
with that occurring in the vertical curves of a pipe line. The distinc
FIG. 106.
tive characteristic of the true siphon is that practically its entire
length lies above the hydraulic grade line. The factor which deter
mines its successful operation is the maximum negative pressure,
which exists at the crown of the siphon. The magnitude of this
negative pressure is, of course, limited by the barometric height and
vapor pressure of the liquid, 17 but it should not be concluded that
such a negative pressure may be obtained in practice; actually dis
ruption of the flow occurs at negative pressures well below the maxi
mum, due to the liberation of entrained gases. Obviously, then, the
factors which change the negative pressure at the crown of the siphon
17 See Appendix V.
BRANCHING PIPES
185
are the factors which are fundamental to its successful operation.
From the approximate pressure grade line it may be concluded directly
that (1) raising the crown of the siphon and (2) lowering the free end
of the pipe both increase this negative pressure and that these, there
fore, are the fundamental variables which determine the success or
failure of siphon operation. Although these conclusions might have
been reached by several applications of the Bernoulli principle the
utility of the hydraulic grade line has been shown again in discovering
the fundamental factors of a problem quickly and efficiently.
44. Branching Pipes. Some of the more complicated problems
of pipeline design involve the flow of fluids in pipes which intersect.
FIG. 107.
The principles involved in problems of this type may be obtained by
a study of pipes which (1) divide and rejoin and (2) lead from regions
of different pressure and meet at a common point.
In pipeline practice, "looping" or laying a line, B, parallel to an
existing pipe line, A, and connected with it (Fig. 107), is a standard
method of increasing the capacity of the line. Here there is an
interesting analogy between fluid flow and the characteristics of a
parallel electric circuit, if head lost is compared with drop in potential
and rate of flow with electric current.
Obviously as flow takes place from point 1 to point 2 through either
pipe A or pipe B the same lost head, &L, is accomplished and from the
continuity principle the sum of the flows in pipes A and B must be the
186
FLUID FLOW IN PIPES
total flow in the single pipe line. Neglecting minor losses, these two
facts allow the writing of the equations
2g
which may be placed in terms of Q for simultaneous solution by
inserting the relations
V A = ^ and V B = ~f
giving
2gd A \ird 2gd B
which would allow calculation of the division of a flow, (?, into two
flows, QA and QB, when the sizes and friction factors of the pipes are
known.
ILLUSTRATIVE PROBLEM
A 12in. water line, in which 12 cfs flow, branches into a 6in. line 1000 ft long
and an 8in. line 2000 ft long which rejoin and continue as a 12in. line. Calculate
the flow through the two branch lines assuming that/ = 0.022 for both of these.
2000' 8"
Q12c.f.s.
<?12c.f,s.
F 8 
1000' 6'
= 5.1
2.87^8
Solving
but
T2"
Q = @ 8 + Q 6 = 12
6  4.89 Cfs 8
7.11 cfs
0.022XX^2!)!
Another engineering example of branching pipes is that typified
by the " threereservoir problem" of Fig. 108 which may be solved
advantageously by application of hydraulic grade line principles.
BRANCHING PIPES
187
Here flow may take place. (1) from reservoir A into reservoirs B and C,
or (2) from reservoir A to C without inflow or outflow from reservoir B,
or (3) from reservoirs A and B into reservoir C. The approximate
hydraulic grade lines representing these conditions are indicated on
FIG. 108.
the figure, and it is obvious that the slopes of these lines and the above
flow conditions are determined by the magnitude of the pressure head
at the junction, 0. Assuming flow to take place from reservoirs A
and B into C, the following equations may be written directly from
inspection of the hydraulic grade lines.
. .
ZA = z H  h /A
w d A
_+*+, hi (&)'
w as <g \ira,B/
and from the continuity principle
Qc
These four simultaneous equations may be solved readily by trial
(if pipe sizes, friction factors, and elevations are known), by assuming a
value of po/w and solving the first three equations for QA, QB, and Qc
The solution of the problem is completed when these values of QA, QB
188 FLUID FLOW IN PIPES
and Qc satisfy the fourth equation, thus allowing prediction of the
rates of flow for a given or existing pipe system.
The above type of solution is not that encountered in design prob
lems where the desired rates of flow are known, the pipeline elevations
fixed by topography and other considerations, and the problem is to
build the most economical pipe system to transmit these flows. In
this case the assumption of a value of po/w immediately fixes the
diameters and, therefore, the initial and operation costs of the pipes.
Various assumptions of p /w may be made and the resulting total
costs plotted agains po/iv. The value of p /w resulting in the mini
mum total cost will determine all the pipe sizes for the most economical
design.
BIBLIOGRAPHY
(See also Bibliography for Chapter IV.)
R. L. DAUGHERTY, Hydraulics, Fourth Edition, McGrawHill Book Co., 1937.
H. W. KING and C. O. WISLER, Hydraulics, Third Edition, John Wiley & Sons, 1933.
G. E. RUSSELL, Textbook on Hydraulics, Fourth Edition, Henry Holt & Co., 1934.
H. W. KING, Handbook of Hydraulics, Third Edition, McGrawHill Book Co., 1939.
A. H. GIBSON, Hydraulics and Its Applications, Fourth Edition, D. Van Nostrand,
1930.
PROBLEMS
252. When 10 cfs of water flow through a 6in. constriction in a 12in. hori
zontal pipe line, the pressure at a point in the pipe is 50 lb/in. 2 , and the head lost
between this point and the constriction is 10 ft. Calculate the pressure in the con
striction.
253. A 2in. nozzle terminates a vertical 6in. pipe line in which water flows down
ward. At a point on the pipe line a pressure gage reads 40 lb/in. 2 If this point is
12 ft above the nozzle tip and the head lost between point and tip is 5 ft, calculate
the rate of flow.
254. A 12in. pipe leaves a reservoir of surface elevation 300 at elevation 250 and
drops to elevation 150, where it terminates in a 3in. nozzle. If the head lost through
line and nozzle is 30 ft, calculate the rate of flow.
255. An 18in. pipe line runs from a reservoir of surface elevation 350 and dis
charges into the atmosphere at elevation 250. If the loss of head in this pipe line
is 90 ft, what flow can be expected?
256. A vertical 6in. pipe leaves a water tank of surface elevation 80. Between
the tank and elevation 40 on the line, 8 ft of head is lost when 2 cfs flow through the
line. If an open piezometer tube is attached to the pipe at elevation 40, what will
be the elevation of the water surface in this tube?
257. What horsepower pump is required to pump 20 cfs of water from a reservoir
of surface elevation 100 to one of surface elevation 250, if in the pump and pipe line
40 ft of head are lost?
258. Through a hydraulic turbine flow 100 cfs of water. On the 42in. inlet pipe
at elevation 145, a pressure gage reads 50 lb/in. 2 On the 60in. discharge pipe at
PROBLEMS
189
elevation 130 a vacuum gage reads 6 in. of mercury. If the total head lost through
pipes and turbine between elevations 145 and 130 is 30 ft, what horsepower may be
expected from the machine?
259. In a 9in. pipe line 5 cfs of water are pumped from a reservoir of surface
elevation 100 over a hill of elevation 165. What horsepower pump is required to
maintain a pressure of 50 lb/in. 2 on the hilltop if the head lost between reservoir
and hilltop is 20 ft?
260. Water flows downward through a 1000ft section of 12in. pipe line running
from elevation 200 to elevation 100. The pressure at elevation 200 is 40 lb/in. 2 , and
the velocity in the line is 10 ft /sec. The head lost in this section is 5 ft. Calculate: (a)
the total energy at elevations 200 and 100, taking datum at sea level; (6) the pressure
at elevation 100; (c) the shear stress at the wall of the pipe, 3 in. from the center
and on the centerline; (d) the lost horsepower in the 1000ft section for this flow.
261. When a liquid flows in a horizontal 6in. pipe, the shear stress at the walls is
1.3 lb/ft. 2 Calculate the pressure drop in 100 ft of this pipe line.
262. If 2 cfs of glycerine at 50 F flow in a 6in. pipe, calculate: (a) the velocity
at the center of the pipe; (6) the loss of head in 100 ft of the pipe by equations 44
and 45 ; (c) the velocity 2 in. from the centerline; (d) the shear stress at the wall; (e) the
shear stress 2 in. from the centerline.
263. Oil of viscosity 0.01 Ib sec/ft 2 and specific gravity 0.90 flows with an average
velocity of 5 ft /sec in a 12in. pipe line. Calculate the shear stress at the wall and
3 in. from the center line.
264. In a section of 2in. pipe line 500 ft long running from elevation 130 to eleva
tion 90 flow 30 gpm of linseed oil at 80 F. If flow is downward and pressure at
elevation 130 is 30 lb/in. 2 , calculate the pressure at elevation
90. Check for laminar flow before calculating this pressure.
265. If 0.25 gpm of oil of specific gravity 0.92 flow in
laminar condition through this vertical 1in. pipe line, calcu
late the viscosity of the oil if the manometer deflection is
10 in.
266. Water flows at 50 F from a reservoir through a 1in.
pipe line 2000 ft long which discharges into the atmosphere
at a point 1 ft below the reservoir surface. Calculate the
flow, assuming it to be laminar and neglecting the velocity
head in the pipe line. Check the assumption of laminar
flow.
267. Glycerine flows through a 2in. horizontal pipe line,
leading from a tank and discharging into the atmosphere.
If the pipe line leaves the tank 20 ft below the liquid surface
and is 100 ft long, calculate the flow when the glycerine has a temperature of
(a) 50 F, (b) 70 F.
268. What horsepower pump is required to pump 50 gpm of linseed oil from a
tank of surface elevation 46 to one of elevation 60 through 1500 ft of 3in. pipe, if the
011 is at (a) 80 F, (b) 100 F?
269. If 180 gpm of water at 70 F flow in a 6in. pipe line having roughness pro
tuberances of average height 0.030 in. and if similar roughness having height 0.015 in.
exists in a 3in. pipe, what flow of linseed oil (100 F) must take place therein for the
friction factors of the two pipes to be the same?
270. In a 12in. pipe line, 15 cfs of water flow upward. At a point on the line at
190 FLUID FLOW IN PIPES
elevation 100, a pressure gage reads 130 lb/in. 2 Calculate the pressure at elevation
150, 2000 ft up the line, assuming the friction factor to be 0.02.
271. Given that 250 gpm of carbon tetrachloride flow downward through 200 ft
of 3in. pipe, from elevation 100 to elevation 40; the pressure at the former point is
30 lb/in. 2 , and that at the latter 50 lb/in. 2 Calculate the friction factor.
272. Calculate the loss of head in 1000 ft of 3in. brass pipe when water at 80 F
flows therein at an average velocity of 10 ft/sec.
273. In. a 2in. pipe line, 25 gpm of glycerine flow at 70 F. Calculate the loss
of head in 200 ft of this pipe.
274. If 100 Ib/min of air flow in a 3in. horizontal clean steel pipe line at t75 lb/in. 2
ab and 80 F, calculate the loss of head and pressure drop in 300 ft of this pipe.
275. If 2 cfs of water flow at 50 F in a clean 6in. castiron pipe line, calculate
the loss of head per 100 ft of line.
276. If 12 cfs of water flow in a 1 2in. spiral riveted pipe at 70 F, calculate the
loss of head in 500 ft of this pipe.
277. Castor oil flows in a 1in. pipe line at 50 F at a velocity of 2 ft /sec. Calcu
late the loss of head in 100 ft of this pipe.
278. A 3in. brass pipe line 100 ft long carries 100 gpm of linseed oil. Calculate
the head loss when the oil is at (a) 80 F, (b) 110 F.
279. Solve the preceding problem when the liquid is castor oil.
280. In a laboratory test 490 Ib/min of water at 60 F flow through a section of
2in. brass pipe 30 ft long. A differential manometer connected to the ends of this
section shows a reading of 14.2 in. If the fluid in the bottom of the manometer has
a specific gravity of 3.20, calculate the friction factor and the Reynolds number.
281. Water flows from a tank through 200 ft of horizontal 2in. brass pipe and
discharges into the atmosphere. If the water surface in the tank is 4 ft above the
pipe, calculate the rate of flow, considering losses due to pipe friction only, when the
water temperature is (a) 50 F, (b) 100 F. (Trial and error solution or by Blasius'
equation for /.)
282. A 1in. clean galvanized pipe 400 ft long leaves a water tank at elevation 30
and discharges into the atmosphere at elevation 27. If the water surface in the tank
has elevation 35 and the water is at 70 F, calculate the rate of flow considering losses
due to pipe friction only. (Trial and error solution.)
283. In a horizontal 8in. clean steel pipe line, 5.0 Ib/sec of superheated steam
flow at an absolute pressure of 250 lb/in. 2 If the steam is superheated 100 F, calcu
late loss of head and loss of pressure in 500 ft of this pipe.
284. Carbon dioxide flows in a 4in. horizontal clean steel pipe line at a velocity
of 10 ft/sec. At a point in the line a pressure gage reads 100 lb/in. 2 and the tem
perature is 100 F. What pressure is lost due to friction in 100 ft of this pipe? Baro
metric pressure 14.7 lb/in. 2
285. What loss of head occurs in 100 ft of a in. galvanized pipe when water flows
therein at 60 F and 30 ft/sec?
286. If 2.5 cfs of water flow in an 18in. clean steel pipe line at 40 F, calculate
(a) the shear stress at the pipe walls and (b) the velocity on the pipe centerline from
Fig. 88 and equation 55.
287. Water flows in a 6in. pipe line at 60 F. If the velocity on the pipe's center
line is 3 ft/sec, what is the rate of flow?
288. Linseed oil flows in an 8in. pipe line at 80 F. The center velocity is
6 ft/sec, Calculate the rate of flow.
PROBLEMS 191
289. Glycerine flows in a 2in. pipe at 50 F and with a center velocity of 8 ft /sec.
Calculate the rate of flow through the line.
290. Carbon dioxide flows in a 3in. pipe at 50 lb/in. 2 ab and 50 F. If the center
velocity is 2 ft/sec, calculate the weight rate of flow.
291. If 250 gpm of water flow at 40 F in a 6in. smooth pipe line, calculate the
approximate thickness of the laminar film, the center velocity, and the velocity V w .
292. Prove that, for smooth pipes having Reynolds numbers below 100,000,
5 58.3
provided that 5 is small.
293. Air flows isothermally in a 3in. pipe line at a velocity of 10 ft /sec, an
absolute pressure of 40 lb/in. 2 , and temperature of 50 F. If pressure is lost by
friction, calculate the velocity where the pressure is 30 lb/in. 2 abs.
294. Through a horizontal 6in. clean castiron pipe line 1000 ft long in which
the temperature is 60 F, 200 Ib/min. of air flow isothermally. If the pressure at the
upstream end of this pipe length is 30 lb/in. 2 abs, calculate the pressure at the down
stream end and the velocities at these two points.
295. Carbon dioxide flows isothermally in a 2in. horizontal clean steel pipe line
and at a certain point the velocity is 60 ft/sec, pressure 60 lb/in. 2 abs, and tempera
ture 80 F. Calculate the pressure and velocity 500 ft down stream from this point.
296. Air at 14.7 lb/in. 2 abs and 60 F flows in a 12 in. by 3 in. horizontal clean
galvanized duct at a velocity of 15 ft /sec. Calculate the loss of head and loss of
pressure per foot of duct.
297. Air at 14.7 lb/in. 2 abs and 60 F flows in a horizontal triangular clean gal
vanized duct having 8in. sides, at a velocity of 12 ft/sec. Calculate the head and
pressure lost per foot of duct.
298. Ten cubic feet of water flow per second in a smooth 9in. square duct at
50 F. Calculate the head lost in 100 ft of this duct.
299. A brick conduit of crosssectional area 10 ft 2 and wetted perimeter 12 ft
carries water at 50 F at a velocity of 8 ft /sec. Calculate the head lost in 200 ft of
this conduit.
300. A semicircular good brick conduit 5 ft in diameter carries water at 70 F
at a velocity of 10 ft/sec. Calculate the loss of head per foot of conduit.
301. A reservoir has surface elevation 200. A 24in. old castiron pipe line 2 miles
long leaves the reservoir at elevation 180 and discharges freely at elevation 100.
Calculate the flow through the line.
302. A new riveted steel 12in. pipe line 1 mile long runs from elevation 350 to
elevation 325. If the pressure at both of these points is 50 lb/in. 2 , calculate the
rate of flow.
303. A new riveted steel 18in. pipe line 1000 ft long runs from elevation 150 to
elevation 200. If the pressure at the former point is 100 lb/in. 2 and at the latter
72 lb/in. 2 , what rate of flow can be expected through the line?
304. A 3 ft square smooth masonry conduit is 3 miles long and connects two
reservoirs having a difference in their surface elevations of 60 ft. What flow can be
expected through the conduit?
305. If 200 cfs of water are to be carried through a smooth wooden pipe line
which leaves a reservoir (surface elevation 290) at elevation 250 and runs 3 miles
192
FLUID FLOW IN PIPES
to elevation 200, where a pressure of 30 lb/in. 2 is to be maintained, what size pipe
line is required?
306. If 60 cfs of water are to be pumped from a reservoir of surface elevation 150
to one of surface elevation 200 through 2 miles of 30in. old castiron pipe, what horse
power pump is required?
307. What diameter smooth masonry pipe is necessary to carry 50 cfs between
two reservoirs of surface elevations 250 and 100 if the pipe line is to be 2 miles long?
308. An extremely smooth 12in pipe line leaves a reservoir of surface elevation
500 at elevation 460. A pressure gage is located on this line at elevation 400 and
1000 feet from the reservoir (measured along the line). Calculate the gage reading
when 10 cfs flow in the line, using: (a) the HazenWilliams formula, (b) Fig. 86.
(Temperature 68 F.)
309. Calculate, by the HazenWilliams method, the head lost in the brick conduit
of Problem 299.
310. If 5 cfs of water flow through a 6in. horizontal pipe which suddenly enlarges
to 12in. diameter, and if the pressure in the smaller pipe is 20 lb/in. 2 , calculate that
in the 12in. pipe, neglecting pipe friction.
311. Estimate the rate of flow of benzene through this
sudden enlargement if the upper gage reads 20 lb/in. 2 and the
lower one 23 lb/in. 2
312. Calculate the manometer reading when 8 cfs of
water flow through this enlargement.
J
10'
313. Solve Problem 312 assuming conical enlargements of 70 and 7.
314. A 4in. pipe suddenly contracts to a 2in. pipe in which the velocity is
10 ft/sec. Calculate the loss of head through the contraction if water is flowing.
315. The velocity of water in a 6in hori
zontal pipe is 3 ft/sec. Calculate the loss of
rhead through a sudden contraction to 2in.
diameter. If the pressure in the 6in. pipe is 50
lb/in. 2 , what is the pressure in the 2in. pipe,
neglecting pipe friction?
316. Calculate the height to which water will
rise in the downstream piezometer tube when 1.0
cfs flows through this contraction.
317. Water flows at 10 ft/sec in a 6in. pipe
line which contracts abruptly to a 3in. pipe.
Calculate the velocity and diameter of the live stream at the contracted section (3,
Fig. 97).
318* If 2 cfs of water leave a reservoir in a 6in. pipe line, calculate the lost head
PROBLEMS
193
caused by the pipe entrance assuming it to be the "reentrant," "sharp," and
"rounded" entrances of Fig. 99.
319. A 90 smooth bend in a 12in. pipe line has a radius of 4 ft. Calculate the
loss of head when 15 cfs of fluid flow through this bend.
320. A horizontal 2in. pipe line leaves a water tank 20 ft below the water sur
face. If this line has a sharp entrance and discharges into the atmosphere, calculate
the rate of flow neglecting and considering the entrance loss, if the pipe length is
(a) 15 ft, (b) 150 ft. Assume /to be 0.025.
321. A 15in. pipe line connects two reservoirs having surface elevations 150 and
90. If the line is 1 mile long and has a sharp entrance, calculate the rate of flow
including and neglecting minor losses. Assume that/ = 0.020.
322. A 12in. horizontal pipe 1000 ft long leaves a reservoir of surface elevation
200 at elevation 180. This line connects to a 6in. pipe 1000 ft long running to eleva
tion 100, where it enters a reservoir of surface elevation 130. Assuming that/ = 0.02,
and neglecting minor losses, calculate the flow through the line and sketch the approx
imate hydraulic grade line, showing its elevation at the contraction.
323. At the rate of 50 gpm, linseed oil is to be pumped through 1000 ft of 2in.
brass pipe line between two tanks having difference of surface elvation 10 ft. Neglect
ing minor losses, what pump horsepower is required if the oil temperature is (a) 80 F
(b) 120 F?
324. A horizontal 2in. brass pipe line leaves (sharp entrance) a water tank 10 ft
below its free surface. At 50 ft from the tank, it suddenly enlarges to a 4in. pipe
which runs 100 ft horizontally to another tank entering it 2 ft below its surface.
Calculate the flow through the line when the water temperature is 68 F, neglecting
no losses. (Trial and error solution.)
325. A pipe line running between two reservoirs having surface elevations of
500 and 300 consists of 1000 ft oMOin, 8in., and 6in. pipe in that order downstream,
connected by sudden contractions. Assuming that/ = 0.023 for all pipes, calculate
the rate of flow and sketch the hydraulic grade line showing all important elevations
(a) including and (b) neglecting the minor losses. Assume that the line has a sharp
entrance.
326. Calculate the rate of flow from this water tank
if the 6in. pipe line has / = 0.02 and is 50 ft long.
327. A 12in. pipe line 1500 ft long leaves a reservoir
of surface elevation 500 at elevation 460 and runs to
elevation 390, where it discharges into the atmosphere.
Calculate the flow and sketch accurately the hydraulic
grade line (assuming that / = 0.022) : (a) for the con
ditions above, and (b) when a 3in. nozzle is attached
to the end of the line, assuming the lost head in the
nozzle to be 5 ft.
328. A 6in. horizontal pipe line 1000 ft long leaves a
reservoir of surface elevation 300 at elevation 250. The
line terminates in a 1 2in. nozzle. Calculate the discharge from the nozzle and the
horsepower available in the nozzle stream, and sketch the hydraulic grade line.
Assume that/ = O.025 and neglect entrance and nozzle loss.
329. A pump close to a reservoir of surface elevation 100 pumps water through
a 6in. pipe line 1500 ft long and discharges it at elevation 200 through a 2in. nozzle.
Calculate the pump horsepower necessary to maintain a pressure of 50 lb/in. 2 behind
Elev. 200
'Elev. 150
194 FLUID FLOW IN PIPES
the nozzle, and sketch accurately the hydraulic grade line, neglecting loss in the
nozzle and taking/ = 0.02.
330. A 24in. pipe line 3000 ft long leaves (sharp entrance) a reservoir of surface
elevation 500 at elevation 450 and runs to a turbine at elevation 200. Water flows
from the turbine through a 36in. vertical pipe ("draft tube") 20 ft long to tailwater
of surface elevation 185. When 30 cfs flow through pipe and turbine, what horse
power is developed? What are the pressures at the entrance and exit of the turbine?
Take/ = 0.02; include exit loss; neglect other minor losses and those within the
turbine.
331. The siphon of Fig. 106 consists of a 2in. hose having/ = 0.025. The crown
of the siphon is 10 ft above and the free end 15 ft below the water level in the tank.
If the hose is 60 ft long and the bend is at its third point, calculate the flow and the
pressure at the crown.
332. What is the maximum flow which may be obtained theoretically in problem
322 when the 6in. and 12in. pipes are interchanged?
333. The horizontal 8in. suction pipe of a pump is 500 ft long and is connected
to a reservoir of surface elevation 300, 10 ft below the water surface. From the pump,
the 6in. discharge pipe runs 2000 ft to a reservoir of surface elevation 420, which
it enters 30 ft below the water surface. Taking/ to be 0.02 for both pipes, calculate
the pump horsepower required to pump 3.0 cfs from the lower reservoir. Sketch
accurately the approximate hydraulic grade line. What is the maximum theoretical
flow which may be pumped through this system: (a) with the 8in. suction pipe,
(b) with a 6in. suction pipe?
334. A 12in. pipe line 2 miles long runs on an even grade between reservoirs of
surface elevations 500 and 400, entering the reservoirs 30 ft below their surfaces.
The flow through the line is inadequate, and a pump is installed at elevation 420 to
increase the capacity of the line. Assuming / as 0.02, what pump horsepower is
required to pump 6.0 cfs downhill through the line? Sketch accurately the approx
imate hydraulic grade line before and after the pump is installed. What is the
maximum theoretical flow which may be obtained through the line?
335. A 24in. pipe line branches into a 12in. and an 18in. pipe, each of which is
1 mile long and rejoin to form a 24in. pipe. If 30 cfs flow in the main pipe, how will
the flow divide? Assume that/ = 0.018 for both branches.
336. A 24in. pipe line carrying 30 cfs divides into 6in., 8in., and 12in. branches,
all of which are the same length and enter the same reservoir below its surface.
Assuming that / = 0.020 for all pipes, how will the flow divide?
337. An 18in. pipe divides into 12in. and 6in. branches which rejoin. If the
6in. branch is 1 mile long, how long must the 12in. branch be for the flow to divide
equally when 10 cfs flow in the 18in. pipe? Assume that/ = 0.019 throughout.
338. A 36in. pipe divides into three 18in. pipes at elevation 400. The 18in.
pipes run to reservoirs which have surface elevations of 300, 200, and 100, these
pipes having respective lengths of 2, 3, and 4 miles. When 42 cfs flow in the 36in.
line, how will the flow divide? Assume that/ = .017 for all pipes.
339. Reservoirs A, B, and C have surface elevations of 500, 400, and 300, respec
tively. A 12in. pipe 1 mile long leaves reservoir A and runs to point at elevation
450. Here the pipe divides and an 8in. pipe 1 mile long runs from to B and a 6in.
pipe l miles long runs from to C. Assuming that/ = 0.02, calculate the flows in
the lines.
PROBLEMS 195
340. A straight 12in. pipe line 3 miles long is laid between two reservoirs of sur
face elevations 500 and 350 entering these reservoirs 30 ft beneath their free surfaces.
To increase the capicity of the line a 12in. line 1.5 miles long is laid from the original
line's midpoint to the lower reservoir. What increase in flow is gained by installing
the new line? Assume that / = 0.02 for all pipes.
341. Three pipes join at a common point at elevation 350. One, a 12in. line
2000 ft long, goes to a reservoir of surface elevation 400; another, a 6in. line 3000 ft
long, goes to a reservoir of surface elevation 500; the third (6in.) runs 1000 ft to
elevation 250, where it discharges into the atmosphere. Assuming that / = 0.02,
calculate the flow in each line.
CHAPTER VII
FLUID FLOW IN OPEN CHANNELS
Openchannel flow embraces that variety of problems which arise
when water flows in natural water courses, regular canals, irrigation
ditches, sewer lines, flumes, etc. a province of paramount importance
to the civil engineer. Although openchannel problems practically
always involve the flow of water, and although the experimental results
used in these problems were obtained by hydraulic tests, modern fluid
mechanics indicates the extent to which these results may be applied
to the flow of other fluids in open channels.
45. Fundamentals. In the problems of pipe lines, as may be seen
from the hydraulic grade line, the pressures in the pipe may vary along
the pipe and depend upon energy losses and the conditions imposed
upon the ends of the line. Open flow, however, is characterized by
FIG: 109.
the fact that pressure conditions are determined by the constant
pressure, usually atmospheric, existing on the entire surface of the
flowing liquid. In general, pressure variations within a liquid flowing
in an open channel are determined by the principles of fluid statics
(Art. 8) unless the flow is sharply divergent, convergent, or curved;
the latter conditions are accompanied by vertical accelerations or
196
FUNDAMENTALS
197
decelerations which disrupt the laws of static pressure distribution
(Art. 15). Therefore, with the foregoing exceptions, the pressure head
equivalent to the pressure at a point in openchannel flow is exactly
equal to the depth of submergence of the point (Fig. 109). This leads
to the conclusion that the hydraulic grade lines for all the streamtubes
which compose openchannel flow lie in the liquid surface.
FIG. 110.
Openchannel flow may be laminar or turbulent, steady or un
steady, "uniform" or "varied," "tranquil" or "rapid." The laminar
flow of liquids in open channels has few practical engineering applica
tions, and the problem of unsteady flow in open channels is an exceed
ingly complex one; the following discussion will therefore be confined
to steady turbulent flow. The definitions of tranquil and rapid flow
will be presented subsequently, but the significance, causes, and limits
198 FLUID FLOW IN OPEN CHANNELS
of uniform and varied flow must now be examined. The meaning of
these terms and the fundamentals of open channel flow may be seen
from a comparison of perfect fluid flow and real fluid flow in similar
channels of unchanging shape leading from reservoirs of the same
surface elevation (Fig. HO). 1 No resistance will be encountered by
the perfect fluid as it flows down the channel, and because of this lack
of resistance it will continually accelerate under the influence of grav
ity. Thus, the velocity of flow in the channel continually increases,
and with this velocity increase, a reduction in flow cross section is
required by the continuity principle. Reduction in flow cross section
is characterized by a decrease in depth of flow ; since the depth of flow
continually "varies" when the forces acting on fluid particles are
unbalanced, this type of fluid motion is termed varied flow.
When real fluid flows in the same channel, motion encounters
resistance forces due to fluid viscosity and channel roughness. Analysis
of the resistance forces originating from these same properties in pipes
has shown the forces to depend upon the velocity of flow (equation 54).
Thus, in the upper end of the channel where motion is slow, resistance
forces are small but the components of gravity forces in the direction
of motion are the same as for the ideal fluid. The resulting unbalanced
forces in the direction of motion bring about acceleration and varied
flow in the upper reaches of the channels. However, with an increase
of velocity, the forces of resistance increase until they finally balance
those caused by gravity. Upon the occurrence of this force balance,
constantvelocity motion is attained, which is characterized by no
change of flow cross section and thus no change in the depth of flow
the depth of flow remains constant, hence the term uniform flow.
Toward the lower end of the channel, pressure forces exceed resistance
forces and varied flow again results.
Obviously, an inequality of the above forces is more probable than
a balance of these forces, and hence varied flow occurs in practice to
a far greater extent than uniform flow. In short channels, for exam
ple, uniformflow conditions may never be attained because of the
long reach of channel necessary for the establishment of uniform flow.
However, in many problems a rigorous treatment of varied flow is not
necessary, and approximate calculations of varied flow may be carried
out with the equations of uniform flow. Solution of the uniformflow
problem forms the basis of openchannelflow calculations.
1 The channel slopes in most of the illustrations in this chapter have been exag
gerated to emphasize their existence. In openchannel practice, slopes are very
seldom encountered which are greater than 1 ft in 100 ft, or 0.01.
UNIFORM FLOW THE CHEZY EQUATION
199
46. Uniform Flow The Chezy Equation. The fundamental
equation for uniform open channel flow may be derived readily by
equating the equal and opposite forces due to gravity and resistance
and applying some of the fundamental notions of fluid mechanics
obtaiiied in the analysis of pipe flow. Consider the uniform flow of
a liquid between sections 1 and 2 of the open channel of Fig. 111.
Horizontal
FIG. 111.
The forces acting upon the mass of fluid, A BCD, contained between
sections 1 and 2 are: (1) the forces of static pressure FI and F%, acting
on the ends of the mass ; (2) the component of weight in the direction
of motion, T^sin a; and (3) the force of resistance, F T , resulting from
the shear stresses, T O , on the bottom and sides of the channel. A
summation of these forces gives
But obviously
and therefore
F l h Wsin a  F 2  F T =
F T = W sin a
But W = Awl, sin a = HL /I, and tan a = 5, in which S is the slope of
the channel bed and of the liquid surface as well. For the small slopes
encountered in openchannel practice
tan a. = sin a = S
Taking T O to be the average shear stress on the sides and bottom
of the channel the resistance force, F T , may be expressed by
200 FLUID FLOW IN OPEN CHANNELS
Substituting these relationships in the above force equation
T pl = AwlS
or
A
TO = W O
p
in which A/p is recognized as the hydraulic radius, R. Therefore
r = wRS (68)
In pipe flow, T O was shown to be given (equation 54) by
in which/, the friction factor, was dependent upon surface roughness
and the Reynolds number, but more dependent upon the magnitude
of the former for highly turbulent flow. The mechanism of real fluid
motion is similar in pipes and open channels, and hence as an approxi
mation it may be concluded that for turbulent flow in open channels
TO = KV 2 (69)
in which K is a coefficient depending primarily upon the roughness
of the channel lining. Substituting this expression in equation 68
KV 2 = wRS
and thus
v = ~ VRS
or, if C =
V=CV~RS (70)
called the "Chezy equation " 2 after the French hydraulician who
established this relationship experimentally in 1775. By applying
the continuity principle, the equation may be placed in terms of Q as
<2  CAVRS (71)
the fundamental equation of uniform flow in open channels.
47. The Chezy Coefficient. Many experiments have been carried
out to determine the magnitude of the Chezy coefficient, C, and its
dependence upon other variables. A theoretical basis for these experi
2 Compare the above derivation of the Chezy equation with that of Art. 41.
THE CHEZY COEFFICIENT 201
mental results may be obtained by the application of some of the
fundamental notions of fluid mechanics in the following manner:
From equation 69
and substituting this expression in the equation for C above,
With g a constant, the Chezy coefficient, C, is thus dependent upon
the friction factor, /, of the channel. Expressing this in a more general
fashion
But from a study of flow in circular pipes which all have the same
shape,
/ = F 2 ( , roughness
In openchannel flow the frictional processes are the same as those in
pipes, but the shapes of open channels are quite different and this
variation in shape must affect the friction factor. Hence, for open
channels
/ = F 3 I , roughness, shape J
Designating roughness by n, and recalling (Art. 40) that the hydraulic
radius, R, characterizes the shape of the section as regards frictional
effects in turbulent flow, this equation may be written
or more generally
/ = F 4 (F, v, n, R)
and, since
C = F 6 (V,v,n,R)
For water flowing in open channels within the usual range of tempera
ture the kinematic viscosity, v, varies little ; taking v as a constant
C  F t (V, n, R)
202 FLUID FLOW IN OPEN CHANNELS
From the Chezy equation
V = F 7 (C, U, S)
Substituting this relationship in the one above,
C = F s (n, R, 5)
Theoretical reasoning thus leads to the conclusion that for the turbu
lent flow of water in open channels the Chezy coefficient is dependent
upon the roughness, hydraulic radius, and slope of the channel.
This conclusion is confirmed by the formula for C proposed by
Ganguillet and Kutter, 8 two Swiss engineers, in 1869. Their formula
was derived from experimental results obtained from hydraulic tests
on artificial and natural channels of all descriptions, ranging in size
from small laboratory channels up to large rivers. The formula has
come to be used widely in this country and abroad, although recently
its accuracy has been criticized because of errors in some of the tests
on which the formula is based. The formula is
in which 5 has no dimensions, R is in feet, and n may be obtained from
the partial list of values of Table IX. Tables and diagrams are avail
able in the hydraulic literature for the solution of the formula. A
graphical method of solution is indicated on Fig. 112.
TABLE IX
VALUES OF THE ROUGHNESS COEFFICIENT, n
Smooth cement, planed timber 0.010
Rough timber, canvas .012
Good ashlar masonry or brickwork .013
Vitrified clay .015
Rubble masonry .017
Firm gravel 0.020
Canals and rivers in good condition 0.025
Canals and rivers in bad condition . 035
8 General Formula for the Uniform Flow of Water in Rivers and Other Channels,
translated by R. Hering and J. C. Trautwine, Jr., Second edition, 1893, John Wiley
& Sons.
THE CHEZY COEFFICIENT
203
Another expression for the Chezy coefficient may be derived from
an analysis by Manning 4 resulting in the formula
for uniform open channel flow. Comparing the Manning and Chezy
formulas,
C = KR 1
in which K is dependent upon roughness and is given by 1.486/w.
Thus, Manning's proposal leads to
r  L486 /?*
C JK.
n
which, because of simplicity and satisfactory accuracy, is increasingly
preferred over the more cumbersome Kutter formula.
3.50
0.75
0.50
.3 .4 .5
1 2345
Hydraulic Radius in Feet
FIG. 112.
20 30 40 50
100
The results of the Kutter and Manning formulas may be easily
compared and presented for practical use by plotting the product
(C X n) against the hydraulic radius, R (Fig. 112). On such a plot,
4 Trans. Inst. Civil Engineers of Ireland, Vol. 20, 1890, p. 161.
204 FLUID FLOW IN OPEN CHANNELS
the Manning formula appears as a single line and the Kutter formula
as a series of curves whose location depends upon slope and roughness.
Another advantage of the Manning formula is that its simplicity
allows it to be inserted into the Chezy equation,
n  L486 x
Q x
giving
Q = AR S (72)
n
and allowing calculations to be carried out without the necessity for
tables or diagrams of C. The equation also shows clearly that an
error in the selection of the roughness coefficient, n, causes an error
of the same magnitude in the calculated quantity of flow, Q. It is
evident from Table IX that experience is essential to the selection of n
and the accurate prediction of rates of flow.
ILLUSTRATIVE PROBLEM
A rectangular channel lined with rubble masonry is 20 ft wide and laid on a
slope of 0.0001. Calculate the depth of uniform flow in this channel when 400 cfs
flow therein
n = 0.017, A = 2Qd, R
20 + 2d
Using the Manning equation
0.017
whence
<20
Solving by trial, d = 8.34 ft
48. Best Hydraulic Cross Section. It is obvious from
Q CAVRS
that for a given area of flow cross section, 4, and for a given slope, 5,
the rate of flow, Q, through a channel of given roughness will be maxi
mum when the hydraulic radius, R, is maximum. It becomes impor
tant, therefore, for best engineering design to proportion the dimensions
BEST HYDRAULIC CROSS SECTION
205
of the channel cross section to give an hydraulic radius which is as
large as is practically possible. From the definition of hydraulic radius
it is obvious that a cross section of maximum hydraulic radius is a
cross section of minimum wetted perimeter, p. Minimum wetted
perimeter means a minimum of lining material, grading, and general
construction work. Hence, a channel cross section having maximum
hydraulic radius not only results in the _.
best hydraulic design but tends toward
a section of minimum cost as well. It,
therefore, becomes of practical interest to
investigate certain channel shapes to dis
cover how they may be proportioned in
order to have maximum hydraulic radii.
For a rectangular cross section of
width b and depth d, having a fixed crosssectional area A, Fig. 113,
the dimensions for the conditions of maximum hydraulic radius may
be calculated by writing a general expression for the hydraulic radius
and obtaining its maximum by differentiating and equating to zero.
From Fig. 113,
FIG. 113.
and
giving
R
b 
R =
b + 2d
A
d
J
Ad
A +2d 2
Differentiating in respect to d and equating to zero
(A + 2d 2 )A  Ad(0
dR
dd
gives
But since A is also given by
(A + 2d 2 ) 2
A = 2d 2
A =bd
206 FLUID FLOW IN OPEN CHANNELS
comparison of these expressions shows that
b = 2d
or, in other words, the breadth of a rectangular channel must be twice
the depth for a condition of maximum hydraulic radius. The magni
tude of the hydraulic radius when maximum may be calculated by
substituting the above expressions in the general equation for hydraulic
radius as follows:
A 2d 2 d
R
2d 2d + 2d 2
Thus, when the hydraulic radius is maximum for a rectangular channel
it is equal to onehalf of the depth of flow.
For the trapezoidal 5 section of Fig. 114, the crosssectional area
is given by
b =  d cot
'//////////////////////////////// d
~t3fe
d co toe
FlG 114 and the wetted perimeter by
p = b + 2d esc a
Substituting the above expression for b
A
p =  d cot a + 2d esc a
d
Thus the hydraulic radius may be stated
p A
d cot a + 2d esc a
d
Differentiating in respect to d and equating to zero, the conditions for
maximum hydraulic radius are obtained, given by
A = d 2 (2 esc a cot a)
6 The angle of side slope, a, is generally limited in an earth canal by the angle
of repose of the soil. If the canal is lined a may have any value.
VARIATION OF VELOCITY AND RATE OF FLOW 207
Substituting this in the general expression for R,
as was obtained in the rectangular channel for conditions of maximum
hydraulic radius.
ILLUSTRATIVE PROBLEM
What are the best dimensions for a rectangular channel which is to carry
400 cfs if the channel is lined with rubble masonry and is laid on a slope of 0.0001 ?
d
n = 0.017, A = 2d*, R = 
0.017
whence
rfi = 364, d = 9.15 ft
I = 2d = 18.30 ft
400 _ 115$ (H) (W (0 . 0001) i
0.017 \b + 2d/
d bd
2 b + 2d
and solving these equations simultaneously
b = 18.30ft d = 9.15 ft
49. Variation of Velocity and Rate of Flow with Depth in Closed
Conduits. Frequently in civil engineering practice, particularly in
sewer problems, closed conduits that do not flow full are used to convey
liquids. Conduits of this type, in which flow does not occur under
pressure, satisfy the definition of open channels and must be con
sidered as such. Openchannel flow in closed conduits possesses
certain special features because of the convergence of the side walls
at the top of the conduit. For sewer design it is essential to have
available the relations of velocity and rate of flow to depth for con
duits of various shapes ; the depth at which the maximum quantity of
flow will occur must be known for capacity calculations, and the
variation of velocity with depth must also be understood since certain
velocities must be maintained in order to transport suspended solid
matter.
208
FLUID FLOW IN OPEN CHANNELS
A simple analysis of these features may be carried out by using the
Manning form of the Chezy equation for rate of flow and velocity
~ 1.486
V =
n
1.486
n
For a conduit of a certain roughness, n, laid on a given slope, *S, it is
obvious that
Q cc AR 1
V oc R l
and that A and R are both functions of the depth d. Hence Q and V
may be plotted against the depth by assuming various values of d, and
calculating the corresponding velocities and rates of flow. By dividing
these velocities and flow rates by those obtained when the conduit
flows full a more generally useful diagram is obtained in which the
velocities and rates of flow are both unity when the conduit flows full.
i.U
.9
.8
.7
.6
<*5
D
.4
.3
.2
.1
C
\
/
/
\
x
x*
j
Q
^
x
/
Qn
y
s
/
^
X
/
/
^
/v
/
s
s
v fun
/
^
^
'
***~
****
^^
) .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0 1.1 l.i
FIG. 115.
A diagram of this type is shown for a circular conduit in Fig. 115;
it gives, for any depth, the ratio of rate of flow and velocity at that
depth to those when the conduit flows full; with rate of flow and
velocity for the full conduit easily obtainable from the Chezy equation,
rates of flow and velocities for a partially full conduit are readily found
with the aid of the diagram.
SPECIFIC ENERGY
209
The depths at which maximum flow and maximum velocity occur
may be obtained directly from the diagram; they are seen to be
0.94D and 0.80J9, respectively, for the conduit of circular cross section.
These depths may be obtained more accurately by stating AR* and
R* in terms of d, differentiating the expressions in respect to d, equat
ing the results to zero, and solving the resulting equations for d.
50. Specific Energy. Many modern problems in openchannel flow
are solved by means of energy calculations. The "specific energy' 1
and its diagram, which were introduced by Bakhmeteff in 1911, have
proved fruitful in the explanation of new and old problems of open
channel flow. Today a knowledge of specific energy fundamentals is
absolutely necessary in coping with the advanced problems of open
FIG. 116.
flow ; these fundamentals and a few of their applications are developed
in the succeeding paragraphs.
Consider the uniform flow shown in Fig. 116. Here, for a given
slope, roughness, shape, and rate of flow, the depth may be calculated
from the Chezy equation. Assuming a uniform velocity distribution,
the Bernoulli equation may be written for a typical streamtube as
Vl
w
2g w Ig *
which indicates that energy is lost as flow occurs. However, if the
sloping channel bottom is taken as a datum plane the energy becomes
b V 2
77 *i ' I
& y ~r T" 
210
FLUID FLOW IN OPEN CHANNELS
which is the same at sections 1 and 2. The term E is called the
" specific energy"; from the figure,
y +  =d
w
and the definition of specific energy may be simplified to
_ , F 2
= d ~T~ 1
(73)
showing that specific energy in openchannel flow is simply the sum of
the depth and the velocity head in the channel.
ILLUSTRATIVE PROBLEM
In a trapezoidal channel having a bottom width of 10 ft, 1000 cfs flow at a
depth of 6 ft. If the side slopes of the channel
are 1 on 3, calculate the specific energy.
A = 10 X 6 + 6 X 18 = 168 ft 2
V = ^2? . 5.95 ft/sec
168
*
= 6
6.55 ft
Jfl
v
?"
s/ssss///////////////////////
L X aJ
^^
In uniform openchannel flow Bernoulli energy is lost as flow takes
place but specific energy remains constant. In varied flow Bernoulli
energy is continually lost, but it will
be seen later that specific energy may
be lost or gained as varied flow occurs.
It is convenient at this point to
deal with flow in a channel of rectangu
lar cross section, whose simple geo
metrical form will allow the use of
simple equations to illustrate the fun
damentals. The principles involved may
be applied to channels of other shape,
but the resulting equations are larger and more difficult to handle.
If a uniform velocity distribution is assumed in the rectangular
channel of Fig. 117, the unit rate of flow, q, through a vertical strip of
1ft width will be given by
2= Vd
FIG. 117.
SPECIFIC ENERGY
211
which will eliminate the width of channel, b, from subsequent equa
tions. Obviously q is related to the total rate of flow, Q, by
Q = bq
From the foregoing equation for g, V = q/d, which may be substi
tuted in the equation for specific energy giving
a more useful expression for specific energy, and relating this quantity
with depth and rate of flow. A thorough understanding of this equa
tion and its physical meaning may be obtained most easily by:
(1) assuming q constant and studying the relation of E and d, and
(2) taking E constant and examining the relation of q and d.
In the flow of a certain quantity of liquid in open channels of
various slopes, it is evident that steep slopes will tend toward high
velocities and small depths, and mild slopes toward low velocities and
large depths. The slopes thus determine the depths, but the depths
in turn determine the specific energy since q is constant in
Plotting of this equation gives the "specific energy diagram'* of
Fig. 118 and introduces some new concepts to openchannel flow.
Region of tranquil flow
and mild slopes
* Specific Energy, E *
FIG. 118. The Specific Energy Diagram.
The specificenergy curve possesses a point of minimum energy at
which the depth is termed the "critical depth." For every specific
energy, E, there are two "alternate" depths at which flow may take
212
FLUID FLOW IN OPEN CHANNELS
place, one greater and one less than the critical depth. If the depth of
flow is greater than the critical the flow is said to be "tranquil" and
the slopes which bring about such flows are designated as "mild"
slopes. The flow is said to be " rapid" if its depth is less than the
critical depth ; rapid flows are caused by "steep" slopes. Uniform flow
at the critical depth will occur when the channel has a "critical slope,"
s c .
Certain general characteristics of open channel flow may be
deduced from the specificenergy curve. In the region of flow near
the critical depth, the depth may change for practically constant
specific energy. Physically this means that, since many depths may
occur for approximately the same specific energy content, flow near
the critical depth will possess a certain instability (which frequently
manifests itself by undulations in the liquid surface). It is also evident
from the curve that a loss of specific energy will be accompanied by
a reduction in depth in tranquil flow, but in rapid flow an increase of
depth is associated with a loss of specific energy.
The relation between depth and rate of flow for constant specific
energy may be obtained by solving equation 74 for q 1 resulting in
q = V2gd 2 E  2gd* (75)
From this equation may be plotted the "gcurve" 6 of Fig. 119. The
physical meaning of such a curve may be determined by assuming a
Since V
Bm h Constant
FIG. 119. ThegCurve.
reservoir of a constant surface elevation which discharges into an
open channel with movable gates at each end. If the velocity in the
6 Originally suggested by Koch. For a summary of his work see A. Koch and
M. Carstanjen, Bewegung des Wassers, 1926.
CRITICALDEPTH RELATIONSHIPS 213
reservoir is neglected the specific energy here will be constant and
equal to the height h throughout manipulation of the gates. If it is
assumed that no energy is lost as flow passes the first gate the fluid
flowing in the channel will possess the constant specific energy h.
By adjusting the gates the depth of liquid in the channel may be varied
between zero and h without changing the specific energy, and during
this variation the relation between rate of flow and depth is given by
the "gcurve" point A occurs when gate 1 is completely closed,
point B when gate 1 is open but gate 2 completely closed, and point C
when both gates are removed and unobstructed flow takes place.
The last condition brings about a maximum rate of flow which occurs
at the critical depth since the point of maximum flow for a given
specific energy content is the same as that of minimum energy for a
given rate of flow. Hence the regions of rapid and tranquil flow may be
indicated on the gcurve as shown.
51. CriticalDepth Relationships. The calculation of critical depth
is necessary to the identification of tranquil and rapid flow, and the
derivation of the equations of critical flow lends a further significance
to the critical depth.
Since critical depth occurs when specific energy is minimum for
a given rate of flow the equations of critical flow may be obtained by
differentiating equation 74 and equating the result to zero. Perform
ing this operation,
dE d
in which d has become d c after differentiation. From this equation
there results
<Z 2 = gd* (76)
which when substituted in the general specificenergy equation gives
E c = d c + 72 =  d c (77)
lgd c 2
Other important criticalflow equations may be readily derived from
equation 76, and these are
q = Vgd* ' (78)
or
0/3
(79)
214 FLUID FLOW IN OPEN CHANNELS
or since Tr ,
q = V c d c
V c = V~& c (80)
Since these equations may also be obtained by the differentiation of
equation 75, the maximum point of the gcurve must correspond to
the minimum point of the specificenergy curve, indicating that the
former point occurs at the critical depth as stated above.
The foregoing equations demonstrate and suggest certain points
concerning critical flow. Equation 79 shows that critical depth
depends only upon the rate of flow in the channel. It is a characteristic
parameter of the flow which may be determined as soon as q is known.
Equation 78 suggests utilizing critical flow as a means of metering
openchannel flow. If critical flow may be caused to exist in a channel
its depth may be measured and the rate of flow calculated.
Equation 80 is the same as another equation of fluid mechanics
which gives the velocity of propagation of a small wave on the surface
of a body of liquid of depth d. The similarity of these equations offers
a rough means of identifying rapid, tranquil, and critical flows in the
field. Since critical flow takes place at a velocity exactly equal to the
velocity of propagation of a small surface wave, such a wave which is
created on the surface of a critical flow cannot progress upstream but
will remain stationary because of the equality of velocities. In tran
quil and rapid flows, velocities are respectively less than and more than
Vgd; thus in tranquil flow small surface waves will progress upstream,
but in rapid flow such waves will be swept downstream. 7
ILLUSTRATIVE PROBLEM
When 400 cfs flow at a depth of 4 ft in a rectangular channel 20 ft wide, is the
flow tranquil or rapid?
~ 3/CW) 2
= xr^ = 2.32 ft
\ 32.2
Since 4 > 2.32 ft, flow is tranquil.
52. Occurrence of Critical Depth. The critical depth will occur
in openchannel flow when a change in channel slope brings about
7 The similarity between wave velocity phenomena in open flow and acoustic
velocity phenomena in gas flow (Art. 22) should be noted.
OCCURRENCE OF CRITICAL DEPTH
215
a change from tranquil to rapid flow or from rapid to tranquil flow.
Two occurrences of critical depth are shown in Fig. 120, where a change
from a mild slope to a steep slope causes the flow to pass through the
critical depth in its smooth transition from tranquil to rapid flow.
An abrupt change from rapid to tranquil flow, the "hydraulic jump;"
Hydraulic
FlG. 120.
occurs when the slope of the channel is changed again to a mild slope ;
here again the flow passes through the critical depth.
At free outfall from a channel of mild slope (Fig. 121), the critical
depth will occur a short distance upstream from the brink. The reason
for this is apparent from the specificenergy diagram (Fig. 118). With
no obstruction to maintain the depth, the
depth may fall from the normal depth of
tranquil flow (existing at a great distance
up the channel) to the critical depth but
may fall no more since further reduction in
depth necessitates a gain in specific energy
and there is no energy supply from which
an energy increase may be drawn. The
fact that the brink depth, d , is less than the critical is no violation
of this principle; here the flow curvatures induce accelerations which
upset the simple laws of specific energy. Rouse 8 has found that for
very mild slopes the brink depth is a fixed proportion of the critical
depth given by
~ = 0.715 or d c = lAd
d c
FIG. 121. Free Overfall.
8 H. Rouse, "Discharge Characteristics of the Free Overfall," Civil Engineering^
Vol. 6, No. 4, p. 257, April, 1936.
216
FLUID FLOW IN OPEN CHANNELS
He proposes using the brink depth as a means of calculating the rate
of flow with the aid of equation 78,
Substituting the above relation in this expression
q  Vg(1.4O 3 = 1.66V7**.*
Q = l.66bV^d '
or
(81)
thus allowing rate of flow to be calculated from measurement of the
brink depth.
Flow at critical depth occurs on the crest of a broadcrested weir
(Fig. 122), owing to circumstances similar to those of the free outfall.
Here, as before,
but, friction losses and velocity in the
reservoir above the weir being neglected,
the specific energy, E c , of the critical flow,
is given by
E c = H
Y/SSSSS///S//////////S///S//S///*
FJG. 122. BroadCrested Weir. But from equation 77
d c =\
and therefore
Substituting above this expression for d c
q =
which may be written for comparison with later weir equations
Q = 0.578 X l
(82)
Neglecting friction loss causes this equation to be somewhat in error,
and tests on broadcrested weirs indicate the coefficient to be from
0.50 to 0.57 (depending upon the weir shape) instead of the above
derived 0.578.
Critical depth is obtained for purposes of flow measurement by
contracting the sides of a channel to form a "venturi flume" or by
VARIED FLOW 217
raising the channel bottom sufficiently (Fig. 123). The latter problem
is essentially one of a submerged broadcrested weir but it presents
a means of measuring openchannel flow without the necessity of an
inconvenient and costly drop in the channel. Difficulties involved in
this method of flow measurement
are due to the critical depth loca
tion changing as the flow picture
changes with varying rates of
flow.
As flow passes through the p IG
critical depth in its transition
from tranquil to rapid flow in the foregoing examples, a " control" is
formed. At a " control" in an open channel the depth of flow is de
termined by the unique relation between depth and rate of flow
(equation 79), and depths upstream from this point are " controlled"
by the critical depth since occurrences downstream from this point
cannot alter the flow above the control. Thus ''controls" in an open
channel are points where the depths may be calculated directly; depths
above and below such controls may be established by the principles
of varied flow.
53. Varied Flow. Calculations on varied flow are based on the
assumption that the loss of energy in a short reach of channel is the
same for varied flow as for a uniform flow having a velocity and
hydraulic radius equal to the numerical averages of the velocities and
hydraulic radii of the sections at the ends of the reach. This assump
tion has never been confirmed by experiment, but errors arising from it
are likely to be small compared to those incurred in the selection of
roughness coefficients. The assumption is undoubtedly more true of
varied flow in which the velocity increases than that in which the
velocity decreases since in the former energy loss is caused by fric
tional effects only whereas in the latter the losses due to eddying
turbulence are added to those of friction.
Considering the varied flow of Fig. 124, in which the changes of
depth and velocity are assumed small,
i
Obviously
i = SJ
and
H L = SI
218
FLUID FLOW IN OPEN CHANNELS
in which S is the slope of the " energy line/' which may be assumed
to be a straight line over short reaches of channel, /. Substituting
these values in the first equation
S l
v\
and solving for /
5  S
(83)
FIG. 124. Varied Flow.
Using now the basic assumption mentioned above, 5 may be calculated
from the Chezy equation for a uniform flow of average velocity and
hydraulic radius. Using the Manning coefficient
whence
in which
and
v 
V av.
s =
n
1.486tf av f
'1 + V 2
2
THE HYDRAULIC JUMP
219
The general method of making variedflow calculations is to start
at a point in the channel where depth and velocity (say d 2 and F 2 ) are
known, assume a depth, d\, slightly different from </ 2 > calculate FI
and S, and solve equation 83 for the length of reach, /. A fairly accu
rate profile of the liquid surface in the channel may be obtained by
this method if the difference between the known and assumed depths
is taken small.
ILLUSTRATIVE PROBLEM
Four hundred cubic feet per second flow in a rectangular channel 20 ft wide,
lined with rubble masonry and laid on a
slope of 0.0001. The channel ends in a free
outfall, and at a point in the channel the
depth is 6.00 ft. How far upstream from this
point will the depth be 6.30 ft?
400
6.30 X 20
400
6.00 X 20
3. 18 ft/sec
3.33 ft/sec
El =
= 6.46 ft
= 6.00
6.17 ft
Fay. =
3.255 ft/sec
3.77 + 3.75
R &v . = * 3.81 ft
/ =
1.486 X (3.81)'
6.46  6.17
0.000233  0.0001
 0.000233
2180ft
54. The Hydraulic Jump. When a transition of flow occurs from
the rapid state to the tranquil state a variedflow phenomenon results
known as the "hydraulic jump" in which the elevation of the liquid
surface increases rather abruptly in the direction of flow. Until the
last few decades the hydraulic jump was considered by many to be a
mysterious and complex phenomenon, but at the present time a com
220
FLUID FLOW IN OPEN CHANNELS
plete theoretical solution of the problem involves only a simple applica
tion of the laws of fluid statics, impulsemomentum, and specific
energy; the results obtained from such a theoretical analysis exhibit
close conformity with the results of experimental observations.
An hydraulic jump in a channel of small slope is shown in Fig. 125.
It is characterized by an increase of depth, 9 a surface roller, eddying
turbulence, and an undulating liquid surface downstream from the
jump. Since the jump can occur only as a change from rapid to tran
quil flow the depth in changing must pass through the critical depth.
In engineering practice the hydraulic jump frequently occurs below
a spillway or sluice gate where velocities are high. From Fig. 125
it is obvious that in passing the jump the fluid loses velocity, and the
FIG. 125. Hydraulic Jump
jump is often used in this capacity as a dissipator of the kinetic energy
of an open flow in order to prevent scour of channel lining.
Between sections 1 and 2 of Fig. 125 there occurs a decrease of
velocity from Vi to V 2 and consequent decrease in the momentum
of the flowing liquid which, according to the impulsemomentum law,
must be caused by forces continually acting on the liquid mass between
these sections. Since the decrease in velocity is in a horizontal direc
tion, the imposed forces are horizontal also and arise (1) from pressure
on the ends of the liquid mass and (2) from frictional effects at the
sides and bottom of the mass. The pressure forces being designated
as FI and /^ , and the friction force as F T , the impulsemomentum law
gives
Fi + FrFi  (Fi V 2 )
9 The jump is not as steep as shown in the figure; the length of jump, /, is usually
4 to 5 times its height, dj.
THE HYDRAULIC JUMP 221
in which F T , resulting from frictional effects over the short distance, /,
is safely negligible. The equation, therefore, reduces to
J7 1? 5 / T/ T7 \
r% "i \ v i v 2)
g
By the principles of fluid statics (Art. 11),
FI = A l wh gl
F 2 = A 2 wh g2
Substituting these relationships and FI = Q/Ai and F 2 = Q/A 2 in
the above equation,
62 ' ~ * ! (84)
which is the basic general equation for the hydraulic jump in channels
of any shape.
Assuming, now, a rectangular channel in order to demonstrate
principles and methods with a minimum of mathematical complexity,
the above equation reduces through substitution of the following
relationships
A i = bdi A 2 = bd 2
hgi == ~^~ ^st ~ ~^~
to
2 ~ 2 " g \d r ~~ d 2
which may be written
 + Y = y + y (85)
which allows the vertical dimensions of the hydraulic jump to be
obtained since if the rate of flow and one depth are known, the other
depth may be calculated.
222
FLUID FLOW IN OPEN CHANNELS
ILLUSTRATIVE PROBLEM
If 400 cfs flow in a rectangular channel 20 ft wide at a depth of 1.00 ft, what
depth will exist after an hydraulic jump has occurred from this flow?
400 ft 3 /sec
q = 20 
* 20 ft
;[
32.2 X (1)
=450ft
Although special expressions for <L\ and d 2 may be obtained 10 from
equation 85, it proves more fruitful to solve the equation by graphical
methods. Taking
it is obvious (1) that, for a given rate of flow, M is a function of d only,
and (2) that solution of the equation is given by
MI = M 2
FIG. 126. Hydraulic Jump and Jkfcurve.
The curve obtained (Fig. 126) from plotting M against d possesses
a minimum value of M at the critical depth and is similar in shape to
the specificenergy diagram. After construction of this curve and with
one depth known, the corresponding, or " conjugate," depth may be
10 Solution of equation 85 gives
* *[ i
* r ^i.
or
THE HYDRAULIC JUMP
223
found by passing a vertical line through the point of known depth.
Since a vertical line is a line of constant M, the intersection of this
line and the other portion of the curve gives a point where MI is equal
FIG. 127.
to M 2 , and allows the conjugate depth and height of jump, dj, to be
taken directly from the plot.
By plotting the Mcurve and specificenergy curve on the same
s>s c
FIG. 128.
diagram, the loss of energy in the jump may be obtained graphically
by the methods of Fig. 127, and from the lost energy the horsepower
consumed in the jump may be calculated by the usual methods.
A simple problem in locating a hydraulic jump is shown in Fig. 128.
224 FLUID FLOW IN OPEN CHANNELS
Here a channel of steep slope in which uniform flow is established dis
charges into a channel of mild slope of sufficient length to contain
uniform flow. For a certain rate of flow the depths d and d 2 may be
calculated from the Chezy equation. From d 2 > d\ may be obtained
from the M curve or by solution of equation 85. Flow in the reach of
channel of length / is of a gradually varied nature, and this length may
be obtained by solutions of equation 83; the larger the difference
between d\ and d, the greater the number of solutions necessary for
an accurate value of /.
BIBLIOGRAPHY
B. A. BAKHMETEFF, Hydraulics of Open Channels, McGrawHill Book Co., 1932.
M. P. O'BRIEN and G. H. HICKOX, Applied Fluid Mechanics, McGrawHill Book Co.,
1937.
B. A. BAKHMETEFF and A. E. MATZKE, "The Hydraulic Jump in Terms of Dynamic
Similarity," Trans. A.S.C.E., Vol. 101, 1936, p. 630.
H. ROUSE, Fluid Mechanics for Hydraulic Engineers, McGrawHill Book Co., 1938.
PROBLEMS
342. Water flows uniformly at a depth of 4 ft in a rectangular flume 10 ft wide,
laid on a slope of 1 ft per 1000 ft. What is the average shear stress at the sides and
bottom of the flume?
343. Calculate the average shear stress over the wetted perimeter of a circular
sewer 10 ft in diameter in which the depth of flow is 3 ft and whose slope is 0.0001.
344. What is the average shear stress over the wetted perimeter of a triangular
flume 8 ft deep and 10 ft wide at the top, when the depth of water flow is 6 ft? The
slope of the flume is 1 in 200.
345. Calculate the friction factor /equivalent to a Chezy coefficient of 120.
346. Calculate the Chezy coefficient which corresponds to a friction factor
/ of 0.030.
347. For an open channel of hydraulic radius 40 ft, value of n 0.017, and slope
0.0001, check the values of (C X n) given in Fig. 112.
348. For an open channel of hydraulic radius 0.8 ft, value of n 0.017, and slope
0.001, check the values of (C X n) given in Fig. 112.
349. What uniform flow will occur in a rectangular planed timber flume 5 ft wide
and having a slope of 0.001 when the depth therein is 3 ft: Using: (a) Manning C\
(6) Kutter C.
350. Calculate the uniform flow in an earthlined (n 0.020) trapezoidal canal
having bottom width 10 ft, sides sloping 1 on 2, laid on a slope of 0.0001, and having
a depth of 6 ft, using: (a) Manning C\ (b) Kutter C.
351. What uniform flow exists in a circular brick conduit 5 ft in diameter when
the depth of flow is 2 ft, if it is laid on a slope of 0.0005, using: (a) Manning C;
(b) Kutter C.
352. What uniform flow occurs in a river of flow cross section 10,000 ft, wetted
perimeter 550 ft, if its slope is 1 in 5000, using: (a) Manning C\ (b) Kutter C?
n is 0.025.
PROBLEMS 225
353. A rough timber flume in the form of an equilateral triangle (apex dowti) of
4ft sides is laid on a slope of 0.01. Calculate the uniform flow which occurs at a
depth of 3 ft, using: (a) Manning C\ (b) Kutter C.
354. What uniform flow will occur in this cross section, if it is laid on a slope of 1
in 2000 and has n  0.017, using (a): Manning C, (b) Kutter C?
loot
i on i
355. What uniform flow will occur in this open channel if it is laid on a slope of
0.0002 and has a value of n of 0.015, using: (a) Manning C; (b) Kutter C?
356. A flume of planed timber has its cross section an isosceles triangle (apex
down) of 8ft base and 6ft altitude. At what depth will 180 cfs flow uniformly in
this flume if it is laid on a slope of 0.01? Use ChezyManning equation 72.
357. At what depth will 150 cfs flow uniformly in a rectangular channel 12 ft wide
lined with rubble masonry and^laid on a slope of 1 in 4000? ChezyManning equa
tion 72.
358. At what depth will 400 cfs flow uniformly in an earthlined (n = 0.025)
trapezoidal canal of base width 15 ft having side slopes 1 on 3, if the canal is laid
on a slope of 1 in 10,000? ChezyManning equation 72.
359. Calculate the depth at which 25 cfs will flow* uniformly in a smooth cement
lined circular conduit 6 ft in diameter, laid on a slope of 1 in 7000. ChezyManning
equation 72.
360. An earthlined (n = 0.020) trapezoidal canal of base width 10 ft and side
slopes 1 on 3 is to carry 100 cfs uniformly at a maximum velocity of 2 ft/sec. What
is the maximum slope that it may have?
361. What slope is necessary to carry 400 cfs uniformly at a depth of 5 ft in a
rectangular channel 12 ft wide, having n = 0.017?
362. A trapezoidal canal of side slopes 1 on 2 and having n = 0.017 is to carry
1300 cfs on a slope of 0.005 at a depth of 5 ft. What base width is required?
363. Rectangular channels of flow cross section 50 ft 2 have dimensions (width
X depth) of (a) 25 ft by 2 ft; (b) 12.5 ft by 4 ft; (c) 10 ft by 5 ft; (d) 5 ft by 10 ft.
Calculate the hydraulic radii of these sections.
364. What are the best dimensions for a rectangular channel having a flow cross
section of 100 ft 2 ?
365. What are the best dimensions of a trapezoidal channel having a flow cross
section of 150 ft 2 and sides sloping at 30?
366. What are the best dimensions for a rectangular rough timber channel to
carry 120 cfs on a slope of 1 in 8000?
367. What are the best dimensions for a trapezoidal canal having side slopes
1 on 3 if it is to carry 1400 cfs on a slope of 0.009? (n = 0.020.)
368. What is the minimum slope at which 200 cfs may be carried uniformly in a
rectangular channel (having a value of n of 0.014) at a velocity of 3 ft/sec?
369. What is the minimum slope at which 1000 cfs may be carried at a velocity
of 2 ft /sec in a trapezoidal canal having n = 0,025 and sides sloping 1 on 4?
226 FLUID FLOW IN OPEN CHANNELS
370. Prove that the best form for a Vshaped openchannel section is one of
vertex angle 90
^ 371. What rate of uniform flow occurs at a depth of 3 in. in a vitrified clay sewer
line of 12in. diameter laid on a slope of 0.003? What is the velocity of this flow?
372. What rate of uniform flow occurs in a 5ft circular brick conduit laid on a
slope of 0.001 when the depth of flow is 3.5 ft? What is the velocity of this flow?
373. At what depth will 800 cfs flow in a circular ashlar masonry conduit 10 ft
in diameter, laid on a slope of 0.006? What is the velocity of flow?
374. Solve problem 359 using the plot of Fig. 115. What is the velocity of flow?
375. Plot curves similar to those of Fig. 115 for an isosceles triangle (apex
up) whose base is equal to its altitude. Find the maximum points of the curves
mathematically.
376. Plot curves similar to those of Fig. 115 for an equilateral triangle (apex
up). Find the maximum points of the curves mathematically.
377. Plot curves similar to those of Fig. 115 for a square laid with diagonal verti
cal. Find the maximum points of the curves mathematically.
378. Calculate the specific energy when 225 cfs flow in a 10ft rectangular channel
at depths of (a) 1.5ft; (b) 3ft; (c) 6ft.
379. Calculate the specific energy when 300 cfs flow at a depth of 4 ft in a trape
zoidal channel having base width 8 ft and sides sloping at 45.
380. What is the specific energy when 60 cfs flow at a depth of 5 ft in a circular
channel 6 ft in diameter?
381. Calculate the specific energy when 100 cfs flow at a depth of 3 ft in a triang
ular (apex down) flume, if the width at the water surface is 4 ft.
382. At what depths may 30 cfs flow in a rectangular channel 6 ft wide if the speci
fic energy is 4 ft?
383. At what depths may 800 cfs flow in a trapezoidal channel of base width
12 ft and sides slopes 1 on 3 if the specific energy is 7 ft?
384. Eight hundred cubic feet per second flow in a rectangular channel of 20 ft
width having n = 0.017. Plot accurately the specificenergy diagram for depths
from to 10 ft, using the same scales for d and E. Determine from the diagram:
(a) the critical depth; (b) the minimum specific energy; (c) the specific energy
when the depth of flow is 7 ft ; (d) the depths when the specific energy is 8 ft. What
type of flow exists when the depth is (e) 2 ft, (/) 6 ft, and what are the channel slopes
necessary to maintain these depths? What type of slopes are these, and (g) what is
the critical slope?
385. Flow occurs in a rectangular channel of 20ft width and has a specific energy
of 10 ft. Plot accurately the gcurve. Determine from the curve: (a) the critical
depth; (b) the maximum rate of flow; (c) the flow at a depth of 8 ft ; (d) the depths
at which a flow of 1000 cfs may exist and the flow condition at these depths.
386. Five hundred cubic feet per second flow in a rectangular channel 15 ft wide
at a depth of 4 ft. Is the flow rapid or tranquil?
387. If 300 cfs flow in a rectangular channel 12 ft wide having n = 0.015 and
laid on a slope of 0.005, is the flow tranquil or rapid?
388. If 400 cfs flow in a rectangular channel 18 ft wide with a velocity of 5 ft/sec,
is the flow tranquil or rapid?
389. Calculate and check the critical depths of problems 384 and 385.
390. What is the maximum flow which may occur in a rectangular channel 8 ft
wide for a specific energy of 5 ft?
PROBLEMS 227
391. An open rectangular channel 5 ft wide and laid on a mild slope ends in a free
outfall. If the brink depth is measured as 0.865 ft, what flow exists in the channel?
392. Calculate the critical depth for 50 cfs flowing in a rectangular channel 8 ft
wide. If this channel is laid on a mild slope and ends in a free outfall, what is the
depth at the brink?
393. What theoretical flow will occur over a broadcrested weir 30 ft long when
the head thereon is 2 ft?
394. The elevation of the crest of a broadcrested weir is 100.00 ft. If the length
of this weir is 12 ft and the flow over it 200 cfs, what is the elevation of the water
surface upstream from the weir?
395. The critical depth is maintained at a point in a rectangular channel 6 ft wide
by building a gentle hump in the bottom of the channel. When the depth over the
hump is 2.20 ft, calculate the flow. Sketch the energy line and water surface showing
all possible vertical dimensions.
396. If 150 cfs flow in a rectangular channel 10 ft wide, laid on a slope of 0.0004
and having n 0.014, what is the minimum height of hump that may be built across
this channel to create critical depth over the hump? Sketch the energy line and water
surface, showing all vertical dimensions. Neglect energy losses.
397. A rectangular channel 12 ft wide is narrowed to 6ft width to cause critical
flow in the contracted section. If the depth in this section is 3 ft, calculate the flow
and the depth in the 12ft section, neglecting energy losses in the transition. Sketch
energy line and water surface, showing all pertinent vertical dimensions.
398. One hundred and fifty cubic feet per second flow in a rectangular channel
10 ft wide having n = 0.014, and laid on a slope of 0.0004. This channel is to be
narrowed to cause critical flow in the contracted section. What is the maximum
width of contracted section which will accomplish this? Neglect energy losses, and
sketch the energy line and water surface, showing vertical dimensions.
399. If 543 cfs flow in a rectangular channel 12 ft wide having n 0.017 laid
on a slope of 0.00228 and ending in a free outfall, calculate and plot the water
surface profile from the brink upstream to the region of uniform flow, taking incre
ments of depth of 0.2 ft. How far from the brink does a depth of 5.0 ft occur?
400. The channel of the preceding problem discharges into a channel of the same
width and roughness, but having a slope of 0.0293. Calculate and plot the water
surface profile downstream to the region of uniform flow, taking decrements of depth
of 0.2 ft. How far fVom the point of slope change does a depth of 3.0 ft exist?
401. In a rectangular channel 12 ft wide having n = 0.017 and laid on a slope of
0.0293, 543 cfs flow uniformly. This channel discharges into one of the same width
and roughness laid on a slope of 0.00228. Calculate and plot the watersurface profile
downstream until a depth of 3.00 ft is reached, using depth increments of 0.2 ft.
How far from the point of slope change does a depth of 2.6 ft exist?
402. Eight hundred cubic feet per second flow in a rectangular channel of 20ft
width. Plot the M curve of hydraulic jumps on the specificenergy diagram of
problem 384. From these curves determine: (a) the depth after a hydraulic jump
has taken place from a depth of 1.5 ft; (b) the height of this jump; (c) the specific
energy before the jump; (d) the specific energy after the jump; (e) the loss of
energy in the jump; (/) the total horsepower lost in the jump.
403. A flow of 200 cfs takes place in a rectangular channel 15 ft wide at a depth
of 0.5 ft. Calculate the depth after a hydraulic jump has taken place from this flow.
228 FLUID FLOW IN OPEN CHANNELS
Calculate FI, FZ, and the change of momentum per second, and check the relation
ship between these terms.
404. If 543 cfs flow in a rectangular channel 12 ft wide having n 0.017 and
laid on a slope of 0.00228, what depth of flow must exist in this channel for a jump
to occur resulting in uniform flow? How far downstream from the point of change of
slope of problem 401 will such a jump be located?
CHAPTER VIII
FLUID MEASUREMENTS
In engineering and industrial practice one of the fluid problems
most frequently encountered by the engineer is the measurement of
many of the fluid characteristics discussed in the foregoing chapters.
Efficient and accurate measurements are also absolutely essential for
correct conclusions in the various fields of fluid research. Whether the
necessity for precise measurements is economic or scientific, the
engineer of today must be well equipped with a knowledge of the
fundamentals and existing methods of measuring various fluid prop
erties and phenomena. It is the purpose of this chapter to indicate
only the basic principles of fluid measurements; the reader will find
available in the engineering literature l the details of installation and
operation of the various measuring devices. Although many of the
following devices frequently appear in engineering practice as appur
tenances in various designs where they are not used for measuring
purposes, a study of them as measuring devices will make obvious their
applications in other capacities.
55. Measurement of Fluid Properties. Of the fluid properties:
density, viscosity, elasticity, surface tension, and vapor pressure,
the engineer is usually called upon to measure only the first two.
Since measurements of elasticity, surface tension, and vapor pressure
are normally made by physicists and chemists, the various experi
mental techniques for measuring these properties will not be discussed
here.
The density of liquids may be determined by the following methods,
listed in approximate order of their accuracy: (1) weighing a known
volume of liquid; (2) hydrostatic weighing; (3) Westphal balance,
(4) hydrometer, and (5) Utube.
To weigh accurately a known volume of liquid a device called
a "pycnometer" is used. This is usually a glass vessel whose weight,
volume, and variation of volume with temperature have been accur
ately determined. If the weight of the empty pycnometer is W\, and
the weight of the pycnometer, when containing a volume V of liquid
1 See bibliography at end of chapter.
229
230
FLUID MEASUREMENTS
at temperature t is TF 2 , the specific weight of the liquid, w t , at this
temperature may be calculated directly from
w,
t
V
Density determination by hydrostatic weighing
consists essentially of weighing a plummet of known
volume (1) in air and (2) in the liquid whosadensity
is to be determined (Fig. 129). If the weight of
the plummet in air is W a , its volume, F, and its
weight when suspended in the liquid, Wi, the
equilibrium of vertical forces on the plummet gives
 W a =
FIG. 129.
and results in
W a 
V
allowing the specific weight, w^ at the temperature / to be calculated
directly.
f Knife edge
e
~TPHnr~
Jl JUl
^Riders
Plummet
>
Thermometer
";
FIG. 130. Westphal Balance.
Like the above method of hydrostatic weighing, the Westphal
balance (Fig. 130) utilizes the buoyant force on a plummet as a measure
of specific gravity. Balancing the scale beam with special riders placed
at special points allows direct and precise reading of specific gravity.
MEASUREMENT OF FLUID PROPERTIES
231
Probably the most common means of obtaining liquid densities is
with the hydrometer (Fig. 131), whose operation is governed by the
fact that a weighted tube will float with different immersions in liquids
of different densities. To create a great variation of immersion for
small density variation and, thus, to give a sensitive instrument,
changes in the immersion of the hydrometer occur along a slender tube,
FIG. 131. Hydrometer.
FIG. 132.
which is graduated to read the specific gravity of the liquid at the
point where the liquid surface intersects the tube.
The unknown density of a liquid 1 may be obtained approximately
from the known density of a liquid 2 (if the liquids are not miscible)
by placing them in an open Utube and measuring the lengths of
liquid columns, l\ and /2 (Fig. 132). From manometer principles,
and thus
This method is not a precise one because the various menisci prevent
accurate measurement of the lengths of the liquid columns.
Viscosity measurements are made with devices known as "viscosi
meters" or "viscometers," which may be classified as "rotational,"
"fallingsphere," or "tube" devices according to their construction
or operation. The operation of all these viscometers depends upon
the existence of laminar flow under certain controlled conditions.
232
FLUID MEASUREMENTS
In general, however, these conditions involve too many complexities
to allow the constants of the viscometer to be calculated theoretically ;
and they are, therefore, usually obtained by calibration with a liquid
of known viscosity. Because of the variation of viscosity with tem
perature all viscometers must be immersed in constanttemperature
baths and provided with thermometers for taking the temperatures
at which the viscosity measurements are made.
Two viscometers of the rotational type are the MacMichael and
Stormer viscometers, whose essentials are shown diagrammatically in
Fig. 133. Both consist of concentric cylinders, the space between
which contains the liquid whose viscosity is to be measured. In the
Rotating
V
Fixed
MacMichael Stormer
FIG. 133. Rotational Viscometers (Schematic).
MacMichael type, the outer cylinder is rotated at constant speed and
the rotational deflection of the inner cylinder (accomplished against
a spring) becomes a measure of the liquid viscosity. In the Stormer
instrument, the inner cylinder is rotated by a fallingweight mechan
ism, and the time necessary for a fixed number of revolutions becomes
a measure of the liquid viscosity.
The measurement of viscosity by the above variables may be
justified by a simplified mechanical analysis, using the dimensions of
Fig. 133. Assuming Ar small, and the peripheral velocity of the
moving cylinder to be V, the torque, !T, is given (neglecting shear
stress on the bottom of the cylinder) by
T = (r2irrh)r = M ~ ^r 2 h
MEASUREMENT OF FLUID PROPERTIES 233
in which
dv = V and dr = Ar
therefore
T = /A 27rr 2 A
Ar
But if 2V = revolutionary speed in revolutions per second, V = 2wrN,
thus
or
in which J is a constant whose magnitude depends on the size, pro
portions, and depth of filling of the viscometer. This equation may
be written
T
it = 
M KN
and since the torsional deflection, 0, of the spring is proportional to
the torque, the equation becomes for the MacMichael viscometer
6
MXi
showing that liquid viscosity, M, may be measured by the torsional
deflection 6 obtained at a speed N. In the Stormer viscometer the
torque, T, is maintained constant by the weight mechanism, and thus
KN
If / is the time necessary for a fixed number of revolutions, obviously
and this time becomes a measure of liquid viscosity.
The fallingsphere type of viscometer is shown in Fig. 134. In this
type of viscometer the time t required for a small sphere to fall at
constant velocity through a distance I in a liquid becomes a direct
measure of the liquid's viscosity. Here again an approximate analysis
234
FLUID MEASUREMENTS
justifies the above. From Stokes' law (Art. 77) the drag D of a sphere
of diameter d, moving under laminar conditions at a velocity V,
through a fluid of infinite extent, is given by
D =
If the proximity of the boundary walls is neglected, this equation may
be applied for an approximate analysis of the viscometer. This drag
force acts upward on the sphere and
acting in the same direction is the
buoyant force, FB, given by
r
#
Bath
FB ~: d wi
6
w Acting downward on the sphere is
its own weight, W, given by
W = d*w.
6
D For constantvelocity motion the
net force on the moving sphere is
zero, giving
D  W + F B =
FIG. 134. FallingSphere Viscometer.
Solving for
K(w 8
or
K(w 8 w{)t
Thus viscosity in this instrument is measured by time of fall but density
of sphere and liquid must be known before viscosity can be calculated.
Two typical tubetype viscometers are the Ostwald and Saybolt
instruments of Fig. 135. Similar to the Ostwald is the Bingham type,
and similar to the Saybolt are the Redwood and Engler viscometers.
All these instruments involve the unsteady laminar flow of a fixed
quantity of liquid through a small tube under fixed head conditions.
The time for the quantity of liquid to pass through the tube becomes
a measure of the kinematic viscosity of the liquid.
MEASUREMENT OF FLUID PROPERTIES
235
The Ostwald viscometer is filled to level A, and the meniscus of
the liquid in the righthand tube is then drawn up to a point above B.
The time for the meniscus to fall from B to C becomes a measure of the
kinematic viscosity. In the Saybolt viscometer the outlet is plugged
and the reservoir filled to level A, the plug is then removed and the
time required to collect a fixed
quantity of liquid in the vessel
B is measured. This time then
becomes a direct measure of the
kinematic viscosity of the liquid.
The relation between time
and kinematic viscosity for the
tube type of viscometer may be
indicated approximately by ap
plying the HagenPoiseuille law
for laminar flow in a circular tube
(Art. 34). The approximation
involves the application of a law
of steady established laminar
motion to a condition of un
steady flow in a tube which is
too short for established laminar
flow to exist and therefore can
not give a complete or perfect
relationship betweenj efflux time
and kinematic viscosity; it will
serve, however, to indicate ele
mentary principles. From equa
tion 46
Saybolt
ostwaid
128ju/
FIG. 135. Tube Viscometers (Schematic)
for steady laminar flow in a
circular tube. But Q = V/t (approximately), in which Vis the vol
ume of liquid collected in time t. Thus,
V
t
and solving for p
128/J
mvi
236 FLUID MEASUREMENTS
The lost head, HL, however, is nearly constant since it is nearly equal
to the imposed head which varies between fixed limits. Since d and /
are constants of the instrument the equation reduces to
fj, Kwt
or
/* = Kpgt = Kipt
thus
 = v = K\t (approximately)
P
The correct empirical equation relating v and t for the Saybolt Uni
versal viscometer is
001935
v in ft 2 /sec = 0.000002365*   (for 100 >/> 32)
in which t is the time in seconds (called " Saybolt seconds"), and in
which the negative term appears as a correction embodying the
neglected factors of the above simplified analysis.
Of the tube viscometers the Saybolt, Engler, and Redwood are
built of metal to rigid specifications and hence may be used without
calibration. Since the dimensions of the glass viscometers such as the
Bingham and Ostwald cannot be perfectly controlled, these instru
ments must be calibrated before viscosity measurements can be made.
56. Measurement of Static Pressure. The accurate measurement
of pressure in a fluid at rest may be accomplished with comparative
FIG. 136. FIG. 137.
ease since it depends only upon the accuracy of the gage or manometer
used to record this pressure and is independent of the details of the
connection between fluid and recording device. To measure the static
MEASUREMENT OF STATIC PRESSURE
237
FIG. 138. Static Tube.
pressure within a moving fluid with perfect accuracy is quite another
matter, however, and depends upon painstaking attention to the
details of the connection between flowing fluid and measuring device.
To measure the static pres
sure in the curved flow of Fig. ^
136, for example, a device must >
be introduced which fits the
streamline picture perfectly and
thus does not disturb the flow at all,
and such a device must contain
a small smooth hole, called a
"piezometer opening," whose
axis is normal to the direction of
motion at the point where the static pressure is to be measured; to
this opening may be connected a manometer or pressure gage to
register the pressure transmitted into the opening. Although meet
ing these requirements perfectly is a virtual impossibility, the attempt
nevertheless illustrates tfie problem involved in the measurement of
static pressure. The above
/fpH^ J device is a practical one for
/ ** l\ use in rectilinear flows (Fig.
137), where the difficulties of
alignment with the flowing
fluid are not so great. In
both cases it is apparent that
the insertion of a solid device
of this type is certain to dis
turb the fluid flow to some
extent, and hence it should be
observed that such devices
should be made as small as
FIG. 139. possible.
A "static tube" (Fig. 138)
may be used for measurement of the static pressure in a flowing fluid.
Such a tube is merely a smooth cylinder carefully aligned with the
flow and having a smooth upstream end. In the side of the cylinder
are piezometer holes drilled radially, or a circumferential slot, through
which pressure is transmitted to a recording device. Since the
introduction of the static tube disturbs the flow and causes the
velocity along its surface to be greater than that in the undisturbed
flow, the pressure transmitted through the piezometer openings must,
I
238
FLUID MEASUREMENTS
according to the Bernoulli principle, be less than the true static
pressure in the undisturbed flow. This error, of course, may be mini
mized by making the tube as small as possible and is usually safely
negligible for most engineering considerations.
The static pressure in the fluid passing over an existing solid surface
(such as a pipe wall or the surface of an object in the
flow, Fig. 139) may be measured by small smooth
piezometer holes drilled normal to the solid surface,
since the surface for each of these conditions "fits"
the flow perfectly, being a streamline of the flow.
Such piezometer openings measure only the local
static pressures at their locations on the solid sur
face and cannot, in general, measure the pressures at
a distance from this surface since such pressures are
obviously different from those at the surface because
of flow curvatures and accelerations. Where no flow
curvatures exist, as in a straight pipe, a wall piezom
eter opening will measure the pressure throughout
FIG. 140. the cross section of pipe in which the piezometer is
Piezometer Ring, located.
Frequently in pipeline practice a large number of
piezometer openings are drilled in the pipe wall at the same cross
section and led into a "piezometer ring" (Fig. 140), whence the pres
sure is transferred to a recording device. The pressure taken from the
piezometer ring is considered more reliable than that obtained from
a single piezometer opening, since probability dictates that the errors
Correct Incorrect
FIG. 141. Measurement of Surface Elevation by Piezometer Columns.
incurred by the inevitable imperfections of single openings will tend to
cancel if numerous openings are used and the results averaged; the
piezometer ring automatically approximates such an average.
57. Measurement of Surface Elevation. The elevation of the sur
face of a liquid at rest may be determined by manometer, piezometer
column, or pressuregage readings (Art. 10).
MEASUREMENT OF SURFACE ELEVATION
239
The same methods may be applied to flowing liquids if the above
precautions in the construction of piezometer openings are followed
and if the piezometer method is used only where the streamlines of the
flow are straight and parallel. Correct and incorrect measurements
of a liquid surface by piezometer openings are illustrated in Fig. 141.
For direct measurements to a liquid surface the hook gage and point
gage (Fig. 142) are common. The hook gage is generally used in a
stilling well connected to the liquid at the point where its surface
elevation is to be measured but may be used directly on the liquid
surface if velocities are not large. To set the point of the hook in the
Hook gage
Point gage
CD
Counter
weight
FIG. 142. Gages for Measure
ment of Surface Elevation.
FIG. 143. Float for
Measurement of Surface
Elevation. (Schematic.)
liquid surface, it is first placed below the surface and then raised until
a small pimple appears on the surface at this condition the point of
the hook is above the liquid surface. When the hook is lowered until
the pimple just disappears, its point is accurately at the same elevation
as the liquid surface. From a graduated scale and vernier on the hook
gage shaft, the surface elevation of the liquid may be read precisely.
A point gage is suitable for swiftly flowing liquids in which the
presence of a hook below the liquid surface would cause local disturb
ances. In measuring, the point gage is lowered until it just contacts
the liquid surface (noted by slight disturbances of the surface) and then
read by scale and vernier located on the gage shaft.
240
FLUID MEASUREMENTS
, Plate
Floats are often used in connection with chronographic waterlevel
recorders for measuring liquidsurface elevations. The arrangement
of such floats is indicated schematically in Fig. 143.
As the liquid level varies, the motion of the cable
is measured on a scale or plotted automatically on a
chronographic record sheet.
An electrical method of liquid surface measure
^ ment has recently been somewhat successful. It
FIG. 144. consists of using a small fixed metal plate, and the
liquid surface as a condenser (Fig. 144). Variation of
liquidsurface elevation varies the capacitance of the condenser, which
may be measured electrically, and after calibration of the device
Stagnation
points v
^E
FlG. 145,
electrical measurements become a measure of surface elevation.
Staff gages give comparatively crude but direct measurements of
liquidsurface elevation. From casual _
observation, the reader is familiar with
their use as tide gages, in the measure
ment of reservoir levels and in registering
the draft of ships.
58. Measurement of Stagnation Pres
sure. The stagnation, 2 or total, pressure,
Po + \pV 2 > may be measured accurately
by placing in the flow a small solid object
having a small piezometer hole at the
stagnation point. The piezometer open
ing may be easily located at the stagna
tion point if the hole is drilled along the
axis of a symmetrical object such as a cylinder, cone, or sphere (Fig.
145). When the axis of the object is aligned with the direction of
* See Art. 19,
FIG. 146. Pitot Tube.
THE PITOT (PITOTSTATIC) TUBE 241
flow the piezometer opening is automatically located at the stagna
tion point and the pressure there may be transferred through the
piezometer opening to a recording device. Theoretically, the upstream
end of solid objects for the above purpose may be of any shape since
the shape of the object does not affect the magnitude of the stagna
tion pressure; in practice, however, the upstream end should be made
convergent (conical or hemispherical) in order to fix the location of
the stagnation point and to cause its location to be insensitive to
small variations in alignment.
A small bent tube, with open end facing upstream, provides an
excellent and simple means of measuring stagnation pressure. Tubes
of this type are called Pi tot tubes after Henri Pi tot, who found (1732)
that, when they were placed in an open flow at a point where the veloc
ity was V (Fig. 146), the liquid in the tube rose above the free surface
of the liquid a distance V*/2g. Obviously, Pitot's results agree with
those obtained by applying the foregoing reasoning on the magnitude
of stagnation pressure.
MEASUREMENT OF VELOCITY
$9. The Pitot (PitotStatic) Tube. From the stagnation pressure
equation
P. =Po + %pV*
or
\ (p. 
V = \ (p.  Po)
it is evident that fluid velocities may be obtained by the measurement
of stagnation pressure, p a , and static pressure, p . It has been shown
that stagnation pressure may be measured easily and accurately by a
Pitot tube and static pressure by various methods, such as tubes, flat
plates, and wall piezometers. Any combination of stagnation and
staticpressuremeasuring devices is called loosely a " Pitot tube " in
most fields of engineering, although Pitot's original device was designed
to measure stagnation pressure only. Aeronautical practice takes
cognizance of this fact by terming a device which measures both
stagnation and static pressures a " Pi totstatic tube."
Pitot tubes may be divided into two classes: (1) those in which
static and stagnation pressure connections are "separate," and
(2) those having "combined" stagnation and staticpressureme^sur
ing devices,
242
FLUID MEASUREMENTS
"Separate" types of Pitot tubes are shown in Fig. 147 as used in
obtaining the velocity profile in a pipe or as an airspeed indicator in
aeronautical practice. Such tubes are simple in construction, but
they cause inconvenience in pipe lines because of the necessity of two
FIG. 147. Pitot Tubes. (Separate.)
pipe connections and the difficulties of obtaining correct static pressure
by a single piezometer opening.
Modern practice favors the "combined" type of Pitot tube, two
types of which (for general and aircraft use) are illustrated in Fig. 148.
Here the static tube jackets the stagnation pressure tube, resulting
PS
FIG. 148. Pitot Tubes. (Combined.)
in a compact, efficient velocitymeasuring device. When connected
to a differential manometer which records the pressure difference,
PS ~~ Pot the manometer reading becomes a direct measure of the
velocity, as may be seen from the Pitot tube equation
V
 (Ps  Po)
P
A static tube has been shown to record a pressure slightly less than
the true static pressure, owing to the increase in velocity along the
THE PITOT (PITOTSTATIC) TUBE
243
tube (Art. 58). This means that the above equation must be modified
by an experimental coefficient, Cj, called the coefficient of the instru
ment, to
in which p' is the pressure measured by the static tube. Since p is
less than p it is obvious that Cj will always be less than unity. How
ever, for most engineering problems the value of Cj may be taken as
1.00 for the conventional types of Pi tot tubes (Fig. 149), since the
American Society of Heating & Ventilating Engineers
t
o o o
National Physical Laboratory (England)
c H
1
Prandtl (Gb'ttingen)
1
^N
\
American Blower Company ^v
FIG. 149. Pitot Tubes (to Scale).
differences between p and p are very small. Prandti^has designed a
Pitot tube in which the difference between p and p is completely
eliminated by ingenious location of the staticpressure opening. The
opening is so located (Fig. 150) that the underpressure caused by the
tube is exactly compensated by the overpressure due to the stagnation
point on the leading edge of the stem, thus giving the true static
pressure at the piezometer opening.
There are many variations on the Pitottube idea resulting in
devices of various shapes and coefficients. Probably the most popular
of these is the Cole Pitometer, a " reversed " type of Pitot tube.
244
FLUID MEASUREMENTS
The Pitometer consists of two similar Pitot tubes, one facing upstream,
the other downstream (Fig. 151). The tube facing upstream measures
the stagnation pressure properly, but the one facing downstream
FIG. 150. Prandtl's Pitot Tube.
measures the pressure in the turbulent wake behind itself, which is
less (Art. 28) than the true static pressure, p . The coefficient of the
Pitometer is, therefore, much less than unity; experiments have shown
it to have values between 0.84 and 0.87.
The advantages of the Pitometer are its
ruggedness and a compactness which
allows it to be slipped easily through a
cock in the wall of a pipe line.
A consideration of velocitymeasur
ing devices is their sensitivity to obli
quity or angle of yaw (Fig. 152). Since
it is always difficult to secure perfect
alignment of tube with flow, it is obviously advantageous to have
a Pitot tube which gives the smallest possible error when perfect
alignment does not exist. The PrandtlPitot tube, designed to be
/Angle of yaw
FIG. 151. Cole Pitometer.
FIG. 152.
insensitive to small angles of yaw, gives a variation of only 1 per cent
in its coefficient at an angle of yaw of 19. For the same percentage
variation in coefficient the American Society of Heating and Ventilat
THE VENTURI TUBE 245
Ing Engineers' Pitot tube may have an angle of yaw of 12, and that
of the National Physical Laboratory only 7. 8
ILLUSTRATIVE PROBLEM
A Pitot tube having a coefficient of 0.98 is placed at the center of a pipe line in
which benzene is flowing. A manometer attached to the Pitot tube contains
mercury and registers a deflection of 3 in. Calculate the velocity at the centerline
of the pipe.
PB  Po = A X 62.4 X 13.55  A X 62.4 X 0.89  197 lb/ft 2
0.98\/64.4[ ) 14.8 ft/sec
\ \0.89 X 62.4/
60. The Venturi Tube. A convergentdivergent tube called a
"Venturi tube" has had some success in aeronautical practice as an
airspeedmeasuring device. A Ven
turi tube is shown in Fig. 153, and
from the familiar Bernoulli principle
it is evident that the pressure differ
ence p p2 created by flow through
FIG. 153. Venturi Tube. the tube is a measure of the velocity
V > Neglecting the losses which
occur in the short distance between sections and 2 of the streamtube
which passes through the Venturi tube, the Bernoulli equation is
and assuming an incompressible fluid
V A
Therefore
or
3 Data from K. G. Merriam and E. R. Spaulding, N.A.C.A. Technical Note 546,
1935.
246
FLUID MEASUREMENTS
in which A is unknown but dependent upon the area A\ and the shape
of the tube. Introducing an experimental coefficient to express the
ratio of A to A i
C = r or A =
Introducing this relationship in the above equation and solving for V
v o= i ,., = V~ (Po ~ p2)
in which Cmust be found by calibration of the tube in an air stream of
known velocity. After calibration the Venturi tube, like the Pi tot tube,
offers a means of measuring a pressure difference and calculating the
FIG. 154. Double Venturi.
velocity which has created it. Venturi tubes have been generally
abandoned as airspeed indicators for aircraft, because of their sensitiv
ity to yaw and their susceptibility
^ to icing difficulties. However,
they are still used in engineering
practice as a source of low pres
sure and frequently appear as
such in the form of the "double
Venturi" of Fig. 154, one of the
flow elements of the standard
carburetor.
61. The PitotVenturi. The
Pitot tube and Venturi principles
are combined in a velocitymeasuring device called the " PitotVenturi"
(Fig. 155), which for the same velocity gives a pressure difference
1
FIG. 155. PitotVenturi.
ANEMOMETERS AND CURRENT METERS 247
(p s p2) larger than that of Pitot or Venturi alone. Here as in the
foregoing examples
Ps  Po = IpVl
Adding these equations gives
and solving for V
Here again an experimentally determined value of C is essential to
velocity measurements with the Pitot Venturi.
62. Anemometers and Current Meters. Mechanical devices of
similar characteristics are utilized in the measurement of velocity in
air and water flow. Those for air are called "anemometers"; those for
water, " current meters." These devices consist essentially of a rotat
ing element whose speed of rotation varies with the velocity of flow and
for which the relation between these variables must be found by cali
bration. Anemometers and current meters fall into two main classes,
depending upon the design of the rotating elements, which may be of
the cup type or vane (propeller) type, as illustrated in Fig. 156.
Anemometers and current meters differ slightly in shape, rugged
ness, and appurtenances because of the different conditions under
which they are used. The cuptype anemometer for the measurement
of wind velocity is usually mounted on a rigid shaft; the vanetype
anemometer is held in the hand while readings are taken. The current
meters are usually suspended in a river or canal by a cable, and hence
must have empennages and weights to hold them in fixed positions in
the flow.
Another type of anemometer which has been very successful in the
field of aeronautical research is the hotwire anemometer, one type of
which is shown diagrammatically in Fig. 157. The device consists of
a fine platinum wire exposed to the velocity V which is to be meas
ured. The fact that various velocities will have various cooling effects
upon the hot wire, which will change its resistance, allows relating by
248
FLUID MEASUREMENTS
Cup Type
N. Y. U.
Anemometers
Vane Type
. Y. U.
Cup Type
N. Y. U.
Current Meters
FIG. 156.
Vane Type
A.S.M.E.
calibration the velocity V and certain electrical measurements.
The hotwire anemometer of Fig. 157 is of the constant voltage type, 4
and during its operation the voltage
across its terminals is maintained con
stant. Variation of velocity will change
the resistance of the wire and, thus,
the ammeter reading; the ammeter
reading thus becomes, after calibration,
a measure of the velocity. The ad
vantage of the hotwire anemometer
lies in the fact that it may be built in
Platinum wire extremely small sizes and so may be
FIG. 157. Hot Wire Anemometer, employed in obtaining the velocity pro
files in boundary layers, etc., where a
Pitot tube cannot be used. It must always be calibrated before use,
and calibration is generally made against Pitottube measurements
of velocity.
4 Constantcurrent and constantresistance types are also used.
VENTURI METERS
249
/Thermometer
Liquid
seal
=
MEASUREMENT OF RATE OF FLOW
63. TotalQuantity Methods. Rate of fluid flow (Q or G) may be
obtained by measurement of the total quantity of fluid collected in
a measured time. Such collections may be made by weight or volume
and are the primary means of measuring fluid flow, but usually such
measurements can be employed for only comparatively small flows
under laboratory conditions.
Measurement of rate of flow by weighing consists of collecting the
flowing liquid in a container placed on a scales and measuring the
weight of liquid accumulated in a
certain time. There are, of course,
many commercial variations of
this method and many automatic
devices are applied to it, but the
principle remains the very simple
one indicated above.
Volumetric measurements of
rate of flow are carried out by
allowing the liquid to collect in a
container whose internal dimen
sions have been accurately deter
mined. By noting the rise in the
liquid surface in a measured time,
or by noting the number of fillings
of the container in a measured
time, the rate of flow may be easily
calculated. The accuracy of volu FIG. 158. Gasometer,
metric measurements is not in
general so high as that of weighing methods because of the larger
number of variables to be measured.
Gases may be measured volumetrically with a " gasometer"
(Fig. 158), a device in which constant pressure and temperature are
maintained while a volume of gas is collected. With pressure and
temperature constant, the rate of rise of the movable top becomes a
measure of the rate of flow into the gasometer, and after pressure and
temperature are noted the rate of flow may easily be calculated.
A correction or automatic compensation must be made for the variable
buoyant force acting on the top due to varying immersion.
\ 64. Venturi Meters. A constriction in a streamtube has been
ekn 5 to cause a pressure variation which is directly related to the rate
6 Articles 21 and 22.
250
FLUID MEASUREMENTS
of flow, and thus is an excellent fluid meter in which rate of flow may be
calculated from pressure measurements. Such constrictions used as
fluid meters are obtained by Venturi meters, nozzles, and orifices.
A Venturi meter is shown in Fig. 159. It consists of a smooth
entrance cone of angle about 20, a short cylindrical section having
diameter J to  of the pipe diameter, and a diffusor cone having a 5 to
7 total angle in order to minimize energy losses. 6 For satisfactory
1 2
10 3 2
10 6 2
5 10* 2 5
Reynolds Number, ~~
FIG. 159. Venturi Meter. 7
operation of the meter, the flow should possess " normal" turbulence
as it passes section 1. To insure this it should be installed after a sec
tion of straight and uniform pipe, free from fittings and other sources of
eddying turbulence, and having a length of at least 30 pipe diameters.
Straightening vanes may also be placed upstream from the meter for
elimination of excessive turbulence of an eddying nature.
See Art. 42.
7 Data from Fluid Meters, Their Theory and Application, Third Edition, A.S.M.E.,
1930. More comprehensive data are available in the Fourth Edition of this
publication.
VENTURI METERS 251
The pressures at the base of the meter (section 1) and at the throat
or constriction (section 2) are obtained by piezometer rings, and the
pressure difference between these points is usually measured by a
differential manometer. An accurate measurement of pressure and
temperature at section 1 will be seen to be necessary in the metering of
gases and vapors, but for liquids the pressure difference between base
and throat of the meter will allow calculation of the rate of flow.
Utilizing now the equations for flow of a perfect fluid through a
constriction in a horizontal streamtube (Arts. 21 and 22):
For incompressible fluids
Q =
For compressible fluids
^2^2 / 2gk
/ . \ o / , \ y
(A^V (P2\
~ W W
But since F 2 = Q/A 2 and F 2 = G/4 2 ^ 2 the equation for velocity
may be written:
For incompressible fluids
F 2 =
For compressible fluids
1 / o^t jjcr *i *ii
1 iV^^^
v , gVi^r^^i *
(h\'(ptf
~ \Ai) \PJ
These velocities will not be obtained with real fluids because of fric
tional resistance and energy losses occurring between sections 1 and 2,
and the above expressions must be corrected by an experimental
coefficient, C V) the " coefficient of velocity," to bring them into con
252 FLUID MEASUREMENTS
formity with reality. When this coefficient is inserted the original
expressions become
CvA* . , ltri ^ z , (g6)
for the real incompressible fluid, and
2gk pi
(%Y /M
~ w w
or
(87)
for the real compressible fluid.
The significance of the coefficient of velocity C v and its relation
to head losses may be better understood if the derivation of equation 86
is considered. For a horizontal pipe and constriction the Bernoulli
equation is
.+!!, fe + Z! + fa
w 2g w 2g
and the equation of continuity
Simultaneous solution of these equations to give equation 86 can
result only when
' 1 .. vl
is inserted in the first equation. In the above equation, K is the minor
loss coefficient for the passage between sections 1 and 2, and it is noted
that K is related to C v by
VENTURI METERS 253
Thus, the lower the value of C v (the greater the difference between
real and perfect fluids), the greater will be the magnitude of K and
the energy loss. Since the energy loss characterized by K results from
frictional effects rather than from eddying turbulence, it may be con
cluded that the variation of K with the Reynolds number will be
similar to that exhibited by the friction factor, /, which decreases as
the Reynolds number increases. Therefore, the variation of C v with
the Reynolds number may be expected to follow a trend opposite to
those of K and /, and increase with increasing Reynolds number, a
fact which is borne out by the experimental results of Fig. 159.
The form of equation 87 makes it too unwieldy for engineering
practice and too difficult of rapid solution, but these inconveniences
may be overcome by the application of the following graphical meth
ods. Equation 87 for the compressible fluid may be placed in the form
of equation 86 if a correction factor, F, 8 is applied. The equation for
compressible fluid flow may therefore be written
YC v
and used conveniently after an expression for Y has been found.
This may be done by equating the above equation 'to
and solving for Y, with the result that
kl
1 
,Aj \pv
ihr (T Pi
Although this calculation seems at first to complicate the problem,
closer inspection indicates that Y is dependent on only three variables :
8 Since the term Y accounts only for the expansion of the gas as it passes from
section 1 to section 2 of the constriction, it is called the "expansion factor."
254
FLUID MEASUREMENTS
the pressure ratio p2/Pi, the area ratio A%/Ai, and the adiabatic con
stant k. This means that Y may be calculated for various values of
these variables once for all, plotted, and thus made readily usable for
engineering calculations. A plot of the expansion factor, F, is given
in Fig. 160 for the solution of problems.
i
= 1.4
1.2
1.00
.98
.96
.94
.92
.90
.88
.86
1.00
.98
.96
.94
.92
.90
.86
1.00 .98 .96 .94 .92 .90 .88 .86 .84 .82 .80
III
1.00
.98
.96
.94
.92
.90
.88
.86
FIG. 160. The Expansion Factor, F.
From the above facts it is observed that the one equation,
YC v A^w l
Cr =
(89)
will allow calculation of fluid flow through a pipeline constriction such
as a Venturi meter, whether the fluid is perfect, real, compressible, or
incompressible.
ILLUSTRATED PROBLEM
Air flows through a 6 in. by 3 in. Venturi meter having a coefficient C v of 0.98.
The gage pressure is 20 lb/in. 2 and the temperature 60 F at the base of the meter,
NOZZLES 255
and the differential manometer registers a deflection of 6 in. of mercury. The
barometric pressure is 14.7 lb/in. 2 Calculate the flow.
k = 1.40, ^ = x . U ' 7 ' . = 0.915
29.92  6
* /20 + 14.7\
V 147 /
. , = 0.25 wi = v r " m = 0.180lb/ft 3
^4i V 6 / 53.3(100 + 460)
From the plot of Fig. 160,
Y  0.95.
0.95 X 0.98 X  X (
Vl  (0.25) 2
G = 3. 3 lib/sec
65. Nozzles. Nozzles are used in engineering practice for the
creation of jets and streams for all purposes as well as for fluid metering ;
when placed in or at the end of a pipe line as metering devices they are
generally termed "flow nozzles." Since a thorough study of flow
nozzles will develop certain general principles which may be applied
to other special problems, the flow nozzle only will be treated here.
Flow nozzles are illustrated in Figs. 161 and 162. They are designed
to be clamped between the flanges of a pipe, generally possess rather
abrupt curvatures of the converging surfaces, terminate in short
cylindrical tips, and are essentially Venturi meters with the diffusor
cone omitted. Since the diffusor cone exists primarily to minimize
the energy losses caused by the meter, it is obvious at once that larger
energy losses will result from flow nozzles than occur in Venturi meters
and that herein lies a disadvantage of the flow nozzle; this disadvan
tage is somewhat offset, however, by the lower initial cost of the flow
nozzle.
Extensive research on flow nozzles, recently sponsored by the
American Society of Mechanical Engineers and the International
Standards Association, has resulted in the accumulation of a large
amount of reliable data on nozzle installation, specifications, and,
256
FLUID MEASUREMENTS
experimental coefficients. Only the barest outline of these results can
be presented here ; the reader is referred to the original papers of these
societies for more detailed information.
The A.S.M.E. "longradius" flow nozzle is shown in Fig. 161.
Section 1 is taken one pipe diameter upstream from the nozzle and
section 2 at the nozzle tip. It has been found that the pressure at the
latter point may be measured successfully by a wall piezometer con
1 Ellipse v . ^_U_ 2
.25
FIG. 161. A.S.M.E. LongRadius Flow Nozzle. 9
nection opposite the nozzle tip which leads fortunately to the simplifi
cation of the nozzle installation since a wall piezometer is easier to
construct than a direct connection to the tip of the nozzle.
The equation derived for the Venturi meter may be applied directly
to the nozzle. This is
YC v A 2 w l I
for compressible fluids and reduces to
Q  C " A *
9 Fluid Meters, Their Tlieory and Application, Fourth Edition, A.S.M.E., 1937.
NOZZLES
257
iVvVVvvs
1.20
1.10
1.00
90
1 .2 .3 .4 5 .6 .7
A*
AI/A,
C
.05
.987
.10
.989
.15
.993
.20
.999
.25
1.007
.30
1.016
35
1.028
.40
1.041
.45
1.059
.50
1.081
55
1.108
.60
1.142
.65
1.183
FIG. 162. I.S.A. (German Standard) Flow Nozzle. 10
for incompressible ones. Values of Y may be obtained from Fig. 160,
and values of C v (over the limited range of tests to date) may be taken
from the plot of Fig. 161.
The I.S.A. (German Standard) nozzle shown to scale in Fig. 162
10 Data from Regelnfur die Durchflussmessung mil genormten Diisen und Blenden,
V.d.I. Verlag, 1935.
258 FLUID MEASUREMENTS
differs from the A.S.M.E. nozzle in shape and in the location of the
piezometer connections, which are made by holes or slots adjacent to
the faces of the nozzle. This method of pressure connection is con
venient in that the nozzle, complete with pressure connections, may be
built as a unit and installed between the flanges of a pipe line without
the necessity of drilling piezometer holes in the pipe.
Although the pressure connections for this nozzle are not made in
the conventional way at pipe and constriction, the flow equations
G _ jrc^, Wl Lfrp*
and
may be applied if it is kept in mind that a nozzle is essentially a means
of securing a regular pressure variation which, with other factors, is
related to the rate of flow. A change in piezometer location, however,
alters the values of Y and necessitates the calculation of a special plot
for this variable (Fig. 162). Another difference between A.S.M.E. and
I.S.A. nozzles is in the definition of the coefficient of the latter as
C =
thus reducing the flow equations to
G = YCA 2 wi
and
Obviously the coefficient C depends upon the area ratio and, through
the coefficient of velocity C VJ upon the Reynolds number as well.
The variation of C with these variables is given in the plots of Fig. 162.
The constancy of C at high Reynolds numbers and the characteristic
decrease of C with decreasing Reynolds numbers should be noted*
ORIFICES
259
ILLUSTRATIVE PROBLEM
An A.S.M.E. longradius flow nozzle of 3in. diameter is installed in a 6in.
water line. The attached differential manometer contains mercury and registers
a deflection of 6 in. Calculate the rate of flow and the loss of head caused by the
nozzle.
From the plot of Fig. 161, C v = 0.991.
0.991 X  X {
4
 1])
Q = 1.01 cfs
Loss of head is composed of f rictional effects between sections 1 and 2
and loss due to turbulence downstream from the nozzle (/fL 2 _ 3 )
j_ 2 )
1.01
= 20.55 ft/sec
(4, 
\Cl
2g
= o.uft
Calculating HL Z _ Z as a sudden enlargement,
_ . (F,  F 3 ) 2 ,_ (20.55  5.14) _
rl>LA _ o ~~ __ ^J . / \J J. L
2 <7 2^
AL = fe!_ 2 + fe 2 _ 8 = 0.11 + 3.70  3.81 ft
66. Orifices. Like nozzles, orifices serve
many purposes in engineering practice
other than the metering of fluid flow, but
the study of the orifice as a metering
device will allow the application of princi
ples to other problems, some of which will
be treated subsequently.
The conventional orifice for use as a
metering device in a pipe line consists of a
concentric squareedged circular hole in a
FIG. 163. SharpEdged Orifice.
thin plate which is clamped between the flanges of the pipe line (Figs.
163 and 164). The orifice differs from the nozzle as to flow character
istics in that the constricted section of flow occurs not within the orifice
but downstream from it owing to the nonaxial direction of fluid
260 FLUID MEASUREMENTS
particles as they approach the orifice. This seems to complicate the
problem since, in the flow equations between pipe and constriction,
YCyAw I
C V A
A% is unknown. At this point another experimental coefficient C c ,
the " coefficient of contraction,'* may be advantageously introduced,
C c being defined by
It is simply the ratio between the unknown area A 2 and the known
area of the orifice A. When this relationship is introduced in the
flow equations, they become
YC v C c Aw l " ~~
(0
and
in which the " orifice coefficient," C, is denned by
C e C,
This reduces the above equations to
and
ORIFICES
261
which differ from the equations for the I.S.A. nozzle only in that A 2
has been replaced by A.
The International Standards Association and the American Society
.98 .96 .94 .92 .90 .88 .86 .84 .82
.80
.75
C.70
.65
.60
. .
A
 .
^.., .
~ ' ii
==
JO
***.
 .
1 r.
.60
.40
. 30
"*,
.
.
N
20
.id
.05
10 4
5 10 5 2
Reynolds Number,
10 6
.70
.68
.66
C 64
.62
.60
/
/
/
/
/
X
'

**^
*
.1 .2 .3 .4 .5
A
A/A]
C
.05
.598
.10
.602
.15
.608
.20
.615
.25
.624
.30
.634
.35
.646
.40
.661
.45
.677
.50
.696
.55
.717
.60
.742
.65
.770
.70
.804
FIG. 164. I.S.A. (German Standard) Orifice. 10
of Mechanical Engineers have Undertaken the standardization of
orifices, approaching the problem in two different ways. The A.S.M.E.
has made available data on orifices for various locations of piezometer
10 See footnote on p. 257.
262
FLUID MEASUREMENTS
connections, but lack of space prevents inclusion of this material here. 11
The International Standards Association proposes the I.S.A.
(German Standard) orifice with only one possible pair of piezometer
connections located as for the I.S.A. nozzle adjacent to the orifice
plate, resulting in the concise presentation of data given in Fig. 164.
The variation of C with the Reynolds number is of some, interest
since it exhibits a trend opposite to that of the nozzle and Venturi
meter. This may be explained from the equation for C
C V C C
C =
and must result from an increase of C c with decreasing Reynolds
number. Since the size of the area
A 2 would be expected to increase
as the flow assumed a more viscous
character (lower Reynolds number),
the coefficient C (which varies
directly with C c ) would be expected
to increase also.
Rounded (bellmouthed) orifices
(which are really short nozzles) are
frequently used in the metering of
gases and vapors. Such an orifice
is shown in Fig. 165, and, for gas
flow with pressure ratio below the
critical, equation 24, Art. 22, modified by a coefficient of velocity,
C v , may be applied directly, giving
G
=
FIG. 165. Rounded Orifice.
VY,
in which the value of C v will be close to unity, 0.995 being a reasonable
coefficient to select if calibration of the orifice is not possible.
In using the orifice of Fig. 165 in metering the flow of a vapor the
methods of Fig. 42, Art. 22, should be followed and a coefficient C v
introduced in equation 25, giving for vapor flow
G = 223.8CU a 4
 / 2
"See Fluid Meters Their Theory and Application, A.S.M.E., Fourth Edition,
1937.
ORIFICES
263
The value of C v is probably close to 0.995, but unless other measure
ments are very precise its inclusion in the equation is usually not
justified.
The orifice is frequently encountered in engineering practice oper
ating under a static head where it may not be used as a metering device
but rather as a special feature in an hydraulic design. 12
The general features of an orifice of the above type may be deter
mined from the study of the submerged orifice of Fig. 166, operating
FIG. 166. Submerged Orifice.
FIG. 167. Orifice Discharging Freely.
under steadyflow conditions. Assuming a perfect fluid and applying
the Bernoulli equation between sections 1 and 2
V\
+ + hi = * 2 + V +
or
V 2 =
 A 2 )
for the perfect fluid. For the real fluid, frictional effects will prevent
the attainment of this velocity and the coefficient of velocity C v must
be introduced, resulting in
V 2
and the rate of flow through the orifice becomes
Q = A 2 V 2 = C V A 2 V2g(h l 
12 For example, as a sluice gate in a dam, Fig. 169.
264
FLUID MEASUREMENTS
As in the pipeline orifice the area A% is unknown but must have a
special relation to the orifice area A, depending upon the shape of the
orifice, velocity of flow, etc. This relation is expressed by the " co
efficient of contraction " C ct defined (as before) by
C c = or A 2 = C C A
A.
which when substituted above gives
Q = C V C C A
 ha)
Orifices and their Coefficients
c v
Sharp
edged
.61
.62
98
Rounded
98
1.00
.98
Short tube
'///f/////A
~ 75
100
Borda
.51
.52
98
FIG. 168.
in which the two coefficients are combined into the " coefficient of
discharge," C, defined by
The above equation may now be written
Q = CA
allowing prediction of the rate of flow for a given difference in surface
elevation, after the coefficient C has been experimentally determined.
When the orifice discharges freely into the atmosphere (Fig. 167),
the head h 2 becomes zero and the equation reduces to 13
Q = CA
18 See Art. 20.
ORIFICES
265
Above these limits of head and
The dependency of the various orifice coefficients upon shape of
orifice is illustrated by Fig. 168. The coefficients given are approxi
mate values for large orifices (d > 1 in.) operating under compara
tively large heads of water (h > 4 ft).
size various experiments have shown
that the coefficients become sub
stantially constant. Experimental
determinations of orifice coefficients
at low heads and for small orifices
indicate that the coefficients vary
with head and orifice size, but the
results of reliable experiments are
so divergent that little can be stated
as to the best values of orifice co
efficients in this range. If orifices
are to be small or to be operated
under low heads and great accuracy
is required they should be calibrated
in place.
A special problem of orifice flow is that of the sluice gate of Fig. 169,
in which contraction can take place on only one side of the jet. Assum
ing a perfect fluid and applying Bernoulli's equation to the typical
stream tube between the liquid surface and contracted section, taking
the base of the structure as datum,
P V\
+ o + A = ^ + 2 + s
w 2g
since the pressure distribution at the contracted section 2, where no
flow curvatures exist, is a static one. Obviously,
FIG. 169. Sluice Gate.

w
and therefore
which gives
h = d
vl
V 2 = V2g(h  d)
for the perfect fluid. For a real fluid
V 2  C 9 V2g(h  d)
266
FLUID MEASUREMENTS
and
or
Q A 2 C V
 d)
Q = C C AC V V2g(&  d)
or, introducing the coefficient of discharge, C = C V C C
Time
Q = CA \ / 2g(h  d)
for the sluice gate. Since the coefficient C will depend upon the head
h, and the gate opening ^4, the relation between these variables must
be determined experimentally before the sluicegate problem can be
treated completely.
Another problem of orifice flow
which frequently arises in engi
neering practice is that of dis
charge from an orifice under
falling head a problem of un
steady flow. With no inflow to
the container of Fig. 170, the free
surface will fall as flow takes place
through the orifice. Thus the head
on the orifice h, and the rate of
flow Q, will vary with time, and
the flow becomes an unsteady 14
one. The time necessary for the
FIG. 170.
Orifice Discharging under Falling Head.
liquid surface to fall from eleva
tion 1 to elevation 2 may be cal
culated by writing the equations of flow for a differential time dt. At
time /, the head on the orifice is h and the rate of flow is therefore
Q = Ca V2&
In time dt the differential volume of fluid, dV, passing from the con
tainer is given by the two expressions
and
dV = Qdt
dV = Adh
which may be equated to give
Qdt =  Adh
"See Art. 16.
FLOW BENDS
267
or by substituting the equation for Q
Ca V2gh dt =  Adh
Solving for dt
dt =
Ca
Integrating between the corresponding limits of t and H
p dt = _
Jt,
h'^dh
gives the elapsed time, tz t\, as
The form of this equation may be simplified by multiplying and
dividing by (h'{ + h^). This gives
"~ h
Ca
+ Ca
or since V, the total volume discharged
in time / 2 ~~ ^i> is given by Q
FIG. 171. Flow Bend.
for a container of uniform horizontal
crosssectional area.
67. Flow Bends. The orifice, noz
zle, and Venturi meter as applied in the
measurement of pipe line flow have been seen to be fundamentally
methods of producing a regular and reproducible pressure difference
which is related to rate of flow. For this reason they are sometimes
called "pressuredifference meters*' or u head meters." Another type
of pressuredifference meter is the "flow bend" which utilizes the
difference between the pressures at the inside and outside of a pipe
bend (Fig. 171) created by centrifugal force as fluid flows through
268 FLUID MEASUREMENTS
the bend. Lansford w has recently obtained the experimental
coefficients of a variety of standard 90 flanged elbows which allows
their use as a successful and economical type of fluid meter. He pro
poses the equation
Po  Pi __ C J^
w 2g
with coefficient Ck ranging between 1.3 and 3.2, the magnitudes
depending upon the size and shape of the flow bend. This equation
may be solved for V
and leads to the flow equation of the familiar form
or, if
C = 7= ,
in which C will have values between 0.56 and 0.88.
68. PitotTube Methods. The rate of flow in pipe lines is fre
quently measured by means of the Pitot tube and Pitometer. These
devices have been seen to be primarily velocitymeasuring instruments
which may be employed in pipe lines to establish the distribution of
velocity; their use in measuring rate of flow is essentially an integra
tion of the product of velocity and area through which the velocity
exists.
One method of obtaining rate of flow from velocity measurements
is to divide the pipe cross section into a number of equal annular areas
(J5, C, D y Fig. 172) and to measure the average velocities through these
areas by placing the velocitymeasuring device at points where these
average velocities are assumed to exist. These points are taken to be
at the midpoints of the areas, i.e., at points where circles divide these
areas in half. This is really assuming the velocity to vary linearly
over the areas considered which (in turbulent flow) is obviously more
18 W. M. Lansford, "The Use of an Elbow in a Pipe Line for Determining the
Flow in the Pipe," Bulletin 289, Eng. Exp. Station, Univ. of Illinois, 1936.
PITOTTUBE METHODS
269
true near the center of the pipe than near the walls; this assumption
does not cause serious errors, however, if a large number of annular
areas are taken. In general, the velocity distribution is not symmetri
K W
V 6 J
Vt*S
FIG. 172.
cal about the pipe centerline, and the average velocity through an
annular area is taken to be the numerical average of the two velocities
measured in this area. Thus (Fig. 172)
V a +V*\A
V 4 \A
3
but the rate of flow, <3, in the line is given by
or by substitution of the above values,
i + V 2 + F 3 + F 4 +
Q
This means that the average velocity in the pipe line is given by
a simple numerical average of the velocities existing at certain special
points on the diameter of the pipe line.
Another method of obtaining rate of flow from velocity distribution
is by graphical integration. From Fig. 173, it is obvious that
vdA
or since dA = 2wr dr
= r>
Jo
v2rdr
270
FLUID MEASUREMENTS
but
thus
2rdr = d(r 2 )
/
vd(r 2 )
This equation suggests plotting velocity against the square of the
radius at which the velocity occurs. The area under the resulting
/R*
vd(r 2 ) and may be obtained
by planimeter or other means. The rate of flow, <2, thus becomes
Q = TT (Area under v vs. r 2 curve)
V
V e
r> R 2
FIG. 173.
The numerical average method is the faster way of obtaining rate
of flow in a pipe line by Pitot tube or Pitometer, and although the
graphical method is theoretically the more accurate in all probability
the two methods will give about the same accuracy in most of their
applications in actual practice.
69. Dilution and Thermal Methods. Dilution methods for measur
ing rate of flow consist essentially of introducing at a steady rate a
concentrated foreign substance to the flow, measuring the concentra
tion of the substance after thorough mixing has taken place, and
calculating from the dilution of the substance the rate of flow which
has brought about this dilution.
DILUTION AND THERMAL METHODS
271
A concentrated salt solution has been used in Europe in applying
this method to the calculation of flow in small mountain streams and
in this country to the calculation of the flow through the turbines of
hydroelectric power plants. If the rate of flow of salt solution into the
unknown flow is Q s , and the concentration of salt in this solution C\
lb/ft 3 , the number of pounds per second of salt added to the unknown
flow is given by Q s C\. If the concentration of salt in the unknown
flow after mixing has occurred is C%, the number of pounds of salt
flowmg in the stream per second is also given by (Q + Q 8 )C2 There
fore,
Q.CI = 02 + <2,)C 2
or
Q s C 2
Since <2s is extremely small compared to Q, it may be eliminated from
the numerator of the expression, giving
Hence by controlling and measuring Q s and obtaining C\ and C% by
titration methods, the unknown rate of flow Q may be found.
Analogous to the above dilution method is the "thermal" method
of flow measurement wherein heat is added at a constant rate to a
flowing fluid and the rate of flow deduced from the temperature rise
caused by this addition of heat. This method, illustrated diagram
matically in Fig. 174, has been applied successfully in measuring the
FIG. 174.
flow of gases in pipe lines. It consists essentially of a resistance coil,
R, and two thermometers, one upstream from the coil, the other at
a point downstream, where turbulent mixing has produced a uniform
temperature across the pipe. If / is the current through the coil in
amperes, and R the resistance of the coil in ohms, the heat added to the
flowing fluid is expressed by 1 2 R watts. The heat added to the flow
in British thermal units per second is, therefore, given by I 2 R/W55.
272
FLUID MEASUREMENTS
If the specific heat of the fluid is c Btu/lb/F, the heat received by
the flow may be expressed as Gc (/ 2 *i) Equating the expressions
for heat supplied and heat received
1055
Gc(t 2 
and solving for G, the weight flow
G =
I 2 R
1055 c(t 2 
Thus by taking measurements of temperature difference and electric
current and knowing the specific heat of the gas and the resistance of
the coil, the rate of flow may be calculated.
70. Salt Velocity Method. An ingenious method of flow measure
ment which has met with success in the measurement of rate of flow
to hydroelectric power plants is the saltvelocity method developed by
Allen and Taylor. 16 In this method a quantity of concentrated salt
solution is introduced suddenly to the flow and the average velocity is
Salt
solution
at high
pressure s
FIG. 175.
obtained by measuring the velocity of the salt solution as it moves
with the flow.
The essential feature of the method (illustrated in simplified form
in Fig. 175) consists essentially of a device for introducing suddenly
the salt solution, and two similar electrodes and circuits. The passage
of the salt charge between the plates of an electrode may be recorded
by a momentary increase in the ammeter reading due to the greater
conductivity of the salt solution. By noting the time t between the
deflections of the two ammeter needles, and knowing the distance /
18 C. M. Allen and E. A. Taylor, "The Salt Velocity Method of Water Measure
ment," Trans A.S.M.E., Vol. 45, 1923, p. 285.
WEIRS
273
between electrodes, the average velocity in the pipe may be calculated
from
H
and the rate of flow by
Q = AV
The details involved in the saltvelocity method are far more complex
than the above statement of principles implies, owing primarily to the
use of a chronographic device to record automatically (1) the variation
of electrical current with the passage of the salt charge and (2) the time
of passage of the charge between electrodes.
71. Weirs. For measuring large and small open flows in field and
laboratory, the weir finds wide application. A weir may be defined in
a general way as "any regular obstruction in open flow over which
Plan
rr
^\\ f~//
UllH
Rectangular contracted weir
'Mil
Rectangular suppressed weir
Elevation
FIG. 176. SharpCrested Weirs.
flow takes place." Thus, for example, the spillway of a dam is a special
type of weir and may be utilized for flow measurement. However,
weirs for measuring purposes are usually of more simple and repro
ducible form, consisting of smooth, vertical, flat plates with upper
edges sharpened. Such weirs, called sharpcrested weirs, appear in
a variety of forms, the most popular of which is the rectangular weir
which has a straight, horizontal crest. Rectangular weirs appeared
originally as notches in a more or less uniform and thin vertical wall
and as such developed contraction of the overfalling sheet (nappe) of
274
FLUID MEASUREMENTS
liquid (Fig. 176), similar to the contraction of the jet from a sharp
edged orifice; because of these nappe contractions at the ends of the
weir, rectangular notches are also called contracted weirs. Obviously
when rectangular weirs extend from wall to wall of an open channel
these contractions are eliminated or suppressed, and for this reason
this type of weir is termed a suppressed weir. Measuring weirs appear
in many shapes other than rectangular, the triangular (Vnotch) and
trapezoidal weir being the more popular of these other types.
The flow of liquid over a weir is at its best an exceedingly complex
problem and one difficult of rigorous theoretical solution. An apprecia
tion for the complexities, however, is necessary to an understanding of
experimental results and the deficiencies of simplified weir formulas.
These complexities may be discovered by considering the flow over the
sharpcrested suppressed weir shown in Fig. 177. Although it is obvi
Stilling
device
Nappe
Atmospheric pressure
beneath nappe
maintained by
adequate ventilation
Roller'
FIG. 177. Weir Flow (Actual).
ous at once that the head H on the weir is the primary factor causing
the flow Q to occur, no simple relationship between these two variables
can be derived for two fundamental reasons: (1) the geometrical form
and (2) the effect of turbulence and frictional processes cannot be
calculated. The more important factors which affect the shape of weir
flow are the head on the weir JEf, the weir height P, and the extent of
ventilation beneath the nappe. Although the effect of these factors
may be found experimentally, there is no simple method of predicting
the flow picture from values of H, P, and pressure beneath the nappe.
The effects of turbulence and friction not only cannot be predicted but
cannot even be isolated for experimental measurement. It may be
noted, however, that frictional resistance at the side walls will affect
the rate of flow to an increasing extent as the channel becomes nar
rower and the weir length, 6, smaller. Fluid turbulence and frictional
processes at the sides and bottom of the approach channel contribute
WEIRS
275
to the velocity distribution in an unknown way. The effects of velocity
distribution on weir flow have been shown by Schoder and Turner 17
to be appreciable, and an effort should be made in all weir installations
to provide a good length of approach channel, with stilling devices
such as racks and screens for the even distribution of turbulence and
the prevention of abnormal velocity distribution. Another influence of
frictional processes is the creation of a periodic, helical secondary flow
immediately upstream from the weir plate, resulting in a vortex
(Fig. 177), which influences the flow in an unknown and unpredictable
way. The free liquid surfaces of weir flow also bring surfacetension
forces into the problem, and these forces, although small, affect the
flow picture appreciably, particularly at low heads and small flows.
r
FIG. 178. Weir Flow (Simplified.)
To derive a simple weir equation in the light of the above complexi
ties will obviously require a vast and artificial simplification of the
problem. Such simplification will lead to an approximate result which
must be corrected by the introduction of experimental coefficients.
To derive a simple weir equation, let it be assumed that (1) velocity
distribution is uniform, (2) that all fluid particles move horizontally
as they pass the weir crest, (3) that the pressure in the nappe is zero,
and (4) that the influence of viscosity, turbulence, secondary flows,
and surface tension may be neglected. These assumptions produce
the flow picture of Fig. 178. Taking section 1 in the approach channel
well upstream from the weir and section 2 slightly downstream from
17 E. W. Schoder and K. B. Turner, "Precise Weir Measurements/ 1 Trans.,
A.S.C.E., Vol. 93, 1929, p. 999.
276 FLUID MEASUREMENTS
the weir crest, Bernoulli's equation may be applied to a typical stream
line to find the velocity v 2 . Taking the streamline 12 as a typical one
which gives
and shows that v 2 depends upon h. Because of the dependence of v 2
upon h> V 2 can be considered to be the average velocity through an area
of only differential height dh, and the flow, dq, through this area may
be written as
dq = v 2 bdh =
Integration of this equation between the indicated limits
/
leads to
g an approximate relationship between Q and H for rectangul
s. If V\ is negligible (as it frequently is), this equation reduces t
the basic flow equation for rectangular weirs.
Into the above equation must be inserted an experimental coeffi
cient C which not only embraces the effects of the various phenomena
which have been disregarded in the above analysis, but may be made
to include the effect of velocity of approach as well. Therefore, real
weir flow may be characterized by the equation
(90)
The coefficient C is essentially a factor which transforms the assumed
weir flow of Fig. 178 into the real weir flow of Fig. 177, and its magni
tude is thus fixed by the most important difference between these
flows, which is obviously the shape of the flow picture. Thus the
coefficient C is in a sense primarily a coefficient of contraction which
expresses the extent of contraction of the true nappe below that
WEIRS 277
assumed in the theoretical analysis. Since the size of the weir coeffi
cient depends primarily on the shape of the flow picture, the effect of
other fluid properties and phenomena may usually be discovered by
examining their influence upon the shape of the flow picture.
Although a dimensional analysis of the weir problem must neces
sarily be incomplete because of the impossibility of including all the
pertinent factors it will provide a rational basis for a graphical com
parison of the weir coefficients proposed by various experimenters.
Neglecting the effect of surface tension, the expressible variables
entering the weir problem may be stated as
Q  F(b, P, H, v, g)
which leads by the methods of dimensional analysis to
and shows by comparison with equation 90 that
P
in which the first ratio is one of linear dimensions and indicates the
general shape of the flow picture. The second ratio is recognized as
a Reynolds number which for water flow is determined primarily
by the size of H since the other terms are substantially constant;
therefore
C = F2 \H'
The dependence of the weir coefficient on head and weir height has
been noted by many experimenters, who have proposed the following
empirical equations for the coefficients of sharpcrested suppressed
weirs
Bazin: 18 C
Frese : lg C =
18 H. Bazin, Annales des ponts et chausstes, 18881898. Summarized by G. W.
Rafter in "On the Flow of Water Over Dams," Trans. A.S.C.E., Vol. 44, p. 220, 1900.
19 F. Frese, " Versuche iiber den Abfluss des Wassers bei vollkommenden Uber
fallen," Zeitschrift des V. d. /., Dec. 20, 1890, Vol. 34, No. 51, p. 1337.
278 FLUID MEASUREMENTS
Swiss Society of Engineers and Architects : w
Rehbock: 21 C
TT
).6035 + 0.0813
0.0036iy
These equations are difficult to analyze, but dimensional analysis has
suggested plotting C against H for various constant values of P/H,
which allows ready comparison of the equations. This has been done
in Fig. 179, which may be used to avoid solution of the equations and
.90
.85
.80
C.75
.70
.65
.60
.1 .2 .3 .4 .5 .6.7.8.91.0 2.0 3.0 4.0 5.06.07.08.0
Head in Feet
FIG. 179. Coefficients for SharpCrested Rectangular Weirs.
to draw some general conclusions on weir coefficients. From this plot
it is immediately evident that: (1) the coefficient tends to increase
with decreasing head and weir height; (2) in spite of precise experi
mental measurements very different weir coefficients are found by
different experimenters; and (3) the divergence of results is small at
20 Code for Measuring Water, Swiss Ingenieur and Architeckten Verein, 1924.
81 Th. Rehbock, "Wassermessung mit scharfkantigen Uberfallwehren," Zeit
schrift des V. d. /., June 15, 1929, Vol. 73, No. 24.
\
Weir Coefficients
Rehbock
Frese 
Swiss Soc.
Bazin 
WEIRS
279
high heads but increases rapidly as low heads are attained. The
Rehbock formula is generally considered the most reliable for the
selection of the coefficients of a weir which cannot be calibrated
in place.
ILLUSTRATIVE PROBLEM
Calculate the rate of flow and velocity of approach when a head of 6 in. exists
on a sharpcrested rectangular suppressed weir 4 ft long and 3 ft high.
'I.
H 0.5
From the plot of Fig. 179, C (according to Rehbock) =* 0.62, approximately.
From formula, C (according to Rehbock) = 0.623.
Q = 0.623 X f X 4 VTg (A)* = 4.73 cfs
4.70
(3 + A)4
= 0.338 ft/sec
H
FIG. 180. Rectangular Contracted Weir.
The rectangular contracted weir (Fig. 180) may be treated by the
method suggested by Francis, 22 who found experimentally that the
end contraction varied directly with the head on the weir and was
equal to onetenth of the head. He reasoned that the end contrac
22 J. B. Francis, Lowell Hydraulic Experiments, Fourth Edition, 1883, Van
Nostrand.
280
FLUID MEASUREMENTS
tions reduced the effective weir length from b to (b 2JEf/10), and
proposed the equation
o / orA
(91)
for rectangular contracted weirs having negligible velocity of approach.
Complete contraction of the nappe is dependent upon sufficient dis
tance between ends of weir and channel walls and will exist if the
dimensions of Fig. 180 are main
tained. If the contracted weir of
Fig. 180 cannot be calibrated, the
selection of a coefficient of 0.62
seems reasonable in the light of
suppressed weir experiments ; such
a weir must be calibrated in place
if great accuracy is desired.
Triangular or Vnotch weirs
(Fig. 181) prove advantageous for
measuring small rates of flow. A simplified analysis of the triangular
weir gives the relationship between head and rate of flow
Q = ^ tan a VYg #*
if velocity of approach is negligible, as it practically always is for weirs
of this type. Introducing the experimental coefficient results in
FIG. 181. Triangular Weir.
Q =
(92)
Early experiments by Thomson 23 on a weir of notch angle (2a) of
90 indicated C to have an average value of 0.593. Experiments by
Barr 24 on the same type of weir made of polished brass led him to
propose
Q 2.48H 2  48
which is equivalent to stating that the coefficient is given by
r _ 0.580
^ ~" rrO.02
23 J. Thomson, "Experiments on Triangular Weirs," British Assoc. Repts., 1861.
24 J. Barr, "Experiments upon the Flow of Water over Triangular Notches,"
Engineering, April 8, IS, 1910.
WEIRS
281
Later King, 25 after experiments on the same type of weir constructed
of rough steel plate, suggested
Q = 2.52H 2  47
as a flow equation for the 90 triangular weir. King's equation leads
to a coefficient given by
0.589
H
.03
A comparison of the results of Thomson, Barr, and King, presented
graphically in Fig. 182, gives striking confirmation of the effects of
details upon weir flow. Roughness or minor obstructions on the weir
.62
.61
.60
.59
.58
.57
56
\\
"X
N
Thomson \
V
/
/King  Re
w
ugh we
ir
X
, "*
 ii
^.
X
.
^Bs
JL
r
'
rr Sn
" i
nooth \
veir '
w
.8 1.0 1.2 1.4 1.6 1.8 2.0
H in Feet
.2 .4 .6
FIG. 182. Coefficients for Triangular Weirs.
plate affect the coefficient of the weir appreciably, by reducing veloci
ties adjacent to the plate, thus reducing the nappe contraction and
increasing the coefficient of the weir.
A trapezoidal weir of ingenious design was proposed by Cipoletti 26
in order to compensate automatically for the end contractions of
a rectangular notch. Cipoletti expanded the contracted weir formula
(91) of Francis to
25 H. W. King, Handbook of Hydraulics, Second Edition, p. 93, McGrawHill
Book Co., 1929.
28 The results of Cipoletti's weir investigations are summarized in Engineering
Record, Vol. 26, No. 11, p. 168, Aug. 13, 1892.
282
FLUID MEASUREMENTS
and considered the negative part as a loss of flow due to the existence
of contraction. He proposed compensating for this loss by cutting
back the corners of the rectangular contracted weir, in effect adding a
half triangular weir at each end. For the proper value of a, the flow
which would be lost by contraction is supplied by the triangular weir
and thus
whence
tan a =
if the two coefficients are assumed equal. Thus to this type of trape
zoidal weir (Fig. 183), the rectangular suppressedweir equation
Q =
FIG. 183. Trapezoidal Weir.
may be applied in which C, determined by Cipoletti, may be taken
as 0.63.
One of the characteristics of
rectangular weirs has been seen
to be the variation of Q with
H 2 > whereas for the triangular
weir Q has been seen to vary
with H*. This leads to the
conclusion that the geometri
cal shape of the weir determines
^ e magnitude of the exponent
an( ^ suggests the possibility of
designing a weir in which Q
FIG. 184. Proportional Weir. varies linearly with H. Such a
weir, termed a " proportional "
or " Sutro " weir, finds wide application in waterlevel control where
a simple relationship between head and rate of flow is desired.
WEIRS
283
Rettger 27 has shown that a proportional weir (Fig. 184) having hori
zontal crest and sides formed of hyperbolas given by the equation
K
results in a theoretical flow equation
and a linear relation between Q and H. Introducing the experimental
weir coefficient, C, the practical equation for the proportional weir
becomes
QCK^VTgH
Broadcrested weirs and spillways occur as overflow devices in
hydraulic structures and are seldom used for measuring purposes;
they are, however, rectangular weirs of special form to which the
v//////////////////////
FIG. 185. BroadCrested Weir.
foregoing rectangular weir equations may be applied. The broad
crested weir (Fig. 185) operates on the criticaldepth principle modified
for friction and flow curvatures and has been treated briefly in Art. 52.
The rectangularweir equation
may be applied in which C ranges from 0.50 to 0.57, depending pri
marily upon the shape of the weir.
27 E. W. Rettger, " A Proportional Flow Weir," Engineering News, Vol. 71, No. 26,
June 25, 1914.
284
FLUID MEASUREMENTS
The "ogee" type of spillway is shown in Fig. 186. Major con
siderations in the design of such a spillway are structural stability
against hydrostatic pressure and other loads and prevention of reduced
FIG. 186. Ogee Spillway.
pressures on the downstream face due to separation of the sheet of
water from this surface. The rectangularweir equation may be
applied to the ogee spillway, the coefficient C ranging from 0.60 to
0.75. The relatively high value
of C may be explained by a com
parison of a sharpcrested weir
(Fig. 187) and an ogee spillway
designed exactly to fit the cur
vature of the lower side of the
nappe of this weir. Obviously,
with a fixed reservoir surface the
flow over the two structures will
be approximately the same, but
the heads for each structure will
be measured from their respec
tive crests and will, therefore, be
different, the head on the weir being greater than the head on the
spillway. If the coefficient of the weir is C Wt the spillway coefficient,
C a , is seen to be the larger if the rates of flow are equated
FIG. 187.
CURRENTMETER MEASUREMENTS
285
giving
72. CurrentMeter Measurements. The construction of a weir
for measuring the flow in large canals, streams, or rivers is impractical
for many obvious reasons; but existing spillways whose coefficients
are known may frequently serve as measuring devices. However, the
standard method of riverflow measurement is to measure the velocity
by means of a current meter '(Art. 62) and integrate the results as for
Pitottube measurements in a pipe line (Art. 68).
.05.25d
FIG. 188. Average Velocity Distribution in a Vertical in Open Flow.
Fundamental to the use of a current meter is a knowledge of the
properties of velocity distribution in open flow. As in pipes, the
velocities are reduced at the banks and bed of the channel, but it must
be realized that in open flow the roughnesses and turbulences are of such
great and irregular magnitudes that the velocitydistribution problem
cannot be placed on the precise basis which it enjoys in pipe flow.
However, from long experience and thousands of measurements, the
United States Geological Survey has established certain average
characteristics of velocity distribution in streams and rivers which
serve as a basis for currentmeter measurements. These characteristics
of velocity distribution in a vertical are shown in Fig. 188 and may be
286
FLUID MEASUREMENTS
amplified by the following statements: (1) the curve may be assumed
parabolic; (2) the location of the maximum velocity is from 0.05 d
to 0.25 d below the water surface; (3) the average velocity occurs at
approximately 0.6 d below the water surface; (4) the average velocity
is approximately 85 per cent of the surface velocity; (5) a more
accurate and reliable means of obtaining the average velocity is by
taking a numerical mean of the velocities at 0.2 d and 0.8 d below the
water surface. The above average values will naturally not apply
perfectly to a particular stream or river, but numerous measurements
with the current meter will tend toward accurate results since devia
tions from the above average values will tend to compensate, thus
giving a greater accuracy than can be obtained in individual measure
ments.
Numerous currentmeter measurements, always required in the
calculation of the flow in a stream or river, are usually taken in the
following manner. A reach of river is selected having a fairly regular
FIG. 189. Division of River Cross Section for Current Meter Measurements.
cross section. This cross section is measured accurately by soundings.
It is then divided into vertical strips of equal width (Fig. 189), the
current meter is suspended, and velocities are measured at the two
tenths and eighttenths points in each vertical (1, 2, 3, etc., Fig. 189).
From these measurements the average velocities (Fi, F, FS, etc.) in
each vertical may be calculated. The average velocity through
each vertical strip is taken as the mean of the average velocities in the
two verticals which bound the strip, and thus the rates of flow (Q\
(?23 etc.) through the strips may be calculated from
2>
and the total flow in the stream may be calculated by totaling the
rates of flow through the various strips.
BIBLIOGRAPHY 287
73. Float Measurements. The velocities of surface floats may
sometimes be used under satisfactory flow conditions to obtain rough
measurements of river flow, but floats of this type are subject to the
vagaries of winds and to local surface currents which may drive them
far off their courses. The reach of river selected for float measure
ments should be straight and uniform and should have a minimum of
surface disturbances; measurements should be taken on a windless
day. The time for the floats to travel a certain distance may be
measured easily, and from this the surface velocities may be computed
and the average velocities approximated by using the relationships of
Fig. 188. It should not be inferred, however, that even under ideal
conditions the accuracy of float measurements is high. This is due to
the abovementioned general difficulties and to the fact that the ratio
of mean velocity to surface velocity, although having an average value
of 0.85, may be as low as 0.80 or as high as 0.95 and quite unpredictable
for a given reach of river.
BIBLIOGRAPHY
DENSITY MEASUREMENT
A.S.M.E. Power Test Codes, Series 1929, Part 16.
VISCOSITY MEASUREMENT
A.S.M.E. Power Test Codes, Series 1929, Part 17.
R. A. DODGE and M. J. THOMPSON, Fluid Mechanics, McGrawHill Book Co., 1937.
E. C. BINGHAM, Fluidity and Plasticity, McGrawHill Book Co., 1922.
MEASUREMENT OF PRESSURE, VELOCITY, AND RATE OF FLOW
Handbuch der experimental Physik, Band IV, Teil 1. Akademische Verlagsgesell
schaft, Leipzig, 1932.
MEASUREMENT OF PRESSURE AND VELOCITY
L. PRANDTL and O. G. TIETJENS. Applied Hydro and Aerodynamics, McGrawHill
Book Co., 1934.
C. M. ALLEN and L. J. HOOPER, "Piezometer Investigation," Trans. A.S.M.E.,
Vol. 54, 1932.
K. H. BEIJ, "Aircraft Speed Instruments," N.A.C.A. Kept. 420, 1932.
MEASUREMENT OF RATE OF FLOW
Fluid Meters, Their Theory and Application, A.S.M.E., 1937.
H. DIEDERICHS and W. C. ANDRAE, Experimental Mechanical Engineering, Vol. I,
John Wiley & Sons, 1930.
VENTURI METER
C. HERSCHEL, "The Venturi Water Meter," Trans. A.S.C.E., Vol. 17, 1887, p. 228.
288 FLUID MEASUREMENTS
WEIRS
H. ROUSE, Fluid Mechanics for Hydraulic Engineers, McGrawHill Book Co., 1938.
J. G. WOODBURN, "Tests of Broad Crested Weirs," Trans. A.S.C.E., Vol. 96, 1932,
p. 387.
CURRENTMETER MEASUREMENTS
W. A. LIDDELL, Stream Gaging, McGrawHill Book Co., 1927.
PROBLEMS
405. A pycnorneter weighs 100 grams when empty and 420 grams when filled
with liquid. If its volume is 200 cc, calculate the specific gravity of the liquid.
406. A plummet weighs 400 grams in air and 300 grams in a liquid. If the
volume of liquid displaced by the plummet is 120 cc, what is the specific gravity of
the liquid?
407. A crude hydrometer consists of a cylinder of f in. diameter and 2in. length
surmounted by a cylindrical tube ^in. in diameter and 8 in. long. Lead shot in the
cylinder brings the hydrometer's total weight to 0.3 oz. What range of specific
gravities may be measured with this hydrometer?
408. To what depth will the bottom of the hydrometer of the preceding problem
sink in a liquid of specific gravity 1.10?
409. Mercury is placed in an open Utube and liquid is poured into one of the
legs. A liquid column 10in. high balances a mercury column 1.5in. high. What is
the specific gravity of the liquid?
410. Water is placed in an open Utube and oil (sp. gr. < 1) is poured into each
leg. The water column is 6in. high, one oil column 3in., and the other 10in. What
is the specific gravity of the oil?
411. A Stormer type viscometer consists of two cylinders, one of 3.0in. outside
diameter, the other of 3.1 in. inside diameter, both 10in. high. A 1lb weight falls
5 ft in 10 sec, its supporting wire unwinding from a spool of 2in. diameter on the
main shaft of the viscometer. If the space between the cylinders is filled with oil
to a depth of 8 in., calculate the viscosity of the oil, neglecting the force on the
bottom of the cylinder.
412. Using the result of problem 29, recalculate problem 411 including the torque
on the bottom of the cylinder, assuming a clearance of 0.05 in. between cylinder
bottoms.
413. A steel sphere (5 = 7.8) \ in. in diameter falls at a constant velocity of
0.3 ft /sec through an oil (S = 0.90). Calculate the viscosity of the oil.
414. What constant speed will be attained by a lead (S = 11.4) sphere 1 in. in
diameter falling freely through an oil of kinematic viscosity, 0.12 ft 2 /sec and specific
gravity 0.95?
415. A Saybolt universal viscometer has tube diameter and length of 0.0693 in.
and 0.482 in., respectively. The internal diameter of the cylindrical reservoir is
1.17 in., and the height from tube outlet to rim of reservoir is 4.92 in. Assuming as
a rough approximation that the loss of head may be taken as the average of the total
heads on the tube outlet at the beginning and end of the run, calculate the relation
ship between v (ft 2 /sec) and t (Saybolt seconds), and compare with the correct
equation relating these quantities.
PROBLEMS 289
416. It takes 80 sec for 60 cc of an oil of specific gravity 0.95 to escape from
a Saybolt viscometer during a routine viscosity test. What is the viscosity of this
oil?
417. The disk of Fig. 137 and a Pitot tube are placed in an air stream aligned
properly with the flow and connected to a Utube containing water. If the difference
of water elevation in the legs of the manometer is 4 in., calculate the air velocity,
assuming w = 0.0763 lb/ft 3 .
418. A Pitotstatic tube on an airplane is connected to a differential manometer
which reads 3 in. of water when flight occurs through still air (14.7 lb/in. 2 and 60 F).
Calculate the speed of the airplane in miles per hour.
419. If the Pitot tube of the preceding problem is connected to a sensitive differ
entialpressure gage which reads 0.08 lb/in. 2 , calculate the speed of the airplane.
420. An airplane is designed to have a top speed of 250 mph when flying through
still air of specific weight 0.0763 lb/ft 3 . What will be the largest pressure difference
recorded between the stagnation and static pressure openings of its Pitot tube?
421. A ijin. smooth nozzle is connected to a horizontal 3in. pipe in which the
pressure is 60 lb/in. 2 Calculate the stagnation pressure in the pipe and in the nozzle
stream.
422. A Pitometer (Ci  0.85) is placed at a point in a water line. If the attached
differential manometer containing mercury and water shows a reading of 5 in., what is
the velocity at the point?
423. A Pitot tube is placed at the center of a 6in. pipe in which carbon tetra
chloride flows at 68 F. The attached differential manometer containing mercury
and carbon tetrachloride shows a difference of 3 in. What flow exists in the line?
424. The pipeline Pitot tube of Fig. 147 is installed on the center of a 12in. water
line and connected to one end of a Utube manometer containing carbon tetrachloride.
The other end of the manometer is connected to the pipe wall. If water fills the
manqmeter tubes above the CCU and the manometer reads 10 in., what is the
velocity at the center of the pipe?
425. A Venturi tube of d\ = \\ in., d% = f in., and C = 0.95 is installed on the
airplane of problem 420. Calculate the pressure difference PQ fa. Calculate the
pressure difference p a pz created by a Pitot Venturi installed on this airplane.
Compare answers.
426. A Venturi tube having C = 0.95 and Ai/A% = 3 has an openended water
manometer connected to section 2. If the manometer reads 12 in. and the open end
is at atmospheric pressure, what is the velocity of the tube through still air of specific
weight 0.0763 lb/ft 3 ?
427. The gasometer of Fig. 158 is used to measure the flow of hydrogen at 65 F.
The internal diameter of the cover is 5 ft, and it rises 1.87 ft in 34.6 sec. The manom
eter reads 16 in. of water, and the barometric pressure is 14.66 lb/in. 2 Calculate
the weight rate of flow. If the counterweight weighs 600 Ib, what is the weight of
the cover?
428. A 12 in. by 6 in. Venturi meter is installed in a horizontal water line. The
pressure gages read 30 lb/in. 2 and 10 lb/in. 2 Calculate the flow, assuming C v = 0.97.
Calculate the loss of head between base and throat of the meter.
429. If the meter of the preceding problem is in a vertical pipe line with throat
2 ft below the base, calculate the rate of flow.
430. Oil (5 = 0.90) flows through a 12 in. by 6 in. horizontal Venturi meter.
The attached differential manometer contains mercury (and oil to the mercury
290 FLUID MEASUREMENTS
surfaces) and shows a difference of 10 in. Calculate the rate of flow if C v = 0.97.
Calculate the loss of head between base and throat of meter. If the cone angle of
the diffusor tube is 7, calculate the total lost head through the meter.
431. If the meter of the preceding problem is in a vertical pipe line with throat
2 ft below the base, calculate the rate of flow.
432. Linseed oil flows through a horizontal 6 in. by 3 in. Venturi meter. What is
the difference in pressure head between base and throat of the meter when 120 gpm
flow at (a) 80 F, (b) 120 F? What is the head loss for these two cases?
433. The maximum flow expected through an 18 in. by 9 in. Ventun meter
installed in an 18in. line is 15 cfs of water at 80 F. How long a manometer is
necessary for this installation if the manometer is to contain mercury?
434. Calculate the weight flow of air through a 4 in. by 2 in. Venturi meter when
the gage pressures at base and throat of meter are 40 lb/in. 2 and 30 lb/in. 2 The
barometer reads 29.5 in. of mercury, the temperature of the air as it enters the meter
is 100 F, and C v 0.985.
435. Carbon dioxide flows through a 6 in. by 2 in. Venturi meter. Gages at base
and throat read 20 lb/in. 2 and 14 lb/in. 2 , and temperature at the base of the meter
is 80 F. Calculate the weight flow, assuming standard barometer and C v = 0.99.
436. Calculate the weight flow in the preceding problem when the throat pressure
gage reads (a) 10 lb/in. 2 , (b) 2 lb/in. 2
437. A 3in. A.S.M.E. longradius flow nozzle is installed in a 6in. water line.
The attached manometer contains mercury and registers a difference of 15 in. Calcu
late the flow through the nozzle. Calculate the head lost by the nozzle installation.
438. A 2in. A.S.M.E. longradius flow nozzle is installed in a 5in. pipe line
where linseed oil is flowing. The attached differential manometer, containing mer
cury, registers a difference of 8 in. Calculate the flow through the nozzle and the
lost head caused by the nozzle installation.
439. If air flows through the pipe and nozzle of the preceding problem, open
mercury manometers at points 1 and 2 show positive gage pressures of 30 in. and
20 in., and the temperature at point 1 is 60 F, calculate the weight rate of flow,
assuming standard barometric pressure.
440. A 1in. fire nozzle has C v = 0.98 and C c = 1.00, and is attached to a 3in.
hose. What flow will occur through the nozzle when the pressure in the hose is
60 lb/in. 2 What is the velocity of the nozzle, stream? What head is lost through
the nozzle? To what height will this stream go, neglecting air friction?
441. Assuming an I.S.A. flow nozzle, calculate the rates of flow in: (a) problem
437; (b) problem 438; (c) problem 439.
442. A 2in. nozzle having C v = 0.98 and C c = 0.90 is attached to a 6in. pipe
line and delivers water to an impulse turbine. The pipe line is 1000 ft long, leaving
a reservoir of surface elevation 450 at elevation 420. The nozzle is at elevation 25.
Assuming a sharp pipe entrance and a friction factor of 0.02, calculate: (a) the flow
through the pipe and nozzle ; (b) the horsepower of the nozzle stream; (c) the horse
power lost in line and nozzle.
443. Calculate the flow through a 3in. I.S.A. orifice installed in a 6in. water line
when the attached manometer containing mercury shows a difference of 12 in.
444. Air flows through a 2in. I.S.A. orifice installed in a 6in. pipe line. A pressure
gage upstream from the orifice reads 30 lb/in. 2 , and a differential manometer con
nected between points 1 and 2 shows a difference of 15 in. of mercury. If the tem
PROBLEMS 291
perature of the air upstream from the orifice is 70 F and the barometer reads 14.3
lb/in. 2 , calculate the weight rate of flow.
445. Air flows through a iin. rounded orifice (C v = 0.99) installed in a 6in. pipe
line. Pressure gages upstream and downstream from the orifice read 70 and 20 lb/in. 2
Calculate the weight flow if the barometer is 14.3 lb/in. 2 and temperature upstream
from the orifice is 100 F.
446. Steam flows through a ^in. rounded orifice, having C v = 0.99, installed
in a 3in. pip^ line, the barometer is 14.5 lb/in. 2 , and the temperature upstream from
the orifice is 400 F. Pressure gages above and below the orifice read 60 and 20 lb/in. 2
Calculate the weight rate of flow, taking (p2/pi) c = 0.55.
447. Water discharges into the atmosphere from a 1.5in. sharpedged orifice
under a 5 ft head. Calculate the rate of flow, diameter of tfie jet at the contracted
section, and velocity at this point.
448. Under a 4.42 ft head, 0.056 cfs of water discharges from a 1in. sharpedged
orifice in a vertical* plane ; 3.30 ft outward horizontally from the contracted section
the jet has dropped 0.65 ft below the centerline of the orifice. Calculate C, C Vt and C c .
449. Water flows from one tank to an adjacent one through a 3in. diameter
sharpedged orifice. The head of water on one side of the orifice is 6 ft and on the
other 2 ft. Taking C c = 0.62 and C y = 0.95, calculate the rate of flow.
450. A 3in. sharpedged orifice discharges vertically upward. At a point 10 ft
above the contracted section, the diameter of the jet is 3 in. Under what head is the
orifice discharging?
451. A Pitot tube is placed in the contracted section of a jet from a 2in. orifice
operating under a 6ft head. The Pitot tube is connected to a piezometer column
in which water stands at a level 1.50 in. below that in the tank. Calculate C v for
the orifice.
452. A sluice gate 4 ft wide is open 3 ft and discharges onto^a horizontal surface.
If the coefficient of contraction is 0.80 and the coefficient of velocity 0.90, calculate
the rate of flow if the upstream water surface is 15 ft above the top of the gate opening.
453. Calculate the rate of flow in the preceding problem if tailwater stands 10 ft
deep over the top of the gate opening.
454. An open vertical cylindrical tank 20 ft high and 5 ft in diameter contains
a valve in the bottom which is connected to a short piece of vertical 3in. pipe which
discharges into the atmosphere at a point 4 ft below the bottom of the tank. If the
tank is full of water and the valve opened, how much time is required to reduce the
water depth to 8 ft? Treat the pipe and valve as a 3in. orifice having C c *= 1.00
and C v = 0.65.
455. A cylindrical tank contains a 1in. orifice 4 in. above the bottom. If the tafk
is 2 ft in diameter and it requires 65 sec for the water depth to drop from 4 f t td 3 ft,
calculate the discharge coefficient of the orifice.
456. How much time is required to drain a full conical water tank (apex down)
10 ft high and having a base diameter of 4 ft through a 2in. orifice (C c = 0.80 and
C v = 0.98) in the apex?
457. Solve the preceding problem assuming a square pyramid with base having
4ft sides.
458. A Vshaped tank is 10 ft long, 5 ft deep, and 4 ft wide at the top. A slot at
the point of the V is 1 in. wide and runs the full length of the tank. How much time
is required to drain the tank from the full condition, assuming that C c 0.85 and
C v 0.98 for the slot.
292 FLUID MEASUREMENTS
459. A flow bend consisting of a 4in. flanged elbow has a coefficient Ck of 1.50.
What flow of water occurs through this bend when the attached manometer (Fig. 171)
contains mercury and shows a difference of 10 in.?
460. A Pitometer (C/ = 0.85) is installed in a 6in. water line. A manometer
containing CCU and water, connected to the Pitometer, shows the following readings
when the tip of the instrument is placed at the points specified for the numerical
average method for calculating rate of flow.
Pitometer location... 1 2 3 C 4 5 6
Manometer readings,
In 1.20 2.04 2.83 3.03 2.89 2.10 1.26
Calculate the rate of flow, pipe coefficient, and the distance from pipe centerline to
station 2.
461. A Pitot tube is placed at various points along a diameter: of a 20in. pipe in
which water is flowing. The pressure difference is measured on a manometer con
taining mercury and water. If the following manometer readings are taken, calculate
the rate of flow and pipe coefficient by the graphical method.
Distance from Pitot
tube location to pipe
centerline, in 2 4 6 8 9 9.5
Manometer reading, in. 6 5.95 5.64 5.07 4.12 3.25 2.45
462. The apparatus of Fig. 174 is installed in an insulated 6in. pipe line in which
air is flowing. A potential of 110 volts is maintained across the 30ohm resistance,
and the thermometers read 100 and 105 F. The gage pressure in the pipe is 50 lb/in. 2
If the specific heat of the air is 0.24 Btu/lb/ F, what weight flow exists in the line?
463. The flow in a brook is measured by the saltdilution method; 0.20 gpm of
salt solution having a concentration of 20 Ib salt/gal are introduced and mix with
the flow. A sample extracted below the mixing point shows a concentration of
0.00008 Ib/gal. Calculate the flow in the brook.
464. The saltvelocity method is to be used on a 24in. pipe line, and electrodes
are installed 100 ft apart. The time between deflection of the ammeter needles is
12.0 sec. Calculate the flow in the line.
465. Calculate the flow over a rectangular sharpcrested suppressed weir 4 ft long
and 3 ft high when the head thereon is 4 in., using the coefficients of (a) Bazin,
(b) Rehbock, (c) Swiss Society, (d) Frese. What is the velocity of approach in (a)?
(Use P/H = oo in finding C.)
466. If the weir of the preceding problem is only 4 in. high, calculate the flows
and velocity of approach.
467. What depth of water must exist behind a rectangular sharpcrested sup
pressed weir 5 ft long and 4 ft high when a flow of 10 cfs passes over it? What is the
velocity of approach? Use Rehbock C.
468. A rectangular channel 18 ft wide carries a flow of 50 cfs. A rectangular
suppressed weir is to be installed near the end of the channel to create a depth of
3 ft upstream from the weir. Taking C  0.62, calculate the necessary weir height.
469. A sharpcrested rectangular contracted weir 6 ft long measures the
outflow from a small pond. If C = 0.623, what is the flow over the weir when the
head is 0.816 ft?
PROBLEMS 293
470. A rectangular contracted weir is to be used to maintain a depth of 4 ft in
a channel 15 ft wide where the flow is 12 cfs. Taking a coefficient of 0.62, what
length and height of weir crest are required?
471. Derive the theoretical flow equation for the triangular weir.
472. Calculate the flow over a smooth sharpcrested triangular weir of 90 notch
angle when operating under a head of 7 in. according to (a) Thomson, (b) Barr.
473. A triangular weir of 90 notch angle is to be used for measuring flows up to
1.5 cfs. What is the minimum depth of notch which will pass this flow?
474. A triangular weir has a 60 notch angle. What is the flow over this weir
under a 9in. head if the coefficient is 0.57?
475. What length of Cipoletti weir is required for a flow up to 20 cfs if the maxi
mum head is limited to 8 in.?
476. Calculate the flow over a Cipoletti weir 12 ft long when the head thereon
is 0.783 ft.
477. If a proportional weir is to be designed for a maximum flow of 5 cfs under
a head of 3 ft, what is the width of its notch 1.5 ft above the crest, taking C = 0.60?
478. A proportional weir is 3 in. wide at a height of 2 ft above the crest. What
rate of flow will occur under a head of 4 ft if C is taken as 0.62?
479. What flow will occur over a spillway of 500ft length when the head thereon
is 4 ft if the coefficient of the spillway is 0.72?
480. A spillway 1000 ft long is found by model experiments to have a coefficient
of 0.68. It has a crest elevation of 100.00. When a flood flow of 50,000 cfs passes
over the spillway what is the elevation of the water surface just upstream from the
crest?
481. A broadcrested weir has a flat crest and a coefficient of 0.55. If this weir
is 20 ft long and the head on it is 1.5 ft, what flow will occur over it? What is the
maximum flow that could be expected if flow were frictionless?
482. A rectangular channel 20 ft wide carries 100 cfs at a depth of 3 ft. What
height of broadcrested suppressed weir must be installed to double the depth?
C  0.56.
483. The following data are collected in a currentmeter measurement at the
river cross section of Fig. 189 which is 60 ft wide at the water surface. Assume
V = 2.22 X (rps), and calculate the flow in the river.
Station.... 0123456789 10 It 12
Depth, ft.. 0.0 3.0 3.2 3.5 3.6 3.7 3.9 4.0 4.4 4.4 4.2 3.5 0.0
Rpm of rotating element
0.2 d 40.0 53.5 58.6 63.0 66.7 61.5 56.3 54.0 52.6 50.0 45.0 ....
0.8 d 30.7 42.8 50.0 54.2 58.8 53.3 49.4 46.5 43.2 40.1 32.5 ....
CHAPTER IX
FLOW ABOUT IMMERSED OBJECTS
Problems involving the forces exerted on a solid body when fluid
flows by it no longer belong exclusively to the aeronautical engineer.
In the design of ship hulls, automobile bodies, and trains, minimizing
fluid resistance or drag has become of increasing importance; in the
design of ship propellers, turbines, and centrifugal pumps, the princi
ples of lift are being applied with increasing success. As these princi
ples find increasingly wide application, it becomes necessary for all
engineers to be familiar with the fundamental mechanics of these
principles. The origins of drag have been discussed briefly in Art. 28,
and it is the purpose of this chapter to expand the treatment of the
subject and to outline the elementary principles of lift as well.
74. Fundamentals and Definitions. In general when fluid flow
occurs about an object which is either unsymmetrical or whose axis
is not aligned with the flow, the velocities on either side of the object
have different magnitudes. From the streamline picture about the
foil of Fig. 190a, it is obvious at once that the velocity is higher on
the upper side of the foil than on the lower, and hence from the Ber
noulli principle the pressure on the upper side is less than on the lower
side. Further examination of the flow picture indicates the pressure
on the upper side of the foil to be less than, and that on the lower side
greater than, the pressure, p , in the undisturbed fluid stream; in
other words, there is a pressure reduction, or * 'suction/* on the upper
side of the foil and in increase of pressure on the lower side. Desig
nating the suction by arrows drawn away from the foil and pressure
increase by arrows drawn toward the foil, the distribution of pressure
over the surface of the foil becomes as shown in Fig. 1906. 1
Along with these pressures which are exerted everywhere normal
to the surface of the foil, there are, of course, tangential shear stresses
T as well, acting on the foil in a downstream direction and resulting
1 Note that the larger pressure diagram on the top of the foil indicates that the
larger part of the force on the foil is contributed by pressure reduction on the upper
side.
294
FUNDAMENTALS AND DEFINITIONS
295
from the frictional effects which exist when fluids flow over solid
boundaries (Art. 26),
The resultant force, F, exerted by fluid on foil by the normal
(pressure) and tangential (frictional) stresses will have a direction as
shown in Fig. 19(k and may be resolved into components parallel and
perpendicular to the direction of the undisturbed velocity, 2 V. The
former component, A is termed the "drag" (force) and the latter
one, L, the "lift" (force) on the foil. Obviously, both these force
components embody the effects of both normal and tangential stresses
exerted by fluid on foil. The effect of normal and tangential stresses
on drag has been discussed in Art. 28 and resulted in the definition of
"profile" and "frictiona!" drag. The effect of tangential stresses
upon the lift force may, however, be safely neglected because the
2 Or, stating this in another way, parallel and perpendicular to the direction of
motion of the foil through still fluid.
296 FLOW ABOUT IMMERSED OBJECTS
tangential stresses r not only are small but also act in a direction
roughly normal to that of the lift force and thus contribute little to it;
thus the lift force exerted on the foil may be safely considered to result
from pressure variation alone.
The width of the foil, , is called the "chord;" its length, 6, per
pendicular to the plane of the paper, the "span;" and the angle a
between chord and direction of the undisturbed velocity, V, the
"angle of attack."
The lift and drag are calculated by
L = C L A ^~ (93)
and
D = C D A P ~ (94)
in which CL and CD are the (dimensionless) lift coefficient and drag
coefficient of the foil, and A the area of the projection of the airfoil on
the plane of the chord. 8 The coefficients of lift and drag may be found
experimentally by windtunnel or flight tests; their magnitudes
obviously depend upon the shape of the foil and (among other vari
ables) upon the angle of attack.
ILLUSTRATIVE PROBLEM
The lift and drag coefficients of a rectangular airfoil of 50f t span and 7ft chord
are 0.6 and 0.05, respectively, when at an angle of attack of 7. Calculate the
horsepower required to drive this airfoil through still air (w  0.0763 lb/ft 3 ) at
150 mph. What lift is obtained when this horsepower is expended?
' 0763 0.00237 slugs/ft*
32.2
150 X 5280
3600
= 220 ft/sec
D = CDA . o.05 X (50 X 7)  1006 Ib
L t
1006 X 220 _
550
8 If the plan form of the foil is rectangular, A = b X c.
DIMENSIONAL ANALYSIS OF THE DRAG PROBLEM 297
75. Dimensional Analysis of the Drag Problem. The general
aspects of fluid resistance on immersed bodies and the properties of
the drag coefficient may be examined to advantage by dimensional
analysis, before considering the physical details of the problem.
Fluid properties />, ft, E
FIG. 191.
The smooth body of Fig. 191, having area 4 A, moves through a
fluid of density p, viscosity ju> and modulus of elasticity E, with a
velocity V. If the drag on the body is D,
and because of dimensional homogeneity,
D = CA a p b n c V d E*
Writing the equations dimensionally
ML _ V (M\ b (M\ (L
J^~ (L) \L 3 ) \LT) \T
the equations of exponents of M, L y and T becomes
M: 1 = b + c + e
L: l = 2a 3b  c + d e
T: 2 = c  d  2e
Solving for 6, d, and a in terms of c and e,
b = 1 c e
d ~ 2  c  2e
4 Any convenient significant area may be used.
298 FLOW ABOUT IMMERSED OBJECTS
Resubstituting these above
and rearranging
D = CA 2 p l  c " e
But, referring to Art. 30,
A*pV V 2 p
= NR and = NC
n E
but (Art. 4)
 = 2
P
Therefore N c = V 2 /c 2 , in which V/c is known as NM, the Mach
number. Substituting these values above, the drag equation may be
written
ApV 2
and by comparison with equation 94
CD =f(N R ,N M ) (95)
This equation indicates: (1) that bodies of the same shape and having
the same alignment with the flow possess the same drag coefficients
if their Reynolds numbers and Mach numbers are the same ; or (2) that
the drag coefficient of bodies of given shape and alignment depend upon
their Reynolds and Mach numbers. Thus, dimensional analysis has,
as in previous problems (ship resistance and pipe friction), opened the
way to a comprehensive treatment of the resistance of immersed bodies
by indicating the dimensionless combinations of variables upon which
the drag coefficient depends.
Although the drag coefficient is theoretically dependent upon both
the Reynolds and Mach numbers simultaneously, this is seldom true
in actual practice because the Reynolds number (containing the effect
of viscosity) affects the size of the drag coefficient only at relatively low
speeds where the Mach number is small and compression of fluid by
body usually negligible; on the other hand, when velocities approach
or exceed that of sound (NM approaching or greater than unity),
compression of fluid by body contributes the major part of the drag
DIMENSIONAL ANALYSIS OF THE DRAG PROBLEM 299
force whereas here the contribution of viscosity is very small. This
reasoning divides the study of drag force into two separate physical
problems, one involving velocities well below the acoustic velocity,
and the other velocities near to or exceeding this velocity. In the first,
the variation in fluid density may be neglected and laminar or turbulent
flow conditions and drag coefficients are governed by the Reynolds
number; this is the field of presentday aerodynamics where the
velocities involved are usually well below the acoustic velocity. The
second case embraces the field of high velocity motion in which flow
is highly turbulent, but in which the Mach number governs flow
patterns and drag coefficients since these result primarily from fluid
compression and not from viscous action. The motion of bullets and
projectiles through air and the motion of airplane propeller tips at
high speeds are governed by compressibility and the Mach number.
A complete treatment of drag includes "drag at low velocities" and
"drag at high velocities." For the first, frictional and pressure forces
must both be considered; they have been seen to result in frictional
drag and profile drag respectively. For drag at high velocities, fric
tional forces are neglected and a study of the topic is one of pressure
effects, which leaves only profile drag to be considered. These three
subjects will be treated in detail in the above order in the succeeding
articles.
ILLUSTRATIVE PROBLEM
An airfoil of 6ft chord moves at 300 mph through still air (14.7 lb/in. 2 abs and
60 F). Calculate the Reynolds and Mach numbers.
Tr 300 X 5280 . .
y =  = 440 ft/sec
3600
p = = 0.00237 slugs/ft 3
32.2
fji = 0.000000375 Ib sec/ft 2
= 440 X 6 X 0.00237 =
0.000000375
Ikp ll
m */ = */
\ \
p 0.00237
 0.393
1120
A X 14.7 X 144
= 1120 ft/sec
300
FLOW ABOUT IMMERSED OBJECTS
76. Frictional Drag. 5 To discover the essential properties of
frictional drag, consider the drag force exerted on one side of a smooth
flat plate aligned with the flow ; such a surface eliminates geometrical
complexities and possesses no profile drag.
As fluid flow occurs over the flat plate of Fig. 192, the viscosity
causes the velocity to be zero at every point on the surface of the plate
whereas the velocity at a very small distance d from this surface is the
undisturbed velocity V. Thus, a very thin layer of fluid, the boundary
layer, containing a velocity gradient, forms over the surface of the
plate; since shear stress in fluids depends upon the existence of a
velocity gradient, it is immediately apparent that resistance effects
are confined to the boundary layer and must, therefore, be dependent
upon the characteristics of this layer.
Turbulent
FIG. 192. Boundary Layers on a Flat Plate.
The layer must start from no thickness at the leading edge of the
plate where viscous action begins and must increase its thickness in
a downstream direction as the increasing viscous action exerted by
increasing plate area extends into the flow and reduces the velocity of
more and more fluid. Near the leading edge of the plate where the
boundary layer is thin and contains small quantities of fluid subject
to high viscous influence, the flow within the layer will be laminar,
but, as the layer becomes thicker and includes more fluid mass, insta
bility results and flow within the layer becomes turbulent. The change
from a laminar to a turbulent boundary layer is not, however, an
abrupt one, but occurs through a transition region in which both
viscous and turbulent action are present; viscosity effects in the
transition region are finally replaced by those of turbulence, and
a wholly turbulent boundary layer results (Fig. 192).
6 The student should restudy Art. 28.
FRICTIONAL DRAG
301
Now consider, for simplicity, the laminar boundary layer (Fig. 193)
having a thickness 6 at a distance x from the leading edge of a plate of
unit width, and apply the impulsemomentum law to an element of
this layer of length, dx. The net applied (drag) force, dD, is given by
dD = rodoc
The flow dQ drawn into the boundary layer in distance dx is of the
order Vd5, and the reduction AFof its velocity is proportional to V.
Thus
JClan
AF~ VddpV
or
dQw
AF = aVddpV
in which a is the factor necessary to make an equality of the above
approximate relations and will be a constant if the velocity profiles
JL i
B+dS
FIG. 193. Laminar Boundary Layer.
throughout the laminar boundary layer are of similar shape. Equating
net force dD and change of momentum per unit time,
and substituting the above values
apV 2 dd
(96)
302 FLOW ABOUT IMMERSED OBJECTS
Now, since the flow is laminar, the shear stress T O is given by
(dv\
r ** **(T)
Wy/Burface
But if the velocity profiles are all of similar shape (dv/dy) Bur t &ce will be
a fixed proportion, /3, of a linear velocity gradient V/d. Therefore,
TO = fti T
d
which may be substituted in equation 96 to give
y
/3p, dx = apV 2 d8
o
/
So
which may be rearranged and integrated
X  s+&
dx = / ddd
giving the relation between x and 5
>IA 2 (97)
thus showing the contour of the laminar boundary layer to be para
bolic in shape. This equation may be rearranged and put in dimen
sionless form by solving it for S/#, giving
(98)
and indicating this ratio to be inversely proportional to the Reynolds
number (Nn) 5 calculated with 8 as the length parameter. Blasius 6
has shown theoretically and experimentally that 2/3/a = 27.0, and
thus the above equation becomes
6 27.0
8 H. Blasius, "Grenzschichten in Fliissigkeiten mit kleiner Reibung," Zeit. f.
Math. w. Physik, Vol. 56, 1908, p. 1.
FRICTIONAL DRAG 303
However, this equation is obviously not convenient for calculation of
6 for a certain numerical value of x, because 5 occurs on both sides of
the equation; it may be placed in more usable form by substituting
for 6, in the Reynolds number, its value obtained in terms of x from
equation 97. This is
a
and substituting in equation 98
20
6 a
* %L J^ZE Vx f>
^ (X pV (JL M
and since
> VXD
^ = V27.0 = 5.20, and letting  = N R
Gi (A
d 5.20
which shows that the shape of the laminar boundary layer an inertia
viscosity phenomenon is dependent, as might have been expected,
only on a Reynolds number.
The drag coefficient and total drag of this flat plate now remain
to be calculated. Expressing the differential drag force dD on the
element of plate of length dx in the conventional way (equation 94),
pV 2
r dx = dD = Cf dx
L
giving
$ (100)
in which c/ is the (frictional) drag coefficient for the element of plate dx.
Since r can be shown to decrease with increasing x, Cf varies with x in
the same way. However, a drag coefficient which differs for every
element of plate is obviously inconvenient, so a more practical average
frictional drag coefficient Cf will be derived for the plate length x, so
that the total drag D for the length x will be given by
304 FLOW ABOUT IMMERSED OBJECTS
But total drag for the plate length x is also given by
C C" pV 2
D = I dD = Cfdx^
JQ J O ^
Equating the above two expressions for D
r
I Cfdx (101)
' Jo
and with this equation C/ may be easily derived after an expression
for / is found. If
r = ft* 7
is substituted in equation 100
and substituting
apV
gives
in which V 2a^3 has been shown by Blasius to have the value 0.664.
Thus
0.664
c f =
and substituting this expression in equation 101
'* 0.664 J
. dx
V~ 1.328 1.328
' "" * IVxp
^
FRICTIONAL DRAG 305
This equation allows the frictional drag on flat plates of length x to be
calculated when the plates are placed in a flow of velocity V, density p,
and viscosity /x, provided that the boundary layer remains in laminar
condition. It should be noted that the results of the above physical
treatment of drag confirm and amplify those of dimensional analysis
which showed that drag coefficients of smooth bodies at low velocities
depend only upon Reynolds numbers (equation 95).
A flat plate with a turbulent boundary layer may be analyzed
physically by methods similar to those above, but the concepts and
mathematics involved are too advanced for inclusion in an elementary
textbook. Prandtl 7 has shown that the equation
0.455
' (logAW 2  68
gives the relation between C/ and NR for a flat plate with turbulent
boundary layer and has also shown that the approximate thickness of
such a turbulent layer is given by
d 0.37
A plot of Cf against NR for smooth flat plates (Fig. 194) with
laminar and turbulent boundary layers bears a striking resemblance
to that of friction factor against Reynolds number for circular pipes
(Fig. 86). However, the critical Reynolds number at which the
laminar boundary layer changes to a turbulent one is not so well
defined as its counterpart in pipe flow, because of flow conditions
which are not so well controlled. With increased initial turbulence
in the approaching fluid flow, earlier breakdown of the laminar bound
ary layer occurs, thus reducing the critical Reynolds number; rough
ening the leading edge of the plate has also been found to decrease the
critical Reynolds number by decreasing the flow stability and causing
earlier breakdown of the laminar layer. A typical equation which
satisfies experimental results in the transition region and which deter
mines the critical Reynolds number is that suggested by Prandtl from
Gebers' tests on a smooth flat plate. This equation is
0.455 1700
2.58
' (log N R )
7 L. Prandtl, Ergebnisse der aerodynamischen Ver$uch$o,nstalt zu Gottingen, IV,
1932, p. 27, R. Qldenbourg.
306
FLOW ABOUT IMMERSED OBJECTS
giving a critical Reynolds number of 530,000. This figure, however,
should be taken only as a typical value ; the range of experimentally
determined critical Reynolds numbers is approximately from 100,000
to 1,000,000.
From the above statements, and from the plot of Fig. 194, there is
obviously great uncertainty in the selection of values of Cf for Rey
nolds numbers of less than 10,000,000. Above this figure there is
ample experimental confirmation of Prandtrs equation.
5 10 9
FIG. 194. Drag Coefficients for Smooth Flat Plates.
ILLUSTRATIVE PROBLEM
A smooth rectangular plate 3 ft wide and 100 ft long moves in the direction of
its length through water (68 F) at 30 ft/sec. Calculate the drag force on the
plate and the thickness of the boundary layer at the trailing edge of the plate.
Vxp 30 X 100 X 1.935
N R = =B _ _ 23
/* 0.000021
0.455
(log 276,500,000) 2  68
pF 2
 0.00186
Total drag (2 sides of plate)  2C f A
D  2 X 0.00186 X (100 X 3)
5 0.37
1.935 X 30 2
972 Ib
100 (276,5QO,000) 2
0.0076, d  0.76 ft
PROFILE DRAG
307
77. Profile Drag. 8 Profile drag has been shown to be that part of
total drag resulting from pressures over the surface of an object and
to be dependent on the formation of a wake behind the object. In
general, when wakes are large profile drag is large, and when wakes
are reduced by streamlining profile drag is reduced also.
Separation point
FIG. 195. Flow about a Sphere at Various Reynolds Numbers.
The properties of profile drag can best be obtained by examining
the details of flow about a blunt object such as a sphere which has an
appreciable and variable wake width, but on which the frictional drag
may be neglected because of the small surface area on which frictional
effects can act.
8 The student should restudy Art. 28.
308 FLOW ABOUT IMMERSED OBJECTS
Even for a blunt object, however, profile drag is not always pre
dominant or frictional drag negligible since flow at very low Reynolds
numbers about a blunt object will close behind the object and no wake
will form (Fig. 195a). Under these conditions total drag is composed
primarily of frictional drag. Stokes 9 has shown that, in laminar flow
at very low Reynolds numbers, where inertia forces may be neglected
and those of viscosity alone considered, the drag of a sphere ot diameter
d, moving at a velocity V through a fluid of viscosity M> is given theo
retically by
D
and this equation has been confirmed by many experiments. The
drag coefficient CD for the sphere under these conditions may be found
by equating the above expression to equation 94
r 4 pV2 * n
Lj)A = OTTJU Va
Taking A as the crosssectional area of the sphere at the center,
A =
4
and substituting this above,
whence
CD ~T ^~ ^ 3717* Fd
c = = 
D Vdp N R
Thus the drag coefficients of spheres at low velocities are dependent
only on the Reynolds number, again confirming the results of the
dimensional analysis of Art. 75.
As the Reynolds number increases, the drag coefficients of spheres
continue to depend only upon the size of this number, and a plot of
experimental results over a large range of Reynolds numbers for
spheres of many sizes, tested in many fluids, gives the single curve of
Fig. 196.
9 G. G. Stokes, Mathematical and Physical Papers, Vol. Ill, p. 55, Cambridge
University Press, 1901.
PROFILE DRAG
309
Up to a Reynolds number of unity, the Stokes equation holds and
the drag coefficient results from frictional effects. As the Reynolds
number is increased to about 1, separation and weak eddies begin to
form, enlarging into a fully developed wake near a Reynolds number
of 1000; in this range the drag coefficient results from a combination
of profile and frictional drag, the latter being of negligible size as a
Reynolds number of 1000 is reached. Above this figure the effects of
friction may be neglected and the drag problem becomes one of profile
drag alone.
100
CD 1.0
.01
0,1 2 4 6 1.0 2 4 6 10 2 4 6 10* 2 4 6 10 3 2 4 6 10 4 2 4 6 10* 2 46 10 6
Reynolds Number, N R  %&.
FIG. 196. Drag Coefficients for Sphere, Disk, and Streamlined Body. 10
The profiledrag coefficient of the sphere is roughly constant from
NR ~ 1000 to NR ~ 250,000 at which point it suddenly drops about
50 per cent and stays practically constant for further increase in the
Reynolds number. In the above range of Reynolds numbers, experi
ments have shown the separation point to be upstream from the mid
point of the sphere, resulting in a relatively wide turbulent wake; the
boundary layer on the surface of the sphere from stagnation point to
separation point has been found to be laminar up to NR ~ 250,000.
At this point, the boundary layer becomes turbulent and the separa
10 Data from L. Prandtl, "Ergebnisse der aerodynamischen Versuchsanstalt zu
Gottingen," Vol. II, R. Oldenbourg, 1923, p. 29, and G. J. Higgins, "Tests of the
N. P. L. Airship Models in the Variable Density Wind Tunnel," N.A.C.A. Tech.
Note No. 264, 1927.
310
FLOW ABOUT IMMERSED OBJECTS
tion point moves to a point downstream from the center of the sphere,
causing a decrease in the width of the wake and consequently a decrease
in the drag coefficient.
The sudden shift of the separation point and decrease in the profile
drag coefficient as the boundary layer changes from laminar to turbu
lent are characteristic of all blunt bodies and may be explained by
examination of the energy properties of the laminar and turbulent
boundary layers of Fig. 197. For comparison let these layers be of
the same thickness and have the same undisturbed velocity V. It is
evident from the velocity profiles that the turbulent layer possesses
greater kinetic energy than the laminar one ; this kinetic energy allows
the flow to continue further around the sphere before friction destroys
this energy, bringing the fluid to rest and causing separation. The
increased energy of the turbulent boundary layer thus brings about the
v ^
Laminar Turbulent
FIG. 197. Velocity Profiles in Boundary Layers.
shift in separation point, the decrease in the width of the wake, and
consequent decrease in drag coefficient.
The change from laminar to turbulent boundary layer on a flat
plate has been seen (Art. 76) to occur at a critical Reynolds number
dependent upon the turbulence of the approaching fluid. This also
occurs with a sphere, and with increased turbulence in the approaching
flow the sudden drop in the drag coefficient curve occurs at a lower
Reynolds number. Thus, a sphere may be used as a relative measure
of turbulence by noting the Reynolds number at which a drag coeffi
cient of 0.30 (see Fig. 196) is obtained.
The change from laminar to turbulent boundary layer has also
been seen (Art. 76) to occur (for the same initial turbulence) at a fixed
distance from the leading edge of a flat plate (or after a fixed length of
boundary layer) for given flow conditions (V, p, and M) Applying
this fact in a qualitative way to elongated bodies (e.g., ellipsoids with
the major axes in the direction of flow), it may be concluded that
breakdown of the laminar boundary layer will occur at a lower Rey
nolds number than 300,000, and conversely that ellipsoids with the
major axis normal to the flow will have breakdown of the laminar layer
PROFILE DRAG
311
and decrease of the drag coefficient at Reynolds numbers greater than
300,000. This trend is borne out by experiments but cannot be
carried to extremes. The limit of ellipsoids with major axis normal
to the flow is one of zero minor axis, i.e., a thin circular disk. The drag
coefficient of such a disk shows practically no variation with the
Reynolds number since the separation point is fixed at the edge of the
disk and cannot shift from this point, regardless of the condition of the
boundary layer. Thus, the width of the wake remains constant, as
does the drag coefficient also. This thought may be usefully gen
eralized and applied to all brusque or very rough objects in a fluid
flow; experiment indicates that such objects have drag coefficients
which vary little with the Reynolds number. 11
10 2 4 6 10 2 2 4 6 10 3 2 4 6 10 4 2 4 6 10 5 2 46 10*
FIG. 198. Drag Coefficients for Circular Cylinders, Flat Plates, and Streamlined
Struts of Infinite Length. 12
The drag coefficients of circular cylinders placed normal to the flow
show characteristics similar to those of spheres. The coefficients
shown in Fig. 198 are for infinitely long cylinders. The drag coeffi
cients of streamlined struts 13 and flat plates of infinite length are also
shown for comparison. The total drags of the flat plate and cylinder
contain negligible frictional drag at ordinary velocities, whereas the
streamlined strut, because of its small turbulent wake, possesses little
11 Cf. relation of friction factor, /, and NR for rough pipes, Fig. 85, and the fact
that the minor loss coefficients of pipe flow show little variation with the Reynolds
number.
12 Data from L. Prandtl, "Ergebnisse der aerodynamischen Versuchsanstalt zu
Gottingen," Vol. II, R. Oldenbourg, 1923, p. 24, and B. A. Bakhmeteff, "Mechanics
of Fluids," Part II, Columbia University Press, 1933, p. 44.
18 The area to be used in the drag equation is the projection of the body on a
plane normal to the direction of flow.
312
FLOW ABOUT IMMERSED OBJECTS
profile drag. The curves are typical of those resulting from tests of
brusque, blunt, and streamlined objects.
ILLUSTRATIVE PROBLEM
What is the drag force on a 6in. diameter smooth sphere when placed in an
airstream (60 F, 14.7 Ib in. 2 ) having a velocity of 30 mph?
T7 30 X 5280 __.,,
y _  36.65 ft/sec
3600
36.65 X A X 0.00237 <__
From the plot of Fig. 196,
CD = 0.49
D = 0.49 X
0.000000375
X
= 0.153 Ib
78. Drag at High Velocities. As objects move at increasing speeds
through compressible fluid the assumption of constant density is less
valid, since higher velocities bring about greater variations of pressure
over the object, these pressures in turn causing changes in the density
of the fluid. At points of low and high
pressure on the surface of the object the
fluid is respectively rarefied and com
pressed. At the stagnation point on the
nose of the object, the increased pressure
compresses the fluid, creating a zone of
dense fluid ahead of the object; the nose
of the body thus moves through a fluid of
increased density and, of course, encoun
ters increased resistance because of this.
An understanding of the effect of high velocity on pressure variation
may be gained by calculating the stagnation pressure on the nose of
an object placed in a high velocity flow of compressible fluid (Fig. 199).
Applying the Euler equation
FIG. 199.
dp
+ VdV 
and integrating between point in the undisturbed flow and the stag
nation point, S.
/" p * Jjt /*Vo
VdV
DRAG AT HIGH VELOCITIES 313
At the stagnation point there is no velocity and V 8 = 0; and if adia
batic compression of the fluid is assumed
P Po '**
P Po
Substituting these values above,
1
F (
<0
VdV
PO^PO .f J o
P
and integrating,
P ^ \[P.*\ k 1 1 ^
Solving for p 8 ,
+
L
but (Art. 4)
Po
and therefore
k
p.*[i + ?
Expanding this expression by the binomial theorem and substituting
c for vkpo/po gives (using the first three terms)
which indicates that the stagnation pressure is always greater than
that of an incompressible fluid and depends not only upon p , V , and
Po but also upon the ratio of undisturbed velocity to acoustic velocity,
the Mach number, NM. Thus
and once again the importance of the Mach number in compressible
fluid flow calculations is observed.
The increase hi stagnation pressure due to compression of the fluid
leads directly to one of the principles of minimizing drag at high
314
FLOW ABOUT IMMERSED OBJECTS
ForK<c
For V>c
FIG. 200.
velocities. Obviously, increased pressure on the nose of the object
contributes directly to profile drag, and to minimize this contribution
the area on which such high pressures act must be reduced to a mini
mum, resulting in pointed rather than blunt noses for highspeed
bodies. At low velocities it has been seen that the shape of the tail
of the body, in determining the
size of the wake, was of primary
importance in determining pro
file drag, whereas the shape of
the nose had little effect. At
higher velocities, where wakes
are fully developed, drag de
pends little upon the tail of the
body and much upon the shape
of the nose. The shapes of the
airship form and projectile of
Fig. 200 are striking examples
of the application of these
principles.
The effect of shape upon the drag coefficients of various projectiles
at high velocities is shown on Fig. 201. In this velocity range,
viscosity has little effect on drag and coefficients will vary primarily
with Mach number as was indicated by the dimensional analysis of
Art. 75. On the plot may be clearly seen (1) the increased effect of
shape of nose on the drag coefficient as the Mach number increases,
and (2) the abrupt increase in the coefficient near NM = 1, where the
velocity equals the acoustic velocity. Here the effects of compressi
bility become pronounced and the nature of the flow changes radically.
For a typical air foil, used as a wing or propeller blade element, the
sudden increase of drag coefficient occurs before NM =1; i.e., the
effects of compressibility are felt before the foil reaches the acoustic
velocity. This condition arises from the fact that local velocities at
certain points on an airfoil are always greater than the velocity of the
foil, and acoustic velocities and serious compressibility effects occur at
these points before the foil itself attains the acoustic velocity.
When bodies travel through a fluid at supersonic (V > c) speeds a
new physical condition arises, which completely changes the nature of
the flow. To investigate this condition briefly, consider the projectile
of Fig. 202, moving at a supersonic speed, V, and let it occupy the
positions 1, 2, 3, and 4, at times t\, t%, 3, and t. At time t\ the nose
of the projectile disturbs the fluid at point 1, and this disturbance
DRAG AT HIGH VELOCITIES
315
progresses through the fluid as a spherical wave with a celerity of
propagation, c. After a time (t t\) has elapsed, the radius of the
spherical wave is c(/ 4 /i). In this same time, however, the projec
tile has moved to point 4; therefore
k
Mach Number, N M ~
FiG. 201. Drag Coefficients for Artillery Projectiles. 14
and the radius of the sphere (by substitution) is cli/V. Similarly the
disturbances which started at points 2 and 3 have (when the projectile
reaches point 4) radii of cl 2 / V and cli/V, respectively. Obviously a
14 F. R. W. Hunt, in The Mechanical Properties of Fluids, p. 341, Blackie and
Sons, 1925.
316
FLOW ABOUT IMMERSED OBJECTS
C. Cranz
FIG. 203. Small Bore Bullet in Flight. 16
18 From C. Cranz, Lehrbuch der Ballistik, Vol. I, B. G. Teubner, Leipzig, 1917.
LIFT BY CHANGE OF MOMENTUM
317
surface tangent to these spherical waves is a conical one with its apex
at the nose of the projectile, and such a surface represents the line of
advance of the aggregation of pressure disturbances; it is a wave
front, or a " shock wave." The " Mach angle," a, may be seen from
the figure to be given by
a = sin"" 1
Since this angle depends only on c and F, the velocity of projectiles
may be obtained from photographs of the wave front by measuring
the Mach angle. A typical photograph of the wave front caused by
a rifle bullet is shown in Fig. 203.
79. Lift by Change of Momentum. The lift force on an unsym
metrical body has been seen to arise from pressure differences caused
by velocity variations over the surface of the body, but this force may
also be associated with the change of fluid momentum caused by the
body. If about the foil of Fig. 204 a rectangle is drawn in such a way
FIG. 204.
that the velocity, FI, entering the lefthand side of the rectangle is
horizontal and the upper and lower boundaries are streamlines, it is
evident at once that the flow Q which enters the rectangle horizontally
at the left is deflected downward, leaving the rectangle with an average
velocity V%. Since there is a continual change in the vertical momen
tum of the fluid within the rectangle, a force must be continually
318 FLOW ABOUT IMMERSED OBJECTS
exerted vertically downward upon this fluid to bring about this change.
This force cannot come from the pressure difference between top and
bottom of the rectangle since this difference is canceled by the dif
ference in elevation of these boundaries. The force must, therefore,
be that exerted vertically downward by the foil on the fluid, the equal
and opposite of the lift force. From the impulsemomentum law
L = (AT),
or
thus giving a physical picture of the relation of momentum change to
lift force.
Such an equation is of some interest in the light of previous applica
tions of the impulsemomentum law and may serve as an exposition of
principles, but to use it for lift calculations is obviously impossible;
the size of the rectangle is unknown, thus preventing the calculation of
Q, and, with unknown magnitudes and directions of velocities at sec
tion 2, AFis also incalculable.
80. Circulation Theory of Lift. Although the foregoing studies of
pressure variation and momentum change have contributed to an
understanding of lift phenomena, a more comprehensive knowledge
may be had by applying the principles of circulation. 16 This was first
done by Kutta (1902) and Joukowski (1906), whose equations were
written for the forces on a body of any shape and whose results were,
therefore, entirely general. Because of the mathematical ability
required to follow their analysis it will not be included here, but the
physical significance of the (Kutta Joukowski) theorem may be seen
by a simple application suggested by Professor Bakhmeteff. 17
The foil of Fig. 205 is a flat plate of chord c and infinite length
from which a section of 1ft length is to be considered. When the plate
is placed in a rectilinear flow (Fig. 205a) of velocity V with its axis
aligned with the flow, there is obviously no lift force since the velocities
and pressures are the same on both sides of the plate.
16 The student should restudy Art. 24.
17 B. A. Bakhmeteff, Mechanics of Fluids, Part II, p. 70, Columbia University
Press, 1933.
CIRCULATION THEORY OF LIFT
319
Now about the plate assume a clockwise circulatory flow (Fig. 205&)
having a circulation T given by
2cv
Traversing a closed curve adjacent to the plate in a clockwise direction,
the length of curve is 2c so that the average velocity along the curve
must be v, if the product of these quantities is to give the above circula
tion. This means that with the circulatory flow there is a velocity v
to the right along the top of the plate and the same velocity v to the
(a)
Po
/K + V
1
2
> ^ _J
 <=
' / """"""""
* v __ v ) >
(c)
FIG. 205.
left along the bottom. Since there is no velocity difference there can
be no pressure difference, and hence no lift force is exerted on the plate.
Now, superpose the two flows, and the flow picture of Fig. 205c
results. The circulatory flow has bent the rectilinear flow upward
at the leading edge of the plate and downward at the trailing edge,
yet the effect on the rectilinear flow at a distance from the plate is
negligible because of the small velocities induced by a vortex at
great distances from its center. The circulatory motion has the same
direction as the rectilinear motion on the top of the plate, but opposes
it along the bottom. Thus, when the two flows are superposed the
320 FLOW ABOUT IMMERSED OBJECTS
average velocity along the top of the plate becomes (V + v) and that
along the bottom (V v). Now, applying the Bernoulli equation
along streamtube 11 between O and the top of the plate and along
streamtube 22 between and the bottom of the plate, letting pT and
ps be the average pressures over the top and bottom respectively.
Po + %PV 2  p T
Equating these,
PBPT =
Simplifying,
ps  PT =
But the lift, L t on a section of 1ft length is given by
L = (p B  PT) o
Therefore
L = %p(Vv)c  (2vc)pV
But
r  2vc
and therefore
L (per foot of span) = FpF
which is the KuttaJojikowski result in its simplest form and indicates
clearly that the combination of velocity and circulation is essential to the
existence of a lift force if either one of these terms is zero there can
be no lift.
Although the foregoing proof is not rigorous owing to the use
of certain average velocities and pressures it nevertheless indicates the
physical essence of the problem ; and in view of the Kutta Joukowski
general treatment it may be applied to airfoils, cylinders, spheres, or
bodies of any shape. The result serves to explain certain familiar
phenomena in which bodies rotating in a viscous fluid create their own
circulation and when exposed to a rectilinear flow are acted upon by
a transverse force. Some examples are the force exerted on the rotating
cylinders of a "rotorship" and the transverse force which causes a
pitched baseball to curve.
81. Origin of Circulation. Although it is not difficult to imagine
a rotating body in a viscous fluid inducing its own circulation, to
explain the origin of circulation about an airfoil, or an element of a
propeller or turbine blade requires knowledge of other principles.
ORIGIN OF CIRCULATION
321
Consider the flow conditions about a typical airfoil as it starts to
move. Before motion begins the circulation about the foil is obviously
zero (Fig. 206a). As motion occurs, the circulation about the foil
tends to remain zero and the "potential" flow of Fig. 2066 tends to be
ro
vo
(a) No Motion.
<b) Potential Motion.
(c) Real FlUijJ Flow.
Starting,
fl
FIG. 206. Development of Circulation about an Airfoil.
set up, but such a flow, which includes a stagnation point near the rear
of the foil and flow around its sharp trailing edge, cannot be maintained
in a real fluid, because of separation caused by viscosity. This poten
tial flow gives way immediately to the flow of Fig. 206c, and in the
process a circulation, F, develops about the foil, and a vortex, the
322
FLOW ABOUT IMMERSED OBJECTS
"starting vortex" (Fig. 206d), is shed from the foil. During the
creation of this vortex, however, the circulation around a closed curve
including and at some distance from the foil is not changed and must
remain zero; thus, from the properties of circulation, the circulation
about, or the strength of, the starting vortex must be equal and
opposite to that about the foil. The, existence of circulation about a
foil is, therefore, dependent upon the creation of the starting vortex;
since the vortex in turn is dependent upon separation and the viscosity
of the fluid it may be observed that circulation and, therefore, lift result
from the existence of fluid viscosity.
View A A
FIG. 207. Airfoil of Finite Length.
82. Foils of Finite Length. When fluid flows about foils of finite
length, flow phenomena result which affect both lift and drag of the
foil ; these phenomena may be understood by further investigation and
application of the foregoing circulation theory of lift.
Since pressure on the bottom of an airfoil is greater than that on
the top, flow will escape from below the foil at the end and flow toward
the top, thus distorting the general flow about the foil, causing fluid to
move inward over the top of the foil and outward over the bottom
(Fig. 207). As the fluid merges at the trailing edge of the wing, a sur
face of discontinuity is set up, and flows above and below this surface
have components of velocity inward and outward as shown. The
tendency for vortices to form from these velocity components is
obvious and in fact this surface of discontinuity is a " sheet of vortices. "
FOILS OF FINITE LENGTH
323
However, such a vortex sheet is unstable and the rotary motions con
tained therein combine to form two large vortices trailing from the tips
of the foil (Fig. 207) ; these are called tip vortices and are often visible
when an airfoil passes through dustladen air.
Since the pressure difference between top and bottom of an airfoil
must reduce to zero at the tips, it is evident that the lift per unit
length of span varies over the span (Fig. 208), being maximum at the
center and reducing to zero at the tips. The total lift of the foil is,
of course, the total force resulting from this lift diagram. Since lift
per unit length of span varies directly with circulation (L = FpF),
a diagram showing distribution of circulation over the span has the
same shape as that of a diagram of lift distribution. The variation of
AZ,
Actual Assumed
FIG. 208. Distribution of Lift and Circulation over an Airfoil of Finite Length.
lift and circulation over the span of an airfoil cannot, of course, be
disregarded in a rigorous treatment of the subject, but such treatment
leads to mathematical and physical complexities which are beyond
the scope of this volume. A simple physical picture may be obtained,
however, from the following analysis in which lift and circulation will
be assumed to be distributed uniformly over the span (Fig. 208).
One of the properties of vortices is that their axes can end only
at solid boundaries. Since there is no solid boundary at the end of the
airfoil, the circulation T cannot stop here, but must continue to exist
about the axes of the tip vortices (Fig. 209). The axes of the tip
vortices extend rearward to the axis of the starting vortex; thus,
according to the theory, the axis of the vortex having circulation T
does not end, but is a closed curve composed of the axes of the airfoil,
tip vortices, and starting vortex. In the real fluid the circulation
324
FLOW ABOUT IMMERSED OBJECTS
persists only about the foil and portions of the tip vortices close to the
foil ; the starting vortex and remainder of the tip vortices are quickly
extinguished by viscosity.
The circulations about the tip vortices induce a downward motion
in the fluid passing over a foil of finite length and in so doing affect
*p
*\
 ^ Axis of
r l starting vortex
FIG. 209. Circulation about an Airfoil of Finite Length.
both lift and drag by changing the effective angle of attack. The
strength of this induced motion will obviously depend upon the
proximity of the tip vortices and thus upon the span of the foil or upon
the ratio of span to chord, b/c, called the aspect ratio of the foil.
An airfoil of finite span is shown at angle of attack a in the hori
zontal flow of Fig. 210. The downward (downwash) velocity induced
FIG. 210.
near the wing by the tip vortices decreases the angle of attack by a
small angle of downwash, , making the effective angle of attack
(a c). This effective angle of attack is that for no induced down
ward velocity or, in other words, it is the angle of attack which would
be obtained if the foil had infinite span and aspect ratio. Calling this
angle of attack ,  /*\
6 < = a 6 (102)
FOILS OF FINITE LENGTH 325
Now treating the foil as one of infinite span at an angle of attack a# ,
the lift LOO exerted on such a foil is by definition normal to the direction
of flow in which it is placed ; therefore L*, is normal to the effective
velocity Foo, and at an angle e with the vertical. The lift, L, on the
foil of finite span is normal to the approaching horizontal velocity V
and is the vertical component of L^. But Loo also has a component
in the direction of the original velocity F, a drag force, >*, called the
induced drag because its existence depends upon the downward veloc
ity induced by the tip vortices. Thus, an additional drag force, D^
must be added to profile and frictional drag in computing the total
drag of a body of finite length about which a circulation exists. Calling
the sum of frictional and profile drag ><, since it is the drag of a foil of
infinite span Cvyhich has no induced drag), the total drag D of a foil of
finite length is given by
D = L>oo + Di (103)
which, by dividing by ApV 2 /2, may be expressed in terms of dimen
sionless drag coefficients as
CD = C Dao + C Di (104)
Thus the drag coefficient, CD, of a body of finite length with circulation
is the sum of the profilefrictional drag coefficient CD^ and the induced
drag coefficient, CD^
From the foregoing statements and Fig. 210, it is evident that
induced drag Di is related to lift L, angle a, and aspect ratio b/c\ the
equations relating these variables are of great practical importance.
Since is small
If the distribution of lift over a wing of finite span is taken 18 to be a
half ellipse (see Fig. 208), it may be shown that
CL (105)
and substituting this value and L = Ci4(pF 2 /2) in the equation
for Di
. (106)
18 An assumption which gives minimum induced drag and conforms well with fact.
326 FLOW ABOUT IMMERSED OBJECTS
Now, expressing Di in terms of the induced drag coefficient
Di  C Di A tj
and substituting this value in equation 106 there results
which relates lift and induced drag through their dimensionless coeffi
cients and shows that induced drag is inversely proportional to aspect
ratio, becoming zero at infinite aspect ratio (infinite span) and increas
ing as aspect ratio and span decrease thus offering mathematical
proof of the foregoing statements on the effect of span, aspect ratio,
and proximity of tip vortices on induced downward velocity and
induced drag.
When the foregoing expressions for and CD; are substituted in
equations 102 and 104, respectively, there result
(108)
vy
and
CD = C D . + 7 (109)
With these equations airfoil data obtained at one aspect ratio may be
converted into corresponding conditions at infinite aspect ratio, and
these data in turn reconverted to foils of any aspect ratio; thus,
extensive testing of the same airfoil at various aspect ratios becomes
unnecessary.
ILLUSTRATED PROBLEM
A rectangular airfoil of 6ft chord and 36ft span has a drag coefficient of 0.0543
and lift coefficient of 0.960 at an angle of attack of 7.2. What are the correspond
ing lift and drag coefficients and angle of attack for a similar wing of aspect ratio 8?
For aspect ratio 8: CL =* 0.960 (No change in lift coefficient.)
Assuming semielliptical lift distribution,
0.0543  0.0489  0.0054
LIFT AND DRAG DIAGRAMS
327
For aspect ratio 8:
C D = 0.0054 + >  0.0421
TT X 8
' = 0.0509 radian = 2.9
For aspect ratio 8:
7.2  2.9 = 4.3
4.3 +^(
7T X 8 \ 27T
83. Lift and Drag Diagrams. The, relation between lift and
induced drag coefficients suggests plotting lift coefficient against drag
Point of zero lift and
minimum drag
( Not always the
same point)
FIG. 211. Polar Diagram for a Typical Airfoil.
coefficient and gives the socalled polar diagram of Fig. 211, which is
used extensively in airplane design. On this diagram equation 107
is a parabola passing through the origin and symmetrical about the
328
FLOW ABOUT IMMERSED OBJECTS
CD axis, the slope of the parabola depending on the aspect ratio.
Since the two curves are for foils of the same aspect ratio, the hori
zontal distance between them is the profilefrictional drag coefficient,
1.6
0.2
.25
FIG. 212. Polar Diagram for 48 ft by 8 ft ClarkY Airfoil. 19 (N/j ^ 6,000,000.)
*', in this way, the polar diagram gives graphical significance to
equation 109. But the diagram does much more than this alone:
"A. Silverstein, "Scale Effect on ClarkY Airfoil Characteristics from N.A.C.A.
FullScale Windtunnel Tests," N.A.C.A. Report 502, 1934.
LIFT AND DRAG DIAGRAMS
329
The important ratio of lift to drag is the slope of a straight line drawn
between origin and the point for which this ratio is to be found ; the
maximum value of this ratio is the slope of a straight line tangent to
i.e n
1.4 
1.2 
1.0 
0.8 
0.6 
0.4 
0.2 
4 8
Angle of Attack, a, Degrees
12
16
0.2 
32
.28
.24
.20
16
.12
.08
.04
FIG. 213. Lift and Drag Coefficients and L/D Ratio for 48 ft by 8 ft ClarkY
Airfoil. 19 (N*~ 6,000,000).
the curve and passing through the origin ; on the diagram are easily
seen also the points of zero lift, minimum drag, and the point ol
maximum lift or "stall," which determines "stalling angle' ' above
which lift no longer continues to increase with angle of attack; the end
19 See footnote on p. 328.
330
FLOW ABOUT IMMERSED OBJECTS
of the upper solid portion of the curve is the point at which the flow
separates from the upper side of the wing, forming a turbulent wake
which increases the profile drag and therefore the drag coefficient,
accompanied by a large drop in lift and lift coefficient because of
increased pressure on the upper side of the wing.
A polar diagram for a ClarkY airfoil obtained in the N.A.C.A.
fullscale wind tunnel is shown in Fig. 212, and anoth'er way of
presenting the same data in Fig. 213. Because of the assumption of
semielliptical lift distribution made in deriving equations 108 and 109
it is to be expected that these equations must be modified for use in
actual practice.
These equations are now written
a = < +
cP' + ' :
C D = Cn + r^ (1 + (7)
and
in which the r and cr are correction factors. Since lift distribution
varies increasingly from the semielliptical one with increasing aspect
ratio these correction factors vary with aspect ratio and increase with
this ratio. Some typical values for r and cr for the ClarkY airfoil are 19
b/c
4
6
8
r
0.030
0.051
0.070
a
0.118
0.176
0.212
The data of Figs. 212 and 213 were obtained at a Reynolds number
of about 6,000,000, and from many foregoing statements it should be
expected that the data will change with changing Reynolds number.
The following trends, which are confirmed by experiment, are of some
interest in the light of foregoing principles. With increasing Reynolds
number the drag coefficient at zero lift decreases ; here the drag coeffi
cient contains predominantly frictional effects and its variation with
Reynolds number is similar to that of the flat plate (Fig. 194). With
increased turbulence, due either to increased initial turbulence or
19 See footnote on p. 328.
PROBLEMS 331
increased Reynolds number, the maximum lift coefficient increases;
in other words, higher angles of attack can be attained without causing
separation. Here the energy of the turbulent boundary layer (Fig. 197)
delays separation, allowing high velocity flow to cling to the upper
side of the foil, causing lower pressures and greater lift.
BIBLIOGRAPHY
L. PRANDTL and O. G. TIETJENS, Fundamentals of Hydro and Aeromechanics,
McGrawHill Book Co,, 1934.
Applied Hydro and Aeromechanics, McGrawHill Book Co., 1934.
L. PRANDTL, The Physics of Solids and Fluids , Part II, Second Edition, Blackie & Son,
1936.
H. GLAUERT, Aerofoil and Airscrew Theory, Cambridge University Press, 1930.
E. G. REID, Applied Wing Theory, McGrawHill Book Co., 1932.
E. A. STALKER, Principles of Flight, Roland Press, 1931.
W. F. DURAND, Aerodynamic Theory, Julius Springer, Berlin, 1934. Six volumes.
C. V. DRYSDALE and Others, The Mechanical Properties of Fluids, Blackie & Son, 1925.
S. GOLDSTEIN, Modern Developments in Fluid Dynamics, Vols. I and II, Oxford
University Press, 1938.
Reports, Technical Memoranda, and Technical Notes of the National Advisory Com
mittee for Aeronautics (N.A.C.A.).
PROBLEMS
484. A rectangular airfoil of 40ft span and 6ft chord has lift and drag coefficients
of 0.5 and 0.04, respectively, when at an angle of attack of 6. Calculate the drag
and horsepower necessary to drive this airfoil at 50, 100, and 150 mph through still
air (40 F and 13.5 lb/in. 2 abs). What lifts are obtained at these speeds?
485. A rectangular airfoil of 30ft span and 6ft chord moves at a certain angle
of attack through still air at 150 mph. Calculate the lift and drag, and the horse
power necessary to drive the airfoil at this speed through air of (a) 14.7 lb/in. 2 and
60 F and (b) 11.5 lb/in. 2 and F. C D = 0.035, C L = 0.46.
486. Calculate the speed and horsepower required for condition (6) of the previous
problem to obtain the lift of condition (a).
487. The drag coefficient of a circular disk when placed normal to the flow is 1.12.
Calculate the force and horsepower necessary to drive a 12in. disk through (a) air
(w = 0.0763 lb/ft 3 ) and (b) water at 30 mph.
488. The drag coefficient of an airship is 0.04 when the area used in the drag for
mula is the power of the volume. Calculate the drag of an airship of this type
having a volume of 500,000 ft 3 when moving at 60 mph through still air (w 0.0763
lb/ft 3 ).
489. A wing model of 5in. chord and 2.5ft span is tested at a certain angle of
attack in a wind tunnel at 60 mph using air at 14.5 lb/in. 2 abs and 70 F. The lift
and drag are found to be 6.0 Ib and 0.4 Ib, respectively. Calculate the lift and
drag coefficient for the wing at this angle of attack.
490. An airplane and an artillery projectile move through still air (14.0 lb/in . 2 ,
40 F), the former at 350 mph, the latter at 1500 ft/sec. Calculate their Mach
numbers.
332 FLOW ABOUT IMMERSED OBJECTS
491. A cylindrical body of 6in. diameter moves through still air (14.7 lb/in. 2 abs
and 60 F) at 500 mph. At what velocity must a geometrically similar body move
through still air (13.0 lb/in. 2 and F) if the two flows are to be completely similar
dynamically? What must be the diameter of this body?
492. A smooth plate 10 ft long and 3 ft wide moves through still air (60 F,
14.7 lb/in. 2 abs) at 5 ft/sec. Assuming the boundary layer to be laminar, calculate
(a) the thickness of the layer at 2, 4, 6, 8, and 10 ft from the leading edge of the plate;
(b) the constants a and /3; (c) the stress, T O at the above points; (d) the*coefficient
/ at the above points; (e) the total drag force on one side of the plate.
493. What is the drag on one side of the plate of the preceding problem if the
boundary layer is turbulent?
494. A flatbottomed scow having a 150 ft by 20 ft bottom is towed through still
water (60 F) at 10 mph. What is the frictional drag force exerted by the water
on the bottom of the scow? How long is the laminar portion of the boundary layer,
using the critical NR of Fig. 194? What is the thickness of this layer at the point
at which it becomes turbulent? What is the thickness of the boundary layer at the
rear end of the bottom of the scow?
495. A streamlined train 400 ft long is to travel at 90 mph. Treating the sides
and top of the train as a smooth flat plate 30 ft wide, calculate the total drag on these
surfaces when the train moves through air at 60 F and 14.7 lb/in. 2 Calculate the
length of the laminar boundary layer and the thickness of this layer where it becomes
turbulent. What is the thickness of the boundary layer at the rear end of the train?
What horsepower must be expended to overcome this resistance?
496. Calculate the drag of a smooth sphere of 12in. diameter in a stream of
standard air at Reynolds numbers of 1, 10, 100, and 1000.
497. Calculate the drag of a smooth sphere of 20in. diameter when placed in an
airstream (60 F, 14.7 lb/in. 2 abs) if the velocity is (a) 20 ft/sec, (b) 28 ft/sec.
498. At what velocity will the sphere of the preceding problem attain the same
drag which it had at a velocity of 20 ft /sec?
499. Estimate the drag on a model of an N.P.L. airship hull of 6in. diameter
which is to be tested in a wind tunnel (14.7 lb/in. 2 , 60 F) at 60 mph.
500. A sphere of 10in. diameter is tested in a wind tunnel (14.7 lb/in. 2 , 60 F)
at 80 mph. At what speed must a 2in. sphere be towed in water (68 F) for these
spheres to have the same drag coefficients? What are the drag forces on these two
spheres?
501. A sphere 1 ft in diameter is towed through water (68 F) at 5 mph. What
size sphere has the same drag coefficient in an airstream (60 F, 14.7 lb/in. 2 ) having
a velocity of 60 mph? Calculate the drags of these spheres.
502. What is the stagnation (gage) pressure of an air stream (14.70 lb/in. 2 abs,
60 F) of velocity 200, 400, and 600 mph considering and neglecting compressibility?
503. At what velocity of air (14.7 lb/in. 2 , 60 F) will an error of 1 per cent be
caused in the dynamic pressure by neglecting compressibility?
504. If the pointed artillery projectile of Fig. 201 is 12 in. in diameter and is to
travel at 2000 ft/sec through air (60 F, 14.7 Ib/in. 2 ), what force is necessary to
propel it?
505. What is the drag of the bluntnosed projectile of Fig. 201 (if its diameter is
3 in.) when it travels at (a) 700 mph, (b) 800 mph through air at 60 F and 14.7
lb/in. 2 ?
PROBLEMS 333
506. Calculate the Mach angle for a bullet moving at 2000 ft/sec through air of
14.5 lb/in. 2 and 100 F.
507. If the Mach angle of the photograph of Fig. 203 is 30 and the bullet is mov
ing through air at 14.0 lb/in. 2 and 50 F, calculate the speed of the bullet.
508. An airfoil of 5ft chord and 30ft span develops a lift of 3000 Ib when moving
through air of specific weight 0.0763 lb/ft 3 at a velocity of 100 mph. What is the
average circulation about the wing?
509. The circulation about a wing of 40ft span and 6ft chord when moving at
150 mph is 700 ft 2 /sec. Calculate the lift on the wing if it moves through still air
at 14.7 lb/in. 2 and 60 F.
510. Derive a general expression for lift coefficient in terms of circulation.
511. If F is the average circulation about a wing per foot of span, calculate the
circulation about the wing at midpoint and quarterpoints of the span, assuming
a semi elliptical lift distribution.
512. A model wing of 5in. chord and 3ft span is tested in a wind tunnel (60 F,
14.5 lb/in. 2 ) at 60 mph, and the lift and drag are found to be 9.00 and 0.460 Ib,
respectively, at an angle of attack of 6.7. Assuming a semielliptical lift distribu
tion, calculate: (a) the lift and drag coefficients ; (b) CD^ (c) CD^] (d) the corres
ponding angle of attack for an airfoil of infinite span ; (e) the corresponding angle of
attack for a foil of this type with aspect ratio 5 ; (/) the lift and drag coefficients at
this aspect ratio.
513. An airfoil of infinite span has lift and drag coefficients of 1.31 and 0.062,
respectively, at an angle of attack of 7.3. Assuming semielliptical lift distribution,
what will be the corresponding coefficients for a foil of the same cross section but
aspect ratio 6? What will be the corresponding angle of attack?
514. From Fig. 212, calculate the lift and drag coefficients for a ClarkY airfoil
of aspect ratio 8, and plot the polar diagram for this airfoil. ,
515. The ClarkY airfoil of Figs. 212 and 213 is to move at 180 mph through air
at 60 F and 14.7 lb/in. 2 Determine the minimum drag, drag at optimum L/D, and
drag at point of maximum lift. Calculate the lift at these points and the horsepower
that must be expended to obtain these lifts.
APPENDIX I
DESCRIPTION AND DIMENSIONS OF SYMBOLS
SYMBOL
A, a
a
b
C, C c , CD, C/, C/, C
C
c
c
A D Pt D f ,
d
d c
E
E
E H
e
F
f
G
g
H
H
h
I
I
K
K L
k
L
I
M
M
m
N
FTLBSEC
DESCRIPTION DIMENSIONS
Area ft 2
Linear acceleration ft /sec 2
Breadths; lengths normal to flow; span
of an airfoil; length of weir; width of
open channel, etc. ft
Various dimensionless coefficients
Chezy coefficient ftVsec
Acoustic velocity ft/sec
Chord of an airfoil ft
Drag forces lb
Diameter ; depth of flow in open channels ft
Critical depth ft
Modulus of elasticity lb/ft 2
Various unit energies ft Ib/lb
Unit heat energy Btu/lb
Height of roughness ft
Force lb
Darcy friction factor
Weight (rate of) flow Ib/sec
Acceleration due to gravity ft /sec 2
Enthalpy Btu/lb
Head on weirs ft
Vertical distance, head ft
Lost head ft
Moment of inertia ft 4
Unit internal energy ft Ib/lb
Various dimensional coefficients
Various dimensionless loss coefficients
Adiabatic constant
Lift force lb
Length; distance along flow; mixing
length ft
Mass lb sec 2 /ft
Expression for graphical solution of
hydraulic jump ft 2
Molecular weight
Revolutionary speed rps
335
336
APPENDIX I
DESCRIPTION AND DIMENSIONS OF SYMBOLS Continued
SYMBOL
DESCRIPTION
FTLBSEC
DIMENSIONS
NF> NM* Afjv*. NR> NW Various dimensionless numbers (force
ratios)
n Distance normal to direction of flow It
P Power ft Ib/sec
P Weir height ft
p Intensity of pressure lb/ft 2
Q Rate of flow ft 3 /sec
q Rate of flow per foot width (in rectangu
lar channels) ft 2 / sec
R Engineering gas constant ft/ F. abs
R Pipe radius; hydraulic radius ft
r Radius ft
S Specific gravity
5 Slope
st Tension stress lb/ft 2
r Absolute temperature F. abs
r Surface tension lb/ft
T Torque ft Ib
/ Time sec
t Thickness ft
/ Temperature F.
u Velocity relative to a moving body ft/sec
V Volume ft 8
V Average velocity, Q/A ft /sec
v Specific volume ft 3 /lb
v Velocity ft/sec
W Weight Ib
w Specific weight; weight density lb/ft 3
Y Expansion factor
Z A gas constant (equation 24) (F. abs)Vsec
z Height above datum ft
a Angle of attack of an airfoil; various
other angles radians
r Circulation ft 2 /sec
6 Thickness of laminar film; thickness of
boundary layer ft
il Efficiency radians
H Coefficient of viscosity Ib sec/ft 2
v Kinematic viscosity ft 2 /sec
p Mass density Ib sec 2 /ft 4
r Shear stress lb/ft 2
w Angular velocity radians/sec
APPENDIX II
SPECIFIC WEIGHT AND DENSITY OF WATER*
Temperature,
F
Specific weight, 2
w, lb/ft 3
Density,
slugs/ft 3 or
Ib sec 2 /ft 4
32
62.42
1.940
39.2
62.45 (max.)
1.941 (max.)
50
62.42
1.940
68
62.32
1.937
86
62.16
1.932
104
61.96
1.926
122
61.71
1.918
140
61.39
1.908
158
61.07
1.898
176
60.68
1.886
194
60.26
1.873
212
59.84
l.?60
1 Handbook of Engineering Fundamentals.
* Taking acceleration due to gravity to be 32.174 ft/sec 2 .
337
APPENDIX III
VELOCITY OF A PRESSURE WAVE
THROUGH A FLUID
Consider fluid at rest in a rigid pipe fitted with a piston at one end
(Fig. 214). This piston is suddenly advanced at a velocity v for a time
dt and sends a pressure disturbance along the pipe at a velocity c.
~fc~Lj !
fe
FIG. 214.
While the piston moves a distance vdt the pressure wave will cover a
distance cdt. Through any time dt, since no mass has been destroyed,
the fluid displaced by the piston must be equal to the gain in the mass
of fluid between sections 1 and 2 due to increased density. Therefore
whence
Apvdt so cdtAdp
(i)
But, according to the law of impulse and momentum, the force
exerted on the mass multiplied by the time the force acts is equal to
the change of momentum accomplished :
thus
Force
Time
Mass
' Adp
dt
cdtAp
Velocity change = v
Adpdt as cdtApv
338
VELOCITY OF A PRESSURE WAVE THROUGH A FLUID 339
or
c = (2)
PP
Multiplying equations 1 and 2,
< a ?
dp
But
Therefore
^ = E
dp p
and substituting in equation 3 and solving for c,
APPENDIX IV 1
VISCOSITY OF LIQUIDS
(M X 10 5 Ib sec/ft 2 )
Tem
perature
o F
Water
Carbon
Tetra
chloride
Benzene
Linseed
Oil
Ethyl
Alcohol
*
Turpentine
32
3.75
2.825
1.892
3.71
4.70
50
2 73
2 38
1 594
3 06
3 73
68
2.10
2 035
1 367
2 51
3 11
86
104
1.667
1 372
1.772
1.56
1.185
1 040
69.2
2.10
1 744
2.68
2 24
122
140
1.148
0.980
1.384
0.928
36.8
1.468
1.238
1.935
158
850
1 116
0.750
1 053
1 52
176
0.746
194
0.662
14.84
212
593
Temperature F
Mercury
Castor Oil
Glycerine
Olive Oil
32
3.475
37
8830
41
7860
50
5060
288.3
57.7
2900
59
3166
224 5
68
68.5
3.235
2060
1735
175.5
77
1362
79 7
1032
86
943
112.8
93.2
3.083
95
661
104
122
158
483
75.9
54.0
25 9
208.4
2.64
1 All physical data from Smithsonian Physical Tables, Eighth Edition, Smith
sonian Insitution, 1933.
340
APPENDIX IV
341
VISCOSITY OF GASES 2
( M X 10 5 in Ib sec/ft 2 )
Tem
perature,
F
Hydrogen
Oxygen
Air
Nitrogen
Carbon
Dioxide
0.01726
0.0372
0.0342
0.0328
0.02775
100
.01982
.04385
.0402
.0383
.0336
200
.0222
.0500
.0457
.0435
.0392
300
.02435
.0555
.0506
.04805
.0443
400
.0264
.0608
.0554
.0524
.04915
500
.0282
.0656
.0598
.0565
.05385
2 Calculated from Sutherland's formula.
^273
in which \J. Q = viscosity at C.
T = absolute temperature C.
C = a constant.
Hydrogen
Oxygen
Air
Nitrogen
Carbon
Dioxide
Mo
0.01812
0.0395
0.0362
0.0347
0.0297
C
72
131
124
110
240
APPENDIX V
CAVITATION
The phenomenon of cavitation has assumed increased importance
in the design and operation of highspeed hydraulic machinery such
as turbines, pumps, and ship propellers. Briefly, it is characterized
by local reduction of pressure to the vapor pressure, formation of a
cavity within the flowing fluid, rapid pitting, and destruction of the
Absolute zero,/'
of pressure
FIG. 215.
parts of the machine in contact with the flowing fluid, losses in the
efficiency of the machine, and serious vibration problems.
The fundamentals of cavitation may be easily observed by a study
of flow through a constriction in the pipe line of Fig. 215. With the
valve partially open, the variation of pressure head through pipe and
342
CAVITATION
343
constriction is given by curve A , the point of lowest pressure occurring
at the point of minimum area, where the velocity is highest. Increased
valve opening produces a condition B, at which the pressure at the
constriction has fallen to the vapor pressure of the liquid. Further
opening of the valve leads to a pressurehead variation C, lowering the
Pitting of surface
FIG. 216.
pressure downstream from the constriction but not changing the
pressure within the constriction. Thus the increased valve opening
cannot increase the velocity and rate of flow through the pipe since at
two points in the flow (in the reservoir and in the constriction) the
pressures are fixed. For pressure variation C a considerable region
downstream from the constriction possesses only the vapor pressure
344 APPENDIX V
of the liquid. Here a cavity forms as shown, the live stream no longer
following the boundary walls of the passage. The cavity contains
a swirling mass of droplets and vapor and, although appearing steady
to the naked eye, actually forms and reforms many times a second.
The formation and disappearance of a single cavity are shown schemat
ically in Fig. 216, and the disappearance of the cavity is the clue to
the destructive action caused by cavitation. The lowpressure* cavity
is swept swiftly downstream into a region of high pressure where it
collapses suddenly, the surrounding liquid rushing in to fill the void.
At the point of disappearance of the cavity the inrushing liquid comes
together, momentarily raising the pressure at a point within the liquid
M. I. T.
Before. After
FIG. 217. Pitting of Brass Plate after 5 Hours' Exposure to Cavitation
(Magnification 10 X).
to a very high value. If the point of collapse of the cavity is in contact
with the boundary wall, the wall receives a blow (Fig. 217) as from
a tiny hammer and its surface is stressed locally beyond its elastic
limit, resulting eventually in fatigue and destruction of the wall
material.
The guiding principles of cavitation prevention obviously are to
maintain pressures relatively high and to provide surfaces having
curvatures gentle enough to prevent separation of the live stream from
them. However, to apply such principles, particularly the latter one,
to problems of design inevitably involves uncertainties because of the
generally complex nature of the flows.
BIBLIOGRAPHY 345
BIBLIOGRAPHY
W. SPANNHAKE, "Cavitation and Its Influence on Hydraulic Turbine Design,"
N.E.L.A., Pub. 222, June, 1932.
L. P. SMITH, "Cavitation on Marine Propellers," Trans. A.S.M.E., July ,1935.
S. L. KERR, "Determination of the Relative Resistance to Cavitation Erosion by
the Vibratory Method," Trans. A.S.M.E., July, 1937.
E. ENGLESSON, " Pitting in Water Turbines," The Engineer, October 17, 1930.
J. M. MOUSSON, "Pitting Resistance of Metals under Cavitation Conditions,"
Trans. A.S.M.E., July, 1937.
INDEX
Absolute pressure, 2526
Absolute temperature, 6
Acceleration, centrifugal, 44
convective, 57
horizontal linear, 42
local, 57
total, 57
vertical linear, 41
Acoustic velocity, 10, 7374, 298, 312
316, 338
Adiabatic constant, 9
for various gases, 9
Adiabatic process, 8
Alternate depths, 211
Anemometer, cup type, 247
hotwire, 248
vane type, 247
Angle of attack, 296
effective, 324
Archimedes' principle, 38
Aspect ratio of a foil, 325
Atmospheric pressure, 24, 26
Barometer, 28
Bernoulli's equation, 60
for compressible fluids, 68
for incompressible fluids, 61
Best hydraulic cross section, 204
Borda orifice, 264
Boundaries, 111
flow over, 110111
Boundary layer, 116, 300
kinetic energy of, 3 10
on flat plates
laminar, 301304
turbulent, 305
Bourdon pressure gage, 25
Boyle's Law, 6, 8
Branching pipes, 185188
Buoyancy, 38
Capillarity, 17, 30
Cauchy number, 127
Cavitation, 63, 184, 342
pitting due to, 63, 344
Center of buoyancy, 39
Center of pressure, 3236
Charles' law, 6
Chezy coefficient, 200203
Chezy equation, 199
Chord of a foil, 296
Circulation, 94
about a foil, 324
origin of, 320
theory of lift, 318
Coefficient of discharge, for nozzles, 258
for orifices under static head, 264
for pipeline orifices, 260
Cole Pitometer, 243, 268
Compressibility, 8
Compressible fluid, flow about immersed
objects, 312316
flow through a constriction in a stream
tube, 7075 '
flow through nozzles, 257258
flow through orifices, 260262
flow through Venturis, 253254
flow with friction in pipes, 160163
stagnation pressure in, 313
Conjugate depths, 222
Continuity, equation of, 58
Contraction, coefficient of, 260, 264
Controls, 217
Critical depth, 211213
occurrence of, 214217
Critical pressure ratio, 72
Critical Reynolds number, 108
Current meter, 247
measurement of river flow with, 285
Curvilinear flow, 92
Cylinders, drag coefficients of, 311
D'Alembert paradox, 1, 121
Density, 4
measurement of, 229231
of water, 337
347
348
INDEX
Dilution methods of flow measurement
270
Dimensional analysis, 131137
of drag of floating objects, 136
of drag of immersed objects, 297
of pipe friction, 147
Dimensions of symbols, 133, 335336
Discontinuity, surface of, 117, 122, 168,
307
Disks, drag coefficients of, 309
Drag, 116
at high velocities, 299, 312317
fractional, 119, 295, 300
induced, 325
profile, 119, 295, 307
Dynamic similarity, see Similarity
Eddy formation, 117118
Eddying turbulence, 118, 122, 168
Elasticity, 8
force, 127
modulus of, 8
Energy, heat, 66
internal, 66
kinetic, 66
loss, see Loss of head
mechanical, 66
potential, 66
pressure, 66
Energy equations, 65, 139
Energy line, 61, 141, 209, 218
Enthalpy, 67
Euler's equation, 59
Expansion factor, 254
Float gage, 239
Float measurements, 287
Floating objects, 38
dimensional analysis of drag of, 136
similarity applied to drag of, 130
Flow bends, 267
Fluid mechanics, development of, 1
Fluid properties, 419
measurement of, 229236
Fluid state, characteristics of, 3
Foils, angle of attack of, 296, 324
aspect ratio of, 325
chord of, 296
circulation about, 318
Foils, drag on, 324331
lift on, 317331
span of, 296
Free overfall, 215
Free surface, measurement of, 238240
Friction factor, 149
in laminar flow, 150
in turbulent flow, 151
plotted against Reynolds number,
150, 153.
Frictional drag, see Drag
Froude number, 127
relation to drag of floating objects, 130,
136
Gage pressure, 26
Gases, equation of state of, 6
gas constants for, 67
modulus of elasticity for, 910
Gasometer, 249
Gravity force, 127
HagenPoiseuille law, 146
Hazen Williams formula, 166167
Head loss, see Loss of head
Heat energy, 66
Hook gage, 239
Hooke's law, 8
Horsepower of fluid machines, 69
Hydraulic grade line, 141, 179184
Hydraulic jump, 219224
location of, 223
solution by Mcurve, 222
Hydraulic radius, 164, 200
Hydraulics, 2
Hydrodynamics, 2
Hydrometer, 229, 231
Ideal fluid, 55
Immersed objects, flow about, 116121,
128, 294331
Impulsemomentum law, 7780
Impulse turbine, 82
Induced drag, see Drag
[nertia force, 127
[nternal energy, 66
isothermal process, 8
Jet, 63
INDEX
349
Kinematic viscosity, 14
Kinetic energy, 66
KuttaJoukowski theorem, 320
Laminar film, 111, 152, 158
Laminar flow, 106
about spheres, 308
between parallel plates, 115
in boundary layers, 116, 301304
in pipes, 107, 143146
similarity for, 129
Lift, 295, 317331
by change of momentum, 317
by circulation theory, 318
coefficient of, 296
Loss of head, 139, 141
in circular pipes due to pipe friction,
for compressible fluids, 160163
for incompressible fluids, 141153,
166
in constrict ions, 252
in hydraulic jump, 223
in noncircular pipes, 164167
in nozzles, 259
in open channels, 199, 209, 218
in pipes, due to bends, 177
due to contractions, 174175
due to enlargements, 170173
due to entrances, 175176
due to exits, 172
due to pipe fittings, 178
Mach angle, 317
Mach number, 298, 313315
Manometers, 2730
Mechanical energy, 66
Mild slope, 211
Minor losses in pipes, 167178
See also Loss of head
Models, see Similarity
Modulus of elasticity, 8
Momentum, 7780
lift by change of, 317
Nozzles, 255259
Openchannel flow, 196224
Chezy coefficient, 200203
Chezy equation. 199
Openchannel flow, critical depth, 211,
213217
hydraulic jump in, 219224
rapid flow, 211, 212, 215,223
specific energy, 209211
tranquil flow, 211, 212, 215
uniform flow, 199208
varied flow, 197, 217219
velocity distribution, 209, 285
Orifices, coefficients for, 264
in pipe lines, 259262
loss of head at, 252
under falling head, 266
under static head, 63, 263265
Perfect fluid, 55
Piezometer, 24
opening, 237
ring, 238
Pipe coefficient, 154
relation to friction factor, 156
Pipes, branching, 185188
flow of compressible fluids in, 160163
head losses in (see Loss of head)
laminar film in, 158159
laminar flow in 106108, 143146
non circular, 164167
pressure grade line for, 179188
Reynolds number for fluid flow in, 107
stress in walls, 40
turbulent flow in, 106108, 150154
Pitometer, 243, 268
Pitot (Pitotstatic) tube, 241245, 268
PitotVenturi, 246
Plates, drag of, 300306, 311
flow between, 113115
Point gage, 239
Polar diagram, 327
Potential energy, 66
Potential head, 61
Potential pressure, 62
Pressure, measurement of, by gage, 2527
by manometer, 2731
by piezometer column, 24
in a flowing fluid, 236
Pressure energy, 66
Pressure force, 127
on submerged curved areas, 3538
on submerged plane areas, 3135
350
INDEX
Pressure grade line, 141, 179184
Pressure head, 24, 61
Pressure wave, velocity of propagation
of, 333339
Profile drag, see Drag
Projectiles, drag coefficients of, 315
Propeller, 88
Pycnometer, 229
Rapid flow, 211
Reaction turbine, 90
Resistance of immersed objects, see Drag
Reynolds' experiment, 106
Reynolds number, 107, 127129, 130,
136
Roughness, 112
in open channels, 200204
in pipes, 150153
relative, 149
Saltvelocity method, 272
Secondary flows, 121
Separation, 116117, 122, 307
Sewer diagram, 208
Shearing stress, 11, 115
in kiminar flow, 108, 146
in pipe flow, 141
in turbulent flow, 109110
on immersed objects, 119, 294
Short tube, 264
Similarity, applied to, drag of floating
objects, 130
drag of immersed objects, 128, 298
pipe flow, 129, 150
dynamic, 126
geometric, 125
kinematic, 126
Siphon, 184
Skin friction, see Drag
Sluice gate, 265
Span of a foil, 296
Specific energy, 209
Specific gravity, 4
of various liquids, 5
Specific voluhie, 4
Specific weight, 4
of water, 337
Spheres, drag of, 307309
Spillway, 284
Stagnation point, 64, 307
Stagnation pressure, in compressible
fluids, 313
in incompressible fluids, 64, 240
Static tube, 237
Steady flow, 55
Steep slope, 211
Streamline, 56
absolute, 57
instantaneous, 56
picture, 56
Streamlined form, 120121
Streamtube, 57
Struts, drag coefficients of, 311
Submerged areas, forces on, 3138
Surface tension, 14
force, 127
Tanks, stress in walls, 40
Torricelli's theorem, 63
Total heat content, 67
Tranquil flow, 211
Transition from laminar to turbulent
flow, about spheres, 309
in boundary layers, 116, 300
in pipes, 106108
Turbulence, eddying, 118
normal, 107, 118
Turbulent flow, 106110
in boundary layers, 116, 305306
in open channels, 197
in pipes, 107, 150154
Uniform flow, 198199
Unsteady flow, 55
orifice under falling head, 266
Vacuum, 26
Vapor flow, 67
through an orifice, 262
through constriction in a Streamtube,
76
Vapor pressure, 18
effect on cavitation, 342
effect on siphon, 184
in barometer tubes, 29
in various liquids, 19
of water, 19
Varied flow, 198, 217
INDEX
351
Velocity, coefficient of, 251252
energy, 66
gradient, 108
head, 61
pressure, 62
Velocity distribution, effect on velocity
head, 157
in boundary layers, 116, 300301
in open flow, 285
in pipes, 154
pipe coefficient, 154
Venturi meter, 249
Venturi tube, 245
double venturi, 246
Viscometers, Bingham, Ostwald, 234
Engler, Redwood, Saybolt, 234
Falling sphere, 234
MacMichael, Stormer, 232
Viscosity, absolute, 11
force, 127
Viscosity, kinematic, 14
measurement of, 231236
of gases, 12, 341
of liquids, 13, 340
Vortex, compound, 98
forced, 94
free, 96
starting, 321
tip, 323
Wake, formation of, 116117, 307
Weber number, 127
Weir, broadcrested, 216283
sharpcrested, proportional, 282
rectangular contracted, 279
rectangular suppressed, 273278
trapezoidal, 281
triangular, 280
spillway, 284
Westphal balance, 229230