SIXmEDTOM
i
MECHANICS
VICTOR LSTREETER
E. BENJAMIN WYLIE
and English conversions and constants
SI
N
4.448
0.3048
_
lib
1 ft
3
1
lb/in
1
_
lb*
32.174 lb m
1 slug
P
SI unit viscosity
/s
= 9802 N/m 3
62.4 lb/ft 3
Pwater
=
1.94 slugs/ft 3
1
=
kcal
lBtu
3.969 Btu
5.38m-N/kg-K
lb/lb* -°R
1000
1
_
kcal
_
(50°F)
kg/m 3
1055 J
kcal
lkg-K
4187 J
1 ft-
2
2
10
-
Btu
N/m =
_
Twater
llb m -°R
1
6894.76
448.83 gal/min
L
0.4536 kg
_
1 ft
14.594 kg
1 slug
1
m
=
1
(39.4°F)
_
±
Digitized by the Internet Archive
in
2012
http://archive.org/details/fluidmechaniOOstre
FLUID
MECHANICS
SIXTH EDITION
FUUD
MECHANICS
VICTOR
L.
STREETER
Professor of Hydraulics
University of Michigan
E.
BENJAMIN WYUE
Professor of Civil Engineering
University of Michigan
McGRAW-HILL BOOK COMPANY
New York
Johannesburg
San Francisco Dusseldorf
London Mexico Montreal
New Delhi Panama Paris Sao Paulo
Singapore Sydney Tokyo Toronto
St.
Louis
Kuala Lumpur
Library of Congress Cataloging in Publication Data
Streeter, Victor Lyle, date
Fluid mechanics.
Includes bibliographical references.
1.
Fluid mechanics.
TA357.S8 1975
ISBN
I.
532
Wylie, E. Benjamin, joint author.
74-9930
0-07-062193-4
FLUID MECHANICS
Copyright
©
1958, 1962, 1966, 1971, 1975
by McGraw-Hill,
Inc.
All rights reserved.
Copyright 1951 by McGraw-Hill, Inc. All rights reserved.
Printed in the United States of America.
No
part of this publication
may
be reproduced,
stored in a retrieval system,
or transmitted, in
any form or by any means,
electronic, mechanical, photocopying,
recording, or otherwise,
without the prior written permission of the publisher.
456789 KPKP 7987
This book was set in Modern 8 A by
Mono of Maryland Incorporated.
and J. W. Maisel,
the designer was Stephen Naab,
the production supervisor was Sam Ratke witch.
New drawings were done by J & R Services, Inc.
The cover was designed by Rafael Hernandez.
The
editors were B. J. Clark
CONTENTS
Preface
PART
Chapter
I
1
XI
FUNDAMENTALS OF FLUID MECHANICS
Fluid Properties
and Definitions
4
1.2
Force, Mass, and Length Units
7
1.3
Viscosity
1.4
Continuum
1.5
Density, Specific Volume, Specific Weight, Specific Gravity,
1.6
Perfect
9
13
Pressure
1.8
Gas
Bulk Modulus of
Vapor Pressure
1.9
Surface Tension
1.7
\J
3
Definition of a Fluid
1.1
Chapter
13
14
Elasticity
17
18
18
Fluid Statics
27
2.1
Pressure at a Point
27
2.2
Basic Equation of Fluid Statics
29
2.3
Units and Scales of Pressure Measurement
33
2.4
Manometers
2.5
Forces on Plane Areas
38
44
2.6
Force Components on Curved Surfaces
53
2.7
Buoyant Force
2.8
Stability of Floating
60
64
2.9
Relative Equilibrium
2
and Submerged Bodies
71
VI
CONTENTS
Fluid-Flow Concepts and Basic Equations
109
3.1
The Concepts
109
3.2
3.10
Volume to Continuity, Energy,
and Momentum
Flow Characteristics; Definitions
Continuity Equation
Euler's Equation of Motion along a Streamline
Reversibility, Irreversibility, and Losses
The Steady-State Energy Equation
Interrelationships between Euler's Equations and the
Thermodynamic Relations
The Bernoulli Equation
Application of the Bernoulli and Energy Equations to
3.11
Applications of the
Chapter
3
3.3
3.4
3.5
3.6
3.7
3.8
3.9
of
System and Control Volume
Application of the Control
Steady Fluid-Flow Situations
3.12
116
121
127
129
130
132
134
140
144
173
4.2
Dimensional Analysis and Dynamic Similitude
Dimensional Homogeneity and Dimensionless Ratios
Dimensions and Units
208
210
4.3
The n Theorem
211
4.4
Discussion of Dimensionless Parameters
223
4.5
Similitude;
Chapter
4
4.1
/
Linear-Momentum Equation
The Moment-of-Momentum Equation
114
Chapter
5
Model Studies
Viscous Effects: Fluid Resistance
5.1
Laminar, Incompressible, Steady Flow between Parallel
5.2
Laminar Flow through Circular Tubes and Circular Annuli
The Reynolds Number
Prandtl Mixing Length; Velocity Distribution in Turbulent
Flow
Rate Processes
Boundary-Layer Concepts
Drag on Immersed Bodies
Resistance to Turbulent Flow in Open and Closed Conduits
Steady Uniform Flow in Open Channels
Steady Incompressible Flow through Simple Pipe Systems
Lubrication Mechanics
5.4
5.5
5.6
5.7
5.8
5.9
5.10
5.11
226
239
241
Plates
5.3
207
249
254
258
262
266
280
286
287
291
309
Compressible Flow
333
6.1
Perfect- Gas Relationships
6.2
Speed of a Sound Wave;
333
340
Chapter
6
Mach Number
CONTENTS
6.3
6.4
6.5
6.6
6.7
6.8
6.9
Chapter
Isentropic Flow
Shock Waves
Fanno and Rayleigh Lines
Adiabatic Flow with Friction in Conduits
Frictionless Flow through Ducts with Heat Transfer
Steady Isothermal Flow in Long Pipelines
Analogy of Shock Waves to Open-Channel Waves
vii
343
350
355
359
366
371
375
Ideal-Fluid Flow
389
Requirements for Ideal-Fluid Flow
The Vector Operator V
Euler's Equation of Motion
Irrotational Flow; Velocity Potential
389
390
7.5
Integration of Euler's Equations; Bernoulli Equation
7.6
Stream Functions; Boundary Conditions
The Flow Net
Three-dimensional Flow
Two-dimensional Flow
403
406
7
7.1
7.2
7.3
7.4
7.7
7.8
7.9
PART
2
Chapter
8
Fluid
Measurement
449
449
451
Pressure Measurement
8.2
Velocity
8.3
Positive-Displacement Meters
8.4
Rate Meters
8.5
Electromagnetic Flow Devices
8.6
Measurement
Measurement
Measurement
8.8
411
414
428
APPLICATIONS OF FLUID MECHANICS
8.1
8.7
396
401
Measurement
of River
of
Flow
Turbulence
of Viscosity
456
457
482
482
482
483
498
498
504
506
9.4
Turbomachinery
Homologous Units; Specific Speed
Elementary Cascade Theory
Theory of Turbomachines
Impulse Turbines
9.5
Reaction Turbines
517
9.6
Pumps and Blowers
9.7
Centrifugal Compressors
9.8
Cavitation
523
530
534
Steady Closed-Conduit Flow
Exponential Pipe-Friction Formulas
544
Chapter
9
9.1
9.2
9.3
Chapter 10
10.1
511
544
CONTENTS
viii
10.2
Hydraulic and Energy Grade Lines
10.3
The Siphon
10.4
Pipes in Series
10.5
Pipes in Parallel
10.6
Branching Pipes
Networks of Pipes
Computer Program
10.7
10.8
10.9
10.10
Chapter
11
for Steady-State Hydraulic Systems
Conduits with Noncircular Cross Sections
Aging of Pipes
Steady Flow
in
Open Channels
Flow
11.1
Classification of
11.2
Best Hydraulic-Channel Cross Sections
Steady Uniform Flow in a Floodway
11.3
11.4
11.5
Gradually Varied Flow
Classification of Surface Profiles
11.8
Control Sections
11.10
Chapter
12
Computer Calculation
of Gradually Varied
Unsteady Flow
in
12.1
Oscillation of Liquid in a
Establishment of Flow
12.3
Surge Control
12.4
Description of the
12.5
Differential
12.6
The
12.7
Boundary Conditions
12.8
Frictionless Positive Surge
12.11
615
619
629
630
Closed Conduits
12.2
12.9
Flow
Transitions
Flow
12.10
592
595
596
600
604
611
614
Hydraulic Jump; Stilling Basins
Energy; Critical Depth
11.7
11.9
590
591
Specific
11.6
547
553
556
560
563
565
569
578
580
U
Tube
Waterhammer Phenomenon
Equations for Calculation of Waterhammer
Method-of-Characteristics Solution
630
644
645
647
649
654
658
Open-Channel Flow
673
Wave in a Rectangular Channel
Frictionless Negative Surge Wave in a Rectangular Channel
674
Flood Routing in Prismatic Channels
Mechanics of Rainfall-Runoff Relations for Sloping Plane
Areas
676
680
687
Appendixes
701
A
Force Systems, Moments, and Centroids
701
B
C
Partial Derivatives
and Total
Physical Properties of Fluids
Differentials
706
711
CONTENTS
D
Notation
719
E Computer Programming
Aids
723
E.l
Quadratures; Numerical Integration by Simpson's
E.2
Parabolic Interpolation
E.3
Solution of Algebraic or Trancendental Equations
Rule
the Bisection
ix
723
724
by
Method
E.4
Solution of Trancendental or Algebraic Equations by
E.5
the Newton-Raphson Method
Runge-Kutta Solution of Differential Equations
Answers to Even-numbered Problems
Index
726
727
728
732
739
PREFACE
The
general pattern of the text
applications.
although
much
the revision
is
unchanged
in this revision in that it
is
first
of the material
is
updated.
The most
noticeable change in
the incorporation of the metric system of units (SI) in the text,
examples, and problems.
emphasis.
is
emphasizing fundamentals and the second
The chapter contents remain the same as in previous editions
divided into two parts, the
The
The SI and Engilsh
units are
now
given equal
generalized control- volume derivation has been improved so
that the limiting procedures are more easily visualized. Several examples
focusing on environmental issues have been included these problems generally
;
some special information or are limited in certain features because
of the normal complexity of natural situations.
Some material that was included in the previous edition has been removed in this revision, primarily in Part 2, while the material on computer
applications has been strengthened. Chapter 8 on flow measurement no
longer carries descriptive material on many devices and the turbomachinery
chapter omits fluid torque converters and fluid complings. The graphical and
algebraic waterhammer solutions have been removed from Chap. 12.
The use of the digital computer in fluid-flow applications is recommended. The addition of a fairly general program to analyze steady liquid
require
flow in piping systems provides the reader with great flexibility in this
Pumps,
pipelines,
networks.
and reservoirs can be treated in simple systems or
The treatment
in
field.
complex
of turbulent flow in pipelines with empirical (ex-
ponential) pipe-flow equations has been added to Chap. 10.
A
more general
xi
PREFACE
xii
program
is
provided for computation of steady gradually varied flow in open
channels.
The unsteady-flow chapter has been reorganized in its treatment of
waterhammer problems, with a larger variety of boundary conditions considered.
Relevant improvements have also been incorporated in the flood-
routing development and accompanying program.
The
first six
chapters form the basis for a
selected materials from Chap. 7
enough material
The
course in fluids, with
2.
Part 2 contains
for a second course, including support for laboratory.
assistance of
Mr. Joel Caves
relating to environmental issues
received from
first
and topics from Part
many
is
in developing
some
of the
gratefully acknowledged, as
is
examples
the advice
reviewers.
L. STREETER
BENJAMIN WYLIE
VICTOR
E.
FLUID
MECHANICS
1
FUNDAMENTALS OF
FLUID MECHANICS
In
the
first
three chapters of Part
1,
the properties of
and the underlying framework of
concepts, definitions, and basic equations for fluid
dynamics are discussed. Dimensionless parameters
fluids, fluid statics,
next introduced, including dimensional analysis
and dynamic similitude. Chapter 5 deals with real
fluids and the introduction of experimental data into
fluid-flow calculations. Compressible flow of both real
and frictionless fluids is then treated. The final chapter
on fundamentals deals with two- and three-dimensional
ideal-fluid flow. The theory has been illustrated with
elementary applications throughout Part 1.
are
1
FLUID PROPERTIES
AND
DEFINITIONS
one of the engineering sciences that form the basis for
subject branches out into various specialties such as
aerodynamics, hydraulic engineering, marine engineering, gas dynamics,
and rate processes. It deals with the statics, kinematics, and dynamics of
fluids, since the motion of a fluid is caused by unbalanced forces exerted upon
it.
Available methods of analysis stem from the application of the following
principles, concepts, and laws: Newton's laws of motion, the first and second
laws of thermodynamics, the principle of conservation of mass, equations of
state relating to fluid properties, Newton's law of viscosity, mixing-length
concepts, and restrictions caused by the presence of boundaries.
In fluid-flow calculations, viscosity and density are the fluid properties
Fluid mechanics
all
engineering.
is
The
most generally encountered; they play the principal roles in open- and closedchannel flow and in flow around immersed bodies. Surface-tension effects
are of importance in the formation of droplets, in flow of small jets, and in
situations where liquid-gas-solid or liquid-liquid-solid interfaces occur, as well
as in the formation of capillary waves. The property of vapor pressure, accounting for changes of phase from liquid to gas, becomes important when
reduced pressures are encountered.
A liquid-fuel-injection system is an example of an engineering problem
which the performance of the product is significantly affected by the properties of the fluid being handled. Fuel is pumped from a storage tank through
a series of fuel lines and spray nozzles. The process is intermittent and occurs
at high speed. It appears reasonable to expect that less force, and less power,
are needed to pump a light-grade, or "thin," oil than a heavy-grade, or " thick,"
oil.
The terms "light" and "heavy" are qualitative terms which describe
how easily the fuel flows. There is a quantitative way of specifying this fluidity
in
FUNDAMENTALS OF FLUID MECHANICS
4
property, and
it
will
be described later in this chapter. Indeed
sary to define a fluid in a rigorous
manner and
to see
it
will
how our fuel
fits
be necesthis defi-
nition.
How the
fuel sprays
from the nozzle
sion determines the drop formation.
will
The
will
be influenced by other properties of the
The
be affected by how surface ten-
actual design of the nozzle passages
liquid.
because fuel is supplied to the
spray nozzles only at specific times during the operating cycle of an engine.
The duration of fuel delivery is carefully regulated. Consequently there are
fluid flow in the lines is intermittent
These pressures can be very high and,
It is possible that when the pressure gets low
momentarily vaporize and interfere with the expected
pulsations of pressure in the system.
surprisingly, also very low.
enough, the fuel
may
performance of the system.
The
pressure pulses are transmitted along the
column of liquid in the fuel lines similarly to sound waves in air. These pressure waves may be in such a phase relationship with one another that the
waves result in momentary pressure peaks, which are many times the expected system pressures. The speed of the pressure waves depends on a property called the bulk modulus.
The
which follow point up the importance of the physical propA number of definitions are also included so that one
about the property, quantity, or assumption being considered.
sections
erties of a liquid or gas.
can be
1.1
A
specific
DEFINITION OF A FLUID
substance that deforms continuously when subjected to a shear
no matter how small that shear stress may be. A shear force is the
fluid is a
stress,
force
component tangent
the surface
is
is
to a surface,
and
this force divided
the average shear stress over the area.
the limiting value of shear force to area as the area
In Fig.
1.1
a substance
is
by the area
of
Shear stress at a point
is
reduced to the point.
placed between two closely spaced parallel
plates, so large that conditions at their edges
may
be neglected.
The lower
and a force F is applied to the upper plate, which exerts a shear
stress F / A on any substance between the plates. A is the area of the upper
plate. When the force F causes the upper plate to move with a steady (nonzero) velocity, no matter how small the magnitude of F, one may conclude
that the substance between the two plates is a fluid.
The fluid in immediate contact with a solid boundary has the same veThis is an exlocity as the boundary; i.e., there is no slip at the boundary.
plate
is fixed,
1
1
S. Goldstein, "Modern Developments in Fluid Dynamics," vol.
University Press, London, 1938.
II,
pp. 676-680, Oxford
FLUID PROPERTIES
Fig. 1.1
Deformation
resulting from
application
AND DEFINITIONS
of
con-
stant shear force.
perimental fact which has been verified in countless tests with various kinds
and boundary materials. The fluid in the area abed flows to the new
moving parallel to the plate and the velocity
u varying uniformly from zero at the stationary plate to U at the upper plate.
Experiments show that other quantities being held constant, F is directly
proportional to A and to U and is inversely proportional to thickness t. In
equation form
of fluids
position ab'c'd, each fluid particle
F =
in
a
which
fi
r
=
The
the proportionality factor and includes the effect of the par-
is
ticular fluid.
= FI A
If r
for the shear stress,
U
M
7
ratio
U/t
is
the angular velocity of line ab, or
deformation of the
fluid, i.e.,
it is
the rate of angular
The angular
the rate of decrease of angle bad.
may also be written du/dy, as both U/t and du/dy express the vechange divided by the distance over which the change occurs. However,
du/dy is more general as it holds for situations in which the angular velocity
and shear stress change with y. The velocity gradient du/dy may also be
visualized as the rate at which one layer moves relative to an adjacent layer.
In differential form,
velocity
locity
T
is
=
du
(1.1.1)
H
the relation between shear stress and rate of angular deformation for one-
dimensional flow of a
fluid.
The
proportionality factor
/x
is
called the viscosity
FUNDAMENTALS OF FLUID MECHANICS
6
of the fluid,
and Eq.
of his "Principia,"
(1.1.1) is
Newton
of his studies of the planets
Newton's law of
viscosity.
In the second book
considered the circular motion of fluids as part
and wrote
Hypothesis
The
resistance arising from the want of lubricity in the parts of a fluid, is, other things
being equal, proportional to the velocity with which the parts of a fluid are separated
from one another.
A
an
plastic substance
initial yield
cannot
fulfill
the definition of a fluid because
it
has
shear stress that must be exceeded to cause a continuous defor-
An elastic substance placed between the two plates would deform a
amount proportional to the force, but not continuously at a definite
rate. A complete vacuum between the plates would not result in a constant
If sand were placed between the
final rate but in an ever-increasing rate.
two plates, dry friction would require a finite force to cause a continuous motion. Thus sand will not satisfy the definition of a fluid.
Fluids may be classified as Newtonian or non-Newtonian. In Newtonian
fluid there is a linear relation between the magnitude of applied shear stress
and the resulting rate of deformation [ju constant in Eq. (1.1.1)], as shown
in Fig. 1.2. In non-Newtonian fluid there is a nonlinear relation between the
magnitude of applied shear stress and the rate of angular deformation. An
mation.
certain
Shear stress r
Fig. 1.2
Rheological diagram.
FLUID PROPERTIES
ideal plastic
du/dy.
A
AND DEFINITIONS
7
has a definite yield stress and a constant linear relation of r to
thixotropic substance, such as printer's ink, has a viscosity that
is
dependent upon the immediately prior angular deformation of the substance
and has a tendency to take a set when at rest. Gases and thin liquids tend to
be Newtonian fluids, while thick, long-chained hydrocarbons may be nonNewtonian.
For purposes of analysis, the assumption is frequently made that a fluid
With zero viscosity the shear stress is always zero, regardless
is nonviscous.
of the motion of the fluid. If the fluid is also considered to be incompressible,
it is then called an ideal fluid and plots as the ordinate in Fig. 1.2.
1.2
FORCE, MASS,
AND LENGTH UNITS
In this text units of the English system and the International System (SI)
are employed.
and the
The basic English units are the pound force (lb) the slug mass,
The basic SI units are the newton force (N), the
(ft).
,
foot length
kilogram mass (kg), and the meter length (m).
These units are consistent in
that the unit of force accelerates the unit of mass
1
unit of length per second
The pound mass (lb TO ) is used in some tabulations of properties
and is related to the slug by 32.174 lb m = 1 slug. The kilogram force (kg/)
is also used in some countries and is related to the newton by 9.806 N = 1 kg/.
per second.
Abbreviations of SI units are written in lowercase (small) letters for
terms
like
hours (h) meters (m) and seconds
,
,
(s)
.
When a unit is named after
a person, the abbreviation (but not the spelled form)
ample, watt (W), pascal (Pa), or newton (N).
is
capitalized; for ex-
Multiples and submultiples
powers of 10 are indicated by prefixes, which are also abbreviated, as in
such familiar combinations as centimeter (cm), for 10 -2 m, and kilogram (kg)
for 10 3 g. Common prefixes are shown in Table 1.1. Note that prefixes may
in
Table
1.1
Selected prefixes
for
powers of 10
in
SI
units
Multiple
SI
Abbre-
SI
Abbre-
prefix
viation
Multiple
prefix
viation
10" 3
milli
micro
nano
M
n
pico
P
10 9
giga
G
10 6
mega
M
10 3
kilo
k
io- 6
10~ 9
lO" 2
centi
c
10 -i2
m
FUNDAMENTALS OF FLUID MECHANICS
8
not be doubled up: The correct form for 10~9 is the prefix n-, as in nanometers;
combinations like millimicro-, formerly acceptable, are no longer to be used.
The pound
of force
is
defined in terms of the pull of gravity, at a speci-
on a given mass of platinum. At standard gravitation, g = 32.174 ft/s 2 the body having a pull of one pound has a mass of one
pound mass. When Newton's second law of motion is written in the form
fied (standard) location,
,
-a
=
F
(1.2.1)
and applied
lib =
to this object falling freely in a
^^32.174
vacuum
at standard conditions
ft/s 2
go
it is
clear that
=
00
32.174 lb„- ft/lb
-s
2
(1.2.2)
Similarly, in SI units
1
k g/
=
—
kg
1
- 9.806
m/s 2
go
and
go
=
9.806 kg -m/kg r
s
2
(1.2.3)
The number g is a constant, independent of location of application of
Newton's law and dependent only on the particular set of units employed.
of a body remains
At any other location than standard gravity, the mass
pull
of
gravity)
varies:
constant, but the weight (force or
M
W
=
M-
(1.2.4)
go
For example, where g = 31.0
,
1A1
10 lb„ weighs
•
.
(10 lb w ) (31.0 ft/s 2 )
32.174 lb m ft/lb
-
The
slug
is
-s
ft/s 2
=
,
9.635 lb
2
a derived unit of mass, defined as the amount of mass that
AND DEFINITIONS
FLUID PROPERTIES
accelerated one foot per second per second
is
by a force
of
9
one pound. For these
Since fluid mechanics
units the constant g is unity; that is, 1 slug ft/lb s 2
is so closely tied to Newton's second law, the slug may be defined as
•
=
lslug
•
.
llb-s 2 /ft
and the consistent
(1.2.5)
set of units slug,
the dimensional constant g Q
.
pound,
foot,
second
may
be used without
Similarly in the SI the kilogram, newton, meter,
and second are related by
1
N=
kg-m/s 2
1
(1.2.6)
and g is not needed. If the pound mass or the kilogram force
dynamical equations, g must be introduced.
On
As they
the inside front cover,
many
is
to be used in
conversions for various units are given.
are presented in the form of dimensionless ratios equal to
1, they can
be used on one side of an equation, as a multiplier; or as a divisor, to convert
units.
1.3
Of
VISCOSITY
all
the fluid properties, viscosity requires the greatest consideration in
the study of fluid flow.
The nature and
characteristics of viscosity are dis-
cussed in this section as well as dimensions and conversion factors for both
absolute and kinematic viscosity.
virtue of which
it
Viscosity
offers resistance to shear.
is
that property of a fluid
Newton's law
of viscosity
by
[Eq.
(1.1.1)] states that for a given rate of angular deformation of fluid the shear
Molasses and tar are examples
have very small viscosities.
The viscosity of a gas increases with temperature, but the viscosity of a
liquid decreases with temperature. The variation in temperature trends can
by explained by examining the causes of viscosity. The resistance of a fluid
to shear depends upon its cohesion and upon its rate of transfer of molecular
momentum. A liquid, with molecules much more closely spaced than a gas,
has cohesive forces much larger than a gas. Cohesion appears to be the predominant cause of viscosity in a liquid, and since cohesion decreases with temperature, the viscosity does likewise. A gas, on the other hand, has very small
stress is directly proportional to the viscosity.
of highly viscous liquids; water
air
Most of its resistance to shear stress is the result
momentum.
rough model of the way in which momentum transfer
cohesive forces.
fer of
and
of the trans-
molecular
As a
an apparent shear
stress,
gives rise to
consider two idealized railroad cars loaded with
FUNDAMENTALS OF FLUID MECHANICS
10
4Mfc^
mmm
Fig. 1.3
fer of
Model illustratingtrans-
momentum.
sponges and on parallel tracks, as in Fig. 1.3. Assume each car has a water
tank and pump, arranged so that the water is directed by nozzles at right
A
B moving to the right,
and being absorbed by the sponges.
Car A will be set in motion owing to the component of the momentum of the
jets which is parallel to the tracks, giving rise to an apparent shear stress
between A and B. Now if A is pumping water back into B at the same rate,
its action tends to slow down B and equal and opposite apparent shear forces
result.
When A and B are both stationary or have the same velocity, the
pumping does not exert an apparent shear stress on either car.
Within fluid there is always a transfer of molecules back and forth across
any fictitious surface drawn in it. When one layer moves relative to an adjacent layer, the molecular transfer of momentum brings momentum from
one side to the other so that an apparent shear stress is set up that resists the
relative motion and tends to equalize the velocities of adjacent layers in a
manner analogous to that of Fig. 1.3. The measure of the motion of one layer
relative to an adjacent layer is du/dy.
Molecular activity gives rise to an apparent shear stress in gases which
is more important than the cohesive forces, and since molecular activity inangles to the track.
with the water from
First, consider
its
nozzles striking
stationary and
A
creases with temperature, the viscosity of a gas also increases with temperature.
For ordinary pressures viscosity is independent of pressure and depends
upon temperature only. For very great pressures, gases and most liquids have
shown erratic variations of viscosity with pressure.
A fluid at rest or in motion so that no layer moves relative to an adjacent
layer will not have apparent shear forces set up, regardless of the viscosity,
because du/dy is zero throughout the fluid. Hence, in the study of fluid statics,
no shear forces can be considered because they do not occur in a static fluid,
and the only stresses remaining are normal stresses, or pressures. This greatly
simplifies the study of fluid statics, since any free body of fluid can have only
gravity forces and normal surface forces acting on it.
The dimensions of viscosity are determined from Newton's law of vis-
FLUID PROPERTIES
cosity [Eq. (1.1.1)].
Solving for the viscosity
AND DEFINITIONS
11
\i
du/dy
and inserting dimensions F, L,
T
and time,
for force, length,
Caaa
u.LT-
t:FL-2
y:L
1
ol\jV\---
s
FL~ T. With the force dimension expressed
Newton's second law of motion, F = MLT~2 the
dimensions of viscosity may be expressed as ML~ T~
The English unit of viscosity (which has no special name) is 1 lb s/ft 2
shows that
/x
in terms of
mass by use
2
has the dimensions
of
,
l
l
.
•
or
1
The cgs unit of viscosity, called the poise
dyn-s/cm2 or 1 g/cm-s. The centipoise (cP) is one one-hundredth of
Water at 68°F has a viscosity of 1.002 cP. The SI unit of viscosity
slug/ft -s (these are identical)
(P), is 1
a poise.
.
meter second or newton-seconds per square meter has no
name. The SI unit is 10 times larger than the poise unit. 1
in kilograms per
Kinematic viscosity
The viscosity m
is
frequently referred to as the absolute viscosity or the dynamic
viscosity to avoid confusing
of viscosity to
p
=
mass
with the kinematic viscosity
it
v
,
which
is
the ratio
density,
-
(1.3.1)
P
The kinematic
ber,
1
which
is
viscosity occurs in
The conversion from the English
1
slug 14.594 kg
ft-s
1
slug
many
VD/v. The dimensions of
1 ft
0.3048
Engl unit viscosity
47.9 unit SI viscosity
m
=
applications, e.g., the
v
are
L2 T~
l
.
Reynolds num-
The English
unit,
unit of viscosity to the SI unit of viscosity
47.9
kg/m«s
is
2
1 ft /s,
FUNDAMENTALS OF FLUID MECHANICS
12
has no special name; the cgs unit, called the stoke (St),
unit of kinematic viscosity
1
To
cm
2
The SI
/s.
m /s.
convert to the English unit of viscosity from the English unit of
kinematic viscosity,
per cubic foot.
tiply
is 1
is 1
2
it is
by the mass density
by the mass density
necessary to multiply
To change
from the stoke,
to the poise
it is
grams per cubic centimeter, which
in
in slugs
necessary to mulis
numerically
equal to the specific gravity.
EXAMPLE
1.1
A
P and
liquid has a viscosity of 0.05
a specific gravity of
Calculate (a) the viscosity in SI units; (6) the kinematic viscosity in
SI units; (c) the viscosity in English units; (d) the kinematic viscosity in
0.85.
stokes.
0.05 g
(G)
^cm,
«
'
(c)
m
- "-
1kg 100 cm
1000
1
W)
f
=
0.005
= -M =
Viscosity
is
perature only.
kg/m-s
X
—
alii
„
47.9
,
3
-
nr
/ff
,
**
•
q
=
X 10- mV s =
5.88
,mVs
0.0001044 slug/ft -s
kg/m-s
0.05g /cm-s
0.85
p
=0.005 kg/m-s
nZT^V,
kg/m
X
0.8o
p
1m
1000 g
g/cm
= AACOC
/s
0.0588 cm22/
3
practically independent of pressure
The kinematic
viscosity of liquids,
and depends upon temand of gases at a given
Charts for the determination of absolute viscosity and kinematic viscosity are given in Appendix
pressure,
is
substantially a function of temperature.
C, Figs. C.l and C.2, respectively.
1
The conversion from the SI
viscosity
lft
V
— (\0.3048m/
i2
1
unit of kinematic viscosity to the English unit of kinematic
is
=
(
i
1
kinematic viscosity unit (SI)
10.764 kinematic viscosity (Engl)
FLUID PROPERTIES
AND DEFINITIONS
13
CONTINUUM
1.4
In dealing with fluid-flow relationships on a mathematical or analytical basis,
necessary to consider that the actual molecular structure is replaced by
a hypothetical continuous medium, called the continuum. For example, velocity at a point in space is indefinite in a molecular medium, as it would be
zero at all times except when a molecule occupied this exact point, and then
it would be the velocity of the molecule and not the mean mass velocity of the
particles in the neighborhood. This dilemma is avoided if one considers velocity at a point to be the average or mass velocity of all molecules surrounding
the point, say, within a small sphere with radius large compared with the
it is
mean
distance between molecules.
mean
distance between molecules
With n molecules per cubic centimeter, the
llz
is of the order n~
cm. Molecular theory,
however, must be used to calculate fluid properties
(e.g.,
viscosity)
which are
associated with molecular motions, but continuum equations can be em-
ployed with the results of molecular calculations.
In rarefied gases, such as the atmosphere at 50 mi above sea level, the
mean free path of the gas to a characteristic length for a body or
conduit is used to distinguish the type of flow. The flow regime is called gas
dynamics for very small values of the ratio, the next regime is called slip flow,
1
ratio of the
and for large values of the ratio
dynamics regime is studied.
The
tion are
it is free
molecule flow. In this text only the gas-
quantities density, specific volume, pressure, velocity,
assumed to vary continuously throughout a
fluid (or
and accelerabe constant).
DENSITY, SPECIFIC VOLUME, SPECIFIC WEIGHT, SPECIFIC
GRAVITY, PRESSURE
1.5
is defined as its mass per unit volume. To define density
mass Am of fluid in a small volume A'U surrounding the point is
divided by AV and the limit is taken as AV becomes a value e 3 in which e is
still large compared with the mean distance between molecules,
The density
p of a fluid
at a point the
p
=
lim
A^<3
—
AV
(1.5.1)
For water at standard pressure (14.7 lb/in2 ) and 39.4°F (4°C), p = 1.94
slugs/ft 3 or 62.43 lb m/ft 3 or 1000 kg/m 3
The specific volume v s is the reciprocal of the density p; that is, it is the
.
1
The mean
free
path
is
the average distance a molecule travels between collisions.
FUNDAMENTALS OF FLUID MECHANICS
14
volume occupied by unit mass
Hence
1
-
=
vs
of fluid.
(1.5.2)
P
The
specific weight
7 of a substance
is its
weight per unit volume.
It
changes with location,
7
=
(1.5.3)
P9
depending upon gravity. It is a convenient property when dealing with fluid
statics or with liquids with a free surface.
The specific gravity S of a substance is the ratio of its weight to the
weight of an equal volume of water at standard conditions. It may also be
expressed as a ratio of its density or specific weight to that of water.
The normal force pushing against a plane area divided by the area is
the average pressure. The pressure at a point is the ratio of normal force to
area as the area approaches a small value enclosing the point. If a fluid exerts
a pressure against the walls of a container, the container will exert a reaction
on the fluid which will be compressive. Liquids can sustain very high compressive pressures, but unless they are extremely pure, they are very weak in
tension. It is for this reason that the absolute pressures used in this book are
never negative, since this would imply that fluid is sustaining a tensile stress.
Pressure has the units force per area and may be pounds per square inch,
pounds per square foot, or newtons per square meter, also called pascals (Pa)
Pressure may also be expressed in terms of an equivalent length of a fluid
column, as shown in Sec. 2.3.
1
1.6
PERFECT GAS
In this treatment, thermodynamic relationships and compressible-fluid-flow
cases have been limited generally to perfect gases.
in this section,
and
its
The
various interrelationships with
perfect gas
is
defined
specific heats are treated
in Sec. 6.1.
The
perfect gas, as used herein,
is
defined as a substance that satisfies
the perfect-gas law
pv 8
1
=
RT
(1.6.1)
These conditions are normally 4°C. In this text standard conditions are considered to be
a temperature of 68°F and a pressure of 30 in Hg abs unless specified otherwise.
AND DEFINITIONS
FLUID PROPERTIES
and that has constant
R
volume,
specific heats,
the gas constant, and
T
p
is
the absolute pressure,
the absolute temperature.
gas must be carefully distinguished from the ideal
fluid.
An
vs
15
the specific
The
perfect
ideal fluid
is
and incompressible. The perfect gas has viscosity and can theredevelop shear stresses, and it is compressible according to Eq. (1.6.1).
Equation (1.6.1) is the equation of state for a perfect gas. It may be
frictionless
fore
written
RT
p =
P
The
units of
(1.6.2)
R can be determined from the equation when the other units
For p in pounds per square foot, p in slugs per cubic foot, and
459.6) in degrees Rankine (°R),
are known.
T
(°F
R =
For p
R =
+
(lb/ft 2 ) (ft 3 /slug-°R)
=
ft-
pounds mass per cubic
in
(lb/ft 2 ) (ft 3 /lb m -°R)
The magnitude
Values of
R
of
R
=
lb/slug
°R
foot,
ft-lb/lb m -°R
in slugs is 32.174 times greater
for several
common
than in pounds mass.
gases are given in Table C.3.
and above the critical temperature
As the pressure increases, the discrepancy
increases and becomes serious near the critical point. The perfect-gas law
encompasses both Charles' law and Boyle's law. Charles' law states that for
constant pressure the volume of a given mass of gas varies as its absolute temReal gases below
critical pressure
tend to obey the perfect-gas law.
perature.
Boyle's law (isothermal law) states that for constant temperature
the density varies directly as the absolute pressure.
units of gas
pV =
is
mv
s
;
The volume V
of
m
mass
hence
mRT
(1.6.3)
Certain simplifications result from writing the perfect-gas law on a mole basis.
A pound mole of gas is the number of pounds mass of gas equal to its molecular
weight;
e.g.,
a pound mole of oxygen
2
is
32 lb TO
.
With
vs
being the volume
per mole, the perfect-gas law becomes
p~v s
if
=
M
MRT
is
the molecular weight.
(1.6.4)
In general,
if
n
is
the
number
of moles of the
FUNDAMENTALS OF FLUID MECHANICS
16
gas in volume V,
nMRT
pV =
since
nM
(1.6.5)
= m. Now, from Avogadro's
law, equal volumes of gases at the
same absolute temperature and pressure have the same
number
of molecules;
hence their masses are proportional to the molecular weights. From Eq.
must be constant, since pV/nT is the same for any(1.6.5) it is seen that
perfect gas. The product
is called the universal gas constant and has a
value depending only upon the units employed. It is
MR
MR
MR
=
The
gas constant
R =
—
1545
ft
-lb/lbm-
R
mole -°R
(1.6.6)
can then be determined from
1545
ft
-lb/lb™ -°R
(1.6.7)
or in slug units,
R =
1545
X S2AU
ft-lb/slug-°R
M
(1.6.8)
In SI units
R =
—— m-N/kg-K
M
8312
(1.6.9)
knowledge of molecular weight leads to the value of R. In Table C.3
Appendix C molecular weights of some common gases are listed.
Additional relationships and definitions used in perfect-gas flow are
introduced in Chaps. 3 and 6.
so that
of
EXAMPLE
1.2
A gas with molecular
weight of 44
is
at a pressure of 13.0 psia
(pounds per square inch absolute) and a temperature of 60°F.
its density in slugs per cubic foot.
From Eq.
g =
(1.6.8)
1545 ^32.174
In 1967 the
name
=li29ftlb/slugOR
of the degree Kelvin (°K)
was changed
to kelvin (K).
Determine
AND DEFINITIONS
FLUID PROPERTIES
Then from Eq.
P
(1.6.2)
(13.0 lb/in2 ) (144 in2 /ft 2 )
= V =
RT
17
o
+
(1129ft.lb/slug- R)(460
=
60°R)
..._
° 00319 SlUg/ft
BULK MODULUS OF ELASTICITY
1.7
In the preceding section the compressibility of a perfect gas is described by
For most purposes a liquid may be considered as incom-
the perfect-gas law.
but for situations involving either sudden or great changes in presbecomes important. Liquid (and gas) compressibility
also becomes important when temperature changes are involved, e.g., free
convection. The compressibility of a liquid is expressed by its bulk modulus
pressible,
sure, its compressibility
If the pressure of a unit volume of liquid is increased by dp,
volume decrease — dV the ratio — dp/dV is the bulk modulus
K. For any volume V of liquid,
of elasticity.
will cause a
elasticity
K
= -
-^-
where
M
To
is
dimensionless,
K
is
expressed in the units of
K
temperatures and pressures,
ordinary
is
it
of
(1.7.1)
dV/V
dV/V
Since
;
the abbreviation for
mega
=
(
=
10
300,000
psi
For water at
p.
=
2068
MN/m
2
,
6
)
gain some idea about the compressibility of water, consider the ap-
plication of 100 psi pressure to
1 ft
3
of water.
When
Eq. (1.7.1)
is
solved for
-dV,
Vdp
=
-dV =
1.0
K
X
100
1
=
ft
3
3000
300,000
Hence, the application of 100 psi to water under ordinary conditions causes
volume to decrease by only 1 part in 3000. As a liquid is compressed, the
its
resistance to further compression increases; therefore
sure.
At 45,000
EXAMPLE
(1000 cm 3
1.3
)
at
A
1
value of
psi the
K
for
K
liquid compressed in a cylinder has a
MN/m
2
and a volume
of 995
cm
3
=
Ap
V- = AV/V
2-1 MN/m
995
f-
2
- 1000/1000
= 200
_
MN/m
,
'
volume
MN/m
at 2
bulk modulus of elasticity?
K
increases with pres-
water has doubled.
n
2
of
2
.
1 liter
What
(1)
is its
FUNDAMENTALS OF FLUID MECHANICS
18
1.8
VAPOR PRESSURE
Liquids evaporate because of molecules escaping from the liquid surface.
The vapor molecules
pressure.
number
If the space
of
exert a partial pressure in the space,
above the liquid
is
known
as vapor
confined, after a sufficient time the
vapor molecules striking the liquid surface and condensing are
number escaping in any interval of time, and equilibrium
Since this phenomenon depends upon molecular activity, which is a
exists.
function of temperature, the vapor pressure of a given fluid depends upon
just equal to the
temperature and increases with it. When the pressure above a liquid equals
the vapor pressure of the liquid, boiling occurs. Boiling of water, for example,
may occur at room temperature if the pressure is reduced sufficiently. At
68°F water has a vapor pressure of 0.339 psi, and mercury has a vapor pressure of 0.0000251 psi.
In many situations involving the flow of liquids it is possible that very
low pressures are produced at certain locations in the system. Under such
circumstances the pressures may be equal to or less than the vapor pressure.
When this occurs, the liquid flashes into vapor. This is the phenomenon of
cavitation.
A rapidly expanding vapor pocket, or cavity, forms, which is
usually swept
away from
its
point of origin and enters regions of the flow where
The cavity collapses. This growth
vapor bubbles affect the operating performance of hydraulic pumps and turbines and can result in erosion of the metal parts in the
the pressure
and decay
is
greater than vapor pressure.
of the
region of cavitation.
1.9
SURFACE TENSION
Capillarity
At the
interface
between a liquid and a
gas, or
two immiscible
liquids, a film
or special layer seems to form on the liquid, apparently owing to attraction
of liquid molecules
below the surface.
It
is
a simple experiment to place a
it is supported there
small needle on a quiet water surface and observe that
by the film.
The formation of this film may be visualized on the basis of surface
energy or work per unit area required to bring the molecules to the surface.
The surface tension is then the stretching force required to form the film,
obtained by dividing the surface-energy term by unit length of the film in
equilibrium. The surface tension of water varies from about 0.005 lb/ft at
68°F to 0.004 lb/ft at 212°F.
Table 1.2.
Surface tensions of other liquids are given in
FLUID PROPERTIES
Table
1.2
Surface tension of
common
AND DEFINITIONS
19
liquids in contact
with air at 68°F (20°C)
Surface tension
Liquid
0.00153
0.00198
0.00183
.0016-0.0022
0.00501
Benzene
Carbon tetrachloride
Kerosene
Water
Mercury
0.0223
0.0289
0.0267
0.0233-0.0321
0.0731
0.0352
0.0269
0.0333
0.5137
0.3926
0.4857
0.0024-0.0026
0016-0.0026
0.0350-0.0379
0.0233-0.0379
In air
In water
In
N/m
lb/ft
Alcohol, ethyl
<r
vacuum
Oil:
Lubricating
Crude
0.
The
is
to increase the pressure within a droplet
of liquid or within a small liquid jet.
For a small spherical droplet of radius
action of surface tension
the internal pressure p necessary to balance the tensile force due to the surface tension <r is calculated in terms of the forces which act on a hemispherical
r
free
body
(see Sec. 2.6),
2a
pirr
2irra
or
r
For the cylindrical liquid
P
jet of radius
r,
the pipe-tension equation applies:
=
Both equations show that the pressure becomes
large for a very small radius
of droplet or cylinder.
Capillary attraction is caused by surface tension and by the relative
value of adhesion between liquid and solid to cohesion of the liquid. A liquid
that wets the solid has a greater adhesion than cohesion.
tension in this case
that
is
partially
is
The
action of surface
to cause the liquid to rise within a small vertical tube
immersed
in
it.
For liquids that do not wet the
tension tends to depress the meniscus in a small vertical tube.
tact angle
between liquid and
solid
is
known, the capillary
solid, surface
When
rise
the con-
can be com-
FUNDAMENTALS OF FLUID MECHANICS
20
h =
Capillary rise or depression,
2
25
r
mm
3
1.0
| 20-
15
5
°' 2
0.02
0.06
0.04
0.08
h = Capillary
0.10
rise or
0.14
0.12
depression,
0.16
0.18
0.20
in.
Fig. 1.4 Capillarity in circular glass tubes. (By permission from R. L.
Daugherty, "Hydraulics," copyright 1937, McGraw-Hill Book Company.)
puted
an assumed shape of the meniscus. Figure 1.4 shows the capillary
water and mercury in circular glass tubes in air.
for
rise for
PROBLEMS
1.1
Classify the substance that has the following rates of deformation and corre-
sponding shear
stresses:
du/dy, rad/s
T,
lb/ft 2
1.2
15
1
3
5
20
30
40
Classify the following substances (maintained at constant temperature)
(a) du/dy, rad/s
r,
(b)
N/m
A
.
6
5
4
4
6
8
6
4
1.1
1.8
2
0.3
0.6
0.9
1.2
2
Newtonian
(Fig. 1.5)
4
0.5
du/dy, rad/s
N/m
1.3
2
du/dy, rad/s
r,
(c)
lb/ft 2
3
liquid flows
The upper
surface
down an
is
inclined plane in a thin sheet of thickness
in contact with air,
which
offers
t
almost no resistance
FLUID PROPERTIES
to the flow. Using Newton's law of viscosity, decide
AND DEFINITIONS
what the value
21
meas-
of du/dy, y
ured normal to the inclined plane, must be at the upper surface. Would a linear variation of
u with y be expected?
1.4
What
1.5
A
When
kinds of rheological materials are paint and grease?
Newtonian
a force of 500
fluid is in the clearance
between a shaft and a concentric
sleeve.
N is applied to the sleeve parallel to the shaft, the sleeve attains a
1500 N force is applied, what speed will the sleeve attain? The
speed of 1 m/s. If
temperature of the sleeve remains constant.
1.6
Determine the weight
1.7
When
in
pounds
of 3 slugs
mass
at a place
where
standard scale weights and a balance are used, a body
=
g
31.7 ft/s2
.
found to be
is
equivalent in pull of gravity to two of the 1-lb scale weights at a location where g =
31.5 ft/s2 What would the body weigh on a correctly calibrated spring balance (for
.
sea level) at this location?
1.8
Determine the value
of proportionality constant go
needed for the following
set of
units: kip (1000 lb), slug, foot, second.
i/1.9
On
another planet, where standard gravity
is
3
m/s 2 what would be
,
the value of
the proportionality constant g in terms of the kilogram force, gram, millimeter, and
second?
1.10
A
correctly calibrated spring scale records the weight of a 51-lb TO
lb at a location
1.11
What
away from
the earth.
What
is
body
as 17.0
the value of g at this location?
Does the weight of a 20-N bag of flour denote a force or the mass of the flour?
the mass of the flour in kilograms? What are the mass and weight of the flour
is
at a location where the gravitational acceleration
is
one-seventh that of the earth's
standard?
1.12
Convert 10.4 SI units of kinematic viscosity to English units
cosity
if
S=
1.13
A
shear stress of 4
formation of
of
dynamic
vis-
0.85.
1
rad/s.
dyn/cm2 causes a Newtonian fluid
What is its viscosity in centipoises?
to
have an angular de-
FUNDAMENTALS OF FLUID MECHANICS
22
3
20
diam
in.
—
1
1b
-
J--
a---
8
~u V=
0-4
ft/sec
in.
Fig. 1.6
1.14
A plate, 0.5 mm distant from a fixed plate, moves at 0.25 m/s and requires a force
per unit area of 2
Pa (N/m2 )
to maintain this speed.
Determine the
fluid viscosity of
the substance between the plates in SI units.
1.15
Determine the viscosity
1.16
A flywheel weighing 500 N has a radius of gyration of 30 cm. When it is rotating
600 rpm,
The
its
mine the
X
J
speed reduces
sleeve length
is
1
of fluid
between shaft and sleeve
rpm/s owing
to fluid viscosity
in Fig. 1.6.
between sleeve and
5 cm, shaft diameter 2 cm, and radial clearance 0.05
mm.
shaft.
Deter-
fluid viscosity.
1.17 A 1-in-diameter steel cylinder 12 in long falls, because of its own weight, at a
uniform rate of 0.5 ft/s inside a tube of slightly larger diameter. A castor-oil film of
constant thickness is between the cylinder and the tube. Determine the clearance
between the tube and the cylinder. The temperature
A
is
100°F.
cm moves
within a cylinder of 5.010 cm. Determine
the percentage decrease in force necessary to
move the piston when the lubricant warms
1.18
piston of diameter 5.000
up from
1.19
to 120°C.
How much
greater
is its
Use
greater
from Fig. C.l, Appendix C.
crude-oil viscosity
is
the viscosity of water at 32°F than at 212°F?
How much
kinematic viscosity for the same temperature range?
1.20
A
1.21
A fluid has a specific gravity of 0.83 and a kinematic viscosity of 3 St. What is its
has a viscosity of 4 cP and a density of 50 lb m /ft 3
matic viscosity in English units and in stokes.
fluid
viscosity in English units
1.22
A
and
.
Determine
its
kine-
in SI units?
body weighing 90 lb with a flat surface area of 2 ft 2 slides down a lubricated
making a 30° angle with the horizontal. For viscosity of 1 P and
inclined plane
body speed
of 3 ft/s, determine the lubricant film thickness.
1.23
What
1.24
Determine the kinematic viscosity
1.25
Calculate the value of the gas constant
ft -lb/lb w
1.26
is
the viscosity of gasoline at 30°C in poises?
of
benzene at 60°F in stokes.
R in SI units,
starting with
R = 1545/M
-°R.
What
is
the specific volume in cubic feet per pound mass and cubic feet per slug
of a substance of specific gravity 0.75?
FLUID PROPERTIES
1.27
What
1.28
The density
(6) specific
A
1.29
What
-
1.30
of a substance
volume, and
(c)
is
2.94
g/cm 3
What
.
(a)
is its
specific gravity,
specific weight?
by F
force, expressed
=
4i
-+-
3j
+ 9k,
acts
upon a square
area, 2
by 2
in,
in
Resolve this force into a normal-force and a shear-force component.
are the pressure and the shear stress?
Repeat the calculations
for F
= — 4i
+
9k.
A
gas at 20°C and 2 kg//cm 2 has a volume of 40
m *N/kg *K.
Determine the density and mass
1
and a gas constant
What
is
the specific weight of air at 60 psia and 90°F?
1.32
What
is
the density of water vapor at 30
1.33
A
N/cm2
ft
2
temperature of 2000 psfa (lb/ft abs) and 600°R, respectively.
volume and specific weight?
kg
of
hydrogen
is
210
abs and 15°C in SI units?
gas with molecular weight 44 has a volume of 4.0
1.0
R=
of the gas.
1.31
1.34
23
the relation between specific volume and specific weight?
is
the xy plane.
3j
AND DEFINITIONS
confined in a volume of 0.1
m
at
3
and a pressure and
What
are its specific
— 40°C. What
is
the
pressure?
^ 1.35 Express the bulk modulus
volume change.
1.36
of elasticity in
For constant bulk modulus
of elasticity,
terms of density change rather than
how does
the density of a liquid vary
with the pressure?
1.37
What
is
the bulk modulus of a liquid that has a density increase of 0.02 percent
for a pressure increase of 1000 lb/ft 2 ?
J
1.38
For
K=
1.39
A
p
=
N/m
of 50,000
2
?
300,000 psi for bulk modulus of elasticity of water what pressure
required to reduce
increased
For a pressure increase
its
volume by
steel container
by 10,000
62.4 lb TO /ft 3 For
.
is
0.5 percent?
in volume 1 percent when the pressure within it is
At standard pressure, 14.7 psia, it holds 1000 lb w water
— 300,000 psi, when it is filled, how many pounds mass water
expands
psi.
K
need be added to increase the pressure to 10,000 psi?
the isothermal bulk modulus for air at 4 kg// cm 2 abs?
1.40
What
1.41
At what pressure can cavitation be expected
is
at the inlet of a
pump
that
is
handling water at 100°F?
1.42
What
is
the pressure within a droplet of water of 0.002-in diameter at 68°F
pressure outside the droplet
1.43
is
is
if
the
standard atmospheric pressure of 14.7 psi?
A small circular jet of mercury 0.1 mm in diameter issues from an opening. What
the pressure difference between the inside and outside of the jet
1.44 Determine the capillary
diameter glass tube.
rise for distilled
when
at 20 °C?
water at 104°F in a circular J-in
FUNDAMENTALS OF FLUID MECHANICS
24
Ring
What
1.45
diameter of glass tube
required
is
if
the capillary effects on the water
within are not to exceed 0.02 in?
Using the data given in Fig.
1.46
two
1.4,
estimate the capillary
rise of
tap water between
parallel glass plates 0.20 in apart.
A method
1.47
of determining the surface tension of a liquid
is
to find the force
needed to pull a platinum wire ring from the surface (Fig. 1.7). Estimate the force
necessary to remove a 2-cm diameter ring from the surface of water at 20°C. Why is
platinum used as the material for the ring?
A
1.48
fluid is a
substance that
any container
(a)
always expands until
(b)
is
(c)
cannot be subjected to shear forces
it fills
practically incompressible
(d)
cannot remain at rest under action of any shear force
(e)
has the same shear stress at a point regardless of
A
1.49
2.0-lb TO object
motion
weighs 1.90 lb on a spring balance. The value of g at this loca-
in feet per second per second,
tion
is,
(a)
30.56
At a
1.50
its
(6)
32.07
32.17
(c)
location where g
=
(of)
33.87
30.00 ft/s2 2.0 slugs
,
is
(e)
none of these answers
equivalent to
how many pounds
mass?
(a)
60.0
(e)
none of these answers
1.51
(a)
The
0.30
(6)
62.4
(c)
64.35
not equivalent units
(d)
weight, in pounds, of 3 slugs on a planet where g
(6)
0.932
(c)
30.00
(d)
96.53
(e)
1.52
Newton's law
(a)
pressure, velocity,
(6)
shear stress and rate of angular deformation in a fluid
(c)
shear stress, temperature, viscosity, and velocity
(d)
pressure, viscosity,
(e)
yield shear stress, rate of angular deformation,
=
10.00 ft/s2
none of these answers
of viscosity relates
and
viscosity
and rate
of angular
is
deformation
and viscosity
FLUID PROPERTIES
FL~2 T
FL-'T'
(b)
1
FLT~
(c)
(a)
can never occur when the
(6)
may
when the
(d)
(e)
can never occur in a frictionless
dyn
-s
/cm
2
FL~ T
2
479
(c)
liquid
at rest
is
momentum
motion
g -s/cm
(c)
is
dyn
(d)
-cra/s2
dyn -s/cm2
(e)
converted to the English unit of viscosity by
(d)
none
(e)
1/p
answers
of these
for kinematic viscosity are
1
L2 T 2
(c)
L 2 T~
(d)
]
L 2 T~ 2
(e)
by
-^
(b)
1/30.48 2
(a) 2
X
10"
5
479
(c)
viscosity of kerosene at
X
3.2
(6)
1CT
5
30.48 2
(d)
2
(c)
90°F
X
10~
(e)
none
of these
second
in square feet per
4
(d)
3.2
X
answers
10~
is
4
none of these answers
The kinematic
second
viscosity of dry air at
25°F and 29.4 psia
in square feet per
is
X
10" 5
(a)
6.89
(e)
none of these answers
1.62
2
p
l
The kinematic
(a)
-s
ML~ T-
(b)
1.59
1.61
2
In converting from the English unit of kinematic viscosity to the stoke, one
1.58
1.60
FLT
(e)
by
The dimensions
multiplies
(e)
g/cm
(b)
W
TT9
1.57
(a)
of
fluid, regardless of its
Viscosity, expressed in poises,
1.56
multiplication
(a)
2
Correct units for dynamic viscosity are
1.55
(°)
FL T
(d)
fluid is at rest
occur owing to cohesion
depend upon molecular interchange
depend upon cohesive forces
(a)
2
Apparent shear forces
Select the incorrect completion.
1.54
(c)
For
fi
2.78
=
0.06
(b)
For m
=
(6)
kg/m
1.0
2.0
1.4
X
X
sp gr
*s,
-4
(c)
=
slug/ft
6.89
X
10" 4
0.60, v in stokes
0.60
(c)
10
10" 4
(d)
0.36
(d)
1.4
X
10" 3
is
(e)
none
of these
answers
the value of n in pound-seconds per square foot
-s,
is
X
10"4
(a)
1.03
(e)
none of these answers
1.63
25
Viscosity has the dimensions
1.53
(a)
AND DEFINITIONS
For
i»
X
=
3X
10~ 7
(6)
10
-4
2.0
10"4
St and p
(a)
5.02
(e)
none of these answers
(6)
X
6.28
X
=
(c)
0.8
10~7
6.21
X
g/cm 3 #4n
,
(c)
7.85
10"4
(d)
6.44
X
slugs per foot-second
X
10" 7
(d)
1.62
X
10~3
is
10" 6
FUNDAMENTALS OF FLUID MECHANICS
26
A
1.64
perfect gas
(a)
has zero viscosity
(6)
has constant viscosity
(d)
satisfies
= RT
(e)
fits
pp
The molecular weight
1.65
degree Rankine
(a)
53.3
(a)
55.2
(b)
1.231
is
incompressible
of a gas
is
The value
28.
R
of
in foot-pounds per slug per
is
The density
1.66
(c)
none of these statements
(6)
1545
(c)
1775
(d)
2
of air at
10°C and 10 kg//cm abs
12.07
(c)
118.4
(d)
none
(e)
of these
answers
in kilograms per cubic
65.0
(e)
meter
is
none of these answers
How many pounds mass of carbon monoxide gas at 20°F and 30 psia is contained
1.67
in a
volume
(a)
0.00453
A
1.68
3
of 4.0 ft ?
0.0203
(b)
(c)
0.652
(d)
2.175
(e)
none
of these
answers
container holds 2.0 lb m air at 120°F and 120 psia. If 3.0 lb m air is added and
is 240°F, the final pressure, in pounds per square inch absolute, is
the final temperature
(a) 300
answers
600
(c)
(d)
The bulk modulus of elasticity K for
1.69
(a)
362.2
(6)
p/p
RTq
(b)
(c)pp
The bulk modulus
1.70
indeterminable
independent of temperature
increases with the pressure
is
has the dimensions of 1/p
(d) is larger when the
is independent of pressure and viscosity
For 70 kg// cm 2 increase
by about
1.71
a)
~sifu
(&)
A pressure of
1.72
The bulk modulus
(a)
-750
(b)
(6)
in pressure the density of
(d)
3
150 psi applied to 10
of elasticity in
750
(c)
( e)
2
ft
3
none
of these
liquid causes a
(d)
more compressible
water has increased,
pounds per square inch
7500
fluid is
75,000
in percent,
answers
volume reduction
of 0.02 ft3
is
(e)
none
of these
answers
Surface tension has the dimensions
1.73
(a)
( c)
"A"
given by
of elasticity
(c)
(
is
none of these answers
(e)
(a)
(e)
none of these
a gas at constant temperature To
pRTQ
(d)
(e)
F
(b)
FL-
1
(c)
FLr2
(d)
Fir 3
(e)
none
of these
answers
.
2
FLUID STATICS
The
science of fluid statics will be treated in
and
its
two parts the study
:
surfaces. Special cases of fluids
moving
as solids are included in the treatment
is no motion
no shear stresses in the
have only normal pressure forces
of statics because of the similarity of forces involved.
of a fluid layer relative to
fluid.
of pressure
variation throughout a fluid and the study of pressure forces on finite
Hence,
all free
an adjacent
Since there
layer, there are
bodies in fluid statics
acting on their surfaces.
2.1
PRESSURE AT A POINT
The average pressure is calculated by dividing the normal force pushing against
a plane area by the area. The pressure at a point is the limit of the ratio of
normal force to area as the area approaches zero size at the point. At a point
a fluid at rest has the same pressure in all directions. This means that an element dA of a very small area, free to rotate about its center when submerged
in a fluid at rest, will
of
it,
have a force
of constant
magnitude acting on either
side
regardless of its orientation.
To demonstrate this, a small wedge-shaped free body of unit width is
taken at the point (x,y) in a fluid at rest (Fig. 2.1). Since there can be no
shear forces, the only forces are the normal surface forces and gravity. So,
the equations of equilibrium in the x and y directions are, respectively,
_
?>r x
__,
2jt y
= px
by
—
ps
ds sin 6
=
= py
dx
—
ps
ds cos 6
—
——
dx by
pa x
=
hxhy
y
—— = bxhy pa
y
=
27
FUNDAMENTALS OF FLUID MECHANICS
28
ySxSy
\PySx
2
Free-body diagram of
Fig. 2.1
wedge-shaped
particle.
which p x p v p s are the average pressures on the three faces, y is the specific
fluid, and p is its density. When the limit is taken as the free
body is reduced to zero size by allowing the inclined face to approach (x,y)
while maintaining the same angle and using the geometric relations
in
,
,
weight of the
bs sin 6
=
by
=
bx
p y bx
—
bs cos
the equations simplify to
px
—
by
The
by
p8
=
p s bx
—
bx by
y
=
term of the second equation is an infinitesimal of higher order of
may be neglected. When divided by by and bx, respectively,
the equations can be combined,
last
smallness and
(2.1.1)
V*
Since
same
is
any arbitrary
angle, this equation proves that the pressure
is
the
Although the proof was
be demonstrated for the three-
in all directions at a point in a static fluid.
carried out for a two-dimensional case,
it
may
dimensional case with the equilibrium equations for a small tetrahedron of
fluid
with three faces in the coordinate planes and the fourth face inclined
arbitrarily.
If the fluid is in
motion so that one layer moves relative to an adjacent
and the normal stresses are, in general, no longer
layer, shear stresses occur
the same in
all
directions at a point.
The pressure is then defined as the average
FLUID STATICS
of
any three mutually perpendicular normal compressive
px
+
Vv
+
29
stresses at a point,
Vz
In a fictitious fluid of zero viscosity, i.e., a frictionless fluid, no shear stresses
can occur for any motion of the fluid, and so at a point the pressure is the
same
in all directions.
2.2
BASIC EQUATION OF FLUID STATICS
Pressure variation
The
forces acting
forces
and body
in a static fluid
on an element of
forces.
fluid at rest, Fig. 2.2, consist of surface
With gravity the only body
force acting,
and by taking
the y axis vertically upward, it is — y bx by bz in the y direction. With pressure
p at its center (x,y,z), the approximate force exerted on the side normal to
the y axis closest to the origin
/
approximately
is
dp by\
bx bz
*y
(
dp Sy
P + -dyJ )Sx8z
y
6x
dp 6y
{p
Fig. 2.2
6x 6y 6z
-dy-2
)SxSz
Rectangular parallelepiped element
of fluid at rest.
FUNDAMENTALS OF FLUID MECHANICS
30
and the
/
force exerted
on the opposite
is
dpSy\
.
bx bz
where by/ 2
is
forces acting
=
bt y
the distance from center to a face normal to
on the element in the y direction gives
bx by bz
—
y bx by
Summing
the
bz
z directions, since
dv
—
bx by
-
bFx =
bF z =
bz
+
dv
—
bx by
act,
bz
dz
The elemental
= \bFx
no body forces
-
dx
SF
y.
dy
For the x and
If
side
force vector 5F
\bFy
the element
is
is
given by
kbF = -(\
+
z
— + \—
+ k — )bxbybz-]y bx by
dy
dz/
bz
\ dx
reduced to zero
size, after
dividing through
by
bx by bz
=
bV,
the expression becomes exact.
—=
—
bV
(i
— + J—
+ k— )p — Jt
dy
dz/
This
is
bV^O
the resultant force per unit volume at a point, which
to zero for a fluid at rest.
V
lim
(2.2.1)
\ dx
The quantity
must be equated
in parentheses is the gradient, called
(del),
dx
dy
dz
and the negative gradient of p,
— Vp, is the vector field f of the surface pressure
force per unit volume,
f
= - Vp
The
f
-
fluid static
j
7
=
(2.2.3)
law of variation of pressure
is
then
(2.2.4)
FLUID STATICS
31
In component form,
*=
£E =
dx
dy
The
=
^
dz
_7
(2.2JS)
partials, for variation in horizontal directions, are
of fluid
two points at the same elevation
at rest have the same pressure.
Since p
is
law, stating that
mass
dp =
one form of Pascal's
same continuous
in the
a function of y only,
-ydy
(2.2.6)
This simple differential equation relates the change of pressure to specific
weight and change of elevation and holds for both compressible and incompressible fluids.
For
is
fluids that
constant,
v = -yy
in
which
pressure
and Eq.
+
be considered homogeneous and incompressible, y
when integrated, becomes
c
the constant of integration.
c is
is
may
(2.2.6),
The
hydrostatic law of variation of
frequently written in the form
p - yh
(2.2.7)
measured vertically downward (h = —y) from a free liquid surthe increase in pressure from that at the free surface. Equation
(2.2.7) may be derived by taking as fluid free body a vertical column of liquid
of finite height h with its upper surface in the free surface. This is left as an
in
which h
and p
face
is
is
exercise for the student.
m
EXAMPLE 2.1 An oceanographer is to design a sea lab 5 high to withstand
submersion to 100 m, measured from sea level to the top of the sea lab. Find
the pressure variation on a side of the container and the pressure on the top
if
the specific gravity of salt water
7
=
1.020
At the
X
9802
top, h
p = yh =
1
=
N/m =
3
100 m, and
MN/m
2
10
1.020.
is
kN/m
3
FUNDAMENTALS OF FLUID MECHANICS
32
measured from the top of the sea lab downward, the pressure variation
If y is
is
p
=
10 (y
+
100)
kN/m
Pressure variation
When
the fluid
is
in
2
a compressible fluid
a perfect gas at rest at constant temperature, from Eq.
(1.6.2)
P
-
=
P
^
(2.2.8)
Po
When
the value of y in Eq. (2.2.6)
tween Eqs.
(2.2.6)
is
replaced
by
pg
and p
is
eliminated be-
(2.2.8),
v
gpo
It
and
must be remembered that
gp/g with g
tween limits
=
if
p
in
is
32.174 lb m ft/lb -s2
•
pounds mass per cubic
If
.
p = p when p =
£*-»£/:?
%
p
,
foot,
then y
=
integration be-
>?\^-
yields
'"
'^!^-
~£ Tl
y-2/o=-^ln^
in
which
p =
In
Po exp
which
is
is
(2.2.10)
the natural logarithm.
-^^\
(-
y
\
Po/gpo /
Then
(2.2.11)
the equation for variation of pressure with elevation in an isothermal
gas.
The atmosphere
frequently
is
assumed to have a constant temperature
FLUID STATICS
gradient, expressed
T =
To
+
33
by
(2.2.12)
fly
= - 0.00357°F/ft (-0.00651°C/m) up
For the standard atmosphere,
The density may be expressed
the stratosphere.
in terms of pressure
and
to
ele-
vation from the perfect-gas law:
"
=
RT
P
=
(2 2 13)
-
R(T +ey)
Substitution into dp
= —pgdy
-
[Eq. (2.2.6)] permits the variables to be
separated and p to be found in terms of y by integration.
EXAMPLE 2.2 Assuming isothermal conditions to prevail in the atmosphere,
compute the pressure and density at 2000 m elevation if p = 10 5 Pa abs,
p
=
kg/m
FromEq.
p - 10
=
3
1.24
/m
78,412
at sea level.
(2.2.11)
2000
2
exp
J
j-
(10 5 N//
m
2
m
/[( 9>806m / s 2)( L2 4kg/m )]
3
)
Pa abs
Then, from Eq. (2.2.8)
P
=
^p=
(1.24
kg/m
3
When
-^L
=
0.972
kg/m
3
compressibility of a liquid in static equilibrium
count, Eqs. (2.2.6)
2.3
)
100,000
p
and
is
taken into ac-
(1.7.1) are utilized.
UNITS AND SCALES OF PRESSURE MEASUREMENT
may
be expressed with reference to any arbitrary datum. The usual
When a pressure is expressed as a difference between its value and a complete vacuum, it is called
an absolute pressure. When it is expressed as a difference between its value
and the local atmospheric pressure, it is called a gage pressure.
The bourdon gage (Fig. 2.3) is typical of the devices used for measuring
Pressure
ones are absolute zero and local atmospheric pressure.
FUNDAMENTALS OF FLUID MECHANICS
34
Fig. 2.3
Bourdon gage. (Crosby
Steam Gage and Valve Co.)
gage pressures. The pressure element is a hollow, curved, flat metallic tube,
closed at one end, with the other end connected to the pressure to be measured.
When
the internal pressure
on a linkage to which
The
dial reads zero
is
is
increased, the tube tends to straighten, pulling
attached a pointer and causing the pointer to move.
when
the inside and outside of the tube are at the same
pressure, regardless of its particular value.
convenient units,
common
The
dial
may
be graduated to any
ones being pounds per square inch, pounds per
square foot, inches of mercury, feet of water, centimeters of mercury, milli-
meters of mercury, and kilograms force per square centimeter. Owing to the
FLUID STATICS
35
measures pressure relative to the pressure
is the local atmosphere.
Figure 2.4 illustrates the data and the relationships of the common
units of pressure measurement. Standard atmospheric pressure is the mean
A
pressure at sea level, 29.92 in Hg (rounded to 30 in for slide-rule work)
pressure expressed in terms of a column of liquid refers to the force per unit
area at the base of the column. The relation for variation of pressure with
altitude in a liquid p = yh [Eq. (2.2.7)] shows the relation between head h,
in length of a fluid column of specific weight y, and the pressure p. In consistent units, p is in pounds per square foot, y in pounds per cubic foot, and
inherent construction of the gage,
medium surrounding
of the
it
the tube, which
.
h in feet or p in pascals, y in newtons per cubic meter, and h in meters. With
its specific gravity S times the
the specific weight of any liquid expressed as
specific
weight of water, Eq. (2.2.7) becomes
= y w Sh
p
(2.3.1)
be taken as 62.4 lb/ft 3 or 9802 N/m 3
the pressure is desired in pounds per square inch, both sides of the
For water y w
When
may
.
equation are divided by 144,
p pai
in
—
62.4
=
Sh = 0A33Sh
which h remains in
feet.
(2.3.2)
1
is measured by a mercury barometer (Fig.
by an aneroid barometer which measures the difference in pressure
Local atmospheric pressure
2.5) or
In Eq. (2.3.2) the standard atmospheric pressure
may
be expressed
in
pounds per square
inch,
62.4
(13J
Tii
when S =
(§)
When
multiplied by 144, the standard atmosphere
62.4 yields 34 ft
2 0. Any of these designations is for the standard atmosphere and may be called one atmosphere, if it is always
understood that it is a standard atmosphere and is measured from absolute zero. These
various designations of a standard atmosphere (Fig. 2.4) are equivalent and provide a
convenient means of converting from one set of units to another. For example, to express
13.6 for mercury.
becomes 2116
100
ft
-Wsince
psi.
H
in
2
lb /ft 2
=
^£-
the
14.7
is
Then 2116 divided by
pounds per square
(14.7)
is
.
H
inch,
43.3 psi
number
of standard
atmospheres and each standard atmosphere
is
14.7
FUNDAMENTALS OF FLUID MECHANICS
36
Standard atmospheric pressure
Local atmospheric pressure
Negative]
suction > Gage pressure
14.7 psi
2116lb/ft
2
vacuum
30 in. mercury
34 ft water
1 atmosphere
760 mm mercury
Local
barometer
reading
Absolute pressure
101,325 Pa
10.34 m water
Absolute zero (complete vacuum)
Fig. 2.4
Units and scales for pressure measurement.
between the atmosphere and an evacuated box or tube, in a manner analogous
bourdon gage except that the tube is evacuated and sealed.
A mercury barometer consists of a glass tube closed at one end, filled
with mercury, and inverted so that the open end is submerged in mercury.
It has a scale arranged so that the height of column R (Fig. 2.5) can be determined. The space above the mercury contains mercury vapor. If the pressure of the mercury vapor h v is given in centimeters of mercury and R is
measured in the same units, the pressure at A may be expressed as
to the
hv
-f-
Fig.
R = K
2.5
Mercury
barometer.
cm Hg
FLUID STATICS
Although h v
a function of temperature,
is
The barometric
temperatures.
it is
37
very small at usual atmospheric
pressure varies with location,
i.e.,
elevation,
and with weather conditions.
In Fig. 2.4 a pressure
may
be located vertically on the chart, which
and to local atmospheric pressure.
dicates its relation to absolute zero
inIf
below the local-atmospheric-pressure line and is referred to gage
datum, it is called negative, suction, or vacuum. For example, the pressure 18 in
Hg abs, as at 1, with barometer reading 29 in, may be expressed as — 11 in Hg,
11 in Hg suction, or 11 in Hg vacuum. It should be noted that
the point
Pabs
=
is
Pbar
I
Pg&ge
To avoid any
that a pressure
is
of the atmosphere,
EXAM PLE
confusion, the convention
is
adopted throughout this text
gage unless specifically marked absolute, with the exception
which
The rate
is
an absolute pressure
unit.
change in the atmosphere with change
of a parcel of air depends on
the density of the parcel relative to the density of the surrounding (ambient)
air. However, as the parcel ascends through the atmosphere, the air pressure
decreases, the parcel expands, and its temperature decreases at a rate known
as the dry adiabatic lapse rate. A firm wants to burn a large quantity of refuse.
It is estimated that the temperature of the smoke plume at 30 ft above the
ground will be 20°F greater than that of the ambient air. For the following
conditions determine what will happen to the smoke.
(a)
At standard atmospheric lapse rate fi = — 0.00357°F per foot and
2.3
in elevation
k =
is
(b)
At an inverted
VdV
J po
The motion
70°.
By combining
(
of temperature
called its lapse rate.
9
f
RJ T
p
lapse rate
(3
=
0.002°F per foot.
Eqs. (2.2.6) and (2.2.13)
dy
+
^
(3y
p_
/
By\ -a/Rfi
po
\
Tj
The relation between pressure and temperature for a mass of gas
panding without heat transfer (isentropic relation, Sec. 6.1) is
ex-
(fc-D/fc
in
which Ti
is
the
solute pressure; k
initial
is
smoke absolute temperature and p the
initial
ab-
the specific heat ratio, 1.4 for air and other diatomic
FUNDAMENTALS OF FLUID MECHANICS
38
Eliminating p/p in the last two equations gives
+
1
tJ
Since the gas will rise until
T =
+
To
temperature
is
equal to the ambient temperature
fiy
the last two equations
a
its
may
be solved for
y.
Let
-1
=
(k
-
l)g/kRfi
+
1
Then
y
-?[©•->]
For
=
fi
= -0.00357°F
R =
per foot,
53. Sg ft lb/slug °R,
•
•
foot,
For the atmospheric temperature inversion
a = -0.2717, and y = 2680 ft.
2.4
MANOMETERS
y
10,570
ft.
a
(3
= 1.994, and
= 0.002°F per
Manometers are devices that employ liquid columns for determining differThe most elementary manometer, usually called a piezom-
ences in pressure.
eter, is
is
illustrated in Fig. 2.6a;
above zero gage.
A glass
it
tube
is
measures the pressure in a liquid when it
mounted vertically so that it is connected
Liquid rises in the tube until equilibrium
then given by the vertical distance h from the
meniscus (liquid surface) to the point where the pressure is to be measured,
expressed in units of length of the liquid in the container. It is obvious that
the piezometer would not work for negative gage pressures, because air would
flow into the container through the tube. It is also impractical for measuring
to the space within the container.
is
reached.
The
pressure
is
large pressures at A, since the vertical tube
the specific gravity of the liquid
would need to be very
S, the pressure at
is
A
is
hS
long.
If
units of length
of water.
For measurement of small negative or positive gage pressures in a liquid
may take the form shown in Fig. 2.66. With this arrangement the
meniscus may come to rest below A as shown. Since the pressure at the menisthe tube
,
FLUID STATICS
(a)
(6)
Fig. 2.6
cus
Ha
is
39
Simple manometers.
zero gage
= —hS
and
since pressure decreases with elevation,
units of length H-2
For greater negative or positive gage pressures a second liquid of greater
employed (Fig. 2.6c). It must be immiscible in the first
fluid, which may now be a gas. If the specific gravity of the fluid at A is *Si
(based on water) and the specific gravity of the manometer liquid is S 2 the
equation for pressure at A may be written thus, starting at either A or the
upper meniscus and proceeding through the manometer,
specific gravity is
,
hA
+
faSi
- hS =
2
unknown
in
which Ka
hi,
h2 are in length units. If
may
is
the
pressure, expressed in length units of water,
A
contains a gas, Si
is
and
generally so small that h 2 Si
be neglected.
A general
procedure should be followed in working
all
manometer prob-
lems:
Start at one end (or
any meniscus
if
the circuit
is
continuous) and write
the pressure there in an appropriate unit (say pounds per square foot)
or in an appropriate
Add
symbol
if it is
unknown.
same
to this the change in pressure, in the
to the next (plus
if
the next meniscus
is
lower,
from one meniscus
minus if higher). (For
unit,
pounds per square foot this is the product of the difference in elevation
in feet and the specific weight of the fluid in pounds per cubic foot.)
FUNDAMENTALS OF FLUID MECHANICS
40
Continue until the other end of the gage (or the starting meniscus) is
reached and equate the expression to the pressure at that point, known
3.
unknown.
or
The
expression will contain one
unknown
a simple manometer or will
for
give a difference in pressures for the differential manometer. In equation form,
Po
~
(t/i
-
-
?/o)7o
(2/2
-
-
2/1)71
in
which y Q y lf
~
(2/3
(2/4
-
2/2)72
2/3)73
'
'
'
-
(2/n
-
2/n-l)7n-l
= Pn
., y n are elevations of each meniscus in length units and 70,
7n-i are specific weights of the fluid columns. The above expres.
.
,
71, 72,
.
.
.,
sion yields the answer in force per unit area
units
by use
A
two points
hyi
—
to other
manometer (Fig. 2.7) determines the difference in presA and B, when the actual pressure at any point in the
system cannot be determined.
to Fig. 2.7a produces
—
and may be converted
of the conversions in Fig. 2.4.
differential
sures at
Pa
-
/1272
+
^373
Application of the procedure outlined above
= Pb
or
Pa
—
Pb
= pB
or
pA
-
Pb = -&i7i
=
^m +
^272
—
^373
Similarly, for Fig. 2.76,
Pa
+
h\y\
No
—
/1272
—
^373
satisfactory to
+
hz y z
7 2 orS 2
S
1
or
7:
M-4
S3
or
y3
Ai
i_.
S,
or 7,
vfey
(a)
Differential
It is
work them out from the general procedure
each case as needed.
Fig. 2.7
^272
formulas for particular manometers should be memorized.
much more
y\
+
S2
or
72
(b)
manometers.
0T
for
FLUID STATICS
If
the pressures at
A
B
and
41
are expressed in length of the water column,
the above results can be written, for Fig. 2.7a,
h,A
—
lis
=
+
hiSi
—
h 2 S2
units of length
h 3 Ss
H
2
Similarly, for Fig. 2.76,
hA
in
-
hB
which
-Mi + h S + hSs
=
2
S 2 and S 3
Si,
,
2
are the applicable specific gravities of the liquids in the
system.
EXAMPLE
2.4
eter liquid
is oil,
termine pa
ing is 730
—
sp gr 0.80; hi
Pb in pascals.
mm Hg,
(a)
,.,
(6)
=
(6) If
H 0) -
(m
hA
- (Am)(l) -
hA
-
hB
- -0.14
Pa
-
Pb
= y(h A -
.
30 cm, h 2
pB
=
=
2
^
= Pb =
7
r
5
hiSn 2 o
-
mH
—
—
cm
kg f 9.806
m)
manom-
=
A
+
in meters of water abs.
h 3 S H2 o
(0.8)
=
H 0)
(m
hB
2
(i^m)(l) =
+
hB
2
=
hB )
o
2
h2 S n
(rife
are water, and the
20 cm, and h s
60 cm. (a) De5 kg//cm 2 and the barometer read-
find the gage pressure at
hA
^5
A and B
In Fig. 2.7a the liquids at
(9802
N/m
N /100 cmV
(
kg f
\
m
)
3
)
(-0.14 m) = -1372 Pa
^^N/m
1
/ 9802
T/
,
3
=
^ m TTH ^
n
50.02
2
—— m \1(13.6)
(730
1000
=
From
fcirt-
50.02
+
9.928
2
abs
=
59.95
mH
/
2
abs
(a)
-
ta.b.
-
0.14
m
=
59.81
mH
Micromanometers
Several types of manometers are on the market for determining very small
differences in pressure or determining large pressure differences precisely.
One type very accurately measures the differences in elevation of two menisci
manometer. By means of small telescopes with horizontal cross hairs
mounted along the tubes on a rack which is raised and lowered by a pinion
of a
FUNDAMENTALS OF FLUID MECHANICS
42
D
C
r
T
A,
Ay
_i_
1
AH
El
^^y^
Fig.
eter
Micromanom-
2.8
two
using
gage
liquids.
and slow-motion screw
so that the cross hairs can be set accurately, the dif-
ference in elevation of menisci (the gage difference) can be read with verniers.
With two gage liquids, immiscible in each other and in the fluid to be
measured, a large gage difference R (Fig. 2.8) can be produced for a small
pressure difference. The heavier gage liquid fills the lower U tube up to 0-0;
then the lighter gage liquid is added to both sides, filling the larger reservoirs
up
The gas
to 1-1.
ment
fills
the space above
1-1.
When the
than at D, the menisci move as indicated in
The volume of liquid displaced in each reservoir equals the displace-
pressure at
Fig. 2.8.
C
or liquid in the system
in the
Ay
A - -a
in
which
A
spectively.
U
is
slightly greater
tube; thus
and a are the
cross-sectional areas of reservoir
The manometer equation may be
and
U
tube, re-
written, starting at C, in force
per unit area,
Pc
+
(fa
+
At/) 7i
+
(
(
k2
-
Ay
+-
J
72
UC2
-
#73
--+
Ay)
72
-
(fa
-
Ay)yi
= pD
FLUID STATICS
43
which y h y 2 and y 3 are the specific weights as indicated in Fig. 2.8. Simplifying and substituting for Ay gives
in
,
-
pc
Pd
= R
73
The quantity
—
72
in brackets
the pressure difference
EXAMPLE
( 1
is
- ~) -
_
Pair
t
1_ =
RT
71- =
a constant for specified gage and fluids; hence,
is
In the micromanometer of Fig.
2.5
mm,
5
(2.4.1)
7
directly proportional to R.
wanted, in pascals, when air
R =
7i
=
is
in the system, $2
1.0,
S3
(0.76
kg/m
3
)
(9.806
m/s 2 )
(0.01)
=
=
1.10,
a/ A
=
is
0.01,
mm Hg.
20°C, and the barometer reads 760
m) (13.6 X 9802 N/m^)
(287N-m/kg-K)(273 + 20K)
(1.205
the pressure difference
2.8,
=
g/m
§
'
0.118
N/m
3
A.
73
-
72
The term
pc
-
(l
-pD
The
jJ
71 (a/A)
=
(0.005
inclined
-
(9802
may
N/m
3
)
be neglected.
m) (1078
N/m
manometer
3
when A and B
)
=
-
0.99)
=
1078
N/m
3
Substituting into Eq. (2.4.1) gives
5.39
(Fig. 2.9)
small differences in gas pressures.
inclined scale,
(1.10
Pa
is
frequently used for measuring
It is adjusted to read zero,
are open.
by moving the
Since the inclined tube requires a
greater displacement of the meniscus for given pressure difference than a vertical tube, it affords greater
Fig. 2.9
Inclined
accuracy in reading the
manometer.
scale.
FUNDAMENTALS OF FLUID MECHANICS
44
Surface tension causes a capillary rise in small tubes. If a U tube is
used with a meniscus in each leg, the surface-tension effects cancel. The
capillary rise is negligible in tubes with a diameter of 0.5 in or greater.
FORCES ON PLANE AREAS
2.5
In the preceding sections variations of pressure throughout a fluid have been
considered. The distributed forces resulting from the action of fluid on a finite
may
area
be conveniently replaced by a resultant force, insofar as external
In this section the magnitude of
reactions to the force system are concerned.
resultant force
tegration,
and
its line
of action (pressure center) are determined
by formula, and by use
by
in-
of the concept of the pressure prism.
Horizontal surfaces
A
plane surface in a horizontal position in a fluid at rest
stant pressure.
fp dA
The magnitude
resultant force.
p dA acting on A are all parallel and in the same sense;
of all such elements yields the magnitude of the
direction is normal to the surface, and toward the surface
Its
To find the line of action of
where the moment of the distributed
point
is
summation
is positive.
area
subjected to a con-
forces
therefore, a scalar
p
is
on one side of the surface
= pf dA = pA
The elemental
if
of the force acting
zero, arbitrary xy axes
is
Fig. 2.10
may
Notation for determin-
ing the line of action of a force.
the resultant,
force about
i.e.,
any
be selected, as in Fig.
the point in the
axis
2.10.
through the
Then, since
FLUID STATICS
the
moment
pAx' =
which
x'
= —
in
axis,
say the y
of the distributed force
axis,
xp dA
/
in
must equal the moment
of the resultant
system about any
45
x' is
the distance from the y axis to the resultant. Since p
xdA =
I
which x
is
constant,
is
x
the distance to the centroid of the area (see Appendix A)
.
Hence,
for a horizontal area subjected to static fluid pressure, the resultant passes
through the centroid of the area.
Inclined surfaces
In Fig. 2.11 a plane surface
the horizontal.
is
taken as the x
The
indicated
is
axis.
The y
The magnitude,
to the liquid,
its
trace A'B'. It
intersection of the plane of the area
axis
is
direction,
is
inclined 0°
and the
from
free surface
taken in the plane of the area, with origin
The xy plane portrays the
0, as shown, in the free surface.
area.
by
and
arbitrary inclined
line of action of the resultant force
due
acting on one side of the area, are sought.
For an element with area 8A as a strip with thickness by with long edges
magnitude of force 8F acting on it is
horizontal, the
= yh 8A = yy
8F = p 8A
Since
all
sin
8A
such elemental forces are
(2.5.1)
parallel, the integral
over the area yields
the magnitude of force F, acting on one side of the area,
F =
fp
dA = y
sin Sjy
dA = y
sin By
A = yhA =
p GA
(2.5.2)
with the relations from Fig. 2.11, y sin = h, and pg = yh, the pressure at
the centroid of the area. In words, the magnitude of force exerted on one
side of a plane area
pressure at
surface
its
submerged
in a liquid
centroid. In this form,
unnecessary.
Any means
it
is
the product of the area and the
should be noted, the presence of a free
for determining the pressure at the cenbe used. The sense of the force is to push against the area if pg is
positive. As all force elements are normal to the surface, the line of action of
the resultant is also normal to the surface. Any surface may be rotated about
troid
is
may
FUNDAMENTALS OF FLUID MECHANICS
46
Notation for force of liquid on one side of a plane-
Fig. 2.11
inclined area.
any
if
axis through its centroid without changing the
magnitude
of the resultant
the total area remains submerged in the static liquid.
Center of pressure
The
line of action of the resultant force
has
its
piercing point in the surface
at a point called the pressure center, with coordinates (x p ,y p )
(Fig. 2.11).
Unlike that for the horizontal surface, the center of pressure of an inclined
surface is not at the centroid. To find the pressure center, the moments of
the resultant x p F, y p F are equated to the moment of the distributed forces
axis, respectively; thus
about the y axis and x
x pF
=
VvF =
/
I
J
A
xp dA
(2.5.3)
ypdA
(2.5.4)
FLUID STATICS
The area element
be 5x
in Eq. (2.5.3) should
and not the
by,
strip
47
shown
in
Fig. 2.11.
Solving for the coordinates of the pressure center results in
Vp
=
xp dA
(2.5.5)
pfypdA
(2.5.6)
In
many
and
applications Eqs. (2.5.5)
(2.5.6)
may be
evaluated most
conveniently through graphical integration; for simple areas they
may be
transformed into general formulas as follows (see Appendix A)
xP
= ———
—
:
yyA
/
xyysmddA =
sin 6 J A
—
xy
/
-
yA
dA =
JA
—
(2.5.7)
A
yA
In Eqs. (A. 10), of Appendix A, and (2.5.7),
xv
=
— +x
(2.5.8)
A
yA
When
=
=
an axis of symmetry
on x = x. Since I xy
may be either positive or negative, the pressure center may lie on either side
of the line x = x. To determine y p by formula, with Eqs. (2.5.2) and (2.5.6),
either of the centroidal axes, x
for the surface,
y,
-
=
-
.
yyysinddA =
In the parallel-axis theorem for
U
=
in
which IG
axis.
Ig
If
Ig
Vp
~-
y, is
I xy vanishes and the pressure center
sin 6 JfA
yyA
x or y
—
yA JfA
y*
moments
dA =
lies
-^-
(2.5.9)
yA
of inertia
+yA
2
Ix
is
is
m
the second
moment
of the area
about
its
horizontal centroidal
eliminated from Eq. (2.5.9),
-
(2.5.10)
or
Vp
-
_
Ig
yA
(2.5.11)
FUNDAMENTALS OF FLUID MECHANICS
48
is always positive; hence, y p — y is always positive, and the pressure center
always below the centroid of the surface. It should be emphasized that y
and y p — y are distances in the plane of the surface.
Ig
is
EXAMPLE
The
2.6
triangular gate
CDE
(Fig. 2.12) is
hinged along
CD
and
opened by a normal force P applied at E. It holds oil, sp gr 0.80, above
it and is open to the atmosphere on its lower side.
Neglecting the weight of
the gate, find (a) the magnitude of force exerted on the gate by integration
and by Eq. (2.5.2) (b) the location of pressure center; (c) the force P needed
to open the gate.
By integration with reference to Fig. 2.12,
(a)
is
;
F =
p dA
/
When
=
y
8,
= y
x
=
sin 6
I
yx dy
= y
0,
and when y =
=
8a
sin 8
x
=
13a
+
13,
xy dy
/
6,
+
y sin
6
I
xy dy
with x varying linearly with y;
thus
+
+
6=
x
=
in
which the coordinates have been substituted to
and b gives
ay
b
b
for a
a
=
Fig. 2.12
b
=
-V
x
Triangular gate.
= i(y-
8)
6
find x in terms of y. Solving
FLUID STATICS
=
Similarly y
—
x
13,
=
y
6;
18,
x
—
0;
F = 7 sin0f[/ (y-S)ydy +
J
(18
Integrating and substituting for y sin
F =
X
62.4
ByEq.
I xy
-Lxy
A =
sin 6
-
Hence
dy~^
leads to
4?/
2
J
+
f 9?/ 2
-
j
1
30
=
|-
=
9734.4 lb
X
62 .4
0.80
as shown, x
X
13
=
2.0,
X
y
0.50
=
X
13.
9734.4 lb
In Eq. (2.5.8),
x
zero owing to
hence x
i.e.,
f \( |'
y)y
y).
_
~\~
_
yA
is
X
With the axes
(6)
_
0.50
-
—
(2.5.2),
F = VgA = yy
Xp
X
0.8
and z = f (18
49
=
xp
=
2.0
symmetry about the
ft.
the pressure center
centroidal axis parallel to the x axis;
In Eq. (2.5.11),
is
0.32
ft
below the centroid, measured in the plane
of the area.
(c)
replaced
P X
6
=
When moments
by the
9734.4
about
CD
are taken
and the action
of the oil is
resultant,
X
2
P =
3244.8 lb
The pressure prism
The concept
of the pressure prism provides another
means
for determining
the magnitude and location of the resultant force on an inclined plane surface.
The volume
of the pressure prism is the
magnitude
resultant force passes through the centroid of the prism.
of the force,
and the
The
is
surface
taken
and its altitude at each point is determined by the
pressure yh laid off to an appropriate scale (Fig. 2.13). Since the pressure
increases linearly with distance from the free surface, the upper surface of the
as the base of the prism,
FUNDAMENTALS OF FLUID MECHANICS
50
Pressure prism.
Fig. 2.13
prism
is in a plane with
an elemental area bA is
8F = yh
its
trace
OM shown in Fig.
The
2.13.
5V
8
force acting
on
(2.5.12)
which is an element of volume of the pressure prism. After integrating, F = V,
the volume of the pressure prism equals the magnitude of the resultant force
acting on one side of the surface.
From Eqs. (2.5.5) and (2.5.6),
;/.
=
xdV
l
-jydV
(2.5.13)
which show that x p y p are distances to the centroid of the pressure prism
[Appendix A, Eq. (A.5)]. Hence, the line of action of the resultant passes
through the centroid of the pressure prism. For some simple areas the pressure prism is more convenient than either integration or formula. For example,
a rectangular area with one edge in the free surface has a wedge-shaped prism.
Its centroid is one-third the altitude from the base; hence, the pressure center
is one-third the altitude from its lower edge.
,
Effects of atmospheric pressure on forces on plane areas
In the discussion of pressure forces the pressure datum was not mentioned.
The
pressures were
below the
computed by p =
free surface.
Therefore, the
or the local atmospheric pressure.
yh, in
which h
is
the vertical distance
datum taken was gage
When
pressure zero,
the opposite side of the surface
is
FLUID STATICS
51
open to the atmosphere, a force is exerted on it by the atmosphere equal to the
product of the atmospheric pressure p and the area, or poA based on absolute
zero as datum. On the liquid side the force is
,
S(po
+
The
effect
yh)
dA = p A
p
A
of the
+
yfh dA
atmosphere acts equally on both sides and in no way
contributes to the resultant force or
its location.
So long as the same pressure datum is selected for all sides of a free body,
the resultant and moment can be determined by constructing a free surface at
pressure zero on this datum and using the above methods.
EXAMPLE
2.7
An
application of pressure forces on plane areas
the design of a gravity dam.
in the base of the
dam
are
computed from the
forces
Figure 2.14 shows a cross section through a concrete
weight of concrete has been taken as 2.57 and y
llOftk
77^^7>
Fig. 2.14
is
given in
The maximum and minimum compressive stresses
Wim
Concrete gravity dam.
is
which act on the dam.
dam where
the specific
the specific weight of water.
FUNDAMENTALS OF FLUID MECHANICS
52
A
dam
1-ft section of
is
considered as a free body; the forces are due to the
and the hydrostatic uplift. Debeyond the scope of this treatment,
but it will be assumed to be one-half the hydrostatic head at the upstream
edge, decreasing linearly to zero at the downstream edge of the dam. Enough
friction or shear stress must be developed at the base of the dam to balance
the thrust due to the water; that is, R x = 5OOO7. The resultant upward force
on the base equals the weight of the dam less the hydrostatic uplift, R v =
6750t + 2625t - 17507 = 76257 lb. The position of R y is such that the free
body is in equilibrium. For moments around 0,
concrete, the water, the foundation pressure,
termining amount of hydrostatic uplift
SM
= Ryx -
=
50007(33.33)
-
is
2625 7 (5)
-
6750 7 (30)
+
17507(23.33)
and
=
x
44.8
ft
It is
customary to assume that the foundation pressure varies linearly
i.e., that the pressure prism is a trapezoid with a
over the base of the dam,
volume equal
*m
"+
Ry
to
<rmiB
=
70
\
thus
7625 7
which o- max <7 m n are the maximum and minimum compressive stresses in
pounds per square foot. The centroid of the pressure prism is at the point
where x = 44.8 ft. By taking moments about O to express the position of
the centroid in terms of o- max and <7 min
in
i
,
,
44g =
<r
min 70
X
^+
((7
ma x
-
cr
min
)^ X
|(70)
(Cmax 4" 0"min)"Tr
Simplifying gives
Om ax
—
11.75<r m n
i
Then
<7
max
= 2IO7 =
When
o-min will
12,500 lb/ft 2
the resultant
falls
(7 min
=
17.17
=
1067 lb/ft 2
within the middle third of the base of the dam,
always be a compressive
stress.
Owing
to the poor tensile properties
FLUID STATICS
good design requires the resultant to
of concrete,
fall
53
within the middle third
of the base.
FORCE COMPONENTS ON CURVED SURFACES
2.6
When
the elemental forces p 8A vary in direction, as in the case of a curved
must be added as vector quantities; i.e., their components in
surface, they
three mutually perpendicular directions are added as scalars, and then the
With two horizontal components at
component all easily computed for a
three components are added vectorially.
right angles
—
—the resultant can be determined.
and with the
curved surface
vertical
The
lines of action of the
components are readily determined, and so the resultant and
can be completely determined.
component
Horizontal
of force
its line
of action
on a curved surface
The horizontal component of pressure force on a curved surface is equal to the
The vertical plane of
projection is normal to the direction of the component. The surface of Fig. 2.15
represents any three-dimensional surface, and 8A an element of its area, its
normal making the angle 6 with the negative x direction. Then
pressure force exerted on a projection of the curved surface.
8FX is
p8A
cos 6
the x component of force exerted on one side of
8
.
Summing up
the x com-
ponents of force over the surface gives
cos
6dA
(2.6.1)
JA
cos 6
p8A
8
A
is
the projection of
8
A
onto a plane perpendicular to
cos e
Fig. 2.15
Horizontal
component of
force on a curved surface.
x.
The element
54
FUNDAMENTALS OF FLUID MECHANICS
Projections of area ele-
Fig. 2.16
ments on opposite sides
of a body.
on the projected area
of force
is
p cos
6 8A,
which
is
also in the x direction.
Projecting each element on a plane perpendicular to x
is
equivalent to pro-
jecting the curved surface as a whole onto the vertical plane. Hence, the force
acting on this projection of the curved surface
is the horizontal component of
on the curved surface, in the direction normal to the plane of
projection. To find the horizontal component at right angles to the x direction, the curved surface is projected onto a vertical plane parallel to x and the
force on the projection is determined.
When the horizontal component of pressure force on a closed body is to
be found, the projection of the curved surface on a vertical plane is always
zero, since on opposite sides of the body the area-element projections have
opposite signs, as indicated in Fig. 2.16. Let a small cylinder of cross section
6A with axis parallel to x intersect the closed body at B and C. If the element
of area of the body cut by the prism at B is 8Ab and at C is 8 Ac, then
force exerted
8A B cos Ob
as cos d c
= —8A C cos
is
negative.
Bc
=
8
Hence, with the pressure the same at each end of the
cylinder,
p 8A B cos 6b
and
+
p 8A C cos
6C
=
similarly for all other area elements.
To
component of force on a curved
system composed of the force components from each area element is required. This is exactly the resultant of
the force on the projected area, since the two force systems have an identical
distribution of elemental horizontal force components. Hence, the pressure
center is located on the projected area by the methods of Sec. 2.5.
find the line of action of a horizontal
surface, the resultant of the parallel force
FLUID STATICS
EXAMPLE
z2 /4
is
+
2.8
2
2/
/4
55
The equation of an ellipsoid of revolution submerged in water
2
£ /9 = 1. The center of the body is located 2 m below the
+
Find the horizontal force components acting on the curved surfirst octant. Consider the xz plane to be horizontal
and y to be positive upward.
The projection of the surface on the yz plane has an area of 7r/4 X 2 X 3
m2 Its centroid is located 2 — 4/37T X 2 below the free surface. Hence
free surface.
face that
is
located in the
m
.
Fx = -
(j-
X
6
X
4
j
(2
-
(2
-
~J 7 =
- (5.425)
3
X
9802
N/m =
3
-53,200
N
Similarly,
F - t
Vertical
f
J
|p J
J
component
of force
7
= -
(3.617)
3
X
9802
N/m =
3
-35,400
N
on a curved surface
component of pressure force on a curved surface is equal to the weight
and extending up to the free surface.
The vertical component of force on a curved surface can be determined by
summing up the vertical components of pressure force on elemental areas 8A
of the surface. In Fig. 2.17 an area element is shown with the force p 8 A acting
normal to it. (Let 6 be the angle the normal to the area element makes with
the vertical.) Then the vertical component of force acting on the area element
The
vertical
of liquid vertically above the curved surface
is
p cos
6 8
,
and the
vertical
Fig. 2.17 Vertical component
force on a curved surface.
component
of
of force
on the curved surface
is
FUNDAMENTALS OF FLUID MECHANICS
given by
-/.' cos 6 dA
When
p
replaced
is
(2.6.2)
by
its
equivalent yh, in which h
area element to the free surface, and
of
8A on a horizontal plane, Eq.
F =
v
7
h cos 6
/
dA = y
A
/
it is
(2.6.2)
is
the distance from the
noted that cos 6
8
A
is
the projection
becomes
dV
(2.6.3)
«>
in which 8V is the volume of the prism of height h and base cos 6 8
or the
volume of liquid vertically above the area element. Integrating gives
,
F = yV
(2.6.4)
v
When
below the curved surface (Fig. 2.18) and the pressure
point, for example, 0, an imaginary or equivalent
free surface s-s can be constructed p/y above 0, so that the product of specific
weight and vertical distance to any point in the tank is the pressure at the
The weight of the imaginary volume of liquid vertically above the
point.
curved surface is then the vertical component of pressure force on the curved
surface. In constructing an imaginary free surface, the imaginary liquid must
be of the same specific weight as the liquid in contact with the curved surface; otherwise, the pressure distribution over the surface will not be correctly represented. With an imaginary liquid above a surface, the pressure
at a point on the curved surface is equal on both sides, but the elemental force
magnitude
Fig. 2.18
the liquid
is
known
is
at
some
Liquid with equivalent
free surface.
FLUID STATICS
57
components in the vertical direction are opposite in sign. Hence, the direction
component is reversed when an imaginary fluid is above
the surface. In some cases a confined liquid may be above the curved surface,
and an imaginary liquid must be added (or subtracted) to determine the free
of the vertical force
surface.
The line of action of the vertical component is determined by equating
moments of the elemental vertical components about a convenient axis with
(Fig. 2.17),
the moment of the resultant force. With the axis at
F
v
in
x
.,/
which x
u
xdV
is
the distance from
to the line of action.
Then, since
F = yV
v
}
xdV
the distance to the centroid of the volume.
Therefore, the line of action of
the vertical force passes through the centroid of the volume, real or imaginary,
that extends above the curved surface
EXAMPLE
2.9
A
up
to the real or imaginary free surface.
cylindrical barrier (Fig. 2.19) holds water as shown.
contact between cylinder and wall
is
smooth.
Considering a
1-ft
The
length of
and (6) the force exerted against the wall.
For equilibrium the weight of the cylinder must equal the vertical
component of force exerted on it by the water. (The imaginary free surface
for CD is at elevation A ) The vertical force on BCD is
cylinder, determine (a) its weight
(a)
.
(?+*•>-=
Fig. 2.19
(2tt
Semifloating body.
+
8) 7
FUNDAMENTALS OF FLUID MECHANICS
58
The
F,
vertical force
on
AB
is
u.--("-f)»--<^
tt)7
Hence, the weight per foot of length
F VBCD
+ F.^ =
(3tt
+
4) 7
= 838
is
lb
The force exerted against the wall is the horizontal force on ABC
(6)
minus the horizontal force on CD. The horizontal components of force on
BC and CD cancel; the projection of BCD on a vertical plane is zero. Hence,
Fh = F Hab =
2T
=
124.81b
since the projected area
jected area
To
is
62.4 lb/ft2
is
2
ft
2
and the pressure
at the centroid of the pro-
.
due to pressure forces, the action of the fluid
be replaced by the two horizontal components and one vertical component acting along their lines of action.
find external reactions
may
Tensile stress
A
in a
pipe
under the action of an internal pressure is in tension around
Assuming that no longitudinal stress occurs, the walls are in
tension, as shown in Fig. 2.20. A section of pipe of unit length is considered,
i.e., the ring between two planes normal to the axis and unit length apart.
its
circular pipe
periphery.
If one-half of this ring is
taken as a free body, the tensions per unit length at
top and bottom are, respectively, Ti and T2 as shown in the figure. The horizontal component of force acts through the pressure center of the projected
,
1 h
Fig. 2.20
1
Tensile stress
in
pipe.
FLUID STATICS
area and
2pr in which p
is
is
the pressure at the centerline and r
is
59
the internal
pipe radius.
may
For high pressures the pressure center
= T2 and
then 7\
T =
pr
(2.6.5)
T is
which
in
stress in
the tensile force per unit length. For wall thickness
the pipe wall
T
= - =
a
be taken at the pipe center;
,
e,
the tensile
is
pr
-
e
(2.6.6)
e
For larger variations in pressure between top and bottom of pipe the
is computed, and two equations are needed,
location of pressure center y
7\
+T
2
=
2rT Y
2pr
-
2pry
which the second equation
in
is
=
the
moment
equation about the lower end of
the free body, neglecting the vertical component of force.
T,
= py
EXAMPLE
T2 =
2.10
A
p(2r
-
y)
4.0-in-ID steel pipe has a
lowable tensile stress of 10,000 lb/in2 what
From Eq.
p
*
=
—
<re
=
\-m wall
the
is
thickness.
maximum
For an
al-
pressure?
(2.6.6)
(10,000 lb/in2 ) (0.25 in)
—
2.0
r
Solving gives
m
,
= 10Knn
1250 lb/in2
Tensile stress in a thin spherical shell
If a thin spherical shell is subjected to an internal pressure, neglecting the
weight of the fluid within the sphere, the stress in its walls can be found by
considering the forces on a free body consisting of a hemisphere cut from the
by a
sphere
vertical plane.
The
fluid
component
of force
normal to the plane
The
acting on the inside of the hemisphere
is
a times the cut wall area
the thickness, must balance the fluid
force;
a
=
hence
pr
—
2e
2irre,
with
e
pwr 2 with
,
r
the radius.
stress
FUNDAMENTALS OF FLUID MECHANICS
60
2.7
BUOYANT FORCE
body by a static fluid in which it is submerged
The buoyant force always acts vertically
or floating
component of the resultant because the
can
no
horizontal
There
be
upward.
or submerged portion of the floatof
body
the
submerged
projection
vertical
zero.
is
always
ing body
The buoyant force on a submerged body is the difference between the
vertical component of pressure force on its underside and the vertical component of pressure force on its upper side. In Fig. 2.21 the upward force on the
bottom is equal to the weight of liquid, real or imaginary, which is vertically
above the surface ABC, indicated by the weight of liquid within ABCEFA.
The downward force on the upper surface equals the weight of liquid ADCEFA
The difference between the two forces is a force, vertically upward, due to the
weight of fluid ABCD that is displaced by the solid. In equation form
The
resultant force exerted on a
is
called the buoyant force.
.
F B = Vy
(2.7.1)
FB
V
the volume of fluid displaced, and y is
the specific weight of fluid. The same formula holds for floating bodies when
V is taken as the volume of liquid displaced. This is evident from inspection
in
which
is
of the floating
the buoyant force,
body
is
in Fig. 2.21.
In Fig. 2.22a, the vertical force exerted on an element of the body in
the form of a vertical prism of cross section bA is
hF B =
(pi
-
in which 8V
Fig. 2.21
bodies.
pi)
is
dA = yhdA = y 5V
the volume of the prism.
Integrating over the complete
Buoyant force on floating and submerged
body
FLUID STATICS
(a)
Fig. 2.22
61
(6)
Vertical force
components on element
of body.
gives
FB =
y f dV
= yV
when 7 is considered constant throughout the volume.
To find the line of action of the buoyant force, moments are taken about
a convenient axis
and are equated to the moment of the resultant; thus,
yjx dV = yVx
in
which x
is
x
or
= — fxdV
the distance from the axis to the line of action.
This equation
volume; hence the buoyant force acts
displaced volume of fluid. This holds for both sub-
yields the distance to the centroid of the
through the centroid of the
merged and
floating bodies.
The
centroid of the displaced volume of fluid
is
called the center of buoyancy.
When
2.22b) the
8F B
in
=
the body floats at the interface of a static two-fluid system (Fig.
buoyant force on a
(p2
-
Pi)
8A = (y 2 h 2
which y h y 2 are the
vertical prism of cross section 8
+
specific
7ifo)
weights of the lighter and heavier
Integrating over the area yields
Fb =
dA
+
7iJ7i!
dA =
is
8A
spectively.
72 J h*
A
72^2
+
71^1
fluids, re-
FUNDAMENTALS OF FLUID MECHANICS
62
where Vi
is
the volume of lighter fluid displaced, and
fluid displaced.
To
V
2 is the volume of heavier
buoyant force, moments
locate the line of action of the
are taken
F Bx =
71 fx
dVi
+
72 Jz
dV 2
or
+
+
yifx dVi
7i^i
dVz
72 fx
+
+
71^1^1
72V 2
7i^i
72^2^2
72^2
which x h x 2 are distances to centroids of volumes *0i, V 2 respectively. The
resultant does not, in general, pass through the centroid of the whole volume.
in
,
In solving a statics problem involving submerged or floating objects,
is generally taken as a free body, and a free-body diagram is drawn.
the object
The
action of the fluid
object
must be shown
is
by the buoyant
replaced
(acting through
its
The weight
force.
center of gravity) as well as
of the
all
other
contact forces.
Weighing an odd-shaped object suspended
sufficient
gravity.
pended and weighted
in
two
Fh F
fluids.
object, are to
1)71
two
different fluids yields
=
are the weights submerged;
^
of equilibrium are written
F
2
+
^72 =
W
i
A
^XCBBBBBB
Kr£Hr£Hru^£H3ir
^r-I-I-Z-ZO£rZ-Z-Z-
^^^^^y^W^y^yyy.
Fig. 2.23
fluid.
71, 72
are
the weight and volume of the
be found.
The equations
+
2
W and V,
the specific weights of the fluids.
Fi
in
data to determine its weight, volume, specific weight, and specific
Figure 2.23 shows two free-body diagrams for the same object sus-
Free-body diagram for body suspended
in
a
FLUID STATICS
:-(°O
Hydrometer
Fig. 2.24
in
water and
in liquid of
63
-AV)Sy
specific
gravity S.
and solved
Ft
D =
w
=
Fiy 2
— F yi
— 7i
2
72
—
A
hydrometer uses the principle of buoyant force to determine specific
72
71
gravities of liquids.
Figure 2.24 shows a hydrometer in two liquids. It has a
stem of prismatic cross section a. Considering the liquid on the left to be
tilled water, S = 1.00, the hydrometer floats in equilibrium when
1)07
in
is
=
W
which
(2.7.2)
V
is
the volume submerged, y
the weight of hydrometer.
The
on the stem to indicate unit
is
the specific weight of water, and
position of the liquid surface
specific gravity S.
When
floated in another liquid, the equation of equilibrium
CUo- AV)Sy =
in
which
Ah =
dis-
AV =
V S -
is
marked
the hydrometer
is
becomes
W
a Ah.
W
1.00
(2.7.3)
Solving for Ah with Eqs. (2.7.2) and (2.7.3) gives
1
(2.7.4)
from which the stem can be marked
off to
read specific gravities.
EXAMPLE 2.11 A piece of ore weighing 0.15 kg/ in air is found to weigh
when submerged in water. What is its volume in cubic centimeters
0.11 kg/
and
its specific
gravity?
FUNDAMENTALS OF FLUID MECHANICS
64
The buoyant
0.15
k*
"££ -
0.00004
W_
=
3
=
cm
40
3
kg,
body
+
be neglected. From Fig. 2.23
(9802
N/ m
^
3
0.15 kg/ 9.806
N/m
air
N/k g/
0.00004
m
=
?5
3
STABILITY OF FLOATING
2.8
A
m
1
9802
7 <U
0.11 kg,
may
^
due to
kg/
1
V =
force
AND SUBMERGED BODIES
floating in a static liquid has vertical stability.
A
small
upward
dis-
an unbalanced
placement decreases the volume
downward force which tends to return the body to its original position. Similarly, a small downward displacement results in a greater buoyant force, which
of liquid displaced, resulting in
upward force.
body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return the body to its original position. It has rotational stability when a restoring couple is set up by any small
causes an unbalanced
A
angular displacement.
Methods for determining rotational stability are developed in the
lowing discussion.
rium.
When
A
a body
body may
is
fol-
float in stable, unstable, or neutral equilib-
in unstable equilibrium,
any small angular displace-
ment sets up a couple that tends to increase the angular displacement. With
the body in neutral equilibrium, any small angular displacement sets up no
couple whatever.
light piece of
Figure 2.25 illustrates the three cases of equilibrium: (a) a
a metal weight at its bottom is stable; (b) when the
wood with
metal weight is at the top, the body is in equilibrium but any slight angular
displacement causes the body to assume the position in a; (c) a homogeneous
sphere or right-circular cylinder is in equilibrium for any angular rotation;
i.e., no couple results from an angular displacement.
(a) Stable
Fig. 2.25
(6) Unstable
Examples
equilibrium.
of stable,
(c) Neutral
unstable, and
neutral
FLUID STATICS
:-:-:(£):
(a)-:-:
Fig. 2.26
A
submerged body.
Rotationally stable
completely submerged object
center of gravity
object
is
65
is
is
rotationally stable only
below the center of buoyancy, as in
Fig. 2.26a.
when
its
When
the
rotated counterclockwise, as in Fig. 2.266, the buoyant force and
weight produce a couple in the clockwise direction.
Normally, when a body is too heavy to float, it submerges and goes down
on the bottom. Although the specific weight of a liquid increases
slightly with depth, the higher pressure tends to cause the liquid to compress
until it rests
the body or to penetrate into pores of solid substances, thus decreasing the
buoyancy of the body. A ship, for example, is sure to go to the bottom once
it is completely submerged, owing to compression of air trapped in its various
parts.
Determination of rotational stability of floating objects
Any
floating object with center of gravity
below
its
center of
buoyancy
(centroid of displaced volume) floats in stable equilibrium, as in Fig. 2.25a.
Certain floating objects, however, are in stable equilibrium
first
when
their center
buoyancy. The stability of prismatic bodies is
considered, followed by an analysis of general floating bodies for small
of gravity is
above the center
of
angles of tip.
Figure 2.27a
tions identical.
is
The
body with
buoyancy is always
a cross section of a
center of
placed volume, which
is
all
other parallel cross sec-
at the centroid of the dis-
at the centroid of the cross-sectional area below liquid
when the body is tipped, as in Fig. 2.276, the center
buoyancy is at the centroid B' of the trapezoid ABCD; the buoyant force
acts upward through B', and the weight acts downward through G, the center
surface in this case. Hence,
of
When the vertical through B' intersects the original
above G, as at M, a restoring couple is produced and the body is
in stable equilibrium. The intersection of the buoyant force and the centerline
is called the metacenter, designated M. When
is above G, the body is stable;
of gravity of the body.
centerline
M
FUNDAMENTALS OF FLUID MECHANICS
66
Stability of a prismatic body.
Fig. 2.27
when below
distance
G,
MG
it is
is
unstable; and
when
at G,
stability of the body.
The
restoring couple
in neutral equilibrium.
it is
called the metacentric height
and
is
The
a direct measure of the
is
WMG sin 6
in
which
the angular displacement and
is
EXAM PLE
In Fig. 2.27 a scow 20
2.12
ft
W the weight of the body.
wide and 60
ft
long has a gross weight
above the water
Find the metacentric height and restoring couple when Ay = 1.0 ft.
The depth of submergence h in the water is
Its center of gravity is 1.0 ft
of 225 short tons (2000 lb).
surface.
h
=
225
X
2000
X
60
X
20
The
6.0 ft
62.4
centroid in the tipped position
is
located with
moments about
BC,
x
=
5
X
20
X
10
+
6
V
=
~
By
X
X
20
20
X
^
X
¥ = 9.46
5X20X| + 2X20X£X5f
6
similar triangles
Ay_ _
6/2
2
~
&P
WP
X
20
AEO
and B'PM,
=
3.03
ft
ft
%*
AB
and
FLUID STATICS
Ay =
Wf _
6r is
1,
6/2
B'P =
10,
-
10
9.46
=
0.54
ft;
then
054_>O0 =
540ft
7.0 ft
£P =
=
67
from the bottom; hence
-
3.03
=
3.97
MP -GP
=
5.40
7.00
ft
and
MG
=
The scow
stable since
is
WMG am
=
225
X
-
MG
2000
=
3.97
is
X
1.43 ft
positive; the righting
1.43
X
—==
=
moment
64,000 lb
is
ft
Nonprismatic cross sections
For a floating object of variable cross section, such as a ship (Fig. 2.28a), a
convenient formula can be developed for determination of metacentric height
very small angles of rotation 0. The horizontal shift in center of buoyancy r
determined by the change in buoyant forces due to the wedge
being submerged, which causes an upward force on the left, and by the other
for
(Fig. 2.286) is
wedge decreasing the buoyant force by an equal amount AF B on the right.
The force system, consisting of the original buoyant force at B and the couple
AF B X s due to the wedges, must have as resultant the equal buoyant force
at B
With moments about B to determine the shift r,
f
.
AFB
Xs = Wr
(2.8.1)
The amount
moments about O,
For an element of area 8A
on the horizontal section through the body at the liquid surface, an element
of volume of the wedge is xd 8 A the buoyant force due to this element is
yxd 8 A; and its moment about O is ydx2 8A, in which 6 is the small angle of
tip in radians. By integrating over the complete original horizontal area at
the liquid surface, the couple is determined to be
of the couple can be determined with
the centerline of the
body
at the liquid surface.
;
AFB
in
Xs =
which /
yd f x2
is
the
dA =
moment
yOI
of inertia of the area
(2.8.2)
about the axis yy (Fig. 2.28a)
FUNDAMENTALS OF FLUID MECHANICS
(6)
Stability relations
Fig. 2.28
in
a body of variable
cross section.
Substitution into Eq. (2.8.1) produces
ySI
in
= Wr = Vyr
which
V
Since
MB sin B
is
the total volume of liquid displaced.
is
very small,
= MBS =
The metacentric
MG
=
height
MB
or
r
is
r
I
6
V
then
MB^GB
or
MG
= -=FGB
(2.8.3)
V
The minus
sign
is
used
if
G
is
above B, the plus sign
if
G
is
below B.
FLUID STATICS
EXAMPLE
2.13
A
69
barge displacing 1000 metric tons has the horizontal cross
section at the waterline
shown
in Fig. 2.29.
Its center of
buoyancy
is
2.0
m
m
below the water
below the water surface, and its center of gravity is 0.5
surface. Determine its metacentric height for rolling (about yy axis) and for
pitching (about xx axis).
GB =
V =
2
-
0.5
=
1.5
m
1000(1000 kg,) (9.806 N/kg,)
9802
N/m
I™ = tt(24 m) (10 m) 3
I xx
= A(10m)(24m)
3
1000 m<
3
+
4(A)
+
2(^)(10m)(6m)
(6
m)
(5
For rolling
MG
=
I
GB =
V
1000
Horizontal cross
Fig. 2.29
section
2250
—
—-
of
waterline.
a
ship at the
1.5
=
0.75
m
m) 3 = 2250
3
+
m
4
(60m (14m) = 23,400m
2
2
)
4
FUNDAMENTALS OF FLUID MECHANICS
70
For pitching
MG
23,400
= - - GB
V
EXAMPLE
A
2.14
1.5
=
21.9
m
with sides
of specific gravity S c floats in a liquid
Find the range of specific-gravity ratios S c /S for it to
homogeneous cube
of specific gravity S.
float
-
1000
vertical.
In Fig. 2.30, b is the length of one edge of the cube. The depth of submergence z is determined by application of the buo}^ant-force equation
b z yS c
in
=
b 2 zyS
which y
is
the specific weight of water.
Solving for depth of submergence
gives
S
The
is
center of
buoyancy
b/2 from the bottom.
GB
b
-
2
Applying Eq.
0-1)
(2.8.3) gives
V
liquid.
z/2 from the bottom, and the center of gravity
z
2
Fig. 2.30
is
Hence
12
Cube
zb 2
floating
in
FLUID STATICS
0.2
0.4
0.6
0.8
71
1.0
S
Plot of
Sc /S
S
/
Fig. 2.31
MG/b.
vs.
or
_
MG
b
b
S\
c
=US -2V-S)
C
When
is
MG
equals zero,
S c /S =
0.212, 0.788.
Substitution shows that
WG
positive for
<
S
-*
O
<
0.212
Figure 2.31
2.9
0.788
<
S
-^
<
1.0
aS
is
a graph of
MG/6
vs.
S
c
/S.
RELATIVE EQUILIBRIUM
is simple to compute, thanks to the
motion such that no layer moves relative
to an adjacent layer, the shear stress is also zero throughout the fluid. A
fluid with a translation at uniform velocity still follows the laws of static
variation of pressure. When a fluid is being accelerated so that no layer moves
relative to an adjacent one, i.e., when the fluid moves as if it were a solid, no
shear stresses occur and variation in pressure can be determined by writing
the equation of motion for an appropriate free body. Two cases are of interest,
a uniform linear acceleration and a uniform rotation about a vertical axis.
When moving thus, the fluid is said to be in relative equilibrium.
In fluid statics the variation of pressure
absence of shear stresses.
For
fluid
FUNDAMENTALS OF FLUID MECHANICS
72
Although relative equilibrium
is
not a fluid-statics phenomenon,
it is
discussed here because of the similarity of the relationships.
Uniform linear acceleration
A
liquid in
an open vessel
is
given a uniform linear acceleration a as in Fig.
After some time the liquid adjusts to the acceleration so that
2.32.
as a solid;
i.e.,
it
moves
the distance between any two fluid particles remains fixed, and
hence no shear stresses occur.
By
selecting a cartesian coordinate
system with y vertically upward and
x such that the acceleration vector a is in the xy plane, the z axis is normal
to a and there is no acceleration component in the z direction. An element
form of a small rectangular parallelepiped having edges bx,
the coordinate axes is taken as a free body, Fig. 2.33. The
center of the element is at (x,y,z), and the pressure there is p. The rate of
change of p with respect to x, y, and z is to be found and is first obtained for
the element. Integration then yields the pressure variation throughout the
fluid. The equation of motion for the x direction is written first:
of fluid in the
by, dz parallel to
2/x =
ma x
(dp
p
F
bx\
J
6x2 )
by
y bz
—
dp bx\
(
) by
[p H
*
dx 2 /
V
bz
= -y
bx by
J bz ax
g
which reduces to
dp _
dx
g
when
bx, by, bz
Fig.
ax
2.32
are allowed to approach zero.
Acceleration
with free surface.
Following the same procedure
FLUID STATICS
73
f.dpdy\ dxSz
d
(p-
S
(
£ i) SySz
.3pix\
dy dz
(>-%¥)**
dy 2
(b)
Uniform linear acceleration of a
Fig. 2.33
for the z direction
(remembering that a z
=
fluid.
because of the choice of axes)
gives
dp
=
~dz
In the vertical direction the weight of the element y bx by
count; thus
(dp —
by\
p
J
dx bz
—
By 2 /
(
I
\
p H
dp by\
]bxhz
dy 2 /
—
y bx by
8z
= y-
8z is
taken into ac-
bx by bz a y
g
Simplifying leads to
Since p
is
a function of position
pendix B)
dp
dp = -^dx
,
dx
+
dp
f-dy
dy
+
dp
^-dz
dz
(x,y,z), its total differential is
(Ap-
FUNDAMENTALS OF FLUID MECHANICS
74
Substituting for the partial differentials gives
- y - dx -
dp =
7
( 1
+-
\
9
which can be integrated
a*
-y —
V =
—
a;
9
To evaluate
= po and
H
\
?!(
i
)
dy
(2.9.1)
9 /
an incompressible
for
a
.
A]y +
i
9
fluid,
c
I
the constant of integration
c,
let x
=
0,
y
=
0,
p = p
then
;
c
Vo-y-x-y[l + -)y
V =
\
9
When
tion
is
av
=
in Eq. (2.9.2).
+
and are
its
9
equa-
Solving Eq. (2.9.2) for y gives
v
v
^*+
;r
+
+
(2-9.3)
/ ,
a y /g)
(
7(1
lines of constant pressure,
dy
(2.9.2)
I
the accelerated incompressible fluid has a free surface,
given by setting p
y
The
9
p
=
const,
have the slope
9
parallel to the free surface.
The
y intercept of the free surface
is
Po
7 (1
+
a y /g)
For an isothermal gas p/y = po/yo with p and p in absolute
Substitution into Eq. (2.9.1) yields, for x = y = 0, p = p y = y
,
To
Po
\
9
units.
,
9 /
or
p
=
po exp
[
L
xa x /g
+
(1
+ (hMy l
Po/7o
{9QA
(2.9.4)
,
J
The compressible equations apply only
to closed containers.
FLUID STATICS
7.5
cm
15
cm
75
>
m
1.2
4.903
m/s 2
m
1.8
15
C
Tank completely
Fig. 2.34
EXAMPLE
filled
cm
f
with liquid.
The tank
in Fig. 2.34 is filled with oil, sp gr 0.8, and accelThere is a small opening in the tank at A Determine the
and C and the acceleration a x required to make the pressure at
2.15
erated as shown.
B
pressure at
B
.
zero.
The planes
a
tan
=
ax
—
=
of constant pressure
4.903
have the slope
= nr
0.5
9.806
g
and at A the pressure
above B hence
zero.
is
The plane through A
passes 0.3
m
vertically
;
=
Vb
(0.3
Similarly,
pc
=
m) (9802
C
(1.425
is
N/m
vertically
m) (9802
3
)
(0.8)
_
.
=
—
1.2
ax
1.8
9.806
N/m
3
)
and
ax
=
J (9.806)
=
6.537
2.352
kN/m
2
below the zero pressure plane a distance 1.425 m, and
For zero pressure at B,
tan
=
m/s2
(0.8)
=
11.174
kN/m
2
FUNDAMENTALS OF FLUID MECHANICS
76
is
2.16 A cubical box, 2 ft on a side, half filled with oil, sp gr 0.90,
accelerated along an inclined plane at an angle of 30° with the horizontal,
as
shown
EXAMPLE
in Fig. 2.35.
Find the slope of
free surface
bottom.
In the coordinate system as indicated in the
=
ax
8.05 cos 30°
The
=
ay
6.98
+
=
figure,
8.05 sin 30°
=
slope of the free surface, from Eq. (2.9.3),
ax
av
6.98 ft/s 2
and the pressure along the
4.02
9
Since tan" 1 0.192
of the box.
=
+
-0.192
10°52' the free surface
ft.
is
32.2
The depth
2 tan 40°52', or 1.73
4.02 ft/s2
parallel to a side
is
inclined 40°52' to the
bottom
on the right-hand side by
from the inclined plane, then
is less
If s is the distance of
A
from the known liquid volume,
.(*+.)-«
or
s
=
0.135.
The xy
coordinates of
A
are
x
=
2 cos 30°
-
0.135 sin 30°
=
1.665
ft
y
=
2 sin 30°
+
0.135 cos 30°
-
1.117
ft
By
substitution in Eq. (2.9.2), the pressure p at the origin
= p -
0.9
Fig. 2.35
tion along
X
62.4
X
6.98
X
Uniform acceleraan inclined plane.
1.665
0.9
32.2
X
1
•
(
obtained,
4.02\
/
62 4
is
+ 3T2
)
1.117
FLUID STATICS
77
or
p
=
Let
90.73 lb/ft 2
t
be the distance along the bottom from the origin; then x
bottom. By using Eq. (2.9.2) again
y
=
0.5t for the
p
=
90.73
which
is
-
=
0.866£,
42.07*
the pressure
t
from
ft
along the bottom.
Uniform rotation about a vertical axis
moving
an axis is called forced-vortex
same angular velocity. This motion
is to be distinguished from free-vortex motion, where each particle moves in
a circular path with a speed varying inversely as the distance from the center.
Free-vortex motion is discussed in Chaps. 7 and 9. A liquid in a container,
when rotated about a vertical axis at constant angular velocity, moves like
a solid after some time interval. No shear stresses exist in the liquid, and the
Rotation of a
fluid,
Every
motion.
as a solid, about
particle of fluid has the
only acceleration that occurs
rotation.
The equation
of
is
motion
directed radially inward toward the axis of
in the vertical direction
that hydrostatic conditions prevail along
at
any point
in the liquid
is
any
on a
free
body shows
vertical line hence, the pressure
;
given by the product of specific weight and vertical
distance from the free surface.
In equation form, Fig. 2.36,
dp
In the equation of motion tangent to the circular path of a particle, the acis zero, and the pressure does not change along the path.
In the equation of motion in the radial (horizontal) direction (Fig. 2.36)
with a free body of length 8r and cross-sectional area 8A, if the pressure at
celeration
r is p,
tion
is
then at the opposite face the pressure
— co 2r; hence
/
p8A - (p
\
dp
\
+ ^8r)5A
dr
)
=
= 7-y u z2 r
dr
g
p
+
(dp/dr)
dr.
The
accelera-
dA
8r yields
8A8ry,
(-co
r)
g
Simplifying and dividing through
dp
^P
-
is
by the volume
of the element
FUNDAMENTALS OF FLUID MECHANICS
78
/^-Zero
pressure
Rotation of fluid about a
Fig. 2.36
vertical axis.
Since p
dp
—
dy
dp =
a function of y and
is
r only,
the total differential dp
is
dp
dr
-\
dy
dr
Substituting for dp/dy and dp/dr results in
—
dp =
y dy
—
-\
co
2
dr
r
(2.9.5)
9
For a liquid (7
=
r
7
a
-w
,.,2
o
~
« const)
72/
integration yields
+c
2
in
(r
which
=
P =
0,
Po
c is
y
=
+
7
the constant of integration. If the value of pressure at the origin
0)
is
p
,
then
c
= p and
(2.9.6)
yy
29
When
lected
the particular horizontal plane (y
(2.9.6) is divided by 7,
=
0) for
which p =
is se-
and Eq.
2
h
coV
= py~~2g~
(2.9.7)
FLUID STATICS
79
which shows that the head, or vertical depth, varies as the square of the radius.
The
surfaces of equal pressure are paraboloids of revolution.
When
a free surface occurs in a container that
is
being rotated, the fluid
volume underneath the paraboloid of revolution is the original fluid volume.
The shape of the paraboloid depends only upon the angular velocity «.
For a circular cylinder rotating about its axis (Fig. 2.37) the rise of
liquid from its vertex to the wall of the cylinder is, from Eq. (2.9.7), u> 2 r 2 /2g.
Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder, the volume of the liquid above the horizontal plane through the
vertex
is
When
the liquid
at rest, this liquid
is
is
also
above the plane through the
vertex to a uniform depth of
2
1 coVo
2
2g
Hence, the liquid
rises
along the walls the same
thereby permitting the vertex to be located
tion are given.
«-r -*
s
^_L^h>:
A'
2a
+
innrLTLnnn
c ^03
Fig. 2.37
Rotation of
circular cylinder
its
axis.
about
when
amount as the center drops,
and depth before rotaco, r
,
FUNDAMENTALS OF FLUID MECHANICS
EXAMPLE
What
=
A
= 200 X
co
=
1
When
in the fluid
Eq. (2.9.6)
coVa
+7—
Po
Then
rA
liquid, sp gr 1.2, is rotated at
the pressure at a point
is
When
Pa
A
2.17
At one point
axis.
m
B 2m higher
Pb
yy
2tt/60
1.5
=
200
m from the axis,
A
than
rpm about a
the pressure
and
1.5
m
is
vertical
70 kPa.
from the axis?
written for the two points,
2
=
rB
is
1
=
2
2
C0 7*£
Po
+7~
20.95 rad/s, 7
=
7 (2/
1.2
X
+
9802
2)
=
11,762
N/m
3
,
and
m
the second equation
is
subtracted from the
first
and the values are sub-
stituted,
70,000
- pB =
(2
m)
(11,762
+
N/m*)
J^l^™'
2 X 9.806 m/s
(20.95/s)'
2
X [lm 2
(1.5
m) 2 ]
Hence
p B = 375.5 kPa
a closed container with no free surface or with a partially exposed free
If
surface
is
rotated uniformly about some vertical axis, an imaginary free sur-
face can be constructed, consisting of a paraboloid of revolution of shape given
by Eq.
(2.9.7).
surface
is
The
vertical distance
For an isothermal gas 7
podp
7o
= —ay
in the fluid to this free
—
=
pyo/po-
Equation
(2.9.5)
becomes
coV
+
p
dr
g
After integration, for p
P
=
-In-
Po
7o
from any point
the pressure head at the point.
-2/
+
= p
,
y
=
0, r
=
0,
—
co
2
r2
2g
Po
p and po must be in absolute pressure units
EXAM PLE
water,
is
2.18
A straight tube 4 ft long,
inclined 30° with the vertical
for the compressible case.
closed at the bottom and filled with
and rotated about a vertical axis through
FLUID STATICS
Rotation of
Fig. 2.38
81
in-
clined tube of liquid about
a vertical axis.
its
midpoint 8.02 rad/s.
the pressure at the
Draw
the paraboloid of zero pressure, and determine
bottom and midpoint
of the tube.
In Fig. 2.38, the zero-pressure paraboloid passes through point A.
the origin
h
=
coV 2
is
taken at the vertex, that
8.02
—
— (2 sin 30°)
2
J
v
64.4
=
which locates the vertex at 0,
the tube is 7 X CD, or
X
62.4
At the midpoint,
Vb
=
p =
0,
If
Eq. (2.9.7) becomes
2
=
2g
4 cos 30°
is,
0.732
X
=
1.0 ft
below A. The pressure at the bottom of
216 lb/ft 2
OB =
62.4
1.0 ft
=
0.732
ft,
and
45.6 lb/ft 2
Fluid pressure forces in relative equilibrium
The magnitude
of the force acting
celerating as a rigid
F = fpdA
on a plane area in contact with a
fluid ac-
body can be obtained by integration over the surface
FUNDAMENTALS OF FLUID MECHANICS
82
The nature
and orientation
of the acceleration
ticular variation of
p over the surface.
of the surface
When
govern the par-
the pressure varies linearly
over the plane surface (linear acceleration), the magnitude of force is given
by the product of pressure at the centroid and area since the volume of the
pressure prism
and
given by pgA.
is
line of action
For nonlinear distributions the magnitude
can be found by integration.
PROBLEMS
2.1/ Prove that the pressure
is
the same in
all
directions at a point in a static fluid for
the three-dimensional case.
The
2.2
container of Fig. 2.39 holds water and air as shown.
A, B, C, and
D
The tube
2.3
in
pounds per square foot and
in Fig. 2.40
is filled
with
meters of water.
Air
'^BZz
T
3
ft
Water
-4-
Air
l
.ft
f
3ft
Water.-
Fig. 2.39
•::«!•*
m
2
Air
B
—
0.E
m
*
Dil
c
il
Fig. 2.40
-*
— SpgrO.85-
oil.
What
is
the pressure at
in pascals?
Determine the pressure at
A
and
B
in
FLUID STATICS
\
A
Air
x
1
Air
\
1 ft
%:=:=>>=i_
Oil
wsp
t
2
c
\
1>
lip
83
gr u.yw
\
ft
V^pS1 ^fecopSX
y-Z
:
=z=z3ft::::::5v
:-z-z-z
-^ D
fc-z-zWa te"r ^z-E-z-z^ii-z1:-:1
Fig. 2.41
Calculate the pressure at A, B, C, and
2.4
of Fig. 2.41 in
pounds per square
inch.
Derive the equations that give the pressure and density at any elevation in a static
conditions are known at one elevation and the temperature gradient ft is
2.5
gas
D
when
known.
By
2.6
Prob.
2.7
a limiting process as
ft
—» 0,
derive the isothermal case from the results of
2.5.
By
use of the results of Prob. 2.5, determine the pressure and density at 5000
elevation
when p =
14.5 psia,
t
=
68°F, and
ft
= — 0.003°F/ft
ft
at elevation 1000 ft
for air.
2.8
For isothermal
the pressure
2j$
is 1
air at
0°C, determine the pressure and density at 10,000
kg// cm 2 abs at sea
In isothermal
air at
80°F what
ft
when
level.
is
the vertical distance for reduction of density
by
10 percent?
2.10
Express a pressure of 8 psi in (a) inches of mercury, (6) feet of water,
(c) feet of
acetylene tetrabromide, sp gr 2.94, (d) pascals.
2.11
A bourdon gage reads 2 psi suction, and the barometer is 29.5 in Hg. Express the
pressure in six other customary ways.
2.12
Express 3 atm in meters of water gage, barometer reading 750
2.13
Bourdon gage
Fig. 2.42
A
inside a pressure
mm.
tank (Fig. 2.42) reads 12
psi.
Another
FUNDAMENTALS OF FLUID MECHANICS
84
bourdon gage B outside the pressure tank and connected with it reads 20 psi, and an
aneroid barometer reads 30 in Hg. What is the absolute pressure measured by A in
inches of mercury?
Determine the heights of columns
2.14
of water; kerosene, sp gr 0.83;
tetrabromide, sp gr 2.94, equivalent to 18
=
In Fig. 2.6a for a reading h
2.15
square inch.
The
A
20 in determine the pressure at
in
pounds per
liquid has a specific gravity of 1.90.
Determine the reading h
2.16
and acetylene
cm Hg.
in Fig. 2.66 for
pa
=
20 kPa suction
if
the liquid
is
kerosene, sp gr 0.83.
In Fig. 2.66 for h
2.17
find
pa
in feet of
S =
In Fig. 2.6c
2.18
=
8 in and barometer reading 29
in,
with water the liquid,
water absolute.
x
meters of mercury gage.
0.86,
If
S2 =
1.0,
h2
=
8.3 cm, hi
the barometer reading
is
=
29.5
17 cm.
in,
what
Find p A in millipa in meters of
is
water absolute?
Gas
2.19
hi
=
5
in,
is
contained in vessel
In Fig. 2.7a Si
2.20
A
of Fig. 2.6c.
determine the pressure at
Compute pa
—
Pb
=
1.0,
S2 -
A
With water the manometer
0.95,
S3 =
1.0,
h=
h2
=
pa~
Pb
= —39 cm H2 0.
30 cm, and h z
=
1
m.
in centimeters of water.
In Prob. 2.20 find the gage difference h 2 for
2.21
and
fluid
in inches of mercury.
2.22 In Fig. 2.76 Si = S* = 0.83, S2 = 13.6, h x = 16 cm, h 2 = 8 cm, and h = 12
cm. (a) Find pa if Pb = 10 psi. (6) For pa = 20 psia and a barometer reading of
720
find pb in meters of water gage.
mm
2.23
Find the gage difference h 2
2.24
In Fig. 2.43,
A
^/
Fig. 2.43
in Prob. 2.22 for
contains water, and the
pa
=
Pb-
manometer
fluid
has a specific gravity
FLUID STATICS
When the left meniscus is at zero on the scale, pa = 10cmH 2 O. Find the read-
of 2.94.
ing of the right meniscus for
2.25
pA
=
The Empire State Building
pounds per square inch
2.26
What
kPa with no adjustment
8
is
1250
What
is
kg/m4 and
A
H
h
is
tube or
the pressure difference in
+
by p = 450
ah, in which a
the distance in meters measured from the free surface?
vertical gas pipe in a building contains gas, p
=
0.002 slug/ft 3 and p
ft
=
cm
H 0.
=
72
2
The
pressure.
6
=
62.4 lb/ft 3 73
;
=
65.5 lb/ft 3 a/ A
;
inclined manometer of Fig.
The diameter of reservoir is
30°, gage fluid sp gr 0.832, find
gage reading
2.30
12
3.0 in
(b) gas
In Fig. 2.8 determine R, the gage difference, for a difference in gas pressure of
2.28
2.29
=
higher, determine the gas
pressure in inches water gage for two cases: (a) gas assumed incompressible and
assumed isothermal. Barometric pressure 34 ft H2O; t = 70°F.
1
scale.
m below the free surface in a fluid that has a
gage in the basement. At the top of the building 800
2
U
water column of the same height?
of a
the pressure at a point 10
is
high.
ft
of the
variable density in kilograms per cubic meter given
2.27
85
R
pa
=
0.01.
when A and B are at the same
and that of the inclined tube J in. For
Pb in pounds per square inch as a function of
2.9 reads zero
2.0 in
—
in feet.
Determine the weight
W that can be sustained by the
100 kg/ force acting on
the piston of Fig. 2.44.
2.31
lift
Neglecting the weight of the container (Fig. 2.45) find (a) the force tending to
,
CD
the circular top
2.32
and
(b) the
compressive load on the pipe wall at A- A.
Find the force of oil on the top surface
is reduced by 1 m.
CD
of Fig. 2.45
if
the liquid level in the
open pipe
2.33
The
upward
container
force
on the plane
2.34
The
shown
on the surface
EF?
has a circular cross section. Determine the
frustum ABCD. What is the downward force
in Fig. 2.46
of the cone
Is this force
equal to the weight of the fluid? Explain.
cylindrical container of Fig. 2.47 weighs 400
with water and supported on the piston, (a)
What
an additional 600-N weight
force
N
is
when empty.
placed on the cylinder,
thie
cylinder? (6) If
will
the water force against the top of the cylinder be increased?
24 cm diam
*
>
4 cm diam
w
100 kg,
M
n Ww/»»»w
m
Oil
w»»///»/A
Fig. 2.44
is
It is filled
exerted on the upper end of
how much
FUNDAMENTALS OF FLUID MECHANICS
86
3ft
<
ft
2.0
3
ft
diam
B
in.
-3
in.
diam
4
diam
—A
ft
-Oil
Sp gr 0.8
Fig. 2.45
2
ft
1
A
diam
2ft
B
C
i ft
D
4
ft
diam
Water
5ft
Fig. 2.46
2.35
A barrel 2 ft
in diameter filled with
water has a vertical pipe of 0.50 in diameter
how many pounds of water must be
attached to the top. Neglecting compressibility,
added to the pipe to exert a force
2.36
A
on the top of the barrel?
vertical right-angled triangular surface has a vertex in the free surface of
liquid (Fig. 2.48)
2.37
of 1000 lb
.
Find the force on one side {a) by integration and
(6)
Determine the magnitude of the force acting on vertical triangle
by integration and (b) by formula.
a
by formula.
ABC
of Fig.
2.49 (a)
2.38
Find the moment about
surface
ABC of Fig.
2.48.
y
=
AB
9000
of the force acting
N/m
3
.
on one
side of the vertical
FLUID STATICS
87
cm diam
24
^ssssssssssssssnw^
r
24 cm
Water:
5
^NSr'.S^.
.WWVVV^
1
0|*-5
cm
777777,
Fig. 2.47
Fig. 2.48
Find the moment about
2.39
ABC
AB of the force acting on one side of the vertical surface
of Fig. 2.49.
Locate a horizontal
2.40
line
pressure force on the surface
Oil
5
(7=55
ft
5ft
^\ Si?
Fig. 2.49
Ib/fT)
below
ABC
is
AB
of Fig. 2.49 such that the
equal above and below the
line.
magnitude
of
FUNDAMENTALS OF FLUID MECHANICS
Fig. 2.50
2.41
Determine the force acting on one side
= 9500 N/m3
2.50.
7
2.42
Calculate the force exerted
shown
in Fig. 2.51.
2.43
Determine the moment at
Fig. 2.51
Fig. 2.52
of the vertical surface
OABCO
of Fig.
.
by water on one
A
side of the vertical annular area
required to hold the gate as
shown
in Fig. 2.52.
FLUID STATICS
Gate 6
wide
89
ft
Fig. 2.53
2.44
If
there
resultant force
water on the other side of the gate (Fig. 2.52) up to A, determine the
due to water on both sides of the gate, including its line of action.
is
2.45 j The shaft of the gate in Fig. 2.53 will fail at a
the
maximum
2.46
value of liquid depth
The dam
of Fig. 2.54 has a strut
AB
every 5 m.
force in the strut, neglecting the weight of the
2.47
moment
of 135
kN
•
m. Determine
h.
Determine the compressive
dam.
Locate the distance of the pressure center below the liquid surface in the
angular area
ABC
of Fig. 2.49
tri-
by integration and by formula.
2.48
By
integration locate the pressure center horizontally in Fig. 2.49.
2.49
By
using the pressure prism, determine the resultant force and location for the
triangle of Fig. 2.48.
2.50
By
2.51
Locate the pressure center for the annular area of Fig. 2.51.
2.52
Locate the pressure center for the gate of Fig. 2.52.
(2. 53
integration, determine the pressure center for Fig. 2.48.
A vertical square area 5 by 5 ft is submerged in water with upper edge 2 ft below the
Locate a horizontal line on the surface of the square such that (a) the force
on the upper portion equals the force on the lower portion and (6) the moment of force
about the line due to the upper portion equals the moment due to the lower portion.
surface.
%%3
Fig. 2.54
mW^
FUNDAMENTALS OF FLUID MECHANICS
90
Fig. 2.55
2.54 An equilateral triangle with one edge in a water surface extends downward at a
45° angle. Locate the pressure center in terms of the length of a side b.
2.55
In Fig. 2.53 develop the expression for y p in terms of
2.56
Locate the pressure center of the vertical area of Fig. 2.50.
2.57
Locate the pressure center for the vertical area of Fig. 2.55.
h.
2.58 Demonstrate the fact that the magnitude of the resultant force on a totally
submerged plane area is unchanged if the area is rotated about an axis through its
centroid.
2.59
The gate
of Fig. 2.56 weighs 300 lb/ft
L€~~~rt
WaterEBJ
^BOBBE^
EHHHHHI-i ^z-z^zzfo;
•fi
--_r-_-_-_-_-_-Z-l-l-T; ';-'d
5
it
Hja
zzizzz^zz-z^zioz^i
-"^-r^.-_-_r-_-_r-_-_-/4
-~---~/? ':,.. "4
i-z-z-z-z-z-z-z-Zt?
<'.:':4
^BBz^^:>;
«__A
Fig. 2.56
:
"
normal to the paper.
Its center of gravity
FLUID STATICS
is
1.5 ft
from the
left face
and
2.0 ft
above the lower
face. It is
the water-surface position for the gate just to start to
91
hinged at 0. Determine
come up.
(Water surface
is
below the hinge.)
2.60
Find h
2.61
Determine the value
maximum
of Prob. 2.59 for the gate just to
to the vertical position shown.
and the force against the stop when
this force
is
a
for the gate of Prob. 2.59.
Determine y
2.62
of h
come up
of Fig. 2.57 so that the flashboards will
tumble when water reaches
their top.
Determine the hinge location y of the rectangular gate of Fig. 2.58 so that
2.63
open when the liquid surface is as shown.
By
2.64
1
its
depth of submergence
is
increased.
Find the magnitude and line of action of force on each side of the gate of
(6) Find the resultant force due to the liquid on both sides of the gate.
Determine F to open the gate if it is uniform and weighs 6000 lb.
(a)
Fig. 2.59.
(c)
will
use of the pressure prism, show that the pressure center approaches the
centroid of an area as
2.65
it
m
IM
/////////////A
Fig. 2.57
Fig. 2.58
FUNDAMENTALS OF FLUID MECHANICS
92
= 54
= 54
16
3
lb/ft
ft
Gate 6
ft
wide
Y777777777777777777777777777777777777777777)
Fig. 2.59
4
m
Fig. 2.60
2.66 For linear stress variation over the base of the dam of Fig. 2.60, {a) locate
where the resultant crosses the base and (6) compute the maximum and minimum
compressive stresses at the base. Neglect hydrostatic
2.67
Work
from 20
2.68
m
at
Prob. 2.66 with the addition that the hydrostatic uplift varies linearly
A
to zero at the toe of the
Find the moment
M at
ft
45V*--------- Water-
m
m
-4 ft-
2
ft
lft
S?5>
Fig. 2.61
dam.
(Fig. 2.61) to hold the gate closed.
MrO
Gate 6
wide
uplift.
FLUID STATICS
93
Fig. 2.62
The gate shown in Fig.
2.69
2.62
is
in equilibrium.
Compute W,
the weight of counter-
weight per meter of width, neglecting the weight of the gate.
Is the gate in stable
equilibrium?
The gate
2.70
as shown.
(a)
is
find
made
of Fig. 2.63 weighs 150 lb/ft
W and
(6)
center of gravity 2
5
ft
>
y////////,
I
¥
/30*
w)
Fig. 2.63
Fig. 2.64
N/m normal to the paper, with its
Find h as a function of 6 for equilibrium
stable equilibrium for any values of 0?
(Fig. 2.64) weighs 2000
m from the hinge at O.
(b) Is the gate in
of the gate,
*
It is in equilibrium
of concrete, sp gr 2.50.
The plane gate
2.71
normal to the page.
arm and brace supporting the counterweight,
determine whether the gate is in stable equilibrium. The weight
Neglecting the weight of the
(a)
FUNDAMENTALS OF FLUID MECHANICS
94
Fig. 2.65
2.72
is
A 15-ft-diameter pressure pipe carries liquid at 150 psi. What pipe-wall thickness
required for
2.73
To
maximum
stress of 10,000 psi?
obtain the same flow area, which pipe system requires the least
pipe or four pipes having half the diameter?
The maximum
steel,
a single
allowable pipe-wall stress
is
the same in each case.
2.74
A
thin-walled hollow sphere 3
allowable stress of 60
2.75
A
MPa determine
2.76
in diameter holds gas at 15 kg// cm 2
the
cylindrical container 8 ft high
with two hoops a foot from each end.
in each
m
minimum
and 4
When
ft in
,
For
wall thickness.
diameter provides for pipe tension
it is filled
with water, what
is
the tension
hoop due to the water?
A
2-cm-diameter
where the pressure
is
steel ball covers
300 kg// cm 2
.
a 1-cm-diameter hole in a pressure chamber
What
force
is
required to
lift
the ball from the
opening?
2.77
If
the horizontal component of force on a curved surface did not equal the force
on a projection
of the surface onto a vertical plane,
what conclusions could you draw
regarding the propulsion of a boat (Fig. 2.65)?
Determine the horizontal component of force acting on the radial gate
and its line of action. (6) Determine the vertical component of force and
its line of action, (c) What force F is required to open the gate, neglecting its weight?
(d) What is the moment about an axis normal to the paper and through point 0?
2.78
(a)
(Fig. 2.66)
Gate 2
w/;;;/;;;;;;;//?^^^^
Fig. 2.66
m
wide
FLUID STATICS
95
Hinge
VZZEZZ2ZZZZZZ&
Gate 4
wide
ft
Oil
sp gr
0.
-Water-
2
X
/777r7777777777V77777777777?77?A
2
m
ft
= 3 rv
S=3.0
Fig. 2.67
Calculate the force
2.79
R=
2.80
R=
2
F
required to hold the gate of Fig. 2.67 in a closed position,
F
required to open or hold closed the gate of Fig. 2.67
ft.
Calculate the force
1.5
when
ft.
2.81
What is R of
2.8£
Find the vertical component
Fig. 2.67 for
no force
F required to hold the gate closed or to open it?
of force
on the curved gate
of Fig. 2.68, including
"~its line of action.
2.83
What
is
the force on the surface whose trace
normal to the paper
2.84
A
is
is
right-circular cylinder with a diameter of 2 ft
Fig. 2.68
OA
of Fig. 2.50?
The
length
3 m.
is
illustrated in Fig. 2.69.
The
FUNDAMENTALS OF FLUID MECHANICS
Cylinder
5
long
ft
Fig. 2.69
pressure, in
2.85
[1
—
2.86
pounds per square
ABC
segment
the
If
4(1
+
as p
=
2p(l
pressure
sin0) 2 ]
+
—
variation
10,
due to flow around the cylinder varies over the
10. Calculate the force on ABC.
foot,
4 sin 2 0)
+
on the cylinder
in
2.69
Fig.
is
2p
p
X
determine the force on the cylinder.
Determine the moment
M to hold the gate of Fig. 2.68,
neglecting
its
weight.
Find the resultant force, including its line of action, acting on the outer surface
quadrant of a spherical shell of radius 60 cm with center at the origin. Its
center is 1 m below the water surface.
2.87
of the first
2.88
The volume
of the ellipsoid given
the area of the ellipse x2 /a2
surface given in
2.89
A log
pushing
it
Example
+
z
/c2
2
=
by x2/
2
1 is irac.
+
2
2
y /b
+ z /c
2
Determine the
2
=
1 is
Iwabc/S, and
vertical force
on the
2.8.
holds the water as shown in Fig. 2.70.
Determine
(a)
the force per foot
against the dam, (b) the weight of the log per foot of length, and (c) its
specific gravity.
2.90
The
cylinder of Fig. 2.71
is filled
with liquid as shown. Find (a) the horizontal
component of force on AB per unit of length, including its line of action, and (6) the
vertical component of force on AB per unit of length, including its line of action.
2.91
The
cylinder.
Fig. 2.70
The
center of gravity of
is
made up from
is
controlled
a circular cylinder and a plate,
by pumping water into or out of the
the empty gate is on the line of symmetry 4 ft from
cylinder gate of Fig. 2.72
hinged at the dam. The gate position
FLUID STATICS
97
Fig. 2.71
the hinge.
feet of
It is in equilibrium
the water surface
A
2.92
when empty
in the position
shown.
water must be added per foot of cylinder to hold the gate in
is
How many
cubic
position
when
its
raised 3 ft?
hydrometer weighs 0.035
N
and has a stem 5 mm
and 1.1.
in diameter.
Compute
the
distance between specific gravity markings 1.0
Design a hydrometer to read
scale is to be 3 in long.
2.93
specific gravities in the
range from 0.80 to 1.10
when the
A
2.94
sphere
1 ft
in diameter, sp gr 1.4,
is
submerged
varying with the depth y below the surface given by p
equilibrium position of the sphere in the liquid.
Repeat the calculations for Prob. 2.94
and a diameter of 1 ft.
2.95
in a liquid
=
2
+
having a density
Determine the
0.1?/.
for a horizontal circular cylinder
with a
specific gravity of 1.4
A cube, 2 ft on an edge, has its lower half of sp gr 1.4 and upper half of sp gr
submerged into a two-layered fluid, the lower sp gr 1.2 and the upper sp gr
Determine the height of the top of the cube above the interface.
2.96
0.6.
It is
0.9.
2.97
3
N
2„9#
in
Determine the density, specific volume, and volume
water and 4 N in oil, sp gr 0.83.
Two
Fig. 2.72
cubes of the same
size,
1
m
3
,
of
an object that weighs
one of sp gr 0.80, the other
of sp gr 1.1,
FUNDAMENTALS OF FLUID MECHANICS
98
Fig. 2.73
are connected by a short wire and placed in water.
above the water surface, and what
is
What
portion of the lighter cube
is
the tension in the wire?
2.99
In Fig. 2.73 the hollow triangular prism
and y
=
is
in equilibrium as
shown when
z
=
1 f
Find the weight of prism per foot of length and z in terms of y for equilibrium. Both liquids are water. Determine the value of y for z = 1.5 ft.
2.100
0.
How many
pounds
having a volume of 0.1
2.101
Two beams, each
in Fig. 2.74.
2.102
m
3
6
ft
Determine the
A wooden
=
kN/m
3
must be attached to a beam
and sp gr 0.65 to cause both to sink in water?
of concrete,
by 12 by 4
y
in,
25
are attached at their ends
specific gravity of
cylinder 60
cm
,
and
float as
shown
each beam.
in diameter, sp gr 0.50, has a concrete cylinder 60
cm
long of the same diameter, sp gr 2.50, attached to one end.
wooden cylinder
2.103
What
so that
it
2.104
Will a
for the
system to
float in
Determine the length of
stable equilibrium with axis vertical.
are the proportions r
will float in
beam
//i of a right-circular cylinder of specific gravity
water with end faces horizontal in stable equilibrium?
10
ft
S
long with square cross section, sp gr 0.75, float in stable
equilibrium in water with two sides horizontal?
2.105
Determine the metacentric height
2. 106
Determine whether the thick-walled cylinder
shown.
_^
Fig. 2.74
of the torus
shown
in Fig. 2.75.
of Fig. 2.76
is
stable in the position
FLUID STATICS
99
Fig. 2.75
r-lft
4ft
=2
ft
2ft
Fig. 2.76
2.107
gen.
A spherical balloon 15 m in diameter is open at the bottom and filled with hydro-
For barometer reading
and the load to hold
2.108
A
it
of
Hg and 20°C, what is the total weight of the balloon
28 in
stationary?
tank of liquid
S =
0.86
accelerated uniformly in a horizontal direction so
is
that the pressure decreases within the liquid
1
psi/ft in the direction of motion. Deter-
mine the acceleration.
2.109
The
makes an angle
free surface of a liquid
accelerated uniformly in a horizontal direction.
2.110
In Fig. 2.77, ax
=
8.05 ft/s2
ay
,
=
of 20° with the horizontal
What
is
Find the imaginary
0.
when
the acceleration?
free liquid surface
and the pressure at B, C, D, and E.
2.111
2.112
In Fig. 2.77, ax
=
In Fig. 2.77, ax
=
0,
ay
=
-16.1
2
8.05 ft/s
,
ay
ft/s2
=
.
Find the pressure at B, C, D, and E.
16.1 ft/s2
.
Find the imaginary
free surface
and the pressure at B, C, D, and E.
2.113
In Fig. 2.78, ax
^lj^ln
and C.
Fig. 2.78, ax
=
=
9.806
4.903
m/s2 a y =
,
m/s2
,
ay
=
0.
Find the pressure at A, B, and C.
4.903
m/s2
.
Find the pressure at A, B,
FUNDAMENTALS OF FLUID MECHANICS
100
Uflj
lft
Oil
Sp
gr 0.8
3
ft
Fig. 2.77
,,a
v
30 cm
i
-a,
-I-I-l
rrv
-Water--------:
1.3
m-
Fig. 2.78
2.115
liquid
A
spills out,
2.116
circular cross-sectional
and accelerated uniformly
tank of
6-ft
depth and
4-ft
in a horizontal direction.
diameter
is filled
If one-third of
with
the liquid
determine the acceleration.
Derive an expression for pressure variation in a constant-temperature gas
undergoing an acceleration ax in the x direction.
2.117
The tube
of Fig. 2.79
the right 8.05 ft/s 2
A. For pa
=
8 psi
,
is filled
with liquid, sp gr 2.40.
draw the imaginary
vacuum determine ax
free surface
When it
is
accelerated to
and determine the pressure at
.
m
2.118 A cubical box 1
on an edge, open at the top and half filled with water, is
placed on an inclined plane making a 30° angle with the horizontal. The box alone
weighs 500
N
and has a coefficient of friction with the plane of 0.30. Determine the
and the angle the free water surface makes with the horizontal.
acceleration of the box
FLUID STATICS
101
r
i ft
L
2
ft
Fig. 2.79
Show that
2.119
the pressure
is
the same in
all
directions at a point in a liquid
moving
as a solid.
A closed box contains two immiscible liquids.
2.120
Prove that when
it is
accelerated
uniformly in the x direction, the interface and zero-pressure surface are parallel.
Verify the statement
2.121
axis that
when a
made
on uniform rotation about a vertical
body, no shear stresses exist in
in Sec. 2.9
fluid rotates in the
manner
of a solid
the fluid.
2.122;
A vessel containing liquid,
sureat one point 2
axis
ft
radially
and with elevation 2
The
2.123
U
ft
sp gr 1.2,
from the axis
higher.
tube of Fig. 2.79
is
)
The
rotated about a vertical axis.
the same as at another point 4
ft
pres-
from the
Calculate the rotational speed.
rotated about a vertical axis 6 in to the right of
at such a speed that the pressure at
2.124
is
is
A
is
zero gage.
What
is
Locate the vertical axis of rotation and the speed of rotation
Fig. 2.79 so that the pressure of liquid at the
midpoint
of the
A
the rotational speed?
U
of the
U tube
tube and at
A
of
are
both zero.
An
2.125
incompressible fluid of density p moving as a solid rotates at speed co about
in the fluid,
an axis inclined at 6° with the vertical. Knowing the pressure at one point
how do you find the pressure at any other point?
A
2.126
right-circular cylinder of radius r
the top and
bottom
2.127
axis.
is
and height h with
with liquid. At what speed must
it
axis vertical
is
open at
rotate so that half the area of the
exposed?
A liquid rotating about a horizontal axis as a solid has a pressure of 10 psi at the
Determine the pressure variation along a vertical line through the axis for density
p and speed
2.128
filled
co.
Determine the equation
for the surfaces of constant pressure for the situation
described in Prob. 2.127.
2.129
Prove by integration that a paraboloid
of revolution has a
volume equal
to
half its circumscribing cylinder.
2.130
A tank containing two immiscible liquids is rotated about a vertical axis.
that the interface has the
same shape as the zero pressure
surface.
Prove
FUNDAMENTALS OF FLUID MECHANICS
102
A
2.131
hollow sphere of radius
axis at speed
ro is filled
Locate the circular
co.
line of
with liquid and rotated about
maximum
its
vertical
pressure.
A gas following the law pp~n = const is rotated about a vertical axis as a
Derive an expression for pressure in a radial direction for speed co, pressure p
2.132
solid.
,
and
density po at a point on the axis.
A
2.133
vessel containing water
16.1 ft/s2
What
.
The
2.134
U
is
is
rotated about a vertical axis with an angular
At the same time the container has a downward
velocity of 50 rad/s.
acceleration of
the equation for a surface of constant pressure?
tube of Fig. 2.79
is
rotated about a vertical axis through
A at such a
A which is
speed that the water in the tube begins to vaporize at the closed end above
What
at 70°F.
is
What would happen
the angular velocity?
if
,
the angular velocity
were increased?
A cubical box 4 ft on an edge is open at the top and filled with water. When it is
2.135
accelerated
upward 8.05
A
cube
ward 8.05
ft/s2
2.136
ft/s2 find the
of
water force on one side of the box.
on an edge is filled with liquid, sp gr 0.65, and is accelerated downFind the resultant force on one side of the cube due to liquid pressure.
1 ft
.
Find the force on side
2.137
magnitude
,
OB
Example
of Fig. 2.35 for the situation described in
2.16.
A
2.138
cylinder 2
ft in
diameter and 6
axis in a horizontal direction 16.1 ft/s2 It
.
pressure along
its
ft
long
is filled
is
accelerated uniformly along
with liquid, y
=
its
50 lb/ft 3 and has a
,
Find the horizontal net force
axis of 10 psi before acceleration starts.
exerted against the liquid in the cylinder.
A
2.139
When
center at
2.140
cm on an edge, has a small opening at the center of its top.
with water and rotated uniformly about a vertical axis through its
closed cube, 30
it is filled
co
rad/s, find the force on a side due to the water in terms of
The normal
stress
is
the same in
all
co.
directions at a point in a fluid
(d)
when the fluid is frictionless
when the fluid is frictionless and incompressible
only when the fluid has zero viscosity and is at rest
when there is no motion of one fluid layer relative to an adjacent
(e)
regardless of the motion of one fluid layer relative to
(a) only
(b)
(c)
only
2.141
The
pressure in the air space above an
The
pressure 5.0
(a)
7.0
2.142
1.03
(c)
9.62
(d)
pressure, in centimeters of
cm manometer
(a)
below the surface of the
8.37
(6)
The
ft
fluid,
(b)
sp gr 2.94,
1.88
(c)
oil
oil,
11.16
layer
an adjacent layer
(sp gr 0.75) surface in a tank
(e)
none
of these
mercury gage, equivalent to 8 cm
(d)
3.06
2
psi.
answers
H
2
is
2.04
is
in feet of water, is
(e)
none of these answers
plus 6
FLUID STATICS
The
2.143
differential equation for pressure variation in a static fluid
(e)
dp
dp
2.144
= — ydy
= —ydp
be written
upward)
(y measured vertically
(a)
may
103
dp
(6)
= — ydy
= —pdp
dy
(c)
(d)
dp
= — pdy
In an isothermal atmosphere, the pressure
(a)
remains constant
(6)
decreases linearly with elevation
(c)
increases exponentially with elevation
(d)
varies in the
(e)
and density remain constant
2.145
same way as the density
Select the correct statement.
always below standard atmospheric pressure.
(a)
Local atmospheric pressure
(b)
Local atmospheric pressure depends upon elevation of locality only.
is
mean
atmospheric pressure at sea
(c)
Standard atmospheric pressure
(d)
A barometer reads the difference between local and standard atmospheric pressure.
(e)
Standard atmospheric pressure
2.146
(a)
10.0 psi, 23.1 ft
10.0 psi, 20.3 ft
(e)
4.33 psi, 10.0
When
- 10.2 m H
(e)
none
2
2
Hg
level.
abs.
Hg
4.33 psi, 10.0 ft
H 0, 20.3 in Hg
H 0, 20.3 in Hg
2
2
(o)
0.075
m Hg
10
kPa
mH
8.91
(c)
suction
is
the same as
abs
2
107
(d)
kPa abs
answers
atm
0.476
13.8 in
10.0 psi, 4.33 ft
(d)
mm Hg,
With the barometer reading 29
(e)
(6)
2
2
of these
(a)
2.149
ft
H 0, 4.91 in Hg
H 0, 23.1 in Hg
H 0, 8.83 in Hg
the barometer reads 730
(a)
2.148
34 in
is
local
Select the three pressures that are equivalent.
(c)
2.147
the
is
(6)
atm
0.493
in
Hg, 7.0 psia
is
equivalent to
7.9 psi suction
(c)
(d)
7.7 psi
abs
In Fig. 2.66 the liquid
is oil,
sp gr 0.80.
When
h
=
2
ft,
the pressure at
A may
be expressed as
(a)
-1.6ftH 2 0abs
{d)
2.5 ft
2.150
h=
(a)
In Fig. 2.6c air
10.14
—
mH
H
(b)
1.6 ft
(e)
none of these answers
(c)
2
contained in the pipe, water
is
A
20 cm. The pressure at
abs
2
(6)
0.2
mH
is
suction
2
the manometer liquid, and
is
vacuum
2
H
1.6 ft
(c)
0.2mH2 O
(d)
4901 Pa
none of these answers
2.151
1.0.
vacuum
2
50 cm, hi
(a)
(e)
H
-3.05
2.152
h = 2.0 ft, h2 =
hA in feet of water is
In Fig. 2.7a,
Then hB
—
(6)
In Fig. 2.7b,
-1.75
h=
1.5
(c)
ft,
h.
1.0
ft,
3.05
=
1.0
h=
hz
ft,
6.25
(d)
ft,
4.0
=
2.0
Si
(e)
ft,
&=
=
0.80,
S2 =
0.65,
£3 =
none of these answers
1.0,
S2 =
3.0,
S3 =
1.0.
104
FUNDAMENTALS OF FLUID MECHANICS
Then p A
—
-1.08
(a)
A
2.153
pounds per square inch
in
Pb
1.52
(b)
0.5
(6)
measuring
7.2
(6)
2.155
A
being
filled
6.3
6.8
(c)
7.3
(d)
to remain at a fixed elevation. 6
with
=
and
R =
(d)
14.4
is
so large that its surface
Used
30°.
1.2
as a simple
ft.
The
(e)
none
7.2
vacuum
oil,
12.5
(c)
When
When
12.9
(c)
it is
accelerated vertically
one
side, 16.1 ft/s
(a)
(6)
—\
When
2.157
mum
(a)
the
(c)
2
,
0.94
(6)
1.125
The magnitude
2.158
of force
,
is
—1
is
none
(e)
of these
pressure in the box of Prob. 2.156
(c)
the other half
upward 4.903 m/s2
accelerated uniformly in a horizontal direction
is
(d)
pressure in meters of water
answers
none of these answers
(e)
the slope of the interface
—\
minimum
14.7
(d)
the box of Prob. 2.155
parallel to
of these
m on each edge, is half filled with water,
1
sp gr 0.75.
11
(6)
manometer
pressure at A, in
is
closed cubical box,
4.9
2.156
answers
none of these answers
(e)
the pressure difference between bottom and top, in kilopascals,
(a)
of these
.
air pressure, it contains water,
inches of water,
(a)
none
(e)
In the inclined manometer of Fig. 2.9 the reservoir
2.154
may be assumed
for
218
(d)
mercury- water manometer has a gage difference of 50 cm (difference in
The difference in pressure, measured in meters of water, is
elevation of menisci)
(a)
8.08
(c)
is
is
answers
zero gage, the maxi-
is
none of these answers
1.31
(d)
on one
side of a circular surface of unit area, with
centroid 10 ft below a free water surface,
1.5
(e)
is
less than IO7
dependent upon orientation of the area
(c) greater than IO7
(d) the product of 7 and the vertical distance from
(e) none of the above
(a)
(b)
A rectangular surface 3
2.159
a free
oil surface,
on one
(a)
side of the surface
38.47
2.160
The
(b)
487
by 4
The
sp gr 0.80.
ft
free surface to pressure center
has the lower 3-ft edge horizontal and 6 ft below
is inclined 30° with the horizontal. The force
surface
is
(c)
51.27
(d)
6O7
(e)
pressure center of the surface of Prob. 2.159
none
is
of these
answers
vertically below the liquid
surface
(a)
10.133
answers
ft
(b)
5.133
ft
(c)
5.067
ft
(d)
5.00
ft
(e)
none of these
FLUID STATICS
The
2.161
pressure center
is
submerged area
(a)
at the centroid of the
(b)
the centroid of the pressure prism
(c)
independent of the orientation of the area
(d)
a point on the line of action of the resultant force
always above the centroid of the area
(e)
What
2.162
the force exerted on the vertical annular area enclosed by concentric
is
circles of radii 1.0
(a)
3t7
and
2.0
9x7
(b)
The
2.163
105
m? The center is 3.0 m below a free water surface, y =
10.257r7
(c)
(d)
Yliry
(e)
none
pressure center for the annular area of Prob. 2.162
of these
sp wt.
answers
below the centroid of
is
the area
m
(a)
A
2.164
0.42
(6)
h/4
The
pressure center
h/Z
(b)
A
2.165
h/2
(c)
vertical gate 4
by 4
about the bottom of the gate
(a)
42.77
577
(&)
The magnitude
2.166
0.44
(c)
m
(d)
0.47
m
none
(e)
of these
vertical triangular area has one side in a free surface, with vertex
Its altitude is h.
(a)
m
(c)
is
answers
downward.
below the free surface
(d)
2h/S
3A/4
(e)
m holds water with free surface at its top. The moment
is
647
(d)
85.37
of the resultant force acting
(e)
none
on both sides
of these
answers
of the gate (Fig. 2.80)
is
(a)
7687
(b)
The
2.167
15937
(c)
I8IO7
(d)
line of action of the resultant force
38207
on both
(e)
none
of these
answers
sides of the gate in Fig. 2.80
above the bottom of the gate
(a)
2.67 ft
answers
Bar.
28
Fig. 2.
in.
Hg
(6)
3.33
ft
(c)
3.68
ft
(d)
4.00
ft
(e)
none
of these
is
FUNDAMENTALS OF FLUID MECHANICS
106
The
2.168
component
horizontal
of force
on a curved surface
(a)
weight of liquid vertically above the curved surface
(6)
weight of liquid retained by the curved surface
(c)
product of pressure at
(d)
force
(e)
scalar
A
2.169
stress of
(a)
on a
sum
diameter
ft in
is
components
to carry water at 200 psi.
the thickness of pipe wall
1.6 in
(b)
The
2.170
centroid and area
of all elemental horizontal
psi,
1.2 in
equal to the
vertical projection of the curved surface
pipe 16
8000
its
is
component
vertical
2.4 in
(c)
For an allowable
tensile
is
(d)
3.2 in
of pressure force
(e)
none
of these
answers
on a submerged curved surface
is
equal to
horizontal
component
(a)
its
(b)
the force on a vertical projection of the curved surface
(c)
the product of pressure at centroid and surface area
(d)
the weight of liquid vertically above the curved surface
(e)
none of the above answers
2.171
The
vertical
component
of force
on the quadrant
of the cylinder
AB
(Fig. 2.81)
is
(a)
224y
2.172
96.57
0>)
The
vertical
(c)
8I7
(d)
42.57
(«)
none of these answers
component of force on the upper half of a horizontal right-circular
and 10 ft long, filled with water, and with a pressure of 0.433
cylinder, 3 ft in diameter
psi at the axis,
(a)
-458
is
1b
-3311b
(6)
(c)
124.8 1b
(d)
1872 1b
(e)
none
of these
answers
2.173
When
A
cylindrical
the barrel
wooden
is filled
barrel is held together by hoops at top and bottom.
with liquid, the ratio of tension in the top hoop to tension in
the bottom hoop, due to the liquid,
(
a)
2
(&)
( c)
1
2
(d)
3
wmmmm
_-_-_-_-_-_-_-_-_-:
r=3
ft
-7
Surface 6
Fig. 2.81
ft
long
is
(e)
none
of these
answers
FLUID STATICS
A
2.174
5-cm-ID pipe with 5-mm wall thickness
megapascals,
tensile stress in the pipe wall, in
4.9
(a)
(c)
of slab
1.6
The
(c)
line of action of
center of gravity of
(6)
centroid of the
(c)
centroid of
centroid of
(e)
centroid of
8.0
(e)
none
of these
Buoyant
(d)
14.4
(e)
the buoyant force acts through the
force
is
body due
the resultant force on a
(c)
the force necessary to maintain equilibrium of a submerged
(d)
a nonvertical force for nonsymmetrical bodies
(e)
equal to the volume of liquid displaced
(a)
(6)
(c)
(d)
(e)
it.
none of these answers
the resultant force acting on a floating
body
answers
feet, is
(a)
A
The
sp gr 0.50, floats in water with a 400-lb load on
(6)
2.178
.
any submerged body
volume of any floating body
the displaced volume of fluid
the volume of fluid vertically above the body
the horizontal projection of the body
(a)
(d)
1 ft,
water at 20 kg// cm2
is
39.2
(d)
submerged, in cubic
6.4
(b)
2.176
2.177
19.6
A slab of wood 4 by 4 by
2.175
The volume
(a)
9.8
(b)
carries
107
to the fluid surrounding
it
body
body
floats in stable equilibrium
when its metacentric height is zero
when its center of gravity is below its center of buoyancy
when GB — I/V is positive and G is above B
when I/V is positive
when the metacenter is above the center of gravity
only
A
2.179
1200
(a)
lb.
closed cubical metal box 3 ft on an edge
Its metacentric height
Oft
(6)
-0.08
when placed
(c)
ft
0.62ft
in
is
oil,
(d)
made
of
uniform sheet and weighs
sp gr 0.90, with sides vertical,
0.78ft
(e)
none
is
of these
answers
2.180
Liquid in a cylinder 10
m
long
The difference
sp wt of liquid is
axis of the cylinder.
in pascals,
(a)
20y
2.181
rigid
if
y
=
(6)
When
200t
(c)
20^7
is
accelerated horizontally 20g
in pressure intensities at the
(d)
2OO7/0
(e)
none
m/s2
along the
ends of the cylinder,
of these
answers
a liquid rotates at constant angular velocity about a vertical axis as a
body, the pressure
(a)
decreases as the square of the radial distance
(6)
increases linearly as the radial distance
(c)
decreases as the square of increase in elevation along
any
vertical line
FUNDAMENTALS OF FLUID MECHANICS
108
any
(d)
varies inversely as the elevation along
(e)
varies as the square of the radial distance
2.182
When
a liquid rotates about a vertical axis as a rigid body so that points on
the axis have the same pressure as points 2
velocity in radians per second
(a)
8.02
(e)
none
2.183
A
11.34
(b)
of these
ft
higher and 2
at the center of the
(c)
64.4
(d)
bottom
one-fourth
open at the top,
is filled
is
value
when
cylinder
zero
indeterminable; insufficient data
(d)
greater than a similar case with water as liquid
(e)
none
A
its
was
answers
forced vortex
(a)
turns in an opposite direction to a free vortex
(b)
always occurs in conjunction with a free vortex
(c)
has the velocity decreasing with the radius
when
the angular
with liquid, sp gr
at such speed that half the liquid spills out.
(c)
2.184
axis,
not determinable from data given
(a)
of these
from the
answers
its vertical axis
(b)
ft
is
right-circular cylinder,
rotated about
vertical line
(d)
occurs
(e)
has the velocity decreasing inversely with the radius
fluid rotates as a solid
full
The
1.2,
and
pressure
3
FLUID-FLOW CONCEPTS AND
BASIC EQUATIONS
The
statics of fluids, treated in the preceding chapter, is
science, specific
almost an exact
weight (or density) being the only quantity that must be
On the other hand, the nature of flow of. a real
very complex. Since the basic laws describing the complete motion
of a fluid are not easily formulated and handled mathematically, recourse to
experimentation is required. By an analysis based on mechanics, thermodynamics, and orderly experimentation, large hydraulic structures and efficient
determined experimentally.
fluid is
fluid
machines have been produced.
This chapter introduces the concepts needed for analysis of fluid motion.
The
basic equations that enable us to predict fluid behavior are stated or de-
and momentum, and the
thermodynamics as applied to steady flow of a perfect
gas. In this chapter the control-volume approach is utilized in the derivation
of the continuity, energy, and momentum equations. Viscous effects, the experimental determination of losses, and the dimensionless presentation of loss
data are presented in Chap. 5 after dimensional analysis has been introduced
rived: these are equations of motion, continuity,
first
in
and second laws
Chap.
4.
of
In general, one-dimensional-flow theory
is
developed in this
where viscous effects
Chapter 6 deals with compressible flow, and Chap. 7
chapter, with applications limited to incompressible cases
do not predominate.
with two- and three-dimensional flow.
3.1
THE CONCEPTS OF SYSTEM AND CONTROL VOLUME
The free-body diagram was used in Chap. 2 as a convenient way to show
forces exerted on some arbitrary fixed mass. This is a special case of a system.
109
FUNDAMENTALS OF FLUID MECHANICS
110
A system refers to a definite mass of material and distinguishes it from all other
matter, called
and
its
surroundings.
may
The boundaries
of a system
form a closed
same
mass during changes in its condition; e.g., a slug of gas may be confined in a
cylinder and be compressed by motion of a piston; the system boundary coinciding with the end of the piston then moves with the piston. The system
may contain an infinitesimal mass or a large finite mass of fluids and solids at
surface,
this surface
vary with time, so that
it
contains the
the will of the investigator.
The law of conservation of mass states that the mass within a system
remains constant with time (disregarding relativity effects). In equation
form
dm
-=
with
(3.1.1)
m the total mass.
Newton's second law of motion
SF =
j
is
usually expressed for a system as
(mv)
(3.1.2)
at
which it must be remembered that m is the constant mass of the system.
2 F refers to the resultant of all external forces acting on the system, including
body forces, such as gravity, and V is the velocity of the center of mass of the
in
system.
A control
volume refers to a region in space and is useful in the analysis
where flow occurs into and out of the space. The boundary of
a control volume is its control surface. The size and shape of the control
volume are entirely arbitrary, but frequently they are made to coincide with
solid boundaries in parts, and in other parts they are drawn normal to the
flow directions as a matter of simplification. By superposition of a uniform
velocity on a system and its surroundings a convenient situation for application of the control volume may sometimes be found, e.g., determination of
sound-wave velocity in a medium. The control-volume concept is used in the
derivation of continuity, momentum, and energy equations, as well as in the
solution of many types of problems. The control volume is also referred to
as an open system.
of situations
Regardless of the nature of the flow,
the following relationships, which
1.
may
all
flow situations are subject to
be expressed in analytic form
Newton's laws of motion, which must hold
instant.
for every particle at every
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
2.
3.
4.
The continuity relationship, i.e., the law of conservation of mass.
The first and second laws of thermodynamics.
Boundary conditions, analytical statements that a real fluid has
velocity relative to a
boundary at a boundary or that
111
zero
frictionless fluids
cannot penetrate a boundary.
Other relations and equations may enter, such as an equation of state
Newton's law of viscosity.
In the derivation that follows the control-volume concept is related to
the system in terms of a general property of the system. It is then applied
specifically to obtain continuity, energy, and linear-momentum relationships.
To formulate the relationship between equations applied to a system
and those applied to a control volume, consider some general flow situation,
Fig. 3.1, in which the velocity of a fluid is given relative to an xyz coordinate
system. At time t consider a certain mass of fluid that is contained within a
system, having the dotted-line boundaries indicated. Also consider a control
or
Control volume
Time
t
+
St
(b)
Inside
Outflow area
(c)
Inflow area
id)
Fig. 3.1
System with
velocity field.
identical control
volume
at
time
t
in
a
FUNDAMENTALS OF FLUID MECHANICS
112
volume, fixed relative to the xyz axes, that exactly coincides with the system
8t the system has moved somewhat since each mass
at time t. At time t
the
velocity
associated with its location.
at
particle moves
+
Let
N be the total amount of some property
(mass, energy,
momentum)
be the amount of this property, per unit
mass, throughout the fluid. The time rate of increase of
for the system is
now formulated in terms of the control volume.
At t + 8t, Fig. 3.16, the system comprises volumes II and III, while at
time t it occupies volume II, Fig. 3.1a. The increase in property
in the
system in time dt is given by
within the system at time
t,
and
let
rj
N
N
Ns y8(+i
in
-
,
N
which dV
= (fnvpdv
BySt
is
+
!mVPdV) t+ 6t - (SnVpdV)
t
the element of volume. Rearrangement, after adding and sub-
tracting
(SiVpdV)
t+ st
to the right, then dividing through
~
^y«+i«
JV 8 ys
t
=
(SnripdV
+
by
dt
leads to
JiVpdV)t+it
~ UnVpdV)
t
dt
dt
(fuir)pdV)t+6t
,
_
(!iypdV)t + 6t
(
j
3)
N within the system
becomes dN /dt. If the
limit is taken as 8t approaches zero for the first term on the right-hand side of
the equation, the first two integrals are the amount of N in the control volume
at t + 8t and the third integral is the amount of N in the control volume at
The term on the left
during time
time
The
t.
dt.
is
the average time rate of increase of
In the limit as
limit
8t
approaches zero,
it
is
d
rjpdV
/
'
dt
the partial being needed as the volume
as
8t
—
>
is
held constant (the control volume)
0.
The next term, which is the time rate of flow of
volume, in the limit, may be written
lim
6t-*o
^ IllVpdV
of
^ t+it
=
dA =
I r)pV •
•'Outflow area
J
7)
Pv cos
a dA
N
out of the control
(3.1.4)
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
in
113
Fig. 3.1c, is the vector representing an area element of the outflow
has a direction normal to the surface-area element of the control
which dA,
area.
It
volume, positive outward; a
is
the angle between the velocity vector and the
elemental area vector.
term of Eq.
Similarly, the last
the control volume,
is,
—
lim°t
«<-»>o
(3.1.3)
,
which
is
the rate of flow of
N into
in the limit,
dA = —
/ rjpV •
Mnflnw
araa
^
Inflow area
/ rip
cos a
dA
(3.1.5)
»
The minus sign is needed as V dA (or cos a) is negative for inflow, Fig. 3. Id.
The last two terms of Eq. (3.1.3), given by Eqs. (3.1.4) and (3.1.5), may
•
be combined into the single term which
trol-volume surface
v
lim
««-*>
is
(!ir>pdV)
/ (SuiVpdV)t + 6t
t+it
\ = /f
1
I
M
\
Where there
U
is
an integral over the complete con-
(cs)
/
no inflow or outflow, V
7/pV
•
,.
aA
J C8
•
dA =
=
f
/
.
r\pv
cos a
dA
J C8
0;
hence the equation can be
evaluated over the whole control surface. 1
Collecting the reorganized terms of Eq. (3.1.3) gives
dN
=
-
dt
f
VP
dV+
f rjpV-dA
(3.1.6)
N within
N within
plus the net rate of efflux of N across
In words, this equation states that the time rate of increase of
a system
is
the control
just equal to the time rate of increase of the property
volume
(fixed relative to xyz)
the control-volume boundary.
Equation (3.1.6) is used throughout this chapter in converting laws and
from the system form to the control-volume form. The system
form, which in effect follows the motion of the particles, is referred to as the
Lagrangian method of analysis; the control-volume approach is called the
Eulerian method of analysis, as it observes flow from a reference system fixed
relative to the control volume.
Since the xyz frame of reference may be given an arbitrary constant
velocity without affecting the dynamics of the system and its surroundings,
Eq. (3.1.6) is valid if the control volume, fixed in size and shape, has a uniform
principles
velocity of translation.
1
This derivation was developed by Professor William Mirsky of the Department of
chanical Engineering, The University of Michigan.
Me-
FUNDAMENTALS OF FLUID MECHANICS
114
3.2 APPLICATION OF THE CONTROL VOLUME
TO CONTINUITY, ENERGY, AND MOMENTUM
In this section the general relation of system and control volume to a property,
3.1, is applied first to continuity, then to energy, and finally
linear
momentum.
In the following sections the uses of equations are
to
brought out and illustrated.
developed in Sec.
Continuity
The
continuity equations are developed from the general principle of con-
servation of mass, Eq. (3.1.1), which states that the mass within a system
remains constant with time;
i.e.,
dm
dt
In Eq. (3.1.6)
mass, or
rj
—
=
f
=
let iV
be the mass of the system m. Then
77
is
the mass per unit
1
pdV
+
J
pV
•
dA
(3.2.1)
In words the continuity equation for a control volume states that the time
mass within a control volume is just equal to the net rate
mass inflow to the control volume. This equation is examined further in
rate of increase of
of
Sec. 3.4.
Energy equation
The
first
law of thermodynamics
to a system
initial
and
minus the work
final states of
for a
form
Q H added
difference in states of the system,
initial to final state,
It is called the internal
energy E.
The
must be a property
first
law in equation
is
Qh-W
The
The
the system.
being independent of the path from
of the system.
system states that the heat
W done by the system depends only upon the
= E*-Ei
internal energy per unit
(3.2.2)
mass
is
called e; hence, applying Eq. (3.1.6),
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
N
= E and
—=-
dE
d
f
pe/p,
by use
,
,
peV
/
7A
-dA
(3.2.3)
J cs
of Eq. (3.2.2)
5Q H
bW
u
dt
—
(
pedV+
dt J cv
dt
or
=
r?
115
— = - J*pedV +
=
dE
d
dt
at
f
,
,
/
/"
peV
•
JA
dh
(3.2.4)
J ra
The work done by the system on its surroundings may be broken into two
work
pr done by pressure forces on the moving boundaries, and
the work Ws done by shear forces such as the torque exerted on a rotating
shaft. The work done by pressure forces in time 5£ is
W
parts: the
bW pT =
By
dtfpy
•
dA
(3.2.5)
use of the definitions of the work terms, Eq. (3.2.4) becomes
SWs
BQh
d
f
f
,
/p
.
(3.2.6)
In the absence of nuclear,
the internal energy
effects,
\
,
e
electrical,
magnetic, and surface-tension
of a pure substance
The
is
the
sum
of potential,
energy u per unit mass
due to molecular spacing and forces (dependent upon p, p, or T)
and
kinetic,
"intrinsic" energies.
intrinsic
is
:
e
=
Q*
Linear
+
g
+
u
(3.2.7)
momentum
equation
Newton's second law for a system, Eq.
(3.1.2), is
used as the basis for finding
by use of Eq. (3.1.6).
the linear-momentum equation for a control volume
Let
N be the linear momentum mV of the system and let
tum per
SF =
unit mass pV/p.
djmV)
dt
=
T
/
dt J M
Then by use
Pv
dV
+
/
J„
of Eqs. (3.1.2)
pVV
-
dA
rj
be the linear momen-
and
(3.1.6)
(3.2.8)
FUNDAMENTALS OF FLUID MECHANICS
116
In words, the resultant force acting on a control volume
is
equal to the time
momentum within the control volume
momentum from the control volume.
rate of increase of linear
efflux of linear
Equations
analysis of
many
(3.2.1),
of the
(3.2.6),
and
problems of
(3.2.8)
plus the net
provide the relationships for
In effect, they provide
mechanics.
fluid
a bridge from the solid-dynamics relations of the system to the convenient
control-volume relations of fluid flow.
Flow characteristics and definitions are next discussed, before the basic
control-volume equations are examined and applied.
3.3
FLOW CHARACTERISTICS; DEFINITIONS
Flow may be
classified in
many
ways, such as turbulent, laminar;
reversible, irreversible; steady, unsteady; uniform,
irrotational.
real, ideal;
nonuniform; rotational,
In this and the following section various types of flow are dis-
tinguished.
Turbulent-flow situations are most prevalent in engineering practice.
In turbulent flow the fluid particles (small molar masses) move in very irregular paths, causing an exchange of
to another in a
manner somewhat
momentum from one portion of the fluid
momentum transfer
similar to the molecular
much larger scale. The fluid particles can range
from very small (say a few thousand molecules) to very large (thousands of cubic feet in a large swirl in a river or in an atmospheric gust) In a
situation in which the flow could be either turbulent or nonturbulent (laminar),
the turbulence sets up greater shear stresses throughout the fluid and causes
more irreversibilities or losses. Also, in turbulent flow, the losses vary as the
square of the velocity, while in laminar flow, they vary as the first power of
described in Sec. 1.3 but on a
in size
.
the velocity
In laminar flow,
fluid particles
move
along smooth paths in laminas, or
with one layer gliding smoothly over an adjacent layer. Laminar flow
is governed by Newton's law of viscosity [Eq. (1.1.1) or extensions of it to
three-dimensional flow], which relates shear stress to rate of angular deforma-
layers,
tion.
In laminar flow, the action of viscosity damps out turbulent tendencies
(see Sec. 5.3 for criteria for laminar flow)
.
Laminar flow
is
not stable in situ-
ations involving combinations of low viscosity, high velocity, or large flow
passages and breaks
down
Newton's law of viscosity
ay
into turbulent flow.
may
An
equation similar in form to
be written for turbulent flow
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
not a fluid property alone but depends upon the
It is called the eddy viscosity.
The
factor
fluid
motion and the density.
77,
many
In
however,
is
117
practical flow situations,
both viscosity and turbulence con-
tribute to the shear stress:
=
r
(m
+
V)
Experimentation
An
(3-3.2)
J"
dy
is
required to determine this type of flow.
and incompressible and should not be conThe assumption of an ideal fluid is helpful
analyzing flow situations involving large expanses of fluids, as in the motion
an airplane or a submarine. A frictionless fluid is nonviscous, and its flow
ideal fluid is frictionless
fused with a perfect gas (Sec. 1.6)
in
of
.
processes are reversible.
The layer of fluid in the immediate neighborhood
ary that has had
its
velocity relative to the
of an actual flow boundboundary affected by viscous shear
called the boundary layer. Boundary layers may be laminar or turbulent,
depending generally upon their length, the viscosity, the velocity of the flow
near them, and the boundary roughness.
Adiabatic flow is that flow of a fluid in which no heat is transferred to
or from the fluid. Reversible adiabatic (frictionless adiabatic) flow is called
is
isentropic flow. 1
To proceed
in
an orderly manner into the analysis of
flow requires a clear understanding of the terminology involved.
the
fluid
Several of
more important technical terms are defined and illustrated in this section.
Steady flow occurs when conditions at any point in the fluid do not change
with the time. For example,
+x
rection indefinitely.
if
the velocity at a certain point
10 ft/s in the
it
(x, y, z
coordinates of the point)
there
no change in density
is
is
remains exactly that amount and in that diThis can be expressed as dv/dt = 0, in which space
direction in steady flow,
p,
is
held constant.
Likewise, in steady flow
pressure p, or temperature T, with time at
any
point; thus
»_o
^=
dt
dt
^=
dt
In turbulent flow, owing to the erratic motion of the fluid particles,
there are always small fluctuations occurring at
any
point.
The
definition for
steady flow must be generalized somewhat to provide for these fluctuations.
To
1
illustrate this,
An
a plot of velocity against time, at some point in turbulent
isentropic process, however, can occur in irreversible flow with the proper
of heat transfer (isentropic = constant entropy).
amount
FUNDAMENTALS OF FLUID MECHANICS
118
Time
Velocity
Fig. 3.2
a
at
point
in
When
the temporal
steady turbulent flow.
flow,
is
given in Fig.
3.2.
mean
velocity
v dt
-IJ
t
by the horizontal line, does not change with the time,
The same generalization applies to density, pressure, temperature, etc., when they are substituted for v in the above formula.
The flow is unsteady when conditions at any point change with the time,
dv/dt 5* 0. Water being pumped through a fixed system at a constant rate
is an example of steady flow. Water being pumped through a fixed system at
an increasing rate is an example of unsteady flow.
Uniform flow occurs when at every point the velocity vector is identical
(in magnitude and direction) for any given instant, or, in equation form,
dv/ds = 0, in which time is held constant and 8s is a displacement in any direction. The equation states that there is no change in the velocity vector
in any direction throughout the fluid at any one instant. It says nothing
indicated in the figure
the flow
is
said to be steady.
about the change in velocity at a point with time.
In flow of a real fluid in an open or closed conduit, the definition of uniform flow may also be extended in most cases even though the velocity vector
at the boundary is always zero. When all parallel cross sections through the
conduit
ai*e
identical
(i.e.,
when the conduit
velocity at each cross section
is
is
prismatic) and the average
the same at any given instant, the flow
is
said
to be uniform.
varies from place to place at any
nonuniform flow. A liquid being pumped through a
long straight pipe has uniform flow. A liquid flowing through a reducing section or through a curved pipe has nonuniform flow.
Examples of steady and unsteady flow and of uniform and nonuni-
Flow such that the velocity vector
instant (dv/ds 9^ 0)
is
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
are: liquid flow
form flow
through a long pipe at a constant rate
119
is
steady
uniform flow; liquid flow through a long pipe at a decreasing rate is unsteady uniform flow; flow through an expanding tube at a constant rate
is steady nonuniform flow; and flow through an expanding tube at an increasing rate
is
unsteady nonuniform flow.
Rotation of a fluid particle about a given axis, say the z axis,
two
as the average angular velocity of
ticle
that are at right angles to each other and to the given axis.
particles within a region
tional flow, or vortex flow.
have rotation about any
If the fluid
is
called irrotational flow. It
is
at rest
and
is frictionless,
is
defined
infinitesimal line elements in the par-
axis,
the flow
If the fluid
is
called rota-
within a region has no rotation, the flow
shown in texts on hydrodynamics that if a fluid
any later motion of this fluid will be irrotational.
is
One-dimensional flow neglects variations or changes in velocity, pressure,
etc.,
transverse to the
main flow
Conditions at a cross section are
direction.
expressed in terms of average values of velocity, density, and other properties.
Flow through a
analysis,
which
of analysis.
pipe, for example,
Many
dimensional.
is
may
usually be characterized as one-
practical problems can be handled
much
by
this
method
of
simpler than two- and three-dimensional methods
In two-dimensional flow
all
particles are
assumed to flow
in parallel
planes along identical paths in each of these planes hence, there are no changes
;
in flow
normal to these planes. The flow
useful
method
sional flow
is
net,
developed in Chap.
for analysis of two-dimensional-flow situations.
7, is
the most
Three-dimen-
the most general flow in which the velocity components u,
v,
w
mutually perpendicular directions are functions of space coordinates and
time x, y, z, and t. Methods of analysis are generally complex mathematically,
and only simple geometrical flow boundaries can be handled.
in
A
streamline
is
a continuous line drawn through the fluid so that
it
has
the direction of the velocity vector at every point. There can be no flow across
a streamline.
Since a particle
instant, its displacement 5S,
of the velocity vector
spectively.
bx
u
_
by
v
_
moves
in the direction of the streamline at
having components
q with components
u,
v,
w
bx, by, dz,
any
has the direction
in the x, y, z directions, re-
Then
bz
w
states that the corresponding
and q have the same
form
6S
components are proportional and hence that
direction.
Expressing the displacements in differential
dx
— = dy
— = dz
U
V
w
3.3.3)
FUNDAMENTALS OF FLUID MECHANICS
120
produces the differential equations of a streamline. Equations (3.3.3) are
two independent equations. Any continuous line that satisfies them is a
streamline.
at
In steady flow, since there is no change in direction of the velocity vector
point, the streamline has a fixed inclination at every point and is, therefixed in space. A particle always moves tangent to the streamline; hence,
any
fore,
in steady flow the path of a particle
direction of the velocity vector at
line
may
shift in
is
a streamline. In unsteady flow, since the
any point may change with time, a stream-
space from instant to instant.
A
particle then follows
one
streamline one instant, another one the next instant, and so on, so that the
path of the particle
may have no
resemblance to any given instantaneous
streamline.
A
dye or smoke
is
frequently injected into a fluid in order to trace
its
subsequent motion. The resulting dye or smoke trails are called streak lines.
In steady flow a streak line is a streamline and the path of a particle.
Streamlines in two-dimensional flow can be obtained by inserting fine,
bright particles (aluminum dust) into the fluid, brilliantly lighting one plane,
and taking a photograph of the streaks made in a short time interval. Tracing
on the picture continuous
lines that
have the direction
of the streaks at every
point portrays the streamlines for either steady or unsteady flow.
In illustration of an incompressible two-dimensional flow, as in Fig.
the streamlines are
drawn
adjacent streamlines
plane of the figure.
must be
is
3.3,
volume flowing between
considered normal to the
so that per unit time the
the same
if
unit depth
is
Hence, when the streamlines are closer together, the ve-
and vice versa. If v is the average velocity between
two adjacent streamlines at some position where they are h apart, the flow
rate Aq is
locity
greater,
Aq = vh
Fig. 3.3
(3.3.4)
Streamlines
for
around a
cylinder between parallel
steady
walls.
flow
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
At any other
position
on the chart where the distance between streamlines
is hi,
the average velocity
lines
drawn,
is
i.e.,
121
=
is Vi
by decreasing
Aq/hi.
By increasing
the
number
of stream-
Aq, in the limiting case the velocity at a point
obtained.
A
the tube made by all the streamlines passing through a
In steady flow it is fixed in space and can have no flow
walls because the velocity vector has no component normal to
stream tube
is
small, closed curve.
through
its
the tube surface.
EXAMPLE
foil
3.1
In two-dimensional, incompressible steady flow around an
the streamlines are
drawn
so that they are
air-
cm apart at a great distance
What is the velocity near the
1
from the airfoil, where the velocity is 40 m/s.
airfoil, where the streamlines are 0.75 cm apart?
(40 m/s) (0.01
m)
(1
m) =
0.40
m /s
3
=
v
0.0075
m
2
and
v
-
3.4
0.40
m /s
3
^^
0.0075 m
,
2
=
_ n m/s
53.3
,
CONTINUITY EQUATION
The use
of Eq. (3.2.1)
is
developed in this section.
First, consider
flow through a portion of the stream tube of Fig. 3.4.
comprises the walls of the stream tube between sections
Control
volume
Fig. 3.4 Steady
stream tube.
flow
through
a
The
1
and
2,
steady
volume
plus the end
control
FUNDAMENTALS OF FLUID MECHANICS
122
areas of sections
(3.2.1) is zero;
f PV
•
1
and
2.
Because the flow
is
steady, the
first
term
of
Eq.
hence
dk =
(3.4.1)
C8
which states that the net mass outflow from the control volume must be zero.
At section 1 the net mass outflow is piVi dAi = —piVi dA h and at section 2
it is p 2 V2
dA 2 = P2V2 dAi. Since there is no flow through the wall of the stream
•
•
tube,
PiVidAi
is
=
p2V 2
dA 2
(3.4.2)
the continuity equation applied to two sections along a stream tube in steady
flow.
For a collection of stream tubes, as in Fig. 3.5, if pi is the average density
1 and p 2 the average density at section 2, then
at section
m =
in
=
/nViAi
which
Vi,
P2V2A2
(3.4.3)
V2 represent average velocities over the cross sections and m is
The average velocity over a cross section is given by
the rate of mass flow.
-4/
If
v
dA
the discharge
Q =
Q
is
defined as
AV
Fig. 3.5 Collection
of
stream
tubes between fixed boundaries.
(3.4.4)
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
the continuity equation
m=
P1Q1
=
may
123
take the form
(3.4.5)
P2Q2
For incompressible, steady flow
Q = A{Vi = A 2 V2
is
(3.4.6)
a useful form of the equation.
For constant-density
f
V
•
flow, steady or unsteady,
dk =
Eq. (3.2.1) becomes
(3.4.7)
r cs
which states that the net volume efflux
volume is filled with liquid at all times)
EXAMPLE
3.2
velocity
3.0 ft/s,
3.0
ft.
At
is
zero (this implies that the control
1 of a pipe system carrying water (Fig. 3.6) the
and the diameter is 2.0 ft. At section 2 the diameter is
Find the discharge and the velocity at section 2.
is
From Eq.
Q = ViAx =
section
(3.4.6)
3.0tt
=
9.42
ft
3
/s
and
For two- and three-dimensional-flow studies, differential expressions of
,
h
,
!
r\
1
r+v
-v,
l
2
II
A,
Pi
1
©
A
2
P2
Fig. 3.6 Control
volume
through series pipes.
for
flow
FUNDAMENTALS OF FLUID MECHANICS
124
J
A
A
r
1
/
pubybz
by
?l
2
—
^ r
bx
Control volume for deriva-
Fig. 3.7
tion of three-dimensional continuity
equation
cartesian
in
co-
ordinates.
the continuity equation must be used. For three-dimensional cartesian coordinates, Eq. (3.2.1)
are u,
v,
applied to the control-volume element bx by bz of Fig. 3.7
(x,y,z),
w, respectively,
the pair of
outward
is
where the velocity components in the x, y, z directions
and p is the density. Consider first the flux through
faces normal to the x direction. On the right-hand face the flux
with center at
is
.
h+
{pu)
(pu)
bx
—\bybz
f]
and u are assumed to vary continuously throughout the fluid.
In the expression, pu by bz is the mass flux through the center face normal to
the x axis. The second term is the rate of increase of mass flux with respect
to x, multiplied by the distance bx/2 to the right-hand face. Similarly on the
since both p
left-hand face the flux into the
pu
\
~
since the step
— (pu) bx by
dx
2~
dx
is
bz
r
volume
is
8z
—bx/2. The net flux out through these two faces
is
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
The other two
directions yield similar expressions hence the net
;
125
mass outflow
is
n
»\
— (pu)
+
dx
—
«J
+—
(pv)
dy
(pw)
dz
bx by bz
J
which takes the place of the right-hand part of Eq.
of Eq. (3.2.1) becomes, for an element,
(3.2. 1)
.
The left-hand
part
dp
bx by
y bz
dt
When these two
expressions are used in Eq. (3.2.1)
,
after dividing
through by
the volume element and taking the limit as bx by bz approaches zero, the continuity equation at a point
£- (pu)
dx
+
j-
B
+dz
(pv)
dy
becomes
(pw)
= -
d
{dt
(3.4.8)
which must hold for every point in the flow, steady or unsteady, compressible
or incompressible.
For incompressible flow, however, it simplifies to
1
^
= °
+?
?
dx + ?
dy
dz
Equations (3.4.8) and (3.4.9)
By
using fixed unit vectors in
tor
V
v =
is
i
=
+
+ iTT
dx
dy
\u
be compactly written in vector notation.
directions,
i,
j,
k,
respectively, the opera-
defined as
k
+
Equation
\v
+
kw
(3 4 10)
-
ir
dz
and the velocity vector q
q
may
x, y, z
is
-
given by
(3.4.11)
(3.4.8) can be derived from Eq. (3.2.1) by application of Gauss' theorem.
See L. Page, "Introduction to Theoretical Physics," 2d ed., pp. 32-36, Van Nostrand,
Princeton, N.J., 1935.
FUNDAMENTALS OF FLUID MECHANICS
126
Then
—
+ k+ —
dx
dy
dz/
i
—
(pu)
dx
—
dy
r)
=
because
V
•
•
i
1,
•
i
\pv
+
k P w)
r)
(pv)
=
j
+
+dz
(pw)
Equation
0, etc.
becomes
(3.4.8)
~
= -
Pq
=
i
r)
+
(\pu
•
)
j
(3.4.12)
Ol
and Eq.
V
•
q
becomes
(3.4.9)
=
(3.4.13)
The dot product V
words
•
q
is
called the divergence of the velocity vector q.
In
the net volume efflux per unit volume at a point and must be zero
it is
for incompressible flow.
See Sec. 7.2 for further discussion of the operator V.
In two-dimensional flow, generally assumed to be in planes parallel to
the xy plane, w = 0, and there is no change with respect to z, so d/dz =
which reduces the three-dimensional equations given for continuity.
EXAMPLE
The velocity
by
3.3
distribution for a two-dimensional incompres-
sible flow is given
x
u =
+
2
x
Show
that
V
y
y
= —
2
x2
it satisfies
+
y
2
continuity.
In two dimensions the continuity equation
du
dv
dx
dy
is,
from Eq.
(3.4.9),
=
Then
du
dx
and
x2
their
+
sum
^
2a2
1
y
2
(x2
+
y
2
2
)
0,
dy
_
2y 2
_J_
x2
+
y
2
does equal zero, satisfying continuity.
(x2
+
y
2
2
)
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
3.5
127
EULER'S EQUATION OF MOTION ALONG A STREAMLINE
In addition to the continuity equation, other general controlling equations are
Euler's equation, Bernoulli's equation, the energy equation, the momentum
equations,
and the
Euler's equation
is
first
and second laws
of
thermodynamics. In
derived in differential form. In Sec. 3.9
obtain Bernoulli's equation.
veloped for steady flow,
The
and some
first
it is
this section
integrated to
law of thermodynamics
is
then de-
of the interrelations of the equations are
an introduction to the second law of thermodynamics.
Euler's
equation is derived for general three-dimensional flow.
7
Chap.
In
flow along a streamline.
it
is
restricted
to
Here
In Fig. 3.8 a prismatic-shaped fluid particle of mass p 8A 8s is moving
along a streamline in the +s direction. To simplify the development of the
equation of motion for this particle it is assumed that the viscosity is zero or
that the fluid is frictionless. This eliminates all shear forces from consideration, leaving as forces to take into consideration the body force due to the
pull of gravity and surface forces on the end areas of the particle. The gravity
force is pg 8 A 8s. On the upstream face the pressure force is p 8 A in the +s
direction; on the downstream face it is \jp + (dp/ds) 8s~] 8 A and acts in the
— s direction. Any forces on the sides of the element are normal to s and do
explored, including
The body-force component
not enter the equation.
— pg 8 A
8m a
8,
8s cos
gives
p8A
(>+j» 8A
'
pgbA
—
pg
8
A
bs
Force components on a
the direction of
the streamline.
Fig. 3.8
fluid particle in
in the s direction is
Substituting into Newton's second law of motion, 2/8
0.
8s cos 6
=
p
8A
8s
as
=
FUNDAMENTALS OF FLUID MECHANICS
128
as
is
the acceleration of the fluid particle along the streamline. Dividing
through by the mass of the
dp
—
+ g cos
1
—
+
as
particle, p
8A
and simplifying leads to
bs,
=
(3.5.1)
p ds
the increase in elevation of the particle for a displacement
bz is
bs.
From
Fig. 3.8,
bz
—=
=
cos
ds
bs
The
v
=
dv
acceleration a s
=
dv
—
ds
=
by
dv
-
dt,
=
v
depends upon
s
and time
t,
dt
of
particle, so
one
may
(3.5.2)
dt
substituting for cos
dv
dv
p ds
ds
ds
dt
the flow
is
dp
dz
dv
p ds
os
ds
2,
motion of a
—
-- + dv
dz
Since p,
in describing the
dv ds
dp
If
t
obtaining
ds dt
By
1
if
dt
dt
1
In general,
dv
H
becomes a function
divide
as
dv/dt.
is
v(s,t),
ds
s
dz
—
and
v
and a s
steady, dv/dt
now
=
0,
in Eq. (3.5.1)
yielding
/rt
are functions of s only, the partials
may
„
..
be replaced by
total derivatives:
— + gdz + vdv
=
(3.5.5)
is one form of Euler's equation of motion and requires three important
assumptions: (1) motion along a streamline, (2) Motionless fluid, and (3)
steady flow. It can be integrated if p is known as a function of p or is constant.
This
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
A
AND LOSSES
REVERSIBILITY, IRREVERSIBILITY,
3.6
process
may be
129
defined as the path of the succession of states through which
the system passes, such as the changes in velocity, elevation, pressure, density,
temperature, etc. The expansion of air in a cylinder as the piston moves out
is an example of a process. Normally,
some change in the surroundings, e.g., displacing it or transferring heat to or from its boundaries. When a process can be made to take
place in such a manner that it can be reversed, i.e., made to return to its orig-
and heat
is
transferred through the walls
the process causes
inal state
is
without a
change in either the system or its surroundings, it
In any actual flow of a real fluid or change in a me-
final
said to be reversible.
chanical system, the effects of viscous friction,
expansion, hysteresis,
etc.,
Coulomb
friction,
unrestrained
prohibit the process from being reversible.
It
is,
however, an ideal to be strived for in design processes, and their efficiency
is
usually defined in terms of their nearness to reversibility.
When
a certain process has a sole effect upon
its
surroundings that
is
have done work on its surroundings. Any actual process is irreversible. The difference between the
amount of work a substance can do by changing from one state to another
state along a path reversibly and the actual work it produces for the same
path is the irreversibility of the process. It may be defined in terms of work
per unit mass or weight or work per unit time. Under certain conditions the
equivalent to the raising of a weight,
it is
said to
irreversibility of a process is referred to as its lost work, 1
work because
to do
referred to, they
of friction
mean
and other
causes.
irreversibility or lost
i.e.,
the loss of ability
In this book when
losses are
work and do not mean an actual
loss of energy.
EXAMPLE 3.4 A hydroelectric plant has a head (difference in elevation of
headwater and tail water) of 50
and a flow of 5 3 /s of water through the
turbine, which rotates at 180 rpm. The torque in the turbine shaft is measured
to be 1.16 X 10 5 N m, and the output of the generator is 2100 kW. Determine
m
m
and the reversible power
the irreversibility, or losses,
m/s 2
The
2,450,500
Ta =
1
g
=
9.8
.
is 50 m N/N. Hence for perfect conyQH = 9802 N/m X 5 m /s X 50 m N/N =
potential energy of the water
version the reversible
The
for the system,
N-m/s =
(1.16
X
10 5
irreversibility
power
2450.5
is
•
3
•
kW. The
rate of
N-m)(\8<f)(27rrad/s) =
through the turbine
Reference to a text on thermodynamics
3
is
is
work by the turbine
2186.5
kW
then 2450.5
advised for a
is
full
kW —
2186.5
kW
=
discussion of these concepts.
FUNDAMENTALS OF FLUID MECHANICS
130
kW or
264
kW 1000 N-m/s
1
=
IkW
5rf/T
9"80^N7^
264
The
irreversibility
-
2186.5
2100
=
5
39m N/N
'
'
through the generator
86.5
is
kW
or
86.5
kW
1000
5m /s
N-m/s
IkW
3
1
Efficiency of the turbine
* =
and
V
10
°
X
50
L 6m
N/m ~
3
9802
rj
m N/N •
t
100
X
5.39
m N/N
~ L76 =
„
•
89
=
SOm-N/N
5°
'
is
efficiency of the generator
=
'
'
22%
g is
rj
96.48%
.
oU
THE STEADY-STATE ENERGY EQUATION
3.7
When
Eq. (3.2.6)
to Fig. 3.9, the
is
applied to steady flow through a control volume similar
volume
integral drops out
2
—— + foi
— + gzi + — + U\ \ piViAi
1
\pi
&Qh
V!
1
I
ot
=
8W
—
—+
S
/
Since the flow
is
and
Ot
steady in this equation,
it
becomes
2
/p 2
[
2
\P2
it is
,
Pi
h gzi
Pi
qH
is
vS
+—
+ U\ = w
2
=
v
+—
+
I
p2
through a control volume.
This
P2V2A2
P2A2V2, getting
u2
the heat added per unit mass of fluid flowing, and
per unit mass of fluid flowing.
I
2
h gz 2
H
\
/
2
pi
B
ui
convenient to divide through by
the mass per second flowing through the system p\A\V\
qH H
v
+—
+
2
h gz 2
is
(3.7.1)
w
s
is
the shaft work
the energy equation for steady flow
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
131
~>W
Control
S
volume
Control volume
volume with
Control
Fig. 3.9
flow across control surface nor-
mal to surface.
The energy equation
(3.7.1)
in differential form, for flow
stream tube (Fig. 3.10) with no shaft work,
d-
+
g dz
+
v
+ du -
dv
dq H
through a
is
=
(3.7.2)
P
Rearranging gives
dp
\-
g dz
-\-
v dv
-\-
du
+
1
pd
For frictionless flow the
Fig. 3.10
dqH
=
(3.7.3)
P
P
sum
of the first three terms equals zero
Steady-stream tube as
control Volume.
from the Euler
FUNDAMENTALS OF FLUID MECHANICS
132
equation (3.5.5) the last three terms are one form of the
for a system,
;
first
law of thermo-
dynamics
+
= pd-
dq H
du
(3.7.4)
P
Now,
*
=
for reversible flow, entropy s per unit
(f
\
/
T
which
is
defined
by
(3.7.5)
*
in
mass
rev
is
Entropy
the absolute temperature.
erty in texts on thermodynamics.
shown
is
In this equation
it
to be a fluid prop-
may have
the units Btu
per slug per degree Rankine or foot-pounds per slug per degree Rankine, as
may
heat
be expressed in foot-pounds
(1
Btu = 778
ft -lb).
In SI units
s is
in kilocalories per kilogram per kelvin or joules per kilogram per kelvin
Since Eq. (3.7.4) is for a frictionless fluid (reversible),
(1 kcal = 4187 J).
dqH can be eliminated from Eqs. (3.7.4) and (3.7.5),
Tds = du
+ pd-
(3.7.6)
P
which
a very important thermodynamic relation.
is
Although
it
was derived
terms are thermodynamic properties, it must
also hold for irreversible-flow cases as well. By use of Eq. (3.7.6) together
with the Clausius inequality and various combinations of Euler's equation
for a reversible process, since all
1
and the
first
law, a clearer understanding of entropy
and
losses is gained.
INTERRELATIONSHIPS BETWEEN EULER'S EQUATIONS AND
THE THERMODYNAMIC RELATIONS
3.8
The
first
law in differential form, from Eq.
(3.7.3),
with shaft work included,
is
dw
s
H
\-
v
dv
-\-
g dz
-\-
du
-\-
pd
P
Substituting for du
dw
s
+
•
+
p d{l/p)
See any text on thermodynamics.
=
(3.8.1)
Eq. (3.7.6) gives
- + vdv + gdz+Tds-dqH
P
1
in
dq H
P
=
(3.8.2)
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
The Clausius
133
inequality states that
ds>^
- T
or
Tds>dq H
Thus
(3.8.3)
T ds —
dqH
a frictionless
>
The
0.
equals sign applies to a reversible process [or
yielding Eq. (3.5.4) with a
fluid,
called losses or irreversibilities
d
= Tds -
(losses)
If
the quantity
(3.8.4)
is
positive in irreversible flow,
and can never be negative.
flow,
work term].
identified as
dq H
seen that d (losses)
it is
is
is
zero in reversible
Substituting Eq. (3.8.4) into Eq. (3.8.2)
yields
dw
dp
s
H
\-
v
dv
+
g dz
+
d (losses)
=
(3.8.5)
p
This is a most important form of the energy equation. In general, the losses
must be determined by experimentation. It implies that some of the available
energy is converted into intrinsic energy during an irreversible process. This
equation, in the absence of the shaft work, differs from Euler's equation by
the loss term only. In integrated form,
— + gzi
=
l
If
work
h
/
J1
is
p
EXAMPLE
z
done on the
Section
negative.
-jr
1 is
+ gz* + w* +
lossesi_ 2
(3.8.6)
volume, as with a pump, then
upstream, and section 2 is downstream.
fluid in the control
w
s
is
The cooling-water plant for a large building is located on a
by a stream, as shown in Fig. 3.11a. The design low-stream
and at this condition the only outflow from the lake is 5 cfs via
3.5
small lake fed
flow
is
5
cfs,
a gated structure near the discharge channel for the cooling-water system.
The temperature of the incoming stream is 80°F. The flow rate of the cooling
system is 4490 gpm, and the building's heat exchanger raises the coolingwater temperature by 10°F. What is the temperature of the cooling water
FUNDAMENTALS OF FLUID MECHANICS
134
Pump house
To
building
From building
-«
Gated spillway
T
<?out'
+ *T
(b)
(a)
Cooling-water system.
Fig. 3.11
recirculated through the lake, neglecting heat losses to the atmosphere
lake bottom,
if
and
these conditions exist for a prolonged period?
A heat balance may be written for the lake,
energy out. Let
T be the
Fig. 3.116,
with energy in
=
7
average temperature of the lake and A7 be the tem-
perature rise through the heat exchanger
QinT in
+
Q HE (T
+
AT) = Q HE T
+
Q in {T
+
A7
W
T
+
7
)
or
5
X
80
+
^490gP"\
449 gpm/1
(T
+
10)
=
cfs
5(!T
+
10)
or
T = 90°F
The temperature
3.9
leaving the lake
is
100°F.
THE BERNOULLI EQUATION
Integration of Eq. (3.5.5) for constant density yields the Bernoulli equation
v*
gz
-r-
-
p
+ -p
z
=
const
(3.9.1)
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
The constant
135
of integration (called the Bernoulli constant) in general varies
from one streamline to another but remains constant along a streamline in
These four assumptions are needed
steady, frictionless, incompressible flow.
and must be kept in mind when applying this equation. Each term has the
dimensions (L/T) 2 or the units square feet (or square meters) per second per
second, which is equivalent to foot-pounds per slug (meter-newtons per kilogram)
:
ft-
lb/slug
-^- = ftVs
=
2
lb-s 2 /ft
as
1
=
slug
lb s 2 /ft. Therefore
•
divided through
is
+—+
z
1
2#
since y
2i
+
=
Eq. (3.9.1)
is
energy per unit mass.
When it
g,
const
(3.9.2)
pg, or
—H
2g
it
=
7
by
=
z2
+—H
2g
y
(3.9.3)
y
can be interpreted as energy per unit weight, or foot-pounds per pound
meter-newtons per newton) This form is particularly convenient for deal-
(or
.
ing with liquid problems with a free surface.
Multiplying Eq. (3.9.1) by p
gives
2
yz
pv
+ —- +
p
=
const
(3.9.4)
which is convenient for gas flow, since elevation changes are frequently unimportant and yz may be dropped out. In this form each term is foot-pounds
per cubic foot (meter-newtons per cubic meter), or energy per unit volume.
Each of the terms of Bernoulli's equation may be interpreted as a form
of energy. In Eq. (3.9.1) the first term is potential energy per unit mass.
With reference to Fig. 3.12 the work needed to lift
newtons a distance
z meters is Wz. The mass of
weight is W/g kg; hence the potential energy
W
WN
per kilogram
Wz
wr
gz
is
FUNDAMENTALS OF FLUID MECHANICS
136
Datum
W
Potential
3.12
Fig.
energy.
The next term, v 2 /2,
2
is dm v /2.
To
is
interpreted as follows.
Kinetic energy of a particle of
place this on a unit mass basis, divide
by 8m; thus v 2 /2
meter-newtons per kilogram kinetic energy.
The last term, p/p, is the flow work or flow energy per unit mass. Flow
work is net work done by the fluid element on its surroundings while it is
flowing. For example, in Fig. 3.13, imagine a turbine consisting of a vaned
unit that rotates as fluid passes through it, exerting a torque on its shaft.
For a small rotation the pressure drop across a vane times the exposed area
of vane is a force on the rotor. When multiplied by the distance from center
of force to axis of the rotor, a torque is obtained. Elemental work done is
mass
is
p 8 A ds by p 8 A ds units of mass of flowing fluid hence, the work per unit
mass is p/p. The three energy terms in Eq. (3.9.1) are referred to as available
;
energy.
By
.Pi
ziH
applying Eq. (3.9.3) to two points on a streamline,
.
h
7
—
vi'
=
2g
h
7
*
—
v2
P2
z2 H
2g
W=
Fig. 3.13
Work done by
tained pressure.
sus-
(3.9.5)
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
137
or
-
Z\
+ pi-jh
22
,
h
2g
This equation shows that
and
is
it is
the difference in potential energy, flow energy,
kinetic energy that actually has significance in the equation.
independent of the particular elevation datum, as
elevation of the
two
points.
Similarly pi/7
—
p 2 /y
is
particular pressure
linear, their
EXAMPLE
datum
3.6
datum
selected.
Water
is
and the velocity
2
ft
is
40.1 ft/s.
Assuming
determine the difference in elevation of the channel
the difference in elevation of floors
is y,
upper water surface to the lower water surface
H
h
2g
21
=
—
2g
7
and
is
not altered
Since the velocity terms are not
ft and a
then flows down a chute into another open channel,
where the depth
—
22
flowing in an open channel at a depth of 4
It
If
—
is fixed.
velocity of 8.02 ft/s.
is
z\
the difference in pres-
sure heads expressed in units of length of the fluid flowing
by the
Thus
the difference in
it is
H
frictionless flow,
floors.
Bernoulli's equation
may
from the
be written
r-22
y
Vi and Vi are average velocities. With gage pressure zero as datum and the
of the lower channel as elevation datum, then z\ = y
4, z 2 = 2,
Vi = 8.02, V 2 = 40.1, pi = p 2 = 0,
+
floor
8.02 2
6iJ
+0 + y + 4
and y = 22
40. I 2
= 6lJ
+°+
2
ft.
Modification of assumptions underlying Bernoulli's equation
Under
special conditions each of the four assumptions underlying Bernoulli's
equation
1.
may
When
be waived.
all streamlines originate from a reservoir, where the energy content
everywhere the same, the constant of integration does not change from
one streamline to another and points 1 and 2 for application of Bernoulli's
is
FUNDAMENTALS OF FLUID MECHANICS
138
equation
may
be selected
arbitrarily,
i.e.,
not necessarily on the same
streamline.
2.
In the flow of a gas, as in a ventilation system, where the change in presis only a small fraction (a few percent) of the absolute pressure, the
sure
3.
4.
gas may be considered incompressible. Equation (3.9.3) may be applied,
with an average specific weight y.
For unsteady flow with gradually changing conditions, e.g., emptying a
reservoir, Bernoulli's equation may be applied without appreciable error.
Bernoulli's equation is of use in analyzing real fluid cases by first neglect-
The resulting equation
then be modified by a coefficient, determined by experiment, which
corrects the theoretical equation so that it conforms to the actual physical
ing viscous shear to obtain theoretical results.
may
In general, losses are handled by use of the energy equation, Eq.
case.
(3.8.6).
EXAMPLE
3.7
(a)
Determine the velocity of efflux from the nozzle in the
(b) Find the discharge through the nozzle.
issues as a cylinder with atmospheric pressure around its
wall of the reservoir of Fig. 3.14.
The jet
The pressure along
(a)
periphery.
all
practical purposes.
its
centerline
Bernoulli's equation
is
is
at atmospheric pressure for
applied between a point on
the water surface and a point downstream from the nozzle,
"
2g
+
P
1
7
.
2<7
7
With the pressure datum as local atmospheric pressure, pi = P2 = 0; with
the elevation datum through point 2, z 2 = 0, ,Z\ = H. The velocity on the
H=4m
10
cm diam
L
Fig. 3.14
reservoir.
Flow through nozzle from
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
surface of the reservoir
V
—
+
is
zero (practically)
;
139
hence
2
+ +H
2
=
+
2g
and
f
= y/2 X
2gtt
9.806
X
4
which states that the velocity of
the surface of the reservoir.
The
(b)
discharge
Q
=
8.86
efflux is
This
m/s
equal to the velocity of free
known
is
fall
from
as Torricelli's theorem.
the product of velocity of efflux and area of
is
stream,
Q = A2V2 =
EXAMPLE
=
7r(0.05m) 2 (8.86 m/s)
3.8
A
0.07
m /s
3
venturi meter, consisting of a converging portion followed
of constant diameter and then a gradually diverging porused to determine rate of flow in a pipe (Fig. 3.15). The diameter at
section 1 is 6.0 in and at section 2 is 4.0 in. Find the discharge through the
by a throat portion
tion, is
pipe
when pi — p 2 = 3 psi and oil, sp gr
From the continuity equation, Eq.
0.90, is flowing.
(3.4.6)
Q = Al V = A 2 V, = ^-V = ^-V,
l
l
36
16
in
which
Q
is
Eq. (3.9.3) for
Pl
-
p2
Pl
-
P2
7
=
3
the discharge (volume per unit time flowing).
zx
X
=
144
=
432 lb/ft 2
vs
432
56.16
2g
Solving for discharge gives
l«1-:
Fig. 3.15
7
or
2g
By
z2 ,
Venturi meter.
Q =
2.20
=
0.90
X
62.4
—— (36
2
= Q2
1
w 2g
cfs.
2
-
=
56.16 lb/ft 3
16 2 )
applying
FUNDAMENTALS OF FLUID MECHANICS
140
3.10 APPLICATION OF THE BERNOULLI AND ENERGY EQUATIONS
TO STEADY FLUID-FLOW SITUATIONS
For an incompressible
2
- + Vi- +
2g
V\
2i
Vi
v<>
which each term now
in
22
y
y
is
+
be inserted
if
be simplified to
losses^
(3.10.1)
energy in foot-pounds per pound or meter-newtons
per newton, including the loss term.
may
may
2
-+-+
2g
=
Eq. (3.8.6)
fluid
The work term has been omitted but
needed.
Kinetic energy correction factor
In dealing with flow situations in open- or closed-channel flow, the so-called
is frequently used.
The whole flow is con-
one-dimensional form of analysis
sidered to be one large stream tube with average velocity
The
tion.
kinetic energy per unit weight given
V
by V /2g,
2
at each cross sec-
however,
is
not the
2
/2g taken over the cross section. It is necessary to compute a correction factor a for V 2 /2g, so that aV 2 /2g is the average kinetic energy per
average of
v
unit weight passing the section.
Referring to Fig. 3.16, the kinetic energy
passing the cross section per unit time
is
— vdA
/.
in
2<7
which yv dA
is
the weight per unit time passing 8A and
energy per unit weight.
yv8A-
Fig. 3.16
ity
Veloc-
distribution
and average
locity.
ve-
By
2
/2g is the kinetic
equating this to the kinetic energy per unit time
v
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
aV /2g,
2
passing the section, in terms of
V
—
yVA
2
a
z
—dA
/
J a %9
2g
By
v
f
= 7
solving for a, the kinetic-energy correction factor,
a=
dA
ifA (f)
(3 10 2)
-
Bernoulli's equation
z\
141
+
H
<*i
—=
becomes
h«2-
z2 H
2g
7
-
(3.10.3)
2g
y
For laminar flow in a pipe, a = 2, as shown in Sec. 5.2. For turbulent flow 1
in a pipe, a varies from about 1.01 to 1.10 and is usually neglected except for
precise work.
EXAMPLE
The
3.9
velocity distribution in turbulent flow in a pipe
is
given
approximately by PrandtFs one-seventh power law,
1/7
j_
= (y\
Vax
\r
/
with y the distance from the pipe wall and
energy correction factor.
The average
V
=
2tt
velocity
V
r
the pipe radius. Find the kinetic-
expressed by
is
rv dr
/
J no
in
which
'V
=
r
=
—
r
2irv m&x
y.
r
J
By
substituting for
r
and
v,
1/7
l
(ro
-
(v\
y)
(
-J
dy
=
irr
2
v max 12
or
1/7
F =
1
«_
V
=
120 /j/Y
98
W
V. L. Streeter,
The Kinetic Energy and Momentum Correction Factors
Open Channels
of
Great Width,
Civ.
Eng. N.Y., vol.
12, no. 4, pp.
for Pipes and
212-213, 1942.
FUNDAMENTALS OF FLUID MECHANICS
142
By
substituting into Eq. (3.10.2)
term losses are
through a system the available energy
decreases in the downstream direction it is available to do work, as in passing
through a water turbine. A plot showing the available energy along a stream
All the terms in the energy equation (3.10.1) except the
For
available energy.
real fluids flowing
;
tube portrays the energy grade line (see Sec. 10.1). A plot of the two terms
+ v/y along a stream tube portrays the hydraulic grade line. The energy
z
grade line always slopes downward in real fluid flow, except at a pump or other
source of energy. Reductions in energy grade line are also referred to as head
losses.
EXAMPLE
The siphon of Fig. 3.17 is filled with water and discharging
Find
the losses from point 1 to point 3 in terms of the velocity
at 2.80 cfs.
2
head V /2g. Find the pressure at point 2 if two-thirds of the losses occur between points 1 and 2.
The energy equation is applied to the control volume between points 1
and 3, with elevation datum at point 3 and gage pressure zero for pressure
datum
3.10
:
V, 2 ^_ Pi
1
,
\-
zi
= TV
p.
,
,
h zz
1
+
,
1
losses
y
2g
y
2g
or
+ +
in
4
= ^-
which the
+ + +
losses
from
1
to 3
—have been expressed as
KVz
2
/2g.
From
the dis-
charge
V3 = J =
A
*§
V
=
and
2
z
/2g
=
8.02 ft/s
7r/y
1.0
ft.
Hence
K
=
3,
and the
losses are
ZV3
2
/2g, or 3 ft-lb/lb.
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
Fig. 3.17
Siphon.
The energy equation applied
2,
with losses 2Vi/2q
+
143
+ 0=
1
=
+-+
2.0
to the control
volume between points
1
and
ft, is
8
+
2
—11
ft
H 0,
7
The pressure
EXAMPLE
at 2
3.11
is
its
or 4.76 psi
The device shown
locity of liquid at point
and
2
1.
It is a
vacuum.
in Fig. 3.18
tube with
its
is
used to determine the ve-
lower end directed upstream
other leg vertical and open to the atmosphere.
The impact
of liquid
against the opening 2 forces liquid to rise in the vertical leg to the height Az
above the
free surface.
Determine the velocity at
Az
1^2:
Fig. 3.18
Pitottube.
£=
1.
FUNDAMENTALS OF FLUID MECHANICS
144
is a stagnation point, where the velocity of the flow is reduced to
This creates an impact pressure, called the dynamic pressure, which
Point 2
zero.
forces the fluid into the vertical leg.
points
2
1
and
Zi
+^+
2g
y
2,
By
writing Bernoulli's equation between
neglecting losses, which are very small,
+^+
=
y
given by the height of fluid above point
Pi/y
is
ing.
p 2 /y
is
given by the manometer as k
+
1
and equals k
substituting these values into the equation, Vi 2 /2g
Vi
ft
of fluid flow-
As, neglecting capillary rise. After
=
Az and
= \/2gAz
This
is
the pitot tube in a simple form.
Examples
3.11
of compressible flow are given in
Chap.
6.
APPLICATIONS OF THE LINEAR-MOMENTUM EQUATION
Newton's second law, the equation
momentum
equation in Sec.
2F = - f pMdV
+
This vector relation
f pvv
may
of motion,
was developed
into the linear-
3.2,
•
dA
(3.11.1)
be applied for any component, say the x direction,
reducing to
2FX = -
M Jfcv P
vx
dV
+
f p^V-
dA
(3.11.2)
J cs
In selecting the arbitrary control volume,
it
is
take the surface normal to the velocity wherever
addition,
if
the velocity
be dispensed with. In
steady flow, the force
is
P 2A 2
V V x2 2
cuts across the flow.
In
with the control surface as shown, and with
is given by Eq. (3.11.2)
acting on the control volume
as
Fx =
it
constant over the surface, the surface integral can
Fig. 3.19,
Fx
generally advantageous to
piAi7i7 ri
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
145
A2
Control volume with uniform
and outflow normal to control
Fig. 3.19
inflow
surface.
or
Fx - pQ(V x2 - V xl
as the
mass per second entering and leaving
When
face,
)
pv 2
pQ =
piQi
=
p 2 (?2.
the velocity varies over a plane cross section of the control sur-
by introduction
may be
is
of a
momentum
correction factor
/?,
the average velocity
utilized, Fig. 3.20,
dA =
/.
Fig. 3.20
(3
P
VA
2
Nonuniform flow
through a control surface.
(3.11.3)
FUNDAMENTALS OF FLUID MECHANICS
146
in
which
jS
is
Solving for
dimensionless.
yields
/3
which is analogous to a, the kinetic-energy correction factor, Eq. (3.10.2).
For laminar flow in a straight round tube, /3 is shown to equal -f in Chap. 5.
It equals 1 for uniform flow and cannot have a value less than 1.
In applying Eq. (3.11.1) or a component equation such as Eq. (3.11.2)
care should be taken to define the control volume and the forces acting on it
clearly. Also the sign of the inflow or outflow term must be carefully evaluated.
The first example is an unsteady one using Eq. (3.11.2) and the general continuity equation (3.2.1).
EXAMPLE
distance
3.12
The
A
jet of
x.
horizontal pipe of Fig. 3.21
with water for the
is filled
constant velocity V\ impinges against the
Fluid frictional force on the pipe wall
is
Determine the equations to analyze
[see Eq. (5.10.2)].
filled
given by tqwDx, with r
=
=
portion.
= pfV 22 /&
this flow condition
Specifiis,
x V 2 = V 2o
0, x
= 20 m/s, D\ = 6 cm, V 2q = 50 cm/s, D 2 = 25 cm, x = 100 m,
= 997.3 kg/m 3 and / = 0.02, find the rate of change of V 2 and x with time.
The continuity and momentum equations are used to analyze this un-
when
initial
conditions are known; that
t
,
.
cally for Vi
p
,
Take
steady-flow problem.
with the two end sections
=
-[pdV+f
° tJ
J
cv
becomes, using A\
d
- l PA
2
x
+
P Ai(Z
P
I
as control
ft
volume the
apart, as shown.
inside surface of the pipe,
The
continuity equation
v.dA
cs
=
-
7rDi2 /4,
x)2
+
A =
2
irD 2 2 /4:,
p(V 2 A 2 -
=
VA
1)
full
over partial length.
1
ot
Fig. 3.21
Jet impact on pipe flowing
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
147
After simplifying,
~ (A
2
-
+ VA -
A,)
2
2
=
FxAx
dt
The momentum equation
2FX = -
dV
I pv x
^J
+
JIcs
Cv
for the horizontal direction x,
pv x V
•
dk
becomes
_ p fV^D
2
x
d
=
which
yMxV + pMl
^
o
dt
_
x)y{]
+ pMYi _
pAiVi2
simplifies to
m*D* + M a
o
(xVi)
_
AiVi
dt
^ + MVj _ AiVi
,
.
dt
As t is the only independent
The continuity equation is
variable, the partials
may
be replaced by
totals.
dx_
dt
VA A —
2
2
FiAj
2
Ai
By expanding the momentum
dV,
equation, and substituting for dx/dt,
= J_ \ A l7l2 _ A2 y 2 _ fffegg
2
dt
zA 2
8
L
+
U*V* -
i 2
it
becomes
Ax7i)'
ii
1
J
These two equations, being nonlinear, can be solved simultaneously by numerical methods, such as Runge-Kutta methods described in Appendix E,
when initial conditions are known. The rate of change of x and
found directly from the equations for the specific problem
dx
— = 0.692 m/s
dt
,
—
dV2
-
=
0.0496
V
2
can be
m/s2
dt
EXAMPLE 3.13 In Fig. 3.22 a fluid jet impinges on a body as shown; the momentum per second of each of the jets is given by M and is the vector located
FUNDAMENTALS OF FLUID MECHANICS
148
Fig. 3.22
Solution
linear-momentum
of
problem by addition
of vectors.
By
at the center of the jets.
body
to hold the
vector addition find the resultant force needed
at rest.
The vector form
of the linear
momentum
equation (3.11.1)
applied to a control volume comprising the fluid bounded
the three dotted cross sections. As the problem
is
to be
is
by the body and
steady, Eq. (3.11.1) reduces
to
SF = Jpvv
•
dA =
ZM
t
out
By
taking
Mi and
M
first,
the vector
Mi —
M
is
the net
momentum
efflux
two vectors, shown graphically on their lines of action. The resultant
of these two vectors is then added to momentum efflux M 2 along its line of
action, to obtain R. R is the momentum efflux over the control surface and
The
is just equal to the force that must be exerted on the control surface.
same force must then be exerted on the body, to resist the control-volume force
on it.
for these
,
EXAMPLE
is
6
flowing,
=
=
ft
=
The reducing bend
Di = 6
=
ft,
D =
40 psi, x
Determine F X)
120°, pi
ft -lb/lb.
ft
3.14
1.
4
2
ft,
of Fig. 3.23 is in a vertical plane.
Q = 300
cfs,
W
=
18,000
lb,
z
=
Water
10
ft,
= 6 ft, and losses through the bend are 0.5 V 22 /2g
Fy and the line of action of the resultant force,
,
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
149
p2 A 2 +pQV{
Fig. 3.23
Forces on a reducing elbow, including the vector
solution.
The
inside surface of the reducing
bend comprises the control-volume
The normal secit.
surface for the portion of the surface with no flow across
tions
By
1
and 2 complete the control
surface.
application of the energy equation, Eq. (3.10.1),
h7
h
=
*i
h-
h
Z2
2g
y
2g
«X14i
IftMi
62.4
64.4
+
A
62.4
lossesi_ 2
»
«*
64.4
64.4
from which p 2 = 4420 lb/ft 2 = 30.7 psi.
To determine Fx Eq. (3.11.2) yields
,
PiAi
40
X
-
p 2 A 2 cos
144tt(3 2 )
6
-
— Fx = pQ(V
2
cos 6
4420tt(2 2 ) cos 120°
-
V\)
- Fx
=
1.935
X
300(23.88 cos 120°
-
10.61)
FUNDAMENTALS OF FLUID MECHANICS
150
Since cos 120°
162,900
Fx =
+
=
27,750
-0.5,
- Fx =
580.5(- 11.94
-
10.61)
203,740 lb
For the y direction
2F„ =
P Q(V y2
Fy —
W - pA
Fy -
18,000
Fv =
78,1001b
2
To
-
- Vyl
2
sin 6
)
= pQV
2
sin
=
4420tt(2 2 ) sin 120°
1.935
X
300
X
23.88 sin 120°
find the line of action of the resultant force, using the
momentum
pQFi = 6160 lb, P QV 2 = 13,860 lb, p x A x = 162,900
lb, p 2A 2 = 55,560 lb.
Combining these vectors and the weight
in Fig.
3.23 yields the final force, 218,000 lb, which must be opposed by Fx and Fy
flux vectors (Fig. 3.23),
W
.
As demonstrated
in
Example
3.14,
a change in direction of a pipeline
causes forces to be exerted on the line unless the bend or elbow
in place.
These forces are due to both
reactions in the turning fluid stream.
pipelines to avoid stress in the pipe in
static pressure in the line
anchored
is
and dynamic
Expansion joints are placed in large
an axial direction, whether caused by
by temperature change. These expansion joints permit relatively free
movement of the line in an axial direction, and hence the static and dynamic
fluid or
forces
must be provided
EXAMPLE
3.15
A
jet of
for at the bends.
water 8
cm
in diameter with a velocity of 40
discharged in a horizontal direction from a nozzle mounted on a boat.
force
of
is
is
What
required to hold the boat stationary?
When the control volume is
momentum is [Eq. (3.11.2)]
Fx = P Q(V x2 - Vxl = PQV )
selected as
997.3
shown
kg/m3 X -
in Fig. 3.24, the net efflux
(0.08
m) 2 (40
m/s)
2
=
The
m/s
force exerted against the boat
is
8021
N in the
x direction.
8021
N
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
mounted
Nozzle
Fig. 3.24
151
on
boat.
EXAM PLE
Find the force exerted by the nozzle on the pipe of
3.16
The
Neglect losses.
fluid is
oil,
=
sp gr 0.85, and pi
100
Fig. 3.25a.
psi.
To determine the discharge, Bernoulli's equation is written for the stream
from section 1 to the downstream end of the nozzle, where the pressure is zero.
Vf
Z
.
+
X
0.85
2g
Since
TV
(100 lb/in2 ) (144 in2 /ft 2 )
z\
=
}
"*"
z2,
62.4 lb/ft 3
and
7 = (Di/D
2
"
2
2)
*8
2g
Vi
(100 lb/in 2 ) (144 in2 /ft 2 )
Yl
{
0.85
2g
X
= 9V h
after substituting,
=
62.4 lb/ft 3
and
Vi
Let
=
V =
14.78 ft/s
Px
2
(Fig. 3.256)
133 ft/s
Q =
14.78 -
X
(
-
X2
)
= 0.725
ft
3
/s
be the force exerted on the liquid control volume by the
nozzle; then, with Eq. (3.11.2),
(100 lb/in2 ) - (3 in) 2
- Px =
(1.935 slugs/ft 3 ) (0.85) (0.725
X
lin.
diam
J_L
:-3.in._diam -_--_!_
"•
—
(a)
Fig. 3.25
Nozzle at the end of a pipe.
(6)
3
ft /s)
(133 ft/s- 14.78 ft/s)
FUNDAMENTALS OF FLUID MECHANICS
152
or P x = 565 lb. The oil exerts a force on the nozzle of 565 lb to the right, and
a tension force of 565 lb is exerted by the nozzle on the pipe.
In
many
situations
an unsteady-flow problem can be converted to a
by superposing a constant velocity upon the system and
its surroundings, i.e., by changing the reference velocity.
The dynamics of
a system and its surroundings are unchanged by the superposition of a constant velocity; hence, pressures and forces are unchanged. In the next flow
steady-flow problem
situation studied, advantage
The momentum theory
The
is
taken of this principle.
for propellers
action of a propeller
is
to change the
momentum of the fluid within which
submerged and thus to develop a thrust that
is used for propulsion. Procannot
designed
according
the
momentum
pellers
be
to
theory, although some
of the relations governing them are made evident by its application. A propeller, with its slipstream and velocity distributions at two sections a fixed
distance from it, is shown in Fig. 3.26. The propeller may be either (1) stationary in a flow as indicated or (2) moving to the left with a velocity Vi
through a stationary fluid since the relative picture is the same. The fluid is
assumed to be frictionless and incompressible.
The flow is undisturbed at section 1 upstream from the propeller and is
it is
Pi
boundary
—
V,
Va
2 3
Fig. 3.26
Propeller
in
a fluid stream.
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
153
it approaches the propeller, owing to the reduced pressure on
upstream side. In passing through the propeller, the fluid has its pressure
increased, which further accelerates the flow and reduces the cross section
at 4. The velocity V does not change across the propeller, from 2 to 3. The
pressure at 1 and 4 is that of the undisturbed fluid, which is also the pressure
along the slipstream boundary.
accelerated as
its
When the momentum
within sections
F
exerted
by the
equation (3.11.2)
is
applied to the control volume
and 4 and the slipstream boundary
1
propeller
tion since the pressure
is
is
of Fig. 3.26, the force
the only external force acting in the axial direc-
everywhere the same on the control surface. There-
fore
F = pQ(V, in
stituting
When
(p,
-
+
since
-
Vx)
= ps-p2
(3.11.6)
Bernoulli's equation
hpVS = p 2
z\
=
z2
=
+
z%
=
Pz-P2 = !p(F 2
4
Eliminating p 3
V =
(3.11.5)
A
is
and between sections 3 and
Pi
p 2 )A
is the area swept over by the propeller blades.
The force on the
must be equal and opposite to the force on the fluid. After subQ = AV and simplifying,
which
propeller
P V(V 4
=
Vi)
—
W
written for the stream between sections
p3
+
|P
7 =
2
In solving for p 3
z*.
p4
and 2
—
+ yvt
p 2 with
,
p\
= p
7i2 )
p 2 in Eqs.
1
4,
4,
(3.11.7)
(3.11.6)
*±*
and
(3.11.7) gives
(3,1.8)
which shows that the velocity through the propeller area is the average of
the velocities upstream and downstream from it.
The useful work per unit time done by a propeller moving through still
fluid is the product of propeller thrust and velocity, i.e.,
Power = FVx =
The power input
P Q(V A
is
- VJVi
(3.11.9)
that required to increase the velocity of fluid from V\ to
FUNDAMENTALS OF FLUID MECHANICS
154
work plus the
Vi, or the useful
kinetic energy per unit time remaining in the
slipstream.
Power input =
^ (F 2
p
7i
4
2
)
= P Q(V, - V )V
1
With the
1
ratio of Eqs. (3.11.9)
efficiency e
+
p
- (V -
and
4
V,) 2
(3.11.10)
2
(3.11.10) used to obtain the theoretical
t,
Vi
27i
(3.11.11)
AV = y 4 — Vi is the increase in slipstream velocity, substituting into
Eq. (3.11.11) produces
If
*'
tvT^
-
(3
which shows that
maximum
efficiency
creases the velocity of slipstream as
is
-
1U2)
obtained with a propeller that inas possible, or for which
little
AF/Fi
is
a minimum.
Owing
to compressibility effects, the efficiency of
an airplane propeller
drops rapidly with speeds above 400 mph. Airplane propellers under optimum
conditions have actual efficiencies close to the theoretical efficiencies, in the
neighborhood of 85 percent. Ship propeller efficiencies__are lesa^around 60
percent, owing to restrictions in diameter.
The windmill can be analyzed by application of the momentum relations.
The jet has its speed reduced, and the diameter of slipstream is increased.
EXAMPLE
N/m
3
,
An
3.17
discharges
Determine
airplane traveling 400
1000
m /s
3
,
x
its
still
two 2.25-m-diameter
air,
y
=
12
propellers.
(a) the theoretical efficiency, (6) the thrust, (c) the pressure dif-
ference across the propellers,
(a)
through
km/h through
XT
Vi
=
and
(d) the theoretical
km 1000 m lh
-———-—---400
1
h
1
km
500m /s
3600
3
(t/4) (2.25 2 )
=
=
s
_ m/s
126
,
111.11
m/s
power required.
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
FromEq.
(3.11.11)
From Eq.
(b)
V = 2V 4
The
F =
N/m
-
126
140.9
from Eq.
is,
m/s
(3.11.5),
12
m /s) (140.9 3
111.11 m/s)
=
36,500
pressure difference, from Eq. (3.11.6),
N/m
'
n
9.806
The
(d)
=
m/s 2
=
V2
=
111.11
3
The
(c)
™
X
2
(1000
9.806
V*-
=
Vi
(3.11.8)
thrust from the propellers
12
155
(36,500
K
'
-
m/s
N)'
is
3
712 (126 m/s) (140.9
theoretical
et
N
(111
-
power
111.11 m/s)
is
U m/B)
^
W
;
1000 N- m/s
0.882
= 4600 N/m2
= 4598
kW
Jet propulsion
The
propeller
is
one form of
jet propulsion in that it creates a jet
doing has a thrust exerted upon
air (initially at rest) is
of fuel; the gases are
it
that
is
and by so
the propelling force. In jet engines,
taken into the engine and burned with a small amount
much higher velocity than in a pro-
then ejected with a
The jet diameter is necessarily smaller than the propeller
For the mechanical energy only, the theoretical efficiency is given
peller slipstream.
slipstream.
by the
sum
work to work input or by useful work divided by the
work and kinetic energy per unit time remaining in the jet. If
ratio of useful
of useful
the mass of fuel burned
is
neglected, the propelling force
F = pQ(V2 - 70 = P Q7 abs
in
which
F abs
(Fig. 3.27)
is
F
[Eq. (3.11.5)]
is
(3.11.13)
the absolute velocity of fluid in the jet and pQ is
The useful work is FVi, in which Vi
the mass per unit time being discharged.
is
the speed of the body.
The
kinetic energy per unit time being discharged
FUNDAMENTALS OF FLUID MECHANICS
156
Fig. 3.27
Walls of flow passages through jet
engines taken as impenetrable part of control
surface for plane when viewed as a steady-state
problem.
in the jet
is
yQV ahs
2
being discharged and
= pQV ahs
/2g
V &hs /2g
2
is
2
/2, since
the theoretical mechanical efficiency
et
output
+
is
Hence,
is
2
X
1
1
pQV, hs V + P QF abs
which
the weight per unit time
FV + pQF abs /2
loss
pQV ahs V
1
is
FVi
output
=
yQ
the kinetic energy per unit weight.
2
/2
1
+
(3.11.14)
T^abs/27!
the same expression as that for efficiency of the propeller.
It is ob-
vious that, other things being equal, Fabs/Vi should be as small as possible.
For a given speed V h the resistance force F is determined by the body and
which it moves; hence, for Fabs in Eq. (3.11.13) to be very small, pQ
must be very large.
An example is the type of propulsion system to be used on a boat. If
the boat requires a force of 400 lb to move it through water at 15 mph, first
a method of jet propulsion can be considered in which water is taken in at the
bow and discharged out the stern by a 100 percent efficient pumping system.
To analyze the propulsion system the problem is converted to steady state
by superposition of the boat speed — Vi on boat and surroundings (Fig. 3.28)
fluid in
Steady-state flow around a boat
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
If a 6-in-diameter jet pipe is used,
=
for Vi
400
=
15
mi/h = 22
1.935Q
=
+
1
3
ft /s,
W2Fx
F a bs =
=
1
+
The horsepower required
FVi
X
400
~ 550 X
550e,
22
_
=
1.935Q
so that
Q =
required
is
With
with
IQQ/tt.
By
use of Eq. (3.11.2),
23.2,
and the
=
65
23.2/44
'
efficiency
is
5%
is
24
0.655
With an 8-in-diameter
400
2
ft/s,
e?-»)
Hence Q = 8.89
*
V =
157
V =
jet pipe,
2
9Q/t, and
(*-)
13.14
3
ft /s,
V =
2
15.72, e
(
=
73.7 percent,
and the horsepower
21.7.
additional enlarging of the jet pipe and the
pumping
more water
The type of
propeller pump.
of
less velocity head, the efficiency can be further increased.
pump
best suited for large flows at small head
Increasing the size of
pump and
jet pipe
is
the axial-flow
would increase weight greatly and
take up useful space in the boat the logical limit is to drop the propeller down
below or behind the boat and thus eliminate the jet pipe, which is the usual
propeller for boats. Jet propulsion of a boat by a jet pipe is practical, however,
in very shallow water where a propeller would be damaged by striking bottom
;
or other obstructions.
To take the weight of fuel into account in jet propulsion of aircraft, let
be the mass of air per unit time and r the ratio of mass of fuel burned to
mass of air. Then (Fig. 3.27), the propulsive force F is
ma
ir
F =
m
air
(F 2
-
Vi)
+
rra air
The second term on the
its
y
right
2
is
the mass of fuel per unit time multiplied by
change in velocity. Rearranging gives
F =
m
air
[72 (l
+ r) - 7J
(3.11.15)
FUNDAMENTALS OF FLUID MECHANICS
158
Defining the mechanical efficiency again as the useful work divided by the
of useful work and kinetic energy remaining gives
sum
et
FVi
=
+
*Ti
and by Eq.
et
rft. ir
"
d+
.
h
2
7i) /2
2
(3.11.15)
=
.
(l+r)(7 -
(8 - lue)
riawMj
2[(l+r) (72/70 -1]
becomes unity for V\ = V2, as the combustion products are
then brought to rest and no kinetic energy remains in the jet.
The
efficiency
EXAMPLE
An
airplane consumes
each 20 lb m air and dis6000 ft/s. Determine the mechanical efficiency for airplane speeds of 1000 and 500 ft/s.
- 6, r - 0.05. From Eq. (3.11.16),
For 1000 ft/s, Vi/Vi =
3.18
charges hot gases from the
tail
1
pipe at
lb m fuel for
7 =
2
UU
e
'
=
(i
+ oUe-i).
-
2[6(1 +0.05)
For 500
"
=
ft/s,
Vi/Vi
(1
+0.05)
+
Vffl-
=
(12 -I)'.
0.05)
°- 287
1]
"
2[12(1
"
-
12,
=
and
154
°-
1]
Jet propulsion of aircraft or missiles
Propulsion through air or water in each case
tion of a jet behind the body.
jet,
turboprop,
ram
jet,
is
caused by reaction to the forma-
The various means
include the propeller, turbo-
and rocket motor, which are
briefly described in the
following paragraphs.
The momentum
relations for a propeller determine that its theoretical
efficiency increases as the speed of the aircraft increases
velocity of the slipstream decreases.
As the speed
and the absolute
of the blade tips approaches
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
159
the speed of sound, however, compressibility effects greatly increase the drag
on the blades and thus decrease the overall efficiency of the propulsion system.
A turbojet is an engine consisting of a compressor, a combustion chamber,
a turbine, and a jet pipe. Air is scooped through the front of the engine and
The
is compressed, and fuel is added and burned with a great excess of air.
air and combustion gases then pass through a gas turbine that drives the compressor. Only a portion of the energy of the hot gases is removed by the turbine, since the only means of propulsion is the issuance of the hot gas through
the jet pipe. The overall efficiency of a jet engine increases with speed of the
Although there is very little information available on propeller
aircraft.
systems near the speed of sound,
it
appears that the overall efficiencies of
the turbojet and propeller systems are about the same at the speed of sound.
The turboprop is a system combining thrust from a propeller with thrust
from the ejection of hot gases. The gas turbine must drive both compressor
and propeller. The proportion of thrust between the propeller and the jet
may be selected arbitrarily by the designer.
The ram jet is a high-speed engine that has neither compressor nor turbine. The ram pressure of the air forces air into the front of the engine, where
some of the kinetic energy is converted into pressure energy by enlarging the
flow cross section. It then enters a combustion chamber, where fuel is burned,
and the air and gases of combustion are ejected through a jet pipe. It is a
supersonic device requiring very high speed for compression of the air. An
intermittent ram jet was used by the Germans in the V-l buzz bomb. Air is
admitted through spring-closed flap valves in the nose. Fuel is ignited to build
up pressure that closes the flap valves and ejects the hot gases as a jet. The
ram pressure then opens the valves in the nose to repeat the cycle. The cyclic
rate is around 40 per second. Such a device must be launched at high speed
to initiate operation of the
ram
jet.
Rocket mechanics
The
rocket motor carries with
that
it
an oxidizing agent to mix with its fuel so
independent of the medium through which
it travels.
In contrast, a gas turbine can eject a mass many times the mass
of fuel it carries because it takes in air to mix with the fuel.
To determine the acceleration of a rocket during flight, Fig. 3.29, it is
convenient to take the control volume as the outer surface of the rocket,
with a plane area normal to the jet across the nozzle exit. The control volume
has a velocity equal to the velocity of the rocket at the instant the analysis
develops a thrust that
it
is
made. Let R be the air resistance, mR the mass of the rocket body, m/ the
mass of fuel, m the rate at which fuel is being burned, and v r the exit-gas veis
FUNDAMENTALS OF FLUID MECHANICS
160
Fig. 3.29 Control
surface for analysis of rocket
acceleration. Frame of
reference has the velocity Vi of the rocket.
V\
locity relative to the rocket.
frame of reference)
V
of reference.
-
P Vy
J
dV
+
is
the actual velocity of the rocket (and of the
the velocity of the rocket relative to the frame
=
dVi/dt
is
the rocket acceleration.
The
equation for the y direction (vertical motion)
PV„V
J
J
is
but dV/dt
zero,
ot J
V
and
momentum
basic linear
2Fy =
is
,
'
rfA
(3.11.17)
'
.
becomes
-R -
+ mf )g
(m R
= -
[(ra*
+ mf )V~\ -
mv r
(3.11.18)
ot
Since
tial
V is
a function of
only, the equation can be written as a total differen-
t
equation
dV _ dV\ _ mv — g(m/
r
dt
The mass
™>f
Mr
dt
—
m>fo
+
+
rns)
(3.11.19)
rrif
with time; for constant burning rate m,
™t> with m/ the initial mass of fuel and oxidizer. Gravity is a
of propellant reduces
—
— R
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
function of
y,
and the
161
depends on the Reynolds number and
air resistance
Mach number (Chap. 4), as well as on the shape and size of the rocket.
By considering the mass of rocket and fuel together [Eq. (3.11.19) ],
thrust
minus the weight and the
riiVr
mass times
its
air resistance just equals
the
the combined
acceleration.
The theoretical efficiency of a rocket motor (based on available energy)
shown to increase with rocket speed. E represents the available energy in
is
When
the propellant per unit mass.
energy
the propellant
E =
converted into kinetic energy;
is
= \/2E. For
locity relative to the rocket, or v r
axes fixed in the earth, the useful power
used up per unit time
is
due to mass
loss
is
2
v r /2, in
v r is
the jet ve-
rocket speed V\ referred to
mv r V\. The
is
ignited, its available
which
rhV^/2
of the
kinetic energy being
unburned propellant
and to the burning rhE, or
=
Available energy input per unit time
The mechanical
=
Vr/Vi
=
—\
Vi2
(3.11.20)
1
efficiency e is
m Vl V2E
m{E + VS/2)
When
/
m[E +
1,
2v r IV,
m
1
the
+
K
(vr/V.y
maximum
absolute velocity of ejected gas
'
=
efficiency e
is
}
'
1 is
obtained.
In this case the
zero.
When the thrust on a vertical rocket is
greater than the total weight plus
mass is continuously reduced. To lift
thrust mv r must exceed its total weight.
resistance, the rocket accelerates.
a rocket off
its
pad,
its
Its
EXAMPLE 3.19 (a) Determine the burning time for a rocket that initially
weighs 500,000 kg/, of which 70 percent is propellant. It consumes fuel at a
constant rate, and its initial thrust is 10 percent greater than its weight.
v r = 3300 m/s.
(b) Considering g constant at 9.8 m/s 2 and the flight to be
vertical
its
without
air resistance, find the
height above ground, and the
From
(a)
mvr
and
m
=
=
350,000 kg
1634.3 kg/s
height
it
will attain.
the thrust relation
l.lWo
=
1634.3 kg/s.
the burning time
speed of the rocket at burnout time,
maximum
is
(550,000 kg/) (9.806 N/kg/)
The
available
mass
=
5,393,300
of propellant
is
N
= m3300
350,000 kg; hence
FUNDAMENTALS OF FLUID MECHANICS
162
FromEq.
(6)
(3.11.19)
-
(1634.3 kg/s) (3300 m/s)
dVi
-
m/s2 ) [350,000 kg
(9.8
(1634.3 kg/s)^
+
=
150,000 kg
dt
+
350,000 kg
-
150,000 kg]
(1634.3 kg/s)«
Simplifying gives
dVi
299.95
305.94
dt
7i
+
=
when
-9.8*
t
=
0,
Vi
9.8*
-
3298.16
_
305.94
t
3298.16 In (305.94
=
0;
t
=
/.214.2
y
214.2, Vi
{
=
117.22
The
=
+
const
^)
1873.24 m/s.
j2
The
"1214.2
height at
329816
/
t
=
/
,.214.2
-
2j
214.2
/
s is
\
^-mSSV*
km
rocket will glide V\ 2 /2g
117,220
-
in (l
F *=- 98
=
t)
*
hence
*--•* -3298.16
when
-
-
m+
24 2
1873 '—
2
X
m
=
ft
higher after burnout, or
296.25
km
9.8
Fixed and moving vanes
The theory of turbomachines is based on the relationships between jets and
The mechanics of transfer of work and energy from fluid jets to moving
vanes.
vanes is studied as an application of the momentum principles. When a free
impinges onto a smooth vane that is curved, as in Fig. 3.30, the jet is de-
jet
flected, its
is
momentum
is
changed, and a force
assumed to flow onto the vane
is
exerted on the vane.
The
in a tangential direction, without shock;
jet
and
furthermore the frictional resistance between jet and vane is neglected. The
is assumed to be uniform throughout the jet upstream and down-
velocity
stream from the vane. Since the
jet is
open to the
air, it
has the same pressure
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
163
o
Free
Fig. 3.30
jet
impinging on smooth, fixed vane.
at each end of the vane.
any,
is
When
the small change in elevation between ends,
neglected, application of Bernoulli's equation shows that the
nitude of the velocity
EXAMPLE
is
unchanged
if
mag-
for fixed vanes.
Find the force exerted on a fixed vane when a jet discharging
2 ft /s water at 150 ft/s is deflected through 135°.
By referring to Fig. 3.30 and by applying Eq. (3.11.2) in the x and y
directions, it is found that
3.20
3
-F
x
= pV
cos e
V A + pV - VoAo)
Fy = P V
(
VoA
sin 6
Hence,
Fx = -
Fy =
(1.935 slugs/ft 3 ) (2
(1.935 slugs/ft 3 ) (2
The
force
and
Fy
3
ft /s)
3
ft /s)
components on the
(150 cos 135°
-
(150 sin 135° ft/s)
fixed
150 ft/s)
= 410
= 990 lb
lb
vane are then equal and opposite to
Fx
.
EXAMPLE
3.21
Fluid issues from a long slot and strikes against a smooth
inclined flat plate (Fig. 3.31).
Determine the division of flow and the force
exerted on the plate, neglecting losses due to impact.
As there are no changes
jet.
The
division of flow
and
after impact,
initial
speed of the
in elevation or pressure before
the magnitude of the velocity leaving
Q h Q2
is
the same as the
can be computed by applying the
momentum
FUNDAMENTALS OF FLUID MECHANICS
164
Two-dimensional jet
impinging on an inclined fixed
plane surface.
Fig. 3.31
equation in the
fluid
by the
equal the
initial
momentum
2F =
S
s direction, parallel to
the plate.
plate in this direction; hence, the final
momentum component
equation for the
f Pv s y
•
dA =
s direction,
= pVoVoA!
+
No force is exerted on the
momentum component must
in the s direction.
from Eq.
The
steady-state
(3.11.2), yields
pVocose(-VoAo)
+ p(-V
)V<>A 2
cs
By
Qi
substituting Qi
—
Q2
=
Qo cos
=
VqAi,
Q2 =
V0A2, and
Q =
^0^.0, it
reduces to
6
and with the continuity equation,
<?i
+
Q2
=
Qo
The two equations can be
Qi
=
The
f (1 + cos
force
F
d)
solved for Qi and Q2,
Q2 =
|°
(1
-
cos 0)
exerted on the plate must be normal to
it.
For the
momentum
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
165
equation normal to the plate, Fig. 3.31,
2F» = f
F = PQ V
P v n \/
•
dk = -F = pF o sin0 (-V QA
)
sin 6
Moving vanes
Turbomachinery utilizes the forces resulting from the motion over moving
No work can be done on or by a fluid that flows over a fixed vane.
When vanes can be displaced, work can be done either on the vane or on the
In Fig. 3.32a a moving vane is shown with fluid flowing onto it tanfluid.
gentially. Forces exerted on the fluid by the vane are indicated by F x and
Fy To analyze the flow the problem is reduced to steady state by superposition of vane velocity u to the left (Fig. 3.326) on both vane and fluid. The
control volume then encloses the fluid in contact with the vane, with its control surface normal to the flow at sections 1 and 2.
Figure 3.32c shows the
polar vector diagram for flow through the vane. The absolute-velocity vectors
originate at the origin 0, and the relative-velocity vector V — U is turned
vanes.
.
C.S.
V -u
A„
(a)
t—
©
(6)
(a) Moving vane, (b) Vane flow viewed as steady-state problem
by superposition of velocity u to the left, (c) Polar vector diagram.
Fig. 3.32
FUNDAMENTALS OF FLUID MECHANICS
166
through the angle
The
ing the vane.
V2 is the final absolute velocity leav— V — u is unchanged in magnitude as
the vane as shown.
6 of
relative velocity v r
it traverses the vane. The mass per unit time is given by pA v r and is not the
mass rate being discharged from the nozzle. If a series of vanes is employed,
as on the periphery of a wheel, arranged so that one or another of the jets
intercept all flow from the nozzle and so that the velocity is substantially u,
then the mass per second is the total mass per second being discharged. Application of Eq. (3.11.2) to the control volume of Fig. 3.326 yields
2FX =
f pv x V
dA = -Fx = p(V -
•
u) cos
[(F - w)A
]
cs
+ p(Vo- w)[-(7o-m)A
]
or
Fx = p(V -
2Fy = f
u) 2 A (l
pv y V
-
-
cos 0)
dA = Fy - p(V -
u) sin d
[(7 - u)A
]
or
Fy = p(V -
u) 2 A Q
These relations are
Fx = pQo(V -
smd
for the single vane.
u) (1
-
cos 0)
For a
series of
Fv = pQo(V -
vanes they become
u) sin 6
EXAMPLE 3.22 Determine for a single moving vane of Fig. 3.33a the force
components due to the water jet and the rate of work done on the vane.
Figure 3.336 is the steady-state reduction with a control volume shown.
The polar vector diagram is shown in Fig. 3.33c. By applying Eq. (3.11.2) in
the x and y directions to the control volume of Fig. 3.336
-F =
x
(997.3
kg/m3
)
(60 m/s) (cos 170°) [(60 m/s) (0.001
+
(997.3
kg/m3 )
(60 m/s)
m
2
)
]
[(-60 m/s)
(0.001
m )]
N
Fx =
3590
Fv =
(997.3
kg/m3 )
(60 m/s) (sin 170°) [(60 m/s) (0.001
m )]
2
=
625
N
2
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
167
4,
120 m/s
o
A
= 0.001 m2
Jet acting on a
Fig. 3.33
The power
uFx =
exerted on the vane
(60 m/s) (3590
EXAMPLE
moving vane.
N) =
is
215.4
kW
Determine the horsepower that can be obtained from a series
curved through 150°, moving 60 ft/s away from a
3.0-cfs water jet having a cross section of 0.03 ft 2 Draw the polar vector diagram and calculate the energy remaining in the jet.
The jet velocity is V = 3/0.03 = 100 ft/s. The steady-state vane control volume is shown in Fig. 3.346 and the polar vector diagram in Fig. 3.34c.
The force on the series of vanes in the x direction is
of vanes
3.23
(Fig. 3.34a),
.
Fx =
(1.935 slugs/ft 3 ) (3
The horsepower
hp =
3
ft /s)
(40 ft/s) (1
-
=
cos 150°)
433 lb
is
(433 lb) (60 ft/s)
550ft.lb/ 8
The components
V 2x = 60-40
=
47 3
-
of absolute velocity leaving the
cos 30°
=
25.4 ft/s
V 2y =
vane
are,
from
=
20 ft/s
40 sin 30°
Fig. 3.34c,
FUNDAMENTALS OF FLUID MECHANICS
168
100
Vr -40ft/
ft/ sec
(a)
(r\
Flow through moving vanes.
Fig. 3.34
and the
V
2
2
=
7T2g
The
exit-velocity
25.4 2
+
20 2
=
7^Z
64.4
head
is
16.2 ft -lb/lb
'
kinetic energy remaining in the jet, in foot-pounds per second,
Q y -± =
3
3
(3 ft /s) (62.4 lb/ft ) (16.2 ft)
=
is
3030ft-lb/s
2#
The
initial kinetic
energy available was
2
3
3
(3 ft /s) (62.4 lb/ft )
100
-—
ft
=
29,030 ft-lb/s
64.4
which
is
the
sum
of the
work done and the energy remaining per second.
When a vane or series of vanes moves toward a jet, work is done by the
vane system on the fluid, thereby increasing the energy of the fluid. Figure
3.35 illustrates this situation; the polar vector diagram shows the exit velocity
to be greater than the entering velocity.
In turbulent flow, losses generally must be determined from experimental
tests on the system or a geometrically similar model of the system. In the
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
Fig. 3.35
Vector diagram for vane doing work on a
following
two
169
jet.
cases, application of the continuity, energy,
and momentum
equations permits the losses to be evaluated analytically.
Losses due to sudden expansion
in a
pipe
losses due to sudden enlargement in a pipeline may be calculated with
both the energy and momentum equations. For steady, incompressible, turbulent flow through the control volume between sections 1 and 2 of the sudden
expansion of Fig. 3.36a, 6, the small shear force exerted on the walls between
The
the two sections
may be
By assuming uniform velocity over the
approached in turbulent flow, application of
neglected.
flow cross sections, which
is
Eq. (3.11.2) produces
PiA 2
At
-
p2 A 2
section
1
=
P
V (V A
2
2
2)
+
fVi(-ViAi)
the radial acceleration of fluid particles in the eddy along the
1__^2_A
1
Pl
A2
p 2 A>
nr
cs^^^m
(b)
(a)
Fig. 3.36
Sudden expansion
in
a pipe.
FUNDAMENTALS OF FLUID MECHANICS
170
surface is small, and so generally a hydrostatic pressure variation occurs across
the section. The energy equation (3.10.1) applied to sections 1 and 2, with the
loss
term
(for
hi, is
a
=
—+-= —+-+
7
2?
2gr
—
2
p 2 ) /t in each equation and equating the
V 2
2
2
2
1
/i?
7
Solving for (pi
7 - VV
1)
7i 2
results give
+ hi
2gr
As F1A1 =
2
2,
^—^^\ -jJ
(Vi
h
7A
- V
2
2)
AA
7i 2 /
2
l
which indicates that the
(3
losses in turbulent flow are proportional to the
-
1L22)
square
of the velocity.
Hydraulic
jump
The hydraulic jump
is
the second application of the basic equations to de-
termine losses due to a turbulent-flow situation.
Under proper conditions a
rapidly flowing stream of liquid in an open channel suddenly changes to a
slowly flowing stream with a larger cross-sectional area and a sudden rise in
elevation of liquid surface.
and
This phenomenon
is
known
as the hydraulic
jump
an example of steady nonuniform flow. In effect, the rapidly flowing
(Fig. 3.37) and converts kinetic energy into potential
energy and losses or irreversibilities. A roller develops on the inclined surface
of the expanding liquid jet and draws air into the liquid. The surface of the
jump is very rough and turbulent, the losses being greater as the jump height
is
liquid jet expands
Fig. 3.37
channel.
Hydraulic
jump
in
a
rectangular
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
171
m
Standing wave.
Fig. 3.38
is
For small heights, the form of the jump changes to a standing wave
greater.
(Fig. 3.38).
The
The jump
relations
discussed further in Sec. 11.4.
is
between the variables
for the hydraulic
jump
in a hori-
momentum, and energy equations. For convenience the width of channel is
taken as unity. The continuity equation (Fig. 3.37) is (Ai = yi, A 2 = y 2 )
zontal rectangular channel are easily obtained
V12/1
of the continuity,
= V2 y 2
The momentum equation
M_
2|f _
—+
2/1
2g
=
—+
2g
which
2/2
is
+ PV
fiY,(yxV»)
and the energy equation
in
by use
1
(-y 1 V
(for points
1)
on the liquid surface)
is
+ hi
hj represents losses
due to the jump. Eliminating
V
2
in the first
two
equations leads to
~r+$j h^^
v!|
(3.11.23)
which the plus sign has been taken before the radical (a negative 2/2 has
no physical significance)
The depths y\ and y 2 are referred to as conjugate
depths. Solving the energy equation for hj and eliminating V\ and V 2 give
in
.
hi
=
^^^
(3.11.24)
The hydraulic jump, which is a very effective device for creating ircommonly used at the ends of chutes or the bottoms of spill-
reversibilities, is
FUNDAMENTALS OF FLUID MECHANICS
172
to destroy much of the kinetic energy in the flow. It is also an effective
mixing chamber, because of the violent agitation that takes place in the
Experimental measurements of hydraulic jumps show that the equaroller.
tions yield the correct value of yi to within 1 percent.
ways
EXAMPLE
If 12
3.24
m /s of water per meter of width flows down a spillway
3
onto a horizontal floor and the velocity is 20 m/s, determine the downstream
depth required to cause a hydraulic jump and the losses in power by the jump
per meter of width
y
12
m /s
20
m/s
2
Substituting into Eq. (3.11.23) gives
!J2
= -0.3
WithEq.
Losses
+
2
Q3
y
X
20*
X
= 7m
9.806
(3.11.24),
(7
=
4
-
X
0.6)
0.6
X
:
=
15.6
m N/N
7
Power/m = yQ (losses) - (9802
EXAMPLE
0.6
3.25
Find the head
celerate the flow of
S =
oil,
N/m
H
3
)
(12
m /s) (15.6 m)
3
—
1.-1,000
i
A
'friction
Fig. 3.39
ft
F,«0
v
"I
?A
Acceleration of liquid
U
|_^
in
1840
in the reservoir of Fig. 3.39
2
0.85, at the rate of 0.5 ft/s
H
*("£)
=
2g
J
a pipe.
kW
needed to ac-
when
the flow
is
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
At 8.02
8.02 ft/s.
entrance
ft/s the steady-state
head on the pipe
is
20
ft.
173
Neglect
loss.
The oil may be considered to be incompressible and to be moving uniformly in the pipeline. By applying Eq. (3.11.2), the last term is zero, as
the net efflux
is
zero,
or
H
=
20
—
0.5
35.52
ft
+
cv
The moment
momentum
equation applied to a control volume,
is
= - f pVdV
MJ
r
=
general unsteady linear
Eq. (3.11.1),
F
X
THE MOMENT-OF-MOMENTUM EQUATION
3.12
The
+
f P VV
•
dk
(3.12.1)
J cs
of a force
F about a point
O
(Fig. 3.40)
is
given by
x F
which is the cross, or vector, product of F and the position vector X of a point
on the line of action of the vector from O. The cross product of two vectors
is a vector at right angles to the plane defined by the first two vectors and
with magnitude
Fr
sin
which
is
of F.
The
the product of
F and
the shortest distance from
O
to the line of action
sense of the final vector follows the right-hand rule.
In Fig. 3.40
If this were
would tend to come up,
the force tends to cause a counterclockwise rotation around O.
a right-hand screw thread turning in this direction,
and so the vector
is
fingers of the right
the
thumb
likewise directed
hand
up out
it
of the paper.
If
one curls the
in the direction the force tends to cause rotation,
yields the direction, or sense, of the vector.
FUNDAMENTALS OF FLUID MECHANICS
174
Fig.
Nota-
3.40
moment
tion for
of a vector.
By
taking
X
x
F,
using Eq. (3.12.1),
rxF=-f P rxvdv+f
The
(
J cs
dt J C v
P r x v) (v
.
dA)
(3.12.2)
is the torque exerted by any forces on the
and terms on the right-hand side represent the rate of change
of moment of momentum within the control volume plus the net efflux of moment of momentum from the control volume. This is the general moment-of-
left-hand side of the equation
control volume,
momentum
equation for a control volume.
It has great value in the analysis
of certain flow problems, e.g., in turbomachinery,
than forces.
When Eq. (3.12.2)
where torques are more
significant
is
applied to a case of flow in the xy plane, with
component of the velocity
the normal component of velocity,
shortest distance to the tangential
3.41,
and
v n is
Fr = T =
t
z
f prv
t
vn
J'*
Fig.
for
flow.
3.41
Notation
two-dimensional
dA
f
+dt J
„„
P rv
t
dV
v t,
r
the
as in Fig.
(3.12.3)
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
Flow through ansuch as a
Fig. 3.42
passage
nular
centrifugal
which
in
175
T
pump
z
is
impeller.
A useful form of Eq.
the torque.
control volume, in steady flow (Fig. 3.42),
Tz =
p2r2Vt 2v n2
/
JA
dA 2 -
/
piriv tlv ni
(3.12.3) applied to
an annular
is
dAi
(3.12.4)
J Ai
2
For complete circular symmetry, where r, p, v and v n are constant over the
and outlet control surfaces, it takes the simple form
t ,
inlet
r.
=
pQKmh -
since (pv n
dA =
(3.12.5)
(rvthl
pQ, the
same
at inlet or outlet.
EXAMPLE 3.26 The sprinkler shown in Fig. 3.43 discharges water upward
and outward from the horizontal plane so that it makes an angle of 0° with
the
t
axis
when
the sprinkler
arm
is
at rest.
It has a constant cross-sectional
and t = 0. The reand discharges q cfs starting with to =
sisting torque due to bearings and seals is the constant T and the moment
of inertia of the rotating empty sprinkler head is I 8
Determine the equation
flow area of
A
,
.
for
co
as a function of time.
Equation (3.12.2)
may
be applied. The control volume
area enclosing the rotating sprinkler head.
that
it
has no
moment
momentum; hence
of change of moment
of
The
inflow
the torque
is
is
the cylindical
along the
—T
due to
axis, so
friction
momentum of sprinkler head
and fluid within the sprinkler head plus the net efflux of moment of momentum
is
equal to the time rate
of
FUNDAMENTALS OF FLUID MECHANICS
176
Fig. 3.43
Plan view of sprinkler and control
surface.
from the control volume. Let
- To -
2
-
Aopcor 2 dr
/
dt J
The
—
(I,
+
ipA or
3
For rotation to
)
=
may
pqr
q/2A
^-° (V r cos
/.
setting du/dt
=
6
-
2
cor )
be used. Simplifying gives
(V r
cos 6
—
ur
)
- T
Vr cos 6 must be greater than T The equation is
coasa function of t. The final value of w is obtained
start, pqr Q
easily integrated to find
by
r
dt
total derivative
ttCO
+
V =
.
in the equation.
m
EXAMPLE 3.27 A turbine discharging 10 3 /s is to be designed so that a
torque of 10,000 N»m is to be exerted on an impeller turning at 200 rpm that
takes all the moment of momentum out of the fluid. At the outer periphery
of the impeller, r
at this location?
=
1
m. What must the tangential component
of velocity
be
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
177
tS:
8
><
in.
12
in.
Rotating jet system.
Fig. 3.44
Equation (3.12.5)
T =
P Q(rv
t
is
) in
in this case, since the outflow has v
T
10,000
pQr
(997.3
EXAMPLE
The
3.28
opening
torque
0.001
is
The
momentum
+
which
=
vri
ft
By
solving for
3
(
1
1.003
m)
sprinkler of Fig. 3.44 discharges 0.01 cfs through each
speed of rotation. The area of each nozzle
its
2
.
has no
moment
of momentum, and no
moment of momentum
of rotation; then the moment
must be
—
leaving
zero.
Let
to
be the speed
is
pQiriVti
va
and
con
v t2
=
are absolute velocities.
Oi
—
cori
=
10
—
Then
to
U.U01
and
Vt2
=
For
Vr2
0?r 2
moment
pQ{nva
and
—
co
=
+
=
of
r 2 v l2 )
v tiL
m/s
exerted on the system externally; hence the
is
pQiTiVa
Va
m /s)
10
fluid entering the sprinkler
of fluid leaving
in
) (
0.
N-m
Neglecting friction, find
nozzle.
of
kg/m
3
=
t
10
—
fa)
momentum
=0
or
to be zero,
10
11.54 rad/s or 110.2 rpm.
-
a
+
J (10
-
fw)
=
FUNDAMENTALS OF FLUID MECHANICS
178
PROBLEMS
A
3.1
m
10
from one water reservoir to another which has its water surface
For a discharge of 0.5 m 3 /s determine the losses in meter-newtons per
pipeline leads
lower.
kilogram and in kilowatts.
3.2
A pump
which
motor running
What
is
located 10
upward a distance
vertically
at rated capacity,
and the lake surface? What
3.3
A
blower delivers 2
water. It
72 percent
is
is
above the surface
ft.
what
the irreversibility of the
is
ft
of 50
is
of a lake expels a jet of
being
If 0.5 cfs is
pumped by
the efficiency of the
motor-pump combination?
pump system when comparing
the irreversibility after the water
m /s air, p —
water
a 5-hp electric
the zenith of the jet
falls to
the lake surface?
1.3 kg/m
at an increase in pressure of 10 cm
Determine the irreversibility of the blower in meterkilowatts, and determine the torque in the shaft if the
3
3
,
efficient.
newtons per kilogram and
in
blower turns at 1800 rpm.
3.4
A three-dimensional velocity distribution is given by u = —x,v=
Find the equation of the streamline through
3.5
this
A
5
—
z.
(2,1,1).
two-dimensional flow can be described by u
= —y/b2
,
v
=
x/a2
2
the flow of an incompressible fluid and that the ellipse x /a2
is
=
2y,w
Verify that
.
+ y /b
2
2
=
1 is
a
streamline.
3.6
Oil
minute)
is
.
expected
flowing in a laminar
The
if
the flow
a flow of 450
is
gpm (gallons per
What losses should be
in a pipeline at the rate of 300
measured at 20
reduced to 200 gpm? (b)
ft-lb/lb™.
What
(a)
losses should
be encountered for
gpm?
In a flow of liquid through a pipeline the losses are 3 kW for average velocity of
6 kW for 3 m/s. What is the nature of the flow?
3.7
2
way
irreversibilities are
m/s and
3.8
When
do the
tripling the flow in a line causes the losses to increase
losses
vary with velocity and what
is
by
7.64 times,
how
the nature of the flow?
In two-dimensional flow around a circular cylinder (Fig. 3.3), the discharge
between streamlines is 0.01 cfs/ft. At a great distance the streamlines are 0.20 in apart,
and at a point near the cylinder they are 0.12 in apart. Calculate the magnitude of the
3.9
velocity at these
3.10
A
two
points.
pipeline carries
oil,
another section the diameter
sp gr 0.86, at
is
V=
2
m/s through 20-cm-ID
pipe.
At
5 cm. Find the velocity at this section and the mass rate
of flow in kilograms per second.
3.11
Hydrogen
section
1
A
is
flowing in a 2.0-in-diameter pipe at the mass rate of 0.03 lb m /s.
the pressure
is
40 psia and
t
=
40°F.
What
is
At
the average velocity?
nozzle with a base diameter of 8 cm and with 2-cm-diameter tip discharges
Derive an expression for the fluid velocity along the axis of the nozzle. Measure
the distance x along the axis from the plane of the larger diameter.
3.12
10
1/s.
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
An
3.13
18-ft-diameter pressure pipe has a velocity of 10 ft/s. After passing through
a reducing bend the flow
the velocity,
how much
pipe per 1000
ft of
is
in a 16-ft-diameter pipe.
If
the losses vary as the square of
greater are they through the 16-ft pipe than through the 18-ft
pipe?
Does the velocity distribution
3.14
179
of Prob. 3.4 for incompressible flow satisfy the
continuity equation?
Does the velocity distribution
3.15
q=
l(fe)+J(fy)
satisfy
3.16
first
+ k(-lte)
the law of mass conservation for incompressible flow?
Consider a cube with 1-m edges parallel to the coordinate axes located in the
quadrant with one corner at the origin. By using the velocity distribution of
Prob. 3.15, find the flow through each face and show that no mass
within the cube
if
the fluid
is
Find the flow (per foot
3.17
is
being accumulated
of constant density.
in the z direction)
through each side of the square with
corners at (0,0), (0,1), (1,1), (1,0) due to
q
= i(16y- 12x)+ j(12y-9a0
and show that continuity
Show
3.18
that the velocity
4x
x2
satisfies
\y
+y
J
2
+y
x2
2
continuity at every point except the origin.
Problem 3.18
3.19
is satisfied.
with magnitude
vT
is
=
a velocity distribution that
Show
4/r.
origin (per unit length in the z direction)
3.20
Perform the operation
3.21
By
is
everywhere radial from the origin
that the flow through each circle concentric with the
V *q
is
the same.
on the velocity vectors
of Probs. 3.15, 3.17,
and
3.18.
introducing the following relationships between cartesian coordinates and
plane polar coordinates, obtain a form of the continuity equation in plane polar
coordinates:
x2
-j-
y
2
=
y
r2
tan0
X
u
=
v T cos 6
—
ve
sin 6
Does the velocity given
v
=
—d =
d dr
dx
dr dx
d 66
j
v T sin 6
+
v$
d$ dx
cos
in Prob. 3.19 satisfy the equation that has
been derived?
FUNDAMENTALS OF FLUID MECHANICS
180
A
3.22
standpipe 20
potential energy
is
ft in
in this
diameter and 50
water
if
ft
the elevation
high
is filled
datum
is
with water.
taken 10
How much
below the base
ft
of
the standpipe?
How much work could be obtained from the water of Prob. 3.22 if run through a
3.23
100 percent efficient turbine that discharged into a reservoir with elevation 30
ft
below
the base of the standpipe?
What
3.24
is
the kinetic-energy flux in meter-newtons per second of 0.01
m /s of
3
oil,
sp gr 0.80, discharging through a 5-cm diameter nozzle?
By
3.25
rise,
3.26
If
is
Show
3.27
ft/s.
the water jet of Prob. 3.25
resistance
is
neglecting air resistance, determine the height a vertical jet of water will
with velocity 40
neglected,
how
is
high will
upward 45° with the horizontal and air
and what is the velocity at its high point?
directed
it
rise
that the work a liquid can do
by
virtue of
its
pressure
is
Jp dV,
3.28 What angle a of jet is required to reach the roof of the building
minimum jet velocity Vo at the nozzle? What is the value of Vq!
The
3.29
v
in
which V
the volume of liquid displaced.
velocity distribution in laminar flow in a pipe
= Fmax [l -
(r/r
is
_v_
2
)
]
For highly turbulent flow the velocity distribution
=
^max
with
given by
Determine the average velocity and the kinetic-energy correction
3.30
of Fig. 3.45
factor.
in a pipe
is
given by
fyW
VV
with y the wall distance and
factor for this flow.
y//////////////////////////////^
k—
Fig. 3.45
25
m
r
the pipe radius. Determine the kinetic-energy correction
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
1 6.1
ft/sec
181
7ft
I
777^7^777777777^
\
Channel
10 ft wide
8n
,
^v \
Fig. 3.46
3.31
The
velocity distribution between
10 -
a
in
which u
normal
What
to,
is
+
two
parallel plates separated
by a distance a
is
20
a\
a/
is the velocity component parallel to the plate and y is measured from, and
the lower plate. Determine the volume rate of flow and the average velocity.
the time rate of flow of kinetic energy between the plates? In what direction
is
the kinetic energy flowing?
3.32
What
is
the efflux of kinetic energy out of the cube given by Prob. 3.16 for the
velocity prescribed in Prob. 3.15?
3.33
Water
is
flowing in a channel, as
mine the two possible depths
3.34
of flow y\
shown in
and yi.
Fig. 3.46. Neglecting all losses, deter-
High- velocity water flows up an inclined plane as shown in Fig. 3.47. Neglecting
calculate the two possible depths of flow at section B.
all losses,
3.35
If
the losses from section
A
to section
B
of Fig. 3.46 are 1.5 ft-lb/lb,
determine
the two possible depths at section B.
3.36
In Fig. 3.47 the situation exists under which each kilogram of water increases in
temperature 0.0006°C due to losses incurred in flowing between
the lower depth of flow at section B.
3.37
Neglecting
50
9.806 m/s
all losses, in
2.5
cm
Channel 2
m
m
I
_Li
wide
B
Fig. 3.47
and B. Determine
Fig. 3.46 the channel narrows in the drop to 6 ft wide at
i
v////////y//A 7777
A
FUNDAMENTALS OF FLUID MECHANICS
182
10
2
Oil
ft
ft
sp gr 0.86
'ZZZZZZZZZZZZZZZZZZZZZL
Fig. 3.48
section B.
For uniform flow across section B, determine the two possible depths of
flow.
3.38
In Fig. 3.47 the channel changes in width from 2
B. For losses of 0.3 m*
N/N
between sections
A
m at section A to 3 m at section
and B,
find the
two
possible depths at
section B.
Some steam locomotives had scoops installed that took water from a tank
between the tracks and lifted it into the water reservoir in the tender. To lift the water
12 ft with a scoop, neglecting all losses, what speed is required? Note: Consider the
locomotive stationary and the water moving toward it, to reduce to a steady-flow
3.39
situation.
3.40
the
In Fig. 3.48
oil
discharges from a "two-dimensional" slot as indicated at
B oil discharges from under a gate onto a floor. Neglecting all losses,
discharges of A and at B per foot of width. Why do they differ?
At
air.
mine the
A
A
into
deter-
water the diameter is 1 m, the pressure 1
m/s. At point B,2m higher than A, the diameter is 0.5 m,
and the pressure 0.2 kg// cm2 Determine the direction of flow.
3.41
At point
in a pipeline carrying
kg// cm 2 and the velocity
,
1
.
3.42
Oil
Neglecting
sp gr 0.75
Fig. 3.49
determine the discharge in Fig. 3.49.
J
Aft
^----Water-
losses,
m
4
in.
diam
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
183
Fig. 3.50
"
Y = 55
T
3
lb/ft
H
6
—
in.
diam
==
+-v
Fig. 3.51
3.43
is
750
For
losses of 0.1
m -N/N, find the velocity at
A
in Fig. 3.50.
Barometer reading
mm Hg.
3.44
The
3.45
For flow
losses in Fig. 3.51 for
of
750
gpm
H=
20
ft
are
3^/2^ ft
in Fig. 3.51, determine
H
= 32
For 1500-gpm flow and
2
system in velocity heads,
/2g.
3.46
•
lb/lb.
H for
What is the discharge?
losses of
ft in Fig. 3.51, calculate
15V2 /2g
ft- lb/lb.
the losses through the
KV
In Fig. 3.52 the losses up to section A are 4Vi 2/2g, and the nozzle losses are
= 8 m.
0.05F2 /2<7. Determine the discharge and the pressure at A.
3.47
H
2
3.48 For pressure at A of 25,000 Pa
mine the discharge and the head H.
3.49
The pumping system shown
D 2 =5
Fig. 3.52
in Fig. 3.52
in Fig. 3.53
cm
with the losses in Prob. 3.47, deter-
must have pressure
of 5 psi in the
184
FUNDAMENTALS OF FLUID MECHANICS
Fig. 3.53
discharge line
when
cavitation
pipe from the reservoir to the
incipient at the
pump
inlet.
Calculate the length of
for this operating condition
if
the loss in this pipe
2
Vi /2g) (0.03 L/D) What horsepower is being supplied to the fluid
What percentage of this power is being used to overcome losses? Ba-
can be expressed as
by the pump?
is
pump
(
.
rometer reads 30 in Hg.
3.50
Neglecting losses and surface-tension effects, derive an equation for the water
surface r of the jet of Fig. 3.54, in terms of y/H.
= 1 m, h 2 = 3 m, Di = 3 m, D2 = 5 m, and the
2.6F2 2 /2gr, with 10 percent of the losses occurring before section 1.
Find the discharge and the pressure at section 1.
3.51
In the siphon of Fig. 3.55, h±
losses to section 2 are
Fig. 3.54
Fig. 3.55
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
185
t
H
J
Water"
U>^
^^
15 cr n diam
Fig. 3.56
3.52
Find the pressure at
3.53
The siphon
diameter to 6
and
of Prob. 3.51
if it is
a stagnation point (velocity zero)
of Fig. 3.17 has a nozzle 6 in long attached at section 3, reducing the
For no
in.
A
losses,
compute the discharge and the pressure
at sections 2
3.
With
3.54
exit velocity
VE
in Prob. 3.53
and
losses
from
1
to 2 of l.7V2 2/2g,
from
2
2 to 3 of 0.9V2 /2g, and through the nozzle 0.06TV/2g, calculate the discharge and the
pressure at sections 2 and 3.
Determine the shaft horsepower
3.55
for
are
H=
10V2 /2g, and
The
3.56
H=
head H p
For
.
3.57
fluid
60
ft
losses of 8
\
2
/2g,
to discharge
pump
the
pump
of Fig. 3.56
determine the discharge and the
losses,
is
is
produced for
H=
200
ft
and Q
=
1000
4V 2/2g,
is
80 percent,
cfs?
exclusive of the turbine.
and runs at 240 rpm. To produce 1000 hp for H =
determine the discharge and torque in the turbine shaft. Draw the energy and
is
90 percent
efficient
hydraulic grade lines.
Neglecting
3.59
10
10.
pump
the hydraulic and energy grade lines.
Losses through the system of Fig. 3.57 are
ft,
pump
the overall efficiency of the system and turbine in Fig. 3.57
If
The turbine
300
(QyH p /550) produced by
horsepower
what horsepower
3.58
efficient
losses, exclusive of
16 m.
and system
Draw
an 80 percent
The system
30 1/s through the system of Fig. 3.56.
ft
losses, find
the discharge through the venturi meter of Fig. 3.58.
diam
Water
Fig. 3.57
FUNDAMENTALS OF FLUID MECHANICS
186
Fig. 3.58
Datum
Fig. 3.59
For the venturi meter and manometer installation shown in Fig. 3.59 derive
an expression relating the volume rate of flow with the manometer reading.
3.60
3.61
With
losses of 0.2Vi 2 /2g
between sections
in gallons per minute.
3.62
In Fig. 3.60 determine
V
for
S=0.8
Fig. 3.60
R=
12
in.
1
and 2
of Fig. 3.58, calculate the flow
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
187
T
H
8
S =
cm diam
1.05
Fig. 3.61
3.63
in
In Fig. 3.61
H=
3.64
Neglecting losses, calculate
3.65
For
in
losses of
terms of
=
6 m, and h
meter-newtons per newton and
5.75 m.
Calculate the discharge and the losses
in watts.
H
in
terms of
0.1# through the nozzle
R
for Fig. 3.62.
what
of Fig. 3.62,
is
the gage difference
R
HI
3.66
A liquid
What
is
3.67
In Fig. 3.63, 100 1/s water flows from section
flows through a long pipeline with losses of 4 ft -lb/lb per 100
ft of
pipe.
the slope of the hydraulic and energy grade lines?
0.4(Fi— F2 ) 2/2<7; p 1
=
75,000 Pa.
Compute p 2 and
1
to section 2 with losses of
plot the energy
and hydraulic
grade lines through the diffuser.
3.68
In an isothermal, reversible flow at 200°F, 2 Btu/s heat
flowing through a control volume.
slug per degree Rankine.
S=3.0
Fig. 3.62
is
added to 14 slugs/s
Calculate the entropy increase in foot-pounds per
FUNDAMENTALS OF FLUID MECHANICS
188
20°
45
cm diam
Fig. 3.63
In isothermal flow of a real
3.69
the temperature at 10°C.
m
through a pipe system the losses are 20
-N/kg
heat transfer from the fluid is required to hold
the entropy change As in meter-newtons per kilogram
fluid
per 100 m, and 0.02 kcal/s per 100
What is
m
per Kelvin of pipe system for 4 kg/s flowing?
Determine the
3.70
momentum
correction factor for the velocity distribution of
Prob. 3.29.
Calculate the average velocity and
3.71
momentum
correction factor for the velocity
distribution in a pipe,
v
^max
-
0"
with y the wall distance and r the pipe radius.
3.72
is
By
3.73
Determine the time rate
(Hint: Consider
3.74
v
+
V v' for v into Eq. (3.11.4) show that > 1. The term
from the average velocity V and can be positive or negative.
introducing
the variation of
all six
of x
momentum
v'
passing out of the cube of Prob. 3.16.
faces of the cube.)
Calculate the ^/-momentum efflux from the figure described in Prob. 3.17 for the
velocity given there.
3.75
If
gravity acts in the negative
force acting
on the
fluid
z direction,
determine the
z
component
of the
within the cube described in Prob. 3.16 for the velocity specified
there.
3.76
Find the y component
of the force acting
3.17 for the velocity given there.
on the control volume given
in Prob.
Consider gravity to be acting in the negative y
direction.
3.77
What
stationary?
3.78
for
What
V =
force
components
Fx F y
,
are required to hold the black box of Fig. 3.64
All pressures are zero gage.
force
20 m/s?
F
(Fig. 3.65)
is
required to hold the plate for
oil flow,
sp gr 0.83,
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
Q =0.7
189
cfs
V= 150
ft/sec
= 1.1
cfs
100 ft/sec
5
cm diam
1
J
L
Fig. 3.65
3.79
How much
by the steady
is
the apparent weight of the tank
jet flow into
full of
water (Fig. 3.66) increased
the tank?
3.80
Does a nozzle on a
3.81
When a jet from a nozzle is used to aid
fire
hose place the hose in tension or in compression?
in
maneuvering a
fireboat,
can more force
be obtained by directing the jet against a solid surface such as a wharf than by allowing
it
to discharge into air?
Area
= lft 2
\
Q!-2cfs
-
Q 2 =2
Water
VZZZZZZZZZZZZZZZZL
Fig. 3.66
3
zT
cfs
FUNDAMENTALS OF FLUID MECHANICS
190
3.82
Work Example
3.83
In the reducing bend of Fig. 3.23,
W=
40,000 kg/,
z
=
3.16 with the flow direction reversed, and compare results.
m, p 2
2
D = 4 m, D = 3 m, 6 = 135°, Q = 50 m /s,
= 1.4 MPa, x= 2.2 m, and losses may be neglected. Find
3
2
x
the force components and the line of action of the force which
must be
resisted
by an
anchor block.
3.84
20
ft
3
/s of water flows through an 18-in-diameter pipeline that contains a
The
horizontal 90° bend.
pressure at the entrance to the bend
force components, parallel
bend
3.85
in place.
Neglect
and normal
losses.
Oil, sp gr 0.83, flows
through a 90° expanding pipe bend from 40- to 60-cm-
diameter pipe. The pressure at the bend entrance
neglected.
For 0.6
10 psi. Determine the
is
to the approach velocity, required to hold the
m /s,
3
is
1.3 kg// cm
2
,
and
losses are to
be
determine the force components (parallel and normal to the
approach velocity) necessary to support the bend.
3.86
Work
Prob. 3.85 with elbow losses of O.QVi 2/2g, with V\ the approach velocity,
and compare
3.87
results.
A 4-in-diameter steam line carries saturated steam at 1400 ft/s velocity.
entrained by the steam at the rate of 0.3 lb/s.
bend
3.88
in place
force
Water
is
required to hold a 90°
is
owing to the entrained water?
Neglecting
losses,
tee (Fig. 3.67) in place.
3.89
What
determine the x and y components of force needed to hold the
The plane of the tee is horizontal.
Determine the net force on the
By noting that the pressure at A and B
on the surface AB.
Is
it
sluice gate
is
shown
in Fig. 3.68.
Neglect
losses.
atmospheric, sketch the pressure distribution
a hydrostatic distribution?
How
is it
related to the force just
calculated?
3.90
The
vertical reducing section
12cfs
Fig. 3.67
shown
in Fig. 3.69 contains
oil,
sp gr 0.86, flowing
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
40
20
191
cfs/ft
ft
H2
1.17
30
ft
ft
wide
Fig. 3.68
upward
at the rate of 0.5
m /s.
3
The
pressure at the larger section
is
1.5 kg// cm2 .
Neglecting losses but including gravity, determine the force on the contraction.
3.91
peller,
Apply the momentum and energy equations to a windmill as if it were a pronoting that the slipstream is slowed down and expands as it passes through the
blades.
Show
that the velocity through the plane of the blades
is
the average of the
downstream and upstream sections. By defining the
(neglecting all losses) as the power output divided by the power
velocities in the slipstream at the
theoretical efficiency
available in an undisturbed jet having the area at the plane of the blades, determine
maximum
the
An
3.92
theoretical efficiency of a windmill.
airplane with propeller diameter of 8.0
0.0022 slug/ft 3 ) at 210
mph. The speed
ft
travels through
still
air
(p
=
through the plane of the propeller is
280 mph relative to the airplane. Calculate (a) the thrust on the plane, (6) the
kinetic energy per second remaining in the slipstream, (c) the theoretical horsepower
required to drive the propeller,
(d)
of air
the propeller efficiency, and
(e)
the pressure
difference across the blades.
A
km/h has a 60-cm-diameter propeller that discharges
Determine the thrust on the boat, the theoretical efficiency
the propulsion system, and the power input to the propeller.
3.93
4.5
of
m
3
boat traveling at 40
/s through its blades.
12 in.diam
18 in.diam
FUNDAMENTALS OF FLUID MECHANICS
192
A ship propeller has a theoretical efficiency of 60 percent.
3.94
and the ship
travels 20
mph, what
is
4
If it is
the thrust developed and what
is
ft in
diameter
the theoretical
horsepower required?
A jet-propelled airplane traveling
3.95
it
m/s
at 550
km/h
1000
takes in 40 kg/s air and discharges
Neglecting the weight of
relative to the airplane.
fuel, find
the thrust
produced.
A
3.96
mph.
jet-propelled airplane travels 700
3 lb m /s of fuel, and develops 8000 lb of thrust.
3.97
What
3.98
A
is
165 lb m /s of
It takes in
What
is
air,
burns
the velocity of the exhaust gas?
the theoretical mechanical efficiency of the jet engine of Prob. 3.96?
boat requires a 1800-kg/ thrust to keep
it
moving
at 25
km/h.
How many
cubic meters of water per second must be taken in and ejected through a 45-cm pipe
What
to maintain this motion?
is
the overall efficiency
the
if
pumping system
is
60
percent efficient?
In Prob. 3.98 what would be the required discharge if water were taken from a
3.99
tank inside the boat and ejected from the stern through 45-cm pipe?
Determine the
3.100
In Fig. 3.70, a
3.101
that the cart
200
is
lb.
is
and the theoretical power necessary to produce a
moving 12 m/s when the propulsive efficiency is 68 percent.
size of jet pipe
thrust of 1000 kg/ on a boat
jet,
p
frictionless
=
3.102
rocket.
3.103
free to
Determine the velocity
directed against the vane.
by a vane through 180°. Assume
The cart weighs
and the distance traveled by the cart 10 s after the jet
2 slugs/ft 3
and
A =
,
is
move
0.02
A rocket burns 260 lb m /s fuel,
How much thrust is produced
ft
deflected
in
2
;
a horizontal direction.
V =
100
ft/s.
ejecting hot gases at 6560 ft/s relative to the
at 1500
What is the mechanical efficiency
m/s relative to the rocket?
and 3000 mph?
of a rocket
moving
at 1200
m/s
that ejects
gas at 1800
3.104
Can
a rocket travel faster than the velocity of ejected gas?
chanical efficiency
to the rocket?
when
it
travels 12,000 ft/s
Is a positive thrust
3.105
In Example 3.19 what
3.106
Neglecting
VQ
is
air resistance,
what velocity would a
ft
'//////////////////////// ////////
Fig. 3.70
What
is
the me-
ejected at 8000 ft/s relative
the thrust just before the completion of burning?
-^
ft
is
developed?
~°
A
and the gas
vertically directed
V-2 rocket
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
attain in 68
s if it starts
ejects gas at v r
3.107
What
3.108
If
=
from
rest, initially
weighs 13,000 kg/, burns 124 kg/s
=
1950 m/s? Consider g
9.8
m/s2
193
fuel,
and
.
altitude has the rocket of Prob. 3.106 reached after 68 s?
the fuel supply
is
exhausted after 68
s
(burnout) what
,
is
the
maximum height
of the rocket of Prob. 3.106?
3.109
What
is
3.110
Draw
the polar vector diagram for a vane, angle
all
doing work on a
6,
Label
jet.
vectors.
3.111
Vo
the thrust of the rocket of Prob. 3.106 after 68 s?
=
Determine the resultant force exerted on the vane of Fig. 3.30. A Q = 0.1 ft2
100 ft/s; 6 = 60°, y = 55 lb/ft 3 How can the line of action be determined?
3.112
;
.
In Fig. 3.31, 45 percent of the flow
is
deflected in one direction.
What
is
the
plate angle 0?
3.113
A
flat
plate
is
moving with
3.114
u
velocity
the expression for power required to
move
At what speed u should the cart
maximum power from the jet?
shown
into a jet, as
Derive
in Fig. 3.71.
the plate.
of Fig. 3.71
move away from
the jet in order to
produce
3.115
3.116
p
=
Q =
If
=
1/s;
p
=
3
,
Vo
=
,
hold the stationary vane of
;
moves in the x direction at u = 40 ft/s, for Q
150 ft/s, what are the force components Fx Fy ?
Q =
kg/m3
10
1/s,
Qi
of Fig. 3.73 find the force
=
3 1/s,
O
=
45°, 0i
=
components
30°,
2
=
2
3
ft /s,
for the following
120°,
V =
10 m/s,
.
3.118
Solve the preceding problem by graphical vector addition.
3.119
At what speed u should the vane
the jet?
=
,
For the flow divider
830
1000
the vane of Fig. 3.72
conditions:
p
80
1.935 slugs/ft
3.117
F x F y needed to
kg/m 3 7 = 100 m/s.
Calculate the force components
Fig. 3.72.
What
should be the angle 8 for
7777777777777777777Z
Fig. 3.71
of Fig. 3.32 travel for
maximum power?
maximum power from
FUNDAMENTALS OF FLUID MECHANICS
194
Fig. 3.72
Draw
3.120
100
ft/s,
u
=
Draw
3.121
40 m/s, u
the polar vector diagram for the moving vane of Fig. 3.32 for Vo
60
ft/s,
and
6
=
the polar vector diagram for the moving vane of Fig. 3.32 for
= -20
What horsepower can
3.122
=
m/s, and 6
vanes (Fig. 3.32) when
water flowing?
=
160°.
V =
150°.
be developed from (a) a single vane and (6) a series of
= 173°, for
10 in2 V = 240 ft/s, u = 80 ft/s, and
AQ =
,
Determine the blade angles 0\ and 2 of Fig. 3.74 so that the flow enters the vane
its leading edge and leaves with no x component of absolute velocity.
3.123
tangent to
Determine the vane angle required to
3.124
deflect the absolute velocity of a jet 120°
(Fig. 3.75).
In Prob. 3.39 for pickup
3.125
force
is
of
30 1/s water at locomotive speed of 60 km/h, what
exerted parallel to the tracks?
Fig. 3.73
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
195
Fig. 3.74
V = 130
ft/sec
Fig. 3.75
3.126
Figure 3.76 shows an
enough
for the fluid velocity near the
orifice called
a Borda mouthpiece.
bottom
of the
The tube
tank to be nearly
zero.
is
long
Calculate
the ratio of the jet area to the tube area.
Determine the
3.127
of liquid, p
pipe,
3.128
g
=
=
irreversibility in foot-pounds per
1.6 slugs/ft
30 ft/s2
3
,
pound mass
3
for 6 ft /s flow
through a sudden expansion from a 12- to 24-in-diameter
.
Air flows through a 60-cm-diameter duct at p = 70 kPa, t = 10°C, V = 60
cm diameter. Considering the gas as incom-
m/s. The duct suddenly expands to 75
pressible, calculate the losses in
meter-newtons per newton of
difference in centimeters of water.
Area*=A»A jet
H
jet
Fig. 3.76
air
and the pressure
FUNDAMENTALS OF FLUID MECHANICS
196
Fig. 3.77
What
3.129
are the losses
when 180
cfs
water discharges from a submerged 48-in-
diameter pipe into a reservoir?
Show
3.130
\/~gy
A jump
3.131
=
yi in
Eq. (3.11.23), the relation
V=
2/2,
V2,
and the
losses in
m
3
/s water at a depth of
meter-newtons per newton, in kilowatts,
occurs in a 6-m-wide channel carrying 15
Determine
30 cm.
and
that in the limiting case, as y\
obtained.
is
in kilocalories per kilogram.
Derive an expression for a hydraulic jump in a channel having an equilateral
3.132
triangle as its cross section (symmetric with the vertical)
Derive Eq. (3.11.24).
3.133
3.134
Assuming no losses through the gate of Fig. 3.77 and neglecting Vo2/2g,
20 ft and yi = 2 ft, find 2/2 and losses through the jump. What is the basis
y =
neglecting Vo2 /2g?
3.135
Under the same assumptions
determine
2/1
—
45
cm and
2/2
=
for
2 m,
2/0.
Under the same assumptions
3.136
as in Prob. 3.134, for
for
as in Prob. 3.134, y
=
22
ft
and 2/2=8
ft.
Find
the discharge per foot.
3.137
For
of 10
m /s,
3
Water
3.138
El
50
losses
down
the spillway of Fig. 3.78 of 2
determine the
is
floor elevation for the
to occur.
flowing through the pipe of Fig. 3.79 with velocity
m
El
Fig. 3.78
m -N/N and discharge per meter
jump
30
m
V=
8.02 ft/s
and
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
16
197
ft
1000
18
ft
in.
diam
>V
Q.
Fig. 3.79
up
losses of 10 ft 'lb/lb
to section
When
1.
the obstruction at the end of the pipe
is
removed, calculate the acceleration of water in the pipe.
Water
3.139
Vi
=
the piping system of Fig. 3.80. At one instant pi = 10 psi, P2 =
is increasing by 3000 gpm/min.
Find the force
fills
10 ft/s, and the flow rate
0,
Fx
required to hold the piping system stationary.
3.140
If in Fig.
Assume
3.141
pump
Q2
3.66
is
1.0 cfs,
what is the vertical force to support the tank?
The tank weighs 20 lb, and water depth is 1 ft.
that overflow has not occurred.
In Fig. 3.42,
r\
=
10 cm,
ti
=
m /s
3
impeller discharging 0.2
16 cm, v
of water.
t
\
=
0,
and
What
v
t
2
=
3
m/s
for a centrifugal
torque must be exerted on the
impeller?
3.142
In a centrifugal
tangential velocity
pump
gpm
400
component
water leaves an 8-in-diameter impeller with a
of 30 ft/s.
It enters the impeller in a radial direction.
pump speed of 1200 rpm and neglecting all losses, determine the torque in the
pump shaft, the horsepower input, and the energy added to the flow in foot-pounds
For
per pound.
3.143 A water turbine at 240 rpm discharges 40 m 3 /s. To produce 40,000 kW, what
must be the tangential component of velocity at the entrance to the impeller at r =
1.6 m? All whirl is taken from the water when it leaves the turbine. Neglect all losses.
What head
4
in.
is
required for the turbine?
Fx
diam
8
I
Y77ZZZZZZZZ2ZZZZZZZZZZZI
If
UZZZZZZZL
^
.1
-j"
m
jzzzzzzzz
10 ft->|«
Fig. 3.80
diam
in.
-«
7
2.1
'
1
|
20
ft
——
>|
FUNDAMENTALS OF FLUID MECHANICS
198
8
K
in.
I
H
-
Fig. 3.81
3.144
The symmetrical
frictionless.
3.145
Determine
What
sprinkler of Fig. 3.81 has a total discharge of 14
its
rpm
gpm and
is
the nozzle tips are J in diameter.
if
torque would be required to hold the sprinkler of Prob. 3.144 stationary?
Total flow 2 1/s water.
3.146
If
there
is
a torque resistance of 0.50 lb
-ft
in the shaft of Prob. 3.144,
what
is its
speed of rotation?
3.147
For torque resistance
of O.Olco 2 in the shaft, determine the speed of rotation of
the sprinkler of Prob. 3.144.
3.148
(a)
(b)
(c)
(d)
(e)
A
reversible process requires that
there be no heat transfer
Newton's law of viscosity be satisfied
temperature of system and surroundings be equal
there be no viscous or Coloumb friction in the system
heat transfer occurs from surroundings to system only
3.149
An
open system implies
(a)
the presence of a free surface
(b)
that a specified mass
(c)
the use of a control volume
(d)
no interchange between system and surroundings
none of the above answers
(e)
3.150
A
is
considered
control volume refers to
(a)
a fixed region in space
(d)
a reversible process only
3.151
1.
Which
(6)
a specified mass
(a)
(c)
an isolated system
a closed system
three of the following are synonymous?
losses
2. irreversibilities
3.
energy losses
4.
available energy losses
5.
drop in hydraulic grade
(a)
1, 2, 3,
(b)
1, 2,
5
line
(c)
1, 2,
4
(d)
2, 3,
4
(a)
3, 4,
5
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
Fig. 3.82
3.152
(a)
(e)
system of Fig. 3.82
Irreversibility of the
9.2 hp
(b) 36.8 hp
none of these answers
3.153
Isentropic flow
11.55
(d)
ft
(a)
irreversible adiabatic flow
reversible adiabatic flow
perfect-gas flow
(b)
is
(a)
steady uniform flow
(b)
uniform flow
(c)
flow which neglects changes in a transverse direction
(d)
restricted to flow in a straight line
(e)
none
The
3.155
answers
continuity equation
(a)
Q = pAv
(b)
piAi
(d)
V
(e)
A1V1
3.156
•
p
=
The
first
=
=
may
take the form
p2 A 2
A2V2
=
pjAiVi
(c)
accounts for
(b)
is
(c)
law of thermodynamics, for steady flow,
is
(d)
is
primarily concerned with heat transfer
(e)
is
restricted in its application to perfect gases
(a)
(d)
Entropy, for reversible flow,
= du + p d(l/p)
ds=dqn/T
(e)
ds
3.158
P2A2V2
all energy entering and leaving a control volume
an energy balance for a specified mass of fluid
an expression of the conservation of linear momentum
(a)
3.157
ideal-fluid flow
(c)
frictionless reversible flow
(e)
One-dimensional flow
of these
ft
is
(d)
3.154
8.45
(c)
is
The equation d
(a)
isentropic flow
(d)
perfect-gas flow
(b)
none
(e)
defined
= T ds
(c)
s
=
u
+
pv s
answers
is
restricted to
reversible flow
none
by the expression
= T dqH
of these
(losses)
(b)
ds
is
of these
(c)
answers
adiabatic flow
199
FUNDAMENTALS OF FLUID MECHANICS
200
3.159
In turbulent flow
(a)
the fluid particles
(b)
cohesion
more
is
momentum
move
in
effective
an orderly manner
than
momentum
transfer in causing shear stress
on a molecular scale only
(d) one lamina of fluid glides smoothly over another
(e) the shear stresses are generally larger than in a similar laminar flow
(c)
3.160
The
transfer
ratio
=
17
is
r/(du/dy) for turbulent flow
is
(a) a physical property of the fluid
(b)
dependent upon the flow and the density
(c)
the viscosity divided by the density
(d)
a function of temperature and pressure of fluid
(e)
independent of the nature of the flow
3.161
Turbulent flow generally occurs for cases involving
(a)
very viscous fluids
(6)
very narrow passages or capillary tubes
(c)
very slow motions
(d)
combinations of (a), (6), and
(e)
none of these answers
3.162
In laminar flow
(a)
experimentation
(b)
Newton's law
(c)
the fluid particles
(d)
the viscosity
(e)
the ratio r/(du/dy) depends upon the flow
3.163
An
required for the simplest flow cases
is
of viscosity applies
is
move
in irregular
and haphazard paths
unimportant
ideal fluid
(a) very viscous
is
one which obeys Newton's law of viscosity
(6)
a useful assumption in problems in conduit flow
(c)
(d) frictionless
3.164
Which
and incompressible
of the following
Newton's law of viscosity
Newton's second law of motion
3.
The continuity equation
4. r
=
(ju
+
77)
(e)
must be
2.
1.
(c)
none
fulfilled
of these answers
by the flow
Velocity at boundary must be zero relative to boundary
6.
Fluid cannot penetrate a boundary
1, 2,
3.165
(a)
3
(b)
1, 3,
6
(c)
2, 3,
5
(d)
2, 3,
6
(e)
Steady flow occurs when
conditions do not change with time at any point
same
at adjacent points at
(6)
conditions are the
(c)
conditions change steadily with the time
(d) dv/dt
is
any
fluid, real or ideal?
du/dy
5.
(a)
of
constant
(e)
dv/ds
is
constant
any instant
2, 4,
5
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
3.166
(a)
(c)
(d)
(e)
201
Uniform flow occurs
(b) when dv/dt is everywhere zero
is steady
when the velocity vector at any point remains constant
when dv/ds =
when the discharge through a curved pipe of constant cross-sectional area
whenever the flow
only
is
constant
3.167
Select the correct practical
example
of steady
nonuniform flow:
water around a ship in a lake
(b)
motion
motion
(c)
steadily increasing flow through a pipe
(d)
steadily decreasing flow through a reducing section
(e)
constant discharge through a long, straight pipe
(a)
3.168
A
of
of a river
around bridge piers
streamline
(a)
is
the line connecting the midpoints of flow cross sections
(b)
is
defined for uniform flow only
(c)
is
drawn normal
(d)
is
always the path of a particle
3.169
to the velocity vector at every point
(e)
is
fixed in space in steady flow
In two-dimensional flow around a cylinder the streamlines are 2 in apart at a
great distance from the cylinder, where the velocity
(a)
75 ft/s
3.170
(a)
400
3.171
(a)
An
(6)
133 ft/s
(c)
100 ft/s. At one point near the
is
The average
cylinder the streamlines are 1.5 in apart.
150 ft/s
(d)
velocity there
200 ft/s
is
(e)
300 ft/s
has a specific gravity of 0.80. Its density in kilograms per cubic meter
oil
414
(6)
800
(c)
(d)
1000
is
25,800
(«)
The continuity equation
requires that Newton's second law of motion be satisfied at every point in the
fluid
work
boundary must be zero
(6)
expresses the relation between energy and
(c)
states that the velocity at a
relative to the
boundary
for a
real fluid
(d)
relates the
(e)
relates
3.172
momentum
mass
per unit volume for two points on a streamline
rate of flow along a stream tube
Water has an average
velocity of 10 ft/s through a 24-in pipe.
through the pipe, in cubic feet per second,
(a)
7.85
(b)
31.42
(c)
3.173
The assumptions about
j dp/p
=
const are that
40
(d)
The
discharge
is
125.68
(e)
none
of these
answers
flow required in deriving the equation gz
it is
(a)
steady, frictionless, incompressible, along a streamline
(6)
uniform, frictionless, along a streamline, p a function of p
+ v /2
2
-\-
FUNDAMENTALS OF FLUID MECHANICS
202
(c)
steady, uniform, incompressible, along a streamline
(d)
steady, frictionless, p a function of p, along a streamline
none of these answers
(e)
The equation
3.174
(a)
m-N/s
(b)
z
The
3.176
(a)
v
3.177
is
/2g
z
(c)
3
(e)
its
m -N/N
sustained pressure
is,
^2gh
(e)
none of these answers
kinetic-energy correction factor
is
expressed by
—
(d)
is
expressed by
—
1
by
/v\2
f
1
expressed
— dA
/
A
—
—
)
JA
W/
f
fv\ z
[
/
A
has the units of velocity head
(6)
v
f
/
1
The
(e)
\^2gh
(d)
v
(c)
3.178
m -N/m
(d)
is
applies to the continuity equation
is
has the units of
capable of doing by virtue of
(a)
(e)
= C
(d) v2/2g
p/y
(c)
(b)
The
liquid
velocity head
/2g
2
pound,
p
2
v
m -N/kg
(c)
The work that a
3.175
+
p/y
N
(6)
in foot-pounds per
(a)
+
z
I
—
)
WJ
JA
dA
dA
kinetic-energy correction factor for the velocity distribution given
by
Fig.
1.1 is
(a)
(6)
3.179
(c)
1
The equation
-J
(d)
2
none of these answers
(e)
2FX = pQ(VXoui — VXia
)
requires the following assumptions
for its derivation:
1.
Velocity constant over the end cross sections
2.
3.
Steady flow
Uniform flow
4.
Compressible
fluid
5. Frictionless fluid
(a)
1,
3.180
2
(6)
1,
The momentum
</>
(e)
5
none
of these
(c)
1,
3
(d)
correction factor
3,
is
5
(e)
2,
4
expressed by
»ilM" <*jL®« <*iJM u
answers
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
The momentum
3.181
correction factor for the velocity distribution given
by
203
Fig. 1.1
is
(b)
(a)
(c)
1
2
(d)
-J
(e)
none
of these
answers
3.182 The velocity over one-third of a cross section is zero and is uniform over the
remaining two-thirds of the area. The momentum correction factor is
(a)
(&)§•
1
(c)
M
f
M
f
none
of these
answers
3.183 The magnitude of the resultant force necessary to hold a 20-cm-diameter 90°
elbow under no-flow conditions when the pressure is 10 kg//cm 2 is, in kilonewtons,
(a)
61.5
A
3.184
43.5
(6)
30.8
(c)
—5
The
psi.
force
component
necessary to hold the elbow in place
—342
A
3.185
none of these answers
12-in-diameter 90° elbow carries water with average velocity of 15 ft/s and
pressure of
(a)
(e)
(d)
223
(b)
(c)
is,
565
in
in the direction of the
approach velocity
pounds,
907
(d)
none of these answers
(e)
5-cm-diameter 180° bend carries a liquid, p = 1000 kg/m3 at 6 m/s at a
The force tending to push the bend off the pipe is, in newtons,
pressure of zero gage.
(a)
70.5
(b)
The
3.186
(c)
141
(d)
515
(e)
none
answers
of these
thickness of wall for a large high-pressure pipeline
is
determined by con-
sideration of
(a)
axial tensile stresses in the pipe
(b)
forces exerted
(c)
forces exerted
(d)
circumferential pipe wall tension
(c)
temperature stresses
Select from the following
3.187
that
jet
by dynamic action at bends
by static and dynamic action
is
deflected
by a
fixed or
list
at bends
the correct assumptions for analyzing flow of a
moving vane:
4.
The momentum of the jet is unchanged.
The absolute speed does not change along the vane.
The fluid flows onto the vane without shock.
The flow from the nozzle is steady.
5.
Friction between jet and vane
6.
The
The
1.
2.
3.
7.
jet leaves
velocity
is
is
neglected.
without velocity.
uniform over the cross section of the
jet before
and
after contacting
the vane.
(a)
1, 3, 4,
3.188
5
(6)
When
the
(6)
no force
6
(c)
3, 4, 5,
6
(d)
3, 4, 5,
7
(e)
a steady jet impinges on a fixed inclined plane surface
momentum
(a)
2, 3, 5,
is
in the direction of the
exerted on the jet
approach velocity
by the vane
is
unchanged
3, 5, 6, 7
FUNDAMENTALS OF FLUID MECHANICS
204
(c)
the flow
divided into parts directly proportional to the angle of inclination of the
is
surface
(d) the speed
reduced for that portion of the
is
jet
turned through more than 90° and
increased for the other portion
(e)
the
momentum component
is
unchanged
parallel to the surface
3.189 A jet with initial velocity of 100 ft/s in the -\-x direction is deflected by a fixed
vane with a blade angle of 120°. The velocity components leaving the vane parallel
to and normal to the approach velocity are
(a)
vx
(d)
vx
= -50, v y = 86.6
= 50, Vy = 86.6
An
3.190
(b)
(e)
vx
vx
=
=
—86.6,
100, v y
vy
=
=
(c)
vx
=
=
50, vy
50
50
sp gr 0.80, discharges 10 kg/s onto a fixed vane that turns the flow
oil jet,
through 90°. The speed of the jet is 30 m/s as it leaves the vane. The force component
on the vane in the direction of the approach velocity is, in newtons,
(a)
424
A
3.191
300
(6)
water
jet
momentum changed
4
A
3.192
jet
212
none of these answers
(e)
having a velocity of 120 ft/s and cross-sectional area 0.05
ft/s in the same direction as the jet. The mass having
per unit time, in slugs per second,
7.74
(b)
(d)
moving 40
flows onto a vane
(a)
240
(c)
(c)
11.61
(d)
15.48
m/s
none
(e)
same direction as the
in the
jet.
2
its
is
The
answers
of these
having a velocity of 30 m/s flows onto a vane, angle 6
velocity of 15
ft
final
=
150°, having a
absolute velocity com-
ponents parallel and normal to the approach velocity are
(a)
vx
(d)
vx
=
=
=
=
2.01, v y
4.39, v y
=
=
13
7.5
(b)
vx
10.6
(e)
none of these answers
7.2, v y
=
vx
(c)
-11,
=
vv
15
A
3.193
vane moves toward a nozzle 30 ft/s, and the jet issuing from the nozzle has a
The vane angle is 6 = 90°. The absolute velocity components of the
leaves the vane, parallel and normal to the undisturbed jet, are
velocity of 40 ft/s.
jet as
it
(a)
vx
(d)
vx
= 10, v y = 10
= —30, Vy — 70
3.194
u
=
(a)
p
0.5
(6)
A
vv
=
10
3
,
(c)
vx
= -30,
vy
=
40
(c)
50
(d)
100
is
(e)
none
of these
answers
moving vanes, u = 50 ft/s, 6 = 90°, intercepts a jet, Q = 1 ft 3 /s,
V = 100 ft/s. The work done on the vanes, in foot-pounds per
is
1875
3.196
The
velocity 20
(a)
= -30,
none of these answers
N is exerted upon a moving blade in the direction of its motion,
30
series of
1.5 slugs/ft
second,
(a)
force of 250
vx
(e)
20 m/s. The power obtained in kilowatts
3.195
=
A
(6)
0.495
(6)
2500
(c)
3750
(d)
7500
(e)
none
of these
answers
kilowatts available in a water jet of cross-sectional area 0.004
m/s
is
(b)
16.0
(c)
17.2
(d)
32
(e)
none
of these
answers
m
2
and
FLUID-FLOW CONCEPTS AND BASIC EQUATIONS
A ship moves through water at 30 ft/s. The velocity of water in the slipstream
3.197
20
behind the boat
is
of the propeller
is,
The
3.198
822
75
(c)
2480
(6)
3300
(c)
A rocket exerts a constant
3.199
86
(d)
none of these answers
(e)
thrust on the ship of Prob. 3.197, in kilograms force,
missile weighs 8 lb
the
and the propeller diameter is 3.0 ft. The theoretical efficiency
ft/s,
in percent,
60
(b)
(a)
(a)
205
downward
and
starts
4963
(d)
is
none of these answers
(e)
horizontal thrust of 40 lb on a missile for 3
from
rest, its
and reduction
acceleration of gravity
s.
the
If
speed at the end of the period, neglecting
weight of the rocket,
in
is,
in feet
per second,
(a)
386
3.200
483
(6)
What
is
580
(c)
600
(d)
(e)
none of these answers
the reduction in weight of the rocket of Prob. 3.199
if
the jet leaves
at 6000 ft/s relative to the rocket?
(a)
0.02 1b
0.04 1b
(6)
0.32 1b
(c)
(d)
0.64 1b
(e)
none of these
answers
3.201
A
stream of
glass tube with a 90°
oil,
bend
directed upward. Oil inside the tube
The
velocity measured
(a)
0.89
3.202
open at both ends.
is
sp gr 0.90, so that one opening
by the tube
0.99
(b)
(c)
is,
1.10
is
5
in
cm
is
It
is
inserted into a flowing
directed upstream and the other
higher than the surface of flowing
is
oil.
meters per second,
(d)
1.40
In Fig. 8.4 the gage difference R' for
vi
(e)
=
none of these answers
5 ft/s,
S =
0.08,
S =
1.2
is,
in
feet,
(a)
0.39
3.203
The
0.62
(b)
(c)
theoretical velocity of
under a head of 4 m,
8.86
is
6.7
(e)
none of these answers
(6)
(d)
oil,
1.17
(e)
none
of these
answers
sp gr 0.75, flowing from an orifice in a reservoir
in meters per second,
(a)
3.204
0.78
(c)
11.8
In which of the following cases
(d)
not determinable from data given
is it
possible for flow to occur
from low pressure
to high pressure?
(a)
flow through a converging section
(c)
flow of a liquid
(e)
impossible in a constant-cross-section conduit
3.205
The head
upward
(6)
adiabatic flow in a horizontal pipe
in a vertical pipe
loss in turbulent flow in a pipe
(a)
varies directly as the velocity
(6)
varies inversely as the square of the velocity
(c)
varies inversely as the square of the diameter
(d)
flow of air
downward
in a pipe
206
FUNDAMENTALS OF FLUID MECHANICS
(d)
depends upon the orientation of the pipe
varies approximately as the square of the velocity
(e)
The
3.206
losses
due to a sudden expansion are expressed by
Vi>- 722
(a)
(6)
—
(d)
(c)
~2g
2g
(Fi- v2 y
•)
%g
3.207
(a)
is
If all losses are neglected,
a
minimum
(a)
ft
(6)
4.55
m
(6)
of a siphon
reservoir only
discharge through the siphon
liquid density
4.5 ft
4.9
summit above upstream
length of the downstream leg
to y
(c)
The depth conjugate
3.209
(a)
of
The depth conjugate
2.32
summit
for the siphon
depends upon height
(c) is independent of the
(d) is independent of the
(e) is independent of the
(b)
3.208
the pressure at the
m
to y
(c)
=
V=
and
1 ft
5.0 ft
(d)
20 ft/s
5.5 ft
(e)
m and V = 8 m/s
7.04 m
(d) 9.16 m
=
is
3
none
of these
answers
is
(e)
none of these
answers
3.210
(a)
The depth conjugate
0.06
ft
(b)
1.46 ft
to y
(c)
=
10
5.06
ft
ft
and
V=
(d)
1
ft/s
10.06
is
ft
(e)
none
of these
answers
3.211
The continuity equation
in ideal-fluid flow
any small volume must be zero
(a)
states that the net rate of inflow into
(6)
states that the energy
is
constant along a streamline
(c)
states that the energy
is
constant everywhere in the fluid
(d)
applies to irrotational flow only
(e)
implies the existence of a velocity potential
4
DIMENSIONAL ANALYSIS
DYNAMIC
AND
SIMILITUDE
Dimensionless parameters significantly deepen our understanding of fluid-flow
phenomena
in a
way which
is
analogous to the case of a hydraulic jack, where
the ratio of piston diameters determines the mechanical advantage, a di-
mensionless
number which
is
independent of the overall
size of
the jack.
They
permit limited experimental results to be applied to situations involving
and often
ferent physical dimensions
different fluid properties.
of dimensional analysis introduced in this chapter plus
the mechanics of the type of flow under study
tion of experimental data.
since one
is
now
make
an understanding of
possible this generaliza-
The consequence of such generalization is manifold,
phenomenon in its entirety and is not
able to describe the
restricted to discussing the specialized experiment that
it is
dif-
The concepts
was performed. Thus,
possible to conduct fewer, although highly selective, experiments to un-
cover the hidden facets of the problem and thereby achieve important savings
in
time and money.
The
an investigation can also be presented to
more compact and meaningful way to faciliEqually important is the fact that through such incisive and
results of
other engineers and scientists in a
tate their use.
uncluttered presentations of information researchers are able to discover
new
and missing areas of knowledge of the problem at hand. This directed
advancement of our understanding of a phenomenon would be impaired if the
features
tools of dimensional analysis
were not available.
In the following chapter,
dealing primarily with viscous effects, one parameter
is
highly significant,
the Reynolds number.
In Chap. 6, dealing with compressible flow, the
Mach number is the most important dimensionless parameter. In Chap. 11,
dealing with open channels, the Froude number has the greatest significance.
viz.,
207
FUNDAMENTALS OF FLUID MECHANICS
208
Many of the dimensionless parameters may be viewed as a ratio of a pair
magnitude indicating the relative importance of
one of the forces with respect to the other. If some forces in a particular flow
situation are very much larger than a few others, it is often possible to neglect
the effect of the smaller forces and treat the phenomenon as though it were
completely determined by the major forces. This means that simpler, although
not necessarily easy, mathematical and experimental procedures can be used
to solve the problem. For situations with several forces of the same magnitude,
such as inertial, viscous, and gravitational forces, special techniques are required. After a discussion of dimensions, dimensional analysis, and dimensionof fluid forces, the relative
parameters, dynamic similitude and model studies are presented.
less
DIMENSIONAL HOMOGENEITY AND DIMENSIONLESS RATIOS
4.1
Solving practical design problems in fluid mechanics usually requires both
theoretical developments
and experimental
quantities into dimensionless parameters
of variables appearing
and
to
make
this
results.
it is
By
grouping significant
possible to reduce the
compact
number
result (equations or data
plots) applicable to all similar situations.
one were to write the equation of motion SF = ma for a fluid particle,
all types of force terms that could act, such as gravity, pressure,
viscous, elastic, and surface-tension forces, an equation of the sum of these
forces equated to ma, the inertial force, would result. As with all physical
equations, each term must have the same dimensions, in this case, force. The
division of each term of the equation by any one of the terms would make the
equation dimensionless. For example, dividing through by the inertial-force
term would yield a sum of dimensionless parameters equated to unity. The
relative size of any one parameter, compared with unity, would indicate its
importance. If one were to divide the force equation through by a different
term, say the viscous-force term, another set of dimensionless parameters
would result. Without experience in the flow case it is difficult to determine
If
including
which parameters will be most useful.
An example of the use of dimensional analysis and its advantages is
given by considering the hydraulic jump, treated in Sec. 3.11. The momen-
tum
72/i
2
2
equation for this case
7?/2
2
2
Viyiy
(Vt --Vi
9
can be rewritten as
7
yi
2
.[i-|'£)']
Fi2 Q
,(
(4.1.1)
DIMENSIONAL ANALYSIS AND DYNAMIC SIMILITUDE
209
Clearly the right-hand side represents the inertial forces and the left-hand
side,
the pressure forces that exist due to gravity.
These two forces are of
equal magnitude since one determines the other in this equation.
Further/2 has the dimensions of force per unit width, and it multiplies a dimensionless number which is specified by the geometry of the hydraulic jump.
more, the term
72/i
2
one divides this equation by the geometric term
If
number
1
—
2/2/2/1
2-|*(l
+ «)
2
02/i
2/1
It
and a
representative of the gravity forces, one has
\
(4.1.2)
2/1/
now clear that the left-hand side is the ratio of the inertia and gravity
is
even though the explicit representation of the forces has been obscured
through the cancellation of terms that are common in both the numerator and
denominator. This ratio is equivalent to a dimensionless parameter, called
the Froude number, which will be discussed in further detail later in this
chapter. It is also interesting to note that this ratio of forces is known once
forces,
the ratio
2/2/2/1 is
what the values
given, regardless of
2/2
and
2/1
are.
From
this
observation one can obtain an appreciation of the increased scope that Eq.
(4.1.2) affords
over Eq. (4.1.1) even though one
is
only a rearrangement of
the other.
In writing the
momentum
equation which led to Eq. (4.1.2) only inertia
problem statement. But other
and viscosity. These were neglected as being small in comparison with gravity and inertia forces; however,
only experience with the phenomenon, or ones similar to it, would justify
such an initial simplification. For example, if viscosity had been included
because one was not sure of the magnitude of its effect, the momentum equation would become
and gravity forces were included in the
original
forces are present, such as surface tension
72/i
2
2
72/2
~
^L
^viscous
=_
yiy
{V2
VI
—
Vl)
2d
with the result that
V1 2
,
9Vi
Fv
2
i
SCOUS y 2
72/i (2/1
-
2/2)
-|»(i + fi)
2
2/1
\
yd
Howis more complete than that given by Eq. (4.1.2)
would show that the second term on the left-hand side is
usually a small fraction of the first term and could be neglected in making
initial tests on a hydraulic jump.
This statement
ever, experiments
.
FUNDAMENTALS OF FLUID MECHANICS
210
In the
equation one can consider the ratio
2/2/2/1 to be a dependent
determined for each of the various values of the force ratios,
Vi 2 /gyi and F V acous /yyi2 which are the independent variables. From the previous discussion it appears that the latter variable plays only a minor role in
determining the values of 2/2/2/1. Nevertheless, if one observed that the ratios
last
variable which
is
i
,
of the forces, Vi2 /gyi
and
FV
i
acoua /yyi
2
,
had the same values
in
two
different
one would expect, on the basis of the last equation, that the values of
would
be the same in the two situations. If the ratio of Vi2 /gyi was the
2/2/2/1
same in the two tests but the ratio F v Bcoua /yy h which has only a minor influence for this case, was not equal, one would conclude that the values of
2/2/2/1 for the two cases would be almost the same.
This is the key to much of what follows. For if one can create in a model
experiment the same geometric and force ratios that occur on the full-scale
unit, then the dimensionless solution for the model is valid also for the prototype. Often, as will be seen, it is not possible to have all the ratios equal in
the model and prototype. Then one attempts to plan the experimentation in
such a way that the dominant force ratios are as nearly equal as possible. The
results obtained with such incomplete modeling are often sufficient to describe
tests,
i
the phenomenon in the detail that
is
desired.
Writing a force equation for a complex situation may not be feasible,
process, dimensional analysis, is then used if one knows the per-
and another
tinent quantities that enter into the problem.
In a given situation several of the forces
may
be of
little significance,
With
two dimensionless parameters
are obtained; one set of experimental data on a geometrically similar model
provides the relationships between parameters holding for all other similar
leaving perhaps two or three forces of the same order of magnitude.
three forces of the
same order
of magnitude,
flow cases.
4.2
DIMENSIONS AND UNITS
The dimensions
of mechanics are
Newton's second law of motion,
F
= ma
force,
mass, length, and time, related by
(4.2.1)
Force and mass units are discussed in Sec. 1.2. For all physical systems, it
would probably be necessary to introduce two more dimensions, one dealing
with electromagnetics and the other with thermal effects. For the compressible work in this text, it is unnecessary to include a thermal unit, as the equations of state link pressure, density, and temperature.
DIMENSIONAL ANALYSIS AND DYNAMIC SIMILITUDE
Table 4.1 Dimensions of physical quantities used
mechanics
Symbol
Quantity-
Length
in fluid
Dimensions
(M,L,T)
L
T
I
Time
Mass
m
M
Force
F
MLT-*
Velocity
V
Acceleration
a
LT-i
LT~*
t
Area
A
L
Discharge
Q
UT-x
Pressure
Ap
Gravity
Density
9
Specific weight
7
M
Dynamic
P
viscosity
Kinematic viscosity
V
Surface tension
Bulk modulus
<T
K
of elasticity
2
ML~
T~ 2
LT-*
l
ML~
ML~ T~
3
2
ML-'TL 2 T~
MT-*
2
1
l
ML-'T-*
Newton's second law of motion in dimensional form
F = MLT-
211
is
2
(4.2.2)
which shows that only three of the dimensions are independent. F is the force
dimension,
the mass dimension, L the length dimension, and T the time
dimension.
One common system employed in dimensional analysis is the
MLT system. Table 4.1 lists some of the quantities used in fluid flow, together
with their symbols and dimensions.
M
4.3
THE n THEOREM
The Buckingham n theorem proves that
1
quantities in
which there are
m
in a physical
problem including n
dimensions, the quantities can be arranged
— m
independent dimensionless parameters. Let Ai, A 2
be the quantities involved, such as pressure, viscosity, velocity,
into
n
quantities are
1
known
E. Buckingham,
A SME,
to be essential to the solution,
Model Experiments and the Form
vol. 37, pp.
263-296, 1915.
,
A
3,
.
etc.
.
.,
An
All the
and hence some func-
of Empirical Equations,
Trans.
FUNDAMENTALS OF FLUID MECHANICS
212
tional relation
must
exist
F(A h A 2 ,Az, ...,A n ) =
If
n
III,
At,
.
.,
.
(4.3.1)
groupings of the quantities
dimensions involved, an equation of the form
..., represent dimensionless
2,
m
then with
/(ni,n 2 ,n 3 ...,n n _ m )
,
= o
Ah A
2,
(4.3.2)
exists.
Proof of the
in Sedov's
book
II
theorem
may be
of determining the
II
parameters
is
among them
ferent dimensions, that contain
them
found in Buckingham's paper, as well as
end of this chapter. The method
to select m of the A quantities, with dif-
listed in the references at the
the
m
dimensions, and to use
as repeating variables 1 together with one of the other
A
quantities for
each II. For example, let A h A 2 A z contain M, L, and T, not necessarily in
each one, but collectively. Then, the first II parameter is made up as
,
=
IIi
A^A^A^A,
(4.3.3)
the second one as
n 2 = AiXiAtf*A 9 s *A4
and
so on, until
lin-m
= AlXn - mA 2Vn -mA Z Zn -mA n
In these equations the exponents are to be determined so that each II is diThe dimensions of the A quantities are substituted, and the
mensionless.
exponents
M,
L,
and T are
set equal to zero respectively.
equations in three unknowns for each
II
These produce three
parameter, so that the
ponents can be determined, and hence the II parameter.
If only two dimensions are involved, then two of the
selected as repeating variables,
and two equations
in the
A
x, y, z
ex-
quantities are
two unknown
ex-
ponents are obtained for each II term.
In many cases the grouping of A terms
is such that the dimensionless
arrangement is evident by inspection. The simplest case is that when two
quantities have the same dimensions, e.g., length, the ratio of these two terms
is
1
the
II
parameter.
m
is essential that no one of the
selected quantities used as repeating variables be
derivable from the other repeating variables.
It
DIMENSIONAL ANALYSIS AND DYNAMIC SIMILITUDE
The procedure
EXAMPLE
to
is
by
best illustrated
213
several examples.
The discharge through a horizontal capillary tube is thought
depend upon the pressure drop per unit length, the diameter, and the vis4.1
Find the form of the equation.
The quantities are listed with their dimensions:
cosity.
Symbol
Dimensions
Q
Ap/l
UT~
ML~ T~
Diameter
D
L
Viscosity
n
ML~
Quantity
Discharge
Pressure drop /length
l
2
l
T~
2
l
Then
f(q,^,Am) =
Three dimensions are used, and with four quantities there
rameter
n = Q^
will
be one
II
pa-
:
(^T D*i»
Substituting in the dimensions gives
n = (L'T-^iML^T-^L^ML-'T- = M°L°T°
1
The exponents of each dimension must be the same on both
tion. With L first,
-
3z!
2*/i
+ zi -
and similarly
for
1
M and T
2/i+l=0
-x 1
22/i
-
1
=
=
sides of the equa-
FUNDAMENTALS OF FLUID MECHANICS
214
from which
x\
=
1,
y\
— — 1,
z\
= — 4, and
D* Ap/l
After solving for Q,
I
/z
from which dimensional analysis yields no information about the numerical
value of the dimensionless constant C; experiment (or analysis) shows that
it is
tt/128 [Eq. (5.2.10a)].
When dimensional analysis is used, the variables in a problem must be
known. In the last example if kinematic viscosity had been used in place of
dynamic viscosity, an incorrect formula would have resulted.
EXAMPLE
4.2
A V-notch
weir
a vertical plate with a notch of angle
is
</>
cut
and placed across an open channel. The liquid in the channel
is backed up and forced to flow through the notch. The discharge Q is some
function of the elevation H of upstream liquid surface above the bottom of the
notch. In addition the discharge depends upon gravity and upon the velocity
of approach V to the weir. Determine the form of discharge equation.
into the top of
A
it
functional relationship
F(Q,H,g,V
,<j>)
=
is dimensionless, hence it
to be grouped into dimensionless parameters.
are
one II parameter. Only two dimensions are used, L and T. If g and
is
<f>
H
is
the repeating variables,
IIi
= H*W"Q = L^iLT-^UT-^
n =
2
H Y Vo
x
2
= L^iLT-^LT-
1
Then
xi
+
?/i
-2yi
+
3
-
-1=0
x2
+
2/2
-2y 2
+
1
=
-1=0
DIMENSIONAL ANALYSIS AND DYNAMIC SIMILITUDE
from which
= — f,
xi
y%
215
= —\,x2 = —\,yi= —J, and
V
Q
or
Q
= o
,</A
f
This
may
be written
<Jk-t)
in
which both /,
Q =
/i
unknown
are
Either experiment or analysis
If
is
required to yield additional information about
/i.
H and
Vo were selected as repeating variables in place of g and
= H^Vovl Q = L^iLT-^L'T-
n =
2
After solving for Q,
^ HWf\ik'*)
the function
Hl
functions.
H *V
x
V2
g
= Lx *(LT-
l
1
)y*LT- 2
Then
xi
+
2/1
+
3
=
x2
-?/i-l =
from which z x
ni
=
~
HW
= — 2,
+
=
-^-2
=
2/1
n = fVo
2
or
W„'fc'*)-°
+
1
2
2/2
= — 1,
x%
n =
3
—
1, 2/2
= —2, and
H
FUNDAMENTALS OF FLUID MECHANICS
216
Since any of the
II
parameters
may be inverted or raised to any power without
affecting their dimensionless status,
Q = VoH%
(&•)
The unknown function f2 has the same parameters as /i, but it could not be
The last form is not very useful, in general, because fre-
the same function.
quently Vo may be neglected with F-notch weirs. This shows that a term
minor importance should not be selected as a repeating variable.
Another method
of determining alternate sets of II parameters
the arbitrary recombination of the
n n n
III,
3,
2,
na =
4
2,
If four
independent
II
would be
parameters
are known, the term
ni ai n 2 a2 iT3 a3 ii4 a4
with the exponents chosen at
n n
first set.
of
II4
3,
will,
would constitute a new
would
set.
yield a
new parameter. Then na
This procedure
may
,
be continued to
find all possible sets.
EXAMPLE
The
4.3
losses
Ah/l per unit length of pipe in turbulent flow
through a smooth pipe depend upon velocity V, diameter D, gravity g, dynamic viscosity p, and density p. With dimensional analysis, determine the
general form of the equation
F
(y
,
Clearly, Ah/l
Ui
=
V,D,p,»,g^
is
a
n
parameter.
If V,
D, and p are repeating variables,
= Vm *D*>-p"n = (LT-1 )*iD'i(ML-*)°iML-1 Txi
+
2/1
—
3zi
1
=
-1=0
-xi
zi
from which
—
xi
+
1
= — 1,
=
yi
= — 1,
z\
= —1.
1
DIMENSIONAL ANALYSIS AND DYNAMIC SIMILITUDE
n = Vx *Dv>p
z
2
x2
-x
+
2/2
—
= {LT- y*Ly*{ML-*) >LT-
32 2
2
+1
-2
2
=
=
=0
22
from which x 2
z
i
*g
217
= — 2,
y2
=
1, 2 2
=
0.
or
F
/TOp
since the
is
II
2
A/A
rt
quantities
may be inverted if desired. The first parameter, VDp/n,
the Reynolds number R, one of the most important of the dimensionless
parameters in fluid mechanics.
the nature of the flow.
The
size of the
Reynolds number determines
It is discussed in Sec. 5.3.
Solving for Ah/l gives
?-«
The usual formula
is
T
T-'< i
EXAMPLE
4.4
A
M,
surface tension
depends upon the velocity V, the denpressure drop Ap, gravity g, viscosity
and bulk modulus of elasticity K. Apply dimensional
fluid-flow situation
sity p, several linear
<r,
dimensions
I,
h,
l2 ,
analysis to these variables to find a set of
F(V, P ,l,l h l 2 ,&p,g,n,a,K)
IT
parameters.
=
As three dimensions are involved, three repeating variables are selected. For
complex situations, V, p, and I are generally helpful. There are seven n pa-
FUNDAMENTALS OF FLUID MECHANICS
218
rameters
:
= Vxl pH zl Ap
n3 = Vx *p v3 *v
n = VX2PV2
n = Vxi PH
n = vx *pH
n =
IIi
i
5
Z2
l
2
z
g
zi
4
z t>K
a
I
-
6
h
By expanding
IIi
the
IT
quantities into dimensions,
= {LT-l )*i(MLr*)viL*iML-l T-*
xi
—
+ zi — 1 =
-2 =
Syi
-xi
+1=0
2/i
from which
x\
= —2,yi= — l,zi =
n = (LT- x >(ML- )y*L
x - Sy + z + 1 =0
-2 =
-z
z
l
2
z
)
2
2
0.
>LT- 2
2
2
=0
y2
from which x 2
= — 2,
=
?/ 2
=
0, 2 2
n = (LT-^iML-^L^ML-'Txz - 3yz + z — 1 =
1.
1
3
z
-ft
2/3
from which
Xz
-1=0
+1=0
= —l,yz= —l,z 3 = — 1.
n = (LT- x *(ML-*)y*L *MT~
=0
X ~ 3?/ + 2
z
l
4
2
)
A
4
4
-2
-a;4
2/4
=
+1=0
from which z 4 = —2,y A
= —1,24= — 1.
DIMENSIONAL ANALYSIS AND DYNAMIC SIMILITUDE
219
n = (LT- )**(ML-3 )*'L**ML- T-2
Zb - 32/s + Z5 - 1 =
-2 =
-x
1
1
6
h
+1=0
3/5
from which
x$
= — 2,
Ap
= — 1,
25
gl
K
n =
^
n'
?/ 5
=
0.
a
H
I
I
n =
°
r2
'
s
and
/
;
/ AP
\P F 2
It is
£L
'
V
'
JL.
7Z P
—
-1—
'
7
2
PZ
'
P^ 2
L
'
^i
'
L\ _ o
k)
convenient to invert some of the parameters and to take the square root
(Ap_
The
2
V*
Vlp
VHp
V
l_
l\
=
Ap/(pV 2 /2), is the pressure coefficient;
the second parameter is the Froude number F; the third is the Reynolds number
R; the fourth is the Weber number W; and the fifth is the Mach number M.
first
parameter, usually written
Hence
After solving for pressure drop
which fh /2 must be determined from analysis or experiment. By selecting
other repeating variables, a different set of IT parameters could be obtained.
in
Figure 5.32
is
a representation of a functional relationship of the type
FUNDAMENTALS OF FLUID MECHANICS
220
just given as
applies to the flow in pipes.
it
are neglected as being unimportant;
the pipe L, and k
is
of the surface roughness of the pipe
Ap
Here the parameters
F,
W, and
M
the pipe diameter D, h is the length of
a dimension which is representative of the effective height
I
is
and
is
given by
e.
Thus
Ki)
~pV 2
The
fact that the pressure drop in the pipeline varies linearly with the length
(i.e.,
doubling the length of pipe doubles the loss in pressure) appears reasonone has
able, so that
Ap
L
/
e
H
Ap
\
°r
pV>-D f4 \ 'Dj
"
P V>(L/D)
/
/4
V
e
\
Dj
The term on the left-hand side is commonly given the notation f/2, as in
Fig. 5.32. The curves shown in this figure have / and R as ordinate and abscissa, respectively, with e/D a parameter which assumes a given value for
each curve. The nature of these curves was determined through experiment.
Such experiments show that when the parameter R is below the value of 2000,
the curves for the various values of e/D coalesce into one. Hence / is independent of e/D, and the result is
all
/ = /i(R)
be predicted in Chap. 5 on the basis of theoretical conremained for an experimental verification of these predictions to indicate the power of the theoretical methods.
This relationship
siderations,
EXAM PLE
but
4.5
will
it
The
airplane propellers
test
thrust due to any one of a family of geometrically similar
is
to be determined experimentally
on a model. By means
from a wind-tunnel
of dimensional analysis find suitable parameters
for plotting test results.
The
thrust
diameter D,
FT
depends upon speed of rotation w, speed of advance V
and speed of sound c. The function
,
air viscosity n, density p,
F(FT ,V ,D,w,tJL,p,c) =
is
to be arranged into four dimensionless parameters, since there are seven
quantities
and three dimensions.
Starting
first
by
selecting
p,
w,
and
D
as
DIMENSIONAL ANALYSIS AND DYNAMIC SIMILITUDE
221
repeating variables,
III
= p*wD^FT = (ML-*)*i(T-
n = p*v Z>
2
22
2
n =
p*w*D*>n
n =
p
3
4
Xi
a>
By
Vi
D
z
= (AfL-
Fo
*c
8
l
)*iL z iMLT- 2
)**(!^- 1 )«'»L**L7
,
-1
= {MIr*)**(T-l )y*L'*ML-l T=
(ML-^iT-^L^LT-
1
1
writing the simultaneous equations in x h y 1} z lf
etc.,
as before
and
solving them,
FT
Vo
c
ju
Solving for the thrust parameter leads to
FT
P
cc
2
^\
pcoD
/Vo
/Vo_ pc*&
2
,
D
4
may
be recombined to obtain other forms, the second
first and second terms, VDp/n, and the
replaced by the first term divided by the third term, V /c thus
Since the parameters
replaced by the product of the
term
is
third
term
is
(Vo
Ft
;
VqDp
VA
pco
Of the dimensionless parameters, the first is probably of the most importance,
The second parameter
it relates speed of advance to speed of rotation.
is a Reynolds number and accounts for viscous effects.
The last parameter,
speed of advance divided by speed of sound, is a Mach number, which would
be important for speeds near or higher than the speed of sound. Reynolds
effects are usually small, so that a plot of F T /pu 2 D A against V /ccD should be
most informative.
since
The
1.
steps in a dimensional analysis
Select the pertinent variables.
may be summarized
This requires some knowledge of the
process.
2.
Write the functional relationships,
F(V,D,p,n,c,H)
=
as follows:
e.g.,
FUNDAMENTALS OF FLUID MECHANICS
222
3.
(Do not make the dependent quantity
These variables should contain all the m dimenOften one variable is chosen because it specifies
the scale, another the kinematic conditions, and in the cases of major
interest in this chapter one variable is chosen which is related to the
forces or mass of the system, for example, D, V, p.
Write the II parameters in terms of unknown exponents, e.g.,
Select the repeating variables.
a repeating variable.)
sions of the problem.
4.
n =
7»iZ>ip*»/i
x
For each of the
5.
that the
6.
sum
= (LT-^Ly^ML-^^ML-'-T
expressions write the equations of the exponents, so
II
of the exponents of each dimension will
be
zero.
Solve the equations simultaneously.
Substitute back into the
7.
the dimensionless
II
II
expressions of step 4 the exponents to obtain
parameters.
Establish the functional relation
8.
/l(lli,n 2 ,n 3
,
... }
Un-m)
or solve for one of
n =/(ni,n
3,
Recombine,
if
2
9.
1
same number
=
II's explicitly:
...,n n _m )
desired, to alter the forms of the
II
parameters, keeping the
of independent parameters.
Alternate formulation of
II
parameters
A
rapid method for obtaining n parameters, developed by Hunsaker and
Rightmire (referenced at end of chapter), uses the repeating variables as
primary quantities and solves for M, L, and T in terms of them. In Example
4.3 the repeating variables are V, D, and p; therefore
V = LT~
D =L
l
p
= ML"
3
(4.3.4)
L =
D
Now, by use
n
T = DV-
1
M
=
PD
of Eqs. (4.3.4),
= ML-'T- = pD'D-'D-W =
1
P
DV
Z
DIMENSIONAL ANALYSIS AND DYNAMIC SIMILITUDE
hence the
II
parameter
Equations (4.3.4)
n
is
be used directly to find the other
n
parameters.
For
2
= LT-2 -
g
may
223
DD-W
2
= V 2D~
l
and
n
gD
9
2
V D~
2
l
V
2
This method does not require the repeated solution
unknowns for each n parameter determination.
4.4
of three equations in three
DISCUSSION OF DIMENSION LESS PARAMETERS
—
The
five dimensionless parameters
pressure coefficient, Reynolds number,
Froude number, Weber number, and Mach number are of importance in
correlating experimental data. They are discussed in this section, with particular emphasis placed on the relation of pressure coefficient to the other
—
parameters.
Pressure coefficient
The pressure
coefficient
When
multiplied
sure.
Ap/(pV 2 /2)
by area, it
is
the ratio of pressure to dynamic pres-
is
the ratio of pressure force to inertial
(pV 2 /2)A would be the force needed to reduce the velocity to zero.
may also be written as Ah/(V 2 /2g) by division by y. For pipe flow the
Darcy-Weisbach equation relates losses hi to length of pipe L, diameter D,
and velocity V by a dimensionless friction factor /
force, as
It
1
kYl
D2g
1
fk-
hl
D " V /2g
2
R
(
^Kd
There are several friction factors in general use. This is the Darcy-Weisbach friction
factor, which is four times the size of the Fanning friction factor, also called /.
FUNDAMENTALS OF FLUID MECHANICS
224
as fL/D is shown to be equal to the pressure coefficient (see Example 4.4).
In pipe flow, gravity has no influence on losses; therefore F may be dropped
to D,
e';
W
Similarly surface tension has no effect, and
out.
liquid flow, compressibility
h
is
not important, and
to roughness height projection
€
M
drops out.
is
dropped.
in the pipe wall,
and k to
For steady
I
may
hence
(4A1)
§-*(*5-D
Pipe-flow problems are discussed in Chaps.
is
5, 6,
and
10.
If compressibility
important,
Compressible-flow problems are studied in Chap.
in
Chap.
8,
V = C
v
6.
With
orifice flow,
studied
y/2gH,
^^^'H'i'D
in
refer
their spacing
which
I
may
(4 4 3)
-
refer to orifice diameter
and h and k
to
-
upstream dimensions.
Viscosity and surface tension are unimportant for large orifices and low-
Mach number effects may be very important for
pressure drops, i.e., Mach numbers approaching unity.
viscosity fluids.
with large
gas flow
In steady, uniform open-channel flow, discussed in Chap. 5, the Ch6zy
formula relates average velocity V, slope of channel S, and hydraulic radius
of cross section R (area of section divided by wetted perimeter) by
v = c
C
is
vrs
= c
4r j;
a coefficient depending upon
(4A4)
size,
shape, and roughness of channel.
Then
2gL
Afr
V*/2g
f *-'<;)
and compressible effects are usually unimportant.
body is expressed by F = CD ApV 2 /2, in which A is a
the body, usually the projection of the body onto a plane nor-
since surface tension
The drag
typical area of
Fona
DIMENSIONAL ANALYSIS AND DYNAMIC SIMILITUDE
mal to the
=
jk-2
flow.
Then F/A
CD=h
The term R
is
225
equivalent to Ap, and
is
(4A6)
(*™'Vl)
related to skin-friction drag due to viscous shear as well as to
form, or profile, drag resulting from separation of the flow streamlines from the
wave drag if there is a free surface; for large Mach numCd may vary more markedly with M than with the other parameters;
body; F
bers
is
related to
the length ratios
may
refer to
shape or roughness of the surface.
The Reynolds number
The Reynolds number VDp/p
is
the ratio of inertial forces to viscous forces.
A critical Reynolds number distinguishes among flow regimes,
such as laminar
boundary layer, or around immersed objects.
The particular value depends upon the situation. In compressible flow, the
Mach number is generally more significant than the Reynolds number.
or turbulent flow in pipes, in the
The Froude number
The Froude number
dynamic (or inertial)
V
2
/gl,
when
by pA, is a ratio of
With free liquid-surface flow the nature
depends upon whether the Froude number is
multiplied and divided
force to weight.
of the flow (rapid 1 or tranquil)
greater or less than unity.
It is useful in calculations of hydraulic
jump, in
design of hydraulic structures, and in ship design.
The Weber number
The Weber number VHp/a is the ratio of inertial
forces (evident when numerator and denominator
forces to surface-tension
are multiplied
by
I).
It
important at gas-liquid or liquid-liquid interfaces and also where these
interfaces are in contact with a boundary.
Surface tension causes small
is
(capillary)
orifices
waves and droplet formation and has an
and weirs at very small heads. The
is shown in Fig. 4.1. To the left
propagation
1
effect of
on discharge of
surface tension on wave
effect
of the curve's
minimum
the
wave
Open-channel flow at depth y is rapid when the flow velocity is greater than the speed
y/gy of an elementary wave in quiet liquid. Tranquil flow occurs when the flow velocity
is less than \/gy.
FUNDAMENTALS OF FLUID MECHANICS
226
Wavelength
Wave speed
Fig. 4.1
vs.
wave-
length for surface waves.
speed
is
controlled
by
surface tension (the waves are called ripples)
minimum
right of the
,
and to the
gravity effects are dominant.
The Mach number
The speed
of
sound
elasticity (Sees. 1.7
T the
in a liquid is written
and
6.2)
y/K/p
ore = y/kRT
absolute temperature, for a perfect gas)
number.
It
is
(k
if
is
K is
V/c or
.
the bulk modulus of
the specific heat ratio and
V / y/K/p is the Mach
By
a measure of the ratio of inertial forces to elastic forces.
squaring V/c and multiplying by pA/2 in numerator and denominator, the
numerator
is
at sonic flow.
the dynamic force and the denominator
It
may
also be
shown
energy of the flow to internal energy of the
relating parameter
4.5
SIMILITUDE;
when
is
the dynamic force
to be a measure of the ratio of kinetic
fluid.
velocities are near or
It is the
above
most important
cor-
local sonic velocities.
MODEL STUDIES
Model studies of proposed hydraulic structures and machines are frequently
undertaken as an aid to the designer. They permit visual observation of the
flow and make it possible to obtain certain numerical data, e.g., calibrations
of weirs and gates, depths of flow, velocity distributions, forces on gates,
l
efficiencies
and capacities
of
pumps and
turbines, pressure distributions,
and
losses.
accurate quantitative data are to be obtained from a model study,
must be dynamic similitude between model and prototype. This simili-
If
there
tude requires (1) that there be exact geometric similitude and (2) that the
dynamic pressures at corresponding points be a constant. The second
requirement may also be expressed as a kinematic similitude, i.e., the stream-
ratio of
lines
must be geometrically
similar.
DIMENSIONAL ANALYSIS AND DYNAMIC SIMILITUDE
227
Geometric similitude extends to the actual surface roughness of model
and prototype. If the model is one-tenth the size of the prototype in every
linear dimension, then the height of roughness projections must be in the
same ratio. For dynamic pressures to be in the same ratio at corresponding
points in model and prototype, the ratios of the various types of forces must
be the same at corresponding points. Hence, for strict dynamic similitude,
the Mach, Reynolds, Froude, and Weber numbers must be the same in both
model and prototype.
Strict fulfillment of these requirements
except with a
1
:
1
forces are of the
scale ratio.
same magnitude.
generally impossible to achieve,
is
Fortunately, in
many
situations only
Discussion of a few cases will
two
of the
make
this
clear.
Wind- and water-tunnel tests
This equipment
is
used to examine the streamlines and the forces that are
induced as the fluid flows past a fully submerged body. The type of test that
is
being conducted and the availability of the equipment determine which
kind of tunnel will be used. Because the kinematic viscosity of water
one-tenth that of
air,
high Reynolds numbers.
a water tunnel
!
is
about
a water tunnel can be used for model studies at relatively
The drag
At very high
effect of various
parachutes was studied in
air velocities the effects of compressibility,
and
Mach number, must
be taken into consideration, and indeed
may be the chief reason for undertaking an investigation. Figure 4.2 shows
a model of an aircraft carrier being tested in a low-speed tunnel to study the
consequently
The model has been inverted
wool tufts can be used to give an
indication of the flow direction. Behind the model there is an apparatus for
sensing the air speed and direction at various locations along an aircraft's
flow pattern around the ship's superstructure.
and suspended from the
ceiling so that the
glide path.
Pipe flow
In steady flow in a pipe, viscous and inertial forces are the only ones of consequence; hence,
number
in
when geometric
similitude
is
observed, the same Reynolds
model and prototype provides dynamic
similitude.
The various
corresponding pressure coefficients are the same.
For testing with fluids
having the same kinematic viscosity in model and prototype, the product,
VD, must be the same. Frequently
models.
this requires
very high velocities in small
FUNDAMENTALS OF FLUID MECHANICS
228
Wind-tunnel tests on an aircraft-carrier superstructure. Model
suspended from ceiling. (Photograph taken in Aeronautical
and Astronautical Laboratories of The University of Michigan for the DynaFig. 4.2
is
inverted and
sciences Corp.)
Open hydraulic
structures
Structures such as spillways, stilling pools, channel transitions, and weirs
generally have forces due to gravity (from changes in elevation of liquid surfaces)
and
inertial forces that are greater
than viscous and turbulent shear
In these cases geometric similitude and the same value of Froude's
number in model and prototype produce a good approximation to dynamic
forces.
similitude; thus
Vf
Since gravity
is
the same, the velocity ratio varies as the square root of the
DIMENSIONAL ANALYSIS AND DYNAMIC SIMILITUDE
scale ratio X
=Vm
Vv
=
l
229
p /lm ,
^
The corresponding times
t
m
for events to take place (as time for passage
through a transition) are related; thus
11
of a particle
—
=
V
t
p
_
m
lP
3
/t p
Force ratios,
Fp
—
ana
t
p
=
yhplp2
ratio
__
t
m
IV
—
—
im V p
V p
The discharge
Qp
=
Q p/Q m
=
t
m
Vx
is
^ 5/2
e.g.,
=
on gates,
Fp /Fm
,
are
X3
yflm^r,
where h
so that
is
the head. In a similar fashion other pertinent ratios can be derived
model
results
can be interpreted as prototype performance.
Figure 4.3 shows a model test conducted to determine the effect of a
breakwater on the wave formation in a harbor.
Ship's resistance
The
motion of a ship through water is composed of pressure
and wave resistance. Model studies are complicated by
the three types of forces that are important, inertia, viscosity, and gravity.
Skin-friction studies should be based on equal Reynolds numbers in model
and prototype, but wave resistance depends upon the Froude number. To
satisfy both requirements, model and prototype must be the same size.
The difficulty is surmounted by using a small model and measuring the
total drag on it when towed. The skin friction is then computed for the model
and subtracted from the total drag. The remainder is stepped up to prototype size by Froude's law, and the prototype skin friction is computed and
added to yield total resistance due to the water. Figure 4.4 shows the dramatic change in the wave profile which resulted from a redesigned bow. From
such tests it is possible to predict through Froude's law the wave formation
and drag that would occur on the prototype.
resistance to
drag, skin friction,
FUNDAMENTALS OF FLUID MECHANICS
230
Fig. 4.3
Model test on a harbor to determine the
(Department of
Civil
effect of a breakwater.
Engineering, The University of Michigan.)
Hydraulic machinery
The moving
parts in a hydraulic machine require an extra parameter to ensure
that the streamline patterns are similar in model and prototype. This parameter
must
relate the throughflow (discharge) to the speed of
For geometrically similar machines
if
moving
parts.
the vector diagrams of velocity enter-
moving parts are similar, the units are homologous; i.e., for
dynamic similitude exists. The Froude number is unimportant, but the Reynolds number effects (called scale effects because it is
impossible to maintain the same Reynolds number in homologous units) may
cause a discrepancy of 2 or 3 percent in efficiency between model and protoing or leaving the
practical purposes
Fig. 4.4
Model tests showing the influence of a bulbous bow on bow-wave
formation. (Department of Naval Architecture and Marine Engineering, The
University of Michigan.)
DIMENSIONAL ANALYSIS AND DYNAMIC SIMILITUDE
M
U OF
r
1
^5.
1007
MODEL
BALLAST
180 KTS
J
3-20-B4
\
j
1
231
i
*
/
Bex
%
•
*"
C
4
•
frfrirfrif-
*?^", J'
;
^Bk.1^
•
11
U
OFM
MOO EL 1007-03
BALLAST
180 KNOTS
FUNDAMENTALS OF FLUID MECHANICS
232
The Mach number
type.
is
also of
importance in axial-flow compressors and
gas turbines.
EXAMPLE
4.6
The valve
coefficients
K
= Ap/(pv 2 /2)
for a
60-cm-diameter
valve are to be determined from tests on a geometrically similar 30-cmdiameter valve using atmospheric air at 80°F. The ranges of tests should be for
flow of water at 70°F at
(VD\
V
(1
"
/min
V
/max
(
=
)
\
610,000
For testing with
v
=
X
(1.8
10- 4
Then the ranges
2
X
10- 5
X
10- 5
Qmin
=
Omax
= 7
X
ft /s)
5
2.5
(0.6
2
ft /s)
=
ranges of airflows
for the prototype valve
m)
is
needed?
is
=
(0.3048 m/ft) 2
1,525,000
80°F
(0.3048 m/ft) 2
=
1.672
X
10" 5
m /s
2
of air velocities are
m /s
Vmin =
=
610,000
=
1,525,000
30.6
m/s
2
(F max )(0.3m)
1.672
m/s)
10-
air at
(7 min )(0.3m)
1.672
X
(1-059
What
to 2.5 m/s.
1
The Reynolds number range
m /s
7 max =
2
2
t (0.3 m) (30.6 m/s)
=
2.16
85
m/s
m /s
3
4
(0.3
m) 2 (85 m/s) =
6.0
m /s
3
PROBLEMS
that Eqs. (3.7.6), (3.9.3), and (3.11.13) are dimensionally homogeneous.
4.1
Show
4.2
Arrange the following groups into dimensionless parameters:
(a)
Ap,p,
V
(b)p,g,V,F
(c)n,F,Ap,t
DIMENSIONAL ANALYSIS AND DYNAMIC SIMILITUDE
By
4.3
(a)
a,
I,
233
inspection, arrange the following groups into dimensionless parameters:
(b) v,
t
I,
A, Q,
(c)
t
(d)
co
K,
a,
A
4.4
Derive the unit of mass consistent with the units inches, minutes, tons.
4.5
In terms of
M,
L, T, determine the dimensions of radians, angular velocity,
moment
power, work, torque, and
4.6
Find the dimensions
4.7
Work Example
of
momentum.
FLT
of the quantities in Prob. 4.5 in the
4.2 using
Q and
system.
H as repeating variables.
them
Using the variables Q, D, AH/I, p, p, g as pertinent to smooth-pipe flow, arrange
as repeating variables.
into dimensionless parameters with Q, p,
4.9
If
4.8
jj.
the shear stress r
is
known
to
depend upon viscosity and rate of angular deforflow, determine the form of Newton's law of
mation du/dy in one-dimensional laminar
viscosity by dimensional reasoning.
4.10
The
variation
Ap
of pressure in static liquids
weight y and elevation difference Az.
By
is
known
to
depend upon
specific
dimensional reasoning determine the form of
the hydrostatic law of variation of pressure.
4.11
When
viscous and surface-tension effects are neglected, the velocity
from a reservoir is thought to depend upon the pressure drop Ap
density p. Determine the form of expression for V.
efflux of liquid
liquid
and
its
V
of
of the
The buoyant force F B on a body is thought to depend upon its volume submerged
and the gravitational body force acting on the fluid. Determine the form of the
4.12
V
buoyant-force equation.
4.13
In a
fluid rotated as a solid
about a vertical axis with angular velocity
pressure rise p in a radial direction depends
Obtain the form of equation for
4.14
In Example
4.3,
upon speed
co,
radius
r,
and
co,
the
fluid density p.
p.
work out two other
sets of dimensionless
parameters by recom-
bination of the dimensionless parameters given.
4.15
Find the dimensionless parameters
of
Example
4.4 using Ap, p,
and I as repeating
variables.
4.16
The Mach number
M for flow of a perfect gas in a pipe depends upon the specific-
heat ratio k (dimensionless)
,
the pressure p, the density p, and the velocity V. Obtain
of the Mach number expression.
by dimensional reasoning the form
4.17
Work
of viscosity
4.18
of
The
T on a disk of radius r that rotates in fluid
and clearance y between disk and fixed plate.
out the scaling ratio for torque
p with angular velocity
model of a spillway for a dam is 3.3 ft/s. For a ratio
what is the velocity at the corresponding point in the
velocity at a point in a
prototype to model of 10:
1,
co
prototype under similar conditions?
FUNDAMENTALS OF FLUID MECHANICS
234
4.19
The power input
the fluid density
p, size
to a pump depends upon the discharge Q, the pressure rise Ap,
D, and efficiency e. Find the expression for power by the use of
dimensional analysis.
The torque
4.20
specific
delivered
by a water turbine depends upon discharge
weight y, angular velocity w, and efficiency
e.
Determine the form
Q, head
of
H
y
equation
for torque.
A model of a venturi meter has linear dimensions one-fifth those of the prototype.
4.21
The prototype operates on water
throat diameter of 60
discharge
cm and
and the model on water at 95°C.
m/s
in the prototype,
For a
what
needed through the model for similitude?
is
The drag F on a
4.22
at 20°C,
a velocity at the throat of 6
density of fluid
an expression
p,
high-velocity projectile depends
acoustic velocity
c,
upon speed
V
of projectile,
diameter of projectile D, and viscosity p.. Develop
for the drag.
is
The wave drag on a model of a ship is 3.52 lb at a speed of 8 ft/s. For a prototype
what will the corresponding speed and wave drag be if the liquid
the same in each case?
4.
24
4.23
fifteen times as long
Determine the
A
4.25
small sphere of radius
of density p
By
r
and viscosity p. The
ft/s.
and density p
tests are
dimensional analysis determine a set
The drag
2I0 m which drop
on a small sphere in
D=
specific gravity of spherical particles,
through air at 33°F at a speed U of 0.3
laminar motion is given by St/jlDU.
force
>
settles at velocity
U in
another liquid
conducted inside vertical tubes of radius r.
of dimensionless parameters to be used in
determining the influence of the tube wall on the settling velocity.
The
4.26
ji
=
0.002 P,
The
how
1.2-m-diameter pipe system carrying gas (p = 40 kg/m3
25 m/s) are to be determined by testing a model with water at 20°C.
losses in a
V=
Y in a
,
laboratory has a water capacity of 75
1/s.
What model
scale should be used,
and
are the results converted into prototype losses?
Ripples have a velocity of propagation that
4.27
is
dependent upon the surface
tension and density of the fluid as well as the wavelength.
By
dimensional analysis
justify the shape of Fig. 4.1 for small wavelengths.
In very deep water the velocity of propagation of waves depends upon the
Upon what
it is independent of this dimension.
4.28
wavelength, but in shallow water
variables does the speed of advance depend for shallow-water waves?
agreement with
4.29
If
this
a vertical circular conduit which
the fluid will attach
2.9)
.
Is Fig. 4.1 in
problem?
itself
is
not flowing
uniformly to the inside wall as
Under these conditions the
full is
it
rotated at high speed,
flows
downward
(see Sec.
radial acceleration of the fluid yields a radial force
and a hydraulic jump can occur on the
suddenly changes. Determine a set of
dimensionless parameters for studying this rotating hydraulic jump.
field
which
is
similar to gravitational attraction,
inside of the tube,
whereby the
fluid thickness
DIMENSIONAL ANALYSIS AND DYNAMIC SIMILITUDE
4.30
A
nant
role.
nearly spherical fluid drop oscillates as
it falls.
235
Surface tension plays a domi-
Determine a meaningful dimensionless parameter
for this natural frequency.
wing are shown in Fig. 5.23. If the wing has a
and drag per foot of length when the wing is operating
7
at zero angle of attack at a Reynolds number, based on the chord length, of 4.5 X 10
in air at 50°F. What force would be on a 1 20 scale model if the tests were conducted in
water at 70°F? What would be the speed of the water? Comment on the desirability
of conducting the model tests in water.
The
4.31
and drag
lift
chord of 10
coefficients for a
determine the
ft,
lift
1
A
4.32
1:5 scale model of a water pumping station piping system
determine overall head
losses.
Air at 25°C,
kg// cm 2 abs
1
is
to be tested to
For a prototype
is
available.
cm/s in a 4-m-diameter section with water at 15°C, determine the air
and quantity needed and how losses determined from the model are converted
velocity of 50
velocity
into prototype losses.
The boat
be made.
(p
=
30 psia,
Cl
coefficient
The
4.34
t
results
scale
=
is
will travel at
90°F)
is
dimensionless.
35
mph
= ClApV2 /2.
Lift
resistance to ascent of a balloon
of a 1: 50 scale
4.35
lift and drag on hydrofoils for a boat are to
through water at 60°F. What velocity of air
required to determine the lift and drag? Note: The lift
Full-scale wind-tunnel tests of the
4.33
model
in water.
is
How would
to be determined
by studying the ascent
such a model study be conducted and the
converted to prototype behavior?
The moment
model
in a
exerted on a submarine
water tunnel.
If
by
its
rudder
is
to be studied with a 1: 20
the torque measured on the model
is
5
N-m
for a
tunnel velocity of 15 m/s, what are the corresponding torque and speed for the prototype?
For two hydraulic machines to be homologous they must (a) be geometri(b) have the same discharge coefficient when viewed as an orifice,
Qi/(Ai y/2gH\) = Q2/ (M \/2gH2) and (c) have the same ratio of peripheral speed
to fluid velocity, wD/(Q/A).
Show that the scaling ratios may be expressed as
Q/ND* = const and H/(ND) 2 = const. N is the rotational speed.
4.36
cally similar,
,
4.37
1
1
By use
of the scaling ratios of Prob. 4.36,
4 model of a centrifugal
pump
determine the head and discharge of a
ft head when turning 240
that produces 20 cf s at 96
rpm. The model operates at 1200 rpm.
4.38
An
\o)D'
incorrect arbitrary
n
recombination of the
II
parameters
'coD/
(a)
\7o'
11
'.oDj
\coD'
/i
'u)D/
FUNDAMENTALS OF FLUID MECHANICS
236
(c)
VcoD
none
(e)
M
com
of these
3
coD/
m
answers
The repeating
4.39
WZ) p
/
variables in a dimensional analysis should
include the dependent variable
have two variables with the same dimensions if possible
(c) exclude one of the dimensions from each variable if possible
(d) include those variables not considered very important factors
(e) satisfy none of these answers
(a)
(b)
Select a
4.40
common dimensionless parameter
(a)
angular velocity
(d)
specific
pressure coefficient
factor
Which
ul
>*
(a)
v
4.43
VDfl
—
UwV
<
(c)\
p
I
The Reynolds number may be
(a)
viscous forces to inertial forces
(6)
viscous forces to gravity forces
(c)
gravity forces to inertial forces
(d)
elastic forces to pressure forces
(e)
none
4.44
The
(a)
^
(e)
none
4.45
of these
{h)
{c)
^T2
The
not a dimensionless parameter:
Darcy-Weisbach
(c)
of a
V
—
gD
Reynolds number?
f
\
(e)
Ap
—
pV
2
defined as the ratio of
answers
pressure coefficient
of these
form
fA\
(d)
specific gravity
Weber number
(e)
of the following has the
tu\
(6)
is
mechanics from the following:
(c)
Froude number
(b)
kinematic viscosity
(d)
4.42
/
kinematic viscosity
none of these answers
(6)
(e)
Select the quantity in the following that
4.41
(a)
weight
in fluid
may
Vv
take the form
(d)Ap
w
answers
pressure coefficient
(a)
viscous forces
(6)
inertial forces
(c)
gravity forces
(d)
surface-tension forces
(e)
elastic-energy forces
is
a ratio of pressure forces to
friction
DIMENSIONAL ANALYSIS AND DYNAMIC SIMILITUDE
How many II
= 0?
4.46
237
parameters are needed to express the function
F(a,V,t,v,L)
(a)
5
4
(b)
3
(c)
2
(d)
(e)
1
Which of the following could be a II parameter of the function F(Q,H,g, V
when Q and g are taken as repeating variables?
4.47
(a)
(e)
QVgW
(b) V /g Q
none of these answers
2
2
2
(c)
Select the situation in which inertial forces
4.48
(a)
flow over a spillway crest
(b)
flow through an open-channel transition
(c)
waves breaking against a sea wall
flow through a long capillary tube
flow through a half-opened valve
(d)
(e)
Which two
4.49
most important
forces are
=
Ql VqR
(d)
Q/g<f>
,<j))
would be unimportant:
in laminar flow
between closely spaced
parallel plates?
(a)
inertial, viscous
(d)
viscous, pressure
A
4.50
[KpQ
PQ
Ap 2
2
I
Q
I,
p,
=
p
is
[p~Q
AplQ
pi
ApQ
I
What velocity of oil,
gravity, pressure
(c)
none of these answers
(e)
dimensionless combination of Ap,
v p
4.51
pressure, inertial
(6)
2
>
p
1.6 slugs/ft 3
,
=
ju
0.20 P,
Ap
must occur
2
I
in a 1-in-diameter
pipe to be dynamically similar to 10 ft/s water velocity at 68°F in a J-in-diameter
tube?
(a)
0.60 ft/s
(e)
none of these answers
4.52
The
(b)
9.6 ft/s
4.0 ft/s
(c)
velocity at a point on a
model dam
corresponding prototype velocity for X
(a)
4.53
X
=
25
(6)
(c)
0.2
(d)
25
crest
is,
0.04
The height of a hydraulic jump in a
The prototype jump height is
(a)
12 ft
less
A
(6)
than 4
2
in
(c)
ft
(e)
was measured
to be 1 m/s.
The
in meters per second,
(e)
stilling
none of these answers
pool was found to be 4.0 in in a model,
not determinable from data given
none of these answers
ship's model, scale 1: 100,
The corresponding prototype wave
(a)
=
60 ft/s
36.
(d)
4.54
5
(d)
10
(6)
100
(c)
1000
had a wave resistance
resistance
(d)
is,
10,000
of 10
N
at its design speed.
in kilonewtons,
(c)
none
of these
answers
FUNDAMENTALS OF FLUID MECHANICS
238
4.55
A
many
times greater would the prototype resistance be
number
1:
in air of the
0.312
(a)
4.56
If
model
5 scale
same temperature and
3.12
(b)
and
Ah
(d)
12.5
(c)
the capillary rise
surface tension a
M=
of a projectile has a drag coefficient of 3.5 at
when
fired at the
How
half the density?
25
(e)
none of these answers
of a liquid in a circular tube of diameter
specific
2.0.
same Mach
weight y, the formula for capillary
D depends upon
rise
could take the
form
(<l)
(d)
M =^y F fe)
(6)
M=
Ah = <l- F
(e)
none of these answers
(
(C)
C
fe)"
Ak = CD
0"
J
REFERENCES
Bridgman, P. W.: "Dimensional Analysis," Yale University Press, New Haven, Conn.,
1931, Paperback Y-82, 1963.
Holt, M.: Dimensional Analysis, sec. 15 in V. L. Streeter (ed.), "Handbook of Fluid
Dynamics," McGraw-Hill, New York, 1961.
Hunsaker, J. C, and B. G. Rightmire: "Engineering Applications of Fluid Mechanics,"
pp. 110, 111, McGraw-Hill, New York, 1947.
Hydraulic Models, ASCE Man. Eng. Pract. 25, 1942.
Ipsen, D. C.: "Units, Dimensions, and Dimensionless Numbers," McGraw-Hill,
New York, 1960.
Langhaar, H. L.: "Dimensional Analysis and Theory of Models," Wiley,
New
York,
1951.
I.: "Similarity and Dimensional Methods
by M. Holt, Academic, New York, 1959.
Sedov, L.
ed.
in
Mechanics," English trans.
5
VISCOUS EFFECTS: FLUID RESISTANCE
In Chap. 3 the basic equations used in the analysis of fluid-flow situations were
The fluid was considered frictionless, or in some cases losses were
assumed or computed without probing into their underlying causes. This
discussed.
chapter deals with real
important. Viscosity
ing fluid;
it is
is
fluids, i.e.,
with situations in which
irreversibilities are
the fluid property that causes shear stresses in a
mov-
also one means by which irreversibilities or losses are developed.
Without viscosity
in a fluid there
laminar, incompressible flow are
is
fluid resistance.
Simple cases of steady,
developed in this chapter, since in these
The concept of the Reynolds number, in-
first
cases the losses can be computed.
troduced in Chap.
no
then further developed. Turbulent-flow shear relationof the Prandtl mixing-length theory and are applied to turbulent velocity distributions. This is followed by boundary-layer
concepts and by drag on immersed bodies. Resistance to steady, uniform, in4, is
ships are introduced
by use
compressible, turbulent flow is then examined for open and closed conduits,
with a section devoted to open channels and to pipe flow. The chapter closes
with a section on lubrication mechanics.
The equations
motion for a real fluid can be developed from consideraon a small element of the fluid, including the shear
stresses generated by fluid motion and viscosity.
The derivation of these
equations, called the Navier-Stokes equations, is beyond the scope of this
of
tion of the forces acting
They are listed, however, for the sake of completeness, and many
developments of this chapter could be made directly from them. First,
Newton's law of viscosity, Eq. (1.1.1), for one-dimensional laminar flow can
treatment.
of the
239
FUNDAMENTALS OF FLUID MECHANICS
240
be generalized to three-dimensional flow (Stokes' law of viscosity)
(du
The
dv\
dw\
(dv
subscript of the shear stress
first
which the
component
stress
is
is
acting.
(dw
du\
the normal direction to the face over
The second
subscript
the direction
is
component.
By limiting the Navier-Stokes equations to incompressible flow, with
gravity the only body force acting (let h be measured vertically upward),
they become
of the stress
-
Id,
—
-
P dx
-
1
-
•
+
(P
yh)
—d (P + yh)^
,
p ay
—
—Id,
+
(p
p dz
in
which
v is
7
.
+
v
+
v
V2 u =
at
dv
=-
o
VH
at
.
2
+ vV w =
N
yh)
— =
the kinematic viscosity, assumed to be constant; d/dt
d
d
u
\-
v
\-
dx
dt
—
dw
dt
w
dy
d
a2
dz
d2
1
dx
2
differen-
d
1
dt
as explained in Sec. 7.3, and the operator
V2 =
is
motion
tiation with respect to the
d
—
du
V2
is
d2
1
dy
2
dz 2
For a nonviscous fluid, the Navier-Stokes equations reduce to the Euler
equations of motion in three dimensions, given by Eqs. (7.3.8), (7.3.9), and
(7.3.10).
vertically
For one-dimensional flow of a
upward the equations reduce
dp
a d2 u
du
dv
d
p dx
p dz 2
dt
dy
dz
1
real fluid in the x direction,
to (u varies with z
.
.
and
t
only)
with
z
VISCOUS EFFECTS: FLUID RESISTANCE
and
241
for steady flow the first equation reduces to
d*u
dp _
" M dz 2
dx
LAMINAR, INCOMPRESSIBLE, STEADY FLOW BETWEEN
PARALLEL PLATES
5.1
The general
case of steady flow between parallel inclined plates
is first
de-
veloped for laminar flow, with the upper plate having a constant velocity
U (Fig. 5.1). Flow between fixed plates is a special case obtained by setting
U =
0.
there
is
In Fig. 5.1 the upper plate moves parallel to the flow direction, and
The flow is analyzed by taking
In steady flow the lamina moves
The equation of motion yields
a pressure variation in the
I
direction.
a thin lamina of unit width as a free body.
at constant velocity u.
pdy
- (pdy +
-£-bl
by\
-
r
hi
+
(t
51
+
j- by dlj
+
y
bl
by sin $
=
Dividing through by the volume of the element and simplifying gives
dr
Jay
=
d
.
(P
J,
dl
+
yh)
(5.1.1)
yblby
sin
6
(r+JftOW
(p+^f
Wh
Flow between inclined parallel plates with the
upper plate in motion.
Fig. 5.1
FUNDAMENTALS OF FLUID MECHANICS
242
which
in
= —dh/dl
sin 6
has been substituted.
Since there
tion in the y direction, the right-hand side of the equation
is
no accelera-
not a function
Integrating with respect to y yields
of y.
t
is
= y-(p
+
+A
yh)
dl
Now,
substitution of Newton's law of viscosity [Eq. (1.1.1)],
du
t
= tidy
for t gives
du
Id
= "
t
ay
j/
\idl
(
+
p
+
yh )y
A
~~
\i
Integrating again with respect to y leads to
u =
—Id— (p + yh)y
2
A
+ -y + B
at
Zfx
ju
which A and
and y =
u =
in
B =
tf
B
are constants of integration.
a,
u = U, and obtain
- ^-|
at
(p
+
7 /i)a
+
2
u =
—~
A
and
T-J, (P
For horizontal
i.e.,
B
+
M ^y -
plates, h
=
0,
2
2/
(5.1.2)
)
C; for no gradient due to pressure or elevation,
For fixed
-f-
plates,
= C and
yh
U =
0,
the velocity has a
and the velocity distribu-
parabolic.
The discharge past a
Eq. (5.1.2) with respect to
Q =
=
—+B
hydrostatic pressure distribution, p
is
them take y
results in
straight-line distribution.
tion
evaluate
n
Zfj,
Eliminating
To
f
udv =
o
In general the
fixed cross section
d
T-lk
maximum
is
obtained by integration of
y:
di
(p
+ yh)a
velocity
°
is
not at the midplane.
(5
'
L3)
VISCOUS EFFECTS: FLUID RESISTANCE
243
p = 20psi
0.24
10
in
ft
p = 12psi
between
Flow
Fig. 5.2
inclined flat plates.
EXAMPLE 5.1 In Fig. 5.2 one plate moves relative to the other as shown.
— 0.80 P; p = 1.7 slugs/ ft Determine the velocity distribution, the dis3
\i
.
and the shear stress exerted on the upper
At the upper point
charge,
p
+
yh
=
(20 lb/in2 ) (144 in2 /ft 2 )
=
3427 lb/ft 2
+
plate.
3
(1.7 slugs/ft ) (32.2 ft/s 2 ) (10 ft)
and at the lower point
p
+
yh
=
12
X
d^jV
_ 1728-
=
3427
lb/ft'
the figure, a
=
(3 ft/s) (y ft)
0.02
=
0.24/12
+
ft
X
Simplifying gives
566?/
-
35,800?/ 2
0.02
ft,
U =
3
(120 lb/ft ) (0.02?/
2
u =
=
10V2ft
dl
u
1728 lb/ft 2
same datum. Hence
to the
From
=
144
ft/s
0.8
P X
-
-3.0
y* ft 2 )
llb-s/ft 2
479
ft/s,
P
and from Eq.
(5.1.2)
FUNDAMENTALS OF FLUID MECHANICS
244
The maximum velocity occurs where du/dy = 0, or y = 0.0079
w max = 2.24 ft/s; the minimum velocity occurs at the upper plate.
The
discharge
and
is
is
0.02
,.0.02
Q =
ft,
udy = 283y 2 - 11,933^
=
0.0177 cfs/ft
•'o
and
downward.
is
To
on the upper
find the shear stress
=
566
-
= -866
71,6002/1
1/— 0.02
plate,
s
-V-0.02
and
r
-,£-^(-866)479
dy
This
is
An
lb/ftf
the fluid shear at the plate; hence the shear force on the plate
lb/ft 2 resisting the
Losses
-1.45
in
motion of the
1.45
is
plate.
laminar flow
expression for the irreversibilities
is
developed for one-dimensional, in-
compressible, steady, laminar flow, in which the equation of motion and the
principle of
work and energy are
in steady flow in a tube,
depth.
The reduction
unit volume,
The
is
in
utilized. There is no increase in kinetic energy
between parallel plates or in film flow at constant
p + yh, which represents work done on the fluid per
converted into irreversibilities by the action of viscous shear.
losses in the length
L
are
Q A(p
+
yh) per unit time.
After examination of the work done on the fluid in one-dimensional
an expression for the losses can be developed. First, the equation of
motion applied to an element (Fig. 5.3) relates the shear stress and change in
p + yh. There is no acceleration; hence 2/* = 0, and
flow,
(p
+
yh)
5y-\p + yh+
^
«*] fy
-
r bx
+
(r
+
j 8y\ hx =
Simplifying gives
djp±y» _
dx
<t
dy
(5 lj4)
.
VISCOUS EFFECTS: FLUID RESISTANCE
245
(r+py)6x
(p
[„.«.«^*..]»
+ yh)6y
Sy
\
6xt8x
Forces on a fluid element.
Fig. 5.3
which implies that the rate of change of p + yh in the x direction must equal
the rate of change of shear in the y direction. Clearly d(p + yh)/dx is independent of y, and dr/dy is independent of x.
The work done per unit time, or power input, to a fluid element (Fig. 5.4)
for one-dimensional flow consists in the power input to the element by p + yh
and by shear stress minus the work per unit time that the element does on
the surrounding fluid, or
(p
+
7fc)
^ + --Jay - |p + 7
(
Power in^
(p
fc
+
5
du by\
by
+
toj
rubx
+
— (ru) by bx — tu bx
dy
—*-tu8x + -j^{tu) 8ydx
+ yh)(u+^^f)8y
[
p + yh
,i<B^U\(u+%%)*y
Power out
tu8x
Fig. 5.4
Work done per
dimensional motion.
unit time
in
a fluid element
in
one-
FUNDAMENTALS OF FLUID MECHANICS
246
Simplifying leads to
—
Net power input
d
Unit volume
= J"
dy
Expanding Eq.
(5.1.5)
TT
..
;
(tu)
d(p
-
u
du
dr
Unit volume
dy
dy
Net power input
:
Unit volume
This power
is
yh)
(5.1.5)
and substituting Eq.
Net power input
With Newton's law
+
ax
d(p
(5.1.4) gives
+
dy
of viscosity,
—
du
=-
= '-T-Ml-r)
r
dy
\dy/
(5- 1 - 7 )
n
m
used up by viscous friction and
is
Integrating the expression over a length
plates,
with Eq. (5.1.2) for
Net power input =
rM
J
I
+
[
Losses
in
=
+
7/1) is
,
dy
= ^Lr
Y
a?L
\
12 M
= —Q
the drop in p
n
-
n
a)
t
\
,
dy
L = Q A(p +
yh)
dl
+
yh in the length L. The expression for
Distribution of velocity, shear,
per unit volume for a round tube.
,
(2y
yields
Energy dissipation
Fig. 5.5
fixed parallel
(5.1.7), gives
r[H(p —
+ yh)
U =
irreversibilities.
L between two
J
Eq. (5.1.3) for
net power input
which A(p
yh)
LT
converted into
and with Eq.
2
)
dl
Q from
U =
/M
—
_ t d(p
Substituting for
du
yh)
dx
and losses
VISCOUS EFFECTS: FLUID RESISTANCE
power input per unit volume [Eq. (5.1.7)]
in a tube.
The
irreversibilities are greatest
tribution of shear stress, velocity,
Fig. 5.5 for a
EXAMPLE
5.2
and
is
247
also applicable to laminar flow
when du/dy
is
losses per unit
greatest.
volume
is
The disshown in
round tube.
A
conveyor-belt device, illustrated in Fig. 5.6,
is
and used to pick up undesirable surface contaminants,
of the sea. Assume the oil film to be thick enough
surface
the
a ship
to be unlimited with respect to the operation of the device.
operate at a steady velocity
mounted on
e.g., oil,
from
for the supply
Assume the belt
to
U and to be long enough for a uniform flow depth
Determine the rate at which oil can be carried up the belt per unit
width, in terms of 0, U, and the oil properties m and y.
A thin lamina of unit width that moves at velocity u is shown in Fig.
5.6. With the free surface as shown on the belt, and for steady flow at constant depth, the end-pressure effects on the lamina cancel. The equation of
motion applied to the element yields
to exist.
-
(t
+ y8yj5l +
When
r
= 7
rdl
- ydy dl sin 6 =
(a
—
-— =
— 7 sin
dy
the shear stress at the surface
sin
dr
or
is
recognized as zero, integration yields
y)
This equation can be combined with Newton's law of viscosity, r
to give
I
du =
Fig. 5.6
/
(a
-
Oil-pickup device.
y)
dy
= —
\x
du/dy,
FUNDAMENTALS OF FLUID MECHANICS
248
or
The
q
=
dn
7 sin
U
u =
(
tf\
flow rate per unit width
7
udy = Ua
I
sin
up the
belt can be determined
by
integration:
a3
Jn
This expression shows the flow rate to vary with a. However, a is still a dependent variable that is not uniquely defined by the above equations. The
actual depth of flow on the belt is controlled by the end conditions.
The depth for maximum flow rate can be obtained by setting the derivative dq/da to zero and solving for the particular a
a
=
.
a
— —y
u»
= (
;
I
\7
To
J
sin 0/
attach some physical significance to this particular depth the in-
may
be considered. If the crest depth A,
then the maximum flow for that
belt velocity and slope will be achieved. If A is physically controlled at a
depth greater than a, more flow will temporarily be supplied by the belt than
can get away at the crest, causing the belt depth to increase and the flow to
decrease correspondingly, until either an equilibrium condition is realized
or A is lowered. Alternatively if A < a, flow off the belt will be less than the
maximum flow up the belt at depth a and the crest depth will increase to a.
At all times it is assumed that an unlimited supply is available at the bottom.
By this reasoning it is seen that a is the only physical flow depth that can exist
on the belt if the crest depth is free to seek its own level. A similar reasoning
fluence of alternative crest depths
Fig. 5.6, is
such that a occurs on the
at the base leads to the
The
V2
sin 0/
1/2
sin 0/
_
I«5L? (
3m
or
\7
conclusion.
discharge, as a function of fluid properties
u (J^J)
\7
same
belt,
9
,'
u»
Y
\7 sin 0/
/2
and
U and 0,
is
given by
VISCOUS EFFECTS: FLUID RESISTANCE
249
LAMINAR FLOW THROUGH CIRCULAR TUBES AND
5.2
CIRCULAR ANNULI
For steady, incompressible, laminar flow through a circular tube or an anis taken as a free body, and
the equation of motion is applied in the I direction, with acceleration equal to
nulus, a cylindrical infinitesimal sleeve (Fig. 5.7)
zero.
From
2tjt 8r
-(
p
the figure,
(
2tit 8r
p
+
2irr Sr
Replacing sin
2irr 8r 81,
!fr +
by
P) +
-7 dl)
\2wr
— dh/dl
81
r
2tjt 81 r
+—
(2irr 81 r) 8r
+
and dividing by the volume
7 27rr
6r 81 sin
of the free body,
gives
7»>+~<">-
Since d(p
+
yh) /dl
is
(5.2.1)
not a function of
r,
the equation
may
be multiplied
y2irr6r8l
Free-body diagram of cylindrical sleeve element for
laminar flow in an inclined circular tube.
Fig. 5.7
=
FUNDAMENTALS OF FLUID MECHANICS
250
by
and integrated with respect to
r 8r
r
(p
iii
+
yh)
+
TT
yielding
r,
= A
(
which A is the constant
must be satisfied when r =
in
For a circular tube
of integration.
0;
hence
A =
for this case.
5 2 2)
-
-
this equation
Substituting
du
dr
note that the minus sign
(u
5.7.
du =
is
is
required to obtain the sign of the r term in Fig.
considered to decrease with
hence du/dr
r,
is
negative.)
AdT
M dr
A
— d— {p J+ yh)r
1
(
2fx
dl
fi
r
Another integration gives
u
+
^T-^Av
4ju at
For the annular
and u =
yh)
--\nr + B
case, to evaluate
when
radius,
r
=
A and
4:fldl
and
Q -
+ T« ^ \
(
for discharge
^
2 Tra d r
= when r = b, the inner tube
When A and B are eliminated,
B, u
a (Fig. 5.8).
/
(p
"
(5.2.3)
/x
r2
+
———
a2
b2
In
In b/a
a\
-
(5.2.4)
r/)
through an annulus (Fig. 5.8),
= -
J | (p + ,»)
[«•
-^-
(^]
(5.2.5)
Circular tube; Hagen-Poiseuille equation
For the circular tube,
A =
u= -"-^jAp +
at
4/z
yh)
in Eq. (5.2.3)
and u =
for r
=
a,
(5.2.6)
VISCOUS EFFECTS: FLUID RESISTANCE
251
Flow through an annulus.
Fig. 5.8
The maximum
velocity w ma x
given for
is
r
=
as
4/x at
Since the velocity distribution
volume
velocity
v
is
is
one-half that of
one-half of the
its
is
a paraboloid of revolution (Fig. 5.5),
circumscribing cylinder; therefore the average
maximum
velocity,
--£*<>++>
The discharge Q
is
(5 2 8)
-
equal to
Vwa2
area,
Q =
also
-
,
«--£!<>+*>
The discharge can
its
(5 2 9)
-
-
be obtained by integration of the velocity u over the
i.e.,
f°
I
2wru dr
''a
For a horizontal tube, h
=
const; writing the pressure drop
Ap
in the length
FUNDAMENTALS OF FLUID MECHANICS
252
L
gives
Ap
dp
=
~
~L
dl
and substituting diameter
—
D leads to
ApirD*
„
Q =
(5.2.10a)
In terms of average velocity,
ApD
2
Equation (5.2.10a) can then be solved
for pressure drop,
which repre-
sents losses per unit volume,
—T^
*V =
The
(5.2.11)
losses are seen to
vary directly as the
viscosity, the length,
and the
charge and to vary inversely as the fourth power of the diameter.
dis-
It should
be noted that tube roughness does not enter into the equations. Equation
(5.2.10a) is known as the Hagen-Poiseuille equation; it was determined experimentally by Hagen in 1839 and independently by Poiseuille in 1840.
The
was made by Wiedemann in 1856.
by Eqs. (5.2.1) to (5.2.10) are not valid near the
the flow enters the pipe from a reservoir through a well-
analytical derivation
The
results as given
entrance of a pipe. If
rounded entrance, the velocity at
tion.
The
to slow
is
velocity
first is
almost uniform over the cross sec-
must be zero at the wall)
As a consequence of continuity the
the central region. The transition length
action of wall shear stress (as the velocity
down
the fluid near the wall.
must then increase
in
U
for the characteristic parabolic velocity distribution to develop
of the
—
is
a function
Reynolds number. Langhaar developed the theoretical formula
1
= 0.058R
which agrees well with observation.
1
H. L. Langhaar, Steady Flow in the Transition Length of a Straight Tube, J. Appl.
Mech., vol.
9,
pp. 55-58, 1942.
VISCOUS EFFECTS: FLUID RESISTANCE
253
p 2 =3kg f/cm2
Flow through an inclined tube.
Fig. 5.9
EXAMPLE
Determine the direction of flow through the tube shown in
which y = 8000 N/m 3 and /z = 0.04 kg/m-s. Find the quantity
liters per second, and calculate the Reynolds number for the flow.
5.3
Fig. 5.9, in
flowing in
At
p
+
yh
section
1
=
(2
=
236.12
k g/ /cm2 ) (9.806 N/kg
<>
kN/m
(3?) +
(8000
N/m
3
)
(5
m)
2
and at section 2
=
p
+
if
the elevation
yh
(3
k g/ /cm 2 ) (9.806 N/kg/)
datum
since the energy
tion)
d
-T
dl
7
than at
/
(p
is
\
/
-
294.18
(5806
the quantity flowing, the expression
——-m
294,180 XT
N/m ,
2
_
5806
AT
N/m
,
10
N/m
)
(x) (0.01
m)«
•
,
7
=
8X0.04 kg/m-s
The average
7=
3
^ ^,
velocity
0.00057
0/4)
m /s
(0.01
is
3
m)
=
;
2
2.
The flow is from 2 to 1
must be the same at each sec-
Substituting into Eq. (5.2.9) gives
V =
kN/m
taken through section
236,120
=
+
)
lm
greater at 2 (kinetic energy
To determine
1.
n
+ yh)
is
Cm
10 °
(
7.26
m/s
0.00057
m /s
3
7
=
0.57 1/s
,
3
is
written
FUNDAMENTALS OF FLUID MECHANICS
254
and the Reynolds number
VDp =
(Sec. 4.4) is
m) (8000 N/m3 )
kg/m-s) (9.806 m/s 2 )
(1.814 m/s) (0.01
(0.04
M
If the Reynolds number had been above 2000, the Hagen-Poiseuille equation
would no longer apply, as discussed in Sec. 5.3.
The
kinetic-energy correction factor a [Eq. (3.10.2)] can be determined
for laminar flow in a tube
V
W m ax
L
by use
of Eqs. (5.2.6)
and
(5.2.7),
W
(5.2.12)
Substituting into the expression for a gives
2irrdr
There
is
twice as
much
=
2
(5.2.13)
energy in the flow as in uniform flow at the same
average velocity.
5.3
THE REYNOLDS NUMBER
Laminar flow is defined as flow in which the fluid moves in layers, or laminas,
one layer gliding smoothly over an adjacent layer with only a molecular
interchange of momentum. Any tendencies toward instability and turbulence
are damped out by viscous shear forces that resist relative motion of adjacent
fluid layers. Turbulent flow, however, has very erratic motion of fluid parThe nature of
ticles, with a violent transverse interchange of momentum.
the flow, i.e., whether laminar or turbulent, and its relative position along a
scale indicating the relative importance of turbulent to laminar tendencies
are indicated by the Reynolds number. The concept of the Reynolds number
and its interpretation are discussed in this section. In Sec. 3.5 an equation
of motion was developed with the assumption that the fluid is frictionless,
i.e., that the viscosity is zero.
More general equations have been developed
that include viscosity, by including shear stresses. These equations (see introduction to this chapter) are complicated, nonlinear, partial differential
equations for which no general solution has been obtained. In the last century
VISCOUS EFFECTS: FLUID RESISTANCE
Fig. 5.10
Reynolds apparatus.
Osborne Reynolds
1
studied these equations to try to determine
different flow situations
Two
1.
They
255
would be
when two
similar.
flow cases are said to be dynamically similar
are geometrically similar,
i.e.,
when
corresponding linear dimensions
have a constant ratio and
2.
The corresponding
streamlines are geometrically similar, or pressures at
corresponding points have a constant ratio.
In considering two geometrically similar flow situations, Reynolds deduced that they would be dynamically similar if the general differential equations describing their flow were identical. By changing the units of mass,
length, and time in one set of equations and determining the condition that
must be satisfied to make them identical to the original equations, Reynolds
found that the dimensionless group ulp/n must be the same for both cases.
Of these, u is a characteristic velocity, I a characteristic length, p the mass
density, and ju the viscosity.
This group, or parameter, is now called the
Reynolds number R,
R =
ulp
(5.3.1)
To determine the significance of the dimensionless group, Reynolds
conducted his experiments on flow of water through glass tubes, illustrated in
Fig. 5.10.
A glass tube was mounted horizontally with one end in a tank
and a valve on the opposite end. A smooth bellmouth entrance was attached
to the upstream end, with a dye jet arranged so that a fine stream of dye could
1
0. Reynolds, An Experimental Investigation of the Circumstances Which Determine
Whether the Motion of Water Shall Be Direct or Sinuous, and of the Laws of Resistance
in Parallel Channels, Trans. R. Soc. Lond., vol. 174, 1883.
FUNDAMENTALS OF FLUID MECHANICS
256
be ejected at any point in front of the bellmouth. Reynolds took the average
velocity V as characteristic velocity and the diameter of tube D as characteristic length, so
that
R = VDp/p.
For small flows the dye stream moved as a straight line through the
was laminar. As the flow rate increased, the Reynolds number increased, since D, p, p were constant, and V was directly protube, showing that the flow
portional to the rate of flow.
With increasing discharge a condition was
reached at which the dye stream wavered and then suddenly broke up and
was diffused throughout the tube. The flow had changed to turbulent flow
with
its
orderly
violent interchange of
movement
able to obtain a value
tor,
momentum that had completely disrupted the
By careful manipulation Reynolds was
of laminar flow.
R =
12,000 before turbulence set
in.
A later investiga-
using Reynolds' original equipment, obtained a value of 40,000
by
allow-
ing the water to stand in the tank for several days before the experiment and
by taking precautions
to avoid vibration of the water or equipment. These
numbers, referred to as the Reynolds upper critical numbers, have no practical
significance in that the ordinary pipe installation has irregularities that cause
turbulent flow at a much smaller value of the Reynolds number.
Starting with turbulent flow in the glass tube, Reynolds found that it
always becomes laminar when the velocity is reduced to make R less than
2000. This is the Reynolds lower critical number for pipe flow and is of practical
importance. With the usual piping installation, the flow will change from
laminar to turbulent in the range of the Reynolds numbers from 2000 to 4000.
For the purpose of this treatment it is assumed that the change occurs at
R = 2000. In laminar flow the losses are directly proportional to the average
velocity, while in turbulent flow the losses are proportional to the velocity to
a power varying from 1.7 to
2.0.
There are many Reynolds numbers in use today in addition to the one
for straight round tubes.
For example, the motion of a sphere through a
fluid may be characterized by UDp/p, in which U is the velocity of sphere,
D is the diameter of sphere, and p and p. are the fluid density and viscosity.
The Reynolds number may be viewed as a ratio of shear stress r due to
turbulence 'to shear stress r„ due to viscosity. By applying the momentum
equation to the flow through an element of area 8 A (Fig. 5.11) the apparent
shear stress due to turbulence can be determined. If v' is the velocity normal
to 8A and u' is the difference in velocity, or the velocity fluctuation, on the
two sides of the area, then, with Eq. (3.11.1) the shear force 8F acting is computed to be
t
,
8F =
in
pv'
which
8A
u'
pv' 8
A
is
the mass per second having
its
momentum changed and
VISCOUS EFFECTS: FLUID RESISTANCE
257
»
iV
Fig. 5.11
Notation
shear stress due to
bulent flov/.
u
f
is
for
tur-
the final velocity minus the
initial velocity in
the
t
T
t
=
=
is
(5.3.2)
stress
due to viscosity
may
be written
*y-
(5.3.3)
which u' is interpreted as the change in velocity in the distance
normal to the velocity. Then the ratio
in
T
t
TV
_
dividing
obtained,
pu'v'
The shear
tv
By
s direction.
through by dA, the shear stress r due to turbulent fluctuations
I,
measured
V'lp
M
has the form of a Reynolds number.
The nature of a given flow of an incompressible fluid is characterized by
Reynolds number. For large values of R one or all of the terms in the numerator are large compared with the denominator. This implies a large expanse of fluid, high velocity, great density, extremely small viscosity, or combinations of these extremes.
The numerator terms are related to inertial
its
forces, or to forces set
up by acceleration
or deceleration of the fluid.
The
de-
nominator term is the cause of viscous shear forces. Thus, the Reynolds number parameter may also be considered as a ratio of inertial to viscous forces.
A large R indicates a highly turbulent flow with losses proportional to the
square of the velocity.
many small
The turbulence may be fine
scale,
composed
of a great
eddies that rapidly convert mechanical energy into irreversibilities
through viscous action; or
it
may
be
large scale, like the
huge vortices and
FUNDAMENTALS OF FLUID MECHANICS
258
swirls in a river or gusts in the atmosphere.
The
large eddies generate smaller
which in turn create fine-scale turbulence. Turbulent flow may be
thought of as a smooth, possibly uniform flow, with a secondary flow superposed on it. A fine-scale turbulent flow has small fluctuations in velocity that
eddies,
The root-mean-square value of the fluctuations
change of sign of the fluctuations are quantitative
measures of turbulence. In general the intensity of turbulence increases as
the Reynolds number increases.
For intermediate values of R both viscous and inertial effects are important, and changes in viscosity change the velocity distribution and the
occur with high frequency.
and the frequency
of
resistance to flow.
For the same R, two geometrically similar closed-conduit systems (one,
have the same ratio of losses to velocity
head. Use of the Reynolds number provides a means for using experimental
results with one fluid for predicting results in a similar case with another fluid.
In addition to the applications of laminar flow shown in this and the
preceding section, the results may also apply to greatly different situations,
because the equations describing the cases are analogous. As an example the
two-dimensional laminar flow between closely spaced plates is called HeleShaw flow. 1 If some of the space is filled between the plates, by use of dye in
the fluid the streamlines for flow around the obstructions are made visible.
These streamlines in laminar flow are the same as the streamlines for similar
flow of a frictionless (irrotational) fluid around the same obstructions. Likewise the two-dimensional frictionless flow cases (Chap. 7) are analogous and
similar to two-dimensional percolation through porous media.
say, twice the size of the other) will
5.4
PRANDTL MIXING LENGTH; VELOCITY DISTRIBUTION
TURBULENT FLOW
IN
Pressure drop and velocity distribution for several cases of laminar flow were
worked out
in the preceding sections. In this section the mixing-length theory
of turbulence
is
developed, including
The apparent shear
r
=
H.
(m
+
J. S.
n)
-r
its
application to several flow situations.
stress in turbulent flow
is
expressed
by [Eq.
(3.3.2)]
(5-4.1)
Hele-Shaw, Investigation of the Nature of the Surface Resistance of Water and
Motion under Certain Experimental Conditions, Trans. Inst. Nav. Archil.,
of Streamline
vol. 40, 1898.
VISCOUS EFFECTS: FLUID RESISTANCE
259
(b)
(a)
Notation for mixing-length theory.
Fig. 5.12
including direct viscous effects.
Prandtl 1 has developed a most useful theory
of turbulence called the mixing-length theory.
In Sec. 5.3 the shear
stress, r,
due to turbulence, was shown to be
Tt
in
puv
which
(5.3.2)
u', v'
are the velocity fluctuations at a point.
In Prandtl's2 theory,
expressions for u' and v' are obtained in terms of a mixing-length distance I
and the velocity gradient du/dy, in which u is the temporal mean velocity
at a point and y is the distance normal to u, usually measured from the boundary. In a gas, one molecule, before striking another, travels an average distance known as the mean free path of the gas. Using this as an analogy (Fig.
5.12a), Prandtl assumed that a particle of fluid is displaced a distance I before
its momentum is changed by the new environment. The fluctuation u' is then
related to
u
,
I
by
~l —
du
dy
which means that the amount of the change in velocity depends upon the
changes in temporal mean velocity at two points distant I apart in the y direction. From the continuity equation, he reasoned that there must be a corre1
2
For an account of the development of turbulence theory the reader is referred to L.
Prandtl, "Essentials of Fluid Dynamics," pp. 105-145, Hafner, New York, 1952.
L. Prandtl, Bericht iiber Untersuchungen zur ausgebildeten Turbulenz, Z. Angew. Math.
Mech., vol.
5,
no. 2, p. 136, 1925.
FUNDAMENTALS OF FLUID MECHANICS
260
between
lation
,
v'
~u ~
f
By
i
I
u'
and
v'
(Fig. 5.126), so that v'
is
proportional to u
f
,
du
-7-
dy
substituting for u' and
v'
in Eq. (5.3.2)
and by
letting
portionality factor, the denning equation for mixing length
I
is
absorb the proobtained
-©*
(5.4.2)
always acts in the sense that causes the velocity distribution to become
When Eq. (5.4.2) is compared with Eq. (3.3.1), it is found
r
more uniform.
that
=
v
But
-
du
Pl
1)
2
(5.4.3)
ay
is
not a fluid property like dynamic viscosity; instead
the density, the velocity gradient, and the mixing length
I.
depends upon
In turbulent flow
77
there is a violent interchange of globules of fluid except at a boundary, or
very near to it, where this interchange is reduced to zero; hence, I must approach zero at a fluid boundary. The particular relationship of I to wall distance y is not given by Prandtl's derivation. Von Karman 1 suggested, after
considering similitude relationships in a turbulent fluid, that
d 2 u/dy 2
in
which
k is
a universal constant in turbulent flow, regardless of the boundary
configuration or value of the Reynolds number.
sometimes referred to as the eddy viscosity, is genmay be considered as a coefficient of momentum
transfer, expressing the transfer of momentum from points where the concentration is high to points where it is lower. It is convenient to utilize a
kinematic eddy viscosity e = 77/p which is a property of the flow alone and is
In turbulent flows,
erally
much
larger than
77,
/x.
It
analogous to kinematic viscosity.
1
T. von K&rm£n, Turbulence and Skin Friction, J. Aeronaut.
Sci., vol. 1, no. 1, p. 1, 1934.
VISCOUS EFFECTS: FLUID RESISTANCE
261
Velocity distributions
The mixing-length concept
is
used to discuss turbulent velocity distributions
and the pipe. For turbulent flow over a smooth plane surface (such as the wind blowing over smooth ground) the shear stress in the
Equation (5.4.1) is applicable, but approaches zero
fluid is constant, say r
at the surface and \x becomes unimportant away from the surface. If
is negligible for the film thickness y = 5, in which \x predominates, Eq. (5.4.1)
becomes
for the flat plate
77
.
77
To
fJL
U
U
p
py
y
The term
<
\A /p
has the dimensions of a velocity and
stress velocity u*.
^=^
u*
(5.4.5)
b
is
called the shear-
Hence
y<6
(5.4.6)
v
shows a linear relation between u and y in the laminar
neglected, and Eq. (5.4.1) produces
film.
For y
>
-"(MS
Since
I
=
ay.
is
has the dimensions of a length and from dimensional consideration
significant linear dimension) assume
,
Substituting into Eq. (5.4.7) and rearranging gives
^
= i^
w*
k
/jl
(5.4.7)
would be proportional to y (the only
I
8,
(5.4.8)
y
and integration leads to
—u = - In y + const
1
w*
It is to
I
be noted that this value of u substituted in Eq.
proportional to y (d 2 u/dy 2
as y increases).
is
(5.4.9)
K
also useful
Equation
when
r
is
is
(5.4.4) also
determines
negative since the velocity gradient decreases
with experiment and, in fact,
because most of the velocity change
(5.4.9) agrees well
a function of
y,
FUNDAMENTALS OF FLUID MECHANICS
262
occurs near the wall where r
is
substantially constant.
It is quite satisfactory
to apply the equation to turbulent flow in pipes.
EXAMPLE
By
5.4
When
—u = Um
w*
=
y
w*
-y
k
r
The
discharge
Virr
2
The
f
=
2?r
r
1
h - In
integration of Eq. (5.4.9) find the relation between the
V
average velocity
,
Vwr
and the maximum velocity u m
u = u m so that
,
2
is
obtained by integrating the velocity over the area,
r °~ s
r
ur dr
/
= 2w
f
°
(
(
u±
um H
In
integration cannot be carried out to y
so small that
1
r-a/
in
2
=
(r
—
y) dy
0, since
the equation holds in
flowing in the laminar zone
+ ainiYi-lYil
which the variable
7=
of integration
is
y/r
Integrating gives
.
Lp_I(iYl + -fi!^-^-I(iYin i + I^Y]l
4 Vo/
2 \r /
2 Vo/
lr
I
Since 5/r
(lim x In x
T7
(,,
it
-y\)
The volume per second
may be neglected. Then
the turbulent zone only.
is
in turbulent flow in a pipe.
7 = u™ ~
L^o
is
=
r
k
J
r
very small, such terms as 5/r and 5/r In
0) thus
r
(5/r
)
JJ
become
1
5 /r
negligible
;
3 w*
o
2 k
—
wm
— V
or
w*
=
3
5"
2/c
In evaluating the constant in Eq. (5.4.9) following the methods of
Bakhmeteff, 1 u = u w the wall velocity, when y = 5. According to Eq. (5.4.6),
,
- = ^ = AT
U
1
(5.4.10)
B. A. Bakhmeteff, "The Mechanics of Turbulent Flow," Princeton University
Princeton, N.J., 1941.
VISCOUS EFFECTS: FLUID RESISTANCE
from which
it is
reasoned that u*b/v should have a
flow changes from laminar to turbulent, since
Substituting u
= uw when
y
=
h
into Eq.
it is
critical
value
263
N at which
a Reynolds number in form.
(5.4.9)
and using Eq.
(5.4.10)
yields
uw
— = JV = - In
1
U*
5
+
const
K
—+
1
Nv
K
U*
= - In
const
Eliminating the constant gives
u
— = - In yu* N
U*
1
1
.
In
\-
K
N
K
V
or
^=
I ln
U*
K
^+4
(5 . 4 . n)
V
A =N —
been found experimentally by plotting u/u*
= 0.417, A = 5.84, but for smooth-wall
pipes NikuradseV experiments yield k = 0.40 and A = 5.5.
Prandtl has developed a convenient exponential velocity-distribution
in
which
against In yu*/v.
(1/k) In iV has
For
flat plates k
formula for turbulent pipe flow,
n
= /y\
u_
um
in
(5.4.12)
\r /
which n varies with the Reynolds number. This empirical equation is valid
For R less than 100,000, n = y, and for
only at some distance from the wall.
greater values of R,
(5.4.11)
and
n
(5.4.12),
decreases.
The
velocity-distribution equations, Eqs.
both have the fault of a nonzero value of du/dy at the
center of the pipe.
EXAMPLE
Find an approximate expression for mixing-length distribufrom PrandtPs one-seventh-power law.
Writing a force balance for steady flow in a round tube (Fig. 5.13) gives
5.5
tion in turbulent flow in a pipe
dp
r
T
dl2
1
J.
Nikuradse, Gesetzmassigkeiten der turbulenten Stromung in glatten Rohren, Ver.
Dtsch. Ing. Forschungsh., vol. 356, 1932.
FUNDAMENTALS OF FLUID MECHANICS
264
+w.^w;^^^^^^^^
lirrblr
ynr
(p+jft8l)irr
1
2
mZ^^^^^^^^.'^^^EMZZ^^M^^ffi^^^^EM^ffl^^^^*
81
Free-body diagram for steady flow through a round tube.
Fig. 5.13
At the wall
dp
r
TO
dT 2
hence
\
r
Solving for
w*
I
r
/
\dy/
gives
Vi -
V/ro
du/dy
From Eq.
u_
Um
=
(5.4.12)
/yy
VO/
the approximate velocity gradient
dy
r
7
Vo/
n
is
obtained,
um
Vo/
x
r
The dimensionless velocity deficiency, (u m — u)/u*, is a function of y/r
only for large Reynolds numbers (Example 5.4) whether the pipe surface is
smooth or rough. From Eq. (5.4.9), evaluating the constant for u = u m when
VISCOUS EFFECTS: FLUID RESISTANCE
=
y
265
gives
r
M
=
il„^-
w*
k
(5.4.13)
y
For rough pipes, the velocity may be assumed to be u w at the wall dis— me, in which e' is a typical height of the roughness projections
and m is a form coefficient depending upon the nature of the roughness. Substituting into Eq. (5.4.13) and eliminating u m /u* between the two equations
tance y w
leads to
— = - In — H
u*
in
k
lnm
u*
e
which the
(5.4.14)
K
two terms on the right-hand
last
side are constant for a given
type of roughness,
-=
-\n^
f
+B
(5.4.15)
In Nikuradse's experiments with sand-roughened pipes constant-size sand
particles (those passing a given screen
and being retained on a
screen) were glued to the inside pipe walls.
sand grains, experiment shows that
k
=
If
0.40,
slightly finer
represents the diameter of
e'
B =
8.48.
RATE PROCESSES
5.5
The
violent interchange of fluid globules in turbulence also tends to transfer
any uneven concentration within the fluid, such as salinity, temperature,
dye coloring, or sediment concentration. Studies indicate that the transfer
coefficient is roughly proportional to, but probably larger than, the eddy viscosity for turbulent diffusions of concentrations other than momentum.
1
If
and
cp
T
is
the temperature,
per unit of mass)
H
H the heat transfer per unit area per unit time,
the specific heat at constant pressure
= -c P v
—=
dy
,
(e.g.,
Btu per unit
of temperature
then
-c pP
2
l
——
(5.5.1)
dy dy
which c p 7j is the eddy conductivity. For transfer of material substances,
such as salinity, dye, or sediment, if C is the concentration per unit volume
in
1
See footnote
1, p.
259.
FUNDAMENTALS OF FLUID MECHANICS
266
pounds
(e.g.,
of salt per cubic foot,
number
of particles per cubic foot)
the rate of transfer per unit area per unit time
c
number
foot per second,
(e.g.,
pounds
and
of salt per square
of sediment particles per square foot per second),
then
= -e
c
c
dC
—
(5.5.2)
dy
and
e c is
proportional to
EXAMPLE
5.6
A tank
e.
agitated so that the kinematic eddy viscosity
If
the
fall
particles
uniform size
be considered constant.
of liquid containing fine solid particles of
is
may
still liquid is v f and the concentration of
measured from the bottom) find the distribution
velocity of the particles in
=
Co at y
is
yo (y
,
throughout the liquid.
By using Eq. (5.5.2) to determine the rate per second carried upward by
turbulence per unit of area at the level y, the amount per second falling across
this surface by settling is equated to it for steady conditions. Those particles
in the height v f above the unit area will fall out in a second, that is, Cvf particles cross the level downward per second per unit area. From Eq. (5.5.2)
of solid particles vertically
—e
c
dC/dy
particles are carried
upward due
to the turbulence
and the higher
concentration below; hence
r f
Cv
dC
= — e —-
dC
or
c
C
dy
=
dy
Integrating gives
In
C =
For
C =
Vf
Co,
y
+
const
y
=
y
--
,
-
C =
Co exp
5.6
BOUNDARY-LAYER CONCEPTS
(y
y
)
In 1904 Prandtl developed the concept of the boundary layer. It provides
an important link between ideal-fluid flow and real-fluid flow. For fluids
1
1
L. Prandtl, Uber Flussigkeitsbewegung bei sehr kleiner Reibung, Verh. Ill Int. Math.Kongr., Heidelb, 1904.
VISCOUS EFFECTS: FLUID RESISTANCE
having relatively small viscosity, the
effect of internal friction
a narrow region surrounding
ciable only in
in a fluid
the fluid boundaries.
is
267
appre-
From
this
hypothesis, the flow outside of the narrow region near the solid boundaries
may be
considered as ideal flow or potential flow. Relations within the bound-
ary-layer region can be
computed from the general equations
for viscous fluids,
but use of the momentum equation permits the developing of approximate
equations for boundary-layer growth and drag. In this section the boundary
layer
is
described and the
flow along a flat plate
is
momentum equation applied to it. Two-dimensional
by means of the momentum relationships for
studied
both the laminar and the turbulent boundary layer. The phenomenon of
separation of the boundary layer and formation of the wake is described.
Description of the boundary layer
When motion
is
started in a fluid having very small viscosity, the flow
is es-
Since the fluid at the
sentially irrotational (Sec. 3.3) in the first instants.
boundaries has zero velocity relative to the boundaries, there
is
a steep ve-
from the boundary into the flow. This velocity gradient in a
real fluid sets up near the boundary shear forces that reduce the flow relative
to the boundary. That fluid layer which has had its velocity affected by the
boundary shear is called the boundary layer. The velocity in the boundary
layer approaches the velocity in the main flow asymptotically. The boundary
layer is very thin at the upstream end of a streamlined body at rest in an
otherwise uniform flow. As this layer moves along the body, the continual
locity gradient
action of shear stress tends to slow
down
additional fluid particles, causing
the thickness of the boundary layer to increase with distance from the up-
stream point. The fluid in the layer is also subjected to a pressure gradient,
determined from the potential flow, that increases the momentum of the layer
if
the pressure decreases downstream and decreases
sure increases
downstream
its
(adverse pressure gradient)
momentum if the
.
The
pres-
flow outside the
boundary layer may also bring momentum into the layer.
For smooth upstream boundaries the boundary layer starts out as a
laminar boundary layer in which the fluid particles move in smooth layers.
As the thickness of the laminar boundary layer increases, it becomes unstable
and finally transforms into a turbulent boundary layer in which the fluid particles move in haphazard paths, although their velocity has been reduced by
the action of viscosity at the boundary. When the boundary layer has become
turbulent, there is still a very thin layer next to the boundary that has laminar
motion.
It is called the laminar sublayer.
Various definitions of boundary-layer thickness
The most
slowing
5
have been suggested.
main flow due to
basic definition refers to the displacement of the
down
of fluid particles in the
boundary zone. This thickness
5i,
called
FUNDAMENTALS OF FLUID MECHANICS
268
Fig.
Definitions
5.14
of
boundary-layer thickness.
the displacement thickness,
USi
= f (U -
is
expressed
by
u) dy
(5.6.1)
Jn
in
which
6 is
that value of y at which u = U in the undisturbed flow. In
= di is drawn so that the shaded areas are equal. This
Fig. 5.14a, the line y
distance is, in itself, not the distance that is strongly affected by the boundary
but is the amount the main flow must be shifted away from the boundary.
In fact, that region is frequently taken as 35i. Another definition, expressed
by Fig. 5.146, is the distance to the point where u/U = 0.99.
Momentum
By
equation applied to the boundary layer
von Kantian' s method the principle of momentum can be applied directly to the boundary layer in steady flow along a flat plate. In Fig.
5.15 a control volume is taken enclosing the fluid above the plate, as shown,
1
following
In the y direction it extends a disundisturbed in the x direction, although
some flow occurs along the upper surface, leaving the control volume.
extending the distance x along the plate.
tance h so great that the velocity
The momentum equation
2FX = -
J
dt J „„
It will
1
pudV
+
puV
J
J
•
is
for the x direction
is
dA
,a
be applied to the case of incompressible steady flow.
The only
force
T. von K&rm&n, On Laminar and Turbulent Friction, Z. Angew. Math. Mech., vol.
pp. 235-236, 1921.
1,
VISCOUS EFFECTS: FLUID RESISTANCE
269
Control surface
Fig. 5.15 Control volume applied to fluid flowing
over one side of a flat plate.
acting is due to drag or shear at the plate, since the pressure is constant
around the periphery of the control volume. For unit widths of plate normal
to the paper,
Drag =
u 2 dy
p f
- P U h + Up
2
f
(U -
u) dy
Jn
Jn
term on the right-hand side of the equation is the efflux of x momenthe second term is the ^-momentum influx through AB.
The integral in the third term is the net volume influx through AB and CD
which, by continuity, must just equal the volume efflux through BC. It is
multiplied by Up to yield ^-momentum efflux through BC. Combining the
The
first
tum from CD, and
integrals gives
Drag =
u(U -
p f
u) dy
(5.6.2)
•'o
The drag D(x) on the
D(x) =
p f
plate
is
in the reverse direction, so that
u{U -u)dy
The drag on the
plate
may
(5.6.3)
also be expressed as
an integral of the shear
stress
along the plate,
D(x) =
f
r
Equating the
dx
last
(5.6.4)
two expressions and then
differentiating with respect to
FUNDAMENTALS OF FLUID MECHANICS
270
x leads to
to
=
d
—
dx
p
which
is
h
f
/
u(U —
u) dy
(5.6.5)
JQ
the
momentum
equation for two-dimensional flow along a
flat plate.
Calculations of boundary-layer growth, in general, are complex and
The parallel-flow cases, laminar
be worked out approximately by use of
require advanced mathematical treatment.
or turbulent, along a flat plate
momentum methods
may
that do not give any detail regarding the velocity dis-
—in fact a velocity distribution must be assumed.
The results can
be shown to agree closely with the more exact approach obtained from general
tribution
viscous-flow differential equations.
For an assumed distribution which satisfies the boundary conditions
and u = U, y = 5, the boundary-layer thickness as well as the
y =
shear at the boundary can be determined. The velocity distribution is assumed
to have the same form at each value of x,
u =
5
0,
= F (l)
when
y
5 is
F(v)
unknown.
= F = \t\~\
T
which
satisfy the
TTO
dd
1
f
/
v
=
\
For the laminar boundary layer Prandtl assumed that
0<y <8
and
boundary conditions.
u\u
F =
Equation
8
1
(5.6.5)
<
y
may
be rewritten
1
and
At the boundary
du
dy
UdF
5
dt)
M)L
=| "?
(5 6 6)
-
-
VISCOUS EFFECTS: FLUID RESISTANCE
Equating the two expressions
271
for r yields
—
dx
= 0.139 P [/2
fju2
b
and rearranging gives
db
b
=
since 5
b
is
If 5
in
pU
a function of x only in this equation.
2
- =
u dx
*—-
10.78
Integrating gives
v
10.78
=
—x+
0, for
=
x
const
0,
the constant of integration
which Rx = Ux/v
is
a Reynolds
leading edge of the plate.
laminar flow shows that
b
is
zero. Solving for b/x leads to
number based on the distance x from the
This equation for boundary-layer thickness in
increases as the square root of the distance from the
leading edge.
Substituting the value of
to
-
5
into Eq. (5.6.6) yields
pV
0.322
The shear
i
\/^-
(5.6.8)
stress varies inversely as the square root of x
three-halves power of the velocity.
width,
The drag on one
and
directly as the
side of the plate, of unit
is
Drag =
f
to
dx
=
0.644
^p~UH
(5.6.9)
Selecting other velocity distributions does not radically alter these results.
The exact
solution,
worked out by Blasius from the general equations of
and 0.664 for Eqs. (5.6.8) and
viscous motion, yields the coefficients 0.332
(5.6.9), respectively.
The drag can be expressed
in
terms of a drag coefficient
CD
times the
FUNDAMENTALS OF FLUID MECHANICS
272
ht/H
r
Transition
I
^
Laminar
^rbulent
Critical
Fig. 5.16
Boundary-layer growth.
(The vertical scale is greatly en-
larged.)
stagnation pressure
P
U
pU
2
/2 and the area of plate
(per unit breadth),
2
Drag = CD —t-
I
in which, for the laminar
CD =
I
boundary
layer,
1.328
(5.6.10)
and Rz =
Ul/v.
When
the Reynolds
number
for the plate reaches a value
between
500,000 and 1,000,000, the boundary layer becomes turbulent. Figure 5.16
indicates the growth and transition from laminar to turbulent boundary layer.
The
critical
Reynolds number depends upon the
initial
turbulence of the fluid
stream, the upstream edge of the plate, and the plate roughness.
Turbulent boundary layer
The momentum equation can be used
to determine turbulent boundary-layer
smooth plate in a manner analogous to the
treatment of the laminar boundary layer. The universal velocity-distribution
law for smooth pipes, Eq. (5.4.11), provides the best basis, but the calculagrowth and shear
stress along a
law. It
and
r
is
is
A simpler
approach is to use Prandtl's one-seventh-power
which y is measured from the wall of the pipe
the pipe radius. Applying it to flat plates produces
tions are involved.
u/u m& =
*.
(y/ro)
1/7
U
=
\8
rj
1'7
117
,
in
VISCOUS EFFECTS: FLUID RESISTANCE
273
and
=
ro
0.0228 P f/ 2
^j
(5.6.11)
in which the latter expression is the shear stress at the wall of a smooth plate
with a turbulent boundary layer. 1 The method used to calculate the laminar
boundary layer gives
= PU2 ^- f
ro
(1
-
r,W) V
CLX Jq
By
W
dr,
=
Ap^
?
2
(5.6.12)
(XX
equating the expressions for shear
boundary-layer thickness
h is
stress,
the differential equation for
obtained,
1/4
5
1/4
=
dh
0.234
-
dx
(0
{
)
After integrating, and then
by assuming that the boundary layer
over the whole length of the plate so that the
initial
conditions x
is
turbulent
=
0, b
=
can be used,
5
=
5'4
0.292 (jj\
Solving for
5
x
gives
°
=
6
it
thickness increases
and
5
layer.
varies
plate r
~
8,
1'4
flat plate, 5 is
eliminated in Eqs.
and
(£-)
(5.6.11)
0.316/R
the drag on a smooth,
(5.6.13),
0.029 P £/ 2
Equation
=
boundary
.
(5.6.11)
f
in the turbulent
,
To determine
=
more rapidly
the thickness increases as z 4/5 but in the laminar boundary layer
as x 112
ro
_ 0.37a;
"
(h)
The
In
0.37a:
xAlb
°- 37
is
(5.6.14)
obtained
(Blasius eq.),
um
~ U.
from
the
following
R = V2roP /ix, and
r
T
pipe
= Mm /1.235.
equations:
To
t
= pfV
2
/8,
transfer to the flat
FUNDAMENTALS OF FLUID MECHANICS
274
The drag
for unit
^
Drag -
f
width on one side of the plate
l
J
,
to da;
=
™„
™ /—Vj
v
0.036 P *7 2 Z
In terms of the drag
/5
-
f
is
O.OSGpUH
[{&
(5.6.15)
coefficient,
CD = 0.072Rr 1/5
(5.6.16)
which R* is the Reynolds number based on the length of plate.
The above equations are valid only for the range in which the Blasius
resistance equation holds. For larger Reynolds numbers in smooth-pipe flow,
the exponent in the velocity-distribution law is reduced. For R = 400,000,
n = |, and for R = 4,000,000, n = £>. The drag law, Eq. (5.6.15), is valid
in
for a range
X
5
<
10 5
Rz
<
10 7
Experiment shows that the drag
is
slightly higher
than
is
predicted
by Eq.
(5.6.16),
CD = 0.074Rr 1/5
(5.6.17)
The boundary
layer is actually laminar along the upstream part of the plate.
Prandtl has subtracted the drag from the equation for the upstream end of
the plate up to the critical Reynolds number and then added the drag as given
by the laminar equation for this portion of the plate, producing the equation
1
CD = 0.074Rr 1/5 -
^
5
X
10 5
<
Ri
<
10 7
(5.6.18)
In Fig. 5.17 a log-log plot of Cd vs. Ri shows the trend of the drag coefficients.
Use of the logarithmic velocity distribution for pipes produces
"0^S
C
in
455
10*<R<<10°
which the constant term has been selected
(5.6.19)
for best
agreement with ex-
perimental results.
1
L. Prandtl,
Uber den Reibungswiderstand stromender Luft,
Goett. Ill Lieferung, 1927.
Result. Aerodyn. Test Inst.,
VISCOUS EFFECTS: FLUID RESISTANCE
275
10"
—
T,.
.
SfSwfenf
^\^
cD
Transition
10" 3
10
s
4
2
6
8
^
io
6
4
2
6
8
10
/
Ri=UI/p
CD = 2£S,
Laminar
Fig. 5.17
Wi
The drag law
transition C|)
-^-^,
R^ R
turbulent
CD =
i
smooth
for
-^
R
1/5
t
plates.
EXAM PLE
5.7
A smooth, flat plate 10 ft wide and 100 ft long is towed through
water at 68°F with a speed of 20 ft/s. Determine the drag on one side
of the plate and the drag on the first 10 ft of the plate.
For the whole plate
still
i
—_
(100
ft)
(20 ft/s) (1.935 slugs/ft 3 )
1.85
0.01
P
slug/ft s
1
479
From Eq.
P
0.455
[log (1.85
The drag on one
X
0.455
10 8 )] 2
side
-
58
8.2675 2
^
which
-
0.00196
58
is
Drag = CD blp -^ = 0.00196 (10
ft)
(100
ft)
l
to the transition
5
0.01
P
slug/ft
479
(20 ft/s) 2
= 760
P
-s
Reynolds number occurs at 5
is
(/oft) (20 ft/s) (1.935 slugs/ft 3
)
1
1<935 slu S s/ft3
lb
2
b is the plate width. If the critical
the length
10 8
(5.6.19)
CD =
in
X
•
X
10 5 k
=
0.27
ft
X
10 5
,
276
FUNDAMENTALS OF FLUID MECHANICS
For the
first
10
ft of
Drag = 0.00274
X
=
the plate, R*
10
X
10
1.85
X ^P- X
X
20 2
10 7
=
,
CD =
0.00274, and
106 lb
Calculation of the turbulent boundary layer over rough plates proceeds
with the rough-pipe tests using sand roughnesses.
At the upstream end of the flat plate, the flow may be laminar; then, in the
turbulent boundary layer, where the boundary layer is still thin and the ratio
in similar fashion, starting
of roughness height to boundary-layer thickness e/8
of fully developed roughness occurs,
and the drag
is
is
significant, the region
proportional to the square
For long plates, this region is followed by a transition region
becomes increasingly smaller, and eventually the plate becomes
hydraulically smooth; i.e., the loss would not be reduced by reducing the
roughness. Prandtl and Schlichting have carried through these calculations,
which are too complicated for reproduction here.
of the velocity.
where
e/8
1
Separation; wake
Along a
flat plate
the boundary layer continues to grow in the downstream
direction, regardless of the length of the plate,
when
the pressure gradient
remains zero. With the pressure decreasing in the downstream direction, as
in a conical reducing section, the boundary layer tends to be reduced in thickness.
For adverse pressure gradients, i.e., with pressure increasing in the downstream direction, the boundary layer thickens rapidly. The adverse gradient
and the boundary shear decrease the momentum
in the
boundary
layer,
and
they both act over a sufficient distance, they cause the boundary layer to
come to rest. This phenomenon is called separation. Figure 5.18a illustrates
if
this case.
tion point,
The boundary
streamline must leave the boundary at the separa-
and downstream from
causes backflow near the wall.
that separates from the boundary
tion
is
to decrease the net
this point the adverse pressure gradient
This region downstream from the streamline
is
amount
known
of flow
as the wake. The effect of separawork that can be done by a fluid
element on the surrounding fluid at the expense of its kinetic energy, with the
is incomplete and flow losses (drag) increase.
Figures 5.186 and 5.18c illustrate actual flow cases, the first with a very small
net result that pressure recovery
adverse pressure gradient, which causes thickening of the boundary layer, and
the second with a large diffuser angle, causing separation and backflow near
the boundaries.
1
Prandtl and H. Schlichting, Das Widerstandsgesetz rauher Platten, Werft, Reederei,
Hafen, p. 1, 1934; see also NACA Tech. Mem. 1218, pt. II.
VISCOUS EFFECTS: FLUID RESISTANCE
Separation point
adverse pressure gradient on
Boundary-layer growth
in a small-angle diffuser. (c) Boundary-layer separation
in a large-angle diffuser. [Parts (b) and (c) from the film
"Fundamentals of Boundary Layers," by the National Committee for Fluid Mechanics Films and the Education Development Center.]
Fig. 5.18
(a)
boundary
layer. Separation, (b)
Effect
of
277
278
FUNDAMENTALS OF FLUID MECHANICS
"^Wake
Fig. 5.19
Streamlined body.
Streamlined bodies (Fig. 5.19) are designed so that the separation point
body as possible. If separation can be
avoided, the boundary layer remains thin and the pressure is almost recovered
occurs as far downstream along the
downstream along the body. The only
drag is due to shear stress in
In the wake, the pressure is not reReduction of w ake size reduces the presloss or
the boundary layer, called skin friction.
covered, and a pressure drag results.
sure drag on a body.
In general, the drag
r
is
caused by both skin friction and
pressure drag.
Flow around a sphere is an excellent example of the effect of separation
on drag. For very small Reynolds numbers, VD/v < 1, the flow is everywhere
nonturbulent, and the drag is referred to as deformation drag. Stokes' law
(Sec. 5.7) gives the drag force for this case. For large Reynolds numbers,
the flow may be considered potential flow except in the boundary layer and
the wake. The boundary layer forms at the forward stagnation point and
is generally laminar.
In the laminar boundary layer, an adverse pressure
Shift in separation point due to induced turbulence: (a) 8.5-in
bowling ball, smooth surface, 25 ft/s entry velocity into water; (b) same
except for 4-in diameter patch of sand on nose. (Official U.S. Navy photograph made at Navy Ordnance Test Station, Pasadena Annex.)
Fig. 5.20
VISCOUS EFFECTS: FLUID RESISTANCE
gradient causes separation
more
readily than in a turbulent
279
boundary
layer,
because of the small amount of momentum brought into the laminar layer.
If separation occurs in the laminar boundary layer, the location is farther
upstream on the sphere than it is when the boundary layer becomes turbulent
first and then separation occurs.
In Fig. 5.20 this is graphically portrayed by the photographs of the two
spheres dropped into water at 25 ft/s. In a, separation occurs in the laminar
boundary layer that forms along the smooth surface and causes a very large
wake with a resulting large pressure drag. In b, the nose of the sphere, roughened by sand glued to it, induced an early transition to turbulent boundary
layer before separation occurred.
The high momentum
transfer in the tur-
bulent boundary layer delayed the separation so that the
tially
wake
is
substan-
reduced, resulting in a total drag on the sphere less than half that oc-
curring in
a.
A plot of drag coefficient against Reynolds number
(Fig. 5.21) for
smooth
spheres shows that the shift to turbulent boundary layer (before separation)
by itself at a sufficiently high Reynolds number, as evidenced by the
sudden drop in drag coefficient. The exact Reynolds number for the sudden
shift depends upon the smoothness of the sphere and the turbulence in the
fluid stream. In fact, the sphere is frequently used as a turbulence meter by
determining the Reynolds number at which the drag coefficient is 0.30, a point
located in the center of the sudden drop (Fig. 5.21). By use of the hot-wire
occurs
10'
10
M
--t
3
10'
10
N
>S
—
A
r
\
/-Dis ks
„.--
^= -
--
/
-i I.
JS
Stokes J
10'
10" 3
Fig. 5.21
10" 2
10" 1
1
Drag coefficients
10
for
..
H--'*- ::-^
V
\n. p"
..J
^Spheres
10
2
10
3
10
4
10
5
spheres and circular disks.
10
6
FUNDAMENTALS OF FLUID MECHANICS
280
anemometer, Dryden 1 has correlated the turbulence level of the fluid stream
to the Reynolds number for the sphere at Cd = 0.30. The greater the turbulence of the fluid stream, the smaller the Reynolds number for shift in separation point.
In Sec.
7.8, for ideal-fluid flow,
equations are developed that permit the
velocity
and pressure to be found at any point
sphere.
In ideal-fluid flow,
in the fluid for flow around a
permitted at the boundary; in addition,
the boundary condition states that the normal component of the velocity
at a
boundary
is
slip is
zero in steady flow. Therefore separation
In ideal-fluid flow, which
is
ruled out of con-
no drag
on a body and no boundary layer. A comparison of Figs. 5.20 and 7.23 shows
the great contrast between ideal- and real-fluid flow around aTbluff body. The
ideal-fluid flow, however, does yield a good representation of velocity and
pressure for the upstream portion of the flow, away from the effects of boundary-layer separation and wake formation.
sideration.
for constant energy, there is
is
DRAG ON IMMERSED BODIES
5.7
The
around bodies are developed in Chap. 7, and
boundary layer, separation, and wake in Sec. 5.6. In this
section drag is defined, some experimental drag coefficients are listed, the effect
of compressibility on drag is discussed, and Stokes' law is presented. Lift
is defined, and the lift and drag coefficients for an airfoil are given.
principles of potential flow
principles of the
100
ic
^4-~
CD
I
\\
l
\
L-
0.1
10"
10
l
10
1
10'
10
io-
4
10
:
10
c
R= UD
Fig. 5.22
1
Drag coefficients for circular cylinders.
H. Dryden, Reduction
of
Turbulence
in
Wind
Tunnels,
NACA
Tech. Rep. 392, 1931.
VISCOUS EFFECTS: FLUID RESISTANCE
Table
281
Typical drag coefficients for various cylinders in two-dimensional
5.1
flow]
CD
Body shape
Circular cylinder
Elliptical cylinder
O
—
—
1.2
cz>
2:1
->
<=>
->
4:1
8:1
Square cylinder
Triangular cylinders
D
—
—
>
Z
120
90°^
o
f
Data from W.
Drag
is
F. Lindsey,
NACA
1.6
10 4 to 1.5
X
4
X
10
10 5
4
10 5
2.5 X
2.5
lOHo
X 10
2 X 10
3.5 X 10
10 5
4
5
4
10 to 10
2.0
120° 1.72
2.15
^90° 1.60
2.20
<]60° 1.39
30"t^>
Semitubular
0.6
0.46
0.32
0.29
0.20
2.0
Reynolds number
4
5
10 4
10 4
10 4
10 4
10 4
10 4
10 5
1.8
10 5
1.0
2.3
4
1.12
4
X
X
10 4
10 4
Tech. Rep. 619, 1938.
defined as the force component, parallel to the relative approach
on the body by the moving fluid. The drag-coefficient curves
and circular disks are shown in Fig. 5.21. In Fig. 5.22 the drag coefficient for an infinitely long circular cylinder (two-dimensional case) is
plotted against the Reynolds number. Like the sphere, this case also has the
sudden shift in separation point. In each case, the drag coefficient Cd is defined by
velocity, exerted
for spheres
Drag = CD A
—
A is the projected area of the body on a plane normal to the flow.
In Table 5.1 typical drag coefficients are shown for several cylinders.
In general, the values given are for the range of Reynolds numbers in which
in which
little with the Reynolds number.
and
drag curve for an airfoil section is shown in Fig. 5.23.
A
Lift is the fluid-force component on a body at right angles to the relative approach velocity. The lift coefficient Cl is defined by
the coefficient changes
typical
Lift
= C LA
lift
—
FUNDAMENTALS OF FLUID MECHANICS
282
2.0
0.40
1.8
0.36
/
1.6
0.32
'
1.4
- <
A
Chord length
J
-
0.28
hS.
sv
1
1.2
s
i
'cL
1
i
0.8
0.24
Q
0.20
£v
0.16
£
o
3
63
O
0.12
w>
2
Q
~*
0.4
0.08
0.2
0.04
-0.2
0.04
-0.4
0.08
-8 -4
Fig. 5.23
4
8
12
16
20
Angle of attack a, degrees
Typical
lift
24
28
32
and drag coefficients
for
an
airfoil.
in
which
A
refers to the
chord length times the wing length for
lift
and drag
for airfoil sections.
Effect of compressibility on drag
To determine drag in high-speed gas flow the effects of compressibility, as
by the Mach number, are more important than the Reynolds number. The Mach number M is defined as the ratio of fluid velocity to velocity of
expressed
sound in the
fluid
medium. When flow
is
at the critical velocity
c, it
has ex-
waves cannot travel
greater than unity, the flow
actly the speed of the sound wave, so that small pressure
upstream. For this condition
supersonic; and
when
M
M =
1.
When M
is
than unity, it is subsonic.
Any small disturbance is propagated with the speed of sound (Sec. 6.2).
For example, a disturbance in still air travels outward as a spherical pressure
wave. When the source of the disturbance moves with a velocity less than c,
as in Fig. 5.24a, the wave travels ahead of the disturbing body and gives the
fluid a chance to adjust itself to the oncoming body. By the time the particle
has moved a distance Vt, the disturbance wave has moved out as far as r = ct
is
is less
VISCOUS EFFECTS: FLUID RESISTANCE
283
from the point 0. As the disturbing body moves along, new spherical waves
are sent out, but in all subsonic cases they are contained within the initial
spherical wave shown. In supersonic motion of a particle (Fig. 5.246) the body
moves faster than the spherical wave emitted from it, yielding a cone-shaped
wavefront with vertex at the body, as shown. The half angle of cone a is
called the
Mach
-
C
=
angle,
sm-^
« =
sin->
The
conical pressure front extends out behind the
wave (Sec. 6.4). There
across a Mach wave.
is
body and
is
called a
Mach
a sudden small change in velocity and pressure
Mach number and becomes
Reynolds number when compressibility effects
become important. In Fig. 5.25 the drag coefficients for four projectiles are
plotted against the Mach number.
For low Mach numbers, a body should be rounded in front, with a blunt
nose and a long-tapering afterbody for minimum drag. For high Mach numbers (0.7 and over), the drag rises very rapidly owing to formation of the
vortices behind the projectile and to formation of the shock waves; the body
should have a tapered nose or thin forward edge. As the Mach numbers increase, the curves tend to drop and to approach a constant value asymptotically. This appears to be due to the fact that the reduction of pressure behind
the projectile is limited to absolute zero, and hence its contribution to the
total drag tends to become constant. The pointed projectile creates a narrower shock front that tends to reduce the limiting value of the drag coefficient.
The drag on
bodies varies greatly with the
relatively independent of the
Fig. 5.24
moving
Wave propagation produced by
at
(a)
subsonic velocity and
(b)
a
particle
supersonic
FUNDAMENTALS OF FLUID MECHANICS
284
—^^^
-*«
0.6
if ^-^
-~
!
^
/
o
b
0.4
""N*
CD
J
J/l
0.2
12
0.0
3
V/c
Drag
Fig. 5.25
coefficients
projectiles as a function
Mach number. {From
L.
for
of the
Prandtl,
Stromungslehre,"
Friedrich Vieweg und Sohn, Brunswick, Germany, 1935.)
"Abriss
der
Stokes' law
The flow of a viscous incompressible fluid around a sphere has been solved
by Stokes for values of the Reynolds number UD/v below 1 The derivation
is beyond the scope of this treatment; the results, however, are of value in
1
.
such problems as the settling of dust particles. Stokes found the drag (force
exerted on the sphere by flow of fluid around it) to be
Drag = QiranU
in
which a
is
the radius of sphere and
fluid at a great distance.
through a
must
iira
1
d
y
To
U the velocity
fluid that is otherwise at rest, the
buoyant force plus the drag force
just equal its weight, or
+
Qwaull
=
of sphere relative to the
find the terminal velocity for a sphere dropping
%ira z y s
G. Stokes, Trans. Camb. Phil. Soc, vol.
8,
1845; vol.
9,
1851.
VISCOUS EFFECTS: FLUID RESISTANCE
in
which 7
is
By
sphere.
the specific weight of liquid and y s
solving for U, the terminal velocity
is
is
the specific weight of the
found to be
U = l-{y -y)
9 n
(5.7.1)
s
The
285
straight-line portion of Fig. 5.21 represents Stokes' law.
The drag
relations
on
particles, as given
by
Stokes' law
and by the
ex-
perimental results of Fig. 5.21, are useful in the design of settling basins for
separating small solid particles from fluids.
Applications include separating
coolants from metal chips in machining operations, desilting river flow,
sanitary engineering applications to treatment of
EXAMPLE
viscosity
ju
=
X
10- 4
-
where y in meters
the base of the stratosphere at 11,000 m.
3.06
is
X
_dy _
f
dt
I0~ 9 y
level.
Estimate the time for these
Neglect air currents and wind
effects.
U =
—dy/dt in Eq. (5.7.1) and recognizing the unit weight
to be much smaller than the unit weight of the solid particles, one has
Writing
dt
Assume the
be expressed by the relationship
measured from sea
particles to reach sea level.
of air
matter 10 jum in
jet aircraft discharges solid particles of
2.5, at
of air, in poises, to
/x
1.78
A
5.8
S =
diameter,
and
raw water and sewage.
~
2 a2 y s
9
= -
'o
m
^f11 ,000
f(1.78
X
10- 4
-
3.06
X
10~ 9 y P)
(5
X
"
86,400
=
where d
[1.78,
s
15.07 d
is
10- 6
3 -° 6X 10 y ]"'
- 3.06X10-VTX
2
the abbreviation for day
IP
1
X
Id
0.1N-s/m2
m)
2
mg
2.5
g
X
9802
N/m
3
dii
y
m
FUNDAMENTALS OF FLUID MECHANICS
286
yAL
Axial forces on
Fig. 5.26
free
body
of fluid
in
RESISTANCE TO TURBULENT FLOW
IN
a
conduit.
5.8
OPEN AND CLOSED
CONDUITS
In steady turbulent incompressible flow in conduits of constant cross section
(steady uniform flow) the wall shear stress varies closely proportional to the
square of the velocity,
xfv
To
in
which
(5.8.1)
For open channels and noncircular
In these
used as the average wall shear stress. Secondary flows 1 occurring
X
is
a dimensionless coefficient.
closed conduits the shear stress
cases, r is
is
not constant over the surface.
in noncircular conduits act to equalize the wall shear stress.
The momentum equation applied
to the control volume (Fig. 5.26) combetween sections 1 and 2, under the assumptions made,
shows no net efflux of momentum, and hence shows equilibrium of forces on
the control volume in the direction of motion
prising the liquid
(pi
in
-
V*)A
which Az
+
yA Az =
t
LP
= L sin 6 and P
is
the wetted perimeter of the conduit,
portion of the perimeter where the wall
liquid surface excluded)
1
Secondary
flows,
.
The
ratio
A/P
i.e.,
the
is
in contact with the fluid (free-
is
called the hydraulic radius
R
of
not wholly understood, are transverse components that cause the main
central flow to spread out into corners or near walls.
VISCOUS EFFECTS: FLUID RESISTANCE
the conduit.
+
Ap
y Az
If pi
=
=
pi
\pV
_
=
r
R
Z
—
Ap,
2
(5.8.2)
~2R
when divided through by
or,
287
y, if
hf
= (Ap
+
y Az) /y are the
losses per unit
weight,
h
-l=
L
in
S
=
^
R2g
which S represents the
losses per unit
weight per unit length. After solving
for V,
V =
J-A V^
- C y/RS
(5.8.3)
'
This is the Chezy formula, in which originally the Ch£zy coefficient C was
thought to be a constant for any size conduit or wall-surface condition. Various
formulas for
C
are
now
For pipes, when X
is
=
//4 and
R =
Z)/4, the
Darcy-Weisbach equation
obtained,
LV
2
=f-—
hf
in
generally used.
which
(5.8.4)
D is the inside pipe diameter.
This equation
may be
applied to open
channels in the form
-4j VRS
(5.8.5)
with values of / determined from pipe experiments.
5.9
STEADY UNIFORM FLOW
IN
OPEN CHANNELS
For incompressible, steady flow at constant depth in a prismatic open channel,
Manning formula is widely used. It can be obtained from the Chezy
formula [Eq. (5.8.3)] by setting
the
C =
—n RV«
(5.9.1)
FUNDAMENTALS OF FLUID MECHANICS
288
so that
r
R 2I3 S
1 '2
(5.9.2)
n
is the Manning
The value of C m
which
V is
formula.
1.49
is
and
1.0 for English
and SI
units, respectively
R
the hydraulic radius (Sec. 5.8),
the losses per unit weight per unit length of channel or the slope of the
the average velocity at a cross section,
and S
bottom
water surface, which is parallel
to be an absolute roughness coefficient, i.e., dependent upon surface roughness only, but it actually depends upon the size and shape of channel cross section in some unknown
manner. Values of the coefficient n, determined by many tests on actual
canals, are given in Table 5.2. Equation (5.9.2) must have consistent English
or SI units as indicated for use with the values in Table 5.2.
of the channel. It
is
The
to the channel bottom.
also the slope of the
coefficient
n was thought
1
Table
5.2
for various
Average values of the Manning roughness factor
boundary materials^
Manning n
Boundary material
Planed wood
0.012
0.013
0.012
0.014
0.015
0.016
0.018
022
0.025
0.025
035
0.029
Unplaned wood
Finished concrete
Unfinished concrete
Cast iron
Brick
Riveted steel
Corrugated metal
.
Rubble
Earth
Earth, with stones or weeds
.
Gravel
f
Work by
the U.S. Bureau of Reclamation and other
government agencies indicates that the Manning roughness factor should be increased (say, 10 to 15 percent)
than about 10 ft. The loss in
due to the roughening of the
surfaces with age, marine and plant growths, deposits,
and the addition of bridge piers as the highway system
is expanded.
for hydraulic radii greater
capacity of large channels
1
To
is
convert the empirical equation in English units to SI units, n is taken to be dimensionthen the constant has dimensions, and (1.49 ft 1/3 /s) (0.3048 m/ft) 1/3 = 1.0 m^/s.
less;
VISCOUS EFFECTS: FLUID RESISTANCE
Fig. 5.27
289
Notation for trape-
zoidal cross section.
When
Eq. (5.9.2)
is
multiplied
by the
cross-sectional area A, the
Mann-
ing formula takes the form
Q =
—n AR
2 '3
S 112
(5.9.3)
When the cross-sectional area is known, any one of the other quantities can
be obtained from Eq. (5.9.3) by direct solution.
EXAMPLE
Determine the discharge for a trapezoidal channel (Fig. 5.27)
= 8 ft and side slopes 1 on 1. The depth is 6 ft, and
the slope of the bottom is 0.0009. The channel has a finished concrete lining.
From Table 5.2, n = 0.012. The area is
5.9
with a bottom width b
A =
8X6 + 6X6
= 84ft
and the wetted perimeter
P=8 + 2X6V2
By
=
2
is
24.96
substituting into Eq. (5.9.3),
1
49
/ 84 \ 2/3
Q =
7T^ 84 (,
W7^>
0.012
\24.96/
area
some instances when the cross-sectional
unknown. Expressions for both the hydraulic radius and the area conthe depth in a form that cannot be solved explicitly.
(
)
(0.0009
1 '2
)
=
703
cfs
Trial solutions are required in
tain
is
EXAMPLE
5.10
required for 4 m /s flow in
m wide with a bottom slope of 0.002?
What depth
planed-wood channel 2
is
3
a rectangular
FUNDAMENTALS OF FLUID MECHANICS
290
If
inEq.
the depth
A =
is y,
2y,
P =
+
2
and n =
2y,
0.012.
By
substituting
(5.9.3),
—
+
/
^2;
4mVs =
2y
(
\2
\ 2/3
0.(
)
2yJ
Simplifying gives
/<»>
=
=
y
°-536
(r+ij"
Assume pirn; then f(y) = 0.63. Assume y =
The correct depth then is about 0.89 m.
0.89 m, then f(y)
=
0.538.
A developer
EXAMPLE 5.11 Riprap problem.
mental regulatory authorities to
has been required by environan open channel to prevent erosion.
The channel is trapezoidal in cross section and has a slope of 0.0009. The
bottom width is 10 ft and side slopes are 2:1 (horizontal to vertical). If he
line
uses roughly spherical rubble (7,
minimum
D
=
135 lb/ft 3 ) for the lining,
5o
Assume the shear that rubble can withstand
r
in
=
- y)D 50
0.040 (y s
which 7
is
S
The
of the rubble that can be used?
is
design flow
described
what
is
is
the
1000
cfs.
by
lb/ft 2
the unit weight of rock and
D
bo
is
the average rock diameter
in feet.
A Manning
n
of 0.03
is
appropriate for the rubble.
To
find the size of
channel, from Eq. (5.9.3)
™
1.49 [2/(10
+
22/)]
6 '»
By trial solution the depth is y =
From Eq.
ro
To
= y RS =
find the
0.040(135
Hence
More
62.4
D
-
D =
50
8.62
ft,
and the hydraulic radius R = 4.84
(5.8.2)
b0
X
4.84
X
0.0009
size for incipient
62.4)7)50
0.0936
=
=
0.272 lb/ft 2
movement
r
=
r
,
and
0.272
ft.
general cases of open-channel flow are considered in Chap. 11.
ft.
VISCOUS EFFECTS: FLUID RESISTANCE
291
STEADY INCOMPRESSIBLE FLOW THROUGH SIMPLE PIPE
SYSTEMS
5.10
Colebrook formula
A force
balance for steady flow (no acceleration) in a pipe (Fig. 5.28) yields
Ap wr
=
2
AL
T 2irr
This simplifies to
n=
r
Av -
fL
(5.10.1)
which holds
(5.8.4)
may
Ap = yh f
for laminar or turbulent
be written
AL
V
flow.
The Darcy-Weisbach equation
=f- P -
2
Eliminating Ap in the two equations and simplifying gives
^°=\/i
F
(5.10.2)
The
stress, friction factor, and average velocity.
be obtained from Eq. (5.4.11) by integrating over
the cross section. Substituting for V in Eq. (5.10.2) and simplifying produces
the equation for friction factor in smooth-pipe flow,
which relates wall shear
V may
average velocity
1
=
Vf"
A +B
s
r
27Tr
s
In
(R v?)
AL
r
o
Apwr\
|
1
1
1
1
I
1
AL
Fig. 5.28
Equilibrium con-
ditions for steady flow in a
pipe.
(5.10.3)
FUNDAMENTALS OF FLUID MECHANICS
292
With the Nikuradse data
1
-i= = 0.86 In (R y/j)
-
for
smooth
pipes, the equation
becomes
0.8
(5.10.4)
VI
For rough pipes in the complete turbulence zone,
=
F2
i/j
in
which
n,
+
\ h)
F
2 is,
—= =
(5,ia5)
i
form and spacing of the roughFor the Nikuradse sand-grain roughness (Fig. 5.31) Eq.
in general, a constant for a given
ness elements.
(5.10.5)
Brln
becomes
1.14
-
0.86 In -^
(5.10.6)
D
V7
The roughness height e for sand-roughened pipes may be used as a
measure of the roughness of commercial pipes. If the value of / is known for
a commercial pipe in the fully developed wall turbulence zone, i.e., large
Reynolds numbers and the loss proportional to the square of the velocity,
the value of e can be computed by Eq. (5.10.6) In the transition region, where
/ depends upon both e/D and R, sand-roughened pipes produce different results from commercial pipes. This is made evident by a graph based on Eqs.
(5.10.4) and (5.10.6) with both sand-roughened and commercial-pipe-test
results shown. Rearranging Eq. (5.10.6) gives
.
-7=
+
0.86 In
^ =
and adding 0.86
In (e/D) to each side of Eq. (5.10.4) leads to
-y=
+
By
selecting 1/ y/f
0.86 In j-
1.14
=
0.86 In
+
(
R
V? ^) ~
0.8
and
In (R y/f e/D)
as abwith slope
+0.86 and rough-pipe-test results in the complete turbulence zone plot as the
horizontal line. Nikuradse sand-roughness-test results plot along the dashed
0.86 In (e/D) as ordinate
scissa (Fig. 5.29), smooth-pipe-test results plot as a straight line
J
J.
Nikuradse, Gesetzmassigkeiten der turbulenten Stromung in glatten Rohren, Ver.
Dtsch. Ing. Forschungsh., vol. 356, 1932.
VISCOUS EFFECTS: FLUID RESISTANCE
293
+3
i
I
/-Nikuradse sand
roughness
/
+2
\
—
—
c
"T
^P"
+
Rough
^-Commercial
pipe
pipe
-1
In
(RV7^)
Colebrook transition function.
Fig. 5.29
line in the transition region,
lower curved
and commercial-pipe-test
results plot along the
line.
The explanation
of the difference in
shape of the
curve of Nikuradse and the commercial roughness curve
sublayer, or laminar film, covers
all
the
artificial
artificial
is
roughness
that the laminar
roughness or allows
it
to
With commercial roughuniformity, small portions extend beyond the
protrude uniformly as the film thickness decreases.
which varies greatly
ness,
film
first,
in
as the film decreases in thickness with increasing Reynolds number.
An empirical transition function for commercial pipes for the region between
smooth pipes and the complete turbulence zone has been developed by Colebrook, 1
—= =
-0.86
V7
which
In
fe/D
+
V3.7
is
the basis for the
2.51
\
(5.10.7)
R VfJ
Moody diagram
(Fig. 5.32).
Pipe flow
In steady incompressible flow in a pipe the irreversibilities are expressed in
terms of a head loss, or drop in hydraulic grade line (Sec. 10.1) The hydraulic
.
1
C. F. Colebrook, Turbulent Flow in Pipes, with Particular Reference to the Transition
Region Between the Smooth and Rough Pipe Laws, J. Inst. Civ. Eng. Lond., vol. 11,
pp. 133-156, 1938-1939.
FUNDAMENTALS OF FLUID MECHANICS
294
1
1
t°
i
B^
Experimental arrangement for determining head loss
Fig. 5.30
in a
—R
1"
pipe.
grade line
is
p/y above the center of the pipe, and if z is the elevation of the
+ p/y is the elevation of a point on the hydraulic
The locus of values of z + p/y along the pipeline gives the hy-
center of the pipe, then z
grade
line.
draulic grade line.
J
1
Losses, or irreversibilities, cause this line to drop in the
The Darcy-Weisbach equation
direction of flow.
(5.8.4)
D2g
generally adopted for pipe-flow calculations.
is
1
hf
is
the head
loss,
or drop in
D and
an average velocity V. hf has the dimension length and is expressed in terms
of foot-pounds per pound or meter-newtons per newton. The friction factor
hydraulic grade
/
is
line, in
the pipe length L, having an inside diameter
a dimensionless factor that
is
required to
make
the equation produce the
measured
measuring the discharge and inside diameter, the average velocity can be computed. The head
loss hf is measured by a differential manometer attached to piezometer opencorrect value for losses. All quantities in Eq. (5.8.4) except / can be
experimentally.
ings at sections
A typical setup is
1
and
2,
distance
shown
L
in Fig. 5.30.
By
apart.
Experimentation shows the following to be true in turbulent flow:
1.
2.
3.
4.
5.
6.
The head loss varies directly as the length of the pipe.
The head loss varies almost as the square of the velocity.
The head loss varies almost inversely as the diameter.
The head loss depends upon the surface roughness of the interior pipe wall.
The head loss depends upon the fluid properties of density and viscosity.
The head loss is independent of the pressure.
The
friction factor /must
be selected so that Eq.
(5.8.4) correctly yields
the head loss; hence, / cannot be a constant but must depend upon velocity V,
diameter D, density p, viscosity /x, and certain characteristics of the wall
See Sec. 10.1 for development of empirical pipe-flow formulas for special uses.
VISCOUS EFFECTS: FLUID RESISTANCE
roughness signified by
e, e',
and m, where
e is
a measure of the
ness projections and has the dimensions of a length,
e'
is
size of
295
the rough-
a measure of the
arrangement or spacing of the roughness elements and also has the dimensions
of a length, and m is a form factor, depending upon the shape of the individual
roughness elements and is dimensionless. The term/, instead of being a simple
constant, turns out to depend
/
upon seven
quantities,
= /(7,ZW,€,e>)
Since /
(5.10.8)
a dimensionless factor,
is
it
must depend upon the grouping
of
For smooth pipe e = e =
m = 0, leaving / dependent upon the first four quantities. They can be arranged in only one way to make them dimensionless, namely, VDp/n, which
is the Reynolds number. For rough pipes the terms e, e' may be made dimensionless by dividing by D. Therefore, in general,
these quantities into dimensionless parameters.
f
-'(?T'3'3'
The proof
pipes a plot of
m
(5109)
)
of this relationship
all
is
left
to experimentation.
For smooth
experimental results shows the functional relationship, sub-
ject to a scattering of
±5
percent.
The
plot of friction factor against the
Reynolds number on a log-log chart is called a Stanton diagram. Blasius 1 was
the first to correlate the smooth-pipe experiments in turbulent flow.
He
presented the results by an empirical formula that is valid up to about R =
100,000. The Blasius formula is
!-*&
<5..0,0,
In rough pipes the term e/D is called the relative roughness. Nikuradse 2
proved the validity of the relative-roughness concept by his tests on sandroughened pipes. He used three sizes of pipes and glued sand grains (e =
diameter of the sand grains) of practically constant size to the interior walls
had the same values of e/D for different pipes. These experiments
show that for one value of e/D the /, R curve is smoothly connected
regardless of the actual pipe diameter. These tests did not permit variation
so that he
(Fig. 5.31)
1
H. Blasius, Das Ahnlichkeitsgesetz bei Reibungsvorgangen in Fliissigkeiten, Ver. Dtsch.
Ing. Forschungsh., vol. 131, 1913.
2
J.
Nikuradse, Stromungsgesetze in rauhen Rohren, Ver. Dtsch. Ing. Forschungsh., vol.
361, 1933.
FUNDAMENTALS OF FLUID MECHANICS
296
0.10
0.09
\\
1
0.08
0.07
1
1
D = 2.412 en
0.06
D = 4.82 cm ^
^».
£=4.87 cm
0.05
\
30
D=9.64cm\
c
D= 2.434 cm
\«
D
D
0.04
b
^IQ
"
Z)
= 2.434 cm x
J.8c
m^
6
~L
1
=
=
j
'hh-ta
0.03
.
O+f "TTT
r
Z)=2.47^
""
D=4.94 cm*
0.02
92 c m
^g e^f**5
\
D
^
Z)=9 .94
e
(
D
1
120
Z)
_€_
1
61.2
=
J_
252
=
1
504
A = J_
D= 9.94 cm
D
1014
1
0.01
10
z
;
10
10
;
10*
VDp
Fig. 5.31
Nikuradse's sand-roughened-pipe tests,
of e'/D or
m
but proved the validity of the equation
one type of roughness.
Because of the extreme complexity of naturally rough surfaces, most of
the advances in understanding the basic relationships have been developed
1
around experiments on artificially roughened pipes. Moody has constructed
factors in clean,
friction
determining
for
charts
one of the most convenient
for pipe-flow
basis
is
the
Fig.
in
5.32,
commercial pipes. This chart, presented
expresses
that
diagram
Stanton
is
The
chart
a
calculations in this chapter.
values
The
number.
Reynolds
the
/ as a function of relative roughness and
experiment
determined
by
are
pipes
of absolute roughness of the commercial
in which / and R are found and substituted into the Colebrook formula, Eq.
for
(5.10.7),
table in
which closely represents natural pipe trends. These are listed in the
the lower left-hand corner of Fig. 5.32. The Colebrook formula pro-
vides the shape of the e/D
1
L. F.
Moody,
=
const curves in the transition region.
Friction Factors for Pipe Flow, Trans.
ASME, November
1944.
VISCOUS EFFECTS: FLUID RESISTANCE
297
E
DO
O
O
=/
JOPBJ
UOIJDIJ
FUNDAMENTALS OF FLUID MECHANICS
298
The
tion.
Apr
=
marked "laminar flow"
straight line
is
the Hagen-Poiseuille equa-
Equation (5.2.106),
2
8nL
may
be transformed into Eq.
_^LV_
V8nL
f
~
yr
~ pD
2
D2g~
(5.8.4)
64
pDV/n
with Ap
LV
D
= yh f and by
solving for h f
,
2
2~g
or
h '- f
LF
2
64L7
2
D2i-RD2i
(5
'
iail)
from which
/
-
64
p"
(5.10.12)
— 1 on a log-log chart,
be used for the solution of laminar-flow problems in pipes. It applies to
all roughnesses, as the head loss in laminar flow is independent of wall roughness. The Reynolds critical number is about 2000, and the critical zone, where
the flow may be either laminar or turbulent, is about 2000 to 4000.
It should be noted that the relative-roughness curves e/D = 0.001 and
smaller approach the smooth-pipe curve for decreasing Reynolds numbers.
This can be explained by the presence of a laminar film at the wall of the pipe
that decreases in thickness as the Reynolds number increases. For certain
ranges of Reynolds numbers in the transition zone, the film completely covers
small roughness projections, and the pipe has a friction factor the same as that
of a smooth pipe. For larger Reynolds numbers, projections protrude through
laminar film, and each projection causes extra turbulence that increases the
head loss. For the zone marked "complete turbulence, rough pipes," the film
thickness is negligible compared with the height of roughness projections, and
each projection contributes fully to the turbulence. Viscosity does not affect
the head loss in this zone, as evidenced by the fact that the friction factor
does not change with the Reynolds number. In this zone the loss follows the
V2 law; i.e., it varies directly as the square of the velocity.
Two auxiliary scales are given along the top of the Moody diagram.
One is for water at 60°F, and the other is for air at standard atmospheric
pressure and 60°F. Since the kinematic viscosity is constant in each case,
the Reynolds number is a function of VD. For these two scales only, D must
be expressed in inches and V in feet per second.
This equation, which plots as a straight line with slope
may
VISCOUS EFFECTS: FLUID RESISTANCE
299
Simple pipe problems
The
three simple pipe-flow cases that are basic to solutions of the
more com-
plex problems are
To find
Given
I.
II.
III.
Q, L, D,
hf
v, e
D,
v, e
Q
h f Q, L,
v, e
D
hf, L,
,
In each of these cases the Darcy-Weisbach equation, the continuity equation,
and the Moody diagram are used to determine the unknown quantity.
In the first case the Reynolds number and the relative roughness are
readily determined from the data given, and hf is found by determining / from
the Moody diagram and substituting into the Darcy-Weisbach equation.
EXAMPLE 5.12 Determine the head loss due to the
= 0.0001 ft /s, through 1000 ft of 8-in-diameter
flow of 2000 gal/min of
2
oil, v
cast-iron pipe.
2000 gal/min
448.8
gal/mm
1 ft
3
/s
VD_
32
12.8 ft/.
0.0001
v
*_
ft
2
/s
V3
}
The relative roughness is e/D = 0.00085/0.667 =
by interpolation, / = 0.024; hence
h
r
>
LV2
= '5
*
- a024
x
1000
-J^~
In the second case,
equation and
ft
V
W#
(12.8 ft/s) 2
_o
= 9L8
0.0013.
Fig. 5.32,
___
,
ft
From
'
lb/lb
and / are unknowns, and the Darcy-Weisbach
Moody diagram must
be used simultaneously to find their
is
of / may be assumed by inspection of
the Moody diagram. Substitution of this trial / into the Darcy-Weisbach
equation produces a trial value of V, from which a trial Reynolds number
is computed.
With the Reynolds number an improved value of / is found
from the Moody diagram. When / has been found correct to two significant
figures, the corresponding V is the value sought and Q is determined by multi-
values.
plying
Since e/D
by the
area.
known, a value
FUNDAMENTALS OF FLUID MECHANICS
300
EXAMPLE 5.13 Water at 15°C flows through a 30-cm-diameter riveted steel
e = 3 mm, with a head loss of 6 m in 300 m.
Determine the flow.
The relative roughness is e/D = 0.003/0.3 = 0.01, and from Fig. 5.32
a trial/ is taken as 0.04. By substituting into Eq. (5.8.4),
pipe,
m
6
nA
0.04
=
from which
m (7 m/s)
m
0.3
2 X 9.806 m/s
2
300
X
V =
2
From Appendix
1.715 m/s.
C,
v
=
1.13
-6
m /s, and so
X
10
=
0.1245
2
_ YD _ CITlWgKOjOm)
1.13
v
From
the
X
10- 6
m /s
2
Moody diagram / =
n
7 =
Q = Aav
0.038,
,mz m)^2 \ (6 m X
\
2
tt(0.15
0.3
and
m)
—m)
(2) (9.806
m/s 2 )
rr^TTTT^;
0.038(300
m /s
3
Z) unknown, there are three unknowns in Eq.
D;
in
two
the
continuity
equation, V, D; and three in the Rey(5.8.4), /,
nolds number equation, V, D, R. The relative roughness is also unknown.
Using the continuity equation to eliminate the velocity in Eq. (5.8.4) and
In the third case, with
V,
in the expression for
h/
=
f
D2g(D
2
R
simplifies the problem.
Equation
(5.8.4)
becomes
T/4) 2
or
h f gw 2
in
which Ci
is
the
known quantity 8LQ 2 /h f gir 2
.
As
VD =
2
4Q/tt from con-
tinuity,
R =
™ = -^D = ^D
4
V
in
which
(5 10 14 )
.
.
TTV
C2
is
the
known quantity 4Q/m>. The
the following procedure:
1
Assume a value
2.
Solve Eq. (5.10.13) fori).
3.
Solve Eq. (5.10.14) for R.
4.
Find the relative roughness e/D.
of
/.
solution
is
now
effected
by
VISCOUS EFFECTS: FLUID RESISTANCE
With R and e/D, look up a new/ from Fig.
Use the new /, and repeat the procedure.
5.
6.
When
7.
all
5.32.
the value of / does not change in the
equations are satisfied and the problem
Normally only one or two
301
is
two
first
significant figures,
solved.
trials are required.
Since standard pipe sizes
by the comNominal standard pipe sizes are |, J, f, J, f, 1, 1J, \\, 2,
2i, 3, 3i, 4, 5, 6, 8, 10, 12, 14, 16, 18, 24, and 30 in. The inside diameters are
larger than the nominal up to 12 in. Above the 12-in size the actual inside
diameter depends upon the "schedule" of the pipe, and manufacturers' tables
should be consulted. Throughout this chapter the nominal size is taken as the
are usually selected, the next larger size of pipe than that given
putation
taken.
is
actual inside diameter.
EXAMPLE
Determine the
5.14
convey 4000
gpm
The discharge
From Eq.
(5.10.13)
D^ =
10,000
8
b
X
75
=
oil, v
0.0001
wrought-iron pipe required to
size of clean
2
ft /s,
10,000
ft
with a head loss of 75
ft -lb/lb.
is
X 8.93
X 32.2 X
7T
2
2
,
/
J
=
M
nj
267.0/
J
and from Eq. (5.10.14)
4
~
X
8.93
If
/
Fig. 5.32,
= 0.019.
maximum
/
Fig. 5.32,
=
/
113,800
D ~
ttO.0001
and from
1
0.02,
=
D
e
=
D =
0.019.
Therefore,
0.00015
1.398
ft.
R =
ft,
81,400,
e/D = 0.00011,
In repeating the procedure,
D =
X
1.382
allowable, an 18-in pipe
is
12
=
16.6 in.
D
If
=
1.382,
and from
R =
82,300,
a 75-ft head loss
is
the
required.
In each of the cases considered, the loss has been expressed in feet of
head or in foot-pounds per pound. For horizontal pipes, this loss shows up as
a gradual reduction in pressure along the line. For nonhorizontal cases, the
energy equation (3.10.1) is applied to the two end sections of the pipe, and
the loss term is included; thus
^T
+ - + zi = ^- + ^ +
Z2
+
h,
(5.10.15)
FUNDAMENTALS OF FLUID MECHANICS
302
which the kinetic-energy correction factors have been taken as unity. The
upstream section is given the subscript 1 and the downstream section the
subscript 2. The total head at section 1 is equal to the sum of the total head
at section 2 and all the head losses between the two sections.
in
EXAMPLE
gravity
is
5.15
In the preceding example, for
0.85, p\
sure at section
=
40
psi, z\
4 ° P8i
X
200
ft,
and
z2
D =
16.6 in,
=
ft,
50
if
the specific
determine the pres-
2.
In Eq. (5.10.15) Vi
0.85
=
0.433 psi/ft
+
= V2 hence
200ft
;
=
_
0.85
P
.„ +
:»i psi/ft
X 0.433
50ft
+
75
ft
and
p 2 = 67.6
psi
Digital-computer solution of simple pipe problems
In solving simple pipe problems by computer, the Colebrook equation (5.10.7)
in place of the Moody diagram, which is its graphical repre-
would be used
C
SCLUTICN CF SIMPLE PIPE PRCBLfcM TC FINJ HLAC LOSS.
C
CCNSISTEM UMTS.
DATA GIVEN IN
REAL K,MU,L
FURMAT<«0
EPS»* ,F8.6, 3H 0=,F7.4,3H Q»,F7.3,3H L=,F8.2,4H MU=
2F9.7,5h RHC=,F8.3,3H G=,F7.3)
F0PMAT(»0
FEAD=',F10.3,3H R = ,FU.l,7H E PS/C = ,Ftt. 6 , 3H F=,F7.4J
3
NAMEL ST/CAT A/EPS tD,U,L, ML ,*HL,G
2
I
1
REAC(5,CMA,eN0=99)
EPS,J,U,l,MU,KHC,G
V=C/(.7€54*D*D)
R=V*0*RK3/MU
K=tPS/D
WRITE(fc,2)
A=. 94* K**. 22 5*. 53 *K
8=ea.*K**.44
C=l.62*K*». 134
F»A*B/R«*C
HEA0=F*L*V*V/(2.*G*0)
h«ITE(fc,3) HEA0,K,K,F
GC TO 1
99 CALL SYSTEM
EMC
CCATA EPS=8.5E-4, 0= 1.333, 0=6. 62. L= 12 34 6. ,MU=2.E-5 , RH0M.935 , G=32.2 CEND
CJATA EPS=.0015,D=0.6,Q=.35,L*500. ,MU= .062 ,RHO=d60. ,G*9.806 , &EN0
EPS=C.CCC850 0= 1.3330 0* 6.O20 L=12346.0C ML*C. 0000200 Rh0=
1.935 G» 32.200
HEAC=
62.752 K=
611769.9 tPS/U=0. 00063 8 F= 0.0194
0=
EPS»0.CC1500
O.toOOO Q*
0.350 L= 500.00 ML=0.062000C RHO* 860.000 G* 9.806
H£AO=
2.178 R«
IJ302.3 EPS/0=0. 002500 F= 0.0334
Fig. 5.33
Computer program,
loss in a single pipe.
data,
and results
for
determination of head
VISCOUS EFFECTS: FLUID RESISTANCE
SCLUTICN OF SIMPLE PIPt PFCBLEM TC FINJ LISCH*RGE.
CATA IN CONSISTENT UMTS
REAL K.fU.L
FCRMATCO ANSWER UCES NUT CONVERGE')
1
EPS = .F8.6.3H 0=iF7.4,6H HEAL= t F 1 0.3 »3H L=tF8.2 f 4H
2 FORMATI'O
2U=tF9.7,5H KHC=,F8.3,3H G=tF7.3t3H F=,F7.4)
3 FOPMATl'O
C = , tF7.3,3H V=,F7.3,3H F=,F/.4,3F R=fFll.l»
NAMEL ST /DAT A/ EPS tO, HE A0,L,MU f RHO,G,F
5 REAC(5,CATA,END=99)
WRITE (6, 2 J EPStJ.h£AO,L,ML,KhCtG,F
C
C
'
303
M
I
=
1
6
K=EPS/C
A^.C94*K**.225*.S3*K
B=88.*K««.44
C=1.62*K**.134
V=SQRT12.*G*i>*HEAU/(F*U)
R=V*D*KFO/MU
Fl = A*t!/R**C
IF(A8S<Fl-F).LT..0CC5) GU TO
1=1*1
.EC.d)
IF
F =
7
GO TO
1
7
8
F1
GO TJ 6
Q=.7d5<t»C*r>*V
WRITE(6,3» Q.V.F.R
GO TO
8
5
WRITEU.l)
GO
TJ
99 CALL
ENO
5
SYSTEM
tCATA EPS=5.f-^,C=3.3 3,L=l6 50. t hEAO=l8.,MU=l.33E-5,RhO=l.S2.G=32.2tF=.02 fcEND
CCATA EPS=.UC18,C=l.,HEAD=4.,L=200.,MU=.COl,rtHO=997.3,G=9.806, CENO
1.920 G= 32.200 F= 0.0200
EPS=0.JCC500 0= 3.3300 HEAJ=
18.0UU L= 165C.00 MU=0.C0O0t33 RHO=
Q=U3.C90 V= 12.985 F= 0.C139 R= 6242214.
EPS = 0.cO800 0= l.JOOO hE/U=
4.000 L= 200.00 MU=0. 0010000 RHO= 997.300 G= 9.806 F= 0.0139
3.192 V= 4. 064 F= 0.C237 R= 4053252.0
C=
Fig. 5.34
Computer program,
charge
a single pipe.
in
Wood
and results
data,
for determination of dis-
has developed an empirical, explicit form of the Colebrook equation which closely approximates it for values of R > 10,000 and
5
1 X 10- < e/D < 0.04.
It is (for k = e/D)
sentation.
+
f
=
a
a
=
0.094/c
bR
The
-
1
c
225
first
+
0.5Sk
b
=
88/c
-
44
c
=
type, for solution of head loss
EPS =
MU
=
1.62/c
0134
(HEAD),
RHO =
is
direct, as
given in
The second type,
for solution for discharge, starts with an assumption for/, Fig. 5.34. The value
of F (Fl) is improved until the criterion (F does not change by 0.0005 in an
iteration) is satisfied.
The third type, for determination of diameter, also
starts with an assumed value of F, Fig. 5.35. If the criterion is not met in
eight iterations, the program moves to the next set of data. The programs
Fig. 5.33.
may be
In the programs
e,
ju,
p.
modified to include minor losses, which are discussed in the remaining
portion of this section.
1
Don
J.
Wood, An
pp. 60, 61.
Explicit Friction Factor Relationship, Civ. Eng.,
December
1966,
FUNDAMENTALS OF FLUID MECHANICS
304
1
2
SCLUTICN CF SIMPLfc PIPE PPGeLtM FCR DIAMtTfcP.
CAT4 IN CONSISTENT
LN11S.
PEAL K.fUiL
FORMATCO ANSWtR DCtS NOT CONVERGE'!
FORMATl'O
EPS= ,F6 .6 , fcH HE AC= ,F10.3 ,3H C*.F7.3,3H L=,FB.2,4H M
•
2U=.F9.7,5H RHO=,F8.3,3H G=,F7.3,3H F=,F7.<»1
FORMATCO
C=',F7.4,3H M,F7.*,3H «=,Fll.l)
NAPELIST/CATA/EPStHEAD.Ut l,MU,KHO,G,F
5 REAC(5,CATA,END=V9)
WRITE 16,2) EFS,HEAC,Q,L,*L,kFu,G,F
3
1=0
6
D=(F*L*C*C/I2.*G*hEAC*.78 54»»2))**.2
K=EPS/D
V=
C/(.7E5<.»D*D)
P=V*C*RhO/MU
A=.C94*K**.225*.53*K
e=
C =
88.*K*«.<.4
1.62*K**.13<»
Fl=A+b/R**C
IF IAUSU 1-FJ.LT..00C5) GO TO
1
=
7
1*1
IFU.EC.8J GO
TO
8
F = F1
7
GC TJ 6
ViRITE(t,3J D,F,R
GC TO 5
e
wRiTEie.u
GO TO 5
°9 CALL SYSTEM
END
tCATA EPS=.CCC8 5,FEAU = 23.,0=7.5,L = 3200.,MU = <».E-<.,RFC=l.69,G»32.2,F=.02 tENO
COATA EPS=.CC15,hEA0=7.,C=.3,L=l000.,MU=.02,RH0=87C..G*9.8C6,F=.C2, GENU
EPS=C.CCC850 HEAO=
23.000 Q*
7.500 L= 3200.00 MU=0.CC0*OOO RHO»
1.690 G* 32.200 F* 0.0200
0= 1.3758 F= 0.0250 R*
29324.
EPS=0.CC15OO FEAL>=
7.0C0 g =
0.300 L* 1C0C.00 MU=0.C2OOCOO RHO= 87C.000 G= 9.806 F= 0.0200
U= C.5029 F= 0.0303 R=
33039.8
Fig. 5.35
Computer program,
ameter
a single pipe.
in
data,
and results
for determination of di-
Minor losses
Those
which occur in pipelines due to bends, elbows, joints, valves,
etc., are called minor losses. This is a misnomer, because in many situations
they are more important than the losses due to pipe friction considered in
the preceding section, but it is the conventional name. In almost all cases the
minor loss is determined by experiment. However, one important exception
is the head loss due to a sudden expansion in a pipeline (Sec. 3.11).
Equation (3.11.22) may also be written
losses
in
which
K
= [.-(|)']*
From Eq.
(5.10.16)
obvious that the head loss varies as the square of
substantially true for all minor losses in turbulent flow.
it is
This is
convenient method of expressing the minor losses in flow
the velocity.
A
(5-10..7,
the coefficient K, usually determined by experiment.
is
by means
of
VISCOUS EFFECTS: FLUID RESISTANCE
^-4
5.36
Fig.
—
305
*t
Sudden
contraction
a
in
pipeline.
loss
and the
If the sudden expansion is from a pipe to a reservoir, D\/D 2 =
becomes Vi 2 /2g; that is, the complete kinetic energy in the flow is con-
verted into thermal energy.
The head
loss
h c due to a sudden contraction in the pipe cross section,
illustrated in Fig. 5.36,
is
subject to the
provided that the amount
sion,
same
analysis as the sudden expan-
of contraction of the jet
is
known. The process
of converting pressure head into velocity head
is very efficient; hence, the
head loss from section 1 to the vena contracta is small compared with the loss
from section to section 2, where velocity head is being reconverted into pressure head. By applying Eq. (3.11.22) to this expansion the head loss is computed to be
1
K
=
2g
VCA = VA
With the continuity equation
coefficient,
i.e.,
the head loss
is
C
2
the area of jet at section
computed
(I
VTA-2
\Ce
J 2g
2
2,
in
which C c
is
the contraction
divided by the area of section
2,
to be
(5.10.18)
The contraction
coefficient
C
c
for water,
determined by Weisbach, 2
is
pre-
sented in the tabulation.
Ai/A
c
1
2
c
x
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0.624
0.632
0.643
0.659
0.681
0.712
0.755
0.813
0.892
1.00
The vena contracta is the section of greatest contraction of the jet.
Julius Weisbach, "Die Experimental-Hydraulik," p. 133, Englehardt, Freiburg, 1855.
FUNDAMENTALS OF FLUID MECHANICS
306
1.2
D2 =
1
/
/
1.5-n
f-
1.0
/
/
0.8
-^=3
//
0.6
c
K
"""l
tl
0.4
v2
1
"\
—
\- <
~^~~~
/
/
\y
0.2
0°
H,-K
20°
40°
60°
80°
100°
(
V
*g
120°
140°
160°
180°
6
Fig. 5.37
Loss coefficients for conical expansions.
The head
loss at the
entrance to a pipeline from a reservoir
is
usually
taken as 0.5V /2g if the opening is square-edged. For well-rounded entrances,
the loss is between 0.01 V 2 /2g and 0.05 V 2 /2g and may usually be neglected.
For re-entrant openings, as with the pipe extending into the reservoir beyond
the wall, the loss is taken as 1.0V 2 /2g, for thin pipe walls.
The head loss due to gradual expansions (including pipe friction over
2
1
the length of the expansion) has been investigated experimentally by Gibson,
whose
Table
results are given in Fig. 5.37.
5.3
Head-loss coefficients
K for
various fittings
Fitting
Globe valve (fully open
Angle valve (fully open)
Swing check valve (fully open)
Gate valve (fully open)
Close return bend
Standard tee
Standard elbow
Medium sweep elbow
Long sweep elbow
A'
10.0
5.0
2.5
0.19
2
1.8
0.9
0.75
60
.
.
The Conversion of Kinetic to Pressure Energy in the Flow of Water Through
Passages Having Divergent Boundaries, Engineering, vol. 93, p. 205, 1912.
A. H. Gibson,
VISCOUS EFFECTS: FLUID RESISTANCE
307
A summary of representative head loss coefficients K for typical fittings,
Company,
is given in Table 5.3.
be expressed in terms of the equivalent length L e of
pipe that has the same head loss in foot-pounds per pound (meter-newtons
per newton) for the same discharge; thus
published by the Crane
Minor
J
D2g
in
may
losses
2g
which
K may refer to
Solving for
L
1
L
e
one minor head
sum
loss or to the
of several losses.
gives
KD
(5.10.19)
e
f
For example,
=
/
if
the minor losses in a 12-in pipeline add to
0.020 for the
=
1/0.020
1000
line,
ft,
then to the actual length of
and
resistance to flow as the
minor
EXAMPLE 5.16 Find the
= 10 m, and determine
The energy equation
+
7
+ 0^-^ +
PV
—
(13.3
2g
H for Q
applied to points
1
=
60
and
2,
+
and
if
X
same
1/s.
including
——
102 m V
1 V
II +
+ --^+/
2
0.15 m2g
2
2
all
the losses,
2g
+
680/)
Globe valve
Standard elbows
30
m
—
60
m
-
Square -edged entrance
Fig. 5.38
Pipeline with minor losses.
Crane Company, Flow of Fluids, Tech. Pap. 409, May, 1942.
V
p-
V% 2
2
2
15 cm-diam clean cast iron pipe
1
20,
discharge through the pipeline in Fig. 5.38 for
the head loss
or
=
=
be added 20
losses.
2
2
2g
ffi
K
written
2
ff i
may
this additional or equivalent length causes the
H
may be
line
2
X
2
0.9
2g
+
10
^
2g
FUNDAMENTALS OF FLUID MECHANICS
308
When
the head
pipe problem.
V
—
given, this problem
is
If
=
/
is
solved as the second type of simple
0.022,
2
10
=
and
2
(13.3
680
X
0.022)
V =
2.63 m/s.
(2.63
m/s) (0.15 m/1.01 /im
2
R =
+
From Appendix
Repeating the procedure gives
The
discharge
Q = V2A2 =
e/D = 0.0017,
From Fig. 5.32, / = 0.023.
m/s, R = 380,000, and / = 0.023.
1.01 /iw 2 /s,
391,000.
2.60
2
Q known,
m /s
1/s
the solution
is
straightforward
3
0.06
(w*)mU*
>
V =
-
v
m/s) (t/4) (0.15 m) 2 = 46
(2.60
Q
=
/s)
C,
is
For the second part, with
v =
A =
2
= 3A0m/s
R =
505 000
/
'
= 0023
and
8
Hl = 9
2
^/
V^nA
9.806 m/s
X
13 3
'
(
2
With equivalent
+
X 0023 =
680
)
lengths [Eq. (5.10.19)] the value of /is approximated,
say / = 0.022. The sum of minor losses
energy at 2 is considered a minor loss,
13.3X0.15
—
=
tt^z
L =
e
m
1706
„
90.7
rtrt
is
K
=
13.3, in
which the kinetic
m
0.022
Hence the
problem
10
m
=
total length of pipe is 90.7
is
solved
L
f
^
If
/
-
by
+L V
D 2g
2
e
0.022,
m/s and Q =
prove L
V =
2
45.6
>
this
,
2.63 m/s,
1/s.
102
=
192.7 m.
The
first
part of the
m
(7 2 m/s) 2
'0.15m 2g m/s2
192.7
_
+
method,
R =
Normally
391,000, and /
it is
=
0.023, then
not necessary to use the
V =
2
new /
2.58
to im-
e.
Minor
may
be neglected in those situations where they comprise
only 5 percent or less of the head losses due to pipe friction. The friction
factor, at best, is subject to about 5 percent error, and it is meaningless to
losses
VISCOUS EFFECTS: FLUID RESISTANCE
309
more than two significant figures. In general, minor losses
be neglected when, on the average, there is a length of 1000 diameters
between each minor loss.
Compressible flow in pipes is treated in Chap. 6. Complex pipe-flow
situations are treated in Chap. 10.
select values to
may
LUBRICATION MECHANICS
5.11
The
effect of viscosity
amined
on flow and
its effects
on head
in the preceding sections of this chapter.
practical importance
is
losses
have been
A laminar-flow case
the hydrodynamic theory of lubrication.
ex-
of great
Simple as-
pects of this theory are developed in this section.
Large forces are developed in small clearances when the surfaces are
and one is in motion so that fluid is "wedged" into the decreasing space. The slipper bearing, which operates on this principle, is illus-
slightly inclined
trated in Fig. 5.39.
same
The
journal bearing (Fig. 5.40) develops
its
force
by the
action, except that the surfaces are curved.
The laminar-flow equations may be used to develop the theory of lubriThe assumption is made that there is no flow out of the ends of the
bearing normal to the plane of Fig. 5.39. From Eq. (5.1.4), which relates
cation.
pressure drop and shear stress, the equation for the force
will
P
that the bearing
worked out, and the drag on the bearing is computed.
Substituting Newton's law of viscosity into Eq. (5.1.4) produces
support
—=
dx
M
is
—
(5.11.1)
dy 2
Since the inclination of the upper portion of the bearing (Fig. 5.39)
/
/
\y
\
\
'%/j\
^mmz<m^
J*
mW/////M^^^
*—
Fig. 5.39
h2
W/M
—L
Sliding bearing.
-
X
is
very
FUNDAMENTALS OF FLUID MECHANICS
310
Journal bear-
5.40
Fig.
ing.
may
be assumed that the velocity distribution is the same as if the
and that p is independent of y. Integrating Eq. (5.11.1)
twice with respect to y, with dp/dx constant, produces
slight, it
plates were parallel
dp
2
f d u
C
dxJ
dy
=
»Jw
_
dv
+A
dp
du
dx
dy
or
and the second time
dp f
f du
ydy = »
—dy
J
J
—
.
The constants
y
=
dptf
dx2
b;
Eliminating
f
—
+B
0.
,
dy
J
of integration
u = V, y
= Ab
+A
A B
,
_
+B
dpyy l
dp
1
or
— - = ^u + Ay + B
are determined from the conditions u
=
0,
Substituting in turn produces
fiU
+B
A and 5 and
=
solving for u results in
.-i*e-.> + *(-9
The discharge Q must be the same
(5.11.2)
at each cross section.
By
integrating over
a typical section, again with dp/dx constant,
t/b
-r
63
dp
(5.11.3)
VISCOUS EFFECTS: FLUID RESISTANCE
Now,
Q cannot vary with x, b may be expressed in terms of x,
= (bi — b2 )/L, and the equation is integrated with
since
which a
in
to determine the pressure distribution.
=
b
311
bi
—
ax,
respect to x
Solving Eq. (5.11.3) for dp/dx pro-
duces
6nU
dp
dx
—
(bi
12»Q
ax) 2
(5.11.4)
—
(bi
ax)
:
Integrating gives
/dp
fdx
dx
,
~
=
dx
f
TT
%t,U /
J
(6i
—
„ f
12 M Q /
ax) 2
J
(6i
dx
—
rr
ax) 3
+C
or
6M Q
6 M /7
=
—
a(bi
ax)
In this equation
—
a(bi
Q and C
c
{
ax) 2
are unknowns. Since the pressure
say zero, at the ends of the bearing, namely, p
the constants can be determined,
Uhb
+
bi
=
0,
x
=
must be the same,
0;
p
=
0,
x
=
L,
6txU
2
__
a(bi
b2
With these values
QnUx{b
-
+
2
6 (&i
+
b2 )
inserted, the equation for pressure distribution
becomes
b2 )
(5.11.5)
b2 )
is positive between x =
and x = L if b > b 2
show the distribution of pressure throughout the
one-dimensional method of analysis the very slight change
This equation shows that p
.
It is plotted in Fig. 5.39 to
With
bearing.
this
in pressure along a vertical line x
The
total force
L
p =
f
pdx =
6fiU
7
^0
P
=
const
is
neglected.
that the bearing will sustain, per unit width,
L
f
is
x(b-b )dx
2
/
^+
^2
&
•/<)
After substituting the value of b in terms of x and performing the integration,
(&i-& 2
2
)
\
b2
bi
+
b2 /
FUNDAMENTALS OF FLUID MECHANICS
312
The drag
by
force
D
required to
move
the lower surface at speed
U
is
ex-
pressed
L
f
D =
L
t
Jo
By
ax
\
—
du
f
= —
/x
I
U=o
dx
dy
Jo
V=0
evaluating du/dy from Eq. (5.11.2), for y
du
b
dp
U
dy
2/x
dx
b
J/=0
=
0,
This value in the integral, along with the value of dp/dx from Eq. (5.11.4),
gives
-
^T^K^ ^)
2fiUL (
bi
&i
62\
3
The maximum load
With
P =
0.16^
6
2>
=
2
£
is
computed with Eq.
(5.11.6)
when
6i
=
-
1L7)
2.26 2
.
this ratio,
2
The
P
(5
0.75^
o
(5.11.8)
2
optimum load
ratio of load to drag for
is
= 0.2l£
(5.11.9)
which can be very large since
EXAMPLE
5.17
A vertical
62
can be very small.
turbine shaft carries a load of 80,000 lb on a thrust
rocker plates, 3 by 9 in, arranged with their long
dimensions radial from the shaft and with their centers on a circle of radius
1.5 ft. The shaft turns at 120 rpm; M = 0.002 lb-s/ft 2
If the plates take the
bearing consisting of 16
flat
.
angle for
maximum
load, neglecting effects of curvature of
path and radial
lubricant flow, find (a) the clearance between rocker plate and fixed plate;
(6) the
torque loss due to the bearing.
(a)
U =
1.5
The load
Since the motion
(W) (2t) =
is
is
considered straight-line,
18.85 ft/s
5000 lb for each
plate,
L =
0.25
which
is
ft
5000/0.75
=
6667 lb for unit
VISCOUS EFFECTS: FLUID RESISTANCE
Hydrostatic
Fig. 5.41
313
lu-
brication by high-pressure
pumping
width.
b2
of oil.
By
-^
=
\l
(6)
solving for the clearance b 2 from Eq. (5.11.8),
,
=
-?
P
0.4
-
The drag due
D - 0.75^ =
X
0.75
0.002
X
For a 9-in
16
X
22.2
X
D =
X
18.85
29.6
X
=
6667
X
X
10- 4
0.75
=
0.002
2.38
plate,
V\
to one rocker plate
h
rocker plates
•25
18.85
X
22.2
X
lO" 4
=
0.0029 in
per foot of width,
is,
0.25
lb.
2.38
_
The torque
loss
due to the 16
is
1.5
= 533
ft -lb
Another form of lubrication, called hydrostatic lubrication, has many
important applications. It involves the continuous high-pressure pumping
1
of oil
under a step bearing, as illustrated in Fig. 5.41. The load may be lifted
starts, which greatly reduces starting fric-
by the lubrication before rotation
tion.
PROBLEMS
5.1
Determine the formulas
tribution for flow in Fig. 5.1
1
for shear stress
when an adverse
on each plate and for the velocity
pressure gradient exists such that
dis-
Q=
0.
For further information on hydrostatic lubrication see D. D. Fuller, Lubrication Mechanics, in V. L. Streeter (ed.), "Handbook of Fluid Dynamics," pp. 22-21 to 22-30,
McGraw-Hill, New York, 1961.
FUNDAMENTALS OF FLUID MECHANICS
314
20
psi
V////////////////A
*
t
u = l poise
2
in.
K
diam
U
F
jszzzzzzzzzzzazzzzzzzzzzzzzzzzm
+
^
u
-
ZZ2//////////////MM//f////tZa
zz
V**-
Radial clearance^
0.002
in.
(a)
(b)
Fig. 5.42
5.2
In Fig.
5.1,
that the shear
is
with
U
positive as shown, find the expression for d(p
zero at the fixed plate.
What
is
-f-
yh)/dl such
the discharge for this case?
In Fig. 5.42a, [7=2 ft/s. Find the rate at which oil is carried into the pressure
chamber by the piston and the shear force and total force F acting on the piston.
5.3
5.4
Determine the force on the piston of Fig. 5.42a due to shear and the leakage from
the pressure chamber for
5.5
Find
F and U in
U=
0.
Fig. 5.42a such that no oil
is
lost
through the clearance from the
pressure chamber.
5.6
Derive an expression for the flow past a fixed cross section
flow between the two
moving
In Fig. 5.426, for pi = p 2 = 1 kg//cm 2
find the shear stress at each plate.
5.7
5.8
Compute
between
5.9
of Fig. 5.426 for
laminar
plates.
the kinetic-energy and
,
U = 2V = 2m/s,a=
momentum
1.5
mm, /x =
0.5 P,
correction factors for laminar flow
fixed parallel plates.
Determine the formula
for angle
for fixed parallel plates so that laminar flow
at constant pressure takes place.
5.10 With a free body, as in Fig. 5.43, for uniform flow of a thin lamina
an inclined plane, show that the velocity distribution is
u=
— (6
2m
Fig. 5.43
2
—
s
2
)
sin
of liquid
down
VISCOUS EFFECTS: FLUID RESISTANCE
315
Fig. 5.44
and that the discharge per unit width
Q=
—¥
is
sin
Derive the velocity distribution of Prob. 5.10 by inserting into the appropriate
5.11
equation prior to Eq. (5.1.2) the condition that the shear at the free surface must be
zero.
In Fig. 5.44, Pl
5.12
=
y = 50 lb/ft 3 and ju =
on the upper plate and
,
a
=
5.14
=
For 6
5.13
3
mm,
The
6
psi,
0.8 P.
p 2 = 8 psi, I = 4 ft, a = 0.006 ft, 6 = 30°, U = 3 ft/s,
Determine the tangential force per square foot exerted
its direction.
90° in Fig. 5.44, what speed
=
pi
p 2 and
,
/x
=
0.2
U
is
belt conveyer (Fig. 5.45) delivers fluid to a reservoir of such a
the velocity on the free-liquid surface on the belt
done by the belt on the
energy to the
5.15
fluid
What
Fig. 5.45
S =
0.83,
fluid in shear, find
how
is
zero.
depth that
By considering only the work
efficient this
device
is
in transferring
fluid.
is
the velocity distribution of the fluid on the belt and the volume rate of
being transported in Prob. 5.14?
'/////////,
required for no discharge?
kg/m«s.
FUNDAMENTALS OF FLUID MECHANICS
316
What
is
section that
is
5.16
A
5.17
the time rate of momentum and kinetic energy passing through a cross
normal to the flow if in Eq. (5.1.3) Q = 0?
film of fluid 0.005 ft thick flows
Determine the
velocity of 2 ft/s.
Determine the
5.18
Water
momentum
down a
fluid viscosity,
fixed vertical surface with a surface
y
=
55 lb/ft 3
.
correction factor for laminar flow in a round tube.
at standard conditions
is flowing laminarly in a tube at pressure pi and
This tube expands to a diameter of 2d\ and pressure p2, and the flow is
again described by Eq. (5.2.6) some distance downstream of the expansion. Determine the force on the tube which results from the expansion.
5.19
diameter
d\.
At what distance r from the
5.20
center of a tube of radius r does the average velocity
occur in laminar flow?
Determine the
5.21
D
maximum
with fluid properties
Show
5.22
/jl
wall shear stress for laminar flow in a tube of diameter
and p given.
that laminar flow between parallel plates
through an annulus for 2 percent accuracy
if
may
the clearance
be used in place of flow
is
no more than 4 percent
of the inner radius.
What
5.23
are the losses per kilogram per meter of tubing for flow of
35°C through 0.6-mm-diameter tube
at
R =
mercury at
1800?
5.24 Determine the shear stress at the wall of a r^-in-diameter tube when water at
80°F flows through it with a velocity of 1 ft/s.
Determine the pressure drop per meter
5.25
/x
=
=
60 cP, sp gr
0.83, at
R =
of
3-mm-ID tubing
for flow of liquid,
100.
Glycerin at 80°F flows through a f-in-diameter pipe with a pressure drop of
5.26
Find the discharge and the Reynolds number.
5 psi/ft.
Calculate the diameter of vertical pipe needed for flow of liquid at
5.27
when the
pressure remains constant,
v
=
Calculate the discharge of the system in Fig. 5.46, neglecting
5.28
through the pipe.
-1
=
55 lb/ft
16
ft
20
ft
diam
-:^=0.1 poise;
Fig. 5.46
R=
1800
1.5/mi2/s.
all losses
except
VISCOUS EFFECTS: FLUID RESISTANCE
317
Fig. 5.47
In Fig. 5.47,
5.29
H
=
0.08 kg/m-s.
liters
H=
10 m,
L =
Find the head
30°,
D=
8
mm, y =
loss per unit length of pipe
kN/m
3
and
and the discharge in
10
,
per minute.
H
5.30
In Fig. 5.47 and Prob. 5.29, find
5.31
Oil,
When
the shear stress at the outer wall
sp gr 0.85,
\i
=
What
is
if
the velocity
is
3 m/s.
0.50 P, flows through an annulus a
is
on the inner tube per foot
the Reynolds
=
0.60
in, b
=
0.30
in.
0.25 lb/ft 2 calculate (a) the pressure drop
,
the discharge in gallons per hour, and (c) the
per foot for a horizontal system, (6)
axial force exerted
5.32
=
20 m,
number
of length.
of flow of 0.3
m /s
3
=
0.27 P,
in a f-in-diameter
tube to
oil,
sp gr 0.86,
ju
through a 45-cm-diameter pipe?
Calculate the flow of crude
5.33
yield a
Determine the velocity
5.34
similar to the flow of 3
5.35
oil,
sp gr 0.86, at
80°F
Reynolds number of 1200.
What
is
of kerosene at
m /s air at
3
the Reynolds
1.4 kg// cm 2
number
90°F in a 3-in pipe to be dynamically
abs and 15°C through a 75-cm duct.
for a sphere 0.004 ft in diameter falling
through
Q Ap by
integra-
water at 80°F at 0.5 ft/s?
5.36
Show
that the power input for laminar flow in a round tube
is
tion of Eq. (5.1.7).
5.37 By use of the one-seventh-power law of velocity distribution w/wmax = (y/fo) 111
determine the mixing-length distribution l/r in terms of y/r from Eq. (5.4.4)
,
A fluid is agitated so that the kinematic eddy viscosity increases linearly from
= 0) at the bottom of the tank to 0.2 m /s at y = 60 cm. For uniform particles
with fall velocities of 30 cm/s in still fluid, find the concentration at y = 30 cm if it is
10 per liter at y = 60 cm.
5.38
zero (y
5.39
2
Plot a curve of e/u*r as a function of y/r using Eq. (5.4.9) for velocity distribu-
tion in a pipe.
FUNDAMENTALS OF FLUID MECHANICS
318
Find the value
5.40
of y/r
in a pipe
where the velocity equals the average velocity
for turbulent flow.
Plot the velocity profiles for Prandtl's exponential velocity formula for values
5.41
n
of
of y, i,
5.42
and
J.
Estimate the skin-friction drag on an airship 300 ft long, average diameter 60
mph traveling through air at 13 psia and 80°F.
ft,
with velocity of 80
The velocity distribution in a boundary layer is given by u/U = 3(?//6) —
2
Show that the displacement thickness of the boundary layer is 8\ = 5/6.
2(y/8)
5.43
.
5.44 Using the velocity distribution u/U = sin (iry/25) determine the equation for
growth of the laminar boundary layer and for shear stress along a smooth flat plate in
,
two-dimensional flow.
5.45
Compare the drag
coefficients that are obtained
with the velocity distributions
given in Probs. 5.43 and 5.44.
5.46
Work
out the equations for growth of the turbulent boundary layer, based on
the exponential law
5.47
Air at 20°C,
How
km/h.
8
u/U =
(y/8)
119
and/ = 0.185/R 1/5
.
(r
=
pf
72/8.)
kg// cm 2 abs flows along a smooth plate with a velocity of 150
1
long does the plate have to be to obtain a boundary-layer thickness of
mm?
5.48
The
walls of a
wind tunnel are sometimes made divergent to
offset the effect of
the boundary layer in reducing the portion of the cross section in which the flow
is
of
constant speed. At what angle must plane walls be set so that the displacement thickness does not encroach
greater than 0.8
What
ft
upon the tunnel's constant-speed cross section at distances
of the wall? Use the data of Prob. 5.47.
from the leading edge
ball, sp gr 3.5, dropped in
P? What would be the terminal velocity for the same-size ball
but with a 7.0 sp gr? How do these results agree with the experiments attributed to
Galileo at the Leaning Tower of Pisa?
5.49
oil,
is
sp gr 0.80,
5.50
fx
=
1
At what speed must a 15-cm sphere
drag of 5
5.51
the terminal velocity of a 2-in-diameter metal
A
spherical balloon contains helium
Balloon and payload weigh 300
Cd =
travel through water at 10°C to
have a
N?
lb.
and ascends through
What diameter
0.21. If the balloon is tethered to the
ground
in a
air at
14 psia, 40°F.
permits ascension at 10 ft/s?
10-mph wind, what
is
the angle
of inclination of the retaining cable?
5.52
How many
30-m-diameter parachutes (Cd = 1.2) should be used to drop a
kN at a terminal speed of 10 m/s through air at 100,000 Pa abs
bulldozer weighing 45
at
20°C?
5.53
An object weighing 300 lb is attached to a circular disk and dropped from a plane.
What diameter
should the disk be to have the object strike the ground at 72 ft/s?
The disk is attached
so that
it is
normal to direction
of motion,
p
=
14.7 psia;
t
=
70°F.
VISCOUS EFFECTS: FLUID RESISTANCE
A
5.54
0.0024 slug/ft
)
it is
.
normal to
in diameter
force
is
held normal to a 100
required to hold
is
km/h
=
airstream (p
at rest?
it
when
plane
its
parallel to the flow
is
and
it.
A semitubular cylinder of 6-in radius with concave side upstream is submerged in
5.56
water flowing 2
A
5.57
On
5.58
plane
Calculate the drag for a cylinder 24
ft/s.
form
projectile of the
km/s through
5.59
What
3
Discuss the origin of the drag on a disk
5.55
when
1
m
circular disk 3
319
=
p
air,
If
c
;
Fig. 5.25,
,
=
300 m/s.
the basis of the discussion of the
often seen before
is
of (a)
kg/m 3
1
an airplane
1
it is
What
Mach
long.
mm in diameter and travels at
108
is
ft
is its
drag?
angle explain
why
a supersonic air-
heard.
mi above the earth passes over an observer and the observer
it has traveled 1.6 mi farther, what is its speed? Sound
does not hear the plane until
velocity
is
1080
ft/s.
What
is its
Mach
angle?
5.60 Give some reason for the discontinuity in the curves of Fig. 5.23 at the angle of
attack of 22°.
5.61 What is the ratio of
attack of 2°?
5.62
mm,
5.63
Determine the
in crude oil at
A
lift
to drag for the airfoil section of Fig. 5.23 for an angle of
settling velocity of small
spherical dust particle at an altitude of 50
atomic explosion. Determine the time
ance with Stokes' law.
Its size
Use isothermal atmosphere
5.64
at
How
20°C
5.65
it
will
mi
is
diameter 0.1
radioactive as a result of an
take to settle to earth
and sp gr are 25
pm
and
2.5.
if it
falls in
accord-
Neglect wind effects.
large a spherical particle of dust, sp gr 2.5, will settle in atmospheric air
What
in obedience to Stokes' law?
The Chezy
A
4.5,
at 0°F.
is
the settling velocity?
coefficient is 127 for flow in a rectangular
with bottom slope of 0.0016.
5.66
metal spheres, sp gr
25°C.
rectangular channel
What
m
1
is
channel 6
ft
wide, 2
ft
deep,
1
m /s.
the discharge?
wide,
Ch6zy C
=
S =
60,
0.0064, carries
3
Determine the velocity.
5.67
5.68
slope
5.69
What
A
is
is
the value of the
rectangular, brick-lined channel 6
ft
factor
wide and 5
n
ft
in Prob. 5.66?
deep carries 210
cfs.
What
required for the channel?
The channel
a slope of 0.0009.
5.70
Manning roughness
A
cross section
What
is
shown
in Fig. 5.48
is
made
of
unplaned wood and has
the discharge?
trapezoidal, unfinished concrete channel carries water at a depth of 6
bottom width is 8 ft and side slope
0.004 what is the discharge?
1
horizontal to l| vertical. For a
ft.
bottom slope
Its
of
FUNDAMENTALS OF FLUID MECHANICS
320
Fig. 5.48
5.71
A
trapezoidal channel with bottom slope 0.003, bottom width 1.2 m, and side
slopes 2 horizontal to
1
vertical carries 6
m /s at a depth of 1.2 m.
3
What is the Manning
roughness factor?
5.72
to
is
What diameter
m /s when its slope
3
5.74
A
5.75
What
5.76
What
S=
0.0049;
A
is
1.
ft
cfs.
is its
of 4
is
m,
side slope 2
The
1
(2 horizontal
required of a semicircular corrugated-metal channel to carry
0.01?
capacity
when
ft in
diameter has a bottom slope
flowing full?
m /s
3
in a gravel trapezoidal
side slopes of 3 horizontal to 1 vertical,
channel with
and bottom slope
km
is
of 0.001.
ft
wide?
0.016.
with a head
loss of 5
is
to be constructed to carry 35
m. The bottom width
the velocity?
5.78
How
does the discharge vary with depth in Fig. 5.49?
5.79
How
does the velocity vary with depth in Fig. 5.49?
Fig. 5.49
on
best velocity for nonscouring
the velocity of flow of 260 cfs in a rectangular channel 12
n=
What
and
the bottom slope required?
trapezoidal channel, brick-lined,
distance of 8
on
is
is
What
Calculate the depth of flow of 60
bottom width
5.77
bottom width 8
semicircular corrugated-metal channel 10
of 0.004.
1
canal,
to be constructed to carry 280
is
,
2.8 ft/s with this material.
5.73
2
A trapezoidal earth
vertical)
1
is
m /s
3
a
4 m, the side slopes
VISCOUS EFFECTS: FLUID RESISTANCE
321
Fig. 5.50
Determine the depth of flow in Fig. 5.49
bottom slope 0.02.
5.80
for discharge of 12 cfs.
It is
made
of
riveted steel with
5.81
Determine the depth y (Fig. 5.50)
5.82
Determine the depth y (Fig. 5.50)
for
for
maximum
maximum
velocity for given
n and
S.
discharge for given n and S.
5.83 A test on 30-cm-diameter pipe with water showed a gage difference of 33 cm on a
mercury-water manometer connected to two piezometer rings 120 m apart. The flow
was 0.23
m /s.
What
3
By using the
5.84
is
the friction factor?
Blasius equation for determination of friction factor, determine the
horsepower per mile required to
lb/ft
3
,
in a
1-cm-diameter pipe,
30
loss is
What
5.87
R=
v
loss per
= 4X
X
3.5
ft in
100
ft of
=
e
X
10
-4
2
ft /s,
y
=
55
(A pipe
is
3
mm,
=
2
number
of 1800.
The
Calculate the discharge in gallons per minute.
tubing.
needed to be "hydraulically smooth" at
smooth when it has the same losses
said to be hydraulically
same
conditions.)
is
the flow through a 3-m-diameter riveted steel
independent of the viscosity of the fluid?
0.03 for
What
m/s
m /s.
Determine the absolute roughness
factor/
5.90
10~~ 5
Above what Reynolds number
5.88
5.89
3.3
kilometer required to maintain a velocity of 4
size galvanized-iron pipe is
10 5 ?
as a smoother pipe under the
pipe,
=
Fluid flows through a ^-in diameter tube at a Reynolds
5.86
head
3.0 cfs liquid, v
through an 18 in pipeline.
Determine the head
5.85
pump
R =
of a 1-ft-diameter pipe that has a friction
1,000,000.
diameter clean galvanized-iron pipe has the same friction factor for
R =
100,000 as a 30-cm-diameter cast-iron pipe?
Under what conditions do the losses in an artificially roughened pipe vary as
some power of the velocity greater than the second?
5.91
5.92
Why does the friction factor increase as the velocity decreases in laminar flow in
a pipe?
FUNDAMENTALS OF FLUID MECHANICS
322
5.93
Look up the
friction factor for
atmospheric
air at
60°F traveling 60
ft/s
through
a 3-ft-diameter galvanized pipe.
5.94
Water
at
20°C
is
to be
pumped through a kilometer of 20 cm diameter wroughtCompute the head loss and power required.
iron pipe at the rate of 60 1/s.
5.95
16,000
ft
3
/min atmospheric
90°F
air at
What
diameter wrought-iron pipe.
is
conveyed 1000
is
ft
through a
4-ft-
the head loss in inches of water?
5.96 What power motor for a fan must be purchased to circulate standard air in a
wind tunnel at 500 km/h? The tunnel is a closed loop, 60 m long, and it can be assumed
to have a constant circular cross section with a 2 m diameter. Assume smooth pipe.
5.97
Must
there be a provision
To what
described in Prob. 5.96?
5.98
2.0 cfs
iron.
If
oil,
each
\x= 0.16
pump
P,
y
=
produces 80
made
to cool the air at
some
section of the tunnel
extent?
54 lb/ft 3
psi,
how
,
is
pumped through a
may
far apart
12-in pipeline of cast
they be placed?
m
long conveys 10 1/s water at 25°C from a
5.99 A 6-cm diameter smooth pipe 150
water main, p = 1.6 MN/m 2 to the top of a building 25
above the main. What
pressure can be maintained at the top of the building?
m
,
5.100
For water at 150°F calculate the discharge
5.101
In Fig. 5.51,
reservoir at the
how much power would be
bottom
for the pipe of Fig. 5.51.
required to
pump
160
gpm from
a
shown?
of the pipe to the reservoir
A 12-mm-diameter commercial steel pipe 15 m long is used to drain an oil tank.
Determine the discharge when the oil level in the tank is 2 m above the exit end of the
5.102
pipe,
5.103
What
5.104
=
m
Two
is
y
0.10 P;
=
For a head
iron
Fig. 5.51
ft
when
loss of 8
260
2-in.-diam
kN/m
3
.
liquid reservoirs are connected
the flow rate
240
wrought
8
ft
by 200
ft of
the difference in elevation
cm water
in a length of
is
200
2-in-diameter smooth tubing.
50
ft?
v
=
0.001
2
ft /s.
m for flow of atmospheric air
VISCOUS EFFECTS: FLUID RESISTANCE
at
15°C through a
1
.25-m-diameter duct,
e
=
1
mm,
323
calculate the flow in cubic meters
per minute.
A
5.105
gas of molecular weight 37 flows through a 24-in-diameter galvanized duct
at 90 psia
and 100°F. The head
slugs per hour?
What
5.106
=
/x
is
0.194
100
loss per
ft of
duct
is
H
2 in
2
What
0.
the flow in
is
mP.
the power per kilometer required for a 70 percent efficient blower to
maintain the flow of Prob. 5.105?
The 100
5.107
lb m/min air required to ventilate a
mine
Neglecting minor
of 12-in-diameter galvanized pipe.
is
admitted through 2000 ft
what head in inches of
losses,
=
water does a blower have to produce to furnish this flow? p
H
= 20 m, L = 150 m,
5.108 In Fig. 5.47
Find the newtons per second flowing.
D=
5 cm,
S=
14 psia;
0.85,
=
p.
=
t
4 cP,
90°F.
=
e
mm.
1
In a process 10,000 lb/h of distilled water at 70°F is conducted through a
smooth tube between two reservoirs having a distance between them of 30 ft and a
5.109
difference in elevation of 4
What
5.110
km
for 1
5.111
e2
=
size of
new
Two
size
cast-iron pipe
with head loss of 2
0.0001
What
ft.
tubing
is
is
needed?
needed to transport 300
types of steel plate, having surface roughnesses of
ft,
have a cost
1/s
water at 25°
m?
differential of 10 percent
more
ei
=
0.0003
smoother
for the
and
With
ft
plate.
an allowable stress in each of 10,000 psi, which plate should be selected to convey 100
cfs water at 200 psi with a head loss of 6 ft/mi?
5.112
An
old pipe 2
m
in
=
30
mm. A
mm. How much
in
pumping
diameter has a roughness of
=
e
12-mm-thick
costs
would
be saved per year per kilometer of pipe for water at 20°C with discharge of 6
m /s?
lining
would reduce the roughness to
e
1
The pumps and motors are 80 percent efficient, and power costs
5.113
Calculate the diameter of
convey 300
5.114
new wood-stave pipe
water at 60°F with a head
cfs
loss of
cent per kilowatthour.
1
in excellent condition
per 1000
1 ft
3
needed to
ft of pipe.
Two oil reservoirs with difference in elevation of 5 m are connected by 300 m of
steel pipe. What size must the pipe be to convey 50 1/s? p = 0.05 kg/m- s,
commercial
7=8 kN/m
5.115
of 3 in
5.116
p
=
1
200
H
2
3
.
cfs air,
=
p
per 1000
Compute
kg// cm
2
,
t
ft.
16 psia,
What
5.118
70°F,
is
to be delivered to a
galvanized pipe
mine with a head
loss
m /niin
air,
needed?
pound due
to flow of 25
3
How
using a 10° conical diffuser?
H in Fig. 5.52 for 125 1/s water at 15°C through commer-
Include minor losses.
In Prob. 5.28 what would be the discharge
the line?
is
20°C, through a sudden expansion from 30- to 90-cm pipe.
Calculate the value of
cial steel pipe.
=
the losses in foot-pounds per
=
much head would be saved by
5.117
t
size
if
Assume a smooth pipe and a well-rounded
a globe valve were inserted into
inlet,
with p
=
1
cP.
FUNDAMENTALS OF FLUID MECHANICS
324
*
H
30
m
30
cm diam
^m
mm
Fig. 5.52
5.119
In Fig. 5.52 for
H=
3 m, calculate the discharge of
through smooth pipe. Include minor
5.120
If
a valve
is
oil,
S =
0.8,
/x
=
7 cP,
losses.
placed in the line in Prob. 5.119 and adjusted to reduce the dis-
charge by one-half, what
is
K for the
valve and what
is its
equivalent length of pipe
at this setting?
5.121
A
water line connecting two reservoirs at 70°F has 5000
steel pipe, three
is
ft of
24-in-diameter
standard elbows, a globe valve, and a re-entrant pipe entrance.
What
the difference in reservoir elevations for 20 cfs?
5.122
Determine the discharge
5.123
Compute the
losses in
in Prob. 5.121
if
power due to flow
the difference in elevation
of 3
m /s
3
is
40
ft.
water through a sudden
contraction from 2- to 1.3-m-diameter pipe.
5.124 What is the equivalent length of 2-in-diameter pipe, / = 0.022, for (a) a reentrant pipe entrance, (6) a sudden expansion from 2 to 4 in diameter, (c) a globe
valve and a standard tee?
Find H in Fig. 5.53 for 200
valve wide open.
5.125
gpm
\l
=
0.1 P,
y = 60
K for the angle valve in Prob. 5.125 for flow of
5.126
Find
5.127
What
is
lb/ft 3 for the angle
,
10 1/s at the same H.
the discharge through the system of Fig. 5.53 for water at 25°C
ff=8m?
210-ft3-in.-diam
Hi
Angle
valve
steel pipe
Fig. 5.53
oil flow,
when
VISCOUS EFFECTS: FLUID RESISTANCE
325
4 ft/sec P
M= 0.80 poise
0.001
in.
Fig. 5.54
Compare the smooth-pipe curve on the Moody diagram with Eq.
5.128
R=
for
10 5 10 6 10 7
,
,
Check the location
5.129
(5.10.4)
.
of line
e/D
=
Moody diagram
0.0002 on the
with Eq.
(5.10.7).
In Eq. (5.10.7) show that when e =
large, it reduces to Eq. (5.10.6)
5.130
R
is
it
reduces to Eq. (5.10.4) and that,
when
very
In Fig. 5.54 the rocker plate has a width of
5.131
bearing will sustain, (b) the drag on the bearing.
Find the
5.132
maximum
1 ft.
Calculate (a) the load the
Assume no
flow normal to the paper.
pressure in the fluid of Prob. 5.131, and determine
its
location.
5.133
Determine the pressure center
5.134
Show
5.135
The shear
for the rocker plate of Prob. 5.131.
that a shaft concentric with a bearing can sustain no load.
stress in a fluid flowing
between two
fixed parallel plates
(a)
is
constant over the cross section
(6)
is
zero at the plates and increases linearly to the midpoint
(c)
varies parabolically across the section
(d)
is
zero at the midplane
(e)
is
none
5.136
The
of these
and varies
linearly with distance
velocity distribution for flow between
(a)
is
constant over the cross section
(6)
is
zero at the plates
(c)
varies parabolically across the section
and increases
(d)
varies as the three-halves
(e)
is
5.137
none
(a)
and
U
Ua/S
5.138
is
power
two
linearly to the
of the distance
fixed parallel plates
midplane
from the midpoint
of these answers
The discharge between two
velocity
from the midplane
answers
and the shear
(b)
Ua/2
parallel plates, distance a apart,
stress is zero at the fixed plate,
(c)
2Ua/S
(d)
Ua
when one has
the
is
(e)
none of these answers
Fluid is in laminar motion between two parallel plates, with one plate in motion,
under the action of a pressure gradient so that the discharge through any fixed
FUNDAMENTALS OF FLUID MECHANICS
326
cross section
zero.
is
The minimum
velocity occurs at a point which
is
distant from the
fixed plate
(a)
a/6
-3*7/4
5.140
The
dp/dx
(d) dp/dx
5.141
(d)
-2U/S
(b)
=
=
The
fi
(e)
minimum
-U/2
(c)
none
velocity
dr/dy
-U/3
(d)
answers
is
(e)
-U/6
stress in one-dimensional
=
(c) dp/dy
dp/dy = dr/dx
none of these answers
(b)
dr/dy
of these
laminar
given by
is
(e)
/x
dr/dx
expression for power input per unit volume to a fluid in one-dimensional
laminar motion in the x direction
(a)
2a/3
between pressure and shear
relation
flow in the x direction
(a)
a/2
(c)
In Prob. 5.138 the value of the
5.139
(a)
a/3
(b)
rdu/dy
(b) r/fj?
(c)
is
\idu/dy
2
(d) r(du/dy)
(e)
none
of these
answers
5.142
When
is in laminar motion at constant depth and flowing down an
measured normal to surface),
liquid
inclined plate (y
(a)
the shear
(c)
r
(d)
the velocity
5.143
=
is
zero throughout the liquid
dr/dy
(b)
The
is
constant throughout the liquid
(e)
is
constant over the cross section
(b)
is
zero at the wall and increases linearly to the center
(c)
varies parabolically across the section
(d)
is
(e)
is
there are no losses
and varies linearly with the radius
none of these answers
zero at the center
When the
pressure drop in a 24-in-diameter pipeline
shear stress in pounds per square foot
(a)
(b)
5.145
at the plate
shear stress in a fluid flowing in a round pipe
(a)
5.144
=
at the surface of the liquid
7.2
(c)
14.4
is
10 psi in 100
is
(d)
720
none of these answers
(e)
In laminar flow through a round tube the discharge varies
(a) linearly as the viscosity
(6)
as the square of the radius
(c)
inversely as the pressure drop
(d)
inversely as the viscosity
(e)
as the cube of the diameter
5.146
When
(a)
-dz/dl
(d)
-d(p
a tube
(b)
+ Pz)/dl
is
inclined, the
-ydz/dl
(e)
term —dp/dl
(c)
-d(p +
is
replaced
-d(p+z)/dl
yz)/dl
by
ft,
the wall
VISCOUS EFFECTS: FLUID RESISTANCE
The upper
5.147
Reynolds number
critical
(a)
important from a design viewpoint
(6)
the
number
at
327
is
which turbulent flow changes to laminar flow
about 2000
(d) not more than 2000
(e) of no practical importance in pipe-flow problems
(c)
5.148
The Reynolds number
O) VD/v
(e)
(b) VDfi/p
none of these answers
5.149
(a)
The lower
200
5.150
ijl
404
5.151
=
0.10 kg/m-s,
808
(b)
(c)
(d)
is
(a)
2460
(6)
(e)
none of these answers
980,000
(d)
40,000
900
(d)
8080
none of these answers
oil,
(c)
(d)
1,178,000
water at 68°F through a 12-in-
14,120,000
is
independent of radial distance from pipe axis
(b)
independent of the shear stress
(c)
zero at the pipe wall
(e)
useful for
(d)
none of these answers
(e)
for 10 cfs discharge of
The Prandtl mixing length
In a
(e)
3-cm-diameter sphere moving 3 m/s through
for a
(a)
5.153
VD/fi
is
The Reynolds number
diameter pipe
5.152
VDp/v
12,000
(c)
The Reynolds number
sp gr 0.90,
(a)
(c)
given by
is
Reynolds number has the value
critical
1200
(6)
for pipe flow
a universal constant
computing laminar-flow problems
fluid
stream of low viscosity
(a)
the effect of viscosity does not appreciably increase the drag on a body
(6)
the potential theory yields the drag force on a body
(c)
the effect of viscosity
(d)
the deformation drag on a body always predominates
(e)
the potential theory contributes nothing of value regarding flow around bodies
5.154
(a)
(6)
(c)
(d)
The
lift
is
limited to a narrow region surrounding a
on a body immersed
in a fluid
stream
body
is
due to buoyant force
always in the opposite direction to gravity
the resultant fluid force on the body
the dynamic fluid-force component exerted on the body normal to the approach
velocity
(e)
the dynamic fluid-force component exerted on the
velocity
body
parallel to the
approach
FUNDAMENTALS OF FLUID MECHANICS
328
5.155
The displacement
thickness of the boundary layer
by boundary shear
(a)
the distance from the boundary affected
(b)
one-half the actual thickness of the boundary layer
(c)
the distance to the point where
(d)
the distance the
(e)
none of these answers
5.156
(a)
(e)
The shear
main flow
Which
e 71
5.158
(b)
The drag
2D/pUH
none of these answers
The average
5.160
(a)
(a)
5.
-q
=
(d)
2
(d)
u/U
velocity divided
(b)
satisfies
the boundary
—
z
(e)
rj
none of these answers
(D = drag)
(d)
pUH/2D
by the maximum
velocity, as given
by the one-
is
( c)
2
(b)
2r)
pUl/2D
(c)
ndu/dy\ v =h
y/8.
17
f
(d)
x 117
(c)
t£u
x llb
(c)
x 112
x 112
(«)
none of these answers
thickness varies as
The turbulent-boundary-layer
l/x 1/5
162
—
77
The laminar-boundary-layer
l/x 1/2
5.161
(&)
(c)
pUl/D
(b)
seventh-power law,
t£o
plate?
coefficient for a flat plate is
(e)
W
is
pdu/dy\ y=o
(c)
flat
cos7iV2
(a)
5.159
of a flat plate
of the following velocity distributions
conditions for flow along a
(a)
boundary
stress at the
0.99
shifted
is
(b) fidu/dy\ y= o
dp/dx
none of these answers
5.157
u/U =
is
(d)
z6/7
(e)
none
of these
answers
thickness varies as
(d)
xAlb
(e)
none of these answers
In flow along a rough plate, the order of flow type from upstream to downstream
is
(a)
laminar, fully developed wall roughness, transition region, hydraulically smooth
(b)
laminar, transition region, hydraulically smooth, fully developed wall roughness
(c)
laminar, hydraulically smooth, transition region, fully developed wall roughness
(d)
laminar, hydraulically smooth, fully developed wall roughness, transition region
(e)
laminar, fully developed wall roughness, hydraulically smooth, transition region
5.163
Separation
is
caused by
(a)
reduction of pressure to vapor pressure
(b)
reduction of pressure gradient to zero
(c)
an adverse pressure gradient
(d)
the boundary-layer thickness reducing to zero
(e)
none of these answers
VISCOUS EFFECTS: FLUID RESISTANCE
5.164
when
Separation occurs
(a)
the cross section of a channel
(b)
the boundary layer comes to rest
(c)
the velocity of sound
(d)
the pressure reaches a
(e)
a valve
5.165
is
is
reduced
reached
minimum
closed
is
The wake
(a)
is
a region of high pressure
(6)
is
the principal cause of skin friction
(c)
always occurs when deformation drag predominates
(d)
always occurs after a separation point
is none of these answers
(e)
5.166
Pressure drag results from
(a)
skin friction
(6)
deformation drag
(c)
breakdown
(d)
occurrence of a
(e)
none of these answers
5.167
A
of potential flow near the forward stagnation point
wake
body with a rounded nose and
(a)
laminar flow
(b)
turbulent subsonic flow
(c)
supersonic flow
(d)
flow at speed of sound
(e)
none of these answers
5.168
A
sudden change
occurs at a Reynolds
(a)
300
(6)
1
5.169
329
The
is
usually best suited for
in position of the separation point in flow
number
(c)
long, tapering tail
of
30,000
effect of compressibility
(d)
3,000,000
(e)
on the drag force
is
none of these answers
to
greatly near the speed of sound
(a)
increase
(6)
decrease
(c)
cause
it
asymptotically to approach a constant value for large
(d)
cause
it
to increase
it
it
around a sphere
about
near the speed of sound
more rapidly than the square
Mach numbers
Mach num-
of the speed at high
bers
(e)
reduce
5.170
it
throughout the whole flow range
The terminal
power
velocity of a small sphere settling in a viscous fluid varies as the
of its diameter
(a)
first
(6)
inverse of the fluid viscosity
(c)
inverse square of the diameter
(d)
inverse of the diameter
(e)
square of the difference in specific weights of solid and fluid
FUNDAMENTALS OF FLUID MECHANICS
330
The
5.171
losses in
power
open-channel flow generally vary as the
of the roughness
(a)
first
(6)
inverse of the roughness
(c)
square of the velocity
(d)
inverse square of the hydraulic radius
(e)
velocity
The most simple form
5.172
(a)
is
steady uniform
(6)
steady nonuniform
(c)
unsteady uniform
(d)
unsteady nonuniform
(e)
gradually varied
In an open channel of great width the hydraulic radius equals
5.173
(a)
computation
of open-channel-flow
y/3
(b)
y/2
(c)
The Manning roughness
5.174
0.020
(a)
0.002
(e)
none of these answers
(b)
(d)
2y/2,
none
(e)
y
of these
answers
coefficient for finished concrete is
0.20
(c)
(d)
dependent upon hydraulic radius
In turbulent flow a rough pipe has the same friction factor as a smooth pipe
5.175
zone of complete turbulence, rough pipes
(a)
in the
(b)
when the friction factor is independent of the Reynolds number
when the roughness projections are much smaller than the thickness
(c)
of the
laminar film
(d)
everywhere
(e)
when the
The
5.176
in the transition
friction factor
is
zone
constant
friction factor in turbulent flow in
smooth pipes depends upon the follow-
ing:
(a)
V, D,
(e)
V L, D, Q,
5.177
(a)
p,
L,
,
f,
5.178
M
(6)
Q, L, Mj p
(c)
V, D,
p, p,
M
(d)
V, D,
M p
,
V
In a given rough pipe, the losses depend upon
V
(6)
/x,
(c)
p
R
(d)
Q
only
(e)
none of these answers
In the complete-turbulence zone, rough pipes,
(a)
rough and smooth pipes have the same friction factor
(6)
the laminar film covers the roughness projections
(c)
the friction factor depends upon Reynolds number only
(d)
the head loss varies as the square of the velocity
(e)
the friction factor
is
independent of the relative roughness
VISCOUS EFFECTS: FLUID RESISTANCE
The
5.179
friction factor for flow of
pipe with a velocity of 5 ft/s
(a)
0.013
0.017
(6)
The procedure
5.180
331
water at 60°F through a 2-ft-diameter cast-iron
is
0.019
(c)
(d)
0.021
to follow in solving for losses
none
(e)
when
Q, L, D,
of these
and
v,
e
answers
are given
is
to
(a)
(b)
(c)
(d)
(e)
assume an/, look up R on Moody diagram, etc.
assume an hf, solve for/, check against R on Moody diagram
assume an/, solve for hf, compute R, etc.
compute R, look up / for e/D, solve for hf
assume an R, compute V, look up /, solve for hf
The procedure to follow
5.181
is
in solving for discharge
when
hf,
L, D,
v,
and
e
are given
to
(a)
(6)
(c)
(d)
(e)
assume an /, compute V, R, e/D, look up /, and repeat if necessary
assume an R, compute/, check e/D, etc.
assume a V, compute R, look up /, compute V again, etc.
solve Darcy-Weisbach for V, compute Q
assume a Q, compute V, R, look up/, etc.
The procedure
5.182
are given
is
to follow in solving for pipe diameter
when
hf,
Q, L,
v,
and
e
to
assume a D, compute V, R, e/D, look up /, and repeat
compute V from continuity, assume an/, solve for D
(c) eliminate V in R and Darcy-Weisbach, using continuity, assume an/, solve for D,
R, look up /, and repeat
(d) assume an R and an e/D, look up /, solve Darcy-Weisbach for V^/D, and solve
simultaneously with continuity for V and D, compute new R, etc.
(e) assume a V, solve for D, R, e/D, look up /, and repeat
(a)
(b)
The
5.183
,
/
due to a sudden contraction are given by
losses
1
F2
\
V22
2
(
1
V
Vo 2
2g
(d)
(Cc
—
—
Vo2
l)
2
(e)
none of these answers
2#
5.184
(a)
(e)
The
losses at the exit of a
2
submerged pipe
(6) 0.05(F /2^)
none of these answers
negligible
5.185
Minor
losses usually
(a)
there are 100
(6)
their loss
is
ft of
may
(c)
in a reservoir are
2
0.5(F /2^)
be neglected when
pipe between special fittings
5 percent or less of the friction loss
(d)
V /2g
2
FUNDAMENTALS OF FLUID MECHANICS
332
there are 500 diameters of pipe between minor losses
(c)
(d)
there are no globe valves in the line
(e)
rough pipe
5.186
used
The length
40
(a)
is
of pipe
200
{b)
=
(/
(c)
0.025) in diameters, equivalent to a globe valve,
300
400
(d)
is
not determinable; insufficient
(e)
data
5.187
(a)
The hydraulic
radius
given by
is
wetted perimeter divided by area
(6)
area divided by square of wetted perimeter
(c)
square root of area
(d)
area divided by wetted perimeter
(e)
none
5.188
(
answers
The hydraulic
a) f
5.189
of these
(6)
2
(c)
radius of a 6
3
(d)
6
by 12 cm
cross section
the same at
is
made
the velocity distribution
(6)
the velocity distribution at any section
(c)
the pressure variation along the bearing
(d)
the shear stress varies linearly between the two surfaces
(e)
the velocity varies linearly between the two surfaces
5.190
A
(a)
0.15
all
4-in-diameter shaft rotates at 240
The
shear stress in an
(6)
1.75
(c)
3.50
oil film,
(d)
centimeters,
that
(a)
of 0.006 in.
in
none of these answers
(e)
In the theory of lubrication the assumption
is
is,
cross sections
the same as
is
is
the same as
rpm
//
=
16.70
if
the plates were parallel
if
the plates were parallel
in a bearing with a radial clearance
0.1 P,
is,
(e)
in
pounds per square
none
of these
answers
foot,
6
COMPRESSIBLE
In Chap.
sidered.
FLOW
viscous incompressible-fluid-flow situations were mainly con-
5,
In this chapter, on compressible flow, one
new
variable enters, the
and one extra equation is available, the equation of state, which
relates pressure and density. The other equations
continuity, momentum,
and the first and second laws of thermodynamics are also needed in the
density,
—
—
analysis of compressible-fluid-flow situations.
In this chapter topics in steady
one-dimensional flow of a perfect gas are discussed.
approach
is
The one-dimensional
limited to those applications in which the velocity
and density
be considered constant over any cross section. When density changes are
gradual and do not change by more than a few percent, the flow may be treated
may
as incompressible with the use of
The
an average density.
following topics are discussed in this chapter: perfect-gas relation-
speed of a sound wave, Mach number, isentropic flow, shock waves,
Fanno and Rayleigh lines, adiabatic flow, flow with heat transfer, isothermal
flow, and the analogy between shock waves and open-channel waves.
ships,
6.1
PERFECT-GAS RELATIONSHIPS
In Sec. 1.6 [Eq. (1.6.2)] a perfect gas
specific heats
V
= pRT
in
which p and
tively, p is
is
defined as a fluid that has constant
and follows the law
(6.1.1)
T
are the absolute pressure
the density, and
R
and absolute temperature, respec-
the gas constant.
In this section specific heats
333
FUNDAMENTALS OF FLUID MECHANICS
334
is introduced and related to specific heats
and the gas constant internal energy and enthalpy are related to temperature
entropy relations are established and the isentropic and reversible poly tropic
are defined; the specific heat ratio
;
;
processes are introduced.
In general, the specific heat
-
..
cv
at constant
volume
is
defined
by
<6.i.2)
(I)
TT)_
y
In words, c v is the amount
by a unit mass of gas to increase its
temperature by one degree when its volume is held constant. In thermodynamic theory it is proved that u is a function only of temperature for a
which u
in
is
the internal energy 1 per unit mass.
of internal-energy increase required
perfect gas.
The
specific
heat
c p at
constant pressure
is
defined
by
(6.1.3)
\eT/ p
= u + p/p. Since p/p
and u is a function only of temperature for a perfect gas, h
depends only on temperature. Many of the common gases, such as water
vapor, hydrogen, oxygen, carbon monoxide, and air, have a fairly small
change in specific heats over the temperature range 500 to 1000°R, and an
intermediate value is taken for their use as perfect gases. Table C.3 of Appendix C lists some common gases with values of specific heats at 80°F.
For perfect gases Eq. (6.1.2) becomes
in
which h
is
equal to
du
=
cv
=
dp
the enthalpy per unit mass given by h
RT
dT
and Eq.
dh
is
(6.1.4)
(6.1.3)
becomes
dT
(6.1.5)
Then, from
h
= u
+
-
= u
+ RT
P
1
The
of
definitions for c v
Eq. (3.2.7)
is
u.
and
cp
are for equilibrium conditions
;
hence the internal energy
e
COMPRESSIBLE FLOW
335
differentiating gives
dh
+ R dT
= du
and substitution
cp
=
cv
of Eqs. (6.1.4)
and
(6.1.5) leads to
+R
(6.1.6)
which is valid for any gas obeying Eq. (6.1.1) (even when cp and c v are changIf cp and c v are given in heat units per unit mass, i.e.,
ing with temperature)
kilocalorie per kilogram per kelvin or Btu per slug per degree Rankine, then
R must have the same units. The conversion factor is 1 Btu = 778 ft -lb
or 1 kcal = 4187 J.
.
The
k
=
specific-heat ratio k is defined as the ratio
^
(6.1.7)
Cv
Solving with Eq. (6.1.6) gives
cp
=
^R
c>
=
^_
(6.X.8)
Entropy relationships
The
law of thermodynamics for a system states that the heat added to a
equal to the work done by the system plus its increase in internal
energy [Eq. (3.7.4)]. In terms of the entropy s the equation takes the form
first
system
is
Tds = du
+ pd~
(3.7.6)
P
which is a relationship between thermodynamic properties and must hold for
all pure substances.
The internal energy change for a perfect gas is
u2
-
Ul
=
cv
(T2
-
Ti)
and the enthalpy change
h-
h=
cp
(T2
-
T{)
(6.1.9)
is
(6.1.10)
FUNDAMENTALS OF FLUID MECHANICS
336
The change
pi
du
+ ^d- =
=
ds
1
1
may
in entropy
cv
dT
—
+ Rn
p
1
1
P
d-
(6.1.11)
p
be obtained from Eqs. (6.1.4) and
(6.1.1).
After integrating,
rp
s2
-
By
s2
si
=
c v In
-*
+ R In -
(6.1.12)
use of Eqs. (6.1.8) and (6.1.1), Eq. (6.1.12) becomes
-
st
= c,lnM(-j
(6.1.13)
or
<»''«
-*-IS©l
s2
and
S2-Si =
cB
[(mi
ln|l~Jl-j
(6.1.15)
These equations are forms of the second law of thermodynamics.
If
the process
is
reversible, ds
=
process should also be adiabatic, dq H
adiabatic process, or s
isentropic.
*i =
Pi*
=
dqH /T, or
=
T ds =
ds =
dqH further,
Thus
0.
;
const; the reversible, adiabatic process
Then, from Eq. (6.1.14) for
s2
=
if
the
for a reversible,
is
therefore
«i,
^
(6.1.16)
k
P2
Equation (6.1.16) combined with the general gas law yields
The enthalpy change
h
-h =
c p (T2
for
- TO =
an isentropic process
cp T,
&-
l\
is
= cT, [(^Y
-
l]
(6.1.18)
COMPRESSIBLE FLOW
The
—
=
and
polytropic process
defined
is
by
const
is
337
(6.1.19)
an approximation to certain actual processes in which p would plot
on log-log paper. This relationship
substantially as a straight line against p
is
work when the polytropic process is revers= fp dV. Heat transfer occurs in a
polytropic process except when n = k, the isentropic case.
frequently used to calculate the
ible,
by substitution
reversible
into the relation
W
EXAMPLE 6.1 Express R in kilocalories per kilogram per
From Table C.3, R = 2077 m-N/kg-K; therefore
R =
EXAMPLE
6.2
and check
in
———
k
1
cp
k
By
kcal
Compute the value
=
0.496 kcal/kg-K
of
R
from the values
and
c p for air
(6.1.8)
=
140— 10
'
'
X
0.240 Btu/lb m -°R
-
0.0686 Btu/lb TO -°R
1.40
converting from Btu to foot-pounds
R =
X
0.0686 Btu/lb m -°R
EXAMPLE
initial
6.3
=
778 ft-lb/Btu
which checks with the value
in
53.3 ft-lb/lb m -°R
Table C.3.
Compute the enthalpy change in 5 kg of oxygen when the
= 130 kPa abs, £i = 10°C, and the final conditions are
conditions are pi
=
500 kPa abs, t 2 = 95°C.
Enthalpy is a function of temperature only.
change for 5 kg oxygen is
p2
H
of k
Table C.3.
FromEq.
R =
*
m-N/kg-K)
(2077
kelvin for helium.
2
-H
l
Xcp (T - T
=
5 kg
=
(5 kg) (0.219
2
By Eq.
(6.1.10) the enthalpy
x)
kcal/kg-K) (95
-
10
K) =
93.08 kcal
FUNDAMENTALS OF FLUID MECHANICS
338
EXAMPLE
Determine the entropy change in 4.0 slugs of water vapor
conditions are pi = 6 psia, h = 110°F, and the final conditions
are p 2 = 40 psia, k = 38°F.
From Eq. (6.1.15) and Table C.3
when the
s2
-
si
6.4
initial
=
+
0.335 In
38 V-"/40\"°'"1
K460
or
S2 - Si= -
(0.271 Btu/lb m -°R) (4.0 slugs) (32.17 lb m /slug)
= -34.7Btu/°R
EXAMPLE
A
6.5
5°C is compressed isentropically
and the work required.
By
kg nitrogen at 1.4 kg//cm 2 abs and
to 3 kg//cm 2 abs. Find the final temperature
cylinder containing 2
Eq. (6.1.17)
v
\
/ O \
(A-l)/ft
=
( -J
(273
+
K)(
5
—
(1.4-1) /l.
=
From
the principle of conservation of energy, the
equal
its
increase in internal energy, since there
tropic process;
u2
—
ui
=
cv
345.6
J
is
K
=
72.6°C
work done on the gas must
no heat transfer in an isen-
i.e.,
(T2
—
= work/kg
Ti) kcal/kg
or
Work =
(2 kg) (0.177 kcal/kg -K) (345.6
EXAMPLE
6.6
-
278 K)
=
23.93 kcal
3.0 slugs of air are involved in a reversible polytropic process
which the initial conditions pi = 12 psia, ^ = 60°F change to p 2 = 20
3
Determine (a) the formula for the process,
psia, and volume V = 1011 ft
work
air,
(c)
the amount of heat transfer, and (d) the
the
the
done
on
(6)
in
.
entropy change.
(a);
V
-£L =
fn =
w
RT
X
12
53.3
X
X U4
32.17(460
+
=
3
0.00194 slugs/ft
*
60)
R was converted to foot-pounds per slug per degree Rankine by multiplying by
COMPRESSIBLE FLOW
Also
32.17.
=
P2
=
Tcrrr
From Eq.
0.002967 slugs/ft
5
(6.1.19)
El _ Vl
n
Pi"
P2
In
In (p 2 /pi)
f|
=
1.20
In (0.002967/0.00194)
In (p 2 /pi)
hence
—=
p
const
1.2
describes the polytropic process.
Work
(b)
of expansion
is
v2
W=
This
PiVi
n
r
/
J Vi
is
pdV
the work done by the gas on
into the integral,
1
—
n
m is the mass of gas. V
=
1011
i^j,
Ui
surroundings.
= p 2V 2 n = pV n
by substituting
if
its
= V (-)
2
I)"
=
2
lOll(ff) 1
/1
-
2
1
=
ft
3
—
n
and
1547
ft 3
Then
_
20
X
144
X
1011
-
12
X
144
X
1548
= _
1-1.2
Hence the work done on the gas
'
is
1,184,000
ft -lb.
Since
339
FUNDAMENTALS OF FLUID MECHANICS
340
From the first law of thermodynamics the heat added minus the
(c)
work done by the gas must equal the increase in internal energy; i.e.,
W
Qh~
= U2 -
Ui
=
cv
m(T -
T{)
2
First
X
20
V2_
T =
2
0.002965
P2R
X
144
X
53.3
566°R
32.17
Then
1,184,000
Qh = —
778
+
0.171
X
32.17 >
— -761 Btu
761
Btu was
s2
—
Si
transferred from the
From Eq.
(d)
mass
of air.
(6.1.14) the entropy
change
is
computed:
-°->°[!(otS)'1--»™
*'•'
and
S2 -
Si
- -0.01420 X
3
X
32.17
= -1.370 Btu/°R
A rough check on the heat transfer can be made
an average temperature
T =
(520
+
566) /2
=
by using Eq. (3.7.5) by using
and by remembering that
,
543,
the losses are zero in a reversible process.
Qh = T(S* -
6.2
Si)
= 543(- 1.386) - -753 Btu
SPEED OF A SOUND WAVE; MACH NUMBER
The speed of a small disturbance in a conduit can be determined by application
momentum equation and the continuity equation. The question is
first raised whether a stationary small change in velocity, pressure, and density
of the
can occur in a channel.
By referring to
written
P
VA =
(
P
+
d P )(V
+ dV)A
Fig. 6.1, the continuity equation
can be
COMPRESSIBLE FLOW
V
V+dV
p
p + dp
p
p
A
A
341
+dp
Fig. 6.1 Steady flow in prismatic
channel with sudden small change
pressure, and density.
in velocity,
in
which
A
is
the cross-sectional area of channel.
The equation can be reduced
to
p
dV
+ V dp
=
When the momentum equation
within the dotted
pA -
(p
+
dp) A
[Eq. (3.11.2) ]
is
applied to the control volume
lines,
=
P
VA(V + dV -
V)
or
= -pVdV
dp
If p
72
dV
=
is
eliminated between the two equations,
dp
(6.2.1)
dp
Hence, a small disturbance or sudden change in conditions in steady flow can
occur only when the particular velocity V = \/dp/dp exists in the conduit.
Now this problem can be converted to the unsteady flow of a small
through
still
velocity
V to
This
is
fluid
the
by superposing on the whole system and
left,
since this in
called the speed of
sound
no way
affects the
c in the
point source would cause a spherical
wave
its
disturbance
surroundings the
dynamics
of the system.
The disturbance from a
emanate, but at some distance
medium.
to
from the source the wavefront would be essentially linear or one-dimensional.
Large disturbances may travel faster than the speed of sound, e.g., a bomb
explosion.
The equation
for speed of
sound
Idp
(6.2.2)
FUNDAMENTALS OF FLUID MECHANICS
342
may be expressed in several useful forms.
can be introduced:
Kin
of elasticity
dV/V
V
which
the volume of fluid subjected to the pressure change dp.
is
dV
dv s
dp
V
vs
p
K may be expressed
K
The bulk modulus
Since
as
= PJP
dp
Then, from Eq.
c
(6.2.2),
= */-
(6.2.3)
This equation applies to liquids as well as gases.
EXAM PLE 6.7 Carbon tetrachloride has a bulk modulus of elasticity of 11 ,460
kg//cm 2 and a density of 1593 kg/m 3 What is the speed of sound in the
.
medium?
IK
c
1
= \/— =
' p
V*
(11,460 kg,/cm») (9.806 N/kg,)
1Kno
/
3
1593 kg/m
i
"
x
"
10
cm 7m =
2
]
840 m/s
Since the pressure and temperature changes due to passage of a sound
wave
are extremely small, the process
is
almost reversible. Also, the relatively
rapid process of passage of the wave, together with the minute temperature
changes,
makes the process almost
adiabatic.
In the
limit, the process
may
be considered to be isentropic,
.
k
pp
=
const
kp
dp
= —
—
dp
p
and
£
(6.2.4)
COMPRESSIBLE FLOW
from the perfect-gas law p
or,
c
343
= pRT,
= y/kRT
(6.2.5)
which shows that the speed of sound in a perfect gas is a function of its absolute temperature only. In flow of gas through a conduit, the speed of sound
generally changes from section to section as the temperature is changed by
density changes and friction effects. In isothermal flow the speed of sound
remains constant.
The Mach number has been defined as the ratio of velocity of a fluid to
the local velocity of sound in the medium,
M = -
(6.2.6)
c
Squaring the
V
which may be interpreted as the
thermal energy, since kinetic energy
and thermal energy is proportional to T. The Mach
Mach number
produces
2
/c 2
,
ratio of kinetic energy of the fluid to its
is
V
proportional to
2
number is a measure of the importance of compressibility. In an incompressible
fluid K is infinite and IVI = 0. For perfect gases
K
= kp
when
(6.2.7)
the compression
EXAM PL E
and
6.8
=
and
c
What is the speed of sound in dry air at
when = — 67°F?
level,
sea level
when
t
= 68°F
t
from Eq.
(6.2.5),
V14 X 32.17 X 53.3(460 + 68)
=
1126 ft/s
=
971 ft/s
in the stratosphere
= y/lA X
6.3
isentropic.
in the stratosphere
At sea
c
is
32.17
X
53.3(460
-
67)
ISENTROPIC FLOW
Frictionless adiabatic, or isentropic, flow
in the flow of real gases. It
nozzles,
is
is
an
ideal that cannot
be reached
approached, however, in flow through transitions,
and venturi meters where friction effects are minor, owing to the short
and heat transfer is minor because the changes that a
distances traveled,
FUNDAMENTALS OF FLUID MECHANICS
344
enough to keep the velocity and temperature
of fluid machines is frequently compared
with the performance assuming isentropic flow. In this section one-dimensional
steady flow of a perfect gas through converging and converging-diverging
particle undergoes are slow
The performance
gradients small. 1
ducts
is
studied.
Some very
general results can be obtained by use of Euler's equation
(3.5.4), neglecting elevation changes,
—=
VdV +
(6.3.1)
P
and the continuity equation
P
AV =
const
(6.3.2)
Differentiating
dp
+
7
dV
pA V and then
dividing through
by pA V gives
dA
+
y T=
From Eq.
VdV +
c
dp
(6.2.2)
2
—
(6.3.3)
obtained and substituted into Eq. (6.3.1), yielding
is
=
(6.3.4)
P
Eliminating dp/p in the last two equations and rearranging give
A/V
dA
A
\
2
ma
The assumptions underlying
frictionless.
(6.3.5)
No
this equation are that the flow
restrictions as to heat transfer
shows that
for subsonic flow
(M <
1),
dA/dV
is
the channel area must decrease for increasing velocity.
M =
is
1
steady and
always negative;
As dA/dV
is
i.e.,
zero for
minimum section or throat
may occur. Also,
flow) dA/dV is positive and
only, the velocity keeps increasing until the
reached, and that
for
is
have been imposed. Equation
is
the only section at which sonic flow
Mach numbers greater than unity
(supersonic
the area must increase for an increase in velocity. Hence to obtain supersonic
steady flow from a fluid at rest in a reservoir,
converging duct and then a diverging duct.
1
H. W. Liepmann and A. Roshko, "Elements
1957.
of
it
must
Gas Dynamics,"
first
pass through a
p. 51, Wiley,
New
York,
COMPRESSIBLE FLOW
When
the analysis
is
restricted to isentropic flow, Eq. (6.1.16)
345
may be
written
= piprV
p
(6.3.6)
Differentiating
V dV +
k
—
u
P
and substituting
k
~2
dP
for
dp
in
Eq. (6.3.1) give
=
Pi*
Integration yields
:
h
k-
2
l Pl
=
: p
kH
const
or
*V
k
Pi_VS
k
Pi
T + *-ift-T + fc^l«
Equation
(6.3.7)
using Eq. (6.1.8)
is
.
when expressed
2
k
V
in
2
f+p^-f
terms of temperature; from p
37)
= pRT
k
2
(6 3 8)
+ t^i**
-
For adiabatic flow from a reservoir where conditions are given by
To, at
-
can be derived from Eq. (3.7.2) for adiabatic flow (dq H = 0)
This avoids the restriction to isentropic flow. This equation
useful
7i
(6
-
po, po,
any other section
V
kR
= j—^ (To-T)
2
In terms of the local
c
2
(k
-
(6.3.9)
Mach number
'o-T)
2
l)A;/er
—
/c
V/c, with
/To
7
1
V7
c
2
= kRT,
\
/
or
p=l + *_i||«'
(6.3.10)
FUNDAMENTALS OF FLUID MECHANICS
346
From
now
Eqs. (6.3.10) and (6.1.17), which
restrict the following equations
to isentropic flow,
N^"-)
*/(*-!)
(6.3.11)
and
i/(*-i)
H>+^»)
(6.3.12)
Flow conditions are termed
there
critical at
V* = \/kRT*. By applying Eqs.
for critical conditions (for k
=
k
p
\k
* -
tn)
when
asterisk.
the velocity
M =
1; c*
=
(6.3.10) to (6.3.12) to the throat section
1.4 in the
0.833
+
T~
the throat section
marked with an
Sonic conditions are
sonic.
is
numerical portion),
1.40
(6.3.13)
1
+
k
=
1.40
(6.3.14)
k
=
1.40
(6.3.15)
1/
= ° 634
These relations show that for airflow, the absolute temperature drops
about 17 percent from reservoir to throat, the critical pressure is 52.8 percent
of the reservoir pressure, and the density is reduced by about 37 percent.
The variation of area with the Mach number for the critical case is
obtained by use of the continuity equation and Eqs. (6.3.10) to (6.3.15).
First
P
AV
in
= P *A*V*
which A*
p*
A_
the minimum, or throat, area.
Then
V*
7T
A*
Now, V* =
Yl
is
(6.3.16)
-
i
(6.3.17)
c*
\T1
= VkRT*, and V =
-
L
III
To
I.
T
=
ctA
1 1+
Ml
=
M
[(fe
^/kRT,
-
so that
l)/2]M
a
1/2
(6.3.18)
[
(/c+l)/2
COMPRESSIBLE FLOW
by use
l+
pop
P
By
and
of Eqs. (6.3.13)
[(*
-
1
[
1
i/(*-i)
(6.3.19)
(*+l)/2
i[
substituting the last
A_
In a similar manner
(6.3.10).
l)/2j*
+ L(k-
two equations
l)/2]M 2
(
into Eq. (6.3.17)
* +1)/2 <*- 1 >
(6.3.20)
|
i*~M|
(fc+l)/2
347
J
which yields the variation of area of duct in terms of Mach number. A /A*
is never less than unity, and for any value greater than unity there will be
two values of Mach number, one less than and one greater than unity. For
gases with k
=
1.40,
1/5 + M
A
2
Eq. (6.3.20) reduces to
V
T*-u\—)
= 140
fc
The maximum mass
flow rate
(6321)
m max
can be expressed in terms of the
throat area and reservoir conditions:
m
r
by Eqs.
(6.3.15)
A*p
and
Ik/
(6.3.13).
2
\
Replacing
Po
by p
/RT
gives
<*+»/<*-»
^--wWst+i)
For
fc
=
m max -
(6 3 22)
'
-
1.40 this reduces to
0.686
—
V/ti
(6.3.23)
o
which shows that the mass flow rate varies linearly as A* and p and varies
inversely as the square root of the absolute temperature.
For subsonic flow throughout a converging-diverging duct, the velocity
must be less than sonic velocity, or M« < 1 with subscript t
indicating the throat section. The mass rate of flow m is obtained from
at the throat
m => va = a
V2
^^©V©H
(6 3 24)
-
-
FUNDAMENTALS OF FLUID MECHANICS
348
which
derived from Eqs. (6.3.9) and (6.3.6) and the perfect-gas law.
is
equation holds for any section and
throat
from Eq.
/
1
subsonic.
is
section,
be applied to the throat section, and for this
(6.3.14),
\k/(k-l)
c,
2
(
may
It
This
applicable as long as the velocity at the
is
V
Po
p is the throat pressure. When the equals sign is used in the expression,
Eq. (6.3.24) reduces to Eq. (6.3.22).
For maximum mass flow rate, the flow downstream from the throat may
be either supersonic or subsonic, depending upon the downstream pressure.
Substituting Eq. (6.3.22) for m in Eq. (6.3.24) and simplifying gives
t
2/
/p\ T
1
/p\
©L -©
(k
=— (*n)
- 1)/k
fc-1/
l
J
\<*+»/<*-i>Ai*Y
2
(6325)
(t)
A may
be taken as the outlet area and p as the outlet pressure. For a given
than unity) there will be two values of p/po between zero and unity,
the upper value for subsonic flow through the diverging duct and the lower
value for supersonic flow through the diverging duct. For all other pressure
ratios less than the upper value complete isentropic flow is impossible and
shock waves form in or just downstream from the diverging duct. They are
A*/ A
(less
briefly discussed in the following section.
Appendix Table C.4
=
fork
1.4.
is quite helpful in solving isentropic flow problems
Equations (6.3.10), (6.3.11), (6.3.12), and (6.3.21) are presented
in tabular form.
EXAMPLE
A preliminary
6.9
ber 3.0 at the exit
t
=
25°C.
Determine
design of a wind tunnel to produce
The mass
desired.
is
kg/s for p
and density at the
m max VRTo ==
At =
0.686p
0.00483
(6)
4.23
1
kg/s
(0.686
m
A =
X
V(287
0.9
m -N/kg -K) (273
kg/ /cm
2
)
of outlet
4.23
X
is
m
2
=
N/kg/ )
(9.806
0.0204
m
kgy/cm 2
and (c) the
,
+.25 K)
(10
determined from Table C.4:
0.00483
Mach num0.9
(6.3.23)
2
The area
=
outlet.
The throat area can be determined from Eq.
(a)
—
A*
is 1
(a) the throat area, (6) the outlet area,
velocity, pressure, temperature,
=
flow rate
2
4
cm /m
2
2
)
COMPRESSIBLE FLOW
From Table C.4
(c)
-
=
- =
0.027
Po
Po
— = 0.357
0.076
i
Po
From
=
349
the gas law
Po
(0.9
=
RT
k g/ /cm2 ) (9.806 N/kg,) (10 4 cm 2 /m 2 )
(287m.N/kg .K)(273
+
25K)
"
,__.__
= L0319
kg/m3
,
hence at the exit
p
=
T =
P
=
0.027
X
0.9
0.357(273
0.076
From
X
kg,/cm 2 = 0.0243 kg,/cm 2
+
K) =
25
1.0319
kg/m =
3
P
A
0.0784
= -166.6°C
kg/m
lkg/s
(0.078
EXAMPLE
ft
K
3
the continuity equation
= mmax =
0.40
106.39
2
6.10
and an
A
kg/m
3
)
(0.0204
m
=
628
2
5
/
m/S
*
)
converging-diverging air duct has a throat cross section of
exit cross section of 1.0 ft 2
.
Reservoir pressure
is
30
psia,
and
Mach numbers and the pressure
range at the exit for isentropic flow. Find the maximum flow rate.
From Table C.4 [Eq. (6.3.21) ]M = 2.44 and 0.24. Each of these values
of Mach number at the exit is for critical conditions; hence the Mach number
temperature
is
60° F. Determine the range of
range for isentropic flow
psia,
is
to 0.24
and the one value
2.44.
From Table C.4 [Eq. (6.3.11)] for M = 2.44, p = 30 X 0.064 = 1.92
and for M = 0.24, p = 30 X 0.961 = 28.83 psia. The downstream pres-
sure range
is
then from 28.83 to 30 psia, and the isolated point is 1.92 psia.
flow rate is determined from Eq. (6.3.23)
The maximum mass
M
0.686
X
- = V53.3X
EXAMPLE
6.11
0.40
A
X
30
32.17(460
X
+
144
W
„
= «L255
rt
.„',«,
^
^^ = ^
,
,
converging-diverging duct in an air line downstream from
a reservoir has a 5-cm-diameter throat. Determine the mass rate of flow
when
FUNDAMENTALS OF FLUID MECHANICS
350
2
p = 8 kg//cm abs,
Po
Po
=
^o
=
(8
7 (0.05
4
xV
=
33°C, and p
kg//cm 2 abs
5
kg/ /cm 2 ) (9.806 N/kg,) (10 4 cm 2 /m 2 )
(287m.N/kg.K)(273
From Eq.
m =
=
t
+
=
=
8 oqq
933
kg/m
'
33K)
(6.3.24)
m) 2
1
(2
at the throat.
/5\ 2/1
4
-
X 8kg/ /cm )(9.806N/kg/ )(10 cm /m )^--^f-J
4
2
2
2
4
[1
-
-
(I)
4'1 4
-
]
1.166 kg/s
SHOCK WAVES
6.4
In one-dimensional flow the only type of shock wave that can occur is a normal
compression shock wave, as illustrated in Fig. 6.2. For a complete discussion
of converging-diverging flow for all
downstream pressure
ranges, 1 oblique
shock waves must be taken into account as they occur at the
In the
exit.
preceding section isentropic flow was shown to occur throughout a convergingdiverging tube for a range of downstream pressures in which the flow was
subsonic throughout and for one downstream pressure for supersonic flow
through the diffuser (diverging portion). In this section the normal shock
in a diffuser is studied, with isentropic flow throughout the tube, except
wave
for the
shock-wave surface.
The shock wave occurs
in supersonic flow
reduces the flow to subsonic flow, as proved in the following section.
very
The
little
thickness, of the order of the molecular
free
has
path of the gas.
controlling equations for adiabatic flow are (Fig. 6.2)
G = - =
piVi
=
V
—+h —
+
fa
Continuit y:
Energy
7i 2
(6.4.1)
P2V2
2
2
ss
which are obtained from Eq.
1
mean
and
It
=
ho
=
(3.7.1) for
V
2
2
+
k
p
(6.4.2)
/c-
no change
1
p
in elevation,
no heat trans-
H. W. Liepmann and A. Roshko, "Elements of Gas Dynamics," Wiley,
1957.
New
York,
COMPRESSIBLE FLOW
Normal
Fig. 6.2
351
compression
shock wave.
and no work done, h = u
fer,
of stagnation enthalpy,
+
i.e., its
p/p
=
cp
T is the enthalpy, and
ho is
the value
value in the reservoir or where the fluid
is
at
Equation (6.4.2) holds for real fluids and is valid both upstream and
downstream from a shock wave. The momentum equation (3.11.2) for a
control volume between sections 1 and 2 becomes
rest.
(Pi
-
P*)A
= P2AV2 2 - piAVS
or
+
Pi
P1V1
=
2
Pi
+
P2V2
(6.4.3)
For given upstream conditions h h p h Vi, pi, the three equations are to
be solved for p 2 P2, V 2 The equation of state for a perfect gas is also available
for use, p = pRT. The value of p 2 is
.
,
1
Vi
/c+
Once p 2
is
OxTY - (k-
l)pj
(6.4.4)
1
determined, by combination of the continuity and
momentum equa-
tions
Pi
+
V
is
2
PiVS = p 2
+
piViVt
readily obtained.
Finally
(6.4.5)
P2 is
obtained from the continuity equation.
FUNDAMENTALS OF FLUID MECHANICS
352
For given upstream conditions, with Mi > 1, the values of p 2 V2 &,
By eliminating V\ and V2 between
2 = V2 /\^kp 2 /p2 exist and M 2 < 1.
Eqs. (6.4.1), (6.4.2), and (6.4.3) the Rankine-Hugoniot equations are ob,
and
,
M
tained
:
P2_ [(fc+l)/(fc-P]Wpi)
"I
(
[(«i+l)/(fc-l)]-pi/ft
Pl
^
and
+
l
P2
[(*+!)/(*-
Pi
V
[(fc+l)/(fc-l)>2 /pi
l)]
+
1
(6.4.7)
V,
p»/pi
These equations, relating conditions on either side of the shock wave, take the
place of the isentropic relation, Eq. (6.1.16), pp~ k
From Eq.
V
2
k
=
const.
(6.4.2), the energy equation,
c* 2
c* 2
p
k
+
1
c* 2
since the equation holds for all points in adiabatic flow without change in
and
elevation,
by Eq.
= \/kp*/p*
c*
P2V2
"-"•is
VV =
l
It
may
<"'- f
satisfied
pi/pi
by use
-'Ks^r +
by Vi =
V
2
of Eq. (6.4.8) leads to
VH
(no shock wave) or by
(6.4.10)
be written
Vi
-;-
=
When
Vi
2
Dividing Eq. (6.4.3)
c* 2
2
V
the velocity of sound.
piVl
and by eliminating p 2 /p2 and
which
is
(6.4.1) gives
(6.4.11)
1
is
greater than
c*,
the upstream
Mach number
is
greater than unity
COMPRESSIBLE FLOW
and
V
c*, and so the final Mach number is less than unity, and
shown in the following section that the process can occur
than
is less
2
vice versa.
353
It is
only from supersonic upstream to subsonic downstream.
By
use of Eq. (6.1.14), together with Eqs. (6.4.4), (6.4.6), and (6.4.7),
for change of entropy across a normal shock wave may be
an expression
Mi and
obtained in terms of
k.
From Eq.
(6.4.4)
-mRS-'-o-»]
Since
=
kpi/pi
-
2fcMi 2
_
P2
c x2
Vi~
(k
-
from Eq.
Vi/ci,
(6.4.12),
1)
(6.4.13)
+
&
and Mi =
1
Placing this value of P2/P1 in Eq. (6.4.7) yields
P2
Mi 2 (/C
=
2
pi
+M
2
x
+
1)
-- 1)
(fc
Substituting these pressure and density ratios into Eq. (6.1.14) gives
—
By
si
=
[2/cMi 2
c v In
-
k
fc+11
substitution of
Mi
>
1
+
1
[" 2
[2
L
+ Mi
2
(/c
-
1)
(6.4.14)
Mi (/c+l)
(/
2
into this equation for the appropriate value of k,
may
be shown to increase across the shock wave, indicating that
the normal shock may proceed from supersonic flow upstream to subsonic
the entropy
Substitution of values of Mi < 1 into Eq. (6.4.14) has no
meaning, since Eq. (6.4.13) yields a negative value of the ratio P2/P1. The
equations for derivation of the gas tables in Table C.5 are developed in the
following treatment. From Eq. (6.3.10), which holds for the adiabatic flow
across the shock wave,
flow downstream.
T
H
k
- 1„
T
k
- 1M
hence
y,
T,
_
+ [(* !+[(*-
1
Since Vi
i)/2]M,'
'
l
1)/2]IVM
= Mi \/kRTi and V2 =
'
M
2
\ZkRT2, use of the
momentum
;
'
equation
FUNDAMENTALS OF FLUID MECHANICS
354
(6.4.3) gives
+ pMMRTi
Vi
= p
2
+
pM*2 kRT2
and
p2
1
Vi
1
+
+
/cMi 2
*
I2
(6.4.16)
2
Now, from the
Mi
Vi
V2T,
(H
continuity equation (6.4.1) and the gas law
Mx
j¥
1
7r^rv = M^i
2
or
m
2
p2
=
2
j¥1
(6A17)
p^t
2
Eliminating T*/Tt and p 2 /p x in Eqs. (6.4.15), (6.4.16), and (6.4.17)
gives
Mi
M
"
2
+
+
1 4- fcMi
fcMi 2
/l
2
>1
1
/cM 2
which can be solved
,2
+
+
- 1)/2]M
1)/2~!M,2 2
4- [(/c
Rfc
!
(6.4.18)
[(/c- l)/2Jfc
M
for
2
nt + 2/(k-i)
"
"
WhenEq.
mined
(6A19)
Da/flb-DJK-i
7
(6.4.19)
is
substituted back into Eq. (6.4.15), T ,/^
is
deter-
in terms of Mi.
To determine the
may write
ratio of stagnation pressures across the
normal shock
wave, one
&-M*
(6A 20)
Vi P1P01
P01
Now
p* =
by use
f
l
+
U+
Pox
By
of Eqs. (6.3.11)
[(fe-l)/2]M,» l
[(*-
and
t/(t
(6.4.13)
^
2
l)/2]Mi»J
fc
(fe
+
-
1)
'
l
1
'
use of Eq. (6.4.19) with Eq. (6.4.21) the stagnation-pressure ratio
expressed in terms of Mi.
In the next section the shock
of
2/clVl!
Fanno and Rayleigh
lines.
wave
is
is
examined further by introduction
COMPRESSIBLE FLOW
EXAMPLE 6.12 If a normal shock wave occurs in the
k = 40°F, Fi = 4500 ft/s, find p 2 &, V2 and t2
From Table C.3, R = 386, k = 1.66, and
psia,
=
Pi
Pl
—
t)\
386
From Eq.
3
0.0000232 slugs/ft
^ '
2
-
(1.66
-
x
1)
144
x
1]
1
(6.4.5)
P
V,-n- -^^
PiVi
FromEq.
=
=
40)
1
317 lb/ft 2 abs
From Eq.
P2
+
=
(6.4.4)
l.oo
=
X
144
32.17(460
x 0.0000232 x 4500
r^m
+ P
=
?2
IX
=
#7\
flow of helium, pi
.
,
,
355
4500
X
~
144
=
0.0000232
2843 ft/s
7
(6.4.1)
Fi
0.000746
V = -^VT
Pl
317
= 4500 -
X
2
4500
=
-° 000367
SlUgS/ft
$843
and
k
= T - 460 = ^- - 460 =
2
6.5
&R
——-— —
0.0000367
X
32.17
X
-
460
= 236°F
386
FANNO AND RAYLEIGH LINES
To examine more
closely the nature of the flow
across a shock wave, where the area
tinuity
may
change in the short distance
be considered constant, the con-
and energy equations are combined
for steady, frictional, adiabatic
By considering upstream conditions fixed, that is, p h Vi, pi, a plot may
be made of all possible conditions at section 2, Fig. 6.2. The lines on such a
plot for constant mass flow G are called Fanno lines. The most revealing plot
flow.
is
that of enthalpy against entropy,
The entropy equation
s-
si
=
c„ln
mi
- -
i.e.,
an hs diagram.
for a perfect gas,
Eq. (6.1.14),
is
(6.5.1)
FUNDAMENTALS OF FLUID MECHANICS
356
The energy equation
Eq. (6.4.2),
=
ho
h .+
(6.5.2)
and the continuity equation
G =
P
=
By
s
cv
T =
+
Si
of state, linking h, p,
ds
is
p,
(6.4.1), is
_
and
V
fc
ha
\G/
-
is
ho
line
/
la
—
or
h
Subsonic
y^Rayleigh
/
line
^-Supersonic
G=pV= constant
'^^Supersonic
*-s
Fig. 6.3
Fanno and Rayleigh
ha
=
h
c
jC,
^<S^
find the conditions for
at the
1
= ^02
'
To
by subscript a values
2
Subsonic^V
.
differentiated with respect to h
1
(6.5.5)
J
kh
^01
p, is
from the four equations,
Fig. 6.3 (not to scale)
indicating
_1
dh
and
-&)<*-!>*]
c v In
shown on
By
Fanno
from Eq.
(6.5.4)
entropy, Eq. (6.5.5)
zero.
in area,
RP
Pi c p
which
no change
(6.5.3)
eliminating p,
=
for
V
The equation
h
with no change in elevation, from
for adiabatic flow
is
lines.
and ds/dh
maximum
+1h
maximum
set equal to
entropy point,
COMPRESSIBLE FLOW
After substituting this into Eq. (6.5.2) to find
h=
=
ha
ha
+
Va
357
,
—
and
Va =
2
-
(k
l)ha
=
(k
-
l)c p Ta
=
(k
-
kR
1)
k
—
Ta = kRTa =
ca2
(6.5.6)
1
Hence the maximum entropy at point a is for M = 1, or sonic conditions. For
h > ha the flow is subsonic, and for h < h a the flow is supersonic. The two
conditions, before and after the shock, must lie on the proper Fanno line for
the area at which the shock wave occurs. The momentum equation was not
used to determine the Fanno line, and so the complete solution is not determined yet.
Rayleigh line
momentum and
Assuming constant upstream conditions and constant
area, Eqs. (6.5.1), (6.5.3), (6.5.4), and (6.4.1) are used to determine the
Rayleigh line. Eliminating V in the continuity and momentum equations
Conditions before and after the shock must also satisfy the
continuity equations.
gives
G
p
2
+-
-
const
= B
(6.5.7)
P
Next eliminating p from
s
=
Si
+c
v
In
—+
Pi
Enthalpy
may
this equation
c v In
P
and the entropy equation gives
(6.5.8)
k
be expressed as a function of p and upstream conditions, from
Eq. (6.5.7)
h
= CpT =
c
V =
Rp
B _<l\
^(
R p\
/
(6 5 9)
.
.
p
The last two equations determine s and h in terms of the parameter p
and plot on the hs diagram as indicated in Fig. 6.3. This is a Rayleigh line.
FUNDAMENTALS OF FLUID MECHANICS
358
The value
maximum
of
by
equations; then
is found by taking ds/dp and dh/dp from the
and equating to zero, using subscript b for maxi-
entropy
division
mum point:
^
Gy[_p {B-Gyph )^-k _
c
=
h
!
dh
Pb
To
satisfy this equation, the
The numerator
not zero.
k
-B
2G*/ Pb
cp
=
that
=
91
— G /p
2
Pb(B
is,
Fanno
M
=
For
1.
line, sonic
flow conditions
^xi
p
b)
b
numerator must be zero and the denominator
set equal to zero yields
p
or
Vhi
= to =
Cb
>
pb
b
this value the
denominator
is
conditions occur at the point of
must be on both
not zero. Again, as with the
maximum entropy.
curves, just before
and
Since the
just after the shock
it must suddenly change from one point of intersection to the other.
The entropy cannot decrease, as no heat is being transferred from the flow,
so that the upstream point must be the intersection with least entropy. In
wave,
all
gases investigated the intersection in the subsonic flow has the greater
Thus the shock occurs from supersonic to subsonic.
The Fanno and Rayleigh lines are of value in analyzing flow
entropy.
These are treated
area ducts.
in Sees. 6.6
and
in constant-
6.7.
Converging-diverging nozzle flow
Following the presentation of Liepmann and Roshko (see references at end
of chapter), the various flow situations for converging-diverging nozzles are
investigated.
Mach number
Equation (6.3.20) gives the relation between area ratio and
By use of Eq.
for isentropic flow throughout the nozzle.
(6.3.11) the area ratio
a* _
A'
P
p*V*
is
1 /2
-i)
=
1.4).
use of the area ratios the distribution of pressure and
now be
may occur.
along a given converging-diverging nozzle can
illustrates the various flow conditions that
pressure
fc
a plot of area ratio vs. pressure ratio and M, good only for isen-
tropic flow (k
By
obtained as a function of pressure ratio
[l- (p/poy*-»' k y'*(p/poy' k
H(/c — l)/2]
C2/(/c + i)]a+i)/2(
v
Figure 6.4
is
is
Mach number
plotted.
If
Figure 6.5
the downstream
p c or greater, isentropic subsonic flow occurs throughout the
COMPRESSIBLE FLOW
M
359
pr
1.0
q
t
0.9
0.5
0.8
0.7
0.6
1.0
h
j
0.5
0.4
0.3
1.5
rs
0.2
2.0
0.1
3.0
0.1
0.2
0.4
0.3
0.6
0.5
A* _
A
0.7
0.8
0.9
1.0
pV
p*V*
Fig. 6.4
Isentropic
relations
for
a
convergingdiverging nozzle (k = 1.4). (By permission, from H. W.
Liepmann and A. Roshko, "Elements of Gas Dynamics,"
John Wiley & Sons, Inc., New York, 1957.)
tube.
If
the pressure
is
at
j,
isentropic flow occurs throughout, with sub-
and supersonic flow downFor downstream pressure between c and /, a shock wave occurs
within the nozzle, as shown for p d For pressure at p f a normal shock wave
occurs at the exit, and for pressures between p f and pj oblique shock waves at
sonic flow to the throat, sonic flow at the throat,
stream.
.
the exit develop.
6.6
ADIABATIC FLOW WITH FRICTION
Gas flow through a pipe or constant-area duct
IN
is
CONDUITS
analyzed in this section sub-
ject to the following assumptions:
1.
Perfect gas (constant specific heats).
2.
Steady, one-dimensional flow.
3.
Adiabatic flow (no heat transfer through walls)
FUNDAMENTALS OF FLUID MECHANICS
360
0.528
1
tm^C
/
Fig. 6.5 Various pressure and Mach number configurations for
flow through a nozzle. (By permission from H. W. Liepmann and A.
Roshko, "Elements of Gas Dynamics," John Wiley & Sons, Inc., New
York, 1957.)
4.
Constant friction factor over length of conduit.
5.
Effective conduit diameter
D
is
four times hydraulic radius
by perimeter)
Elevation changes are unimportant compared with
No work added to or extracted from the flow.
(cross-
sectioned area divided
6.
7.
The
controlling equations are continuity, energy,
equation of state. The Fanno
was
it
for constant area
line,
momentum, and the
developed in Sec. 6.5 and shown in Fig.
6.3,
and used the continuity and energy equations; hence,
applies to adiabatic flow in a duct of constant area.
upstream end of the duct
A particle of gas at the
may
be represented by a point on the appropriate
for proper stagnation enthalpy ho and mass flow rate G per unit
Fanno
line
area.
As the
particle
moves downstream,
its
properties change, owing to
such that the entropy always increases in adiabatic
Thus the point representing these properties moves along the Fanno
friction or irreversibilities
flow.
friction effects.
COMPRESSIBLE FLOW
Table
6.1
Supersonic
flow
Subsonic
Property
Velocity
flow
V
Decreases
Increases
Mach number M
Increases
Decreases
Pressure p
Decreases
Increases
Temperature
T
Decreases
Increases
Density p
Stagnation enthalpy
Decreases
Increases
Constant
Constant
Entropy
Increases
Increases
line
361
toward the
maximum
s
point,
verging-diverging nozzle, the flow
where
may
M
=
the duct is fed by a conbe supersonic; the velocity
subsonic at the upstream end,
1.
If
originally
must then decrease downstream. If the flow is
the velocity must increase in the downstream direction.
For exactly one length of pipe, depending upon upstream conditions, the
flow is just sonic (M = 1) at the downstream end. For shorter lengths of
have reached sonic conditions at the outlet, but for
must be shock waves (and possibly choking)
if supersonic and choking effects if subsonic.
Choking means that the mass
flow rate specified cannot take place in this situation and less flow will occur.
Table 6.1 indicates the trends in properties of a gas in adiabatic flow through a
constant-area duct, as can be shown from the equations in this section.
The gas cannot change gradually from subsonic to supersonic or vice
pipe, the flow will not
longer lengths of pipe, there
versa in a constant-area duct.
The momentum equation must now
include the effects of wall shear
stress
and
is
conveniently written for a segment of duct of length 8x (Fig. 6.6)
pA -
(p
+
j- 8x
A -
t ttD 8x
=
J
r iTD8x
>+£
h)A
P A-
~v+£«.
V|«-&t->|
Fig. 6.6
of
Notation for application
equation.
momentum
P
VA (v +
—
8x
- V)
FUNDAMENTALS OF FLUID MECHANICS
362
Upon
simplification,
+ pVdV
=
dp
+
By
use of Eq. (5.10.2), r
jfdx
(6.6.1)
= pfV2 /S,
in
which /is the Darcy-Weisbach
friction
factor,
fpV 2
+ *'-£— dx +
dp
pV dV =
(6.6.2)
For constant /, or average value over the length of reach, this equation
can be transformed into an equation for a; as a function of Mach number.
By dividing Eq. (6.6.2) by p,
f pV
—
—
2D p
h
p
each term
V
pV
2
dp
2
=
is
dx H
,*ao\
dV =
(6.6.3)
p
now developed
in terms of
M. By
definition
V/c
= M,
£
(6.6.4)
M
(6.6.5)
P
or
pV 2
=
2
V
for the
middle term of the
momentum
equation.
Rearranging Eq. (6.6.4)
gives
— dV = kM
2
p
Now
h
=
^V
to express
h
+
V
(6.6.6)
dV/V
V
—
M, from the energy equation,
2
2
j=
in terms of
cp
T
+
(6.6.7)
Differentiating gives
cp
dT
+ V dV
=
(6.6.8)
COMPRESSIBLE FLOW
V = WkRT yields
2
Dividing through by
CpJ_dT
dV _
~
+ V
EhW
T
Since
/R = k/(k
f=
cp
-
-M*(fc
T=
2
—
1),
f
1)
(6.6.9)
F = M kRT
2
2
Differentiating
2
and dividing by the equation give
m +Y
Eliminating
dT/T
(6 6 10)
-
in Eqs. (6.6.9)
and
(6.6.10)
and simplifying lead
V
l(k- 1)/2]M
2
+
(6.6.11)
dV/V
from Eq.
(6.6.6), yielding
J dV =t(k-l)/2W+l
And
finally, to
(6 6 12)
-
express dp/p in terms of M, from p
= pRT and G =
pV = GRT
pV,
(6.6.13)
dV
~~f~~V
Equations (6.6.9) and (6.6.11) are used to eliminate
p
'
differentiation
dp_dT
djP_
to
1
which permits elimination of
p
-
dM/M
dV
By
363
=
(fe
-
1)M 2
+
[(/c- 1)/2]M 2
Equations
1
+1
dT/T and dV/V:
dm
l
;
IVI
(6.6.5), (6.6.12),
and
(6.6.14) are
now
substituted into the
FUNDAMENTALS OF FLUID MECHANICS
364
momentum
L dx =
D
equation (6.6.3). After rearranging,
-M)
2(1
kM*{t(k- 1)/2]M +1}
2
_
2M
_
k
+
~kM*~
M
1
+
M{[(/c- 1)/2]M 2
k
which can be integrated
= M,
directly.
D~
ln
By
(6
'
615)
lj
using the limits x
=
0,
M =M
,
x
=
I,
M
AM'Jmo
1/1
Up
i
For
=
/c
fl
-
[(£-i)/2]M
2fc
+ 1 [YMoY
+ "aT ln
fii)
Kb)
1\
1.4, this
5/1
fc
2
(*
+
(6A16)
lJ Mo
- i)M +
2
FWT2]
1
\
.
r/IVIo\
,
2
M +
2
5 1
M
=
"d"
5/1
7
There
is
W
(6617)
reduces to
M is less than M
M = 1 and k = 1.4,
If Mo is greater than 1,
cannot be less than 1, and
not be greater than 1. For the limiting condition
/L max
2 1
\
~
V
6M02
+ f ln M7T^
some evidence
1
fe
if
1,
= L4
to indicate that friction factors
can-
(6A19)
may be
smaller in
supersonic flow.
EXAMPLE
6.13
for flow of air,
From Eq.
0.02
max
0.05
"
from which
1
maximum length of 5-cm-ID pipe, / = 0.02
Mach number at the entrance to the pipe is 0.30.
Determine the
when
the
(6.6.19)
5/J_
7 V0.3
2
L max =
\
,
7
6
X
0.30
0.30 2
2
+
5
13.25 m.
H. Keenan and E. P. Neumann, Measurements of Friction in a Pipe for Subsonic and
Supersonic Flow of Air, J. Appl. Mech., vol. 13, no. 2, p. A-91, 1946.
J.
COMPRESSIBLE FLOW
The
pressure, velocity,
form
tegral
terms of the
in
365
and temperature may also be expressed in inMach number. To simplify the equations that
follow they will be integrated from upstream conditions to conditions at
M =
From Eq.
indicated by p*, V*, and T*.
1,
1)M„
-*4 k - k+
2
+
(6.6.14)
2
(6.6.20)
1
Pi
From Eq.
y*
(6.6.11)
Uk -
i
FromEqs.
dT
T
when
T* _ (k
-
(6.6.21)
(6.6.11)
M,/M
--(*-:L)
which,
2
1
and
(6.6.9)
1
+
1)M„*
k+
Vo"
N
[(*- 1)/2]M
+
2
1
integrated, yields
l)IVIo
2
+
2
~~
k
To
+
(6.6.22)
\
/ = 0.020, has air at 14.7 psia and at t =
60°F flowing at the upstream end with Mach number 3.0. Determine L max
p* V*, T*, and values of p{, V'Q T' and L at IVI = 2.0.
EXAMPLE
6.14
A
4.0 in-ID pipe,
,
,
FromEq.
°- 02
r
Lmax
m~
0.333
from which
is
=
,
(6.6.19)
5/1 Hi
vy
Lmax =
6X32
ei
+ fin
3 +
in
1)
'
,
'
'
8.69
2
'
ft.
5
If the flow originated at
M =
2,
the length
L max
given by the same equation
—
6X2
+fln^^
2
0.02
Lmax = f(i-
from which
L max =
1)
5.08
ft.
Hence the length from the upstream
where
IVI
The
V =
=
2
is
8.69
-
5.08
=
3.61
velocity at the entrance
VkRTM
= y/lA X
53.3
X
section at
M =
3 to the section
ft.
is
32.17(460
+
60)
X
3
=
3352 ft/s
FUNDAMENTALS OF FLUID MECHANICS
366
From
Eqs. (6.6.20) to (6.6.22)
^-s^^—
_Z!_
2354
3354
7* _
So p*
X
now
i
1
X
0.4
6.7
+
22
2
=
-
M =
2 the
same equa-
:
2.45
2.4
0.4
/r~
X
22
+
2
_
"
3
=
2787
=°- 6124
2 4
,
X
22
+
2
2.4
=
3
,
<~
2
T'Q
So p{
7
V* = 1707 ft/s, T* = 1213°R. For
solved for p'0) V' and T'
0.4
_
_
~
2
~
TT^ 2 V
1213
_
^ r
0.509
67.4 psia,
67.4
1707
37
+
32
2.4
tions are
VT
=
2.4
"
=
2
-
*
3 \
-
+
32
/
0.4
~
520
X
0.4
I
~
27.5 psia, V'
2
FRICTIONLESS FLOW
ft/s,
and T' = 809°R.
THROUGH DUCTS WITH HEAT
TRANSFER
The steady
flow of a perfect gas (with constant specific heats) through a con-
stant-area duct
work
is
is
considered in this section.
done on or by the
The appropriate equations
Continuity:
Momentum:
+
is
neglected,
and no
for analysis of this case are
G = - = pV
p
Friction
flow.
pV 2 =
const
(6.7.1)
(6.7.2)
COMPRESSIBLE FLOW
367
7 - VJ
2
Energy
qH
:
=
fa
=
cp
2
—h+
7 - Vf
2
(T2
(^02
- T +
2
x)
—
(6.7.3)
^01
Tqi and To2 are the isentropic stagnation temperatures, i.e., the temperature
produced at a section by bringing the flow isentropically to rest.
The Rayleigh line, obtained from the solution of momentum and continuity for a constant cross section by neglecting friction, is very helpful in
examining the flow. First, eliminating V in Eqs. (6.7.1) and (6.7.2) gives
G
+—
2
p
=
const
(6.7.4)
P
which is Eq. (6.5.7). Equations (6.5.8) and (6.5.9) express the entropy s
and enthalpy h in terms of the parameter p for the assumptions of this section,
as in Fig. 6.7.
Since
by Eq.
(3.8.4), for
no
losses,
entropy can increase only when heat
added, the properties of the gas must change as indicated in Fig. 6.7,
is
moving toward the maximum entropy point as heat is added. At the maximum
s point there is no change in entropy for a small change in h, and isentropic
conditions apply to the point.
is
given by
c
= y/dp/dp
The speed
by Eq.
as given
u/l
G=pV= constant
Fig. 6.7
Rayleigh
line.
of
sound under isentropic conditions
(6.2.2).
From Eq.
(6.7.4),
by
differ-
FUNDAMENTALS OF FLUID MECHANICS
368
entiation,
dp
p
2
Hence
using Eq. (6.7.1).
\/dp/dp
and
also
M
=
1,
at the
heat
1,
is
and
Mach number
just the proper
if
added, choking results
mass rate
to reduce the
an increase
in the
point of the Rayleigh
s
or sonic conditions prevail.
supersonic flow causes the
M =
maximum
V =
of heat to
toward
becomes 1. If more
and conditions at the upstream end are altered
amount
of flow.
of the flow to decrease
of heat
is
The addition
Mach number toward
transfer causes choking with
The addition
line,
added,
M
of heat to subsonic flow causes
M=
1, and again, too much heat
an upstream adjustment of mass flow rate to a
smaller value.
From Eq.
pressure
is
(6.7.3) it is
noted that the increase in isentropic stagnation
a measure of the heat added.
From V =
2
M
2
kRT, p = pRT, and
continuity,
pV = GRT
pV = kpM
and
2
Now, from the momentum
Pi
+
/cpiMi 2
= p2
+
2
equation,
kp 2 N\z 2
and
1
Pl
p.
=
+
/cIVM
YTm
(67
2
Writing this equation for the limiting case p 2
p
>*
+k
+ /cM
=
p* when
M =
2
1
-
5)
gives
1
1
(6.7.6)
2
with p the pressure at any point in the duct where M is the corresponding
Mach number. For the subsonic case, with M increasing to the right (Fig. 6.7)
p must decrease, and for the supersonic case, as M decreases toward the right,
p must
increase.
To
develop the other pertinent relations, the energy equation (6.7.3)
kR
m
kR
m
V
2
is
COMPRESSIBLE FLOW
in
T
which
is
the isentropic stagnation temperature and
temperature at the same section.
through by
|2 =
and
+
1
kRT /(kY
(*
-
Applying
T
369
the free-stream
this to section 1, after dividing
I), yields
1)
^
(6.7.7)
1)
"£
(6.7.8)
for section 2
|=
+
1
(*
-
Dividing Eq. (6.7.7) by Eq. (6.7.8) gives
To_i
=
T
Ta
The
K 2+
2
2
ratio
+
(h-
l)!^ 2
-
1)M 2 2
(fc
(6.7.9)
TV ^2 is determined in terms of the Mach numbers as follows. From
the perfect-gas law, pi
= piRTh p = piRT2
2
,
£-**
T
2
From
(6.7.10)
Vi Pi
continuity pi/ pi
VkRT!
= Vi/V2 and by
,
definition,
-\/kRT2
so that
V
2
M ^T
2
2
and
Now substituting
Eqs. (6.7.5) and (6.7.11) into Eq. (6.7.10) and simplifying
gives
_ /Mi
T ~ \M 2
Tx
1
2
1
+ fcM \
+ ZcMxV
2
2
2
(6.7.12)
FUNDAMENTALS OF FLUID MECHANICS
370
This equation substituted into Eq. (6.7.9) leads to
T01 _
/IVIx 1
7*02
\M
1
2
+
+
/cMA 2
When
this equation
M =
1
2
7-0
_
2
fcMxV 2
is
+
+
M»(fe
+
Tt
(6 7,13)
-
2
(k
2
applied to the downstream section where
and the subscripts
"
- 1)M^
- 1)M
(fc
for the
Tw = T* and
upstream section are dropped, the
result
is
l)[2+ (fe-l)M']
(l
+
(0./.14)
/cM 2 ) 2
All the necessary equations for determination of frictionless flow with
now
heat transfer in a constant-area duct are
unit mass
is
equations
is
EXAMPLE
=
given by qH
— T
c p (T*
for
)
Heat
available.
M
=
transfer per
at the exit.
1
Use
of the
illustrated in the following example.
Air at Vi
6.15
=
300
=
p
ft/s,
How much
40 psia,
= 60°F
t
flows into a
heat transfer per unit mass
is needed
Determine pressure, temperature, and velocity
and at the section where M = 0.70.
4.0-in-diameter duct.
for sonic conditions at the exit?
at the exit
300
Vi
1
~ \ZkRTx ~
The
X
53.3
X
32.17(460
+
"
60)
'
isentropic stagnation temperature at the entrance,
T01 = tJi
The
\/l-4
+
—^ mA = 520(1 + 0.2 X 0.268
isentropic stagnation temperature at the exit,
T
(k
+
+ /cM
[2 +
1)M 2
2
2
(l
527(1
)
(Jb
1)M 2 ]
"
)
from Eq.
+
(6.7.7), is
= 527°R
2
1.4
X
0.268 2 (2
-
527)
2.4
from Eq.
X
(6.7.14),
0.268 2 ) 2
+ 0.4 X
0.268 2 )
= 1827°R
The heat
qH
=
cp
transfer per slug of air flowing
(T*
-
Toi)
=
0.24
X
is
32.17(1827
=
10,023 Btu/slug
is
COMPRESSIBLE FLOW
The
pressure at the exit, Eq. (6.7.6),
I _i_ £|V|2
= v ,,
p*
=
—40
(1
+
X
1.4
is
=
0.268 2 )
and the temperature, from Eq.
(6.7.12),
m f 1 + /cM l =
r* = T
[jkTm\
/l
+
(
2.4X0.268
2
2
At the
y* =
= ^/kRf* = Vl-4 X
c*
,
= p
X
1-4
18.34 psia
0.268 2
V
)
-
1522
°
R
exit,
At the
v
520
371
section
where
fe+1
18.35
TTm*
From Eq.
L
M
=
i
+
i.4
=
53.3
0.7,
X
32.17
X
from Eq.
X 2.4 =
x o.7*
1522
=
1911 ft/s
(6.7.6),
.
26
'
1
psla
(6.7.12)
1
+ kW
\1
J
+
1.4
X
0.7 2 /
and
F =
MVkRT
=
0.7
Vl-4 X
53.3
X
32.17
X
1509
=
1332 ft/s
The
trends in flow properties are shown in Table 6.2.
For curves and tables tabulating the various equations, consult the
books by Cambel and Jennings, Keenan and Kaye, and Shapiro, listed in the
references at the end of this chapter.
6.8
STEADY ISOTHERMAL FLOW
IN
LONG PIPELINES
In the analysis of isothermal flow of a perfect gas through long ducts, neither
Fanno nor Rayleigh line is applicable, since the Fanno line applies to
and the Rayleigh line to frictionless flow. An analysis somewhat similar to those of the previous two sections is carried out to show the
the
adiabatic flow
trend in properties with
Mach number.
FUNDAMENTALS OF FLUID MECHANICS
372
Table
6.2
Trends
flow properties
in
Heating
M >
Property
Pressure p
Velocity
V
Cooling
M <
1
M >
1
M <
1
1
Increases
Decreases
Decreases
Increases
Decreases
Increases
Increases
Decreases
Isentropic stagnation
T
temperature
Density p
Temperature
T
Increases
Increases
Decreases
Decreases
Increases
Decreases
Decreases
Increases
Increases
Increases for
Decreases
Decreases for
M <
The appropriate equations
Momentum
M <
1/y/k
Decreases for
Increases for
M >
M >
1/y/k
l/y/k
are
oV
oV
—
—p + ^:
dx + — dV
2D p
dv
[Eq. (6.6.3)]:
1/Vfc
2
f
=
(6.8.1)
p
Equation of state
:
P
-
=
—
— = dp
dp
const
V
P
r,
•
rr
=
pV
•
Continuity:
dV
= ——
dp
const
—
T = T
(6.8.3)
V
P
Energy [Eq. (6.7.7)]:
(6.8.2)
P
[l
(k
+
~
1}
M
2
(6.8.4)
J
in
which
T
is
the isentropic stagnation temperature at the section where the
free-stream static temperature
is
T and
Stagnation pressure [Eq. (6.3.11)]:
in
which p
is
the
p
Mach number
= P
/
(
V =
definitions
M = \/kRTM
and the above equations
dV
dM
V " M
dW
2M
2
M
\ */<*-«
k — 1
+ -^^M Y
v
the pressure (at the section of p and
the velocity to zero isentropically.
From
1
is
2
)
M)
(i
s.:
f
obtained by reducing
COMPRESSIBLE FLOW
nV
V dV
c2
p
RT
RT
RT
v
Substituting into the
dp_dp_
V
373
"
P
equation gives
ldW_
_
~
~V "
'
The
may
dV
momentum
dx
differential
M
2
/cM 2
==
"
2
1
-
/cM
positive in the
is
fdx
2
(6.8.6)
2D
downstream
direction,
conclude that the trends in properties vary according as
velocity
M <
than l/\/k. For
or greater
and
so one
than
l/\/k, the pressure and density decrease and
is less
increase, with the opposite trends for M > 1/y/Jc;
always approaches l/\/k, in place of unity for
and Mach number
Mach number
hence, the
M
adiabatic flow in pipelines.
To determine the direction of heat transfer differentiate Eq.
and then divide by it, remembering that T is constant:
(6.8.4)
?=
(6 8 7)
^W
Eliminating
dM
dT
2
-
(1
-
in this equation
k(k
T "
M2
2
- 1)M
/cM )[2
+
4
(6.8.6) gives
fdx
-
(Jb
and Eq.
-
1)M 2 ]
(6.8.8)
D
which shows that the isentropic stagnation temperature increases for M <
l/\/k, indicating that heat is transferred to the fluid. For
> 1/y/k heat
transfer is from the fluid.
M
From
dp
_
p
2
-
2
+
Eqs. (6.8.5) and (6.8.6)
(k
+
1)M 2
(k
-
1)M
2
fdx
fcM 2
/cM
-
2
1
2D
Table 6.3 shows the trends of
fluid properties.
By integration of the various Eqs. (6.8.6) in terms
Mach number is found. The last two terms yield
f
rL m „
dL
l/V-k
dx
=
-x f
k JM
_
—
(1
fcM2)
ivR
dM
of
M, the change with
FUNDAMENTALS OF FLUID MECHANICS
374
Table
Trends
6.3
in fluid properties for
isothermal flow
M <
Property
Pressure p
Density p
Velocity
M > 1/y/k
subsonic or supersonic
1/VJfc
subsonic
V
Mach number M
T
Stagnation temperature
Decreases
Increases
Decreases
Increases
Increases
Decreases
Increases
Decreases
Decreases
Increases
Stagnation pressure p
Decreases
Increases for
Decreases for
M < \/2/(k + 1)
M > y/2/{k + 1)
or
f
1
—
frMs
(6.8.10)
in
which
LmtLX
as before, represents the
,
maximum length
lengths choking occurs, and the mass rate
is
decreased.
of duct.
To
For greater
find the pressure
change,
~
r 1,v
i
L
k
dW
M
"'
2./m
V
v
2
and
(6.8.11)
V
The
superscript
*'
indicates conditions at
M =
1/y/k, and
M
and p represent
values at any upstream section.
EXAMPLE
6.16
M =
Helium enters a 10-cm-ID pipe from a converging-diverging
= 14kN/m2 abs, T = 225 K. Determine for isothermal
flow (a) the maximum length of pipe for no choking, (6) the downstream
conditions, and (c) the length from the exit to the section where M = 1.0.
/ = 0.016.
From Eq. (6.8.10) for k = 1.66
(a)
nozzle at
0.016L B
0.1
m
from which
1.30,
1
-
p
1.66
1.66
Lmax =
X
X
1.3 s
1.3 2
2.425 m.
+
In (1.66
X
1.3 2 )
COMPRESSIBLE FLOW
From Eq.
(6)
p
*t
= p^/kM =
v*t
dy
dV
f
-Iff
x
(6.8.11)
kN/m Vl^6
2
14
The Mach number
at the exit
M
1.3
=
23.45
l/\/lM =
is
i/VIj2
~V~2J hM
Jv
375
or
2
kN/m
0.756.
y*t
i
F
V^M
2
abs
From
Eqs. (6.8.6)
At the upstream section
V = M y/kRT =
V1-66X 2077X225 =
1.3
1145
m/s
and
V
m/s-
1145
'
FromEq.
(c)
0.016 r/
L' m«
=
-r—
m
m/s
M
1,
(6.8.10) for
—
-
683.6
1.66
=
nn
+ In 1.66
,
,
i
1.66
0.1
or L'max
6.9
1
=
=
M
0.683 m.
=
1
occurs 0.683
m from the exit.
ANALOGY OF SHOCK WAVES TO OPEN-CHANNEL WAVES
Both the oblique and normal shock waves
in
open-channel flow.
which y
such that
An
in a gas have their counterpart
elementary surface wave has a speed in still liquid
the depth in a wide, open channel.
of \/gy, in
is
channel
V = V =
is
When
flow in the
c
y/gy, the Froude number is unity and flow is
a small disturbance cannot be propagated upstream.
be critical; i.e.,
analogous to sonic flow at the throat of a tube, with Mach number
unity. For liquid velocities greater than Vc = \/gy the Froude number is
greater than unity and the velocity is supercritical, analogous to supersonic
gas flow. Changes in depth are analogous to changes in density in gas flow.
The continuity equation in an open channel of constant width is
said to
This
Vy =
is
const
and the continuity equation
for compressible flow in a
tube of constant cross
FUNDAMENTALS OF FLUID MECHANICS
376
section
Vp =
is
const
Compressible fluid density p and open-channel depth y are analogous.
The same analogy is also present in the energy equation. The energy
equation for a horizontal open channel of constant width, neglecting friction,
is
—+y=
const
After differentiating,
V dV +
By
g dy
=
substitution from
V dV + V
2
c
V =
c
y/gy to eliminate
g,
—=
y
which
to be
is
compared with the energy equation
for compressible flow
[Eq. (6.3.4)]
V dV +
c
2
dp
-
=
P
The two
critical velocities
V
c
and
c
are analogous, and, hence, y
and
p are
analogous.
By
applying the
momentum
equation to a small depth change in hori-
zontal open-channel flow, and to a sudden density change in compressible
flow, the* density
analogous.
In
and the open-channel depth can again be shown to be
the analogy is between the Froude number and the
effect,
Mach number.
Analogous to the normal shock wave
is
the hydraulic jump, which causes
a sudden change in velocity and depth, and a change in Froude
number from
greater than unity to less than unity. Analogous to the oblique shock and
waves in gas flow are oblique liquid waves produced in a channel
by changes in the direction of the channel walls or by changes in floor elevation.
A body placed in an open channel with flow at Froude number greater
than unity causes waves on the surface that are analogous to shock and
rarefaction waves on a similar (two-dimensional) body in a supersonic wind
rarefaction
COMPRESSIBLE FLOW
377
Changes to greater depth are analogous to compression shock, and
Shallow water tanks, called
tanks, have been used to study supersonic flow situations.
tunnel.
changes to lesser depth to rarefaction waves.
ripple
PROBLEMS
had its temperature increased 2°C
an insulated constant-volume chamber. Deter-
3 kg of a perfect gas, molecular weight 36,
6.1
when
6.4
of
mine
c„
kJ
and
cp
6.2
A
gas of molecular weight 48 has c p
6.3
Calculate the specific heat ratio k for Probs. 6.1 and 6.2.
6.4
The enthalpy
work was done on
in
it
.
of a gas
is
increased
stant pressure, and the internal energy
is
maintained constant and heat
6.5
h
is
—
by
is
0.372.
0.4
What
for this gas?
is c v
Btu/lb m °R when heat
-
increased by 0.3 Btu/lb TO
= 5°C
=
to vi
30
kN/m
abs,
k
=
at con-
=
14
kN/m2 abs,
170°C.
6.6
Calculate the entropy change in Prob. 6.5.
6.7
From Eq.
(6.1.13)
added
added. Calculate the molecular weight.
Calculate the enthalpy change of 2 kg carbon monoxide from pi
2
is
°R when the volume
-
and the perfect-gas law, derive the equation
of state for
isentropic flow.
6.8
to
k
6.9
Compute the enthalpy change
=
100°F
in
In an isentropic process
Work
1
What
solute pressure doubled.
6.10
per slug for helium from
t\
=
0°F, pi
=
15 psia
an isentropic process.
kg oxygen with a volume
is
of 100
1
at
15°C has
its
ab-
the final temperature?
out the expression for density change with temperature for a reversible
polytropic process.
6.11
Hydrogen at 40
psia,
polytropic process with
6.12
A
n
30°F, has
=
1.20.
its
temperature increased to 120°F by a reversible
Calculate the final pressure.
gas has a density decrease of 10 percent in a reversible polytropic process
when the temperature
decreases from 45 to 5°C.
Compute
the exponent n for the
process.
6.13
A projectile moves through water at 80°F
at 3000 ft/s.
What
is its
Mach num-
ber?
If an airplane travels at 1350 km/h at sea level, p = 101 kPa abs, t = 20°C,
and at the same speed in the stratosphere where t = — 55 °C, how much greater is the
6.14
Mach number
6.15
What
is
in the latter case?
the speed of sound through hydrogen at 80°F?
FUNDAMENTALS OF FLUID MECHANICS
378
7~r
V+dV*"
y+dy
I
I
Fig. 6.8
Derive the equation for speed of a small liquid wave in an open channel by using
6.16
the methods of Sec. 6.2 for determination of speed of sound (Fig. 6.8)
By
6.17
V dV +
using the energy equation
— + d(losses)
=
P
the continuity equation
in a pipe the velocity
const,
and
c
in the
=
\/dp/dp, show that for subsonic flow
downstream
direction.
= 40 psia, t = 90°F,
which brings the velocity to zero.
Isentropic flow of air occurs at a section of a pipe where p
6.18
V=
and
pV =
must increase
537
ft/s.
An
object
is
immersed
in the flow
What
are the temperature and pressure at the stagnation point?
6.19
What
is
the
Mach number
for the flow of Prob. 6.18?
How do the temperature and pressure at the stagnation point
compare with reservoir conditions?
6.20
in isentropic flow
6.21
Air flows from a reservoir at 70°C, 7 atm. Assuming isentropic flow, calculate
the velocity, temperature, pressure, and density at a section where
6.22
Oxygen flows from a
section where the velocity
the
Mach number,
reservoir
is
pressure,
600
where po
ft/s.
=
100 psia,
to
—
M=
0.60.
60°F, to a 6-in-diameter
Calculate the mass rate of flow (isentropic) and
and temperature
at the 6-in section.
Helium discharges from a J-in-diameter converging nozzle at its maximum rate
2
p = 4 kg//cm abs, t = 25°C. What restrictions are placed
on the downstream pressure? Calculate the mass flow rate and velocity of the gas at the
6.23
for reservoir conditions of
nozzle.
6.24
Air in a reservoir at 350 psia,
a converging-diverging nozzle.
6.25
What must
For
t
=
290°F, flows through a 2-in-diameter throat in
M=
1
at the throat, calculate p, p,
50
sonic
there.
M=
2.4?
Nitrogen in sonic flow at a 25-mm-diameter throat section has a pressure of
kN/m
6.27
T
be the velocity, pressure, density, temperature, and diameter at a
cross section of the nozzle of Prob. 6.24 where
6.26
and
2
abs,
What
and
is
t
= — 20°C.
the
Determine the mass flow
Mach number
in subsonic flow?
for Prob. 6.26 at a
rate.
40-mm-diameter section
in super-
COMPRESSIBLE FLOW
6.28
What diameter
throat section
monoxide from a reservoir where p
6.29
which
needed for
300
psia,
A supersonic nozzle is to be designed
20
is
— 85°C.
6.30
is
=
cm
t
=
critical flow of 0.5 lb TO /s
carbon
100 °F?
for airflow with
diameter and has a pressure of 7
in
379
M=
kN/m2
3 at the exit section,
abs and temperature of
Calculate the throat area and reservoir conditions.
In Prob. 6.29 calculate the diameter of cross section for
M=
1.5, 2.0,
and
2.5.
For reservoir conditions of po = 150 psia, to = 120°F, air flows through a converging-diverging tube with a 3.0 in-diameter throat with a maximum Mach number of
6.31
0.80.
Determine the mass rate
perature at the exit where
6.32
M=
of flow
and the diameter, pressure,
velocity,
and tem-
0.50.
Calculate the exit velocity and the mass rate of flow of nitrogen from a reservoir
where p
=
4 atm,
t
=
25°C, through a converging nozzle of 5
cm
diameter discharging
to atmosphere.
6.33
of
Reduce Eq.
(6.3.25) to its
form
for airflow.
Plot p/po vs.
A* J A
for the range
p/po from 0.98 to 0.02.
A*/ A =
6.34
By utilizing the plot of
6.35
In a converging-diverging duct in supersonic flow of hydrogen, the throat
diameter
is
2.0 in.
Prob. 6.33, find the two pressure ratios for
Determine the pressure
ducts where the diameter
is
2.25
ratios p/po in the converging
in.
Nozzle exit plane
7//////(/////M
©!
Reservoir
P=P
©
yff/////////////m
Nozzle schematic
Distance along nozzle axis
(b)
Fig. 6.9
0.50.
and diverging
FUNDAMENTALS OF FLUID MECHANICS
380
A shock wave occurs in a duct carrying air where the upstream Mach number is
and upstream temperature and pressure are 15°C and 20 kN/m 2 abs. Calculate
the Mach number, pressure, temperature, and velocity after the shock wave.
6.36
2.0
Show
6.37
6.38
tu
=
that entropy has increased across the shock
wave
of Prob. 6.36.
Conditions immediately before a normal shock wave in airflow are p M = 6 psia,
Vu = 1800 ft/s. Find M w M d pd and t d where the subscript d refers to
100°F,
,
,
,
,
conditions just downstream from the shock wave.
6.39
wave
6.40
For
in
A =
ft
2
in Prob. 6.38, calculate the
entropy increase across the shock
Btu per second per degree Rankine.
From Eqs.
(cf. Fig. 6.96)
Are these
.
From Eqs.
Fig. 6.96.
(6.3.1), (6.3.4),
(De Laval)
gent-divergent
6.41
0.16
Do
(6.3.1)
,
and
(6.3.5)
nozzle, point
differentials zero
(6.3.4)
not consider
,
and
1
deduce that at the throat of a converdp = 0, dp =
for M 9^ 1
of Fig. 6.9a,
f or
M=
1?
Explain.
(6.3.5) justify the slopes of the curves
shown
in
EFG.
For the nozzle described below, plot curves ADB and AEC (Fig. 6.96) ( SuggesDetermine only one intermediate point. Use section VI.) The reservoir has air
at 300 kPa abs and 40°C when sonic conditions are obtained at the throat.
6.42
.
tions:
Section
I
II
III
V
IV
VI
VII
VIII
IX
X
exit
Distance downstream
from throat,
A/A* (A* =
6.43
cm
cm 2
27
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
1.030 1.050 1.100 1.133 1.168 1.200 1.239 1.269 1.310 1.345
)
Using the data from Prob. 6.42 determine pz/po when a normal shock wave occurs
at section VI.
6.44 Could a flow discontinuity occur at section VI of Prob. 6.42 so that the flow path
would be described by ADFG of Fig. 6.96? (Hint: Determine the entropy changes.)
6.45
What
is
pz/po
=
p\
and p u
(Hint: p d
when a normal shock wave
is
pz for isentropic flow
up
occurs just inside the nozzle exit?
to section
6.46
Suggest what might occur just outside the nozzle
which
is
above that
the receiver, point
C
for
of
VI
of Prob. 6.42.)
if there is a receiver pressure p 4
which the gas flows isentropically throughout the nozzle into
Fig. 6.96, but below that for which a normal shock is possible
at the nozzle exit (cf. Prob. 6.45)
Speculate on what occurs within and without the nozzle
below that corresponding to point C of Fig. 6.96.
6.47
is
if
the receiver pressure
COMPRESSIBLE FLOW
Show, from the equations
6.48
decrease in
real,
of Sec. 6.6, that temperature, pressure,
381
and density
adiabatic duct flow for subsonic conditions and increase for supersonic
conditions.
What
6.49
enters at
and leaves
3.0
of the duct length
is
M=
Determine the
6.51
maximum
10-cm-diameter duct, /
200 m/s, p = 2 kg// cm 2 abs.
What minimum
6.52
=
M=
is
needed when oxygen
0.4
M=
and leaves at
M=
What
0.6.
portion
0.5?
length, without choking, for the adiabatic flow of air
when upstream
0.025,
What
are the pressure
duct
size insulated
The upstream temperature
ft?
0.018,
2.0?
required for the flow to occur at
in a
1000
at
Air enters an insulated pipe at
6.50
=
length of 4-in-diameter insulated duct, /
M =
is
is
conditions are
and temperature
t
=
V=
50°C,
at the exit?
required to transport 2 lb w /s nitrogen
80°F, and the velocity there
is
200
ft/s.
/
=
0.020.
6.53
Find the upstream and downstream pressures in Prob.
6.54
What
through 6
p
=
6.55
inlet
1
is
the
rate of flow of air from a reservoir,
m of insulated 25-mm-diameter pipe, / =
kg//cm2
=
15°C,
In frictionless oxygen flow through a duct the following conditions prevail at
outlet: Vi = 300 ft/s; h = 80°F; M 2 = 0.4. Find the heat added per slug
and
ratio pi/p2-
In frictionless air the flow through a 10-cm-diameter duct 0.15 kg/s enters at
=
0°C, p = 7 kN/m 2 abs.
without choking the flow?
How much heat, in
6.57
Frictionless flow through a duct with heat transfer causes the
t
t
0.020, discharging to atmosphere?
abs.
and the pressure
6.56
maximum mass
6.52.
decrease from 2 to 1.75. k
=
1.4.
kilocalories per kilogram, can be
Determine the temperature,
added
Mach number
velocity, pressure,
to
and
density ratios.
6.58
In Prob. 6.57 the duct
is
2 in square, pi «= 15 psia,
and Vy
=
2000
ft/s.
Calculate
the mass rate of flow for air flowing.
6.59
How much heat must
increase
6.60
from 2 to 2.8
Oxygen
at
V =
l
be transferred per kilogram to cause the
in a frictionless duct carrying air?
525 m/s, p
=
How much
diameter frictionless duct.
80
kN/m
2
abs,
Y\
t
=
Mach number
to
500 m/s.
= - 10°C
heat transfer per kilogram
is
flows in a 5-cm-
needed for sonic
conditions at the exit?
6.61
Prove the density, pressure, and velocity trends given in Sec. 6.8 in the table
of trends in flow properties.
6.62
Apply the
first
law
of
thermodynamics, Eq. (3.7.1), to isothermal flow of a
and develop an expression for the heat added per
perfect gas in a horizontal pipeline,
slug flowing.
FUNDAMENTALS OF FLUID MECHANICS
382
Air
6.63
/
=
is
mum
flowing at constant temperature through a 3-in-diameter horizontal pipe,
=
At the entrance Vi
0.02.
300
pipe length for this flow, and
ft/s,
t
=
how much
120°F, p x = 30 psia. What is the maxiheat is transferred to the air per pound
mass?
6.64
Air at 15°C flows through a 25-mm-diameter pipe at constant temperature.
the entrance V\
=
60 m/s, and at the exit
72 =
90 m/s. /
=
0.016.
What
is
At
the length
of the pipe?
6.65
If
the pressure at the entrance of the pipe of Prob. 6.64
pressure at the exit and
what
is
is
1.5
atm, what
is
the
the heat transfer to the pipe per second?
Hydrogen enters a pipe from a converging nozzle at M = 1, p — 2 psia, t = 0°F.
Determine for isothermal flow the maximum length of pipe, in diameters, and the pressure change over this length. / = 0.016.
6.66
6.67 Oxygen flows at constant temperature of 20°C from a pressure tank, p = 130
atm, through 10 ft of 3-mm-ID tubing to another tank where p = 110 atm. / = 0.016.
Determine the mass rate of flow.
6.68
In isothermal flow of nitrogen at 80°F, 2 lb m /s
tubing, /
6.69
=
Specific heat at constant
^
dr)
6.70
HI
\
is
\du)
,
any
of these
N
h— h
<•>££
p
6.71
volume
defined
^
is
to be transferred 100 ft from a
What
160 psia.
,
^
\d77
7N
is
the
minimum
Aw+
is
none
of these
not given
by
A(p/p)
(rf)
answers
For a perfect
gas, the
enthalpy
always increases owing to losses
depends upon the pressure only
(c) depends upon the temperature only
(d) may increase while the internal energy decreases
(e) satisfies none of these answers
(a)
(6)
6.72
The
following classes of substances
may
be considered perfect gases:
saturated steam, water vapor, and air
(a)
ideal fluids
(c)
fluids
(d)
water vapor, hydrogen, and nitrogen at low pressure
none of these answers
(e)
(b)
size
by
Specific heat at constant pressure, for a perfect gas,
<•>*
(e)
=
—
200 psia to a tank where p
0.016, that is needed?
tank where p
with a constant bulk modulus of elasticity
answers
COMPRESSIBLE FLOW
6.73
cp
and
=
cp /c v
(a)
k
(e)
none
cv
are related
=
k
(b)
of these
by
k
(c)
cp c v
If c p
=
0.30 Rtu/lb m -°R and k
cv
equals
(a)
0.582
6.75
If
1452
(6)
cp
=
0.30
kilogram per kelvin
(a)
0.075
R=
6.76
(a)
1.2
(a)
6.78
•
(d)
=
cp
cv
k
1.66, in
—
foot-pounds per slug per degree
7500
1.33,
(e)
none
of these
answers
the gas constant in kilocalories per
is
lb/lb ro
°R and
-
specific heat ratio is
(»)l
none
answers
+ 5^
The entropy change
cp
=
1.66
(c)
ft/Cp
0.399
(c)
1.33
of these
c v /cp
(d)
kcal/kg-K and k
—— V1
(e)
ft
(6)
The
6.77
62
=
4524
(c)
0.099
(6)
=
answers
Fahrenheit,
6.74
(d)
0.699
0.279 Btu/lb TO °F.
-
(d)
1.89
none
(e)
The
none
(e)
of these
of these
answers
given by
(e)
-+fl
(d)
7-^-7=
1
c„
for a perfect gas
— c /R
v
is
(Aq H/T) Te
always positive
(d)
a thermodynamic property depending upon temperature and pressure
(e)
a function of internal energy only
(a)
(b)
(c)
(d)
(e)
6.80
An
(b)
isentropic process
always
and adiabatic
and isothermal
frictionless and adiabatic
frictionless and irreversible
none of these answers
irreversible
The
relation p
=
const p k holds only for those processes that are
reversible poly tropic
(6)
isentropic
(c)
frictionless isothermal
(d)
adiabatic irreversible
(e)
none of these answers
The
reversible poly tropic process
(a)
adiabatic frictionless
(b)
given by p/p
given by ppk
(c)
(c)
reversible
(a)
6.81
a function of temperature only
is
=
=
const
const
answers
isentropic exponent k
(a)
6.79
383
is
-v
is
FUNDAMENTALS OF FLUID MECHANICS
384
(d)
(e)
given by p/p n = const
none of these answers
A
6.82
reversible polytropic process could be given
1
7
(a)
J2
(e)
-1
/piY
fer
1
!
,
n
vi
/Pi\
s.(fiY
w
...
W
.
/vA n ~
x
(of=(-r
^2
\P2/
\pl/
P2
T
.
by
(d)
f=(fiY
T2
\P2/
none of these answers
In a reversible polytropic process
6.83
(a)
some heat
(b)
the entropy remains constant
(c)
the enthalpy remains constant
transfer occurs
(d) the internal energy remains constant
(e)
the temperature remains constant
The
6.84
differential equation for energy in isentropic flow
dp+d( P V>) =
/
(a)
x
dV
/
(6)
may
take the form
dA
=
—V +-+—
A
dp
p
(c)
2VdV+
— =0
(d)
VdV+ — =0
9
(e)
none of these answers
Select the expression that does not give the speed of a
6.85
(a)
P
VkRT
The speed
6.86
(b)
Vkpjp
of a
sound wave
(c)
(d)
yf&pf&p
in a gas is
sound wave:
VWp
the speed of flow in an open channel
(6)
the speed of an elementary wave in an open channel
(c)
the change in depth in an open channel
(d)
the speed of a disturbance traveling upstream in moving liquid
(e)
none of these answers
The speed
of
sound
\TKfp
analogous to
(a)
6.87
(e)
in water, in feet per second,
under ordinary conditions
about
(a)
6.88
460
(b)
The speed
1100
of
(c)
sound
in
4600
11,000
(e)
none of these answers
an ideal gas varies directly as
(a)
the density
(d)
the bulk modulus of elasticity
(6)
(d)
the absolute pressure
(e)
none
(c)
the absolute temperature
of these answers
is
COMPRESSIBLE FLOW
385
Select the correct statement regarding frictionless flow:
6.89
(a)
In diverging conduits the velocity always decreases.
(6)
The
(c)
In supersonic flow the area decreases for increasing velocity.
(d)
Sonic velocity cannot be exceeded at the throat of a converging-diverging tube.
(e)
At Mach
velocity
is
always sonic at the throat of a converging-diverging tube.
zero the velocity
is
sonic.
In isentropic flow, the temperature
6.90
(a)
cannot exceed the reservoir temperature
(6)
cannot drop, then increase again downstream
Mach number
Mach number only
(c)
is
independent of the
(d)
is
a function of
(e)
remains constant in duct flow
The
6.91
(a)
0.528
(b)
0.634
(c)
0.833
(d)
1.0
carbon monoxide
(e)
none
of these
is
answers
Select the correct statement regarding flow through a converging-diverging tube.
6.92
(a)
critical pressure ratio for isentropic flow of
When
the
Mach number
at exit
is
greater than unity no shock
wave has developed
in the tube.
(6)
When
the critical pressure ratio
is
exceeded, the
Mach number
at the throat
is
greater than unity.
(c)
For sonic velocity at the throat, one and only one pressure or velocity can occur at
downstream location.
The Mach number at the throat is always unity.
The density increases in the downstream direction throughout the converging
a given
(d)
(e)
portion of the tube.
In a normal shock wave in one-dimensional flow the
6.93
and density increase
and temperature increase
temperature, and density increase
density, and momentum per unit time increase
(a)
velocity, pressure,
(6)
pressure, density,
(c)
velocity,
(d)
pressure,
(e)
entropy remains constant
A
6.94
normal shock wave
reversible
(a)
is
(b)
may
(c)
is
irreversible
(d)
is
isentropic
(e)
is
none
6.95
A
occur in a converging tube
of these answers
normal shock wave
(a)
an elementary wave
(b)
the hydraulic
jump
in
is
analogous to
still
liquid
FUNDAMENTALS OF FLUID MECHANICS
386
<
(c)
open-channel conditions with F
(d)
flow of liquid through an expanding nozzle
(e)
none of these answers
1
Across a normal shock wave in a converging-diverging nozzle for adiabatic flow
6.96
the following relationships are valid
(a)
continuity and energy equations, equation of state, isentropic relationship
(b)
energy and
(c)
continuity, energy,
(d)
equation of state, isentropic relationship,
momentum
equations, equation of state, isentropic relationship
and momentum equations; equation
momentum
of state
equation, mass-conservation
principle
(e)
none
of these answers
Across a normal shock wave there
6.97
(a)
p,
M,
s
(d)
p,
M
no change
A
6.98
;
(6)
Fanno
line
p, s; decrease in
in s
is
(e)
momentum and
(b)
(c)
energy and continuity
momentum and energy
(d)
momentum,
(e)
none
A
6.99
continuity,
Rayleigh line
is
s,
M
and energy
developed from the following equations:
momentum and
energy and continuity
(c)
momentum and
(d)
momentum,
(e)
none of these answers
continuity
energy
continuity,
and energy
Select the correct statement regarding a
Two
p\ decrease in
M, T
answers
(6)
(a)
(c)
in
continuity
(a)
6.100
M
an increase
developed from the following equations:
(a)
of these
p,
is
points having the
same value
of
Fanno
or Rayleigh line:
entropy represent conditions before and
after a shock wave.
(b)
pV
is
held constant along the
line.
Mach number
always increases with entropy.
(d) The subsonic portion of the curve is at higher enthalpy than the supersonic portion.
(e) Mach 1 is located at the maximum enthalpy point.
(c)
6.101
Choking
(a)
a valve
(b)
a,
is
in pipe flow
means that
closed in the line
restriction in flow area occurs
COMPRESSIBLE FLOW
(c)
the specified mass flow rate cannot occur
(d)
shock waves always occur
(e)
supersonic flow occurs somewhere in the line
In subsonic adiabatic flow with friction in a pipe
6.102
(a)
(c)
(e)
M, s increase; p, T, p decrease
M,
s increase; V, T, p decrease
p,
T, V, s increase; M, p, p decrease
V,
p, V,
(b)
M,
(d) p,
M
increase; T, p decrease
s increase;
V, T, p decrease
In supersonic adiabatic flow with friction in a pipe
6.103
(a)
V,
(c)
p,
(e)
p,
M, s increase; p, T, p decrease
M, s increase; V, T, p decrease
p, s increase; V, M, T decrease
(6)
p, T, s increase; p, V,
(d)
p, T, p, s increase; V,
M
M
Adding heat
Adding heat
(c)
Cooling supersonic flow decreases the
(d)
The Fanno line is valuable in analyzing the flow.
The isentropic stagnation temperature remains constant along the
6.105
to supersonic flow increases the
to subsonic flow increases the
V
1
(b)
V,
T
(d)
V,
p,
p, p,
(e)
To, V,
6.106
M<
increase; p, T, To decrease
(a)
(c)
increase; V, To decrease
p increase; p,
T
V
increases; p, p, T, To decrease
V
increase; T,
T
decrease
p, p,
V, T, To increase; p, p decrease
^decrease
V
increase;
1
T, To decrease
(b)
p,
(d)
p, p increase; V, T,
p,
M>
T
decrease
In steady, isothermal flow in long pipelines, the significant value of
6.107
determining trends in flow properties
1/k
6.108
(b)
l/y/k
(c)
1
M
for
is
(d)
VI
(e)
k
Select the correct trends in fluid properties for isothermal flow in ducts for
0.5:
(a)
V
(c)
V,
(e)
increase; p, p decrease
increase; p, p,
Select the correct trends for cooling in frictionless duct flow
(e)
M<
T
T
decrease
(c)
(a)
pipe.
Select the correct trends in flow properties for frictionless duct flow with heat
transferred to the pipe
(a)
decrease
Mach number.
Mach number.
Mach number.
(6)
(e)
decrease
Select the correct statement regarding frictionless duct flow with heat transfer:
6.104
(a)
387
M, To, p, p p decrease
M, To increase; p, p p decrease
V, M, po, To increase; p, p decrease
increases;
,
,
(b)
V,M
(d)
V,
T
increase; To, p, po, p decrease
increase;
M,
p, po, p decrease
388
FUNDAMENTALS OF FLUID MECHANICS
REFERENCES
Cambel, A. B., and B. H. Jennings: "Gas Dynamics," McGraw-Hill, New York, 1958.
Keenan, J. H., and J. Kaye: "Gas Tables," Wiley, New York, 1948.
Liepmann, H. W., and A. Roshko: "Elements of Gas Dynamics," Wiley, New York,
1957.
Owczarek,
J.
A. "Fundamentals of Gas Dynamics," International Textbook, Scranton,
:
Pa., 1964.
Shapiro, A. H.:
vol. 1,
"The Dynamics and Thermodynamics
Ronald,
New
York, 1953.
of Compressible Fluid
Flow,"
7
IDEAL-FLUID
FLOW
In the preceding chapters most of the relationships have been developed
i.e., flow in which the average velocity at each
used and variations across the section are neglected. Many
design problems in fluid flow, however, require more exact knowledge of
for one-dimensional flow,
cross section
velocity
is
and pressure
distributions, such as in flow over curved boundaries
along an airplane wing, through the passages of a
over the crest of a dam.
An understanding
of two-
pump
or compressor, or
and three-dimensional flow
much broader
approach to many real fluid-flow situations. There are also analogies that
permit the same methods to apply to flow through porous media.
In this chapter the principles of irrotational flow of an ideal fluid are
developed and applied to elementary flow cases. After the flow requirements
of a nonviscous, incompressible fluid provides the student with a
are established, the vector operator
and the velocity potential
is
V is introduced, Euler's equation is derived
defined.
Euler's equation
is
then integrated to
obtain Bernoulli's equation, and stream functions and boundary conditions
are developed.
7.1
Flow
cases are then studied in three
and two dimensions.
REQUIREMENTS FOR IDEAL-FLUID FLOW
The Prandtl hypothesis,
Sec. 5.6, states that, for fluids of
effects of viscosity are appreciable
low viscosity, the
only in a narrow region surrounding the
For incompressible-flow situations in which the boundary
may be applied to flow of a real fluid to a
satisfactory degree of approximation. Converging or accelerating flow situations generally have thin boundary layers, but decelerating flow may have
fluid boundaries.
layer remains thin, ideal-fluid results
389
FUNDAMENTALS OF FLUID MECHANICS
390
separation of the boundary layer and development of a large
wake that
is
difficult to predict analytically.
An
ideal fluid
The continuity
1.
dv
du
—
—
+
H
dx
dy
dw
must
satisfy the following requirements
equation, Sec. 3.4, divq
=
0,
or
=
dz
Newton's second law of motion at every point at every instant.
Neither penetration of fluid into nor gaps between fluid and boundary
at any solid boundary.
2.
3.
in addition to requirements 1, 2, and 3, the assumption of irrotational
made, the resulting fluid motion closely resembles real-fluid motion for
fluids of low viscosity, outside boundary layers.
Using the above conditions, the application of Newton's second law to a
fluid particle leads to the Euler equation, which, together with the assumption of irrotational flow, can be integrated to obtain the Bernoulli equation.
The unknowns in a fluid-flow situation with given boundaries are velocity
and pressure at every point. Unfortunately, in most cases it is impossible to
proceed directly to equations for velocity and pressure distribution from the
boundary conditions.
If,
flow
is
7.2
THE VECTOR OPERATOR v
The vector operator V, which may act on a vector
or
may
act on a scalar function,
is
most useful
as a scalar or vector product
in developing ideal-fluid-flow
theory.
Let
VU
=
U
be the quantity acted upon by the operator.
The operator
V
is
by
defined
lim
-f^UdS
(7.2.1)
VJa
U may be interpreted as
-a,
Consider a small volume
V
x
a,
where a
with surface
is
any
S and
unit vector in the direction of the outwardly
element
dS
(Fig. 7.1).
vector, or as a scalar, say
surface element dS.
drawn normal n
This definition of the operator
develop the concepts of gradient, divergence, and
curl.
is
Ill is
<j>.
a
of the surface
now examined
to
IDEAL-FLUID FLOW
Fig. 7.1
vector
ri!
391
Notation for unit
normal to area
elements.
When U
grad
=
<f>
V<f>
a scalar, say
is
=
—
lim
To
interpret grad
<f>,
J
the gradient of
(7.2.2)
the volume element
and the other end area
o-
+ hr
(Fig. 7.2)
/
rii<£
Jf\i4>
dS =
Js
rii (
•o_>o
<f>
\
—
do dn dn
in the surface
is
no change in
dS over
dn do
dn
J
side of Eq. (7.2.2)
=
rii
—
dn
^constant
scalar 0.
Surfaces
of
in surfaces parallel to the
<t>)
dn
<P+wan
Fig. 7.2
<f>
end
the curved surface of the element vanishes.
— dn -
+
and the right-hand
lim
is taken as a small prism of crossend area in the surface <t>(x,y,z) = c
const
As there
.
symmetry,
is
a
sectional area dS, of height dn, with one
*
<f>
n^ dS
/
V
1Un
<f>,
constant
becomes
by
Then
faces,
FUNDAMENTALS OF FLUID MECHANICS
392
and
grad
in
=
<f>
V<f>
= n
—
d<f>
1
(7.2.3)
dn
which
rii is
the unit vector, drawn normal to the surface over which
By
interpreting
Let
obtained.
divq =
V
•
q
U
U
<f>
grad is a vector.
as the scalar (dot) product with V, the divergence
constant, positive in the direction of increasing
<f>;
is
<f>
is
be «q; then
= lim-Jri!
lim - / n
x
q dS
•
TUo U J S
(7.2.4)
This expression has been used (in somewhat different form) in deriving the
is the volume flux per unit volume
general continuity equation in Sec. 3.4. It
at a point
The
and
curl
is
V
a scalar.
x q is a more
difficult
concept that deals with the
vorticity
or rotation of a fluid element:
= V
curlq
With
x q
=
lim
-
/
reference to Fig. 7.3,
surface element
dS
x q
rii
rii
x q
dS
is
(7.2.5)
the velocity component tangent to the
at a point, since the vector product
is
a vector at right
angles to the plane of the two constituent vectors, with magnitude q sin
=
Then
x q dS
8,
an elemental vector that is the product of
tangential velocity and surface area element. Summing up over the surface,
as wi
Fig. 7.3
curl
of
vector.
1.
rii
for
Notation
the velocity
is
IDEAL-FLUID FLOW
then dividing by the volume, with the limit taken as
V—
»
0, yields
393
the curl q
at a point.
A special type of fluid motion is examined to demonstrate the connection
between curl and rotation. Let a small circular cylinder of fluid be rotating
about its axis as if it were a solid (Fig. 7.4), with angular velocity w, which is
a vector parallel to the axis of rotation. The radius of the cylinder is r and
tli x q at every point on the curved surface is a vector parallel
I.
the length
to the axis having the
Hi x q
is
magnitude
x q
111
dS =
co
•*s
cor.
Over the end areas the vector
rlr
/
da
=
=
Ir
da,
2
2irr lto
•'o
Equation
curl
=
Then, since ds
tributes nothing to the curl.
J
q
equal and opposite at corresponding points on each end and con-
(7.2.5)
q = lim
—-
now
yields
2irrH<&
=
2co
u*o TrrH
showing that
for solid-body rotation the curl of the velocity at a point is
If one considers the pure translation of a small
then the curl q is always zero. As any rigid-body
motion is a combination of a translation and a rotation, the curl of the velocity
vector is always twice the rotation vector.
A fluid, however, not only may translate and rotate but also may deform.
twice the rotation vector.
element moving as a
The
solid,
definition of curl
Fig. 7.4
Small fluid
cylinder rotating as a
solid.
q
applies,
and hence the
rotation of a fluid at a point
FUNDAMENTALS OF FLUID MECHANICS
394
by
is
defined
to
= icurlq = }(V
When
<o
=
x q)
(7.2.6)
throughout certain portions of a
described as irrotational.
The vorticity vector
similar to the velocity vector q.
the motion there
fluid,
and vortex tubes, comprising the vortex
vorticity vector,
small closed curve, follow certain continuity principles;
vorticity
by area
of the tube
= V -(V
div (curlq)
The operator V
lines
viz.,
through a
the product of
must remain constant along the vortex
=
x q)
is
q has certain characteristics
Vortex lines are everywhere tangent to the
curl
tube, or
0.
acts like a vector but
must be applied
to a scalar or a
vector to have physical significance.
Scalar components of vector relationships
Any
vector can be decomposed into three components along mutually per-
pendicular axes, say the
The component
x, y, z axes.
nitude and sign (sense) are needed to specify
it;
is
fx
a scalar, as only mag-
= —3
indicates the x
component of a vector f acting in the —x direction.
The vector is expressed in terms of its scalar components by use
fixed unit vectors,
a
=
The
+ ja +
\a x
y
i,
j,
respectively:
unit vectors combine as follows:
The
•
x, y, z axes,
ka z
j.j=j.j=kk =
jxk =
ixj = k
a
k parallel to the
b
=«=
=
kxi=j
i
scalar product of
(\ax
ax b x
+ \ay + ka
+ ciyby + a b
z)
z
The vector product
a x b
ij=jk=ki=0
l
of
(io.
=
\(ay b z
y
—
two vectors a
•
+
(\b x
\b v
•
+
b
etc.
is
k& 2 )
z
two vectors a x b
+ \a + ka
=
= — ixk
z)
x
(\b x
a z b y ) -f \{a 9bx
—
is
+ \bv + kb
ax b z )
z)
+ k(ax b —
y
ay b x )
of the
IDEAL-FLUID FLOW
It is conveniently written in
J
k
dx
Oy
az
bx
by
bz
1
a x b
=
To
which a
a
•
find the scalar
is
any
= a
V<f>
vector.
Mi
first
consider a
•
V0
(Fig. 7.5), in
Eq. (7.2.3)
— = a cos —
dn
d<j>
•
determinant form
components of V0,
By
395
d<t>
dn
as
the angle between a and
is
and
fli
of a corresponds to a change in n, given
a cos
n\
=
by da
1.
A
cos 6
change da in magnitude
dn; hence
=
= a—
—
dn
da
and
a
•
V0 =
a
60
—
(7.2.7)
da
The
V0 =
scalar
V0
60
V0 =
k
d<f>
dx
Fig. 7.5
components of
J
Change
corresponding to
normal direction.
•
of
vector a
change
in
are
•
V0 =
60
FUNDAMENTALS OF FLUID MECHANICS
396
and
V0 =
dd>
.
— + k—
dz
dd>
.
l— +J
dx
•
d
d
+J
,
,
^
The
v
-
q =
=
(7.2.8)
v
dy
The operator V,
v =
dd>
.
^
'
in terms of its scalar components,
d
+
i
k
^
<
V
scalar product, say
i
(
is
£ +j i +k l)-
q,
•
7 2 9>
-
-
becomes
(iu+j!
'
+M
dv
dw
—
+
+
dx
dy
dz
du
(7.2.10)
as in Sec. 3.4.
The vector product V x
i
.
The
— + — + k -)
j
/div
dv\
\dy
dz)
.
7.3
x
(\
u
components,
+ \v +
is
kw)
/du
dw\
/dv
du\
\dz
dx)
\dx
dy)
quantities in parentheses are vorticity components, which are twice the
value of rotation components
V
q, in scalar
x q
=
i2w x
+ \2uy +
co x ,
co y , co z ,
and
so
k2co,
(7.2.12)
EULER'S EQUATION OF MOTION
In Sec.
3.5, Euler's
along a streamline.
equation was derived for steady flow of a frictionless fluid
The assumption
is
made
here that the flow
is frictionless,
and a continuum is assumed. Newton's second law of motion is applied to a
fluid particle of mass pdV.
Three terms enter: the body force, the surface
force, and mass times acceleration. Let F be the body force (such as gravity)
per unit mass acting on the particle. Then Fp 8V is the body-force vector. The
surface force, from the preceding section, is — J s t\ip dS if the fluid is friction-
IDEAL-FLUID FLOW
nonviscous, so that only normal forces act.
less or
term
p
is
The mass-times-acceleration
5V dq/dt. Assembling these terms gives
-
Fp 5V
397
ds
flip
=
J
dq
p
8V
dt
Now, dividing through by the mass
613—>0 yields
=
n lP dS
of the element
and taking the
limit as
P
dt
Use
F
of the operator
V
leads to
^P = §
-
This
(7.3.1)
dt
p
is
Euler's equation of motion in vector notation.
product of each term with
then
i,
j,
then
k,
By
forming the scalar
the following scalar component
equations are obtained:
_iJ_P = du
x
in
p dx
_idP =
dv
p by
dt
dt
z
_ldp =
dw
p dz
dt
Z
are the body-force components per unit mass. The accelerabe expanded. In general u = u(x,y,z,t), and so (see Appendix
which X, Y,
tion terms
Y
may
B)
du =
du
—
dx H
du
dx
dy
du
du
dy
dz H
-\
dz
dt
dt
For du/dt to be the acceleration component of a particle in the x direction,
the
x, y, z
du
may
ax
=
du
—
dt
=
moving
coordinates of the
be divided by
du dx
dt,
du dy
du dz
H
1
dx
dy dt
dt
dx
—
dt
v
=
dy
—
dt
w =
du
1
dz dt
But
u =
particle
yielding
dz
—
dt
dt
become functions
of time,
and
FUNDAMENTALS OF FLUID MECHANICS
398
and
du
=
at
u+v
du
du
dx
dy
du
+w
dz
du
+
(7>3<3)
dt
Similarly
dv
—
= u
dv
dv
dy
dw
— = u dw
dx
(F
X
dv
H
(7.3.4)J
K
dz
dw
\-v
dy
dt
If the
dv
\-w—
\-v
dx
dt
dt
dw
Vw
dw
(7.3.5)
1
dz
extraneous force
dt
is conservative,
it
may be
derived from a potential
= -gradfi):
=
dfi
dSl
Y =
dy
dx
In particular,
Z =
if
gravity
is
dSl
(7.3.6)
dz
the only body force acting,
12
=
gh,
with h a direction
measured vertically upward; thus
X=-g—
dx
Y=-g—
dy
Remembering that
p
is
Z=-g—
dz
(7.3.7)
constant for an ideal
fluid,
substituting Eqs. (7.3.3)
to (7.3.7) into Eqs. (7.3.2) gives
Id,
_
+ v- + w — + + yh)=u---(v
dt
dy
dz
dx
p dx
du
du
du
mnn
du
dv
dv
Id,
dv
dv
---(p
+ ih)=u- + v- + w- + -
dx
p dy
Id
---(p +
p dz
yk)
dy
dz
,
(7.3.8)
(7.3.9)
dt
dw
dw
dw
dw
=M—
+ V—
+ W—
+—
dz
dx
dy
,
m
_ ...
(7.3.10)
dt
The first three terms on the right-hand sides of the equations are convectiveacceleration terms, depending upon changes of velocity with space. The last
term is the local acceleration, depending upon velocity change with time at a
point.
IDEAL-FLUID FLOW
399
two-dimensional flow
Natural coordinates
in
Euler's equations in
two dimensions are obtained from the general-component
w = and d/dz = 0; thus
equations by setting
1
d
p dx
7
Id,
p dy
N
\V "T yh)
(P
+
7^)
= u
du
dy
dv
dt
dv
\~
dx
(7.3.11)
1
dx
=,w
du
du
h v
v
dv
(7.3.12)
1
dy
dt
By
taking particular directions for the x and y axes, they can be reduced to a
form that makes them easier to understand. If the x axis, called the s axis,
is taken parallel to the velocity vector at a point (Fig. 7.6) it is then tangent
to the streamline through the point. The y axis, called the n axis, is drawn
toward the center of curvature of the streamline. The velocity component u
is v s and the component v is v n
As v n is zero at the point, Eq. (7.3.11) becomes
,
.
Id
-—
+
(p
p ds
dv s
yh)
= Vs—ds
+
—
dv s
(7.3.13)
dt
Although v n is zero at the point (s,n), its rates of change with respect to
and t are not necessarily zero. Equation (7.3.12) becomes
-—^
1
p
dn
Fig. 7.6
,
JN
+ yh)
.
(p
=
vs
—+—
dv n
dv n
ds
dt
Notation for natural co-
ordinates.
s
(7.3.14)
FUNDAMENTALS OF FLUID MECHANICS
400
When
the velocity at
changes from zero to
and at
s
5v n
from similar triangles
.
With
s
r
+
8s
along the streamline
is
considered, v n
the radius of curvature of the streamline at
s,
(Fig. 7.6),
8s
8v n
dv n
vs
r
vs
ds
r
Substituting into Eq. (7.3.14) gives
-1± (, + *)-* + §
dn
p
r
(7.3.15)
dt
For steady flow of an incompressible
be written
may
---d
1
(p
p ds
fluid
Eqs. (7.3.11) and (7.3.15)
d
+
yh)
+
yh)
"
ds
(7.3.16)
(?)
and
d
1
p
,
dn
(P
2
Vs
=
(7.3.17)
r
Equation (7.3.16) can be integrated with respect to s to produce Eq. (3.9.1),
with the constant of integration varying with n, that is, from one streamline to
another. Equation (7.3.17) shows how pressure head varies across stream-
With
lines.
EXAM PL E
vs
and
known
r
functions of n, Eq. (7.3.17) can be integrated.
A container of liquid is rotated with angular velocity
co about a
Determine the variation of pressure in the liquid.
the radial distance, measured inwardly, n = —r, dn — —dr, and
7.1
vertical axis as a solid.
n
v8
=
cor.
1
(P
is
By
+
integrating Eq. (7.3.17),
7
N
yh)
= —
f
2
r2
dr
(-
J
o
P
co
r
or
1
- (p
p
+
yh)
=
qfra
—— +
2
const
const
IDEAL-FLUID FLOW
To
v
evaluate the constant,
=
Po
-
yh
p
= p when
r
and h =
=
0,
then
—
co
+
p
if
2
r2
which shows that the pressure
hydrostatic along a vertical line and in-
is
creases as the square of the radius. Integration of Eq. (7.3.16)
pressure
is
401
constant for a given h and
v8
,
that
is,
shows that the
along a streamline.
IRROTATIONAL FLOW; VELOCITY POTENTIAL
7.4
In this section
it is
shown that the assumption
the existence of a velocity potential.
tion of a conservative
The
body
of irrotational flow leads to
By use of these relations and the assump-
force, the
Euler equations can be integrated.
individual particles of a frictionless incompressible fluid initially at
cannot be caused to rotate. This can be visualized by considering a small
body of fluid in the shape of a sphere. Surface forces act normal to its
surface, since the fluid is frictionless, and therefore act through the center of
the sphere. Similarly the body force acts at the mass center. Hence no torque
can be exerted on the sphere, and it remains without rotation. Likewise, once
an ideal fluid has rotation, there is no way of altering it, as no torque can be
exerted on an elementary sphere of the fluid.
By assuming that the fluid has no rotation, i.e., it is irrotational, curl q =
0, or from Eq. (7.2.11)
rest
free
dv
—
du
= —
dw
dx
dy
dy
dv
du
dz
dz
—
=:
dw
—
=
(7.4.1)
dx
These restrictions on the velocity must hold at every point (except special
singular points or lines)
two-dimensional flow.
u dx
is
+
v
.
The
first
equation
is
the irrotational condition for
It is the condition that the differential expression
dy
exact, say
udx
+
v
dy
The minus
= —d<f>=
sign
is
dy
dx
dx
arbitrary;
(7.4.2)
dy
it is
to
a convention that causes the value of
By comparing terms in Eq. (7.4.2),
decrease in the direction of the velocity.
(f>
FUNDAMENTALS OF FLUID MECHANICS
402
u =
= — d<j>/dy.
—d<t>/dx, v
flow, of a function
direction
is
This proves the existence, in two-dimensional
such that its negative derivative with respect to any
<f>
the velocity component in that direction.
strated for three-dimensional flow.
q
=
is
equivalent to
=
-grad</>
can also be demon-
-Vcj>
d<f>
u =
It
In vector form,
d<f>
=
v
(7.4.3)
dx
w =
dy
The assumption
d(f>
(7.4.4)
K
J
dz
of a velocity potential
is
equivalent to the assumption of
irrotational flow, as
-V
(-grad0) =
curl
because
du _
dy
V
v =
x
0.
d 2 <f)
dv
dx dy
dx
=
proving dv/dx
x
v</>
This
is
=
(7.4.5)
shown from Eq.
(7.4.4)
by
cross-differentiation:
d 2 cf>
_
dy dx
du/dy,
etc.
Substitution of Eqs. (7.4.4) into the continuity equation
du
dw
dv
1
=
1
dx
dy
dz
yields
<f>
d 2 <f>
d 2 <f>
dx 2
dy 2
dz 2
d2
In vector form this
V
•
and
Any
-V
q
=
is
written
function
is
V0 = -V
V2
<f>
fluid-flow case.
=
0.
2
=
(7.4.7)
Equation
(7.4.6) or (7.4.7) is the
that satisfies the Laplace equation
As there are an
infinite
number
is
Laplace equation.
a possible irrotational
of solutions to the Laplace
equation, each of which satisfies certain flow boundaries, the
is
main problem
the selection of the proper function for the particular flow case.
IDEAL-FLUID FLOW
Because
<j>
linear equation,
appears to the
and the sum
fa are solutions of
V 2 fa =
Eq. (7.4.6)
power in each term of Eq. (7.4.6)
two solutions is also a solution; e.g., if
first
of
then fa
+
fa
is
a solution; thus
a solution
if
C
403
,
it is
fa
a
and
V 2 fa =
then
V 2 (<£i
+
Similarly
if
+
= V2fa
fa)
V 2 fa =
fa is a solution
Cfa
is
is
constant.
INTEGRATION OF EULER'S EQUATIONS; BERNOULLI
EQUATION
7.5
Equation
(7.3.8)
can be rearranged so that every term contains a partial
From Eq. (7.4.1)
derivative with respect to x.
du
dv
d V2
du
dw
dy
dx
dx 2
dz
dx
and from Eq.
d
d<f>
dt
dx
dt
a
As u 2
d_
dx
u2
v2
w
2
2
2
J
+
v2
+
iv
2
=
2
q
,
2
dt)
the square of the speed,
e + *+«-«)..
d_
dy
d_( v
c-
+
dx 2
d<f>\
Similarly for the y and z directions,
dz
2
these substitutions into Eq. (7.3.8) and rearranging give
(V
dx \p
w
(7.4.4)
du
Making
d
* + 1 - *)
=
FUNDAMENTALS OF FLUID MECHANICS
404
The
quantities within the parentheses are the
Equation
(7.5.3).
(7.5.1)
since the derivative with respect to x
show that the quantity
tion of
only, say F(t)
t
+ gh + \--Z
Z
dt
p
In steady flow
+
-
+^
2
gh
p
The
=
d<t>/dt
is
same
states that the quantity
is
zero.
not a function of y or
in Eqs. (7.5.1) to
not a function of
is
x,
Similarly the other equations
z.
Therefore
it
can be a func-
:
F(t)
=
(7.5.4)
and F(t) becomes a constant E:
= E
(7.5.5)
available energy
is
everywhere constant throughout the
fluid.
This
is
Bernoulli's equation for an irrotational fluid.
The
pressure term can be separated into two parts, the hydrostatic
pressure p s and the dynamic pressure p d
Eq. (7.5.5) gives
,
Ps
,
.
gh-\
Pd
gh
p
= p
s
+
p d Inserting in
.
2
2
two terms may be written
first
-f-
so that
~ = E
p
p
The
q
\~
1
,
—
= -
p
p
+
(ps
yh)
with h measured vertically upward.
The
expression
expresses the hydrostatic law of variation of pressure.
be included
in the constant E.
is
a constant, since
These two terms
it
may
After dropping the subscript on the dynamic
pressure, there remains
-
+£
= E
(7.5.6)
This simple equation permits the variation in pressure to be determined if
is known or vice versa. Assuming both the speed q Q and the dynamic
the speed
pressure p to be
p
2
p
known
2
at one point,
IDEAL-FLUID FLOW
405
or
-*+f [-©']
A submarine moves through water at a speed of 30 ft/s. At
7.2
on the submarine 5 ft above the nose, the velocity of the submarine
relative to the water is 50 ft/s. Determine the dynamic pressure difference
between this point and the nose, and determine the difference in total pressure
between the two points.
If the submarine is stationary and the water is moving past it, the velocBy selecting the
ity at the nose is zero, and the velocity at A is 50 ft/s.
dynamic pressure at infinity as zero, from Eq. (7.5.6)
EXAMPLE
a point
A
— = 450
QQ2
2
Q
+-
E =
=
2
ft
-lb/slug
2
For the nose
1
= E = 450
v
= 450 X
1.935
= 870 lb/ft 2
p
A
for point
E_,_*_«0-^
2
p
and
p
=
1.935
Therefore the difference in dynamic pressure
-1548
-
The
to point
gh A
870
= -2418
(?-?)=
2 /
\
and
h
-\
lb/ft 2
by applying Eq.
to the nose n,
—=
gh n
h
-\
z
p
—
J
p
Hence
PA-p n
=
p
(^ -
= -2740
pfci
lb/ft 2
lb/ft<
is
difference in total pressure can be obtained
A
-1548
2
2
+
gn *
~ g^ =
)
1.935
(-5g -
y)
(7.5.5)
FUNDAMENTALS OF FLUID MECHANICS
406
It
may
also
be reasoned that the actual pressure difference varies by 57 from
A is 5 ft above the nose, or —2418 —
the dynamic pressure difference since
5
X
= -2740 lb/ft
62.4
2
.
STREAM FUNCTIONS; BOUNDARY CONDITIONS
7.6
Two
stream functions are defined: one for two-dimensional flow, where all
motion are parallel to a fixed plane, say the xy plane, and the flow is
identical in each of these planes, and the other for three-dimensional flow
with axial symmetry, i.e., all flow lines are in planes intersecting the same
line or axis, and the flow is identical in each of these planes.
lines of
Two-dimensional stream function
If
A,
P
represent two points in one of the flow planes,
(Fig. 7.7)
lines
,
and
ACP,
if
ABP
e.g.,
the xy plane
the plane has unit thickness, the rate of flow across any two
must be the same
if
the density
is
constant and no fluid
created or destroyed within the region, as a consequence of continuity.
is
Now,
a fixed point and
P
a movable point, the flow rate across any line
connecting the two points
is
a function of the position of P.
if
is
A
\f/,
is
and
right to left
is
If this function
taken as a sign convention that it denotes the flow rate from
as the observer views the line from A looking toward P, then
if it is
defined as the stream function.
If fa, fa represent the values of
7.8), respectively,
then fa
—
fa
Fig. 7.7
Fluid region showing
the positive flow direction used
in the definition of a stream
function.
is
stream function at points Pi, P2 (Fig.
PiP 2 and is independent of
the flow across
IDEAL-FLUID
between
Fig. 7.8
Flow
points
a fluid region.
in
the location of
of fa, ^2
The
A Taking another point
u
in the place of
.
viz.,
the flow across OA.
A
changes the values
Then \p is indeterminate
an arbitrary constant.
velocity components u, v in the x, y directions can be obtained from
the stream function.
left, is
407
two
by the same amount,
to the extent of
FLOW
— u by,
In Fig. 7.9a, the flow
b\p
across
AP =
by,
from right to
or
b^
=
(7.6.1)
dy
and similarly
_
b$
_ ty
8x
dx
(7.6.2)
P
.
vr
,8r
P
T
I
;hr p
'
(b)
(a)
Fig. 7.9
velocity
Selection of path to
components
to
show
relation of
stream function.
FUNDAMENTALS OF FLUID MECHANICS
408
In words, the partial derivative of the stream function with respect to any
direction gives the velocity
=
vr
from
1
ty
—
r
do
—
to that direc-
ve
—
= ty
dr
Fig. 7.96.
When
^i
component +90° (counterclockwise)
In plane polar coordinates
tion.
^2
given by
as there
\p
P h P 2 of Fig. 7.8 lie on the same streamline,
no flow across a streamline. Hence, a streamline is
comparing Eqs. (7.4.4) with Eqs. (7.6.1) and (7.6.2),
the two points
=
=
const.
is
By
d0
d\p
d<j>
dxf/
dx
dy
oy
ox
These are the Cauchy-Riemann equations.
By Eqs. (7.6.3) a stream function can be found
for each velocity potenthe velocity potential satisfies the Laplace equation the stream
If
tial.
function also satisfies
it.
Hence, the stream function
may
be considered as
velocity potential for another flow case.
Stokes' stream function for axially symmetric flow
In any one of the planes through the axis of symmetry
A
such that
fixed
is
and
P is variable. Draw
select
two points A,
through the surface generated by rotating AP about the axis
a function of the position of P. Let this function be 2^, and
symmetry be the x axis of a cartesian system of reference. Then
of x and to, where
co
is
=
vy
+
2
£
let
\f/
is
the axis of
a function
2
the distance from
To
P
AP. The flow
of symmetry is
a line connecting
P to the x axis. The surfaces ^ =
find the relation
between
\p
const are stream surfaces.
and the velocity components
f
u, v parallel
and the cD axis (perpendicular to the x axis), respectively, a
employed similar to that for two-dimensional flow. Let PP'
be an infinitesimal step first parallel to to and then to x that is, PP' = 5co and
then PP' = 8x. The resulting relations between stream function and velocity
are given by
to the x axis
procedure
is
;
— 27rco
5co
u =
2it
5\f/
and
2-7rco
dx
v'
=
2tt
5\J/
IDEAL-FLUID FLOW
Solving for u,
v'
gives
1#
u =
co
dco
co
(7.6.4)
dx
The same sign convention is used as in the two-dimensional case.
The relations between stream function and potential function
d<f>
409
1 dip
d(f>
dco
dco
are
1#
(7.6.5)
dx
co
co
dx
In three-dimensional flow with axial symmetry,
or
\J/
L T~
3
has the dimensions
l
,
volume per unit time.
The stream function
is
used for flow about bodies of revolution that are
frequently expressed most readily in spherical polar coordinates. Let
needed because of axial
=
2ttt sin 6 8r v e
r
and 6 be the polar angle; the meridian angle
symmetry. Referring to Fig. 7.10a and b,
distance from the origin
2-/r 8\f/
— 2wr sin d r 56 v =
r
2
be the
is not
w5\f/
from which
1
d$
dxfr
(7.6.6)
Ve
r 2 sin 6
r sin 6 dr
dd
and
dxfr
—
sin 6 66
dr
1
—dr =
— sin 6 —
(7.6.7)
dd
56
4^
It {
/
rfh~~2^
\e
(a)
Fig. 7.10
Displacement
A
(b)
of
P
to
show the
nents and Stokes' stream function.
relation
between velocity compo-
FUNDAMENTALS OF FLUID MECHANICS
410
7^/////////ZZ.
Notation for boundary
Fig. 7.11
condition at a fixed boundary.
These expressions are useful in dealing with flow about spheres,
and disks and through apertures.
ellipsoids,
Boundary conditions
At a fixed boundary the velocity component normal to the boundary must
be zero at every point on the boundary (Fig. 7.11)
q
•
rii
is
tli
=
(7.6.8)
a unit vector normal to the boundary.
In scalar notation this
is
easily
expressed in terms of the velocity potential
(7.6.9)
dn
at all points
on the boundary.
For a moving boundary
(Fig. 7.12),
where
the boundary point has the velocity V, the fluid-velocity component normal
to the
boundary must equal the velocity of the boundary normal to the bound-
ary; thus
q
•
rii
= V
•
rii
q-V
Fig. 7.12
Notation for boundary
condition at a moving boundary.
(7.6.10)
IDEAL-FLUID FLOW
411
or
(q
-
V)
ni
•
For two
=
(7.6.11)
fluids in contact, a
the pressure
viz.,
dynamical boundary condition
must be continuous
A stream surface in steady flow
for a
7.7
may
boundary and
is
required;
across the interface.
(fixed boundaries) satisfies the condition
be taken as a
solid
boundary.
THE FLOW NET
In two-dimensional flow the flow net
is
of great benefit
;
it is
taken up in this
section.
The
line
line
<f>(x,y)
along which the value of
Since velocity
v&
given by
d<t>
=
v s in
—
As ^
As
,.
<£
any direction
= — hm
ds
=
const
is
called
an equipotential
line.
It is
a
(the velocity potential) does not change.
s is
given by
A<f>
two closely spaced points on an equipotential line, the
no component in the direction defined by the line through
the two points. In the limit as As —>
this proves that there is no velocity
component tangent to an equipotential line and, therefore, the velocity
vector must be everywhere normal to an equipotential line (except at singular
points where the velocity is zero or infinite)
The line \p(x,y) = const is a streamline and is everywhere tangent to
the velocity vector. Streamlines and equipotential lines are therefore orthogand
A<f>
is
zero for
velocity vector has
onal;
is
i.e.,
they intersect at right angles, except at singular points. A flow net
of a family of equipotential lines and a corresponding family of
composed
streamlines with the constants varying in arithmetical progression.
It is
customary to let the change in constant between adjacent equipotential lines
and adjacent streamlines be the same, for example, Ac. In Fig. 7.13, if the
distance between streamlines is An and the distance between equipotential
lines is As at some small region in the flow net, the approximate velocity v s
is then given in terms of the spacing of the equipotential lines [Eq. (7.4.4) ],
A<f>
— Ac
Ac
As
As
As
FUNDAMENTALS OF FLUID MECHANICS
412
Elements
Fig. 7.13
of a
flow
net.
or in terms of the spacing of streamlines [Eqs. (7.6.1) and (7.6.2)],
Ad/
Ac
An
An
These expressions are approximate when Ac is finite, but when Ac becomes
very small, the expressions become exact and yield velocity at a point. As
both velocities referred to are the same, the equations show that As = An,
or that the flow net consists of an orthogonal grid that reduces to perfect
squares in the limit as the grid size approaches zero.
Once a flow net has been found by any means to satisfy the boundary
conditions and to form an orthogonal net reducing to perfect squares in the
limit as the
number
of lines
is
increased, the flow net
is
the only solution for
the particular boundaries, as uniqueness theorems in hydrodynamics prove.
In steady flow when the boundaries are stationary, the boundaries themselves
become part of the flow net, as they are streamlines. The problem of finding
the flow net to satisfy given fixed boundaries may be considered purely as a
graphical exercise, i.e., the construction of an orthogonal system of lines that
compose the boundaries and reduce to perfect squares in the limit as the number of lines increases. This is one of the practical methods employed in twox
dimensional-flow analysis, although
it
usually requires many attempts and much
erasing.
Another practical method
fixed boundaries
is
the
of obtaining a flow net for a particular set of
electrical analogy.
The boundaries in a model are formed
out of strips of nonconducting material mounted on a
surface,
and the end equipotential
lines are
flat nonconducting
formed out of a conducting strip,
IDEAL-FLUID FLOW
413
e.g., brass or copper. An electrolyte (conducting liquid) is placed at uniform
depth in the flow space and a voltage potential applied to the two end conducting strips. By means of a probe and a voltmeter, lines with constant drop in
voltage from one end are
mapped out and
plotted.
These are equipotential
By reversing the process and making the flow boundaries out of conduct-
lines.
ing material
and the end equipotential
lines
from nonconducting material,
the streamlines are mapped.
A special conducting paper, called Teledeltos paper, may be used in place
tank with an electrolyte. Silver ink is used to form a conducting strip
or line having constant voltage. One cuts the paper to the size and shape
needed, places the constant-voltage lines on the paper with a heavy line of
silver ink, then marks the intermediate points of constant voltage directly
on the paper, using the same circuits as with an electrolyte.
The relaxation method numerically determines the value of potential
function at points throughout the flow, usually located at the intersections
of a square grid. The Laplace equation is written as a difference equation,
and it is shown that the value of potential function at a grid point is the
average of the four values at the neighboring grid points. Near the boundaries
special formulas are required. With values known at the boundaries, each
grid point is computed based on the assumed values at the neighboring grid
points; then these values are improved by repeating the process until the
changes are within the desired accuracy. This method is particularly convenient for solution with high-speed digital computers.
of a
1
Use of the flow net
After a flow net for a given boundary configuration has been obtained,
be used for
all
necessary to
is
point.
it
may
irrotational flows with geometrically similar boundaries.
know
It
the velocity at a single point and the pressure at one
Then, by use of the flow net, the velocity can be determined at every
Application of the Bernoulli equation [Eq. (7.5.7)] produces
other point.
If the velocity is known, e.g., at A (Fig. 7.13), An
from the adjacent lines. Then Ac tt An v s tt As v s With
the constant Ac determined for the whole grid in this manner, measurement
of As or An at any other point permits the velocity to be computed there,
the dynamic pressure.
or As can be scaled
Ac
us
^^
As
X
C.-S.-
—
.
Ac
—
An
Yih, Ideal-Fluid Flow, p. 4-67 in V. L.
New York, 1961.
Dynamics," McGraw-Hill,
Streeter
(ed.),
"Handbook
of
Fluid
FUNDAMENTALS OF FLUID MECHANICS
414
The concepts underlying the flow net have been developed for irrotational
flow of an ideal
fluid.
Because of the similarity of
differential equations
describing groundwater flow and irrotational flow, the flow net can also be
used to determine streamlines and lines of constant piezometric head (h + p/y)
for percolation through homogeneous porous media. The flow cases of Sec. 7.9
may
then be interpreted in terms of the very rotational, viscous flow at extremely small velocities through a porous medium.
THREE-DIMENSIONAL FLOW
7.8
Because of space limitations only a few three-dimensional cases are conThey are sources and sinks, the doublet, and uniform flow singly
or combined.
sidered.
Three-dimensional sources and sinks
A
source in three-dimensional flow
uniform rate in
resembling
it
flow patterns.
directions.
all
in nature,
The
is
a point from which fluid issues at a
It is entirely fictitious, as there is
but that does not reduce
strength of the source
its
nothing
usefulness in obtaining
m is the rate of flow passing through
any surface enclosing the source.
As the flow is outward and is uniform in all directions, the velocity, a
distance r from the source, is the strength divided by the area of the sphere
through the point with center at the source, or
m
Since v r
=
—d<f>/dr
and
ve
=
0, it
follows that dcfr/dd
=
0,
and the velocity
potential can be found.
d<j>
m
dr
4tit 2
n r LU.
id
ell
m
(7.8.1)
<t>
47JT
A negative source is a sink.
and there disappear.
Fluid
is
assumed to flow uniformly into a sink
IDEAL-FLUID FLOW
415
Three-dimensional doublets
A
combination of a source and a sink of equal
which are allowed to approach each other in such a manner that
the product of their strength and the distance between them remains a condoublet, or double source, is a
strength,
stant in the limit.
In Fig. 7.14, a source of strength m is located at (a ft) and a sink of the
at (— afl).
Since each satisfies the Laplace equation, their
same strength
sum
also satisfies
m/1
1\
4tt \ri
r2 /
it:
(7.8.2)
</>
By
the law of sines and Fig. 7.14,
n
sin
2a
r2
sin
2
as the angle
n =
r2
From Eq.
^>
=
—m
r2
4tf
(-a,0)
Fig. 7.14
sin (0i
0i
2a
—
2)
between
r2
and
a (sin
0i
—
-
sin i (0i
r\
sin
cos
2)
2 sin \ (0i
at
P
is 0i
2.
-
2)
cos \
2)
(0i
—
Solving for
2a cos
2)
\ (0i
—
—
^ (0i
cos \
(0i
+
-
2)
2)
(7.8.2)
—
2am
ri
rx r2
cos ^
47rrir 2
Auxiliary
for
cos ^
(0i
+
—
(a,0)
(0,0)
systems used
(0i
coordinate
Rankine body.
^2)
2)
M cos
rir2
cos
\ (0i
+
2)
\ (0i
—
2)
r2
2)
—
ri
gives
FUNDAMENTALS OF FLUID MECHANICS
416
Streamlines and equipotential lines for a three-dimen-
Fig. 7.15
sional doublet.
In the limit as a approaches zero,
= -o
r
which
cos 6
2
=
0i
=
6,
r2
= n =
r,
and
(7.8.3)
the velocity potential for a doublet 1 at the origin with axis in the
is
Equation (7.8.3) can be converted into the stream func(7.6.7). The stream function is
positive x direction.
tion
by Eqs.
_ _
£ = _ Msin^
f
3
(?
g4)
r
Streamlines and equipotential lines for the doublet are drawn in Fig. 7.15.
1
L.
M. Milne-Thompson,
1938.
"Theoretical Hydrodynamics," p. 414, Macmillan, London,
IDEAL-FLUID FLOW
Source
The
in a
417
uniform stream
radial velocity v r
due to a source at the origin
m
(7 8 1}
* = 4
Vr
-
-
is
m
d<f>
Vr
~
~
'
dr
"
which,
when
47rr 2
by the
multiplied
gives the strength m.
surface area of the sphere concentric with
it,
Since the flow from the source has axial symmetry,
Stokes' stream function
is
defined.
For spherical polar coordinates, from
Eqs. (7.6.7),
—
dxf/
= —
.
Q
sin 6
—
=
—
66
d<t>
d\J/
66
dr
WithEq.
.
r 2 sin 6
—
d<f>
dr
(7.8.1),
—=
m
—86 =
dxl/
d\I/
dr
.
sin^
4tt
Integrating gives
$ =
— Cos
TYl
(7.8.5)
6
4?r
the stream function for a source at the origin. Equipotential lines and streamlines are
shown
in Fig. 7.16 for constant increments of
<f>
and
\f/.
A uniform stream of fluid having a velocity U in the negative x direction
given by
throughout space
is
-»=-U
dx
^=
doo
Integrating gives
4 =
Ux=Ur cos 6
(7.8.6)
FUNDAMENTALS OF FLUID MECHANICS
418
Streamlines and equipotential
Fig. 7.16
lines
for a source.
The stream
function
U
The
Ur
is
found in the same manner as above to be
2
sm
flow network
2
is
6
(7.8.7)
shown
in Fig. 7.17.
Combining the uniform flow and the source flow, which may be accomby adding the two velocity potentials and the two stream functions,
plished
gives
4>
=
771
z
h
Ur
— cos
The
resulting flow
is
6
sin 2
-\
(7.8.8)
2
47T
were added
Uv^
771
^ =
cos 6
47JT
everywhere the same as
if
the separate velocity vectors
for each point in space.
A stagnation point is a point in the fluid where the velocity is zero.
The
conditions for stagnation point, where spherical polar coordinates are used
and when the flow has
vr
=
=
dr
vd
axial
symmetry, are
=
=
r
dd
IDEAL-FLUID FLOW
419
y=6U
^=5U
xf/=MJ
=3U
rf
= 2U
^=U
y^=0
xf/
= -V
^ = -2U
^=~3U
^ = -AU
= -5U
^=-6U
if
Fig. 7.17
form flow
Use
uni-
of these expressions with Eqs. (7.8.8) gives
m
4tit2
U sin 6 =
U cos 6 =
which are
6
Streamlines and equipotential lines for
in negative x direction.
satisfied
/
=
by only one point
in space, viz.,
m
U
Substituting this point back into the stream function gives
is
the stream surface through the stagnation point.
surface
is
found from Eqs.
2irU
cos 6 H
m
r 2 sin 2 6
=
1
\f/
=
m/Air, which
The equation
of this
(7.8.8)
(7.8.9)
FUNDAMENTALS OF FLUID MECHANICS
420
The flow under consideration is steady, as the velocity potential does not
change with the time. Therefore, any stream surface satisfies the conditions
for a boundary the velocity component normal to the stream surface in steady
flow is always zero. Since stream surfaces through stagnation points usually
split the flow, they are frequently the most interesting possible boundary.
This stream surface is plotted in Fig. 7.18. Substituting w = r sin in Eq.
(7.8.9), the distance of a point (r,6) from the x axis is given by
:
m
(1
2^U
-
cos0)
which shows that
cD
maximum value \/m/irU as 6 approaches r, that is,
Hence, w = \/m/irU is an asymptotic surface to
surface. Equation (7.8.9) may be expressed in the form
has a
as r approaches infinity.
the dividing stream
H^
m
6
SeC
(7.8.10)
2
from which the surface
half-body, as
The
in
it
is
easily plotted.
pressure at any point,
i.e.,
which the dynamic pressure at
Fig. 7.18
Such a
figure of revolution is called a
extends to negative infinity, surrounding the negative x
infinity is
taken as
Streamlines and equipotential lines
for a half-body.
axis.
the dynamic pressure from Eq. (7.5.7)
zero,
q
is
,
is
the speed at
IDEAL-FLUID FLOW
any point. Evaluating q from Eqs.
q
\drj
r
2
m
+
\ddj
421
(7.8.8) gives
mil
2
2
167r r
cos
27rr 2
2
and
(m
p TT
m
cos
2
\
(7 8
>-i*feF-iSw)
-
from which the pressure can be found
a singular point.
pressure
3m
/
m
2
for
Eq. (7.8.10)
given in terms of
is
p
p
When
r
for
any point except the
is
origin,
-
n)
which
is
substituted into Eq. (7.8.11), the
any point on the
half body; thus
\
=Mi^-^)
(7 8 12)
-
This shows that the dynamic pressure approaches zero as
r
increases
-
down-
stream along the body.
Source and sink of equal strength
A
in a
source of strength m, located at
point
P
uniform stream; Rankine bodies
(cr-,0),
has the velocity potential at any
given by
m
4*
= -7—
47Tfi
in
which
n
is
the distance from (a,0) to P, as shown in Fig. 7.14.
the potential function for a sink of strength
<t>2
m at
(— afl)
Similarly,
is
m
—
= —
~
47rr2
Since both fa and
<f> 2
satisfy the Laplace equation, their
sum
will also
be a
solution,
m
/l
1\
4tt \ri
r2 /
Because r h r 2 are measured from different points, this expression must be
handled differently from the usual algebraic equation.
The stream functions for the source and sink may also be added to give
FUNDAMENTALS OF FLUID MECHANICS
422
the stream function for the combined flow,
$ =
—m (cos
-
0i
cos02 )
(7.8.14)
47T
The stream
7.19,
which
values of
<f>
=
<f>
and equipotential surfaces take the form shown in Fig.
by taking constant
surfaces
plotted from Eqs. (7.8.13) and (7.8.14)
is
and
\J/.
Superposing a uniform flow of velocity U in the negative x direction,
= \U&> 2 the potential and stream functions for source and sink
Ux,
,
\f/
of equal strength in a uniform flow (in direction of source to sink) are
= Ux
+
—m
47T
Vi
r2 /
ml'
4tt
2
f = iUr
Lv
sin 2 6
11
1
(x
—
a)
(cos
-\
2
+
ft
-
co
2
cos
47T
Fig. 7.19
Streamlines and equi-
potential lines for a source
sink of equal strength.
and
V(x +
ft)
a) 2
+
co
2
J
(7.8.16)
IDEAL-FLUID FLOW
As any stream
surface
may
be taken as a
solid
boundary
423
in steady flow,
the location of a closed surface for this flow case will represent flow of a uniform
stream around a body. Examining the stream function, for x > a and di =
= 0. For x < -a and 6X = 2 = B = ir, = 0. Therefore, ^ =
2 = 0, yp
\J/
must be the dividing streamline,
since the x axis
equation of the dividing streamline
+
m
(cos
6i
—
cos
2)
is,
is
from Eq.
the axis of symmetry.
The
(7.8.16),
=
(7.8.17)
2irU
which w = r sin is the distance of a point on the dividing stream surface
from the x axis. Since cos 0i and cos 02 are never greater than unity, cD cannot
exceed y/m/irU, which shows that the surface is closed and hence can be
replaced by a solid body of exactly the same shape. By changing the signs of
m and U the flow is reversed and the body should change end for end. From
in
Eq. (7.8.17) it is seen that the equation is unaltered; hence, the body has
symmetry with respect to the plane x = 0. It is necessarily a body of revolution because of axial
To
x axis,
it is
From Eq.
known
that the velocity
x2
—
a2
+
Ux
since
x
—
Fig. 7.20
of the equations.
is
D
a
r2
x
+
Rankine body.
a
(Fig. 7.20),
which must be on the
along the x axis
(7.8.15) the velocity potential
ma
2ir
symmetry
locate the stagnation points C,
<f>
x
for points
(it is
a streamline)
on the x axis
is
given by
FUNDAMENTALS OF FLUID MECHANICS
424
Differentiating with respect to x and setting the result equal to zero yield
d<f>x
TT
u
~
T*-
where x
is
maxo
= °
7w=*>*
(7 8 18)
-
-
the x coordinate of the stagnation point. This gives the point C (x 0)
The half-breadth h is determined as follows. From Fig. 7.20.,
,
(a trial solution)
0i
=
ir
—
a
.
= a
62
which
in
a
cos a
Vh +
2
a2
Substituting into Eq. (7.8.17) gives
m
rU
a
y/h
2
+
a2
from which h can be determined (also by trial solution)
Eliminating m/U between Eqs. (7.8.18) and (7.8.19) leads to
—
=
= -h
U
Xo
a
2
a
Vh +
2
a2
The value of a can be obtained for a predetermined body (xo, h specified)
Hence, U can be given any positive value, and the pressure and velocity
distribution can be determined.
In determining the velocity at points throughout the region it is convenient to find the velocity at each point due to each component of the flow, i.e.,
due to the source, the sink, and the uniform flow, separately, and add the
components graphically or by cD and x components.
Bodies obtained from source-sink combinations with uniform flow are
called Rankine bodies.
Translation of a sphere
The
1
an
infinite fluid
velocity potential for a solid
at rest
1.
in
must
moving through an
satisfy the following conditions
The Laplace equation Vfy =
everywhere except at singular points.
G. G. Stokes, "Mathematical and Physical Papers," vol.
sity Press,
London, 1880.
infinite fluid otherwise
1
i
1,
pp. 38-43, Cambridge Univer-
IDEAL-FLUID FLOW
425
The fluid must remain at rest at infinity; hence, the space derivatives of
must vanish at infinity.
The boundary conditions at the surface of the solid must be satisfied.
2.
3.
<f>
For a sphere of radius a with center at the origin moving with velocity U
x direction, the velocity of the surface normal to itself is
U cos 0, from Fig. 7.21. The fluid velocity normal to the surface is —d(f>/dr;
hence the boundary condition is
in the positive
-
dd>
-Z
= u
cos 6
dr
The
velocity potential for the doublet [Eq. (7.8.3)]
\i
cos
r
satisfies
V 2 <£ =
for
any constant value of /*. Substituting it into the boundary
condition gives
d<f>
dr
=
—r cos
2fJL
6
which is satisfied
components,
= U
for r
cos 6
=
a
if
n
=
£/a3 /2. It
and
~dr
r ~dd
Fig. 7.21
Sphere translating
the positive x direction.
in
may also be noted that the velocity
FUNDAMENTALS OF FLUID MECHANICS
426
are zero at infinity.
Therefore,
Ua?
COS0
(7.8.20)
2r 2
the conditions for translation of a sphere in an infinite fluid. This
one of unsteady flow, solved for the instant when the center of the sphere
Because this equation has been specialized for a particular
is at the origin.
instant, the pressure distribution cannot be found from it by use of Eq. (7.5.7)
satisfies all
case
is
Streamlines and equipotential lines for the sphere are shown in Fig. 7.22.
The stream function
Ua*
.
for this flow case is
,
Steady flow of an
(7.8.21)
infinite fluid
around a sphere
The unsteady-flow case in the preceding section can be converted into a steadyby superposing upon the flow a uniform stream of magnitude U in
= Ux = Ur cos to the potenthe negative x direction. To prove this, add
flow case
<j>
tial
function [Eq. (7.8.20)]; thus
W
4>
2r 2
cos
+
Ur
cos
Streamlines and equilines
for
sphere
a
moving through fluid.
Fig. 7.22
potential
(7.8.22)
IDEAL-FLUID FLOW
Fig. 7.23 Streamlines and equipotential lines for
form flow about a sphere at rest.
The stream function corresponding
—
Ua?
Ur2
.
sin2
H
2r
to this
427
uni-
is
.
sm
2
(7.8.23)
2
Then from Eq. (7.8.23), \p =
when =
and when r = a. Hence, the
stream surface \p =
is the sphere r = a, which may be taken as a solid,
fixed boundary. Streamlines and equipotential lines are shown in Fig. 7.23.
Perhaps
it
should be mentioned that the equations give a flow pattern for
the interior portion of the sphere as well.
No
fluid passes
through the surface
of the sphere, however.
The
velocity at
any point on the surface
of the sphere
is
Ictyl
r
deL a
q
"
The stagnation points are at 6 = 0, 6 = ir. The maximum velocity ft/ occurs
= 7r/2. The dynamic pressure distribution over the surface of the sphere
at 6
is
V
=
pU
2
—tt
-fsin
2
0)
(7.8.24)
dynamic pressure of zero at infinity.
In Sees. 5.6 and 5.7 the flow around a sphere in a real fluid is discussed.
Figure 5.20 shows the actual wake formation due to laminar and turbulent
boundary-layer separation. The drag on a sphere in real fluids is given by
for
FUNDAMENTALS OF FLUID MECHANICS
428
the drag-coefficient data in Fig. 5.21. In the ideal-fluid-flow case the boundary
If the x component of force exerted on
dynamic
pressure
is
obtained
from Eq. (7.8.24) by integration,
the sphere by
no
will
found
to
zero.
There
is
drag
on a body in ideal-fluid flow
it
be
be
energy
is
constant
and
everywhere
the high velocity of fluid
because the
obtained by flow over the front portion to the largest cross section is completely converted back into pressure in flowing over the downstream portion.
condition does not permit separation.
7.9
TWO-DIMENSIONAL FLOW
Two
simple flow cases that
may be interpreted for flow along straight boundexamined; then the source, vortex, doublet, uniform flow, and
flow around a cylinder, with and without circulation, are discussed.
aries are first
Flow around a corner
The
4>
potential function
= A(x*-y>)
has as
\p
its
stream function
— 2Axy = Ar 2
Fig. 7.24
around
90°
sin 20
Flow
net
bend.
for
flow
IDEAL-FLUID FLOW
a = 225°
and
are polar coordinates.
It is plotted for equal-increment
in Fig. 7.24.
and
Conditions at the origin are not defined,
a stagnation point. As any of the streamlines may be taken as fixed
which
r
changes in
as
= 45°
Flow net for flow along two inclined surfaces.
Fig. 7.25
in
oc
429
it is
<j>
\J/
may
be taken as walls, yielding flow into a 90°
corner. The equipotential lines are hyperbolas having axes coincident with
the coordinate axes and asymptotes given by y = ±x. The streamlines are
rectangular hyperbolas, having y = ±x as axes and the coordinate axes
as asymptotes. From the polar form of the stream function it is noted that the
= 0.
two lines =
and = w/2 are the streamline
This case may be generalized to yield flow around a corner with angle a.
By examining
boundaries, the plus axes
\f/
4>
= Ar irja
it
is
ird
Ar rla
cos
noted the streamline
flow nets are
shown
\f/
=
7T0
sin
is
and 6 = a.
now given by =
a = 225° and a = 45°.
Two
in Fig. 7.25, for the cases
Source
A line
imagined to flow uniformly
in all directions at right angles to it, is a source. It appears as a point in the
customary two-dimensional flow diagram. The total flow per unit time per
normal to the xy plane, from which
unit length of line
is
fluid is
called the strength of the source.
As the flow
is
in radial
lines from the source, the velocity a distance r from the source is determined
by the strength divided by the flow area of the cylinder, or 2Trn/2irr, in which
FUNDAMENTALS OF FLUID MECHANICS
430
the strength
Then, since by Eq.
is 2-k^.
(7.4.4) the velocity in
any direction
is
given by the negative derivative of the velocity potential with respect to the
direction,
d<f>
dr
dd
and
<t>
= —
is
the velocity potential, in which In indicates the natural logarithm and r
ijl
In r
the distance from the source.
<f>
satisfies
is
the Laplace equation
two dimensions.
in
The
streamlines are radial lines from the source,
r
~dr
From
* =
i.e.,
dd
the second equation
-/*0
Lines of constant
A
This value of
sink
is
Fig. 7.26
vortex.
<f>
(equipotential lines)
and constant
a negative source, a line into which fluid
Flow
net
for
source
or
is
\f/
are
shown
flowing.
in Fig. 7.26.
IDEAL-FLUID FLOW
431
Vortex
In examining the flow case given by selecting the stream function for the
source as a velocity potential,
<t>
= —nd
\f/
which also
=
\i
In r
the Laplace equation,
satisfies
it is
and the streamlines are
lines are radial lines
tangential direction only, since
d<f>/dr
=
0.
It
seen that the equipotential
circles.
The
velocity
is
in a
is
ia</)_M
q "
r 36
since r 66
is
r
the length element in the tangential direction.
In referring to Fig. 7.27, the flow along a closed curve
tion.
The
flow along an element of the curve
is
is
called the circula-
defined as the product of the
length element 5s of the curve and the component of the velocity tangent to
the curve, q cos
V =
q cos
a.
Hence the
ads =
Jo
is
<f>
constant.
The value
of the circulation is the strength of the vortex.
By selecting any circular path of radius
q = n/r, and ds = r d6; hence,
r
=
is
= —fi6 is for the vortex
velocity distribution given by the equation
such that the circulation around any closed path that contains the
is
vortex
C
n
The
and
T around a closed path
ds
q
/
J
circulation
r
/
q cos
ads =
Jn
r
2lt
I
-
Jn
r
6s
Fig. 7.27
Notation
for
definition of circulation.
r d6
=
27Tju
r
to determine the circulation,
a
=
0°,
FUNDAMENTALS OF FLUID MECHANICS
432
At the point
=
r
0, q
point. Figure 7.26
=
p/r goes to infinity hence, this point
;
shows the equipotential
lines
is
called a singular
and streamlines
for the vortex.
Doublet
The two-dimensional doublet
is
defined as the limiting case as a source
and
sink of equal strength approach each other so that the product of their strength
and the distance between them remains a constant 2irfi. y. is called the strength
of the doublet. The axis of the doublet is from the sink toward the source,
i.e., the line along which they approach each other.
In Fig. 7.28 a source is located at (a,0) and a sink of equal strength at
( — a,0)
The velocity potential for both, at some point P, is
.
<f>
= —m
with n,
2irm
for
ri
is
In
-f
m In r
2
measured from source and
r2
sink, respectively, to the point P.
the strength of source and sink.
2am =
and
n
ju,
may
r2
To take
must be altered. The terms
the form of the expression for
be expressed in terms of the polar coordinates r, 6 by the cosine
<f>
law, as follows
r22
=
r2
-
+
a2
+
a2
+
O
a
2ar cos
2ar cos
6
=
=
r
2
r2
L0 5
a
Si
Fig. 7.28
tion
of
doublet.
So
Notation for derivaa
Thus
the limit as a approaches zero
two-dimensional
1
1
+
(-
+
(-)
-
2 - cos o\
+
2 - cos d\
J
IDEAL-FLUID FLOW
Rewriting the expression for
= -
^ (In n 2
<f>
with these relations gives
+
= -^|lnr2
In r2 2 )
433
In
|~1
+ ("Y -
2 " cosfll
-lnr -lnri + (^Y + 2%os0l}
2
By
,
the series expression,
* =
x2
3
xA
+ »)-*--+--- +
,
N
In (1
m
f/ a
Y
--2|W-
a;
^ a
2
COSe
;
2
1
IY<A
-2LW-
~ a
2
COSS
7
.
2
(;)
2
-^cos«J-...-[Q +
:©,+i
2^co S «]
M-i[©,+ -M+-}
After simplifying,
</>
= 2am
h
|
r
Now,
if
ii
/aV
cos 6
[cos
2am =
cos
\r/
n and
if
r
4
/a\ cos
\rj
the limit
is
6
2
4 /a\ cos 3
3
r
\rj
+
]
taken as a approaches zero,
cos0
which
is the velocity potential for a two-dimensional doublet at the origin,
with axis in the -\-x direction.
Using the relations
d<j>
dr
_
ld\p
1
d<f>
_ d£
r dd
r d$
dr
FUNDAMENTALS OF FLUID MECHANICS
434
gives for the doublet
=
—
dd
dxf/
m cos 8
—
r
dr
d\p
= -n sin
.
6
r2
After integrating,
if,
/x
=
sin0
r
is
the stream function for the doublet.
The equations
are
a2
+
y-
$ =
x2
+
y
2
Rearranging gives
(-£)
Fig. 7.29
for the
,+
,
**£
" + (» +
S)
-£
Equipotential lines and streamlines
two-dimensional doublet.
in cartesian coordinates
IDEAL-FLUID FLOW
435
constant <£ are circles through the origin with centers on the
and the streamlines are circles through the origin with centers on
the y axis, as shown in Fig. 7.29. The origin is a singular point where the
The
x
lines of
axis,
velocity goes to infinity.
Uniform flow
Uniform flow
<t>
= Ux
—x
in the
direction,
u =
— U,
expressed
is
by
= Uy
\p
In polar coordinates,
<t>
= Ur
cos 6
\f/
= Ur
sin 6
Flow around a circular cylinder
of the flow due to a doublet and a uniform flow results in flow
around a circular cylinder; thus
The addition
<j>
= Ur
fjL
cos
cos
= rr
Ur sin 6
.
H
yp
m sin
r
r
\f/
As a streamline
given by
in steady flow
is
a possible boundary, the streamline
= Ois
= (ur which
Ur
-
If this
M
6
-J
sin 6
is
satisfied
-
=
value
is r
by
=
6
a,
=
0,
ir,
which
is
or
by the value
of r that
makes
a circular cylinder, then
= Ua?
and the streamline
\p
=
is
the z axis and the circle
r
=
a.
The
potential
and
stream functions for uniform flow around a circular cylinder of radius o are,
FUNDAMENTALS OF FLUID MECHANICS
436
by
<f>
substitution of the value of
= U
for the
a2 \
/
ir H
cos
lines for this case are
The
= U
yp
J
uniform flow in the
velocity at
ju,
a2 \
/
\r
sin
J
— x direction. The equipotential lines and stream-
shown in Fig. 7.30.
any point in the flow can be obtained from
velocity potential or the stream function.
velocity
is
necessarily tangential
U
The
of
velocity
2U
Eq.
+
(l
\
at 6
is
—
which holds
I
is
either the
the surface of the cylinder the
expressed
by
d\f//dr for r
=
a;
thus
sin
r=a
= 0, ir and has maximum values
For the dynamic pressure zero at infinity, with
zero (stagnation point) at $
7r/2,
(7.5.7) for
= 2U
-]sin0
r2 /
and
On
p
=
37r/2.
0,
q
any point
for
=
U,
in the plane except the origin.
For points on the
cylinder,
p
=
p
- f/2(i
_
4sin 2
The maximum
Fig. 7.30
0)
pressure,
which occurs at the stagnation points,
Equipotential lines and
streamlines
for
circular cylinder.
flow
around a
is
pU
2
/2;
IDEAL-FLUID FLOW
437
and the minimum pressure, at 6 = v/2, Sir/2, is — SpU 2 /2. The points of
dynamic pressure are given by sin = ±J, or 6 = ±tt/6, ± 5w/6. A
cylindrical pitot-static tube is made by providing three openings in a cylinder,
and ±30°, as the difference in pressure between and ±30° is the dyat
namic pressure pU 2 /2.
The drag on the cylinder is shown to be zero by integration of the
x component of the pressure force over the cylinder; thus
zero
r
Drag =
/
pa cos
Similarly, the
lift
6 dd
force
=
——
r
pau 2
/
(1-4 sin
on the cylinder
is
2
0)
cos
dd
=
zero.
Flow around a circular cylinder with circulation
The addition
of a vortex to the doublet
around a circular cylinder with
4>
-K)
The
results in flow
r
cos 6
-*('-t)
2^
distances
around a
\f/
=
In r
sin 6 H
2tt
r = a, and, at great
=
from the origin, the velocity remains u
—U, showing that flow
circular cylinder is maintained with addition of the vortex. Some
streamline
(
of the streamlines are
The
T/2ir) In a
shown
is
the circular cylinder
in Fig. 7.31.
velocity at the surface of the cylinder, necessarily tangent to the
Streamlines
Fig. 7.31
around
and the uniform flow
circulation,
a
circulation.
circular
for
flow
cylinder with
FUNDAMENTALS OF FLUID MECHANICS
438
cylinder,
is
r
2?7sin0H
dr
2ira
Stagnation points occur when q
sin 6
=
0; that
is,
r
=
4*Ua
When the circulation is 4:vUa, the two stagnation points coincide at r = a,
= — 7r/2. For larger circulation, the stagnation point moves away from the
6
cylinder.
The
pressure at the surface of the cylinder
The drag again
is zero.
The
lift,
is
however, becomes
2"
Lift
= —
f
/
pa
sin 6 dd
•'o
-
-
*f f b
2 sin
(
+
sb)1 sin
•*
pC7r
showing that the lift is directly proportional to the density of fluid, the approach velocity U, and the circulation I\ This thrust, which acts at right
The
angles to the approach velocity, is referred to as the Magnus effect.
Flettner rotor ship was designed to utilize this principle by mounting circular
cylinders with axes vertical on the ship and then mechanically rotating the
cylinders to provide circulation.
Airflow around the rotors produces the
thrust at right angles to the relative wind direction. The close spacing of
streamlines along the upper side of Fig. 7.31 indicates that the velocity is
high there and that the pressure must then be correspondingly low.
The theoretical flow around a circular cylinder with circulation can be
transformed into flow around an airfoil with the same circulation and the
1
same
lift.
The
shape. It
1
by producing a circulation around it due to its
can be shown that the lift is pUT for any cylinder in two-dimensional
airfoil
develops
its lift
1
V. L. Streeter, "Fluid Dynamics," pp. 137-155, McGraw-Hill,
New
York, 1948.
IDEAL-FLUID FLOW
flow.
The
439
angle of inclination of the airfoil relative to the approach velocity
(angle of attack) greatly affects the circulation.
For large angles of attack,
the flow does not follow the wing profile, and the theory breaks down.
It should be mentioned that all two-dimensional ideal-fluid-flow cases
be conveniently handled by complex-variable theory and by a system of
conformal mapping, which transforms a flow net from one configuration to
another by a suitable complex-variable mapping function.
may
EXAM PL E
1
A source with strength 0.2 m /s -m and
3
7.3
a vortex with strength
Determine the equation for velocity potential
What are the velocity components at x = 1 m, y =
m /s are located at the origin.
2
and stream function.
0.5
m?
The
<t>
-
=
velocity potential for the source
2
is
m 2//s
2
l
In
r
2tt
and the corresponding stream function
*
0.2
=
~
The
<t>
mVS
6
Yir
velocity potential for the vortex
= -
—
6
is
m /s
2
and the corresponding stream function
4/
— In
=
is
is
m /s
2
r
2tt
Adding the respective functions gives
4>
= ~
- ( 0.1 In
7T
The
At
r
+-
\
radial
^ =
and
)
-
2/
- (0.10
-
Jin r)
7T
and tangential velocity components are
_ 0* -
_JL
dr
lOirr
~
(1,0.5), r
= a/1 2
V$
+
_ 1
~
0.5 2
r
=
60 _
J_
dd
2irr
~
1.117
m,
vr
=
0.0285 m/s,
ve
=
0.143 m/s.
FUNDAMENTALS OF FLUID MECHANICS
440
PROBLEMS
7.1
Compute
the gradient of the following two-dimensional scalar functions:
= -2 In
(a)
+
(x 2
7.2
Compute
7.3
Compute the
7.4
Forq =
y
2
(6)
)
= Ux + Vy
the divergence of the gradients of
l(x
+
+ \(y + 2) + k(z +
2
y)
<j>
found
curl of the gradients of
2
2/
=
(c)
2sy
found in Prob.
7.1.
in Prob. 7.1.
+z
2
)
components
find the
of rotation
at (2,2,2).
7.5
Derive the equation of continuity for two-dimensional flow in polar coordinates
efflux from a small polar element to zero (Fig. 7.32)
It is
by equating the net
C^r
Vr
IdVe
dr
r
r
7.6
y
2
+
7.7
.
_
d$
The x component
z
2
.
is u
component
of velocity
Find the simplest
z
=
x2
+z +
2
and the y component
5,
is v
=
of velocity that satisfies continuity.
A velocity potential in two-dimensional flow is
(f>
=
y
+
x2
—
if.
Find the stream
function for this flow.
7.8
The two-dimensional stream function
for a flow
is \p
=
9
+
6x
—
4y
+
7xy.
Find the velocity potential.
7.9
Derive the partial differential equations relating
<f>
and
\f/
for two-dimensional
~flow in plane polar coordinates.
7.10
From
the continuity equation in polar coordinates in Prob.
7.5,
derive the
Laplace equation in the same coordinate system.
7.11
Does the function
In three-dimensional flow
7.12
By
=
is it
l/r satisfy the Laplace equation in two dimensions?
satisfied?
use of the equations developed in Prob. 7.9 find the two-dimensional stream
function for
Fig. 7.32
<f>
<f>
=
In
r.
IDEAL-FLUID FLOW
Find the Stokes stream function for
7.13
For the Stokes stream function
7.14
In Prob. 7.14 what
7.15
r
=
7.16
its
=
6
1,
and
r
=
1,
d
is
\f/
=
<j>
=
441
1/r.
2
9r sin2
0,
find
</>
in cartesian coordinates.
the discharge between stream surfaces through the points
=
tt/4?
Write the boundary conditions for steady flow around a sphere, of radius
a,
at
surface and at infinity.
7.17
A
velocity
circular cylinder of radius a has its center at the origin
and
V in the y direction.
terms of
satisfied at its surface
7.18
20 cfs
and at
Write the boundary condition
in
is
translating with
<f>
that
is
to be
infinity.
A source of strength 30 cfs is located at the origin, and another source of strength
is located at (1,0,0).
Find the velocity components u, v, w at (—1,0,0) and
(1,1,1).
7.19
If
the dynamic pressure
calculate the
7.20
dynamic pressure
is
zero at infinity in Prob. 7.18, for p
(—
at
1,0,0)
and
A source of strength m at the origin and a uniform flow of 5 m/s are combined in
By
1
m /s located 1 m from a plane barrier.
3
ft
long and 2
ft
thick in a transverse direction.
A source of strength 10 cfs at
(1,0,0)
combined with a uniform flow of 25
Rankine body formed by this flow.
are
7.24
the
A sphere of radius
—x
and p
Obtain the velocity
Equations are wanted for flow of a uniform stream of 12 ft/s around a Rankine
body 4
7.23
.
use of symmetry obtain the velocity potential for a three-dimensional sink
of strength
7.22
2.00 slugs/ft 3
(1,1,1)
three-dimensional flow so that a stagnation point occurs at (1,0,0)
potential and stream function for this flow case.
7.21
=
=
1
and a sink
ft/s in the
of the
same strength at ( — 1,0,0)
Determine the size of
— x direction.
m, with center at the origin, has a uniform flow of 6 m/s in
it.
At (1.25,0,0) the dynamic pressure is 500 N/m 2
direction flowing around
1000
kg/m3 Find
.
the equation for pressure distribution over the surface of the
sphere.
7.25
By
integration over the surface of the sphere of Prob. 7.24
on the sphere
is
V^=
Show that if two stream
0for^ = fc + &.
7. 27
Show that if u\, vi and u2
7.26
and 02 which
ponents are u =
<pi
7.28
show that the drag
zero.
,
functions
v2 are
\pi
and
+
u 2 and
v
=
both satisfy the Laplace equation,
the velocity components of two velocity potentials
satisfy the Laplace equation,
Ui
\p 2
then for
<£
=
<£i
+ $2 the velocity com-
Vi~\- v2 .
A two-dimensional source is located at (1,0)
and another one of the same strength
and (1,1).
draw the velocity components by adding the
at (—1,0). Construct the velocity vector at (0,0), (0,1), (0,-1), (0,-2),
(Hint: Using the results of Prob. 7.27,
FUNDAMENTALS OF FLUID MECHANICS
442
individual velocity components induced at the point in question
without regard to the other, due to
Determine the velocity potential
7.29
by each
source,
strength and location.)
its
for a source located at (1,0)
Write the equa-
.
tion for the velocity potential for the source system described in Prob. 7.28.
Draw
7.30
from
a set of streamlines for each of the sources described in Prob. 7.28 and
diagram construct the streamlines
this
of the sources
for the combined flow. (Hint: For each
draw streamlines separated by an angle of 7r/6. Finally combine the
intersection points of those rays for
which
\f/i
+
xf/ 2
is
constant.)
7.31
Does the line x =
form a line in the flow field described in Prob. 7.28 for
which there is no velocity component normal to it? Is this line a streamline? Could this
line be the trace of a solid plane lamina that was submerged in the flow? Does the
velocity potential determined in Prob. 7.29 describe the flow in the region x >
for
a source located at a distance of unity from a plane wall? Justify your answers.
Determine the equation for the velocity on the line x =
for the flow described
Find an equation for the pressure on the surface whose trace is x = 0.
What is the force on one side of this plane due to the source located at distance of
unity from it? Water is the fluid.
7.32
in Prob. 7.28.
7.33
In two-dimensional flow what
7.34
By
lines for the flow
By
7.35
is
the nature of the flow given by
<j>
=
7x
+
2 In r?
using a method similar to that suggested in Prob. 7.30 draw the potential
given in Prob. 7.33.
using the suggestion in Prob. 7.30 draw a flow net for a flow consisting of a
source and a vortex which are located at the origin. Use the same value for
p,
in
both
the source and the vortex.
A
7.36
is located at (—1,0), and a sink of twice the
For dynamic pressure at the origin of 100 lb/ft 2 p = 1.8
find the velocity and dynamic pressure at (0,1) and (1,1).
source discharging 20 cfs/ft
strength
slugs/ft 3
is
,
located at (2,0)
.
,
Select the strength of doublet needed to portray a uniform flow of 20
7.37
around a cylinder
of radius 2
m/s
m.
Develop the equations for flow around a Rankine cylinder formed by a source,
an equal sink, and a uniform flow. If 2a is the distance between source and sink, their
strength is 2tt/i, and U is the uniform velocity, develop an equation for length of the
7.38
body.
7.39
p
=
•
A
circular cylinder 8 ft in diameter rotates at 500
0.002 slug/ft 3 moving at 400 ft/s, what
,
is
the
lift
rpm.
When
in
an airstream,
force per foot of cylinder, assum-
ing 90 percent efficiency in developing circulation from the rotation?
7.40
An
unsteady-flow case
may
be transformed into a steady-flow case
(a)
regardless of the nature of the problem
(6)
when two bodies are moving toward each other in an infinite
when an unsymmetrical body is rotating in an infinite fluid
(c)
fluid
IDEAL-FLUID FLOW
(d)
when a
(e)
under no circumstances
single
body
translates in an infinite fluid
Select the value of
7.41
+
that satisfies continuity.
(a)
x2
(e)
none of these answers
y
The
7.42
2
(b)
sin x
(c)
In (x
-\-
(d)
y)
x-\- y
units for Euler's equations of motion are given
force per unit
(6)
velocity
(c)
energy per unit weight
(d)
force per unit weight
(e)
none
answers
of these
Euler's equations of motion can be integrated
(a)
the continuity equation
(6)
the fluid
(c)
a velocity potential exists and the density
(d)
the flow
(e)
the fluid
is
is
is
by
mass
(a)
7.43
when
it is
assumed that
satisfied
is
incompressible
rotational
is
constant
and incompressible
nonviscous
Euler's equations of motion are a mathematical statement that at every point
7.44
mass inflow equals rate of mass outflow
mass equals acceleration
(a)
rate of
(b)
force per unit
(c)
the energy does not change with the time
(d)
Newton's third law
(e)
the fluid
of
momentum
motion holds
is
constant
In irrotational flow of an ideal
7.45
(a)
a velocity potential exists
(6)
all
(c)
the motion must be uniform
(d)
the flow
(e)
the velocity
particles
A
7.46
is
must move
must be
(b)
is
fluid
in straight lines
always steady
must be
function
(a)
</>
zero at a
boundary
that satisfies the Laplace equation
linear in x and y
a possible case of rotational fluid flow
(c)
does not necessarily satisfy the continuity equation
(d)
is
a possible fluid-flow case
(e)
is
none
7.47
If
</>i
of these
and
<f>i
answers
are each solutions of the Laplace equation, which of the following
also a solution?
(a)
443
4>i
—
2</>2
(b) 0i02
(c)
0i/02
(d)
2
<f>i
(e)
none
of these
answers
is
FUNDAMENTALS OF FLUID MECHANICS
444
Select the relation that
7.48
du/dy
(d) du/dy
(a)
The
7.49
+ dv/dx =
=
dv/dx
must hold
(e)
=
du/dx
(6)
none
if
the flow
is
dv/dy
irrotational.
d 2 u/dx 2
(c)
(c)
when the speed
(d)
the energy
(e)
the net flow rate into any small region must be zero
is
is
is
constant along a streamline but
The Stokes stream function
applies to
three-dimensional ideal-fluid-flow cases
all
ideal (nonviscous) fluids only
(c)
irrotational flow only
(d)
cases of axial
(e)
none
symmetry
of these cases
The Stokes stream function has the valued = 1 at the origin and the value
The discharge through the surface between these points is
7.51
2 at (1,1,1)
.
w
(b)
1
7.52
(c)
2t
(d) d<j>/dx
=
=
4
(d)
Select the relation that
(a) dcf>/dx
(e)
must hold
=
in two-dimensional, irrotational flow:
-dxjz/dy
d\fr/dy
(b)
d<j>/dx
d(j)/dy
(e)
none of these answers
(a)
is
constant along an equipotential surface
(b)
is
constant along a streamline
defined for irrotational flow only
(c)
is
(d)
relates velocity
(e)
is
(a)
(6)
(c)
(d)
(e)
(c)
d<f>/dy
=
ty/dx
and pressure
none of these answers
In two-dimensional flow
\fy
discharge between the two points
7.55
none of these answers
The two-dimensional stream function
7.53
(e)
vary across streamlines
constant throughout the fluid
(b)
(a)
may
increases, the pressure increases
(a)
7.54
=
constant along a streamline
the velocity
the energy
(a)
2
Bernoulli equation in steady ideal-fluid flow states that
(b)
^ =
2
of these answers
(a)
7.50
+ d v/dy
from left to right
(b)
none of these answers
=
4
ft
/s at (0,2) and
(c)
2 cfs/ft
for steady flow of
an
is
=
(d)
ideal fluid
zero at the boundary
component normal to the boundary is zero
velocity component tangent to the boundary is zero
boundary surface must be stationary
continuity equation must be satisfied
velocity
velocity
xf/
2
ft
2
/s at (0,1).
is
4tt cfs/ft
The boundary condition
2
1/tt cfs/ft
is
that the
The
IDEAL-FLUID FLOW
An
7.56
equipotential surface
(a)
has no velocity component tangent to
(6)
is
composed
it
of streamlines
(c)
is
a stream surface
(d)
is
a surface of constant dynamic pressure
(e)
is
none of these answers
A
7.57
source in two-dimensional flow
(a)
is
a point from which fluid
(6)
is
a line from which fluid
angles to
is
is
imagined to flow outward uniformly in all directions
imagined to flow uniformly in all directions at right
it
(c)
has a strength defined as the speed at unit radius
(d)
has streamlines that are concentric
(e)
has a velocity potential independent of the radius
7.58
445
circles
The two-dimensional vortex
(a)
has a strength given by the circulation around a path enclosing the vortex
(b)
has radial streamlines
(c)
has a zero circulation around
(d)
has a velocity distribution that varies directly as the radial distance from the vortex
(e)
creates a velocity distribution that has rotation throughout the fluid
it
2
APPLICATIONS OF FLUID
MECHANICS
In Part 1 the fundamental concepts and equations have
been developed and illustrated by many examples and
simple applications. Fluid resistance, dimensional
analysis, compressible flow, and ideal-fluid flow have
been presented. In Part 2 several of the important
fields of application of fluid mechanics are explored:
measurement of flow, turbomachinery, closed-conduit
and open-channel flow, and unsteady flow.
8
MEASUREMENT
FLUID
Fluid measurements include the determination of pressure, velocity, dis
charge, shock waves, density gradients, turbulence, and viscosity.
are
many ways
these measurements
gravimetric, volumetric,
measurements
electronic,
may
be taken,
electromagnetic,
e.g.,
and
indirect,
optical.
Direct
volume or
Indirect methods
for discharge consist in the determination of the
weight of fluid that passes a section in a given time interval.
of discharge
There
direct,
measurement require the determination
of head, difference in
and with these commethods are the gravimetric or volumetric determinations, in which the weight or volume is measured by weigh
scales or by a calibrated tank for a time interval that is measured by a stop
pressure, or velocity at several points in a cross section,
puting the discharge.
The most
precise
watch.
Pressure and velocity measurements are
first
undertaken in
by positive-displacement meters, rate meters,
ment, and turbulence and viscosity measurement.
followed
8.1
this chapter,
river-flow measure-
PRESSURE MEASUREMENT
The measurement
of pressure
is
required in
many
devices that determine the
velocity of a fluid stream or its rate of flow, because of the relationship be-
tween velocity and pressure given by the energy equation. The static pressure
of a fluid in motion is its pressure when the velocity is undisturbed by the
measurement. Figure 8.1 indicates one method of measuring static pressure,
the piezometer opening.
variation
is
When
the flow
is parallel,
as indicated, the pressure
hydrostatic normal to the streamlines; hence,
by measuring the
449
APPLICATIONS OF FLUID MECHANICS
450
y////////////////^^^^^
mzzmmzmmzmi wzzymmmmmmym,.
Piezometer
Fig. 8.1
ment
opening
for
measure-
of static pressure.
pressure at the wall, the pressure at any other point in the cross section can
be determined. The piezometer opening should be small, with length of opening at least twice its diameter, and should be normal to the surface, with no
burrs at its edges because small eddies form and distort the measurement.
A small amount of rounding of the opening is permissible. Any slight mis-
alignment or roughness at the opening
therefore,
it is
into a piezometer ring.
the reading
is
may
cause errors in measurement;
advisable to use several piezometer openings connected together
When the surface is rough in the vicinity of the opening,
For small irregularities it may be possible to smooth
unreliable.
the surface around the opening.
For rough surfaces, the static tube (Fig. 8.2) may be used. It consists
is directed upstream with the end closed. It has radial holes in
the cylindrical portion downstream from the nose. The flow is presumed to be
moving by the openings as if it were undisturbed. There are disturbances,
however, due to both the nose and the right-angled leg that is normal to the
of a tube that
Fig. 8.2
Static tube.
FLUID
flow.
If it
451
The static tube should be calibrated, as it may read too high or too low.
does not read true static pressure, the discrepancy Ah normally varies as
the square of the velocity of flow
by the tube;
i.e.,
—
v
Ah = C
in
MEASUREMENT
which
2
C
is
velocity are
determined by towing the tube in
known or by inserting it into
still
fluid
where pressure and
a smooth pipe that contains a piezom-
eter ring.
Such tubes are
relatively insensitive to the
Mach numbers below
Reynolds number and to
Their alignment with the flow is not critical,
so that an error of but a few percent is to be expected for a yaw misalignment
unity.
of 15°.
The piezometric opening may lead to a bourdon gage, a manometer, a
micromanometer, or an electronic transducer. The transducers depend upon
very small deformations of a diaphragm due to pressure change to create an
electronic signal. The principle may be that of a strain gage and a Wheatstone
bridge circuit, or it may rely on motion in a differential transformer, a capacitance chamber, or the piezoelectric behavior of a crystal under stress.
8.2
VELOCITY MEASUREMENT
Since determining velocity at a
number
of points in a cross section permits
evaluating the discharge, velocity measurement
measuring
flow.
particle takes to
is an important phase of
by measuring the time an identifiable
move a known distance. This is done whenever it is con-
Velocity can be found
venient or necessary.
This technique has been developed to study flow in
regions which are so small that the normal flow would be greatly disturbed
and perhaps disappear
locity.
A
if
an instrument were introduced to measure the ve-
made available, and by means
and a powerful microscope the very minute impurities in the
transparent viewing region must be
of a strong light
fluid can be photographed with a high-speed motion-picture camera.
From
such motion pictures the velocity of the particles, and therefore the velocity
can be determined.
Normally, however, a device is used which does not measure velocity
directly but yields a measurable quantity that can be related to velocity.
The pitot tube operates on such a principle and is one of the most accurate
of the fluid in a small region,
measuring velocity. In Fig. 8.3 a glass tube or hypodermic needle
is used to measure the velocity v in an open channel.
The tube opening is directed upstream so that the fluid flows into the opening
methods
of
with a right-angled bend
APPLICATIONS OF FLUID MECHANICS
452
m
A7i
v
A
1
Simple
Fig. 8.3
pitot tube.
up
until the pressure builds
velocity against
it.
streamline through
in the tube sufficiently to withstand the impact of
Directly in front of the opening the fluid
1
leads to the point
2,
is
at rest.
called the stagnation point,
The
where
and there divides and passes around the tube. The pressure
the liquid column within the tube. Bernoulli's equation,
applied between points 1 and 2, produces
the fluid
at 2
is
—+—
2g
is
at rest,
known from
—
=
= ho+ Ah
y
y
since both points are at the
same
elevation.
As pi/y =
ho,
the equation re-
duces to
Ah
(8.2.1)
= \/2g Ah
(8.2.2)
2<7
or
v
Practically,
The
it is
very
difficult to
read the height Ah from a free surface.
which is also referred
two parts, the static
pressure h and the dynamic pressure Ah, expressed in length of a column of
the flowing fluid (Fig. 8.3) The dynamic pressure is related to velocity head
pitot tube measures the stagnation pressure,
to as the total pressure.
The
total pressure
is
composed
of
.
by Eq.
(8.2.1).
By combining the static-pressure measurement and the total-pressure
measurement, i.e., measuring each and connecting to opposite ends of a difFigure 8.4
ferential manometer, the dynamic pressure head is obtained.
illustrates one arrangement. Bernoulli's equation applied from 1 to 2 is
_
2g
_i_
Vl
y
_
¥l
y
(8.2.3)
FLUID
MEASUREMENT
453
///////////////////////////////////////////////,
*m
m
Spgr=S
WWWWWMWWA
7777777777Z7777777777\
0=
Sp gr=S
Use of pitot tube and
Fig. 8.4
opening for measuring velocity.
The equation
for the pressure
piezometer
through the manometer, in feet of water,
is
-S + kS + R'So -
(k
+
R')S =
-S
7
y
Simplifying gives
vuzli = R
'(!-')
Substituting for (p2
—
(8.2.4)
pi)/y in Eq. (8.2.3) and solving for
-V**(f-i)
v results in
(8.2.5)
The pitot tube is also insensitive to flow alignment, and an error of only a
few percent occurs if the tube has a yaw misalignment of less than 15°.
The static tube and pitot tube may be combined into one instrument,
Analyzing this system in a manner similar
called a pitot-static tube (Fig. 8.5)
to that in Fig. 8.4 shows that the same relations hold; Eq. (8.2.5) expresses
.
the velocity, but the uncertainty in the measurement of static pressure requires a corrective coefficient
,
A
=
C^'(!°-l)
C
to be applied:
(8.2.6)
particular form of pitot-static tube with a blunt nose, the Prandtl tube,
APPLICATIONS OF FLUID MECHANICS
454
Y//////////////////////////////////////,
m^
-:-:-:-:2^
Pitot-statictube.
Fig. 8.5
has been designed so that the disturbances due to nose and leg cancel, leaving
C = 1 in the equation. For other pitot-static tubes the constant C must be
determined by calibration.
The current meter (Fig. 8.6) is used to measure the velocity of liquid
flow in open channels. The cups are shaped so that the drag varies with
orientation, causing a relatively slow rotation. With an electric circuit and
headphones, an audible signal
The number
is
detected for a fixed
of signals in a given time
are calibrated
is
by towing them through
number
of revolutions.
a function of the velocity.
liquid at
known
speeds.
The meters
For measuring
high-velocity flow a current meter with a propeller as rotating element
as
it
is
used,
offers less resistance to the flow.
Air velocities are measured with cup-type or vane (propeller)
eters (Fig. 8.7)
which drive generators indicating
counters indicating the
By
number
anemom-
air velocity directly or drive
of revolutions.
designing the vanes so that they have very low inertia, employing
and optical tachometers which effectively take no power
to drive them, anemometers can be made to read very low air velocities. They
can be sensitive enough to measure the convection air currents which the
human body causes by its heat emission to the atmosphere.
precision bearings
Velocity
measurement
in
compressible flow
tube may be used for velocity determinations in compresIn Fig. 8.5 the velocity reduction from free-stream velocity at 1
to zero at 2 takes place very rapidly without significant heat transfer, and
friction plays a very small part, so that the compression may be assumed to
be isentropic. Applying Eq. (6.3.7) to points 1 and 2 of Fig. 8.5 with V2 =
The
pitot-static
sible flow.
FLUID
MEASUREMENT
455
gives
?- 1^(5- 9 -£i <*-*>- **<£-)
The
substitution of c p
is
(&-i)/fc
from Eq.
~\
(6.1.8).
r
<-"
Equation (6.1.17) then gives
/^\(fc-i)/fc"
?-4(r -']=-[-©]
(8.2.8)
may be obtained from the side openings of the pitot
and the stagnation pressure may be obtained from the impact opening
leading to a simple manometer, or p 2 — Pi may be found from the differential
manometer. If the tube is not designed so that true static pressure is measured,
it must be calibrated and the true static pressure computed.
The
static pressure p\
tube,
Fig. 8.6
{W. and
Price current meter.
L. E. Gurley.)
APPLICATIONS OF FLUID MECHANICS
456
Fig. 8.7
Air
anemometer.
(Taylor Instrument Co.)
Gas velocities may be measured with a hot-wire anemometer, which works
on the principle that the resistance to flow through a fine platinum wire is a
function of cooling due to gas flow around it. Cooled film sensors are also used
for gas flow and have been adapted to liquid flow.
8.3
POSITIVE-DISPLACEMENT METERS
One volumetric meter, the positive-displacement meter, has pistons or partiwhich are displaced by the flowing fluid and a counting mechanism that
records the number of displacements in any convenient unit, such as gallons
tions
or cubic feet.
A common meter is the disk meter, or wobble meter (Fig. 8.8), used on
many domestic water-distribution systems. The disk oscillates in a passageway so that a known volume of fluid moves through the meter for each oscillation. A stem normal to the disk operates a gear train, which in turn operates
a counter. In good condition, these meters are accurate to within 1 percent.
When they are worn, the error may be very large for small flows, such as those
caused by a leaky faucet.
FLUID
457
Disk meter. (Neptune Meter Co.)
Fig. 8.8
The
flow of domestic gas at low pressure
inflow to one end of the
it is
chamber
in
which
is
The
metric meter with a traveling partition.
valving,
MEASUREMENT
it
usually measured
partition
operates,
is
by a voluby gas
displaced
and then, by a change
displaced to the opposite end of the chamber.
The
in
oscillations
operate a counting mechanism.
Oil flow or high-pressure gas flow in a pipeline
a rotary meter in which cups or vanes
is
frequently measured
by
move about an annular opening and
volume of fluid for each revolution. Radial or axial pistons
arranged so that the volume of a continuous flow through them is
displace a fixed
may be
determined by rotations of a shaft.
Positive-displacement meters normally have no timing equipment that
measures the rate of flow. The rate of steady flow may be determined with a
stop
watch to record the time
8.4
RATE METERS
A
rate meter is
for displacement of a given
volume
of fluid.
a device that determines, generally by a single measurement,
the quantity (weight or volume) per unit time that passes a given cross section.
eter
Included
and
among
rate meters are the
weir, discussed in this section.
orifice, nozzle,
venturi meter, rotam-
APPLICATIONS OF FLUID MECHANICS
458
Orifice in a reservoir
An
orifice may be used for measuring the rate of flow out of a reservoir or
through a pipe. An orifice in a reservoir or tank may be in the wall or in the
bottom. It is an opening, usually round, through which the fluid flows, as in
Fig. 8.9. It may be square-edged, as shown, or rounded, as in Fig. 3.14. The
area of the orifice
is
the area of the opening.
With the square-edged
orifice,
the fluid jet contracts during a short distance of about one-half diameter
downstream from the opening. The portion
the wall cannot
make
of the flow that approaches along
a right-angled turn at the opening and therefore main-
tains a radial velocity component that reduces the jet area. The cross section
where the contraction is greatest is called the vena contracta. The streamlines
are parallel throughout the jet at this section, and the pressure is atmospheric.
The head H on the orifice is measured from the center of the orifice to
the free surface. The head is assumed to be held constant. Bernoulli's equation applied from a point 1 on the free surface to the center of the vena contracta, point 2, with local atmospheric pressure as datum and point 2 as elevation datum, neglecting losses, is written
—
2g
H
h
si
=
—
2g
y
H
h
22
y
Inserting the values gives
+ +#
= -^
+ +
2g
Fig. 8.9
Orifice in a reservoir.
FLUID
MEASUREMENT
459
or
F2 = y/2gH
(8.4.1)
This is only the theoretical velocity because the losses between the two points
were neglected. The ratio of the actual velocity Va to the theoretical velocity
V
t
C,
is
=
called the velocity coefficient
C
v
;
that
is,
j
(8.4.2)
t
Hence
Via
= C V2g~H
(8.4.3)
v
The
Q a from the orifice is the product of the actual veand the area of the jet. The ratio of jet area A 2 at
orifice A is symbolized by another coefficient, called
actual discharge
locity at the vena contracta
vena contracta to area of
the coefficient of contraction
C =
c
C
c
:
^
(8.4.4)
The area
at the vena contracta
is
CA
C
.
The
actual discharge
Qa = C C A s/2gH
V
It is
is
thus
(8.4.5)
C
customary to combine the two
coefficients into a discharge coefficient Cd,
Cd = C V C C
(8.4.6)
from which
Q a = CdA
V2gH
(8.4.7)
There is no way to compute the losses between points 1 and 2; hence,
C v must be determined experimentally. It varies from 0.95 to 0.99 for the
square-edged or rounded
one, the
amount
orifice.
For most
such as the square-edged
and test results must be
orifices,
of contraction cannot be computed,
There are several methods for obtaining one or more of the coefficients.
area A the head H, and the discharge Q a (by gravimetric or
Determination of either
volumetric means), Cd is obtained from Eq. (8.4.7)
used.
By measuring
,
.
APPLICATIONS OF FLUID MECHANICS
460
C
v
or
C
c
methods
1.
]
then permits determination of the other by Eq. (8.4.6).
Several
follow.
By measuring
Trajectory method.
the position of a point on the trajec-
tory of the free jet downstream from the vena contracta (Fig. 8.9) the actual
velocity Va can be determined if air resistance is neglected. The x com-
ponent of velocity does not change; therefore, Va t = x in which t is the
time for a fluid particle to travel from the vena contracta to point 3. The
time for a particle to drop a distance y under the action of gravity when
it has no initial velocity in that direction is expressed by y = gt 2 /2.
After t is eliminated in the two relations,
,
Va =
So
VW
determined by Eq. (8.4.1) the ratio Va /V = C v is known.
Direct measuring of Va With a pitot tube placed at the vena contracta,
the actual velocity Va is determined.
With
V
2i
t
.
Direct measuring of jet diameter.
jet at the vena contracta
may
With
outside calipers, the diameter of
be approximated. This
is not a precise measurement and in general is less satisfactory than the other methods.
Use of momentum equation. When the reservoir is small enough to be
suspended on knife-edges, as in Fig. 8.10, it is possible to determine the
force
F
that creates the
closed, the
tank
is
orifice discharging,
and opposite
weights
momentum
leveled
a force
in the jet. With the orifice opening
by adding or subtracting weights. With the
creates the momentum in the jet and an equal
force F' acts against the tank.
W the tank
is
again leveled.
Weights
Fig. 8.10
Momentum method
determination of C v and Cc
.
for
From
By
addition of sufficient
the figure,
F = Wxo/y
r
.
With
FLUID
the
momentum
yF
V
Wx
™
\
9
as
Vxin
Losses
°
QayVa
g
2/o
Va
in orifice
Va
and
zero
is
measured,
461
equation,
9l (v
._
MEASUREMENT
the final velocity.
is
unknown
the only
is
Since the actual discharge
is
in the equation.
flow
The head loss in flow through an orifice is determined by applying the energy
equation with a loss term for the distance between points 1 and 2 (Fig. 8.9),
V
—
la
>
Pl
h
h
zi
TV
—
=
2g
y
2g
p
—
+
2
h
+ losses
zi
y
Substituting the values for this case gives
Losses
in
or
=
H-
which Eq.
V2a and C v
EXAMPLE
lb
ft
^
A 3-in diameter orifice under a head
8.1
C
v
ft.
of 16.0 ft discharges
2000
,
,
loss.
V
theoretical velocity
2t
velocity
is
2
16
=
t
is
32.08 ft/s
determined from the trajectory.
ft is
-£-4J^J and the velocity
Xo= V2a t
H and
trajectory
V = V%gH = V64.4 X
4
(8.4.8)
was determined by measuring x = 15.62
Determine C v C c Cd, the head loss per unit weight, and
The
s.
a drop of 4.0
The actual
l)
has been used to obtain the losses in terms of
(8.4.3)
the horsepower
The
^ (^ -
.
water in 32.6
for
= H(l - Cf) =
is
0-49S,
expressed
V2a =
=
]^
0.498
by
31.4 ft/s
The time
to drop
APPLICATIONS OF FLUID MECHANICS
462
Then
31.4
CV
0.98
~ Vu
The
32.08
actual discharge
Qa
is
2000
Qa
62.4
With Eq.
w
X
0.984 cfs
32.6
(8.6.7)
0.984
=
A y/2gH
(ir/64)
=
X
16
0.98 2 )
=
V64.4
0.625
Hence, from Eq. (8.4.6),
C
Cd
0.625
C,
0.98
0.638
The head
Loss
loss,
from Eq.
= H(l - C?) =
The horsepower
0.63
550
X
X
2000
=
(8.4.8), is
16(1
-
0.63
ft
-lb/lb
loss is
0.070 hp
32.6
The Borda mouthpiece (Fig. 8.11), a short, thin- walled tube about one
diameter long that projects into the reservoir (re-entrant), permits appli-
F
=
yHA
Control
volume
Fig. 8.11
The
mouthpiece.
Borda
MEASUREMENT
463
equation, which yields one relation between
C and
FLUID
momentum
cation of the
Cd-
The
velocity along the wall of the tank
the pressure distribution
is
v
almost zero at all points; hence,
hydrostatic. With the component of force exerted
is
on the liquid by the tank parallel to the axis of the tube, there is an unbalanced
The final velocity is V2a the initial
force due to the opening, which is yHA
.
velocity
is
and Q a
zero,
yHA = Q a 1
is
,
the actual discharge.
Then
Vi
and
Q a = CdA Q
V2a = C ^2gH
V2gH
Q a and V2a and
Substituting for
In the
v
simplifying lead to
orifice situations considered,
has been assumed to be held constant.
the liquid surface in the reservoir
An
unsteady-flow case of some prac-
tical interest is
that of determining the time to lower the reservoir surface a
given distance.
Theoretically, Bernoulli's equation applies only to steady
flow,
but
if
the reservoir surface drops slowly enough, the error from using
The volume discharged from the orifice
is negligible.
which must just equal the reduction in volume in the reservoir in the same time increment (Fig. 8.12), A R (— by), in which A R is the
area of liquid surface at height y above the orifice. Equating the two expresBernoulli's equation
in
1
time
8t is
Q
^fl^
8t,
f
h
pSMmmM
1
Fig. 8.12
ing head.
Hi
Notation
for
fall-
APPLICATIONS OF FLUID MECHANICS
464
sions gives
-A R 8y
QSt =
Solving for
t
=
t
8t
and integrating between the
AR
t
=
yi,
t
and y = y2f
=
orifice
Q
discharge
f
1
=
dy
Q
"yi
The
limits y
yield
is
CdA
-\/2gy.
After substitution for Q,
y2
cdA V%g Kx
When A R
is known as a function of y, the integral can be evaluated.
Conwith other SI or English units, t is in seconds. For the special case of a
tank with constant cross section,
sistent
m
Ar
y/Tg
C A ^t=
(
2Ai
y~ 112 dy
Cd A V^g
d
EXAMPLE
8.2
(Vyi -
Vwd
A tank has a horizontal cross-sectional area
of 2
m
2
at the ele-
and the area varies linearly with elevation so that it is
2
1 m at a horizontal cross section 3 m above the orifice. For a 10-cm-diameter
orifice, Cd = 0.65, compute the time in seconds to lower the surface from 2.5
to 1 m above the orifice.
vation of the
AR =
2
-
|
orifice,
m
2
and
t
=
0.657r(0.05 2 )\/2
X
9.806
f
(2
-|)^-^ = 73.7
Venturi meter
The venturi meter
is
used to measure the rate of flow in a pipe. It
a casting (Fig. 8.13) consisting of an upstream section which
as the pipe, has a bronze liner,
is
and contains a piezometer ring
is
generally
the same size
for
measuring
FLUID
MEASUREMENT
Venturi meter.
Fig. 8.13
static pressure
;
a converging conical section a cylindrical throat with a bronze
;
liner containing a
piezometer ring; and a gradually diverging conical section
leading to a cylindrical section the size of the pipe.
A differential manometer
attached to the two piezometer rings. The size of a venturi meter
is
465
is
specified
by the pipe and throat diameter; e.g., a 6 by 4 in venturi meter fits a 6-indiameter pipe and has a 4-in-diameter throat. For accurate results the venturi
meter should be preceded by at least 10 diameters of straight pipe. In the
flow from the pipe to the throat, the velocity is greatly increased and the pres-
The amount of discharge in incompressible
shown to be a function of the manometer reading.
The pressures at the upstream section and throat are actual pressures,
and the velocities from Bernoulli's equation without a loss term are theoretical
sure correspondingly decreased.
flow
is
When
velocities.
losses are considered in the
are actual velocities.
First,
energy equation, the velocities
(i.e., without a head-
with the Bernoulli equation
term) the theoretical velocity at the throat is obtained. Then by multiby the velocity coefficient C v the actual velocity is obtained. The
actual velocity times the actual area of the throat determines the actual dis-
loss
plying this
From
charge.
V
,
Fig. 8.13
V2t
lt
h h
H
y
+
1
P2
(8.4.9)
7
which elevation datum is taken through point 2. Vi and V2 are average
and 2, respectively; hence, a h a 2 are assumed to be
2
unity. With the continuity equation ViDi 2 = V2 D 2
in
velocities at sections 1
,
I
1l
2g
2
== VI /DA
2g
\dJ
4
(8.4.10)
,
APPLICATIONS OF FLUID MECHANICS
466
which holds
for either the actual velocities or the theoretical velocities.
tion (8.4.9)
may be
solved for
V
£[-©1- -^ +
h
v
Equa-
2 t,
and
v
F- =
J
2g[h
V
i
+
(pt
_
-
P2 )7t]
(8- 4 - 11 )
WDl y
Introducing the velocity coefficient
^
t/
/
/ggPH-
/t j
gives
2gQ+(?>i-P2)/7]
After multiplying by
n
Via = C,Vu
«-^-V
i_
the actual discharge
.A 2 ,
~
(pi
,
Q
is
Q , 10
.
determined to be
? 2 )/y]
Qi10
(8A13)
,
.
(ft /A)*
The gage difference #' may now be related to the pressure difference by writing
the equation for the manometer. In units of length of water (Si
S
gravity of flowing fluid and
-Si+
(h
+k+
the specific gravity of
R')Si- R'So -
y
kSi
=
is
manometer
the specific
liquid)
- Si
y
Simplifying gives
h
+ *=J±
By
=
R>
(|
-
l)
(8.4.14)
substituting into Eq. (8.4.13),
-
Q =
......
CUWfaR'jSp/S,
^^/^
which
is
1)
(8-4-15)
the venturi-meter equation for incompressible flow.
C =
The contraction
be noted that h has dropped out
of the equation. The discharge depends upon the gage difference R regardless
coefficient
is
unity; hence,
v
C<*.
It should
r
FLUID
MEASUREMENT
467
1.00
0.99
0.98
0.97
0.96
0.95
/
0.94
10 4
1.5
2
4
3
10 5
8
5 6
1.5
2
3
4
5 6
8 10
(
Reynolds number -^—^-
Coefficient C v for venturi meters. ("Fluid
Meters: Their Theory and Application," 5th ed., American
Society of Mechanical Engineers, 1956.)
Fig. 8.14
of the orientation of the venturi meter;
whether
it is
horizontal, vertical, or
same equation holds.
C v is determined by calibration, i.e., by measuring the discharge and the
gage difference and solving for C v which is usually plotted against the Reyinclined, exactly the
,
nolds number. Experimental results for venturi meters are given in Fig. 8.14.
tolerances
shown by the dotted
be selected so that
bers for which
The
it is
lines.
its coefficient is
D /Di
from 0.25 to 0.75 within the
venturi meter should
constant over the range of Reynolds num-
JThey are applicable to diameter ratios
2
Where feasible, a
to be used.
may
be slightly greater than unity for venturi meters
mean that there are no losses
but results from neglecting the kinetic-energy correction factors a h a 2 in the
Bernoulli equation. Generally ai is greater than a 2 since the reducing section
coefficient
that are unusually smooth inside. This does not
make the velocity distribution uniform across section 2.
The venturi meter has a low overall loss, due to the gradually expanding
acts to
conical section,
in reconverting the high kinetic energy at the
which aids
throat into pressure energy.
change between sections
1
The
and
loss is
about 10 to 15 percent of the head
2.
Venturi meter for compressible flow
The
theoretical flow of a compressible fluid through a venturi meter
stantially isentropic
and
is
is
sub-
obtained from Eqs. (6.3.2), (6.3.6), and (6.3.7).
APPLICATIONS OF FLUID MECHANICS
468
When
multiplied
l
m
2
'
The
by C v the
,
L2k/(k
-
l)>lPl(?>2/?>l)
^
i-
velocity coefficient
velocity coefficient,
is
(p2/pi)
2/fc
[l
-
Vh (A t /A
it
yields for
mass flow rate
(p»/ Pl ) <*-»/*]
(8.4.16)
1
y
the same as for liquid flow. Equation (8.4.13)
reduced to horizontal flow
when
and modified by insertion of an expansion factor
can be applied to compressible flow
m
= CvYAi
£
2piAp
(A/ A)
(8.4.17)
4
Y
can be found by solving Eqs. (8.4.17) and (8.4.16) and is shown to be a
k, P2/P1, and A 2 /Ai.
Values of Y are plotted in Fig. 8.15 for
=
k
1.40; hence, by the use of Eq. (8.4.17) and Fig. 8.15 compressible flow
can be computed for a venturi meter.
function of
Squar e edged
orifice
-
01
*-1.4
.04
I
H
^0.6
^ ^
^X
0.9
.
Zs
c •75
na
^0.8
Dt/Dt-
0.8
fe
Ve nturi me er
or nozzle
0.7
*=1.4
^
0.6
0.5
1.0
0.70
0.90
>M
Fig. 8.15
Expansion factors.
0.60
FLUID
MEASUREMENT
469
Flow nozzle
The ISA (Instrument Society of America) flow nozzle
is shown in Fig. 8.16. It has no contraction
flow nozzle)
(originally the
VDI
of the jet other
than
that of the nozzle opening; therefore the contraction coefficient
is
unity.
Equations (8.4.13) and (8.4.15) hold equally well for the flow nozzle.
For a horizontal pipe (h = 0), Eq. (8.4.13) may be written
Q = CA 2
2Ap
(8.4.18)
P
in
which
C=
C
y/l
v
(8.4.19)
- (A/A)'
m
An
1.20
^1
061
1.18
1.16
060
1.14
1.12
0.55
1.10
nsn
1.08
1.06
0.45
040
1.04
/
1.02
y
1.00
0.98
035
*^
0.30
/^
'//V,V
0.96
0.94
0.20
010
U.Ob
y
// V
0.92
10«
10
5
10*
mi
Fig. 8.16
ISA (VDI) flow nozzle and discharge
coefficients. (Ref. 11 in
NACA
Tech.
Mem.
952.)
APPLICATIONS OF FLUID MECHANICS
470
and Ap =
(8.4.18)
pi
—
The value
p%.
When the
.
of coefficient
C
in Fig. 8.16 is for use in
coefficient given in the figure
is
to be used,
it is
Eq
important
that the dimensions shown be closely adhered to, particularly in the location
(two methods shown) for measuring pressure
At least 10 diameters of straight pipe should precede the nozzle.
The flow nozzle is less costly than the venturi meter. It has the dis-
of the piezometer openings
drop.
advantage that the overall losses are much higher because of the lack of
guidance of jet downstream from the nozzle opening.
is found by Eq. (8.4.17) and Fig.
For other values of specific-heat ratio k, Eq. (8.4.16) may
Compressible flow through a nozzle
8.15
if
k
=
1.4.
be used.
EXAMPLE
Determine the flow through a 6-in-diameter water line that
The mercury-water differential manometer has a gage difference of 10 in. Water temperature is 60°F.
8.3
contains a 4-in diameter flow nozzle.
From the data given, S = 13.6,
- 0.0873 ft p = 1.938 slugs/ft 3
2
x/36
,
,
&
m
=
=
1.0,
- H = 0.833 ft, A =
X 10" lb s/ft Substitu-
Bf
2.359
2
5
•
2
.
ting Eq. (8.4.19) intoEq. (8.4.15) gives
Q = CA 2
From
Mf-0
Fig. 8.16, for
A
2
/Ai
=
=
=
C
2
(-J)
of the curves applies; hence,
assume that the horizontal region
1.056; then compute the flow and the
0.444,
Reynolds number.
Q =
1.056
X
0.0873 ^/64.4
X
0.833
(y^ -
=
1.0
2.40 cfs
J
Then
7l
=
T=
Ai
^
=
12 21 ft/s
'
7T/16
and
R=
V}D}E=
n
12.21X1.938
2
X
2.359
X
The chart shows the value
cfs.
lO" 5
of
C
to be correct; therefore, the discharge
is
2.40
FLUID
MEASUREMENT
471
Orifice in a pipe
The square-edged orifice in a pipe (Fig. 8.17) causes a contraction of the jet
downstream from the orifice opening. For incompressible flow Bernoulli's
equation applied from section 1 to the jet at its vena contracta, section 2, is
y
2g
2g
y
The continuity equation
Co
= A 2 /A
After eliminating
and by solving
2t
Vu
Vu
for
,
V
2t
the result
~ ViTh~
2 9(Pi
J
\1-C
Fig. 8.17
and
V
2t
with the contraction coefficient
(8.4.20)
4
4
v "
relates
,
2
C
(A)/A) 4
Orifice in a pipe.
is
APPLICATIONS OF FLUID MECHANICS
472
C
Multiplying by
T/
V2a
r
- Cv
I
to obtain the actual velocity at the vena contracta gives
v
-
2(pi
V2)7p~
Vi- C >(Do/D y
1
and, finally multiplying by the area of the
jet,
CA
C
,
produces the actual
dis-
charge Q,
in
which Cd =
CC
V
C
.
In terms of the gage difference R' Eq. (8.4.21) becomes
,
Because of the difficulty in determining the two
simplified formula is generally used, Eq. (8.4.18),
Q = CA
coefficients separately,
J—
v
a
(8.4.23)
P
or its equivalent,
Q = CA \\2gR
f
(^ -
l)
(8.4.24)
C are given in Fig. 8.18 for the VDI orifice.
Experimental values of expansion factor, for k = 1.4, are given in Fig.
Equation (8.4.23) for actual mass flow rate in compressible flow becomes
Values of
8.15.
m = CYA
The
\/2pi
Ap
location of the pressure taps
(8.4.25)
is
usually specified so that an orifice
can be installed in a conduit and used with
forming a calibration at the site.
sufficient
accuracy without per-
Elbow meter
The elbow meter
for incompressible flow is one of the simplest flow-rate measPiezometer openings on the inside and on the outside of the
elbow are connected to a differential manometer. Because of centrifugal force
uring devices.
MEASUREMENT
FLUID
v
0.82
^s
Ai
70
0.80
^^
0.78
>30°
473
0.76
o
e
2
o
Ar
<u
JL_
0.74
I
*•
0.72
*-^0.02D,
0.70
-
's
\^^
%
0.68
60
O.bO
v^
^O.ID
0.40
0.66
L
C03D,J
0.03Z),
2
=
CA
0.64
Ap
"^•^
0.30
0.62
0.20
**»"""
P
010
0.60
0.0b
4-10
3
10
4
5-10
4
10
5
5-10
5
f
10
Mi
Fig. 8.18
VDI orifice and discharge coefficients. (Ref.
Mem.
Tech.
11 in
NACA
952.)
at the bend, the difference in pressures
is
related to the discharge.
A straight
calming length should precede the elbow, and, for accurate results, the meter
should be calibrated in place. 1 As most pipelines have an elbow, it may be
used as the meter. After calibration the results are as reliable as with a venturi
meter or a flow nozzle.
Rotameter
The rotameter
is a variable-area meter that consists of an enlargmetering
"float" (actually heavier than the liquid)
and a
displaced upward by the upward flow of fluid through the tube. The
graduated to read the flow directly. Notches in the float cause it to
(Fig. 8.19)
ing transparent tube
that
is
tube
is
rotate
and thus maintain a central position
in the tube.
The
greater the flow,
the higher the position the float assumes.
Weirs
Open-channel flow
may
be measured by a weir, which
channel that causes the liquid to back up behind
1
W. M.
Lansford,
an obstruction in the
and to flow over it or
is
of an Elbow in a Pipe Line for Determining the Rate of Flow
Eng. Exp. Stn. Bull 289, December 1936.
The Use
in a Pipe, Univ. Ill
it
APPLICATIONS OF FLUID MECHANICS
474
Outlet
Metering
float
8.19
Fig.
(Fischer
&
through
it.
flow
is
Rotameter.
Porter Co.)
By measuring
the height of upstream liquid surface, the rate of
determined. Weirs constructed from a sheet of metal or other material
so that the jet, or nappe, springs free as
it
leaves the upstream face are called
sharp-crested weirs. Other weirs such as the broad-crested weir support the flow
in a longitudinal direction.
The sharp-crested rectangular weir (Fig. 8.20) has a horizontal crest.
The nappe is contracted at top and bottom as shown. An equation for discharge can be derived
if
the contractions are neglected. Without contractions
the flow appears as in Fig. 8.21.
The nappe has
parallel streamlines with
atmospheric pressure throughout.
Bernoulli's equation applied
#+0+0= -+H-y+0
2g
between
1
and 2
is
FLUID
MEASUREMENT
475
^ttg^3l
y7yzzzzzzz^zzz-z-zzzzzzzzzz
>_j^ZZ^Z}Z^Z-Z-Z-Z-ZZ-Z-Z-Z-Z-
Sharp-crested rectangular weir.
Fig. 8.20
which the velocity head at section
in
v
1 is
neglected.
By
solving for
v,
= \/2gy
The
Q =
t
theoretical discharge
f
v
Q
t
is
dA = f vLdy = ^2g L f
y 1 '* dy
=
f yj2g
LH^
Experiment shows that the exponent of H is
too great. The contractions and losses reduce the actual discharge to about 62 percent of the theoretical, or
in
which
L
is
the width of weir.
correct but that the coefficient
3.33L# 3
'2
is
English units
Q =
(8.4.26)
1MLH*'
When
2
SI units
the weir does not extend completely across the width of the
wm
Fig. 8.21
Weir
contractions.
nappe
without
APPLICATIONS OF FLUID MECHANICS
476
m
H.
0.1H-4W-
Weir with end contractions.
Fig. 8.22
channel,
it
has end contractions, illustrated in Fig. 8.22.
tion for the reduction of flow
for each
-HKO.ltf'
is
An
empirical correc-
accomplished by subtracting 0.1H from
end contraction. The weir
in Fig. 8.20
is
said to have its
L
end contrac-
tions suppressed.
The head
H
measured upstream from the weir a sufficient distance
A hook gage mounted in a stilling pot connected to a piezometer opening determines the water-surface elevation from
which the head is determined.
When the height P of weir (Fig. 8.20) is small, the velocity head at 1
cannot be neglected. A correction may be added to the head,
is
to avoid the surface contraction.
(
Q = CL
Fig. 8.23
yi\m
v*\
(8.4.27)
V-notch weir.
FLUID
MEASUREMENT
477
in which V is velocity and a is greater than unity, usually taken around 1.4,
which accounts for the nonuniform velocity distribution. Equation (8.4.27)
must be solved for Q by trial since Q and V are both unknown. As a first trial,
the term aV 2 /2g may be neglected to approximate Q; then with this trial discharge a value of V is computed, since
Q
L(P
+ H)
For small discharges the F-notch weir is particularly convenient. The
is neglected, and the theoretical discharge is computed (Fig. 8.23) as follows.
The velocity at depth y is v = y/2gy, and the theoretical discharge is
contraction of the nappe
Qt
By
= J vdA =
j
vx dy
similar triangles, x
may
be related to
y,
L
H-y H
After substituting for v and
Q<=V2~9^f
L/H
Expressing
L
—
=
2H
V
m (H -
x,
y)
dy
=
A V^g I tf
in terms of the angle
<j>
of the
5 '2
V
notch gives
tan -
<f>
2
Hence,
Qt
=
AV^tan^tf^
The exponent in the equation is approximately correct, but the coefficient
must be reduced by about 42 percent because of the neglected contractions.
An
approximate equation for a 90° V-notch weir
2.50# 2
-
50
English units
-
50
SI units
Q =
is
(8.4.28)
1.38#2
APPLICATIONS OF FLUID MECHANICS
478
Experiments show that the coefficient is increased by roughening the upstream,
side of the weir plate, which causes the boundary layer to grow thicker. The
greater amount of slow-moving liquid near the wall is more easily turned, and
hence there is less contraction of the nappe.
The broad-crested weir
(Fig. 8.24a) supports the
hydrostatic at section
sure variation
is
tween points
and 2 can be used
1
2.
nappe so that the pres-
Bernoulli's equation applied be-
to find the velocity v2 at height
z,
neglecting
the velocity of approach,
H+ +
In solving for
v2
z
2
- + z+(y-z)
Do
=
v2
= y/2g(H -
,
y)
drops out; hence,
constant at section
v2 is
2.
For a weir of width
to the plane of the figure, the theoretical discharge
Q =
V2
Ly = Ly ^/2g(H
A plot
is
of
Q
-
(8.4.29)
y)
as abscissa against the depth y as ordinate, for constant
given in Fig. 8.246.
The depth
shown
is
mum discharge, by the following reasoning.
A gate or other obstruction placed at
pletely stop the flow b}'
pass section 3 (holding
and the discharge
curve.
By
L normal
is
is,
making y
section 3 of Fig. 8.24a can
= H. Now,
H constant)
for example, as
,
H,
to be that which yields the maxi-
if
a small flow
the depth y becomes a
is
com-
permitted to
little less
than
H
shown by point a on the depth-discharge
further lifting of the gate or obstruction at section 3 the discharge-
depth relationship follows the upper portion of the curve until the maximum
discharge is reached. Any additional removal of downstream obstructions,
however, has no effect upon the discharge, because the velocity of flow at
Fig. 8.24
Broad-crested weir.
FLUID
section 2
in
is
y/gy, which
liquid of
still
depth
MEASUREMENT
479
exactly the speed an elementary wave can travel
Hence, the effect of any additional lowering of the
is
y.
downstream surface elevation cannot travel upstream to affect further the
This depth y,
y, and the discharge occurs at the maximum value.
called the critical depth, is discussed in Sec. 11.4. The speed of an elementary
wave is derived in Sec. 12.10.
By taking dQ/dy and with the result set equal to zero, for constant H,
value of
dQ
2g
and solving
for y gives
Inserting the value of H, that
v2
gives
v*
=
3?//2, into
the equation for velocity
-s/gy
and substituting the value
SmLH*'
Qt
is,
2
of y into Eq. (8.4.29) leads to
English units
=
(8.4.30)
1.705L# 3
'2
SI units
Experiments show that
[3.03ZJ/ 3
Q =
'
2
for a well-rounded
upstream edge the discharge
is
Engiish units
(8.4.31)
1.67
which
SI units
within 2 percent of the theoretical value.
is
itself to
LHW
discharge at the
maximum
The
flow, therefore, adjusts
rate.
Since viscosity and surface tension have a minor effect on the discharge
coefficients of weirs, a weir
should be calibrated with the liquid that
it
will
measure.
EXAMPLE
8.4
Tests on a 60° V-notch weir yield the following values of head
H on the weir and discharge Q:
H,it
0.345
0.356
0.456
0.537
0.568
0.594
0.619
0.635
0.654
0.665
Q, cfs
0.107
0.110
0.205
0.303
0.350
0.400
0.435
0.460
0.490
0.520
APPLICATIONS OF FLUID MECHANICS
480
By means
of the theory of least squares, determine the constants in
Q = CH m
for this weir.
By
taking the logarithm of each side of the equation
lnQ = lnC
it is
+ mlnH
Y = B+
or
mX
noted that the best values of B and m are needed for a straight line through
when plotted on log-log paper.
By the theory of least squares, the best straight line through the data
the data
points
is
minimum value
the one yielding a
cal displacements of each point
Z
F =
s.-
2
=
2Q/i
-
(B
+
from the
of the
sums
line; or,
from
of the squares of vertiFig. 8.25,
mxi)j
t=i
n
is
the
number
To minimize
F,
two equations
dF
—
=
of experimental points.
dF/dB and dF/dm are taken and set equal
two unknowns B and m, as follows:
to zero, yielding
in the
= 2Z[>- (B
+
«*<)]( -1)
from which
(1)
»-X
Fig. 8.25
Log- log plot of
for V-notch weir.
Q
vs.
H
FLUID
MEASUREMENT
481
and
dF
—
=
= 22[^ - (B
dm
+ mx,)^-^)
or
XXiyi
m=
-
BXxi
-
mXxi 2 =
(2)
Solving Eqs. (1) and (2) for
m gives
— ^Vi/n
r~
— Xxi/n
—
XXiyi/^Xi
—
Xxf/'EXi
B =
Xyi
mUxi
n
The logarithms of Qi and H may be looked up, and the summations taken
by desk calculator. This problem, however, is very easily handled by digital
computer with the following program:
{
READ M,LH,LQ
DIMENSION
Q(10),H(10)
NAMELIST/DIN/Q,H,N,M,C
READ (5,DIN,END = 99)
X=.0
Y = .0
XX =.0
XY=.0
DO 1I = 1,N
LH = ALOG(H(I))
LQ = ALOG(Q(I))
X=X+LH
Y=Y+LQ
XX = XX+LH**2
1
XY = XY+LH*LQ
M= (XY/X-Y/N)/(XX/X-X/N)
B= (Y-M*X)/N
C = EXP(B)
99
WRITE (6,DIN)
CALL SYSTEM
END
N=
10,Q(1)
=
H(l) =
For the data of
.107, .110, .205, .303, .35,
.4,
.435, .46, .49, .52
.345, .356, .456, .537, .568, .594, .619, .635, .654, .665
this problem,
m=
2.437,
C =
1.395.
482
APPLICATIONS OF FLUID MECHANICS
8.5
ELECTROMAGNETIC FLOW DEVICES
field is set up across a nonconducting tube and a conducting
through the tube, an induced voltage is produced across the flow
which can be measured if electrodes are embedded in the tube walls. 1 The
If a
magnetic
fluid flows
voltage
a linear function of the volume rate passing through the tube.
is
Either an ac or a dc
A
at the electrodes.
may be used, with a corresponding signal generated
disadvantage of the method is the small signal received
field
and the large amount of amplification needed.
measure the flow in blood vessels.
8.6
The device has been used
to
MEASUREMENT OF RIVER FLOW
Daily records of the discharge of rivers over long periods of time are essential
to economic planning for utilization of their water resources or protection
against floods.
The
daily
measurement
of discharge
distribution over a cross section of the river
is
by determining
costly.
To avoid
this
velocity
and
still
obtain daily records, control sections are established where the river channel
change in bottom or sides of the stream bed. The
control section is frequently at a break in slope of the river bottom where it
becomes steeper downstream.
A gage rod is mounted at the control section so that the elevation of
water surface is determined by reading the waterline on the rod; in some
installations float-controlled recording gages keep a continuous record of
river elevation. A gage height-discharge curve is established by taking currentmeter measurements from time to time as the river discharge changes and
plotting the resulting discharge against the gage height.
With a stable control section the gage height-discharge curve changes
very little, and current-meter measurements are infrequent. For unstable
control sections the curve changes continuously, and discharge measurements
must be made every few days to maintain an accurate curve.
Daily readings of gage height produce a daily record of the river disis
stable,
with
i.e.,
little
charge.
8.7
MEASUREMENT OF TURBULENCE
Turbulence
a characteristic of the flow. It affects the calibration of measurupon heat transfer, evaporation,
is
ing instruments and has an important effect
diffusion,
1
and many other phenomena connected with
H. G. Elrod,
Trans.
Jr.,
ASME,
An
and R. R. Fouse,
vol. 74, p. 589,
May
1952.
fluid
movement.
Investigation of Electromagnetic Flowmeters,
FLUID
Time
Fig. 8.26
MEASUREMENT
483
t
Turbulent fluctuations
in
direction of flow.
by two
and the
components at a point are constant. If these mean values are u, v, w and the velocity components at an instant are u, v, w, the fluctuations are given by u'
Turbulence
is
generally specified
mean
velocity
w', in
v',
u = u
+
The
is
quantities, the size
In steady flow the temporal
intensity of the fluctuations.
u'
v
root
=
v
w = w
-{- v'
mean square
of
+
w'
measured values
of the fluctuations (Fig. 8.26)
a measure of the intensity of the turbulence.
VWT
These are \/(u
f
2
)
,
\/(v') 2
,
2
.
The
an average measure of the size of eddy, or
two
When
velocity measuring instruments (hot-wire
size of the fluctuation is
vortex, in the flow.
anemometers) are placed adjacent to each other in a flow, the velocity fluctuations are correlated, i.e., they tend to change in unison. Separating these
instruments reduces this correlation. The distance between instruments for
zero correlation is a measure of the size of the fluctuation. Another method
for determining turbulence
8.8
(1)
discussed in Sec. 5.6.
MEASUREMENT OF VISCOSITY
The treatment
for
is
of fluid
measurement
determining viscosity.
by use
of
equation; (3)
Viscosity
is
concluded with a discussion of methods
may be measured
in a
number
of
ways
Newton's law of viscosity; (2) by use of the Hagen-Poiseuille
by methods that require calibration with fluids of known vis-
cosity.
By measurement
of the velocity gradient
du/dy and the shear
stress r,
APPLICATIONS OF FLUID MECHANICS
484
Newton's law of viscosity [Eq.
in
T
=
(1.1.1)],
du
(8.8.1)
fX
dy
the dynamic or absolute viscosity can be computed. This is the most basic
method as it determines all other quantities in the denning equation for viscosity. By means of a cylinder that rotates at a known speed with respect to
an inner concentric stationary cylinder, du/dy
is
determined.
By
measure-
ment of torque on the stationary cylinder, the shear stress can be computed.
The ratio of shear stress to rate of change of velocity expresses the viscosity.
A schematic view of a concentric-cylinder viscometer is shown in Fig.
8.27. When the speed of rotation is N rpm and the radius is r2 ft, the fluid velocity at the surface of the outer cylinder is 27rr2 iV/60. With clearance b ft
du
dy
2irr2
N
606
«
r2
The torque Tc on the inner cylinder is measThe equation is based on b
ured by a torsion wire from which it is suspended. By attaching a disk to the
wire, its rotation can be determined by a fixed pointer. If the torque due to
'/////.
-1
IffiES
Fig.
8.27
wmm
Concentric-
cylinder viscometer.
.
FLUID
fluid
below the bottom of the inner cylinder
is
MEASUREMENT
485
neglected, the shear stress
is
7
27rri
2
/i
Substituting into Eq. (8.8.1) and solving for the viscosity give
157y>
(8.8.2)
W^hN
When the clearance a is
is
appreciable,
it
so small that the torque contribution from the bottom
can be calculated in terms of the viscosity.
Referring to Fig. 8.28,
8T = rr8A =
— r 8r 86
cor
rpL
a
in
which the velocity change is cor in the distance a ft. Integrating over the
and letting co = 27riV/60 lead to
circular area of the disk
Td
=
^~N
f
The torque due
f
r*drd9
to disk
=
-£-
Nn*
(8.8.3)
and cylinder must equal the torque
wire, so that
a60
156
Fig.
8.28
Notation for
determination of torque
on a disk.
15
\4a
b
)
T
in the torsion
APPLICATIONS OF FLUID MECHANICS
486
in
which
all
known
quantities are
must be laminar
except
The
n.
for Eqs. (8.8.2) to (8.8.4) to
be
flow between the surfaces
valid.
Often the geometry of the inner cylinder is altered to eliminate the torque
which acts on the lower surface. If the bottom surface of the inner cylinder
is made concave, a pocket of air will be trapped between the bottom surface
of the inner cylinder and the fluid in the rotating outer cup. A well-designed
cup and a careful filling procedure will ensure the condition whereby the torque
measured will consist of that produced in the annulus between the two cylinders and a minute amount resulting from the action of the air on the bottom
Naturally the viscometer must be provided with a temperaturesurface.
controlled bath and a variable-speed drive which can be carefully regulated.
Such design refinements are needed in order to obtain the rheological diagrams
(cf. Fig. 1.2) for the fluid under test.
The measurement of all quantities in the Hagen-Poiseuille equation,
except
/z,
by a
suitable experimental arrangement,
for determination of viscosity.
distance
is
tion after
A
required for the fluid to develop
it
is
another basic method
setup as in Fig. 8.29
its
may
enters the tube therefore, the head or pressure
by some means
be used.
Some
characteristic velocity distribu-
;
The volume V
must be measured
can be measwhere the reservoir surface is held at a constant level. This
yields Q, and, by determining y, Ap can be computed. Then with L and D
known, from Eq. (5.2.10a),
at a point along the tube.
ured over a time
of flow
t
ApirD 4
12SQL
An adaptation
of the capillary tube for industrial purposes
viscometer (Fig. 8.30).
Fig. 8.29
Determination
through a capillary tube.
A
short capillary tube
of
viscosity
by
is
flow
utilized,
is
the Saybolt
and the time
is
FLUID
MEASUREMENT
487
Overflow
Constanttemperature 3t-z-z-
Er
-----
bath
:-!-=;
Schematic
Fig. 8.30
view
of
Saybolt viscometer.
cm
3
measured
for
The time
in seconds
60
of fluid to flow
is
through the tube under a falling head.
This device measures kinematic
the Saybolt reading.
from a rearrangement of Eq. (5.2.10a). When Ap = pgh,
and
when
the terms are separated that are the same regardless of
V/t,
viscosity, evident
Q =
the
fluid,
ghwD 4
128
Ci
VL
Although the head h varies during the test, it varies over the same range for
all liquids and the terms on the right-hand side may be considered as a con;
stant of the particular instrument.
v
Since
/x/p
=
v,
the kinematic viscosity
is
= dt
which shows that the kinematic viscosity varies directly as the time t. The
capillary tube is quite short, and so the velocity distribution is not established. The flow tends to enter uniformly, and then, owing to viscous drag
at the walls, to slow down there and speed up in the center region. A correction
in the above equation is needed, which is of the form C/t; hence
v
=
C
+-
2
Cit
APPLICATIONS OF FLUID MECHANICS
488
The approximate
relationship
between viscosity and Saybolt seconds
is
ex-
pressed by
=
v
-
0.0022*
^—
t
which v is in stokes and t in seconds.
For measuring viscosity there are many other industrial methods that
generally have to be calibrated for each special case to convert to the absolute
in
One
units.
consists of several tubes containing "standard" liquids of
graduated viscosities with a
ball to fall the length of the
steel ball in
known
each of the tubes. The time for the
tube depends upon the viscosity of the
placing the test sample in a similar tube,
its
by comparison with the other tubes.
The flow of a fluid in a capillary tube
is
liquid.
By
viscosity can be approximated
the basis for viscometers of the
Oswald-Cannon-Fenske or Ubbelohde type. In essence the viscometer is a
U tube, one leg of which is a fine capillary tube connected to a reservoir above.
The tube is held vertically, and a known quantity of fluid is placed in the reservoir and allowed to flow by gravity through the capillary. The time is recorded for the free surface in the reservoir to fall between two scribed marks.
A
calibration constant for each instrument takes into account the variation
from the standard, the bore's uniformity, entrance condue to the falling head during the 1- to
2-min test. Various bore sizes can be obtained to cover a wide range of viscosities.
Exact procedures for carrying out the tests are contained in the
standards of the American Society for Testing and Materials.
of the capillary's bore
ditions,
and the
slight unsteadiness
PROBLEMS
A
8.1
liquid
tube (Fig. 8.2) indicates a static pressure that
static
is
is
0.12 psi too low
when
flowing at 8 ft/s. Calculate the correction to be applied to the indicated pres-
sure for the liquid flowing at 18 ft/s.
Four piezometer openings
8.2
in the
same
cross section of a cast-iron pipe indicate the
following pressures for simultaneous readings: 4.30, 4.26, 4.24, 3.7
cm Hg. What
value
should be taken for the pressure?
A
8.3
7
=
,
ju
=
0.65 P,
Ah =
is
inserted into a small stream of flowing
1.5 in, h
=
5
in.
What
is
oil,
the velocity at point 1?
A stationary body immersed in a river has a maximum pressure of 69 kPa exerted
8.4
on
simple pitot tube (Fig. 8.3)
55 lb/ft 3
it
at a distance of 5.4
depth.
m below the free surface.
Calculate the river velocity at this
MEASUREMENT
FLUID
8.5
From
8.6
In Fig. 8.4
For R'
=
Fig. 8.4 derive the equation for velocity at
air is flowing
R =
of air
1
For
between
A
1.47 in
16 psia,
=
t
40°F) and water
in the
is
manometer.
(p = 101 kPa abs, h = 5°C) and mercury is in the ma20 cm, calculate the velocity at 1 (a) for isentropic compression
air is flowing
nometer.
8.8
=
1.
1.2 in, calculate the velocity of air.
In Fig. 8.4
8.7
(p
489
f
and 2 and
(6) for air considered incompressible.
pitot-static tube directed into a 12 ft/s water stream has a gage difference of
on a water-mercury differential manometer. Determine the coefficient for the
tube.
A
8.9
pitot-static tube,
manometer when
C=
1.12,
has a gage difference of
directed into a water stream.
cm on
1
a water-mercury
Calculate the velocity.
A pitot-static tube of the Prandtl type has the following value of gage difference
8.10
R' for the radial distance from center of a 3-ft-diameter pipe
r,
0.0
0.3
0.6
0.9
1.2
1.48
4.00
3.91
3.76
3.46
3.02
2.40
ft
R', in
Water
flowing,
is
and the manometer
fluid
has a specific gravity of 2.93. Calculate the
discharge.
What would
8.11
be the gage difference on a water-nitrogen manometer for flow o
nitrogen at 200 m/s, using a pitot-static tube?
and corresponding temperature 25°C. True
Measurements
8.12
the static pressure
is
in
The
static pressure
static pressure
is
1.75 kg// cm 2 ab
measured by the tube-
is
an airstream indicate that the stagnation pressure is 15 psia>
is 102°F. Determine the
10 psia, and the stagnation temperature
temperature and velocity of the airstream.
0.5 kg/s nitrogen flows through a 5-cm-diameter tube with stagnation tempera-
8.13
Find the velocity and
ture of 38°C and undisturbed temperature of 20°C.
static
and
stagnation pressures.
8.14
A
tion.
Calculate the flow in gallons per minute for 86.5 oscillations per minute.
8.15
A
disk meter has a volumetric displacement of 1.73 in3 for one complete oscilla-
disk water meter with volumetric displacement of 40
quires 470 oscillations per minute to pass 0.32 1/s
pass 2.57
A
8.16
32.4
s.
8.17
liters
8.18
1/s.
Calculate the percent error, or
volumetric tank 4
What
is
ft in
and 3840
slip, in
diameter and 5
ft
cm 3
per oscillation re-
oscillations per
minute to
the meter.
high was
filled
with
oil in
16
min
the average discharge in gallons per minute?
A weigh tank receives 75 N liquid,
sp gr 0.86, in 14.9
s.
What
is
the flow rate in
per minute?
Determine the equation
small orifice with head of 15
ft
for trajectory of a jet discharging horizontally
and velocity
coefficient of 0.96.
Neglect
from a
air resistance.
APPLICATIONS OF FLUID MECHANICS
490
Fig. 8.31
An
8.19
orifice of
It discharges
i/o
=
1.23
8.20
area 30
N
in a vertical plate
m. Determine C v C c
,
H and a.
,
=
has a head of
1.1
m of
oil,
Trajectory measurements yield Xo
s.
sp gr 0.91.
=
2.25
m,
Cd.
maximum
Neglect
In Fig. 8.31, for a
8.21
cm 2
of oil in 79.3
Calculate Y, the
terms of
Cv
6790
rise of
a jet from an inclined plate (Fig. 8.31), in
losses.
45°,
Y=
0.48F. Neglecting
air resistance of
the
jet, find
for the orifice.
Show
8.22
4Y(H —
that the locus of
Y) when
8.23
A
head.
The
difference
A
points of the jet of Fig. 8.31
3-in-diameter orifice discharges 64
is
The manometer
R'
=
8.25
N
3.35
ft.
10-cm-diameter
liquid
is
ft
3
liquid, sp gr 1.07, in 82.2 s
Determine C v C c and
,
,
orifice
to resist impact of
Compute
Air,
2
,
psi
ft:
in.
diam
spgr0.92
Q-0.74
Fig. 8.32
2
under a
discharges 44.6 1/s water under a head of 2.75 m.
downstream from the vena
the jet. Find Cd C v and C c
3
:-I-:0il,
X =
9-ft
Cd.
,
A flat
contracta requires a force of
.
the discharge from the tank shown in Fig. 8.32.
6
given by
is
plate held normal to the jet just
320
is
determined by a pitot-static tube with coacetylene tetrabromide, sp gr 2.96, and the gage
velocity at the vena contracta
efficient 1.0.
8.24
maximum
losses are neglected.
FLUID
MEASUREMENT
491
Air
30 kN/m 2
:r-_-__-_-^--_--_-;
Air
^-_-_-_j-_-_-_-
20 kN/m
2
m
2
Water
10
cm diam
Water
Cd =0.85
Fig. 8.33
For
8.26
C =
v
0.96 in Fig. 8.32, calculate the losses in foot-pounds per
pound and
in
foot-pounds per second.
8.27
Calculate the discharge through the orifice of Fig. 8.33.
8.28
For
8.29
A
C =
v
Cd,
is
jet at the
and Cc
A
8.30
newton and
in watts.
4-in-diameter orifice discharges 1.60 cfs liquid under a head of 11.8
diameter of
C„
0.93 in Fig. 8.33, determine the losses in joules per
vena contracta
found by calipering to be 3.47
is
ft.
The
Calculate
in.
.
Borda mouthpiece 5 cm
in
What
diameter has a discharge coefficient of 0.51.
the diameter of the issuing jet?
8.31
A 3-in-diameter orifice, Cd = 0.82, is placed in the bottom of a vertical tank that
ft. How long does it take to draw the surface down from 8 to 6 ft?
has a diameter of 4
Select the size of orifice that permits a tank of horizontal cross section 1.5
8.32
to have the liquid surface
Cd =
orifice.
8.33
at the rate of 18
ft
above the
orifice in 83.7
Select a reservoir of such size
s.
m
head on the
Calculate the discharge coefficient.
and shape that the
In Fig. 8.34 the truncated cone has an angle 6
draw the
8.36
for 3.35
0.63.
over a 3-m distance for flow through a 10-cm-diameter
8.35
cm/s
2
A 4-in-diameter orifice in the side of a 6-ft-diameter tank draws the surface down
from 8 to 4
8.34
drawn down
m
liquid surface
down from
y
=
12
ft
to y
=
=
4
liquid surface drops 1
orifice.
60°.
Cd
How
=
m/min
0.74.
long does
it
take to
ft?
Calculate the dimensions of a tank such that the surface velocity varies inversely
When
as the distance from the centerline of an orifice draining the tank.
30 cm, the velocity of
fall of
the surface
is
3 cm/s; orifice diameter
8.37
Determine the time required to
8.38
How long does it take to raise the water surface of Fig.
surface
is
is
the head
1.25 cm, Cd
raise the right-hand surface of Fig. 8.35
8.36 2
a large reservoir of constant water-surface elevation.
=
is
0.66.
by 2
ft.
m? The left-hand
APPLICATIONS OF FLUID MECHANICS
492
Fig. 8.34
8.39
Show
that for incompressible flow the losses per unit weight of fluid between the
upstream section and throat
of a venturi
meter are KV<?/2g
if
tf=[0/C„) 2 -l][l-(D2/A) 4]
8.40 A 200 by 100 in venturi meter carries water at 80°F. A water-air
manometer has a gage difference of 2.4 in. What is the discharge?
differential
What is the pressure difference between the upstream section and
cm horizontal venturi meter carrying 50 1/s water at 48°C?
throat of a
8.41
16 by 8
A 12 by 6 in venturi meter is mounted in a vertical pipe with the flow upward.
= 0.1 P, flows through the pipe. The throat section is 4 in
2000 gpm oil, sp gr 0.80,
above the upstream section. What is pi — y?2?
8.42
/jl
8.43
Air flows through a venturi meter in a 5-cm-diameter pipe having a throat
calculate the
—
0.97.
For p x
flowing.
—
=
C =
mass per second
diameter of 3 cm,
v
I
6
ft
1
:
~~
3
in.
diam
Cd =0.90
Afl=16ft
2
Fig. 8.35
^
*S4fl=20ft
:
=
830 kPa abs, h
=
15°C, p 2
=
550
kPa
abs,
FLUID
T
5
m
MEASUREMENT
493
diam
m
3
20 cm diam
Cd =0.83
Fig. 8.36
Oxygen, pi
8.44
=
40
a pressure drop of 6
h = 120°F, flows through a 1 by \ in venturi meter with
Find the mass per second flowing and the throat velocity.
psia,
psi.
ISA
Air flows through a 8-cm-diameter
8.45
flow nozzle in a 12-cm-diameter pipe.
=
5°C, and a differential manometer with liquid, sp gr 2.93,
has a gage difference of 0.8
m when connected between the pressure taps. Calculate the
=
px
2
1.5 kg// cm abs,
mass rate
A
8.46
h
of flow.
2.5-in-diameter
What
diameter pipe.
ISA
nozzle
is
used to measure flow of water at 40°F in a 6-in-
gage difference on a water-mercury manometer
is
required for
300 gpm?
Determine the discharge in a 30-cm-diameter line with a 20-cm-diameter VDI
water at 20°C when the gage difference is 30 cm on an acetylene tetrabromide
(sp gr 2.94) -water differential manometer.
8.47
orifice for
A J-in-diameter VDI orifice is installed in a 1-in-diameter pipe carrying nitrogen
120 psia, h = 120°F. For a pressure drop of 20 psi across the orifice, calculate
8.48
=
at pi
the mass flow rate.
Air at
8.49
1
atm,
t
=
diameter square-edged
21 °C flows through a 1-m-square duct that contains a 50-cm-
orifice.
With a head
loss of 8
H
cm
2
across the orifice,
compute
the flow in cubic feet per minute.
A
8.50
7
=
6-in-diameter
52 lb/ ft 3
.
An
VDI
orifice is installed in
oil-air differential
manometer
a 12-in-diameter
is
oil line,
fx
=
6 cP,
used. For a gage difference of 22 in
determine the flow rate in gallons per minute.
8.51
1.3
A
rectangular sharp-crested weir 4
m high.
m
long with end contractions suppressed
Determine the discharge when the head
L = 10ft,P
10 ft. P =
H=
8.52
In Fig.
Fie. 8.20,
8.20.
weir.
C =
8.53
A rectangular sharp-crested weir with end
should
flow?
it
1.8
ft,
0.80
ft.
is
is
25 cm.
Estimate the discharge over the
3.33.
contractions
is
1.5
m long. How high
m for 0.45 m /s
be placed in a channel to maintain an upstream depth of 2.25
3
APPLICATIONS OF FLUID MECHANICS
494
Determine the head on a 60° V-notch weir
8.54
H=
Tests on a 90° V-notch weir gave the following results:
8.55
cfs;
for discharge of 170 1/s.
H=
1.35
ft,
Q =
5.28
Determine the formula
cfs.
0.60
ft,
Q =
0.685
for the weir.
8.56 A sharp-crested rectangular weir 3 ft long with end contractions suppressed and
a 90° V-notch weir are placed in the same weir box, with the vertex of the 90° V-notch
Determine the head on the V-notch weir
weir 6 in below the rectangular weir crest.
when the discharges are equal and (b) when the rectangular weir
greatest amount above the discharge of the F-notch weir.
(a)
A broad-crested weir 5 ft high and
8.57
What head
A
What
torque
=
P?
M
0.8
a
=
cm
circular disk 20
The
8.59
is
=
160 rpm.
0.05 in;
n=
What
the viscosity?
is
8.60
With the apparatus
discharged in
The
1
is
rpm when
of Fig. 8.29,
6.0 in.
mm from
a
flat plate.
the clearance contains
D=
What
is
0.5
The torque
mm, L =
(e)
none of these answers
static
(b)
tube
is
density
velocity
(c)
m,
1
is
24 lb- in
H=
the viscosity in poises?
piezoelectric properties of quartz are used to
temperature
A
=
2.8 in; h
h 30 min.
(a)
8.62
in diameter has a clearance of 0.3
required to rotate the disk 800
speed
8.61
long has a well-rounded upstream corner.
oil,
concentric-cylinder viscometer (Fig. 8.27) has the following dimensions:
0.012 in; b
cm 3 was
ft
its
required for a flow of 100 cfs?
is
8.58
10
discharges
when the
0.75 m, and 60
S =
0.83.
measure
(d)
pressure
used to measure
(a)
the pressure in a static fluid
(c)
the total pressure
(d)
(b)
the velocity in a flowing stream
the dynamic pressure
(e)
the undisturbed fluid
pressure
8.63
A
piezometer opening
is
used to measure
(a)
the pressure in a static fluid
(c)
the total pressure
(d)
(6)
the velocity in a flowing stream
the dynamic pressure
(e)
the undisturbed fluid
pressure
8.64
The simple
pitot tube measures the
(a)
static pressure
(6)
dynamic pressure
(c)
total pressure
(d)
velocity at the stagnation point
(e)
difference in total
8.65
A
and dynamic pressure
(C = 1) is used to measure air speeds. With water
manometer and a gage difference of 3 in, the air speed for 7 = 0.0624
pitot-static tube
differential
in feet per second, is
(a)
4.01
(b)
15.8
(c)
24.06
(d)
127
(e)
none
of these
answers
in the
lb/ft 3
,
FLUID
8.66
The
(a)
static pressure
dynamic pressure
(c)
total pressure
(d)
difference in static
(e)
difference in total
8.67
(a)
(c)
(d)
(e)
8.68
(a)
(b)
(c)
(d)
(e)
8.69
known flowing gas can be determined from measurement
of
and stagnation pressure only
(6) velocity and stagnation pressure only
and dynamic pressure only
velocity and stagnation temperature only
none of these answers
static
The
velocity of a
known
flowing gas
may
be determined from measurement of
and stagnation pressure only
and temperature only
static and stagnation temperature only
stagnation temperature and stagnation pressure only
none of these answers
static
static pressure
The
hot-wire anemometer
pressure in gases
pressure in liquids
(c)
wind
(d)
gas velocities
(e)
liquid discharges
A
is
used to measure
velocities at airports
piston-type displacement meter has a volume displacement of 35
revolution of
8.71
of a
velocity
(a)
(a)
and dynamic pressure
and dynamic pressure
The temperature
(b)
8.70
495
measures
pitot-static tube
(b)
MEASUREMENT
1.87
Water
its shaft.
(b)
4.6
The discharge
35
(c)
for a pipeline
in liters per
(d)
40.34
minute
for 1000
rpm
cm 3
per
is
none of these answers
(e)
was diverted into a weigh tank for exactly 10 min. The
was 4765 lb. The average flow rate in gallons per minute
increased weight in the tank
was
(a)
8.72
66.1
(6)
8.73
(c)
7.95
0.13
(d)
none of these answers
(e)
A rectangular tank with cross-sectional area of 8 m
by a steady flow
(a)
57.1
14.44
Which
of liquid for 12 min.
(6)
867
(c)
of the following
(a)
current meter
(d)
pitot tube
(b)
(e)
901
disk meter
was
filled
to a depth of
The
rate of flow in liters per second
(d)
6471
measuring instruments
venturi meter
2
(c)
(e)
is
was
none of these answers
a rate meter?
hot-wire anemometer
1 .3
m
APPLICATIONS OF FLUID MECHANICS
496
The
8.74
voir
is
(a)
C
v
V2gH
A
8.75
actual velocity at the vena contracta for flow through an orifice from a reser-
expressed by
fluid jet discharging
vena contracta.
(a)
1.31
CC
(a)
C
(a)
V
J
(d)
Cd V2gH
(c)
from a 2-in-diameter
(c)
0.875
0.766
(d)
Cc Cd
C v Cd
(c)
Mir"
C
^T~¥
1
)
\/y
A
(b)
(c)
y
0.96
(6)
orifice is
Cd /C c
1/
\/y
(d)
\Iy
rate, the area of reservoir
Ar must
none of these answers
(e)
1/s
under a head of 3 m.
is
0.97
The discharge
200,000
(a)
through an
answers
Mirw-i)
5-cm-diameter Borda mouthpiece discharges 7.68
(a)
8.81
(e)
of these
none of these answers
(e)
velocity coefficient
(a)
none
it
H
The
8.80
its
losses in orifice flow are
For a liquid surface to lower at a constant
vary with head y on the orifice, as
8.79
has a diameter 1.75 in at
(e)
Cd /C v
(d)
8.78
(a)
orifice
C v Va
(e)
coefficient of contraction is
1.14
(6)
VW
(d)
ratio of actual discharge to theoretical discharge
V
The
8.77
The
(6)
The
8.76
CcVtgH
(b)
(c)
0.98
coefficient for a 4
(d)
0.99
by 2
in venturi
(e)
none
of these
answers
meter at a Reynolds number of
is
0.95
(6)
0.96
(c)
0.973
(d)
0.983
(e)
0.992
Select the correct statement:
The discharge through a venturi meter depends upon Ap only and
is
independent
of orientation of the meter.
(6)
A venturi meter with
the flow
(c)
is
vertically
a given gage difference R' discharges at a greater rate
downward through
it
than when the flow
is
vertically
when
upward.
For a given pressure difference the equations show that the discharge of gas is
when compressibility is taken into account than when
greater through a venturi meter
it is
neglected.
(d)
The
The
(e)
coefficient of contraction of a venturi
overall loss
with the same
D
2 is
is
meter
is
unity.
the same in a given pipeline whether a venturi meter or a nozzle
used.
FLUID
The expansion
8.82
(a)
k, yoJvi,
factor
(c)
(d)
k,
depends upon
R, and A 2 /Ai
none of these answers
(e)
The discharge through a V-notch weir
8.83
H~
(a)
l/2
H
(b)
1'2
(c)
H
312
(d)
varies as
H
bl2
(e)
none of these answers
The discharge of a rectangular sharp-crested weir with end contractions
than for the same weir with end contractions suppressed by
8.84
(a)
5%
(e)
none
8.85
(b)
10%
of these
(c)
15%
(d)
(a)
t
A homemade
=
46
Ci
=
s.
no fixed percentage
viscometer of the Saybolt type
is
calibrated
known kinematic viscosity. For v = 0.461 St,
The coefficients d, C2 in v = C\t -\- C2 /t are
0.005
(6)
C2 =-2.3
(e)
none
Ci
=
0.0044
C2 =3.6
of these
is less
answers
with liquids of
St,
497
A 2/Ai
and
R, p 2 /p\, and A 2 I A x
k, R, and P2/P1
(b)
Y
MEASUREMENT
(c)
d=
t
=
by two measurements
97 s and for v = 0.18
d=
0.00317
C2 =1.55
C2 =
14.95
A
Bibliography,"
0.0046
(d)
answers
REFERENCES
Dowden, R. Rosemary: "Fluid Flow Measurement:
BHRA
Fluid
Engineering, Cranfield, Bedford, England, 1972.
American Society
ASME
of
Mechanical Engineers: "Fluid Meters," 6th ed., New York, 1971.
Its Measurement and Control in Science and Industry,
Symposium on Flow,
Pittsburgh,
May
9-14, 1971.
9
TURBOMACHINERY
To
turn a fluid stream or change the magnitude of
forces be applied.
momentum,
When
its
velocity requires that
a moving vane deflects a fluid jet and changes
its
between vane and jet and work is done by
displacement of the vane. Turbomachines make use of this principle: the
axial and centrifugal pumps, blowers, and compressors, by continuously
doing work on the fluid, add to its energy; the impulse, Francis, and propeller
turbines and steam and gas turbines continuously extract energy from the
fluid and convert it into torque on a rotating shaft the fluid coupling and the
torque converter, each consisting of a pump and a turbine built together make
use of the fluid to transmit power smoothly. Designing efficient turbomachines
utilizes both theory and experimentation. A good design of given size and speed
may be readily adapted to other speeds and other geometrically similar sizes
forces are exerted
;
by
application of the theory of scaled models, as outlined in Sec. 4.5.
Similarity relationships are
tion of homologous units
and
first
discussed in this chapter
specific speed.
by considera-
Elementary cascade theory
next taken up, before considering the theory of turbomachines.
bines and
pumps are then considered,
The chapter closes with a
compressors.
9.1
HOMOLOGOUS
followed
by blowers and
Water
is
tur-
centrifugal
discussion of cavitation.
UNITS; SPECIFIC SPEED
In utilizing scaled models in designing turbomachines, geometric similitude
is
required as well as geometrically similar velocity vector diagrams at entrance
to, or exit
from, the impellers.
Viscous effects must, unfortunately, be ne-
two above conditions and
have equal Reynolds numbers in model and prototype. Two geometrically
glected, as
498
it is
generally impossible to satisfy the
TURBOMACHINERY
similar units having similar velocity vector diagrams are homologous.
499
They will
have geometrically similar streamlines.
The velocity vector diagram in Fig. 9.1 at exit from a pump impeller
can be used to formulate the condition for similar streamline patterns. The
blade angle is /?, u is the peripheral speed of the impeller at the end of the
vane or blade, v is the velocity of fluid relative to the vane, and V is the absolute
velocity leaving the impeller, the vector sum of u and v; V r is the radial
component of V and is proportional to the discharge a is the angle which the
absolute velocity makes with u, the tangential direction. According to geometric similitude, /3 must be the same for two units, and for similar streamlines
a must also be the same in each case.
It is convenient to express the fact that a is to be the same in any of a
series of turbomachines, called homologous units, by relating the speed of
rotation N, the impeller diameter (or other characteristic dimension) D,
and the flow rate Q. For constant a, V r is proportional to V (V r = V sin a)
and u is proportional to V r Hence the conditions for constant aina homologous series of units may be expressed as
also
;
.
Vr
—
=
const
u
The discharge Q
is
proportional to
is
D
proportional to
2
.
The speed
VD
r
2
,
since
of rotation
any
cross-sectional flow area
N is proportional to u/D. When
these values are inserted,
^
=
const
(9.1.1)
expresses the condition in which geometrically similar units are homologous.
The
Fig. 9.1
pump
discharge
Q through homologous
units can be related to head
Velocity vector diagram for exit from a
impeller.
H
APPLICATIONS OF FLUID MECHANICS
500
and a representative
cross-sectional area
A
by the
orifice
formula
Q = CdA V^gH
which
in
Cd
,
the discharge coefficient, varies slightly with the Reynolds
number and so causes a small change in efficiency with size in a homologous
series. The change in discharge with the Reynolds number is referred to as
scale effect. The smaller machines, having smaller hydraulic radii of passages,
have lower Reynolds numbers and correspondingly higher friction factors;
hence they are less efficient. The change in efficiency from model to prototype
may
be from 1 to 4 percent. However, in the homologous theory, the scale
must be neglected, and so an empirical correction for change in efficiency
with size is used [see Eq. (9.5.1) ]. As A
D 2 the discharge equation may be
effect
~
Q
d
=
Vh
2
const
Eliminating
H
ND
2
=
(9.1.2)
Q between
Eqs. (9.1.1) and (9.1.2) gives
const
(9.1.3)
2
Equations
and
(9.1.1)
and speed.
are most useful in determining performance
from those of a homologous unit of different size
(9.1.3)
characteristics for one unit
1
,
1
Application of dimensional analysis is illuminating. The variables appearing to be pertinent to the flow relations for similar units would be F(H,Q,N,D,g) = 0. There are
and D may be selected as the repeating variables,
two dimensions involved, L and T;
N
yielding
(-9-,»,JL.).
\ND D WD J
3
'
'
Solving for
H
=
H gives
>—)
if—
\ND* NWJ
Df,
'
Experiment shows that the second dimensionless parameter actually occurs to the power
—1
;
hence
AW,
„
I
Q\
gH
,
(
Q
\
3
characteristic curve for a pump in dimensionless form is the plot ofQ/ND as abscissa
H/(N 2 2 /g) as ordinate. This curve, obtained from tests on one unit of the
series, then applies to all homologous units, and may be converted to the usual charand D. As power is proportional to
acteristic curve by selecting desired values of
The
D
against
N
yQH, the dimensionless power term
7
Q
7 iv5"3
H
N D*/g
2
=
power
P
N D*
3
is
TURBOMACHINERY
EXAMPLE
A prototype
9.1
test of a mixed-flow
pump
501
with a 72-in-diameter
discharge opening, operating at 225 rpm, resulted in the following characteristics
H,
:
Q, cfs
ft
e,
200
228
256
280
303
60
57.5
55
52.5
50
%
H,
47.5
45
42.5
40
37.5
69
75
80
83.7
86
What size and synchronous
to produce 200 cfs at 60
teristic
Qi
H
ND
2
ft
e,
%
87.3
88
87.4
86.3
84.4
330
345
362
382
396
H,
ft
35
32.5
30
27.5
25
Q, cfs
e,%
411
82
425
438
449
459
79
75
71
66.5
speed (60 Hz) of homologous pump should be used
head at point of best efficiency? Find the charac-
curves for this case.
Subscript
=
Q, cfs
ft
345, e
=
1 refers to the 72-in pump.
For best efficiency Hi
88 percent. With Eqs. (9.1.1) and (9.1.3),
Hi
Q
ND*
N^D?
2
=
45,
Qi
NxDi*
or
ND
2
200
45
60
2
225
2
X
72
After solving for
345
ND*
2
225
X
72 3
N and D,
D=
A = 366rpm
r
51.1 in
nearest synchronous speed (3600 divided by number of pairs of poles)
360 rpm. To maintain the desired head of 60 ft, a new D is necessary. Its
size can be computed:
The
is
D
=
The
VU X m X 72 = 52
in
discharge at best efficiency
Q = ^"777 =
NiDi
3
which
is
345
slightly
X
is
mm* =
then
208
more capacity than
cfs
required.
With
N
=
360 and
D =
52,
APPLICATIONS OF FLUID MECHANICS
502
equations for transforming the corresponding values of
can be obtained:
H
and Q
for
any
efficiency
H
-
JET,
(^f-J
-
J5Ti(*»
X
If)
2
-
1.335ffi
and
iV7)
J
= Qi(fM)(«) =
Q =
Qi
The
characteristics of the
H,
ft
80
76.7
73.4
70
66.7
3
JV1D1 3
Q, cfs
121
138
155
169
183
e,
%
69
75
80
83.7
86
0.603Q 1
new pump
H,
ft
are
Q, cfs
e,%
H,
200
208
219
87.3
88
87.4
86.3
84.4
46.7
43.4
40
36.7
33.4
63.5
60
56.7
53.5
231
50
239
ft
Q, cfs
*,%
248
257
264
82
79
75
271
71
277
66.5
The efficiency of the 52-in pump might be a fraction of a percent less
than that of the 72-in pump, as the hydraulic radii of flow passages are smaller,
so that the Reynolds number would be less.
Specific speed
The
specific
speed of a homologous unit
is
a constant widely used in selecting
the type of unit and in preliminary design.
for a
pump and
It is usually defined differently
a turbine.
N
The specific speed s of a homologous series of pumps is defined as the
speed of some one unit of the series of such a size that it delivers unit discharge
at unit head. It is obtained as follows. Eliminating D in Eqs. (9.1.1) and
(9.1.3)
N_VQ
#3/4
and rearranging give
const
By definition of specific speed,
(9.1.4)
the constant
is
N
s,
the speed of a unit for
Q =
1,
TURBOMACHINERY
H
=
l;
-££
N. =
The
i.e.,
503
specific
(9.1.5)
usually denned for the point of best efficiency,
and head that is most efficient.
a homologous series of turbines is defined as the
speed of a series
is
for the speed, discharge,
The
specific
speed of
speed of a unit of the series of such a size that it produces unit horsepower
with unit head. Since power P is proportional to QH,
—=
const
The terms
(9.1.6)
D and Q may be eliminated from Eqs.
(9.1.1), (9.1.3),
and
(9.1.6)
to produce
N VP
-^iT =
const
(9.1.7)
For unit power and unit head the constant of Eq.
or the specific speed
*
-
N
s
(9.1.7)
becomes the speed,
of the series, so that
*g
<«.8)
The specific speed of a unit required for a given discharge and head can
be estimated from Eqs. (9.1.5) and (9.1.8). For pumps handling large discharges at low heads a high specific speed is indicated for a high-head turbine
producing relatively low power (small discharge) the specific speed is low.
Experience has shown that for best efficiency one particular type of pump or
;
turbine as usually indicated for a given specific speed.
Because Eqs. (9.1.5) and (9.1.8) are not dimensionally correct (7 and g
have been included in the constant term) the value of specific speed depends
,
on the units involved. For example,
in the
United States
Q
is
commonly
ex-
pressed in gallons per minute, millions of gallons per day, or cubic feet per
second when referring to specific speeds of pumps.
Centrifugal pumps have low specific speeds; mixed-flow
medium
specific
speeds; and axial-flow
Impulse turbines have low
specific speeds;
pumps have high
specific speeds; Francis turbines
and propeller turbines have high
pumps have
specific
speeds.
have medium
specific speeds.
APPLICATIONS OF FLUID MECHANICS
504
Fig. 9.2
9.2
Simple cascade system.
ELEMENTARY CASCADE THEORY
do work on a fluid or extract work from it in a conit flow through a series of moving (and possibly
fixed) vanes. By examination of flow through a series of similar blades or
vanes, called a cascade, some of the requirements of an efficient system can
be developed. Consider, first, flow through the simple fixed cascade system
of Fig. 9.2. The velocity vector representing the fluid has been turned through
an angle by the presence of the cascade system. A force has been exerted on
the fluid, but (neglecting friction effects and turbulence) no work is done on
the fluid. Section 3.11 deals with forces on a single vane.
Since turbomachines are rotational devices, the cascade system may be
arranged symmetrically around the periphery of a circle, as in Fig. 9.3. If
the fluid approaches the fixed cascade in a radial direction, it has its moment
of momentum changed from zero to a value dependent upon the mass per
Turbomachines
either
tinuous manner by having
Fig. 9.3 Cascade
arranged
on the periphery of a cir-
cular cylinder.
TURBOMACHINERY
Fig. 9.4
505
Moving cascade
within a fixed cascade.
unit time flowing, the tangential
component
of velocity
V
t
developed, and
the radius, from Eq. (3.12.5),
T = P QrV
(9.2.1)
t
Again, no work
done by the fixed-vane system.
series of vanes (Fig. 9.4) rotating within the
fixed- vane system at a speed co.
For efficient operation of the system it is
important that the fluid flow onto the moving vanes with the least disturbance,
i.e., in a tangential manner, as illustrated in Fig. 9.5.
When the relative
Consider
velocity
shown
is
is
now another
not tangent to the blade at
The
its
entrance, separation
may
occur, as
tend to increase rapidly (about as the square)
with angle from the tangential and radically impair the efficiency of the
machine. Separation also frequently occurs when the approaching relative
in Fig. 9.6.
velocity
is
losses
tangential to the vane, owing to curvature of the vanes or to ex-
pansion of the flow passages, which causes the boundary layer to thicken and
Fig. 9.5
Relative
tangent to blade.
velocity
APPLICATIONS OF FLUID MECHANICS
506
Flow separation, or
Fig. 9.6
shock, from blade with relative velocity not tangent to
leading edge.
to rest. These losses are called shock or turbulence losses. When the fluid
from the moving cascade, it will generally have its velocity altered in
both magnitude and direction, thereby changing its moment of momentum
and either doing work on the cascade or having work done on it by the moving
cascade. In the case of a turbine it is desired to have the fluid leave with no
moment of momentum. An old saying in turbine design is "have the fluid
enter without shock and leave without velocity."
Turbomachinery design requires the proper arrangement and shaping
of passages and vanes so that the purpose of the design can be met most
efficiently. The particular design depends upon the purpose of the machine,
the amount of work to be done per unit mass of fluid, and the fluid density.
come
exits
9.3
THEORY OF TURBOMACHINES
Turbines extract useful work from
fluid energy;
and pumps, blowers, and
turbocompressors add energy to fluids by means of a runner consisting of
vanes rigidly attached to a shaft. Since the only displacement of the vanes is
work is done by the displacement of the tangential
components of force on the runner. The radial components of force on the
runner have no displacement in a radial direction and, hence, can do no work.
In turbomachine theory, friction is neglected, and the fluid is assumed to
have perfect guidance through the machine, i.e., an infinite number of thin
vanes, and so the relative velocity of the fluid is always tangent to the vane.
This yields circular symmetry and permits the moment-of-momentum equation, Sec. 3.12, to take the simple form of Eq. (3.12.5), for steady flow,
in the tangential direction,
T = PQL(rV )ont- (r70m]
in
which
T
t
(9.3.1)
is
the torque acting on the fluid within the control volume (Fig.
TURBOMACHINERY
507
Steady flow through control volume
Fig. 9.7
with circular symmetry.
9.7)
and pQ(rV
t
)
out
and pQ(rV
t
) in
represent the
moment
of
momentum leav-
ing and entering the control volume, respectively.
The
polar vector diagram
ships (Fig. 9.8), with subscript
fluid.
V
is
is
1
generally used in studying vane relation-
for entering fluid
and subscript 2
for exiting
the absolute fluid velocity, u the peripheral velocity of the runner,
The absolute velocities V, u
and the relative velocity connects them as shown. V u is
designated as the component of absolute velocity in the tangential direction.
a is the angle the absolute velocity V makes with the peripheral velocity u,
and (3 is the angle the relative velocity makes with —u, or it is the blade angle.
and
v
the fluid velocity relative to the runner.
are laid off from 0,
Entrance
Fig. 9.8
Polar vector diagrams.
APPLICATIONS OF FLUID MECHANICS
508
is assumed. V r is the absolute velocity component normal
In this notation Eq. (9.3.1) becomes
as perfect guidance
to the periphery.
T =
pQ(r 2 V 2 cos a 2
= pQ(r 2 V u2 -
—
riVi cos en)
riVui)
= m(r2 V u2 - nV u i)
(9.3.2)
The mass per unit time flowing is m = pQ = (pQ) 0U = (pQ)m- In the form
above, when T is positive, the fluid moment of momentum increases through
the control volume, as for a pump. For T negative, moment of momentum
t
of the fluid
is
decreased, as for a turbine runner.
When T =
0,
as in passages
where there are no vanes,
rV u = const
This
is
with the tangential component of velocity varying
free-vortex motion,
inversely with radius.
It is discussed in Sec. 7.9
and compared with the
forced vortex in Sec. 2.9.
EXAMPLE
9.2
The wicket
gates of Fig. 9.9 are turned so that the flow
makes
an angle of 45° with a radial line at section 1, where the speed is 2.5 m/s. Determine the magnitude of tangential velocity component V u over section 2.
Since no torque is exerted on the flow between sections 1 and 2, the
Wicket
Fig. 9.9
Schematic view
of propeller turbine.
TURBOMACHINERY
moment
Vur =
of
=
is
constant and the motion follows the free-vortex law
const
At section
Vui
momentum
509
1
2.5 cos 45°
-
1.77
m/s
Then
V ul n =
(1.77
m/s)
(1.2
m) =
2.124
m /s
2
Across section 2
V* =
m /s
r m
2
2.124
At the hub
2.124/0.6
=
Vu =
Head and energy
By
2.124/0.225
=
Vu =
9.44 m/s, and at the outer edge
3.54 m/s.
relations
multiplying Eq. (9.3.2) by the rotational speed
(rad/s) of runner,
co
Tu = P Q(ur 2 V u2 - conV u i) = pQ(u 2 V u2 - uiV u i)
For no
H
is
losses the
Qy
potential energy per unit weight.
QyH,
Tec
in
which
H
is
the
pump
head.
is
Q Ap = QyH,
in
which
the weight per unit time and
H the
power available from a turbine
the head on the runner, since
(9.3.3)
Similarly a
is
pump
runner produces work
The power exchange
= QyH
is
(9.3.4)
Solving for H, using Eq. (9.3.3) to eliminate T, gives
H
=
U2Vu2
~
UlVul
(9.3.5)
g
For turbines the sign is reversed in Eq. (9.3.5)
For pumps the actual head p produced is
H
Hp
=
eh
H
=
H - HL
,
(9.3.6)
,
APPLICATIONS OF FLUID MECHANICS
510
and
for turbines the actual
head
H
t
is
TT
H
t
=
-
H + HL
=
(9.3.7)
eh
in
which
the hydraulic efficiency of the machine and
e h is
HL
represents
all
The overall efficiency of the machines
by friction caused by fluid between
the internal fluid losses in the machine.
further reduced by bearing friction,
runner and housing, and by leakage or flow that passes around the runner
without going through it. These losses do not affect the head relations.
is
Pumps
are generally designed so that the angular
entering the runner (impeller)
H
=
momentum
of fluid
Then
zero.
is
-^^
U
(9.3.8)
9
momentum is zero at the exit section
Turbines are designed so that the angular
of the runner for conditions at best efficiency; hence,
H
=
^^
(9.3.9)
In writing the energy equation for a pump, with Eqs. (9.3.5) and (9.3.6)
_ U2V2 cos
a<i
—
i/iFi cos ai
_
9
for
which
it is
total energy.
v relative
assumed that
With the
to the runner,
(Fig. 9.8) by the
ui
2
+
Vi 2
-
all
streamlines through the
relations
among
pump have
the same
the absolute velocity V, the velocity
and the velocity u
of runner,
from the vector diagrams
law of cosines,
2uiVi cos ai
=
V!
2
u 22
Eliminating the absolute velocities
+V -
Vh V
2
2
2
2U2V2 cos a 2
in these relations
=
and
v2
2
in Eq. (9.3.10)
gives
m
uj-ut _ tf-tf _ £_,, _
2g
2g
y
_
(93
n)
TURBOMACHINERY
511
or
V2
Z2
29
The
L\2<7
)
7
"
~
(2"
+~+
(9 3 12)
Zl
'
losses are the difference in centrifugal head, (u 22
head change in the relative
fromEq.
(9.3.11),
^^ +
- *
22
flow.
For no
loss,
—
2
ui )/2g,
=
in the
(9.3.13
2#
2g
With no flow through the runner,
vi,
v2
are zero,
and the head
rise is as ex-
pressed in the relative equilibrium relationships [Eq. (2.9.7)].
head
and
the increase in pressure head,
is
7
occurs, the
'
)]
rise is
When
flow
equal to the centrifugal head minus the difference in
relative velocity heads.
For the case of a turbine, exactly the same equations
result.
9.3 A centrifugal pump with a 24-in-diameter impeller runs at
rpm.
The water enters without whirl, and a 2 = 60°. The actual head
1800
produced by the pump is 50 ft. Find its hydraulic efficiency when V 2 = 20 ft/s.
From Eq. (9.3.8) the theoretical head is
EXAMPLE
__
H =
U2V2 cos a 2
=
1800
X
g
The
actual head
50
**
- 5^6 =
9.4
85
'
is
50.0
ft;
X 20 X
X 32.2
2tt
60
0.50
=
ro „
58.6
.
ft
hence, the hydraulic efficiency
is
4%
IMPULSE TURBINES
The impulse turbine
is
one in which
all
available energy of the flow
is
con-
verted by a nozzle into kinetic energy at atmospheric pressure before the fluid
contacts the
moving
blades.
Losses occur in flow from the reservoir through
may be computed from pipe friction data. At the base of the nozzle the available energy,
the pressure pipe (penstock) to the base of the nozzle, which
or total head,
ha
=
^ + ^f
7
2#
is
(9.4.1)
APPLICATIONS OF FLUID MECHANICS
512
Headwater
-
Energy grade
"
Pressure pipe
*
l)
= C
x/2gh a
v
The head
ha~
With C v the nozzle
Fig. 9.10.
V2 = C
^¥W
V
2
is
(9.4.2)
2
v
ha
=
ka (l
- C
*)
(9.4.3)
efficiency of the nozzle is
Vi/2g
C
v
*h a
=
ha
The
9.12)
coefficient, the jet velocity
lost in the nozzle is
— = ha- C
and the
Tail water
Impulse turbine system.
Fig. 9.10
from
(g)
jet,
which
d
(9.4.4)
with velocity V2, strikes double-cupped buckets (Figs. 9.11 and
the flow and turn the relative velocity through the angle $
split
(Fig. 9.12).
The
F =
x component of
pQ(v r
—
iv
momentum
(1
changed by
(Fig. 9.12)
cos 0)
and the power exerted on the vanes
Fu = pQuv r
is
-
cos0)
is
(9.4.5)
To maximize the power, theoretically, 6 = 180°, and uv r must be a maximum;
is, u(V 2 — u) must be a maximum.
By differentiating with respect to u
that
TURBOMACHINERY
513
Southern California Edison, Big Creek 2A, 1948, 8§-in-diameter jet
impulse buckets and disk in process of being reamed; 56,000 hp, 2200 ft
head, 300 rpm. (Allis-Chalmers Mfg. Co.)
Fig. 9.11
and equating to
(7,-t*)
u =
V /2.
2
zero,
+ t*(-l) =0
After making these substitutions into Eq. (9.4.5)
Fu-pQjfa-
Fig. 9.12
y)(l-
-1) - yQ
Flow through bucket.
7
2
2
(9.4.6)
29
APPLICATIONS OF FLUID MECHANICS
514
energy of the jet. The velocity diagram
shows that the absolute velocity leaving the vanes is zero.
Practically, when vanes are arranged on the periphery of a wheel (Fig.
9.11), the fluid must retain enough velocity to move out of the way of the
which accounts
for the total kinetic
for these values
following bucket.
The
Most
of the practical impulse turbines are Pelton wheels.
two and turned in a horizontal plane, and half is discharged
from each side to avoid any unbalanced thrust on the shaft. There are losses
due to the splitter and to friction between jet and bucket surface, which make
the most economical speed somewhat less than V 2 /2. It is expressed in terms
jet is split in
of the speed factor
vm
* -
For most
(9
efficient
turbine operation
speed as shown in Table
9.1.
<f>
-
47)
has been found to depend upon specific
The angle
6 of
the bucket
is
usually 173 to 176°.
and the diameter of the wheel is D at the centerline of the buckets, it has been found in practice that the diameter ratio D/d
should be about 54/iV, (ft, hp, rpm), or 206/iV (m, kW, rpm) for maximum
diameter of the
If the
jet is d,
s
efficiency.
In the majority of installations only one jet
is
used, which discharges
horizontally against the lower periphery of the wheel as
The wheel speed
shown
in Fig. 9.10.
carefully regulated for the generation of electric power.
is
A governor operates a needle valve that controls the jet discharge by changing
Table
9.1
Dependence of
<f>
on specific
speed*
Specific speed
(m,
kW, rpm)
7.62
11.42
15.24
19.05
22.86
26.65
(ft,
N
s
hp,
2
3
4
5
6
7
rpm)
4
0.47
0.46
0.45
0.44
0.433
0.425
Modified from J. W. Daily, Hydraulic
H.
(ed.),
Machinery,
Rouse
in
"Engineering Hydraulics," p. 943,
Wiley, New York, 1950.
TURBOMACHINERY
its
So
area.
V
515
remains practically constant for a wide range of positions of
2
the needle valve.
The efficiency of the power conversion drops off rapidly with change in
head (which changes V2 ), as is evident when power is plotted against V2
for constant u in Eq. (9.4.5)
The wheel operates in atmospheric air although
it is enclosed by a housing.
It is therefore essential that the wheel be placed
above the maximum floodwater level of the river into which it discharges.
The head from nozzle to tail water is wasted. Because of their inefficiency at
other than the design head and because of the wasted head, Pelton wheels
usually are employed for high heads, e.g., from 200
to more than 1 km. For
high heads, the efficiency of the complete installation, from headwater to
tail water, may be in the high 80s.
Impulse wheels with a single nozzle are most efficient in the specific
speed range of 2 to 6, when P is in horsepower,
is in feet, and N is in revolu.
m
H
Multiple nozzle units are designed in the specific speed
tions per minute.
range of 6 to
EXAMPLE
12.
A
9.4
Pelton wheel
The water
cm
is
to be selected to drive a generator at 600
and has a velocity of 100 m/s.
With the blade angle at 170°, the ratio of vane speed to initial jet speed at
0.47, and neglecting losses, determine (a) diameter of wheel to centerline of
buckets (vanes), (6) power developed, and (c) kinetic energy per newton
rpm.
remaining in the
The
(a)
u = 0.47
X
jet is 7.5
in diameter
fluid.
peripheral speed of the wheel
100
=
is
47 m/s
Then
—
600 /
- I 2tt
D\
- =
47 m/s
J
(b)
(997.3
From Eq.
kg/m3 ) j
or
D =
From
m
(9.4.5), the power, in kilowatts, is
(0.075
m) 2 (100 m /s)
(47 m/s) (100
x
(c)
1.495
[1
-
-
computed to be
47 m/s)
Fig. 3.32, the absolute velocity
kW
^^
1
(-0.9848)]
=
2170
kW
components leaving the vane
APPLICATIONS OF FLUID MECHANICS
516
are
Vx =
(100
-
47) (-0.9848)
Vv =
(100
-
47) (0.1736)
+
=
= -5.2 m/s
47
9.2
m/s
The
kinetic energy remaining in the jet
5.2 2
+ 9.2
2
X
is
2
=
m -N/N
5.69
9.806
EXAMPLE 9.5 A small impulse wheel is to be used to drive a generator for
60-Hz power. The head is 300 ft, and the discharge 1.40 cfs. Determine the
diameter of the wheel at the centerline of the buckets and the speed of the
wheel. C v = 0.98. Assume efficiency of 80 percent.
The power is
X
yQHe
Pn =
62.4
Taking a
value of
1.4
-Jw=
N
XT
trial
N H7=-
—\/P
5 '*
S
=
4
=
X 300 X
^o
N
X
=
oo
38
,
-
,
lhp
of 4 gives
s
5'4
300
7=^
0.80
=
o
_ rpm
809
\/38.1
For 60-Hz power the speed must be 3600 divided by the number of pairs of
For five pairs of poles the speed would be 3600/5 =
720 rpm, and for four pairs of poles 3600/4 = 900 rpm. The closer speed 720
is selected, although some engineers prefer an even number of pairs of poles
poles in the generator.
in the generator.
Ns =
NVP
~H^
For
N
u =
<t>
s
=
Then
720
=
V3^2 = 356
300"
3.56,
take
\/2gH =
4>
0.455
=
0.455,
V
2
and
co
=
W
2»
=
75.4 rad/s
X
32.2
X
300
=
63.2 ft/s
TURBOMACHINERY
The
=
w
peripheral speed
—
^
o>D
=
Z>
2
2
of the jet
v
Q
77
V
0.98
2
201
— =
d
The
».
d
are related
1.676
ft
=
20.1 in
is
y/2
obtained from the jet velocity V2; thus
X
32.2
X
300
X
7^
136
144
=
!-
482
in2
«*
=
136 ft/s
Sa
1.482
= \
= \fc^77 =
\ 7T
>0.7854
—
/
1-374 in
ratio Z)/d is
= 1A«
14.6
1.375
desired diameter ratio for best efficiency
54
iV s
which
9.5
co
1.40
=
Hence the diameter
-
and
2 X 63.2
2u
—
= —— — =
75.4
V = C y/2qH =
=
D
co
The diameter d
a
u and
517
is
is
_ _5i _ 15a5
3.56
satisfactory.
Hence the wheel diameter
is
20.1 in
and speed 720 rpm.
REACTION TURBINES
In the reaction turbine a portion of the energy of the fluid
kinetic energy
by the
fluid's
is
converted into
passing through adjustable gates (Fig. 9.13)
before entering the runner, and the remainder of the conversion takes place
through the runner. All passages are filled with liquid, including the passage
(draft tube) from the runner to the downstream liquid surface. The static
fluid pressure occurs on both sides of the vanes and hence does no work. The
work done is entirely due to the conversion to kinetic energy.
The reaction turbine is quite different from the impulse turbine, discussed in Sec. 9.4. In an impulse turbine all the available energy of the fluid
is converted into kinetic energy by a nozzle that forms a free jet. The energy
is then taken from the jet by suitable flow through moving vanes. The vanes
are partly filled, with the jet open to the atmosphere throughout its travel
through the runner.
In contrast, in the reaction turbine the kinetic energy is appreciable as
the fluid leaves the runner and enters the draft tube. The function of the
APPLICATIONS OF FLUID MECHANICS
518
Fig. 9.13
Stay ring and wicket gates for reaction turbine. (Allis-Chalmers
Mfg. Co.)
draft tube
is
to reconvert the kinetic energy to flow energy
by a gradual
ex-
pansion of the flow cross section. Application of the energy equation between
the two ends of the draft tube shows that the action of the tube
is
to reduce
upstream end to less than atmospheric pressure, thus
increasing the effective head across the runner to the difference in elevation
between headwater and tail water, less losses.
By referring to Fig. 9.14, the energy equation from 1 to 2 yields
the pressure at
^ +
Vi 2
2flf
The
v\
^+-=
its
+
+ +
7
losses include friction plus velocity
head
loss at the exit
from the draft
tube, both of which are quite small; hence
(9.5.1)
7
2g
TURBOMACHINERY
Fig. 9.14
519
Draft tube.
shows that considerable vacuum
is
produced at section
increases the head across the turbine runner.
which
1,
The turbine
effectively
setting
must not
be too high, or cavitation occurs in the runner and draft tube (see Sec.
9.8).
EXAMPLE 9.6 A turbine has a velocity of 6 m/s at the entrance to the draft
tube and a velocity of 1.2 m/s at its exit. For friction losses of 0.1
and a
tail water 5
below the entrance to the draft tube, find the pressure head at
the entrance.
m
m
From Eq.
Pi
£i
(9.5.1)
62
= _5
T
1
2
7
X
9.806
1.2 2
2X
x
1_
9.806
o.l
= -6.66
m
as the kinetic energy at the exit from the draft tube
head of 6.66
m
is
is lost.
Hence a suction
produced by the presence of the draft tube.
There are two forms of the reaction turbine in common
use, the Francis
turbine (Fig. 9.15) and the propeller (axial-flow) turbine (Fig. 9.16). In both,
all
passages flow
ing the
moment
full,
of
and energy is converted
to useful
momentum of the liquid. The flow
work
by changthrough the
entirely
passes
first
wicket gates, which impart a tangential and a radially inward velocity to the
fluid.
A
space between the wicket gates and the runner permits the flow to
close behind the gates
and move
as a free vortex, without external torque
being applied.
In the Francis turbine (Fig. 9.15) the fluid enters the runner so that the
The radial
is tangent to the leading edge of the vanes.
component is gradually changed to an axial component, and the tangential
component is reduced as the fluid traverses the vane, so that at the runner
relative velocity
exit the flow is axial
The
with very
little
whirl (tangential component) remaining.
pressure has been reduced to less than atmospheric, and most of the
remaining kinetic energy
is
reconverted to flow energy by the time
it
dis-
APPLICATIONS OF FLUID MECHANICS
520
44
in.
diam
Turbine data
150,000 hp
330 ft hd
120 rpm
Fig. 9.15
port
News
Francis turbine for Grand Coulee, Columbia Basin Project. (NewShipbuilding and Dry Dock Co.)
charges from the draft tube. The Francis turbine is best suited to medium
head installations from 80 to 600 ft (25 to 180 m) and has an efficiency between
90 and 95 percent for the larger installations. Francis turbines are designed in
the specific speed range of 10 to 110 (ft, hp, rpm) or 40 to 420 (m, kW, rpm)
with best efficiency in the range 40 to 60 (ft, hp, rpm) or 150 to 230 (m, kW,
rpm).
In the propeller turbine (Fig. 9.9), after passing through the wicket
gates, the flow moves as a free vortex and has its radial component changed to
axial component by guidance from the fixed housing. The moment of momentum is constant, and the tangential component of velocity is increased through
the reduction in radius. The blades are few in number, relatively flat, with
very little curvature, and placed so that the relative flow entering the runner
is tangential to the leading edge of the blade. The relative velocity is high, as
with the Pelton wheel, and changes slightly in traversing the blade. The
velocity diagrams in Fig. 9.17 show how the tangential component of velocity
is reduced.
Propeller turbines are made with blades that pivot around the
hub, thus permitting the blade angle to be adjusted for different gate openings
and for changes in head. They are particularly suited for low-head installations,
up to 30 m, and have top efficiencies around 94 percent. Axial-flow turbines
TURBOMACHINERY
521
Fig. 9.16 Field view of installation of runner of 24 500-hp
100-rpm, 41-ft-head, Kaplan adjustable-runner hydraulic
turbine. Box Canyon Project, Public Utility District No. 1
f
of
Pend
operation
Oreille County,
in 1955.
Washington. Plant placed
f
in
(Allis-Chalmers Mfg. Co.)
are designed in the specific speed range of 100 to 210 (ft, hp, rpm) or 380 to
800 (m, kW, rpm) with best efficiency from 120 to 160 (ft, hp, rpm) or 460
to 610 (m, kW, rpm)
The windmill is a form of axial-flow turbine. Since it has no fixed vanes to
give an initial tangential component to the airstream, it must impart the tangential component to the air with the moving vanes. The airstream expands
in passing through the vanes with a reduction in its axial velocity.
APPLICATIONS OF FLUID MECHANICS
522
u,=cor
q
Velocity diagrams for entrance
Fig. 9.17
and
exit of a propeller turbine blade at fixed radial
distance.
EXAMPLE
Assuming uniform
9.7
axial velocity over section 2 of Fig. 9.9
and
using the data of Example 9.2, determine the angle of the leading edge of the
=
propeller at r
At
=
r
0.225, 0.45,
u =
W(2t) (0.225) =
At
=
r
r
-
0.6
The
(0.6
=
(0.45)
m,
5.66
Vu =
m/s
(0.6)
=
240 rpm.
m/s
m/s
Vu =
4.72
m/s
15.06
m/s
Vu =
3.54
m/s
discharge through the turbine
m)
(2.4
m)
(*) (2.5
8
tt(0.6 2
-
=
Moody
1
is,
m/s) (cos 45°)
8.24
from section
=
8
1,
m /s
3
is
m/s
0.225 2 )
Figure 9.18 shows the
1
9.44
11.3
Hence, the axial velocity at section 2
Va =
for a propeller speed of
m,
u = W(2tt)
Q =
0.6
0.45 m,
u = W(2tt)
At
and
0.225 m,
initial
vane angle
for the three positions.
has developed a formula to estimate the efficiency of a unit of
Lewis F. Moody, The Propeller Type Turbine, Trans.
ASCE,
vol. 89, p. 628, 1926.
TURBOMACHINERY
523
0,-144°26'
u = 15.06 m/s
O
Va =8.24
Va =8.24m/s
Ka =8.24m/s
Vu =4.72m/s
r= 0.225
Fig. 9.18
m
r=0.45
m
Vu =3.54m/s
r=0.60
m
Velocity diagrams for angle of leading edge of a propeller turbine
\
blade.
a homologous series of turbines when the efficiency- of one of the series
is
known
€
in
=
/DA
1
which
9.6
1
"
-d-eOf^)
e\
(0.5.2)
and D\ are usually the
efficiency
and the diameter
of a model.
PUMPS AND BLOWERS
Pumps add energy
to liquids and blowers to gases. The procedure for designthem is the same for both, except when the density is appreciably increased.
Turbopumps and -blowers are radial-flow, axial-flow, or a combination of the
ing
Fixed
blades
Fig. 9.19
Axial-flow
soll-Rand Co.)
pump.
(Inger-
m/s
APPLICATIONS OF FLUID MECHANICS
524
Fixed
blades
Fig.
9.20
pump.
A
mixed-flow
(Ingersoll-Rand Co.)
two, called mixed-flow.
For high heads the radial (centrifugal) pump,
quently with two or more stages (two or more impellers in series),
adapted.
For large flows under small heads the axial-flow
(Fig. 9.19) is best suited.
The mixed-flow pump
pump
is
fre-
best
or blower
(Fig. 9.20) is used for
medium
head and medium discharge.
The equations developed in Sec. 9.2 apply just as well to pumps and
The usual centrifugal pump has a suction, or inlet,
blowers as to turbines.
Fig. 9.21
Velocity relationships for flow through a
centrifugal-pump impeller.
TURBOMACHINERY
525
Sectional elevation of Eagle Mountain and Haypumps, Colorado River Aqueduct. (Worthington
Fig. 9.22
field
Corp.)
pipe leading to the center of the impeller, a radial outward-flow runner,
as in Fig. 9.21,
and a
to the discharge pipe.
collection pipe or spiral casing that guides the fluid
Ordinarily,
stage units in which the flow
is less
is
no
fixed
vanes are used, except for multi-
and the additional
relatively small
fluid friction
than the additional gain in conversion of kinetic energy to pressure
energy upon leaving the impeller.
Centrifugal
Fig. 9.23
Corp.)
Impeller types used
///a Mixflo
in
pumps and
Propeller
rotation
blowers. (Worthington
APPLICATIONS OF FLUID MECHANICS
526
1,000
800 Positive
displace
600
500 .merit S
pumps
400
300
.
N.
80
60
50
40
E
T,
c8
00
E
\
\,
*
-
S
\3
j\
\ \
^-X
\\
ft''\.
\\
\.
s
\
X
c
jf
Wff
eep
v /ell turbin e pi jmps,
\
w
f
Nv
\
>
v
^
s
30
\
*s
\
\\
s
:
s
<
\
20
-
rf*
Vjvfc
3oN .)x$wm
-
<
c
\.*6
\W4.%*\
V
)i
ifyvftfflfflfy
i i
Wk
mSr\
J
6
\
m
w
t
3
2
4
\ X
Propeller
pumps
^
V
\-
3
2
10
4
6
;
2
3
4
6
10'
U. S. gallons per
Fig. 9.24
\
\_
6
10'
10
\
.
-Jk
4
IIIH
5L
41
3
v
\
,
//f
^
2
m: <ed flow
pu mps
<MW%
\,/
\\
s
?
^s^p
S
\ WW*
gS I ''^''''tyflfuffiy'
jj
10
•*
k
V
v^
S
s
\
Vffl,
\y
-^
<&\
V
200
a>
« $Wffl(ffl%*
>v
.
10
:
minute
Chart for selection of type of pump. (Fairbanks, Morse
&
Co.)
Figure 9.22 shows a sectional elevation of a large centrifugal pump.
For lower heads and greater discharges (relatively) the impellers vary as
shown in Fig. 9.23, from high head at left to low head at right with the axialflow impeller. The specific speed increases from left to right. A chart for
determining the types of pump for best efficiency
is
given in Fig. 9.24 for water.
Centrifugal and mixed-flow pumps are designed in the specific speed
range 500 to 6500 and axial pumps from 5000 to 11,000; speed is expressed in
revolutions per minute, discharge in gallons per minute, and head in feet.
Characteristic curves showing head, efficiency,
and brake horsepower as
a function of discharge for a typical centrifugal pump with backward-curved
vanes are given in Fig. 9.25. Pumps are not so efficient as turbines, in general,
owing to the inherently high
losses that result
from conversion of kinetic
energy into flow energy.
Theoretical head-discharge curve
A
theoretical head-discharge curve
and the vector diagrams
V
2
COS
C* 2
= V U2 =
U2
of Fig. 9.8.
— V
rz
COt ft
may
be obtained by use of Eq. (9.3.8)
the exit diagram of Fig. 9.8
From
TURBOMACHINERY
527
1
100
He ad
£ 80
o
*>
*a3
%
60
30
^
-o
M
40
^V
.0)
40
^2
52S
I
20 g
O
10
20
CD
n
200
400
600
800
1000
1200
1400
1600
1800
2000
Gallons per minute
Fig. 9.25
Characteristic curves for typical centrifugal
pump;
10-in
impeller, 1750 rpm. (Ingersoll-Rand Co.)
From
is
the discharge,
if 62 is
the width of the impeller at
r2
and vane thickness
neglected,
Q =
2irr2b2V T 2
Eliminating
V
r2
and substituting these
last
two equations into Eq.
(9.3.8)
give
XI
=
U22
u 2 Q cot
For a given
The usual
&
(9.6.1)
27TT2&20
g
pump and
speed,
H varies linearly with Q,
design of centrifugal
/?2
pump
>90°
H
Q
Fig.
9.26
Theoretical
charge curves.
head-dis-
has
&<
90°,
as
shown
in Fig. 9.26.
which gives decreasing
APPLICATIONS OF FLUID MECHANICS
528
Fig. 9.27
Effect of circulatory flow.
head with increasing discharge. For blades radial at the exit, /3 2 = 90° and
is independent of discharge.
For blades curved forward,
02 > 90°, and the head rises with discharge.
the theoretical head
Actual head-discharge curve
By
subtracting head losses from the theoretical head-discharge curve, the
actual head-discharge curve
is
is
obtained.
not an actual loss but a failure of the
the relative velocity with angle
(infinite
number
had an angle
f
&
The most important subtraction
number of blades to impart
finite
of the blades.
of blades) the fluid actually
is
Without perfect guidance
discharged as
if
the blades
than fa (Fig. 9.27) for the same discharge. This
inability of the blades to impart proper guidance reduces V u i and hence
decreases the actual head produced. This is called circulatory flow and is
shown in Fig. 9.28. Fluid friction in flow through the fixed and moving passages
f3 2
which
is less
causes losses that are proportional to the square of the discharge.
shown
in Fig. 9.28.
The
final
head
They
are
loss to consider is that of turbulence, the
due to improper relative-velocity angle at the blade inlet. The pump can
be designed for one discharge (at a given speed) at which the relative velocity
is tangent to the blade at the inlet.
This is the point of best efficiency, and
shock or turbulence losses are negligible. For other discharges the loss varies
loss
HSS2£*A
head
lire
Turbulent
Fig. 9.28
ships.
Head-discharge
relation-
TURBOMACHINERY
529
about as the square of the discrepancy in approach angle, as shown in Fig. 9.28.
The final lower line then represents the actual head-discharge curve. Shutoff
head is usually about u 22 /2g, or half of the theoretical shutoff head.
In addition to the head losses and reductions, pumps and blowers have
torque losses due to bearing- and packing-friction and disk-friction losses
from the fluid between the moving impeller and housing. Internal leakage is
also an important power loss, in that fluid which has passed through the
impeller, with its energy increased, escapes through clearances and flows
back to the suction side of the impeller.
EXAMPLE 9.8 A centrifugal water pump has an impeller (Fig. 9.21) with
= 12 in, n = 4 in, fa = 20°, fa = 10°. The impeller is 2 in wide at r = n
and f in wide at r = r 2 For 1800 rpm, neglecting losses and vane thickness,
determine (a) the discharge for shockless entrance when a\ = 90°; (6) a 2 and
r2
.
H;
the theoretical head
(c)
the horsepower required; and (d) the pressure rise
through the impeller.
The
(a)
wi
HW*)
=
peripheral speeds are
=
(I)
The vector diagrams
u 2 = 3wi = 188.5
62.8 ft/s
are
shown
known, the entrance diagram
hence
Q =
22.85(x) (J)
At the
(b)
7,2
=
7.97
2tt
By
X
X
12
(A) =
determined, Vi
u\
and the angles a h
U\ tan 20°
V
22.85 ft/s;
it,
the vector
is
r2
20.3 ft/s
line distance
-v, 2
a2
r2
from
=20.3
\^^
*— K2-73.5*—
F
—
u 2 = 188.5
>-
Exit
Fig. 9.29
fa
=
0.75
drawing w 2 (Fig. 9.29) and a parallel
Entrance
=
7.97 cfs
exit the radial velocity
=
With
in Fig. 9.29.
is
ft/s
Vector diagrams for entrance and exit of
pump
impeller.
—
530
APPLICATIONS OF FLUID MECHANICS
triangle
is
=
v u2
a2
=
determined when
=
20.3 cot 10°
20 3
—
-
tan- 1
=
/3 2
laid off.
is
Vu2 =
115 ft/s
V =
15°26'
2
Thus
188.5
-
115
20.3 esc 15°26'
=
=
73.5 ft/s
76.2 ft/s
io.O
From Eq.
ti
—
(9.3.8)
u2 V2 cos a 2
=
u 2 V u2
9
,
hp =
By
(d)
—X
188.5
73.5
=
430
ft
32.2
Q
_
.
(c)
=
QyH =
W
7.97
X
62.4
X
430
=
tE6
ooo
388
applying the energy equation from entrance to exit of the
impeller, including the energy
H
added (elevation change across impeller
is
neglected),
v_i
H+
29
T
2g
T
V2
y
and
Vi
-
Vi
=
430
+
7
22.85 2
-
64.4
76.2 2
348
ft
64.4
or
2> 2
9.7
-
Pi
=
348
X
0.433
=
151 psi
CENTRIFUGAL COMPRESSORS
Centrifugal compressors operate according to the
same
principles as turbo-
machines for liquids. It is important for the fluid to enter the impeller without
shock, i.e., with the relative velocity tangent to the blade. Work is done on
the gas by rotation of the vanes, the moment-of-momentum equation relating
torque to production of tangential velocity. At the impeller exit the highvelocity gas must have its kinetic energy converted in part to flow energy by
suitable expanding flow passages.
the gas) the actual work
For adiabatic compression (no cooling of
wa of compression per unit mass is compared with the
TURBOMACHINERY
531
work
wth per unit mass to compress the gas to the same pressure isentropically.
For cooled compressors the work w h is based on the isothermal work of comt
pression to the
v
same pressure
Hence
as the actual case.
—
W
=
(9.7.1)
a
is
the formula for efficiency of a compressor.
The
formula for compression of a perfect gas
efficiency
developed for
is
the adiabatic compressor, assuming no internal leakage in the machine,
no short-circuiting of high-pressure
impeller.
up
T th
back to the low-pressure end
i.e.,
of the
Centrifugal compressors are usually multistage, with pressure ratios
From
to 3 across a single stage.
with
(9.3.2)
fluid
the
moment-of-momentum equation
inlet absolute velocity radial, a\
=
90°, the theoretical torque
is
mV u2 r
T th =
(9.7.2)
2
in which m is the mass per unit time being compressed, V u2
component of the absolute velocity leaving the impeller, and
radius at exit.
The
actual applied torque
Ta
is
is
the tangential
r 2 is
the impeller
greater than the theoretical
torque by the torque losses due to bearing and packing friction plus disk
hence
friction;
T th = Talm
(9.7.3)
the mechanical efficiency of the compressor.
In addition to the torque losses, there are irreversibilities due to flow
through the machine. The actual work of compression through the adiabatic
machine is obtained from the steady-flow energy equation (3.7.1), neglecting
elevation changes and replacing u + p/p by h,
if
r]
m
is
y _
-=—
2
-w
a
The
=
isentropic
y*
+
work
h2
-h
(9.7.4)
of compression can
be obtained from Eq.
(3.7.1)
in
differential form, neglecting the z terms,
-dwa = VdV
V
+ d- + du
=
VdV +
p
The
last
two terms are equal to
T ds
dp
1
P
P
— + pd- + du
from Eq.
(3.7.6)
which
is
zero for isen-
APPLICATIONS OF FLUID MECHANICS
532
tropic flow, so that
—
dv
-dwa = VdV +
(9.7.5)
p
By
=
integrating for p/p k
— Wth =
J?
const between sections
k
+
k
-
_ 72
2
(iL _
(A
-w _
_
77
may now
efficiency
th
(V 2tt?
~
~"
-wa
since A
=
-Wa =
—m —m
c p T.
2,
Pi)
~ l)/fc
-
t-'+W®
The
and
Pi)
1 \p2th
r/M
1
i
(97
1
-
6)
]
be written
- VV/2 + c,ri[(p,/pi)<^>/» (TV - W/2 + c p (T2a - 7\)
1]
{
•
}
In terms of Eqs. (9.7.2) and (9.7.3)
=
7-
T)
=
=
m
Tim
(9.7.8)
7)
m
then
Use
of this equation
is
made
in the following example.
EXAMPLE
9.9 An adiabatic turbocompressor has blades that are radial at the
15-cm-diameter impeller. It is compressing 0.5 kg/s air at 1 kg//cm 2
abs, t = 15°C, to 3 kg//cm 2 abs. The entrance area is 60 cm2 and the exit
area is 35 cm2 77 = 0.75; t] m = 0.90. Determine the rotational speed of the
exit of its
,
,
impeller and the actual temperature of air at the exit.
The density
Pl
=
pi
RTr
at the inlet
9.806
X
10*
is
N/m
2
(287J/kg.K)(273+15K)
= L186 kg/m
TURBOMACHINERY
and the velocity at the entrance
m
"
1
is
0.5 kg/s
_
~
_
(0.006
m
The
theoretical density at the exit
is
P2th
=
Pl4i
=
(-)
pi
kg/m
(1.186
1.186
3
)
X
31'1
-
4
=
and the theoretical velocity at the
V" h =
m
-^A t
0.5
2.60
X
For radial vanes at the
V
=
==
/S
2
)
2.60
kg/m3
exit is
=
54
Vu =
u%
'
945m/S
0.0035
exit,
i
=
cor 2 .
From Eq.
(9.7.9)
LXPi
l
—
533
0.90 T
I
(0.24
X
4187) (273
+
0AI1A
15) (S
-
+
1)
54.945 2
-
70.26 2 !
I
and u 2 = 359.56 m/s. Then
u2
359.56
r2
0.075
= 4794
rad/s
and
N
= „(^) =
The
w 22
.
wa =
It is
—
X
theoretical
-w
w th =
7]
4794
tft
=
— = 45,781 rpm
work Wth is the term in the brackets in the expression
X 10 6 m-N/kg. Then from Eq. (9.7.1)
for
0.1058
X
10 5
1.058
r-zT
0.75
at
liMlx/inl
= -1.411
X 10 m-N/kg
Since the kinetic energy term
5
is
/i
small, Eq. (9.7.4) can
be solved for h 2
—
h\
APPLICATIONS OF FLUID MECHANICS
534
and a
trial solution effected,
-h
h2
As a
=
first
c p (T2a
-
T{)
=
approximation,
1.411
let
X
10 5
V2a = V
2t h
+
——
=
54.945; then
temperature the density at the exit is 2.387 kg/m 3 and the velocity
is 59.85 m/s. Insertion of this value in place of 54.945 reduces the temperature
to T2a = 429.1 K.
For
this
,
CAVITATION
9.8
When
a liquid flows into a region where
pressure,
boils
it
and vapor pockets develop
its
pressure
is
reduced to vapor
The vapor bubbles
in the liquid.
are carried along with the liquid until a region of higher pressure
where they suddenly
collapse.
This process
is
called cavitation.
is
reached,
If the
vapor
bubbles are near to (or in contact with) a solid boundary when they collapse,
the forces exerted by the liquid rushing into the cavities create very high
localized pressures that cause pitting of the solid surface.
The phenomenon
accompanied by noise and vibrations that have been described as similar to
gravel going through a centrifugal pump.
is
In a flowing liquid, the cavitation parameter a
the susceptibility of the system to cavitate.
a
is
useful in characterizing
It is defined
by
-=^
= V
(9.8.1)
V
J
P 7 /2
2
in
which p
is
the absolute pressure at the point of interest, p v
is
the vapor
pressure of the liquid, p is the density of the liquid, and V is the undisturbed,
or reference, velocity. The cavitation parameter is a form of pressure coefficient.
In two geometrically similar systems, they would be equally likely to
same degree of cavitation for the same value of <r.
(7 = 0, the pressure is reduced to vapor pressure and boiling should
cavitate or would have the
When
occur.
Tests
made on
chemically pure liquids show that they will sustain high
tensile stresses, of the order of
thousands of pounds per square inch, which
TURBOMACHINERY
is
in contradiction to the concept of cavities forming
to vapor pressure.
pressure
Since there
is
when
reached with commercial or technical liquids,
is
pressure
generally spontaneous boiling
it is
is
535
reduced
when vapor
generally ac-
cepted that nuclei must be present around which the vapor bubbles form and
grow. The nature of the nuclei is not thoroughly understood, but they may
be microscopic dust particles or other contaminants, which are widely dispersed through technical liquids.
Cavitation bubbles may form on nuclei, grow, then move into an area of
higher pressure and collapse, all in a few thousandths of a second in flow
within a turbomachine. In aerated water the bubbles have been photographed
as they move through several oscillations, but this phenomenon does not
seem to occur in nonaerated liquids. Surface tension of the vapor bubbles
appears to be an important property accounting for the high-pressure pulses
resulting from collapse of a vapor bubble.
Recent experiments indicate
pressures of the order of 200,000 psi based on the analysis of strain waves in a
photoelastic specimen exposed to cavitation. 1 Pressures of this magnitude
appear to be reasonable, in line with the observed mechanical damage caused
by
cavitation.
The formation and
collapse of great
numbers
of bubbles
subject that surface to intense local stressing, which appears to
surface
by
fatigue.
Some
on a surface
damage the
ductile materials withstand battering for a period,
called the incubation period, before
damage is noticeable, while brittle materials
may lose weight immediately. There may be certain electrochemical,
and thermal
corrosive,
which hasten the deterioration of exposed surfaces. Rheingans 2 has collected a series of measurements made by magnetostrictionoscillator tests, showing weight losses of various metals used in hydraulic
machines (see Table 9.2).
Protection against cavitation should start with the hydraulic design of
the system in order to avoid the low pressures if practicable. Otherwise, use
of special cavitation-resistant materials or coatings may be effective. Small
amounts of air entrained into water systems have markedly reduced cavitation
damage, and recent studies indicate that cathodic protection is helpful.
The formation of vapor cavities decreases the useful channel space for
liquid and thus decreases the efficiency of a fluid machine. Cavitation causes
effects
three undesirable conditions: lowered efficiency,
damage
to flow passages,
and noise and vibrations. Curved vanes are particularly susceptible to cavitation on their convex sides and may have localized areas where cavitation causes
pitting or failure. Since all turbomachinery and ship propellers and many
1
2
G. W. Sutton, A Photoelastic Study of Strain Waves Caused by Cavitation, J. Appl.
Mech., vol. 24, pt. 3, pp. 340-348, 1957.
W. J. Rheingans, Selecting Materials to Avoid Cavitation Damage, Mater. Des. Eng. t
pp. 102-106, 1958.
APPLICATIONS OF FLUID MECHANICS
536
Table 9.2
machines
Weight
loss in materials
used
in
hydraulic
Weight
Alloy
Rolled
7%
layers,
17%
Ni)
6.0
Hot-rolled stainless steel
(26%
Cr,
13% Ni)
Tempered, rolled stainless steel (12% Cr)
8.0
9.0
13.0
20.0
80.0
97.0
98.0
105.0
124.0
156.0
224.0
Cast stainless steel (18% Cr, 8% Ni)
Cast stainless steel (12% Cr)
Cast manganese bronze
Welded mild steel
Plate steel
Cast
steel
Aluminum
Brass
Cast iron
f
X
§
mg
0.6
3.2
5.8
stellitef
Welded aluminum bronzej
Cast aluminum bronze§
Welded stainless steel (two
Cr,
loss
after 2 h,
This material is not suitable for ordinary use, in spite of
its high resistance, because of its high cost and difficulty
in machining.
Ampco-Trode 200: 83 % Cu, 10.3% Al, 5.8%
Ampco 20: 83.1% Cu, 12.4% Al, 4.1% Fe.
Fe.
hydraulic structures are subject to cavitation, special attention must be given
to
it
in their design.
A cavitation
and
index
a' is useful in
the proper selection of turbomachinery
The
in its location with respect to suction or tail-water elevation.
minimum
pressure in a
pump
or turbine generally occurs along the convex
side of vanes near the suction side of the impeller.
point of
minimum
downstream
In Fig. 9.30,
if e is
the
between e and the
between the two points, may be
pressure, Bernoulli's equation applied
liquid surface, neglecting losses
written
7
2g
in
which p a
y
is
the atmospheric pressure and p e the absolute pressure. For
the
e, the pressure must be equal to or less than p v
cavitation to occur at
,
TURBOMACHINERY
Turbine or
Fig. 9.30
vapor pressure.
a
,
V
=
2
Pa
e
If
-
= p
setting,
v,
- jH
yH
Pv
2gH
is
pe
pump
s
(9.8.2)
the ratio of energy available at
The
kinetic energy.
537
ratio a'
is
e
to total energy
H
,
since the only energy
a cavitation index or number.
The
is
critical
may be determined by a test on a model of the homologous series.
For cavitationless performance, the suction setting s for an impeller installation must be so fixed that the resulting value of a is greater than that of o- c
value a c
H
.
indicate a <j c = 0.10. A homologous
where p a = 13 psi and p v = 0.50 psi
and is to pump water against a head of 80 ft. What is the maximum permissible
suction head?
Solving Eq. (9.8.2) for H s and substituting the values of cr c H, p a
and p v give
EXAMPLE
unit
9.10
Tests on a
pump model
to be installed at a location
is
,
H> _
Pj^j^ _
a
,
_ 13-050 _
H
0.433
7
The
less
the value of
H
s,
0WX8O=
2Q8
,
ft
the greater the value of the plant
and the greater
a',
the assurance against cavitation.
The
tion of
net positive suction head
minimum
(NPSH)
is
frequently used in the specifica-
suction conditions for a turbomachine.
NPSH =
^=
A
run on the machine to determine the
test is
Va
-
Pv
-
It is
yHs
(9.8.3)
maximum
value of
H
s
for
APPLICATIONS OF FLUID MECHANICS
538
operation of the machine with no impairment of efficiency and without objec-
Then, from this test NPSH is calculated from Eq.
machine where the suction lift is less than 8
as found from Eq. (9.8.3), is then acceptable. Note that
8 is positive when
the suction reservoir is below the turbomachine, as in Fig. 9.30.
A suction specific speed S for homologous units may be formulated. By
tionable noise or damage.
Any
(9.8.3).
H
setting of this
,
H
elimination of
V
NPSH = -1
2
e
D
~ Q^-
2g
S
is
two equations
in the
e
2
D
2
Q =
-^-
ND
A
const
z
e
obtained,
N VQ
S =
(NPSH)^
When
-
different units of a series are operating
equal values of
is
(9 8 4)
S
under cavitating conditions,
indicate a similar degree of cavitation.
not present, the equation
is
-
When
cavitation
not valid.
PROBLEMS
By
9.1
9.2
A
pump
How
centrifugal
pump
load increases.
is
P = yQH
use of Eqs. (9.1.1) and (9.1.3) together with
the homologous relationship for
is
P
in
for power, develop
terms of speed and diameter.
driven by an induction motor that reduces in speed as the
A test determines several sets of values of N, Q, H for the pump.
a characteristic curve of the
pump
for a constant speed
determined from these
data?
9.3
What
is
the specific speed of the
pump
of
Example
9.1 at its point of best effi-
ciency?
9.4
plot several points from the characteristics of the
are they not exactly on the
9.5
pump
Plot the dimensionless characteristic curve of the
same curve
Determine the
pump
of
Example
size
of Example 9.1. On this
new (52-in) pump. Why
same curve?
and synchronous speed
9.1 that will
produce 3
of a
m /s at
3
100
pump homologous
m head
to the 72-in
at its point of best
effi-
ciency.
9.6
Develop the characteristic curve for a homologous
and 1800 rpm.
pump
of the series of
Example
9.1 for 18-in diameter discharge
9.7
A pump with a 20-cm-diameter impeller discharges
head at
its
point of best efficiency.
What
is its
specific
100 1/s at 1140
speed?
rpm and
10
m
TURBOMACHINERY
539
9.8 A hydroelectric site has a head of 300 ft and an average discharge of 400 cfs.
For a generator speed of 200 rpm, what specific speed turbine is needed? Assume an
efficiency of 92 percent.
N
A model turbine, s = 36, with a 14-in-diameter impeller develops 27 hp at a
head of 44 ft and an efficiency of 86 percent. What are the discharge and speed of the
model?
9.9
What size and synchronous speed of homologous unit of Prob. 9.9 would be
needed to discharge 600 cfs at 260 ft of head?
9.10
22
9.11
m /s
3
component
in
water flowing through the fixed vanes of a turbine has a tangential
m/s at a radius of 1.25 m. The impeller, turning at 180 rpm, discharges
of 2
What
an axial direction.
torque
In Prob. 9.11, neglecting
9.12
exerted on the impeller?
is
losses,
what
is
the head on the turbine?
A generator with speed N = 240 rpm is to be used with a turbine at a site where
m and Q = 8 m /s. Neglecting losses, what tangential component must be
given to the water at r = 1 m by the fixed vanes? What torque is exerted on the impeller? How much horsepower is produced?
9.13
H=
A
9.14
3
120
site for
a Pelton wheel has a steady flow of 2 cfs with a nozzle velocity of 240
With a blade angle
ft/s.
of 174°,
diameter of wheel, (b) the speed,
and C v
(c)
=
0.98, for
60-Hz power, determine (a) the
the horsepower, (d) the energy remaining in the
water. Neglect losses.
An
9.15
impulse wheel
120 m, and
e
=
Q =
75
is
to be used to develop 50-Hz
Determine the diameter
1/s.
of the
site
where
H=
speed.
C =
0.97;
power at a
wheel and
its
v
82 percent.
At what angle should the wicket gates of a turbine be set to extract 12,000 hp
from a flow of 900 cfs? The diameter of the opening just inside the wicket gates is 12 ft,
and the height is 3 ft. The turbine runs at 200 rpm, and flow leaves the runner in an
9.16
axial direction.
For a given setting of wicket gates
with the discharge?
9.17
9.18
Assuming constant
how
axial velocity just
does the
moment
of
momentum vary
above the runner of the propeller turbine
components if the hub radius is 1 ft
of Prob. 9.16, calculate the tangential-velocity
and the outer radius
is
3
ft.
Determine the vane angles ft, fa for entrance and exit from the propeller turbine
no angular momentum remains in the flow. (Compute the angles
inner radius, outer radius, and midpoint.)
9.19
of Prob. 9.18 so that
for
9.20
Neglecting
9.21
The hydraulic
80 m.
What
is
losses,
what
is
the head on the turbine of Prob. 9.16?
efficiency of a turbine is 95 percent,
the actual head required?
and
its
theoretical
head
is
APPLICATIONS OF FLUID MECHANICS
540
©
Fig. 9.31
A
9.22
turbine model test with 25-cm-diameter impeller showed an efficiency of 90
What
percent.
A
9.23
efficiency could be expected
turbine draft tube (Fig. 9.31) expands from 6 to 18
the velocity
is
ft
diameter.
At
section
1
and barometric pressure of 32 ft of
incipient cavitation (pressure equal to vapor pressure at
30 ft/s for vapor pressure of
Determine h s
water.
from a 120-cm-diameter impeller?
for
1 ft
section 1).
Construct a theoretical head-discharge curve for the following specifications of a
= 1200 rpm,
pump: ri = 5 cm, r2 = 10 cm, bi = 2.5 cm, 6 2 = 2 cm,
9.24
N
centrifugal
and
=
ft
30°.
an impeller r\ = 2.5 in, 61 = If in,
width at n and r 2 respecJ
tively)
Neglect thickness of vanes. For 1800 rpm, calculate (a) the design discharge
for no prerotation of entering fluid, (6) a 2 and the theoretical head at point of best
efficiency, and (c) for hydraulic efficiency of 85 percent and overall efficiency of 78
percent, the actual head produced, losses in foot-pounds per pound, and brake horse-
A
9.25
r2
=
pump
centrifugal water
4.5 in, 6 2
=
in, ft
=
(Fig. 9.21) has
30°, ft
=
45°
(61,
6 2 are impeller
,
.
power.
A
9.26
61
=
centrifugal
5 cm, b 2
=
pump
3 cm, ft
has an impeller with dimensions
=
ft
=
30°.
For a discharge
of
r*i
=
7.5 cm, r 2
vanes compute (a) the speed,
(b) the head, (c) the torque, (d) the
pressure rise across impeller.
Neglect
9.27
3
A
in, b 2
centrifugal water
=
1
in, ft
=
60°
«i
=
15 cm,
power, and
(e)
the
90°.
pump with impeller dimensions r± = 2 in, r2 = 5 in, 61 =
to pump 5 cfs at 64 ft head. Determine (a) ft, (6) the
is
and (d) the pressure rise across the impeller.
and assume no shock at the entrance. a\ = 90°.
speed, (c) the horsepower,
losses,
losses.
=
55 1/s and shockless entry to
Neglect
Select values of r h r2 ft, ft, b h and b 2 of a centrifugal impeller to take 30 1/s
water from a 10-cm-diameter suction line and increase its energy by 12 m-N/N.
= 1200 rpm; on = 90°. Neglect losses.
9.28
N
,
TURBOMACHINERY
541
has blade angles ft = ft; h = 262 = 1 in; r x = r2 /3 = 2 in. For a
head of 95.2 ft at a discharge at best efficiency of 1.052 cfs, determine the
blade angles and speed of the pump. Neglect thickness of vanes and assume perfect
guidance. (Hint: Write down every relation you know connecting ft, ft, 61, b2 ri, r2
from the two velocity vector diagrams, and by
U\, U2,
h, Q, Vrz, VU 2, Vi, co, and
substitution reduce to one unknown.)
A pump
9.29
theoretical
,
H
,
N
t
A mercury-water differential manometer, R' = 65 cm, is connected from the
10-cm-diameter suction pipe to the 8-cm-diameter discharge pipe of a pump. The
9.30
cm below
centerline of the suction pipe
is
calculate the head developed
by the pump.
The
9.31
30
impeller for a blower (Fig. 9.32)
turns at 1200 rpm. For 10,000
exit blade angles (ai
=
ft
3
/min
air,
7
the discharge pipe. For
is
=
Q —
60 1/s water,
18 in wide. It has straight blades and
0.08 lb/ ft 3 calculate (a) entrance
,
90°), (6) the head produced in inches of water,
and
(c)
and
the
theoretical horsepower required.
An air blower is to be designed to produce pressure of 10 cm H 2 when operating
7 = 11.5 N/m 3 r2 = Lin; ft = ft width of impeller is 10 cm;ai= 90°.
9.32
at 3600 rpm.
Find
;
;
ri.
=
9.33
In Prob. 9.32 when ft
9.34
Develop the equation for
Vm
r
9.35
(V 2 th2 —
2
u2Uv\
Vi
2
30°, calculate the discharge in cubic meters per minute.
efficiency of a cooled compressor,
pi
p2 \
pi
pj
Find the rotational speed
in
Example
9.9 for a cooled compressor, using results
of Prob. 9.34, with the actual air temperature at exit 15°C.
9.36
p
=
What
is
the cavitation parameter at a point in flowing water where
2 psia and the velocity
Fig. 9.32
is
40 ft/s?
t
=
68°F,
APPLICATIONS OF FLUID MECHANICS
542
A turbine with <rc = 0.08 is to be installed at a site where = 60 and a water
barometer stands at 8.3 m. What is the maximum permissible impeller setting above
tail water?
H
9.37
Two
9.38
units are homologous
when they
are geometrically similar
(a)
similar streamlines
(b)
the same Reynolds
(c)
the same efficiency
(d)
the same Froude
(e)
none of these answers
The
9.39
following
(a)
H/ND* =
(c)
P/QH =
(d)
N VQ/H =
const;
const;
312
The
9.40
two
specific
Q/N*D2 =
const
H/N D =
2
2
number
number
N
const;
(b)
y/P/Hzl * =
pump
speed of a
Q/D
2
V# =
const;
is
const
(e)
of unit size with unit discharge at unit
of a unit
head
power for unit head
delivers unit discharge at unit head
delivers unit discharge at unit power
requires unit
it
(c)
of such a size that
it
(d)
of such a size that
it
(e)
none of these answers
impulse turbine
(c)
always operates submerged
makes use of a draft tube
is most suited for low-head
(d)
converts pressure head into velocity head throughout the vanes
(e)
operates by initial complete conversion to kinetic energy
(6)
A
9.42
const
none of these answers
denned as the speed
of such a size that
(a)
H/N*D =
const
(b)
An
and have
relationships are necessary for homologous units:
(a)
9.41
m
installations
Pelton 24-in-diameter wheel turns at 400 rpm. Select from the following the
head, in feet, best suited for this wheel:
(a)
9.43
(a)
9.44
7
(b)
A
30
120
(c)
shaft transmits 150
26.2
What
(b)
250
torque
is
(c)
170
(d)
kW at 600
2390
(e)
480
rpm. The torque
(d)
4780
required to give 100 cfs water a
has a tangential velocity of 10 ft/s at a distance of 6
(a)
(e)
116 lb-ft
(b) 1935
none of these answers
lb- ft
(c)
6000
lb- ft
(e)
in
newton-meters
none of these answers
moment
ft
is
of
momentum
so that
it
from the axis?
(d)
11,610 lb-
ft
m
in flowing through
The moment of momentum of water is reduced by 27,100 N
vanes on a shaft turning 400 rpm. The power developed on the shaft is, in kilowatts,
9.45
(a)
(e)
•
181.5
(b) 1134
none of these answers
(c)
10,800
(d)
not determinable; insufficient data
TURBOMACHINERY
momentum
Liquid moving with constant angular
9.46
has a tangential velocity of
from the axis of rotation. The tangential velocity 5
4.0 ft/s 10 ft
543
ft
from the axis
is,
in feet per second,
(a)
2
4
(6)
A
9.47
8
(c)
2750
=
(a)
52.6
W
(e)
9.51
2275
70.7
(d)
105.3
(b)
132
(c)
o)2
=
=
90°;
The
=
Ui cot ft
(d) riVi
(6)
=
a head of 7.5
m
and with an
in kilowatts,
is,
none of these answers
(e)
efficiency of
80 percent, for
rro,~
W
T ro/»
2
P 7 /2
is
pump
(e)
none of these answers
vector diagrams.
Vu2 = u2 — Vr2 cot ft
r2 Vr2
cavitation parameter
2
P 7 /2
none
vi
r2/vv
165
(d)
Select the correct relationship for
(a) ai
9.50
3
100 ft/s,
9.49
(c)
(c)
m /s under
The power developed
The head developed by a pump with hydraulic
V2 = 60 ft/s, a 2 = 45°, a x = 90°, is
9.48
U2
2500
(b)
none of these answers
(e)
reaction-type turbine discharges 34
overall efficiency of 91 percent.
(a)
16
(d)
(e)
defined
( C)
none of these answers
by
T ,o
2 /^
7^ /2
( rf )
pV2/2
of these answers
Cavitation
(a)
high velocity
(d)
low pressure
is
caused by
(6)
(e)
low barometric pressure
low velocity
(c)
high pressure
REFERENCES
Church, A. H.: "Centrifugal Pumps and Blowers," Wiley, New York, 1944.
Daily, J. W.: Hydraulic Machinery, in H. Rouse (ed.), "Engineering Hydraulics,"
Wiley, New York, 1950.
Eisenberg, P., and
M.
P. Tulin: Cavitation, sec. 12 in V. L. Streeter (ed.),
"Handbook
Dynamics," McGraw-Hill, New York, 1961.
Moody, L. F.: Hydraulic Machinery, in C. V. Davis (ed.), "Handbook of Applied
Hydraulics," 2d ed., McGraw-Hill, New York, 1952.
Norrie, D. H.: "An Introduction to Incompressible Flow Machines," American
of Fluid
Elsevier,
New
York, 1963.
Stepanoff, A. J.: "Centrifugal and Axial
Flow Pumps," Wiley,
New
York, 1948.
New York,
Wislicenus, G. F.: "Fluid Mechanics of Turbomachinery," McGraw-Hill,
1947.
10
STEADY CLOSED-CONDUIT FLOW
The
basic procedures for solving problems in incompressible steady flow in
closed conduits are presented in Sec. 5.10, where simple pipe-flow situations
are discussed, including losses due to change in cross section or direction of
The
flow.
great majority of practical problems deal with turbulent flow, and
The
velocity distributions in turbulent pipe flow are discussed in Sec. 5.4.
Darcy-Weisbach equation
is
introduced in Chap. 5 to relate frictional losses
to flow rate in pipes, with the friction factor determined from the
gram.
cial
A number of exponential
and
friction
industrial applications are discussed in this chapter.
hydraulic and energy grade lines in solving problems
ticular applications are developed.
Moody
dia-
formulas commonly used in commeris
The use
of the
reiterated before par-
Complex flow problems
are investigated,
including hydraulic systems that incorporate various different elements such
as pumps and piping networks. The use of the digital computer in anatysis
and design becomes particularly relevant when multielement systems are
being investigated.
10.1
EXPONENTIAL PIPE-FRICTION FORMULAS
Industrial pipe-friction formulas are usually empirical, of the form
%
\
-
in
which hf/L
grade
544
line)
,
Q
is
the head loss per unit length of pipe (slope of the energy
the discharge, and
D
the inside pipe diameter.
The
resistance
STEADY CLOSED-CONDUIT FLOW
R
coefficient
is
a function of pipe roughness only.
R
exponents and coefficient
is
developed, and
In
diameters.
its
is
An
R =
it
normally limited to a range of Reynolds numbers and
range of applicability such an equation is convenient, and
it is
1
—
equation with specified
valid only for the fluid viscosity for which
nomographs are often used to aid problem solution.
The Hazen-Williams formula for flow of water
through pipes is of this form, with R given by
R =
545
at ordinary temperatures
4.727
-—
Qn -
with n
=
1.852,
English units
(10.1.2)
SI units
(10.1.3)
m =
140
4.8704,
and C dependent upon roughness
as follows:
120
extremely smooth, straight pipes; asbestos-cement
very smooth pipes; concrete; new cast iron
wood stave; new welded steel
110
vitrified clay;
100
cast iron after years of use
130
new
riveted steel
95
riveted steel after years of use
60 to 80
old pipes in
bad condition
One can develop a special-purpose formula for a particular application
by using the Darcy-Weisbach equation and friction factors from the Moody
diagram or, alternatively, by using experimental data if available. If P sets
of values of D and Q are available and h f /L values are obtained, either from
an experimental program or by use of the Darcy-Weisbach equation, the reThe
sistance coefficient and exponents in Eq. (10.1.1) can be evaluated.
method of least squares is used to find R, n, and m in Eq. (10.1.1) as follows:
h
f
In— =
,
R+
In
n
In
Q —
m In D
Li
Let Zi
Z = B
{
1
=
In (hfi/L),
+
nXi
B =
In R,
X
{
=
In
Q and
it
F,
=
In
D
t ;
- mYi
H. W. King and E. F. Brater, "Handbook of Hydraulics,"
York, 1954.
then
(10.1.4)
p. 6-11,
McGraw-Hill,
New
:
APPLICATIONS OF FLUID MECHANICS
546
The
Eq. (10.1.4), Z i} is computed from the Darcy-Weisbach equaor it is an experimental value. The right side of the equaevaluated with the corresponding values of Qi and Di and represents
left side of
Di and Q iy
tion at
tion
is
the value of In (h f /L) as calculated from Eq. (10.1.1).
and
of R, n,
left
and
The unknown values
m that provide the best match between the P sets of values on the
right side of Eq. (10.1.4) are determined
by minimizing the sum
of
the squares of the differences
p
p
££
F =
The value
which
F „
—
=
dn
„
a function of the three
is
unknowns B,
n,
and
m for the
be minimized when
set, will
<>
-t:
(10.1.5)
{
of F,
given data
dF
=
dB
Z(Zi- B -nX + mYiY
=
2
dF
—
=
dm
These three conditions provide the three principal equations
- PB - n^Xi + mZYi =
2Zi
ZZiXi
- BXXi - nXXi + mXXiYi =
ZZiYi
- BZYi - nZXiYi + mXY'* =
in
2
which the indices on the summation symbol have been dropped but are
A
understood.
simultaneous solution of the three equations provides the
values of B, n, and m, and since
R =
e
B Eq.
(10.1.1)
,
is
determined.
The
fol-
lowing example shows a calculation of the constants in Eq. (10.1.1) by use
of the computer program in Fig. 10.1.
EXAMPLE 10.1 For the Reynolds number range, RMIN = 50,000 to
= 106 diameter range DMIN = 0.667 ft to DMAX = 2.5 ft, and
for pipe roughness, EPS = 0.0005 ft, and kinematic viscosity, VNU =
RMAX
,
X
1.217
10~ 5
Four
2
ft /s,
determine R,
(KDIV =
the limiting values of
n,
and
m in Eq.
4) equally spaced values of
R and D. Then
(10.1.1).
R and
of
D are taken within
since
Q = -DRv
4
16 sets
(PP =
16) of data,
D
and
Q, are available to yield 16 values of/,
by
STEADY CLOSED-CONDUIT FLOW
1
547
DIMENSICN 0(6) ,RE(6>
NAMEL 1ST/DIN/VNU,EPS,K01V ,CrtAX,DN IN, RUN, &M AX , G , EK , EN ,R
READ(5,DIN>
00 1 I*1,KDIV
D(I)=OMIN+(I-l )*(DMAX-OMIN)/ IKDIV-1)
RE( I)=RPIN*< I-l)*(RfAX-RHINJ/(KOlV-i)
DATA J,SX,SY,SZ,SZX,$XX,SXY,SZY,SYY/C,.0,.Q,.0,.0,.0,.O,.C,.0/
CUN=l./<.7et><i**2*2*G)
00 2 I=1,K0IV
Y*ALOG(C(I)l
AW=.094*(EPS/C 1))* « .22 5
BW=88.*<EPS/C<1) )**.««
C*=l.62*(EPS/CU >*«.134
00 2 Il=l,K0IV
1
)
J=J*1
F=AW+Brf*RE II )**(-Cfc
Q=.7854*C(l )*RE( UMVNU
HL=F*CUN*G*»2/C( I)**5
Z=ALUG(HL)
X~ALOG(C)
SX=SX+X
sz=sz*z
SY=SY*Y
SZX=SZX*Z*X
(
2
)
SXX=SXX*X*X
SXY=SXY*X*Y
SZY=SZY*Z*V
SYY=SYY*Y*Y
PP=KDI V*KOIV
Gl=SXX/SX-SX/PP
C*=SXY/SY-SXX/SX
C5=SXY/SX-StY/SY
C6=SZX/SX-SZY/SY
EM=(C6/C4-(SZ/PP-SZX/SX)/Cl>/USY/PP-<>Y/SX)/Cl-C5/C4)
EN=-(C6*C5*EM) /C4
R=EXP( SZ-SX*EN*SY*EM)/PP)
WRITE(6,UIN)
(
END
COIN VNJ=l.2l7E-b,EPS=.0CC5,KDI
RHIN = .5E5,G=32. 174 &tNC
Computer
Fig. 10.1
Example
V=
<t ,
program
0MAX = 2.
5
,
DM IN= . 66
7,
exponential
for
RM AX=
1
E6
,
equation,
10.1.
Wood
approximation for friction factors, and 16 values of h//L
from the Darcy-Weisbach equation. The computed results from the program
use of the
1
are
R =
0.000542
If
n = 1.908
an equation
in all input
is
wanted
m =
5.041
in SI units, it
data to the program.
is
only necessary to use SI units
These data include kinematic viscosity,
diameters, absolute pipe roughness, and gravitational acceleration.
10.2
HYDRAULIC AND ENERGY GRADE LINES
The concepts
of hydraulic
complex flow problems.
1
See p. 302.
and energy grade
If,
more
term p/y
lines are useful in analyzing
at each point along a pipe system, the
APPLICATIONS OF FLUID MECHANICS
548
determined and plotted as a vertical distance above the center of the pipe,
is the hydraulic grade line.
More generally, the plot
of the two terms
is
the locus of end points
V
7
+z
,
for the flow, as ordinates, against length along the pipe as abscissas, produces
the hydraulic grade
which liquid would
ings in the line.
is
line.
The hydraulic grade
When
the pressure in the line
negative and the hydraulic grade line
The energy grade
line is the locus of heights to
tubes connected to piezometer open-
rise in vertical glass
is
is less
below the
than atmospheric, p/y
pipeline.
line is a line joining a series of points
available energy in foot-pounds per
pound
for
marking the
each point along the pipe as
ordinate, plotted against distance along the pipe as the abscissa.
It consists
of the plot of
V
p
+-+
2
for
*
y
2g
each point along the line. By definition, the energy grade line is always
above the hydraulic grade line a distance of V 2 /2g, neglecting the
vertically
kinetic-energy correction factor.
The hydraulic and energy grade
lines are
shown
pipeline containing a square-edged entrance, a valve,
To
of the line.
necessary
including
head
when the
and a nozzle
reservoir surface
is
at the
end
given,
it is
to apply the energy equation from the reservoir to the exit,
minor losses as well as pipe friction, and to solve for the velocity
Then, to find the elevation of hydraulic grade line at any point,
first
all
V /2g.
2
construct these lines
in Fig. 10.2 for a simple
120
ft
6-in.-diam pipe
B^C
K=10
Arbitrary
/= 0.020
datum
-?
Loss due to nozzle 0.10
Fig. 10.2
Hydraulic and energy grade lines.
-3-in.-diam
nozzle
2g
STEADY CLOSED-CONDUIT FLOW
549
is applied from the reservoir to that point, including all
between the two points. The equation is solved for p/y + z, which is
plotted above the arbitrary datum. To find the energy grade line at the same
point the equation is solved for V2 /2g + p/y + z, which is plotted above the
arbitrary datum.
The reservoir surface is the hydraulic grade line and is also the energy
the energy equation
losses
grade
F
At the square-edged entrance the energy grade
line.
drops by
line
/2g because of the loss there, and the hydraulic grade line drops 1.5V 2 /2g.
This is made obvious by applying the energy equation between the reservoir
0.5
2
surface
and a point
just
V
—
+
V
2
H+ +
=
downstream from the pipe entrance:
2
2g
Solving for
Z
+
z
+
2
p
—
+ -+0.5
2g
y
p/y,
V
H -1.5-
2
V
P
7
=
2g
The head
shows the drop of 1.5V 2 /2g.
not actually occur at the entrance
diameters of pipe downstream.
EXAMPLE
due to the sudden entrance does
but over a distance of 10 or more
customary to show it at the fitting.
loss
itself,
It is
Determine the elevation of hydraulic and energy grade lines
and E of Fig. 10.2.
If the arbitrary datum is selected as centerline of the pipe, both grade
lines start at elevation 60 ft. First, solving for the velocity head is accomplished by applying the energy equation from the reservoir to E,
10.2
at points A, B, C, D,
60
V
V
200 V
1 V
+ -— + 0.020-— — + 10— +
2
+
+
2g
=
0.50 2g
2g
the continuity equation,
VE =
— (16 +
16
From
60
2
2
2
E
= -^
+ +
+8+
\
10
+
X
V
2
/2g
=
reservoir to
1.66
A
+ +
=
2g
2
2g
47. After simplifying,
0.1)
=
36.1
—
Applying the energy equation
ft.
gives
V
p
—
+^+
2g
7
V
—
2
2
60
V
—
E
0.10
2g
2g
and
2
z
+
0.5
2g
for the portion
from the
APPLICATIONS OF FLUID MECHANICS
550
Hence the hydraulic grade
+
=
z
60
-
— = 60 -
1.5
The energy grade
2
—+
*
A
is
1.5
X
59.17
ft
=
1.66
57.51 ft
2g
a
V
line at
p
7
+
2g
A
line for
+
=
57.51
=
V
p
—
+^+
is
=
1.66
For£,
V
80 V
—
+ 0.02 X — —
2
60
+ +
+
z
0.5
t
2g
2
2
2g
0.5 2g
and
V
+
60
z
-
(1.5
the energy grade line
+
is
=
3.2)1.66
at 52.19
+
52.19
1.66
=
ft
53.85
ft.
Across the valve the hydraulic grade line drops by 10V 2 /2g, or 16.6 ft.
Hence, at C the energy and hydraulic grade lines are at 37.25 ft and 35.59
ft,
respectively.
At point D,
+
T
2g
=
7
+ +
z
[
200 \
ia5
+
\
002 x KIR )
0.50/
V
2
T
2g
and
V
+
= 60 -
z
19.5
with the energy grade
At point
grade line
TV
—
=
E
X
1.66
=
27.6 ft
line at 27.6
+
1.66
the hydraulic grade line
=
is
29.96
ft.
at zero elevation,
and the energy
is
V
—
=
2
16
2#
16
X
1.66
=
26.6
ft
2g
The
hydraulic gradient
is
the slope of the hydraulic grade line
if
the con-
STEADY CLOSED-CONDUIT FLOW
duit
d(z
horizontal; otherwise,
is
551
it is
+ v/y)
dL
The
energy gradient
zontal; otherwise,
d(z
+
y/y
+
is
the slope of the energy grade line
if
the conduit
is
hori-
it is
V*/2g)
dL
In
many
situations involving long pipelines the
minor
losses
may be
may
neglected (when less than 5 percent of the pipe friction losses) or they
be included as equivalent lengths of pipe which are added to actual length in
solving the problem. For these situations the value of the velocity head V 2 /2g
2
is small compared with f(L/D) V /2g and is neglected.
In this special but very
common
case,
when minor effects are neglected,
The single grade line,
the energy and hydraulic grade lines are superposed.
shown
in Fig. 10.3,
is
commonly
referred to as the hydraulic grade line.
No
change in hydraulic grade line is shown for minor losses. For these situations
with long pipelines the hydraulic gradient becomes h f /L, with h f given by the
Darcy-Weisbach equation
hf
= fJ
or
by Eq.
'
(10.2.1)
D2g
(10.1.1).
Flow (except through a pump)
is
always in the direction
of decreasing energy grade line.
Pumps add energy
to the flow, a fact
which
may be
expressed in the
energy equation either by including a negative loss or by stating the energy
per unit weight added as a positive term on the upstream side of the equation.
Fig. 10.3
Hydraulic grade line for long pipelines
where minor losses are neglected or included as
equivalent lengths of pipe.
APPLICATIONS OF FLUID MECHANICS
552
Loss due to bends and
in vertical
friction
section
Energy grade
line
Negative pressure head
because hydraulic grade
If line is below pipe
Loss due to bends and
in vertical
friction
section
Energy
grade line
Hydraulic grade
Fig. 10.4
Hydraulic and energy grade lines for a system with
line'
pump and
siption.
The hydraulic grade
line rises sharply at a
and energy grade
draulic
lines for a
true slope of the grade lines can be
EXAMPLE
percent
is
pump. Figure
system with a
shown only
10.4
pump and
shows the hya siphon.
The
for horizontal lines.
10.3 A pump with a shaft input of 7.5 kW and an efficiency of 70
3
connected in a waterline carrying 0.1
/s. The pump has a 15-cm-
m
diameter suction line and a 12-cm-diameter discharge line. The suction line
enters the pump 1
below the discharge line. For a suction pressure of 70
kN/m 2 calculate the pressure at the discharge flange and the rise in the hy-
m
,
draulic grade line across the
If the
the fluid power added
QyE =
pump.
energy added in meter-newtons per newton
7500
X
0.70
is
symbolized by E,
is
or
E=
7500
Q
1
X
7
9gQ2
=
5.356
m
Applying the energy equation from suction flange to discharge flange
STEADY CLOSED-CONDUIT FLOW
553
gives
1^ +
in
+
=
which the subscripts
From
respectively.
F
5.356
y
2g
01X4
«=oT^
=
5
-
s
^+^+
1
2g
y
and d
refer to the suction
and discharge conditions,
the continuity equation
F
66m/s
oi y
4
<=oT^
=
8
-
84m/s
Solving for p d gives
5.66 2
Va
y
=
and pd = 89.6
(?)
8.84 2
70,000
2~^80^ + ^80^ +
kN/m
2
.
The
£. 9 145 +1
.
7
5 356
'
rise in
-
^^80o "
= 9145
*
hydraulic grade line
-»
9802
m
is
-3.004m
In this example much of the energy was added
and the hydraulic grade line rises only 3.004
of 5.356 m.
in the
form
of kinetic energy,
m for a rise of energy grade line
A
turbine takes energy from the flow and causes a sharp drop in both
the energy and the hydraulic grade lines.
of fluid
10.3
A
may
be treated as a
loss in
lines.
THE SIPHON
closed conduit, arranged as in Fig. 10.5, which
tion higher than its free surface
is
The energy removed per unit weight
computing grade
a siphon.
It
lifts
and then discharges
has certain limitations in
summit s.
Assuming that the siphon flows
its
the liquid to an eleva-
at a lower elevation
performance due to the low presit
sures that occur near the
full, with a continuous liquid column
throughout the siphon, the application of the energy equation for the portion
from 1 to 2 produces the equation
2g^
2g^ J D2g
APPLICATIONS OF FLUID MECHANICS
554
H
Siphon.
Fig. 10.5
K
which
in
the
is
sum
of all the minor-loss coefficients.
Factoring out the
velocity head gives
+ K + f±)
= Yl( 1
H
which
is
(10.3.1)
solved in the same fashion as the simple pipe problems of the
or second type.
With the discharge known, the
ward, but the solution for velocity with
assuming an
The
V
2
H
is
first
straightfor-
H given is a trial solution started by
/.
pressure at the
between
for the portion
=
solution for
1
summit s is found by applying the energy equation
and s after Eq. (10.3.1) is solved. It is
V
2
+ & + * + X'f
f
2g
2g
v*
y
VV
+f~
D
2
2g
which K is the sum of the minor-loss coefficients between the two points
is the length of conduit upstream from s. Solving for the pressure gives
and
f
in
V
V
= -y* - ir
2g
2
ps
—
7
(+-+ 1)
decreases with y 3 and
If the solution of the equation should be a value of p s /y equal to or
which shows that the pressure
V
2
/2g.
less
1
(10.3.2)
is
negative and that
than the vapor pressure 1 of the
liquid,
it
then Eq. (10.3.1)
A
is
not valid be-
liquid boils when its pressure is reduced to its vapor pressure. The vapor pressure is a
function of temperature for a particular liquid. Water has a vapor pressure of 0.203 ft
abs at 212°F (100°C).
abs at 32°F (0°C), and 33.91 ft (10.33 m)
(0.0619 m)
2
2
See Appendix C.
H
H
STEADY CLOSED-CONDUIT FLOW
555
cause the vaporization of portions of the fluid column invalidates the incom-
assumption used in deriving the energy equation.
Although Eq. (10.3.1) is not valid for this case, theoretically there will
be a discharge so long as y s plus the vapor pressure is less than local atmospheric pressure expressed in length of the fluid column. When Eq. (10.3.2)
yields a pressure less than vapor pressure at s, the pressure at s may be taken
as vapor pressure. Then, with this pressure known, Eq. (10.3.2) is solved for
V 2 /2g, and the discharge is obtained therefrom. It is assumed that air does
not enter the siphon at 2 and break at s the vacuum that produces the flow.
pressibility
Practically a siphon does not
summit
work
satisfactorily
when
the pressure in-
vapor pressure. Air and other gases come out
of solution at the low pressures and collect at the summit, thus reducing the
length of the right-hand column of liquid that produces the low pressure at
the summit. Large siphons that operate continuously have vacuum pumps to
remove the gases at the summits.
The lowest pressure may not occur at the summit but somewhere downstream from that point, because friction and minor losses may reduce the
pressure more than the decrease in elevation increases pressure.
tensity at the
EXAMPLE
equal to
10.4
its
is
close to
Neglecting minor losses and considering the length of pipe
horizontal distance, determine the point of
minimum
pressure
in the siphon of Fig. 10.6.
When minor losses are neglected, the kinetic-energy term V
Then the hydraulic grade
neglected also.
two
x
liquid surfaces.
= — 40 m
y
The equation
y
=
0.0414a;
+
4
5.656
is
usually
m
m
y
substitution into y
= mx
and
is,
by
x
=
56.57
=
+
8
m
b,
m
pressure occurs where the distance between hydraulic grade
Hydraulic
grade
Fig. 10.6
/2g
line is a straight line connecting the
Coordinates of two points on the line are
of the line
The minimum
V^^
=
2
line
y
H
'"
7
^**
Siphon connecting two reservoirs.
APPLICATIONS OF FLUID MECHANICS
556
line
To
maximum,
and pipe
is
a
0.0025a: 2
-
0.0414a:
find
minimum
-
5.656
p/y, set d(p/y)/dx
— 5.827 m of fluid flowing.
=
0,
which yields x
=
8.28,
The minimum point occurs where the
and p/y =
slopes of the
pipe and of the hydraulic grade line are equal.
PIPES IN SERIES
10.4
When two
pipes of different sizes or roughnesses are connected so that fluid
flows through one pipe
nected in
series.
A
and then through the
desired for a given discharge or the discharge
trated in Fig. 10.7.
other, they are said to
wanted
for a given
Applying the energy equation from
A
be con-
H may be
typical series-pipe problem, in which the head
H,
is illus-
to B, including all
losses, gives
2g
Dx 2g
2g
^hD
in
which the subscripts refer to the two pipes. The
from pipe 2. With the continuity equation
last
item
at exit
VJV
V
2
is
-
TW
eliminated from the equations, so that
-&>+f+HI)7+f(I;Ht)'
Fig. 10.7
Pipes connected
in
series.
is
2
2g
+
2g
the head loss
STEADY CLOSED-CONDUIT FLOW
557
H
Q
H
Fig. 10.8
Plot of calculated
selected values of Q.
For known lengths and
H
2 /i
which
in
is
^ (d + C
=
ft,
C
2,
+
reduces to
sizes of pipes this
C3/2)
(10.4.1)
C3 are known.
readily computed,
for
With the discharge
and the/'s maj' be looked up
given, the Reynolds
in the
number
Moody diagram. Then
H
H
is found by direct substitution.
With
given, V h /1, /2 are unknowns in
Eq. (10.4.1). By assuming values of /1 and/2 (they may be assumed equal),
a trial V\ is found from which trial Reynolds numbers are determined and
values of /1, /2 looked up. With these new values, a better Vi is computed from
Eq. (10.4.1). Since / varies so slightly with the Reynolds number, the trial
The same procedures apply
solution converges very rapidly.
two pipes
In place of the assumption of /1 and /2 when
tion
may
be utilized in which several values of
corresponding values of
10.8.
for
more than
in series.
H
Q
H is given, a graphical soluassumed
are
in turn,
and the
are calculated and plotted against Q, as in Fig.
By connecting the points with
Q for the given value of H.
a smooth curve,
it is
easy to read off the
proper
EXAMPLE 10.5 In Fig.
= 800 ft, D 2 = 3 ft,
U
10.7,
€2
=
=
0.5,
U
=
0.001
ft, v
=
0.00001
K
e
1000
ft,
A=2
2
ft /s,
and
ft, €1
=
H
20
=
termine the discharge through the system.
From
the energy equation,
2
20
=
Ji- {0.5
+/,
w) +
[i
-
2
(I) ]
2
+/
2(
ir)(i) 4
+
4
(i)
:
0.005
ft.
ft,
De-
APPLICATIONS OF FLUID MECHANICS
558
After simplifying,
—
=
20
-
(1.01
From ei/A =
sumed
/1
=
for the
0.025
+
0.0025,
solving for
Ri
=
+
€2
52.6/2 )
/D 2 =
and
0.00033,
Fig. 5.32 values of
fa are
as-
complete turbulence range,
=
f2
By
500/i
Vh
^^7 -
0.015
R2 =
1,898,000
'
0.00001
=
with these values, V\
^
4 2
:
9.49 ft/s,
3
0.00001
and from Fig. 5.32, fx = 0.025, f2 =
and Q = 9.46tt = 29.8 cfs.
0.016.
=
V =
2
4.21 ft/s,
1,263,000
'
'
By solving for V
x
again,
V =
x
9.46,
Equivalent pipes
Series pipes can be solved by the method of equivalent lengths. Two pipe
systems are said to be equivalent when the same head loss produces the same
discharge in both systems. From Eq. (10.2.1)
Li
;i
fl
and
Q1
2
A(AV4)
=
2
2^
/1L1
8&
A
+g
5
2
for a second pipe
/2L2
8Q 2 2
For the two pipes to be equivalent,
hfi
=
h f2
Qi
= Q2
After equating h fl
f2 L 2
fiLi
D16
"
W
=
h f2 and simplifying,
STEADY CLOSED-CONDUIT FLOW
Solving for
L
2
559
gives
(DA*
= lJj[
u
ft
which determines the length of a second pipe to be equivalent to that of the
For example, to replace 300
first pipe.
of 25-cm pipe with an equivalent
length of 15-cm pipe, the values of fa and fa must be approximated by select-
m
ing a discharge within the range intended for the pipes.
fa
=
Say
fa
=
0.020,
0.018; then
—— —
0.018 \25/
0.020 /15\
300
(
5
)
=
25.9
m
m
m
For these assumed conditions 25.9
of 15-cm pipe is equivalent to 300
25-cm pipe.
Hypothetically two or more pipes composing a system may also be replaced by a pipe which has the same discharge for the same overall head loss.
of
EXAMPLE
pipe
for
1,
ft -
and
Solve Example 10.5 by means of equivalent pipes.
by expressing the minor losses in terms of equivalent lengths,
10.6
First,
0.5
+
[1
-
(flij
=
0.809
L el =
^ ^^
=
=
65
ft
for pipe 2,
*_1
Le2
=
^
=
L£3
2 00ft
0.015
fa
The values of fa, fa are selected for the fully turbulent range as an approximaThe problem is now reduced to 1065 ft of 2-ft pipe and 1000 ft of 3-ft
tion.
pipe.
By
expressing the 3-ft pipe in terms of an equivalent length of 2-ft pipe,
by Eq. (10.4.2)
By
adding to the
2-ft pipe,
of finding the discharge
the problem
through 1065
is
+
reduced to the simple pipe problem
79 = 1144 ft of 2-ft diameter pipe,
APPLICATIONS OF FLUID MECHANICS
560
€
=
0.005
=/
20
for a
ft,
1144
V
2
2g
head
loss of
20
ft,
2
0.025, V = 9.5 ft/s and R = 9.5 X 2/0.00001 =
e/D = 0.0025, / = 0.025 and Q = 9.5tt = 29.9 cfs.
With / =
For
PIPES IN PARALLEL
10.5
A combination of two or more pipes connected as in Fig.
is
1,900,000.
among
divided
the pipes and then
is
joined again,
is
10.9, so
that the flow
a parallel-pipe system.
In series pipes the same fluid flows through all the pipes, and the head losses
are cumulative; however, in parallel pipes the head losses are the same in
any
of the lines,
and the discharges are cumulative.
it is assumed that the minor
In analyzing parallel-pipe systems,
are added into the lengths of each pipe as equivalent lengths.
From
losses
Fig. 10.9
the conditions to be satisfied are
h/i
=
=
h f2
= Pa
h f3
y
Q =
in
Qi
+
Q2
which z A
,
+
zB
(-
zA
(?
+ -)
(10.5.1)
Qz
are elevations of points
A and B, and Q is the discharge through
the approach pipe or the exit pipe.
Two
at
A
types of problems occur: (1) with elevation of hydraulic grade line
known, to find the discharge Q; (2) with Q known, to find the dis-
B
and
and the head loss.
nesses are assumed to be known.
tribution of flow
The
Fig. 10.9
first
type
is,
Sizes of pipe, fluid properties,
and rough-
in effect, the solution of simple pipe problems for dis-
Parallel-pipe system.
STEADY CLOSED-CONDUIT FLOW
561
charge since the head loss
is the drop in hydraulic grade line. These discharges
added to determine the total discharge.
The second type of problem is more complex, as neither the head loss
nor the discharge for any one pipe is known. The recommended procedure is
are
as follows:
1.
Assume a discharge Q[ through pipe
1.
2.
Solve for h'fl using the assumed discharge.
3.
Using
4.
With the three
,
f
Q 2 Qi
h'A , find
Q
given
,
discharges for a
common head
up among the pipes
is split
in the
loss, now assume that the
same proportion as Q[, Q'2 Qi;
,
thus
Qi=
lh *-**
Check the correctness
computed Q h Q 2 Q 3
5.
q
>=§ q
of these discharges
(io 5 2)
-
by computing
-
h/ v h/ 2 h/ 3 for the
,
.
,
This procedure works for any number of pipes.
By
judicious choice of
obtained by estimating the percent of the total flow through the system
that should pass through pipe 1 (based on diameter, length, and roughness)
Q[,
Eq. (10.5.2) produces values that check within a few percent, which
within the range of accuracy of the friction factors.
is
well
EXAMPLE 10.7 In Fig. 10.9, U = 3000 ft, Dx = 1 ft, ei = 0.001 ft; L 2 = 2000
D 2 = 8 in, e 2 = 0.0001 ft; L 3 = 4000 ft, Z> 8 = 16 in, e3 = 0.0008 ft; p = 2.00
slugs/ft v = 0.00003 ft /s, Pa = 80 psi, z A = 100 ft, z B = 80 ft. For a total
ft,
3
2
,
determine flow through each pipe and the pressure at B.
Assume Q[ = 3 cfs; then V[ = 3.82, R[ = 3.82 X 1/0.00003 = 127,000,
flow of 12
cfs,
ei/A = 0.001,/; =
^=
J
0.022
0.022,
and
^ ^-^
3
X
3
14.97 ft
64.4
1.0
For pipe 2
^Ki
2
14
.
97=/'
0.667
Then e 2 /D 2 =
|
X
2<7
0.00015.
1/0.00003
=
Assume/^ = 0.020; then
89,000,/;*
=
0.019,
F2 =
V = 4.01 ft/s, R^ =
Q = 1.44 cfs.
2
4.11 ft/s,
2
4.01
X
APPLICATIONS OF FLUID MECHANICS
562
For pipe 3
VIL2
= Jn,4000
M
14.97
1.333 2g
Then
/D 3 =
es
=
1.333/0.00003
The
=
2Q'
Assume /£ =
0.0006.
178,000, /J
=
total discharge for the
3.00
+
1.44
X
12
+
5.60
=
0.020; then V'z
0.020, Qi
=
5.60
=
4.01 ft/s, R£
=
4.01
X
cfs.
assumed conditions
is
10.04 cfs
Hence
Qi
=
——
3.00
=
3.58 cfs
fv
&
Q =
——
2
10.04
1.44
X
12
=
1.72 cfs
10.04
ah
= 7^7 x
10.04
12
Check the values
=
6 70 cfs
-
of h h h 2 h 3
,
:
o ro
Vi
=
—- =
7J7
V =
2
4.46
Rx
=
152,000
fx
=
0.021
h fl
=
20.4
4.93
R2 =
109,200
f2
=
0.019
h f2
=
21.6 ft
4.80
R3 =
213,000
ft
=
0.019
h ft
=
20.4
4
—72- =
1
7T/9
V =
3
——
=
ft
ft
47T/9
f2 is about midway between 0.018 and 0.019.
would be 20.4 ft.
To
Pa
V zA
7
find
=
pB
Pb
h zB
7
or
p£
y
,
= 80X144
62.4
+
hf
If
0.018 had been selected, h2
STEADY CLOSED-CONDUIT FLOW
which the average head
in
183.5
X
Vb
32.2
=
was taken. Then
81.8 psi
144
BRANCHING PIPES
10.6
A
X
2
loss
563
is shown in Fig. 10.10.
In this situation the
wanted when the reservoir elevations are given.
The sizes and types of pipes and fluid properties are assumed known. The
Darcy-Weisbach equation must be satisfied for each pipe, and the continuity
equation must be satisfied. It takes the form that the flow into the junction
J must just equal the flow out of the junction. Flow must be out of the highest
reservoir and into the lowest; hence, the continuity equation may be either
simple branching-pipe system
flow through each pipe
Qi
If
& + Qz
=
or
is
Qi
+
Q2 =
Qz
the elevation of hydraulic grade line at the junction
of the intermediate reservoir, flow
is
into it
;
but
if
is
above the elevation
the elevation of hydraulic
J is below the intermediate reservoir, the flow is out of it. Minor
be expressed as equivalent lengths and added to the actual lengths
grade line at
losses
may
of pipe.
The
solution
is
equation.
tion,
which
Fig. 10.10
effected
by assuming an
elevation of hydraulic grade line
Qi, Q 2 Q3 and substituting into the continuity
If the flow into the junction is too great, a higher grade-line eleva-
at the junction, then
will
computing
,
reduce the inflow and increase the outflow,
Three interconnected reservoirs.
is
assumed.
APPLICATIONS OF FLUID MECHANICS
564
EXAMPLE
In Fig. 10.10, find the discharges for water at 20°C and with
10.8
the following pipe data and reservoir elevations: Li = 3000 m, Di = 1 m,
= 600 m, D 2 = 0.45 m, e 2 /D 2 = 0.002; L 3 = 1000 m,
€i/A = 0.0002;
Z> 3 = 0.6 m, € 3 /Z) 3 = 0.001; zx = 30 m, u = 18 m, z 3 = 9 m.
U
Assume
3000
V
y
600
V
=
5
2
= f*7^rzir
0.45 2gr
1000
14=/
3
V
Assume
0.278
+
zj
/i
=
0.014
V =
1.75
m/s
ft
=
1.380
m /s
f*
=
°- 024
72 =
1.75
m/s
Q2 =
0.278
m /s
/3
=
0.020
7 =
2.87
m/s
Q =
0.811
m /s
1
3
3
2
3
7tt0.60 2g
-
Then
23 m.
2
so that the inflow
1.380
=
pj/y
2
-j-
7
/i
+
zj
-
is
greater than the outflow
0.811
pj/y
=
3000 7i
—
——
3
=
0.291
24.6 m.
3
3
by
m /s
3
Then
2
5.4
=
/x
600
7
7T7Z IT
0.45 2g
15 6
=
^7
IT
0.60 2g
The
inflow
-
-
pj/y
=
/•
0.015
Vi
=
1.534
m/s
Q =
1.205
m /s
h =
0.024
7 =
2.011
m/s
Q =
0.320
m /s
=
°- 020
F3
=
3 029
m/s
Q3 =
°- 856
mVs
by 0.029
m /s.
1
is still
24.8 m,
3
2
2
=h
6 6
=
/x
2g
1
/»
greater
Q =
l
1.183,
Q2 =
2
3
0.325,
-
2
By extrapolating
Q 3 = 0.862 m /s.
3
linearly, zj
+
3
In pumping from one reservoir to two or more other reservoirs, as in
Fig. 10.11, the characteristics of the
the
pump
runs at constant speed,
suitable procedure
1.
2.
is
pump must be known. Assuming
its
that
head depends upon the discharge.
A
as follows:
Assume a discharge through the pump.
Compute the hydraulic-grade-line elevation
pump.
at the suction side of the
STEADY CLOSED-CONDUIT FLOW
Fig. 10.11
Pumping from one
565
reservoir to
two other reservoirs.
3.
From
the
pump
characteristic curve find the
suction hydraulic grade
Compute drop
head produced and add
it
to
line.
J and determine
in hydraulic grade line to the junction
elevation of hydraulic grade line there.
For
this elevation,
If flow into
is
J
compute flow into
reservoirs 2
equals flow out of J, the problem
too great, assume less flow through the
This procedure
is
easily plotted
and
is
3.
solved.
pump and
If flow into
J
repeat the procedure.
on a graph, so that the intersection of two
elevations vs. flow curves yields the answer.
More complex branching-pipe problems may be solved with a similar
approach by beginning with a trial solution. However, the network-analysis
procedure in the next section is recommended for multibranch systems as
well as for multi-parallel-loop systems. Such problems are most easily handled
with a digital computer.
10.7
NETWORKS OF PIPES
Interconnected pipes through which the flow to a given outlet
several circuits are called a network of pipes, in
through
electric networks.
many ways
may come from
analogous to flow
Problems on these in general are complicated and
require trial solutions in which the elementary circuits are balanced in turn
until all conditions for the flow are satisfied.
The
1.
2.
3.
The
following conditions
must be
satisfied in
a network of pipes
algebraic sum of the pressure drops around each circuit must be zero.
Flow into each junction must equal flow out of the junction.
The Darcy-Weisbach equation, or equivalent exponential friction formula,
APPLICATIONS OF FLUID MECHANICS
566
Pipe network.
Fig. 10.12
must be satisfied for each pipe; i.e., the proper relation between head
loss and discharge must be maintained for each pipe.
The
first
condition states that the pressure drop between any two points
in the circuit, for example,
through the pipe
AG
A
G (Fig. 10.12), must be the same
AFEDG. The second condition is
and
or through
whether
the con-
tinuity equation.
Since
it is
impractical to solve network problems analytically, methods
of successive approximations are utilized.
in
which flows are assumed
A
junction.
for
The Hardy Cross method
one
at every
then computed in turn
each pipe so that continuity
correction to the flow in each circuit
is
1
is
is satisfied
and applied to bring the circuits into closer balance.
Minor losses are included as equivalent lengths in each pipe. Exponential
equations are commonly used, in the form h f = rQ n where r = RL/D m in
,
Eq. (10.1.1) The value of r is a constant in each pipeline (unless the DarcyWeisbach equation is used) and is determined in advance of the loop-balancing
procedure. The corrective term is obtained as follows.
For any pipe in which Q is an assumed initial discharge
.
+ AQ
Q = Q
where Q
is
the correct discharge and
hf
= rQ n = r(Q
If
AQ
1
is
(10.7.1)
+
AQ) n = r(Q
small compared with
Hardy Cross, Analysis
286, November 1936.
of
Flow
in
Q
n
,
AQ
is
the correction.
+ nQo^AQ +
all
terms of the
Networks
•
•
Then
for
each pipe,
•)
series after the
second
of Conduits or Conductors, Univ.
III.
may
Bull.
STEADY CLOSED-CONDUIT FLOW
Now
be dropped.
2h,
in
= 2rQ Q
which
for a circuit,
= 2rQ
1
I""
|
567
\
Q |—
+ AQ 2m
*
Q h" =
1
|
AQ has been taken out of the summation as it is the same for all pipes
and absolute-value signs have been added to account for the
summation around the circuit. The last equation is solved for
in the circuit
direction of
AQ
in
each circuit in the network
2rQo
I
Qo
n_1
l
2rn Qo h"
1
|
When AQ
is
applied to each pipe in a circuit in accordance with Eq. (10.7.1),
the directional sense
tion
1.
is
important;
i.e., it
adds to flows in the clockwise direc-
and subtracts from flows in the counterclockwise direction.
Steps in an arithmetic procedure may be itemized as follows:
Assume the
best distribution of flows that satisfies continuity
by
careful
examination of the network.
2.
Compute the head
= rQ Qn
in each pipe. Compute the net head loss
around each elementary circuit: Xhf = SrQo Qo n_1 (should be zero for
a balanced circuit)
Compute for each circuit Snr Qo n_1
Evaluate the corrective flow AQ in each circuit by use of Eq. (10.7.2).
Compute the revised flows in each pipe by use of Eq. (10.7.1).
Repeat the procedure, beginning with the revised flows, until the desired
accuracy is obtained.
loss hf
1
3.
-
|
4.
5.
6.
|
The values
|
both numerator and denominator; hence, values
proportional to the actual r may be used to find the distribution. Similarly,
the apportionment of flows may be expressed as a percent of the actual flows.
To find a particular head loss, the actual values of r and Q must be used after
the distribution has been determined.
EXAMPLE
is
of r occur in
The
10.9
distribution of flow through the network of Fig. 10.13
desired for the inflows
given the value
and outflows
as given.
For simplicity n has been
2.0.
The assumed distribution is shown in diagram a. At the upper left the
term SrQ Qo n_1 is computed for the lower circuit number 1. Next to the
diagram on the left is the computation of Snr Q n_1 for the same circuit.
The same format is used for the second circuit in the upper right of the figure.
1
|
|
|
APPLICATIONS OF FLUID MECHANICS
568
70
2
35
2
•30
2
x2=
xl=
9800
x4
3600
7425
AQ
57
17
1225
=
1
2x70x2 = 280
2x35x1= 70
2x30x4=240
x2= 6500
2
xl =
289
<<
2x57x2 = 228
)
r
^
=5
20
2x17x1=
2x43x4 = 344
-611
606
\s
43
#
-30
^X
r=4
AQ^Itt-l
20^x5=2000
-17 2 xl
34
43 x4=-7400
2
21
2
2
2
42
x2= 6740
xl=
441
2x21x1=
r=
2x42x4=336
131
610
=2
=5
2x20x5 = 200
2x17x1=
2x30x1=
34
60
294
811
AQ 2 -f£«-3
^
5
~TT
42
x4=-7050
290
30
(b)
^
2x58x2 = 232
= -289
xl=-900
606
58
2x15x5=150
1325
290
AQ,
= -13
=2
2
1125
-1325
590
7425
590
2
15^x5=
-35 2 xl = -1225 2x35x1= 70
-35 2 xl = -1225 2x35x1= 70
1*
Sf
42
r
=
17^x5= 1444
2x17x5 = 170
-21 2 xl'
-
441
2x21x1=
42
-33 2 x2:
-1089
2x33x1=
66
278
-86
—
*-
r=
Fig. 10.13
The
4
30
AQ
^i = -6TT
(c)
Solution of flow distribution
2
86
278
.
a simple network.
in
+
=
20 and
Diagram b gives the distribution
after both circuits have been corrected once. Diagram c shows the values
correct to within about 1 percent of the distribution, which is more accurate
correction for the top horizontal pipe
for the diagonal as 35
+ — 13) —
(
5
=
is
determined as 15
5
17.
than the exponential equations for head loss.
For networks larger than the previous example or for networks that
contain multiple reservoirs, supply pumps, or booster pumps, the Hardy
Cross loop-balancing method may be programmed for numerical solution on
a digital computer. Such a program is provided in the next section.
A number of more general methods 1-3 are available, primarily based upon
1
R.
Epp and A. G. Fowler, Efficient Code for Steady-State Flows
ASCE, vol. 96, no. HY1, pp. 43-56, January 1970.
in
Networks, J. Hydraul.
Div.
2
3
Uri Shamir and C. D. D. Howard, Water Distribution Systems Analysis, J. Hydraul.
Div., ASCE, vol. 94, no. HY1, pp. 219-234, January 1968.
Michael A. Stoner, A New Way to Design Natural Gas Systems, Pipe Line Ind., vol.
32, no. 2, pp. 38-42, 1970.
STEADY CLOSED-CONDUIT FLOW
569
Hardy Cross loop-balancing
or node-balancing schemes.
In the more
normally modeled with a set of simultaneous
equations which are solved by the Newton-Raphson method.
Some programmed solutions 2,3 are very useful as design tools since pipe sizes or rough-
the
general methods the system
is
nesses
may be treated as unknowns in addition to junction pressures and flows.
10.8
COMPUTER PROGRAM FOR STEADY-STATE HYDRAULIC
SYSTEMS
Hydraulic systems that contain components different from pipelines can be
handled by replacing the component with an equivalent length of pipeline.
When
is a pump, special consideration is needed.
more than one fixed hydraulic-grade-line ele-
the additional component
Also, in systems that contain
must be introduced.
For systems with multiple fixed-pressure-head elevations, Fig. 10.14,
pseudo loops are created to account for the unknown outflows and inflows at
the reservoirs and to satisfy continuity conditions during balancing. A pseudo
loop is created by using an imaginary pipeline that interconnects each pair
of fixed pressure levels. These imaginary pipelines carry no flow but maintain
a fixed drop in the hydraulic grade line equal to the difference in elevation of
the reservoirs. If head drop is considered positive in an assumed positive direction in the imaginary pipe, then the correction in loop 3, Fig. 10.14, is
vation, a special artifice
AQ
150
S
-
135
nn
|
nQ,
Qa h"
\
1
QA
+
n~l
-
\
nri
\
&
nft
|
&
\«~ l
©
3
Fig. 10.14
Sample network.
AQ^r\EI
n 7m
APPLICATIONS OF FLUID MECHANICS
570
This correction
1 and 4 only.
If additional real pipelines
would be adjusted accordingly during each
applied to pipes
is
existed in a pseudo loop, each
loop-balancing iteration.
A pump in a system may be considered as a flow element with a negative
head
head rise that corresponds to the flow through the unit.
pump-head-discharge curve, element 8 in Fig. 10.14, were expressed by
a cubic equation
loss equal to the
If the
H
+A
= A
where
be
A
is
135
X
Q8
+
AM + A Q
3
the shutoff head of the pump, then the correction in loop 4 would
-
-
117
+
(A
nn Q 5
A,Q,
|-i
|
This correction
The
3
8
-
+
(At
AM + A&i)
+
2A 2 Qs
applied to pipe 5 and to
is
FORTRAN IV program in Fig.
+
is
r5
Q5
1
&
1
l""
3^ 3 Q 82 )
pump 8 in the loop.
may be used to analyze a wide
10.15
variety of liquid steady-state pipe-flow problems.
balancing method
+
used. Pipeline flows described
The Hardy-Cross loopby the Hazen- Williams
equation or laminar or turbulent flows analyzed with the Darcy-Weisbach
equation can be handled; multiple reservoirs or fixed pressure
sprinkler system, can be analyzed;
pumps can be treated.
levels, as in
a
and systems with booster pumps or supply
Either English or SI units
may be used by proper speci-
fication of input data.
A network is visualized as a combination of elements that are interconnected at junctions. These elements may include pipelines, pumps, and
imaginary elements which are used to create pseudo loops in multiple-reservoir
systems. All minor losses are handled by estimating equivalent lengths and
adding them onto the actual pipe lengths. Each element in the system is
numbered up
to a
consecutively.
A
maximum
in the arithmetic solution,
that continuity
tion in a
Any
is
is
and not necessarily
assigned to each element, and, as
an estimated flow
satisfied at
pump must be
solution with
of 100, without duplication
positive flow direction
is assigned to each element such
each junction. The assigned positive flow direc-
in the intended direction of
backward flow through a pump
is
normal
invalid.
pump
operation.
The
flow direc-
tion in the imaginary element that creates a pseudo loop indicates only the
direction of fixed positive
head drop, since the flow must be zero in
this ele-
may represent the termination of a single element
or the intersection of many elements, is numbered up to a maximum of 100,
without duplication and not necessarily sequentially. An outflow or inflow at
ment. Each junction, which
a junction
is
defined during the assignment of initial element flows.
HARDY CROSS LCOP BALANCING INCLUDING MULTIPLE RESERVOIRS L PLPPS(PU)
H*ZEN-WILLIAMS(HW) CR C ARCY-WE Se ACHt DW ) KAY BE USED FOR PIPES
ENGLISH(EN) OR SI UNITS(SI) MAY BE USED, POSITIVE DH IN ELEMENT IS FEAD OROP
DIMENSICN ITY(4) ,IC(2),ITYPE( 100) , ELE
500 ) , ND ( 500 , Q( IOC ,H 100
2,S(20), IXI24C)
DATA ITY/•^w•,•DW•,•PS•,•PU•/,IE/ , CC , /,ID/ , EN•, , SI•/
CO 12 J=1,50C
IC
IF (J.LE.ICC )ITYPE(J>=5
IF (J.LE.IOO) H(J)=-100G.
IF (J.LE.240) IX( J} =
IND(J>=C
12
C READ PARAMETERS FOR PRCfiLEK, ANO ELEMENT CATA
READ
5,15,END=99) N T, KK TOL , VNU ,DEF
FORMAT (A2 , 1 8 ,F 10.4 F10. 7 , F 1 0.5 I
15
C
C
I
C
M
I
)
)
(
,
(
,
(NT.EC.IC(2))G0 TO 20
WRITE (6,18) VNU
FORMAT (• ENGLISH UMTS SPECIFIED, VISCOSITY IN
UNITS=4.727
C=32.174
IF
18
20
21
22
24
32
33
FiO. 7)
'
(
IF (NT.EQ.ITY(NTY))CC TO 33
CONTINUE
ITYPE(I)=NTY
KP=5*(I-1)*1
(41, 42, 53,64) ,NTY
IF (X3.EC.0.)X3=DEF
3**
ELEM(KP)=UNITS*Xl/
EX=1.852
(
GO TO 4
42
• ,
VNU
FORMAT (• SI UNITS SPECIFIED, VISCOSITY IN M**2/SEC= ,F 10.7
UNITS = 1C674
G=9.806
WRITE (6,24) TCL.KK
FORMAT (• CESIREO FLCW TCLERANLE= • ,F5. 3 , • NO. OF I TERAT ICN S* • , I 5//
2' PIPE
CICFS CR M**3/S) L(FT OR
Hm C CR EPS')
C FT OR M)
READ (5,30) NT , I ,CQ ,X1 ,X2 ,X3 ,X4 ,X5
FORMAT ( A2 , 3X , I 5 , 3F 10. 3 ,F 10. 5,2F10. 3)
IF (NT. EC. IE) GO TO 68
Q(I)=QQ
CO 32 NTY=1,4
GO TO
41
T**2/S EC*
GO TO 22
WRITE (6,21)
)
26
30
F
IF
1 . 8
52*X2**4.8 7C4
3
(X3.EQ.C.)X3=0EF
EX = 2.
ELEM(KP)=Xl/(2.*G«X2**5*.7854«.7854)
ELEM(<P»l)=l./(.7E5 4*X2*VNU)
ED=X3/X2
ELE M KP *2
(
43
45
(
GO
53
55
)
=
.
094 *EC*«. 22 5*. 53 *ED
ELEM(KP*3)=8e.*EJ**.44
ELEM(KP*4)=l.62*EC*«.l34
WRITE (6,45) I,Q( I),X1,X2,X3
FORMAT
5 F 1 8 . 3 ,F 1 . 1 F 1 2. 3 , F 1 4. 5
EN=EX-1.
I
,
,
TO 26
ELEM(KP)=Xl
WRITE (6,55) I, XI
FORMAT (15,
RESEKVCIR ELEV OlFFERfcNC E= • , Fl 0. 2
•
GO
TO
26
ELEM(KP)=X2
ELEM(KP*3)=(X5-3.* (X4-X3 )-X2)/(6.*Xl*«3)
ELEM(KP*2)=(X4-2.*X3*X2> /( 2. *X1 **2 )-E LEM( KP*3 *3. *X
ELEM{KP+1)=(X3-X2)/X1-ELEM(K.P*2)*X1-ELEM(KP*3)*X1*XI
WRITE (6,66)
,X1,X2,X3, X4,X5, ELEM( Kf «J-l ,J = 1 ,4)
FORMAT (15,' PUMP CURVE, Dg** t F7.a«' h = , 4F 8. 1/ 5X
66
2' COEF IN PUPP EQ=»t4*ll.31
64
)
)
I
•
GO TO 26
I
READ LOOP
68
7C
INDEXING CATA,
12=11*14
READ (5,75) NT , INC
FORMAT (A2,2X,15I4)
(
75
INO=NO.
P
I
PES ,P I Ft ,
P
I
PE
,
E TC
.
CLOCKW ISE* ,CC-
11=1
(
I ) ,
I
=
I
1
,
I
2
INT. EC. IE) GO TC 78
11=12*1
GO TO 7C
78
IF
(Il.EC.l) GC TC 140
WRITE (6,79)
UNO! Ilil-ltll)
79
FORMAT (• I NC= / ( 1 5 I 4 ) )
C BALANCE ALL LCCPS
CC 130 K=1,KK
CDQ=0.
IP=1
IF
'
Fig. 10.15
(Continued on pages 572 and 573)
571
)»
8C
81
82
83
(
,
)
>
INOIIP)
IF(IL.EQ.O) GC TO 124
DH=0.
HDC=0.
00 110 J=1,I1
I=IND(lP*Jl
IF (I) 61*110*62
Il =
SUl=-l.
1 = -I
GO TO 8 3
S(J) = l.
NTY*ITYPE(II
KP=5*(l-l)*i
GO TO (91, 92, IC3, 104), NTY
91
R=ELEM(KP)
GO TO
92
93
9 5
REY=ELEM(KP+1)*A8S(Q( I))
IF(REY.LT.l.) REY=1.
IF (REY-2000.1 93,94,94
R=ELEM(KPM64./REY
GO TO
95
94
R=ELEM(KPI*(ELEMKP*2)*ELEM(KP*3l/REY**ELfcM(KP*4li
95
DH=DH+S( J)*R*C( I *AE S (Q I I **EN
FOQ=HDQ*EX*R*AES(Q( I II** EN
)
GO TO
103
(
)
110
DH=DH*S(JI*ELEM(KPI
GO TO 11C
DH=DH-S(Jl*(ELEM(KP)*G(I)*(ELEM(KP*l)*t.(I * ELEM( KP + 2 *C II *
2ELEM(KP*3))II
FDQ=HDU-(ELEM(KP*1I»2.*ELEM(KP*2I*Q(I I *3. *ELEM I KP* J )*C( IM*2)
110 CONTINUE
IF (ABS(HCG).LT..CCCl) FCG=1.
00=-OH/HOO
DDQ=oou**esi6gi
104
)
DO 120 J»lill
1= IABSI INCf IF*JI )
IF
( ITYPEI I). EC. 3)
CII
>
= G(
I
I
GO TO 120
II*SIJ)*O0
12C CONTINUE
IP=IP*I 1*1
GO TO 8C
124 WRITE 16,1251 K,ODC
125 FORMAT |« ITERATICN NC. 1 ,!^.*
IF (DDQ.LT.TCL) GC TC 140
13C CCNTINUE
140 WRITE (6,1411
LAI FORMAT !• ELEMENT
FLOW I
CO 150 1=1, IOC
SUM OF FLOW CORRECT IONS=«
,F 10 .4
NTY=ITYPE( II
GO TU
142,142, 150, 142, 1501, NTY
142 WRITE (6,1431 I,Q(
143 FORMAT (I5.F1C.31
15C CCNTINUE
X= JUNC ELEMENT , JUNC. ELEM , JUNC , ETC.
C REAC CATA FOR FGL CCMFLTATICN,
152 SEAD (5,155) NT,K,HF
155 FORMAT (A2, I8,FIC3)
(
I
I
IF (NT. EG. IE) GO TO 160
H(K } = HH
GO TO 152
160 11=1
162 12=11*14
REAO ( 5.75INT, IX(K I ,K=I 1,12)
IF (NT. EG. IE) GO TO 17C
11=12*1
GO TO 162
17C WRITE (6,171) I IXII ), 1=1,11)
171 FORMAT !
IX=«/(15I4)>
IP=1
18C CO 200 J=l,238,2
(
(J.EQ.l) U=IX(IF)
I=IX( IP»J)
N=IX( P* J* 1
IF(I) lei, 199,182
181 SS=-1.
IF
I
I=-I
GO TO 16 3
ie2 ss=i.
183 NTY=I TYPE(
KP=5*( - 1
1
GO TU ( 184,165,139, 19C, 199), NTY
I
I
Fig. 10.15
572
»
STEADY CLOSED-CONDUIT FLOW
573
184 R=ELEM(KP)
GO TO iee
165 REY*ELEP(KP*1)*ABS(C(I))
(REY.LT.l.) REY=1.
(REY-2000.) 186,167,167
ie6 R=ELEM(KP)*64./REV
IF
IF
GO TO
188
R=ELEM(KPM<ELEM(KP + 2)+ELEM(K.P*3)/R£V**ELEM(KP + 4)
188 HINl = HtUI-SS*R*«(I)*ABSiaiin**EN
18 7
)
GO TO iSS
189 H(N )=H( l)-SS*ELEM(KP)
GO TO 199
19C H(N) = H( Ii)*SS*IELEP(KP)*Qin*(ELEM(KP*l)+l.(I)*(ELEM(KP*2)»Gm
2ELEMIKP + 3)
1S9 IF
IF
20C
)J
I
(IXl J*IP«-3).EC.C»
(
GC TC 210
1X(J»IP+2).EJ.C) GO TO 205
Il=N
2C5 IP=IP*J*3
GO TO
ISO
210 WRI TE (6,215)
215 FORMAT l« JUNCTION
00 220 N=l,100
HEAC)
IF
(H(N).EC.-IOOG.) GC TO ZZO
WRITE (6,143) fSH(N)
22C CONTINUE
9S
GO TO
STOP
END
Fig. 10.15
first
1C
FORTRAN program
for hydraulic systems.
The operation of the program is best visualized in two major parts: the
performs the balancing of each loop in the system successively and then
manner until the sum of all loop-flow corrections is less
At the end of this balancing process the element
computed and printed. The second part of an analysis involves the
repeats in an iterative
than a specified tolerance.
flows are
computation of the hydraulic-grade-line elevations at junctions in the system.
Each of these parts requires a special indexing of the system configuration
in the input data. The indexing of the system loops for balancing is placed in
the vector IND. A series of integer values identifies each loop sequentially by
the number of elements in the loop followed by the element number of each
element in the loop. The directional sense of flow in each element is identified
by a positive element number for the clockwise direction and a negative element number for counterclockwise. The second part of the program requires an
identification of one or more junctions with known heads. Then a series of junction and element numbers indexes a continuous path through the system to all
junctions where the hydraulic grade line is wanted. The path may be broken at
any point by an integer zero followed by a new junction where the head is
known. These data are stored in the vector IX by a junction number where the
head is known followed by a contiguous element number and junction number.
Again the positive element number is used in the assigned flow direction, and
the negative element number is used when tracing a path against the assigned
element-flow direction. Any continuous path may be broken by inserting a
zero; then a new path is begun with a new initial junction, an element, and
574
APPLICATIONS OF FLUID MECHANICS
a node,
etc.
computed
All junction hydraulic-grade-line elevations that are
are printed.
As shown below, the type of each element is identified in the input data,
and each element is identified in the program by the assignment of a unique
numerical value in the vector
ITYPE.
Element
Data
Program
Hazen- Williams pipeline
Darcy-Weisbach pipeline
Pseudo element
HW
DW
2
PS
3
Pump
PU
4
The
1
physical data associated with each element are entered on separate cards.
In the program the physical data that describe
stored in the vector
an example
ELEM,
with
all
elements in the system are
five locations reserved for
each element. As
of the position of storage of element information, the data per-
number 13 are located in positions 61 to 65 in ELEM.
Data preparation for the program is best visualized in four steps, as
shown in Fig. 10.16 and described below. Formated input is used, as shown
in Fig. 10.16 and the program.
taining to element
Step
Parameter description card
1:
The type
of unit to
be used in the analysis
for English units or SI for the SI units.
number
is
An
defined
by the characters
integer defines the
of iterations to be allowed during the balancing scheme.
EN
maximum
An accept-
is set for the sum of the absolute values of the corrections in
each loop during each iteration. The liquid kinematic viscosity must be speci-
able tolerance
Darcy-Weisbach equation
fied if the
Williams equation
or
if
is
used for pipeline
losses.
used, a default value for the coefficient
the Darcy-Weisbach equation
is
If
the Hazen-
C may be
defined,
used, a default value for absolute pipe
may
be defined. If the default value is used on the parameter card,
need not be placed on the element cards; however, if it is, the element data
roughness
it
is
override the default value.
Step
2:
Element cards
Each element
quire either
in the
HW
or
system requires a separate card.
DW
Pipeline elements re-
to indicate the equation for the problem solution,
the element number, the estimated flow, the length, the inside diameter, and
(if the default value is not used) either the Hazen- Williams coefficient or the
pipe roughness for the Darcy-Weisbach equation.
Pump
elements require
STEADY CLOSED-CONDUIT FLOW
575
/
Junction Element Junction Element
(A2, 2X, 1514)
etc.
•HV6 Head
calculation
path cards
Junction Head
(A2, 18, F10.3)
IVa Junction elevation
cards
No. of
elements
in loop
PS
Element Element Element
,.„ _ w ,-«.-,
(A2, 2X, 1514)
No. of
elements
in loop
etc
III
Loop index cards
Element
number
Element
PL)
Flow
number
HW
Element
FIc
Length
nUmb6r
DW
/EN
(A2, 3X, 15, 3F10.3, F10.5,
Number
Tolerance
of
iterations
or
Element cards
\
Kinematic
Default
v/iscosity
HW —
C
or
SI
DW-
e
<A2, 18, F10.4, F10.7, F10.5)
I
Fig. 10.16
PU
Parameter card
Data cards for Hardy Cross program,
to indicate the element type, the element number, the estimated flow, a
AQ at which values of pump head are specified, and four values
head from the pump-characteristic curve beginning at shutoff head and at
equal flow intervals of AQ. The pseudo element for the pseudo loop requires
PS to indicate the type, the element number, a zero or blank for the flow, and
a difference in elevation between the interconnected fixed-pressure-head levels
with head drop positive. The end of the element data is indicated by a card
with "&&" in the first two columns.
flow increment
of
Step
3:
Loop index cards
These data are supplied with 15 integer numbers per card in the following
number of elements in a loop (maximum of 20) followed by the
element number of each element in the loop with a negative sign to indicate
order: the
counterclockwise flow direction.
are defined.
umns
1
and
The end
2.
This information
of step 3 data
is
indicated
is
repeated until
all
by a card with "&&"
loops
in col-
APPLICATIONS OF FLUID MECHANICS
576
Step
Head-calculation cards
4:
Junctions with fixed elevations are identified on separate cards by giving the
number and the
junction
more
or
hydraulic-grade-line elevation.
of these cards followed
by a card with "&&"
There must be one
two columns
in the first
to indicate the end of this type of data.
The path to be followed in computing the hydraulic-grade-line elevations
by supplying 15 integer values per card in the following order:
specified
is
2
.001
.12
.03
3
.0
4
.03
.03
SI
30
HW
I
HW
HW
HW
HW
5
PS
6
PS
PU
7
.000001
ICO.
toe.
300.
50C.
40c.
300.
.3
.15
.6
.3
.3
15.
18.
.06
8
30.
.C3
&£
117.
5
4
8
5
LL
SI UMTS SPECIFIED, VISCCS1TY IN M**2/SEC= C.CC00010
DESIREO FLCW TOLER ANCE'O .00 1 NO. OF ITERATIONS'
30
PIPE
LIFT (JR M)
D(FT OR M)
HW C OR EPS
JICFS CR M**3/S)
1
0.120
600.0
0.300
100.00000
2
C.C30
300.0
0.150
100.00000
3
0.000
500.0
0.6O0
100.00000
4
400.0
C.030
0.300
ICO. 00000
5
0.C30
300.0
C.300
100.00000
RESERVOIR ELEV DIFFERENCE'
6
15.00
7 RESERVOIR ELEV DIFFERENCE=
18.00
8 PUMP CURVE, CQ=
C.C30 H =
30.0
29.0
26.0
20.0
-555.556 -6172.636
COEF IN PUMP EQ=
30.000
-11. Ill
INC
=
eoocccoocococcc
2
2
I
-3
3
4
-5
3
3
6
-4
-1
3
5
7
c
ITERATION
ITERATION
ITERATION
ITERATION
ITERATION
ELEMENT
1
2
2
4
5
NO.
NO.
NO.
NO.
NO.
FLCW
0.143
SUM CF FLCW CORRECTICNS=
SUM OF FLOW CORRECTICNS=
SLK CF FLCW CORRECTIONS'
4 SUM OF FLOW CORRECTIONS'
5 SUM CF FLCW CORRECTIONS'
0.1385
1
2
0. 1040
3
0.C372
0.0034
C.C0C6
-0.C34
O.C27
0.080
O.0S4
O.C67
58422114300OCOO
E
IX'
C
JUNCTION HfcAC
137.811
1
2
150. C44
3
4
135. C44
137.797
117. OCC
5
Fig. 10.17
Program input and output
for
Example
10.11.
STEADY CLOSED-CONDUIT FLOW
number where
577
known, element number (with a negative
assumed flow direction) junction number, etc. If one wants a new path to begin at a junction different from the
last listed junction, a single zero is added, followed by a junction where the
head is known, element number, junction number, etc. The end of step 4
data is indicated by a card with "&&" in columns 1 and 2.
junction
the head
is
sign to indicate a path opposite to the
,
EXAM PLE 10.10 The program in Fig. 10.15 is used to solve the network problem displayed in Fig. 10.14. The pump data are as follows:
Q,
m /s
3
H,m
30
0.03
0.06
0.09
29
26
20
The Hazen- Williams
pipeline coefficient for all pipes
is
100.
Figure 10.17 dis-
plays the input data and the computer output for this problem.
Figures 10.18 to 10.20 give input data for three systems which can be
solved with this program.
El
El
500
480
;£/ 1.-5000.-1.25
a
3
Q(ft /s)-L(ft)-D(ft)
EN
HW
L.
.C0001
3C00.
2CO0.
5000.
6.
0.
0.
-80.
-20.
6.
5.
Hh
HW
PU
PS
PS
7.
120.
1.5
1.
1.25
120.
113.
99.
LI
-2
-1
-6
LL
ACC.
7
LL
7
Fig. 10.18
6
11
Input data for branching pipe system
units with Hazen-Williams formula.
in
English
APPLICATIONS OF FLUID MECHANICS
578
v
520
El
JL
B,
-
©f>
yS
L4-4000-1
,
/
0.-3000-0.8^(8)
®
EI500
^zo.aooo.o.e
[J]
00
1
®\
1
d
6
o
o
©i
CO
O
O
O
I
/
El 4 [
J5
Vm3/s°-
^K
SI
DM
DW
OW
CM
DM
DM
CM
DM
DM
DM
DM
PS
PS
a
v
30
.0C2
1
3.
1.6
2
1
.5
.5
0.
0.
5
7
8
1.4
1.4
9
10
1.4
^JH
® 1
1.4
1.
C.8
0.8
0.5
0.3
C.8
0.6
1.0
1.
0.6
0.6
-45.
0.
c.
20.
&G
5
I
-4
5
2
4
m 3/s
00C5
<.ccc.
(.000.
<»coc.
(»CC0.
C.
11
12
13
14
-
3(m3/s)-L(m)-D(m)
:cooo 12
3CCC.
3tC0.
JCC0.
3CCC.
3CCC.
30CC.
3C0C.
1.0
3
4
m
4000 ai
/<$r
\
0.5-3000-0.3
©
m
/ 0.5-3000-0-5
3
4
13
4
14
7
-9
-1
2
3
3
4
9
10
11
4
5
C
-2
2-7
8
&6
520.
1
£&
1
1
2
2
9
Fig. 10.19
7-8
3-11
6
12
8
Input data for hydraulic system. SI units and
Darcy-Weisbach equation.
10.9
CONDUITS WITH NONCIRCULAR CROSS SECTIONS
In this chapter so
far,
only circular pipes have been considered.
sections that are noncircular, the
if
the term
D
Darcy-Weisbach equation
may
For cross
be applied
can be interpreted in terms of the section. The concept of the
R permits circular and noncircular sections to be treated in
hydraulic radius
STEADY CLOSED-CONDUIT FLOW
579
500
El
—-~~
w~
'30.-4000-2.5
®f
s
©
®m QP
X\©
©
10.-40001.5
X
[H yf
rump/"
15 cfs
\
10.-4000-1.5
\_J/
\
\
\
5 ftVs
5.-4000-1.5
in
t—
in
o
o
o
*t
®{
3
Q(ft /s>-I(f0-D(ft
©1
\
\\®
o
o
o
d
\
©
m,
"
/
5
ft
3
10.-4000- 1.5
Q0
©
[91
10,4000-1.5
\^J/
EN
HM
HW
HW
HM
HW
HW
HW
HW
HW
PU
PU
PU
PS
10
.CCC01
4C00.
4C00.
40C0.
4000.
4CCC.
4C00.
4CCC.
4CCC.
4ccc.
.02
30.
30
1
10.
2
3
5.
5.
C.
4
5
6
10.
IC.
10.
15.
IC.
IC.
7
8
9
12
13
1*
7.
7.
7.
5.
C.
15
/^rSH!
®\
El
Hr/pumpp^ri_V
S
5,4000-1.5
"V
/s
\
@
-3-
©
i—
ft
400
^
3
/s
120.
2.5
1.5
1.5
1.5
1.5
1.5
1.5
1.5
1.5
100.
110.
150.
S3.
1C3.
143.
79.
89.
129.
55.
65.
1C5.
100.
se
-i
8
2
12
2
-2
-1
5
-7 - 13
-6
-9
e
15
2
3
12
4
5
9
4
7
14
8
6
5
8
8
4
3
-e
10 -13
11
14
£G
500.
40C.
1
8
e&
-
1
1
!>
12
2
-
-7
CL
Fig. 10.20
Input for booster-pump system.
the same manner.
divided
by the
R =
area
perimeter
The hydraulic
wetted perimeter.
_
ttZ)
2
/4
irD
_
radius
is
denned as the cross-sectional area
Hence, for a circular section,
D
(10.9.1)
4
and the diameter is equivalent to 4#. Assuming that the diameter may be
by 4R in the Darcy-Weisbach equation, in the Reynolds number,
replaced
APPLICATIONS OF FLUID MECHANICS
580
and
h
>
in the relative roughness,
=f
L V
~
V4Rp
_
2
R =
UiTg
e
e
D
(10 9 2)
= 4R
-
-
may
be handled in a similar manner. The Moody diain Eqs. (10.9.2) cannot be expected
to hold for odd-shaped sections but should give reasonable values for square,
Noncircular sections
gram
The assumptions
applies as before.
and
oval, triangular,
EXAMPLE
similar types of sections.
Determine the head loss in inches of water required for
/min of air at 60°F and 14.7 psia through a rectangular galvanized-iron section 2 ft wide, 1 ft high, and 200 ft long.
10.11
flow of 10,000
R =
i
=
0.333
10,00 °
V =
60
ft
X
-
3
-^ 4R
ft
0.00038
74#" =
83.3 ft/s
X
83.3
4
X
4
=
1330
2
/ = 0.017
Then
L V
~ ^4R 2g ~
2
f
The
specific
0.017
4
X
weight of
X
200 83.3 2
0.333
In inches of water the head
275
X
0.0762
X
12
=
,
^
"
=
(14.7
y
air. is
2 5
64.4
X
t
144)/ (53.3
X
520)
-
0.0762 lb/ft 3
.
loss is
.
4.04 in
62.4
10.10
AGING OF PIPES
The Moody diagram, with the values of absolute roughness shown there, is
for new, clean pipe. With use, pipes become rougher, owing to corrosion, incrustations, and deposition of material on the pipe walls. The speed with
which the
friction factor
changes with time depends greatly on the
fluid
being
STEADY CLOSED-CONDUIT FLOW
handled. Colebrook and White 1 found that the absolute roughness
e
581
increases
linearly with time,
=
€
in
e
+
at
which
e
10.10.1)
is
the absolute roughness of the
required to determine
new
surface.
Tests on a pipe are
a.
The time variation of the Hazen-Williams coefficient has been summarized graphically 2 for water-distribution systems in seven major U.S. cities.
Although it is not a linear variation, the range of values for the average rate
of decline in C may typically be between 0.5 and 2 per year, with the larger
values generally applicable in the
sure
way
through
first
years following installation.
The only
that accurate coefficients can be obtained for older water mains
is
field tests.
PROBLEMS
10.1
Sketch the hydraulic and energy grade lines for Fig. 10.21.
10.2
Calculate the value of
Prob. 10.1
10.3
is
10.4
24
ft.
for the valve of Fig. 10.21 so that the discharge of
reduced by one-half. Sketch the hydraulic and energy grade
Compute
energy grade
K
H=
the discharge of the system in Fig. 10.22.
Draw
lines.
the hydraulic and
lines.
What head
is
needed
in Fig. 10.22 to
produce a discharge of 0.3
m /s?
3
commercial
steel pipe
Fig. 10.21
1
2
C. F. Colebrook and C. M. White, The Reduction of Carrying Capacity of Pipes with
Age, J. Inst. Civ. Eng. Lond., 1937.
W. D. Hudson, Computerized Pipeline Design, Transp. Eng. J. ASCE, vol. 99, no. TE1,
1973.
APPLICATIONS OF FLUID MECHANICS
582
—
r~
2
S=0.88
M = 0.04
60 m
30 cm
diam
30 m
20 cm
diam
Oil
-I-
m
poise
F
cm diam
45
Square edge
Smooth pipe
Fig. 10.22
Calculate the discharge through the siphon of Fig. 10.23 with the conical diffuser
10.5
removed.
H=
4
ft.
Calculate the discharge in the siphon of Fig. 10.23 for
10.6
minimum
Find the discharge through the siphon of Fig. 10.24.
10.7
minimum
Estimate the
8
What
is
the
the pressure at
A1
ft.
What
is
pressure in the system.
Neglecting minor losses other than the valve, sketch the hydraulic grade line for
10.8
The globe valve has a
Fig. 10.25.
maximum
What
is
the
eter reading
is
29.5 in
10.9
10.10
Two
300 m, Di
Q=
H=
pressure in the system?
0.1
10.11
height of point
m /s
20 cm;
A
K=
4.5.
(Fig. 10.25) for
L 2 = 360 m,
by three clean cast-iron pipes in series; L\ =
30 cm; L 3 = 1200 m, D 3 = 45 cm. When
D =
2
water at 70°F, determine the difference in elevation of the reservoirs.
Solve Prob. 10.10 by the method of equivalent lengths.
Standard elbow
\D
6ft
4-20
8
ft
ft
Water
60° F
12
ft
8-in.-diam
steel pipe
-I
Fig. 10.23
no cavitation? Barom-
Hg.
reservoirs are connected
=
3
loss coefficient
k-1.5
D
STEADY CLOSED-CONDUIT FLOW
rn^
1.8
m
583
Close return bend
10-cm-diam
smooth pipe
6
Oil,
S =
H -
m
3.6
m
L
cm
15
0.8
5cm-diam
N-s/m 2
0.01
nozzle,
Loss=0.1^Fig. 10.24
For a difference
10.12
in elevation of 10
m in Prob. 10.10, find the discharge by use of
the Hazen-Williams equation.
of
By use of the program in Fig. 10.1, determine the exponential formula for flow
water at 21 °C through clean cast-iron pipelines ranging in diameter from 10 to
30
cm
10.13
for the
What
10.14
500
ft
=
pipe
is
10.16
ft?
100
ft of
What
10 5
.
is required to convey 100 gpm kerosene at 90°F
There are a valve and other minor losses with total K of 7.6.
Air at atmospheric pressure and 60°F
The upstream pipe
0.06) in series.
pipe. Neglect
X
diameter smooth pipe
with a head of 16
10.15
(e
Reynolds number range 10 5 to 5
is
400
is
carried through
two horizontal pipes
24 in diameter, and the downstream
ft of
36 in diameter. Estimate the equivalent length of 18-in
minor
(e
=
0.003)
losses.
pressure drop, in inches of water,
is
required for flow of 6000 cfm in
Prob. 10.15? Include losses due to sudden expansion.
10.17
Z>i
=
Two
pipes are connected in parallel between two reservoirs; L\
1.2-m-diameter old cast-iron pipe,
For a difference
in elevation of 3.6
C=
100;
Water
Fig. 10.25
at
60°F
D =
2
1
=
2500 m,
m, C
m determine the total flow of water at 20°C.
Globe valve
8-in.-diam smooth pipe
L 2 = 2500 m,
=
90.
APPLICATIONS OF FLUID MECHANICS
584
Water 60° F
Pump A
H
Qcfa
it
Pump B
e%
H
70
!t
e%
Qcfs
80
60
2.00
59
70
2.60
54
55
2.56
70
60
3.94
70
50
3.03
76
50
4.96
80
45
3.45
78
40
5.70
73
40
3.82
76.3
30
6.14
60
35
4.11
72
20
6.24
40
30
4.48
65
25
4.59
56.5
20
4.73
42
Fig. 10.26
For 4.5
10.18
m /s
3
flow in the system of Prob. 10.17, determine the difference in
elevation of reservoir surfaces.
Three smooth tubes are connected in parallel: L\ = 40 ft, Di = J in; L 2 = 60
L 3 = 50 ft, D z = f in. For total flow of 30 gpm oil, y = 55 lb/ft 3 m
0.65 P, what is the drop in hydraulic grade line between junctions?
10.19
D =
2
1 in;
=
=
,
Determine the discharge of the system of Fig. 10.26 for L = 2000
and H = 25 ft, with the pump A characteristics given.
10.20
e
ft,
D=
18
in,
L = 4000
ft,
ft,
0.0015,
10.21
D=
Determine the discharge through the system of Fig. 10.26
smooth pipe, H = 40 ft, with pump B characteristics.
for
24-in
10.22
Construct a head-discharge-efficiency table for
connected in
pumps A and B
(Fig. 10.26)
pumps A and B
(Fig. 10.26)
series.
10.23 Construct a head-discharge-efficiency table for
connected in parallel.
pumps A and
series;
Find the discharge through the system
5000 ft of 12-in clean cast-iron pipe, H
10.25
Determine the horsepower needed to drive pumps
10.26
Find the discharge through the system of Fig. 10.26 for pumps
5000 ft of 18-in steel pipe, H = 30 ft.
10.24
parallel;
of Fig. 10.26 for
=
100
B
in
ft.
A
and
B
in Prob. 10.24.
A
and
B
in
STEADY CLOSED-CONDUIT FLOW
585
•
t
H
60
°>S
120
90
m
€
m
8 cm diam
= 0.009 m
10
cm diam
€=0.012
M-
m
Fig. 10.27
Determine the horsepower needed to drive the pumps
10.27
For
10.28
sp gr
=
H=
m
12
in Fig. 10.27, find the discharge
in Prob. 10.26.
through each pipe,
/z
=
8 cP;
0.9.
H in Fig.
m /s flowing.
3
=
=
10.29
Find
10.30
system
Find the equivalent length of 12-in-diameter clean cast-iron pipe to replace the
of Fig. 10.28. For H = 30 ft, what is the discharge?
10.31
With
10.27 for 0.03
\i
5 cP; sp gr
velocity of 4 ft/s in the 8-in-diameter pipe of Fig. 10.28 calculate the
flow through the system and the head
H required.
10.32
In Fig. 10.29 find the flow through the system when the
10.33
If
the
pump
and the elevation
10.34
Find
The pump
Q A and Q B
Fig. 10.28
0.9.
.
of Fig. 10.29
is
pump
is
removed.
delivering 3 cfs toward J, find the flow into
A and B
of the hydraulic grade line at /.
is
adding 7500
W
fluid
power to the flow (toward J)
in Fig. 10.29.
-
.
APPLICATIONS OF FLUID MECHANICS
586
100'
El
El
^—
80'
—
_
-
|
^~^n
El
50'
——
1
—
<*
o\
\
•
i<py
-
^cpA
El
WOCTipj
^=0.04'
€
^0.04'
r-^
Fig. 10.29
10.35
With pump
A
of Fig. 10.26 in the
system of Fig. 10.29, find
QA
,
Qb, and the
elevation of the hydraulic grade line at J.
10.36 With pump B of Fig. 10.26 in the system
and the elevation of the hydraulic grade line at J.
10.37
pump
For flow
of
1 cfs
into
B of Fig.
efficiency of 70 percent,
10.29,
of Fig. 10.29, find the flow into
what head
how much power
is
is
B
produced by the pump? For
required?
10.38
Find the flow through the system of Fig. 10.30 for no
10.39
(a)
pump
in the system.
With pumps A and B of Fig. 10.26 in parallel in the system of Fig. 10.30,
and D and the elevation of the hydraulic grade line at J\ and Ji.
find the flow into B, C,
(b)
this
Assume the pipes in Fig. 10.30
problem by use of the program
10.40
n=
are all cast iron.
Prepare the data for solution to
in Fig. 10.15.
Calculate the flow through each of the pipes of the network shown in Fig. 10.31
2.
10.41
Determine the flow through each
10.42
By
line of Fig. 10.32.
n
=
2.
use of the program in Fig. 10.15, solve Prob. 10.35.
EI90mi£
^0.025
Fig. 10.30
f= 0.025
l
STEADY CLOSED-CONDUIT FLOW
587
25
<?
75
100
Fig. 10.31
10.43
(b)
By use of the program
Fig. 10.19; (c)
in Fig. 10.15 solve the
problems given in (a) Fig. 10.18;
Fig. 10.20.
10.44 Determine the slope of the hydraulic grade line for flow
80°F through a rectangular 18 by 6 in galvanized-iron conduit.
10.45
What
size
square conduit
of hydraulic grade line of 0.001?
10.46
Calculate the discharge of
sheet-metal conduit
10.47
A
A
e
=
is
10.49
(a)
50
0.003.
30
air at
ft/s.
water at 60°F with slope
S
=
loss is
0.85,
/x
60 cm.
=
e
4 cP, through 30
=
1 ft
on a
side,
Calculate the slope of the hydraulic grade
Estimate the head
loss
per 1000
m of 5 by 10 cm
0.00015 m.
its
conveys 6
cfs
line.
absolute roughness doubled
ft for
a flow of 15 cfs
when the
25 years old.
An
18-in-diameter pipe has an / of 0.020
In 10 years /
10.50
oil,
when the head
cfs
atmospheric
V=
0.003.
clean 24-in-diameter cast-iron water pipe has
in 5 years of service.
pipe
needed to convey 10
=
duct, with cross section an equilateral triangle
water at 60°F.
10.48
is
e
of
=
0.029 for
V=
The hydraulic grade
Fig. 10.32
5 ft/s water flow of 60°F.
line is
always above the energy grade line
-
when new for
3 ft/s. Find / for 4 ft/s at end of 20 years.
(6)
always above the closed conduit
:
APPLICATIONS OF FLUID MECHANICS
588
(c)
always sloping downward
(d)
the velocity head below the energy grade line
(e)
upward
in the direction of flow
when
in direction of flow
pipe
is
inclined
downward
10.51 In solving a series-pipe problem for discharge, the energy equation is used along
with the continuity equation to obtain an expression that contains a V 2/2g and /i, J2,
etc. The next step in the solution is to assume
(a)
Q
(b)
V
(c)
One pipe system
10.52
R
is
(d)/i,/2 ...
h,
Q
10.53
Q
L,
(b)
none
of these quantities
said to be equivalent to another pipe system
following two quantities are the
(a)
(e)
,
(c)
when the
same
L,
D
(d)
f,
D
(«)
V,
D
In parallel-pipe problems
(a)
the head losses through each pipe are added to obtain the total head loss
(6)
the discharge
is
the same through
(c)
the head loss
is
the same through each pipe
(d)
a direct solution gives the flow through each pipe
(e)
a trial solution
10.54
analytically
(6)
by
by
by
by
(d)
(e)
the pipes
when
the total flow
is
known
not needed
Branching-pipe problems are solved
(a)
(c)
is
all
by using as many equations as unknowns
loss is the same through each pipe
assuming the head
equivalent lengths
assuming a distribution which satisfies continuity and computing a correction
assuming the elevation of hydraulic grade line at the junction point and trying
to satisfy continuity
10.55
In networks of pipes
(a)
the head loss around each elementary circuit must be zero
(6)
the (horsepower) loss in
(c)
the elevation of hydraulic grade line
(d)
elementary circuits are replaced by equivalent pipes
(e)
friction factors are
10.56
The
all circuits is
assumed
is
the same
assumed
for each pipe
following quantities are computed
by using 4R
noncircular sections:
(a)
velocity, relative roughness
head loss
Reynolds number, relative roughness, head
(d) velocity, Reynolds number, friction factor
(e) none of these answers
(b)
(c)
for each junction
velocity,
loss
in place of diameter for
STEADY CLOSED-CONDUIT FLOW
10.57
(a)
Experiments show that in the aging
of pipes
the friction factor increases linearly with time
(6)
a pipe becomes smoother with use
(c)
the absolute roughness increases linearly with time
(d)
no appreciable trends can be found
(e)
the absolute roughness decreases with time
589
11
STEADY FLOW
A
IN
OPEN CHANNELS
broad coverage of topics in open-channel flow has been selected for this
Steady uniform flow was discussed in Sec. 5.9, and application of
equation to the hydraulic jump in Sec. 3.11. Weirs were
momentum
the
introduced in Sec. 8.4. In this chapter open-channel flow is first classified
and then the shape of optimum canal cross sections is discussed, followed
by a section on flow through a floodway. The hydraulic jump and its application to stilling basins is then treated, followed by a discussion of specific
energy and critical depth which leads into gradually varied flow. Water-surface
profiles are classified and related to channel control sections. Transitions are
next discussed, with one special application to the critical-depth meter.
The mechanics of flow in open channels is more complicated than closedconduit flow owing to the presence of a free surface. The hydraulic grade line
coincides with the free surface, and, in general, its position is unknown.
For laminar flow to occur, the cross section must be extremely small,
One
the velocity very small, or the kinematic viscosity extremely high.
example of laminar flow is given by a thin film of liquid flowing down an
inclined or vertical plane.
This case is treated by the methods developed
in Chap. 5 (see Prob. 5.10). Pipe flow has a lower critical Reynolds number
of 2000, and this same value may be applied to an open channel when the
diameter D is replaced by 4i£. R is the hydraulic radius, which is defined
as the cross-sectional area of the channel divided by the wetted perimeter.
In the range of Reynolds number, based on R in place of D, R = VR/v < 500
flow is laminar, 500 < R < 2000 flow is transitional and may be either laminar
or turbulent, and R > 2000 flow is generally turbulent.
Most open-channel flows are turbulent, usually with water as the liquid.
The methods for analyzing open-channel flow are not developed to the extent
chapter.
590
STEADY FLOW
IN
OPEN CHANNELS
591
of those for closed conduits. The equations in use assume complete turbulence,
with the head loss proportional to the square of the velocity. Although
practically all data on open-channel flow have been obtained from experiments
on the flow of water, the equations should yield reasonable values for other
low viscosity. The material in this chapter applies to turbulent flow
liquids of
only.
11.1
CLASSIFICATION OF FLOW
Open-channel flow occurs in a large variety of forms, from flow of water
field during a hard rain to the flow at constant
depth through a large prismatic channel. It may be classified as steady or
unsteady, uniform or nonuniform. Steady uniform flow occurs in very long
over the surface of a plowed
where terminal
due to turbulent flow
potential energy due to the uniform
the channel. The depth for steady
inclined channels of constant cross section, in those regions
velocity
is
has been reached,
exactly supplied
i.e.,
where the head
by the reduction
in
decrease in elevation of the bottom of
uniform flow
loss
called the normal depth.
In steady uniform flow the discharge
everywhere constant along the length of the
channel. Several equations are in common use for determining the relations
between the average velocity, the shape of the cross section, its size and roughness, and the slope, or inclination, of the channel bottom (Sec. 5.9).
Steady nonuniform flow occurs in any irregular channel in which the
discharge does not change with the time; it also occurs in regular channels
when the flow depth and hence the average velocity change from one cross
section to another. For gradual changes in depth or section, called gradually
varied flow, methods are available, by numerical integration or step-by-step
means, for computing flow depths for known discharge, channel dimensions
and roughness, and given conditions at one cross section. For those reaches of
a channel where pronounced changes in velocity and depth occur in a short
distance, as in a transition from one cross section to another, model studies
are frequently made. The hydraulic jump is one example of steady nonuniform
is
is
constant, and the depth
flow;
it is
is
discussed in Sees. 3.11 and 11.4.
Unsteady
Unsteady uniform flow rarely occurs in open-channel flow.
nonuniform flow is common but is difficult to analyze. Wave motion is an
example of this type of flow, and its analysis is complex when friction is taken
into account.
Positive
and negative surge waves
in a rectangular channel are
analyzed, neglecting effects of friction, in Sees. 12.9 and 12.10.
The routing of unsteady flood flows in channels is discussed in Sec. 12.10.
Flow is also classified as tranquil or rapid. When flow occurs at low velocities
so that a small disturbance can travel upstream
and thus change up-
APPLICATIONS OF FLUID MECHANICS
592
stream conditions, it is said to be tranquil flow (F < 1). Conditions upstream are affected by downstream conditions, and the flow is controlled by
1
When
the downstream conditions.
flow occurs at such high velocities that
a small disturbance, such as an elementary wave
flow
described as shooting or rapid
is
(
F
>
1)
is swept downstream, the
Small changes in downstream
.
conditions do not effect any change in upstream conditions; hence, the flow
by upstream
is
controlled
is
just equal to the velocity of
(F
critical
=
conditions.
When
flow
such that
is
an elementary wave, the flow
its
velocity
said to be
1).
The terms
subcritical
and
supercritical are also used to classify flow
velocities. Subcritical refers to tranquil flow at velocities less
and
is
supercritical corresponds to rapid flows
when
than
critical,
velocities are greater
than
critical.
Velocity distribution
The
boundary must be
velocity at a solid
zero,
and
generally increases with distance from the boundaries.
does not occur at the free surface but
of 0.05 to 0.25 of the depth.
is
in open-channel flow it
The maximum
The average
velocity along a vertical line
sometimes determined by measuring the velocity at 0.6
more
reliable
method
is
of the depth,
to take the average of the velocities at 0.2
of the depth, according to
velocity
usually below the free surface a distance
measurements
is
but a
and 0.8
of the U.S. Geological Survey.
BEST HYDRAULIC-CHANNEL CROSS SECTIONS
11.2
For the cross section of channel for conveying a given discharge for given
and roughness factor, some shapes are more efficient than others. In
general, when a channel is constructed, the excavation, and possibly the
slope
lining,
must be paid
and so both
wetted perimeter or
Manning formula
1
is
From the Manning formula it is shown that when the
a minimum, the wetted perimeter
lining
dimensions of
Q =
for.
is also a minimum,
and excavation approach their minimum value for the same
channel. The best hydraulic section is one that has the least
area of cross section
—AR^S
n
its
equivalent, the least area for the type of section.
The
is
1 '2
See Sec. 4.4 for definition and discussion of the Froude number
(11.2.1)
F.
STEADY FLOW
in
Q
which
is
the discharge (L 3 /T),
A
OPEN CHANNELS
IN
the cross-sectional flow area,
593
R
(area
by wetted perimeter P) the hydraulic radius, S the slope of energy
grade line, n the Manning roughness factor (Table 5.2), Cm an empirical
constant (U lz /T) equal to 1.49 in English units and to 1.0 in SI units. With
Q, n, and S known, Eq. (11.2.1) may be written
divided
A =
in
is
cP 2
'b
which
(11.2.2)
c is
known.
To
a minimum.
(Fig. 11.1)
A =
-
(P
P =
b
2y)y
by elimination
This equation shows that
+
and
2y,
= cP
of
P
is
a
minimum when A
find the best hydraulic section for a rectangular channel
A =
Then
by.
2'b
The value
b.
of y
is
sought for which
P
is
a
minimum.
Differentiating with respect to y gives
2
)» +
(SSetting
b
=
*-*-*—f
dP/dy =
gives
P =
4y, or since
P =
b
+
2y,
2y
(11.2.3)
Therefore, the depth
is
one-half the
bottom width, independent
of the size
of rectangular section.
To
P=
b
find the best hydraulic trapezoidal section (Fig. 11.2)
+
11.1
+m
2
After eliminating b and
.
J
b
-.
Fig.
2y y/l
Rectangular
cross section.
A
A =
by
+
in these equations
my
2
,
and
APPLICATIONS OF FLUID MECHANICS
594
Trapezoidal cross section.
Fig. 11.2
Eq. (11.2.2),
A =
By
by
+ my
holding
2
m
= (P -
2y \/l
+ m )y + my
2
2
= cP
2'5
(11.2.4)
constant and by differentiating with respect to
y,
dP/dy
is
set equal to zero; thus
P=
\y \/\
+m —
2
2my
(11.2.5)
Again, by holding y constant, Eq. (11.2.4)
and dP/dm is set equal to zero, producing
is
differentiated with respect to
m
T
2m
Vl + m
2
After solving for m,
m
and
P =
V3
after substituting for
2y/3y
=
b
V3
2
—y
m in Eq.
A =
o
(11.2.5),
V3 y 2
(11.2.6)
b = P/S and hence the sloping sides have the same length
As tan-1 m = 30°, the best hydraulic section is one-half a
which shows that
as the bottom.
hexagon. For trapezoidal sections with
wet earth will stand) Eq.
depth ratio.
The
semicircle
cross sections.
is
(11.2.5)
is
m specified
(maximum
slope at which
used to find the best bottom-width-to-
the best hydraulic section of
all
possible open-channel
STEADY FLOW
EXAM PLE
11.1
Determine the dimensions
of the
IN
OPEN CHANNELS
595
most economical trapezoidal
brick-lined channel to carry 8000 cfs with a slope of 0.0004.
WithEq.
A
*= P=
(11.2.6),
y
5
and by substituting
1
8000 =
49
5iie
v§2'8
into Eq. (11.2.1),
AA
2/3
(I)
Vooooi
or
y
8 '*
= 3930
and from Eq.
11.3
A
=
22.3
ft
(11.2.6), b
=
y
25.8
ft.
STEADY UNIFORM FLOW
practical open-channel
IN
A FLOODWAY
problem of importance
charge through a floodway (Fig. 11.3).
rougher than the river channel, and
is
the computation of dis-
In general the flood way
is
much
depth (and hydraulic radius) is
much less. The slope of energy grade line must be the same for both portions.
The discharge for each portion is determined separately, using the dashed
line of Fig. 11.3 as the separation line for the two sections (but not as solid
boundary), and then the discharges are added to determine the total capacity
its
of the system.
Since both portions have the same slope, the discharge
Fig. 11.3
Floodway cross section,
may
be expressed
APPLICATIONS OF FLUID MECHANICS
596
Qi
VS
= Kt
O2
=
K V&
2
or
Q=
(Kt
+ K^VS
in
which the value of
K
=
ft
(11.3.1)
K
is
AW*
from Manning's formula and is a function of depth only for a given channel
with fixed roughness. By computing Ki and 2 for different elevations of water
surface, their sum may be taken and plotted against elevation. From this
plot it is easy to determine the slope of energy grade line for a given depth and
K
discharge from Eq. (11.3.1).
HYDRAULIC JUMP; STILLING BASINS
11.4
The
relations
among
the variables
Vh
y\,
V
2,
y 2 for a hydraulic
jump
of determining the conjugate depths for a given discharge
method. The
2/i
and
7/2
momentum
(Fig. 11.4)
— -—
is,
= Pq(V
2
equation applied to the free body
for unit
Vi)
width (Viyi
= V
2
y2
=
to occur
Another way
in a horizontal rectangular channel are developed in Sec. 3.11.
is
the
of liquid
F
+
M
between
q),
= pVty* - P V^ yi
Rearranging gives
72/r
+
pVS yi
Fig. 11.4
W+
pV22 y2
Hydraulic jump
in
horizontal rectangular channel.
(11.4.1)
STEADY FLOW
IN
OPEN CHANNELS
597
or
Fi
+
in
which
2
F
is
M
+
Mi = F
(11.4.2)
2
M the momentum
M
+ for a given discharge q
the hydrostatic force at the section and
per second passing the section.
By
F
writing
is
per unit width
F
+
M
=
72/2
2
a plot
q
=
is
+
made
10 cfs/ft.
pf
(11.4.3)
y
of
M
F+
Any
having the same value of
of y for
minimum F
y and setting d(F
+
as abscissa against
?/
as ordinate, Fig. 11.5, for
at
two points
+ M; hence, they are conjugate depths.
The value
vertical line intersecting the curve cuts
+
F
M [by differentiation of Eq.
M)/dy
equal to zero]
it
(11.4.3) with respect to
is
'•=©""
(11.4.4)
The jump must always occur from a depth
greater than this value.
This depth
is
the
the following section to be the depth of
400
800
1200
1600
F+M
Fig. 11.5
F+ M curve for hydraulic jump.
less
than
critical
minimum
2000
this value to a
depth, which
energy.
is
depth
shown
in
Therefore, the
APPLICATIONS OF FLUID MECHANICS
598
jump always
energy
occurs from rapid flow to tranquil flow.
lost in the
jump prevents any
The
fact that available
possibility of its suddenly
changing
from the higher conjugate depth to the lower conjugate depth.
The conjugate depths are directly related to the Froude numbers before
and after the jump,
F,
=
is
—
m
F2
=
—
(11.4.5)
9V2
From
the continuity equation
Vihj x 2
=
= V 2 y2 2 = ghy2
glriy^
Z
2
or
Fi2/!
3
-
F 2 y2 *
FromEq.
(11.4.6)
(11.4.1)
gyJ
\
QVil
\
Substituting from Eqs. (11.4.5) and (11.4.6) gives
(l
+
2F1)
The value
Fr2/3 =
(l
+
2F2 ) F2 - 2
of F2 in terms of Fi
is
/3
(11.4.7)
obtained from the hydraulic-jump equation
(3.11.23),
2
By
*\2/
g
2/i
^
m
Eqs. (11.4.5) and (11.4.6)
F2 =
—
8Fl
,
(11.4.8)
These equations apply only to a rectangular section.
The Froude number before the jump is always greater than unity, and
after the jump it is always less than unity.
STEADY FLOW
Stilling
A
OPEN CHANNELS
IN
599
basins
basin
stilling
a structure for dissipating available energy of flow below
is
a spillway, outlet works, chute, or canal structure. In the majority of existing
installations a hydraulic
the energy dissipator.
jump
is
housed within the
This discussion
is
stilling
basin and used as
limited to rectangular basins with
some cases to save exwork by personnel of the
Bureau of Reclamation classified the hydraulic jump as an effective energy
dissipator in terms of the Froude number Fi(Vi 2 /gyi) entering the basin as
horizontal floors although sloping floors are used in
An
cavation.
authoritative and comprehensive
1
follows
At
Fi
=
At
Fi
is
=
3 to
6.
=
Near
Fi
=
Pre-jump.
There
is
is
only a slight difference in con-
3 a series of small rollers develops.
The water
uniform, and the head loss
fairly
length of pool
At
Standing wave.
to 3.
1
jugate depths.
surface
is
low.
is
quite smooth, the velocity
No
baffles required
if
proper
provided.
Oscillating action of entering jet, from bottom
Each oscillation produces a large wave of irregular
period that can travel downstream for miles and damage earth banks and
riprap. If possible, it is advantageous to avoid this range of Froude numbers
Fi
6 to 20.
Transition.
of basin to surface.
in stilling-basin design.
At
=
20 to 80. Range of good jumps. The jump
is well balanced, and the
Energy absorption (irreversibilities) ranges from 45
to 70 percent. Baffles and sills may be utilized to reduce length of basin.
At Fi = 80 upward. Effective but rough. Energy dissipation up to 85 percent.
Other types of stilling basins may be more economical.
Fi
action
is
at its best.
Baffle blocks are frequently used at the entrance to a basin to corrugate
the flow.
They
are usually regularly spaced with gaps about equal to block
employed at the
downstream end of a basin to aid in holding the jump within the basin and
to permit some shortening of the basin.
The basin should be paved with high-quality concrete to prevent erosion
and cavitation damage. No irregularities in floor or training walls should
be permitted. The length of the jump, about 61/2, should be within the paved
basin, with good riprap downstream if the material is easily eroded.
widths.
1
Sills,
either triangular or dentated, are frequently
Research Study on Stilling Basins, Energy Dissipators, and Associated Appurtenances,
Progress Report II, U.S. Bur. Reclam. Hydraul. Lab. Rep. Hyd-399, Denver, June 1,
1955. In this report the Froude number was denned as V/\/gy.
APPLICATIONS OF FLUID MECHANICS
600
A hydraulic jump occurs downstream from a 15-m-wide
The depth is 1.5 m, and the velocity is 20 m/s. Determine (a) the
Froude number and the Froude number corresponding to the conjugate depth;
(6) the depth and velocity after the jump; and (c) the power dissipated by
EXAMPLE
11.2
sluice gate.
the jump.
IV
20 2
'^^r^^ur
(a)
FromEq.
27 2
-
(11.4.8)
X 27.2
VI + 8 X 27.2 8
F2
(
F2
(6)
=
—=
1)3
=
°-° 831
V y = V lVl =
0.0831
2
2
1.5
X
20
=
30
Then
Wand
30
V =
2
X
X
0.0831
=
2.90 m/s, y2
From Eq.
(c)
h,
9.806
^=
- (* -
The power
(3.11.24), the
(10 -34
4
42/i2/ 2
X
dissipated
1.5
head
~ 1 -5 ° )3 =
X 10.34
loss hj in the
11.13
jump
is
m- N/N
is
Power = yQhj = 9802
11.5
10.34 m.
X
15
X
30
X
11.13
=
49,093
kW
SPECIFIC ENERGY; CRITICAL DEPTH
The energy per
E
with elevation datum taken as the bottom of
It is a convenient quantity to use
in studying open-channel flow and was introduced by Bakhmeteff in 1911.
the channel
It
is
£=
is
unit weight
called the specific energy.
plotted vertically above the channel floor,
J/
+
J
(H.5.1)
STEADY FLOW
Fig. 11.6
Example
of
IN
OPEN CHANNELS
601
specific
energy.
A plot of
specific
energy for a particular case
tangular channel, in which q
E=
y
+
is
is
shown
in Fig. 11.6.
the discharge per unit width, with
In a rec-
Vy =
q,
(11.5.2)
2gy
2
It is of interest to note
how
the specific energy varies with the depth for a
For small values of y the curve goes to
while for large values of y the velocity-head term
= y asymptotically.
is negligible and the curve approaches the 45° line
The specific energy has a minimum value below which the given q cannot
constant discharge (Fig. 11.7).
infinity along the
E
axis,
E
occur.
The value
of y for
minimum
E
12
3
is
obtained by setting dE/dy equal to
2g
1
^^m%^%
4
5
6
E
Fig. 11.7 Specific energy required for flow of a given
discharge at various depths.
APPLICATIONS OF FLUID MECHANICS
602
zero,
from Eq.
dy
(11.5.2), holding q constant,
gy
3
or
>-w
The depth
for
(11.5.3)
minimum
energy y c
is
called critical depth.
Eliminating q 2 in
Eqs. (11.5.2) and (11.5.3) gives
Emm
(11.5.4)
2i)C
Vc
showing that the critical depth
E in
is
two-thirds of the specific energy. Eliminating
Eqs. (11.5.1) and (11.5.4) gives
V = VoVc
(11.5.5)
c
The
velocity of flow at critical condition
V
c
at the critical condition
is
to determi ne the
occur for a given specific energy.
The
which was used in Sec.
Another method of arriving
is s/gijc,
8.4 in connection with the broad-crested weir.
maximum
discharge q that could
resulting equations are the
same as
Eqs. (11.5.3) to (11.5.5).
For nonrectangular cross
sections, as illustrated in Fig. 11.8, the specific-
Specific energy for a
nonrectangular section.
Fig. 11.8
STEADY FLOW
IN
OPEN CHANNELS
603
energy equation takes the form
J-ir + jgi
in
which
dE_
A
is
the cross-sectional area.
To
find the critical depth,
Q^dA
_
()
di.5.6)
1
gA* dy
dy
From
Fig. 11.8, the relation
between dA and dy
is
expressed
by
dA = Tdy
in
T
which
is
the width of the cross section at the liquid surface.
With
this
relation,
£r.-i
di.5.7)
The
critical
and
(11.5.7) gives
#=
2/c
+
depth must satisfy
for irregular sections,
on
A =
11.3
energy occurs when the velocity head
(11.5.7)
may
+|
be solved by
trial
=
1.
Determine the critical depth for 300 cfs flowing in a trapbottom width 8 ft and side slopes 1 horizontal to 2 vertical
2).
82/
in Eqs. (11.5.6)
by plotting
ezoidal channel with
(1
minimum
A/T. Equation
depth occurs for that value of y which makes f(y)
EXAMPLE
Q
(11.5.8)
one-half the average depth
Critical
Eliminating
^
This equation shows that the
is
this equation.
T =
8
+
2/
APPLICATIONS OF FLUID MECHANICS
604
Hence
By
+ y)
+ 2/V2)
300 2 (8
=
f(y)
32.2(%
+
2795(8
8
(82/
+
2/)
0.52/
2
:
)
trial
The
y
=
2
4
3
3.2
3.24
3.26
3.30
3.28
f(y)
=
4.8
0.52
1.33
1.08
1.03
1.02
0.975
0.997
critical
depth
is
3.28
ft.
In uniform flow in an open channel, the energy grade
line slopes downshowing a steady decrease
in available energy.
The specific energy, however, remains constant along
the channel, since y -f V 2 /2g does not change.
In nonuniform flow, the
energy grade line always slopes downward, or the available energy is decreased.
The specific energy may either increase or decrease, depending upon the slope
of the channel bottom, the discharge, the depth of flow, properties of the
cross section, and channel roughness. In Fig. 11.6 the specific energy increases
during flow down the steep portion of the channel and decreases along the
ward
parallel to the
bottom
of the channel, thus
horizontal channel floor.
The specific-energy and critical-depth relationships are essential
in study-
ing gradually varied flow and in determining control sections in open-channel
flow.
The head
F+
loss in a hydraulic
jump
is
easily displayed
by drawing the
M
curve (Fig. 11.5) and the specific-energy curve (Fig. 11.7) to the
same vertical scale for the same discharge. Conjugate depths exist where any
given vertical line intersects the
F
+
depth may be observed to be always
sponding lower conjugate depth.
11.6
(if
at
less
The specific energy at the upper
than the
specific
energy at the corre-
GRADUALLY VARIED FLOW
Gradually varied flow
area,
M curve.
is
steady nonuniform flow of a special
class.
The depth
roughness, bottom slope, and hydraulic radius change very slowly
all)
along the channel. The basic assumption required
loss rate at
a given section
is
given by the
is
that the head-
Manning formula
for the
same
depth and discharge, regardless of trends in the depth. Solving Eq. (11.2.1)
STEADY FLOW
for the
head
AE
=
which S
in
is
OPEN CHANNELS
605
per unit length of channel produces
loss
AL
IN
nQ V
/
\Cm ARW/
now
(11.6.1)
the slope of the energy grade line
sine of the angle the energy grade line
makes with the
varied flow the slopes of energy grade
line,
or,
more
specifically,
the
horizontal. In gradually
hydraulic grade
Computations of gradually varied flow
line,
and bottom
may
be carried out
Horizontal
either by the standard-step method or by numerical integration.
channels of great width are treated as a special case that may be integrated.
are
all different.
Standard-step method
Applying the energy equation between two sections a
finite
distance
AL
apart, Fig. 11.9, including the loss term, gives
V1
2
2
7i
-^
+
So
AL
+
Vi
=
2
-r
+
2/2
+
S AL
(11.6.2)
2g
2g
Solving for the length of reach gives
AL =
If
(Fi 2
- V *)/2g +
2
S-
conditions are
wanted a distance
known
AL
yi
-y
2
(11.6.3)
So
at one section,
away, a
e.g.,
section
1,
trial solution is required.
and the depth y 2
The procedure
is
is
as
follows
1.
Assume a depth y 2 then compute
A V
2.
For the assumed y 2 find an average
y,
;
Fig. 11.9
Gradually varied flow.
2,
2.
P, and
A
for the reach [for prismatic
APPLICATIONS OF FLUID MECHANICS
606
=
channels y
compute
+
(2/1
/2 with
2/2)
A
R
and
computed
3.
Substitute in Eq. (11.6.3) to compute AL.
4.
If
AL
EXAMPLE 11.4 At
10 m, mi = 2, 2/1 =
m
n =
0.035.
A =
l
Pi
=
*
61
-
section
of a canal the cross section
1
7 m, and at section
higher than at section
+
=
b2
1,
mi?/!
22/1
1!
2
=
vV
X
10
and
Assume
2
+
1
=
A =
2
=
6.9
15
X
6.9
15
+
2
is
trapezoidal, 61
=
downstream 200 m, the bottom is
and m2 = 3. Q = 200 m 3 /s,
2
X
72
=
168
m
10
+
2
X
7
\/2 2
+
2.
2
Vi
1
=
=
41.3
M=
1.19
m/s
m
- 00004
-
since section 2
2/2
.
15 m,
+
7
Since the bottom has an adverse slope,
tion,
2,
Determine the depth of water at section
+
612/1
depth] and
not correct, assume a new y 2 and repeat the procedure.
is
0.08
for this
S.
i.e., it is
larger than section
is
rising in the
downstream
probably
1, 2/2 is
less
direc-
than
2/1.
m; then
+
3
X
=
6.9 2
246
m
V = U% =
2
2
0.813
m/s
and
P =
2
The average
X
6.9
A =
VlO =
58.6
m
207 and average wetted perimeter
find an average hydraulic radius for the reach,
R =
P =
4.14 m.
50.0 are used to
Then
V_
0.035X200
_ / nQ V _ /
S \CnAR™) - U.O X 207 X 4.lW "
Substituting into Eq. (11.6.3) gives
(1.19 2
-
0.813 2 )/(2
X
0.000172
A
+7-6.9 _
+ 0.0004
larger y2 for example, 6.92
closer to the actual length.
,
9.806)
m, would bring the computed value
of length
STEADY FLOW
IN
OPEN CHANNELS
607
Numerical-integration method
A more
satisfactory procedure, particularly for flow through channels having
a constant shape of cross section and constant bottom slope,
is to obtain a
then to perform the integration
When AL is considered as an infinitesimal in Fig. 11.9, the
numerically.
rate of change of available energy is equal to the rate of head loss —AE/AL
differential equation in
terms of y and
L and
given by Eq. (11.6.1), or
£S+.-«+.)--(s&?
in
of
which Zq — SoL
bottom at L =
is
0,
the elevation of bottom of channel at L,
and L
is
measured positive
in the
zq is
the elevation
downstream
direction.
After performing the differentiation,
_Zg +4 _±.( *U'
dL
dL
\C AR
(U.6.5)
21 */
m
g
Using the continuity equation
VA = Q
and expressing dA =
which
T dy,
in
leads to
T is the liquid-surface width of the cross
section, gives
dV
VTdy _
A dL~
_
dL~
Substituting for
T !k
4- Q
^
°
and solving
for
dL
_
So
_
QT dy
A dL
2
V in
Eq. (11.6.5) yields
_ -^ =
dL
dL
(
nQ
V
\CmARwJ
gives
i - qr/gA'
- (nQ/CVW)*
dy
(1L6
.
6)
APPLICATIONS OF FLUID MECHANICS
608
After integrating,
-i
1 - Q*T/gA*
- (nQ/CrnARwy
So
dy
(11.6.7)
L is the distance between the two sections having depths y and y
When the numerator of the integrand is zero, critical flow prevails;
there is no change in L for a change in y (neglecting curvature of the flow and
in
which
x
nonhydrostatic pressure distribution at this section)
.
Since this
is
2.
not a case
of gradual change in depth, the equations are not accurate near critical depth.
When
the denominator of the integrand
is zero, uniform flow prevails and
no change in depth along the channel. The flow is at normal depth.
For a channel of prismatic cross section, constant n and So, the integrand
becomes a function of y only,
there
is
F(y)
m
i
So
-
Q>
W
- (nQ/CnAR219
)'
and the equation can be integrated numerically by plotting F(y)
against y as abscissa.
values of y
is
The area under the curve
the length
L
between the
(Fig. 11.10)
as ordinate
between two
sections, since
1/2
-/
F(y) dy
EXAMPLE
11.5
A
trapezoidal channel, b
0.001, carries 1000 cfs. If the depth
is
10
-
ft
10
ft,
m=
at section
1,
1,
n -
0.014,
£ =
determine the water-
surface profile for the next 2000 ft downstream.
To determine whether
the depth increases or decreases, the slope of
F(y)
Area=L
y2
y,
Fig. 11.10
y
Numerical integra-
tion of equation for gradually
varied flow.
STEADY FLOW
energy grade line at section
A =
by
P=
b
+ my
2
=
10
+
2y
y/m
2
+
200
-—
=
5.24
ft
X
1
38.2
OPEN CHANNELS
computed [using Eq.
+1X
10
=
1 is
IN
10 2
=
200
ft
609
(11.6.1)]:
2
ft
and
R =
38.2
Then
V
0.014X1000
/
„
S =
( l.49 X
X
200
=
5.
nMnndn
000
° 243
S < S hence, the specific energy is
can be accomplished only by increasing the depth downstream. Substituting into Eq. (11.6.7) gives
The depth
is
increasing,
and
f
~
The
v
J 10
-
l
greater than critical, and
X WT/A*
3.105
0.001
;
this
-
88.3/A 2 ^ 4
y
/3
following table evaluates the terms in the integrand
10 3
y
10.0
10.5
11.0
11.5
12.0
X
Numer-
Denomi-
A
P
R
T
ator
nator
200
215.2
232
247.2
264
38.2
39.8
41.1
42.5
43.9
5.24
5.41
5.64
5.82
6.01
30
0.8836
0.9037
0.9204
0.9323
0.9426
757
800
836
862
884
The
31
32
33
34
integral fF(y) dy can be evaluated
the area under
it
between y
=
by
F(y)
L
1167
1129
1101
1082
1067
574
1131
1677
2214
plotting the curve
10 and the following values of
y.
and taking
As F(y)
does not vary greatly in this example, the average of F(y) may be used for
each reach and when it is multiplied by Ay, the length of reach is obtained.
Between y = 10 and y =
1167
+
-h
1129
X
0.50
=
10.5
574
—
+
)
APPLICATIONS OF FLUID MECHANICS
610
Between y =
1129
+
1101
1
2
10.5
X
and y = 11.0
n rrt
0.50
rr ^
-
557
and so on. Five points on the water surface are known, so that it can be plotted.
Horizontal channels of great width
For channels of great width the hydraulic radius equals the depth; and for
So = 0; hence, Eq. (11.6.7) can be simplified.
The width may be considered as unity; that is, T = 1, Q = q and A = y
horizontal channel floors,
}
R =
L
y; thus
(1L6
--L*tiw**
i (—y (y im - ^
EXAMPLE
i3/3
t( y
)
From Eq.
x
0.7,
= -
(11.6.9),
and with
—
(
13 \0.015
——
+
4
X
Critical
yc
2/c
=
-
\g
)
=
4/3
n69
-
)
(
-
)
X
with x replacing
X
0.7
J
10.5/
(y
^
+
(y
Ky
413
340*/
10.5
0.7 13 ' 3 ;)
-
0.7 4 / 3 )
4'3
/
(
10.5 2
Y
)
V9.806/
/3
-
o o.
2.24
m
L
as distance
m /s,
-
1 *'*
v
—
9.30?/ 3/3
=
15
of 0.7
m. Find the equa-
0.015.
depth occurs [Eq. (11.5.3)] at
/<A 1/3
(
q
=
(
^— J
9.806 V0.015/
= -209.3 -
-^
m/s and a depth
n =
tion for the water-surface profile,
yi
4/3
(2/
After contracting below a sluice gate water flows onto a wide
11.6
horizontal floor with a velocity of 15
=
8)
performing the integration,
or, after
l = -
"
2
from section
1,
where
STEADY FLOW
IN
OPEN CHANNELS
611
The depth must increase downstream, since the specific energy decreases,
and the depth must move toward the critical value for less specific energy.
The equation does not hold near the
critical depth because of vertical accelerahave been neglected in the derivation of gradually varied flow. If
the channel is long enough for critical depth to be attained before the end of
the channel, the high-velocity flow downstream from the gate may be drowned
tions that
or a
jump may
occur.
must begin with
The
water-surface calculation for the subcritical flow
depth at the downstream end of the channel.
The computation of water-surface profiles with the aid of a digital com-
puter
critical
discussed after the various types of gradually varied flow profiles
is
are classified.
CLASSIFICATION OF SURFACE PROFILES
11.7
A study
has
of Eq. (11.6.7) reveals
definite characteristics.
its
many
types of surface profiles, each of which
The bottom
slope
is
classified as adverse,
and steep] and, in general, the flow can be above the
normal depth or below the normal depth, and it can be above critical depth
horizontal, mild, critical,
or below critical depth.
The various
used are
A very
assumed in the reduced equations which follow, with R = y.
profiles are plotted in Fig. 11.11; the procedures
discussed for the various classifications in the following paragraphs.
wide channel
is
Adverse slope profiles
When
the channel bottom rises in the direction of flow (So
resulting surface profiles are said to be adverse. There
is
is
negative), the
no normal depth, but
may be either below critical depth or above critical depth. Below
depth the numerator is negative, and Eq. (11.6.6) has the form
the flow
critical
where C\ and Ci are positive constants. Here F(y) is positive and the depth
increases downstream. This curve is labeled A 3 and shown in Fig. 11.11. For
depths greater than critical depth, the numerator is positive, and F(y) is
negative; i.e., the depth decreases in the downstream direction. For y very
large, dL/dy = 1 /So, which is a horizontal asymptote for the curve. At y = y c
dL/dy is 0, and the curve is perpendicular to the critical-depth line. This
,
curve
is
labeled
A
2.
612
APPLICATIONS OF FLUID MECHANICS
A-
Adverse
Horizontal
Mild
Horizontal
Horizontal
Critical
Steep
The various
Fig. 11.11
typical liquid-surface profiles.
Horizontal slope profiles
For a horizontal channel S = 0, the normal depth is infinite and flow may be
below critical depth or above critical depth. The equation has the form
either
dL = -Cy
l
i*(y*
- d)
dy
dL/dy is positive, and the depth increases downstream.
For y greater than critical (H 2 curve) dL/dy is negative, and
the depth decreases downstream. These equations are integrable analytically
for very wide channels.
For y
less
than
It is labeled
H
critical,
3.
,
Mild slope profiles
A
mild slope is one on which the normal flow is tranquil, i.e., where normal
depth y is greater than critical depth. Three profiles may occur, Mi, M 2 M 3
,
,
STEADY FLOW
IN
OPEN CHANNELS
613
depth above normal, below normal and above critical, and below critical,
For the M x curve, dL/dy is positive and approaches l/S for
very large y; hence, the Mi curve has a horizontal asymptote downstream.
As the denominator approaches zero as y approaches y the normal depth is
an asymptote at the upstream end of the curve. Thus, dL/dy is negative for
the M 2 curve, with the upstream asymptote the normal depth, and dL/dy =
at critical. The M 3 curve has an increasing depth downstream, as shown.
for
respectively.
,
Critical slope profiles
When
the normal depth and the critical depth are equal, the resulting profiles
are labeled Ci
and C 3
for
depth above and below
critical, respectively.
The
equation has the form
i
\-bif
,
with both numerator and denominator positive for Ci and negative for C3.
Therefore the depth increases downstream for both.
For large y, dL/dy
approaches 1/*S hence, a horizontal line is an asymptote. The value of dL/dy
;
at critical depth
also
is
0.9/
;
hence, curve Ci
is
convex upward.
Curve
C
3
is
convex upward, as shown.
Steep slope profiles
When
the normal flow
is
rapid in a channel (normal depth less than critical
depth), the resulting profiles
Si,
S2 S 3
are referred to as steep profiles: Si
,
above the normal and critical, S2 between critical and normal, and S 3
below normal depth. For curve Si both numerator and denominator are
positive, and the depth increases downstream approaching a horizontal
asymptote. For curve S 2 the numerator is negative, and the denominator
positive but approaching zero at y = y
The curve approaches the normal
depth asymptotically. The S 3 curve has a positive dL/dy as both numerator
and denominator are negative. It plots as shown on Fig. 11.11.
It should be noted that a given channel may be classified as mild for one
discharge, critical for another discharge, and steep for a third discharge,
since normal depth and critical depth depend upon different functions of the
is
.
discharge.
section.
The use
of the various surface profiles
is
discussed in the next
APPLICATIONS OF FLUID MECHANICS
614
11.8
A
CONTROL SECTIONS
small change in downstream conditions cannot be relayed upstream
when
downstream conditions do
not control the flow. All rapid flows are controlled by upstream conditions,
and computations of surface profiles must be started at the upstream end of a
the depth
is critical
or less than critical; hence,
channel.
Tranquil flows are affected by small changes in downstream conditions
and therefore are controlled by them. Tranquil-flow computations must start
at the downstream end of a reach and be carried upstream.
Control sections occur at entrances and exits to channels and at changes
under certain conditions. A gate in a channel can be a control
both the upstream and downstream reaches. Three control sections are
illustrated in Fig. 11.12. In a the flow passes through critical at the entrance
to a channel, and depth can be computed there for a given discharge. The
channel is steep therefore, computations proceed downstream. In b a change
in channel slope from mild to steep causes the flow to pass through critical at
the break in grade. Computations proceed both upstream and downstream
from the control section at the break in grade. In c a gate in a horizontal
channel provides control both upstream and downstream from it. The various
in channel slopes,
for
;
curves are labeled according to the classification in Fig. 11.11.
The hydraulic jump occurs whenever the
momentum
conditions required
by the
In Fig. 11.13, liquid issues from under a
gate in rapid flow along a horizontal channel. If the channel were short
enough, the flow could discharge over the end of the channel as an H 3 curve.
equation are
satisfied.
With a longer channel, however, the jump
occurs,
mmmmmmmmzm^
Fig. 11.12
Channel control sections.
and the
resulting profile
STEADY FLOW
IN
OPEN CHANNELS
615
/-Depth conjugate
/
tO
^3
^^^^^^^^^^^^^^^^^
Fig. 11.13
jump
Hydraulic
between
two
control
sections.
H and H
jump
In computing
computed, starting at the
gate (contraction coefficient must be known) and proceeding downstream
until it is clear that the depth will reach critical before the end of the channel
is reached. Then the H 2 curve is computed, starting with critical depth at the
end of the channel and proceeding upstream. The depths conjugate to those
along H 3 are computed and plotted as shown. The intersection of the conjugate-depth curve and the H 2 curve locates the position of the jump. The channel
may be so long that the H 2 curve is everywhere greater than the depth conjugate to H 3
A drowned jump then occurs, with H 2 extending to the gate.
consists of pieces of
these profiles for a
3
known
2
curves with the
discharge, the
H
3
curve
in between.
is
.
All sketches are
drawn
to a greatly exaggerated vertical scale, since
usual channels have small bottom slopes.
11.9
COMPUTER CALCULATION OF GRADUALLY VARIED FLOW
In Sec. 11.6 the standard-step and numerical-integration methods of computing water-surface profiles were introduced.
The
repetitious calculation in the
handled by digital computer. The program, listed in
Fig. 11.14, calculates the steady gradually varied water-surface profile in
any prismatic rectangular, symmetric trapezoidal or triangular channel.
The concepts of physical control sections in a channel must be understood in
latter
method
is
easily
order to use the program successfully.
Input data include the specification of the system of units (SI or ENGin the first columns of the first data card, followed by the channel
dimensions, discharge, and water-surface control depth on the second card.
If the control depth is left blank or set to zero in data, it is automatically
assumed to be the critical depth in the program. For subcritical flow the
control is downstream, and distances are measured in the upstream direction.
LISH)
)
,
1
)
WATER SURFACE PROFILE IN RECT, TRAPEZOIDAL OK TRIANGULAR CHANNEL.
XL=LENGTH, B=6JT WIDTH, Z=SIDE SLOPE , RN=NANNI NG N,SO=80T SLOPE, ti*FLOW,
YCCNT=CCNTRCL DEPTH. IF YCCNT=J. IN CATA, YCLNT IS StT EUCAL TC YC.
CATA ISI/'SI'/
AREA(YY)=YY*(e*Z*YY
PER(YYi=R*2.*YY*SCRT( l.+Z*Z)
YCRIT( YY) = I. -£**( B*-2.*Z*YY)/(G*A«EA(YY>**3)
YN0RM(YY) = l.-0*U*CCN/(AREA(YY)**3.333/PfcR(YY)**l.-»3 3)
DL( YYMYCRIUYY) / YNCRM YY )*SO)
FPM(YY) = CAM*(YY*YY'»(3'».5*Z*YY/3.)*U*CV(G*AREA(YY) ))
ENERGY (YY)=YY«-C*w/(2.*G*ARfcA(YY)**2)
READ (5,7,E.\D=99) 1 LNI T , XL , 6, Z ,PN ,SO, I YCCNT
FORMAT (A2/<: 10.1,3F10.4,F1C.6,2F10.3)
IF (IUNIT.EC.ISI) GC TC 10
GAM=62.4
)
(
(
5
7
G=32.2
CCN=(RN/1.466)**2/SC
WR ITE (6,9)
FORMAT (• ENGLISH
GO TO 12
9
UMTS
1
)
G=9.806
GAM=9802.
CCN=RN**2/S0
WRITE (6,11)
FORMAT (• SI UNITS'
1
WRITE (6,13) XL,*,B,Z,RN,SO
1
FORMAT (/• CHANNEL LENGTH* 1 fFlO.lt • DISCHARGE *',F12.3/
3
2' d=',F8.2,' Z=',F6.3,' RN=',F6.*,» SC=',F9.6)
NN=30
DETERMINATION OF CRITICAL AND NORMAL CEPTHa
IC
)
UP=30.
CN = 0.
YC=15.
DO ZO
14
1=1, 15
IF
(YCRIT(YC))
DN=YC
GO TO
14,21,15
20
15 UP=YC
2C YC=(UP*CN'M.5
21
(YCCNT. ECO.) YCCMT=YC
(SO.LE.C.) GO TC 33
UP=40.
IF
IF
CN = 0.
YN=20.
DO 30
IF
2 3
1
=
1,15
(YNOM(YN)) 23,31,24
DN=YN
GO TO 3C
24 UP=YN
3C YN=(UP*DN)*.5
31 WRITE
6,32) YN,YC
32 FORMAT (/• NCPMAL DEPTH=
GO TO 35
(
F7.3," CRITICAL DEPTH*' ,F7.3
33 YN=3.*YC
34 FORMAT (/•
CRITICAL DEPTH*' , F7. 3
WRITE (6,34) YC
IF
(YN.LT.YC) GJ TO 50
/"ILC, ADVERSE, OR hCRIZCNTAL CHANNEL, YN. GT. YC
IF
(YCUNT .LT.YC) GO TO 45
SUOCRITICAL FLCrf, YCCNT .GE. YC
SIGN=-1.
DY=(YCONT-YN) «.9?8/NN
WRITE (6,44) YCLNT
44 FORMAT (/• CCNTPOL IS LCWNSTREAf-, CtPTH = ,F7.
35
'
GO
TO
6C
SUPERCRITICAL FLCw
45
SIGN=l.
CY=(YC-YCCNT)/NN
WRITE (6,46) YCCNT
46 FORMAT (/• CCNTPCL IS LPSTKEAM, DEPTH* • ,F7.
30 TO 60
STEEP CHANNEL, YN .LT. YC
50 IF (YCONT.LE.YC) GC TC 55
SL8CRITICAL FLCw, YCCNT .CT.YC
SIGN=-l.
DY=(YCONT-YC)/NN
WRITE
(6,44) YCCNT
GC
6C
TO
Fig. 11.14
616
3)
3
STEADY FLOW
C
OPEN CHANNELS
617
YCCNT .LE. YC
SIGN=1.
KN=NN*2
IJY= (YN-YCCNT)*.<J98/NN
rfRITE (6,46) YCCNT
SL-PERCRIT ICAL FLCw,
55
6C
SL=0.
Y=YCOMT
E=ENER3Y( Y)
FM=FPM(Y)
wRITF (6,62)
62 FORMAT (//•
CISTANCE
WRITE (6,64) Sl,Y,E,FP
64 FORMAT (5X.FIC. ,2F 1 2 .3 , F 13.0
hATER SURFACE PRCFILE (ALCULATION
'
C
IN
CO
8J
1
= 1,
NN,
DtPTH
ENERGY
F*M«)
)
2
Y2=YC0NT*SIGN*CY«( i* I)
CX=OY*(DL(Y)*OL( Y2)»4.*CL(YCCNT»SIGN*I»0Y)
Sl=SL*DX
IF(SL.GT.XL) GC TC 62
Ml.
Y=Y2
E=ENERGY(Y)
FM^FPM(Y)
IF
l.EC.NN-l.ANO.SL.LT.C.) SL=XL
PC WRITE (6,64)SL,Y,E,FM
(
GO TJ
62
5
Y=Y2-SIGN*2.*CY*(SL-XL)/UX
E=ENERGY(Y
)
FH*FPM( Y)
WRITE
6,6<t)XL,Y,E,FM
(
<J9
GO TO
STOP
ENO
5
ENCLISF UNITS, EXAMPLE 11.7
toCC.
8.
0.8
ENCL ISH UNITS
16CC.
0.012
C.02S
873.
CO
8.
C.8
0.C12
C.CC02
873.
2.023
8.
C.6
J.C12
C.C002
673.
6.C
ENGLISH UNITS
1800.
Fig. 11.14
FORTRAN program
for water-surface profiles.
For supercritical flow the control depth
is
upstream, and distances are mea-
sured in the downstream direction.
The program begins with
several line functions to compute the various
and functions in the problem. After the necessary data input,
critical depth is computed, followed by the normal-depth calculation if normal
depth exists. The bisection method, Appendix E.3, is used in these calculations. The type of profile is then categorized, and finally the water-surface
profile, specific energy, and F +
are calculated and printed. Simpson's
rule is used in the integration for the water-surface profile, Appendix E.l.
The program can be utilized for other channel sections, such as circular
or parabolic, by simply changing the line functions at the beginning.
variables
M
EXAMPLE
11.7
A
trapezoidal channel,
B =
bottom slopes. The upstream portion is 600
8
ft,
ft long,
side slope
S =
=
0.025,
0.8, has two
and the down-
stream portion, 1800 ft long, S = 0.0002, n = 0.012. A discharge of 873 cfs
enters at critical depth from a reservoir at the upstream end, and at the downstream end of the system the water depth is 6 ft. Determine the water-surface
profiles throughout the system, including jump location.
e
C
APPLICATIONS OF FLUID MECHANICS
618
Yc
= 5.86
ft
:>-:-"-
f
4.13
Fig. 11.15
Yn
ft
= 10.49
ft
Solution to Example 11.7 as obtained from computer.
Three separate
sets of data,
shown
in Fig. 11.14, are
the results used to plot the solution as shown in Fig. 11.15.
needed to obtain
The
first set for
the steep upstream channel has a control depth equal to zero since
it will be
depth in the program. The second set is for
the supercritical flow in the mild channel. It begins at a control depth equal
to the end depth from the upstream channel and computes the water surface
downstream to the critical depth. The third set of data used the 6-ft down-
at automatically
assumed
critical
ENGLISH UNITS
E73.000
1800. C CISCHARGE =
CHANNEL LENGTH*
8.00 Z= C .800 RN=0.0120 S0= 0.CC0200
B=
5.861
NORHAL DEPTH= I 0.495 CRITICAL DEPTH*
3.C23
CCNTPOL IS UPST «EAP, OEFTM
F*M
ENERGY
DEPTH
DISTANCE
49635
14.954
3.023
O.C
46627
13.4
78
3.212
83.3
44C43
3.401
12.301
165.0
41821
3.591
11.356
2 44.9
10.596
J9910
3.780
322.4
38271
9.965
3.969
397.0
468.4
4.15 8
9.493
36671
4.347
535.6
9.101
35683
598.6
4.536
8.789
34666
A. 726
8.544
656.2
33860
4.915
8.356
707.7
33190
752.4
8.215
32662
5.1C4
789.1
8.115
32267
5.293.
31994
5. 462
8.C49
816.9
8.012
31835
834.6
5.672
31784
8.001
840.9
5.t6l
ENGLISH LNITS
6" 3. DOC
CHANNEL LENGTH=
1800.0 DISCHARGE =
B=
8.0J Z= C .8CC RN=0.0120 S0= 0.CCC200
NORMAL DEPTH= 1 0.495 CRITICAL CEPTH=
5.e6l
CCNTPCL IS DCKN STREAM, CEPTH= 6.J0O
DISTANCE
DEPTH
ENERGY
F*M
31811
O.C
3.UC6
6. CCO
32044
34.7
6.299
8.053
6.5«8
32504
8.14C
112.7
33174
243.7
b.897
8.259
439.5
34046
7.196
8.4C4
715.8
7.495
8.571
351C8
1092.
7.794
8.756
36354
1597.9
a.c<3
8.9 56
37 7 79
38241
1800.
8.ie3
9.C19
Fig. 11.16
Computer output, Example
11.7.
STEADY FLOW
OPEN CHANNELS
IN
619
stream depth as the control depth and computes in the upstream direction.
Figure 11.16 shows the computer output from the last two data sets. The
jump
last
is
located
two data
by
finding the position of equal
F
M from the output of the
+
sets.
TRANSITIONS
11.10
At entrances
to channels
and at changes
in cross section
and bottom
slope,
the structure that conducts the liquid from the upstream section to the
section
is
profile in
a transition. Its purpose
such a manner that
is
to change the shape of flow
minimum
losses result.
A transition for tranquil
flow from a rectangular channel to a trapezoidal channel
Applying the energy equation from section
11.17.
—+
yi
=
—+
2/2
+z+E
1
new
and surface
is
illustrated in Fig.
to section 2 gives
(11.10.1)
x
2g
2g
In general, the sections and depths are determined by other considerations,
By good
z must be determined for the expected available energy loss E x
and
.
with slowly tapering walls and flooring with no sudden changes
in cross-sectional area, the losses can be held to about one-tenth the difference
design,
i.e.,
between velocity heads
for accelerated flow
Elevation
Fig. 11.17 Transition from rectangular channel to trapezoidal
channel for tranquil flow.
and to about three-tenths the
APPLICATIONS OF FLUID MECHANICS
620
difference
between velocity heads
mechanics
is
EXAMPLE
11.8
In Fig.
rectangular section
6
ft
rise
is
8
11.17,
400
x
W=
—
400
—
- = 6.25
64
=
For rapid
flow,
wave
flows through the transition; the
cfs
wide; and
ft
=
2/1
8
The
ft.
wide at the bottom with side slopes 1:1, and y2
z in the bottom through the transition.
V =
V*
for retarded flow.
required in designing the transitions. 1
0.61
A =
2
=
trapezoidal section
7.5
ft.
is
Determine the
ft 2
101.25
2g
V =
-L
=
2
3.95
101.25
E =
0.24
x
2<7
—
/TV
2
400
0.3
(
\2flf
v \
) =
2
2
0.11
2gJ
After substituting into Eq. (11.10.1),
=
2
0.61
+
8
-
0.24
-
7.5
-
0.11
=
0.76
ft
The critical-depth meter 2 is an excellent device for measuring discharge
an open channel. The relationships for determination of discharge are
worked out for a rectangular channel of constant width, Fig. 11.18, with a
raised floor over a reach of channel about 3y c long. The raised floor is of such
height that the restricted section becomes a control section with critical
velocity occurring over it. By measuring only the upstream depth y h the
discharge per foot of width is accurately determined. Applying the energy
equation from section 1 to the critical section (exact location unimportant),
in
"
Fig. 11.18
1
2
L
%
Critical-depth meter.
A. T. Ippen, Channel Transitions and Controls, in H. Rouse (ed.), "Engineering Hydraulics," Wiley, New York, 1950.
H. W. King and E. F. Brater, "Handbook of Hydraulics," 5th ed., pp. 8-14 to 8-16,
McGraw-Hill,
New
York, 1963.
STEADY FLOW
IN
OPEN CHANNELS
621
including the transition-loss term, gives
W+
-z2#
=
yi
TV
+
z
+ —- +
yc
— /7
1
FA
—
2g/
2
C
1
I
\2^
10
2^
Since
^c
,
2
in
which
2£c is
2/i
+
—=
1.1
FromEq.
Vc
TV
_
#
c
the specific energy at critical depth,
+
z
1.033#c
(11.10.2)
(11.5.3)
= %EC =
(?-\
(11.10.3)
E
In Eqs. (11.10.2) and (11.10.3)
solved for
g
= 0.517^ ( yi -
Since #
q
=
=
eliminated and the resulting equation
/
I
The equation
y
is
1.1
yj
—
2
55
solved by
\ 3/2
—,
trial.
(11.10.4)
J
As
may
?/i
and
2 are
known and
is
small,
the two
it
#'s
first
are the
same the equation
the width of channel are known, a chart or table
Q
for
any
y±.
the right-hand
be neglected for an approximate q.
larger than the approximate q may be substituted on the right-
When
side.
+
-z+
x
term containing q
A value a little
z
Vi can be eliminated,
Fij/i,
0.5 17^ 2
hand
is
c
q,
may
is
solved.
Once
z
and
be prepared yielding
Experiments indicate that accuracy within 2 to 3 percent
may
be expected.
With
tranquil flow a
rapid flow a
jump
jump occurs downstream from the meter and with
occurs upstream from the meter.
APPLICATIONS OF FLUID MECHANICS
622
m
EXAMPLE
2/i
q
is
=
11.9 In a critical-depth meter 2
wide with
measured to be 0.75 m. Find the discharge.
0.517 (9.806 1
'2
)
As a second approximation
q
=
/
0.517 (9.806 1
'2
=
(0.45 3 / 2 )
) f
=
/
0.517 (9.806 1/2 ) I 0.45
0.50,
\ 3/2
+ -j— X
+
\
m the depth
0.3
2
55
0.45
and as a third approximation,
q
be
=
m /s
0.489
let q
z
0.5 2
=
0.512
J
m /s
2
0.513,
—— X
55
0.513 2
9.806
V
/2
=
)
0.513
m /s
2
/
Then
Q =
2
X
0.513
=
1.026
m /s
3
PROBLEMS
11.1
Show that for laminar flow
11.2
down an
to be ensured
per unit width cannot be greater than 500i\
inclined surface, the discharge
(See Prob. 5.10.)
Calculate the depth of laminar flow of water at 70°F down a plane surface making
of 30° with the horizontal for the lower critical Reynolds number.
(See
an angle
Prob. 5.10.)
11.3
Calculate the depth of turbulent flow at
R = VR/v = 500
20°C down a plane surface making an angle 6 of 30° with the
formula, n = 0.01; S = sin 6.
11.4
A
rectangular channel
is
n
=
lined with galvanized iron,
of
metal needed for each 100
11.5
A
to carry 1.2
0.011,
what
m of channel?
for flow of
horizontal.
m /s at a slope of 0.009.
3
is
the
minimum number
water at
Use Manning's
If
the channel
of square
is
meters
Neglect freeboard.
trapezoidal channel, with side slopes 2 on
1
(2 horizontal to
1 vertical), is
to
carry 600 cfs with a bottom slope of 0.0009. Determine the bottom width, depth, and
velocity for the best hydraulic section,
11.6
=
0.025.
A trapezoidal channel made out of brick, with bottom width 6 ft and with bottom
slope 0.001,
for the least
11.7
n
What
is
to carry 600 cfs.
number
What
should the side slopes and depth of channel be
of bricks?
radius semicircular corrugated-metal channel
is
needed to convey 2.5
m /s
3
STEADY FLOW
-*|
40
400
ft
IN
OPEN CHANNELS
623
ft
Fig. 11.19
a distance of
km
1
with a head
m? Can you
loss of 2
find another cross section that
requires less perimeter?
11.8 Determine the best hydraulic trapezoidal section to convey 85
bottom slope of 0.001. The lining is finished concrete.
11.10
4
ft,
with a
11.11
For 7000
S =
cfs flow in
0.0009 and y
=
8
ft.
when the depth over the floodway
the section of Fig. 11.19
calculate the energy gradient.
For 25,000
the floodway
cfs flow
when the
11.12
Draw an F
11.13
Draw
through the section of Fig. 11.19, find the depth of flow in
slope of the energy grade line
0.0004.
is
+ M curve for 2.5 m /s per meter of width.
3
the specific-energy curve for 2.5
chart as Prob. 11.12.
0.5
3
Calculate the discharge through the channel and floodway of Fig. 11.19 for
11.9
steady uniform flow, with
is
m /s
What
is
m /s
3
per meter of width on the same
the energy loss in a
jump whose upstream depth
is
m?
11.14
Prepare a plot of Eq. (11.4.7)
11.15
With
11.16
q
=
100 cfs/ft and
Fi
=
12,
determine v h
Determine the two depths having a
specific
yi,
and the conjugate depth y2
energy of 6
ft for 1
.
m /s per meter
3
of width.
11.17
What
is
the critical depth for flow of 18 cfs per foot of width?
11.18
What
is
the critical depth for flow of 0.3
m /s through the cross section of Fig.
3
5.49?
Determine the critical depth for flow of 8.5 m 3 /s through a trapezoidal channel
with a bottom width of 2.5 m and side slopes of 1 on 1.
11.19
11.20
An unfinished
concrete rectangular channel 12
ft
wide has a slope of 0.0009.
It
and has a depth of 7 ft at one section. By using the step method and
taking one step only, compute the depth 1000 ft downstream.
carries
480
11.21
Solve Prob. 11.20 by taking two equal steps.
cfs
What
is
the classification of this
water-surface profile?
11.22
A very wide gate
(Fig. 11.20) admits water to a horizontal channel. Considering
APPLICATIONS OF FLUID MECHANICS
624
Gate
:v=o
20
Cc = 0.86
C-0.96
ft:
tt
o
I
Horizontal
floor,
n=0.015
Fig. 11.20
the pressure distribution hydrostatic at section
the discharge per foot of width
11.23
is
If
65.2
when y =
3.0
0,
compute the depth at section
of Fig. 11.20 is 2 ft and the discharge per foot
compute the water-surface curve downstream from the gate.
the depth at section
cfs,
11.24
Draw
11.25
If
and
ft.
of
width
the curve of conjugate depths for the surface profile of Prob. 11.23.
the very wide channel in Fig. 11.20 extends downstream 2000
ft
and then has
a sudden dropoff, compute the flow profile upstream from the end of the channel for
q
=
65.2 cfs/ft
in the channel.
11.27
is
for gradually varied flow.
Using the results of Probs. 11.24 and 11.25, determine the position of a hydraulic
11.26
jump
by integrating the equation
(a)
is 0.6 m, and the velocity
compute the depth at the downstream end of the
by use of the computer program given in Fig. 11.14,
In Fig. 11.21 the depth downstream from the gate
12 m/s. For a very wide channel,
adverse slope. (6) Solve part (a)
or write a similar program to obtain the solution.
Sketch (without computation) and label
11.28
can be obtained from Fig. 11.22 by varying
for 2 2
11.29
<
Z\,
all
z h z2 ,
the liquid-surface profiles that
and the lengths
of the channels
with a steep, inclined channel.
In Fig. 11.22 determine the possible combinations of control sections for various
6.012
Fig. 11.21
STEADY FLOW
IN
OPEN CHANNELS
>
z2 ,
with the inclined channel
values of z h z2 and various channel lengths for
always steep.
Sketch the various liquid-surface
11.30
obtained by varying channel length for
z2
zx
profiles
>
other discharge.
Use the computer program
11.32
to locate the hydraulic
flow of
1
m /s,
3
stream depth
n
is
=
0.8
jump
for
one discharge and steep for an-
to be critical?
in Fig. 11.14 or a similar
program you have written
in a 90° triangular channel, 0.5
0.015, So
=
0.001.
for Fig. 11.22
Z\.
Show an example of a channel that is mild
What discharge is required for it
11.31
and control sections
625
The upstream depth
km
is
long, that carries a
0.2
m, and the down-
m.
Design a transition from a trapezoidal section, 8 ft bottom width and side
on 1, depth 4 ft, to a rectangular section, 6 ft wide and 6 ft deep, for a flow of
250 cfs. The transition is to be 20 ft long, and the loss is one-tenth the difference between velocity heads. Show the bottom profile, and do not make any sudden changes
11.33
slopes
1
in cross-sectional area.
from a rectangular channel, 2.6 m wide and 2 m deep, to a trapebottom width 4 m and side slopes 2 on 1, with depth 1.3 m has a loss
3
four-tenths the difference between velocity heads. The discharge is 5.6 m /s. Determine the difference between elevations of channel bottoms.
11.34
A transition
zoidal channel,
11.35
A critical-depth meter 20 ft wide has a rise in bottom of 2.0 ft.
depth of 3.52
11.36
ft
With flow approaching a
ber of 10, what
11.37
is
the
critical-depth
minimum amount
meter
site at
6
m/s and a Froude num-
the floor must be raised?
In open-channel flow
(a) the hydraulic grade line
is
always parallel to the energy grade
(b)
the energy grade line coincides with the free surface
(c)
the energy and hydraulic grade lines coincide
(d) the hydraulic grade line can never rise
(e)
For an upstream
determine the flow through the meter.
the hydraulic grade line and free surface coincide
line
APPLICATIONS OF FLUID MECHANICS
626
11.38
Gradually varied flow
(a) steady uniform flow
(6)
(c)
unsteady uniform flow
(e)
none
11.39
(a)
(d)
of these
unsteady nonuniform flow
Tranquil flow must always occur
(6)
(e)
below normal depth
on adverse slopes
(c)
above
an adverse channel
(c)
in
(e)
in a steep channel
Flow
jump
(6)
in a
mild channel
at critical depth occurs
when
changes in upstream resistance alter downstream conditions
(6)
the specific energy
(c)
any change in depth requires more specific energy
the normal depth and critical depth coincide for a channel
the velocity is given by \/2gy
(e)
11.42
The
=
depth)
(a)
y=2b
y
11.43
The
is
a
maximum
for a given discharge
best hydraulic rectangular cross section occurs
(b)
y=b
(c)
y
=
6/2
best hydraulic canal cross section
(a)
the least expensive canal cross section
(6)
the section with
(c)
the section that has a
(d)
the one that
(e)
none of these answers
11.44
(a)
(c)
(e)
depth
in a horizontal channel
(d)
(a)
(d)
critical
Supercritical flow can never occur
(a) directly after a hydraulic
11.41
steady nonuniform flow
(d)
answers
above normal depth
below critical depth
11.40
is
is
y
an M3 curve to an Mi curve
an S3 curve to an Si curve
below critical depth to above
=
62
(e)
(6
y
=
=
bottom width,
6/5
defined as
minimum roughness coefficient
maximum area for a given
has a minimum perimeter
The hydraulic jump always
flow
occurs from(6)
(d)
an H 3 curve to an H 2 curve
below normal depth to above normal depth
critical
-r-r Horizontal
Fig. 11.23
(d)
when
depth
STEADY FLOW
IN
OPEN CHANNELS
627
Fig. 11.24
11.45
(a)
depth in a rectangular channel
Critical
Vn
11.46
VWy
(b)
(a)
Q T/gA* =
(e)
none of these answers
The
QT*/gA
(6)
1
specific
2
11.48
3
(6)
(d)
2
=
1
4.26
11.50
tions of
5.43
(c)
12.02
H2
(d)
9.86
The number
zi,
z2
,
The
3
loss
(Vi-
expressed by
Q A*/gT 2 =
(e)
(c)
(d)
22.15
(e)
H3
(d)
A2
of channel in Fig. 11.24
4
(d)
V2)'
(6)
(a)
(b)
(c)
(d)
(e)
m/s y
ft lb/lb.
•
critical-depth
=
1
m,
in
The discharge
none of these answers
in Fig. 11.23
5
(e)
0.1
W-
is (zi
9^ &i)
6
is
about
(Vi- V2Y
V22 )
(c)
0.3
29
(e)
A
4.43
1
is
classified as
A3
(e)
2<7
none of these answers
2<7
11.52
7=
Q 2 /gA* =
is
through a diverging transition
0.1
0.3
(d)
1
none of these answers
from flow under the gate
2<7
(d)
lls
of different possible surface profiles that can occur for
and length
(b)
17
(c)
(c)
Wg)
is
profile resulting
(6)
2
11.51
(b)
The
(a) Hi
(a)
is
The minimum possible specific energy for a flow is 2.475
11.49
(a)
M
2
(c)
per foot of width, in cubic feet per second,
(a)
VqTg
energy for the flow expressed by
meter-newtons per newton,
(a)
by
depth in a nonrectangular channel
Critical
2
11.47
Vgy
(c)
expressed
is
meter
measures the depth at the critical section
is always preceded by a hydraulic jump
must have a tranquil flow immediately upstream
always has a hydraulic jump downstream
always has a hydraulic jump associated with
it
any
varia-
628
APPLICATIONS OF FLUID MECHANICS
REFERENCES
New York,
New York, 1959.
New York, 1966.
Bakhmeteff, B. A.: "Hydraulics of Open Channels," McGraw-Hill,
Chow, V.
T.: "Open-channel Hydraulics," McGraw-Hill,
Henderson, F. M.: "Open Channel Flow," Macmillan,
1932.
12
UNSTEADY FLOW
Up
to this point practically
all
flow cases examined have been for steady flow
or have been reducible to a steady-flow situation.
larger
equipment
is
As technology advances and
constructed or higher speeds employed, the problems of
hydraulic transients become increasingly important.
The hydraulic
tran-
but produce excessive noise,
fatigue, pitting due to cavitation, and disruption of normal control of circuits.
Owing to the inherent period of certain systems of pipes, resonant vibrations
may be incurred which can be destructive.
Hydraulic-transient analysis deals with the calculation of pressures and
velocities during an unsteady-state mode of operation of a system. This may
be caused by adjustment of a valve in a piping system, stopping a pump, or
innumerable other possible changes in system operation.
The analysis of unsteady flow is much more complex than that of steady
sients not only cause dangerously high pressures
Another independent variable enters, time, and equations may be
than ordinary differential equations. The
digital computer is ideally suited to the solution of such problems because of
its large storage capacity and its ability to operate at very high computing
rates. This chapter is in two parts: the first part dealing with closed-conduit
transients, and the second part with open-channel unsteady flow.
flow.
partial differential equations rather
U
by its application to
and the establishment of flow
in a system. Equations are next developed for cases with more severe changes
in velocity that require consideration of liquid compressibility and pipe-wall
elasticity (usually called waterhammer)
The open-channel cases are the
positive and negative surge waves in a frictionless prismatic channel with
instantaneous gate changes, the more general case of flood routing through a
First, oscillation of
pipelines
and
a
tube
is
studied, followed
reservoirs, the use of surge tanks,
.
629
APPLICATIONS OF FLUID MECHANICS
630
prismatic channel with friction, and flow over plane inclined surfaces due to
rainfall.
FLOW
IN
CLOSED CONDUITS
The unsteady
flow cases in closed conduits are treated as one-dimensional
The equation
of motion or the unsteady
and the unsteady continuity equation
takes special forms. With nonlinear resistance terms for friction and other
effects, the differential equations are frequently solved by numerical methods,
usually by digital computer.
distributed-parameter problems.
linear-momentum equation
is
used,
OSCILLATION OF LIQUID
12.1
Three cases of
IN
oscillations of liquid in
frictionless liquid,
(2)
A U TUBE
a simple
U
tube are of interest: (1)
laminar resistance, and (3) turbulent resistance.
Frictionless liquid
For the
frictionless case, Euler's equation of
(3.5.3) ]
may
be applied.
dv
dv
dp
dz
-—
+ g- + v- + 1
p ds
ds
When
ds
sections
grated from
Vi
~
Vi
+
P
1
1
to
(Fig. 12.1)
^-.O+^ + 0>
= p 2 and
=
pi
g(z2
-z = -L^
v\
and the equation
v 2 ; also
dv/dt
is
= O
L
is
inte-
independent of
(12.1.1)
s,
hence
(12.1.2)
1)
which
is
for incompressible flow
But
in
=
dt
and 2 are designated
2,
motion in unsteady form [Eq.
It is
the length of liquid column.
By
changing the elevation datum
to the equilibrium position through the menisci, g{z 2
—
zi)
=
2gz; since v is
UNSTEADY FLOW
Oscillation
Fig. 12.1
uid
a
in
dt
2
The
=
only, dv/dt
t
dv
~
liq-
U tube.
a function of
d2 z
of
631
may
be written
dv/dt, or d 2 z/dt 2
,
2g
dt~~
~L
Z
(12.1.3)
general solution of this equation
Ci cos
*/—
t
r
+
C2
C
^
sin a/
is
t
which Ci and C2 are arbitrary constants of integration. The solution is
by differentiating twice and substituting into the differential
equation. To evaluate the constants, if z — Z and dz/dt =
when t = 0,
then Ci = Z and C2 = 0, or
in
readily checked
z
= Z
cos
fa
*/—
(12.1.4)
t
This equation defines a simple harmonic motion of a meniscus, with a period
for a complete oscillation of 2ir y/L/2g. Velocity of the column may be obtained
by
differentiating z with respect to
t.
EXAMPLE 12.1 A frictionless fluid column 2.18 m long
z = 0.5 m. Find (a) the maximum value of z, (b)
when
and
(c)
the period.
has a speed of 2
the
m/s
maximum speed,
APPLICATIONS OF FLUID MECHANICS
632
Differentiating Eq. (12.1.4), after substituting for L, gives
(a)
dz
„
= -3Zsm3*
—
.
dt
If
h
0.5
the time
is
= Z
cos
when
z
=
and dz/dt =
0.5
-2 = -SZ sin 3*i
3*i
Dividing the second equation by the
tan
or
=
0.5/ (cos
0.309 s, sin 3*i = 0.8, and cos 3*i =
= 0.833 m, the maximum value of z.
The maximum speed occurs when
0.927 rad, k
=
J—
=
0.6.
Then Z =
0.5/0.6
3Z =
or
1,
The period
(c)
2tt
=
3*i)
(b)
sin 3*
equation gives
first
= |
3*i
3*i
2,
3
X
0.833
-
2.466
m/s
is
= 2.094 s
2g
Laminar resistance
When
a shear stress r at the wall of the tube resists motion of the liquid
column, it may be introduced into the Euler equation of motion along a streamline (Fig. 3.8)
by the mass
1
dp
dz
+ ^7 +
"f
ds
p ds
This equation
tion
is
made
The
.
resistance in length 5s
of the particle
dv
T
is
dv
pA
5s, it is
4r
„
=
-^
pD
f
t ttD
8s.
and Eq.
After dividing through
(12.1.1)
good
for either laminar or turbulent resistance.
+
'
that the frictional resistance in unsteady flow
tube
becomes
/„« „ -x
(12 L5)
7,
dt
stress at the wall of a
=
is
/I>,
+
ds
the steady flow at the same velocity.
ro
4r
From
is
The assump-
the same as for
the Poiseuille equation the shear
is
(12.1.6)
UNSTEADY FLOW
633
After making the substitution for r in Eq. (12.1.5) and integrating with respect to
s
as before,
__ =
S2pvL
dv
_ Zl)+LT _
+
.
,
,
ff(Z2
=
Setting 2gz
g(z 2
—
„
changing to total derivatives, and replacing
Zi),
v
by
dz/dt give
d2z
%2v dz
2g
M + Wlt + L* =
°
the column
is
In
effect,
(
12 17)
-
assumed to have the average velocity dz/dt at any
cross section.
By
z
=
Cie at
substitution
+ Ce
bt
2
can be shown to be the general solution of Eq. (12.1.7) provided that
,
32?
2g
S2v
,
2g
and C2 are arbitrary constants of integration that are determined by given
values of z and dz/dt at a given time. To keep a and b distinct, since the equations defining them are identical, they are taken with opposite signs before
the radical term in solution of the quadratics; thus
C\
//16A
16?
2
2g
//16A
16?
L
S
2g
L
To
simplify the formulas,
m =
16?
n =
w
//16A
if
2
Aw)
"
2g
l
then
z
=
Ci exp
When
the
(
— mt +
initial
nt)
+
condition
C2 exp — mt —
nt)
(
is
taken that
t
=
0, z
=
0,
dz/dt
= V
,
then by
APPLICATIONS OF FLUID MECHANICS
634
substitution Ci
= — C2 and
,
= de-mt (e nt -
z
e~ nt )
(12.1.8)
Since
_
p nt
p-nt
= smh
nt
2
Eq. (12.1.8) becomes
=
z
By
dz
—
2Cie~ mt sinh nt
differentiating with respect to
=
2Ci(
— me~ mt sinh nt +
and setting
= V
for
<
ne~ m cosh n£)
*
=
gives
- 2Cin
Fo
=
since sinh
z
rf-s/d^
t,
=
V
—
n
e
~ mt
and cosh
0=1. Then
sinh nt
(12.1.9)
This equation gives the displacement
z of
one meniscus of the column as a
function of time, starting with the meniscus at
with velocity
Two
16*
^
V
z
=
when
t
=
0,
and
rising
.
principal cases 1 are to be considered.
When
J2g
is so great that the motion is damped out
becoming negative, Fig. 12.2 (m/n = 2). The
time Jo for maximum z to occur is found by differentiating z [Eq. ( 12. 1.9) ] with
n
is
a real number and the viscosity
in a partial cycle with z never
1
A
third case,
lQv/D 2
sulting oscillation
is
=
mt The re\/2g/L, must be treated separately, yielding z = V te~
and is a limiting case of IQv/D 2 > \/2g/L-
for a partial cycle only
.
UNSTEADY FLOW
635
0.6
0.5
\ m
i
0.4
0.3
\
0.2
zn
\
Vn
m
0.1
—
*"
^*>
;i
l
;
\
l\
t
\
7
8
9
10
nt
n't
liquid in a
meniscus as a function
U tube with laminar resistance.
respect to
t
Position of
Fig. 12.2
—
=
=
and equating
—
n
dt
(
of
time for oscillation of
to zero,
—me~ mt sinh nt
+
ne~ mt cosh nt)
or
tanh
nt
= —
(12.1.10)
7H
Substitution of this value of
t
into Eq. (12.1.9) yields the
maximum
displace-
ment Z
Z =
o_ _ /m
s/m? — n2 \m
The second
16*>
I
-
n \ m/2n
+
n)
case,
when
JL_
/m
*2g\m
-
n \ m/2n
+ nJ
(12.1.11)
APPLICATIONS OF FLUID MECHANICS
636
results in a negative term, within the radical.
,
V-|?-(£M-@) -«
in
which
= \/—l and n' is a real number.
*
Replacing n by
in' in
Eq.
(
12.1.9)
produces the real function
z
= —,
e~ mt sinh in't
=
—
-
e~ mt sin n't
(12.1.12)
n
in
since
sin n't
=
- sinh
in't
i
The
motion of
resulting
plitude, as
mum
or
dz/dt
=
shown
minimum
0,
tannic =
z is
an
oscillation
about
z
displacement
=
m/n' = \.
obtained from Eq.
in Fig. 12.2 for the case
is
with decreasing amThe time t of maxi(12.1.12) by equating
producing
n
—
(12.1.13)
m
There are an indefinite number of values of t satisfying this expression, corall the maximum and minimum positions of a meniscus. By
substitution of t into Eq. (12.1.12)
responding with
/
/ „
\/n' 2
+
,
m?o
eX P
I
m
",
n'\
tan_1
_
)
= V
EXAMPLE
12.2
A
1.0-in-diameter
m
n'\
\L
I
exp - - tan- Jmj
*2g
\ n'
1
(
)
(12.1.14)
tube contains oil, v = 1 X 10~ 4 ft 2 /s,
Applying air pressure to one of the tubes
U
with a total column length of 120 in.
makes the gage difference 16 in. By quickly releasing the air pressure the oil
column is free to oscillate. Find the maximum velocity, the maximum Reynolds number, and the equation for position of one meniscus z, in terms of time.
The assumption is made that the flow is laminar, and the Reynolds num-
UNSTEADY FLOW
The constants
ber will be computed on this basis.
16*>
m = -— =
m and n are
16 X 10————
= 0.2302
4
//16A 2
n =
637
V(^J
-
2g
I
=
i
V
'
23022
"
X
= V-l
~^—
2
32.2
.
2.527
=
fi.527
or
=
n'
2.527
Equations (12.1.12) to (12.1.14) apply to this case, as the liquid will
above and below z = 0. The oscillation starts from the maximum
position, that is, Z = 0.667 ft. By use of Eq. (12.1.14) the velocity (fictitious)
when z = at time t before the maximum is determined to be
oscillate
z„
y°
=
fa
Vr
exp
(m
t
n'\
tan
\?AA
„
°' 667
m) =
Vic"
/0.2302
exp
\Tm
,
tan
2.527 \
oara)
1.935 ft/s
and
"'
tan n
*
=
m
in
=
z
= Z at
t
=
0.
+
0.586)] sin 2.527(*
The maximum
Differentiating with respect to
dz
V = - =
^ro.
°' 586 S
into Eq. (12.1.12)
0.766 exp [-0.2302(£
which
'
5^ =
,
tan
IEZ7
Hence by substitution
z
2 527
1
=
t0
+
0.586)
velocity (actual) occurs for
t
>
0.
to obtain the expression for velocity,
t
-0.1763 exp [-0.2302(*
+
0.586)] sin 2.57(*
+
0.586)
dt
+
1.935 exp
[-0.2302 (t
Differentiating again with respect to
mum V produce
tan2.527(Z
+ 0.586)
= -0.1837
t
+
0.586)] cos 2.527 (t
and equating
+ 0.586)
to zero to obtain maxi-
APPLICATIONS OF FLUID MECHANICS
638
The
solution in the second quadrant should produce the desired
= 0.584 s. Substituting this time into the expression
V = —1.48 ft/s. The corresponding Reynolds number is
t
VD
R =
=
1.48
(A X
10 4 )
=
for
maximum,
V
produces
1234
V
hence the assumption of laminar resistance
is justified.
Turbulent resistance
In the majority of practical cases of oscillation, or surge, in pipe systems there
is
With
turbulent resistance.
large pipes
and tunnels the Reynolds number
when the
is
very near to zero. The
assumption of fluid resistance proportional to the square of the average veIt closely approximates true conditions, although
locity is made (constant /)
it yields too small a resistance for slow motions, in which case resistance is
almost negligible. The equations will be developed for / = const for oscillation within a simple U tube. This case will then be extended to include oscillation of flow within a pipe or tunnel between two reservoirs, taking into account
the minor losses. The assumption is again made that resistance in unsteady
large except for those time periods
velocity
is
.
flow
is
given by steady-flow resistance at the same velocity.
Using Eq. (5.10.2) to substitute for
1
P
dp
dz
dv
dv
—
0— + »- + - + —
+
ds
ds
2D
fv
y
ds
When
first
dv
Since
2
=
this equation is integrated
=
v is
Eq. (12.1.5) leads to
(12.1.15);
V
dt
term drops out as the
out as dv/ds
r in
0; the fourth
fv
from section
limits are
and
p
fifth
=
1
to section 2 (Fig. 12.1), the
in each case
;
the third term drops
terms are independent of
s,
hence
2
a function of
t
only the partial
may be
replaced with the total de-
rivative
S+s>-m+*-°
The
absolute-value sign on the velocity term
(12
is
-
U6)
needed so that the resistance
UNSTEADY FLOW
opposes the velocity, whether positive or negative.
d2 z
f dz
dz
2
2D~dt
dt
dt
This
is
20
+ /z
By
expressing v
=
639
=
dz/dt,
(12.1.17)
a nonlinear differential equation because of the ^-squared term.
It
can be integrated once with respect to t, but no closed solution is known
It is easily handled by the Runge-Kutta methods
(Appendix E) with the digital computer when initial conditions are known:
t - t
dz/dt = 0. Much can be learned from Eq. (12.1.17), however,
z = z
by restricting the motion to the — z direction; thus
for
the second integration.
,
,
2
dt
~ 2D
2
2a
+ -^
L
\dt)
=
(12.1.18)
The equation may be
integrated once, 1 producing
0'=^ (+!)+in
which
for dz/dt
C
C is the
= 0,
±gD 2
PL_^
-_
constant of integration.
+ _j exp
1
To
evaluate the constant,
if z
=
zm
^__j
and
ey^K-KfK-^]
By
V
substitution of
~
dz
d 2z
~dt
~dt
dpdz
dp
~
2
dt
~
" P
dp
'
dz dt
Jz
then
dp
P
dz
f
2D V
tgz
2
x L
,
n
**'*>. For
This equation can be made exact by multiplying by the integrating factor e
3d
Equations,"
Differential
"Elementary
Rainville,
D.
Earl
see
the detailed method
ed., Macmillan, New York, 1964.
APPLICATIONS OF FLUID MECHANICS
640
Although
this equation
cannot be integrated again, numerical integration of
t.
The equation, however, can
particular situations yields z as a function of
be used to determine the magnitude of successive oscillations. At the instants
of maximum or minimum z, say zm and 2 m +i, respectively, dz/dt = 0, and Eq.
(12.1.20) simplifies to
/-
/2m\
V + -d)
.
exp
/
fZm\
V"
T)
=
I.
l
\
fZm+l\
fzm+ l\
+ -^) exp (
.
/loiOIN
(12L21)
V "d"J
Since Eq. (12.1.18) the original equation, holds only for decreasing z, z m must
be positive and z m+ i negative. To find zm +2 the other meniscus could be con,
sidered
and
Zm+i as a positive
number
substituted into the left-hand side of
the equation to determine a minus z m+ 2 in place of zm +i on the right-hand side
of the equation.
EXAMPLE 12.3 A U tube
has a maximum oscillation
tion of the surface
FromEq.
(l
=
consisting of 50-cm-diameter pipe with /
(Fig. 12.1) of zm
=
0.03
6 m. Find the minimum posi-
and the following maximum.
(12.1.21)
+
^^) -P (-5^f- )=
+
0.06zm+ i) exp (-0.06zw+ i)
6
+ 0.06^,)
(1
exp
(-0.06^)
or
(1
which
(1
+
is
satisfied
=
0.9488
by z m +i = —4.84 m. Using z m = 4.84
0.06z w+ i) exp
(-O.O&w+i)
+
-
(1
=
0.9651
0.06
X
m in Eq.
4.84) exp
(-0.06
(12.1.21),
X
4.84)
which is satisfied by zm+ i = —4.05 m. Hence, the minimum water surface
z = —4.84 m, and the next maximum is z = 4.05 m.
= fz/D, then
Equation (12.1.21) can be solved graphically. If
is
(f>
F(4>)
=
(1
+
(12.1.22)
<t>)e-+
which is conveniently plotted with F(<j>) as ordinate and both — and +<£
are found as
on the same abscissa scale (Fig. 12.3). Successive values of
indicated by the dashed stepped line.
<f>
<\>
UNSTEADY FLOW
641
1.0
0.8
\
0.6
F(<t>)
\-<t>
0.4
0.2
0.4
0.2
+
Fig. 12.3
0.8
0.6
(j>,
or
1.0
-0
F(tf>) = (1+ 0)e~*
maximum and minimum
Graphical solution of
yielding successive
displacements.
Although
z
cannot be found as a function of
given as a function of
by equating dV 2 /dz =
z,
since
V =
dz/dt.
t
from Eq.
(12.1.20),
The maximum value
to find its position
z'
;
of
V is
V
is
found
thus
dz
After solving for
z'
'—-Hi+Is)
and
after substituting
back into Eq. (12.1.20),
(12.1.23)
Oscillation of two reservoirs
The equation
same
for oscillation of
two
as that for oscillation of a
U
reservoirs connected
by a
pipeline
is
the
tube, except for value of constant terms.
APPLICATIONS OF FLUID MECHANICS
642
Ai
A2
^^
j;*2
-
-%
-
i
Oscillation
Fig. 12.4
of
two
res-
ervoirs.
If z\
and
represent displacements of the reservoir surfaces from their equi-
z2
librium positions (Fig. 12.4), and
ticle
if z
within the connecting pipe from
zA =
ziAi
=
z2
A
represents displacement of a water parits
equilibrium position,
2
which A\ and A 2 are the reservoir areas, assumed to be constant in this
Taking into account minor losses in the system by using the
length
equivalent
L e of pipe and fittings plus other minor losses, the Euler
in
derivation.
equation with resistance included
-yA(zi
+
z2 )
+
for z decreasing.
d?z_
j_U
2
2D L
dt
yAfL
2gD \dt/
iih
is
yAL dH
g
dt 2
After simplifying
/dz\ 2
gA(\
1
\
(12.1.24)
After comparing with Eq. (12.1.18), /
gA(l/Ai
+
V
D
J
L
1/A 2 )/L. InEq.
(12.1.22)
is
replaced
by fL e /L, and 2g/L by
UNSTEADY FLOW
643
EXAMPLE 12.4 In Fig. 12.4 a valve is opened suddenly in the pipeline when
L = 2000 ft, A = 200 ft2 A 2 = 300 ft D = 3.0 ft, / = 0.024,
Zl = 40 ft.
losses
are 3.50V /2g. Determine the subsequent maximum negaand minor
2
x
,
,
2
tive
and positive surges
in the reservoir A\.
The equivalent length
^
™
minor
losses
is
= 438ft
0.024
/
+
Then L = 2000
e
X
40
ZlAi
A
= 2438 and
438
200
2.25tt
The corresponding
* =
of
f~ =
<f>
is
=
0.024 (ffltt)(Uf*)
11.04
and
=
F(4>)
which
is
=
F(<f>)
which
+
(1
satisfied
+
(1
=
4>)e-+
=
I)*" 1
is satisfied
by
-1 X
<*>LZ)
fL e
and
for
<j>
=
0.593
2m
0.024
X
0.024
fc
X
=
=
+
(1
2000
X
X
3
0.000193
4)<r*
-0.593. The values of z m
X
3
= -102.6
2438
0.593,
=
60.9
2438
The corresponding
*-
=
2000
Zm
'
0.736
</>
11.04)e- 1104
— 1.0. Then
^
by
+
(1
values of
z\
are
^- -102.6 X^=
-3.63ft
are, for
<f>
= -1,
APPLICATIONS OF FLUID MECHANICS
644
and
*i
=
12.2
=
X——
zUU
2.25tt
60.9
2.15 ft
ESTABLISHMENT OF FLOW
The problem of determination of time for flow to become established in a pipewhen a valve is suddenly opened is easily handled when friction and minor
line
losses are
is
taken into account. After a valve
available to accelerate the flow in the
head
creases, the accelerating
is
is
opened
is
first
H
the head
but as the velocity inand minor losses. If L e
(Fig. 12.5)
,
instants,
reduced by friction
the equivalent length of the pipe system, the final velocity
V
is
given
by
application of the energy equation
L
H ~
-ff >Yl
(12.2.1)
D2g
The equation
yA
of
motion
is
yALdV
«-'§S
dt
g
Solving for dt and rearranging, with Eq. (12.2.1), give
LTV
J
~
gH
v
dV
t
J
VJ
V
2
H
X
Fig. 12.5
Notation for establishment of flow.
UNSTEADY FLOW
645
After integration,
LV
The
V
+V
V
velocity
(
approaches
V
time for
V
asymptotically;
V
to attain the value
mathematically
i.e.,
Practically, for
.
V
12 2 2 )
it
-
-
takes
to reach 0.99
takes
LF„1
gH 2
=
<
.
ln
infinite
V
V
p V7
=
,
^
L9_9
LV_
0.01
gH
must be determined by taking minor
does not contain
L
losses into account,
but Eq. (12.2.2)
e.
EXAM PLE 12.5 In Fig. 12.5 the minor losses are WV /2g, f = 0.030, L = 3000
m, D = 2.4 m, and H = 20 m. Determine the time, after the sudden opening
2
of a valve, for velocity to attain nine-tenths the final velocity.
L =
e
3000
FromEq.
7 =
°
=
\2gHD
<~Jl7
3000
X
19.612
12.3
The
2 4
•
= 3128
m
U.Uo
(12.2.1)
/19.612
=
V
V =
Substituting
t
X
16
+
0.9
3.17,
X
20
X
2.4
=
_ m/s
o
3 17
'
.
0.030X3128
V
1.90
In
20
X
into Eq. (12.2.2) gives
=
^
o
71.3 s
0.10
SURGE CONTROL
oscillation of flow in pipelines,
when
compressibility effects are not im-
For sudden deceleration of flow due to closure
of the flow passage, compressibility of the liquid and elasticity of the pipe
walls must be considered; this phenomenon, known as waterhammer, is disportant,
is
referred to as surge.
cussed in Sees. 12.4 to 12.7. Oscillations in a
As one means
of eliminating
U tube are special cases of surge.
waterhammer, provision
is
made
to permit the
APPLICATIONS OF FLUID MECHANICS
Fig. 12.6
Surge tank on a long pipeline.
liquid to surge into a tank (Fig. 12.6) The valve at the end of a pipeline may
be controlled by a turbine governor, and may rapidly stop the flow if the
.
generator loses
its
To
load.
destroy
all
momentum
in the long pipe
system
quickly would require high pressure, which in turn would require a very costly
With a surge tank as near the valve as feasible, although surge will
occur between the reservoir and surge tank, development of high pressure in
pipeline.
this reach
is
prevented.
It
is still
necessary to design the pipeline between
surge tank and valve to withstand waterhammer.
Surge tanks
may
be
not to overflow (unless a spillway
air to enter the pipeline.
and differential. The
and must be large enough
classified as simple, orifice,
simple surge tank has an unrestricted opening into
It
must
is
it
provided) or not to be emptied, allowing
also
be of a
size
that will not fluctuate in
resonance with the governor action on the valve. The period of oscillation of
a simple surge tank
The
is
relatively long.
surge tank has a restricted opening, or
orifice, between pipeand tank and hence allows more rapid pressure changes in the pipeline
than the simple surge tank. The more rapid pressure change causes a more
rapid adjustment of flow to the new valve setting, and losses through the
orifice aid in dissipating excess available energy resulting from valve closure.
The differential surge tank (Fig. 12.7) is in effect a combination of an
In
orifice surge tank and a simple surge tank of small cross-sectional area.
case of rapid valve opening a limited amount of liquid is directly available
orifice
line
Fig. 12.7
Differential surge tank.
UNSTEADY FLOW
647
from the central riser, and flow from the large tank supplements this flow.
For sudden valve closures the central riser may be designed so that it overflows into the outside tank.
Surge tanks operating under
air pressure are utilized in certain
stances, e.g., after a reciprocating
pump.
They
circum-
are generally uneconomical
for large pipelines.
Detailed analysis of surge tanks entails a numerical integration of the
equation of motion for the liquid in the pipeline, taking into account the
particular rate of valve closure, together with the continuity equation. The
particular type of surge tank to be selected for a given situation depends upon
a detailed study of the economics of the pipeline system. High-speed digital
computers are most helpful in their design.
Another means of controlling surge and waterhammer is to supply a
quick-opening bypass valve that opens when the control valve closes. The
quick-opening valve has a controlled slow closure at such a rate that excessive
pressure is not developed in the line. The bypass valve wastes liquid, however, and does not provide relief from surge due to opening of the control
valve or starting of a pump.
The
following sections on closed-conduit flow take into account com-
pressibility of the liquid
and
elasticity of the pipe walls.
Waterhammer
cal-
may be
accomplished in several ways; the characteristics method,
recommended for general use in computer solutions, is presented here.
culations
12.4
WATERHAMMER PHENOMENON
DESCRIPTION OF THE
Waterhammer may occur in a closed conduit flowing full when there is either
a retardation or acceleration of the flow, such as with the change in opening
of a valve in the line.
carried out
conduit
rigid.
When
the changes are gradual, the calculations
a valve
side of the valve
is
is
be
rapidly closed in a pipeline during flow, the
is
to decrease the velocity of flow.
side of the valve the pressure
is
reduced, and a
wave
pressure travels downstream, which also reduces the velocity.
is
may
and the
reduced.
action of this pressure pulse
downstream
liquid incompressible
This increases the head on the upstream
and causes a pulse of high pressure to be propagated upstream.
flow through the valve
The
If
by surge methods, considering the
If
On
the
of lowered
the closure
rapid enough and the steady pressure low enough, a vapor pocket
formed downstream from the valve. When this occurs, the cavity
tually collapse and produce a high-pressure wave downstream.
may be
will even-
Before undertaking the derivation of equations for solution of water-
hammer, a description of the sequence of events following sudden closure
of a valve at the downstream end of a pipe leading from a reservoir (Fig.
12.8) is given. Friction is neglected in this case. At the instant of valve closure
APPLICATIONS OF FLUID MECHANICS
648
a~+
—
h
Ji
1
'
^^0
1
V
7/iA
_
v=-v
4
V=0
n
0<t<^a
2L
L<
t<
~ a
a
(a)
(b)
a
—
n^o
v=-v
—
v
a
v=o
[
—
a
Fig. 12.8
=
Sequence
of events for
a
(d)
(c)
one cycle
of
sudden closure
of a valve.
compressed and brought to rest, and the
As soon as the first layer is compressed
the process is repeated for the next layer. The fluid upstream from the valve
continues to move downstream with undiminished speed until successive
layers have been compressed back to the source. The high pressure moves
upstream as a wave, bringing the fluid to rest as it passes, compressing it, and
expanding the pipe. When the wave reaches the upstream end of the pipe
(t = L/a s), all the fluid is under the extra head h, all the momentum has
been lost, and all the- kinetic energy has been converted into elastic energy.
There is an unbalanced condition at the upstream (reservoir) end at the
instant of arrival of the pressure wave, as the reservoir pressure is unchanged.
The fluid starts to flow 'backward, beginning at the upstream end. This flow
(t
0) the fluid nearest the valve
pipe wall
is
is
stretched (Fig. 12.8a).
returns the pressure to the value which
was normal before
wall returns to normal, and the fluid has a velocity
V
closure, the pipe
in the
backward
sense.
This process of conversion travels downstream toward the valve at the speed
of sound a in the pipe. At the instant 2L/a the wave arrives at the valve, pressures are back to normal along the pipe,
and the velocity
is
everywhere
V
Q
in
the backward direction.
Since the valve is closed, no fluid is available to maintain the flow at the
valve and a low pressure develops ( — h) such that the fluid is brought to rest.
UNSTEADY FLOW
649
This low-pressure wave travels upstream at speed a and everywhere brings
rest, causes it to expand because of the lower pressure, and allows
the pipe walls to contract. (If the static pressure in the pipe is not sufficiently
high to sustain head —h above vapor pressure, the liquid vaporizes in part
the fluid to
and continues to move backward over a longer period of time.)
At the instant the negative pressure wave arrives at the upstream end of
the pipe, 3L/a s after closure, the fluid is at rest but uniformly at head — h
less than before closure. This leaves an unbalanced condition at the reservoir,
and fluid flows into the pipe, acquiring a velocity V forward and returning
the pipe and fluid to normal conditions as the wave progresses downstream at
speed a. At the instant this wave reaches the valve, conditions are exactly
the same as at the instant of closure, 4L/a s earlier.
This process is then repeated every 4L/a s. The action of fluid friction
and imperfect elasticity of fluid and pipe wall, neglected heretofore, is to damp
out the vibration and eventually cause the fluid to come permanently to rest.
Closure of a valve in less than 2L/a is called rapid closure; slow closure refers
to times of closure greater than 2L/a.
The sequence of events taking place in a pipe may be compared with the
sudden stopping of a freight train when the engine hits an immovable object.
The car behind the engine compresses the spring in its forward coupling and
stops as it exerts a force against the engine, and each car in turn keeps moving
at its original speed until the preceding one suddenly comes to rest. When the
caboose
is
at rest all the energy
is
stored in compressing the coupling springs
The caboose has an unbalanced
force exerted on it, and
an unbalanced force on the
next car, setting it in backward motion. This action proceeds as a wave toward
the engine, causing each car to move at its original speed in a backward direction.
If the engine is immovable, the car next to it is stopped by a tensile
force in the coupling between it and the engine, analogous to the low-pressure
wave in waterhammer. The process repeats itself car by car until the train
is again at rest, with all couplings in tension. The caboose is then acted upon
by the unbalanced tensile force in its coupling and is set into forward motion,
(neglecting losses).
starts to
move backward, which
followed in turn
all
cars are in
by the
in turn causes
rest of the cars.
motion as before the
When
this
original impact.
wave reaches the engine
Then the whole cycle is
repeated again. Friction acts to reduce the energy to zero in a very few cycles.
12.5
DIFFERENTIAL EQUATIONS FOR CALCULATION OF
WATERHAMMER
Two
basic mechanics equations are applied to a short segment of fluid in a
pipe to obtain the differential equations for transient flow: Newton's second
law of motion and the continuity equation. The dependent variables are the
APPLICATIONS OF FLUID MECHANICS
650
H
elevation of hydraulic grade line
velocity
V
at a cross section.
above a fixed datum and the average
The independent
variables are distance x along
the pipe measured from the upstream end and time
V = V (x,t)
t;
hence,
H
=
H(x,t),
not taken into account in this derivation.
For pipelines with expansion joints it does not enter into the derivation. Friction
Poisson's ratio effect
.
is
considered to be proportional to the square of the velocity.
is
Equation of motion
The
element between two parallel planes 5a: apart, normal to the pipe
taken as a free body for application of Newton's second law of motion
in the axial direction (Fig. 12.9). In equation form
fluid
axis, is
d
r
pA — \pA
1
(pA) 8x
H
dx
L
+
J
p
dA
—
dx
bx
+
yA
8x sin 6
—
tqttD bx
dV
= pAbx—dt
Dividing through by the mass of the element pA bx and simplifying give
dp
--—
+
1
p dx
Fig. 12.9
motion.
.
4r
81110--- =
pD
—
dV
at
Free-body diagram for derivation of equation of
(12.5.1)
UNSTEADY FLOW
Introducing the hydraulic-grade-line elevation, from p
/dH
dp
dx
= pg(H —
z),
651
leads to
dz\
^fe^s)
(12A2)
But
—=
— sin 6
dx
and
dV
dH
—
—=
9—
+
+
pD
dx
4t„
(12.5.3)
dt
flow, r = pfV 2 /8 [Eq. (5.10.2)].
The assumption is
that the friction factor in unsteady flow is the same as in steady flow.
For steady turbulent
made
Hence, the equation of motion becomes
dH
dV
fV
+ a+
'*r
Since friction
2
22>
= °
(12 5 4)
-
must oppose the motion,
V
2
is
written as
V V
\
|
-
to provide the
proper sign. After expanding the acceleration term,
L*
=
-^ =0
dH
dV
dV fV 7
9—
+ V—
+ — + 2D
dx
dx
J
Tr
dt
The equation
tinuity Li
9
,„
I
(12.5.5)
by Li to distinguish
next derived.
indicated
is
which
is
it
from the equation of con-
Equation of continuity
The unsteady
continuity equation (3.2.1)
is
applied to the control volume of
Fig. 12.10,
-
d
j- (pAV) bx
dx
in
which dx
= -
(pA
(12.5.6)
to)
dt
is
not a function of
t.
Expanding the equation and dividing
:
APPLICATIONS OF FLUID MECHANICS
652
pAV+fx
Control
Fig. 12.10
volume
for
{
P AV)bx
derivation
of
continuity equation.
through by the mass pA dx give
V dA
A dx
The
first
1
dA
A
dt
V
dp
I
dp
dV
(12.5.7)
p dx
p dt
two terms are the
dx
total derivative 1 (1/A) dA/dt,
and the next two
terms the total derivative (1/p) dp/dt, yielding
dA
1
A
H
dt
The
dV
\dp
1
p dt
first
=0
rt
(12.5.8)
dx
term deals with the
elasticity of the pipe wall
and
its
rate of deforma-
tion with pressure; the second term takes into account the compressibility of
For the wall elasticity the rate of change of tensile force per unit
is (D/2) dp/dt; when divided by the wall thickness t',
it is the rate of change of unit stress (D/2t') dp/dt; when this is divided by
Young's modulus of elasticity for the wall material, the rate of increase of
the liquid.
length (Fig. 12.11)
unit strain
is
obtained, (D/2t'E) dp/dt.
D/2, the rate of radial extension
is
perimeter irD the rate of area increase
dA
dt
D dpD
2t'Edt 2
See Appendix B.
D
After multiplying this by the radius
obtained;
is
finally,
obtained
by multiplying by the
UNSTEADY FLOW
Tenpipe
12.11
Fig.
653
sile force in
wall.
and hence
D_dy
\_dA
A~di ~
From
K
the definition of bulk modulus of elasticity of fluid (Chap. 1),
dp
= -
dV/V
and the rate
dp
dt~ Kdt
1
dp/p
change of density divided by density yields
(12.5.10)
Eqs. (12.5.9) and (12.5.10), Eq. (12.5.8) becomes
—-
Kdt\
It
is
a'
=
in
of
= ~^-
dp
1
p
By
(12.5.9)
7e~cU
Et'
(12.5.11)
dx
convenient to express the constants in this equation in the form
K/p
1+
which
(12.5.12)
(K/E)(D/t') Cl
d
is
unity for the pipeline with expansion joints. Equation (12.5.11)
now becomes
I dp
- -ypat
+a
2
dV
=
—
dx
(12.5.13)
APPLICATIONS OF FLUID MECHANICS
654
Now,
since
dp
ydp
=
p
z)
(Fig. 12.10),
dp
Hy
dx
dt
= pg(H —
dt
\dx
™\dt
dx)
dt)
H
The change of p with respect to x or t is very much less than the change of
with respect to x or t, and so p was considered constant in the preceding equation. If the pipe is at rest, dz/dt = 0, and dz/dx = —sin 0; hence
I
dp
(dH
jr
\dx
p dt
and Eq.
U
=
\
a2
,
dH
,
/
(12.5.13)
dV
TT
\-V
g dx
dt
becomes
dH
dH
—
+—+V
dx
sin 6
-
(12.5.14)
dt
which is the continuity equation for a compressible liquid in an elastic pipe.
L\ and L 2 provide two nonlinear partial differential equations in V and H in
terms of the independent variables x and t. No general solution to these equations is known, but they can be solved by the method of characteristics for a
convenient finite-difference solution with the digital computer.
12.6
THE METHOD-OF-CHARACTERISTICS SOLUTION
Equations L\ and L 2 in the preceding section contain two unknowns. These
XL 2
equations may be combined with an unknown multiplier as L = L\
+
Any two
all
in
real, distinct
values of X yield two equations in
the physics of the original two equations L\ and
any
solution.
It
may happen that
ticular values of X are found.
L, with
L 2 and may
replace
great simplification will result
L\ and
L2
.
V and H that contain
if
them
two par-
are substituted into the equation for
some rearrangement.
[dH
/Ty
N
dHl
[dV/ Tr
a2 \
dV~\
+ V Smd +
This expression
is
arranged so that the
first
fv\v\
J
\
-^=0
(12.6.1)
term in brackets would be the
UNSTEADY FLOW
total derivative
dH/dt
655
if
dx
(12.6.2)
and the second term
dx
—=
in brackets
would be dV/dt
if
2
-a
V+
dt
(12.6.3)
g\
since
dH
dHdx
dt
dx dt
+
.
dH
dV
dt
from
calculus.
7+
\g
= 7
dV dx
dx
dt
dt
dV
'
dt
Equations (12.6.2) and (12.6.3) must be equivalent,
+-
(12.6.4)
Solving for X gives
(12.6.5)
9
Therefore, two real, distinct values of X have been found that convert the
two
partial differential equations into a pair of total differential equations
by Eqs. (12.6.2) and (12.6.3).
These equations, with X substituted, become
restricted
dH
adV
dt
g dt
Tr
a/7
.
7
1
(12.6.6)
2gD
C+
dx
= V
+
a
(12.6.7)
dt
dH
adV
dt
g dt
7 sin $ + Tr
.
afV\V\
'
'
=
n
(12.6.8)
2gD
\c-
dx
Tt
=v ~ a
To understand
(12.6.9)
the significance of these four equations,
it is
convenient
APPLICATIONS OF FLUID MECHANICS
656
it
C*
L
Fig. 12.12
character-
xt plot of
along which solution
obtained.
istics
is
to consider the solution being carried out on an xt plot (Fig. 12.12)
V
H
.
Consider
known at the two known locations R and S in the figure.
The curve labeled C+ is a plot of Eq. (12.6.7) Eq. (12.6.6) is valid only along
a C + characteristic. The curve labeled C~ is a plot of Eq. ( 12.6.9) Eq. ( 12.6.8)
that
and
are
;
;
C~ characteristic. Each equation, (12.6.6) and (12.6.8),
contains two unknowns for a known point on its characteristic, but at the
intersection of C+ and C~ at P both equations may be solved to yield V P and
H P At this point Eqs. (12.6.7) and (12.6.9) may also be solved for x and t.
Hence the solution is carried out along the characteristics, starting from known
conditions and by finding new intersections so that heads and velocities are
is
valid only along a
.
found for later times.
In waterhammer calculations in metal pipes, the subject of this treatment, V is very small compared with a and may be dropped from Eqs. (12.6.7)
and (12.6.9). The characteristic lines are now straight, with slopes ±a. a
is the speed with which the pressure-pulse wave is transmitted along the pipe.
The pipe is considered to be made up of
equal reaches (Fig. 12.13), and H
and V are initially known at each of the dividing sections. The solution to
the waterhammer problem can then be carried out at the intersections of the
characteristic lines, as shown by the solid dots. It is to be noted that the solution can be carried over only a limited region, unless information is given at
x =
and x = L, of some external condition as a function of time. (See under
N
"Boundary conditions/' page
658.)
From the grid of Fig. 12.13, it is seen that the time step of the calculation is At = Ax/a. By using the grid, x and are known at each intersection,
and Eqs. (12.6.7) and (12.6.9) need be considered no further. By multiplying
Eq. (12.6.6) by dt and integrating along the C+ characteristic, Fig. 12.13,
t
HPi -
tf«
+
— (Q
Pi
Qi-i)
+
Qi- .iAi
.
sm
+
fAx— „
-
-
2gDA'
Qi-i
Qi-
=
(12.6.10)
UNSTEADY FLOW
657
1
At
Ax
Ax
Ax
Ax
Ax
At
HPl
cy
At
\c-
ft+l
3-1
JV+1
i+l
Rectangular
Fig. 12.13
grid
solution
for
of
char-
acteristics equations.
where a first-order approximation has been used in the integration of the
two terms and, for convenience, the equation has been written in terms of
charge
Hpi
.
Q
in place of velocity V.
Hw .^ iQpi
last
dis-
Similarly for Eq. (12.6.8)
. Qi+l)+
9^^
e
.llL Qi+1
=
Q, +1
(12.6.11)
By
adding the
HPi
=
last
0.5 !#;_!
two equations, Qp
+ H i+1 +
-^-
(Q t-_i
i
is
eliminated:
- Q i+i) -
^-^ (Qi-i + Q
i+ i)
/Arc
2gDA'
(C*-i|Q<-i|-Q<4!|Q.-+*l)
(12.6.12)
With HPi known, either Eq. (12.6.10) or (12.6.11) can be used to solve for
With N reaches in a pipe there are N + 1 sections along the pipe, inQp
cluding the two ends. The unknowns Hp and Qp are determined at each section 2 to N by using Eq. (12.6.12) and either (12.6.10) or (12.6.11). The two
variables at each of the two ends of the pipeline are determined by using the
..
t
appropriate one of Eqs. (12.6.10) or (12.6.11) together with the equation for
the boundary condition. When the variables
P and Q P have been evaluated
at each section in the pipeline, time is incremented by A£ and the procedure
H
is
repeated.
are replaced
At the end of each time step the values of H and Q
by the newly computed values of H P and Qp.
at each section
APPLICATIONS OF FLUID MECHANICS
658
Boundary conditions
At the upstream end of a pipe, Eq. (12.6.11) for the C~ characteristic provides
one equation in the two unknowns QPl and Pl (Fig. 12.14a). One condition
is needed exterior to the pipe to relate the pipeline response to the boundarycondition behavior. This condition may be a constant value of one of the
variables, such as a constant-head reservoir, a specified variation of one of
the variables as a function of time, an algebraic relationship between the two
variables, or a relationship in the form of a differential equation. Some bound-
H
ary conditions
may involve additional variables, e.g., pump speed in the
pump connected at the upstream end of the line. In this
of a centrifugal
case
case
two independent equations must be available to combine with Eq. (12.6.11)
to solve for the three unknowns at each time step. The simplest boundary
condition is one in which one of the variables is given as a function of time.
A
direct solution of Eq. (12.6.11) for the other variable at each time step
provides a complete solution of the interaction of the fluid in the pipeline and
the particular boundary.
This includes the appropriate reflection and trans-
mission of transient pressure and flow waves that arrive at the pipe end.
At the downstream end
of the pipe (Fig. 12.146), Eq. (12.6.10) for the
two variables HPn+1 and Qp N+v
needed that either specifies one of the variables to
be constant or a known function of time or provides a relationship between
the variables in algebraic or differential equation form. The simplest end
condition is one in which a variable is held constant, as at a dead end where
Qp.y+i is zero. Then Eq. (12.6.10) provides a direct solution for HPrf+1 at each
C+
characteristic provides one equation in the
One
external condition
is
time step.
A
computer program
is
provided for solution of the next example, which
Complex systems
involves simple boundary conditions on a single pipeline.
can be visualized as a combination of single pipelines that are handled as
described above, with boundary conditions at the pipe ends to transfer the
transient response from one pipeline to another and to provide interaction
with the system terminal conditions. Thus it may be noted that a compli-
7=iV
AMI
1
(a) Upstream end
Fig. 12.14
(b) Downstream end
Boundary conditions.
UNSTEADY FLOW
XL =
Fig. 12.15
Example
600 m,
D
= 0.5
659
m
12.6.
common solution procedure
with a systematic coverage of each
terminal and interconnection point in the system. The primary focus in the
treatment of a variety of transient liquid-flow problems is on the handling of
boundary conditions, which are discussed in the next section.
cated system can be treated by a combination of a
for the interior of each pipeline, together
BASIC WATERFAMMER PROGRAM. WAVE ON RESERVOIR UPSTREAM, OEAC END D.S.
OIMENSICN HP(2i),gP(2l),H(21),w-'21)
READ (5, 15,END=99) A, XL ,0 ,F ,G,Q0, h J OH CMfc G A ,TMAX , N , I PR
10
FORMAT (3F10.2.3F10.4/4F10.2/2I5)
15
NS = N-M
R=F*XL/(2.*G*D**5*.78 54**2*N)
CH = A/(G«. 7854*0*0)
C
,
,
DT*XL/(MM
T = 0.
K=0
CU 20 1=1, NS
HI I )=H0-(I-1)*R*U0*C0
C(I)=CG
WRITE (6,25) A,XL,D,F,G,QC,H0,OH,CMEGA,TMAX,DT,rv,IPR
25
FCRMATC A, XL,
L F= • ,2F€. i , 2F8.4/' G, Co, HO £ 0H=» , 2F8. 3,F8. 1
HEADS
2F8.3/* CMEGA, TMAX, CT=' , F8.3 ,F8. 1, F8. 3/ • N £ lPRs',214//'
TIMt X/L=
3 AND CISChARGES ALGNG THE PIPE*//'
0.
.2
20
4
.4
.6
.8
1.')
WRITE (6,35) T , (Hi I , 1 = 1 ,NS ,2 ) , (0 ( I ) , 1= 1, NS ,2
FORMAT
1H0 ,F 7.3 , 5H
H= , 6F8. 2/ 10X, 3H Q=,6F8.3)
40
T=T*DT
IF ( T.GT.TfAX) GO TC 10
K=K*1
COMPUTATION OF INTERIOR PCINTS
30
35
C
)
(
DO 50 1=2,
C
CP=H(I-l)+0(I-l)*(CH-R*ABSiq(I-l)))
HP (I >=.5*(CP*MI*i)*Q( 1*1 )*(R*AES(Q( 1*1) )-0>
50
GP(I)=(CP-HP(I))/CH
«CUNC*RY CCNCITIONS
HP(l)=HC»CH*SIN(OMEGA*T)
0P(l) = (hP( l)-H(2)-U(2)*(R*ABS(J(2) )-CH) )/CH
QP(NS)=0.
HP(NS)=MN)*Q(N)*(Ch-R*ABS(C(N) ))
)
DO 60 1=1, NS
60
H(I)=HP( I)
0(I)=QP(I)
IF IK/IFR*IPR-K) 40,30,40
99
STOP
END
12CC.
ICC.
1C
600.
3
.5
018
9.806
3.1416
2
Fig. 12.16
Basic
waterhammer program.
0.
APPLICATIONS OF FLUID MECHANICS
660
EXAMPLE
end
12.6
Determine the transient response
surface
is oscillating,
grade-line elevation,
XL,
G,
:o,
£
IPfU
is
H = H + DH
DH
10
C
t
2
HEADS AND CISCHAROE i ALCNG ThE PIPE
.2
.4
TIME X/L =
0.
H=
100.00
O.OOJ
100.00
ICO. 00
0.000
0.000
C.000
Q=
H=
O.ICO
100.93
100. cc
1C0.0C
0.000
0.001
C.000
C=
H=
101.76
IOC. 93
ICO. GO
0.200
Q0.003
0.001
C.OOC
0.300
H=
101.76
100.93
102.43
=
0.004
0.003
0.001
O.fCQ
H=
102.85
102.43
1C1.76
0.004
0.005
0.003
c*
H=
C.500
103.00 102.85
1C2.43
0.C05
0.005
C.0C4
c=
H=
0.600
102.6 5
103. OC
lC2.d5
0.0C5
0.005
C.0G5
fi«
C.7C0
102.43
102.85
H=
U3.00
0.005
0.C04
C.0C5
Q=
H=
0.800
101.76
102.43
102.85
=
0.003
0.004
C.0C5
H=
100.93
103.35
101.76
C.9C0
=
C.O01
0.CC3
0.0G2
H=
100.00
101.85
103.53
1.000
-0.00C -0.000 -COCO
Q=
H=
99.07
101.76
103.35
I. 100
-0.002
0- -0.CO4 -0.003
H=
102. 6d
98.24 100.57
1.200
-0.008 -0.007 -C.005
G»
=
H
101.15
97.57
99.33
1. 300
-0.012 -0.01C -0.008
G=
H=
97.15
98.15
99.33
1.400
=
-0.014 -O.C13 -C.O10
h=
97.00
97.15
97.57
1.500
-0.014 -0.014 -0.012
Q=
=
1.600
H
96.43
So. 06
97.15
=
-0.C14 -0.014 -C.J12
H=
97.57
96.06
94.93
1.700
Q=
-0.012 -0.012 -C.Oil
H=
1.800
98.24
96.07
94.30
-0.C08. -0.009 -C.0C9
C=
=
h
1.900
99.07
96.47
93.29
-0.004 -0.006 -0.005
U=
H=
92.93
2.00C
1C0. 00
96.29
C.000
C.000
0.000
U*
H=
2. ICC
100.93
96.48
S3. 29
0.00b
O.0C5
O.C07
Q=
h=
2.200
101.76
97.93
S4.30
0.014
0.U12
C.009
G=
h=
S6.78
2.300
102.43
99.58
Q-»
0.019
0.018
C.014
=
2. 400
H
99.58
1C2.85 101.28
0.023
0.C21
c=
C018
2."500
H=
103.00 102.85
102.^2
0=
0.023
0.C24
C.C19
H=
2.600
102.85 10H.14
U5.C3
0.023
0.022
0.019
G=
2.700
H=
102.43
105.03
1C7.14
0.C19
0.019
C.017
Q=
2.6CC
H=
101.76
105.42
1C8.55
=
C.C14
U.C15
C.014
H=
2.9CC
100.93
105.29
110.05
=
C.C07
0.008
C.007
Fig. 12.17
H
F=
1200.0
6C0.0
5000 0.0180
CH =
9.806
O.OOC
100.0
3.000
CT=
3. 142
6.C
0. 050
L
ho
CMEGA, TMAX,
N
in a single horizontal dead
connected to a reservoir in which the water
sin ut, where
is the steady hydraulicthe amplitude of the wave, and co the frequency.
pipeline, Fig. 12.15, that
Computer
solution to
.6
100.00
0.000
100.00
0.000
100.00
0.000
100.00
0.000
100.93
0.001
101.76
0.003
102.43
C.004
102.85
0.005
1C3.93
0.003
IC4.62
o.oo<i
1C1.65
0.000
1C4.62
-0.002
IC3.93
-C.O03
102.85
-C.0C5
iGC.57
-0.007
.e
10C.CC
O.OOC
ICC. CO
C.OCC
100.00
0.000
IOC.CC
0.000
IOC.CC
O.OCO
IOC. 93
0.CC1
101.76
0.0C3
103.35
0.002
104.62
O.0C2
105.43
0.0C1
1C5.71
COCO
105.43
-O.OCl
1C4.62
-0.CC2
103.25
-0.0C2
101.76
-0.0C3
98. ZH
99. C7
-0.008
96.07
-0.009
94.30
-0.009
9^.15
-0.007
9C.77
-0.003
90.30
0.000
90.77
0.003
-0.0C4
96.47
-0.CC6
93.29
-0.005
90.77
-O.0C3
89.15
-O.0C2
88.59
O.CCC
92. 15
0.007
9<».30
0.009
97.93
0.012
101. 76
0.014
105.42
0.315
1C8.55
C.014
111.77
0.010
113.84
0.005
Example
89. 15
u.0C2
9C.77
0.003
93.29
O.0C5
96.47
CCC6
100.93
CCC7
105.29
0.0C8
11C.C5
C.CC7
113.84
CCC5
116.27
C0C3
12.6.
1.
100.00
0.000
100.00
0.000
100.00
0.000
100.00
0.000
1CC.CC
0.000
100.00
0.000
101.85
0.000
103.53
0.000
104.85
O.OOC
105.71
0.000
106.00
0.000
105.71
0.000
104.85
0.000
103.53
0.000
101.65
0.000
ICO. 00
0.000
96.29
o.ooo
92.95
0.000
90.30
0.000
88.59
O.OOC
88.01
O.OCO
88.59
0.000
90.30
0.000
92.95
O.OCO
90.29
0.000
100.00
0.000
105.56
0.000
110.57
C.000
114.55
O.CCC
117.10
0.000
UNSTEADY FLOW
m
661
H
=
Other conditions include a = 1200 m/s, / = 0.018, Q = 0.0 3 /s,
= 10 reaches in pipeline,
100 m, DH = 3 m, co = 3.1416 rad/s, T max = 6 s,
and IPR = 2 time increments between printout. The value of g is also needed
N
and in this case it is 9.806 m/s 2 since SI units are used.
The computer program in Fig. 12.16 is used to solve the problem. The
important steps in the program include (1) data input, (2) setup of initial
conditions, (3) printout, (4) compute Hp and Qp at interior points, (5) compute Up and Qp at the boundaries, (6) substitute HP and Qp into H and Q at
each section, and (7) return to step 3. The formated input data are shown in
Fig. 12.16, and the computed output appears in Fig. 12.17. It is of interest
in the input data,
to follow the wave of the pressure pulse through the pipe. Since the excitation
happens to be at the pipeline natural period, a pressure-head amplification may also be noted.
BOUNDARY CONDITIONS
12.7
The term boundary
may be
condition refers to the end condition on each pipeline.
a system terminal at a reservoir, valve,
may
be a pipeline
a pump
or a storage volume. In each of the many options at the downstream end of
the pipe the equation along the C+ characteristic is used to interface with the
particular end condition. It is convenient to write Eq. (12.6.10) in an abIt
etc.,
or
it
connection to another pipeline or a different type of element,
breviated form for section
HPn8
downstream end,
= CP - CH Qp N8
= a/gA
where Ch
known
CP =
NS at the
e.g.,
is
(12.7.1)
a constant for the pipeline and Cp
is
a combination of
quantities at each time step
HN +
QN (ch -
^ sin 9 - ^j
|
QN
\j
Similarly at the upstream end of the pipe the equation along the
(12.7.2)
C~ charac-
teristic is
used to relate the pipeline behavior to the end condition. Equation
(12.6.11)
may
HPl
= Cm
+
be expressed
CH QPl
(12.7.3)
with
Cm = H, -
Q*
(cH
+ ^ sin -
±^
|
Q2
(12.7.4)
1)
APPLICATIONS OF FLUID MECHANICS
662
A few common boundary conditions follow, and in each case either Eq.
or (12.7.3)
(12.7.1)
used to represent the pipeline response.
is
Valve at downstream end
For steady-state flow through the valve, considered as an
orifice,
(Cd A v )oV2gHo
Qo=
with
Q
(12.7.5)
the steady-state flow,
H
the head across the valve, and (Cd A v )
area of the opening times the discharge coefficient.
the
For another opening, in
general,
Qp = Cd A,y/2gH P
(12.7.6)
Dividing the second equation by the
Qp =
first
and rearranging give
-^tVH~p
(12.7.7)
which r is the dimensionless valve opening, r = 1 for steady flow Q and
for the closed position of the valve. When the subhead drop H and t =
script NS is added to the variables Qp and Hp in Eq. (12.7.7) and this equa-
in
,
tion
is
solved simultaneously with Eq. (12.7.1), the value of
mined as a function
Qp N8
is
deter-
of the valve position including the transient response
from
the pipeline
The corresponding value
of
HpNS
can be determined from either Eq. (12.7.1)
or (12.7.7).
Minor loss
In some problems it may be important to use the energy equation at the
boundary and to include a minor loss. The pipe entrance from a reservoir is
discussed as an example. Figure 12.18, left shows the energy and hydraulic
grade lines for flow into the pipe. The energy equation between the reservoir
UNSTEADY FLOW
Energy grade
^C
Energy grade
line
[
H,
1
VP
HR
Hydraulic gradeline
line
ig
u
_L_.
line
^ Hydraulic grade
y2
663
2
2
Fig. 12.18
surface
Minor
loss.
and section
1
Qp?
in the pipeline
,
is
QpS
(12.7.9)
Simultaneous solution of this equation with Eq. (12.7.3) yields
gC„A*
Qp =
-kt~i +
which
is
^
(9C»A>Y
VUrv
valid as long as
Fig. 12.18, right,
H Pl
=
2gA>
{Hr
+
Hr — Cm
HR
,
and a
-
Cm)
positive.
is
zn
(127
For flow into the
direct solution for
QPl
is
-
10)
reservoir,
possible
by use
of Eq. (12.7.3).
EXAMPLE 12.7 Consider a single horizontal pipeline with reservoir upstream
and an orifice a short distance (10 diameters) downstream from the entrance.
At the downstream end a valve is to be closed linearly in 1.5 s. The steady flow
Qo discharges to the atmosphere at the valve. Write the equations for both
boundary conditions so they are ready to be programmed. The kinetic-energy
term should be included at the upstream orifice. Assume the loss coefficient to
be the same for flow in either direction.
The steady-state head loss at the valve must first be determined by
writing the energy equation for the system.
head
H
H °= H
K
e
is
just
upstream from the valve
If
HR
is
the reservoir head, the
is
«-M i+K +K+ f)
<
the entrance-loss coefficient and
K is the orifice-loss coefficient.
APPLICATIONS OF FLUID MECHANICS
664
For the downstream boundary,
Ch -
cp =
If
t
9a
hn +
<
-^
Qn (ch
2
\q n
\)
1.5,
Qp NS =
If
r
t
=
>
+
\ \2~)
1.5,
QPn8 =
0.0
#p„ s - CP
0.0
For the upstream boundary, when flow
h +
J"
Qp
Qp
n +
*
+k Qf
*
is
negative, the energy equation gives
*
The second term on the right-hand
side
is
the exit loss from the pipe.
By
solving simultaneously with Eq. (12.7.3)
>=K-V\k)
in
which
CM =
H 2
for
QPl
Wpi "
which
is
1
+K +
c
valid only
f^
2
1
flow, the
if
5
it
HR —
energy equation yields
gives
C
V
2 A*(ffB
—
-
Cm)
Cm]
/
+ g1+K.
+ \\l + K ~+k)
l+K. + K
K)
+
,
if
C = gC«A
Q> 1)
valid only for negative flow; hence
(12.7.3),
Cs
_
"
is
For positive
Combined with Eq.
)2gA'<
K
Q 2 (ch -
The equation
negative.
(HR -Cj
//ftV
C,
Qp
HR —
e
Cm >
0.
Cm must be
UNSTEADY FLOW
665
Junction of two or more pipes
At a connection of pipelines of different properties, the continuity equation
must be satisfied at each instant of time, and a common hydraulic-grade-line
elevation may be assumed at the end of each pipe. These statements implicitly assume that there is no storage at the junction, and they also neglect
all
minor
effects.
In multipipe systems
it is
necessary either to use double-
number and the
second to the pipe section number, or to use continuous sectioning in the entire
system. If the former scheme is used to handle the three-pipe junction in
subscript notation, the
first
subscript referring to the pipe
and Eqs. (12.7.1) and (12.7.3) are written in the following form,
provides a simple solution for the common head
summation
a
Fig. 12.19
QPl.NS
=
Hpi, ns
Chi
HP1,NS
Chi
—Qpz.i
=
Hp ltNS
-_
n _
-
+
4-
Cp
l
Ch\
Cmi
Ch
2
Cm%
Chz
Chz
yn
ZQp
+
I
-HTJPltNS 2 —- +
CPl
77^H\
y^H
CM2
+7,
tff 2
Cm
—
3
h
C#3
or
Hpi -"
= CpJChi
+ Cm /Ch +
2
2
Cmz/Ch
3
.
-
I(ijc7)
With the common head computed, the equations above can be used
termine the flow in each pipe.
Fig. 12.19
Pipeline
_
1
(12 7
junction.
v
n)
1
-
to de-
APPLICATIONS OF FLUID MECHANICS
666
Valve
in line
A
valve or orifice between two different pipelines or within a given line must
be treated simultaneously with the contiguous end sections of the pipelines.
It is assumed that the orifice equation (12.7.7) is valid at any instant for the
control volume shown in Fig. 12.20. This assumption neglects any inertia
effects in accelerating or decelerating flow through the valve opening and also
implies that there is a constant volume of fluid within the indicated control
volume. At any instant the flow rate in the end sections are equal, Qp llNS =
Qp 2i1 and the orifice equation for positive flow, written with the subscript
notation to identify the pipeline as well as the section, becomes
,
Qor
Qp,.i
VH
= Qpuss = -T^P
AH
where
is
at the flow
Pl Ka
,
- HPltl
(12.7.12)
the steady-state drop in hydraulic grade line across the valve
Q when
r
=
When Eqs.
1.
(12.7.1)
and
(12.7.3) are written
with
the same notation
Hp llNS = CPl —
HP2A
= CMi
(12.7.13)
ChiQpi, ns
+ CBiQPatl
(12.7.14)
and are combined with Eq.
Qpi.ss
2
+
CA (CHl
In this equation
tion
+
(12.7.12), a quadratic equation results:
Ch )Qpu N8
2
C =
4
t2 Q
2
/AH
~
.
C<(CPl
The
- C M2 =
)
solution for flow in the positive direc-
is
Qpu.s
=
f
[- (CBl + CH2 +
)
yj(C Hl
+
CW + 4(
%
CM2)
4
]
(12.7.15)
®
u- Control volume
Q
jjj
Orifice or valve
Fig. 12.20
Valve
in line.
©
UNSTEADY FLOW
667
Equations (12.7.13) and (12.7.14) may be used to find the hydraulic grade
line when the flow is known. For flow in the negative direction the orifice
equation must be written
Qpi,\
=
Qor
=
Qpi,ns
,-7jr
— HPuNS
-\/Hp 2tl
(12.7.16)
When
the equations that are valid along the characteristic
with this equation, the solution is
=
Qn,„s
f
Cm
(Cpi
+ CH - ^(C„ + C„^-4
[(CBl
,)
lines are
')
(12.7.17)
i
c<
]
Examining the
This equation yields a valid answer only for negative flow.
equation shows that a negative result
Eq. (12.7.15)
Cm <
Multipara
I
lei
if
CPl - Cm >
0,
2
and Eq.
if
Cpj
— CV <
0-
Thus,
used
if
CPl -
2
(12.7.17)
is
pipes
a number of identical parallel pipes exist in a system, as in a cooling-water
condenser, they
to the
is
used
possible only
0.
2
If
is
is
combined
may be correctly treated as one element with a flow area equal
sum of the
areas of the individual pipes but with a resistance factor that
appropriate to the individual pipe diameters.
resistance
f8 and
D
s
term
in Eqs. (12.7.2)
and
(12.7.4)
Thus the multiplier for the
2
where
/, Ax/2gD A T
becomes
refer to the smaller or individual pipes while
flow area.
As long
as the individual pipelines are the
s
At
,
refers to the total
same
in every respect,
the transient conditions in each line will be identical and this special element,
which accommodates the
total flow, will accurately portray the physical be-
havior of this portion of the system.
be appropriate to the smaller-diameter
Centrifugal
pump
with speed
The
pressure-pulse
wave speed must
lines.
known
If a pump is operating at a constant speed, or if the unit is started and the
pump and motor come up to speed in a known manner, the interaction of the
pump and the fluid in the connecting pipelines can be handled by a fairly
simple boundary condition.
(9.1.3),
when the
The homologous
transient behavior of a given
conditions, Eqs.
pump
is
(9.1.1)
investigated,
and
may
be
B
APPLICATIONS OF FLUID MECHANICS
668
written
H
j— =
const
where
H
is
—Q
=
may be
is
expressed
pump and N is the speed. If the pump
by a parabola, then in the homologous form
written
^-A + A^ +
The
(12.7.18)
the head rise across the
characteristic curve
it
const
J
d)'
pump
When Eq.
compressibility of the fluid in the
pared with the
in Fig. 12.21,
Hps>1 When
rest of the system.
= N*B,
this equation is
charge
may be
n
QplNS
"
c
" Hl
(12.7.19)
is
negligible
applied to the
com-
pump
takes the form
it
H PltNS
assumed to be
is
+ B NQPltNS +
2
BsQPl ,„ s *
combined with Eqs.
(12.7.13)
(12.7.20)
and
(12.7.14), the dis-
determined as
+
cH2 - b 2n
2*3
r
r
"
+
L
1
wara + Cp
1
(CHl
i
+ cH2 -
-Cir,)
T
Ny
J
2
/,
|
J
(12.7.21)
During a pump start-up a
variation.
If the
pump-curve constants. Also
reservoir, the equation
along the
C+
may
pump
often assumed for the speed
N may be combined with the
characteristic in the suction pipeline.
(of
:
Pump
Fig. 12.21
is
if the pump is operating directly from a suction
be simplified by the elimination of the equation
!_L
^— Q
is
constant,
linear speed rise
speed of the
Centrifugal
pump.
UNSTEADY FLOW
669
EXAMPLE 12.8 Develop the necessary boundary-condition equations for the
pump in Fig. 12.22. The pump is to be started with a linear speed rise to N R
in
U
A
s.
check valve exists in the discharge pipe. The
initial
no-flow steady-
head on the downstream side of the check valve is He. For a steady flow
of Qo there is a loss of AH across the open check valve. Assume the check
valve opens instantaneously when the pump has developed enough head to
state
Hc
exceed
.
The equation
valve (after the
HP
= N*B
X
for the hydraulic grade line across the
check valve
is
open)
+ B NQP + B Qp> 2
3
pump and
check
is
^
where C 4 = r 2 Q /AH and N is the speed. When Eq. (12.7.3) is introduced
on the left side of this equation, the quadratic may be solved to determine the
flow:
CH - B 2N
The equations
for the
H -
Qi
2
N B < Hc
X
If iV 2£i
HP
-
Cm)
DA2 \Q2\J
t
<
t
t
>
U
2
Q P =0
Qp
(iV 2£i
<
[N R
If
1/CO
f
boundary condition are
\Ch
"0
N=
4 (£3-
_
a
CH =
Cm -
f
is
,
and
> Hc
defined
= Cm
+
Hp = He
,
by the above
ChQp
solution to the quadratic equation,
and
APPLICATIONS OF FLUID MECHANICS
670
/
'
^ Final
hydraulic
gradeline
/
Initial
Example
Fig. 12.22
gradient
12.8.
Accumulator
Many
different types of air or gas accumulators are used to help reduce pres-
sure transients in liquid systems.
In an analysis of the behavior of the ac-
cumulator shown
is visualized as being the same
any
instant. It is assumed to be
throughout the indicated control volume at
frictionless and inertialess. The gas is assumed to follow the reversible polyin Fig. 12.23 the pressure
tropic relation
HAV»
= C
where
HA
heads,
V
is
(12.7.22)
is
the absolute head equal to the gage plus barometric pressure
the gas volume, n
is
the polytropic exponent, and
®
i
Fig. 12.23
Simple accumulator.
«*,
C
is
a constant.
UNSTEADY FLOW
The
derivative of this equation with respect to time leads to
dHA
nC dV
V n+1
dt
The
671
dt
continuity equation applied to the control volume in Fig. 12.23 yields
— = Qp
- QPl
2
dt
When
these two equations are combined and placed in finite-difference form
for the time increment At,
HP
H+
=
C*(Qi,ns
-
Q2.1)
C 6 = 0.5nC At/V n+1
where
+
Cs(QPl NS
,
- QPitl
In this equation
.
it is
(12.7.23)
)
assumed that the volume V
volume during
at the beginning of the time step adequately represents the gas
the time step.
The new volume
is
computed before proceeding
to the next
time step by using a finite-difference representation of the above continuity
Equation (12.7.23) can be combined with Eqs. (12.7.13) and
unknowns Qp llNS Qp ,u and Hp. For those
cases where the change in V is significant during At, a second-order method
should be used.
An accumulator with inertia and friction is shown in Fig. 12.24. The
equation of motion written for connecting line 3 yields
equation.
(12.7.14) to evaluate the three
7
3
(
+H
Hp
V
HP4 + H
2
fU
4
ZW
2
2
,
Qz\Q*\ \
2g
"
yA L z dV
)
z
g
z
dt
or
HP -HPi
= C7
+
CsQp 3
(12.7.24)
where
C8 =
—
~gA
2/
3
and
C7 =
At
Equation (12.7.23) written
yields
Hpt = Hi
+
CqQz
+
CqQp 3
fL
H -H + j^-^Qs
DzgAz
A
for the air
2
\Qz\- CS QS
volume with the notation
in Fig. 12.24
APPLICATIONS OF FLUID MECHANICS
672
Gas
K~M
Qp
®
®
Fig. 12.24
The
friction
and
inertia.
continuity equation at the bottom of pipeline 3 yields
Qpi.ns
The
Accumulator with
=
five
Qp2,i
+
Qpz
unknowns may be determined
at each time step
solution of Eqs. (12.7.13), (12.7.14), (12.7.24),
and the
by a simultaneous
two equations.
last
Vapor-column separation
If the pressure level at.
a section in a pipeline drops below vapor pressure for
the liquid, vaporization occurs and a vapor pocket forms.
When this happens,
vapor pressure at the section and the flows must
be computed using this pressure. When the flows have been computed, a local
continuity balance may be used to identify the size of cavity. The vapor
column may grow and subsequently collapse, giving rise to a substantial overpressure. The calculation of the pressure rise at the collapse of a vapor cavity
is one of the most important calculations in many transient water-flow problems. An accurate description of the phenomenon is beyond the scope of this
treatment; however, the following example provides a reasonable description
of the behavior and yields a conservative design, in that pressure levels are
likely to be less than predicted with the model.
the pressure level
is
fixed at
UNSTEADY FLOW
C
C
C
C
C
673
COLUMN SEPARATION STATEMENTS FOR INTERNAL SECTIONS
ASSUMPTIONS: 1) COLUMN OPENS ONLY AT SECTIONS, 2) CAV„ VOLUME
SMALL COMPARED WITH LIQUID VOLUME IN REACH, 3) NO NEGATIVE
PRESSURES, 4) WAVE SPEED REMAINS CONSTANT. HV=VAPOR PRESSURE
IN UNITS OF LENGTH OF FLUID, GAGE.
DO 11 1=2,
VCAV(I)=.0
ICAV(I)=0
11
DO 34 1=2,
CP=H(I-1)-K) (1-1) *(CH-R*ABS (Q (1-1)
IF (ICAV(I+1) .EQ.1)CM=H(I+1)-QPX(I+1)*(CH-R*ABS(QPX(I+1-)))
IF (ICAV(I+1) .EQ.0)CM=H(I+1)-Q(I+1)*(CH-R*ABS(Q(I+1)))
IF (ICAV(I) .EQ.l) GO TO 33
HP(I)=.5*(CP+CM)
IF ((HP(I)-EL(I)) .LT.HV) GO TO 32
)
31
)
QP(I)=(CP-HP(I))/CH
GO TO 34
HP(I)=EL(I)+HV
ICAV(I)=1
QPX(I)=Q(I)
QP(I)=(HP(I)-CM)/CH
QPP(I)=(CP-HP(I) )/CH
VCAV(I)=VCAV(I)+.5*DT*(Q(I)-KJP(I)-QPP(I)-QPX(I))
QPX(I)=QPP(I)
IF(VCAV(I) .GT.OJ GO TO 34
ICAV(I)=0
VCAV(I)=.0
HP(I)=.5*(CP+CM)
QP(I)=(CP-HP(I))/CH
CONTINUE
32
33
34
Fig. 12.25
Column-separation statements.
EXAMPLE
12.9
Prepare a set of
FORTRAN
recognize and to calculate vapor cavities at
statements that will be able to
all interior
points in a single pipe-
line.
The
set of statements
is
shown
in Fig. 12.25.
Two
to identify the existence of a vapor pocket at a section:
the volume of the vapor cavity, and
if
no cavity
exists at the section
ICAV(I)
and
1 if
is
variables are defined
VCAV(I)
identifies
an integer that has the value
a cavity
exists.
The comments
in
Fig. 12.25 set forth the assumptions; the first three statements are inserted
above the time-increment iteration loop; and the
replace
all
interior-point calculations in the pipeline.
at each section
is
accumulated, and when
it
rest of the statements
The
size of
the cavity
collapses, the pressure rise is
com-
puted by use of the normal equations along the characteristic lines. When a
cavity exists, the flows at the upstream and downstream sides of the cavity
are identified as QPP(I) and QP(I) at the current time and QPX(I) and
Q(I) at the previous time, respectively.
1
OPEN-CHANNEL FLOW
In general, open-channel transients are more complex to handle than closedconduit transients. Surface-wave motion is an example of open-channel and
APPLICATIONS OF FLUID MECHANICS
674
unsteady flow. The subject is too vast to attempt to cover as part of a chapter.
few special topics are discussed that use about the same approach as the
A
waterhammer equations frictionless positive and negative surge waves,
routing, and a case of rainfall and runoff from a plane area.
:
FRICTIONLESS POSITIVE SURGE WAVE
12.8
IN
flood
A RECTANGULAR
CHANNEL
In this section the surge wave resulting from a sudden change in flow (due to
a gate or other mechanism) that increases the depth
channel
assumed, and friction
is
is
is
studied.
Fig. 12.26 shortly after a sudden, partial closure of a gate.
analyzed by reducing
it
A rectangular
is shown in
The problem is
12.27. The con-
Such a situation
neglected.
to a steady-state problem, as in Fig.
tinuity equation yields, per unit width,
(Vi
+
= (V2
c)yi
+
c)y 2
(12.8.1)
and the momentum equation for the control volume
stress on the floor, per unit width, is
1
-
2
(2/i
2
2/2 )
= -
2
By
ViiVt
+
c)(V 2
+ c-V 1
1
—
2,
neglecting shear
(12.8.2)
c)
g
elimination of
V
2
in the last
two equations,
11/2
*46
i+
9
(12.8.3)
In this form the speed of an elementary wave
proach 2/1, yielding
Vi
+
c
= \/gy
m.
—v u
Fig. 12.26
Positive surge
obtained by letting y 2 ap-
(12.8.4)
v.
y?
i
gular channel.
is
wave
in
a rectan-
UNSTEADY FLOW
675
Surge problem reduced to a steadyproblem by superposition of surge
Fig. 12.27
state
velocity.
For propagation through still liquid Vi — 0, and the wave speed is c = \/gy
when the problem is converted back to the unsteady form by superposition
>
V =
of
-c.
In general, Eqs. (12.8.1) and (12.8.2) have to be solved by trial. The
in the two equations
hydraulic- jump formula results from setting c =
(3.11.23).
EXAMPLE
A
12.10
rectangular channel 3
m /s,
m wide and 2 m deep,
m
discharging
3
/s at the downstream
suddenly has the discharge reduced to 12
end. Compute the height and speed of the surge wave.
V x = 3, yi = 2, V2 y2 = 4. With Eqs. (12.8.1) and (12.8.2),
3
18
6
=
4
+
c(y2
Eliminating
yi
-
4
=
c
-
and
4
9.806
and
2)
V
2
yi
-
=
4
|^|
(c
+
3) (3
-
7,)
gives
(^-2 + 3)(3 -^)
or
/j/2
2\
Ky2
V32/2-4/
^
)U
=
After solving for y 2 by
_±_ =
9.806
trial,
y2
=
2.75
m. Hence
V =
2
4/2.75
=
1.455 m/s.
.
APPLICATIONS OF FLUID MECHANICS
676
The
c
height of surge
2
=
—
2
=
2/2-2
:
wave
=
is
2.667
0.75 m,
and the speed
is
m/s
0.75
FRICTIONLESS NEGATIVE SURGE WAVE
12.9
wave
of the
IN
A
RECTANGULAR CHANNEL
The negative
surge
liquid surface.
wave appears
and lowering of a
downstream from a gate
as a gradual flattening
It occurs, for example, in a channel
being closed or upstream from a gate that
is being opened. Its propaaccomplished by a series of elementary negative waves superposed
on the existing velocity, each wave traveling at less speed than the one at
next greater depth. Application of the momentum equation and the con-
that
is
gation
is
tinuity equation to a small depth change produces simple differential expressions relating
wave speed
c,
velocity V,
and depth
y.
Integration of the equa-
tions yields liquid-surface profile as a function of time,
and velocity as a funcand time (x and t)
tion of depth or as a function of position along the channel
The
assumed to be frictionless, and vertical accelerations are neglected.
In Fig. 12.28a an elementary disturbance is indicated in which the flow
upstream has been slightly reduced. For application of the momentum and
continuity equations it is convenient to reduce the motion to a steady one,
fluid is
by imposing a uniform
as in Fig. 12.286,
c
y-6y :-:-:-:-z-:-r^^:-:
mmmmmmmmmm?.
(a)
CV
^
^^^v j^^^^v^
(b)
Fig. 12.28
Elementary wave.
velocity c to the
left.
The
continuity
UNSTEADY FLOW
equation
(V
or,
(c
is
-bV -c)(y-by)
= (V -
by neglecting the product
-
677
V) by
c)y
of small quantities,
= ybV
(12.9.1)
The momentum equation produces
(yI
l
ty)
- y-y =
2
2
y
- {V
Z
-
c)y[y
-
c
- (F-sy-c)]
g
After simplifying,
=
by
c
- V
bV
(12.9.2)
9
Equating bV/by in Eqs.
c
(12.9.1)
and
(12.9.2) gives
- V = ±\/gy
(12.9.3)
or
c
= V
d=
\/gy
The speed of an elementary wave in still liquid at depth y is \/gy and with
wave travels at the speed \/gy relative to the flowing liquid.
flow the
Eliminating
c
from Eqs.
(12.9.1)
and
(12.9.2) gives
*y
dy
and integrating leads
V = ±2
\/gy
+
to
const
For a negative wave forming downstream from a gate,
the plus sign, after an instantaneous partial closure, V =
V =
2
-y/gy'o
+
const
Fig. 12.29,
by using
y and
V when y =
,
APPLICATIONS OF FLUID MECHANICS
678
Negative
Fig. 12.29
wave
gate
after
closure.
After eliminating the constant,
7 =
-
7„
The wave
ct,
= (VQ
x
- v5)
+x
travels in the
V -
2
(12.9.4)
direction, so that
^gy +
motion occurs at
If the gate
by x =
yfg (ViJ
V^ =
+
= V
c
2
t
=
3
V^
(12.9.5)
the liquid-surface position
0,
is
expressed
or
-
2
+ s Vgy)t
-s/ijfo
(12.9.6)
Eliminating y from Eqs. (12.9.5) and (12.9.6) gives
V
7o
2i
3
3*
which
is
(12.9.7)
Vgy*
the velocity in terms of x and
EXAMPLE
12.11
t.
In Fig. 12.29 find the Froude number of the undisturbed
is just zero when the gate is suddenly
flow such that the depth y\ at the gate
closed.
For
V =
20
ft/s, find
the liquid-surface equation.
V\ =
In Eq. (12.9.4), with V
when
It is required that
t
=
0.
V =
2
gyo
or
F
=
=
0,
—=
y
4
yx
=
=
0,
at x
=
for
any time
after
UNSTEADY FLOW
V =
For
20,
V1
—=
2
=
Vo
20 2
-
-r
±g
By
3.11ft
±g
Eq. (12.9.6)
=
x
679
The
(20
-
2
V32.2
liquid surface
is
X
3.11
+
=
3 -v/32.2j/)«
17.04
V?/
a parabola with vertex at the origin and surface concave
upward.
EXAMPLE
12.12
In Fig. 12.29 the gate is partially closed at the instant t =
is reduced by 50 percent.
V = 6 m/s, y = 3 m. Find
so that the discharge
Vi,
and the surface
2/1,
The new
q
6
X
3
-y-
-
=
n
9
By
Eq. (12.9.4)
7i
=
-
6
2
x
=
-
2.12 m.
(6
-
2
which holds
Dam
=
is
T7
Fi2/i
V^806 V3 - V^)
(
Then V\ and
yi
profile.
discharge
?/i
are found
The
by
trial
from the
liquid-surface equation,
V% + 3 V^/)*
or
for the range of values of y
a;
two equations, V\
from Eq. (12.9.6), is
last
-
(9.39
V^ -
=
4.24 m/s,
4.84)^
between 2.12 and 3 m.
break
An idealized dam-break water-surface profile,
Eqs. (12.9.4) to (12.9.7).
From a
Fig. 12.30,
can be obtained from
channel with depth
frictionless, horizontal
on one side of a gate and no water on the other side of the gate,
in
suddenly removed. Vertical accelerations are neglected. V =
the equations, and y varies from y to 0. The velocity at any section, from
Eq. (12.9.4), is
of water y
the gate
is
v = -2 Vg (Vvo - Vv)
(
12.9.8)
APPLICATIONS OF FLUID MECHANICS
680
Dam-break
Fig. 12.30
profile.
always in the downstream direction.
The
water-surface profile
is,
from Eq.
(12.9.6),
s
= ($VqV -
At x =
y
0,
section x
2 VoVo)t
=
=
4?/ /9,
is,
(12.9.9)
the depth remains constant and the velocity past the
from Eq.
also independent of time.
(12.9.8),
The
leading edge of the
height and moves downstream at
V =
c
wave feathers out to zero
The water surface is
= —2\/gy~
.
a parabola with vertex at the leading edge, concave upward.
With an
actual
or wall of water, to
dam
break, ground roughness causes a positive surge,
move downstream;
i.e.,
the feathered edge
is
retarded
by
friction.
12.10
FLOOD ROUTING
IN
PRISMATIC CHANNELS
In the two preceding sections instantaneous changes in frictionless rectangular
channels were considered.
In this section friction
is
taken into account, and
may
change gradually at inlet or outlet sections. With prismatic
channels the area is a function of depth of flow. The assumption is made that
the channel slope a is small enough for cos a to approximate 1 and hydrostatic
conditions
conditions to prevail along any vertical line in the fluid.
function of time
may be added
downstream section
to the flow or taken from
Any known
it
at the
flow as a
upstream or
of the channel.
In Fig. 12.31 an element of the flow is taken as control volume, the x
direction is taken parallel to the bottom of the channel, and the depth y is
measured normal to the bottom. The unsteady-momentum equation (3.11.2)
UNSTEADY FLOW
681
^Ax
dx
yA^Ax
dx
>
^yAAx sin
r
a
PAx
^^^^^^^^^^^^3^^^^
—»-Ux|—
Control volume for application of unequation.
Fig. 12.31
steady
is
-
momentum
applied,
-^-
y Ax
A -
t
P
Ax
dx
+
yA Ax
=
a
sin
—
(pV A)
dx
2
Expanding and dividing through by the mass
dy
dV
V
—
-^--gsma + 2V— + +
dx
A
R
dx
to
g
TT
.
P
P
is
2
dA
dx
+-
VdA
+
+ A~—
dt
(pAV Ax)
dt
of the element
the wetted perimeter of the cross section, and
The
Ax
dV
—
=
pA Ax
gives
(12.10.1)
dt
R
is
the hydraulic radius.
continuity equation (3.2.1) applied to the control volume of Fig.
12.31 yields
-
~ (pAV)
Ax =
dx
d
-
(
P
A
Ax)
dt
After expanding and dividing by the mass of the element,
VdA
dA
dV
-—+-—
+— =
A
1
A
dx
dt
,
,_
12.10.2
dx
Equation (12.10.1) may be simplified by subtracting Eq. (12.10.2), when
multiplied by V, from it
Ll
=
d
g
JL
dx
Z Z=
d
+ ll_ gsina+V dx +
pR
Equation (12.10.2)
may
(12 10 .3)
.
dt
be written
+ °!L-0
U „A*L+V*
T dx
dx
dt
d
(12.10.4)
APPLICATIONS OF FLUID MECHANICS
682
since
dA
=
dAdy
=
dx
dy dx
dx
T
with
method
dy
the top width.
L\ and
of characteristics.
Idx \
L
2
are
now
in suitable
Combining, as in Sec.
TJ
ldx\
dt]
+
form
for solution
by the
12.6,
X/
dt\
^- gdna =
(12.10.5)
pti
For the term within the
dx
=
+
dt
and
for the
dx
—
=
„
V+
first
pair of brackets to be a total derivative
\A
Y
term within the second pair of brackets to be a total derivative
q
-
X
dt
After equating the two latter expressions and solving for
X
=
db
X,
(12.10.6)
yJY
and
^ = V±J±
V T
(12.10.7)
v
dt
For a rectangular cross section
wave (surface wave)
speed of a
it
is
should be noted that this expression for
given by
^=v±Vov
(12-10.8)
dt
Equation (12.10.5) reduces to
=
5 + *? + ^-^ina
pR
dt
dt
subject to Eqs. (12.10.6) and (12.10.7).
(12-10.9)
UNSTEADY FLOW
For small slopes
may be
sin
a
«S
expressed as gS, where
where S
,
S
is
is
683
the bottom slope, and r /pR
the slope of the energy grade line as de-
fined by the Manning or Chezy equation. This assumes that viscous losses in
unsteady flow are described in the same manner as losses in steady flow at the
same depth and discharge. After making these substitutions and those of the
values of A from Eq. (12.10.6), the four differential equations become, in
finite-difference form,
VB
—
H
yR )
(yp
+
g(SR
-
So) At
=
(12.10.10)
Cr
c+
—
XP
Xr
= {Vr
Vp-Vs--
+
Cr) At
-
ys)
cs )
At
(yP
(12.10.11)
+
g(S s
- S
)
M
=
(12.10.12)
cs
c-
—
Xp
xs
= (V s —
(12.10.13)
where
IgA
(12.10.14)
and the subscripts
S
R and S indicate evaluation of the quantity at points R and
in Fig. 12.32.
assumed that initially y and V are known at equiAx apart. To find yp and VP at one of the
sections at the new time t + At, the variables y, V, and c must be evaluated
at R and S. This is accomplished by linear interpolation from known values
at A, C, and B. Then Eqs. (12.10.10) and (12.10.12) may be solved for VP
In Fig. 12.32
it is
distant sections along the channel,
Ax P
Ax
At
Ax
Ax
Ax
C
Ax
S B
R
Solution
distance interval.
Fig. 12.32
for
specified
time
and
APPLICATIONS OF FLUID MECHANICS
684
and yP
The
.
Vc — Vr
Y
VA
V7 c — F"
where
CC
cc
~
—
=
=
6
°R
linear interpolation for
—
xc
—*Ax
Xr
=
_H— Xr
Xp
=
Ax
At/ Ax and use has been
= 0(VR
+
VR
is
n .__
9(
expressed
+
Vr
made
by the proportion
c*
(12.10.15)
of Eq. (12.10.11).
Similarly
(12.10.16)
cR )
cA
Simultaneous solution of these two equations yields
+
e(-VcCA + CcVA)
VA + c c - Ca)\
6(Vc — ^T~T
Vc
Vr
i
-f
1
miom
(12.10.17)
/,/t/
i
The value
of cR can be determined
section
from (12.10.14).
tjr
from Eq. (12.10.16) and for the rectangular
For the general cross section it is easier to find
yR from
VC
yc
~
-
VR
= 6(Vr
+
(12.10.18)
cr )
va
by using the computed values
of c R fromEq. (12.10.14).
of
VR
and
cR ;
then find a new compatible value
For subcritical flow, which is most common for flood-routing analysis,
S lies between C and B in Fig. 12.32. A similar linear-interpolation
procedure using Eq. (12.10.13) in place of Eq. (12.10.11) leads to
point
~ 6(Vcc B -e(V c - V B Vc
1
=
cc
ccV B )
cc
+
cb)
+ VsO(cc - cb)
\ + e{cc-c B
(12.10.20)
)
ys
=
yc
+
e(V s
-
c B ) (yc
-
(12.10.21)
yB )
Again in the rectangular section Eqs. (12.10.19), (12.10.20) and (12.10.14)
are used, and for the general cross section the above three equations are used,
with a final value of c s determined from Eq. (12.10.14) using y s from Eq.
In supercritical flow, which is not covered here, the control of the
(12.10.21)
unsteady flow shifts to the upstream end only, and point S lies to the left of
,
.
point C.
With the
variables determined at points
R
and
S, Eqs. (12.10.10)
and
UNSTEADY FLOW
and Vp\
(12.10.12) are solved simultaneously for yp
»yp =
+
lysCB
:
cr
+
cs
+
yR c s
~
R
cR c s
L
{
VP = VR -g y^-~^ -
g At (SR
S
M
-
(SR
- S8
(12.10.22)
)
JJ
g
-
685
(12.10.23)
So)
cR
For the end sections of the channel Eqs. (12.10.10) and (12.10.12) each
two unknowns VP yP which, together with a
known condition at the end, permits the solution to be carried forward. This
analysis is limited to subcritical flow where one boundary condition is needed
at each end of the channel.
In solving the equations it is absolutely essential that R and S always
yield one linear equation in
AB
stay within
EXAMPLE
(Fig. 12.32)
A
12.13
0.8 horizontal to
slope
time
;
otherwise the solution method
m
trapezoidal channel 3000
1
,
,
long, 7
m
is
wide, side slopes
Manning roughness, and
vertical, 0.016
unstable.
0.001
bottom
discharging under steady uniform flow conditions at y n = 1.8 m. At
= the flow begins to increase linearly to a flow of 60 3 /s in 20 min.
is
t
m
The flow then decreases linearly to
boundary condition
Q = Cw (y
-
is
15
m /s in another 10 min.
3
The downstream
given as a gage-height discharge curve of the form
1 -5
2/do)
.
A
computer program is presented in Fig. 12.33 for finding the velocity
and depth in the channel. The discharge Q at steady uniform flow is first
computed using the given normal depth in Manning's equation. The upstream
boundary is
Q= 1Qm +
Qf
4
tp
~ QM
—
-h)
(t
4
t
F
t>t F
^Qf
where Q M =
60,
QF =
15,
h
Z
x
the side slope.
=
1200
s,
h =
1800
s.
By
continuity the dis-
B is the bottom
A nonlinear equation results which must be solved
charge in a trapezoidal section
width and
k<t<
t\
is
equal to (By
+
Ziy 2 ) V, where
simultaneously with the C" compatibility equation, Eq. (12.10.12) Newton's
Similarly the downstream
is used in the solution (see Appendix E).
gage-height discharge equation, when combined with the continuity equa.
method
FLCCD ROlTlhG IN TRAPEZCICAL CHANNEL. D.S. BOUNDARY Q*=CW*( Y-YOO 1**1.5
LINEAR RISE 6 FALL OF INFLOW HYORUGRAPH, FLOW AT YN AT INITIAL T,QM AT
Tl f ANC UF AT TF. WIDTH B, SIDE SLOPE Zl, ROUGHNESS RN, SLOPE SO.
DIMENSICN Y(2l),V(2l),YP(ill,VP(2l),C(21), 1CI2I
CATA ID/'EN' ,'SI'/
AREA(YY)*YY*(B*YY*Z1)
PER(YY)*B*2. *YY*SCRT(Z1*ZW1.)
CEL(YY)=SQRT(G*YY*(B*YY*Z1)/(B*2.*ZI*YY)1
SLOPE YY,VV)=(VV*RN/CMA)**2*(PfcR(YY)/(YY*(B*YY*Z 1) ))**l. 3333
READ (5,15,END=99)XL,B,Z1,RN,S0,CW YN. CM C F T 1 TF, THAX NT,N, IPR
10
FOPMAT(3FlO.3,2Fl0.5,FlC.2/toFl0.2/A2,3X,2I5)
15
C
C
C
.
IF
(NT. EC. 10(2)1
,
.
,
,
GO TO 16
CMA=1.49
G*32.17«
GO TO 17
CMA=i.
G*9.8C6
17
WRITE (6*18) NT , XL B Z I ,RN S J C M, YN , CM OF T 1 TF TMAX N IPP
FCRMAT(1H0,A2, UNITS SPECIFIED'/' XL*',F9.i,' B*',F8.2,' Zl»'
18
2F8.2,' PN=',F6.4,' S0=',F7.5,' CW*',F7.2/« YN*',F8.2,» CM*',F8.2
3,' 0F=',F8.2,' Tl,TF,fc TMAX=' , 3F8 .1 ,' SEC'/'
N*',I5,» IPR*', 151
16
,
,
,
,
,
,
,
,
,
,
'
Q0=CMA*AREA(YN)**1.6666*SURT(S0)/(RN*PER(YN)**C6666)
C6=(CM-C0)/T1
C7=(UF-CM)/(TF-Tl)
YOO=YN-(C0/Crf>**(2./3.*
NS=N*i
VO=QO/AREA(YN)
CC 20 1*1, NS
vm*vo
20
C(I)=CEL(YN)
YP(I)=YN
Y(I)=YN
OX*=XL/N
Q*go
T»C.
K»0
CT=.9*CX/(V0»C(l)>
WRITE (6,251 QCYOO
FORMAT (• QO*',F10.2,' Y0C=',F8.3/' TIME IN MINUTES'//'
25
.4
.6
.8
.0
.2
2
L.S.C X/L=
TM=T/6C.
30
35
40
Cn = AREMY(NS))*V(NS)
WRITE (6,35) TM,C>,( V( I ) ,
FCPMATUH ,F9. 3.F9.2, 3H
I
*
tf=
1,NS ,2)
,
(
Y<
1 1
,oF 8. 3/19X, 3H
, 1
=1
Y=
,
TIME
1.
0')
,NS,2 ) ,QW
6F8.3 ,F8.2)
T*T»OT
K = K*1
IF (T.GT.TMAX) GO TC 10
IX=0
43
DXX=0.
DO 45 1=1, NS
CXI = (V( I)*C(I))*OT
IF (CXI.GT.DXX) OXX=DXI
IF (OXI.GT.OX.AND.IX.EJ.O) GO TO 43
GO TO 45
T=T-.1*CT
CT=.9*CT
IX = 1
CONTINUE
TH=OT/DX
INTERIOR PCINTS
45
C
00 50 1*2,
CA=C(I)-C(I-1I
VR=(V(I )*TH*(CU)*V(I-l)-V(I)*C(I-l)))/(l.*TH*(V(I)-V(I-l)*CAI)
CR=(C( I)-VR*TH*CA)/(l.*TH*CA)
YR = Y(I )-TH*(VR*CR)*(Y( U-YII-U )
CR=CEL(Y«)
SR=SLOPE(YR,VR)
CB=C(I)-C(I*1)
VS=(V(I )-TH*(V(I)*C(I*l)-C(II*V(IU)))/(l.-TH«(V( I )-VU *1 1-CB)
CS=(C( I)*VS«TH*CB)/(1.*TH*CB)
YS=Y(I)»TH*(VS-CS)*(Y( I)-Y(I»1)
CS=CEL(YS)
YP(I )=(YS*CR*YR*CS*CR*CS*((VR-VS)/G-DT*(SR-SLCPE(YS,VS))))/(CR*CS)
VP(I )=VR-G*( (YP(I )-YR)/CR*LT*lSR-SO))
50
UPSTREAM BCUNCARY
CB=C(l)-C(2)
VS=(V(l)-TH*(V(l)*C(2)-C(l)*V(2)))/(l.-TH*(V(l)-V(2)-CBI)
CS=(C( l)*VS*TH*C8)/( l.*TH*C8)
YS=Y(1)*TH*(VS-CS)*(Y(1)-Y(2))
IF (T.LT.T1) U=U0*C6*T
)
C
Fig. 12.33
686
UNSTEADY FLOW
687
(T.GE.T1.AND.T.LT.TF) U=t.M*C7*( T-T I
(T.GT.TF) Q=QF
IF
IF
C2=G/CfcL(YS>
Cf=VS-C2*YS-G*DT*< SLOPE (YS,VS)-SO>
E=CM*Zl*C2*e
DU 60 j=l,3
F=Q-YP{ll*(YP(l)*(YP(i)*C2*Zl*tJ*CM*8)
60
YP(l)=YP(n*F/(YP(l)*(3.*C2*Zl*YP<l)«-2.*E}*CM*8J
VP«1)=CK*C2*YP(1)
COhNSTREAP eCLNCARY
CA=C(NS)-C(N)
VR=(V(NS )*7K*(CINS)*V(N)-V(NS)*C(N) J)/ll.*-TH + lV(NS)-V(N)+CA)l
CR = (CUS)-VR*TH*CA) /< 1 .+TI-*CA)
YR=YCNS)-TH*(VR*CR) *<Y(NS)-Y(N1)
C
C4 = G/CEUYR)
CP=VR*C4*YR-G*DT*(SLCPE4YP,VR)-SOJ
E=-CP*Z1*B*CA
CU 70
JM,3
H=YP{NS)-YOO
F=YP{NS)*<YP(NS)*(YP(NS)*C4*Z1*E)-B*CP)*C*****1.5
YP(IMS)=YP(NS)*F/(YP(NS)*(-YP(NS)*C4*Zl*3.-t*2.)*B*CP-1.5*CW*W**.5)
VP{NS)=CP-C4*YP(NS)
70
DC 80 1=1, NS
vm=vp(i)
C(II=CEL(YP(I)J
YII)=YP(I1
IF (0XX.LT.0.8*DX) CT=l.l5*UT
IF (K/IPR*IPR.EJ.K) GO TO 30
80
GO TO 40
STOP
99
ENO
ooc.
1.8
7.
.8
6C.
1!>.
10
SI
Fig. 12.33
tion
.016
1200.
.0010
20.
1300.
2400.
2
FORTRAN
and Eq.
IV
program
for flood routing.
(12.10.10), yields a nonlinear equation
iteratively using
which
is
also solved
Newton's method.
are given in Fig. 12.34. It may be noted that formatted
used in the program and that either SI or English units may be used
by specifying SI or EN in the first two columns of the last card of input data.
Computed results
input
is
The two
integers
on
this card specify the
number
of reaches the channel is
divided into and the number of time-increment iterations between printout
of
computed
results.
The program
rectangular, or trapezoidal sections.
is
able to handle prismatic triangular,
It also contains
an adjustment
of the
the time increment during the computations in order to minimize the
size of
approximation due to interpolations. Some caution should be used in applyprogram to continually increasing flows or sudden flows, as no provision is made for handling hydraulic bores.
ing the
MECHANICS OF RAINFALL-RUNOFF RELATIONS FOR
SLOPING PLANE AREAS
12.11
An
1
problem is that of the relation beand runoff on a plane sloping surface. Percolation rate may be
interesting, simplified characteristics
tween
rainfall
1
M. Henderson, and R. A. Wooding, Overland Flow and Groundwater Flow from a
Steady Rainfall of Finite Duration, J. Geophys. Res., vol. 69, no. 8, 1964.
F.
APPLICATIONS OF FLUID MECHANICS
iSI
UMTS SPECIFIED
XL=
YN=
N=
CC=
TIME
C.80 RN= O.OloO St1=0.0010 3 CW= 2 3.00
1800 .0 240
1200 .0
15.00 Tl. TF,C TMAX*
7.00 11 =
8=
3C00.
60.00 QF=
1.8C CM=
2
10 IPR=
0. 322
35. S2 Y00=
IN
MINLTES
L.S.C X/L=
35.92 v=
TINE
C.000
Y=
V=
Y=
1.441
37.65
2.681
39.39 V=
4.322
Y=
V=
Y=
41.12
5.762
42.86 V=
7.203
44.59
V=
46.33
Y=
V=
Y-
8.643
Y=
-8.06 V=
10.C84
Y=
11.525
49. 8C
V=
Y=
12.965
51.53 V=
14.334
53.18 V=
15.430
Y=
V=
Y=
54.74
Y=
V=
16.927
56. 3C
18.223
57.86 V=
Y=
Y=
19.520
59.42
20.816
Y=
56.33 V=
V=
Y=
22.113
50.49 V=
23.409
44.66
Y=
V=
24.706
38.82
26.002
32.99
27.15
Y=
V=
Y=
V=
Y=
V=
28.595
21.32
Y=
V=
29.892
15.49 V=
27.299
Y=
Y=
31.188
15. OC
V=
32.465
15.00
Y=
V=
33.7ei
15.00
35.C73
15.00
36.375
15.00
37.671
15. OC
Y=
V=
Y=
V=
Y=
V =
Y=
V=
Y=
Computer
Fig. 12.34
.0
.2
.4
.6
2.364
1.800
2.432
1.830
2.488
1.865
2.536
1.903
2.577
1.944
2.615
1.985
2.649
2. 028
2.681
2.071
2.711
2.114
2.739
2.156
2.764
2.304
1.600
2.364
1.800
2.411
1.820
2.^60
2.364
1.600
2.364
1.800
2.364
1.600
2.396
1.614
2.437
1.8J6
2.478
1.865
2.516
2.364
2.1W
2.7ae
2.235
2.810
2.272
2.832
2.3C9
2.853
2.346
2.739
2.321
2.561
2.242
2.390
2.144
2.219
2.C29
2.041
1.898
1.847
1.750
1.628
1.564
1.368
1.395
1.415
1.316
1.486
1.261
1.546
1.217
1.597
1.182
1.637
1.156
1.670
1.136
1.649
2.5C5
1.862
2.546
1.919
2.585
1.956
2.621
1.998
2.654
2.G40
2.686
2. 082
2.714
2.122
2.740
2.160
2.764
2.198
2.786
2.236
2.810
2.274
2.832
2.311
2.759
2.305
2.618
2.250
2.460
2.176
2.340
2.066
2.195
1.982
2.C44
1.665
1.897
2.55o
1.933
2.592
1.971
2.627
2.011
2.658
2.049
2.686
2.037
2.713
2.124
2. 73?
2.
162
2.764
l,20J
2.788
l.Zid
2.811
2.276
2.766
2.233
2.657
2.246
2.544
2.
19*
2.428
2.124
^.lO*
2.04^
1.735
184
1.946
1. 701
2.034
I
-fad
I
1.592
1.633
1.486
1.636
1.413
l.tw
I.J53
1.665
1.303
1.680
l.zt>L
solution of
2.
1.
64
3
1.915
1.728
1.822
l.b2o
1.733
1.545
1.763
1.473
1.759
1.42
Example
I.
ECO
2.364
i.eoo
2.264
1.800
2.364
i.eoo
2.386
1.6C9
2.419
1.627
2.454
1.850
2.492
l.87e
2.529
1.911
2.565
1.946
2.59t
1.931
2.629
2.C17
2.658
2.C53
2.636
2.C9C
2.712
2.126
2.738
2.165
2.763
2.203
2.767
2.241
2.762
2.257
2.681
2.238
2.589
2.201
2.493
2.149
2.392
2.C84
2.287
2.CC8
2. 179
1.923
2.C67
1.628
1.975
1.736
1.920
1.656
1.687
1.586
.8
2.364
1.800
2.364
1.800
2.364
1.800
2.364
1.800
2.364
1.800
2.379
1.806
2.4C4
1.819
2.435
1.839
2.468
1.663
2.502
1.892
2.534
1.923
2.564
1.955
2.593
1.989
2.622
2.024
2.650
2.061
2.676
2.098
2.702
2.136
2.728
2.174
2.752
2.212
2.7*3
2.235
2.683
2.229
2.610
2.206
2.532
2.167
2.452
2.115
2.370
2.C51
2.286
1.978
2.199
1.896
2.121
1.811
2.065
1.733
,0
SEC
1.
2.364
1.800
2.364
1.800
2.364
1.800
2.364
1.800
2.364
1.800
2.364
1.800
2.369
1.806
2.381
1.819
2.398
1.839
2.418
1.864
2.441
1.892
2.465
1.921
2.490
1.953
2.517
1.987
2.543
2.022
2.570
2.058
2.597
2.095
2.624
2.133
2.65C
2.171
2.676
2.2 09
2.691
2.231
2.663
2.220
2.663
2.190
2.633
2.146
2.594
2.091
2.547
2.027
2.492
1.955
2.429
l.b77
2.363
1.798
35.92
35.92
35.92
35.92
35.92
35.92
36.13
36.63
37.35
38.27
39.32
40.44
41.66
42.95
44.32
45.74
47.21
48.72
50.25
51.81
52.73
52.28
51.04
49.25
38.77
35.86
12.13.
subtracted from rainfall rate to yield more meaningful results.
sional flow
sin 6 tt
S
.
is
assumed
(Fig. 12.35)
One-dimen1 and
on a mild slope so that cos
^
UNSTEADY FLOW
Control volume for continuity and
Fig. 12.35
equations for
and
rainfall
The momentum
-
yy Ax So
r
Ax =
dx
After replacing r
momentum
runoff.
equation, for an element of unit width, yields
dy
—
Ax -
yy
689
p
—d (V
2
y)
dx
by ySy, where S
Ax
d
+ p -dt
{Vy) Ax
the slope of the energy grade
is
line,
and
simplifying
dy
+
gy^
dx
The
—
(Vy)
dx
-
gyS
gyS
dV
dy
dV
dy
+ 2Vy— + y— + v^
+v*f
dx
dx
dt
dt
=
o
(12.11.1)
continuity equation yields
Ao:
=
z;
Ao;
Aa:
dt
or
y
h
dx
V
^o
1
dx
=
(12.11.2)
dt
Multiplying Eq. (12.11.2) by
»«
QVTdx
gyS
-
gySo
+
Vy
XT
subtracting from Eq. (12.11.1) lead to
-IT
—+ — +
j-.
+
V and
»>
y
dx
-
dt
v
V =
(12.11.3)
APPLICATIONS OF FLUID MECHANICS
690
or
(by
S = S -l-r\dx
VdV --ldV
+ -—+
+
dx
dt
g
g
—V\
v
(12.11.4)
)
gy /
For one class of problems the quantity in parenthesis
*So and may be dropped:
is
small compared with
SttS
Manning's equation
If
q
= Vy =
or,
q
(12.11.5)
by
—n VSo~y
is
used to describe frictional losses
5,z
(12.11.6)
generalizing, for resistance formulas other than
Manning,
= ay m
The
(12.11.7)
continuity equation in terms of q
is
+ £-*
?
dx
(12.H.8)
dt
These
last
two equations, one
a characteristic.
dQ
=
—
dqdy
dx
ay dx
From Eq.
m
= may™'
i
1
algebraic,
one
differential,
may
be solved along
(12.11.7)
dy
—
no
11 (^
(12.11.9)
dx
which, combined with Eq. (12.11.8), yields
may m-'
—+—
dx
Now,
v
(12.11.10)
if
— = may™dt
=
dt
1
(12.11.11)
UNSTEADY FLOW
691
Eq. (12.11.10) becomes
dy
=
v
(12.11.12)
which is valid along the
For constant rate
xt
path given by Eq. (12.11.11).
from t = 0, Eq. (12.11.12) inte-
of rainfall starting
grates to
=
y
(12.11.13)
vol
which states that along any
By
linearly.
xt characteristic
the depth of liquid increases
substituting Eq. (12.11.13) into Eq. (12.11.11) and integrating,
dx
—
= ma(vQt) m~
l
dt
or
=
x
x
+
aVo
m ~H n
(12.11.14)
with £ the starting point on the characteristic for t = 0. For x = 0, the line
plots on the xt plane of Fig. 12.36a as shown. Since all the characteristics
issuing from the t =
line are parallel at a given t, and since y = v t along
each characteristic, the build-up on the plane occurs with the surface parallel
to the plane, as indicated in Fig. 12.366.
mined from Eq.
=
x
av m ~H m
(12.11.14) with x
= -
(v
t)
m
=
The
steady-state profile
is
deter-
0,
ay m
=
Vo
or
ay s m
Xs
(12.11.15)
{
Vo
The time required
ts
=
L
/
is
by Eq.
(12.11.14) for x
=
L, x
=
1/m
(12.11.16)
-)
(
which
\
for steady state is given
also the
time at which
maximum
depth and flow begin to occur. The
APPLICATIONS OF FLUID MECHANICS
692
ti
h-
Steady state x s = -32k
^J*""*-— -^
—^
r-
Build-up
y
= v
o
l
(a) Characteristics on xt plane
water surface on physical plane.
Fig. 12.36
and
(b)
discharge at the downstream end of the plane
q x= L
= ay m =
a(v
t)
m
<
t
<
ts
is
(12.11.17)
with steady-state discharge
q x =L
= Lv =
a(v
ts
m
)
t
>
ts
(12.11.18)
UNSTEADY FLOW
vQ
693
For subsidence of the runoff following the cessation of the rainfall, let
= 0, for t > to > t8 Then dy/dt = 0, which shows y to be a constant along
.
the characteristic
dx
—
=
amy™-
1
(12.11.19)
Figure 12.36a shows the characteristics drawn from conditions for steady
tQ
The profile conditions as a function of time may be found by
integration of Eq. (12.11.19)
state at time
x
=
xs
+
where x s
.
amy
is
s
m ~ l (t
-
t
(12.11.20)
)
the distance to the steady-state depth y s
.
By
substituting Eq.
(12.11.15)
x
=
ay s m
h
amy
= ay m
Since q
s
,
s
m - l (t
-
t
(12.11.21)
)
the discharge during subsidence
may
be determined at any
position x,
DM
x
= -
am + am(-)
J
Vo
For x
(t-
I
=
to)
>
U
(12.11.22)
t>
to
(12.11.23)
t
L,
(m— l)/w
=
-
+ am (-)
(
(t
]
-to)
from which the subsidence hydrograph at the downstream end can be calculated. The discharge hydrograph for the complete rainfall period is schematically
EXAMPLE
shown
in Fig. 12.37 for
A
t
>
ts
.
200 yd square on a slope of 0.0016 is
Resistance is given by Manning's
0.025. Determine the hydrograph for this storm from this area.
12.14
paved parking
lot
subjected to rain of 2 in/h for 30 min.
n =
2.0
V
°
"
12
X
1
3600
~
t/S
21,600
Cm ^
T VS = 0^25" X V° mm = 2A
1.49
«
=
°
APPLICATIONS OF FLUID MECHANICS
694
Fig. 12.37
Hydrograph
from an inclined plane.
for
From Eq.
m=
ts
(12.11.16), for
FromEq.
Q =
8.596
Qmax
=
t
-
"
1488
V,2.4 X
t)
X
m
= 600 X
'
2.4 (0.00004630 5/3
<
1488
600
-
36Q
t
0.218Q 2
The hydrograph
20
'
5
is
t
<
(12.11.23), for q
U
>
t
=
<
1488
<
1800
Q/600,
1800
plotted in Fig. 12.38.
40
60
100
80
Time, min
Fig. 12.38
s
0.0000463 2 3 /
10-^ 5/3
16.67 cfs
From Eq.
\3/5
(12.11.17)
600a{v
=
f,
600
wV
~
runoff
Runoff hydrograph for Example
12.14.
UNSTEADY FLOW
695
PROBLEMS
Determine the period
12.1
cross-sectional area
A U
12.2
2.4
is
mum
fluid velocity
12.3
A liquid, v =
is
Neglect
.
tube containing alcohol
equilibrium position of 5.0
The
in.
and the period
0.002
U
of oscillation of a
cm2
oscillating
is
total
ft /s, is in
1
of water.
The
maximum
with
column length
of oscillation.
2
tube containing \
friction.
40
is
Neglect
displacement from
Determine the maxi-
in.
friction.
U tube. The total liquid column
a 0.50-in-diameter
when the column is at
determine the time for one meniscus to move to within 1.0 in of its equilibrium
70 in long.
rest,
one meniscus
If
is
15 in above the other meniscus
position.
Develop the equations
12.4
when lQv/D 2 = y/2g/L.
A U
12.5
for
motion
Suggestion:
U tube for laminar resistance
of a liquid in a
mt
Try
z
=
e~
+c
(ci
2 t)
tube contains liquid oscillating with a velocity 2 m/s at the instant the
Find the time to the instant the menisci are next
menisci are at the same elevation.
at the
L =
A
12.6
1
same elevation and determine the velocity then,
v
=
10 /mi 2 /s,
D=
0.6 cm,
75 cm.
mi
10-ft-diameter horizontal tunnel has 10-ft-diameter vertical shafts spaced
When
apart.
a depth of 50
ft in
valves are closed isolating this reach of tunnel, the water surges to
one shaft when
it is
20
ft in
the other shaft.
For/
=
0.022 find the
height of the next two surges.
Two standpipes 6 m
12.7
/
=
0.020,
other one
and minor
in
diameter are connected by 900
losses are 4.5 velocity heads.
when a valve is rapidly opened
One
in the pipeline.
m of 2.5-m-diameter pipe;
9 m above the
reservoir level
Find the
is
maximum fluctuation
in water level in the standpipe.
m
12.8 A valve is quickly opened in a pipe 1200
long, D = 0.6 m, with a 0.3-mdiameter nozzle on the downstream end. Minor losses are 472 /2<7, with V the velocity
= 9 m. Find the time to attain 95 percent of the steady-state
in the pipe, / = 0.024,
H
discharge.
12.9
A
D=
3.0
(K =
globe valve
ft,
/
=
0.018,
10) at the end of a pipe 2000 ft long
minor
losses are
2V2 /2g, and
H=
75
ft.
is
How
rapidly opened.
long does
it
take
for the discharge to attain 80 percent of its steady-state value?
A steel
12.10
thickness.
pipeline with expansion joints
When
Benzine
12.11
it is
(K =
g-in wall thickness.
ft,
D=
4
ft, t'
150,000
psi,
S =
90
cm
in
diameter and has a 1-cm wall
0.88) flows through f-in-ID steel tubing with
Determine the speed
Determine the
12.12
3000
is
carrying water, determine the speed of a pressure wave.
=
maximum
of a pressure
wave.
time for rapid valve closure on a pipeline:
§ in steel pipe,
V =
10
ft/s,
L =
water flowing.
12.13 A valve is closed in 5 s at the downstream end of a 3000-m pipeline carrying
water at 2 m/s. a = 1000 m/s. What is the peak pressure developed by the closure?
APPLICATIONS OF FLUID MECHANICS
696
12.14
Determine the length
12.15
A
valve
closed at the
is
only one-third of the line
of the time
2L/a
of pipe in Prob. 12.13 subjected to the
is it
downstream end
subjected to
is
peak pressure.
of a pipeline in such a
maximum
manner that
During what proportion
pressure.
closed?
A pipe, L = 2000 m, a = 1000 m/s, has a valve on
m/s and h = 20 m. It closes in three increments, spaced
=
12.16
its
downstream end,
2.5
1 s
apart, each area reduc-
tion being one-third of the original opening. Find the pressure at the gate
midpoint
Vq
=
A v /A
pipeline,
6 ft/s and ho
1
vo
t,s
L = 2000
=
100
ft.
ft, a = 4000 ft/s, has a valve at its downstream end,
Determine the pressure at the valve for closure:
0.75
0.60
0.45
0.30
0.15
0.5
1.0
1.5
2.0
2.5
3.0
In Prob. 12.17 determine the peak pressure at the valve for uniform area reduc-
12.18
tion in 3.0
s.
Find the
12.19
12.17
and at the
of the pipeline at 1-s intervals for 5 s after initial closure.
A
12.17
vq
maximum
area reduction for |-s intervals for the pipeline of Prob.
when the maximum head
at the valve
not to exceed 160
is
ft.
Derive the characteristics-method solution for waterhammer with the pressure
12.20
p and the discharge Q as dependent
variables.
Alter the reservoir boundary condition in Example 12.6 to include the minor
and velocity head at the pipe inlet. Assume a square-edged pipe entrance.
12.21
loss
Develop a single-pipeline waterhammer program to handle a valve closure at
12.22
The valve
the downstream end of the pipe with a reservoir at the upstream end.
closure
m=
300
is
3.2;
given by r
=
L =
ft,
5743.5
(1
—
a
=
t/t c )
3927
m where
ft/s,
is
tc
,
D=
4
the time of closure and
ft,
/
=
0.019,
V =
is
3.6 ft/s,
6.2
and
s,
and
H =
ft.
12.23
In Prob. 12.22 place a wave on the reservoir at a period of 1.95
s
and obtain a
solution with the aid of a computer.
12.24
Develop the computer program to solve Example 12.8 using your own particular
input data.
12.25
T
t,
The valve
1
S
closure data for the series system
shown
0.73
0.5
0.31
0.16
0.05
0.0
1
2
3
4