Whitaker - Introduction to Fluid Mechanics

lntroduction to Fluid Mechanics

STEPHEN WHITAKER Professor of Chemica/ Engineering University of California at Davis

ROBERT E. KRIEGER PUBLISHING COMPANY MALABAR, FLORIDA

Original Edition 1968 Reprint Edition 1981 w /corrections Printed and Published by ROBERT E. KRIEGER PUBLISHING COMPANY, INC. KRIEGER ORIVE MALABAR, FLORIDA 32950

Preface

Copyright© 1968 by PRENTICE HALL, INC. Transferred to Stephen Whitaker 1976 Reprinted by arrangement with author

All rights reserved. No part of this boolc may be reproduced in any form or by a•y electronic or mechanical means including information slorage and retrieval systems without permission in writing from the publisher.

Reprint. Originally published: Englewood Cliffs, N. J. : Prentice-Hall (Prentice-Hall international series in the physical and chemical engineering sciences. lncludes bibliographical references and index. l. Fluid dynamics. l. Title. 11. Series: PrenticeHall international series in the physical and chemical engineering sciences. 532'.05 81-1620 (QA91l.WS8 1981) ISBN 0-89874-337-0 AACR2

This book is intended for use in an The student is expected to have com¡:] and to be familiar with ordinary diffe multiple integrals, Taylor series, and The book is based primarily on in which the author participated wh each department (with the excepti advanced undergraduate courses in to provide a rigorous foundation · eight chapters are the result of ex can be covered satisfactorily in app used in either a three-unit semester e Chapters 9, 10, and 11 were adde who may wish to use the book in stances it may be necessary to delete s certain sections must be covered if s These sections are marked with an Vector notation is used freely th~ elegance or rigor but simply because in a form which attempts to connec A variety of people contributed of this text; they have the autho , Professor John C. Slattery of Nort text rests largely on an endless ser· the problems of teaching fluid mee

10 9 8 7 6 S

Davis, California

Printed in the United S tates of America

Library of Congress Cataloeinlln Publieation Data Whitaker, Stephen. lntroduction to Fluid Mechanics.

.J. A..\'-oQ.~\o Oc~oa.. \~f~O.. Uo..\l~

s.

SuiMQ..' 1, the behavior is the Bingham model parameters and are given in Tables 1.5-3 and 4. I torces PARAMETERSr.

Weight,

m, lbr-sec"/ft 0.194 O.J50 0.089 0.418

1

n (dimensionless) 0.566 0.171 0.52 0.575

Llahtfoot. Tr11111p~~rt I'Mrfi>IMIIII (New York: Jolm Wlley

P;

TTr2 -

Po

rrr 2

t

Areo

t

Areo

to

Interior pressure

Ambient pressure

to

(j

2rrr0

t Perimeter t Surface tension

Fl¡. 1.5-7. Force balance on a spherical drop.

9. A. B. Metzner, Advances in Chemical Engineering (New York: Academic Press, 1960), Vol. l. JO. R. B. Bird, W. E. Stewart, andE. N. Lightfoot, Transport Phenomena (New York: John Wiley and Sons, lnc., 1960), Chaps. J, 3. Jl. A. Fredrickson, Principies and Applications of Rheology (Englewood Cliffs, N.J.: Prentice-Hall, Inc., 1964).

22

lntroduction

p, is given by

where p 0 r0

= ambient pressure

2a P;=Po+ro

Chap. 1

Sec. 1.6

Vectors

(1.5-10)

= radius of the drop

For an arbitrary surface it can be shown12 that the pressure "jump" or discontinuity is (1.5-11)

where r1 and r 2 are the principal radii of curvature. There are sorne fascinating and important situations where surface tension effects domínate the fluid motion. These cases are best illustrated by the brilliant movie prepared by Professor L. Trefethen, 13 which is both educational and entertaining.

Vapor pressure Under sorne conditions, the fluid pressure can become less than the vapor pressure,t p1111 , and cavitation may occur. The most common cases are in high-speed pumps and in the wake of a bluff body, such as that shown in Fig. 1.5-8. The cavitation number, Nca, is defined as, N ca-PoPv' IJ (1.5-12) 1 2 "[PUo and is an important parameter in establishing the probability that cavitation will occur. Figure 1.5-8 shows a missile which has been launched under water and has just reached the surface. The cavitation numbers range from 1.10 to 0.054, and the results indicate that we can expect severe cavitation for this type of flow when Nca is less than 0.10. Since cavitation can drastically alter the operating characteristics of a centrifuga! pump14 and lead to serious corrosion, knowledge of the vapor pressure of a fluid is an important factor in pump design and operation.

*1.6 Vectors The task of describing the complex world that surrounds us in terms of abstract symbo/s is an immense one. In addressing ourselves to this problem, 12. R. Aris, Vectors, Tensors, and the Basic Equations of Fluid Mechanics (Englewood Cliffs, N.J.: Prentice-Hall, lnc., 1962), Chap. 10. 13. L. Trefethen, "Surface Tension in Fluid Mechanics," distributed by Educational Services, lnc., 47 Galen St., Watertown, Mass. 02172. t The vapor pressure is the pressure exerted by the vapor in equilibrium with the liquid. Thus, the vapor pressure of water at lOO"F is 0.065 atm, and at 212"F it is 1.00 atm. 14. R. H. Sabersky and A. J. Acosta, Fluid Flow-A First Course in Fluid Mechanics (New York: The Macmillan Company, 1964), Chap. 10.

p.= 1.00 atm u. Nca

= 44.2 ft/sec = 1.10

Po u. N ca

Fig. 1.5-8. Cavitation behind courtesy of U.S. Naval Ordnan cussion see NAVWEPS Report Missiles, by J. G. Waugh and G.

it is wise to use symbols which con which relate this information to t Inasmuch as velocity is a quantity vector), we choose a notation whi velocity as v. Instead of using bold just as well place an arrow over th< v*, etc. Use of boldface type is th t The use of boldface type to indicate

lntroduction

2a

Po + -

ro

Chap. 1

Sec. 1.6

Vectors

23

(1.5-10)

shown12 that the pressure "jump" or (1.5-11)

There are sorne fascisurface tension effects domínate the by the brilliant movie prepared both educational and entertaining.

can become less than the vapor The most common cases are in a bluff body, such as that shown in is defined as, (1.5-12) ~v"''"u•¡;

the probability that cavitation which has been launched under The cavitation numbers range from we can expect severe cavitation for 0.10. Since cavitation can drastically centrifuga! pump 14 and lead to serious of a fluid is an important factor

Po = 1.00 atm u. Nca

= =

44.2 ft/sec 1.10

p.= 0.51 atm u. = 47.8 ft/sec Nca = 0.47

p. = 0.10 atm u.= 55.5 ft / sec Nca

=

0.054

Fig. 1.5-8. Cavitation behind an underwater missile. Photographs courtesy of U.S. Naval Ordnance, Pasadena, Calif. For a detailed discussion see NAVWEPS Report 7735, Part 1, Water-Exit Behavior of Missiles, by J . G. Waugh and G. W. Stubstad.

world that surrounds us in terms of addressing ourselves to this problem, Equations of Fluid Mechanics (Englewood 10. Mechanics," distributed by Educational 02172. by the vapor in equilibrium with the liquid. 0.065 atm, and at 212°F it is 1.00 atm. Flow-A First Course in Fluid Mechanics Chap. 10.

it is wise to use symbols which contain as much information as possible and which relate this information to the physical world as closely as possible. Inasmuch as velocity is a quantity having both magnitude and direction (a vector), we choose a notation which symbolizes this fact and represent the velocity as v. Instead of using boldface typet to represent a vector, we could just as well place an arrow over the symbol, v, or mark it with an asterisk, v*, etc. Use of boldface type is the traditional method of communicating

t The use of boldface type to indicate vectors is known as Gibbs notation.

24

lntroduction

Chap. 1

information, and, like Pavlov's dogs, we respond toa symbol in boldface type wíth the thought, "lt's a vector." Our physical notion of a vector is that of a directed line segment, such as that iUustrated in Fig. 1.6-1. The vector v may be represented in terms ofits

z

Sec. 1.6

Asirle from the physical

has certain definite transfo

two-dimensional form beca three-dimensional rotation sions is shown in Fig. 1.6-2, y'

Vz

---------

Vectors

y

V

X

Fig. 1.6-J. A vector in rectangular Cartesian coordinates.

three scalar components

v.,, V11 ,

and

v = iv.,

Vz

Fig. 1.6-2

by the relationship

+ jv + kvz

11 (1.6-1) where i,j, and k are the three unit base vectors. The three scalar components are the projections of v on the three coordinate axes. The scalar (or dot) product between two vectors, A and B, is defined as

A •B

=

AB cos ()

(1.6-2)

where A and B are the magnitudes of the two vectors, and () is the angle between them. The scalar product of an arbitrary vector and a unit vector yields the projection of the vector on a Iine defined by the unit vector. For example, V•

i

= V COS ()""' = V.,

(1.6-3)

x,y-coordinate system, and th associated with the transform

l. 0.,.,,, angle between the 2. 011.,,, angle between the .

3. ()""''' angle between the 4. ()1/ll,, angle between the .

The subscripts of () obviously helpful in constructing a com for vectors. By the constructio: V~= t

and by simple geometrical cons

wh(!re ()""' is the angle between the vector v and the x-axis.

ot=

lntroduction

Chap. 1

respond to a symbol in boldface type that of a directed line segment, such as v may be represented in terms of its

Sec. 1.6

Vectors

25

Aside from the physical significance of a directed line segment, a vector

has certain definite transformation properties. These are best examined in two-dimensional form because the algebraic effort associated with a fully three-dimensional rotation can obscure the analysis. A vector in two dimensions is shown in Fig. 1.6-2, and the scalar components are noted for both the y

y'

~

- - - - - - - - - eJ. - -

yy 1/

V

co"'

_..

«¡ \

__./V

1 1 1

\ \

X

1

1 1

Fig. 1.6-2. Transformation of a vector.

Cartesian coordinates.

x,y-coordinate system, and the rotated coordinates, x', y'. The four angles associated with the transformation are given by the following expressions: (1.6-1) vectors. The three scalar components coordinate axes. The scalar (or dot) is defined as (1.6-2) the two vectors, and () is the angle an arbitrary vector and a unit vector a line defined by the unit vector. For (1.6-3) v and the x-axis.

l.

0.,.,., angle

2. 011.,·, 3. ()ZII., 4. 01111 ·,

between angle between angle between angle between

the x-coordinate the y-coordinate the x-coordinate the y-coordinate

axis axis axis axis

and and and and

the x'-coordinate the x'-coordinate the y'-coordinate the y'-coordinate

axis; axis; axis; axis.

The subscripts of () obviously have a definite meaning, and they will be helpful in constructing a compact formulation of the transformation law for vectors. By the construction shown in Fig. 1.6-2 we may write, V~ =

V., COS ().,.,,

+V

11

COS

ex

(1.6-4)

and by simple geometrical considerations, the angle ex may be expressed (1.6-5)

26

lntroduction

Chap. 1

Substitution of Eq. 1.6-5 into Eq. 1.6-4 yields An expression for

v~

v; = v., cos O.,.,. + VIl cos o""'' may be derived in a similar manner.

(1.6-6a)

(1.6-6b) The extension of this result to three dimensions is straightforward but tedious; we shall list only the results.

v; = v., cosO.,.,. + V v; = V., COS 0., + V v; = v., cosO., •. + V

11

cos

11

(J""',

1111 •

+ v. coso•.,. + v. COS 0. + v. coso...

(1.6-7a) (1.6-7b)

01111 • 11 • (1.6-7c) 11 cos 011 •• The terms, cosO.,.,,, cos 0.,,,, etc., are called the direction cosines for the rotation, x, y, z---+- x', y', z'. We note in these equations an ordered arrangement of subscripts, the first always identical to the subscript on the velocity component in the x, y, z system, and the second always identical to the subscript on the velocity in the x', y', z' system. This order will allow us to express Eqs. 1.6-7 in a much more compact forro if we make use of index notation rather than Gibbs notation. 11 •

11

COS

lndex notation

Traditionally, we denote the three coordinate axes of a rectangular Cartesian coordinate system as x, y, and z, and the three scalar components of the velocity vector as v.,, v11 , and v•. It would be just as satisfactory to denote the coordinate axes as x 1 , x 2 , and x 3 , and the scalar components as V¡, v2 , artd v3 . Similarly, the base vectors can be expressed as e< 1 l _e(2)

= =

i j

e< 3 , =k

(1.6-Sa) (1.6-Sb) (1.6-Sc)

Using this notation, we may express the vector vas (1.6-9) v = iv., + jv11 + kv. = e(l,v1 + e< 2 ,v2 + e< 3 ,v3 Use of numerical subscripts allows us to express a vector in terms of its scalar components and base vectors as 3

v = !e(i)v;

Vectors

Index notation is most usef ofvectors. For example, the Eqs. 1.6-7 as V~= V1 C

v; = v., cos 0., + v cos 0 11 •

Sec. 1.6

(1.6-10)

i= l

Further simplification results if we accept the convention that repeated subscripts (or indices, as the subscripts i are called) are summed from 1 to 3. This is called the summation convention, and allows us to write Eq. 1.6-10 as (1.6-10

V~= V1 CO V~= V1 CO

where it is understood that (') coordinate system. The : Eqs. 1.6-12 toa more compa

Although i was used as the r k,j, n, etc.--could have been index may be represented by single, or unrepeated, index is from 1 to 3. Using the con 1.6-13 further by writing

Certainly Eq. 1.6-14 is a far n presented in Eqs. 1.6-12; how is not to produce a "slick" m stress (Chap. 4), in Newton's energy equation (Chap. 7), w because it saves us from gettin In Gibbs notation, the ve index notation may be used t remembered here is that the relative to the x,y,z-coordinat components just as well by Gibbs notation is that by tra of a directed quantity, while three numbers. The former i while the Iatter is extremely u If we jump ahead briefiy viscous dissipation to interna!

t T represents the total stress t sans serif type whenever possible.

lntroduction

Chap. 1

(1.6-6a)

Sec. 1.6

27

Vectors

Index notation is most useful in dealing directly with the scalar components of vectors. For example, the use of numerical subscripts allows us to express Eqs. 1.6-7 as (1.6-12a) + Va cos ea (1.6-12b) V~ = V1 COS e12 + V2 COS e22 + Va COS ea 2 (1.6-12c) v~ = v cos e a+ v cos e a+ Va cos e where it is understood that the second subscript on e refers to the primed v{

(1.6-6b) dimensions is straightforward but

= v1 cos en + v2 cos e21 1

e + v. cos e•.,. cos 0 + v. cos e. cos el/•. + v. cose... cos

(1.6-7a) (1.6-7b) 1111 • 11 • (1.6-7c) called the direction cosines for the these equations an ordered arrangeto the subscript on the velocity the second always identical to the z' system. This order will allow us to l.llrrln,rt form if we make use of index 11 .,.

coordinate axes of a rectangular z, and the three scalar components It would be just as satisfactory to and Xa, and the scalar components as can be expressed as =i =j =k

(1.6-8a) (1.6-8b) (1.6-8c)

vector vas

+ e< 2lv2 + e(a)Va

(1.6-9) express a vector in terms of its scalar

e to be. given by (2.2-17) 4> = -(xg., + yg11 + zg.)

-Vtfo =

(¡~ + j~ + k~)(xg., +

ax

= ig.,

ay

az

+ jgll + kg.

(2.2-18)

This method of representing gis possible only because gis a constant vector. Equation 2.2-13 may be written in terms of tf>. (2.2-11) O= pVtf>

+ Vp

(2.2-19)

If the density is constant, this equation may be written as

a partial derivative, and Eq. 2.2-12

(2.2-12)

(2.2-13)

(2.2-14)

O= V(ptf>

+ p)

(2.2-15)

OX¡

gradient of the pressure and denoted V is more informative and will represent the gravity vector in terms , tfo. This function is called the it represents the gravitational

(2.2-20)

Since this result holds everywhere in the fluid, the term ptf> + p must be a constant, and we may write (2.2-21) p=C-ptf> Here we ha ve integrated the equations of fluid statics, and C is the constant of integration. Use of the potential function to obtain the integrated form of Eq. 2.2-13 is satisfying from the mathematical point of view, but the solution of this equation deserves a treatment more closely connected to the student's notion of static fluid systems. For the system shown in Fig. 2.2-2, the gravity vector is g= -kg

this result would be expressed, i--

yg11 + zg.)

=g

~)]}

op

(2.2-16)

it follows that minus the gradient of 4> is equal to g.

z-surfaces

surfaces is such that an "x-surface" is and the unit outwardly directed Eq. 2.2-10 by Llx .iy ~ and take O, we obtain i(

-41

Fluid Statlcs

potential energy of the fluid. The gravity vector g may be expressed

Forces on the (2.2-10) y-surfaces

(kp Llx Lly)

Sec. 2.2

(2.2-22)

where g is the magnitude of the gravity vector. Substitution of Eq. 2.2-22 into Eq. 2.2-13 yields the three scalar equations

op =o

(2.2-23a)

op =o

(2.2-23b)

ox

oy

op

- =

oz

-pg

(2.2-23c)

42

Fluid Statia and One-Dimensional Laminar Flow

Chap. 2

These three scalar components of the original vector equation are quite obviously expressed by the index notation of Eq. 2.2-15, where g 1 = g 2 =O and g1 = -g. lntegra,~on of Eq. 2.2-23c gives

p = -pgz

+ C(x,y)

A single boundary condition is needed in order to determine the constant e and thus specify the pressure everywhere in the fluid. lt is obtained by recognizing that the pressure at z = L is the atmospheric pressure p 0 • We may write

p=po,

z=L

(2.2-26)

the application of which to Eq. 2.2-25 yields,

PI•-L =Po= -pgL+ e

(2.2-27)

and the constant of integration is

e=p0

+ pgL

+ pg(L -

vector is a continuous function. Both ti postulate discussed in Sec. 1.1. Boundat of the second rule-i.e., pressure is a e

2.3

Barometers

A barometer is a device for mea atmosphere. The one illustrated in Fi at one end and immersed in the bar obtained by filling the tube with the closing the open end (the thumb does 1 pool of barometer fluid. The fluid v; pressure there is the vapor pressure p.,p. For mercury the vapor pressure at room temperature is approximately 3 x 10-t .a tm and may be considered to be zero. The differential equation for the pressure is

op

oz = -pagg

(2.2-28)

Substitution of e into Eq. 2.2-25 yields the final expression for the pressure. P =Po

Barometers

(2.2-24)

where the constant of integration may be a function of x and y; however, Eqs. 2.2-23a and b indicate that the pressure is neither a function of x nor y, and we write Eq. 2.2-24 (2.2-25) p= -pgz+ e

B.C. 1:

Sec. 2.3

z)

(2.2-29)

This problem is such a simple one that the student can easily have missed the significance of the various steps; therefore, a review will be helpful. l. The linear momentum principie and Eq. 2.2-8 were applied to a differential volume element to develop the equations of fluid statics. 2. The equations were solved to yield Eq. 2.2-25. 3. A boundary condition was specified and applied to Eq. 2.2-25 to obtain an expression for the pressure.

lt may not seem so to the student, but the final step in this general process is usually the most troublesome. The differential equations of motion can be derived once and for all, and methods of solution (for those cases which can be solved) are neatly tabulated in mathematics and fluid mechanics texts; however, boundary conditions are specified mainly on the basis of physical intuition, and thus present a more diffi.cult problem. Things are not quite as bad as they may seem, for there are two rules which guide us in specifying boundary conditions: velocity is a continuous function; the stress

(2.3-1)

which is integrated to give

p = - Pa 1gz

+e

(2.3-2)

The boundary condition is

B.C.1:

p =p,p ~o,

z = h (2.3-3)

the application of which to Eq. 2.3-2 gives p = Pugg(h - z) (2.3-4) Since the pressure at z = O is the atmospheric pressure, p0 , we write

Po = Paggh

(2.3-5)

Because of the universal use of mere pressure, it is often reported in terms o pressure varies from day to day, the,

t Taking the pressure to be continuous n liquid interface.

and One-Oimenslonal Laminar Flow

Chap. 2

the original vector equation are quite of Eq. 2.2-15, where g1 = g 2 =O gives

Sec. 2.3

43

Barometers

vector is a continuous function. Both these ideas follow from the continuum postula te discussed in Sec. 1.1. Boundary condition 1 results from application of the second rule-i.e., pressure is a continuous function for fluids at rest.t

(2.2-24)

be a function of x and y; however, pressure is neither a function of x nor y, (2.2-25)

in order to determine the constant e in the fluid. lt is obtained by L is the atmospheric pressure p 0 • We

z=L

(2.2-26)

= -pgL+ e

(2.2-27)

2.3

Barometers

A barometer is a device for measuring the absolute pressure of the atmosphere. The one illustrated in Fig. 2.3-1 consists of a single tube closed at one end and immersed in the barometer fluid. Such a system may be obtained by filling the tube with the barometer fluid (usually mercury), closing the open end (the thumb does nicely), and immersing the tube in the pool of barometer fluid. The fluid vaporizes in the closed end, and the pressure there is the vapor pressure vp p.,,. For mercury the vapor pressure at room temperature is approximately z =h 3 x 10-e atm and may be considered to be zero. The differential equation for the pressure is (2.3-1)

(2.2-28)

the final expression for the pressure.

which is integrated to give p = - PHggz

(2.2-29)

that the student can easily have missed therefore, a review will be helpful. and Eq. 2.2-8 were applied to a develop the equations of fluid statics. Eq. 2.2-25. and applied to Eq. 2.2-25 to the final step in this general process differential equations of motion can of solution (for those cases which in mathematics and fluid mechanics are specified mainly on the basis of a more difficult problem. Things are there are two rules which guide us in is a continuous function; the stress

+e

(2.3-2)

The boundary condition is B.C.l:

p

=p.,,~

o,

z=h

(2.3-3)

z=O

the application of which to Eq. 2.3-2 gives p = PHgg(h - z) (2.3-4) Since the pressure at z = O is the atmospheric pressure, p 0 , we write

Po= PHggh

Fig. 2.3-1. Mercury barometer.

(2.3-5)

Because of the universal use of mercury barometers to measure atmospheric pressure, it is often reported in terms of h, or inches of mercury. Atmospheric pressure varies from day to day, the average being 29.92 in. Hg. This value t Taking the pressure to be continuous neglects any effect of surface tension at the gasIiquid interface.

Fluid Statics and One-Dimensional Laminar Flow

Chap. 2

is referred toas one standard atmosphere, and we may use Eq. 2.3-5 to determine that p0 (standard) = 14.696lbr/in. 2

Sec. 2.-4

Manometers

Both these conditions are derived from continuous function.t Solution of the tw p¡

This quantity is often written 14.696 psia (pounds per square inch absolute).

2.4

= -:p¡gz

P2 =

-p~Z

Application of boundary condition 1 giv

Manometers

Manometers are devices which make use of columns of liquid to determine pressure differences. The simplest type is the U-tube illustrated in Fig. 2.4-1. The manometer fluid must be immiscible with the fluid in the / /

Pllz-111 = Po = or

C1 =Po+ and the pressure in fluid 1 is

Open tothe otmosphere

P1 =Po+ P~ Application of boundary condition 2 giv

P2!z=ll1 -

Pressure tonk

=-

P2Kh1

+ C2 = P1l

or

-z=hz

and the pressure in fluid 2 is

P2 = Po

- - z=h 1

+ p~(h1 -

z

The gauge press.ure, pg, is defined as the pheric pressure; therefore, the gauge pre _ _ _ _ __ _ z=O

pg = (p2 -Po) = p~(h¡

Fl¡.l.4-l. U-tube manometer.

tank, and its density, p1 , must be greater than the density of the fluid in the tank, p 2 , or there will be a tendency for it to replace the fluid in the tank. The equations for the pressure in the two fluids are op¡

-

= -p¡g

(2.4-1)

aP2 = -p 2g

(2.4-2)

oz oz

The boundary conditions for this system are

If the fluid in the tank is a gas, the usually be much larger than that of the pressure is pg ~ p¡g(h Manometer calculation

The fluid in the tank is a gas and a me the pressure. The readings on the man01

= 3.25 · ft} h3 =5.17ft

h¡

B.C. 1:

P1 = Po•

z = h3

(2.4-3)

B.C. 2:

P2 = P1•

z = h¡

(2.4-4)

or (h

t Interfacial tensions are obviously being ne

and One-Dimensional Laminar Flow

Chap. 2

and we may use Eq. 2.3-5 to deter-

= 14.696lb1/in.2

Sec. 2. p2, and the gauge J~ressure ís (2.4-12) Manometer calculation

The fluid in the tank is a gas and a mercury manometer is used to measure the pressure. The readings on the manometer are

ft}

h¡ = 3·25 ha = 5.i 7 ft

or

(h

a-

h) 1

= 1.92 tit

t Interfacial tensions are obviously being neglected here.

Fluid Statics and One-Dimensional Laminar Flow

Chap. 2

Sec. 2.5

Forces on Submerged Plane Surfaces

z_=L----------

The density of mercury is 13.5 gfcm3, and the gauge pressure is p 11 = {(13.5 g/cm )(32.2 ft/sec~(1.92 ft)}

3

Original terms

{(~) (

X

453.6 g

3 2 lbr sec ) (2.54 cm) (12 in.)} 32.2lbm ft in. ft

z=t--------------· df= -npdA

Conversion factors

~

= (13.5)(1.92)(2.54)3(12) = 11.2lb /in.2 ~53~ f

To distinguish gauge pressure from absolute pressure (psia) we write

Fig. l.S-1. Surface force a

p11 = 11.2 psig (pounds per square inch gauge)

*2.5

The net surface force exerted on the gat by the liquid and by the surrounding atn

Forces on Submerged Plane Surfaces

Fnet = -

Now that we have learned to determine the pressure field for a fluid at rest, we can direct our attention to the second problem noted in Sec. 2.2, i.e., the calculation of forces on submerged surfaces. According to our previous definition, the force which the surroundings exert on a system is force per unit area } {acting on the system

Jnp d . dA

¡

Sec. 2.8

One-Dimensional Laminar Flo

Equation 2.8-6 is called the stress eq in terms of the velocity gradient if

(2.8-3)

Jllf,.(t)

This result is identical to that for static ftuids, but in this case the stress vector has both normal (pressure) and tangential (shear) components. We are only interested in the z-component of Eq. 2.8-3, which may be applied to the differential volume to yield Forces on the z-surfaces

(2.8-4) Forces on the r-surfaces

A very definite sigo convention is used with the shear stress, -rr:· On the surface having an outwardly directed normal in the positive r-direction, the shear stress -rr• is represented as acting in the positive z-direction. On the surface having an outwardly directed normal in the negative r-direction, the shear stress is represented as acting in the negative z-direction. The actual direction of the shear stress will depend on the specific problem under investigation. For example, ifthe flow in the tube were reversed, the direction of the shear stress could certainly be reversed. In formulating problems, we must adhere to the sigo convention that a positive shear stress acts in the positive coordinate direction on positive surfaces,t and in the negative coordinate direction on negative surfaces. This scheme is necessary to insure that shear stresses on opposite sides of a differentially thin shell of fluid act in opposite directions, which is, of course, intuitively obvious. A proof will be given in Chap. 4 where stress is studied in detail. Dividing Eq. 2.84 by 21rt1r!1z and taking the limits !1r-+ O and !1z-+ O, we get

The relationship between the shea flow is very similar to the relations flow. In Chap. 5, Newton's law of the present, we must regard Eq. relationship between the shear stres circular tu~. If the viscosity 1-' is constant, w to yield

op

-=p

oz

The solution of this equation req the functional dependence of v, a1 fluid particle must move in a straig this flow is constant, the volumetri must remain constant (which follo implies conservation of volume, · conclude that the velocity v, is not that the ftow is axisymmetric,t v, i1 V

The pressure will be a function of it will certainly depend on z. Thu

p=

0 = lim [-r(PI:H• - PI•)] /::,.z

.,. .... o

(2.8-5)

op

which immediately leads to the differential equation

op + -1 -a (r-rr,)

O= - -

oz r or

However, the effect of gravity wil and the pressure gradient will dependence of opfoz may be expr

oz (2.8-6)

t By positive surfaces we mean the surfaces having normal vectors pointing in the positive coordinate directions.

Examining Eq. 2.8-8 in light of t side is only a function of z while K.ceping in mind that r and z ar

t See Birkhoff's plausible intuitive h

One-Dimensional Laminar Flow

Chap. l Sec. 2.8

of a constant vector is zero, Eq. 2.8-1

One-Dimensional Laminar Flows

59

Equation 2.8-6 is called the stress equation of motion, and it may be expressed in terms of the velocity gradient if Newton's law of viscosity is used.

(2.8-3) (2.8-7) fluids, but in this case the stress vector (shear) components. We are only . 2.8-3, which may be applied to the Forces on the z-surfaces

(2.8-4) Forces on the r-surfaces

with the shear stress, T rz· On the normal in the positive r-direction, the in the positive z-direction. On normal in the negative r-direction, in the negative z-direction. The actual on the specific problem under in the tube were reversed, the direction reversed. In formulating problems, that a positive shear stress acts in the surfaces,t and in the negative This scheme is necessary to sides of a differentially thin shell of is, of course, intuitively obvious. A is studied in detail. taking the limits Ar ~ O and Az

~O,

The relationship between the shear stress and the velocity gradient for this flow is very similar to the relationship given in Chap. 1 for plane Couette flow. In Chap. 5, Newton's law of viscosity will be examined in detail; for the present, we must regard Eq. 2.8-7 as an experimentally determined relationship between the shear stress and the velocity gradient for flow in a circular tu~. lf the viscosity /-' is constant, we may substitute Eq. 2.8-7 into Eq. 2.8-6 to yield (2.8-8) The solution of this equation requires sorne careful arguments regarding the functional dependence of v, and p. Because v11 and vr are zero,t each fluid particle must move in a straight line along the tube. If the density for this flow is constant, the volumetric flow rate across any section of the tube must remain constant (which follows from the fact that conservation of mass implies conservation of volume, if the density is constant), and we can conclude that the velocity v. is nota function of z. If, in addition, we assume that the flow is axisymmetric,t v. is only a function of r. v.

equation

= p(r, O, z)

op =f(z)

having normal vectors pointing in the

(2.8-10)

However, the effect of gravity will be the same everywhere along the tube, and the pressure gradient will not depend on r and O. The functional dependence of opfoz may be expressed as

oz (2.8-6)

(2.8-9)

The pressure will be a function of r and O owing to gravitational effects, and it will certainly depend on z. Thus, p

(2.8-5)

= v.(r)

(2.8-11)

Examining Eq. 2.8-8 in light of these arguments, we see that the left-hand side is only a function of z while the right-hand side is only a function of r. Keeping in mind that r and z are independent variables, we conclude that

t See Birkhoff's plausible intuitive hypothesis IV, Sec. 1.3.

60

Fluid Sutics and One-Dimensional Laminar Flow

Chap. 2

both sides of Eq. 2.8-8 must be constant,t and we write !-' ! !!:._ r dr

(r dv,) dr

= _ !lp

(2.8-12)

L

Here the partial derivatives have been replaced by total derivatives, inasmuch as v. is only a function of r, and the pressure gradient opfoz has been replaced by -(!lp/L), where (2.8-13) Multiplication ofEq. 2.8-12 by r dr, division by ¡,t, and integration give

dv,

r dr

= -

(!lp) L 2¡,t + e r

2

(2.8-14)

1

Dividing by r and integrating again, we have an expression for the velocity profile.

v,

,2 - + e In r + e (!lp) L 4¡,t

= - -

1

2

(2.8-15)

The boundary conditions for this flow are

B.C. 1:

v, is finite for O :::;; r :::;; r0

B.C. 2:

v.

=o,

r ='o

(2.8-16a)

viz r-ro

Eq. 2.8-18 is a very useful result, if the dimensions of the system, the g The volumetric flow rate is given b

º fr:., =

o o

This relationship between the volu pressure gradient !lpfL is called th hydraulician, Hagen, and the ph established this result in the early p We see by Eq. 2.8-19 that the flo pipe radius; thus, a small error in error in the flow rate or pressure fluid, specification of the pressure d must be made on an economic bas cost of the pipe is small but the cos constantly be confronted with prob economic constraint be placed on tll Other quantities of interest can velocity, Vz,mu• occurs at r =O and v.,max

2

~+e2

One-Dimensional Laminar

(2.8-16b)

Both these conditions are derived from the idea that the fluid is a continuum; the velocity is therefore a continuous function. The first boundary condition is somewhat unusual in that the velocity is not specified at sorne point; it is simply restricted to a finite value. Boundary conditions of this type are often found in the analysis of fluid motion, heat transfer, and mass transfer. Because the statement is so obvious, it is sometimes overlooked by students unfamiliar with the solution of boundary value problems. Application of boundary condition 1 requires that e1 be zero, beca use In r - - oo as r - O, and boundary condition 2 yields an expression for e2,

!lp) , = 0 =( -L- 4¡,t

Sec. 2.8

The same conduits are often used to costs be minimized and piping layout to know how long it will take before conduit arrives at the exit. The ma the "breakthrough" time. The aver metric flow rate divided by the croSl (v.)

= _Q__

7Tr~

(2.8-17)

The average velocity is often a use since the extent of a chemical reacf in the reactor. This time is called tll

(2.8-18)

average resid

The velocity may now be given as

indicating that the velocity profile is parabolic.

Application of the Hagen-Poiseuill therefore, we must be sure that the

t The arguments presented here are mainly qualitative and should be accepted in that context. A quantitative treatment of this problem is given in Chap. 5.

t Throughout the text, angular brack area averages.

One-Dimensional Laminar Flow

t

Chap. 2

and we write (2.8-12)

replaced by total derivatives, inasmuch gradient opfoz has been replaced (2.8-13) by ¡.t, and integration give (2.8-14) we have an expression for the velocity (2.8-15)

(2.8-16a) (2.8-16b) the idea that the fluid is a continuum; !lutlcuon. The first boundary condition is not specified at sorne point; it is conditions of this type are often heat transfer, and mass transfer. is sometimes overlooked by students value problems. Application of be zero, because In r ---. - oo as r ---. O, ex~:;¡reS:SIOID for C2,

Sec. 2.8

61

One-Dimensional Laminar Flows

Eq. 2.8-18 is a very useful result, for it allows us to calculate the flow rate if the dimensions of the system, the pressure drop, and the viscosity are given. The volumetric flow rate is given by

Q=

fo 2I"o Vzr dr dO = -7TT~(/:l.p) 8¡.t L ro

(2.8-19)

This relationship between the volumetric flow rate, the radius r 0 and the pressure gradient LlpfL is called the Hagen-Poiseuille law in honor of the hydraulician, Hagen, and the physician, Poiseuille, who experimentally established this result in the early part of the nineteenth century. We see by Eq. 2.8-19 that the flow rate is very sensitive to changes in the pipe radius; thus, a small error in sizing a pipe can lead to an_appreciable error in the flow rate or pressure drop. For a given flow rate and a given fluid, specification of the pressure drop (i.e., the pump size) and the pipe size must be made on an economic basis. If a small-diameter pipe is used, the cost of the pipe is small but the cost of the pump is large. An engineer will constantly be confronted with problems of this type which require that an economic constraint be placed on the derived result. Other quantities ofinterest can be derived from Eq. 2.8-18. The ma:>fJmum velocity, Vz,max• occurs at r =O and has the value

(!l.p)

r~1-' L (2.8-20) 4 The same conduits are often used to carry different fluids in order that piping costs be minimized and piping layouts simplified. In such cases it is important to know how long it will take before the new fluid being pumped through the conduit arrives at the exit. The maximum velocity can be used to calculate the "breakthrough" time. The average velocity (vz)t is defined as the volumetric flow rate divided by the cross-sectional area; therefore, Vs,max =

(vz) =

_Q_ = 7Tr~

(

'~) (Llp)

(2.8-17)

L The average velocity is often a useful quantity for sizing pipeline reactors, since the extent of a chemical reaction depends on the time the fluid spends in the reactor. This time is called the average residence time and is given by

(2.8-18)

. L average res1.dence hme = (v:)

8¡.t

Application of the Hagen-Poiseuille law is limited to steady, laminar flow; therefore, we must be sure that the flow under consideration is not turbulent. qualitative and should be accepted in that is given in Chap. 5.

t Throughout the text, angular brackets (()) will be used to denote both volume and area averages.

Fluid Statics and One-Dimensional Laminar Flow

62

Chap. 2

Calculation of the fl.ow rate should always be accompanied by a determination of the Reynolds number, _ p(v.)D N Re¡t

~r (r-rr.) = -e~:) r

X

1

1 }{

10 dynefcm

e

Orfllnal tcrm

To complete the solution, we R.eynolds number is given by

(2.8-22)

r

u

If we wish the answer in terms

(2.8-21)

(tlp)!: + C¡ L 2

= -

T

O..Oimenslonal Laminar

/:ip = { 1.91

to be sure that it is Iess than the critica! value, 2100. Knowledge of the stress distribution for laminar fl.ow in a tube is sometimes desired. Rewriting Eq. 2.8-6 in the form

and integrating, we have

lec. 2.1

Because the shear stress must always be finite, we may write, B.C. 1:

Tr•

is finite for O :::::;: r :::::;: r 0

Thus, the constant of integration C1 is zero, and the shear stress is a linear function of r.

-(/j.p)!: L 2

=

T u

The fact that T r• is negative i.s in keeping with our intuition, because the shear stress acting on the r-surface having a positive outwardly directed normal must certainly act in the negative z-direction if the fluid is flowing in the positive z-direction. Sample calculation of pressure drop

Oil having a density of 0.87 gfcm 3 and a viscosity of 1.3 poises is to be pumped through a smooth pipe lOO ft long with an inner diameter of 2.0 in. We wish to calculate the pressure drop for a volumetric flow rate of 1000 ft3/hr. Rearranging Eq. 2.8-19 gives

tlp =

Carrying out the computation,

(2.8-23)

8p.LQ/7Tr~

1 Thus, the fl.ow is laminar. Non-Newtonlan flow betwee1

Figure 2.8-3 illustrates the fl. parallel plates, jnclined at an profile is indicative of a pseu flow to be one-dimensional an Eq. 2.8-3 to the differential vol

O=

(pgr~:) !:u !:ly !:lz

(2.8-24)

+ {p !:ly tlzlr~: -

Substituting the appropriate values and the necessary conversion factors, we get !:l = {(8)(1.3 poises)(lOOft)(1000fe/hr)} P (3.14)(1.0 in.)4

Dividing by Ax !:iy /:iz and taki

Original tcrms

X { (1

2

dyne-~ecfcm ) (12 in.) ( p01se

4

ft

1 hr ) } 3600 sec

O= pgs·

Conversion factors

tlp = 1.91

X

106 dynefcm 2

+IiAw-+

d One-Dimensional Laminar Flow

Chap. 2

be accompanied by a determination

S.. 2.1

8

= {t.91 X

2 }{

10 dynefcm

c·

:y!~-'lbt) c-~~J} = 27.6lbr/in.

25

1

Con.ualon facton

Ori,UW term

To complete the solution, we need to verify that the flow is laminar. The R.eynolds num.ber is given by _ p(v.)D _ 4pQ N Re1-' '"D¡.¡

(2.8-21)

(2.8-22)

63

If we wish the answer in tenns of pounds-force per square inch, we write dp

value, 2100. for laminar flow in a tu beis sometimes

On....Oimenslonal Laminar Flowa

3

N Re

_ {(4X0.87 gfcmSXlOOO ft fhr)} (3.14X2.0 in.X1.3 poises) Ori,UW terma

be finite, we may write, O::;; r::;; r 0

8

X {(

zero, and the shear stress is a linear

1 hr ) (12 in.) 3600 sec ft

(2-~4 cm) ( 2

tn.

1 poise ) } g/cm-sec

Con.uaiOil facton

Carrying out the computation, we have (2.8-23) NRe =

with our intuition, beca use the shear a positive outwardly directed normal if the fluid is flowing in the posi-

and a viscosity of 1.3 poises is to be long with an inner diameter of 2.0 in. for a volumetric flow rate of 1000

1320

Thus, the flow is laminar. Non-Newtonlan flow between two flat plates Figure 2.8-3 illustrates the flow of a non-Newtonian fluid between two parallel plates, ~nclined at an angle () from the horizontal. The velocity profile is indicative of a pseudoplastic "power-law" fluid. Assuming the flow to be one-dimensional and laminar, we apply the x-component of Eq. 2.8-3 to the differential volum.e shown in Fig. 2.8-3 to obtain

O = (pg#e) dx dy dz (2.8-24)

+ {p dy dzlz -

Body force Forces on the x-suñaces

p dy dzl~.u}

+ {-Twzdxdzlw + 1'wzdxdzlw+Aw}

(2.8-25)

Forces on the y-suñaces

Dividing by dx ay dz and taking the limit as before, we get

• () - lim PI~Áz O = pgsm .U-+0

Plz

dx (2.8-26)

64

Fluid Statics and One-Dimensional Laminar Flow

Chap. 2

S.C. 2.8

On•Dimenslonal Laminar Flc

wc note that the left-hand side is in -r,. is independent of x, thus botl constant. Integration is then straig T

vz

=-

(fl.p, L

The boundary condition that we integration is that the shear stress

B.C. 1:

The easiest way to demonstrate th 2.8-3 to the differential volume sl

y=O

~

Fig. 2.8-3. Flow between two flat plates.

where g., has been replaced by g sin the limiting process gives us

o=

e.

pg sin

Each limit represents a deriva ti ve, and

e - -op + -oT'J/X ox oy

(2.8-27)

We may also apply the y-component of Eq. 2.8-3 to this flow to obtain

o=

-

pg cos

e - .é)p -oy

(2.8-28)

stresses at y = +fl.y and y = - j following the assumption of symrr

lntegration of Eq. 2.8-28 gives an expression for the pressuret p(x, y)

= - pgy cos e + C(x)

(2.8-29)

which leads to the conclusion that opfox is independent of y. Rearranging Eq. 2.8-27 into the form,

op -ox

. e a'TIIX pgsm = oy

Flg. 2.11-4. Shear s1

(2.8-30)

t Note that integration of a partial derivative yields a "constant of integration," which may be a function of the other independent variables.

Evaluation of the terms in Eq. 2.8

O = (pg 111)2 fl.x fl.;

+ 2p !l.y fl.zl.

One-Dimensional Laminar Flow

Chap. 2

S.C.. 2.8

65

On•Dimenslonal laminar Flows

we note that the left-hand side is independent of y. For one-dimensional ftow ,...., is independent of x, thus both sides of Eq. 2.8-30 must be equal to a constant. Integration is then straightforward, and we obtain (2.8-31) The boundary condition that we wish to apply to evaluate the constant of integration is that the shear stress is zero at the plane of symmetry, y= O.

B.C. 1:

1"wz

= O,

Y= O

(2.8-32)

The easiest way to demonstrate the validity of this condition is to apply Eq. 2.8-3 to the differential volume shown in Fig. 2.8-4. Note that the shear

two flat plates.

Each limit represents a derivative, and

éJp

éJTyz

--+éJx éJy

(2.8-27)

of Eq. 2.8-3 to this flow to obtain (2.8-28)

Fl¡. 1.8-4. Shear stress at a plane of sym.metry.

stresses at y= +~y and y= -~y are both actingin the same direction following the assumption of symmetry about the plane y = O, i.e., (2.8-29)

éJx is independent of y. Rearranging

(2.8-33) Evaluation of the terms in Eq. 2.8-3 gives

O = (pgz)2 ~X ~y ~z (2.8-30)

+ 2p ~y ~zlz -

Body force Forces on the

2p ~y ~zlz+4z x-suñaces

Forces on the

y-surfaces

(2.8-34)

66

Fluid Statics and One-Dimensional Laminar Flow

Chap. 2

Ifwe Jet dx and dz remain finite and allow dy to go to zero, Eq. 2.8-34 reduces to 2rul.=o dx dz = O

(2.8-35)

and it follows that the shear stress at the plane of symmetry must be zero. Thus, C1 is zero and the stress distribution is Tu=

-(d: + pgsin o)y

(2.8-36)

Sec. 2.8

One-Dimensional Laminar Flows

The volumetric flow rate per unit wid Eq. 2.8-43. dp -H

q=

(~) ( 1:_ 2n + 1

The velocity profile is somewhat dimensionless by dividing by the avell

v.,

The shear stress for a power-Iaw fluid in one-dimensional flow is given by (see Eq. 1.5-9) (2.8-37)

U.,= (v.,)

(2n

= -;;-

where Y= 2yfh, a dimensionless di puted from Eq. 2.8-45, are shown in

Since the flow is symmetric about y = O, we need only treat the region O L. y L. h/ 2. Noting that the velocity gradient will always be negative in this region, we can write

~;"' =

-1 ~;"' ,

(2.8-38)

Combination of Eqs. 2.8-36, 2.8-37, and 2.8-38 yields Tu=

-m

1~;"'In= -(d: +

pgsin

o)y

(2.8-39)

Dividing by -m and raising both sides to the 1/n power,

dp dv., = ( L 1dy 1

+

pg sin

0)

1

/n ylfn

m

(2.8-40)

Using Eq. 2.8-38 yields

e;.,)

dp

= - (L

+

pg sin

m

0)

1

/n

i'"

(2.8-41)

This equation may be integrated, subject to the boundary condition

B.C.l:

h y=2

(2.8-42)

Fl1. 2.8-5. Velocity profiles for ~ flat plates.

to obtain the velocity distribution

t The total volumetric flow rate (in cub (2.8-43)

Q, while the volumetric flow rate per unit 1 represented by q.

~ Whenever possible, capital letters w

and One-Dimensional Laminar Flow

Chap. 2

allow dy to go to zero, Eq. 2.8-34 reduces

dz =O

Sec. 2.8

67

One-Dimensional Laminar Flows

The volumetric flow rate per unit width qt is readily obtained by integrating Eq. 2.8-43.

(2.8-35)

at the plane of symmetry must be zero. "bution is (2.8-36) in one-dimensional flow is given by

(2.8-37)

(2.8-44) The velocity profile is somewhat easier to examine if the velocity is made dimensionless by dividing by the average velocity.!

U :e=~= (2n (v:e)

+ 1)(1n+1

(2.8-45)

y 1, the apparent viscosity increases with increasing shear rate; thus, the fluid appears to be more viscous near the walls, and a larger proportion of the flow takes place in the central region of the channel. Fluids which behave in this manner are called dilatant, and are not nearly as common as pseudoplastic fluids . If n = 1, the fluid is Newtonian and we replace m with p,. The velocity profile is parabolic, and the volumetric flow rate per unit width is given by q

= -h3 (!!:.p - + pg sin () )

Fla. Atmospherlc pressure

(2.8-47)

12t-t L

This type of flow is sometimes referred to as "plane" Poiseuille flow. The methods presented in this section have allowed us to solve sorne rather important practica! problems without recourse to detailed analysis; however, this approach is only useful when the streamlines are straight. Much work remains to be done before we can handle more complex flows.

PROBLEMS 2.1 . If the density of fluid 1 is 62.4lb m/ft 3 and the density of fluid 2 is 136.8 lb m/ft, determine the gas pressure in the tank shown in Fig. 2-1. Assume that the density of the gas in the tank is negligible compared to the two manometer fluids. Ans.: 20.4 psia.

2-2. A simple U-tube manometer can be used to determine the specific gravity y of fluids which are more dense than water by the arrangement illustrated in Fig. 2-2. Derive an expression for y in terms of z1, z2, and z3 •

o Fla. 2-2 2-3. For ftuids with a density close to, gravity is best determined in the pression for y in terms of z1, z1, z Pt

A.ns.: y = PI =

(z4

-

zs) - (za (z¡ - zs)

and One-Dimensional Laminar Flow

Chap. 2

69

Problema

's law of viscosity, we see that the

~Atmospheric pressure

(2.8-46)

decreases with increasing shear rate, Since the shear rate is zero at the wall, the apparent viscosity becomes the flow resembles a case where there Eq. 2.8-44, that as n becomes small, the sensitive to the channel depth, h. In we must be extremely careful when for a smal! error in geometry can at a given pressure drop. tcn~as,es with increasing shear rate; thus, near the walls, and a larger proportion region of the channel. Fluids which and are not nearly as common as

2 Fl¡. 2-1

we replace m with p. The velocity flow rate per unit width is given by

+ pgsin o)

Atmospherlc /pressure~

(2.8-47) Less dense fluid

to as "plane" Poiseuille flow. have allowed us to solve sorne recourse to detailed analysis; the streamlines are straight. Much handle more complex flows.

Woter

LEMS and the density offluid 2 is l36.8Ibm/ft, shown in Fig. 2-l. Assume that the compared to the two manometer

used to determine the specific gravity y water by the arrangement illustrated in in terms of z1, z2, and z3 •

Fl¡.l-l

2-3. For fluids with a density close to, but less than, _that_ of water, t~e specific gravity is best determined in the system shown m Flg. 2-3. Denve an expression for y in terms of Z¡, z1, z8 and z,. P1

Ans.: Y = PI =

(z4

-

za) - (z8 (zl - za)

-

zJ

Problema

70

Fluid Statics and One-Dimensional Laminar Flow

Chap. 2

2-4. If the air can be treated as an ideal gas, and the temperature of the atmosphere varíes linearly with height above the earth, T

=

T0

-

a.z

derive an expression for the pressure as a function of z neglecting fluid motion in the atmosphere and the rotation of the earth. "

2-5. Two fluids are confined by a hinged gate as shown in Fig. 2-5. If the lower fluid is water and the upper fluid has a specific gravity of 0.8, determine the moment per unit width about point "A."

Ans.: Moment per unit width is 69.4 x 104 lbr-ft/ft. z Oil

L.

y=0.8

Fla.l-s

2-6. A wooden sphere (p = 58 lbm/ft3) is floating atan air-water interface. What fraction of the sphere is submerged? Solve by first deriving Archimedes' principie for an arbitrary body located at a fluid-fluid interface. 2-7. A wooden plank of density 46lbm/ft3 is anchored in a submerged surface as illustrated in Fig. 2-7. If the plank is 1 in. thick and 10 in. wide, what is thc moment about point "A"? Ans.: Moment is 571br-ft.

2-8. Determine both the horizontal and vertical components of the force per unit width exerted by the fluid on the curved gate shown in Fig. 2-8. Use Eq. 2.5-2 to determine these forces, and deduce from the result that the y-component is simply the weight of the fluid above the gate, while the x-component is the average pressure exerted on the gate times the projected area. In ordcr that the units remain consistent, take {J = 1 ft 112• 2-9. Work Prob. 2-8 for the case where the density is a linear function of y. P = Po - 11p

where Ap

lbm/ft3

(L) 6ft

10 62.4lbm/ft3 y= ft Ans.:f., = 896lbr/ft; [ 11 = -2960 lbr/ft. =

Po =

2-10. The spherical tank shown in Fig. 2indicated by the mercury manometer is 2000 lbm. If30 bolts are used to bol perbolt? 2-11. A weather balloon 1Oft in diamete rcsult obtained in Prob. 2-4 with o the balloon will risc. Assume tha is constant at 10ft, and the mass Given: Ralr = 0.0253 atm ft3/lbm•

and One-Dimensional Laminar Flow

Chap. 2

Problema

71 y

gas, and the temperature of the atmosthe earth,

L.

as a function of z neglecting fluid motion of the earth. gate as shown in Fig. 2-5. If the lower has a specific gravity of 0.8, determine the "A." x 104 lbr-ft/ft.

z

L.

=0.8

Fl¡.l-7

L.

is floating at an air-water interface. What ? Solve by first deriving Archimedes' at a fluid-fluid interface. is anchored in a submerged surface as is 1 in. thick and 10 in. wide, what is the

vertical components of the force per unit gate shown in Fig. 2-8. Use Eq. deduce from the result that the y-comabove the gate, while the x-component gate times the projected area. In order

p=

1 ftl/2,

the density is a linear function of y. - llp

(L)

Ibr/ft.

6ft

Fl¡.l-1

2-10. The spherical tank shown in Fig. 2-10 contains both water and oil ata pressure indicated by the mercury manometer. The mass of each half of the spherical shell is 2000 lbm. If 30 bolts are used to hold the two sections together, what is the force perbolt? 2-11. Aweatherballoon lOftindiameterisfilled with 7.0 lbm ofhelium. Using the result obtained in Prob. 2-4 with a: = 0.005°K/ft, determine to what altitude the balloon will rise. Assume that T0 = 298°K, the diameter of the balloon is constant at 1O ft, and the mass of the balloon is 1O lb m:. Given: Ralr = 0.0253 atm ft 3/1bm 0 R, RHe = 0.182 atm ftll/lbm 0 R.

Chap. 2

Fluid Statics and One-Dimensional Laminar Flow

72

Problems

d

Fig. 2-10 Fluid 2

2-12. Determine the thickness t of the concrete dam shown in Fig. 2-12, which is required to produce a zero moment about point B. Assume that the hydrostatic head on the bottom of the dam varies linear! y from 80 ft atA to zero at B.

Ans.: t = 34.1 ft.

1'1 .

, .. ·.•

·.. ~··..:·; ~·~·:

Water

,.

..

. ..

Fig.

2-14. The center of stress (often referred t systems) is defined as the positio? still give the same torque as the d1st1 position vector locating the center ol above definition is

· : ·.· 9

i' X F =

;.:: :. ·. ' ¿ · ..

.. Coherete ·:. ~ : y=25 .'

80ft

j

.

. ...

.

.. 6 •

.

. .

...

. . ·. 4

F'

where

..

/

Fig. 2-12

2-13. The micromanometer illustrated in Fig. 2-13 is a useful device for accurately measuring small pressure differences. If the densities of the two manometer fluids are nearly the same ( p1 ""' p2), measurable val ues of the distance d can be obtained for very small values of the pressure difference, PB - PA· Letting A 1 be the cross-sectiona1 area of the reservoirs, and A 2 be the cross-sectional area of the connecting tu be, derive an expression for PB - pA in terms of P1• p2 , g, A 1 , A 2 , and d.

Use these equations to loca te the ce1 when the density is constant. 2-15. Prove the following: (a) A • (B x C) = B • (C X A) = (b) A X (B X C) = -(B X C) X

Given that

ixj=k=-

jxk=i=-

kxi=i=-

and One-Dimensional Laminar Flow

Chap. 2

73

Problems

Fluid 2

concrete dam shown in Fig. 2-12, which is about point B. Assume that the hydrodam varies linearly from 80ft atA to zero

. , ••. d

...

. ••

•

• A

2-14. The center of stress (often referred toas the center of pressure for hydrostatic systems) is defined as the position at which total force could be applied and still give the same torque as the distributed force system. Ifwe define i' as the position vector locating the center of stress, the mathematical equivalen! of the above definition is

f •

•

Fig. 2-13

•.

: ~·· .~;;, :_: ~::_

i' X F

.-.: :.. ·•... 9 ·

f

= r

X

ten> dA

A

Concrete ·

~ · r = 2.s -~. ~

.

• ¡J

••..

.

.

;·_:

dA

,4. ••

A

•

/

Use these equations to loca te the cen!er of stress for the gate shown in Fig. 2-8 when the density is constant. 2-15. Prove the following :

Fig. 2-13 is a useful device for accurately If the densities of the two manometer measurable val ues of the distan ce d can be pressure ditference, PB - PA· Letting reservoirs, and A 2 be the cross-sectional an expression for PB - pA in terms of

(a) A · (B x C) = B · (C x A) = C ·(A x B) (b) A X (B X C)

-(B x C) x A

=

=

B(A · C) - C(A • B)

Given that

i

X

j

j

X

k

i

k

X

k

=

-j X i,

=

i

=

-k

=

j

=

-i

=

X X

i

X

i

=

0

j,

jxj=O

k,

kxk=O

74

Fluid Statlca and One-Dimenalonal Laminar Flow

2-16. Demonstrate that the cross product, A x B, can be written in

term~

Chap. l

of the

determinant

k

j

A X8

=

A.., A" A. B.., B" B,

2-17. Show that A • (B x A) is zero.

Kinematics

2-18. Show that the cross product rules may be expressed in index notation as e< 1> x e11 , ~ O, e< 1> x e(i,

=

i

=

j

±e(kl•

i "- j ,¡. k ,¡. i

2-19. Prove that V X (Vp) = 0

thus, the curl of the gradient of a scalar is zero. To do this, expand the term V x (Vp) and regroup the scalar components. 2-20. Determine the velocity profile for laminar flow of a Newtonian fluid in the annular region illustrated in Fig. 2-20.

2-21. Derive the flow-rate, pressure-drop relationship for the flow of a power-law

fluid through a circular tu be. The relationship between shear stress and shear rate is given by 1 = 1dv, ,..Tr• m dr dr

(dv,)

for one-dimensional laminar flow in a tube.

In the previous chapter, we anal pressure field and surfa.ce forces .ror st2 profiles for one-dimens10nallammar fi tum principie was applied to a diffe1 appropriate differential equations. Th more general differential equations e possible only because the acceleration 1 was zero. To analyze the complex fi01 we must be capable of describing tt Kinematics is the description of moti how this motion is brought about. 1 represents a key step toward our obje tions of motion. In addition, sorne of formulation of the macroscopic balan The student is encouraged not to e as "mathematical gymnastics" unrela final result, for the derived differential balances (Chap. 7) cannot be appliec fidence unless the derivation is unde understood unless the mathematical mastered.

One-Dimenslonal Laminar Flow

A

Chap. 2

x B, can be written in term~ of tbe k

3

Kinematics be expressed in index notation as j

=j i#j#k-:Fi

is zero. To do this, expand the term

~"""u'''""'~' for the flow of a power-Jaw , .... vuo'"'~'

between shear stress and sbear

In the previous chapter, we analyzed the problem of determining the pressure field and surface forces for static fluids, and we determined velocity profiles for one-dimensionallaminar flows. In both cases, the linear momentum principie was applied to a differential volume element to derive the appropriate differential equations. These were, in fact, special forros of the more general differential equations of motion, and the derivations were possible only because the acceleration term in the linear momentum equation was zero. To analyze the complex flows commonly encountered in practice, we must be capable of describing the fluid motion in a precise manner. Kinematics is the description of motion per se, and it takes no account of how this motion is brought about. The material presented in this chapter represents a key step toward our objective of deriving the differential equations of motion. In addition, sorne of the derived results must be used in the formulation of the macroscopic balances treated in Chap. 7. The student is encouraged not to dismiss these preliminary developments as "mathematical gymnastics" unrelated to the practical application of the final result, for the derived differential equations (Chap. 5) or the macroscopic balances (Chap. 7) cannot be applied to practical problems with any confidence unless the derivation is understood. And the derivation cannot be understood unless the mathematical tools used in the development are mastered. 75

Kinematics

76

*3.1

Chap. 3

Material and Spatial Coordinates

Our objective in this section is to develop an understanding of material coordinares and their relationship to spatial coordinates, so that we can adequately describe the motion of a fluid. The term "spatial coordinates" refers to a fixed rectangular coordinate system in which all points may be located. The existence of such a coordinate system follows from the assumption that the space is Euclidean. There are two possible methods of locating or identifying a "particle" of fluid, which may be defined as a differential material volume element, d"Ym(t), i.e. it is a vanishingly small volume of fluid which contains the same material at all times. At sorne time, t, we may designate the position of a fluid particle in terms of its spatial coordinates-x, y, and z. In this manner, all fluid particles may be located in terms of the spatial coordinates and time. At sorne reference time, chosen as t = O for convenience, the position of any fluid particle can be specified as

x=X, At sorne other time, t

z=Z,

y= Y,

at

Sec. 3.2

Time Oerlvatlves

material coordinates do not represent a deforms with the fluid; they simply i coordinates-namely, those which the fl -time, t = O. Since no two fluid particles at the same time, the material posit! particle. Equation 3.1-2 may be writte ' r

= r(l

which simply expresses the fact that the are a function of time and the materia The time rate of change of the spatil particle is the velocity of that particle. with the material coordinates held con In keeping with the nomenclature to 1:)

t =O

> O, the position of this particle may be expressed

as

*3.2 Time Derivatives

t

X =

X

+ f (~:) dt

(3.1-1a)

o

(3.1-lb)

(3.1-1c)

We may put this result in more compact vector form by multiplying these three equations by i, j, and k, respectively, and then adding to obtain

The time derivative of an arbitral! given by dS = lim [S(t dt AC-oO

where the spatial dependence of S is \l S is a function of time only, this defin However, if Sisa function ofthe spati by Eq. 3.2-1 is not well defined until s' in space at which S is measured for ti As our first example, we shall e through space, letting x, represent tl illustrated in Fig. 3.2-1. According to

t

r =R

+

J(~:)

dt

(3.1-2)

o

dx, = lim [x,( dt At-oO

= Vz, theve

+ jy + kz R = iX +iY + kZ

where r = ix

We shall call r the spatia/ position vector, because it locates the fluid particle inspace, while R will be called the material position vector, because it represents the coordinates used to "tag" or identify a given particle. Notice that the

x-directi<

t Describing the fluid motion in terms e referred to as the Lagrangian method, while tb to as tbe Eulerian method.

Kinematics

Chap. 3

develop an understanding of material spatia/ coordinates, so that we can fluid. The term "spatial coordinates" system in which all points may be system follows from the assumpare two possible methods of locating may be defined as a differential it is a vanishingly small volume of at all times. At sorne time, t, we may in terms ofits spatial coordinates-x, may be located in terms of the reference time, chosen as t = O for particle can be specified as

Sec. 3.2

n

Time Derlvatlves

material coordinates do not represent a coordinate system which moves and deforms with the fluid; they simply represent a specific set of spatial /coordinates-namely, those which the fluid particle occupied at the reference O. Since no two fluid particles can occupy the same spatial position · time, t at the same time, the material position vector uniquely defines a fluid particle. Equation 3.1-2 may be written as

=

r = r(R, t)

(3.1-3)

which simply expresses the fact that the spatial coordinates of a fluid particle are a function of time and the material coordinates.t The time rate of change ofthe spatial position vector for a particular fluid particle is the velocity of that particle. Since this time derivative is evaluated with the material coordinates held constant, it is called a material derivative. In keeping with the nomenclature to be established in Sec. 3.2, we write

(~:)a=~:=

(3.1-4)

T

at t =O of this particle may be expressed

(3.1-1a)

(3.1-1b)

(3.1-1c)

•3.2 Time Derivatives The time derivative of an arbitrary scalar function, S= S(x, y, z, t), is given by

dS dt

=

1im [S(t

+ dt)- S(t)J

(3.2-1)

dt

At-oO

where the spatial dependence of S is understood. For the special case where Sisa function of time only, this definition of the derivative is unambiguous. However, if Sisa function ofthe spatial coordinates, the derivative expressed by Eq. 3.2-1 is not well defined until sorne statement is made about the point in space at which S is measured for the two times, t and t dt. As our first example, we shall consider a system of particles moving through space, letting x 11 represent the x coordinate of the pth particle, as illustrated in Fig. 3.2-1. According to Eq. 3.2-1, we write

+

vector form by multiplying these , and then adding to obtain

(3.1-2)

dx 11 = 1im dt

At-oO

=

because it locates the fluid particle position vector, beca use it represents fy a given particle. Notice that the

[X (t + dt) 11

X 11(t)J

dt

(3.2-2)

v.,, the velocity of the pth particle in the x-direction

t Describing the ftuid motion in terms of the material coordinates and time is often referred to as the Lagrangúzn method, while the use of spatial coordinates and time is referred to as the Eulerúzn method.

78

Kinematics

Chap. 3

il meaningless until we specify where, in 1

z

aample, we measured the temperature ever, the temperature of a fluid may be t a thermocouple or sorne other device. constant for the limiting process, we Wli

ar _ (d!'\

1 y

Ot

dt }r

_lim [~ At_,O

/ /

.........1 / 1 1

1 /-___ ,-:::.

..............

........

/

/

~~----~~----~~~-------------+X

xP(t)

xP(t + fl f)

Fi¡. 3.1-1. Motion of a particle in space.

The derivative given by Eq. 3.2-2 is the time rate of change of a scalar, which is measured as we move with the particle-i.e., we might imagine an observer riding on the particle and continuously observing his position along the x-coordinate. This type of derivative, encountered previously in the study of particle dynamics, is called a material derivative and denoted by

Dxv = (dxv) = lim [xv(t Dt dt R M .... O

+ Llt) -

xv(t)J

Llt

(3.2_3) Material coordinates held constant

In this particular example, there is no question about the type of derivative we obtain by the limiting process indicated by Eq. 3.2-l. The time derivative of any quantity associated with a particle is necessarily a material derivative, since the quantity can only be measured by an observer or a device which moves with the particle. For example, if we evaluate the time rate of change of the temperature of the pth particle,

dTv = lim [Tv(t dt At_,O

+ Llt) -

Tv(t)J

Llt

(3.2-4)

it is obviously a material derivative. Consider now the difficulties encountered in specifying the time rate of change of the temperature of a fluid. The deriva ti ve

dT = lim [T(t dt At_,O

+ Llt) Llt

T(t)J

(3.2-5)

where oTfot is called the partial deriva to measure the temperature of a fluid the fluid. Under such conditions the 1 constant, and the material derivative lJ held constant, the temperature of a sin~ the temperature of a succession of fiuJ constant. Thus far we have indicated that the f takes on special meaning when either t held constant. lf neither is held cons if either the function depends only 01 at which the function is measured has 1l Before we present specific equations it will be helpful to give examples of ti sidering a skin diver measuring the wat The temperature may be a function of tij time t. If the diver anchors himself at se of change of temperature that he meast

(~n.

lf he allows himself to drift with the cw of change of temperature, the materia] ,

If the skin diver ís energetic and moves ; the total derivative, dTfdt, which depend velocity w, and the partial derivative wit We would now like to formulate thel temperature is a function of the spatial e this functional dependence as T= T(1

Kinematics

Chap. 3

lec. 3.2

nme O.rlvatlvea

79

ia meaningless until we specify where, in space, T is measured. In the previous eumple, we measured the temperature while moving with a particle; however, the temperature of a fluid may be measured at a point fixed in space by a thermocouple or sorne other device. lf the spatial coordinates are held constant for the limiting process, we write

ar = (d!'\ = lim [T(t at dt }r At-+0

........ / 1 1 1

___ )~~,_..,

____ / /

H.f)

of a particle in space.

time rate of change of a scalar, which .e., we might imagine an observer observing his position along the encountered previously in the study of derivative and denoted by xv(t o[

+ Llt) Llt

Xv(t)J

!it

(3.2_6) Spatial coordillateo beld C:ODIIaDI

/

..,.. /

+ At)- T{t)J

(3 _2 _ ) 3 Material coordinatcs held constant

question about the type of derivative by Eq. 3.2-l. The time derivative is necessarily a material derivative, by an observer or a device which if we evaluate the time rate of change

where ar¡at is called the partía/ derivative. It is possible, but not practical, to measure the temperature of a fluid with sorne device which moves with the fluid. Under such conditions the material coordinates would be held constant, and the material derivative DT/ Dt would be detemiined. lf R is held constant, the temperature of a single fluid particle is measured, whereas the temperature of a succession of fluid particles is measured if r is held constant. Thus far we have indicated that the time rate of change of a scalar function takes on special meaning when either the spatial or material coordinates are held constant. lf neither is held constant, the derivative is meaningful only if either the function depends only on time or the velocity of tlie point at which the function is measured has been specified. Before we present specific equations for the material and total derivatives, it will be helpful to give examples of the three types of derivatives by considering a skin diver measuring the water temperature in the Big Sur River. The temperature may be a function of the spatial coordinates, x, y, and z, and time t. lf the diver anchors himself at sorne point in the river, the time rate of change of temperature that he measures is given by

(3.2-7) If he allows himself to drift with the current while he measures the time rate of change of temperature, the material derivative is determined.

(3.2-8) (3.2-4)

in specifying the time rate of The derivative

+

(3.2-5)

If the skin diver is energetic and moves about with a velocity w he measures the total derivative, dTfdt, which depends on the spatial variations of T, the velocity w, and the partial derivative with respect to time. We would now like to formulate these ideas in mathematical terms. The temperature is a function of the spatial coordinates and time, and we express this functional dependence as T= T(r, t) (3.2-9)

Sec. 3.2

lf the material coordinates are to be held constant, we express the spatial coordinates in terms of R and t by means of Eq. 3.1-3. T

=

T(r, t)

=

(3.2-10)

T[r(R, t), t]

= T[x(R, t), y(R, t), z(R, t), t]

Time Derivativa

Returning now to Eq. 3.2-ll, coordinates holding R constant a vector v, and the derivative of the ten deriva tive. DT = (o!'\

Holding R constant, we differentiate with respect to time and get

Dt

(:~)a- ~~- (~~ (::)a + (~~ (:na + (~D (::)a + (~n.

dG

=

(oG) dT + (oG) oT "·" oV

dV T,n

+ (oG) on

(3.2-13)

dn T,v

There is sorne curious notation associated with thermodynamics which is rarely encountered elsewhere-i.e., the use of subscripts on the partial derivatives to indicate which variables are being held constant. In sorne respects, this system is superfl.uous, because a partial derivative implies that "all other independent variables are held constant" during the limiting process. However, in thermodynamics there is a wide choice of independent variables, and the subscripts are used as a reminder that the independent variables are (in this example) T, V, and n. If T, V, and n are functions of time, Eq. 3.2-13 may be divided by dt to obtain dG _ dt

(oG) oT

V,n

(d!'\ dt }

+ (oG) oV

V,n

+ (oG)

(dV) dt .

on

T,v

(dn) dt

(

3.2_14)

Under these conditions, the functional dependence of G could have been expressed as G = G[T(t), V(t), r(t)] (3.2-15) which is very similar to the functional dependence of T in Eq. 3.2-10 when R is held constant. The only difference is that the temperature, T, depends explicitly upon time in addition to the implicit time dependence via x(t), y(t), and z(t).

J

+ v.,(~

a

Using vector notation, we can expr€

DT =~

(3.2-11) where the chain rule has been used to obtain the first three terms. This step always presents sorne difficulties, and it may be helpful to consider it further. If the student has had a course in thermodynamics, he is familiar with functions of several variables. For example, consider the Gibbs free energy, which may be expressed as a function of the temperature, volume, and number of moles, n. (3.2-12) G = G(T, V,n) The total derivative is

ot

Dt

Noticing the repetition of x, y, an quite naturally leads us to the use convention

If the location of the point at with time owing to the motion of t no longer held constant. Instead of

(~n =

(:n

+ (~~ (~t

where dxfdt, dyfdt, and dzfdt are which describes the diver's motion. dT dt

This result is the most general t Equation 3.2-17 is justa special case derivative (i.e., w = 0). The sky-dlver

As an illustration of the applica diver who has mistakenly strapped a onto bis wrist. His error makes th to gather sorne interesting meteorolo ture is independent of time and dec an altitude of 10,000 ft. Thus, we

T=

Klnematlcs

Chap. l

be held constant, we express the spatial means of Eq. 3.1-3.

11r(R, t), t]

(3.2-10}

t),y(R, t), z(R, t), t]

S.c. l.l

Returning now to Eq. 3.2-11, we note that the derivatives of the spatial coordinates holding R constant are the components of the fluid velocity vector v, and the derivative ofthe temperature holding r constant is the partial derivative.

(o!'\J + v:l)(o!'\ + vv(o!'\ + v.(o!\ ox} oy} oz}

DT = Dt ot

with respect to time and get

(~D (~;)R + (~D (~;)R + (~n. to obtain the first three terms. This step it may be helpful to consider it further. in thermodynamics, he is familiar with example, consider the Gibbs free energy, of the temperature, volume, and (3.2-12)

dV + (aG) dn On

(3.2-13)

T,v

with thermodynamics which is , the use of subscripts on the partial bies are being held constant. In sorne because a partial derivative implies that constant" during the limiting process. a wide choice of independent variables, that the independent variables are (in and n are functions of time, Eq. 3.2-13

P-'V"'"'""

oG) (dv) + (aG) (dn) (3.2_14) OV dt . On dt .

V n

.

T"

dependence of G could have been V(t), r(t)]

(3.2-15}

dependence of Tin Eq. 3.2-10 when R is that the temperature, T, depends the implicit time dependence via x(t),

(3.2-16)

Using vector notation, we can express Eq. 3.2-16 as DT = oT + Dt ot

(3.2-11)

,n

81

Time Derivativa

v·VT

(3.2-17)

Noticing the repetition of x, y, and z in the last three terms of Eq. 3.2-16 quite naturally 'leads us to the use of index notation and the summation convention

~~ = (~D + vi (~~

(3.2-18)

If the location of the point at which the temperature is measured varies with time owing to the motion of the skin diver, the material coordinates are no longer held constant. Instead of Eq. 3.2-11, we write

(~D = (~D + (~~

e:)

+

(~~ (~n

+

(~v (~:)

(3.2-19}

where dxfdt, dy/dt, and dzfdt are the components of the velocity vector w which describes the diver's motion. In vector form, Eq. 3.2-19 becomes dT oT -=-+w·VT dt

ot

(3.2-20)

This result is the most general type of time derivative we shall encounter. Equation 3.2-17 isjust a special case (i.e., w = v) ofEq. 3.2-20, as is the partial derivative (i.e., w = 0).

The sky...c:tlver As an illustration of the application of Eq. 3.2-20, let us consider a sky diver who has mistakenly strapped a dial thermometer instead ofhis altimeter onto his wrist. His error makes the jump more precarious but allows him to gather sorne interesting meteorological data. Assume that the air temperature is independent of time and decreases linearly between ground level and an altitude of 10,000 ft. Thus, we may represent the temperature as

T= T0

-

ru

82

Kinematia

where

tX

=

5

X

Chap. J

10-3°F/ft, and Eq. 3.2-20 reduces to dT - = dt

-WtX

•

(3.2-21)

In the free-fall stage, the sky diver's velocity is about 200 mph. Hence,

dT = -{(-200mi)(5 x 10_ dt hr ft

3

oF)}{(528~ft)( mt

hr )} =1. 47 oF/ sec 3600 sec

and he experiences a rapid rise in temperature. After the parachute is opened, the velocity decreases to about 5 mph, and the rate of temperature rise is reduced to only 0.037°F /sec.

S.C. J.l

Time Derivativa

Here we ·see that the acceleration Ct) dV,

J

D.t

(3.4-4)

Klnematlcs

90

Chap. 3

Since the limits of integration on the first two integrals are the same, they can be combined to give

~Js dV = lim dt

4t-+O 7'"a(t)

J

[S(t

~t

L=

w

~t) dVn -

[ Vu(4tl

J

S(t

The volume dV is given .by the leng

+ ~t) dV1]

dV=

v 1 •]

P

where

4>(x) '

3

a

= 0.075 Ibm/ft = 1 x I0- 5/ft 2

b

=

Po

3-5. The Leibnitz rule for differentia1 functions of the independent varia

1

X

then

10--4/ft

and the velocity of an airplane flying through the atmosphere is given by, w

=

iw.,

+ jw11 + kw.

where w., = 300 mph w11 = 20mph w. = -20mph how would the air density measured at the airplane change with time if the coordinates of the plane are x = y = O, and z = 15,000 ft? Ans.: 4.9 x

lo-~> Ibm/ft3-sec.

d4> dx

1 Jb-Direction

) +-1 OTer OT"] - T-98 +-

1 --(TTrr [ r or

r oiJ

r

oz

(a)

(e)

(b)

In the next section, the viscous stress tensor "' will be represented in terms of the rate of strain tensor d for a Newtonian fluid. The relationship is given by

op

'f

- oz

= 2,ud + [(K -

i,u)V • v]l

(5.1-lSa)

or (e)

T¡¡

=

2p d¡¡

+ [ (K-~ ,U) (~~J ~íi

J

(5.1-18b)

134

The Differential Equations of Motion

Chap. S

where we refer to ¡.t and K properly as the shear coefficient of viscosity and the bu/k coefficient of viscosity, respectively. In practice, we refer to ¡.t and K as the "viscosity" and the "bulk viscosity," respectively. The rate of strain tensor is given in terms of the velocity gradients ast

d¡;

= !(éJv,

+ OV¡)

ox,

(5.1-19)

2 OX¡ A detailed derivation of this result is presented in Sec. 5.2 followed by a qualitative discussion of the rate of strain in Sec. 5.3. The effort required to understand the development is large compared to the benefits reaped from such study; therefore, many students may wish to skip these sections and go directly on to Sec. 5.4. They may do this without detriment to the study of subsequent material; however, any student with a good background in solid mechanics can cover this material very quickly and should do so.

Sec. 5.2

Newton's Law of Viscosity

in this section will be difficult to unders student. A mathematical description is applications and a qualitative description The rate of strain

In order to determine the rate of str consider a materialline element of length trated in Fig. 5.2-1, P and Q are materia

z

5.2 Newton's Law of Viscosity Newton's law of viscosity is based on a linear relationship between the viscous stress and the rate of strain. In Chap. 1, Newton's law of viscosity for a simple one-dimensional flow was introduced in the form of a single component of the stress tensor being equal to the coefficient of shear viscosity times a velocity gradient,

r

(5.2-1) Although this expression may appeal to the student's intuition, it is not at all obvious t~at dv.,fdy is a measure of the rate of strain for a fluid in plane shear flow, ev_9) though the units (sec-1) are satisfactory. Gaining sorne understanding ofthe rate of strain and its relationship to the viscous stress is undoubtedly one of the most difficult problems in fluid mechanics; in fact, more than 100 years elapsed between Newton's statement in 1686 that, "The resistance arising from want of lubricity in the parts of a fluid is, other things being equal, proportional to the velocity with which the parts of the fluid are separated from one another." and the derivation of the Navier-Stokes equations by Navier (1821), Poisson (1831), and Stokes (1845). 1 lt is only natural, then, that the ideas discussed

t This relationship should be reminiscent of that between the strain tensor and gradients of the displacement vector encountered in solid mechanics. The scalar components of á11 in rectangular, cylindrical, and spherical coordinates are listed in Table 5.2-1, and the relationship between the scalar components of the viscous stress tensor and the velocity gradients are given in Tables 5.2-2, 5.2-3, and 5.2-4 for Newtonian fiuids. l. P. F. Neményi, "The Main Concepts and Ideas ofFluid Dynamics in Their Historical Development," Archive for the History of Exact Sciences, 1962, 2:52.

X

Flg. 5.2-1.

Materi

element connecting them. If the fluid element would be translated and rotated However, if the line element were length deformed and the rate of change of len rate of deformation. W e note that at any point in space, of line elements, each having a differ tangent vector to the line element

A.,= li As-+

Differential Equations of Motion

Chap. S

the shear coefficient of viscosity and the In ·practice, we refer to p, and K as ," respectively. The rate of strain gradients ast

+ ovi) i

ox,

Sec. 5.2

135

in this section will be difficult to understand, requiring extra effort by the student. A rnathernatical description is presented first, followed by sorne applications and a qualitative description in Sec. 5.3. The rate of strain

(5.1-19)

is presented in Sec. 5.2 followed by a in Sec. 5.3. The effort required to compared to the benefits reaped from wish to skip these sections and go this without detriment to the study of with a good background in solid quickly and should do so.

Newton's Law of Viscosity

In order to determine the rate of strain at sorne point in space, let us consider a rnaterialline element of length !J.s embedded in the fluid. As illustrated in Fig. 5.2-1, P and Q are material points and !J.s is the rnaterialline z

o on a linear relationship between the Chap. l, Newton's law of viscosity introduced in the form of a single to the coefficient of shear viscosity

Material line element, ~s

(5.2-1)

the student's intuition, it is not at all rate of strain for a fluid~· n lane shear ¡austa1cto:rv. Gaining sorne nderstandto the viscous stress · ndoubtedly fluid mechanics; in fact, more than l"u,;uu;Ju in 1686 that,

X

Flg. 5.2-1. Materialline element.

equations by Navier (1821), Poisson then, that the ideas discussed that between the strain tensor and gradients mechanics. The scalar components of eoordiDtates are listed in Table 5.2-1, and the the viscous stress tensor and the velocity 5.2-4 for Newtonian fluids. Ideas of Fluid Dynamics in Their Historical Sciences, 1962, 2:52.

element connecting them. If the fluid were rnoving as a solid body, the line elernent would be translated and rotated, but the length would never change. However, if the line element were lengthened or shortened, the fluid would be deformed and the rate of change of length of !J.s would be a rneasure of the rate of deformation. We note that at any point in space, we could visualize an infinite number of line elements, each having a different direction A1, where .A, is the unit tangent vector to the line elernent

A; = lim 4s-+O

(!J.' •) !J.s

(5.2-2)

The Differential Equations of Motion

136

Chap. S

In solid mechanics, the word "strain" denotes a change in length per unit length. We will follow this nomenclature and call the rate of change in length per unit length the "rate of strain." We remember that each line element passing through a particular point in space may be undergoing a different rate of strain, depending on its direction; therefore, we define the rate of strain in the following manner: the rate of } strain in the {). -direction 1

{1 D

.

= hm - - (t..s) As -o

}

S.C. 5.1

Newton's Law of Viscosity

Remembering that the material deriva ti we obtain

!(_1 )!!._ (A.f2) . 2 As Dt 2

Now, t.v1 represents the change in th and a Taylor series expansion of v, ab~

(5.2-3)

• (~a

t.s Dt

t.v.=

Following Fig. 5.2-1, we write (5.2-4) where we understand that this equation only holds in limit as t.s-+- O. We may rewrite Eq. 5.2-3 in terms of the square of t.s to obtain the ratein of . { (1) strain the} = hm - ( - 1 2) -D (t..s 2)} {A;-direction As-o 2 t.s Dt

.Here, we have dropped the higher ord limit t.s-+- O. Noting that !:u1 = t.r1, form

l(l)D

(5.2-5)

Substitution of this result into Eq. 5.2

In the following steps, it will be understood that we intend to take the Iimit t.s- O. By Eq. 5.2-4 we obtain

!(_1 )!!.__ (t..s2) = !(_12){!!.__ (t.r . t.r)\ t.s Dt t..s Dt ' '

2

j

2

(5.2-6)

J

2

Differentiation of the product yields

1){!!.__ (t..r;) t.r; + t.r; !!.__ (t..r;)} !(_1)!!.__ (t..s2) = !(2 t..s 2 Dt 2 t..s 2 Dt Dt 1(1)D

1{

Noting that

the rate of } strain in the {). -direction 1

Note that our result here is very simih the stress in a continuum. In that cas by a tensor Tii, and the stress acting direction n1 was given by

T.... =

(5.2-8)

In this development, we find that the ' by the tensor ov,fox1, and the rate of S jn the direction ).1 is given by Eq. 5.2-1 The tensor ov,fox1 may be rearr conjugatet ovifox, to obtain

OV¡ = !(OV¡ oxi 2 oxi

(5.2-9)

we write .P_(t..r;) Dt

=

Dr;(Q) _ Dr;(P)

Dt

Dt

=

t..(Dr;) Dt

=

(5.2-7)

By symmetry, we may write the right-hand side as

D (t..r;) } - - 2 - (t.s 2) = - 2 t.r;2 t.s Dt t.s Dt

2

- -2 -(t..s)= 2 t..s Dt

(5.2-10)

+ ovi) OX;

t It may be helpful to remember from the that tbe atraln In lbe} { A,rtirection

and Eq. 5.2-8 takes the form (5.2-11)

where u, is the deplacement vector. t The transpose of a matrix is obtained b this operation is applied toa tensor, the resu

Differential Equations of Motion

Chap. S

" denotes a change in length per unit and call the rate of change in /ength We remember that each line element may be undergoing a different rate therefore, we define the rate of strain

lim

{_!_ !!_ (~s)}

&s~o ~s

(5.2-3)

Dt

S.C. 5.2

Newton'a Law of Vlscoslty

137

Remembering that the material deriva ti ve ofthe position vector is the velocity, we obtain (5.2-12)

Now, dv, represents the change in the velocity vector between Q and P, and a Taylor series expansion of v, about the point P gives

(av.)

dv, = -• dx 1

(5.2-13)

OX¡

(5.2-4)

only holds in limit as square of ~s to obtain

~s-+-

O. We

.Here, we have dropped the higher order terms in anticipation of taking the limit ~s-+- O. Noting that dx1 = dr1, we may rearrange Eq. 5.2-12 in the form

l

(-1)!!_ (~s2) = (av,ox )(~r,) (~r

2 ds 2 Dt

(5.2-5)

~s

1

~s

1)

(5.2-14)

Substitution of this result into Eq. 5.2-5 and use of the limit ds-+- O yieldt that we inte d to take the limit

)A.,A.¡ {!~:a:t: ~~e} (av, ox =

.l.¡-direction

(5.2-6)

Note that our result here is very similar to that obtained when we analyzed the stress in a continuum. In that case, the "state of stress" was determined by a tensor T,1, and the stress acting on a plane having a normal n, in the direction n1 was given by (5.2-16)

(5.2-7)

side as (5.2-8)

- r;(P)

In this development, we find that the "state of rate of str.ain" is determined by the tensor ov,jox;. and the rate of strain on a surface having a normal A., jn the direction A.1 is given by Eq. 5.2-15. The tensor ov,Jox1 may be rearranged by adding and subtracting its conjugatet ov1Jox, to obtain

(5.2-9) (5.2-10)

(5.2-15)

i

(5.2-17)

t lt may be helpful to remember from the small deformation theory of solid mechanics that tbe s.train in tbe} {..1.¡-du:ecuon

_!_{~'· ~(D'')} ~s Dt 2

'

(5.2-11)

= ( iJu,)

iJx ).). 1

• 1

where u, is the deplacement vector. t The transpose of a matrix is obtained by interchanging the rows and columns. When this operation is applied to a tensor, the result is called the conjugate tensor.

138

The Differential Equations of Motion

Chap. S

Sec. 5.2

Newton's Law of Viscosity

The first term on the right-hand side is called the rate of strain tensor, d¡;, and is symmetric.

d¡;

=

!(OV¡ + OV;) 2 OX;

(5.2-18)

OX¡ .

Table 5. SCALAR CoMPONENTS OF THE

Rectangular Coordinates (x, y, z)

(d;; = d;;) The components of d;; for rectangular, cylindrical, and spherical coordinates are listed in Table 5.2-1. The second term in Eq. 5.2-17 is called the vorticity tensor, 0¡1, and is skew symmetric

o .. = !(ov¡ _ ov;) " 2 OX ; (O;;= -0;;)

(5.2-19)

= d;;A;A; + O;;A;A;

1 ove des = -r -ao d..

(5.2-21) and in the second term on the right-hand si de we relabel i as j and j as i to obtain O;;A;A; = !O;; A; A; + !O;;A;A¡ (5.2-22) Changing the order of multiplication by A; and A; allows us to express this resultas (5.2-23) Because O;; is skew symmetric, we know that (O;; and the proof is obtained. O;;A;A; = O

+ O;;) is identically zero (5.2-24)

Our expression for the rate of strain is now given entirely in terms of the rate of strain tensor the rate of } strain in the {1.;-direction

Cylindrical Coordinates (r, O, z)

(5.2-20)

where the unit tangent vector A; is completely arbitrary. We can show the last term on the right-hand si de of Eq. 5.2-20 to be identically zero by carrying out the double summation and suitably regrouping the terms; however, the proof is somewhat easier if we use the technique of relabeling the repeated or dummy índices. We first split the term into two parts,

= d;;A;A;

(5.2-25)

av. oz •

d.. =

OX¡

We now wish to show that the vorticity tensor O;; makes no contribution to the rate of strain. Later we shall show that the components of this tensor represent the rotational motion of the fluid. Returning to Eq. 5.2-15 and making use of the definitions given by Eqs. 5.2-18 and 5.2-19, we obtain the rate of } strain in the { 1.;-direction

d., =d.

=

v,

+ -r •

ds. =

av. oz'

Spherical Coordinates (r, O, r/>)

dr 8

=

d8 r

= ~2

[r !_ (~) or r

1 ov8 d88=-r

v, ao +-. r

d••

1

OV.

= (r sin-O-a,¡,

V,

V9 COt

0)

+-r + -r- '

Newton's law of vlscosity We begin our task of formulatin by assuming that the viscous stress te tensor. We then restrict ourselves to rate of strain to obtain Newton's la similar procedure in treating solids-i. of the strain tensor, and then restrict pendence on the strain to obtain Hoo

Differential Equations of Motion

Chap. 5

Sec. 5.2

139

Newton's Law of Viscosity

is called the rate of strain tensor, d; 1,

Table 5.2-1 SCALAR CoMPONENTS OF THE RATE OF STRAIN TENSOR

i

+ av.) ' ax;

(5.2-18)

Rectangular Coordinates (x, y, z)

, cylindrical, and spherical coordinates term in Eq. 5.2-17 is called the vorticity

(5.2-19) d..

tensor n;; makes no contribution to that the components of this tensor fluid. use of the definitions given by Eqs.

=

ov. oz '

Cylindrical Coordinates (r, 8, z)

/\

1 OVe

dee=-r o8

(5.2-20) d..

,..,.,,.,.,.,.,," arbitrary. We can show the last to be identically zero by carrying out the terms; however, the proof of relabeling the repeated or dummy parts, (5.2-21) si de we relabel i as j and j as i to

that (.Q;;

=

v,

+-, r

ov. oz'

d., = d.. =

Spherical Coordinates (r, 8, )

dre = der = !2 [' .!_ (~) + !r ov,J or r o8 1 OVe

d88=-r o8

Vr

+-. r

1

8) '

(5.2-22)

ov.p v, ve cot _( d··----+-+-,sin 8 o,¡, r r

(5.2-23)

Newton's law of viscosity

+ il;;) is identically zero (5.2-24)

now given entirely in terms of the rate (5.2-25)

ov,) 21 (ov. a,: + oz

We begin our task of formulating a constitutive equation for fluids by assuming that the viscous stress tensor is a function of the rate of strain tensor. We then restrict ourselves to a linear, isotropic dependence on the rate of strain to obtain Newton's law of viscosity. We generally follow a similar procedure in treating solids-i.e., assume the stress tensor is a function of the strain tensor, and then restrict the analysis to a linear, isotropic dependence on the strain to obtain Hooke's law.

140

The Differential Equatlons of Motion

Chap. 5

S.C. 5.2

Newton'1 Law of VIICOIIty

of Eq. 5.2-6 subject to the restriction th

Our starting point is, then, (5.2-26) Thus, we have a tensor function of a tensor as our first step in formulating a constitutive equation. This idea is undoubtedly new to the student, and it is best to back up a bit to begin at the beginning. We shall consider first a sca/ar function of a scalar, (5.2-27) p = f(T)

.,.,1

is given by the following six equations:

We have indicated here that the density is a function of the spatial coordinates, X¡, x 1 , x 8 (or x, y, z). If p were independent of one of the coordinates, we could plot it as a function of x, to obtain a surface in 3-space. Derivatives are understood as the slopes of tangents to this surface and integrals as volumes under the surface. We might ask, "What of a vector function of a vector?" (5.2-29) Certainly this idea is readily acceptable and simply states that each scalar component of the velocity depends on the spatial coordinates.

= / 1(X¡, x 1, x 8) v1 = h(x¡, x 2, x 8)

v1

V8

= /

3(X¡,

x1, x8)

or v.

(5.2-30a)

or

= v.(x, y, z) v. = v.(x, y, z)

(5.2-30b)

or v. = v.(x, y, z)

(5.2-30c)

Although we may feel quite comfortable with such function, we cannot draw a "picture" of it easily, so we must rely on mathematical formalism. Now let us return to our tensor function of a tensor and state more explicitly what this terminology means. In words, we could say that each component of .,.,1 is a function of all the components of d11 • Symbolically, we may write (5.2-3la) Tu = fn (dw du, daa, du, d2B• du) "u= / 11(d11 , d22 , d 83, d12 , d18, d 81)

etc.

+ C1rtiw + C1ad•• + C11d,. + c.adw + c..c~•• +

.,.,. = Cud,.

.,." = .,.•• =

We have indicated here that the density is a function of the temperature. This function is easily plotted as p versus T; derivatives are readily identified as the slope of this curve and integrals as the area under this curve. Next we shall consider a scalar function of a vector, (5.2-28)

=O when

.,... = .,.•• =

.,... =

C11d,. + C1rtiw + Cud•• + C,1d,. + C,adw + C,ad•• + Cud,. + Caadn + C,ad., + C11d,. + C1 rtiw + C,ad•• +

We see that the general linear relationshi Buried in this array is the special form of in Sec. 1.5 for a simple shearing flow-i

"··=p which is simply a special form of Eq. S. described in Sec. 1.5 we can deduce that

c.. = We can now show that by restrictin tropic, only 2 of the 36 coefficients in of isotropy for Eq. 5.2-26 is expressed

.,.;1 =/,

where .,.;1 and d¡1 are the viscous stress rotated coordinate system x 1 - x~. Eq relationship between .,.,1 and d11 is in idea of isotropy is sometimes difficult further before we apply this restriction t Let us consider two observers locatec! respectively. Each observer is going ~ cxperiment on the same body of fluid. ol 1tress tensor and the rate of strain tens~ will reduce to

(5.2-Jlb)

Owing to the symmetry of both "H and dii, we have only six equations representing the dependeó.ce of "'~ on the six distinct values of d11 • Having established what we mean by a tensor function of a tensor, we need only write down the linear form of this relationship and require it to be lsotropic to obtain Newton's Iaw of viscosity. The most general linear form

and .,.:,. = e~ d:W,

in the

Now, if the fluid is isotropic we ex ayatem to obtain the same dependence t A ICalar zero in a tensor equation shoul

Differential Equations of Motion

Chap. 5

S.C. 5.2

Newton'a Law of Vlscoslty

141

of Eq. 5.2-6 subject to the restriction thatt (5.2-26)

tensor as our first step in formulating a new to the student, and it is beginning. We shall consider first a (5.2-27) is a function of the temperature. T; derivatives are readily identifi.ed as the area under this curve. (5.2-28)

is a function of the spatial coordinates, eoc:nd,ent of one of the coordinates, we a surface in 3-space. Derivatives to this surface and integrals as volask, "What of a vector function of a

dxs)

(5.2-29)

and simply states that each scalar the spatial coordinates.

= vix, y, z) v11 = v11(x, y, z) v. = v.(x, y, z)

or vf/e

(5.2-30a)

or

(5.2-30b)

or

(5.2-30c)

with such function, we cannot draw on mathematical formalism. function of a tensor and state mqre In words, we could say that each the components of d11 • Symbolically, (5.2-31a) etc.

(5.2-31b)

d11 , we have only six equations reprcdistinct values of d11 • by a tensor function of a tensor, wc of this relationship and require it to be ¡,¡Qi'~!liltv, Thc most general linear form

.,.,1

=O

when d11

=O

(5.2-32)

is given by the foUowing six equations:

+ C1rd., + C1ad•• + Cudn + C1r,d,. + C1,d.f/e '~"n = Clld• + C.rd., + C,d•• + CtAdn + C dv• + C,.d" '~"•• = C11d.,. + C1rd., + C,d•• + C"d" + C d,. + C,.d. .,... = C,ld.,. + C,sdr, + C.,d•• + C"d" + C.,d,. + C.,d. .,.,. = C61d.. + C6adn + C6,d•• + Cudn + C d,. + C.,d. .,... = C, 1d.,. + c,r14, + c,,d•• + c"d" + C86dv• + c ..d.f/e .,..,. = Cud.,.

(5.2-33a)

11

(5.2-33b)

81

(5.2-33c)

66

(5.2-33d) (5.2-33e) (5.2-33f)

We see that the general linear relationship gives rise to 36 unknown constants. Buried in this array is the special form of Newton's law of viscosity discussed in Sec. 1.5 for a simple shearing flow-i.e., dvf/e .,..f/e=p dy

(5.2-34)

which is simply a special form of Eq. 5.2-33d. From experiments of the type described in Sec. 1.5 we can deduce that c.. = 2p (5.2-35) We can now show that by restricting this linear relationship to be isotropic, only 2 of the 36 coefficients in Eqs. 5.2-33 are distinct. The condition of isotropy for Eq. 5.2-26 is expressed as '1":¡ = f,¡(d~,) (5.2-36) where .,.;1 and d;1 are the viscous stress and rate of strain tensors in sorne Equation 5.2-36 states that the functional rotated coordinate system 1 relationship between .,.,1 and d11 is independent of the coordinate system. The idea of isotropy is somctimes difficult to grasp, and we should explore it further before we apply this restriction to simplifying Eqs. 5.2-33. Let us consider two observers located in the 1 and coordinate systems, respectively. Each observer is going to perform a simple shearing flow experiment on the samc body of fluid. Only a single component of thc viscoua stress tensor and the rate of strain tensor will be nonzero; thus, Eqs. 5.2-33 wiU reduce to in thc x 1-coordinate system (5.2-37) and .,.:,_ = e~ d:,_, in thc x;-coordinate system (5.2-38)

x x;.

x

x;

Now, if thc fluid is isotropic we expect thc observcr in the x;-coordinate l)'ltem to obtain thc same dcpendencc of stress upon rate of strain as tbe t A ecaJar zero in a tensor equation ahould be read u the null temor.

142

The Differential Equations of Motion

observer in the x,-coordinate system. Thus, C~, beco mes

Chap. S

= C", and Eq. 5.2-38 (5.2-39)

which is justa special case of Eq. 5.2-36. We are now in a position to examine Eqs. 5.2-33 for a series of coordinate transformations, for which we will require that the Cii remain constant. The transformed values of -r,1 and 1 will be related to the original values by Eqs. 5.2-40 -r;¡ = CX¡~r:CX¡¡'TIJ:¡ (5.2-40a) a;i = ¡ Ol.i/r.Cltjlakl (5.2-40b) Z 1Z

a.

Sec. 5.2

Newton's Law of Viscosity

We now return to examine the first of Eq that the C¡¡ terms are invariant; thus, 5.2-33a becomes

Substitution of the results from Eqs. 5.2

1

1

1

1

/)(

1 1

/

/

1 / 1 // 1 // 1 1 ,"' Y----------~----------Y

X

Fl¡. 5.1-l. Coordinate transformation.

First, let us examine a coordinate transformation consisting of a 180° rotation about the z-axis as shown in Fig. 5.2-2. The direction cosines cxm,. for this transformation are given byt cx11 = -1 cx13 =O CX12 = 0 cx21 =O cx 28 =O CX22 = -1 cxal =O cxa2 =O CXaa = 1 (5.2-41)

Comparing this result with Eq. 5.2-33a,

By examining the remaining Eqs. 5.2-33, also be zero, and the array of coefficien

We may now apply these results in forming the double sum indicated in Eqs. 5.2-40. For -r~ 1 , we write T~l = cxucxu-r~~:¡

Noting that only cxw cx22, and cx38 are nonzero, we quickly find that T~1 = ex u CXuTn = -( -1 ) 2-r11 =

Tbe remaining terms for

T 11

-r;1 and a;1 are cals_ulated readily, and we find 1

-r.,., = -r.,., 'T;II 1

'T ••

-r.,=-r.,ll ' 1

= 'TIIII

'TI/1 = -TIIS

= -r••

-r•' ., = --r.:x:

and a~z = azz

a;"= a"" a~.=

a••

(5.2-42)

a~=a.,

a;.= -a". a~z = -a.z

(5.2-43)

t Remember that according to Eq. 5.2-40, the first index on Ot refers to the x/-coordinate system, and the second index refers to the x,-coordinate system. Thus, Otu is the cosine of the angle between the x'-axis and the y-axis.

C11

C12

C1a

G

c21

C22

c2a

G

Cal

Ca2

Caa

c41

c,2

C,a

(j

o o

o o

o o

o o

Finding the other zeros in this array is a straightforward but rather tedious task, and we will only outline briefiy the remaining steps. For the rotaqon shown in Fig. 5.2-3, the array is reduced further to Cu

c12

c13

c21

C22

C2a

Cal

Ca2

Caa

o o o

o o o

o o o

o o o e" o o

o o o o e55 o

o o o o o Css

(5.2-48)

Differential Equations of Motlon

C~,

=

Chap. S

Cu, and Eq. 5.2-38

(5.2-39) Eqs. 5.2-33 for a series of coordinate that the C11 remain constant. The be related to the original values by Eqs. 5.2-40 (5.2-40a) T¡¡ = tXj~~:Cl¡¡Tic¡ d¡¡ - ¡ Otj/t.Otjldkl (5.2-40b) First, let us examine a coordinate transformation consisting of a 180° rotation about the z-axis as shown in Fig. 5.2-2. The direction cosines cx:mn for this transformation are given byt cx:11 = -1 Cl12 =o Cl13 =o Cl22 = -1 Cl23 =o + r sin 8 ati = -V(p + P4>) = -V[(p- Po)+ P4>l

(5.5-2)

where p 0 is a constant generally taken to be atmospheric pressure. This rearrangement is possible only if the density is constant-i.e., for incompressible flows. Substitution of Eq. 5.5-2 into Eq. 5.4-10 gives

p(~: + v· Vv)

=

(5.5-3)

-V[(p- Po)+ p4>] +,u V2v

For any given flow there may be several possible choices for the characteristic length and velocity. In closed-conduit flow, the characteristic length is traditionally chosen to be four times the hydraulic radius, R 11 , and the characteristic velocity is taken to be the average velocity, (vs>· The hydraulic radius is defined as the cross-sectional area divided by the wetted perimeter.

R = cross-sectional area 11

(5.5-4)

wetted perimeter

For a pipe, the hydraulic radius is equal to one-fourth the diameter,

(1T~2) (5.5-5)

R,.=--1TD

D 4 principie behind dimensional variables (v and p) are functions of:

For the present we will simply designate the cbaracteristic length and velocity by L and u0 , and multiply every term in Eq. 5.5-3 by L/ pu~ to obtain

and t); differential equations (p, ,u, and g); boundary conditions.

o(;J + (..!.) ·(LV)(.!.)

~~'>"''~..··E>

form, we can reduce the number from three to one, and show that for two-phase flows or free-surface represent the gravity vector as the (5.5-1)

é)r~o)

Uo

Uo

= -(LV)(p- {

PUo

0

+ 1> + (_l!:_)(r,2V )(..!.) 2

Uo

2

)

puoL

u0

(5.5-6) Each term in parentheses in Eq. 5.5-6 is dimensionless; to work with a less cumbersome equation, we need to define sorne dimensionless variables. In defining these variables, it is helpful to choose symbols which clearly represent the quantity in question. Insofar as it is possible, we shall try to do so by using the capital letter or script letter associated with the particular dimensional quantity. Often, such a choice would lead to confusion, as it would if

160

The Differential Equations of Motion

Chap. 5

Sec. 5.5

Dimensional Analysis

V were used to represent the dimensionless velocity, or if T were used to represent the dimensionless time. Under these circumstances, sorne other variable is chosen to represent the dimensionless variable. The dimensionless variables in Eq. 5.5-6 are defined as U =

(:J

0 =

c~o)

{dlmenslonleas veloclty}

{dimensionleastime}

V = LV {dlmensionleas vec:tor operator}t p = N

(p - Po) pu~

{dlmensionleas preasure}

_ PUoL {dlmensionleas parameter called the}

Re -

[JJ

--

=p

p.

Reynolds number

+ 1!._2 {dimensionleas preasure whlch lncludes} the force term Uo

body

Use of these dimensionless variables allows us to write Eq. 5.5-6 as

1 oU +U. VU =-V&'+ - - V2U

o0

Flg. 5.5-1. Sudden

(5.5-7)

NRe

For incompressible flows, the dimensionless continuity equation takes the form (5.5-8) So far .we have reduced the number of parameters in the differential equations from three (p., p, g) to one (NRe); however, both U and &' will depend on the parameters that appear in the boundary conditions, and we need to examine two specific cases to indicate how the boundary conditions are treated in dimensional analysis. From these two examples, we shall see that the Froude number enters the solution as a parameter in the normal stress condition at a fluid-fluid interface. For confined flows, the dimensionless velocity and pressure, U and &', are independent of the Froude number. Sudden contraction in a pipeline

As an example of flow in a closed conduit, we consider the sudden contraction in a pipeline such as that illustrated in Fig. 5.5-l. We assume that the Reynolds number is less than 2100; the flow is therefore laminar, and the velocity profile is essentially parabolic at sorne distance from the t Lack of a suitable symbol for the dimensionless "del" operator compels us to use thc samc symbol for both the dimcnsionless and thc dimensional form.

sudden contraction. For practica] the mathematical point of view w parabolic at z = ± oo. While it is possible to solve tn thus determine the pressure variati sive to investigate this effect experiJJ pressure variation. To obtain the n mínimum amount of experimenta dimensionless form. The differe dimensionless form and we must J form. The boundary conditions a

=

B.C.l:

v,

B.C.2:

v. =o,

B.C. 3:

v. =O,

B.C. 4:

v, = 2(v,)I

2(v,){l- 4 r

= T¡, 2

(~~

t Numerical methods have been use found in J. E. Fromm, "A Method for Flows" (Los AJamos, Calif.: Los Alam

Differential Equations of Motion

Chap. S

Sec. 5.5

161

Dimensional Analysis

velocity, or if T were used to these circumstances, sorne other ~~~vUJll;;~~ variable. The dimensionless

~+oo

Grovity Flg. 5.5-1. Sudden contraction in a pipeline.

(5.5-7) continuity equation takes the (5.5-8) of parameters in the differential (NRe); however, both U and 9 will in the boundary conditions, and we how the boundary conditions these two examples, we shall see as a parameter in the normal stress confined flows, the dimensionless independent of the Froude number.

conduit, we consider the sudden illustrated in Fig. 5.5-1. We assume 2100; the fiow is therefore laminar, ua•au•Jll'-' at sorne distance from the énsi{ 1 -

B.C. 2: B.C. 3:

v. =o,

r =r¡,

v, =O,

r

B.C.4:

v, = 2(v.)¡

(~~T1- 4(~JJ

= r2 ,

-00

O>z:::::-oo +oo:::::z;;:::O z = +oo

(5.5-9a) (5.5-9b) (5.5-9c) (5.5-9d)

t Numerical methods have been used with success on such problems. Details may be found in J. E. Fromm, "A Method for Computing Nonsteady, Incompressible, Viscous Flows" (Los Alamos, Calif.: Los Alamos Scientific Laboratory, 1963), LA-2910.

162

The Differential Equations of Motion

Chap. S

Sec. 5.5

where we ha ve made use of conservation of mass (or vo1ume in this case of incompres~ib1e fl.ow) 1r

D~ (v.)1 = 4

D~ (v. ) 2 4

1r

U,= 2(1- 4R2 ),

B.C. 2':

U,

B.C. 3':

= O, u. = o,

= t, O> Z R = t&lD• 00

B.C. 4':

) =

~

-

and the pressure becomes

Z = -oo

R

For the configuration shown in Fig.

(5.5-10)

in formu1ating boundary condition 4. Choosing (v, )1 as the characteristic ve1ocity :md D 1 as the characteristic 1ength 1eads to the following dimensionless boundary conditions,

B.C. 1':

Dimensional Analysis

(5.5-11a)

¿ - oo

(5.5-11b)

z

(5.5-11c)

¿

Z=

o +oo ¿

Here we see that gravity simply gives term, pgz cos Ot, and does not infl.uen wou1d seek on1y to determine the fu Racing sloop hull design

(5.5-lld)

To illustrate how the Froude nu1 1et us assume that we are assigned t hull shape for an America's Cup defe s1oop is a subt1e and comp1ex task gen sailor-designers; however, sorne use obtained by towing models through < Fig. 5.5-2. The model is a scaled-do dimension-such as the 1ength, beam.

where

R = ..!._ DI

Z=.!.... DI ÉlD= D2t

DI

On the basis ofEqs. 5.5-7, 5.5-8, and 5.5-11, we may express the functiona1 dependence of U and & as U= U(R, (), Z, 0, NRe• ÉlD)

(5.5-12)

f!P = f!P(R, (), Z, 0, NRe• ÉlD)

(5.5-13)

Confining our attention to the pressure term, we express the time average and area average ast (5.5-14) where the dependence on R, (), and 0 is removed by the averaging process. This resu1t indicates that the dimension1ess pressure depends on Z and the two parameters, NRe and ÉlD. Note that one ofthese parameters carne from the differentia1 equation, whi1e the second entered the solution via a boundary condition. Actual experiments will be concerned with the dimensional pressure p, which may be expressed as

Position of the free surface is z, z, (x, y)

=

(5.5-15) t A script 9t with an appropriate subscript will be used throughout the text to denote a dimensionless ratio of two like quantities. t Time averages will be discussed in Chap. 6 and will always be denoted with the overbar ().

Fig. 5.5-2. M

Chap. S Sec. 5.5

163

Dimensional Analysis

of mass (or volume in this case of For the configuration shown in Fig. 5.5-1 , the scala r (c/J ) is given by (cjJ )

(5.5-10)

=

-gz sinn:

(5 .5-16)

and the pressure becomes . Choosing (v.) 1 as the characteristic leads to the following dimension(5.5-11 a) (5.5-11b)

(p)

= p(üz)U(Z,

NRe• !J?n)

+ pgz

sinn: + Po

(5.5-17)

Here we see that gravity simply gives rise to an additive hydrostatic pressure term, pgz cos ex, and does not influence the velocity U. An experimental study would seek only to determine the function f(Z, N Re• !Ji n).

(5.5-11c) Z

= + oo

Racing sloop hull design (5.5-lld) To illustrate how the Froude number enters into free surface problems, Jet us assume that we are assigned the job of determining the best possible hull shape for an America's Cup defender. The design of a hull for a racing sloop is a subtle and complex task generally undertaken by wise and venerable sailor-designers; however, sorne useful information about hull design can be obtained by towing models through a testing tank, such as that illustrated in Fig. 5.5-2. The model is a scaled-down version of a proposed hull, and every dimension- such as the Jength, beam, etc.-is related to the dimensions ofthe

5.5-11, we may express the functional Towing mechonism

(5.5-12) (5.5-13) term, we express the time average (5.5-14) is removed by the averaging process. ess pressure depends on Z and the one of these parameters carne from entered the solution via a boundary with the dimensional pressure p,

Model hull Position of the free surfoce is Z5 z5 (x, y)

=

z

L.

(5.5-15) will be used throughout the text to denote 6 and will always be denoted with the Fig. S.S-2. Model towing tank.

Sec. 5.5

164

The Differential Equations of Motion

Dimensional Analysis

Chap. S

The Froude number is defined ast

actual hull by (5.5-18) where ~ L < l. Thus, the model is geometrically similar to the actual hull. Our experiment rnight consist of towing severa! models, all having the same scale factor ~ L> through the tank to determine which produces the least drag ata given speed. Presumably we know approximately at what speed the actual hull will be moving, and we know the physical properties of the fluid (salt water) through which it will move. The question to be answered is: What should be the conditions in the tank so that the tests are dynamically similar to the actual motion? The answer is that the Reynolds number and Froude number must be the same; our objec.tive is to prove it. lt is easiest to view this problem as if the hull were fixed in space and the fluid were flowing past it with a uniform velocity. Under these conditions the boundary conditions are

x,y, z-.. ±oo

B.C. 1:

(5.5-19a)

(i.e., the flow is undisturbed far from the hull, the origin of the coordinate system being on the hull), and B.C. 2 :

V=

a(x,y, z) =O

0,

z = z,(x,y)

p =po,

(5.5-19c)

This last boundary condition indicates that the fluid pressure is equal to the atmospheric pressure at the free surface, designated by z,(x, y). This boundary condition neglects viscous and surface tension effects, but this is a reasonable assumption. Rewriting the boundary conditions in dimensionless form, we get

=

B.C. 1':

U

B.C. 2':

U = O,

B. C. 3':

f!JJ

=

-i,

4>

u 02

X , Y, Z--..

± oo

(5.5-20a)

A( X, Y, Z)

= O

(5.5-20b)

Z = Z,(X, Y)

1 , z= z, (o:,!ll

(5.5-20c)

The function 4> is given by

4> = gz

(5.5-21)

and to incorporate the dimensionless variable Z, we multiply and divide 4> by the characteristic length L , and rewrite boundary condition 3' as B.C. 3":

f!JJ

=

(~~)z.

Z

=

z

f!JJ = - , NFr

B.C. 3":

On the basis of Eqs. 5.5-7 and 5.5-8 a Eqs. 5.5-20a and b and Eq. 5.5-24, the is represented by

U= U[X, Y,Z, 0, A1

f!JJ

=

f!JJ[X, Y, Z, 0, A

Since the models are geometrically si function A( X, Y, Z) is identical for bot flow to be dynarnically similar, we req Froude number be the same; thus, _ N Re-

(5.5-19b)

Here, a(x, y, z) is a function which describes the surface of the hull; thus, boundary condition 2 indicates that the tluid velocity is zero on the surface of the hull. B.C. 3:

thus yielding the final form of boundar

(u L) 0

11

actUI

and 2

NF -

u ) --º. r - ( gL actu

When these conditions are subjected t we find and 11model

= 9i

The condition on the velocity is easil kinematic viscosity is difficult if not iJ cases. Since ~ L is necessarily much sm of the fluid in the towing tank must b than the kinematic viscosity of water. and in practice we can maintain eithe ' number constant but not both. Under intuition are required to determine the ance of a hull.

t Most often the Froude number is defiii Eq. 5.5-23 arises naturally from the dimensio

Z,(X, Y)

(5.5-22)

be used throughout this text.

Sec. 5.5 Differential Equations of Motion

165

Dimensional Analysis

Chap. S

The Froude number is defined ast (5.5-18) similar to the actual hull. severa! models, al! having the to determine which produces the we know approximately at what speed know the physical properties of the The question to be answered is: so that the tests are dynamically is that the Reynolds number and objec.tive is to prove it. if the hull were fixed in space and the velocity. Under these conditions (5.5-19a) the hull, the origin of the coordinate

NFr

B.C. 3" :

Z

that the fluid pressure is equal to the designated by z.(x, y). This boundary effects, but this is a reasonable vu•uuoJu~ in dimensionless form, we get (5.5-20a) Y,Z) =O Z

= z.(X,

(5.5-20b) Y)

gz

(5.5-20c)

(5.5-21) Z, we multiply and divide tfo boundary condition 3' as

Z = Z,(X, Y)

(5.5-22)

(5.5-23)

gL

= Z,(X, Y)

(5.5-24)

On the basis of Eqs. 5.5-7 and 5.5-8 and the boundary conditions given by Eqs. 5.5-20a and b and Eq. 5.5-24, the functional dependence of U and & is represented by

e, A(X, Y, Z), NRe• NFr] Y, Z, e, A(X, Y, Z), NR e• NFrl

U

=

U[X, Y, Z,

(5.5-25)

&

=

&[X,

(5.5-26)

Since the models are geometrically similar to the proposed actual hull, the function A(X, Y, Z) is identical for both. For the actual flow and the model flow to be dynamically similar, we require that the Reynolds number and the Froude number be the same; thus, · N

_ Re -

(5.5-19c)

u2 _Q,

thus yielding the final forro of boundary condition 3,

(5.5-19b) the surface of the hull; thus, ñuid velocity is zero on the surface

=

(u L)

_(u L\ 0

0

v

actual -

v }model

(5.5-27)

and

NF- -u~) - (u~) r gL actual-- gL model

(

(5.5-28)

When these conditions are subjected to the restriction given by Eq. 5.5-18, we find (5.5-29) and 3/2 (5.5-30) Vmodel = :71-L Vactual /M)

The condition on the velocity is easily met; however, the condition on the kinematic viscosity is difficult if not impossible to meet for most practica! cases. Since ~Lis necessarily much smaller than one, the kinematic viscosity of the fluid in the towing tank must be one to two orders of magnitude Iess than the kinematic viscosity of water. Such a fluid is rather difficult to find, and in practice we can maintain either the Reynolds number or the Froude number constant but not both. U nder such conditions considerable skill and intuition are required to determine the effect of a free surface on the performance of a hull.

t Most often the Froude number is defined as u0 /VgL; however, the form given by Eq. S.S-23 arises naturally from the dimensionless form of the equations of motion and will be used throughout this text.

166

The Differential Equations of Motion

Chap. S Se c. 5.6

The treatment of dimensional analysis given here is brief in comparison to the usefulness of the method. A more thorough discussion is available elsewhere. 2 - 3

For uniform acceleration,

Dv -=a Dt ' and

•5.6 Applications of the Differential Equations of Motion

In light of Eqs. 5.6-4 and 5.6-5, Eq. 5

There are only four classes of problems for which we may easily obtain analytic solutions of the equations of motion: fluids at rest or moving at a constant velocity; uniformly accelerated flows; one-dimensionallaminar flow; and irrotational flow. With the aid of advanced analytical or numerical methods, practically all laminar flows are susceptible to analysis; however, these four types of flow represent the limits of the undergraduate at this time. Although irrotational flow theory is a key tool in the analysis of flow around immersed bodies, and wave propagation in open channel flow, it will not be treated in this text. A readable account of this particular area of fluid · mechanics is given by Temple, 4 although the classic in the field is still the detailed treatment of Milne-Thomson. 6 lf the fluid velocity is constant,

0= -Vp

for uniform acceleration. Equation except that the gravity vector has bee three scalar forms of Eq. 5.6-6 by tald respectively,

op

where the constant vector bis zero if the fluid is at rest. Substitution of Eq. 5.6-1 into Eq. 5.4-10 yields the previously derived result,

+ pg

ox

op

-=

oy

op

oz

Noting that the total differential of th

dp =

op dx +

ox

we may use Eqs. 5.6-7 to obtain dp = p[i • (g - a) dx

+j ·

Integration of Eq. 5.6-9 between x 0 , Yo the general expression for the pressur

Uniformly accelerated flow Uniform acceleration means that each element of the fluid experiences the same acceleration; thus, v is a function of time but not position. Making use of the material derivative, we may write Eq. 5.4-10 as

Dv ~p-=-Vp+pg+¡.tV-v Dt

.

PJ

-=p

(5.6-2)

The solution of this equation was treated in detail in Chap. 2; we shall not discuss it further here.

.

- = pl

(5 .6-1)

v=b

O= -Vp

Appl ications of the Differential E

p = Po

+ p [ A.,(x -

x0)

where p = p(x, y, z)

Po = p(xo, Yo• Zo) (5.6-3)

2. G. Birkhoff, Hydrodynamics, A Study in Logic, Fact and Similitude (Princeton, N.J.: Princeton University Press, 1960). 3. P. W. Bridgman, Dimensional Analysis, (New Haven, Conn.: Yale University Press, 1931). 4. G. Temple, An Introduction to Fluid Dynamics (New York : Oxford Press, 1958). 5. L. M. Milne-Thomson, Theoretica/ Hydrodynamics, 4th ed. (New York : The Macmillan Company, 1960).

A.,= i · (g- a) A 11

= j · (g- a)

Az =k • (g- a)

If the pressure p 0 is known or specifie( where in the system. Ifthe system con1 is constant, Eq. 5.6-10 may be used t(

Chap. S Sec. 5.6

given here is brief in comparison thorough discussion is available

Applications of the Differential Equations of Motion

167

For uniform acceleration,

Dv Dt

-=a

a constant vector

'

(5.6-4)

and V 2 v =O

(5.6-5)

In Iight of Eqs. 5.6-4 and 5.6-5, Eq. 5.6-3 reduces to for which we may easily obtain motion: fluids at rest or moving at a flows; one-dimensionallaminar flow; of advanced analytical or numerical are susceptible to analysis; however, of the undergraduate at this time. is a key too! in the analysis of flow vv•"l'."·uvu in open channel flow, it will account of this particular area of fluid the classic in the field is still the

O= -Vp

op= p i.· ) + l o-r~;> + o-r~;>] oz r or r o() oz

(6.3-6)

Turbulent Flow

198

Chap. 6

Sec. 6.3

A Qualitative Description of Turb~

Following the analysis for laminar flow, we make four assumptions: l. 2.

vr = v8 = O, the time-averaged flow is one-dimensional; av.foz = o, following from assumption 1 and the time-averaged con-

tinuity equation; 3. f~';>, f~';>, and -r~;> are independent of () and z; 4. g. =o. Equation 6.3-5 immediately reduces to

(aozft)

= 1 ~ (r-r~;'>)

(6.3-7)

r or

Once again we argue that the right-hand si de of Eq. 6.3-7 is independent of z while the left-hand side is independent of r. We conclude that both sides are equal to a constant and write

- t.p = 1 !!._ (r-r) L

r dr

(6.3-8)

rz

1.0

_(f)

Fig. 6.3-4. Variation of intensity o

-,-,

Trr = - p v, Vr

where ó.p is the time-average pressure tube. We have again neglected entran result. Integration and application of t

B.C. 1 : f~;'> is fini

yields an equation for the total time-av 11> 11>

~ 11>

.,....

o .e

Thus, the total shear stress distribution pressure drop. Experimental studies of the turbulei

(/)

- (t)

Trz

0 .6

0.4

0.2

-----rtr0 Fl¡. 6.3-3. Total shear stress and turbulent shear stress for ftow in a tube.

o

=-

have been made by a number of inve shown in Fig. 6.3-3. The variation of ii Fig. 6.3-4. We find from these two fi¡

Turbulent Flow

Chap. 6

Sec. 6.3

A Qualitative Description of Turbulent Flow

199

, we make four assumptions: flow is one-dimensional; 1 and the time-averaged con-

uu'f'"'vu

of Oand z;

(6.3-7) side of Eq. 6.3-7 is independent of z of r. We conclude that both sides are

(6.3-8)

1.0

- - - - - r;r0 Fig. 6.3-4. Variation of intensity of turbulence with radial position.

where !J.p is the time-average pressure difference and L is the length of the tube. We have again neglected entrance and end effects in obtaining this · result. Integration and application of the boundary condition

B. C. 1:

-r;;>

is finite for

O $;; r

$;;

r0

(6.3-9)

yields an equation for the total time-average shear stress fCTJ rz

=

f

rz

+ fCrztl = -

(tlp) !_ L 2

(6.3-10)

Thus, the total shear stress distribution is determined by measurement of the pressure drop. Experimental studies of the turbulent shear stress, (6.3-11)

turbulent shear stress for flow

have been made by a number of investigators, and illustrative results are shown in Fig. 6.3-3. The variation of intensity of turbulence is illustrated in Fig. 6.3-4. We find from these two figures that the turbulent shear stress

200

Turbulent Flow

Chap. 6

Sec. 6.4

The Eddy Viscoslty

predomina tes in the core region and falls to zero at the wall. We also discover that the maximum intensity of turbulence and turbulent shear stress is near r/r0 = 0.9. We conclude that turbulence is generated near the wall and the intensity falls off toward the center of the tube. In the central region, the generating force (shear deformation) decreases and viscous forces tend to reduce the intensity.

6.4 The Eddy Viscosity An interesting intuitive description ofthe turbulent stress may be obtained by proposing a constitutive equation for -t< 1> similar to Newton's law of viscosity. In Chap. 5 the constitutive equation for an incompressible Newtonian fluid was stated as (6.4-1) ~ = 2,ud The time-average form is (6.4-2) i = 2,ud By analogy (albeit poor), we might write

consider the steady (on the average), o 6.4-1. If we assume that discrete "lum one position to another and still retai then the net rate of momentum transfe

(6.4-3)

thus defining the turbulent or "eddy" viscosity, ,u. The total time-averaged stress tensor is now given by (6.4-4)

The eddy viscosity is not a very useful quantity for analysis, but it helps to develop a qualitative feel for the nature of turbulent flow. The constitutive equation, Eq. 6.4-1, is a postulated relationship between the stress and the rate of deformation, and experiments show that ,u is indeed a constant for a large class of fluids. On the other hand, ,u is nota constant but depends on position and on the intensity of turbulence. In this section, we shall make use of an empirical expression for ,u to analyze the turbulent velocity profile in a tube. Prandtl's mixing length theory Prandtl's mixing length theory 2 was one of the first attempts at a semitheoretical analysis of turbulent flow. The arguments leading to its development are not very precise, but they are useful in developing a qualitative understanding of the mechanism of turbulent momentum transfer. We now 2. L. Prandtl, "Investigations on Turbulent Flow," Zamm, 1925, S :136.

x-momentum transferred {across a y-surface

where the mass M of the two fluid pa incompressible. The turbulent shear s which this transfer takes place, and the . occurs. Thust,

{momen force per) {unit area = unit tin

and we write

where tl.t is the average time required f1 the average area associated with the t1 and tl.A are dependent on the structure tude and frequency of the velocity fluct of a Taylor series expansion for v., give

v.,¡

~+l

~v.,l 11

t This expression follcws from the linear tion is not obvious, and the student must wa! is derived in Chap. 7 before he can understan

Turbulent Flow

Chap. 6

Sec. 6.4

201

The Eddy Viscosity

to zero at the wall. We also discover and turbulent shear stress is near is generated near the wall and the the tube. In the central region, the decreases and viscous forces tend to

ofthe turbulent stress may be obtained for i' 1l similar to Newton's law of ve equation for an incompressible 2,ud

-

2,ud

Flg. 6.4-1. Turbulent momentum transfer.

(6.4-1) (6.4-2)

consider the steady (on the average), one-dimensional ftow illustrated in Fig. 6.4-1. Ifwe assume that discrete "lumps" or particles offtuid can move from one position to another and still retain their momenta during this motion, then the net rate of momentum transfer for the process shown in the figure is

(6.4-3) x-momentum transferred} {across a y-surface

(6.4-4) quantity for analysis, but it helps to of turbulent flow. The constitutive ·onship between the stress and the that ,u is indeed a constant for a fL' 1l is nota constant but depends on In this section, we shall make use the turbulent velocity profile in

one of the first attempts at a semiarguments leading to its developuseful in developing a qualitative Ient momentum transfer. We now Flow," Zamm, 1925,5:136.

=

Mv1

_

Mv1

"' v+l

"' 11

(6.4-5)

where the mass M of the two fluid particles must be the same if the ftow is incompressible. The turbulent shear stress, :;~~. will depend on the rate at which this transfer takes place, and the y-surface area over which the transfer occurs. Thust, fo~ce per} = { umt area

{m~m~ntum tran~ferred per}

umt time per umt area

.

(6.4-6)

and we write M ( -~ -r-W 11.,=-- v.,

!l. t !l. A

v+l

- 1) -v.,

(6.4-7)

11

where !l.t is the average time required for the process to take place and !l.A is the average area associated with the transfer. lt is easy to imagine that !l.t and !l.A are dependent on the structure of the turbulence, i.e., on the magnitude and frequency of the velocity ftuctuations. Making use of the first term of a Taylor series expansion for v., gives,

v.,l v+l ~ v.,¡ + 11

t

dv) (--= dy

(6.4-8)

t This expression follcws from the linear momentum principie; however, the connection is not obvious, and the student must wait until the macroscopic momentum balance is derived in Chap. 7 before he can understand this result clearly.

202

Turbulent Flow

Chap. 6

and Eq. 6.4-7 becomes .¡: 11

=

"'

Mt (dv.,)

~t~A dy

Turbulent Flow in a Tube

6.5 Turbulent Flow in a Tube (6.4-9)

In this section, we will apply the problem of turbulent fl.ow in a tube. 1 gravity is neglected, the stress equatio

We can carry the analysis further by writing the mass M as M=p~V

Sec. 6.5

(6.4-10) .¡:= 2P.

[( ov,)

2

(~

ove

~

a¡:+ roO+,

a (ve +P. [ ror -r

av,

ov,

+P. (¡; + or Spherical Coordinates r, O,

ll>

=

2P-

(~ove ~)· or + r ao + r [( av,)'

V· (pfv) dV

r.(t)

f

V· (v • T) dV- dV r.(t) r.(t)

t This analysis is given in Secs. 10.1 and 10.2.

o

D

2 - (~ pv) =Dt

THE DISSIPATION FUN

This derivation has been somewhat more abstract than our previous efforts and it may be helpful to discuss this result before going on to the derivation of the macroscopic balance. We started with the stress equations of motion, which representa balance between the time rate of change of momentum per unit volume and the forces per unit volume. Forming the scalar product of the velocity and the time rate of change of momentum gives rise to the time rate of change of kinetic energy, while the scalar product between the velocity anda force (force times distance per unit time) naturally gives rise to the rate of work. The body force, pg, gave rise to a term representing the rate of work done by gravity, while the surface force V · T gave rise to two rate of work terms. The second term represents the irreversible rate of surface work, which is converted to thermal energy. If the rate of surface work is negligible, Eq. 7.3-18 indicates a balance between kinetic energy and potential energy. The student may recall experiments in beginning physics courses designed to demonstrate that kinetic and potential energy were conserved provided "frictional" effects were small. Expressions for in rectangular, cylindrical, and spherical coordinates are given in Table 7.3-1 for Newtonian fluids. There, we note that is composed entirely of squared terms and is therefore always positive. Forming the integral of Eq. 7.3-18 over the volume "Ya(t) yields the macroscopic balance,

:t (!

Here, we have made use of

unit ·rume

= -V· (pfv) +V· (v • T)-

Dt

The Macroscopic Mechanical E

Sec. 7.3

(7.3-19) sin O a ( +P. [ -,-aó '

ic Balances-lnertial Effects

Chap. 7

clearly only when the principie of detaiJ.t Substitution of Eqs. 7.3-4, the differential mechanical energy

Sec. 7.3

225

The Macroscopic Mechanical Energy Balance

Here, we have made use of

(7.3-20) Rate of work done

by surface forces

u ·rume

per ni t

+ V· (v • T) -



(7.3-18)

since V · v = O. Application of the general transport theorem to the first integral in Eq. 7.3-19, and of the divergence theorem to the second, third,

i

done

Table 7.3-1 el> FOR

Rate of conversion

unit

to thermal encrgy

THE DISSIPATION FUNCTION

per unit volume

abstract than our previous efforts t before going on to the derivation of motion, which represent a balance per unit volume and the forces product of the velocity and the time to the time rate of change of kinetic the velocity and a force (force times rise to the rate of work. The body ng the rate of work done by gravity, two rate of work terms. The second surface work, which is converted to Eq. 7.3-18 indicates a balance . The student may recall experito demonstrate that kinetic and "frictional" effects were small. rica!, and spherical coordinates are There, we note that is composed always positive. over the volume "f/a(t) yields the

Rectangular coordina tes x, y, z 2

2

(

(

]

ov. ov.) ov, ov.) ov, ov.)'] + .u [( ox + oy + oy + oz + ox + oz 2

2

(

(

Cylindrical Coordinates r, O, z 2

cl>=2.u

0Vr) [ ( a;+

(

1

ov

Vr) +a; ov,) ] 2

2

0

(

~ar¡+;:-

+.u [ ra,.o (v-; ) 8

ovr

1(oOovr)~J +.u [1~ (ov,) ov oO + ( a;2

+~

ov,)"

+.u ( oz +a; Spherical Coordinates r, O,

-

JV· (pr/Jv) dV

ov,)+ ov.)+ ov,) [(ox oy oz 2

cl>=2.u

el> - 2.u

dV=-

NEWTONIAN FLUIDS



2 ovr)" + (1 ov0 Vr)" + ( -1- -ovq, + -Vr + v0 cot 0) ] + [ ( or r oO r r sin o orf> r r

-

'f'".(t)

dV-

I

(7.3-19)

dV

'f'".(t)

sin O o ( vq, ) 1 ov8] • + .U [ -r- oO sin O + r sin Oaj;

0 )]

2

226

Macroscopic Balances-lnertial Effects

Chap. 7

Sec. 7.3

The Macroscopic Mechanical E

and fourth integrals, gives

~t

J

(t pv ) dV + 2

r .w

=

J(t

(

2

pv ) (v - w) • n dA

.ot.w

J

J

J

.ot.w

.ot.w

r.w

(v • T) · n dA -

pcfov • n dA -

(7.3-21)

p._j_t..J~

dV

1 1

Once again, there are sorne very definite reasons for transforming the volume integrals to area integrals. For example, if the gravitation term were left as it first appeared in Eq. 7.3-3, the macroscopic balance would contain an integral of the type

f

L

v • pg dV

r.(t)

Evaluation of this integral would require knowledge of v everywhere in the system. However, expressing the gravity vector as the divergence of a scalar and applying the divergence theorem yield the area integral in Eq. 7.3-21, which we may evaluate with reasonable accuracy in terms of distance above a reference plane and average velocities.t Following our development of the macroscopic mass and momentum balances, we again wish to separate the total area daCt) into the area of entrances and exits, fixed solid surfaces, and moving solid surfaces. Thus,

da(t)

=

A.(t) Area of entrances and exits + A.(t) Area of solid moving surfaces

+A.

(7.3-22)

w = O on A.(t)

2

..ot.w

(v • T) · n = v represents the rate at which the surro If we define Was the rate at which tn on the control volume, t then

w) • n dA =

J(tpv )(v 2

.A,I

and the total area integral takes the f

J(

v • T) · n dA ' :;,t.w

Here, we have made use of the fact tt of viscous dissipation to interna! ener

and we write

J(tpv )(v -

In attacking the first term on the r· remember that t(o) is the force per un the system. Thus, the term

Area of so lid fixed surfaces

An illustrative control volume "Ya(t) is shown in Fig. 7.3-1, and we now wish toexamine each term inEq. 7.3-21 to see what kind of simplifications or special interpretations we can make in terms of the three different areas. The first term (from left to right) remains unchanged. In the second term, we note that v = w =O on A. v -

Fig. 7.3·1. Movi1

w) • n dA

(7.3-23)

E= 11

A.(t)

t If we sought an exact solution, we would ha ve to know v everywhere, in which case there would be no point in using the macroscopic balance and no advantage in using the area integral. However, we seek approximate solutions to a great many problems, and in those cases the area integral has great preference.

t The dot over this term is used as a r per second or sorne equivalent.

Chap. 7

Sec. 7.3

227

The Macroscopic Mechanical Energy Balance

( Jprf>v • n dA - J



(7.3-21)

dV

reasons for transforming the volume if the gravitation term were Ieft as balance would contain an

knowledge of v everywhere in the vector as the divergence of a scalar yield the area integral in Eq. 7.3-21, accuracy in terms of distance above

t macroscopic mass and momentum total area da(t) into the area of and moving solid surfaces. Thus, of entrances and exits of so lid moving surfaces

(7.3-22)

Doshed line indicotes

Fig. 7.3-1. Moving control volume,

In attacking the first term on the right-hand side of Eq. 7.3-21, we must remember that t is the force per unit area that the surroundings exert on the system. Thus, the term

(v • T) · n = v • (T · n)

on

on A 8(t)

J(tpv )(v 2

W=

J

(7.3-24)

V • f¡a)

(7.3-25)

dA

A,(t)

and the total area integral takes the form

J

J

:W.(t)

A,( t)

(v • T) · n dA =

A,

= v • t(a)

represents the rate at which the surroundings do work on the control volume. If we define W as the rate at which the solid moving surfaces are doing work on the control volume, t then

of solid fixed surfaces

is shown in Fig. 7.3-1, and we now to see what kind of simplifications or terms of the three different areas. The In the second term, we note

~(f)

v • t dA

+W

(7.3-26)

Here, we have made use of the fact that v = O on A •. Defining the total rate of viscous dissipation to interna! energy as

w) • n dA

(7.3-23)

ha veto know v everywhere, in which case · balance and no advantage in using the solutions to a great many problems, and in

t The dot over this term is used as a reminder that the units are foot pounds force per second or sorne equivalent.

228

Chap. 7

Macroscopic Balances-lnertial Effects

and incorporating Eqs. 7.3-23 and 7.3-26, we can write Eq. 7.3-21 Time rate of change of kinetic energy of the control volume

~t J(! pv

2 )

Net flux of kinetic energy leaving the control volume

dV +

f' a(t)

Jl

pv\v- w) • n dA

and invoke the order of magnitude a (v'l

--, v' ·T

J

= v • t). If the streamlines are curved, from hydrostatic according to the of control volumes will always be so that the surface forces may be

235

Sudden Expansion in a Pipeline

-V&J

+ - 1-V 2U Nne

(7.4-26)

When the Reynolds number becomes very large, it might seem that the viscous terms would be negligible, but this is not necessarily so. As an example, Eq. 7.4-26 for steady fiow in a pipe reduces to

o= -V&J + - 1-V 2U

(7.4-27)

Nne

and the viscous terms are equal to the pressure terms regardless of what the Reynolds number might be. Thus, a high Reynolds number does not necessarily mean negligible viscous effects. In general, we will find that viscous effects can be neglected when (7.4-28) If the Reynolds number is large and U • VU = 0(V2U)

to steady turbulent flows, we again .4-17 ast

+ p') + pg(z + z') = c.

(7.4-22)

term and taking the time average, we

(7.4-29)

the inequality will be satisfied. This latter condition requires that the rate of change of velocity in the direction of fiow is comparable to the rate of change of velocity perpendicular to the fiow. The best way to gain sorne insight regarding the importance of viscous effects is to solve sorne problems and compare the results with experiment, for there are no hard and fast rules to guide us in making these approximations.

(7.4-23) tude argument that

PART 11-APPLICATIONS

(7.4-24) 's equation becomes pgz

=e,

7.5 (7.4-25)

we need to consider under what terms in the equations of motion to be the time average of

Sudden Expansion in a Pipeline

ov'fot is zero.

In this example, we wish to calculate the pressure change caused by a sudden expansion in a pipeline, such as that shown in Fig. 7.5-1. The dashed line indicates the surface of the control volume to which we wish to apply the macroscopic mass, momentum, and mechanical energy balances. The points a and b indica te the segment of the center streamline to which we will apply Bernoulli's equation. The flow is steady, turbulent and incompressible.

236

Macroscopic Balances-lnertial Effects

"O"

"1"

Chap. 7

"2"

Sec. 7.5

Sudden Expansion in a Pipeline

vector is given by

1

1

In all these examples, we will dispense should be used with turbulent flows, a adjustments necessary to account pro tution of Eq. 7.5-4 into Eq. 7.5-3, and the unit vector k, yield

J pv,v ·odA= A,

Fig. 7.5-1. Sudden expansion in a pipeline.

Mass balance

For a fixed control volume and steady, incompressible flow, we start with Eq. 7.1-11

J

pv • n dA= O

(7.5-1)

A,

which quickly reduces to (7.5-2)

where we have made use of the fact th1 placed far enough up.stream and d dimensional at these pomts; however, t be independent of the placement of Oa correct location is in sorne respects a the disturbance propagate upstream one go before the flow is on~.-dime~si "What is the flow topology? Askmg assumptions made in solving the protl 2 is very large,t the assumption of neg The area integral on the left-hand s at the entrance and exit of the control right-hand side must be evaluated 8 • k = O along the walls of the pipe, a area averages yields

or

2 - p(v,) 0

=

4D~ + p(v,)s11

1TD2

(P)o7

2

?:

1T

+ (p)¡4

Momentum balance

We start with Eq. 7.2-11

J

(pv)v • n dA=

A,

There is a very important point ~), for experience has s this term. The reason appears to be a the analysis of macroscopic balance entrances and exits and completely f4 ever the symbol .91 appears on an int each and every portion of the control entrances or exits.

( 11f4)(D:-

J

pg dV

r

+

J

t dA

(7.5-3)

d

and assume that viscous and turbulent stresses are small,t so that the stress

t The assumption of negligible viscous effects is to be made throughout this chapter. Turbulent stresses are identically zero at solid surfaces and are generally considered to be small compared to the pressure at entrances and exits.

t The distance between O and 1 is negligi

Sec. 7.5

"2"

237

Sudden Expansion in a Pipeline

vector is given by

1

1

t(n)

(7.5-4)

= -np

In all these examples, we will dispense with the time-average symbol, which should be used with turbulent flows, and require the student to remember the adjustments necessary to account properly for the time averaging. Substitution of Eq. 7.5-4 into Eq. 7.5-3, and formation of the scalar product with the unit vector k, yield

J A,

y, incompressible flow, we start with dA =0

(7.5-1)

(7.5-2)

pv.v • n dA = -

J

pk • n dA

(7.5-5)

.!#

where we have made use of the fact that k • g = O. The planes at O and 2 are placed far enough upstream and downstream so that the flow is onedimensional at these points; however, the answer obtained in this example will be independent of the placement of Oand 2. The purpose in placing them in a correct location is in sorne respects a device for forcing the questions: "Does the disturbance propagate upstream at all ?" "How far downstream must one go before the flow is one-dimensional ?" or, in the language of Sec. 1.3, "What is the ftow topology ?" Asking these questions helps to emphasize the assumptions made in solving the problem, for if the distance between 1 and 2 is very large,t the assumption of negligible viscous forces is unsatisfactory. The area integral on the left-hand side of Eq. 7.5-5 is to be evaluated only at the entrance and exit of the control volume, while the area integral on the right-hand side must be evaluated over the entire surface. Noting that n. k = O along the walls of the pipe, and expressing the integrals in terms of area averages yields

(7.5-6)

There is a very important point to be made regarding the term (p)1

f

dV +

('1T/4)(D:- Dl:), for experience has shown that students will often neglect t dA

(7.5-3)

.o/

stresses are small,t so that the stress is to be made throughout this chapter. surfaces and are generally considered to be exits.

this term. The reason appears to be associated with a general error made in the analysis of macroscopic balance problems-i.e., focusing attention on entrances and exits and completely forgetting about other surfaces. Whenever the symbol .91 appears on an integral, the student is urged to examine each and every portion of the control surface, especially those which are not entrances or exits.

t The distance between O and 1 is negligible except for ftows where Nxe < 1.

238

Macroscopic Balances-lnertial Effects

Chap. 7

(7.5-7)

we must eliminate (p )1 from Eq. 7.5-6. If the distance between O and 1 is negligible and the pressure variation over the annular region is hydrostatic, it is reasonable to make the approximation

(p)l

=

(7.5-8)

(p )o

and Eq. 7.5-6 can be rearranged in the forro 2

Ap

=

p(v~) 2 p(v~)o(~:) -

Sudden Expansion in a Pipeline

The assumptions for this result follow.

If we wish to develop an expression for the pressure differencet

Ap = (p )o - (p )2

Sec. 7.5

(7.5-9)

F or turbulent flows, the velocity profile is nearly flat; thus, the average of the square may be approximated by the square of the average. We shall investigate this approximation in more detaillater; for now, we simply write, (7.5-10)

l. Viscous effects were neglectedconsidered negligible. 2. The average pressure on the annu the average pressure at point O. 3. The velocity profiles were assumec Next, we will attack this problem using t will allow us to obtain a solution for AP different from those listed above.

Mechanical energy balance We start this analysis with the stead the mechanical energy balance

f!

and make use of the macroscopic mass balance to arrange Eq. 7.5-9 in the forro

2

pv v • n dA

A,

(7.5-11)

f

v · t

Sec. 7.8

tcnl =

-np 0

+t

where t(';,> is the stress in the solid in ambient pressure. The stress produced by atmos to the area integral on the right-ha balance yields - pvib where AP is the cross-sectional area Sin ce t{';,¡A P is the force exerted on t and opposite to the force exerted o and Eq. 7.8-14 is equivalent to Eq. the proper choice of a control vol u for in this case viscous effects need The effect of ambient pressure We saw in the Iast example tha made no contribution to the area int balance. Since we will encounter tH is worthwhile now to draw sorne ge is a constant p 0 , the stress vector m t(n)

=

where t,!> will be referred to as the n written in terms of the total pressUJ

1

t(n) = 1

------1

L

1 1 1 1 1 1 1

1

1

1

1 1 1 1

1

1

l

Grovity

1 1

L-------- --

-

_______ J

1

"3"

Fig. 7.8-2. Plane jet impinging on a flat plate.

and the net stress vector is then gi Cross sect1onol oreo= Ap

* = -n

t(n)

where

p

the gauge pressure. In the absence o terms of the gauge pressure

*

t(o)

The area integral of the stress vect Eq. 7.8-15 to give

roscopic Balances-lnertial Effects

Chap. 7

that the velocity is zero at the solid that

Sec. 7.8

volume, the stress vector becomes ttn)

surface of the plate

(7.8-8)

the surface of the plate

(7.8-9)

and assuming that u., is zero at points 7.8-1 to obtain bL - (p)piatebL

(7.8-10)

is specified as flat, the force F., that the = p0 bL

+ pvibh¡

(7.8-11)

the back side of the plate producing a on the plate becomes (7.8-12)

= pvibh1

ng this result is to apply the momentum in Fig. 7.8-2. For that control

-------, 1 1

257

Applications of the Momentum Balance

=

=

t 10 ¡

-npo,

-np 0

everywhere except at the solid surface

+ t 10*¡

at the solid surface

(7.8-13a) (7.8-l3b)

where t;';.> is the stress in the solid in excess of that arising from the constant ambient pressure. The stress produced by atmospheric pressure will contribute nothing to the area integral on the right-hand side of Eq. 7.8-1, and the momentum balance yields - pvibh¡ = i · t~n>Av (7.8-14) where AP is the cross-sectional area of the solid section supporting the plate. Since t;';.>AP is the force exerted on the plate by the solid support, it is equal and opposite to the force exerted on the plate by the impinging jet of liq u id, and Eq. 7.8-14 is equivalent to Eq. 7.8-12. The point of this analysis is that the proper choice of a control volume can greatly simplify the computation, for in this case viscous effects need never be considered. The effect of ambient pressure

We saw in the last example that the ambient pressure, being a constant, made no contribution to the area integral ofthe stress vector in the momentum balance. Since we will encounter this problem in subsequent applications, it is worthwhile now to draw sorne general conclusions. Ifthe ambient pressure is a constant p 0 , the stress vector may be written as t(n)

=

- np0

t

=

- np

+ t 1:¡

(7.8-15) where t1!> will be referred toas the net stress t•ector. The stress vector may be written in terms of the total pressure p and the viscous stress tensor, 't.

+ n · 't

(7.8-16)

and the net stress vector is then given by Cross sect1onol oreo= Ap

ft:> = -

where

n(p - Po)

+ n · 't

(7.8-17)

P- Po = Pu

- - - _ _ _ _ .J

the gauge pressure. In the absence of viscous stresses, the net stress is given in terms of the gauge pressure * = - npu t(n) (7 .8-18) The area integral of the stress vector in Eq. 7.2-10 may be written in terms of Eq. 7.8-15 to give

J

ftnl

mpinging on a flat plate.

d.(t)

dA = -

J

np0 dA

.-1.(1)

+ Jt~¡ dA

+ (p !p(v~- v~) + (Pa!p(v~ - v~)

/

/ // /

/

2 -

If we neglect gravitational effects an is atmospheric, Eqs. 7.8-36 yield

// /

Moving Control Volumes and U

We still require one more equati we have ex hausted the tools at our di · equation always gave the same res however, in this case the difference our aid. lf we apply Bernoulli's eqm lines indicated in Fig. 7.8-4, we obta

1

1 Streomlines

Sec. 7.9

/

V¡=

This result is consistent with Eq. 7. information by using Bernoulli's e everywhere, and Eqs. 7.8-32 and 7.8

/ /

/ / / /

'

/

pv~(A 2 - A 1

/

/

A¡=

V /

The areas and volumetric flow rates a by Eqs. 7.8-38 and 7.8-39,

Fig. 7.8-v, • n dA -

through the turbine driving the compressor and expands in the exhaust nozzle. lt leaves the nozzle close to atmospheric pressure and at a higher velocity than the entering air. If we assume steady flow, the momentum balance in the x-direction may be written

Ev i·

this equation inasmuch as a moving actual work done is given by (7.9-20)

W is most easily determined by

J(pv,)v, •

=Ji· t~l

dA

(7.9-21)

d

Assuming that the pressure at the surface of the control volume is atmospheric, the net stress t~l results only from the stress on the motor mount (presumably the fuelline would not be subjected to any stresses). If we take the velocity profiles atO and 1 to be flat, Eq. 7.9-21 reduces to -p0 v~A 0

+ p viA

=F., (7.9-22) We notice there is no contribution to the momentum equation from the fuel added to the system, because the x-component of the fuel velocity vector is zero at the surface of the control volume. The mass balance requires that 1

+ m, =

1

(7.9-23) where m1 is the mass flow rate of the fuel. Making use of Eq. 7.9-23 allows us to write the force F., as PoVoAo

moving through still air at a constant represented as the engine fixed in a velocity v0 . As the air enters the rises. The compressor further into the combustion chamber where ace. The combustion mixture passes

n dA

Ae

F.,

=

p0 v0 A 0 [v 1(1

PtVtAt

+ ex) -

v0 ]

where

m,

cx=--

PoVoAo

(7.9-24)

266

Macroscopic Balances-lnertial Effecu

Chap. 7

Because of the addition of fuel and the combustion, the exhaust velocity v1 is always much greater than the entrance velocity v0 and F,. is positive-i.e., the force acts in the positive x-direction. We remember that the definition of the stress vector required that it represent the force per unit area exerted by the surroundings on the control volume, thus, the force that the engine exerts on the air frame is the negative of F:c, and we see that the "thrust" of the engine is in the proper direction. At this point, the problem has not yet been solved because we must determine v1 to obtain a relationship between the thrust and velocity. In practice then, the engineer must make use of the principie of conservation of energy and knowledge of the combustion process to predict the temperature ofthe exhaust gases. Once he knows the temperature, he may use an equation of state to determine the density p1 and Eq. 7.9-23 to evaluate v1 •

Lx

~t Jpv dV + J pv(v r.(t)

w) • n

A,

we neglect gravitational effects and no volume is constant to obtain

A,(t)

In solving this problem, we will wish to the force F; thus, we might first take tl:l unit vector i to obtain

V

A¡

r---------------"1"

1 1 1 1 1 1 1

1 1

Ao

from the point of view of an observer macroscopic mass, momentum and point of view of an observer moving Eqs. 7.9-6, 7.9-8, and 7.9-19. We shall to discuss the first to gain sorne · treating moving control volume proble Starting with Eq. 7.2-10,

pv(v- w)

We now wish to extend the problem examined in the previous section to the case where the curved vane is moving with a constant velocity. This system is illustrated in Fig. 7.9-2.t We may attack this problem in two ways: z

1

Moving Control Volumes and U

J

jet impinging on a moving vane

1

Sec. 7.9

1 1 1 1 1 1

1

I"Q"

1 1 1 1

.,_wo

í - l_ _

Fig. 7.9-2. Jet impinging on a moving vane.

t Note carefully that the jet leaving the vane is drawn as it would be seen by a moving observer. Here we have made use of the intuitive idea that Vr should be tangent to the vane.

W e can now start to see sorne of ti momentum entering the control volm terms describing the momentum leavil note first that the fluid velocity vector that the angle {3 be calculated. Todo 1 vr, of the fluid leaving the jet. If we ca very shortly that the task of computi beco mes cumbersome; however if the a direction either parallel or perpendic difficulty in calculating the momentu If we now switch our attack on 1 moving with the control volume, we s tational effects to obtain

Making use of Bernoulli's equation al system, Eq. 7.9-5, we find that the m vr is a constant at points O and l. 1

Chap. 7

combustion, the exhaust velocity v1 velocity v0 and F., is positive-i.e., We remember that the definition of the force per unit area exerted thus, the force that the engine of F"'' and we see that the "thrust" of yet been solved because we must between the thrust and velocity. In use of the principie of conservation of · process to predict the temperature temperature, he may use an equation Eq. 7.9-23 to evaluate v1 •

Sec. 7.9

from the point of view of an observer fixed in space and applying the general macroscopic mass, momentum and mechanical energy balances; or from the point of view of an observer moving with the control volume and applying Eqs. 7.9-6, 7.9-8, and 7.9-19. We shall use the second approach, but we need to discuss the first to gain sorne insight into the difficulties encountered in treating moving control volume problems. Starting with Eq. 7.2-10,

~t

f

pv dV

+

r.

f

pv(v - w) • n dA =

f

pg dV

+

r.

A.(t)

f

t7n> dA

(7.9-25)

.JIJ1.(t)

we neglect gravitational effects and note that the momentum of the control volume is constant to obtain

f pv(v examined in the previous section to · with a constant velocity. This may attack this problem in two ways:

267

Moving Control Volumes and Unsteady Flow Problems

w) • n dA

=F

(7.9-26)

A.(t)

In solving this problem, we will wish to determine the x- and z-éomponents of the force F; thus, we might first take the scalar product of Eq. 7.9-26 with the unit vector i to obtain

V

-pv0(v 0

-

w0)A 0

+ p(v 1 cos {J)(v1 sin {3)( ¿1 sm

)

e

=F.,

(7.9-27)

We can now start to see sorne of the difficulties in this approach. The momentum entering the control volume is easily calculated; however, the terms describing the momentum leaving are not so easy to work with. We note first that the fluid velocity vector v is not tangent to the vane, requiring that the angle {3 be calculated. Todo so we must know the relative velocity, vr, of the fluid leaving the jet. If we carry this problem much further, we see very shortly that the task of computing the momentum leaving the system beco mes cumbersome; however if the fluid stream left the control volume in a direction either parallel or perpendicular to the vector w, we would ha ve no difficulty in calculating the momentum flux. If we now switch our attack on this problem to a frame of reference moving with the control volume, we start with Eq. 7.9-8 and neglect gravitational effects to obtain

f

(pYr)vr • n dA

=F

(7.9-28)

Ae

on a moving vane. is drawn as it would be seen by a moving idea that vr should be tangent to the vane.

Making use of Bernoulli's equation along a streamline in a moving coordina te system, Eq. 7.9-5, we find that the magnitude of the relative velocity vector vr is a constant at points O and l. Using this fact and taking the scalar

268

Macroscopic Balances-lnertial Effects

Chap. 7

Sec. 7.9

Moving Control Volumes and Unste

product of Eq. 7.9-28 with i and k, we get - pv~A 0 + p( v, cos O)v,A 1 = F z p (v, sin O) v,A 1 = Fz

(7.9-29) (7.9-30)

where A 1 is the cross-sectional area of the jet and not the area of the exit, which is A1/sin O. The mass balance gives

r-------------------

1

1 1 1

(7.9-31)

Ao =Al

~~--------~--~

1

Fx---~

and if we write the magnitude of the relative velocity vector as (7.9-32) the components of the force exerted by the vane on the jet are p(vo _ Wo)2Ao ( 1 _cosO)= -Fz Force exerted by the ( 7 .9 _33) vane on the jet

Force exerted by the vane on the jet

(7.9-34)

The power - W generated by the action of the jet on the vane is the scalar product of the force -F times the velocity w. power = p(v0

-

2

w0) woA 0(1 -cosO)

Moving water scoop

Figure 7.9-3 illustrates a water scoop used by a fast-moving train to pick up water from a trough set between the tracks. We would like to know the rate at which the train takes on water and the drag on the train resulting from this process. The width of the scoop is b and the thickness of the water film is h. In this example, we will view the problem as observers fixed in space and apply Eqs. 7.1-7 and 7.2-10. In applying the macroscopic mass balance

· 1"'0 (1)

Fl¡. 7.9-3. Movin¡

(7.9-35)

Note that if O = O, no force is exerted on the vane and no power is generated. If O = 1r-i.e., the vane directs the jet back upon itself-the power generated is a maximum. If we compare Eqs. 7.9-27 and 7.9-35, we can see that a rather large amount of algebraic effort must be expended to reduce Eq. 7.9-27 to the simple form given by Eq. 7.9-35. lt is apparent that visualizing the problem in terms of a moving coordinate system leads to an easy solution.

~t Jp dV + Jp(v- w) • n dA = O

1/í1/í//////////////lí//ll///lí///llll/í!/1~

(7.9-36)

A.(t)

we note that the fluid velocity v is equal to the velocity of the control volume w everywhere except in the water, where v = O. The time rate of change of

the mass of the control volume results .tbus, Eq. 7.9-36 yields d -(pVn 1 o)dt

or rate at which water is taken { onto the train

Forming the scalar product of the mac unit vector i gives

ti_ dt

f

pvz dV

1"'.(t)

+

f

pvz(V -

.A.(t)

The velocity of the water in the movin enter the tank at a velocity greater tha the tank must be w0 , and the first ter becomes

ic Balances-lnertial Effects

Chap. 7

Sec. 7.9

Moving Control Volumes and Unsteady Flow Problems

269

L.

(7.9-29) (7.9-30) the jet and not the area of the exit,

,---------------------------------,

(7.9-31) Fx-----1~

velocity vector as (7.9-32) the vane on the jet are Force exerted by the (7.9-33) vane on the jet Force exerted by the (7.9-34) vane on the jet

Water

of the jet on the vane is the scalar

1/í11111/í11111111/í//Í/111111111111111111111/11//lí11111/1//l/1111/í/í/1/i1/í/111//, Fl1. 7.9-3. Moving water scoop.

1 -cosO)

(7.9-35)

the vane and no power is generated. upon itself-the power generated

the mass of the control volume results only from the accumulation of water; -thus, Eq. 7.9-36 yields

we can see that a rather large to reduce Eq. 7.9-27 to the ~n·•,..,.·nt that visualizing the problem leads to an easy solution.

d -(pVa o)- pw0 hb =O

dt

lvn••nt1••11

used by a fast-moving train to pick tracks. We would like to know the the drag on the train resulting from b and the thickness of the water film blem as observers fixed in space and the macroscopic mass balance - w) • ndA =O

(7.9-36)

to the velocity of the control volume v = O. The time rate of change of

(7.9-37)

•

or rate at which } water is tak~n = pwJzb { onto the tram

Forming the scalar product of the macroscopic momentum balance with the unit vector i gives

ti_ dt

J

pv., dV

+

7".(1)

J

pv.,(v - w) • n dA =

..t.(t)

Ji •~~>dA

(7.9-38)

d.(t)

The velocity of the water in the moving tank is not constant because it must enter the tank ata velocity greater than w0 ; however, the average velocity in the tank must be w0 , and the first term on the left-hand side of Eq. 7.9-38 becomes -d dt

J

pv., dV = -d (p Va 1 oW0) dt

7".(1)

(7.9-39)

270

Macroscopic Balances-lnertial Effects

Chap. 7

Sec. 7.9

Moving Control Volumes and U

The momentum flux term is identically zero, since either V=W

or

v.,

=o

everywhere on the control surface. Ifwe neglect the small increase in pressure in the water film, the net stress acting on the system is the result of only the applied force F.,, and Eq. 7.9-38 yields

,-----------------1 1 1

:

d - (p vH.oWo) = F"' dt

Compress

1

(7.9-40)

Because w0 is a constant, we may use the mass balance to obtain the final result,t (7.9-41)

This force must be equal and opposite to the force the water exerts on the scoop. It is exerted on the curved portion of the scoop where the water is suddenly accelerated from rest to a velocity somewhat greater than w0 : For a train moving ata high velocity, inertial effects probably predominate, and Eq. 7.9-41 is a reasonable solution. On the other hand, velocity gradients at the leading edge of the scoop will be large and viscous effects could be important if a very accurate answer is desired. The accelerating water tank

Another application of the mass and momentum balances for moving control volumes is the accelerating water tank shown in Fig. 7.9-4. Inasmuch as the jet leaves the tank in a direction parallel to the velocity of the control volume, we will have no difficulty in determining the momentum flux term; however, if the jet issued from the tank in sorne direction other than that shown, the problem would be a great deal more complicated. The tank contains a supply of compressed air regulated so that the flow rate leaving the tank through the nozzle is constant; thus,

~ (pVH.o) = -m

dt

(7.9-42)

where VH

m= a positive constant = the volume of water in the tan k •0

t This result can be deduced intuitively from the linear momentum principie since the force Fz is just equal to the time rate of change of momentum; i.e., (pw 0 )(w0hb) = momentum per unit volume x volume per unit time.

Fig. 7.9-4. Accelen

We assume:

l. No friction occurs between the 2. The velocity profile in the emerg1

The mass of the control volume, exclud · the water in the control volume at macroscopic mass balance gives p(v- w) • nln where A 0 is the area of the nozzle. momentum balance with the unit vecto effects in the surrounding air, we get

~t

f

pv., dV

+

f

pv.,

r.(tJ A, We now assume that the velocity e is w. This is not strictly correct beca nozzle; however, if the amount of fiui through the nozzle is small, it may be ~ use of Eq. 7.9-43 and the above assum d - (Mw 0 dt

+ pVH oWo 2

Sec. 7.9

271

Moving Control Volumes and Unsteady Flow Problems

zero, since either

neglect the small increase in pressure on the system is the result of only the

--------------------------------,1

1 1

1

1

1

1

1

:

Compressed oir

1

1

(7.9-40) i - - - - • wo(fl

the mass balance to obtain the final (7.9-41) of the scoop where the water is somewhat greater than w0 ." For effects probably predominate, and the other hand, velocity gradients at be large and viscous effects could be desired.

and momentum balances for moving tank shown in Fig. 7.9-4. lnasmuch parallel to the velocity of the control determining the momentum flux ter m; in sorne direction other than that deal more complicated. lJJauit:~~c:u air regulated so that the flow is constant; thus, (7.9-42)

from the linear momentum principie since the ofmomentum; i.e., (pw 0)(w0hb) = momen-

Fig. 7.9-4. Accelerating water tank.

We assume: l. No friction occurs between the wheels and the solid surface. 2. The velocity profile in the emerging jet is flat . The mass of the control volume, excluding the water, is M, and the volume of the water in the control volume at zero time is Vo. Application of the macroscopic mass balance gives (7.9-43) p(v - w) . nJnozzle Ao = m where A 0 is the area of the nozzle. Forming the scalar product of the momentum balance with the unit vector i and neglecting viscous and inertial effects in the surrounding air, we get

~t

f

f

(7.9-44) pvx dV + pvx(v - w) • n dA = O r .(t) A,~[z- !(zn

(7 .10-1 O)

This result indicates that the pressure increases continuously from (p ) 0 at the entrance ofthe channel to (p ) 0 p ( v., )~ at the end ofthe channel. Differentialmacroscopic balances present no serious difficulties provided we can recognize when they must be used.

+

PROBLEMS o 7-1.

Make use of the continuity equation and the material derivative to show that the stress equations of motion given by Eq. 7.2-3 are identical to those given by Eq. 4.4-13.

7-2. If

f

(7.10-4)

v · ndA = O

A.,(t)

and the flow is incompressible, show that the macroscopic mechanical energy balance may be written as

divide by A llx to obtain

(p)Jz+&z (p)J.,

A

2p(v.,)~( 1 _ ~) = d(p)

-p(v!>A. + p(v!>Aiz+&z =

Qo

Use of Eq. 7.10-8 allows us to write Eq. 7.10-7 in the form

(7.10-2) so that the flat velocity profile will be neglected. We now wish to determine the pressure as a function of x, and we therefore need to develop a differential-macroscopic balance in the x-direction. We do so by applying Eq. 7.2-11 to the control volume shown in Fig. 7.10-2. Neglecting viscous effects and dotting Eq. 7.2-11 with i, we get

=

(7.10-5) m the y- and z-

~

fG

pv

)

dV

+

JG J

pv

2

)

(v - w). n dA

A , (l)

fa(t)

=

(7.10-6)

2

J

V •

t(n) dA -

A , (t)

prpV • n dA

+ W-

Év

d"(l)

7-3. A hydraulic brake that consists of a cylindrical ram displacing fluid from a slightly larger cylinder is shown in Fig. 7-3. The velocity of the ram is specified as u0 and we wish to know the force F required to maintain this =-

d(p)

(7.10-7)

dx

motion. Neglect viscous effects and use both the momentum and the mechanical energy balances to obtain two answers. State carefully the assumptions that must be made in order to obtain a solution. Ans: The mechanical energy balance gives

(7.10-8) F

=

pu~( 1T Di/4) 2[(D 1 / D0) 2 - 1F

276

Macroscopic Balances-lnertial Effects

Chap. 7

Problems

Fi Flg. 7-l. The hydraulic ram.

7-4. A jet of water issues from a 1-in. I.D. nozzle at a velocity of 30 ft/sec and strikes a flat plate moving away from the jet at 17 ft/sec, as illustrated in Fig. 7-4. What force does the water jet exert on the plate?

7-6. A very simple device for pumping 7-6. The fluid to be pumped is (instead of the force supplied b sec 0 ndary fluid issuing through th place vía viscous effects, and the

"1''

17ft/sec

l

Gravity

Fig. 7-4

7-5. Water issues from a pipe imbedded in a concrete wall as shown in Fig. 7-5. lfthe velocity leaving the pipe is 20 ft/sec and the cross-sectional area is 1 in. 2, what are the components of the resultant force exerted on the pipe by the wall? Assume that the pressure drop from the wall to the end of the pipe is negligible. Ans: F.,= 0.72Ib1 , F11

=

2.70 lb1

Fig

energy balance cannot be neglectec be neglected if the flow is turbule pump can be obtained with the mo1 the pressure rise between points 1 nozzle area A 0 while maintainin¡ Assume that the secondary fluid is

7-7. The Egyptians reportedly used wa 7-7. The radius r 0 of the circular the bottom of the bowl. Determin' quired if the depth of the liquid is

F

Fig. 7-5 hydraulic ram.

I.D. nozzle at a velocity of 30 ft/sec and the jet at 17 ft/sec, as illustrated in Fig. exert on the plate?

• 7-6. A very simple device for pumping fluids is the ejector pump illustrated in Fig. 7-6. The fluid to be pumped is called the primary fluid, and momentum (instead of the force supplied by mechanical pumps) is supplied by the secondary fluid issuing through the nozzle. The momentum exchange takes place vía viscous effects, and the viscous dissipation term in the mechanical "f' 01

r

----

Area= A

:

--1--

17ft/sec

l

Grcvity

7-4

in a concrete wall as shown in Fig. 7-5. ft/sec and the cross-sectional area is 1 in. 2, force exerted on the pipe by the from the wall to the end of the pipe is

Fig. 7-6

energy balance cannot be neglected. However, the viscous surface forces can be neglected if the flow is turbulent and a satisfactory design of an ejector pump can be obtained with the momentum balance. Derive an expression for the pressure rise between points 1 and 2, and note the effect of reducing the nozzle area A 0 while maintaining the secondary flow rate Q0 constant. Assume that the secondary fluid is the same as the primary fluid. • 7-7. The Egyptians reportedly u sed water clocks similar to that illustrated in Fig. 7-7. The radius r 0 of the circular bowl is a function of z, the distance from the bottom of the bowl. Determine the functional dependence of r0 on z required if the depth of the liquid is to be a linear function of time.

Macroscopic Balances-lnertial Effects

278

Chap. 7

Problems

7-8. Ifthe valve on the pipe shown in Fi elbow will decrease suddenly. The 1 for very short times when the velo effects are negligible. Determine p Hint: Two scalar components of ti Ans: p*

=

p0 + pgL 1 L 2/(L 1 + L 2)

7-9. A standard method for determinin of measuring the pressure differenc Fig. 7-9. Derive an expression for

1"""""'""'-'-'-~"'"'~" Areo

= A1

'

Flg. 7-7

Fi

í

y

Lx

L1

and 'the pressure difference measu the relative merits of the momentu is much superior to the other for t

7-10. What is the resultant force at the C< the pipe in Fig. 7-10, because ofth velocity in the 2-in. l. D. section is 7 lbm, and the interior volume is : 7-11. The device illustrated in Fig. 7-11 mine the resultant horizontal forc leaving fluid streams. Given: PA PB

p•

1 1

-f---

gauge pressure at A gauge pressure at B gauge pressure at e L2 Flg. 7·8

=

15 psig 20 psig

=

o

=

The average velocity atA is 10 ft/s• Ans: Fx

=

-28 lb1

7-8. Ifthe valve on the pipe shown in Fig. 7-8 is opened rapidly, the pressure at the elbow will decrease suddenly. The problem is a complex one but can be solved for very short times when the velocity may be set equal to zero and viscous etfects are negligible. Determine p* for short times after the valve is opened. Hint : Two scalar components of the momentum equation must be used. Ans: p* = p0 + pgL 1 L 2 /(L 1 + L 2) t 7-9. A standard method for determining turbulent flow rates in conduits consists

of measuring the pressure ditference across a flow nozzle such as that shown in Fig. 7-9. Derive an expression for the volumetric flow Q in terms of A 1 , A2 ,

Areo = A1

A 2 =oreo

T

7-7

Fig. 7-9

and ·the pressure ditference measured by the manometer. Consider carefully the relative merits of the momentum and mechanical energy balances, for one is much superior to the other for this particular analysis.

L,

E--- -

L2

- --

7-1 O. What is the resultant force at the connection between the nozzle assemb1y and the pipe in Fig. 7-1 O, beca use of the water issuing from the jet? The average velocity in the 2-in. l. D. section is 1Oft/sec, the mass of the nozzle assembly is 7 lbm, and the interior volume is 38 in.3

- --

-.-¡

7-11. The device illustrated in Fig. 7-11 is used to mix two miscible fluids. Determine the resultant horizontal force on this device owing to the entering and Ieaving fluid streams. Given: PA = 70 lbm/ft 3 PB = 59 lbm/ft 3 gauge pressure at A = 15 psig gauge pressure at B = 20 psig gauge pressure at e = o The average velocity atA is 10 ft/sec; at B, it is 15 ft/sec. Ans: Fx = -28 lb1

280

Macroscopic Balances-lnertial Effecu

Chap. 7

Problems

L. Grovity

Flg. 7-10

rate is the same in both cases.) E funétion of O, and assume that the m over the surface of the exit as indi Fig. 7-13b for assumption (b). momentum flux (a) Ans: momentum flux (b)

=

1 ~l1 (1 -

2

8 0=1in. Flg. 7-11

7-12. Water is being pumped into a tank as shown in Fig. 7-12 ata rate of 8 ft 3/sec. lfthe diameter ofthe pipe is 6 in. and the inside diameter ofthe tank is 30 in., what is the force required to hold the tank in position for h equal 24 in .. excluding the gravitational force acting on the tank itself? 7-13. In solving macroscopic balance problems, we always attempt to place the suñace of the control vo1ume so that the streamlines are parallel and the momentum flux is easily determined. As an example of the error that may occur if the streamlines are not parallel, compute the momentum flux at the exit of a circular diverging section assuming (a) the stream1ines are straight lines converging at point O, and (b) the streamlines are parallel. (Note: the magnitude ofthe ve1ocity vector must be adjusted so that the volumetric flow

(b)

Fig.

<

lcrclScc,Dic Balances-lnertial Effects

Chap. 7

281

Problems

L.

t h(t)

-t

12in.

~~~--~=---~_l Fig. 7-12 7-10

rate is the same in both cases.) Express the ratio of momentum fluxes as a function of O, and assume that the magnitude of the velocity vector is constant over the surface of the exit as indicated in Fig. 7-13a for assumption (a) and Fig. 7-13b for assumption (b). Ans:

momentum flux (a) momentum flux (b)

-l¡ - J

1 cos2 O In (secO) 2 (1 - cos 0) 2

= -

0=2in.

7-11

shown in Fig. 7-12 ata rate of 8 ft 3/sec. the inside diameter of the tank is 30 in., the tank in position for h equal 24 in., on the tank itself? we always attempt to place the the streamlines are parallel and the As an example of the error that may compute the momentum flux at the assuming (a) the streamlines are straight the streamlines are parallel. (Note: the be adjusted so that the volumetric flow

"o"~-:_--1!__----.................... ,,

Fig. 7-ll

282

Macroscopic Balances-lnertial Effects

Chap. 7

• 7-14. Fluid distributing systems often consist of an array of perforated pipes providing discrete distribution at any number of points. A single perforation in a pipe is shown in Fig. 7-14 . lf the diameter of the hole is smaller than the pipe-wall thickness, the jet is very nearly perpendicular to the main stream. Assuming that the volumetric ftow rate through the perforation can be expressed ast Q 0 = A 0 v 2(p 1 - p 0 ) / p, where Q 0 is the volumetric flow rate through the hole, A 0 is the area of the hole, and p 0 is the ambient pressure, use the momentum and mechanical energy balances to obtain an expression for the pressure ditference, p 1 - p 2 • Neglect viscous etfects and explain wh y you obtain two ditferent answers .

Problems

Air at 8 psig

8ft _::::;>

Air at 11 psig

---.,.:e.-

F

1 in.

Fig. 7-14

v0 = 25 ftlsec 7-15 . Calculate the discharge rate from the upper reservoir to the lower reservoir in Fig. 7-15. Take the fluid to be water and neglect viscous etfects.

Ans: Q

=

-.-t_______,

0.18 ft 3 /sec.

7-16. Does a converging nozzle on a garden hose place the hose in tension or compression at the junction between the hose and the nozzle? Don 't guess-analyze. • 7-17. The inclined flat plate in Fig. 7-17 is moving toward the plane jet at 10 ft/sec, and the jet velocity (relative toa fixed reference frame) is 25 ft¡sec . lf the jet is 1 in. thick, determine the force per unit width exerted on the plate by the stream of water.

Ans: fx

=

F

-106 lb11ft

• 7-18. Two miscible turbulent streams of densities p1 and p2 are flowing in a wide, rectangular duct separated by a thin plate. Neglecting viscous effects, use the momentum balance to calculate the pressure change in the mixing region in terms of (v.)¡, (v. )2 , h1 , and h2 . Assume the pressure is uniform across the channel and the velocity profiles of the unmixed and completely mixed streams are flat.

t This is an approximate version of a formula to

be derived in Chap. 8.

Fl1. 7-18. Mixing

Balances-lnertial Effects

Chap. 7

of an array of perforated pipes number of points. A single perforation diameter of the hole is smaller than the nearly perpendicular to the main stream. rate through the perforation can be p, where Q0 is the volumetric flow rate hole, and p0 is the ambient pressure, use rgy balances to obtain an expression for eglect viscous effects and explain why you

283

Problems

Air ot 8 psig

8ft __::::>

Nozzle diometer

Air ot 11 psig

= 2in

Fig. 7-15

1 in.

v0 = 25 ft/sec the upper reservoir to the lower reservoir and neglect viscous effects.

--~-------------

hose place the hose in tension or compresand the nozzle? Don't guess-analyze. is moving toward the plane jet at 10 ft/sec, reference frame) is 25 n¡sec. If the jet is it width exerted on the plate by the stream Fig. 7-17

f densities p1 and p2 are ftowing in a wide, in plate. Neglecting viscous effects, use the he pressure change in the mixing region in ~ssume the pressure is uniform across the ~ of the unmixed and completely mixed

~rmula

to be derived in Chap. 8.

Flg. 7-18. Mixing of two turbulent streams.

284

Macroscopic Balances-lnertial Effects

Chap. 7

7-19. A laminar jet ofwater issues from a horizontal nozzle as shown in Fig. 7-19. The velocity profile of the jet as it emerges from the nozzle is parabolic, but the profile becomes flat sorne distance from the nozzle owing to viscous effects. Neglect the effect of gravity and the ambient air and derive an expression for the final area A 1 . Ans: A 1/A 0

Macroscopic j Visco

=! Nozzle oreo, A 0

--------Fluid jet

orea, A, Fig. 7-19. The laminar jet.

7-20. Use the mechanical energy balance (rather than Bemoulli's equation) to show that v1 = v2 = v3 for the flow shown in Fig. 7.8-4. 7-21. Solve the problem of the moving water scoop illustrated in Fig. 7.9-3 by using Eq. 7.9-8 instead of Eq. 7.2-10. 7-22. Derive Eq. 7.10-6 by integrating the x-direction stress equation of motion over the y,z-surface. Neglect viscous effects.

In Chap. 6, we presented a qualit showed that the time-averaged stress as the instantaneous stress equations equations could not be used to det because the dependence of the turb velocity is unknown. Empirical exp length equation, may be used to detc cannot be applied to any arbitrary gc In solving turbulent flow problem: are important, we will always require of this chapter is the formulation e experimental data and the applicatior viscous effects must be considered.

*8.1

Friction Factors-Deflnit

In examining the momentum at 7.2-10 and 7.3-27), we see that knowJ,

Jten> dA

a

""·(t)

2

Chap. 7

horizontal nozzle as shown in Fig. 7-19. from the nozzle is parabolic, but from the nozzle owing to viscous and the ambient air and derive an

Macroscopic Balances: Viscous Effects

8

than Bernoulli's equation) to show in Fig. 7.8-4. scoop illustrated in Fig. 7.9-3 by x-direction stress equation of motion effects.

In Chap. 6, we presented a qualitative description of turbulent ftow and showed that the time-averaged stress equations of motion too k the same forro as the instantaneous stress equations of motion. However, the time-averaged equations could not be used to determine time-averaged velocity profiles, because the dependence of the turbulent stress i on the time-averaged velocity is unknown. Empirical expressions, such as the Prandtl mixing length equation, may be used to determine velocity profiles; however, they cannot be applied to any arbitrary geometry and their usefulness is limited. In solving turbulent ftow problems under conditions where viscous effects are important, we will always require sorne experimental data. The objective of this chapter is the formulation of a consistent method of interpreting experimental data and the application of this result to sorne problems where viscous effects must be considered.

*8.1

Friction Factors-Definition

In examining the momentum and mechanical energy balances (Eqs. 7.2-10 and 7.3-27), we see that knowledge of the integrals

f

t, and p represents p. Splitting the area integral on the right-hand side of Eq. 8.1-2 into the area of entrances and exits A.(t) and the area of solid surfaces (both fixed and moving) A, + A,(t), we get

A,(t)

= -A

A,+

.llfo(t)

-;'"G(t)

·J

t'

Because A will be orthogonal to th so lid surfaces, F n is given by

Fn= n • T dA

(8.1-4)

A,+A,(t)

t In the design of airfoils, an engineer is naturally interested in a lift force, the force the fluid exerts on the solid in a direction perpendicular to A..

for conduits having a constant eros only for mathematically smooth co for real conduits.

Balances-Viscous Effects

Chap. 8

solutions to these equations. In neglected and these integrals were were obtained for a variety of r.•otru·t .. n to cases for which we knew In accounting for viscous effects, and the dissipation integral to and flow rates. the solid surfaces that it contacts. In in the component of this force in the be helpful to define a unit vector A flow (8.1-1) that the fluid exerts on the solid, which is especially convenient for Eqs. 7.3-7, 7.8-15, and 7.8-20 theorem for a scalar, we get w) • n dA

(8.1-2)

pcp]

Sec. 8.1

Friction Factors-Definition

For convenience, the total drag force is split into a form force and a friction force Frorm

=A

J ·f

n[(p- p0 )

+ pcp] dA

Frrlction =

-A

(8.1-6)

n • 't' dA

A,+A,(t)

The drag force contains the negative of the Iast two terms in Eq. 8.1-3, beca use it is defi.ned as a force the fluid exerts on the so/id, while the terms in Eq. 8.1-3 represent the force which the surroundings exert on the fluid. Because ofthe way in which the drag force has been defi.ned, the normal vector n in Eqs. 8.1-4 through 8.1-6 is directed from the fluid into the solid. The separation of the drag force into a form force and a friction force is made because the former depends mainly on inertial effects and is roughly proportional to pu~, while the latter depends on viscous effects and is roughly proportional to ¡.¿u0 • The word "form" results from the fact that this portion of the drag force is greatly influenced by the geometry of the solid surface, while the word "friction" indica tes that this portion of the drag force depends primarily on the area of the solid surface. The drag force for flow in closed conduits is generally represented and correlated in terms of a dimensionless friction factor f defi.ned by

+ n • 't'} dA

f= ___!j¿_

(8.1-7)

A*KE*

be kept in mind that for turbulent p. Splitting the area integral on area of entrances and exits A.(t) and moving) A. + A.(t), we get {-n[(p- p0)

+ pcp] + n · 't'} dA

(t)

dA+

f

(8.1-3) D•'t'dA

(8.1-5)

A,+A,(t)

where A* KE*

=

a characteristic area

= a characteristic kinetic energy per unit volume

There are many important processes dealing with the flow of fluids through conduits of constant cross-sectional area. For such conduits the characteristic kinetic energy per unit volume and characteristic area are generally given by (8.1-8) KE* = tp(z\) 2 A*

= t wetted surface

(8.1-9)

A,+A,(t)

force which the fluid exerts on the

f

] dA - A • n • 't' dA

Fn

(8.1-4)

A,+A,(t)

interested in a lift force, the force pc:nlncuJtai to A.

Because A will be orthogonal to the outwardly directed unit normal at the so lid surfaces, Fn is given by

= -A

-J n •

't'

dA

(8.1-10)

A,

for conduits having a constant cross-sectiona1 area. Equation 8.1-1 O is valid only for mathematically smooth conduits and thus is only an approximation for real conduits.

288

Macroscopic Balances-Viscous Effects

Chap. 8

Frl~ion

Sec. 8.1

Factors-Definltion

Assuming that the stress at the wall · the numerator of Eq. 8.1-12 as Z=O

-A

Z=L

.f

n • 'T dA= -A;

A,

The wall shear stress

---,\

D

7'0

was previous

and we may use this definition to ex)

!= This result simply indicates that the dimensionless wall shear stress. The ·momentum balance Fig. 8.1-1. Flow in a circular tube.

Asan example, we will apply these ideas to flow in the circular tu be shown in Fig. 8.1-1. The characteristic area is given by (8.1-11)

A*= !rrDL

and substitution of Eqs. 8.1-8, 8.1-1 O, and 8.1-11 into Eq. 8.1-7 allows us to express the friction factor as -A

F

f = __ D_ = _ A*KE*

J

f{

-n[(p- Po)+ p]

f

(!rrDL)(tp(v.i)

1}

(8.1 13a)

= nr = n 2 = n 8 = O at the tube wall

(8.1-13b)

n3 = n. =O

(8.1-13c)

A1 = Ar =O

(8.1-14a)

A2 = Ae =O

(8.1-14b)

A3 = A• = 1

(8.1-14c)

n[(p

Here, we have assumed that the v identica1, so that the momentum fl1 Forming the scalar product with A a Eq. 8.1-18 to FD =A

·f

""'·

and A has on1y one nonzero component.

+ n ·'

+ ""'·

(8.1-12)

For a mathematically smooth circular tube, there is only one nonzero component of the outwardly directed normal at the so lid surface, n1

O=

.4-¡

n • 'T dA

__:A~•!....,-_ _

It will be helpful at this point to a shown in Fig. 8.1-1. A rather care presented here, and future discussiom that these points are understood. F 8.1-1, the momentum balance reduc~

-n[(p- Po)

Dividing by A*KE*, we obtain th factor:

ic Balances-Viscous Effects

Chap. 8

Sec. 8.1

289

Frl~lon Factors-Definition

Assuming that the stress at the wall is independent of () and z, we may write the numerator of Eq. 8.1-12 as

-A

Z=L

.f

(8.1-15)

n · 't" dA= -Ai(n;T;i)TTDL= -TrzTTDL

A,

The wall shear stress

T0

was previously defined in Sec. 6.5 as at

r

=

(8.1-16)

r0

and we may use this definition to express the friction factor as

4r0

f =

(8.1-17)

tp(!\)2

This result simply indicates that the friction factor may be interpreted as a dimensionless wall shear stress. The momentum balance a circular tube.

to flow in the circular tu be shown given by (8.1-11)

It will be helpful at this point to apply the momentum balance to the flow shown in Fig. 8.1-1. A rather careful analysis of the stress terms will be presented here, and future discussions of the momentum balance will presume that these points are understood. For the control volume indicated in Fig. 8.1-1, the momentum balance reduces to

8.1-11 into Eq. 8.1-7 allows us to

O=

f

{-n[(p- p0)

+ ptf>] + D•'t"} dA

f-

+

(8.1-12)

n[(p- p0)

+ ptf>] dA+

J

n • 't"dA

(8.1-18)

A,

tube, there is only one nonzero at the solid surface, (8.1 13a)

Here, we have assumed that the velocity profiles at points 1 and 2 are identical, so that the momentum flux term in Eq. 8.1-3 is identically zero. Forming the scalar product with A and using the definition of Fn, we reduce Eq. 8.1-18 to

(8.1-13b) (8.1-13c)

Fn

=A

.J

-n[(p- p0)

+

ptf>] dA

+A

.J

n • 't" dA

(8.1-19)

A.

(8.1-14a) (8.1-14b) (8.1-14c)

Dividing by A* KE*, we obtain the following expression for the friction factor:

f =

i(

&'1

-

&'2) -+-

[A

J

n1 • 't" dA

Aentrance

J

+A

n 2 • 't" dA

.dexlt

J1A *KE* (8.1-20)

Macroscopic Balances-Viscous Effecu

290

Chap. 8

Friction Factors-Definition

and Eq. 8.1-20 may be written as

In obtaining this result we have used the definition

fJJ = ((p) - Po)

Sec. 8.1

+ p(cp)

(8.1-21)

f

D =

L ( &\

!-p(t\)2

and the area integral over the entrance and exit has been represented as two separate integrals. The stre~s terms are best treated in terms of index notation,

- !/.J -

1 A* Kl

If the structure of the turbulence does 1 stress terms will cancel and we are left

(8.1-22) where

Thus, the friction factor may also be in drop.

Remembering that the turbulent stress was given in Sec. 6.2 as -(t)

T;;

=

1

Dimensional analysis for the

1

-pV;V;

and the time-averaged viscous stress is given by f;;

Before examining the experimental of dimensional analysis to determine friction factor. Writing out Eq. 8.1-12

= !!:.(ov; + ov;) 2 OX;

OX;

we may write Eq. 8.1-22 as

A· (n ·

frictio~

~) = A;n;~(~:: + ~;) - pv;v~J

sr(-~~

(8.1-23)

oo

J=---

We now wish to evaluate this quantity at the entrance and exit where n has only one nonzero component n1 = nr =O } n2

=

n8

=

O

(8.1-24a) at the entrance and exit

n3 = n. = ±1

(t1TD1

Forming the dimensionless variables

Z=

(8.1-24b) (8.1-24c)

R=

Carrying out the summations indicated in Eq. 8.1-23, and making use of Eqs. 8.1-14 and 8.1-24, we get

~ (n · ~) = ± (~ oz av.

-~ - pv.v.

(8.1-25)

Ü= •

In obtaining this result, we have assumed that the velocity field is described by

Nne=

1\ •

Vr

=

Ve

v.

= O,

v;, v~, v;

-=1=-

1

-::j=.

)

O

allows us to express Eq. 8.1-29 as

O

The time-averaged velocity in the z-direction will be independent of z for all points downstream of the entrance region (i.e., in the region of onedimensional flow); thus,

ov. =o oz

(8.1-26)

!=

(~)(~)(Ll { o

This result indicates immediately that Reynolds number, but we must know

Balances-Viscous Efrects

Chap. 8

Sec. 8.1

291

Friction Factors-Definition

and Eq. 8.1-20 may be written as (8.1-21) exit has been represented as two

f = D(/JlJ L vr¡-

OJl)

vr2

1 (< -;-;) ( -;-;))7TD A*KE* pv.v. 1 - pv.v. 2

4

-

2

(8.1-27)

If the structure of the turbulence does not change from 1 to 2, the turbulent stress terms will cancel and we are left with the result

of index notation, (8.1-28)

(8.1-22)

Thus, the friction factor may also be interpreted as a dimensionless pressure drop.

given in Sec. 6.2 as

Dimensional analysis for the friction factor

Before examining the experimental values off, it will be wise to make use of dimensional analysis to determine what parameters will infiuence the friction factor. Writing out Eq. 8.1-12 gives

+ :;) -

pv;v;J

(8.1-23) (8.1-29)

the entrance and exit where n has (8.1-24a) entrance and exit

Forming the dimensionless variables

Z=_::

(8.1-24b)

D

(8.1-24c)

in Eq. 8.1-23, and making use of

R=.!:.

-,') -ov. - pv.v.

ü =!!L

D

(8.1-25)

•

the velocity field is described by

_ p(v.)D NRe---

# allows us to express Eq. 8.1-29 as will be independent of z for all (i.e., in the region of one(8.1-26)

2

1 ) JL/DJ "(- oܕ) f= (~)(!l.)(7T

L

NRe

o

o

oR

R d8 dZ

(8.1-30)

R=l/2

This result indica tes immediately that f depends u pon the ratio L/ D and the Reynolds number, but we must know upon what parameters oü.foR depends

292

Macroscopic Balances-Viscous Effects

Chap. 8

to complete the investigation. The continuity equation and the NavierStokes equations in dimensionless form are given by

V·U=O

au +u. vu = ae

(8.1-31)

-VfJJ

+

- 1- V' 2U

(8.1-32)

NRe

Sec. 8.2

Friction Factors-Experimental

lf the velocity profile is fully develope be independent of Z. lntegration wit yield the factor L/ D, which just cano entrance length for turbulent ftow is number and is on the order of 50 tub we may neglect the effect of the entra

The boundary conditions for this ftow may be expressed in both dimensional and dimensionless form as follows:

B. C. 1:

V=

0,

f=f(NRe)

*8.2

Friction Factors: Experi

(8.1-33)

U=O, B.C. 2:

V

=

V1(r,

U

=

U 1(R, (), e),

Z=0

(), t),

B.C. 3:

(8.1-34)

Z=O

z=L L

(8.1-35)

Z=D B.C. 4:

p = Pl = fJJ¡,

[JJ

pcp¡,

z= O

(8.1-36)

z=o

Although we cannot solve Eqs. 8.1-31 and 8.1-32 for U and fJJ, we can state that these two dependent variables are functions of the independent variables (R, (), Z, e), the parameters in the differential equations (NRe), and the parameters in the boundary conditions (L/ D). Thus, we can write

U=

u(R, (), Z, e, NRe• ~)

(8.1-37a)

=

[JJ(R, (), Z, e, N Re• ~)

(8.1-37b)

[JJ

Ifthe time-average ftow is steady, Ü will be independent ofthe dimensionless time e, and the gradient at the wall may be expressed as

oü. 1 = oR R=l / 2

function of

(e, Z, NRe• .!:) D

(8.1-38)

In Eq. 8.1-30, the () and Z dependence will be eliminated by integration, and the functional dependence of the friction factor is (8.1-39)

If the friction factor is measured u values of L/ D, we soon discover that represented by a single plot off vers performed the first experimental ~tu we immediately wonder, "Where dtd As is often the case with a mathematic the boundary conditions. In Eq. 8.1 zero at r = D/2, and while this seems true only for a mathematically smootl written

B.C. 1':

V=

O,

r

where the roughness function e(O, z) l that describes the roughness of the co magnitude of e is of the order of 0.001 "small effect," but for turbulent ftow i do not lead to small effects." This ex~ it is a classic example of a plausib erroneous conclusion. The early work by Darcy and other experimental study of the friction facto tubes. The rough tubes were obtaine< possible with sand of a definite grain si Nikuradse obtained a roughness ce encountered in commercially availab plotted as the friction factor versus th results are characterized by the relati•

1. H. P. G. Darcy; for an account, see 1 (New York: Dover Publications, Inc., 1963), 2. J. Nikuradse, VDI-Forschungsh. 361, A original paper is available as NACA Tech. M

Chap. 8

and the Navier(8.1-31) (8.1-32)

Sec. 8.2

293

Friction Factors-Experimental

lf the velocity profile is fully developed at Z = O, the velocity gradient will be independent of Z. Integration with respect to Z in Eq. 8.1-30 will then yield the factor L/ D, which just cancels the multiplying factor, D/ L. The entrance length for turbulent flow is nearly independent of the Reynolds number and is on the order of 50 tu be diameters. For large values of L/ D, we may neglect the effect of the entrance region and write

be expressed in both dimensional

F or large L/ D

*8.2

(8.1-40)

Friction Factors: Experimental

(8.1-33)

z=O Z=O Z=

(8.1-34)

L L

Z=-

(8.1-35)

D

z=O

(8.1-36)

8.1-32 for U and ~. we can state ~mct1011s of the independent variables equations (NRe), and the D). Thus, we can write (8.1-37a)

(8.1-37b) independent of the dimensionless be expressed as

(o. z, NRe• ~)

(8.1-38)

be eliminated by integration, and factor is (8.1-39)

If the f¡iction factor is measured using a variety of pipes all having large values of L/ D, we soon discover that the experimental values off cannot be represented by a single plot off versus N Re as Eq. 8.1-40 implies. Darcy 1 performed the first experimental studies leading to this conclusion, and we immediately wonder, "Where did our dimensional analysis go wrong ?" As is often the case with a mathematical analysis, the trouble líes with one of the boundary conditions. In Eq. 8.1-33 we specified that the velocity was zero at r = D/2, and while this seems like quite a reasonable statement, it is true only for a mathematically smooth tube. In actual fact, we should have written B.C. 1':

V=

0,

r

=

lD + e(O, z)

(8.2-1)

where the roughness function e(O, z) is sorne unknown function of O and z that describes the roughness of the conduit. For most commercial pipes the magnitude of e is of the order of 0.001 of the pipe diameter. Surely this is a "small effect," but for turbulent flow in pipes we will find that "small causes do not lead to small effects." This example should be remembered well, for it is a classic example of a plausible intuitive hypothesis leading to an erroneous conclusion. The early work by Darcy and others led Nikuradse 2 to carry out a detailed experimental study of the friction factor for smooth and artificially roughened tubes. The rough tubes were obtained by covering the surface as tightly as possible with sand of a definite grain size glued onto the wall. In this manner, Nikuradse obtained a roughness considerably more uniform than that encountered in· commercially available pipe. The experimental results are plotted as the friction factor versus the Reynolds number in Fig. 8.2-1. The results are characterized by the relative roughness parameter, ef D, where e l. H. P. G. Darcy; for an account, see H. Rouse and S. Ince, History of Hydraulics (New York: Dover Publications, Inc., 1963), p. 170. 2. J. Nikuradse, VDI-Forschungsh. 361, Ausgabe B, Band 4, 1933. A translation of the original paper is available as NACA Tech. Mem. 1292, 1950.

294

Macroscopic Balances-Viscous Effects

12xl0-~

1 9 8

4

1

•

1~ \

c/0=9 . 86xl0~4

r. '

o

= 1.98xlo- 3 • 3 . 97xlo- 3 6

o

•

=8.34xl0·3 = 1.63xlo- 2 3.34xlo- 2,

.

X

l}/0 =

Chap. 8

~-8~ JJ .-= 4

(commerciolly rough)_

Sec. 8.2

Friction Factors-Experimental

This equation is an indication that the to affect severely the parabolic veloc' pipes (say, efD > 0.10), we could ce laminar friction factor; however, this e If we use Eq. 8.1-17, in conjunction wi stress, we obtain 8J.L (v.)

To=--

Eq. 5.6-~ ~~ ~

3

1\

2.5

\

D

_... _. o;li

Eq. 8.2 /8 -. . loo...

2.0xlo-2

r'"'

!"" llfti!ll

.... fll

~rx::­ .5,~~~~--~~~~--~~~~,~~~m-.~~~Krr~~-~~

indicating that the drag force is propo and the velocity. Under these conditio the friction force; hence,

1

FD

· 2 ~~~--+---~~~--+-~~~~~-.7 8*.2~-~9~-~i~~a~~~

1.0 x lo-2LW,..L---J-__-!---J-!--L--+--+-+~~~~:=;:=f:¿:L.:~ 4 6 8103 2 4 6 8 104 2 4 6 8 105 2 4 6 8 106 2

=

Frriction•

Critica! region

NflA>=u0 0/v

Fig. 8.2-1. Friction factor versus Reynolds number for sand-roughened tu bes.

is the height of the sand grain. Because all the tu bes were roughened in the same manner-i.e., the sand grains were placed as close together as possiblethe single parameter ef D was sufficient to characterize the roughness. The student should give sorne serious thought to these results, for they are a clear indication that we can easily perform an apparently rigorous analysis, which in fact neglects important effects. lt is quite common in engineering analysis to treat surfaces as mathematically smooth and impose boundary conditions of the type given by boundary condition l. Generally, such an approach is satisfactory (as it is for laminar flow); however, for turbulent flow in tubes, a small effect such as wall roughness can lead to a striking effect in the pressure drop-flow rate relationship. In the absence of experimental studies, the average investigator is not likely to formulate the boundary condition as indicated by boundary condition 1'; thus, we must be constantly on the alert for flaws and limitations in a mathematical analysis. Several aspects of this friction factor plot must be discussed before we continue with the results for commercial pipes. Laminar region

We note first that all the data in the laminar flow region fall on a single line, the equation for which was derived in Chap. 5,

f =

64

N Re

(8.2-2)

Pipe roughness is not a factor number, and the transition to turb critica! Reynolds number may depen the pipe, upstream conditions such as spurious disturbances such as buildin there is a lower bound for the criti however, recent studies 3 have shown t to Reynolds numbers of 2 X 104 by flow free of disturbances. In practic disturbance and the transition to tur region between NRe = 2100 and NRe · it is here that the transition is comple and in the critica! region the flow turbulent regimes. Transition and rough-pipe regions

For smooth tubes there is only o the two previously mentioned-i.e., flow existing for NRe > 4 X 103 • H the flow continues to change as the R1 the relative roughness is less than O. smooth pipe curve for a region in w 3. R. J. Leite, J. Fluid Mech., 1959, 5:81

ic Balances-Viscous Effects

Chap. 8

Sec. 8.2

295

Friction Factors-Experimental

This equation is an indication that the values of ef D were never large enough to affect severely the parabolic velocity profile. By using extremely rough pipes (say, ef D > 0.10), we could certainly find an effect of ef D on the laminar friction factor; however, this case is not of great practica! importance. If we use Eq. 8.1-17, in conjunction with Eq. 8.2-2, to express the wall shear stress, we obtain 8JJ. (v.) To= - - D

for laminar fiow

(8.2-3)

indicating that the drag force is proportional to the product of the viscosity and the velocity. Under these conditions, the drag force results entirely from the friction force; hence, FD 2

=

Ffriction•

for laminar flow

(8.2-4)

Critica! region

=u0 0/v number for sand-roughened

all the tubes were roughened in the placed as close together as possibleto characterize the roughness. thought to these results, for they perform an apparently rigorous effects. lt is quite common in mathematically smooth and impose by boundary condition l. Generally, is for laminar flow); however, for as wall roughness can lead to a rate relationship. In the absence of is not likely to formulate the condition 1'; thus, we must be ns in a mathematical analysis. must be discussed before we

laminar flow region fall on a single in Chap. 5, 64 (8.2-2)

Pipe roughness is not a factor in determining the critica! Reynolds number, and the transition to turbulent flow starts at NRe = 2100. The critica! Reynolds number may depend strongly on the inlet conditions to the pipe, upstream conditions such as valves and bends, and the presence of spurious disturbances such as building vibrations. lt is well established that there is a lower bound for the critica! Reynolds number of about 2100; however, recent studies 3 have shown that laminar flow can be máintained up to Reynolds numbers of 2 x 104 by taking extreme care to ~eep the inlet flow free of disturbances. In practica! cases, there are numerous sources of disturbance and the transition to turbulent flow starts at Nfi.e = 2100. The region between NRe = 2100 and NRe = 4000 is called the critica! region and it is here that the transition is completed. The transition is ,not a sharp one, and in the critica! region the flow alternates between the laminar and turbulent regimes. Transition and rough-pipe regions

For smooth tubes there is only one more region of flow in addition to the two previously mentioned-i.e., the region of fully developed turbulent flow existing for NRe > 4 x 103 . However, for rough pipes the nature of the flow continues to change as the Reynolds number is increased. Provided the relative roughness is less than 0.01, the friction factor curve follows the smooth pipe curve for a region in which the rough pipe could be considered 3. R. J. Leite, J. Fluid Mech., 1959, 5:81.

296

Macroscopic Balances-Viscous Effects

Chap. 8

hydraulica/ly smooth. As the Reynolds number is increased, each curve eventually departs from the smooth pipe curve, progresses through a transition region, and finally reaches a constant value depending only on the relative roughness. This latter region will be called the rough-pipe region offlow, and the zone between the smooth-pipe and rough-pipe regions will be called the transition region.

Sec. 8.2

Friction Factors-Experimental

numbers, the major source of resista turbulent eddies with the wall rough compared to the form drag. In the 11 factor is constant, and Eq. 8.1-17 i1 proportional to the density times the T0

Low Reynolds number

=

(~) p(v.) ~n 2

Comparing Eqs. 8.2-3 and 8.2-5, we s momentum transport exist in pipe flo the other of inertial effects. Also, it i. the formula developed in Chap. 7 fo1 which indicated that force exert {on the pla

where v and A are the velocity and parison,. we might interpret Eq. 8.2force per unit exerted on the { protrusions

There is obviously sorne similarity be Commercial pipes

Flg. 8.2-2. Qualitative deséription of wall effects in turbulent ftow.

We can explain the existence of the transition and rough-pipe regions only in a qualitative manner; however, the explanation will help to explain the nature of turbulent pipe flow and should be valuable for this reason alone. At low Reynolds numbers the laminar sublayer is generally thicker than the average roughness e, and the wall shear stress To consists primarily of viscous stresses. As the Reyno!ds number is increased, the energy of the turbulent eddies increases, and they penetrate more closely to the wall. This process causes the thickness of the laminar sublayer to decrease, and the irregularities of the tube wall begin to protrude through the laminar sublayer, as Fig. 8.2-2 illustrates. When the turbulent eddies come into contact with these protrusions, the interaction of the fluid and solid changes. At high Reynolds

Pipes and tubes used in engineer smooth nor be considered rough in roughened pipes. The roughness in < it subject to direct measurement. H1 to any given pipe if the friction fac Moody 4 has made an extensive inve~ cial pipes, the results of which appe< widely used in engineering design Values of the relative roughness are pipe. When using these charts, an engi of e/ D are only approximate and th relative roughness for any given typ< probable variation injfor smooth t1 ± 10 per cent is to be expected for co 4. L. F. Moody, "Friction Factors for

Chap. 8

number is increased, each curve curve, progresses through a transivalue depending only on the relative the rough-pipe region offlow, and · regions will be called the

Sec. 8.2

297

Friction Factors-Experimental

numbers, the major source of resistance to flow is the interaction of the turbulent eddies with the wall roughness and friction drag becomes small compared to the form drag. In the rough-pipe region of flow, the friction factor is constant, and Eq. 8.1-17 indica tes that the wall shear stress is proportional to the density times the velocity squared. _

7

o-

(/_)

8

(

P v.

)2

In the rough-pipe region of turbulent flow

(8.2-5)

Comparing Eqs. 8.2-3 and 8.2-5, we see that two very distinct mechanisms of momentum transport exist in pipe flow-one a result of viscous effects and the other of inertial effects. Also, it is of interest to compare Eq. 8.2-5 with the formula developed in Chap. 7 for the force exerted by a jet on a plate, which indicated that force exerted} {on the plate

=

v2 A

(8.2-6)

P

where v and A are the velocity and area of the jet, respectively. For comparison, we might interpret Eq. 8.2-5 as force per unit area} exerted .on the {protrus10ns

(f)

= -

p(v.)

2

(8.2-7)

8

There is obviously sorne similarity between the two flows. Commercial pipes

in

and rough-pipe regions only will help to explain the be valuable for this reason alone. is generally thicker than the -r0 consists primarily of viscous the energy of the turbulent closely to the wall. This process to decrease, and the irregularities the laminar sublayer, as Fig. come into contact with these solid changes. At high Reynolds

l'lJJiamtuu•u

Pipes and tubes used in engineering practice can neither be regarded as smooth nor be considered rough in the same sense as Nikuradse's sandroughened pipes. The roughness in commercial pipes is not uniform, nor is it subject to direct measurement. However, a value of ef D can be assigned to any given pipe if the friction factor is known in the rough-pipe region. Moody 4 has made an extensive investigation of friction factors for commercial pipes, the results of which appear in Fig. 8.2-3. This particular chart is widely used in engineering design and is often called the Moody chart. Values of the relative roughness are given in Fig. 8.2-4 for various types of pipe. When using these charts, an engineer must keep in mind that the values of ef D are only approximate and there may be significant variations in the relative roughness for ány given type of pipe. Moody has indicated that the probable variation inJfor smooth tubing is ±5 per cent, anda variation of ± 1Oper cent is to be expected for commercial steel pipe. Corrosion may also 4. L. F. Moody, "Friction Factors for Pipe Flow," Trans. ASME, 1944, 66:671.

298

Macroscopic Balances-Viscous Effects

Chap. 8

Sec. 8.2

Friction Factors-Experimental Pipe diamete

04 06 0 .2 0.3 0.5· 0.81

1\

Riveted steel

1/

\,)

1"'-111. 11 ro. Concrete

"

N

~

l"i

;::.

"

Woo d~ 1'\stave

·~

-;;

·¡:¡

~ ~ 8 •

a:



e

en

'O

oe

>.

Q)

a::

o

...o .S "'...o t;

~

= o

]

,.;

,:.

•..

¡¡:

Pipe diomet1

Fig. 8.2-4. Relative roughr

J

JO¡:lO~

UO!¡:lDJ.:::I

cause large variations in ef D with tir analyzing pipelines which have been in time. There are two useful empirical form

5. C. F. Colebrook and C. M. White,"" Pipes with Age," J. /nst. Civ. Engrs. (London)

Balances-Viscous Effects

Chap. 8

Sec. 8.2

299

Friction Factors-Experimental Pipe dio meter in feet, O 5

0.1

0.4 0.6 0 .2 0.3 0.5 0.81

2

3456810

25 20

o.o 1a§gga!ffi~lnm~nmtEf1Ei±tlno.o7

0.04k: 0.03 ro. 0 .02

0 .06

[\

0 .05

40 60 100

300

Pipe diomeler in inches, O

Fig. 8.2-4. Relative roughness for commercial pipes.

cause large variations in ef D with time,5 and we must be cautious when analyzing pipelines which have been in operation for an extended period of time. There are two useful empirical formulas giving the friction factor in terms 5. C. F. Colebrook and C. M. White, "The Reduction of the Carrying Capacity of Pipes with Age," J. Inst. Civ. Engrs. (London), 1937, 7:99.

300

Macroscopic Balances-Viscous Effects

Chap. 8

of the Reynolds number for smooth tu bes. For Reynolds numbers less than 105 , Blasius6 gives the following equation

f =

0.316Nfi~ 4

For smooth tubes

(8.2-8)

Sec. 8.2

Reynolds number can then be determi1 assumption for f may be examined. If second trial must be made. In the follo type of calculation.

This equation is plotted in Fig. 8.2-1 and is in excellent agreement with Nikuradse's data for smooth tu bes provided N Re :::;:; 105. For higher Reynolds numbers, the Prandtl equation7 is useful:

JJ1 = 2.0 log (Jj N Re) -

0.8

For smooth tubes

(8.2-9)

This equation is also plotted in Fig. 8.2-1, and shows good agreement with the experimental data. The derivation of Eq. 8.2-9 is semitheoretical and should give satisfactory results for arbitrarily high Reynolds numbers. The Prandtl equation requires a trial-and-error solution to determine J; however, if the Blasius equation is used to obtain the first estímate, the final value can generally be determined with only one additional calculation. A useful equation for determining the friction factor in the transition and rough-pipe regions has been developed by Colebrook. 8 2.51 ) JJ = -2.0 log (e/DD + NRe,J] 1

For transition and rough-pipe regions

Friction Factors-Experimental

Determination of the flow rate for a pressure drop

Let us determine the volumetric fic 750-ft length of 4-in. diameter (nominal drop of 23.5lbcfin. 2 We shall assume d tional effects need not be considered. 1 D

=

L= 750ft

= 0.95 centi¡ p = 62.4 lbm/f

p.

t:J.p

= 23.5 lbr/in

.E_=

6. H. Blasius, "The Law of Similarity for Frictional Processes in Fluids," Forsch. Arb. lngr.-Wesen (Berlin, 1913), 131:361. 7. L. Prandtl, "The Mechanics of Viscous Fluids," in Aerodynamic Theory, F. W. Durand, ed. (Berlin: Springer-Verlag, 1935), 3:143. 8. C. F. Colebrook, "Turbulent Flow in Pipes, with Particular Reference to the Transition Region between the Smooth and Rough-Pipe Laws," J. lnst. Civ. Engrs. (London), 1938, 11:133.

0.0004

D

(8.2-10)

This empírica! relationship is, in fact, the basis for the curves shown in Fig. 8.2-3. lt requires a trial-and-error solution for f; however, such procedures are straightforward and this equation may prove useful if a digital computer is used for pipe-sizing calculations. Problems dealing with turbulent flow in pipes generally fall into two categories: determination of the pressure drop, given the flow rate, the physical properties of the fluid, and the geometry of the system (i.e., the length, diameter and relative roughness of the pipe); or determination of the flow rate, given the pressure drop, the physical properties of the fluid, and the geometry of the system. The first case is straightforward because the Reynolds number can be calculated and the friction factor determined immediately. The second type of problem requires that an initial guess be made for f, allowing us to calculate the average velocity by Eq. 8.1-28. The

4.03 in. (a

We may rearrange Eq. 8.1-28 to yield

and the Reynolds number becomes _ p(v,)D _ N Rep. For this example, we obtain 2

N

1 { [(2X23.5 lbr/in. ) Re - ,Jj (0.95 cen

--

x [ ( centipoise-ft se1

0.672 X 10-3 ]1>¡ or

4.

NRe=-

Now we need only find the value ofjJ equation and the relationship between The mínimum value off in this case

Balances-Viscous Effects

Chap. 8

For Reynolds numbers less than

(8.2-8)

Sec. 8.2

Reynolds number can then be determined, and the accuracy of the initial assumption for f may be examined. If the two values differ significantly, a second tria! must be made. In the following example, we consider this type type of calculation.

and is in excellent agreement with N Re ::;; 105. For higher Reynolds

For smooth tubes

(8.2-9)

2-1, and shows good agreement with of Eq. 8.2-9 is semitheoretical and "ly high Reynolds numbers. The solution to determine f; however, n the first estímate, the final value one additional calculation. friction factor in the transition and by Colebrook.8 2.51 )

NReJJ

For transition and rough-pipe regions

301

Frlctlon Factors-Experimental

Determination of the flow rate for a given pressure drop Let us determine the volumetric flow rate of water at 75°F through a 750-ft length of 4-in. diameter (nominal) commercial steel pipe for a pressure drop of 23.5lbr/in. 2 We shall assume the pipe is horizontal so that gravitational effects need not be considered. lt is given that

D = 4.03 in. (actual diameter) L= 750ft p

= 0.95 centipoise

p = 62.4 lbm/ft 3 l:lp = 23.5 lbrfin. 2 !_

D

(8.2-10)

= 0.0004

We may rearrange Eq. 8.1-28 to yield the basis for the curves shown in solution for f; however, such promay prove useful if a digital

1 (21:1p D) Jj ---¡;L

112

(vz) =

and the Reynolds number becomes in pipes generally fall into two drop, given the flow rate, the the geometry of the system (i.e., of the pipe); or determination the physical properties of the fluid, rst case is straightforward because and the friction factor deterproblem requires that an initial guess averagevelocity by Eq. 8.1-28. The Frictional Processes in Fluids," Forsch. Fluids," in Aerodynamic Theory, F. W.

N

- p(v.) D - _!._ (21:lpp Da)I/2 Re-

p

-Ji

p2L

For this example, we obtain 3

N Re=~ { [(2X23.5 lbr/in.~62.4lbm/ft X4.03 in.)

-J¡

3 ]

(0.95 centipoise) 2(750 ft)

(__.!!__) J}

2

x [ ( centipoise-ft sec ) (32.2 lbmft) 0.672 X 10-3 lbm lbr sec2 12 in.

1 2 '

or

_ 4.13 N Re-

X

Ji

104

43.

with Particular Reference to the TransiLaws," J. Jnst. Civ. Engrs. (London),

Now we need only find the value ofjfor ejD = 0.0004 that will satisfy this equation and the relationship between f and N Re given in the Moody chart. The mínimum value off in this case is 0.0 16, which yields a value for the

Macroscopic Balances-Viscous Effects

302

Chap. 8

Reynolds number of N

- 4.13 X 104 - 3 26 105 .Jo.016 - . x

Re -

Examination of the friction factor chart for s/ D 3.26 x 105 gives != 0.0178

=

0.0004 and N Re

=

Sec. 8.2

Friction Factors-Experimental

pipe sizes is available in handbooks, 9 of facing page) will be cited here for a The variation in inside diameter is s designing any real system. Throughout dimensions of any system are assumed t~ Flow in closed conduits of noncircul

which in turn yields the second approximation for the Reynolds number, 4.13 X 104 = 3.08 X 105 .Jo.0178 Returning to the chart, we see that a Reynolds number of 3.08 x 105 gives a friction factor of 0.0178 for this particular relative roughness. Having established the Reynolds number, we may now determine the volumetric flow rate, Q. NRe

=

2

(7T~ )(v.) = (NR~:D/l)

Q=

= [(3.08 x 105)(3.14)(4.03 in.)(0.95 centipoise)J (4)(62.4Ihm/fe) 3

X

=

0.672 X 10- lbm ) ( ft )] [( centipoise-ft sec 12 in.

0.83 fejsec

Nominal pipe diameters

Closed conduits of noncircular although they have received much les laminar flow, each noncircular conduit problem, for the result from one case e and Katz11 list severa! solutions of t dimensional laminar flow in noncircula Bateman12 give a more thorough discu severa! solutions. Since an engineer i. pressure drop-flow rate relationship, t suggested ·by Gaydon and NuttaP 3 is of odd-shaped conduits. The method re effort to determine the pressure drop-f per cent. Turbulent flow has been studied by rectangular and circular notched condUJ in Fig. 8.2-5. For each conduit, the fr·

J=At

It is unfortunate but true that the nominal pipe diameter is not the actual pipe diameter; we must take this into account when specifying pipe diameters. In the previous example, the actual diameter of a 4-in. commercial steel pipe was 4.03 in. The difference between the nominal and actual diameters depends upon the diameter and the so-called schedule number of the pipe. The latter is simply a measure of the wall thickness and is therefore an indication of the pressure that the pipe can safely withstand. Information on Nominal Pipe Size, in.

Outside Diameter, in.

6

6.62

Schedule No.

SS lOS 40ST,40S

soxs, sos 120 160

XX

Inside Diameter, in. 6.41 6.36 6.06 5.76 5.50 5.19 4.90

where A* and .KE* are given by Eqs number is given by 4 NRe=9. Chemical Engineers' Handbook, 4th ed. lnc., 1963), Sec. 6. 10. L. S. Marks, ed. Mechanical Engineers Book Company, Inc., 1958). 11. J. G. Knudsen and D. L. Katz, Flu1 McGraw-Hill Book Company, Inc., 1958), Cl 12. H. L. Dryden, F. D. Murnaghan, a Dover Publications, Inc., 1956), Chap. 2. 13. F. A. Gaydon and H. Nuttal, "Viscou Cross Sections," Trans. ASME Ser. E 81, 19 14. L. Schiller, Z. Angew. Math. Mech., 1 15. J. Nikuradse, Ingr.-Arch., 1930, 1:3 16. L. Prandtl, Proc. lntern. Congr. Appl.

Sec. 8.2

ef D =

0.0004 and N Re =

303

Friction Factors-Experimental

pipe sizes is available in handbooks, 9 - 10 and only a single example (see foot of facing page) will be cited here for a 6-in. diameter steel pipe. The variation in inside diameter is significant and must be considered in designing any real system. Throughout the remainder of this text the nominal dimensions of any system are assumed to be identical to the actual dimensions. Flow in closed conduits of noncircular cross section

v,. .... ~.UV'U 4

l0

= 3.08

for the Reynolds number, X

105

Reynolds number of 3.08 x 105 gives particular relative roughness. Having may now determine the volumetric

Closed conduits of noncircular cross section are occasionally used, although they have received much Iess attention than circular tubes. For laminar flow, each noncircular conduit must be Iooked upon as a separate problem, for the result from one case cannot be extended to others. Knudsen and Katz11 Iist severa! solutions of the Navier-Stokes equations for onedimensional laminar flow in noncircular conduits. Dryden, Murnaghan, and Bateman12 give a more thorough discussion of these problems and present severa! solutions. Since an engineer is generally concerned with only the pressure drop-flow rate relationship, the approximate method of solution suggested by Gaydon and NuttaP 3 is of great value in obtaining solutions for odd-shaped conduits. The method requires only a small computational effort to determine the pressure drop-flow rate relationship to within a few per cent. Turbulent flow has been studied by severa! investigators14- 16 for triangular, rectangular and circular notched conduits; the experimental results are shown in Fig. 8.2-5. For each conduit, the friction factor is defined by

FD f= A*KE* nominal pipe diameter is not the actual specifying pipe diameters. ,.,uu....,... of a 4-in. commercial steel pipe nominal and actual diameters so-called schedule number of the pipe. wall thickness and is therefore an indican safely withstand. Information on l""~·v .. •cu when

Schedule No.

lnside Diameter, in.

5S lOS 40ST,40S 80XS, 80S 120

6.41 6.36

160 XX

6.06 5.76 5.50 5.19 4.90

(8.2-11)

where A* and KE* are given by Eqs. 8.1-8 and 8.1-9, and the Reynolds number is given by _ 4p(v,)Rh N Re-

(8.2-12)

¡;, 9. Chemical Engineers' Handbook, 4th ed. (New York: McGraw-Hill Book Company, lnc., 1963), Sec. 6. 10. L. S. Marks, ed. Mechanical Engineers Handbook, 6th ed. (New York: McGraw-Hill Book Company, Inc., 1958). 11. J. G. Knudsen and D. L. Katz, Fluid Dynamics and Heat Transfer (New York: McGraw-Hill Book Company, lnc., 1958), Chap. 4. 12. H. L. Dryden, F. D. Murnaghan, and H. Bateman, Hydrodynamics (New York: Dover Publications, lnc., 1956), Chap. 2. 13. F. A. Gaydon and H. Nuttal, "Viscous Flow Through Tubes of Multiply Connected Cross Sections," Trans. ASME Ser. E 81, 1959, 4:573. 14. L. Schiller, Z. Angew. Math. Mech., 1923, 3:2. 15. J. Nikuradse, Ingr.-Arch., 1930, 1 :306. 16. L. Prandtl, Proc. Jntern. Congr. Appl. Mech., Zürich, 1927.

304

Macroscopic Balances-Viscous Effects

'

0. 1

""

O. 1 0. 08

["-.

O. 06

1"" O. 01

O. 01 O. 008

r

~

lb..

,_'

o N1kuradse . Re f. 15

O. 004

O. 001

'<

.....

•

Schl11er , Ref . 14 1

4.0

_)---¡

1

2.0

Eq 8. 1-8

~ 'So ~o· ·o

~ 1--:-:::-.,_ equ 11 atera¡ J \

Co

./'

-o-_ ~-1f

kC!i~ ~~!~

""" ~"~''''"'•'lJt~

O. 006

10.0 8.0 6.0

J 6-38 Eq . 5.

"k~ " "'\ ~

Chap. 8

1

oy

1 r.... ~!r le._1

-o-:. :!!_tched

--.,__In.

"'

r-<

1"--

~

! f

~ r-<

1--

"

t'-....: ~ ~"""-

~¡...

1.0 0.8 o. 6

r--

o. 4

o. 2

o.1

10

2

4

6 81Q2

2

4

6

11103

2

Fig. 8.2-6. Drag coefficien

1 1

1

The characteristic area A* for imme projected area in the direction of fio Flg. 8.2-5. Friction factors for noncircular conduits.

A*-

where Rh is the hydraulic radius. The results illustrated in Fig. 8.2-5 indicate excellent agreement with the Blasius equation; in addition, the values in the laminar fl.ow region fall on the line given by

TTD2

4 '

A*= DL,

The characteristic kinetic energy is ta velocity far removed from the imme and 8.2-15, the drag force acting on

(8.2-13)

FD=eD(

These results must not be construed as verification of the friction factor chart for all shapes of closed conduits, for it is well known that laminar ftows may deviate significantly from Eq. 8.2-13. However, we may conclude that if the shape of the conduit is not far removed from circular the friction factor chart can be used to obtain :.atisfactory results. Drag coefficients for spheres and cylinders

The friction factor for solid bodies immersed in a flowing fluid is traditionally referred to as a dimensionless drag coefficient and defined by

e D-

__fJ¿_

A*KE*

(8.2-14)

and the drag force acting on a cylindl

FD= eD

Experimental values of eD for spher Dimensional analysis would again 1 function of Reynolds number and reJa has not been studied extensively bec with reducing the drag on immersed 1 of accomplishing this end.t

t This is not always the case, for somet drag. This phenomenon, along with the c detail in Chap. 11.

10.0 8.0 6.0

Chap. 8

4.0 2.0 Co

1.0 0 .8 0.6

"'

~

' ""'

~"

....

Circular cylinder

"

0.4

Sphere

_....,

t---

0 .2

o.1

10

\.11 2

4

6 81Q2

2

4

6 8¡Q3

2

4

6 8

104

2

4

6 8

105

2

4

6 8¡Q6

NRe

Fig. 8.2-6. Drag coefficients for spheres and cylindl:rs. 6

8

The characteristic area A* for immersed bodies is generally taken as the projected area in the direction of ftow; thus, for noncircular conduits.

A* =

results illustrated in Fig. 8.2-5 indicate · ; in addition, the values in the 64

(8.2-13)

verification of the friction factor chart is well known that laminar ftows may However, we may conclude that if the from circular the friction factor chart

2

D

for a sphere

(8.2-15a)

A* = DL,

for a cylinder

(8.2-15b)

7T

4 '

The characteristic kinetic energy is taken to be !pu;,, where U 00 is the fluid velocity far removed from the immersed body. On the basis of Eqs. 8.2-14 and 8.2-15, the drag force acting on a sphere is (8.2-16) and the drag force acting on a cylinder of length L and diameter D is (8.2-17)

immersed in a flowing fluid is tradidrag coefficient and defined by

Experimental values of eD for spheres and cylinders are shown in Fig. 8.2-6. Dimensional analysis would again lead us to the conclusion that en is a function of Reynolds number and relative roughness. The effect of roughness has not been studied extensively beca use engineers are generally concerned with reducing the drag on immersed bodies; keeping them smooth is a means of accomplishing this end.t

(8.2-14)

t This is not always the case, for sometimes increasing the roughness can decrease the drag. This phenomenon, along with the curves shown in Fig. 8.2-6, will be discussed in detail in Chap. 11. 305

306

Macroscopic Balances-Viscous Effects

Chap. 8

Sec. 8.3

Pipeline Systems

For Reynolds numbers less than 1.0, the inertial terms in the NavierStokes equation can be neglected under certain circumstances, which leads to a linear set of equations first solved by StokesY The drag coefficient for this condition is - 24 (8.2-18) en-

T

NRe

and the drag force acting on the sphere is given by

FD =

31TflU 00 D

2 00'

(8.2-19)

Equation 8.2-19 is known as Stokes law and has been used extensively by engineers to describe the motion of solid spheres moving through gases and Iiquids. The discussion here has been quite brief; however, a detailed treatment of flow around immersed bodies is given in Chap. 11 .

l_ 2'

8.3

Pipeline Systems

One ofthe standard problems that an engineer may encounter is the design of a piping system such as the one illustrated in Fig. 8.3-1. In the design of a new system, the flow rate and the physical properties of the fluid are generally given, and the engineer must determine the pipe size and the power requirement for the pump. The final solution will naturally be subject to certain economic constraints beyond the scope of this text. The main economic problem hinges on the fact that the cost of pumps and pipes depend on the size and that the final design of the system should satisfy the physical conditions at a minimum cost. The mechanical energy balance is the most suitable macroscopic balance w use in solving pipeline problems, because the troublesome viscous effects are nicely lumped into a positive dissipation term. If we restrict ourselves to steady flow, fixed control surfaces at entrances and exits, and incompressible fluids, the mechanical energy balance (Eq. 7.3-27) takes the form

f

Net outftow of kinetic energy

2

f -f

(lpv )v • n dA=

T

20'--

Pump flg. 8.3-1. 1

If we now take the positive z-coord gravity vector, the potential energy fl

In addition, we will consider control entrances and exits is normal to the e

f V·

t(n)

dA

A.

prf>v·ndA

A.

dA=

A.

Rate of work done on the system at the entrances and exits Rate of work done by gravitational forces, or the net outftow of potential energy

V • t n~ b

re

00

5.3D ----

99% -

laminar

J. '

turbulent

00

To gain sorne familiarity with the orders of magnitude of these times, we could say for turbulent flowt

/oo = O(0.005) task beca use of the complex varianumber. A numerical solution make an interesting class project),

D (v.)

= 0 (l ft) = O(1 ftjsec)

Under these conditions, the time would be (8.4-11)

199

%

=

= (8.4-12) will not be valid at short times when if the major portion of the transient region of flow, the approximation of Eq. 8.4-12, and application of the

0 =0

(8.4-13)

(8.4-14)

o[

(5.3)(1 ft) min (0.005)(1 ftjsec) 60 sec

J

0[1.77 min]

From this order of magnitude calculation we see that the transient time only becomes significant for large diameter tubes and low flow rates. Remember that this analysis assumed that the Iength of the pipe was long compared to the entrance length and is therefore not valid for short pipes. Oscillating systems-the U-tube manometer Various fluid systems are subject to upsets which may give rise to oscillatory motion. Actual processes may become extremely complex, and large analogue or digital computers may be required to determine the fluid motion even approximately. The purpose of studying the U-tube manometer is to gain sorne insight into the important aspects of oscillatory systems. In

t Read 0(0.005) as "on the order of magnitude of 0.005."

324

Macroscopic Balances-Viscous Effects

Chap. 8

Sec. 8.4

Unsteady Flow in Closed Conduits

addition, this example is of sorne interest because it can easily be studied experimentally, thus introducing the student to the errors incurred in an approximate solution. The U-tube manometer to be analyzed is shown in Fig. 8.4-2. Once again, the differential equations describing the motion are too complex to be solved, and the mechanical energy balance is required. We start with the complete equation and note that W is zero to obtain

For the moving control volume i'a(t) that the gas-liquid interface is horizon 8.4-2. Actually, it is not because the interface will be as shown in the inset. Considering Eq. 8.4-15, we note th1

~t J(!pv

and the convective transport term on energy balance is identically zero. If vi! are neglected, we may assume that

2 )

J

dV +

i'"0 (t)

2

(!pv )(v - w) • n dA

A,(t)

=

J

V•

t(n)

dA-

A,< O

J

pgzv • n dA -

Ev

(8.4-15)

.91' aW

'

__("' Gas pressures suddenly equilibrated

( - - - - -- - -

V=W

V•

~n> hert =

and the first term on the right-hand assumption is open to question, becaus depends upon whether it is advancing solve the detailed problem associated v assume it is ftat and make the assump

T ho

_________________ l __ _

Final position of manometer fluid

1 - - - i + - - lnitial pasition of manometer fluid

1

z=O

¡

If the flow in the manometer were t 8.4-17 could be replaced by average term could be expressed as in the pr ftow in a manometer is likely to be la mate analysis for laminar flow. Assur one-dimensional, we write v(r, t)

Manometer liquid

Advancing liquid

"''""' '"'"'"" 11111 Liquid

Receding liquid

u

Fig. 8.4-2. U-tube manometer.

Liquid

= 2(v)l

Here, v(r, t) represents the component Expressing the transient velocity profil is a common technique (and one whi obtaining approximate solutions to eng method depends on the variations witl velocity profile is never far from par we shall explore the validity of this as! The average velocity (v) in Eq. 8.4 defined as positive when the flow is fr< V•

n heft

=

and the position of the two interfaces Z

= -h heft•

ic Balances-Viscous Effects

Chap. 8

because it can easily be studied to the errors incurred in an

Sec. 8.-4

For the moving control volume "Ya(t) we ·choose the fluid itself and assume that the gas-liquid interface is horizontal and well defined as shown in Fig. 8.4-2. Actually, it is not because the fluid adheres to the tube wall, and the interface will be as shown in the inset. Considering Eq. 8.4-15, we note that

f

~a(t)

on

V=W

dA-

325

Unsteady Flow in Closed Conduits

and the convective transport term on the left-hand side of the mechanical oenergy balance is identically zero. If viscous effects at the gas-liquid interface .are neglected, we may assume that pgzv • n dA-

E" (8.4-15)

""·(t) Gas pressures suddenly equilibrated

T ho

______ l __ _

¡

z=O

J of

V•

t.:n)

u L iquid

=

-V •

t

lrlght

(8.4-16)

and the first term on the right-hand side of Eq. 8.4-15 is also zero. This assumption is open to question, beca use the nature of the interface obviously depends upon whether it is advancing or receding. However, we must either solve the detailed problem associated with the moving interface, or we must assume it is flat and make the assumption indicated by Eq. 8.4-16 to obtain

~f (!pv

2

)

dV

= -

f

pgzv • n

dA - E,

(8.4-17)

Jll'G(t)

'j'"G(t)

If the flow in the manometer were turbulent, the various ve1ocities in Eq. 8.4-17 could be replaced by average velocities and the energy dissipation term could be expressed as in the previous example. However, oscillating flow in a manometer is likely to be laminar, and we must devise an approximate analysis for laminar flow. Assuming the flow to be quasi-steady and one-dimensional, we write v(r, t)

Receding liquid

hert

(~)]

= 2(v)[ 1 -

(8.4-18)

Here, v(r, t) represents the component of the velocity vector along the tube. Expressing the transient velocity profile in terms of the steady state solution is a common technique (and one which should be used with caution) for obtaining approximate solutions to engineering problems. The success ofthis method depends on the variations with time being small enough so that the velocity profile is never far from parabolic. After obtaining the solution, we shall explore the validity of this assumption. The average velocity (v) in Eq. 8.4-18 is a function of time, and will be defined as positive when the flow is from left to right. Since V•

n hen

=

-V •

n lrlght

(8.4-19)

and the position of the two interfaces is given by, Z

= -h hert>

z=

+h lrtght

(8.4-20)

326

Macroscopic Balances-Viscous Effects

Chap. 8

Sec. 8.4

Unsteady Fl 0 w in Closed Conduits

time 0 as,

We may simplify the area integral fn Eq. 8.4-17 to obtain

~1(4p7TL(v)2f" [1 - (~rJ r dr) = - 2pgh(v)7Tr~ - J

h ho

H=-

2

dV r-.(t) .

o

(8.4-21)

0 =t

From Table 7.3-1, we find the dissipation function to be

= 2~-'[av~ + or

(!r ovao + ~)r + (av.) oz 2

+ ~-'[ (::r) + (:~·)

r

2

]

r

+ ~-'[;(:;·) +

(::8)

r

= v, and reduces to

16J,L(v)2 (~)

2

=

where

In terms of the dimensionless variables,

In this case, there is only one component ofvelocity, v. 4>

d H + 2{3 dH d02 dG

2

8

+ ~-'[' :,(;) + ~(:;r)

and Eq. 8.4-24 reduces to

(8.4-22)

Substituting this result into Eq. 8.4-21 and carrying out the integration, we get (8.4-23)

B.C. 1':

H= 1,

B.C. 2':

dH =O d0 '

The solution of this second-order, ordin by assuming solutions of the form

H= which when substituted into Eq. 8.4-26 eme(m2

Because the time rate of change of height of the gas-liquid interface is equal to the average velocity,

dh - = (v) dt

2

r~ dt

2L

(8.4-24)

=

B.C. 2:

dh dt

h0 ,

=0

m1

= -({3 +

m2

=

-({3-

Provided m 1 and m2 are distinct (i.e., {3

This equation is to be solved subject to the boundary conditions h

+ 2{3m ·

Thus, m has two possible values, given

dh+ (6v)dh + (3g)h = 0 B.C. 1:

This equation will be satisfied, and E differential equation, if m is chosen so t m2

We may rearrange Eq. 8.4-23 to obtain

dt 2

+ 2{3¡

t=O

(8.4-25a)

=o

(8.4-25b)

H

=

C¡emle

However, if the roots of Eq. 8.4-30 are ,

t

It will be helpful to put the differential equation and the boundary conditions in dimensionless form. To do so, we define a dimensionless distance H and

H

=

(C1

+

26. L. R. Ford, Differential Equations (N Inc., 1933), p. 70.

Chap. 8

Sec. 8.4

time

. 8.4-17 to obtain

327

Unsteady Flpw in Closed Conduits

e as, h

H=-

-2pgh(v)1Tr~- J dV

=

ho

(8.4-21)

E>=t

7".(1)

!3i

~-u

and Eq. 8.4-24 reduces to

+

(~:·n

d2H

)r

+ ~t[;(~~·) + (~:8)

r

d82

+ 2{3 dH + H = O

(8.4-26)

dE>

where

fJ =

f2L r~ ~Ji

3v

In terms of the dimensionless variables, the boundary conditions are ofvelocity, v.

= v, and reduces to (8.4-22)

and carrying out the integration, we

B.C. 1':

H= 1,

e =O

(8.4-27a)

B.C. 2::

dH =O dE> ,

8=0

(8.4-27b)

The solution of this second-order, ordinary, differential equation is obtained by assuming solutions of the form

H= eme (8.4-23) of the gas-liquid interface is equal

(8.4-28)

which when substituted into Eq. 8.4-26 gives

em 9 (m 2

+ 2flm + 1) = O

(8.4-29)

This equation will be satisfied, and Eq. 8.4-28 will be a solution of the differential equation, if m is chosen so that

+ 2flm + 1 = O

m2

(8.4-30)

Thus, m has two possible values, given by m1

(8.4-24)

m2 = -({3Provided m1 and m 2 are distinct (i.e.,

the boundary conditions

=o

(8.4-25a)

t =o

(8.4-25b)

t

= -({3 + .Jf32 - 1)

H

=

.Jf3

fJ =1=

C1 em1 e

2

-

1)

(8.4-31a) (8.4-31b)

1), the general solution is

+ C emsE> 2

{8.4-32)

However, if the roots of Eq. 8.4-30 are not distinct, the solution is 26

and the boundary conditions a dimensionless distance H and

H

=

(C1

+ C2 8)e-e

(8.4-33)

26. L. R. Ford, Differential Equations (New York: McGraw-Hill Book Company, lnc., 1933), p. 70.

328

Macroscopic Balances-Viscous Effects

Chap. 8 Sec. 8.4

Unsteady Flow in Closed Conduits

The solution is again given by complex function containing bo part independently satisfies the conditions, and takes the forro

This result indica tes that H takes which decrease exponentially wi damped." -i.OL------------:8::--------

-

Fig. 8.4-3. Position of the gas-liquid interface for an oscillating manometer.

The motion of the fluid in the manometer may have three distinct modes depending on the value of {3. The three types of motion are illustrated graphically in Fig. 8.4-3. We may discuss them as follows .

l. {3 > l. For this case, both m1 and m 2 are real negative numbers, and the constants of integration in Eq. 8.4-27 are readily determined to yield (m 2emt9 _ m1em28) H=~::.:.-----2..::...~ (8.4-34) (m 2 - m1) This type of motion is called "over damped," and the fluid approaches its equilibrium position exponentially. 2. {3 = l. Here, the roots are indistinct, and the solution is obtained by application of the boundary conditions to Eq. 8.4-33, yielding

H

=

(1

+ 0)e-9

(8.4-35)

The fluid motion is similar to the previous case in that the equilibrium position is approached exponentially. However, this solution represents the most rapid approach to equilibrium without "overshooting" the equilibrium position. Such motion is called "critically damped." 3. {3 < l. Under these conditions, the roots become complex, and we must take special care in obtaining a solution. We write the roots as

{3)t

- (p

+ pcf>)2]

(8.5-7)

where Pm is the density of the manome Experimental studies indicate that accurate for engineering practice, anda to give

Q = CaA2

A plot of Ca versus Reynolds numbe~ meter with (A 2/A 1) = 0.25. We see tha sis is good because Ca is nearly unity. we should not expect the analysis to ho velocity profiles are no longer flat. A profile into aC)

1 -

(p

+ pcf>)2]

p[l - (A2/At)2]

(8.5-8)

Now, 4> is a linear function in z; therefore, the average value is the centerline value, and (8.5-9) However, the pressure at point 2 is not a linear function of z because the streamlines are curved,t and strictly speaking we cannot replace the average pressure by the pressure at the centerline. A more detailed analysis might

t Does this suggest a differential-macroscopic balance as a possibility? ! See Sec. 7.4, Eq. 7.4-21.

{2(;;: .J~

/'

0.90

/

V 0.80

1/

2

4 681Ql 2

4

NRe=

Fl1. 8.5-:Z. Venturi ~

Chap. 8

The mechanical energy balance is well for steady, incompressible flow, the

Sec. 8.5

include an estimate of the effect of the curved streamlines on the average pressure; however, we shall neglect this effect and replace the average pressure by the centerline pressure to obtain

Q- A

(8.5-1)

(8.5-2)

333

Flow Rate Measurement

2

J

2(p¡ - P2) p[1 - (A 2/A1) 2]

(8.5-10)

If this formula is expressed in terms of the height of the manometer fluid, z 2 - z1 yields (8.5-11)

(8.5-3) viscous effects are negligible and write (8.5-4)

where Pm is the density of the manometer fluid. Experimental studies indicate that Eq. 8.5-11 is not always sufficiently accurate for engineering practice, and a discharge coefficient e, is introduced to give (8.5-12)

into Eq. 7.3-31, and assuming flat

- (vMp)2A2 - (t/>)1(v)1A 1 ]

(8.5-5)

(8.5-6)

A plot of e, versus Reynolds number is shown in Fig. 8.5-2 for a Venturi meter with (A 2/A 1) = 0.25. We see that at high Reynolds numbers the analysis is good because e, is nearly unity. For Reynolds numbers less than 2300, we should not expect the analysis to hold because the flow is laminar and the velocity profiles are no longer flat. A more carefulanalysis, takingthevelocity profile ·into account, can lead to the conclusion that e, should decrease rapidly as laminar flow is approached; however this problem will be left as an exercise for the student.

(8.5-7)

1.00

~v

(v)1 and obtain an expression for

0.90

(8.5-8) the average value is the centerline (8.5-9) ~~"""'F.

we cannot replace the average A more detailed analysis might

balance as a possibility?

/

eti 0 .80

V

~

lo-"

V

V

V

2

4 6 8 103 2

4 6 810" 2

4 6 8 1dl

Nrt.= (v) 1D1/V Fl¡. 8.5·1. Venturi discharge coefficient.

334

Macroscopic Balances-v 1scous Effects

Chap. 8

Sec. 8.5

Flow Rate Measurement

Orífice meter

The sharp-edged orífice shown in Fig. 8.5-3 is another device for accelerating the flow and obtaining a measurable pressure difference related to the flow rate. The orífice is often used as a metering device in preference to a Venturi meter because of its versatility and low cost of construction and installation. Because the area of the jet at point 3 (the vena contracta) is

t---.... !'--0 .80

!'--t---.... t-. t--r--..,

Control volume

"1''

"2"

1 1 1 1

0.60 2

r--- ~J

... r--..t- ¡... t--r--.., t-11 ... l-..

4

6 8

4

10

2

4 6

Fig. 8.5-4. Discharge coefficien '

Our analysis would indicate that

Ca.=~

'V [1- (

Flg. 8.5-3. Sharp-edged orifice.

unknown, the solution to this flow problem will not be as accurate as that for the Venturi meter. Calibration of orífice meters is therefore a necessity, and, in general, they are considered less reliable than Venturi meters. If we express the area of the jet at the vena contracta as (8.5-13)

where A2 is the area of the orífice and Ce is the contraction coefficient, the flow rate is given by _ CA

Q-

e

J

2

2(p¡ - Pa) p[1 - (CeA2/A1) 2]

(8.5-14)

Here we have applied the mechanical energy balance, neglected viscous effects, and replaced average pressures with centerline pressures. Because the contraction coefficient is a function of the ratio of areas A2/ A1 , the equation is generally written in terms of a discharge coefficient

Q = Ca.A 2~ 2(p 1 - p3 )/p

(8.5-15)

Because Ce changes with A2/A 1 it is ratl with experimental results. However, as coefficient should equal the contraction e simplified. Figure 8.5-4 shows sorne exp1 sharp-edged orífice. For small values o approximately 0.6, a reasonable value again the analysis is only satisfactory at Flow nozzle

A flow nozzle, such as that illustrat difficulty of an unknown contraction cot parallel as the fluid emerges from the nm is applied as before to yield

Sec. 8.5

Flow Rate Measurement

335

¡-..... r-...

8.5-3 is another device for acceleratpressure difference related to the a metering device in preference to a and Iow cost of construction and jet at point 3 (the vena contracta) is

0 .80

0.70

1·~

1'-.

~

0 .60

¡... ..............

r--

0 .40

........

0.60 2

...... t- """'¡... ..._¡........

0.20 0.10 0.05

......

4 6 8104

2

4 6 8105

2

4

6

81o6

2

NR• =1D,fv Fig. 8.5-4. Discharge coefficients for a sharp-edged orifict;.

Our analysis would indicate that (8.5-16)

meters is therefore a necessity, and, than Venturi meters. (8.5-13)

c. is the contraction

Because c. changes with A 2/A 1 it is rather difficult to compare our analysis with experimental results. However, as A 2/A 1 becomes small, the discharge coefficient should equal the contraction coefficient, and matters are somewhat simplified. Figure 8.5-4 shows sorne experimental values of C,¡.for a standard sharp-edged orífice. For small values of A 2/AI> the discharge coefficient is approximately 0.6, a reasonable value of the contraction coeffi.cient. Once again the analysis is only satisfactory at high Reynolds numbers.

coefficient, the Flow nozzle

(8.5-14) energy balance, neglected viscous with centerline pressures. Because the the ratio of areas A 2/A 1 , the equation coefficient (8.5-15)

A flow nozzle, such as that illustrated in Fig. 8.5-5, does not offer the difficulty of an unknown contraction coeffi.cient, because the streamlines are parallel as the fluid emerges from the nozzle. The mechanical energy balance is applied as before to yield

(8.5-17)

Macroscopic Balances-Viscous Effects

336

"2"

"1"

Chap. 8

Sec. 8.5

Flow Rate Measurement

The experimental values of Cd shown is in error by about 10 per cent or less, area to the duct area, A 2/A 1 • As the n erated more severely, and inertial effec the Reynolds number is greater tha1 viscous dissipation begins to show i decreases.

1 1

Pitot-static tube

The pitot-static tube shown in Fi velocities ata point; it may be used t pro file is flat or if the en tire profile is m at point 2 which is the stagnation poin holes are drilled around the peripher_)l static pressure. The pressure at point 2 difference between these two pressure and relatéd to the fluid velocity. In thi the pressure along two streamlines rat

Fig. 8.5-5. Flow nozzle.

1.00

A)A,Lb1.~

í1

1

11 1 1_1 1 -0.50

~ """"

A2/A1 A2l;A 11=0.65

~

u

~~

0 .90 f-A 2 !A 1 = 0 .30

0.80

2

Th 4

6

8104 2

4

6

8105 2

4

6

81o6 2

l_

NRe =D,Iv Fl1. 8.5-6. Discharge coefficients for a flow nozzle.

Fig. 8.5-7. Pit

Sec. 8.5

Flow Rate Measurement

337

The experimental values of Cá shown in Fig. 8.5-6 indicate that the analysis is in error by about 1Oper cent or less, depending u pon the ratio of the nozzle area to the duct area, A 2/A 1 • As the ratio becomes smaller the fluid is accelerated more severely, and inertial effects very definitely predominate, provided the Reynolds number is greater than 105 • For lower Reynolds numbers, viscous dissipation begins to show its effect and the discharge coefficient decreases. Pitot-static tube The pitot-static tube shown in Fig. 8.5-7 is a device for measuring fluid velocities ata point; it may be used to determine the flow rate if the velocity profile is flat or if the en tire profile is meas u red. The tu be has a small opening at point 2 which is the stagnation point, i.e., the velocity is zero. A series of holes are drilled around the periphery of the tube at point 3 to measure the static pressure. The pressure at point 2 is called the dynamic pressure and the difference between these two pressures may be measured by the manometer and related to the fluid velocity. In this particular system, we wish to compute the pressure along two streamlines rather than over a control surface, and we Flow nozzle. Boundory loyer (thickness is exoggeroted)

"1"

"2"

·~========::::-

~.

4 6

e105

2

4 6

e o6 1

Th 2

1__

=D11v

for a flow nozzle.

Fig. 8.5-7. Pitot-static tube.

338

Macroscopic Balances-Viscous Effects

Chap. 8

shall use Bernoulli's equation. Application of Eq. 7.4-17 to the streamline between 1 and 2 yieldst PI

Since v2

=

+ !pv~ + pgzi =

P2

+ tpv~ + pgz2

(8.5-18)

O, the pressure at point 2 is P2

= PI

+ !pv~ + pg(zi -

Z2)

(8.5-19)

Sec. 8.5

Flow Rate Measurement

Sharp-crested weir

Flow rates in open channels may the channel over which the fluid m monly used weirs, but we will consi Fig. 8.5-8. It represents a rather e sence of experience and knowledge,

Along the streamline from points 1 to 3 we write PI

+ fpv~ + pgzi =

Pa

+ !pv~ + pgza

(8.5-20)

To continue this analysis, we must know something about the flow field around the tube. We need not guess the flow topology, for experimental and theoretical studies have been carried out 27 indicating there is a thin region (called the boundary !ayer) around the tube where the velocity changes rapidly from the free stream value to zero. At high Reynolds numbers, the boundary 1ayer thickness is small and the pressure difference across it is negligible. Under these conditions, the pressure at point 3 is very nearly equal to the pressure at the static hole at the surface of the Pitot-tube. We take the streamline from points 1 to 3 to be outside the boundary !ayer; thus, (8.5-21) and the pressure at the static hole is given by Pa = PI

+ pg(zi -

Za)

(8.5-22)

Substituting Eq. 8.5-19 into Eq. 8.5-22 and solving for the velocity, we get vi

= J1.[(P2 - Pa) + pg(z2 - za)J .

(8.5-23)

//7//////§//ff//,ij¡

Fig. 8.5-8. S

p

As in the other systems studied in this section, the analysis is not exact and Pitot-static tu bes must be calibrated in terms of the equation, vi=

e J2(p2;

Pa)

(8.5-24)

The gravitation terms have been dropped because they are always negligible. If a Pitot-static tube is properly designed, thf" coefficient e can be very close to unity. The implication is not that the analysis is exact; it simply indica tes that severa1 errors have cancelled. t Remember that Bernoulli's equation neglects viscous effects. 27. H. Schlichting, Boundary Layer Theory, 4th ed. (New York: McGraw-Hill Book Company, Inc., 1955).

attack the problem. Both the mon would seem to be suitable tools fo equation and lea ve the application o Along any streamline we may w p

+ !pv

provided viscous effects are negligi the thickness of the jet at the weir the weir will be small. U nder th specified by applying Eq. 8.5-25 at

ic Balances-Viscous Effects

Chap. 8

tion of Eq. 7.4-17 to the streamline (8.5-18)

(8.5-19)

Pa

+ tpvi + pgza

Sec. 8.5

Flow Rate Measurement

339

Sharp-crested weir

Flow rates in open channels may be measured by a weir, an obstruction in the channel over which the fluid must ftow. There are severa! types of commonly used weirs, but we will consider only the sharp-crested weir shown in Fig. 8.5-8. It represents a rather complicated ftow problem, and in the absence of experience and knowledge, it is not at all obvious how we should

(8.5-20)

know something about the ftow field flow topology, for experimental and out27 indicating there is a thin region the tube where the velocity changes zero. At high Reynolds numbers, the the pressure difference across it is pressure at point 3 is very nearly at the surface of the Pitot-tube. We to be outside the boundary !ayer; thus, (8.5-21)

(8.5-22) and solving for the velocity, we get

(8.5-23) section, the analysis is not exact and terms of the equation, (8.5-24) because they are always negligible. thP coeffi.cient C can be very close analysis is exact; it simply indicates viscous effects. 4th ed. (New York : McGraw-Hill Book

Fig. 8.5-8. Sharp-crested weir.

attack the problem. Both the momentum and mechanical energy balance would seem to be suitable tools for the analysis, but we will use Bernoulli's equation and Ieave the application ofthe macroscopic balances asan exercise. Along any streamline we may write p

+ tpv 2 + pgz =c.

(8.5-25)

provided viscous effects are negligible. If the channel is deep compared to the thickness of the jet at the weir (H ';J> h), the velocity far upstream from the weir will be small. Under these conditions, the constant c. may be specified by applying Eq. 8.5-25 at sorne point far upstream from the weir.

C, = p

+ pgz

(8.5-26)

Macroscopic Balances-Viscous Effects

340

Chap. 8

Since the velocity in the z-direction is negligible there, the pressure is just the hydrostatic pressure, (8.5-27) P =Po pg(H - z)

+

and we see that every streamline has the same constant given by (8.5-28)

C, =Po+ pgH We may now write Eq. 8.5-25 as

p

+ l pv + pgz =Po + 2

pgH

= ..}2g(H- z)

PRO

(8.5-30)

The volumetric flow rate over the weir is given by z ~ L+h

f

Q= b

v ·i

dz

While experiments must be carried 01 this analysis does have value in that i rate on (H - L). In Chap. 9, we w show how practica! flow rate measuri theory, thus providing a more reliab The results presented in this sectio on fluid metering 28- 30 to obtain both and a thorough discussion of the m tions.

(8.5-29)

At the crest of the weir the fluid pressure is unknown; however, the pressure at both the top and bottom of the jet is atmospheric, and the pressure within the liquid should not be too far removed from this value. Applying Eq. 8.5-29 to the streamlines at the crest of the weir and setting p = p 0 , we find

v

Problems

(8.5-31)

z= L

where bis the width ofthe weir. While Eq. 8.5-30 may be a satisfactory expression for the magnitude of the velocity vector, it does not provide us with the x-component of the velocity vector. If we make the approximation (8.5-32)

8-1. To determine the pressure drop-fl<

bundle used in heat transfer e configurations similar to that shown in Fig. 8-1. The tubes are arranged in a triangular pattern with the mean flow directed along the diagonal of the square. Describe how you would interpret the experimental results obtained i such a system-i.e., how would you define the friction factor, what parameters would f be a function of, etc.? Neglect wall effects and assume the tubes are infinitel)i long. 8-2. Derive an integral expression for tube shown in Fig. 8-2. Express

Eq. 8.5-30 may be substituted into Eq. 8.5-31 and the integration carried out to yield (8.5-33) Q = fb..}2g{[H- L]3 i 2 - [H- (L + h)] 3i 2} In practice, the height of the weir L is known, and the depth far upstream H is measured with a depth gauge ; however, the position of the free surface at the weir remains asan unknown. Traditionally, we assume H- (L

+ h)

~

H- L

Q

EJ A1

---

-----.:

(8.5-34)

Flg. S.l. Flow iJ

and introduce a discharge coefficient to obtain (8.5-35) We expect the discharge coefficient to be less than unity, and experiments indicate that (8.5-36)

28. " Flowmeter Computational Handt lication, New York, 1961. 29. " Fluid Meters : Their Theory and New York, 1959. 30. H. W. King and E. F. Brater, Ham Book Company, Inc., 1963).

Balances-Viscous Effects

Chap. 8

negligible there, the pressure is just the (8.5-27)

pg(H- z) same constant given by

(8.5-28)

=Po+ pgH

Problems

341

While experiments must be carried out to determine the discharge coefficient, this analysis does ha ve value in that it predicts the correct dependence of flow rate on (H- L). In Chap. 9, we will analyze the broad-crested weir, and show how practica! flow rate measuring devices can be altered to fit available theory, thus providing a more reliable method of measurement. The results presented in this section are brief; we must refer to handbooks on fluid metering 28 - 30 to obtain both accurate values for discharge coefficients and a thorough discussion of the many metering devices and their applications.

(8.5-29)

is unknown; however, the pressure atmospheric, and the pressure within from this value. Applying Eq. the weir and setting p = p 0 , we find (8.5-30)

(8.5-31) Eq. 8.5-30 may be a satisfactory expresvector, it does not provide us with the we make the approximation (8.5-32) . 8.5-31 and the integration carried out

PROBLEMS 8-1. To determine the pressure drop-flow rate relationship for flow through a tu be bundle used in heat transfer equipment, experiments are performed on configurations similar to that Tube diometer is O shown in Fig. 8-1. The tubes are arranged in a triangular pattern ¿/'¿/'##/$#$!'#/#/& with the mean flow directed along o o o o the diagonal of the square. Describe how you would interpret the Mean o o o experimental results obtained in - - flow such a system-i.e., how would o o o o you define the friction factor, what o o o o o parameters would f be a function of, etc.? Neglect wall effects and //7//////////////////////////// assume the tubes are infinitely Flg. 8-1. Flow through a tube bundle. long. 8-2. Derive an integral expression for the pressure drop in the expanding circular tu be shown in Fig. 8-2. Express !lp as a function of Q, A1 , A2, and f The r

- [H- (L

+ h)]al2}

(8.5-33)

, 1

1

known, and the depth far upstream H the position of the free surface at iiuium1a11 , we assume (8.5-34)

1 1

1 1 1

L,

1

A~~-----------------------'~,,~ Az

Flg. 8-l. Flow in an expanding tube.

_ L)312

(8.5-35)

be less than unity, and experiments

(8.5-36)

28. "Flowmeter Computational Handbook," Am. Soc. Mech. Engrs. Research Pub/ication, New York, 1961.

29. "Fluid Meters: Their Theory and Application," Am. Soc. Mech. Engrs. Paper, New York, 1959. 30. H. W. King andE. F. Brater, Handbook of Hydraulics (New York: McGraw-Hill Book Company, lnc., 1963).

342

Macroscopic Balances-Viscous Effects

cross-sectional area is given by A = A1

+ (A 2

-

Chap. 8

Problems

)(i)

A1

and yo u are to assume that f is constant and the flow is turbulent. 8-3. Water is pumped through a 1-in. diameter pipe 100ft long at arate of 120 in. 3 /sec. Neglecting entrance effects, calculate the pressure drop for a relative roughness ej D of 0.002. 8-4. Calculate the pressure drop for the conditions in Prob. 8-3 if the pipe is replaced by a square duct, 1 in. on a side. 8-5. Crop-dusting of the artichoke fields around Castroville, California, is traditionally done by pilots flying at an altitude of 17 ft. If the particles of insecticide are approximately spherical with an average diameter of 10- 2 cm, anda density of 127lbm/ft, how long does it take them to fall to the ground? If the nearest population center is 1 mi from the fields, what is the maximum wind velocity (directed from the field toward the town) that can be tolerated before the dusting operation must be stopped? 8-6. The flow rate through a flow nozzle was given in Sec. 8.5 by 2(p¡ - P2)

Pipe diometer =0 0 Surge tonk diometer =fJ

Flg. 8-7. Use of a surge t

an approximate solution for the t~nk h'(t) may be obtained. Det maximum damping, and comme Hint: Assume a solution for h'(t h'(t)

where the discharge coefficient Cd had to be determined experimentally. Repeat the derivation retaining the viscous dissipation term in the analysis. Explain what experiments you would perform, and how you would correlate the data to obtain a satisfactory method of predicting flow rates by a flow nozzle. The assumption of flat velocity pro files is a reasonably good one and should be incorporated in your analysis. 8-7. The pump illustrated in Fig. 8-7 has an output given by Q = Q0

If aQ ~ Qo

the friction factor may be taken as a constan t. In the absence of a surge tank, the fluctuations occurring at the process are identical to those at the pump and the process is therefore difficult to control. You are asked to design a surge tank which will reduce the amplitude of the fluctuations by a factor of 10. Set up the problem in a general fashion, locating a surge tank of diameter D1 at a distance rxL (O : : :; rx ~ 1) from the pum p. Neglect minor losses and inertial effects. An exact solution is difficult to obtain; however, if the depth in the surge tank is represented by a steady term and a time-dependent term =

8-8. Apply the momentum and mecha for the flow rate over the shar fully list the assumptions that mu

+ aQ sin wt

where Q 0 is the time averaged flow rate, and aQ represents the amplitude of the volumetric flow rate variations . The frequency of these variations is w.

h

and make use of the approximat1

h0

+ h'(t)

where h'(t)

~

h0

8-9. Determine the water fiow rate fa the siphon shown in Fig. 8-9. Th inlet to the siphon is placed 10 1 below the water leve! and the ow let is even with the bottom of th tank. The pipe diameter is 1 ft, th totallength of the piping is 450 f¡ and the relative roughness is 0.0~ The fittings in the lineare standar 90° elbows. Assume that the levf of the water in the tank i constan t.

Balances-Viscous Effects

Chap. 8

Problems

343

pipe 100ft long at a rate of 120 calculate the pressure drop for a relative conditions in Prob. 8-3 if the pipe is si de. around Castroville, California, is traaltitude of 17 ft. If the particles of with an average diameter of w- 2 cm, does it take them to fall to the ground? · from the fields, what is the maximum toward the town) that can be tolerated stopped? was given in Sec. 8.5 by 2(p¡ - P2)

p[1 - (A 2/A 1) 2]

had to be determined experimentally. viscous dissipation term in the analysis. perform, and how you would correlate of predicting flow rates by a ftow profiles is a reasonably good one and

r

-a)L.---..l•l

and the ftow is turbulent.

\\ ;f'0 Pipe diometer = 0 0 Surge tonk diometer

=0 1

Flg. 8-7. Use of a surge tank to damp pump functions.

an approximate solution for the amplitude of the fluctuations in the surge tank h'(t) may be obtained. Determine the value of oc that will provide the maximum damping, and comment on the effect of increasing the ratio D 1/ D 0 • Hint: Assume a solution for h'(t) of the form h'(t) =a,. sin (wt - 8)

and make use of the approximation

y¡,

=

y-¡;0

(t + ~ h'(t)) ho

8-8. Apply the momentum and mechanical energy balances to derive an expression for the flow rate over the sharp-crested weir discussed in Sec. 8.5. Carefully list the assumptions that must be made to obtain a solution in each case.

+ aQ sin wt and aQ represents the amplitude of The frequency of these variations is w.

In the absence of a surge tan k, are identical to those at the pump to control. You are asked to design a of the ftuctuations by a factor ~r locating a surge tank of diameter the pump. Neglect minor losses and to obtain; however, if the depth in term and a time-dependent term

8-9. Determine the water flow rate for the siphon shown in Fig. 8-9. The inlet to the siphon is placed 1O ft below the water level and the outlet is even with the bottom of the tank. The pipe diameter is 1 ft, the totallength of the piping is 450 ft, and the relative roughness is 0.02. The fittings in the lineare standard 90° elbows. Assume that the level of the water in the tank is constan t. Flg. 8-9. Flow in a siphon.

Macroscopic Balances-Viscous Effects

Chap. 8

Problems

8-10. Explain how you might estimate the totalloss owing to the presencc of a flow nozzle in a pipeline. Hint: Apply both the momentum and mechanical energy balances to a suitable control volume. 8-11. Determine the cros~-sectional area of the jet illustrated in Fig. 8-11 as a function of z. The area of the rounded orifice is A 0 , and you are to assurne that the velocity profile everywhere in the jet is flat. Treat the flow as quasi-steady. Neglect viscous effects and sol ve the problem using both Bernoulli's equation and the momentum balance.

r

p = 68.5 lbm/ft J.L

Ll.t:z::z::~:::z::¡~¡____z =o

¡

:

1

I¡J,II:J.IIr.t.liJ.II~11r.t.IU,II¡.¡,I.J • 1 2

A1 =0.10 ft Flow

!

Sudden contr 6in. I.D. to

5' .L_

Seo wot

Flg. 8-13. Flo

"2"

1

'J .

centipoise

h

8-12. Water at 70°F flows through a Venturi meter as shown in Fig. 8-12. Cavitation occurs when the local pressure falls below the vapor pressure of the liquid, allowing vapor bubbles to form. The vapor Flg. 8-11. Flow through a rounded orifice in a tank. pressure of water at 70°F is 0.36 psia. Calculate the maximum flow rate that this meter can handle without cavitation. "1"

= 1.2

3

:

~~~¡¡wwuu~

1

! i

1

~~·~~

1

i

A2=0.025 ft

/1 (/ti(/! 1111111.1..

2

_......·.:.r:, _ _ 1/T~--

"~mnmm~~

Fig. 8-12. Flow through a Venturi meter.

8-13. Figure 8-13 shows a pipeline system for pumping sea water into a reservoir. The Venturi meter has a discharge coefficient of 0.96 anda throat area onehalf that of the pipe. The pipes are 4-in. and 6-in. I.D. commercial steel, and the manometer fluid is mercury. Determine the horsepower requirement of the pump and the volumetric flow rate. The density of sea water may be taken as 68.5 lbm/ft 3 and the viscosity as 1.2 centipoise. 8-14. A i-in. plastic sphere (p = 38 lbm/ft 3) is released from the bottom of a deep lake. What is the terminal velocity? 8-15. Calculate the horsepower delivered by the pump to maintain a steady flow ofwater in the system shown in Fig. 8-15. All piping is 6-in. I.D. commercia1

Flg. 8-15. Flo·

Balances-Viscous Effecu

Chap. 8

Problems

345

totalloss owing to the presencc of a flow the momentum and mechanical energy

r

p =68 .5 1bm/fl 3 JL

= 1.2

centipoise

Pump

h

L~p::z:ZJ-z=o

Sudden controction from 6in. I.D. lo 4in . I.D.

Flg. 8-11. Flow through a rounded orifice

in a tank.

5' .L...

Seo water Flg. 8-13. Flow in a pipeline.

for pumping sea water into a reservoir. eoeffic.ient of 0.96 and a throat area oneand 6-in. I.D. commercial steel, and the horsepower requirement of rate. The density of sea water may be as 1.2 centipoise. is released from the bottom of a deep the pump to maintain a steady flow

5. AII piping is 6-in . I.D. commercial

Flg. 8-15. Flow in a pipeline.

Macroscopic Balances-Viscous Effects

346

Chap. 8

steel, and all fittings are 90° standard elbows. Assume a value of 0.7 for the contraction coefficient of the 4-in. diameter orifice in the si de of the tan k. 8-16. Electrical transmission towers are placed at 1000-ft intervals, and i-in. diameter cables are strung between them. Determine the drag force on a single cable between two towers if the wind velocity is 60 mph. 8-17. When 15 ft 3 /sec ofwater flow in a 12-in. I.D. pipeline, 85 hp are lost in viscous dissipation for every 1000 ft of pipe. For these conditions, calcula te the head loss, friction factor, and shear stress at the pipe wall. 8-18. Calculate the flow rate of water from the tank illustrated in Fig. 8-18.

l

Water

1 in. 1.0. smooth tubing

lOOft

k----------250ft---------·~! Fig. 8-18

8-19. Calculate the smallest flow rate required to keep the pipeline shown in Fig. 8-19 running full. The pipe is 1-in. I.D. commercial steel. Neglect minor losses and keep in mind the phenomenon of cavitation.

r-30'----j

t

50'

t

t

Water at 70° F

LL

l

100'

j

\

)._ Pump

Flg. 8-19. Flow in a pipeline.

Problems

8-20. Calculate the flow rate through tht submerged orifice in Fig. 8-20 The orifice diameter is 2 in. Esti mate the discharge coefficient.

8-21. Derive an expression for the vol· umetric flow rate in a Ventu meter (see Sec. 8.5), assuming th velocity profile is always paraboli Compare your result with th experimental values of the dis charge coefficient shown in Fig 8.5-2.

ic Balances-Viscous Effects

Chap. 8

elbows. Assume a value of 0.7 for the orífice in the side ofthe tank. at 1000-ft intervals, and i-in. diamDetermine the drag force on a single velocity is 60 mph. .I.D. pipeline, 85 hp are lost in viscous For these conditions, calculate the head at the pipe wall. the tank illustrated in Fig. 8-18.

1 in. /.D. smooth tubing

250ft _ _ _ _.., 8-18

to keep the pipeline shown in Fig. I.D. commercial steel. Neglect minor of cavitation.

t 50'

t

r-30'4

l

lOO'

Problems

8-20. Calcula te the flow rate through the submerged orífice in Fig. 8-20. The orífice diameter is 2 in. Estímate the discharge coefficient.

8-21. Derive an expression for the volumetric flow rate in a Venturi meter (see Sec. 8.5), assuming the velocity pro file is always parabolic. Compare your result with the experimental values of the discharge coefficient shown in Fig. 8.5-2.

347

Air ot 4 psig

Atm. press.

5'

j__ Fig. 8-20. Flow

orifice.

through a subrnerged

Sec. 9.1

Open Channel Flow

9

Uniform Flow

the curvatu re of the free surfacet is sm and rapidly varied ftows, where the cu In the study of ftow in closed con~ fiow rate, pressure drop, and geometry will seek a relationship between the geometry [Q, h(x), b(x), 'l)(x)]. We sha terms of the macroscopic mass, mome and then go on to consider the flow ir This willlead us to equations defining shall analyze rapidly varied ftows, suc

9.1

Uniform Flow

Consider the ftow in a rectangula sluice gate, such as that illustrated in gate, the ftow will be nonuniform ar

The study of open channel flow is primarily directed toward applications in irrigation, ftood control, and water suppty. The analysis of ftow in open channels is more complex than the analysis of flow in closed conduits for two reasons : the existen ce of a free surface complicates the analysis; and the actual channels encountered in practice are comparatively complex. A large portian of the ftow in closed conduits is restricted to pipe flow, where the diameter and relative roughness describe the geometry fairly well. However, in dealing with flow in open channels we may encounter anything from a carefully constructed aquaduct to a grass-lined irrigation ditch. The flow characteristics of the former are fairly well known; however, those of the latter must be studied experimentally for every individual case if accurate results are desired. The study of open channel ftow could easily be the subject of an entire course, and in a single chapter we can only hope to comment on the most important aspects. These ftows may be classified as either steady or unsteady, uniform or nonuniform. With the exception of the solitary shallow-water wave studied in Sec. 9.3, al! the ftows investigated in this chapter will be steady. A uniform ftow is one for which the fluid depth h above the channel bed is constant. Nonuniform flows can be further classified into gradually varied flows, where 348

_j~___ ---·-Fig. 9.1·1. Formati

distance downstream, a balance .bet will be achieved and the flow will be a representative rectangular channel indicate a rather curious situation. B wetted surface is the only force opp leads us to expect the maximum velo

t The term "free surface" denotes an that the tangential stress is zero and the no

Sec. 9.1

9

Uniform Flow

349

the curvature of the free surfacet is small compared to the depth of the fluid, and rapidly varied flows, where the curvature is comparable to the fluid depth. In the study of flow in closed conduits, we sought equations relating the flow rate, pressure drop, and geometry (Q, !l.p, L, r0 ), while in this chapter we will seek a relationship between the flow rate, fluid depth, and channel geometry [Q, h(x), b(x), r¡(x)]. We shall first analyze a simple uniform flow in terms of the macroscopic mass, momentum, and mechanical energy balances, and then go on to consider the flow in a channel with a changing geometry. This willlead us to equations defining gradually varied flows, after which we shall analyze rapidly varied flows, such as the hydraulic jump.

9.1

Uniform Flow

Consider the flow in a rectangular channel formed downstream from a sluice gate, such as that illustrated in Fig. 9.1-1. Just downstream from the gate, the flow will be nonuniform and rapidly varying; ho~ever, at sorne

· y directed toward applications supply. The analysis of flow in open · of flow in closed conduits for two complicates the analysis; and the are comparatively complex. A large is restricted to pipe flow, where the the geometry fairly well. However~ we may encounter anything from a lined irrigation ditch. The flow well known; however, those of the for every individual case if accurate annel flow could easily be the subject r we can only hope to comment on ther steady or unsteady, uniform or solitary shallow-water wave studied · chapter will be steady. A uniform abo ve the channel bed is constan t. into gradually varied flows, where

_j)___ ----------------------Fig. 9.1-1. Formation of a uniform flow.

distance downstream, a balance .between gravitational and viscous forces will be achieved and the flow will become uniform. The velocity profile for a representative rectangular channel is shown in Fig. 9.1-2, and the curves indicate a rather curious situation. Because the drag force occurring at the wetted surface is the only force opposing the gravitational force, intuition leads us to expect the maximum velocity to occur at the centerline on the free

t The term "free surface" denotes an air-water interface, and has the characteristic that the tangential stress is zero and the normal stress is - np 0 •

Open Channel Flow

350

Chap. 9

M ximurri velocity

Sec. 9.1

Uniform Flow

Time-averaging of all the equation. because the open channel fiows of p turbulent. The general macroscopic 1

:J

pv dV

?'"a (t)

z h

+

J

pv(v - w) •

A, (t)

can be dotted with i and applied to t

Because the fiow is uniform, (v!)2

Unes of constont velocity ( Vx)

and Eq. 9.1-6 reduces toa balance o Fl¡. 9.1-l. Velocity profiles in a rectangular channel.

O = pgzAL surface; however, the maximum velocity is located below the free surface. This anomaly appears to result from the secondary flow (indicated in Fig. 9.1-2), which carries the low velocity (vz) fluid upward along the sides and then out over the free surface.

Mass balance

dt

f +f p

dV

p(v - w) • n dA

A,

?'"4 (t)

.

=

O

Assuming that the shear stress at th should be for uniform fiow), Eq. 9.1

(9.1-1)

Defining an average wall shear

stres~

l '~'o=--

J

v·ndA =O

S,

we obtain

O= pg sin (9.1-2)

A,

Application of Eq. 9.1-2 to the control volume illustrated in Fig. 9.1-1 yields the obvious result (9.1-3)

= A1 =

d

(pg)l

and simplifying it for a fixed control volume and incompressible fiow, we obtain

Since A2

f [-

where S is 'the wetted perimeter, b -1 that

Starting with the general macroscopic mass balance,

f!..

+

bh, we obtain

(9.1-4)

Following the procedure used in tC arrange this equation in dimensionle Dimensionless wall shear stress

t The use of the net stress vector is a inasmuch as t 1:, is zero at a free surface.

Open Channel Flow

Chap. 9

Sec. 9.1

Uniform Flow

351

Time-averaging of all the equations used in this chapter is understood, because the open channel flows of practica! importance are almost always turbulent. The general macroscopic momentum balancet

:J

pv dV

7"• (t)

z ¡¡

+

J

pv(v- w) • n dA

=

A. (t)

J

+ t~>

7" • (t)

.9/• (t)

pg dV

J

(9.1-5)

dA

can be dotted with i and applied to the fixed control volume to yield p(v! )2 A - p(v! )1 A

=

pg.,AL

f

+ t~> • i dA

(9.1-6)

.9/

Because the flow is uniform,

(v.~:.. )

(9.1-7) and Eq. 9.1-6 reduces to a balance of surface forces and body forces,

a rectangular channel.

O = pg.,AL located below the free surface. This (indicated in Fig. 9.1-2), along the sides and then out

+ f [-i · npg + i · (n • T)] dA

(9.1-8)

.9/

Assuming that the shear stress at the solid walls is independent of x (as it should be for uniform flow), Eq. 9.1-8 becomes (9.1-9) S

where S is ·the wetted perimeter, b that

+ 2h.

(pg)]

(9.1-1)

=

Once again uniform flow requires

(pg)2

(9.1-10)

Defining an average wall shear stress as (9.1-11)

and incompressible flow, we we obtain

=0

(9.1-2)

O= pg sin O AL-

-r0 SL

(9.1-12)

Following the procedure used in the analysis of closed conduits flow, we arrange this equation in dimensionless form to obtain illustrated in Fig. 9.1-1 yields

Dimensionless wall shear stress

Dimensionless body force

(9.1-3) (9.1-13) (9.1-4)

t The use of the net stress vector is a convenience in analyzing open channel ftow inasmuch as t(:) is zero at a free surface.

Open Channel Flow

Chap. 9

where the hydraulic radius is given by R~

cross-sectional area

bh

wetted perimeter

S

= ----:---:---

n2

116 R~13

Uniform Flow

Flow rate-channel depth calculati uniform flow

Severa! empírica! expressions exist for the friction factor, f The most popular is the work of Manning, 1 which we may write in the form

f =

Sec. 9.1

If the geometry and fluid depth fo of the flow rate is straightforward. If and solve for (v:z), we get

(9.1-14)

The coefficient n is a measure of the roughness, and values for severa! types of surfaces are Iisted in Table 9.1-1. Since f is dimensionless, we must be Table 9.1-1 T ABLE OF ROUGHNESS COEFFICIENTS na Type of Channel

8g sin()

bh

Q=

R:'

116n 2

n, ft' /6

If the volumetric flow rate and channe

Artificial Channels of Uniform Cross Section Sirles and bottorn Iined with well-planed tirnber evenly laid . . . . . . . . . . . . . . . . . . . . Neat cernen! plaster, srnoothest pipes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Cernen! plaster (3 cernen! to 1 sand), srnooth iron pipes . . . . . . . . . . . . . . . . . . . . . . Unplaned tirnber evenly laid, ordinary iron pipes .. . .... .. . .. ........ ... .. .... Ashlar rnasonry, best brickwork, well-laid sewer pipeb . . . . . . . . . . . . . . . . . . . . . . . . Average brickwork, foul planks, foul iron pipes, ordinary sewer pipes after average uneven settlernent and average fouling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Good rubble rnasonry, concrete laid in rough forrns, poor brickwork, heavily incrusted iron pipes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

From the channel geometry and fluil volumetric flow rate. For a rectangul

0.009 0.010 0.011 0.012 0.013

is to be determined, we use a trialchannel; we multiply Eq. 9.1-15 by bJ R~¡=

and rearrange to obtain

0.015

h=![(2h+b

0.017

Channels Subject to Nonuniforrnity of Cross Section

If we assume a value of h, we can th value of h and test the validity of th1

Excellent clean canals in firrn grave!, of fairly uniforrn section; rough rubble, "drying paving" . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.020 Ordinary earth canals and rivers in good order, free frorn large stones and heavy weeds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.025 Canals and rivers with rnany stones and weeds . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.03-0.04

yield close agreement between the a¡ fluid depth is determined. More con channel, Iead to a more tedious tria! is straightforward.

a Data fro m L. Marks and T. Baumeister, eds. , Marks' Mechanica/ Englneers' Handbook . 6th ed. (New York : McGraw-Hill Book Company.Inc. 1958). pp. 3- 81. A more tho rough list of values.is given in Ref. 2. b This last value should be used for the previous categories, if doubt exists asto the exccllencc of construction and the maintenance free from slime, rust, or other growths and deposits.

careful to remember that n has units of ftl/ 6 • Note that the value off given by Eq. 9.1-14 is independent of the Reynolds number, indicating that it is only valid for the " rough-pipe" region of flow where inertial effects predominate. A detailed discussion of the drag force is given by Chow.2 l. R. Manning, "On the Flow of Water in Open Channels and Pipes," Trans. Inst. Civ. Engrs. Jreland, 1891,20 :161. 2. V. T. Chow, Open Channel Hy drau/ics (New York : McGraw-Hill Book Cornpany, Inc., 1959), Chap. l.

1

Mechanical energy balance

If the flow is incompressible and may use the results of Prob. 7-2 to 1

:t fr.w(!pv~

dV

+j

= f V • ttn) dA A.,(t)

A

-

Open Channel Flow

Chap. 9

Sec. 9.1

353

Uniform Flow

Flow rate-channel depth calculations for uniform flow

bh

S

for the friction factor, f The most we may write in the form 2

16~ R~ 13

If the geometry and fluid depth for a channel are given, the determination ofthe flow rate is straightforward. Ifwe substitute Eq. 9.1-14 into Eq. 9.1-13 and solve for (v.,), we get

(9.1-14)

uuJ~nn.c:::~s.

Since f

and values for severa! types is dimensionless, we must be

8g sin

··· ··b ····· ··· ······· ····· ·· ·. .

····· ·· ·· ···· ·· ······ ·· ··· ... .

0.015

forms, poor brickwork, heavily

uniform section; rough rubble, 0.020 , free from large stones and heavy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.025 . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.03-0.04 Mechan/cal Englneers· Handbook . 6th ed. (New York : thorough list of values.is gi ven in Ref. 2. if doubt exists asto the excellence of construction and deposits.

ft 116 . Note that the value off given number, indicating that it is of flow where inertial effects predrag force is given by Chow. 2 Open Channels and Pipes," Trans. Inst. Civ. (New York : McGraw-Hill Book Company,

(9. 1-15)

eR:'

3

rectangular channel

(9.1-16)

If the volumetric flow rate and channel geometry are given and the fluid depth is to be determined, we use a trial-and-error solution. For a rectangular channel, we multiply Eq. 9.1-15 by bh, express the hydraulic radius as

and rearrange to obtain

1[

h = - (2h

b

0.017

······· ·· ···· · ··· ···· ... o·· · ···

413

116n2

116n 2

n, ft 1 i 6

pipe .. .. . ..... .. . . ... . . .. . . . ordinary sewer pipes after average

eR 11

From the channel geometry and fluid depth, we may readily compute the volumetric flow rate. For a rectangular channel,

Q = bh

0.009 0.010 0.011 0.012 0.013

8g sin

v·ndA

Here we have made use of the fact that

Q = (v,. )bh = constant

(9.2-18)

Differentiating Eq. 9.2-18 or Eq. 9.2-5 to obtain

bh d(v,.) + (v,.)b dh dx dx

f

(9.2-15)

carrying out the indicated integration and differentiation, and using the approximationt (9.2-16) we get d(v,.) d (bh p(v,.)bh - = pg,.bh - pg.dx dx 2

NFr)- NFr(~)b

+ (v,.)h db = dx

O

(9.2-19)

t Chow (Ref. 2, p. 28) indicates that Eq. 9.2-16 may be in error by as muchas 20 per cent; hence, any study of real systems must incorporate the necessary correction factor.

A.

=

A0

Applying this result to the differenti collecting the terms in Eq. 9.1-19, we

Open Channel Flow

Chap. 9

Eq. 9.2-10 is easily overlooked, for our forces at the entrance and exit of the

Sec. 9.2

Gradually Varied Flow

359

and applying this result to Eq. 9.2-17, we can obtain the following dimensionless form of the differential-macroscopic momentum balance, dh[g• _ (v.,?J dx g gh

(9.2-11)

_ (v.,) (~) db =

2

gh

b dx

g., _ T 0 S g pgbh

(9.2-20)

For practica! purposes, gradually varied fiows are restricted to cases where O is less than 10°, and under these conditions g. =coso~ 1 g

· uD -g., = sm

(9.2-12)

g

~

tan

= - (dr¡) dx

{)

which leads to balance yields dh (1 - NFr)dx h(x)

)

tlxf Pu dz - S tlx To

(9.2-13)

-o, we get dz

+ (~~)

NFr(~)db = b dx

(dr¡ dx

+

(9.2-21)

ToS )

pgbh

We have replaced (v.,)2/gh with the Froude number, so that it becomes clear the surface profile depends on whether NFr is greater or less than one. Flows for which NFr = 1 are called critica/ fiows; NFr > 1 represents a supercritica/ jlow and NFr < 1 a subcritica/ fiow.

h(x)

f

Pu dz -

ToS

(9.2-14)

o

Mechanical energy balance

Starting with Eq. 9.1-19, we express the kinetic energy flux termas

J

(9.2-15) and differentiation, and using the (9.2-16)

2

(ipv )v • n dA = ip(v!>AI

"'

Using Eq. 9.1-20 to express 4> and writing the gauge pressure as

Pu (9.2-17)

(9.2-22)

- ip(v!>AI x+4x

A,

=

pg[(h - z) cos O]

(9.2-23)

we can combine the pressure and gravitational terms in Eq. 9.1-19

-JPuV • n dA - Jptf>v • n dA = - pgJ(r¡ + h cos 6)v • n dA (9.2-18)

A,

A,

(9.2-24)

A,

Applying this result to the differential volume element in Fig. 9.2-2, and collecting the terms in Eq. 9.1-19, we get db + (v.,)h=O dx

(9.2-19)

.2·16 may be in error by as muchas 20 per the necessary correction factor.

iP(AI x+4x

- (v;)AI ) = - pg[(r¡ x

+ h cos O)(v.,)AI x+4x

- (r¡

+ h cos O)(v.,)AIJ

-

Ev

(9.2-25)

Open Channel Flow

360

Chap. 9

Dividing by !::u and taking the limit ó.x -- O, we have the differential macroscopic mechanical energy balance

iP .!!._ (A) = - pg .!!._ [(11 dx

dx

+ h cos e)(v.,)A] - e,

(É")

Gradually Varied Flow

to eliminate the velocity gives

(9.2-26)

where • 1'1me,=

Sec. 9.2

(9.2-27)

Az-o Ó.x

Assurning that (9.2-28)

For a given volumetric flow rate and s h. Exluding negative and complex valu values of h or none. The solutions are for each value of Q2/2gb 2 there is a mil is no solution for h. By inspection of comes Iarge the fluid depth may tend t

and using Eqs. 9.2-18 and 9.2-19, we can rearrange Eq. 9.2-26 to obtain

!!_ [ (v.,)2 dx

2g

+ h] = -

(d1] dx

+ ~)

(9.2-29)

pgQ

6

providing e is less than 10°. Referring to Eq. 9.1-26, we see that the viscous dissipation per unit length e, for uniform ftow is given by

~=~= pgQ

sin

e=

- d7]

pgQL

for uniform flow

(9.2-30)

dx

4

lf the viscous dissipation for gradually varied flow is very nearly the same as for uniform flow, we may expect the right-hand side of Eq. 9.2-29 to be small. Integration of Eq. 9.2-29 gives 2

(v.,) 2g

+h=

-J

5

(d1] dx

+ ~) dx + constant

h

(9.2-31)

pgQ

which is generally written

(v.l

--

2g

+ h-E Bp

(9.2-32)

E 8P is known as the specific energy, and it should be very nearly constant for gradually varied flow, although the changes in the geometry and roughness of the channel will certainly lead to small variations. Ifthe flow rate and channel width are given and values for ,.o are available, either Eqs. 9.2-21 and 9.2-18 or Eqs. 9.2-29 and 9.2-18 can be used to solve for the two unknowns, (v.,) and h. In general, the process requires numerical integration, although analytic methods may be used to obtain approximate solutions for simple cases. The mechanical energy balance is not as powerful a too! for solving open channel ftow problems as it was for solving closed conduit problems; however, it can be used to good advantage to illustrate sorne important characteristics of open channel ftow. Combination of Eqs. 9.2-18 and 9.2-32

Fig. 9.2-3. Fluid depth as a

We may determine the value of h fe the derivative dEsp/dh equal to zero a1 dEsp = 1 _ dh

This condition corresponds to what i energy, fluid depth, and velocity all a:

Open Channel Flow --+-

[('17

Chap. 9

O, we have the differential macro-

+ h cos O)(v.,)A] - e,

Sec. 9.2

Gradually Varied Flow

361

to eliminate the velocity gives

º2

- -2 2

(9.2-26)

2gb h

+ h = E sp

(9.2-33)

For a given volumetric flow rate and specific energy, Eq. 9.2-33 is a cubic in

(!:)

(9.2-27)

(9.2-28) rearrange Eq. 9.2-26 to obtain

h. Exluding negative and complex values of h, this equation yields either two

values of h or none. The solutions are shown in Fig. 9.2-3, where we see that for each value of Q2/2gb 2 there is a mínimum value of Esv below which there is no solution for h. By inspection of Eq. 9.2-33, we can see that as E sp becomes large the fluid depth may tend toward either zero or Ew

(9.2-29) Eq. 9.1-26, we see that the viscous flow is given by

= - d17

for uniform flow

(9.2-30)

dx

flow is very nearly the same as side ofEq. 9.2-29 to be small.

riL-Jitar:tu

~) pgQ

dx

+ constant

h

(9.2-31)

(9.2-32) should be very nearly constant for in the geometry and roughness of

Esp

Fig. 9.2-3. Fluid depth as a function of specific energy.

given and values for r 0 are available, and 9.2-18 can be used to solve the process requires numerical be used to obtain approximate as powerful a too! for solving open solving closed conduit problems; to illustrate sorne important · of Eqs. 9.2-18 and 9.2-32

We may determine the value of h for the mínimum value of E sp by setting the derivative dE5 vfdh egua! to zero and solving for h. (9.2-34) This condition corresponds to what is called critica! flow, and the specific energy, fluid depth, and velocity all assume their critica! values. Expressing

362

Open Channel Flow

Chap. 9

Sec. 9.2

Gradually Varied Flow

Eq. 9.2-34 in terms of the velocity, we find the critica! depth h. to be h = (v.,)~ e

(9.2-35)

g

----

and the critica! velocity is (9.2-36)

Substitution of these results into Eq. 9.2-33 indicates that the critica! specific energy is always three halves the critica! depth (9.2-37)

In terms of the Froude number, we note that Eq. 9.2-35 gives Flg. 9.2-4. Establishm

for critica! flow

(9.2-38)

There are perhaps two reasons why the condition NFr = 1 is designated as the critica! ftow. Ifwe remember that Eqs. 9.2-29, 9.2-30, and 9.2-31 indicated the specific energy was only approximately constant, we can certainly expect small variations in Esp from changes in geometry and channel roughness. At the critica! condition, the slope of h versus E 8 P is infinite; hence, small variations in E 8 P lead to large variations in h and wave motion is likely to occur. For this reason, designers try to avoid such ftows. A second reason for designating NFr = 1 as the critica! ftow is that it separa tes two fairly distinct ftow regimes. When NFr < 1, the fiow is subcritica/ or tranquil and disturbances can travel upstream; thus, a change in downstream conditions can alter the ftow upstream. When NFr > 1, the ftow is supercritica/ or rapid, and downstream disturbances are not propagated upstream. The speed at which a disturbance is propagated will be discussed in Sec. 9.3.

As an example of the application of the equations developed in this section, we will investigate the gradually varied fiow indicated in Fig. 9.2-4 to determine the length required for establishing uniform ftow. This example will also serve to introduce the student to the variety of surface profiles he may encounter for any particular fiow system. We start with the differential momentum balance, expressing the Froude number in terms of the volumetric fiowrate and restricting the analysis to a rectangular channel of constant width.

2 3

r0

= 14.5p

If we further restrict the analysis to ver¡ the approximations

Under these conditions, we can combir

dh(1- ~)=sil 2 3 dx

gb h

We have replaced -dr¡fdx with sin O. T simplified somewhat by noting that

Establishment of uniform flow

dh(l~)- -(dr¡ + ~) dx gb h dx pgbh

Assuming that the Manning formula ftows, we can combine Eqs. 9.1-13 and! of the volumetric ftow rate to obtain

º2 (gbº2h

gb 2 =

2

because Q 2 fgb 2h~ is the Froude number • to one. Equation 9.2-41 can now be an

~~[1- (~n =sin o

If we designate hnt as the fluid depth f (9.2-39)

t The subscript n is used because the de¡ called the normal depth.

Open Channel Flow

Chap. 9

Sec. 9.2

Gradually Varied Flow

363

find the critica! depth he to be (v.,)~

(9.2-35)

g

-- --

(9.2-36)

indicates that the critica! specific depth (9.2-37)

that Eq. 9.2-35 gives Fig. 9.2--4. Establishment of uniform ftow.

(9.2-38)

the condition NFr = 1 is designated as s. 9.2-29, 9.2-30, and 9.2-31 indicated constant, we can certainly expect geometry and channel roughness. At versus Esp is infinite; hence, small in h and wave motion is likely to avoid such flows. = l as the critical flow is that it When NFr < l, the flow is subtravel upstream; thus, a change in flow upstream. When NFr > 1, the disturbances are not propagated is propagated will be discussed in

Assuming that the Manning formula can be applied to gradually varied flows, we can combine Eqs. 9.1-13 and 9.1-14 and express the result in terms of the volumetric flow rate to obtain To

=

Q2n2

(9.2-40)

14.5p 2 2 1/ 3 b h R,.

If we further restrict the analysis to very wide channels, b the approximations

> h, we can make

R,.~h s~b

Under these conditions, we can combine Eqs. 9.2-39 and 9.2-40 to find

dh( 1 -

-

dx

Q2 )

~

gb h

. Q2n2 = sm () - 14.5 2 1013

gb h

(9.2-41)

We have replaced -dr¡/dx with sin O. The left-hand side of this result may be simplified somewhat by noting that of the equations developed in this varied flow indicated in Fig. 9.2-4 · uniform flow. This example to the variety of surface profiles he system. balance, expressing the Froude and restricting the analysis to a

-(dr¡ dx

+ ~) pgbh

Q2 -

gb2 -

(L)ha- ha gb2h~

e -

e

(9.2-42)

because Q 2jgb 2h~ is the Froude number for critica! flow and is therefore equal to one. Equation 9.2-41 can now be arranged in the form

dh[1(he)3] =sin ()(1- 14.5Q2n2 ) dx h gb h 1 sin () 2 10 3

(9.2-43)

If we designate hnt as the fluid depth for uniform flow (h-+ hn as x-+ co), (9.2-39)

t The subscript n is used because the depth under uniform ftow conditions is often called the normal depth.

364

Open Channel Flow

Chap. 9

Sec. 9.2

we may rearrange Eq. 9.1-16 to obtain (9.2-44) Our original differential momentum balance may now be written in compact forro dh dx [ 1 -

(hhe)3] = sin e[1 - (hhn)10/3]

(9.2-45)

which is to be sol ved subject to a boundary condition of the type B.C. 1:

x=O

(9.2-46) Critic

Surface profiles

Before going on to obtain an approximate solution to Eq. 9.2-45, we will discuss the possible surface profiles this equation may describe. Our purpose is not to acquaint the student with the nine specific surface profiles associated with the flow downstream from a sluice gate, but rather to drive borne the point that flows in open channels can assume many possible forros. In his discussion of gradually varied flow, Chow (Ref. 2, p. 222) lists severa! dozen commonly encountered surface profiles. Thus, in analyzing flow in open channels it is extremely important to have knowledge of experimentally observed surface profiles, for it is especially difficult to "guess the flow topology" if the various possibilities are unknown. For a given volumetric flow in a given channel, it should be intuitively obvious that the normal depth hn will decrease with increasing slope. This idea is expressed quantitatively in Eq. 9.2-44 and leads us to classify channels as mild, i.e., having a slope small enough so that the normal depth is greater than the critica! depth, (hn > he), critica/ (hn = he), and steep (hn < he). These conditions are illustrated in Fig. 9.2-5, and the flow at the critica! slope is wavy or undular in agreement with our previous comment on the instability of critica! flow. We now wish to examine Eq. 9.2-45 for mild, critica!, and steep channels downstream and upstream from a sluice gate. For mi!d channels, hn > he, and we must consider three possibilities in the gradually varied flow region. Ml. h > hn >he, subcritical flow; M2. hn > h > he, subcritical flow; M3. hn > he > h, supercritical flow.

Steep

Fl¡. 9.2-5. Classifi

Examining Eq. 9.2-45, we find the foil Ml. dhfdx M2. dhfdx M3. dhfdx

> O, backwater curve; < O, drawdown curve; > O, backwater curve.

The condition that h increases with in called a backwater curve, while a dect The three profiles for mild channels are approaches the critica} depth, the profil as Eq. 9.2-45 indicates, the flow cann< The Ml backwater curve is the typc dam (or the sluice gate shown in Fig curve is important to reservoir desig

Open Channel Flow

Chap. 9

Sec. 9.2

365

Gradually Varied Flow

---

(9.2-44) nce may now be written in compact

(9.2-45) condition of the type

x=O

(9.2-46)

solution to Eq. 9.2-45, we will equation may describe. Our purpose specific surface profiles associated gate, but rather to drive home the assume many possible forms. In his (Ref. 2, p. 222) lists severa! dozen Thus, in analyzing flow in open have knowledge of experimentally difficult to "guess the flow unknown. channel, it should be intuitively decrease with increasing slope. This and Ieads us to classify channels so that the normal depth is greater (hn = he), and steep (hn < he). and the flow at the critica! slope previous comment on the instability Eq. 9.2-45 for mild, critica!, and from a sluice gate. consider three possibilities in the

VAIIllla'""

........ ....

.........

Steep

.............

........ .............................

Fl1. 9.1-5. Classification of channels.

Examining Eq. 9.2-45, we find the following restrictions on dhfdx: Ml. dhfdx M2. dhfdx M3. dhfdx

> O, backwater curve; < O, drawdown curve; > O, backwater curve.

The condition that h increases with increasing distance down the channel is called a backwater curve, while a decrease in h is called a drawdown curve. The three profiles for mild channels are illustrated in Fig. 9.2-6. As the depth approaches the critica! depth, the profile is drawn with a dashed line, because, as Eq. 9.2-45 indicates, the flow cannot be gradually varied as h --+-he. The Ml backwater curve is the type of profile that exists upstream from a dam (or the sluice gate shown in Fig. 9.2-4). Knowledge of the backwater curve is important to reservoir design, because the profile líes above the

366

Open Channel Flow

---------~---- ~--

MI

-..._....._

-------~

----

-- ----

---------------__------- ---.

__

-_-:~

Chap. 9

......

--

............_.............._

Sec. 9.2

Gradually Varied Flow

The C 1 and C2 backwater curves o< may exhibit hydraulic jump. The C3 ba from a sluice gate, and a mild, or undull supercritical to critica! flow.

......__---- , - - - - '

~ --·- -----... --h .. = he, subcritical ftow; C2. h11 = he = h, critica! ftow; C3. h 11 = he > h, supercritical ftow. Examination of Eq. 9.2-45 for this channel indicates: Cl. dhfdx >O, backwater curve; C2. dhfdx > O,t backwater curve; C3. dhfdx > O, backwater curve. The three profiles for critica! channels are illustrated in Fig. 9.2-7. t This result is easily obtained by application of I'Hospital's rule to Eq. 9.2-45; however, the actual existence of such a profile is open to question.

__

------- ---~, , -

Fig. 9.2-7. Surface profilc

F or steep channels, he > h11 , and th SI. h > he> h.., subcritical flow; S2. he > h > h.., supercritical flow; S3. he > h 11 > h, supercritical flow. These conditions lead to the following SI. dhfdx > O, backwater curvt:; S2. dhfdx < O, drawdown curve; S3. dh fdx > O, backwater curve.

These surface profiles are illustrated in The Sl backwater curve is the type t The S2 and S3 profiles can occur downs1 being similar to that illustrated in Fig.

Open Channel Flow

Chap. 9

Sec. 9.2

Gradually Varied Flow

367

The Cl and C2 backwater curves occur upstream from a dam, and both may exhibit hydraulic jump. The C3 backwater curve may exist downstream from a sluice gate, and a mild, or undular, jump occurs at the transition from supercritical to critica! flow.

for mild channels.

flooding behind a dam may extend of the water in the reservoir. from a sluice gate. As the transition from supercritical to subís known as a hydraulic jump. occur for the configuration shown in a change in channel slope from mild again we have three possibilities in the

Fig. 9.2-7. Surface profiles for critica! channels.

For steep channels, he> hm and the possible types of flow follow: SI. h >he > hm subcritical flow; S2. he > h > hm supercritical flow; S3. he > hn > h, supercritical flow. These conditions Iead to the following surface profiles:

indicates:

are illustrated in Fig. 9.2-7. ofi'Hospital's rule to Eq. 9.2-45; howto question.

SI. dhfdx > O, backwater curvt:; S2. dhfdx < O, drawdown curve; S3. dhfdx > O, backwater curve. These surface profiles are illustrated in Fig. 9.2-8. The Sl backwater curve is the type that may occur upstream from a dam. The S2 and S3 profiles can occur downstream from a sluice gate, the S2 profile being similar to that illustrated in Fig. 9.2-5.

368

Open Channel Flow

Chap. 9

Sec. 9.2

Gradually Varled Flow

allows us to write Eq. 9.2-45 as

To extractan analytic solution from mation ][10

If t < H < 2, this approximation w however, it is not excessive when coro in the use ofEqs. 9.2-16 and 9.2-28 a factor. Simplifying Eq. 9.2-47, separatin

where

e is the constant of integrat

B.C. 1':

H=H0

The integral in Eq. 9.2-49 is available condition 1', is

Fig. 9.2-8. Surface profi.les for steep channels.

Once again, the point to be made in this discussion is that a variety of surface profiles are possible for a given geometrical configuration and it is necessary to have sorne idea what to expect before plunging into an analysis. Returning now to Eq. 9.2-45, we wish to extractan approximate solution for the approach to uniform flow downstream from a sluice gate. The fl.ow at the sluice gate cannot be classified as gradually varied; however, at a short distance downstream fwm the gate the variations in fluid depth are more gradual and Eq. 9.2-45 is satisfactory. Defining the dimensionless quantities H

= -h ,

dimensionless fluid depth

hn

X = x sin O hn

dimensionless length

'

fJl = he , ratio of critical to normal depth hn

+ j_ [tan_1 (2H + 1)

J3

J3

F or practica! purposes, we can assu when h is within l per cent of h.,. or ll required to achieve uniform ftow · designated by L,.

Entrance lengths

For mild channels, only the M3 gate, and a hydraulic jump occurs ir type, 9l < 1, and the maximum entr because the term in braces ({ }) is al

3. H. B. Dwight, Tables of Integrals a. Macmillan Company, 1961), p. 42.

Open Channel Flow

Chap. 9

Sec. 9.2

Gradually Varied Flow

369

allows us to write Eq. 9.2-45 as 3

dH( 1 - Ha [Jt ) dX =

(

1 ) 1 - Hto¡a

(9.2-47)

To extractan analytic solution from this equation we must make the approximation ¡pota~ na (9.2-48) If i < H < 2, this approximation will lead to an error of about 25 per cent; however, it is not excessive when compared to the percentage of error incurred in the use of Eqs. 9.2-16 and 9.2-28 and the Manning formula for the friction factor. Simplifying Eq. 9.2-47, separating variables, and integrating, we get

f(~3-=-~ ) 3

where

dH = X

e is the constant of integration to

B.C. 1':

+e

(9.2-49)

be determined by application of

X= O

(9.2-50)

The integral in Eq. 9.2-49 is available, 3 and the result, after applying boundary condition 1', is for steep channels.

H2)

Ho)2]

(H- Ho) +(&la- 1){In [( 1 + H + (16 1 + H 0 + H~ 1 - H

(9.2-51) in this discussion is that a variety of geometrical configuration and it is before plunging into an analysis. sh to extractan approximate solution m from a sluice gate. The flow gradually varied; however, at a short variations in fluid depth are more Defining the dimensionless quantities

For practica! purposes, we can assume that uniform flow has been reached when h is within 1 per cent of h .. or H is within 1 per cent of unity. The length required to achieve uniform flow will be caiied the entrance length and designated by L •.

Entrance lengths For mild channels, only the M3 profile occurs downstream from a sluice gate, anda hydraulic jump occurs in the neighborhood of H = 81. For this type, ,tjt < 1, and the maximum entrance length will be obtained for 91--+ 1 because the term in braces ({}) is always positive. Letting 81 = 1 to obtain

to normal depth

3. H. B. Dwight, Tables of Integrals and Other Mathematica/ Data (New York: The Macmillan Company, 1961), p. 42.

Open Channel Flow

370

Chap. 9

Sec. 9.3

The Solitary Wave

an upper bound gives

L = (0.99- H 0)h., sin 8 • and an order of magnitude estimate of the entrance length is

(9.2-52)

L.=o(~) sm O

(9.2-53)

Since h., is generally on the order of feet, and sin O may be 0.01 or less, the entrance length is on the order of 100ft or more for mild channels. For critical channels, fJt = l and the entrance length is given by Eq. 9.2-52. For steep channds, we see from Fig. 9.2-8 that there are two possible profiles leading to a uniform flow.: S2. fJt S3. fJt

~

H ~ 1;

~

1

~

H.

For these two limiting cases we will take H 0 ~ 1 and H 0 obtain; S2. L.= (h.,fsin O) [1.44fft3 S3. L. = (h.,fsin O) [2.02fft3

+ (1.01 + 0.99].

- H 0)];

< 1, respectively, to (9.2-54a) (9.2-54b)

In bath cases, the en trance length can be hundreds of feet long, and it should be apparent that uniform flow is the exception and varied flow the rule for most practical cases.

9.3 The Solitary Wave To understand sorne ofthe phenomena encountered in open channel flow, we must determine the speed at which a disturbance to the surface profile moves. We first consider a disturbance caused by a movable wall which boun4s an initially quiescent body of water. This process is illustrated in Fig. 9.3-1, and a sketch of the control volume to be u sed in the analysis is shown in Fig. 9.3-2. We assume that the wave profile is preserved and choose a differential control volume that moves with sorne specific portion ofthe wave. Note that the control volume is not a material volume, for it is the profile which moves, not the water. In this respect, Watts' has written that: "Buddhism has frequently compared the course of time to the apparent motion of a wave, wherein the actual water only moves up and down, creating the illusion of a 'piece' of water moving over the surface." 4. A. W. Watts, The Way ofZen (New York: Pantheon Books, Inc., 1957).

Flc. 9.3-1. Gener:

r------- - - flh

ho

h(x}

~"""'"~'' z

Lx

Flc. 9.3-2. Moving control vol

Open Channel Flow

Chap. 9

Sec. 9.3

371

The Solitary Wave

---

(9.2-52)

(9.2-53)

' and sin () may be 0.01 or Iess, the or more for mild channels. the entrance Iength is given by Eq. 9.2-8 that there are two possible

H0

~

1 and H0

~

1, respectively, to (9.2-54a) (9.2-54b)

hundreds of feet long, and it should and varied flow the rule for

Flg. 9.3-J. Generation of a solitary wave.

encountered in open channel flow, a disturbance to the surface profile caused by a movable wall which . This process is illustrated in Fig. to be used in the analysis is shown profile is preserved and choose a sorne specific portion ofthe wave. material volume, for it is the profile Watts' has written that:

Control volume

r----------flh

1----c ho

h(x)

the course of time to the apparent water only moves up and down, moving over the surface." : Panthcon Books, Inc., 1957).

Flg. 9.3-1. Moving control volume for analysis of the solitary wave.

372

Open Channel Flow

Chap. 9

Sec. 9.3

The Solltary Wave

Neglecting viscous effects and app, in Fig. 9.3-2, we bave

Mass balance

d

We begin the analysis with the macroscopic mass balance

~t Jp dV +Jp{v 7'"0 (t)

p dx

(9.3-1)

w) • n dA = O

Expressing the gauge pressure ast

A,(t)

pg=

and assume that w is a constant vector such that

w.,=c,

and carrying out the integration it (9.3-2)

w11 = w. =O

The mass in the control volume is a constant so that Eq. 9.3-1 immediately reduces to

J

p(v - w) • n dA

I.,+Az

+

A,W

J

p{v - w) • n dA

1

=O

d [(v.,(v., dx

p-

Assuming that the velocity profil

(9.3-3)

((v.,)-e)

A,W

Here we have used the fact that

Use of Eq. 9.3-7 to eliminate d(v.,

(v - w) • n

= O

h(z)

(v.,- e) dz

o

1

z+Az

-J

ll(z)

(v., - e) dz

o

1

.,

=O

(9.3-5)

d

dx [((v.,) - c)h] =O

(9.3-6)

+ (v.,) dh =

(9.3-7)

which we may also write, dx

e dh

dx

Momentum balance

The momentum of the fluid in the control volume will be a constant, so the x-component of Eq. 9.1-5 becomes

J pv.,(v- w) • n dA= Ji· [-np., + n ·~]dA ..t,(t)

.w

JI(

or

e=

Provided the amplitude of the wa depth !1h ~ h0, the fluid velocity (v.,) ~e, and Eq. 9.3-14 reduces

e

where the density has been removed because the flow is incompressible. Dividing by !1x, expressing the integrals in terms of average velocities, and taking the limit !1x --+ O, we get ·

h d(v.,) dx

(e-

(9.3-4)

at the free surface

Note that Eq. 9.3-4 does not imply that the control volume moves with the fluid (w = v), it simply means that the control surface always coincides with the free surface. Evaluation of the terms in Eq. 9.3-3 gives

f

[(v.,(v.,-

(9.3-8)

The result of our analysis app tions-i.e., the wave profile is pt speed of a particular point on tt constant. However, our analysis of waves m channels, for observ faster than the leading or trailing in front of the crest. However, if over the wave, and a reasonable :

t This equation assumes that the hJ the variations caused by fluid motion i t There is sorne tradition associate• time rate of change of position of the w word speed for v = .,¡;:-;: However, misinterpret. 5. J. J. Stoker, "The Formation o 1 :l.

Open Channel Flow

Chap. 9

Sec. 9.3

373

The Solltary Wave

Neglecting viscous effects and applying this equation to the control volume in Fig. 9.3-2, we bave /0(.,)

ros,:oo:tc mass balance

p ddx [(v.,(v., - e))h] = - ddx (9.3-1)

(9.3-9)

o

Expressing the gauge pressure ast

pg = (9.3-2)

Jpg dz

(9.3-10)

pg[h(x) - z]

and carrying out the integration in Eq. 9.3-9, we find

p ~ [(v.,(v.,- e))h] dx

stant so that Eq. 9.3-1 immediately

= -pgh dh

dx

(9.3-11)

Assuming that the velocity profile is flat, and using Eq. 9.3-6, we get (9.3-3) ((v.,)- e)h d(v.,)

= -gh dh

dx

dx

(9.3-12)

Use of Eq. 9.3-7 to eliminate d(v.,)fdx leads us to (9.3-4) the control volume moves with the surface always coincides with in Eq. 9.3-3 gives (9.3-5)

( e-

(v.,) ) 2

e= (v.,) ±

(9.3-6)

(9.3-7)

(9.3-13)

../gh(x)

(9.3-14)

Provided the amplitude of the wave is small compared to the quiescent fluid depth !l.h ~ h0 , the fluid velocity will be small compared to the wave speed~ (v.,) ~e, and Eq. 9.3-14 reduces to

e= the flow is incompressible. in terms of average velocities, and

= gh(x)

or

±../gh(x)

The result of our analysis apparently contradicts one of the first assumptions-i.e., the wave profile is preserved- for Eq. 9.3-15 indicates that the speed of a particular point on the profile depends upon h(x), which is not constant. However, our analysis does predict qualitatively the real behavior of waves m channels, for observation5 indicates that the wave crest travels faster than the leading or trailing edge and that the wave tends to get steeper in front of the crest. However, if !l.h ~ h0 , the speed will change only slightly over the wave, and a reasonable approximation for small amplitude waves is

e=±~ ntrol volume will be a constant, so

• [-np,

b

+ n •-r) dA

(9.3-8)

(9.3-15)

(9.3-16)

t This equation assumes that the hydrostatic pressure variations are large compared to the variations caused by fluid motion in the z-direction. t There is sorne tradition associated with the use of the word ce/erity to describe the time rate of change of position of the wave profile, reserving the word velocity for v and the word speed for v = VV:V. However, the term wave speed or wave velocity is difficult to misinterpret. 5. J. J. Stoker, "The Formation of Breakers and Bores," Comm. Appl. Math., 1948 1 :l.

37-4

Open Channel Flow

Chap. 9

We must clarify the assumption associated with Eq. 9.3-10. Expressing the pressure as hydrostatic is satisfactory only if pv!fh0 is negligible compared to pg (inertial effects are therefore small compared to gravitational effects). A careful study of wave motion• indicates that this assumption is satisfied when the wave length A. is long compared to ~he fluid depth. Such waves are often called shallow-water waves, and the wave speed is given correctly by Eq. 9.3-16. Waves for which the wave length is less than the fluid depth are called gravity waves, and their speed is given by

e= ±

/gi,

-J2;

gravity waves (A< h,}

Sec. 9.3

The Solltary Wave

.fih-

(9.3-17)

We see that gravity waves move ata slower velocity than shallow-water waves. If the wave length is very short-say, less than 1 cm-the wave velocity is governed by surface tension effects and these waves are called ripp/es. Their wave speed is given by e --

±

!5'"pA.(f '

ripples (A < 1 cm}

(9.3-18)

Both Eqs. 9.3-17 and 9.3-18 are restricted to the condition that the wave amplitude is small compared to the wave length, Ah < A.; thus, the results are applicable only for small disturbances. An important point to be cleared up before we discuss the result derived for shallow-water waves is the sign ± associated with the wave speed. There is nothing in the analysis which tells us whether the wave generated in Fig. 9.3-1 will travel to the left or to the right, but there is little doubt in our minds that it will move to the right. One might ask, "If our analysis is correct, why shouldn't it predict the direction in which the wave moves ?" The answer is that one must investigate carefully the actual generation phenomena in order to predict the direction of wave propagation. Stoker7 has presented a thorough treatment of this problem indicating that a disturbance always propagates into the "region of quiet," i.e., the undisturbed region. In our problem the undisturbed region is to the right of the movable wall; hence, the wave moves to the right. An analysis of wave propagation in a flowing channel is similar to the development presented here and Eq. 9.3-14 is again obtained with the velocity (v,.) being the normal flow ve1ocity. It is thus indicated that a disturbance plus the stream may propagate downstream at a velocity equal to velocity, and upstream ata velocity equal to minus the stream velocity. This situation is illustrated in Fig. 9.3-3 where a disturbance has been created by momentarily immersing a gate into the stream.

Jih

Jih

6. L. M. Milne-Thomson, Theoretica/ Hydrodynamics, 4th ed. (New York: The Macmillan Company, 1960}, Chap. 14. 7. 1. 1. Stoker, Water Waves (New York: Interscience Publishers, lnc., 1957), p. 303.

Fig. 9.3-3. Propagati

The interesting case to consi is given by Eq. 9.2-36 as Thus, for critical flow a disturb nel; for subcritical flow (such a• may propagate upstream; and ~ carried downstream. Note tha~ waves because they have a small may propagate upstream in a e¡

Thus, if the fluid depth is Ü(fe Such small waves are rapidly that shallow-water waves, grav· in supercritical flow.

Open Channel Flow

Chap. 9

Sec. 9.3

375

The Solitary Wave

with Eq. 9.3-10. Expressing the if pv!fho is negligible compared to to gravitational effects). A this assumption is satisfied when fluid depth. Such waves are often . speed is given correctly by Eq. 1s less than the fluid depth are called

.¡g¡; - 1, supercritical ftow; < 1, subcritical flow.

t This statement assumes that we do not have supercritical flow on a mild or critica) channel.

Open Channel Flow

378

L

Chap. 9

Sec. 9.'1

Flow Over Bumps, Crests, an

Supercriticol flow

-------------------1~:..-------~bcriticol

Flow over a crest flow

We now wish to examine the ¡:¡ in Fig. 9.4-2. Water flows from a reservoir formed by an adjustable take a chronological approach.

~

Z=H

Z=L----------Flg. 9.4-1. flow over a bump.

For case 1, Eqs. 9.4-1 and 9.4-4 indicate that dh

dx

>o

'

dh 0 dx '

on the upstream side

(9.4-6a)

on the downstream side

(9.4-6b)

Thus, for subcritical flow our analysis indicates that the profile dips down over the bump as indicated by the dashed curve in Fig. 9.4-1. This conclusion, of course, violates our assumption that h and Toare constant in the neighborhood of the jump. However, if h decreases and -r 0 increases owing to the increased velocity, the inequalities given by Eqs. 9.4-4 still hold, and we conclude that the profile rises over a bump when the flow is supercritical and dips when the ftow is subcritical. We must note· that this discussion is very qualitative in nature, and the behavior of the surface profile in the neighborhood of a bump is best understood in terms of experimental observation. The analysis is put in its proper place if we simply state that it does not contradict the experimental observation.

Let us consider first the case ftow occurs and the..water in the 1 so that a small fiow takes place, t profile will occur at the crest. F increase the ftow rate until the crit1 ftow has been subcritical and c1 However, once critica! flow is rea propagated upstream, and further out the downstream portion of th' tion is illustrated in Fig. 9.4-3, wh the crest, the critica! fiow at the c1 of the crest. The broad-chested weir

A broad-crested weir is shown as ftow rate measuring devices, a viously discussed sharp-crested v analysis of this ftow is identical t1 where the application of Bernoull

p + lpv2

•

Open Channel Flow

Chap. 9

Sec. 9.4

379

Flow Over Bumps, Crests, and Weirs

Flow over a crest We now wish to examine the problem of flow over a crest, as illustrated in Fig. 9.4-2. Water flows from a reservoir over the crest and into a second reservoir formed by an adjustable gate. In examining this problem, we shall take a chronological approach. Z=H----------------~L_-------------_

Z=L----------- h

(9.4-Sa)

(9.4-Sb) Fig. 9.4-2. Flow over a crest in a reservoir.

up over the bump, and our ~~lJLuuiul;ouu of the bump may be

(9.4-6a)

wnstream side

(9.4-6b)

that the profile dips down over Fig. 9.4-1. This conclusion, of are constant in the neighborhood increases owing to the increased still hold, and we conclude that supercritical and dips when the discussion is very qualitative in in the neighborhood of a bump The analysis is put in not contradict the experimental

Let us consider first the case where the control gate is raised so that no flow occurs and the...water in the reservoir is quiescent. If we Jower the gate so that a small flow takes place, the flow will be subcritical anda dip in the profile will occur at the crest. Further Iowering of the control gate will increase the flow rate until the critica! velocity is reached. Up to this time the flow has been subcritical and controlled by the downstream conditions. However, once critica{ flow is reached small disturbances can no longer be propagated upstream, and further lowering of the control gate simply washes out the downstream portion of the dip in the pro file at the crest. This situation is illustrated in Fig. 9.4-3, which shows the subcritical flow upstream of the crest, the critica{ flow at the crest, and the supercritical flow downstream of the crest. The broad-chested weir A broad-crested weir is shown in Fig. 9.4-4. Broad-crested weirs are used as flow rate measuring devices, and have a certain advantage over the previously discussed sharp-crested weir (Sec. 8.5). The initial portion of the analysis of this flow is identical to that presented for the sharp-crested weir where the application of Bernoulli's equation to any streamline gave us

p + lpv 2 + pgz = p0

+ pgH

(9.4-7)

380

Open Channel Flow

Chap. 9

Subcriticol flow

Sec. 9.4

Flow Over Bumps, Crests, and

The broad-crested weir has its grea reached the critica! value. Under ti

Critico! flow

V . e

and Eq. 9.4-9 is written as

J2i

Ve=

Supercriticol flow

We may now eliminate h., solve for rate,

Q=b~

'

Flg. 9.4-3. Flow over a crest in a reservoir.

Over the central portion of the weir, the streamlines are straight and the pressure is hydrostatic, (9.4-8) p =Po pg(he L - z)

+

+

This result ts similar in form to t however, in deriving this equation coefficieut to correct the assumpti been able to specify the fluid depth The results obtained in this se flows and negligible viscous effects. should be regarded only as estímate

and the velocity in Eq. 9.4-7 is given byt

v

= J2g .Jcn- L)- h

Calculatiorl of flow rate (9.4-9)

This equation could be used to determine the volumetric ftow rate over a broad-crested weir if JI and h were measured with depth gauges. Because the streamlines are straight, the volumetric ftow rate per unit width is

q = h.fii

.Jcn- L)- h

(9.4-10)

For a value of 11 L-= 0.525 unit width, and the fluid depth on By Eq. 9.4 -13 the volumetric fio

q = .Jg'[i(H- L))312 = or

q=l. Because the flow on the weir will b

q =vA=

Jih.

and. sol ve for he:

he= i(H

Experiments described by Doering conditions, q= 1

he= O Flg. 9........ Broad-crested weir.

t Remembcr that Bernoulli's equation neglects viscous effects; therefore, it is not unreasonable that Eq. 9.4-9 indicates a ftat velocity profile.

Thus, the actual flow rate is within the measured fluid depth on the wei value. In view of the simplificatio between measured and calculated f1

8. H. A. Doeringsfeld and C. L. Bark

Open Channel Flow

Chap. 9

Sec. 9.4

Flow Over Bumps, Crests, and Weirs

381

The broad-crested weir has its greatest utility when the flow at the weir has reached the critica! value. Under these conditions, the velocity is given by

.fihc

(9.4-11)

J2i ~(H- L)- h.

(9.4-12)

v.

=

and Eq . 9.4-9 is written as

v. = ulic jump

We may now eliminate h., solve for rate,

v.,

and obtain the total volumetric flow (9.4-13)

L- z)

(9.4-8)

This result IS similar in form to that obtained for the sharp-crested weir; however, in deriving this equation we have not had to introduce a discharge coefficieut to correct the assumption of a flat velocity profile, and we have been able to specify the fluid depth at the weir. The results obtained in this section are restricted to gradually varied flows and negligible viscous effects. Calculations made using these equations should be regarded only as estimates. Calculatiorl of flow rate

(9.4-9)

the volumetric flow rate over a with depth gauges. Because rate per unit width is L)- h

(9.4-10)

For a value of H - L = 0.525 ft, determine the volumetric flow rate per unit width, and the fluid depth on a broad-crested weir. By Eq. 9.4-13 the volumetric flow rate per unit width is given by

q = .Ji[i(H- L)]312 = ~32.2 ft/sec 2[(i)(0.525 ft)] 312 or

q = 1.17 ft 3/ft-sec Because the flow on the weir will be critica!, we may express q as

q =vA= Jg'h;h. = }.g[i(H- L)]a12 and. sol ve for h.: h. = i(H - L)

=

0.35 ft

Experiments described by Doeringsfeld and BarkerS indicate that for these conditions, q = 1.08 ft 3/ft-sec

h.= 0.28 ft weir.

viscous effects; therefore, it is not profile.

Thus, the actual flow rate is within 8 per cent of the calculated value, while the measured fluid depth on the weir is 20 per cent smaller than the calculated value. In view of the simplifications made in the analyses, the agreement between measured and calculated flow rates is remarkable. 8. H. A. Doeringsfeld and C. L. Barker, Trans. A.S.C.E., 1941, 106:934-46.

Open Channel Flow

382

9.5

Chap. 9

Sec. 9.5

Hydraulic Jump

after the jump. Expressing the left-h and evaluating the right-hand side, '

Hydraulic Jump

A hydraulic jump occurs at a transition from supercritical to subcritical flow. This transition may be brought about by a change in the slope of the channel, such as the jump illustrated in Fig. 9.4-3, or it may occur when a supercritical flow is formed on a rnild channel by a sluice gate. This situation is illustrated in Fig. 9.2-4. In analyzing the hydraulic jump, we will consider the flow shown in Fig. 9.5-1. As shown, it consists of a supercritical flow coming off a steep channel onto a mild channel. On the mild channel, the depth gradually increases and then undergoes a sudden change, which is called an hydraulic jump. We will, in the course of this analysis, show that the jump may be lo=ocf>+v·Vcf> Dt ot

(10.1-11)

Since the gravitational potential e/> is independent of time, ocf>fot = O and we may combine Eqs. 10.1-11 and 10.1-lO·to obtain

(10.1-6)

De/> Dt

pg•v= - p at which work is done on the body 1-5 represents the rate at which work show shortly that this term can be of the potential energy of the body, Jeft-hand side of Eq. 10.1-5~ This been encountered in thermodynamics (10.1-7) by surface forces. of the Reynolds transport theorem p DS dV

Dt

f

pg · v dV

+

Substitution of this result into Eq. 10.1-9 yields the differential energy equation in terms of internal, kinetic, and potential energy. Time ratc of changc or intcrnal. klnctic and potcntial cncrgy pcr unit volume

D 2 p - (e+ tv

Dt

Ratc of hcat transfcr pcr unlt volume

Ratc of surfacc work pcr unlt volume

+e/>)= -V· q +V· (T • v)

(10.1-13)

We will refer to this resultas the total energy equation. In Sec. 10.2, we will show how the thermal energy equation may be derived by subtracting the mechanica/ energy equation from Eq. 10.1-13. In the analysis of compressible flows, the total energy equation is the most useful forro of the principie of conservation of energy. In this brief introduction to compressible flow, we will only use the macroscopic total energy balance, and Eq. 10.1-13 must be integrated. The integration is complicated by the presence of the density p outside the material derivative; we must remedy thi.'s difficulty before proceeding. Letting O represent the total energy per unit mass, (10.1-14) n =e+ tv2 cP

+

f

(n • T) • v dA (10.1-8)

derivative.

(10.1-12)

and expanding the material derivative in Eq. 10.1-13, we get

an

p - + pv ·Vil= -V· q +V· (T • v)

ot

(10.1-15)

Chap. 10

Sec. 10.2

The Thermal Ener¡y Equation ant

Noting that n [(opfot) +V· pv] is zero by the continuity equation, we may add this term to the left-hand side of Eq. 10.1-15 to obtain a form suitable for integration.

Adding

a~d subtracting p/ p from O yic

Compressible Flow

394

(10.1-16)

dt Carrying out the integration over the arbitrary moving volume "Ya(t), and applying the divergence theorem to the second, third, and fourth integrals, we get

+ J(pvO) • n dA=

(pO) dV

1'"0 (!)

d

0

-

f d

4

(l)

(10.1-17)

do(l)

Applying the general transport theorem, and noting that the convective energy flux term will be zero except over the area A.(t), we obtain the macroscopic total energy balance

~t

f

pO dV

+

f

,V.(t)

f

pO(v- w) • n dA=- q • n dA+ Jt

q • n dA+ Jt)v

If we neglect variations of h, v1, and restrict oprselves to a system having write Eq. 10.1-24 as A(h

+ !v1 +

where m is the mass flow rate, and A exit conditions (where v • n is positive] Y· n is negative). With the mass, momentum, and en ready to analyze the propagation of a " and begin our investigation of compress we should tie up sorne loose ends re encountered in the derivation of the me section, therefore, we will derive the tl the viscous dissipation does indeed apJ1

10.2 The Thermal Energy Equa the Entropy Equation Our objective in this section is to p energy equations in proper perspectiv We start the analysis with the total en D

p -(e + lv') Dt

= -V· •

+ PI• Y+

Chap. 10

Sec. 10.2

by the continuity equation, we may Eq. 10.1-15 to obtain a form suitable

Adding

Compressible Flow

(10.1-16)

-V· q +V· (T · v)

arbitrary rnoving volume "Ya(t), and second, third, and fourth integrals,

-f

q • n dA+

.ofoW

Jt(n) •V dA .ofoW

q • n dA+

..,• (t)

Jt(n) •

V

dA

(10.1-18)

JI(. (t)

area integra1st over A,(t) and A,(t)

-f

q • n dA

..,.w

a~d subtracting p/ p from O yields

fp(

h

+ tV1 + cfo- E)v •n dA= Q + W + Jt1• 1 • vdA p

~

(10.1-22)

~

where the integral representing the rate of heat transfer to the system has been replaced by Q. As we have noted in severa! previous examples, the stress at entrances and exits may be approximated by

Substitution of this expression for the stress vector into Eq. 10.1-22leads to a cancellation of terrns and we obtain

J

p(h

+ lv1 + cfo)v • n dA = Q + W

(10.1-24)

..t. If we neglect variations of h, v1 , and cfo across the entrances and exits and restrict ourselves to a system having one entrance and one exit, we may write Eq. 10.1-24 as

(10.1-25)

+ W + f t(n) •V dA ..t.< t)

(10.1-19) this text always refers to the rate of rnoving surfaces. In thermodynamics the useful work, while the work done (the 1ast terrn in Eq. 10.1-19) is referred general forrn of the macroscopic total ourselves in this chapter to fixed and steady state conditions. Under to for fixed control surfaces at entr.ances and exits and steady state conditions

(10.1-20) this result in terrns of the enthalpy per

(19.1-21)

where m is the mass ftow rate, and ti indicates the difference between the exit conditions (where v • n is positive) and the entrance conditions (where v • n is negative). With the mass, momentum, and energy balances at our disposal, we are ready to analyze the propagation of a weak sonic disturbance (a sound wave) and begin our investigation of compressible ftows. However, before doing so, we should tie up sorne loose ends regarding the viscous dissipation terrn encountered in the derivation of the mechanical energy equation. In the next section, therefore, we will derive the therrnal energy equation to show that the viscous dissipation does indeed appear as thennal energy.

10.2 The Thermal Energy Equation and the Entropy Equation Our objective in this section is to put the total, therrnal, and mechanical energy equations in proper perspective, and to derive the entropy equation. We start the analysis with the total energy equation,

D Dt

p - (e surface, A •.

395

(10.1-23) (10.1-17)

and noting that the convective energy area A,(t), we obtain the macroscopic

-f

The Thennal Enercy Equation and the Entropy Equation

+ lv') = -V • q + PI• v +V· (T • v)

IOtaiCMr~JequalioD

(10.2-1)

Compressible Flow

396

Chap. 10

the stress equation of motion,

Dv = pg Dt

p-

+V·T

(10.2-2)

and the continuity equation

~: + V · (pv) =

O

(10.2-3)

Forming the scalar product of Eq. 10.2-2 with the velocity v yieldst p

!!_ (!v 2 ) = pg • v + v ·(V · T) Dt

(10.2-4)

The last term on the right-hand side may be expressed ast

v • (V · T)

=

V · (T · v) - Vv : T

mechanic:al energy equation (10.2-6) = pg. v + V · (T · v) - Vv : T Dt Subtracting this result from the total energy equation yields the thermal energy equation:

+ Vv : T

thermal energy equation

temperatures, or temperature profil studied in detail in courses in heat tr The total energy equation is gen temperature and kinetic energy are t course, for compressible flows. We must be careful to remember derived from the laws of mechanics, equations are two distinct equations. under special restrictions, can dege equation and the mechanical energy regarding the role of the fundament2 Entropy equation

p !!_ (!v2)

De = -V • q Dt

The Thermal Energy Equation a

(10.2-5)

Substitution of this expression into Eq. 10.2-4 yields the mechanical energy equation for compressible flow:

p-

Sec. 10.2

(

l 0.2-7)

In the next two sections, we wi entropy) processes. lt is therefore im so that we know what we mean by that the ·interna! energy may be sp?.ci

e= we may take the material derivative

De = Dt

(oe) os

D~

p

Dt

T = - pi+

De . = 1p Dt

"t'

p-

the last term on the right-hand side of Eq. 10.2-7 may be expressed as§

=

-pV · v + )

(10.2-15)

398

Compressible Flow

Cflap. 10

Sec. 10.3

The Speed of ound

We now wish to determine the time rate of change of the entropy of a material volume. Rearranging the right-hand side of Eq. 10.2-15 and forming the integral, we obtain

f pD;dV=-f v.(~)dv+f (-q·~T +~)dv D 7'".,. T 7'".,.(ll T T

(10.2-16)

7'"... O

or

(10.5-5) x 1 and x 2 gives

J -f 2

1

dh T

1

2

dp pT

(10.5-6)

be treated as perfect gases, we may

(10.5-7) Eq. 10.5-4, and we need only find the shock. lt is easiest to do so perfect gas law. Rearranging Eq. from x 1 to x 2 , we obtain

(10.5-8) 10.4-23)

Our analysis indicates that only transitions that decrease the Mach number are possible. The analysis does not indicate that such transitions do occur, nor does it indica te that the transition should be from supersonic to subsonic.t lt only states that such a transition is possible, and if it occurs, the entropy increases. From the analysis of isentropic fiows we know that there are certain conditions (i.e., a back pressure in the "forbidden region") which disallow the possibility of an isentropic fiow. With this information in hand, one might ask, "Why isn't there a smooth, nonisentropic transition from supersonic to subsonic fiow?" The answer to this question is apparently unavailable at present. We can only state that transitions from supersonic to subsonic fiow occur abruptly. PROBLEMS

10-l. Integrate Eq. 10.1-19 with respect to time to obtain the traditional form of the first law of thermodynamics for open systems. Carefully explain the meaning of each term. 10-2. Starting with Eq. 10.2-15, derive the general macroscopic entropy balance for a control volume, '"~'"a(t).

(10.5-9)

10-3. Derive Eq. 10.4-7 by integrating the x-direction equation of motion over the cross-sectional area A(x)-i.e.,

J

the pressure ratio as

i · p [:;

AW

(10.5-10)

t

+

JdA = - Ji ·

v • Vv

Vp dA

+

AW

Thus, we could have the situation M 1 > M 2 > 1.

J

i · pg dA

AW

+

p.

J

i · V 2v dA

AW

420

Compressible Flow

Chap. 10

Make the same simplifying assumptions as in Sec. 10.4, and use the Leibnitz rule (see Prob. 3-5) for interchanging integration and differentiation. 10-4. For a given value of T o, show that an upper limit exists for the velocity given by (v., )

<

..J~ y-=-} ~

10-5. A perfect gas at T1 and p¡ in a large reservoir flows isentropically through a convergent nozzle of area A2 into the atmosphere where the pressure is p 0 • Derive expressions for the mass flow rate mtaking into account the phenomenon of choke flow. (This type of calculation would be a key step in the design of a safety valve for a high-pressure chemical reactor.)

Flow Around lmmersed Bodies

10-6. If ata shock such as that shown in Fig. 10.5-1, M 1 = 3.0, T1 = 1000°R and p 1 = 0.2 atm, find the Mach number and pressure after the shock. Assume 1' = 1.5. 10-7. For isentropic flow in a divergent channel, the density decreases 14 per cent (p2 = 0.86p1 ) for a 10 per cent increase in the cross-sectional area (A 2 = 1.10 A1). Calculate the change in velocity between these two points. 10-8. A stream of air flows ata Mach number of 0.8. lf a Pitot tube is used to measure the velocity, compute the error incurred by the use of the incompressible form of Bernoulli's equation. 10-9. Air ftowing in a duct enters a nozzle at a velocity of 5000 ft/sec, a pressure of 100 psia, and a temperature of 800°F. The nozzle is designed to increase the pressure isentropically to 125 psia. If the mass ftow rate is 1O lbm/sec, what is the downstream cross sectional area of the nozzle? 10-1 O. Air is contained in a tan k at 200 psia and 7rF. lf the flow of a ir through the outlet valve is approximated as isentropic and one-dimensional, what is the maximum flow rate into the surrounding atmosphere? Take the crosssectional area at the valve as 1.0 in. 2 and the ambient pressure as 14.7 psia. What would the ftow rate be if the pressure in the tank were increased to 400 psia?

The objectives of this chapter are to ena associated with ftow around imm a rational explanation of the drag shown in Fig. 8.2-6 in Chap. 8. Bou thus giving sorne practice in setting U balances. We shall also use boundary performing an order of magnitude am

11.1

Descri ption of Flow

The study of ftow around imme subject to discuss with any degree of the fact that there are three distinct, inftuence the force exerted by the ftui the ftow past a thin airfoil at a zero a number. At high Reynolds numbers, except in a thin region close to the outside this thin region is governed a called irrotational. As the name impli

4

Compressible Flow

Chap. 10

as in Sec. 10.4, and use the Leibnitz integration and differentiation. an upper limit exists for the velocity

reservoir flows isentropically through a atmosphere where the pressure is p 0 • rate mtaking into account the phenomcalculation would be a key step in the chemical reactor.)

Flow Around Immersed Bodies

11

10.5-1, M 1 = 3.0, T1 = 1000•R and and pressure after the shock. Assume the density decreases 14 per cent in the cross-sectional area (A 2 = between these two points. of 0.8. If a Pitot tube is used to error incurred by the use of the incomat a velocity of 5000 ft/sec, a presof 800°F. The nozzle is designed to to 125 psia. lf the mass ftow rate is 10 sectional area of the nozzle? and n•F. If the flow of air through the and one-dimensional, what is the atmosphere? Take the crossand the ambient pressure as 14.7 psia. pressure in the tank were increased to

The objectives of this chapter are to introduce the student to the phenomena associated with ftow around immersed bodies and to provide him with a rational explanation of the drag coefficient-Reynolds number curves shown in Fig. 8.2-6 in Chap. 8. Boundary layer theory will be introduced, thus giving sorne practice in setting up and solving differential-macroscopic balances. We shall also use boundary layer theory to explore the method of performing an order of magnitude analysis.

11.1

Description of Flow

The study of ftow around immersed bodies is an enormously difficult subject to discuss with any degree of exactness. The difficulty results from the fact that there are three distinct, and fairly complex regions of ftow that inftuence the force exerted by the fluid on the body. Figure ll.l-1 illustrates the flow past a thin airfoil at a zero angle of inclination and a high Reynolds number. At high Reynolds numbers, inertial effects predominate everywhere except in a thin region close to the surface of the solid body. The ftow outside this thin region is governed almost entirely by inertial effects and is called irrotational. As the name implies, fluid elements in this region undergo -421

Flow Around lmmersed Bodies

Chap. 11

Sec. 11.2

The Suddenly Accelerated Flat Plac Fluid extends

y

-oo~--c=========~

t sO, the plote is stoti t >O, the veiocity of ti

Fl¡. 11.2-1. The suddenly acc lrrotationol

Fl¡. 11.1-1. Flow regimes around an immersed body.

translation but do not rotate; therefore, there is no shear deformation and viscous effects are negligible. A great deal is known about irrotational flow, because in the absence of viscous effects the equations of motion can be solved by fairly simple mathematical techniques. While the study of irrotational flow is well within the student's capabilities, it will be treated only · qualitatively here, and we will concentrate our attention on the boundary layer. The wake is extremely difficult to treat theoretically, and again we will restrict ourselves to qualitative comments regarding this region. Near the solid surface, viscous effects become important, and this flow region is called the boundary layer. The equations of motion here reduce to what are known as the Prandtl boundary layer equations. In the next section, we willlay the ground work for both the derivation and the solution of these equations.

the area of fluid mechanics. The systern Fig. 11.2-1. We start with the two-dimensiona1

. p-+v-+v-(ov~ OV~ ov.,) ot ~ ax ay ovtl ovtl ov") = - 1¡ p (- + V~- + V at ax ay , 11

11 -

and the continuity equation

OV~

ox

+~

é

Since the plate is infinite in the x-directi and p are not functions of x. The conl to

ovtl = oy

and application of the boundary condi1

11.2 The Suddenly Accelerated Flat Plate As an introduction to the analysis of boundary layers we will examine the fluid motion that results when an infinitely wide, infinitely long, flat p1ate is suddenly accelerated from rest to a constant velocity, u0 • We may rightly ask, "Why study a hypothetical flow that can never be realized in practice ?" The answer is that by performing the analysis we will gain intuition-i.e., we may improve our skill at guessing the flow topology. In addition, the equations we will solve are identical to those describing transient heat and mass transfer into a semi-infinite medium; thus, the analysis has value beyond

B.C. 1: V"= O, leads us to the conclusion that V11 is zero the equations of motion reduce to p

ov.,) (a¡ =¡

o=-(~~

We need solve only the first ofthese te

Flow Around lmmersed Bodies

Chap. 11

Sec. 11.2

-423

The Suddenly Accelerated Flat Plate

Fluid extends to y =oo y

-oo~--~c:::::::::::::::~:::I::::x::~:::Jr---~+oo t sO, the piote is stotionory t >O, the velocity of the plote is o constont, u0 Flg. 11.2-1. The suddenly accelerated flat plate.

the area of fluid mechanics. The system under consideration is illustrated in Fig. 11.2-1. . We start with the two-dimensional fonp. of the Navier-Stokes equations there is no shear deformation and is known about irrotational flow, the equations of motion can be techniques. While the study of ircapabilities, it will be treated only our attention on the boundary treat theoretically, and again we will regarding this region. become important, and this flow equations of motion here reduce to layer equations. In the next both the derivation and the solution

Flat Plate boundary layers we will examine the wide, infinitely long, flat plate is eonsta111t velocity, u0 • We may rightly can never be realized in practice ?" analysis we will gain intuition-i.e., the flow topology. In addition, the those describing transient heat and ; thus, the analysis has value beyond

ov., 11 ov.,) --'--+,u-+. op (o 2v.,2 02v.,) (otav., + v-+v"' ox oy - ax ox oy2 p(avti + v., ovti + vti ovti) = - op- pg + .u(éJ2v112+ éJ2v11) ot ox oy oy ox oy2 p--

(11.2-1)

(11.2-2)

and the continuity equation (11.2-3)

1,

Since the plate is infinite in the x-direction, we can argue logically that v.,, v and p are not functions of x. The continuity equation immediately reduces to

ovti =o oy

(11.2-4)

and application of the boundary condition

B.C. 1 :

1

v = O,

y=

O

(11.2-5)

11

leads us to the conclusion that v is zero everywhere. Under these conditions, the equations of motion reduce to

2 P(ov.,) =,u (o v.,) ot ol o=-(~~)- pg

(11.2-6) (11.2-7)

We need solve only the first ofthese to determine the velocity as a function

Flow Around lmmersed Bodies

Chap. 11

of y and t. Defining the dirnensionless velocity Uz as U

=

z

Vz

(11.2-8)

Uo

Sec. 11.2

The Suddenly Accelerated Flat Plat

At this point we can see that setting a sirnplifying our equation. Substituting setting a = 1, we get

we rnay write Eq. 11.2-6 in the forrn

(~~z)bytb-1 =

(o Uz) 2

oUZ ='V (11.2-9) ot ol which we rnust solve subject to the following boundary conditions B.C. 2: Uz =O, t =O, O ::;;y::;; oo (11.2-10)

= 1,

B.C. 3:

Uz

B.C. 4:

Uz =O,

t >O,

y= O

(11.2-11)

y= oo,

t ¿ O

(11.2-12)

The boundary condition B.C. 2 is often referred toas an initial condition since it specifies the velocity field at sorne initial time. The boundary condition B.C. 3 indicates that for times greater than zero the plate is rnoving at a constant velocity, thus an infinite acceleration is required. While this is physically irnpossible, it provides us with sorne insight as to how disturbances originating at salid surfaces are propogated into the surrounding fluid. Equation 11.2-9 is a partial differential equation. Until now such equations have been considered beyond the scope of this text; however, we can solve this one without expending too rnuch effort, and in the process we can becorne acquainted with a fairly general rnethod of solving partial differential equations. A solution rnay be obtained by rneans of Laplace transforrns, but here we will use a technique known as a similarity solution. Thus, we define a new variable 1J in terrns of powers of y and t, (11.2-13) where a and b are as yet undeterrnined constants, and we seek a solution of the forrn (11.2-14) Uz(y, 1) = Uz[1J(y, !)] If we find that Uz can, indeed, be represented in terrns of the single variable rJ, we will have been successful. lf we cannot represent Uz in terrns of 1], we rnust try other rnethods of solution. We note first that the two partial differentials in Eq. 11.2-9 rnay be written oUz ot

=

ot

oUz = d Uz (01]) 2

ol

2

drJ 2

2

2

ay

1] =

yt

Equation 11.2-16 reduces to

(:~x) = - ~ Putting the boundary conditions in B.C. 2':

Uz= O,

B.C. 3':

uz =

B.C.4' :

1, Uz= O,

Here we see that boundary conditions E terrns of the transformation of variables h We solve the differential equation t P=

-~

~

where

P=·

Separating variables and integrating gi1 lnP

=-

or

(dUz) 01J = (dUz) byatb-1 drJ

If we could sornehow elirninate y and would be reduced to solving an ord· accornplish this by setting b = -t, ther

(11.2-15a)

d11

au.,_ e

dr¡2 -

+ dUz (d 1J)

1

2

drJ

d Uz 1 b 2 =-(ayat) d1] 2

Forrnation of the indefinite integral giv

ol

+ dUz -[a(a- 1)y

(11.2-15b)

a-2 b

d1J

t]

Flow Around lmmersed Bodies

Chap. 11

velocity U., as (11.2-8)

Sec. 11.2

425

The Suddenly Accelerated Flat Plate

At this point we can see that setting a = 1 will eliminate one term, thus simplifying our equation. Substituting Eqs. 11.2-15 into Eq. 11.2-9, and setting a = 1, we get 2

dU"')bytb-t = ( dr¡

(11.2-9) boundary conditions

o :S:: y

:S::

00

y=O t¿O

(11.2-10) (11.2-11) (11.2-12)

11

(ddr¡U"')t 2

b

(11.2-16)

If we could somehow eliminate y and t from this equation, our problem would be reduced to solving an ordinary differential equation. We can accomplish this by setting b = -!, thereby defining r¡ as,

r¡ =

(11.2-17)

yr-t /2

Equation 11.2-16 reduces to 2

rred to as an initial condition since it time. The boundary condition B.C. 3 the plate is moving at a constant velocWhile this is physically impossible, how disturbances originating at solid fluid. equation. Until now such the scope of this text; however, we much effort, and in the process we general method of solving partial be obtained by means of Laplace · known as a similarity so/ution. of powers of y and t,

2

U.,)

dUx) = _ 2v(d ( dr¡ r¡ dr¡ 2

(11.2-18)

Putting the boundary conditions in terms of r¡, we find B.C. 3':

U.,=O, U.,= 1,

r¡ = 00 r¡=O

B.C.4':

U.,=O,

r¡ =

B.C. 2':

00

(11.2-19) (11.2-20) (11.2-21)

Here we see that boundary conditions B.C. 2' and B.C. 4' become identical in terms of the transformation of variables from y and t to 11· We solve the differential equation by first reducing the order to obtain p

(11.2-13)

= _

2v(dP) r¡ dr¡

(11.2-22)

where P=dU., dr¡

(11.2-14) L""~·~~·~-u

in terms of the single variable represent U., in terms of r¡, we

Separating variables and integrating gives

In P

¡trerentíals in Eq. 11.2-9 may be written

(~~"')

byatl>-

1

= -

2

!l._ 4v

+ C1

or (11.2-15a)

dU.,- C' e·· -·"14• d172 -

(11.2-23)

1

Formation of the indefinite integral gives

dU., -[a(a - 1)ya-2 tb ) dr¡

(11.2-15b)

f

~

u"' = e;

o

e-r"t 4 • d-r

+ c2

(11.2-24)

Flow Around lmmened Boclia

Chap. ••

Sec. ll.l

The Suddenly Accelerated Flat Plate

where -r is the dummy variable of integration. It will be convenient to define a new dummy variable Eby T

E=allowing us to write Eq. 11.2-24 as

---A~

#

~rv4.

u;l: =

c~J e-~~ dE+ C

2

u.

(11.2-25)

o

By boundary condition 3', C2 = 1, and application of boundary condition 2' leads to the result

o

(11.2-26)

Flg. 11.1-1. Velocity profil accelerated ftat plate.

The definite integral in Eq. 11.2-26 is known to be equa1 to expression for the velocity becomes

U:1:=

1 -2-

f

J;,o

.¡;;.¡2, and our Approximate solution

~¡v'4v

e-~ 2

dE

(11.2-27)

The latter term in this equation is known as the error function, abbreviated erf(r¡/../4v). Tabulated values of the error fl,lnction are available, 1 and our final expression is

U;l: =

1- erf ~

(11.2-28)

'Y 4vt

This result is plotted in Fig. 11.2-2. The key point of this analysis is that U,. reaches 99 per cent of the undisturbed value for y/../ 4vt = 1.8, and we can use this fact to define a "boundary layer" thickness () as

() = 3.6J;i

(11.2-29) For values of y larger than (), ·the velocity is, for all practica} purposes, zero. We may also interpret Eq. 11.2-29 by saying that the distance the disturbancc penetrates into the quiescent fluid is proportional to the square root of time t, and the kinematic viscosity, v. This result agrees with our intuition, which should tell us that a large viscosity leads toa rapid propagation of a disturbancc, and a zero value of the viscosity would not allow the disturbance to be fclt anywhere in the fluid. 1. M. Abramowitz and l. A. Stegun, eds., Handbook of Mathematical FunctiolfS (Washington, D.C.: National Bureau of Standards, 1964).

We will now solve this problem agai compare the results with the exact s~h "Why solve a problem wit~ an approxtm is already available-espectally when ~he any real situation ?" Aside fr?m keepmg that by comparing an approxtmate result sorne insight into the validity o~ th.e ¡ it provides an interesting apphcatton balance. The control volume for the approx1 l1.2-3. The upper boundary is located y-direction. We start w~th the complet and dot it with i to obtam

~~ f pvz dV +f pvz(v f

0

(1)

A,W

-1

We again impose the restriction that the v of x, and make the approximation (A) t

A. 1:

Flow Around lmmersed Bodia

Chap. 1 ,

Sec. 11.2

427

The Suddenly Accelerated Flat Plate

1---'....----i---- Exact solution t-----i - - - Approximate solution

(11.2-25) application of boundary condition 2'

o

o

(11.2-26)

Fls. 11.2-l. Velocity profiles for the suddenly accelerated flat plate.

known to be equal to

J;¡2, and our Approximate solution

(11.2-27) as the error function, abbreviated ft,mction are available, 1 and our (11.2-28) key point of this analysis is that U. value for y/../ 4vt = 1.8, and we ca: thickness ~ as (11.2-29) is, for all practica! purposes, zero. that the distance the disturbance to the square root of time t, agrees with our intuition, which a rapid propagation of a disturbance, not allow the disturbance to be felt

We will now solve this problem again, by an approximate method, and compare the results with the exact solution. The student might well ask, "Why solve a problem with an approximate method when the exact solution is already available--especially when the problem is somewhat removed from any real situation ?" Aside from keeping idle minds busy, the answer must be that by comparing an approximate result with the exact solution, we can gain sorne insight into the validity of the approximate method. In addition, it provides an interesting application of the macroscopic momentum balance. The control volume for the approximate solution is illustrated in Fig. 11.2-3. The upper boundary is located at y = ~(t) and is moving in the y-direction. We start with the complete macroscopic momentum balance and dot it with i to obtain

~~ Jpv., dV +Jpv.,(v f.w

J

i • t dA

(11.2-30)

-"'.w

We again impose the restriction that the velocity and pressure are independent of x, and make the approximation (A) that

Handbook of Mathematical FunctioiU 1964).

w) • n dA =

A.w

A. 1:

v.,

=o,

y ;;>8(t)

(11.2-31)

428

Flow Around lmmersed Bodies

Chap. JI

r:---------~ 6(f)

J -oo-1

1

1 ~

1 --;

1

1

1

1

The Suddenly Accelerated Flat Plate

We now make the assumption tha1 allowing us to express the velocity as

L.

1

Sec. 11.2

= v,

v.,(y, t)

A.2:

Note that this equation is comparable to

--•+00 7}

= .l

!5(

"----L ----.4

We further assume that the functional represented by a polynomial.

Flg. 11.2-l. Control volume for the suddenly accelerated ftat plate.

Of course, we know this approximation is very reasonable on the basis of the exact solution. A little thought will indicate that all the momentum flux terms are zero, and Eq. 11.2-30 reduces to L

L

dt ff pv., dy dx =f [r 6(1)

E_

o o

o

11 .,,

11=6(t)

r .,¡ Jdx 11

(11.2-32)

11=0

Substitution of Newton's law of viscosity for r 11., yields

v., = a + b(~)

A.3:

d JLJ d d JL{(ov.,). dt- o o pVz y X = ¡..t o -oy

-

1/=6(1)

(ov.,). }d oy v- o X

(11.2-33)

By approximation 1, we have specified v., = O for y ~ d(t). If we require the derivative to be a continuous function, then approximation 1 naturally leads us to the condition,t

(ov.,joy)

(!:"') =o,

v,.= O, ov.. _ o oy- , v,. = u

1

(11.2-34)

and Eq. 11.2-33 reduces to

(an intuit1

y= c5(t)

(continuit

y=O

(a bonafi

oyz- o, 0 Vz

-

y

=o

(from the

Equations 11.2-38 may be used to deter: obtain

6(1)

~ J pv., dy = dr o

y = d(t)

These three equations allow us to deten Eq. 11.2-37. We could obtain more derivatives of v,. to be zero at y = !5; h1 regarding the velocity profil~ ca~ be ?bu Examining Eq. 11.2-1, keepmg m nund we quickly conclude that 2

y= d(t)

ce

To determine the constants in the po . information regarding the velocity profil

0,

6 1

+

-¡..t

(ov.,) oy

(11.2-35) 11= 0

Here we have used the fact that v., is independent of x ; thus, the integration with respect to x can be dropped. This result states that, the force exerted by the plate on the fluid is equal to the time rate of change of the momentum of the fluid. tOne can also argue that all higher derivatives are zero at y

=