Solution manual fundamentals of fluid mechanics (4th edition)

Instructor's Manual to ACcolnpany
FOURTH EDITION
Fundamentals
BRUCE R. MUNSON
DONALD F. YOUNG
Department of Aerospace Engineering and Engineering Mechanics
John Wiley & Sons, Inc.
THEODORE H. OKIISHI
Department of Mechanical Engineering
Iowa State University
Ames, Iowa, USA
New York Chichester Brisbane Toronto Singapore

TABLE OF CONTENTS
INTRODUCTION ...................................................................................................................1
COMPUTER PROBLEMS .................................................................................................... 2
Standard Programs-File Names and Use .................................................................... 2
SOLUTIONS
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Chapter 10
Chapter 11
Chapter 12
Appendix A
Introduction...................................................................................... 1-1
Fluid Statics...................................................................................... 2-1
Elementary Fluid Dynamics-Bernoulli Equation .......................... 3-1
Fluid Kinematics.............................................................................. 4-1
Finite Control Volume Analysis....................................................... 5-1
Differential Analysis of Fluid Flow ................................................. 6-1
Similitude, Dimensional Analysis, and Modeling ........................... 7-1
Viscous Pipe Flow............................................ ................................ 8-1
Flow Over Immersed Bodies ........................................................... 9-1
Open-Channel Flow....................................................................... 10-1
Compressible Flow......................................................................... 11-1
Turbomachines ............................................................................... 12-1
Listing of Standard Programs.......................................................... A-I

INTRODUCTION
This manual contains solutions to the problems presented at the end ofthe chapters in the
Fourth Edition ofFUNDAMENTALS OF FLUID MECHANICS. It is our intention that
the material in this manual be used as an aid in the teaching of the course. We feel quite
strongly that problem solving is an essential ingredient in the process of understanding
the variety of interesting concepts involved in fluid mechanics. This solutions manual is
structured to enhance the learning process.
Approximately 1220 problems are solved in a complete, detailed fashion with (in most
cases) one problem per page. The problem statements and figures are included with the
problem solutions to provide an easier and clearer understanding of the solution
procedure. Except where a greater accuracy is warranted, all intermediate calculations
and answers are given to three significant figures.
Unless otherwise indicated in the problem statement, values of fluid properties used in
the solutions are those given in the tables on the inside of the front cover of the text.
Other fluid properties and necessary conversion factors are found in the tables of Chapter
I or in the appendices.
Some of the problems [those designed with an (*)] are intended to be solved with the aid
of a programmable calculator or a computer. The solutions for each of these problems
are presented in essentially the same format as for the non-computer problems. Where
appropriate a graph of the results is also included. Further details concerning the
computer and their solutions can be found in the following section entitled Computer
Problems.
In most chapters there are several problems [those designated with a (t)] that are "open-
ended" problems and require critical thinking in that to work them one must make
various assumptions and provide necessary data. There is not a unique answer to these
problems. Since there are various ways that one may approach many of these problems
and since specific values of data need to be assumed, looked up, or approximated, we
have not included solutions to these problems in the manual. Providing solutions, we
feel, would be counter to the rational for having these problems-we want students to
realize that in the real world problems are not necessarily uniquely formulated to a have a
specific answer.
One of the new features of the Fourth Edition of FUNDAMENTALS OF FLUID
MECHANICS is the inclusion of new problems which refer to the fluid video
segments contained in the E-book CD. These problems are clearly identified in the
problem statement. Although it is not necessary to use the CD to solve these "video-
related" problems, it is hoped that the use of the CD will help students relate the analysis
and solution ofthe problem to actual fluid mechanics phenomena.

Another new feature of the Fourth Edition is the inclusion of laboratory-related
problems. In most chapters the last few problems are based on actual data from simple
laboratory experiments. These problems are clearly identified by the "click here" words
in the problem statement. This allows the user of the E-book CD to link to the complete
problem statement and the EXCEL data for the problem. Copies of the problem
statement, the original data, the EXCEL spread sheet calculations, and the resulting
graphs are given in this solution manual.
Considerable effort has been put forth to develop appropriate problems and to present
their solutions in a manner that we feel is helpful to both instructors and students. Any
comments or suggestions as to how we can improve this material are most welcome.
COMPUTER PROBLEMS
As noted, problems designated with an (*) in the text are intended to be solved with the
aid of a programmable calculator or computer. These problems typically involve
solutions requiring repetitive calculations, iterative procedures, curve fitting, numerical
integration, etc. Knowledge of advanced numerical techniques is not required. Solutions
to all computer problems are included in the solutions manual. Although programs for
many of these problems are written in the BASIC programming language, there are
obviously several other math-solver or spreadsheet programs that can be used.
A number of the solutions require the use of the same program, such as a program 'for
curve fitting, or a numerical integration program, and these "standard" programs are
included. For those requiring use of one of the standard programs, there is a statement in
the problem solution which simply indicates the standard program used to solve the
problem. A list of these standard programs, with their file names, follow. The actual
programs are given in the appendix. Most of the standard programs are, of course,
readily available in other math-solver or spreadsheet programs, and the student can
simply use the programs with which they are most familiar.
Standard Programs-File Names and Use
EXPFIT.BAS
LINREG l.BAS
LINREG2.BAS
POLREG.BAS
POWERl.BAS
Curve Fitting
Determines the least squares fit for a function of the form
y=aebx
y=bx
y=a+bx
y =do+dJX +d2x2 +d3x3 +...
y=axb

SIMPSON.BAS
TRAPEZOLBAS
Numerical Integration
Calculates the value of a definite integral over an odd num-
ber of equally spaced points using Simpson's rule
Calculates the value of a definite integral using the
Trapezoidal Rule
Miscellaneous
COLEBROO.BAS Determines the friction factor for laminar or turbulent pipe
flow with the Reynolds number and relative roughness
specified (for turbulent flow the Colebrook formula, Eq.
8.35, is used)
CUBIC.BAS Determines the real roots of a cubic equation
FAN_RAY.BAS Calculates Fanno or Rayleigh flow parameters for an ideal
gas with constant specific heat ratio (k> 1) for entered
Mach number
ISENTROP.BAS Calculates one-dimensional isentropic flow parameters for
an ideal gas with constant specific heat ration (k>1) for
entered Mach number
SHOCK.BAS Calculates normal-shock flow parameters for an ideal gas
with constant specific heat ratio (k>1) for entered upstream
Mach number (Ma)
3

t. t I
1..1 Detennine the dimensions. in both the FLT system and
the MLT system, for (a) the product of mass times velocity,
(b) the product of force times volume. and (c:) kinetic energy
divided by area,
mASS
Sinee.
;( ve/oc;'& .:. (;VI ) (L 7-
1
) -
F .:. M L T-.2
( b) ./oree J( Y&/I/ml! - F L 3
Fr
=
_ (ML T-2.)(L3) _ /'1L ifT-Z.
(~ ) J::,;'e/:'G e ner.!~
t:l reL
/- I
-2.
/'1T

/'2
( 0.)
1.2 Verify the dim~nsions, in both the FLT
and MLT~ystems..ofthe folioWing quantities which
appear in Table 1.1: (a) angular velocity, (b) en-
ergy, (c) moment of inertia (area), (d) power,
and (e) pressure.
= a1'19tI //1r c/'spkce/?'J()~';' ..!.
-time
(.b) e he1'"1:J ~ C.a.;aci +!J 01 b~cJ!1 1-0 do w()rk
Since. Wt?/'"K =I()rce;( d/sl-tll1tt:..)
~nerJ!J ; F L
tJr ~if;, F _' /11 LT- 2
e.nerj tj ~ (MI- T -2)(L) == M L 2 T-2
cc) /7l{pmfl1t 0/ inerlltt.~V'ea.) = sec~l?d /nl'Jme/}f D/ t:lff?l
. (1.:2-)(L~) =. LIf
+-()rce ,
= - ~
. L- 2
=F.£
LZ.
J..---------- - - - - - - - - - - - - - - - - - - - - - - -
/-2.

1.3
. ~ Verify the dimensions, in both the FLT system and the
MLT system, of the following quantities which appear in Table
1.1: (a) acceleration, (b) stress, (c) moment of a force, (d) vol-
ume, and (e) work.
acc-e/ero.:tt'tJl1 :::: Ve.JDC.I+~ .:=
+/me
~ t-r-< ./C)Yce F.
eS5 = == L;" -
0. rea..
(C) /?1t:J/)')t"l1i ,,{ (£ (-kyce = .force.K dlsftln('~ .-: 1= L
=f/1LT-VL ...: I1L
2
T-
Z
(a) volume Oen~f-h) 3.-:. L3
- -
(e) Work - !=L
/- '3

I
II
I
i
II
I
/''1 I
ra..)
(b)
(C)
/.5 I
1.4 If P is a force and x a length, what are
the dimensions (in the FLT system) of (a) dPI
dx, (b) tf'Pldx and (c) JP dx?
dP . p- . != L- 2
- -- -- - -dJC L
d3
.f F 1= L-3. .-:::r - -
dx:. 3 L3
jPdx .
PL--"'
1.5 If p is a pressure, V a velocity, and p a fluid density,
what are the dimensions (in the MLT system) of (a) pip, (b)
pVp, and (c) p/pV2
?
1> _
--f
(a. )
.
--
f.1L-'T-Z. .
-(ML -3) (LT-I
) Z
-
'--_ _ _ _ _ _ _ _ _._..........._____________________......J
I-~

/. ID I 1.6 If V is a velocity, fa length, and I a fluid
property having dimensions of UT-I, which of
the followingcombinationsare dimensionless: (a)
vr", (b) VC/', (e) V'" (d) VIM
(L T -'j(L)f1. z r)
,
L~ T-1 mol dlm.nsienle,s)(a.) V J. -zJ .:.. -
(1:, ) v.R (Lr')(L) . LOr" ( dimension /ess)- -V (L'2. T I)
(C! ) V 2
-z) - (L T-) "(L •r -I
) ~ L~r3 (oof dimfnsl'oIl!ess)-
(d) V (LT- 1) . -l.
(not dlfnension!e>s )- - L
).11 {L )(L' r ')
j· 7 I 1.7 Dimensionless combinations of quan-
tities (commonly called dimensionless parame-
ters) play an important role in fluid mechanics.
Make up five possible dimensionless parameters
by using combinations of some of the quantities
listed in Table 1.1.
Some possible e"Qmpl e~ :
(L r2)(T)uC(eIe r,,-/-'M " f 1m e • . L"T"- -
ve/OCI f '1 (L rlJ -
frefllenc'j ;( hme - (rl){r) ..:. TO
(ve!oci+!j) 2.
• (LT -I)'" ,
L"T= ",
/ t'179f!? x. <lea/uP/1M (L)( L r'-)
force " -lime , (F)(r) (j=){T) :. F"i"TO
= -
(1'7zr:J(Lrj/771Y/n en rum (11 LT -~
deMif-') " velocil-j " len-P'4 --' (Mr3)(LT-I}(d •
M'L"T'- =
d'fnllr>1i< visUJ~if:J Mr' 7-1
1- 5

/.~ I
118 The force, P, that is exerted on a spher-
ical particle moving slowly through a liquid is
given by the equation
P = 37CJlDV
where Jl is a fluid property (viscosity) having di-
mensions of FL -2T, D is the particle diameter,
and V is the particle velocity. What are the di-
mensions of the constant, 37C? Would you classify
this equation as a general homogeneous equa-
tion?
.p =- 37T;<D V
[f] -'- [3rr][pc:Lr][L][L r~
[F] == [:?7TJ [pJ
,
••
37T /.5 d/men.510I1Je~s, C(nd -the ~2t1a!-/{)I1·
15 tt 1 ene Y'(J/ hl/rn()~eneOU5 efJtAa..f-/on. yes.
/- ~

/. 'I I
t. /0 I
According to information found in an old hydraulics
book, the energy loss per unit weight of fluid flowing through
a nozzle connected to a hose can be estimated by the formula
h = (0.04 to 0.09){D/ d)4V2 /2g
where h is the energy loss per unit weight, D the hose diameter,
d the nozzle tip diameter, V the fluid velocity in the hose, and
g the acceleration of gravity. Do you think this equation is valid
in any system of units? Explain.
~ = (O.OLf 1-0 ('). {) 9) (.!J)If ~~
2.J
gn= [D.O~ 1-. O.O~ [tJ[i] [~:J[t]
[L J== [O.OLf -1-0 0,07] [LJ
Since eac.h hrf}z li-t the e$tt.a..f./~h must: n4t1e the
:Slll71e d;'mel1$/tJh5 -the Cf!)I1~"'lfi I-erm (~. ~'f ~ ~. ~tj) rnusf
I
/;~ climfns/f.,hless. Thus1 the e$ti/{,t/~H /.5 a. !J~n(lY~ I
h~mo1enet!Jvs €lttA..-6I4;;' .fh{(.i: IS 1I11//c/ IH CiI1.!! ~!:f5Iem
~f Un ,·f..:5. Yes.
1.10 The pressure difference, Ap, across a
partial blockage in an artery (called a stenosis) is
approximated by the equation
cosity (FL -~T), p the blood density (ML -3), D
the artery diameter, Ao the area of the unob-
structed artery. and A I the area of the stenosis.
Determine the dimensions of the constants K,.
and K". Would this equation be valid in any sys-
tem of units?
pV (All )2 1
.1p = K! D + K" A I - 1 pV-
where V is the blood velocity, Jl the blood vis-
Since eac.h -terM mv.st h~lJe. the same dimensions;
k'v Cll'ld Ku are dirnen5ionJe-:'5. Thu~.1 fhe efuafltJJI/
IS (;( ttener~1 h()f71~jel1eO"s e~ ua.l-;tJv, -tnCI'/- w{)uld be
va/ic/ t'n Cfn!! C()tJ5isffnt sfjsl-em of U)1jf5. yes.
/-7

I. / / J I.II Assume that the speed of sound, c, in a fluid depends
on an elastic modulus, Eu, with dimensions FL ~2, and the fluid
density, p, in the form c = (Eu)"(p)h. If this is to be a dimen-
sionally homogeneous equation, what are the values for a and
h? Is your result consistent with the standard formula for the
speed of sound? (See Eq. 1.19.)
0)
FPr ~ d)J11eY1~/Of1tt/I'1 h(!)mt1ef1eDIJ5 -€$ad
'
{!)'J1 ea.ch +erm
In the etua.t,bJ-f fntlS1- haf/(. -fJu 5f1/)/e dlmeY15JO#.s, Thtl5,
-/ne Y"'9Jtf hand ~/de ()f. P~l OJ mus+ h~ve the dlmenslPA,s
of- L 7-'. There /dYe)
a-tb==o
2.},=-1
.ta -f If b = - I
(i:1> sa-/-1 's.f." C6"t/, ',,,()~ "n r)
(.£. :!iJ1-,~ I-y ~Y1dJ/o'" "" L)
a. =L tlnt! /:; =- ).Z. 2.
So -tn..-f. c = ~i0: 1
Thb re.5u 1+ /s ~nsisl-f"r/AI;-!/1 the, sblltlt/J'p ~rIl1U/A -kr 17te
:5peed ()j2- 5DUJlJd. YeS.
1- 'j

I, /2. I
1.12 A formula for estimating the volume rate of flow, Q.
over the spillway of a dam is
Q = C v28B (H + V2/2g)3/2
where C is a constant. g the acceleration of gravity. B the
spillway width. H the depth of water passing over the spillway.
and V the velocity of water just upstream of the dam. Would
this equation be valid in any system of units? Explain.
5/~ce ea.c;" I:errn ,i1 ~e .e.Su.Lf/~H rnus-t- ha.ve +he
SQ/7Ie dimellsi{)l/s -the ~11.sb1l/i C VI must:- he
cilmeI15/!)/J )e~s. Thtls; -tnt!.. .et(f~tltJH is a ~-ene r-a I
htPl1IP,e/ledJt(J egUa.,tIOJl -1'n¢,f WOf,{ /~ be. vt).. //d I»
411'1 e4)A~/sl:ent Set: of (,Iilif.s. Ye~.

/. / if I 1.14- Make use of Table 1.3 to express the
following quantities in SI units: (a) 10.2 in.lmin,
(b) 4.81 slugs, (c) 3.02lb, (d) 73.1 ft/s2, (e) 0.0234
lb·s/ft2
•
(c>-) 1t),2 :;;'1 - (;0. 2 ;,;J(Z,S*;t/O-",:'.) ( ~;;n)
-3 /W1
- i-. 'a2. .;c It) s = tf. 32. T
[h) If. 9/ S/fAl'= ('I:?/ sill!> ) (;.'f$f';< I () sju~) = 70, 2 ).ff
(~) 3. tJ:L /b::: (3.~Z /b ) ( If. If'If f1 ).=: /3. If AI
Cd) 73. J :Efi :
ce) CJ, tJ23'1
Ib·s
(0. ~Z3'f ITt.) ('/,7.?1;tIO
N· -':
)~ ",.,1-
ff~ lb. s
-ft'l-
I, /2 N·s
- M'J'l.
1-/0

/./.5' I 1.15 Make use of Table 1.4 to express the
following quantities in BG units: (a) 14.2 km,
(b) 8.14 N/m3, (c) 1.61 kg/m (d) 0.0320 N·m/s,
(e) 5.67 mm/hr.
(b) o !!..o.llf. ,11'I'f 3 " (g. 'If ~ ) (~3U;(/O·3 ':3)= 5'. IF)( 10'2 Pt.
,,",,3
( -3 SJUjS)
lI. CfIff) )(. /0 W =
~
~~
(d) 0.0320
N-1'H1
(~, 0j 20 N~I1f1 ) (7, 371P;( /V-I
il-·Ib
)- - -- oS
S
N·/'M
--2 .{.f·/b oS
2.3b)(JD- -oS
1-1-to
- s: 17 )1.10
-...5
/-11

/. /(0 I 1.lG Make use of Appendix A to express the
following quantities in SI units: (a) 160 acre, (b)
742 Btu, (c) 240 miles, (d) 79.1 hp, (e) 60.3 OF.
IfpO a. ere
(6) 7tf2 137U = 6'1-2 sru) (.°£,;</03
J.)= 7.g3X/~5J
BTU
C~) .2LjO int.' = (;'''10 tni ) (;'''Oq;(./(;.3 1"YY1,)::: 38iDX/oS"t?11
I'n1L
Cd) 71. / hp 0: (7'i'./ hp ) (7.'f5"7 X /02. (;{;) '"
(e) Tc = l' ~1).3 - 32) '= /5.7 "C::
k = /5",7 f) ( -r 273 ::::), gr 1<
1-/2

/./7 I
1.17 Clouds can weigh thousands of pounds due to their
liquid water content. Often this content is measured in grams
per cubic meter (glm3). Assume that a cumulus cloud occupies
a volume of one cubic kilometer, and its liquid water content
is 0.2 glm3. (a) What is the volume of this cloud in cubic
miles? (b) How much does the water in the cloud weigh in
pounds?
1M1= 3.281 U
(;0'/111.1) (g, Z8'1 ~)
J
-
(£2!b >fIb) t:)3
0,2 £j 0 nn,,3
(h) %J == 0X -Vol"rn~
d' =: jJd = {0.2 ;'3){!D-l
;g.)(r.8/ ;)=f. UU/iJ-;;J
"lJ =- (I. '((,,2 ;(JD-3;;', ) (10 1;m3) = /. '(~2 X IDI, N
= (I. "t,z X /D (.N ) (:1., 2tf8 x/D-
1
-J& ) :::: ~, If! XJO
S
fh
1- 13

1.18
(a)
1.18 For Table 1.3 verify the conversion re-
lationships for: (a) area, (b) density, (c) velocity,
and (d) specific weight. Use the basic conversion
relationships: 1 ft = 0.3048 m; lib = 4.4482 N;
and 1 slug = 14.594 kg.
I it1..: (/ .ft'")f(a 301f.>') 2/1?1 ,,-] = 0, () q29{) /H1 ~
L I-i ~
Thus) rnu//-'/0 -ft2
bJ 9. '2'i{) £ - 2. +0 t!trJnvfrf
fo /ffI :2..
II;) /
Thus) mu/fipJ'j slugs/.ft.3 b!:J 57 IS-If E of 2. ;'0 CtJl'Jtlfrl
-to Ie? //I'n ~
(I!) / If- = (/ fj )(~. 30'/; jJ)~
Thus.) muillpl!) Ills bIJ 3.0'le f - / -1-0 cOl1vert
-I:0 /t11 /s.
(d) I JIz - (I !l ') (If. 't'l12 !!..)[ I Ii 3 l
If3 - l' -It3 ) l~. /j, ( 0, "3()Iff)3/W1 3 J
IV
-= /57, / ;;;;
TfJlAS) m IA If/pI:; /b/R ~ b!:J /. 5'7/ };-t 2 -10 t'e>ntlfY't
fo #/;m3 4
/-/if

/,/9 .J -- -
1..1q For Table 1.4 verify the conversion re-
lationships for: (a) acceleration, (b) density.
(c) pressure. and (d) volume f1owrate. Use the
basic conversion relationships: 1 m = 3.2808 ft;
1 N = 0.22481 lb; and 1 kg = 0.068521 slug.
(a)
(b)
Thus) m""/+ipllj
tt/ .J.t/.5 J..
I ~ ~ = (I ~3 ') (0. oft> f/5:L/ slugs) [ 1m,3 J1111 ~ "" ( T; (3. ZFO~)3-f1:: 3
- I 040 x /0- 3 S lu ~~
. 1 f-t3
Th ~S.i m ul.f.ipJ'1 ~J/tt113 h,!j /. qLfo E-3 to ~J1t/fri.
-1:0 S /u~/.ft 3.
(C) I Ji :: (I !:!. ) (O,2.2lfgl ~)f I (M1. l/'I't1 ? tn1 2. N l (3. lfOg) 2. ft 1.J
-.2. Ik
"=' '2. ()gr i. ID f.t1-
Thu5) m/,.{lfip/~ N/rrn l b~ ;;'.Ogq E-l fo ~~n()fYt
1::-0 /h/ f.t :L,
(d) / 73
== (I ~)[cg,1.KOS/~:l= 35". 3/ fr'
ThUS) rn f.,( It ifl':J 1»1
3
/5 b~ 3. 531 E+ I -1:.0 rlOl1Vfyt
+(/ ft 3
/s.
------------~~- --------
/-/5

/.2..0 J
(()...)
1.20 Water flows from a large drainage pipe at a rate of
1200 gal/min. What is this volume rate of flow in (a) m
3
/s. (b)
liters/min. and (c) ft3
/s?
f./owrat e =
-:2
757 ;<. 10
/i'Y7.3
.5
(b) Since / Ii fer = / [) -3t1"/1 ~
/lowrfLte= (7.57 ;'/6-
2
~.3)(/o3///.er.5)({Po.s)
S /H1 3 /'1?1/11
(C ) I I()W r (I. +. e. = (7 S 7)( J()- ~if3 ) (3 S3 I X J0
-I'tJ
:: 2. ~ 7 s
-
I-/~

1,,;2 / 1.2 , A tank of oil has a mass of 3 0 slugs.
(a) Determine its weight in pounds and in new-
tons at the earth's surface. (b) What would be its
mass (in slugs) and its weight (in pounds) if lo-
cated on the moon's surface where the gravita-
tional attraction is approximately one-sixth that
at the earth's surface?
( t(.) w.ei9h i- .: ~. as.5 )(. 3
;,:2 2
= (3 0 5 /uqs ) ( 32.2 ;:)== _o/~r;, 16
- (30 shillS) ('t. Sf 14 )("I.E! -f,,)-= ,/Z'foN
( b) /h') 4 s.s = 3 () 5 J/A 9 S ( /n1 ASS dtJts t}IJt- dep~;1d t!)1'1
JY'~ vihfitJl1ll / a ffrtu..J-if!)11 )
w.eijhi = (30 s/uqS ) (32.~:Ef..) /fa/ /b
1.22 A certain object weighs 300 N at the earth's surface.
Detennine the mass of the object (in kilograms) and its weight
(in newtons) when located on a planet with an acceleration of
gravity equal to 4.0 ft/S2.
9, 8/
'I: () ftIs :J.
)
- (3tJ.(P Jj. ) ( if. 0 ~) ((), 30'fg ;;)
= 37.3 N
1-1:1

1.23 An important dimensionless parameter
in certain types of fluid flow problems is the Froude
number defined as Vlv'g'ii, where V is a velocity,
g the acceleration of gravity, and ra length. De-
termine the value of the Froude number for V =
10 ft/s, g = 32.2 ft/s2
, and r = 2 ft. Recalculate
the Froude number using SI units for V, g, and
e. Explain the significance of the results of these
calculations.
In B6 tI/lits /
;0 /.25"
In JI uni-t-s:
V:: (to ft )(~. '3IJJfr ~):: 3.06 TS ft
~;: 1',:g I ~
~ ::: (~+t:) (0. "3 04-g ~ ):: O. bI0 t'l?1
-Fe
v
y!~
=
Th e. Va /lle D I a.
in cle;enciel7i of
1.25--
d im-et1sjt'Jn less parl!met ev
-the un i t ~1 sl-em.
1-/8
IS

/, '25
1.2 4- The specific weight of a certain liquid is
85.3 lb/ft3
• Determine its density and specific
gravity.
d" g5.3
Ii?
s Iu 9.5-
;0 -= - .ft-3 2.&'5
1 '32,2 .pc
f-t3
5.2.
fJ 2.~5
5/,,?.5
k-i
56= -
I. @ f~c
/. fi- S/W9S
If;l.O -..ft.~
1.25 A hydrometer is used to measure the specific gravity
of liquids. (See Video V2.6.) For a certain liquid a hydrometer
reading indicates a specific gravity of 1.15. What is the liquid's
density and specific weight? Express your answer in SI units.
5G
(J
-=
~D@'" °C
//5 - f- /o/)o .k;'
1m 3
1.37
f== (I. /5) (I ()r;O :'3) 1150 ~h)?3
1- /q

/.2 10 1
l. 2~ An open, rigid-walled, cylindrical tank contains 4 ft3
of water at 40 of. Over a 24-hour period of time the water
temperature varies from 40 of to 90 of. Make use of the data
in Appendix B to determine how much the volume of water will
change. For a tank diameter of 2 ft. would the corresponding
change in water depth be very noticeable? Explain.
/)1QSS of w~l:er = -V X t
Wheve ¥ /s the {/oh{rne and! 1he. deI15rfr:1. J/J1Ce.. -the.
rnA$~ mU$1- Yefl1111M ~l1sfa)1i (/5 the -iempera.-tuye ehf/flqeJ
-tfx iJ := -tI-)( ~
'fcc /'io p '1~. (f~ (> (I )
Ff~'11 ra6)e B. J
/Hzo ~ r~"F = I. 13/ s~
TherekYt) (nil'! E$- (/) /' .:!':!i~)
+ = ( if/t.3)( I, 9'1" .f-c3 If DIFb -Pi.]
1p() I. y"j J ~:;:3
Thus; the, I~crell~ In Vt) lumt: ,:s .3
'I: 1/Ji L - If. "00 -== 0. 0/ i I:, .pt
The chtlltfe Jil Wl..fey cle;1Jt" 41) .i<J "o/tI~j fo
il-V- O. OJ;~ ..ft3
-3
Ai.:= area :=. == 5, '12 xlD +t=~. ()7/~in.
71 (7f-1:) 2-
Lf-
7h'5 ~/lJ4I/ e-hl(Hge In def1h would n~.J. he iJel"!1
tJ()l-lcet:l/;/~. AI0,
,4 S/;1hf/lj d.:PkY(~i. Vfi!,,(! for' .l1) f-I/;II b~ "h.fa;HfA
If ~I'~c"f,i ("J(I;hf lJ!-wphr Jr Iur,r fflilJey 1ltQIf4t11s/-J-!1'
11J1~ 'J du e -10 t'h.e /rtc.t tho! 1Jtele is SIP/II e IIHcem,id]l
117- -!itt! fi,Jlr1h ~/;1;ln(~111 /'9l1Y'e of 1Jte..re. +tv" 1It//l(es,l lit'!
ff;.(J ~()//,('h~Jt 'S SPfls/fl';~..fl':J 7}"j unc..ryitlin-J.t;.
/-20

/,27? I
1.2 ~ A liquid when poured into a graduated
cylinder is found to weigh '8 N when occupying a
volume of 500 ml (milliliters). Determine its spe-
cific weight, density, and specific gravity.
w~i~ht gN
(=-
1/0/ tllYJ e
f=
/ra;<. /~
:3 JL
?! /1113
-
c;. 1.81 hH
57..
.= 10. a
- J. ~3 x /0
3
-k ~
1»1 3
f
1 ~'f.
/. b3 x / D ;m .?l
/. to 3-
~o@
JfDC
/0
3 ..fEg.
;m3
S6
/- 2/

/,2Cj 1 I.'2.q The information on a can of pop indicates that the can
contains 355 mL. The mass of a full can of pop is 0.369 kg
while an empty can weighs 0.153 N. Determine the specific
weight, density, and specific gravity of the pop and compare
your results with the corresponding values for water at 20°C.
Express your results in SI units.
y=
v~/JlJf ()+- I-/UIC;
{/p/um~ t:J/ .fltlt'c/
(/ )
-h~/ we/fltf = maSS x 9- = ~.3tf J# )(r.JI ~)::: d: 62 II
wf,jhf ~f GIn:' C/. /53 IV
/
r. / -3L) / -3 /YYI3) -(. 3
Vp/l(l'n~ ~ "'-/1114::- c:i'S5oX I~ (/0 T =- ~:,-S-x/tJ /YYI
Th u~ I-r~1t7 E%. (/)
3. "Z /II - ~. 153 II - Cf77o!::!
tf7 7013
r.8J~
;rna
oS;&. ~
rtf~ 1'm.3 - (J. 99/'
j~o~ y
/J?'13
Ulaler ~t 20·C (see. ~.J/e B. 2 J~ ApjJfHd,X J])
v - '17g'13.'3 · /.J :: f'/t. 2 ~ . 56 = 0. qqg2.
oJ+z. iJ - /J11) (Jt.z. ~ 1')n3 )
jj etJll11l"n;;" ~f 1AlS~ Jltl/IIR.I £" lOll/IV with 1ht)s~
~y 11te P(J); sh()w.s 1Jt~.j iJ;~ ~~C;-hC 1A.)(',jhf~
c/tnS,fYI ifl1d ~eClhc' :Jr/tv,f, cf- iJre i",P are. all
Sl'jhflfj Jp l<Jer 1JJlfn ine ~rre;;t~nd/~ J/tJlllfS Ibr UJt:der.
/-22.

/.30*1 1.30* The variation in the density of water, p,
with temperature, T, in the range 20°C :$ T :$
60°C, is given in the following table.
Density (kg/m') 1998.21997.11995.71994.11992.21990.21988.1
Temperature (0C) I 20 I 25 I 30 I 35 1 40 1 45 1 50
Use these data to determine an empirical equa-
tion of the form p = c, + C2T + C3T1 which can
be used to predict the density over the range
indicated. Compare the predicted values with
the data given. What is the density of water at
.42.1°C?
To S()/ve 1h:S pr()~Jem use POLRF6.
***************************************************
** This program determines the least squares fit. **
** for any order polynomial of the form: **
** y = dO + dl*x + d2*x
A
2 + d3*x~3 + ... **
***************************************************
Enter number of terms in the polynomial: 3
Enter number of data points: 7
Enter data points (X , Yl
? 20,998.2
? 25,997.1
? 30,995.7
? 35,994- .. 1
? tiO,992.2
? 1±5.990.2
The coefficients of the polynomial are:
d2 = -4,.0953E-03
d1 = -5.3332E-02
dO = +1.0009E+03
X
+2.0000E+01
+2.5000E+01
+3.0000E+01
+3.5000E+01
+1±.0000E+01
+1±.5000E+01
+5.0000E+01
Y
+9.9820E+02
+9.9710E+02
+9.9570E+02
+9.91±10E+02
+9.9220E+02
+9.9020E+02
+9.8810E+02
Y(predicted)
+9.9825E+02
+9.9706E+02
+9.9566E+02
+9.91±07E+02
+9.9226E+02
+9.9026E+02
+9.8805E+02
Tn US)
f=== /00/ - O. OG"333 T - 0.00'1095 T:J.
!Vote tl14t f (pJ'ecl'~fed) ~ l'n 9()OO Q9reemfl1t w;'1h f (gJ~~h).
At r = '1-2. / "C)
! = /00/- O.~S333 (Jf.Z. / DC) - (J. {)O tj.O?S (1f.2./ cc)
~
/-23

/,32 I
1.32 The density ofoxygen contained in a tank is 2.0 kg/m3
when the temperature is 25°C. Determine the gage pressure of
the gas if the atmospheric pressure is 97 kPa.
p=f)/U = (.2.0 #!.)(.)51.8 Ie~k.) [r.wer ,m)Kj
- /5'5 i Pa. (4 bS )
-p (JCd/e): -1;, I - 1:. I :: /g5 J.~ - rt71e ~ = 5? k ~fibS 4.rm
/.33 I
J.33 Some experiments are being conducted in a laboratory
in which the air temperature is 27°C. and the atmospheric
pressure is 14.3 psia. Determine the density of the air. Express
your answers in slugs/ft3 and in kglm3.
P=fJRT
Tempera.fllYe,
- 0.00222
/ :: /0. Of) 222 SlIl9.s) (s. /S¥ X If)
2
!,,g! ) ::( c. R a .5LuS.l
--:r£~
1.14 1r<!33In"
"

l.3 If A closed tank having a volume of 2 fe is filled with
0.30 lb of a gas. A pressure gage attached to the tank reads 12
psi when the gas temperat.ure is 80 of. There is some Question
as to whether the gas in the tank is oxygen or helium. Which
do you think it is? Explain how you arrived at your answer.
W~/ji, t = tJ. go IJ,
~)( (lo/ume (';2.2. ~) (z. ft3)
-3 /
1=
Sin ce.
~
-1=
rr4'1'J?
ttinc1
Thus;
pre.sStlre ~~JUMFd
T =- (tf/J (>;: + If.b~) llR
(2/,,7
;0:::
Ta ble I. 7 R :: /. 5"'S"1f X J~
;€ = I, 2 if.Z X It) If Ii· J.j,
0511/9 ' (),R.
./r()1'}1 Ff.(J J I;:' -!he
I 7./Z
::- 155'f X/~ 3
he /11,lm
,tJ - 7. /2..
r- - /, 21f2 X/I) If
3
lor-
~y
9tif
s/tl1..:5
/t3
1~~x/o St.(9S
.;:-t;3
( J2 T I't: 7 ) fS/~
b( ~ / if-: 7 f.J'/a )
Ii .ft, //,,(jJ j '"that:
7 12
R.
O)(jgel1
he/lto'n.
Is f.9XY'l/n
= *5'i.x M-3 !/u~
H3
A- ~mJJIIYJ51J1}
of -!7te '1q5
6/ 1he.5e va/lies w/ fl, -tJ"e tlC/:t(o/ df!1>/~
/rl the -bani. /nell C.1I~...s 7h/I t 1it e
9t1S /?1 uS i- be
1- 2 5
(I)

1.3G A tire having a volume of 3 fe contains
air at a gage pressure of 26 psi and a temperature
of 70 oF. Determine the density of the air and the
weight of the air contained in the tire.
t== R.T
~1P!l20 .,. /It, 7.f! ) (Ilflf In.2.)
.: Jh. I J1 2. .ft:"L
(/7/~ h,/)' ) li(7~d;:+'ffPO)"R
5hlj'~ II ~
-3 I
~, "If)( /D s~
wei!JH ::- ! 3- ;( ",,,Iume =(t..1fIf x/0
3
s!,,:) ('32.2 ::) ( ~.f.t~
1- 2f.s.

/.37 I
1..'07 A rigid tank contains air at a pressure of 90 psia and
a temperature of 60 oF. By how much will the pressure increase
as the temperature is increased to 110°F?
-P::tRT
J=oy a
V~/vme
/!rIPt7I
Y'lr/4 c.losed Jan./( -the "';, rnpS5 4nd
4Y~ 'DI1"iR~Z. ,;fO 1= ~n5i:4nt-. Thus",
ct. /. f (W/~ R etPI'I5rq"t)
-P, _ FL
-T, 7;..
wht'f'e -A ~ fit) psia-) r;:: bO
Ii
r -J- Jflt,D - S2.~ c ~.1
"nil 7i:: //()oF-+Jf6o = s-'lO~. FY'~pr ct. (j)
( I )
-b = 7i.. ~ = /S7tJ
eJ
l<) (flJf~t.A.) ='18'. 7 OSLo..
r2. 7; 0 ( 5z()-;e I .
l-l7

I. 3i -'II: I
"J .3X Develop a computer program for calculating the
density of an ideal gas when the gas pressure in pascals (abs).
the temperature in degrees Celsius. and the gas constant in
J/kg· K are specified.
;::';;1" lin ,dtlt/ ~115
1::/RT
~
1= "kr
t.)htY~ t iJ tlb.$l)/u/-(l. 1'1"!'SS"~) R 1h~ 9"S (}PIIS J-f,(1'1 i: I ClI'1~ 7
is tlbsl)l",,~ -lempRrtlwre. Thus" ,-I 1'ht!. t~mJ~f-I4.t:UY~
{s / J1 4)C -Inti'!
T = ~ { + 273. JS
,4 spreadshp8 t (exCEL] PY'()jY4h1 ~,... C/4leulai-lIlj fJ j.o/jfJLt)s.
This program calculates the density of an ideal gas I
when the absolute pressure in Pascals, the temperature I
in degrees C, and the gas constant in J/kg-K are specified.
To use, replace current values with desired values of
temperature, pressure, and gas constant. i
I
A 8 C D !
--+--------+--------~~~--~----~----~
Pressure, Temperature, Gas constant,. Density, :
Pa °C J/kg· K I kg/m
3
I
1.01 E+05 15 286.9 1.23 Row 10
Formula:
~----~--------~-------;
=A1 0/«81 0+273.15)*C1 0)
1
~xt¥rnp/e" taJcuLa..f-e I ~r P= 2.~o~ P()..) trhljJfrl.i:ure. -
~O·CJ t1'1~ R:: 2..97 J/~. I~ I
A 8 C D I
Pressure, Temperature, Gas constant, Density,
Pa °C J/kg-K
2.00E+05 i 20 287 2.38 Row 10
1-2.~

I. 3f-- I
':'1.3l) Repeat Problem 1.38 for the case in which the
pressure is given in psi (gage). the temperature in degrees
Fahrenheit. and the gas constant in ft·lb/slug,oR.
F(J)Y ql? /c/ea/ 9qS
1=fRT
/.)::: .::t, 1<;
whtY.f!. p /.s a bs()/utt prf.5SUYl'j t:i1'1I( T IS Q"~~/lJ.fe +emptrai:tlre.
ThIl..5 I f ~empYa-bo"e IH ";:: tfn" PYt'5~uve ,ft os£. -the" . -a
/ / r I J In.
T = tI;= -r if5~. ~7 411r.1 f':: [ p(1"£) -r t.ihtt(psia.) x14LfJ;t~
.4 spreAdsheet. (t:XCEU pr()!f4m ~y C4/fL{J~t-Jir." f lallows.
This program calculates the density of an ideal gas~:~-~-~--------------~------~----+------4------~----~
when the gage pressure in psi, the atmospheric
pressure in psia, the temperature in degrees F, and
the gas constant in ft.lb/slug.deg R are specified.
To use, replace current values with desired values of
gage pressure, atmospheric pressure, temperature,
and gas constant.
A B C D I
Pressure, Temperature, Gas constant, Atm. Pressure, '
psi I of fi Ib/slug.oF psia
o ! 59 1716 14.7
E
Density,
slugs/ft
3
0.00238
i
i
Row 12
~---l-------+_ _ _ _--+------il Formula:
i=((A12+D12)*144)/((C12)*(B12+459.67))
I
J?J(" rn)J Je' ell/,uL~ -te fJ ~y P=LfOPJi.) ifmprrIJ ture =/~~()t;
fi.J:"" = 1'1-.7 P5L'(ij tind R= /7JI:, .fJ.t.lb/.sJU~'''~ ,
A I B I C i D j
f-:::---- .
Pressure, Temperature, Gas constant, I Atm. Pressure, !
psi I of ft Ib/slug of psia
40 100 1716 14.7
J - Zer
E I
Density, i
slugs/ft3
0.00820 Row 12~ ___ _

/. '10 I
I.LfO Make use of the data in Appendix B to determine the
dynamic viscosity of mercury at 75 of. Express your answer in
BG units.
/-30

/. ~I
1.4 J One type of capillary-tube viscometer is shown in
Video V1.3 and in Fig. PI ~( . For this device the liquid to
be tested is drawn into the tube to a level above the top
etched line. The time is then obtained for the liquid to drain
to the bottom etched line. The kinematic viscosity, v, in m2
/s
is then obtained from the equation v = KR4
t where K is a
constant, R is the radius of the capillary tube in mm, and t
is the drain time in seconds. When glycerin at 200
e is used
as a calibration fluid in a particular viscometer the drain time
is 1,430 s. When a liquid having a density of 970 kg/m3
is
tested in the same viscometer the drain time is 900 s. What
is the dynamic viscosity of this liquid?
Glass
strengthening
bridge
Capillary ---lr-+-.-li"
tube
• FIGURE P1.41
~y ~/tI~er/.n @
!. / r X /1)-) hn"l-Is ::
20D[ 7J= !JCfxIP-~1s
••
Uc R Ij.) 0, ~30 s)
k R4-= 8. ~ 2 X If) -7 /}?12-1s 2..
v=
=
IJ zu/d win, t::. rODs
(3. $ ~ i/O-7 /n1 "2./s2.) (90 t) 5 )
-r-z/
(970 --k#~3) (7. If 'I x/0-If fn1% )=
D. 727 ~Im-S
= u,727
1-3/

I. ¥2 I
J042 The viscosity of a soft drink was determined by using
a capillary tube viscometer similar to that shown in Fig. P1.41
and Vidl'O V1.3. For this device the kinematic viscosity, v, is
directly proportional to the time, I, that it takes for a given
amount of liquid to flow through a small capillary tube. That
is, II = KI. The following data were obtained from regular pop
and diet pop. The corresponding measured specific gravities
are also given. Based on these data, by what percent is the
absolute viscosity, J-l, of regular pop greater than that of diet
pop?
I(S)
sa
Regular pop
377.8
1.044
Diet pop
300.3
1.003
- tJ1< /OD
}-32.

1.13 I
1.43 The time, t, it takes to pour a liquid from a con-
tainer depends on several factors, including the kinematic
viscosity. ", of the liquid. (See Video V1.l.) In some labo-
ratory tests various oils having the same density but differ-
ent viscosities were poured at a fixed tipping rate from small
150 ml beakers. The time required to pour 100 ml of the
oil was measured. and it was found that an approximate
equation for the pouring time in seconds was t = I + 9 X
102" + 8 X I03
,,2 with" in m2
/s. (a) Is this a general ho-
mogeneous equation? Explain. (b) Compare the time it
would take to pour 100 ml of SAE 30 oil from a 150 ml
beaker at O°C to the corresponding time at a temperature of
60°C. Make use of Fig. B.2 in Appendix B for viscosity
data.
(a..) -I:. =: / -t 'I J( /o"l.-u T 9 X/OS -v 2-
fT] == [i ] + [tf;<JoV[~J-t [3x/oJ] [-¥.]
5/~c~ each +rn11 ;'n +he egutL.f:lbJ1 !1?"fs-t hftlle -t-he stlme
d/1rJl'''''tM5 -tJte ~IJ51-o",1-.s a..?petlr/n~ /rl 1J1e efllLa.:I:,clI
.I ' . / -3mu 51- have dllnen ~/pif..s l e. [ ]
[)] :;; [TJ [11.>< JD 1-J ~ [.1::] D~ XID3J.:. -b-
7htl..5) w; 1h a. c.hol1"~ I;' Ut1/fs /he J/,,/tI~ "I 7h~
C(J)115-fz1l1i5 wPt//tI e..l1l1l1fe qntl -t7J/~ I S 11~t:- tt j-enent /
hornojen e(J)tI.J. ..(2gaa-i'£;;J . ;1.0.
(j;) Pr~m Ta /;Ie 8.2 /n Appel1d1 x B
(~r SAE:3'tJ ();/ @) O°c) -z/= 2. 3 )( jtJ -,3/Yn 2./s
(-for 5A-E~!) ()// @ ~DC) -V= 'I: ~ x /o-s/m2-1s
@ (00° C
l7f£. (; )
i::- I T- C/Xj// (2.3X/D-~)-t
:3.1/ s
1+
I, 0'1- 5
)-33
(I)

I. Lf4 I
/. 'is J
1.44 The viscosity of a certain fluid is 5 x
1O-~ poise. Determine its viscosity in both SI and
BG units.
/= (5.>L/o-I(.,P~i~e)(IO-' ~~)=
p~/$e
(/n el Frpl7? ~ 6/e /. If
~ -l
~ ::: (5 X- /D- .!:!.:.!.. ) ( :/., o~q )(./ 0
. /1'11 2
1.4S" The kinematic viscosity of oxygen at 20°C
and a pressure of 150 kPa (abs) is 0.104 stokes.
Determine the dynamic viscosity of oxygen at this
temperature and pressure.
--
vm'2.
-.s

Rate of
*1.46 Auids for which the shearing stress, T, is not linearly
related to the rate of shearing strain, 1', are designated as non-
Newtonian fluids. Such fluids are commonplace and can exhibit
unusual behavior as shown in Video V1.4. Some experimental
data obtained for a particular non-Newtonian fluid at 80 of are
shown below.
T(lb/ft2)-.J~ 2.11 1 7.82 I. 18.5 L31.7 I
l' (S-I) I 01 50 100 150 I 200
Plot these data and fit a second-order polynomial to the data using
a suitable graphing program. What is the apparent viscosity of
this fluid when the rate of shearing strain is 70 s-I? Is this
apparent viscosity larger or smaller than that for water at the
same temperature?
Shearing
shearing
strain, 1/s
o
stress,
Ib/sq ft
o
=40 'r - 0 OO~8 i ±-q.Oill5-¥~..
! 30+--~!--~i--~i--~·~,--~i
~...' 20 +__+--_--+1_--:.1/fC-----1!----i
50
100
150
200
2.11
7.82
18.5
31.7 I
I

U; 1../ i
g' 10 +----l---j-~--i~--t----rl------l
.~ 0 ....- .....~~---+1---t---1i---------i1:.
U)
o 50 100 150 200 250
Rate of shearing strain, 1/s
~----------------------
-G'" 1/0•.5
/'lPf)f Table. S.I I~ A-PfHAti,x B) fl+z.O@$ODF =1,7Q/XID +1;'-)
q,,~ 5/l1ct.. WA.-tev is a. Newt:r;nl41f riu/d 171J~ l/a/ut. 1.5
/11~/eprl1dt"t ,,{ i . TheiS", 17J~ t(l1KI1f)WfI nO!1-Newl:.t>Jltt1i-t
/-/('ui/ hilS a mu(;.h /t'{Y''1fY VI:( J~ e. .
J-3S

/.47 I 1A7 Water flows near a flat surface and some measure-
ments of the water velocity. u. parallel to the surface. at different
heights, y, above the surface are obtained. At the surface y = O.
After an analysis of the data. the lab technician reports that the
velocity distribution in the range 0 < Y < 0.1 ft is given by
the equation
u = 0.81 + 9.2y + 4.1 x lOV
with u in ftls when y is in ft. (a) Do you think that this equation
would be valid in any system of units? Explain. (b) Do you
think this equation is correct? Explain. You may want to look
at Vicko 1.2 to help you arrive at your answer.
:J .3
(a) U= ~.a/ + q.2 :J -r tf..1 >(/0 !J
[Lr-]= ~.8il1-0.2][L] +- ft.JX}D~[L3J
E'ach ferm ,A 1he ezua.;fJ~1I rn liS f h4 V! the ~(Jrne dl fYlfUI5I!)IIS.
Thus,) the ~/J~t"J1t &. f / m1l51: hAlle dln1tl1~'IJII~ "f. /.... T-;
r.2. dl':"'(AS/Pj,~ cf T-~ t:I/ld 11-. / )( 1t;.3 dl",eA.5/PlIs D f L-2 T~I
Slh,t!. 7?te. t~I1.si::4I1b /11 '1h~ ~!U4.tr~JI hf(~e c/lmpIIS/~;'.s '1J1elr
Yll/ues JtI/I/ Ch(Jl1f~ If)/tn a. Chll119E (rl /,411;f.5. No.
(b) E1Ji4tJP/IJ ~lIl1npt he t!lJrrRd ;5/nCti! a.t fj=o /A.:: ~.8J tils )
a. ntJlI-,eYb J/A/we w}",h wDulel V{'D.laJ~ .fne. '~o-sJlp'l
e(JJl1d;i:/~)I. NIJ-t t.Prrec-t.-!
1-314

1.'11 I
1.-+1-1 Calculate the Reynolds numbers for the flow of water
and for air through a 4-mm-diameter tube, if the mean velocity
is 3 m/s and the temperature is 30°C in both cases (see Example
1.4). Assume the air is at standard atmospheric pressure.
Ftf)t" wILier a...t 3 olf>C (-Ir()rn Ta~/e 8,2 ,h A-ppel1dJi B)~
~#
I::: 9 '15'. 7 ;;;; 3
Re
_ ~ V D
/
_ (rqs: 7 1!~J (3 Lf) ( t'. 00 Lf ,'YY1)
=10°00
7. 975' ;( IO-1f H,S
rrn :a.
/;r at ~ ~-I: 30' C ( 4~m Ta, ble 8. If /n 4pperJdi)( B) :
f ::: t. I (,fi :!. jA-::: I. 11. ;< ID- S :s....
Re = 752
1-37

/,.tIc:! I 1. qq For airat standard atmospheric pressure
the values of the constants that appear in the
Sutherland equation (Eq. 1.10) are C = 1.458 x
10-"kg/(m·s·KI:) and 5 = 110.4 K. Use these
values to predict the viscosity of air at 10 °C and
90 °C and compare with values given in Table 8.4
in Appendix B.
3
( TT.
=
T-tS
3
T'-
T -t" /J o. If I<
T= /O ·e. = lo 'e -t" .),7 S.Jt,- =
, %
fl. "fF8 ;(10- ) ( :1.83./51<) -5' N
1.71.5" 10 .,=
). 1'3. 15' k .,. /I O.1f "".,'"
From Table B.It)/-'
T = '10'C = '10'C -t" ;'7~. W : 31. -g. IS k )
(
3/,
_ (f.If!JgXIO-') 3(,~./!ik.) ;Z
3&, :1. I~- k. r 1I O. If
-5"
== .2.13)(.10 NoS
-:;;;;,
Frc;m
/ -3~

/.~o-tl'
1.5r)* Use the values of viscosity of air given
in Table B.4 at temperatures of 0, 20, 40, 60, 80,
and 100°C to determine the constants C and S
which appear in the Sutherland equation (Eq.
1.10). Compare your results with the values given
in Problem l.lf'f. (Hint: Rewrite the equation in
the form
T 312 = (!)T + S
Ii C C
and plot 'P'2/Ii versus T. From the slope and in-
tercept of this curve C and S can be obtained.)
(J)
T (k) I- (/V'S//tI1l.) T~ [J<~(ljJ.,.s)]
o
0;.0
'fi>
60
80
I()~
A- plot
,173. IS"
J./i3./6
313.1!i
3~3. /;-
3S3./~
373.lb
0/ V.s.
I. 7/ ;C If) - b- :2. 6'f~ ~/0 fj
/. 'iZ X10 -6' z.7sf X If) 8
-(i
~. 963)L /0 8I.f?;('JD
/, Cj 7 ;( If) -tJ- 3.037 X 10 g
-:,- l. ;; ~" X 1() 8J.o 7 ;( If)
-5 3. 322;( It) 82../7.xJO
T I~ Sh()Wn b<-/ow.'_3~
1jP-
3.Si.JD
8
':i::-:_~~==~-==-~ ~~~Fi~ ~:~..~~~;:~:j~~~=J~=::'~~-~J~~:::::::r .
:.~.:~~::-~:~:::-:.: ~-=~h:- .'.:...: =:.-:::. -=:=::d:'::~ =:-ir .::!::
:::==;~ :~::=::~=: ;::.:!~:: ~:~~:~.~::1::= ::~=1~>~: =:::..• 'j_' ..
=:-~~~:~~-=:::~~~:: =~::=:~::::- ~:==- :_='::-: : f <
3.0 X/{~f: ~~;-~cc;.~ ~~~,--: ~==-r~ .
·:1:·: ::'::=:=::~ :::: -:~/~l~:::1"'-::;" :.: :~:.:- ---...~: :-
!::::~ 7~· :.:t:::. J : .(- -:::!::
::j- (£-.-~; .... ::::l .•.•.. :~.-l·· ....;:.: -::-:-

/. 50 jIi I (C~11'i )
5/~le. the dt-tA. P/Dt a..s flff approXJlt1l.te .rfrtli9Jtt
ES' (/) C1i11 be. refrt!S'f111 kd "''I flh e8"'4..ti()~ of
form
!f::: b x. -t a..
wher-I! !J/V T3~ ) X"V T) .b ""' lie I tll1 d
To obffllH a 411d j, use LJNRFG J. po
a.N .sIc.
K**************************************************
** This program determines the least squares f it. **
** for a function of the form y = a + b * x **
***************************************************
Number of points: 6
Input X, Y
.) '273.15,2. 640E8
? 293.15!2.758E8
'? 313.15,2.963E8
,:"J 333 .. 15,3.087E8
? 353.15,3.206E8
? 373.15,3.322E8
a = +7.~~1E+07
b = +6.969E+05
X Y
+2.7315E+02 +2.6~00E+08
+2.9315E+02 +2.7580E+08
+3.1315E+02 +2.9630E+08
+3.3315E+02 +3.0870E+08
+3.5315E+02 +3.2060E+08
+3.7315E+02 +3.3220E+08
Y(predicted)
+2.6~76E+08
+2.7869E+08
+2.9263E+08
+3.0657E+08
+3.2051E+08
+3.3~~~E+08
2 =a. =
7
7. JiJf/ X ID
C
Qi/(i 1her-(~fe
s= ID7 /(
Th,Se. [.It/lues .kl' C f/11t1 5 4te In 91)t)d tl1f'femfllt
tv/ii? il4/tltS rivtl1 il1 Problem /. '11 .

/.5'/ I
1.51 The viscosity of a fluid plays a very important role in
determining how a fluid flows. (See Vieko V1.1.) The value of
the viscosity depends not only on the specific fluid but also on
the fluid temperature. Some experiments show that when a
liquid, under the action of a constant driving pressure, is forced
with a low velocity, V, through a small horizontal tube, the
velocity is given by the equation V = K/,.,.. In this equation K
is a constant for a given tube and pressure, and JJ is the dynamic
viscosity. For a particular liquid ofinterest, the viscosity is given
by Andrade's equation (Eq. 1.11) with D = 5 X 1O-7 lb • s/ft2
and B =4000 oR. By what percentage will the velocity increase
as the liquid temperature is increased from 40 of to 100°F?
Assume all other factors remain constant.
I<
-)AIfoo
I<
': [b' _~~IOD)J-'DOo IJ
-'1
5~lD e
ell
(2.)
(3)

/.52#
1.52.* Use the value of the viscosity of water
given in Table B.2 at temperatures of 0, 20, 40,
60, 80, and 100 DC to determine the constants D
and B which appear in Andrade's equation (Eq.
1.11). Calculate the value of the viscosity at 50 DC
and compare with the value given in Table B.2.
(Hint: Rewrite the equation in the form
1
In Jl = (B) T + In D
and plot In Jl versus 11T. From the slope and
intercept of this curve Band D can be obtained.
If a nonlinear curve fitting program is available
the constants can be obtained directly from Eq.
1.11 without rewriting the equation.)
£"8"4 it;;" 1.11 DIn be t.Jr;flf'i1 Ih the .{;,rm
Ih;" ::- (/3) / of In..D
tina w/th 1he cI~ia ~J'Om 746ft!. 8.2 "
T ("() T(k) I/T(K) it (J./.sk 1.)
:173. IS 3. b"l ;tID
-.3
I. 7K 7.x'If)-J0
3. 'III xIf) -.3
-3
.10 ;'~3.1; /. (!)Ol x10
/.Ib '3 I 3. IG' .1.1f3 x1D-
3 6:.. 5"29 ;(I~ -1(0
~6 333./6
-3 ~ ~'G'"Xlo-if3. ooz xlO
yo 35'3. I£" t2.152, .x' 10-
3 3. S"1f. 7 x/0- If
-.I -~
I ()o 373. ),!,- ~. 'R~ .rlf)
2.81;-,;(10
A- plot of In!- liT
.
shf){'()n be/f)w:VS. IS
( Col1t)
/-'1'2
(I)
It} ~
- t.. "3Z 7
- I.. t/ob
- 7. -33 Y.
-7. ~70
- 7. c,'f'f
-8.J7lf

/. 52 ~ I (C£J" It)
A
51,,'ce the dfti~ plat as '11/ tlflr()x,mll te sl,.Ai,1J, i
£1, (J) (/117 b~ ".!~d .fo refyr.sfrli 1hese ddt/.,
To t)btlJ/H B /In'f' f) lise k-X?FI T,
***************************************************
** This program determines the least squares fit **
** for a function of t.he form y = a * e' b*x **
***************************************************
Number of points: 6
Input X, Y
? 3.661E-3,1.787E-3
? 3.411E-3,1.002E-3
? 3.193E-3,6.529E-4
? 3.002E-3,4.665E-4
? 2.832E-3,3.5~7E-~
? 2.680E-3,2.818E-4
a = +1.767E-06
b +1.870E+03
X
+3.6610E-03
+:3.4110E-03
+3.1930E-03
+3.0020E-03
+2.8320E-03
+2.6800E-03
Y
+1.7870E-03
+1.0020E-03
+6.5290E-04-
+4.6650E-04
+3.5470E-04
+2.8180E-04
Y(predictedl
+1.6629E-03
+1.04-18E-03
+6.9298E-0,*
+4.84-82E-04
+3.5277E-0l,t
+2.6548E-04
-, I 2-
D=~:: I. 7' 7 X /D N·S /1')1
~d 3
13 -= b = I, f'J~ i< /0 /(
So i}ttrt. I!'10
-6 -T
~:: /.7~7 x/a e
Ai SOO{ (323,)5"1<»)
-,
;<= I. 7'7 ;(. /()
1370
e iJ23, )b-
-
1-'13
S.7~x)o
-it-
N.S//P1~

I. 53 I
1.5 ~ Crude oil having a viscosity of 9.52 X 10-4
Ib·s/fe
is contained between parallel plates. The bottom plate is fixed
and upper plate moves when a force P is applied (see Fig. 1.3).
If the distance between the two plates is 0.1 in., what value of
P is required to translate the plate with a velocity of 3 ftls? The
effective area of the upper plate is 200 in.2
I-'ll.{

/. 54
1.54 As shown in Video V1.2, the "no slip" condition
means that a fluid "sticks" to a solid surface. This is true for
both fixed and moving surfaces. Let two layers of t1uid be
dragged along by the motion of an upper plate as shown in Fig.
Pl.54. The bottom plate is stationary. The top fluid puts a shear
stress on the upper plate, and the lower fluid puts a shear stress
on the botton plate. Determine the ratio of these two shear
stresses. Fluid 1
Fluid 2
I-- 3 rnIs --i
f-02m/s..j
• FIGURE P1.54
n,r .fj{,lid I
'1j" h (~t f_" (6.L1 ~)(J;;Op JUr1TACt.
iJl = 0.4 N • 51m2
N20-
1m1.-
~r +I"'lei. 1. mt
1;.~A(~) = (0.2 ~)( :.o:~) =
bo-(h,,,, sur(.,,,
T-b,p ~141""+'Ct
;"";>-
( "O~fI' ~14 r...(.,(.,

1.55 There are many fluids that exhibit non-Newtonian
behavior (see for example Video VI.4). For a given fluid the
distinction between Newtonian and non-Newtonian behavior is
usually based on measurements of shear stress and rate of
shearing strain. Assume that the viscosity of blood is to be
determined by measurements of shear stress, T, and rate of
shearing strain, du/dy, obtained from a small blood sample
tested in a suitable viscometer. Based on the data given below
determine if the blood is a Newtonian or non-Newtonian fluid.
Explain how you arrived at your answer.
T(N/m2
) 0.04 0.12 2.10
du/dy ~-I) 2.25 4.50 11.25 22.5 450
Foy a- Net.AJJ:()mQI1 .P/u/c/ in( ra.C/o of -t ilJ du/dfj 15 ~
C(!hISi::.tll1t:. ;=;:'1' -th~ tiAta., 9/Vt' 11
?- (lV,s/h1~) O. ~/78 P.OI3~ o. ~//)71 ~.()()Io (/.0067 ~.()Q5F O.CIJ,5() (),()()'f-7
dull,;
Th~ ra /:./0 IS noi. tJ.. ~f1si""i ""~ de're(Js~r q s the Ya.te t!;f shear/".,g
stYII/n /n,yel/~l'!. Thll~ thiS F/w/d (;/tKJd) l,j ~ /7()I1- lIeu//;ol1l';' -flt(/d.
A- p/oi of 7hf! cltt(:a. .£S .sh/)(,Qb bt!/ow. ;:PI" A. f/ewI:rUII~H 71u,C/ 1J1~
C-/.tl"'II-e WIJ)I/I,( b~ .a si:y~",h t /J~e UJI17t I( IllJpt!! (If / f() /.
:-;< ,I ~- ~ ~;i:',; .~.... - ., " !
'I' ,
'" ..• ,1' II .': ·:::T::::F~::-:-::~:::;V:~:;:'~#,>:> ••" ..
1 - i:-:;; ~~:~!:~: <:: ::: ::::..: :: ..,~~:j:':':~ ~ ~::~; ::::t?~~ :::~:~~ ~.:: :::::::::::: ;::: ;::IC~=: :
• ,>.' :;'t'•.• ~;::" _ ... ;.~:: ,,:,:,:,,~:::::, :;,;.;:;,,::::: ::, ~j~,,,:, :1':•. '
i.-j I l"~::i::lll I : I
'--.... L - _-'-!,_ '.' ! I II : r i i ' I .1: I
. - ' 1 - - -If ._+-- -:'1'--1-1-f--,H I 'H ; ,
': ! i i : I ,1 i, ' " I,. I
/0.0 ICO.(]

1.56 A 40-lb, 0.8-ft-diameter, I-ft-tall cylindrical tank
slides slowly down a ramp with a constant speed of 0.1 ftls as
shown in Fig. P1.56. The uniform-thickness oil layer on the
ramp has a viscosity of 0.2 lb . S/ft2. Determine the angle, 8,
of the ramp.
• FIGURE P1.56
(I)
)
LJ heve V l~ the. Ve.IDC.d"'1 of- -b(A..,k..
aVI ~ b"~ Tn j(:..k:nt5S f)f t) i I Ia. 'jt..,..
#
t= (0.2 ~)( :.1 r)'=..
.DOZ
F'V't?m l?~ .CJ)
WO Ib) ~"YJ f7 - (t{) ~2..)(:q:)(O.8..ft)2
SI nf) = 6. J2.5'1
& = ,. Z2.
D
/-1.f7

/.57 I
I. '57 A piston having a diameter of 5.48 in. and a length of
9.50 in. slides downward with a velocity V through a vertical
pipe. The downward motion is resisted by an oil film between
the piston and the pipe wall. The film thickness is 0.002 in., and
the cylinder weighs 0.5 lb. Estimate V if the oil viscosity is
0.016 Ib·s/ft~. Assume the velocity distribution in the gap is
linear.
2:f rVer " ... 1
=D 't~
nUS.) 1
OW:. ~A
tA ~
wkev! A = rrDi
i
aVId ,I)- (ve 1t'>C.:~) _
frL= t- -( +ilm1hlc.l::lle.5s)
.56 -th~t
1»= (I'- t )(1TDj)
~
TI'W ~
'¥ ~ P-

i
f- D ~

/.5'8 I
1.f) & A Newtonian fluid having a specific gravity of 0.92
and a kinematic viscosity of 4 X 10-4
m2
/s flows past a
fixed surface. Due to the no-slip condition, the velocity at
the fixed surface is zero (as shown in Video V1.2), and the
velocity profile near the surface is shown in Fig. PI ~g. De- ~ _~L _ l(l)3
termine the magnitude and direction of the shearing stress U - 2 0 2 0
developed on the plate. Express your answer in terms of U
and 0, with U and 0 expressed in units of meters· per sec-
ond and meters, respectively.
?- 5(,/J'"loc~
(~:o)
dt.{
d!J
@ J=-O)
• FIGURE P1.f>'B
I U
)'
I

r~_---i
u
~--i
'----i
~
Io

151 When a viscous fluid flows past a thin sharp-edged
plate, a thin layer adjacent to the plate surface develops in which
the velocity, u, changes rapidly from zero to the approach ve-
locity, U, in a small distance, 8. This layer is called a boundary
fayer. The thickness of this layer increases with the distance x
along the plate as shown in Fig. PI.59. Assume that u = U y/8
and 8 = 3.5 Vvx/U where v is the kinematic viscosity of the
fluid. Determine an expression for the force (drag) that would
be developed on one side of the plate of length f and width b.
Express your answer in terms of f, b, v, and p, where p is the
fluid density.
U
r---.
f--
f--
'----
~
Plate
width == b
f- Boundary layer
-.1I==U /
' ~
I
· _---r--- ~- ---~.---I-
_--- e 8 ~11 == U~
_- __ ~ t (5
""----=.::=---L--IL---_____ x
..;
tJ her~ dA---
(I )
Clntl
l/
-J
~3. (f)
t{nd
IJ - 0,571 bf V-zJ1.U 3
I-50

1.601'
0.08
14.43
The coordinate Y is measured normal to the sur-
face and u is the velocity parallel to the surface.
(a) Assume the velocity distribution is of the form
u = CIY + C2Y)
and use a standard curve-fitting technique to de-
termine the constants CI and C1• (b) Make use
of the results of part (a) to determine the mag-
nitude of the shearing stress at the wall (y = 0)
and at Y = 0.05 ft.
(~) Use n()nh;'e4r yejrl'.ss/~1'J progf'YIf'fl) such a,s SAS- NLJN,)
10 ()btllil'J u;eIHc/fl1i::s (, lind C.1,.. Th;'.j pr(!)9fY1m pr{)duce.s
let/si s1"nrt's est/males ~I /he. ,PIIY'lIl71ei:trs of II /?~111J11~4r
(~)
rtl&r/el. /=by iJle dttia.. JiVfl1)
C = 153 s-',
.
SIJ1ce) du
1:=~ d;
/f ~/J/)U/s 1114t
r=;- (~ t 3 C;z. :; l )
Thus) at- the wal! (~=())
Ai
/-51

1.6 I The viscosity of liquids can be measured
through the use of a rotating cylinder viscometer
of the type illustrated in Fig. Pl.61. In this device
the outer cylinder is fixed and the inner cylinder
is rotated with an angular velocit)-" w. The torque
:, required to develop w is measured and the vis-
. cosity is calculated from these two measurements.
Develop an equation relating fl, w, 5", C) Ro and
Ri • Neglect end effects and assume the velocity
distribution in the gap is linear.
Tor'tue; d r, due. +() ~he"t;l1j sms.s
t::J11 /nneJ- C!j/Jnc/fr I~ e!tltd..fr,
d '7: rr::. T dA
whi're. c/It = ~. de) 1.. Thus)
2-
d'T= ~. J Ttit;
{J nd /-vrfflie. regtI /re d to rtJfa I:e
,nne", c'1/lntler i,S 2JT
J= 1</-1 ride()
- .2 TT R.t.''- J. r
~~~
FIGURE P1.61
top View
(J. "'" C'j Ilndrr leMi fi.J )
POI' a Iln'ell!' ve/oc./+:; distyibtl'l'/on In fhe gap
T=/-
R'WL
~ 7i R,~}.t tV
Ro-RI.'
/-5'.2...

/.bZ I
1.62 The space between two 6-in. long concentric cylinders
is filled with glycerin (viscosity = 8.5 X 10-3
Ib·s/ft2
). The
inner cylinder has a radius of 3 in. and the gap width between
cylinders is 0.1 in. Determine the torque and the power required
to rotate the inner cylinder at 180 revImin. The outer cylinder
is fixed. Assume the velocity distribution in the gap to be linear.
Prl)/'/em /. " (, )
T = 02'ff R,.3))A- W
:eo - /Ct..'
(;80 !!.!. )(eillTmilt
~ )(1 mlh)= blT
rev '0s
W=
vad
s
.:l.7T (i£ft)3(-A ft:)(s,s)(/a-
3
!Jt.)(67T 0/) =
( ~ -ft)
120
0, qIf 'f It·l)'
S/f]ce pouJey =
f()wer': (~tJif'fft'/h)(67T r;d) == 178 ~.Ib
/- '53

I. (P3 1.63 One type of rotating cylinder viscometer, called a
Stormer viscometer, uses a falling weight, 'lV, to cause the cyl-
inder to rotate with an angular velocity, w, as illustrated in Fig.
PI.6.3. For this device the viscosity, J.L, of the liquid is related
to 'lV and w through [he equation 'lV = KJ.Lw, where K is a
constant that depends only on the geometry (including the liquid
depth) of the viscometer. The value of K is usually determined
by using a calibration liquid (a liquid of known viscosity).
(a) Some data for a particular Stormer viscometer, obtained
using glycerin at 20°C as a calibration liquid, are given
below. Plot values of the weight as ordinates and values
of the angular velocity as abscissae. Draw the best curve
through the plotted points and determine K for the vis-
cometer.
'lV (lb) 2.20
5.49w (rev/s)
• FIGURE P1.63
(b) A liquid of unknown viscosity is placed in the same
viscometer used in part (a), and the data given below
are obtained. Determine the viscosity of this liquid.
'lV (lb) I 0.04 0.11 0.22 I 0.33 I 0.44
w (rev/s) 0.72 1.89 3.73 5.44 7.42
( Cl) 5;;'re ~: K)4u.J -th~ ~/()fe ~I -the czJ tis.
IJ
If)' =
We/b)
s/t:Jpe =
tv C~V)
SO fha..i ~ (16.s )
k=
.5 0 jJe YeV
/<-(~)
fi,y -rJ,e :J/~~en~ dai:a. (.see p/Dt: ~11 f')(?x/; page)
('/?Q.5rd ,t)11 a /ul.ri S,!ftllres ,f,:t Df the d~~) I.J
S/CJj)e (J/,/ceni1):= O,.?9R J~~
5In'~ U tjl'lcerln) =: 3.13X//}_zJb·s -tnel1
IF. it. ..
1<=
(),g9S Ib...5
/'?v
tu
Fixed outer
cylinder
UlriJe
(I)
(h) ;:P". the. un klJ~tVf1 filii (J d~-k.. Gee. IJ/t)i GJI1 11t't.-t f4~~) the
sJ()P~ (b"~e# tf}Jf ~ legst, $8"II"S ~it ~I tH~ d/l,tA,.) IS
S/Db.# (tll1KdtJuJl1 rill/g) = O,CJ6o/ 11:1'.$
r- ~v
/-51.f

I, (P3 (~l1lt )
Thu~ /rpm E"r.l/)
I S/tJff!!
I (lIlJillPlIJlI fluid) = 1
,.0
Ib·5
tJ. tJ(p ~/ -rev
/2., 7 H,2.
rev
/-5,

1.6Y* The following torque-angular velocity
data were obtained with a rotating cylinder vis-
cometer of the type described in Problem 1.61.
Torque (ft-lb) 13.1 26.0 39.5 52.7 64.9 78.6
Angular
velocity (rad/s) 1.0 2.0 3.0 4.0 5.0 6.0
For this viscometer Ro = 2.50 in., Ri = 2.45 in..
and r= 5.00 in. Make use of these data and a
standard curve-fitting program to determine the
viscosity ofthe liquid contained in the viscometer.
The
-the
.fz,r~lIeJ ~ J'J
.e$Ua. /;/~11 )
Jl'e/a.bed .J-o -the tlnJU/dY (/(!/t:Jcil-!1.; UJ.)
3
:;,7T R,: )J t.U
~--
l!o - ftL'
(see
(1I'1d
so/tJl-,{)~ ..J.t> Problem I. hI, ). Thus) /"y Ii flx'ed ,et:Jmef,.J
a. 'lIt/en /l1~CC).s;-f!:J J E~ ,/1) /05 ot 1h~ ~f'rn
y=hx ( !1 rv'J Qni1 x' r.." W )
If. ~,,~tr,1'J i .fl/Utl/ .fc
j, = :zTr ~.:1).!=
IS
)2'(.
WI ih LIAlfrF6 I.
**************************************************~
** for a function of the form y = b * x **
***************************************************
Number of points: 6
Input X, Y
? 1.0.13.1
'? 2.0,26 .. 0
~) 3.0,39.5
'? 4.0.52.7
? 5.0,640.9
'? 6 .. 0,78.6
b = +1.308E+Ol -Ft,Jb'5
X
+1.0000E+OO
+2.0000E+OO
+3.0000E+OO
+4,.OOOOE+OO
+5.0000E+OO
+6.0000E+OO
Y Y(predicted)
+1.3100E+Ol +1.3082E+Ol
+2.6000E+Ol +2.6165E+01
+3.9500E+Ol +3.921±7E+01
+5.2700E+Ol +5.2330E+Ol
+6.4,900E+01 +6.51±12E+01
+7.8600E+Ol +7.81±95E+Ol
(C()I?'t )
/-5~
(/)
().)

(emit)
! = (b) ( ~. - f?.: ')
.27T ~.3,R
find w/th -/he daf&" g/~tl1)
r;g Of ft.f/;'$ ) (.? 5"0 - :I.. LJ.S- ft-)
/ = ' 1'2. _ rJ. 1f-5 IJ,·s
~lT (~.'1~ ft) J (~It) -Fel.
12- 12.
/-57

I. 'S-
1.65 A 12-in.-diameter circular plate is placed over a fixed
bottom plate with a O.l-in. gap between the two plates filled
with glycerin as shown in Fig. PI.6S. Determine the torque
required to rotate the circular plate slowly at 2 rpm. Assume
that the velocity distribution in the gap is linear and that the
shear stress on the edge of the rotating plate is negligible.
ili FIGURE P1.65
kYO'lUi) dCiJ J c1u~ h s hellYJ~1j sfrfSSl'.J
pn pI4+~ l.s~ e1Uq / -1-"
do;= ,. tdA
~ ntrt dIt.: 2.11" yo dr, Thel'5J
5;nce
cJ C7J: "" T Z.".. rd '"
"(=ufl'!r.r tlr
o
T,: f<- ~ ) 1/fl.t/ "r fA.
lIe/"'Jf.t J'5tY; h",-I-U/H (SlefijHye)
-:: D. 0772 /1:. ·/t
I-58
Rotating plate
0.1 In. gap
ellA _ V 'rW
d;-I~T

I. ~ 7 J
1.(,1 A rigid-walled cubical container is completely filled
with water at 40 of and sealed. The water is then heated
to 100 oF. Determine the pressure that develops in the container
when the water reaches this higher temperature. Assume that
the volume of the container remains constant and the value of
the bulk modulus of the water remains constant and equal to
300,000 psi.
5/~ce 1h~ t..Jd..frr tnf4S) YfMAll1S ttPl1s~fJ
~.1I::1 (..pl71.l¥-)
~~ /tJO D
+ /j 1If)/Ume t1"~ 4-1' iJ C.hlll1'1t2 111 ve)/um e 1"1 tva/-!'/'"
'/ir, I, / z.
sJuf,.S
/, l''fb .ft.J
I. '/27 5'::'
dp
-& = - d¥
V ~
7h II~)
-/
If ~I/IJ UJS U.l/1J.. d1l- ::; 4-v' 411 t:I. PI» -.:::= fJp 1hAf- 1hf CJtIlH'I€
/11 />Y'e> 5 ",y~ r-e'gwred. ..f-r> ~nJ;,.es.s the Iva..-tey bltJt. Ie i-h
tJ /1'/1/11.) 110 I~ ft1e... ,'.1
t1f= - r;tJ~/~~()f~L'){-O.OO~75)
3 .
2.o3iJD pst..

/,iD't I
1.(o~ In a test to determine the bulk modulus
of a liquid it was found that as the absolute pres-
sure was changed from 15 to 3000 psi the volume
decreased from 10.240 to 10.138 in.3
Determine
the bulk modulus for this liquid.
d~ z A ¥ -= I~, e1JfO - I~, 13~ :
Ib
rJ9i'S ;;,.
( C', 1t)2 In,7)
/", ;l'l0 in.'
1.tJ1 Calculate the speed of sound in mt,s for
(a) gasoline, (b) mercury, and (c) seawater.
(a) f:by 7(is~l/ne,'
I 3
t:J, jOJ, Ii'l,
( Eg. I,/Q)
I, Lf5 ~I'm
s

T 1.70 Air is enclosed by a rigid cylinder con-
taining a piston. A pressure gage attached to the
cylinder indicates an initial reading of 25psi. De-
termine the reading on the gage when the piston
has compressed the air to one-third its original
volume. Assume the compression process to be
isothermal and the local atmospheric pressure to
be 14.7 psi.
PC>t' i~othermlJ/ ~&)mpY'ess/t:Jl1)
-Pl.'
..f',-.
Pf =
= -fE
~.f
f!t f;.
~. ~
Where (.'~ /ni.J-,i.d state.. UI1A
f """ f/nAI ~illie.. .
0/nc.e f=
m4SS
)J/t)ltll11 e.
1'h~reloye
1;. =(3)[(:15 f- / 'I. 7) p:s L' ftJ6s)) :::
t (p-;e) =(i/9- 1'1. ~f'i :-

It 7 / J
h,y
So
1.1 I Often the assumption is made that the flow of a certain
fluid can be considered as incompressible flow if the density of
the fluid changes by less than 2%. If air is flowing through a
tube such that the air pressure at one section is 9.0 psi and at a
downstream section it is 8.6 psi at the same temperature. do you
think that this flow could be considered an imcompressible
flow? Support your answer with the necessary calculations. As-
sume standard atmospheric pressure.
1"0 therMo / CHfiA'It 111 den~;I!:J
-/;), :-
P2-
- -f, ~
1},4.-t I, ?z..-- ----- ~f,
)( I D0
TAus
I

1.72. J
1.72. Oxygen at 30°C and 300 kPa absolute pressure ex-
pands isotherrnalIy to an absolute pressure of lZ0 kPa. Deter-
mine the final density of the gas.
For /sofh~rma/ ex.ftlI15i()YJ ) ::t =t!LJl'Jsft/tJt:
~
~.
~~ u)),ere
,
;'rllt,4/ s fai-eL ...
t"'"- - -~. !j I- "V .fInII / st:A..ie.
SIJ 1h4t
Pi1#
- 3.8/ ~3/J?"I

/,73 J
For
C( J1 c/
~r
1.73 Natural gas at 70 OF and standard atmospheric pres-
sure of 14.7 psi is compressed isentropically to a new absolute
pressure of 70 psi. Determine the final density and temperature
of the gas.
;'sen irop'c- c~m?re~S/()11 , --P = ~t!/ns tQl1t
;O~
~. J}
tVhere ,,; 'V ;'n;';';'/ 6~te" ::t
cou/- 1,*p.-A
.f 'V .f.J'ntl/ sta.te .L -f-
~I
=
51) -thA-f
-,3
'I.2S )(Jf) S h~f5
-ft:'a
, ~ )
-for IIJ I I/l.
7f
..... (7c; 7ii:a. ) ( If tf -:t:;"A-
-
ft.R -
('I.25 ;I. / ()- ~ Shtf..s ){3.O'If;( j 03 ~b IJ: )
h3 sh",.liI~
- 7(P5 (J)R-
71: 7fD 5 oR - /f-IPt) - 305 of

/.7 if
1.7ll-- Compare the isentropic bulk modulus of
air at 101 kPa (abs) with that of water at the same
pressure.
J:C;r a l r ( E''I. /,) 7 ))
£ y ~ I<. f =(/,'i-o ) (10/ x 1t'3~ ) =
/=(;r tva:te". ( Tr,; bJeo /, /, )
E,,:: :1.)6;< It> " ,q
Thu.5)
Ev (WIJ.,-ter) _
£v (cur)
9
~, 15" ;< II) Pa
I, Ifl X. 1~5'"/}

I. 75" '4' J 1.7.5'* Devel~p a -computer program for cal-
culating the final gage pressure of gas when the
initial gage pressure, initial and final volumes,
atmospheric pressure, and the type of process
(isothermal or isentropic) are specified. Use BG
units. Check your program against the results ob-
tained for Problem 1.70.
r-oY C/!)11? pye5SIol1 e1 Y ex..pQI1JIOII)
?... = e04stoni.
!
wheye h=/ -ky isotho-mal process) and It::: .Jj'e,;{tc. helL-/: va.!:'"
lOr 1.st'l1frt'Jllc proc.ess. Thus J
~. = !i:.
;:.-* /;.-P.
where /.'/1; In/ha'/ ~k.te I .f''V IiH~/ .str;le) So 1J1Ii't
if : (-J,:) "-f:: (/ )
1hel1
ml1ss
t.: Vt!)/~lI1e
~ = Vt·
~. ~
w he¥"e v;,.) ~) tire
Thus) Ir~m S~ /1)
{
'It. )-k
t.Jhe Y'e
CaM be.
l T -A ::'..f.g a:eM Vf (~! T 1:.t"" )
fh (! SWhSChpi 3 reI-Frs 1::6 JaJe /J Y"e sst/re
w y-;He!? as
if, = (~) -I. (~'j -t tt"J -~t""
(c~n It )
( 2 )

175 itt I
100 cls
110 print "*********************************************************"
120 print "** This program calculates the final gage pressure of **"
130 print "** an ideal gas when the initial gage pressure in psi, **"
1""0 print "** the initial volume, the final volume, the **"
150 print "** atmospheric pressure in psi, and the type of **"
160 print "** process (isothermal or isentropic) are specified **"
170 print "*********************************************************"
180 print
190 input "Enter initial gage pressure in psi, Pi = ",p
200 input "Enter initial volume, Vi = lI,vi
210 input "Enter final volume, Vf = ",vf
220 input "Enter atmospheric pressure in psi, Patm = ",patm
230 pabsi=p+patm
240 print:print "Enter type of process"
250 print "0 Isothermal"
260 print "1 : Isentropic"
270 input pt
280 print
290 k=l
300 if pt=l then input "Enter specific heat ratio, k = ",k
310 pabsf=pabsi*(vi/vf)~k
320 pf=pabsf-patm
330 print
3l.!-O print using "The final gage pressure of the gas
is Pf = +#.####~~~~ psi";pf
*********************************************************
** This program calculates the final gage pressure of **
** an ideal gas when the initial gage pressure in psi. **
** the initial volume, the final volume, the **
** atmospheric pressure in psi, and the type of **
** process (isothermal or isentropic) are specified **
*********************************************************
Enter initial gage pressure in psi, Pi =ZS
Enter initial volume, Vi = 1
Enter final volume, Vf = 0.3333
Ent.er atmospheric pressure in psi, Patm = 1"".7
Enter type of process
o Isothermal
1 Isentropic
? 0
The final gage pressure of the gas lS Pf = +1.0~/E+02 psi

I. 7 G:. I
1.7<; An important dimensionless parameter concerned
with very high speed flow is the Mach number, defined as Vic,
where V is the speed of the object such as an airplane or
projectile, and c is the speed of sound in the fluid surrounding
the object. For a projectile traveling at 800 mph through air at
50 of and standard atmospheric pressure, what is the value of
the Mach number?
Thu.s
~.h)e. B.-3 In
~al'r ~ 50-F
!v1I.G'r1 numblY -
-.
y
Co
LO{P

1,77l
I. 11 Jet airliners typically fly at altitudes between approx-
imately 0 to 40,000 ft. Make use of the data in Appendix C to
show on a graph how the speed of sound varies over this range.
c = VIeR.r
7
(EZ' I· 2o)
t:;r 4< ::: lifO Clf1d If=- 17/t, 1i·1/,
s/,,!! .t)~
C= '19. tJ YT(~)
1
Fr~m 1a6/t: C. / In 4ppend':x C at tI/1 (J / -fi.f.udf! til
T= S'I. ~ C) of- !fl;o : 5J9°~ .50 -thl.;/;
::: I / /!. .f.i:-s
5;'17/;/tfl'" Cd /CU/4,tltPIfS CiJn De mtule -hr (JiJ,n~ dlf/ltlt!t'~
tin' -the ,es~/f/;'! 1raJh is :J"/1()W)f b~/ow.
Altitude, ft Temp.•o F Temp.," R c, fUs
0 59 519 1116
5000 41.17 501.17 1097
10000 23.36 483.36 1077
15000 5.55 465.55 1057
20000 -12.26 447.74 1037
25000 -30.05 429.95 1016
30000 -47.83 412.17 995
35000 -65.61 394.39 973
40000 -69.7 390.3 968
1120 r--,.-----,---;---;----;----:---:-~__,
1100 -~...... I
~1080 I~ I
;; I I' 1
g1060 . L'" "-
(f) 1040 t---t---t---t--..-""-~--:-----1f---J!---L-...---l
~1020 .t--r--+--+I_--+-~-~--+I---!--l
~1000 .r----+---+--_+_I- 1 -1, ----+~~,__LI-_+_I____i
(f) I 1.' I.
980 t----;---;---t---t---t---+---.......;.......:""~~---1
I' 1 I r----960 +-.---+----'----..:.---..:.---;--~_~---1
o 5000 10000' 15000 20000 25000 30000 35000 40000
Altitude, ft
o +-t

I. 73 I
1.7 R When a fluid flows through a sharp bend. low pres-
sures may develop in localized regions of the bend. Estimate
the minimum absolute pressure (in psi) that can develop without '
causing cavitation if the fluid is water at 160 OF.
cC/J/i.faflon mtJ'1 (pee",,. whtn fhe I~cq/ pY{"$Stlfe e~t",ls the
(/t2for ,res'Sure. !=Or waiel" ai- /(p() df' (lj.~,." ?;bJl8,j ,~Apptw:l/J(B)
Thus/
i = if. 7Lf pSI,' (IIbS)
r
1.79 Estimate the minimum absolute pressure (in pascals)
that can be developed at the inlet of a pump to avoid cavitation
if the fluid is carbon tetrachloride at 20 °c.
Cavi i-A.I/!)11 rnp'1 (peetI r when fhe stlciion P;'(J$stlJle
at- -tnt!.. pum,P inlet etttlA/.s the 1/a.,Pcr' fJY'es$ure.
t;r ClJrhtm tei'('ac.J"Joy;d~ t2 t 2.0
IJ
C ...n = /3 ~ R. (g,!;s)
IV
rn ; n /m J,I J?1 !f".Rssure /3 .Ie Pa. (4.6S)
/-70

I, So J
I .~D When water at 90°C flows through a converging sec-
tion of pipe, the pressure is reduced in the direction of flow.
Estimate the minimum absolute pressure that can develop with-
out causing cavitation. Express your answer in both BG and S1
units.
('C/vif4tl{)" nUl'j cc.CClr /n 111e Ct'''J/er9'/~~ sec..-i-Idn ~ pile whel1
-rhe. pr~55«;,e tEf;aolS -th~ va.~J' fYe.J5'tlre. ;:-r/)/7'I 74lie B. 2 I;' I+!'ff"c/J( fj'
~r wA,ter at 9~ °C.I 1;:: 70. / -h Po.... (ql,,,). Thu~
/.31 I
.
minimum pre~suv~ ::' 7()./ --k.?c.. (q/'5) 1/1 sr tln,fs.
86 Hnifs
I11lnlmuM .P'fSJare =f;().JxJ~3::.. )(/. lj5"1; )(/J- ~fl,.{)
::: /0, 2 psI.a
1.8/ A partially filled closed tank contains ethyl
alcohol at 68 OF. If the air above the alcohol is
evacuated what is the minimum absolute pressure
that develops in the evacuated space?
f.iz I
1.8Z Estimate the excess pressure inside a rain drop having
a diameter of 3 mm.
().oo/5 /1n
/-7f

I. 113
I. r~ A 12-mm diameter jet of water discharges vertically
into the atmosphere. Due to surface tension the pressure inside
the jet will be slightly higher than the surrounding atmospheric
pressure. Determine this difference in pressure.
"Ft;/' erp,iJ/bri/l"" fspe IIjure ).;
1(z~Ii/: cr(zJI.)
SO -rnA i
-t==
12 >' it; -.3 ~
2:
= 12. 2 Ii
1-72..
1'1V ex,t'SS f rfSSU re
Sur-Hlle -ftHSIDIl ~'(,e:- cr 2. £~

/,8'-1
1. 'a Y. As shown in Vidl'O V1.5, surface tension forces
can be strong enough to allow a double-edge steel razor
blade to "float" on water, but a single-edge blade will sink.
Assume that the surface tension forces act at an angle erel-
ative to the water surface as shown in Fig. PI ~~. (a) The
mass of the double-edge blade is 0.64 x 10-3
kg, and the
total length of its sides is 206 mm. Determine the value of
erequired to maintain equilibrium between the blade weight
and the resultant surface tension force. (b) The mass of the
single-edge blade is 2.61 x 10-3
kg, and the total length of
its sides is 154 mm. Explain why this blade sinks. Support
your answer with the necessary calculations.
Surface tension
force
• FIGURE p1.<64
T
(a. ) L FV€r+t '4I
::.0 ~
tow
~
o.
VW=Ts/n8
Luheye ttJ :: (Y'(l X
blade ~ Ql;1d T:::- a- ><. Jenfn, of. slqes
( (), 10'1- ;( 10-3-ka ) (U I I'tr./~.) = fr. 3~ ;( }O-2 .Jt.)(IJ, ZO~ /In ) 5'111 e
:sin e- =- o. Lf-15
9 = :J... 4-.5
0
(b) For slnrle-edtje blade
'2J = /yrl MAde X d- " (~2.I.1 x: 10-
3
-ka.J ('1. ~J I'M/~')
:: ().DZ~1s, N
uYld
T 5111 e :: (vx. JenJ1n of. /,lode ) ~I;' f7
::- (7.3LJ.x/o-Z
Ntm) (O.15LtM1) '51Y1 G
= O. 0 I J 3 '5/n e
r n t>Y'aer +O~ hjq de +0 "-J./Da.i It ~ -< T "SIn e.
"StYlet.. rma)(Jf1'lUfn1 Value JoY" ~Ine IS J'+- followsI
that '1.<.J > T St'n e and 'Sin9/<!-eciJe hlade w; II si"k.
/-73

I. 8'5 I
(a.)
1.1'5 To measure the water depth in a large open tank with
opaque walls, an open vertical glass tube is attached to the side
of the tank. The height of the water column in the tube is then
used as a measure of the depth of water in the tank. (a) For
a true water depth in the tank of 3 ft, make use of Eg. 1.22 (with
() = 0°) to determine the percent error due to capillarity as the
diameter of the glass tube is changed. Assume a water
temperature of 80 oF. Show your results on a graph of percent
error versus tube diameter, D, in the range 0.1 in. < D < 1.0 in.
(b) If you want the error to be less than 1%, what is the smallest
tube diameter allowed?
The e;(ce~s he'jh t I h) CtlfA~ed bt
h= zo-~~
'oR
1h~ .sur~(t. ien>/~~ ,1
(E'Z. J.ll.)
f:ipy- tr:: 0 D tv;f;, b =.z.R.
h = 'fO-
rD
PY'If)I?1 7i.J,/~ B./ln A-ppendl)( B /Dy- WtJ.-tev ~t
0-= Jf.9Ix//J-3/bj.Pt Clnd r= 'Z.2z. Jb/k~
Th~~ .fr.9m 1:1. (I)
h~f):: tf (Jf,tjI x:IO-J -!)
{(e>Z.lZ. ~,) D(,n.)
I 2. I~, /';-1::
~l~ error =. h ~J )f.. 100
.5'.ft;
froW! eq.l 1-) 1ha.~
-3
fl.1 q )( J0
.D ( I'n.)
D -3
f)/0 eyr"y = 3.7&f x J0
x ID0
3 D(J'n,)
A- plot. t>.f ~ eY'r~r V-(i"S(,/S t"'be C/'t:ll11et:er IS
ShtJWI1 "n 111t /1f~t I"a'je,
( C!L;,/t. )
/-7Lf
0)
( 3 )

/. 8S- I ( Ccr/t.)
Diameter % Error
of tube, in.
0.1 1.26
0.15 0.84 1.50
!
,
0.2 0.63
I
I i
,
I
I I
... 1
0.3 0.42 0 1.00 1
I
: ...
'{ I
I
i
-
0.4 0.32 i ...W
,
I i i :
0.5 0.25 I ~ 0.50
,
, 0 I
~
!
I
0.6 0.21 i I
,
,
I 0.00
i ..... i
0.7 0.18 i ,
!
0.8 0.16 i 0 0.2 0.4 0.6 0.8 1 1.2
0.9 0.14 I Tube diameter, in.
1 0.13
I
"
--
Values obtained
from Eq. (3)
(1) For /ofo eyrpy ;;'PII1 £Z. (3)
J=
t). /2b
/)(,'rJ.)
D-= ~./2.1D
.In.
/-7S

1.H6 Under the right conditions, it is possible, due to surface
tension. to have metal objects float on water. (See Video V 5.)
Consider placinf, a short length of a small diameter steel (sp.
wt. = 490 lb/ft) rod on a surface of water. What is the
maximum diameter that the rod can have before it will sink?
Assume that the surface tension forces act vertically upward.
Note: A standard paper clip has a diameter of 0.036 in. Partially
unfold a paper clip and see if you can get it to float on water.
Do the results of this experiment support your analysis?
o. 0 ~ J '+ (n.
grr
cri.. rrL
I
-3 r.L
5 II X' I0 ;-"'l..
S/nc-e ~ ;st.andArd ~I:ee/ paptr c)lf hAS ~
d/~met;('r ~f ". ~3" il1') wh/ch IS Jess fr..a11
O.O'/'/- /n.) It sJ1()~J~ f/~4.t. A- ~/mpj~ e)l../Jtrirnmi
iP f / I Vof v,'f ~ 111I.S • Ye s .
J-7f6

J.37 I
I. g$ I
1.~7 An open. clean glass tube. having a diameter of 3 mm.
is inserted vertically into a dish of mercury at 20°C. How far
will the column of mercury in the tube be depressed?
2 (}C&S e
?rR
1. gB An open 2-mm-diameter tube is inserted
into a pan of ethyl alcohol and a similar 4-mm-
diameter tube is inserted into a pan of watef. In
which tube will the height of the rise of the fluid
column due to capillary actton be the greatest?
Assume the angle of contact is the same for both
tubes.
( ~g. j. 22 )
-3
3.00 X ID 1m
3. 0 0 1)')1 t'YY1
(Eg. j,22.)
.J,. (C// t~hp/)
~ (tva tel")
U (~dtph()/) '0(WA if,) (If 1m"" )
a-(WA. tf'I") !"" (,,/tCh"/) ~ IWIIW1
= (;.2.81-/0-'). ~)('/.r{)Xlo3~3)(#MAI'IIA)
( 7. 3lf)( JD-~ f;, ) (7.tlf X }f)3~3) (;).M1~ )
(J, 7g7
/-77

1.~~ * The capillary rise in a tube depends on
the cleanliness of both the fluid and the tube.
Typically, values of h are less than those predicted
by Eq. 1.22 using values of (J and efor clean fluids
and tubes. Some measurements of the height, h,
a water column rises in a vertical open tube of
diameter, d, are given below. The water was tap
water at a temperature of 60 of and no particular
effort was made to clean the glass tube. Fit a curve
l7o/n t.~. I. ')."L.
to these data and estimate the value of the prod-
uct (J cos e. If it is assumed that (J has the value
given in Table 1.5 what is the value of e? If it is
assumed that eis equal to 0° what is the value of
(J?
d (in.) 10.3 I 0.25 I 0.20 I 0.15 10.10 I 0.05
h (in.) 0.133 0.165 0.198 0.273 0.421 0.796
-P. = 2 O-d-U:;S e(-k); 'f(j'C:S e(-f)
d=l12. Thus; £j.(J} 1..5 ()t the. /cYm
i,: b d'
b= d'::: J...
d
The ~I/ S /-III1't.; b) C41J b.c ()j,~/~et/ b'1 4 //I1t'l.Y least
szuare''s fL't 6f 1J1.e, 911/"11 d...-ba... (J.. Ql1d lid).
If.0
'f~
!Po
80
120
J.'fO
( CD!) t)
/-7'6
-P. (ft)
O.O!l~8
(). () 13 ?:;-
O. () /65'0
(). t)z27S'
(). b ~5"oK
(),0""33
(Z)

(C4;I1't)
To "btfll~ b 11.5 e. LJNREG 1 r
***************************************************
** for a function of the form y = b * x: **
***************************************************
Number of points: 6
Input X, Y
? 4-0,0.01108
'7 4-8,0.01375
? 60,0.01650
? 80,0.02275
? 120,0.03508
? 24-0,0.06633
C.L, 2.-
b = +2. 799E-04- rt,
X
+4-.0000E+01
+4-.8000E+01
+6.0000E+01
+8.0000E+01
+1.2000E+02
+2.4-000E+02
Y
+1.1080E-02
+1. 3750E-02
+1.6500E-02
+2.2750E-02
+3.5080E-02
+6.6330E-02
Thus,
rr e~Se = h a
If
Y(predicted)
+1.1195E-02
+1.34-34-E-02
+1.6792E-02
+2.2390E-02
+3.3584-E-02
+6.7169E-02
_ (,<..799 x)0· If It "L.)~z. If ~J)
'i-
.3 lob
= 1-. 37 X jo .ft
II
.3
/J,/fi 1hen0--= So {)3 Jt /0
I
if: "17X/lJ
-.) fk
Cc>J e :: -Fe =- o.g~r
-s. tJ3X /o-J .1.l!
.fi:
alld
~ = J 1.70
If B=o
0
-rhfl1 Cos a = /.0 ClI1 d
3 J..E 3 /.6'f,37X)O
rr= .pt- ::' if, 37 XJO
--..
,ft
/.0

1.90 Fluid Characterization by Use of a Stormer Viscometer
Objective: As discussed in Section 1.6, some fluids can be classified as Newtonian flu-
ids; others are non-Newtonian. The purpose of this experiment is to determine the shearing
stress versus rate of strain characteristics of various liquids and, thus, to classify them as
Newtonian or non-Newtonian fluids.
Equipment: Stormer viscometer containing a stationary outer cylinder and a rotating,
concentric inner cylinder (see Fig. P1.90); stop watch; drive weights for the viscometer; three
different liquids (silicone oil, Latex paint, and corn syrup).
Experimental Procedure: Fill the gap between the inner and outer cylinders with one of
the three fluids to be tested. Select an appropriate drive weight (of mass m) and attach it to the
end of the cord that wraps around the drum to which the inner cylinder is fastened. Release
the brake mechanism to allow the inner cylinder to start to rotate. (The outer cylinder remains
stationary.) After the cylinder has reached its steady-state angular velocity, measure the amount
of time, t, that it takes the inner cylinder to rotate N revolutions. Repeat the measurements us-
ing various drive weights. Repeat the entire procedure for the other fluids to be tested.
Calculations: For each of the three fluids tested, convert the mass, m, of the drive weight
to its weight, W = mg, where g is the acceleration of gravity. Also determine the angular ve-
locity of the inner cylinder, w = Nit.
Graph: For each fluid tested, plot the drive weight, W, as ordinates and angular velocity,
w, as abscissas. Draw a best fit curve through the data.
Results: Note that for the flow geometry of this experiment, the weight, W, is propor-
tional to the shearing stress, T, on the inner cylinder. This is true because with constant an-
gular velocity, the torque produced by the viscous shear stress on the cylinder is equal to the
torque produced by the weight (weight times the appropriate moment arm). Also, the angu-
lar velocity, w, is proportional to the rate of strain, duldy. This is true because the velocity
gradient in the fluid is proportional to the inner cylinder surface speed (which is proportional
to its angular velocity) divided by the width of the gap between the cylinders. Based on your
graphs, classify each of the three fluids as to whether they are Newtonian, shear thickening,
or shear thinning (see Fig. 1.5).
Data: To proceed, print this page for reference when you work the problem and click hl're
to bring up an EXCEL page with the data for this problem.
Rotating inner cylinder
Outer cylinder
Fluid
Ii FIGURE P1.90
(cOn 't )
/- go

/.9'0 I
Solution for Problem 1.90: Fluid Characterization by Use of a Stormer Viscometer
m, kg N, revs t, s co, revls W,N From the graphs:
Silicone oil is Newtonian
Silicone Oil Data Corn Syrup is Newtonian
0.02 4 59.3 0.07 0.20 Latex paint is shear thinning
0.05 12 66.0 0.18 0.49
0.10 24 64.2 0.37 0.98
0.15 20 35.0 0.57 1.47 co = Nit
0.20 24 31.7 0.76 1.96
0.25 30 31.0 0.97 2.45 W=mg
0.30 20 17.4 1.15 2.94
0.35 25 18.8 1.33 3.43
0.40 40 26.0 1.54 3.92
Corn Syrup Data
0.05 1 28.2 0.04 0.49
0.10 2 27.5 0.07 0.98
0.20 4 27.2 0.15 1.96
0.40 8 25.7 0.31 3.92
Latex Paint Data
0.02 2 32.7 0.06 0.20
0.03 2 20.2 0.10 0.29
0.04 5 32.2 0.16 0.39
0.05 10 47.3 0.21 0.49
0.06 10 37.2 0.27 0.59
0.07 10 29.8 0.34 0.69
0.08 10 24.6 0.41 0.78
0.09 10 20.1 0.50 0.88
0.10 20 34.0 0.59 0.98
/- 9I

!. 'to ( c~I") t )
Problem 1.90
Weight, W, vs Angular Velocity, 0)
for
Silicone Oil
Problem 1.90
for
Corn Syrup
4.50
4.00
3.50
3.00
··..--~~~~---·---------~l
~·-=~I
4.50 ..,--.----
4.00
3.50
3.00
z 2.50
~ 2.00
1.50
1.00
0.50
-----------1
W=2.5~ 0) I--_ .._----------------j
I
-- ---------------..-----------1
z 2.50
~ 2.00
1.50
1.00
·W = 12.80)------------_._---+---_. ---_..- - j
0.50 +-~'-------~------~---------.-- ~ ..
0.00 4----,..------;----,-----1 0.00 +-----r----r----,--------;
0.00 0.50 1.00 1.50
OJ, rev/s
1.20
1.00
2.00 0.00
Problem 1.90
for
Latex Paint
0.10 0.20
OJ, rev/s
-'1---..-----1
0.80
z
+-------~----~-~----1
~
0.60
DAD
---,1tfI""'------- ~-- ----------- ----- - --.-j
I------ ..._-
W =1046600°
707
0.20
0.00
0.00 0.20 DAD 0.60 0.80
00 rev/s
J- 8'2.
0.30 0.40

I. 'f /
1.91 Capillary Thbe Viscometer
Objective; The flowrate of a viscous fluid through a small diameter (capillary) tube is a
function of the viscosity of the fluid. For the flow geometry shown in Fig. P1.91, the kine-
matic viscosity, v, is inversely proportional to the flowrate, Q. That is, v = KIQ, where K is
the calibration constant for the particular device. The purpose of this experiment is to deter-
mine the value of K and to use it to determine the kinematic viscosity of water as a function
of temperature.
Equipment: Constant temperature water tank, capillary tube, thermometer, stop watch,
graduated cylinder.
Experimental Procedure: Adjust the water temperature to 15.6D
C and determine the
flowrate through the capillary tube by measuring the time, t, it takes to collect a volume, V,
of water in a small graduated cylinder. Repeat the measurements for various water temper-
atures, T. Be sure that the water depth, h, in the tank is the same for each trial. Since the
flowrate is a function of the depth (as well as viscosity), the value of K obtained will be valid
for only that value of h.
Calculations; For each temperature tested, determine the flowrate, Q = Vlt. Use the data
for the 15.6D
C water to determine the calibration constant, K, for this device. That is, K = vQ,
where the kinematic viscosity for 15.6°C water is given in Table 1.5 and Q is the measured
flowrate at this temperature. Use this value of K and your other data to determine the vis-
cosity of water as a function of temperature.
Graph: Plot the experimentally determined kinematic viscosity, v, as ordinates and tem-
perature, T, as abscissas.
Results: On the same graph, plot the standard viscosity-temperature data obtained from
Table B.2.
Data: To proceed, print this page for reference when you work the problem and click hu!'
Th
1
IIil FIGURE P1.91
/-93

I. f I
Solution for Problem 1.91: Capillary Tube Viscometer
V,ml t, s T, deg C Q, mils v, m"2/s
9.2 19.8 15.6 OA65 1.12E-06
9.7 15.8 26.3 0.614 8A9E-07
9.2 16.8 21.3 0.548 9.51 E-07
9.1 21.3 12.3 OA27 1.22E-06
9.2 13.1 34.3 0.702 7A2E-07
9A 10.1 50A 0.931 5.60E-07
9.1 8.9 58.1 1.022 5.10E-07
v =KlQ K, m"2 ml/s"2 v (at 15.6 deg C), m"2/s
5.21 E-07 1.12E-06
K =v Q =1.12E-6 m"2/s * OA65 mils =5.21 E-7 m"2 ml/s"2
Problem 1.91
Viscosity, v, vs Temperature, T
1.5E-06 ,......----------------~
I!
1.0E-06 -I------'k-----~- ---------1
From Table B.2
T, deg C v, m"2/s
10 1.31 E-06
20 1.00E-06
30 8.01 E-07
40 6.58E-07
50 5.53E-07
60 4.75E-07
~<
!
i
! • Experimental I
E
5.0E-07 -I------~-----c--------'~~-------l
O.OE+OO -I------,..------,------r-----i
o 20 40
T, deg C
/- 8"1
60 80
[-- From Table B.2: '

.2. I I
2.1 The water level in an open standpipe is 80 ft above the
ground. What is the static pressure at a fire hydrant that is con-
nected to the standpipe and located at ground level? Express
your answer in psi.
1>= ~~ -T-Po
Since.. -the ~.ftt"dPI'pe J~ o,Pen I~ =0 I
2.2. I
l' = ~;;. J.;. ~'!J ) (8() .ft)(/I ItL )
'T"l. ""~/),."
2.2 Blood pressure is usually given as a ratio of the
maximum pressure (systolic pressure) to the minimum
pressure (diastolic pressure). As shown in Video '2.1, such
pressures are commonly measured with a mercury mano-
meter. A typical value for this ratio for a human would be
120170, where the pressures are in mm Hg. (a) What would
these pressures be in pascals? (b) If your car tire was
inflated to 120 mm Hg. would it be sufficient for normal
driving?
t= !'h
(a.) For /.20 ~ HJ : p= (I-3J X /O~ ) (0, /:J,Otm) =/~. O~ Pa.
l' = (;33 X f03;!. XC;, 070,"",) '" r.31 ~J?.
5/~ce a.
is n"t
1'= (/t..O;( I03!.)(f./fSIJXfb-'I ::~~ )
::- ~,~2. fjL
-t'jp/ca / -1-,r-e pve~s uye 105 30-35 -p5(,,~ I Z()ItH",.. tlJ
~u f+; dJ el1 t -fOr /Jpr!114/ drI VI!1j .
~-I

2.:J I
2.3 What pressure, expressed in pascals, will
a skin diver be subjected to at a depth of YOm
in seawater?
t : ~~ -to ~
4f- t1, e 1Ju t'ff., e A =(.) StJ thllt
3 ] N
~::' (10 • Ix. /o.-it.. )(lfom-a) = 40,/-;(. 10 -"L - '1()'1"t/ir~, ~
2.4 The two open tanks shown in Fig. P2.4 have the same
bottom area, A, but different shapes. When the depth, h, of a
liquid in the two tanks is the same, the pressure on the bottom
of the two tanks will be the same in accordance with Eq. 2.7.
However, the weight of the liquid in each of the tanks is dif-
ferent. How do you account for this apparent paradox?
-.- .. .. c _ ..,_
/ /
Area = A Area = A
lv'~'"''''
I; .....,.,
14. I
I,
J?'
•
;=1)y fl!~ 1t/IfK ~/flt tJ7e. I ;1Cj,ned wa //05,; 1h( pressure ()n the
b" ~m i.s tlwt! fD fh~ wei,/I'! ".f -fl,e /JiJl1Q Ii The t-olumn
t/J;'~(.f& ~iJ~tle -Ih~ bf)/Iotl1 as rlt~""11 iJ't 1h( cI(/SH~d Ild~ 111
71tt h11("~, 7h(s iJ the 'Am! w~'i)Jf AS thAI- ;'1' The /:(lI1K.
un"1it 1'ht ~fyAi9l1f 5idts. 711 uS1
-fh~ PY'(S$I(/,'~ 01'1 7h( b~1I-01Yl
t;.f- i71~ ..f-wo +tt.llt~ Ij the 54f'ne. 711~ e:tdcl.f'PIt#/ iv-t11hi
In -the bl1it. WJ7"h th~ Ihc..JIH~R wAils 1:S j(,(pporl:ect blj th~
Il1clJl~1ed w{.(lIs) t!l5 11IusfrA.hef lit 1ht' f,8tO·-e.

.2. ~
2.5 Bourdon gages (see Video V2.2 and Fig. 2.13) are
commonly used to measure pressure. When such a gage is
attached to the closed water tank of Fig. P2. ~ the gage reads
5 psi. What is the absolute air pressure in the tank? Assume
standard atmospheric pressure of 14.7 psi.
f=: r-h + R,
..{) - liz k)6;. -))
II-ale (Ii. 4t) - lair
../h _
/0/ . _
'(111'
12 in.
Bourdon gage
6 in.
• FIGURE P2.5
In. 2-
-.f!tl-
1;1/; - 1'1-3 1~t.'tZ.
l.ti Bathyscaphes are capable of submerging to great depths
in the ocean. What is the pressure at a depth of 5 km, assum-
ing that seawater has a constant specific weight of 10.1 kN/m3?
Express your answer in pascals and psi.
P=-~J. +~
A-f -the :5urhtce 10 =0 ~o !hAt
1>=Ul)·J i. JD
S
;!; )(5" )(./0
3
/WI) =SO. S
A/seJ)
f = (so. 5
= 50.S" MP~
7 320 p~iJ
Air

~. 7 j
2.7 For the great depths that may be en-
countered in the ocean the compressibility of sea-
water may become an important consideration.
(a) Assume that the bulk modulus for seawater
is constant and derive a relationship between
pressure and depth which takes into account the
change in fluid density with depth. (b) Make use
(aJ
Thus)
So th..t
~::: -~ =-('3
~ :: - 1- dr
6
/e-:'df =
~
.so -inti/:
1>-
of part (a) to determine the pressure at a depth
of 6 km assuming seawater has a bulk modulus
of 2.3 x 109 Pa, and a density of 1030 kg/m3 at
the surface. Compare this result with that ob-
tained by assuming a constant density of 1030
kg/m3.
( £g. 2. If)
(I)
be hre..
(Eg. 1.1'3 J
a.t 1'=0

~. 7 I (C~n'i)
(b) Frt91?? f~rt ~) )
-p = - Ev Jh, (/ - t. ::1. )
So -thAt at A:: ,~~
~= - (:J..3 x IO".!tr .-m2.
)in [I - (;'03X;()3};3)(r'd'/~)(b)(llhtt)J
; . 3 ;<. I tJ '1 ...!:!..
""' :r.
= &'1. Lf M P'I-
(c) h,.,. ~I1S tant dens N-!J
I
p= (/~ = 13- ~ =(I. ~3 ~/D3~)(tj,f/ ~)("~I/~)
- fo6. t, ;'1~
2.8 Blood pressure is commonly measured with a cuff
placed around the arm, with the cuff pressure (which is a mea-
sure of the arterial blood pressure) indicated with a mercury
manometer (see Video 2.1). A typical value for the maximum
value of blood pressure (systolic pressure) is 120 mm Hg. Why
wouldn't it be simpler, and cheaper; to use water in the manome-
ter rather than mercury? Explain and support your answer with
the necessary calculations.
-p = ~It
J; V" /2, 0 /YK 1M I-J, : t -= >fh
:::: (/33 X Jb),!~ )(CJ. JZ 0 Mt )
- //,..() 4eP...
?; (> /, .till;' 1J'/lj pY.t'$~/.I~ f.IJ i'h1 A. wa..-I:ey Col""., 11
3 tV
I', () X /b ;;;":a.
- J, It. 3 tWl
Q (J 3 IV
" ,,()Xlb ;;;;;
Thus) il tv~bY W(I/f!. «Sid In -h1< 'Ttlan()l1?f..k..t- -th ~
V"etlAit-ftl CO/U"''' htI9h/:.s w()u/d be t()tJ hl~h anti.
1m'Y~C,-ht41. N'o.
,,2.-5'

2, Ii Two hemispherical shells are bolted together as shown
in Fig. P2.9 . The resulting spherical container, which weighs
400 lb, is filled with mercury and supported by a cable as shown.
The container is vented at the top. If eight bolts are symmetri-
cally located around the circumference, what is the vertical
force that each bolt must carry?
'1 "... hwce In ~ne h"JI:.
p.-y frf-,surt. at mld-p}t/he
A,-.; CI r-fA ai rnl d-p/~ne..
AI AI we.lgh i li In-ert.ur!1 1Y1 h()#-fPfI1 hoiI
I+! 0';:' She II
~ "'" w-eijJJt af j,tlt/Dnt hlJJf ()f sheJ/
/;1' ett"'jibri,,m/
L ~e"'hctl/ = 0
Thus)
gIX f Ii -r W#-j 1- Ws
Cable
Sphere diameter = 3 ft
• FIGURE p2.Cf
- O"u/f)(fD~ + '04*)(flJ~ T ~ ('too /J,)
=(8lf'l !,X¥")(f)(3ft) ~ ~lf7~,)(~)(f)(Htf+ J,oo Ib
ferlO 1.6

)./0 I
2.10 Develop an expression for the pressure
variation in a liquid in which the specific weight
increases with depth, h, as y = Kh + Yo, where
K is a constant and Yo is the specific weight at the
free surface.
:!1-_vdr - Q
( Eg. .2.4-)
Le t -R. :: 1:0- Z
So -(heel: d~ :-cli:
tind
Qh&i
v

d.;1 ~ I
2.11.* In a certain liquid at rest, measurements
of the specific weight at various depths show the
following variation:
h (ft) Y (lb/ft3)
o 70
10 76
20 84
30 91
40 97
50 102
where
CQI) ht. ;".teyrIl1e4 Ilvmprlcltl/y
I
(Lont)
60 107
70 110
80 112
90 114
100 115
The depth, h = 0, corresponds to a free surface
at atmospheric pressure. Determine, through nu-
merical integration of Eq. 2.4, the corresponding
variation in pressure, and show the results on a
plot of pressure (in psf) versus depth (in feet).
D
/ 1 1 / / / / / 1 ' 1 '
(I)
depth -l,'. £!l.Illtl£J" (1)
t/.J/ns 171ft ~//~/I1J 1)f"oJrAIJ1.
(!Vof~: The. n"m,y/cil I /;'.ft"Y4;'1()~ elfl1 "Iso h< (1c(t!)rnp/lsju~4
1hroufh f'e;fq.J.e~ ~.se. of '/1Ie pY~9J1'''m T.R,lfPb"ro/).

~.J/ jIk
100 cls
110
120
print "*************************************************"
print "** This program integrates Eq. 2.4, numerically **"
130 print "** using the trapezoidal rule to obtain the **"
14,0 print B** pressure at different depths **"
150
160
print If*************************************************"
print
170
180
190
200
210
220
230
24,0
250
260
270
280
290
300
310
dim p(11),gamma(11)
n=11
dh=10
p(1)=0
for i=l to n
read gamma(il
next i
data 70,76,84,,91,97,102,107,110,112,114,,115
for i=2 to n
s=(gamma(1)+gamma(i) )/2
im1=i-l
for j=2 to im1
s=s+gamma(j)
ne)':t j
p(i)=dh*s
320 next i
33C '
34,0 'Print the results
350 print
360 print If h (ft) Pressure (psf)"
370 for i=l to n
380 print using "###.# #####.#"; (i-1l*dh,p(i)
390 next i
Th ~ fa 1,1( j~fed f'esu/.J-$
~rye.!p()ncl/".7 P/()t of
be/"wI 1/011, tJlI1h
liS. dfpTh.
~***********************************************~
Lt This program integrates Eq. 2.4, numerically **
** using the trapezoidal rule to obtain the **
** pressure at different depths **
**~c *** ****' *** *********' *********'**'** >I: * >I: ******'*'*'*'**
*10·
1.2
1.-1 (ft) Pressure (psf)
0.0 0.0
10.0 730.0
1. o·
20.0 1530.0
:30.0 2405.0 ~ O.B
40.0 334,5.0
.!;
50.0 4,34,0.0
Co
• O. B
60.0 5385.0 m
Co.
:J
70.0 64,70.0 •a
80.0 7580.0 ~ 0.4
Q.
90.0 8710.0
100.0 9855.0 0.2
0.0
the..
0 20 40 60
Oapth, h (ft)
60 100

II
~.)2
"2. .12. The basic elements of a hydraulic press are shown in
Fig. P2.12. The plunger has an area of 1 in.2
, and a force, FI ,
can be applied to the plunger through a lever mechanism having
a mechanical advantage of 8 to 1. If the large piston has an area
of 150 in?, what load, F2, can be raised by a force of 30 lb
applied to the lever'? Neglect the hydrostatic pressure variation.
Plunger
/Hydraulic fluid ""-F]
If. j..(')~(E. o/. ,3.o)b o.ff>Jj~cI -Ir:, the. Jevn- V'esl-{/f.s P1
t:A. P/un7fr" ~jI'ce,) F;) ~f Fj = (?)(:6.o) =2lfo II:; .
'SIn ce. F; :.. tAl QI1t/ P-;...= plr2 whe~e ? Is "The..
,Pt"'e SS{,{ r'e COtel 1// ('01 rI.42., co·~ 1i1e t{V'e4S ()f -the..
pJuh 1f r 4 ncJ PJ5 foJ?) i'e~fec.tJt/eI!1. SInce... ? I S elMS ~n t
-thi"'ou 9htJu t 7he chqm be-yo )
{:,
:: 1=2-
-1+/ A2..
:50 1h,,-/;
F: = 6- t=; (/~~ I~.~) (2'10 Jb)-:2- ,401 / /11.
2.13 A 0.3-m-diameter pipe is connected to a 0.02-m-
diameter pipe and both are rigidly held in place. Both pipes are
horizontal with pistons at each end. If the space between the
pistons is filled with water, what force will have to be applied
to the larger piston to balance a force of 80 N applied to the
smaller piston? Neglect friction.
~ ~ fA-I
!=s. = f A2
ThuS)
--
~-IO
.....
3Gj 000 1.1:,
i
I.~

2. IS- I
2.15 What would be the barometric pressure
reading, in mm Hg, at an elevation of 4 km in
the U.S. standard atmosphere? (Refer to Table
C.2 in Appendix C.)
Itt C( n eleva.-tJt)1'/ 01 If J.m J P= ~. /~ (p X/0 S-- :!2-
Ta b /e C.:;' l'n /1.Ppen dI xC) . S I H (. e.
2. J(, I
1:> ==- ~~
If AI
..:P... =- ~. J" ~ :I. 'D -::;-2-
d' J3~;(,/~3..!:L
2.1 ~ An absolute pressure of 7 psia corre-
sponds to what gage pressure for standard at-
mospheric pressure of 14.7 psia?
AH'3
( Ir~m
- 71'.)(.«- )Lf.7/:lf.: ::: -7.7 PSt:
2-11

2./7*
*2.17 A Bourdon gage (see Fig. 2.13 and Video V2.2) is
often used to measure pressure. One way to calibrate this type
of gage is to use the arangement shown in Fig. P2.17a. The
container is filled with a liquid and a weight, W, placed on one
side with the gage on the other side. The weight acting on the
liquid through a O.4-in.-diameter opening creates a pressure that
is transmitted to the gage. This arrangement, with a series of
weights, can be used to determine what a change in the dial
movement, (J, in Fig. P2.17b, corresponds to in terms of a
change in pressure. For a particular gage, some data are given
below. Based on a plot of these data, determine the relationship
between (J and the pressure, p, where p is measured in psi?
W (lb) 0 1.04 2.00 3.23 4.05 5.24 6.31
(J (deg.) 0 20 40 60 80 100 120
Bourdon Gage 'IV
(b) (a)
• FIGURE P2.17
7rt, ~(J") (I)
So
Theta, deg. W,lb
0 0.00
20 1.04
40 2.00
60 3.23
80 4.05
100 5.24
120 6.31
( IN hfY~ P Jj I;" p-,£)
=o.()SZ2lii
fi,-."m R"'I' (j)
1> (~,;)
7.Cfb
-p (f'~t.) .:- o.'II, e
W =0.0522 e8.00
I,Q 6.00
~ 4.00
'Q)
~ 2.00
0.00
0 50 100
Theta, degrees
:L-1'2.
]
150

2.18 For an atmospheric pressure of 101 kPa
(abs) determine the heights of the fluid columns
in barometers containing one of the following liq-
uids: (a) mercury, (b) water. and (c) ethyl alco-
hol. Calculate the heights including the effect of
vapor pressure, and compare the results with
those obtained neglecting vapor pressure. Do
these results support the widespread use of mer-
cury for barometers? Why?
(1h C./luhn1 va.f~Y fYt'St;I/Ye.)
r(A.brn) : yJ.. + fJ'I/"
w heJl'e. ? "V /,!A ftJY" 't'~.ssuY'e.
f =
3 /'I -I N
( WI thou f VttfDY' f Y~5.sUY"e )
f (A. 6;J/'1) .. d'h
'3 tv
(It) F~". mtY'cur'j:
/OIx.JtJ - -I.(PXJD-l.."",,:z. t111
133)( It) 3..J:L
/I't?~
101 )(.ID ;;~
J"3 3)(. 10"! .J!..
(M.3
= 0,7511W1
3 N J 3 }J
It
/o/x./b - l . - 1.77xIt; .Ji. /01 'J. I D -;;-~
t.b) (;r /.VA.-ter: = ;?H AM 1.
~ =
'i. 30 /. )0 3 J:!..,.,.,3
1. fo X/0 3
J:L
/'tIA3
= I(). J /)'Y1
= /0,3 rm
(() Por erf" J 3# 311
ttJ/;<.ID -"""i -5.'1J<'/~ - .. /01 )I.. Jb.J.!:!...
~ .{a /Coho/ : ;:
I/'H "."..,
~
1'171"1..
7. 7 If- .x. I 0 3 .!:L 7. 71f X ID3 31..311'H3
- I ;J.. "3 1??'1 - /3,OI1?1-
For mereu r!) btlY'tlmeiers -fhe effect. 01 IItLlt/Y PWSflI1"e
is nf'r//qi bfe / tlnd t:h~ v€$u/red he/9hi 0/- The n1er(.11'!/
Co/urn n /S renstPnt:lb/e.
;;. -ra
Ilt1

:I. I q I 2. II:! Aneroid barometers can be used to mea-
sure changes in altitude. Ifa barometer reads 30.1
in. Hg at one elevation, what has been the change
in altitude in meters when the barometer reading
is 28.3 in. Hg? Assume a standard atmosphere,
and that Eq. 2.12 is applicable over the range of
altitudes of interest.
..5u10 Irac. t £1. (2.) /re;m £~. (J) -k ~bta (n)
FDr
U//th
-fhen
l~-r( = ~ UfJ1- (rt)r]
Tq,=2.89K) (5 = f!). 00 "So ~ ) -P~
I:: tt, '!/ ~
"
R. = 2?7 .::!- ) Ct I'J Iisa.. Ie;. If(
&. =
(:'8'7 *fl< ) (~. 00#;57) ~ )
d
-p,:: 011-,# J,., =
f:;. =!~ ~ ::
.?".pm E'fJ3)
9.tl ~
52.
==
( f'g. :2. J:J. )
/(:;) Je Pc. J
:: f!J. I q 0
~ - 1: ::
).f~ /( [( lol ~p~ )o.Ii~ ( qo.i. .v~y,/qOJ
~ J ~ O~~5tJ !fn I() I .t PL 101 -It.. Pa.
- 54-3 rm-
(Il
(2 )
[] )

2.20
(a)
2.20 Pikes Peak near Denver, Colorado has
an elevation of 14,110 ft. (a) Determine the pres-
sure at this elevation, based on Eq. 2.12. (b) If
the air is assumed to have a constant specific
weight of 0.07647 Ib/ft3, what would the pressure
be at this altitude? (c) If the_~ir is assumed to
have a constant temperature of 59 OF what would
the pressure be at this elevation? For all three
cases assume standard atmospheric conditions at
sea level (see Table 2.1).
Ihen
(abs)--
L/; ) -p -= ~ - tlJ,.
J.I/~. z ~l. - (t). 67blf7 f;3)(;~ /lo.f.t)
- IO/.fo J:!. (o.bsJ
ft1-
[(. )

2.21 2.21 Equation 2.12 provides the relationship
. between pressure and elevation in the atmo-
sphere for those regions in which the temperature
varies linearly with elevation. Derive this equa-
tion and verify the value of the pressure given in
r~bf.lec~ in Appendix C fOfr an;~evation of 5
d fJ _ ~ dl:
T - - R T
~ -l-J
Le t 1;~t .ft,,,. -},.::.O ) {Nf j"". 7:2 ~ £ I (;(11 d T= To.. -rs7: .
Thus) ,p
Iti: 1:, di-
=- -
/? ?;., "7 i
~
t:PY"
L
and -h. tiJ11
c.>
1::
£. ~ [- ~ 1m (iA-~~i- =~
~
::' .1 h (J _ &-r)
Rr4 TA
/()5a rithrn 0/ h~tn $1 atd D-t e3L( a.:tIO·H
~
-P == ~ (I - ~)~
s. 'f0 X I () If !!.1
mr
~ [e. (To.-(3<) -hTJ
':JleLdu
(~t.2.,2.)
_L-~~.-.-.--.=-.=.--.~---_--.J
1.-I<&.

2.22
2.2'2 As shown in Fig. 2.6 for the U.S. stan-
dard atmosphere, the troposphere extends to an
altitude of 11 km where the pressure is 22.6 kPa
(abs). In the next layer, called the stratosphere,
the temperature remains constant at - 56.5 0c.
Determine the pressure and density in this layer
at an altitude of 15 km. Assume g = 9.77 m/s2
in your calculations. Compare your results with
those given in Table C.2 in Appendix C.
h =22. ~ ..lPa,1i )
-r;:: -5 '.5 ·C -r ;L 73. 15" = :; I (p. " 5' k .
- / ~ I I -k Pa.
_ ~77 70,..)(IS-x If/1M - 11;(. ,//WI )J
[ (,1S''l tfk )(:l /1.. /, 5"/< )
3 IV
1")./.x.JtJ ::;;'l..
Lb/e. C.2 J~ A-ppeJ!1&//x. C) h : J;1, II ~ pC(.rJ;.
L-/7
and

:<.23 +-1
2.23* Under normal conditions the tempera-
ture of the atmosphere decreases with increasing
elevation. In some situations, however, a tem-
perature inversion may exist so that the air tem-
perature increases with elevation. A series of
temperature probes on a mountain give the ele-
vation-temperature data shown in the table be-
low. If the barometric pressure at the base of the
mountain is 12.1 psia, determine by means of
numerical integration the pressure at the top of
the mountain.
Elevation (ft)
5000
5500
6000
6400
7100
7400
8200
8600
9200
9900
Temperature CF)
50.1 (base)
55.2
60.3
62.6
67.0
68.4
70.0
69.5
68.0
67.1 (top)
W; tb -the. .frmftral",Yt. d~b... '9W(11 fhe. ;tJ.J.e1rl/ 'H B~ . 2. '1 Ca i1
be eV/J/u4Iell J111h1fnc"l1~ tls/llj rteAPE"iol.
*****************************************************
** This program performs numerical integratior~ **
** over a set of points using the Trapezoidal Rule **
*****************************************************
Enter data points (X , Y) /Jeff :
'? 5000,1.962E-3
7 5500,l.942E-3
'? 6000,l.923E-3
? 6y,OO,l.915E-3
'? 7100.1.899E-3
? 7{,J.OO.l.894E-3
? 8200.1. 888E-3
? 8600,l.890E-3
'? 9200,l.895E-3
7 9900,1.898E-3
The approximate value of the integral is: +9.34S2E+00
--
SODoft
So fhllt ( w/th j :: 32.2. 4/52. tlh4 R= 17/1, It., Jj,/sJu~. 42)<. )
111 ft ~ - (~Z. 2 tt.)(r. 3 ~ t:) - - O. 175"i.f. (I)
11 17/ to ft:·/bI sIll1--;< ( COl? t. )
~-Ig

2.231-1 ( COI71; )
:t.f Ie//()IUS -h--PIJ1 1;'1,(/) /() J1H 'A = /2.J PoJ L'a. 7).,lIf
- tJ,17S'1f.
~ = (1'2. I ,s/a) e == 10, Z PSI.'a.

- Closed valve
2.2 Y A U-tube manometer is connected to a closed tank
containing air and water as shown in Fig. P2.2~. At the closed
end of the manometer the air pressure is 16 psia. Determine the
reading on the pressure gage for a differential reading of 4 ft
on the manometer. Express your answer in psi (gage). Assume
standard atmospheric pressure, and neglect the weight of the air
columns in the manometer.
I-- Air pressure = 16 psia
~
Water
Gage fluid
(y =90 Ibfft3
)
• FIG U REP 2 •2f-
I~ Ib .!.!!:. 2..)
/i(p ~ 1.. - /'t. 7 -:- 1.. '0/fLf rJ-l-
~ I~. 'y,. J f -n..
+ ( (p Z. ¥ -!ft:3 )( ~ f-t)
T2 ft
,1
t
Pressure
gage
( I 16)( /It 2. ) - '/-. b7 DSl.
(J) 72.. -Ft."- Ju..u • 2. r.TT'I?,
--

Z.2.5" J
I 2.2.5' A closed cylindrical tank filled with water has a hem-
ispherical dome and is connected to an inverted piping system
as shown in Fig. P2.2S. The liquid in the top part of the piping
system has a specific gravity of 0.8, and the remaining parts of
i the system are filled with water. If the pressure gage reading at
A is 60 kPa, determine: (a) the pressure in pipe B, and (b) the
pressure head, in millimeters of mercury, at the top ofthe dome
(point C).
Hemispherical dome
c /
Water
~.
I /SG=O.8
4m
+3m
-1-2m
-
* RJ
I
!
Water
(a.) 1;; -I- [S G,)( ~:l.()) [3 ~) + ~o (L/W1) :: 1>B
(b )
1>8 = ("o.J.p.. + (O.a)(r.II(//!.)(3",,)+ (r.c~~1D3;;'.)(z1'M)
= /03 ~Pa,
~ :: -p - ~ (3hY1)it 2,0
- k0 .kPa. - (9. ~DXID
3
.;~ ) (:! '"" )-
5f). ~
3 N
- )( JD -- 1I?1'10
..pc: 3 N
-It Bo.t. xH -;;'l-
= - :::
tI,230m!
!'jJ -
/'33)( J0 3
.!:Lj 11H'3
= t>. 230 ml (la 3
fW1,!:, ) = 2 3 0 1)1"1 /WI
ht'
2-2./

2.2.("
Manometer fluid
2.26 For the stationary fluid shown in Fig. P2.26, the pres-
sure at point B is 20 kPa greater than at point A. Determine the
specific weight of the manometer fluid.
Density = 1500 kglm3
• FIGURE P2.26
Le-/; r-",.,:: spe,l!,,, w-e/;ht t>11Y1~I'I()metel' fl"t"
dA = (SG)( ~ ) (,.) = (1.2) (IODD '*~)('l.81!f.. )1.Jr. D f!' 'I-c. ..".
:: II} i()()..!!
hr' 3
(fa :: ~ 3- = (iSt>o #!. )(Cj .81 ;,.) == JLfj 7()() ! 3
-1=>13 - tA- = - ~ (2hYI);- ~ (2m1) -t rB (~IW)
t.(n<ib J !!.~ :: - (I') 8DI;;.,)('2.JIK) -+ 'Cr... (2,..) -t (I~,100;'3)(1....)
I'm
IV
'O~ = 7/ J DO ;;3
2. -2<.

Z. 2.7 I
o1'3"= 2 psi
2.7..7 A U-tube mercury manometer is connected to a closed
pressurized tank as illustrated in Fig. P2.27. If the air pressure
is 2 psi, determine the differential reading, h. The specific
weight of the air is negligible.
Air
Water
t,~ f ~ 1 -
'1
~ (it + If (t) ::
#,.0
fa/r
~ =
'0JI.% 0 (If!-/J =('2.¥f4J )(If ./'.;)
(/5. I,)({.2.lfAJ) - tz'If/c.~.J - ~:LO
2.-2. 3
T
h
~
Mercury (SG =13.6) - '------'
- tJ. 517 Ii
-'-,I
2ft
t2ft
-t--I
2ft
~

2.28
2.28 A suction cup is used to support a plate of weight W
as shown in Fig. P2.28. For the conditions shown, determine
W.
.. FIGURE P2.28
FOr etu; I; hrlUM 0+ f.o'rces ~I plo.;h-e.
~ ~ ~,A. £1 }
WheY'fo A, 1$ ar~A of. c.up aVltl P, (s &( ne9a...4:ve pr-tssur"e.
FY()(n manomet-er .QZ ULJ·/O·v :
~ - ~z.D (1. ~ H) t (sG) ('0~t.0 ) ( (j. ~ .f-i:) .:: 0
t, ~ oil-tO [I. ~ ft - (1I)(0.4.ft)]
" btlt ~Juft - (E){Mft)]
-= - Cfcr ~ ~~
Tht.ts/ /-Yom t=~. c. J}
OW " (qu :")(Tt) ( O. 5'.f.r) 2. =: 1g. ~ IJ,
2-2..,-/

2.29 A piston having a cross-sectional area of 3 ft2 and neg-
ligible weight is located in a cylinder containing oil (SG = 0.9)
as shown in Fig. P2.29. The cylinder is connected to a pres-
surized tank containing water and oil. A force, P, holds the pis-
ton in place. (a) Determine the required value of the force, P. T
(b) Determine the pressure head, expressed in feet of water, act- 2 ft
ing on the tank bottom. +2ft
t3 ft
L
Tank bottom
For .eblA~IL},rlilf'"
• FIGURE P2.29
( a...)
-p,t+=p (J )
I I . I _ 1
...C)nev~ "P1~.5 presslAV'~ ac..-tlnq f>~ p/~ror1. l-t'mat1t!1M-eber-
(b ")
e$l.l aJ-LtP J1 &J i 1/-e.J.
1)1 T o~i I (Sft) - ~l' J ( 2 H.):: ~I'r
.so iYzl4..t
P, = 11,r - ~~ J (s+1:) t ~; J (2 ft-)
" (s ~.~)(I¥*~:)- (0. q) (~1.~~J)(5-H) T (0. q)(~2.~~J(1-tt)
5~~ ~.-
Th us fyf)m E'C£. () )
) b ( 2.) I l, to 0 1P=(5"5"2 k~) 3 fi -
~bo&~ :- "RL'r -1- O~2.0 (3 +t) -1- ~6; J ('t-fi)
"(5' ~:-)(i~'t~:)·d":2.4-~3 ) (3 H)+~.'1)~1.~~~~ft)
Ib
- J 13D J;:a- Ib
I 3D "h:a. :::
2.-25'

),31
2.31 The mercury manometer of Fig. P2.3
indicates a differential reading of 0.30 m when
the pressure in pipe A is 50 mm Hg vacuum.
Determine the pressure in pipe B.
O. m
Water
~r
0.30 m
Mercury L .
FIGURE P2.3
-fa r 'O.il (6.1>.... T O. 3D.. ) - tJf3 (0.3",,) - !u.o (0.1."",) : ~
Wh €Y'l. -R:: - 6'JI. ( IJ, fJ3() hrI )
,4 ~
Oil
1'{:, = - r~ (~, P30 ... ) - ~i I ((), Vfi""') +!U; (~, 3",,) ... (fu,/D./!;"",;
=- - (/;3 t )(~,~30"") - (g9b~)(t1~~~) +(/~3~)~~MI) +
(r.80 ~y6.}~I&1)

2,32.
2.•U For the inclined-tube manometer of Fig. P2.32 the
pressure in pipe A is 0.6 psi. The fluid in both pipes A and B
is water, and the gage fluid in the manometer has a specific
gravity of 2.6. What is the pressure in pipe B corresponding to
the differential reading shown? -.-
3 in.
+
FIGURE P2.32
-t !#,20 (IiH:) - ~f ~ ft:) Sih3tJ° - cr~t> (J~ ft) :: 18
'0;1 /oJ the ~fec/f,c /Nfl"}, t ~ the. ;tqJe .flu/d)
P,e :: Pj1. - 4'#1- ({ ft) .silt :30 ~
Water
" (0.6 /t,..)(N't!ji )-(j.b)({,2/f~,)a ft)(O.5-) :32.3
:: 32.3 Jbl.yt3-/I'I1I- til.·la· .= 0.22'-f ?s,:
2. 3~ Compartments A and B of the tank shown in Fig.
P2.3~ are closed and filled with air and a liquid with a specific
gravIty e~ual to 0.6. Determine the manometer reading, h, if the
ba:ometnc pressure is 14.7 psia and the pressure gage reads 0.5
pSI. The effect of the weight of the air is negligible.
A
Air
Water
Liquid
B (SG = 0.6)
AI> - 'rllJh) -+ f.u (h) -(- 1;.; (~.J.ft) :::0
~ = ~/r f- ~t (tJ./ .pi)
~l 0 - ~l /
IOpen
----.L
0.1 ft
-t
Mercury (SG = 13.6)
= ( o. 5 ~.). )(Ill-If !ft.:)-t {; ~.,)(t,1.1f- ~3) (~.J ft)
(,2. lf .!Jt3 - (().'){~Z·'f!lt3)
:: ~. 2i' it
2.-2.7

2.34-
2.31} Small differences in gas pressures are
commonly measured with a micromanometer of
the type illustrated in Fig. P2.31f. This device con-
sists of two large reservoirs each having a cross-
sectional area, A" which are filled with a liquid
having a specific weight, YI, and connected by a
V-tube of cross-sectional area, A" containing a
liquid of specific weight, Y2' When a differential
gas pressure, PI - P2' is applied a differential
reading, h, develops. It is desired to have this
reading sufficiently large (so that it can be easily
read) for small pressure differentials. Determine
the relationship between h and PI - P2 when the
area ratio A,IAr is small, and show that the dif-
ferential reading, h, can be magnified by making
the difference in specific weights, Y2 - i'1, small.
Assume that initially (withpI =P2) the fluid levels
in the two reservoirs are equal.
I;'ih~/
l-e lit! I
J~;'h4/ I,ve/
/Dr grlit' 1-/,11'd
1~
__II
,
i.
When ei dl'fffnn/;Ja / pYt'$5Ure) It-Ii) I~ a.pp)/eH
Vf5ervpj,. drt),P' b~ A c/J~kllrce.J Alll tU,(/ ritjJri
111~ /7) til') " In et: fI- e$"Il t.i~)n be~4'me.s
Th
__________~!_ 12
FIGURE P2.3Lt-
_ _ _ J
'1.
i
~l..
_1_1
- -- - r. At
we ~$$tI',H. tIt"t; lell~1 111
l-eve/ r-is,s b~ Ah. Thus,
~ ~ ~ (1, + -It - i Ja.) - d'2 ~ - ~, (-R., T ~), ) = +2
11 - f2 :: ~ J. - d'/ ..£. -t- ~ (2 A-I.. ) OJ
S/nee -the /, *f) J:dJ /"
LJl A ::- -A 41r :2. 1:
1he rntln6)meter /:Ire /ncomfY't'ssi /')e;>
o r- :z J h _ At
T - A)-
4 15 $/"nIl)J /henAi-· :l~h«~ / tift term in EZ.(J)
be ne,/Jec:'ted, Thus)
-f:: - +2 :: (~ - " )l
Cind /tlr1e vt}ltI'RS pi h
d/f-.feY'fnt/a1s if 0;1,. - 6',
.
IJ

2.3S
2.J5 The cyclindrical tank with hemispherical ends shown
in Fig. P2.35 contains a volatile liquid and its vapor. The liq-
uid density is 800 kg/m3, and its vapor density is negligible.
The pressure in the vapor is 120 kPa Cabs), and the atmospheric
pressure is 101 kPa Cabs). Determine: (a) the gage pressure read-
ing on the pressure gage; and (b) the height, h. of the mercury
manometer.
t1 m
t-1m
l
• FIGURE P2.35
Open
T
h
1
Mercury
(a) Le-{; dL =~p. who 6 f LiUJ'~ =(goo ~3 )(~.8J;~): 1850;;3
QVI~
i? (,119e.)::: IZO .kPo- C~bs) - J0 / ,kPet (ttb$): I q ~ Po.
va-poY'
Thus)
~A5~ ::
=
-FVAPdY' + ~ (ll'M)
~'l )(b3;;'~ 1" (ll35"D~~ )(fIM~
2 ~, ~ -k p~
{=O.ZOL/tM

2.3' J
2.3'- Determine the elevation difference, Ah, between the
water levels in the two open tanks shown in Fig. P2.36.
.2.37 I
2:31 Water, oil, and salt water fill a tube as shown in Fig.
P2.37. Determine the pressure at point I (inside the closed tube).
" /
Oil density
= 1.20 slugs/ft3__
iI-- 2-in. diameter
-lOin. diameter
~t-
3 'ft
1.1'--- Salt water,
'-- -~f------'--- ~ (l) SG = 1.20
2ft
[~:~'~_C _Lei}
(SG)$4(+ ~U2.0 (3Ii) -t ~"I (3Ii) i' ~2.0 (2..rt) =0
W4+Vr
-p-I
~ = (/.20 W. 2.'f ~. )(3,f1:) - (;. 20 !.f/:)(az.2';."{JIt)- ('~lfkJ(z.ft)
= _ I ~ I iJ:,
. .f·e·
2-30

2.38 An air-filled, hemispherical shell is at-
tached to the ocean floor at a depth of 10 m as,
shown in Fig. P2.31. A mercury barometer lo-
cated inside the shell reads 765 mm Hg, and a
mercury U-tube manometer designed to give the
outside water pressure indicates a differential
reading of 735 mm Hg as illustrated. Based on
these data what is the atmospheric pressure at the
ocean surface?
I735 mm
Ocean surface
FIGURE P2.38
Let: Pa.. "'" &lb~c/ute ~/j,. pr"esS(Jre Ins/de. shell = 1fuJ. (~, 7(P5'1'W'l )
/;'fMl -v Sit( rlae. e at:1Y/1 f)sphent. fY'{,sSU re
~w- 11 Sft't:.iIi'c. uN, ;'11 t 0 f Selttc.J4.:/;eyo
f~~
.Jo 1h4t
~ihn= t - (f ( / (), 'b ~) -t ~ ( (), 7" 5"~ )
Sw- d
-- (N3 :f!.)ftJ.7'b-_) - (;I!J.I ~)(;t>. 3b"") T ~~3 ~)(".13f...)
-;;'-31

2.3'7* Both ends of the U-tube mercury ma-
nometer of Fig. P2.3Cf are initially open to the
atmosphere and under standard atmospheric
pressure. When the valve at the top of the right
leg is open the level of mercury below the valve
is hi' After the valve is closed, air pressure is
applied to the left leg. Determine the relationship
between the differential reading on the manom-
eter and the applied gage pressure, pg. Show on
a plot how the differential reading varies with Pg
for hi = 25, 50, 75, and 100 mm over the range
o::s Pg ::s 300 kPa. Assume that the temperature
of the trapped air remains constant.
114/(/( C /~seeJ. e'lttr4 a. pyeSSlJre I fj I
-A _ 0. .tJI..:' .h
'1 #1 rtt.
/j~:
FIGURE P2.3~
.
an,! -t. 4f'e :JItJe fyeSSUYfs. /;1' IS() therm4 J
~ tra.pped aIr
.:::E..:: ~t1.5 .fqII i
f .
~11.5ft1l1t 4' r mas~
(I)
tJheY'e -V- Is (f/r II~/"me, / l' ,:S 4bS~/l(k /vt'SSI//I'el (J;ul " lIJ1d -r
Y'elfr 10 in/hal Clnd ft~iAI sfttff.5/ r-f~ft'c.hllelfj. TIHtS)
t if =-a:/:1I'1 t'
(~ -r f,.t», )~ ( 2)
for 41J!' IrltlPul /11 r;g~i leJ J
if.:" ~. (Ar~4. D/ I-~b< )
/. t.
£tj,(Zj CQI1 btl 1t.I,.;ffen 4.5
t :: titm [
,£~.
- I]~. _ An
L. ::z..
( 3)
B$ ./3) /nflJ Es. (J)
./)h =: -t[p; t- -/;./-""
l-32

eX.fv~s~ed J ~ 1]" ( f"JI'YM
-?; ;- ~f"",) IJh + 2.1 t.e
<1Hg. a-u;
Qt)d -!he r()o.f.s cf 1hJ.S g"adr~II'c. e~IJ({i:";1( ave.
E ~Ii' tid) n (it) Cflll b-e.
(fj-R.)).- (t1.J.~+
(-R. i-
t..
~ -r ~i"") + - r,---(~-L-'-.,.-f;.-#-of--~-a.-f~-)-2.--2-18-i-,; (s )
,.1 ~a V ~ ~3 b'Ml~
J; eV~)1(1Jk IJ~ -the. ne.14-J.1 v< SJ1H IS t1J ~J 5/~'e. .A h=() h.,. f1 =o.
,4- pr"Jl'tI!Y1 /tJr ~mluIt", iJh (}s ~ htn't/~;' of 1 f-y 114 'l'/oU..!
J.i !"/Iows (WItH ;1i,.."" /O/.,QP" qn' ~= 133:.B.Nk3 )'
100 cis
110 print
120 print
130 print
140 print
145 print
150 print
160 print
"***************************************************"
"** This program calculates the lower root of a **"
"** quadratic equation to give Dh (in m) for a **"
"** range of gage pressure, Pg (in kPa) , and [or **"
"** a set of different initial heights, hi (in m) **"
"***************************************************11
162 dim dh(S)
164 patm=101
166 ghg=133
170 print
Dh(hi=0.100)
" Pg Dh(hi=O.OOO)
180 for pg=O to 300 step 30
190 for i=O to S
195 hi=(i-1)*0.025
200 a=hi+(pg+patm)/(2*ghg)
210 dh(i):::a-(a~2-2*pg*hi/ghg)A.5
220 next i
Dh(hi=0.025)
230 print using 11####.# ###.##### ###.#####
## # , ### It #It ; pg , dh ( 1 ) ,dh ( 2 ) ,dh ( 3 ) ,dh ( 4 ) , dh ( 5 )
240 next pg
Dh(hi=0.050) Dh(hi=0.07S)

.1
I
(C.()//t )
~**************************************************
** This program calculates the lower root of a **
** quadratic equation to give Dh (in m) for a **
** range of gage pressure, Pg (in kPa). and for **
** a set of different initial heights, hi (in m) **
***************************************************
Pg Dhlhi=O.OOO) Dhlhi=0.025) Dhlhi=0.050) Dhlhi=0.075) Dh(hi=0.100)
0.0
30.0
60.0
90.0
120.0
150.0
180.0
210.0
2'*0.0
270.0
300.0
0.00000 0.00000 -0.00000 0.00000
0.00000 0.01101 0.02120 0.0306'*
0.00000 0.01816 0.03538 0.05170
0.00000 0.02313 0.0,*539 0.06681
0.00000 0.02678 0.05280 0.07807
0.00000 0.02956 0.058,*7 0.08673
0.00000 0.03175 0.06295 0.09359
0.00000 0.03353 0.06657 0.09913
0.00000 0.03'*99 0.06956 0.10370
0.00000 0.03621 0.07205 0.10753
0.00000 0.03725 0.07'*18 0.11078
0.20p-----------------~-----------------------------------,
- 0 - - hi-O.OOO m
- 0 - - hl-0.025 m
- b . - - hl-0.050 m
- 0 - - hl-0.075 m
-+ -- hl-0.l00 m
0.15~----------------~
~
~
0.10
~ 0.05
III
J:
0.00000
0.03938
0.06716
0.08739
0.10258
0.11'*33
0.12365
0.13119
0.137'*1
0.1,*262
0.1,*70'*
O.OO~~~~---e----~----~--~~--~~---&----~----e----t
-0.05L-______~________~__~____________~________________~
o 100 200 300
Gage Pressure, Pg (kPa)

2.40 A O.02-m-diameter manometer tube is connected to a
6-m-diameter full tank as shown in Fig. P2.40. Detennine the
density of the unknown liquid in the tank.
SG = 1.10
4m
~
" __!IiIII-¥
Specific weight
= 25.0 kN/m3
II FIGURE P2.40
LeI:: 0':: ~R 14.71:. cl- UI7k.nrJwY/ .fluid tUt~
0-;. :: rI,/O)(f. 80 x /D3) ':' If). 8)( Jo3 N/~3,
Thus,;
1~ -t d' (1",,) - (zS.>W
3
;;.) (~",,) - 0; (3",,) :: t;
r= 8. ~ ;.. 1D3 .!!.3
/WI
/J - cr
t - d-
3 N
_ J ~. q .x 10 -;;-3
2..-.3S-
= /'f30 ~/WI 3

2.4 A 6-in.-diameter piston is located within a cylinder
which is connected to a ~-in.-diameter inclined-tube manometer
as shown in Fig. P2.4. The fluid in the cylinde.· and the ma-
nometer is oil (specific weight = 59 Ib/ft3
). When a weight "W
is placed on the top of the cylinder the fluid level in the ma-
nometer tube rises from point (1) to (2). How heavy is the
weight? Assume that the change in position of the piston is
negligible.
• FIGURE P2.41
tv/in pi$t::I?"'1
m4111?me iel'"
~/t)l1e let:- py",ss,,"G ()11 hce ~f f,·stol1 =
-Rg"'Lt/~'1 b~c~""e.s
~ - a:: , 1.., sil? 30° .::: 0
p d/l
~ incyot'IZ>8d -toetdded 'py.,SStI r~
I W
~ ':: ~ + A-p
an p( m~1/14'me.j;el'"..egk~t"~J1 hecl!>m es
PI" - 4il (-'-, + ~ It) ~JH '3ofJ =0
51,;( bI-t&tci E Z.(J) -fr~m Eg ,0) -k o.biol~
1:0'_ f. - J:'J (.J ft.) ~/;' ~() ~ :.0
r f '" /..
(!:II"
;../
~;J (;tIt) S/H 3D to
- '::.
A,
SO thlll-
W &9 ~3)(-{ ft )(()/~)-
'I:( ~ It)"-If ,2
tine!
AI = ;1.,Qo /1,
I
A wh e~e
p
tI t fJ~+OH )
(I)
(2 )

2.4 Z The manometer fluid in the manometer of Fig. P2.4Z
has a specific gravity of 3.46. Pipes A and B both contain water.
If the pressure in pipe A is decreased by 1.3 psi and the pressure Tin pipe B increases by 0.9 psi, determine the new differential
reading of the manometer.
M=l-t-2c("
-.l Gage fluid
(sG = 3.46)
• FIG U REP 2.42.
~I" +;,~ in;·fJill clPnli1l1NltJf:m :
0)~ r<tz0 (;l) r ~f (2.) - d;h~ (, ) == ~
where ~/I 1t'l1jtn,j I'(JA~ I;' R. tVhen ~ dec.re~st'.s fa ~ I Q H~
1:, in~re/(se.5 +0 ~ I 11Je heljhfs Df t'he .flllid Caltlmns eh4n'je
ts 'hown &Ji1 .fi:;w~e. h>)I- -tne ./;ht1/ ~()11./;9j,(rtll:l()·H "
p; + ~() (cl.-a.) -J- Jjf (J-r;;'~) -~o (/+a.) = pl!J'
J~ bI:.~ct 1;g.c"2) fum EZ' (l) -10 bb..bn'J1
~ -1;/ 1- ~~/It} - ~f (.,,,) r ~zo (q )" Pa - Pta'
( ?13 -18I ) - (~ -p~)
t{. =
"" ( (f~ () - (fIt)
( 2)
Since I
, J
~f = ~ lfb ~:J.OfA. -~ :: I. B f5L. / ~-fB :- - 0. q fSi.) tinA
( q )b)(t /n.~ )
- t/ ;;;,'l. Ilflf h~ - (/, 3 -/t,.~)(IV-t;*~ )q,.= - /. rJ3 It
rJ. ('2. If };1) ( I - 3. if~ )
CI/Jc/ 1hfre"r~
~h = :2..ft + ;. 4: -- ~.ft +:J. (1.03 H) - -'I.Ob It

2.43 Determine the ratio of areas, AllA2 , of
the two manometer legs of Fig. P2.43 if a change
in pressure in pipe B of 0.5 psi gives a corre-
sponding change of 1 in. in the level of the mer-
cury in the right leg. The pressure in pipe A does
not change. Area = Al
.hI:
~ercury
FIGURE P2.43
For -the.. /n;t-141 C(!)nll'rwrlJf/~" (Sf'e H1lfre) :
~ + lJ: (t.. +.411,. ) -!:/ (iJh,') -!,. rt): t (/)rT "'LD' .. " . , . . (), J ' /3
when 18 inc~~ps~s the Y'/9I1t ("Iu",,, /rills A d;si~""e )a. J 4HI-i
111e lef-t t~/umn nje.s Ct cI'sl-t:/I'ue) h, Sin(e. 1ne. I/o/lime tjf. 11te
/ I;; /lId must: YemA/h ~tf)I4S/:;4I1t .A h = Az (4.. e>r At Q...
7) ) rt, A; ::: h
FoY' -the. .finAl (Onh;"r4t/~H) w/"ht /,y,ssure In f3 eJlltll -b, P/3 J :
fl4 or d;, ( ~, -r- lJ hI.' -,,) - ~ (J h.-A. - b) - d:" (t.'+~) =..J.. I
"LO ~'()/ t. rl3
Su bfr", t t:iZ. tJ) &171 EZ, (2) -fo ohio;",
I
- 0t:J.o (1:,) 7' ~;. (" +h) - d;/, (If'{) ::: 1:>,8 - "3
b = (PR/ -~13) - I~ (I/{) t ?f,,;j (tAo)
!~ - ~~o
P/l/-f.e =o. S" psi tl~tI a. = / ,'/1.),-t ~/IDWS thai
b : (tJ.5 ft,..)(I'fIf!li: )-r'l7)t, )(,i It) + (tJ.gj('H ~,)M,cr)
'Ktt7 It, - b'Z. If ~
= O. O~71/ I-t
:: /t.7
{z.j

2.44 The inclined differential manometer of
Fig. P2.44 contains carbon tetrachloride. Initially
the pressure differential between pipes A and B,
which contain a brine (SG = 1.1), is zero as
illustrated in the figure. It is desired that the ma-
nometer give a differential reading of 12 in. (mea-
sured along the inclined tube) for a pressure
differential of 0.1 psi. Determine the required
angle of inclination, 8.
FIGURE P2.44-
, I
tdhtl1 1:;,. - P/3 's in c.reAsed I::c ~ - Pa me le/i Cr:;/qmn IR//s t<.
dlj'.f.an'e tt ClI1 P the r/9}rt Co/tlIY'n II'/Jes ~ dishtl"lce. b t:lj~l1f
) J
111~ Inclined ~be- 4$ .sh~UIi? /n .f,9£.{l"e. J::;v -tn,'.J +-/Yl4J ~nfi:juf'tJ.t/(;)1:
1ftI + ~, (I,,. +fA.) - ~c1l~ (It t-.b ./" e) - lrbJ( -bs,." e)=+~
1,/- Psi -I- (<1j,.. - 0;.'/11- )(a -I- h fJlh t9) ::: 0
Th e eI,:P1erent/oj read/nJ) 1J h) tI )!)f19 111e -tube is
~h=
Thus,; +rt)rn El.{))
Thus )
--h I '
'# -'Pia + ( ~~ - 1',,1If ) ( Ah 'S /n ~ ) = 0
- (~'- Pra')
(I)
- (,), £fbI-

2.45" Determine the new differential reading
along the inclined leg of the mercury manometer
of Fig. P2.45", if the pressure in pipe A is de-
creased 10 kPa and the pressure in pipe B remains
unchanged. The fluid in A has a specific gravity
of 0.9 and the fluid in B is water.
~". fh~ /n;/-/q/ C(J)n h,~~a tl(;rl
FIGURE P2.4S
~ + i' (0. t) + t'Jt (().05 5111 30") - ~20 (0. ()8) ~ ~
A A 3
wh(;'"( fill /enftns eire l'n m, tJ),e YI ~ clec.l"'eases /e/-t CI::J/Utnl1
rnlP V-PS up A. ch$-hnce; a.) QJ'1e1 I"l!J hi ~''''ml1 moves dlPWM
t:t c/1'.5-k01Ce... J~) as .sh~Wh /" fIJIIJI"~, Ft:>r the ·f.ti'1ll/ t~4tj"NI.. 6/~ ~ "
(I)
14I.. ~ (tJ, J - a. ~/n 30 oj + 6"IJ., ( " ",;, JO' + O. oS sin 30' -/' «) -
!: (~.08 Ta.) = 1, l2.)
~~O 8
IS the !'lew !Y'~5S"'Y'e In t If4!. A.
EZ,C2.) from Eg.{J) 1:.0 ob-N/~
a. =
~A S/H 3DO - ;Yft~ (5//1 30'TI) ;- O~O
I
fA -fA:: /0 ltPa..
a.. -
~N
- 10 --:
M1
( 0, q) (tt. SJ ~) (0, 5" ) - (J3 3 ~ )(0,~-TI) of-
0, 051./0 1??1
New d/ffel"en-t14/ V'eudln1) 4h) me45tAY'eJ a/{)n1 /Mc..J1I1eJ +t.tl"e 1',5
e'l I{a. I to Cl
o ~ 1- = + 0, 05 4- a.n Sin '30 0
o,OSJ.jo /I'YI -r t).05M1 i" O. D5¥O tn1 = O,Z12 IW1
(!:l,S'

2.4~ Determine the change in the elevation
of the mercury in the left leg of the manometer
of Fig. P2.4f> as a result of an increase in pressure
of 5 psi in pipe A while the pressure in pipe B
remains constant.
1 .
2 m.
diameter
FIGURE P2.4~
-fA -r 'tJ.l~O (-f!.) - 6'#3- (.1;. SIn 30°) - ~il (~) = ~ (I )
t..Uhet'~ Id/ /el1~1h.s 4re /n f.t. IJhen ~ /nc.retlses +0 1Jt' -the
Ie.ft column ~/Is blj -the. dt's.J..q;tce; (1,) and 1ne rlJhi ~/"mJ1
m~l/e.s Up -me d/s~n~e. b) t:lS sh~ttll'J ;11 11Ie Irj",I"~. /;y 71l<.
( , J
-rlnal t!4'I1.h!}HI""i;-JiJn.'
*: I + a;: ( 7£-t 4.. ) - ~ (a + ~ :5/n 30D-r b oS/11 3CJ 0 ) -
n "~o 3
~ 'I (!3. - b $JH '30
6
) ~ 1:.()/ 1:1. !3
( 2)
Subtract ~b' (I) tr()rn FZ.ll) .f.o ()~iJ1
.p~ - ~ "1- ~~o (a.) - ~3 (li-rb5/~306) T ~i/ (bS/1') 3c~ =0
Since -the I/()/ume 0/ IJZ'u'"d rnusf ~.e ~s.ftzl',.t A, a. = A,z b)
(i /n.) ~ a- = (; j n. ) '))
.b ::. If 4-
- ~/-j;A ) :.
a.. = o#:lO - !Ut (3) -+ 4,;/ (2. )
= (!), 30 y. .pi: (down)
( 3)

2.4-1* Water initially fills the funnel and its
connecting tube as shown in Fig. P2.47. Oil
(SG = 0.85) is poured into the funnel until it
reaches a level h > H12 as indicated. Determine
and plot the value of the rise in the water level
in the tube, e, as a function of h for HI2 ~ h ~
H, with H = D = 2 ft and d = 0.1 ft.
rD-jd
T______ H
H -----, I
h=2 v'--.l
~ Water
Initial
{= ~(lJ:r)-AD)T-1oOed
FIGURE P2.4-1
Tht V()/lime cf w~trY' mu.st b€ t!Dl7ser/le~ (1111{ 1iJeyekt'eI
4 ('2.)
Final
OJ
.E d"< /J - E (DI.:)L it _ 7r (D~)';
f .,(. - 3 "'2.2. 3:z. c (2)
A-l~())
12:: Pt..' =
Do t/HU J) . .:: 12
- ~H ff/:z.
t. 2-
( 3)
find !i.1·t2·) CAn be. w"iffen ct..s
D2H
(-P)~Jo
3
3d'f. - - -8
toY' H = 2 -Pt I D =2 f t. I d = (). I It ) 411"
E~. (I) hec~d1es
;. :: 0. ~51t 1- O. IS ~ - /D
(If )

2.
3 (t;, 1ft) P =
be~m's
(2 ft)2 (1. f6)
8
1:
(1_ ().~3J. ) 3 (S)
IuI1(,t/~n ~{ -f. /.o/I()w~.
100 cl=:
110 print
print
print
print.
print
print
1(****************************************************"
"** This program solves iteratively a system of **11
"** equations to calculate the elevation 1 (in ft) **"
"** range of heights h (in ft) **"
"****************************************************"
120
130
14,0
150
160
165
166
print It h (ft) 1 (ft)"
167
170
180
I for h=l 1=0
print using" ###. ### ###. ###"; 1. ,0.
for h=1.10 to 2.01 step 0.10
1=0.0
190 las=l
200 hO=(1-0.03*las)~(1/3)
210 1=0.85*h+0.15*hO-1
220 if abs(l-las/l»O.OOl then goto 190
230 print using" ###.### ###.###";h,l
24,0 next h
To. bu/Ctted dtl:l:a.. find 4(, plot of tAe d4.ta Clre .sh"IlJII heJou).
****************************************************
** This program solves iteratively a system of **
** equations to calculate the elevation 1 (in ftl **
** range of heights h (in ft) **
****************************************************
h (ft) 1 (ft) 1.0~------------------------------------------------,
1. 000 0.000
1.100 0.085
1.200 0.170
1.300 0.255
0.8
1.4,00 0.339
1.500 0.4,24,
1. 600 0.509 o.e
1.700 0.59lJ. -...
1. 800 0.679
..-1.900 0.76lJ. ....
2.000 0.84,9 0.41
0.2
1.2 1.4 i.e 1.8 2.0
h (ft)
).-¥3

2.1.f8 Concrete is poured into the forms as shown in Fig.
P2.4e to produce a set of steps. Determine the weight of the
sandbag needed to keep the bottomless forms from lifting off
the. ground. The weight of the forms is 85 lb, and the specific
weight of the concrete is 150 Ib/ft3•
i-3ft-1
--==L 10 In. tread
W FIGURE P2.1f8
FrfJm the. Iree- bfJJ'1 - dt'a.:JYI/m
([3:LP1=O
~+-~+'Ui-tA::;o
tJhere .'
(I)
~ = we'flJt of 5t!Jldbtrg
'1d" ~ lVu;ld of- C())lcreb
PJ1 : tve&Itt (/{ *l'Ins
Ph : ?r~jII~ uft;"1 6"l!Dm ~lIrl4(i! due if) ~lfcyeie
A ::: I:Irelt ~I btP/folr) sur/ga
?rpm 1h~ d~ flU/A.'
"
J~
~1J
i~A
1vc-- (/~P !t3)(~f ~HCf'~) . ' .
=-;; Jl!)/ L'L) [(I/);i1.f,z'f',h.) + (/0/".)(16",.) 7- fi()",.)(~ tn.J)
(IS-O Us l.37T; . ~
:n:. / f!-ff .!11·
.f-C.~
Thus) fr~m F!. OJ
~ = (jOt;~,.) (7.5"H.2
) - ISOt) It - '65/.J
~ 1&,5 1.J,

2A9 A square 3 m X 3 m gate is ·}ocated in the 45° slop-
ing side of a dam. Some measurements indicate that the resul-
tant force of the water on the gate is 500 kN. (a) Determine the
pressure at the bottom of the gate. (b) Show on a sketch where
this force acts.
(a) ~':. ir'hc.A
SDO-k. N ::- (~. ~D ~) (-/tc.)(5'1nl JC3,,")
~ c.. =: S. 10 7/YYt ~
* = ~ (~(. + I.SI'tY )( 5111 Il-S·)
bofh,ft'
f
_ (ct. Kl> :~ ) ( ~.(..1 ~ + I.Sht sin Lf~)
(j, )
= G,".O~ = ~'.D~g
tl1'""'
lj :
1.;1(.("
+ Yc..R.
~c. A
':1c., = ~c:. 5'.fgJIM g.02 NY1:
-s~'~ ~ S~ sin 4-5°
..L )312. (3",,)(3M
f 8,D2,*, :' ().l> q3Siv.. +8.0 Z,w.
~It =
(8.021ft" ) (31M 'i.5/r&4 )
== <3.11 IWo
/
/
/
~
1.5"I'tI + 6. 0935 ~ = .5"'i ('In ~m
top of. ~ cs.1:!

2.6D I
2 SO An inverted O.l-m-diameter circular cylinder is par-
tially filled with water and held in place as shown in Fig. P2.50.
A force of20 N is needed to pull the flat plate from the cylinder.
Determine the air pressure within the cylinder. The plate is not
fastened to the cylinder and has negligible mass.
h>y e ZU'I //bY/um
2. F:-erfJCa I .:: 0
-Ph -+ ZtJ N =~
r-0.lm~
Air
TWater 0.2 m
Plate
1
F= 20 N
...pA
.t.4l~lt!*l
-p- ( N.u -that. pr'es~tlre fYJtJJt.
be a '~u"I-/~/ pre.~fllye.J
A-)~ (J .I
(t). ltttf )
~. -+ ~ = 1>aIr 1J.z,~
Th{Js) ZtJ IV (iJ~ j. /~of.!!.. ) (t;.2M')
-faIr = r~.j,,") ~
/1')1.3
= Lf 5' /0 .!!.. - - If. 5" / .J. ~
/1112.
2-'1~

d.?! I
2.5/ A large, open tank contains water and is connected to
a 6-ft diameter conduit as shown in Fig. P2.5) . A circular plug
is used to seal the conduit. Determine the magnitude, direction,
and location of the force of the water on the plug.
wheYt!
:: G-ft) If + /"L Ii
(/2 It) 11 flU)"
f;:~:,::,~~~-::::~:::,;: T
:'::", 1i:9_J..J..
f
-,t_"'-f..,.l~;:;:P_IUg_-.[r,-- ' L::;"';'! ...1..
: ," I
• FIGURE P2.S
-- 12./ r R
{he ';;rce 01 .).~ ZfJo/J:, acfs /21'i.fi: b!/t:J1d the
wp,i:ey Stlrhrce t:lh d IS pet/end, Ci;(/p r k fh~ pJllf
surface ~5 5h&(//)1.
;.- '17

2.52
2.52 A homogeneous. 4-ft-wide, 8-ft-Iong rectangular gate
weighing 800 lb is held in place by a horizontal flexible cable
as shown in Fig. P2.S2. Water acts against the gate which is
hinged at point A. Friction in the hinge is negligible. Determine
the tension in the cable.
Water
Th f,{!JJ
'i< =U2,lf!J)("4Jr;,;'h·){tftdft)
=3390 If,
To / tJ~4,te Fte )
~Ie ::
I. 1l C.
-r '1, where. '1e. =31i:IJ, A
So ihd-t: :3
Cable
117. ::
I~ (ifl-t)(I.,ft)
-I- 3 f-I:: - if.0 II:
tin d
(g h) (,ftJlIffl.)
ef"i//6;illlYl )
210)1 :::'0
T (3Ii)rs/~ 6~' = 'iJ ['1.r;t)(~()Sbb()) + ~ (2 +t:)
(sbb/b)('f.ft)~S"D~)+ ea'lb Ij,) (2ft.)
T= (g .ft ) (~t'11 boo)
/3S{) /J.,
2-'18

2.54- An area in the form of an isosceles tri-
angle with a base width of 6 ft and an altitude of
8 ft lies in the plane forming one wall of a tank
which contains a liquid having a specific weight
of 79.8 lb/ft). The side slopes upward making an
angle of 60° with the horizontal. The base of the:
triangle is horizontal and the vertex is above the
base. Determine the resultant force the fluid ex-
erts on the area when the fluid depth is 20 ft above '
the base of the triangular area. Show, with the
aid of a sketch. where the center of pressure is
located.
~c :(/f~D')ft -&)tt
.:: JO. t.r3 ft
.-f - U S/~ jp~ 0
~ - -rC PY'~5SI1Jl'e.
~ ~ ~J., A = (]If. 8 ~3f.,l M 3 t-t) S I~ t.01(~){"h d H)
- 33;QooJb
a{../-s 1hr~ufh
c/l, fql1~e t:J{ ~ t:) Ii - ;'p, t. Ii :: ;.If'1 /-1:
$liJ (Po~
~I- i'Ae. .fr/~if,le. as 5h~U,)n in ~Jetc.h.
;;. -Lf'f

2.55 Solve Problem 2.Stt if the isosceles tri-
angle is replaced with a right triangle having the
same base width and altitude.
1
- - -
FI< = 3~) 9()o/b
~ 1= J, If q Ii
(see se>/ut-i,;;, ~
Prt;j,len? 2. '5"4 )
L';jc.
'Ie.- It
_ ( to It) :J.( t fi) 2. -=
-
(Eg. 2,2.0)
2, ~ '7 -Pt
The 4rce. J ~ J e:.tcl-s "/h;'''''ff h '1'hc cen.f-er t:J/ 1'1"eSStlrc w/tu
t,4)(!')rdi"",l-e.r ~~ = :l, (J 7 f t: Q#f~ .!JJ = .:J, 'I- 9 f-t (5 -t!e. .ske I-c.h ) ,

z.s~
2.56 A tanker truck carries water, and the cross section of
the truck's tank is shown in Fig. P2.S6. Determine the magni-
tude of the force of the water against the vertical front end of
the tank.
Open
• FIGURE P2.56
PR,:: '(f he. A
Brea./( area. It'lto 3 P& ....i:-s
a s shewn. Few' a~a.. ([) :
i=k ='0 h~ At. I
I I
:: ~2.lf k3)(1)(~.ct)({)(2+1: ,( lfi:)
- 333 /1,
5/'(I(t F,~ = F~ 1'J"el1 1=" ::: 3-33 11,
I 3 ~3
~r tfret< @:
~ ~ ?> hi! Pt.,."2.. z..
=(~2. Lt tt3')(1H)('t-+t f. trt-t)
t:: ~ L F" .,. ~
A. "I 'Z..
:. 2. (333Ib) -t 02.4- ~~) (z.f.t) (~H ,c. 't-H.)
- 2~ fDb Ib
2.-5"/

2.57
2.'57 Two square gates close two openings in a conduit con-
nected to an open tank of water as shown in Fig. P2.57. When
the water depth, h, reaches 5 m it is desired that both gates open
at the same time. Determine the weight of the homogeneous
horizontal gate and the horizontal force, R, acting on the vertical
gate that is required to keep the gates closed until this depth is
reached. The weight of the vertical gate is negligible, and both
gates are hinged at one end as shown. Friction in the hinges is
negligible.
T Horizontal gate, 4m x 4m
h
1
Water
R
Vertical gate,
4m x4mL-________________________
~
_
),,,1"17(:)11111 /
L MJ.J =0
1ha.i
OW = .pA ~Jure
~~m
..-6 =~ (2.h!'Jr /.It. (J
P /j tnt' tvp.-br fr~Sr/lye ~11 fh~
SI/f"face .
11141:- N
1v =(9!OO;;3 )(z~) (tf/M J< Ji.M) 314 -I<N
!;r tlfy-t/ c~d rttie)
fk =I'A~A Luhey~ ht
:, 7hn
Se> 7'ha. t.
1;.:: (?;p~1-3)(7trn )('l-tm X Jf~)
=- IIt)o.JGN
To /1/ettk FIC-
U L~c.
J~ -=. Y"A -+~, =
~ (ifltH ) (If/11( ) ;
~-----~ ~
(7""" J(If,.,., , IfI1H )
~ u/'II'br/um
2: MIi :::'0 So 1Yz41:.
R=
(;/ IJ~ I<IV ) (rAn - 7.1 'IIIIK )
J..-S2..
r'ff"!!--R.
11'W1 = 7.1'l1M4
Hinge

2.$8-
2.58 The rigid gate, OAB, of Fig. P2S8 is hinged at 0 and
rests against a rigid support at B. What minimum horizontal
force, P, is required to hold the gate closed if its width is 3 m?
Neglect the weight of the gate and friction in the hinge The
back of the gate is exposed to the atmosphere.
F. =Yh A tJheyf:. he. = 5""1
, ',I I
~ =(qfpo!.. ) (5;w, )('ftm X 3 1m)
/J'Y13
- s'i'8xl{)SN
h = 7M-1
Cl.
~o "'fhAt
;;.:: (1fOO ;;3 )(1/h1)(2~ X ~~)
- 'I. 12 ;( /O~N
To
Open to atmosphere 1
3m
Water-
il-' 0/.';,', .. ·/t
-Hi~-. -
4m
';':.:'-':-. -.:- ::;.-:..: B A
. - .. -. .'. p:z-.-::;-.:r;::,.~_--1..._p
".~_:~.~::./-:;-~-:::<::'r-2
m--1
• FIG U REP 2.58 o~ ;
o III---~"'" ()1
,
p
/oc",-if:. 0) / ( 3
1~ = Ixc
ie,
/ 2. ~""" ) ( if;ffl )
-t - -+I
!:fc A, (S""m?) ('1M .(-i;Jn)
I
The -krce Ii at:--Is 4i 17re Cenhr 0/ 1J,e
L H =-0
0
F; (6: 2.6, 7~ - ~~) -r Ii (J".,) =:
SD 1714.i
p=
2.-53

01.5'1
'2.'S'f The massless, 4-ft-wide gate shown in Fig. P2.59 piv-
ots about the frictionless hinge O. It is held in place by the 2000
lb counterweight, W. Determine the water depth, h. Gate -+-III
For' ezu"~'br;"m)
::EM =()
()
h :: .b.C 2.
f;. d = %) (3A:)
So -;-;'4 i
( 2~&J() Ii; ) (3 H)
(0/.1 ~)(!J...~) It It)
z, 2-
-It.3,= (3){Z. 0(1)<> It ) (3 f+)
(~Z.¥~,Yf )('rR)
i == 5'. lift/:
Pivot 0
Width =4 ft

2.60* A 200-lb homogeneous gate of 10-ft.
width and 5-ft length is hinged at point A and
held in place by a 12-ft-Iong brace as shown in
Fig. P2.60. As the bottom of the brace is moved
to the right, the water level remains at the top of
the gate. The line of action of the force that the
brace exerts on the gate is along the brace. (a)
Plot the magnitude of the force exerted on the
gate by the brace as a function of the angle of
the gate, 0, for 0 :5 0 :5 90°. (b) Repeat the
calculations for the case in which the weight of!
the gate is negligible. Comment on the results as
0--+0.
(a.) HY tn, free- j,()dy • dlR~rllm 0/ tlte
f}4 -fe (St*~ rl,lfl'tl))
Z t=;. =0
00 thAt
FIGURE P2.60
'}. (*) + q,v (f et>se) ~ (1=; UJ.5 ~)(1 ,I~ e) .,. (Fa jl;HJ)(jf()J a) (/)
Also"
J. ~/h e -=- L .5/n tf; ( as.slllfl/~ ~ h/l11e. t:{ ntl end /) f
brllle II + .slllIIe elfIN. 'h~11 )
..511, ~ = J. ~/;' r;
L
f:.R.:: o!'c A ~
r (J ~/~'~ )(,Q w)
.
w hert! W- I.J the sa~e wulln. Thus) I:!. (II Clln b.e Wf, /fflt (,{~
'6({'J(sl.9) w- T OW! CBJ {J "" Fa 1 ( (ss ~ J/~() l' J/' " ('.{)s e)
.50 that
(
V' A 2. _) <AI ( (:zw-) I'AII e ~ ~
~ = 0 ,(G.> WjSI"B + '2 ~.s e -= t;~ -
Few
F,.,=
8
~s ¢; 51"!? T 51 HtP Cl;s 9
'O-::'fDZ.'f 161ft)) 1=5ft/
(6,2.If !:))(Sft)l(10jt) hi" (; r
&;
=
( CDn't )
~ ~() () -hi"G f' I () 0
(r)s 9 frill /} + ~/H 4
( Z )
( J)

.J
~.bO 1-1 (~,,'t )
SInce 51;' t/; :: ~ 511"} e
/....
SII1r/ :: S-
S/~e-12..
c( h d. ~j/ ~ ?liI!11 l) I ? t!iJh ~~
Clift b( II.5f'd .fD tlel-ermHIe... Fe
proJftlm foY' C/J IUtllllulJ ~ 4.s It
100 cis
f:: s-ft L= /2 It)
r/e.J.en11lI'lt4. 711 tiS, £3' (3)
~Y' t{ JJi/!J.1 &. A ~"'Pu.,/;fY
I-uI1C.N()·11 of e- ~/j()l().s •
110 print n**********************************************************"
120 print "** Variation of the resultant Fb as a function of theta **n
130 print n**********************************************************n
1/,r0 print
150 print n Theta (deg) Fb (lbs) (w=100 lbs) Fb (lbs) (w=O lbs)"
160 pi=4.0*atn(1.0)
170 for theta=pi/2 to pi/36 step -pi/36
180 sph=5/12*sin(theta)
190 phi=atn(sph/(1-sph~2)A(O.5))
200 fbl={2600*tan{theta)+100)/(cos(phi)*tan(theta)+sin(phi))
210 fb2=2600*tan(theta)/(cos(phi)*tan(theta)+sin(phi))
220 print using It ###.# ####.# ####.#";theta*180/p
i,fbl,fb2
230 next theta
Tabu/flied data.. CIne! a. ,lot of the data. af'e .!IJlfrJ ~I"}
rhe h//6'/A.)lh~ pa..je.
(b) h;r OW=O lEg. (3) reduces t~
-- G!;s ¢ frJn& -r 5/)1 ¢'
(/11(/ the jC/me ,ProJY'lIm t:lS WIIS u.sed In f4rita)
(w/·tn w set: fS"4/ -t-o "'!freJ) (!.All be used ~ t!)bt4/h
£' ((.s t:I ftlnt;t/~h ~f e. T4J:,t(14~~ data.. qllt/ ~
l!J
f/r;t t!)f the d~i£ ~Ye J/;fYl ~11 the .f,p//!')/P1i1!J p~~e.

OJ
.n
.....
ri
LL
**********************************************************
** Variation of the resultant Fb as a function of theta **
**********************************************************
Theta (deg) Fb (lbs) (w=100 lbs) Fb (Ibs) (w=O lbs)
90.0 2860.1 2860.1
85.0 2757.4: 274:8.1
80.0 2659.4: :264:1.5
75.0 2567.0 :254:0.9
70.0 24:80.9 24:4:6.7
65.0 24:01. 6 2359.2
60.0 2329.4: 2278.8
55.0 2264:.8 2205.4:
50.0 2208.0 2139.0
4:5.0 2159.6 2079.6
4:0.0 2120.0 2027.1
35.0 2090.0 1981.2
30.0 2071. 3 194:1.9
2:,.0 2066.4: 1909.0
20.0 2081.1 1882.2
15.0 2128.8 1861.6
10.0 224:9.8 184:7.0
5.0 264:6.3 1838.2
* 10
3
4.5
Legend
G o w-100 lbs
h 1:. w-O lbs
4.0
3.5

3.0

2.5

2.0
o 20 40 60 BO
Theta. deg
( t.(}1)'t)
.2-57

A-s r; ~ 0 me /la/lit eJ{ 17.3 CIIJ?"~ det~Y1mn e,,( !rtJlI1 E!. (~))
2 (Pot) Clll1 eF-,:
e tps ¢ 1T!1I f) rS/",
I f "/I~II)J thAt.
U.sf -- V1- sli/'~ ":: pi/-- hS:)~/"6
Th US tis B ~ 0
I
/ + ..£'/2.
:: I f 11-0 J.b
Pht{'S/{II//'1 IJ/I~ Y'lIStili mellllS '/h,t: ~r' ~ =.()I the lJtl/"e () / ~
is Indeferft,,'n~ft! I but ft,y tll1~ "//n*, sn",f/ I, Vlllue. ~I e I ~
WII/ QfPt()lIc.h / ? J/.O /b.

2.til An open tank has a vertical partition and on one side
contains gasoline with a density p = 700 kg/m3
at a depth of
4 m, as shown in Fig. P2.61. A rectangular gate that is 4 m high
and 2 m wide and hinged at one end is located in the partition.
Water is slowly added to the empty side of the tank. At what
depth, h. will the gate start to open?
F=Rj -::. ~,. ..t" ~ A.~
WneV't '} refers +0 :J4S6i1l1 e .
• FIGURE P2.61
~;=(700 ~.)(q.8/ ~)(ZNO) (tf,.. ~ z.,..)
== 11 0 X 10J N == II 0 k N
t- -:. 0. )" w- 1+l4-
~l.cr Lv'"
LU heV'e. 1..0. If e .(ers toWcc....-h}l' .
F; ;: ('t.(?O)c'ID3N)(.h)(2~~h)
Rw- /)tt :i 2.
Wher~ n f~ dtpTh ~t wo...~V'.
~kr": (~.g0 ~ Ir/)~4
h:>.... e~u',ll bfl ~m )
2 M .t :'0
So -/11«,t { l
~"" 1.",,::' 1=/26 if! w d" .Q... = '3 a"01 1, =3 N't
Tn11.5) ( q. 8DX I03
) (h'1) [ ~ ) =(I I 0 ~ I()
3
N) ( - N')
~"'cA h:: 3. ~5 fYYl
wh,c.~ I~ th<. ,~~h~~ vAlwf -h,v h.

1Water 6m
2.'-2 A gate having the shape shown in Fig. P2.62 is located
in the vertical side of an open tank containing water. The gate
is mounted on a horizontal shaft. (a) When the water level is at
the top of the gate, determine the magnitude of the fluid force
on the rectangular portion of the gate above the shaft and the
magnitude of the fluid force on the semicircular portion of the
gate below the shaft. (b) For this same fluid depth determine
the moment of the force acting on the semicircular portion of
the gate with respect to an axis which coincides with the shaft.
~l
~
(aJ
(h)
r:; r rec.-t.al1'1l1/a,. f()rfu,'II)
(f):: J-he A ~he f" At = ~ ""'
r
S" "fna.t:
(Fi<)y::(ri~b';.3)(3"""')(~~;('"",,) - /~6,t)~N
I :: ~I'M + 'fie
he 31i
Side view
of gate
:= ~f'M -t If (~m,) = 1. Z1 M1
31T
/ 6 J 0 J? N
7.3~1"n1
ThUfJ, tn ~mt/I i ~;fh reSfec.1 -4 s h~.f.f:.J M)
11 =(~) )( (7. 3' I'M - ". ao ~ )
5,
:: (;O/D xJD3N ) (i. 3'1'M)
I. 37 X /~ I:. Iv· /WI=

2.63 A 6 ft X 6 ft square gate is free to pivot about the fric-
tionless hinge shown in Fig. P2.63. In general, a force, P, is
needed to keep the gate from rotating. Detennine the depth, h,
for the situation when P = O.
1=;,.,. eO,;,I,'enum
21'1).J. =0
ThllJ ~r p= () ~ /NO" fd
J
hlft/( to,Pass tnrPJl9h 1ne
/J1n9.e) ,', e,) fJl?:::?' s.f.t + h
.
Qnl{ WIt;,
t:J(e
y", A-
!:f~ -= h T 3 f.I::
• FIGURE P2.63
3.sR+h
_ -A ('~tJ ((,. ft)) -+-
(I, t 3ft) (tft,t. b-H)
6.5 it:. -
h= 3. ()t> -tt
2.-(,.'

2.64 A thin 4-ft-wide, right-angle gate with negligible mass
is free to pivot about a frictionless hinge at point 0, as shown
in Fig. P2.64. The horizontal portion of the gate covers a I-ft-
diameter drain pipe which contains air at atmospheric pressure.
Determine the minimum water depth, h, at which the gate will
pivot to allow water to flow into the pipe.
POt' egu;/,'hrJum
:2 1'10 =0
I=:: >< ..J. :: I=R )(12. (I )
~J I z.
17< ='rhe A-,J I
=(~z.lf~,)(f) (If-Pc ~ h)
== 12.5' h'l-
Width = 4 ft
Right-angle gate
Hinge
.. FIGURE P2.64
I=R,
1"
)., t:='J(L
~
OJ!'
1,,~
~)ll.~
I;y -the t:,rt~ ~n -tnt! h~rJ~pl1ta J ~YI::J~YI 0 ~ ~ sa,l::e
(whu..lt JJ hajq~ced b~ P~tS5"H/·t. ~~ bDTh 5Jaej e"cepi
lo~ tnf. t.lre~ f)f 'lrre. p/pe)
t=;~: tfhtr;.)(lh)2::: aZ.Lf:3)(h)(![)(!1i)).
::: Lf'l.O h
Thus, Pr8m ~~. 0) /AJ/'f;. ).., ~ g dnPl 1.=3t1:z
U1.5h;2)(f) = r;.1.t>J,)(3+t)
h::: /. rFfi
pipe

2.~s
2.65 The specific weight, 'Y, of the static liquid layer shown
in Fig. P2.65 increases linearly with depth. At the free surface
'Y = 70 lb/ft3, and at the bottom of the layer 'Y = 95 lb/ft3
•
Make use of Eq. 2.4 to determine the pressure at the bottom of
the layer.
=951b1ft3
z=o
• FIGURE P2.65
dp _
-di: -~
( ~~. 2..lf)
0= 'f5- 12.£' 2-
-/nat () c=-z.
fl' = irs-12.H)d:t
1b.1trJ17I 6
1t,lhm=- [r,~- It'~-r9. 2.
z ()
- - ['1Sb') - t... '1.5' {2)'j-
t /~5"
J.b
=- .ft,..
hot"fl,m

2.'~* An open rectangular settling tank con-
tains a liquid suspension that at a given time has
a specific weight that.varies approximately with
depth according to the following data:
2.0
1.4
2.8
., ?
".-3.6
12.3
12.7
12.9
13.0
13.1
II (m)
o
0.4
0.8
1.2
1.6
10.0
10.1
10.2
10.6
11.3
The depth II = 0 corresponds to the free surface.
Determine. by means of numerical integration,
the magnitude and location of the resultant force
that the liquid suspension exerts on a vertical wall
of the tank that is 6 m wide. The depth of fluid
in the tank is 3.6 m.
'17
r-_ - - - T~~
1-:,.....,-:01'''-'1'-""'"-:,/-:,/,,I-:-j""'-'-:-_L-:-'L-:-h-,.!'L
The mAA",IslI/t. 01 71te -lIt/Itt ~~ce fi., . /,..... ,
(!/In k "ff;III1' ", SlimIY/ I111 1i1~ d,I=HY'fIlf.,it/
~"'ces a,chhlj fJlI Pte. h~I'I'o"'';'/ ~l-rlp ShtPWII
,,,; 1I1e. frrt.ll'e. Thus)
f ().-t lot
/I J..I
Fk=jd{-bj1>Jh OJ
() 0
Whel'"e ? 'oJ the presslire at (j.eft;, ~ .
To iii,''' P wc uSe. Gr 2.'1
cjj; =-K
d't
((HI IN,:;'" d'i= -dh {
?(A.)~ 1~ d-P.
L
(Z )
E1IJ 4.t,0I1 (2)
PrOfj'yl"m ti111{
CIJIt be If} JeJrllff~ 1/" meY'I~q / f;, "':/t" 111(! kIf()uJIit"
11.51;'9 11ze. Vlly,'u·holl I; (f MJITh h :;llIel1.
100 cls
110 print
120 print
130 print
1lJ,0 print
150 print
160 print
"*************************************************"
"** This program integrates Eq. 2.4 numerically **"
"** using the t:r:apezoidal rule to obtain the **"
"** pressure at different depths **"
"************:t:************************************"
170 dim p(10),gamma(10)
180 n=10
190 dh=0.4
200 p(l)=O
210 for i=l to n
220 read gamma(i)
230 next 1
2lJ,0 data 10.0,10.1,10.2,10.6,11.3,12.3,12.7,12.9,13.0,13.1
250 for i=2 to n
260 s=(gamma(1)+gamma(i))/2
270 iml=i-1 ( Ct>l'Jt )

280 for j=2 to im1
290 s=s+ganuna(j)
300 next j
310 p(i)=dh*s
320 next i
330 '
3~0 'Print the results
350 print.
360 print" h (m) Pressure (kPa)"
370 for i=1 to n
380 print using "###.# ###.##";(i-1)*dh,p(i)
390 next i
K************************************************
** This program integrates Eq. 2.~ numerically **
** using the trapezoidal rule to obtain the **
** pressure at different depths **
*************************************************
h (m)
0.0
O.l±
0.8
1.2
1.6
2.0
2.4
2.8
'J OJ...J • .:....
3.6
Pressure (kPal
0.00
l±.02
B.08
12.2~
16.62
21.34-
26.3~
31. 46
36.6~
41.86
~gfl4. t/(;11 (I J CilI1 ,,()/AJ b4 Ihk!hlkJ lJ("(mer;c,,li~ US/'n~ TIGItPEfO/
*****************************************************
** This program performs numerical integrat.ion **
*****************************************************
Enter data points (X , y)
? 0.0,0.00
? O.4,~.02
? 0.8,8.08
? 1.2,12.2~
? 1.6,16.62
? 2.0,21.34
? 2.4,26.34
7 2.8,31.46
? 3.2.36.64
? 3.6,41.86
The approximate value of the integral is: +7.1068E+Ol
( Ci>!1 'i )
2-(ps

(Cd)tJl )
w;-n II
j,?d~ =7/.07~
() A'H
~ = (~hrI ) ( 7). 07 #).: t.j z{, --k Iv'
.
To /"c4k lie. sum h1~l'I'1fl1.f:s (I/'(Jlli a.x.I'.s !;;rmed b!l l"trj'S~C.t:'()H of
IIfl'h{~/ Wf/I II~~ I(/;"H Sf,( Y' /Au;,. rn"s,
~ f,e = bItI> d -f. (.3)
o
The il?.fejY4f)I ~f CAli be. dekY""~t4 41111 Eo . (3) /~l-ef,.,,/~4
"l/mer"clI//~ us/nj T,eA-PErIJ/. TIIJ,"/"#I r~.!tI/1s ql'e ,iflflt k/"w.
~****************************************************
** This program performs numerical integration **
*****************************************************
Ent.er number of data points: 10
Enter data points (X , Y) Note: YtV (}f? 0.0 to. 000 ~
') O.y,.1.608
:1 O.8,6,L16lJ.
'? 1.2,1"',688
? 1.6.26.592
? 2.0,l:t-2.680
? 2.4,63.216
? 2.8,88.088
7 3.2.117.24:8
? 2.6.150.696
The approximate value of the integralls: +1.7y,37E+02
TIlliS) IN/tA H
J-I-fd-t. " / 7'1-. If -J, /(
0
if /DI/()WJ 1r~1J1 Ii,!. (3) 1'h1.t
,£ =
h l~I"h { fD 1m )(/7'1: If ,AI)
--/? fi
?,,,, .A)/
The Y'e.suJitlni ~f'te. (j c.ls 2. '1-1, MI below fluId
:::
2. 'fie 1m
sur-fa c.e ,

2.~"1 The inclined face AD of the tank of Fig.
P2.•1 is a plane surface containing a gate ABC,
which is hinged along line Be. The shape of the
gate is shol1 in the plan view. If the tank contains
water, determine the magnitude of the force that
the water exerts on the gate.
{.() he re.
i'
A =[).;<.'d!j , (". -H1""')
=j;{{)I/?d;/ = ff({f}T=~'A,'
o
I f~' /'1 I, I
'de ,4 = t:f d A:: .t. !:J,I(. d:J
o c)
I
I '1, .::
lind
J'n ce. {c. = 'fG $,1'1 30
0
)
Water
""'
=
x'
FIGURE P2. (p7
/'
./
~ = ~2.1f ~~ )(3"ti:)(J/~30tJ)( ~".prl.) _ Sfet /b
y'
y' = 4(x,)2
Plan of
gate

~2m-j
1
2.68 Dams can vary from very large structures with curved
faces holding back water to great depths, as shown in Video
V2.3, to relatively small structures with plane faces as shown
in Fig. P2.68. Assume that the concrete dam shown in Fig. P2.68
weighs 23.6 kN/m3 and rests on a solid foundation. Determine
the minimum coefficient of friction between the dam and the
foundation required to keep the dam from sliding at the water
depth shown. You do not need to consider possible uplift along
the base. Base your analysis on a unit length of the dam.
. 5 m
4m
If-~---6 m----"!
FIGURE P2.1o<B
~ := K/,(! A
where .4 - ( If ""'" ) (I )
:S/].t SI.3°
/=Or eZ1I;//6r/ufYI )
z: F;<. =0
or
fie SII1 s/.3 II =F; = )/11
/f-/jt!);
2.. FfJ :. 0
.:50 1h()./;
N =C>v<../ -t- !=Je C6S 51.3
0
where.
't<I:: (d"'~n~l"ei-e. )( V()/r.lm~ DI- UDrlCrei:e)
Thus)
N = (~:3. ~ ~~ )(20 tm 3) '1 (I()() Ie. N) C~s S"I. 3 0 = s 3 If -k N
(/~O-RN) Sin Sf. 3 0
-=
5'""'3 'f ..It.. N

2.b'f "" Water backs up behind a concrete dam
as shown in Fig. P2.~~. Leakage under the foun-
dation gives a pressure distribution under the dam
as indicated. If the water depth, h, is too great,
the dam will topple over about its toe (point A).
For the dimensions given, determine the maxi-
mum water depth for the following widths of the
dam: r = 20, 30:- 40, 50, and 60 ft. Base your
analysis on a unit length of the dam. The specific
weight of the concrete is 150 Ib/ft3.
/- hfe- b~d~- dla~r({m ot the dct~ 15
~howl"l in -the +t~CAV'e. a.t 1J1e I'i~lti)w;,e",.e.:
~I = '6;"- (lev l.(nl'+ lel?~11t)
F. = (CS~'" '({~T ) Q
3
:z.
F. = 'c5 (~T) (~r ') = ~-a "2-,.'2. 7 Sihe
Z 5i"e
-t
'12,=- I (~,. )~,:: 3 3 SI'w8
PB = 'Yh
l·
FIGURE P2.lDQ
+elt') G=
·':13 ~~/{Jer the. pfeSSwve dlS.fV',J,,,,,~,o~ On The. 10 ~~ :
sW/
~ =1(~-ti)J
Summ In~ mt)meY~ a~ot.(.i AJ
~~ ~3 = FI (~) + f!c (~~)
I
180 ft
( It.'
~----------------~--------------~~~ '--------------------------------~
'2..- f.:. ?

( CD" 't )
5 0 f;,4 t .J. )
1f3 = Fr (.. ) .,. ~ (tft
Fj
5,,6>1-, '.fl(j.1f)~ of exl/I'fSSIDlf5
j. ( *-r 7~ )
-i-r-1t
Fe>I( eguiiij,.,./"W1 of fhe dqm I L: fi1A:: 0 I So 1h~t
'J ':1, - ';v' (;ft) - (=2. '12. + ~ t:13 =0
and w'Jth 'If::: fe,2.lflb/ft3
) crt. = 150Iklft-', (/1111 ;'T':./tJ oft) then:
F, = 3/.2 i 2. ~= {gf!)ooJ. F-:. = 312. 0 ':I ::I 10/3
:z. Slit G :z si", 9
F..:: 3/. 2. (.f. ~,0) .i.3
'13 ~ P. (if.,. 1J():r (2.~ -t-I(Jj J)..
-R.1"-R..r 3(...(t-t IO)
Su h:st,/£l.tJDJ( 01 1h~se ex..preSS/O/LS I;,k l:Z. (J) ~ I eJds )
(3/. do.1 )t~ ) - ((,,""O~) (; Y-) - (3~~:){-~~~~)
r ['3/.2 aT/O);] [~~t:::~QJ=0
Thus;, .for tL t:j"UH .P. I B C411 be de,t.ermuucJ /YOHf the
c,4>rld,f/{;H -hl11~=?o/,f1 Qnd Ef,(2) ~()JlluJ kV h.
OJ
A CiPmtufty pl"0J)"qm for detfnn/lll1iy h ~v t:t 7,l/el1 ). ~J/()WS
( C~t1t )
:2-70

J. 6Cf /1& I
100 c12
110 print "*******************************************************"
120 print "1I:* This program solves a cubic equation to determine **"
130 print "** the maximum water, h, depth for a series of dam **"
110,0 print. "** widths, 1 **"
150 print "*******************************************************"
160 print.
170 print" Dam width, 1 (ft) Maximum depth, h (ft)"
180 for 1=20 to 60 step 10
190 theta=atn(80/1)
200 h=O
210 hp=h
220 h=(3/31.2*(3896*1-2+10y,OO/(sin(theta)-2)-20.8*1-2*hp) )-(1/3)
230 if abs(l-hp/h)O.OOl goto 210
210,0 print using" ##.# ##.#";l,h
250 next. 1
~r the dtlm w/dths s!..f'c.I'f,Pd The fI14X Imum WAifY depth.s
ClY'e glYfn be/"u). ,t10t( i'}Jat .fr:,j, -the +1<10 /aY'~e~t dtll?1
/AI/tf1"hs -the tvaier wt)"'/,f ~vPJrf/~tV fh~ d~m h~fa't' IC lV~l"q'
+Otf/~ .
¥******************************************************
),.. This program solves a cubic equation t.o determine ct.1I
:1<* the maximum water, h, depth for a series of dam **
** 'i"idths. 1 **
*******************************************************
D~n width, 1 (ft)
20.0
30.0
40.0
50.0
60.0
Maximum depth. h (ft)
48.2
61.1
71. 8
81. 1
89.2
2-71

:;.70
2.70 A 4-m-long curved gate is located in the side of a
reservoir containing water as shown in Fig. P2.70. Determine
the magnitude of the horizontal and vertical components of the
force of the water on the gate. Will this force pass through point
A? Explain.
J~II' E'~ u; I, 'bY'/un'I I
L rJ(. ::.0
r:;.:: ~ ~ l(!,c'- A:& = '( {'11ft+J.!1rC)(1,-" .:~..)
f:u =(9. HD ~ )(7shfI. ) (12 /WI ,.) = ~8'2. -Ie tJ
Slmll~.,.It-:J)
L F~ =-0
whe/l"f '
II FIGURE P2.70
r, = [3' (6~)](311M xLfht1) = (Cf.FO/lt/J )(6~)(/2tm")
/IH'3
~ = ~ 11- = (9. eo ~)(q i1 M4
3
)
Thus} f:,," (r.80 ~)[7l ",,3 .,.'l7f ",,3] " 1113 ).IJ
I~
--~
(/V/).;.~: !;Ne ~I W4/;fr tOn ,j4~ INtil be. tt>ppes/!c I~ dll'~cf-~i,., +0)
111111- ~h~wn tDn h9ul"e.
The. d';'ecbtJ"n c/ 4// C/;lleYf"'';'/~/ Ior~es etch;'f ~'1 th~
CtlJI'vetl SUY'~c.e l.s ~r,Pl'''''41&U'4/1' ~ St.lr/ttce./ ctl'1d "t11f/l'(ft,"'f!!)
in! yes III 14111 /1'1 ",.sl- ?IH5 i11rlP"''1h the ';'+~Y.rt'~I-I"';, 01 all 7/lelf!.
~yce$ which I~ ~i; tt!)Jnf A-. Ye s.
2..-72

2·71
Z.il The air pressure in the top of the two liter pop bottle
shown in Vidt'o V2,4 and Fig. P2.7 I is 40 psi, and the pop depth
is lOin. The bottom of the bottle has an irregular shape with a
diameter of 4.3 in. (a) If the bottle cap has a diameter of I in.
what is magnitude of the axial force required to hold the cap
in place? (b) Determine the force needed to secure the bottom
2 inches of the bottle to its cylindrical sides. For this calcula-
tion assume the effect of the weight of the pop is negligible. (c)
By how much does the weight of the pop increase the pressure
2 inches above the bottom? Assume the pop has the same spe-
cific weight as that of water.
1 in. diameter
Pair =40 psi
4.3 in. diameter
• FIG U REP 2 .1 I
(a.J 3I. Lf /j,
L F. =0
VfY+1 c..J.
F. d ::. F = (pr'e~surt @ 2,;". QbolllZ h,HoWl)'51 ~ I
X. (A yeo... )
- (Lfo ~: )(.~) ('1.3 iny ~------------------
= 58 11,
(c)
lnCYea.5e. U1 pY'e~suY'.e crue .J.o w~'9ht ::
e$S 11cH1 010 0 f tU r pr'eSSu ye).
J-73

).- 72
1.n Hoover Dam (see Vidl'O 1.3) is the highest arch-
gravity type of dam in the United States. A cross section of the
dam is shown in Fig. P2.72(a). The walls of the canyon in
which the dam is located are sloped, and just upstream of the
dam the vertical plane shown in Figure P2.72(b) approximately
represents the cross section of the water acting on the dam.
Use this vertical cross section to estimate the resultant hori-
zontal force of the water on the dam, and show here this force
acts.
1..---- 880 ft - - - 1
." . :~,.;t£.'i··~~iAiW':h'i?t.;.· y
~6601t-......,
(a)
Ii FIGURE P2.72
- 290 It ..
(b)
!3re~t tlvea /'rJfD S pa~.fs (I,S ~ho{J)n.
17:;y aYea.. / .'
liz! = /)hA=~2.1(. 'by~ )(715".f-I:'ji){t.qS-Ii)[7/fift)
::: l.:r7 .x/O
q
/.6
}:?,y arell 3 : Fo -:: FI? =157X If)if J.b
".3 I
For- Clrea. Z:
/=,e~~ (0),,- A~ = u'1.iI !t3){l;.)(7Isft)(290 1i)(7Jsfi)
- if. l3 X / D'f /b
~ -t ~ -t Fo : /'57;(JD ~I.J -I- 'I: ~3/ J~ f /;; -t /.!;-7x JD9
Job
J z. 1'.3
7, 77 ;( JD' /j,
Since -the.. /7J()mefl /
~f the. dam t7l us t
F;.,) 17<l..) ~1It;{ he3
)
01 111(. rt's/J/1zurt ~yce a~lIf 'the. b4.se
be ega-aJ -fz, iht! rn(!)m~Y1 f..s due. tt:>
11- .fi,//oI{)S 1h~t
(Ce;n '-/: )

.2, 72 I (~tJl'} 'i )
Pi xci" ~ (1)615ft;) T F?/-Df7;5"H) t ~ r;)(7/5"H)
a::: (/..f7;( Il/ h)(~)[715"It)+(;f. LJilbqli'fJ.)(?J~f-f.) .ff'/.57t /DqliE)(71OIt)
7.77 XJ'/J' i);
- LJ Ob ·f-/;
(hils) 1h~ r~.>ul/rll1.f:- hfJ;'''I()t1fal -hvc~ ~11 -tJ,e dam /J)
7 77 )( I() 9)1:, acJ-/nj -'fot It (,I.P Irm? fh( bl/5t!
t>f 1he dqm a/Pili 17te tl~is 0f ~'f/11fl1e 11''1 t>/ the "rea.

2.73
, . -"
2.73 A plug in the bottom of a pressurized
tank is conical in shape as shown in Fig. P2.73.
The air pressure is 50 kPa and the liquid in the
tank has a specific weight of 27 kNIm3
• Determine
the magnitude, direction, and line of action of
the force exerted on the curved surface of the
cone within the tank due to the 50-kPa pressure
and the liquid.
Fo r e$u.;;I '/:n"/um )
L f:verbI c.a / =- 0
50 in"t
F;. = -?a,irA 1- '.v'
Wher{ F" 10$ The. force. The CdI"( e~frl:s
ot. the .flu/ci,
Als~)
~ . A = (50!1i )(-r;)(d:1)
flO' 2.
= (50 -.le.P4. )(~)(I. J~£rrt1)=' S ~,LI·-k.N
CHI PI
Thus)
75'. 'f,k, N5'1.4- ..k..N +
50 kPa
Air
I Liquid
3m
kL..--- ~
FIGURE P2.73
I
ilNr-t'l---,d.
.fon 30·:= ~
I
d = :L -I:~" 30 •.: /. /55'IIM
/Z8--kN
and -Ine. it' "'c. I£) .., 111e. c...on e. h1(5 a. rn t:t,n I .f.1A e/-e. ~.f.. /2. 8'--k IV
and I~ d,,.~c.I-~,{ vey.flclI/l!:1 d~W}1W4Y'~1 Q/~I'ff the C-!Jne. ~)(.;~.
2-7'-

2.7J.f.
2.7If A 12-in.-diameter pipe contains a gas under a pressure
of 140 psi. If the pipe wall thickness is i-in., what is the average
circumferential stress developed in the pipe wall?
i=;" e6u/ }/}/"),1"1'um
()f the. pipc.) )
(ieI' 4 un;! /"'fllt
Or
·At/~ ~
rr =
1?.
( I 'fa Jh
)(G. I~. )-I'I? .~
- I / .'"'Ii /11, )
.
= 33~o pst.
2.-77

J.7E
2.75 The concrete (specific weight = 150
Ib/ft3
) seawall of Fig. P2.75 has a curved surface
and restrains seawater at a depth of 24 ft. The
trace of the surface is a parabola as illustrated.
Determine the moment of the fluid force (per unit
length) with respect to an axis through the toe
(point A).
The CompDnen+s 01 tne fllJ/d .ft,rce (Ie ·/-JiI,
(!)I') The (pllli fire If t:ll1d W (J S S h()WI1
(HI -!-he fljuY't. where ..
F;= ?f~(.A == ('If.OIt3)(;J.~t)(2'tliJ{l+t)
= J~Lf()oJ;
.4-150 )
W -= ~-if-
Clnd
To d~.feY'rHJne -¥ -hne/ IIrea B CD. Thus)
(s e~ .It9lire .fo ,,;,l1f) ~
/J. =- !;~'/-!1 ) dJi. = ((,2; -".;;;x"') dx
o Jo '
:: [~ Jt x - 0,;x'~lQ)to
t(11 d WI th XI:: rl20 A =)
175'" H:2.
-p.L = .4;< I f-l: -- '7~- h 3
---15 ft ·1
FIGURE P2.75
T
:l'f-Ij
1
Jt./) =~
( /'I()';'~: All l-tn(ths I~ II: )
So 1}ud:
Thus) I. /j,
<tv = (to If. (; 7t:l ) ( 1'15 It 3) = I~ Zt!)O lob
To /~cllie Cft1ff'()/~/ 0/ A : J&"
'-<. A =- f::dA = i1~'f-'J) x dx .jj.1¥x - a, 2/) d>< =
o
;t. =(.
/~ (~) t _ (). 2 (Vj"Z;,) ¥-
t
J7S
4.11 tt

2.7b A cylindrical tank with its axis horizontal
has a diameter of 2.0 m and a length of 4.0 m.
The ends of the tank are vertical planes. A ver-
tical, O.l-m-diameter pipe is connected to the top
of the tank. The tank and the pipe are filled with
ethyl alcohol to a level of 1.5 m above the top of
the tank. Determine the resultant force of the
alcohol on one end of the tank and show where
it acts.
1
r~r------''------.
2..0tfll
whe~e ie = /. !;J-1tY) 1- I. /)/ffIt = 2. 5'"." ~
SO -tnA t
fR ::- (7,7if~ ) (2,5"~)(;:)(-2. OAM) '-= ~~. j ~ JI
AIs())
( ;}. S-NY! )(r;)(Z~)?
Thus) the restl//q"t "wee hilS q rn4'Jl1lfude "I 6~. 8-k)/
a11v{ ~C.f.s of a ellS fttl1ce "f ~ -':f~ = :<. /,O/M - :l. S"D hI1 = O. /O[)/'IH
bejpw ~en feY' 0 I fRill< e h c/ L/.)O/;.
2.-7'1

~. 77
2.17 If the tank ends in Problem 2.7b are
hemispherical. what is the magnitude of the re-
sultant horizontal force of the alcohol on one of
the curved ends?
~or e 3";)/j,,.u;'m J
;;;, = J:;." // (U., ;;,w,,)
SInce.
SC/me
s(>/u 11,,0 (." hOn)Mfe, I !ere",
4S .f,:,,.. PNblem 2.1'0.
:< ./8
2.7 & Imagine the tank of Problem 2.7(. split
by a horizontal plane. Determine the magnitude
of the resultant force of the alcohol on the bottom
half of the tank.
(onsJde, t>. fne- b(K/!rdl;'f~ Df
Dojl.,m 11.1/ 01- "",,,); [Sp, H~~Y'<")
where:
1> ~ fr~'suY< D/ ~/u;tI CJH h";~M/.1
1./ ~ l1...e. .1 I1.N,MI" I pl'H"-
OW ~ wel9i1f of. volu",. ./ l/"i4 IH
h./f' of !<rill:.
~ 1
- I
1
I
I=f ~ f'£'Sulk,,1 ~".ce e"",y,t.tI by *,,,1: 0" 1'1.,4
Ii". .eSu;;'bNvm (rf'Hr -10 "nohlt", 2.80 /;, #"i< d"ne_SicnS J,
fi: jA 'l'W = d' (I.~""./"")(:l"" d",,) + ?r [U;)(2~)(~...)J
(-7.71t ~ )(~O",,3T ~7T,,")) = 203J,.IV
Tht-lS, fo~(e t>{ alcohol DM +P.1k - 203 -k IV dlYe~+ed
verhcqll'j doUJn1()4 rd.

l.llf A closed tank is filled with .wat~r and has a 4-ft-
diameter hemispherical dome as shown m Fig. P2.79. A U-I~be
manometer is connected 10 the tank. Determine the vemcal
force of the water on the dome if the differential manometer
reading is 7 ft and the air pressure al the upper end of the ma-
nometer is 12.6 psi.
~/" epidi/; r,uJ'YI,
2- F 1 I ~over-nCa
;So tit..t
Fp: -plr - 1v
Fe, -I~
.
l' I':'
Fr"lf)m 1'he m {(nPIli e teY')
-the de>me ex.ri-.
pl"e H""... af -n..
~ -I ~f (7.ft.) - d"J./." (If ft) = P
So -thAi;
t-pA
(I )
Oil 1h. 1-0/4
h,.u of. 171( dt>mf.
f = (;2./. ~.~ )(J'l'f ~: ) + (~.)('2¥~.)(71t)-(iZ. ¥~.)Htt)
Ib
= Z ~ 8~ k'"
fr4>m F'f, .r;) w) n, 1/0IufYI••1 Sphfr. = :r(d/~m(~/Y' ).3
Fe, = f2 Ho ~~ V;;Ylfftl - t [:-(iHf)J (~2. 4~.)
35: /00 Jb)
The -Crt<. 1hot- 1he v.rl-I<..I ~r(. -tn.+-
3S; /00 Ib t.::~",)(",e!.rl~~_~"!!.h_7h=,,=-....!!d.~om~e,--I~
2-V 1

2..80
2.S0 If the bottom of a pop bottle similar to that shown in
Fig. P2.71 and in Vid(>() V2,4 were changed so that it was hemi-
spherical, as in Fig. P2.80, what would be the magnitude, line
of action, and direction of the resultant force acting on the hemi-
spherical bottom? The air pressure in the top of the bottle is 40
psi. and the pop has approximately the same specific gravity as
that of water. Assume that the volume of pop remains at 2 liters.
4.3-in. diameter
11 FIGURE P2.80
t;Y'(<( ~ W<t,"-i. o/- pcp SlAfp"r.J:rd b~ bo}h,r>1 t k'f"Ct!
dU( i;o Qlj,- ft"tssur~
WeI ;h t 0 ~ pdp =?r )( v0 I~mof c I pof (II
Pdp 1 .f.t" )
V()/UhI(' = z. I, f.ey.s == (2..x. J0- ?'m'I 3) x (.3S3( x I{) ;;;) : O. 07oj, ..f.t3
Thus1
~11'l E"1.(I)
W.el!hi- #/- )'IP/, : (6z.,.R,)(()·~7~b Pt
9
) = Lf. '+J /b
FDYc~ c1/1e to til". pr~JS/lre :::- .J.... x pr() '~c..1Pt;I aveA a.{...-/;11" c) , • L l.J.
YH'It1"pheYt(a/ OOJl1!)1')'1
= (LtD ~.~)(r){~.31·n.)2-
::: 58 J Jb
Re~~1 t.At1t force. = Lf.'l-I J;, + 5~J Jj, - b-2b II,
Tn e rt'su) tQIf t: ~Y(' t. /5 dItee.ted ()t vtica J1lf down lOtIrd)
CI n d due .J.o slfmme tY'11
/t aC':f3 ()Il 1Jt(. hem/5pheriC41
bt!' mm a/~n~ -/)Ie. verft'c~d t2J(.jj ()fhle bot+fe.

2.31
2.81 Three gates of negligible weight are used to hold back
water in a channel of width h as shown in Fig. P2.81 . The force
of the gate against the block for gate (b) is R. Detennine (in
tenns of R) the force against the blocks for the other two gates.
(a) (b)
Fi>r C4$e (.J:, ) • FIG U REP 2 . 81
~ =(rhe A:: d'l4 )(hxj,j = ~b
411' 'J/(:: -f: J..
Thus;
:FMJ.I=o
:>0 "/h,d hR = (~ l ) Ii
(c)
i R:: ( ~ It. )( }::J, )
_ "Yhlb
R..-- 0)
3
For case (el.) On free- bt)c? - c/utfr41?J s),owl1 ft.!1
F;.=tY~.b ( f-rrJh1 1fI~"Vt') 4114
IJI?::- --t 1
t:?11"
~.:: 6)( -Val
':
J' [1T~)~~)J
:: 7TJh2
j,
lie,
Ihu$;
Z /vIH =0
~D th"t
'&J (~ - ~;) 1- !k~~) ~h-
2.-3.3
.......--.....
T
F/<.,
Fa

2.%/ I ((.~nr.)
.tf I>II~IU.J 1k,i
Fa :: :r),:Lb (t. 3f~)
FrtJl1'I ~f' OJ J-h2.£ =-5R./ th".s
F/3 = 1.17R.
j;r C!ttse (C).J -kl" -the !ree- bal~ - dt".J'rl//J1 .sh~Jl4l1t)
-the /f:,Y(f! r:. tJl'7 1h~ eu",.ve~ sec,.,T::It;11 p~s.Jes 1h;"u,Jr..e,
fh e hI Yl1e ~n If( 1ht'I"e-hYe d()es /1ld. ~l1fr":J/.(k -h
f;, <. m "me" i ~Y()III1I1 /I. ()n b~fltul1 />4rl ()/ JIlI:e
lif..: ~he It ::: t (~Jfh)(f x b) :: -ph2.b
he
t!fc
-d:(h)f-z
h).3
ah- -t-!f(; 1/
(~)(-f xb)
-Lf
28 h
3~
Thus
) 2}1U=0
~(it~j~~h
iJr f8~(4r426)~:) =fir6'///,
Fr6Jm j;:'~. rJ) J'h"},h;;; 5 eI -fh u.>
Fa =-;,e = !), irS/<.
4~
I )./.'"
I--~
1,
"""
I7e
!fR.1.
1J
-- ~z~
,
Fa

2,12 I
2. g2. A 3 ft X 3 ft X 3 ft wooden cube (specific weight =
37 Ib/ft3
) floats in a tank of water. How much of the cube
extends above the water surface? If the tank were pressurized
so that the air pressure at the water surface was increased to 1.0
psi, how much of the cube would extend above the water sur-
face? Explain how you arrived at your answer.
h y e$U~ /, 'br/U n1)
-n;z:Frrr+IC'1fJ : (J
So -thAt --_--+-1"W~ -~ i~
~ - FI3
ThUS)
(57 i!'t3 ) (3 .ft)3:(2,1. If1t3)(3It-h)(3ft i3{t)
{= j, 22 -!-/;
1:1 -the (IIi- pressure til- the W~';"r sur/lice IIIt.Yf4!fs
/:;'1 / P~/ tntr( WI IIIR bt 11() Chllnfe 1/1- /h( he,!Ai
~f the ttlb~ ((1111/" tht. IdAter ..5'tlrhct. The Il1trellst'tI
frt'$j",r~ /z:,r("~ ()/1 tAt -t-~f 0/ 'i7te e"b~ ,; bp/4lfCeA b'l
tin "',tiP I -force 0}1 the btFmm b/ 1ht ttt6e 051 flee Iht'
Jurhc~ fft5SIIYt '$ rfih,j/Jf;jff;d -11lr~"!h~lJ,i 1lJe .f/,,;'ri .
2 -8"S-
3.?l:-h
--=k

2.83 The homogeneous timber AB of Fig. P2.83 is 0.15 m
by 0.35 m in cross section. Determine the specific weight of the
timber and the tension in the rope.
W= ~ Jf WheV'( ~ is tHe.
~fe,jfj,- w(l~ht af The -h'mh.eY'
al-lp( -tf' /~ ;fs lIa/"me Thus;
/J:; }( (0. J5~ )( ~.3' /It4 ~ /01lM)
= ()"Z, t
~or e~u'd /brJum )
IM =0
A-
So /ha.t
• FIG U REP 2 . 83
Vi (103,1WI ) c'os0( = Fe (8:, )Cos J..
Cit'"
(t?,5"Z, ~)(S"1'fYI) = (P·IfU) d'Uz.o )C yo ~ )
~ =(,?,Jf2t»( UD ~3 ) ( ~ "" 1
(tl. 7"1.;-) ( 5" h" )
:50 trt,d:
T = ~ - 1/= 6,lfzorrnJ
)('1.80 ~~) - (o.{j]"tn13)(~.27~,) = 1ZLflV

Z.iJ.f. I
2.R4 When the Tucurui dam was constructed in northern
Brazil, the lake that was created covered a large forest of valu-
able hardwood trees. It was found that even after 15 years un-
derwater the trees were perfectly preserved and underwater log-
ging was started. During the logging process a tree is selected,
trimmed, and anchored with ropes to prevent it from shooting
to the surface like a missile when cut. Assume that a typical
large tree can be approximated as a truncated cone with a base
diameter of 8 ft, a top diameter of 2 ft, and a height of 100ft.
Determine the resultant vertical force that the ropes must resist
when the completely submerged tree is cut. The specific grav-
ity of the wood is approximately 0.6.
h;Y esfA /!J ],rJU(YI J
Z. p: :::.0
ver·hctC J
so -tnI., -I::
T== Fa-'W
/7;y a. fruflcabd ~()J1e)
0/"me:' j.b (IJ ZT Yj Yi. -t Jl'z.3. )
Whe~ : r; ~ bl(S( Y'tJ..d I us
Y'l. :: i:cp rll dI u.s
h ~ heIgh i.
(/)
2v 'V we,~ It t.
Fa;.l bt.t,,~an I: ft;rCt!
Thus) ¥. =(-rr) (IDbh)[Ctt.fl:/+ ('fit J( 1ft) 1- Uul]
+ve~ 3
_ 2200 -Prj
0: J( 4'- =- (CJ.,){t,z.1f ~~)(l.Z-ooft.~ =81- If,,ol~'L~ +~~ n ) I
7Y~e
E'l. {J)
T == / g~ I)()O / b - 9A) Lj()a Ii,

2.86 An inverted test tube partially filled with air floats in
a plastic water-filled soft drink bottle as shown in Video V2.S
and Fig. P2.86. The amount of air in the tube has been adjusted
so that it just floats. The bottle cap is securely fastened. A slight
squeezing of the plastic bottle will cause the test tube to sink
to the bottom of the bottle. Explain this phenomenon.
-~~Air
Test tube
Water
Plastic bottle
• FIGURE P2.g("
" :' • .L--_--..lI..,•
. .......
, .'

2.87 The hydrometer shown in Vidl'O '2.6 and Fig. P2.S7
has a mass of 0.045 kg and the cross-sectional area of its stem
is 290 mm2
: Determine the distance between graduations (on
the stem) for specific gravities of 1.00 and 0.90.
When ine h!Jc/~metfY i.s 1/"t1.tll1'i ;';'s we/rAt; w)
fiJl'lI/!he butP!1lt"i .f"Y'~e. Q~e e$u(J/ ~/J1C~
2: F: ::;0
v(yii (4 /
h,r IJw"t/ wi1h 5G -= o. q
I )
~I =W
• FIGURE P2.~7
01"
(SG, ) (~zo)-t7 ( J.Jhfre t ;t. @ If·C)=w I+3.D JJ,. 0
~Jml J4rJIj ) ft,.,. Ih~~d w/th Sq1,. -= /.0)
(SG~ ) (~~O )~ = W
ClI'1t1 ~wbfr~c.t/r1~ '1}1, e$uah6?l1s !:J~eJd.s
-if t:- V..! - 'W 1
J :l. (SGJ('O'tI-:z.o) (SC71.)(Y~~O)
S/~C~ -t1 -it = LJ~ .45
-(-~_/_-[ S~, - tGJAs ¥~hCl ')
_ (tJ."lfsJ1 )(r·E'JTa- J [
(290 Jf. I()-,"" ~('I. 81 ;< 10:J!; ) -0-.cr-
-2
I. 72 ;(. I0 I't'Y = /7, Z "" m1
Let
QI14
A "" s+em (ll"ftt.
s
¥ N jubmp~ed
vt>}t..Ime..

2.SS An L-shaped rigid gate is hinged at one end and is lo-
cated between partitions in an open tank containing water as
shown in Fig. P2.88. A block of concrete (y = 150 Ib/fi') is to
be hung from the horizontal portion of the gate. Determine the
required volume of the block so that the re-action of the gate on
the partition at A is zero when the water depth is 2 ft above the
hinge. The gate is 2 ft wide with a negligible weight, anrl t.he
hinge is smooth.
• FIGURE P2.8a
r-;y esu;/d",U/11)
I. MU =-0
.56 ~1:
~ J-1 -t ~ J..2- ~ TJ..1.
where,'
(I)
'1= ~h,-f, :: ('2,~~,) (.3t1:)(zA4=~ z.hJ
: 7 Lf-9 1.1,
J:;.. :: ?rhi!Z ,4z =~2. ~ ~3) ('t4t) (Zft"- If:tJ
T
~I
1=1
- fCf8 1.1, J.. () 'j
L:f = r,<.c.. -t ~ :: 17.. 7.+t (2.fi) -t.3 H. = 3.1/ +t.
RI ~c.. A c:. (.3.ft) (2.~t I. z. ft)
.Q = ~~ - z. +'.f .= j. JJ fi - z. t-t. :: . -ttI I
J.l. = /H
Thlls) fr,nn e:a~. (J J
(7'f'f IJ,) (1./1 fC) -t- (t?f8 J1,) (j ft) = T (I fiJ
T= 18.30 11
f;r ~ncre~ hJocJt..)
s" th,t-
L t=""V(lo't. c. 1.:.0
0 'r ~c. ~ Tl' Fec..
d: -Y: -= I B30 I~ l' 6'"u, Jfc..
" '- J-.{)
I gj 0 Jb
41:=c..
2.-10
T

2. 81 I
2.89 When a hydrometer (see Fig. P2.87 and Video V2.6)
having a stem diameter of 0.30 in. is placed in water, the stem
protrudes 3.15 in. above the water surface. If the water is re-
placed with a liquid having a specific gravity of 1.10, how much
of the stem would protrude above the liquid surface? The hy-
drometer weighs 0.042 lb.
Whln f'/,( h~drometfY IS I-J,,4.'l:Jr,!J
/1-5 weI1ht.) 'h.J) J,i hAi4J1teei. ht;1
-fhe bu/)~I(" t- k;yc. e) r-13 , J=o r
e~ /Ai JIbn um I
~FL =0
Vfy/-t~al
IhJl$1 ~y fA)A,l-ev
~=~
(~t-o)-v;- = 1;) (I )
whevt!. -yt Is tne. JuJ,merged volume.. w(1l1 1l1e new
J~iu'd
(s 6) ( 5Uz,o ) 1{ :,ew
C~mb,nJn~ cIs. (IJ 11I1I((Z) wJin ~
(rfJ.2.() ) -r:- ::: (SQ ) ('ritz,() ) 4f2-
~ = -¥,
~ 56
1-1/
U)

J. '/0
2.90 The thin-walled. I-m-diameter tank of
Fig. P2.90 is closed at one end and has a mass of
90 kg. The open end of the tank is lowered into
the water and held in the position shown by a
steel block having a density of 7840 kg/m3. As-
sume that the air that is trapped in the tank is
compressed at a constant temperature. Deter-
mine: (a) the reading on the pressure gage at the
top of the tank. and (b) the volume of the steel
block.
Water
Air
3.0 m
~
FIGURE P2.90
Let 1f-:: At ~
(Jhd
1;:: ~ (-£.-lJ. to ) 't" ~~th» {wlter~ «II
T}lIASJ
Smce.
-+..-+/'..
~ = A ~ ::- l. "
t-
f:}.
fL' =- fat_ lfilf~ -b'-. = A--t:(3 )l..
-{ 1 ;- (p.-I-; _0.,,){ -
./..11
9. go -:;; ~ 1:1-l1l'i :: ItJl ~ PI).. I
=0
~ 2. + ( /()/-iPIL.. ) G
3 ('0 I .4..p~ ')
n - 0.' /Itt 1t-
9.8/J ~hf
or
1. :l -t- Cf.7 I ~ - .3 0.1 :: 0
1.71 -r V(r.7} )1+ ~ (~{),9) ::' :2. S-3hYl
:L
Th us.l +tt»rJ ~6' (/ )
if. (1tA.9<) ::'
.2. -12
U)

(b) F;Y' e'6l1; /,'hY" ~,.". 0 I i'/11it. (s@e Ir(~- b~df:J -
c/La 9Y'II m ) J
T:: ~ At -<W-e
wheY'~ ~ N ·hr(l1~ we/'111 t / t'I J;/ #( .k ,...
5h~ / 6/()c,k
Thus)
-¥,-=s
T
3
- 0, 208 I'm
),,-93
y

;;, 'II ~J
2.91 * An inverted hollow cone is pushed into
the water as is shown in Fig. P2.91. Determine
the distance, C, that the water rises in the cone as
a function of the depth, d, of the lower edge of
the cone. Plot the results for 0 :;; d :;; H, when
H is equal to 1 m. Assume the temperature of
the air within the cone remains constant.
! T·'2 /! l~
Le
H
d w",,· cr.~
Open end ~
-----+----
f-R--I
FIGURE P2.9
I-emfJe~4.l:uye Cf;tr1pt'es>/o~ 0 I
.
tJ, e. " I r tell11,1J1 fire. O!Jne,.J
~. Vi = ~ -v; (I)
whev-e ¥ 15 the. (1/; V,,/"'lne. ClIfU " an" I n! Ie", +r, ,;;,.fuj/ and,
.fln/ll :s--ta ~ J respecltve/!J . .:t".f ~D flow.} 111.f (.s('t! f'JI-(~):
1:..:: ;{ilM
~ = ~ (d-,fl) + fa.~
11 = 7fR'H if =- 'J(H/)~>(H-.l) : :r:Y(H-Jl
Th(,is, .from Et. (J )
t (7r;<211) = [6' (d-.L) + tim<] :rP:t(1+-.1) 3
41-""" .3
wh/c.h 5ImfJ,·he.s -1-0
1 = d - 1'1-- [(!-i/-I]
WheY'e 1. all" d 4Ve
I-t ~(!JmflilfY ji:Jt"tJJNlm
t:> f- c/.
J~ I"n1 e..f.eY' S .
k,//OIlJS ~r Cti/c.J(IaII~
I
( C(f)11 t )
(.3 )

~.91 "I
:00
110
120
130
135
1'"'0
150
160
170
180
185
190
200
210
220
230
( C~1i't )
print "**********************************************"
print "** This program solves iteratively a fourt.h **"
print "** order equation to give the water rise, **"
print "** 1, as a function of t.he depth. ,j **"
print "******************************************t:***II
print.
print" Depth, d (m) Water rlse, 1 (m)"
for d=O.O to 1.01 step 0.1
1=0.0
if d=O then goto 220
lp=l
1=1-((d-lp)/10.3+1)-(-1/3)
if abs(l-lp/l)O.OOl goto 190
print using" #.###
next d
#.####":d.l
*********************************************~
** This program solves iteratively a fourth **
** order equation to give the water rise. **
** 1, as a function of the depth, d **
**********************************************
:§:
.....
.;
In
·M
L
L
<lJ
.jJ
'"~
Depth, d (m)
0.000
0.100
0.200
0.300
"10-2
O. '.,!-OO
0.500
0.600
0.700
0.800
0.900
1.000
Water rise, 1 (m)
0.0000
0.0031
0.0062
0.0092
0.0122
0.0152
0.0182
0.0211
0.0239
0.0268
0.0296
3~----------------------------------------------~
//
2
1
o ./
0.0 0.2 0 . .:1 0.6 0.8 1.0
Depth, d (m)

~.92. I
2.92 An open container of oil rests on the
flatbed of a truck that is traveling along a hori-
zontal road at55 mi/hr. As the truck slows uni-
formly to a complete stop in 5 s, what will be the
slope of the oil surface during the period of con-
stant deceleration?
:: d:t :::
cJ~
.//1'111/ we/~c"~ - I~' ~/.f J ve/~,;..J::,
-61,;, ~ I;'h ,./Il, I
o -
9. rl ~ + C>
O,50Z

( 6 )
2.93 A 5-gal. cylindrical open container with a bottom area
of 120 in.2
is filled with glycerin and rests on the floor of an
elevator. (a) Determine the fluid pressure at the bottom of the
container when the elevator has an upward acceleration of 3
ft/s2. (b) What resultant force does the container exert on the
floor of the elevator during this acceleration? The weight of the
container is negligible. (Note: ] gal = 23] in.3 )
T{
xl ..po
illillI' I, '11
-a. A=VDll-tl'Yle..
-t (120;~,1.) : (b~tAI)(~~'al;I1~)
{, = C/. ~ 3 1'11.
iJ, - f (~ -t Qi ) -{
- (::1.4-11 sluq.s)(3 2 2 £E 1"" 3~t )( 1.'3 ft)T ft~ , $4 52. 12.
=
Fr~m frte- bcd!! -dla9NlfYJ or CtJntaI~e r)
~: 1J, A
LEJ( Jb.) ( · "1)(' +-I.' )- /2~ Ih, -
- ~i.rft'l. I 't4-"~:-
1Ff
:: 57. 4- 11
Thus; tcrle 101- ClPYJ-I:alnty f JOt} r
. S/.If. Ib dOWhWQY'cJ.. •
~'"
J~
2 -'17

2.qIf- An open rectangular tank 1 m wide and
2 m long contains gasoline to a depth of 1 m. If
the height of the tank sides is 1.5 m, what is the,
maximum horizontal acceleration (along the long
axis of the tank) that can develop before the gas-
oline would begin to spill?
-
T
t.y.4
---
------
--d~ _
cd;, <
I. 5 IMt - J. 0 ,4.t1
JIN
== - O.5'C> -L '-______----1
(S{!e +lj~re).
d -l: =
dtJ
T

2•.95 If the tank of Problem 2.'ftt slides down
a frictionless plane that is inclined at 30° with the
horizontal, determine the angle the free surface·
makes with the horizontal.
Fre>m NewtM~ J mel Jaw I
I
L F~ I =' 1m a!J
S/~'e, -tne. ()11/:; ~Ne In -th~ ::; ~ dl;ec.:l-toJi
I:; the. e{)m!()l1ff,.i 01 kJel9J1f (rrr.;)sjI18)
( trr1 J) Sin ~ -::. IYY1 a. ~
s" 1h~i
, .
a~ -= - a!J SJn~
=
- -
(4y,d loY' G Z ~D 0
d=1:
- =d~
I
- ~ Sln'Z6>
I • I •
/- 2: 5111 ~O
+an Q= O. 7~ t.f
f= 37.4-°
(I:~. 2.28)
-=
m us~ ot ~k ChId
'}aso 111"1 e..

2. Cf~ A closed cylindrical tank that is 8 ft in diameter and
24 ft long is completely filled with gasoline. The tank, with its
long axis horizontal, is pulled by a truck along a horizontal
surface. Determine the pressure difference between the ends
(along the long axis of the tank) when the truck undergoes an
acceleration of 5 ft/S2.
1:. - J; ::2. I
=-
I·
(2t. 2. 2S) C~:
0)
- t tl~ (2."1- k)
- (1,3:2 s';/; )(s ~: )~". ft)
I/J
- Jse -1-t7..
2-100
,2'f .ft .-1,.
Oht1.)
"'7
a - S'-t-i:
!1 - Tz..

2.91 The open U-tube of Fig. P2.0!1 is par-
tially filled with a liquid. When this device is ac-
celerated with a horizontal acceleration, a, a
differential reading, h, develops between the ma-
nometer legs which are spaced a distance r. apart.
Determine the relationship between a, r, and h.
d.:r a..'::J
-d~ ~T C(i:
51!Jce)
ar-. ~
- "7- ..
d!J
+he" -~ a.
=y: 3-tCJ
Or
a.~
~ - -J.
a. =a~
hFIGURE P2.''1
..2-10/

2. q'ii' An open I-m-diameter tank contains
water at a depth of 0.1 m when at rest. As the
tank is rotated about its vertical axis the center
of the fluid surface is depressed. At what angular
velocity will the bottom of the tank first be ex-
posed? No water is spilled from the tank.
F;;r f;~e surface W I in -{ =0 0.-1: r =-0 J
{= w"Z.l-~
,.,
The VOi"fme. of .fluid I~ Y'ok-t'n~ 1-qt1,k is ~Vel1 1':1
~" f:l:Y -i1 dy -= :>.rr w" (~3dr = rrw~fC-,!
D ~~ Jo '13-
S/nce. -the. J'n ,'-I-t'tJ 1)0/ume) ~. = rrR'2. hl.·) rn uS 1.: e~ ua I the +1n4J
1r ~'2!!:. ~.: 1r/?'2.~~
Iff}
CU = -~ If ~!" -, =- - ( 1f{9.?I-f,. )(0,7",,) I
V (O. S/tn ) 7-
2..-/02.
raJ/0.5 _
5

2. qcr The V-tube of Fig. P2.Q'1 is partially
filled with water and rotates around the axis a-
a. Determine the angular velocity that will cause
the water to start to vaporize at the bottom of
the tube (point A).
fY'e SSWI! 111 D..
-the. 12 DI.l .. i:.~ t:J 11. )
the (!.4)()Y'mnA.fe oS '1si-e1Yl ~h()wn)
--p=o a. 1:: 1"'= ,+Ii? 0111/ z:-
2(,+ )2-
e.(!;n5t~l1't = -. ~ IA) Ii It ~ ~ (~~.ft) _
'l.
At pO/flt A) r=o 411'" i-={J I (0111
-R = - f~l. t ~
A /8
If fA = vapo'y prfSSlAy-e - O. "Z~-~ psC:a...) ~Y'
a
FIGURE P2.Gj~
U)
fit". (0. 25'1. pH· - l't.7P$" )(I 'l-1f.iJ~~).: -:l. ago :.~ ftqr)
w=
- 141 ~s
=
.t..-/03

d.. }OO I
2.100 The U-tube of Fig. P2.100 contains mer-
cury and rotates about the off-center axis a-a. At
rest, the depth of mercury in each leg is 150 mm
as illustrated. Determine the angular velocity for
which the difference in heights between the two
legs is 75 mm.
The e DU.a..t/~H of. 1h~ free => Ur/4ce.
1.=
6.)"2.,...2-
-t ~tfs-/:q~ i
,;1..3
pfl~SIYl1
Thus} w2.
( ~1._ ~~ )l:A -18
- fJ~- - -:1.,
150 mm
t ~
-220 m:-:/.....·~1·-.....,·190 mm
I
I
a
FIGURE P2.100
-tnrou9h A 4 nd 8 1.5
[ E 'h .2. "32 )
1
= -
v
l ('I. 'if f:a.) rD, 07Shtt)
(0, 220hrl)2. - (0. O'l{)",,)~
yad
to, ()If -
5
2-/bLj

:2./0/ I
2.10 I A closed, OA-m-diameter cylindrical
tank is completely filled with oil (SG = 0.9) and
rotates about its vertical longitudinal axis with an
angular velocity of 40 rad/s. Determine the dif-
ference in pressure just under the vessel cover
between a point on the circumference and a point
on the axis.
Pre~s"y~ In 4 rtJta.tIJ1, .flt-lltl /1(1 nes
/11 aCC"f'drNICe witl! -th~ £jU4-iJtOJ1}
A t- - --~
" - I . /
.tJti)2.;- 2..
P= (~ - d-i' -t ~l1s.f.toli
_ f-tJ 1. (t; '1. _ J-. 1)
:l. B,4-
- ({), 'I )(10' ~~) ('If) .~:,d)2 01. )2. ]J (0.2h11 - 0
:L
~g, ~ kP~

2. /02
2.102 Force Needed to Open a Submerged Gate
Objective: A gate, hinged at the top, covers a hole in the side of a water filled tank as
shown in Fig. P2.102 and is held against the tank by the water pressure. The purpose of this
experiment is to compare the theoretical force needed to open the gate to the experimentally
measured force.
Equipment: Rectangular tank with a rectangular hole in its side; gate that covers the hole
and is hinged at the top; force transducer to measure the force needed to open the gate; ruler
to measure the water depth.
Experimental Procedure: Measure the height, H, and width, b, of the hole in the tank
and the distance, L, from the hinge to the point of application of the force, F, that opens
the gate. Fill the tank with water to a depth h above the bottom of the gate. Use the force
transducer to detennine the force, F, needed to slowly open the gate. Repeat the force mea-
surements for various water depths.
Calculations: For arbitrary water depths, h, detennine the theoretical force, F, needed to
open the gate by equating the moment about the hinge from the water force on the gate to
the moment produced by the applied force, F.
Graph: Plot the experimentally detennined force, F, needed to open the gate as ordinates
and the water depth, h, as abscissas.
Results: On the same graph, plot the theoretical force as a function of water depth.
Data: To proceed, print this page for reference when you work the problem and click heft'
If!i FiGURE P2.. 102
(Con t)

~. / ()Z (ef!JrJ't )
Solution for Problem 2.102: Force Needed to Open a Submerged Gate
L, in. H, in. b, in. y, Ib/W'3 Ixc, ft"4
5.5 6.0 4.0 62.4 0.003472
h, in. F,lb F1,lb Yr - Yc, ft d,ft F,lb
21.1 10.1 15.69 0.0138 0.264 9.03
18.5 8.9 13.43 0.0161 0.266 7.80
16.2 7.6 11.44 0.0189 0.269 6.71
14.5 6.7 9.97 0.0217 0.272 5.91
12.8 5.8 8.49 0.0255 0.276 5.11
11.1 4.7 7.02 0.0309 0.281 4.30
10.1 4.3 6.15 0.0352 0.285 3.83
7.4 2.9 3.81 0.0568 0.307 2.55
Since h > H, A =H*b =constant and Ixc =b*H"3/12 =constant.
F =F1*d/L, where F1 =y*(h - H/2)*A, d =H/2 + (Yr - Yc), and Yr - Yc =Ixc/(h - H/2)*A
.c
Problem 2.102
Force, F, vs Water Depth, h
12 ...,....-----------..,...----.----,
10 ---.---.--.-.--'.---.~--- ---- -I
8
-II
-------1LL. 6 - j - - - - - - " - -
-I!
4-----~--·--~.LV-
2 .-------..,-- -------1
i
O+----t-----+---+---...,.------I
o 5 10 15 20 25
h, in.
ft- J 07
- - Theoretical
-a- Experimental

2./03
2.103 Hydrostatic Force on a Submerged Rectangle
Objective: A quarter-circle block with a vertical rectangular end is attached to a balance
beam as shown in Fig. P2.103. Water in the tank puts a hydrostatic pressure force on the
block which causes a clockwise moment about the pivot point. This moment is balanced by
the counterclockwise moment produced by the weight placed at the end of the balance beam.
The purpose of this experiment is to determine the weight, W, needed to balance the beam
as a function of the water depth, h.
Equipment: Balance beam with an attached quarter-circle, rectangular cross-section block;
pivot point directly above the vertical end of the beam to support the beam; tank; weights; ruler.
Experimental Procedure: Measure the inner radius, R1, outer radius, R2, and width, h,
of the block. Measure the length, L, of the moment arm between the pivot point and the
weight. Adjust the counter weight on the beam so that the beam is level when there is no
weight on the beam and no water in the tank. Hang a known mass, m, on the beam and ad-
just the water level, h, in the tank so that the beam again becomes level. Repeat with differ-
ent masses and water depths.
Calculations: For a given water depth, h, determine the hydrostatic pressure force,
FR = yhcA, on the vertical end of the block. Also determine the point of action of this force,
a distance YR - Yc below the centroid of the area. Note that the equations for FR and YR - Yc
are different when the water level is below the end of the block (h < R2 - R1) than when
it is above the end of the block (h > R2 - Rl).
For a given water depth, determine the theoretical weight needed to balance the beam
by summing moments about the pivot point. Note that both FR and W produce a moment.
However, because the curved sides of the block are circular arcs centered about the pivot
point, the pressure forces on the curved sides of the block (which act normal to the sides)
do not produce any moment about the pivot point. Thus the forces on the curved sides do
not enter into the moment equation.
Graph: Plot the experimentally determined weight, W, as ordinates and the water depth,
h, as abscissas.
Result: On the same graph plot the theoretical weight as a function of water depth.
Data: To proceed, print this page for reference when you work the problem and click here
Pivot point
Quarter-circle block
Counter
weight
)..-/03
III FIGURE P2.103

2. / P3 J
Solution for Problem 2.103: Hydrostatic Force on a Submerged Rectangle
R1, in. R2, in. L, in. b, in. g, ftlsA2 y, Ib/ftA3
5.0 9.0 12.0 3.0 32.2 62.4
Experimental Theoretical
m, kg h, in. W,lb FR,lb Yr -Yc, ft d,ft W,lb
0.00 0.00 0.00 0.00 0.750 0.000
0.02 1.11 0.04 0.07 0.719 0.048
0.04 1.58 0.09 0.14 0.706 0.095
0.06 1.92 0.13 0.20 0.697 0.139
0.10 2.51 0.22 0.34 0.680 0.232
0.12 2.76 0.26 0.41 0.673 0.278
0.14 2.99 0.31 0.48 0.667 0.323
0.16 3.20 0.35 0.55 0.661 0.367
0.18 3.41 0.40 0.63 0.655 0.413
0.20 3.60 0.44 0.70 0.650 0.456
0.22 3.80 0.48 0.78 0.644 0.504
0.24 3.99 0.53 0.86 0.639 0.551
0.26 4.17 0.57 0.94 0.0512 0.634 0.597
0.28 4.33 0.62 1.01 0.0476 0.631 0.637
0.30 4.50 0.66 1.08 0.0444 0.628 0.680
0.35 4.95 0.77 1.28 0.0376 0.621 0.794
0.40 5.39 0.88 1.47 0.0328 0.616 0.905
0.45 5.83 0.99 1.66 0.0290 0.612 1.016
0.50 6.27 1.10 1.85 0.0260 0.609 1.127
0.55 6.70 1.21 2.04 0.0236 0.607 1.236
W =32.2 ftlsA2 * (m kg * 6.825E-2 slug/kg) Sum moments about pivot to give W*L =FR*d
For h < R2 - R1:
FR=y*(h/2)*h*b
d =R2 - (h/3)
Forh>R2-R1:
FR = y*(h - (R2 - R1)/2)*(R2 - R1)*b
d = R2 - (R2 - R1 )/2 + (Yr - Yc)
Yr -Yc =Ixc/hc*A
Ixc = b*(R2 - R1 )A3/12 = 0.000771 ftA4
hc=h - (R2 - R1)/2
A =b*(R2 -R1)
l-IOCf

2. /03
1.2
Problem 2.103
Weight, W, vs Water Depth, h
1.0 -1------1------+----------.----
.c 0.8 +-___-+---____----i--_-+_---+-____
~ i0.6 --------------- .-.---~~----.----- ---I
I--I0.4
0.2 ~---------i---~,---~-l
0.0 -J-=:::::::::::..---.,.------+----t---------.;
0.0 2.0 4.0 6.0 8.0
h, in.
'1- II 0
- Theoretical
• Experimental

2. /o/f
2.104 Vertical Uplift Force on an Open-Bottom Box
with Slanted Sides
Objective: When a box or form as shown in Fig. P2.104 is filled with a liquid, the ver-
tical force of the liquid on the box tends to lift it off the surface upon which it sits, thus al-
lowing the liquid to drain from the box. The purpose of this experiment is to determine the
minimum weight, W, needed to keep the box from lifting off the surface.
Equipment: An open-bottom box that has vertical side walls and slanted end walls;
weights; ruler; scale.
Experimental Procedure: Determine the weight, Wbox, of the empty box and measure
its length, L, width, b, wall thickness, t, and the angle of the ends, e. Set the box on a smooth
surface and place a known mass, m, on it. Slowly fill the box with water and note the depth,
h, at which the net upward water force is equal to the total weight, W + Wbox, where W = mg.
This condition will be obvious because the friction force between the box and the surface on
which it sits will be zero and the box will "float" effortlessly along the surface. Repeat for
various masses and water levels.
Calculations: For an arbitrary water depth, h, determine the theoretical weight, W, needed
to maintain equilibrium with no contact force between the box and the surface below it. This
can be done by equating the total weight, W + Wbox, to the net vertical hydrostatic pressure
force on the box. Calculate this vertical pressure force for two different situations. (1) As-
sume the vertical pressure force is the vertical component of the pressure forces acting on
the slanted ends of the box. (2) Assume the vertical upward force is that from part (1) plus
the pressure force acting under the sides and ends of the box because of the finite thickness,
t, of the box walls. This additional pressure force is assumed to be due to an average pres-
sure of Pavg = -yh/2 acting on the "foot print" area of the box walls.
Graph: Plot the experimentally determined total weight, W + Wbox, as ordinates and the
water depth, h, as abscissas.
Results: On the same graph plot two theoretical total weight verses water depth curves-
one involving only the slanted-end pressure force, and the other including the slanted end
and the finite-thickness wall pressure forces.
Data: To proceed, print this page for reference when you work the problem and click hac
F .,/ Footprint of box
i t
1
L
1 ~
it III FIGURE P2.104
( ~Dn t )
~-III

2./()/f I ( Cf)r/t)
Solution for Problem 2.104: Vertical Uplift Force on an Open-Bottom Box with Slanted Sides
8, deg L, in. b, in. t, in.
45 10.3 4.0 0.25
Experimental
m, kg h, in. W + Wbox, Ib h, in.
0.00 2.06 0.942' 0.00
0.05 2.23 1.052 0.25
0.10 2.42 1.162 0.50
0.15 2.53 1.272 0.75
0.20 2.67 1.382 1.00
0.25 2.81 1.491 1.25
0.30 2.94 1.601 1.50
0.35 3.06 1.711 1.75
0.40 3.16 1.821 2.00
2.25
2.50
2.75
3.00
3.25
W =g*m =32.2 ftlsA2 * (m kg * 6.825E-2 slug/kg)
Theory 1. Including only the slanted-end pressure force:
W + Wbox = y*Vol
Vol =b*h*h
Wbox,lb y, Ib/ftA3
0.942 62.4
Theory 1 Theory 2
W + Wbox, Ib Pavg, Ib/ftA2 W + Wbox, Ib
0.000 0.00 0.000
0.009 0.65 0.047
0.036 1.30 0.111
0.081 1.95 0.194
0.144 2.60 0.295
0.226 3.25 0.414
0.325 3.90 0.551
0.442 4.55 0.706
0.578 5.20 0.879
0.731 5.85 1.070
0.903 6.50 1.279
1.092 7.15 1.506
1.300 7.80 1.752
1.526 8.45 2.015
Theory 2. Including the slanted-end pressure force and the finite-thickness wall pressure force:
W + Wbox =y*Vol + Pavg*A
Pavg =0.5*y*h
A =(b + 2*t)*(L + 2*tlsin8) - b*L =8.33 in.A2 = 0.0579 ftA2
2- /11.

2. /{)l/- I
I
I
Problem 2.104
f
Total Weight, W + Wbox, vs Water Depth, h
2.5 .
I
Ii
2.0
i
-~-------~--~--------r-------j
/ I
~'. I • Experimental
/.
.c 1:.1
- 1.5- , ,,-><
0 !
i I Theory 1.c
3:
, I
(slanted ends.. !
+
"
i only)
1.0 I
1---------------- ------ -~----i
3: / I
i - . - . Theory 2.
1
/ I (slanted ends
/ I and bottom
0.5
/ I edge)
;
I" , I
0.0
" , ' , , I
0 1 2 3 4
I
I
I
h, in.
I
I
~- 1/3

2, j O~
2.105 Air Pad Lift Force
Objective: As shown in Fig. P2.105, it is possible to lift objects by use of an air pad con-
sisting of an inverted box that is pressurized by an air supply. If the pressure within the box
is large enough, the box will lift slightly off the surface, air will flow under its edges, and
there will be very little frictional force between the box and the surface. The purpose of this
experiment is to detennine the lifting force, W, as a function of pressure, p, within the box.
Equipment: Inverted rectangular box; air supply; weights; manometer.
Experimental Procedure: Connect the air source and the manometer to the inverted
square box. Detennine the weight, Wbo,,' of the square box and measure its length and width,
L, and the wall thickness, t. Set the inverted box on a smooth surface and place a known
mass, m, on it. Increase the air flowrate until the box lifts off the surface slightly and "floats"
with negligible frictional force. Record the manometer reading, h, under these conditions.
Repeat the measurements with various masses.
Calculations: Determine the theoretical weight that can be lifted by the air pad by equat-
ing the total weight, W + Wbox, to the net vertical pressure force on the box. Here W = mg.
Calculate this pressure force for two different situations. (1) Assume the pressure force is
equal to the area of the box, A = L2, times the pressure, p = 'Ymh, within the box, where I'm
is the specific weight of the manometer fluid. (2) Assume that the net pressure force is that
from part (1) plus the pressure force acting under the edges of the box because of the finite
thickness, t, of the box walls. This additional pressure force is assumed to be due to an av-
erage pressure of Pavg = 'Ymh/2 acting on the "foot print" area of the box walls, 4t(L + t).
Graph: Plot the experimentally determined total weight, W + Wbox, as ordinates and the
pressure within the box, P, as abscissas.
Results: On the same graph, plot two theoretical total weight verses pressure curves-
one involving only the pressure times box area pressure force, and the other including the
pressure times box area and the finite-thickness wall pressure forces.
Weight
Th
1
IN-----------L-----------m
~-/JLf

2. I/)5" I
Solution for Problem 2.105: Air Pad Lift Force
L, in. t, in. Wbox,lb YH20, Ib/ft"3
7.5 0.25 1.25 62.4
Experiment Theory 1 Theory 2
m, kg h, in. W + Wbox, Ib p,lb/ft"2 W + Wbox, Ib W + Wbox, Ib
0.0 0.54 1.25 2.81 1.10 1.17
0.1 0.64 1.47 3.33 1.30 1.39
0.2 0.74 1.69 3.85 1.50 1.61
0.3 0.82 1.91 4.26 1.67 1.78
0.4 0.94 2.13 4.89 1.91 2.04
0.5 1.04 2.35 5.41 2.11 2.26
0.6 1.12 2.57 5.82 2.28 2.43
0.7 1.23 2.79 6.40 2.50 2.67
0.8 1.32 3.01 6.86 2.68 2.87
0.9 1.42 3.23 7.38 2.88 3.08
1.0 1.52 3.45 7.90 3.09 3.30
1.1 1.63 3.67 8.48 3.31 3.54
1.2 1.72 3.89 8.94 3.49 3.73
1.3 1.83 4.11 9.52 3.72 3.97
1.4 1.96 4.33 10.19 3.98 4.26
1.5 2.06 4.55 10.71 4.18 4.47
1.6 2.12 4.77 11.02 4.31 4.60
1.7 2.23 4.99 11.60 4.53 4.84
1.8 2.32 5.21 12.06 4.71 5.04
W =g*m =32.2 ftls"2 * (m kg * 6.825E-2 slug/kg)
Theory 1. Involving only the pressure times the box area:
W + Wbox = p*L"2
P = YH2o*h
Theory 2. Involving the pressure times the box area plus the average pressure times the edge area:
W + Wbox = p*L"2 + (p/2)*((L + 2t)"2 - L"2)

2. lOS-
Problem 2.105
Total Weight, W + Wbox, vs Pressure, p
•5 ----j
I4
I
• Experimental
:9
x
- Theory 1 (box area0
J
~ 3 -------~--
I
only):
.~+ Theory 2 (box area
3: i~
1
plus edge area)
2
I!
•
1
1
i
0
0 5 10 15
p,lb/ft"2

3./
3.1 Water flows steadily through the vari-
able area horizontal pipe shown in Fig. P3.1. The
velocity is given by V = 10(1 + x)i ft/s, where
x is in feet. Viscous effects are neglected. (a) De-
termine the pressure gradient, iJplax, (as a func-
tion of x) needed to produce this flow. (b) If the
pressure at section (1) is 50 psi, determine the
,pressure at (2) by: (i) integration of the pressure
gradient obtained in (a); (ii) application of the
Bernoulli equation.
(1)
f---- x
FIGURE P3.1
(a) -C'sine -M- ::: ev1¥ hut "e-::o Qntl V::IO(/+X) fils
~ = - f Vf¥ 0r ¥x-::: -f vPx- :::: -f (/0(I +X)) (10 )
ThtJs) #- ~ -1,9'f!f(}(lo!f)'J.(/+x) ) with X if) feet
.: - 19'1( /+ x) Ira '
fo- X" ==3
(b)(l) i == -19/f(/iX) so fhai Jo/t == -:19LfJ{I-fX)dx
I,::sops, XI == 0
1-- /b '1
or ~ == SO pSJ' - /'1'1 (.3 +.;})@ ( -)-'f;~i") = SO -/0. I :::3Q,9E/
///) + I 01/2. ~!7 _~ +.1- 1/2. }1:7 ,jh ~ -7
l." fJl 2" r VI -+0 ~I -1'2. :2.. ('Y2. +(/ ~2. "" or Wlr ~/-~:J.
A. .::: fJ; +-ff (V,2_ ~L) where V;; 10 (1 of 0) == loll
'vi ::: /0 (/+.3) ::: '/-0#
Thvs; "
_ ..J.. I/. OLL .s~) II ,. LL 2.)JL ( / ·ft~ )
'2.-EO pSI, +:z. (••7T 773 "/ tj 0 -,fl s::Z 71f'/-;":/. = 3 P, PPJ/
3-/

3.2 I
3.2 Repeat Problem 3.1 if the pipe is vertical with the flow down,
,
.~
:.
W
~
N -
(a) - asine - M- = f v*f with e:::: - qOO and v=io(JtX)!I
if = -p V¥S + If {j r *;%-f Vif +0 -: - f (10(J +x)) (J0) +0'
ThvsJ -*::: -L9~ ~:"(IO#)2(f+X) + 62.¥ ~
- -/9~ (IT x) .... 62.Jf #p
J
wNh X in feet
(b) (i) if =: -/9'1 (t +x) + If)..'! .so t60-£ ~ '"J;;;/fX)+62.1'Jdx
f, :;SOptl/ XJ:::O
f) . (..3:z.) Ih ( I ft l. ) lb ( I If 2-
or T2. = .5 0PSII - 19'1 .3 +~ fi'- i~ifii." + 6:L.'f (3) ff2. /'1-'1 i/J,'J.)
= .50 -/0,/+/, g ;:= 4-/.2fJS/
(il) f1 +J..p~:J.. +'2i =;2.of dp~2.+!"22. PI' w/I/J '&/::: 0,) Z2. :::-311
alld 10/::: /f'J(1 +0) :;:; /0# L{::: /O(ff3) =-~IJ.:fiJ J
f{ :::1/ +i. f (14"'-~2.) - 022-
= SO!s/ +i (I.?/f :jp-)(J01_ lftJ'") - t2.Y. #3(-.3 ft)
:: ,1./ /. 'lfJS/
3-Z

..3..3
3.3 An incompressible fluid with density p flows steadily
past the object shown in Video V3.3 and Fig. P3.3. The fluid
velocity along the horizontal dividing streamline
(-00 ::s; x ::s; -a) is found to be V = Vo (1 + a/x). where a is
the radius of curvature of the front of the object and Vo is the
upstream velocity. (a) Determine the pressure gradient along
this streamline. (b) If the upstream pressure is Po. integrate the
pressure gradient to obtain the pressure p(x) for -00 ::s; x ::s; -a.
(c) Show from the result of part (b) that the pressure at the stag-
nation point (x = -a) is Po + pV~/2. as expected from the
Bernoulli equation.
• FIGURE P3.3
rQ) t-= - ~V.g. where v;:Va (J. +f)
ThvsJ 41 == g::-
or
VA fL·
x2. .
¥s=: ~=-e~(J~+)(-4l)::
x
faltj {f,. + ?)J~(b)
f X
{dp =f #dx =
po
or
P-fo =: eaVo2.[- f
-00
I a I- 2: XI
X
-00
(c) From part (h)) when X=-a
1'/ ::; 1'. - fit 11.
2
[- i qf&]
xc-a
Frt)m the Bern()vll/ eqvatiDn
where IV, =V =:: Yo (J +f-la») :: 0
)(~-a
-rhll~ '" =: PrJ +ifVo.2. 4S expected.
3-3
.x

3.5 I
3.6 I
3.4 What pressure gradient along the
streamline, dpfds, is required to accelerate water
in a horizontal pipe at a rate of30 mfs2?
where
ThvsJ
e:::o and
VPs- ::: a.s :::
M- ::: - ('as:;:: - 999.!;!:s (30~) :;:: -30Poo(~)/m
or
~ = - 30.0 kPa/mdS .
3.5 At a given location the air speed is 20 m/s and the
pressure gradient along the streamline is 100 N/m3
• Estimate
the air speed at a point 0.5 m further along the streamline.
If neqlecf qrQv/fy 1$ :: - D vlY or) I S I C)S
or ~v N
- = -Ioo~ /(1 23 l!!L )(,-o!!!.)(jS m 3 • m3 s
Thus)
I
= .... '1-.07 -:s
JV~ ¥S 6S - (-if,07-}) (O,511?) = -2.V3.qz) so/hat V+crV:::20~ -2.03:
or V~/i.o';
3.6 What pressure gradient along the streamline, dp/ds, is
required to accelerate water upward in a vertical pipe at a rate
of 30 ft/S2? What is the answer if the flow is downward?
where
Ti1f1S, for /Jp flow
e == qo ~ for flP Ilow }
e-= _qOO for down flow)
and vjf: ~ ::: 3.0 fi
~ - - 62,'1-(1) Ih - 1.9'1 ~~(30 it)c}.s - W s.2-
o.n" for Jown f/()I;f/
- -12.o.6(-!fp.)/ff :::-0. 83S:e;f
d.(l Ib .sJup(. ft)7.S = -6).,'1(-I) lP - /,9# ff.6 3°7
3-if

3.7 ,
3.7 Consider a compressible fluid for which
the pressure and density are related by pip" =
Co, where n and Co are constants. Integrate the
equation of motion along the streamline, Eq. 3.6,
to obtain the "Bernoulli equation" for this com-
pressible flow as [n/(n - l)]plp + V 2
12 +
gz = constant.
~f +.f- +~ r = constant 010n9 a stream line
or
ThlJs;
S ~ - nr n -I
n
n -I
3-.5
so that

3.8 The Bernoulli equation is valid for steady, inviscid, in-
compressible flows with constant acceleration of gravity. Con-
sider flow on a planet where the acceleration of gravity varies
with height so that g = go - cz, where go and z are constants.
Integrate "F = rna" along a streamline to obtain the equivalent
of the Bernoulli equation for this flow.
From Z b&::: 6m as one obfains
df + f pd(V2) of- 0til where 't::: f'J
(see EI.(.3.$)
ThusJ
(I)
(2.) S
Jm
rlp + rJ. (f pV2) + P(go - c1) d1 ~O J or by infe9r4fin9 from (I) to(:;.):
(2.) (1) ~
S"f +Sd(tfV'-) ~ f>f(~()-C2)J;z=:O
W til (/)
or
f{ -p, +f p(v:- ~~) .,. P'}o(~,.-l,) - fpc (2').2.-z/-J =0
Thvs,
fJ + t pV2. +fi, z - f fC 2~ -; cons/alii a/oil' a .sfreamline.
3-6

3.9 J
3.9 Consider a compressible liquid that has a constant bulk
modulus. Integrate "F = rna" along a streamline to obtain the
equivalent of the Bernoulli equation for this flow. Assume
steady, inviscid flow.
From Ef. .3.6
dp +fpdrv1.) +odz~O
Th-us} (Alon, Q sfream}ine;
where
and
£.,.1f +:i-pd(V''J +fgrie::o or
(:: e9
dp::£~ 1f INhere
Ell ::: hulK 1Y}()t/"lvs ::: cOMI4nt
(see E~. 1.13)
£'V ~ + d(f Vtl.) +~ dl :: 0 which can. he infe9rate d he~ween
e belween pomfs (I) ana (2.) T() '1ille
f,. v,. 12.
£v { ~ T fdaV~)+ f9d,Z ~O
f' V, 1/
or
"Ev[fr2 -i] t t[V,.2..- '/,1.] +~[Zl-Za ~O
Hence:
?z - ~ +r-:= cunsl.nf 410"'1 Q slreamline

3,10
3.10 Water flows around the vertical two-di-
mensional bend with circular streamlines and
constant velocity as shown in Fig. P3.1O. If the
pressure is 40 kPa at point (1), determine the
pressures at points (2) and (3). -Assume that the
velocity profile is uniform as indicated.
I4m
FIGURE P3.10
-a' d~ _y, _ e'C
Tn on - "1( wNh ~::/
(1)
OlJrP V == /om/.s
,hilS; w/fh '1< -::; 6 - /}
alJrJ,
wlfh f; :: '1-01<14 4nd fl,3
()r
Ilz. =<2..,0 kPIJJ
== ~m S/J ::: 'If) kP~ - 9.I()XIO:J~ (;;'/1/)
-999 -t!s (10-1)2. /1 (-9-)
or
1.3 ::: -20. / kPiN
3-8

.3.12 I
3.12. Water in a container and air in a tornado flow in hor-
izontal circular streamlines of radius r and speed Vas shown
in Video V3.2 and Fig. P3.12..Determine the radial pressure
gradient. ap/ar. needed for the following situations: (a) The
fluid is water with r = 3 in. and V = 0.8 ftls. (b) The fluid
is air with r = 300 ft and V = 200 mph.
'14~--+-x
-
• FIGURE P3.J2
Fo"r ourved sfreamJ/ne,s,
- ~ '" ~+ if ~ J or wilh ~ =0 (horilDnf4! sfrea/lllines)J
and fn =- fr- this becDmes
!Ie :: iJ-"(;/.r r
CA) With r::: J ff and V= 0,8 #- und wafer (p =/,9'f ~)J
r!£ /.9'1- *¥(O,81})2. s//J9s = ~,97 Jb3J,'r :: (Ii fI) :: ~. 97 fI··.s~ f/
Jt
(6) W'i-Ih r =300fT and V=- zoo tnph( 88 oS h ) =t,omp
and air ( f:: 0.00238 .:!~~s )J .
2 '13 fts
~ (11:)2 I /brkr ~ O,OO:2.3a~ 2'13..$. == 0.681 slv9,S2. 068/
d:r 300 f1 ft"·,s ::, W
3-'1
1?:::rJ

.3./3
3.B As shown in Fig. P3.13 and Video '3.2. the swirling
motion of a liquid can cause a depression in the free surface.
Assume that an inviscid liquid in a tank with an R = 1.0 ft ra-
dius is rotated sufficiently to produce a free surface that is
h = 2.0 ft below the liquid at the edge of the tank at a position
r = 0.5 ft from the center of the tank. Also assume that the liq-
uid velocity is given by V = K/r. where K is a constant. (a)
Show that h = j(2 [(1/,-2) - (1/R2)]/(2g). (b) Determine the
value of K for this problem.
v:L
(a) - t" f"R Of' #,:,:
Thvs (fo (R J
, j clp == eK'- ~ ~f
F r
Bvi fo::: 0hand f:::O at r 0/1 the free stJrface.
ThvsJ ,.
oh:: - ef[p -f,.J or sInce
• FIGURE P3.13
(I)
(b) With h:: 2 f( R:: Iif I and r == o.s1/ £qn. (jJ 9/ve.s
_ K2. [I
2 ff - 2 (32..2 fils>') lJo.stW - (f ~)L ]
or
K::: 6.55 #2.
3-/0

3./1f
3.1# Water flows from the faucet on the first floor of the
building shown in Fig. P3.11f with a maximum velocity of 20
ft/s. For steady inviscid flow, determine the maximum water
velocity from the basement faucet and from the faucet on the
second floor (assume each floor is 12 ft tall).
and
or
(3)
(I)
' =20 ft/s f~"
~=======r=-
12 ft
• FIGURE P3.11f
wIth /J ::fll =() (froe Jet)
a/ld VI=20 # J 2/ =-If. If
2'.3 ::: /6 ItoJ;l 2. 02
(2 oS) + if It ::: ~ + /b H
2. (32 . Z ¥:J 2- (32.2 !h.)
Of' ~ ="';').02. _2(.32..2)(lJ..) == ,; _ 373' ImjJoss/b/el #0 //tJlJ/
from .secf)/Jd {/rJf)fI felVest.

3.15
3.15 Water flows from a pop bottle that has holes in it
as shown in Video V3.5 and Fig. P3.1S. Two streams com-
ing from holes located distances hi and hl below the free
surface intersect at a distance L from the side of the bottle.
If viscous effects are ne~ligible and the flow is quasi-steady,
show that L = 2(hl~)1 . Compare this result with experi- ~-O
mental data measured from the paused video for which the
holes are 2 inches apart.
• FIGURE P3.1.5
For sfeady /nviscid I/o~ Ihe velocities of the horizonfa I jets of water
at pOinfs 0) and (;z.) are obtained from ihe Bernovlli erv41ion 4S:
fo +-J:f Vr/".f0Zo::: A+-1f ~z.+o2, ~12. +if1Iz:L+!'Z:J. where
fJo~f,==f2.::0; Vo==Oj zo==h2.j 2, :::.hz.-h,; and 22.::0 X=VI:
Z
ThusJ
~ =r2~h,'" and ~::y2!h: (I) ~2= h_t~t2
Once olltside the fqnKJ qrtJvify is the on~ force on
fhe fluid parTicles. Hence the hOf'iiontQ J compone"f
of ve/ocify remlJillS cons/antJ bvi fhe parlicJes
o.........----x
accelerate downward wah fhe acceJert.lfion ct.f 9rav/1y. ThusI for a {II/itA
par/ide Ihat exiled fhe lank t seconds Q90 at z;:h /-1 fol/()w.s theff
X::: Vt and 2::: h-±~ l~ T % h:z..:: Z
By e/il1Jind1in9 " ihi.s 9/ves
the paflide pafh ((,~ e. the
shape a/the VltAler jet Q.s
Z =:. h- g X2
iV'
Thvs" the shapes()/ fhe
flllo Water jets are as shown
in the fi9J1re.
Z= 0 -=~~----~--------x
3-/2

.3,J5 I(c on'I)
The sfreams intersect when X-L tJl1d ~::Z3.. 71Jl),sJ fromEf"s.(l.)tJnd(.3~
(h -h )_LL'l. - _1-L2.
2. I ~ ~2. - 2 ~2.
L=12(~-h'~ W-1;L
From Eqn. 0) J
I / I J _ .J. [-1- _.J..]-v,2. - ~2. :: 2Jh, - 21h~ - 2, hi h2. =
-rhvsJ
by comh/nilJi ["'f/ls. (If) and (s) we ohtain
L: Jz(kh,)j (h..-h,) = 2yh h.
Y 9' 7 21 hi h2. 1··
3-/3
fhz.. -hi)
2j n,h:;.
(S)

.3./6
3.16 A 100 ftJs jet of air flows past a ball as shown in
Video V3.1 and Fig. P3.16. When the ball is not centered
in the jet, the air velocity is greater on the side of the ball
near the jet center [point (1)] than it is on the other side of v1 = 110
the balI [point (2)]. Determine the pressure difference,
P2 - PI' across the ball if VI = 140 ftJs and V2 = 110 ftJs.
Neglect gravity and viscous effects.
The Bernoulli e9uafion from
point (3) fo (2.) and (1/-) to (I) with
9rQvily neq/ecfed 'lives
• FIGURE P3.16
I 2 ..L 2. J.. 1/2 1. v,"{J.3 "':2f ~ ::; p:;. + 2 f> V:z. tllld f'f of 2. PVII- =:: P, -#- ,. PI
V= 100 IUs
(Lf)
But {1.3 -== 1'1:::0 and ~.::- ~
Thvs} e 'len fholJ.~h poinls (I) and (~) are nul 0" the same streamline}
I ,,2 -L /2
fl +:i f V, :- ~ of 2. f V2.
3-/~

3,/8 I
3. 18 A fire hose nozzle has a diameter of H
in. According to some fire codes, the nozzle must
be capable of delivering at least 2.50 gallmin. If
the nozzle is attached to a 3-in.-diameter hose,
what pressure must be maintained just upstream
of the nozzle to deliver this fiowrate?
Thus;
fJl:;:: 4-[V;.~-v:2-] where
alld
.so iAqt w/IA f ~ f
/J = i (/.9~ ~;P)[80.7). -1I.3/f2.]~
::: 6/90 J1L == Jf.3.0 psif12.
3"'/5

3.19 Water flowing from the O.75-in.-diameter outlet
shown in Video V8.6 and Fig. P3.19 rises 2.8 inches above
the outlet. Determine the flowrate.
The flowr~te is Q::: A, V; 1 where trom
the Bernoulli efvalion
I?J- + .M: +E ::: I!!= +~ .,.E2.
o ~1 I t 2.1
• FIGURE P3.19
ThtJ~ with' fl =:~ ~ ~I ~ ~ ~ 0 we Obtain
V; ={212;;. = {2(.32.2fi/.s:l.) (2,8/12)ft ::: 3.88ffl.s
So that 2. .3
Q::: AI ~ = I(O~~SH) (3,88 ¥) :: 0,0//9 f

3.20
3.20 Pop (with the same properties as water) flows from a
4--in. diameter pop container that contains three holes as shown in
Fig. P3.20 (see Video .t5). The diameter of each fluid stream is
0.15 in., and the distance between holes is 2 in. If viscous effects
are negligible and quasi-steady conditions are assumed, detennine
the time at which the pop stops draining from the top hole.
Assume the pop surface is 2 in. above the top hole when t = O.
Compare your results with the time you measure from the video.
Q:: Q/ + Q,. .,. Q.a ::: -lJr -df
where /J.::: ~·A· :: Y2ah· A·,(1 ,~ (/' I J,
(i /If/J 2.".1)
or L
and
t AT [ dh
::: A, Y~1 ((h +Vh+L +fh+2J.)
o L
• FIGURE P3.20
III :: fJ:r. :: A.3 :: f (o/IIi)'}.
=/,227 X /0-'11/2-
AT:: ~((fn)'-;: 0.087.3 fl2.
where i is the time /1
lake fur fhe fros .s()rI~e
fo re40h fhe IIf/Df'hole
(h~o),
_ O. OB 7311'"( dh
- (1.22.7~10"fJ'J. )K2.)(.3:J., 2. ff/s'-)]~t (Yh +Yh fL +1(h+2.L )
o
-rhll~ L
j :: 88.7 (dh where L:: -ki-l == 0.1667 ff
J«(h .fYh+L- +YhfZL)
o
Nofe: Will; LIf) leef) flJl'.r crvtJJfiOf) 9illes t in seCl)l'Idr.
Since fherB is flO closed form
3-/7

3.1- 0 I(con 'i )
The nvtrJerica/ valve of lhe i"fe9raI is obfailu}d by vsin9 the
trapezoidal rule since the closeJ form analytical so/uil'on
is nof 9/ven jn illfe9f'al fahies. The EXcel- spreadsheelv.red
for this is 9irJefJ be/ow.
S
L J
, :;; ,88.7 f(h) Jh where [(h)::: (VE -I-WL ./-1(hiii.)
o
:=88.7[ih(f,.t{+,)(hit/-h,)] «(88.7 frd~.,tom1= /0.75
h, in. h,ft f(h) , 11ft1/2 (1/2)*(fi + fi+1)*(hi+1 - hi), ft112
0.0 0.0000 1.015 0.00804 1
0.1 0.0083 0.914 0.00743 2
0.2 0.0167 0.870 0.00711 3
0.3 0.0250 0.837 0.00686 4
0.4 0.0333 0.810 0.00665 5
0.5 0.0417 0.786 0.00646 6
0.6 0.0500 0.764 0.00629 7
0.7 0.0583 0.745 0.00614 8
0.8 0.0667 0.728 0.00600 9
0.9 0.0750 0.712 0.00587 10
1.0 0.0833 0.697 0.00575 11
1.1 0.0917 0.684 0.00564 12
1.2 0.1000 0.671 0.00554 13
1.3 0.1083 0.659 0.00544 14
1.4 0.1167 0.647 0.00535 15
1.5 0.1250 0.637 0.00526 16
1.6 0.1333 0.627 0.00518 17
1.7 0.1417 0.617 0.00510 18
1.8 0.1500 0.608 0.00503 19
1.9 0.1583 0.599 0.00496 20
2.0 0.1667 0.591 21
Sum of column =integral = 0.12011
Thus, t =88.7*0.12011 =10.7 s
3-/8

J.21 Water flowing from a pipe or a tank is acted upon by
gravity and follows a curved trajectory as shown in Fig. P3.21
and Videos V3.5 and V4.3. A simple flow meter can be con-
structed as shown in Fig. P3.21. A point gage mounted a dis-
tance L from the end of the horizontal pipe is adjusted to indi-
cate that the top of the water stream is a distance x below the
outlet of the pipe. Show that the flowrate from this pipe of di-
ameter D is given by Q = 'TT'D2Lgl/2j(2S/2 XI/2).
~I'--L - - - . I...
• FIGURE P3.21
The only force aclin9 ofallY w4fer parfiele ill -Ihe free
Jet is Ihal d()e 10 9ravily - the parfic/e:S wel9hf.
Thv.sJ for the X- YQxes shown
i) :: a ana ~~:::o whicA for a parhcle .sfarfi»"
",fl.. tf dP' I
af xr:: y~ 0 ttl I:::0 9ive
X;::: t1-t2. and yo:: Vi
£//mt'nale l to 9ive fhe wafer Irajec7ury as
x~{~(~t or x==r~t:
ThvsJ
wifh y::::1.. :
Vc V-/x·'"Land
Q:::AV:= fj)"v :: 7T D~L Vj/(251
:J, Vi)
3-/9

.3 ."1. I
3.22 A person holds her hand out of an open
car window while the car drives through still air
at 65 mph. Under standard atmospheric condi-
tions, what is the maximum pressure on her hand?
What would be the maximum pressure if the "car"
were an Indy 500 racer traveling 220 mph?
2. 2
.I!.!-+XL +Z = h. ~ V2. ..J. Z
r 2j I ~ 2.1 2.
~ =Z2
,/;::6StrJph (88fj ) =«5.3!1-
6om,"17=0) 4 =-0
or Ii =- ~ (2.38X/O-
3
~~ig)(95.3#/·= /0.8 ~
If V;:: 2.20mph ( B8 ~) ::: 3 ~3 !f fhen
60mph J
3.23 T
12. :::: f (2.38 X/O-.3 ~:fi)(3:L3 fj) 2.= /2.'1 ~
3.23 A differential pressure gage attached to a Pitot-static
tube (see Video '3.4) is calibrated to give speed rather than the
difference between the stagnation and static pressures. The cal-
ibration is done so that the speed indicated on the gage is the
actual fluid speed if the fluid flowing past the Pitot-static tube
is air at standard sea level conditions. Assume the same device
is used in water and the gage indicates a speed of 200 knots.
Detennine the water speed.
Af :: tpV2.
In air J Afail'::: f (0. 00'-38 ~~~.s ){2ooknofs)f/.
In w4fer AIl ::: ..1 (/ 9/f S/()fJJ.) ( V)2.i rW41er 2-' ff3
so thai w/lh A Ali- ::A fwater J
t(O,OOJ-31 )(ZtJO/' ::: i (I.9~) V.2
or
V= 7.0/ knofs

,3,2Jf I
3.,.. A 40-mph wind blowing past your house speeds up as
it flows up and over the roof. If elevation effects are negligible,
determine (a) the pressure at the point on the roof where the
speed is 60 mph if the pressure in the free stream blowing to-
ward your house is 14.7 psia. Would this effect tend to push
the roof down against the house, or would it tend to lift the
roof? (b) Determine the pressure on a window facing the wind
if the window is assumed to be a stagnation point.
a) Thvs, from (JJ to (:1-) :
I, +i p~:z. =/2. + 1:P~2.
b) From (I) +(13): Since ~'='O"
(J~ == PI +1pU
Z
or
~V2.
(2.)
-o (3)
(J.J -P, :: -tpv,:L=- f (0,00').38 :~"/f) (S8.7il/s;2 ::: ~/O ~
3-2.1
v,
~
(I)

.3.2.S
.~.25 Water flows steadily downward through the pipe
shown in Fig. P3.2S. Viscous effects are negligible, and the pres-
sure gage indicates the pressure is zero at point (I). Detennine
the flowrate and the pressure at point (2).
A/so)
f=. +Z -I- ]'z.2. =- .fL ~z ... ~£
• FIGURE P3.2S
Jr 2. ~I l: '~,.
where f,:;: 0 4nrJ sInce II, ~,42 Ii {QI/olIIs fha-r Vz. # V,
TlJv~
2:J. - 2, ~ - if 0 r l; ~ -2 ff
(Jr
12 :: -2. ft (62,tf Jjp) :: -/25 ~
3-22

3,26
3.2~ Small-diameter, high-pressure liquid jets can be used
to cut various materials as shown in Fig. P3.26. If viscous ef-
fects are negligible, estimate the pressure needed to produce a
O.lO-mm-diameter water jet with a speed of 700 m/s. Deter-
mine the flowrate.
:/:L .{2. (J~ ~ 2-
r +~I +Z, = V + $ .,.;$J. where ~ -zO J 2, ~ Z2.J and fJ2- =0
Thll.s ~ == t f ~2. = t ev,.2.=t ('9'1-i)(70()~)2 = 2.9-5')(/0
5
.lsJ.
Also)
Q =~ A.z = 700 ~[¥(/O-i'1h)2] = S.$() x /o-~.!fi
3-23

3.2-7
3.27 Air is drawn into a wind tunnel used for testing auto-
mobiles as shown in Fig. P3.27. (a) Determine the manometer
reading. h, when the velocity in the test section is 60 mph. Note
that there is a I-in. column of oil on the water in the manome-
ter. (b) Determine the difference between the stagnation pres-
sure on the front of the automobile and the pressure in the test
section.
(I)
~
~
•
~
60 mph ('-)
~
• FIGURE P3.27
(a)
(b) f2. 1- Z. ~ Jt ::: k .f Z.?l .f-!2.
7 z,.,. 4' -,
where
Z:L :::Z.J 41114 ~::'()
Thvs.J 2-
ty ~1; = If or
Wind tunnel
A-(J2. ~ f f 142.::: -f(0.00]..38 -§.~~.s )(eNfj)2 == 9.:z.z1-.
~
-'"
-....
"'"

.3.1-8 I
3.28 A loon is a diving bird equally at home
"flying" in the air or water. What swimming ve-
locity under water will produce a dynamic pres-
sure equal to that when it flies in the air at
40 mph?
or
3-25

3.29 A large open tank contains a layer of oil floating on
water as shown in Fig. P3.29. The flow is steady and inviscid.
(a) Determine the height. h. to which the water will rise. (b)
Determine the water velocity in the pipe. (c) Determine the pres-
sure in the horizontal pipe.
(c)
3-26
r4m
p

.3.30
3.3.0 Water flows through the pipe contrac-
tion shown in Fig. P3.30. For the given O.2-m
difference in manometer level,determine the flow-
rate as a function of the diameter of the small
pipe, D.
I I
0.2 m
FIGURE P3.30
./!L + 1,2. +Z, =.if + V.' + Z,1 or w/lh z/ =ih ani 11=0
~ ~? zi
~ =1;>'1 (fi/') ,
butf!, '" ~k and /':z. ~ rh2 so that 17-12. '" ((fir-h.) =0.').0-
Thlls,
~ =rrZ-?-o.-;""/ ="';2.1(0.').)'
3-27

3.31
3.31 Water flows through the pipe contrac-
tion shown in Fig. P3.31. For the given O.2-m
difference in the manometer level, determine the
flowrate as a function of the diameter of the small
pipe, D.
or
2-
Q_ 0.0156/)
- Y(0.1)'1 - l/'{
or ~ ==
3-28
•0.2 m
t
FIGURE P3.31
0.2. (2-$)
[(!ff)'"-I]
1
1/ = (0.1) V.VI D I

3.3.2
3.32 Water flows through the pipe contraction shown in
Fig. P3.3-2. For the given O.2-m difference in the manometer
level, determine the flowra1;e as a function-of the diameter of
the small pipe, D.
12- +X1t Z :: {J:J,. -1-]1: +£.
it :LI I .., ~ 2-
where Z,:: Z2. t4nrJ ~ -:::0.
ThtJSJ
k +ff = br z.,f r
Bi/t
IJ!-:::x and t'f::: D,2m +x Sf) -Ihat
Il'"
X +¥ :: O.2m +X or
+
0.2 m
• FIGURE P3.32
-------- kV, == f2-1 (0.')..",) = (2 ('1.81 /§i)(0. :1-/11)) '2.::: 1.98-fL
ThusJ
Q::; II, Vt ::: f (o.linl- (I.?!.p.) ~ 0.01,56!f-3 for olJy D
3-1-9

3.33 J
3.33 The.speed of an airplane through the air is obtained
by use of a Pitot-static tube that measures the difference be-
tween the stagnation and static pressures. (See Video V3.4.)
Rather than indicating this pressure difference (psi or N/m2)
directly, the indicator is calibrated in speed (mph or knots).
This calibration is done using the density of standard sea
level air. Thus, the air speed displayed (termed the indicated
air speed) is the actual air speed only at standard sea level
conditions. If the aircraft is flying at an altitude of 20,000
ft and the indicated air speed is 220 knots, what is the ac-
tual air speed?
~~ooo
or
3-30
V
•p
(3)
•
...... (4)
r--
v.==VI
PI := f
V2 ::: 0
(1)
~.•
(2)

3.3Lf
3.3'1 Streams of water from two tanks im-
pinge upon each other as shown in Fig. ·P3.3'1. If
viscous effects are negligible and point A is a stag-
nation point, determine the height h.
.()- ~~ z. - ~/I .xl ~
-II( +').1 + 2. ,,- T + ~I' +.c/J
ThusJ
h +2.0II ::: .If; f- 2 () If
or
h == i!f
II/so;
fJ, 'h2. ~ + VtJ..2.+Z
4" + '-I +Z, .:: i' ').j 1'1
TAv.sJ
.If::: 'f+z, -219
h =- If+z, - Z.19 ~
1h
I
1Free jets
20 ft
1
FIGURE P3.3Jf
where Ii.=-0.) ~:::- 0 J 2',.:::- h+~()rl
~ =0.1 a/Jd ~ = 2.() If
3-31
t8ft
~
(I)

3.35 I
:US A O.15-m-diameter pipe discharges into a a.IO-m-di-
ameter pipe. Detennine the velocity head in each pipe if they
are carrying 0.12 ro3
/s of kerosene.
I, "..Q.
m'O./Z-S m
- 1£( • = 6.79$I /II
and
Thvs,
'I' O,/sm)
",3
0.12. ;s-
f (o.lom)'
" 15.271}
/" (';.7?f!1/
v, _ -:"::;':':"'-£"f- " 2. 35m.
"1 - 2. (9.9I-f;,)
alld
v.' (ls.'Z.7!}l
2, "2(9;8/~) =
II."111
3-n

3.36 J
3.36 Water flows upward through a variable
area pipe with a constant flowrate, Q, as shown
in Fig. P3.36. If viscous effects are negligible,
determine the diameter, D(z), in terms of D if
the pressure is to remain constant throughout the
pipe. That is, p(z) = Pl.
ICf'I
.1
Z~_D(Z)!
l-Dl_L
FIGURE P3.36 ..... I J (1)
or will; I' =fJ; and z, =0 tQ
2. 2-
V, -V =2<;i!
and V=-f
Thus)
('f~)2. (JfQ)2..TiD/- - TtI)2- ~ 2pE
or
..l... I
DII- -11I
3-33

..3.37 I
3.37 Water flows steadily with negligible viscous effects
through the pipe shown in Fig. P3.37. Determine the diameter,
D, of the pipe at the outlet (a free jet) if the velocity there is V = 20 ftls *
20 ft/s.
• FIGURE P3.37
or
V, == 8.83 !1-
BtJf AI Vj =A:l ~ So fhtff
.1tD'- (8 i3 H) = F/). 2.(2.0 if)'f-I • S r:.z. oS
or ~
4. =(8~~3) ,.(fffI) == 0.0831 ff:: O. '197 in.
rOpen
115 ft :::; h
-111.5I,. do""'"
fen

3,38 J
3.38 The circular stream of water from a faucet is observed
to taper from a diameter of 20 rnm to 10 IJ1m in a distance of
50 cm. Determine the flowrate.
r
:! ":;) (/J
I!!- + ~2. +Z = -1'2._ + v,,2. +E
~ ')..~ I T '-1 2
where A ;::f2, ': 0 J Z,-- =0 J Z, == o. son
and
II - Q v.: - ()VI - 7f; J 2. - liz
~~ ,
0.5f) m
L.Q
II:J. fii2,(%t+21i!, =("*J'or Q= [(.;~*~)r=
, t I -(/J:J./A)'J.
or Since
112 (- A. 2.
- ) we obfRinA,-o, I
Q- A 1I2jz/ _ 11.( )2.[ 2(9.eI'f;.)(O,SOfIJ1
2
- z f J -(D.zID,)7J - 'I- 0,010n, ,_( 0.0/0 ~If
f), fJ2.0 )
_¥ m3
=2,5Lf x/a S
3-35
D =o.O:Lom,
('-) D ;:: O. 0/0",
2.

3.3 'I
3.3Q Water is siphoned from the tank shown
in Fig. P3.39. The water barometer indicates a
reading of 30.2 ft. Determine the maximum value
of h allowed without cavitation occurring. Note
that the pressure of the vapor in the closed end
of the barometer equals the vapor pressure.
2.-
r30.2 ft
Closed end
(0)
3 in.
diameter
FIGURE P3..3Q
5 in. diameter
ft t Vi +Z
r 2.1 I
Thv,s,
where jJ/ -=0 J V; ::;-0J fJz =Ivaplr
Z, =°,22 ::0 ft
o - fVlltPfJl' + '(,,2. + 6 ff
- 0' 21
bllf 10 +.30, '). II r ::11
lIe/)c8.;
o= -30.:2. if + ~.t of 6 II
:1-1
Tf;vsJ
l{:: 3 q..!' !I-
Since ~ 11.3 = Vz A;z ,
or
'vi = 1'I-,21J
HiJwever)
or .since ft::::{Ji/(/.por J f/lapor =-30,').. It
~
All VJZ LJ.a V:~ f - +z, ::: ./L::!. + - +28
o zi ~ ~
or ~ = Y2-?h I
ThtJ.sJ
I if. 2 f1 = ';2.(32.Zli ) h ;/s s~
or h::::3.13 ff
3-36

.3. 'f0
3.40 An inviscid fluid flows steadily along the stagnation
streamline shown in Fig. P3.40 and Vidt'H l3_l, starting with
speed Vo far upstream of the object. Upon leaving the stagna-
tion point, point (I), the fluid speed along the surface of the ob-
ject is assumed to be given by V"" 2 Vo sin 8, where (I is the
angle indicated. At what angular position, 61, should a hole be
drilled to glve a pressure difference of PI - P2 = pV~/2? Grav-
ity is negligible.
Vo
----;;-(0)
• FIGURE P3.40
.3.'fI
~ 11.'" ~ IlL +~ v.L
I.+:z.f· "'/t+2.PV/ ::f.. :z.f~
where V, =0
Thvs,
f, - P.. d 'ip(v,.'" - W'J ~ d(' ~L
sO fh"f if
f, -f1,. =' drv/ then v.. "V.
Th"lis:
~ '" 2 V. sih ~ '" Vo
II () __ 30'
ntlJc8; L =
or . " ..LStn~==2.
.l.41 A eelain vacuum cleaner can create a vacuum of2 kPa
just inside the hose. What is the velocity of me air inside the
hose?
fl +f p~ 2. "f:z. +t pV,:
where II ::0, ~ '" 0 .so fhal
~ !I.'"f",=-:z.f2.
flenceJ
3.JL .J. k ~
-2.XIO 1112. ~ -2.(1.~3!J3) ~
3-37
OJ
///// 1/ / ///111/7
•

3.4f2 J
3,113 I
3.42 Water from a faucet fills a 16-oz glass (volume = 28.9
in.3
) in 10 s. If the diameter of the jet leaving the faucet is 0.60
in., what is the diameter of the jet when it strikes the water
surface in the glass which is positioned 14 in. below the faucet?
= 0.852 it...s
HenceJ __------------------~--,1:1. 2 -
( tf )(111- fl + (0. BS2.s) )
Vz. = 2 3.2..:2.-;2. n:. T 2 (3.2..2 {tao)
But,
11, ~ % 112 Vz. SI) fhal
= 8.7/ Ii.s
or ~ F+ k
D - (Ji) f) - (0.BS27)2-(0 LO ,·n) :: 0.188 il'l.2.- V,2. '/ - 8.1111 .0 •
.$
0.2 m
3.43 A smooth plastic, lO-m-long garden hose with an in- : (I) ;;~
shown in Fig. P3.43. If viscous effects are neglected, what is '/h ~/" /////////////~ 1-1
side diameter of 20 mm is used to drain a wading pool as is :.;;~::;:: C"" 0.2*3 ~
the flowrate from the pool?
./!!- + '0:L +i!
~ zg I
Thus}
~=f2~(Z,-Z2)i -
- 2 ~Q.!1l- .7.s
or
FIGURE P3.¥3
where 11:r/3.. == 0 J ~I :: O. ')../1'1
2 :: - O.23m and ~I =02. .I
k
(2. (9.8/ ~)(O,2 m - (- O.'].3hJ)~ 2.
Q = fJ2. ~ =-f(o.02-0ml·(2.90!P-) = '1.1/ X/O'f./ff-3

3. iflf I 3.11-'1- Carbon dioxide flows at a rate of 1.5 ft3/ s from a 3-
in. pipe in which the pressure and temperature are 20 psi (gage)
and 120 of into a 1.5-in. pipe. If viscous effects are neglected
and incompressible conditions are assumed, determine the pres-
sure in the smaller pipe.
Tillis,
f:;. :::; fJ, + i f (~ 2. - ~'")
Ib I I" If,.:: 2. ero 7[" - .53. f[z. == 2.J 82-71f2.
(Jr
~ ::; 19. 63 psi 9Q ge
3-39
o·(I) ::_---.....:.-=c:.:..~):....;,o
D - ':I,,,' D ;: I,S/I').I-~ . 2.

3. 45 I
r--
3.1f5 Oil of specific gravity 0.83 flows in the .
pipe shown in Fig. P3.IfS If viscous effects are
neglected, what is the flowrate?
14 in:= h
Water~ ~ (tAl
~ -. f1in
i
TVt ll--;a=o.83 )1
tl or- ~ (.liT ()1FIGURE P3. Jf.5
112. 1) V/,
LJ, +..!.L +:z ~ L .... -1--::2
7" ~i ~I = T -r~ ~2
Thus,
V;- _1'1 -I':l.
21 - )"
but,
fl., :: ~ +rJ :::;If +oj
and
fJ:z.:: 't(J, +h) - i}nh -ffll-
Thv!'J
where z, ~ Z:;. and V, ~ 0
(I)
ft -fz :: ( ~ -l')h (2)
Comhine £,05. (I) aIJrk (2.) 10 obtain ,h
.1 ' ./ 1 (.62,.'ffl3 )'.!i:r/l
v.. =V2~V·-") = v2.!(1-1) h =12(32·1.t/-.)~O.83(6';fff.rJ 6.."1
or
Vz = 2.lo!f
ThtJ-s,
Q =1l2. ~ =-f(!r.n)2.(z.lo!j) =0.183 ~
"'?-ifo

3.'f6
.3.16 Water flows steadily from a large open tank and dis-
charges into the atmosphere through a 3-in.-diameter pipe as
shown in Fig. P3.46. Determine the diameter, d, in the narrowed
section of the pipe at A if the pressure gages at A and B indicate
the same pressure.
p~ +~ e~2 +~Zlf == f:J. +if'6.2. +i'z~ J where Z2.:: 0 and1:l.~O
Thus.! since 14 =1'"
13 +t e~2 + i'r.1I- == f: pVs2. 0)
HOllleve~ fl +ie~2..,. r2, := fL +ipli2.+~E2..1 where It -::'11-~ ~~Z2.
so that #0
t.pv,.2.;:~i!, or Vz. ""Y2fz,' =y;.,;; =(2.(32..2 ~~}(16 fI)]~ 32.di/.s
Bvl
f.3 + 1:P~2. t rZ.3 ~ f~ +i p{2. f r22. "hers ~ ~ ~ since 112. ~1l.3
ThllsJ
1.3::; -rZ3:: -(16+Q)fI(6,..Jl-lbltl3
) ~ -/Sto l/,lfl:J. (2)
From Efs. OJ ana (,.): ..
-1,s&0 jtJ. f i (I.9'1- ~') '42. := 1::(1,9'1- ::~9f) (:l21/i})2-
or
~ ;:: ~t./ 11/0$
Since A~ ~ =,42. V,.
¥a~~ =1/):1.2. v,.
or
d '" L>.. V~' '" (3io.)
if follows 1h~f
3,..1111.r
-~1t--I.~,/ = 2.5 0 in.~.I T /0$
3-if/

3,'1-7 I
3.47 Determine the flowrate through the pipe in Fig. P3.47.
m= 2.20-s
Tho.sJ
FIGURE P3,,1f7
where z, =- Z2. and l.'2.:::0
Q==II, ~ = -: (0,08 m)2. ('1. .20 !f) ::: 0.0/11 if

.3. '1-8
3.48 Water flows steadily with negligible viscous effects
through the pipe shown in Fig. P3.48. It is known that the 4-
In. diameter section of thin-walled tubing will collapse if the
pressure within it becomes less than 10 psi below atmospheric
pressure. Determine the maximum value that h can have with-
out causing collapse of the tubing.
or
h= /,3/ fl
• FIGURE P3.4B

3 . Ifq I
3. ljq For the pipe enlargement shown in Fig. P3.lf'1. the
pressures at sections (1) and (2) are 56.3 and 58.2 psi, respec-
tively. Detennine the weight flow rate (Ibis) of the gasoline in
the pipe.
or
~( 1-(4;f) =
or
FIGURE P3./fQ
Vt~2.I.'f!} tUld ~=!l,~=:e/';f'f(21.'f~/) '" 0.'190 if
Thvs,
O'Q = If2.S"#-, (0.'1-90.[/) = ZO.8!E.
==.$=
<21

3.50 I
3.50 Water is pumped from a lake through an 8-in. pipe at
a rate of 10 (13/ 5. If viscous effects are negligible, what is the
pressure in the suction pipe (the pipe between the lake and the
pump) at an elevalion 6 ft above the lake?
ill e 0 J ilL " 6.0 II
'I (/o ij!)
1T(l£fIY
3 -'15
= 28.6!f

.3.5/
b =width =0.06 m
3.SJ Air flows through a Venturi channel of rectangular Free jet
cross section as shown in Video V3.6 and Fig. P3.51 The .(1) Air
constant width of the channel is 0.06 m and the height at the ~ ·(If)
exit is 0.04 m. Compressibility and viscous effects are neg-
ligible. (a) Determine the flowrate when water is drawn up 004 13'::: 0.10 m
0.10 m in a small tube attached to the static pressure tap at . m 1
the throat where the channel height is 0.02 m. (b) Determine
the channel height, hz, at section (2) where, for the same
flowrate as in part (a), the water is drawn up 0.05 m. (c) De-
termine the pressure needed at section (I) to produce this Water
flow. • FIG U REP 3 . 51
(n
0.04 m
~2. Ih 2 .3Ji
(2.) (b) t;+;.; := 1-+i.f Whel'€ A =' OJ I:;. =-~.oJ2. := U()X'J!.3 (o.asml
= - lI90 :z.
II m mFrom parr (a) I V.f::: 23.,:s
ThllsJ
£1n. (~) becomes
- Jf90 .f-,. '1,.2 (~3.1 ~)2. II 3/ r m--.,...-- + ::: or V2::: D • .,j s
1'1..0/3 2(q.81!;,) 2 (Q,8'fi)
8 uf Vi AA ::: V'I,44' .so ihai
(36 ..s.!JL) (o.o6m)h,. r:: (u.,!-) (O.06m) (0. oIf.m) or h;. ::O.02.53m
LJI h:z. fJ!t. ~ :z.
(3) (c) AIsoJ -r-I- Jj :: ? +-ii where fJ/f -:::() and AI ~ ::t9~ ~
Bvf since /)1:::: (0.0'1-111)( 0.0&,,) ::: lip then V; ~ ~ alia EfP. (3) r/ve..s
{JI ::: f/f e 0

.3.52
3.52 An inviscid, incompressible liquid flows steadily from
the large pressurized tank shown in Fig. P.3.S2. The velocity at
the exit is 40 ftls. Determine the specific gravity of the liquid
in the tank.
• FIGURE P3.52

3.53
3.53 Air (assumed frictionless and incompressible) flows
steadily through the device shown in Fig. P3.53. The exit ve-
locity is 100 ft/s. and the differential pressure across the noz-
zle is 6Ib/ft2
• (a) Determine the reading. H, for the water-filled
manometer attached to the Pitot tube. (b) Determine the diam-
eter, d. of the nozzle.
• FIGURE P3.S3
(a) fJl +rz, +fpV;2-:: A+Z:). ft:pv,.:z.
where
~ ::::z2. J " :: () J ona l{::: ()
Thus;
p,. -::: 1:p~t. J /;vi /;1. :::: dH.1c1 1/ so ij,aT
(62,1/- ~) H=: i (().00:1-38 ~if) (100 #iJ. or fI == O. /9/ ff
HenceJ
61f-,. +t(O,OO').38sX'f)(I/-If."'Ij.)~;: 1(0.00:;"30:'$)~.2
or £i
~::= 83.7 s
so th4f with III/ V¥ -::: Q4' :' ~ J
-fd~(83.7#) ::: 0.79s
or
d:::; o. /09 ff
3-¥1

,3.5 if
3.54 The center pivot irrigation system shown in Fig. P3.54
is to provide uniform watering of the entire circular field. Water
flows through the common supply pipe and out through 10
evenly spaced nozzles. Water from each nozzle is to cover a
strip 30 feet wide as indicated. If viscous effects are negligible,
determine the diameter of each nozzle, di , i = 1 to 10, in terms
of the diameter, d!(), of the nozzle at the outer end of the arm.
Supply pipe Nozzle
• It I , I • I ! I I I ! I tV~ • I r I I I
o r, r2. T3 r;..., r,' fa r, r,o
ai fr,·2.- r·,"]4_ - t I-I
rJ,o - /7/00
J
These reslJ/fs are r;il/en
in the f.hle.
= (rl.·2.-t;...:)
(3fJ02. -'-70"")
I.
I
2
.3
11-
5
6
7
a
q
10
3-'f9
r: H4
30
60
90
,,-0
/50
0.229
0.3'17
0.51.3
().607
0.688
/80 0.7&/
210 0.82.7
').11-0 o.seq
270 O,Q'l-6
300 /,00
(Z)
(3)

x=L x-----"x = 0
3.5.5 Air flows steadily through a converging-diverging ;;Q~~l~~~~~;;;::lZ~r~=:;;:,;;..;n~ 0
re. tctangular channel of constant width as shown in Fig. P3.55 ;::;;,:
and Video V3.6. The height of the channel at the exit and the .£.....
exit velocity are Ho and Vo, respectively. The channel is to be
shaped so that the distance, d, that water is drawn up into tubes
attached to static pressure taps along the channel wall is lin-
ear with distance along the channel. That is, d = (dmax/L) x,
where L is the channel length and dmax is the maximum water
depth (at the minimum channel height; x = L). Determine the
height, H(x), as a function of x and the other important para-
meters.
III FIGURE P3.55
1-+ z'( -1-1. Pvz. ~ If) 1- Zo 0 4- t (lVoz. wners p::: air dells/I,
where
-Z cEo J {Jo::: 0 J fJ::: - ~LO rJ. =- ~D
7hvsJ
},If J ...L V2. I I .z.
- 0H,.o "''ltIX X + :2. P ::: 2: f Yo
Bvf
AV A I} V- Ao V, - flo II J.1.. f:::: 0 Vo J () r -,CJ 0 - 7T Yo .s0 In"
- ~.~ ~rflX +tp(!J;v.t«i ev,"
or
H/Ho vs xlL
. --.
Q
::J:
-::J:
1 0.8 0.6 0.4 0.2 o
xlL
3-50

*.3,56 I
*3.56 Air flows through a horizontal pipe of variable di-
ameter, D = D(x), at a rate of 1.5 ft3/S. The static pressure
distribution obtained from a set of 12 static pressure taps along
the pipe wall is as shown below. Plot the pipe shape, D(x), if
the diameter at x = 0 is 1, 2, or 3 in. Neglect viscous and
compressibility effects.
x (in.)
0
I
2
3
4
5
6
p (in. H2O) x (in.) p (in. H2O)
fI~ f
-1.00
ail' p 1)0.72 7 0.44
D. (f-P. D-V J t0.16 8 0.51
-0.96 9 0.65 L _J
10 0.78 .- -.....
-0.31
J..-x ........0.27 11 0.90
_:Ih0.39 12 1.00
water
rN.z.O
If+ ¥(+zo =- f +lj+Z J where Zo"::Z
n~ £
V= 1'i2
+ 2('9-/)' 'llilh v.::: Q =1.5 ~ - 1.91 .£l whereD.",fI
V 0 P D ~ :It D2. - D2 S J 0
If 0 0
and _ 62.If -Ita .
fo-f= ~.ao (ho-h) - 12.11}. (1m. -h) c: s.20(J-h) ~Jw,:jh h"'in.
LJ '/'L -3. 11' 11
nenceJ
WI II p:::2.38 X/O pf3 we oOTain
jk ~uV= [(.J,iJ.)2.+ 10.'+ (/-h) 2 =[3.&5 -l- '1-370 (j-h) :2. (J)
D: 2.38 x/,,3 Do'"
AlsoJ
AV:::Q or IllV:::(f so fhtrl
D- [ if~]~ -[¥(J'#fq~- 1.38;- 7l'V ,... 7T.$ - VV J or when comhined wilh Et{,{/J
t382 .
D-= [3 65 ~l~ H where Do'V ffJ h,.,/1). (3.)
Dl t'l370(J-h)J J
Plot D=/)(X) with Do:= ifJ +J and f fl) vsiIJ9 fhe va/lies of /,::/'(X)
fro/TJ the lahle. Nofe : h i.s the SQIIJ(; a I/(J On. liz. 0)'1 in Ihe tab/e.
AII EXCEL pro9r4m W~J' used If) ob/fA/1i fhe fol/oUl/II, re.rvl/u.
(con'-fJ
3-5/

~3,.s6 r (con 'f)
x, in. 0, ft (Do = 1/4 ft) 0, ft (Do = 1/6 ft) 0, ft (Do = 1/12 ft) p, in. H2O
0 0.24996 0.16664 0.08332 1.00
1 0.20277 0.15733 0.08299 0.72
2 0.16776 0.14435 0.08234 0.16
3 0.13999 0.12870 0.08112 -0.96
4 0.15299 0.13667 0.08182 -0.31
5 0.17245 0.14649 0.08247 0.27
6 0.17841 0.14902 0.08260 0.39
7 0.18123 0.15015 0.08266 0.44
8 0.18558 0.15179 0.08274 0.51
9 0.19616 0.15537 0.08291 0.65
10 0.20944 0.15911 0.08306 0.78
11 0.22710 0.16300 0.08320 0.90
12 0.24996 0.16664 0.08332 1.00
,
Dvsx
0.30 - j
I
I
i
0.25 ~--- I
·------19
: ~I
I ' I
"
, ,I
,._._J,
I
i
0.20
..
i
...t-~ I
'".. ... I.. .. ••_ .. -1- .-.. ... : I
- - - 0 =1/4 ft '1
= ,....' .,.
1.......... " ..... . _-!'- -~
0.15 -------,~-.;;.~~ .",- - - H _ _ • ~ - - -------.1
- -0 =1/6 ft Ic ,
'-": ! - 0 =1/12 ft
I
0.10 ---.-~.--~+------ ~,
!
0.05 I II --1
i
, i
iI
I
0.00 I I
,
I
0 2 4 6 8 10 12I
I
x, in.I
I
I
3-052-

3.57 I
357 The vent on the lank shown in Fig. P3.57 is closed
and the tank pressurized to incr~se the f1owrate. What pres-
sure, PI' is needed to produce twice the flowrate of that when
the vent is open'?
FIGURE P3.57
.
~~'------'Wi/;' ihe veil! open:
• A. 1-:'b +~ +2 =J,;. +2.. -I-&Z
r 2.! ' r Z?
TIIVS,
11,'
z =-L or, Zj
To hove dovUe fhe f/gwrafe /'liM Ihe veil! clfMed (j'l 10):
•
./.J- u, ==..Y:. where fot'th;.s uue 1Iz.: 2. (:z.s./ffi) ~ .>0.8 II, ~ 6
ThVJ}
Il (SO.8!t/
(< ~',J,) t/ofl == fl
t2.'1''fP 2.(.32.2 Si)
or Ih .
It == /87671'- = 1:l.OpS,
3-53

- 3.581
3.58 Water flows steadily through the large
tanks shown in Fig. P3..sS. Determine the water
depth, hA •
For sfeady (low)
QI/- ~!9~ ~ w/fA
Also;
-1.1-+ ~:L +z
r ')..1 I
.so fhaf
V.2. =,f2fh//
Thus}
A
1i2.~ ==cr~ or f(O.03hJi";2(9.81f;.)~j = O.O/J-31}3
3-5'1-

3.5'1 Air at 80 of and 14.7 psia flows into the tank shown
in Fig. P3.59. Determine the flowrate in ft3
/s.lb/s. and slugs/so
Assume incompressible flow.
3-55
0.6-in.
Q
dja~eter
(I) +
• ~ .(2.)
t
/ 0.5 in. Hg vacuum
~ump
FIGURE P3.Sq

3.60 I
3. 60 Water flows from a large tank as shown in Fig. P3.6o..
Atmospheric pressure is 14.5 psia and the vapor pressure is 1.60
psia. If viscous effects are neglected. at what height, h. will
~avitation begin? To avoid cavitation should the'value of DI be
mcrea~ed or decreased'! To avoid cavitation should the value of
D~ be increased or decreased? Explain.
'J (0)
.-.-•.-:::-'.'•.'.'.'-:-' :j:-
FIGURE P3. 60
where f10 ; 1'1.5psio. ,/'t ~ 1.60 pSia,
E.; h, Z, ~o, and V. ~O
How.ver,
A, V, "11~ V, Or 11 =( z:r~
where
wah
(D.)'" .= - hD,
( ) Ih( in?)IIf.5 -./.60 In? 1'f1fW-
'h [ .. • ,1
6:z..'f N' e::J -'J
From £y. (J) il is seen fhaf h increases in increasing D,
and. c1ecreo.in9 D:z.. Thvs, fo avoid cavilaf;.n (,;~. to have
h slnall enot/9h) D, should be increq.sec1 and 4. decreased.
3-56
(I)
(2)
(3)

3.61 Water flows into the sink shown in Fig. P3.61 at a rate
of 2 gal!min. If the drain is closed, the water will eventually
flow through the overflow drain holes rather than over the edge
of the sink. How many OA-in.-diameter drain holes are needed
to ensure that the water does not overflow the sink? Neglect
viscous effects.
=fy +if+%2. , where It"0, V, =0, Inti. .21"ti fJ..=0
V1 .. V2,~,·=[2(32..2. Jj) (1+/~2.ff)]~ =2 .5i' ff.
Also)
Q ::: nA2. 14 :: n ~ fd; V:z. J where n:: IJlllflher ofPfJles reqlJired,l
Q2.:= o.'I in; and. C;:::c,nfraction coel.
= 0.61 (see ri'1.3.19)
ThlJs~ with
Q=2 w,( I ""'" ) (~31 in.
J
) (_ Ifl~ )::: ~ ~tx/O".J fi.!111111 60.s 19GJ 17'-8 m.3 , .& )
3-57

3.62
3.(i2 What pressure, PI' is needed to produce a flowrate of
0.09 ft3
/s from the tank shown in Fig. P3.62?
If+:i+ Z2. " 1-+{f +j!~
ThusJ
11 +Ooh
o
where
or
FIGURE P3.62
where -/'2.::f1 +~ hJ fJ3 -: 0
"Z2 :: 3.6 fl J Z3 =0
and ~::tO
31.8-¥
lg Ib 3 I •8 ~ )2. [( £:!. 2-
o(2~ - Z2. ) - t"h == (1./ (6J..1f 7i;» 2.(32.. 2. ~~)
or
- '12.5 ~ (2,0{-0
fJl == 7'1-6 ~ - 5./8 ps/
3-S8

3.63 ,
3.63 Laboratories containing dangerous materials are often
kept at a pressure slightly less than ambient pressure so that
contaminants can be filtered through an exhaust system rather
than leaked through cracks around doors, etc. If the pressure in
such a room is 0.1 in. of water below that of the surrounding
rooms, with what velocity will air enter the room through an
opening? Assume viscous effects are negligible.
I { viseoV.s e(feefs tJ,,~ neqligihleI '
~~ v,~o
V2. (1)rL (I)
-........-.-..----.-..--.
/71111111/77
I v.2. ..L 1/2. h V,I, +'2 P,:: I,. + ,.. f v,. J '" ere I ~ 0 4fJd It -/2 :: i'azo h
or
p, - f:J.::: (~fI)('2.1f {t3) ;;0.520 ~
ThusJ
~ ==[ 2 (prp,.J ] ~== [ 2 (0.520 Ib/I/1.) ]~-:= 20.9 Ji
2. e ] O.OO:J.39SIIJ1s/113 .s
3-SQ

3. 61f I
3.6/f Water is siphoned from the tank shown
in Fig. P3.6'f. Determine the ftowrate from the
tank and the pressures at points (1), (2), and (3)
if viscous effects are negligible.
(1)
•
FIGURE P3.6.1f
fJo of 'v"2. + =z :: -(1,- + ~.2.. +2. f ' r (I)
~ ~~.L.-o r ~ " Of' 6 :: ~'I;; ~
For t';:: 5 and 1'0:::0 J Va =-0) fJs =::0 flus become.s
i!.0 = :;2.+Es or Vs"";2 ~ ( Eo -Z.s)' =1,--2-(3-.2.-2_~:~:--)-(3-ff--"')I
-,-L = /3.'1 lI n/Js)
Q:: /}s 'Is ::: .;-(-Ii:fl)'L(I.3,q fj) = 0.303 ~
Fro/IJ £r.(I) with l-::/ a/Jd t{:::o J ~ = o(Zo-Z,)
~ =(62,Lf ffa)(8 (I) :7l9?fA
Fro/1J E'[ II) wifh /=2 J LO =:Ij+ i,.j. E:z.
where 1J2. ~ ::: flo!' ~ ~ ~'-
SJ/Jce f}2::' 115 if f()llows that L{:::~ or ;; = 2; =20-Z,5
Thus}
~2. ~!1.
1- =Z!o -22. - :2.1 =Eo-%:z. - ,.; =Zo -22. -(Zo-Es )
=i! -2or :5 2-
11 :: t(zs -~2.) =(62,1- -/fo)(s if) =3/2-ffo.
From Er{ (J) wifli t''''3J
20:: Jf+ ;:f+Z.3
where fla ~ ::lis ~ ~2. ~ 2.
Since 1).3:::115 if fo/lows thai ~ t:: Vs or ~; =- 2..1 =&0 -Zs
ThvsJ 2. 2-
:fi; = Z!~ -2'3 - ~ : Zo-E3 - ;; :: %. -2.3 -(Zo-Es)
:; Zs - z,3
o~ =t(zs -%3) =(62.'1' ~~)(-.3fI) :0 -/B7~
3-60

3.65 J
3.65 Redo Problem 3. 61f if a l-in.-diameter
nozzle is placed at the end of the tube.
~) (1)
•
fJo 4,2. _ ..t2. /,.7. • r . _ / S (I)
i' + 2.7 +Zo - ~ +?-9 +%1 Ter t - ,/°°)
For l ~ 5 2. and -/b =0 J !0:::0 J fJ.s =0 fhls becomes
:Eo =- y~ or ~ =- 12.9{~.-:Z5)1= {2.(3Zo2fj-.J (3ff/
Thus ::: /.3.'/ 4-
J 11' (1)2. ii) _ :£:L3Q =A.s ~ == T -;z:ff (/3.9 oS - O.07S8 oS
From E'f(!) w/fh i=/ and it;::o J A::: a(zo-Z,)
=(6Z./f-#r3)(8f-l) == '1991-2-
Fro/h £i/o (J) wilh i ~2 J i!o:: ~ +-¥!i+ ~2.
where 112 V:z. :: A5 ~
Since A:z. == (~y2.A5 = (t)~.s =LJ.As iT follows that
J _ .L ,/ J!:l _J-(-1-1/)2. _ J ~
2
_ -L ( z )V.z - ¥ V.s or 2~ - 29 If ~ - 7D~ - /6 Zo - oS
Thus) 2-
fJ:;. = Zo -Z2 - ~ -:= 2'0 - Z2. - -!r(:ZO-Z5)= afl-!t(3ft)
~ tI == 7. 81 ff
or Ih ) Ib
-Ii =(62 JI- ff.i) (7.81 PI- ::. 'f$8 7P-
From fro (I) w/lh i =3 J 20 == :t; oj. ~;+Z.3
where f}.3 ~ e.As Vs ~2 ~2. /
or s/nce 113.=!J2. then ~::: ~. and 2; = 2.~ = 16 (zo-2~J
Thu,s)
~ ~2. .-l.. J
-; ~ L O- Z3- 2$ :::Zo-~3- /6 (Zo-Z5) = -76 (3{-i) =-/~ PI
or Jl;.. 3 Ih
fj == (62.Jf fP.) (-Tl"H)= - II. 77fi.
_~-61

3.66 Determine the manometer reading, h,
for the flow shown in Fig. P3.66.
Thus)
f, :::: f:;.
I T0.37 m
h
1 .. . 1--I--H--~
a08m - ... - Free
. ...-~ ----,-_ M'·-~ jet
dlameter(J} ,---
'c:::=:=l==== I ....0.05 m diameter
FIGURE P3.66
where Z, =:Ez , ~ -:: 01 and
Vz = 0
!loweVer, f1:::: O'h and 122.::: 0' (0.37m)
so Ihot
h = 0.37 m
3 -6z.

3.67 I
3.67 The specific gravity of the manometer fluitl shown in
Fig. P3.67 is 1.07. Determine the volume flowrate, Q, if the
flow is inviscid and incompressible and the flowing fluid is
(a) water, (b) gasoline, or (c) air at standard conditions.
fluid
kN
O'J m3
m3
Q)s
fa) Wc.i-ler 9.80
-.3
/.06 X /0
(b) ~a.so//ne 6,61 .3. 02. X /0-.3
(e)
. /2 X/o-
3
O.UBair
3-63
t0.09-m
diameter
~
(f) (2.)
.-.,t::::;'"
£1 o.~ m
-~-10 mm
t
20 mm=h
J
4,..--' .... ...,.j
~
FIGURE P3. 67

3.68
3.68 IP-4 fuel (SC = 0.77) flows through the
Venturi meter shown in Fig. P3.68 with a velocity
of 15 ftls in the 6-in. pipe. If viscous effects are
negligible determine the elevation, h, of the fuel
in the open tube connected to the throat of the
Venturi meter.
v = 15 fUs
3.69
FIGURE 1'3.68
where Z/::O J Z2. :::: .f}f-l, (I)
and If::: 15 {-I/.s
or
1!f=-7.S3 ff
Buf if=-h so that h== Z 53 ff
3.69 Repeat Problem 3.68 if the flowing fluid
is water rather than IP-4 fuel.
Note from the solution fo Prob lem 3.68 fhai fhe
value of 0 is /Jot needed, Thus; h =7. S3 ff for
efther wafer 0 r J P- if rIJeI.
3-6Lf

3.70
3.70 Air at standard conditions flows through the cylin-
drical drying stack shown in Fig. P3.70. If viscous effects are
negligible and the inclined water-filled manometer reading is
20 mm as indicated, detennine the flowrate.
FIGURE P3.10,
However) fJz +~J2. +Ymh =={1- '0(1-h-~) where h =(-zomm)sil)l.
or PI;12. ""(~ -I)h +i (2)
By comhinilJg £rt's~(!) and (2)
1;/2.=(!p- -/)h
or ~ 9.8tJ X/0
3
,*3
II, = 2 ~(~ - J) h _ 2 (9. 8/ ~) /2 .o=!b
VI /5 --------~--/~5~~--~-------
- 2.35 ~
Thus}
Q::: AI 'It ==.:IfD/-I/; ==f(2",)2.(2,35.r;) -
====
3-65

.3.71
3.71 Water, considered an inviscid. incompressible fluid,
flows steadily as shown in Fig. P3.71. Determine h.
• FIGURE P3.71
f, -I- l~ +l-pv,2. :::f:z. f rzz.+l:pV/1.
IIi
where ~ '"0 I %;L::.3 (I, Va '"0, and ~ '" -1;:= ;;f/J'" = S. 09 f1
oS'
IhvsJ
1'1 + t(1,9'1 !'ff;)(£.()9 fjl' := 12. +62.oil-if, (3 f.I)
or
" - f" =: / t:J. -Jf,. (I)
B,,1 from fhe h14f1omelerJ
(JJ -¥rJ-l3flJ .J. a(h+/.) :::Iz
or
f,-d2,1f -fp (3ft) ~62.Jf ~ h =/,3-
lienee,
f, :::: f% + le7 - 62.'fh wlll~1J whslJ C()/JIbilJDd wilh £t. II) ,ive.r
It..+ 117 -J2.'fh - f1~ -::/ bJ.
Qr
h= OJI-ooff
3-66

3. 72. I
3.72 Determine the flowrate through the submerged orifice
shown in Fig. P3.72 if the contraction coefficient is Cc = 0.63.
3-67
1 (/)
~ ::----:::::--_-:..:-:---::=:= :..::=::-
--- --:-::=:=::----:..:-:-:-::=:
4ft
1('2.) (.3)
=f='f3-in. 2ft
diameter ... ·f
FIGURE P3. 7~
6ft
/
where fJl:::O J ~:::O J zl::.'1f~
Z2.=OJ qnd 'l =2 ff

3.73
3.7;3 Determine the flowrate through the
Venturi meter shown in Fig. P3.73 if ideal con-
ditions exist.
.£J..+ ~l+Z =.:b:. + v;.2 +Z
'0 2.i I K' 2-9 2.
ThlJs)
tl+ (~rtv;
r '2-~
'Y = 9.1 kN/m3
FIGURE P3.73
(73S -s50)kPa
2(9.8/~) (9.1~)" m == ZI,5!}-
3-68
I - I /9 him )If
t- 3/mm

3.7'1- I
3.74 For what flowrate through the Venturi meter of Prob.
3.73 will cavitation begin if PI ., 275 kPa gage, atmospheric
pressure is 101 kPa (abs), and the vapor pressure is 3.6 kPa
(ab')?
(I) .£!.. + WI.+.i? =.:IJ,. + It.. z
r ~'J ' r 2.7 2
Thvs, wl/h A, l'J' ;: 112 Vt.
or
Vz " 30.6 -!}
Thvs,

Q t '---.)~,k'----1I
--:)31rm.cn .(') 19mm/
')' _ 9.1 kN/m3/
where Z," 1!:2. ) ~ =-3.6 KI'~
and 11 =(pH/OI)kfQ(Qbsl
= 376 kPa labs)
Q .". 2. v,"" ' )" m) -3 m 3
=/12 v.. =vLl" :z. = -:y:- (0.01'1 m (30.65 =8.68 X/O "S
3-69

3.75
3.75 What diameter orifice hole, d, is needed if under ideal
conditions the ftowrate through the orifice meter of Fig. P3.7S
is to be 30 gal/min of seawater with PI - P2 = 2.37 Ib/in.2?
The contraction coefficient is assumed to be 0.63.
FIGURE P3.75
(.3.06 .fi) + 2. (32 .2 .£i) 2.37 X11I-'11p·2. ( Ib)
.s oS 2. 6'1:0 J.li. ft3
or
v,. == IB~8 fj
ThvsJ since
Q = 112. Vz = ~ f d2. ~ if {QI/ows fh47
d. =[ 'IQ ~Ji=[ ¥ x 0.0668 .~·e J~ o.08'f7ff := 1.016 //1,
7l'Cc V,. J 17' (O,63)(l8,Bf)
3-70

.3.76
3.76 An ancient device for measuring time i~
shown in Fig. P3.76. The axisymmetric vessel is
shaped so that the water level falls at a constant
rate. Determine the shape of the vessel, R :::J
R(z), if the water level is to decrease at a rate of
0.10 m/hr and the drain hole is 5.0 mm in di- I
ameter. The device is to operate for 12 hr without
needing refilling. Make a scale drawing of the
shape of the vessel.
-!l~~J.o-mm diameter
~, l~
,
FIGURE P3.76
jf the flow l.s assumed fo
he tj/ItJ.oS i - .sIeQrly,
z, =z./ and Z2- ~o
Thus)
42. = 'vj2. t z which} if ~« ~ (i, e. R» s, OhJhJ~ become.s
Z? 2?
'4 =y2IJZ'
Since AI Vt "Az ~ and l{ = I~I = 0./ T,.(-3,/6~~.s)
we ob1ain == 2.. 7B X /0-
5
~
1T R2 (2,78 X10
5
.p-) =-f-(o.oosm)2.J2 ('1.8J1£;.) Z J
where Rand z are ~m
Thu.s,
B= o.qqB zu;'
or
z. m R.m
0 0 Vessel Radius, R, vs Water Level, Z
0.02 0.375
0.05 0.472 1.0
0.12 0.587
0.22 0.683
0.8
0.32 0.751
0.42 0.803
0.60.52 0.847 E
0.62 0.886 N
0.72 0.919 0.4
0.82 0.950
0.92 0.977 0.2
1.02 1.003
1.12 1.027 0.0
1.22 1.049 0.0 0.2 0.4 0.6 0.8
R,m
3-71
1.0

3.7B* A spherical tank of diameter D has a
drain hole of diameter d at its bottom. A vent at
the top of the tank maintains atmospheric pres-
sure within the tank. The flow is quasisteady and
inviscid and the tank is full of water initially. De-
termine the water depth as a function of time,
/! = h(t), and plot graphs of h(t) for tank diam-
eters of 1, 5, 10, and 20 ft if d = 1 in.
~t~ZI = 1+ ~i+i!z
where fJ; ='0 J fJz" 0 J z/ "'h, Zz ='0 and
Thvs) .
Vz=12.jh' which when cumbined IIIdh At~=' Az ~ ?IPes
-II/I/t =11:.y7.1l or - rr r:1.4f = ¥dzY:l.jh' (I)
where R:1.:= rZ +(h -R)' ~
wdh R,=4- =radivs of lank h-R w:;
Thvs) r=vR'--(h-R;'-' so flJflt £s.{/) beco~i9.s
-[R'--(h-RJ'J ;t '"4; 1/2.1h
or
(h3
/z, _ 2Bh li) dh = d2.fi Jt
h t
Yh%.-2Rh'1.)dh = dZ.pfdt
2R 0
whid elln 6e infe9rafed from
the il'll/tfil limo aAd deplh (t=o,
h'"2.R) fg dII or6drdry f;m~ aM-
depll! (UI) as
or . d2.1/2i
1;(H~i_(2.Rt2)_::R(h3/z._(2Rf2) = :7 t (2)
Use d= ,i ff (wi j=32.2 ~ and pM !J=h(I) fu/'
valves of R = o..s J 2.S J..s:; Q/ld III II
!Vole: .It is easier to solve £r. (2.) as t s t(h) rafhe r
fhl/n h =hit)
3-72.

J.71A. I (con'/.)
Results of an EXCEL Program to calculate h(t) from Eqn. (2):
D=1ft D=5ft D = 10 ft D = 20 ft
t, s h,ft t, S h,ft t, s h,ft t, S h,ft
0.00 1.000 0 5.000 0 10.00 0 20
0.09 0.950 5 4.750 28 9.50 158 19
0.35 0.900 19 4.500 110 9.00 620 18
0.77 0.850 43 4.250 242 8.50 1370 17
1.34 0.800 75 4.000 422 8.00 2390 16
2.05 0.750 114 3.750 647 7.50 3661 15
2.89 0.700 161 3.500 913 7.00 5163 14
3.84 0.650 215 3.250 1216 6.50 6876 13
4.91 0.600 274 3.000 1552 6.00 8778 12
6.06 0.550 339 2.750 1917 5.50 10846 11
7.30 0.500 408 2.500 2308 5.00 13055 10
8.60 0.450 481 2.250 2718 4.50 15376 9
9.94 0.400 556 2.000 3143 4.00 17782 8
11.31 0.350 632 1.750 3577 3.50 20237 7
12.69 0.300 710 1.500 4014 3.00 22706 6
14.06 0.250 786 1.250 4445 2.50 25144 5
15.37 0.200 859 1.000 4862 2.00 27502 4
16.61 0.150 929 0.750 5253 1.50 29714 3
17.72 0.100 990 0.500 5603 1.00 31695 2
18.62 0.050 1041 0.250 5889 0.50 33311 1
19.14 0.000 1070 0.000 6053 0.00 34239 0
3-7.3

1.0
0.9
0.8
0.7
0.6
Water Depth vs Time
D=1ft
----t--
I
!
--~-
-=_ 0.5
.c
0.4
0.3
0.2
0.1
0.0
o 5 10
t,s
Water Depth vs Time
D =10 ft
15 20
10 ,----~-----,-----r--,
9
8 -1---__-
7
6
5
4
3
2 +------~--~-~--+-_4
o +-------~----~r_-----4--~
o 2000 4000 6000
t,s
3-79
Water Depth vs Time
D=5ft
5,----:--~-----,--..,..........-...,.----.
:-i-=:.---"<=~~-=~--=:~---'----i--~---l3 +---t---T---+----r---r-----+----1
-= 3+---~--~--~---+-------~
.c
2 +-_~--4-~~-~_--+--~
2 +----+----'-
O+---+--+--+---+---;-..L--!
o 200 400 600 800 1000 1200
t,s
Water Depth vs Time
D =20 ft
20,------..,..........----~---------~
! I18 -t---~-. ----:-----1
~:~--~--~~----~:--~~
, I
12 +--~--'<+-- -+----1
-= I
.c 10 --I,
:~--~--~--~--~2 I
--I
O+----~----_r--~~~~
o 10000 20000 30000 40000
t,s

3.7q* An inexpensive timer is to be made from
a funnel as indicated in Fig. P3.7C(. The funnel is
filled to the top with water and the plug is re-
moved at time t = 0 to allow the water to run
. out. Marks are to be placed on the wall of the
funnel indicating the time in 15-s intervals, from
oto 3 min (at which time the funnel becomes
empty). If the funnel outlet has a diameter of d
= 0.1 in., draw to scale the funnel with the timing
marks for funnels with angles of 8 = 30,45, and
60°. Repeat the problem if the diameter is
changed to 0.05 in.
.£L+ VI2. +Z = -J2.,. + V:+2
~ 2'1 I r ~9 ;J.
Plug
FIGURE P3.7q
I(I) r- R--..t
where "':::0 J ~ =: 0 J :if =- oJ
22 ::: OJ and ~:::!*-11 « ~ h
ThIlS, if R» -t -l ~~ --.J
{ ~ ';2tjh' which when combined wl1h A~ :::A~ li 9ille.s
-IJ,-# :::A2./zgh' or -7TR
2
1!f =fd:;i~9h'
where R =h ftlne
Th!1sJ £'1' (I) become.s - h2. tan1-e ~ == /-V2.gh'
(I)
or
h
9
/
2
dh =~"f{8 di which Ct.1II De infeprqied From h=ho
III) of t~o as
h t
(h3/3.'d'L =_ d1.Yii Sdl: 2- [5~ 54] _ d2{ij
J. n l/- tlJn"8 or 1f h - h. - - 'f flJll~() t
ho 0
ThvsJ
. .2/5
h [h5/2 S d2. V2i"'t]= 0 - 8 Ttln2.e Since h ==0 wnen
. }, if (ol/ows thaTJ
(2)
t= 3m//)
== /80.s
5A sd2.1(2(32.2 !J) (180s) .
ho ~. = . + 28 which when combined
8 an ~ w/fh £'1' (2.) gives
h
:::[ S d212(32'~#:l./iIBOS)J (I _--L)Z/S
8 tan2.8 . J 180
or
rJ. ¥/.s t 2-,1; (3)
h = /5.2 ( fan e) (1-/eO) where h- ft J d~nJ and t~s
(con'i)
3-75

The results of an EXCEL Program using Eqn. (3) to calculate h as a function of t are shown below.
The time interval markings for the sixfunnels are shown in the figures on the following page.
d=0.1 in.,8=30deg d = 0.1 in., e== 45 deg d = 0.1 in., e== 45 deg
t, s h,ft t, s h,ft t, s h,ft
0 0.512 0 0.330 0 0.213
15 0.495 15 0.319 15 0.205
30 0.476 30 0.307 30 0.198
45 0.456 45 0.294 45 0.190
60 0.435 60 0.281 60 0.181
75 0.413 75 0.266 75 0.171
90 0.388 90 0.250 90 0.161
105 0.361 105 0.232 105 0.150
120 0.330 120 0.213 120 0.137
135 0.294 135 0.190 135 0.122
150 0.250 150 0.161 150 0.104
165 0.190 165 0.122 165 0.079
180 0.000 180 0.000 180 0.000
d = 0.05 in., e== 30 deg d = 0.05 in., e== 45 deg d = 0.05 in., e== 45 deg
t, s h,ft t, s h,ft t, s h,ft
0 0.294 0 0.190 0 0.122
15 0.284 15 0.183 15 0.118
30 0.273 30 0.176 30 0.114
45 0.262 45 0.169 45 0.109
60 0.250 60 0.161 60 0.104
75 0.237 75 0.153 75 0.098
90 0.223 90 0.144 90 0.093
105 0.207 105 0.134 105 0.086
120 0.190 120 0.122 120 0.079
135 0.169 135 0.109 135 0.070
150 0.144 150 0.093 150 0.060
165 0.109 165 0.070 165 0.045
180 0.000 180 0.000 180 0.000
(colli)
3-76

six funnels (d ~ O.oS in. or
! Z i i _ I : Y ; , ;. -h.." i ! : ' -c-+
1-+-+-i--1C-+-i--+-#:-++-r+7!V~!+-t-+-'r-+-t1 -i:;.'",':?-+-t-J-----r-t-';- ~. --l-+- H---+-.L-+
I: ,I I i/' , .v, I / , -H-+-~t-:.'. ,: ~+- +++'.' I l.i I' v, ' , I .Y ~-t-i. " .' . .
0
1 .----rrl if v, , I/~. I~ , .. , I,
• /, IV 1/111 II I 'I
I ' , : i [/
I ii' , I ; ~ / ' : : + :+r-~'+'~ j-L
~..i-H y-: vr· I ! : I ' I ~ ~.'.. '• ...l.. f-C--,.'1-~-...:......:iT,.
~;- I!./'
3-77

3.80
3.80 The surface area, A, of the pond shown
in Fig. P3. SO varies with the water depth, h, as
shown in the table. At time t = 0 a valve is opened
and the pond is allowed to drain through a pipe
of diameter D. If viscous effects are negligible
and quasisteady conditions are assumed, plot the
water depth as a function of time from when the
valve is opened (t = 0) until the pond is drained
for pipe diameters of D = 0.5, 1.0, 1.5, 2.0, 2.5,
and 3.0 ft. Assume h = 18 ft at t = O.
h (ft)
o
2
4
6
8
10
12
14
16
18
FIGURE P3.80
A [acres (1 acre = 43,560 ft2)]
o
0.3
0.5
0.8
0.9
1.1
1.5
1.8
2.4
2.8
D
where -Ii =0 J 12.::: 0 J z; =hJ 22 =-311
and ~ ;: - j <.< ~
-2-
Note: t ~ D
which when combined with A~ =112. ~
(con/f)
3-78
OJ

( cO/J'I)
An EXCEL Program using a trapezoidal integration approzimation was used to calculate the results
shown below.
D =0.5 ft D =1.0 ft D =1.5 ft D =2.0 ft
h,ft A, acres A, ft2 t,s t, s t,s t,s
18 2.8 121968 0 0 0 0
16 2.4 104544 32181 8045 3576 2011
14 1.8 78408 59530 14882 6614 3721
12 1.5 65340 82354 20589 9150 5147
10 1.1 47916 101536 25384 11282 6346
8 0.9 39204 117506 29377 13056 7344
6 0.8 34848 132412 33103 14712 8276
4 0.5 21780 145035 36259 16115 9065
2 0.3 13068 153988 38497 17110 9624
0 0 0 157704 39426 17523 9857
The graph for D =1 ft is shown below. The shape of the curve is the same for any D.
20
18
16
14
12
~ 10
.c
~~
Water Depth vs Time
for D =1 ft
~
~....
I~
~.....
~
D =2.5 ft D =3.0 ft
t,s t, s
0 0
1287 894
2381 1654
3294 2288
4061 2820
4700 3264
5296 3678
5801 4029
6160 4277
6308 4381
8
6
4
2
o
~
'..
1...
o 10000 20000 30000 40000
t, s

3,8 I I
J.8' Water flows through the branching pipe shown in Fig.
P3.81. If viscous effects are negligible, determine the pressure
at section (2) and the pressure at section (3).
10 (7-):
A3 =0.035 m
2
~_------, ;:3 = 10 m
(3)
(2)
V2 =14 m/s
A2 =0.03 m2
where Z,:: ~2. -= 0 Qllri
or s
(1,. ::: 2,52 XIO .f-,. :::: 2S2 kP.
Alon9 fhe sfreamline from (0 10 (3):
~1 ~'Il ~&
t +if +Z, = 7' t 4 +2,3 where Since rt, =Q,. +Q3 then (I)
Q3 :::: A.J v., =~ ..Q;l =CV, - At. V2. Sf) thai
~::: Q,:A1.V2. _ 1111% -O.03","(/¥1II1,s) =16,t,.!R-
A.3 - 0,035 /)'J2.. ~
TIllIS) £'1' flJ becomes (wtlh ZJ ~O J Z,J <# 1011J)
.300 X10
3
#/.,2 + flo 1II/.s)'J. , _ {J3
9.801.,03 /'11m3 2 (tl. 81rnl.s2.} - 9.f}ox/ulNlm3
or
I.J::: 1,lifX/O
S
f;. :::: JIJfkPa
3-80
+ (JI•.tWJIs)'J.. +-Iom
2-('1,f)1mIs')

3.6:2.. Water flows through the horizontal branching pipe
shown in Fig. P3.82.at a rate of 10 ft3
/s. If viscous effects are
negligible, detennine the water speed at section (2), the pressure
at section (3), and the ftowrate at section (4).
From (I) 10 (2.):
From (!) +0 (3) ;
Also}
(Jo~l _
2(32..2 f{) -
-
Q~ :::: Q, - Q,. - Q3 ::: Q, - II:;. { - A3 ~
A3 =0.2 ft2
V3 = 20 ft/s
"------..-
or Ih .
~ ::: 1/50 ff'- ::: 7.98PSI
or 3
() IP I'J. ( tf ) f'.L2.f fi) - 3, 97 LstU{1f ::: lOs- - o.o7f 2.9.0-:s - 0.2 l. l20 oS

3.83
3.83 Water flows from a large tank through
a large pipe that splits into two smaller pipes as
shown in Fig. P3.83. If viscous effects are negli-
gible, determine the flowrate from the tank and
the pressure at point (1).
T7m
L
0.02 m
FIGURE P3.,B3
t:; +Iff Zo '" :If 4- lJ':.,.4. where A"0 I 1'2. ,,0 I Vo '"0 J zb" 7/1)
n~ ~d~=~
J
Ik ::';21 (Zo -Z20)' =f2. (9.8Ilfo.) (7- ~)ml ::::: 7, 67.1f
!Similarly
~ '~ Ir+-7-7-(:2-0
--Z-~---,)I = ,;2 ('1.8/ fA) (7m)' = //. 7.1f
ThvsJ
Q:: Qz.'" Q3 :: flt;2. -I- fD;~:J.
or
Q:: f[(O.03h1)"(7.67~) .J.(o.02.m)~(1I.7.p.)J ::: 9./fJx/o-.3.!J!
Also)
-h. t Vg=z'f' 70 - .fl. + Vt'- +Z uJhere 2/:::: 0 and
r ~I .z; - r ~'i 'I "':" 3 .l!J!
or /,:: Q == 9.I()xlo- s =- ~63.1!L
I -;z;; -5(0. 0.5 11J)2 • .s
_~[z - V,2] n 3 J:L. I (Jf.63!f-)2.] .¢ AI
f1 - 0 0 2Q =7.fo)l./O m3 7m - (0 0(.') =s:7?X/O 4. T 2 h8Js~ m
or
f1 ::: 5 7. 9 kPo.
3-81-

3,8Lf
3•.8/f Water flows through the horizontal Y-fitting shown in
Fig. P3.'6/f. If the flowrate and pressure in pipe (1) are QI = Ql-"
2.3 ft3/S and PI = 50 Ib/in.2• determine the pressures, P2 and
P3' in pipes (2) and (3) under the assumption that the flowrate
divides evenly between pipes (2) and (3).
FIGURE P3.9/f ~(2)
Q2
3-83
20 ft

3.85 r
3.85 Water flows from the pipe shown in Fig.
P3.85 as a free jet and strikes a circular flat plate.
The flow geometry shown is axisymmetrical. De-
termine the flowrate and the manometer reading,
H.
where
7l4.h I
11. D2. V2
If I
Hencel
Eq. 0) qive.s
t H
1-0.1 m -1- . I
I Diameter 1 -II(2)
--~~r«~=~ I 0.4 mm
0.2 m
i i,
III (/)O.Ol-m
. 'J 111r diameter
PIPe/blJ
tQ
FIGURE P3.85
(J.6ov"f
J
·:: ~2. ~ 2,(fl.81-!k) (0. 2./IJ) or v" = I.Sq.lf
.so fhtrf
r; .: ~1 v,. = 71(0,/PJ)(t;.X!O-IIIh ) (I.sql}) :: 2.00X/O-'".p-3
3- 8Jf
(I)

3.86
3. 86 Air, assumed incompressible and inviscid, flows into
the outdoor cooking grill through nine holes of O.40-in. diam-
eter as shown in Fig. P3.96 . If a flowrate of 40 in.3/s into the
grill is required to maintain the correct cooking conditions, de-
termine the pressure within the grill near the holes.
Q = qA.z v.z
Thvs~
~ = Q
,.. 9~
A/soJ
where
-¥Q
- 971D,2.
:2
FIGURE P3.86
9 holes, each
DAD-in. diameter
£!. + 'Vi2.+ -=z =..ti:. + V2.'2. -1-:1 where LJ =0 z::Z and ~::o)f' Z lJ ~I K' "-9 2. r I J I ") I
Thlls,
f:;. =- f PV: = - i (2.38 x/o
3
, .s};r-)(2..Qlf!j.) 2.=- /.03 x/o-2.#1.
3-85

3.87
3.87 A conical plug is used to regulate the air
flow from the pipe shown in Fig. P3.87 . The air
leaves the edge of the cone with a uniform thick-
ness of 0.02 m. If viscous effects are negligible
and the flowrate is 0.50 m3
/s, determine the pres-
sure within the pipe.
Q '" 0.50 m3/s
--I~~ -~+-
FIGURE P3.87
v

3.8a r3. 98 An air cushion vehicle is supported by
forcing air into the chamber created by a skirt
around the periphery of the vehicle as shown in
Fig. P3.Be. The air escapes through the 3-in.
clearance between the lower end of the skirt and
the ground (or water). Assume the vehicle weighs
10,000 lb and is essentially rectangular in shape,
30 bySO ft. The volume of the chamber is large
enough so that the kinetic energy of the air within
the chamber is negligible. Determine the f1ow-
rate, Q, needed to support the vehicle. If the
To svpporf fhe load
Also}
ground clearance were reduced to 2 in., what
f10wrate would be needed? If the vehicle weight
were reduced to 5000 Ib and the ground clearance
maintained at 3 in., what f10wrate would be
needed?
Fan Q
,J ( / Vehicle
Skirt .rr ~~~ 11 ..",
~ 'If .¥ - .'M L~
'///////////////////////////////////////////////t//////7,1/,0
3 in.
FIGURE P3.89
where W= vehlf;!e weigh!
and Ao= (30·f/) (50!+) ::: 1500 fll.
.I!E... + to:L+.z. :::: -A- + ,42. +% where 4.:::0 J ~()::OJ andZ()-;:22.
t 2g 0 r ~1 2 F~
so fha+ ~/_';26' /_ 2.W
V2. - P or V2. - ,qfJf
Wah h::: fjrovnrl clearance if follows fha-l
Q=fJ2 ~ == 2h (L+ b) ~ where L=50 tI and b:: 30il
Thus) I '
Q=2h(soff +30f.l)i(JsOOfI1)(:'~8X/O-3 ~). fI.i
or
Q:::: IIq.8 hYw' ¥ where h.-v fJ and W-/6
Thus) If h =/~ ff and W=/g 001) /6 J the/} q:::: 3000 !f
if h ;: ~ ff and W=/~ooo I~ J fhe/) Q::: 2.000 .if
======
and if h = -Ii ff and 'II =5J OOO /i:J J the/) Q= 2/20 .fj-~

3.8Q
3. e'l A small card is placed on top of a spool as shown in
Fig. P3.eQ . It is not possible to blow the card off the spool by
blowing air through the hole in the center of the spool. The
harder one blows, the harder the card "sticks" to the spool. In
fact, by blowing hard enough it is possible to keep the card
against the spool with the spool turned upside down. (Note: It
may be necessary to use a thumb tack to prevent the card from
sliding from the spool.) Explain this phenomenon.
• FIG U REP 3 •eq
As fhe air flows r4dially ovfward in fhe 90P he/ween the cord and
fhe spool if .slow.s down since fhe Flow area ilJcreases will? r; the
radial dis/once from fhe center. That ISJ
(J = 2-71' r h VJ or V= ').!hr (see the fi'jvre). ,_1 rexif ~ (I)
I r.. I exit
If visco[).s effecf.r are nfJ/ imporfanfJ h I I VI
then :2.. r1// /! /#.: +X :::: cOl/slanl ;::: {JexH -I- 'lexif
r '-I- (f 29'
/ for since lex/I:::O (a free jel) if Q
follows fhal 2. 2. Z _ 2._ Q 2.[ J _ J..
{J '" if('iex/I - V ) ) where from Er- (/) ~Xlt V -t2rrh) ~t ri]
Bvf rex/I> r so fhat f ~ 0, !here /s a Vaclltlm wlthi/) fhe 9a/.
The card is svcked 1J9oi/ls! fhe spool, Tl;e harrier one blows fhrIJ{)flA
fhe sfJ()o/(/atyBr())) fhe /rJrger fhe V(JCIIIIII'IJ IJnd fhe harder fh~ CQrd is
held rJ9aiIJsi the spool.
3-88

3.1:(0 I
sz £ech !t=CHlI2=f
3.90 Water flows over a weir plate (see Video V 10.7) which t'~(/f ~-Q~-:--'-"'" f * I - , - '~'-:J
has a parabolic opening as shown in Fig. P3.90. That is, the 'Z 2)::- H
opening in th~ weir plate ~as a width CHIP, where C is a con- t,' ,,~ ] _f /
stant. Detennme the funcbonal dependence of the flowrate on D f. '" "tAR
the head, Q = Q(H). ZeD
Q== fLl dll where IJ. is t.t {()nc1;OA () rh.
Thai is, frl1m Ijf +1/#2, '" IF#.%j+'E:z.
or 2. U1-
(H-~/)+lj+z,::O +0/ +(II-h)
Tnvs) __
II = Vl,h+Jlt2. ~ f'J-,:h il ~ is "s/lla//
IJ
wilh /If =H-Z'J t{ =ll
If::o (,'free jei,
alld, ZJ. :::II-h
Also
dA):# C fftJ,r (i. e. dll:: 0 dl fflr Z${). till;&cffiifIr ;~II) So fhai
bH J
Q=f'~I'1/h cWrJl where h:::H-z.
p~o H
ThlJs) f¥ =: c1};jf1-Z/l--2--""1; dl
J
where
H 0 pH
S1{iiHI71i:: f[(z--!)YHZ -Zo." +(~t.sin-I[(Z--f)/(H/J.)fl
o zaO
whjGh reduces If)!
~::: ¥ Vii H2. Thai is Q- H:L
AIfernative/yJ Q::: Vt9 where the QveraJe ve/Qc/IJ isfJrPjXll'liPIlII!
fo 1if(t·.e. V'" Vi-,,#) IJlJti fhe 1,14'/ flow area., is pl'rJfJOn/olJa/
fo H~ (t·. e. A- HN (C HI'..) ::: cH.3h. ). TlJv~
Q'V ¥2-~H- (CH*) :: CV,.i' H2-
That Is) ~'V HZ 4S ohfained above.

3.91 A weir (see Video VI0.7) of trapezoidal cross section
is used to measure the flowrate in a channel as shown in Fig.
P3.91. If the flowrate is Qo when H = e/2, what flowrate is ex-
pected when H = e?
FIGURE P3.Q(
Q == II V where if t'.s eXp'ecferl Mai Vis (J. function of the head.J fl.
Thai is I V-,J2,Hi
Also, from fhegeomelry A=fH(t+J,.) where~::=.t+2.Hff/173o"
ThusJ
A::: /I(i + H11/1'}.300) so fha1 T
Q== C; 1'"1'"(J.f H tan30G
) H~/:1. where q ;s a con.stanf
Lei Qo= f/oWfa1e when. jJ::: l-
and Ql::: {Iowrale when H=1
Thus, 3,t
Qo _ C;p:j (1-+- f ftln 300)(i);" _ (J + ifan300)
q; - c,Vii (P. 4- I, fan 31l) (£)3/2. - (I +tQn 30°) (2~/2.) :; O,2.Sq

3.92 Water flows down the sloping ramp
shown in Fig. P3. q2. with negligible viscous ef-
fects. The flow is uniform at sections (1) and (2).
For the conditions given show that three solutions
for the downstream depth, h2 , are obtained by
use of the Bernoulli and continuity equations.
However, show that only two of these solutions
are realistic. Determine these values.
:£L + Vl +:r = -Ii + V2.2. +Z2
~ 21 I 1 29
fJ/so) A, V; ::A:t ~
or H
V -..hLv, - (/N)(IO.s):; /0
2 - h,. I - h2. h2.
Thus) Ect. (I) becomes
Jj. 2. ( /0 )2.
( 10 ;s) +3 ff _ hi.
2 (32.2 ¥:z.) - 2(32.2 ~J
or
6'f. if h:-2 93 h:~ 100 -::: 0
FIGURE P3.Q2.
(I)
where f' =0 , ~ -::0 J Z, :: 3 ill
and Z2 =n2
By v.sin, a root findin~ pro9ram the fhree roofs to this o()bic
erllalion ~re fovnd fo b8;
h:;. = 0.630 f1
h2 = if.IJf8 fl
Or
h2 = a ne9tt1il/e roof C/early if is not po.ssihle (physicql/y)
If) hlJve h2. <0 -rhus, h,.:::: 0.630 H or
h2. = Jf.'f8 ft

3.Q3 J
3. q3 The flowrate in a water channel is some-
times determined by use of a device called a Ven-
turi flume. As shown in Fig. P3.Q,3 , this device
consists simply of a hump on the bottom of the
channel. If the water surface dips a distance of
0.07 rn for the conditions shown, what is the
flowrate per width of the channel? Assume
the velocity is uniform and viscous effects are
negligible.
T0.2 m
FIGURE P3.q3
w/fh"11 =0 J fJ2. =0 J E, == 1.2mJ
ond 22 ::: /.. 2m - 0.07117 = /. /.9m
= I, 29 ~
or [(1.2.9)2.-I] w'"= 2. (q.8/~) (1,2. - /.13) m
or 1 =I,'I38~
Hence,
Cf =h, ~ =(1. 'I38.f) (1,2m);:: I. 73 ¥
(I)

3.9'f Water flows in a rectangular channel that
is 2.0 m wide as shown in Fig. P3. qlf . The up-
stream depth is 70 mm. The water surface rises
40 mm as it passes over a portion where the chan-
nel bottom rises 10 mm. If viscous effects are
negligible, what is the ftowrate?
Also) A,W:::A:l~
or hiI I, O.07m / II
V2::: h:t I = 0,/0 m VJ == 0·7-'(1
FIGURE P3. q'f
where f'l:::0, 12.:::0 , Z,:::: O.07R'1, (J)
and Z2.;:; (o,()/ +o,/o)rn ~ O,lIm
Thus, &t. (I) become.s
[J-O.72.]/,'-=2(Q.81*)(O,II-o.o7)m or y,== /.2'1~
Hence}
Q=A/~;:: (O,07m)(2.0m) (J.2lf!f):: O.17q.
3-93

3.QS I
3.95 Water flows under the inclined sluice gate shown in ~'~~'..-:~'''' - ... --'--~
1
(1)
Fig. P3.95. Determine the flowrate if the gate is 8 ft wide.
ThllsJ
~+6H = Jt.t'f-f"-I ,..,.
B(IT A, V, :: II,. V2-) Qr
/_ A, _ 611" II
Y2. - 1f1 VI - 7iT v, :: 6 v,
Hence., £". tI) become.s
Jt" +6ff ;;: f1Y~Y/·t IIf
'-, "1
6ft
j
wher~ 11 # (J I fo ::I) J Z, ~ IfT,
alit/. Zz. # I ff
~1_/]~2 ::: 2 (32.Z.!f,.)(6-I)ff Q ~ ;;:;.3.03#
HenceJ .1
Q:::./1, V, :: ~H (et~)(.3.0.3!j.) =I~S.!f
(/)

3.96 Water flows in a vertical pipe of 0.15-m
diameter at a rate of 0.2 m'/s and a pressure of
200 kPa at an elevation of 25 m. Determine the
velocity head and pressure head at elevations of
20 and 55 m.
",g
V= il. == 0. 2 -::s m 1/
LI:a 2. =//,3 -..5 =Y'o :::- ~
n If (O,15m) ....
IJI p()int (0):
~ == (1J,31fl == 6,5/m
d
'"I 2(Q,8I/fo)
an ~_ ,h2 A)/ ~2.
.u:.!L+-L+z ::~+....l..+z
't 29 0 ~ ~D I
orA _ 200 ~ ( ,
r - 9.80!y:£ + 25 - 20) m
m3
== 25.'1 In
D=0.15", -
~ =~5tn • (I)
(1= 200kPa
70 =20m • (0)

3.t17 Draw the energy line and hydraulic
grade line for the flow shown in Problem 3.6/f.
~) (1)
•
FIGURE P3.6'1-
For /lJviscid {low w/IIJ no pfllnjJs or furbines} Me e/Jer9Y line (EL)
is hori:zo/Jla/ I af tJlJ elevation 01 the free s(Jr~7ce. T/;e /;ytirduhc
grade line (H6L) is ()ne ve/oc/I; head lowe!',) ~welJ w/IJ file fJipe
outief. Since Ihe Iluid ve/ocily is cOlis/tint Ihr1)v9h,vr lIJe pipe
wilh -fi= :3 r~ the fd//ow/IJJ is Qbfained:
3-96

3.98 I 'V (0)
..:----:::;:-----:----~::-- --
3.98 Draw the energy line and the hydraulic
grade line for the flow of Problem 3.60
FIGURE P3.60
For il)lI/scid flow wilh no ptJmps or turbines, the energy line
is horj2()nfo~ a disflJnce /; above fhe ovf/et . From Problem 3.60
we obf"in h :: /. 7'1 fl..
The hydrolJlic qrade line is ~~ below fhe energy line, sl"rftng
af Ihe free surface where Vo::: 0 and ending at fhe pipe exil
where fJ2. =0 and ;f=h. IN poinl (I) the pressure he(ui
i.s fJ'/l = (2..98 - 1'1-.5) jff.:z. ( /~:;A?-) /62Jf Ia =- 2.6.S I~
and £/ =0. 2-
In the ~;n. pipe Vj =/12. ~/Il.1 =(t) ~ so fIJaf
V:z. If ~2. 'I- If
-.¥.-J =( D:l.,) ~ =(.A) h =(..£-)(/.7'1 fO = O. //2 (I
29 D3 ~9 D3 If
The corresponding EL and HGL are drawl) 10 scale below,
~
li:: 0./11 II
~
-----Jr~~' --;:=-_=-=_-==-~Ener91 LliJe
V (0)
4(3) , .l/). - 'Cf) Z! 1:: 0
)
1 , t_ _ _ _ _ ' I ,,,2-
pipe eenlerline: : 2:!. =h=1.7'1 It
I I ZQ, I ,
I I
I ', ,I ,
I ,
I
,
HydralJlic Grorie --:
:1-= -26.afj
I
ILine (HGL) :, ,
, I
I ,
, ,
I ,
, '
, r
I ,
, I
r ,
I ,
!..J--1
3-'17
(EL)

3. qq J
2-in.-diameter
hOS~
l
3. qq Draw the energy line and hydraulic
grade line for the flow shown in Problem 3.65. J,lln.
~~ .
2 ft (0) '/
7. ~ ·(3)t- t=-=::-J:§:j::e:-=
--L B ft
(S), (If)
(1)
•
For il'Jlliscid flow wilh IJIJ lumps or furbine.s) fhe e/)~r}' li/)e (£1-)
is. horizo/J/tll, af (J/) elevafio/J ()llhe free sur/dee. TlJe~hy/rdv/Ie
9rade line (/l6L) ;0$ one veloc'fJ heod lower. SInce 2
Ys ;: 3 It
if follows fhal fhe !leL pa.rses Il)/'f)u,h tlJe f.i;; u( fhlIlOZ2/8.
I//so, since ~ '" ~:s =(*r~ if Mlol/IS Ihal
~.7- D)'" ~2 II- 3
-;.r =(l); zg = (t) (3fl) :: III ff . T/;r()u,houf the pipe
Ihe velocil, head is c()ns/olll Sf) fh4f the fQ/lolllilJg /.S obTtJ/IJeri:

3.10'0 * Water flows up the ramp shown in Fig.
P3.100 with negligible viscous losses. The up-
stream depth and velocity are maintained at hi
= 0.3 m and VI = 6 m/s. Plot a graph of the
downstream depth, h2' as a function of the ramp
height, H, for 0 s:,.H s:, 2 m. Note that for each
value of H there are three solutions, not all of
which are realistic.
FIGURE P3.100
III + ~2. + Z I =.:t!2: + V'%'2. + i!.:;..
o 2.q ~ 2q
(I)
where fJI ~O J 1'2. ~O J 2/ ::: o.3m,
I//.soJ At11 :::,42. U. .so fhat
(Jlui z,2.::: fI+I;2.
J - fJ, II - (O,31TJ)(6~) - ~ 1~II..JeJ"'e h2,"''''V2. - ~ Yj - n2. - h2. ",rl", .
ThvsJ £r- (I) become~
(1.8 )'J.
:; +0.3 m = ~ + (f'~h3.) or wdli L{ =' b~ J
(61}i' +2. (9.81 ':2.) (0.3 - f/- h,Jm =(-~~ )'J-q;
which CdlJ be writ/en as ~
h:-(2_13.5 -R)h: +0.//'5/ =0
For 0 ~ f1~ 2m solve Ert. (2) {or h2.
Ra-lher than SO/Vifl1 a Cf/b/c e9{)ahon for h,. (?llIe II)J one
con directly solve for 1/f9/llen 171-)' From £r. (1-) :
(2)
1/=2./3.5 _l _ 0,/651 (3)
"2. h"J.2..
fJ 9raph of £'1, (2) or(3) /s 9/l/en on the {pi/ow/f)! pa98.
(con'-I)
3-qq

The results of an EXCEL Program to calculate H for given values of h2 are shown below.
h2' m H,m
0.3 0.001
0.4 0.703
0.5 0.975
0.6 1.076 2.2
0.7 1.098
0.8 1.077 2.0
0.9 1.031
1.0 0.970 1.8
1.1 0.899
1.2 0.820 1.6
1.3 0.737
1.4 0.651 1.4
1.5 0.562
1.6 0.471 E 1.2
1.7 0.378 N
1.8 0.284
.c 1.0
1.9 0.189
2.0 0.094 0.8
2.1 -0.002
0.6
0.4
0.2
0.0
Water Depth vs Elevation Change
I+--~~---~---------- .~--------1
-~I
----I
I
I
t====~:::::::::~=---...-=::::::--~--~---t-
0.0 0.2 0.4
I 1
-------- ------ --·----r -----1
J i
I I
0.6 0.8
J.oqg
1.0,' 1.2
H,m
For H~ I,OQ8 m fhere are no realJ posilil/e roofs of Ef(. (2.).
ThaiisJ for fhe (jive/J upstream condilions (V; == 6 f- and hI::: O.3/n)
We must have 11<: /,098 m. II w()IJld nol be possible fo hove
the flow 90 vp aramp of 9rB41er hei9)rI fhalJ fhis w/-/huv/
/ncreas/f)1} e/lher I1lJnd/or hi . The two P(J,s.s;;/g waler de,fJ;.r
for a 9ive/J HtJfQ plo/leri be/ow.
3-100

3, /01 I
3.101 Pressure Distribution between 1vo Circular Plates
Objective: According to the Bernoulli equation, a change in velocity can cause a change
in pressure. Also, for an incompressible flow, a change in flow area causes a change in ve-
locity. The purpose of this experiment is to determine the pressure distribution caused by air
flowing radially outward in the gap between two closely spaced flat plates as shown in
Fig. P3.101.
Equipment: Air supply with a flow meter; two circular flat plates with static pressure
taps at various radial locations from the center of the plates; spacers to maintain a gap of
height b between the plates; manometer; barometer; thermometer.
Experimental Procedure: Measure the radius, R, of the plates and the gap width, b,
between them. Adjust the air supply to provide the desired, constant flowrate, Q, through the
inlet pipe and the gap between the flat plates. Attach the manometer to the static pressure
tap located a radial distance r from the center of the plates and record the manometer read-
ing, h. Repeat the pressure measurements (for the same Q) at different radial locations. Record
the barometer reading, Hatm, in inches of mercury and the air temperature, T, so that the air
density can be calculated by use of the perfect gas law.
Calculations: Use the manometer readings to obtain the experimentally determined pres-
sure distribution, p = p(r), within the gap. That is, p = -"Imh, where "1m is the specific
weight of the manometer fluid. Also use the Bernoulli equation (Ph + V2
/2g = constant)
and the continuity equation (AV = constant, where A = 27Trb) to determine the theoretical
pressure distribution within the gap between the plates. Note that the flow at the edge of the
plates (r = R) is a free jet (p = 0). Also note that an increase in r causes an increase in A,
a decrease in V, and an increase in p.
Graph: Plot the experimentally measured pressure head, ph, in feet of air as ordinates
and radial location, r, as abscissas.
Results: On the same graph, plot the tJleoretical pressure head distribution as a function
of radial location.
r- _ 1
hI
I
/
circular plates r-r ----- II1II T
1________---___II-Water
-1
1
• ~ir.-,.__vBlli·_-'"
~-----R-----4~1
tQ
(con'/: )
3-/01
• FIGURE P3.101

3,/0 I
r
I
I
I
(conJl)
Solution for Problem 3.101: Pressure Distribuition between Two Circular Plates
Q, ft"3/s R, in. b, in. Halm, in. Hg T, deg F YH2D, Ib/ft"3
0.879 5.0 0.125 29.09 83
Experiment Theory
r, in. h, in. ply, ft V,ftls
0.7 -9.05 -663.75 220.8
1.0 -6.02 -441.52 161.2
1.5 -2.02 -148.15 107.4
2.0 -0.96 -70.41 80.6
2.5 -0.48 -35.20 64.5
3.0 -0.24 -17.60 53.7
3.5 -0.13 -9.53 46.0
4.0 -0.03 -2.20 40.3
4.5 -0.01 -0.73 35.8
5.0 0.00 0.00 32.2
P =Palm/RT where
Palm =YHg*Halm =847Iblft"3*(29.09/12ft) =20531b/ft"2
R = 1716 ft Ib/slug deg R
T =83 + 460 =543 deg R
Thus, P =0.00220 slug/ft"3 and Y= p*g =0.00220*32.2 =0.0709 Ib/ft"3
ply =YH2D*h/y
v = Q/(27trb) = 0.879 ft"s/(2*3.1415*(O.125/12)ft*r)
Problem 3.101
62.4
ply, ft
-740.7
-387.2
-163.1
-84.7
-48.4
-28.7
-16.8
-9.1
-3.8
0.0
Pressure Head, ply, vs Radial Position, r
-200
4::
-;.. -400
-Q.
-600
41
f
-800
0.0
/!
2.0
r, in.
:
,
,
I
I
4.0 6.0
• Experimental
- Theoretical
IL-______________________________________________________
3-/02
I

3. 102,
3.102 Calibration of a Nozzle Flow Meter
Objective: As shown in Section 3.6.3 of the text, the volumetric f]owrate, Q, of a given
fluid through a nozzle flow meter is proportional to the square root of the pressure drop
across the meter. Thus, Q = Khl/2, where K is the meter calibration constant and h is the
manometer reading that measures the pressure drop across the meter (see Fig. P3.102). The
purpose of this experiment is to determine the value of K for a given nozzle flow meter.
Equipment: Pipe with a nozzle flow meter; variable speed fan; exit nozzle to produce a
uniform jet of air; Pitot static tube; manometers; barometer; thermometer.
Experimental Procedure: Adjust the fan speed control to give the desired flowrate, Q.
Record the flow meter manometer reading, h, and the Pitot tube manometer reading, H. Re-
peat the measurements for various fan settings (i.e., flowrates). Record the nozzle exit di-
ameter, d. Record the barometer reading, Hatm, in inches of mercury and the air temperature,
T, so that the air density can be calculated from the perfect gal law.
Calculations: For each fan setting determine the f]owrate, Q = VA, where V and A are
the air velocity at the exit and the nozzle exit area, respectively. The velocity, V, can be de-
termined by using the Bernoulli equation and the Pitot tube manometer data, H (see Equa-
tion 3.16).
Graph: Plot flowrate, Q, as ordinates and flow meter manometer reading, h, as abscissas
on a log-log graph. Draw the best-fit straight line with a slope of Yz through the data.
Results: Use your data to determine the calibration constant, K, in the flow meter equa-
tion Q = Kh1
(2.
Flow meter
maometer
Air
meter
Pitot tube
Water
Exit noule
Pitot static
tube
3-/03
• FIGURE P3.102

Solution for Problem 3.102: Calibration of a Nozzle Flow Meter
d, in. Hatm, in. Hg T, deg F
1.169 29.01 75
h, in. H, in. ~P, Ib/ftA2 V, fUs Q, ftA3/s
11.6 5.6 29.1 162 1.20
11.1 5.4 28.1 159 1.18
10.7 5.2 27.0 156 1.16
10.1 4.9 25.5 151 1.13
9.6 4.7 24.4 148 1.10
8.8 4.3 22.4 142 1.06
7.9 3.9 20.3 135 1.00
7.2 3.6 18.7 130 0.97
6.1 3.1 16.1 120 0.90
5.4 2.7 14.0 112 0.84
4.5 2.3 12.0 104 0.77
3.8 2.0 10.4 97 0.72
2.9 1.5 7.8 84 0.62
2.1 1.1 5.7 72 0.53
1.0 0.6 3.1 53 0.39
P =Patm/RT where
Patm = YHg*Hatm = 847 Ib/ftA3*(29.01/12 ft) = 2048 Ib/ft"2
R =1716 ft Ib/slug deg R
T =75 + 460 =535 deg R
Thus, P = 0.00223 slug/ft"3
V = (2*~p/p)1/2
Q =AV where
A = nd2
/4 = n*(1.169/12 ft)"2/4 = 7.45E-3 ftA2
From the graph, Q = K h1/2 = 0.358 h1/2 where Q is in ft
3
/s and h is in in.
Thus, K =0.358 ft3
/(s*in.
1/2
)
3-/01/-

I
I
Problem 3.102
Flow Rate, Q, vs Manometer Reading, h
i
I
' ,I', i
" i
i
iii
I !,' !
, I :!
i : I
.v i i
i ! I Iii I '1
111
;1
'
I I, ! I' I'
I II iii I ill:0.1 +-------~~~~~----~~--~~~~
1 10 100
h, in.
3-/05
• Experimental I

3,103
3.103 Pressure Distribution in a Two-Dimensional Channel
Objective: According to the Bernoulli equation, a change in velocity can cause a change
in pressure. Also, for an incompressible flow, a change in flow area causes a change in ve-
locity. The purpose of this experiment is to determine the pressure distribution caused by air
flowing within a two-dimensional, variable area channel as shown in Fig. P3.103.
Equipment: Air supply with a flow meter; two-dimensional channel with one curved side
and one flat side; static pressure taps at various locations along both walls of the channel;
ruler; manometer; barometer; thermometer.
Experimental Procedure: Measure the constant width, b, of the channel and the chan-
nel height, y, as a function of distance, x, along the channel. Adjust the air supply to provide
the desired, constant flowrate, Q, through the channel. Attach the manometer to the static
pressure tap located a distance, x, from the origin and record the manometer reading, h. Re-
peat the pressure measurements (for the same Q) at various locations on both the flat and
the curved sides of the channel. Record the barometer reading, Halm, in inches of mercury
and the air temperature, T, so that the air density can be calculated by use of the perfect gas
law.
Calculations: Use the manometer readings, h, to calculate the pressure within the channel,
P = "Imh, where "1m is the specific weight of the manometer fluid. Convert this pressure into
the pressure head, ph, where "I = gp is the specific weight of air. Also use the Bernoulli
equation (Ph + V2/2g = constant) and the continuity equation (AV = Q, where A = yb)
to determine the theoretical pressure distribution within the channel. Note that the air leaves
the end of the channel (x = L) as a free jet (p = 0).
Graph: Plot the experimentally determined pressure head, ph, as ordinates and the dis-
tance along the channel, x, as abscissas. There will be two curves-one for the curved side
of the channel and another for the flat side.
Results: On the same graph, plot the theoretical pressure distribution within the channel.
Static pressure taps
~--~+-----L----------~'I
• FIGURE P3.103
3-106

(c.on'/)
o ution for Problem 3.103: Pressure Distribution in a Two-Dimensional Channel
b, in. Q, ft"3/s Halm, in. Hg T, deg F L, in.
2.0 1.32 28.96 71 21.75
Experimental
x, in. y, in. h, in. h, in. ply, ft ply, ft
flat side curved side flat side curved side
0.75 2.00 0.28 0.31 20.2 22.3
2.50 2.00 0.21 0.37 15.1 26.6
4.00 1.28 -0.42 0.03 -30.2 2.3
4.63 1.05 -0.77 -1.63 -55.5 -117.4
5.38 1.05 -1.01 -1.05 -72.7 -75.6
8.14 1.29 -0.63 -0.62 -45.4 -44.7
10.75 1.54 -0.32 -0.31 -23.0 -22.3
13.25 1.77 -0.15 -0.15 -10.8 -10.8
15.78 2.00 -0.05 0.00 -3.6 0.0
21.75 2.00 0.00 0.00 0.0 0.0
P = Palm/RT where
Palm = YHg*Halm = 847 Ib/ft"3*(28.96/12 ft) = 2044 Ib/ft"2
T = 71 + 460 =531 deg R
Thus, P =0.00224 slug/ft"3 and y = p*g =0.00224 slug/ft"3*(32.2 ftlsll2) = 0.0722 Ib/ft"3
ply =YH20*h/y
Theoretical:
ply =Vex/12g - V2
/2g where
V = Q/A = Q/(b*y) and
Vex~ = Q/Aex~ = (1.32 ftIl3/s/)*(2 *2/144 ft1l2) =47.5 ftls
Problem 3.103
Pressure Head, ply, vs Distance, x
40
Theory
ply, ft
0.0
0.0
-50.5
-92.2
-92.2
-49.2
-24.1
-9.7
0.0
0.0
20
0
-~--J
.~~--:-~~~~~~
-20
= -40
.:i:-
-60Q. ~~--------j-----------------r=-=JI
-80 .--~~~--t-------T-------r-~~-- J
!
-100 ~•......- Experimental, flat side
-120 -1--------+-------+----- - .. - Experimental, curved side
--Theoretical
-140
0 5 10 15 20 25
x, in.
3-/07

.3, /oLf
3.104 Sluice Gate Flowrate
Objective: The flowrate of water under a sluice gate as shown in Fig. P3.104 is a func-
tion of the water depths upstream and downstream of the gate. The purpose of this experi-
ment is to compare the theoretical flowrate with the experimentally determined flowrate.
Equipment: Flow channel with pump and control valve to provide the desired flowrate
in the channel; sluice gate; point gage to measure water depth; float; stop watch.
Experimental Procedure: Adjust the vertical position of the sluice gate so that the
bottom of the gate is the desired distance, a, above the channel bottom. Measure the width,
b, of the channel (which is equal to the width of the gate). Turn on the pump and adjust the
control valve to produce the desired water depth upstream of the sluice gate. Insert a float
into the water upstream of the gate and measure the water velocity, VI> by recording the time,
t, it takes the float to travel a distance L. That is, VI =Lit. Use a point gage to measure the
water depth, z> upstream of the gate. Adjust the control valve to produce various water depths
upstream of the gate and repeat the measurements.
Calculations: For each water depth used, determine the flowrate, Q, under the sluice gate
by using the continuity equation Q = AI VI = b ZI VI' Use the Bernoulli and continuity equa-
tions to determine the theoretical flowrate under the sluice gate (see Equation 3.21). For these
calculations assume that the water depth downstream of the gate, Z2, remains at 61% of the
distance between the channel bottom and the bottom of the gate. That is, Z2 = O.61a.
Graph: Plot the experimentally determined flowrate, Q, as ordinates and the water depth,
ZI> upstream of the gate as abscissas.
Results: On the same graph, plot the theoretical flowrate as a function of water depth up-
stream of the gate.
Data: To proceed, print this page for reference when you work the problem and click III're
• FIGURE P3.104
3-10 8

3./0/f I (conJ/)
Solution for problem 3.104: Sluice Gate Flowrate
a, in. b, in. L, ft
1.2 6.0 4.0
Experimental Theoretical
Zl, ft t, S V1, ftls Q, ft"3/s
0.183 4.2 0.952 0.087
0.267 5.0 0.800 0.107
0.343 5.2 0.769 0.132
0.453 6.2 0.645 0.146
0.569 6.4 0.625 0.178
0.725 7.0 0.571 0.207
0.877 8.6 0.465 0.204
Experimental:
V1 = Ut
Q = V1bz1
Theoretical:
Q = b*Z23/2*(2*g)1I2*[((Zl/z2) -1)/(1 - (z2/z1)2)]1/2
where
Z2 = 0.61*a
Problem 3.104
Flow Rate, Q, vs Depth, Z1
1.00 -r=======:r====r==:r==r==+=+==l=+=I
tn
-CO')
< 0.10
=d
0.01
i I
i I T
1 I
', 1 i !
1
I
I ! ,i •
--------+--------'-----1 --t,~-
i : ! , ;
.-.
•
0.1
!
I
I
I
I
I
I
I ,i.-~
1 i~l! i
~~l i
,
,
!
; I ; ! I
I
I i
-r
II
!
i r
r I
i
I
, I
1 -I I I
I
I ! i
I
,
1 I
! I I
i
1
I , !
I
! !
i
i i
i
I
II
i I1
1 1
Z1' ft
3 -/0'1
1
Q, ft"3/s
0.091
0.114
0.132
0.155
0.175
0.200
0.222
• Experimental
- Theoretical
I

/1./
4.1 1}te velocity. field.of a flow is given by V =
(3y + 2)i + (x - 8)j + 5zk ft/s, where x, y, and z are in feet.
Determine the fluid speed at the origin (x = y:::;: Z = 0) and
on the y axis (x = z = 0).
IJ..::= 3Yf2. J v=x-8, W=Sz
Thl/s) af X= y:::r ~O V=1«2.+ V2. +W,,' = J2~+(-Sl ' = e.2s1j
and Of) the line x;:Z =OJ
V=/(3y +2)2. +(_8)").1 :::: J~9-y-2.-t-/2-y-+-68-" #- where y--ft
if. '1-
4.2 A flow can be visualized by plotting the velocity
field as velocity vectors at representative locations in the
flow as shown in Video V4.1 and Fig. E4.1. Consider the
velocity field given in polar coordinates by vr = -lOlr
and Vo = 101r. This flow approximates a fluid swirling into
a sink as shown in Fig. P4.2. Plot the velocity field at
locations given by r = I, 2, and 3 with e= 0, 30, 60, and
90 deg.
W;1h tV";;:: -/olr and N8 ::: lo/r then
V-=Jrv;.'-+IV92. ::; I(-Io/r)% +(Jo/r)1 = I'I-;!.'I-
The QI191e ex hetllleen the radial J/re{;/ion and
the velocify vee/or is r;iven hy
filii lolr
tan eX = -AIr ;:: _(-Iolr) =I
Thvs ()( = ~s 0 for anv r B
• FIGURE P4.2
o
J / J
~·.e. the velocify vector is alway oriente') /fS"re/4live 1orodia/ lines)
3 8=60
Nofe: VIS
independ'snt
of 9.
V=II/.III 1ftr::: I
If-I
Vr

#.3 r
4.3 The velocity field of a flow is given by
V = 20yl(."2 + y2)l'2i - '20xl(x2 + )'2)1121 fils,
where x and yare in feet. Determine the fluid
speed at points along the x axis; along the y axis.
Also,
fan e=-tr-=
-2.0X
(xl. +y2.)J..2.
20Y
or (x'J. t y2.) Ji.
fane == - f
ThusJ
for (xJy) =(sJ 0)
fan e::: -00 or e=- qoo
For (~y) =(5,5)
tQn8=-/ or B=-'fsl!)
for (XJY) =(0,.5)
fan () = 0 or B== ot)
What is the angle between the velocity vector and
the x axis at points (x, y) = (5,0), (5, 5), and
(0, 5)?
y
2 off.;s
(0,5)
(5,0)
~~----~--=-....;"".-- X
2.0 ff
,s
~v
u
4.4 The x and y components of a velocity field are given
by u = x - y and v = x2
y - 8. Detennine the location of any
stagnation points in the flow field. That is, at what point(s) is
the velocity zero?
v== 0 provided fhQf bolh u=0 and v::: o.
ThIJs1
f).:: X- Y:::0 or X :=y anti V =x~- B ::: 0 or x2y == 8
By combinin9 ob/ain XJ
:- 8 or X::: 2. Since X:::.y N follow..! fh41
y=2. also. Thl/.s (x) y) == (2.,2.)
Jf-2

~,5
45 The x and y components of velocity for a two-dimen-
sional flow are u = 3 ft/s and v = 9x 2
ftls. where x is in feet.
Determine the equation for the streamlines and graph repre-
sentative streamlines in the upper half plane.
Ii = 3 and If:::: 9x2. so fhai Sfre4f)1//nes are 9/ven hy
4¥ ~ -f~ 2/ ::3x2 Or Jdy =:J.3x2.dx
ThvsJ
y:: X.3 +CJ where c is ac()llsfanT.
Repre.renfalirJe .sfream/ines correspondil'l9 70 differenT va/VB.! of
care shoVln he/ow.
Streamlines y =x3
+ C
100
50 -+-I-'--I------+--
c=/OO C=.5o
I
I
-J
i
I
I
c= 0
c= -so
o ~~~--r_~r--,--~--~~~--~~~~
-5 -4 -3 -2 -1 o 1 2 3 4 5
x
~-.3

~. 6 I 4. 6 Show that the streamlines for a flow whose
velocity components are u = c(x2- y2) and v =
- 2cxy, where c is a constant, are given by the
equation x2y - y3/3 = constant. At which point
(points) is the flow parallel to the y axis? At which
point .(points) is the fluid stationary?
U=C(X2..-y2.) J V=-2cxy
Sireamlines given ny y"{(x) Qre -such fhal *"f
Consider +he {une/ion x'l.y - -f = t:.ofJsf, (I)
!VoTe; If i.s nof easy fo wr/le fhis eXf/iviily as y~ f(x)
HoweverJ we can differentials Ef. (JJ ff) 9ille
2xyr/x +x'-dy - y'1.rly :;;: 0 I or
(x2.._y2.) dy +2xy dx =0
Thvs.l fhe //ne.s /n fhe x-y plane given by £1, (I) hdl/O a slope
dL == - ;XY) or for allY consIanI C J lIx:: -(2. ~xy ) =L.(IX (X _y'1. C X _y2. U
3
~,e. fhe fun6fion x2y - ~ =consf. represenfs the .sfreafllines
of +he 9iv81J flow.
The flow is par4/Je/ fo ihe x- axis wASil *::0J or v=:O.
rlus OCC{)rs whelJ eifher X=0 or y:::O J t:e-.; fhe X-ax/s Of
. the y-qX~
The flow i.s para/IeJ fo The y-axi.s when *=: 00 J or IJ.. ::::0.
This occurs whotJ X =±Y
The livid has z.ero ve/oc/I)' at x::y =0

~. 7 I 4. 7 The velocity field of a flow is given by
U =- - V~y/(x~ + y2)112 and v = Vox/(x2 +
y2)II2, where Vo is a constant. Where in the flow
field is the speed equal to Vo? Determine equation
of the streamlines and discuss the various char-
acteristics of this flow.
11 = - v. y V = V, X oS 0 fh41
o (Xl. +y"J.'fo- 0 (X2-+yl.)V2.
V ;:_IU2. -+- V2. := . Yo +X = t,. [/2- (y2. '-) 1~
V (Xl. + y2.) 0
Thus, V=~ fhrou?houl the entire {low fielt1
Sfream lines IJro 9iveIJ by
i1... =L = 2- or - ydy = xdx which can be infe9rafeddx IJ. -y
fo give X2. t y2. =cons!.
ThusJ fhe flvid {low wifh circular .sfreamliIJes and the
speed is constanT fhrollghout.
,+-5

.If.B
4.8 Water flows from a rotating lawn sprinkler as shown
in Video V4.6 and Figure P4.8. The end of the sprinkler arm
moves with a speed of wR, where w = 10 radls is the an-
gular velocity of the sprinkler arm and R = 0.5 ft is its ra-
dius. The water exits the nozzle with a speed of V = 10 ftls
• FIGURE P4.S
(a) Wafer leave.s the nozzle wilh Q Ve10ClfY 0 f V= If) fIls af all QI191e of ,gOO
relalive 10 the radial direvlion - for an obserJler ri,)/nfj on the sprinKler 4rb1.
Th/s is fhe rel41ille velocity. II.! shown ill the .skefch, fhe sprillk/PI 4rll1 has
a c;"c/lmf~enci41 veloc/Iy 0 f RfP -::: O. b fI (10 rad/s) =: s His. TlJe IJbsolv/e
ve10cilyJ Va; 4.!' observed hy aperson sfanJilJr. on -Ihe lawn is the vee/or
SlJm of re/t'J1/ve velocily and the nO"j"Jle ve/oeily.
From fhe 9(;omell'Y of tne fi9vre :
, .i a.n0( = 10 Sin 30 -.5 ::::. 0
10 Ct'J~300
Tho! i.s ex, == 0
(,',e,) the ahsolvfe wafer
velov/ly is in the r4rJial
rJ.irecfion. Since there is
no force ac+i"1 on fhe wafer
o
after iT leavesJ "he W41er p4f'h'c/e-s
)
w;:;/o
coni/nile f~ move in The raJialdirec,lion.
R=o,s
Th~ the palh1ine..r are .s1rai9hf raJiallines.
(b) TAe sh4pe of the water stream «1 a 1irJen installf (t:e"a ~fJ4P .shot"of
fhe W4fer) can he ob-lained 4S follolVs, COf).rider the wafer .sfream emaIJaliA9
from +he end af-lhe nO"/.i/e al r::R (md (}::oat -lime /;=:()
(con'i)
'/-6

-¥. 8 (con II)
A particle in -Ihi.s sfream that lefl from -Ihe no~~/e t seconds 490 d'd
so when the nozzle wasaf f) ~ wi. SinGe /'0 /.t::0
the parth/e.s in sfrai1h1J ratii4/ pAths w/lh Va~.""
speed ~ (.see pa,.{ (al)) fhis parliole is ~"t ser;o~Js 070
af a dis/once of r = R+Va { from .Jhe / ;-~e=wl
orl9tn. ... ~ ~ i=o
ThllsJ
the s1reanJ sho.pe js )fJ)
r = R +Va t and et:: fA) t J or by eli,.,,;ncdiJ19 i
r::Rf(-i5-)e
For the qiven data wi1h Va -;:: V&0.s30' =00 1j-)(.os30
0
::: 8. Db ~ (seeparl(4J)
CI/Jd I.U == If) rad/s this hecomes
r = o.s +0. 866 f:) J where r,v ff alld ()~ rad.
'his .sfream shape i.s plofled beJoU/.
---2
------------------~--~--~---xJ~
OJ
#-7

~.9
'''4.9 Consider a ball thrown with initial speed Vo at an angle
of 0 as shown in Fig. P4.9a. As discussed in beginning physics.
if friction is negligible the path that the ball takes is given by
y = (tan O)x - [g/(2 V0
2
COS
2
O)Jx2
That is. y =CIX + c&x 2, where Cl and C2 are constants. The path
is a parabola. The pathline for a stream of water leaving a small
nozzle is shown in Fig. P4.9b and Video '4.3. The coordinates
for this water stream are given in the following table. (a) Use
the given data to determine appropriate values for C1 and C2 in
the above equation and, thus, show that these water particles
also follow a parabolic pathline. (b) Use your values of c, and
C2 to determine the speed of the water, Vo. leaving the nozzle.
)'
x, in. y, in.
o 0
0.25 0.13
0.50 0.16
0.75 0.13
1.0 0.00
1.25 -0.20
1.50 -0.53
)'
1.75 -0.90
2.00 -1.43
- ........... ,
,
(a)
(b)
, x
,,

"'-, ",
., '
, '

• FIGURE P4.9
An EXeEL Pttt79f!tJl!J W4S v.red to p/of the x-y d4itl d/Jd
to fif a .seco"J order curve to fhe data. Tile results are .showfJ he/oUl.
y vs x for Water Stream
0.04 -,-----.,-----.,---------l
0.02 -~~~~~ ~J
o I
-0.02 ..-~-.~-..~-~--~--PIIIo~-.---~-~-... i
:::~ -0.04
>-
--~--~-.~---~--~~~~
-0.06 +----~--!------,-------!---~~
-0.08 ~-----"C -8.4987x
2
+ 0.7n 15x
-0.1
-0.12 +----.,..-----,---------,.--+-----1
o 0.05 0.1
x, ft
C,:::: O. 7//5 :::: 14nB
and
or e::: 3S,~ I
c,. ::: -8. 'f?11 =:
0""
2. .32.. ~
'10 = 2. (8.'f187) cos'"(.3S,rJ
Thll~ '10 -::: 1.69 fj
JI.-e

'I-~ 10 I
4. t0 The x and y components of a velocity field are given
by u = :l-y and v = -xy2. Determine the equation for the
streamlines of this flow and compare with those in Example
4.2. Is the flow in this problem the same as that in Example
4.2? Explain.
Stream Jines are 9t"ven by ~J:: == J!. ::; _ X y2. __ .X-
d J X U x2.y - X
or .-t :: - ~ which cal} he iIJ1e9raied as:
y x
s*=-J~ Thus, In y = -Iny +c) where C /s ~ CIJI7.ff4nl.
Thfls) Xy :::C
No fe: These streofllline.s are the So.lhS shape (same Ilf/ow paifenn
as in £xample ~ 2 - but fhe veJoc/fy fields are dt'lferel'r/.
HoweverJ lhe ralios -£ are the same:
.;f.:; _ X y2. :: _t
w x"Y x
and
~ := (Volt) (-y) := _ X-
u (voll) (x) x

4.11 In addition to the customary horizontal
velocity components of the air in the atmosphere
(the "wind"), there often are vertical air currents
(thermals) caused by buoyant effects due to un-
even heating of the air as indicated in Fig. P4.12.
Assume that the velocity field in a certain region
is approximated by u = Un' v = Va (1 - y/h) for
o< y < h, and u = un' v = 0 for y > h. Plot
the shape of the streamline that passes through
the origin for values of lio/vo = 0.5, 1, and 2.
y
/'"
/
~ II
J.
I
I
o
FIGURE P4.12
".-
.so fhai streamlines
x
=~fdx
o
/
X
Nofe: The lower limds () f infe1raJiol)
(x=o) y:::o) insure fhal fhis
eqvaflon is for fhe streamline
fhrov'Ih fhe ori9in.
Thi.s sfream I//JB
.c
->.
X ==-h (~) In (J - i) i~ ploffed befow.
y/h vs xlh
1 .. _p'. -~ .- •• _M -:.. 'I
~ .... , : -.-'-""-
...r ..... t'"'"' I
O8 +-------+--#~-,--- I . . , . , ,
. .. .......r.., ,
,# ,/
, /'
----j--r-0.6
0.4 : , / "~~--+--------'---~--I
I
-11
0.2 -~~~
0~--~----1---~-----r----r---~
o 0.5 1 1.5
xlh
2 2.5 3
Jf-/O
-uO/vO=2
- -uOlvO = 1
- - - uO/vO = 0.5

4.13* Repeat Problem 4.12 using the same in-
formation except that u = UoY/II for 0 ::; y ::; II
rather than u = U(). Use values of uo/vo = 0,0.1,
0.2, 0.4, 0.6, 0.8, and 1.0.
Uo
~
y
/
0
/7f'
/
.."....
II
A
I
I
,
ll= UhY , V= Va (I-i) tor o<y<h .so fhaf .sfreafh//ne.s
for y '"h are given by
1>:- =..::L = vo(J-{) = Vo (h-y) or wtth X::::O when y=o
(J x IJ. U 1. Ua YD h
Y IX{ (h~y) fiy = ~ dx This ifJ1e'lf'ales 117 'live
-~ -h In(h-y):h In (h) = -to; X or f=r&.,)[ln~)- iJ
Thi.s sTreamline is plolled he/ow for 0 ~ "* ~ I J wilh
~ = 0 J 0.1 J 0.2.J O. 'I-J O. /)I o.BJ Qnri I. 0 The VIJ Ities were
calctJ/oted ana plotted ()SifJ? 41) £XC£L PrtJ'1J'i1J11.
y/h vs xlh
1 _ D,/
0.9 - ~~-"~=---
O.C
-----
0.8 --- ---------- I
0.7 ~-I--
0.6
to.5 -l--'--l---1.f-,~
,,
--- ------------- -- ---I
I--~--1
!
-j
0.4 -P.I-'-J(-q--~--~- -~~~~~---~-
0.3 -WJIj'---~~--~__i____--~~~~_l______- -~--~-----------------:
,
0.2
i
- --~~----~-.----
i0.1 ------~-------- ----;
iO+-----------T-----------~----------r_--------~
a 0.5 1
xlh
'1-//
1.5 2
//
X

'f.II/- I
"".15 I
4.14 A velocity field is given by u = cx2
and
v = cy2, where c is a constant. Determine the x
and y components of the acceleration. At what
point (points) in the flow field is the acceleration
zero?
dlJ. dU dlJ, 2. .3
Ox =rr +l1.Tx +VTy - (CX'")(')..CX) ::;2c X
and
ay =1f +u.¥X+v:; :: (cy2.)(2-c.y) =2C2.y 3
Thus) a=axl+ayJ =0 af (X1Y) -::(0)0)
.1.1.5 A three-dimensional velocity field is given by u = x 2
,
V = -2xy, and w = x + y. Determine the acceleration vector.
JU Lt1 JU O!J,
QX = If + U DX +~19 +jJI n
=X2.(2X) ~ 2X3
J/IT dAl. aN:. JP:.
Cly ::: rr +It d)( +IV ry of)lf 1i
- X%.(-2y) +(-2.XY){-).X) -= 2X'y
Q1. =1f +IJ¥f+1'1'1'1-fiJI W
=x2.(J) +(-2-xy)0) '=- xz.-:J.xy
IhvsJ
~ 3/1 .1 A 2 I'
11. - 2 X t + 2. X YI +(x ...2XY) k

If. ./7
-4.11 The velocity of air in the diverging pipe shown in Fig.
P4.17 is given by VI = 4t ftfsand Vo = 21 ftfs, where t is in
seconds. (a) Determine the local acceleration at points (I) and
(2). (b) Is the average convective acceleration between these
two points negative, zero, or positive? Explain.
b) convecfive acceleration a/on9 the pipe :::: U#-
where IJ. >0. IN aIJr lime' ~ < Vt . r/;ps he/ween OJ a"d(:J.)J J .... J
JI£ ~ V2.- ~ /0
JX"'" 1 ~
/lel1cc", 11. ¥X c::: 0 or the tlver41e conl/eet/ve acceleration
is neq4live.

If. {8 f
4. Ia Water flows through a constant d!arneter pipe with a
unifonn velocity given by V = (8/t + 5)j mis, where t is in
seconds. Detennine the acceleration at time t = 1, 2, and lOs.
~
~V .- ..... s
a:::Tf + V.vV WiIh U == 0 J V -(T +.5)':
thi.s becomes
......110 ( cV ~V)" _ dV A _ _ ~ 1 m
a = -rr + vJy J - n J - t2. J S'-
ThlJs, a=-81./fi af t=/.s
a=-2.0J.f.. af t=2 oS
and
a=-O.OBJ~ at t::::/O..s
4. Iq When a valve is opened, the velocity of water in a
certain pipe is given by u = 10(1 - e-'), v = 0, and w = 0,
where u is in ft/sand t is in seconds. Detennine the maximum
velocity and maximum acceleration of the water.
/
w-=oJ

~,zo
~- ---_._- '-_...-
4.20* Water flows through a pipe with V = t (s) u (ft/s) t (s) u (ft/s)
u(t)i where the approximate measured values of 0 0
u(t) are shown in the table. Plot the acceleration 1.8 1.7
as a function of time for 0 ~ t ~ 20 s. Plot the 3.1 3.2
acceleration as a function of time if all of the 4.0 3.8
values of u(t) are increased by a factor of 2; by 5.5 4.6
a factor of 5. 6.9 5.8
8.1 6.3
10.0 7.1
~
a= ~v +V'VVelt u=u(t) v=:OJ )
thi.s becomes
~ (~U. ~IJ.)I
a:: Tf+urx l
~ tL 1./1
=rr /J - 1Jlor "'x - ~t
11.2 8.1
12.3 8.4
13.9 8.3
15.0 8.1
16.4 7.9
17.5 7.0
18.4 6.6
20.0 5.7
w=-o
The IJ. ~U(f) ,rttph Qnd fhe hesf [// clIbic erlJatiofJ .shown helolJl
was ploNeJ ilsifl9 In £XCEJ.. Pr091'4RJ.
Thvs w/lh IJ.::: -0.0012.-/
3
- O,OI/2tJ. +0,97.5" if fo//ows 11141J
fix := 1~ :::: -0.0031 flo - O,01.Z.lft +0.97.51 jt J where tl'VS
This acce/fJI'4/ion is a/so p/ofleri he/QUI,
Note Ih4-f if UincreMes by fA fAvlol' 01 Kti, e, K:::2. ,,, K:::S)J
Ihe ()'caeleralion J tI)(. .:= IfJ does also.
Water Speed vs Time Acceleration vs Time
u =-0.001213
- 0.0112r + 0.97561
5
9 4
B - 3
7 2
N
1
~ 0
;c
-11"11
6
~ 5
:; 4
3 -2
2 -3
1 -4
0 -5
0 5 10 15 20 0 5 10 15
t, s t, S
(I)
20

Jf.2.1 I
4.2 The fluid velocity along the x axis shown
in Fig. P4.21 changes from 6 mls at point A to
IS mls at point B. It is also known that the ve-
locity is a linear function of distance along the
streamline. Determine the acceleration at points
A, B, and C. Assume steady flow.
VA = 6 m/s
~
VB = 19 m/s
~
• • •A C B
x
too5m~ "'//// ,///////"" I
,. 0.1 m----l
FIGURE P4.2.1
~
.- oV - -.
a = 1T +V-VV U==U(X) v=O and w=O) J
fhi.s becomes
a=(!t +u ~J)t = u~~ t (I)
Since U is a /ineqr ftlnclion ()f X J (j,= c,X +c2, where fhe
cons/an/.s C, J C1. are given as: I.JA " 6 :: Cz.
and ltD = 19 =0,/ C, .r C2.,
)
m 0 r C, ;:;: J20 J C2.:: 6.
ThusJ (). =( /20 X +6 s with x~m
From Efj. (I)
a= U ~~ L -= (/20X +6 ) P. (/20 :5) t
or
for XII =:0 J ~= 7~O r1i
for X8-::: 0,0.5m) aB
= IJftiOtp.
and.
l' / ~ 6 Amrf!Jr Xc.== 0, m J ac
=- 21 0 l':S'J-
Lf-/6

4.2'1 When a fluid flows into a round pipe as shown in Fig.
P4.22, viscous effects may cause the velocity profile to change
from a-uniform profile (V = Voi) at the entrance of the pipe to
a parabolic profile {V = 2Vo [1 - (r/R)2]i} atx = e. Velocity
profiles for various values of x are as indicated in the figure.
Use this graph to show that a fluid particle moving along the
centerline (r = 0) experiences an acceleration, but a particle
close to the edge of the pipe (r ""'" R) experiences a deceleration.
Does a particle traveling along the line r = 0.5 R experience
an acceleration or deceleration, or both? Explain.
..,).
DV ..,). ~
a=F +V'VV
we obtain
r
R ~~~==:-,.....----x = 0
x = fl4
0.75 R
,
,,~x =fl2
I~
i I~~'x = 3Cl4
I I ,
I I I x = t
I I
0.5 R
0.25 R
CL I IO~--~----~~--~--~~--~u
(b)
FIGURE P4.2 Z
v=o and w=OJ J
~ ~ aU dU dU
a:= Ox t where ax == Jf + uTi := UR
a)
b)
c)
U
Along the centerline (r =0) we finri li >0 and ~ >0
ThusJ Qx >0 on r::: O.
/Veo.r the pipe wall (r ~ R) we find Ii >0 btA -if<0 (i.e.; fhe
ve/oc/ly changes from (j.::: Va fo u.c: Vo (J.s X increase!;.)
Thv.s J ax <0 for r ~ R,
For r:: f R we r/nd u >0 and fx>0 'leaf fhe pipe
en/ranee) buf #..c.0 elsewhere. This is indicofed /n fhe
t;9IJre beJow.
Th{)~ for r =1- ~JJ ~ J
ax >0 neQr the en/raneeI
-I -
I
r== iR
I
Ox <0 elsewhere
o~~x
¥x->o ~ <0
x = 0
(a)
tf -/7
and
x = I'

y f-o------el2-----l·1
4.23 As a valve is opened, water flows through
the diffuser shown in Fig. P4.23 at an increasing
flowrate so that the velocity along the centerline
is given by V = ui = Vo(1 - reI) (1 - xII' )i,
where uo, c, and eare constants. Determine the
acceleration as a function of x and t. If Vo =
10 ftls and e= 5 ft, what value of c (other than
c = 0) is needed to make the acceleration zero
for any x at t = 1 s? Explain how the acceleration
can be zero if the flowrate is increasing with time.
u
1
u = -VoG - e-et)
L__~~~~~_~2_~~~
~
Q =iY.. +V-VV~t
this becomes
~ (~U JM)/ 1
a::: rt t u 1)( t == Cl~ t
I fax s: 0 for any X af t:: 1 s
[c i(;t _ .:;. (/_ e-ct )2J :: 0
-c 10 ( _C)2
ce -5/-e =0
----i~~
u = VOG - e- c0
~
FIGURE P4.23
v::OJ J
and w::O
we mils! htlve
with Vo::: /0 Q/Jd1.:::5
For fhe above condl'fionJ fhe /()co,/I/ccelerafio/J (ft- >0) is
precisely Do/alJced hy fhe Colll/eelitlc decelero/ion (it :~ <.0),
The {Iowrote if)crea.re.s wilh limeJ /;vi fhe I/Vid flows 10 an
area of lower ve/oc/Iy.
4-/8
x

4.24 A,fluid flows along the x axis with a velocity given by
V = (x/t)i, where x is in feet and t in seconds. (a) Plot the
speed for 0 :s x:s 10 ft and t = 3 s. (b) Plot the speed for x =
7 ft and 2 :s t :s 4 s. (c) Detennine the local and convective
acceleration. (d) Show that the acceleration of any fluid particle
in the flow is zero. (e) Explain physically how the velocity of
a particle in this unsteady flow remains constant throughout itf.
motion.
(0) U. -;;; : 1j so at t:: 3.s J Ii:;: f !f
(b) For X == 7 If U:;.2 --.sfl
J t
JU x d JU X(')_ X
(e) It = --p. an Urx =T T - /;'"
tf
tJ,J fp.s
3
2.
0
3
U, fp5
2-
t =3.s
0 10
X= 7 f+
Fi9,2
~
(d) For Qny fluid particle a=1¥+V,VV
whic.n with v:;: 0 J W -::: 0 becomes
...... (dlJ. (j(J,)J' _ ( X + X ) -1 0a :- "IT +uTx l - -72- 72- { :=
°0
(e) The {Jarfie/c.s flow Inlo area-s of hl9het ve/oc/ly (see r;.~./),
hoi af 0"1 9iven loca/Ion fhe veJooifj i.s de6rea.sifJ9 i/J Ilme
(see Fig-2J. For The 9iven velo&/fy field the local qnd
COhflBr;fil/C acceleratio"s ore e'lv41 and Oll()s,ie) 1iflifJ9
'2 e/,O acee/era/ion fhl'otJ9n ~vl.

4.25 A hydraulic jump is a rather sudden change in depth
of a liquid layer as it flows in an open channel as shown in
Fig. P4.25 and Vitleo V10.6. In a relatively short distance
(thickness = e) the liquid depth changes from ZI to Z2' with a
corresponding change in velocity from VI to V2• If VI =
1.20 ftls, V2 = 0.30 ft/s, and e= 0.02 ft, estimate the average
deceleration of the liquid as it flows across the hydraulic jump.
How many g's deceleration does this represent?
Hydraulic jump
.
1'1 FIGURE P4.25
.-S
JV .> - -- A
Ii -:: T{ of V·vV SO w/fA V:: IJ.(XJi
.- '" JIJ, 1
a =Ill. ::t(,TXtJ X
Will1 ouf knDWin9 fhe acftJa/ ve/oc/Iy Ji.flr;/'u/i()11 J t( == IJ(X~
the (Jcceleraliol) can he approximated tis
tlx ~ fA .w~ i (V,+~) (~i"') :;;; 1(1.2.0 +0,30) if
= -33.,{!z.
( O. 30 -1.;).0)!
O.Oloft
ThllsJ
- /.05

'I-:k6 I
-t.U. A fluid panicle flowing along a stagnation stream·
line. as shown in Vick,) Y4.:: and Fig. P4.26. slows down as it Stagnation pornt. t::: 0
approaches the stagnation point. Measurements of the dye flow ~~~J(f~~¥ FI Id It I
in the video indicate that the location of a panicle starting on i,; ~:tt~" ";.i;oro. ~ v / U PC! Ie e
the stagnation streamline a distance s "" 0.6 ft upstream of the 'tfi~ s
stagnationpoint at ( "" 0 is given approximately by s =O.6e- o.s" ,-
where t is in seconds and s isin ft. (a) Determine the speed of • FI G U REP 4 . 2 6
a fluid particle as a function o(time, Vpanicl.Jt), as it flows along
the sleamline. (b) Determine the speed of the fluid as a func·
tion of position along the streamline. V"" V(s). (e) Determine
the fluid acceleration along the streamline as a function of po-
sition. a, :::: a,(s).
(,,) Vlilh s" 0.6 e-o.s
i
if follows Ihat
II ds () _O.Sf _o.st~ ~"J "-rr ~ 0.6 -o.s e "-0.3 eP',.TfC e ,,~
His
(b) From pari (a),
[
_0.51] _o.sl
V" (-0.5) 0_6 e where .s"0.6 e
ThvsI
v~ (-0..»[5], or V= -0.s5 [lis whero s-ff
dV
(e) For sfeadl {lOlli, I?r" V 4i
Thvs, wilh V= -0.5s pnd ~ « - O,S,
Qs " (-o.s 5) (-0.5) " 0.255 fils' where $- H
NoIe : For S >0, Os iJ pari/lve - the parficlt:r acee/eralion is /0 /he ri'llrl.
Si"ce fhe paMic/e is mavifltj fa the left. aplJS//ive as fof' fj,is ca.re
implies -fhal Ihe p",.fide is dece/era/in,! (qS ,Im~f be for Ihis
sfa?nalion point lJow1.
'f-:J./

¥.271
4~1.7 A nozzle is designed to accelerate the fluid from VI to
V2 in a linear fashion. That is, V = ax + b. where a and b are
constants. If the flow is constant with VI = 10 m/ s at XI = 0
and V2 - 25 m/ s al X2 = I m. delennine the local acceleralion.
the conveclive acceleration, and the acceleration of the fluid at
poinls (I) and (2).
Wiih u =ax +b ) V=O J and w=o
can be wriHen as
h IV,
a= Ox t were Ox = /J.1X •
fhe accelerafion
(I)
Since /J. = ~ = IO.p af x=0 and Ii- =v.. =25.p. al x =1 we oMain
10 = 0 +b
25 = a +b So thai a= 15 and b=10
Thai i.s, /J. =(!5X+10).If ; where x- m , so th.i from EZ'O)
ax = (/5X+ IO)f (15'f) = (t1.5XTISO) f,.
'/-22

4.21.) Repeat Problem 4.27 with the assumption that the
flow is not steady, but at the time when VI = 10 m/s and
V2 = 25 mis, it is known that aVl/ar = 20 m/s2 and
aV2/ar = 60 m/s2
•
~
f I f
.- JV ~ ~
Wi h U=fJ.(x)l:) J v::: oJ
and w=:0 the acce era ,'on a =Tf +V·vV
can be wriffen a.s
a:: Ox f where Ox:::: #- +Ii ¥X J w/lh /.J.:::; a(f)x +btfJ. (I)
III fhe given lime (fc:io) u:: ~ =IO.qz. ~.d X ==0 (JnJ IJ == ~ ==25 11 aTX:=/m
Thus) 10 = 0 + bOo)
25 :: aCto) t ha,,) so fhai aCt,,)::: /s tJlld !JII,) == 10
II/sf) af {:::to J #- == ~~~L :: 20 ~ at X=:(}
~/J. JV2. _ f !l!... f X-I MI "'If,. / Iand Tt = -rr -0 0 .s:J- a - moe: I 'lese Qre OC"
(Jcce /er"f ions (J f time t=to
The convective accelerofion al X::O (Ef-0) i.s
u¥x= (oxtb) (a) =(15(O)+IO)~/5 ~):::ISO?
vJ'h//e a f '1= / if i.s
u~ :: (/5 (I) +JO)!J-(ls f) == 375ii
The fllJid accelerofion ,d t:::to Is
a= (#- +I.l ~)t == (2.0+ ISO)'( ~ = 170t ~ af x::()
and
a=( 60+ 37s )t ';2. ::: 'f.9st-jt af X'::lm

9.30
-1.30 An incompressible fluid flows past a turbine blade as v
shown in Fig. P4.30a and Vidt'o '4.5. Far upstream and down-
stream of the blade the velocity is Vo. Measurements show that 1.5 Vo
the velocity of the fluid along streamline A-F near the blade is
as indicated in Fig. P4.30b. Sketch the streamwise component
of acceleration, as, as a function of distance, s, along the stream- Vo
line. Discuss the important characteristics of your result.
0.5 Vo
0
A B C D E
A .I'
1 .~
(b)---1 Vo
s
(a)
II FIGURE P4.30
tA.s :: Vis where froRJ the fit/fire of V =VIs) The (vl/cfion
1!has the foJ/olNin1 Sh4I'S.
F
.s
F
The ffwd decelerates from 1/ 10 C) 4cce/er4les {rum c If} IJ) and
fhe decelerates a94JI; from [) 10 F. the nel acceler4li(}17 (rpm
Atf) F is zero (i. 8/ ~::: V, =- VF ),

4.31 * Air flows steadily through a variable
area pipe with a velocity of V = u(x)i ft/s, where
the approximate measured values of u(x) are given
in the table. Plot the acceleration as a function
of x for 0 :S X :S 12 in. Plot the acceleration if
the flowrate is increased by a factor of N (i.e.,
the values of u are increased by a factor of N),
for N = 2, 4, 10.
x (in.) u (fils) x (in.) u (ft/s)
o
1
2
3
4
5
6
10.0
10.2
13.0
20.1
28.3
28.4
25.8
7
8
9
10
11
12
13
-
20.1
17.4
13.5
11.9
10.3
10.0
10.0
Since /). ~U(x) J v==o) and w=o /1 follows fha! (j= :: +v·vV
simplil/e.s fo 11 = ax t where (Ix -::: U ¥X (I)
The valves U are given in fhe faMe ; the correspondin9 values
of ~~ C4n be tJbt"ined by an approximale numerical diFreren-fialion o.s
qiven in Pro9rom P'f#31 shoJ¥fJ he/ow,
No Ie fhat since (/x ~ f). #- if fo//ows thol and increose in ve/ocIl,
from u. 10 NU Increases Ihe occelerafiol) from Ox If) ;V2dx
2-00 cls
110 open "prnl! for output as #1
120 dim u(l~), n(~)
125 u(l)=lO.O : u(2)=10.2 : u(3)=13.0 : u(~)=20.1 : u(5)=28.3
130 u(6)=28.~ : u(7)=25.S : u(S)=20.1 : u(9)=17.~ : u(10)=13.5
135 u(11)=11.9 : u(12)=10.3 : u(13)=10.0 : u(1~)=10.0
1~0 n(l)=l : n(2)=2 : n(3)=~ : n(~)=10
150 print#l, "************************************************"
160 print#l, "** This program calculates the acceleration **"
170 print#l, "** as a function of position. **"
180 print.# 1, "************************************************"
200 print#l, " I!
210 for i = 1 to ~
220 print#l, II "
230 print.#l, using "For N = ##";n(i)
2~0 print#l, 11 x, in. u, ft/s a, ft/s2"
300 for j = 1 to 13
310 a = n(i)-2*((u(j+l) + u(j))/2)*{(u(j+l) - u(j))/(1/12))
320 uavg = (u(j+ll + u(j))/2
330 x = j - 0.5
3~0 print#l, using "###.# ###.# +#.###----";x,uavg,a
350 next j
360 next i
Nole fhQf aHhou9h Ihe velocIty· da/aJ U :: fJ. (X)/ o/fJerJrs fo he
qulfe "smoolh '; the occelerat;on resvll J ox:::: I.i ~ I is somewhaT
irrerju/ar (especia/~ for x> 7 in.).

~***********************************************
** This program calculates the acceleration **
** as a function of position. **
************************************************
For N = 1
x, in.
0.5
1.5
2. :>
3.5
4.5
5.5
6.5
7.5
8.5
9.5
10.5
11. 5
12.5
3000
2.000
aXJ
£l1.s2- 1000
0
-1000
-2~
u, ft/s,
10.1
11. 6
16.5
24,.2
28.3
27.1
23.0
18.8
15.4,
12.7
11.1
10.1
10.0
30
u"
ff/.s
2.0
10
a, ft/s2
+2.424E+01
+3.898E+02
+1.410E+03
+2.381E+03
+3.402E+01
-8.455E+02
-1.570E+03
-6.075E+02
-7.231E+02
-2.438E+02
-2.131E+02
-3.654,E+01
+O.OOOE+OO
5
For N = 10
x. In.
0.5
1.5
2.5
3.5
4.5
5.5
6.5
7.5
8.5
9.5
10.5
11. 5
12.5
u, ft/s,
10.1
11.6
16.5
24,.2
28.3
27.1
23.0
18.8
15.4,
12.7
11.1
10.1
10.0
10 Xl in.
a, ft/s2
+2.4,2l"E+03
+3.898E+04,
+1.l,,10E+05
+2.381E+05
+3.l,,02E+03
-8.l"55E+0l,,
-1.570E+05
-6.075E+04,
-7.231E+0l"
-2.438E+0l"
-2.131E+0'"
-3.654E+03
+O.OOOE+OO
15
N=F
- -- -/-.---------,---------------- -:-------~-----,------~--
-.
f ---------.-------~'--.------~.----.----- ---:----- --- --------.- --------- ---
/.
~~---+-'------~,....------'J5--Xi-in;
___~~~/O-.----________,______

'1.32. J 4.32. Assume the temperature of the exhaust
in an exhaust pipe can be approximated by T =
To(1 + ae-bx)[l + c cos(wt)], where To =
100 °e, a = 3, b = 0.03 m-1
, c = 0.05, and w
= 100 rad/s. If the exhaust speed is a constant~
mis, determine the time rate of change of temp-
erature of the fluid particles at x = 0 and x =
4 m when t = o.
Since u-== 2 T v;O and w=O if fol/ow.s +h~+'V J J
DT 'OT ~ T oT ~T IJT dT oT 'dT
Dt =rr+ V'V =rt+lJ.rxf-V 1y tWn-=-rr+UdX
ThusI
gr=To (I +a ibx)(_c fA) sin(4Ji)+u.T,; (t+c c.o.s(wt»)(-ab e-b:x)
When t;:: 0 :
%f = - ab IJ 7; (I +c) e-
bx
J or w/llJ fhe given data.,
%f== -(3) (0.03 ;h)( 2.~)(100 DC) (/+0.05) i o.o3X
=-/B.q e-O
•
03X
:C J where x,vm
Thus, Iff-;- /B.q:j- at X;::O J t=:o
and
DT °c
Df = -/6.8 s af x;: Lf »J J t =0
/f-27

4.33* As is indicated in Fig. P4.33, the speed
of exhaust in a car's exhaust pipe varies in time
and distance because of the periodic nature of the.
engine's operation and the damping effect with;
distance from the engine. Assume that the speed.
is given by V = Vo[1 + ae-bx
sin(wt)], where·
Vo = 8 fps. a = 0.05. b = 0.2 ft-I, and w = 50
rad/s. Calculate and plot the fluid acceleration at
x = 0, 1, 2. 3, 4, and 5 ft for 0 :5 t:5 n/25 s.
v = vorl + ae- bx sin(wtl)
FIGURE P4.33
Since fj,=U(X,i) v=o and w=o if fo/low.5 thaf.- J J
~ aV _.-v J h DUo aua = Tf +V•V = Qx I. , were Ox:: Tf +UIX
ThIJS, w/lh U:: Vo [I +a e-
bX
Sin (wt)] Ef. (/) 9ives
Ox = Va aw e-bxco.s(wt) +Vo[I+ a e-bx
.sin(fAJi)] Va a. (-b) e-bxsin(wt)
= Vo a ebX
[ tQ COoS ((pt) - Vo b Sin (4JtJ (J ta e-bX
sin (wi »)]
W/lh Va:: 8!j J a;:; 0.05 , b =:: 0.2. ft J and I» =50 !.1d
fhis becomes
(I)
ax == O. If e-o.2.X [so cos(sot) - 1.6 sin (Eot) (J +0.0,5 eo.2x
sin (sot))J f}a. PJ
where t~.s and x- ff
PI07 tlx from £'1- (:J.) (or o:s t ~ if s w/lh X =0) J) 2.,3, if, tlnrl51!,
AIJ £xce/ Prorflam Wa.r vs-eci 10 (;4/clI/4Ie ax frolll £r (~). The re.rvlfs
ore shown en Ih9 /Jext p4,e,
(con 'I)

'I: 33- {con 'I)
Acceleration at various x locations, fUs"2
t, s x=Oft x=1ft x=2ft x= 3 ft x=4ft x=5ft
0.000 20.00 16.37 13.41 10.98 8.99 7.36
0.005 19.22 15.73 12.88 10.55 8.64 7.07
0.010 17.24 14.11 11.56 9.46 7.75 6.34
0.015 14.18 11.61 9.51 7.79 6.38 5.22
0.020 10.24 8.39 6.87 5.63 4.61 3.77
0.025 5.67 4.65 3.81 3.12 2.55 2.09
0.030 0.74 0.61 0.51 0.42 0.34 0.28
0.035 -4.23 -3.46 -2.83 -2.31 -1.89 -1.55
0.040 -8.93 -7.31 -5.98 -4.90 -4.01 -3.28
0.045 -13.08 -10.71 -8.76 -7.17 -5.87 -4.81
0.050 -16.42 -13.44 -11.00 -9.01 -7.37 -6.04
0.055 -18.73 -15.34 -12.56 -10.28 -8.42 -6.89
0.060 -19.89 -16.29 -13.33 -10.92 -8.94 -7.32
0.065 -19.81 -16.22 -13.28 -10.87 -8.90 -7.29
0.070 -18.51 -15.15 -12.41 -10.16 -8.32 -6.81
0.075 -16.06 -13.14 -10.76 -8.81 -7.21 -5.90
0.080 -12.61 -10.32 -8.45 -6.91 -5.66 -4.63
0.085 -8.37 -6.85 -5.61 -4.59 -3.76 -3.07
0.090 -3.62 -2.96 -2.42 -1.98 -1.62 -1.32
0.095 1.36 1.12 0.92 0.75 0.62 0.51
0.100 6.26 5.13 4.20 3.44 2.82 2.31
0.105 10.77 8.82 7.22 5.92 4.84 3.97
0.110 14.61 11.96 9.80 8.02 6.57 5.38
0.115 17.54 14.36 11.76 9.63 7.88 6.45
0.120 19.38 15.87 12.99 10.64 8.71 7.13
0.125 20.01 16.38 13.41 10.98 8.99 7.36
Acceleration, ax, vs Time, t
20
15
10
5
N
tn
¢:! 0
~
co
-5 -.~----'---
-10 ~-----.--------------
-15
-20+-----~----~----~--~--r_----~----~
0.00 0.02 0.04 0.06 0.08 0.10 0.12
t, S
1/--29

4.34 A gas flows along the x-axis with a speed of V = 5x sure for a fluid particle flowing past x = I. (c) Explain without
m/s and a pressure of p = IOx2
N/m2
, where x is in meters. using any equations why the answers to parts (a) and (b) are
(a) Determine the time rate of change of pressure at the fixed different.
location x = I. (b) Determine the time rate of change of pres-
a) Since fJ:::: /0 X
2
/i follows IhQf *= 0 for a1/ x.
b) 'Wifh fj,:::5x /tr:::O )P=:O and ll==jox2.d follows fha.fJ ~ J r-
~1 ::*+u~ +/Ir ~ +jJr ~ ::: 11.* ::(SXlf)(2oX1.) == IOOX2. ~/.s
-rhvs ~fl == /00 ;;(6
Xc:/m
c) For this s learJy flow fhe press/Jre at apoinT is cO/Jsiallf (p4rl(",)~ blli
1he pressure fOr a 9i1l811 parfie Ie chafJ?eJ; with lime (,~flf (bJ) heca/)se
fhe pafl1(~/e flow.: into a hir;hel' prossfire re9io/J.
4.35
4.35 The temperature distribution in a fluid is given by T =
lOx + 5y, where x and y are the horizontal and vertical coor-
dinates in meters and T is in degrees centigrade. Determine the
time ra.te of chan~e of temperature of a fluid particle traveling
(a) honzontally WIth u = 20 mis, v = 0 or (b) vertically with
u = 0, v = 20 m/s.
DT ~T JT JT h Jf
Dt == rr +IJ IX +V dY J were 1T :::: 0
ThIlS, if IJ,::;2o'f and v~OJ then ~ :::u{f:::(2.o.t;-)(IO !.#r) =:200!f
and '" T 7 iJC PC
If u=0 anJ v=: 207) fhel1 ft-;; vfr :::: (:LO!)(5'-;;;) -:::jOIJ ~
..U6 At the top of its trajectory, the stream of water shown
in Fig. P4.36 and Yid,'o V4.3 flows with a horizontal velocity
of 1.80 ft/s. The radius of curvature of its streamline at that
point is approximately 0.10 ft. Determine the normal compo-
nent of acceleration at that location.
2.ij2.
a: == VI = (1.8 ) = 3 2.~
n 1? o,/off s2.·
1./-30
ffi =0.10 It
m FIGURE P4.36

~37
4.37 As shown in Video V4.2 and Fig. P431, a flying
airplane produces swirling flow near the end of its wings.
In certain circumstances this flow can be approximated by
the velocity field u = -Ky/(;r? + y2) and v = Kx/(;r? + l),
where K is a constant depending on various parameters as-
sociated with the airplane (Le., its weight, speed, etc.) and
x and yare measured from the center of the swirl. (a) Show • FIG U REP 4 .37
that for this flow the velocity is inversely proportional to the
distance from the origin. That is, V = KI(;r? + yl)ln. (b)
Show that the streamlines are circles.
or
Jir J
V-=
(b) Sfreamline.s are (piJen by
Kx
(x'".f y2.)
-Ky
(x 1
+y'")
which when inferraled qives
t y1. ::: - f X1. +C'.J where t; is I-i cOlJrfalJ/.
or
X2. -l- y"J, :: Con.sfanf
1f-3/
_ x
-y

'1-.38
4.3){ Assume that the streamlines for the wingtip vortices
from an airplane (see Fig. P4.37 and Vidt'(j V4.1) can be ap-
proximated by circles of radius r and that the speed is V = K/r.
where K is a constant. Determine the streamline acceleration.
as. and the normal acceleration. an. for this flow.
• FIGURE P4.13
a - V!li-~ - as where SlfJce V-J:- r J
if~(J
ThvsJ
f4.s -:: £
IJ/so
J 2-
(Klr):1 ::: KQ _ 'i _
-3"-Y[- -r r
-

4.3 q A fluid flows past a sphere with an upstream velocity
of Va = 40 mls as shown in Fig. P4.3Q. From a more advanced
theory it is found that the speed of the fluid along the front part
of the sphere is V = ~Va sin e. Determine the streamwise and
normal components of acceleration at point A if the radius of
the sphere is a = 0.20 m.
FIGURE P4.3Q
v= ~ Va sin e== -j: (~O ';) sine == 60 sine f (I)
2. (6 . 0)2. m1
a == X == 0 Sin 1./-0 SJ::::: 7/f'fO ~
n 1( O.2m s
and
~v . ) ~v JV ~v ~e
CIs ::: V15 == (60 smf) dS J where J.5 ::: Ie TI
From £~,(J) J j~ :: io cos fJ
h +dB - J
/lIsa s::: a8 =0.1- e I1J J where e-raJ, J so f Q r.s - ~
Thus, for B='fOD
as:: (60 sinIfOO,T)(6Dco.sIfOO~)(o.~m) ::: 886o~

4.40* For flow past a sphere as discussed iI}
Problem 4.39, plot a graph of the streamwise ac-
celeration, as> the normal acceleration, a,,, and
the magnitude of the acceleration as a function
of 0 for 0 ::; 0 ::; 90° with Vo = 50 fils and a =
0.1, 1.0, and 10 ft. Repeat for Vo = 5 fils. At
what point is the acceleration a maximum; a min-
imum?
_r _(tVa sinet-_ 9 '102
• 2.8
an -1'(- a Jfa Sin (I)
dV ~V de oV 3
and G.s = VTs =V18 E J where T8:: '".2 Vo cosfJ and .s:::a()
or gjl =..L
ThllsJ ~5 a
as = (fVa sine)(tYo co.s8) f = ff sinO cos e (2)
lienee The ma9ndIJde of the acee/era/ion i.s
I i 9Vc 2. I . 'I i 9 ~ 2. J i
IaI = yan2.+a.: = Jl.ao Ysin e +sin2.8 (,os2.8 = If aD sine Vsir/e +co.s~e
or 2-
(3) urI = ¥.i sinB Thvs, Ia/min = 0 tlf 8=0 J la/mIX= ~ af fj =90
0
lin bxcel Pr09fafn"W4s wed to caJGt6'i4iB tl.s a/J 4/ld C{ frllm
£flns. (fJ/)'~ afJa (1'J. The l'eJ()/ls are shown ie/(JJ4I~ 7lJe resilils tor
ofher V()'/V8s Od'e simIlar ,llhe f4,,/pr Vo'"/a is 4CG()vnieri /I/",
The folroWif9 derfa is f~f Vo:::5 fils 1 Q::: / ff
e, deg an, ftls2 as, ftls2 a, ftls2
o 0.0 0.0 0.0
5 0.4 4.9 4.9
10 1.7 9.6 9.8
15 3.8 14.1 14.6
20 6.6 18.1 19.2
25 10.0 21.5 23.8
30 14.1 24.4 28.1
35 18.5 26.4 32.3
40 23.2 27,7 36.2
45 28.1 28.1 39.8
50 33.0 27.7 43.1
55 37.7 26.4 46.1
60 42.2 24.4 48.7
65 46.2 21.5 51.0
70 49.7 18.1 52.9
75 52,5 14.1 54.3
80 54.6 9.6 55.4
85 55.8 4.9 56.0
90 56.3 0.0 56.3
Acceleration vs Angular position
50 T---~~-4--~~~---~
N
I/)
¢:! 40
o 10 20 30 40 50 60 70 80 90
e,deg
Va ~ 5 fils
a :: IN
1-normal accel, ftsQ2 i
1- -streamwise accel, '
fUsA
2
- - - accel, ftlsA2

If. Iff r
4.4 J A fluid flows past a circular cylinder of
radius a with an upstream speed of Vo as shown
in Fig. P4.41. A more advanced theory indicates
that if viscous effects are negligible, the velocity
of the fluid along the surface of the cylinder is
given by V = 2Vo sin e. Determine the streamline
and normal components of acceleration on the
surface of the cylinder as a function of Va' a, and
e.
a - V2
_ (2. Vo.sin8/- _ 'tVQ.2. ",·n2.n
n- f( - Q - a ~ r:7
and ========
6V V dV ~e
0.s= V().f= ~B Ts J where ~~:::: 2 Yo cos e
lJB I
or rs =-L-
ThusJ "
as::: (2'Vo sine)(ZVo cos(i)+ = ~ sinO cose
~.tj2" I 4.4,2* Use the results of Problem 4.41 to plot
graphs of as and an for 0 :5 e:5 90° with Vo =
10 m/s and a = 0.01, 0.10, 1.0, and 10.0 m.
v
-
FIGURE P4.41
and .s =: a8
Fro/?? Problem if. 'II J C1n:= If.cJIo:l Sifle C/nd t1.s :: "al4,/I. sine cos().
These results with Vc,-.::/o§! and a == 0.0/ J 0./0) /.0.1 and /0. Om
are plotted below.
a =0.01 m a =0.10 m a =1.0 m a =10 m a =0.10 m a =0.10 m a =1.0 m a =10 m
e, deg as, fUs
2
as, fUs
2
as, fUs
2
as, fUs
2
an, fUs
2
an. fUs
2
an, fUs
2
an, fUs2
0 0 0 0 0.00 0 0 0 0.00
5 3473 347 35 3.47 304 30 3 0.30
10 6840 684 68 6.84 1206 121 12 1.21
15 10000 1000 100 10.00 2679 268 27 2.68
20 12856 1286 129 12.86 4679 468 47 4.68
25 15321 1532 153 15.32 7144 714 71 7.14
30 17321 1732 173 17.32 10000 1000 100 10.00
35 18794 1879 188 18.79 13160 1316 132 13.16
40 19696 1970 197 19.70 16527 1653 165 16.53
45 20000 2000 200 20.00 20000 2000 200 20.00
50 19696 1970 197 19.70 23473 2347 235 23.47
55 18794 1879 188 18.79 26840 2684 268 26.84
60 17321 1732 173 17.32 30000 3000 300 30.00
65 15321 1532 153 15.32 32856 3286 329 32.86
70 12856 1286 129 12.86 35321 3532 353 35.32
75 10000 1000 100 10.00 37321 3732 373 37.32
80 6840 684 68 6.84 38794 3879 388 38.79
85 3473 347 35 3.47 39696 3970 397 39.70
90 0 0 0 0.00 40000 4000 400 40.00
(con't,
11--35

N
.Ie
E
(CO!) 'I)
100000 ES==:::E==t~
10000
1000
100
10
T----:-. --:---±~
I ,
T·-----·--~- . .
--~;---'------~-
~.....---r-+:...,I------t---t---i
"
o 50
e,deg
100
---a =0.01 m
- - - a = 0.10 m
- - - - - a = 1.0 m
- - - -a =10. m
N
.Ie
E
100000 B=:====:E===l
1000 t~_=~~~'=-=-=~~~=-,.-]
I ' •
---a = 0.01 m j
i- • I ---a=0.1m
: , i
- - - - - a = 1.0 m
----a=10m
100
10
f-.r'--C/~~~~-~~~-
~-~.-~
:1 .
f--+<---_-;-~_+______i_----+----i--
II
r
o 50 100
e, deg

4.4.3 Determine the x and y components of
acceleration for the flow given in Problem 4.6. If
c > 0, is the particle at point x = xo > 0 and
y = 0 accelerating or decelerating? Explain.
Repeat if Xo < O.
Since 11.== c(x"J._y 2.) and v= -2 cxy if fo//ow.s fhaf
~ A A ha = 0xt +(JyJ J were
Clx ==1t +u-¥x + v-W == c(x2--y')(~cx) +(-.2cxy)(-2CY)
or
Ox =2c2
X (x2.+y2)
and
Oy == ~ + u1f +v¥y- =c(x2.-y2.) (-2CY) +(-2cxy)(-2cx)
0"
Qy == 2 c2. y( x"+y 2.)
For X=Xo and y::::O We obfain:
lJ.:::cx2. vr::O
d
o J
an
Ox= 2C2.x: J Oy =0
ThllsJ wilh c>o and Xo>O ;i foJJows thai u >0 J Ox >OJ i.e.~ the
flviJ l.s acceleratinq.
With C>0 and xo~o if foJlo~.s fhat u> 0 J ax<o j t',8'J fhe
f/uirA is rJece/era1if1g.
4-37

-fM Water flows through the curved hose shown in Fig.
P4.44 with an increasing speed of V = lOt ft/s, where t is in
seconds. For t = 2 s determine (a) the component of acceler-
ation along the streamline, (b) the component of acceleration
normal to the streamline, and (c) the net acceleration (magnitude
and direction).
a) h f ~ =0J lJ dS
fdr a/I t.
2-
b) On::: f
or
an} ::- .5"(")'- == 2.0if,.
t=2.;
c) I}{ t ::= 2 s Ii = Cis f + all n=- /0; .,. 2() n~
or
J
~J [2. "]~ [1. 2.]~ ffa == Q.s + all ::: 10 +20 == 2 2.-i' ?
and
e::arcfo,n(M-) ;63.¥fJ
tj-38
o:.JI. =20 It
1

4.45 Water flows steadily through the funnel shown in Fig.
P4.45. Throughout most of the funnel the flow is approximately
radial (along rays from 0) with a velocity of V = clr2, where
r is the radial coordinate and c is a constant. If the velocity is
0.4 mls when r = 0.1 m, determine the acceleration at points
A and B.
ThvsJ 2.
_ (C )( 2..C) 2CQs-- ]"'2. -? :::: rs
At point A!( -.3 m3 )2.
_ 2. Lfx/o s :::: 3 2. 0 l!!-
a.s - (O.lm).b 's:L
o
FIGURE P4.4S
'1(-:::00 (i.e) fhe sfreamline.s
arB sfraight )
o,I"m
0,11n
O,lm
2. 2-
rOB= (0,1) +(0.06)
:: O. /167m
o

4.4·6 Water flows through the slit at the bottom of a two-
dimensional water trough as shown in Fig. P4.46. Throughout
most of the trough the flow is approximately radial (along rays
from 0) with a velocity of V = clr, where r is the radial co-
ordinate and c is a constant. If the velocity is 0.04 mls when
r = 0.1 m, determine the acceleration at points A and B.
~ A. J h V2.
a :: an n +lls .s J were CAn= 1? =: 0 since
lJ ~v ~v h V C
n/so) a.s::: VT.S = - V1FI Were ::: -;:
FIGURE )'4.46
1?:=OD (,,~e~ the sfreamJine.s
are sfraiqhf)
Since V== o.oq.!J- when r:::. O.lm if follow.s 1hat -.3
( -3:z. V 'f-x/o
c::: Vr == o.oIfIp)(O.lm)::: 'fx/o -r, or :::: r !!l. whert~ r rJ TnoS J c;;
Thvs,
as = -( ~ )0;1)= f-
IN poin! 1/: -.3 m2. 2-
(J ~ (lfxlO ~) = 3.13 x/o-5 4
~ (O.Bm't oS
At point B: -.3 m2. ~
(ifx/o :$") -3 m
~::; (0. '}..rn) 3 = 2. oox /0 :sa.

thlf7 I
4.,47 Air flows from a pipe into the region
between two parallel circular disks as shown in
Fig. P4.1f7. The fluid velocity in the gap between
the disks is closely approximated by V = VoRlr,
where R is the radius of the disk, r is the radial
coordinate, and Vo is the fluid velocity at the edge
of the disk. Determine the acceleration for r =
1, 2, or 3 ft if Vo = 5 ft/s and R = 3 ft.
DX5
7
R
'I
r
~vo
~,r ~
~ t~PiPe
V
FIGURE P4. 47
- J A h V
2
0a =an n + Qs S I were Cln :::: 'It :: since 1(::00 (t', eJ +he sireamJine.s
are .straighT. )
Also, as = V¥S ::: V¥r J where V:::: V~R
S/nce Va:: 5*and R:=3f.J J V== Irs ~ J where rrvff
ThvsI ( f+)"( )2-
_(YoR)(_ VoR) __ VrfR2
__ 5':$ 3£1.. :=_ '),.2,s
a.s - r /,,2 - r 3 - , ;3 fl3 r3
fJl r:: / ff J (l.s == - 225 !is.
Af r=3 ff J Q.s = -8,33~
4-JfI

4. '1-8 Air flows from a pipe into the region between a cir-
cular disk and a cone as shown in Fig. P4.49. The fluid velocity
in the gap between the disk and the cone is closely approximated
by V = VaR21r2, where R is the radius of the disk, r is the
radial coordinate, and Va is the fluid velocity at the edge of the
disk. Determine the acceleration for,. = 0.5 and 2 ft if Va =
5 ftlsand R = 2 ft.
!
r----R-----+<.!
FIGURE P4.4S
- A A h V
2
•
a ::::an n +CI.s.s , were Cln == "1( -:: 0 smce
AJ - aV - av h V- b R2.so) O.s - Vr.s - V1"F ,were ---;:r
'f<::co (e:e'.l fhe .s+ream/ine.s
are .sfraiqhi)
Thvs
~(~) (_ 2.Vo R2.) =_ 2. '"~RI/-
as r2. r3
r-5
lit r ::o.sHJ O.s =-2.5)600 ~
::: -
.fl 2. If
2-(5 s) (20ft) __ 800 .fi
.s .cl"s - r5 s
r TT
. where rNf-j
4-42.

4.4Q Water flows through a duct of square cross section as
shown in Fig. P4.4Qwith a constant, uniform velocity of V =
20 m/s. Consider fluid particles that lie along line A-B at time
t = O. Determine the position of these particles, denoted by line
A'-B', when t = 0.20 s. Use the volume of fluid in the region
between lines A-B and A'-B' to determine the flowrate in the
duct. Repeat the problem for fluid particles originally along line
C-D; along line £-F. Compare your three answers.
B B' D F
A A' C E
FIGURE P4.4Q
'1-.50
Since Vis cons/ani Ii) time and sptlceJ all parfic/e.s on line IJB
move a disfal/ce f -:: VAt :d(:2o-!;) (o.2.s) =Lf/J'J from t ==0 10 f~ O.2.s
Thv-s, the volume of ABA'O' l.s ~8A~' = (o.Stn)~(LfIn) :::/.00 m
3
so thQt
Q= ~8t~1 = 1.00 m
3
::: S.O ~3
0.2s $
Simdar/y from t==o fot==o.2..s fhe {Ivirl a/ol/9 JilJes CD and EF
move fo C'/)' and E'r;respecfiveV. II/so, ~DC~/:::: ~F$~I ::: ~BJ981
so fhat weobfat'n Q::: ft :::5.fJ!f re9ardlcss which line we consiclel'.
20 m/s B
4.50 Repeat Problem 4.4 qif the velocity pro-
file is linear fromlO to 20 mls across the duct as
shown in Fig. P4.50.
10 m/s A'
FIGURE P4.S0
From t==0 fo I:~ 0.2$ the parficle imtially ai B fravels a distance
1B ::: Ve At ::: (20.!p- )(o.2.s) ='f-m as Sholll/} wht'/e one af II/ravels a
Jisiance If/ ~ ~ At r:::~o-¥) (o.'1.s) == ~m. Since the '1e/ocdy prutlle
is linear i/ne 198 remaiIJSsfraI9h.. blll ..h11.s..;..10Iinell.ll~ TAvsJ
!he VOIllfll8 of flvid cros.silJ9 the in//;4/ IIIIe III ;'s
t:SB1A' ::: i (iA of J8 ) II ::; i:(2 m +'1-111) (0. Sin/.;: 0.7£ ",3 t'
so that __F"-----ft
IJ:: ~BB~I = O.7S m
J
::: 3.75 d
If - .6t O.1-s oS
F.Or any cvrved line £F (which /noves fo £'FI) E £'
~FF~' :: ~BB'R' sO thai the same I/oillme f/oUlrtJie) Q, is obfained
for allY l/nB cunsidered.
lJ-'f3

?': 5 I I r~---------~Control surface
1- -~.!"r_~~: /Sluice gate I
"'.51 In the region just downstream ~f a slui~e ?at~, the w~- '~~i: Vb = 3 ftlS~1
t~r may develop, a r~ver~e flow reg~on as IS. mdlcated m,' ..•.. ..' ~-------:-:::i=--7-~ ,.J ~~'"i=~.---': ~
FIg. P4.51 and Vldl'O 141.:-. The velocIty profile IS assumed to I ., ''VV.. ..".~. ~.. A~ ..' 1.8 ft
consist of two uniform regions, one with velocity V" = 10 fps , ' ~ ~': ' .' !
and the other with V/J = 3 fps. Determine the net flowrate of ................ '. '. --..... ·ct,..;:::~ ......· 1.2 ft
water across. the porti?n of the control surface at section (2) if .... - r7~/. --",. ~ ;,-:t-?,;
the channel IS 20 ft wIde. (1) (2)
Q-
FIGURE P4.5'
= (10 if) (J.2.ft)(:J.olf) -(3!f )(J,Bff)(;'()ffJ
_ /32 fI.3- oS
Va = 10 ftls
L{.. s 2. /4.52 At time t = 0 the valve on an initially as p = p.,(1 - e-bl
), where b is a constant, de-
empty (perfect vacuum, p = 0) tank is opened termine the time rate of change of mass within
and air rushes in. If the tank has a volume of -Vo the tank.
and the density of air within the tank increases
FOr t~o J p::. Po [I - ebt
] so fhai M::: mtJss of air /n tanK
cJM -bt = eVa :::f'oVo [I-e-
bt
]
Thvs} df == po~ b e

4.5'1- :
4.5+ Air enters an elbow with a uniform speed of 10 mls
as shown in Fig. P4.S4. At the exit of the elbow the velocity
profile is not uniform. In fact, there is a region of separation or
reverse flow. The fixed control volume ABeD coincides with
the system at time t = O. Make a sketch to indicate (a) the
system at time t = 0.01 sand (b) the fluid that has entered and
exited the control volume in that time period.
Control volume
From i =0 to t:::: O.O/s parlic/es IIJ BJ ~ DJ
and. E /!J()ve the
followin9 distances:
~::: ~ dt -:: (10.!Jl- )(o.o/s) ::0./m :: OD
JB
-:: Va Ji:: (5!j) (o,o/.s) =0.05 m
6c~ '!cit:: (J5~) (o.OJS) =0./517) J and
aE =0
Thv.sJ
fluid. a/on9 lines /JIJ 1J1Jd. BEe ori9in(J/~ moves 10 lines
fill;' ana B'E'c' shown be/()w,
O./m
-1 ~
D ,:_
I.. 'I J)I ...... -..;;
~
---- .sysfem at 1::::0
- - - - system at i :::0.0/£
//// {Iu/J -IhaT exHed con1rol
volume
c
15 m/s
flvid tho" entered. cunlrol
volvme
4-Jf5

'1-.55 J
.f.55 A layer of oil flows down a vertical plate as shown in
Fig. P4.55 with a velocity of V = (Vo/h2
) (2Ju - .xl) j where
Vo and h are constants. (a) Show that the fluid sticks to the plate
and that the shear stress at the edge of the layer (x = h) is zero.
(b) Determine the flowrate across surfaceAB. Assume the width
of the plate is b. (Note: The velocity profile for laminar flow
in a pipe has a similar shape. See Vidt.'o V6.(t.)
a) fir ==~(2h)(-X2.)
Thus)
/1TI =i£-(0 - 0) == 0 and
X==O
'/ =,# 1i/=JJ{r[2h -2xj '"' ()
x=:h x=h x=h
Plate ~
A----~---~~~-~-----B
Oil
1+----,.- h - -
f/ence) fhe flvid sficKs 10 the p/tJfe and there is no shear
slress al .Jhe free S{)rf4c.e.
x=n h
b) IilIiB '" JIlT dlJ:= J/If" bJx =f.}a (2hx _X2) ba'x
x=o 0
or
h
Q - Vo b[h 2. J x3 / - 1::. I hb118 - V X -:3 - 3 Vo
()

If. 56
4.56 Water flows in the branching pipe shown in Fig. P4.S6
with uniform velocity at each inlet and outlet. The fixed control
volume indicated coincides with the system at time t = 20 s.
Make a sketch to indicate (a) the boundary of the system at time
t = 20.2 s, (b) the fluid that left the control volume during that
0.2-s interval, and (c) the fluid that entered the control volume V3 = 2.5 m/s
during that time interval.
- - - Control volume
FIGURE P4.56
Since V/.s consf4nf J the flvrd trovels a disitmce i == VJt 117
fime At, Thus J -'t:::: Vt of :: (2.J}!-) ('20. -2.0)05 = O./ftn
12..= V:l. Jt = (1 -P) (zo, - ')..o)s :::: O,2m
and 1,3;;: ~ U = (2.5!f)(').0. - ')..o)..s = 0.50 m
The sy.sfem of f == 2-O.2.~ and fhe {Iuid fhal has eIJlercd or
exiled. the conlrol volume are indicafed in fhe {''lure below.
- - - confroJ volume
------- -- sysfem at i=20,2s
V1 = 2 m/s

*S7
..LSi Two liquids with different densities and viscosities fill
the gap between parallel plates as shown in Fig. P4.57. The bot-
tom plate is fixed; the top plate moves with a speed of 2 ft/s.
The velocity profile consists of two linear segements as indi-
cated. The fixed control volume ABeD coincides with the sys-
tem at time t = O. Make a sketch to indicate (a) the system at
time t = 0.1 s and (b) the fluid that has entered and exited the
. control volume in that time period.
II FIGURE P4.57
The flv/J af y ~ -0.'1-11 (Ihe hoflom piaIe ) relJlaifls slafiP/JdflY.
AI y:::0 fhe flvid speed is I,.s fflr .$'f) -fhaf af fjlbe t ~ 0,10$ ;/ A4s
fYJDved 10 -Ihe rifJhl 4 d;slance X:: Vt:: /.s #(o,/s) ::: o.lsll. In
Ihe sahJe I;",e periOd Ihe fop p/IJ'/e anrllhe fluid sff)ci If) il has
/hoI/sri a di.rt4fJCe X~ 2. !f (OJls) -;:: 0.2 If. Since fhe velOCity prof/Ie
is piec,ewire //neat) Ihe enJ.r ()f the sysfeM 1I1t///hove.,jt7 IhtJr
Jines- At; 4"rJ, Be ~emain straiqh-l, This isilJdicated in the.skeloh below.
y
41# ~2H
-:1 rAl ID IDI
rlIJidfh4+ AI!" r--~~-= ---~-~--~ f/viJfhq,t
entered .---r_~/: ex/fed co"'~1
con1(IfJI V/: ' vo/Vti/e
VO/fllYJe ,X
r // l-r. (f .-JI I-~; 0, IS ' 0, IS ft
B~; --= -:....=- -=-=-- =-= --=-- -=-=- --=-- d Co ~ CI
- - - cOlliro/ vO/llf11e
----- - - - - - sysfem af I ~o,J.s

~,se
4.5 Q Water is squirted from a syringe with a speed of V =
5 m/s by pushing in the plunger with a speed of Vp =
0.03 m/s as shown in Fig. P4.S8. The surface of the deforming
control volume consists of the sides and end of the cylinder and
the end of the plunger. The system consists of the water in the
syringe at t = 0 when the plunger is at section (1) as shown.
Make a sketch to indicate the control surface and the system
when t = 0.5 s.
.. ".' ~ ..'V = 5 m/s
(1) I
r--O.08 m--J
FIGURE P4.E9
DIJriIJ9 the i == 0.5s lime irderval fhe p/lJnger moves ~:::: Vp Jt :::O.OJ5/r).
and fhe waferiIJ/lio/1y af Ihe exd moves 12.::: V6i ::::2..!5m. The
corresponrJifJ9 confrol surfaces and systems af t=:O t1nd t= O.ES
S hOWl) in fhe fl9vre be/rJU/.
'
" ..... ,: ....," .. 110.. ..
r.'::.,-:.::.~~.:.~.;t!!.. =:.-~L.~;"T"':'"",'T.-•• ":'":-:-••-..........-.-••-.-"-••-.-.,'-',-"j.,..-:-:••""".-'•......,~.,.....-.~••~~-:-.~....."
l
:::· .....::.,:/J. .,' '...... ~ ..•..... :,..................... : ....,
-"~~:....:. fI) I.. 2 ..5 m ~
O.06.5m
O.OBm
control Vo/()me af t= 0.5.$
system 41 f=0.55

/1-, s9
4.59 Water enters a 5-ft-wide. loft-deep channel as shown
in Fig. P4.59. Across the inlet the water velocity is 6 ft/s in
the center portion of the channel and I ft/s in the remainder
of it. Farther downstream the water flows at a uniform 2 ft/s
velocity across the entire channel. The fixed control volume
ABeD coincides with the system at time t = O. Make a sketch
to indicate (a) the system at time t = 0.5 s and (b) the fluid
that has entered and exited the control volume in that time
period.
6fUsl----
", .".......................... Control surface
II FIGURE P4.59
DtJrin9 the l :::0,5 s I/m8 1,,/el'lItJ/ the Ilv/rJ ff;a! was a/()119
//n8 Be Af lime t=o h4J' m()vea I() /he rl?111III dirfaJlce
J. ;: Vi::: 2 f1 ((j,ss) ::: / If, Si/IJilri"IyJ porfions oflhe
fluid a/on9 Iills AD have fHPved J.::: 1#(o,s-.sJ ::= (),sll
()nd 1.::: b!/- (0,£s) ~ ~ fl. TIJ/s 4.5'.[(//118$ fhe /!! and,
6 ij flvid streallJs d() !'Jof nl/x Dr IAlcrmillf/R t:/vril'J91lJe
(),£ s time lidel'va/. See ftf;tll'e be/ow.
A ~J B 8'
I::::---=-=- =-=-- -=-= - ~ --:-:. - =-=-- - -r..:.-;~:
•• I - .. ,
• "1 ' , t Iff
flfJid tha14f;: ~r' "',:----..I. t _ _ _ _ _ _ _ _ ~ .. I
enfered. ~ :;fr:·:..:...~ '. '.' ~ f/vid fH41
coniroJ 1:-:",. ':- ,~ ,.- ""'1 ;.' .'-"~- exiler./, cOfJ/~/
Vo/u,"e f>.:-------- r···': VOIV/H8
.-..:· ....... o.,sH " 'I

:"'1 l' .',1.. , J • ~ I
• 1 I
• ::.= _________ - - - - - - - - - - _":'_1 I
1> -D J - - - - - - -c c
_ - - fixed con/I'o/ volt/me
___________ sysfem af 1::0.5s
4--50

'1-.60 1
4,60 Water flows through the 2-m-wide rectangular channel
shown in Fig. P4.60 with a uniform velocity of 3 m/s. (a) Di-
rectly integrate Eq. 4.16 with b = I to determine the mass
flowrate (kg/s) across section CD of the control volume. (b)
Repeat part (a) with b = 1/p, where p is the density. Explain
the physical interpretation of the answer to part (b).
a) Bow == S rbv·~ rAil
cSouf
A
~V= 3 m/s B
-" A
Wilh b=/ and V·n == VC()s{)
Bf = SoVcos f) dA =-~Vcose SJ,/}
Ihis becomes /;
c
O/) c/)' CD
=pVc~B ACJ) J where IIcD ::: 1, (2"')
_(0,5Ih)( )- - ').fYJ
CfJ,r8
b) With b::: J/p £f'O) becomes
Bout ~Sv·~ dfl =JVcosf) dlJ::: Vcose lie/)
CD Gl>
m3
=(3~)cosf) (~();8)n/· :: 3.00-:S
D
r+C
Control surface
/ ) D (I)
I vi _...
~n
Wilh b'" Vp = r~) = !:~ if {,fj,w.s fhal '8 ~ V,//lJfle'
vol •
(i.e:; b =-!a.r,s) .so fha! f V'ntill ::: B,uI refJI'8seIJIs fhe f/O/{)1fI9
f/ollr""fe C",3/s) frolll fhe cOld!'(;! 1f)/PIfIO.
ii-51

'f.(, I I
4.61 The wind blows across a field with an appro;(imatc
velocity profile as shown in Fig. P4.6 1. Use Eq. 4.16 wi th the
parameter b equal to the velocity to determine the momentum
flowrale across the vertical surface A- B. which is of unit depth
inlo the paper.
o
15 ft/s
• FIGURE P4.61
y"20fi
- f ~Vt)[MH'] (lWely
y.O
Buf, V= ;~y -# for o~y~loff(i.•., V;o,,+y,Oj V='S'~alrlo)
and V; Is1f For y",o(l
Thus, ,0 2D ' 10 ,.
- fr[ J(1fyfdy+SostJy] = fF[2.1.s fl + 2~ YI ]
o 10 0 0
-
=0.00238 S;7s
s
[750 :( +2250 ~'],
If-52

5·/
S.l Water flows into a sink as shown in Vidt'H V5.1 and
Fig. P5.1 at a rate of 2 gallons per minute. Determine the av-
erage velocity through each of the three 0.4 in. diameter over-
flow holes if the drain is closed and the water level in the sink
remains constant.
Three O.4-in. diameter
overflow Q= 2 gal/min
• FIGURE PS.1
for the confrD/ vo/vme indt'caTeJ,
0..00'1-'1-£ lj-.1
1i/,70 s
5-1

5.2 Various types of attachments can be used with the shop
vac shown in Video V5.:. Two such attachments are shown in
Fig. PS.2-a nozzle and a brush. The flowrate is I ftl/s. (a)
Determine the average velocity through the nozzle entrance, V•.
(b) Assume the air enters the brush attachment in a radial di·
rection all around the brush with a velocity profile that varies
linearly from 0 to Vb along the length of the bristles as shown
in the figure..D~temtine the value of Vb'
(a) rlJ ='41z
Thvs,
II, V, =' Qz
so
Of'
v" = "fS. 8 .fj
• FIGURE PS.2
.fl'Q" =' / s lind Q3': ~ 113 where
~ = ayerage velocify at (3) = ±Vb
113 =' 7TlJ, h3
Thvs,
-} v;, [17 (ftFil(# fil] = / i/, Of'
Yo =' 2M #-
5 - '2-..

5.3 Water flows into a rain gutter on a house as shown in
Fig. PS.3 and in Vidl'O '1.0.3 at a rate of 0.0040 ft3
/s per foot
of length of the gutter. At the beginning of the gutter (x = 0)
the water depth is zero. (a) If the water flows with a velocity
of 1.0 ftls throughout the entire gutter, determine an equation
for the water depth, h, as a function of location, x. (b) At what
location will the gutter overflow?
(a) Ft7V the, (Cl1M VO/lAme. .shown IV) -the sketch abdVe
Q in == Qo~f
(O.ooLJ-O ~/1)(x if)
~o h:: 0.012 X
So .L f+ _ ~, 012 X
I)"
x ...
- 10.g f+
h :: 2. ff
J~
5'_ 3

5. if I
504 . Air ~ows steadily between two cross sections in a long.
straight section of O.l-m inside-diameter pipe. The static tem-
perature and pressure at each section are indicated in Fig. P5.4.
If the average air velocity at section (1) is 205 mis, determine
the average air velocity at section (2).
~b ;/
Section (1)
PI = 77 kPa (abs)
TI = 268 K
VI =205 rnIs
II FIGURE PS.4
P2 = 45 kPa (abs)
T2 = 240 K
This C/;1aly.fis Is sl-"'i/(J~ ~ flte. dJ1e ()f' Cxtll11pJe 5:2.
For steady /kw i>elweeY7 $ecl-/tJns (I) and (2)
WI,.. - I'hI
or
.oA V :: ~ A. V
/.). ,2;'" 1/ I I
Thu5
--V=
2..
AS$um/"!J t-haf ul1der +J,e UJi1dihilns of f~i.s
behllj/e~ a ~ Cin ie/eal gt/5' we U$e Ihe
etjutrHoYl of date (E~. /.8) 10 get-
(I)
,Pl7Jblel11/ ClIT
('deal 9a5
~~~ !i~.
,., ~ P T
(2.)
~ .,. f
Cowrb/n/nJ Etjs. I anti 2. 411&1 observ/~ Mal
we. gef
V =z.
p, ~ V,
'1- 7j
3/ t.f !!!.
5
:. (11k fa.(abJ)] ('2'1-0 J< )
[ ,/5 --APt;. (u105 )](268 k)
(-Z(}5 ~)

5.5
!').:') The wind blows through a 7 ft x 10 ft garage door
opening with a speed of 5 ft/s as shown in Fig. P5.5. Deter-
mine the average speed, V, of the air through the two 3 ft X 4 ft
openings in the windows.
OS I Ti16 ft . .~ !",: 10 ft ts.ft/'
1~~1:30'
5.6
II-·-----22 f t - - - - - - !
• FIGURE PS.S
roy- sfeady
Q.9a~e
ad"Y'
or A V -- t1 V + A, V
'Jt1tr1fe 1'Jf?Y'Mt%/ -It, w~"drJw iJlmeifMI
ac)(J r ,1U'a.9e. 4"OY'
$0 hie ()..~je ~~eJ) v: df fhe ~/r
Aa/t.aAe. VMt"",,~ +r, III1'"4.'It boY'
V:'~ tI =
5.6 A hydroelectric turbine passes If million gal/min
through its blades. If the average velocity of the flow in the
circular cross-section conduit leadin2: to the turbine is not to
exceed 30 ft/s. determine the minim~m allowable diameter of
the conduit.
I=or IncompreSSible flow through fhe conduif a#d fur6ine.
A vCtJndJlif C(HId1.1if
( /, 6~ )
Lj. ) ( 4 )( /0 min
ana
dCtJMu/f
-~oYldui-f
d :: I~rs ff
Cf)ndui f
5-5

5.7 I
5.7 The cross-sectional area of the test section of a large
water tunnel is 100 ft2. For a test velocity of 50 ftls, what vol-
ume flowrate capacity in gal/min is needed?
~~AV
Q:= OO(){fJ.)(50 £!)(7.~P: ~ )1/ / )S ff-] (t"O ~
m/n
5-7

5.8 A hydraulic jump (see Video ,10.5) is in place down-
stream from a spill-way as indicated in Fig. P5.8. Upstream of
the jump, the depth of the stream is 0.6 ft and ~he average stream
velocity is 18 ft/s. Just downstream of the Jump, the average
stream velocity is 3.4 ft/s. Calculate the depth of the stream, h,
just downstream of the jump.
·' ,
• FIGURE
(2.)
,I
'h ~ 3.4ft/s
Ft,y 5feady ,;" coYJlfpfesj;6le flow belween sech'()n~ (/ )QJ/1d(Z)
Q, = 0.2.
Or
-AV, ,=
Thus
~ hi
and
-
h =2.
~ h. _
l{
( /8 !j.) (0. 61+) =
(3,tf ~-f)
5-$
3, /8 -ff

5. q A water jet pump (see Fig. P5. q) in-
volves a jet cross section area of 0.01 m2
, and a
jet velocity of 30 m/s. The jet is surrounded by
entrained water. The total cross section area as-
sociated with the jet and entrained streams is
0.075 m2
• These two fluid streams leave the pump
thoroughly mixed with an average velocity of 6
.m/s through a cross section area of 0.075 m2
•
Determine the pumping rate (i.e., the entrained
fluid flowrate) involved in liters/s.
conf"ol V"/U1't1.e,
V3 .:: 6 m/s
V, ::
r----'--;f---;P--~. 30 m/s
jet I
su+io,,~) I
FIGURE P5. q
For sfeady in compre$sib/e flow -I-hl'ough +he.. COfJ~/ volume
0-, + Q.2. = QJ
or
- QV, AI + :7.
Thus
Q2
- 'V;A3 - V,A,
/50 trIer)
s

5./0
5.10 Water enters a cylindrical tank through
two pipes at rates of 250 and 100 gal/min (see
Fig. PS.lO). If the level of the water in the tank
remains constant, calculate the average velocity
of the flow leaving the tank through an 8-in.-
inside diameter pipe.
Section (2)
Q2 =
250 gal/min~
--------
Section (3)
For :'feady and /11 compress/hIe flow
0.3 = QJ + Q2,.
FIGURE P5.10
or--
/
5./1
f+
/
1;d2
-"3
L.f
5.11 At cruise conditions, air flows into a jet
engine at a steady rate of 65 lbm/s. Fuel enters
the engine at a steady rate of(!J.60 Ibm/s. The
average velocity of the exhaust gases is 1500 ft/s
relative to the engine. If the engine exhaust ef-
fective cross section area is 3.5 ftl, estimate the
density of the exhaust gases in Ibm/ft3.
For sfeody flow
or
Ij A3 ~ ::: m-tfYI
I 2.
Thus
~ -
,
111, +- m..2,
250 ~I'J'I1')(231 !:!:..3 V.i. Vi.)/ gal A60 ~AIz ;n.
J?'11;, If
~ fuel in
I~
I exhoIJ.S-I
~~==~~-~-~-~-~!OS6
conh-o I
VtJlume
65 ~ +- O. CO !?
(3. 5 ~+2) (/500 fJ )
(J"f
5'- 10

5.IL
5.12 Air at standard atmospheric conditions
is drawn into a compressor at the steady rate of
30 m3/min.The compressor pressure ratio, PexiJ
Pinle!' is 10 to 1. Through the compressor pipn
remains constant with n = 1.4. If the average
velocity in the compressor discharge pipe is not
to exceed 30 mis, calculate the minimum dis-
charge pipe diameter required.
FoY' sfeady flow
, .
m,.... =f}?
I
or
- (Lf!A V - ;::2 ~ 2- I
50
Q,
d2. :=
1r
~-
~
fil1ally
d ::: 0.004 m
2
--
5, / I
seclion(t) (inlef)
, --'r--------
• I
----.I, secl/on(Z) I
I c'mf~1 (eX)f),:
, VtJ!u/tJe. ~i-----1I~--7
compressor
di.tc/1drj€.
pipe
, :3
(;'yo; 30 ~- PlII'/
/0 --'- ( )
~ 30!j-
60~
min

5·/3
Two rivers merge to form a larger river as shown in
Fig. PS.13. At a location downstream from the junction (before
the two streams completely merge), the nonuniform velocity
profile is as shown. Determine the value of V.
U$e fhe.. CI/YJ1Y?J1 VO/vln1C 5h~wn wit1J1~ brol<ey, //ne.s
IYJ fne ske I-c-t-t ab(}Ve.. {tie nole fha! m::: f A V {;Ir'Jc1
~ ~ ((/Y)s~(Vtvf/on of masS' fY/YJl1pJe vve..
gel-
m I +VYl 2
._ m -
3 - Y11 t- tv1
().fv v
oA v t fA ~ .. fA 0.2 V + f A V{ I I z. 2 0.1 II I
v
v ~ 3·63 ti
=- 5
5-/2
=(50ftJ(3ff)(3f! t(iM! {Slf)ftft)
('Yo.ft){6ff)(o,'6)t (7()ff}{iH)

5./tt
5.14 Oil having a specific gravity of 0.9 is pumped as
illustrated in Fig. P5.14 with a water jet pump (see Vi(ko V3.6).
The water volume flowrate is 2 m3
/s. The water and oil mix-
ture has an average specific gravity of 0.95. Calculate the rate,
in m3
/s, at which the pump moves oil.
{or sfeadJ flow
.
m+m, A.
Or
~ ~ + t Qz. = Ij Q3
Section (1)
Section (2)
tOil (SG = 0.9)
FIGURE P5.1tf
(I)
~
Water
and
oil
mix
(SG = 0.95)
A/50 Since fhe wakv and oil /11ay be. cOh.$t'dered Ii, clJrnlre»/61e.
I
Q" -f Qz = Q 3 (2)
Comb/nt'nJ £9.$. / a/l7d z we (jef
I? Q, + /} Qz If ({), f 0( 2 )
or
Q + SGzQ, = 5G J
(Of f Qz)I
Qnd
Q, ( 1- 5G3 )
Q'2. -
SG3
- 5G 2-
Thus
(1- ;3)(/ _0.95) J
~
:: ::: 2.00 :!!.
s
O.9S - 0,90
5- /3

5·/~ I
Air at standard conditions enters the compressor
shown in Fig. P5.15 at a rate of 10 ft3
/ s. It leaves the tank
through a 1.2-in.-diameter pipe with a density of 0.0035
slugs/ft
3
and a uniform speed of 700 ft/s. (a) Determine the
rate (slugs/s) at which the mass of air in the tank is increasing
or decreasing. (b) Determine the average time rate of change of
air density within the tank.
.,.----- - --
-~1.2 In.
! _ 700 ftls
- - - - -r b.0035 slugs!ft3
Tank volume = 20 ft3
use the.. CtJrl-lrvl vtJ/umf:. W; ff"n the hyo/< ~YJ //nc5.
(oj ~ +ne c(/y}se. '(VCi -h'dY) of m~>s prj-1"7 [ I ~/-e we !}ef
(h)
DMJ'(! _
Dt-
f)iJ1WJ =
[)t
oMs-ys -
Pt-
So
~ :::
,
_ ~ ~/}CJ..-L) A V
01At- c,~ In rOLAi Out (Jur
0.00'15£ SIll!! -
'" "'Yea.!J'!}
s
O{f ~r~ ) = V De o. ooLf56 ~/Lf!J
$,s :::
Dt .s
Ot
0.00 LfS{ r!J.J 6.00'fS£
s!t{.J -tf
J" .s - 2·ZPXI0
- - -
51'1J
-Of
~>fI
20 ff'? I'f~ >
5-/'1-

5.16 An appropriate turbulent pipe flow ve-
locity profile is
V = lie (R ; 'rni
where lie = centerline velocity, , = local radius,
R = pipe radius, and i = unit vector along pipe
centerline. Determine the ratio of average veloc-
ity, u, to centerline velocity lie for (a) 11 = 4; (b)
n = 6; (c) n =8; (d) n = 10
For'" any CYOSS sec lion area
Ii? = ,..oAu - J,P v. ndA
A
A/50
~ .A
V. n
~A
; V. l = uc (R,/)
.J..
11
cross seclion
ayec<
for a lIni~rm/y dislribufed
I f< u (R- t:) *Z1rrdr
dA == 2rrr dr
densify/ ~ over Cltet/ A
o C R /
u. -
and
R ..1-
u
Z f f -; )(~)d(;)
2 n7.
- - -Uc 2n2.+ 3n + Io _
n fA.
lie
if- 0.71 f
6 0.79/
10 0.866
5-/5

5,/7
5.17 The velocity and temperature profiles
for one circular cross section in laminar pipe flow
of air with heat transfer are
where the unit vector i is along the pIpe aXIs,
and
The 1'h~~J f/OWYfJ/e 15
.
=/t -l. /'
m v·n dA
A
T = Te [ 1 + i(;r -~ (;)]
The subscript c refers to centerline value. r =
local radius, R = pipe radius and T = local tem-
perature. Show how you would evaluate the mass
flowrate through this cross section area.
/?;r (Ur acft;y as tin ideq/ !ftlJ'
f
P
- -RJIIlr
f~y a.., Ciy~/qj" cr()S5, feelloY! oJ'ea.
dA -- 217'rdr
-
ThU5 /?
fY1 - fir U
C
[I - f;)2] Zfr r dr
- () II,;,
and ror a ul1ikm!J til;h--i6uIerI sIalic
~ I< '1
h1 =21ft~ J[I - (;) ]
~i,-r::: fJ [ Iff(I/--
If we
Ii ~
fhen
=
5-/6
ieft)
;(;)7
O.7(!)/
pYe5SLlye

5 .I~ * To measure the mass flowrate of air
through a 6-in.-inside diameter pipe, local veloc-
ity data are collected at different radii from the
pipe axis (see Table). Determine the mass flow-
rate corresponding to the data listed below.
r (in.) Axial Velocity (ft/s)
o 30
0.2 29.71
0.4 29.39
0.6 29.06
0.8 28.70
1.0 28.31
1.2 27.89
1.4 27.42
1.6 26.90
1.8 26.32
2.0 25.64
2.2 24.84
2.4 23.84
2.6 22.50
2.8 20.38
2.9 18.45
2.95 16.71
2.98 14.66
3.00 0
The W1 a>> flowvlJ.k Is calCM.LaIe.d. w/tt.
R
In ::: I ;OfA 21ir-dr :
o
R ::
p= 0.00238 slu9
f+3
U:: 1~CtJ.{ ax.la! ve.{dci+y ifl ff
S'
r :: I oCd.I yadiu.s ,;' j".
and f ~ 11 rdr /5 (. va./u.a. leo! IJUfltU;c.a.tI~ w;fI1 fk ~rJIJ"d.().1 'rule
()
wi/t1. une1tA.M J;"~v().ls. Thf!- COY/ltp/A,~ PW)(IYa/V. u.s~d ro Jollie /his
l'flJblern IS II~ led On the Y1ex. t- pa.je. .
(con't)
5-/7

Ii. J~ I (coi'1'f)
100 CL:-:;
110
1.2' 0
130
-1 1 r.
1.~U
150
PRINT
F)F~ I 17r
F'F~II~T
FPIKT
PF'INT
" t::.t -* ;t:.:t *:t: *t: *-* :j:: *-:t)< *:*--:t:>I::t: **.i'- :t:i *t* :t,;r:t :t.* .~.- i-*;I; i: ***:t·t ***_:.t i, t"
"of.-;f: Thit:. program c:omputes the ma:=;;s flow rac:e ;t,,~"
,,** for problem 5,16 using the trapezoidal rule it"
")::+: applied to unequal intervah:.. *t"
":t.***.********** i:***:t;*:*****:t** ***:t**i~ t.:;j::t::t.*******:Lt.:t"
160 PF:IIH
170 DIN U(19), R(19)
180 '
190 'Initialize the variables
200 N = 19
210 RHO = .00238
220 P I = 4! * ATN ( l! )
230 FOR I = 1 TO N
240 READ R <I ), U( I )
250 R(I) = R(I) / 12!
260 NEXT I
270 DATA 0,0, 30.00,
280 DATA 0.8, 28.70,
290 DA~A 1,6, 26.90,
300 DATA 2.4, 23.84,
310 DA~A 2,95,16.71,
0.2, 29.71,
1.0, 28.31,
I.e, 26.32,
2.6, 22.50,
2.ge,14.66,
O.
1.
2.
"Co.
.:::) .
4, 29. 3:;, ~
'"'Co, 27. 89,
0, ?h
"- ...J • 64,
8 J 20. 38,
0, 00. 00
:~:3C '~::'.!=,rnpute integral using trapezoidal rule
340 FOR I = 2 TO N
O. 6, 29. 06
1
4, 27. 42....
? ':I
24. e-4-'-' . '-',
2. 9, 16. 71
350 SUM = SUM~(U(I-l)*R(I-l)+U(I)*R(I»*{R(I)-R(I-l)/2!
·:,6;) NEXT I
:::',,(,0 Jl1DOT = F~HO * 2! * PI * SUM
3130 '
3g0 'Print the results
400 PRINT
410 PRINT USING "The mass flow rate is ##.#### slugs/s"; MDOT
*************************************************** This program computes the mass flow rate **
** for pr oblem 5. 18 usi ug the trapezo idal rule **
** applied to unequal intervals. **
*************************************************
The mass flow rate is 0.0114 slugs/s
5-/8

5·/9
As shown in Fig. P5.19, at the entrance to a 3-ft-wide
channel the velocity distribution is uniform with a velocity V.
Further downstream the velocity profile is given by u = 4' -
2l. where u is in ft/s and y is in ft. Determine the value ~f V.
Use the U/YJ~/ V~/ume iJ101/cttkd by /he hY1J/<.eYl //YJe.J J/J
if, e sk-elvh ahwe .
~ ~ CIfYJ.J'-ervt( -h'or-- of /'I?ClSS pYj'y}c/pJe
CLI
= Q1.. I ff
'1AI = f LA dfJ flly-2y':J b dy
A2. (J
/ If
V (iJ.7Sff) b y fi!2_ 2y311>= fb 11 3
- -3 /) :3 s
V - 4 - /.78' Ft
-3{~.7§) oS

5.20
Section (2)
5.20 Flow of a viscous fluid over a flat plate
surface results in the development of a region of
reduced velocity adjacent to the wetted surface
as depicted in Fig. P5.20. This region of reduced
flow is called a boundary layer. At the leading
edge of the plate, the velocity profile may be
considered uniformly distributed with a value U.
All along the outer edge of the boundary layer,
the fluid velocity component parallel to the plate
surface is also U. If the x direction velocity profile
at section (2) is
Section (1)
u
develop an expression for the volume flowrate
through the edge of the boundary layer from the
leading edge to a location downstream at x where
the boundary layer thickness is b.
From +he
II()w fhJ"()u.1h
~ have.
./1'7 -::;..
6/
::::
Ovtd fhus
Q
61
t:
Conservt) lioY! of- """ttSS
rhe con/nJ / voIt.l ",., C
.
1~
-.l 'I. dtn - /. n A
2
A2
WIdth ()f ~c p/afe
1 7JJ. b
8
5-Z0
~u
·1
FIGURE PS.20
prlnc.iple. C<.ppl/ed fr; fk.
sitown ,;., fhc /igul'e
Outer edge
of
boundary
layer

5.22.
5.22 How long would it take to fill a cylin-
drical shaped swimming pool having a diameter
ofIf) m to a depth of 1.5 m with water from a
garden hose if the fiowrate is 1.0 liter/s?
defu.rm;IIJ
c.onfrt; I
V() il,,( fYt.e
From appl/cah'on of fhe c,onservo,h'oVi CJf J?!1.tlS$" principle
Iv fhe. COrllro1 volume COl'7hlY-J 'ny wak-r only Cf S $hown ,;"
we have
r -l /
+ J(' V. n dA
cs
== 0
FOr Incontpres.si6le flow
Thus
-t ==
or
2-
trJ) fa
::=-----
if Q.
11 (IDYl") 1. (;.5"",) (lOCO ~)
tf (1. 0 I;e~) (?600tf )
5"- 2. J
1

5.2.3 J
5.23 The Hoover Dam backs up the Colorado River and cre-
ates Lake Mead, which is approximately 115 miles long and has
a surface area of approximately 225 square miles. (See jdl'o
'2.3.) If during flood conditions the Colorado River flows into
the lake at a rate of 45,000 cfs and the outflow from the dam is
8,000 cfs, how many feet per 24-hour day will the lake level rise?
Far the cont('0 I volt/lYle .shoWfJ:
min - moul == kf~ d-1I
, cVwafer
or since m::::pQ. )
Qin - Qouf - k(II/ok. h) '""Joke 1b.
Thvs, d.h" Q.ut - Qin " ('1s,ooo:8,odo) .fjJ z =5. 90X/Oi
.if
err fI/lJ,t.e. 22S mi (52.90{t.) N
= .s.'lOX/O-61f- (3,;pO()ir) ( :ttf.lf;,) :::: 0.5/0idy
5-22.

1
de foYdII n!J.
5.2 Lf Stonn sewer backup causes your basement to flood at l/Jrrfl71/ llolume fhaf
the steady rate of 1 in. of depth per hour. The basement floor r ClJnfriin5 water
area is 1500 ft2
. What capacity (gal/min) pump would you rent -- - - J - - - - - - , 1/1 /1to (a) keep the water accumulated in your basement at a constant / '
level until the stonn sewer is blocked off, (b) reduce the water / ' / I h
accumulation in your basement at a rate of 3 in./hr even while / / ) _ _ _ _ _ _ _ .t. / _ -:.
the backup problem exists? J - - -/- - - - - - -, /
I / / I "t~t Cji "~f/~oY ayea -= AI / - ~ /
I _ _ _ _ _ ~l/
How nut flow In
For a aefwmil1j COn/nJI Vd/('(me fhaf u;n!l1/J"S fire WtI/ey
OIlf!Y fhe baS"ewtenf floor- (.see Skef.ch ClbJve}) The
Conserva-hfJYI 01- maS5 priJ1c;,Ie. (Etg. 5.17) /eaC/s fr;
t- 1,;O~' nciA -- 0
cs
or Con5ft:ln1 flt(id c:leJ1Sily aYlc( ar'ea, (A)
A dh Q. -r QDwf ~O
- II?
d-t
(I)
F;y pari ct .I Etj. / leads fr;
(J6u.f :;:; Qin
ro eva/utile ()Ii,) ~ use fl{ / wIth Q~wf =:0. 1Ju,s"
Q,;" - A dh (/500 f/') (/ /h) (--L ) 125 f+1- = ::;
-d-t hr 12 Ih. hr-
and Ff
Qokf
::; (125 fl)(7.'18 :J~)(',-!-- .) "' /5:6 JE
hy {f~ ~tJ ~ min
hI"
FoY parf b E~./ yieldsoJ
,
!;
j

5.2.5 I
. 5.25 A hypodermic syringe (see Fig. P5.2?)
is used to apply a vaccine. If the plunger is moved
forward at the steady rate of 20 mmls and if vac-
cine leaks pass the plunger at 0.1 of the volume
fiowrate out the needle opening, calculate the
average velocity of the needle exit flow. The in-
side diameters of the syringe and the needle are
20 mm and 0.7 mm.
I
FIGURE ps.25
USing a delDr"'/~ C()n/YrJ/ LIP/lime QJ1d fhe COnservClI/tJn ()f
/J1(1 Sf !JI'/;'c.iple. ( Etj. 5'.17) as ot/II/ned In €x.ohlple, S. 5';
w~ tJ61r:lln (see £1' 3 ,,{ Ex Qmple s: g)
-;0 A, ~ + ;0 Q2. of' I'QleaJ:.
1:
Since ;;::
~ obm/~
C/)nJ/rJnf) Q :. o. / Q ..
leak.. ..
~ £1. 1
1.1 ~~ = A, ~
or
:: (;, ) ~ -(1) I?V:z..
- - d! -;; =A.1 1./
and
V; :: IJf. & m
-5
0
1
( Z 0 yrJl'k) (z0 n,,,,,Is )
( d. 7 ~,., ) .. (I. I )
(;)
(tot}o :~)
I
,
J

5.271
5.17 It takes you 1 min to fill your car's fuel tank with 13.5
gallons of gasoline. What is the approximate average velcoity
of the gasoline leaving the nozzle at this pump?
v A - ~
hOJ31e f10JJ/e
2.
Clhtl
AnOJ31e.
:: 7!.!h()'J1/~
If
,..
So == {I'~. 5' )(JI)(12 )
--
2.
VI1P}3ie
(7T~!r/L(7.'l-8)(bd)
v == 17. if 1"1-
n()1Jle S
5-25
..

5,2.6 A gas flows steadily through a duct of
varying cross section area. If the gas density is
assumed to be uniformly distributed at any cross
section, show that the conservation of mass prin-
ciple leads to
dp + dV + dA = 0
p V A
where p = gas density, V = average speed of
gas, and A = cross section area.
Foy a sfeady,l oJl1e - dIMensioA(J/ /i~~ fhe CIJnSer-va/ion 01 PI(J'>]
pYiJ1tift1! /elJd.$ -I7J £~. 5./2 or
;aAV = Consfa",d
7hus
cl(fAV) :::0
Or d{J AV + ;aA d 'V T I'dA V - 0 (I)
P/v/d'1f ~. / by ",A V we obfl1/~
~ +
dV
+ dA 0- -;<1 V A
5"'- 26

5.2.9 A IO-mm diameter jet of water is deflected by
a homogeneous rectangular -block (15 mm by 200 mm by
100 mm) that weighs 6 N as shown in Video V5.4 and Fig.
P5; 2.9 Determine the minimum volume nowrate needed to
tip the block.
From fhe free body diaqram
of fhe block when if is ready
10 lip Z i110 =: oJ ()r
Rx JRx ::: WIII' where Rx is
the fore iha+ the wafer pllts
on fhe block.
ThllsJ
Rx -==
(0..0/5 )
W/W - 6N ~", ~O,90N
tRx ~ o.OSom
For fhe conff'o/ vollJme shown Ihe X"-componerr! of the m()/l/8I1fllttl
eqv"lion
Sup V'n dlJ c Z rx,
cs
becomes
bloc~

5.JO
Water enters the horizontal, circular cross-sectional,
sudden contraction nozzle sketched in Fig. P5.30 at section (I)
with a uniformly distributed velocity of 25 ftls and a pressure
of 75 psi. The water exits from the nozzle into the atmosphere
at section (2) where the uniformly distributed velocity is 100
ft/s. Determine the axial component of the anchoring force re-
quired to hold the contraction in place.
r::f:9 ,.- -fh;s plDh/em we incll1de. I;' the C();,~/ V()/Uftllf!.
fhe hOJ}/e as ~// as Me wpf~1" al 4M ins!-zrvtl 6~~en sechiJns
(/)al"ld(Z)cu Ihdicaf~ /n th~ sl::ekh above. The hOY'i}lhfal
.foyce s ac,f/Yl9 ()YJ +he coll1fey,fJ of the UJn-/-nJ1 vaJu.~ a,.t shown
In fhe sk.e fCh _ Note f~rA+ ihe.. CJfWl()S pheriL foy-,eJ CAY/eel (JIA. f
()V1d aye no!- sh()wYl. AppliCA.htJr. of -fhe. hOYI]tJYlftJ.1 tJ'r x-
d/"ecfid~ cOMpaneYlf of +I1e / /jlfeat }?IUYkeYlfu~ eqtJ.a.fion (Gr- >.22)
fo fhe.. flow fhY()U111 fhis ~J11nJ1 v()luYHe yieLds
P.A -1= - vA' I A r2 l
Frorvt iN... tfrt'lSf/rv~I/,,1') af I'YIQ-s.r er~n~ ( 1:::).5. /2) we tJbltt/~
m-- fllA. AJ =to 41 A,.
1bu.~ f1- (I) n1t:!y b~ 6e~S~rI".s
m(t.4" - ~1):= p'A, - ~ -Ii1I"l-
(I)
Or ;t ~ 2.
F =. /',11, -f{A1
+- tYJ 6,,,-U,) = f, 1"-1-/)' - r;, 1£f}1. -flu,TI'p, (u2 -fA,)
1/ 'I ..., If
and F. ~ rr;;!l):!(~.j_ 016- f'llf!,uznIz5 £t ) Jr(3,:..j IO() ft- 25ft rIb.~
A {II' 'i C -ft'/ l~ .s.-l tj"" (N'f ,;.,.?- SI"1.
11
f,.~
~:=~ Ih
5-28

5.3 I A nozzle is attached to a vertical pipe and discharges
water into the atmosphere as shown in Fig. PS.31. When the
discharge is 0.1 m3
j s, the gage pressure at the flange is 40 kPa.
Determine the vertical component of the anchoring force re-
quired to hold the nozzle in place. The nozzle has a weight of
200 N, and the volume of water in the nozzle is 0.012 m3
• Is
the anchoring force directed upward or downward?
to.lom3/S
• FIGURE PS.3i
~ Cr:J')1 serl/(J..h·/JI" tJI "" a.rJ ~e D~I"Jn
Q).. ~ Q,
(1,
A').
(I)
(2)

~. ':' ~DON ~ 'ZOON -ill.6 N - 0 N -: 'f82 N downwttrCf

5.32-
5.3 Z Determine the magnitude and direction
of the x and y components of the anchoring force
required to hold in place the horizontal 180
0
el-
bow and nozzle combination shown in Fig. P5.31..
Also determine the magnitude and direction of
the x and y components of the reaction force
exerted by the 1800
elbow and nozzle on the flow-
ing water.
,
Section (2)
12 in.
FJGUI~E 1>5.32 PI = 15 psi
VI = 5 ftls
Fo~ defet'm;~/''' fhe X and y dl;ec.fi~Y} c(J¥Vtpf)nenff
x
fone a. Coi11J..D I 1/f)/uJne fhr;t f con llI;nf ';'he elbow.l 1'10}j/e pnd MIt/fer
/:JelweeYl S'ecfioYls (/ JAnti (z.) is used. Th C oJ'dnJ/ v"/l(l11e a;1(1 f'/'e
fOY'ces in vIJ/ved QI't. 5htJWh In the s/t:.~fci1. above. /tj?p/,"cern'Dr, 01
the y cI/recl-;on c.()mptJn~n+ of -file lJ~eclY YYJ()YJ!Iel"t.ftA~ quafltJJI)
(E1' ~. 22) leads -/-t;
r: = 0
1-Y
Ap,IIc.(J.+iDVI 01 -fh~ Xf:kt'ecf/()n CIJ~p~neyd 01- +k linea)! mOWlei'lful'n
e~fA aA'oYl YIelds
- LAI f U A - U f M A A F{I 2 1.:2 -::: P I - A + f.. A'2..I ~ X. 2
From fA e C(n,.,serllaf,'on of WUlSS e1ud-f,'on
rn = !Ju/ A, :; fJlA2. A'2,
1hUj &t. I May be eXl'fej~ed (N~
-P'fA (lA, of U.) :: p, A, - ~," + f,A2
~Vd
F :;
A/x
Als6 ~
ThUS
F
AX,
f~, A/w,-tIJ.,) f P,A, t P.A2 =t u, : o,'(IA,tfA.) +p.
9. 2
(con't)
5-31
II)
(Z)

5.32. (con'-/; )
,
r (is lJ. ) 11' (J2 ,•.J
{n.'l. if
Ftly defwMJY1;11j fhe x aYlq y cOMf~nenlJ of fhe yetlcii())'J -hY(;e
tit COn fro/ v() IUMe fhaf ClJytfa,i-u (1;1Iy the. Wit k,y- between sec/lam
(I) fNt;td (A.,) ;f us~d. Ap"lklAh'6-h of fhe y dlrecht)Y} CiJM-pIJt1(Jnf
of fhe. hhe4Y VUOJ1-f~;, fuWl e&JUtX.·fion yieLdJ
I?., := 0
AfP /;caf;'Oh of fhe X d/yecf/ol'J cOl"l-'lptJnenl of -the. II;';ea v
hi}() /)If e k fuJ'V/ fYJua f/(Ji1 leadf fr;
Or
=- /i9~ /b
Section (2)
x
Section (1)
5-32-

5·33 1
Water flows as two free jets from the tee attached to
the pipe shown in Fig. P5.33. The exit speed is IS m/s. If
viscous effects and gravity are negligible, determine the x and
y components of the force that the pipe exerts on the tee.
Use fne Lo)'lfyl)1 vollAYhe.. s-h()wn.
Foy fhe X - c.offlf()Y1enf of fhe -force. eX~l'kcl by +he pIpe.
OYJ +he fee we. lAS'~ fhe x- c.oYhplJneni r:;f the l/neCir
i'hO m-e. nItAYVI .€1/Aa.ll(Jy).
- ~(' l{AJ -I- ~f~1J3 :: f,AJ
- IfA.J- ~f,A,""fl3)+ f
"().~r.;t..)Al-(-;;. ~)43- ~-,43) -J- ~
=p A -t- F /1)
I I )( l'
9t1t:je
To 1e.f. '1 we LASe cOYlservah'rm 01- mass
61, = Q 2 +- ~3
6y A, ~ -= liz V2. 1- 11"1 V,3
50 v.::; A.l. V:z.. + A~ IIJ
= (0, 3M2.) (IS" m1 ) +- (0. S"'')(It;" m/.s) ::: /2 1'Yl1
I AI /~~
To eJ f-/.-necle. f/;fjQff! we. l-1.{e 8~Y'I1(hll/i J' ~~6~ ~ f/(IW belwr£." (1)4t101(2..)
- 72 ()()() N:. FI J(
So F":: 72/~OON ~
.x
Fw /1u.. .Y ~!~ ff I1u ~ exekfecJ /:;f hr-t-, PiPe Oh flu. lee, ,ve L).5~ #V
Y (IYYtMc"f Itl- h..L !;~e6N" hlnntA11w.-" erah ,., ..fe, 9~f
.1',2. ~ II). ,4 2 :: 5
(JSJXq'f9!:!)(l5;)(D.1"'~) ~ {,~'fOoN t "Fj
5'-33

(r) (,mltya,1
.j. 10 psi ~ VOt..cme.
5.3'1- Water flows through a horizontal bend and discharges ... -- - .......
into the atmosphere as shown in Fig. P5.3'1. When the pressure P, A I
gage reads 10 psi, the resultant x-direction anchoring force, FAx' Q =? ~ •~
in the horizontal plane required to hold the bend in place is !
shown on the figure. Determine the flowrate through the bend I i=.======::::'1
"~l- FA, =1440 Ib
Water
and the y direction anchoring force, FA" required to hold the Area =0.2 ft 2
bend in pJace. The flow is not frictionless.
I. "FA,=?
Area =O. 1 ft 2 " ~
(2.)'if:;(
/~
• FIGURE P5.3tj-
A uh1 .fl..D/ voluMe!. -f1.taf 'O,,~iJ1S -H,e. 1oe.V1d OlH& -tkt wakr wi+k;Y
the loeHd be:fwee.n "S«'tlOl'1oS (I) a~d (2.) as s/"OI.JI1 J~ -H1~ 51<.e.1-'/'"
aho~ ''S uS'-ea. Applic~no~ 01- file. x - direc+ion <J:)MpOyf.~t-
t> f -HICo IirJt.<1r ,.t'0i) en tu,.., eti.. tAa:n011 uieIds 0
-0 J ~Qge
v. cos CfSo f) a. -=. DA - F' + J1A cos '1-5 0
I/o I rI I Ax. ,1i) 1.- u, pQ
w',H"
~, - ~ al-1d V : Q
- -A, a.
Az.
e.~. J loe'b~(.s
1- 0
~-a.r Qrc:OsLfS
AI A
.. ~ A - FI I
AxL
or +Cy po.r+ Co.)
Q-::
5-31.f

ft,v ~r+ (h) We ltS't. 1{e. y'" dtre.c+;oll C!.O'M.poYlf!"1f of fhe
, irl eo. y IA-0 MtLl1 fuWI ~ IA.I:f.-He", -ft, qe+
• C>
R. Sin
•
fQ.J= ~
~ ~/n q., f ~ - '-+5
-AY A,.
()f
'1..
Q
os I'VI ifS
0
t= :.
fAY
A2.
a.~a 1.
(? +t') . • (I.9'1 S~)~ Ib.~'"
)" 'l'i Ib
."I ~ oS I '" 'i-S"
F -- $I ""'1. -FtAy (O,Ol f+") --

Thrust, vector control is a new technique that can be
u~ed to greatly :mprove t,he maneuverability of military fighter
aJrc:aft. It consIsts of USIng a set of vanes in the exit of a jet
engIne ~o deflect, the, exhaust gases as shown in Fig, PS,3S, (a)
DetermIne the pI,tchIng moment (the moment tending to rotate
the nose of t?~ aJrc,raf: up) about the aircraft's mass center (cg)
for the condItIOns IndIcated in the figure, (b) By how much is
the thrus~ (force along the centerline of the aircraft) reduced for
:he case IndIcated compared to normal flight when the exhaust
IS parallel to the centerline?
Vane
- -- i" - -Pj';= 0 - -
!lim = !liout = 16 slugs/s
-_.....
ro.r petrI (0,) WL CA-ff'J the.. (p/'Yfp()nenf of file J')?()hlt!nf-of-tJ1(JMen~'"
e7~1/'tIYJ fl,af ;'r pe."pend,'udar -/tJ the plane of i1Je ske}~h of
fhe alrcraff -k the. c()nfe.",ff 01 fhe. cvn-!n,/ Vo/",~e .show,? -k 9f!i
i V >/;' () m - j . V. h,. = pifc~ ),14 J'))otne"f
olAf ()c.. f O/A f In '''I II? ./
(2 0 oFf) (/~OO fj) ~/;' 8'(i6 Jfr.£j _1,.. p(){) fJJf'fr) "rrd.""j ",•..e~f
/ SILl,. ff-
/1,. $2..
I sluy.lt
/J. s~
bh IStJ() _ 1000 1. fl. /b -:: pilej'J'y ~P1~hl
. I"
~ p~r(b) ~ apply !he /U)Y/3()YI/R1 Cdn7{'(jY)enf t)/ /he I/YJeoY'
yYJ 0 'I'YIeYl Iu ~ Ilg1A.t:c. f/vYl -ft; ~ (/m ~Ir t)f htL U/n ~/ V()/U hi ~ Iv 9d
.fa
T1wus+
(/~(}O if)(UH/-W '1(6 1l-J
)
=
5-36
/ ~//,(i' If
16 . .5'

5,,36 The thrust developed to propel the jet ski shown in
Video V9.7 and Fig. PS:3G is a result of water pumped
through the vehicle and exiting as a high-speed water jet.
For the conditions shown in the figure, what flowrate is
needed to produce a 300 lb thrust? Assume the inlet and out-
let jets of water are free jets.
For the cordrol volume ii1d/cafed.
the x- componenf of ine fJ10menlllm
erlJ~h()n
Supg'n dlJ ::: ~ Fx becomes
cs
• FIG U REP 5 .'36
(~ Co.s 3 011 ) f (- lit )III of Vz P(+V2.L42 ::: Rx
where We have asslJtned ihai I::0 on the ell/ire controls/Jrl4ce
and fhai -fhe ext'l/fJ9 wafer jet is hori~(Jnlal.
W/fh m::: eli,~ == fA,. V2 £1. (I) beaJlll6S
Rx -; m(V2 - ~ cos e) =- f Vt AI (v" - 1 cos.30")
A/so) A, ~{ ; A2. ~ so fhal
I. AI ~ • 7- S in." ,I I
V
z s= - 112. = v, = 2.60 v,
*(3,5 in.)2-
By comb,'nin9 frs. (/) 4nd tJ.) ;.
Rx ~ pV/'A, (2,60-cos30·)
or Jfz.
:::r .Joo,!}; l =22 7 11
~ l(l.9¥ ":~)(iJrl')(2.60-co.s30')J '.s
Thvs)
Q :::AI V, =: ~/:~ft~)(:;'2.7 ¥);: 3, 9'f-¥
5-37
V2.
(1.)
(I)
(2)

5.:37
5.37 Water is sprayed radially outward over
1800
as indicated in Fig. PS.3 The jet sheet is
in the horizontal plane. If the jet velocity at the
nozzle exit iszo ft ls, determine the direction and
magnitude of the resultant horizontal anchoring H' _"
force required to hold the nozzle in place.
FIGURE P5.37
The Lo~1
(I) a..rI (Z)
y dlie elion
volu,.,e includes ik ""'3/e anti waler i:Jdwren secfionJ
<lJ /hd;""kd I;' ike sk.ehJ., abcve. Ap,I;c..lio" d' fl.e
eo""'pMtnf o.c lite j;neay 1'>10""'''1..,., ~a6'"" ·iel"s
or F
A,y
()
F :: 0
~,y
ApI'IiCAnM ()f fhe
eg uo.lio'1 le.df n,
or
and
'1-3 16
F
A, x

5.3~
5.3B A circular plate having a diameter of 300
mm is held perpendicular to an axisymmetric hor-
izontal jet of air having a velocity of 40 mls and
a diameter of 80 mm as shown in Fig. psS8. A
hole at the center of the plate results in a dis-
charge jet of air having a velocity of 40 mls and
a diameter of 20 mm. Determine the horizontal
component of force required to hold the plate
stationary.
FIGURE PS.3E
-t--( ~ Plate
I
';;+""':"":;::L.:!..JJ ~ 40 m/s
The U)/J!Yo1 lIoluYHe Cl)nfa/n 5 fhe p/ak l./JIui {low,;' cor a.> Inri/caler!
Ih -the f/<.efdt ab"ve - Appliudion of I-h-e. hOfljtJJI1fa/ t.JY X
d,;ecJio/l1 cOMpoYlenf of -fi1~ //Y/ear mOMen-lum ~ua.f;~ yIelds
- u, f u A + IA f I.L AI I l 2 . 2 -
or

5. ?=I A sheet of water of unjfonn thickness (h = om m)
fl ows from the device shown in Fig. p5.39. The waler enters
venicall)' through the inlet pipe and exits horizontally with a
speed that varies linearly from 0 to iO m/s along the 0.2 m F
length of the slit. Detennine the y component ofanchoring force AY
necessary to hold this device stationary. ~
;~~--,~
I - - ...
I
,
Q
l
--- - ... ",,""
• FIGURE P53~
A ",~t",1 volune. tMt ",n+"'i"s he. bOl( poyf,,,... ot l1,c. dev;,e
G<Vd ttoe. wtl1e'l' ,'" -1,e bo>< o;s ,hown In -Ine ~ke.itJ" above. 'I' u,~ ,
ArpliC<l.fi." of ttoe y_dl~+io" tOWlpone"f of "'... Ii_y "'OIIe.nJu""
e'b.....""" ~ielJ.~ 0 'l.
-' .
F = I 1) ~V , r. dll ~ D ,,' hd ...
Av fI '
5t;+ 0
T~<. var'a.+:o" of v- ..;#, x .. li"'e<t. "y
o.t.
f (<;O)() ilx =
o
or
FAY :(q~~~y-so
-
s-~o

5.40 The results of a wind tunnel test to de-
termine the drag on a body (see Fig. P5.40) are
summarized below. The upstream [section (l)J
velocity isuniform at 100 ft /s. The static pressures
are given byPI = PI = 14.7 psia. The downstream
velocity distribution which is symmetrical about
the centerline is given by
change in the direction normal to the paper. Cal-
culate the drag force (reaction force in x direc-
tion) exerted on the air by the body per unit
length normal to the plane of the sketch .
Vz - 100 IUs
u ~ 100 - 30 ( 1 -if) lyl s3 ft
u ~ 100 l yl> 3ft 3 It I
Section 11 - - 1:'-=::where u is the velocity in ft lsand y is the distance - ~ - - - - - i-
on either side of the centerline in feet (see Fig. LutrlrDJ..til..",. C::3PS.42). A ssume that the body shape does not FIGURE PS.40 Section (2 ~
The (017 &.1 volume. conm/;'i"j Plr Mfy PS Shown In fAe ry"ye if lIfed ,
AppliCAfion d f/,. X dlftcl-;O" componenf of Ihe linea, ""me"fuM
e.<j"",fion iead! * 3 ff
-V;r1!,A,+ ZfufUdy
o 3 II
2.
-Rx
= (I)ZfI[100 - 30(1- f)] dy
To de k.,.mlne. fh e di;1tJ",u h ~ fhe c.#hscrvlAh'o,. 01 .-.'lt1H l!flA.tA.h;"'" is
!Applied ,,,!weel) sed/onf (I) a"d(z)-as loll"",
ph -u;
Thuf
h
3ft
= zJfJ u. d'j
()
3fi
; ~ [fOO - 30(1- })7 "1
, 0
". , fI')h = ( z ) {2 55 :;-
or
5.1 fl
(100 !!)(I il)
Then fr.,., lj' J .
R"r.m731J1"JJ )(too fJ){5_1f+ill/, )(iff )
x -II'/ (-'1"3- if
s'
- l (o,002JfJ/"!,) ( 7Z1, if00 [!') Ii I. )
ffj ) sl. { ~;I.f.1' (!
.'17, 1 Ib P"'r ff of Je~ nor"'41 fo fhc pl.ne of Ihe skeIGh
s- 'II

So 4/
5.41 The hydraulic dredge shown in Fig. P5.41 is used to
dredge sand from a river bottom. Esti.mate the thrust needed
from the propeller to hold the boat statlOnary. Assume the spe-
cific gravity of the sand/water mixture is SG = 1.2.
• FIGURE PS.41
lJ.rilJj ~ (,1Ih~1 vptut11e fh4'vl>? by ~ hi?;~ I,'v,e In
/t..L- s/::.elct, a6;ve W~ lASe fht., hoy1foJ?W oY X CllwtftIYJenf
of ~ IlfJea-r /YlrM1enft..vm e~lA.P.hin" ~ gel-
"2
~ =~/J V V ::: f. (sq) 7I5!2 V V CO.J .fa fJ
.2. 2 .2. :2.X Hz 0 2. l.f .2. 2-
whe!re ~eefi~ / is' where IlfIW en/t,yf Ik C-tM-h-s/ vp/ume
verh·ca.llj tA.i?d secf/(/)1 2 iJ v.;),eye flctw J~ve.s- k ~I
v"kn,e. aT h'1 t:l/}Jle of .?() 0 ~ ~ hor-'Jtr;lhl oIiYec-h~.
N~1e That Me'l'~ is" no fu·y-t]tI>?kl d.io/'eC/h~ //Y1e~
#thne;,1u~ Flow a1 5'€c-fiQn I .
F ::: 1/.9'1- s/uffl(J,l1) 7T (.2. Iffro f!)(so :~) tlJs70o/3..)>< " fi$ / f - . s ( f/-: $lw!J
F.. = {6SIJ /6 s2-
){

5.42. Water flows vertically upward in a cir-
cular cross section pipe as shown in Fig. P5.4Z.
At section (1), the velocity profile over the cross
section area is uniform. At section (2), the ve-
locity profile is
V = We (R ; ryn k
where V = local velocity vector, We = centerline
velocity in the axial direction, R = pipe radius,
and r = radius from pipe axis. Develop an expres-
sion for the fluid pressure drop that occurs be-
tween sections (1) and (2).
Section (2)
h
Section (1)
FIGURE PS.4 Z.
The anCl!ysh' for fhls pmb/em IS fin;i/ar 7z, Ihe one of Exa~e. 5:/?
The (onfml volUMe confaln5 fhe fiuid only belw~eYJ -sechorJs(/)tJnd{z.)
as Ind/cllIter/ in the .ske·h;h. Apf//cafion ~ fhe verI/col Or ~
cOl'Ylf"nenf of fhe jt'neor mohfen~m qua/itJYI leads ~
R
+ j{JJ.,f W
z
21"1 rdr ::: p A - If 1- r.. A - W.. " r 2.2.. w
o R .l.. Z.
P,- P. = ~~ _pw,-" + f.~1f[{we ('~~I)]rdr -t f" (I)
The we;~/d of fhe I/IKJ/w I"" -f-he confYDI volu,n,e may 6e exp,/(;ffed as
w~ =9fAh
The value of w£ may be obfa,y,yJ frOi'VI fhL col'lseyvalion of ma.S'.f
e~tJ4.fJoY1 (,(s -/O/IOVJ5 l
I? 7
f~ AI ::: JftU~ (,<?~) 2rrrdr
or
(2)
10 evalu.afe
t>( = R-r
-R.
fhey,
dt>( = -dr
R.. (If )

s,lf2. I(con'f)
and ~ -!. 0 1.. 1
~ ('7/;"7ydr -= - / C{ 7(1_1><) R riC<'
C.omb/n,'.", ~s. 2. a",«1 5" we () 1::,.1-.../"
u.J = €.E- tvc "19 J
Thu.5 frlJWJ &J. /
R 2
~-P.. ;::;:~-fWI+
2. lTRl.
fo ev()/t,ux/-e 1-he.
Thus
R 2
[(;;r/rd~
o
and ~. 6 beCIJmes
or-
'19 2.-
= - R
1'1'-/
(6)
/Vofe fhaf In [tJntrAsf Iv -fhe restllf- Of Exttl?'lp/e s:13, only
a very $"mtl// porf/on of' -the pyessuYf! drtJj7 if clue fo
a c,ht11?ge t"n fh~ mtJmen-hon //ow befWeeh seci/onf /

£#3 I
5.43 In a laminar pipe flow that is fully de-
veloped, the axial velocity profile is parabolic,
that is,
average velocity, u, with the axial direction mo-
mentum flowrate calculated with the nonuniform
velocity distribution taken into account.
as is illustrated in Fig. P5.43. Compare the axial
direction momentum flowrate calculated with the
I ,
,,
j
I
!,...........
"I( 'u ......I ,
I

~uc
~~
.~
JI
-FIGURE PS.43
The axial d,;ecfitJl1 fr/orwen-furn IIowrafe. btlsed OJI/ a un/101m velocity
pyolde wifh u =: U /s
M~ :::: ufu A = /Jrl"1'rR1.
x Uf/ifOYIJ1 {
J
1he axial direcfion ff/()l1teJ1TUh1 rlow~afe ba!Jed On the fjOfJ-
un; fOYM fltlra btJl/e-.t<. ve locify pr~fi/e if I 2-
M~nOI1- =jurUZ'1l hi,- " (1 u~271' R' 0-r;)J(I)ff)
J IJl7ifr;rm 0 ~L
MFx nol'} -
J (Jl)ifaYn?
U MAti u ~ vve use -fhe
tl5 fo/I~ws
To ob-fal"n a ye/~·lionfhip befween
con serva/-l'on of mass e~lt.ahdYt
fU 1TR' ~ f hR' Uc; [rt-({Jj(;) d(:)
-U -- lAc,
-Z
if
-3
Mi=;( nOn-
J
U"ifm",
--
5-'1-5

~ 5.44 For the pipe (6-in.-inside diameter) air
flow data of Problem 5./8, calculate the rate of
flow of axial direction momentum. How large
would the error be if the average axial velocity
were used to calculate axial direction momentum
flow?
F.) :. tJ. 002'31' >(1.(9
I H:1
Vi::: /OCA./ ve/()ci/y I ~f (~ Table 61' /'r1Jb!eu,., s: /6 )
r ::. local radius; in. ( .fr~ Table. of I'mb~ s./B)
I? ': pipe YtJdiu.,JJ "3 in.
J~.- /·....1""_,,"'/ r~J~ 1-r -,,/ e / 1_.-1 • / . J. Ltn<- ""-"1'''' VI CIt, 1$ va u.tA.f'Ct;If by nbtl11eY,CI( l¥lfCfY~n~
~
f-/5!ny ft,c. frate.fdic/Ift/ you/e WI·!t-t 4negl1a.) ,;'/e-yVIJ.!;.7}tG
Lt/nI.(J1A. k¥ PYlJfYd-Wl lA5ed. fw ftJ/s )7v.y,o/f.. /S 1/J1e'/ 6n /'he,
//lex f pt1Jje. The ve.$ L( If of »'1 I,.v",e,y;'c.czJ t'J1leqvahon aJ1d s~11A h~J7 of
Etj. / is
MF -:: 0.29'1
x
MFx
wl-te>/e
v =
( MF - MF)x )( /f7p
/VI~

conti: )
100 CLS
110 PE I NT ":t:::f.*;t:,.t* *****:~t:**;,j *.:j*::t~.* **-*:******i:t J>I':,tt.*:.;;f*t::t :tJt* ::1'::t. *,.
120 PE I KT .. i.* This program computes the axial-direct ion ;j;;'."
1::".0 PR I NT "** momentum f lovr rate for problem 5. 44; usi ng :i';:>i:"
140 FRnn tI.t:* the trapezoidal rule appl ied to unequal **"
150 PRINT "** intervals. til!
160 FR I NT "t:i:*::t::t*******::t;j.*****:t:*********i.*;j~;;j.**********:l:>::**"
170 PRINT
180
190
DIM U (9), R (19)
,
200 'Initialize the variables
210 1{ == 19
220 RHO = .00238
2:~,0 PI = 4! ): ATN (1! )
240 FOR I = 1 TO N
250 READ R<U, UU)
2~'.')O P'I:i = R<I) / 121
270 NEXT 1
IJft..T A
T,}" 'T /
..,.../M.. 1. ..t;.
0.0,
t-'. C
"./ . ./ ~
-) ..'
-<-,0,
::".4,
30.00,
2e..70,
26,90,
2:3.2-4,
250 FOR I = 2 TO N
~:: '! (; ~~E)~'T 1
:. ,:~. (:~'.:::: X. = RHO ,; ::;.
,- ,
0.2,
1.0,
'1 Q
...L.,U,
2. e.j
,
2fJ. 71,
2e.31,
26.32,
22. ::.0,
0.4,
1 ':)
-l.. '-'" t
'2. (),
s~ p,,--,. '- ,
.3.0,
29.39,
27.69,
25.64,
00.00
0.6,
1. 4,
'11~~tJt~t*~****1***tt**~~*******~*:t:****~
~* Thi~ program computes the axial-direction **
~* momentum flow rate for problem 5.44 using **
tt the trapezoidal rule applied to unequal **
*~*****************************************t***
The axial-direction momentum flow
rate is 0.284 slug-ft/s2
5'-If?
27.42
24.. =,4
16.71

Consider unsteady flow in the constant diameter, hor-
iZontal pipe shown in Fig. P5.45. The velocity is uniform
thro!:ghout the entire pipe, but it is a function of time: V =
u(t) i. Use the x component of the unsteady momentum equation
to determine the pressure difference PI - P2' Discuss how this
result is related to F, = mat"
II (I)
," .: I
I ---------------tDI
I I
: p =density : I
-- -------(1) (2)
--x
U5/11} the C-tJn-/rD1 VIJ/umf(. shpwYl. I~ fhe 5kel-ch and etffJ/y,'IJ fhe
x- U)t?1(J()t'J eYlf of /he unstead) //YletfY' h/()n1eY)..J-.q/'Yl 8fl,fetl)~ -fo; !1Je
Contentr of ft;/ s c.v fN~ gel-
or
~ JflA.dil f f rIA ~. ;:.d1/ - 2 rx
")-t CV (J
d (f~ 7r pl., )
df: 'f
"fA 2.Ar I I :::. I' u7.2./1" CI~~t.un;'':} ~ = tt, ().t -eI/f/Yy
{Y/c/,tmlesJ f;fJW
5-48

s.1./-6
SAt) The propeller on a swamp boat produces a jet of
It;r having a diameter of 3 ft as illustrated in Fig. P5.46. The
ambient air temperature is 80 of, and the axial velocity of the
flow is 85 ftls relative to the boat. What propulsive forces are
produced by the propeller when the boat is stationary and when
the boat moves forward with a constant velocity of 20 ftls?
3 It diameter
.. FIGURE PS.46
F;.., ~ sla..J.IonaY!1 btJa-t +he hoyi]ontlA J Uwtl'~ tJ/ fhe h~e(),Y
y11trme;,fu ...... e~/,4aJ"/m ().PI'/led -It; /tu l6?~ ,,{ IItJ cmM VD/Un1e
S htJWn Ii-; ~ ~":..ei'(,h A,J,cve yields
FJ ::: w, ~ - VI)::.bA V V 51nce V« V+"'YUS-t- <. V2 r 2 ~ :z. I 2
.Iff, 2. / /1, V. in~ ) '2 ~
F = /LA .I'):i~ (I'I.7rn2..1/lf¥.f~...J 'TT'{JfI)('I5.f-1
fj,YUf-i- ( R-r 'I- 15-3.3 -fl. I: )(5'1/)'7<) If /j~.). ibpt.{f)
.. /~. 'R (- ,J,.S:l./
F =//7 Ii;
+),.,.uff
ft?Y fhe lo()tAf Wf()vi, fd.,tNf,VYr/. J/V}'rt.. ~ ffuJ 0/ 2tJ If,{; -the
.5~e. (.~ftb-{ Id/u~e fhfIWn )n the S-kelc.J/ aINJV~ is u.red)
h~ IttL ~/a,li~ V(,ltrt,:,'1 W is h~ /;"fJI¥~'
~ ~ hIJV/]dvrk/ ~ ()/ fj,~ /J~eA¥ Yh(fWJe;,~
e~ ~ gel-
F. :: 1'1/W - w) := ~II W (w -w) :: ..E-7?-~:lV{(~-W)
-IhYIH..}- ( I 2. , I ~ 2 .'1 I R r <.j
Fw tIu- yn()v;~ h~ ~ ~ gS- ft tw,,( V,j =20ft anti .>0
F = 8,/.2 /6
';hY'Laf -

5". if 71
A free jet of fluid strikes a wedge as shown in Fig.
P5.47. Of the total flow, a portion is deflected 30°; the remainder
is not deflected. The horizontal and vertical components of force
needed to hold the wedge stationary are FH and Fv, respectively.
Gravity is negligible, and the fluid speed remains constant. De-
termine the force ratio. FHIFv'
The h()rjt1nfa / QYJd. ve-y-h'Ul/ (tJn..,fJOneYJh of /he lInea ~
m 0 men fu ffJ ~M 17(;n cut OfP//ed f1J -I'hi. [ffn /evrh of ft,e
CPYJ~I vO/l.Il'Yle ~hdwn fu fief
- V, f' V, Aj +- ~ f ~ A:J. t- ~ CbS ?O~ ~ AJ : - ~ (/)
- ~ f/~:?O·~~ AJ
~ ~ (2)
)-mweve¥"' v -= v - II: .... V SeJ eo>. (l)a n d{2) beUfrYJe/ 2.. - I P
V.2.".o (Ill -t A3 &Of. ~(). - II,) -- - Fit
V It' A3 flY) 30 tJ ~ -- FI/
~ _ A2. +Iq t()$" $0 "-11,
Fv AJ $ /;" 3 () 0
~ C(}YIS( Y'VCt h~ of yY1 cus we.. ge f
G(I ::;. ~.2. t- blJ
or
anPl
AJ:::' Az. r /lJ
~b i:"/~ €,f . (?) aAt-d (tf) twt-- qe f
.4, -I- /;7 CQ.J JIJ 0 - Ill.. - //1
113 >,1;' JO i)
('3)
fJ 3 {COS?OO -I _
-~-------
, 0
A »,-, 3IJ
7
0.27
--
The ne'jCihve 5/tjY1 Indht:l/el' ~f ~ if ~wn r-~~ /Mn
ul CiS 5"how}? I;' the .Jl:..eM~
5-50

S.lJ8 Water flows from a two-dimensional open channel and
is diverted by an inclined plate as illustrated in Fig. P5. When
the velocity at section (1) is 10 ft/s, wh~t horizontal force (per
unit width) is required to hold the plate in position? At section
(1) the pressure distribution is hydrostatic, and the fluid acts as
a free jet at section (2). Neglect friction.
2.13 1b
--
5- 51
• FIG U REP 5 .'f~
.......""'"l.Oft

5.5'0
5.50 A vertical. circular cross section jet of
air strikes a conical deflector as indicated in Fig.
P5.50. A vertical anchoring force of 0.1 N is re-
quired to hold the deflector in the place. Deter-
mine the mass (kg) of the deflector. The mag-
nitude of velocity of the air remains constant.
v = 30 m/s
conhol
'Jolu;Ye
Sec·tiM (I )
FlGURE PS.50
To deteYmlne. fh~ Y>'fas.5 of fhe UJn/c,(,1/ de.fiecJ'()'r we use fJ,e.
SIP.lionaY"y / )1dl"} - d~fuy»1/~ c.unfnJ/ VtJ/Unl~ ~hdWJ1. In l1te sl:.e kJ"
fA bdlle . /tp,P//CIIl h~n of fite v~h'clA / a'/yec ndn C/)}IJI?pr;nenf of
the /1';-,e.tJ. yo IntJl'H en ~"'" t!jtlal/o¥! ( c1· 5'. 2.2) fp -/he ChlIeHIr (J f
Ih/$ LfI)1/n;/ VtJ/Ul"He y/e/JJ
Ii! (- v: of t-;. UJS 30 0) ::: - 'A - W
Cone
Or
Wc~nt' - ~l1ne J - M (~- ~ UJf f()~ - ; :;: f~~(/{-J{CPJJO~-~ (;)
Hd/N"€,vev-
l!;' ::
~
tint:{
).
A; :: 711),
If
Thll) E~. / Ct2J1t be O<Iyc5s~d as
m -= f frO,"). I{ (~- ~ CbS 10 D) - ~
t.tJne.. fj' .9
CJy
frJ ':= f2? A-J)ClJne. /'I'fJ
ancJ
IYI ::- 1).108 *1
CMe
5-52

5:51 I
Water flows from a large tank into a dish as shown in
Fig. PS.SI. (a) If at the instant shown the tank and the water in
it weigh WI Ib, what is the tension, TI , in the cable supporting
the tank? (b) If at the instant shown the dish and the water in
it weigh W2 Ib, what is the force, F2 , needed to support the dish?
Pifr ~ ~) we ~j?I'I !he Y~rh'c.a" J
c~mptJneYlt of the linef/ Y
m(JI'l'1eh htlM ~/Aa~~ Iv /he lWJ1en1J
of tdnfrtrf vr;/tlme A) C~;J -Iv Je-I
(j)
Td qef /,¥flue 01 ~ill /Ne Of!",!
!,exnfJu/Jj f erJ.tCk~~ -h; ~ fttMI ff(/WJ .'f'N. (yee StA-YhttL ~! ltv.
IN&< -kv J;' /hL -1?tY11<, -k ~ -Ia~J, ewi Ief fr; gef
Vu",! 0- Vz9"A = VCz)(J Z.2- 'Ii)O() fl.) - ZS· f f!
Then ~ i,.(1) we. gef 2
- (Z' 't f:)(''1,{ i/fJ{z., If If) 7T/ / it) =T, - W,
/ ~/u). /-1
16 . .5'"
~n-fr::> M /;"-/0 ~
ve YHca / CVMj'I{/';JC/h./ ()(
pi CV
B
b qef
~- IN'z
C~ cV8
;; ere- f V1';.fv we 1A5-e ~l"no""IJ; 's €-1l-tWt$"/:rn be-Iwee-n free furtee of
1tV~ II') c~k. +v '~e su¥/-e,u 0-1- I#a.k- I;' di.fh Iv ,ej
~;-,~ = Vz. fj (hlJ -t-ho) - Z(?2. z. (i,,){IO 1-1 rl2,-H) -: J 7-6 PCI$ j
~ ,,;,/.,., we L{S' fn,.,... t<M ~.J..,., a! ""'<if5, ';",;'", ~ ';'6..,.ff ~I-!..f
c~ c~ ..fa..,/c..
50 .fr-v.-.. G CO • (2) ~ 7(;f
(
37. (, tf)/J.lj'l ~ YzS".if ~ ) 1Tt·1 ff/(I /~1.J -r;. -W2-5 C' -frJJl" .s ) 'I '>/v9 ..ff
One) F ;:" W + 14· 7 Ib2.. '"l.. _
5"-53

S.52 I
5.52 Air flows into the atmosphere from a nozzle and
strikes a vertical plate as shown in Fig. PS.S2. A horizontal force
of IZ N is required to hold the plate in place. Determine the
reading on the pressure gage. Assume the flow to be incom-
pressible and frictionless.
t
~:~=(oz,).1 I
• -=-.- ..... • 4- 9N
( .(/) I~; ~
V=.==:f====::::7 Area =0.003 m2
, I
{3}7'"
Area = 0.01 m2
• FIG U REP 5.52.
To de~"'M·It'e. +ke. sk+ic:. 'f'Ie. pr~SSLlrE.. at sfz:..t;OY (I) "",f! .f(rs+
Cone:; i dey fi..c. frtC'Hofless (}.Vo. i~C.cmpv~ssi bte. flow of air ~
(1) -tz, (2) . The BernolA I"I e~(.A.a.+10..-. fv.r ~'lS Ho",", "s
.:::f 0 5Q,c1..
£: +- V~ -= r!.~- + _Vl
~ ~ "Z
W'<.. VO~ ,+tw.+ VI aV'(;.l Vt (lV-I{ iYlked b~ ~e ~oV1+iyu"i+y
((.t!VSe.No..1-tDy of ~oss) e~~.a:tioV
Q I ': Q'2., ~r 'VI :: A2.V2-
C.o~Io:V;V~ E,s. l Gtt'O 'l. vI"- okl..,ir"
i -t (~~ V'-)L _
'3.
V'-
-<-
2.
To d.-e~"V)IVe V1... W<. use +i1c ;Y1eav iYoWeyt~~ eolA.A,,"c,,",
~'f" +h~ -fltJN ~ .2) ~(3). FISY -tL-vt C,.6Y-tn1 "oM.me.
s~ft.-V~J o..b:>ve +he l;~e.ay )"Yot"'ewtullV' py'Y'lc.·I~Ie. ':jeds
or
( 12. N
57 F"
-'5
((!on' + )

5:52. 1 (um'+)
NdW, w;#-1 9· 3 'l.
P. " l~
S--55

5.SLf I
5.54- Two water jets of equal size and speed strike each
other as shown in Fig. P5.5~. Determine the speed, V, and di-
rection, 0, of the resulting combined jet. Gravity is negligible.
ex ~~~~1'/. '/
,-------" ,/"
V2 =10ftlS--':b Io.lft T /.'"
------~ 1
• FIG U REP 5 . stj
90
0
,11 r
I I
I 0.1 ft ,
,I-
t'Vj=lOft/s
,
For the. c.orrty~l vch.u)e skow~
mOMeyd~M e-~ ~:tjo'" for +he
1l-te. X dl fe.C:f,'oV
lrt -i1t. s«.ef-"h aloove +he. l&.rea.(
x o.Vld y dl~c+tot.s are I for
- V7. ~ V2. A'l. + (V CfJS e ) ~ VA = 0
£1t1ct ~y ~.: y dLr~c:hoV
- VI ~ VI A,-+-(V "SlV'l. e) -eVf :' C
Also .for coV<;erva+1l>v of vYO.ss we. ~le
PIV( AI -t pV1. "'1. - ~V~ =c
F-roFV E~s. I ard 2 We.. ~e+
:l-
V'l. ~I.. =' CJ) s e = O)t- ~
J~ AI ~ e 1.
~o e -= c.O(1 V:A-z, = cot_I~(10 !;)1r (~t~ -=
.1. II. '10 '"
'41 n I ( 0 f~) 1t' (~l.ft)
'i
N bWI U) M ~ r in~ E~s. I). C1.tO '3 INt q€.+
- V~ A, 1'" V osin e (V I AI +"2 AIJ -= 0
V ::
V .::
v=
5'-56
v

S. 55
5. S£ Assuming frictionless, incompressible,
one-dimensional flow of water through the hor-
izontal tee connection sketched in Fig. P5. s5 ,
estimate values of the x and y components of the
force exerted by the tee Jon the water. Each pipe
has an inside diameter of 1 m.
z
V1 = 6 m/s
Pl = 200 kPa
FIGURE P5.55
Section (2)
We COJl'l U$e fhe x and y CCIYVL(ltJnel1'/-r of Ik I/neay YY1()YY1eYl-ft"YVI
eCjtAa-fitTYI (&J. 5. 22) To defen-'Yfine fhe x and Y ct:JYVl{J()YJenls ()f
-fhe. 'feacf7'oYl foyce exeyfed. by -!-he wa.fer Oh -the.. fee. Fw
-fhe.. cor.+rol vo tUYJ1e coYtfA/~jn.J wafey I~ the, fee, E$' f.22 leAds
(2)
1/ ::: ll.1
V3 -
1T!J.~
Li'
(con 'I: )
$'-57

S,5h
5. Water is added to the tank shown in Fig. P5. 56 through
a vertical pipe to maintain a constant (water) level. The tank is
placed on a horizontal plane which has a frictionless surface.
Determine the horizontal force, F, required to hold the tank
stationary. Neglect all losses.
i
l' ~ _C0!lst~nt. --
'-+)C , water level I
1 -:-:~--~-:~i- -~--~~:;::;:~-';:-
Jet area (2.) ..! '-.+1.m '. -
=1250 mm2
',lm
Jet area
F~
..i =625 mm2
Frictionless I
surface i !
• FIG U REP 5 . 5.6
of -the ljreAr WOWlCY-tuM -e~(AClfi ov)
vo luM(. Sk.tt.~J dloove vJ~ ~e..f
,
A?py'It"~ -the. x- dlV~,tioil' c.oMPOVe.I+
to the. U>~~ws 0+ +""- (.O~~l
V,~VI~'- V2..~Y'l-A"l. -:: F (I)
,
USlh'3 8errOIAl','~ ~uo.fjon ~ d.~~c.r~1oe.. -the .ffidio~l-ess -flow
frbM -tltC'- c.o'~-hwt WGkY" SU1~'E:. level +0 +nCo .flow l~o..Vl~~ o.t
s.m,1io~cz, (l) and (z) VJe... alo~in
V?. "; [z~ h"-
<l.vtcl
I/I=~
~1o;vll'l'3 ~5. l) 1. dVld. '3 we qet
F = ~C3 ~, fA. - L~h2.~ A2.
or
F = D ~
(2.)
/'

5.57 I
Water flows steadily into and out of a tank that sits on
frictionless wheels as shown in Fig. PS.S7. Determine the di-
ameter D so that the tank remains motionless if F = O.
I .........
1.L-_ _.,....llL.
(/f'0
I
1
,
I
'"I
I F
~
AffJ/Y1nJ fhe hfFfljQVlIa/ l{fVl1p(}ntVlf
pI f1tl I/Ylea4 Wlornen..f.um -e,u&&-I/f/h
fn -hu. [tin ~f.s 0 f !Itt. (gy,.fn; /
'~'~;..:.o..~";;;""'~3-)
(ltD ---if
Vd tUY'YJe s h(JWYI I;' the we qef:
f VtV.;dll :
cs
Vd~= V pI---I-VdL.
I 2. J
A'}t:un/ fl»c.e. Vz. =~ -= VlJh we gef
V;cl7.. = ~p7,..f ~ ti1.
~o/c.IVJj ~f ~$. (I) ~ (z)1-ork w<. ~kvk.
If V3 L.. V; eS' (/) CfAYlJI'Io-f I:x 5iJ.HrteJ
e~.~) CtU-t be 5df,j.fl~
If e8 .(I ) '-t4t.- I..t sall.rh'ecl
(2)
e~. (J.) CaJIIYl,f be $t$--H.r~e,j
et. (j ) ~vo. k ,$a. ~!he,{ w:/t. j) -;; 0
~. (7-) C. (:( 1'1 ~ ~-hj/je"l J/l/'tt. p:: 0
lIs-;-~ ~[)=O
L?. _ h wt..u..!' f k.L.Je f Jf) fhal
~ V'S ~ "'"
Vj -; vzgh ::: l-j
5'-60

of each is atmospheric. and the flow is incompressible. The
. . . ' " contents of each device is not known. When released. which
I h
TIhe four de~lces shown In .Flg. P5.5~ res.t on fnlctlOn. devices will move to the right and which to the left? Explain.
ess w ee s, are restncted to move In the x directIOn on y anc
are initially held stationary. The pressure at the inlets and outlets (I) ,-- - - ~ ~I
~
We (),pply jt.e. horijoYlfal
c()mPOYlenf. of f1.,e II~ e~ ~
Y>1 Om e,.., /urn eg/,f(.f IltJn f, /he
("OJ1kj1f$ ~f fl,e. CdYlh-;;/ I/IJluYlfe
(bfl);:'en liYles) aY'ld delel"n1J;'e the.
5en$e of fll e tll'l cJ" ~;t1J ..fo.taL ~.
;If F. Is /n fhe d"~cJ-i()n.,
Shown I~ the .fke~l,es, nulion
will be. f.o {he /e fl. .If ~ if
I
I
(Z),~~.__::=;:_-=::J'~~L-'
(oj
(cJ
, I
. -'
(eI)
/n ~ d{Y€.cf/(Jn ()flPIJS; Ie fr; {-Jaaf
riJhf . .J..F FA:O.; there- IS 1'10
Fr,-y- ske tel, (a.)
shown.! the. Y??o~;'Y) is' n fhe
)'(J'rJ jonI-,. / l?1(rhOn.
- v, ~v, AI - ~ ~ ~ Al. = ~
5(ncf:. ~ is fo f1-,e lei f / mIJh'on /J -/tJ fhe. YO;J},!.
PlN s/r:.e kh (b)
-~f'~,A,+ ~f~A2.=:F
().nol {y()n/) Ulnset"vah'trn of YY'Ja>J
fJ V, A, :::"p ~ 4l..
~I'ki )"tnee V '> V rhe" FLI If -/7J 11.e /eff a'JPI 1"I1(J~~",JIoIl.e~;flu..I 2.; n ,/
Fov s~~ ~) (nD~: flow i.r ,;".ft, c va:f (t )
-l1p V, A I = ~
a~d ~ ;.1 h ft,e lelf- f, nro hon /s h lJ,e njAf.
FIN' 5Ke. f-c~ (d)
-v,p~~ + ~ p~,.42. =F;;
and ~ UJYI.JerVtA 170 n of ~a.fS
,.0 V, 4J ~ f'V2, A2.
{And ~ <. "2.
so ~ if -10 fie r/!/'t- lAne! ~Pf.~r'J /$ /r; the Jeff.
5-61

~. 5'1 I
Water discharges into the atmosphere through the dc-
vice shown in Fig. PS.59. Determine the x component of force
al the flange required to hold the device in place. Neglect the
effect of gravity and friction.
To ,,,,,,,/~k fht ;(-d,.-<ch~~
",-,clo.,,'''3 {o". "fU'Ye,) to hold f/,e
device In flac¢ / the x- d/.-edio'1
CO>'>1I'OYI,,,1- of the 1,'.,,,,1" mommlu..,
egl/ali."" is fAJed 0" h.e (ft,k-..iJ
fj,e skt-kJ., +0 .I.fa;n;
-V,,,,VA -VflV4 + 112.2.1. V, cos'~'f' V A = - F. +f. 1/, 3 3 1/ "
To dekTI'M/ne V .1A LJ C J,
J I Jn ~ IJn5€rvAT7 • ..,
7-D ./,fa,';., :
Q.,=>;<" T Q,}
~ II, =Ii A, f >J AJ
30 fl~,~______
(20 ~f;:O. 8r~') ~f(J~t)(f/'ff'f')t- v,. (b.3h')
f.
V3
=S ff
S
fY7f'r-> E3 ' / we ,e!-
_ ('20 V)(I.U it}(2O¥j{d.Zff')
F -=).'1'+0 Ib
A
1>-62
,,,
(i )

S.60
5. 60 A vertical jet of water leaves a nozzle at
a speed of 10 mls and a diameter of 20 mm. It
suspends a plate having a mass of 1.S kg as in-
dicated in Fig. PS,bO . What is the vertical dis-
tance h?
FIGURE P5.60·
To de/eYYJlllne. fhe ve~H(.eI1 cI/sfttnce h we apply fk verhl.al tilfecho""!
cowtpbY1enf tJl fhe. II~eIlY I"n()YHehfu".,., e'6u().h·~h (61' >.2Z) fD ft,e
wafe r I;" fhe con-fn,/ voluY11t2- shown 1i1 fhe .s-kefct a.bove.. Thlls,
- R -;()O -if - - V A V J. 2-
~ r .7 wale,.. - I f I J == -;0 l{ ~~
tt
7Ae verfico,/ reaciion force of lhe ;/ole on Jhe kI~/er is
efu,o./ in 1n49nilude 10 lite l,fKiIighl of Ik pJII.le) ()r
RJ :::" 9 !11p/a le = ('1. $1 !f) ( I. 5 ~J:: I if. 7 N
(J)
A/050./ !he. wBiyh f of M£ wQ.ler willl/n I/J£ cfl)1IYol volt/meJ fJ.9 ~41e.r J
;.s 17~1ligiiJle I and fAe /?It.Z.fS flow role is
m=:jJ~ l{ == ;; Ao Yo -::( 999 ~ )f (/J. o:J. hi)'- ( 10 ;) ;:: 3, /3 l!f
Thull Elf / becames
- /~. 7 II ::: - ~ m or v,:: 11f-.7 N - If.70 tp
, 3,/3.1t1s-
From lite Berl1lJtI// £tjpg!itJn (Er;.3.7) Wi! h9ve
~ + i.fl v,,:l+ J1zo :::: -f1 .J. 1:(/ 'v; ~ 0'~ I wAe.r8 /'0 -::It :.0
20 1# 0 J 2, :::,.1
T/;I/~
1fU2~ ifJl1~ h
or .Ji/lce l' ~~?
L I (~,. /I1) I (:z.. :z..) m~
If =:21 0 - I =::"2.(q.81!f{J /0 -- JI.. 70 ~ =:: 3. q 7 /J'l

5.6'1 Exhaust (assumed to have the properties of stan-
dard air) leaves the 4-ft diameter chimney shown in Video
VS.3 and Fig. PS." ,."ith a speed of 6 ftls. Because of the
wind, after a few diameters downstream the exhaust flows
in a horizontal direction with the speed of the wind, IS ftls.
Determine the horizontal component of the force that the
blowing wind puts on the exhaust gases.
• FIG U REP 5.61
For -the control volume indic4fed fhe x-Gomponen! ()r the
momenlvm eCI'Jafi~f7
s:u pV·/1 dll = ~ Fx become.s
cs
~ pVz- Ill- === Rx J where Rx ".s the nef hOl'i'l.()IJio./ force
fha1 -fhe winrh pvis 0" -foe exhaust 9a.ses.
ThtJs
RxJ~ m:;. V~ Whel'B m2
::: e!9" 4 == ~ A, V, (i. e. m,:::: '+',. )
or m~::: (0,00238 sJ?-)I1J(lfHtJr611) ::: 0, /79 ~~~
HenCB.!
Rx '" O.1793J?(Jsf}),: 2,6'1 0f.!1:= 2.tQ/iJ
S-6tf
15 Itls
..-

5· 62- I
5. 62. Air discharges from a 2-in.-diameter nozzle and
strikes a curved vane, which is in a vertical plane as shown in
Fig. P5. b2.. A stagnation tube connected to a water V-tube
manometer is located in the free airjet. Determine the horizontal
component of the force that the air jet exerts on the vane. Ne-
glect the weight of the air and all friction.
• FIGURE P5.~~
Note. +hat we 'Iqnoyc. --l,<. ef~+ of o~ost>h(,ri c.. Pf.f'SSUfC On ~h(. lal~~
0+ R ;n OUf SOlu1iOVl below a~ck lASe. ~~e. pye~SLtfe.. Is 'In~c~6 in
)( L ' .
Ex.~mpe. 5, 10 J '#(. ~~Sphc":l to ?y~ ~~uye -rvrc.e. MOJ-j ree.d c()ns I"~r~+ll>""
whe", id.t~ti~iY'l~ rto.L+ioV +orc.e!,. For ~e Qty .f'low·It~ ihY¢~~h+he
C-Ot+rol votu,,",e., ~ ~~ aoove) +he. X- d'f~C-He>'" COMpot'le.Ii of +he.
I',neo.'(' M()mc.r'1u~ ~(AO.,fi on 'I~
- VI P ,vI AI - V'l. eos 30
0~. V2,.Al.. =- R;(.Qlf Qr
IfP/~tioY' 0-( 8eroololll IS ~W)..p.oY 11r t-he. flow fn>M
(,) -tv ('2..) y'lec:is
V2, =VI
Ihen) +ron'l +he. C()",c;erV'a.-i1on of Mo.S~ t>Y'Ylc.iple
A,V, :: AI...V't.
We lAse m t3c.rnotAll', e..rlAAfioY ~dl~ -+0 oJ::,~~1' +he
.fDlloWl·t'~ e,~+lOY fz,y ~ S~~~(7Y -hAbe decele.~~Oh
1
~ + VI :::
~ (lr '2..
For +he Vfo."'o me.kr'"I w.t-h. +h-c. e.IAct1io,""
Pt~ + h CS' -
fY'~t'O WQ..e,y
h 0'. - 1:1Ytt;trg o.lr s+o.~
( eDV)'i )
(I)
(~)
(~)
(s)
( G,)

5·62.. I (COn't)
R ==
x.
~ =x
2.9f, b
--
ThiS is the .f-ov'ce -eKe..rkd. ~~ -hc. vane CY --the. .f.lovl,~ ~~V'.
The. .fu.ru2. -ex.uktl ?~ th.t. ·tlc,w1~'1 <:1 txer+s on 1i1e VCl.n~
~ ~~ ~y ~7 I'-Mo4. b~ Ofposi k iYl d ;~c:nCTY 00 -Ihe rl1h-f)
5"- 66

5.6S
5.6 S A 3-in.-diameter horizontal jet of water
strikes a flat plate as indicated in Fig. P5,65. De-
termine the jet velocity if a lO-lb horizontal force
is required to: (a) hold the plate stationary; (b)
allow the plate to move at a constant speed of 10
ft/s to the right.
(al (bl
FIGURE "5.65
The COI1/-rr;/ volume shoWn In The skefch is used. The
sfah'oYlPYJ plale CAfe if. cPI1J'ldeYea fi;ff. IIpfllicaltan ~I
the htJYlj()nft?1 OY x- dlf'ec.iltJY1 cOYJ/If1~nent of fj,e /;i1eor
tyJ()Me yt /u,yy, e~ua/j(7}/l y /e/dj
Oy
lA,
Thus
()nd f.,(,
plate
the x- dIYec,fioY]
or
and
- -~ XJ
5- 67
tit fP.eed V= 10 ffr I :s /
-the lin eery hI~hZehluh-t erUIl/;OY'j
-== ZOo Z ff
==== J'
/)10 VI;'" platt!

5,66 A Pelton wheel vane directs a horizontal. circular
cross-sectional jet of water symmetrically as indicated in Fig.
PS.66 and Vidtm V5.4. The jet leaves the nozzle with a veloc-
ity of 100 ft/s. Determine the x direction component of an-
choring force required to (a) hold the vane stationary. (b) con-
fine the speed of the vane to a value of 10 ft/s to the right.
The fluid speed magnitude remains constant along the vane .,;
surface. D = 1 m.
(a) (b)
FIGURE P5..6.6
(A) To deferm,nG -fJ,e. )(- dl;ecli~n cpl'H{J4J1ehf of qJ1c.h~Y'I~ ~e !7!fUI'i-etl iP
hlJ/d fhe V4J1{! S"1ei1/onlJry JIVe use file S'lRh'dJ1~'y on"/rrJ/ v~/JtM(!
skwn tU?IJV~ tlJtd th~ x- d,yeCndn ct?m;;~l1ehf of- the //heo.- m,Jlfle"lu"",
e,MA:n#1I1 (£1' Ii· 2"2 ) • l"hu~
~ :: m(tft ~I.IJS 'Is ') :::j7AJ ~ ('1 t~ ~Jljl"1 :: I' ~p/I{~ of ~t;.5'1~'
F. :: Iii Ib
"(6) 70 de!e,YYn,;'e fhe X'- d,;e&h'Oh CIJ"VJ()nel'1f or /lJ1cJ,fH'~ ~e ",(HIed
11J U)wh'ne .fife. Vane.. fr; ()l Chlsnrl1f s/eed of It) fr";" IRe Yl9"'f w-e
life ~ Clhrlrol V()/ume. Ynov/nf/ fo ftt~ Y/1hf WJ'fh t{ $,ud pf /0 ~f"
and me x- d/ved/on comp4J1od ()f tHe. I/;r.u.r MOJll1.tJ1/ztm ejlo"h',,,
IrN- &t f YtlJ1.rlitn·"?j clh1fy,1 vO/~Ylte (£1' $".21). lJut~"
,: ::. ~ A, ltV (vv of" WCIJ$ Y.rD) & p 7r~). W /W t W U.f ¥.fD) {t}
Ii J I l. / (it' ' (' I 1. /
We JI16-k 11tA-1-
/'1-1 'I,
s- 68

'i.h7 I
5.67 How much power is transferred to the moving vane of
Problem 5.66?
P"wer ~ E V where frolf} Prohlem s. 66 Fa -::. II/bibA ) n
PDWer :::
( l/.fb b) ( ID ft")
(
t;S'o ~+.lb
~. hp J
':2 2.6S hp
~- 69

5.68 Water enters a rotating lawn sprinkler
through its base at the steady rate of 16 gal/min
as shown in Fig. P5.68'. The exit cross section area
of each of the two nozzles is 0.04 in.2
and the flow
leaving each nozzle is tangential. The radius from
the axis of rotation to the centerline of each noz-
zle is 8 in. (a) Determine the resisting torque
required to hold the sprinkler head stationary.
(b) Determine the resisting torque associated with
the sprinkler rotating with a constant speed of
500 rev/min. (c) Determine the angular velocity
of the sprinkler if no resisting torque is applied.
Th;s if f/~;I"y -It> GK'amfJle 5: 17.
! r= 8 ij./NOZZleexit .--t;-- area = O.04m.
2
-- -/,.,..----- ---/ /-- I .......
1/ '
V ~
~~ ~
" .//j
...-///
--- -- -::-:--t(sio.fit1nd.YJ
t......... ClJtrfl7Jl IItJlu~e.
@=16 gal/min
--- - I
~
FIGURE PS.,8
(a) To dekv»tlne fhe ~5"ir/Jy,f 1t;Yi Jt~ r~tlli~d fz> ;u4d M~ 5jJn'hk/er-
head sfafl~l1t:1'Y we U$f:.- The. m"JIIt.~f - tJl-;;noJ1teJllfu"" ~f"Iue
etlA.alidn {EI' ~. 50}. i'Jul5,.
= 6'/./7 t!oS
-r. ::- z.96 ff.16
$/uil-I
(;)
(b) T() dehYm,ne the r~5iJh'~1 -I-oY'ltl~ a>$dd~f.ed w/ft,.. tt spYIVJkiet' ~ttd
of 5"t1~ ye.v we. a.5e c1· / attll;'. fhn.veVeyI (/ot'ifA ro/7tI/t/YJ Wf.. /,ave.
ml;')
(Z)
FI7Y W We use
(16 !~a()(Ill'! ;;~~)
,..
{,'1./7 ff.bL -=
W c - J
l..
2 AI1()J}~ (2){ ". ()'f ,;"....) (7. '1R 9,4 , I ~ ~.L )
F/~ ....,;.,exi,,"
(GLJI1' f )
5',70

S:6B I (cOJ1'f)
r~y Z{ /,Ale ('(fe
= (9 In.) (soo ~ ) (i7r ftt )
(;2 I;") (6tJ~ )off ;Jfm
Th{JJ with £1' 2. (,N~ htMe.
V. :: 6'1./7 If - "llf.91 ff =- 2f/- 2{ ff-
~2. T .J .J"
C/Jlfti wilt, E't. / W~ tJ6"kt,;"
T
- (to 9'1 f.!!J.!)(I~ f.."/ Y8 In.j{zf. zt t)(/ iJ!.. )
- e-z lJcin ./ • sll4,. fI-
,$/11llf (~'fK 9tf I) (bIJ ~) /12. i" ') .sa..
alo'U/
rf2 him ( H
/. 3~ f-l. /h
CC) To de/e",m/YJe fAt.. ~n.I~/ar veto~/fy ~f- I1rL >jJ}'/nlder /f ~ Y~/>h~
f()y~ tl.e !s p-p/I/ed we. U.${! the ~/Y1tA-h(;YI ()f CtjS. / aJlut 2
fl; () bID. Ii?
tf ::W
).. 2.-
~
W2
(6'1.17 ~/)(IZ if)w = =:. ":::.
yo..... (S,;'.)
7he Yl;ftJy ~1MU.I IV.I i.r fI, Ci.J
N -::::(?I.? YQd) (b()~) = 920 ~
s /(277 ~d) -======A=,=''''
rev
5-7/
9~·J
ypd
-s

5.69
S.t19 Five liters/s of water enters the rotor shown in Video
V5.5 and Fig. P5.69 along the axis of rotation. The cross-
sectional area of each of the three nozzle exits normal to the
relative velocity is 18 mm2
• How large is the resisting torque
required to hold the rotor stationary? How fast will the rotor
spin steadily if the resisting torque is reduced to zero and (a)
() = 0°, (b) () = 30°, (c) () = 60°?
II FIGURE P5.69
7b de/erm/l'Je fh~ flJ~u~ "'~UI"ed -/zJ }u,/d ..fhe. YrJl»y fiP./7i;)Ylaq
we use, fhe. ~oJ+tenf - ,,1-- m()JIH~Iu;+1 rorf"G I!.j;UIJtIJ'un (1:1' s:So)
-10 obfa/~
;- - ~ y V t115!)
Shlll'f - ~uI- duf-
We nl)f~ fHaT
J1I1 :: p t:2.
3A""'"1le
eXl!
£1.$. IJ Z. aJlld 3 we
2
r' Q ~t co.> 6'
3 AI'I6JJle
exit
(I)
(it)
To riekY"w,ine fhe Y7Jtrrr fAJ19/,1Itty veltJvify t1>s()c/tJtf~d wifh 3ero
oSha!f iDrfJ/lfe Wl. ~!IJ In use. fhe m()~eJ?f- of - fflomel'1luW'l -/Dt"ffJe
e~J,(ah'oV1 ( €'j. ~. So) +t> "hm,n.l it,)$ l7~e w /t?t hJIRhol'7)
~hah-:: Y)1 ~"f (tv 1)1A.7 CbS e qt.tf- ) (5)
W-e nofe tha.f
(7)
(cOYl'f)
5-72

5.69 I (c~n'f)
r;half :::
I = Zt:Jo AI. not
s"4.11
Fr~ £1" w~ obJ,un ~ fl.t:?H:: tJ
w = If I~/ (OJf 3()")(/PI?P ~),. -= IbtJ YAel
3(18~~L-){ IPI}~ ~)((J.S"..) -- .$
(c.) F7Jy e == 60 d wt use £" 'I Iv ge.f
r & (9ff!1)(s ~/(t).'-'")(Co! to') {tPtJtI :,.:/(f1'/;")
py
~4fi ~3 ~ f>
(/IJIJIJ ~~y~ (3) (IYn-.n,~)
r :: III N.~
Jhaff
~ t'j. f we 16hun -fp.., ~ L =- ()
'firr 2-
W ~ (S ~)(UJ6tJ~ (IPPd ~
(3J(13ht,.,~J(lOPcJ ~) (d~~;)
;-73

5.7 J A water turbine wheel rotates at the rate of 50 rpm in
the direction shown in Fig. P5.7J. The inner radius, T2, of the
blade row is 2 ft, and the outer radius, r I , is 4 ft. The absolute
velocity vector at the turbine rotor entrance makes an angle of
20° with the tangential direction. The inlet blade angle is 60°
relative to the tangential direction. The blade outlet angle is
120°. The flowrate is 20 ft3/ s. For the flow tangent to the rotor
blade surface at inlet and outlet, determine an appropriate con-
stant blade height, b, and the corresponding power available at
the rotor shaft.
since
Q. II:: 211;-; b ~,I
+hen the blade /'el,kl-, bI 1.5
b
The
a
:Iffr; ~ I
51101f P()WeY'"
•
vv:.~tlH
lief (NIt
powef" ~uah'tJn (£1.5'·53) .
•
-vtt;.haff .::.
) 15"
Section (1) Section (2)
• FIGURE PS.7!
b1 f-
j1.
OJ
nef (Jt.ri 1/ .,
(Ina the Use of 11+ 'I 0,. wi It, l{ V~ 2. d el'end.s 0" wire/her
Ve 2 /s- opposi Ie. 7-0 or i" +he. 5"ante d/rechon as V; respecfively.
/
To defermine the value of ~,I we use. the velocify fYiong/( af
se c. Non (I). Thus we hallf:I
V.R,I
V,
Wi~ fhe ve/~ci1y /Y/(Jn9Ie we.. have
~,I V.
V= R~ I
+
hn 2()(J
,
fAn 60 0
However
V; - r;w
(con't )
5-74

5".7/_1 (c()n't)
thus fq.]
V ::
leads fr;
r,w
R,/ / -L _ I ')
( hn 2.0~ fDn 60,/
tViff" ~9 ,I we tJbh'n
b ~ (to #1)
/- rOd)(if ff) (5() rpm) ( 271' ;;;
:::
= tJ. O~25 if
211"('1/1 )(9.651 9")
For fhe blade lIe/ocifl'es 1i1 E~. 2 we ref
7J = r, (J.) == ('I Tn (so YPWt) ( 2TT ~) = 20.9'1 If
I (60 .s: ) .s
"'11"
1.l:z. = J: w = ('Zff) (5 0 t'p~) ( 2-71' ~)
- --------------~'~~~---== ~.~7 ti
60L s
HI/I?
Foy ~I J we we fhe, velrJc/Iy iYj(Jnv1e af secf/(}n (I) -h; ~blzi,;'
~ , : : ' ~I I _ " .
6,1 T ~ 2~. ~l f f
I fz:tn 20 0 mh 2(} t1 of'
F-or- V. we COnJirucf The se.cfiol? (2.) ve/(h:,fltj fy/1I1'10/(: ske.~kd 6elfMI
~ 2 ./
(Ve, 2. ntrt h salle)
L{
(AYld We yeahie -fhaf
V
8,2.
VB 2 :: ~ Z fan ~() 0 - t{I I
From cOf/servaHon of Mass
'II.: : V. A,:::
R,2. /(, I A
2
(con'f)
(It )
5-75

50 wi Hz £~. 'T we obmln
V~ 2. -= {ltf.3 ~f)1-rJJ1 30
0
_ /&.'17 ft == 0.673 ;f
Finally w/111 €r. 2 we (}b~/~
vir ~ I!,ti'l slufffi.o if) )!(zfJ.71f £1 )1;.652"£rl-f{O.'f7 fJ:)'/O.67JB)}/i !it- )
shall- (," 1-+1/, s If co S /(4. S/ s I· s"(j{ sllI1.ft
net ou.t J~
Or
W ": '2.. /8 X I() 'I ft. II:;
s)1tI.H -
net 4ltf
s
a~c{
'I
1.. iN XII) If· 110
-W - s -::: '39. 6 hp
sna.If
nef"-o/,tf s~o ff./b
s. hI'
5-76

5.72 An incompressible fluid flows outward
through a blower as indicated in Fig. PS.72. The
shaft torque involved, T'haft' is estimated with the
following relationship:
Tshafl = t1U2VO.2
where t11 = mass flowrate through the blower,
'2 = outer radius of blower, and VO•2 = tangential
component of absolute fluid velocity leaving the
blower. State the flow conditions that make this
formula valid.
I - i11~ ~)2S-i1a It
.....- _...
FIGURE PS.1Z
(;)
valId J1I1a..y 6e Ide",.l/h·ed 6y UJ>?1ffJl'I"J ~&J' I with -IJ,e
c(JY14fJ~neJ1f of Ee. s· 'f2.. Thast! cohdi -n(n,,$ ar~
/Ix/a I
a.. s tah'oJ1.IJl;':.Y and h~J? - de/w~/~ CQ'y1~1 volume (see .rtelc~ a6uv&)
". slea.dJ - I;' - flit! - MeQI'I ,clow
c. ne&I'j/6Ie She4Y sIY~s.! 1oY9UL w;~ ve>Iee-f ~ tlx/s IJ{ ttJ/r../JOh
V ::: 0
0,1d.
e. no
sftt!>~e~
f. uJ1/ flJy~ d /s fy/huh'on of t,.;, 2
I
~77

5.73 VR2 = 30 ftls
5.73 The radial component of velocity of water leaving the
centrifugal pump sketched in Fig. PS.73 is 30 ft/s. The mag-
nitude of the absolute velocity at the pump exit is 60 ft/s. The
fluid enters the pump rotor radially. Calculate the shaft work
required per unit mass flowing through the pump.
• FIGURE P5.73
The stah'o}fcuy
Sk~l-ch abtJve
aMt! 11011- d~/PYMlry ct)J1irrJI vt)/tlYJt~ shown /~ -IJ,e
i.s used. /0 ttebYmln-R (he shlJlf wOy/<.. fer
1""1_ r, Id~ C~"'" U~I £4. s: S"~. TJII~.JWlt/5J
$ntlrt I"""" l«,. ""'L II
lIni! ./
= 1J Vw.$/'tlfl- 2. 9) 2-
7}te bIt/de ~".ud tJ;. cd;r be ()6M,ned (lS ..fiJI/(}w>,
(I)
I )
V ::: C W ::;; (0.5 ff )(2.000 y!!' )fTT ~)/l- )::: JD5 If
"l- ~ ~/;' yeV (~O...E.- s
The. fangeJ1hd ve/tJti/-yJ ~ I CAli k dbl,J1net/ (/J fp/;;/.,;~
~2 = (v:- v~...)i =' [(6off/-(30~t/l ~ 52
ThuJI
.frn.n £$' I
= (105 :1)( ~2 t;lf60
~78
{f
-s
ff·/6
s/uJ

5.74 A fan (see Fig. P5.74-) has a bladed rotor
of 12-in.-outside diameter and 5-in.-inside di-
ameter and runs at 1725 rpm. The width of each
rotor blade is 1 in. from blade inlet to outlet. The
volume fiowrate is steady at 230 fe/min and the
absolute velocity of the air at blade inlet, VI' is
purely radial. The blade discharge angle is 30°
measured with respect to the tangential direction
at the outside diameter of the rotor. (a) What
would be a reasonable blade inlet angle (mea- "
I
sured with respect to the tangential direction at " .... ,c I
....
the inside diameter of the rotor)? (b) Find the .;;:--._--- I
power required to run the fan. FIGURE PS.74 1p-I in.--J I-
The ~lah(;Ylary and nOn - delrrrm/~J confn;/ VtJ!ui11e .fjzDWYl In -the
skel-ch above. is UJed. 10 de~n1/h-e. a retrst)JlUf61e. blt:lde
Ih/ef tJ~1/e we aSSume fhat the 6/adl. $htJl1/d ft,e /-t;;n~Mf fo
-Ihe ve/afive ve/ocJfy ttl lite Iy,/~f. The.. Inlef ve/tJcl/-y /Y/an!/e
/> JkekhuJ belttw.
WI
(;)
Now
Q
= 35./ f!Q::. --A, S
== YoJ = (2.5 11].)
I .
(12. In)
ff
7hu; wifh E~. / fI-
e = +aM -t{J§·I J) 7= 'f3 0
, (37-" 51)]
(Con'c)

5: 7/f I (con It )
.
The power ye~~"y.ed) ~J,tlll) may be tJb-blird wi/tJ &1- 5".53.
Thus
.
W:::mVV
fha/l- 1.. z. ~ "L
(2)
VV1IiH -!Iow-ra-!e / rl17-J YYlay k ~bIt;l~i a$ hl/lIWf_
Also
th< 1- '" jl Q =' (2-31X 10-
3
S/"ffJ (z"i0 tfY-'- );; q.12 xmJ~
ifJ / fIIlh). to.!... S
111,/)
u.. : ~w '" (6 In. ) (172 ~ J£f J{27T fl! )'" 90."3 t!.
(/2 In.) (6tJ .ff-: ) .s
H /Hln
The value. of ~I ~ may be. o6-/r:(Jnet/ by cOh-rld'Ylj .fh~ lIe/«:i7
friPYJ1/e -fpy -Ihe. flow letW'l1j -lite j?J~ af 5ec;hdn(~_ The
ye!/ftfNc velocily at /he Yl)~ exjf- i> UMsJdereti -IT; be. IoWffld
-Iv the-- hhde fheY'e _ !he yafw exif flrJ/N vtl()~ifj -IY/aY/le is
skel-d1eJ belfIN.

-
5.7S An axial flow gasoline pump (see Fig.
P5.75) consists of a rotating row of blades (rotor)
followed downstream by a stationary row of
blades (stator). The gasoline enters the rotor ax-
ially (without any angular momentum) with an
absolute velocity of 3 m/s. The rotor blade inlet
and exit angles are 60° and 45° from the axial
direction. The pump annulus passage cross sec-
tion area is constant. Consider the flow as being
tangent to the blades involved. Sketch velocity
triangles for flow just upstream and downstream
of the rotor and just downstream of the stator
where the flow is axial. How much energy is added
to each kilogram of gasoline?
c~m+-ml
VIJ /Utl1e ~5C
I I
I I mean radius blade
60
0
j£ sections
I I
I
:7U:~J I coYlfrol
I I VDlume
I l
)
FIGURE P5.7S
lhe veioc-i-f'l l'IiI1YJj/e." f;;r {low jusf (Jff~tM11 ~ /he, y,,1-tJy Is
SKefrfrled belaJ lOY fhe arithmefic mean Yael/uS.
WI
V-"3rn,- -s
With 'he TriorJj!e., We con~(ude fhai
W :; V, (3 ;:-) 6 m::. =J
cos 606
s
Co> 66"
ay,d
v;= 5,2
5"- 81

5.75 -coy/Ii)
The ve-I()ciiy -/-vldYJg/e I()r (I(}w jllsf c/()/)JJ1.JIYetlM pf fhe Y()1-or is
ske hhed b~/(NJ f()y -the tJyifhmelic, JlY/etlYi Yadws, ~
t'i7UJn1fJm5ib/~ flow ~)2.:::' ~. f:or /J1eon ratift.l.f t~ l{=Z{.
Thus lor relafive lIow faYJgeJllf fr; fhe blade Wt obkiln iAe
ve(rJcify -Iv/angle.- .>kefd7ed belM.
W2.
~lI
With fhe fY/an,/e we undude thaI-
Vn., = U - W -:; TJ- v: 1anl/§D::: 5,2 m_h m)-M~'1/= 2.2 ?Z'
~~ ~ @JZ 2 J2 s l- $ S
5"- 82

5.7~
5.7" A sketch of the arithmetic mean radius blade sections
of an axial-flow water turbine stage is shown in Fig. P5.76. The
rotor speed is 1000 rpm. (a) Sketch and label velocity triangles
for the flow entering and leaving the rotor row. Use V for ab-
solute velocity, W for relative velocity, and U for blade veloc-
ity. Assume flow enters and leaves each blade row at the blade
angles shown. (b) Calculate the work per unit mass delivered
at the shaft.
I /oll.(jIre
J ~76~Ufe (0,.+11)1 lOhl~e
/ / ~I I Blade sectIons
, I mean radius
I I
I/ 'A' at the arith~etic
FIGURE PS.7b 45¥' 17.50
J I
The ve/oci.fy -/Y-/dr},f/es foy fhe.. Tk>w enleYl?J and fhe flow
leav,h9 fh e vofrJr Yow cd -!he dJ'lfhmelic.- metlh Ytlc!tu.J (1ye JKelched
below.
1J2.
Ai fhe ari~lJtefic m~a;1 vCid/uJ the bltlde ve/oclfy ) ~ is)
V:: t{ rw (fD ,;".) I DIJO re.", (Z'Tt' ~)
II:: = nt/~ Yev
:: ?:? '3I f!I
(12 In.) ('02-: ) s
If )lHlh
W;t4, th~ vel~cify fYl'aYl!/Ie.-
~>in 7()fJ -= VgI
f,Y fh~ flow lnHn-IYtJ the Y7JIuy Wl- cmclwle fIdf
" I
V, (;()5 70 = ~ I
W I §irl LfS II =V~ ,- u
tV I Co> '1-5" ;: 'Ix, I
(Con 'e )
5"'- 83
(I)
(2..)
('3)
('I)

5. 76 (con '7; )
Fro}1l1 fhe rtlfio tJ{ ~5 . 3 /AMI if we. Obf71'~
V.X,,
which when um6/necl wift;. Elf. I ~d 2. y/e/ds
or
(J
1af'llfS =
=
v U)S 70"
I
u
[5/i1 70
0
_
=- ~'7.fofl-
.s
TheYl
.j' S1i17i' =(f7' ~/) 'S,;" 70 () =- SZ,3 IfVBt
::
.>
I
V tj (.f)$ 7(/ -= fa7.b ~+) (()5 70
0
:- z" '1
fl-
= -~/ I
S
()nd
Vx I (~'1, q ~)
WI = 12,'-I !!,
- ::
cos fs D
U'Js Ifr () s
W/th fl1~ ve/ocify -/-Y/OYl9/e fllY fhe fl()w leavlnJ fhe r-olrH we.
conclude ihaf
. ()
I - V - W S-1Y1 ¥sv~ l - 2. 2-
I
{ $,;' «z.
Vz. (,0$ O{z. =
_ Zq, Cf {l-
S
(conI t )
(5)

ThUJ {yom G~. 5
= '12. 'I f-f
.$
oJIId fr-uwt ~. fLJ
V ::: U _W ~1Y, Iff; 0 = 52.3 f! _(ttl,'! f.f) ~/~lf) 4:. 22.l{-!J
(},'2- ").. 2. $ S .J
The valio or £15. 7 aMd! y/e/dJ
.x; = -fan-I (Vt9I'-) -= fru-,-!(2.2.4 it)J = 37 0
Vx, ~ ( Z'i, 7 ~+;J
aJ1d -IYl7Y'n E,/. 7
v - Ve, z
:z.
We CLJn U5~ Etg. S:S'I fz; C'a/ClAltlk fhe WdYf,. per u~;f
mass deJ/veyed ().f /he s-ha.ff. 71zt-1I
W :::: - TJV + 7J V
5htl.fl- J ~ I ~ fJ. ZI
::f(5"2, 3 ft)(n.3 ~+) f (5"z.3 f f)(z~.'/fI-~/t !.L )
5 J ) s1Jl flUj. (f
5).
::: _ :3130 fl. Ib
5/uy
5- 85

5.77 Sketch the velocity triangles for the flows entering and
leaving the rOlor of the turbine·type flow meter shown in Fig.
PS.77. Show how rotor angular velocity is proponional to aver-
age fluid velocity.
,..'"
• FIG U RE P 5 . 77 (CoutICsy or EG&G Flow Tech.
nology, Inc.)
F"" '" 5edjt7l-> of {he /uYbj'lG Mo.de. af r",<I/u6 f/ fl.e b/tff/~
J17fN<S t-~"fhl!> w i /I, '" vdQc.i/y V = rrAj. 1k ve/(Jei!;t
Irl"'''9/e! hltJ.y be fkl.heo( as ShtM>'> .
W,
V,
v
v
V ~2.J
/.1s'''J IE$. 5".'50 IN· 'lei
- V -+-1...- r W
XJ
2.
50
cu -
fJ-86

5".78
S.7f:3 By using velocity triangles for flow upstream (1) and
downstream (2) of a turbomachine rotor, prove that the shaft
work in per unit mass flowing through the rotor is
V~ - Vi + U~ - Uf + Wr - ~
Wshaft = 2
net in
where V = absolute flow velocity magnitude, W = relative
flow velocity magnitude, and U = blade speed.
Any .fef or veJo(ify fnaJ19/e.- fw ,rIM fhYI/lA;/' t{ /z,tvbt'YJltPch/fe ft)/Py
YOw wtJlAld 9/Jle.. fh~ Jdme I'e~u.lt. We IA.Sc ~ l;r/dnt/es Df 10,. ?S. 77.
Fr()/'YI fhe ,'rile! Ilow velocify fYl'aYJ9/e w4Z qef
':2. l . " l .
V. '=>. V -)/KI V D II I v,
tlnd
1. 2.
': W - (~ +u) :J ~ I I
'l. '1. ).
AI - V - 2 Uv - U~ I I e I
c~(,/Y)J nJ

£Cjr. / t:IM-d z c.vc ohfaly)
if ~ I
'2.. _ v l _ til
- I I
I I
Z
V
~2
FrO)IVI -fh t!. outJe f flow v~/pt;fy fy/an9/e. we- gef
'2 'l
Vy ::: V -
~ 2.. .2.
and
(COy/It)
5- 81
(;)
(2)
('f)
(s)/

? 78 I . COY) 'f)
C(YI/yt /n'n /Y/J Etj~. 'I tA nd S' we. b6h/~
lj'-W"" T
"l-
Ull
l{ (,)
- :;?. 2-
~ 87..
ZI
;t:-()y the ~ef of veltJcify -Iv,'a119/es
uJ - U!/ f ~~~
(7)
sh4f+ I ~I I
he! /n
Corn b/YJ,fij ~1s. 3/ h ClL-vtd 7 (AJ(.. ok-/-,; I;'
V 2
_ V 2-
l. 2- 2. W"2-
-t V-U +W-
W5htfff-
::: ;). / :a.. ~ I 2-
hel Ii-? 2
5"-88

5.70/
5.71* Summarized below are air flow data for Upstream of Rotor Downstream of Rotor
flow across a low-speed axial flow fan. Calculate
the change in rate of flow of axial direction an-
gular momentum across this rotor and evaluate
the shaft power input involved. The inner and
outer radii of the fan annulus are 142 and 203
mm. The rotor speed is 2400 rpm.
1ItL chA.llje lit
across the.
~FAMx
ov
f}Jhey'~
Radius
(mm)
142
148
169
173
185
197
203
Absolute
Axial Tangential
Velocity Velocity
(m/s) (m/s)
0 0
32.03 0
32.03 0
32.04 0
32.03 0
31.09 0
0 0
r t»td (
.
~nnu II,(S Ih1/e¥ ()nd t1tt. /ey rad'-;
(, f)
Absolute
Axial Tangential
Velocity Velocity
(m/s) (m/s)
0 0
32.28 12.64
32.37 12.24
31.78 11.91
31.50 11.35
29.64 11.66
0 0
(/ )
~ anti ~ aye local y-a.dii af secfil)n (2) dll.lll1s/tla-11-! ()f 1aJ1 y,hy
ClAd 5c,c+io~ (;) I.If$~ of ~ ~iur
~ OVId. V~ I ()Yf!., local aj,5tJluk k"le)1h4! vdlJcJly al secftdJ7J(2)IJIUiU)
~2 I
. a~ (1)(4/ a-x.i(J.1 vt-/()(iffes aJ Sec piJn.1 (2) a-Itd (I)
lIx'/L aMd VX, J
As 5Wj1t!,JktJ. bj t~. 5·'15
7: ~.tJ FAMJ(
Jlu/ff
(2)
and Gj. 2- IS
!ttL -Iror'Jt)l'dA. 1
eva {ua./(tA nUYY/fJY,"{,Il,/l witr... a. U/mlufw fYP<jYdll? Maf (,iht~J
rukwi/'t, tAJ1e~Yl inWet/J. & pri7Y~ fisi ~yf.d
(e.su.lfs att!, on fhA.. Yle)C-f paJe .
.
WrA4ff- =- ~Illf fA.)
:3 ;'s .eN1t tlA.ojuI 6~
.
~h((ll) ;f ev~ 4-takD/ w'-f(., Gj. 5. '17. ThUi.)
( 3)
IIu.. ~ujt,./ tJr;r;Y~ I/)/e~ on bu
(um If )
5"- 8'1

i
(Con 't)
100
110 PR HIT ., :+. *tt*******:t;*.******************:*********:*********:.t:"
120 PRINT ":rt This program computes the change in rate of *:t"
130 PRINT ">1* axial-direction angular momentum and power**"
140 PRINT "** input for problem 5.79 using the trapezoidal **"
150 PRINT "** rule applied to unequal intervals. **"
160 PR I NT "i' t::!.~~.t*****~~*********:t:*.*:t:* :j;~;~;~;**:t,*t*******i:*;t:*****i:"
170 PRINT
180 DIM UXU (19), UTU (19), UXD (9), UTD (9), R <19 )
190 '
200 , Initialize the variables
210 N = 7
220 RHO = 1 ")0
.L • .:::.. ,.~
230 PI = 4! * ATNO! )
240 RPM = 2400!
250 FOR I = 1 TO N
260 PEAD R (1) , UXU (I) , UTU (I) , UXD (I) , UTD ( I )
270 RO) = R (1) ,/ 1000!
2130 NEXT I
29U DATA 142.0, 00. 00, 00.00, 00,00, 00,00
300 DATA 143.0, 32,0:::-" 00.00, 32,28, 12.64
310 DATA 169.0, 32.03, 00,00, 32.37, 12.24
320 DATA 173.0, 32.04, 00.00, 31. 7,13, 11. 91
:3~~ 0 DATA lot=:~) -~ , 0, 32.03, 00.00, 31. 50, 11.35
340 DATA 197. 0, 31. 09, 00.00, 29.64, 11.66
,3'50 DATA 2():3.0, 00.00, 00.00, 00.00, 00.00
~wo 'Compute integral u:sing trapezoidal rule
380 .:3UJ{U :;:;: O!
:3(~' (I SU]lfD = O!
400 FOR I = 2 TO N
410 TEMPU=UTU(IJ:t:UXU(I)*R(I)A2+UTU(I-1)*UXU(I-l)tRCI-l)A2
420 TEMPD=UTD(I).tUXDrI)*RCI) ft
2+UTD(I-l)*UXDCI-1):t.R(I-1)-2
430 BUNU BUMU + TEMPU * (RCI) ReI - 1» / 2!
44(1 SUMD = ,sT):MD + TEMPD * (R(I) - R(I - 1) / 2!
450 NEXT I
460 MFXU - RHO * 2! * PI * SUMU
470 }{:FXD = RHO * 2!t PI * SUMD
480 POWER =:: 01FXD - MFXU) * 2! .t PI * RPM / (60! * 1000!)
490 '
500 'Print the results
510 PRINT
520 PRINT USING "The shaft torque is ##.## N-m"; MFXD - MFXU
5:30 PRINT USING "The power input is ##.## KWH; POWER
**************************************************
** This program computes the change in rate of **
** axial-dire..::tioYl angular momentum and power )j~*
** input for problem 5.7~ using the trapezoidal **
** rule applied to unequal intervals. **
********.t*****~***.t*******************:t:*******.t***
I The shaft torque is 4.79 N-m
L..The pm',er input is 1.20 KW
5"-'10

10
S.l:{() Air enters a radial blower with zero angular momen-
tum. It leaves with an absolute tangential velocity, VB' of 200
ftls. The rotor blade speed at rotor exit is 170 ft/s. If the stag-
nation pressure rise across the rotor is 0.4 psi, calculate the loss
of available energy across the rotor and the rotor efficiency.
+
nef In
(I)
~
!he. ~haff w()y/::... 11"1 j lNSh4
1:1- CAh be ()6~/~e?l wilt-..- the" V/.1tA11eVlf-of-
tlt:'f ,},
MCJYI1eJ1fuYYl wcJvl,- e~ lA.a:f,'1IY (. E~. s·r-<t-). 7hM.s,
uJS;'ulH
:= v;,uf- V
~ dkr (Z)
net- In
c~ndnY')J'""J trJ ·
14-J,tt/2 /~df -ID
loss - ~ I';. - Po auf r tr V
" " i71A-f- t'1,t,J
;0
loss -=
(lJ1,cI
loss 1t9tJo rf. It
5"- 9/

5.81 Water enters a pump impeller radially.
It leaves the impeller with a tangential component
of absolute velocity of 10 m/s. The impeller exit
diameter is 60 mm and the impeller speed is 1800
rpm. If the stagnation pressure rise across the
impeller is 45 kPa, determine the loss of available
energy across the impeller and the hydraulic ef-
ficiency of the pump.
The antl/y.fi$ of EXtfI,nf'le ~. Z 7 /5 a/pi/cable 10 S()/V)~ #til
py()bleVl1' !.JS1i1J €~. ~ &f Exq"".pJe 501.7 we tJb/aln
actual ir)'fp/ Py($$U~ Yise ({&-Yr;$f I),.~fk""
uv -2 &;2
H()lIIe1If/r"
~) (17T ,.,.4)J
(1900
~ ::: t;.UJ = (60 yY''''' )
Yw =5.66 ~
(2)(!()()O ~) (bo.!-. ) .5
}'?'lIn
Thus
(5.66 J)(!Os'jf~,~", ~ t '5)(10 AI /
/0$5 ::
- ;;i)(f1f~~)s2-
/055 ::: / /. 6 /II. ~
-
7<J
Fronz &1' 5' of EX~k 5. Z 7 we tJbM,;"
{)C~aJ foia/ f1"t'eJj(Jf( ti.re aCYlJJ I~pe/I""
( ;::
to

5.82. Water enters an axial-flow turbine rotor with an ab-
solute velocity tangential component, VII' of 15 ft/s. The cor-
responding blade velocity, U, is 50 ft/s. The water leaves the
rotor blade row with no angular momentum. If the stagnation
pressure drop across the turbine is 12 psi, determine the hy-
draulic efficiency of the turbine.
To defeo'J1lne fhe ef{/c/ency of f),e, fuy/:;/ne We use
acfua/ Nod~- ou.f
'1 ocfua/ //II()r/<.. our of loss
The. achttl-I Wf.!'rK o('vf; w,ft,af{ } Ir
fief ouf
momenf - of - rnomenfuYl'l Wt7Yf:.
v(f J~I
(I)
(2 )
To
rofvv use
fhe
fhe
I()5) of a vtf.f/la6/e.. eneYO'r ac.Y()sr the
eYJ~ j e~ lAcL/-;'tJy, (61' 5:?2) ~ ~6hln
L ~ ~ n~,Juf
loss = -I- ~;.. - ~Uf -I- 9(~/" h~~o + ~h(1fl-
2 1 - l'Jef ,;.,
p.-I! f", I J1 411, IJu. V t/ .Ii? 6j I;.
V· tI .In ~ II?
=
7.J V . 1- /055
III ~ In
a.nd
5-'13
(3)

5.$13 j
.
.ff3ctd
CJ)...t..I
5.83 An inward flow radial turbine (see Fig. P5.83) in-
volves a nozzle angle, 0:'1' of 60° and an inlet rotor tip speed,
Ul' of 30 ft/s. The ratio of rotor inlet to outlet diameters is 2.0.
The radial component of velocity remains constant at 20 ft/s
through the rotor and the flow leaving the rotor at section (2) is
without angular momentum. If the flowing fluid is water and
the stagnation pressure drop across the rotor is 16 psi, determine
the loss of available energy across the rotor and the hydraulic
efficiency involved.
IOfct,ae
---:o-:~~-~ ~.} ~,~
~ I
: ,,~'-1IiIIg1!
,----- -
• FIGURE PS.83
An analyJf.~ //ke fhe One of ~)(tlmf/e 5.Z8 IJ/(juld be liff~flyjok
fw 50/V;'!j fhis flYbh/eJ'J1. ~j"ce t:4 fur6ine is Inwlv~d lit fh/s
pJ1)b1eJ1t'1.1 wshaff :: - W$half and fnm" ~. I ()f EXtJI")flJe ~U
nd i" hel(Jut
we Cdn c.())1c/ude that
/055
If) $5 -=.:
s itlfnali()'" pf"e.s.5'I'1~ dfPl' acyO~s yofuy
r'
= - V 1/J ~ I
51?A(P1~Ht1YI pt(.s.sU'l( dY'Pf o.CrtJSf YrJIuy
;0
(r)
10 c/elwTJ'1J~e" -fh(. V()/{"(e o-f ~ I we eX'tlm/~e file.- ve/dclfy fy/ol1j/e
ft;y fn(.. I/(JW '€~/ny fhe, ~ fhal- iJ JJ::.ekhuJ k/vw
hvm fhe velOcity fr;'tll1,fJe we Obltil!1

From we, ob1-al;'
It/55 (;6 f:.~)(/'I'Il,':)
(t 1'1 f!:J$)
{f"3
= I'I!J fh ,,,
tv$htlff f J655
hel ()uf
Oy In ofhe,y wWti.J / lite
re.Jt.l/ir /n J'ju/~1 l/VlI'Yk
sla?l'u/jJ'fFYl piU.> u re dJ0/J acrou ihe. roJuy
;0
Slayifa.'6'on preISu Y!. dyojO PlU'~.r -Ihe. n:drJy
alu:/ /0$.5 I/f dt/tJ/'/a6/e ~er,/ .
7hl/ 5 &f meaA1~:fiA.1 e ff;c-ieYlC;Y If
UJShtlff
l1er OlAf
=

/he
WOr/c.
ol1d
5.8 if An inward flow radial turbine (see Fig. P5.83) in-
volves a nozzle angle, 0: 1, of 60° and an inlet rotor tip speed of
30 ft/s. The ratio of rotor inlet to outlet diameters is 2.0. The
radial component of velocity remains constant at 20 ft/s through
the rotor and the flow leaving the rotor at section (2) is without
angular momentum. If the flowing fluid is air and the static
pressure drop across the rotor is 0.0 I psi, determine the loss
of available energy across the rotor and the rotor aerodynamic
efficiency.
105$ P,-~
slJa{f wor1.., WSJulf~
,,(!f ,;,
e~~~ CEi. 5".5'1/).
wS/-'Q.ff_ = -VI/.:
l1ef ,;,
I ~I
-= W:ha (.1-
J?~ t DI1..!-
C~61""/~ ~s. I t:tA-fd z y/elds
/t7$5 - P.- P-a.
r
v,2_ ~~
UI- I ~ I
/l 2.
~I,
~I,
(Coy/f)
5- 96

con'f)
With the velocifJ /-Y/an9Ie.. /life
== I/O .fl-~ :: (-Zo ~ )
s
and
v :: V SIn b() 0 -:: ('fo -H) SIn ~() 0 :: 3'f.~Lf it~ I I S S
5iYl c e the flow /eav/~ fhe YD/w is Yael/a/ I theY!
V. =- l/ = 20 f+
~ R,L S
~ £1.3 we- tJ6~/;"
lOSS (b, 0 I /!!-.)1/'1'1 i!J....)Ie -= I". (I' f+~
(1a1X10" i.J.1j ')
ffi/
fl().~
lhe t.- {{/c/ency may he o!?h:lIi1t:.d wilt,
1 ::
(lclvtai woY/L Qui
aclvtal wov/£.- o-wt + /()S$
oy

S·es I
5.S5 How much available energy is lost during the process
shown in Video V5.7?
All of tA~ pore"IitJ.1 eneY9!! /0;1- I;' I'YJOI/ly ~ fhe
fr;, of Ihe ~y ~ the. bo7ldh?_

What is the size of the head loss that is needed to raise
the temperature of water by 1°F?
in
c(r - r)~lAl })J
h =77f IfL -=
S"-qq
Ib~ . It ;j7f f.f. Ik
lb. 57..) . Bn.)

5.B1
S.K7 A 100-ft-wide river with a flowrate of 2400 ft3/s flows
over a rock pile as shown in Fig. PS.87. Determine the direc-
tion of flow and the head loss associated with the flow across
the rock pile.
IIlI FIGURE P5.87
To dele.rrn,;'e Hte d/'~.c~ of ~ we will Q$.rLinte p. dJ~cIJ8n
:/
u.r~ -fivL eA?~ e~~ (Gfo 5":tt'l) aJ'u;/ C'QlclAlQk the. >
head /bS,J. .Ir I/t.L /vaal 11)5$ is ?()fi~·ve'" OUy aSJ'tlmec/ c4~o)"
t'f ft~ if ~CI: I..f fJ.,e hR.atl /lJf~ if J?ryal,~ whjlh iJ J1td
fJAyfic~/0 ;tJ()ss;ble.,J OrA¥' a.!stJ""et/ d/~c.-hlrn (JI }Iqw iJ w~.
50.; tlSfllm/~ {he HdlN is ~ Y&Af -k lei; (/Y' ~
POI'" r (i) -Iv po/ni- (2.) I~ !he .5'kehl, I1bcv<./ we gef
fA sin} .JJ"(JIf wDrL
N(IIN
0 - ~
-A,
(Jl'ld
V2,
-::: ~
AL
So
=(:Zlf(}() fr3)
.(Lj fi) (in: ff)
-::: ('].lf~o ft )
(2/-1-:(100ff)
- G f+
- r
c 12 fr
S
l..
_ (If+)
~_/oo

5". S8
5J:{S If a t-hp motor is required by a ventilating fan to pro-
duce a 24-in. stream of air having a velocity of 40 ftls as shown
in Fig. P5.88, estimate (a) the efficiency of the fan and (b) the
thrust of the supporting member on the conduit enclosing the
fan.
1m FIGURE PS.8S
40 ftls
--+-
(0.) The s~/uf;d"h -IZJ fJ,if fM] df ht.t poblew, iJ It"k.e b:dMJ'1e s:),'1.
We. lAse-
7 = ~~<tf{- - /"fS
IPs-h,r/!.
fr, ca;/Ctf.l4'/~' fJ,e. rd41 e .f17·c/eYlc).
We use fl,e ~ e~a,,;,rnr11.5;S2.)-Pr f/(IW -fA~Uj'h the
~-h-t VI//ume fJ.elel,ed abow.. 10 c,/A/cl1la,!e fh<. iPJS II!
f.oll~5"
~ + '/,.2 -I- a c -=- !i. + ~~+9r of WfJ,tJlft loss
fJ 2" .I l. r' ':l- I nef in
8fA f !{ :: R a-rJ 2-..z -: ~J j .j:: 0) wfJ,~1 == ~
~ ~f~ m
fUse} ~ :: fA~~ = L ~clJ. V;z...
I?T 'I
So
~ 101

5.88 I (c.orj'+)
'Itt- 1-/./6 _ 2 i/.? II. /6
Ibn-, 1~Yt?
/0;5 -
50 ~~ _ 1f!,2 fU'f'f
7
Ihl"" 16_
-
¥Y ..p.~
1/'/!tO
Frrr
-I'././b
= /9.2
- ~·~b
(/:;) We uSe fhe hOY)Jdn/a} ctJYl'IptJnehf of fhe //YJea yo
YY)Qyyle,n ~ etjvtt:~,;h~ ~ e valuak /I....L anch()Y);'J
fWa reJu ,.;e4 Iz; h()/d ih( 1ztv? In place
F = Vii?AX 2.
h-~ pf).;{f (fA.,)
fo
2.
n:, = L 1T dv L{ =
f<T 4
. 9.t!-1 Ibmm = s
('10 ft)~'II 'f')F -,
Ax
(~2.2 (k.f+ )
. Ih. $').
5-/02
= /1.7 II;
-

Air flows past an object in a pipe of 2-m diameter and
exits as a free jet as shown in Fig. P5.89. The velocity and
pressure upstream are uniform at 10 m/s and 50 N/m2
, respec-
tively. At the pipe exit the velocity is nonuniform as indicated.
The shear stress along the pipe wall is negligible. (a) Determine
the head loss associated with a particle as it flows from the
uniform velocity upstream ofthe object to a location in the wake
at the exit plane of the pipe. (b) Determine the force that the air
puts on the object.
U!___ .___________C!:)~.~
1_ . Air l
1 1 , , , - ....... " . 1
•- : "I(K,t T l ~4 m/s
~I_ 2m-dia. ':~W~kel-m,dia
' _ I ~ t l 1
I I ,-".. '« 12m/s
•r-:::: __~___ -- - ---_-J
' Exit/"
I' = 50 N/m2 V= 10 m/s
(rA) T(') de/-fH--m/(Je fhe /O$} rlJfle~eel h, el flu/d po~/i"k (}{s;1 !loWJ
f,--qn,a (I) -10 ~ I()u"h~n I~ lite VII()*e ,,-I- (2.) we (J.fJply fht..
ene49J e$uaA'on (€f' ~. 8l/) -fr; thaI- pay/7'de floIN' 10 'Ie!:
i'; v/ .-1' = r.;V,'1- fj' r vJ£~; h (t)
fi ~;g ~ T i9 I /~ - L
()'"r
"3-
J,L =
f,
+ ~
-'I -2J
e;...,~
(5"q ~~)
hL ::
('7-~)
-
+
'1
v,
-;}.J
"l-
"1.
(If 1)(10 ; )
-
2 (9.i';i) ~(9,'1 ;'~)
ffVA) +
7'~
.$
(coY'/f)
5"-103
- $'. '15 n7
--

avu::J
1/ =lIt) IVr'lx _
5-IOif

5.90 Oil (SG = 0.9) flows downward through a vertical
pipe contraction as shown in Fig. P5.90. If the mercury
manometer reading, h, is 100 mm, determine the volume
flowrate for frictionless flow. Is the actual flowrate more or l~ss
than the frictionless value? Explain.
~- f..
;0
COMb/I'JJ~
v:;I.
Oy
_ h( $G/fJ
- 9 5(;011
J _ ('OD ",m)'t
']()O PtM
aJ1d .fnJn" ~f' I wC ha~
1Y (Oo/",,) 1- (S. 7-9;') :. O. o¥z ~ 3
~ s
(tf )
AcAv.~ / I-I(Jw rill<- wdl-1.IJ j,!. Ie.>.! fhol'! fhe.. {y/c,!-/"I',1esJ Iv Iwc /.e-l.4MJe-
!he /05.5 WO lAId. be.. frt..a ky- livtl'l fhe. }e-nJ a""unl t-t-;.eP( a/'ov( .
5-105

5.91 . A~ incompressible liquid flows steadily along the pipe
shown 10 FIg. P5.91. Detennine the direction of flow and the
head loss over the 6-m length of pipe.
• FIGURE P5.91
A.sSllme f/()'-'V /Y(Jm (I) 1-0 (2) a I'td
efut((-,'t:)n (£1- S:BI./) h gef hY- 11te,
c(}n~1 volume .>howl1:
f{
-
0
145
-r}. - f, +
"I
1,1.5e fhe. enedj'J
CQnfe-l?h of fJ,e.
h --~
p, _ ~ + Z - 'l:- -::. '3tNt - I. 0 Yn _ I, S"." = tJ. 5 ~
I J.
(f (f
5_106

5.'1Z I
5.92. A siphon is used to draw water at 70°F from a large
container as indicated in Fig. PS.9Z. The inside diameter of the
siphon line is 1 in. and the pipe centerline rises 3 ft above the
essentially constant water level in the tank. Show that by vary-
ing the length of the siphon below the water level, h, the rate
of flow through the siphon can be changed. Assuming friction-
less flow, determine the maximum flowrate possible through
the siphon. The limiting condition is the occurrence of cavita-
tion in the siphon. Will the actual maximum flow be more or
less than the frictionless value? Explain.
Q ::: Aa Vs -=
'1
frY8
Lf
VB'
To 01,1-0 I~ VB' we apply -I-he ene,,) e~lAah'()n
A anrJ B t'n fh e skefrh above -Iv ()bftA/n
or
and
or
j: v.' jOJO..P.. 1- 9Z~ == ~ 1- .!.. f-
2 " , 0 2
2
~ ::; 9{~ -~e) - los}
1
0
!&/I rhlf
"leI in
c
FIGURE PS.92..
(I)
(3)
('f)
Vc == Z(q, 11 !!:!)(-3ff )f030'/s ",,) 1- (101000 /'I -122B N )
s.. ~ - ii '" n,-z lh~ ::; 9.0'18 !?!
(999. 7 ~)(/ ~ ) s
(c.on't) ,.,J 1t,,;:,
~-I07

b. q1... (con If )
z.
Q = Ac ~ = ~c ~
if
we have. hv Ik- mtLX/Y'1/Um fitJW~~ -fhyoVfJA ih~ 5//hoY1 ;
1. 2-
Q. ':: '!YO /Y1.) (f).1()ft In) (tJ,Olf8 P1') "; ~.)8.kIIJ-:J ~3
If (Iif'! Ii,. 7.) fi S s
7'P'
With E1f. J tlJ'lrj ¥ we Co~clkde Mal Pny los! WOWd pt.f Ir;
/ovJ-e,.r the. va//l..e of V /11 Ik SJf;hIJYJ OYld I'1w"f maJ..e. fJ"e
Oi/IvtJ max/mum f/owmk wi#! fy/chon /e.fs fft4l'1 IJ,e ~axl"mlJl'VJ
flowr~1e wi/11out {Y/cfi't1n.
5-/08

5.93 A water siphon having a constant inside
diameter of 3 in. is arranged as shown in Fig.
P5.c;g. If the friction loss between A and B is
O.5V2/2, where V is the velocity of flow in the
siphon, determine the flowrate involved.
10 c/elel'hJlne -fhe /Iowrak I Q/ we Use
Q .: AV = 1t11. V
J/
3 in.
FIGURE 1)5.93
(/ )
4ft
i 12 ft
4ft
1
To obfo/~ V w~ apply fhe eneyfY e~lAAj/on (E~. '.i2.) 6elweeYl
po/~ Ir A anti 0 I;' IAL J/aJd a &0lie. • 1J-..MSI
/fOr v/+ F, ;4:4:fJfl 7- ~hoff
nef- ,-.,
los.!
tJr
~ J -Ill - (). '8 V
].
2
v = 9(~A - ~II) - f6. 9 ff- = -0, q o.q s
OVId wifl1 £Ij. I
'2
{I 0.1
~ -=
1Y ('3/n .) ;f) O.l3 {fl::::
~ (I'll/- '!:....) J
ff'~
£"-Ioq

5,95 J
5.95 Water flows through a vertical pipe as is
indicated in Fig. P5.95. Is the flow up or down
in the pipe? Explain.
r(AA
r
'-"-i'~
• I
I I
I I
H I I
1::I I
.:::r!:a!i;;;;
"""""Mercury
~
h
I
FIGURE P5.95
The- ~onIY()1 vtJ/ume 5hown 11'1 fhe skefc-h ah()V~ i.! uset/.
Fov Sf~t2.d'tl ;"nCQM.fJlessible flow dowJ'!wa,rd .fnw.. (AJ fD (8)
we Dbfelin ~ ft. s.n
2 1
Ps 1- ~ f ~ r ~ F;:; + ~ 1- g~A - losS
r' 2:' fI to 2 A 8
~ C~.5eYII(lL·l(}}'1 t)f Wlfi>$ we (onv/llfde fhaf
~ = V8
Tnu s fi..urn ~,. I
loss = q H +
A 8
J./tJWever fhe f?1til'JtJmefer ej1Aah'un (5ee Sec-lion 2.6) y/e/ds
~;;; - g[ h(l- S't,)-H}
[JJhic--h is a. ne$tlfive. ~t(al1lif! .5/nce SG
lij
~ /3.' . /I ne.Jt:i.f/ve
j() S ~ is rJ(}f fJhysi Cd /17 pt1ssJ6/e So -fh~ flow /'nMsf be
vpwllyd ~ 8 -k A. Fur ~wtJ/rd fldW fAe tlbt1ve anti;SiJ
leads -Iv
B It;5SA =!Jh ( 5"GHi - 1 )
wh;ch , of p()Ji five lIlY/d fh~Yc 1rH--t. pAj'si CA 1ft p"ea.5tJn.a6/e.
5'- 110

5.96 A fire hose nozzle is designed to deliver
water that will rise If 0 m vertically. Calculate
the stagnation pressure required at the nozzle
inlet if: (a) no loss is assumed; (b) a loss of 30
N'm/kg is assumed.
1b dele¥mlhe fhe- rlaiJllan't7n ptt:sS~ af ~ MJJk Ih/ef we.
aSJ ul'J1e ilia! fhe. 5ktfMt:l/;'gy, ples.fUle aJ- htL ;1()]}/e exif if
& Same af fl"e. >1a5Y141ftfr' tY'tJJute a/ fh~ nOJ]k ;h/ef Clnd we
apply fhe eJ'% ~J e,.,.".alJPYJ (f,. S:i'l) -/7; Ik -f!tJw ~ h
)203Jlc ex/t iv t1u.. i1It~o(;muYJ-t e/~ vt;"nift" of IN. w~ ~IAN -,0 ~el
~ -.:=. ?f l1 ~ + 17{I0$$ ) (I )
(a) ~ ;110 IO~J ., ~~. / /eods -Iv
p :=
(9. '(D 1<AIl'l(J "" ) -=- "?q1- -kN =S "2 -k..PC<..
0
n,1 ",-z.
(b) Fw /()SS - 30 tv. 1'1
&t.! y;'eldf
*7
.)
s:.lll

5.17 For the 1800
elbow and nozzle flow shown
in Fig. P5. q7, determine the loss in available
energy from section (1) to section (2). How much
additional available energy is lost from section
(2) to where the water comes to rest?
6 in.
12 in.
.j:'" (Irrt,.,I
- - ... V6h.tme
"
:YLI x
I
J
Pl = 15 psi
Vl = 5 fUs ~ u.---l.~----,~~
Section (1)
FIGURE P5. q7
fOr so/v/ny fhe II'sf fClyf of fhi.J jlT()bleYn,l the [(};rfrol 1/()/tI~e
&h(}wn t'n fhe sl::eh;h ahtJlle 15 lIser-1.. To defe1"nriYJe the /1J~5
acct!n1.p()lIYI7 flow fYl'/YJ1 sech'P>1 I it; Sec/ion 2. E9' 5". 71 Cdn 6eUJfd
CiS' .jfJ//OW5.
~- P-z.
-;0 '2
5',';'ce x - y C.()oyd,;'",ks
15 hot,'},,, fa / a",eI 2, - ~2. : o· A/so.; ?Z'::' ~h.,:: 0 ?'si.
Fr{frlA fh (! C-I)-1se rva,n,,1'l of JI'Ylflf! PJ'/;u./ple vv~ (..QYIc/lAtie Iha f
ThU5
oY
11 :::2.
(loSJ:;..
/055
I ;l.
v,f;:j~
2. %
{(I -(iJJZ
::
::;
- - 5/U!
For fhe second ptWT- tJ{ th/S' ,roh/en., we c~n5;de" .Jhe Ilow 01 p..
flu/d f",yh'c.le. ~ Jec -hOYl 2. fo t:f .$/nk o~ mf;, 4'£8- ~ 79 /f!Atis -1-0
jtJss
2. tt
Nofe. T1ta:f
7hllJ
5"- 112

5•.1g An automobile engine will work best
when the back pressure at the exhaust manifold,
engine block interface is minimized. Show how
reduction of losses in the exhaust manifold, pip-
ing, and muffler will also reduce the back pres-
sure. How could losses in the exhaust system be
reduced? What primarily limits the minimization
of exhaust system losses?
We. an1y -f~ e)1~J ~~'I/Y) ( Eb· ~. I)) h f/1L lJo~ ~ Me.
en7111e bloc/<. I ex-haus t YJ14J1Jift;ld I~/er-kta fp IN- ex.l1auff J!lJ~
eXif fo get
i;"r; :::; ~t.tf- f -r ;;(toss ) (I )
Wifh E$. J lIVe Y(,ducfJ'oYl of loss /)1 !he -exfJa/,o-J.
5y>j-&w1. resulh In a.. ItJlN'er valtle of
P;;, OJlld fh" J l11e enf'he.
b().cJe.- p1ers[JYe. . LtJsse J In ftt(.. exh~.ff .J)'f/4", C(J)A-IcI 6~
re.tiu.ced by el/wlI·Y14. rh' !J mVt7Y los.J UI Wt(1t)}1 ~n/f .fue/' ILJ /he
Caj-~lyh'c UlYJ I/"W fer- and ike M/,(ffler ()J /~ ()f~ done
In race CiJrs. However, noise Qi1d emiJ.r/dns le!ps 1~/i(/)1 //m;fs
fhe e.x. k YL f 10 withh /"hiJ ,k/Vlt:t of /~.fl ~cf,'()1'J can OCC/..IIY"
In con venh'onaJ veh/cler. S~P1.C IOSI rec!.MciJOn can aflo 6(("("('v
by ((JIl hjUY'":J !he exhatlJ f .fffJen, P/{J'ny wilh few 6eHd.s
OJ'fd ~I'YlJjJYiak area clis~"6ul"()nf. Howetl~,
ve..T",nyefr/ el'rh of~ /et:ulJ Iv behci.r aJ?d t-UYJ'}.J" I" fhe PIPI"
(Ina CO.rff /;mif the exhxd of ofJf/"Yl/j/~ C/J"eC( diJIn·bu//d)'Js.
5 ... / /.3

5. 91 Water flows vertically upward in a cir-
cular cross section pipe. At section (1), the ve-
locity profile over the cross section are·a is uni-
form. At section (2), the velocity profile is
(
R - ,)117 .
V = We -R- k
where V = local velocity vector, We = centerline
velocity in the axial direction, R = pipe inside
radius, and, , = radius from pipe axis. Develop
an expression for the loss in available energy be-
·_£.s~c--fj,n. ('2.)
I
I
I
I
I I
-f~sedj~I'(i)
fl 6!AI
Fay- deI-e¥WJ/~;~ /~Sf we use. -tlte, eneYJJ eg/l'lafh'on M 110YJ-
(jYli-/Qym fld'WS J €g. S.i7- ThuS' /
~-!{
-2 -:to
loss 0(, ~ - t(2. ~
J(~/- ~)
:::-
+-,/J -t
Z
(; )
~ UfrISeyv4.hfly, of J.'nti. sf (tj. S.l3) I,tk kve
II Is()
/
ha lie
I
/If secho..t2)i
C(' == /.0
/
ve/o(ilJS/nce -fhc.. pyt)-h'le. af .rec/'oJ'l(I) /$ un/Iov"",.
we. sollie
c:r: :::J.-
7J.u...s ~1' /-'
loss -
~1 ' S'.3£
/.00
Yields
~-~
jO
(sec. sol£( nOn fw p>ob/~ 5./25(c »tvYIoI db/a'n
_"2-
CJ· O~ V; 1-!} ( ~I - :Z"2. )
2.
5. 100 5, J(Jo Discuss the causes of loss of available energy in a
fluid flow.
50me cc("uses of loss of Q.Vd.lllA.b/,e; fYVIf!H9't In a fluid fJ(jW
,. fj,ic17'01'1
z. heal trCln.!/w acft)ss t1 km,-t-rt1Iure. d/fM8Ha.,
'3. /law I1.CYOSJ' fA Shock.,
5- fiJI-
I
i

s:10 I
Consider the flow shown in Fig. PS.91. If the flowing
fluid is water, determine the axial (along the pipe) and normal
(perpendicular to the pipe) components of force that the pipe
puts on the fluid in the 6-m section shown.
0.1 m
[)5/~j f1,£ C-fhlm / v(}/umt. ,$41wn bt .:1 _.:
hrD~~ I;ne.s we '~l'Jfl/Y ~e aXJ~1 &Inri - ---:: c'
t
iI
3m
I
,
11.(y'rn1tI/ (.QM!(JneYI h tJI.-/he I,'",ea yo n-,()meHIurVJ e-rlA..Q;hfn.. -Ir; get-: .
Z ~ :: IJ $i;'ce. t1,e~ i.l no mflY"le,.!u"", flow I~ !Itt 1"1(/~nfA/ rlJlech;..,
PY1d 2 F ::. 0 ,fl;"c,e the flow i.s 4.$suI"'JeeJ. f~l/tf de",eUJp~P/ a ~CI the
So
A he f- ~Y'1(JIM-lf- (Jf Qx/~I J.,,'~cf/(J n ~'r11enlf.1lN1 {Io IN 0{,.#1 of
& CV if ~e~
I? - Wcos 8 ': 0 OV ~: Wco.rB
N 2
W = n! " "4A.R = )' ~d '1 -f!.IX1/~l'--(17.'",)"'&..'.';= '162 tJ
_ l o r
fJ :. SI';, ';::. /'1.;-,
f<N .:: (ifb 2N )(~f /9.S') == ~ N
f!W- ~ 4X-/~J cJ;~th~
~ Al. + RA of W>';' (7- ~ II, = 0 00;
R :: DI} - D A 1- WSJ;' (J =If:-f1.)11 .f- W>I:' t9
}q r, I r-z l. ;
~ the- ,..,1( ~omefe,y ~ad/~.J
f{'= >h)... ~r;( f: =- d' h)
fJ,us
I1-P2, = ~ (h;- h2 )
Pi.".,c/
~ =t/(h,-h2
)f) -(Wfln Irt.SO)
~ =r.cx1o':')(7.0 ","- o.s:Jtrt·I"'J~ (r62 Nj(.f;;'j9-5°j
R~ = 32 N
5- }/5

5./02
5.102 Water flows steadily down the inclined
pipe as indicated in Fig. P5.102. Determine the
following: (a) The difference in pressure PI - P2'
(b) The loss per unit mass between sections (1)
and (2). (c) The net axial force exerted by the
pipe wall on the flowing water between sections
(1) and (2).
w/UfYte
t6 in.
*
Mercury
(fA) The. d;ffeyeYlce In pr'eSSuf'e.; ~- P2, J J')1~y be. obfa/~ed fY~ -the.
WlanOYHeter (see Jecf/oVl ~. 6) with tAe fiLtid slllt:Nc.J e~lAll.fiol1
p.- P- :. - 6' f(s it) S/~ 3(JD 1- (6 in.) ] + ~ (6 I~ .)
or- 2. H2 o l " (/,;Z};) ~~ (,;).f~)
D P. := -42.4 Ik)fSff)Si"J11,o-f (0.50) 7+ ~3.6)(62..'f'&)((J.5.ff)= 2]7'1,1/ - 2.. '" f.I i j f-l'J -f+3
and
(h)
P,- P, = 2~7.lJ2. I ': /.65 f S /'
I ) . f+' (lw~~
The IOSf per ".,if ,"7,;) be!weeh Sedi."I (J) IimI(2) "'''1 be 06iJ~1'4
wlft, EZ' ~. 71. Thu5
R -= - 77. 2 110
)(
5'"- II 6

s:103
y
5.103 Water flows through a 2-ft-diameter pipe arranged
horizontally in a circular arc as shown in Fig. P5.IO 3. If the
pipe discharges to the atmosphere (p = 14.7 psia), determine
the x and y components of the resultant force exerted by the
water on the piping between sections (I) and (2). The .stea?y
flowrate is 3000 ft3/min. The loss in pressure due to flUId fflc-
tion between sections (I) and (2) is 25 psi.
Section (2)
Section (1)
FIGURE P5.103
TO defeYYJlJJne fhe X 4nd Y c(}ht{Jonenh of lite teru/fan"; -£1'C~ exerlep/
"y ihe. w().fe Y On fhe. p/YJJ~ be~eYJ sechdn (I) and(2.) we lAS e the
X and. y C()yY!po}1eHIJ 01 -lite. I,j"eay mOYnenfu"1 e$uaf/OY/ (Etj £Z2) .
fi';r fhe. CtJnlrtJl vo It/me Con-I-~,'ni~!he w(:('/ey I'? the p;r klwtey,
secf/~YJ (,) 4nd (2) / £1' 22 le~d.f fo
~
R><-=-~It-V;fQ. =-RlI~ -!{f1~
and ¥
Ry -= r:....4,4 7- ~ f{} (2)
7he resu /!-aMi IoYCe c.ompo)"Jenl:r In Elf. / C'R1t1d 2. att exeykd by-/JrG
pipe ('h1 -jh~ vvCt~. l/-t~ Yl.!tdktYll fir.u. of wah" lin p/,e if ~/ltd In
mayni ludt hi-fi o,,,aldr:. ;'" dl;~cf/oYl.
To de{oymln~ PI we use. the e11e,.-~y ~ua lIun) Efj. ~ Kl. 711.UJ/
~ = ~ (loSS) ':. :25 fS i
Also
~ ~ ()
A,
::
().nd
Rx =_11.,850 110
(J..J.1d the. x dlt'~c-6~n GtNYtptJYlenf of ft,e, for-c.e. exe,!ed &7 ~
watw On -/1t.t p/"e bd~eey> se..ch'~J (/) tlMrJ (2.) Js +/2,850 110.
(con'f)
5- 1/7

.oS ,J 03 I (GOII ' t )
{)JiM t;~. 2 vvc. ob-/-4t"n
~ =(1).92 If) (ql{. r!!!1!) lJooOft)t ~ .) J ~ 15'10//;
Y s . ffl / ( ~ no.. ~/"J'::)(lo ~
CU7Ij the y- d J;let;H~Y) UJmpoYJenf of In-e- mil?
fwee eXf!/¥kd b'l the. walw ~Vl the... pIpe be~ sechdn.J
( I) aY! d (2) t5 IS ifd lb.
~ -//8

s. /041
S.lOLf When fluid flows through an abrupt ex-
pansion as indicated in Fig. P5.10~, the loss in
available energy across the expansion, loss eXl is
often expressed as
loss = 1 _....!. _I
(
A)Z VZ
ex Az 2
where Al = cross section area upstream of ex-
pansion, Az = cross section area downstream of
expansion, and VI = velocity of flow upstream
of expansion. Derive this relationship.
f
I
~_ _~. I
~~ {j ,'~ I -() II I V I
Sectiln(1) II / ;'
(1t+(t'",a~ '.1 S:ctiln(2)
10c.afi()(I ~f
.u.dio"., (I) FIGURE PS.I0if
Apply/nJ fhe ene,.,'1 e'tIA~h'()Yl (r=IJ· ).!Z) fo the fl()w ~ Jech'qn(J)
to secH()Y/ (2) we. olofrA/n
/OS5 ::: ~-P.. +
ex -
;0 I I
APply/)'},J f~e ax/a.
c~ IAA-HtJ Y) (~~. 5· Z 2.)
fr-gyv, sec -hoY! (I) -Iv
,. l
1-11
-~
2-
d,~echfJ n C. Oi'VtptJne nf
W ~e f!iA.,'d ctP'l.frv~ eo!
se c. fr()yt ('G) we. 010fa I~
(I)
()f f/'e /;~etlY" mIJYhel?/z"""
'n fhe. UJh-h--D1 voluYl1t
Rx f AAI - P.. A). ~ - ~ f AI l-j + l( fAl. 112 (2 )
NOw, ;f we.. consider sec.+io'VI (J) as tJCCUYY'lry af f;,e evrd
of the. 5ma Iler di(lW/e~ p/pe (-the be91;'J1J~ of fhe /(A'rger
d/(ll'11eiw pipe) uS 1~d.iCA.1-ed In -IAe.. skid. above I £r I fnll
yields fhe C~tJn.J/oy, /oss ~nd ES' 2. 1oe~.J
Rx + P, A2. - P'1. A" - - ~ t'AV, -t- ~I' ..42 11:1. (3)
Nofe th~f- wife, sec.flon (I) ptJsih'tJY/ed at- /1,e, QJ?d of fhe Jma.lleY'
d/4me/e¥ flfe~ .t; ().dJ ovev- ar-e~ .A2.. Also, b(!CtUt.Je of fj,e
jef HdW fy~ fl1e.. S"Ynalle~ d/a.~t-eY' pl;"e- Info f),e /orfe-r
d/(lAeff!Al" p/,o-e) the Va./La. of ;fx /AI/II fpe sYna/1 eJ?4k,jh (.PwIf'4~tA
ft; ft,e. olher ie"-m.s- 1;-, E<e. J 11t~ f ~ can dY()jO Rx' F~ E;.3
~ - P"2.
-;0
L.UYYlh/YJJ.,
joss
eo(
::
~
V ~- V A,
2. I
A2,
(f)
5"-1;9

FrI/WJ CtJY?.Jervah'pn of }'YJIIISJ (&I. s: /i) we have
II = I/. II,:l. J _
A.".
~h/1'117 f1S. > ill nA 6 we ,er 1
/OJSex = ~(A~/ ~11)f _l{_'2.__~_~_A_~~)
t:Jr
f ·/OHe)( ~
~ 2((21 ) 2 A,
Z til. liz.
(). Jll() '2..
/oss~x . '= ~{(- ~.)2- A"2,
5./05
5.105 Near the downstream end of a river spillway, a hy-
draulic jump often forms, as illustrated in Fig. P5.105 and Viu,'()
V10.5. The velocity of the channel flow is reduced abruptly
across the jump. Using the conservation of mass and linear mo-
mentum principles, derive the following expression for h2,
h2 = - i + (iJ+ 2VgIhJ
The loss of available energy across the jump can also be de-
termined if energy conservation is considered. Derive the loss
expression
2
+-1
it,')}
section{I }
FIGURE (15.105
A'I'//ca;f/oi'/ of !'he. h()~Jjoj!J-/r,r1 cOWLl'lJnenf of fhe. //neay hllJn,el'1iuw.
etjt<~h'(JYI (Ef' >.22.) -fo the. wale,.... In fhe.. (dYl~1 Vo!un-tf! -hw.
see-tim U) Iv se ch'(JY7 (2) /ea.dr fo / -&- Un;f w/dlh ~f' FIII'W/
h2. j
- RX + '0_, _ r h~ = - l{l'h, ~ + I{ J:) h:L ~ (I )
~ ;: r
(con'-f )
5".. /2-0
we dYf)p Rx
dlolaIn
(:;;.)

5105 (COy) I t )
/ and 2. we ololtJI;'
/; ]_ 2 V I
- 9( (&)-1
hi
(
2. ( Zv'l.
!:~) + h~) __,
h, 11, 9"',
= 0 (3)
2.
Jtll'YII /IS! :: g
'Ih"2.hJ
5=-/21

5".JOb
5.1 % Two water jets collide and form one homogeneous
jet as shown in Fig. pS.I06. (a) Determine the speed, V, and
direction, 0, of the combined jet. (b) Determine the head loss
for a fluid particle flowing from (1) to (3), from (2) to (3).
Gravity is negligible.
1'1 = 4 m/s
• FIG U REP 5.106
f:Dr +he wakr .flOWing throll9h the. tOntYl) volume -sk(.ttY!!td o.ioov€, the
x. - tAYcA 'i- dlt(.dior' GOYYpofetn ()l -1,e. lir'u'f rYOMe",h,tM ~cAo.tiOl- ate
- ~~~ V2. A2. -t "1>CfJS (8 ~J3 Af =0 (I)
Q..VO
- V, ~VI AI + v~ Sin e fV1 A3= C (2)
H-oy) -the. <..Ons-enb-hOt' of ~"S s p.,.inc..~pe we. qd"
- ~)J AI - ~V2.A"l. + PV3A-?, -= 0 (~)
COY'~"I)'',,,a f=a..s. l aVId 2 We.. Obktl.-, "1. r.' ICJ. 'l.
JP d ' ) ~ITYI)
+o.V e = v;-AI = Vl - I (~ ~ '
-:7r7A ~ :. -----:...-- -= 0 ,';os b
V'). '1 V ndl. ( I' ~ '2..
~ " 10 i-) 'IT~l.m)
'1
So
NOW) c.ofYI~;n;n'3 f:~s. l o..nc:l:; we. "Ie.t
- J~~A2.t V;Cos e (~j,Al T f V'J. "2.) ==- 0
( c..on'i )

(4 )
(C;)
(6)
anc!
felk of (05$ :;:: S"22 -S
~#" 123

5".1 D7
5.107 The pumper truck shown in Fig. P5.107 is to deliver
1.5 ft3/s to a maximum elevation of 60 ft above the hydrant.
The pressure at the 4-in. diameter outlet of the hydrant is 1U psi.
If head losses are negligibly small. detennine the power that
the pump must add to the water.
Hydrant
II FIGURE P5.107
h = ~."3 f"l-
s
.
W.rhtlf.f
her in
s- /2/f

5:}08
S.lOS What is the maximum possible power output of the
hydroelectric turbine shown in Fig. P5.108?
6 mls
Turbine
III FIGURE P5.108
lOr IIDw .(;.()fY1 sech'()n(t)-h; sechol'tfz.), yt.5.i2. y/e/cls
J.;. :2f- 92}. : !5 + '12 +- Q ~ t W~hq{t - loss (; )
r . - J O . , . / I
f ~ 1'1(1 I;"
5/nce P = I? =/!,.--L/ 2. r,vn,)
w =-wfhq(r SiJqH
7htl5
and
•
W;i1alf
IJe{ ou f
llIaXihluhl
,
-W5haH
hef bfA..f
.
W;haff
hef ()IA./-
mil.x/,."£tJNI
J1ef in nel .,d
'2..
9(i, - ~2..) - y~ - /t;'sf
2
,
MW =
sltt,1f
lief ou.f
~/Jxirnl.t"'"

5.109 Estimate the power in hp needed to drive the main
pu~p of the.l~ge-scale water tunnel shown in Fig. P5.109. The
design condition head loss is specified as 14 ft of water for a
flowrate of 4900 ft3
ts.
Test
section
II FIGURE P5.109
7he Sf) It,{ -hi;VJ (Jf fhif prob IeI't1 is fimi /a r ~ !he Ohe
of t;x~fJ/e. i'. 6. L00I'I~ p.rtJunc/ iJ,~. wafer tul1l?e/
~ &iy Cn;of.f I"et//jan t7F- the. -iunY)e/ bacl -It> 1/,e
5ame, cl7JSS -S"t;Jcn"rfr, we. UYlc!ude WI;'j ~fhe. ener,fJ
e.f (,url/tr;,) GS' ~ BIf
hs = hL..
The y? ~ E8· 5: t>S
=J, r C<f
.
W
S"hq/f
nef I;'
5-125

5, I/O
5.11 0 Water is supplied at ISO ft3 /sand 60 psi
to a hydraulic turbine through a 3-ft-inside di-
ameter inlet pipe as indicated in Fig. PS.ll'"Q. The
turbine discharge pipe has a 4-ft-inside diameter.
The static pressure at section (2), 10 ft below the
turbine inlet, is 10 in. Hg vacuum. If the turbine
develops 2S00 hp, determine the rate of loss of
available energy between sections (1) and (2).
Section (1)
P.1. = 60 psi_
Q = 150 ft3/s
VI = 3 ft
...,...~~
10 ft
L P2 = 10 in. Hg
·vacuum
D2 = 4 ft
~
(I )
2-
=(z /. 22 If) (3 If) :: II. '1'1 t!
s (Lf 1+1 .5
FrlJW1 Gf. I
power I()s!
(
5'50 f+. /10 )
S.hr
-t' p2.1- f! )~(J f1- J(I 110 _
. s~ slu!. If )
- 2500hp
sa.
pllWe¥ t()ss": 30 I hp
5"-/26

5, /1/ 5.111 A steam turbine receives steam having
a static pressure, PI, of 400 psia, an enthalpy, hI,
of 1407 Btu/lbm, and a velocity, VI' of 100 ft/s.
The steam leaves the turbine as a mixture of vapor
and liquid having an enthalpy, h2' of 1098 Btul
Ibm, a pressure, P2' of 2 psia, and a velocity, V2,
of 200 ft/s. If the flow through the turbine is
essentially adiabatic and the change in elevation
of the steam is negligible, calculate: (a) the actual
work output per unit mass of steam; (b) the ef-
ficiency of the turbine if the ideal work output is
467 Btu/lbm.
(tl) This pmblem If fimi/ay ~ Ex.am/,/e 5:21.
=
11: /'1-07 ~ _ /Ofl KIw f
2. l.
(loa fj) - (200 f!)
/,,~ Ib~
2, ('2.'). Ibm. ft)(771 f.f.,j,)II, . s 2. flhA
= 308 '!i::-
=/hm
(b) /I reaS()fllJ,hle e,(!-/ciency 15 /he rtlfit) of ac&aJ worJ. ()ulpwf
it; Jd&.ll I/V'lJ rk out-puf ()y
J()B 8!!3
___/~"I'f'>;..- X 100 - 6610
'It? 8&
Jb;..
5-17-7

5//Z I
5.11 '2. A centrifugal air compressor stage op-
erates between an inlet stagnation pressure of
14.7 psia and an exit stagnation pressure of 60
psia. The inlet stagnation temperature is 80 oF. If
the loss of total pressure through the compressor
stage associated with irreversible flow phenom-
ena is 10 psi, calculate the actual and ideal stag-
nation temperature rise through the compressor.
Calculate the ratio of ideal to actual temperature
rise to obtain efficiency.
We aS5ume -II, 0..1- -the air compressdr Opera re.J tJdia6afkal/y.
fin ideal Compl"e~$;"Y1 pyoces~ is fl"/c fitlnleSf IAl1d (;l.d/abalk
()//Id thus ~cc.oyd/~ +0 Eg. 5".101) if is IlL' ct/J15fdM.f e;?1YOfl1
Or /5e//l /Y~pic.. pYOC. e5S. Wi fh £9' 5". It)! we p.,1$t) c.I)JI'! (//ude
fhaf an aclu~ I ad"abpn'c. CVWt/J;IC$$/()YJ pY()ceSJ witt, rl"/c.h'tJYJ
IJ1tJsf IMVO/VC- aM en~y In~YCIl.>e. tJJ1 femfJe,yr,tMl -141:/nJpy
cdo/d/Ylate.;; fhf!., idt!/A I al'ltl /Ae-fJ.t.a1 C~~I/~$,ti()'yI PYPC.ti!CS
()PI'ecw a5 l~dicak<J In -the. ~I<f!fd" 6eltIW. A/~o shMn i5
ft-r(.. /0 jJfi J~$ I~ S ~f"1al"()Yl pY~S$ uYt! dlA~ ~ fYic,IJ()J'I.
70psitl.- :: ,Po' '::
___...,.~_~;......~_o!f:JA. ~ ~ 2, tfchtJ ~2, ideAl
"2 4cl-wJ.I, ,
-----.,01'--...-.--7; 2. id~(J.1
J I
I
I
S--1z'8

5. //'f
5.11q* Total head-rise values measured for air
flowing across a fan are listed below as a function
of volume flowrate.
Q (m3
/s)
o
0.14
0.28
0.42
0.57
0.71
0.85
0.99
1.13
1.27
Total Head Rise
(mm H20)
79
79
76
67
65
70
76
79
75
64
Determine the flowrate that will result when this
fan is connected to a piping system whose loss in
total head is described by loss = KLQ2 when: (a)
KL = 49 mm H20/(ml/s)2; (b) KL = 91 mm H201
(ml/s)2; (c) KL = 140 mm H20/(m3/s)2.
7he II~wy-ak
of !ttL Ufrn6/i?af/tJYI of a.. -IaJl1. tN pump an4 IL uJ1l1eckd
f)fJll1J sys~ IS qe/~)IJIJI'y]ed by /J1t.. /nwsecl-/",., of /h.e
fan Oy pump head y/u. tis· V/J(Uhle.. fl6NY'tLk- utrv{.. ~l1.tJI I1te
S'ls~ /t/H vs I/olul'l'le. .f/(jyo/ yak... uo've. 7~ dehm/J1e I1u
flO'Wya.k Y'esu/~ Wh~ !It.t.. ~ of !tz,/,J 'p~b~ /{ ~nYleck~
f?J Il1.L ft,l"e~ [( aJ(b) ~(c)] p/P)i?y sys~s / Ik. )j.,/eysec-f/Ph.J
<:J f fkt. III111'2} .sy5'~ / tJ S S vs. b( W YII'f..J ()i1 ~ flu. fan
-roW hc~d yise... v.s. ~ UuYlle fil- weye de/wY11iJ1Cd
wirh I1u U/V'J-tfJlA./e¥ pn;rY~ liJk,l On ~ ~?/IIII'W)~ ftifeJ.
A p~/yn()mia I /etl,rf fjuwye.s 4lyl/e. f/I ()f ~ /tJ..6ula..k.d
cL~ /5 used· 1k- jY1/e-rsech'()"¥I pd Jj;,f> wey~ dejeyJl"1/~ee:l
1# ittt I1v.. Ne.w+oYJ - I(O<.f/hJdYI f-e. c.-~ ;'; 11/ e .
(UJn't)
~-/30

5.//1
100 CLS
110 PR ~NT "****************.***********t;************************:t*"
120 PRINT "*:t, This program determines the intersection of the **"
130 PRINT "** head loss and head rise curves for problem 5. 11lf *:r."
140 PRINT "** A least square fit polynomial of the form: :t.*"
150 PRINT "** Y = dO + d1*x + d2*x A
2 + d3*x~3 + ... t:*"
160 PRINT ":t:.t: is used to describe the head rise dat.a. **"
170 PRINT "*****************)!'************************************"
180 PRINT
190 DIM B(21), DC2l), S(21), XClO1) , WOOl), YOOl), FOOl)
200 DIM ERRF (01), PJ (101), PJMl (101), YBAR ClO1)
210 '
220 'intialize t.he variables
230 NTERMS = 8: NTERMSAVE = NTERMS
240 NPOINT = 10
250 INPUT "Enter the head loss coefficient."; KL
260 PRINT
270 FOR I = 1 TO NPOINT
280 READ XC!), YO)
290 W(n = 1
300 F(l) = YO)
310 NEXT I
320 DATA 0.00, 79.0, 0.14, 79.0, 0.28, 76.0, 0.42, 67.0
330 DATA 0.57 , 65.0, 0.71, 70.0, 0.85, 76.0, 0.99, 79.0
340 DATA 1.13,75.0,1.27,64.0
350 PRINT "The polynomial fit to t.he head ris:;e dat.a is of ordey-";
.360 PRINT U'::~ING "##"j NTERMS - 1
370 '
380 'determine the polynomial coefficients
390 PRINT "The coefficients of the polynomial are:"
410 FCI) = FCI) - D(NTERMS + 1) * XCI) - (NTERMS)
420 IfEXT I
430 FOR J = 1 TO NTERMS
440 BU) = 0
450
4()O
470
DO )
E~ (J )
NEXT
= 0
= 0
J
480 C(U = 0
D (1) D (1) + F (1)
*500
~,10 EO) = B (1) + X (1) .*
520
5::::0
S(1)
NEXT
=
I
S Cl) + WO)
540 DO) = DO) / S(1)
560 ERRF(l) = Fer) - D(l)
570 NEXT I
W(l)
W(1)
580 IF NTERMS = 1 THEN GOTO 850
590 B(l) = B(l) / S(1)
6 1 () F!J M1 ( I) = 1
620 PJ(l) = X<l) - B<l)
630 NEXT I
(un 't)
~-.I31

(~n't)
640 FOR J = 2 TO NTERMS
660 P = PJ(l) *' WO)
670 D(J) = DeJ) + ERRF(I) * P
680 P = P * P]CI)
690 B(J) = B(J) + X(I) * P
700 seJ) = S(J) + P
710 NEXT I
720 D(J) = D(J) / S(J)
740 ERRFCI) = ERRF(I) - D(J) * pel)
7e:-.;0 NEXT I
760 IF J = NTERMS THEN GOTO 850
770 B(J) = B(J) / S(J)
780 C(J) = se]) / S(J - 1)
790 FOR I =- 1 TO NPOINT
800 P = PJel)
810 PJO) = (XCI> - B(J» * PJ(1) - C(J) * PJM1(l)
820 PJM1(I) = P
8,30 NEXT I
840 NEXT J
850 PRINT USING" d# = +#.####-"~~"; NTERMS - 1; D(NTERMS)
860 NTERMS = NTERMS - 1
370 IF NTERMS ) 0 THEN GOTO 400
e.r:,o '
890 'determine the intersection using the
900 'Newton-Raphson method
910 qNF = 1!
~)20 QN = QNP
930 F == O!
940
950
960
970
980
990
FP =-
FOR
F ==
NEXT
FOR
FP =
ot
I =
F +
I
I =
FP
1000 NEXT I
1 TO NTERMSAVE
D <l)
*' QN
,.
(}
2 TO NTERMSAVE
+ I
* D (1)
* QN
1010 F = KL * QN ft 2 - F
1020 FP = 2! * KL * QN - FP
1030 QNP = QN - F / FP
STEP 1
- 1)
STEP 1
.,
(l -
1040 IF (ABSCQNP - QN) > .0001) THEN GOTO 920
1050 F = O!
1060 FOR I = 1 TO NTERMSAVE STEP 1
1070 F = F + D ( I) * QN ft (I - 1)
H)80 NEXT I
1090 PRINT
1100 PRINT USING "Head loss coefficient:
1110 PRINT USING "Volume flow rate ----:
1120 PRINT USING "Operating head ------.
(COn'f )
5"-1.32
###.##"; KL
##.### m~3/s"; QN
### rom of H20"; F

5./1'-1 (Con't)
***************************************************~**
** This program determines the intersection of the **
**: head loss and head rise curves for problem 5. 11'1 ~:*
** A least square fit polynomial of the form: **
** y = dO + dl*x + d2*x-2 + d3*x-3 + ... **
** is used to describe the head rise data. **
******************************************************
(e) Enter the head loss coefficient? 140.
The polynomial fit to the head rise data is of order 7
~he coefficients of the polynomial are:
d7 = -1.7369E+03
d6 = +8.2623E+03
d5
d4
d2
d?
dl
dO
-
=
=
=
=-
=
-1. 5353E+04
+1. 3788E+04
-5. 9543E+03
+1.0551E+03
-6. 2329E+Ol
+7. 8983E+Ol
Head loss coefficient:
Volume flow rate ----:
Operating head ------.
140.00
Q.705 m"3/s
70 rum of H20
******************************************************
** head loss and head rise curves for problem 5.114 **
*t y = dO + d1*x + d2*x-2 + d3*x~3 + .. , **
**~*******************t*******************************
(b) Enter the head loss coefficient? 91.
d7 = -1. 7369E+03
d6 - +e..2623E+03
d5 = --·1 . 5353E+ 04
d4 = +1. '3788E+04
-' r,
- -5. 9543E+03c.':"
C:2 = +1. 0551E+03
dl = -6. 2329E+01
dO -. +7. 8983E+Ol
Volume flow rate ----;
Operating head
91. 00
0.928 m-S/s
78 rom of H20
(COJ'7't )
~-13.3

(Con If )
******************************************************
** head loss and head rise curves for problem 5.11Q **
** y = dO + dl*x + d2tx-2 + d3*x
A
3 + ... **
*****~************************************************
(al Enter the head loss coefficient? 49.
d7 = -1. 73~59E+03
d6 = -+8. 2623E+03
d5 = -1.5353E+04
d4· - +1.3788E+04
d3 = --5. 9543E+03
(i~~ = +1. 0551E+03
ell -- -6. 2329E+Ol
dO = +7. 8983E+Ol
Volume flow rate ----:
49.00
1.203 m~3/s
71 mm of H20
~-/31/-

5'.115"
~ lJ)
Water is pumped from the tank shown in Fig. J=~ -~1~ I
PS .1ISa. The head loss is known to be 1.2 V2 /2g, where V is I I
the average velocity in the pipe. According to the pump man-: 6 m I
ufacturer, the relationship between the pump head and the ftow- J LJ (
rate is as shown in Fig. PS .IISb: h" = 20 - 2000 Q2, where, : I 2.)
hi} is in meters and Q is in m'/s. Determine the ftowrate, Q. I - .- - - -,
1_ ,-? - .. i.......t
- - - Pump 0.07 m
VIe wanf Iv 1:.'11"" !he ftowrak b2.
'FrN-!'he- C/Jnl-mJ Vd /41"t1e. f hfJ1/11Y7 )
a/p/,·u/f.htJn ()f /he.. fYherJ:;
(a)
ej1AfA.hlgy. (G~, ).gl() y/ei£i.r, 0
£+ V~2- f- 2-z =7.I- ~ ~I 1-1-h;
(( 2J 6':2J
~ nd 'Z.
h =h ::: 20 - 20a? as p
Since ~::: V2. A1. w<- h~ ..fn,." et z
"l,.
h =: !..:.!: (£)'1 Z9 A
().J/}~ U¥nb:n'j ~.L (/~(])~d. ('1) I/Jt!
).
..L (!E.) -I- -t =~ -J- 2 ~ - 2,.000
2.J Az 2. I
S--/3S
''',
/",
Ii" =20-2000Q2 ,.
'.
oO':-------::--':-:----.i
0.05 0,10
Q, m3/s
(h)
(z)

5.116 Water flows by gravity from one lake to another as
sketched in Fig. PS.116 at the steady rate of 80 gpm. What is
the loss in available energy associated with this flow? If this
same amount of loss is associated with pumping the fluid from
the lower lake to the higher one at the same flowrate, estimate
the amount of pumping power required.
or 1/1 tJ
FIGURE PS.116
~-------:::::---T--------------------------------I
: 5.1/ 7
5.11 -, A !-hp motor is required by an air ven-
tilating fan to produce a 24-in.-diameter stream
of air having a uniform speed of 40 ft/s. Deter-
mine the aerodynamic efficiency of the fan.
The Qeyt)djJ'1dnt/c. efFic./e"yzc,!/ or fhe fan" '( I /5
1 = "dept ?~wey yejl4lrld
acl"tA1 Pd1Alev Ye.IU,;eJ/
~ fAcill1."'/ .rhflff p"wer resu/yM.) WAc.fl.(.::</) IS O.7S hI' .
The. ;dr.al ~h(iff I'0JA/er rtjw;&:/" vV;ClealJ i.1 ()~/;,ed ~ ES' ,.32 hY /low
Wifhou.f loss tl.CYDS$ fire fal1. Thus
• • 2. 1. '2. v3
lN7,.1D_1 :: I'M ~uf :: I'A V J ~uf = I'?r..!!ou+ ~ut =/ tlFr:Jf. 2 ,ut 'ilT 2. - 'f l.
4r VV;ded :: tl. '135 hr
~-136

5".118
5.119 Water is pumped from a tank, point 0), to the top
of a water plant aerator, point (2), as shown in Video VS.S
and Fig. P5.118nt a rate of 3.0 ft3
/s. (a) Determine the power
that the pump adds to the water if the head loss from (1) to
(2) where V2 = a is 4 ft. (b) Determine the head loss from
(2) to the bottom of the aerator column, point (3), if the av-
erage velocity at (3) is V3 = 2 ftls.
Aerator column
• FIGURE P5.118
(a) The enerqyeCfvof/on {rom {/J to (:;.)
~ +JZ + z +h - h ~ P:L +li2. of ~~ 2j I P 'J. 7 z1 J.
w,-fh
A :::~ :: 11 ~ ~ ::: 0 9/ves
hp == hi- +z2.- Z, ::: iff! f(;O+3)ff-slf == 12ff
Thvs, the pump power ;'s J
~ == 't(Jh.s ~ 62.'1- *(3!f ) (JJ-H) :: 22"'15 tt;Ib (5510~)
= '/-.08 hp S
5"- /37

~.IJq
(I)
The turbine shown in Fig. PS.119 develops 100 hp (:=--"-r~"~~~"--:::-- J
L
/(/ (,V,. tCVB
when the flowrate of water is 20 fto/s. If all losses are negligible, I 1 P3 1'4
determi~e (a) the elevation h, (b) the pressure difference across I h I 0 0
the turbme. and (c) the flowrate expected if the turbine were J 1 1 / .:: ::-- /
removed. - - - ~ .' '.' - ......(2)
(C)
I
l
[;"-/38
______ 12..in·_ (-aJ,' T ~~ i~
~ : . J
~---=.l..
- -,' - - - -:I' - -/(
-
- 2 S-·5 li-
..J
- - Free jet
(/)

5. /zo J t Pl = 50 psia
V2 = 35 ftls
-7 1 . . - - - - - - .
Section (2) L
Section (3)--1 .6~
5.120 A liquid enters a fluid machine at sec-
tions (1) and (2) and leaves at section (3) as shown
in Fig. P5.12.0, The density of the fluid is constant
at 2 slugs/ft3
• All of the flow occurs in a horizontal
plane and is frictionless and adiabatic. For the
above-mentioned and additional conditions in-
dicated in Fig. 5.120, determine the amount of
shaft power involved.
-..r;--Section (1)
P3 = 14.7 psia
V3 = 45 ftls
A3 = 5 in.2
Pl = 80 psia
V1 = 15 ftls
Al = 30 in.2 FIGURE PS.12D
Fi:J~ fhe iricfiOY1/tSJ rAJIlri. (JtI/abah'~ flaw +hyoUj"
&j5. 5'.64J S. 65 and S.7' lead -h>
-/-his IlbO(J WlllCh/ne
vir :: m3 (!l + ~2) _ n1 (fl. + I{
2
) +
S'htlff ;0 2. / I fO i )
nef- in
(I)
S/nce
• V . v . v ( . . ) v . v . '" . (., V)rnI UJ - YYJ
2
U).. - YY13 ()3 = m1-+m3 <AI - m.l. U).- n1J ~0;: Yn2 lA,-iA2.
IIf secfioYl (3)
+ n, (u-u ) :: 03 I '2.
~-13'1

5./21
5.121 Water is to be moved from one large
reservoir to another at a higher elevation as in-
dicated in Fig. P5.121. The loss in available en-
ergy associated with 2.5 ft3
/s being pumped from
sections (1) to (2) is 61V 2
/2 where V is the av-
erage velocity of water in the 8-in.-inside diam-
eter piping involved. Determine the amount of
shaft power required.
Section (2)
HGURE PS.121
flow .(y.IIYn Sech'oh (t) -Iv secft'ol'J (2) E~. 'i.iz leads
FrtJVJ.-t the
I :::
.
W
s-liIlH
ner i"
v()/ul'He
bl -=
If
=
flowrale
~
;;:-i/- =
-Lf
loss],: ;at;[!J(~-~) f ", fj (I)
w-e obhl'n
(2.5 {t7)
11-/!.i:!... )z¥ ( 12 117.
fT
£-1'1-0
7.162 fJ:.
.J

~.12 z..
Oil (SC = 0.88) flows in an inclined pipe at a rate
of S ft3/Sas shown in Fig. PS.l22. If the differential reading in
the mercury manometer is 3 ft, calculate the power that the
pump supplies to the oil if head losses are negligible.
~ =
~-Jlf/
12 in.
S" IIi'
.r
--1-
;(: ft)
-.
ZS'·5
(2)
H
--+I3ft
L
_"':::t .. ,::r·,.~r
1'1-
J
6 in.

5.124 I 5.12'-1 The velocity profile in a turbulent pipe
flow may be approximated with the expression
E. = (~)lln
Uc R
where u = local velocity in the axial direction,
Uc = centerline velocity in the axial direction,
R = pipe inner radius from pipe axis, r =
local radius from pipe axis, and n = constant.
Determine the kinetic energy coefficient, a, for:
(a) n = 5; (b) n = 6; (c) n = 7; (d) n = 8; (e)
n = 9; (f) n = 10.
ror -the kinetic. eneryy ctJelhci~l1~ ~ we ma, use Efj. s:3, . 711/,'($,
I< 1 1/ ) fr- 1 I .L
rx = ;: 1f'U 27r;-r)y = 2 b tlJ(f ,,~)= 2 Uc!,f-i)rf)d(f)
IX ::: /.08
(C) For n = 7
(d);=w n= r
IJ( == 1·()5
(e) I=tJv. n::: q
0(' ::: /.0"1
-3
fA
(f) ~n=l()
tr ::::"/·~J
5"-/.'1-2
-3{.(
(Z)
(~)
(5)

5.125 A small fan moves air at a mass flowrate
of 0.004 lbm/s. Upstream of the· fan, the pipe
diameter is 2.5 in., the flow is laminar, the ve-
locity distribution is parabolic, and the kinetic
energy coefficient, (11, is equal to 2.0. Down-
stream of the fan, the pipe diameter is 1 in., the
flow is turbulent, the velocity profile is quite flat,
and the kinetic energy coefficient, (12' is equal to
1.08. If the rise in static pressure across the fan
is 0.015 psi and the fan shaft draws 0.00024 hp,
compare the value of loss calculated: (a) assuming
uniform velocity distributions; (b) considering ac-
tual velocity distributions.
(Jml
loss ::: 3.36 fl-· Ib
5-/Lf3

5.126 Force from a Jet of Air Deflecled by a Flat Plate
Objective: A jet of a fluid sU'iking a flat plate as shown in Fig. P5.126 exerts a force on
the plale. It is the equal and opposite force of the plate on the fluid that causes the fluid mo-
mentum change that accompanies such a flow. The purpose of this experiment is to compare
the theoretical force on the plate with the experimentally measured force.
Equipment: Air source with an adjustable flowratc and a flow meter; nozzle to produce
a unifonn air jet; balance beam with an attached flat plate; weights; barometer; thennometer.
Experimental Procedure: Adjust the coumer weight so that the beam is level when
there is no mass, m, on the beam and no flow through the nozzle. Measure the diameter. d,
of the nozzle outlet. Record the barometer reading, Hoam, in inches of mercury and the air
temperature, T, so that the air density can be calculated by use of the perfect gas law. Place
a known mass, m, on the flat plate and adjust the fan speed control to produce the necessary
flowrate, Q, to make the balance beam level again. The flowrate is related to the flow meter
manometer reading. II, by the equation Q = 0.358 1Il/1, where Q is in f(l/s and h is in inches
of water. Repeat Ihe measun:ments for various masses on the plate.
Calculations: For each flowrale, Q, calculate Ihe weight, W = mg, needed to balance the
beam and use the continuity equation, Q "" VA, to determine the velocity, V, at the nozzle
exit. Use the momentum equation for this problem, W = pV 2A, to determine the theoretical
relationship between velocity and weight.
Graph: Plot the experimentally measured force on the plate, W, as ordinates and air speed,
V, as abscissas.
Results: On Ihe same graph. plot the theoretical force as a function of air speed.
Data: To proceed, print this page for reference when you work the problem and dick ,,~/'(!
to bring up an EXCEL page with Ihe data for this problem.
II FIGURE PS .126
(C!on't)
5- I'f'f

(COJ1'-t )
Solution for Problem 5.126: Force from a Jet of Air Deflected by a Flat Plate
d, in. Hatm, in. Hg T, deg F Q = 0.358 h"0.5, with Q in cfs and h in inches of water
1.174 29.25 70
Experimental
m, kg h, in. Q, ft"3/s V,ftls m, slug
0.010 0.54 0.263 35.0 0.00069
0.020 1.08 0.372 49.5 0.00137
0.030 1.52 0.441 58.7 0.00206
0.040 2.18 0.529 70.3 0.00274
0.050 2.72 0.590 78.5 0.00343
0.060 3.25 0.645 85.8 0.00411
0.070 3.81 0.699 92.9 0.00480
0.080 4.32 0.744 98.9 0.00548
0.090 4.92 0.794 105.6 0.00617
0.100 5.46 0.837 111.2 0.00685
0.150 8.13 1.021 135.7 0.01028
0.200 10.85 1.179 156.8 0.01370
0.250 13.72 1.326 176.3 0.01713
Experimental:
V =Q/Awhere
A = nd
2
/4 =n*(1.174/12 ft)"2/4 = 7.52E-3 ft"2
W=mg
Theoretical:
W = pV
2
Awhere
p = Patm/RT with
Patm =YHg*Hatm =847 Ib/ftJ3*(29.25/12 ft) =2065 Ib/ftJ2
T =70 + 460 =530 deg R
Thus, p =0.00227 slug/ft"3
(con't)
5 - /JfS
W,lb
0.022
0.044
0.066
0.088
0.110
0.132
0.154
0.177
0.199
0.221
0.331
0.441
0.552
Theoretical
W,lb
0.021
0.042
0.059
0.084
0.105
0.126
0.147
0.167
0.190
0.211
0.315
0.420
0.531

s:/2-' (Con' -t)
..0
0.6
Problem 5.126
Weight, W, vs Velocity, V
i
O5 --.--~-- -------~--~---~-.-------~-,-
.
• I
~--I
I
I
. ,
0.4 -----~--i
!
!
•~ 0.3
i
I0.2 --------------~---~--~-_r_----------- --1
0.1 J,
0.0 +,- - - - - r - - - _ r - - - - - - - - , - - - - - - - - - j
o 50 100 150 200
V, fUs
5"-/'16
I • Experimental
1-Theoretical

5 .12/
5.127 Pressure Distribution on a Flat Plate Due to the
Deflection of an Air Jet
Objective: In order to deflect a jet of air as shown in Fig. P5.127, the flat plate must push
against the air with a sufficient force to change the momentum of the air. This causes an in·
crease in pressure on the plate. The purpose of this experiment is to measure the pressure
distribution on the plate and to compare the resultant pressure force to that needed, accord-
ing to the momentum equation, to deflect the air.
Equipment:
flat plate with
thermometer.
Air supply with a flow meter; noule 10 produce a unifonn jet of air; circular
static pressure laps at various radial locations; manometer; barometer;
Experimental Procedure: Measure the diameters of the plate, D, and the nozzle exit,
d, and the radial locations, r, of the various static pressure taps on the plate. Carefully cen-
ter the plate over the noull! exit and adjust the air f1owrate, Q, to the desired conSlaIll value.
Record the static pressure tap manometer readings, h, at various radial locations, r, from the
centcr of the plate. Record the barometer reading, HOlm' in inches of mercury and the air tem-
perature, T, so that the air density can be calculated by use of the perfect gas law.
Calculations: Usc the manometer readings, h, to delennine the pressure on the plate as
a function of location, r. That is, calculate p = "Ym h, where "Ym is the specific weight of the
manometer fluid.
G raph: Plot pressure, p, as ordinates and radial location, r, as abscissas.
Results: Use the experimentally detennined pressure distribution to detennine the net
pressure force, F, that the air jet puts on the plate. That is, numerically or graphically iIlle-
grate the pressure data to obtain a value for F = IP dA = fp (217"r dr), where the limits of
the integration are over the entire plate, from r = 0 to r "" D/2. Compare this force obtained
from the pressure measurements to that obtained from the momentum equation for this now,
F "" pylA, where V and A are the velocity and area of the jet, respectively.
Data: To proceed, print this page for reference when you work the problem and did..' "('re
II F IGURE P5.127
(con 't )

57 /2.7 I (c()n 't )
Solution for Problem 5.127: Pressure Distribution on a Flat Plate due to the Deflection of an Air Jet
D, in. d, in. Halm, in. Hg T, deg F Q. ftA3/s
8.0 1.174 29.25 77 1.41
r, in. h, in. p, Ib/ftA2 p, Ib/in."2
0.00 6.62 34.42 0.2391
0.39 5.92 30.78 0.2138
0.79 3.04 15.81 0.1098
1.24 0.55 2.86 0.0199
1.59 0.19 0.99 0.0069
2.04 0.13 0.68 0.0047
2.41 0.09 0.47 0.0033
2.85 0.05 0.26 0.0018
3.23 0.03 0.16 0.0011
3.67 0.00 0.00 0.0000
P = YH2o*h
P =Palm/RT where
Palm =YHg*Halm = 847 Ib/ftA3*(29.25/12 ft) = 20651b/ftA2
T = 77 + 460 =537 deg R
Th us, P =0.00224 slug/ftA3
Using the trapezoidal rule for integration
p*r, Ib/in.
0.0000
0.0834
0.0867
0.0246
0.0109
0.0096
0.0078
0.0051
0.0035
0.0000
Fexp =2n*0.5*L:J TO 9[(prj +pr;+1 )*(ri+1 - rj)] = 2n*0.5*0.189 = 0.594 Ib
Theory:
F = pV
2
A where
A =nd
2
/4 =n*(1.174/12 ft)"2/4 =0.00752 ftA2
V =Q/A =(1.41 ftA3/s)/(0.00752 ft"2) = 188 ftls
Thus,
Flh =0.00224 slug/ftA3*(188 ftls)"2*(0.00752 ftA2) = 0.595 Ib
5-IJf8
pr;+pr;+1 r;+1 - rj
1 0.0834 0.39
2 0.1701 0.40
3 0.1114 0.45
4 0.0355 0.35
5 0.0205 0.45
6 0.0174 0.37
7 0.0130 0.44
8 0.0086 0.38
9 0.0035 0.44

>'127
40
35
30
N 25
<
~
20-.c
ci: 15
10
5
0
0
0.10
0.08
c: 0.06:::
:£
{ 0.04
0.02
0.00
Problem 5.127
Pressure, p, vs Radial Location, r
.-.--~----~------~~-- ------_._---------j
.----~..-------------------~- --- --_.__ ..-.- - ------- ----- - -----------..j
I- - - - -- --------.......------------------------.-.------- --- ---i
I---'-------~---~--'------~~I !--.- Experimental i
1
0 1
--~~
2 3
r, in.
Problem 5.127
Pressure Times Distance, p*r,
vs
Radial Location, r
------~--~.---.~--- ..-
2 3
r, in.
5"- IJfCf
I------4
i
----1
i
i
4
4
- - - - - - --
1--.-Experimental 1

5.129
5.128 Force from a Jet of Water Deflected by a Vane
Objective: A jet of a fluid striking a vane as shown in Fig. PS.128 exerts a force on the
vane. It is the equal and opposite force of the vane on the fluid that causes the fluid mo-
mentum change that accompanies such a flow. The purpose of this experiment is to compare
the theoretical force on the vane with the experimentally measured force.
Equipment: Water source; nozzle to produce a uniform jet of water; vanes to deflect the
water jet; weigh tank to collect a known amount of water in a measured time period; stop
watch; force balance system.
Experimental Procedure: Measure the outlet diameter, d, of the nozzle. Fasten the
8 = 90 degree vane to its support and adjust the balance spring to give a zero reading when
there is no weight, W, on the platform and no flow through the nozzle. Place a known mass,
m, on the platform and adjust the control valve on the pump to provide the necessary flowrate
from the nozzle to return the platform to a zero reading. Determine the flowrate by collect-
ing a known weight of water, WwaleP in the weigh tank during a measured amount of time,
t. Repeat the measurements for various masses, m. Repeat the experiment using a 8 = 180
degree vane.
Calculations: For each data set, determine the weight, W = mg, on the platform and the
volume flowrate, Q = Wwate'/('}'t), through the nozzle. Determine the exit velocity from the
nozzle, V, by using Q = VA. Use the momentum equation to determine the theoretical weight
that can be supported by the water jet as a function of Vand 8.
Graph: For each vane, plot the experimentally determined weight, W, as ordinates and
the water velocity, V, as abscissas.
Results: On the same graph plot the theoretical weight as a function of velocity for each
vane.
Data: To proceed, print this page for reference when you work the problem and dick herl'
~ FIGURE PS.123
(con't)
5- ISO

5:123 I (COfl't)
Solution for Problem 5.128: Force from a Jet of Water Deflected by a Vane
d, in.
0.40
m, kg Wwater,lb
Data for 8 = 90 deg:
0.02 7.71
0.07 8.66
0.17 8.87
0.12 8.92
0.22 9.66
Data for 8 = 180 deg:
0.05 6.81
0.10 9.02
0.20 8.84
0.25 7.88
0.30 8.86
0.35 7.97
0.40 6.37
W=mg
Q = Wwate/(y*t)
V = Q/Awhere
t, s
29.8
18.2
10.1
12.6
10.6
24.5
20.8
13.2
10.9
11.1
9.5
7.6
Experimental
m, slug W,lb Q, ftA3/s
0.0014 0.044 0.0041
0.0048 0.154 0.0076
0.0116 0.375 0.0141
0.0082 0.265 0.0113
0.0151 0.485 0.0146
0.0034 0.110 0.0045
0.0069 0.221 0.0069
0.0137 0.441 0.0107
0.0171 0.552 0.0116
0.0206 0.662 0.0128
0.0240 0.772 0.0134
0.0274 0.883 0.0134
A = nd
2
/4 = n*(0.40/12 ft)A2/4 = 0.000873 ftA2
Theoretical:
W = pV2A for 8 = 90 deg
and
W = 2pV2A for 8 = 180 deg
(Con'~)
5-/5/
Theoretical
V, ftIs W,lb
4.7 0.038
8.7 0.129
16.1 0.440
13.0 0.286
16.7 0.474
5.1 0.088
8.0 0.215
12.3 0.512
13.3 0.597
14.7 0.727
15.4 0.803
15.4 0.802

5:/28'
..c
CDn't)
Problem 5.128
Weight, W, vs Velocity, V
1.0 -,-----~------------1
0.9
0.8
0.7 --
0.6
--- ------~~----------------- - -----r;~-------
~ 0.5
0.4
0.3
0.2
0.1
• I
-.#-+-- ----..rr-----~-~-~__J
-~-~~---~~cF----~~---'----:~---j
!~--.-----.-- ----- --------1
o.0 J-....-I!~=--..,----r-------t------j
o 5 10 15 20
V, ftls
5""-1$2
• Experimental, 90 deg
• Experimental, 180 deg
Theoretical, 90 deg
Theoretical, 180 deg

5.129 Force of a Flowing Fluid on a Pipe Elbow
Objective: When a fluid flows through an elbow in a pipe system as shown in Fig. P5.129,
the fluid's momentum is changed as the fluid changes direction. Thus, the elbow must put a
force on the fluid. Similarly, there must be an external force on the elbow to keep it in place.
The purpose of this experiment is to compare the theoretical vertical component of force
needed to hold an elbow in place with the experimentally measured force.
Equipment: Variable speed fan; Pitot static tube; air speed indicator; air duct and 90-
degree elbow; scale; barometer; thermometer.
Experimental Procedure: Measure the diameter, d, of the air duct and adjust the scale
to read zero when the elbow rests on it and there is no flow through it. Note that the duct is
connected to the fan outlet by a pivot mechanism that is essentially friction free. Record the
barometer reading, Hatm, in inches of mercury and the air temperature, T, so that the air den-
sity can be calculated by use of the perfect gas law. Adjust the variable speed fan to give the
desired flowrate. Record the velocity, V, in the pipe as given by the Pitot static tube which
is connected to an air speed indicator that reads directly in feet per minute. Record the force,
F, indicated on the scale at this air speed. Repeat the measurements for various air speeds.
Obtain data for two types of elbows: (1) a long radius elbow and (2) a mitered elbow (see
Figs. 8.30 and 8.31).
Calculations: For a given air speed, V, use the momentum equation to calculate the the-
oretical vertical force, F = pV2
A, needed to hold the elbow stationary.
Graph: Plot the experimentally measured force, F, as ordinates and the air speed, V, as
abscissas.
Results: On the same graph, plot the theoretical force as a function of air speed.
Data: To proceed, print this page for reference when you work the problem and click /tat'
Centrifugal fan
II FIGURE PS.129
5"-/53

s:/2 9 I (COli 'f )
Solution for Problem 5.129: Force of a Flowing Fluid on a Pipe Elbow
d, in. Hatm, in. Hg T, deg F
8.0 29.07 73
Experiment Theory
V, ftImin F,lb V, ftIs V, ftIs Fth,lb
Long Radius Elbow Data 0 0
0 0 0.0 5.0 0.02
1200 0.38 20.0 10.0 0.08
1420 0.51 23.7 15.0 0.18
1800 0.79 30.0 20.0 0.31
2160 1.05 36.0 25.0 0.49
2440 1.38 40.7 30.0 0.70
2700 1.65 45.0 35.0 0.96
2900 1.91 48.3 40.0 1.25
3100 2.19 51.7 45.0 1.58
3520 2.83 58.7 50.0 1.95
3750 3.12 62.5 55.0 2.36
3950 3.38 65.8 60.0 2.81
65.0 3.30
Mitered Elbow Data
1400 0.30 23.3
1780 0.55 29.7
2000 0.74 33.3
2300 1.12 38.3
2630 1.44 43.8
2900 1.72 48.3
3150 2.06 52.5
3360 2.38 56.0
3550 2.62 59.2
3620 2.74 60.3
P =Patm/RT where
Patm = YHg*Hatm = 847 Ib/W3*(29.07/12ft) =2052 Ib/W2
T =73 + 460 = 533 deg R
Thus, P = 0.00224 slug/W3
A = rrd1l2/4 = rr*(8/12)1I2/4 = 0.349 ftll2
(CtJl1'-/:)
5"- 15"1-

Problem 5.129
Force, F, vs Velocity, V
4.0 ------~,---,------
.c
3.5
3,0
2.5
-_ 2.0
LL
1.5
1.0
0.5
----.......--- ··---------1
I
O.a -t-..e::::.---,------'""1r-----i------l
a 20 40
V, ftls
60 80
5"-/55
- Theoretical
• Experimental: Long
radius elbow
A Experimental: Mitered
elbow

'-./ J 6.1 The velocity in a certain two-dimen-
sional flow field is given by the equation
V = 2xti - 2ytj
where the velocity is in ft/s when x, y, and tare
in feet and seconds, respectively. Determine
expressions for the local and convective compo-
nents of acceleration in the x and y directions.
What is the magnitude and direction of the ve-
locity and the acceleration at the point x = y =
2 ft at the time t = 01
FroM eJGj>rt'$S 14J" ft,.,. II~ I"c; 1-:; ,)
5:/nce
a" (Ci!Jnv) : (,{ Ju. +1/' Pll.
!);< ~
- ,+xl:2.
S/miJell" /'J)
pi/" u ~-t v-q
a~ :: Tt:+ t9;( J~
CO'1 PI
a; (local) = 1.!! =- - 2!:J
at-
= (Z.X'.1;) (2.t) -t r- Z!:Jt)(o)
a., (C~nj/.) .: u~t-1r~ - (2xt)(o) -+- (- 2}jt)(-l-t-)
l);<. d !J -
= If lj t-2..
At x.-:.1.j=2ft Clnd t~o
U = 2. (2)(0) =0
So -that
~
V==o
a~ = .2;<. + il-xt'2 =
Cl
j
= -2J + 'f.!Jt 2- .:
a:= 41: - LfJ +-tis"
v- : - 2. (2 ) ((}) .:: 0
2 (r.) + tt (2) (0) = Lf .ftls z.
_ ,2 (2) -t t (2) (0) ::: - If R Is 2.

6.2 Repeat Problem 6.1 if the flow field is
described by the equation
V = 3(x2 - y2)i - 6xyj
where the velocity is in ft/s when x and yare in
feet.
and
a;( (tDnv) = [.(.~~ +r~ -:: 3{;tl:.!:J2.)(b,t)-t (-6~.!J)r-b.!J)
= / g(X. 3 + X J 2.)
qnd
a..'J (/OCtJ/):: ~; - 0
a.j
(~nJl): U ffT r rf == 3 (i2_!J"l.) (-6.!f) -t (-h)(!J) (-,I.. )
== Je(.x2
':J ~ !i 3)
At K=!1 = / ft (/hd i:~o
fA. = 3 [O/-ut] =0 v-:: -b (,)(i) = - Ie
Qnd
~ A
V= -&,j
ai = /3 (x 3
+x :/,) = 18> [(03 ..,. (/)(t)~ :: gtlt/.s·
a!J = IS( x~-+y3) =18[{/)2{1) -t{n
3
] = 3{, -ftJs~
,,-z

6••~ The velocity in a certain flow field is given by the equa-
tion
v = xi + X2zJ + Yzk
Detennine the expressions for the three rectangular components
of acceleration.
From e)(.frt'j>/~Y1 /t,1" ve/()'l"t!1) fA. -:: ><
Sinc.e
a.~ ::
-!hen
a~ ::
5t'mt/arl'j./
0+ ( >< )(1) ,. (i~)( 0 ) +~r:)(O)
X
--
a - olr
:;- t>t +
and a. - 0 +
~
~w + fA.. ~ -J- 'lr!J!F 1" W' Jur
p~ ox 8~ J~
o 1- ( X ) (0) + (X2~) (~) + 6~) (fj )
2-
xLi'· T :1 r

(t:L)
(j,)
6.4 The three components of velocity in a
flow field are given by
u = x 2 + y2 + Z2
U = xy + yz + Z2
W = -3xz - z2/2 + 4
(a) Determine the volumetric dilatation rate, and
interpret the results. (b) Determine an expression
for the rotation vector. Is this an irrotational flow
field?
V()/I.{me.fr/c. d,/t<la.flon v-a.fe = J(.I. Jir
;a;: + Jy T
Thus) {"Y fle/()c,fy tdJ/'rJjJtt7l'1fl'll-s f/ien
I/()I"tn~InCo d;/a.ta.t,b" rate -: 2.t. -r (x.,.r) -t- (- 3x. -7:) .= 0
This YfSU/t. /nd,ctl+es ih~1: 1heyt! I~ nD chtllfl/e I;' -the
Volume. e>,f. a.. +Itlid elemfl1i (I~ it rn~lIeJ -hm dJl1e.
/I)Cltt'OIJ to (lI1IJ1her.
F'IPm .Gss.
'lIven:
I
W ::
~
:i (;;- ~;) =; (!f - 2!J) = - 1:
(1;--::)= -1 [0 - (y-tZr)J " - (!J: H)W;(. =
CU!f --
I
:l
I
-;;.. (~ _ ~):: .!... [2i - (-3z.)1 = sz~oi:. p;(;L J
- r-; +7:) i +
S-fo ~
2. J
~
SI'nt.e W /.5 nat )fr~ elleffjwheve -the fl{)w field
I ~Y'oia-t';'1111J. N~ .
.
/s nCJt

6.5 Detennine an expression for the vorticity of the flow
field described by
v = - xy3 i + y4j
Is the flow irrotational?
FY'CJm e;<"fr"t~s,i;n -h,Y ve/tJci-l-!j) tl =- X!J3J
tr= !:J'f) ond w-= o}
C( n cl tv ;' rA
: 1- (~_ JJr)W;e. :2.. Jy or
W
11
~ ~ ( ~ - ~~)
W = 1- (Jtr_~)
~ ;L ax. 0 !:j
/i Iolll)ws inA-t
CJ~l )
- z [CO) r + (0) j + (f. ~f1") kJ
--
.-.l.
S/nce :f
/~ hot.
,'.5 not jer() evef'fjwhere -f;,e
/;/,()t4",6'(,I1I1I, NtJ.
(£t ~. /3)
(po. 6. 11f)
(EZ·k./2.)

t:.. ~ I 6.6 A one-dimensional flow is described by
the velocity field
u = ay + by2
v = w = 0
where a and b are constants. Is the flow irrota-
tional? For what combination ofconstants (if any)
will the rate of angular deformation as given by
Eq. 6.18 be zero?
i=tP1' /~rottl.~/6,~q/ ./-/~w-
---"'
W '=0 ,
tJ1r) _- -0
of:
Thu.5) c;) /5 n~i 7'fY't) ellfr!1w/'ere (If/Pi fife f'/tJu/"
is no i irr~l:-tL~itPn(J/, NtJ .
Since.. (~m I:i$. t,.18)
• oU- JL.4
~:: ~ t- J!;j
(I: ~//~w.s .fr:,,, hie.. IIt/I)(./-/:y c/iIfn'hlli.;()J1 71J1fJ1 1114.1:
i=- a + zbfj
Thus) 1her-e tire no
et(,{q/ -h> /~r()) -uud.
~f :to N~ne.
va. /1{ ts tJ/ t4 1111(/ 6
•wil/ rlv~ a- =-0
( eJt~~fi: k11t
.r;,r 4// I)a/wt'.s

6.7 For incompressible fluids the volumetric
dilatation rate must be zero; that is, V . V = O.
For what combination of constants a, b, c, and e
can the velocity components
u = ax + by
v = ex + ey
w=O
be used to describe an incompressible flow field?
For- CI n ine(f)mp tess //'Je /I,,/t/)
Ju. Jtr ~1Jt-o
~;( -t- Tj"" o~-
lv/In -the Ve /C;c./-/!1 d,.!J'.f,.;bl('t;{)~ rI i/el1
~tr
=e;'1
a. +e = 0
oW- _
- -0
~t

6.8 An incompressible viscous fluid is placed
between two large parallel plates as shown in Fig.
P6.8. The bottom plate is fixed and the upper
plate moves with a constant velocity, U. For these
conditions the velocity distribution between the
plates is linear, and can be expressed as
u = ux.b
Determine: (a) the volumetric dilatation rate, (b)
the rotation vector, (c) the vorticity, and (d) the
rate of angular deformation.
(a) (/e>Jtlmett/c d//a.. t~tt4?JI f'"11 te
(h) ,t:;,,. (/e/~c./ry d,'.s I,;hI.( f,'eJ11
'"~
Wi! -/;..UJ~
~ ~ +
I
'11 11 1:11 j
4ntl
.J. (~_ dU )tv = :2- =~ oJ< 0 'J
Thus; .-'
~ V ~w -- -2b
~
~
'"CC)
-S :: 24) = U -{
-.b
• otr cJu
cd. ) "0 :: - -rp;<. J!:J
Thus)
(' = [r
-b
u
Moving
plate
Fixed
plate
FIGURE P6.S
~i" cW- .=.O()fj 41:.
--
V
-zb
(~t b,le)

6.9 A viscous fluid is contained in the space between con-
centric cylinders. The inner wall is fixed, and the outer wall ro-
tates with an angular velocity w. (See Fig. P6.9a and Video
V6.1.) Assume that the velocity distribution in the gap is linear
as illustrated in Fig. P6.9b. For the small rectangular element
shown in Fig. P6.9b, determine the rate of change of the right
angle 'Y due to the fluid motion. Express your answer in terms
of ro, ri' and w.
Foy the. lInear d,.sfy/buf,o"n
t<...: - r; co !J
~-Y"','
.5c thai:
d"!. - _ V;w
J!j Yo-Y;;
alllf S/"te r= 0
The nejal-iII~
rfjlJi- anjle
sifn
(a) (b)
• FIGURE PS8

~./O I
fl.l!) Some velocity measurements in a three-dimensional
incompressible flow field indicate that u = 6xy2 and v =
- 4y2Z. There is some conflicting data for the velocity com-
ponent in the z direction. One set of data indicates that w =
4yz2 and the other set indicates that w = 4yz2 - 6y2Z. Which
set do you think is correct? Explain.
To ;5A..t,54 -Ine ~"-t1;1I,"1!1 e~ u.~.fItD~
Ju air ~w-
u..,. 75 1" Tt. .= 0
J"lnce dU :; ts,!J 2- IIi1 II Jzr= - 3'1 i
~x ~fJ
-tlIfl1 IttJtn b3' 0 )
fS,:; '2._ 9fJ % + oW- ::'0
Jt-
Thus) 2t,U-
-~i: -
Ff "A-f",iJ CZ) (!fin be
fdw- :::
~r
w-=
8'fJr -~d2.
inkJra-kd w/fh Y~S~ct h i i-c ()j,flJl)'
J8!1 1: d:J - f '!J1.d~ + +(;I.; !1)
'ffj1: 2 _ ":J2.-J: 1- '(Yo) ~)
The :5et ~f dA-bo. (tvJ'-n-, kt,'J) =0)
w- ::: 4/J:r 2. - ~!:J l.:z
wouitA tlflfllY -10 he fne ~ff'ec,t se i::.
('-10
(/)
c. Z )

b. JI I
6.11 The velocity components of an incompressible, two-
dimensional velocity field are given by the equations
u = 2xy
v = x2 _ y2
Show that the flow is irrotational and satisfies conservation of
mass.
II -the t:.UJo-d/mfnSIr;I111/ .{jew 15 1~~()I::a..be,ntI/)
UJ =".J... (~V- _ Ju.)= 0
:e ,;J. ~;< 17 ':J
F;;r 1J1e. ve/()Cit, d/~ fri /:)/1'1-1';/1 f' ven.l
Ct 11 c/ 71te IS
-r; Sllt;.s!::t ttP;1.serJ/ail4J1f "I mass I
eJu. Jtr_
~-;. "t" P!J-()
-then

'./2.. I
6.12 For each of the following stream functions, with units
of m2
/s, determine the magnitude and the angle the velocity
vector makes with the x-axis at x = I m, y = 2 m. Locate any
stagnation points in the flow field.
(a) '" = X)'
(b) '" = -2x
2
+ Y
~f ~~ ~fY'e/lrn fl.4 Hc..:hiPJ1 J
tr-= - ~
d'X
l{':: dt =x V-=-~ ::-~
~, . ~~
Ai: x::: l.-m) 'f ~ J-,.",,) I i -4/1IJuJ.J ~.f::
Th~sI
>Co
-~
L - z.
C"a n 9 ":
5Jn!c '" =0 ",I:.){:: C> ClI1" v:. 0 a--t ~::o/ tt j+a9n4f../~;'
p()/n I: (!)~C.IAY.s at:. X =- ':J .::. ().
(j,) Ft'y
u. -= 0 cf -= J hM 1r.:: - ~ ~ '!-x
TtJ j D~
A1:- )(. = jlWl} I:J;:: 2-/I't1 I it ~11f)W5 fhA-/. tA.:: I ~ (.f"d 1r= ~ 0/

'-. , '3 I
6.U The stream function for a certain incompressible flow
field is
'" = lOy + e-Y sin x
Is this an irrotational flow field? Justify your answer with the
necessary calculations.
~'r 1'h~ /f"w -tD hfC l;r()-ttLt:lo~a I (S4'e l:g. 1../2.) J
W =-L (~ - ~)-o~ 2. ~/. d!?-
alt~ /z,r f;,~ S.f:rerun .f.U)1c. t.,';rl (J I'veI'!
ThUj
)
~() --/J1a;t
tA.- =- ~ = /0 - e-~/';~
-:1
11"":: - Jtf :: - e ~s ;(
7i
-':1
OlA.
. alr_e .sIn ~
=- - -~j ~;(.
-!j )
1 ( -~
W ::' e jJ~ l- e sint.
~ 1-
-~ .
e SIJIJ ~
='0
Jln"e Wl;:O I -r;1J':s JS an J'rroi:(L+r()~t:t1 flow -I,eid. Yes.

~./'f I
6.14 The stream function for an incompressible, two-
dimensional flow field is
IjJ = ay2 - bx
where a and b are constants. Is this an irrotational flow? Explain.
Ql1d #r the slrellln ~1'I'£~J1
IA.:: ()If
~
-
u-::: - ij: =~;(.
Thus;
S /n t. e w~ 1:- 0
(t{n/-es~ a =0) .
G CL j
b
.
7 HI(J1 J

lO./5 I
6.15 The velocity components for an incom-
pressible, plane flow are
Vr = A,-l + Br-2 cos f)
Vo = B,-2 sin 0
where A and B are constants. Determine the cor-
responding stream function.
Fr~m 1he de/;;'/i,blt of th~ StYi'II/11 Iwnc,6/~n J
..,,... _ ..!.. dip .,,... :: _ _dip
v~ - r ~ Vt; Or
So tha.t. +OY' fhe velocity c/Js'ft;'I7J,(.it~1/ fll/en,
I cJl; -I -2
;:. Ft9 :- A- r T B I- ~s e
J~
-2.
; - /3}- S;'11 &
;r;,.
Ii?Ie9n1fe I:'j'()) w/-ht r-eSf~t-t: to e to
Jd¥ :: J{A+ 8 }--/ e.~ s e ) da +
S/~/Jar01 /nte'lrllte E~,l Z} u/rlh Y'l'~fec-i +0
fdif =- - faj--2sil1l!} d/' + I'..fB)
If:: 13 j- - 'sin CI 1- ~ (t))
Thw5) .fo .sa,tIS.f.!1 both ./;"gs. (.3) 1(l1d/lf)
t.f:: Ae+ /3~-I.5/I1C1 +C
where C ~ an Clrb /trA r.!:J c~n.5itll1 t.
~b-b,/~
.f. (r)
I
(/)
( 2.)

t,.I' I 6.1G, For a certain two-dimensional flow field
u = 0
v = V
(a) What are the corresponding radial and tan-
gential velocity components? (b) Determine the
corresponding stream function expressed in
Cartesian coordinates and in cylindrical polar co-
ordinates.
(()..) Ai- an arh i fyay'J pfJ,ni P
(.5~e ti'llAY'e)
Vr=V~/na
ve = V ~s B
Cb) Since.
u= Jif v-= -~:::v=0
d':J oX
/t f"Ilow thai fII 15 n()t ~ f/,fnc I:ltJ~
Lf=-Vx: -fC
l) f. I;j
wher-e. C
.
Ii r b/IrA1"':1 UJ1siol1i .is lin
IIlsrJ) tui"ht .x.::: Y t.t!;S B
'1= - V f"UJS& fC

6.17 Make use of the control volume shown
in Fig. P6.17 to derive the continuity equation in
cylindrical coordinates (Eq. 6.33 in text).
~ ffo'¥ + f-V·:' 0',4 "'0
CV cs
J~ f~t/-tr
"......
-Cp-
Clnd
y
8
z
Volume element
has thickness dz
x
FIGURE P6.17
( .E'S, &,,)'1)
(I)
f~y,~ dA - net i"ll.te o! mNSS ()ui//()w fhr()II9h
Cs
FrPR1 h'lwn 4 f ~J;h t.
Ne t: rLtc 01 m~~s
()U-tf/DW /n r- dJ';-e,..i/(J~ =
( /J r,: ...,. '4.f3 q.,JI' v,.of tk)d(; d1: .
r, f)y 2 J.I' .".
- {t ~ -~ r)(r- t)dGdl
oS /'(Y'ffu.es cf ~nff"()/ J/()/wme
( f V/. r d::1- t)(I- or t )de d~
C2.)

~,/7 (con i)
FrfJl11 !t9~"{ at- ttfht.'
Net Y'a I:e ~/ /nASS
~utfl"/N ,;., t) - d/ree:t";JJ =
(;~ r ~ ~e) drdr
-fv;; - ;~~ ~~ )drdil-
(fife - dfife dB)cJl'"dr:
~~~____ 09 ~
aI'/J{; ely d~d~
(J$
I'lfPl'n hitlYe at: rl9lJt:
!Ve.t f"11 te (;/ tnQS.J
{)u-t-R~u; In r - el;~c....j.ldH
/IJV"; ~ ()f"i- ~) rd&df
(I r Ji 2-
- (/1'j. - of"i dl:- ) rd~ til'
)2: ;l,
(;t~ ydrd{)d~
(:)r
y
5u"sl-/-j.j,tI-'~1j ~f Gjs. to inY'/'< (fl.) ,"nl-o £1. "./9 'j;eIJs
¥t r dydf)Jr -r Jr:r f- eli" 11'6 dt +!v;. dY'd&d:&
-+ a/v[; drdf)d2 + 0 f~ rdy-dedr :0
)9 J~
or
De. wr/ He1'1 ,q.s
%.,. -/: fi (r~lIj.) + -f: ;j~Ilj;) ..
,,-/ g
( 3)
('I)

~,/8
6.18 It is proposed that a two-dimensional,
incompressible flow field be described by the ve-
locity components
u = Ay
v = Bx
where A and B are both positive constants. (a)
Will the continuity equation be satisfied? (b) Is
the flow irrotational? (c) Determine the equation
for the streamlines and show a sketch of the
streamline that passes through the origin. Indi-
cate the direction of flow along this streamline.
(u) 10 :5otisry the Ct'JI1III1U/1y e~uai'';J1
J u...,. d II"" =-0
p;( J';J
Sin ee /1:;r 1J1e ve IDtif.y c/''s-';''''/b,,/-ti;/J 1i1I~11
)
&..!:!. - 0 q,x =-0
t>,K. - tJ~
the ~l1hiJl,/t1 e.!",a. &;'J1 ls 5 at;/sfietl. Yes,
(b) :t.n ~I"der .ft,,.. the .fl(!)w +D be 1~f'~t:flb~i1111 wz.-=Oj IAJheY'{!
w~~l(t-~~) (~%-b,/Z)
tf-'/3 ~ =Aa;(. - d r:1
I
wi = ;L
(I3-A)
Thus; f/~w will ()fJ/Y be ;rr{)l:atl()~t:// ,'f
(!) 4/~n9 a. s.fre/lI'l1/;;'e
dy _ (/
d~ - fA.
S" that for fhe tle/Deit!! dl;fn'bu!tPH
ttltie" cI!I = B x.
dZ fi!:1
(/11" 1htre fe,re
'1 c/!J =-J)(. d x
.t:11.ferrlll/!:)h 'fields
t./ 2 - /.3 ,,(,z+ C
J - -;r
,4=-8.
slope:: r1

(p.jq I
6. Iq In a certainsteady, two-dimensional flow
field the fluid density varies linearly with respect
to the coordinate x; that is, p = Ax where A i~
a constant. If the x component of velocity u i&
given by the equation u = y, determine aq
expression for u.
For p.. vCll"it:l/;/e. del1~;.J.'1 f/~t.V)
J fcou.) ~ J (fv-)
~x. J!:f
(t<..:: C4~)(J) =AX!J
11- /t;/llws -thllt
Th (As)
In Ie'Irille
otlA.J .. Ay
Jt'fV-J :: -AfJ
t1b
eg .(;) wifh r~sl'ec.1: tD !f 10 ~61rt/"
jdrtV-) = -jAjd'J + fIx)
With ! =A-x
where
v=
v- _- - ~ z. + f{)I.)
:2.,;(,
~-2..0
( I )

(p, 2,0 I
y, ft
6.20 In a two-dimensional, incompressible flow
field, the x component of velocity is given by thl"
equation II = 2x. (a) Determine the correspond-
ing equation for the y component of velocity if
v = 0 along the x axis. (b) For this flow field
what is the magnitude of the average velocity of
the fluid crossing the surface OA of Fig. P6.2D!
Assume that the velocities are in ft/s when x and
yare in feet.
1.0 ~ ________ A
(a) ro s~.J/s~ the t.D"tl;'U~.f!J -RgU4 b()~
dlA Jtr_
'?X -r Jb -0
SInce. '
~~ =2-
I t ~/I"ULS thft-l.
CPoA :
1-7I
I ~8
.f ~DG :80' I 1.0 x, ft
FIGURE P6.20
(con~'der (i.,
un/t tnJcJ:.lllSS =1ft)
JLr=_Z (I,)
t?!1
rn.f.e11'~6p;' "I E"1,fJ) With resrc.-t. to :t 'j/eJdJ
1r= - 2:; + f{~)
rf '2r=O ~/tP"9 X-AxIS (f/:'O) 7hen /-(;1.) =0 So thd
1/= - :2..:J
(h) 1;, ~al';;4 C4?11.st!YIIA-k"o'f ~f m~s.3
%,4 :: 0;.~ - 4113 (,,~~ !t7t1~)
oft-
tt/()n'j A/J U =~ (,) = 2 S .50 1h4.. t.
(j)AI3 = U AA8 = (.2.. ItIs ) (/ ,ct)(/,ct) =
/f/0I19 t!)13
Thus)
'1/" =0 Sa
~,tI = {/J;/3 ::
{Jt;1J
Y;v =
a f"elt.
()A
ih~t (/)Ot3 = CJ ,
2
-Ft3
-5
+-1:3
2 s=
v:r +t'J.
/. If./ +-1:-...
S

cD. ;2/ I
S/nce
6.21 The radial velocity component in an incompressible,
two-dimensional flow field (v: = 0) is
vr = 2r + 3r2
sin ()
Detennine the corresponding tangential velocity component,
vo' required to satisfy conservation of mass.
J.. ~ (rtlj,. ) / J ~
r ~ r -r -;. gB +
-v; ::.tJ )
.
oU{; :
de
Pllc/ wIth
r v:. ::' :<. r 2 + 31"3 05/11 er
l"f fr;//ows 1htt t
~ (r 'J/j,.) -
Tr -
Thu~ E$. (j) be,-",-"es
~v: =- - (IIr f qr 2:S/11 e)
E$Ufli:I4?I1(l) Ctin be ;"'-teff'tLieq' with r~.5rc.t
Jdvg = - f (IfI-' f- qrz. si" e) dB r
VB::' - tfre - 9 ,...zcos ~ + .;(y.)
!(r) I"; an Hncle ter/rJ,npd Ittl1ctle;'H ~/ Y".
'-.2 Z.
("I)

6.22 The stream function for an incompres-
sible flow field is given by the equation
IjI = 3x2y - y3
where the stream function has the units of m2
/s
with x and y in meters. (a) Sketch the stream-
line(s) passing through the origin. (b) Determine
the rate of flow across the straight path AB shown
. in Fig. P6.2].
(a..) Li~,s fJ/ 6Pns/:t?lJ I: If ore str(1/1If}/Ii7es.
FOr- tf: 3;( 2; - 1 ~ tht! slre/lllllJ;'e
fX'ssinf thrDllfA 1h~ ()I'if,n (~:=~ !I=())
has &t. vpl,,~ cf=o, Tfltl~ -the
e$tlat-ldJl1 .{()Y fHl!.. S-f"flll11/iIlPS ihmllfk
1he ()f' 19In I~
() :: ax z!J _'j3
j= zv3x
y, m
1.0 B
If.:0
A
1.0 x, m
FIGURE P6.22
If=o-----7(----"':""-":"'-' X
4- ~fel:c"h ()f these sfrellm/Jl1e.s 15 sJU)Wi1 il1 1Jfe ;;jllr~,
(b)
4?=Y8-~
,4t 13 .x. =(») ':1:: IMI .so tha..t
lfs :: B(o) ~/) - (1)3 = - I Mf~/s (per tlnd WJdtA)
Ai
3(1)'(0) - (0) 3 -- 6
Ihus)
The. ne1al-l~(' sI9n l~d,C4k.s inll';' the Iltjw /s IT()I'YJ
rl ,ht .f.o Jett Q S We /fPtJK ./rpm A Ie E
t-23

~.23 I
6.23 The streamlines in a certain incompres-
sible, two-dimensional flow field are all concentric
circles so that v, = O. Determine the stream func-
tion for(a) Vo = Ar and for (b) Vo = A,-l, where
A is a constant.
Prd)m "the. de-h/1;fltPll 0/ -the .5.fr~(Jm knc.t'()'11
)
.J..a~= y Je v-: -= - ~
(;; Jr
So /hilt tt)/f1t
and 1l1prej,,,.e
;f '/:'//t)tuS 1/.~-t ~ :::'0
~ = f(r)
(A) ht ~ = A;-
dlf :-Ar
Jr
EIl1) with 1'I'.s/,ct t:. l-
fd", = -jArdY
'f = - A:'2. + ~ (~)
JIoweVf') .sill~e (jl is 1/6t A. htl1C.i./()1I lJl &1 It- ,t;//"U/.! thL t
t.f = - t!f2.f C
Where. C I~ tin Qrbifral''1 Uf1sMl"rl:,
Cb) SJfn"/tlY'i':J) -k.,. ~ = A- y-I
fdtf- == - fA-r-'dr
or
tf = - A 1/1 Y + C
('-24
( I )

6.24-* The stream function for an incompres-
sible, two-dimensional flow field is
'II = 3x2
y + y
For this flow field plot several streamlines.
The 12.~ ua.:tltJlJ Icy a ~.frellm/il1~ ,:S -h,fll1d'/''1 ~fh~ tjJ=~lJsizJl/i
/n ffJe ejua.f::lon -Idr 1JIe offlllllll -func.t:j~l1. Thusl
lew -t11e
9/11el1 ~fre/l111 ./uHtittJ#
Lf = 3i
4
y +-b
./I:: ~11t)u)J inai t1Je. eStJd..tl(f)JJ c I loS
-':'00 cls
t/J
'j= 1 + 3x"2
.
kJh-ere VIIY'I()/,LS Ct;IJ,st:""t 11r,III'.s erNI be Q~.sJ1ned
+0 t)hto/" tl fbm,'/.!t "f ,jrrf'lIm//lJes.. I-J- Py",NIn?
-rhe XI '1 eoord/n;ks ,,{ 1/'4Y'1(;~ Sfreqm/;d/ZS "?;//f)II/S
110 print "****************************************************"
120 print "** This program calculates the X,y poin+.s for **"
130 print "** various streamlines **"
150 print "****************************************************"
160 print
162 dim y(4,)
Iv If-;
{"yo cqJc" 14. 1:,~
165 print " x y(Psi=l) y(Psi=2) y(Psi==3) y(Psi=4-)"
170 for x=-10 to 10
180 for psi=l to 4-
190 y(psi)=psi/(1+3*x R
2)
200 next psi
210 print using "###.# ###.#### ###.#### ###.#### ###.####";x,y(1) ,y(
2) ,y(3) ,y(4,)
220 next x
( C&>I'/t )

Tabu/alea res",Its -h'l' if =/; 21 ~ If ol1d a... fltrt .sh"uJ/~" the
..:51-f"ell/l1I1~~ 4J1'~ 71?-t'11 ..6tluw.
~***************************************************
:7: ylPsi=l) y(Psi=2) y(Psi=3) y(Psi=L,.
-5.0 0.0132 0.0263 0.0395 0.0526
-L,..5 0.0162 0.032L,. 0.0L,.86 0.06L,.8
-4.0 0.020L,. 0.0L,.08 0.0612 0.0816
-3.5 0.0265 0.0530 0.079:) 0.1060
-:3.0 0.0357 0.0714 0.1071 0.1~29
-2.:'- 0.0506 0.1013 0.1519 0.202:)
-2.0 0.0769 0.1538 0.2308 0.3077
-1. 5 0.1290 0.2581 0.3871 0.5161
-1.0 0.2500 0.5000 0.7500 1.0000
-0.5 0.571~ 1.1~29 1. 71~3 2.2857
0.0 1. 0000 2.0000 3.0000 ~.OOOO
0.5 0.571~ 1.1429 1.71L,.3 2.2857
1.0 0.2:;,00 0.5000 0.7500 1.0000
1.5 0.1290 0.2581 0.3871 0.5161
2.0 0.0769 0.1538 0.2308 0.3077
2.5 0.0506 0.1013 0.1:)19 0.2025
:'. a 0.0357 0.071~ 0.1071 0.1~29
3.5 0.0265 0.0530 0.0795 0.1060
! r)
Lj,. '- 0.020~ 0.0~08 0.0612 0.0816
~. :' 0.0162 0.0324 0.0~86 O.06~8
:).0 0.0132 0.0263 0.0395 0.0526
5~----------~----------------------------------~Legend
.~--E)Ps1-1
" 6 Ps1-2
4 • • Ps1-3
~---+I Ps1-4
1
o
-1L-~__~~~____~__~__~~~__~~~____~______~
-6 -4 -2 0 2 4 6
x

6.25* The stream function for an incompres-
sible, two-dimensional flow field is
'I' = 2rl sin 38
For this flow field plot several streamlines for
o~ (J ~ n13.
Tile e$u..b~n ;;" . <t st,~qm/;;'e "' HII"d b!l sellin.!
/n t'he "illlt&t''' .,4:>1' 111e sfr~"m Alndlc" . TflIIs~
'lIven sfrfl1lJ1 Itt/utt;,,,
If= 2;-30$';" 3~
file fft"ai:IOI1 cf ... . strellm/lli.,
r=//f )~
(Z ~i. 3 B
I t 1<>//(lI<IS IJ
where VPrt"ll.5 t"",f,,,,i VII/lies CQ/1 be I/J5/fjlled-lt,
1-0 obb"j., a ibm//!! ()r shwm/Jites. ;1- ~r()Jrqm ,I"()I"
epic/daMI)' f"lJe ",.!J ~"rp;,~ks (t./;'"re JC=yc~se "",/
'j= /-5/"19) o-f yarlt'UJ sftepm//II~ ~//ows .
100 cls
110 print " ~* '~***~*********************************************',
1 20 print "-..>1' This program calculates the x .y po ints for **"
130 print "** various streamlines **"
150 print "~****** *** **************************************** ** "
160 print
162 dim psi(4). ,:C4J,y(4)
161.1 print" Psi"'1 Psi=5 Psi=10 Psi;20 "
165 prin~" x y x y x y x y"
166 pi=l.I*atn( l.O)
1 67 date 1 . 5.10 .2 0
168 fo!:' i=1 t o 4
1 69 read psi(i)
170 n ext i
1 75 for theta=pi/ l 80 t o 59*pi/ l SO step Pi/45
180 for i""1 to J,.
182 r"'(psi(il/(2*sin(3 *thetaJ))-(1/3)
185 x ( i J"'r*cos (theta J
190 y Ci)=r*sinltheta l
200 next i
210 print u sins " 1t.UIt M.UM II.UM *,UII II.UIt it,UM II.UII
x( 1 ) .y( 1) ,x C2) , y(2) , x (3 ) ,y ( 3 ) .x ( 4 ) ,y(t... )
220 next theta
It.UIt" :

( C{)t1 '-L )
Tab,,/a.f(d Y-l'.sllih ~r
The ~ fl'"ellm/J;je,.s Cire
If ::: I; S; /~ 2CJ
91tifJJ bellJw.
~***************************************************
** This program cal culates the x, y point~o foy *)f'
** various streamlines **
****************************************************
Psi=l Psi=5 Psi=10 Psi=20
x y x y x y x y
2.122 0.037 3.628 0.063 /,),.571 0.080 5.759 0.101
1.2/,),1 0.109 2.122 0.186 2.673 O.23/')' 3.368 0.29::
1.020 0.162 1. 744 0.276 2.197 0.348 2.769 0.439
0.902 0.208 1.543 0.356 1.9/,),4 0.4/,),9 2.450 0.566
0.826 0.252 1. 412 0.432 1. 779 0.544 2.241 0.685
0.770 0.296 1.317 0.505 1.659 0.637 2.090 0.802
0.728 0.339 1.244 0.580 1. 568 0.731 1. 975 0.921
0.695 0.385 1.188 0.658 1.496 0.829 1.885 1.04:)
0.668 0./,),34 1.143 0.742 1. 440 0.935 1.814 1.178
0.649 0.489 1.109 0.836 1.397 1.053 1.760 1.327
0.635 0.552 1. 086 0.944 1. 368 1.190 1. 724 1.499
0.630 0.630 1.077 1. 077 1. 357 1.357 1.710 1.710
0.638 0.734 1.090 1.254 1. 374 1. 580 1. 731 1.991
0.672 0.892 1.150 1. 526 1. 449 1.923 1.825 2.422
0.802 1. 235 1. 372 2.112 1.728 2.662 2.178 3.353
5
Legend
0 oPsi=1
0 a Ps i=5
4 ~ l>Psi=10
0 e> Ps 1=20
-1~~~~__~~~__~~~~__~~~~__~~~__~~~~__~~~
o 1 2 3 4 5 6
x

6.2~ A two-dimensional flow field for a non-
viscous, incompressible fluid is described by the
velocity components
u = Vo + 2y
v = 0
where Vo is a constant. Ifthe pressure at the origin
(Fig. P6.2') isPo, determine an expression for the
pressure at (a) point A, and (b) point B. Explain
clearly how you obtained your answer. Assume
the units are consistent and body forces may be
neglected.
Chec.K ~ ;U. ;'{ ,?'/()uJ is Ir'r~t(J.:I:r()~A/. 51Y1c'e
y
B(O,})'
A(1,O)
Po x
FIGURE P6.2~
W ::.J.. (!.Jt' _ d LI ) ( /3"'.~.I2.)
2: J- ~;(. ~ b
an d lor fne rive" (le/~ctf'J c!tSfrlbU.bl()#.J ~::::.{) 1111# j~ =Z I
; f fr; II~W.J thfi I- tU~:j:. O. Sinee I/~w i.5 nat /'rr()tait{;;14/
CLJl1noi ().ffJ/'1 the. i3ernou.lli .f!gU(:I:I~1I bet.Wtel1 an3 fw{) poin1s
In the. f/()w lie/d.
(a)5/I1~e 1/=0, 1"he ~Y'19;11 and p()/ni A t'/Y'e {)IJ the
5.ame. .5 frelJln //"e.. Th us)
..po "t- ~ 2, = -P14
;r :L~ 4'
At the Dr;'r1n
f.r~m E~. 0)
~~~
(.b) ~/~ t 15 Ij /Jot ~11 St'lh1e ~rrY?II/'11j/lle as tJJ"/9In ~o
aff)::; Ber/')tJul// e!"lLt'~# betw~el1 13 pnd tJ. 10
Use the '1- wm~J1ent ~ Gu.lfY'; .e~tlA,t:'tJl1'):
L) ~ -!...! = -<J [JIl" ..,..u Jlf 7/)tr fAY Jtr)
r- (f!J ; ':J r pi pX + ; 'J l' ) t J
1fJ =0 )
=0
c,- 2'1
(I)

, ~,l7 J
b.Zl In a certain two-dimensional flow field the velocity is
constant with components u = -4 ft/s and u = - 2 l'tis.
Determine the corresponding stream function and velocity po-
tential for this flow field. Sketch the equipotential line 4J = 0
which passes through the origin of the coordinate system.
k= *"" -t11e lie /"Clf-rt t~I7If'J"flfh J/vell
~ =-i{-
p':;
~ _ '7
J;< - .(,..
£'1' (;) /AI;rh Yt'5f~Gt ti, b +0 OhtrUh
jdIf ::- f- 1/ d!1 7- ~ (,-)
0)
{2 J
If =- If ~ -(- +; (x) ( .3 )
iI/feJr4i:e e1.(Z) /AJ ifh res/ni. .Jo)( .j." "b1711h
fd'l .:: I Z d;l. + .{ (!J)
If = Z X. -r I&. (j )
ThU5) -h Sa I:I'S 1'1 /;ith £ss, (3) a11d (If)
If,: 2;( -Lf!J -f-C
wher'e C IS till flrbi 1rtJY.!J ~51:t;I1i.
tf(J)H1 1he deli'''t'.f'47;' 0./ the lIe/()cif!J ffJ-/;tlfhQj
fA.. '= @j v= II~~ p~
00 fh~t /r;, the lIe/f)CI'nj Mf11POflfttt.s 'lIvefI
E-i =: - 4-
J;<.
!.-4> = - z
J':;
{,-30
(5")
(~)

t..27
In.f-efrio k E~ .(5 ) (,IJ /tn '('eoSfet f.j.y, x .f,; 0 /,f//I it
fd4 = f- ~dx + ~(!J)
~ ::- - It-x t !ar!J)
k+efYllk 13;, (It,) w/h re5fee..i b J -k aj,.j",;,
fdp =J-z d:; -t 1Ij.(~)
OY'
9:: - 2.) -r 4('1.)
Thus) -k 'Sa f:/'5.f!:t b()P, E$S, (7 J tlJtti {a)
¢=-/ft.-Zjt-C
t.Jhfte C l:S (1;1 aY'blfr'4Y; ~slo!1i.
jJl1C.e, ihe .e!lIlfp,LPHf,q/ 11I1e) rI~a) flfs.se.5 1hr()1I9h Pte
()r"91i1 ()C:.y=o)) ihel1 C.:::. 0 ;'it E"f,fr) So 7hlff -file
e1litt1n~H ()f 71t( ¢;:::o egfJ'Jeftlff'4/ /111' IS
2 fJ =: - 'Ix.
~=-,Z)(
,,-31
( 7)

C:..Zg I
6. 2.~ The velocity potential for a given two-
dimensional flow field is
¢ = (!)x3
- 5X."2
Show that the continuity equation is satisfied and
determine the corresponding stream function.
to sai:ls/!1 fhe ~"-t/I1,,,if!:1 ~0UIL t/(~i1 )
~ -r ~=0
fi,y -t;,e- ?/~t'1'I tle/~c,'..fy f'ctel1tJdil.;
u= ~ =(3)({))C2.- S !JJ.. ~ Sx 2
-5!:J2
- IOx.~
Since
fA. =' ~ ::- S';{2. _ S"!:J l.
, P!J .
I/'}I-e!rl/I'h~ With re.s'pe~i.f() ~ 9Jlle.Jand
//11- : if5";( "-- 5'!/) d!1
~ =- S (x1 - }J) + !, ()f )
51ml'Jarl, ) J...,,.. - _ J == - lOX w
v - ~X ..J
til'ltl Inff,rAitl);' wi"tl, re.s;evt to x. q J i/e..s
If::: 5';( 2!:J 7- -f;z. (!J)
10 sa-t-';/!; both Ets, OJ qn~ (2),)
tf -- S x 2
!J - f.:J 3 -f C
0)
( <)

6. z'i Determine the stream function corre-
sponding to the velocity potential
¢ = x3
- 3xi
Sketch the streamline If! = 0, which passes through
the origin.
fA.. = q.J: :: ~ ¢; =- 3X 2_ 3!:J ~
J'j ox
Znlet""it! tVI n, (,PS/~C.:I; -h !1 -t() tJb';"1;-;
!dI.f : fr3;1.2- 3fj 1..) dJ
..5!;"';/lIr0 J
1/.- - ij: ~ 'J.J. = - Iex!:J
pi ~!j
and 1~I-e'lrAfllJ', with Y'eJ/Pc,t 1:. x. f1/~'/ds
jd If ::- fix!! d;(
If -= 3x 2!/ t !;.fjJ
To 'stl.ti.sfJ h(!)"/h GiS' (I) 11114 (Z)
Lf:- 3x"J.y _ :/3 + {'
(2.)
wheY'e C t.s
passes 1hr~tlfh
4h tlY'bilrlfll"fJ et)lIsi7Il1t. '51;'C~ -the ~+re4m/;l1e 0=0
1J1e. Drlflit (x.:=())y::o) t"t 1:,/I~tUS th~t- C=o qi?~
"3
tf= 3;(':; -!:f (3)
The. -e~uQ,bol1 0/ the stream line
p(J..sS/n~ tnrotA.:Jh the ()Y'lgl;' I~ /o/.{l1d b'.t
5f!H;'n:; If=o IH £~.r3) -1-0
ij/eltl
fJ=!-V3x
,4 ske icJ, ~I ilJe If =0 sfrellm);I/(;S
4/'e ~hotVl1 IH -the lijwye..
b-33
r=c
x

630 A certain flow field is described by the stream func-
tion
r/J =A (J + B r sin (J
where A and B are positive constants. Determine the corre-
sponding velocity potential and locate any stagnation points in
this flow field.
1h - .!. Jtf - d ¢ - .e + IS ~S er- r;n;--;n:,- r
:t"'nkJrai:e. Win, rf'.s~",f ~ ~ ~ ~bl-olj,
fl ~ f 1-t 8 ~.s(?) dr
tf ~ A /11 r t B yo ~.s~ -t- ~ (~)
S/mj'JAr/~ )
;<J r' _ _ J If _ .J. ~ ¢ = - B51n8
VI? - Tr - ;. ~
Or
aH~ (
f; ~ =- ) 8 r-~in l} d ()
tj:; /3 J- ~s~ + t; 0-)
To ~a,i-'s.f!j bOin EJS. {2) ttH PI. {If}
tf ::-;.J. /n r + B r-et;s e +C
fA)hey~ C /~ tin arb/.fr~Y.!:J ~nsi:(JI1I-.
(2)
(If )
51:a.'I14./:"18/~ ~/;,-h 6CC.~~ U)"er~ Vj.:;() al1ll ~.:o .
PrtP/11 ~r. (3) ~=-o IJ..-t: G:o a~",< () =-rr. t=r~1?1
~!. (/) wi'fh 9=0
1). =-$ 1- /j
A /1 •
So thlJ,i 7J;. : 0 h>r' r =: -"8 . /'1TJ£l.}ev~v.; ~/11"! A- 411d []
t:(t'<l.. b~Th p",joh,i.e, ~".$'frlYJ -Is "n1l~ Yes,,/-/:. il1d,ca.1-e.s a..
fie!},,+-, v~ va/"e ~I' r 1AJ/11i.J, /oS I'}()t de fined.
Ii--t t9:; 7T
17f,. =1 + e
~o -thAt ry.. .:::: 0
sIT;..111Ai-T~JI f¥'llit
e==-lT
o (!C UY$ a,t;
,A.
aHA r="Jj

~, 31
6.3 r It is known that the velocity distribution
for two-dimensional flow of a viscous fluid be-
tween wide parallel plates (Fig. P6.3 ) is para-
bolic; that is
with v = O. Determinoe, if possible;o'the corre-
sponding stream function and velocity potential.
- - -- -
To de-t-ernll;,e -!he sireQI>1 hnc/.JoJ1 let
~=~ = [{ [I-(t)~]
an'; 111/efrg.fe WIn, YIZSf~C.1: .fv.!:f 1-0
fdY! :: f~[,-{t)jd,
If = 0; [ :J - 3~: ]f J; (;()
FIGURE P6.31
J'f7/= - =a
lX
) I.f Ij net A. I'UJ1c. t-/O~ /t).{ X f /t) -/hit t
[{!1 [1- ; (-/)Y+ c
where C /s an ar/n.ftor!:1 6!JJ1stal1t.
Tc de.termJYJe. ihe /I~/tXrl"!:J ,()ifJ1-/;/~/ let:
fA. = ~f :: 7{ [I - ({)6)
I~ief"..tlie wif1t Y'e~fect -/;0 x 1:0 ob-l:(lil1
PI = J 7:{ [, - (%f-) d;<
1 == L{ [;< - £i)6;e] + i. t)
~,..- or/; - 0 -
v - d!J -
_.2D(X!1 + Jo-fz(.!J}
h'J. dJ
.x. tlnd~. Thus) "there 15 not q ve/oc./"ft1 ,Poi:.ent/q/ 17111"/:
desct"lues fl1/j -'c/otd ("17te .fl()w IS /J(Ji l'rrc;ill..-btJnql) .
====================~
6-35

6.32, The velocity potential for a certain inviscid flow field
is
cP = - (3.ry - l)
where cP has the units of ft2
/ s when x and yare in feet. Deter-
mine the pressure difference (in psi) between the points (1, 2)
and (4, 4), where the coordinates are in feet, if the fluid is water
and elevation changes are negligible.
5/;' Cf!. 1n~ 111)4.1 -/ie /4 1..5
is /'rro &a.t-II)H(J 1 01'1d '111~
tin (j .fwo pDlni:s. lit", s"
de.sC!·jbut by A. I/e/{J"-fr; pot:ellt';'/ -fhe /I'H~-
13~".n~"//i -eJ/,la..t,tD~ ~11 be appl,;d beiweei1
3+ Yt'2. = OJ
J- 2..#-
fA..':: 0 t/J : -" x. '"P;t. ..J
At )(.= / -I-i I 'J = 2 /-i:
U = - ~ {, )(2.) = -/2.f:t:
I s
'V'"r :: - 3 (I) 2+ 3(t) '3..:. '1~
I S
00 thll.. i:: V;"2..: 4/'''' 'Pi 2.::: (-/2 1)z.. T('f +:)2..= 2- Z£"(7)2.
A-t x=- ¥ ft J 'J:::' if ft
U2,= -b{If){Lf) =- - ft:. £:#oS
1-i :: -3(tf)4-r 3{JI)'l.=o
So 1h4.i ~ 2..:" (_ f~ ~) ~

(p.33
6.33 Consider the incompressible, two-di-
mensional flow of a nonviscous fluid between the
boundaries shown in Fig. P6.33. The velocity po-
tential for this flow field is
<p = x1 - y2
(a) Determine the corresponding stream func-
tion. (b) What is the relationship between the
discharge. q. (per unit width normal to plane of
paper) passing between the walls and the coor-
dinates Xf' Yf of any point on the curved wall?
Neglect body forces.
(a.)
y'

/
"r 1/;=0
FIGURE P6.33
To qet.errm·"e tf 1~t-e1rll k (# /h re.sfu,.1: .;.. !J .;.. ob-blIH
SIMi/III" /'1.1
So thAt
fdtf = p.)(d!:J
'f:: Z.x!J + "Ix)
J If _ ~$
'1r~ - ~ - o!J -;: - Z ':J
Jdif:- !Z!:J dx
If == 2..x.!J + hlj)
To sa./;i.si!;t 6"th E$s. (/) '1111/ 11..)
tf=:LXj+ C
z
x
(I)
(2 I
where
C :: 0
~ IS tln ArlJilrl'lf'fj ttf)II-f-/:qHt. S,nt.e tf=O a/"nJ !f=O
tll1e{
C.b) The
the
if =- Z)C'.:J
dischaY'~e) b) pqssin9 1hrf)U94 tll1fj stlr~ce
+WD UlQlls) stich as .48 (see nfllre) I /:S
t ~ ~ -911
F1 .l3)) ~ =0
(3)

(a..> -r;
F;;y
So
6.3 Lf The stream function for a two-dimensional, nonvis-
cous, incompressible flow field is given by the expression
!/J = -2(x - y)
where the stream function has the units of ft2
/ s with x and y in .
feet. (a) Is the continuity equation satisfied? (b) Is the flow field
irrotational? If so, determine the corresponding velocity poten-
tial. (c) Determine the pressure gradient in the horizontal x di-
rection at the point x = 2 ft, Y = 2 ft.
sa.t/s4 ,.;ne C{)n IIi,,"fl1 eSti a.-/;IrdI'J)
J fA -t- a.J!' =- 0
p.JC. Ii !1
iJJe :5l-rellh1 knc.-tI4Jn
.
91,;~n)
qJf - 2- fr v-= -l!t =2u= - aX.P!j s
.It:
s
anc/ 7Jte ~111"/~tJ/';'!1 elulJ.i/~J1
( fl/ofe: 0J,e" ~ //(')w .field /5
-the ~l'1il;'u/-f!J egua.iliJ;t /,$
1.5 's4tls/;ed. Yes.
dt:-hht>d h!:J A .sfretlm IUnc.t:I()·J!
(lIWtHIS /del7l-lc~,;1.:t .saill;·//ed.)
(h) S/~{~
(/11" ~fA
- '='0
;;tr
alj a;<. .:: CJ
;t- 4//aws 1hat CJJ =a Clnd the .(/()(,U -hidcl 15 iY'r~ia-ti()1f4/. Yes,
~ =
tA-= 04 = 2-
~;<.
Clnd 'l1'.fefr4t:,olJ !f/idcb
CJhere
(c) Wi1n
¢ =2{x +!J)+ C
C /s an "Y'bl.fran;j C/Pl?si:"d.
the .x-a;(t'~ h~Y;1/)11"~ /) d~ =0 J anv(
- ~ -= /J (IA. 0'" T v- ~ )
lJ;<: (~ J';('''~ 11
anll a.i: X= 2. ~t:) , =2.+-1; ~f = ~ [z ~r (0) T- 2 ~t-(0)J

c;,. 35" I
6.35 In a certain steady, two-dimensional flow
field the fluid may be assumed to be ideal and
the weight of the fluid (specific weight = 50
Ib/ft3
) is the only body force. The x component
of velocity is known to be u = 6x which gives
the velocity in ftls when x is measured in feet,
and the y component of velocity is known to be
a function of only y. The y axis is vertical, and
at the origin the velocity is zero. (8) Determine
the y component of velocity so that the continuity
equation is satisfied. (b) Can the difference in
pressures between the points x = 1 ft, y = 1 ft
and x = 1 ft, y = 4 ft be determined from the
Bernoulli equation? If so, determine the value in
Ib/ft2. If not, explain why not.
( a.) To saiisly the (!t),rl:,~tljf!J eglla1:U)# )
Ju. ocr_t:)
~x T ~ -
17 u ::: ~ i t .t~//{)141.$ 71111i,,;(.
;v-
J ~ =-t,
(!q11 bi! /J,.J.efrAfe,f w~1h re.ste~t t:~ !J -Ie :J"eJ~
1/= - ~J + ~ (;()
Since 11" IS I7tJt a. HtIfCi:.If!i" ,,( x Rnd JJ ~er() at The. OrJif/ H)
-rr=-6 ':J
,.b) The I3frn"u.JJi Rgu4f/oJt Ct/n he QPpJ/~~ hel:wten 4"!J +Wo POln+':'
/f -I-he ·f--/"w l.s J yo/"'(') i It i:t011"J. :sJJ''lCe..
t.v=J.(~V-_~) (Eq.h,JZ)
i- :2.. Jx " 'j ()
( 2)

SinCe
(Pith
6.3~ The velocity potential for a certain inviscid, incom-
pressible flow field is given by the equation
4J = 2x2
y - (ih,3
where 4J has the units of m2
js when x and yare in meters.
Detennine the pressure at the point x = 2 m, y = 2 m if the
pressure at x = 1 m, y = 1 m is 200 kPa. Elevation changes
can be neglected and the fluid is water.
1h~ ,fJ{)w i..s I ;Y'() -64. 61f)ila J)
.p, Yj2. .fJ2. {.2..
- + :: - .,.
J- - r '-}.2.;
)'~it11 II; J 'J JveJ1 J
~c$ 7. 'Z.
~:: Z;{ -2!J
Ai fDI;' f / Ie t )(. =. I h7I
If (1)(1) tf~
u.. - '::
I - S
CI n cI
V 2..::- ('I ~ ),. ::
"/
f()/~ t :2
IA :: J.f (z)(z.) -
~
(/" p)~ -
(:oul
!J =. 11m ~()
v-: :- 3.
~ (I) -
I
/1112..
-$'&
,uA"Z.
-s .....
::. &'0. I ~ Pa..
tn,i
:2.. (J)z = 0

6.37 (a) Determine the velocity potential and the stream
function for a steady, uniform, incompressible, inviscid, two-
dimensional flow that makes an angle of 300
with the horizon-
tal x-axis. (b) Detennine an expression for the pressure gradi-
ent in the vertical y direction. What is the physical interpretation
of this result?
(ct,) Pr'l?7 1::$5_ (,..80 anol /'.8/
(bJ
tf ::- C/ (.>< t4sol. -+-:J :s I;' 01..) (!='J. ~, 36)
(iI/Pi ~r ot.::: 30"
tf = 0- (;< OJS1o" + ~ S/11300) ~ V-(~,8'b;( + ~.5'h~!1)
Sin"/111' /!jI
tf =v (!f U;sd - x. sin~) (Ef. ,,80
t/ h a. fpr 0(.:; ;-"c
If '" v ('J ~jJiJ'- >< J /" :11/) .. l/ (~.ffU,J - O. S()(),( )
SJ~'{'
U. :: l:; cfi ~t'lt11l 1,r..::
, ~ ~'::J
If ~11~tI)J 1nd
1A.::~.f"V- Cll'ttl V-:::. ~. 5"00 r;-
p:;,.f)I?J itte t=IA fev -R. ~ tl.1L-tJix I ~ "the IIfr1-/CA. J .!J- d I y~dtolt
1) t:1 - £.1 -:: [) (0 IT+ u ~Jr -r V- e.r + JAr E) (£='9, '.Slb)
( if ~ J ~ I dt 4~.x P!1 d r 6
anI! hi I in 1r:=' &;115.fz1l, t Clud #1):: - ~
*~ -tJ
or
/fJJ~ 'r'f1jI.(JI:: /Yld,C4.J.e.s jh4/- 1he pres.suve dJshi hU.J-,D~ 1.5
h'1d"~s-taI-(G ,Th/~ I~ nDi ~ ~l1r/,Y/$ln, Vt'.5u/-I: .sInce.
-tJ, ~ ~ frn~/,(II; e$U¢I-/~11 Inti, c.al-e..s ~ f II- 1Ittrt.. loS
no ehfll1'1e. 111 Vc. J~'-/:;'.!::J 1h~ eltlll-1pt' 111 pv-e.fS"Y~ Ii
Slmpl'J ~jtl e ..fa */he. We 19h f of fh(. ,fluId.; L: Po) a..
h fj drt>.sta..J.ti. v4 y, a.1-tt>n'

(a)
6. '38 The streamlines for an incompressible,
inviscid, two-dimensional flow field are all con-
centric circles and the velocity varies directly with
the distance from the common center of the
streamlines; that is
Vo = Kr
where K is a constant. (a) For this rotational flow
determine, if possible, the stream function. (b)
Can the pressure difference between the origin
and any other point be determined from the Ber-
noulli equation? Explain.
~=
_ J If = k y-
17","
En.f-e? ra te 1£,.lI} w/1h re.sfec.t.
/dt{; :- - for dr
"y-
1<: r"
if :: fIe)- -r;z.
Sinct:. _ ( ~ If
11 - ? u; =0
/t Iol/()ws 1ha:t t.p J.5 ntJ1.. a. fu11 C. t /tJJ1 ()I a a;,d 1heye~ye.
~=
- J<;-6. +c
""wher-e C
.
arb "+va. V:J Cton S1:411 t.IS an
(j,) The Ilew Js l"()ta..tl()~IIJ ohd Therefore fne 13erl1t/uJJi
e~ U~&/DI1 CA nt10t be o..pp);'ed betWteh --the.. oY/~/i1 a h d
q 11 ~ pOInt J SI h ce These fo/nb QV€ }1{Yt 6)J1 The
Same ~fr"etlm //l1e. Nt).
(I?e.fer 1-0 d'SCtI$S'DI1 t'fS.sOCICl·l:t-tI tu/fh der/~IJ.I:I~·h
olE;Z. it>. ~7. )

'.31 I
6. 39 The velocity potential
¢ = -k(x2 - y2) (k = constant)
may be used to represent the flow against an in-
finite plane boundary as illustrated in Fig. P6.~t
For flow in the vicinity of a stagnation point it is
frequently assumed that the pressure gradient
along the surface is of the form
ap = Ax
ax
where A is a constant. Use the given velocity'
potential to show that this is true.
Fe>y the IIe/()c. /-J.!J poi:en.f,QJ 7' vefJ
LA.. == ~cP - -2.~)!.
a;. -
11"::-
o~
= - 2. '" !:J-~b
Clnc! tne. S~JtJA.ft~11 pt!),~i () CCtlrs
;:';y 'this -6fea, ':J.- 1-kJO - dlmt/l.s/~II/
y
.._._---- ---_.-
FIGURE P6. ~9
at the
...~r'.j In
/It;w
/
X
u)
(Z)
-ij. =-t (IA, ~ 1" r~)
a~ 1))< tJ!:J
(£:. ~>/o.)
Qnq (lIon? -the. SUyfac.e. (:;=.o)
?l. : /JU ~
1J.x. I dX.
v-= () So
t-;..t)m E1. U ) u ': - Z--k:X.
au.: -z/{
p;(.
qnP( Gg .(3) beu;me.s
~ :: fi) (- Z Ie)/ ) r-z~) - IfJe G,x
PX I
where
1J,a.i
(..3 )

'.'fO Water flows through a two-dimensional diffuser hav-
ing a 20° expansion angle as shown in Fig. P6.40. Assume that
the flow in the diffuser can be treated as a radial flow emanating
from a source at the origin O. (a) If the velocity at the entrance
is 20 m/s, determine an expression for the pressure gradient
along the diffuser walls. (b) What is the pressure rise between
the entrance and exit?
11 =2.0 ~~ 5(> "/hILi:

~
2m
~----- ,. ~
- -"",,~,
o -~~ Flow
:~ran~7 m .
~/Exlt
1
(Sf'e Ta'ie ,. J )
1??'1 -= 2. rry. 7rj,.:: 2. rr ("2.hH) (2t!; ? )
Pr~i'Yl /he. /3fYI/~ul/,' ~ lJ.4.;/:J()·}1
-p -+ it 1"';-"2.. = ~(Pl1slr1l1t:.
~ = -~ v;. ~ tJ;.
"y I ~ r
:50
(I)
' .,,..,.,.", -1-1. 0 v;. ==
JJnc.e VI'":; ~1T'Y Inel1 ~ Y'

6.4 I An idealfliilcffiows betweenthe-inciined
walls of a two-dimensional channel into a sink
located at the origin (Fig. P6.41). The velocity
potential for this flow field is
m
¢ = -In'
2n
where m is a constant. (8) Determine the cor-
responding stream function. Note that the value
of the stream function along the wall OA is zero.
(b) Determine the equation of the streamline
passing through the point B, located at x = 1,
Y = 4.
(et)
r n fe~t"rt .J.e. ~g ./1) w/fh r-e.s/ec.t:
fe/if = f ~ it;:l7T
to ~
t>Y'
/t11~ + ~ f;")if= -27T
5/nee.
-~
I
qj = 0
~= - --Pi'" t- oB
Lf
,
n()iis A tunc.+;~/1 01 r- So
if:: ~ ~C.:;,,,.
where C ~
~I1SiP"t . If/~&)J,j ~
-So ~a-t
C
n?1
=--C.
qh"
(~~
I
)/jJ.: /1"n --&:.
"
Ir, ~6io/11
eS .f'l)
o
FIGURE P6.4
bectJl1Us
!p:=.o ;;'1' ~ ~ It.3
A
x
( .b) At 8 I-tJn ~:o;- So PtII. i: ({) .:- /. :3 3 rd d . J:;.I!)/YJ ~ g.t3 )
.
f11e I/Idw€ of tf Possil'1f -t;,Y~w'lh ih/5 ?(),hi 1.5
if ~ t?'n (1.33 - ~ ).:- D. ~1fS"olYl1
211 .
( I )
(1. )
(..3 )
4,nd there~re fI1e eg(J~biJ,J1~.f 1he. sirellll1hlJe. plIsslnJ 1hrtJL(jh 13
IS (). 0 "SO 11'11 =/)?1 (~~ - i )
(fj/) Ie.' It CtlIJ be seefl /rom '&1. (3) 1htl.t fAt! ,;Irflllf1lilJfS
tlY'e II/I ~"""'1I9I1t };heJ f'k~.sl'n~ 1hrlJtI&11 -!he "yiglit. )

6.42. It is suggested that the velocity potential
for the flow of an incompressible, nonviscous,
two-dimensional flow along the wall shown in Fig.
P6.4Z is
¢ = r~/3 cos !8
Is this a suitable velocity potential for flow along
the wall? Explain.
:tl this ;5 et suifaJ,ie ~ -fiJe C/!)rre.sp()ndln~ tfJ must hAVe. et u;IIsT:t!l1i
I/o/we fI/on~ the 1U1/1/ (s/nce -the tvll/I tnust tt:)rl"'''~}'~l1d --h a sfr~A/IJ/i"e).
-...L ~ d¢ - If- ~ If
Vj - r JB =- ;r;. - "3 r ~~.s"3 & (I)
:1:nk1t'Aie I:iS.tJ) w/tIJ reSfect 1:a B 10 ohtolH
jdif:: f .!j f-"~c..S : e
where C
A-JotJ? one
1ne other
c.~n5 1::11n i
t~)'feI'rhq,j
If:: !- +£/.8 ,JIb ~ B T ~ tr) (2)
/I ~. I,
:: - ~ r o5J J:j :t:. e3 .3
If
if = r -; .5;'1'3 : S + +;. (8) (3)
both E'Is, (2) an4 (.J)
If:: f 1f~ SIn : B + {
is an (/rb;~rlJr!1 e()nsbtllt.
se~b~n ~I fJu! wl//I; t) =0) Qnd tj;::: C. A-/()nf
set..-bo'n e::: V a nil tp =C. Th us) tp hils ~
va/we. a/f)!'''} ihe WtI/J and the 1/vel1 j)e/~c.;-h;
Ct/n b.e used -k y~tYese" t f~~ a/tJ"1 the Will/. Yes.

ft" if.3 I
6.43 As illustrated in Fig. P6.43 a tornado can be approx-
imated by a free vortex of strength r for r > Ret where R, is
the radius of the core. Velocity measurements at points A and
B indicate that VA = 125 ft/s and VB = 60 ft/s. Determine the
distance from point A to the center of the tornado. Why can the
free vortex model not be used to approximate the tornado
throughout the flow field (r 2: O)?
t } fA R x
" /
.....- --- / I ,"".. I
" //~
"...... ./
----,.,.,....
• FIGURE P6.43
Thus) ILl:
(:011( a.-t
There /r:;re J
.~/nce
fA
13
)
~:
) ~:
/ ZS' fiJ .::.
/zt5" it Sf)
.5 )
(Po H
-:r ) ..50
Go 'B
"k -fA =/{)() l-i
/1- f~//t9IPJ ~t
J2.5" 'A ::: I> 0 (/~O +fA )
fA. ~ '11.3 It
ina i Ie =/zS' fA
that: /G:: 60;-/3
7h e hee vorte.(. C/ln/Joi b~ tlsed .j." CljJpro,(i"'41:.e t<. /:or)'UIII'o
1nrou'lhoJl.t the /-ltJUJ .fle/d Since at:- r=o The
t/e/o~ifrt b~~me5 /;'f;';,/fe.

6.44 The velocity distribution in a horizontal, two-dimen-
sional bend through which an ideal fluid flows can be approx-
imated with a free vortex as shown in Fig. P6.44-. Show how
the discharge (per unit width normal to plane of paper) through
the channel can be expressed as
q = cJ¥-
where Ap = PB - PA' Determine the value of the constant C
for the bend dimensions given.
(1 = 0.5 m
b =0.9 m
/
• FIGURE PS.44
1< k
~A = a:. ~S: 7:
fhe. 13fl'o/)~f,t/ It; ebafi"-6u:;'n
1~ .,. ~; = P13 +-
~ 2..~ d-
~
-ve.e-2-J.
~ ±t (~A" - 7Ijg8'- ) = ;;; I("l. ( (J.I,.
=
or 7-
1<. ::
.111 .J!.
tt.
Thus
-1t
r"l.
( ;~ - i~))
tJp= (~ -f: )1-
-kOf'"
.b
( ' ' ) % iff(.tLt - - ;;; - i~ I'- a..
Clncl 1he~eh"~
rYt ::
C
J,
VTh tJ,'1 hI'f
. V2 jn ~ (),S' hwt
/,Q/th
c=
r~
=- ~,S"oo m'I
I
V(ft;...), /
--bl. - -{o, ~",,/"
b-'f~

6.45" When water discharges from a tank through an
opening in its bottom, a vortex may form with a curved sur-
face profile as shown in Fig. P6.4S'and Video V6.2. Assume
that the velocity distribution in the vortex is the same as that
for a free vortex. At the same time the water is being dis-
charged from !he tanlc ~t point A it is desired to discharge a
small quantity of water through the pipe B. As the discharge
through A is increased, the strength of the vortex, as indi-
cated by its circulation, is increased. Determine the maxi-
mum strength that the vortex can have in order that no air
is sucked in at B. Express your answer in terms of the cir-
culation. Assume that the fluid level in the tank at a large
distance from the opening at A remains constant and viscous
effects are negligible.
,. b )
(14-
8.,,-l.r'Z..J
• FIG U REP 6 .4-S
A-/r w/Jj be.
/hws)
_ J/ I ....L ~ Jh 7..~ =- / f' 1 .fr,". ;- =- 2 it.sUCJC.et /I1r;~ P'fe. "'" en J::... -r-r:
Ir I = /01 -s

6•. /fb The streamlines in a particular two-di-
mensional flow field are all concentric circles, as
shown in Fig. P6.1f{.. The velocity is given by the
equation VII = wr where w is the angular velocity
of the rotating mass of fluid. Determine the cir-
culation around .thep~~!!.~_~CD~____..._.....
Pri- d-;r=
ABeD
__~B
FIGURE P6.Lf:b
- f ~ j, df) +f-v;. dl- +f~ adt; + Iv;.fli'-
CD lJAA-13 /3,
Sln~e 7/j. =0 4ntl ~= CUr J E~. U) beOJ)me.s
fez fe,(7:: tv i/' d~ +0 + t<) a..l. c/e r{)
~I ~L
(I)

t6.~7-
4 fils
~ ------------
t
6.47 Water flows over a flat surface at 4 ftls as shown in
Fig. P6.47. A pump draws off water through a narrow slit at a
volume rate of 0.1 fe/s per foot length of the slit. Assume that
the fluid is incompressible and inviscid and can be represented
by the combination of a uniform flow and a sink. Locate the
stagnation point on the wall (point A) and determine the equa-
tion for the stagnation streamline. How far above the surface,
H, must the fluid be so that it does not get sucked into the slit?
H •
..--------~..- ..-----t-I$~
0.1 ttl/s
(per foot of length of slit)
• FIGURE PS.47
t.f= t.f"'I)I ;;,f"In
7- If:sl~k. - Ur SIj,e- !!!e
2.7/
Hew
Thu.5; I JIp m1- V CfJsB-?If :' '"F dCJ - 2rr~
Cind
- ~tp - -US/~e1/;=I) ~ Y" -
1I-/c)fJ? the willi ~:: 0 ) ClI1r/ the s.fa9IJAiltf)H p"J/1i (jcc('{I"S
wheY'{: 7/j. =0 J .so 1111 t ..f~m Beg. t'/.)
t; ::
.s
/YY1
- -
Ql1d the .Jta91J~t,ol} fo/ni IS pn the teJlII/ t!), 007'f~ +t.
-bo 1he f"lfl1t cl S//t.
( CfPl1 '-L )
('-5/·
0)
(2..)

t..47 I ( COl? 'i )
The (/Q/&1e c/ If Itt the. si:fltll1llitOJ1 f~,~i (r = ~.t)()7~~ It) e :'0°)
IS i.e,..o (E"'g.1) :so thAi the- e~t.iai:'4)H ~/ the cjk'1I1'1t/to~
sfr~t1mhi1~ i,s
Sinle y=- rSln (; 1he. .eJIi it. t/~1J 01 the. ~*,9/1I1-bP" sf"'~J1",J/1Ie
.
can b£ W/I'I ffe 11 as
!:J= :::Va
P/u/e/ aboVe the. ;5i-dfn~I:'~11 sfrellml1l1e w;11 luJi b.( sucked 1~+lJ
.:5/''t. Tlte tn4;(imlAm c/,~tt1h,e.1 HJ
~y -Inc sl-otf/llhd,,,, s-trel/mJIi1e
t;~Ctlr..s as c9 ~ 7T :s 0 ~ t
tJ.2 tt2.s --H=
Z (If '{~)
(Noi~: AI/ th~ ./It//d be/f)W -tne s-lA9nl( ttDI1 s+f"ellml,;'e must
111r()tl9h the. .s//t. Thus) f,."m t4J#S~rJ/4t'4JH ~I mASs
/-IV =- ,f-/()W ,,iii) ~ /;i
~, ~ 250 .ft
'f~.$
wh ic..A e-hec.k:.s WI'fh 71te. tll1swer a J,~lIe • )
'-52

"'.!.f8 Consider two sources having equal strengths located
along the x axis at x = 0 and x = 2 m, and a sink located on
the y axis at y = 2 m. Determine the magnitude and direction
of the fluid velocity at x = 5 m and y = 0 due to this combi-
nation if the flowrate from each of the sources is 0.5 m3
/ s per
m and the flowrate into the sink is 1.0 m3
/s per m.
At pel~i. Pr IlI~H1 The )C.-tH.U A.~ j.,,:: SMA
Th( Ve.oc..;+-I:U dl.l" -h Th~ -two SOI4 VC.eJ
e:t~~ 11te !:I1~c.. a.v( a.s ,G,llow.s:
SOuVce i
{VY1 } A= ::
/#tl.
0,5 "5=
2lT (r- z"",) ?rr (S'IW -l"""J
':1
=
FoY' The ~ i)1<-
(1r~ )A -=
fYV
W hev( y-= ~'(2./M)"l.;{s3l- = ~ 2~ ;;.,.
2..rY"
;)0 -i'hD.lt. "I.
C>.OZq~: ~1.0 ~
(V-y ) A =
':
2iT 'rv:' 1M
Th~s, a.-i A n~ ht>,(,I'~t>",~J veoc.;~ ~VYpo)'e ...tl (A.j .~
LA.. -= 0.6 S"'i ~ + 0.0"2.(05 ~ ~, (0. O"2.q~') ~
=- O.Of'i~
o.VI~ -tne ve v+ic~1 ve Oc.~+1j ~YYfo~e,,",i) 11", LS
v= 2. (O.02.qt)C!:! = O.OlO~ ~V2q S ..
T~e V€.0 c. +, (I..t A ~ nevety"~
,
VA: fLt~-t1rl = ~ ( o. 0 4-q ~';1.+ (0. 0 II 0 ~) 2-
O. olg5 ~ ~Lf-c

6. ~ 9 The velocity potential for a spiral vortex
flow is given by ¢ = (f/2n) (j - (rn/2n) In r.
where f and In areconstants. Show that the angle.
a, between the velocity vector and the radial di-
rection is constant throughout the flow field (see
Fig. P6.~n.
HW me I/e /0'/"::1 p~ 1::e11til:!I 'II IIf H J
~d ~
~'1=- :'
~I- ~ .".;-
Sln"ce --100
A
IVIV·e = CtJscl.y
oord ~ -1 "'
V :- r;. er + ~ eB
-then
~
""V· eJ-
Cbs~: =
IVI
ThUS) ;;, .,. a. rand
a.. Ct;J1 st:an t .
y
::c
FIc;URE P6.1.f ~
I or/> f7:
r 0& ::- 2.7rr
7/;.
Y1/j.1. -I- VB1. 1
=
/+
(..i2 ) 1-
.2..7T,...
(_ ;,...) 1.
angle 01.
.
IS

~.5"O I
6.50 For a free vortex (see Vidt'tI '(1.2) determine an ex-
pression for the pressure gradient (a) along a streamline, and
(b) normal to a streamline. Assume the streamline is in a hor·
izontal plane. and express your answer in terms of the circu-
lation.
1r. " .L &lIP = 0
r r ~B
5,;',,,, fne .fy~~ 114r&" refre,fnH ,ql1 ';r"td'M"/ Ilow
..f-/d4, -me I3fY'm~II; f?l''(L-t'o;,
1: + X.: t i: = i!J11ls7:t1l1 f:. (j )
J- z...~
's /lei,#( ht.'!:w,-el1 an, -I:wo poi".b.
(a.) ,4/,1/1 u .ffru",I,Atf; (r-=- wnsf""i), ~ )j ~"sfq.1-
all'" 1).=0 $0 -tIIaf from £'3.(11 w,111
:z. i!,p/ls/-tt"f- .-In.., press"" ,J C_:;h:lltf) I..'.e.)
£..e ': 0
.5.
-
ot9-
"lretlml,i1~ wi'" 10-=0 ({nil t-= cpns1:J,.f
~2. + i: -= ~"s.Jq"",
2-~
(,-S5"

~.s/
6.51 Potential flow against a flat plate (Fig.
P6.51 a) can be described with the stream function
'1/ = Axy
where A is a constant. This type of flow is com-
monly called a "stagnation point" flow since it
can be used to describe the flow in the vicinity of
y
0
/'
x
(a)
-fiGURE PGSI
lj;= A-x ~
..,.. /h1
ITt;;
the stagnation point at O. By adding a source of
strength, m, at 0, stagnation point flow against
a flat plate with a "bump" is obtained as illus-
trated in Fig. P6.51 b. Determine the relationship
between the bump height, h, the constant, A; and
the source strength. m.
y
[ X
Source
(b)
- A }--lS/11 28 + (!!J
B- ;z.. ~7T
For 1J,e. bump the ~t:(l.1nlt tlOh po,'"i will "cct/y Ai
.>G= 0) ~=/,. (B= ~) f-=-P..). FDy -the ~'L1e H ot.retlm IuYlC- tl(11)
, J if
A-r ~~ z.!I
qn
V;=?ae = oJ. (' 1 )
:z.1T'r
(JIn'
J/fv::: - AI- :)111 2.8-e or-
Th e lOIn /;) B: r) r- =- t,) w/// be A .:;-t:a1I1tft f/~;, pOlni: Ir
v;.. =0 ~ /)''Ue v;;.:::a 0. i: -htJ~ jJo/n t:. Th us) Ir!Jm £'1. (J )
o ::: A h C"$ 7T +

'.52 I
6.5Z The combination of a unifonn flow and a source
can be used to describe flow around a streamlined body
called a half-body. (See Video V6.3.) Assume that a certain
body has the shape of a half-body with a thickness of 0.5 m.
If this body is placed in an air stream moving at 15 mis,
what source strength is required to simulate flow around the
body?
wheye
b=
EZ·{,·9'1
b=
rn 15
7.SCJ
2rrb
Sf-renJth) (JHd iheve-/Cye
(/51f)(t',:;)
/h12.
-S

(P.5'3
6.53 A body having the general shape of a
half-body is placed in a stream of fluid. At a great
distance upstream the velocity is U as shown in
Fig. P6.S3. Show how a measurement of the dif·
ferential pressure between the stagnation point
and point A can be used to predict the free-stream
velocity, U. Express the pressure differential in
terms of U and fluid density. Neglect body forces
and assume that the fluid is nonviscous and in-
compressible.
!3ernrJulh' esuQ,.fl{)~ bei:ween
A- ~ ()iPTxlI;,
1;1:a.J = ~ 7" ~t ~2.
u
-----
ItIsf»
I.: 2. ='
It
V 2. (/ + 2.- ~ e,,,s e +
A
ql1d
b (7T-B)
j-=
05,-;., e
A t P()/~t A e.:. JT
2-
So 7h1J. t
b(rr-I) 7Tb
t:= :::-
2.
A SliJ r-:z..
b 2-
- .:
'h 7T
y
FIGURE P6.S3
!=.>');:2-
If
5u. /:'51:: i tLl. t/(;11 IJ f [;b.(2) ,rl.f-c) E j. {,.. 10 J :flf feb
~:2.: V- 7. ( / + 0 + =iz.)
Clnd 1n~yt krt! /rtJl11 J;g . (/)
(I)
(~'t. './01 )
(E'g . tt..J()" )
( :l )
-F]b.j = fA + i t T/ 7.. (/ + :;...) - ~ r tJ.7p3t V '-
Thus)

6.54 One end of a pond has a shoreline that resembles a
half-body as shown in Fig. P6.S4. A vertical porous pipe is lo-
cated near the end of the pond so that water can be pumped
out. When water is pumped at the rate of 0.08 m3
/s through a
3-m-long pipe, what will be the velocity at point A? Hint: Con-
sider the flow inside a half-body. (See Video V6.3.)
15 m----J
• FIGURE P6.sLf
lud.f -Io(!)d~ )
tf -= V- r-SJh B "f"
.::50 -that.
~~
V .51j"BVB =- JI-
:::
QIIP(
dfj; ~
'1/.'= 1.. ::: 77 &;S e .,...- :tvJ.-l- I" ae
Thus 41: p~J;' t A e.::t:J /-.:: IS'_
)/ I
l..-B .:: 0
AN(
1/:.=r ~: V+ ~TT(JS)
!=or 4 f/()w ra te
Source ~ /-reng'/h J~
then
V=
Z1T (S,,")
- ¥ ;IJ4f
{g,37 )( 10 S +
--
(l2g. (p, 97)
q,,~
( /)
/OI'JJ f Ipe ) the.
(I:g. '.99)
-It M4
(P.37 xl/) 7

6•.5.5* For the half-body described in Section
6.6.1 show on a plot how the magnitude of the
velocity on the surface, V., varies as a function
of the distance, s (measured along the surface),
from the stagnation point. Use the dimensionless
variables V,I U and slb where U and b are defined
in Fig. 6.24.
On 1JJ~ Stlr/ttce, 6/ the hflJf-b()dy
.b (7r-e)
/-= S/h e
d5 =: 1((d;<) 2.-1 {d.J).1
oj, - TbU
wIth X ::. r t~s !} ol1d !1 = Y' 5,;, 8-. It ~1/(JuJ.s ihpt.
d;(.: y- (- Sill b) db -t C(),s (} til"
db = V' (~se) dB of- S/nB d~
ClitP there~ye
ds :: Vr '2.(de) 2. +-(dyo) l.
, .
"V
~ds -:: r2-f- (~)" de
Let .::s~= sib qncl y1' : 1/6 So 7hA. t
, ~
ds;;' = V{rI'F+(UjZ de
r-r~m E Z. c.. /fJO
dr~_ 5111 e r (rr- B) C6'J 19
-de S/n'17
Th IA S) 'f}u.. tire /e12jfh s+ IS "Jlit 11 by
.5~= //;;1<)'"1- {~/- de
1T
O~cP='7T.
(I)
( 2. )
(3)

(e&r/t)
The. ve/oc;~ .I ~ I (')11 -!he. .s",.fp(~ ~f. th~ hal/-b~J!1 CItH b~
(!)b#//;"" #Om £'1. "./()/ WI"'I'Hl'11 111 11t~ ft,rm
V~= tis D-r 2. I!fJs!, + I ] ~- =- [p,.) l-
V j---110
ThusI ~Y' a.. 7Jv't'fJ e ) ".-1' 0111 he ~6kllle~ fr~", tf. t. /~O)
S"" .frpl'II el. (3 ) J tlif II V #- fj..Ptn Ef .flr). ,4- fY"()9f'(/ffI /t,y-
CtlJ~lAla.J-/nJ V* tls a iul'1c.t.Jl)it fJ/ S'" f,;//()u}s. (Not.~:.I;,
1}1e pYf),rlllfl V"" 1.5 c/e~J9n.kd 4.5 V' 41111 S'" QS S.)
100 c:1""
110 print "*******************************************************"
120 print "** This program calculates the velocity distribution **"
130 print "** over the arc-length of a half body **"
150 print "*******************************************************11
160 print
170 dim th(18),r(18),s(18),intgd(18),v(18)
171 pi=~.O*atn(1.0)
180 n=18
190 dth=pi/18
200 s(l)=O.
210 for i=l to n
220 th(i)=pi-(i-1)*dth
222 if i>l then goto 230
22~ r(i)=1.
226 drdth=O.
228 goto 236
230 r(i)=(pi-th(i))/sin(th(i))
232 drdth=-(sin(th(i))+(pi-th(i))*cos(th(i) ))/sin(th(i))~2
236 intgd(i)=(r(i)~2+drdth~2)~0.5
238 v(i)=(1+2*cos(th(i) )/r(i)+1/r(i)~2)~.5
240 next i
250 for i=2 to n
260 sum=(intgd(1)+intgd(i))/2
270 iml=i-l
280 for j=2 to iml
290 sum=sum+intgd(j)
300 next j
310 s(i)=dth*sum
320 next i
330 print" Theta Arc-length Velocity"
340 for i=l to n
350 print using" ###.# ###.#### ###.####";180/pi*th(i),s(i),v(i)
360 next i
(con't )

to.55 W'-I (~I1t.)
TObU/ofed da-ta.. (Jlla A. fie-/:. ~f -the cf4-itA. tire gnifH be/tJl().
::>
~
>-
~
...c
U
0
~
.,
>
Gl
Gl
III
~
c::
0
...c
Gl
c::
III
E
...c
0
******************************************************~
** This program calculate::-. the velocity distribution **
** over the arc-length of a half body **
*******************************************************
..s V5Theta Arc-length b Velocit.'}T
180.0 0.0000 ) 0.0000 ) U
170.0 0.1751 0.1739
160.0 0.3527 0.3444
150.0 0.5352 0.5078
140.0 0.7255 0.6611
130.0 0.9269 0.8013
120.0 1.1437 0.9257
110.0 1.3811 1. 0322
100.0 1.6464 1. 1192
90.0 1. 9495 1.1854
80.0 2.3052 1.2306
70.0 2.7366 1.254:7
60.0 3.2814: 1.2588
50.0 4,.0079 1. 24,42
40.0 5.0539 1. 2134
30.0 6.74,87 1.1693
20.0 10.14,19 1.1159
10.0 21. 54,87 1.0577
1.5
.......
r
1.01-
0.5 h
0.0
-0.5~____~____I~ __~~~ __~~ __~~ __~~~ ____~i____~____~
o 5 10 15 20 25
Dimensionless erc-length. sIb

~ ".5'''' Consider a uniform flow with velocity V in the pos-
itive x-direction combined with two free vortices of equal
strength located along the y-axis. Let one vortex located at y =
a be a clockwise vortex (1jJ = K In r) and the other at y = - a
be a counterclockwise vortex, where K is a positive constant. It
can be shown by plotting streamlines that for ValK < 2 the
streamline IjJ = 0 forms a closed contour, as shown in Fig.
P6.56. Thus, this combination can be used to represent flow
around a family of bodies (called Kelvin ovals). Show, with the
aid of a graph, how the dimensionless height, HIa, varies with
the parameter ValK in the range 0.3 < ValK < 1.75.
y
--- H
U
------L-l----t:--",.L1---x
----
--_7/
(I)

e ,&. (/) Illul
(.!j -t j) 2-
(-* -I)).
yk: H~. tttUP..f,f)fl (I)
( 2.)
t;,y A :Jlec;';;~1fJI va/1ft ~I Va./k 1=1.L2) ~11 1)(: S'tJ/tleA b'1 ~ +r/~j
tfnll err~r :)o/(.('h{;11 /0 t'6-klfl HI),. 5IPme -itJ/;ulak4 till/ties Alit( the
~6?rre~ff)I1I"i1j' tjl"R,lJ1J tJre $hpWIJ bel/9/.(). 3
Ua/K
0.30
0.50
0.75
1.00
1.50
1.75
H/a
2.65
2.09
1.74
1.54
1.32
1.25
2.5
2
H/a
"1 .-.- - -- ---- -- --- -- - - -..__ .__________ _
0.5 - - - --- -----------1 - -- -------1-
O,------~I------~I------~I-____~o 0.5 1
Ua/K
15 2

ro.57 I
6.57 A Rankine oval is formed by combining
a source-sink pair, each having a strength of 36
ft2
/s, and separated by a distance of 12 ft along
the x axis, with a uniform velocity of 10 ftls (in
the positive x direction). Determine the length
and thickness of the oval.
I/.,
). :: [1r~~ + I J~a...
{ ~ [( ~) •_ I ] tt<~ [2.CT;':") : ]::
a..
(Et b./07)
( E'b. ~. /0' )
/3, / Ii
Th e fh /ck.neS5/ 2..l) Ctln be de term/ned frf!)fYJ Fg. '.IOC, .h:J
+rla'/ '111'" fl'"Y'fJ)"'. AS$"fme I/Q/"Ie ~.... -/"/It.. Cfl1d CtPmlflY'e
w;'th rl9II t A4nd Side ~f E$. to. It) ~. (See -& J,1e. J,(, IDw. )
-:. ~ [(4)'-IJfah r2. (!>.l~) fJ
C!J. ZSo O. 2~9
0.25"'1 t).2'Z
~. 25"2 o. 2S"I.
O. 25"3 0.25"0 ~ use
Thus) ~ ~ D.ZS3a.,
al1c/ -thickness - 2J. = 2. (".ft)(~. 25"3) - 3. a'l- -fi:- -

cauf
6.58* Make use of Eqs. 6.107 and 6.109 to
construct a table showing how (ila, hla, and rJh
for Rankine ovals depend on the parameter n Val
m. Plot rlh versus n Valm and describe how this
plot could be used to obtain the required values
of m and a for a Rankine oval having a specific
value of rand h when placed in a uniform fluid
stream of velocity, V.
w h{)Y~ the Itl1fl1t ()f. the hody I~ Z), tlnd the W/dth IS z~.
/r;r tt J /vtl1 /l'p/ue e>f 'IT 7/tt, //1'11, F'f. (P. /07 Cfln b~ Jt!)jtJed
/r,y .i/a / dlU{ Ef. ,./tJ/f CIII1 be .s~/i/fd ("SJ~.f 4n ~ceY'l-ft:Iff)d
jJYtJcetlu~) kr ~/~. The Yllt-/o ~/-i Mn 1Jtel1 be dekrmll1f4.
Ii- jJr011"1I1')1 iDy Ck/~"lgl/111 t/a) -A/a., "Nd l/J as t1 htnc-I:I()~
of 7T VtJ.-/11?1 ~//()IVS.
:00 c15
2.. 10 pc:int, "**~: *:+ ****** ************************* *:j:***:t. ******:t: ****~,: '" I'
:L2'J print. "** This program calculates l/a, h/a, and l/h as a **"
130 p:r:'int "** f1.lnct,ie,n of pi*U*a/m fOT Rankine ovals :tic"
150 print. II ** *** ***** **** *********** *** *>l:******ll: **:k ****:**:)1:0+ *'*:t:*** *'''
160 print
162 print "pi*U*a/m l/a h/a 1/h"
168 data 10.0.5.0,1.0,0.5.0.1.0.05,0.01
170 for i=l to 7
172 st.art.=O. 001
175 read a
180 la=(1!a+1)~.5
190 for has=start to 10.0 step 0.0001
210 ha=O.5*(has
A
2-1)*tan(2*a*has)
220 if abs(1-has/ha)<O.002 and ha>O then goto 230
222 next has
230 lh=la/ha
250 print using "##.#### ##.#### ##.#### ##.####";a,la,ha,lh
255 start=ha
260 next i

Tabu/pted dabJ.. a"d I( pi,,-/:. ()f- lit. as a. .ffl11~t/~n Df 7T'Tr~/m1 tire
g/l/fl1 be/otV.
10
1
r******************************************************
**' This program calculates lla, h/a. and l/h as a **
1'* function of pi*U*a/m for Rankine ovals **
*******************************************************
!:i:tU*a/m
10.0000
-2
10
5.0000
1.0000
0.5000
0.1000
0.0500
0.0100
l/a
1.01,,88
1. 0951"
1.411,,2
1. 7321
3.3166
1".5826
10.01,,99
-1
10
h/a l/h
0.11,,27 7.3,*83
0.2632 1".1623
0.8601" 1.61,,37
1.301,,2 1.3281
3.1022 1.0691
1".1,,227 1.0362
9.9538 1.0096
For tA. Rfll1KIIJe O{/4! iAJl1H 1. tinA i. :spec.;!J(~d the IoJJf)wJn~ s.J..eps
f!6;U/d 6~ ~//OWftl It; det:.fJl'InI!1e. /Yn and 4..:
(J) /7;.". a. 9Jvfl1 1.1.Jr. d~i:.f~/7J'I1·e 7J,e reSIJIN'd value t:Jf 7l[r1Z)m
from -nte. gl"aph.
(2.) tI~/I1.1 111/~ J/p/ue ~f 7rD"~ //1'11 ~h/cLtla;i:e 1~
(....3) W/1h 'the. VII/we of .ellt. c1eiermJ~e~1 tlnd ,R.
the '!I4/we !)f a..
+-1'{)1I1 Eg. ,. /" 7.
:Jj>f'C,'tl p'd) dei:.frI11Ji1e
LIf) 0;171 7T77a/,m Clnd
/.5 -1r110WJ1/ Q 11 d /r;r
a.. de tf}l'm,He'!.; the /,It} lue 0 f u1m
~ jlVfJ1 1/ the I/p/U(! 0-1 t?11 /s f;.xe4.

6.5"'f Assume that the flow around the long
circular cylinder of Fig. P6.59 is nonviscous and
incompressible. Two pressures, PI and P2, are
measured on the surface of the cylinder, as illus-
trated. It is proposed that the free-stream veloc-
ity. U. can be related to the pressure difference
/)./) = fJl - fJ2 by the equation
u~c~
where p is the fluid density. Determine the value
of the constant C. Neglect body forces.
r:,2::
.2..
=
si:a.5n4. tlon
-P'2. + ~1.,
7 ~
u y
-
- FIGURE P6.59
-J: ;0 (3 U:t) = -j-!, l! <.
So
u: )ffy~. «r;¥·
1hey~~t"e
c= V-f
(I)

6.60 An ideal fluid flows past an infinitely long
semicircular "hump" located along a plane
boundary as shown in Fig. P6.60. Far from the
hump the velocity field is uniform, and the pres-
sure is Po. (a) Determin~expressions for the max-
imum and minimum values of the pressure along
the hump, and indicate where these points are
located. Express your answer in terms of p, U,
andPo. (b) Ifthe solid surface is the'll =0 stream-
line, determine the equation of the streamline
passing through the point () = n/2, r = 2a.
( ~) ()1'1 ine suv-1ace cf' ihe hump)
1;. = 1:," ;/l/ 2 (I - J{. si.'-a)
r
FIGURE P6.60
V, Po
--
----
The. max.IYnU/?? pY"e.sst-ft"'e Dec.fAY'S 0hef'e $1;" 8::0) ~Y" a..-t t9.:q 71;
(/11 d 4.-t 1'hese.. f~;n-l=s
1;(rm ~IJ(.) ~ 10 + ~! V'- (ILt t9 =(J 0-#/ 7T )
. .
rn, n I rn III rn f~t5S u Y'~
A t t71{5 !Dln t
1: (mlln ) :-
:.s
(4.-1: e::: r)
(.b) ;:;r un/form f/r;w In 1h( J1e,14h~e ..JC- d/t'Pc.b()~ )
If:: - V r (I - f:~~).rI ~ a
d;5C"SS/~J11 as.s(!)~,ate'" w,·f;., the dey/vA-flO;' ~I £g. ~.JI:;').
~-v-r(l-
ttl -a..) ._ SII,B
r'-

6.61 Water flows around a 6-ft diameter bridge pier with a
velocity of 12 ftls. Estimate the force (per unit length) that the
water exerts on the pier. Assume that the flow can be approxi-
mated as an ideal fluid flow around the front half of the cylin-
der, but due to flow separation (see Video V6.4), the average
pressure on the rear half is constant and approximately equal
to 1/2 the pressure at point A (see Fig. P6.61).
u= 12ftls ~
•
~----
-----.. FIGURE P6.61
~I? C{
hJ. '.2.~ t"t !-o11"lJ/! 7nA.-t -t:he drttj
.sec..i::/~J#J (~twei'h 6)=0 C/J;'Jj( ,9=""J
of tA Clrcw/(lY clj/lnt/fY' t.3 71,,1('110<."'1 17te. ~JU4.i:-/~'11
DrP.J ~ ~ =-1t ep~t9 ",dB
C)
;::;1" 1;,(:, -k>l'ce 6)n me .j;.P4-t. hq/f ~ ~e c.ylln/fr <Per III'I//: lei1flh)
r::;.::- - z j; ~s~ ~d t9
( 1T/z
OJ
Dill!. t1t{e /::0 SrrnlJ'Jel-rlj ~ =tJ. rrt)/n e;:( 6:.. lib
f.s = ~ -t ;!u~ (;- If .51;' 2-!) )
d.l1P< sinCe V-J(.. (Ire. tPl1/~ II1.fert's/-PII I", 1lt~ -brce ci(.(e 6-0
rEt. 6,J16)
fh~ //PUlIJ,j ;:/,,/d h/of! (VIII I'f!t 1b'=6. 7hus h~hn 5!.f))
f
7r /
1-;1 = - 2. -1;1/2. (J - 't .5/~ 1.f) ) cpse v... de
'IT '11lz
i t-t;.sBdB == :5/"P17r "-I
'1f/, 7T/1.-
i1T5/~'Z-f) ~~l7dB - jf~3f}lIT .: -j
1!/z- . 7T~
Chid
( Z)
('-70

~. 'I I {(!or;'i )
It f,;1/()IIIS +r()/?1 i:?!.(2.) -inat
_ .,0 U-za.
t: ~ (-XI .3
lV()f~ -1J",f 1h~ neja.I:JI/~ slid /nciJcJi:-er %,.+ 1I1~ w«lry J~ ac&all.!/
'P4IFn.:; It DI1 1J!e. c..rlOftlPy (,frpl1t h4/f) I'; tHe upst:re/lm c/1J'f'c:bf)I"'I.
I-I-e>wevt~ (,Jhe It the -e .f.fe~1:.. of- t'lte Y'eIiY holf' of the ClfIJIfAel'" 6 tAJcPI1
Inib aC~ilJ1i (lit ct YPtd Flu,'d) -then fu,'ll be A net drtlj in til, dlYer;hD/f oI-f/lJlJJ.
The prtS~I(JI'~ A.I: 1ht -I:Dp D/ fh.f (!~kndfY' (pt>J~t ,40) /J fl'/tK ""
P.s =: ~ + 1; 7)"2. (J - Lf -S/I1'l.{;) {ig. •. IJ~
a/,(&( WI n. {) =71'Iz..
-PI, :: -A - ; f V' ~
SInce. A-o~ - 3 1-
PA- :; - ""i! if
lJi)fo( 1),/-r 1ne. /J@Jllt/iI~ P(~.5S{,4Y.f w,·l/ '1JJt.. 4. D()..s/~f,J~ ~
~n~ / ~
/:; : - ~ >< prf>Jt'"ft.d aVE'~ = - ~ ;(.. 1..a.. (I)
~t Z L
SC f'n../;. '1. ~ 'Z.
t- ::: .1 f 1/ (2.a.){I) :: 1: fD u-
Ii<7.. If
f:;. = ~I +~l-
= _ p. V-~ +
3
- ffrJ').~
[,.u I ~ -Ine datA fJ I ~e11 )
t-; =7; (J.1~ ~:)(t2 ;~) 2(3h) =
fo-11

6.62* Consider the steady potential flow around
the circular cylinder shown in Fig. 6.26. Show on
a plot the variation of the magnitude 'Of the di-
mensionless fluid velocity, VI V, along the posi-
tive y axis. At what distance, yla (along the y
axis), is the velocity within 1% of the free-stream
velocity?
-:~
I.J./on, -the !1-I1;(JS
Df 1I1e. ~e/I!)c.;ff1) V,
1I'j..=O So th4t The m4'}fJrl-~de.
i...J e!/IAI -fa /118/' J I;;Ut!..
- V- (/ + ;-:) SIn b (Pt. ~.JIG')
(a=¥:) r;::f;j)
100
v
v
/ + /+
/
-(-1:)~
110 print "******-*- **- *- *****-* ***- *********' *:t:*- *- *- *-*- *- ** ***********:1'''
120 print "** This program calculates t.he velocit.:,r profile **n
130 print fI*:j( on the +y-axis for flow ar()und a cylinder **"
1~0 print. !! **:t:or. *:0: ***********- **** *):***:I::;j: **- **;~ **- ****- *** **)1: **;1: *:t."
150 print
155 print " y/a
160 for ya=1.0 to 10.0
170 u=1+1/ya~2
180 print using "##.##
190 next ya
V/U"
#.####";ya,u
( UJ:'J t )
t. -72.

(C4'JI'/t)
da,ip- (j ntI a. fl"t ().f /he data.
S(Jfl1 fr"l?1 fh'J~ ye.su/-h 1h4i ",..
'1 > ICJ
a.
~*************************************************
**' Thi:: program calculat.es the velocits profile **
** on the +y-axis for flow around a cylinder **
**************************************************
y/a V/U
1. 00 2.0000
2.00 1.2500
3.00 1.1111
4:.00 1.0625
:).00 1.04:00
E,.OO 1.0278
7.00 1.0204:
8.00 1.0156
9.00 1. 0123
10.00 1. 0100
2.0~--------------------------------------------------------,
1.8
1.6
1.4
1.2
1 3 5 7 9 11
y/a

6.63 The velocity potential for a cylinder (Fig.
P6.63) rotating in a uniform stream of fluid is
¢ = Ur (1 + ;:) cos 0 + ;n ()
where r is the circulation. For what value of the
circulation will the stagnation point be located at:
(a) point A, (b) point B?
(a.)
(.I:, ) II I • /3 e. - 371'
/"f' r; ~/l1t .) S~j - 2.. J
(1
(':: ¥7TlIa S;'n ~
u
~
y
FIGURE P6.63
x

6. b Lf A fixed circular cylinder of infinite length
is placed in a steady, uniform stream of an in-
compressible, nonviscous fluid. Assume that the
flow is irrotational. Prove that the drag on the
cylinder is zero. Neglect body forces.
:J.rr
D"fl~ : F; : - f. 1's eeS(J ~ de
fs :: ~ + -it [/2(1 - tf S/~~f))
- [
27T
f. Css t7 dB =
2.7r
f SIj, ~~ ~se dB ~
C>
. 3 J2.
lT
s,~ (;} =0
o
if f.o JjDW5 ih(l.-t.
Drlt j =0
( Eg. ,. "I, )

- - _.
6. ~S Repeat Problem 6.~4 for a rotating cyl-
inder for which the stream function and velocity
potential are given by Eqs. 6.119 and 6.120, re-
spectively. Verify that the lift is not zero and can
be expressed by Eq. 6.124.
27T
Dr-tl') '" F;,. " - [ t CfJSI9 a. dt9
1.2.(' . ~~ 2f1s/nB
t = -fo + ~ITI (/- if 5111 l7 of- 7r~ TJ
I
,ll
C!JS e
u
lrr
de '" sin 17]. =0
2'Jr
1. :;I·M"'~ Cilsede- "
z1/
ICt)58 51;' 9 de- =
o
d: +0IIows fh--l:
DY'a.~ = 0
• 3 J2T
~ '="0
3 0

,.17"
[ 5/·~e de =
J~:iH3ede .-
£)
"jJ
is/~"e de "
J-t !cIJoW5 1h().-t.
~·-71

6."~ A source of strength m is located a dis-
~ tance etrom a vertical solid wall as shown in Fig.
P6.G,b. The velocity potential for this incompres-
sible, irrotational flow is given by
m
<P = 4n {In[(x - e)2 + y2]
+ In[(x + e)2 + y2]}
(a) Show that there is no flow through the wall.
(b) Determine the velocity distribution along the
wall. (c) Determine the pressure distribution
along the wall, assuming p = po far from the
source. Neglect the effect of the fluid weight on
the pressure.
(A) k = ~rJ
~x.
S/~Ce.
Ix 1., [ ex-;./"+ ~.])
QYltA.
~ f.n [l;<.1-1 ) ~ j ~J
It 10 /J()WS 1tll. t
::
=
U: - +
Mt [2.ex-.e )
Jf7r (X-).)2 rfJ 'l.
~e-i
FIGURE P6.l,.~
2 ( x-~ )
(~_). ) ~ -t ~ "J..
2. ()(11.)
-ex-rQ ) ';l. +-.!:J'l..
2 C;{ -t..e )
J(x+~ ) 2.+ 'j 2-
o
Thus theye I
IS
)
nD i/tJw ihr-ough -the Wa IJ •
(b) Tn! l/eJ~c~+j o/f!)IJJ /AJa 1/) V. :. 7Tw- Slhce u.. =o. A-l.so
1r=
~tfi
dlj
U)I'th the.
. tie 1t:)'/~ ftJ leI''':''~ /Cln d (il v'f? '"
t1?1
[ 4 ~ z~ J7/=
'fTr (x-}) '2-r!:J 2-
+
(x+J)l. +~ ~
(c.~J1't )
'-7B
( /)

( ~) !=;y Ir~m ih(!. Sou Y'C e.l 1;) =f;; Q h d V~ (). Th usJ
-PO _ -Pur + V;
7 - 7 2.d
whey-e. fw- l".s the fyeS5Uye ai the 1.J4/}) .s~ 1h4i
PJ,r:: ~ - -1! VIu-'J.
( 2)

6.t.7 A long porous pipe runs parallel to a
horizontal plane surface as shown in Fig. P6'(.1.
The longitudinal axis of the pipe is perpendicular
to the plane of the paper. Water flows radially
from the pipe at a rate of 0.5 IT ft3
/s per foot of
pipe. Determine the difference in pressure (in lb/
ft2) between point B and point A. The flow from
the pipe may be approximated by a two-dimen-
sional source. Hint: To develop the stream func-
tion or velocity potential for this type of flow,
place (symmetrically) another equal source on the
other side of the wall. With this combination
there is no flow across the x-axis, and this axis
can be replaced with a solid boundary. This tech-
nique is called the method of images.
sou.,.,e )
--
/s rnea.suJI'ed ;".I!)fYJ
.sh~wn /n fi9ure
~ 2n 1"''2-
.".".
-fne SOl.{r,e. W,"th the. C.tJOJl'dIl1~/e
r?.: X"2.+ {'j_3)2.
C/ lui
t'2.: / ()Wtt" sourre )
so -inA.t ~r --the. C!t9mblnt'c/ S(U(Y"ces
and
/1:.
l'= ~ f..en [x 2 + (!j-3 )~J ... in [x'+ 6+3)'J]
u= ()X
() ~ [ ;("Z + (~-:3) 2.] =
2.><
)(2 -+ (f;j-3) l-
~
2 iYl [ )("2..-t-l~+3 )1-] =
2)(
)(2.+ (:;'1"3) 1-;;;<
/0//"w.s -thai
WII/I) '1 =0, 1/ =0 and
Vw- = u = n;; (-X-'2.lf
--r-i..q- )
( Con 't )

At f()Jn t A) x.:. Jj. It) find wJ1'h /'YY1 =- 0.57' -¥2.)
O. SIT t;ta. [ Lf (~Pt.) J_~ :ft
VW-A = -
Jf- 7T (If- -Pi:) '1. + 9-PI:. 2. .2 5" S
At- p{)/n t B) x. =0 ) an d
V'/(r/3 =0
irtJm the /3ern()ulji
~
1>13 Vt.crB::
-+ -
5 :L~
Or
J..L .2.
~ - ~ = 2 d VWA
I Jb) .2-
:: (~2.IfH3 (3. fE) - O. OOG> 2t> psf
:L (32. 2 ~ ) 25'S

y
source
(a)
6.68 At a certain point at the beach, the coast line makes a
right angle bend as shown in Fig. 6.68a. The flow of salt wa-
ter in this bend can be approximated by the potential flow of
an incompressible fluid in a right angle comer. (a) Show that
the stream function for this flow is I/J =A r2 sin 28, where A is
a positive constant. (b) A fresh water reservoir is located in the
comer. The salt water is to be kept away from the reservoir to
avoid any possible seepage of salt water into the fresh water
(Fig. 6.68b). The fresh water source can be approximated as a
line source having a strength m, where m is the volume rate of
flow (per unit length) emanating from the source. Detennine m
if the salt water is not to get closer than a distance L to the cor-
ner. Hint: Find the value of m (in tenns of A and L) so that a
stagnation point occurs at y = L. (c) The streamline passing
through the stagnation point would represent the line dividing
the fresh water from the salt water. Plot this streamline.
• FIGURE P6.68
(a) f;r 1n~ 1lve~ $f-rtam fun(.:f/~A)
~ =A- r 2$;"1 '2 e9-
a/tin, ().=() if=-o ~,( B-=-rr/2, lP=o ,
(b)
Thus) -f/,(, ('A~~ G=o (JHP. B,. 17'/2. (!,(JI'I be replac.eeA
lv/Tn. a. "5&/;" b~Jllu'"y't 4/(!)A1 4Jh,;;'h the Sfrefll?f
-Iunc.-tItP~ r»1tJJr bt ~n$hnf. 7}J/5 b"u.;;tdf(Y"~
(J:;)
.j;;rmJ ~ Y"lint tlJllfJe. Allit 1'nff'e,/rJle Ink! sf-reAm
ful1(...I-,D /I Cfln b" U$~~ fD yefr~.sfnf ft4~ JJ4' ~
YO/ill! qrtfle ~rl')er.
.
5'/1'l'~
11::::' 1r r
a..b t9:: -rr/'l,..
1;-: :: 2 A r Uls7T =- ZA-r
r
To Cr-t(L ~fL Ct
J~I;- "lrr =
1')?1
-Zvr
~ 1-1t7"lVbf)H
11';.1
(~tp" 't )
b-B2..

( CfJl1't)
2. hL
_ /n1
- -2.7rL
tjJ= A-r2.s/~'2.B -t g ()
CIna. IA-Jl' th rrn =: If rrA-L"l.
ljJ= ,A-;-2Sln 28 T lAL!'e
The ~4/ue "l /jJ 41- the. SJ.afnaf/~~ ~In·t (r=L) G::: tr/z ) J~
'&/ A L2 $1'" 7T '1- 2.41..'"("1£)
2-
= ,4L 1T
Th'-lJ) -the €!"UviIDI' .;,,,. 1he ~fre(unJltj.e, pa~sln, inv"~UfJ.t
1n~ sh911aA-/~i4 p~/;'f Ij
,4L'2.11 =: A- rs/n 2B t- 2.A L2 e
V
,I f- _ 7T-28
f.,.. - -
r - L - St'11 2.8
~Y'" pj~H-;'l1j let.
, I
X. = r- CDS a
t:lnlA ~ pjr;z ().f- In~ d/I/;d,ltj .sfY"e4mJ/~<t.. ;r~m
J?~.cn I~ sh() wn ~1'1 17u. ~J1DW/~~ pa.,j e.

Theta(deg) Theta(rad)
10 0.175
20 0.349
30 0.524
40 0.698
50 0.873
60 1.047
70 1.222
80 1.396
90 1.571
r/L
2.857
1.950
1.555
1.331
1.191
1.100
1.042
1.010
1.000
x'
2.814
1.832
1.347
1.020
0.765
0.550
0.356
0.175
0.000
y'
0.496
0.667
0.778
0.856
0.912
0.952
0.979
0.995
1.000
IStreamline I
~ .~~ -.----.,.-:'-----,r-
j.-;:::::::,:==..,..----,-!,-----,
. t----F--~_j~~~~~----~---ti---j
0.80 .f-----+_!,------=i~-_--±r--.---_--+-----r--------j
"» 0.60 t----t------j------t---=~--iiiiiiiOO;~:.:__1
0.40 +------l-----I----_!__--__+----+__--_l
0.20 +------l-----I~--_!__--__+----+__--_l
I
I~O.OO
I .00 0.50 1.00 1.50 2.00 2.50 3,00
! Source X'
i J'L_______________________________________

~. bet I
6. ,~ The two-dimensional velocity field for
an incompressible, Newtonian fluid is described
by the relationship
V = (12xy2 - 6x3)1 + (18x2y - 4y3)j
where the velocity has units of mls when x and
yare in meters. Determine the stresses a:m ay)',
and Tn at the point x = 0.5 m, y = 1.0 m if
pressu~e at this point is 6 kPa and the fluid is
glycerin at 20 °e. Show these stresses on a sketch.
(j" :: - -F -r 2.;« l.!:>t){ ~ ,,(
(T, - -p-r~ ~()'"
~~ ~ O~
txJ ~ f (OU 2)tr)
ay'" ~J<
(Eg. b,lz5'a. )
( '=g. b, J15' 1.,)
( ~O. ~.I'Z.£'d)
Fo.,. ih~ 9/(,/(11 lIe/t!>';+:; d;sf~ibui/~J1) With X::.t1),S"",. (JI1J j::1.0M-1:
dU '2 4 'Z. ~ /
- : : 12!J -/~;( ::: 12 (/'0) -Ie (b,s) ::: 7.StJ s-
~x
~.:
d!:J
~=
~)(
36X::;:
,:1.if (~,S)(I,D)
3b (~,S)(I,o)
I
:: 1;)..6 S
I
- If, (j s
~ ()"' ==
alj
C) 2 2
/dX -IZ!:f
Thus) .f.o~ and /"-:: /,5'0 ~: )
+ 2 (I, 5"'() ~) ( 7, ~() f ) :: -S. r8 ..Ie fa.
cr'J~ ::
tj = (1.5'D
L X

6.70 Typical inviscid flow solutions for flow around
bodies indicate that the fluid flows smoothly around the
body, even for blunt bodies as shown in Video V6.-1. How-
ever, experience reveals that due to the presence of viscos-
ity, the main flow may actually separate from the body cre-
ating a wake behind the body. As discussed in a later section
(Section 9.2.6), whether or not separation takes place de-
pends on the pressure gradient along the surface of the body,
as calculated by inviscid flow theory. If the pressure de-
creases in the direction of flow (a favorable pressure gradi-
ent), no separation will occur. However, if the pressure in-
creases in the direction of flow (an adverse pressure
gradient), separation may occur. For the circular cylinder of
Fig. P6.70 placed in a uniform stream with velocity, U, de-
termine an expression for the pressure gradient in the di-
rection flow on the surface of the cylinder. For what range
of values for the angle ewill an adverse pressure gradient
occur?
rYP/71 Eeg. ~. /I ro
-u
--------• FIG U REP 6 .J0
11= -to + i fVl(i- if ;1'ln1.B)
Since el/J
p?Slh;e.,
adverse fr~.s>nre J rdQle",,-t
t3lljPB) /r -!oj/Pills hom Ei.f/)
I
Hils /17 1J;e rAl1fe tJf 2. rt'cJ #r 1/11 tJdJlerse
p('e.sst/~ 1I'IU//I'I1/. Th/J 1"t/f1/fe e87rrRspl1ti.I -Jo
Y'etlr half "I 171e C!f/ltlder.
e
( I)

'.71 I
,
6.7' For a two-dimensional incompressible
flow in the x-y plane show that the z component
of the vorticity, C, varies in accordance with the
equation
DC = v V'2[.
Dt -<
What is the physical interpretation of this equa-
tion for a nonviscous fluid? Hint: This vorticity
transport equation can be derived from the Na-
vier-Stokes equations by differentiating and elim-
inating the pressure between Eqs. 6.127a and
6.127b.
~~. 1../27b
f
(1.)
b, fleyel1tJ~te Eg.") W,.th Ye'feci +0 l:J cliol E ~ . tZ) wi-h
V'.esfe,:i: fc) )( ) a""pA $Ubtl"'U.t, G:fJI) .frfJ", E~.lZ) * ,,~-I:'IIH
;J (JV- dtr Jtr) d (JU dIA. JI.(.)ax 'it -t l.(. ~x .,. V ~!J - a!1 ~"t + u ax + v- d ~ -
~ [ d (~1. tr ~• V-) d ( ~ 't (,.( d'1. '-<. )J"1 ]X. ~X'L T ~-z.. - ~ o}('L -t d':J'L (3)
'8~ cJe//","/-,()i1 (s~e li'~. '.17)
f
~v- ;;I..L
~:~-r-;
Re-wr;te Et·(.3} ~ r;j,-/;g/h
C> (~tr_~)+ d ( ~v-
- u. 'fX ~x-
~t: ,X Pfj
7[ ~7.. (o~ dU
)
at· ax. - J; +
b-87
d L< ) -+ & e'" (j,,)v- '""i a>< - ~ -
d!:1 -
ol. (~_J'<)J
~!J 'I.
oJ< t3'J ( If)

(P, 7/ 1
:5/nce each -fer-rn
. pqyen thRSI'S
.
t01 In Erg. (If) lor
it iol/()ws Iha.t
d~ afr ~ fiL h.(~ d'f. )
+ tA. -r7.J'" = +~ (S)
- ;0 J.x2.. J.!:J 2.-
- J!}tit a"><.
The Ieli f:J icJe of ~ Z.(5) CtV1 be expyessed a. s f'slJe' G!.lf.S)
the ~teyab:;r D () IS 1hf! mtJfenid
Dt
Df~ t.J here
-Dt
def"iva tille . The rl~h1:; haJ1 d s;eJ t. 0 I E ~ . (5) ~h b-e
ex.pYessed as
-v V2. ~
where -V =)L~ So thai E''!. tG") ~"11 b~ tdr/fltH I¥~
nfz :: -V l7~
Dt .z
Fo y a. 1'Jt:n{//~C.()u..s .{/uid J -zj=a J QJ1c/ 111 Thu Cq.s-t!
D f1:-
- -=0
Df:
Th us -{O.,.. a.. fwo- ttllmpI151OA4J +/~w I){ On Ji1~ompt~.s5aJ/p
) J
lIon I//SC/J(J.s ,f-/"I(jI -In e. c.hal1/1€ /H the. v();-iJ~/r':1 t>f "-
·/-Iu,'d j)ru·t,·cJe as t't f?1 () lies ih Y'~"9h the.. !..jaw It'eld
is -,.fl"o.

~.72. I
6.72 The velocity of a fluid particle moving
along a horizontal streamline that coincides with
the x axis in a plane, two-dimensional incom-
pressible flow field was experimentally found to
be described by the equation u = x2
• Along this
streamline determine an expression for: (a) the
rate of change of the v-component ofvelocity with
respect to y; (b) the acceleration of the particle;
and (c) the pressure gradient in the x direction.
The fluid is Newtonian.
(~) From -the C4Jni./~ul'-f!:t e8ua.tl~~)
c;u ~V'"'-o
ox. "1' ~ -
.so tn~ t: tv in u. = )( :z..
?;tr _~ = -;LX
~:: oJ<
A/so) ~g .II) Cal-( btl ;".f.@9Y'4ftM wit;, res.ft>ct 1:.0!:J +0 OJ,-i-4/~
Jel 11'" =- J- z~ d!1
1r::: - z)(:; + f-l)l.)
SIHce the X-fiX/J is 0. .s/-Ytlll'l1/Jhe) r=o 4/"11, "this o;tiJ (J/I1~
theV'ekl"t. f (;<)::: 0 Jo -tit. t:
'1r = -zx';j
(b)
(/)

'.73 I
6. 13 Two horizontal, infinite, parallel plates are spaced
a distance b apart. A viscous liquid is contained between the
plates. The bottom plate is fixed and the upper plate moves
parallel to the bottom plate with a velocity U. Because of
the no-slip boundary condition (see Video V6.5), the liquid
motion is caused by the liquid being dragged along by the
moving boundary. There is no pressure gradient in the di-
rection of flow. Note that this is a so-called simple Couetle
flow discussed in Section 6.9.2. (a) Start with the Navier-
Stokes equations and determine the velocity distribution be-
tween the plates. (b) Determine an expression for the
flow rate passing between the plates (for a unit width). Ex-
press your answer in terms of band U.
(a.. )
7htl~ ~r ~.,f'O f'''.5..fIlY~ ~YdQ/;'''t
)2.k, _
7JJz -0
..50 1hd
U::: C, j -t Cz.
1/
>-
T" I Z
Z t I i ' I t I , It
b
1 ~ 77 7 7 I 7 I I I n-X
J:/)(~d pJ..-le.
At 'j=D fA. ='0 and ,.f -h//t)/.IJS 1h~t
V
at: ::J =-~ u.-= V and C/ ==];
TheYekreJ
fA.. ¥!J
(1:,)
b
V f~fu (I) d:J J; ':J d::;c;:: =
0
0
where '6 is 111e. flDwrate
E~f -- j, 2. 0
-
per unit- WIdth
Vb
2..

r;.71f I
(k)
6.74 Oil (SAE 30) at 15.6 °C flows steadily between fixed,
horizontal, parallel plates. The pressure drop per unit length
along the channel is 20 kPa/m, and the distance between the
plates is 4mm. The flow is laminar. Determine: (a) the volume
rate of flow (per meter of width), (b) the magnitude and direc-
tion of the shearing stress acting on the bottom plate, and (c)
the velocity along the centerline of the channel.
_ (tJ/,i. ~tr)
1-~)( - I ~ -r ~;<. (Eg. b.J2!i"J.)
Sln'~ I op
all1d
;f
and
At-
(C )
lA.. :- - -.......
').)A aX
v-=o
iO//f!)wS thai
~k I eE (Z!1 )
Ot..r -=0
- - ~ ~x a.xa':1
-theY'e kwe
Tj,x. :: ~ ('1)
the boll-om pia ie) '1 =-,,£. ) (/11 d S il1 C(! ~
~x.
......:..
=: 61' (-R,):: (ZO;<I03.1Y.-a)(Z'(JO~'/YYI)L'jJ< ;. /WI
-If ~)
(2.81 xlO ~
(Z)(Z)( IO-3M-{ ')
:: - ¥J
- D. 105 ~

fD·75 I
6.75 Two fixed, horizontal, parallel plates are spaced 0.2
in. apart. A viscous liquid (J-L = 8 x 1O-3
lb· s/fr-, SG = 0.9)
flows between the plates with a mean velocity of 0.7 ft/s. De-
termine the pressure drop per unit length in the direction of
flow. What is the maximum velocity in the channel?
Tn r.(5)
~i ..
'3 (P-b) I 5 tt::: 2: 0.7 5 = .0 S
= 1.If2 ~z. p.,,..It::

6.710 A layer of viscous liquid of constant
thickness (no velocity perpendicular to plate)
flows steadily down an infinite, inclined plane.
Determine, by means of the Navier-Stokes equa-
tions, the relationship between the thickness of
the layer and the discharge per unit width. The
flow is laminar, and assume air resistance is neg-
ligible so that the shearing stress at the free sur-
face is zero.
3)1. = 3 S/11 d...
w/th the UJoyd'l14.t~ ~'1s/-em Sh"Wh ,it the hjUye
7/':0 t..cr::O and #()fl1 1l1e eLJl1tinuif!J e.tJtl4:i:IOH ~ =0
J J () ox. Thus)
froM -the )(.- ~m'p'Mfl1t of the lilli/lty -S.f.tJke.s ~!UlLtl()H~ (Et. '.I27a..))
() = _ i j + 1'1} Slit eJ/ +j dZu
fJ.x. (f d!J 2-
A-/~o) sJj.,{~ there /S 4(. free slwface there CtiIfMt.
jYrJdlent Ih the X-dlY'ecf/~H So -h.c.i: E..e -= 0
C
L · . 1 L , ax-qh De. wrt 7T~11 I/lS
6:,-1'3
(I)
be a. pre.ssf.lY-e,.
Qut/ E'I. (I)
( Z)

(P.77
6.77 A viscous, incompressible fluid flows be-;
tween the two infinite, vertical, parallel plates of
Fig. P6.11. Determine, by use of the Navier-
Stokes equations, an expression for the pressure
gradient in the direction of flow. Express your
answer in terms of the mean velocity. Assume
that the flow is laminar, steady, and uniform.
Direction of flow
~h+h~
FIGURE P6.17
With the ~f'dtn4~e .j,/.stem ~hf)UJn t<. =0) W- =0 C(lId .frt!Jm 111e
UJni/nu/f!:1 ez(Aa.I::/~J1 ;:; = o· Th liS) rr"m 11te !/-UJmjJDJ1bl'"i
of -the Nllllley-SlDk~.s -ejtlal-ID;"~ (EI· ~.1Z.1h)) tul1Jt ~:J -= -J)
Jp cJ 2 tr
6 = -h -f3 +;«- d;<'2.
S/I1C~ 'fhe ,"~~stlye is I'}()t a h(J1(:.-b~n 01 .x.) £'1' 1/) C/l11
be wl"iHeJ1 IJs
d 2 v- _ .p
- --tlfi'L /-
(t.Jhe~e P = #+;03) ol1d /nff?rp/-en
dt.r:: P X T C,
d.JL r
J:;1!)f)1 s'f/7J/nelr~ ~r=c o.t: x = (J .j~
of 1£,.(2) ~if1/d.s
Thu5 w/1'J-,)
v (me~Yl veJocj~ ) 9ivel1
<!- = J..E -I.. "2.
V= 2:h 3 /'-
'!hai
~
'4!:J -
(I)
( Z)

r..78
6.78 A fluid of density p flows steadily down-
ward between the two vertical infinite, parallel
plates shown in the figurefor Problem 6:17. The
flow is fully developed and laminar. Make use of
the Navier-Stokes equation to determine the re-
lationship between the discharge and the other
parameters involved, for the case in which the
change in pressure along the channel is zero.
See.
Z=
Problem to. 8 3
'2. p-l3
3 ~
where ! J5 the ci'scha"ge pel" "1I11 t.vidth
.f = !.e +/,q. Thus)
P':J ~
~ +/J
"to
CJp
TY =
!=b~ ~ =0
':;
- -
3
- -2.
3~g
2.. -It. J
11. -fJ
(lIo/:e: Jj,(! ne!a·h;,~ SI'!), iJ1dlC4ks -fh4t 1Ite d'l"e"tJ(/n ~( IIf)W
l?1us t: b.fl d"wnw~lI·eI .f.o create. a IeY'O fY't'ssl.J~~ jradJlfrlt.)

6.79 Due to the no-slip condition, as a solid is pulled out
of a viscous liquid some of the liquid is also pulled along as
described in Example 6.9 and shown in Videu V6.S. Based on
the results given in Example 6.9, show on a dimensionless plot
the velocity distribution in the fluid film (vIVo vs. x/h) when the
average film velocity, V, is 10% of the belt velocity, Vo.
l'VeJ1 b'J the ~tu4:h~)I
en
r",.,,,, e1.(3)
(J J:..~ ~ 2.7
~VD
tlnll eZ. (If) ~''U1 W wr, HfH a.S
V.~ J. 3S(f.)1_ 2·7 [~ ) + J
A pia+- ~ f. -tn(! I/~ /Oc./-hJ c/I'flt-vi 6/.(.J-/~·1'I
xlh vNo
1.000 I
0.744 I 1.500 ,---,----,---r--...,.--,----,-..,---,----.----.
! i
0.514 i I
0.312 1.000 f'... ! !
~ I
0.136 <>0 0.500 ~
-0.013 lr-....t-..,. i
( 3)
(Lf' )
o
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.000 -1--+--+-+--~___~"...__-I---t--4----l
-0.134 r-I~~:...-;Lj.
-0.229 -0.500 -I--......!-I--L._.L...--L-~!----l_..l.---L---L--I
1
a
-0.337
-0.350
0.5
x1h
-0.296 l_~ ~-~~------------'
~.,........,.-...,....."....--,
Calculated from
Eq. (5)

G,.FO I
6.80 An incompressible, viscous fluid IS
placed between horizontal, infinite, parallel
plates as is shown in Fig. P6.80. The two plates
move in opposite directions with constant veloc-
ities, VI and V2 , as shown. The pressure gradient
in the x direction is zero and the only body force
is due to the fluid weight. Use the Navier-Stokes
equations to derive an expression for the velocity
distribution between the plates. Assume laminar
flow.
FIGURE P6.80
RPr fne :steelh iod c~l1di .J-IDt1~)
Sa 171tl.t the.. X.-6Pmf~l1p"t 01
( EZ' ~. /2'71<.) reduces -1::0
v =" w-:: 0 tit:.£) 1:111" ! ::0) ,) p,l(. I J(.,)
d zf.,( .:. 0
d!J 2.
In.f-e1rll 6,p;' t!J.f- Gr(I) 7Ie/cis
U:- ~ J of ("2.
:J= 0) u.. =--L{
c =- - VJ. ~
the Natlle~ - S-iDkfJ e'l"lttJol1S
F;; r 'J =. b) fA.. ':: V, ~o 1ha t
V;=C,h-~
c =I
(u; -t u;. ) V
u..= t;J - "b
u)
( 2.)

u
6.81 Two immiscible, incompressible, viscous
fluids having the same densities but different vis-
cosities are contained between two infinite, hor-
izontal, parallel plates (Fig. P6.E,J . The bottom
plate is fixed and the upper plate moves with a
constant velocity U. Determine the velocity at the
interface. Express your answer in terms of V, 111 ~
and 112' The motion of the fluid is caused entirely
by the movement of the upper plate; that is, there
is no pressure gradient in the x direction. The
fluid velocity and shearing stress is continuous
across the interface between the two fluids. As-
sume laminar flow.
Fixed
plate
FIGURE P6.81
/=r;r the. spec/hed C~lId;'tloIf.SJ 7/::: 0; w-:::0.l if. :'0j QHIiI Ix. =oJ -Se>
'/hat the x- ~mf()nMt of th~ IVallier-5fDkes -esaa:l:loHs (E?fJ. ',/Z711)
-hI- e/they fht! up~Y tJr IOIVIY laffer vedtlces t.o
dzu
- ':0 (I)
d!J J,.
I.nk~rA:ti"lt ~I ES. (/) fj/e/ds
u=A-~ -f8
fA.) h;c h '1 Jiles
r", the uppey
tJt~ lIe/f)c/-ht dt.sfn~ktIDN I it
/4.reY' at- 'j:= 2 -'.) '" =Z7
B -: V - A, (2.1,)
e/fhev JalJ"'Y".
.so /ha.:t
I
W her~ the S'IbSC. "'pi J re fers 10 11te U.f'f~v la.Yfy',
POI' the lowe#' latter- CIt fj::oJ u=o ..so thai
B =02-
(..)her~ the. '5L1bu,JI'lpt Z Jl'eier.s +0 the JO£Oty JaY-iV'. Thus)
U I -:: AI ( ':J - 2 t..) +- 7J
tlnu
A'2, ~fA =2.
AI: 'J='{ I U, ::: U2 Jo 1ha-l:
AI (~-Z.h) + U .:: AI. --{
",... u
A:z - -A +
~I
(c'()n 't )
( z )

( CI)I1'i )
S/~(e 1h(. lIeJ()C"-ht chs'frJ'blll:lDh IS /'~elll' '11 e~c.h Jtl~er
he 5he~Y'/n, ~fYi.s.s
"f-.. ( J u. .,. Ii tr):: .u. .i!::
L':Jx -jA. J ~)x /- d!J
Is ~l1si:ql1i 111r()l.(fIHt)I(,f. eac.h lA-iffY'. hr the "'fpl'Y J~'ttr
L; ~ /-1 A,
a J1 eI laf' the )tJWdy /tl'1fY'
Li ":! jJ,. A2.
",... =- ~ Co,.J.,a -I:£., £.z. v lit
PI A, -=)tz. AI-
ft, - l:!-~
liz. - ~,
u

6.82 The viscous, incomp.r0:Jslb1e flow between the paral-
lel plates shown in Fig. P6.8~ is caused by both the motion of
the bottom plate and a pressure gradient, iJp/iJx. As noted in
Section 6.9.2, an important dimensionless parameter for this
type of problem is P = -(b2
12 p.U) (iJp/iJx) where p. is the fluid
viscosity. Make a plot of the dimensionless velocity distribu-
tion (similar to that shown in Fig. 6.31b) for P = 3. For this
case where does the maximum velocity occur? u
• FIGURE P6.82
I (dP) 2 /'I
U ': ~ ~ ':f t ...., 'J -t ~L
At t.< :' V Jo "'1}t1A..i: Cz. ::- v: A -I:
C, :: - .J. (lj") b _ ~
/)..1" ~;< / b
(~ )(!J "- b:;) 7 [,r (1 - ~)
u - i.~ (~)(Y)(!LI) '::J
t I
r; -ttr ,;( b J, b
,
SII1C<)
h"'- ('?i')P=-iirr ~
1:1 . (I) Can b< wr/ffel'1 as.
u.
-p(t)(l-)- ~
t I
- - ];...
f7
Fixed
plate
( 2.)
IJ- p/oi 6/ -this VI! / ~(..;+rt chsf(; b/.( -tIC ';" loy P=- 3
/$ sh()wn. On -h1t!. .ft, //()u) I n', p4.J<! .
,-100
(I)

u/U
1
1.17
1.28
1.33
1.32
1.25
1.12
0.93
0.68
0.37
o
y/b
o
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.2 ..,--.,--,...---,----,..--,----r--,--;--,--,-I----r--,---'I,--;-1
1
- ,
1 t:tl~~j;~t:r:['JI=JI=JI=I!=rI=Li
]
0.8 ~i I! I i
It~:~1 I!
~ O.6 +---+-+-+----+----+-----j---I----l----tl---i~.....~d~-----r--T!-----j
0.4 +-+--+-+-+--+-+-+--1--+,-+-----+----i--1'l}l-t-I, --j
, 1
0.2 +-+----+--+--+----+--+---+---t----t--t-__-t-~--r-::~-,r--:-I--j
, 0-l--...I....-...I....-.L.....-.L.....-j---!l.--J----1--L---'-=:;.l..---,----,---,--;
o 1.50.5 1
Calculated
from Eq. (2)
with P =3.
u/U
L-____________________________________ i
To de. krrrlJ~.c. whp,,< fh{ )114)(1I1U011 V€ )0(, I ~ DC'" C..O"
c/,/ft.r-tl1tI4·+-t!.. E'!. (2) Cll1d .sei eS~AJ ~ 'ty~. Thus"
d (1A-1v) 0[1. (J. ) - 1] - J. .= 0
. ~-..L J,a .b b
~'J
ClntA. WJ 1-;, P = 3
c{(fA./v) = _ 3 [i (2 i-1ij-±= ~
dey
So -InlVr
~ .... 1
"I"" 3
&'-/0/

u= 0.02 ftls
•
6,83 A viscous fluid (specific weight = SO Ib/ft3
; viscos-
ity = 0.03 lb . s/ft2) is contained between two infinite, hori-
zontal parallel plates as shown in Fig. P6.S3. The fluid moves
between the plates under the action of a pressure gradient, and
the upper plate moves with a velocity U while the bottom plate
is fixed. A V-tube manometer connected between two points
along the bottom indicates a differential reading of 0.1 in. If the '~~_. m_BB~ Fixed~ .1= • plate
upper plate moves with a velocity of 0.02 ft/s, at what distance
from the bottom plate does the maximum velocity in the gap
between the two plates occur? Assume laminar flow.
• FIGURE P6.83
MCI;(llnt.f/YI ve/oc,1t:t ltI/// 4'JCCllr at cllS-h}fce ~ntf tJl1t~e
Thus)
For rnanofYItte.,. (St~ h~lJrt 1:0 r"IhtJ)
~+ ~f Ah - '(1f .An = f1.
t,-1'2. ~ ('0}t - ~ )Ah
=(J ()O it11
gD l:! ) ( 6. J 111.) _ Ih
.ft'3 I ' -O.I~7-~2.~. .ft.
+t-
A-Is0 J
.p, -R'l.
1'-'
JP O. /(P7 4=t;2. lb::
= = (). ~3LJ H3ox. - ( (01')1.~ )12. ,'01.
n
I. 0 I~'
12. ,'..,.
+ ~
Z
r = 100 Iblff
( J)

6•.S1+- A vertical shaft passes through a bearing and is lubri-
cated with an oil having a viscosity of 0.2 N·s/m2
as shown in
Fig. P6. g'r. Assume that the flow characteristics in the gap be-
tween the shaft and bearing are the smne as those for laminar
flow between infinite parallel plates with zero pressure gradient
in the direction of flow. Estimate the torque required to over-
come viscous resistance when the shaft is turning at 80 revImin.
Shaft
75 mm
~"'i"
• FIG U REP 6.84-
The -bOY~ ue due. +0 'yce dr ac.+'~j
~)I (). d~++eyet,.hal ~vea.) dA-=- vi.Rc1e J
.1.5 C$e E' +1;ijll'.iC! Il.t VI·'~t.)
cI~ ~ y. dF :: y.'J. r.1 d8.J l l.
().Iheve t IS tne. ~ neAyil1j styes s. Thu S J
1oJT"
u=
id8 = :1.11" r/ t ;
lJ.:t.
b
Thus) from EZ.II J
J= 2iTr/"r;«-¥)f. =
= ),11 (0. 015m1Y{0.2
0.355 N· rm
dF== "t'dA
= Tr.·2dBL
(I )

6.85 A viscous fluid is contained between two long con-
centric cylinders. The geometry of the system is such that the
flow between the cylinders is approximately the same as the
laminar flow between two infinite parallel plates. (a) Determine
an expression for the torque required to rotate the outer cylin-
der with an angular velocity w. The inner cylinder is fixed. Ex-
press your answer in terms of the geometry of the system, the
viscosity of the fluid, and the angular velocity. (b) For a small
rectangular element located at the fixed wall determine an ex-
pression for the rate of angular deformation of this element.
(See Video V6.1 and Fig. P6.9.)
Lx
de
1 '" c.~Jihdf( Jen,1h
t .... sheanng s+ress
(ill) The -torgue W~'c.~ must be. o.ppl,fa +v out(lv CLfIJ~dtf 10 ovevco",e 1k.t
kY,e du~ 1v tk~ S~~4v(n, J1-l"'ts.s IS (see .r,9·u~)
d~:: ~ d F = Y" (t ~ .Q de) = ~~ 1:"..Q de
ITT"
J" 'fb' tJ [ de
I.n the 3().P
t:: Lt d..Y: :::
F' db
1::: YQ - Y"l.' ) 11
G Z. (J) -hia+-
2. '-'" n
::: ;2:".,. ro l!..
+0Hows
(I )

~.fS I
(j,) FYtJln E'fj. t,,18
P_~ 01.(
~ - "K '+ 0'1
/i;~ 1h( jJt1ear' J,ifr;bu .j./~'1J
u.= - r;~
Y.'-y-.() L
~l-{ __ V-
Fb" b
I/nll 6 V.::.()
v
--.b
__ Z/~
J,
/h~ lIe!a..f/,t~ ~J1n indtC4h.! -/n"t;- iJt~ ()rt9J11~/
rl911t ClIt9/~ .Ihf)IINI I;' Fitj, P', 'f J, jJ IhCretl..slh~.

6. 8~* Oil (SAE 30) flows between parallel
plates spaced 5 mm apart. The bottom plate is
fixed but the upper plate moves with a velocity
of 0.2 mls in the positive x direction. The pressure
gradient is 60 kPa/m, and is negative. Compute
the velocity at various points across the channel
and show the results on a plot. Assume laminar
flow.
The ve/o(;/t-!1 d/sfr/i1U";'OH £5 'iIi/eM b~ the egt{~bt)~
u = V J+ ¥~~) (:;2_ b!1)
u=
so 1ha i
me 9111iw dA:UI
(~.:J. ;')
~ +
(tJ. ()t)SI'm )
u = 'fO!:f + 7.!f;<. It; 'I- ( O. O()S!:f - :/J)
tv/lh U. 111
CCI/ tilJ~1-,;'.1
100 cis
fm/5 whfn !:J i~ I;' 1)t1.
t( as 4 I-ttl1c;bl)d bf
If pr0tjY4m
f:J fO//()lQs.
110 print ,,******* *** **** ******** ** **** ***:+ *************** :t-* *II
120 print II ** This program cal culates t.he velocit.y prof il e * ill
130 print ,,** for Couette flow **"
140 print "**************************************************"
150 print
155 print." y U(y)"
160 for y=O.O to 0.0051 step 0.0005
170 u=40*y+78900*(0.005*y-y~2)
180 print using "#.#### #.####";y,u
190 next "}t

( CtPn't. )
[;,,/;II/ated dak. 4Hd A. p/tJi ()f- the dlLia. a;e !}/vf'J1 j,e/()w.
*************************************************w
** This program calculates the velocity profile **
** for Couette flow **
**************************************************
y
0.0000
0.0005
0.0010
0.0015
0.0020
0.0025
0.0030
0.0035
0.0040
0.0045
0.0050
*10-
3
u(y)
0.0000
0.1975
0.3556
0.4742
0.5534
0.5931
0.5934
0.5542
0.4756
0.3575
0.2000
5r-------------------~~----------------________------____
4
3
E
2
1
0.1 0.2 0.3 0.4 0.5 0.6
u (m/s)
b-lo7

6•.~ 7 Consider a steady, laminar Bow through
a straight horizontal tube having the constant el-
liptical cross section given by the equation:
x2 y2
a2 + b2 = 1
The streamlines are all straight and parallel. In-
vestigate the possibility of using an equation for
the z component of velocity of the form
(
X2 y2)w=A 1----
a2
b2
as an exact solution to this problem. With this
velocity distribution what is the relationship be-
tween the pressure gradient along the tube and
the volume fiowrate through the tube?
Fr(Pin '/he descl"I,iJt/It 1)/ -the pJ"()b/{'m,) U=O) V-':0
l
Ix- =0) ur:j: I(t),
C/hd the. C4?l1ilnU; f~ egtilt tl,,;n li?tl;~~~J' -II1,d Jur.:::.o. w;f1.t 11use
CbI1t/;f/~;'nJ' The iE -CtPmp~pl1t of 1ke NIJII/e}l"-J~".s .e!(J,d,()H.s (Ef. h, IZ 7c.)
rePtltes -16 "Z
Thus) the
~11 d,' +,f!JIt
J!, _ (Lur1- J "Zur)Ji: ~)' ~;(Z. ~J ~
)(,'2.. ';/"2..
- + - =Ial: 6'3,
, .
pr()t~secl veJoc,'-ht d'SfYI btl I-IO~
5 /J1 ce 011 the bt)LlHdtli'fj
( I)
0": A ~ -1':-- r)" A[i- (~: ;-~~» =A D-oj)=()
Th/s V'~.sLlJi 1~(";att'J 7hllt the PY'tJfos('cl lIe/{)cir!1 d/~f~l:)jrho~
Carr b.e used as tt .s~J£li.JbM. Subst/tu 6til1 ~f the ve/oci ~
d',S"fr;l?t/fJoH II1'-1-lJ Et. t.J) ql'/es ihe YeJa-b~sHj be.J-w~eY1
the Pf'.e>StlY.(! 1Ylltltf;'t; #) (/11# 111e Lle/oci'1tj. S;;;C'f!.,)
d2
W-::- _;(,4 d2
W-"" _ Z,4
P,x'z a 7.. o!J'Z - .bz.
I t ~llows 1H4i:
~ - - l AJ<. (-f. + j,~ )
J-t /
( coni)
G-108
(2)

QV1d 1Y1ere fr:, roe
,4:
!="rom EZ·(2)
~ =-
~t
zeD
1ra...b
Lf# C)
Tr4b
T
h
-t--
(-jy. i-~)
d4=-di.d':J
I I I

6. ~ g A fluid is initially at rest between two
horizontal, infinite, parallel plates. A constant
pressure gradient in a direction parallel to the
plates is suddenly applied and the fluid starts to
move. Determine the appropriate differential
equation(s), initial condition, and boundary con-
ditions that govern this type of flow. You need
not solve the equation(s).
D/';:kYPII~,(/j e$ua.ildJl1S OYt! -hte Sf/me.. QS £$5. 1..1291 6./3~ ql1~
~. /;J J e.t"ce,Pt 7Hc t- :; :# 0 (.5,;'a:.. iHe .f'14L.) 1.5 ~J1sl-ea",).
IhvsJ
£1. '. IZ1 r?/ws-t 'nclwde tHe. IOCQI aCce/enJt:lo~ i:t'r""'.J
~ ) Q HII 1Jte ,/ov(lY'l1ln f d,.,ch~J4hlf'j -ef""iJt9Xs tlre:
( x- d/Ye(.-htJ~)
J:n,+,q'j &;11t/J'..j.,p#
/3f)t/J1r/II,.!} C4'JnQi.fioll5 :
0::-
0:
k=o
u=o
_ Jp
d!}
-~
dZ.
.ft:,y-
h>r
6:,-/10
- fi-
i:a ~r ,I'll !:1
!f=~~ -k>y t ~(j •

j
6.89 [t is known that the velocity distribution for steady,
laminar now in circular tubes (either horizontal or vertical)
is parabolic. (See Vidl'Cl "'6.6.) Consider a IO-mm diameter
horizontal tube through which ethyl alcohol is flowing with
a steady mean vclocity 0.15 mls. (a) Would you expect the
vclocity distribution to be parabolic in this ease? Explain.
(b) What is the pressure drop per unit length along the tube?
(a) Chec.K
Re =
flow /; )IJm/;'"r (} Hd
PIIY" /JoFc.. y~j .
= g (/.I~ ></0-
3
yt,~)(t? /~f')
(0. o;,lJ"") "
IVS7 / -, p.er ml
;m
'--1 1/

~. '10
6.90 A simple flow system to be used for steady flow
tests consists of a constant head tank connected to a length of
4-mm-diameter tubing as shown in Fig. P6.90. The liquid has
a viscosity of 0.015 N . s/m2
, a density of 1200 kg/m3
, and dis-
charges into the atmosphere with a mean velocity of 2 m/s. (a)
Verify that the flow will be laminar. (b) The flow is fully de-
veloped in the last 3 m of the tube. What is the pressure at the
pressure gage? (c) What is the magnitude of the wall shearing
stress, Tn' in the fully developed region?
(C<. )
(h)
(c )
Check:. Re'fIJ~/"s hum ber-
Re::: !- V (2~) ::
/'"
/7;r /qmlfJRY /-/f)W,;
v= J!..2.Ae
K)4 ,.l.
SInce fJt= 1: -1, :: 1:- 0
? = J'1t vi
J ..e ~
=
~____IIIIIiiI_m
3m---I.!.Diameter = 4 mm
• FIGURE P6.90
(1;3- 1;, /sz)
fs.ef! I/J'wr-e )
B' ( 0.0/5 ~:) (2 '; ) (3/IH) ::- /~o.le. ~
(0.0:' If I'M) 1
7Ij. :'0 I 50 tha. t
4he/ w/fh v;,"'i::.z V / tJhf"-e. V is the me4n ve/oci"l:'J
Lrr = .< y/ (- f"1-)
ThuS) o.t: the. wall) r= F2}
I(~" --~ ': 1- If
l Lri-)w./I - R
(2 ~ ) ((). 0 Ifj ~.. ')
(0.00 ij. IYI1 )
Z
f&.-1/2
N
= ~O.O-z.m'I

1--'_'1_,_ ......1 6.91 A highly viscous Newtonian liquid (p = 1,300 kglm3;
J.L = 6·0 N . s/m2) is contained in a long, vertical, 150-mm di-
ameter tube. Initially the liquid is at rest but when a valve at
the bottom of the tube is opened flow commences. Although
the flow is slowly changing with time, at any instant the ve-
locity distribution is parabolic, that is, the flow is quasi-steady.
(See Vidt-o V6.6.) Some measurements show that the average
velocity, V, is changing in accordance with the equation
V = 0.1 t, with V in mls when t is in seconds. (a) Show on a
plot the velocity distribution (v: vs. r) at t = 2 s, where v: is the
velocity and r is the radius from the center of the tube. (b) Ver-
ify that the flow is laminar at this instant.
(a) [;.,,- ~rAj,~II' veloci-fJ dl;.fy"bH-h~ II
(b)
3: = J- (f)1-
't;q).
:5,'""C! ~IIK:: 2. V
~=tV[I-(~Y"]
V= 6.1t ) 0.-1: i=- 2.s V = O. 2~
15"~~111'1 :::- 1StmM1 . Thus) ct. (J)
2..
~= 2. (6,2 r)D-("'~':""J"]
1i- ': o. Lf (I - 178 r).)
A- pl{)t (;f -hi;.; ve/~(,;frt d/~Jy"butl()'"
Vz (m/s)
0.000
0.100
0.185
0.256
0.313
0.356
0.384
0.400
0.384
0.356
0.313
0.256
0.256
0.185
0.100
0.000
r(m)
0.075
0.065
0.055
0.045
0.035
0.025
0.015
o
-0.015
-0.025
-0.035
-0.045
-0.045
-0.055
-0.065
-0.075
o - f-Y D -
/'e- 7 -
0.1 I I
0.08 ~
0.06 ,..,~,
0.04 "
0.02 ,
~ 0 t-------t--T------t----t-------I
-0.02 t----t--T---+-~)--t-______j
iV, !
-0.04 i~ I '
~ , I
-0.06~! ! i
~:.:ooo 0;" o~oo o~oo o~oo o~oo I
vz(m/s) ~
6,. () NJ
h1'I~
== '.5 «2..J{)t; ( Flow J".s 14m/naY')
(I)

6. ~ 2 (a) Show that for Poiseuille flow in a
tube of radius R the magnitude of the wall shear-
ing stress, TTl> can be obtained from the relation-
ship
4JlQ
I(Trz)wall! = nR3
for a Newtonian fluid of viscosity Jl. The volume
rate of flow is Q. (b) Determine the magnitude
of the wall shearing stress for a fluid having a
viscosity of 0.003 N·s/m2
flowing with an average
velocity of 100 mmls in a 2-mm-diameter tube.
( Q..)
(b )
/]- J~
(..r~ -:: /' pr
Vi- = ~t" [I -(~]']
a~d V 1.5 1H~ rn el#n ve Jocl+'"J ) /f /o//{)II)S
0lJ.-= _ 'fVr
~r- /C2.
ThUS) a. t tHe u/(',// {}- ==-Je) ;
( "....) - _ 'f~V
Lrr willi - Ie
CIII d ";" "ftt cP:: Tit 2 V
(0-~L./I I=
= /.20 Fa.
If (0.003 ~ )(0. Joe> !!f)
( 0. 00 2. ,.,.. )
"Z.

6.'J3 An incompressible, Newtonian fluid
flows steadily between two infinitely long, con-
centric cylinders as shown in Fig. P6.~3 . The
outer cylinder is fixed, but the inner cylinder
moves with a longitudinal velocity Vo as shown.
For what value of Vo will the drag on the inner
cylinder be zero? Assume that the flow is laminar,
axisymmetric, and fully developed.
Fixed wall"-,..
G'tua 6,;'1:/ t. ./Jf7 J wh'Gh WIIS 'ec/e/~ptd ~Y' fl"w 111 clrcuilfY -lubes)
CA.pplles JJ1 The. flnnu/lir reJI()n. ThUll
~ = ;. (~) r~ + C, in J- + C% (t)
With .b()"i1dtJr~ U)H(J/f;()J1S I y. =~) -v; =0) tlnd yo ::. rz·) ~.:: ~ )
It fe>1/()w.s tnAt:
SUh frac.t
The
( ~ ) ~ 2 + c, i Y1 t;; T Cz
(U) 1.0 Z. + Co,in 1-';.7- C.
~~ '{; ..-
E~ . (Z) 1',./J11'1 E$ J3) Ie CJbl:~u~
~" ?Of)(r,.z- r;') + c, 2~
C '::::
I
w,'j/ be reY'o
L -:: M. (~T JVi-)ri- I Jr J y
COIf:/ w;fJ, v;. ;: a J if I"/I,,ws tha i
L :::)k J Vi:
rr / or
(c.~n 't. )
( 3)

~.93 I
D,t~/eY{1ntJ~l:e £$. OJ w;tn NSfec.t t:cJ
J ~ _ -L (£..e)I-''''' ~
~ - ;1.1<- 0 r J-
So 1h1l t ~i r.: r..
(Z-'ttr.,,~&(~:)r.. +
,
ThusJ
1/1 "rdey ./0r ihe c/Y'II? .fD be ~eY'o.I
-L (ll. ) r. + ~ - ~ (#)(~.7.-y;; 2.)
021' ; z- /..

6.'14- An infinitely long, solid, vertical cylin-
der ot radius R is located in an infinite mass of
an incompressible fluid. Start with the Navier-
Stokes equation in the () direction and derive an
expression for the velocity distribution for the
steady flow case in which the cylinder is rotating
about a fixed axis with a constant angular velocity
w. You need not consider body forces. Assume
that the flow is axisymmetric and the fluid is at
rest at infinity.
p"t" this (//)W he/cf) v;.=0)
I ~(I-l/j-)
-z.; =0; Pi'll! frt)h1 the
-r al--
I d~ ~l-i-
+ - - + -=0
r pC; ~z..
t"t 10IIt)w.s '1hI(. t.
e
.
I hus) the Ntll//fY- S~kes efua-tlt/H 111 the e-d/}~ec.t/l)~ (Eg.6.1ZJb)
loy s tflld~ Ht)w Y.ftluce.s io
I )~ [f d f J~)
o = - ;. ae +~ 1== h- l'" av- -
Due. fo the S'lmme tl''1 tJ/ 1he k"lV )
o-p ::::"0
dB--
-L~
(r ~?)- VB =0
r Jf-o /-'1..
.so tJu. t
ti)Zt/j; I Jv;, _ ~
~",2. 1--;;;: }-~ =0
S,n·ce ~ /05 4 iul1c.i'f!)·Jf "f ~n/!1 r.l E $ ,(J) Cllf1
eA./"'~.ssed 4~ t1n ~ rd/)'JA1"!1 dl Ife"/'Inq/ eI tllJ til)J"J
V'e-tvY'I '#f'J1 t1S
d l/(:; +
dl--
.!!.. (~) =0
dr T
"V&----- -J-
I- d ~ -t- ~.: C1 r
ely.
( CCn 't )
b-117
( I)
(2.)
(.4 )

~. 9'1 1
E$lIflt:/f) ~ t.J) ~n be e)(.pY'~.ss(?d 4£
d (;-I./j;) _
c, y
.405 I- -~ eO) ~ ~ 0) (s,;'ce f/tI~d IS ai r~si O-t 1~/;;lIf~)
So 1ha i C, ;: o. T h US)
.;Jince a i
"'i~ Cz.
ve':: ~
;-=-R v: -= RtV) /f/ a

6. qs A viscous fluid is contained between two
infinitely long vertical concentric cylinders. The
outer cylinder has a radius '0 and rotates with an
angular velocity w. The inner cylinder is fixed and
has a radius 'i' Make use of the Navier-Stokes
equations to obtain an exact solution for the ve-
locity distribution in the gap. Assume that the
flow in the gap is axisymmetric (neither velocity
nor pressure are functions of angular position ()
within gap) and that there are no velocity com-
ponents other than the tangential component.
The only body force is the weight.
( See PY'(),bJeh1
W,th The. hOUY1dIlYf:J
r =:. r: V; :: Y'. U)
0; C7 ()
~j.fd~ f,;Jt.5 t- =Yi,' ) ve.= 0) 4J.1 til
(s~e fij£ll'e ~Y' hOia.t:IOI1)) it:
+-ff)1?1 ~ b' C1) tnat .'
D::
c, r;..
.2..
"f-
(, Y'"
+-~w=
.2..
2W
/-
-r:z..
o
C:2. :::
/-
.so 1n1l.:t
r-tU
1/:::-
G;
I -
.....:a.
t.
1";1-
0
or"
v:=(7
rW [4 I
(1- %2-)
eZ.
1"".''-
C:J.,
-to
,- //7
2-
r.. ~
I.
1- (/- !i.~ )
h"()
r/]1-2-
(I)
kl/()ws.

/;1'
6.96 For flow between concentric cylinders, with the outer
cylinder rotating at an angular velocity w and the inner cylin-
der fixed, it is commonly assumed that the tangential velocity
(ve) distribution in the gap between the cylinders is linear. Based
on the exact solution to this problem (see Problem 6.95) the ve-
locity distribution in the gap is not linear. For an outer cylinder
with radius ro = 2.00 in. and an inner cylinder with radius rj =
1.80 in., show, with the aid of a plot, how the dimensionless
velocity distribution, ve/row, varies with the dimensionless ra-
dial position, r/ro' for the exact and approximate solutions.
fh~ e~a(,.J S/)/U.l:,t~·11 (see PrIJf,Jem i" 15')
rtc.J
[I !lJ~ :: -(/ - ~.I.) f-ao
1=b~
ClIJe! It. n~n d/fY1frlsi~n,,/ ~f"1"n
~':(lr1119
....
[I -.. ~
- ..
f;AJ (1- !.:':) V't> Yo
...."
rf)Y r;.:: I,g 0 In· 1/ MAt;; := 2... Ot) /n . S~I11~ k;u J4. fpP{ VII Juel I'JI( II
" jY'A.'ph t:1l"e sh~WiJ beID"V. Npte ~£.J; -h:tY~ IJ 1:tt-.1e •
d;f./e'fll('e he-/;w~eJ1 the eJC.ac.t Q"d a.ppf'l'ilml.fe .sc/U:tlt9fl.S,'(;'Y' thIS
srntJ/ JA..p wid-t'Jt. FtJr 1111 ,rA&'"C4i purft'Jt.s ho11t :5tJ1u.r'/f)Jl/s HII
on 1/te SJhf)e C.UVlltl shpwl'J.
Linear Exact
",/r.w Yo/row
0.000 0.000
0.125 0.131
0.250 0.260
0.375 0.387
0.500 0.512
0.625 0.637
0.750 0.759
0.875 0.880
1.000 1.000
r/ro
0.900
0.913
0.925
0.938
0.950
0.963
0.975
0.988
1.000
:::::.~_._..__.__1___~__~
a960 -- .------:/-------------
~O.9401--~------.-.-
0.920 /-------- .. --'---'--
0.900 y - - - - - f - - - - - -
0.000 0.500
Vf) Ir;,w
1.000
(I)
(2)

().97 A viscous liquid (p. = 0.0121b . s/ft2
, p = 1.79
slugs/fe) flows through the annular space between two hori-
zontal, fixed, concentric cylinders. If the radius of the inner
cylinder is 1.5 in. and the radius of the outer cylinder is 2.5 in.,
what is the pressure drop along the axis of the annulus per foot
when the volume flowrate is 0.14 feIs?
Check Re'flll';ds iJIIl1'Ibtr 1-0 JeternufJe
R ~ I..V iJ"(! )"-
Whev-e v::-Di. :: :l. (';; - ft·) Q/I1 d
1T (~).-ft..1.)
:J. (1,7t; s!:/;)(t). ILl £13)Thus)
=
tr (~.()/2 I~~ ) (.?Sln, + :-SI#.)
12. 111 •
3'1,Q < 2100
h
S/~ce the Re'f,,~Jd$ l7L1mb!y IJ w!/I b(,/ow Z/oo -the flow I~
/Qm /J1~'" 411 d
So
41' ==
J
-- 33.1 ~h
ft2.
fey It
fr.-,i,
(£6. 6.1~6)
~( ~.S/;'~ (JIS/~ ,2J2-12 I~· ) 12..J.!l' )
-:rot ft:
.2.51/'1'
1,5'1'11.

The
6. '1'6> * Plot the velocity profile for the fluid
flowing in the annular space described in Problem
P6.Q7. Determine from the plot the radius at
which the maximum velocity occurs and compare
with the value predicted from Eq. 6.157.
veloc/f!J ch~ fy; btl.tlt)J1
'" the
- -!. f.ti»[;< 2
~- ~ or r - ~ +
annulus
1":.2_ y; 2-
' (/
.1n Yo
-r,.f..
From ProbJem t .Q7
~: - ~ = - J.a. 'I- !.k.
tJz J. tt 3
.
hI; the -Rgtl~,",dj,IS 9,yfJ1
Jm 1:]f;
(Et, "ISS)
Thus w/iJ, J,t. = o. 016 I~ ,s/~i 2. r·:: I, 5" /11. 1//1 tI Y'() =2. S' I'".) /-. I J. J
If fo//()ttlS "1hAi
~= - (:u.'I i!:3) rr~ (Yft): (¥..ft)':.M~)ln ~ J
if (d. /)/1. ~ ) l I:z. ) n 2..~ !;{.ft
"-L 1.5
wh(JY~
~ 111 .fils w;th J- JH ft. A fY'c9rtl!n .loy
c~/Ctl /~t-/ng ~ as .,Ltll1c. t:1(/~ af r .
the. Y'An'l~~ /"
1,5 ft L.. 1--= ~,ft k//()WS.- 12.1'2.
:00 cls
110 print "**************************************************"
120 print "** This program calculates the velocit.y profile **"
130 print "** for flow in an annulus **"
1~0 print. "**** ** ******* *** **** **** *** * **** ****** **>I: *** *** ***"
150 print
155 print II r (ft) v (ft/s)"
160 for r=1.S/12 to 2.501/12 step 0.1/18
170 v=-~~~*(r-2-0.0~3403-0.05~38*log(r/0.208333))
180 print. using "##. ### ##. ###"; r, v
190 next r
I
( Ct?JI) i )
'--/22-

'.1.& .110. (eO!?!)
Tahu}qted da.ttL ~l1d
FrdJrn 1hp.5e dat:tt. 1--1;
.JS
?/tJ t a.f the d" ia. ()r~ f/IJI(Jn beJ!)f.().
speJ1 1hllt -lite. f'n"XllYlttm lJeJoc.I'i!:J
oCCuvs a.t-
~. /"5 ft
..-
**************************************************
** This program calculates t.he velocity profile **
** for flow in an annulus **
**************************************************
I
r (ft) v (ft/s) ( '~I) t)
0.125 0.000 0.169 1.53l±
0.131 0.l±19 0.175 1.l±64
0.136 0.768 0.181 1.3lj.l
0.14,2 1.048 0.l86 l.169
0.147 1. 265 0.192 0.947
0.153 1. 419 0.197 0.678
0.158 1. 514 0.203 0.362
0.164 1. 552 0.208 0.000
0.23~------------------------------------------------------~
0.21 11111/1'''' ,/ ~
0.19
.::. 0.17
L
0.15
0.13
Hat/mum
ve1ot,tlj OC('LO'~
a.t r= O./("SPt
O.11L-~__~__~__~__~~__~__~______~~__~__~______~~
0.0 0.5 1.0 1.5 2.0
v (ft/s)
/;.-/'2-3

6. q,,)4c As is shown by Eq. 6.150 the pressure
gradient for laminar flow through a tube of con-
stant radius is given by the expression:
zlC rr·1
1
0
.
2
1
0
.
3
1
0
.4 r·5
r·6
1
0
.
7
1
0
.
8
r·9
11.0
R(z)IRo 1.00 0.73 0.67 0.65 0.67 0.80 0.80 0.71 0.73 0.77 1.00
iJp 811Q
- = - -
For a tube whose radius is changing very grad-
ually, such as the one illustrated in Fig. P6. qq ,
it is expected that this equation can be used to
approximate the pressure change along the tube
if the actual radius, R(z), is used at each cross
section. The following measurements were ob-
tained along a particular tube.
Compare the pressure drop over the length efor
this nonuniform tube with one having the con-
stant radius Ro. Hint: To solve this problem you
will need to numerically integrate the equation
for the pressure gradient given above.
/'//////.
,. i .,
FIGURE P6.'1 Cf
(fhe prl'.sslIY'e dY'tJ,) ; t
K!4Jf.ft[R(rT'tdi
;l.Jt..:;: r/J. a" II Ie "" =;e / Ro I
I
!Jf>= g;;C/J).'jrR")-'fdr.-#.
7rR ¥
() 0
(4)!),sitJrli ra.d;us t:ube. (oJt'e bS. 6./S/))
~A Cp}.
1r'F'() If
So tha.i
LI P(/1()nl/()J ~rtn -tube)_
fj f (tln,term I:llbe)
This it?te'jrtlJ Con be. etltl/uated numfY'Jct:/l& USII1.fj
SIft1PjON Cll1d -the da1;tt, ~/veJ1.
/"

(~n't )
~*************************************************
** This program perf01..1l1S numerical integration *>1'
** over a set a set of an odd number of equally **
** spaced point,s using Simpson I s Rule **
**************************************************
Enter number of data points: 11 _¥
Enter data points (X , Y) Ncte: X/V t¥" and '(,.., (RJt..)
? 0.0,1.00
? 0.1,3.52
7 O.2.~.96
? 0.3,5.60
7 O.4~4.96
? 0.5,2.'-1:'-1:
? 0.6.2.'-1:'-1:
? 0.7.3.9'-1:
? 0.8.3.52
? 0.9,2.8'-1:
? 1.0,1.00
The approximate value of the integral 1S: +3.5707E+00
LJf (non IIn'#'rm -f"b~) =
Ap (un,ftJrh1 i::uj,~)
3,57

6,. /00
6.100 Show how Eq. 6.155 is obtained.
;:;y -/It)tt) lit ~11 Ql1l1tt/ tlS) 1': .::c>
1:
ttl: r== Y:~
t:l11 pt
1i=D a.t r= y.(". Thll~ /rtJl'h Er. ~.11/-7
t;'" L (l/?-) '1;2. r
ft< tJz.
C; In t; l' Cl.
I) ..:: ..1 (le ) r:.2
t-
tf)A. d i: {.
el 111 Ii 1" Cz.
t:ll1a ~oJIIlitf ~r- (!, tI)1~
'z. tJt! hnlle
(I)C .::
I
-L I~) (z. .2.)
- ~ L~£ fJ - ~.
111 (~. )
l.2.. )
~ -~ /n f;,
In (~. )
( z )
'1.: -,,{) / yot,.'l:l. ]
/n(~.) n ~
I~

~. I()I I
6.101 A wire of diameter d is stretched along the centerline
of a pipe of diameter D. For a given pressure drop per unit
length of pipe, by how much does the presence of the wire
reduce the ftowrate if (a) diD = 0.1; (b) diD = 0.01?
The
//)c) 4 tk,d:. ';;'y ~ =0 ( h 0 w/re)
cp = 711(; 'rI::.p
?r';uJ.
0:;yre.rjJt)J1ds ~ PO/Jet/Illes Ltw (Eg. 6. / b'1) .
(I)
(z.. )
~ =o. IJ E 1,c l) '7/(/~5 '2..
if=- 7T~If. AI' [I _(tJ.l) '1-+ [t - CO. J) 91 == 6.';7'f
$)A1. (n (o. J) )
Th us) ~'y 1ne stirn e f:.,p +he -f/Ol()r~b:. l:.s reduced b::J
r(o tedttcl:t~1'1 I" t?.=- (;- ~.S7¥).>C )tJfJ ::= 'f2. b %
(J:J) St~ a(ty~ .,t;;v 1; -=-~. () I E1. (2.) 91 tJ-l'.j
cp; rid y.~I'[, _ (/).01) 1',... V- (~.01)9 ~7=- D.]S3
g;uJ. J11 (0,0/) j
t:>t reductIon /11 tjJ:.- (; - (j.7 F3) oX /00 :- :2.1,7 0/0
;tit; ie 1Hlrl- t;,e lYeJfH(€ of
-tlt~ kJe Cef1ffrJJ~e hII~ t:i
et/P 11 ~ t/f V1 S/'hfl/I unIe Q!()jIff
~';II; tC~Jllt e~c.-t t!Jn tke f/otJmde
b ~/ 2.7

7. /
7.1 The Reynolds number, pVD/I1-, is a very important pa-
rameter in fluid mechanics. Verify that the Reynolds number is
dimensionless, using both the FLT system and the MLT system
for basic dimensions, and determine its value for water (at 70
0c) flowing at a velocity of 2 m/s through a 2-in.-diameter pipe.
fJ-VD
(FL-If r 2 )(I. T-)(L) .
Re'111()/tis num bit' =
= -~ FL- 2
r
. (M L- 3)(LT-,)(LJ
- ML_1,-'
/=;;;-- wa..t:er at 10 tJc) ~ = ;V.s
-/m 2
LJ = Cj77. g ..k,.
r /YYI~
(Ta b/e 13. 2 ,,, AppendlJ( B) .
7-1
-
Qnd
rOL~T~
M~LD TD

7.2 I
7.2 What are the dimensions of density,
pressure, specific weight, surface tension, and dy-
namic viscosity in (a) the FLT system, and (b)
the MLT system? Compare your results with
those given in Table 1.1 in Chapter 1.
=
.../arce. .
p= fr~S5UYe:" area.. .::: L.~
~ = spec; lid. we/rift .::
tv(liltj .£ -
Wlllt vi/tune - 1-3
-
-2-
HL T ...:. /-1
L..3 T'1-Z
(a) 1ft the. FtT ~'t.skmJ
. J= L-If T 2-
t::
1'::' t=L- 2
&; FL- 3
r:T= FL- J
;"
-.
FL- 2 T
7-2
(j,) /11 the.
-2
MLT -. ;1.1
L T2..
/. -2.' _2.
. ()/ILT ;;L .- -- -T -I
!VJL r .5Lf.sffm)
I~ HL- 3
p; ft1L-' r-2
~= ML- 2
r- 2
0-::- M T - 2.
/-= ML-'T- I
M
-LT

7,3 1
7.3 For the flow of a thin film of a liquid with a depth h
and a free surface. two im,E2rtant dimensionless parameters are
the Froude number. V/Vgh. and the Weber number, pV2
h/cr.
Determine the value of these two parameters for glycerin
(at 20°C) flowing with a velocity of0.7 mls at a depth of 3 mm.
V
/WI
6,7 "5
If. 08. - --
'f;h
i -
j (r.81 Cf-z)(O,OtJ31M)
Je )~
f-V2.h - (/"2fDo !!-a )(CJ. 7 ~ ~. ~t)3 trn )
-- -, -,2. "
7,Lf I
(t:?-)
0- .33 )I.)D -
tm
7. Lt The Mach number for a body moving
through a fluid with velocity V is defined as V/c,
where c is the speed of sound in the fluid. This
dimensionless parameter is usually considered to
be important in fluid dynamics problems when its
value exceeds 0.3. What would be the velocity of
a body at a Mach number of 0.3 if the fluid is:
(a) air at standard atmospheric pressure and 20
°e, and (b) water at the same temperature and
pressure?
V :: (),3
c
!=or tllr tit- 20De
J
c =3Lf.3. 3
So 111a. t
!!!1
S
V-= tJ. 3 (3'f3. 3 T) ::
( Tq6J~
/03
I'm
-S
2 9. 3
e. if J;' //ppfnix /3 )
Pol" WA-ter at to DC /'1-8'/
t!!1.
(Ta.6/e 8.2
.
/t,pfrld/~8)c.b ) c :: S 111
I
.:s () 1;Iat
V=~, 3 (/'f~/ c;t) - 'flfLf ff!1.
s
7-3

7.S I
7.5 At a sudden contraction in a pipe the
diameter changes from Dj to D2• The pressure
drop, !::.p, which develops across the contraction
is a function of DJ and D2, as well as the velocity,
V, in the larger pipe, and the fluid density, p, and
viscosity, /1. Use DJ , V, and /1 as repeating vari-
ables to determine a suitable set of dimensionless
parameters. Why would it be incorrect to include
the velocity in the smaller pipe as an additional
variable?
n&m the. pt.' -tlJe~rtm) &,- 3 = ~ dlmfl1"M/ess fClYl/metfYS
l1) ~ t/Md? as refe~/;/n1 variables. Tl1us)
a: b C
1Tj = iJp q V~
(;=L-'-)(L ) a- (L T-) h ( PL -J. T) c~ j:~L"T0
I-rC=O
-2 +~,../:,-:2c=o
- j, + C ='0
1t ~ jJ"w.s ih"i a. :£ / ~ .h:: - J,) C:: -/ ) 1/11 PI 1J1fY'~f"r~
/JI. /),
TT.= t::..
I V/,
? == FC
2
r
reg"in''''' Use
{, heck. d"';)fn.sltJfJs u~/n, 1/4L 7 ~'Iskl11 :
IJ.t DJ ~ (ML-1r-Z)(L) .:... M"LoTo
·v)<-' (L I-i) ( HL-JT-)
.'. ok.
((, .b c
11;:: '11), V;a
L (L)Q. (LT-jb (FL- 2
r) c.: rOL~rf)
c=o
'i'" ~ +b -).C=o
-.b + c ':0
(.=-1 I h=c» c=o
7r.. = Dz.
2 D,
7--'1

7. s I ( CC)/I t: )
7Tz 1:S Obv/~usJ'1 cilmfnslol'I/ess.
'{?,y 7T3.' ((. b C
7T3~ /,£1 VI-
(F L- 1fT2) (L) Q. ( L7 -') b ( F1..-2. T) C~ poL~T C
I+G =-0
-If +a -r j, -2C. =0
:l-b -t- C = 0 (Ie" T)
11 ~J/f)tIIS thai et = I) h = I; c =- -/ f/HP( 1h-eJ'e{o~e.
77:= t-D,V
3 ~
. .
Chec,/c JlmpnSIIJIJS us/nJ fv1 1..T .5lfskm .'
4/), 11 (Mi.-3 )(L.)(LT-' )
( . - -d MDI/TD
~ ML-' T-/
., tJ/t..,
Th US)
~Yrhn~/f!;J egU(J..;tt"·J'} )
-rr ~_ 7r 2..
V 7j. D, - ~ ~ D2
wheY't Vs /s 1he ve/l't:.;t!J I ~ the jtn4 J/ey pife.. Since
Vc =(!j )2. V.5 /)z.
lis ,j (lot inrie(Jf}1deni of ~ I P2) I1J1A J/ til1d 1heYe~tt!
.sh"uId no t be /11c,luded. Q~ tJYI /l1depeY1deKi lJav,able.
7-5

7.6 Water sloshes back and forth in a tank as shown in Fig.
P7.6. The frequency of sloshing, w, is assumed to be a function
of the acceleration of gravity, g, the average depth of the wa-
ter, h, and the length of the tank, e. Develop a suitable set of
dimensionless parameters for this problem using g and eas re-
peating variables.
. T-'w=
• FIGURE P7.6
FY-"rn 1/,t. p,' +11 eo rem) 4 - 2 -=. Z d J'me11's Ibn less
paY"'!'€t.eYs Y'e!",'red. Use 3- ann L a.s Y'.ep~(J.·h;'~
vaY'ltJ/JJe.s) Thll5) .b
"'" = W?a..1
and (T"'I)(L T-2)Q.(L)b::!; LO T6
(,f;,r L)
(r;,~ r)
z+ h)/~ws b =y'z a"d 1here/r,yeI
Chec..k
Fo yo "TTl.: J q IJ b
7r'}.. := ?1 d )(..
L (LT-t.) a:.(L.).b == L()TtJ
/ + a+b = t;
- 'l tL '::'0
, Dk• ¥

7.7 I
7.7 It is desired to determine the wave height when wind
blows across a lake. The wave height, H, is assumed to be a
function of the wind speed, V, the water density, p, the air den-
sity, Po' the water depth, d, the distance from the shore, e, and
the acceleration of gravity, g, as shown in Fig. P7.7. Use d, V,
and p as repeating variables to determine a suitable set of pi
terms that could be used to describe this problem.
___..~+_H_··_··_·~t_____________~_·~~~~_V_.~
'1
#= L V'= LT-1 f -= FL-"r2
fa.: FL-
If
rl- d::' L
F;."111 the 1/ fh et>YftJ1 7- 3 = Lf f/J
d J ~ 411"I 4.s fete4.-/;;'lIj VPrt'a6/es ,
tT, = II d a /I b; C
.ferms ret(; u/rpti. Use
Thus)
(L ) ( L J(;. (L T-I) b (Ft - 'fT ~) C = F ()L" TO
~=o
I + a... rb - 'Ie =-0
- b -f"Z.C =0
( hI" F)
(~r L )
(.{".,. T)
Cl=-/ 1,=0 C=6 anti ThftelCteoJ J I
1r~J:t.
I d
oj, '/J~{lSi.:J dJMtns/on/ps...5.
tTl.::: fa. a4. tr bt C
[FL -'I r7)tl..) fA. (LT-1 ) b (FL -IfT~) t: = POL6 TO
I -f C = 0
- If -t- a. -t-b - 4- c =0
2-b-r2C ==0
((!6)n 'i)
7-7
(~r' r )
(full' '-)
(,fr,'(" T)

7. 7 I ( c~I1'i)
til'll' e::{S J;r 7T/) a =-I) b= 0) C=o ,50 That.
-rr - ).
1';3 - cr
/;,. 7T'f: 11"'f:;- d da.f J, /' c
{LT--z)(L)4.(LT-'jh (pL-lf r2-)c=. fDL()T"
C'.::. 0
/ -r A. ~h - t.t- c. =0
-2. -b "1"2(.-=0
C :: 0
)
H _ ).. (Ia. ft ~)
d - r f') d..) V'"
7-3
(.for t)
(-hf' L)
(f;r T)
I 01<, ,

:-:--:s--:::;:S-:-:-~-::-::::- t
7.S Water flows over a dam as illustrated in Fig. P7.~ .
Assume the flowrate, q, per unit length along the dam depends
on the head, H, width, b, acceleration of gravity, g, fluid density,
p, and fluid viscosity, J.L. Develop a suitable set ofdimensionless
parameters for this problem using b, g, and p as repeating
variables.
'f ~ L2. , - ' Jf:: L j,..:. L ;= L ,-2
FrtJn? -the IJL the()rem 1,-3= 3 pi
H
1
1=
tfrmJ
f---b-l
• FIGURE P7.~
Pi -'111- f=- FL-2.r
regUJY'f'd. Use..
bJ 31
Vlnd I as relea-ifh:J vt/n 'af;/es Th(Js~
and
77f= ~ b aJ~(c
(L Z T -)(L ) et( L T-2.) b (FL- lIt:2.) Co .-:
c=-O
2 -f a... -t-b -LfC =0
- / - z.b -I-L C '::0
'J
'p3~ J ~
pOL 0 TO
Check d/~en~/~M us,;'j 1'-1 L T .5't:,f.em.'
t:£ /, C
77;.-- JI 6 j !
(L)(L) ~ (LT-z).J(,t=t-'fT-a.j c. ~ FOL oro
C=o
I-t- ~ rb -tfC =0
- 2-b -rl. C =0
a =-1) b ::())
11: = 1-12.. b
1'.5 () b1/J~~.50 dt'm !14.5IPI') Jess
7-Cf
(.for ;:)
(:ky L)
(lor T)
, ()/<
(J::,r r)
thy L)
(.ky T)

7.8 J (~I1'i )
a. " Co
113=)<- b j f
(FL -2. T) (J-) #I. (J- T-j /,(;=-L-If-T 2) c --= pOLO TO
/ -I-C =0
-). 1" t{ -t.b -tfc-=o
/ - 2b-r2 c. =0
:t.t- 10IltP4IJ thai a =- ~) ~ =- 1z.)
n= L
3 I/Iz/'1
c= -J
I
o//rnel1~/i;f1J u5iny ML T s"'f~.J-em.'
?: ~ (I1L-'r-' )
f:,9Iz
if'lz..
f - (L)3/z(I..T-V'IZ (M[3)
(,k,- F)
(J;y L)
(oCy T)
11
()I tJ)-o ,
- L. " ())(

I
7. 'f I
7. 'l The pressure rise, !:J.p, across a pump can
be expressed as
tJ.p = f (D, p, ro, Q)
where D is the impeller diameter. p the fluid den·
sity, (j) the rotational speed, and Q the flowrate.
Determine a suitable set of dimensionless param-
eters.
-2
II .fellows 1h~t tZ = -2) b' -/ , C. =-2 ) qnt/ 111trehte
_ 4p
. . Tli- D'(o4J'
CheclL. dlmel15l()11S IlSin, NI-T slfskm .-
t1p Mf. -'T-'"
-
D,/,tcJ' (L) Z (J..,J:')(r~·
n; = rP D"'/4J c
(L 3r)(L.),''(;:r¥r)b(rJC: rLOT"
.b=o
31''' - ~b =0
-1+2/'-C.=0
:U f",11.ws 1hai a. =-3 b=0, c" -/ 4n4 1h'",/(,reI ,
- cp
. 1r;z - D34J
Checir dimenSIOns 1151;', Mt..T ~'fsJ..em "
CD _ L"T-' _ == NDLDFo
D0p,) (L)3 (r-')
7-1/
.'. bl<.
(;;,~ ,t)
(f"y '-)
(Tor T)
.: OK.

7. /0 I
7. 10 The drag, ~, on a washer shaped plate
placed normal to a stream of fluid can be ex-
pressed as
ill = f(d" dh V, fl, p)
where d, is the outer diameter, d2 the inner di-
ameter, V the fluid velocity, f1 the fluid viscosity,
and p the fluid density. Some experiments are to
be performed in a wind tunnel to determine the
drag. What dimensionless parameters would you
use to organize these data?
~ == F "-, -= L dz. :d: LV':: L T-
1
)A =- FL_aT
rrlJ)tn tJte. p/ fhet7r'rm J
, - ~ =3 fi Hl"l71s Y',!~,;'ed. lise
d,) V; tlno I "" ~ft'a,6/11~ V4ntl"/e~. Thus)
Al tl "C'rr;.: pV d, j/ t
(F) (L) (4 (L r-j b(FL-'trz) c: = F"L ~Tfj
/-t-C=O
a.+h -Ifc=o
-.b-+2.C=o
(/;,,,.. F)
(-ICY' L..)
(.fey T)
11: r;,//PI(I$ ihlli /<..=-2.; },=-2, C ='-/ ;
~
1T:~
I
0/ VI'
ChecK qlmfl1SI;'ns "'SlfJ.!) NtT ~lfs-t-em :
71..'.z.
~ MLT
-2
::::::
(L) 2. (LT-') (I1L-3)d/' VZ;
~ b c
712 = d;L d, V !
(L )(L) tZ (LT -') b lfL -1fT 2) c =
C1 =0
/-t-a + b - 1Jc. =0
.b +:J..c ':;0
-
7-12
tJl1t/ 1here'ye
NIJttJTO . ()I<.
- ..

7. 10 I
.zt -Iol/e;lVs 1h1lt t:l. :: - I; .b =ole::0 I tI/I t/ 1heY'e/:;"e.
tJh Icn IS
fOr 7?3:
17:=~2. d,
ObV/()VS/; tiltnf'I1.5I(J/Jlpss .
IrC .::a
- J. -t t( -t./J -ifc ::0
/ -.b -+ 2.c =lJ
11:=.3
(~r )=")
(.j;,r L)
("'yo T)
.: ~k
d, Vj1
II)
SIhC(!; I-Vdl I~ a. sirJl1d£trtl dl ~pn~/~l1less PtlrqmebY'
I"-
(Re!;//J"lds num bt!}") I E3. (I) AJould rnOYe Comm t)l11':1 be.
ex-tressed (J 5
(Z)
As h,y AS cfJmfl1,sIOl1et J /Jllq !t1Si.s Ij ~f1C'erl1pd I ,=;s. (J)
til? tI (Z) qre .fJgu/llalen t" .
?-/3

7. 1/ I
7.11 Under certain conditions, wind blowing past a rectan- .
gular speed limit sign can cause the sign to oscillate with a fre-
quency w. (See Fig. P7.11 and Video VY.6.) Assume that w is
a function of the sign width, h, sign height, h, wind velocity, V,
air density, p, and an elastic constant, k, for the supporting pole.
The constant, k, has dimensions of FL. Develop a suitable set
of pi terms for this problem.
• FIGURE P7.11
tv: r- I
b:.L -4t=L V== LT-
1
f= r-L-'tr7. .,k';FL
PY'611?1 1'nt. p/ 1I1eorem '-J = 3 pi hn?lj YeZUlyptJ/. Use
b.J ~ tin,!;; 11.5 repea.-t,Hfj INIYlable...s. Thu;~
tJ; ::- tv j, a. V f;C
(T -I) (L)tO.(L r-jh(rL "''fT~ c: Pi- "TtJ
C=:.o
a-t J, - i.fC =0
- I - j, i"lC. =0
(tor r)
(/:,y L)
(Joy T)
t t /-oIJIf}IJI..$ #fa a: IJ j,:. -I) C:. 0 ) CIIf a 1htre ~re.
wb
1Ti= V
C he("k c/"YlfI/S/()I1.5: ( (
fA) b T-1) L) ....:... L() TO
V-':: (L,J)
,: 01<..
7T; - h j,Ow vb
/
C
(L) (L)(J. (LT-1)b (FL-'fr2-)(.=. pOLO TO
(! =0
J -t a. +b- ¥c. =0
- .b "i"Zc.. =0
7-1'f
(~y F)
(~Y' L)
(,(oJ" T)

7 1/ I (tt!)l1t)
/=;r TT3:
1T3:: ~ J/~' Vbf c:.
(r-L)(L)Q..{L7-0 h ( FL-'Ir).) C =toL0 TiJ
/-I-C!. =0
l-t a...,. b - Jfc.. =a
-1 -r Zc. =0
:Ii ~/jf)Wj 1;J"i a-= - 3) .b: -l) c:::- - I) ~J1A 1here(ore.
-k
173 =.b'V'1
Ch~CK r/;mel'l~PII.s ~ Slh..J ML T .s'l~klt1f I'
o
- fi1 ~ L ~ T :. Dk.
-4< . HL'1. T - Z
),3V'/~ = (L')(LT-,)2(HL-3)
7-1S-

7.12- I
1.12. The flowrate, Q, of water in an open channel is as-
sumed to be a function ofthe cross-sectional area of the channel,
A, the height of the roughness of the channel surface, c, the
acceleration of gravity, g, and the slope, So' of the hill on which
the channel sits. Put this relationship into dimensionless form.
(p::' L3 T-1
FYO/YI the ?'" 1he~yeWI I S--"2:. '3 P/ ffrrns Y'e0/J/~r". US~
,4 tlnli 9 as rffJ(Jul;'~ Vt(v/ab Je..s Thus)
11j = ([J ,44
J6
Check
(i ~ r-I)(I..") I(. (L r-1.) "=- L()TO
E+Z4-tb=c
- I - 2 J, =D
a. :' - 5"/'f J b:: - I/z. 1 a,,1 1Ite"'~ fo,re.
~ - --=-.(/)......""",,==-
1- A5"/'fff
I -t2.&L-tb =0
- 2..b = D
1J,p f a ':= - 1/4,) b = (:)) Ii 11 ~ 1'ltfre f;,1I'e
1T. - e2- -
.: 01<
(toy L)
(rfr~ T)
r.A
IJ t>bv/~I4.5J:J d,me"s/t!JJ1/eS'J. Th( 17l1vd f,' ~m Ij
-SC) Jj dun tl1SJ~IIJfSS. 7ft us)
7-/(II

7. /3 1 I.U Because of surface tension, it is possible, with care,
to support an object heavier than water on the water surface as
shown in Fig. P7.!3. (See Video V1.S.) The maximum thick-
ness, h, of a square ofmaterial that can be supported is assumed
to be a function of the length of the side of the square, e, the
density of the material, p, the acceleration of gravity, g, and the
surface tension of the liquid, 0'. Develop a suitable set of di-
mensionless parameters for this problem.
I' {' -I
~);fit~~8 i{Fm>4');~I~1
T
~=f(.J. 1) ..,0-)) () (/')
-l=' L l='L ;= 1=I--'t,2- d= Lr-1
Fy~rn the p/ 1he~Y'em) b-- B::. Z fi .f.fr/Y1J Y'R11J1rnl...
t) ffJ (lI1d I (Is retea,6;~~ vl/f'iflbJe..s. Thus)
1Ti -:: ),. .1 a.1- 6; C
(t.. )(L)~(l..T-~) b(FL-If r2.) c_' rfJLD(-
C! ::: 0
'-t-~-+,b-,+c.. =0
- 2.b +2.. c.. .:= ()
a=-') j:.o) e:&J,
-It
flt=7
(/1111 171tJrt"re
wAlch /5 ~j,JlJ~II.sJ!:1 citinf,,~i!JIJ!1sS .
/=by 7Tz....' ~ IJ c:.
7T"J.- = rr t d !
(FL-')(L) 4..(L T--~ b (FL-'fr 2.) Co = j='tiO I
/ -+ C := 0
-/ +tJ... -rb - 'Ie =0
_ 2.~ -r '2C =0
a:-z. )
l,-:-/ c.=-/ t/htl there{;r<.) )
rr
j.~t
If'!'J;'~ /vtt.. 7 ~'Is/e1l1 f
0- ' ( fo1 7- "L.. )
J, 2#f =(L7.)0. 1-Z
){ML- ~ )
-f:= cf (t-f-it )
7-17
.-:. M~L 0 TP
..

7.Jlf
7.J4 As shown in Fig. P7.14 and Video V504, a jet of liq-
uid directed against a block can tip over the block. Assume that
the velocity, V, needed to tip over the block is a function of the
fluid density, p, the diameter of the jet, D, the weight of the
block, 'lV, the width of the block, b, and the distance, d, be-
tween the jet and the bottom of the block. (a) Determine a set
of dimensionless parameters for this problem. Form the di-
mensionless parameters by inspection. (b) Use the momentum
equation to detennine an equation for V in terms of the other
variables. (c) Compare the results of parts (a) and (b).
• FIGURE P7.14
(a) V =-f (t) DJ ~) b; d )
V; t.. ~-I f =- r-L-1f T -a. D:: L t:p,J =f bd: L cJ -= L
p;.1)f11 1nt. ?/ 1htJ)rem J '-3 :: 3 p,' -krrns r~tIJJY'ed .
B/f j';S?eVr/DI1 h:,y 11, (tDI1Hlil/;'..1 V)
11i = V D f:G '~ (L 7-')(/...) ({r:~~T~) "," rot!'
Check US/IfJ I1LT:
Vb Y.l
1
=
H,r 71., J~~
b....
11". =:z- d
aha lor 17j
cJ
rr,3:: D
til1lt b,,"In 1T~ tll1'{ tT) tire oJ:, JlI ~u.s 1':7
{.hI t=;r Irnpend/n1 flPplnj aY~I(;1f' 0
L MIJ =e>
50 171(1,,.1: h
t= cl =t:tJ (-;.)
7-18
c/;men~/lJl1less .
( I )
o

7. Itf I ( 6PJ1 ~ )
.
f:y-IP"" m",m~l1ftlm ~"~Jr/eya,t/~ t{51;'~ tJ,,~ CV sh(Jwlt
C9 !fu ;.;. dA =- L F"
f V2,A = F
7hH~j /rPTn I='1. {J.}
(;V2.A)(d) = ~ ({)
$;0 i114, t
V= V~ (j,) •
2.f,4 d
" nil /A/11h ,4 =..,.,.fir D Z
11= /21v b
r ( 7Tf'd-D-1
(') Front part Ca.)
... Ire;;;-' A (i E.)V - V~ '1 d) D
1?"f .£2.) Clll1 b£ IN yoI ifr11 a s
{ 2. }
£.3)
.t:f h//oltJs b'f &JlI1flJr'Jn.J E,/s. (2) tfn,;;{ /.)) 1VJtJ.t
f (~) ~ ) = (~)(-ff)'
So 1/t..,t 1(~) ~) /.s actua ":1 inrJetflf("J~nt tJ f -i .
7-1'1

7. /s
7.15 A viscous fluid is poured onto a horizontal plate as
shown in Fig. P7.r5". Assume that the time, t, required for the
fluid to flow a certain distance, d, along the plate is a function
of the volume of fluid poured, ¥, acceleration of gravity, g.
fluid density, P. and fluid viscosity. fL. Determine an appropriate
set of pi terms to describe this process. Form the pi terms by
inspection.
I---d--j
• FIG U REP 7 •1%S
i:-= T
t=rpm -the pt,' 7he"rem
B:; /n.5f~dtt3;j/ f;r 70
;.= J...T- z
!;;: FL.-
lf
r2. !-~ FL-2. r
6, - 3 = 3 pi t.errns rejVI"retJ..
(('~11i:4Iitll1.7 -t) :
II,
_ (T)(I-T-7L) L == pOLdro
,', ~/:.
CL ) ~.z.
1T;=iv'J
7-20
/

7. Ie, I
7.It; Assume that the drag, ID, on an aircraft
flying at supersonic speeds is a function of its
velocity, V, fluid density, p, speed of sound, c,
and a series oflengths, ell ... ,ei , which describe
the geometry of the aircraft. Develop a set of pi
terms that could be used to investigate experi-
mentally how the drag is affected by the various
factors listed. Form the pi terms by inspection.
oJ.:: f ('0/.1 C) )1) ., " Y/., )
V::;L. 7-1 I == FL-If T 'Z Co == LT-'
r-rtJtn the pi thetJI'etn) ("/--fL)-3 = /+i pi ffl"tns Y"f'J"/re~ where
L' IS the ntlmhtr of /en91h itrmJ (t'= ~ ; 3) etc. ).
Blj /J1Sfectl(;H) fer 7T; (Ct)J1fq/111)1~ otJ) .-
1T. - ,(J -=- F :::: !=~L 0 TO
I - I V2..J,/, - (FI--IfTl.)t1.r-'j 2.fLj 2-
Check. us/nj MLT:
~ Mt..T
_2.
rM/.-3) ( LT-I
) z. (2) ~
Fbf ~ (c~nl-(jin';'j c) .
Co ~
7Ti. = V Dr C
Cll1d both (lYe abl/lows/'1 dllnens/on Jess.
/i;r tI / / () the,. ft' .feyms c,tPl1 kI/~/iJ!J J.L'
7T.' =- t,'
t J/~
t)/<..
til1d these tefm5 /~I1()/III;'-' 1he 1; CI re obJl,otl~ Iy dimens1011 Jess.
Thus J
wheyt:.
7- 21

7, J7 I
7.11 When a fluid flows slowly past a vertical plate of
height h and width b (see Fig. P7.17), pressure develops on the
face of the plate. Assume that the pressure, p, at the midpoint
of the plate is a function of plate height and width, the approach
velocity. V, and the fluid viscosity, /.L. Make use of dimensional
anaJysis to determine how the pressure, p, will change when the
fluid velocity, V, is doubled.
1'= .f(~) b) tI))<--)
t= ~c2. J..':: L b-= L l!:: Lr-I
~= /=C' T
FrfJfJ? the pt' 1heOYfm 5- 3 .::. z. fi' +trl11J
8'1 1;"'pec,t:/~:/II.1 h,,.. m (dd?11 1-4,iuh, -;):
1Ji = ~-t. ~ (FL-
z
)(L) .:..
V)A-- Ct.. ,I)(rl--:Z r)
Check. IA$/~.! fitf.., T :
1'-4.. _ fHL _lr--a.)(L) -.: M"L()TtJ
V)<-- - (L 7-1 ) ( Mt.-',-')
ff;r ~ (CtP,rltJ/'1I;'J b) :
frJ- = i
whIch JJ obvifJ('(sb dlm·e~~~oll/tS.5. Tn u~,
ti~ef({)
V;«-
So
Fr~m 13"1. tJ) if ,f:,;I"WJ 1),a1: -kr A. ?J v(~ re&flefr;
and VI~t~5Ifr/ if 1At! v(»eI(I+'1/~ Ii dou~/ed -t1te..
l...rlS5l1fe) r: wal he dCtlb/fd.
Plate Width =Ii
, CJK..
0)

7;3 1
7.1 ~ The pressure drop, Ap, along a straight
pipe of diameter D has been experimentally stud-
ied, and it is observed that for laminar flow of a
given fluid and pipe, the pressure drop varies di-
rectly with the distance, e, between pressure taps.
Assume that Ap is a function of D and e, the
velocity, V, and the fluid viscosity, fl.. Use di-
mensional analysis to deduce how the pressure
drop varies with pipe diameter.
fJ I' =- f. (/)/ J) ~)t)
At=- I=L-'J.. D-' L }-' L V.; LT-' 1-; FL-'-r
Frt)m 1he p/ fhet)~pmJ s-3 =.l pI.' 17rms ye$tlIYed.
Bt{ II1:5fec.l-u;H) -foy -n; (CtP)1.Jqil1lhj A.p):
7r= /l.f>J) ::!: (I=L-
2
)(L) --.: rOL~To
I /' V (FL- 2 T) (LT-')
ChecK. uSln~ ML T :
Af b ~ (I,/}L""7-2
)(L) . 11 6 LO TD
!'" Y (Mt..-'r-')(LT-')
.: 01<.
FDy -n; (CtPnltlil),n,')'
1Tz =i
Oblli()usJlj dlmeY/51()J1less. Thus;
LJfb =1(1.
D
)
~V
shfrll1fJ1t af 1Jte pY'()/'iem) LJp oC 1
rnus i be t).f. the ~rl'Y1
4f D == k 1..
)'IV D
(I)
So 1J14i
whfll'e I{ IS c50me tons/:QJ1t. It 'thus JcJ/ow~ 1hot
I
-D'2.
for a. !lIVen lteloci~.
7-23

7.11 I
I. [q The viscosity, /-L, of a liquid can be measured by de-
. termining the time, t, it takes for a sphere of diameter, d, to
settle slowly through a distance, e, in a vertical cylinder of
diameter, D, containing the liquid (see Fig. P7.19). Assume that
t = fee, d, D, /-L, ~'Y)
where ~'Y is the difference in specific weights between the
sphere and the liquid. Use dimensional analysis to show how t
is related to /-L, and describe how such an apparatus might be
used to measure viscosity.
rYllfl1 1ht f/ 1heOYflf1) ,- 3::, 3
t y 7Ti (tlPn hi11IIi.J 1: ) :
1T: - i: L1ad. .
,- ,)A. -
{,) (f:t..-a)(L)
( FL-2. T )
Che,k. USlhj MI-T: -t Ard -!
~
f;,r 11;. (CSJI11rti/11;'j D).'
11'1. =~
(T ) (N C 2T-2.~IL) ~
( "-1. L-'T-' )
"bVIDUS).!1 dlmf"si~I1J~5s
(~fPn ft/IJlIH.J J.) :
1T - L.3- t:i-
()6111~U6/~ dlmtl1s/o,,/es.5. Thus,
-I::lJoa_ rf,(O ft)- r d.;-
r d..
If. hied tjeemef,. ~
-/:Llad.= C
~
C J:s q e.,11 :,-/-aM:i) ~ r
/Cylinder
{ ~~~ ,>-<Gf'"
J
Sphere
)A- == ~ !J /I -f ::: {!, Mt
0.. ~"s.f-a",..+ ~1" a f.1 ~ eJ. 3e()'l1 e fr.!:J. Tit ~sI .fr,.".
)A. = ~ Dr-/:
Ilt.e e,1I,s.f.II/1-t (, talt be at.termlt1'(,p{ b'J ~a/,'b¥'aJ'f)')( w,1lt A
~/Ul" ,/ ~iI~lJ)n Ih5eP5rh. {P, Th C, ~ h,wII fh( VIS·COS,'-#, ~ I
D1/1(r fltI' CD Ok-t. hL dt-/:eYlrJt/I'e '" ih Y'OtJj h 4 IYl eQsII. llem P4
~I- 1'1u. -f.IMt! -t: Ii! t.pnjuncf,f)'x L<Jl7h. Et. (J) ,
7-24-

7.2.0 I
1.40 A cylinder with a diameter, D, floats upright in a liquid
as shown in Fig. P7.20. When the cylinder is displaced slightly
along its vertical axis it will oscillate about its equilibrium po-
sition with a frequency, w. Assume that this frequency is a func-
tion of the diameter, D, the mass of the cylinder, m, and the
specific weight, y, of the liquid. Determine, with the aid of
dimensional analysis, how the frequency is related to these var-
iables. If the mass of the cylinder were increased, would the
frequency increase or decrease?
. -1
w::.T D=L
Fr~'" 1/1e p/ 1heD~~W1, 4 -3 =
8'J I;' 'f'ec.;f-rD~tJ .
Check..
71:=I
S/~Ce- -there
. ",,1, / p/ -/-en"" ;~ P~IIDWS 1h14:t:JJ
W
~:: C-D J'"
where C Is it. ~"s1-aHf . Thus)
tv= CJ) vr
i'
Fr~m p,i$ reSHli /f ~1/~klJ tnoi:. /f tn1 ,j incy-ease"
tv w/Jl decrease.
7-2~
/Cylinder
C diameter =f)
,: OJ(

7.:lJ/f: I
V, ftls
*7.21 The pressure drop, ap, over a certain length of hori-
zontal pipe is assumed to be a function of the velocity, V, of the
fluid in the pipe, the pipe diameter, D, and the fluid density and
viscosity, p and p,. (a) Show that this flow can be described in di-
mensionless form as a "pressure coefficient," Cp = ap/(0.5 pVl)
that depends on the Reynolds number, Re = pVD/p,. (b) The fol-
lowing data were obtained in an experiment involving a fluid with
p = 2 slugs/ft3, P, = 2 X 1O-31b· s/ft2, and D = 0.1 ft. Plot a
dimensionless graph and use a power law equation to determine
the functional relationship between the pressure coefficient and
the Reynolds number.
------j------_._._.__.._-_.
(c) What are the limitations on the applicability of your equa-
tion obtained in part (b)?
(a. ) A r=: +- (~ DJ f; f< )
I.Jp= PL-1. Y=LT-
I
D=L
FY'1'1I fh( p/ theDremJ 5"- 3 ::
8~ ';,spec....hc if .fc>r 7T,.)
-rr. = A. P -==
I /,v:z.
Che(.k U'SIII'j 11LT ~'I.5.J-er11 :
Ap ML-1r-Z.
IV,," == (Hc3) (LT-j l- -
3
11
17
20
fJ VD -
TT2 = ~
(PL-lf,2.)(LT-1
){Lj _
( FL.-Z. T )
us,;" MLT S'lsiem :
I-VP::: (ML-a) (LT-I
) (L).= H"L/)T~
,#-- ( /vi L- ' 7-1
)
Thus) J1p
-IV2.
192
704
1088
1280
:. Ok..
:. OJ::.
~
" i..s aJit unk.hOUJH .{u/lfc.;hoi1 1
a ..{(J.Gl::or
o.G"" Ct:ln b~ inc.-Iude d 111 Tr, (/1 deS/re,) $0
Thw,S/
(,uhel"t. Cp
ntlt11 bey,
AL _ .1 (~V.b)
aStV2. - 't' ~..
Lp - c)(R~)
,..; tnt. pre~sur.e ~eff,c.I;' t tll1d Re 1lr-e. Re~"d/J..s
(Co c>n !t.)

7. 2 I I (CDn 'z. )
(b) fA$1';1 inl!. da.-/::a '7Jlie n )
.6.P _ _.d~p---:----:-_
c;.=6,SfV'- - (6.5){2. 1;,r) V' -
eli-II( (~~)( ) r / (",
t€ :: /:.YJ:> =- 2 ok 3- Y (D. -M:..i _
e)A- 1...>< 10-.1 I~'),..
Ta~U/4-k-~
1J,e 4..t-a
v, ftls
3
11
17
20
valtleJ
tire
Ll p, psf
192
704
1090
1280
/01" Cp C/J!fP. Re
Show H be/,u; .
Re Cp
300 21.3
1100 5.82
1700 3.77
2000 3.20
(DD V
.....
cQ) 25.0 -r---------~---__.
• Cp = 638 Re-1
.
00
;g 20.0 f-----'-----'--------..-
:g 0. 15.0 +---~'-------------l
~ () 10.0 f----~.........~~--.. -... ----------1
~ 5.0 1-----~----=----iiOiiiiiii,;~~----....
£ 0.0 +----,---...,.-----,.---...,.-----1
o 500 1000 1500 2000 2500
Reynolds number, Re
{hlL ?()u.JtV /t(w Y'e /d,o'"Sh,f ,j
~3'8
c =~ Re
(C) BII.sep "11 1he tI~rJ46Jej Ust-d 411,{ 1'h~ rille" aa..f:a. ~
1nL ern;/y./ca I y-e/...·/-,~II.sh, i') £$. (j)..) WI)u./~ (!)"J'J b4!
a pp/" c.. "I... ,.,,- 1h... ReIf1101"$ h f.I {'II ~/Y rA 11'i&.
300 < Re. 5... 200D
ND1:.e.: 41-thPI(tlh -me efu£t/~11 I71JfhC he.. VI//,&I ""b/de
-fh/J rtJJ19€1 re1/,(!-h .s-hflu/I( /lot- I:u.. ex. t:/'"#..fJdlal:e'
loe,~"1f 1he rlll'lfe ()f d~ta used.
?-1.7
[1)

7.22 The height, h, that a liquid will rise in
a capillary tube is a function of the tube diameter,
D, the specific weight of the liquid, y, and the
surface tension, (J. Perform a dimensional anal-
ysis using both the FLT and MLT systems for
basic dimensions. Note: The results should ob-
viously be the same regardless of the system of
dimensions used. If your analysis indicates oth-
erwise, go back and check your work giving
particular attention to the required number of
reference dimensions.
Us/;" P/...T ~fjstem :
-i -= L .D == L ~= PL-3
CT= /=t--
I
FrtJm The P": the()rem I 1./--2 = .z pL'tertnJ
8'1 InSfec..ttDII) for iTj (~ni4i/1liJ, -l) :
11i=:f
/.5 OJ,V'Dt/S!y dt'/nfnS/onJess.
( cDnt:olnln, ;r tlnd rr) :
FL -/
{= ~ (a-a;.. )
USlfl'~ ULT ~'fJkm .-
-1,. =L D-' L ~= ML-
2
;-2. 0-= M r-2.
/.J/tJ,o~,), '!here a.fftllY'5 f() be .3 reierel1((!.. dl;'fI1S/~~) tJJ1/Y
,t reierFJ1ce dlmeflslol1-5 dYe act",ol/!I Yt'$tlIYfd ( Land M T-z)
1-0 de5cr/~e 'the i/qnqbJes, 8,/ inspec.i't!)"n,l hr 77; Csfle 41"ye)
tt=:l.I .D
CU1C,( ~Y' 7Tz (~ittIHI"" d- (JJ1t1 (j) .'
0- = /VI T - 2. _ _ H DL ~TD
~= OJ)2. (ML- Z T-z."j(L) l
7-28

7:2g I
7.23 The speed of sound in a gas, c, is a func-
tion of the gas pressure, p, and density, p. De-
termine, with the aid of dimensional analysis,
how the velocity is related to the pressure and
density. Be careful when you decide on how many
reference dimensions are required.
c= f' (?;f)
fJ:l: FL-'2.
/flf/J()tl9i1 1he~ ~'p'"r.; to k 3 reference c/Jmfl1~II)IJS
(wjJlch would I~c!tcat~ thac rhl'r~ tlre ilo foss/Me. pi I:eyrns);
011/'1 2 re/erel1ce. d,rn.fI1SII)II.s (LT-1 (/11&/ FJ...- 2
) aYe
aciU4//'1 resulted since _,/2-
t= (rt.- 2
)(LT
Thu~ fr~m 1Jte pl' 1Jzel)~m) 3 -2 =/ Pl' term re~ulred.
8'1 In~pecl::l{;11 .' ~ '..0 _I) 2 r~ -2,1 -,)-~
1T. = ~ == (1-T UF L / (LTd: riJLt) TO
J -p PI..-'2.
Che~k U51J1J NLT.'
(LT- 1)2.(M/..-3
): MDLtJT D
ML-'7- 2
onJIj / fL' term)
~= C
where C J~
1>
a CtJHstani.
c=
c= Clff
where. C, JS a. CliI1S,j;qHi (4= rC ) .
, OJ<

7. 2. 4 Hr -'I- I. 2.L/ The pressure rise, t1p = P2 - PI' across the abrupt
expansion of Fig. P7.24 through which a liquid is flowing can
be expressed as
where A 1 and A2 are the upstream and downstream cross-sec-
tional areas, respectively, p is the fluid density, and VI is the
upstream velocity. Some experimental data obtained with A2 =
1.25 ft2, VI = 5.00 ft/s. and using water with p = 1.94 slugs/ft
3
are given in the following table:
AI (ft2
) 0.10 I 0.25 0.37 0.52 I 0.61
10.3 I 11.6 I 12.3t1p (lb/ft2
) I 3.25 I 7.85 1
Plot the results of these tests using suitable dimensionless
parameters. With the aid of a standard curve fitting program
determine a general equation for t1p and use this equation to
predict t1p for water flowing through an abrupt expansion with
an area ratio A 1IA2 = 0.35 at a velocity VI = 3.75 ftl s.
/
PI
o
Vi ........
IArea = Al
IArea = A2
tJf= FL-
2
AJ
== L1. A;L::: L1. f::' FL-'1 r"2. V, -= L7-1
p;.~m iJ,e p" fJ1e~l'(tn) s- 3 = 2. pi ffrmJ '("e~tI/¥'et1.
B~ /,,~pec.tlbJl /;;r ~ (t:£;n.fail1lhg fl f) .'
AE _ Fi..-2.
7r.=I Iv;:t (FL-IfTl.)(Lr-,)l-
ChecK
. MLT:USJI1!j
.-
~ - M L-/ T-2.
- M~L~TO
I~~ (1-11..-3)(L r-J) 1-
{;y 7T2- (tlllftli4lrlj A-, tiff" Az
~;
I r, J.
) :
tI~/;'j -the daia. 1JI/el11 ,+ ~/I()JI/.5 tha.i
LlPR~2. o.ofp7 0,",2 0.212 o.2J~ 0.2.5'4-
A,/Al. O.ORO O.2()o ().lq~ 6.~lb O.4-F~
( r!41/ i)
. ~I<.

0.3 . I I i
0.25 ·····--i------~----I --f-.- ·
.,- 0.2. I .__ / ____~I-.- ...
i:J... 0.15 . ·7------------------------------- --
0.1 IV ,-----
0.05 I--I-------~------- ------------L------------
I '
o-f------!------!,f----+,----+-----I,
o
The (!urlle
7'V1J11 7)r~
0.1 0.2 0.3 0.4 0.5
Pi2
dl't:ltvn 1"'" 111 e ~ ylt ph
.e~IiIt.I:I"1f
~ - / J0 /,4/ )1. +
fl-j ~ - -. {:4z. I. 0 7 ( 4'z. ) - 0, 0 103
7- 31

7.2.5
7.2;- A liquid flows with a velocity V through -
a hole in the side of a large tank. Assume that
V (m/s) 3.13 4.43 5.42 6.25 7.00
V = f(h, g, p, 0")
where h is the depth of fluid above the hole, g is
the acceleration of gravity, p the fluid density,
and 0" the surface tension. The following data
were obtained by changing h and measuring V,
with a fluid having a density = 103 kg/m3 and
surface tension = 0.074 N/m.
h (m) 0.50 1.00 1.50 2.00 2.50
Plot these data by using appropriate dimension-
less variables. Could any of the original variables
have been omitted?
V= LT-' -A =L ;.; LT-2. t= FL-"'r'Z- 0":: /="L- 1
Fr~m 1/1e pt' the()/"em I S- 3 =2 p" .feJl'/'J'1S f"f'ifllY'fd.
B'1 iYJS;eCt./~H lOr 7T; (~/,-t(Jlnl.l1j V):
.Y- _ /.. r-I
- 1. 0 TO
77f = '{if {t-T -~ 'I,. (L) ~
/=r;r TT2. (eon /rJ;'lInfj t IIlfd rr) .' 1-
IT: ::- /-3-1.2. == (Ft..- If
r2.)(/.T-'1.)(L) ..:. rOLo TO
:z. (/ r-L-/
Check tAStnj
v
V3..A..
;::;,,. fit€. data.. ~J~en:
f#*z.lrr 3."3 J)(.104-
Vl'fih /. If I
~
I·
__ .-••j
-: -j
v -
IML-3j(LT-') (L J_2. == /l1"LD TO
M 7-2.
13.3 )(10 t' 2.'1.3 X10 If S3.0 x10;
I. '1-1 /. 'f I I. ifl
. . :~~=J.:
J'2. 'I xIf) 'I'
/.Cf/
-
rjT
I.P~--.. +-~~~~~~~~~-+~~~~~~~~~~
The 9rr~ph 411d +Able shDw- thaI: VIVjh I{J l~clepfndfHi elf fJ#J.)./rr. Thus"
-the. vtlnil6/es ~ anti rr ~,,/d hI/lie hfen ~mdft-c/.

7 2~ 1
7. 2 ~ The time, t, it takes to pour a certain volume of liq-
uid from a cylindrical container depends on several factors,
including the viscosity of the liquid. (See Video V1.1.) As-
sume that for very viscous liquids the time it takes to pour
out 2/3 of the initial volume depends on the initial liquid
depth, e, the cylinder diameter, D, the liquid viscosity, J1.,
and the liquid specific weight, )'. The data shown in the fol-
lowing table were obtained in the laboratory. For these tests
e= 45 mm, D = 67 mm, and), = 9.60 kN/m3
. (a) Perform
a dimensional analysis and based on the data given, deter-
mine if variables used for this problem appear to be correct.
Explain how you arrived at your answer. (b) If possible, de-
termine an equation relating the pouring time and viscosity
for the cylinder and liquids used in these tests. If it is not
possible, indicate what additional information is needed.
I I 17 39 61 107
t(s) IS 23 53 83 145
i = ((1) ~ jA) ~)
J..=L ])=L ?-= FL
w2
r
FYtJrn -the pt' meoYern
80/ Jh~Fec..--/-tt:;"/1) -/'0;- 7J;
77: = t- tr [)
5-3 == 2 p/ ffrms Y'efftl.,yed.
( CCJI1.J-t/;l1ln!J 1:)
/ )A--
(T)(FL- 3)(L) . pOLoT D
(!=L-~ r)
Check {,/SII1,j M L T
+;r-D
~
~'I~1-e1l1 :
=- (T)(ML-2. r-2)(L) _ Me. L0 TO
(M L-' /-1)
Wh/ch
.
C) bv /o{.{ 5 ~ dlmeJ1 j/lJnless Th"s)Is .
t-tf D cf (; )- (/ )
)<-
.1-- I.j.S-/TIl/iH
(!()IISI-oll-t) .For -th(! dt1.,~ == o. Is,72 ( a.J/v(n 75- '7/fHhf1
h~m Eg. (/)
i:lr.D
?=
7- 33
e£)115-kl11+ IT- 101/005
the dt'l.:ta. 9 I v~ Y1 :

7.2Co J
?i77 ? 70 871f 375 g72
Since ~ J~ -e.~sen1:.IQ.IJ.!J CtPns1::ant ~lIey 1he. Y'an~e of-
1ne eJt-peYJin en-ba. J da.,i-a.. the VtlYla6/es 1/6f4 ~Y' the pYtJ/;/em
a.. p peay b be a6)Yrec.i:.
Lb) The So Tha:i
i: = 1.31. h
WI fiJ -I:- I~ seuMds f.<Jhen /-- JJ t'n tJl1''f...J tJ{ lV,s//Wt~
NtJ-I:.~ 1ha.t f)US r~s.f-Yl~~d ~zud/~JI /j 0'111 1/a.I/d
.for ~;0:::~. (,7t,; [) = ~ 7/111-*1) and 6-~ 9:/pt}4eN//fn 3 Uil n
:L)B t>.f- 1J1e J/JI'+U!J' / I/o iwme be,n.J pour@d.
7-3'-f

7,.;2.7 I
Q (m3
/s) Ap (N/m2
)
3.6 x 10-6
1.1 X 104
4.9 X 10-6
1.5 X 104
6.3 X 10-6
1.9 X 104
7.9 X 10-6
2.4 X 104
9.8 X 10-6
3.0 X 104
7.27 The pressure drop per unit length, Apr,
for the flow of blood through a horizontal small
diameter tube is a function of the volume rate of
flow, Q, the diameter, D, and the blood viscosity,
jl. For a series of tests in which d = 2 mm, and
jl = 0.004 N's/m2
, the following data were ob-
tained, where the Ap listed was measured over,
the length, e= 300 mm.
Perform a dimensional analysis for this problem,
and make use of the data given to determine a
general relationship between Apr and Q (one that
is valid for other values of D, ~ and 11).
/' =- FL.-2.r
Fr()tn the ft' fhe"Yftn J /.f -3 = / pi tfym reslilyed.
81f In;pec.tlon ~ ~ LlAo D' _ (Fr3) (L) Y
I ~ cP (PL-'Z r) ( L3 7-')
MLT:
.Ll~ Dif
~Q
S/nce -there IS
wheve
(ML-1T-Z)(L)'f ::.
(M'- -lr-I)(Lsr-')
aJ1 d fnere ft,re
L1i1 Dlf
/0
U&ln,9 the. da.& IH
'fo." ~, 7 40. / itO. If 'fo.7
.' OJ<.
-~ .6p
=/,33 x If) Q

"7.28 As shown in Fig. 2.26, Fig. P7.28, and Video '2.7,
a rectangular barge floats in a stable configuration provided the
distance between the center of gravity, CG, of the object (boat
and load) and the center of buoyancy, C, is less than a certain
amount, H. If this distance is greater than H the boat will tip
over. Assume H is a function of the boat's width, b, length, e,
and draft, h. (a) Put this relationship into dimensionless form.
(b) The results of a set of experiments with a model barge with
a width of 1.0 m is shown in the table. Plot this data in di-
mensionless form and determine a power-law equation relating
the dimensionless parameters.
ft= i(b;J-)~)
e,m h,m
2.0 0.10
4.0 0.10
2.0 0.20
4.0 0.20
2.0 0.35
4.0 0.35
iii FIGURE P7.28
H,m
0.833
0.833
0.417
0.417
0.238
0.238
(r"l?1 the fL' 1h eoyrm) '-/-/ =~ p,' krl71s rP~l4;ye,f. 91;1
/ I? .$pec.f,o n "
1=~(i)4)
A/I ~.f. 1h~ p/ krms tjye obv/ousl!:! dJ mens/on/esS.
(/:,) /="0". the dL.J:.a fivel1 1-4bltl"ktl va/U(ls for H-/b) l1/bJ
~t1~ .{Ib
ay~ ShbtJIlt btlow. I
h/b
0.10
0.10
0.20
0.20
0.35
0.35
H/b fib
0.833 2.0
0.833 4.0
0.417 2.0
0.417 4.0
0.238 2.0
0.238 4.0
All J~sf"cA·,()'1I
"~t iI-ep~nd
Is obl-ttll:'~,(
,----------------~-------- -
I
I
0.8 I
..c 0.6
--I 0.4
0.2
o
0.00
"-
H/b = 0.0833 (h/br1OO
" --
--
0.10 0.20 0.30 0.40
i h/b
~------------------------
61- 1He.H d"i-A. reI/eli 15 1H1l.-t H-/b d"e.s
fJ~ "l./b; (.f',,) 111-t "Sl1l?1f Ua luE' Df. H-/b
ICY' cI,'Flerf'''' t (/a {wtJ$ IJ I- .l. Ii, .Thus,
41'1 d 4{)t1f t;,~ plDt 01 ihe da.ttl.) lAS U1~
.e$u,4,. .f/D YJ
0... pllwev- law-
1+ ('-Pc.)-I."0
h =0.0933 1:

7.2,Q A fluid flows through the horizontal curved pipe of
Fig. P7."2.Q with a velocity V. The pressure drop, !1p, between
the entrance and the exit to the bend is thought to be a function
of the velocity, bend radius, R, pipe diameter, D, and fluid den-
sity, p. The data shown in the following table were obtained in
the laboratory. For these tests p = 2.0 slugs/ft3
, R = 0.5 ft,
and D = 0.1 ft. Perform a dimensional analysis and based on
the data given, determine if the variables used for this problem
appear to be correct. Explain how you arrived at your answer.
:;£;::;ft2 ) I,~:~ I ~:~ ! !:~ I ~:;
v - >L---_I--.
- ~-,,--""'---
• FIG U REP 71. 2 ~
Prt9m 1Jte pi thepl'em) 5 - 3 = 2. I" 1::('1"/71 S ret"'/r(#~;
130/ ItJ$;e~l1) .(;;y -u; (C4'nt:III#lny tip) :
..6p _ (Fl- Z )
1Tj = ,,4 V 2. (~L-IfT1.) (LT-) Z
C,heck.. USII1? NI..T s'fs/-em .
~ f' _ (M'- - I T-I. )
(1'11..-') (I..T-tj Z
" ~/,
Hr 1Tz. (~n£-~iH,il7 R PI1P /) )
D
1T;..=R
which iJ ~bv;'o{,/f;,/~ diM fI1SIOII/P.5,S_ ThtlS)
/jQ - ~ (-#);? j/ 2. - 'f' r-
D _ (). /.11:: _ /
--s
;;t. (j) tv J'"'h-1 ~ ~ CPH St:qH i
~1?.s-l:-t1l1i . ~«/eve~ #1" the
(I)
C,rls&:tn# J. Thu51
/0 /J"WJ -IrzJ
C4?1'1~ brJt i / 1:';- f,:;1/()f().S 4 pi·, -the
'/he jJYD/,ltm tJlre n~i &f'f'ec.-F·
I
7-37

7. 30 I
1.3() The water flowrate, Q, in an open rectangular channel
can be measured by placing a plate across the channel as shown
in Fig. P7.30. This type of a device is called a weir. The height
of the water, H, above the weir crest is referred to as the head
and can be used to determine the flowrate through the channel.
Assume that Q is a function of the head, H, the channel width,
b, and the acceleration of gravity, g. Determine a suitable set of
dimensionless variables for this problem.
Q :: +(H; b) ~)
h:i L d = i 7-
2
I- b
V
Weir plate
p/ the~yefYll '1- 2 '::: 2. p/ +(ftnj ri'~UJypJ.
In 'spe d-/~/f .J:,y- 10 (t...~ntot'J11 n, 4'):
4J :=. L3 T-1
HSIzJIJz. (L )5""/z. (L T-z) 1)2-
712.. (CAIt HI#/l19 /;)
tr = -A;z. f+
7-38
-I
T
H
l

7. 3/ I
J.51 From theoretical considerations it is known that for
the weir described in Problem 7.30 the ftowrate, Q, must be
directly proportional to the channel width, b. In some laboratory
tests it was determined that if b = 3 ft and H = 4 in., then
Q = 1.96 ft3/ s. Based on these limited data, determine a general
equation for the ftowrate over this type of weir.
7.30)
(j)
-
-- c (J;)H5"/z..J'Iz.
where C J:.s tt CtPn sftvtt f. Th (,lSI -ft,y he dd4. 9'Ilf)1
/. 91, if3 .
c --
1he Jen&Y'4 / egIia:t:zPI1 J.5
({J == 6. 5"'1t b Y&ft3 ]
(/)

7.32 l
7.31 SAE 30 oil at 60 OF is pumped through a 3-ft-diame-
ter pipeline at a rate of 6400 gal/min. A model of this pipeline
is to be designed using a 3-in.-diameter pipe and water at 60 OF
as the working fluid. To maintain Reynolds number similarity
between these two systems, what fluid velocity will be required
in the model?
Pol" Re~I'J~/ds num btl' s/m;/tlr/f!:J;I
VhH D.-m V D:::
Or
and
+hen
Thus/
~-
v= r(sft) z.
ff-()m £r.lJ )
- 2.02 fts
I. = (1.2/ X- /0-
5
{t; (3 ftl (2.02 £t)
m1 ( If. 5"x /0 - 3 if.)( ?z. ft) oS
7-t.io
(I)
I '-I. 3 I-t3
-S
-z.
- ~.52 ;C/o

7·33 I
7.3 3 Glycerin at 20°C flows with a velocity of 4 m/s
through a 30-mm-diameter tube. A model of this system is to
be developed using standard air as the model fluid. The air ve-
locity is to be 2 m/s. What tube diameter is required for the
model if dynamic similarity is to be maintained between model
and prototype?
FOr d'lntlliJ/c.. simi/tTr/fy/ i'he Reyno/cis numblY' rnu.st. be
the S time (;,r mode I aJ.1d prototype.. Th usJ
v~ Dtm Vb
-
~ v
So thai
DhY - V- ~ D
(I, '16 XI()-!J'"-;. 2. ) (tf f1)
(O,/)3~/1f1)- -
V ~ ( t. /'I X /0" 3!l!1.2.) (z ~-)
.s s
.. 3
- CJ. 73IP ~/D I"WJ .. O. 73 fe, /'WI I'm-
7-Lf!

7.3Lf I
7.3 if The drag characteristics of a torpedo are to be studied
in a water tunnel using a 1:5 scale model. The tunnel operates
with freshwater at 20 °e, whereas the prototype torpedo is to
be used in seawater at 15.6 0c. To correctly simulate the be-
havior of the prototype moving with a velocity of 30 mis, what
velocity is required in the water tunnel?
FOr d'lnQ"',"6 s/m; /tlr/ ty the Re'l,,~/tlJ number must be fhe.
I
SeNne hI'" m~r/e/ ClI1c1 prtJioiype. Thus J
Vmr D":} _ 'i!l
~ 1/
~=-V411D V
/)?1 V ~
5t'nce) ~ (Warey @ ~()oc )::: /. ()Olf ;(./o-ID 11)12.6 (7db/e B. 2))
-V (SeplVaier @ 15:boC):: /./7~/()""6-111~ (r;6/e /.~)J lind
.D/])/M =5 I;t follows thai
( _, ,.",,2.)
V::: /. 00 II- x. I/) ""5
M (/. 17 X IIJ -6 '; 2. )
/2 Cf

7.35 I
7.J5 The design of a river model is to be based on Froude
number similarity. and a river depth of 3 m is to correspond
to a model depth of 50 mm. Under these conditions what is
the prototype velocity corresponding to a model velocity of
1.2 m/s?
Pi:;y. I=;."ude 17/,(1n j,pr sI'm ~ J~Y'if!J )
V.;m V
=
V!/WI ctw. V#-tJ..
Wheve d . the fflAIC; depth . 7h (.{ S JI~
v: I/f{ ~
flirt d4H
Cll1d w,'1h ! ::j'WI
V= Vt ~ =~(. ~)0. O.5() Mt • Z 5
7-'13
= 1.30~

7.3(0 I
7.3 ~ For a certain fluid flow problem it is known that both
the Froude number and the Weber number are important di-
mensionless parameters. If the problem is to be studied by using
a 1: 15 scale model, detennine the required surface tension scale
if the density scale is equal to 1. The model and prototype op-
erate in the same gravitational field.
t=br dlfnqm/e- S;lru'/f/I";'+!J)
Vim ::"
V
V?rm )., VjJ-
ay,d
fm, V'; )~
=
f- V'-J·
0;", cr
To saflsly F;.t:JlJde HumhtY' .s/~/'Jlly/-J~ ( w~Th
V~ _
f¥--V
a=J~ ))
et..,d inere.-h,re !Dr Webev nlAmher .5 '/ mJ'lt:? Y/-f!J
"~(-r)f=~(;r:: ~ (~)~ &
(J V f
a;,
-()
Thus) w'rth JIWI /1. ~ I/I~ and ~ /f= I J

7. 37 I
7.37 The fluid dynamic characteristics of an airplane flying
at 240 mph at 10,000 ftare to be investigated with the aid of a
1:20 scale model. If the model tests are to be performed in a
wind tunnel using standard air, what is tht" required air velocity
in the wind tunnel? Is this a reaHstic vefocity?
For d~J14m"~ sim;14;i-lJ,I -the. ReYfJ()/d5 nUn? hiY' rn us.f ~ the.
SaMe. #>Y' model tlnd pY'~ioi'J.pe. ThUS)
(/)
jJJ1Cf)
3. 5'3Lf.x JO-
7
~ -3 I
(TaMe C.I)/-~ . /= /, tsb x/0 s ut..s
ftl. )
h J
-7
~ -3A =3.711- x JD Ib.s I = 2, 3r~)f) slui' (~kJe 1.7)"" ft~ )
1m +i'
'h,1I< ),,/1,," =1..0 J ,+ follows /r()m liZ·O) -thai
- 3750 mph
No I If / s not 4 rea//stti, ve/()c/f::; - muc.h fDa h/811 .
7-'1s

7.38 If an airplane travels at a speed of 1120
km/hr at an altitude of 15 km, what is the re-
quired speed at an altitude of~ km to satisfy Mach
number similarity? Assume the air properties cor-
respond to those for the U.S. standard atmo-
sphere.
( V ) _ (V) (I)
C J5"~'»1 - C 8 ~~
Th e ~fRfcl ~ f SoufJd ~lJn be CA/cedafe'" ~()m 1he Rgtltl:'='!J#
~ :: V-'~T (13 ~, /' Zo)
qHP{ -ft,r til y.) I<.:j, ~()) R:: z 86, '1 J/"*;, I<.
It-/: /~~"'" olh I-ude I
T= - s" ~o °C + 273.16 - J./&',7 k (Table (,2)
aHti (4'1. 8 ~~
T= - 3b.9,/ t:JC f ,;J.73,)~ - 23',21<. (Iab/~ (l)
Thus) a, t. I~'*1Wf ojI-,I",dt!.
C =l/-r;-.11-0-)(-2.-g~-.ri-.d---)-(:;.-/,-,7-k-) =
/;-6."", if -Ie#-'/<
Ci11d () f: ~ .h. I1YI
r-------------------------
C(j ~ =...1(/'tfo) (21,.1~J' ) (13&,21<) ::
()~.,., J 1Il;./<
308 ~s
From £1' (J)
~.h~ =
~
(
308-
= ~ ) 1120 {M( )
.2.'15"~ (I h...
5
1170 ~hr

7.t;o I
7.40 The lift and drag developed on a hydro-
foil are to be detennined through wind tunnel
tests using standard air. If full scale tests are to
be run, what is the required wind tunnel velocity
corresponding to a hydrofoil velocity in seawater
of 15 mph? Assume Reynolds number similarity
is required.
For ?e'fn()/dJ number s;m "Jllr/f~ J
Vh)1 .i~ =- rt
~ 7/
IAJhere J. I~ Some cnoY'pcten'Sf,c, IeYJJth of the hfjdf'Dfoi /.
Thus)
a Ilt/
~= -u"
m1 V
tv /"ht ..,e/11tn =I
-z4,
7:l Y
1. V
),.",
(.,cull scale kst)
_ fl,5"7 x /0-If !j'1..)
- (/,z' x 10-5" .ft'&.)
.s
7-"17
(/5 mph)

7.'1/
Ca)
7.41 A 1/50 scale model is to be used in a towing tank
to study the water motion near the bottom of a shallow chan-
nel as a large barge passes over. (See Vidl'O V7.7.) Assume
that the model is operated in accordance with the Froude
number criteria for dynamic similitude. The prototype barge
moves at a typical speed of 15 knots. (a) At what speed (in
ftls) should the model be towed? (b) Near the bo.ttom of the
model channel a small particle is found to move 0.15 ft in
one second so that the fluid velocity at that point is ap-
proximately 0.15 ftls. Determine the velocity at the corre-
sponding point in the prototype channel.
(1) :51~ C-e. f-rom t;;~. (I)
V~O ~V"" - V1~-1 =-V
S0 ~a.-i:-
V= '/sO (0.15 ~) = /. O~ V

7. ¥2 J
7.4'Z. AI: 40 scale model of a ship is to be
tested in a towing tank. Determine the required
kinematic viscosity of the model fluid so that both
the Reynolds number and the Froude number are
the same for model and prototype. Assume the
prototype fluid to be seawater at 60 oF. Could any
of the liquids with viscosities given in Fig. B.2 in
Appendix B be used as the model fluid?
As dISCIISS(?c/ /n Sec./;/()11 7. e.3 if) rnQJ~ talh both ReynoJd.J
nurn ber- 411d Fr""r/e number SlInllq n'7-;;
-V ].
; : (AJ.) ~
3
( -!.. ) --;: ( -" tWI z..)Jro /, /7 x/I) s
- if /Y11 ~
If. 6 J. .>G JD oS
!VI). The VIJ lues tJf y /c;- the
aYe a/I much /t1YfO'" thlll1 Ine.
?-'fr
I,Z't/lds
re~ lJ t'rer)
tfIt/t'n /It Fi9. B, 2
va/ue.

7. '-13 I
1.Lf.3 A solid block in the shape of a cube rests partially
submerged on the bottom of a river as shown in Fig. P7.43. The
drag, q]j, on the block depends on the river depth, d, the block
dimension, h, the stream velocity, V, the fluid density, p, and
the acceleration of gravity, g. (a) Perform a dimensional anal-
ysis for this problem. (b) The drag is to be determined from a
model study using a length scale of 1/5. What model velocity
should be used to predict the drag on the prototype located in
a river with a velocity of 9 ft/s? Water is to be used for the
model fluid. Determine the expected prototype drag in terms of
the model drag.
(a..) r1J =.f. (d.) ~) V) fJ J )
i) == f d= L -l:: L V= )..7-
1
1= t=L-lfr'1-
/=1""111 the fi -fh-e~Y.flY1) ~-3 =3 p/ 1-~"I?1~ r~1"I;p~, 1/;,1(
().. d"n·.eI1$/~111J1 an4/151J 'lie/tis
#h.= 1(~) ~)
(b) For- ~;m'l lari+!1 be f-w~ell tn "de I arrti prDfIJ -t:!:/pe
d"",_ d 4", fA VI'M .JL-- -Y1,Ak +. ::-
WItH d"" V8~
Thus dMA. -/,."," - (
)
-- X -5d
and
V-FY77V
Yfr9~ Jf. 02 -r:v: - =1IfJ-
The fr'~dl~-I-JDn elua.-t1fPn J".$
Jj = ~m1
fV1. hl !'WI V,,; h~
7- so

7. /1.11- I
rtit)
7.44 The drag on a 2-m-diameter satellite
dish due to an 80 km/hr wind is to be determined
through a wind tunnel test using a geometrically
similar O.4-m-diameter model dish. Assume stan-
dard air for both model and prototype. (a) At
what air speed should the model test be run? (b)
With all similarity conditions satisfied, the mea-
sured drag on the model was determined to be
170 N. What is the predicted drag on the pro-
totype dish?
I=rtJhI E~. 7.19) Re'lJ1fJ/ris number slm;14r1Tfj
~ lJ.... - VD
-v....... -z/
. Y'eglJiYfl'i. Thus,IS
whey!. D I.J the dt'sh clJtlrneteY". It ~//"WJ 1f14t
~:~.12. V1m 7/ ~
and w/n, ~ /v = /
~ =( 2. /WI )/~~ ~-m) =".,., ~. J.jIYYI ( ~ h Y'
(g~ ~~ )2.
hI"
17D tV
(tVote Th_+ I!) :.,/j1Yr! Ii-, -IJ,ti.s pJ'tJb/eWl) 5/~(e I....fJm The. CJJtI(ltf,,,'yj
of Re'lI'1()Jds nwm My S/ffll·/tll',.-ffj) v'llv,;= D; ID~ ThiS IS nrt
rrue 111 ,ln~ra /. )
7-S/

7. '15" I
7.45 The pressure drop between the entrance
and exit of a 150-mm-diameter 90° elbow,
through which ethyl alcohol at 20°C is flowing,
is to be determined with a geometrically similar
model. The velocity of the alcohol is 5 m/s. The
model fluid is to be water at 20°C, and the model
velocity is limited to 10 m/s. (8) What is the re-
quired diameter of the model elbow to maintain
dynamic similarity? (b) A measured pressure
drop of2.0 kPa in the model will correspond to
what prototype value?
Foy //()W In a. c/(Jsed C6ndtJ/f;
De,PtfldfHt pi I-e,m = f (-i', ; J 7!) (Eg. 7. 10)
T-;r 1J,/,J fJIlY/-lcu/lJr Py()b/fn? the deppndent vAYJahJe. I~ The
pJl'~.sSfA~ c!r'~f) Jj1>J So -!hilt
Deoey, cleft t pi f~rWJ .: 1J.fJ
r IV2.
,4/so, the. ~h(JY'(Jcifn~f;c, lel1f/h hV flaw through a. q{)D e/b()w IS the
c/lame-H. v) 0 J So 1ha-/;
lJ.f _ J. (£ ,oVD)
IV:z. - r D) I~
d'lMIIIlJIc.. :!J/~/"jarl+!1)
;'11H ~ j)~ = &
~~ ~
VAH D"" =: 'i.P
v
:: If 9. i 1m IW1
( Con Ii )
(/;/;/e B. 2) 4Hd
I
(lib),- /. ~)/ So thAt

(b) W,''/H 1ne same Reynolds number fer /nodel and pr~iol.fjpe.J
aJ1cJ tv ,'tH jeomefr.h 'sl m;)4 y;-I:J (tv hIC ~ Impllec, ThAt
fAM /Dtm = t / D) -t11e.,
A,p1)M ~ ~
(.) 2. ..0 y'"1/1')' VMt {
7-53

7¥6 I
7.4 ~ For a certain model study involving a
1:5'scale model it is known that Froude number
similarity must be maintained. The possibility of
cavitation is also to be investigated, and it is as-
sumed that the cavitation number must be the
same for model and prototype. The prototype
fluid is water at 30°C, and the model fluid is water
at 70 0c. If the prototype operates at an ambient
pressure of 101 kPa (abs), what is the required
ambient pressure for the model system?
for naurie. number ~/m;J(JI'I+'J)
V-*, V
-:
VJfW "i~ ViT
$0 -hrAt (WIt;.. j:'Mt)
V; = V!;
I="tPr C(l{//I-~f/~1I n()mbpy slmtlay,fr:J )
(-p". - ?v- ttf'l = (1-;. - p~ )
-t(hI V""l. if V4.
.f.o J/ow-s ~t 2-
( & V.,.. (AJ lJ )
1'.".- tv-)Itt = ~ VJ. r.,. -, v
It
m,k/J1~ use ~f '""1. ll)
(-fr - P... L... = 7- It- (tj.-fJ-)
afJfl
SD,2 -A Pet (a/'o5)
7-51(-
OJ
( Z )

7.0/-7
7.47 As shown in Fig. P7.47, a thin, flat plate containing a
series of holes is to be placed in a pipe to filter out any parti-
cles in the liquid flowing through the pipe. There is some con-
cern about the large pressure drop that may develop across the
plate, and it is proposed to study this problem with a geomet-
rically similar model. The following data apply.
(a) Assuming that the pressure drop, Ap, depends on the vari-
ables listed above, use dimensional analysis to develop a suit-
able set of dimensionless parameters for this problem.
(b) Determine values for the model indicated in the list above
with a question mark. What will be the pressure drop scale,
APm/Ap?
Prototype
d-hole diameter = 1.0 mm
D-pipe diameter = 50 mm
,u-viscosity = 0.002 N . s/m2
p-density = 1000 kglm3
V-velocity = 0.1 m1s to 2 m1s
•
Model
d=?
D = 10mm
,u = 0.002 N . s/m2
p = 1000 kg/m3
V=?
• FIGURE P7.47
fa) Llp=/{d;D)t;f;V)
J.p=- p L-2. do:: L D=- L f<:: FC2.T t= FL-'tT '2. V= LT-
I
Fr~rJ1 the. pi theoJIPm, 6,- 3= 3 pi +--et-rnS Y'e~UI rrd; tI ncJ a..
dlinen.5li)l14/ ~1')41'f.si.s ';jltld.s
~ = ~ (.!L fX!»fJV:z. r D) ~
(b) Foy ~/~,I~Y;+!J)
cl"... cI
D~:: 15
w in 1n~ fAa,./:a
cJ : 0= d
".,... 0
O. lOO""'hH
~ }J0 wS ThtA.i.
:: ~- V
D.50Dlf fr~ ID.O~

7. 'IS ,
7.4 8 At a large fish hatchery the fish are reared in open, ation' of gravity, and Q is the discharge through the tank.
water-filled tanks. Each tank is approximately square in (a) Determine a suitable set of dimensionless parameters for
shape with curved comers, and the walls are smooth. To cre- this problem and the prediction equation for the velocity. If
ate motion in the tanks, water is supplied through a pipe at water is to be used for the model, can all of the similarity
the edge of the tank. The water is drained from the tank requirements be satisfied? Explain and support your answer
through an opening at the center. (See Video V7.3.) A model with the necessary calculations. (b) If the flowrate into the
with a length scale of I: 13 is to be used to determine the full-sized tank is 250 gpm, determine the required value for
velocity, V, at various locations within the tank. Assume that the model discharge assuming Froude number similarity.
V = f (e, ej
, p, j.L, g, Q) where eis some characteristic length What model depth will correspond to a depth of 32 in. in
such as the tank width, ej represents a series of other perti- the full-sized tank?
nent lengths, such as inlet pipe diameter, fluid depth, etc., p
is the fluid density, J.L is the fluid viscosity, g is the acceler-
Fr()h7 'f), e.. pi thf'ore", I 7 - 3 ~ If
dl/neI151;P/14/ anll/';Jsi..J fj/~/d.s
V.12. _ (1.1.: Q z.
q - ~ T )}.:r:J )
Thus I 1h~ ~irn;Jtlf,'f!1 r-egulrernellfs are
)/.'/YYI :. ~. (J)'ht2.::. CD 2.
J-m. " 1-;d-m J.:>J
pyedt C.ht)J, etu.~/;iC) H I.J
V'n2- V; oj 2.
;::. _ 111' 1M
Clhd
Q - QM-i 0
J:r()(n -fhe Jllst sirn//4ra-~ re2t1/remetit 0 017;"
(j)nn == -e ~ )~ ~
Q ft??1)1M1 I- J.
Ir~pt 1he cSec.,r1JJ1d SI'm',/4rl '-J-!1 Y-l''jtl l remeJ1.i With
0"" =(b Fz.Cp .J-
5//?ce --these iwo yeg ul'remett b aY~ /11 et)J1rJ/~ t I't-
IoJ/~ws /n(i= 1he s;'m;/4f,'-!:J V-e~tllremell+S CtU1J1ot
b-e 5~'= 15he'd. 1/()
(COI11: )
7-510

Ge(!)me t Y"J C. SIin /I(jrl f!t r-ez fA I YeJ -t;,f).-t
.1""'" :: !:.i.I.~ ).
() I'" j.'#H _ ,R.~ _.-L
r j.. J3
SO -fn~1: a/I /eY1;1hs sca.Je as 1Jte lel1f th S~tJ.Je . Thus)
(defth )17J(}del ::: ( IF)(depTh~y.-b,-t!Jpe
= (/~) (:12 in.) = 2I ¥-b In.
7- 5 7

f--d----!
7.49 The pressure rise, Ap, across a blast wave, as shown
in Fig. P7.49 and Video V11.5, is assumed to be a function of
the amount of energy released in the explosion, E, the air den-
sity, p, the speed of sound, c, and the distance from the blast,
d. (a) Put this relationship in dimensionless fonn. (b) Consider
two blasts: the prototype blast with energy release E and a model
blast with 11l000th the energy release (Em = 0.001 E). At what
distance from the model blast will the pressure rise be the same
as that at a distance of 1 mile from the prototype blast?
Ii FIGURE P7.49
(a) /J P= .f( E') f) c) d)
fjf == FL- 2. E';' FL f == f:L-'I T~
Fyt)/I1 the p/ theorem I S-3:: 2 PI
a d /me"s/tJI'/4 I qnal'tSls !tIe Ids
..1P (E)jJc2.. = CP /,c2.d3
(j,) iiI" ~/~aanry I
E"... = E
~h)I C~ d;, f c.."Z.d3
/4I/'fh ~::.! J c~ =- C I If
d J::: '=~ cl 3
;n, 1:
E"., IE ::: ~.OtJ I tfll1 d cI ::
3 3
0"., =(c,. ()t> I ) (I ;tHi )
d = 0. /OtJlYtZL,11M
Ju/'th 1'hi.5 ~/ml'J~rl'+~ re~/n'l"ernfl1t
p/ed/~fro'" .I2JuatliJ'i I~
.Ll -Pht1 =-
f~ c,,:
Ap
-fJC'"
-th~re hre.
LJf~= L1 P
d ~.:: ~, 100 11111.:
?-S8

7. So I
7, !;-/) The drag, qv, on a sphere located in a pipe through
which a fluid is flowing is to be determined experimentally (see
Fig. P7.50). Assume that the drag is a function of the sphere
diameter, d, the pipe diameter, D, the fluid velocity, V, and the
fluid density, p. (a) What dimensionless parameters would you
use for this problem? (b) Some experiments using water indi-
cate that for d = 0.2 in., D = 0.5 in., and V = 2 ft/s, the drag
is 1.5 X 10-3
lb. If possible, estimate the drag on a sphere
located in a 2-ft-diameter pipe through which water is flowing
with a velocity of 6 ft/s. The sphere diameter is such that ge-
ometric similarity is maintained. If it is not possible, explain
why not.
-v
----
I. Sphere· d
G~
r
1
Wf)h1 the fi fhfl)Yetn) 5"-~: 2 ft' ffrhJJ r"gu/ld'd, t/hd a.
dll'n el1 ~;'tJl1d / 411 ~ / '1 "j (11 ~ / d,J
,&
(j,) The s/;rn'/",./f.7 rf~"ll'emt'lI-1: /J
d'm _ d
- - Df)-*1
O. "2 I)", =
(). 5"'/n. Z -I./:
d = 0, e ~-I:. (I"e'jlllf('d dltfhleffr).
7 hus, the Pf!'dlc.tle/n
E
7-S"f

7. SI 1
(0..)
(1)
7.51 Flow patterns that develop as winds blow past a
vehicle, such as a train, are often studied in low-speed en-
vironmental (meteorological) wind tunnels. (See Video
V7.S.) Typically, the air velocities in these tunnels are in the
range of 0.1 mls to 30 mls. Consider a cross wind blowing
past a train locomotive. Assume that the local wind veloc-
ity, Y, is a function of the approaching wind velocity (at
some distance from the locomotive), U, the locomotive
length, e, height, h, and width, b, the air density, p, and the
air viscosity, J-L. (a) Establish the similarity requirements and
prediction equation for a model to be used in the wind
tunnel to study the air velocity, Y, around the locomotive.
(b) If the model is to be used for cross winds gusting to
U = 25 mis, explain why it is not practical to maintain
Reynolds number similarity for a typical length scale 1:50.
V -: .p (~ ,1.1 J,; b) ~ jJ-)
-It 1. 11 =' r-L-1.rV==LT-1
V= LT-1
J.:' L J" -= L J, == L f= FL T fV' r-
From 1he. pi. tneoV'em 7 - 3 .:: '+ fl' -i-noms Y'egu'-y~tI. I Qrt pC, A-
d /meYls/~yJa I aMaI~ ~J:S ' !j' eIds
~ <: 4( ~ ).t )0/)T~ us The ~jl1l" 14y,:f'j regUI Y'fm ~11.+.s aye.
;.;' _ j. h~ =.! t:'tK 11M! Tl~:= t- h l/
TIWf -;,., --It,.. h ~jG==..,..====-)4__
T~e. pred;".J-loJ.. €ZJ)..~-/'Ii>~ 1".5
V - VM
V=-1Z.
5/~ce 1he den,j~ ant< VljC"SI~ at- tHe Ci/'r f/OWIHfj 4rouI1d The
1rg/,., Cind -the (Ur In 'hH Wlj1'd tunnel kJ~u/~ b< pYac.1-,"c.4l1!t
-tn~ same (I'WJ~! )),*~p..)) /t +olloWJ +rom tn"e IIHi
5Jm/lt(Y/+; vezu/yemen:i.. CWn,cl-1 ,J --f1.u ~ey"oldj nurn},.er)
111 ~t If ::('-A ) U
/Wt -ttA4of
nu~) w/-fh
V::- 2. S /1'n /s
e::( /eJ1j1n Scale. of 1:50 C/;1(:/ w;1h
~ ::- (sO) (Z§mf/s) = ~ Z§~ ~/S
Th/~ ye'SlJlrf'd mfJael t/e/~;-f!J ,,5 rnuch hl9he.r lhal1
CaJ1 be Ct. c h ~ ; v -e a' / h '171 e !AI",id i:. I,.( 11;-1e. / Cin d
'1heve-hY(' ,"/- Is ntrt PytAc..+/~/ Iv /Yr1l1lrl';"/~ Re'tho/d..s
l1ul111JtI"" Slfnl"/arl+Y. Th< v~J&II;eel rn od(l./ v-e!oc'+Ij 1.5
+t90 h 19 h . I

7.52 I
1.5l An orifice flowmeter uses a pressure drop measure-
ment to determine the flowrate through a pipe. A particular or-
ifice flowmeter, when tested in the laboratory, yielded a pressure
drop of 8 psi for a flow of 2.9 fe /s through a 6-in. pipe. For a
geometrically similar system using the same fluid with a 24-in.
pipe, what is the required flow if similarity between the two
systems is to be maintained? What is the corresponding pressure
drop?
A-5sllm~ cp = +(4p) d) D) f;;')
Wh rre.' Q'" /iDlUya,te.:' L~r-') LJ p"'" pressure dr~?::: t=C2
)
d Ao ~rl Ii(~ c/t'lfIhlef;,r-': L) D"", I'lpe c/,itlnetfr'= L )
!N .r/~/d df"~i-I:J:::: FL-'lr",,; I/Iul )AN +lwlO vis~sH!:J =- FL -2 T.
fr~f}1 1/Ie p/ the~re",,) ~ - 3 ::: 8 p/ -trttrlS re$uiyed I tlHa tt
d;'mt'n5/~~lIj IIl1aJ'1~/.s f1it/JJ
t.2. = t(Pi !Jl D't )
)lD /)J)<2.
d
D
p""", -
-d
.Ll f>.,... D': !-'"1 = ~P D'"t
D-w. _
--
D
),1,..,,1. ?<Z
~t=- f)t)1{-!?)/~) AI..,
),=)1.", I ! =/~
tJt =(iff) ~AI.. - ff;.fri,!si): o. 5'p~ psi
~ CJ~ ~ ~
/~ Dm, ;fA-.D
Q=(*"f~j{)(?~
=(i)4J"" =- (If )(2. f .fP) =- p. tIt

7.53 During a storm, a snow drift is formed behind some
bushes as shown in Fig. P7.53 and Video '9.4. Assume that
the height of the drift, h, is a function of the number of inches
of snow deposited by the storm, d, the height of the bush. H.
the width of the bush. b, the wind speed, V, the acceleration of
gravity, g, the air density, p. the specific weight of the snow,
'Ys' and the porosity of the bush, TJ. Note that porosity is defined
as percent open area of the bush. (a) Determine a suitable set
of dimensionless variables for this problem. (b) A storm with
30 mph winds deposits 16 in. of snow having a specific weight
of 5.0 Ib/ft3
• A half-sized scale model bush is to be used to in-
vestigate the drifting behind the bush. If the air density is the
same for the model and the storm, determine the required spe-
cific weight of the model snow, the required wind speed for the
model, and the number of inches of model snow to be deposited.
ra) ~ :: +(d) H J bJ ~
Bush
~ & . Dritt 1
.~!
• FIGURE P7.S3
3-) f) IS) 'Z )
-It -= L d= L /+~ L h=L V -' LT-
1
d= L T-2.
t == FC'f-T "2. ir".$~ FL-3 1= rDL~TO
FV'tJ/11 1h4! pi. 1'heare"",) Cf-3 =t. 1'/ .J.trnu reZ~/;e~1 tlH&./
a. dlme'hSltJnaf dna/'1siJ 'jle'J.s
~ =~(1;)"~) 9s /rJi,) '1)
4/so) Vhh V
- =
Vd,,"~ V"g.H
So 'h1J fAltr" g~= ~ t:{ut;! #1tI4 / )J.
:::1.'l-
V11f =ri7
V = V(d)"' (3(),-,..ph) := J./. ;1.~pJ,
a'ncl
d,1t" £t
Ji-1t'J'I ~
d~:(W) d ({ ) (I~ in.) :: ~. 00 ~n.
-

7. S ~ As iIlustrated in Video V7.2, models are commonly
used to study the dispersion of a gaseous pollutant from an
exhaust stack located near a building complex. Similarity re-
quirements for the pollutant source involve the following in-
dependent variables: the stack gas speed, V. the wind speed,
U, the density of the atmospheric air, p, the difference in
densities between the air and the stack gas, p - PJ' the ac-
celeration of gravity, g, the kinematic viscosity of the stack
gas, II... and the stack diameter, D. (a) Based on these vari-
ables, determine a suitable set of similarity requirements for
modeling the pollutant source. (b) For this type of model a
typical length scale might be 1:200. If the same fluids were
used in model and prototype, would the similarity require-
ments be satisfied? Explain and support your answer with
the necessary calculations.
(a.) 5/" Ce. v= L i-I EJ'; L T -I ;;= FL-1f T 2 f-f.s ::i FC'+T4
if ~ 1.. ,-2. ~:!:1'/'7 -I D= L) I f ~ //t)UlJ. ~r~/7t the pi.
(};)
-theoytm --thAi 7-3 Jf- j:>;' ffrrn; 4r~ Yi'p/,I/ypd. /+ dU':'fI1SIt)n~/
CinA/1~/.s fjlf,ld...s .:::!...) ~) 2:2.) Clnd ~ tfs a.. ~5j,;)t!
I r . L ~L. ~ j.0.J • I I' . J J.I 1_.
se t I:) r- pI ~rrn5. / rJ"'~ 'In~ 511111/111'1 r!:l retJIJll'emfl17::l toOl/I« 0::: :
~ ~ II I/",. D~ = V.D Vitti 2 =- L Z
0-$),.., _(I-A)
Z7,m Y ~M1"11 Y;:-DItIf dD ~ - I'
~r
D",.,., ::: and V£1rtt =- Ys 1ne ~eC&>I1" j/~/'/lIrl'-h:t- Z()OD
ret"':rt"'1'1'71. 10:> Un, _ -VS"" D := Zoo (see abt)ve)
- - D/mV I/s ..
ThJS re5t-t/J:: ~/7.f/I;;1-J
:fI till'!" y/f'1 re%Uiref'Yl erll:/
re~ UI remen+S (!OJ1f/pC be
&;n drf-Ion 5 , Iv'tJ.
W / 1h 1I11t-t In>m -j;, f. S ect:J;f,(
lind 1hfrej,ye. 17I~ j/hn,')lIr,'ft:J
5~.f:./jj,e4 UHt/er 1h'f .s+A.fe4

7.:rt
7.55 The drag on a small, completely sub-
merged solid body having a characteristic length
of 2.5 mm and moving with a velocity of 10 m/s
through water is to be determined with the aid
of a model. The length scale is to be 50, which
indicates that the model is to be larger than the
prototype. Investigate the possibility of using
either an unpressurized wind tunnel or a water
tunnel for this study. Determine the required ve-
t locity in both the wind and water tunnels, and
the relationship between the model drag and the
prototype drag for both systems. Would either
type of test facility be suitable for this study?
As c/ehl"I'J.J.frll~rd IJIl Pg. 7./fl fo,y .f/~w (J rlJul1d /mmfrst"d b()cJ/~,
Re'Jn~/d.s numbty s/m;/"1"1 =4 /J f".fl tllY'e" J() ihd
~j.-, _ v~
- --tiM! Y
~ = ~1 YAt'! 7/ J._
J:f fYlodel ./-esls IIYf YUh In unpr~.S'.slo,,/'t4 wlnt-1 -hlill1tlI 1kel1
~ (sfqildllJl'd (Jlr) .:: /. 'fb X. 16-f,-/WIYs I ~J1d 1J (~pttY'): /. /2 XIO-'l'fHl/s J
So 1hlrl: - l.
(/.I.,,~ X /()-!J a!!. ) (/ ) /
Vm. = (1.12;( ID-~ ~....l~ (IO ~)=2,1. I~ (+., wIHII 1,,..,/)
r.{ mode! fest-s t:(re /1'101 Irl water t.unnel WIth. -14 =~ ihel'1
VIWI:: () )(l() )(10 7) = ~. 200 T (~~ w.. fe#' funnel)
SIY1C.e ~ J~ Y'iIJ5~illlbJe J~ both CAstS J
e;1I1er 1he. Wmd +ulInel
or 111e. Wa fer .fttJlIl e/ Cou/d b~ wSed .
?v/I;. '1etPl11efl'lc' 411t{ d'll1l1lfJl~ .s-;hllil/f/-f~1 /t ~I/~U'.s thllt
dJ _ ~"."
-I V';R '- ~ ~J.Jt1If2-
~:: ..L.J:..2. 1"~
rJJhi" ~ v,..." J..ff41.
2.13
0, Lf LUI-

7.'5b The drag characteristics for a newly de-
signed automobile having a maximum character-
istic length of 20 ft are to be determined through
a model study. The characteristics at both low
speed (approximately 20 mph) and high speed (90
mph) are of interest. For a series of projected
model tests an unpressurized wind tunnel that will
accommodate a model with a maximum charac-
teristic length of 4 ft is to be used. Determine the
range of air velocities that would be required for
the wind tunnel if Reynolds number similarity is
desired. Are the velocities suitable? Explain.
,Rel(IIt>ld~ lJumb.tr s;ml'/4f11!J;
!1" VIIPI ~~ = J- V..R
)A~ r
V'/m =-7;tt V
SlfUe. the. W1I1t1 ..fulllfel 1,$ Uh/!,ftSSII}'I1ft/ the 1/1 rpr~Pfrh'e.s w//J be
aP!r(Jxirn~telt; -I1te St1I?1t. ~I'" medel 1"l1d p~/()f!lpe. TI1/1SI €'$. e/)
feciuce.; -f..o
tll1c{ #~ IRe dll~ ?llIe""
~ :: (20 It) J! = 5" V
",., ( tf -It)
Ttterefr;rf) al IoU) Sfted
/I""., :: S (ZtJ mph) =- /o~ mph
aHa af 11t9h 6,~ed
~ :: S (q~ IWIP),) : tfb() trrtp h
00 fhg-f the mode / vel!)c in, Y'~11ge /s /Oornp;' fo 'f.>7J mph
OJ
,41 i/Je 1111,/1 vel()clfzt IH the WJ~tI 1-uJ1fJel e(f)/)1pr~ss~b/)'+!J ,,,f the
Qir W()u!tI. .6ft1r,t -to be~me 411 lin'(JY'k~t titclr;r/ 14JherftlS
Cbmpressllllllf,; l.r /Jot: /n7'p~yfy,l1t -fi,y 1fte pYbfoftjpe. nlt/S, ine
ht'ql?er l/e/~cii-!1 y;oe!tl1retl ~r 1ke ff1()tie/ wf)IIld /Jot be $UJ14j,/e.
"v~.

7. 57 I
7.57 If the unpressurized wind tunnel of
Problem 7. 5~ were replaced with a tunnel in
which the air can be pressurized isothermally to
8 atm (abs), what range of air velocities would
be required to maintain Reynolds number simi-
larity for the same prototype velocities given in
Problem 7.'S{,,? For the pressurized tunnel the
maximum characteristic model length that can be
accommodated is 2 ft, whereas the maximum
characteristic prototype length remains at 20 ft.
Re'fIJ(>Ic/s nIIm};!y SI'm; /IInfy .;
11m ~ J":!. :: t-E
)A~ ~
~ ::~ & .l. V
Ib1 fnt'! ~ j..,..
/=PI'" lin Ide,,} 9t:1s ) --;:::I't€ T I til'ld .{;,y ;'s~thfrf)J41 t!.f)/npf"e5S/~H
:1=. =~11S I:ni1 t
f'
~=J:..
tl'fYJ fJ
L-£
tt7'1 - ft'l?'1
( ass tJml;'7 )J1YI -:::.; )
I) - .:f...J;.. V
~ - 1'1f'OI:4
/1 )
where. p /.s 'l,'6-mf!)~phfr'~ pr-e.sswY'<' (PY't'SStll"e at:
0{f'Y"Ilf-eS ) J and ~ Is !f?.5SUY'f t:>f ~jJf't'ss('d fli}-
tJ)""h PY'ototljpq
In the Wll1d -ftt11t1el.
ror tm=af>
~ == (.J...)(zo It) V=- /.2;- V
/11'1 f ( z .ft)
ThusJ 4t /fJl.J .sIted
~ ::. I, Z~ (2~ mph) = Z 6' /W1 ph
CH1t1 'If hlfjh s?~ed
V"",:: /. "25"" (90/J11ph) ::- 112,5"tmph
There ~rel 1)1(. r'e~(/Irett tnfJdel ve/f)cl1:J rnl1ge /~
dis mph ..fr:, //2.5"trnph.

?s~ I
7. Sg The drag characteristics of an airplane
are to be determined by model tests in a wind
tunnel operated at an absolute pressure of 1300
kPa. If the prototype is to cruise in standard air
at 385 km/hr, and the corresponding speed of the
model is not to differ by more than 20% from this
(so that compressibility effects may be ignored),
what range of length scales may be used if Rey-
nolds number similarity is to be maintained? As-
sume the viscosity of air is unaffected by pressure,
and the temperature of the air in the tunnel is
equal to the temperature of the air in which the
airplane will fly.
l1/,fhJl1ty slm;1f/f'lf!1 )
~ Y:" ,R~ ~ I-v.1
So th,,1:
,)1-i11 /-
i,," ~ ~ k 1:.
): ~~~
fir fin Icle~J "as) ? =;,e~ IJnd w/1H ~s.f.tI/Ji femppnr/uYe. J
:E.. = e,,,.sfa11 t
f
or -p _ -4-
~- 1m,
(fntl 5~ .(J) CttI1 he 411'/#e11 tI.s (tvi1H )Atttt :::)A )
);fIIf _ 1;) V
) - t.. ~
t;y the dll~ '1J;;el1
J.4f4 - f;o/~fa.) V
-T- O500 k Pa. ) ~
al1d w/fH ~ = (Ii: (). z) VI /7 .fr;/loliJ5. 7hfJi
j~ (/oJ-IGfl ) J
- -1- - (/3()().k fa. ) ( / j; (), 2)
CJ, Ot,/f 7 to t). 0'17I ,
(f)

7. SCf I
1 Stf Wind blowing past a flag causes it to "flutter in the
breeze." The frequency of this fluttering, w, is assumed to be
a function of the wind speed, V, the air density, p, the acceler-
ation of gravity, g, the length of the flag, .e, and the "area den-
sity," PA' (with dimensions of ML -2) of the flag material. It is
desired to predict the flutter frequency of a large e = 40 ft flag
in a V = 30 ft/s wind. To do this a model flag with e = 4 ft
is to be tested in a wind tunnel. (a) Determine the required area
density of the model flag material if the large flag has PA =
0.006 slugs/ft2. (b) What wind tunnel velocity is required for
testing the model? (c) If the model flag flutters at 6 Hz, predict
the frequency for the large flag.
W= 7-1 V= L7-
1
1= MC 3
!=' /...T-1. l;;'L fA':: NL-2.
j:Y'~f)1 the. ?/ 1he~rem) ~-3 = 3 f/ +erl11s re$tIlYfJ.} t1l1d a.
d/ln et1 ~/PII~/ tin,,/'1'IS ttI.eI"~
W Vj = cf (~ J )
Slhl//Ilf'I1-Y
~~
~ .J/t'n
Vtrn l
Vdr;r,)~ V8.R.
gl'm =d
!",,:: Vf V ::
W,1"h fhe simi I~r/1-!:I
.
s~.:t I ~+I pi(t) reIUJremPt11=s 7H(' pr"ecl,'ct./~i1
e!tl.Q.':t:J~M j5
Lv V-f= tu." V~
aht1
jo fn~-/;
W - Vi- J V~ ,4! r.!t:E 1(~ II,) = I. ~0 f-l)=- !~ J- ~ ~o .f-.f

I, (Po I
7. fDO River models are used to study many different
types of flow situations. (See, for example, Video V7.6.) A
certain small river has an average width and depth of 60 ft
and 4 ft, respectively, and carries water at a flowrate of 700
ft3jS. A model is to be designed based on Froude number
similarity so that the discharge scale is 11250. At what depth
and flowrate would the model operate?
fi;y P.rpud~ n 1-1111 J,~Y 5/;"'I'IlIrJf.!l
V/??1
=Vj~~~ Vi). -,
wherll 1. IJ SOn1t c,naY/lc. t-en s'1-,C 1('HJt;, ) I1I1K /1117], ! ""' ;:: j
ah to{ for- ()...
e"rr-t's/o/1dlit'!
£mt.=. (). I J 0
t
prof:" -I:!f~ dep1lt t;f If f-t the.
model dep1J1 (s
J.t'm ::: ((J.II 0) (If H) == O.If-'f{).f-t

7. ,/ I
(a.J
7. r..l As winds blow past buildings, complex flow pat-
terns can develop due to various factors such as flow sepa-
ration and interactions between adjacent buildings. (See
Video V7.4.) Assume that the local gage pressure, p, at a
particular location on a building is a function of the air den-
sity, p, the wind speed, V, some characteristic length, e, and
all other pertinent lengths, ei• needed to characterize the
geometry of the building or building complex. (a) Deter-
mine a suitable set of dimensionless parameters that can be
used to study the pressure distribution. (b) An eight-story
building that is 100 ft tall is to be modeled in a wind tun-
nel. If a length scale of 1:300 is to be used, how tall should
the model building be? (c) How will a measured pressure in
the model be related to the corresponding prototype pres-
sure? Assume the same air density in model and prototype.
Based on the assumed variables. does the model wind speed
have to be equal to the prototype wind speed? Explain.
1>= rL-·
FY~fl1 -rj,~ P"
dl n,-en SI;I'1IC./
f =FL-'fr 2. V -= L T - ( 1 :.- L Y,,' ::1
1]1eDref"YI/ s-3= 2 -p,: kY"l71s ve~",,'red, Ql1d c..
Ci 11 4. J1 S 1:S 'J J ~ I OJ
I'~Z = ~ (f: )
georne fy/c, 'S1'm,'l a rj'+j
j.~ _ 1
-ti,m ).,:
J.1)rt .. J-~'hff
T - J.i
t"f- {"IIDWS ThAt 4/1 P(Y't/~fl4t lenfth~ aY( Scaled
le~ffh Sca/(. f.".,. /j., Thus, WITh j;ttl I;. = /aoo
l'mode.1 h-eicd,t -= .ill.Ji:- :: O.3~3 +toJ '300
f'Wt =f ( 1
-P~.lf~) f~
Wd1t 1Jr( ~e1: of. '1' ~e r1 Va r/a hieJ 1hfY(!' IS no ve8~;ref1l~t
foil" f'he V~JDC.·I~ sca./t!; VlYI4jV) 5fJ fk~ rnod~ WI~e.{ -speed.
dDe~ npt hail~ -to be .e~u~ ilt... tJu prI)-/:oiypre. WU1~
opee&. No .
7- 70

7. ~2 j
7.62 A nscale model is to be used in a towing
tank to determine the drag on the hull of a ship.
The model is operated in accordance with the
Froude number criteria for dynamic similitude.
The prototype ship is designed to cruise at 18
knots. At what velocity (in mls) should the model
be towed? Under these conditions what will be
the ratio of the prototype drag to the model dra~?
Assume the water in the towing tank to have the
same properties as those for the prototype and
that shear drag is negligible.
Foy FR.oude n tim blr- s·lm~ jar; +!;j )
Ym, = V
V!hO ~"., vF
Qnp( IN,Tn ?=-im,
/ )1 ~)- a!!.
w,·th
~:Vfv-- Vi I /~ ~lIdfs (0, S/J.f.1{- ",,~t - /. s/ s
Frowde lIt1mber ~im i /4 f ,+, tlltd 9tfPllleffli SI ",dan f'j I 1hel1
ptJ :: ~I)M
~ ~loJ:fy'~J.~
J,j -the dra.9.
~=.t.
otJ~ fm,
5Jiu:e. I =/fWI and V/Vmt::" V)/-fM4 I ,·t ~JJ()/JJj thai
4:(t) (fJJ = (J) ( 5"0)
3
'" /. 2 ~X 105'
M'1
7-7/

7.64 Assume that the wall shear stress, 'Tw, created when a
fluid flows through a pipe (see Fig. P7.64a) depends on the pipe
diameter, D, the flowrate, Q, the fluid density, p, and the kine-
matic viscosity, v. Some model tests run in a laboratory using
water in a O.2-ft-diameter pipe yield the 'Tw vs. Q data shown in
Fig. 7.64b. Perform a dimensional analysis and use the model
data to predict the wall shear stress in a O.3-ft-diameter pipe
through which water flows at the rate of 1.5 fefs.
LLu- -= +(DJ cr ) f) -V)
lw--=- FL-'- D= L 4>::: L
3
,-'
FY'{)f)1 1he pi tr. eOytM) 5- 3=2.
ct c/,menS,bJ'l4 I fUt allj'SI'..r fjl 'e'd..s
;cp~~ == 1(!J)
SIin//fI r,'+~
w:>~
(al
'::: 0.7 ;,
J3 0.6 , ,
J 0.5 ----1-.-__._+._.__ -L_. __.
o 0.5 1 1.5 2
Flowrate. Q. ft/s
(bl
• FIGURE P7.64

7. ~f) 7.65" The pressure rise, f!.p, across a centrifugal pump of a
given shape (see Fig. P7.6Sa) can be expressed as 8r-------------------------~
Model data
Q
f!.p = I(D, w, p, Q)
where D is the impeller diameter, w the angular velocity of the
impeller, p the fluid density, and Q the volume rate of flow
through the pump. A model pump having a diameter of 8 in. is
tested in the laboratory using water. When operated at an an-
gular velocity of 407T rad/s the model pressure rise as a function
of Q is shown in Fig. P7.6S'b. Use this curve to predict the
pressure rise across a geometrically similar pump (prototype)
for a prototype flowrate of 6 ft3
/s. The prototype has a diameter
of 12 in. and operates at an angular velocity of 607T rad/s. The
prototype fluid is also water.
6!---------··~-------~=
Wm = 40!l" rad/s
Dm =8 in.
°O~----~----~~~~~----~0.5 1.0 1.5 2.0
• FIG U REP 7.65
Centrifugal pump
(0)
Jp = j ( D/~.) f.J a )
. -/
(",,).::: TtJ t ~ f '-- 2.- D -' L
h-If)tn t:-he PI.: 1:hetJytlYl/ S-~:
1= FL-'tTl. Q'= L3 T-J
.2 Pi t!'rrns retII/yeti) 11114 t4
t:ift~/'fsi5 &' e/tis
LIe
Itt)11>l-
Ye!VI r-emt'lt t I..J
~~ =: cp
;;;-£3 tV D3
'*" ~
a;ui jpy 1;,e dd£ ;,/iJev
tj),., =- (folT tf')( flit,):3( I. .pP) _ I. /'1 {!J
Qm (ft 3/s)
(b)
( ~7T ~d) /2./n. .5
..5
Fr()1J1 tHe 1ntlJ, (r;,. P7.bS';) .t1 /:m :: .s:S'o ?.5l.: ~y til =/. / 'i ;tJ. fitliS)
LJp ::
1-73

7.~ (P I
7. " <c, Start with the two-dimensional conti-
nuity equation and the Navier-Stokes equations
(Eqs. 7.35, 7.36, and 7.37) and verify the non-
dimensional forms of these equations (Eqs. 7.38,
7.41, and 7.42).
Jt< Jtr=o
J;( r PIj
;0 (~-t U ~ .,. 7r ~)O, - ~ 1")A (~~ 1" ~;~) (liS. ? 36)
If) (Jrr+ t< U- -r viJ!:) = - ~ -,0 II -riA- (~-r JY} (e~. 7. 37)
I _ P~ ~" d ':1 a!1 f ~ / 0,( l. ~:; 1.) ,
4s In(/ic.a/-PII 111 Sec.·ltoN 7.10 let
k """= ..!:!:.. ?i14:. :JC -p"..; 1:.
V V 1;,
.;. X fA 11=__ _1:1 ..J....L
X = ~ v.J C7"-= t-~
The /llit/,,'US 1r((i1.sk~IYM.j./~'n..s t4h be mille as /c/JClJIs:
au. _ J(Vu.~) Jx'" V Ju-i-
ax. - ~;(, /4. d x. :: --;. Jx'-tIr
1f11,{ j'/ln ;/(11" &)
J v- _ ~ J I/#- ~ .: J:: !..!!:.f!. 0 1.1" _ 1::lYv.
;2 -.). I;<*" J'j.J. Jyf. ;; - ~ o!J+
,4/so) J "2 fA. _ v d (0U -It) ~It -: Y- J ZU I-
p,(2. - -X U.,. a;.. a~ 12. ~x1tZ-
Clnc/ jim/iffr/~/
J7- V- _ 1/ J2.v-¥ 27.(A. _ 1/ 0'1.", i'.
tJx2. - F ~L p;2. - Ji J!j'loZ-
fEy 17te local acce/fy~-b~·If.J
dUo J rvu4') Jtl. = V Ju"
rJt:: Jt--f.. it ~ Jt+
til/it ~/m,'k,yo)'11
JU- _ V J ]/-J.
n - ~ ;-c~

7. ~ ~ I (~11 Z)
roY' -the. pY'e.ssure I-erm.s,)
ij::, _ ~ Pa f;J.. d.JG~ ::
J.x. - ~x. aX.
t/hd s/m;/Ilr/y)
li:.:: A li:..jJ.
o!J j. d!J*
5Ub.'Jf,i-uf,il1 eJ/ 1h~ I/tlrI()~ ftrms J
-(xpresspo/ In Hrms ill
tnt! dlln~I1.5;{)I?J-es.r vtJY'lobJe.s, Crill h< IIltlde I~t-~ the (!)Y'I'gJ;'ql
d, fleVfl1h4/ eglAa.t:I()AS (£IS . 7. 3~~ 7.3~,; ~1'1'" 7. 37) -1-0
tjie/ tI Ef~. 7· 38) 7· ~'J 1/(111{ 7. 'fo. To o/'.ftt"j the hh~ /
!orin for ES5. 7. 'fJ tll-ul 7. 'f2 dlV, de each krm
hy IvYi.

r 7.'-7 I
I. Ie.7 A viscous fluid is contained between wide, parallel
plates spaced a distance h apart as shown in Fig. P7.67. The
upper plate is fixed, and the bottom plate oscillates harmonically
with a velocity amplitude U and frequency w. The differential
equation for the velocity distribution between the plates is
au a2
u
p at = J.L ay2
where u is the velocity, t is time, and p and J.L are fluid density
and viscosity, respectively. Rewrite this equation in a suitable
nondimensional form using h, U, and was reference parameters.
Le i:
~=
O!j
02U _
~t-
u.
-VI
7-7~
(
(
I'
f
) Fixed pi ate
--....11
h
l x
• II = UCOS WI
TJ' ou*vW-
0-(;'*

7. ~8 I
7. ~8 The deflection of the cantilever beam of
Fig. P7.bi is governed by the differential equation
d2
y
EI dx2 = P(x - C)
where E is the modulus of elasticity and I is the
moment of inertia of the beam cross section. The
boundary conditions are y = 0 at x = 0 and
dy /dx = 0 at x = O. (a) Rewrite the equation
and boundary conditions in dimensionless form
using the beam length, r~ as the reference length.
(b) Based on the results of part (aYwhaiire the
similarity requirements and the prediction equa':
rion for a model to predict deflections?
( (,L J Le i. J1- =1- and ;(.14 =
;
did dt (J.b~) d)(. ~ _
d~ - - -d )(..,.. d;<.
Co( iii&{
)(.
.so
X
J. dt1~
dt,JJr (-i )
rJz!:1
= E. (!li*)!.£'" =...L rJ'!f'I--d.x, l.. J;e"* dx.l/- dx ).. dx" z.
OV'
(.b )
re~ UI veme,.:tr et Y'f.
--
X,w. _ ~
J.... - )...
7-77
-
FIGURE P7.1D8
7"ha.i
:: ~d)(,~
hec.Dme..s

7. fo'f I
7. fo9 A liquid is contained in a pipe that is
closed at one end as shown in Fig. P7."~ . Initially
the liquid is at rest, but if the end is suddenly
opened the liquid starts to move. Assume the
pressure PI remains constant. The differential
equation that describes the resulting motion of
the liquid is
auz = PI + (a2uz+ ! auz)
P at e JL ar2 r ar
where Uz is the velocity at any radial location, r,
and t is time. Rewrite this equation in dimen-
sionless form using the liquid density, p, the vis-
cosity, jl, and the pipe radius, R, as reference
parameters.
_ t ,/sed
r
'_P_l_ _ _ _ _--1-_-----. End initia!l~
rLz ~vz R V
T
1-1.---1----1,1
Le t rlf =..t ) t1-: ± ) tll1t1 ~*= ~ Whe~~ rR , ~ V
IS -?~me ~m'JI~lftt/1'J.S of the p4rtimekYS /)/-I QM~ R hal/ll1? ihe
dlmeI1SI()~ ~f t.lme) tJn~ V I~ J~mt Ct;mlJ/ntt/~~ cf the sqme
p4fllmeter.s hHtllht 1?,e d/rnfrlsl!;/LS t>f tt., tle/()c/7Tj. Let:
1:=/-,R?=. (pl.-If r"L) (L) ~ r
/" FL-2. T
alit! JA. -2 T I
V=...£- ~ I=i. d ~ L T-
fJ R. ( F/..-'f r2.) (L)
W,1h these d/~ei1S;~l1less IIlJyltd,ies .-
a~: a(v~~ ~~:- V J~" (...!..): ff)!}!:. )J~ "_ (e)~ dV;""
dt Jt-l- at W 1: ~,R. (1£2. ~t:1ft - fJ 1<3 Jtf'
~ = ~ (V'Zi-tt-) or·: v ~-I- fJ.): (~ )(!.) p~-Jt .:- A Jtfr+
ar ar-to. df- J}--i- ( i< f>~ fo Jyo$. (';€,. 0Y'""
a7tJi: _~ ~ (JVjI)k! = ~ d 21it~ /1) :: ~ JZ1Jj~
ayol. - 1',e2 ,It#- --r;;:-. Jr I'Rz ~ rj.2. (~ ;o;e.3 () r1' z.
The ()l"IflJ14J d,'.fkrfntlq/ e$tlflf:to'n ClJM n~w b~ ex.p'l'tSSfd aoS
[; ~r:f.] ~;: = :. + ~ (f;e3)J(~';:: +?~)
7-78

7. 70 I
7.70 An incompressible fluid is contained be-
tween two infinite parallel plates as illustrated in
Fig. P7.70. Under the influence of a harmonically
varying pressure gradient in the x direction, the
fluid oscillates harmonically with a frequency w.
The differential equation describing the fluid mo-
tion is
au a2
u
p - = x cos wt + f.l -
at ay2
where X is the amplitude of the pressure gradient.
Express this equation in nondimensional form us-
ing hand w as reference parameters.
h YL~u
-f- x
h
FIGURE P7.70
Let ~j(= f I ti-:. 4Ji , c/l111 ,,+::: {~ So 1h4.i '
~: ~ (J.tu '" -I-~j.._- i ~""(w) = IttU'-~.f.
dr ,; 'C""' t3t tu Jt:1c J r It
&IA. _ ~ fA fA) ().:"'") J.!1" ~ -I,tv ~ ~ (J) =- UJ ~~
;!:J - J :;-11- ~ - djf. ~ J!:1'*'
dZt-< :: t.U d ( Pu"*)lit-'Ie. UJ () Zu 1'. LL) = tU dZu-11
Jj l. ; !J f. a!:1~ J!:J. 0J~ 2.. {h -::t: b!J;' z.
The OY'J9/f1lt! d,/nY.fJ1ttal egIAlJ.tJtPH Ct(11 naw b.e -e.x.pytssed a~
[;-AtJj ~~: ~ X c:~s tt' + ~-i!] ~>.~z

7.7/ I
7.7 I A viscous fluid flows through a vertical. square chan-
nel as shown in Fig. P7.71. The velocity w can be expressed as
w = f(x, y. b, J.L. Y. V, aplaz)
where J.L is the fluid viscosity, y the fluid specific weight, V the
mean velocity, and apIaz the pressure gradient in the zdirection.
(a) Use dimensional analysis to find a suitable set of dimen-
sionless variables and parameters for this problem. (b) The dif-
ferential equation governing the fluid motion for this problem
is
ap (a2
w a2
w)-= -y+J.L -+-~
az ax2
ay
Write this equation in a suitable dimensionless form. and show
that the similarity requirements obtained from this analysis are
the same as those resulting from the dimensional analysis of
part (a).
(tL)
I - b - - I
• FIG U REP 7 . 71
w= LT-
1
H()t'Y1 1he
tll1t1lf1s/~
w-= { ( J<.) !i; JJ.;)<.1 0.1 V" if. )
;(= L J=- L b=' L ? == P1..-;. T cY=PL-
3 V-:'L--1 .?E...:.. c::- - 3
- I Jt: - I L
(.b)
p(.: 1hel)i'e~ ?-3 == S J'i i-f>rms ye~IJly.etL I tlnd A.
y/e/Pi..s
~
j/
w~= r) x*= ~ ) (jJl.= t
dw_ d(VW~)())t.Jj.= V~1f
~X - ~;<.. oX J: h<.+
Let So
JZW- .=
gXl-
V d /dWi-) h(" _
]; Ot:l- ( ~;< 1i:- JX -
5,hit'IIIY Iy.-' ~~tv V ~2W-J/c.
);; ~ = T... J.'}f:. 2-
&1'/'1);,,,) d;l{eyel1-6iJ; efud,oJl can nov b~ -f'xjJressp,<
Ul. -= -J- + W](~1' + ~~)
)c t h2. ~;<,JJ. 2- g~*' 1.-
lhe.
hZ
qp = _ (rbL. -r (»2W :lc -r ~"2~)
)tV Ji!: I'V gx"'z. g~+2-
7-80
(I)
as
(2)

7.71 1 ((1;/1't )
E1' (2) /~titcRk thai
j, ('/;* -¥ cr j, 2. .b.2. J P )
kr+-= .if == 'r / ~) )All) ~V n (.3)
If/ 1'h()1I! h 1'hu, r.e';WJ-C d~~.s t1D~ all'eAr' i:v mtt.1-cM ih~
eJlldf~" "j,Ioo',l1erl b1 tltmfl1S'M~( tl"""{'1~Jj (E'j,. I)) 1t.<
las I:. 1-Wo PI +fYI11S /11 £7.t/) !4,z b~ ~m blAe4 ftJ y/~/tI
(-1. q)11Iz.2.) = ~ 2lf
rlz: (~J/ ~ V Jt:
:50 -/7ta.. f £,.!)) e.11J1 II b.se b~ wr/ 1feJ1 a.s
(
X f1 ?r/; ~ b"2...il )
-r= 1 b) b.J ~V ) ,.ft y ~t ( If)
ttl! d 1h/~ Y~>u./f IS -lite $time IJS </hllf-
the ~/'ml'/C{y/~ Y'eJtI;Yempl')-/s I~d,cq)-e~ /:;7
Ii re -the sam e .
/11 £9 J3). 7htl5~
£t.J. (3) 4-M.P( (If)

7.72.
7.72 Flow from a Tank
Objective: When the drain hole in the bottom of the tank shown in Fig. P7.72 is opened,
the liquid will drain out at a rate which is a function of many parameters. The purpose of
this experiment is to measure the liquid depth, h, as a function of time, t, for two geometri-
cally similar tanks and to learn how dimensional analysis can be of use in situations such as
this.
Equipment: Two geometrically similar cylindrical tanks; stop watch; thermometer; ruler.
Experimental Procedure: Make appropriate measurements to show that the two tanks
are geometrically similar. That is, show that the large tank is twice the size of the small tank
(twice the height; twice the diameter; twice the hole diameter in the bottom). Fill the large
tank with cold water of a known temperature, T, and determine the water depth, h, in the
tank as a function of time, t, after the drain hole is opened. Thus, obtain h =h(t). Note that
t ranges from t = 0 when h = H (where H is the initial depth of the water), to t = tfinal then
the tank is completely drained (h = 0). Repeat the measurements using the small tank with
the same temperature water. To ensure geometric similarity, the initial water level in the small
tank must be one-half of what it was in the large tank. Repeat the experiment for each tank
with hot water. Thus you will have a total of four sets of h(t) data.
Calculations: Assume that the depth, h, of water in the tank is a function of its initial
depth, H, the diameter of the tank, D, the diameter of the drain hole in the bottom of the
tank, d, the time, t, after the drain is opened, the acceleration of gravity, g, and the fluid den-
sity, p, and viscosity, J.L. Develop a suitable set of dimensionless parameters for this problem
using H, g, and p as repeating variables. Use t as the dependent parameter. For each of the
four conditions tested, calculate the dimensionless time, tgl/2/Hl/2, as a function of the di-
mensionless depth, h/H.
Graph: On a single graph, plot the depth, h, as ordinates and time, t, as abscissas for each
of the four sets of data.
Results: On another graph, plot the dimensionless water depth, h/H, as a function of di-
mensionless time, tgl
/
2
/H1
/
2
, for each of the four sets of data. Based on your results, com-
ment on the importance of density and viscosity for your experiment and on the usefulness
of dimensional analysis.
Data: To proceed, print this page for reference when you work the problem and click her~>
rH
L
iii! FIGURE P7.72
7-- 82-

7.72. I
Solution for Problem 7.72: Flow from a Tank
H for big tank, in. H for small tank, in.
16.0 8.0
h, in. t, s
tg1/2/H1/2 h/H
Big Tank with T =57 deg C
16.0 0.0 0.0 1.000
12.0 9.2 45.2 0.750
8.0 20.0 98.3 0.500
4.0 33.8 166.1 0.250
0.0 57.0 280.1 0.000
Big Tank with T =20 deg C
16.0 0.0 0.0 1.000
12.0 9.0 44.2 0.750
8.0 20.3 99.a 0.500
4.0 33.0 162.2 0.250
0.0 57.2 281.1 0.000
Small Tank with T =57 deg C
8.0 0.0 0.0 1.000
7.0 3.1 21.5 0.875
5.0 9.5 66.0 0.625
3.0 18.2 126.5 0.375
1.0 30.1 209.2 0.125
0.0 41.4 287.7 0.000
Small Tank with T =20 deg C
8.0 0.0 0.0 1.000
7.0 3.0 20.8 0.875
5.0 10.0 69.5 0.625
3.0 18.1 125.8 0.375
1.0 32.5 225.9 0.125
0.0 43.0 298.8 0.000
7-83

t:
..c:
( ~,,'t )
18
16
14
12
10
8
6
4
Problem 7.22
Water depth, h, vs time, t
-------~--~-==-----=~J
~..-~ ~-··-I
. I
-------~:------~
2 +----~-..:----~~--_r--.---1
0
0 20
i
40 60 80
t, S
Problem 7.72
Dimensionless Depth, h/H,
vs
Dimensionless Time, t*(g/H)"O.5
1.20 -,--.------------,
I
1.00
: I----·-·-·~-------------~~----·--I
I--I0.80
- Big tank, T = 57 deg C
• Big tank, T =20 deg C
--.-. Small tank, T = 57 deg C
X Small tank, T =20 deg C
• Big tank, T = 57 deg C
_ I
• Big tank, T = 20 deg C
- .. - Small tank, T =57 deg C
I
J:
:c 0.60
i
0.40 ----------'-- I
I0.20 -l-----+--~,.,----+-----l
I
I
0.00 +----+-----1--__'*----1
o 100 300 400
t*(g/H)"O.5
?-9'1
X Small tank, T = 20 deg C

7.73 Vortex Shedding from a Circular Cylinder
Objective: Under certain conditions, the flow of fluid past a circular cylinder will pro-
duce a Karman vortex street behind the cylinder. As shown in Fig. P7.73, this vortex street
consists of a set of vortices (swirls) that are shed alternately from opposite sides of the cylin-
der and then swept downstream with the fluid. The purpose of this experiment is to deter-
mine the shedding frequency, w cycles (vortices) per second, of these vortices as a function
of the Reynolds number, Re, and to compare the measured results with published data.
Equipment: Water channel with an adjustable flowrate; flow meter; set of four different
diameter cylinders; dye injection system; stopwatch.
Experimental Procedure: Insert a cylinder of diameter D into the holder on the bot-
tom of the water channel. Adjust the control valve and the downstream gate on the channel
to produce the desired flowrate, Q, and velocity, V. Make sure that the flow-straightening
screens (not shown in the figure) are in place to reduce unwanted turbulence in the flowing
water. Measure the width, b, of the channel and the depth, y, of the water in the channel so
that the water velocity in the channel, V = Q/(by), can be determined. Carefully adjust the
control valve on the dye injection system to inject a thin stream of dye slightly upstream of
the cylinder. By viewing down onto the top of the water channel, observe the vortex shed-
ding and measure the time, t, that it takes for N vortices to be shed from the cylinder. For a
given velocity, repeat the experiment for different diameter cylinders. Repeat the experiment
using different velocities. Measure the water temperature so that the viscosity can be looked
up in Table B.l.
Calculations: For each of your data sets calculate the vortex shedding frequency,
w = Nit, which is expressed as vortices (or cycles) per second. Also calculate the dimen-
sionless frequency called the Strouhl number, St = wDIV, and the Reynolds number,
Re = pVDI/-L.
Graph: On a single graph, plot the vortex shedding frequency, w, as ordinates and the
water velocity, V, as abscissas for each of the four cylinders you tested. On another graph,
plot the Strouhl number as ordinates and the Reynolds number as abscissas for each of the
four sets of data.
11& FIGURE P7.73
S/de IIJeW

7.73 I
Results: On your Strouhl number verses Reynolds number graph, plot the results taken
from the literature and shown in the following table.
St Re
0 <50
0.16 100
0.18 150
0.19 200
0.20 300
0.21 400
0.21 600
0.21 800
Solution for Problem 7.73: Vortex Shedding from a Circular Cylinder
T, deg F b, ft
70 0.50
Data from Literature
Q, ftA
3/s y, ft D,ft N t, s 00, cycles/s V, ftls Re 8t Re 8t
0.036 0.82 0.0202 10.0 13.2 0.758 0.0878 169 0.174 50 0.00
0.036 0.82 0.0314 10.0 19.9 0.503 0.0878 263 0.180 100 0.16
0.036 0.82 0.0421 10.0 24.5 0.408 0.0878 352 0.196 150 0.18
0.036 0.82 0.0518 10.0 30.1 0.332 0.0878 433 0.196 200 0.19
300 0.20
400 0.21
0.062 0.79 0.0202 10.0 6.3 1.587 0.1570 302 0.204 600 0.21
0.062 0.79 0.0314 10.0 9.6 1.042 0.1570 469 0.208 800 0.21
0.062 0.79 0.0421 10.0 12.5 0.800 0.1570 629 0.215
0.062 0.79 0.0518 10.0 15.1 0.662 0.1570 774 0.219
0.029 0.86 0.0202 10.0 19.2 0.521 0.0674 130 0.156
0.029 0.86 0.0314 10.0 28.2 0.355 0.0674 202 0.165
0.029 0.86 0.0421 10.0 33.1 0.302 0.0674 270 0.189
0.029 0.86 0.0518 10.0 36.7 0.272 0.0674 333 0.209
0.018 0.92 0.0202 10.0 31.2 0.321 0.0391 75 0.165
0.018 0.92 0.0314 10.0 41.3 0.242 0.0391 117 0.194
0.018 0.92 0.0421 10.0 52.2 0.192 0.0391 157 0.206
0.018 0.92 0.0518 10.0 65.3 0.153 0.0391 193 0.203
00 = Nit
V =Q/(by)
8t =ooDN and Re =DV/v, where
v = 1.052E-5 ftA
2/s
7-8'

7.73 ( C<Jn't)
Problem 7.73
Shedding Frequency, ro, vs Velocity, V
1.S --r-------;-----~-----------,
1.6
1.4
1.2
..- ..__._-----_. __._-.._------_.----------/-
--I
~ ,
~ 1.0 ----~~--___:;iI""""'-------J
u I
~ O.S +-~---------~-+--------;;;,;L------t-:...------l
8 0.6 +-------'-----/--~-___;;;;O..=-_::7'~---.~
I0.4 -----------------I
0.2 ~~-=----..----.-------..-------- ------
0.0 - f - - - - - - ; - - - - - - i - - - - - - , - - - - - - j
0.00
0.25
0.20
0.15
-
0.05 0.10
V, ftls
0.15
Problem 7.73
Strouhl Number, St,
vs
Reynolds Number, Re
. I_._.__._---_._---,---_._----------
!
j
0.20
-+-0 =0.0202 ft
_0 =0.0314 ft
""""-0 =0.0421 ft
___ 0 =0.0518 ft
en
0.10
II
---1
~-~-,--_4--~--~~
! • Experimental I, I
1_Data from literature I !
0.05
0.00
0 200 400
Re
600 SOO
7-87
I
t
1000
I
I

7.7'/-
7.74 Head Loss across a Valve
Objective: A valve in a pipeline like that shown in Fig. P7.74 acts like a variable resis-
tor in an electrical circuit. The amount of resistance or head loss across a valve depends on
the amount that the valve is open. The purpose of this experiment is to determine the head
loss characteristics of a valve by measuring the pressure drop, 6.p, across the valve as a func-
tion of f1owrate, Q, and to learn how dimensional analysis can be of use in situations such
as this.
Equipment: Air supply with flow meter; valve connected to a pipe; manometer connected
to a static pressure tap upstream of the valve; barometer; thermometer.
Experimental Procedure: Measure the pipe diameter, D. Record the barometer read-
ing, H.tm , in inches of mercury and the air temperature, T, so that the air density can be cal-
culated by use of the perfect gas law. Completely close the valve and then open it N turns
from its closed position. Adjust the air supply to provide the desired f1owrate, Q, of air through
the valve. Record the manometer reading, h, so that the pressure drop, Ap, across the valve
can be determined. Repeat the measurements for various f1owrates. Repeat the experiment
for various valve settings, N, ranging from barely open to wide open.
Calculations: For each data set calculate the average velocity in the pipe, V = Q/A, where
A = 7TD2
/4 is the pipe area. Also calculate the pressure drop across the valve, 6.p = T'mh,
where T'm is the specific weight of the manometer fluid. For each data set also calculate the
loss coefficient, KL, where the head loss is given by hL = 6.p/y = KL V2/2g and T' is the
specific weight of the flowing air.
Graph: On a single graph, plot the pressure drop, 6.p, as ordinates and the f1owrate, Q,
as abscissas for each of the valve settings, N, tested.
Results: On another graph, plot the loss coefficient, KL, as a function of valve setting, N,
for all of the data sets.
Data: To proceed, print this page for reference when you work the problem and click II/'re
Th
1
Water --""--',,
,-:--Free jet II FIGURE P7.74

Solution for Problem 7.74: Head Loss across a Valve
D, in. Hatm, in. Hg T, deg F
0.81 28.7 70
h, in. Q, ft"3/s ~p, Ib/ft"2 V,ftis
N =2 Turns Open Data
9.20 0.235 47.8 65.7
6.50 0.195 33.8 54.5
5.04 0.169 26.2 47.2
9.40 0.479 48.9 133.9
6.33 0.386 32.9 107.9
5.01 0.341 26.1 95.3
3.62 0.289 18.8 80.8
1.92 0.214 10.0 59.8
9.35 0.827 48.6 231.1
7.65 0.767 39.8 214.3
6.01 0.691 31.3 193.1
4.32 0.578 22.5 161.5
3.24 0.504 16.8 140.8
2.62 0.456 13.6 127.4
1.85 0.391 9.6 109.3
0.98 0.283 5.1 79.1
3.03 0.897 15.8 250.7
2.37 0.799 12.3 223.3
1.79 0.701 9.3 195.9
1.39 0.618 7.2 172.7
0.97 0.517 5.0 144.5
0.64 0.426 3.3 119.0
6.p =YH20*h
KL =6.p/(pV
2
/2) where
V =Q/A =Q/(n*D2/4)
and
p = Patm/RT where
Patm =YHg*Hatm =847 Ib/ft"3*(28.7/12 ft) =20261b/ft"2
T =70 + 460 = 530 deg R
Thus, p =0.00223 slug/ft"3
N KL
2 9.95
2 10.21
2 10.54
3 2.45
3 2.54
3 2.57
3 2.59
3 2.50
4 0.816
4 0.777
4 0.752
4 0.772
4 0.762
4 0.752
4 0.723
4 0.731
5 0.225
5 0.222
5 0.218
5 0.217
5 0.217
5 0.211

60
50
40N
<
~
30.c
c:
<l
20
10
0
12
10
8
...J
6~
4
2
0
Problem 7.74
Pressure Drop, ~p, vs Flowrate, Q
--------~-
.. ~
--+-N=2
___ N =3
-.-N = 4
--e-N=5
+-----__~--~~----~~~4----~
0
- - - -
0
0.2 0.4 0.6 0.8
Q, ftA
3/s
Problem 7.74
Loss Coefficient, KL,
vs
Number of Turns Open, N
,----------------.
I,·_------1
. 2 3 4 5
N
I--+-N=2
I.N=3
I&N=4
I---e-N=5

7.75
7.75 Calibration of a Rotameter
Objective: The flowrate, Q, through a rotameter can be determined from the scale read-
ing, SR, which indicates the vertical position of the float within the tapered tube of the ro-
tameter as shown in Fig. P7.7S. Clearly, for a given scale reading, the flowrate depends on
the density of the flowing fluid. The purpose of this experiment is to calibrate a rotameter
so that it can be used for both water and air.
Equipment: Rotameter, air supply with a calibrated flow meter, water supply, weighing
scale, stop watch, thermometer, barometer.
Experimental Procedure: Connect the rotameter to the water supply and adjust the
flowrate, Q, to the desired value. Record the scale reading, SR, on the rotameter and mea-
sure the flowrate by collecting a given weight, W, of water that passes through the rotame-
ter in a given time, t. Repeat for several flow rates.
Connect the rotameter to the air supply and adjust the flowrate to the desired value as
indicated by the flow meter. Record the scale reading on the rotameter. Repeat for several
flowrates. Record the barometer reading, Halm, in inches of mercury and the air temperature,
T, so that the air density can be calculated by use of the perfect gas law.
Calculations: For the water portion of the experiment, use the weight, W, and time, t,
data to determine the volumetric flowrate, Q = Wht. The equilibrium position of the float
is a result of a balance between the fluid drag force on the float, the weight of the float, and
the buoyant force on the float. Thus, a typical dimensionless flowrate can be written as
Q/[d(p/Vg(PI - p))l/2], where d is the diameter of the float, V is the volume of the float, g
is the acceleration of gravity, P is the fluid density, and PI is the float density. Determine this
dimensionless flowrate for each condition tested.
Graph: On a single graph, plot the flowrate, Q, as ordinates and scale reading, SR, as ab-
scissas for both the water and air data.
Results: On another graph, plot the dimensionless flowrate as a function of scale reading
for both the water and air data. Note that the scale reading is a percent of full scale and,
hence, is a dimensionless quantity. Based on your results, comment on the usefulness of di-
mensional analysis.
Data: To proceed, print this page for reference when you work the problem and click hue
Scale
reading
o
t
Float
~ FiGURE P7.75

7.75 I
Solution for Problem 7.75: Calibration of a Rotameter
d, in. V, in.A3 PI, slug/W3 Hatm , in. T, deg F
1.40 1.50 15.1 29.05 78
Air Flow Data
SR Q, W3/s (Q/d)[p/(Vg(PrP))]1/2
14.6 0.229 0.142
21.5 0.321 0.200
28.1 0.413 0.257
33.6 0.491 0.305
39.2 0.564 0.351
44.8 0.644 0.400
50.2 0.714 0.444
55.9 0.798 0.496
63.1 0.888 0.552
68.6 0.973 0.605
73.5 1.05 0.653
76.2 1.08 0.671
Water Flow Data
SR W,lb t, s Q, W3/s (Q/d)[p/(Vg(Prp))]1/2
13.1 6.52 19.9 0.0053 0.103
18.5 8.01 17.7 0.0073 0.143
24.2 7.02 10.4 0.0108 0.213
28.2 7.81 10.1 0.0124 0.244
37.1 8.20 8.4 0.0156 0.308
45.7 9.21 7.5 0.0197 0.387
52.6 8.19 5.7 0.0230 0.453
P =Patm/RT where
Patm = YHg*Hatm = 847Ib/W3*(29.05/12 ft) = 2050 IblftA2
T = 78 + 460 = 538 deg R
Thus, P =0.00222 slug/ftA3
7-'12

7. 75"
.!!M
<
-=a
!::!~
~
rcQ.
-Cl
>:::::-
..s
-"t:I
-0
-
Problem 7.75
Flowrate, Q, vs Scale Reading, SR
1
0.1
0.01
0.001
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
10 100
SR
Problem 7.75
Dimensionless Flowrate vs Scale Reading
-r--~-----~---~--~'----i
-------------
---JIP------ - ---~---1
,-~----~-----jI
i
--;---~--------I
-I-~~__'_~~~C---~~_+_~~~-----:
)
'~_W""------------'-'------~-_j
i,
~------ '--~--------------~-i
0.0 -t----,.-----..,----;-----t-----
o 20 40 60 80 100
SR
1-93
I-+-Air I
i--waterll
-+-Air
--Water

a, I I
8.1 Rainwater runoff from a parking lot flows through
a 3-ft-diameter pipe, completely filling it. Whether flow in
a pipe is laminar or turbulent depends on the value of the
Reynolds number. (See Video V8.1.) Would you expect the
flow to be laminar or turbulent? Support your answer with
appropriate calculations.
Re == ~ = ~D If Re >t/-ooo fhe flow is iI/rbI/lenT, The
correspondinq velocify is
5 .fi~
V- Re 11 - ('1-000)( 1."J XIO- oS ) _ 0 0161 Ii
- D - 3 ft - , oS
Mo.sf likely Ihe veloc/fy will be greQter Ihan thisJ
t:8'J furbfJ/en! flruv.
8,3 I
8.3 The flow of water in a 3-mm-diameter pipe is to remain
laminar. Plot a graph of the maximum fiowrate allowed as a
function of temperature for 0 < T < 100°C.
VD
Re:= -V ~ 2.100 J where
~ is gil/en by
V- Q - J.l.f1.
-7f - ll'D2For laminar flow
Thu.s} fhe maximum
Re ::: (fk)D = ifQ
11 71'11 D =2./00 J
or Q= ~/OO 71' 1//)
if
or
Q == 20/00 :(O,003m)p = 7!9S f/ J where 1/-=::~ Qlld r;~¥
WHh valves of 1/ {rom TabJe B,z we Obtain
T, deg C v, m
2
/s
1.79E-06
1.00E-06
6.S8E-07
4.7SE-07
3.6SE-07
2.90E-07
a, m
3
/s
8.86E-06
4.9SE-06
3.26E-06
2.3SE-06
1.81E-06
1.44E-06
o
20
40
60
80
100
Flowrate vs Temperature
1.E-OS -,----,--------------,
M$1 :::~~:+~-"'----"~-----------~---~--~--'~----~JE i
ci 4.E-06 ------- --- ---- - ------ -.--------.-.-. --- - -- -- -- . - -- - j
j
2.E-06 -1,
I
1O.E+OO
0 20 40 60 80 100
T, deg C
8-1

8.9 I
SA Air at 100 of flows at standard atmospheric pressure in
a pipe at a rate of 0.08 Ibis. Determine the maximum diame-
ter allowed if the flow is to be turbulent.
fY!inimllhl Re::: e~D for tvrbvlent flow is Re :::/f000.
or wHh (1161 )D
V= Ii) = ¥Q Re:: efi"@ ;::: ~p Q ::: '1-000
7f 11' D2. } )l rr'p D
lienee}
Q-== lfO~O:J.i D (I)
Given oQ::; O.OB~ J where O'::'}p and p:: fr
ThIJ,s, .&
== OJ!-. 7 x/Jflf fir.) = 0 002.20 s/()q.s
p (/716 H'ib )('l-to+ 100)'P, . ft.3
s/IJ,.fiR
so fhat Jh .u3
_ O. 08 S ::: /. /3 .JL
Q- (..32. 2fi)(000220S/UlJ,S) s
. s~· ft:S
LJ h -7 Ib'S (
nenceJ
wif Ji':= 3.'1# 1./0 fl20 see Table B~3)J £rt. (I) 9ives
_ if eQ _ If (O,OO2.20~) (/. 13 ~) - fi
D- qOOO 11'f'- - iJooo"" (3-<>'f'l XI 0-7 '~t) - 2.27

8.5 I
8.5 Carbon dioxide at 20°C and a pressure
of 550 kPa (abs) flows in a pipe at a rate of 0.04
N/s. Determine the maximum diameter allowed
if the flow is to be turbulent.
For turbulent flowJ Re =fJ~D > 'fOOO J where Q=: VII ::*D2.V
or _JfeQ D _ f.e(i _
Re - 1l}i D2. - 1l'fJ'D - ~ooo
Thvs, _ II- eQ II 5
D- 11-00011'1'- J where ifG = 0.04-:s and j1. =/.'1- X/O- 'fn~ (Table I.B)
Hencs JL ( I )
J _ Lf. (0.0'1 oS) 9.S/f..)
D- 4000"" (J.Jf7 X10-5 !f;;f) =0.0883 m
8-3

e.6
8.6 It takes 20 seconds for 0.5 cubic inch of water to
flow through the 0.046-in. diameter tube of the capillary tube
viscometer shown in Videu V1.3 and Fig. PS.6. Is the flow
in the tube laminar or turbulent? Explain.
If Re = Y,9 < ')../00 the {fow /.$ /t~rnina~
.oS
'O,5/f'J. /
( 12;f1./fl)3 20..$
'f(~i~JI,E.f4)'-
where
V:: R :: 1//1:_
A JI D2.
If
or f+
V=I. 25 :s {2-
Thvs wifh 11:::. /,1../ X lOS ~ (see Table I.s)J
R
1.25 !1 ( °iO:AU)
- = 396 -< 2./00
e - /,2.1 x/f.5 reoS
The flow is laminar.
8-'1-
• FIGURE P8.~

8.7 I
'1,.7 To cool a given room it is necessary to supply 5 fe/s
of air through an 8-in.-diameter pipe. Approximately how long
is the entrance length in this pipe?
Ie
V:: % = ; (!ffY = 1'1.3 !f Thus, wilh t/=J. 57XI0-'1if(see 74Me I. 7)
R VD /If.3~ (~ tf) J.L 1/ . f h I Ie ;;: 7 ;;: -/{. fJ.~ =6OJ 700 > if 000 so me 1/0'1/ /s IJr /) en .
1.57xI() .!j-
Hence}
~ ~ '-A'I ReY6
J
or ~ ~ 'fJI (6q 700)~(ft) ::: 18~ If ff
e-.5

8,8 I 8. 8 The wall shear stress in a fully developed
flow portion of a 12-in.-diameter pipe carrying
water is 1.85 Ib/ft2. Determine the pressure gra-
dient, ap/ax, where x is in the flow direction, if
the pipe is (a) horizontal, (b) vertical with flow
up, or (c) vertical with flow down.
In genel"o/, AfJ -Ii sine == ')..r{
Thus) willi 7".: 7; at r:: -#: tlnd ~:: --f fA,s bec()mes
M J/-'IW It'. f)
W == - D - (J Sih
a) For a hori'ionla/ pipe ()~o
~ = _ JfiW .. _ .II- (J,BS -#-2.) = _ Z ¥o ~
1'X D I fl ' H
b) For verlica! flow up e::: qOO
Ul.. __ 'fj _}to _ _ tI- (1, 65 jJ 6" LL 16 69. Ib
dX - U - I It - ~,7' 1/3 == - •8 fi3
and
c) For verfiCt~! flow down e:::_qoD
ail.. __ Lf'lW + l'I == _ 'f a.as ~) +62 tf JA. = 55 0 .&
1X - D O l ff . ff3 I fl3
8-6

8 ,9 I 8.9 The pressure drop needed to force water
through a horizontal l-in.-diameter pipe is 0.60
psi for every 12-ft length of pipe. Determine the
shear stress on the pipe wall. Determine the shear
stress at distances 0.3 and 0.5 in. away from the
pipe wall.
8./°1
For a hori:zontaI pipe 1= ~rc- or '(= -f ~
Thus, b
_ (O.6X/Jf~.pp) _ ..!k. ,.,
'( - ~ 'J.(J').. If) - 3,6 r FfL J where r it
Hence]
'J- (o,S) /b'w :: 3.6 7Z =- 0./5 W
and wifh r=(o.s-O.3)in. =O.2in'J
'( == S.6 ( 9,~ ) :: 0.06 ft2.
FinallyJ w/fh r =: (O,E -O,S)'17. ::< 0 in. r =- 0
KHI Repeat Problem 8.9 if the pipe is on a 20° hill. Is the
flow up or down the hill? Explain.
For a pipe ()/J a hilI ¥.= 2.;: + 0sine J where e== ±zoo
/lsslJfJ'Je fhe flow /s uphd/ ; e·= +20° Ih
-rL ~ LJ~ . 31 '>- J.. (05 )[0.6xllPI- rp. lb· ;u!7I flUS, I:: 2. L7 - j smf)j or Iw:: 2 /rfl I~ If -t2·'fff3SIIJ J
Or Ih
fw ::: - 0.29£ H2o Since we fnv.s1 have ?W >0) fhe flow IJ7vstnQt
be uphill.
Assume fhe flow is riownh,11: f:)= -Zo' .JA
-rL L [~ . ,7 r[0.11 XIII-If FlJ. Ih· 6JI nilsJ '{ r::: 2. :£ - i sJnf)j or '(:: 2: 12. If +t2,/ffj3 stn20
=1'1. 3 r ~ J where r ~ ff . The
lienee, wilh r =E rj- T, ow is downhill
fw = /~3(1;t) == 0,596 ~
l#iJh r = (0.5 - 0.3)in. :: 0.2. in.)
(= /Jf.3 (~;) = 0,238 #i
Wilh r =(0.5-0.S)i/J. ::0 '( =0J

e,ll I
8.11 Water flows in a constant diameter pipe
with the following conditions measured: At sec-
tion (a) po = 32.4 psi and Za = 56.8 ft; at section
(b) Pb = 29.7 psi and Zb = 68.2 ft. Is the flow
from (a) to (b) or from (b) to (a)? Explain.
(a)
iJSSfJme the {Jow is uphill. Thus) Ij- + ¥i+~4 = 'I-t ¥i+Zb +0.
(b)
or wilh JIg =~ ) .
h = Ib +:z - I'i> - Z = (32.1psi-2rt. 7psi)(;IfIff!:) +56 eff -682ft
'L '1 a 4' b 6ZJf !A. ' .
or ~
hi =-5./7ff<o, which is impossible. ThllsJ
fheflowjsdownhi/l,frorn(bHo(Q).
8-8

x.J:2 Water flows downhill through a 3-in.-diameter steel
pipe. The slope of the hill is such that for each mile (5280 ft)
of horizontal distance. the change in elevation is az ft. Deter-
mine the maximum value of az if the flow is to remain lami-
nar and the pressure all along the pipe is constant.
or
.tlZ ~.JsilJfJ ~ - (s~eoff))( (-I,f¥- K/()-s) :::: O. / ()2. If
8-q

8. /.3 I 8. 13 Some fluids behave as a non-Newtonian
power-law fluid characterized by r = - C(dill
dr)", where n = 1, 3, 5, and so on, and C is a
constant. (If n = 1, the fluid is the customary
Newtonian fluid.) For flow in a round pipe of a
diameter D, integrate the force balance equation
(Eq. 8.3) to obtain the velocity profile
u(r) = -n 1 r(II"'1) 11 _ _
(a)1111 [ (D)lll-lI nJ
(n + 1) 2ee 2
For any FlUid i.f == 2-l so thai with (=:;-c(1fr)n we ob14in
Af=-2f(djflor #;:=-(/fp)*r-h ~
or -L
-sdu :::!(-2ti)jr*dr which iIJfe9rafes foqive
lL=-(2~1t(n~J) r(lW.) t-C, , where 0 is a conAanl.
.The fluid sficks to fhe pipe so thaf {). = 0 at r =: ~ •
Hence1 from Eq. (I)
, (Ll)1; n ( D(n,t')
C, = 2C~ (ntJ) 2"J
so fhot -L.
;:: --'1- (- Af )nf- ~ (lL)flWlJtJ. (n+J) 2cL r -f- 2.
..NOTe: Since we tire considering onl,y oJd inleger valves lor
n we ct/IJ vse fhe felt;t ffJqt it"
(1Ir)n= - KJ where 1<> oJ TPe/J ~::- K~
sO fhal ¥r<0.
8-10
(/)

8.IJf'" I
8.14* For the flow discussed in Problem 8.1S.
plot the dimensionless velocity profile u/V<, where
Vc is the centerline velocity (at r = 0), as a func-
tion of the dimensionless radial coordinate r I(D/
2), where D is the pipe diameter. Consider values
of n = 1, 3, 5, and 7.
From Problem 8./3,
Ii. (r) := -IL ( Ll{J )hfr(lJril) +(JJ..)(CW)] (I)
(n+1J 2 ic; 2. I (nil)
Let Yc = (). (r=o) , or Yc;:: (~~I) (2£1 Jii(~J-n (2)
Nofe: For ?- == c(M)nwdh djf<0 and!) on odd inlegerJ to hove
( >0 J we tnllsf hove C~o. ThlJs, from £q. (2.) J Vc >0 as if musT.
By dividing £~. (J) by £1.. (1.) we obiQin
u. [r](CW)
Vc = I - (~) r
This resuH i.s ploHed below for n =J) 3J 5; t1nd~ wtfh 0 ~ (~) ~ I.
An EXCEL pr0'll'am was vsed 10 do the calev/41ion.r alkiploif/fl9.
n = 1 n=3 n=5 n=7
r/(O/2) uNc uNc uNc uNc
0 1 1 1 1
0.05 0.998 0.982 0.973 0.967
0.1 0.990 0.954 0.937 0.928
0.15 0.978 0.920 0.897 0.886
0.2 0.960 0.883 0.855 0.841
0.25 0.938 0.843 0.811 0.795
0.3 0.910 0.799 0.764 0.747
0.35 0.878 0.753 0.716 0.699
0.4 0.840 0.705 0.667 0.649
0.45 0.798 0.655 0.616 0.599
0.5 0.750 0.603 0.565 0.547
0.55 0.698 0.549 0.512 0.495
0.6 0.640 0.494 0.458 0.442
0.65 0.578 0.437 0.404 0.389
0.7 0.510 0.378 0.348 0.335
0.75 0.438 0.319 0.292 0.280
0.8 0.360 0.257 0.235 0.225
0.85 0.278 0.195 0.177 0.170
0.9 0.190 0.131 0.119 0.113
0.95 0.097 0.066 0.060 0.057
1 0.000 0.000 0.000 0.000
(con'f)
9-11

r/(D/2) vs uNc
1
0.8
0.6
-a-1::
0.4
0.2 --- -~~----~------~~
a +-----~------r-----~----_+----~
a 0.2 0.4 0.6 0.8 1
8-/2
--n=1
---n=3
··--·n=5
---·n = 7

f.IS I
K J5 A fluid of density p = 1000 kglm3
and viscosity
J.L = 0.30 N • s/m2
flows steadily down a vertical O.IO-m-
diameter pipe and exits as a free jet from the lower end. De-
termine the maximum' pressure allowed in the pipe at a loca-
tion 10m above the pipe exit if the flow is to be laminar.
Re ::: 2100 for maxilf/flm pres.rfJl'e.
Th(}J'.; k
2/00 ::: eVlJ _ 1000~ Y(O,/I1})
po 030 #:!.• PJ~
or
V= 6.30lJ
Buf for Jam/par flowJ
Jom
O.lm
--
V:: (~e -!'J sine) /)2. where lJ eo/1m J~/()IJI 4IJd &:: -po'
.3:z.,P J I I JI
t:::PIO:::: ~1/olW.3
(LlP - ~I/O.j. {/()m)silJt-P()'))(()'/m):J.
..32 (O.3() ~) C10m)",
ThlJ~
6,30 f :::
So fhal
Af ~ -3, 7,XI 0 IfJf ::: -.3 7. t kfa,
11/'-
~-13

e./6
~.16 Water is pumped steadily from one large, open tank
to another at the same elevation as shown in Fig. PB.16. De-
termine the maximum power the pump can add to the water if
the flow is to remain laminar.
Length = 100 ft
Diameter = 0.1 ft
vII FIGURE PS.16

8./7 I
8.17 Glycerin at 20°C flows upward in a ver-
tical 75-mm-diameter pipe with a centerline ve;
locity of 1.0 m/s. Determine the head loss and
pressure drop in a lO-m length of the pipe.
For /qmillfJr Flow in fJ pipe,
P=I2.60!itn3
Il=/. 50 N·S
r' m'J..
V=- overQge velocify =f ~qX :: i{J~):: 0.5.1f J =/Om
Thus k
o '_ .eVD _ (J2607,?)(O.S1j) (0. 07£ In) = 31.5 < 2100
ne - p - I 50 N·s
• m2.
The flow i.s /(Jlflil}tJr so ih4f
-- ~ D-=O.07Sm
V= (AfJ-K'I.sinB)J)Z h e=no8
32~1 J were 7 .:...... ej) 2,
ThlJ,sJ ( N•.5)( I l!l. L.<:::
.ALl == 3.2
D
JI.!V -I- ~i ;:.32 1.50'/ii2. lom),o,s.s) +('I.8/~){/').60~)(!Otn)
r (0.07.5m)2.· III
= 1.66 x/Os -J,'J.. J or ~f::: /66 kPQ
/I/S(), 2. ~2
t;+z, + ~ ::: "F +Z2. f ;; +hi.. J or wNh ~ -= ~ J 22.- "Z/;::J, and
PI ={J2. +AfJ this 9ille.s
A /,61)(/0
05
1,.
hL== 4f -1. = (9. 8111;)(1260 :a) -/0", = 3, ¥3 m
8-/5

8./8 I
8,19 I
8./8 A fluid flows through a horizontal O.l-in.-diameter
pipe. When the Reynolds number is 1500, the head loss over a
20-ft length of the pipe is 6.4 ft. Determine the fluid velocity.
_.l::C.:hi. - f D 'J-, J where since Re::: /500 <: ZlfJO fhe Flow
ioS /4111ina,. ,
ThV$ f;: 6~/Re =t'l//.5oo;::; O,O~27 .so thaiJ
6, If fI :: 0,01/-2.7 ;LO ·Ff
(0,/ //'}./4)
or J:l
V::: 2,01 oS
X.J q A viscous fluid flows in a 0.1O-m-diameter pipe such
that its velocity measured 0.012 m away from the pipe wall is
0.9 m/s. Ifthe flow is laminar, detennine the centerline velocity
and the flowrate.
For /omif/or flow in a pipe
u(r) = Vc[I-(-¥-lJ J where D=O./m aIJd IJ.:::O.8~ cd
-rJ.. r == °1m - O. O/2m == O.03Sm
I ,,{lSI 2. ~
hi _ 1/ [1_ (2 (o.o3Sm))] II - I Sqf!L
O.B S - Vc - . O./Om or Vc - " s
so t/Jai
Q-= 11/'V:::*D'J.rO.5 'Ie) ~ fro. 1mi'(0.5)(1.89 1')== 7. 'fZx 10-
3
.If.
g-/6

8.20 J I--I~- - - 0.5 m -!
8.2.0 Oil (specific weight = 8900 N/m viscosity = 0.10
N·s/m2
) flows through a horizontal 23-mm-diameter tube as
shown in Fig. P8.20. A differential U-tube manometer is used
to measure the pressure drop along the tube. Detrmine the range
of values for h for laminar flow.
~•.-Oi_1_ _ _ _f..L-23~jn_m__,. (2) ~""12
t~
h
tor Iamii'J4r flow Re ~ 2./00; or e.:f- 5 )./00
where e::::r~. Thv~ flJe milJiI!JvlIJ his h::: 0 (no 1/(11) and
the maximvm h is for Re::: ~/oo.
Hence,
2100 ::
( a'lOONlm') v(o O').3m)9.8/IfJ /.s2. •
HenC8J
0.5/11
A P -::It -f:J. -::: 0.0305 0.0'-3 HI
Dr
A fl ::: 30J
-¥OD "y1m2-
From m4nomeler equalions:
" of 0(Hofh) - SG~Hzo h- t H:::f:;. J or
Ap::{J,-/2.::: (SG ~2.0 -r)h
Thvs
h
= J. 3 OJ If. 00 N/m" ::: O.50qm
(7 (9 8ooN/m3) - 8900/1//113)
Hence o ~h ~ 0.50Q!YI
8-/7

8.2/ J
S.21 A fluid flows in a smooth pipe with a Reynolds num-
ber of 6000. By what percent would the head loss be reduced
if the flow could be maintained as laminar flow rather than the
expected turbulent flow?
For eilher laminar or furhu/eni flow
1;£
hL ~ f D Z9 .- ThvsJ w/lh Ihe same ~"/), 4nd9
hL/(J1b _ {I,m
"hL-IVI'b - +Ivrh
If the flow is laminar
If the flow is furbf)lenf with Re :::to()() (lAd "* :::(),
then fro", fhe ltlooJ)' charf (Fi,. 8,2()) ~vrb:: 0,035
ThvsJ
h_ L /fJ'" ::: O. 0101 ::. O•.JI 16
hl.-I
lJrh
0.035
The heqd/oss would be redvced Dy
(hLturh -hL/41l1 )/h/.tU,.~ == /-0.'186 If:! O,S/If, or 5/,11-70

8.22 I
SG = 0.87_[1
- el). :::==--='--'-
S.22 Oil ofSG = 0.87 and a kinematic viscosity v = 2.2 x
JO - 4 m:!/ s flows through the vertical pipe shown in Fig. PB.22.
at a rate of 4 X 10- 4 m3/ s. Determine the manometer reading.
h.
20 mm
-- -P. c::4 m
FIGURE PS.22.
SG = 1.3
(I)
(2.)
From manomeTer consideralions
p, +ahI - %h+~h2. :: f2. 1 where ~:: SGtrlHJ.O:: 1.3(fl.811$)=12.7/f ~
ThusJ
and h=h- h:J.+jJ or h,.+h, :: hfi
I,-P,. =Ap ::;-'O(h1 +h) +rm h::(om- o)h - 01. (3)
Combine Eqs. (~) and (3) fg qive
Lf3.7 ~ = (l2.7J1.-8.53)1$ h- (8.5.3 ~)(1-IIJ)
or
h:: 18.5 m

8.23 J
8.23 Determine the manometer reading, h, for Problem 8.22
if the flow is up rather than down the pipe. Note: The manometer
reading will be reversed.
SG = 0.87_[1
- (2}.::=~_
20 mm
- -ics4m
FIGURE PS.23
-r-
1 -
h,
~Cr'
tQ .. 'SG = 1.3
•2rm
Fr()m /flan()mefer considerafion.s
III - oh, -I- d;n h- rhz =-III J where Om == s~ ~,.o =/.3(p.er ~)=/2.71f ~
and 172 =i+h-h, or h2-+ hi =J+h
Th(Js,
(1,-1,. -== Af == ~(h2.+h,)- ~mh = -(d;,,- a)h +d'i
Combine Ef(s. (2) and (3) if) 9ive -
/I/.q ~ =-(J2.7'!-8.~3)~h +8.S.3~ (lfllJ)
or
h=-/8.5m
{2.)e '--
Nofe: Since h<o the man()llJeler is displaced -;:::=-
/n the direcTion 0flfXJS,-le Ihal sbowlI in fhe (I). ~
ofl9inal ft'yvre.
8-20
1'S.5m
J
(I)
(z)
(3)

8,25
8.24 For Problem 8.22, what fiowrate (mag-
nitude and direction) will cause h = O?
From Prob. 8,22, AI=- (%-(f)h-(f'l
Thvs) with h=0) Af::: - (fi af}d
Q= 71' (-~J +rJ,)/)1f =0
/2.8 j/.i =
Note fhqf ~fJ :J:0J but Q=O .since Af+rJ:::o
8.25 The kinetic energy coefficient, a, is de-
fined in Eq. 5.86. Show that its value for a power-
law turbulent velocity profile (Eq. 8.31) is given
by a = (n + 1)3(2n + 1)3/[4n4(1l + 3)(2n + 3)].
dll
_ eJu 3rJIJ
From Err,S, 86) 0( - AVV2 where V== overtJ'je velot;ify, /}=7TR/
k P :l /
QlJd - 1/ [/ ..t:..Jn Fi £; I 8 if v- 2.n Yc
U = Vc - R • rom xample . I - (n+l)(zn+/J
Thvs wilh dl/ = 2. 71'r d,.
i-3d. R s } ~
(0)
IX = : v/'J where JaJ,j/l=2 71'S~tr)-j ]"I'rdr =271'R"V,/S[I-y]ydy
where y::: ~ , r=O y~o (I)
Let x::: J- Y so thai y = I-x and dy =-dX
J
Hence 1.3 0 ~ J.:: 3 +/
J S[1_y]=11y dy =-Jx~(J-X)dx =J(Xn
-X7f )dx
y=o x~1 0
n ~ n
- n+-3 X - 2n+3
X=I
2n+31X n
x=o
(ntll(1.fJ+l'1
'l-n+'(n+3) (2I)f3)
(2.)

8.26
~.Z6 As shown in Video V8.3 and Fig. P8.26, the velocity
profile for laminar flow in a pipe is quite different from that for
turbulent flow. With laminar flow the velocity profile is para-
bolic; with turbulent flow at Re = 10,000 the velocity profile
can be approximated by the power-law profile shown in the fig- .!..
ure. (a) For laminar flow, determine at what radialloaction you R
would place a Pitot tube if it is to measure the average veloc-
1.0l<r----_L
Turbulent with Re = 10,000
.!!... = 11 - .!..1115
Vc R
ity in the pipe. (b) Repeat part (a) for turbulent flow with
Re = 10,000.
0.5
_______ -1- ______ _
a 0.5 1.0
/I
V;
• FIG U REP 8.2.6
For lam/nor or furbv/enl (lowJ
R
Q ~ fJV::: 1TR'J-V:: fUdl] =[U(~llrdr) =27Tfurdr
r::O
a) LQtrJinor flow: R 2. 2- 2-
7TR"V,. 271Vc, Jr[l-(i/}:lr == 271'f~- ~] == 71'4 Yc
Thus) V==t{ 0 For U::: V::: ~ fhe ervttfiOI1 for -% 9;ves
i == i =' I - (~)""I or (f)"'" t Thus,!' "'Vr R r; O.707R
b) TlJrb {Jlen! fJow R I Vi
7TR1
V: 27TVc Sr[l-ftolr == 27TR2.'Vc J(.f)[J-(f)] 5d(f)
o 0
LeT y:= I - (f) .so -fha-f (f) :: /-y and d.(f) ;: - Jy
ThlJ's >:=0 I
7T R"" V'" 271R2.Vc f(I-y) y I(s (-tiy) := 21!R2.f .f(YIi- - yVs) 1)1
_/ 0
y- [.5 S 2. (').C)::: 21lR'" Vc '6 - It J= 271R Vc 7&
Or V= if Vc hr U::: v':: 1* the e'!vQfitJn fof' ~ 9ives
~ := #- '" [1- f t.s or f '" O.7SO so fh4"/ t' == O.7S0 R

8.27 I
8,27 Water at 80 of flows in a 6-in.-diameter pipe with a
ftowrate of 2.0 cfs. What is the approximate velocity at a dis-
tance 2.0 in. away from the wall? Determine the centerline
velocity.
V
R 2.0" PI VD (JO.2.{i) (Aft) -E
= II = *(.f2.fO" ::; /0.2 s so fhaf Re::; r :: 9.2-6)/0-6if =S.5/ X/rr
- .J..
The flow is /()rbu/enf with ~::: (J- t )n) where n';; 8.3 (see Fiy. S.I )
Thvs, (see Example B.If)
V :J.n'- 7-(8.3)'-
Vc ::= (n+l)('-fJtJ) :::: (8.3.1-/)(2X8.3+/) : O.8Jf2
or _ IO.2~ ff
Ve - O.BJt.2. = /2./ :s -L
.L fi.( .8.3
/llso) af r= 3ifJ.-2.0in. =/.Oil'),) u::::Vc(I-f)fI:/:2.1s J-;~~~)= J/.S#

8.28
8.26 During a heavy rainstonn, water from a parking lot
completely fills an 18-in.-diameter, smooth, concrete stonn
sewer. If the flowrate is 10 ft3
/s, detennine the pressure drop
in a 100-ft horizontal section of the pipe. Repeat the problem if
there is a 2-ft change in elevation of the pipe per 100 ft of its
length.
(b) W/{h f/fJW Vfh/I/ ~- Z/ : 2 ff so fhat
£jf ::(62ll-!t3)(t.H)( /tf.~~:'''1 +O. 266/~1. ::: /'/3 psi
(c) Wi-Ih flow downht11 .F2.- Z, :-2ft s() fhQf
.lJ{J:;:: (62·~ffo)(-1.fI)(II{.'J~?-)+ 0.266 *'= :=: - 0.10/ psi

8.zq r
8.30 I
8.29 Carbon dioxide at a temperature of 0 °C
and a pressure of 600 kPa (abs) flows through a
horizontal 40-mm-diameter pipe with an average
velocity of 2 m/s. Determine the friction factor
if the pressure drop is 235 N/m2
per lO-m length
of pipe.
8.3:0 Water flows through a 6-in.-diameter horizontal pipe
at a rate of 2.0 cfs and a pressure drop of 4.2 psi per 100 ft of
pipe. Detennine the friction factor.
8-2.5
=0.0300

8.31 I
8.31 Air flows through the 0.108-in.-diameter, 24-in.-long
tube shown in Fig. P8.3J. Detennine the friction factor if the
flowrate is Q = 0.00191 cfs when h = 1.70 in. Compare your
results with the expression f = 64/Re. Is the flow laminar or
turbulent?
-
--
T
h
7,- .l..
./"77; Water
I 24 in.
(I) I i --I
Air • 0.108 in.
-I
t V
_r---
(2)
• FIGURE P8.~1
..LJ. /,2 ~2. £ V2.
L.L ""...!.L +Z = 7'J2. +~ +Z2. + f75 -
i' 2p I r~? 2p
(I)
where 'Z,:: 2'2 ; fJ2. =: 0 J ~::O and
f, =Oq~o h = (62.1f.!ffi)(ftFI) =8. 8'1 ~
A~~ ~
V v. !i O,OOlql oS
::: 2. = A = f (0'118 ff)2.
= 30.0 {i
HenceJ Eq. OJ becomes
fJl = -fpV2(1 + f-Z-)
or
8. e'l- *- =;. (0.002.38 S~~)[If f (-o.~o~~Jl30.0~)" or f=0.03 2.~
Also Re = .p!
or R _ (~Ii) (30,0 y)
e - 1.57)( IO~~ ~2. = /72-0< '2.100.J -fhe fl()w is lamifJar
Note~
47-e ::: 6'1- :: 0.0372
1'1 1720

8.33 J
8...33 Detennine the thickness of the viscous sublayer in a
smooth 8-in.-diameter pipe if the Reynolds number is 25,000.
f 51/ h .. ('fMt)lf '>- D1 f1. I ~O.s = U. J IN eref~2 =- e and £w:: 'I- • Si~ce LlfJ == D "i"pV
we obfain fw = ea and u.4
== yJI V
Thvs,
.51/ .511D r _ SD
65 = ""ff' = 1Jr ) or °05 - D 1tr (I)
V'8 V Va vD l1e Ye
From Fi9.8.20} for a smooth pipe with Re:: 2.5)( IO~ f == 0,02'1
Thus) from £,{.tl)
6 = sf§' (R.~-I) = O.OO2.Jf3 fI
~ 2,5XIO'" Vo.O")./f
8-27

B.3#- I
8.34 Water at 60 OF flows through a 6-in.-di-
ameter pipe with an average velocity of 15 ft/s.
Approximately what is the height of the largest
roughness element allowed if this pipe is to be
classified as smooth?
Lei h= rov9hness heiqhf. Thos} h=4 J where O:S::: 5;
with u~=( "; )~ and?;::: %ie. Since AI' -== f-!ieV:l. we obflJin
'>- - pfV2 .,--IT V(w - 8 or U -Va
For a smoofh pipe wilh Re =ift::: {/sfj. ){j.:;~ = 6.19x/Os we obtain
f
//J./X/O -
from h9' 8,2 0 ::: 0.0125 .s
ThV.5, U· = (- O.;IZ5t·(/s!j) =' O. 593 ~
or (-.5.fi.z,)
6'=.£it. = .5 /,2/X/O oS -= /'02x/ri~ff
:s U· o.sq,3 ~
8-28

8.3.5 J
8.35 A 70-ft-long, 0.5-in.-diameter hose with a roughness
of e = 0.0009 ft is fastened to a water faucet where the pressure
is PI. Detennine PI if there is no nozzle attached and the average
velocity in the hose is 6 ft/s. Neglect minor losses and elevation
changes.
D=0.5 in. V
~~~:------~~~(~
(I) .£=70ft
0)
0.000'1 If = 2.16X/o-z.
(~fJ)
= 2.07 X/O¥ we obfain f= 0.052.
Hence} from Eft. (j)
fJ, '" (0.052) (;;'ff) ~ {1.91f ~"1f)(6#/''" 3050 ~ ~21.2.ps;
/2. ff
8- 2. 'I

8.36 I
8.36 Repeat Problem 8.35if there is a nozzle
of diameter 0.25 in. attached to the end of the
hose.
v : D=o.5 in.
a-+- ~(I) (~)
1.=70 ff D =: 0.').5 in..2.
Jff+*+II = Pr+fi+Z2. +(IiiI where Z6h I V,=V;61f, fz=O
and V,. ::: Y!& =V. (.Q..) -'6f1 )(z/":::2.'1-11lil I D~ ~ s .s
Thus,
It =(V/·+{-t V:L)~ =-t p(v,.:/.+fl V'") (I)
r · 'Ih £. o.oooqff -2-
rrom Fi,- 8.2·0 WI D = (0.5 ft) =2.1& x/a
and ~
Vb (6 f1)(0.s 0) ~
Re=-= oS -rr =2.071</0 weobfaif) (:::O.052
1/ /.2/x10-5
!F
Hence) from £'1. flJ
1', := i (I.9'1 S~~$)«2.'f fj-/ +(0.052) (~:) (6!1f) =36oq I$. '"2.5.1.j}.
Note: To lI'Jainfain the same flowrale wdh fhe noy-zleaHocherJQS compared
fo fhat wlthouj fhe nO"i2/e (see Probe 8.35) the pressure mvsf be
increased from 2/.2 psi 10 25.1ps,:
~-30

8.37 "J
8.37* The following equation is sometimes used
in place of the Colebrook equation (Eq. 8.35):
1.325
f = [In[(e/3.7D) + (5.74/Reo.9
)]F
iteration procedure to obtain f. Plot a graph of
the percent difference in f as given by this equa-
tion and the original Colebrook equation for Rey-
nolds numbers in the range of validity of the above
equation, with e/D = 10-4
•
for 10-6
< e/D < 10-2
and 5000 < Re < 10+8
(Ref. 22, pg. 220). An advantage of this equation
is that given Re and e/D, it does not require an
Lei ..of ==
fapp-f
fapp
J where fap,:- approximafe reSfJ/f ohI4/iJe.d (rum
/.:3 25
..J..... =-2.0 log[~ +M. J1(f 3.7/) Re1(f
ThvsJ with t == IO-1f these become
fa = . /.3'-5 2-
pp [I (2 10X/05
-t .s.7~ )11
and n, ReoA
~
W= -2.0 loq [2.7()x/O-.s -I- t:Vr]
3 8 I I I /1 f J00 ( fqpp - {)
For Sx/O ~ Re ~ 10 J ca cv ate an" p07 IOOA:: f J
where Fapp ana f are obtained from £qs. (I) and (2.)
Proqram P8·fI=.37 shown be/ok! Wq-s used f()r !he cQ/cu/afions.
:'00 cis
120 print "**************************************************"
130 print ,,** This program calculates the difference **"
1i!-0 print "** between the friction factor given by the **"
150 print "** Colebrook equation and that given by the **11
160 print "** approximate formula provided. The Cole- **"
170 print "** brook result is determined by an iterative **"
180 print "** routine. **"
190 print "**************************************************"
200 rr = lE-i!-
210 Re = 2500
220 print "Re f faprox f - faprox, %"
230 for i = 1 to 16
2i!-0 Re = Re*2
250 faprox = 1.325/(log(rr/3.7 +
8-3/
(.1)

8.37· I(con't)
260 fp = faprox
270 goto 290
280 fp = f
290 f = 1/(-2.0*log(rr/3.7 + 2.51/(Re*fp~0.5))/log(10) )~2
300 if abs(l - f/fp) > 0.000001 then goto 280
310 diff = ((f - faprox)/f)*100
320 J,?rint using "#.###~~~~ #.###### #.###### +#.###~~~~II;Re,f,faprox,
diff
~30 next i
~0
-><0
**************************************************
** This program calculates the difference **
** between the friction factor given by the **
** Colebrook equation and that given by the **
** approximate formula provided. The Cole- **
** brook result is determined by an iterative **
** routine. **
**************************************************
0.6
0.4
0.2
0
-0.2
Re f faprox f - faprox, %
5.000E+03 0.037505 0.037961 -1.216E+00
1.000E+0~ 0.031037 0.031138 -3.233E-01
2.000E+0~ 0.026101 0.02605~ +1.809E-01
~.OOOE+O~ 0.022286 0.022196 +~.017E-01
8.000E+0~ 0.019319 0.0192~1 +~.0~7E-01
1.600E+05 0.017026 0.016985 +2.397E-01
3.200E+05 0.015290 0.015295 -3.227E-02
6.~00E+05 0.01~032 0.01~077 -3.176E-01
1.280E+06 0.013179 0.0132~6 -5.091E-01
2.560E+06 0.0126~3 0.012713 -5.513E-01
5.120E+06 0.012332 0.012391 -~.7~8E-01
1.02~E+07 0.012162 0.01220~ -3.~99E-01
2.0~8E+07 0.012072 0.012100 -2.298E-01
~.096E+07 0.012027 0.0120~3 -1.362E-01
8.192E+07 0.012003 0.012012 -7.108E-02
1.638E+08 0.011992 0.011995 -2.875E-02
(f - fapprox)1f vs Re
!
/ ~
t J r ..-.
! V ~
j,;"
VQ. -0.4
l-<~Co
III
-0.6 ...'to-
'to- -O.S
- -1
-1.2
-1.4
1.E+03 1.E+04 1.E+05 1.E+06 1.E+07 1.E+OS 1.E+09
Re
8-32.

8,38 T
6.38 Water flows at a rate of 10 gallons per minute in a
new horizontal O.75-in.-diameter galvanized iron pipe. Deter-
mine the pressure gradient, 6.p/ e, along the pipe.
Q:= /0 ~ (JmJE..)(7.3/ in})( 19
4
/--"3) -::: 0.0223 Itmm 60S / fa I 172'/IJ. .s
Thvs ~
V
- J Ji _ 0.02.2.3 s _ .tt
- II - :tI. (0.75 )2. - 7.')..
7
;s
J{- -r;:-H
Now, for a horizon/IAI pipe
Af:: f-tt pV2. where si"ce
Re :: til:: 7.27J} (~2.fl) ~-If S f+ ::: 3. 76}( /0
y 1.2/1.10--:s
and.
S::. ::: O.-Ooosft ::: O.Oog
D e;-".sII)
if follolNS from Fi'l' e.20 fhal f:: 0.037
ThvSJ
2-
E£:;; 0.0.37 (J.'I'fs/f/9S Ifl.3) (7.27(I/s) ::
1 (f!;?ffl) (2.)
II; ( I-N~
3 o. ¥ 1{3 --'11'1 in.~)
= 0.2/1 psi/II
~-33

8. if! I
8.41 Air at standard temperature and pressure flows through
a l-in.-diameter galvanized iron pipe with an average velocity
of 8 ft/s. What length of pipe produces a head loss equivalent
to (a) a flanged 90° elbow, (b) a wide-open angle valve, or
(c) a sharp-edged entrance?
KL D .I L D VD ( 8!sf)(I ~ H) .3 -L. <J L
t. ::: -f where 'lid" ne:: --:n-::: -"".f:f! =Lf.25x If) TnlJS WI1'1
e f ' y I.S7XIO .I
£. 0.0005 (f .s ( .
75 =(//12. H) :: O.~06 (see Table 8./) we oblain f:: 0, 0'15 Fi9·8.20)
Thus t. :: kJ.. (/2- f-l) =1.85ZKi. or 0) 90'elbo'l} : k;. :: 0.3 or i~(J::: o.S56Fl
J ef oO-¥S (, -
. b) qlohe valve: K:: 2. orie'l =3.70 If
c) sharp en/ral)ce: KJ.::: 0,5 Of is'f,=o.q2.6If
8.4-211 18.42* Water at 40°C flows through drawn tub- 10-' m.1/s. In your solution obtain the friction
ings with diameters of 0.025, 0.050, or 0.075 m. factor from the Colebrook formula.
Plot the head loss in each meter length of pipe
for flowrates between 5 x 10-4
m3
/s and 50 x
hi=f-!s!i ' where V=:!l~2 Thus, hL= fl ~(::r/= ~~~'-
orwdhi=lm, 2. ~
h 8 {(Jm)Q. h frt h /tI3 (I)
L-.".2.(q.g/~)D5 or 'J. ::O.082.6r;s , UI ere hJ."'In,D,vIllJQ"'s
For artWn IUbil1tJ &~ 0.0015 ",m ~ 1.5XIO-
6
m
co -6
or -B':: I.S~ 10 ) where D~m (2.)
f)lso} Re:: y# = 1T~~ = -¥ Q or fie =1.9'1-x/0
6
2 (3)
1f(6.S8XIf)-7~) D J .
Nofe: The lhil'JillJU/fJ f?e OCCbrS for @1II1i'J and D/lJllx. Thus,
Relhin =/Jllfx/o" ~.~~~If = /2/100 The flow is 7urbv/cnt over the
rfln98 01 pqrameler.s considered.
Hence, from Ef. (8.35)
J- == -2,0Ioq[.-f- + 2..5 /] or with £.:: /.5 x/o-'m
Vf ,3.7D fie 'ff
-L = -2 0 100 [If.o.sx/o-
7
+ 2.5 J 1 (fl.)
1ff ' J D Re vr]
ThvsJ
for .5 xIO-~~ :s. Q ~ .so x10-'1f-3 and w/lh D=: o.0 2.5 rn I
O. OSO m J or 0.075 mJ
de fermine Re frofll £1. (3); f from £'1' ('S'~
B-.3Lf

8,1f2" j (conll)
and hI... frOfh El(.O). These resvlls are calculafed and plotted
below. (h. -::hJ.. (QJ D), See Pro9ram P8# '1-2 sh()lIIn be/ow.
100 cIs
110 open "prn" for output as #1
120 print#l, "**********************************************"
130 print#l, "** This program calculates the head loss **"
1~0 print#l, "** as a function of flowrate and pipe dia- **"
150 print#l, "** meter, using an iterative scheme to **"
160 print#l, "** determine the friction factor from the **"
170 print#l, "** Colebrook foxTrlula. **"
180 print#l, "**********************************************"
200 D = 0
210 for i = 1 to 3
220 D = D + 0.025
230 rr = 1.5E-6/D
2~0 print#l, " "
250 print#l, using "For D = #.###:!t m with e/D = #.##:!t~~~~l!;D,rr
260 print#l, " Q. m3/s Re f hL. m"
270 Q = 0
280 for j = 1 to 10
285 f = 0.02
290 Q = Q + 5.0E-4
300 Re = 1.94E+6*Q/D
310 fp = f
320 f = 1/(-2.0*10g(rr/3.7+2.51/(Re*fp~0.5) )/log(10) )~2
330 if abs(l - f/fp) > 0.0001 then goto 310
340 h = 0.0826*f*Q~2/D~5
350 print#l, using II #.#:!t#~~~~ #.:!t##~~~~ #.#### #.##:!t~~~~II;Q,Re,f,h
360 next. J
370 next i
**********************************************
** This program calculates the head loss **
** as a function of flowrate and pipe dia- **
** meter, using an iterative scheme to **
** determine the friction factor from the **
** Colebrook formula. **
**********************************************
For D = 0.0250 m with
Q, m3/s Re
5.000E-04 3.880E+04
1.000E-03 7.760E+0~
1.500E-03 1.164E+05
2.000E-03 1.552E+05
2.500E-03 1.940E+05
3.000E-03 2.328E+05
3.500E-03 2.716E+05
4.000E-03 3.104E+05
4.500E-03 3.492E+05
5.000E-03 3.880E+05
e/D = 6.000E-0:,
f hL, m
0.0223 4.718E-02
0.0193 1.629E-01
0.0178 3.384E-Ol
0.0169 5.702E-01
0.0162 8.563E-Ol
0.0157 1.195E+00
0.0153 1.586E+00
0.0150 2.028E+OO
0.0147 2.520E+00
0.0145 3.062E+OO
8-35

8. ifZ"
3
.J 2
1:
enUl
o
.J
C
oc(
III
J:
1
o
Q, m3/s Re
5.000E-0~ 1.9~OE+0~
1.000E-03 3.880E+0~
1.500E-03 5.820E+0~
2.000E-03 7.760E+0~
2.500E-03 9.700E+0~
3.000E-03 1.16~E+05
3.500E-03 1.358E+05
~.000E-03 1.552E+05
~.500E-03 1.7~6E+05
5.000E-03 1.9~OE+05
Q. m3/s Re
5.000E-0~ 1.293E+0~
1.000E-03 2.587E+0~
1.500E-03 3.880E+0~
2.000E-03
2.500E-03
3.000E-03
3.500E-03
~.OOOE-03
4:.500E-03
5.000E-03
5.173E+0~
6.~67E+O~
7.760E+0~
9.053E+0~
1.035E+05
1.16~E+05
1. 293E+05
e/D = 3.000E-05
f hL. m
0.0261 1.727E-03
0.0222 5.873E-03
0.0203 1.208E-02
0.0191 2.021E-02
0.0183 3.017E-02
0.0176 ~.189E-02
0.0171 5.532E-02
0.0167 7.0~2E-02
0.0163 8.717E-02
0.0160 1.055E-01
e/D = 2.000E-05
f hL. m
0.0289 2.516E-0~
0.02~~ 8.~83E-0~
0.0222 1.738E-03
0.0208 2.897E-03
0.0198 ~.313E-03
0.0191 5.975E-03
0.0185 7.876E-03
0.0180 1.001E-02
0.0176 1.237E-02
0.0172 1.~95E-02
PROBLEM PSIlI"f-2.
D=O,0.50m
D= 0.075 m
-1~__~____~______~__~____________________________~~
o 1 2 3 5
''10-
3
FLOW RATE. Q (M3/S)
NoIe fhe sfl'on9 dependence of hI. on D.
g-36

g, if] J
8.43 Air at standard temperature and pressure flows at a rate
of 7.0 cfs through a horizontal, galvanized iron duct that has a
rectangular cross-sectional shape of 12 in. by 6 in. Estimate the
pressure drop per 200 ft of duct.
For a horjzontq/ dvcf Af::: 'thL = fi "ipV2.J
where V=~
or _ . 7 #3 _ fi _ Vf)h
V- (J~in.)(6in,)( I n~ 2.) -I.If.o sand Re" - T
'fh /If'fIl1,
WI D .= !td = #- (O.5f/1.) = 0.66711
?J P (2+/)fI
Thus, .f:/.)I. !)
Re = (I'/-. 0 oS [O.
667ffJ. = &.9SXIOif
h /.57 XIO'f.p.2.
E. O,QOQStt
AlsoJ for 90 /I/QlJiled /ron E~ 0 ..0005 fI) or 25h::: 0.6671/ :: 0.000750
From ~'9' 8.20 we ohtain f= 0.02.2.7
Thus) from q. (/) w,fh i =20()/fJ
AP = (0. 022.7)~:6~ f (2..38XIO-.3 s~!!J) (III-.o!f)2. == I.Sq Ita =0.0 I/O pSI
8-37
(/)

8,JfJf
~.44 Water flows at a rate of 2.0 fels in an old, rusty 6-in.-
diameter pipe that has a relative roughness of 0.010. It is pro-
posed that by inserting a smooth plastic liner with an inside di-
ameter of 5 in. into the old pipe as shown in Fig. PS.44, the
pressure drop per mile can be reduced. Is it true that the lined
pipe can carry the required 2.0 ft3
/s at a lower pressure drop
than in the old pipe? Support your answer with appropriate cal-
culations.
Old New
ill FIGURE PS.44
yes", fhe /Jew11;&n4S 41()wer At
=
8-38

8,Jf6
8.:46 To conserve water and energy. a "flow
reducer" is installed in the shower head as shown
in Fig. P8A6. If the pressure at point (1) remains
constant and all losses except for that in the "flow
reducer" are neglected, determine the value of
the loss coefficient (based on the velocity in the
pipe) of the "flow reducer" if its presence is to
reduce the flowrate by a factor of 2. Neglect grav-
ity.
8-3q
Flow reducer washer
FIGURE P8.46
50 holes of
diameter 0.05 in.

j 8.lf7
SA7 Water flows at a rate of 0.040 m3
js in a 0.12-m-
diameter pipe that contains a sudden contraction to a 0.06-m-
diameter pipe. Determine the pressure drop across the contrac-
tion section. How much of this pressure difference is due to
losses and how much is due to kinetic energy changes?
(
(
D, =0.12m
1
• - - 3
(0 61=0 O'ff!!..
D:z. =0.06 f17
(
'(2.)
, Sf
LJJ +v,:z. + ~:lo Vl +l/ .{2. J.. _ "::t ( )
¥ 2j z, = !If +:lj- +%2. nL :;.'} wI/ere z, -~;.. I
and",3 m3
~ ~* =;(00;;),,- ::3.5'f-P- J v.. =!=~O'f'Tf = IJM~'I- • m If. O.o6m
Thus) wifh -t==(~~)2.-( ~.~::)2= 0.25 we Ob/flin from Fie;. 8.30
KJ.= O. 'fO
Hence, from £~, 0)
{J,-(J:z = -tf{KLIS."-+ ~'--V,S] =f(999 ~)[o.'10 (11f.1 :f"+(11f.Iq.l-(3.5'ff~
or .l N -3 N
fJl-/~::: 3Q.7 X/O fi2- + 93,OX/u- fii :: 133 kPQ
This represents a 3 q. 7 /rPo drop from losses Qnd a 93,0 kPa drojJ
rif)e fo on incretlse in kinelic ener9,.
8-'10

8. Jfq I
t -::=:=::------ ----::::=::-
- 3 f t - 2 ft at t = 0 i-3ft-
!--:j:j:::::::---';;';;-::;::::-'~---25 f t -
8•.4Q At time t = 0 the level of water in tank
A shown in Fig. P8AQ is 2 ft above that in tank
B. Plot the elevation of the water in tank A as a
function of time until the free surfaces in both
tanks are at the same elevation. Assume qua-
sisteady conditions-that is. the steady pipe flow
equations are assumed valid at any time, even
though the flowrate does change (slowly) in time.
Neglect minor losses. Note: Verify and use the
fact that the flow is laminar.
~__B____~~~~c=====~__~AL-~
"'-o.l-in.-diameter, gal. iron
FIGURE P8.4Q
Thus, Er.s, (J.) and (If) 9ive
2~/-ho = 1f!:-iIi= 3;'$1V,or bl uS;/)9 Ef#!3)
2 Z -h = _(~)2. 32ft J dz/, 0 D 'I D'J. at
Lef FE z,-iJ:- so fhed ~ =11- and £q. (05) bec()mes
(
'Dr)2 3 2jJi dF
2. r = - D r D'J. (fF
(con)/J
8-1.f{

or 0( dE +F==0 where ex. = 16# I. (Dr )2.(ff J lD2. D
ThvsJ fdF .r()( F = -j dt or 0( In F=- t + C) where C:::consTfJlJf
f/ence, t. .
F C -(~) TI- f . ho C -(tlo<) 'lh lh . .J. I _J-/-
= e I no IS} Z/ - T::: e WI r. /, e Inn/tJ CO'KJI1/~n
z/ =ho when t =0 J or C::: ~
Thu.s z -.!:k == l1.!. e-O/tX)
J I 2. :2-
or
2/:::: if-[J+e-(i/o()] /Vole; As l:'-"OlJJ
Zl-+~
For the cond/liof/.S 9iven} ho:: 2 ff tJnd
/6(2.3/fX/0
5
~)(ZSN)(~N )2. 5
«== {62.'f~)(~r1r (WD =2.801</0.s
Hence~ t
Z - J +e-(;..flX/O:S) h ff d 1
I - ) were Zt.tV an r; ,..,..s
This resu/f is ploNed be/ow. (Nofe: lim z, = Iff)
l"'Ob
0.5 +--+--------1f-------I---+--t----t---t-----r---t----.
O+---~--~~--_+--~--~--~--+_--~~
o 20qooo 40QOOO 60QOOO BOqOOO 1J
OOQOOO
t, sec

8.50~
8.50* Repeat Problem 8.4Q if the pipe diam-
eter is changed to 0.1 ft rather than 0.1 in. Note:
The flow may not be laminar for this case.
(con'/J
8-tf3
f
2 ft at t = 0
l
O.l-ft-diameter, gal. iron
FIGURE P8.·4Q
(I)
(2.)
(0. 25Lflj)2.
( it) or Zt=(i+O.00382)fI
2 32.:z.S.1.
75
£:::: 0.005 {f :; o. os
o.lff
(see Table B.O

Hence, _~
I J [ -3 3.0'f-X}O ] 11vr = -2.0 09 1.35·xIO T vYf J where V-v .s
Solve (t:e. ifJfe9rtlle) £r. (ff) for Z, $2(f)J siarfin9 wdh in/fiat
cOl1ri;lion Zt::: 2 ff 41 t::: o. Ohfa/n f {which is a {()ncl/oll of i
becavse V= - t'/oo ~ (Ef. (3)) ,:r Q runc hon () rt) from £'1' (s).
ProqrQm P8NEO shown he/ow WQS /}Sed 10 oldfll" fhe results.
100 cls
120 print #1, " * * * * * * * * ** * * * * * ** * * * * * * * * * * ** * * ** * ** * * * * * * * * * * * "
130 print#l, "** This program calculates the water depth **"
1""0 print#l, "** as a function of time. The friction **"
150 print#l, "** factor is obtained by itteration from **"
160 print#l, "** the Colebrook formula. **"
170 print#l, "***********************************************"
200 print#l, " "
210 t = 0
220 VP = 0
230 rr = 0.005
2""0 dz = 0.05
250 print#l, " z, ft dz/dt, ft/s f t, sIt
260 for i = 1 to 20
270 z = 2 - (i-1)*dz
280 f = 0.02
300 dzdt = -7.98E-""*(z - 1)AO.5/f AO.5
310 V = -900*dzdt
330 Re = 8.26E+3*V
3""0 fp = f
350 f = 1/(-2.0*log(rr/3.7 + 2.51/(Re*fp
A
O.5))/log(10) )A2
360 if abs(l - f/fp) > 0.001 then goto 3""0
380 if abs(l - VP/V) < 0.01 then goto ""00
385 VP = V
390 goto 300
""00 t = t - dz/dzdt
'*05 zn = z - dz
(5)
""10 print#l, using" #.#### +##.##### #.#### +#.##AAAA";zn,dzdt,f,t.
""20 next. i

8.sot I (con'l)
2.0
1.9
1.8
1.7
1.6
;:
1.5~
N
1.4
1.3
1.2
1.1
1.0
***********************************************
** This program calculates the water depth **
** as a function of time. The friction **
** factor is obtained by itteration from **
** the Colebrook formula. **
***********************************************
'" ft dz/dt, ft/s f t, s'-' ,
1.9500 -0.00440 0.0329 +1.14E+01
1.9000 -0.00429 0.0329 +2.30E+01
1.8500 -0.00417 0.0330 +3.50E+01
1.8000 -0.00405 0.0331 +4.74E+01
1.7500 -0.00392 0.0332 +6.01E+01
1.7000 -0.00379 0.0333 +7.33E+01
1.6500 -0.00366 0.0334 +8.70E+01
1. 6000 -0.00352 0.0335 +1.01E+02
1. 5500 -0.00337 0.0336 +1.16E+02
1.5000 -0.00322 0.0337 +1.32E+02
1. 4500 -0.00307 0.0339 +1.48E+02
1.4000 -0.00290 0.0341 +1.65E+02
1.3500 -0.00273 0.0343 +1.83E+02
1.3000 -0.00254 0.0345 +2.03E+02
1. 2500 -0.00234, 0.0348 +2.24,E+02
1.2000 -0.00213 0.0352 +2.4,8E+02
1.1500 -0.00189 0.0358 +2.74E+02
1.1000 -0.00162 0.0365 +3.05E+02
1.0500 -0.00130 0.0377 +3.44E+02
1.0000 -0.00089 0.0402 +4.00E+02
---------o 100 200 300 400
t, sec
8-t;-5

8.5/
8..5/ As shown in Fig. P8.5/ , water flows from one tank
to another through a short pipe whose length is n times the
pipe diameter. Head losses occur in the pipe and at the en-
trance and exit. (See Video VS.4.) Determine the maximum
value of 11 if the major loss is to be no more than 10% of
the minor loss and the friction factor is 0.02.
If hL .=lo'ZhL..~ ) thenhJ4Jor mlfJor
• FIGURE P8.51
1 v'" - v2. 1- _ £KJ.
10 f D 2., - '£1 1j or D- 10 f
where 2: K =: K 1 + k: =O. e+I :: I. eL IJ. enrf'4nce 'L ex/I
-rhvsj
wifh f:: 0.02 and 1:n/) £r. {/J hecomes
nD /.8
- -D /0 (0,02..)
Of'
n - q
--
(I)

8.52 J
8.S 2. Gasoline flows in a smooth pipe of 40-
mm diameter at a rate of 0.001 m3/s. If it were
possible to prevent turbulence from occurring,
what would be the ratio of the head loss for the
actual turbulent flow compared to that if it were
laminar flow?
Lei ( )/ denote fhe turbulent flo/IJ and {~ the /lJ/fJlfJor flolll,
- /, r 1. yt
Thv.5J hif - ft D:1-9 olld h/.J ::: ~ D 21 (I)
where V= I; _ 1/ _ Q _ O.OO/~3 m
VI: - VJ - 7f - 17'( )2. = 0, 79{-:s
7j O.()lfm
Fro", Table I. 6 f =6eo~ QndJ1. =3./X/O-
1I
~~ So thaf
Re = eVD = (680~)(O.796~)(o.O'fm) = 6.98X/O'f
p. .3 / x/0-'1- !!.:!.• 1112-
lielJce} ,from h9. 8.20) for a smoolh pipe ft:: 0.0/92
while for lominar flow £'1 = 4! ~ "6 '16~ ~ = 9.lbX/ O-
1f
"e • 8x/O
Thus) from Eft fl)
hl.t _ ft _ O.Olq:;.
- - - - 'I- =2/.0
hLJ ~ 9.1tXIO-

8.53 l
8.54-1
8.53 A 3-ft-diameter duct is used to carry
ventilating air into a vehicular tunnel at a rate of
9000 ft3/min. Tests show that the pressure drop
is 1.5 in. of water per 1500 ft of duct. What is the
approximate size of the equivalent roughness of
the surface of the duct?
(/)
8.54 Natural gas (p = 0.0044 slugs/ft3 and
v = 5.2 X 10-5
ft2/s) is pumped through a hor-
izontal 6-in.-diameter cast-iron pipe at a rate of
800 lb/hr. If the pressure at section (1) is 50 psi
(abs), determine the pressure at section (2) 8 mi
downstream if the flow is assumed incompressi-
ble. Is the incompressible assumption reasona-
ble? Explain.
2. 2. lr11+~ +~;::- '1+¥j+Z3. +fD~ J where Z/~~Z and ~=-~
/II. J!; Ib 0 2:1.2 i.!L fl3
so} 'tQ==800 LIIu:-)=O.Z2Z-- orQ= ' . s =/,57-
sr 3600s ;s ~ (32.2~)('f.'hlo-3.f!J!1)
Thvs V=!i.. == 1.57 !f3 = 800 li {f3
} II .1l(.k.-)2. ..s
~. ,,-if lt l6 )
Wlfh Re c: e'L~ = V
ll
D = (8.00$ )-~7f). = 7.69x/o'l-and (Fr()/J'J Tahle 8, t)
.r- 5.2X/O
£ o.oooasff 'I' f oS ..,-1.. t' r ' , I
75::: (6/12ff) -,:Qool7 weOl) tlln == o.o2lf5 I "vsJ rro/)'J t='!,(/./
j) ;: 1J - rJ...L D V2.= 50 Ib - (O.02~5) (Smi)(S).8ofj,.) La,. IfXI0.3~)(8.00!:t)27/2. TI D 2 r Ii?- o.s ff 2. C • 1f3 oS
::: SOfJsi -:;.qI ~ = (so -1...o2)psi or -A.;::- ~8. 0 psi
Nole: pj-~ - 2.02psi =0 OLLOL1. a Lf..O 7. cL/lnoe in pressure.
- I, - SOfJS;' .,. T.J n "/
Since f ==pRTJ
w/fh T essenlia/ly co,;s/Q/Jf.J IJ small choIJge ill I
qives (J SIIJ4// ChIJb96 il) p. TiJtI.t Ihe /l')cfJlIJ,ore.ss//;/e 4.IsfJPJ,tiofJ is valid.

8.55· J 8.55* Water flows in a 20-mm-diameter gal-
vanized iron pipe with average velocities between
0.01 and 10.0 m/s. Plot the head loss per meter
of pipe length over this velocity range.
.l. v~
h z:: f D '),.9 J or wilh i =/m qnd [) == O.02.0 m
hi ;= ftt.:,.",)2~~8/~ J Of' hi = 2.SS f V2.
I//so} Re ::: X,f.: V (0._0:117; or Re =1.79x/Oil- V
r /.I2X/0 4L }
FOr this pipe) *:: O~':::"III = 7.5 )(/0-
3
(see Table 8. I ) $0 Ihat
fhe Colebrook /ortrJlJ/o hecomes (£f{. 8. 3S)
.J... ~ [£, 2.£ ] I r -3 2. 5 ]
vr ::: -2.0 °9 i% + Revt or w:: -2.0 logJ!.03XIO + Pte1fT
Nole: II Re ~ 2100 the (low is lominqr qnd f::: ~
If I?e ~ '1000 Ihe flow is f{)rhlJlenl and f is ohftlined from $"r.t:J).
For 2/00< fie <'fOoo rI;..s ngf clear which value of f 10 f)se, For
simplici& a.s.svllJe lamiIJtJr flow is tr)oiIJlainerl up fo Re ='10()().
From £0, (:J.) V= I ¥~oooS' == 0.22.3 ~ when /?e -::'1000.
],) .7 x/a
Thil~ {or 0.0/ ~ V~ O.223~ obf4in flo a/ld rfrollt £tts.(1.)PlJdC'I)
(I)
(2.)
(.3)
(¢)
and hL from £1'(/)' FOr O.~23< v~ /0-1 06111il) Re and f fro/l/ Ef.r. (2.) and('I).
100
110
120
130
14,0
150
160
170
200
210
220
230
300
310
320
330
34,0
350
360
4,00
4,10
4,20
The lIalue.s of h. 4re CQ/c()/Qfed and plolled he/ow (see pro9l'fJ/YJ
P8#55).
cls
open "prn" for output as #1
print.# 1, II **************************** *********** ********* If
print#l, "** This program calculates the head loss in **"
print.#l, "** pipe as a function of velocity. The **"
print#l, "** friction factor is obtained by the Cole- **"
print#l, "** formula. **"
print# 1, "********************************* *************:** ff
rr == 0.0075
V = 0.005
print#=l, If "
print#l, " VI m/s Re f hL, m"
V == 2*V
Re = 1.79E+4,*V
f = 64,/Re
if Re < 4,000 then goto 4,00
fp == f
f == 1/(-2.0*log(rr/3.7+2.51/(Re*fp~0.5) )/log(10) )A2
if abs(l - f/fp) > 0.0001 then goto 34,0
h == 2.55*f*V~2
print#l, using" ###.###
if V < 10 then goto 300
#.###A~A~
(con'l)
#.### #. ### A~. A ~"; V, Re, f, h

E
.:..c
************************************************
** This program calculates the head loss in **
** pipe as a function of velocity. The **
** friction factor is obtained by the Cole- **
** formula. **
************************************************
I
1.E+OO ~~§ill~!~II~~)lfIJII
/
1.E-01 §'III~tml/~mll - - Turbulent flow
I) - - ~ Laminar flow
I"
1.E-04 .4,::1"_L......I...J..~JJ.l.I-.....J......J....J...I...J..j.""'""+_................................."'"'i
0.01 0.1 1 10
V, m/s
8- SO

8.56 I 8.56 A fluid flows through a smooth horizon-
tal 2-m-long tube of diameter 2 mm with an av-
erage velocity of 2.1 m/s. Determine the head
loss and the pressure drop if the fluid is (a) air,
(b) water, or (c) mercury.
R,;c
, where hi -::: {7i"1 J Z, == &2 J and ~=:-V2.
or hi ==22S{ m 0)
tt) air l.¥6x/o-s 287 lQl11in.r 61/- 0 3 ro 21{e= .2.2 ~. /2.0
b) wafer 1./210.'0-
6
3150 ..,turbfJlem 0.0'1-0'1- 9.09 q800
c) mere-vry 1./05 )./0-
7
36,5"00 furbulent 0.021.0 '1:95 /33,000
602-
B,9/X/o'"
6.58Xl05
8.57 I 8.57 Air at standard temperature and pres-
sure flows through a horizontal 2 ft by 1.3 ft rec-
tangular galvanized iron duct with a flowrate of
8.2 cfs. Determine the pressure drop in inches of
water per 200-ft length of duct.
Jr" _(J~ ~2. .l~2
~ +1-1 +&, - T +~I +Zz. ~ f lJ
h
Z-1 ' where z/= z~ (lI1d ~ =~
Also D:;!M.. = ~(2f1)(1.3fl) =I. 576 ff
and) 11 P 3 z[:Zf/ +1.3f.1.]
1£
V= (J) = 8.20$ =3/511
1f (2f1)(1.3t/) . S
ThtJsJ Ir1'. '"f ~ f pV', where {or 9o/r/Qniled iron Eo = 0.0005 {f (robleS.I)
Hence, -t.. '" f~~:O:/' '" o. f)00317 and Reh~ '{.P,I = (I.S7tffJ!:;:/J"'31600n ' , /.S7X/O.$ J
so from Fit). a:to, f=0,01-5
Thus / )(2.00If , I ( -3 §!J!s) I 11)2 . JbJ ft -f:J.. ==lO.02.S 1.57&F1J2: 2.38X/O ~ ,3,/S.$ =O,037'ffP
or wdh fJ,-f:z. :::: ON;.o hJ
.IJL
h= f1-;ta = 0,::':'1-/12. == 6.00XIO-~ff =0.00720/17. of wafer
;,f) f1~
~-51

8.58 I 8.58 Air flows through a rectangular galvanized iron duct
of size 0.30 m by 0.15 m at a rate of 0.068 m3
/s. Determine
the head loss in 12 m of this duct.
h ::::r1 r where D. ::=.!J 0:: 4(O.3m)(o./sln) ::= 0.2 m
L Dh 2.7 J ~ r- 2.[a.3m to.ISm]
Clnd m~ . JIL
V~(Q = O.068-r ::: /5/1l1.. I//s D = Vl>h.::: (I.5I oS) (o.1.m) ::= 20700
7f (0.3m)(O,15ni) . oS oJ neh T /.Jf6 X/O-.$".oe J
ond from Table a.1J .s
i:. ::= o,lsx/o-.3m = 7.Sx/o-~ lienee.! from Fi,. 8,~O F=O,027
Dh 0.21'11
so fhat
h ::::(0.027)(12.111 .,
L O.2m)
{/.5/f-i
2(q.a/~
= O./88m
6.5 q I 8. S9 Air at standard conditions flows through
a horizontall ft by 1.5 ft rectangular wooden duct
at a rate of 5000 ft3/min. Determine the head loss,
pressure drop, and power supplied by the fan to
overcome the flow resistance in 500 ft of the duct.
h -f1. r h V-.!i - (5000fln)(-1!W-) =55.6 lisT
L - Dh 2.9- J 141 ere - II - (J ff) (1.5 fI)
!tf1 _ {l (I fO(I.SrI) _ ,
and Dh == P - 2[1Ft t/.5ff] - /.2 ff
II/so} D VDh (£5.6!j) (/.2. f-I) n .5 and f,..,,'IIJ -r..4"~ 8,/
neh::z - = ,.t.l~ =7.2.5 X/O Tt v. '~O't7
11 1.57 x10- .lj-
£ ~ 0,0006 f-I to 0,,·003 If. Use al') ''avera1c'J £.::= 0.001 8 fI so -!hat
~ 0.0018 ff Ii r L' 8"0 fDh = /.2.1-1 =0.0015 I flIJ-SJ Trom r'9"~ ::: o.o'-zJ or
h ( EOO ff _, (5S.61Jf ff
'i.= (0,022) /,2(f ) .2(3.2.2£) = 'f~O
For this horitonial pipe IJ!- +!i2
+z,:: 11+If+Z:;. f hL} where :Z,;:2:;.
and V,;: v,. .
Thus} fl-P:J.::: O'hL :::::(7.65X/O-2.. tJp)(l/-'f()f/) == 33.7 ~~%.;: O.23'1-pSi
P =rQ hL,: 6( (,,-f:t ) ,:(sooo!/:~~;~b)(33.7 It>)=(2810 ~1r55~~j]
or
p =- s.// hI'
8-52

8.60 I
8. 60 When the valve is closed the pressure
throughout the horizontal pipe shown in Fig. P8.60
is 400 kPa, and the water level in the dosed surge
chamber is h = 0.4 m. If the valve is fully opened
and the pressure at point (1) remains 400 kPa,
determine the new level of the water in the surge
chamber. Assume the friction factor is f = 0.02
and the fittings are threaded fittings.
Closed
f,....--Surge chamber
Angle -.-
valve 0.5 m T D = 0.02 m
5? ! 1 ~Tee ~
(2)
~C:::r=:::::::::::=::::!~.J=======~====::{ (1)
• .IC,3) t IVfH!-'- 5 m--f+·---8 m' - - - - I .
FIGURE P8.60
'f+¥i+z,::; 1!j-+1i+z.z+(fifZKL)¥i J wherez,=Z:l.J ~::~':l:::'O
~ 2 I. V"
Thus) If = ~ f(flS +L~)'1 wilh v:: ~ .
Wi.Jh ~:: 2 for 4n an9/e valve ond K::: o.q for The fee (see Table 8.2.)
we obtain
'1-00 Jsjf. == I [I +(O.02)(-(B+S)/TJ +2.+0.91 V2.
9.80.1$ 2 ('1.81!jsJ O.021h -; j
or
V== 6,88%
Tho.!) fJ.3 is de1ermined from
v,~ 4s y;- J rJf+i'g +2,::: "'f -I- ').1 +Z.3 +(fD +~ K):z.g J where 2,==Z.3 ond ~=O
f)Jso} ~ =V HenceJ
'1 == .t;-f (J +fl +£Kt.)¥j J where j,== em and K::o
Thus) .! y-'J. l 2-
(13 = II -(J +fD )21°':: f,-{/+f"5)teV
== '-foo kP - (I +-(O~02.)(-a~: )~ (qqq~) (6.89lJl/'= 'fOO/rP-2.13X,cf.!. m m m
orfJ;a::: /e7 kp
Thus} LJ~ = '1-00 kRJ with the I/q/ve closed wheh h:::: O·lf-m .
,-. I'IJlr
Q(1ri ;;.3 ::::187 /{pq wilh the valve open qnri h=:hD "asM:::: /J74.s.s of air ill surge chamber -::: f ¥;::~Ql)sfalJt, I-VA.....~
where ¥ '= II (o.5m-h) ond fJ::: pl?~or p=if O.5m ~
Thus, wdh ( )c rienoli1J9 fhe closed valve condition) ~ (a)
Pc ~ ==fo"Va J or 41J(O.5 -0.'1-) == RII~ /I(o.5-h,) IIssume ~~To
Or n c D xk
0.11(;::: (o,s-ho)lfJ J where pc:::: 'foo/r~-ohc ='Iool<fJ4 -?9~(o.'f/IJ)
and (J,=:J87kPa-tho =- 187/{P4 -r;.eot/fh,= (lB7-9.sho)kPq =39tkPQ
Th()sJ 0.1 (39t) ==(o.5-ho)(l87-9.8ho)J or hD:::: 0.285 m
g-53

8.61
8.61 What horsepower is added to water to pump it verti-
cally through a 200-ft-long, I.O-in.-diameter drawn tubing at a
rate of 0.060 ft3
/ s if the pressures at the inlet and outlet are the
same?
8.62
8.62 Water flows from a lake as is shown in
Fig. P8. 62 at a rate of 4.0 cfs. Is the device inside
the building a pump or a turbine? Explain and
determine the horsepower of the device. Neglect
all minor losses and assume the friction factor is
0.025.
(Il
525 ft
. ('J.)
0)

8.63
8.63 Repeat Problem 8.62. if the flowrate is
1.0 cfs.
8-55
525 ft
495 ft
300-fi-long.
O.4-fl-diameter

8.6 If
8.64 At a ski resort water at 40 of is pumped through a
3-in.-diameter, 2000-ft-Iong steel pipe from a. pond at an ele-
vation of 4286 ft to a snow-making machine at an elevation of
4623 ft at a rate of 0.26 ft3
/s. If it is necessary to maintain a
pressure of 180 psi at the snow-making machine, detennine the
horsepower added to the water by the pump. Neglect minor
losses.
8-56

8,65 I
8.65 Water flows through the screen in the
pipe shown in Fig. P8.65 as indicated. Determine
the loss coefficient for the screen.
~-S7
t -.-1 6 in. "-
...l.--il-t , "'SG = 3.2

8,66
8.66 Water flows steadily through the 0.75-in. diameter
galvanized iron pipe system shown in Video VS.6 and Fig.
PS.66 at a rate of 0.020 cfs. Your boss suggests that friction
losses in the straight pipe sections are negligible compared
to losses in the threaded elbows and fittings of the system.
Do you agree or disagree with your boss? Support your an-
swer with appropriate calculations.
6 in. length
elbows
4 in. length
Closed ball
J. v2.
Major loss ~ { If q. where
j::(6+6+'t+j)in. ==/7//I.~ D:::O,7S/n. • FIG U REP 8 .6~.
anrJ. 0, 0.2 ~ f-l
V=~:. 1f(0. 7S//;l.)''f.1, :. 6,$'2 :s
Th{)sJ wi1h H ( O.7S' r )
R =~ == 6,5)'--;s ~it ::: 3.37X10'1- and
e P J. 2/ x. /0-.5 P
t. 0, ooos ff -3 ( ..,.. hI) 11' ( C". 8 )
D:: (0.7,s;4) = 8x/o see fa Ie 8,1 we OD70ln see T'9. .'2.0
-rr .1 JC' _ /7 ill. JC' V'-
f :::; 0.038 .so fhai f D~? - 0.038 0.7.5/11. ~p = O.·~61 J:j (I)
Al.so) V" V" V2.
Minor loss = r:KJ..'! :::[2 (J.s), + 2 of 0,)05' ] Zj ::: 5,1.5' ~ (:l.)
o /iee / / 'L A ( ~. 2-
90 elbow rcdvcer wHo :1. =:" !!'O/~) ::0,6'f
7f; O.iSln·
(see h1' 8.26)
-rhvs from £'(.s. (I) and (j.) :J ,.
major loss o. 861 fimino!' loss - s,/s ~ == 0./67 :: It,? %'
:J-f
Prob(),bly dl.saZ/'Be wll/J boss hec4v.re pipe frict/(J1l /S ab()!)!
17% of other losses.
8-58

8.67 I
X.C)7 Because of a worn-out washer in a kitchen sink faucet.
water drips at a steady rate even through the faucet is "turned
off." Readings from a water meter of the type shown in Yidl'O
'S.7 indicate that during a one-week time period when the
homeowners were away. 200 gallons of water dripped from the
faucet. (a) If the pressure within the 0.50-in-diameter pipe is
50 psi. determine the loss coefficient for the leaky faucet. (b)
What length of the pipe would be needed to produce a head
loss equivalent to the leaky faucet?
(a)
f/ence;
(50 ffi.. )(IIf'f il/)
10." VIi::: IC
62.'1 ~ i
(O,032.1fUl"
2. (.32,2. /t)-
or I
K ::: 7. 01 x/O
'L.
(b)
(0, O.3:J. If #-)(W-li) . .
-5 ffl· ;' //2 ..:::: 2-/()() Sf) the flow Is lam/liarI,l/XIO _
S'
g-sq

8.68 I
S.6X Assume a car's exhaust system can be approximated
as 14 ft of 0.1 25-ft-diameter cast-iron pipe with the equivalent
of six 90° flanged elbows and a muffler. (See Vilko VB.S.) The
muffler acts as a resistor with a loss coefficient of KL = 8.5.
Determine the pressure at the beginning of the exhaust system
if the flowrate is 0.10 cfs, the temperature is 250 of, and the
exhaust has the same properties as air.
8-60
d'0)

8.6Q I 8.69 Air is to flow through a smooth hori-
zontal rectangular duct at a rate of 100 m3
/s with
a pressure drop of not more than 40 mm of water
per 50 m of duct. If the aspect ratio (width to
height) is 3 to 1, determine the size of the duct.
Reh
Triol and error so/lJfi()n of Efts. (2.~ (3)) (If) for fJ
ReJ and V:
(I)
(2.)
(3)
(If)
/lss{)me f~ 0,02 Sf) thai o.01- ~ o. 017:J.h
S
or h :::/.03/IJ. ;:rom Ef. (:;~
Reh::: .3·t:-;,q6=3.32 x/ot which fro/fJ h1' a2.o~ive.s F=O.009t :10.02
Assume f =0.0096 which ,ives h=0, 890m, Thll~ Reh :: 3. etr-xlr!
or f;:; 0,0093 r/ 0,009t
/lsslJme f 0:;: 0,0093 or h:::: O.8fPfm. 7IJIIS ReI.. = 3.87x/0
6
of f::: 0.009..3) .) n J
which ofjrees ~Vllh fhe (JsslJmed va/uB.
Thus, Ihe dflci is h::0.88'1 m /;, 3h::: 2. £5m iIJ s/j.e.
8-61

8,70
8.70 Repeat Problem 3.14- if all head losses are included.
The pipes are I-in. copper pipes with regular flanged fittings.
The faucets are globe valves.
Bill from 74h/e 8.1, -5-::
olJd .s;"ce' fi
R =.lJJL::: (ik.f.I) (2.0$ )
e l' /.2.1 x10-5
IF"
f= 0.0/65
• FIGURE P3.1"'"
.5x/o-'ff -5
--:-:---:--- = 6.0 x/0
(JU)
=/.,.38 X 10
5
if {oliows from hr. 8. 20 IhaT
Helice.! from £y- {/)J If::: 7f1.,. 6. ')../(8/1-(0.0/0$) +11)1-1 =83,9 II
or 11 ::::(83 • 9 (-I) (t2.'f~)( Jlf'!!:'~)=3~,¥-,.ri
Now; O.fsllnJe I, I'ema/IIS ihe S4me re14rdle.s-s wh/c.h fallcel is ojJen.
This is e.s.relrlra/f irlle if fhe Sf)I'I/Y line /e4dilJf IP the pifJes ;.s re/41iflefy
/01'98 compared 10 the fJj;es IIJ Ihe h{)lIce.

S.70 1(con/f)
(0) Open h4se/lJenf favcef:
v,:z. .A:J ~2. ). 2-
Ij!- -I-;:jof %, :: if +~ +Z.3 +(f D t z:KJ.)zi )
AJ :: 0 v,:-~ :: V Ie-::: 93.9ft i:::511 IJl1d(f".J J .) (J0 .J ~
z::K :: ~o' t KL fjlohe ::: (),3 +10 :;r /0.3
sl'l1w V6/1/8
ThvsJ
V2.
83.9f/" -sfl +(ftZJ J-/o,3);.(31.2¥5
Of' 2.
5730 = (60 f f /0. 3)V1
whel'e v~ 1j OJ
£ -5
II/so 1)::: 6.0x10 ana
n DV (f,:fI) V 3 . L (I
"e::: T = 1 2/1./0-5it = t.89xlD V wnere v~:s (2)
• sl.
TrifJ/ IJlJd erl'or .so/t/tion: IIssume f:: IJ, OldS as be/()f!e.
Thvsj
from Ef, II)
,..------
V= .5 730) = 22. s!f QIJJ frolJ1 £o.(~)
(60)(0.0/&S. +/0.3 J
Re =6,8'1 x/rl (22.~) :: /..5.5 x/oS so fhqf from Fit;. fl. 20;
f:: o.aIDS, This afre8s with tlJ~ (J.r~vmed va/de ~
Hencej V=22. ~ {i
(h) Open 2nd floo! fflvcel: I/s 4bove (excepl /,elweel1 pOilJ7.s (I) IJlJd (If)J
Zi' -f (f'£ +L /0r;'"f wheN Zi' '" If/II, 1"/fI~ 1f'"83.9't
and
£ KI. ::: ~/ee +~ 90' + ~ 91,;8 =0.2 +0.3'" /0 ~ /(},E
elboll/ vallie
1! 'V' 2.
HenoeJ
19+({(i!) +/0.5)2(31..2.) =83.9) or /f180==(2.29{t/o,s)V (i)
where VNli
/JsstJI1ISf==O.O/7S. Fr()1J1 £r. (3)j V::/Z()# ,so {~omE?(:;..) s
Re -::: /1/7XjO-'. TlJVS.l (rom Fi'9, 8,2-0 w//h f:- b'()X/()-~ f= 0,d17o$
whic;~ 41/'ees Wtlh the (JofS()lIIerJ ralfle.
lIence.l V=17. o!J.
8-63

8.71 I
8.71 Water at 40 of flows through the coils of
the heat exchanger as shown in Fig. P8.71 at a
rate of 0.9 gal/min. Determine the pressure drop
between the inlet and outlet of the horizontal
device.
lienee) {rom Fif}. 8.;1.0
f= 0,0#/
fllld from £f· (/)
r--I'--18 in'-~'I(J)
A------------~.~.---Q
("+-----------..,.,.
(
(
(
J
)
Threaded 180 0
return bend
(2)
• •
' " O.5-in. copper pipe (drawn tubing)
It -flo = ( O.OJfl ( :;t.f/)+/0.5) (l ){/.9~ .sir'P)(!.lf7!t /"
1"-
of
f, -f2. = 96.B~ = O.32SP$;
8-6'-1-

8.72 I
Threaded 1"--30 ft
elbow (2) '
lotft 3-in.-diameter," )h--.....,~--!8.72 Water at 40 of is pumped from a lake as shown in Fig. 7.(1} -I f = 0.02 Pump
PB.72.. What is the maximum flowrate possible without cavita- ~~=~++b===tion occurring? - -- --:-::=:-::;.- ::;:::~--------:-:~:::::;
20 ft
l
FIGURE PS.72.
2. 1. y."~
if+~2fZI = If-I- ~ -I- Z2 -I-(f-o +r;K)~? ' where 2, ~OJ Z~'" Illfl, (I)
/J;:; 1'l.7~(abs)J ~ =0) V2 =VJ
CI/Jd fro/h fahle 8./ f3. =0.12./7 {ffilabs) =/7.52 ~
Thus} with the 9illen f=O.02 we obtai;' from£f{.(I)
( .& )I If in"!') JIL 2-
/1f.7 in'-. cI'f 1fi - 17.52 pt2. =- loff +(O.02( SOH) + I + 1.5+ o~e) V
62 ~ ./1L A ff 2(32.2,tp,
• ffl 1'- S !
where we have (Jsed K;.::: 0, e for the enff'tfnceJ K;.::/.5 for the 90'elbOW
(see 09.8,,.2- tlnd Tahle 8.2.)
Thus) V= 1,/-,116fj so -fhal Qc:llv=*(lifl)2.(JJf.If6~) ~ 0.7/0 if
8-65

8.73 Q,.,
Nozzle tip diameter,., 0.30 in. 0.010 cfs
8.73 The i-in.-diameter hose shown in Fig. PS.73can with-
stand a maximum pressure of 200 psi without rupturing. Deter-
mine the maximum length, e, allowed if the friction factor is
0.022 and the flowrate is 0.010 cfs. Neglect minor losses.
Water
L-j
D ,., 0.50. ;".~A~ 10ft
(I)~I j
LL_~=:[p~um./p FIGURE P8.73
~ +~2. +~I := If+if+Z;. +rl ¥; J where ~ ::oJ $;z.:: /off, f1t =:;'OfJjJSl:
_ _.!!. _ 0.01 ¥ F+ _ g _ 0. 01#3 _ fJ.
fo-0J ~ - II, - f (~fI") = 7.33 s) fIlJ' ~ - 112.- f{~ ft)~ - 20.'1- oS
ThvsJ
klilh 1=0.022 £". (/J hecomes (vsin9 V=VJ)
(I)
(2oolJi!-)(I'f'f~) + f7.33Jj),- _ (2.0.lf.fj)').
62.'1-14"3 2(3:1..2%) - 2(3.2.2~)
(
i ) (7.33 !})2..
,*,/0f/+O.022 0.5 2(. ,.Ii )
rrfl 31. S:J.
or
1::: 1012. ff
q =
Nozzle tip diameter = 0.30 in. 0.010 cfs
8.74 The hose shown in Fig. PS.73 will collapse if the pres- D 0 50 in ~.4{r'
sure within it is lower than 10 psi below atmospheric pressure. Water- =. . ..< ••
Determine the maximum length, L, allowed if the friction factor 10ft
is 0.015 aod the flowra!e is 0.010 cfs. Neglect miDorlosses. .:~ ne;,"..L~~' j
::z. 2 Pump FIGURE P8.73
-If+ vL'+21 = if+¥i+Z.1 +{iii, where It 0:::0, Vi::OJ :2, ::311J (I)
'"I I;' _ 6i 0.01 1/3
_ f1 _
Z2.:::: 0 J fJ~;::; -I01h!- J olJd ~ - If :;: 1l (o.s ~'). - 7.33 s - V,. If n:FI
ThvsJ
wilh r:: 0.0/s £it. (I) becomes
3rf '" (-1Of!.){tIflf·j~) +(1 f{O.OI.5J( L )) (7.33 ~)2.
62.¥ iJh tJ ..rgFI 2. (32.2. :g)TO l'k S
or
L = 8'1.0 fl

8.7S
(l)
~'75 The pump shown in Fig. PS.75 deli'Vers·a. hF.mi of 250 ft
to the water. Determine the power that the pump adds. to the wa-
ter. The difference in elevation of the two ponds is 200 ft.
• FIGURE P8.75
or ~
(667 f +/2.8) V ::3],1-0
/llso Re== iV/J = (1.9~~)V(O.7$tl)
J ).t 2.3'1-X/()-s I~~
or
, Re :::; 6.22 x/o'fV
and {rom Fi9- B.2tJ:
f
8-66
v
Pipe length =500 ft
Pipe diameter = 0.75 ft
Pipe roughne55 =0

8,76
8.76 As shown in Video V8.6 and Fig. P8.76, water
"bubbles up" 3 in. above the exit of the vertical pipe attached
to three horizontal pipe segments. The total length of the
O.75-in.-diameter galvanized iron pipe between point (1) and
the exit is 21 inches. Determine the pressure needed at point
(1) to produce this flow.
• FIG U REP B. 76
~ +~2 +Z -h == b +~2. +Z2-
~ 2j l : l It ~j
where 2, =0 J 12.-::0) ~ ~o Thlls)
£!- h~:z.
( I) r :;: Z:z. + ~ - f? Where V,::: ~ ~ V
Wilh no head loss from (3) to (2) and 12. :::f,g :: ~ ;::-0 lUll. oblaill
~:z.
-q: +1,3 '" Z.. J or ~ =:{~j(Z2. -Z3) =.y2(32.7.!t~) ( ;%:f.I) '" J{;OIg.
ThvsJ ft (0.1SIf
Re== Y.lJ.:= '9J2. = 'I,Ols _-rr -, =2. 07 XIO'l-
1/ 11 1.2./XIO 5 .s "
and.
£ ::::: 0.00005 ff =0.008 (see /o,h/e /0, I) J so fh4f (see F;9' F/.2lJ)
D (~)fi
f == 0,039 :z.
1 V 2-
Also) hL = f D -rj 1- r:~ ~ where l: KJ. :: 3 (I.s) =:: ¥. S
Henc~ £0. (I) hecomesJ 1 2. 2-
*'"z.. +[ft +~ k.t] ~ - i1 where V, "v
~ ~ -/~fI +[O.03Q :;"II~. +'1:S-I] ~'l:OIY.X) =(0.583 +1.I'I7)ff
~ ~ 0.75 /n.' 2 32..2. 52-
= 1.73 tf
ihllsJ
I, ~ (62.'f~ )(I. 73 ft) := 108 1f-2- ~ 0.7.50 ps/
8-67

8.71
X.i7 The pressure at section (2) shown in Fig. PS.77 is not
to fall below 60 psi when the flowrate from the tank varies from
oto 1.0 cfs and the branch line is shut off. Determine the mini-
mum height, it, of the water tank under the assumption that (a) mi-
nor losses are negligible, (b) minor losses are not negligible.
FIGURE P8.77
All pipe is 6-in.-diameter plastic
(tID == 0). flanged fittings
/';
6 ft •• e~~::!:@iWW"i ~:1 ~i~~
1 l-:i~~[b .1, 900 ft~90° elbows
.ift-t-f+~I == £t+-ii+Z2. +(f-t t&K)¥j. J where It=:~ ~=OJ z,=16f11hJ
ond Z2,==o ThvsJ
wilh V::~
/6 +h = I} +¥;+(fl +&I<t. )t;2 . !Vole: h. must he no less IhQn tnt/t wiTh
LJ • =:: 6ons; (J/Jd ~ == I cls orrr~(1Im r- fJ..i mIx J
I J' - v- &- - I oS - S. 00 Ji.
Vj - - 7[ - *(-Ar~)" - . , .s
lienee) IJ.. in~ it 2-
h== -1611 + (6°7?)(~f1i=) +(1 +fr h+6+600 +9(0 ) +Elf;.) (-'.09 oS)
62. 'fw:~ 7f 2.(32.2 f,,)or ,,-
h=/~2..5+(I+f(JS~~5+h)tLK)(o.'I-(}2.) fI) where h"'f/ OJ
With -!=0 (Jlld Re == YI- ::: (s.o9#)ff:/J) == 2.10X/O s we ohio/I)
I.ZIXIO oS
f= 0.015.5 (see hl-- 8.20)
0) Ne9/ect minDr lo.sses (£K=:O):
From EfJI)
h=/Z2~&-(/ +(O.OIG!» (15:'~+h)) (O.l}02)
or h = IJf3 ff
b) Inclvde m/nor losses:
£ If;. == Jf;.enlrtJlJce +/5 ~e/IJ(Jw + ~/ee ::: 0.5 +/5 (0,3) +0.2 =S. 2
(see TobIe 8:2.J q,s.rUfIIC f/1J1J9Cd
Thus} from Ef. (JJ
fif/iIJ95 )
h :::122~5+(J +(0.0/55)( /~o:th) +s.2.)(0.'1-02.)
or
h= I#-6 ft
Note: For fhis cqse minor losses ar~ no! VerI i"'floridlJl.
8-68

8,78 I
FIGURE PS.71
S.78 Repeat Problem 8.77 with the assumption that the
branch line is open so that half of the flow from the tank goes
into the branch, and half continues in the main line. All pipe is 6-in.-diameter plastic
(flD =: 0). flanged fittings
b) Inc/tide minor /().Sses :
£1<;. =: ~L ,J +/.5~ IL_ = 0.5 t Is(0.3) =.s.0 (see Table 8,2,' (/SSV/He
d
q emr4nce 'l.i!ICJ(IW " ",
M ~~~)
'& IV..L -::: f(; ::: 0,2
o 'l.fee
From £f,t:;..)
h:=12-2..5+(/ +(o.0Io5S)('60t +h)+ 5. O)(O.tl-Ol) +(1800(0.01705) +0.2) (0./0/)
0.5
or
h= 137ft
Nole :For fhis cqse milJor losses are nol veri imporfllnl.

8,79
3.43 A smooth plastic, lO-m-Iong garden hose with an in-
. . side diameter of 20 mm is used to drain a wading pool as is
Repeat Problem 3.43 If head losses are mcluded. shown in Fig. P3.43. If viscous effects are neglected, what is
the ftowrate from the pool?
0.2 m
f.t:. +Jr +2 :::: & + ~ +Z +(fJ.. 4- &~ ):t? ').1 I ~ 1i '3- D II. '1
where It 0=/,.#0.1 ~ =0, 22. =0J ~ lIS a'f3th dnd liz. =V
ThIl$J wifh £~. ::: ~tI,,1 ::: 0.9 £e.(/) beco/l11J.r
V" /0/11
O. If3 IIJ ~ 2 (9.81.,:5(I +f o.02.m+ O. 8)
or ,..
8. 'fJf = (/,8 +500 f ) V where V~.p-
J
AlsooJ Re::: PY. ::: (o~ O~m) V = 1.7Q X/O/!- V
V 1,/:J.x/(j61£
Trial and error so/vtion:
f} sSlJlhe f ::::o. 02. or {rpm £ r. (,.)
V- - 0 811-6!!L
r 8./f~ ]~
- (1.8+500(0.02.)) - , s
so from £r. (3) J Re::: 1.7Q x/o'(0,8'1-6) ::: 1.5/XjO'l
Wilh this Ae valve and f:::o we obtain {rom the
Moody charI (Fi9. 8,:JO)) f;:: 0.02.7 wh/ch is /Jot fhe
assumed 'la/vB. Tht/s fry 49aiIJ.
/lsSI/I»8 {=-O.027 or fro", £tts, (z)ana (3~
V= O.1Lf2. If 4nd Re:;: 1.33 x/o~ 71JtI~ {rpm fhe h10()o/
chari f::: 0.028 ::J: f). 0').7
lI.ssvme f::: 0.02.8 W~i(;A 9it/es V~O,731 1JL) ((eo; /.3/ XI{JJ~
ond from the Mu()dy charf) f ~ 0, 028.1 f/;e t/s.flllYJed '14ft/f).
Hence V:::O.731 'f ~nr1J 3
11' 2 hi -If f1I
qt=: ,I}V== 7j (O.OZOfh) (O.731 T) -;:: 2.30X 10 S
8-70
0.23 m
(I)
(2.)
(3)

8.80
~.so The exhaust from your car's engine flows through a
complex pipe system as shown in Fig. P8.80 and ith~(j K5.
Assume that the pressure drop through this system is ApI when
the engine is idling at 1000 rpm at a stop sign. Estimate the
pressure drop (in terms of API) with the engine at 3000 rpm
, when you are driving on the highway. List all assumptions you Intake manifold
made to arrive at your answer. II FIG U REP 8 . 8 0
For sfeady {fowl
If t i!, +fi-4. '" ~ H2o t ~.
J.. V1
!lssvm8 :E,::: Z.z and ~ c ~ Sf) f1741 will, ~ == [ f D +,,]iI
Alld ilf=: It -12- we o~-J4i/J
J V2. 1. 1~I;:; 0'4 :::: 0(f]) +J()"ij = i f V (f J) +K)
lIeneeJ
A~()f)() _ t BoDo ~0~0 ( f31t1Q f +I<L.)
A I/oOf) - :i. flo" V,D~ ( f,tJtJo 1-I- ~)
/JsJ'pme eooo ::: ~(J(J() anJ ~()()O::: f.]()"(1 ((:e. filJcJ~endelll of /(e )
rnvs 1, (
4b:" _( 'vj... "" .f ~
Ll{JIOOO - ~ooo j Re
Buf V-=:: ~ where Qis a.r.rtll)1eri pl'OjJol'fi()lIdJ/ If) ef)?i/J~ r;;Pl.
T/JP,1 is ~O()o;:;.J ~otJt) Sf) fhai
~fJOOO ~ (3)1... ~ q
~f/Of)O =
8-7/
Exhau
(2.)

8.8/ I
s.~1 Water flows from a large open tank, through a 50-ft-long,
O.lO-ft-diameter pipe and exits with a velocity of 5 ftls when the
water level in the tank is 10ft above the pipe exit. The sum of the
minor loss coefficients for the pipe system is 12. Determine the
new water level needed in the tank if the velocity is to remain
5 ftls when 20 ft of the pipe is removed (Le., when the length is
reduced to 30 ft). The minor loss coefficients remain the same.
Of'
h::: 8.02-ff
8-71-
-u (lJ I
-:;: ,
I
,f I V=Vz. h:J
LV2I ..... 1
r
(~

x.Sl Water is to flow at a rate of 3.5 ft3/S in a horizontal
aluminum pipe (8 = 5 X 10-6ft). The inlet and outlet pres-
sures are 65 psi and 30 psi, respectively, and the pipe length is
500 ft. Determine the diameter of this water pipe.
I!L ~r. -& ~:a. 1 '1
r +L/ +iI == ; +22 +;.1+fD ~ ; ~:::Z:J. J V, ::: ~ ::V
where 3,5 ~
V = *'" ~/)~ = * where D~(f, V~ fI/.s
v
Thv~
~
f.1
/'- :::{-tIi Or
Ih I jn~ ) ( '1-.111) 2-
(6S - 30) 7i~ lllflf 1fi _f §OfL 'Jif
6 ~ /1> - J) 2(32-.2)
which simplifies to
2. 1fS
D=/./38 {liS -,
Also from Table 8,/ .£- ;:: -!Dx/
!..JJ J
(MfA
R ::; e.,VD ;:: 1.9# (~)D ()r Re::: 3.70X/O
S
e -p ')...3/f 't 10-'& D
(I)
Tri4/ and erl'fJl'so/vfion : -¥ IIAK/lQWns (~ j; ReJ
r); 're9v4Iio/)s ((/)J(~
(.J),.41Jt:A /)food,
Ch4rf (F;1~ 8.').0))
Assume f -== 0.02 so from £'1_ (I) D::: O.S'-o rf. TIll/sol frolf} Efs. (l)alJelliJ
Re ~ 7,// x/osand i ~'1.1x/P-~ so froll! 0,.8'],()J f: D. 012.8 r/;().02.
AssUl1}e f..0.0/:2.8 whir;h 9iv()S /) =:0. If71 II) Re ~ 7. 77X/O~
qnd -t:::I,/x/(}-~ TI)(/~ from Ft'? 8.2-~ f:: 0.0128 1V1}J~h
Q9rees w,-Ih the tJsslII1Jed vallie.
TlJv~ /) ~ 0.'1-71 f1

8.83 J
8.&3 Water flows downward through a vertical smooth
pipe. When the flowrate is 0.5 ft3
/ s there is no change in pres-
sure along the pipe. Determine the diameter of the pipe.
r=
m '1
V ~ 1
(I)
(:J.)
(3)
Triol find er~/' so/ufioh : .3 UlJtlJlJWIJS (D~ ReJ f) and 3 ef/(/41i()lJs
((~t3~ ~lJd Fi9. 8.20)
Assume (;:0.02. so from £f(. (~ /)~o,/t'fl (llld frum
ftt.(3).) Re::: 3.18 xlOs. T/Jvs~ from Fii. 8,").0.1 f:::. O.O/If :I O. 02
Assume f:~ o. O//f .so fh4f D;: o./ssfl and !Ie :::3.1f2 XIO~
7hvs" from Fi9' 8.~~ f-O.llf which cheo~.I wilh fhe
Assvmea Vo./1I8.
Thus D:: 0./55 fIJ

8.8/f
8..8'" As shown in Fig. P8.8f, a standard household wa-
ter meter is incorporated into a lawn irrigation system to
measure the volume of water applied to the lawn. Note that
these meters measure volume, not volume flowrate. (See
Video V8.7.) With an upstream pressure of PI = 50 psi the
meter registered that 120 ft3 of water was delivered to the
lawn during an "on" cycle. Estimate the upstream pressure,
PI' needed if it is desired to have 150 ft3 delivered during
an "on" cycle. List any assumptions needed to arrive at youl
answer.
• FIG U REP 8 .84
The ener9Y €9tJalion {Of' fhis flow i.!
~ ~ z
~~lf+z,-[fJ+L~]~ ~*~~.Ll:l
where Z, =:: ~:J. J f:L:;-o J ~ ~ ~ and V2. =: 4; ~
/hvs from £,. (I) 2-
1'1 J frV,'- [ft +r~ +(4&) -/]
Bof Q:::II, ~ ::: ¥J where VIS fh~ VO/1I1110 01 kldler so/II/od
dvrifl9 a/) '1ln"cycle and t is fhe /elJ9fh ollhe cyc/~.
Fof' tl 9iven sysfem 2:k;. is indepenJofJl 01 Q. S'/JI,lt1l'ly./ fot' l/Jif/9
Re pipe flow) f is ifJde{JendcnJ of Re ( or ~). Thv~
[f1 +Z kJ. of(! )2- /J is ~on.sf(J"IJ liJdependf}/J! ufQ.
Hence, from £'1' (;')J if fhe len,fh ofthe cycle is cOlJsianfJ
{JIt,ft = ~pv,~t =[ v,),r. ]~=r~E' ):2.;0 (.!flot::I. 5"63
It )17.0113
2" P'" ~:z.o lV, )/;10 " I UJ
or
f,) :;; 1.£6311;) =:& /.SI3(sopsi) ~ 78./fJSi
150 /1.0
8-7.5
(I)

8.85
X.SS When water flows from the tank shown in Fig. P8.85,
the water depth in the tank as a function of time is as indicated.
Determine the cross-sectional area of the tank. The total length
of the 0.60-in.-diameter pipe is 20 ft, and the friction factor is
0.03. The loss coefficients are: 0.50 for the entrance, 1.5 for
each elbow. and 10 for the valve. --Z-................-.......,............"....,.,
!Ii FiGURE PS.S5
8-7t
100 200 300
1.5

8.96
KS6 Water flows through a 2-in.-diameter pipe with a ve-
locity of 15 ftls as shown in Fig. PS.S6. The relative roughness
of the pipe is 0.004, and the loss coefficient for the exit is 1.0.
Determine the height, h, to which the water rises in the piezome-
ter tube.
OJ
h= /6.S If
Open
Ih
15 ftls
(I)t-I.--8 ft----i
• FIGURE PS.S6

8.e7
H.H7 Water flows from a large tank that sits on frictionless ~- - - I) - - -
wheels as shown in Fig. P8.87. The pipe has a diameter of 0.50
m and a roughness of 9.2 X 10-5 m. The loss coefficient for
the filter is 8; other minor losses are negligible. The tank and
the first 50-m section of the pipe are bolted to the last 75-m
section of the pipe which is clamped firmly to the floor. De-
termine the tension in the bolts.
{OJ
(I)
(2.)
or
v- /9,9
- Vq f ~50f
IIlso Re:: ~ :::
or
Re == ~.~tx/os V
and 5
s:: 9.2 XIO m ::: l.i/fXli'f
D O.ShI
rnvs from Ihe AI(jot1j cnarl:
f ~*~J.Jlfxlril'
Re
11 FIGURE P8.87
(con'f)
~--.-----' (~ h
boH
CfJntro/ YDJlltne

8.87 I(c,on'tJ
T/'/0/4fJd error srJ/vliof) :
Assume f:: 0.0:;' ..£l V::: s.2-?;- !E-Re :=2. ~X/06!:!!.. I::: 0, o/~ :I (),()2-
flSSPIlI9 f:: 0.01'f'-!!'!' V::.s: 60qL .!!:!- Rf) ~ 2.SX/OI (.1). {::: O. f)/~
7i7~ V== SIIO.fL
/I/so ~
~z ~~
f; +:i!.3 +i!; -4J
-2. := f!; ~ Zz +~
where
Z~:::~~ J V:z..:::'~ IJn~ f,.. ~() Sf) fl1al
1'.3_ =: h -;:: (f b. -I- k;)J[.1 ;-(0.0/.,. (?E!L)-I8) (S.6o.fl)2 == 1£,/1l'J
'7 'LJ-2. /) ll. z,! f). SAl 2 (9.11i!!..)
or ,r2.
3# .s-AI
{1== p,aox/o Iii (/~./fII):;; I.se X/O AJi
TIJv~ from Erp (0):
FboH -;:: (iJ 113 +PA.l ~2-
=p.~x/o$f.):;(O.ShI)2- +(9P9tJ)1j-(o.,smll-S'.IIJ11-J1.
:: 3./(JXIO~N + 6,ISX/O.Jjr/
8-79

8.88
8.88 Water flows through two sections of the
vertical pipe shown in Fig. P8.8 B . The bellows
connection cannot support any force in the ver-
tical direction. The O.4-ft-diameter pipe weighs
0.2 lb/ft and the friction factor is assumed to be
0.02. At what velocity will the force, F, required
to hold the pipe be zero?
From Ihe m'OmenTum e9uafion appn'ed 10 fhe
con/rol volume indicofed
I,II, - ~:l0 - Wpipe =rh (V,.-~) ::::0 ~i,!ce VI:: Vl
Thvs .;:; WH~o+WPif' = '0ill, + I (:t=)
J 1/ AI II,
or~ == 0/ +(O.20 lt)P :: 0/ +1.59i where f) NiA 1"'//
I!'j ~ ( 0.Jffl)2 J 1// n )
,t;/soJ
4t-+¥g2HI = 't+.r:;1U;J. +f /;liJ where 12::0,
1 =l{ ~ V, 2, =0, (Jnd 2,-:::J
ThlJs) f) = };tZ +fl..LfV~TI Q 2. D 2.. J
Pipe weighs
0.20 Ib/tt
D = 0.40 ft
FIGURE PS.BS
or when comhined wifh the ()br;Ve force DC/lance resvll
11 :::: oj+f-t ip V2. = 0'1 +1.591 i
TL I· fDVV2. v- _I 2D(/.S9) - 2(O.JI.)(/.S9) ==5.7'3fiIfJOiIS, -Tii ==/.59 or - Y pf - (1.9'1)(0.02.) = oS
Nofe: Tilis answer is indepenrJelll of the pipe /enqthJ 1.
8-80

8.8q I
7(1)
60-mm-diameter, t ------:-:--:-----
8.Sg The pump shown in Fig. PS.89 adds 25 kW to the 30-m-long pipe; I..
water and causes a flowrate of 0.04 m3/s. Determine the flow- 40-mm-diameter f=/ 0.016 ___rJ~
no~zle ~.
rate expected if the pump is removed from the system. Assume / J .-
f = 0.016 for either case and neglect minor losses. ----~===========:t,~ "£"-L--_.--l
(2.) Pump
FIGURE P8.S9
8-81

8.90 I 8.90 A certain process requires 2.3 cfs of water
to be delivered at a pressure of 30 psi. This water
comes from a large diameter supply main in which
the pressure remains at 60 psi. If the galvanized
iron pipe connecting the two locations is 200 ft
long and contains six threaded 90° elbows, de-
termine the pipe diameter. Elevation differences
are negligible.
v.'" b. ~2. £ V'" .
++29 .fZ/:: ; +~+l:t +(fD+r~)~ J where~==30fJs, J I, ~ 60fJsi,
o::z = 7. ~=O It == V= !l.:: 2.3 fJ ::: ~ £i wifh DNff
~I 2. J I t ' " fj.!r D'" D2. S J
If
Thvs,
l1-f2. ::: (f~ +L")fp V2-
orr60 -30) ~(PI-If #i)=(ff{( 2~Oft)+6(I.S) +0.5) (~lIf-1jf·(t.)(1.9'f~)
where we hove used
2: f( :: 6 KelhtJw +"elllrlJhce = 6(I.s) -1-0.5
Thus,
Jl.tJ, Lf :: (J +/~Ol) iff
A/so, _ VlJ _ ('i!) D:: 2.9..3!f~ D .5' I
Re - y - 11 1.2IXIO-!;tpD' or Ile :: 2.Jf1.X/O IS
and from TahIr: 8.1
e o.ooo.sN
75 == D
Finol/y) from Fi9' 8.7-0'·
Trial and error so/vfion of
Eqs. (I~ (:J.); (JJ, and('I) for
f
fJ DI ! J and Re. D
(I)
Normally if is e4siesf fo 9uess a va/V6 off, cQ/cv/afe ~ efc, In this cose
(becoIJSe ofminor losse), Eq,{/) is no! eosy to IJse in fhis fashion, TIlus) Qsstme
DJ
calcfJlafe r(Ef{. (/))J lie (£", (2.))J ond f (£f. (.1)). Look up f;1) Fi7~ 8.20
(cr· (~)) IJlJd c()",pare wifh fhal fro", Eft· (I),
/issII",e D=o./ffi. T/;/IS, f=O.00557, Re:6.05Xlo
s
J
f;;:o.oo/:l.S
or from Fi9' 8.:Z 0 f:: 0.021 -:f: 0.00557
Assume D=O.5f-l j f=o.OSSIJ
Re =II;8/f/<IO: .g-:::O.OO/orf=O.0203¢O.OSSI
Assume D:: D.'isff. F:: 0.02 -s'3 fie::$,38"'0$ -DE = 0.0011/ or f == 0.02.05 :to.Ol/l-.3
J J J
/J~siJme D=O.llifff; f::: O. 0197J Re =,s..sox'0: !-:::D. 0 0//'1-0/'" f=O.0205t:O.OlQ7
INfer e/JofJ9h trifJ/5 ()hloifJ D= O. If'/-2 fi
Nofe ; If H9. 8.20 (Ef-('I)) ;0$ replaced hJ1he Colehrook erll4lio/J
fhis prOblem cOllld be solved efJ.II'Ij will; acfIJ'J,PtJler..

8.Q' The turbine shown in Fig. P8,QI de-
velops 400 kW. Determine the ftowrate if (a) head
losses are negligible or (b) head loss due to fric-
tion in the pipe is considered. Assume f = 0.02.
Note: There may be more than one solution or
there may be no solution to this problem. 120 m of 0.30-m-diameter
cast iron pipe
FIGURE P8.'U
'J!-+¥J fZ, ::: iff ri+Z2- +ft¥; -l-~ I where I,;:~ =0; z,c20tnJ
Z ::: 0 Thus z = XI +{lr+1-. (/)2. I, "]..9- D 2.9- ,,.,.
a)#eq/ecf head losses (f==o): 3 N.m
_ vl +l. "" L:= ¥oo x/o -:s 52.0
Z, - 2j ''7' J where hr rQ (q.80x/rrP.3)!j(Jm/ ~ ;: T m
Thu,s, /2.
V2 .52..0 ,,3 I 0
~om=2(9.81:;") + ~ or Y2. -392v,.+/020= (2.)
De/ermine Ihe roofs of fh,-s cuble erVfJlion. Lei v,.3-392~ f/O')..O=F
kIanf F:::0. Note Ihol ~ = 3V/'-392 so fhaf ~ :::0 aI ~;: r//.'1l}-
A/so} F::: 1020 when ~=0J F::; ~ 007 wheh V,.::: -II. ¥J and
r = - /9t7 whe/) ~:::I/.Jf. IJs ~- ()b J r-0(). F
lis ~ - - 00; F- -ctJ. This liJformaliof)
Indicales Ihere 4re IUlo posilive real rools
(.see Ihe fi9ure). The nefolive rool hlJs no ~
physical meQlJifJ9' So/ulion of £'1' (1-) 9ives -1967 .---
Vi = 2. 6s ~ Or ~::: 18.3 ~ Thlls, Q-:: 1i:J. liz. := *(JtntV;.
or 3 3
Q= 2.08!f or Q=/~Jf~
b) Inclvde heQd loss (f ;:0.02): From E'(. II) V='h,t2 = ~ (.~)2.= ~ (~.~m)
o~o", =(lfO.o2/:/~om )(JJ.J)~) Vl + 52.0 m ::::.//./~
0.3 m 2 (q, 81-$.) V~
n~ 3 3
~ -O.398~ +1.03'1=0 Lei G= ~ -O.3?B'Vz +1. 03/f j de1ermiIJ8
v,. fhot 9ives 6=0. lis above) G-±ooa,s v,.-tooj ~~ ::::3~2..-0.3?8=O
for Vi =:J:. O. 36Jfj G=/. 03'1- for v,. :::0) an d G::1.13 for ~.::' - O.36oS' I
G=O.Q37 of V,,== +0,36'1: Thus, fhe qr4fJh Df G G
looks QS shown, A cuhic e'lIJClfiol} has al most
two min. or max. IJs ,shown, there isno pos/live
real roof, The Flow clJlJn'" occtJr (mvsf have P<'f(}{)~W) -0.361f

8.92. I
K.n A fan is to produce a constant air speed of 40 m/s
throughout the pipe loop shown in Fig. PS.92. The 3-m-diam-
eter pipes are smooth. and each of the four 90-degree elbows
has a loss coefficient of 0.30. Determine the power that the fan
adds to the air.
i'" 20 m
(;1: --f--
110 m
~J
(1).::{21
_(Fan
~I
-- 'i'-
ii2 FIGURE P 8.92
../
-- I~v= 40 mls
~
D=3m- :.-
~

8.9.3 ,
KL exit =.0 . V
g.~3 Water is circulated from a large tank, through a filter,
and back to the tank as shown in Fig. PS.93. The power added
to the water by the pump is 200 ft·lb/s. Determine the ftowrate
through the filter.
KL elbow = 1.5
~2. 2. V2
1-+Z; +#+hp '" If ~Zz. +41+(fl +f 1V.,.)1j
where
,,-:::fJ,. J ~ =- ~ =-0" and Z,:: 23-
AJsoJ ~::: r()hp Dr
200 f1.:1! 1/-08h - ,s: -
p - 62.,~ (¥(O.lfO"2)V - V
Thus £0, (I) becfltnts
J T 2
'1-08 ( 200 ff I. )) V ¢ '
V = 0,111 f +,0.8+05(1.£) -I- /2 +6 +/ 2(32.2~}
or
3 13./.3
V ~ (f +0.01365)
1:: eVD =- 1.1",,5wrv!1){O./If)
-p 2 a~ x/o& Ii"is.
• fll.
or Re::: B290V
Tri.1 and error so/vl/o/J:
flssum~ r::: 0.0'1. From £,. (2)J V::: 6.26 # ;from Ft. (3~
Re :: 5.'2. 0 x/0 ~. Thv~ from Fi~. e.20.1 f:::: O. 039 =I: 0.0/1
/Jssllme (:::O,o3Q or V:: 6. 29.if. IJnd Re :::S.21 X/O~and f -:::O.O:Jf
J
(Check)
Tf,v~ Q:::/I V::: "'(O.lf/l(6.29~) =o.O.y.9~-!f3
(I)
(2)
(3)

8.94 Water is to be moved from a large, closed tank in
which the air pressure is 20 psi into a large, open tank through
2000 ft of smooth pipe at the rate of 3 ft3
Is. The fluid level in
the open tank is 150 ft below that in the closed tank. Determine
the required diameter of the pipe. Neglect minor losses. .
(IJ
1=2000H
(I)
T
1501+
~ (2.)
_ P(~)D _ 1.9~ (3.82) or Re == 3./~'JIO~
- P - 2.3¥'tIO-s /) I
Trial lind error solufio!) :
IIssume f=O.02 S~ from Ey. (1.)" /):; O.S'fo II lind from £".(3)
Re =:.£.87X/O
S
, Thv~ fr(J/II F". 8.UJ(w/lh -i-.=:o) f::o.ol.3 :;0.02.
(2.)
!JssulIJe f=:0.0/3 wh/cA 'lives lJ ~ O. 'f9.5 II f(e:: 6..1fQ lOs QlJd f::o.ol2..5J J
Assume f :O.OI2.S, ~o D=O. ~91f-lJ He ~6Jf~,(/OsJ f ~ O.O/'-S (checJu)
Thvs D ::: O.'1-9/ (IJ
g-86

8.,95 I
8..95 Rainwater flows through the galvanized
iron downspout shown in Fig. P8.'15 at a rate of
0.006 m3
/s. Determine the size of the downspout
cross section if it is a rectangle with an aspect
ratio of 1.7 to 1 and it is completely filled with
water. Neglect the velocity of the water in the
gutter at the free surface and the head loss as-
sociated with the elbow.
70 mm
L
T~. ~.
Ifm
LI"li'~lW#Hi$l'W""''> ~. 3m---l·1

8.Q6 *1
8.96 * Repeat Problem 8.QS
spout is circular.
if the down-
f = f-2.0 JO'J[¥."tX/~-6+ 3.6
WO-9
D] J
2
Solve £rs. (t) and(7) iferqlive/y. Sfflrl "11Th Q.ss/Jhlod vfllues
D;;: 0./ J f= 0.02. 0 From El(. (I) o61qill Q /Jew Dvalve 0 WI/h Mis
new DcQ/cvlole 0 /Jew f vqlve from fro (7). Re,PfJ4t suc.h CQ/cp/otir¥ls
unfit the nih lind IJ_/.sf I/IJ/ves s4iisr, The CfJIJV8ry8nce crderiQn
) 1- i!L J < 0.001 and II - ~ }<O,OOI
fn-I ~I
8-88
(2.)
(3)
(5)
(6)
(7)

Pro9ram P8# 96 .BIJS show/) below kills used 10 solve
£ro5. (6) QlJd(7) 405 indictied above 10 9/~ve D=o. OJf~Sflla/lJ
f= 0.02.78
100 cIs
110 print "*****************************************************"
120 print "** This program determines the friction factor, f, **"
130 print "** and the diameter, D, solving iteratively **"
1l±0 print "** Colebrook's equat.ion **"
170 print "*****************************************************II
180 print
190 f=0.D2
200 d=O.l
210 dp=d
220 fp=f
230 d=( (f+0.1429*dp)/195600)-0.2
240 f=1/(-2.0*log(0.OOOOl±05!d+O.000368*d/fp~.5)/log(10))-2
250 if abs(l-f/fp)O.OOl or abs(l-d/dp)O.OOl then goto 210
260 print.
270 print uSlng "The friction factor is f = +#.####----";f
280 print. uSlng " The diameter is D = +#.####----";d
****************************************************~
** This program determines the friction factor, f, **
** and the dia~meter, D, solving iteratively **
** Colebrook's equation **
*****************************************************
The friction factor is f = +2.7842E-02
The diameter is D = +4.l±518E-02

g,q7 1
8.97 Air, assumed incompressible, flows through the two
pipes shown in Fig. P8.97. Determine the flowrate if minor
losses are neglected and the friction factor in each pipe is 0.015.
Determine the flowrate if the 0.5-in.-diameter pipe were re-
placed by a l-in.-diameter pipe. Comment on the assumption I
of incompressibility.
p = 0.5 psi
~F(0) I.
1 in. 0.50 in.
~ m ! (2.) (.!)
t
.1.20 ft
__t20 ft- 'II
~,
FIGURE P8.97
{?a ~ fJ.j_ :: /~. 7p.Ji . ;:: 0.967
lJo l'i (0.5 f/1f.7)PSI
The floJV is neflr/j JiJcompressi!JJe.
3-QO

8,178 *1 p = 0.5 psi
*8.98 Repeat Problem 8.97 if the pipes are galvanized iron
and the friction factors are not known a priori.
TiD= 150°F 1 in. 0.50 in.
* (I) t (2.) (.l)
(0) I. _ t lJ.-- t 201:._........-..1-20 It-- J.
~
FIGURE P8.Qg
I
I
and
and
Re == ~DI Re == ~ ~ where fr()m 'able 8,3I 1/ J 2. V J 1(.4
·7~ ['.12.
11 == ..l! = '1.181./0 {fl. =2.00X/O·¥.I-L-
fo O.0020q ~ .s
ffS
Hence} Re == (O,2.sV,J (-kfl.) =/O/f V,
1 I 2. OOX /0-11-]£ ~
ano ( I .s
o == V2. Vf H) =2 08 ~
nez 2.OOx/o·iff:J:
S I [ £ 2 5/ ]
For lurhulenf Flow £'(. 8.35 9iV~s vr =-2.0 Jog .3~ i- R~yr
By comhinifJ9 £fs. ('1-) fhrough (8) we obtajn
I I [ -3 ',1111</0·2.]
W, ==-2.0/09 I, 62xl0 + V2.1{f;
and, [-3 I 1-1 X10-
2
]
w: =-2.0 Jog 3,2.lf.x/o t . '4W;
(c,onJl)
(2.)
(3)
(If)
(s)
(7)
(8)
(q)
(10)

B.98 f , (conI!.)
Solve E"q,s.(.3)) (9)J Qnd (jo) {or fhe unknf)w/Js ~J h.J and Ih. (see below).
If D, == D2 J then ~ ~ ~ J ~ :: -& since -t; == i2. == 0.006 J qnd
Re =Re = V:z. D:J. = V:t (7i:ft~ ='1-16 ~
I 2 1/ 2.00)(/0-"']£
..s
Thlls, Ef. (I) hecomes
tPo =ie~~[~(el;:~) +IJ
or
(o.s1::,.)(JJflf ~ ) :: i (0.00209 ¥f;){:l.[f2. (10
;:) +I]
Hencel
6, eq x10
9
= ~~ [ '180 -& +1] (II)
Also) frohl El(. (8)
.-L =-2.0 lo~ ~.62x/0-3 + 6,03)(/0-
3
] (/2)
vr;. L V:t w:
Solve £'(s. (II) ond (/7-) for &. tJnd ~ (see he/f)w)
Nole: S/nce {J:::: pRT il fo/lolAls +hal
.fi._(,ln)_hJi. ' 1 ' ffo - ( A) -]10 73 If we (J.S.SVtrJe 7; =70 (/ prohfJbly willnf) heJ
RTo
buf rt should be Q reasof)oble apprOXillJaTi()n) fhen
~.!I :;: IJt- = ( Ilf. 7fJ~i :::: OJI67 The flow is neqrly incompressible.
0 , () 0.5 tllf.7 psi
Pro9rIJIiJ P8# qg .shaWl) be/ow 'lidoS used fo Dbl';1J the f()//f)lIIiIJ9
resfJ/ls :
0) W,-Ih fhe fwo different diall1efer pifJes ; Q::: 0, 07~6 .f
b) With fhe sill'lle pipe: Q=: 0.339 #3
100 cls
110 print 11 *****************************************************f1
120 print "** This program determines the frict.ion factors, **"
130 print "** f1 and f2, and the velocity V, solving **f1
140 print "** iteratively Colebrook's equation **"
170 print "*****************************************************"
180 print
190 f1=0.002
200 f2=0.002
210 f1p=f1
220 f2p=f2

230
240
245
250
260
265
270
275
280
290
300
380
385
390
400
420
430
445
450
470
480
490
500
v=(68900/(15*fl+480*f2+1))~0.5
fl=1/(-2.0*log(O.00162+0.0241/(v*flp-.5) )/log(10))-2
f2=1/(-2.0*log(0.00324+0.0121/(v*f2p-.5) )/log(10) )-2
if abs(l-fl/flpl)O.OOl or abs(1-f2/f2p»0.OOl then goto 210
print
print "For the CB2'.e of unequal diameter pipes: II
print. using "Tbe friction factors are fl = +#.####'·~'·-I!;f1
print using" and f2 = +#.####----If;f2
print. using" The velocity is V = +#.lHt##··--- ft/S";V
Q = 3.14159*(0.5/12)-2*v/4
print using" The flowrate is Q = +#.####---- ft3/slf;Q
print
print.
print "For the case of equal diameter pipes:"
f2=0.002
f2p=f2
v= (689001 (480*f2+1) ) -0.5
f2=1/(-2.0*log(0.00162+0.00603/(v*f2p-0.5) )/log(10) )-2
if abs(1-f2/f2p»O.001 then goto 420
print using" The friction factor is f2 = +#.####----";f2
print using" The velocity 1S V - +#.####---" ft/s";v
Q=3.14159*(1/12)-2*v/4
print using If The flowrate is Q = +#.###---- ft3/s";Q
k****************************************************
;j:* This program determines the friction factors. **
** f1 and f2. and the velocity V, solving **
** iteratively Colebrook!s equation **
*****************************************************
For the case of unequal diameter pipes:
The friction factors are f1 +4.2508E-02
and f2 = +4.4593E-02
The velocity is V = +5.4682E+01 ft/s
The flowrate is Q = +7.4561E-02 ft3/s
For the case of equal diameter pipes:
The friction factor is f2 = +3.5069E-02
The velocity is V = +6.2160E+01 ft/s
The flowrate is Q = +3. 390E-Ol ft.3/s

9.100 I Elevation = 15 m
_~(A> Diameter of each pipe = 0.10 m
----;!--~-
8.100 With the valve closed, water flows from tank A to
tank B as shown in Fig. P8.100. What is the flowrate into tank
B when the valve is opened to allow water to flow into tank C
also? Neglect all minor losses and assume that the friction fac-
tor is 0.02 for all pipes.
·A
1..0.0..._
(/)
FIGURE P8.100
Q, == C¥2 +Q3 where QI' =11£ V:. == f 4~~' J i:: ~2Jl
ThvsJ
since /)J:: /)2:: /).3 iI fo/lows fhol
~ == l{ +~ (I)
Also, for flvid flowin9 from II loB,
&.+Yl+z~ =& +:!i+Ze tf, 1L;!L.2 +t j. v: where LJ,q: A78 == 0,~ 21 I' 21 I D, 2., 2 D,. 'J.~ J Tn y~
~ == ~ =0) ZA ::/5 ~2.rJnd Ze==O'
Thus z = ,,1L.:iL +t !.~.2. or OJ
) A I D, ~, :t D:t 2j ~
15m == . (0,02) [(801'h)~3.+(ifO/1})~2]
(0.1/11) (2)(1.8/$.)
Hence,
J8,Jf == ,12- +0,5 ~2. J l'l N ~ (.3)
SiIYJ'/rJl'/yJ lor fluid {/owin9 from IJ fo C,
tfJ/J +~+Z. ::: f&. + ~+z. +£.! }IL.:J. +£ h li2. where LlLJ ::AJC:::(),
T 21 /i r 2? c I D, 2.1 3 /).3 ~1 J Vb yc
~r VccO 219 =/.5m, IJnd Zc= 0
J j ~2. b ~2.
ThIJS, :Z/J == £~ i1 +b~ '"9 (~)
B , r (J J ( ) r I r 12. V:_£ b vIy comptJrth9 k.rs. J., ana ~ we Tina T2 D; 1:j - .3 D3 29
or since &=~ and 4.:::D~ J
~ 142.=~ }6.2 Thvs) ~o ~2.= 7..5~.2 or ~ ==1.369 ~ (sJ
Solve £9s. (IJJ
(3); IJnd (S) for 11J ~ J find II.J •
From £fl' OJ and(s); ~ == I, 3t9 ~ +~ == 2,369 v.s. and from Ef.(3)
fB.if =(2.369~)'). +().5 (1.369'4)2 or 'V.J = I. t7t '!- ~ =1.319(l.t71f)
J
and ~ =2.1-9 f +/. 67& r = If. 00~ =2. 7- 911-
~~ 3
<V2. :::1i.2 Vz == *"(0,1mi' (2.2.9.p) ::: 0.0/80 ':

8.101 -I Elevation = 15 m
F-'¥~d(A) Diameter of each pipe = 0.10 m
,'::A Elevations = 0'-- ---:-
LBOc-++-""40 m+;:'~CS) ee) -J/)
*8.101 Repeat Problem 8.100 if the friction factors are not
known, but the pipes are steel pipes.
E=75=m::;:;;(3~)~~"f'Y>l-:~.:::"'1_~C
FIGURE P8.100
Q/ =Q2. +Q3 where Q/ ::;At'l{' :: f4·~Vt· I /::~J./3
Thus, silJce D,:: D2 ::D.3 if fo/lows fhot
~=~f~ ro
Ais0) for fluid fJowin9 from It to BJ
f +¥it ~/I ='1 t ¥$+28 t t! ¥i +(;I. t ~iJ where IA ::14 ::0,
?i:: '4 =0 211 ::::/Sml
a!)d.&o:::O
J J. 2 I. J.
Thvs} Z/I :: r; D' ~ -I- -& f li or (2.)
I 2, :2 1-1 }
/5", =[f,f~.~:nw+M~::)'6·]2(q.~/~)
or
0.368 == ~ ~2 +0.5 f:J. ~2. (3)
SinJl/fJr/y, for flvid flowing from 1110CJ
ik v,..,z _ tPc_ + ~2.+ =z r la .M: + r iJ~:l L
?1' +~ +Z/i - T 2.1 ~c 4- TJ D,'" ~ D; '2-1') wnere (111::fJc:: 0,
~ =Vc ::: oJ ~A= IS 111) lI!)d Zc==(}
IL f, 11. ~2. r J~ Y.j2
I nusJ ZIJ = I D iii +TJ 7i :z.! ('1-)
I ~ :3, Z
By cu"'parill9 EfS' (2) afJd (If) we find fz i ~ = &-! ¥i
or since D2 II:: D~ J
. ~
f2 J2. Vzl.::: -& i~ &2. Thlls} "f()~ V/" ::: 75 ~ ~z or ~= I.3&9( t)~3 (S)
From £q,8./35 J::-2..oJo9[! + 2..SI] where from TolJle 8.1
{,. J 3.7 Re vr
£:: O. OJ.f.-S /1')hJ so fhol for each pipeJ E =o.o~.s mill = 11:5 XIO~
VlJ 7S '00111111 •
II/so} Re = V or for i::~2}3
Re. =V~'Di = ~. (O,/m) = 8.93X/O""'i'
l tI /.12 x/o-I>f I
(con't)

(6) (7)(~
.I ;
Solve 6 e'll/(JtiolJs for b (J/)KIJoMlIJS; £rs, O)J (,g) J (&) /6),(7)) (Jnd(8) for
~ J &I &I ~ J ~J ~. Ir/a/ olJd error so/pi/on tiS fo//pw.s:
From £'(. (s)) ~ =' 0.73 0(-t;t-~,which when combined "lith
EC(. (J) g/ves
~ =[I to.730(~)-'i] ~ (q)
ThllsJ
by combini/}! £rtsI (3) and (q) we obfqin
0.368 ='f~[I+O.730(t;;rto.S fz] v,,2
or
~ -f 0.368 ] ~ (10)
2.- ~ [I +0.730(4;Y']2.+0.5 f2-
fI/s~ from £'1' (I)J
(II)
So/vfion method!o)GlJess valves of "J fl.; and & (1I9ooJ siQrfiIJ9
valve IS fne /Q/'7e Re valvB fDr i:::9.sx/o¥-, or "=-&=£-=o.OI7~
h) Ca/clJ/ale ~) ~J alJd Va fro/IJ £ys. ('I); (/0)1 dl/d (1)/ c) Ct,jc(}/4fe
t,J F2 J qnd~ frmEts. (6~ (7~ (8) j d) Compare the new ~. IV/lh fhe
previous ones j e) If IIPi 9t1oc/ elloVfA tl9reellleflfJ refJeal wt"-Ih the
new ~. as fhe 9()ess~
Program P8#/OI .shown below Wf}-S used ..fo ca/c()/aie the fO//OIViIJrI
resulfs:
Q m3 1»3 m3
J:: 0.03.3/ S J Q2::: 0.0193 T J l(j::: O.0/38T

8./01 II I (conJi)
100 cls
120 print fI************************************************"
130 print "** This program calculates the flowrates in **"
1l;0 print "** the three pipes using the Colebrook form- **"
150 print 11** ula to determine the friction factors. **"
160 print H** An i terat.ion scheme is used. **"
170 print "****** *********************** ***** ******** ****** II
200 dim f(3), fp(3). V(3). VP(3). Re(3)
210 for i = 1 to 3
220 f(i) = 0.017
230 VP(i) = 0
2/,00 next i
250 rr = /,o.5E-/,o
260 print II II
270 print "pipe no. Re f V. m/s Q. m3/s"
300 del=O
305 V(2)=(0.368/(f(1)*(1+0.730*(f(2)/f(3) ) ft O.5) ft 2+0.5*f(2)) ) ft
O.5
310 V(1)=(1+0.730*(f(2)/f(3) ) ft
O.5)*V(2)
320 V(3)=V(1)-V(2)
330 for i = 1 to 3
3/,00 fp(i)=f(i)
350 Re(i)=8.93E+/,o*V(i)
360 if Re(i)(2100 then goto /,000
370 f(i)=1/(-2.0*log(rr/3.7+2.51/(Re(i)*fp(i) ft O.5) )/log(10) )ft2
380 if abs(l-fp(i)/f(i) )0.001 then goto 3/,00
390 goto /,010
/,000 f(i)=6/,o/Re(i)
410 del=del+abs(l-VP(i)/V(i))
iJ20 next i
500 if del<O.OOl then goto 600
510 for i = 1 to 3
520 VP(i)=V(i)
530 next. i
540 goto 300
600 for i = 1 to 3
610 Q=(3.14159*0.1~2/4)*V(i)
620 print using" ## #.##ftftftft #.#### ##.### #.##~ftft~";i.Re(i),f(i)~
(V( i) ,Q
630 next. i
************************************************
** This program calculat.es the flowrates il"} **
** the three pipes using the Colebrook form- **
** ula to determine the friction factors. **
** An iteration scheme is used. **
************************************************
pipe
1
2
3
no. Re
3.76E+05
2.19E+05
1.57E+05
f
0.0176
0.018/,0
0.0190
V, m/s
/,0.211
2./,051
1. 760
Q, m3/s
3.31E-02
1. 93E-02
1.38E-02

8.102
8.102. The three water-filled tanks shown in Fig.
P8.102 are connected by pipes as indicated. If
minor losses are neglected, determine the flow-
rate in each pipe.
D = 0.08 m
f = 400 m
f = 0.020
FIGURE P8.102
Assume fhe fluid flow.s from IJ 10 Band /110 c. TIJl/sJ ()I: Q,. +Q3
or f (o./m)" ~ ::: f(O.OBhJ/v,. +*(O.0811ii'~
Thus, ~ = O,6Jf ~ rO.6~ ~
For fluid flowiIJ9 from /I foB with fA: /,8 ~O and ~ ~ ~ :::-0)
Z =Z + t i, .:it +~ .&. Yl'/I 8 I DI 2.1 2. D~ '1
or (200/1)) V/2.
60m- 2 0111 = (0..0105) O,/m 2(fl.8J.ff's)
HenceJ
+( )(200h) ) V/
0.02J) o,oem Z(q.811f,J
LfO =/,529 ~2. +2.55 ~2.
Simdqr/y, for fluid f/owinfJ from II fa C.wilh fJA ::fc =0 C/lJcI ~::: ~ ~OJ
Z =2 +~k.it + f b. YJ.'-1J C I D, 2., 3 D3 ,.,
or '-
60", :: (0, 01.5)(-2oam) V,
O,lm 2.(fl.B/~')
)(
/f.0()IIJ) V;+(0.020 o.oem ( Il!.)
2. 9. 810$2.
HenceJ
60 =- /,,529 V/ +5.10 ~2
Solve Ers. (/)J(z)lond(.3) for 11,) ~14I1dl(g. From£fs.(J)and(3):
60= 1..5'29 (o.tJTi(v,. +~/ +.5./0 ~:ZJ or 95.8 = (V~ +l4l+B./If ~2.
Su6lr(Jcf £'1' (2.) from Ef. (3) :
60-'1-0 = .s,/0~:Z + 2.55 ,42 or ~ =12~'- -7.8'1- I
ThvsJ from £I(s. (If) fJnd (s): 8,//f 1/./+(j2.vi_7.ei +'V.3 )2.- 95.8 :::-0
This C4fJ be sillJplified fo
I ~ I 2-
2 ~ l' .2.~ -7. 81f := 103.6 -I/./Jf ~ Square both sirles and
rearrange 10 qive ~If - /9,63 ~2 + Q2. 5 =0 which C4n he solved
by fhe qlltJdr4/ic formfJla fo give
~2.= Iq.63:±JI9,~":'-tf(q2..5)' ;: /1.77 or 7.86 Thvs ~::3.4l3f
or ~ == 2.BO.!f
(I)
(2.)
(3)
(If)
(5)
(6)

8./02' (con'i)
Nofe: The value ~=9.'f3: is not a solufion af the orifjino/ e'lufJlionsJ
E,(s. (I), (2.), (lnd (3). Wah this' valve fhe rl9hf hand siJe of EfJ6)
is neqaliv8 (i.e, /0.3.6 - II./If ~3.:; 103.6- //.IIf (3.11-3)2. =- 2.'1;5), lis
seen from the lefl hand side of E".(6)J -fhi.s etlAIl"; be. This ex/ro
root WQS inTroduced by squariIJ9 Eq. (i),
Thus} ~= fJ3~=f(O.oBm)'"(2.fJO!f) ;:;0,0IJfI.;-3
lJ/soJ
from Eq, (.3) ~
6o=I,S:J.9V/'+S.JO(2.80)" or ~.::: 3,62f}
1l 2. ",3
or Q/:: AI ~ =;yo (O.JOm) (.3.62 f) = O.02.B'f-:s
olJd from £'1' (0:
3.62:: 0.6'/- ~ +0.6'1(2.80) or Vt. =2.86f
W 3
Q2. =fiz ~ ::::.1J(0.08m/ (~,86 ~) =O.0/'1-3.!J

8./03
Free jets
=5 in. Q
2
K.IIJ3 Water is pumped from a lake, into a large pressur-
ized tank, and out through two pipes as shown in Fig. PS.I03.
The pump head is hp = 45 + 27.5Q - 54Q2, where hp is in
feet and Q (the total flowrate through the pump) is in fe/so Mi-
nor losses and gravity are negligible, and the friction factor in
each pipe is 0.02. Determine the flowrates through each of the ,
pipes, QI' and Q2'
f----1200 It ----I.j (2.)
(I)
(a)
('1-)
II FIGURE P8.103
/+0.02. (1ooH/(6112.1-I))
---:'---:ll~/-'-----:~) = O. 703 'vi/ +0.02 (/2fJO rT/ (SII1. If)
T/]vsJ
Q:z. A Vi ff,fnt(0. 703V,) r'" _,., LL nD!":) J./. -'-
-- == z.::: 6 ~ 0 I.f:z. - v. TU u uri .s6 {naT
({I II, V, f (12#) ""
Q = (fl of c(:z. ::: Q, +O. '1-11f¥1 ::: /, '1-11fYl
A/.t0J
wi-lh no /O.r.t8s (rr;m Ihe lake 10 the f41J~
If of fi+hp = It +ir S~ 11141 wf!h Ie::l) al/d Y.<' ~ ::"~
h, =:: f? W/JldJ CAli De r;()/'IIb//lerJ w/I/J -4(I) ff) five
h :: JZ[/+£l]fJ,.., I /)J WlW6 ~ '" l, :: .1Z (~ • "'..5. o91/i
I '" ,,.11)
and hp::: 'l-S J-2.7.S' -S'f~:z. so fhal Elf (f:) ber;ollle.r
'$'~ +:1.7.£ ( /.IfN~/)- ~If. (I.'PI(t,/ = :r;:.~([!.fQ. 02. llj ]or I~
/31r.r/·-~(}.9QI -~Sr:O wht'clJ /;4,S the so/v/;'61l
1:>/ = I/-o.9.t ii'l().f)"+'f(~)('3,j 0 76 II
J
I br. ,J.' ,/. h
'-r, (/3 J) - • 3 S or - O.7V '().W{I/r;II 'tiS IJO
2 phYJ"/&4/II1Caf}iIJf)
IIIsoJ
t¥2.. ::: O.'MJ Cl ::: O,'HJ((),7P31j3) :::: 0.37.2. !J3
8-100

BJOIf J
X.IO-l A 2-in.-diameter orifice plate is inserted in a 3-in.-
diameter pipe. If the water flowrate through the pipe is 0.90 cfs,
determine the pressure difference indicated by a manometer
attached to the flow meter.
v , t
.-.( ~=3jn.
1..1.
d==2in.
IT
Q _ 1L _ 2. in. _ 2
to' - D - 3 in. -"3 J
Ji3
Q:: O. 70 s J and
fJ -lld0- If
V~1l~ = O.7-!f =ltf.26 Ii
If D f (f,. rf)~ ~
6
=2.'1Sx/o
~-IOJ

8./0S I
8.105 Air to ventilate an underground mine flows through
a large 2-m-diameter pipe. A crude flowrate meter is constructed
by placing a sheet metal" washer" between two sections of the
pipe. Estimate the flowrate if the hole in the sheet metal has a
diameter of 1.6 m and the pressure difference across the sheet
metal is 8.0 mm of water.
or
Q == 29.5 Co -¥
DV (2m) V
AisOJ Re ::: -:;;- = ---=-----";"----=----rv
" /./fb X/O-S 1!L-
and d
~; -D = 1..6m
2.0 hi.
:::0.8
Trial and errol' so/tJlion:
.s
or
3
Assume G,;::: 0.6/ so fhQf from Eq.O), Q:::29.5 (0.61):: /8.0..!p-
m3
Ll v- Q - Jao-:s 5 ./!1..
nenceJ - 7f - Il ( . )2. ::. ..73 oS
JI. 2.01l'J.
Froth £q'(~)J Re -:::1.37x/Os (5.73) ::' Z 85 x/OS
This Re and @ 9ive Co::: 0,6/ (see Fi1.8.'1-.1) wh/ch "1rees
w/fh fhe ps.!vmed valve.
m3
ThfJS; Q-::: /8.0 r!
8-102
0)
(2.)

9./06 T
8./07 I
8.106 Gasoline flows through a 35-mm-diameter pipe at a
rate of 0.0032 m3
/s. Detennine the pressure drop across a flow
nozzle placed in the line if the nozzle diameter is 20 mm.
1.'1 -c Ll /2(-p,-f>l)' h t:J~.d. = 20 mm = 057/ IJ = lld2 (I)
'¥ - n nn p(J-~"') ,were t" D 35mm . 7 n Jf
-From Toble /.6 0.:: 680 ~ and }l =3./x/0'1- ~':' m 3
ThuS1 Re :::: ejD where V=~ == O,00321F = 3.33 111
fh f J ~D .ff(O,o3smt s
so OR = (68of!&)(3.33f)(o.035 m) = 2.56 x/Os
e 3.1 xIO-1f'!!.-:fm
lieneel from Fi9- 8.4.3 J en == OJI86
Froth E'{. (J) I
O. 0032 ~ =(O.986).1j (o.02.0m)" (. ;,if-fo) If)
690""-1 J -0.57/
or '''-
11 -fJ,-= 3.2'fX/O¥.g.'1. = 32JfkPa
8.107 Air at 200 of and 60 psia flows in a 4-
in.-diameter pipe at a rate ofO.S21b/s. Determine
the pressure at the 2-in-diameter throat of a Ven-
turi meter placed in the pipe.
8-/03

8. /OPJ 8.108 A 50-mm-diameter nozzle is installed at
the end of a 80-mm-diameter pipe through which
air flows. A manometer attached to the static
pressure tap just upstream from the nozzle in-
dicates a pressure of 7.3 mm of water. Determine
the ftowrate.
v
Q C. 11 2. (PI-e;).L 50mm d
J:: n nn pO _~If) I Wnere ~ = 80mm = 0.:625 an
..! h::7.3 mm H,.o
f
1,-/2== 0H oh =(q800~ )(7.3XIO-
3
m) ==7/.5J:{,"
Thus, with lin;:: Ifd~" m
Q=C. 1l(O. 050ml" ~ (7/.5~)
n If (/IZ3~)(J-0.62.SII)
or 3
Q:: 0.0230 en IJssiJ/1'Je Cn=0.97 so that Q= O.0223J}
and V=JL = O.0223-f-3 :: 'f.'1'f/ll
fD'" 1J(O,o8m)'" oS
or ~
Re :: JLQ = (If.lI/f oS )(O.o8m) :; ~,Jf3Xlo'l Wilh fhe Re and awe oblain
'Y !.lftx10-.5 !:~ . "'
Cn = 0.963 :/:0.97 (fhe ass{)med value) (See Flq.8.Jf3)
3
ThusJ assume en:: 0.963 So fhai Q = 0.0230 (0.963):; O.022/.if
J JlL3
ana V= 0.02.2/ $ = If: 'fO J!l
f(O.OBml S
C'L '- C . I./'th R (Jf.Jfo~)(0.oBm2 ~ b1 ' C.neCK n L WI e== /.~6xJO-.s.!§3 =2./fIX/O we 0 10111 n-;::: 0.963
from Fi9- 8.11-3 (checks)
TL ~ m3
I nUs, l';;:: O. 02. 2/ T
8-10'1-

.... .."rQ/r
: --1-
8./09 J
8.109 A 2.S-in.-diameter nozzle meter is installed in a
3.8-in.-diameter pipe that carries water at 160 of. If the inverted
air-water U-tube manometer used to measure the pressure dif-
ference across the meter indicates a reading of 3.1 ft, determine
the flowrate.
h=3.lff .
,
V I f . (I) ... +
- ..} D=3.Bm.· • d=2 5J'n
() ! ~f' ,
Q=Cnlln 2({1-PtJ.) where ~=Ji = 2.5(n, =0.658 (I)
e{l-~'1) J D 3.8 m,
C 7:ll 8 slv9:!, -& Jb's t,rom IQole .J: P=I.896 fl3 JjI-=8.32X/ O -fP. sofha
Re:= £VD =(J.eQ641¥) v(flfI)
ft- 8.32x/o-6
Jti!
or ~ 1i
Re :: 7. 2.2x10 VJ where V.... 0$ (2.)
f)/SOI wilh (} =:: 1fD2V £1. (JJ bec()mes (usi"9 {1-(J:;.:;: 0'h):
.1l(~ fI)2. V= C .1l(3:§.ff)~[ 2 (32.2.~)(1.8q6 ¥;f)(3.lf-l) J~
If /2 n If /2 (1.8Q6 o$il105)(1-O.658/f)
or
V::. 6.78 en (3)
Trial and error solo/ion ().Sif}9 Fi9' 8,/f3 for en-::Gil (ReI ~:::a6s8) ;
lissome en=0.Q9 From £1,(3) V= 6.78 (0.99):: 6.11ij
From Ef(. (2) Re;: 7.?..2.XIOII-(6.7/~) ='f.8/f"t/o
s which from
F'9. 8.tt7 9ives en=-0.99 (checks wifh fI.uumed vqllJe)
Thf)~J V=6.71# Ql1d Q=~llVr:::f(~"lft)2(6.7/If):: O.5281j-3
8-/05

8,110 J
8.110 Water flows through the Venturi meter *.(I) (2.)! . Q '
shown in Fig. P8.110. The specific gravity of the 6 re • 3 In. ~ I
manometer fluid is 1.52. Determine the flowrate. ~ 1---------1/
1~_:2 in. ~tm
t 'SO = 1.52
FIGURE P8.110
Q::: C II Z({1-"fo.)· h ~ == JL == 3 in, - 0 5
v T p(J-(3'1-) , were " D 6in. - .
Also,
f' +'II = f~ +h'(l-h) +~(SG)h or f,-I2.:= o(SG-J)h = P9(SG-I)h
HenceI ,
/) - G II /2 eg(S6-nh
L' - v T p(J-~If) or
Q= C l£(~f.ll [ 2(3:;..2fA)(J'E2-J)(lifl)]~
v If /2. [ (1- 0.511 )
Thus,
Q= 0, I/qB Cv Assume Cv';:. 0,9£1 so tha+ Q;:. 0.1198 (O.98)=O.1J71j3
Hencs 1i.~
J@ ~H7~ li
V;;:"2. :::; 1l (6 )1. :::; 0.596 oS so thaI
IfD 'I- jiH
Re = 'tQ == (O.5q6!f)(~FI) :::; 2.'1-6 X/O'l-
'Y 1.2/XIO
s
#-
From Fi9' 8.'15 cd this Re J
Cv~O.96 :/=0.98.1 fhe assumed value.
Hehce ass ume Cv =0.96 orJ J
Q'" O. /Ifl8 (0.96) '" 0.11s1f3 and V=':('1)'" ::0.586 y
(
6 If 1'1.
There{o,.°e R - 0.5S6 nJ -2 Lf2.XIOIf In I h R 8 ¥sJ e - /. 2J x 10-.£ - . so r.r rom '9, ' J
Cv ~ 0.96 Checks 'IIdh l).r-svlfJed valve.
Hence) Q::: 0, I IS ¥3
8-/06

B.II/ I
X. j II If the fluid flowing in Problem 8.110 were air, what
would the flowrate be? Would compressibility effects be im-
portant? Explain.
*6'tPl (l,) .13;0 ~ ')
~,." ~ ,----_---.J/1. ~_:2 in. ~tm
t 'sa = 1.52
FIGURE P8.110
or
Q:= 5. elfCv .fl.'! (:2.)
fl3 ~ S.61 oS fi.
As-s(}me Cv=0.96, or Q= .s.61'S so thai V=!fD2 =f (f,.Jfl -= 28,6 s
Hence. 1:1 ('.9- )
D _ VlJ - (l8.6o$) p.fl - CJ it Fi r fhFIe - - - -'l-Rl -7,1/ X/O rom r/f]- SA'S is qive.sv I,S7x/O :s
Cl~O.975 or Q=5.8Jf(O.97S)== .5.69!!.3
Nofe: Wah v= "*=:/1-)". "'29.0 f/- we oh14in
'I- 1'-
Re = 29.0(1£) = 9 2IfX/O"" T/J/)SJ from Flo. 8.'»'5 c.v~ O.97S
J/,57)(/0-'1 " 7
which 49rees wdh /he Qss()/lled
valve.
g-/07

8./12
8.112 Water flows through the orifice meter shown in
Fig. P8.112 at a rate of 0.10 cfs. If d = 0.1 ft. determine the
value of h.
I h
~
FIGURE PS.1l2
Q='CoA.y ~t':p{J where e=*=' ~/; =0.6, f1t -/~4h =e~;'
II/so, -lL _ 0.10#3 _ it
V-.g/)2. - 14(Ii.Hi - /f.58.s so thor
Re == Y!2 == (Jf.58fJ)(k~t) =6.3/XIO~ Hence) from Fi.9
· .8Jfl) Co::::O.6/6
1/ /.2.lx/fs ¥
Therefore} from E1. {/J!
O.IO.[j-3 =(0.616) ¥(0.Iff?" 2 P(32.21iJh or h==5.. 771;
pu-0.6/1-) TJ
9,/13
8.113 Water flows through the orifice meter shown in
Fig. P8.112 at a rate of 0.10 cfs. If h = 3.8 ft. determine the
value of d.
(f)
Q== r L1 2{-(J,-(Jz.) h ar:& tl =_d_ . - - It''h =: h (j)
V()11o e(J - (J~) J were :' 15 I~ ft J fJl f:J., - 0 . f'J
and, V= Si... = O.IO~ =~.58It R :: ~::: ('!.ssY~(Aff) == 6.3/X10'f
.:JlD2. 1l(.!.f~)2. S ) e ".. I 21XI0 ..s1L~ # I~ ~ • s
Trial and error sO/lJfion: From Ef· O) I
x.
o,if == c. 11.(d ff)'2.f 2 e(32.zYa)(3.BfI)J2. orQOOB/If= CQ d2.. he d~N Ct.)
..s 0 If f (/-~If) • . { J_ ~'f ) W 1'e I
/Jssvme ~= O.6} or d= I~~ =: ~ (0.6)== 0, I off. rhv~ from £rt.(:t-)
Co = 0,7£9. !loweverJ
(rom Fi9,8.91 for fhis He and ~J Co=O.6IS rFO.7S9
Assume ~ =0, 65, or d== -h(0.65) = 0, J08(-I. From £".(;.) Co = 0.6 33.
Fro", F;,- e.III J Co = 0.618 :# 0.633. .
A,ssume ~ =0.67) or d== 12.(o.67) ::: 0.112 fl. From £f' (,,) Co = 0, SBo
From Fi9' 8.'fl J Co.::0,6/9 :;:0.S90 ThlJs, d~ 0,109 (1
8-108

e.II'I- I I h =1.6 (f
K 114 Water flows through the OIifice meter shown in
Fig. PS.1 12 such that h = 1.6 ft with d = 1.5 in. Detennine the
flowrate.
~
FIGURE PS.1l2
Q c. Ll 2(f1-(h.) h d /.5ill. d }oth h= ono p(J-~'1) J were p;: 15 = 2. in. :;:; 0.75 an /1-1:1.-:: 0 ;: e1
Thvs, it. ~
Q=c 1l(1,5 )2.[ 2 e(32.2~){J.6H)]
o I/- ITff f (/- 0.7£11")
or
Q=O./S/ CO . 1Z. (I)
Also Re::: ~ = V(7i.f1) ;:: 1.3BJ<IO
If
VJ
~he/'e V=!-/J2.;;;: /f5.Bt:'l (2.)
, 'V 1,'J.IX/fT fL2 -: 'Y
s
Trial olld error so/urion;
IIssvme Co =O.6j or from £".(OJ ~ =- 0.151 (0.6)::: 0.0906 £j3
HenceJ
from Eft. (2.), V= '15.8(0.0906)::: Jf.15 and Re =5,73X/Olf
From Fi9. e,If I wifh this He and ~) Co;;;: 0,62 :F 0,6 (the qssvmed Vt,J/ue)
lissome Co= 0,62 or Q=-O./51(o.6J.)::. 0.09361j3, Th/lS V:::~S.B (0.0936)
or V=if.29!j and Re -:::,s.92X/rI, !Tom Fi9' 8.'1-/ J Co =O.62; Me
()ssflmed VtI/lJe.
fJ!lienee, r;.:: 0,0936 s
8-IOQ

8.lIS
S.I 15 The scale reading on the rotameter shown in Fig.
P8.115 and Video VS.6 (also see Fig. 8.46) is directly propor-
tional to the volumetric flowrate. With a scale reading of 2.6
the water bubbles up approximately 3 in. How far will it bub-
ble up if the scale reading is 5.0?
I!:L +Z -tYL7.-h ::: 1J2.~.J.Z +~
jt I 2{f rl ~ z. 2.?
t
~ FIGURE PS.115
where .
I, ::I,.~IJ J Z, ::: OJ ~ ~()~. s() thai w/lh no /o.r.res (hI. ~(J~
(0
It/2.
"£j ::= Zz
For the f'Of4meter
Thv.r
J /) KtSR
~::-ft~T
r¥ ::: KI(SR WhC/'9 SR ~sca/e reat.iilkl (,fAd
J{is a CQIJ.S141J1.
so th4 i whell c()lfJ/;i/J~J w/IA Elf. (/~
,l<2.(SB)2. :::£ K2.(l./~'- _ .l
Al-(z!-) ,. or 1;(~,) - (r;.fI)
By Jil/iJiA~ 11J6S8 fWfJ e(I'/4ti()ll~
(s.o/- == -f-- or f; == O.92s" If =//'/117.
(2.&)7- (r;.ff)
8-110

8.116 I
8.116 Friction Factor for Laminar and Transitional Pipe Flow
Objective: Theoretically, the friction factor, f, for laminar pipe flow is given by
f = 641 Re, where the Reynolds number, Re = pVDIIL, is based on the average velocity, V,
within the pipe and the pipe diameter, D. Also, the flow is normally laminar for Re < 2100.
The purpose of this experiment is to use the device shown in Fig. PS.116 to investigate these
two properties.
Equipment: Small diameter metal tubes (pipes), air supply with flow regulator, rotame-
ter flow meter, manometer.
Experimental Procedure: Attach a tube of length L and diameter D to the plenum. Ad-
just the flow regulator to obtain the desired flowrate as measured by the rotameter. Record
the manometer reading, h, so that the pressure difference between the plenum (tank) and the
free jet at the end of the tube can be determined. Repeat for several different flowrates and
tube diameters. Record the barometer reading, Hbar, in inches of mercury and the air tem-
perature, T, so that the air density can be calculated by use of the perfect gas law.
Calculations: For each of the data sets determine the pressure difference, Ap ='Ymh,
between the plenum pressure and the free jet pressure. Here "'1m is the specific weight of
the manometer fluid. Use the energy equation, Eq. 5.84, to determine the friction factor, f
Assume the loss coefficient for the pipe entrance is KL = 0.8. Also calculate the Reynolds
number, Re, for each data set.
Graph: On a log-log graph, plot the experimentally determined friction factor,/' as ordi-
nates and the Reynolds number, Re, as abscissas.
Results: On the same graph, plot the theoretical friction factor for laminar flow,
f = 64/Re, as a function of the Reynolds number. Based on the experimental data, determine
the maximum value of the Reynolds number for which the flow in these pipes is laminar.
Data: To proceed, print this page for reference when you work the problem and elide hen'
II FIGURE PS.116
e-/II

8./16 I (CDIJ'f)
Solution for Problem 8.116: Friction Factor for Laminar and Transitional Pipe Flow
L, in. Hatm, in. Hg T, deg F
24 28.9 73
Theoretical
h, in. a, ml/min a, cfs v, fps Re f Re f
D =0.108 in. Data 100 0.6400
7.5 6600 0.003887 61.11 3202 0.0341 2100 0.0305
6.75 6200 0.003652 57.40 3008 0.0349
6.26 6000 0.003534 55.55 2911 0.0345
5.54 5650 0.003328 52.31 2741 0.0344
4.66 5150 0.003033 47.68 2499 0.0349
4.29 5000 0.002945 46.29 2426 0.0339
3.92 4860 0.002863 45.00 2358 0.0325
3.48 4600 0.002709 42.59 2232 0.0322
3.21 4500 0.002651 41.66 2183 0.0307
2.34 3700 0.002179 34.26 1795 0.0338
1.86 2900 0.001708 26.85 1407 0.0461
1.11 1800 0.001060 16.67 873 0.0758
0.63 1100 0.000648 10.18 534 0.1194
D =0.046 in. Data
9.52 560 0.000330 28.58 638 0.1007
7.68 475 0.000280 24.24 541 0.1134
7.08 425 0.000250 21.69 484 0.1311
5.26 315 0.000186 16.08 359 0.1785
3.39 221 0.000130 11.28 252 0.2348
2.61 165 0.000097 8.42 188 0.3256
D =0.063 in. Data
4.58 925 0.000545 25.17 770 0.0838
3.32 680 0.000401 18.50 566 0.1140
2.51 530 0.000312 14.42 441 0.1431
1.48 325 0.000191 8.84 270 0.2270
0.86 190 0.000112 5.17 158 0.3893
P = Patm/RT where
Patm =YH20*Hatm =847 Ib/ftJl3*(28.9/12 ft) = 2040 Ib/ftJl2
T =73 + 460 =533 deg R
Thus, P =0.00223 slug/ftJl3 and Y=p*g =0.0718 Ib/ftJl3
Also, ).l =3.83E-7 Ib s/ftJl2
Theoretical for laminar flow: f =64/Re =64/(pDV/).l)
/!p/y =(fUD + KL + 1)(VJl2/2g) where KL = entrance loss coefficient =0.8 and V =a/(nDJl2/4)
~-//2-

8.//6
1.00
-
0.10
Problem 8.116
Friction Factor, f, vs Reynolds Number, Re
~:-=:- -=~i:-:== ! f! i : :~
-+---~~-~~:--.-~.-~--'-----'-----l. . I
- -.-----.-- ------~---'b.-"-----
.------.. - .. --------~:r-
- -- ..---------------.;......-~-
..- ..--------,--~--'--'
-------~------_r_~--+--i---""'---.
: ' I ,
-- --------4------+--'--+---------"~__.;-~->___+-+--+--+ • Experimental, D =0.108 in.
• Experimental, D = 0.046 in.
• Experimental, D = 0.063 in.
--Theoretical, laminar
0.01 -l-----------'--"-r------~---~""___1
100 1,000
Re
10,000
8-/1.3

8,117
8.117 Calibration of an Orifice Meter and a Venturi Meter
Objective: Because of various real-world, nonideal conditions, neither orifice meters nor
Venturi meters operate exactly as predicted by a simple theoretical analysis. The purpose of
this experiment is to use the device shown in Fig. P8.117 to calibrate an orifice meter and a
Venturi meter.
Equipment: Water tank with sight gage, pump, Venturi meter, orifice meter, manometers.
Experimental Procedure: Determine the pipe diameter, D, and the throat diameter, d,
for the flow meters. Note that each meter has the same values of D and d. Make sure that the
tubes connecting the manometers to the flow meters do not contain any unwanted air bubbles.
This can be verified by noting that the manometer readings, hy, and ho, are zero when the sys-
tem is full of water and the flowrate, Q, is zero. Tum on the pump and adjust the valve to
give the desired flowrate. Record the time, t, it takes for a given volume, V, of water to be
pumped from the tank. The volume can be determined from using the sight gage on the tank.
At this flowrate record the manometer readings. Repeat for several different flowrates.
Calculations: For each data set determine the volumetric flowrate, Q = VIt, and the pres-
sure differences across each meter, IIp = 'Ymh, where "1m is the specific weight ofthe manome-
ter fluid. Use the flow meter equations (see Section 8.6.1) to determine the orifice discharge
coefficient, Co, and the Venturi discharge coefficient, Cy, for these meters.
Graph: On a log-log graph, plot flowrate, Q, as ordinates and pressure difference, IIp, as
abscissas.
Result: On the same graph, plot the ideal flowrate, Qideal (see Eq. 8.37), as a function of
pressure difference.
.. FIGURE P8.117
8-/ltf

8.//7 T(con)" )
Solution for Problem 8.117: Calibration of an Orifice Meter and a Venturi Meter
d, in. 0, in. V, gallons
Ideal
0.625 1.025 2.00
C = 1
t, s ho, in. hVI in. L1PoI Ib/W2 L1PvI Ib/ft"2 A, ft
3
/s Co Cv L1P, Ib/ftA2
27.0 9.3 3.8 48.4 19.8 0.0099 0.611 0.956
13.2 37.1 14.5 192.9 75.4 0.0203 0.626 1.001
34.2 5.5 1.9 28.6 9.9 0.0078 0.627 1.067
16.6 23.9 10.1 124.3 52.5 0.0161 0.620 0.953
12.0 43.2 18.1 224.6 94.1 0.0223 0.638 0.985
11.7 51.3 21.7 266.8 112.8 0.0229 0.600 0.923
15.4 27.9 11.2 145.1 58.2 0.0174 0.618 0.976
25.1 10.1 4.2 52.5 21.8 0.0107 0.631 0.978
20.4 14.7 6.2 76.4 32.2 0.0131 0.643 0.990
17.3 21.4 8.7 111.3 45.2 0.0155 0.629 0.986
15.7 26.7 11.2 138.8 58.2 0.0170 0.620 0.957
Average discharge coefficient: 0.624 0.979
a = V gal/t s x (231 in."3/gal)x(1 ft"3/1728 in."3)
L1P = YH2o*h = 62.4 Ib/ftA3 *h ft
Ov =A2/[1 - (A2/A1)"2]"0.5*Cv*(2*g*L1p)YH2o)"0.5
and
0 0 =A2/[1 - (A2/A1 )"2]"0.5*Co*(2*g*L1PoIYH20)"0.5
where
A1 = 1t 0"2/4 = 1t (1.025/12 ft)"2/4 = 0.00573 ft"2
and
A2 = 1t d"2/4 = 1t (0.625/12 ft)"2/4 = 0.00213 ftA2
Problem 8.117
orifice venturi
Flow Rate, Q, vs Pressure Difference, ~p
iii....,. J••: , :'
I : i : :
i : Iii
iii! ! I
! I:
i i!i I I '. i I
0.001 +-_--'-_..!...i---L:.....I-~'..!...:'-tl!---'----.!..--'---'-...........-.W....{
10 100
~PI Ib/ftA2
1000
8-/IS
• Experimental, orifice
• Experimental, venturi
- Theory, C =1
18.0
75.5
11.2
47.7
91.4
96.1
55.5
20.9
31.6
44.0
53.4

8.//8
8.118 Flow from a Tank through a Pipe System
Objective: The rate of flow of water from a tank is a function of the pipe system used
to drain the tank. The purpose of this experiment is to use a pipe system as shown in
Fig. P8.118 to investigate the importance of major and minor head losses in a typical pipe
flow situation.
Equipment: Water tank; various lengths of galvanized iron pipe; various threaded pipe
fittings (valves, elbows, etc.); pipe wrenches; stop watch; thermometer.
Experimental Procedure: Use the pipe segments and pipe fittings to construct a suit-
able pipeline through which the tank water may flow into a floor drain. Measure the pipe di-
ameter, D, and the various pipe lengths and note the various valves and fittings used. Mea-
sure the elevation difference, H, between the bottom of the tank and the outlet of the pipe.
Also determine the cross-sectional area of the tank, Atank• Fill the tank with water and record
the water temperature, T. With the pipeline valve wide open, measure the water depth, h, in
the tank as a function of time, t, as the tank drains.
Calculations: Calculate the experimentally determined flowrate, Qex' from the tank as
Qex = -AU!nk dh/dt, where the time rate of change of water depth, dh/dt, is obtained from
the slope of the h versus t graph. Select a typical water depth, hI> for this calculation.
Graph: Plot the water depth, h, in the tank as ordinates and time, t, as abscissas.
Results: For the pipe system used in this experiment, use the energy equation to calculate
the theoretical flowrate, Qth, based on three different assumptions. Use the same typical water
depth, hI> for the theoretical calculations as was used in determining Qex. First, calculate Qth
under the assumption that aJ1losses are negligible. Second, calculate Qth if only major losses
(pipe friction) are important. Third, calculate Qth if both major and minor losses are important.
Data: To proceed, print this page for reference when you work the problem and clu:k hen'
D
Floor
.. FIGURE P8.118
8-1/6

Solution for Problem 8.118: Flow from a Tank Through a Pipe System
The pipe is galvanized iron with threaded fittings.
The system contains:
D, in.
0.595
one sharp edged entrance
one fully open globe valve
two 45-deg elbows
four 90-deg elbows
Atank, ft"2
0.654
H,ft
1.00
Total pipe length, in.
135
T, deg F
71
h, ft t, s
1.00 0
0.90 13
0.80 26
0.70 40
0.60 54
0.50 67
0.40 81
Experimental: Qex = -(dh/dt)*Atank = -(0.0074 ftls)*(0.654 ft"2) = 0.00484 ft"3/s
Theoretical with no losses: Qth = V2*A2, where when h = 0.90 ft
V2= (2g*(h + H))"0.5 = (2*32.2*(0.9 +1.0))"0.5 = 11.06 ftls
and with A2 = rcD"2/4 = rc*(0.595/12 ft)"2/4 = 0.00193 ft"2
Qth = 0.00193 ft"2*(11.06 ftls) = 0.0213 ft"3/s
Theoretical with major losses: Qth = V2*A2, where the energy equation gives
h + H = V//2g(1 + fLID), where again use h = 0.90 ft and f is a function of Re and £ID
Thus, with h =0.90 ft,
1.9 = (V//64.4)*(1 + f*135/0.595), or
122.4 =V/*(1 + 227f)
Re = V2D/v = V2*(0.595/12 ft)/(1.04E-5 ft"2/s) = 4768*V2
and
£ID =0.0005 ftI(0.595/12 ft) = 0.0101
Trial and error solution: Guess f, solve for V2, calculate Re, obtain new f from Moody chart
The solution is: f =0.041, V2= 3.44 ftls, Re = 16,430
Qth = 0.00193 ft"2*(3.44 ftls) = 0.00664 ft"3/s
Theoretical with major and minor losses: The energy equation gives
h + H = (1 + fLID +L.KdV//2g
where L.KL = 0.5 + 10 + 2*0.4 + 4*1.5 = 17.3
Thus, with h =0.9 ft
1.9 =(V//64.4)*(17.3 + f*135/0.595), or
122.4 = V/*(17.3 + 227f)
Trial and error solution gives: f =0.42, V2= 2.14 ftls, Re =10,200
Qth = 0.00193 ft"2*(2.14 ftls) = 0.00413 ft"3/s
(con J-/)
8-117

~-------------------------------------------------
I
1.2
1.0
0.8
~ 0.6
.c:
0.4
0.2
0.0
Problem 8.118
Water Depth, h, vs Time, S
-6c-----+----~----_r___----~-1
I-------~~~~----1
I
I
I i .
h = -q.0074t if 0.9965!
!
0 20 40 60 80 100
t, S
8-//8
I • Experimental I I
I-Linear h vs t II

8.//9
8.119 Flow of Water Pumped from a Tank and
through a Pipe System
Objective: The rate of flow of water pumped from a tank is a function of the pump prop-
erties and of the pipe system used. The purpose of this experiment is to use a pump and pipe
system as shown schematically in Fig. PB.119 to investigate the rate at which the water is
pumped from the tank.
Equipment: Water tank; centrifugal pump; various lengths of galvanized iron pipe; various
threaded pipe fittings (valves, elbows, unions, etc.); pipe wrenches; stop watch; thermometer.
Experimental Procedure: Use the pipe segments and pipe fittings to construct a suit-
able pipeline through which the tank water may be pumped into a sink. Measure the pipe di-
ameter, D, and the various pipe lengths and note the various valves and fittings used. Mea-
sure the elevation difference, H, between the bottom of the tank and the outlet of the pipe.
Also determine the cross-sectional area of the tank, Atank• Fill the tank with water and record
the water temperature, T. With the pipeline valves wide open, measure the water depth, h, in
the tank as a function of time, t, as water is pumped from the tank.
Calculations: Calculate the experimentally determined flowrate, Qex' from the tank as
Qex = -Atank dhldt, where the time rate of change of water depth, dhldt, is obtained from
the slope of the h versus t graph.
Graph: Plot the water depth, h, in the tank as ordinates and time, t, as abscissas.
Results: For the pipe system used in this experiment, use the energy equation to calcu-
late the pump head, hp, needed to in order to produce a given flowrate, Q. For these calcu-
lations include all major and minor losses in the pipe system. Plot the system curve (i.e.,
pump head as ordinates and flowrate as abscissas) based on the results of these calculations.
On the same graph, plot the pump curve (i.e., hp as a function of Q) as supplied by the pump
manufacturer. For the pump used this curve is given by
hp = -2.44 X 105
Q2 + 51.0 Q - 12.5
where Q is in feIs and hp is in ft. From the intersection of the system curve and the pump
curve, determine the theoretical flowrate that the pump should provide for the pipe system
used.
Data: To proceed, print this page for reference when you work the problem and dick here
• FIGURE PB.119

Solution for Problem 8.119: Flowrate of Water Pumped from a Tank and Through a Pipe System
The pipe is galvanized iron with threaded fittings.
The system contains:
0, in.
0.625
h, in.
25
24
23
22
21
20
19
18
17
16
15
one sharp entrance
eight 90-deg elbows
two 45-deg elbows
two globe valves
one union
Atank, ft"2 H,ft
0.647 3.50
t, s
0
7.6
16.1
25.2
32.3
40.8
48.9
57.7
65.7
74.9
82.7
Experimental:
Total pipe length, in.
242
Pump equation
hp, ft
12.50
12.31
11.63
10.46
8.80
6.66
4.02
0.90
T, deg F
62
Q, ft"3/s
0.000
0.001
0.002
0.003
0.004
0.005
0.006
0.007
Qex =-Atank*(dh/dt) where from the graph, dh/dt =-0.1204 in.ls
Thus,
Qex = -(0.647 ft"2)*(-0.1204/12 ft/s) = 0.00669 ft"3/s
Theoretical:
The energy equation gives
h +hp - hL = H +V2/2g, where
System equation
V,ft/s Re f hp, ft
0.00 o 2
0.47 2070 0.0309 2.16
0.94 4140 0.0490 2.73
1.41 6210 0.0470 3.62
1.88 8281 0.0450 4.84
2.35 10351 0.0430 6.37
2.81 12421 0.0425 8.27
3.28 14491 0.0420 10.50
hL = (fLlO + LKd*V2/2g = (f"(242 in.lO.625 in.) + 0.5 + 8*1.5 + 2*0.4 + 2*10 + 0.08)*V
2
/2g
= (387*f + 33.4)*V2/(2*32.2) =(6.01*f + 0.519)*V
2
Thus, with h =18 in. =1.5 ft,
hp =H - h + hL + V2/2g =3.5 - 1.5 + (6.01*f + 0.519)*V
2
+ V
2
/(64.4)
or
hp =2.0 + ( 6.01*f + 0.535)*V
2
But V =Q/A =Q/(rc02/4) =Q/(rc*(0.625/12 ft)"2/4) =469*Q
Thus, the system equation is
hp = 2.0 + ( 6.01 *f + 0.535)*(469*Q)2 = 2.0 + (1.32E+6*f + 1.18E+5)*Q2
Also, obtain f from the Moody chart with
Re =VO/v =V*(0.625/12 ft)/(1.18E-5 ft"2/s) =4414*V
c/O = 0.0005 ft/(0.625/12 ft) = 0.0096
From the graph, the pump and system equations intersect at Qth = 0.0051 ft"3/s
( ('0,,'1)
8-120

8./19
C
.£
30
25
20
15
10
Problem 8.119
Water Depth, h, vs Time, t
h=-O.1204tf+-24.9 ;
5
0
0 20 40 60 80 100
t, S
Problem 8.119
Pump Head, hp, vs Flowrate, Q
i • ' L'!
8 +--l--~I--i'-+-+~.f-.----'Il-+-¥-.·-+--+-crr-+--'---+
. i'/ I
..Jf' ! i I
I ! I I ! ! _l !
1 ; i
"'1o +---J.--'--i--'--~---r_,..J,J_-,-1. -r--'--........-f
0.000 0.002 0.004 0.006 0.008
Q, ftJ3/s
8-/"-1
• Experimental
-Linear h vs t
-+-Pump curve
-+-System curve

a, }20 I
8.120 Pressure Distribution in the Entrance Region of a Pipe
Objective: The pressure distribution in the entrance region of a pipe is different than that
in the fully developed portion of the pipe. The purpose of this experiment is to use an ap-
paratus, as shown in Fig. P8.l20, to detennine the pressure distribution and the head loss in
the pipe entrance region.
Equipment: Air supply with flow meter, pipe with static pressure taps, manometer, ruler,
barometer, thennometer.
Experimental Procedure: Measure the diameter, D, and length, L, of the pipe and the
distance, x, from the pipe inlet to the various static pressure taps. Adjust the flowrate, Q, to
the desired value. Record the manometer readings, h, at the various distances from the pipe
entrance. Record the barometer reading, HbaI> in inches of mercury and the air temperature,
T, so that the air density can be calculated by use of the perfect gas law.
Calculations: Detennine the average velocity, V = Q/A, in the pipe and the pressure
P = 'Ymh at the various locations, x, along the pipe. Here 'Ym is the specific weight of the
manometer fluid.
Graph: Plot the pressure, p, within the pipe as ordinates and the axial location, x, as
abscissas.
RESULT: Use the graph to detennine the entrance length, Le, for the pipe. This can be
done by noting the approximate location at which the pressure distribution becomes linear
with distance along the pipe (i.e., where dp/dx becomes constant). Use the experimental data
to detennine the friction factor for fully developed flow in this pipe. Also detennine the en-
trance loss coefficient, KLent•
rh
~
iIii FIGURE P8.120
~-I2.2.

Solution for Problem 8.120: Pressure Distribution in the Entrance Region of a Pipe
0, in. L, in. Q, ft"3/s Hatm, in. Hg T, deg F
0.74 50 0.481 29.7 75
x, in. h, in. p, Ib/ft"2
0 9.98 51.9
1 7.21 37.5
2 6.61 34.4
4 6.19 32.2
6 5.82 30.3
10 5.15 26.8
15 4.23 22.0
20 3.64 18.9
30 2.28 11.9
40 1.09 5.7
50 0 0.0
P = Patm/RT where
Patm = YHg*Hatm = 847Ib/ft"3*(29.7/12 ft) = 2096Ib/ft"2
T = 75 + 460 = 535 deg R
v = Q/A =(0.481 ft"3/s)/(n*(O.74/12 ft)12/4) = 161 ftls
p = YH2o*h
From the graph, the p vs x results are linear after (approximately) x =15 in. Thus, Le =15 in.
For the fully developed flow portion, dp/dx = -fpVI2/20 and from the graph dp/dx = -0.635 (Ib/ft12)/in.
Thus,
f = 0.635 (Ib/ft12)/in.*2*0.74 in./(0.00228 slugs/ft13*(161 ftls)12) = 0.0159
From the entrance to the exit of the pipe Pent = (KL + fUO)pVI2/2
Thus,
KL = 2Pent/(pVI2) -fUO= 2*51.9Ib/ft12/(O.00228 slugs/ft13*(161 ftls)12) - 0.0159*50in./0.74 in.
=0.'682
Results: Le =15 in.; f =0.0159, and KL = 0.682.
~-1').3

N
<
~
-.c
c.
L~__
60
Problem 8.120
Pressure, p, vs Axial Location, x
50 -j~------:-----~-------1
40
•
~ • Experimental
•30 --+-
• t:. fully developed flow
20 - Linear (fully developed
flow)
10
0
0 20 40 60
x, in.
8-/1-'1-

8.I2.J I
8.121 Power Loss in a Coiled Pipe
Objective: The amount of power, P, dissipated in a pipe depends on the head loss, hL'
and the flowrate, Q. The purpose of this experiment is to use an apparatus as shown in
Fig. PS.12l to determine the power loss in a coiled pipe and to determine how the coiling
of the pipe affects the power loss.
Equipment: Air supply with a flow meter; flexible pipe that can be used either as a
straight pipe or formed into a coil; manometer; barometer; thermometer.
Experimental Procedure: Straighten the pipe and fasten it to the air supply exit. Mea-
sure the diameter, D, and length, L, of the pipe. Adjust the flowrate, Q, to the desired value
and determine the manometer reading, h. Repeat the measurements for various flowrates.
Form the pipe into a coil of diameter d and repeat the flowrate-pressure measurements. Record
the barometer reading, Hbar, in inches of mercury and the air temperature, T, so that the air
density can be calculated by use of the perfect gas law.
Calculations: Use the manometer data to determine the pressure drop, IIp ="Imh,
and head loss, hL = IIp/y, as a function of flowrate, Q, for both the straight and coiled
pipes. Here "1m is the specific weight of the manometer fluid and "I is the specific weight of
the flowing air. Also calculate the power loss, P ="IQhL' for both the straight and coiled
pipes.
Graph: Plot head loss, hL' as ordinates and flowrate, Q, as abscissas.
Results: On a log-log graph, plot the power loss, P, as a function of flowrate for both the
straight and coiled pipes. Determine the best-fit straight lines through the data.
Data: To proceed, print this page for reference when you work the problem and click /rae
1
Air supply
h
k-Manometer
- T
1m 'I '
r ...." ~)Ol ed pipe
-
- Free jet
-- • FIGURE PS.121
8-/25

Solution for Problem 8.121: Power Loss in a Coiled Pipe
D, in. L,ft Hatm, in. Hg T, deg F
1.44 18 29.9 80
h, in. Q, ft"3/s ~p, Ib/ft"2 hL' ft
Straight Pipe Data (d = infinity)
10 1.19 52.0 709
8 1.06 41.6 568
6 0.913 31.2 426
4 0.731 20.8 284
2 0.505 10.4 142
Coiled Pipe Data (d = 8 in.)
10 0.835 52.0 709
8 0.745 41.6 568
6 0.641 31.2 426
4 0.517 20.8 284
2 0.357 10.4 142
~p = YH20h where YH20 = 62.4 Ib/ft"3
hL = ~p/y where Y=gp
P = Patm/RT where
Palm = YHg'"Hatm = 847Ib/ft"3'"(29.9/12 ft) = 2110 Ib/ft"2
T = 80 +460 =540 deg R
Thus, P= 0.00228 slug/ft"3 and Y= 0.0733 Ib/ft"3
P = (yQhJft Ib/s'"(1 hp/550 ft IbIs)
e-I2.6
P, hp
0.1125
0.0802
0.0518
0.0276
0.0095
0.0789
0.0563
0.0364
0.0196
0.0068

8.11-/
Problem 8.121
Head Loss, hL• vs Flowrate, Q
800
700
600
500
..
f, •
L
1/
~ 400
, /
/ /.s::.
c..
.r:.
0:
300
200
100
a
a
) -/
rf ('
0.5 1
Q, ftA
3/s
Problem 8.121
Power, P, vs Flowrate, Q
1.000 -,.------~-----...,
0.100 +------>;Q;;----t1--------;
(coiled)
P = 0.0679Q2.87
(straight)
0.010 +-----,...--1---------;
0.001 +-------+------.......
0.1 10
Q, ftA
3/s
~-/27
1.5
~Straight pipe
_Coiled pipe
• Straight Pipe
• Coiled Pipe
- Power (Straight
Pipe)
- - - Power (Coiled
Pipe)

q,/
9.1 Assume that water flowing past the equi-
lateral triangular bar shown in Fig. P9.1 produces
the pressure distributions indicated. Determine
the lift and drag on the bar and the corresponding
lift and drag coefficients (based on frontal area).
Neglect shear forces.
U = 5 ft/s
~
p = 0.5 pU2
b = length = 4- ft
FIGURE P9.1
olJ= {fJ cose dll +f~ sin e JII, where ~ == 0
Thu.s,
rIY= {fJ ~os e dll +Jf cose dA +Sf cos eJ/I
I 2. .3
::: 2 JfJ cos 60° dll - JfJ dll == 2 (o.S~lI:z.) cos 60
0
iIJ
1 3 -(-0.25 f {j'-) i b
or
p = -0.25 pU2
otJ= 0,5 pU"ib (I)
.so fhat JJ= O.S (I.f/I} .s~~.s )(5 ~+ /"(o.lft)(LfftJ == 9.70 /b
Becavse of syrnmefry of fhe object} ;t:::~
A/so} from El(. {/J
C - ~ - ~sfY1b
D - i p7J'-A - f pY2.Jb
and .since ;t::: 0
C cI- 0
"L =-f pTlIJ =-
=;,00

9.2 r
9.2 Fluid flows past the two-dimensional bar shown in Fig.
P9.2. The pressures on the ends of the bar are as shown, and
the average shear stress on the top and bottom of the bar is Tavg.
Assume that the drag due to pressure is equal to the drag due
to viscous effects. (a) Determine Tava in terms of the dynamic
pressure, pU2
/2. (b) Determine the d~ag coefficient for this ob-
ject.
u ..
Width = b
p= +pu2
~I
I.
a) elf:;:: {ric/ion drfA1 =2 7;'19 (/oh b) == 20 hb~1I1
anrt
f' = -0.2(+pU2
)
r avg
;/-- ir avg
1011 .,
cJ,.=pf'e.$.Svre Jr41:::: -I: pyz.(hh) -(--I; pU'"(O.2)) (hh) :: 1.2 (1. py2..)(bh)
TlJv~ if 4 ~~f fhsA
I y1
20 hh faV! = 1.2 (bh) '£ P
or
tv, ::: 0.06 (fp7J~)
h) t#= 4 +4~ Cll
~ pu2.lJ:: Cll
tp1J2.bh
Thvs~
2,0/76 fall, +/.2 (ipy2) (bh) =CD dpU"hh
or
20 (0.06) (fpU2
) +1.2(ifU~) -eel) (-feV:I.)
ThvSj
eLl:::: 2. if0

9..3'" I
9.3* The pressure distribution on the 1-m-
diameter circular disk Fig. P9.3 is given in the
table below. Determine the drag on the disk.
Cll L(2.)
~......
~-
FIGURE P9.3
D
r::::r
of}== ffrill - ff rilJ = Jf (271'r dr) -11 : D2.J
si/Jce dlt:: 27Trdr
I 2. r=o
ThlJSJ 0.5 m 0 •.5
of) == 271'{ f r dr - (-Sif,.)fOmJ.) := Z7TSf r dr oJ- 3.93 kN
o 0 where LjAJ ~ r.-vIn
r /11}
Evu/lJale fhe /nle9ral numerically tlsiR9 the fOUowiIJ9 In/of/ranJ:
fr J kN/m r (m) p (kN/m2)
o
0.05
0./0
0.15
0.20
0.25
0.30
0.35
O.JfO
O.¥S
o
O.")../tf
0.'106
0.558
0.62-0
O.6QS
0.71/
0.662-
O.S6~
0 ..333
o
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
4.34
4.28
4.06
3.72
3.10
2.78
2.37
1.89
1.41
0.74
0.0
0.50 0.000 0.5
By /lSln9 fhe pr09rlJH1 SIMPSON. OilS we ohlain S{Jr dt' = O.2LJ/
~************************************************* 0
** This program performs numerical integration ·t"
** over a set a set of an odd number of equally **
** spaced point.s using Simpson f s Rule :t::t
**************************************************
Enter data points (X . y)
? O.OO.O.OOCI ,.... 0.4-5.0.333
'; 0.05.0.:14 r) 0.50.0.000
? 0.10.0.4-0E
'? O.15~O_55P
? O.20.0.62C
? 0.25,0.695
? O.30~O.711
The approximate value of the integral 15: +2.4-083E-Ol
? O~35.0.662
·~I O.4:0.D.564
q-3

9.11- I
br:Widfh
9.4 The pressure distribution on a cylinder is approximated
by the two straight line segments shown in Fig. P9.4. Determine
the drag coefficient for the cylinder. Neglect shear forces.
~ eJr
o I I e deg
< 90 180 '
" I I
P I I
, I I
f-- : :
I I
I I
_pu2 ______ '1 I
8'I!: 2,,,,, .".
,fJ-fpC()S8dlJ::fpcosB (brd9) =2.Sf- cos() (brJJe (J)
where
8cO 0
fJ :: - PyZ for of~9~" find
fJ =: -#:(,172.[1-#9] f(JrO~8~ f (,:e. fJ::: 1fU2.if f)::()
Thus) fJ ::: -f U"if er=: f)
1T 1f' 1f
2.( 2. I 2.
Sf cosf) df)~ -pU JC(Js(jd9::: ~pV sine :: eV (JJ
1V'2 ~ 7Th
and.
~ ~ ~
{ fJ CQS() tie =TPV'f[I -Ie]c()sede:: i-fV2.[siIl 9 -*(cos9 f9sin8)}
0 0 0
=t pU2.[I-~ (f) -(-1;)] :: t('u~[ ~ -2] (.1)
Th/l~ fr()111 Frs. (I), (2), IJlld (3)
tb=2br Iic()s8 de = 2 hr[ifu2.{fr-2) +p U
Z
] =ipv'('-;,hr)()
so thai
9-If

Linear distribution
9.5 Repeat Problem 9.1 if the object is a cone (made by
rotating the equilateral triangle about the horizontal axis through
its tip) rather than a triangular bar.
u = 5 ft/s
p= 0.5 pU2
• FIGURE P9.1
By symmelry the fifl IS 2ero. Thvs; tf. ::0 and CL::0
A/so ,jJ= r/{ron+ +'tear where
4ronf :: SfJ C()So dll and dlJ;::; ~rrrdx
Thv-S.J wilh fJ ==o.sf U2. ( J-lOX)
i.e. fJ/ == o,spU2.
xc:O
and
fli :: 0 J we hove
b =length =2 ft
x~~/ 0/
0.1 ft
X:::O.11iJ
~J •
4ronl '"f0.5 PU2. (J-/OX) 271(X cole) dx =27rCOSR.60'(0.5 pT72.)f(X-IO)t)dx
X=O
or 0,'
~ronf '" 27r(0.S/(o.spU")[ 4
2
_ ~xsl,o" 4runf ='0.00130q eU"
o
Also)
4ear':::: _(o.1.SpTJ2.) *(0.1)' =0,00/96 eU2
.so thaI
o:9=4ronl +tiJ,.c4r -::: (0.00130'1 -1-0.00/9/) p7J2.=O.00327 f7J2.
-so w/fh p=/,9'1 s;;rs and U:: 5 #J
tx9::= 0,00327 (/,?/fJ(5)'- :: 0./s9 II;
9-5

Q.6 I
9.(; A 17-ft-long kayak moves with a speed of 5 ftls (see
Video V9.2). Would a boundary layer type flow be developed
along the sides of the boat? Explain.
Re = Pf- J SO wdh 1:::/7ff Ql1d TJ:: 5!1 and 00'F w4fer
w/.Jh 1/= j.Z/x10-
5
1f2 we halle
Re- (J7ff)(Sfj.) - 7.02Xj06
- I.Z/x/o5
if -
Since Re ~ }OOO is offen assumed +0 be the lower Jim/I
for bOlJndary layer fype flow J if is clear thai a b()undary
loyer wov/d develop l1/on9 the sides of the kayak. ~.

9.7 ,
9,7 Typical values of the Reynolds number for various an-
imals moving through air or water are listed below. For which
cases is inertia of the fluid important? For which cases do vis-
cous effects dominate? For which cases would the flow be lam-
inar; turbulent? Explain.
Animal
(a) large whale
(b) flying duck
(c) large dragonfly
(d) invertebrate larva
(e) bacterium
Speed
10 mls
20 mls
7 mls
1 mmls
0.01 mmls
IlJerlia important if Re ~ / (i.e whale, d{}ckI dra9()nfl;)
VisctJl4 effecTs arJlfJin41e if Re~1 (l.e larva I haefer/vliI)
BOlJnrilJfl), layer flow becomes lurhllleni for Re ()/J -Ibe
order of lOS'10 /o~ (t',e, whale anapsrh4ps Ihe dlJ&~)
The flow w()u/d be /amilJQr for fhe dr4?OllfifJ /1J1'1I1J, 4nd
haGTer/llh1 anJ perhQp5 Me duck.
9-7
Re
300.000.000
300.000
30.000
0.3
0.00003

'l. 9 I
9.9 Approximately how fast can the wind
blow past a O.25-in.-diameter twig if viscous ef-
fects are to be of importance throughout the en-
tire flow field (i.e., Re < 1)? Explain. Repeat for
a O.004-in.-diameter hair and a 6-ft-diameter
smokestack.
UDRe:=7 </ or 7J < b if viscous effecfs are fo be ImporlQnf
For sfandard air
Thu-s}
fhrouqhoul fhe flow.
1/ -= /. 5 7 x/o-If 1-2
-If
U
1.57XIO
< D J where D l.s fhe diamefer in feef.
object DJ
ff yllJ S
fwi'] 2,OBXI02. 7, S'I- X/0.3
hair 3.331-10'" O.if7/
smokesfack 6 2.62. x/os
9-e

9./0 I
9.10 A viscous fluid flows past a flat plate such
that the boundary layer thickness at a distance
1.3 m from the leading edge is 12 mm. Determine
the boundary layer thickness at distances of 0.20,
2.0. and 20 m from the leading edge. Assume
laminar flow.
For laminor flow 0::: c1(i J where C i.s Q cOllslQn7.
Thus)
C
- L _ 12XI0.3m
- ...Iv' - {i:3m = 0.0105
rX 1.3m
or 6 = 0.0105 rx where X~m o~tnJ
'1,// I
x) m 6J m 6) mm
0.2. 0.00'-170 If. 70
2.0 O.OI'fB /'f. 8
20.0 O.OJf70 Jf7.0
9•.17 If the upstream velocity of the flow in
Problem 9.10 is U = 1.5 mis, determine the ki-
nematic viscosity of the fluid.
.m TJJ2.
For /tlmifJfJr flow 6= sVT )or j/ = :;.s X
Thus, m 3 '2.
(I.5-r)(J2 Xlo-m) 6 -6m2-
1/ = 2-5 (/,3m) = .6SX/0 S

9./2 I
9.12 Water flows past a flat plate with an upstream velocity
of U = 0.02 m/s. Determine the water velocity a distance of
J0 mm from the plate at distances of x = J.5 m and x = 15 m
from the leading edge.
From the BJasivs sO/III/on For bovndary laygr flOk! on a flal plate}
I
U = ufr,,) J where YI J fhe simi/drily variable) I'~
17 = yllx I . Vq/ves of f(J7) are "iI/en in TabJ~ 9.1.
Ux (O.02./f) (J5m) 5
SInce Rex = ~ = J./2.x/tJA ~ == 2..68 x/O is Jess fhan fhe
crifical Rex =5)(/05
if fol//ws thai the boundarv Jailer {/Pli/ is labJilJar.cr J / F
/II X,: /. 5 m andy:: /ox/0-
3
m We obfain:
n -(lox 0-.3 ) O. 02 ~ - IOql
'f, - '/ /'n (J./2X/f6..!!l!) (I.sm) -.
oS
Linear /nferpo/afion from Tahle q,/ 9ives:
I (O.3Q38 - O.2.6'1?l
f = 0.26'1-7 -I- (J.2 -0.8) (/.Oq/ -0.8) :::: O.3EQ
lIence,
U, == [j ('(il) =(O.oZ1f) (0.359);:; 0.00718.t;
SihJilarly J ot Xl. ::: 15m and y:: /ox/f
3
m we ohfain:
.JJ1.
I -3) 0.02,..s
T!z ::llox/O rn (I.I2.Xlo-6:J!f) (Js/h) = 0.3'15
s
L/I)etlr ;nferpo/afi()n {rohl Table 9.1 q/ves :
f 1= 0.0 + (0./328 -0.0) (0.3'1.5 - 0.0) -= 0, /IJfS
(0.8 -O,¥)
lienee,
tlz ~ ufrlJ2,) =(0.02~)(0.II'fS)= O,002.2Q.tp
'1-/0

9. /3" T
"'9,13 A Pitot tube connected to a water-filled U-tube ma-
nometer is used to measure the total pressure within a boundary
layer. Based on the data given in the table below, determine the
boundary layer thickness, 8, the displacement thickness, 8*, and
the momentum thickness, 8.
From fhe Bernoulli eflJaliofJ) ~/Ih
faif' .c::< PH,20 H follow.s thai
PI +-t fQ,r v,2. == 12 +i eq,-,. ~'J.J where
V, -= {j, I ~ ~ 0 J fl,:::0 J and '2.:::'tNJ.o h
Thus} I¥ I. II I
U = 2 ~zoh = 2l9BOO pjJ)h m
Pair /.23~
or 11L m
tJ..= /26.2. Vh J where h-m} tJ~s
For Y> 26.8mf11 we see that h='11.0/11111
ThvsJ
U= /~ 6.Z/(O.OJfI/ = 25.55.p.
For y::: 23.6mm J U::: /2-6.2 /(O.OJl.05)
== 25.'1-0 .p-
or U 25,.1/<)
V;;; 2.5..5.5 =O. 99If
Thvs 6 ~ 23.6 mmJ
The displacemeni fhickness J 6~ is
oD
l =S(1-11- )dy or Since
o
~ = /2.6.21/h ='1-. 9~''i( Ih. hecome.sU 25..5.5 ,n IS
0,0268 In
/=' {(I-'f.9!fl/h)dy
y=o
y (mm)
o
2.1
4.3
6.4
10.7
15.0
19.3
23.6
26.8
29.3
32.7
h (mm)
o
10.6
21.1
25.6
32.5
36.9
39.4
40.5
41.0
41.0
41.0
Numerical infeqrafion of fhe tabulafed dafa qJves 6-- ~/8X/O-3m
(See nex+ PtJge.)
9-1/

IJ/so} the momenlvm fhicknes.s) (iJJ is
~ aOU8m
f!J =fV(J - -V-) Iy '" 4i. 9/f ~1(h(J-1f-.9'f fh) dy .
o y~o
Nt/merical infe~rlJlion of the fabulafed JatfA gives @::: 2.23xI0-
3
m
Use proqrafYI T!(IIPEIOlio ;rde9rqle the Infe9rands fo611/afed be/olJl:
YJ m ( 1- 'f.?'ffh) 1h(/-If,9ifYii')
0 I 0
0.002.1 0.1f9/ 0.0506
0.00'1-3 0.282 O. O~/O
0.006'1- 0.210 0.0335
0.0107 Q./Oq 0.01Q7
0.0/50 0,0511 0.0098/
O.OIQ,3 O.OI9Jf 0. 00386
0.0236 0.0058'1 0.001/8
0.02.68 0 0
0.02.93 0 0
0.0327 0 0
****************************************************~
*****************************************************
? 0,1
? 040021,0.4:91
? 0.00f,j,3,O.282
'? 0.0064:-.0.210
? 0.0107,0.109
'? 0.0150,0.0511,
? 0.0193,0.019f,j,
? 0.0236,0.00584:-
? 0.0268,0
The approximate value of the integral
0.0268 -
Th"s~ S(/-Jf.9Jftfh)dy==o.OOlfI8
o
(con't)
'1-12
l" c·
'" . +f,j,.1777E-03

****************************************************~:
*****************************************************
? 0,0
'? 0.0021,0.0506
? 0.0043,0.0410
? 0.0064,0.0335
? 0.0107,0.0197
? 0.0150,0.00981
? 0.0193,0.00386
? 0.0236,0.00118
"0.0268,0
The approximate value of the integral is: +4.5206E-04
0.0'-'
Thv.sJ
SVh (J-q..qJf-Yh),y =~.51.X/o"fI.
o
or ¥ - 3
~ =Jf.9~ x~,S2 '1.10- =2..23 x/O In
~-I3

u=
1 m/s
-+-
Streamline A-B
A ------
9.14 Because of the velocity deficit, U - /l,
in the boundary layer, the streamlines for flow
past a flat plate are not exactly parallel to the
plate. This deviation can be determined by use
of the displacement thickness, 15*. For air blowing
past the flat plate shown in Fig. P9.14, plot the
streamline A-B that passes through the edge of
the boundary layer (y = 158 at x = C) at point B.
That is. plot y = y(x) for streamline A-B. As-
sume laminar boundary layer flow.
--- - " Edge of bOlJndary layer
/_ f = 4 m------H.!
FIGURE P9.14
5 . R U£ (If) (JfIn) s 5 X105 fh b J
mee ej:::: T:::: /.Jft X/O-OS..f!!J- =2.7Jf X /0 <.:: J e ounat/ry
5
layer .flow remain.5 laminar alonq fhe enlire p'lafe. flelJce,
6=sj ~. or 68
;5 [ u'm~o;-1')(tf"')] ~ O.0382-m
The flowrQle aJrried hy fhe acfual Dovndtlry
loyer is by de{'l/dion ef/val 10 thaf carried by
a vniffJrm velOCity w/lh Me plare d/sp/qced.
by an aRlotJnf 6~ Since there is no flow
fhrouf/h the plate or sfreamline II-BJ
QA :::QB J or UYI/::= (68 -6;) U
where 6'=1.7211';;· ,
u
i,----""
or(. [ (/.'l-6XJ(r.5:fflj{lfm)]~ 0 0
°8=/·72.1 IT J=' J315m
u
~6
-1---
(I)
Thus, ~
y,q == 68 - dB -= O. 03B2-m - O.O/315m :: 0.025/m : sfream//ne
lienee, for ooy x-lOCI/lion A S[": Y)~
QA == Q or [/>A::: U(y -6 ) Y-~7 _ L~
or y :::: IA +6.J= ~ t 1.72/ / v:I ~"""r~-r-r""""7/'r'7""':)r-r-Tj"7")/"rrr
_5m'J.. ]i;;:. O.OUI m t 1.7').1 [(/.~x/o -:s) Xm = 0.025/ +6.58 X/031x' mI
- /.If where X""m
0.04 _._-,
0.035 --.---------~ --~---- ------.---.-- -1
E o~o~; --------1~ o~o~~ =~-=~----------:-----=-~-==~-~-===~=~IO.01 ---------------~-----------~----------------~---. ----------- ----- -----{
0.005
i
- l
Or--------r~------~------~------~
o 1 x, m 2 3 4
9-/Jf

9.15 Air enters a square duct through a I-ft
opening as is shown in Fig. P9.15. Because the
boundary layer displacement thickness increases
in the direction of flow, it is necessary to increase
the cross-sectional size of the duct if a constant
U = :2 ftls velocity is to be maintained outside
the boundary layer. Plot a graph ofthe duct size.
d, as a function of x for 0 :s x :s 10 ft if U is to
remain constant. Assume laminar flow.
FIGURE P9.15
For incompressihle flow
and
Q0 == (}(x) where Qo::: flolVrafe info !he dudn3
:::: (JAo ::: (2. fJ) (Jf/1) :::2 -:s
Q(x):: [J/II
Thus,
where fJ:: (d - 2. 6')2. /s Ihe effective areq of Ihe
dlJ~f (1J//owiIJIj for fhe decrefl.sed f/owrale if) fhe
bOllndary layer).
Qo :::: U(d -26·)2 or d= / fI f 2t5*) (0
where f1/X' [r. -'1.f:J..l. J~
6'= /.721V-1!- = I. 71.1 1.57~1;.'" )X 0: 0.0/52 fX {~ where xNf/
Hence, from Ef. (I)
d= / + 0.030'1 rx ff
For example) d:: / IIIJI x:::O ond d::: l.oQ6 fl of x ~ /0 ff.
1.10
1.08
1.06
~ 1.04't:J
1.02
1.00
d vs x
---.--------.~~-
-I
0.98 -1-,---;---------r-------,----,------j
o 2 4 6 8 10
x, ft
9-/S

9.16 I
9.£6 A smooth fiat plate of length e = 6 m and width b =
4 m is placed in water with an upstream velocity of U = 0.5
m/s. Detennine the boundary layer thickness and the wall shear
stress at the center and the trailing edge of the plate. Assume a
laminar boundary layer.
2. -,
6::: 5/1/TJX •==-5 (/./2.x/o-t) X == 7.¥8 x/o-
3
Vx mJ where x~m
0.5 oS
and 3 fi)"J? 3
fw =0.332 U ~v-ef· =0.332. (o,s1}-)Y2.
(999.15,;3) (1.12. XIO-3
!!.;f,.) .
X
O./2Lf. N h- rx m'2. J W ere X~m
ThvsJ
af X=3m 6=7,lf8X/(j.31(3::: D.O/30m
">- _ 0,/,.1/ _ 0 07/6 J!.
'IV' - f3' -' In'-
while of X =6 m o= 7,118 X/031(6' = O· 0183 m
0./2'1- N
?w:: 16' = 0.0506 m"
'1-16

9./7
9.17 An atmospheric boundary layer is formed
when the wind blows over the earth's surface.
Typically, such velocity profiles can be written as
a power law: u = ay", where the constants a and
n depend on the roughness of the terrain. As is
indicated in Fig. P9.17, typical values are 11 =
0.40 for urban areas, n = 0.28 for woodland or
suburban areas, and n = 0.16 for flat open coun-
try (Ref. 23). (a) If the velocity is 20 ftls at the
bottom of the sail on your boat (y = 4 ft), what
is the velocity at the top of the mast (y = 30 ft)?
(b) If the average velocity is 10 mph on the tenth
floor of an urban building, what is the average
velocity on the sixtieth floor?
>.
(n) U -- C yO./6 he' J f'" J were IS a cons/tin
FIGURE P9.17
Thvs} tJ2. (V)O
•J
6
ff (30If)0.16 1i
U, =r~ or Uz.;: 20 S lIff == 27.6 S1
(b)
~ o.~ -
lJ. = C Y J where C is t1 consTanf
ThlJs U (Y; )0.tlo ( 60 )o.l!
, ;; = ~ Or U2. = lomph /0 == 2.0.s mfJh
q-/7

9.18 A 30-story office building (each story is 12 ft tall) is
.built in a suburban industrial park. Plot the dynamic pressure,
pu
2
/2, as a function of elevation if the wind blows at hurricane
strength (75 mph) at the top ofthe building. Use the atmospheric
boundary layer infonnation of Problem 9.11.
From F/q. P9./7 fhe houndary layer
0.28 C O. '2.9 hU /'V Y ) Or u.:: Y J were
ve10cdy prof/Ie is (Jlven by
C is a consf4nf.
Thus) .JL ::::(L)0.26
()./ X
or 028
U= /JO(3rO) .fj. where y~ ff
Hence}
t pu~.= i(2.38X,03~¥4)f,o(~~or;r
a
o
o
n
n
o X= 301..12
=360ff
o
o
or 0.5 6 ,,--7J~~-T-'i""7--;~rr'7-"
-t. f u'J.= /'1: tf (k) fr:L J where Y'" If
Th/.s /.5 plolfed in the fi9l1re below.
400~----------~------------~-----------,
350
3 0 0 - ------~--- --------------- --- -----
250 --------------------
=~ 200
>-
150
100 -------------------"-~-~------ -------------
50 ~~-------------------- - - - - - -
o+-~~--------~------------r_----------~
o 5 10 15
,/-18

9.19 The typical shape of small cumulous
clouds is as indicated in Fig. P9.19. Based on
boundary layer ideas, explain why it is clear that
the wind is blowing from right to left as indicated.
FIGURE P9./9
lis indicafed in h~. P9./7) beCIJilSe ofthe afmosphmc bOtJndary
layer the velocily of fne wind 'IM.rally increases wiM ajf;/pde. ThlAf,
fhe fop porlioM of a cloud fravels Fosfer Ihun ifs base - Ihe cloi/ds
fend fo "lip " fOlllard fhe direclion of the wind. Tha! isJ Ihe wind
is from ri9M fo leff.

9.2.0 I
9.20 Show that by writing the velocity in terms
of the similarity variable '7 and the function f(,})
the momentum equation for boundary layer flow
on a flat plate (Eq. 9.9) can be written as the
ordinary differential equation given by Eq. 9.14.
The 90verninq e'lvtJl/ot}s CI/'Ie
dl/. +- U - 0 d (I)
IX ~y - tin
U#- + v~ = r 41t (2)
COIJsirler /).= r; f(n) and v=(%:%)~(1 r'-f) where ( /=1" (2.1)
and;; =(lxi'-y
ThvsJ JJ7 , ~ --% .1. II JJ'l 1fiJ'-k
rx = -2. fV- yx =-2. X ond Ty = rJi X :z (.1)
,so fhaJ "
~= !x (Uf') =utf = U~¥X =-i-¥i!f'l (If)
and ~ ~ .;,
If ::;(r;})2it (17 f '- f) = (fJJ~("f" i- fl- f')f9- =(¥ff)'-'1 f"Yfx-~
°fr :: i Jf1'/ {'I (5)
~y
-rh{)~ by tJsi/J9 £1s.('I-)tJIJd (s) we see fhai .Errl) is .sqj;Slied fol'
any funcfion (tif). .k
II/so, ~ =1#-#- =In(ur')[Yf£li] :: (rJ'/)~ f" (6)
and a'"U =('.1.)Js..Lf.1 =( U3)~ fill u/ ::: J.t:. fill1"y2 vx ~y ~x ry 1/X (7)
Thus) by lISiIJ9 Efs. (2./)) (t)) fJlJd (7) w/lh £r. (2.) we ohltli/)
(U f') (~-¥J? {") +(1f¥)Y,.(1t f '- f) (!i)~ [" =vllf'"
which simplifies 10 :
2. fill- ff"=o
From El(. (2.1) the bovndary conditiofl.S at y=O (i.e. 1=0) beC()me
u.::0 =Ufco) aIJd V =0 =(!!f)~(0 (to) - fro))
That is J fro):::o and f'roJlllo
Similarly us y-II' 0() (l.e" l?-..otJ) we re'lvlre 1).'-' u. r/Jv~ from
J I
£1,(2..1) f - / as ".-.00.

9.21 "I 9.21* Integrate the Blasius equation (Eq. 9.14)
numerically to determine the boundary layer pro-
file for laminar flow past a flat plate. Compare
your results with those of Table 9.1.
Solve the fo/lowin9 third ()rder differenfiaJ ef/llafion by a
nvmerical in fe9fafion feGhniqve:
III 1/
Z f + f f -::0 with boundary condil/oHS
f:: f'=o at '1=0 and f'-I as 7'/- 00 (( )/: In )
Wrile this third order ef/IJtJfion (/.f 3 lIi'sf order e'ltlalio/lS (,l/Jd /J.se
a RVfJ94 - kuffa IIU/fl81','C4/ techniqfJe TO i/Jf~rale fhem. 7h1/~ /e f
>1 == f J X~ f I::. Y.z J ~ I ::: f ": ~ J dlUI }) /= {"/::: - t f{'/:: -i~13
Thai is:
/
}j=~
X./:: ~ ond
~/=_}jY3h·
These can De tlfJl'roximqted q.s
A)j = ~ AI? J .A~ = ~ Ai? J find A)j :: (-J1Y.J/2.)Ar;
start wilh Yt ~y,.=O af 1;::0. ;?sstlllle ~ == c afi; =() (w/;el'e C is
s()me 9ivell cIJI)J'rtJllf) IJlJd',hfe!r*le ff) J;:: IX) II by yt.:: X(0) +2;AX··LJI;
I ;;
If Y2 (00) :f / (t:ej
f't()()) ofI) tlt.(jlJ.jt the /la/tie 01 c(t:o; f'lo)) a/Jd
fry Q9a/n. Tile IwofJlJifJl bOllllldl')' Jl4/ve jJ/'IJb/e/fJ (~;e) f(o)=:rro)~ 0
and f 't¢) ==I) is sp/ved by l/er41/()/J fiG IJIJ //I/iial Jla/()e problem
r,:e) fto) =: ('to):/J) fto) ~ c).
Pro9ram P8#J.' show/J below was vsed for fhe c4/c()/atiI}As. rAe
f/n4/vallle 01 C :: 0.332 and f/Je Jle/()c/"fy 'prollie J U-:: V f'r}J),
rJ9ree very well w/lh fhe sfQfJJfJrJ valves ~;velJ in Tab /e 9.1
lOO cls
110 open "prn ll
for output as #1
120 print "******************************,..********************"130 print H** This program integrates the boundary layer **"
14:0 print "** equation (Blasius equation) for a flat. plat.e **"
150 print "** using a Eunga-Kutta type routine. The user **"
160 print "** must specify an initial value of f'! (0) :=;,0 **"
170 print "** the boundary condition 'at infinity' (f!! '= **"

9. 21 41 (con't)
180 print "** 1) 1S satisfied. **"
190 print "***************************************************"
200 print " "
210 print If "
300 print "Input a value for f" (0)"
310 input c
320 print "Input stepsize and number of stepslf
330 input dx, n
335 print "Input how often to print output (number of steps)"
336 input. nn
340 print" eta f f! f! '"
350 yl = 0
360 y2 = 0
370 y3 = c
380 x = 0
385 m = 0
390 for 1 = 1 to n
395 m = m +1
400 x = x + dx
410 yl = yl + y2*dx
420 y2 = y2 + y3*dx
430 y3 = y3 - (y1*y3/2)*dx
435 if m ( nn goto 450
440 print using" ##.#### +#.##~~~~ +#.##~~~~ +#.##~~~~";x,yl,y2,y3
445 m = 0
450 next. i
460 got.o 210
t**************************************************
** This program integrates the boundary layer **
** equation (Blasius equation) for a flat plate **
** using a Runga-Kutta type routine. The user *)f'
** must specify an initial value of f" (0) so **
** the boundary condition 'at infinity' (f' I '= **
** 1) is satisfied. **
***************************************************
=~pu~ a value for f" (0)
? 0.332
Input stepsize and number of steps
? 0.01 700
Input how often to print output (number
? 50
eta f f' f' ,
O. :,000 +4.07E-02 +1.66E-01 +3.31E-Ol
1. 0000 +1. 64E-Ol +3.30E-Ol +3.23E-Ol
1.5000 +3.68E-01 +4.87E-01 +3.03E-01
2.0000 +6.47E-Ol +6.30E-Ol +2.67E-01
2.5000 +9.93E-01 +7.52E-01 +2.17E-01
3.0000 +1.39E+00 +8.47E-Ol +1.61E-01
3.5000 +1. 83E+OO +9.14E-01 +1.07E-01
4.0000 +2.30E+00 +9.56E-Ol +6.38E-02
4.5000 +2.79E+00 +9.80E-Ol +3.36E-02
5.0000 +3.28E+00 +9.92E-Ol +1.56E-02
5.5000 +3.78E+00 +9.97E-01 +6.41E-03
6.0000 +4.28E+00 +9.99E-01 +2.32E-03
6.5001 +4.78E+00 +1.00E+00 +7.36E-04
of steps)
~
e~
r-
6 ~
Lfr- 1
/
2. .7
ov. If'
7.0001 +5.28E+00 +1.00E+00 +2.06E-04 o 0.5 I

9.22 An airplane flies at a speed of 400 mph at an altitude
of 10,000 ft. If the boundary layers on the wing surfaces be-
have as those on a flat plate, estimate the extent of laminar
boundary layer flow along the wing. Assume a transitional
Reynolds number of Rexcr = 5 x lOs. If the airplane maintains
its 400-mph speed but descends to sea level elevation, will the
portion of the wing covered by a laminar boundary layer
increase or decrease compared with its value at 10,000 ft?
Explain.
R UXcr h U=/fOO h( Ihr )(S2BOff) =587.£i
eXcr = 11 ,were mp 3600.5 mi Ib .s
-1 '5
d f T II - p - 3.S3#-XIO -w:-(Jf} rom I aD e C. I) 11 - f - -3 §.l~
/,7S6'X/O ft3
= 2.01 X/O-¥ 1j-2.
lienee) wifh RexC,. = 5X/O
S
}
X
cr
':: 1/ Rexcr _ (2.0/ x/oJ/. fj2.) (SxIDS)
V 587 i!
:: 0./71 If
s
IN sea- level:
(b) ReX
cr
== VlI
Xcr
J where
and
u== 'fOO mph ( J hr ) (5
2
9,0 ft) :: 587i±3600S ml S
1/ == 1.57 x /0" 1'/2.
Hence, -J/. Hl 5
X - 11 Rear _ (1,57 X/0 -::s )(5 xJO ) _ 0 I 'lJ'[ fl
cr - Tr - ft - ,..::1.... ,
v 587 -:s ===
The laminar boundary layer occtJPt"es the ft'rsf o.13Jfff of Ihe
winq at seQ Jeve I and (from par! (a) rrbove) fhe f/rsf O. 171 if
af an alfilude of /~ 000 ff. This is dve mainly f" fhe lower dellsilt
(iarger kinematic. viscosi1y ). The d),!7amic visco.sifies are approximaiely
fhe same.

9,2/f I
9.24 A laminar boundary layer velocity profile is approxi-
mated by u/U = [2 - (y/o)](y/o) for y:s; 0, and u = U for
y > o. (a) Show that this profile satisfies the appropriate bound-
ary conditions. (b) Use the momentum integral equation to de-
termine the boundary layer thickness, 0 = o(x).
ra) t:::9(1) == 2Y_12. where Y:: yld
TlJvsJ #:1 -= 0 as if mllstJ -#-/ =2 - j -:; / 01' IJ -:: lJ()Iy:::J
y=O y~J
4.$ /1 /J1wI. J,
A/so, rr =U[f -~] so 1114t a1J~ 77[f -r] ~O
y~J
(b) From the momon/vI'» infer;rlAl (j1fJafiof),
I d
o= 2-i~,P)(, where c, ~ (, (l-fo)JY and c~ ~ 1r)
1:: 0
T/;t/s) I
c, '" ((:lY-Y') (J-zY+Y'")JY = !(2Y-Sy1+lfy3
-Y'l)dY
o 0
-I-S-I-/--#- =~- :3 ~ /-5'
and
CL
=: (2 -2yj
so -fhat
Ire0
J = / 2 ~2.) 1/X '"' / 3O;X
r:s- 7l
fie f/cS; w/fh Rex:::: 1ff J
<5 _ -{.30 _ S.'hP
-X - VRex - 1/f/.e)(
9-2'1-

9. z-s I
9.25 A laminar boundary layer velocity pro-
file is approximated by the two straight-line seg-
ments indicated in Fig. P9.25. Use the momen-
tum integral equation to determine the boundary
layer thickness, t5 = t5(x), and wall shear stress,
T" = T,,(X). Compare these results with those in
Table 9.2.
y
0
012
Compare these results To fhose in Table q.2..
/1
71
/l
I
I
I
[7
I I
I I
! I ,U
0 2U U
3
FIGURE P9.25

9,261t I
9.26* An assumed dimensionless laminar
boundary layer profile for flow past a flat plate is
given in the table below. Use the momentum in-
tegral equation to determine b = b(x). Compare
your result with the exact Blasius solution result
(see Table 9.2).
From the momentum infe9/'al eqllof/on
J ' df = 2. Cz. 1/X where C =1#3-}
o ve, J ~ dt
1=0
and I
C, -::S9( }-~)dY with 1f ::~(Y)and y=f
o
ylb
o
0.080
0.16
0.24
0.32
0.40
0.48
0.56
0.64
0.72
0.80
0.88
0.96
1.00
ulU
o
0.133
0.265
0.394
0.517
0.630
0.729
0.811
0.876
0.923
0.956
0.976
0.988
1.000
The I/Q/oe of Cz.ClJn he appr(Jximafed tiS
A(~) 0./.13
C2.~ A (f) -= 0.080::; I. 66
:t-cO
and Ihe VQ/IJe of ~ C(JII be oblo/ned from IJPhJ81'1&a1 ~
infe9r41ion (pro9rtlfn Tf(flPElO])
Y ,#(1-9)
o
0.8
0./6
O.ZIf
0.32-
O.IfO
O.Jf8
0.56
o.61f
0.7,.
().eo
0.88
0.91
/.00
o
0.115
0./'.5
0.2.39
0.2.50
0,233
0./98
0./53
O.joq
0.07/
O. 0If2
0.02.3
0.0/2-
o
fI.s ilJdictried helow;
I
C, -= J~(J-VJdY ~ 0,/3/ so thaf
o I
6=[- 21/X (f.66) J1: - 5 03 (11X)1:U (0./31) - . u
or <5 _ 5.03 h eV}.
x --V Re;< J were Rex::: p
Nole: The Blasius sollliion h•.s ~ nols.03
1~***************************************************
*****************************************************
Enter number of data points: 1'-1,
Enter dat.a points (X , y)
? 0.00,0.000 ? 0.56,0.153
? 0.08,0.115 ? 0.6'-1,,0.109
'? 0.16,0.195 ? 0.72,0.071
? 0.2'-1,,0.239 ? 0.80,0.0'-1,2
? 0.32,0.250 ? 0.88,0.023
? 0.40,0.233 ? 0.96,0.012
? 0.'-1,8,0.198 ? 1.00,0.000
The approximate value of the integral is: +1.3096E-Ol

9.27* I 9.2.7* For a fluid of specific gravity SG = 0.86
flowing past a flat plate with an upstream velocity
of U = 5 mis, the wall shear stress on a flat plate
was determined to be as indicated in the table
below. Use the momentum integral equation to
determine the boundary layer momentum thick-
ness, 8 = 8(x). Assume 8 = 0 at the leading
edge, x = O.
S/nce 7W = f TJ:l.~ ;t fol1owtS that d@ = ?W d
eU2 )(
which can be irdeqrorec1 fo give (t}s//Jq @=O af x=0)
X x
(j) '" ~~2 fIw dx =(o.e6)(1O~~)(5P-)'2. ~ Iw dx
or x
-5 ( N
(j) -== if. 65 x10 J fw dx J where (f),.., mJ x ~ rn J and fw..v ;,;.
o
For 0 ~ x6 2.0 m, iIJfe9r4fe E'{. (I) fo defermine @os a {()ncl/on
of x. To do soJ we need fhe vaItJe of 7,; af x=:0) whic;h is !Jof 9itlell
in the fable. Theorel/callyJ fw ~ 40 aT the /eadif)}. For our pvr,os8~
based On the exfrapolaTed curve be/olllJ
assume ~ =: 22 -ffi at X:: 0
'tw VS X
25
i
20 . -
'" 15E
-z
i 10~
Or---~----~----~--~I
0.0 0.5 1.0 1.5 2.0
x,m
x (m)
o
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
Pr09r4rn P91127 showl1below was {}Sed for fhe calcvlafio/Js.
(con'l)
13.4
9.25
7.68
6.51
5.89
6.57
6.75
6.23
5.92
5.26
(I)

9.27,1. I (con 't)
100 cls
120 print# 1, "******** ********************** ******************** tf
130 print#l, "** This program calculates the momentum **"
140 print#l, "** boundary layer thickness as a function of **"
150 print#l, "** x from the given wall shear streSfj distri- **"
160 print.#l. n** but.ion. **n
170 print#l, H**************************************************"
200 dim t.au (11)
210 tau(1)=22.0 : tau(2l=13.4 : tau(3)=9.25 : tau(4)=7.68
220 tau(S'=6.51 : tau(6)=5.89 : tau(7)=6.57 : tau(S)=6.75
230 tau(9)=6.23 : tau(10)=5.92 : tau(11)=5.26
240 print#l. " If
250 print#l, 11 x, m momentum thickness, m"
260 for l = 1 to 11
270 x = 0.2*(i-1)
280 if l = 1 then goto 400
290 intgrl = 0
300 for j = 1 to i-I
310 intgrl = intgrl + 0.5*0.2*(tau(j+1) + tau(j»
320 next. j
330 theta = 4.65E-5*intgrl
400 print#l, using" ##.### #.*##~~~A";x.t.heta
410 next l
~*************************************************
** This program calculates the momentum **
** boundary layer thickness as a function of **
** x from the given wall shear stress distri- **
** bution. **
**************************************************
x. m momentum thickness, m
0.000 O.OOOE+OO
0.200 1.646E-04
0.400 2.699E-04
0.600 3.487E-04
0.800 4.146E-04
1. 000 4.723E-04
1.200 5.302E-04
1.400 5.922E-04
1.600 6.525E-04
1.800 7.090E-04
2.000 7.610E-04
0.0008
0.0006
I
~I
E
0.0004
~
0.0002
0
0.0 0.5 1.0 1.5
x,m
2.0
I
I
I
~------------------------------------------------------------I
'1-28

9.Z8
9.28 The square flat plate shown in Fig. P9.28a is cut into
four equal-sized prices and arranged as shown in Fig. P9.28b. ...!!..
Determine the ratio of the drag on the original plate [case (a)]
to the drag on the plates in the configuration shown in (b).
Assume laminar boundary flow. Explain your answer physi-
cally.
Te
1
.. FIGURE P9.28
By comparinq £'ts. CO and (:J.) we see fhaf
4a = 2.0 4h
1
• • • • •~e/4'
-------1..1 T
In cose (b) fhe bOIJndary ItJyer on fhe reor plale i.s fhicAel' Iholl on
fhe fronl plate. lIellce fhe shear sfre.tS is /e.ss I)" fhe r{](Jf pfafe fhall
if is on fhal p/afe In confl9vrafion (a)) 9,vin, /es.s dray for c4!e (b)
fhan for cqse CO)J eve" fholJ9h the fola/ areas are the Sqme.

9.ZQ I
9.29 A plate is oriented parallel to the free
stream as is indicated in Fig. 9.2.9. If the boundary
layer flow is laminar, determine the ratio of the
drag for case (aHa that for case(bl. Explain your
answer physically.
u
~ PLATE
(a) I(b)
FIGURE P9.29
For case (a):
JCI. _..L T/2.C /I h c. - 1.329 - 1.3)..8 d IJ 02-
<Vfa - 2- P Df were 'Df - VRetI - YuJ,' an == Lf"
Y-hvsJ
V
'1a =feU" ~('fJ') := 2.56 pU% y:v;% (I)
For case (6)
4,,:= t PU"CDf /I where c.... = (ffijJJ and ,4:'f12
UT U(tflJ
Thus, 11
PI == ...La U2. 1.32.eW ('fill.) =...L (2 56 OU~ 1fV f)~) (2.)
oUfb 2. y4{fi i :,t 2.. ~
From E~s. (J) and (2.) we see thai
4a = Z
cftfb =
The shear sfress decrease.s wdh disfance from thrl /eadifl9 erlfje
of fhe plafe (i.e.}fhe fhickenifl9 of the bOlJnriary/tJ!er). ThIJS.I eve/J Ihofl9h
fhe plate !Area /J the same for case (aJ or (b)J the QVfJra9fJ shear
,stress (and the drat;) is 9realer for ca.se (a),

fl. 30 I
9.30 If the drag on one side of a flat plate
parallel to the upstream flow is S) when the up-
stream velocity is U, what will the drag be when
the upstream velocity is 2U; or U/2? Assume
laminar flow.
and
9-31
C ::: 1,328
Vf J~i I

9.3/ I
9.31 Air flows past a parabolic-shaped flat plate oriented
parallel to the free stream shown in Fig. P9.31. Integrate the
wall shear stress over the plate to determine the friction drag
on one side of the plate. Assume laminar flow.
Treat each .stripof thickness dy and /e/J9lh i=i(y) liS a SRJf/I/I
f/Q/ plate wilh rlr49 dtb where for lal1linllr flow
,,jj= CD, tpU
2
JA with rill ~ I tly olla CD1 " ~ =
ThllsJ
/.32 B 2 1r::::: 3..i 1/h
db", V0/ t eU I rly =O.66-!t- ff'f' U rt tiy
But 1:::: If _y2 so fh41
+2-
r r 3/2 V ·,/) ::::JrlrB =: J0.66'1- ~p' U Jf-y~ dy
1.328
ret!
/Vole: The un/Is ()/J the inTftJftI/
are HlIa (i.e. 2 rtd: L ~/.a)
9-32

9.32 I
9.32 It is often assumed that "sharp objects
can cut through the air better than blunt ones."
Based on this assumption, the drag on the object
shown in Fig. P9.32 should be less when the wind
blows from right to left than when it blows from
left to right. Experiments show that the opposite
is true. Explain.
U'?
'. FIGURE P9.32.
!l si9IJiliconf porfion of fhe drq9 on an ohject cqn he from
fhe relaliflely low pressure developed in Ihe wake re9ion hehind Ihe
objecf. By maxin9 fhe obJeci sfrean,!;/Jed (i. e./ (luI#! fro/)} left fo
ri9ht) nol rl9hf 10 Iell /f) Ille ahove !J9f)re) botJndary loyer separation
/s avoided I.IlJd (J relafively fhill woke lVill; I()w drtJ9 is O/)"/tlifled.
Whefher fhe front of fhe obJec1 is ~h41'j) /I or ''/;/u"r does noll.llfecf
the cOl)lr/{;(}/iofJ to fhe dra9 frulIJ the frunf pal'7 of Ihe body -of
/eqsf not CIS mvclJ CI fhe widfh of fhe wake affecfs fhe dr4g.
'1-33

9,33
9.33 Two small holes are drilled opposite each other in a
circular cylinder as shown in Fig. P9.33. Thus, when air flows
past the cylinder, air wil1 circulate through the interior of the
cylinder at a rate of Q = K (Pi - P2). where the constant K de-
pends on the geometry of the passage connecting the two holes.
It is assumed that the flow around the cylinder is not affected
by either the presence of the two holes or the small flowrate
through the passage. Let Qo denote the flowrate when e= o.
Plot a graph of Q/Qo as a function of efor 0 s; es; TT/2 if (a)
the flow is inviscid, and (b) if the boundary layer on the cylin-
der is turbulent (see Fig. 9.17c for pressure data).
...-~......,..........~,.,
-'
u
. ~
(0)
;?'
",'"
....~---......~......~-.~...........................~.,.
(0)
• FIGURE P9.33
For invi.rciJ {Jow :
fJ-lJf)
C ::: r T' == / - 'I-.si!) 2.(}
'f tpu"
ThtJs) Q -= K(II -f:J. ) ::: K[r"-fo) -(f,--fo)]
=K[-tpU'-(I-Jfsl,,2.(}) - t.(JU2.(J-tf.sin'J.(e+1r))]
nul si,/O e sirl"(f) +71')
Hence Q:: 0 fir if/viscid flow "J
/Vote! !his is 10 he eXfecred
becafAre 01the .rYlI/mell'l~/
pl'8.sS fJrB dislr/blllioll.
(6) For a ftJrbtllenT hotlndary layer =
Q::: K((1, -(12) ::: K[(P,-Po) - (/2.-fo)] ::: tpllzK[ef, - c,:z.]
where efl is for () CllJd c,:J. is fol' 18o-e dey,
Ohta,n ~ 141a fr()m [;1- 9.17
Noie; C
f
:;; / fur ().:: () 4/Jd C, z~,1f fur B:::/It)Ie!
TfuI.r Q0 == QI ::; if717.K[ / - (-0, t,t)] :: /. If ( tl!J1-K)
J e~o
,so fhaf
.R:: teu'k[cp,-ce:a.] _ CPI -Cf2.
Qo I. If (1:pTf'''K) - I,If
The resvlls are f4/;vlaled 4J'1d;Ioiletl be;PW.
(c()n'-I)

9,3,3 (con If)
e,deg 180 - e, deg cp1 cp2 0/00
0 180 1.00 -0.40 1.00
15 165 0.70 -0.40 0.79
30 150 0.00 -0.42 0.30
45 135 -0.90 -0.43 -0.34
60 120 -1.70 -0.45 -0.89
75 105 -2.10 -1.30 -0.57
90 90 -1.90 -1.90 0.00
0/00 vs 8
1.0
0.8
0.6
i_____ 1
i
0.4
0.20
a 0.0
- --~--------.---~--~~.-:-~~~-~-
a -0.2----------~~~----:-~~-~-
-0.4 --~. ---------------
- 0 . 6 - . ------ ----~--------~~-
-0.8 - -------,-------------
-1.0 +------,-------r-----j------r-----.-----i
o 15 30 45 60 75 90 1
8, deg
9-3S

9.3'1-
9.34 Water flows past a triangular flat plate oriented paral-
lel to the free stream as shown in Fig. P9.34. Integrate the wall
shear stress over the plate to determine the friction drag on one
side of the plate. Assume laminar boundary layer flow. U =0.2 m/s
--.... l.Om
• FIGURE P9.34 IY
.b~ {7W dl/ where Iw::: 0.332. V
34
ftf.
Thvs
' " '" 10•33.2//4.Ve,u.~{*tillx==o.S y=O,s-x
=0.332 7l~Vpp (2)f f rf.f}x
XeO y=o
=0.332. U
3
/7- P"f;;- (2)f D'~'YiX dx
x=o O,S
:: 0.332 7l4.VfjA (2.J[o.~('-)xt - ix~i
o
dA
x=o
= O. 66'f(O.2l})3/,. 999!,Ml.fU/o3~.;)[Vo.S -y.(0.5)3;" ]
or
JJ == 0, 02.96 N
')-36

9,35
9..35 A three-bladed helicopter blade rotates at 200 rpm. If
each blade is 12 ft long and 1.5 ft wide, estimate the torque
needed to overcome the friction on the blades if they act as flat
plates.
Let dill=fOrYIlJe {rom fhe dr49 on area elemenf dll
or
dlll= (4op +cbbcilom ) Y==z(tPU
2
CDf dll)y
where
[J = t»y 4/Jd fof' laminar floW *
1.328
CDr= !/~ with ReD= UJ 0
Rei .,lI
Thvs)
dM =Dy2 1.329 1 U~l~ ~ d,,; (llL~ Y dy = 1.328 P 11 Y r
'fh 11 l~or WI
U=Q)Y
~ ..k ~ s.
el/Y/::: 1.~2e ptJ) "3.J ~ V ~ Y~ tiy ~
=(1.326)(2 38XIO-3 ~ ,h(.200 rc..v)( Imifl)(2.lI'rad)l(~ 5F/)-!'(1 S7Xlo¥1il.)i. ~ J
, ~~ 'L min 60.s I rev ~ . , .s 1'"'11
or
5~
d/YJ =O.OJ25 Y 2. dy ff·lb J where YN fI
Thus) fhe net f()r9ve for the fhree hlades
~ ~
!YJ = 3Sd/J'J == 3 (0.0 /2E ).fy-tdy == 3(0.012.5 )(-f)(12.)~
or
M= 6'1-.1 1f,IIJ
Nofe: The forf{ve covld be greate,. ,'f Ihe bOfJi'Jdal'y layer is fvrhlJ/enl.
r.eV
*III fhe lip y ==,~ fl so fhaf TJ== (J) Y=t~:op)(2-11 f!tv)(J2 fl)~ 2.5/ 1)
JEt tnln
and. Re - 25/ -s ) (1.£11) 6 ,L' h 1 JL JL
~tip - /,57X/O-'I- f: l. =2.'I-OX/O, 'Alnlc is 9reaTer rnan rne
oS
ori/ical valve. Thl/~ the bOlJndIJry is t(Jrbf}/ent al the lip and laminar at
the huh. FbrsiIiJP//t:.Ity ass()me a laminar b()undary layer fhr()v,h()v1.
9-37

9..36 I
9.36 A ceiling fan consists of five blades of 0.80-m length
and 0.1 O-m width which rotate at 100 rpm. Estimate the torque
needed to overcome the friction on the blades if they act as flat
plates.
Lei dllJ= 1orq{)efromfhe drfJ9 01} elemenf dA
dA of fhe hlade IP ,,~ U .1
or _ A + _ J.. 2. ) ( ... .....dy Ji =O,lm
d/IJ - (oV.top 4oHom)Y-2(2. pUC/)1 d~ Y E -JT 0'
where {} =tily and tJJ=/oo..£!).. ( liYlilJ)( 2-'!trod) Y -flO dF 'J
mm 60S I rev
or(j.}::: 10,'17 ::J. L== 0.8m
The fr/()ximvPJ R~ will ()cc()r at t:inf OJ where y:;:L ();
Re == UR. = wLR = (10.'1-7 CfL)(O.8m) (a.lm) = S.7~X/OJI.
:1/ 1/ 1/ 1.1}6xI0-.5 .Di!oS
ThlJs, af a/I poinfs on fhe blade Rex < Rexcr = SXIO
s
Clnd fhe f/uw is JlJlfJil1or.
C = 1.329 = 1. 3 )'sW
Df ~ ReJ, I ifui
so fhrrl from Elf. CI)
W 3~
d/YJ =rr;2 'HfJ!' (i dy)y == 1.32.8 f U 2 Vll t ydy ,hv/ Wit/; Tf::t(Jy
d/Yl::: /.328 f!JJ~ Yvi' y"% dy ,
= /.31.8 (1.Z3~) (10.'1-7 o/l~-g,'.¥,{x/o-
s
.()(O./IIIt" /4Jy
or
~
d/l1 =0,0669 Y 2. dy N'm J where YAJ'"
ThtJsJ
the nef fOfYjve O/) the four hlades is 0.8
S S
o.aM s ( 2) k-lM= 5 d#J == 5 0.0669 Y ~oy - 5 (0.06t9) "7 y 2-
or
y=O 0
/'1:::; 0,0 Jf38 N'/IJ
9-38

9.37
~
9.37 As shown in Video V9.2 and Fig. P9.37a, a kayak
is a relatively streamlined object. As a first approximation
in calculating the drag on a kayak, assume that the kayak
acts as if it were a smooth flat plate 17 ft long and 2 ft wide.
Determine the drag as a function of speed and compare your
results with the measured values given in Fig. P9.37b. Com-
ment on reasons why the two sets of values may differ.
For a fled plate cIf~ fpU'-CD,II where (I)
JJ :: /711(2 ff) ::: 3/f ff:;' tlnd cIJ1 is a {tinct/of) of Ret::: 11J
Thvs
Re J::;- J7 N TJ it = /JI-())( /06 7J (2-)
:e I. 2/Xlo
5
ft ~ 6
Consider I ~ rJ!E 8ft J or /,ifOX/O ~Re.J ~/.12xJ07
From Fi9' 9,/5 we see thaf in this ReJ rtJ"ge fhe houndary layer
flow i.s in fhe fransitional r0/19c. ThllsJ
from To.h/e 1.3
GDI
:; O.lf.5.5/(/09 Rel)2.&8 - /700/Rel (3)
By cOlnbt'nin, Ef(s. (0.1 (2); anti(3);
J) :::: -! ( I. q'fs.J.~9f )*U 'J.
CDf (3'1-f/J.) or
rIJ= 33.0 V" [0.",.5"5/(101 (1.'10I./iU»)'~1700/(/.'fOX/o617)J (II)
The reslJll.r from fhi,s erilarion tire ploffed he10w,
8~----~----~----~----_.
7l, ff/s rf}J Ib
I o.oQe6
2 O,/fIO
.3 o.9oq
if 1.58
S 2.'1-2-rYIf>Q5 dvalves
61------+-----~-----+---+--j
6 .3.L/-3
7 'I-.sq
8 ,5,90
Ib 41---------+------r----~~--~
..---.- theory
(E .('I)
2~----+-----~~--_+~~~
• FIG U REP 9 .37
2 4 6 8
Kayak speed U. ftfs

9.38 I
9.38 A sphere of diameter D and density Ps falls at a steady
rate through a liquid of density p and viscosity J.L. If the Rey-
nolds number, Re = pDU/ J.L, is less than 1, show that the vis-
cosity can be determined from J.L = gD2(ps - p)/18 U.
For .steady flow E F; ::0 tu
"'- diameTer D
orof) + FE == W J where Ii::bVOytJRt force = f~11= f1(f)77'( ~ t
W:: wei,hf =Ps gV::: flJ (4) Tf(~t
Thvs}
and IJ == drtJ7 ::: CD f f '%D2) or siIJce Re< I
rf} =37l'Dll,P
31l'DfJl-" + f9(.j-)1!(-B:)3 =fs9(l)7T(~)3
which CC/II be reorf'IJngecJ fo ,ive
= g D2.(e.s-p)
f IBTl

9. 3 q I 9.39 Determine the drag on a small circular
9. Jf() ,
disk of 0.01 ft diameter moving 0.01 ftls through
oil with a specific gravity of 0.87 and a viscosity
10,000 times that of water. The disk is oriented
normal to the upstream velocity. By what percent
is the drag reduced if the disk is oriented parallel
to the flow?
o/}= CD ~ pU:J/1 J where e~ (o.e7)(J.9~-§.~~):: /'688 ~~ (I)
and p=IO')tHz,O::: Il1(z.311XIOs 1:;~)-:::o.:z.3~ It;
Thvs 0 :: UD:: eIli) = O.&88,#!fp)(O.Oll)(o,OJf-lJ _ 7."'/ I -If«/
J ne 11 P Ih's - . ~ x 0
so tho! the low Re d41Q of TaMe ?,1f 0.23/f-w:
is valid.
FOr fhe df.sx normal fo fhe flow) r;::: 2~: = 2.O.~-.# = 2~300
so thai frolll £~Itl) 7.2./ XIO
/}= 28,300(t )(1. t8B S~})(O,Olfj)'-f (O.OlfJ)2. ~ I. sax/o-If Ib
If Ihe disk i.s PQraJ/e/ fo the flow, ~::: 1~~6 so thai
~ c. (13,6)
pqral/el ;;::. Dpar41/81 = 7[; = 0.667 a 33.3 7. reduction
JJ"OI'1IIQ/ CD"or/YIQ/ ( 2iiel/) J
9.40 For small Reynolds number flows the drag coefficient
of an object is given by a constant divided by the Reynolds
number (see Table 9.4). Thus, as the Reynolds number tends to
zero, the drag coefficient becomes infinitely large. Does this
mean that for small velocities (hence, small Reynolds numbers)
the drag is very large? Explain.
For a 9iven ohjecf CD ~ ~e (where the 1/4/118 orCdepends
on the shape of the ()bject)J provided Re ~ I, TIllIS" (loS
Re - OJ- CD - ca.
lIowev(J~
~ ..L 2. C L ~ f'r
rU = CD 2. pU /I ;:(e!D.) ~ eU II"'" C/
TIJqt iSJ
as lJ--';; (t:e. Re-O)I Ihen 0#- U
TAvsJ
dfJes C/) - O t ) l')Je4A fhaT ~-o f No.
9-'11

9. 'l-f
9.41 Compare the rise velocity of an i-in.-
diameter air bubble in water to the fall velocity
of an Hn.-diameter water drop in air. Assume
each to behave as a solid sphere.
diQ. D== O./25in.
(a) air bubble in wafer: For sfeady rise L:r;r; 0
or
Fe = W+tb' }where ct=JrQ9 = CD {pif2.f J):1.
W= we'jhf ~ ~ir-V :=; Oaifl ~71'( g)3
~J D3
Fa== buoyanf force = dN;LO-V = ~o !fJl(T)
However} since rair.eC.<. ~,.O if f()//o'Us fhal W<'< F8
or
F8 =d7
Hence J CJ p i17'(Jt)3== CD f fY2. f D2. or U=/I/- Dg ·= 4 (!2!.f¥)f1(31,
3 CD 3 CD
or 0.669
U= _rr-'
yeD
¥) where CD -= C/)(Re) and Re::: u.v/)
or (0115)
R = -TcH U = 861 Ue -5tJ~
I,l/XIO -:t-
(J)
(2.)
From Fig. 9,ll :
Tria Iand error solufion for v:fI.s.svm~ CD j obtai!)
U {r()tn E'{. (OJ Re from Ef. {:J.)j check CD (3)
from Elf, (3).1 the 9raph,
Re IIS.sUII/8 Cb =1--.-. Y;O.66,1---fI- Re=576 -- ClJ =0.5:1: J
Assume CD:fJ,S--U-=O/I/ftl -- Re :: '615 - CD= 0,5 (checks)
Thv$J U= O.9¥-6 .g.
(b) wafer drop in air: Since ~il'« (fH:I.() ) F8 <:<: W
ThusJ W=~ J or 4. 0 !If-(l}? = cD teu2.fIJ2.
2. J.
orU_/4D olt.oI J /I (~R)(6Z,'/-") ]~ /9,1 Jj. (Jf) all'
S feD L3(2.38)(IO-3~~j})CD VC; ~
fUso} 0 /').5 t
Re = UD = ( '17.. f!) ~ = 66.3 U (5) U
"11 1.5 7xlO 'flf
Trial and error solution of£'1s. (If), (&)) and 9raph 13) :
Il.s.sume CD:: O•.s---. U-= 27. f) fj - Re ::: /790 - CD:::; 0, if "1= o,s
Assume C/):: O,tf - U::: 30.2 P.- Re =2000- C/)::: OJ,l (checks)
Th [J = 30,2 ff
Nofe; Becavse of fhe 9raph (Fij. Po 2.1) the tiS J .s
answers are nof ()ccvrafe to fhree sI91J1//canl N9l1res•

9.42. A 38.1-mm-diameter, O.024S-N table tennis ball is re-
leased from the bottom of a swimming pool. With what velocity
does it rise to the surface? Assume it has reached its terminal
velocity.
tU dia, D=- 38./mm
-,
For sfeady rise 21&=0 (
or
Fa = W+,/J) where /J-:;drag =CD i fu2.fD2.
VI =: weigh! =- O. 02't5 N
tFB } water
~W f'jf
TJ1
~ :::: bl/oyanf force = oJ{- ='t if371 (~)2
or
CD y2 = o~1/-.55 J where U~ ~ (J)
A/so, Re =f{P
or
R
- U(O.038Im) {l V rn
e - J./2X/o-6J1J.! = 3,1./-0 x/O '(/ J where ~-:s
..s
(2.)
Fino//y, from Fi~. '1.21; Gill "--v-- (3)
Re
Trial tlnd error so/ufion : /I.s.sume CD j obiain TJ from Ef. 0)) Re from
E'l~(J.) j check C/) from £1' tJ) J the 9r4ph.,
A.ssf)m~ CD =0.5 -.-. U= O.Q599 - Re = 3.2JfX/O'"- Cl) =O,If '*0,5
Il.s.sume CD =OJ; - U= I. 06 ~ ---.. Re::: 3.62.1./ol.f-.-. CD =0. ~(checks)
ThlJs, V= 1.06.p
Nofe : 8eco/J.se of the qraph (Fig, 9.2.1) the answer,s are nof
accllrate fo three sI9ni{icanf fi9lJres.

9.44 A hot air" balloon roughly spherical in
shape has a volume of 70,000 ft3 and a weight of
500 Ib (including passengers, basket, balloon fab-
ric. etc.). If the outside air temperature is 80 of
and the temperature within the balloon is 165 of,
estimate the rate at which it will rise under steady
state conditions if the atmospheric pressure is 14.7
psi.
For .sfeady rise ~ F; =0) or Fs ;:: W+/!
where
o!) :::drag ::: CD t.f U
2
: D2.
~ :: 6110yanl force == 0'¥
and.
W= lofal welJhf =.500 Ih +O;n .y
volIJme 11
( Jb ) { ")2-
!Vow D= -13- ~ I'f,7-w. I). In· = 0,002-29 slllfM
RT {!715 s ,,;.• )('16oteo)OR fI.:I
0= p~ ::: (0.00229 .stjf)(32~2~.)::: O. 0736 ¥p
and Jb • 2. Ii
0; = .1i:. == (11f.77ii?-){t2.%) (32.2 .s~)
In R"Tin (1715~!;~R)('f.60+/65)'R
Thus wifh 11::: 7xIO'lfl3: ~~f)3
, or D= 51.1 ft we obtaIn
JJ= CDiro.002- 29) U2.*(51., )'1.
;:: 2.36 CtJ rJ2. /b) where U~.g.
_ L Ib- 0.0036 N.i
/'Iole: Since fhe ba//oon i.s
open at the hoftoll1J
fhe
pres.spre within fhe Da//ODn
i~ lJeal'/}, Me same QJ ill~
o()fsir/e.
Ills0)
W;:: Soo Ib +(0,0636 PJ..i)( 7~000 fi3) ::: 9952/1>
ano rr-
~;:::(O.(J736f!P)(7o"OOOff.3) =SIS2Ib Tht/sJ Fa::: W+rf} 9ive.s
2. 2-
5/52 /b ::: 1'-952 Ih + 2.36 Co TJ or CD TJ ::: Blf.7
fJlsoJ Re ... !II- . '
(J)
or 51./ ff U' .s
Re ::: -~.fi!" =3.2.SX/O U
J,S7X/O ~ , I~ ~
ond from [;7- 9.21 CD~
(:;.)
(3)
Re
Tr;al 4nd error solufion: IIssv!1J8 CDj ohrQ/t} (/ from £i.OJ) Re from Ef{.t:J.);
check C/) from £,.(3)J the qraph.
lJ.s..stJme CD::: 0•.5 -+ U::: 13.0 #- ~ Re :::1.J..23 x/o6
---. Cb ::: o.2.¥;t O~S
lJ.sst/me CD =0.2.'1-' V ::/8.8 ft - Re = 6.11 X/06 - CD :=0.30:/= O.2~
Assume CD:: D.30~ U::::/6.8§. ~ Re ;5.'16 x/06 ~ ~ :: 0.30 (checks)

9. If5
9.45 A 500-N cube of specific gravity SG = 1.8 falls
through water at a constant speed U. Determine U if the
cube faBs (a) as oriented in Fig. P9.45a, (b) as oriented in
Fig. P9.45b.
(a) (b)
• FIG U REP 9.45
For steaay fa II) :£ F;;; ma:::O
or
W== oI!+Fa, where W::: wel9hf :::500 N
Fa:: buoyanl force::: 0 D3
Qfld of}::: f f UCD A -:;; dr4fj
But 3 .3
W~ 0; D3::; SG 0D3
, or 500 N ::: I. fI ('I, eox 10 -j.) D
ThvsJ
D== 0.30.5 m Sf) that from £~. {/J
teD
.5ooN:: i(qqq~)UCD (O.305m)% +('l80 X/0
3!3) (o,305m)3
or
U:z.CD ::: '1-.78 where V,.. l}-
(a) for case fa) CD::: 0.80 (see F/". q.2Cf)
Hence.l U:: ( ~:;~ )1/,. ~ 2.Jfif if-
(b) For case (b) CD:::: /.0£
(
'1-.78 )"2. WI
Hence) U::: 1.0.5 ~ 2.137"
Q-'f5
(I)

9.46 The 5 X 10-6
kg dandelion seed shown in Fig. P9.46
settles through the air with a constant speed of 0.15 m/s. De-
termine the drag coefficient for this object.
For sfeady faJ/if19 41 tl cQl/stani .fl'eed,
rfj c VI or /IJ?-:: C/) i p7J2.IJ
ThIJs;
• FIGURE P9.46
sx/O-6kj (Q.II-fi) ~~ ({)(1.23i1a)(O.lst)1. f(O,OlflJl)2.

9,Jl7
9.41 A 22 in. by 34 in. speed limit sign is supported on
a 3-in. wide, 5-ft-Iong pole. Estimate the bending moment
in the pole at ground level when a 30-mph wind blows
against the sign. (See Video V9.6.) List any assumptions
used in your calculations.
For e9va//br/tltfJ I LJ IY/q ::: 0 or
(YJB = 2.,.5' fl a9f' of (s +H)ff 4) where
4 .: dra1 on the pole and d& :: dray on 1Hfsir/? 3S1
From Fic/. 9.28 wifli ilD.:0,/ for the si7rJJ
i!,p~ D
CDs::: /, q 1
I---l-..... ~
17 tn.
T
(/)
From ~7' 9./9 if the po.rf aots as tf srvare rod .....,1-1
with shal'p cornel'S cnp == 2.2. Th/J~ wilh 17== 3 0nJl'h ;:: 'f'f tt I
t:l :: -J-DT!:z.C ,,==...L '0 001-3~ §,/vf;!..) (tf'fii)2(19) (~).. (9Jf)ff2.) -22 III
~ ~ r 'j)s I'1s 2. I , f{3 ..r' / ~ -¥- - . I/O
and
~f =1: W2.C/)P lip =' i(o.oo:m ;'1;-) ('f'ff1. t(2..2.) ( ~ (S)ft) " 6.3'f/l,
-rhtJs; from Eq.,(;) ~
MB :::; ~.sff (b.31f/b) +(s+!J)ff{22,716):::: /62 ff-Ib
9-'1-7

9.48 Determine the moment needed at the
base of ZO-m-tall, 0.12-m-diameter flag pole to
keep it in place in a 20 m/s wind.
FOr ervi/iDriomI fYJ =l,jJ where
Jj=c. i y2.iD
• I> P UD (2.0'1-)(0.12 m) .s
Since Re =-:r::= -5 2. ==!.6'fx/0 J
/.lf6 XIO !}
thai CD == /.2
Thus, &0= /.2 (f)(I.~ ~3)(ZO'f)2.(2()11I)(O.I2.IIJ):;; 70aN
HenceJ from £r· (/)
M:;; ~ (708N)::: ?,080;Vi?}
2.
(I)

9,1ft( I
~}.49 Repeat Problem 9.48 if a 2-m by 2.5-m flag is attached
to the top of the pole. See Fig. 9.30 for drag coefficient data
for flags.
For eqtJi/ibritJm, /YJ =4tbi+(1, - ~)~2.
where ~ =20 m) 12.. = 2.5 mJ (Inri D2 ::: 2m.
From the so/v-lion +0 Problelh q, Jf8 ) ~I ~::
IJ/so}
2. . : . 1 1 12. 2.5
/}2.., =CD t pTJ 12. D2 ) where from FIg. 9.30 WllfJ n == T :;: /.2S'
2-
we obfoin ClJ = 0,08.
Thus}
082
== o.o8(1)(I.1-3~)(20':)~(2IEhJ)(21T1) = 98,'111
By combinifJ9 £'1s. (/), (2.)) and (3) we ohfain
/'1:: 7,080 /V.", +(20m - 1m) (98,';' N) = B)qSO N'm
(I)
(2.)

9,SO
'1' 9.50 During a flood, a 30-ft-talJ, lS-ft-wide tree is up-
rooted, swept downstream, and lodged against a bridge pillar
as shown in Fig. P9.S0 and Vjdl~O V7.(j. Estimate the force that
the tree puts on the bridge pillar. Assume the tree is half sub-
merged and the river is flowing at 8 ft/s. See Fig. 9.30 for drag
coefficient data.
8 !tis
.--+-
.. FIGURE P9.50
Force of free on bridge ::: dr(J1 on free c r/J: c" fell'lJ
where lJ== 8 H&-
/lS,Sl)me fhe free is .rha;ed as a/} e//ifue anti -t I~ /n fhe 1114/91',
7htJsJ
A :;;(dJ *(30 It) (/S fl) :: /771-12.
FrDfJ1 Ft9' 9,3(} tis the wind pasIafree ;f)cr~a.re~ I/Jc clrllf ctJe!l;c/~/J1
decre.'IIsBS (fhe leaves "(Did /;P,(;X P
). 1I.r.rv",e //;e.rafJJe Inif)! l1o//BIJ.r III
wafet 411d (/.re c;::; 0.1..0,
7h{).JJ h
rx9'::: o,,,o(f)(/,9~.!;')(eP/· (!77f1:L) -::: 2Z00//;
9-£0

9.s/ I
9.51 If for a given vehicle it takes ZO hp to
overcome aerodynamic drag while being driven
at 65 mph, estimate the horsepower required at
"15 mph.
9-SI

9..52. I
9.52 Two bicycle racers ride 30 km/hr through
still air. By what percentage is the power required
to overcome aerodynamic drag for the second
cyclist reduced ifshe drafts closely behind the first
cyclist rather than riding alongside her? Neglect
any forces other than aerodynamic drag. (See Fig.
9.30).
~D == power when nof dl'offing == Volta :: UCDJ/IJ f py"A= q,NDipV~
dnd
3
Po = power when dl'oflilJfj :: CD -ipu A
D
From Fi9' 9. 30 CD == 0,88 Q/Jd q, == 0,50
Thus,
WI) D
- 0.88 -O,5Q. =0,'1.3)",) i.e',) a 1,L3,27, decrease
0,88

9.S'f I
9.54 On a day without any wind, your car consumes x gal-
lons of gasoline when you drive at a constant speed, U, from
point A to point B and back to point A. Assume that you repeat
the journey, driving at the same speed, on another day when
there is a steady wind blowing from B to A. Would you expect
your fuel consumption to be less than, equal to, or greater than
x gallons for this windy round-trip? Support your answer with
appropriate analysis.
1rlp wHh fhe lafrJer pOW81' lOST aV8 f() qerodYI)/Jmic drdf willllJ'e fhe
mO.It 94.f. Lef ( ), mean np winJu tlnti ( )2. mean //wine/II,
0) No wino!:
4 :: c/) -tpu"'l; fof' ~ofh 11.-.B 41ld B~II
Thtl~
-;;~ p()wer ::: 1J~ :;: 1:p7l}[)/l
(2) Wind (7Jw-'= wilJd speed; as.sV/IIe ~<:7J):
,j~ ::: CIJ tf ttlf7lwfll for /1-+8
/)2. == Col) 1.P(ll"'~J2.fi fDt' B~II
TlJIJ.s
I ~
11 ~ tp(U+1lw) 7Jcj)1I fot' I}~B
~ :: i (J (rJ-7Jwt 7J ~ IJ for 8-,./1
).
B
£ner'lf f).sed ::: ~( where ,,::: filJ/B III 1Q froll/ /I~B PI' !~Il
ThvsJ
E, == 2- (te7J3c/JIJ)i
(Jfld
..1.. 2. .L 2. I
£2. == 2 f (77-171) UC/) Il t -I- 1. p(TJ-Uw) UCb/l r
::
t. e. mf)/,fi fvel needeJ whelJ Windy
q-53
-- I :z. <: I
/ +(71w/7J)

9..5.5
9.55 A 2S-ton (SO,OOO-lb) truck coasts down a steep 7%
mountain grade without brakes, as shown in Fig. P9.SS. The
truck's ultimate steady-state speed, V, is determined by a bal-
ance between weight, rolling resistance, and aerodynamic drag.
Assume that the rolling resistance for a truck on concrete is 1.2%
of the weight and the drag coefficient is 0.96 for a truck with-
out an air deflector, but 0.70 if it has an air deflector (see Fig. v
P9.S6 and Video VIJ.S). Determine V for these two situations. ---
• FIGURE P9.55
For cOlJs/ant .rp~eJJ & Fx :mtlx ::: 0
Of'
Wsi»e::: a9 -/- F
where (J == Qrc fan (i/o) ::: J/.. 00 tie,
!)
'I.
) tiJ= tf 712.c/) /I
TrUCk width =10 ft
Thvs
£0, ~00 fA (sin 'f.oode,) =t (0.001-38 :./;~) lJ'-C/) (/2 N1%fl)+0.0/2 (5'~000 /b)
or 1.
3 'f9a 16 = O,/~3 UcJ) +600 Ii
(aJ 1rC/) ::. 0.9t J fhen II=/If£!- =:- qi//meA
(b) If c/j::: O,7() J then U::: /70 ji =/lbPlpn

9..56
9.56 As shown in Video V9.S and Fig. P9.56, the aerody-
namic drag on a truck can be reduced by the use of appropri-
ate air deflectors. A reduction in drag coefficient from
CD = 0.96 to CD = 0.70 corresponds to a reduction of how
many horsepoer needed at a highway speed of 65 mph?
(a) CD =0.70
-p :::poWer == gO' where
2-
rIJ == f pU CD A
Thus) tJ. P ;;: reJl1cliolJ if) power
==1b-~
::: t p7J3
A[CDb - C~a]
W;Ih 71::: 65'mph ::: 95 .3 .ft;sJ
(b) CD =0.96
• FIGURE P9.56
b'P ;::. i (O,00238!.}:J-!.) (9S.3.fi)3(JOff)(Jlf/) [0.96 -0.70]
;::. 32;; 100 ¥ (£i-ohb) ::: .58.if hp
oS'
9-S5

9.57
9.57 The structure shown in Fig. P9.57 consists of three;
cylindrical support posts to which an elliptical flat-plate sign is
attached. Estimate the drag on the structure when a 50 mph
wind blows against it.
where If we a..r.r(/llJe the s;911 is (Jf) el/~eJ
/l:: f (;off) (5f1) ~ 39.3 fl2.
IJ'J. ~ 0.6 ff (15ft)::: 9.00 1-12-
IIJ .:: 0, aft (;0511) ::: /2.,0 Ifl. /lAd
At.;: l(f(lsfI) == IS; off)"
From Fif. fl, z?J f()r tI Ihin disc ~/::: /,/
For -fhe cylindricaJ po.ri obl4/n Cb fr()m Fif-9.'J-/ Cis ~ (7J:::S()IHIA::: 7.3.3.f!.)
Re ~ ~ :: 13.3 ~!OI6~) == 2.8X/().S' --+ C ::: 0.6
2. l' /,S7X/() i'1p. /)~
Sinl/lar/YJ
Re3
::;: .31 7x/Os~ C.o ~ 0. s
.5 .J
{(elf =:Jf.7X/o - Cb~ =: ().'J-S
ThvsJ
from £f. (/):
rff=i-(O.O()2.33 ~:f.t )(73.3P.tf!./ (39.3ft) +O. Irp.ol-!j +O.S(!1. ff2)+o.2-S (lsf:,l2)]
::= 3 7~ /iJ
9-56

9.SY As shown in Video V9.5 and Fig. P9.S9, a vertical
wind tunnel can be used for skydiving practice. Estimate the
vertical wind speed needed if a ISO-lb person is to be able to
"float" motionless when the person (a) curls up as in a crouch-
ing position or (b) lies flat. See Fig. 9.30 for appropriate drag
coefficient data.
For eC((J///hritJm cond/Iiof},s
W=rIJ =CD t f TJ2.A • FIG U REP 9.59
fls.sume
ThlJ,s}
w tv
W::: /60 Ih and C/)/}:: q fl:/. (see Fi'9. 9.30)
/601b == (f)(O.00238S1pS) U'l-(9f1'-) where U..... fj
or
U-( ff)( Iml )(360 0..$) h- 122 5 S28of+ I hn :::83.2",1'
Nofe : If fhe skydiver /~flr/erJ lip il1fo ahaIr, fhen Call?:. 2.,slf
(see Fi9.9.30) and (J= /s8mph
9-S7

9. 6o'It
9.60* The helium-filled balloon shown in Fig.
P9.60 is to be used as a wind speed indicator. The
specific weight of the helium is y = 0.01lIb/ft3
,
the weight of the balloon material is 0.20 lb, and
the weight of the anchoring cable is negligible.
Plot a graph of f) as a function of U for 1 sUs
50 mph. Would this be an effective device over
the range of U indicated? Explain.
For- fhe balloon fo remain stationary
2: Fx ~ 0 and 2:Fy =0 rb
u
FIGURE P9.60
Thlls) IJ = T cos e or T == c()..s e
I4nrJ. ~ == W+TsilJ e +k1Je pJP
which comhine 10 qiV8 Ie . . bl= nSlon I/J CQ e
Fa .::: Wt c!J ftJ178 + ~e
Sui W.= 0,2 /6 ) F8 =eq"V =(7. 6£ x/ti2.~) ¥(-fH)3:::: o.32o'llb
and ~e == ONe V = (0.0 II ~) 'i:{l (1:H)3:: 0.01,16/ /6
ThLJs) £f/I (I) becomes
o.320'l1i; = 0,2 /b +,j} f4/JB +O.oi.jt/ II:,
or
of!fan e:: O. 07l(-3 /6
100 cis
120 print#l, "***************************************************"
130 print#l, "** This program calculates the angle of the **"
1'"'0 print#l, "** cable as a function of the velocit.y of the **"
150 print#l, "** air. Values of the drag coefficient are **"
160 print#l, "** obtained from Fig. 9.23 as a function of **"
170 print#l, "** Reynolds number. **'1
(con'!:.)
Q-S8
(I)
(2)
(I/-)

I I
I!
i i
9.60' (con'I;)
180 printfl:l, "***************************************************"
190 printfl:l, " II
200 pi = 4*atn(1)
210 printfl:l," U, mph CD theta, deg"
220 for i = 1 to 10
230 print " "
240 input "For a velocity of (in mph) U=",U
250 Re = 1.27E4*(88*U/60)
260 print using "the Reynolds number is Re=#.###~~~'-II;Re
270 input "Enter the drag coefficient: CD=".CD
280 theta = (atn(19.9/(CD*(88*U/60)~2)))*180/pi
290 print using "For U = fl:fl:#.#fl: mph theta = ##.###fl: deg";U,theta
300 print#l, using II ####.# ##.### #fl:#.##fl:";U,CD,theta
310 next i
, ***************************************************
** This program calculates the angle of the **
** cable as a function of the velocity of the **
** air. Values of the drag coefficient are **
** obtained from Fig. 9.23 as a function of **
** Reynolds number. **
***************************************************
U. mph CD theta, deg
1.0 0.400 87.524
2.0 0.420 79.707
5.0 0.540 34.421
10.0 0.550 9.548
15.0 0.330 7.102
20.0 0.100 13.022
25.0 0.080 10.482
30.0 0.090 6.516
40.0 0.120 2.759
50.0 0.160 1.325
--------------------------------------------------,
evs U
90 1
80
70 1
60+--~--~~------+-------~------~------~
~ 50+--4-----~------+_------~------~------~
! 40+---~--~------_+------_4--------~----~
30+---~--~-------+------~--------~----~
20~-~~--~----~--~--~
./ ""--
10+-------~----~~--~~_4--------~----~
~ ./ """'-I---
O~--~----~--~~==~==~
o 10 20 30 40 50
U,mph
q-sq
lNofe :oBe:cause of the sudden
chon98 in C/) when Ihe
boundar! layer DecrJfII9S
furhu/slll (at ahout /S mph),
the evs UeUl've i.s hi9hly
non -linear. 1/J loe" for So/l#
valves of e there is more fhfJA
one po-ss/bie va/fie 01 Y. If
would /Jot wo"k well rJ,f aINiAd
speeJ indica/of' ,n Ih;s f'4IJfR.

9.61
9.61 A 2-in.-diameter cork sphere (specific weight = 13
Ib/ft3
) is attached to the bottom of a river with a thin cable, as
is illustrated in Fig. P9.61. If the sphere has a drag coefficient
of 0.5, determine the river velocity. Both the drag on the cable
and its weight are negligible.
r:Fx = 0 or t.#-::: TC(Js6o'
and
ZFy::: 0 or Fa -W=Tslnlo'
(I)
(1.)
Since T '60·
11 ::Vo/Vl1Js fJf cork== -!llr3
, -
60°
= /71'(Ji fl)3:; 2.1-2 t/o-.1 f+
3
if f()llows frf)/II Ff. (2.) thai
4.10 V- ¥:fJl'lt. V :: 7si/l bi
l
J or
(62../f -13)#" (j..'l-2 x/o-J
JfI' t:: .,-si",rI
TlJvs~ T::: 0./3 aIh
From £'1' (I) I ~::: CD feu%1J ::;; 7 c()Sto() where /I'lI!1Tr
2
ThvsJ
1/
D ~ k
U=[ 2. T (;(J.s6~ ] =[ :2 (O.l3sIIJ)(;t7S6o
fJ
] 2_ ft
CD e7rr 0.5 (1,'I1f ~:¥)1f(;tft)'I. - 2 .S.5-:s
Q-60

9.12.
9.62 Two smooth spheres are attached to a thin rod that is
free to rotate in the horizontal plane about point 0 as shown in
Fig. P9.62. The rod is held stationary until the air speed reaches
50 ftjs. Which direction will the rod rotate (clockwise or coun-
terclockwise) when the holding force is released? Explain your
answer.
Lef 4 and 4 del/ote fhe dra9
forces on fhe fell and ri9h f
~pher8.sJ respeclively. If 4 >~
fhe rod will To/ale covllier CIDG~t;ise.
_ I 7Js :z
4. -£;J. 2. f ~ and a&J? ~ CJ)R t e71 IIR Sf) thai
:r~
1u
../O.7-ft diameter
~ CDt( IJA. wlJere CL) =:: CJ) (Re J. (J)
; ( =:: CbJ. IIJ.
#OIi0 UD. O.0023S
sllJ ~(50#)(I,Sf/) 5
Re::: e.. ;:= _ 1$ =:/1.771./0
1..# 3.7'1'/./0 fii
and. =:: D UDR __ 0.002-38 -!fp(SO~)(O.7f./) .5
De l- "1 ~ ::; 2,23X 10
!Tj R fJ' 3.1'1-1./0 ,.;!
Thv~ frrJlII Fif- B.21 Or 8.25J for smooth spheres
CJ)L =0.06 alld ~R::: 0.5 $0 thai Eq.f/) 'lives
11 2. 2.
D8p. CDR Lf DR 0.5 (O.7{~)
.....tJ == .1l 12 : O.06(J.5f1)1 == /.81
(17/.. C/)L -¥- v/..
Since 4 >4 Ihe r()d will rotafe counter clockwise.
/lole; II11hOUfh Ihe rip,1 sphere /s S llJaller fhan Ihe Ielf, Ifhas
more dra'j hSCfJlI..re il n4S a It/lYe rJr41 coefficielJf (/al1li/JfJr
boundary layer, wiJe waKe). The /4r,9 sphere has a smaller
drtJ1 coefficient (lurbulenl bounJ4ry layerJ narl'pw waf.e).
9-61

Q.63 I
9.63 A radio antenna on a car consists of a
circular cylinder t in. in diameter and 4 ft long.
Determine the bending moment at the base of
the antenna if the car is driven 55 mph through
still air.
FOr e9ut//nrivm J g /YJ():: 0 J or fYIo = lrlJ
,dia- D
V=5S mph
- - ff ltd}==80.7 - 1
oS 2.=2ff
where of)= CD fpV2.f1 (J)
5 . R - Jll. - (80,7#) VB ft - I 07 X/0'1
/l7ce e - .y - -I/o ft'- -, )
c:;,slRx
/,57 x10 :s-
it follows from Fi9. q. 2/ fhol C~ ~ I. 3
lienee 01} = /..3(t)(0.002.38 =!.~~5 )(BO.71j)'J. (if (1) (~ fI)
Thus) Mo = (2 fI) (0.8'10£) = /.68 Ff •Jb
9. 65 Estimate the wind force on your hand
when you hold it out of your car window while
driving 55 mph. Repeat your calculations if you
were to hold your hand out of the window of an
airplane flying 550 mph.
/Yl()
Ii
b= CDterf2./I J where U=(ssmph)( ~~m;h) = 80.7!t
Ass/)f/)c your hand 1'.5 'fin_ by bill, i/) si~e dnd acl.s like
a t~in disc wifh CD ~ 1.1 (see Pi-g.9.zq).
Thvs)
rIJ=(J,J)(~)(O,00238)(eo.7fj)').(11f1)(&N) == /, 1f2 Ih
If your hand is norma} 10 the fhe lif! force is zero.
For U =550 mph == e07 Y (i,e') a /0 fold increase in rJ) fhe
draq will increase by (J factor of 100 (i.e~ 08- U2
)J or JJ= /'f2/b
Note: We have assumed fh4t CL) is not a funcl/on of v. That /sJ
if is /Jof a fvn&lion of either Re:::: ~ or /Y/fJ -:: *".
9-62

?, 66 It I 9.66* Let 0>0 be the power required to fly a
particular airplane at 500 mph at sea-level con-
ditions. Plot a graph of the ratio 0>/0>0' where 0> is
the power required at a speed of U, for 500 mph
=:; U =:; 3000 mph at altitudes of sea level, 10,000
ft, 20,000 ft, and 30,000 ft. Assume that the drag
- coefficient for the aircraft behaves similarly to
that of the sharp-pointed ogive indicated in Fig.
9.2f. 'P t eV3C A C 3
-p== Urb ::; CD i pV
3
fJ .sO fhaf -::;:}:;::; ..L eJ1!C
D
A - e D (-JL) (I)
(88 ft ) ro 2. 00 Do - PoCDo Va
Nodw1
Yo=(s00mph) (6()m~h) = 733~
an
/Ylo =¥- where C =1k RTi so fhat Co -:: /1.If (17/6/it '~~R )('1-60+ 5'1)oRi
or Ii. "-/ 9
M = J&. = 733 ~ == 0 656 = / //71f
1'lfJo Co 1//7 11 ..s
HenceJ
from Fi9. 9,ZtfJ
eno= 0,2 so fhal £". (f) hecOf!J8.5
1> _ eCD (- U )3 or -p _ -6 C V.3 h
% - (0.00238 sllJ~)(O.2) 733 if 70 - 5,33 x/o r D J were {:l.)
f f S . U,., Ii and
fils 0, CD =Cb(fYla) from h~. 9.2'1- CDI F- e~ s;:r- (3J
where
M- U ~
Q - Vf.JI. (1716~J~~t/l)(Jf6() t T) 'R'
or
Ma = U I where TJN fj and T'" 'F (If)
,; 2 Lf 00 ('1-60 +T) J
ThosJ for fhe givel) a/fi1ude obfain Tand pinTable C.I. Se leef
bOOmph ~ U~ 3,()()Omph (i.e~ 73.3!f ~ U:: 'I-~oo1j )J determine /!Ja
from £'1,,('1) J C[) from Ef(J3J (the graph)) and ~ frblll Eq. (~J. The
reslI/fs are ploHeri be/olll.
0) IN /~OO() ffJ p::: 1.76 x/o-
3
WI J T = :2 3. Lj. "F
Thus, i:,::; 9,98 x/o9 C
b
V3
(J Tt
Ma= 10eo
b) 11-1 2~oooflJ P=/.27 x/o-
3S/7!- J 7= -/2.3 'F
Thvs :e -9 3
~ :: 6.77 X/0 Cb U
Nf _ U
I'IQ - IOIf()
9-63

9.66 4 I (conll)
and
c) IN 3~OOO ff J e~ 8.9/x/o-If. sf:¥-J T =-tf7.8 fir
ThusJ ~::: 'I,7S X/O-9CD y3
M- UI'-I(J.- q'ls
Proqratn P91166 s/70Wh belolV Wqs flsed for the CQ/cfJ/4Iiol)s. Cb vs (;14 rit/fa
WQS f4keIJ fro", Fi9J ?2'1-.
100 cIs
120 dim M(17), C(17), A(3), AA(3)
130 print#l, "*************************************************"
ll,o.O print.#l, "** This program calculates the power rat.io **"
150 print#l, "** at different altitudes of flight for **"
160 print#l, "** flight speeds from 500 to 3000 mph. **"
170 print#l, "*************************************************"
200 C(1)=0.19 : C(2)=0.23 : C(3)=0.l,o.2 : C(l,o.)=0.55 : C(5)=0.52
210 C(6)=0.49 : C(7)=0.46 ! C(8)=0.43 : C(9)=0.40 : C(10)=0.38
220 C(11)=0.37 : C(12)=0.36 : C(13)=0.35 : C(ll,o.)=0.35
230 C(15)=0.34 : C(16)=0.34 : C(17)=0.34
240 A(1)=1080 : A(2)=10l,o.O : A(3)=995
250 AA(1)=9.38E-9 : AA(2)=6.77E-9 : AA(3)=4.75E-9
260 for i = 1 to 17
270 M(i) = 0.25*(i+l)
280 next i
300 for i = 1 to 3
310 z = 10000*i
320 print#l, " "
330 print#l, using "For an altitude of z = ######. ftl!;z
3l,o.O print.#l. " U, mph Ma CD P/Po"
350 U = 0
360 for j = 1 to 6
370 B = 0
380 U = U + 500/60*88
390 Ma = U/A(i)
~oo for k = 1 to 17
410 if B = 1 then goto 450
~20 if Ma > M(k) then goto l,o.50
430 CD = C(k-1) + (C(k) -C(k-l) )*(Ma - M(k-l))/(M(k) - M(k-l))
44·0 B = 1
450 next k
460 ratio = AA(i)*CD*U
ft
3
465 UM = U*60/88
470 print#l, using" #####.# #.#### #.#### #.###ft"ftft!l;UM,Ma,CD,ratio
l,o.80 next. j
l,o.90 next i
(con II;)
9-6'1

?# 664~ (con'-lJ
o
a..
it
~~~*********************************************~
** This program calculates the power ratio **
** at different altitudes of flight for **
** flight speeds from 500 to 3000 mph. **
*************************************************
For an altitude of z =
U, mph Ma CD
500.0 0.6790 0.2186
1000.0 1.3580 0.5370
1500.0 2.0370 0./,r556
2000.0 2.7160 0.3827
2500.0 3.3951 0.35/,r2
3000.0 /,r.07/,rl 0.3/,r00
U, mph Ma CD
500.0 0.7051 0.2228
1000.0 1./,rl03 0.5308
1500.0 2.115/,r 0./,r/,r62
2000.0 2.8205 0.3772
2500.0 3.5256 0.3500
3000.0 /,r.2308 0.3/,r00
U, mph Ma CD
500.0 0.7370 0.2279
1000.0 1./,r7/,r0 0.5231
1500.0 2.2111 0./,r3/,r7
2000.0 2.9/,r81 0.3721
2500.0 3.6851 0.3500
3000.0 /,r./,r221 0.3/,r00
100
f------ ---~,-- .
100 1000
U, mph
10000 ft
PIPo
8.088E-Ol
1.589E+Ol
/,r.550E+Ol
9.061E+Ol
1. 638E+02
2.717E+02
20000 ft
PIPo
5.9/,r9E-Ol
1. 13/,rE+Ol
3.216E+Ol
6./,r/,r5E+Ol
1.168E+02
1.961E+02
30000 ft
PIPo
/,r.270E-Ol
7.839E+00
2.198E+Ol
/,r./,r61E+Ol
8.196E+Ol
1.376E+02
--z =10,000 ft
- - - z = 20,000 ft
- - - - - z =30,000 ft
10000
9-65

9,~7 J
1).67 A 0.50-m-diameter meteor streaks through the earth's
atmosphere with a speed of 1800 mls at an altitude of
20,000 m where the air density is 9 X 10-2 kg/m3 and the speed
of sound is 300 m/s. The specific gravity of the meteor is 7.65.
Use the data in Fig. 9.24 to determine the rate at which the me-
teor is decelerating.
z:: r:::ma or tb= mq
TIJvs,
1:pref) f 1/- == fm .;1/(.g)3fA Phi ~ SG PN~o
where If. (/) 3 [ l!. 1 'I- (0.5111)3 I.
m ~ Pm -:3 11 ~) ::: 7,65 (1000 /113) J r 71 --r == £0/ ~9
IIlso
j U
/fIa ::: C" =
9-66

9.68
9.68 A 30-ft-tall tower is constructed ofequal I-ft segments
as is indicated in Fig. P9.68. Each of the four sides is similar.
Estimate the drag on the tower when a 75-mph wind blows
against it. 4~===~~.... ....
fu
2 in.flssllme no inferference be/ween the
fronf and back portions of the lower.
A/so, neq/Bd fhe cJrfHI on the sider of FIGURE P9.68
Thus) from £r.. (t)
L
t
-t
(I)
,j} = 30(l.(O.OO2.3FJ s~~s )(110~l)[(J.q8)(2.)(Iff)(-kH)+(18)(kH)( et~~8V2rt)
t (1.82.) (z){/f/)(k H)+(1.rO(J.f+)( 8 t8/~812 r+)J
or
,j} = BSq Ib
9-67

9.69
9.()9 A 2-in.-diameter sphere weighing 0.141b is suspended
by the jet of air shown in Fig. P9.69 and Vidl:o V3.1. The drag
coefficient for the sphere is 0.5. Determine the reading on the
pressure gage if friction and gravity effects can be neglected for
the flow between the pressure gage and the nozzle exit.
9-68
Area = 0.6 ft2
Area = 0.3 ft2
t
Air

Q.70
9.70 The United Nations Building in New York is approx-
imately 87.5-m wide and I54-m tall. (a) Determine the drag on
this building if the drag coefficient is 1.3 and the wind speed is
a uniform 20 m/s. (b) Repeat your calculations if the velocity
profile against the building is a typical profile for an urban area
(see Problem 9.17) and the wind speed halfway up the building
is 20 m/s.
(C/) of) == CD f f v'lJ == /.3 (i )(1.23 ~)(2oft(15'1m)(87.SIfJ)
or
JJ== /1,3/ x/0
6
N = '1-.3/ MN
(b) For ~n vrDQ/J areo, U;::. C yo.1.}
Thv.sJ wdh u=2.0 lJ- af y-:: -b: : 77fJJ
we obfain
y
T
h
1h==l.5ifm
C = :,.11- == 3.52. J or f). == 3.52 yO.4 wilh UIV 1} J Y'"In
The 10141 drtl9 is y= /s'!-
cb = fdeD =,fCD ip u.
2
dlJ = -freD{(3.s2. yO.If/- (87.5) dy
y::o
or 15~
.()=f(J.23) (1.3) (3.Sz/(87.S)S/,8dy = 967C~ ) (f5'1r= '/;/7 ~/06 N
Thvs) 0
rf} == 'f. /7 MN
9-61

0)
(2.)
9.72 When the 0.9-lb box kite shown in Fig. P9.72 is flown
in a 20 ft/s wind, the tension in the string, which is at a 30° an-
gle relative to the ground. is 3.0 lb. (a) Determine the lift and
drag coefficients for the kite based on the frontal area of 6.0 ft2.
(b) If the wind speed increased to 30 ft/s. would the kite rise
or fall? That is, would the 30° angle shown in the figure in-
crease or decrease? Assume the lift and drag coefficients re-
main the same. Support your answer with appropriate calcula-
tions.
u= 20 ftls
•
/r':31b
t
• FIGURE P9.72
(a) z::Fx :::If) Ox =: 0
Thfl.fJ
tiJ
e/) == t('U2-A
of'
e/)::: O,i)/(J
/llso;
0'" ~::: Tcos30fJ
:::(316JC()s30o :::: 2.IOI/;
zry =mtly=:O or
!hils; t
C.l. ==ff TJ'''I1 =~
or
G.:: 0.8'1-0
2Jfo/6
9-70
-r=3Ih W= o.911J

9.73 I
9.73 A regulation football is 6.78 in. in diameter and
weighs 0.91 lb. If its drag coefficient is CD = 0.2, detennine
its deceleration if it has a speed of 20 ftls at the top of its
trajectory.
0.0238 Ii> ff::: O.Sil-1 ~
0,0283 S/fJr;s oS
9-71

fl. 7'1-
9.74 Explain how the drag on a given smokestack could be
the same in a 2 mph wind as in a 4 mph wind. Assume the val-
ues of p and J..L are the same for each case.
/)= CD i f U'J./J :::: CD £pU
z
1l'Dt
LeI ( )1 denofe conJilions with lJ= Jl}-
and ( )1. with U= 2. !p-
7/;lIsJ with p, =f:l 1 fo hove /J;::: ciA We hove
C ' u,2 A r J 2. IJ C ~ 2-
D/2ft I rrDL =V[)z2f2. U1. lTDL or DJl1 :::CD:;. Uz
Thai i.sJ
CD1
:::: LfC/)2 where CD, and C/)2- ore fvncfions of Re == ~D tJs shown
Since 11 =u- and ~ =±Uz if foJ/oU/s fhat
Rei = 0,5 Re2.
From F;~, 9,1-1 tq) we cqn deTermine CD
a value of Re2
such fhat Re, -::: 0. S Rez
and CD1 =- Lf CD:;. j hence the drd9s
are e'ltJo/ eve"fhou9h the velocHies
in Fi9vre 9,21 (a),
J.
CD2,.== 0. 25
are /JIJerlla J. This occurs beCOIJ.J'8
• _ _ _ _ 1 _ _ _
of fhe s()dden drop in ClJ as the
bo()ndafY Jqyer Decomes tlJf1hu/(}n1.
q-72

9.77
Radius =0.845 in.
9.77 A strong wind can blow a golf ball off the tee by
pivoting it about point I as shown in Fig. P9.77. Determine the
wind speed necessary to do this.
-- Weight = 0.0992 Ib
When the ball is ahout 10 be blown
from Ihe fee fhe free bOdy dia9rtlm
is 4S showlI. flence.; hy .full/min, IfJD/lJ8111s
ObO()1 (I):
[J~ ~ 0 J or Wi == tJr
Tnvs)
(0. O'!f/2 /6) ( o.2.oin.) = IJ (0.821 i".)
or
08= o. 0292/iJ I where /Ie ~ fflf21Tr2 r=O.8IfsilJ.
-ri. D.B2/in.
IfIUS) . 2-
Il C 1...1. £(}fS) V%r1°.89SIIJ)O. O'}.'I-~/O;: D 2. (0.00138 If' In i{!;. n '
~T1 x,:::O.lOln.
or
C/) U2. =: 130S1
where u-P. (I)
For a sphere4- C/) ==~(Re) (see [;'9- 9./..5) where (2.)
R == ~ - (O.00233S'''9S/FlJJU(2.(().8~5)Il2.r.l)
e p - 3.1f7X/0700',slfl3.)
01'
Re ~ 9tt rJ.; where V~!!- (.1)
Trial and erN!' s()/fliion:
IlSSf)hJ e C/) =: O. If oS0 Intll frf)/IJ £1- (/)J lJ:::S l /!) and
fro/l} £o.{,1) Re::: 9bl (~7./) -::! s.S 2 X/O ¥. TAvs fro/J?1 J J
Fl9' 9.],5, Cj; -;:: O. 2.5 :f (), ~Il Try 41(J//;.
!Js.su/lJe Cb =: 0.1.2 so fh4t V::: 77,() #- tllJti Re:: 7,1f'rX/O~
T/JlIs.; Fro/IJ r;,. 9.~5J Cb -;:: 0.22 ChecKs.
lienee; U;:; 77,tJ #-

9.78 I 9.78 An airplane tows a banner that is b =
0.8 m tall and e= 25 m long at a speed of 150
km/hr. If the drag coefficient based on the area
bEis Co = 0.06, estimate the power required to
tow the banner. Compare the drag force on the
banner with that on a rigid nat plate of the same
size. Which has the larger drag force and why'!
P = rbV J where oft= CD ±fy'J/I w/I/; f}::: bi.
Thus J
wilh CD=O.06tJnd lJ:::~.50~)( 3}0'::.5)( 'f~O: ):::~1.7.p
fhis 9ive.s
-p = (0,06) (±)(1, 2.3 ~)('fl. 7.!fl (O.Bm)(-;.Sm) ;: 5 3, S xl 0.3 W :: S 3,5 kW
For a r'9id fla-l plate
P = rlJ7J = 2 CD 1fV~bi (fhe f4c1or of fwo is needed hectlf)Se the
drQ9 coefficjenf is bClSed QIJ file dra1 QIl ()ne
side ()f Ihe pIaIf) )
iI,'fh R V}, ('l-1.7,f-)(25hJ) 1 IJ·.f'
Wj e/J ~ -u':= UI' -$ m£ = 7, /~X/O We ()oraln rrom.e" k-rD X/O oS
FI9J 9./s a va/tie of CD:::: o.O();).s for a smooTh pia/e.
Th()s)
'P == 2- (0.002.5)( J..)(1.231;) (If!. 7-lJ!-i(o. 8m)(2-.5",) = ¥-.lft xIO.3W :: /f. f~ AW
For +he flal p/4le cQ.!'e the d1'4! is re/alillefJ sll/al/ becfJlJSe /1 IS
doe elJf"f'e~ 10 sheqr(viscotJ,s) forces. /)V{) 10 fhe I'/flvller/1J9
h
of 1);61
banner, (/ 900d porl/o!} of /4 JrtJ1 (4!}ril;elJce power) is (J reS'()/1 01
preS-SlIf'8 forces . II /'.t /lui (/s slreomh"ned tiS a 1'I"tiel /hIpIale.
9-7'1-

9.7Q I
9.80 I
9.19 By appropriate streamlining the drag coefficient for an
airplane is reduced by 12% while the frontal area remains the
same. For the same power output, by what percentage is the
flight speed increased?
p :;: g TJJ WI;ere rfJ ::: CD f f 7Jit
Lei ( )0 denole the ori,inlJI conlijllrtlfion t:fnJ ( )05 the slream/ied one.
Thvs J w/lh r;:: 11 we ohfa/n
eDt) feo Tl;;2.flo Yo == CDsi f!s (!./I1.s Us or w/-/IJ t90 ~lJs ) po -::f~
. 3 . 3 tl [c.]~ C. k
Tfo Cao == u.s c[)~ ThllsJ T:::: cl>P. =[ iJo J3=I, 0'135
... Vo Ds CDo -O,/2CD(')
t:e'J
a '1-.35 h speed Increqse
Note: 'P~ U3Cl> .so/hal JfJ==3V2.GbJU.fU3JCj). ThvsJ
w/IJJ
JfJ=O this give.s t5U = - f Jet) =- -0;2 =-+0.01/- == 4-%
J TJ CD
9.BO The dirigible Akron had a length of 239
m and a maximum diameter of 40.2 m. Estimate
the power required at its maximum speed of 135
km/hr if the drag coefficient based on frontal area
is 0.060.
P:;alJU J
where ob'=Cf)teU'll
Thus) with [J:::(/3.5~)( 31oh;s)( ',Okhlm)= 37.5.lf-
P = CD -tpU3
f}
=(0.o60)(i)(J.2.3 !;;-3)(37.S!p)3Cf) (lfo.2tn/- == 2.'1- 7 X/06 k~:2.
or
p = 2.1- 7 x 10
3
AW (/.3/f.~)= 3310 hp
9-7S

9.8/ I
9.81 Estimate the power needed to overcome the aerody-
namic drag of a person who runs at a rate of 100 yds in lOs in
still air. Repeat the calculations if the race is run into a 20-mph
headwind; a 20-mph tailwind. Explain.
9-76

Q.83 I
9•.83 A fishnet consists of 0.10-in.-diameter
strings tied into squares 4 in. per side. Estimate
the force needed to tow a 15 ft by 30 ft section
of this net through seawater at 5 ft/s.
o.lo-in.-dia. cylinder
ifin. /
Jfin·/.V /
~/ /'.
IU~ecfjon1/ / / /
The net cQn be freafed as one lon9 ....1-~--.!'-/~--t---t---1~
0,/0 -/n. -diameTer .circular cylinder wlfh
ell= C; f [JUl./} J where U::: 5 #- . Each I fl' .sectirJn Qr Ihe net
conTains 6 feel {}f sfri171 (donoi C()IJnt fhe edges fwice). ThPJ.; f/;e
fola/ sfri/Jy lell,lh ;0$ 0fJlioximalely 1=(6ff2o) (/5 f-l)(30f/) = 2700 Ii
f)lso, since p= /,99 s~¥ and 1/:::: /.26 x/o-.s ~2. (see Table 1,5)
Re:::: Vb
0::: rs1J)( q,;.fI) = 33/0
1,26 x10-.51J!
ThlJs}
/J==(I,I)(1)(1.99%'1S)(sfj/{~ft)(2700ft):::: 6/61b
9-77

9.84 An iceberg floats with approximately,
of its volume in the air as is shown in Fig. P9.84.
ITthe-wInd velocity is U and the water is station-
ary, estimate the speed at which the wind forces
the iceberg through the water.
11
- ---.~--------~--
u
• 7 volume in air
FIGURE P9.84
Lei ( ~ denofe fhe portion of the /cebel"fJ in fhe air
dnd ( )w Ihal porlio/) in Me w4ler.
Thv.s} ~::: t-v and Vw::= f VJ where 11:: voItlme 0 f fhe iceber9
For sfeady mofion) 4::: 08w J where 4 -= ~q i ~(U - rh')2.Aa
and A _ C .1. 0 11 Z Ll
GVw- 'Dw2. W Vb nw
wlln
i1 ::speed ot fhe iceber9
9-78
(I)

9.85 I
'1.86 )
9$5 A Piper Cub airplane has a gross weight
of 1750 Ib, a cruising speed of 115 mph, and a
wing area of 179 ft2. Determine the lift coefficient
of this airplane for these conditions.
9.86 A light aircraft with a wing area of200 ft2
and a weight
of 2000 Ib has a lift coefficient of 0040 and a drag coefficient
of 0.05. Determine the power required to maintain level flight.
For eqlJi/;hrium t~ W=20 00/b == ~ ifU~1l
or .sIll 2-
2000 /b =(0.'100) f. (0.00238 -;/f) U (zoo ff')
Hence)
TJ =1'IS!i
Also; p::: power =: It Tl) where
rIJ= e/) i:pV'-;; =(O,OS)i(O,00238E~¥)(/'1$fil·(2.oofl2)~ 250/1;
/Vofe; Thi.s valve of rbC()vld he ohlailJed frfJm
~ == ~ := f!:. :: (),1f() :: 8 fJr d9 ~ ~ ::: ?:..,ooo//; ~ :;.50 II;
/IV (7./ c{) o.oS.J 0 8
ThvsJ
JJ /f.!.:Ul.( /h£ )P ~ 250/lJ (1'1-5 s )::: 3.b3 x/O ~ sso {(.J1J :: 65.9 hp
-:r
9-79

9,17
9.87 As shown in Video V9.9 and Fig. P9.87, a spoiler
is used on race cars to produce a negative lift, thereby giv-
ing a better tractive force. The lift coefficient for the airfoil
shown is CL = 1.1 and the coefficient of friction between
the wheels and the pavement is 0.6. At a speed of 200 mph,
by how much would use of the spoiler increase the maxi-
mum tractive force that could be generated between the
wheels and ground? Assume the air speed past the spoiler
equals the car speed and that the airfoil acts directly over
the drive wheels.
Tracfive force = Fz ~Ji~
where j1. ~ coef{j'c,ienirI{net/on w:: O. 6
Thf)~)
.Af; :::;fl A~ =fti!.; where LlF;. ,".r fhe
increase in fracl/ve force d{)e fo the rdowlJwardJ //I'f.
Hence will; lJ:::; 200 hJnh:: 293 fils) ,- J
b =spoiler length =4 ft
JW
t= ipU
2
CL IJ =i(o.OO').3e SK¥-)(2Q31j/·(;'/){J.5ff)(lfffJ-:= 67¥/b;
and
~ S::= 0,6 (67~/b)-:: ,!-os /6
9-30

9.98
9.88 The wings of old airplanes are often strengthened by
the use of wires that provided cross-bracing as shown in
Fig. P9.88.1f the drag coefficient for the wings was 0.020 (based
on the planform area). determine the ratio of the drag from the
wire bracing to that from the wings.
Speed: 70 mph
Wing area: 148 ft 2
Wire: length =160 ft
diameter = 0.05 in.
• FIG U R E P9.88
so -Ih'rf
andsillce
Re - Jl!2 -- 1/ -
( 88# )( 0.0.5 )
(7olYlph) -,om'" --n:f.I
I. ~7 xIO-JI. J:L2.oS
:::: 2720.
From Fi9' 9.21 } w/lh Re =: 271-0 we obTain
Hence)
clJwire
rx9lJ1ifJ9
(1.0)(0. 667f-lZ.)
- (0.02) (IJf8 f/2.) :: 0.225 J or
9-e/
22.570

9.89 I
9.81) The jet engines on a Boeing 757 must develop a cer-
tain amount of power to propel the airplane through the air with
a speed of 570 mph at a cruising altitude of 35,000 ft. By what
percent must the power be increased if the same airplane were
to maintain its 570 mph flight speed at sea level?
-fJ:::: p()wer :::,fJ7J = ir1]3Cb
1/
Lef ( )0 and ( )3$ den()le cOllrl/II()M af.seo, lei/el &llJri 3.s;()()() ftrCfJflC!illof)
Tnvsj z?;:: 77;. .sf) fh41
..L 71.
3
2. po 0 c/)t)lio Sf) IIA:, ~/lJ.5 411d c/Jo ~~3') J f/;ell
df.;s ~ C.o.s./~~
7j
- -11.s-
~ or w,Yh () Vq/vOs from iaiJ/e cJM ~ r ,
16 ::::
~s
~
~ __ 0.00238 ~ 22 32'1.. tI-f · .I'D e~ ~ =8, = h~~M
13s- 0.0007.38 stqf

9.90 J
9.90 A wing generates a lift 5£ when moving through sea-
level air with a velocity U. How fast must the wing move
through the air at an altitude of 35,000 ft with the same lift
coefficient if it is to generate the same lift?
- 2.38)(/0 :7I!
)~ ~ -3
S
/IJIJ )Ji
~eQ leve/- 7.38X/O""1/* q•• level
l{¥oofl = .J. 80 r{eQ level

9.9/* When air flows past the airfoil shown
in Fig. P9.91 the velocity just outside the bound-
ary layer, U, is as indicated. Estimate the lift coef-
ficient for these conditions.
If shear forces are negligible
~ = Jf cos e dlJJ where fhe +siqn is used on
fhe lower surface j - sI9n on upper surf(lce. IIlso,
fJ = 1'0 +ipV2.- -feu.'-· Since the
x
c
FIGURE P9.9/
f--
I
r-
0.8 1.0
c.onlrib"fe fo the liFt, we sei f. ~O f. u -Ie -7.ZZz:z=-
consfcmf aim().spheric preSSllre doe.s not ~
ThvsJ
fJ =fpy2.[J - (~tJ So fh4t J~a dA
1.= f#f(,u1
[I-(tt] cos8dR ~. '(~
However, dll~ j Js where j: will9 leI/II/II ~
or cose dll:= cose ids = i rk
Hence I
£=-fpu
2
j[t -(~):L]idx +ffV2.f[I-(~)2-J 1dX
upper lower
or
t :: f f rl"l f[(%)2 - (%)~ ]dx
upper lower
II/so; since G;.= f f'~2A = i e~2tC I wnere C: chord len9th
if fo/lows thaf
I
CL =j[(-fff' -(-~)2- ]d(i-) Thi.s Infe9ral is obfained bynVQJcriCQ/
.!.::O upper fOlller infeqrafio/J of fhe d4ta 9ir1en in fheh9vre.c
The fo//OW/tJ9 f4h/e of Jaf4 is oJ/a/ncd ;
(con'£)
9-H'f

9.9/ "1 (conli)
x
[(~~:per- (~)~werJ-c
0 0
0.05 0.65
0./0 0.52
0./5 0.'1/
0.20 0.36
0.2.5 0.32
0.30 O. '2.q
0.35 0.2'/-
O.lfo 0.'2.0
0.Jl.5 0./8
D.SO D.17
0 ..55 O./S
0.60 O./~
0.6..5 0,12.
0.70 0.1/
O.7E O.Oq
0.80 0. 07
0.85 0.05
0.90 0.03
0/1.5 0.01
/.00 0
By lJ,sinJ the pro9ram TRAP£~OI wHh the ahove valves of Ihe
infcqrIJluJ we ohf4in
*****************************************************
*****************************************************
Enter data points (X , Yl
? 0,0 ? 0.55,0.15
? 0.05,0.65 ? 0.60,0.11,t.
? 0.10,0.52 ? 0.65,0.12
7 0.15,0.lxl '? 0.70,0.11
'? 0.20,0.36 ? 0.75,0.09
? 0.25,0.32 ? 0.80,0.07
? 0.30,0.29 ? 0.85,0.05
? 0.35,0.2lx ? 0.90,0.03
? 0.I,t.O,0.20 ? 0.95,0.01
? 0.1,t.5,0.18 ? 1.00,0
? 0.50,0.17
The approximate value of the integral is: +2.0550E-Ol
Thus) ~:::: 0.206

9.93
9,91f
9.93 A Boeing 747 aircraft weighing 580,000
lb when loaded with fuel and 100 passengers takes
off with an airspeed of 140 mph. With the same
configuration (i.e., angle of attack, flap settings,
etc.) what is its takeoff speed if it is loaded with
372 passengers. Assume each passenger with lug-
gage weighs 200 lb.
For sfeady fli9ht "£::: CL ifV~A -::: W
Lei ( )'00 denofe conditions wifh /00 pdsset/ljers
ond ( )372 wilh 372 pqs.renrer.s. ThuS; w,lh ~/OO'" ~37.2 J
~oo ;:: 11372. J (}nd 800:::: ('37.2 Ef· (I) 91ves
(I)
til0 0 U;:o u == u ~r58~ooo+(372 -100) (2.00)J Ib]~ III TJ _
= -2.- or '.312. 700 580, 000 Ib J WI '100- JlfOlfl/J1J
;/.37.2 ~7:Z
Th/)~J ~12:: /'f6 mph
9.94 Show that for unpowered night (for which
the lift, drag, and weight forces are in equilib-
rium) the glide slope angle, e, is given hy
tanH = CnICt.•
FOr steady vn~wered fh9hf
L Fx =0 9lve.s b= Wsil}8
tllld
~ Fy == 0 91i1es t ::: W&Dsf)
ThlJs}
b_
y- WsinG _.J. ()
Wt Os e - 1(JIJ J
f
Cb
an e=Ct.
9-86
v~
Te

9. q5 J
9.96 I
9.'J5 If the lift coefficient for a Boeing 777 aircraft is 15
times greater than its drag coefficient, can it glide from an al-
titude of 30,000 ft to an airport 80 mi away if it loses power
from its engines? Explain. (See Problem 9.94.)
From ProbleIn 9, qifJ
Hence,
I or d=if;S)</05{1
I
/5
~ooorl
/5 J
=8S.21J7/
lienee; the plane can q/ide 80m,:
9.96 On its final approach to the airport an
airplane flies on a flight path that is 3.00
relative
to the horizontal. What lift-to-drag ratio is needed
d
if the airplane is to land with its engines idled -.;:::!J
back to zero power? (See Problem 9.94.) ~ _ y-.pr
/??~1~-;~J
From Prohlem 9.9if J
fan e= CD
C'-
c'- 9-==/,/
CD
9-87

9.97
9.97 A sail plane with a lift-to-drag ratio of 25 flies with
a speed of 50 mph. It maintains or increases its altitude by flying
in thennals, columns of vertically rising air produced by buoy-
ancy effects of nonunifonnly heated air. What vertical air speed
is needed if the sail plane is to maintain a constant altitude?
WJ1h no vertical air h/o·l;on Ihe sailp/ane w!.IJ/d
91ide wilh a slope Qn9/e OJ where since L F =0
tIf = 1.11sine (Jnd t =Wcos{;. Helice) .!It::: 1I/si,,() == tane
~ ~ Wc~B U
or since cIJ-::: ffV C/) /I and ~
t = 1: PY
4
Cl II if follow.s Ihat tan B=-*. Therefore in sf/I/
4ir the s4i/p/one w()vld 10os8 alfilllrJe ctf arole ()f Usin()) where
()== fan)( g~):::fan't1.sj:=2,29 ~ lienee, an upward wind 0 f(so"'ph) sin 2.29"
=: 2,00 mph wI/I a/low hori2onlo/ fl'9hl.
9-ea

9Jl& I
9.98 Over the years there has been a dramatic increase in
the flight speed (U) and altitude (h), weight ('W), and wing
loading ('WIA = weight divided by wing area) of aircraft. Use
the data given in the table below to determine the lift coeffi-
cient for each of the aircraft listed. v __
Aircraft Year 'W,lb U, mph 'WIA, lblit 2
h, ft
..................................................................................................................................................................
Wright Ayer 1903 750 35 1.5 0
Douglas DC-3 1935 25,000 180 25.0 10,000
Douglas DC-6 1947 105,000 315 72.0 15,000
Boeing 747 1970 800,000 570 150.0 30,000
~ W _ 2 (.W)
Ct. ::: .jf 7J'-,tJ -
1. f u2j} - p77L
rr
Tnt/oSJ
eJ s/V1.$/{f3 '" ffls w}IJ. /61(12-
Wri9hf F7yer 3Jl -.3
2. X/o SZ3 ~.s
DC-.3 I. 7t 'tJO-
3
2t/f. 2. S". 0
DC-6 /,sox/o
3
'It2- 72-, ()
7¥7 8.9/x/o-¥ 83t ISO
9-e9
ct..
o.¥--&o
O.'ff)9
O.f.SI
OJI-J2

9,99 J
9.99 The landing speed of an airplane such as the Space
Shuttle is dependent on the air density. (See Video V9.1.)
By what percent must the landing speed be increased on a
day when the temperature is 110 deg F compared to a day
when it is 50 deg F? Assume the atmospheric pressure re-
mains constant.
'ThlJS)
('1-60 +/10)
(~60 +£0) == 1.//76
IT ., == VI, // 76 lloo :. 1,0572 77;0- or a 5.727. IlJcre4Se110 ~
9-Qo

9,/00 J
'.100 Commercial airliners normally cruise at relatively
high altitudes (30,000 to 35,000 ft). Discuss how flying at this
high altitude (rather than 10,000 ft, for example) can save fuel
costs.
FOr /eve I Fli9ht W:::aircrafl wei'lht =~::: ~ 1pu'"jJ
T/;/JsJ for given ~ C; J tI/ldf} Ihe dYIJ41J1/C pressure
is COllstaniJ /ndependenf of a/h'fvde. Thalls
.1. V 2) - .L 7/2.) or lJ - (-J14t1t1~)~U
2.. P 10., 0001; - 2 e 4~OfJ() ff J :3~fJOO - f.3rJ,;OODj I~OOO
Hence ~o OO() > q"O()O
.I ~ '" :a-
li/so,} s/I)ce flJe drill ;'S t#~c/) -fpV/I II f()//()JII..$ fhal
,/) =C/).J:pll2.I/) =C/JfplJ'II) slIJce tpT{:fJI(J =f:e~(}tJl
.3";,OOD .10""(JI I~QIO -;
HenceJ fhe o/rcrarl can fly fosler af hif};f) a/l/ltJr/ed wllh file
.same (JlJlotJlJf of dr49 (4~,oll ::~~"()o)
9-9/

9.102
9.102. Repeated controversy regarding the
ability of a baseball to curve appeared in the lit-
erature for years. According to a test (Life, July
27, 1953) a baseball (assume the diameter is 2.9
in. and weight 5.25 oz) spinning 1400 rpm while
traveling 43 mph was observed to follow a path
with an 800-ft horizontal radius of curvature.
Based on the data of Fig. 9.39 do you agree with
this test result? Explain.
For steady IJ')olion a/oIJ! Ihe curved
path ZJ F,. = m Or
o r . 2 . 1(::: 800 f-l
£ =mJ{ ::: ¥#J where V={lf3mphJ(:os~:h)=63.1~
Thu~ ( ~62SIb)(63.1 ~),._
~ - (32.2 iO(800 f+) - 0.0507 lb
BUT)
~ = CL
-J: fy2.fJ = ~ t Py2.fD2.
or
_ 9 t _ e (o.OSo71J,)
CL- 7l'pTli!.D2- - 1l'(O.OO2.38-?'fI)(6.3.I{fi"(#(1/- =0.:2.33
From h.,. ?39 wHh
D ( If rev)( Jmin )(2.11rad)(2-.9 U)
PL- = 1 OOMfii -;;o.s Jrev n :; 0.28/ We ohft:lin
2. 7J 2.. (63.Ii}) .
CL ~ O.OB which is less Ihal) IhB t; =0.23.3 in £,,_ (I).
(I)
lienee (J smorJlh sphere would not curve QS mvch (IS indicated, but
perhaps a rotJ9h ball (i.e'J (J/le w/fh seams) wov/d#
9-92

'I, /0.3
9.103 Boundary Layer on a Flat Plate
Objective: A boundary layer is formed on a flat plate when air blows past the plate. The
thickness, 8, of the boundary layer increases with distance, x, from the leading edge of the
plate. The purpose of this experiment is to use an apparatus, as shown in Fig. P9.l03, to mea-
sure the boundary layer thickness.
Equipment: Wind tunnel; flat plate; boundary layer mouse consisting of ten Pitot tubes
positioned at various heights, y, above the flat plate; inclined multiple manometer; measur-
ing calipers; barometer, thermometer.
Experimental Procedure: Position the tips of the Pitot tubes of the boundary layer
mouse a known distance, x, downstream from the leading edge of the plate. Use calipers to
determine the distance, y, between each Pitot tube and the plate. Fasten the tubing from each
Pitot tube to the inclined multiple manometer and determine the angle of inclination, e, of
the manometer board. Adjust the wind tunnel speed, U, to the desired value and record the
manometer readings, L. Move the boundary layer mouse to a new distance, x, downstream
from the leading edge of the plate and repeat the measurements. Record the barometer read-
ing, Hbar, in inches of mercury and the air temperature, T, so that the air density can be cal-
culated by use of the perfect gas law.
Calculations: For each distance, x, from the leading edge, use the manometer data to de-
termine the air speed, u, as a function of distance, y, above the plate (see Eq. 3.13). That is,
obtain u = u(y) at various x locations. Note that both the wind tunnel test section and the
open end of the manometer tubes are at atmospheric pressure.
Graph: Plot speed, u, as ordinates and distance from the plate, y, as abscissas for each
location, x, tested.
Results: Use the u = u(y) results to determine the approximate boundary layer thickness
as a function of distance, 8 = 8(x). Plot a graph of boundary layer thickness as a function
of distance from the leading edge. Note that the air flow within the wind tunnel is quite tur-
bulent so that the measured boundary layer thickness is not expected to match the theoreti-
cal laminar boundary layer thickness given by the Blassius solution (see Eq. 9.15).
Data: To proceed, print this page for reference when you work the problem and click hen'
Boundary layer mouse
/Pitot tubes
-.::::-
Flat plate
Water-E........~
9-9.3
• FIGURE P9.103

Solution for Problem 9.103: Boundary Layer on a Flat Plate
8, deg
25
y, in.
Halm, in. Hg
29.09
L, in.
Data for x = 7.75 in.
0.020 0.20
0.035 0.35
0.044 0.48
0.060 0.70
0.096 0.95
0.110 1.06
0.138 1.21
0.178 1.44
0.230 1.70
0.270 1.85
Data for x =5.75 in.
0.020 0.20
0.035 0.42
0.044 0.50
0.060 0.71
0.096 0.98
0.110 1.06
0.138 1.30
0.178 1.54
0.230 1.76
0.270 1.88
pu
2
/2 = YH20*L sin8
where
p =Palm/RT where
T, deg F
80
u, ftls
19.9
26.3
30.8
37.2
43.4
45.8
48.9
53.4
58.0
60.5
19.9
28.8
31.5
37.5
44.0
45.8
50.7
55.2
59.0
61.0
YH20, Ib/ft"3
62.4
y, in. L, in.
Data for x =3.75 in.
0.020 0.15
0.035 0.35
0.044 0.45
0.060 0.71
0.096 1.20
0.110 1.30
0.138 1.56
0.178 1.77
0.230 1.95
0.270 2.00
Data for x = 1.75 in.
0.020 0.20
0.035 0.50
0.044 0.68
0.060 0.90
0.096 1.51
0.110 1.70
0.138 1.90
0.178 1.95
0.230 2.00
0.270 2.00
Palm = YH20*Halm = 847Ib/ft"3*(29.09/12 ft) = 20531b/ft"2
T =80 + 460 = 540 deg R
Thus, p = 0.00222 slug/ft"3
Approximate boundary layer thickness as obtained from the graph:
x, in. 0, in.
1.75 0.15
3.75 0.20
5.75 0.27
7.75 0.30
9-9'1-
u, ftls
17.2
26.3
29.8
37.5
48.7
50.7
55.6
59.2
62.1
62.9
19.9
31.5
36.7
42.2
54.7
58.0
61.3
62.1
62.9
62.9

c:
>-
Problem 9.103
Velocity, U, vs Distance, y
0.30 ,-----,----~---,------,I
0.25
0.20
0.15
0.10
0.05
0.00
0.35
0.30
0.25
I----II--+-~--~-I
I
---~-------~~-------_H'__iI_+_-----1
o
- -I-------------i--------I+--f--4-----1
20
II
j
I
I~~.-:;~-----------~----I
I
40 60
U, ftls
Problem 9.103
Boundary Layer thickness, 8,
vs
Distance from Leading Edge, x
I
80
----"--~----+----~~___1
-+-x =7.75 in.
__ x = 5.75 in.
--.- x =3.75 in.
-e-x = 1.75 in.
c: 0.20
r-O~ 0.15
- ------'---- ~~~~~~
I
I--.---------~------ ---.------~-------_i
• Approximate boundary layer
thickness
- Best fit power-law curve
8 = OJ 12x°.48 ; '---~_______---.J
0.10 - ---~-----------------------------~
I
O.05 +=----t-----~---t----;----__t__
0.00 +-----~-----r-----T----_,----~
o 2 4 6 8 10
x, in.
9-9.5"

9./otf.
9.104 Pressure Distribution on a Circular Cylinder
Objective: Viscous effect within the boundary layer on a circular cylinder cause bound-
ary layer separation, thereby causing the pressure distribution on the rear half of the cylin-
der to be different than that on the front half. The purpose of this experiment is to use an ap-
paratus, as shown in Fig. P9.104, to determine the pressure distribution on a circular cylinder.
Equipment: Wind tunnel; circular cylinder with 18 static pressure taps arranged equally
from the front to the back of the cylinder; inclined multiple manometer; barometer;
thermometer.
Experimental Procedure: Mount the circular cylinder in the wind tunnel so that a sta-
tic pressure tap points directly upstream. Measure the angle, /3, of the inclined manometer.
Adjust the wind tunnel fan speed to give the desired free stream speed, U, in the test sec-
tion. Attach the tubes from the static pressure taps to the mUltiple manometer and record the
manometer readings, L, as a function of angular position, O. Record the barometer reading,
Hbar, in inches of mercury and the air temperature, T, so that the air density can be calcu-
lated by use of the perfect gas law.
Calculations: Use the data to determine the pressure coefficient, Cp = (p - Po)/(pU2
/2),
as a function of position, O. Here Po =0 is the static pressure upstream of the cylinder in
the free stream of the wind tunnel, and p = 'YmL sinf3 is the pressure on the surface of the
cylinder.
Graph: Plot the pressure coefficient, Cp' as ordinates and the angular location, e, as
abscissas.
Results: On the same graph, plot the theoretical pressure coefficient, Cp = 1 - 4 sin2
0,
obtained from ideal (inviscid) theory (see Section 6.6.3).
Water .. FIGURE P9.104

Solution for Problem 9.104: Pressure Distribution on a Circular Cylinder
~, deg Hatm, in. Hg T, deg F U, tUs
25 29.97 75 47.9
Experiment
e, deg L, in. p, Ib/ft"2 Cp
0 1.2 2.64 1.00
10 1.1 2.42 0.92
20 0.7 1.54 0.58
30 0.1 0.22 0.08
40 -0.6 -1.32 -0.50
50 -1.6 -3.52 -1.33
60 -2.4 -5.27 -2.00
70 -3.1 -6.81 -2.58
80 -3.0 -6.59 -2.50
90 -2.7 -5.93 -2.25
100 -2.7 -5.93 -2.25
110 -2.6 -5.71 -2.17
120 -2.6 -5.71 -2.17
130 -2.6 -5.71 -2.17
140 -2.6 -5.71 -2.17
150 -2.6 -5.71 -2.17
160 -2.7 -5.93 -2.25
170 -2.7 -5.93 -2.25
180 -2.8 -6.15 -2.33
P =YH20*L sin~
P = Patm/RT where
Patm = YHg *Hatm = 847 Ib/ft"3*(29.97/12 ft) = 2115 Ib/ft"2
T =75 + 460 =535 deg R
9-97
Theory
Cp
1.00
0.88
0.53
0.00
-0.65
-1.35
-2.00
-2.53
-2.88
-3.00
-2.88
-2.53
-2.00
-1.35
-0.65
0.00
0.53
0.88
1.00

1.0
0.5
0.0
-0.5
Co
-1.0u
-1.5
-2.0
-2.5
-3.0
0
Problem 9.104
Pressure Coefficient, Cp, vs Angle, e
30 60 90 120 150 180
e,deg
• Experimental
- Theoretical (inviscid
flow)

/0, I I
/0.21
/0.3 J
10.1 Water flows at a depth of 2 ft in a lO-ft-wide channel.
Determine the flowrate if the flow is critical.
10.2 The flowrate per unit width in a wide
channel is q = 2.3 m2
/s. Is the flow subcritical or
supercritical if the depth is (a) 0.2 m, (b) O.Sm,
or (c) 2.5 m?
v= -AQ =:tl :: -i- so fho f Fr ::: ~ = if
yb Y vg Y yfiY
or m~
Fi - 2..3-:9 = 0.73# h
r - t .l!l. ' ~ S/2 I were y-m
19.81 s" y Y
y, m Fr flow fype
a) 0.2-
b) 0.8
c) 2.5
8.21
1,03
0.186
supercrificQ/
supercrifical
slJhcri/icol
10.3 Water flows in a canal at a depth of 2.8
ft and a velocity of 5.3 ft/s. Will waves produced
by throwing a stick into the canal travel both
upstream and downstream, or will they aJl be
washed downstream? Explain.
C=V~Y =j(32.2!J)(2.8ff)' ==9..5'O¥
ThlJs, with V::5.3 ~ < c fhe wove CQI) frl1vel upsfream Cl9tJiIJSi
fhe c{)rrent. Re/alive to fhe sfretJlh bOI1K fhe wave fravels fJps1re4m
wi~h velocity t- V = 9.sofj -5.3 !+ =¥.2 !/-.1f also frove/.s dOlllfJsfream.
Nofe: Fr = ¥:: :~!l =0.558 j fhe flow i~ slIberifical. .
Waves travel both vp-rfreCfff} 4IJd downstream.
/0-/

/o.Jf-
10.4 Consider waves made by dropping objects (one after
another from a fixed location) into a stream of depth y that is
moving with speed Vas shown in Fig. PlOA (see Video '9.1).
The circular wave crests that are produced travel with speed
c = (gy)l/2 relative to the moving water. Thus, as the circular
waves are washed downstream, their diameters increase and the
center of each circle is fixed relative to the moving water. (a)
Show that if the flow is supercritical, lines tangent to the waves
generate a wedge of half-angle a/2 = arcsin(l/Fr), where • FIG U REP 1 0 . 4
Fr = V/(gy) 1/2 is the Froude number. (b) Discuss what happens
to the wave pattern when the flow is subcritical, Fr < 1.
v
(a) In a lime inferval of i slf)ce the ohject h,1fhe wafer (allJi11141i4ff)J
fhe Wave)J the cef)fer 0 r{he wave has heel) swepl J()U'fNfreClfYJ a
d,'.s14f)Ce Vi and fhe W4Ve has eXjlallJerJ. If) 6e adis/af/ce ct frpm
ifs eenfer. Tn/s i-r .showlI 1/J1/Je f/ytJre be/ow. !Vole fhal Vi >c-t
if V>c (,·.e, Fr>/).
....... - -- ./
01' f::: arcsin (//Fr)
ct
vt
_ c
-1'
(6) If Fr </ the 4b()V8 result 9ive.s s"rf ~~ Wht"ch is' jPJ!~J'.r/b/e.
For Fr</ The fol/ow/flr; w~ve pal/ern wfllJ/d re.rt//f. Tnere is I')()
IlweJre UproJfloeJ.
/ -,,,-I ci
U J
'" Vi /
Vi <. ct if
Fr<l
'-. - -
10-2

/0.5
10.5 Waves on the surface of a tank are observed to travel
at a speed of 2 m/s. How fast would these waves travel if (a)
the tank were in an elevator accelerating upward at a rate of 4
m/s2
, (b) the tank accelerates horizontally at a rate of9.81 m/s2,
(c) the tank were aboard the orbiting Space Shuttle. Explain.
Since c:;; 1{iy-' Hfollows That the fank aep+h ~
_ c2. (2 ~)~
Y- - = ::: O. ¥o8 m
'I Q.81f;.
(a) If ihe rank accelerafe.s upward wifh acceleration 0. J Jhe effective
acceLe.rcrlion of qrav/fy is g
e
ff -:: 9+a '= (9.8/ +tt)f! -= 13.81fi.
ThusJ
C =:: V-ge-(r--'y'" == 1('3.8/If,. )(O./f()8m) == 2,37f-
(b) If the tank accelera1es hori2onla/& 'II/Iii
acceleration a the effeciive acceleration isJ
jefr ~ / i~+a2. ':;: /9.8/2.+'1.8/": =/3.87.r;:..
Thus)
,...---------.
c::: (13.87!f;.)(o,If08m) = 2.381}
(c) In orhif ~err -::0 (weI9hfless) .so c:!!.....
10-3
a

10.6 I
10,7 I
10.6 In flowing from section (I) to section (2) along an
open channel, the water depth decreases by a factor of two and
the Froude number changes from a subcritical value of 0.5 to a
supercritical value of 3.0. Determine the channel width at (2) if
it is 12 ft wide at (I).
10.7 Obser'ations at a shallow sandy beach
sho' that even though the waves several hundred
yards out from the shore are not parallel to the
beach, the waves often "break" on the beach
/~w,,,c,,st
~Cll
nearly parallel to the shore as is indicated in Fig.
PIO. 7. Explain this behavior based on the wave
speed c = (gy)v:.
Ocean (2.).' . . - -
-
........--""H»-jH!>!>~C·7g'7
..;.. :.... M~ Bea~h'" . .' •... '.
FIGURE PIO.7
Since C =",,9YI il {OJ/Ollis fhaf c, > C2 becQlJse of the fact lhat YJ >)t .
There forel
a~ fhe waves move, fhat porfion in Ihe deeper water
fends fo ~afch up /I wilh fhat porliDn closer fo shore in the shallower
wafer. The wqve cresf Je/ld.s 10 become more nearly pIJra/le I fo Ihe
shore line. The waves "break" on fhe shore 4S if The wind were blolllin9
normal 10 fhe shore.
ID-J.f

/0.8 I
'10.9 ,
10.8 Waves on the surface of a tank containing water are
observed to move with a velocity of 1.8 m/s. If the water is
replaced by mercury, with aU other conditions the same, deter-
mine the wave speed expected. Determine the wave speed if the
tank were in a laboratory on the surface of a planet where the
acceleration of gravity is 4 times that on earth.
Since C =(gi if foJlows fhat the walle speed is independenf of
fhe f/vid density. ThusJ CH;J.o :: CH, ::: /.8!f on eqrlh.
However) on 'he planet .t
-/ ·-fge4rfh) ( )~- ( 9PI(Jnet)~ .!z:
Cplanef - ~/anefY - gearfh fjplanef/ - gearlh (~eQrth y)
or
Cpt/l1Iet = ( Lf )~Cearfh =('f)~(1.8!f) = 3.60 If for wafer or mercury
10.9 Often when an earthquake shifts a seg-
ment of the ocean floor, a relatively small am-
plitude wave of 'ery long wavelength is produced.
Such waves go unnoticed as they move across the
open ocean: only when they approach the shore
do they become dangerous (a tsunami or "tidal
wa'e"), Determine the wave speed if the wave-
length. A. is 6000 ft and the ocean depth is 15,000
ft,
From £'(, /0,11-: C =[~ fqnh(2.JX )]~
or k
C = [ (.92.Z[l,.)(6oooff) f h(-2.17'(J5~oooft))~1 ~- /75 H
2"" Qn 6000 fl ~ s
/0-5

la,to
10.10 A bicyclist rides through a 3-in. deep puddle of wa-
ter as shown in Video V.lO.l and Fig. PIO.IO If the angle
made by the V-shaped wave pattern'procfuc-ed by the-1ront
wheel is observed to be 40 deg, estimate the speed of the
bike through the puddle. Hint: Make a sketch of the current
location of the bike wheel relative to where it was !::..t sec-
onds ago. Also indicate on this sketcr the current location
of the wave that the wheel made !::"t seconds ago. Recall that
the wave moves radially outward in all directions with speed
c relative to the stationary water.
AI fime t ::0 the froni whee I wos
af point (OJ. AI fhe Gvrrent limeJ
t:;LJ. t J Ihe wheel has traveled (J.
dis/an ce d:: V.~i and is atpoint (j),
Ai t:rne t==A( a wave prodvced
by the wheel when NWas of (0)
wi If he a diS1ance CAt from (0)
as indicafed i/l-lhe fi9vre.
Waves prodvced 01 variOf}.s /i,."e.s
(from t:; 0 -/0 i =:At) hy the from
wheel will form fA V-shaped Wo,ve
as shown ill fhe sec()/Jd fJ'rttJfIe
(proviJed V>c jsvperor,.fical
bike speeJ),
c
I
I

CA.i
• FIG U REP 1 0.10
wave proallced
01 (0) when t=0
V
(/)
~
C ~ fiY '" [32.2 ~ (~H)Ji:::-; 2..e~.p
or
V=
sinf
c
where
ThtJsJ
V= 2.f'ffj
sln 20rl
=: 8.30#
10-6

/0.1/ J
IO.1l Water flows in a rectangular channel with a flowrate
per unit width ofq = 2.5 m2
Is. Plot the specific energy diagram
for this flow. Detennine the two possible depths of flow if
E = 2.5 m.
E- y + ~"
- 21 y"
Thus, plot
£ ::: y .,. 0.3/'1 wAel'e £ N my'" my" J ) .
0.'- ~ ((2 sJ11!fJ.)~
Nole; Yc =(T) = . ~ ::: 0. 86 0 111
d
IF Q.8Ist.
01} 3 _ 3
Emili =3.' Yt: - 2: (0. 860m) ::: 1.29m
yvs E
2.5
/2 ~'"
/)1.5
/'
VE
1/'/'/~
1
//
n~/.~
l-':: 0.8,0
~
0.5
," Ell t"-......
,,
",
0 IL
0 0.5 1 1.5 2
E,m
0.3/9
For £ =2.5 mJ Ef. (I) is 2.5::: Y + yi
or y3 -2..5y2 +0.3/9 ::: 0
2.5 3
The roo/.s ff) fhi.s e'!volion ore y::: 2.'1-5 , 0.338 I tJnd - 0.335
Thus, y =2...#5 m or y ::: 0.386 IIJ
/0-1

/0./2.
10.12 Water flows radially outward on a horizontal round
disk as is shown in Video VIO.6 and Fig. PlO.12. (a) Show that
the specific energy can be written in terms of the flowrate, Q,
the radial distance from the axis of symmetry, r, and the fluid
depth, y, as ( Q)2 1
E=y+ - -
27r1" 2gy2
(b) For a constant flowrate, sketch the specific energy diagram.
Recall Fig. 10.7, but note that for the present case,. is a variable.
Explain the important characteristics of your sketch. (c) Based
on the results of Part (b), show that the water depth increases
in the flow direction if the flow is subcritical, but that it
decreases in the flow direction if the flow is supercritical.
'----J/ "/ ./
~( /r '4
I
/
" /
I--r-
r--~
• FIGURE P10.12
(a) The specific elJergy i.s E. ~ y + t-gJ.) where V:= ~ ::: Q
21lry
Thus,
( Q)2. I
£=y+ 21Tr 2fjY'"
"-12.
(b) Lei ~ ;;;2":r so Ihaf E=Y f ~yt. ",hid, is Ihe slime tis For
fINO d/mensi()(JQ/ Flow wifh ff s:: f hei119 replaced by ¥. I!pwever;
for fWo dimensional flow 't is co/ul4nfj for radial flow 'i Is &l
variable since r Varies. Bvf ~ vs YC/JrVes for cons/ani rwovla
look as shown below (Fi9~ /0,7). / ~ubc'i1;CQI
(e) Front fhe Bernoult· e'lvalion Y / / /-.J'l.
£, ;::£2. or £=-cons/QlJt for this flow. ,.
Consider suhcr/fic4/ flow -point
II. For olAf/ow r incrB4se.s so Ih,,1
f decre4ses. Thv.s since £::c.oMl.J
the flow 90es {f'()HJ sfetfe ~ If) II~;
the depth increQsc.s. F()r sUb - E
crilic.a/ in(10111 r decreases J ~ increasesJ
E, =E2,
fhe flow 'lues fro/ll II, 10 11.1) and the depth decreases.
For supercr/IiC41 Flow f is true. 'ThIlSJ
oulfoll! increases fJ decreases ~j or
fr()fn B, fo 82- decreasin, depl-h.
Supel'cr/llca,/ inflow from B, fa 8.3-
increasifl9 depth.
/0-8
inflow
oliff/ow
stlhcrifiCIII
depfh
decreases
depth
increases
supercriliccl
depth
incre4ses
deplh
decreases

/0,/3- I
10.13* Water flows in a rectangular channel
with a specific energy of E = 5 ft. If the flowrate
per unit width is q = 30 ftl/s, determine the two
possible flow depths and the corresponding Froude
-
numbers. Plot the specific energy diagram for this
flow. Repeat the problem for E = 1, 2. 3. and 4
ft.
(0)
(I)
(2)
y, fI (s()bcrificalJ r;. YJ ff fsvpercrdical) Fr
I
2
.3
if
.5
no .solufion poss,"J,l
I
.J
if. 21 0.6/2. 2.26 /,56
s ( 1. ~ I(30 If:L)2] ~
Note Ihai Emili ='2 Yt: -J whers Yc:: f) = 32.2 {! = 3, 03 ff
Thlls E. =- ! (3. 0311)= ~s.s ft, mlfJ ...
As shown on the 9rllph be/fI~J there are noposifive real roots of Ct{. (I)
if £ < £mi,,::: '1.S5 rl.
7~-r--~~~--~~--~-r--~~--
...j"6+-~--~~~--~-+~+--+--~~~
/V
5+-~--~~~--~"'~/+--+--~~~~~
1; ...... /
It: 4 -t--+---+---+-~~-A--+--+-----+--.J---I---I
......... 1,/>f ;:.3.o3H
~3+-~--~~~-~~--+--+--~~~~~
/ " EI7./~ =Krs f.I
2-t--+--,,~--~4--4~~_~___--+~--~~~--~
1 ./ r--r-+---L-.I
~"
o II
o 1 2 3 4 5 6 7 8 9 10 11
E,ft
/o-q

/ D.llf
10.14 Water flows in a rectangular channel at
a rate of q = 20 cfs/ft. When a Pitot tube is placed
in the stream, water in the tube rises to a level
of 4.5 ft above the channel bottom. Determine
the two possible flow depths in the channel. Il-
lustrate this flow on a specific energy diagram.
= ~.S -y or y.3 -If.Sy2+6.2! =0 I where y-{/
The rOOTs ()I this gfl'lt1IiOIJ are y ~ 9. I'fJ
I. If1-, lind - I. 0~
Thv-s,
y.: ~./'l-II or y;:1.'1-2 If
/0-/ 0
(I)

/0./5
10.15 Water flows in a 5-ft-wide rectangular channel with
a flowrate of Q = 30 f~/s and an upstream depth ofYl = 2.5 ft
as is shown in Fig. PIO.15. Determine the flow depth and the
surface elevation at section (2).
FIGURE PIO.15
A ~2. IJ~ ~2
7 f t? fZ, =-I ti1+Z2} where ~~I1.=:0J z,='t-=21IJ Z2=o.2rl+Y~J
-= Q = (30!f) = fJ. :' Q ::: 30 i}.J -.A.
V, tii (2.£+)(5fI) 3 S J IJlJd ~ 7& (5 If) Y1. - Y:z.
Thvs,
( Ii)2. (1. fi)2...
3 .s +2 ff = y~ oS • f 0.2ff f ~
2(32.2 ~) 2. (32.2'f{)
or 3 . 2 •
~ -1.9/fy,. to.5Sf/=O wl"chhQsroofs Y1.='I.77'1-I
O.632.1
11IJd-o.632.
!Vole: Fr::jL = 3!f =O.37/f < I
I v;y; K32.2~J(2fnJ~
If Y2.=O.63:J. J
fheh Fi;,>/. This
cQlJnol he silJce there ,".s no "6vPlpl
bel.,een (tJ anti(2.) of whi&n critical
COfldihfJ"S CfllI Oa&Vr.
Thus} Y1. =I, 77~ II (1I1d 22. =1.97'1 ff
10-1/
y
.
, • " (I) y,:: 2 ft
, • (2.) Y,2,:: I. 77'1-fI
• I
,
E

/0,/6
10.16 Repeat Problem 10.15 if the upstream
depth is YI = 0.5 ft.
FIGURE PIO.16
Thvs} ~ ::0.528 f1 olld :&3. =:: O. 72.8 ft
10-/2

/0.174 I
ZB = O.2e -x2
10.17* Water flows O-er the bump in the bot-
tom of the rectangular channel shown in Fig. PIO.17
with a flowrate per unit width of q = 4 me/s. The
channel bottom contour is given by ZB = O.2e-x
:.
where ZB and x are in meters. The water depth
far upstream of the bump is Yl = 2 m. Plot a
graph of the water depth. y = y(x), and the sur-
face elevation, Z = z(x). for -4 m s; x s; 4 m.
Assume one-dimensional flow. FIGURE PIO.I7
'1.2. V2.
Ii+i1+2/ :: -f +ij +ZJ where fJl o:::f ==0) ~:::~:: 2m J 22.:: y+28
-x2. v,!t ~ Iff- .., I a. If
or Z = Yt 0.2 e ,:::: " .:: :: 2!:r tlnQ V=.Jj.. :: -J ,,'2-1h J Y Y
TL ( 11!..)2. (!l m)" x:2
I nlls, 2 oS +2 m .:: y:S -+ Y+0.2 e-
2. (9,81;l) 2(fI.&//fi)
or 2
y3 _( 2.2.0 -0.2 e-X
) yZ+ 0.8IS::0 where y-m
Solve for y with - ~ ~ X~ Ifm
/Vole: Fr: ~; - = 2 f- = 0"1-52. ""
I 'iy, [(f/.8/~){21n~Ji.
ThiJs, the flow will remQin slJhcrilicQ/
fhrof},holJf-lhe largesl roof of Er. (I)
wdl be the correc.1 one.
y
/
1/
Use proJrarn CUBIC fo solve For yct) froth Ef. (fJ (/Qrpesf roof)
~
and fhen z = y to.2 e-x for -'IRJ ~ x~ If""
_xl. -x2.
X}m - (2.20 -0.2 e ) y Z== Y+O.2 e Jm
"! 'f.O - 2.200 2.000 2.000
;t .3.5 - 2/2;.00 2.000 2.000
± 3.0 - 2.200 2.000 2.000
t 2...r - 2.199 / .995 J .q,/s
±2.0 - 2,196 1,990 1,99'1-
!:. I.S - 2,179 /.961/ /,990
!:. /.0 - 2./26 1.900 1.979
~o.s - 2. Olf'/- /.71/0 1.9Jf6
0 - 2,000 1.72. 7 1.92.7
/0-/3

/0.17"/ (cantO
The ahove resolls are plotfed in the 1ra,h De/ow,
I
i
yvs X I
2.00
L
~
1.95
V1.90
/E_ 1.85
/>-
1.80
1.75
/i
V
1.70
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
x,m
zvs x
I 2
L.----
t-
1.99
L1.98
jV
1.97
E / I
_ 1.96
/N
1.95 1/
1.94 J
/1.93 .-'
i
1.92
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
x,m
/0-1'1

/0.184
*10.18 Repeat Problem 10.17 jf the upstream depth is
O. 4 m.
O2
_.2
:n = . e
FIGURE PIO.IS
'VJ2 LJ V:J..
4'-+ 21 +2, :: ~+'ij. +zJ where 1t:::/~OJ 2,::Y;:::0,Jfm J z,.= y+ZB
2. '1-42
or Z:z. :: y +0.2. e-
x
J ~;:: t::: o.lim ::: lolJlJ dnd V:::!f- ::: 1-
Thl/s, (IOr/-)2. +O,Jf, = (if !f)2 +Y -1-0.2 e-x:J..
2 (fl.811f)'l. In 2 {'lBlf)2
or 1
y3 _(5.50 - 0.2. e-x )y2 +O. 815;;0 where ylWm
Solve for y wilh -7'~ X~ /,t m
~ IO.!!l
Nofe: Fr ==--1- == ~ =5.0.5>/
I Yj» [('l.8/~) (o,'Tmil~
ThvsJ
the flolI! will remoin svpercrilical
fhroll,hotJf --fhe shQ//esf posit'vB roof of
f,(.O) will he Ihe correcf one.
y
/
(joSe prOf/ram CVBIC 10 solve fOry(x)frofIJ Fr.O) (..sma//esf positive
K2.
roof) alld then ~:: y +0,2 e- for -tf.m ~ X ~ Jfm.
2- 2.
X)rn - (5,SO-0.2 e-x ) y 2;; y -I-a2 e-X
m
J
-!. if. 0 -5 ..500 0..',.000 O. 'fooo
t .3.5 - S.500 O. 'fOOO O. Jfooo
±3.0 - 5 • .sOf) O. 'fOOO O. JfOOO
±2.5 - 5 . '1-99 0.3998 O. I..fOO~
±2.0 - S. '1-96 0.3999 O. if. 036
±1•.5 - S .479 O.Jfo06 0.lf~/7
±/.0 - s. 'f2.6 O. J/OZB O. Lj 76Lf
±O.S -.s. 3/flf 0, '1-063 0: 5 6 21
0 - .S'. 300 0.11-082 0.6082
/0-1.5
(J)

/o,/8~ I (conI')
The ahove resv/ls are plolfed on the qrllfJh iJe/ow.
zvs x
0.65 ......------.------------....-------.
/0-/6

VI Im77r-fflTflZTA3
10.1 q Water in a rectangular channel flows
into a gradual contraction section as is indicated
in Fig. PlO.lq. If the ftowrate is Q = 25 ft'/s and
the upstream depth is YI = 2 ft, determine the
downstream depth, Y2.
~ bl'" 4 ft b2 = 3 ft ~ V2
. 1 ~wkvvvmp:?2Z2ZZZZZ7azz?:
FIGURE PIO.,q
Top view
Side view
or
'h.3
- 2./5~2 +/.077 =0 which htJs roofs Y1.::I. 82 8J
O.9~6J and -0.62.3
1,
. I1
Note: F0 ~vi>' ~~ 3.~ s ]k~ 0.390..:.1
~YI (32.2.$ )(2.fI) 7-
Since Ihere is no relotive lIIinilllVIf/
area he/ween OJ and (2.) where critical
fJow can occur if Fo/lows fhot Fr:z.~ I
also. ThllsJ
IIis nof pOSS/b/6 10
have Y:z.:: O.9¥t
Thvs) Y1.::: 1.828 ff
/0-/7
y
(I)

10.2.0
10.20 Sketch the specific energy diagram for
the flow of Problem 10./'1 and indicate its im-
portant characteristics. Note that qi =;6 q2' FIGURE PIO.20
'] l:_-~J~~L~-_:~--- .. ,i (2)_________---'O.-'-'-'-~
~. Yl 12 -+'2
>;//fTm?//7/7/7'/////1»//7//7/7////7// t'T/T/Jd//
(1) (2)
_ Q _ 25 !/-J =6. 25 -.sfl~
Cf,- T - Jff+
or fil.~
E
- + (6. 25.s ) _ + 0,607
- Y 2(j2,2~)y1. or E - Y y:z.
!i. 2.S .J:ti fi2.
For the b2-=3 II chQ!1IJe) J ,/1. = b1. = 3H =B.33 .s
(I)
or fi2. 2-
E= t (8,3~) 1,077
Y 2(32.,2~J y2. or E:= Y + y2.
Noh; Yc =(-r)~ sofhaf Yc,~( f ~ = ( (~~Z;17l=I.067R
(JJ
and k 2 k
.x =(~)3= (-(8;33~)2.):3= 1,2.92 f-l
Cz. Ij 32.2. %
A/s~ Emin =-: Yc J or Emilll =f(I.067{f)::: 1.600ft
£/1IilJz. := f (1.1-12 It)= 1.938f/
The specific ene"9Y ditJ9ro/fJ.S (EfS. (I) IJlJd(J.) tire p/offed he/w:
.. .. .. ~- ~ j 1 I I
~, ---------r------
; I •
2. ' '
I ff Jemf). ''')
~ ~I~
/0-/8

10.21
10.21 Repeat Problem 10. Iq if the upstream
depth is Yl = 0.5 ft. Assume that there are no
losses between sections (1) and (2).
Top vie<
FIGURE PIO.2,'
----___......:01,....:...(2_)_
Y2 ~'2
>7??>tW?7Tffff/?/T77ff/?l??Tff7ffffff/J/t77?Tff/J/
(1) (2)
or 3 2 '
Yz - 2.93 Y2 + /. 077 :: 0 which h,s roots Yz-::: 2. 79, 0.699, ond- o.SSE
AI f . Fi v, 12..S!!IVO e' r,::: - == = 3.12>/
Vii' [(32.2{~J(O.5ff~~
Since fhere is no relafive milJilllvl1I
arCQ between (JJ qnd(2) where cri/icfJi
Flow COli occur if follows fhal Fr2 > I
also. ThvsJ
if is noT poss/6le 10
hove ~ :::2.7 q- (fhe s{)bcrifiCIJIroo/).
Thus, ~ =O. 6?~If
/o-/q
y
/
E

/0.7-1. I
10.22. Water flows in a rectangular channel with a flowrate
per unit width of q = 1.5 m2
/ s and a depth of 0.5 m at section
(1). The head loss between sections (1) and (2) is 0.03 m. Plot
the specific energy diagram for this flow and locate states (1)
and (2) on this diagram. Is it possible to have a head loss of
0.06 m? Explain.
(I) (~)
1.1 • t ~ • t
~ ---~
Y' t7777777777777?7
1.6 --,---...,.....----r--....,.----,--.,.--...,....-...,.....-.....,
"'/1.4 -t---t---+---+-----j--+---+--~V~---t
~:/1.2 -t---t---+---+-----j--+------::-..f"V'----F-----j----t
~;
1.0 -r---t---t----t--I-----:;;...rl""/'----...--/---+-----f-----I
E ..,.......
~ 0.8 -t---t---t----t--~~-/-A----I---'-h-.,9---I
>. ..,.f'f' 1. ,/EJ. F O.q.
0.6 +---t---t---:,...,-F---f--~---'f-cv:-.A-----I-----1I-----t
"" ... '" ""min= ).9Ia-':';;'r--..... E; =0.9.9
0.4 -r------r--...--:....f----t---='---'-'--t--'--'---'--+----"""oo...._d_:-:::..-.:..--:..~--:...-_~--I
... '" -0.2 i-~...~--+--+---j--+---+--_+_-_I
'" I1;'"0.0 ~--+--+---+--_+_--!---!---+--__!
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6
E,m
10-20
Note: If hL.:: 0.06111
wilh £, =O.9SQm so
thaf £2.::£, - 0,06,
fhen
£,. ==0.eqq III <: Emil')
ThusJ
if is not possible
to hOlle hL-:::o.o6 wifh
fhe qivclJ ~ andYt .

/0.2.3 I
10.23 Water flows in a horizontal rectangular
channel with a flowrate per unit width of q = 10
ft2/s and a depth of 1.0 ft at the downstream
section (2). The head loss between section (1)
upstream and section (2) is 0.2 ft. Plot the specific
energy diagram for this flow and locate states (1)
and (2) on this diagram.
(1..)
OJ f ·.,1~ Y, ~-~
77)777777777 )777
yvs E
3.0 --r----~--....,------r-----r"-----,--,-I'~,'I""V"'"7'""-.....,
1;"1"/2.5 -i------+-----+-----+----+--~I'~--+--t---------I
",,'/V,.'2.0 -i------+-----+-----+------:I'.-F----/:7'-i---t__-______I
;:= ,,'1',' I/E,.~;IJ :: 2.1' II
~ 1.5 -i------+-----+---~I'~--~.~~---t---------I
>- 1,.,,,,,,,,, ~/E2. =. ,s51f
1.0 -i------+---",..F----+---~--____f"I....______;;::::-:--+__-______I
",'" ~~~~----~
",' l£ =2..7sH, I
0.5 -i----~--__+_--__+----t--~r___------'---+__-______I
'"'",,
1,.,'"0.0 -¥---...f---+----+----+-----i-----+------1
o 0.5 1 1.5 2 2.5 3 3.5
E,ft
10-').'

10.'1..../1 I
10.2+ Water flows in a horizontal, rectangular channel with
an initial depth of 2 ft and initial velocity of 12 ft/s. Determine
the depth downstream if losses are negligible. Note that there
may be more than one solution. Repeat the problem if the initial t
depth remains the same, but the initial velocity is 6 ft/s.
, ~ ~~ pi - - /2-
V.=12 fp.s ~ =2 t1' - - -~ - -
I 77'7117771711/717777J7//
b;::widfh
which has 3 roofs; ()ne ne9alive (no phY.Jic41
rneanin9J1 one is ~ =:2. ff (no change in depth),
and ~ :::: 3,51 If (an increasei" deplh).
If ~:::611 the/) v.: == (6 fj)(2.11) ~ .ll:.
I .s) 2. Yz Y1
and £'1.'(/) becomes
2 ( /2. )2-
2 + 6 - / + 2i_ .
2(32.2.) - J'2 2 (32,2.)
or
3 2-
X. -2.56 Y:z. +2-.2-'1- ~O The p()s/live real roQ/s are
~ c: 2 fl J or ~ ::: /,38 ff (a dec.rease ill deplh)
10-2.2

/0.25
10.25 A smooth transition section connects
two rectangular channels as shown in Fig. PlO.2S.
The channel width increases from 6.0 to 7.0 ft
and the water surface elevation is the same in
each channel. If the upstream depth of flow is 3.0
ft, determine 11, the amount the channel bed needs
to be raised across the transition section to main-
tain the same surface elevation.
~'''''''-''''TTh.'''·
""''''I'''''''''''''''
6ft 7ft
&",L",,,,~ 1~~,,~~,,~~~~
FIGURE PIO.2S Top view
(I) S7
.L), 1/1. l!f V2 h Side view
-l.k!- +..!L f Z = ... f J ~ ~ WI ere 4 ~ Ll.. ::: 0 (J/Jd ~,::: 22,
r Zj' Z, 3. J '-' r~
ThIlS, ~;::"~ or
*=t lienee, 11, :::112. or (6 ff)(3(t) =(7ff)(3f1-h)
or
h:::; 0.'129 ff
Nofe: ~/:::; f, =f IJ/Jd '/2.::; ~ :::: ~ < rt,
Md ~2 ~
£, :::; Ii +21 IJnd £2.::: ~ +"iff Thv~ .since ~ c: ~ II fo/lows
fhaf ~ -Ez :::; ~ -~
The correspoIJdin9 specific eIJer9Y di(J9ralll is qs ille/kateri be/oJP=
/
y //
~ - ~ - ~ - - - - - - /' - • (J)
~ -------"/./
/
/
/
/
/
fj.,
/
LL....-_ _ _~---'--_ i:2
/0-23
(2.)
•

/0.26 I
f(x)10.26 Water flows over a bump of height
h = h(x) on the bottom of a wide rectangular
channel as is indicated in Fig. P10.26. If energy
losses are negligible, show that the slope of the
water surface is given by dy/dx = -(dh/dx)/
[1 - (V2/gy)], where V = V(x) and y = y(x)
are the local velocity and depth of flow. Com-
ment on the sign (i.e., <0, = 0, or >0) of dy/dx
relative to the sign of dh/dx.
-.2.,=,",,"-=-,,-=-= __ ~ _ =-_ __
VI .:--=-=-----:-r=-----c-=-- ~) t---
~ Yl JT
-{ ~~7/~/r~.... ~~f
h(x)
FIGURE PIO.26
For any fwo points on Ihe free s()rface:
-If!- +- ift &, :: ~-I- ¥i.,.$.2 I where fJ, -:::1.2:::()J Z, :: M) (u,d~2 =hf Y2
Thvs, JC+h +y == cOfJslonf so tha+ hy diflerehlialillq
').~
:V't dh fit. =0 (I)
').9 Tx (JX
Also, for cOl/servoliDA 01h/qss
Vj >1 :: Vy or V~ t Y ~ ;:;0 or ~:: - f 1x (Z)
Combine £'10$. (I)(I/Jd (2.);
V (-J!. dy ) +dh +d.x -:::0 "dy _ -(df)
~ y dX d1 di J or OJ - (/_(fy'))
/lole: If Fr ==ffy <: I j . fhen ~ (/nd ¥X hove the OpPosite sijn
9
If Fr > I the" dA and!U V· dll ---
J Tx dX -..- :;:<0 Fr </
have fhe SQllle sifjlJ. ~
-x di>o
Fr>/
/0-')../1-

/0.1-7 j
10.27 Integrate the differential equation ob-
tained in Problem 10.26 to determine the "draw-
down" distance, e= e(x), indicated in Fig. P10.26.
Comment on your results.
From Prohlem /0.26 :
or
1. 2 1J'2 &
y.3_(y,t 26 -h)Y:Zt(v~; )::0 (I)
Obtain y== y(x) fro/}') £'1' (jJ and fhen 1=J(x) from y,:: h +y +1
or j:: YJ -h-y
Nofe: £'1' (I) is r/fJlhill9 ",ore Ihon Inc Bern(){)//i efV4//o/) :
y;+y, " 1;+y+h wilh V,,~ so Ih4t
~2 (¥J.2 l 1
2; +}j::: ').~ t y+h wnkh sim/Jlilie.s 70 [r.(I),
/0-25

/0.28 I
10.28 Determine the maximum depth in a 3-m-wide rec-
tangular channel if the flow is to be supercritical with a flowrate
of Q = 60 m3
/s.
3
_ Q _ 601]- _ 20
V- 7f - (3m) y - y , where y:depfh
A!.so, V (Vl.) 6.3Q
Fr '" YjY - f9.81;)y]%'" y3Az /Vole: Fr decrt!Qses as y increases.
Thvs wdh Fr:/ y =(6.3q)~ == :).'1I1-/fJ
J J
10-26

/ o. 30 I
J0.30 The following data are taken from measurements on
a river: A = 200 ft2, P = 80 ft, and So = 0.015 ftlSO ft. Deter-
mine the average shear stress on the wetted perimeter of this
channel.
rr- Y D ( h D A 200//2. -"
rW == 0 nh~o J were "I, == p == eo(+ ;:: 2.$01"1-
anri
ThvsJ
?W:::: 62.,'f#S (2,s()11) (0. o()030) :::: O,O/fD~1ff:L
/0.31 1
10.31 A viscous oil flows down a wide plate with a uniform
depth of 8 mm and an average velocity of 50 mm/s. The plate
is on a 3° hill and the specific gravity of the oil is 0.85. Deter-
mine the average shear stress between the oil and the plate.
t~ b::plate widlh - .
77; 7 77 77; ,;;, 7777,/ ;; 7 ;;"1
y= Shim
lw == 't~ So J where 0:: O. fJS rH,.o =0.95 (QgOOU;3) = 8330 ~3
For a wide flal pIa/e) /)::: hy and p:: b so fhat Rh'::';:: y~ 8XIO-
3
p/
A/SOl S(J:; sin3 ' so fhaf
/0"'27

10.32 The following data are obtained for a
particular reach of the Provo River in Utah:
A = 183 ft~. free-surface width = 55 ft, average
depth = 3.3 ft. R" = 3.22 ft, V = 6.56 ft/s, length
of reach = 116 ft. and elevation drop of reach =
1.04 ft. Determine the (a) average shear stress
on the wetted perimeter, (b) the Manning coef-
ficient, n, and (c) the Froude number of the flow.
X" D S h 1.0JfrfaJ 7;=Ol)h'o } were 50 = 116t/- =0.00897
Thus} ?;, = (62.Jftfs)(3.2.2 ft )(O.OOgQ7) = I. BO~
b) rt::: ~ A Rh2.;3 So%. :: /I V J where K::: l.if?
Thvs} = I.lf9 R:/.3s;:. :: (t.Lf9) (3. ').2./"/s(0. OOeq7)~_
n V 6.56 - 0.Oi.f69
c) Fr =y{yy =r ;;~6!f)]~ '" 0.636 < I (subcrifico{)
j (32.2 05
3.3ft 2.
10-2-8

/0.33 I Center board
/ J
r----:-----::-:----..::.~---:---
bl2
10.33 By what percent is the flowrate reduced
in the rectangular channel shown in Fig. PIO.33
because of the addition of the thin center board?
All surfaces are of the same material.
1~-bI2---<O{fl
W7m~~~~~
/0.3 if I
Q~ ~ fJ Rh~ S/i
Wi/houl fhe centerboard
1034 Water flows in an unfinished concrete channel at a
rate of 30 m3
/s. What flowrate can be expected if the concrete
were finished and the depth remains constant?
I. b---+l·1
FIGURE PIO.33
(I)
(2.)
Q=-*f} R;/3 S~ Lei ()f denote finished j ( )u denote vnfinished.
Thus) since fJu == Af ) Rhu -:: Rhf and SOil -::: S"-F J if fo/lows fhal
Q )( fJ o.~ ti
Ll == nu. U 11/1 IJ Sou _ !!L.
Q.f Ji. II R~ s~ - nunf .f hf Of
From Table /0./ nu::: O.D/if J nf :: 0.012-
or
Q ::.!1JL n = O.O/if (30 l!L
l
) = 35.0.11l
f nf fu 0.0/2.s .s

10.3.5
10.35 The great Kings River flume in Fresno County, Cal-
ifornia, was used from 1890 to 1923 to carry logs from an el-
evation of 4500 ft where trees were cut to an elevation of 300
ft at the railhead. The flume was 54 miles long, constructed of
wood, and had a V-cross section as indicated in Fig. PlO.35. It
is claimed that logs would travel the length of the flume in 15
hours. Do you agree with this claim? Provide appropriate cal-
culations to support your answer.
• FIGURE P10.35
i ::: dis/alice frQl/e/ed ~ ~119 I: • TlJI/~
~ ::= 1. ~ (S~ mi) (5l eo fll",,') == S, 2c9 1i.
lot t (/s hr) (3o(JO S /hf') S
De/ermine the t/veratj8 wafer velocily; ~ (Inri cPPlfaf'O ,I w/lh 1I)j~' ~Of'
Vz:: ~ Rh1-~1f'fo ) where }(.::/.If?; 11:! t (J {il.) ~ O,S f/~J p::: 2 II
_H. I R - fi. - o.s f-l ~ - 0 2 ,.. f'.l
$0 7fl(J1 h - p - ~ #- - , ~ T
Also;
~ z (~500 -300) H
So :: T::: (5/fmi) (52 eo filmi) = o. O/lf7
ThvsJ
wtlh n::: o. 0/2 (see 74h/e /0. /J planed wood)J
V= /.119 (OI2.5)~ 1o.01lf7 =.6.97 Ii0,0/2 ~
Nole: V;s slif/1rl1y J4rger than 0"1. TIJ/I~ the claim Clt/eal'S If) be
curr(J(;t. Ves.
10-30

/0.36 I
1036 Water lows in a river with a speed of 3 ft/s. The river
is a clean, straight natural channel, 400 ft wide with a nearly
uniform 3-ft depth. Is the slope of this river greater than or less
than the average slope of the Mississippi River which drops a
distance of 1475 ft in its 2552-mi length? Support your answer
with appropriate calculations.
(I) V:::: ~ Rh~ rs; J where X=:/JI~
V=.3ff/s, Y=3ff J b=: t;.ofJ1f.l l/::iJy:: /20()f/~ P:::b+2Y~'f(J6rl
TiJfls ~
D J II /2-()o/l /J 1'1
I1h := P:::: I/-otff =2./tTT
IIfroJ
(rom TaIJ/e /0//) n~O.03 SrJ 11;41 fl'fJIYI /5."1- (I):
..3 c I. Jf9 ( '). ,9t)~ rs:0,Q3 "
Of'
So = O. 000858
ihB aVerIJ9B lJIIi.rs'ss;ili .s/ofe /:s
10-31

/0.37
6or---------------------~
10.37 At a particular location the cross sec-
tion of the Columbia River is as indicated in Fig.
P1037. If on a day without wind it takes 5 min
to float 0.5 mi along the river, which drops 0.46
ft in that distance, determine the value of the
Manning coefficient. n.
QO~--u40~O--~80~O~~1~20~O~~=-~·
Width, ft
FIGURE PIO.37
c: . (0,5mi)(5280 if,.) == 8 8 £t
[rom the qlven dalo) v::: ( ,)( S) '.s .
5/11m 60mrn
From fhe lIlanning ettlJlJli()n,
V K ~ ~ Q~H
== n Rh So J where ><~/Jf?, so= (I ~ ')(5 JJ)
o A o. ml 280mi
IJlJd !lfJ :::: p .
ilppr()ximaie !J and P fro", fhe fi9vre a.s
II ~ tby -;: -J: (I 7oo(1)(lflfff) :::: 3~¥()(){l2
and
P~/8ooff Thvs i?Jh~ 37.,JfO~/J2 =zQeff
J 1800
lieneeJ from £1- (I) ;
1.~9 26 ~
8. e :: n-(20. s) (0.000/7#)
or
n:: 0.0/69
IO~32
=0.000/7"1- (J)
)

10.38 ,
10.38 If the free surface of the Columbia River
shown in Fig. PlO.37were 20 ft above the bottom
rather than 44 ft, as is indicated in the figure, how
long would it take to float the O.5-mi stretch con-
sidered in Problem 1O.37? Assume the elevation
change remains 0.46 ft.
Width, It
FIGURE PIO.31
Lei ( )20 denofe fhe 20 Hdepth and ( )#~ fhe 'PI- rf depth.
Thvs,
X 213 J.i
~o =n,.o Rh"o so~o
and $020 ::: S,,¥¥
HenceJ
'" 0 =( Rh20 )2.h From
V~Jf Rhl#l-
the Nivre I/~'f ~ f by =i:(J700!+) (tf'ffl)
:: 37 'fOOfl2
~If ~ /800 Ff J
@d ~ 2.
A:J.o';:: dby:: i (/S50-Ifoo)(20)ff :: 1~5ooff
F;.o ~ /5S0 - 600~ /oooff
(I)
~ ~ d
-E!......::( II.S f+) ::: O.b7~ so fhal wilh j= Vi (Jnd 1,.u=:Pp; :O..5nll;
Vito 20.8 f1
1 - ( .{,,,,) t - 5 milJ - Z ¥ 2 .
L9.0 - V:10 #~ - 0.67'1- - ~ 111m
10-33

/0.39 I
lO.3Q Rainwater runoff from a 200-ft by SOO-ft parking lot
is to drain through a circular concrete pipe that is laid on a slope
of3 ft/mi. Determine the pipe diameter if it is to be full with
a steady rainfall of I.S in./hr.
K 7.; ~
Q;: n ARh:!I So ~ J where
A = *D1- and
= 0.000568
From Table /0./, n =0.0/2 .
AIso} ~ =Alof r J where r ~ rainf411 rale ::: 1.5 fr
,
Thus, . , N Jh rl3
Q:: (').oOH)(sooff)(l.s fr)(Ii1nJ( 360;.s) ==3.lf7.!j-
HenceJ from Elf' 0):
3~tl1~ :'~tt;~ (:qD'") ( 1lt)~ (0. 000568) ~
or
0= 1.6Lfff
/O-.3Jf
(I)

/O.lf~
10.40 To prevent weeds from growing in a clean earthen-
lined canal, it is recommended that the velocity be no less than
2.5 ftls. For the symmetrical canal shown in Fig. P10.40, de-
termine the minimum slope needed.
x ~ k D IIV== n Rh So" J where X~/.'1'1 and nh ;:: l'
II :: t (Iff.I +12. f+ ) (.3 H) :::: 2If fll. Qnd p:::: rt ff +:2 (5[.0 =/ITff
Thus) R = 2.'Tf/2 =171'1(1
h II/- fl •
From Table 10.1 n = 0.022J
_ /.1f9 ( "4 ~
2.s - 0.022 /.7/Jf) So or
so lhQf Ef.(J) qives (with V:: 2.5 !J)
SD ~ 0.00066'+
10. Jf I
10.41 The smooth concrete-lined symmetrical channel
shown in Video V10.3 and Fig. PI0.40 carries water from the
silt-laden Colorado River. If the velocity must be 4.0 ftls to
prevent the silt from settling out (and eventually clogging the
channel), determine the minimum slope needed.
V J< R2h 5 V2. h d R A::: 7f h 0 J Were K::I. If? an h:: p
FIGURE PIO.40
fJ ::: t (if fI +12 fO (.1ff) == 2 Jf f+2. anrJ P=if ff f 2. (5 f+) =Ilf ff
Th 21/- H2o
liS, Rh = Ilfft = I. 7/Jf ff
From Table 10./) n::: 0.012 so fhal £'j.(J)qives (with V=If.fJ)
1.11-9 ~ ~ _
If..O ;:: 0.0/2 0.7/J-f) So or So - 0.000505
10-35
(I)
(I)

10.'1-'2.
10.4Z The symmetrical channel shown in Fig.
PlO.40 is dug in sandy loam soil with 11 = 0.020.
For such surface material it is recommended that
to prevent scouring of the surface the average
velocity be no more than 1.75 ftls. Determine the
maximum slope allowed.
10.'13
FIGURE PIO.40
x ~ ~ A
V=rrRh So ) where K=I.J.f9 and Rh=P
A ;:: t (Lf ff t 12H)(3 ff) :;:: Z If ff 2. and P== If fl +2 (5 fI) -::; 1if ff
Thus) R :: 2lfff
a
=I.7IJf{1
h IJfff ff
'Wilh n == 0.020 and V= 1.7.5 T fli.aJ qives
/.7.5 =6:~:o (J.711/'/~ S~ or So :: 0.0002 6q
10.43 The flowrate in the clay-lined channel (n = 0.025)
shown in Fig. PlO.43 is to be 300 ft3
/s. To prevent erosion of
the sides, the velocity must not exceed 5 ft/s. For this maximum
velocity, determine the width of the bottom, b, and the slope, So.
v== '*J where A=f [b t ( b+i, +12)] y
FIGURE PIO.43
wilh ~ == h~:~o ::3. '1-6 fl
and J,. = ta::So =2 ff 3
Thus) 5 11 = 300!f or b::. 27..3 ft
s fEb +(b+.3.'l-6fff:ZU5J2fO J ===
V
K %/,a ~
A/so} :::: n Rh So } where "1.==/.'19 and from Table /0./) n::. 0.025
From Efj.tO} A;:: i [2 (27. 3 to .,. 3. '1-6 ff +2 rtJ(2 ff) :.60.0ff2.
Also} 2ff 2. ff ..
P== b +J~ .,.J.u :: 27.3 ff + - SOD + . LU 0 =3'1-.1 ff~ T Sin SIn ,oJ
Thus} R :: A. = 60.0£12. :: I. 76 ff so fhaf ££1. (2.) becomes
h p 3lf.1 ff • G
_1,11-9 24 ~ _
5 - 0.02..5 (1.76) So I or So- 0.00331
10-36
(I)
(I)

10.Jf4- J
10.44 A trapezoidal channel with a bottom
width of 3.0 m and sides with a slope of 2: 1
(horizontal: vertical) is lined with fine gravel
(n = 0.020) and is to carry 10 m3
/s. Can this
channel be built with a slope of So = 0.00010 if
it is necessary to keep the velocity below 0.75
mls to prevent scouring of the bottom? Explain.
3
Determine Vwilh ~== 10./f and So :: 0.00010,
Q= ~ AR;hVS;; J where /l= fy[3+(.3+lfY)]::: 2Y2.+3y
f/nd Rh
:: JJ wdh p:: 3 t 2 (1/5'y)
Thvs, f 2]~ I
J0::: --L (zy:L+3y) 2. y r 3 y (o.oooJ)J4.
0.02 3 t 2. V5 Y
or ~
(2. yz.,.3y) :3
20 :: ( -v:r )~ wht"cn COli he wrdlen QS
3 +1 ;.s 'y
2 ( O.lI-
:J.y .f-3y-6.03 3fl1/Ey) ==0 =F(y)
Solve (hy frial and error) E'l' OJ for y:
YJ m F(y) 0.1 -
'2.20 -O,'f.Sq
2.2.2 -0.2.6'1 fry) -
2,2.Jf -0,077 ..
-2.26 0,1/7
o I I I ~-
~ 2.20 2.2~"f 2.2.6
From fhe 9raph '118 see
fhal y;:; 2. 25 m -0,6
~.
2-
lIenceJ
from £'1' (0) II ::: 2. (2.2..5) + 3(2.2.5) == /6.9 IJJ 2-
so that m3
V== l:: JOs ::: 0,592 f/6.9m2.
ihlls" V< 0.7.5.f so fhat scofJrin, will nul occur,
/0-37
(0)
0)

10.4-5J
10.45 Water flows in a 2-m-diameter finished
concrete pipe so that it is completely full and the
pressure is constant all along the pipe. Ifthe slope
is So = 0.005, determine the flowrate by using
open-channel flow methods. Compare this result
with that obtained by using pipe flow methods of
Chapter 8.
For open channel flow Q== ~ f} Rh~ S/~· J where x:::: J
A/soJ A=fD2
==1j(zmt::::3.1'fm:J. and P::::"D=6.28m so thai
R _.1i - 3 .IIf /17'- - 0 5
h - P - 6.2.8 m - . m
HenceJ
wifh n:::: 0.012 For finished concrete (see Table 10./)
Q I ~(~ 111
3
( )
=0.012. (3.1'1-) (0.5) 0,005) 2. :::: 11,7 -:s open channel
For pipe flok! with cons/ani pressure:
t£J- t JL.2 +z, =b +Yl+Z:J. +f!t~ 2, ~ 21 '
where PI == f~ and It;' = ~
Thus, with z/ - 2:;. == JSo •
1rf So :::: f D ').g
or
fV'-::: 2.tjDSo :: Z(9,8/~.)(2.m)(O.005)
F'rom Fig. B. 2.2 J (or smooth concrete ~ == 1.5 x/o-If-
AI 0 - Y.Q - V(2111) - 6V
SOl ne - 11 - 1.I2.XIO'm..2.. - 1.79x/o
,s
and from fhe Moody chari (Fi9' 8.23) :
Solve £'15. (I), (2), tJlJd (3) for ~ VI Re : f
ils.sul!1e f= 0,015 so fhqi from E,.(J)
V::::[ 0'/96]~ = 3 61l!1.
0,01.5 '..s
Re
or .' .
Re.: I. 79x/O~(3,61):: 6J1-6x/06
Thu~
f:::: 0.013 :I 0.015. /lsslJme f c 0.013 I
from £f.t.1) (t1foldy c.hqrf)
V [ 0.196] J.2 801
or = 0,013 =.3.8 .s
(2.)
Z~=o
(2.)
$0 fhat Re ==/, 7?XI0
6
(3. 88) =6.'15 x/o6
T/;/I~J from £1. (3) f:: 0.0/3 (check.s
wHh fhe t/.t.svlIIsd vo/ve) lienceJ V=..3. 88'; ()r
3
Q=!9V.::: !(2.111)" (.J.88~) :' 12,2 1f3 (pipe flow) :::, II. 7~ (open c.hannel
flow)
10-38

IO·Jf7 I
lO.4i Because of neglect, an irrigation canal has become
weedy and the maximum flowrate possible is only 90% of the
desired flowrate. Would removing the weeds, thus making the
surface gravel, allow the canal to carry the desired flowrate?
Support your answer with appropriate calculations.
Lei ( )", and (~ denofe weedy Qnd r;ravel co"difIPf)SJ re.sle6 //l/ely.
TlJv.JJ
Q _ X ~ 1r:::-
(I) w- /1w!lw ~w ,s"", and
(1.) Q., = f 1 Rny"4~
wJw./wdh !he Ch4M9/ M/J flU!«111 J 4,,/R",J (lnd -5;,,,,~s;"
lIe/Jce) /;y diViJ;f}~ £,(:;.)/;1 Ff(/J :
Q .-1-
J ::::: !!L = !jf- ) WAel'B f",fJjf/ ";"/;/0 /().j: /)w::: O.03(); I? ::: f},().2.,s-
Qw *~ 1
ThtJs)
('J :: (). 03 0 I) == /, 2 ta.1(, 0.Orli ~£.&I IV
Evt ~w:: 0.9 41 • rJ So fH4+lftJeslre
CV, :: If 2 (0, qQ"esil'eti) = /, 0~ qriesif'84
YeSJ
;1 WOIIIJ worA.
/0-39

10·f8 I
(I)
10.48 An open channel of square cross section had a
flowrate of 80 ft3
/s when first used. After extended use, the
channel became half-filled with silt. Determine the flowrate for
this silted condition. Assume the Manning coefficient is the
same for all the surfaces.
LeI ( }n alJd ( ).r denoTe

/0,501
10.49 A circular, finished concrete culvert is to carry a dis-
charge of 50 ft3
/s on a slope of 0.00IO. It is to flow not more
than half-full. The culvert pipes are available from the manu-
facturer with diameters that are multiples of I ft. Determine the
smallest suitable culvert diameter.
x 2..-1 k
Q=n ARh 50"') where X= l.'f9 ) S/)= 0.001) anrJ(frollJ TabJe /0.1)
n=0.0/2-
For a circular pipe half fuJI /I::: i D'"J p::: f Dso that Rh =~ =*Thus, 50 ::;o~o~~ (*D'')(~)''h(o.ooJ)~) or D::.5.21ff
To make sure if i.s not more thon holf (vII use fhe 6 ff diameler pipe.
10.50 A rectangular unfinished concrete
channel of 28-ft-width is laid on a slope of 8
ft/mi. Determine the flow depth and Froude
number of the flow if the flowrate is 400 fe/so
¥ I
Q j(AR2~s~ h 8U:::: n h 0 J were k= I.'f'lJ So = .52BOff = 0.00Is/51
Qnd
from Table /0./ n= 0.01,/-
AI A f
A 2By
so) =28 Y and p= 2Y+2B SO fha Rh :::: P :::2y +28
Thus) '100 == /,lI" ( 2By )~/3(2SY)(O OO/5/S)~O,O/If 2y+ZB •
or
Y
5'/,9
O.5QJf = -....:......--
(y t/lli/3
Hencel 5./.i
O.tf58(yt/if)-Y 2. :::0 !fiFfy)
Trial and error soJufion fay Ffy)=:O
From Ihe 9rQph r=O when y= 2.23
ThvsJ
y:: 2..23 ff
F
2.2
0,5 ...
. y:2.23
ol---...L..-I /-4-....L-_....I1_ -
2.3 Y
V- Q Jl-OO¥ 6 VI Jt
- If :: (2BfO{2.23ff) = .,I.s
so thof -0.5 I-
£i.
Fi 1.. 6. JfI .s 0
r=V'lY =[(32.2¥a)(2.:l3ff)J~ = .7.56
IO-LfI

/O.SI I
0)
10.51 A 10-ft-wide rectangular channel is built to bypass a
dam so that fish can swim upstream during their migration. Dur-
ing normal conditions when the water depth is 4 ft, the water
velocity is 5 ft/s. Determine the velocity during a flood when
the water depth is 8 ft.
Lef ( ~ and ( 1- denule f/Ol'nJa I af/a flood ctJlJd/fiol/s, f'cspecfillely.
ThvsJ
II k D 1,./3.fr;:: Clf)d
Vn ::: n;; f1hlJ ySolI
/ )( R2-4 Ir;:-'
Vf z:: -n; hf
v~{
where nn ~ nf J ~on -::; S()I and
lin :: I (J If(if fIJ::: ~ (JII~ J 11;:: 10If(&f-IJ ::; fo til.
Pn =10 If f 2 ('ffT) =Ii If~ fJ ::: lofT+2(111) ;:2.of-l
A 112-
ThVSJ
Rn/}:;- t-::: j;ff ::: 2..:J.:J. II
anJ _ II 8()t/L
~r - 71" :z.IIf '" 3. on.;
flence~ d/v/Je Eq(:J.) by £q(jJ t() ()~/4in:
Jf _(Rhl )~~ ( .3.oe# )~~
Vn - Rpn - 2:).~ff - 1.2/f
so that
~ ;= I.;z.~ ~ ::: I. 2'f(s ij) ::: 6,).2-#
to-If2-

/0.52 J
10.52 An engineer is to design a channel lined with planed
wood to carry water at a ftowrate of 2 m3
Is on a slope of 10
m/800 m. The channel cross section can be either a 90° triangle
or a rectangle with a cross section twice as wide as its depth.
Which would require less wood and by what percent?
Qr: -H- AR;hS/i
ana ( ),.. denofe the
(I)
Let ( )t denote fhe frianguJar cross -sec/ion
rectangular- cross-section
3 10
ThvsJ 4Jr:=tVl =2.!J- J ~,.. =.s;,t = 800
and nr= nl So thai £q. (I) give.s
IT]
1-2Yr~
~!;: 11ArRh,.. =III Rh: J where Rh :: p (2.)
Hence, ~
iJ,.:=2.Yr:z. J p,. =lfYr S(7 fhal Rhr := ~t =~ Yr
Also)
Ai =i(2)i)X:= ~~ J '1 = 2 (1{iyt ) so tho! Rhl = 2~ .
Thus) from Ef.(').) :
2 ( I)~ ,.(J)% '2Yr "2Yr ::::)1 2."12 M J or Yr = 0.707 Yt
The amotlfJf of wood i.s proporliolJfll fo fhe welled perimeter, P.
Since
!l= 21/i'Yt == zl'I'y;. = /.00
P,. if Yr q.( 0.107 ) Yt
the friqnq/e requires lhe same amount of wood as 'he recTangle

/0.£3
10.53 The two channels shown in Fig. PIO.53 are laid on
the same slope and lined with the same material. When these
channels are flowing half full, in which one will the flowrate
be the greatest? Show any calculations needed to obtain your
answer.
0)
(~)
/O-1./-'f
• FIGURE P10.53

10.5 If
10.54 Water flows in a channel with an equi-
lateral triangle cross section as shown' in Fig.
PIO.54. For a given Manning coefficient, n, and
channel slope, determine the depth that gives the
maximum flowrate.
FIGURE P]()..5+
There foreJ .'%
Q _ K J ( 2,)[ J-hy - y:J.. ]:3 ~
- 7f fOI1~o' lhy-y 2(h + y) So
cos60°
'---I
1- 2.h
- fan 60°
b =2lh-y)
fan 60°
/J _ Y
~.s - Sin600
For fhe !rJQxi/1JVIfJ {/owrale J ~ -= 0 J which is e'{lJivQ/eni fo
dE ( : l )51.3
Or = OJ where F7y) == ( 2hy - Yo)~ Vpon cli!ferelJlitJlioIJ
y+ hco.s60
and simp/ilicrrlion !his 9ill8S
S ( Y+hcos6otl) (h-y) - (2.hY- y") =0
or
+!y2. +(.5hcos60() -.3h)y - 5 h2.Cos60tl
:: 0
which cqn he wrilfen as
e({-t:.- (-f) - 5 =0
Y _ J;t '~I+-'1--(-8)-(5---"),
Hence} h - /6 = -0.73/ or + 0.856
The ne9afive ro01 has no physical meQnin9'
Thus} y =O. 856 h
10 -'1-5

/0.55 (
10.55 At what depth will 50 ft:1/s of water flow in a 6-ft-
wide rectangular channel lined with rubble masonry set on a
slope of I ft in 500 ft? Is a hydraulic jump possible under these
conditions? Explain.
/. If.1 ~
Q:: on ARh vs, where
LJ /y 0 A _ 6r'-
n ~ 0 J 1111:: P - "ly+6 J
anfJ.
n:: 0,01-5 (see Table. 10,/)
Thvs) '-,.1
60 -c: 1.1f9 (6Y)[ 6y ] '3(0 002.)~0,02.5 :2. y+ 6 '
Whic.h becomes
y5/3
:= (2y+ 6)'-/3 (O.9~g)
l 6ft j
.~~--nc: 0.02.5
The Irial and error solvfion 70 this e9tJal;~1J IS
Y= 2.S3fl
-rl... V- !l 5 0 ft~ 3 f" /.I f/VSJ - IJ = 6 (2.$"3) ffl::: · ')..9 TIS
so thai
V
Fr ::~~
v'J y
3.29 ff4
S/nce Fr < I if /s not poss/hle 10 have a hytlr4V//c .ifl~l.
/0-'1-6

10.56 I
10.56 . Water flows in t~e s~mmetrical, unfinished concrete ~~21 ~':'
~~~~~~~alsc:,~f~/i~~~t~~~~~i~~~~e6n~tr:;::~;[u~~~:;:~~ .'V==:<-:-o-: --,g'oo'--':--:.;::;:::-:-::.;:-:-=-=--~.
of concrete needed to line each 1000 ft of the channel. 35 0'._ ~ _ _ .• ,0
- - , - _.••." ., •.••.• d .. I .••.•• " ' ; " .;... 4-in.-thick
I. 10ft .1 concrete lining
FIGURE PIO.56
X fA %
Q::: 7f fJ Rh 3 So % J where from Table /0.1 n;: 0.01'1-
A/SOl
A= t(~ +/off)y = t (20{ff2.86Y)Y
or
A==(Jo +1.¥Jy)y
~-.A.- II
h - P - (joll 102/1.)
or R _(JOt/.'f.3Y)Y
h - (Jot3,lfBy)
HenceJ with x:: /.~p £ 1{. (I) becomes
~ 'i ;t~
~/off--lV
'" =/011 +). fQ~3.sfl
12 =sin~sO :: I. 7¥ Y
r
( "J~~ ~J20= I, lf9 (10 t/.IfJy)y IO-t/·JfJY,y (11-.2.) 2
O.O/If (jo+3.If By) 2000
or
(10 y +1.#-.3y2.)S SA
/tf,eqo = (JO+3.~8Y)2. J or /22 (JO+.3/fSy) -(Joy+!'Jf.3yZ) 2=O=F(y)
Solve (frial qnd error) for F(y):::O
v F,
/.60 185
200""
F
(I)
1.6S 1.f2.S
1.70 -III
}oo
yo;:. I. 66Jf ff
ThIJsJ
y -= 1.66'f fl
v-= volume of concrefe per Jjoooff
.:: (P +Lf ff) ( 'JOOO £1) ( ~ fI)
O~L-------'~~--~'
1.65 ~y /,701.60
- /00 I- ~
where p= loft +2/,. :: /ofl +2. (I.7Jf)(I.tl'fff)== 15.8ft
HenceJ
11 =(15.8 {f +Lffl){IJOOO f·I)(~H) = 6}600 {fJ( k~~~) = 2-~'f yd3
lo-'t7

10.57J
10.57 Determine the critical depth for a flow
of 200 m'/s through a rectangular channel of 10-
m width. If the water flows 3.8 m deep, is the
flow supercritical? Explain.
i 1
Q= *fJRh1:1 S/2. where for crilicQ/ flow Fr =/ or V= 1iY[!L3
Thvs, wifh V" *"~~;In""i " 2/ '; we hqve
¥- :::VQ.81 y' or Y=3.tf'fI'J7
If y=3.8m J fhen V:= f.~ =S.26.f-
The fJow i.s slJbcrilicQ/.
lO-tf-8
Fi V S.:z.6 ~
and r = V9y =[(q.81 p,.)(3.8111)]~
=0.862

10.58 I
10.58 Water flows in a rectangular, brick-lined aqueduct of
width 1.2 m at a rate of 73,000 m3
/day. Determine the water
depth if the change in elevation over the 16-km length of this
channel is 9.6 m.
X ~/3.~
Q== 1) f) Rh vSo where (f)
X= I,) n::' o.o/s (see TalJle /0.1)J and
Q
",3 14':l Jhr ) m3
;" 7~ 000 j,y ( ').I/-hr) ( 36 oos z:: O. BIfS s-
Also) So = (9.6h1)/(/6x/o3m) ::: 0.0006
A
DA 1·2.Y_
~ /. ')..Y and Ilh ~ P = i"y+I.1..
Thvs Ea.(!) becomes:
~ I' "'A
-L I /.2.Y...] ~----------0,8J1.S :: 0.0/5 (I.').y) ';y+I,,. VO,QOfJ6
of ~
_ [If" Y ] '.3
O. ~3/ - Yr:2ytl.2
IJ fria / and error soluiion of Ihi.s eCI'Ja/ion 9jl)es
Y= 0.86/ hi
lo-~q
t
Y
l

/0.59 J
10.5'1 A smooth steel water slide at an amusement park is
of semicircular cross section with a diameter of 2.5 ft. The slide
descends a vertical distance of 35 ft in its 420 ft length. Ifpumps
supply water to the slide at a rate of 6 cfs, determine the depth
of flow. Ne£!Ject the effects of the curves and bends of the slide.
r 2.
5
!f 1
~f~~~~ ..
.'....:.::~.:.~./~y
k 2A k 35 ff te •Q::: n Ii Rh 3 So" J where k;::I.'f9J SD:: J/.2oFt :: 0.08331
Q= 6.0 .s
and from Table /0./ n == 0.012
D2. .
Also (see £xomple /0.5) IJ == '8 (f) -smB) and
J R _ D(9-sin8) where D::: 2.5 ff
h- '1-8 J
Thus,
I D
94
[ (e -sinef4
j .J
Q= *So~ 8 (1/)"4 e~ J where eN
rQ
6 ,
or 8~ s/:
6 0 :: 1.1f? (O.0833)~ (2.5) [ (() -sillS) ~]
• 0.012. 8 (lfl/.3 fji l3
lIeneeJ
2. ~ : z . . )5
O.2.q3f)~=(e-si"B) JI 0.02528 -(e-sme ==O=F(e)
Trit,j (Jnd error so/ufi()() for F(e) =0
eJ~d F .
1.50 O.02tf7 0.02 ~
1.57 O.oolqS F
0,01 ~
1.60 -O.OJ3.5
Ie:::1.57¥
0 J I j • I
1.50 1.55
~60 e
-0.01 l-
•
-0.02. I-
Thvs, f)=(1.57/f rq,j)( J~or:?) = ,qO.2- 0
or since
y:;:. ~ (/- cos (!)) if follows fhol
y=(¥fI)(J- co.s( ~)) =0.369 ff
/0-50

10.60
10.60 Two canals join to form a larger canal as shown
in Video VI0.2 and Fig. PI O.6Q. Each of the three rectan-
gular canals is lined with the same material and has the same
bottom slope. The water depth in each is to be 2 m. Deter-
mine the width of the merged canal, b. Explain physically
(i.e., without using any equations) why it is expected that
the width of the merged canal is less than the combined
widths of the two original canals (Le., b < 4 m + 8 m =
12 m).
II FIGURE P10.60
Q.3 = Q/ + Q2 where for &':::1,2,3
Q{' :: ~. A,. Rh~~ VSOl~
ThvsJ
K fI R1;j 1r;-' X ~ 1r;::-' }< 2.4 t~
n3 l3 h~ r50,3 :: "1i1. A1. Rh, rSo,- + If; AI RIl, rSo, (t)
But n,:::n2 ::: n3 and SOl::: SOl.. ~ SO.3 .$0 thaf Eq. (J) become.s
A R2/3 :::!l R='-h -I- II R2.h (2)
3 h3 2. h,. I hi
where :z.
AI::: 2m (4111)::: 8"l J F( ~ (2f2+Jf).t8m so -fhtat Rh, ~ f. r: : ; ::: 1m
A:z..::: 2m (em):::/6tn'" Fi::: (2-+2+B):#/:Lm So +ha1 Rh =~:: //,,,," -=-/,333m
J 2. ;~ 12-111
and.
/).3 == 2 b /112-) f1, :: (~+2+b) ==(b+¥)m so rhal Rh3 == t :: (!~9-)
ThvsJ
Er.(2) becomes
!~J"/.3 ~ ~(2b)L(b+~) == /6 (1.333) .} 8(J) ::: 27. 7'-
or ~A
b5
/3 =8.63 (6+'+) 3 (3)
A tria I and error or e9/Jaiion so/vel' so/vrion 10 £ft- (3) 9/ves
b= 10,6& m
I { Ihe two ori'lin4/ canals mer'jed If) form a/2m wide cafJQ~ fhe
wafer depfh flJov/d he less fhlJlJ 2m heC4t1.se wdhovl the fwo w~/Is thele
would be I~ frief/on force hold the wa1erb4CK. ThllsJ to lJIallllain
+he 2m depfh W~f}1tJ.rl have b<'/:Lm.
1 fJ. 1=1=1 g.3
j..--8rn~1 ~/:J.m-
/0-51

10.61 ..
2ft----..ol
l10.61 * Water flows in the painted steel rec-
tangular channel with rounded corners shown in
Fig. PlO.61. The bottom siope is 1 ft/200 ft. Plot
a graph of flowrate as a function of water depth.
for 0 :s y :s 1 ft with corner radii of r = 0, 0.2,
0.4, 0.6, 0.8, and 1.0 ft.
'=---=::=:=:::=;=---~----~--:=:-:-=:-:-=::::=:--y-=r_=::_=:-=-_-_=-_:_=::::dc_- Ir
FIGUnE "10.61
Q=*fJ Rh sji, where X= I. Iff J from 7Qb/e /0./ n::: o.O/~J 4nol
Jff
So= 200ft = 0.005
(Q) AsslJme y ~r :
,,....- - 2.fI -----t
1-.-i
Thus) A=2(y-r) +r(2-Zr)+ i 7Tr2.
or A=2Y-(2.-1[)r2.
and p=2 (y-r) +(').-Zr)+ llr
or p:: 2y -(Jf-Tf)r +2
HenceJ
wi!h Rh :: I Ef(s. (JJ, ('J.)J lind (3) give
1.1:: I.Jf9 A~3 -'- (0 005)~
¥ 0.011f p~'
or ~
y-r
.,.
(I)
(2.)
(3)
[2Y-(2.-¥)r2.] r ff3
Q= 7.53 ["Y-('1-1I)r+2.]J~ TOr r~y:!JJ where r-fl,y,.ft/;",s (If)
(b) Assume Y ~ r:
Thus, A= AJ +f}2. +A.3
From Example /O.S J 'IIilh D=2r
A,+A3= (Zrf(9-sin8) where fJNrad and
e
cosl = r;y
Hence) A= .f(e-sin8) of- (2-2.r)y
AIs0
1
p= 2-2r <lop' ... P3 ) where
from Example /0.5) R+~ = (2i)9 = re
ThusJ
p:: 2- 2r tr8 =~ +(9 -1.)r
By combing £'1.5. (OJ (5)J and (6) We obtain:
() - J.~9 A5.I.s I (0 005)./i
q I 0.0111 ~ ,
•
: r:z.> y
.
+
I
• (3)
.I
(5)
or ['2.. J5~-f(9 -sine) +(2-2.I")Y
Q=7.53 2A for O~y6r where r",ff y,..fI (7)
[2.+- (8-Z) r];3 Q'" fJ-3J
and ~ =2 co£'( r;Y)"" ;ad
(con/i)
IO-.5~

10.61 .. , (&onJt)
For r=OJo.2JO.I/-JO.6Jo.8Jand Iff pl()f Q:::Q(y) from either
ft{. (#oj or Eq. (7) for ()~ y~ I ft. Pro9ram P/o#~ I show" De/()W
wtU used 10 calclJlate fhe .tall1jJle re.rf)//s .J'howfJ.
:"00 cls
120 print# 1, "************************ *** *** ***** ** ***** *****"
130 print#l, "** This program calculates the flowrate as **"
1l,r0 print#l, "** a function of depth for various values **rt
150 print#l, "** of the radius of curvature of the **"
160 print#l, "** corners. **"
170 print#l, "***********************************************"
180 r = -0.2
190 pi = l,r*atn(l)
200 for i = 1 to 6
220 1.' = 1.' + 0.2
230 print#l, " "
2l,r0 print#l, using "With 1.' = ##.## ftrt;r
260 print#l, " y, ft Q, ft3/s"
280 y = -0.1 + 0.00001
290 for j = 1 to 11
295 y = y + ('.1
300 if y < = r then goto 500
320 0 = 7.53*(2*y-(2-pi/2)*r*r)ft(5/3)/(2*y-(l,r-pi)*r+2)ft(2/3)
3l,r0 got.o 600
500 th = 2*atn((r*r-(r-y) ft 2) ft O.5/(r-y))
5200= 7.53*(r*r*(th-sin(th))/2+(2-2*rl*y)ft(S/3)/(2+(th-2)*r) ft(2/3l
600 print#l, using" ##.## #.###~ft~~";y,O
620 next j
6l,r0 next i
***********************************************
** This program calculates the flowrate as **
** a function of depth for various values **
** of the radius of curvature of the **
** corners. **
***********************************************
T,tJith r =
y, ft
0.00
0.10
0.20
0.30
0.l,r0
0.50
0.60
0.70
0.80
0.90
1. 00
-0.00 ft
0, ft3/s
6.987E-08
3.0l,r5E-Ol
9.123E-Ol
1.700E+00
2.613E+00
3.620E+00
l,r.699E+00
5.835E+00
7.017E+00
8.236E+00
9.l,r87E+00
With r =
y, ft
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1. 00
0.60 ft
0, ft3/s
2.806E-08
1. 79l,rE-Ol
6.220E-01
1.281E+00
2.122E+00
3.107E+00
l,r.198E+OO
5.357E+00
6.566E+00
7.815E+00
9.096E+00
10-53
With 1.'
y, ft
0.00
0.10
0.20
0.30
0.l,r0
0.50
0.60
0.70
0.80
0.90
1. 00
1. 00 ft
0, ft3h:.
1.589E-I0
7.158E-02
3.112E-Ol
7.2l,rl,rE-Ol
1.305E+00
2.0l,rlE+00
2.918E+00
3.919E+00
5.022E+00
6.207E+00
7.l,r51E+00

Qvsy
10
9 /.
8
7
6
I/) --r= 0.00 ft
-..,
t:: 5
d
4
- - - r =0.60 ft
- - - - - .r = 1.00 ft I
/
v", ,,'
j; ",
3
2
1
0 ~
~ ............. .--...-:.-:--... " ...
0.0 0.2 0.4 0.6 0.8 1.0
i
I
I y, ft .
I
i
/O-EJ.f.

10.61.* Water flows in the fiberglass (n = 0.014)
triangular channel with a round bottom shown in
Fig. PlO.62. The channel slope is 0.1 m/90 m.
Plot a graph of flowrate as a function of water
depth for 0 :0:; Y :0:; 0.50 m with bottom radii of
r = 0,0.25, 0.50, 0.75, and 1.0 m.
~~-1
r 0.5 m
(b) I/sslJme y~ 1,
" 90
0
/
/V"
FIGURE PIO.62.
A-= III +Al , where from £X(lIfJp/e 10.5 with er:: 1/
I j
A = (2r)2.(1l - sin 1I) _ (TT-2) r2. .~-------,F
I B 2. 2.- ~ I
Also, for the frapezoidal area
iJ2. (see fiqvres):
1t2 =If1/2r +2[Yf(l'i-I)r]][y~(J-k)r]
= [Y+(~-I)r][y-(J-*)rJ
ThIJsJ
11=[y+(~-J)r][y-(J-~)r] + (7T~2.) r2
Also}
P = 11 t ~J where R= "fr qnd ~ = 2 (YZ)[Y-(l-vi-)r]
Thus,
p: rr +212[Y-(J-y})r]
10-55
(.3)

/0.624 r (con Ii)
Thus, wifh Rh :: 1- Elf. (I) hecome.s
f:) I 115/3 I ( ~
{ = O.OJJf n p~ 0.00//J)
or 115/3
Q= 2,38 p~/3
Hence) for y~ (I-;=})r calculafe 6i from £'1' (5) J lAIdh t1and P from
frs. (3) and (If),
(5)
Thtls; p/o-lQ~Q(Y) for O~y~O.51h wdhr=O 0.2050.50 O,7S/,Om
.L J I J J J
wI/ere .S"/:
Q 81a (e - sin e):3 . h () r-It f (..La) ::: 0.7SO r e'-/3 wil CQ.J.a -::.:...r- i y;!f 1- y,Jr
and 5~
b) Q=2. 38 112,~P ....3
with f) ::[Y +(/i-/)r][Y-(J- &)r] + (1T~2.) rand
p =fr +2y£[y-(J-~ )r] if y>(t-Vr)r
These restI/t.s are co/cfJla/cd and plolted below usifl9 PrOrrqm P/O.#62.
100 cls
120 pr int.# 1, "****************************** ***********'* '* '* '* '*:t:"
130 print#1, "** This program calculates the flowrate in **"
lLrO prir.t#l. t!** a vee shaped open channel with a rounded Oli"
150 print#1, "** bottom. '*:t!1
160 pr int.# 1, "* * ** * * ** * ** ** ** ** **** *** ** * * ** ** ** * *** * * * ** * * * *"
190 pi = Lr*atn(1)
195 r2 = 2"0.5
200 r = -0.25
210 for i = 1 to 5
220 r = r + 0.25
225 l)rint#1, " "
230 print#l, using "l>Jith r = ##.## m";r
21,00 print#1, " y, m Q, m3/s"
250 y = -0.05 + 0.000001
260 for j = 1 to 11
270 y = y + 0.05
2S0 if y < = (1 - 1/rZ)*r then goto 1,000
300 A = (y+(3/r2-1)*r)*(y-(1-1/r2)*r) + (pi-2)*r*r/1±
320 P = pi*r/2 + 2*r2*(Y - (1 - 1/r2)*r)
31,00 Q = 2.3S*A~(5/3)/P~(2/3)
360 goto 500
1,000 th = 2*atn( (r*r - (r - y)-2)~0.5/(r-y))
1,020 Q = 0.750*r-(S/3)*(th - sir.(th))~(5/3)/th-(2/3)
500 print#1, using" ##.### #.###----";y,Q
510 next j
520 next i
/0-56

/O.62~
***********************************************
** This program calculates the flowrate in **
** a vee shaped open channel with a rounded **
** bottom. *Jl'
***********************************************
With r = 0.00 m
y, m Q. m3/s
0.000 1. 194-E-16
O.OSO 4-.038E-04-
0.100 2.564-E-03
0.lS0 7.559E-03
0.200 1.628E-02
0.2S0 2.952E-02
0.300 4-.800E-02
0.3S0 7.24-0E-02
0.4-00 1.034-E-Ol
0.4S0 1.4-1SE-Ol
0.500 1.874E-Ol
0,6 ---~~'--'-'-----
0,51-----
0.'+ -----------
With r =
y, m
0.000
0.050
0.100
0.150
0.200
0.250
0.300
0.350
0.4-00
0.4-S0
O.SOO
f
O,Sm
t
0.3 ------------------------ --- ____ C_~~·~~-~-"
0.2 i--------------------------------
1.00 m
Q. m3/s
3.4-36E-13
5.120E-03
2.263E-02
5.361E-02
9.839E-02
1.S69E-Ol
2.290E-Ol
3.138E-Ol
4.113E-Ol
S.21SE-Ol
6.448E-Ol
0,' ------~f
O.5m
t
0
0 0.1 0.3 O.Jf 0,5
10-57

/0,63
10.63 The cross section of an ancient Roman aqueduct is
drawn to scale in Fig. PlO.63. When it was new the channel
was essentially rectangular and for a fiowrate of 100,000
m3
/day, the water depth was as indicated. Archeological evi-
dence indicates that after many years of use, calcium carbonate
deposits on the sides and bottom modified the shape to that
shown in the figure. Estimate the fiowrate for the modified shape
if the slope and surface roughness did not change.
I I ! I
a 0.2 0.4 0.6 0.8 1.0 m
II FIGURE P10.63
where from measuremenls on the fi9/Jre
Ao'X 0, 6m (I. 25hJ) -:: 0,75,,/ and
R Ao O,751'n2. 0 306
ho = Pc ~/.25m +2.(0,6",)::' m
7nIJsJ
from £'/_ (jJ:
_ )i '".<5 1r:-1.16 - no (O.7S)(O,3()6) ,SOD
or
f.rs: = 3.1f1
frlodified:
J< 2./3 1'-;:-
Qm == n", flm Rhm VSom
from fhe fi9vreJ
Am -;::. I m( /.2111/ O.3b1) -;::: 0.75m2.
and.
R - Alii".. O.7S m'J. :: O.2B3m
11m - P,n"'" (0, SSm +/.Im +IhI)
10-58
'. . : .' .
.. . '
surface
Original water
surface
Calcium carbonate
deposits
~O.6m~
ori9inai
o.ssm
modified
/.2-hJ
a.3m

10.61/-
·~------6m--------4
10.6'" The smooth concrete-lined channel shown in Fig.
PlO.64 is built on a slope of 2 m/krn. Determine the flowrate
if the depth is y = 1.5 m.
0.5 61=__> 3 m------1
y''''
,.... - - •• 0 • • , •• ',,,, ••••
FIGURE PIO.64
Q
_ L IlD~4S~ L _ _ 2m
- n nnh 0 J Wnere X-I) So- looom == 0,002.) and from TaMe jO./
n= 0.012-
Wdh Y =1,5mJ ,4= (.3m) (o.,sRJ) +f. (3m fom) (J,Ohl ) == 6/11'-
ond ok
p= /,Smf.3m fO.Emf (1'-+32
)2. m =8.J6m
ThlJs) ~ = ~ =B.~;; == 0.735 "', tJlld Ef. (I) 9ives
Q= 0.;12. (6)(0. 735//
3
(0.002)~ = 19.2-tf
/0.6.5
10.65 Determine the flow depth for the chan-
nel shown in Fig. PIO.64 if the flowrate is 15
m3
/s.
(J) .
Q== ~ fJR;-1 SO~) where X=J J So:= 3~:m =0.003) and froh17aJJle/(J,/ 1I=0.01J..
A/so, fl:: 3y'" f [3(y-O..5)](Y-O.5)= ~ y2..f-;' Y+ ~ ,....~3(_Y_-O_.5_).,-;.--r-__--.
tJnd ~ ~ 2.]k :Y-O.5 Tp= yt3to,.5t
L
(y-t) +q(y-f) 2. :+
:: y +.3.5 of- '{iO'(y- 0.5) :: ~/6 y +/.92 L..01-.5_f1l_~
HenceJ wifh Rh =J and rr::/S 1f-3 we obt4in
1 2. .5/3 I k
15 =1'1 0/2 (/.5y fl.Ey +a37S) ( )"A (0,003)'"v. Jf.//'yt/.l:J. '3
or 0 I/- 0.1
2.01f(lf.16y+I.92)· -1.sy:J. -/•.sy-O.37S=O :;:F(y) F
Tria/and error solufion for F(Y)~ 0 0.05
Y F
1.20 0.085.5
1.22 O.OOIfI
/.2'1- --0,0786
o +----'--_~---L-_
Thus) y ~ J.22m
Nofe: Since y < 1.5m the wafer-o.os
does nof conrad the Jef-f
verrica/ wall -0.1
/o-Sq
Y

/O.66~
10.66* The cross section of a creek vaHey is
shown in Fig. PlO.66. Plot a graph of ftowrate as
a function of depth, y, for 0 :s y :s 10ft. The
slope is 5 ftlmi.
z = 530 ft
30Y-120
yt1-6
Hence} ~ ~
Q:::: ,.lf9 (8Y)[~ (O,OOOqIf7)~ t I. 1/-9 (3 0Y-I1-D) [ 30r-IJ.O] (o.OOO9'f7~0,03 yt/2J 0.05 yt2.6·
or_ ySI.3 (Y-/f)5!z, .3
Q- LfB.9 ( 12.)2/.3 + 265 ( )'"A for 'f~ y~effJ where Q~~yt yt26 !I
Y-8 y-l/
(c) For y~ 8 ft :
Q= Q, +4>2 tQ3+ QI/-
with n,=o.o3, n2.=o.o5In.3=O.03.sJn~=O.15 eN
Also, A,:= 8y J 1J2. = 30(Y-9) I A-i =90(Y- 8) , find 11/1- =50 (y-e)
gild
R,=e+etJ.f=2o R= 3()tJf=3tf fj=90 +(y-8)= y+82 ond
J ) /10 '
~=50t(y-8)= ytif2 .so thol wilh Rni
=11,: and
(c.onJ-t)
fO-60
(3)
y-s

Q.:: Lntfj /I. Rh~ S ~ == I.n'f!./I: RJt~/8 ( o.OOOqtj.7)1£:; O.Oll-59 fl. D. l.h
t t t l Q ~ lilt n; , ''th'
ThilSI SA 51 [ ~
Q
_ O.Olf5q (8)') ~ + O.(JIfsq [30(y-/f)] + o.o~sq QO(y -8)]
- 0.03 2.0'h 0.05 3'f""'3 0.035 (y+82.),,13
+ O.0'l-5Q [.50(Y-B)]5/
3
0./5 (y tlf2.)'4
or
~ .s~ (y-8)5/.3 (y- e)~
Q::. 6.6'1-y +2S.3(y-If) +2370 (y+BZt/3 + 208 (y+'I2.)~ ForY~&J
Q~1f
For o~ y~'O plol Q::Q(y) from £'(..5. (2.)J (3)J O{l (If')'
ProqraPJ P/01166 sh()wlI be/ow wa.; used fo ca/cv/oie the res()/ls,
100 cls
110 open "prn" fOT output as #1
120 print#l, "*************************************************"
130 print#l, "** This program calculates the flowrate in **"
1'*0 print#l, "** the creek at various depths of flow. **"
150 print#l, "*************************************************"
160 Y = -0.5
170 print#l, " "
180 print#l, " y, ft
190 for i = 1 to 21
200 y = y +0.5
Q, cfs"
210 if y < = '* then goto 300
230 if y < = 8 then goto '*00
2'*0 goto 500
300 Q = 30.8*y~(5/3)/(y + ,*)~(2/3)
310 goto 600
,*00 Q = ,*8.9*y~(5/3)/(y + 12)~(2/3)+265*(y - ,*)~(5/3)/(y + 26)"(2/3)
'*10 goto 600
500 a = 5/3
510 b = (y-8)~a
520 Q=6.6,**y~a+25.3*(y-,*)~a+2370*b/(y+82)~(2/3)+208*b/(y+,*2)~(2/3)
600 print#l, using" ###.# #.###~-~~";y,Q
700 next i
*************************************************
** This program calculates the flowrate in **
** the creek at various depths of flow. **
*************************************************
y, ft Q, cfs
0.0 O.OOOE+OO 5.5 1. 765E+02
0.5 3.559E+00 6.0 2.2,*5E+02
1.0 1.053E+Ol 6.5 2.781E+02
1.5 1. 9,*3E+Ol 7.0 3.366E+02
2.0 2.961E+Ol 7.5 3.997E+02
2.5 ,*.072E+Ol 8.0 ,*.669E+02
3.0 5.252E+Ol 8.5 5.872E+02
3.5 6.,*86E+Ol 9.0 7.607E+02
,*.0 7.761E+Ol 9.5 9.755E+02
,*.5 1.011E+02 10.0 1.226E+03
5.0 1.350E+02
(conJi)
10-61
('1-)

/O.66~ (conli)
The flowrafe os a funcfion of de plh is pJo/fed beJow.
------ ---------i--~----<--~--,----~ - - ----r- -
1200~------~--------~--~----~-------4--------~
- -~---.------- ___._~ -.-.---~~----_+--------<. -------·-l~---. ~- ------
-----_~~__~===~E~~==---=~===3~~~~==--~--. ~J=~~_=~:=_=:=_~~-~~l -_~--_=_
, ,
--.------------+--- - - --.,------ ---~'-- ----------i----<---- -- .----
- -- .-----. - - -- ____J.________ ----"..-- --.---r-------------- --~:~-- ------- -_.
SOD
------ ------------i---~--~,~---+--------+------~-- ~
-------~--+--. --:~-
----~~--==--== ==--1--- =----=-=-1 --+-----;----
600 ~------~--------_4--------_+----~--~~------+
-- - - ------ -1--------'---
-------1---- ----------
----------+-------~----i---------- -
~oo~-------+--------~--~----+---~-+~---------+-/_--+_________L _____j
- - - - - - - - - " - - - - - - - 0 - - - _•• ----_.---t
------:- ------j
-~~-----+I-'~--I, t
200r--------+--~----+---~--~----~--4---~--~
"-----. -f---~-- --- - - ---~-.. ----.-
-----1------
10-62.

10.67 I
·,1--'-----6 m - - - - - 4
10.67 Repeat Problem 10.64 if the surfaces are smooth con-
crete as is indicated except for the diagonal surface, which is
gravelly with n = 0.025.
:.1
l.d~:loo::"~-.-,'-"-.,-.,-(j)(j)~,--..--l"''---®---l... /cref e
i~~ /
0.5 ~·~t.=.=.=....,....,~._m=.=..=..:j
FIGURE PIO.64
Q K"/: ~ )( LJ RZh ~
=Q,+Q2, = tij II,Rhl
:3 50" + n;, f72.. h~ So J where X=IJ 50::: 0.0021
(fJ
nJ=:o.ozs,(lnd from TabJe jO./ n,.=o,OI2.
A/SOI II,:: t(I.Om)(3f11)== 1.50 hi" J P, =(1.0 2.+ 3.02.)~ == .3,/6m
D ,4, I.SOm"
or Ilh =: - = = O.if.7S In
rJ
J R 3.16 hi
an
A2 =(.3I17)(J,5m)= 'f.Sn/ J ~ =QSm +3117 +1.5", =sm
D 112. 'I. Sin'"
or nh,,::: Ii = 5 In =0,90hi
flenc8J
from Eq. (/):
I '"..3 Js. I '"h ~
Q= 0,02.5(1.50)(0.'1-7.5) (o.oo2J + 0.012. (#.5)(0.90) (0.002)
or 3
Q= 17.3 f-
3
Note: With al/.svrlflce.s concrefeJ Q= 18.z.!f (see Problem /O,&lf).
10-63

10.68* Water flows through the storm sewer
shown in Fig. PlO.68. The slope of the bottom is
2 m/400 m. Plot a graph of the flowrate as a
function of depth for 0 ~ y ~ 1.7 m. On the same
graph plot the flowrate expected if the entire sur-
face were lined with material similar to that of a
clay tile.
t------2.5 m----l"
T
-- - - ------------- - --:---:-:---:-:-:--.---:---:----
--'-l-----'«~.v-c,aytile
FIGURE PIO.68
(a) For 0 ~ y -==O.5m: The flow is Ihe SQl11e as fhal in 4 circiJlar pipe.
Rubble
masonry
Thf)sJ from EX4mple /0.5 wilh D=/ fI1} X=/} and n =O.O/~ (7Q6/e /0.1);
J< k D% (e- sinel.1.3 I ').111 1i (1)fV3 (8-sinell.3
Q= n So:z. B ('ft~ e2/.s =O.OIJl. ('l-oom) 8 (Jf?13 ei /3
or ~ 5~
Q=0 251 (f) -SinS) .m! where fj"'rad
. e2/ 3 S J
and 8 = 2 cos-,(o.s-y)
O.S
(b) For Y~ O,Sm: LI4--- 2, 5m
y-O.5 (2.)
Q X 11 D"~ ~ 'Ih r ~-----,- -----,= n; n, nh, SO 2'J W' n, = o.o/~} j.!m..l (J) ..~m-l
II,:: ¥(0.5tn)'-=o.39311J2., !?=7r(0.S)=/, 57m Sf) thaf
R -==.AL =O.3Q3111:L =0.250 m
hi P, 1.57m
ThIJs}
(;,= o.~IJI.(O.3q3)(O.2.50t.&(If~O )'1. :; O.787.!f3
Also,
.K 2~ li.
Q2 = n" fJ2. Rh2. So ) wilh n1 = 0.025 (see Toh/e /0./)
A2. = (2,5m) (y-o.s) =2.5 Y -1.2.5 and 8. =2. (y-o..5) T 2(~)=2Y t 0,5
lienee, wilh Rh2. = *} E'(. (;;..) hecolTle.s
Q'= ~ ( _ )5/3 , (_2._)~ _
2. 002.5 2.5Y 1.25 ( )~ ~oo - /3,0
• 2y+4S
Therefore) s
Q::! 0.787 +/3.0 (Y-O.5) ~
(2. yt 0.5)'''/3
3
1f for y ~ O,Sm
(y_o.S)514
(2ytO.S)%
Plot Q= Q(y) for O:;y ~ 1.71'11 /)SJn9 Eqs. (I) and (3),
(codi)
/O-6Lf
(I)
(2.)
(3)

If fhe en/ire surfqce were lined with ",c,feria} with n, =n2 = o.o/if,
Etfn. (I) would remain valid. The coefficielJl NJ3. 0/
1
in Eq. (.3) wovlJ
become /3.0(0.025) =23.2 For this caselrO.D/Jf . •
( l~ 3
Q=0.787 + 23.2 y~0.5 ~ -sm for y~ 0.5h1 (4)
(2Y+O.5) 1.3
Thi.s re.sulf i.s also plo-lted U.~. Q) frorn £r, {/J for o~ YS:O'S"I and
Q from £'{. (If) for 0.5 4! y:5 1.7/11). See Proqram P/0#68 below.
100 cls
120 print#l, "*************************************************"
130 print#l, "** This program calculates the flowrate in **"
140 print#l, "** the channel as a function of depth. **"
150 print#l, "*************************************************"
160 dim a(2)
170 a(1) = 13.0
180 a(2) = 23.2
190 for i = 1 to 2
200 print#l. " "
210 if i = 1 then goto 260
220 print#l. "With n = 0.014 for the entire channel"
230 goto 280
260 print#l, "With n = 0.025 for part of the chann8l"
280 y = -.J. + 0.00001
290 print#l, " y, m Q, m3/s"
300 fo~ j = 1 to 18
320 y = y + 0.1
340 if y < = 0.5 goto 500
360 Q = 0.787 + a(i)*(y - 0.5)~(5/3)/(2*y + 0.5)~(2/3)
380 goto 600
500 th = 2*atn(((0.5~2 - (0.5 - y)-2)-0.5)/(0.5 - y))
5200= 0.251*(th - sin(th))-(5/3)/th-(2/3)
600 print#l, using" ##.# #.###~---";y,Q
610 next. j
620 next i
~************************************************
** This program calculates_the flowrate in **
~* the channel as a function of depth. **
*************************************************
With n
y, m
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
= 0.025 for part of the channel
Q, m3/s
7.552E-l1
3.293E-02
1.381E-01
3.089E-01
5.315E-01
7.870E-01
9.837E--01
1. 367E+OD
1.853E+00
10-65

/0,68· (cadi)
10
5
0.9
1.0
1.1
1.2
1.3
loY,
1.5
1.6
1.7
With n
y, m
0.0
0.1
0.2
0.3
O.L!
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.L!
1.5
1.6
1.7
2.y,07E+OO
3.010E+00
3.6L!9E+OO
L!.315E+00
5.003E+00
5.708E+00
6.L!26E+00
7.157E+00
7.897E+00
= 0.01~ for the entire channel
Q. m3/s
7.552E-11
3.293E-02
1. 381E-01
3.089E-01
5.315E-01
7.870E-01
1.138E+00
1.822E+00
2.689E+00
3.678E+00
L!.75L!E+00
5.89y,E+00
7.083E+00
8.310E+00
9.568E+00
1.085E+01
1. 215E+01
1. 3L!8E+01
I
- _...• _.... .1...
in::O.Ol'!- for
_. _.__..____._._.__.________~e.l!flre. channeL...
o,l/- O,B
y, m
/0-66
n= 0,025
for top porfion
of channel .
/.6

10.6q Determine the flowrate for the symmetrical channel
shown in Fig. PlOAO if the bottom is smooth concrete and the
sides are weedy. The bottom slope is So = 0.001.
Q Q
K .1.A k.
:::: QJ +<?,. +-Q3 =Q, +2Q2 I where l :::: n;. fI/ Rht.3 SD:I. with){ ~/, 1.11
Also, AI =(3(1)(11-(1) :::/2 f/2. J 1J2 ::. f(3;I)(lffl) :: 6ff~J P, :::~/~ 4nd ~ =5f1J
fha! D - A, - J2.ff2. -3 ff nnd 0 - A,. - 6 fl2. -/1) ffso nh, - If - If fl - .. nh2 - 7{ - E ff - .~
HenceJ
wilh n,::: 0,0/2 and n,.:: 0.030 (see Table 10./) we obfain:
Q= ~:~;2 (/2) (3/,4 (O,OOJ)""4-(2)~:~~ (6) {J.2),"/3 (0.00014 = IIq ¥3
/0-67

/0.70 I
10.70 Water in a rectangular painted steel
channel of width b = 1 ft and depth)' is to flow
at critical conditions, Fr = 1. Plot a graph of the
critical slope, Son as a function of y for 0.05 ft :5
Y :5 5 ft. What is the maximum slope allowed if
critical flow is not to occur regardless of the
depth?
I ~ II--- /R --~..I
V 1<. zh Ji h= 7f Rh So J were X== J.lfq and from Table 10.1 n= O.O/if.
Also) Rh =J.: .,.01 and with Fr==m ==/ J V= 1fiY
Thv.5J ~
IA'! ( Y )l~ ~ [(2y+/)'fJ 3V32.2. Y := o.O/Sf zytl So: or Soc = 0.0028Jf Y
EquafioIJ (J) is ploHed below. To de/ermine fhe mini/lJlJ/IJ critical oS lope
set daS;c == 0 . That is:
-2h 3
dSoc _(J.) 0 If)[ (ZY+J)IJ] fif(2ytJ) (2)Y -(2Y+J)lJj =0
~ - 3 0.0028 Y y2
Thus, y == -t so fhfff frol11 £C(. OJ
S := 0,0028'1 [ (t,+1)'1 ]~ = 0.007.5 7
oCmin "6
If 50
<0.00757 critical fJoWl cannot occur aiany dep-th.
The followinq volves are ohfained from Er·OJ. !Vole fhal
J [ (2YtJ)I/o]~Jim So == O.0028tf ill) Y = oa and lim So =- 00
y_O c y~O y_DO C
10-68

1--/0_,7_°---1 (con V)
Soc vs y
0.045
0.040
0.035
0.030
0
0.025
0
Cf)
0.020
0.015
0.010
I
I
--~-~~'---'--~~-~~. . ~"~
~~-~--~ ~---"-l
0.005 -~--.- ---------- --·-~--l
I
0.000
0.0 1.0 2.0 3.0 4.0 5.0
y, ft
SOc vs y (expanded scale)
0.0090 .,.----,....---,--....,-----r---....,-----~------__n
0.0088
t
.___________~ _ _ _~'_~~ _ _ ___+__--~~--_-- ____. _ _ _ -----0-----
0.0086
0.0084
g 0.0082 -C/)
-- ~0.0080
.~=~=~~-~~
0.0074 +------r-----r------r-----.-------f
0.0078
0.0076 ---------;----~I;;;;;;;;--
0.0 0.1 0.2 0.3 0.4 0.5
y, ft
/0-6 q

10.7' 1
10.71 Water flows in a rectangular channel of
width b and depth y with a Froude number of
unity. The slope, Son of the channel needed to
produce this critical flow is a function of y. Show
that as y --') 00 the slope becomes proportional to
y (i.e., Soc = CIY, where CI is a constant) and
that as Y --') 0 the slope becomes proportional to
-113 (. S C / 113 h C· )Y I.e., oc = 2 Y ,w ere ~ IS a constant.
Show that the channel with an aspect ratio of
b/y = 6 gives the minimum value of Sllr'
l4--b ..
f
r
(/)
lis y -0) (ZVb)"- f so Iha1
S (- n9 ) (J1)l - c,.Oc --. X b'll.l Y - yY.i
To de/ermine fhe minimtlm Soc J calculafe ~~c =0 from £q.(J):
dS -2.;3 3 J!-
oe =( ng ) (1-J[(2 ytb)"'] ['1(2Yfb) (2.) Y- (2ytb) ] == 0
dY X b'l/.3 3 Y Y2.
or 3
(2y +b) [BY - (;;.y tb)];O
Thvs)
y= ~
/0-70

10.72 I
10.73 J
10.72 Water flows in a rectangular channel with a bottom
slope of 4.2 ft/mi and a head loss of 2.3 ft/mi. At a section
where the depth is 5.8 ft and the average velocity 5.9 ftls, does
the flow depth increase or decrease in the direction of flow?
Explain.
her S =hL - 2.3ft s = '1-.2 ft
J w e f T - .5280# J 0 52-80H,
and 1T= ~:::: 5.9 g. ::Q Jf32
(g y}li [(32.2.~)(5.8fl)J ~ .
Thus)
(2.3 -~.z)
dlx - Is;eo t =-0. OOOIf'lZ <: 0 The flow depthdecreQse.s in flokl direclittJ.
dx - O.lf32
There is less heQa JrJSS fhan chonle in e/evotiof) for fhis .suhcrifico/ f/uw.
The flUid speeds up a/ld qefs shallower.
_'V , '(3)
10.79 Water flows in the river shown in Fig.
PIO.73 with a uniform bottom slope. The total
head at each section is measured by using Pitot
tubes as indicated. Determine the value of
dy/dx at a location where the Froude number is
0.357.
~(--l-):"- -------------' J. 1-1(4)
~ " - (~--------------- -- ~
~?77;77i~~~~"d/,J (. (2)
Zl = 620.1 ft lZ3 = 628.3 ft
x2 - Xl = 4100 ft
Z2 = 618.7 ft
Z4 = 625.0 ft
FIGURE PIO.7S
dy Sf -so J where from/he fi'lvre S =~ = 23 -i!~ :::. (628.3-62.5.0)(+
(]X=: I-Fl' f J X,-X:z. ¥/ooff
or ,... =8 os 10-'" d S .::: Zl-~:l = (620,J-t/8.7){f =3 li/X/OL{.
..::J; • x an 0 R If/oof! .-,.
Thus,
dy
;rx=
10-11

10. 7"1- I
! I
10.7/f Repeat Problem 10.73 if the Froude number is 2.75.
/0.7S I
ft:::: 5f - So I where from fhe fi9vre
dx 1- Frl
S -8 -If J$ 2/-Z~
or f - .05 x/o tina (J= J =
ThlJs,
21 = 620.1 It
23 = 628.3 ft
X2 - X] = 4100ft
"2 = 618.7 It
Z4 = 625.0 It
FIGURE PIO.73
d -it 'f /-'11 = 8.051./0 -3. Ix 0 __ 7. 07X 10-5
dx /-(2,75)2
10.75 Assume that the conditions given in Fig.
PlO.73 are as indicated except that the value of
Z4 is not known. Determine the value of Z4 if the
flow is uniform depth.
:5:b ~rZ t . ,/ ///////)'/7
2] = 620.1 ft ~
23 = 628.3 It z2 = 618.7 It
For uniform flow
Thus,
X2 - _'I = 4100 It
FIGURE PIO.73
~:::: Sf - S~ =0 or S~ = So
" /-Fr J T
Z3 - '2'1- _ 2i -Zz (
J - i J or z~ == 23 +Z2. - 2, = 628.3 +6/8.7-620./) ff
or
%'1 == 626.9 ff
10-12

10.76 I
10.76 A 2.0-ft standing wave is produced at the bottom of
the rectangular channel in an amusement park water ride. If the
water depth upstream of the wave is estimated to be 1.5 ft,
determine how fast the boat is traveling when it passes through
this standing wave (hydraulic jump) for its final "splash."
t £~,Off t
-!'!. I.S If =)1 , Y2.
7.f777 77 7777711t
*:::dr-I+/1 +8 Fr,'" ]
or / 2.0 U+1.5if) ::: ..J..[_I +jl+8 Fr, '" ]
~. I.sfl 2. I V,
Thus, Fr,::: /,971 or since FiJ = ~f})'I
10. 77 I
~ ::: Fr, #l = /.97 (32.2!t) (1.5 ff)
10.77 The water depths upstream and down-
stream of a hydraulic jump are 0.3 and 1.2 m,
respectively. Determine the upstream velocity and
the power dissipated if the channel is 50 m wide.
y2. I. 2m I r. / fi 2.'] .Y, .::: 0.3 m ==2"[1 + / +8 r, or Fti =3./6 Thus, sInce
if follows that V,:: (.3.16)[(Q.8J?) (O.3/h)] ~ == 5. t;2 !f
The power dissipafed is qiven by
-p:: lQh where !:E:.:: /-~ -I- Fr/' (I - (4-t)r;.. J Yt Y, 2. 12
0=(0.31>7)[1- ~.~: + (3~6)1(1-tt:::n] = O.SO'fm
Also, Q= A/~ :: Ytbl1:: (o.3m) (SO/H) (5.1f2!f-) :0: S/.31}3
Thll.sl
;1)::'(9.8 ~)(81.3.t;-3)(O.50¥fn):: ~Ol k~'m = '101 kIN
/0-73

/0.78 I
10.78 Under appropriate conditions, water flowing from a
faucet, onto a flat plate, and over the edge of the plate can pro-
duce a circular hydraulic jump as shown in Fig. P10.78 and
'idl!i1 '10.6. Consider a situation where a jump forms 3.0 in.
from the center of the plate with depths upstream and down-
stream of the jump of 0.05. in. and 0.20 in., respectively. De-
termine the flowrate from the faucet.
For a hyriratJ/ic jump:
~ ::= i[-I +1{i+8Fr/· ] or
..
-0J"~
, / ' ,..,....
i
II FIGURE P10.78
== f £1 +Y1+N F;?-]0.2.0111.
o.osin.
.so fhai Frl ::: 3.1t
ThIJsJ
V, = 3,161.32.·:z.f{(o,os//;l.)11 ::: I. /b!J.
.-,~3in. 4/J~mp (2)
I fI)i····· pL ...•...
··~.l:,:::;;;;:,:,::;:;::i::*::;:;:::::.:::::;:::::+'t;;:::;;;;;;:::·.':H'···
0.05 in. 0.20 in.
a/Jd ff3
Q== AI ~ ~ 271R, Ii ~ :: 271(ff. f.I){qi~5 11) (/.It {t) ::: 0.00759 s
IO-7Jf

/0.79 I
.10.79 In order to have a hydraulic jump, the flow upstream
'of the jump must be supercritical. This implies that a wave made
by a disturbance upstream of the jump cannot travel upstream; it
gets washed downstream (see Video VI0.(;). Show that for a hy-
draulicjump in arectangular channel, the Froude number upstream,
- Frl' and the Froude number downstream, Fr2' are related by
-2 8 Frf
Fro -
2 - [(1 + 8 Frf)l/2 - 1]3
Plot Fr2 as a function of Frl and show that the flow downstream
of a jump is subcritical.
2.
Fr2
2
" [I 8 Fr: y Thi.s resuH iJ p/o/-/ed below.
J +8Fr/ - J
;>77Y7777T777;;;
(I)
(z)
(3)
, 3.5 T-------------------I
FrJ Fr2.
0.5 2.26
/ I
2 O.S'f7
3 0.1/10
if O.33Q !
5 0.296
I
3.0
2.5 ~---
2.0 ---.~
.!'
LL
1.5 - -- - -.-- ~
1.0
0.5 ---.
-- -~- - - .--
:
I
I0.0 -I-------,-----r---..,....-----,----i
0.0 1.0 2.0 3.0 4.0 5.0
Fr1
!
i
I
Nole; To hQve ajump we
mils! have FIf> I. FrOIl?
the 9roph FI3. ~I if ~ '>/.
I
~----------------------------------~
NOTe; Fr/ =/ rp"ve.s Frz =/. IJ/so (;QI1 show from £9' (3)
dFr2./rlFr, <0. Hence" Fr2<1 for Q JUMP,
/0-75
that

/O.BO I
10.80 Water flows in a 2-ft-wide rectangular
channel at a rate of 10 ft3
/s. If the water depth
downstream of a hydraulic jump is 2.5 ft, deter-
mine (a) the water depth upstream of the jump,
(b) the upstream and downstream Froude num-
bers, and (c) the head loss across the jump.
Q i13 I " f:l.
(a) < - - 10 ;s - 2 0 Ii so fhai Fr2.= V:z. ~ = ~ .s
2. -7[; - (2.5f1)(:;.ff) - • S (g.y,) ,. [(3;2..2~,.)(2.5ft)J~
or
Fr2, =: 0.223 Obfain Fri from fhe reslJH of Problem /0.7Q : ..
Fr:2. :: 8 Fr,z or (0.223) 2r(J +8 Fr,2.)~_1]3- 8 F~/:z. =0 == f( Fr,)
:2 [(JtBFr,2.)V2.-1]3 LI
Triol and error solviion for f =0 ~
f Fr:::8.10/5 -Fr, f
8.0 -7,3q
8,/ - 0.26
8,2 7.25
o I V I Fr
8.0 .1 8.2. I
Thvs, Fr," 8.10 =(v, )V~ , where
l,
IL3 gY,
II - - JO s - 5 fh fv, - I - (2. ff)}'I - ~ so "
-s f-
8./0= (ft)
(32.2. Y/)~
Hence,
Yt = 0,228 Hand .1:: 0.;29 = 2.1.9#
(b) From pari (a) Ff) = 8./0} Fr2, == 0.223
(c) A/so,
hL ::: r;[I - 2l, + !if(J -(it)]=O.22Bff[l- ~;;8 + (8~O)(J-( o-;~ey·)J
or
hL= 5./£ f+
Ir Or could use ~:: f [-I +~ J+B Fi/] wilh Y:z. =2.sfl so fhat
/1 :a. I I 2. 2.
(I)
5 +y. = ~-Vr-/-+8-Fi-f--:->2.' /Vow wdh Fr,2.:: J1.. = CV (by,)) :: (JO/(:z.X))
I I I .) J 'JYt 'J ~ 32.2 Yt I
or L. 2 _ 0.776 we ~bf4in
rr, - v3 J
1/ .k
5+}j a:: y~ [I -1-8(-°·;:6)]" .By sqlJQri"9 both sirles ofJdsill1ph!JilJy we
aMain X:L+2 ,5X -0.62-/;:0 which 9ive.s )1.:0. 228 {las ahove.
/0-76

/0. eI I
10.82 ]
t / f V~
10.81 A hydraulic jump at the base of a spillway of a dam
is such that the depths upstream and downstream of the jump
are 0.90 and 3.6 m, respectively (see Vidt'o V10.5). If the spill-
way is 10m wide, what is the flowrate over the spillway? ~ -}'J::O.9m Y2.':::3.6rn
777p 7 7777}7 / 7 /7
Y2. I [I 2,'] 3.6m I [J £]Y, ='2 -} +v 1+8Fr, J or o.qm == 2: -I +- 1+8Fi;
Hence) ~::: 3. 16 J hur Fr, = (fJ~Yi so that
~ = 3.16 [(9.81 qi)(o.'I/11~Y2-::: 9.3Q !}
ThuSJ
3
Q=IJ, V, :::: by! U=(I o.Om)(o.QI7l)('I.3Ql}) = 8.tt,5 ~
10.82 Determine the head loss and power dis-
sipated by the hydraulic jump of Problem 10.81.
b =100m
hL= Y,[I-~ +1f?(J-(~t)] I where from~::: ;':~=i[-H-JI+8F"'{I.'J
Hence} Fr, =3.16 so fhat
h :::(0 q )I1- 3.6/11 -#- (3.16 )~(I _(o.QI1l)?)J::: I 51
L • m O/lm 2. 3.6m J .. m
.J,. [ ,,%.AIso) 'fJ::: rQhi. J where v,::: (9Y)2. 0::: (q.81:',.)(O.qm~ (3.16)~ Q.3Q:
HenceJ
7'= (9.80~)UO.qf)J)(/Oom)(q.3q~)](J.51 m}= 12J500kN;111 == /2.J500kW
10-77

/0.8.3
10.8.3 Water flowing radially outward along a circular plate
forms a circular hydraulic jump as is shown in Fig. PIO.83a.
This is shown easily by holding a dinner plate under the faucet
of the kitchen sink (see Video VI0.ll). (a) Sketch a typical spe-
cific energy diagram for this flow (see Problem 10.12) and lo-
cate points I, 2, 3, and 4 on the diagram. (b) Which of the wa-
ter depth profiles shown in Fig. PlO.83b represents the actual
situati~n? Explain.
, (1) (2) (3) (4)
(a)
-__---'I
-
(b)
iii FIGURE P10.83
From Problem /0./2 fhe. specific ener9Y dio'lram for this radial flow is
shown below.
y
/
.~Uj,~~iJical
lncreas~flI' ' ,
~, ••.. • .supercriticaI
• . G
/. /fJcre4Sln9 m
/increasin9 r/
E"
Up.slream of fhe jump fhe flow mv.sf be Slipercr/fica I so (I) and (2) are
localed, Enerpy is consef'teJ- £, =E.2., The deplh decreases from (01rJ(21
1nthe jump enerqy decreases - £.3 ~£'I- < £~. The flow /s stJbcrdt"co/
dOlJlns-freo.m of fha jfJmp and fhe deplh jncreases. (See 4he G1hove qraphJ
Th/)~ fhe flow is like fhe {o//owin,:
/O-7g

IO.Slf I
10.84 Water flows in a wide finished concrete
channel as is shown in Fig. PlO.84 such that a
hydraulic jump occurs at the transition of the
change in slope of the channel bottom. If the
upstream Froude number and depth are 4.0 and
0.2 ft, respectively, determine the slopes up-
stream, SOl' and downstream, S02' of the jump to
maintain uniform flows in those regions. The jump
I can be treated as a jump on a horizontal surface.
FIGURE PIO.8Jf
For uniform flow V== *R;1.3 5,;1-',. J where X-::;/·tf.9 and for a wide
channel R = A::: yb ~ Ysince b»y ~
h P 2.y+ h i ; _ I
From Table /0., n::: 0,0/2 ~I..-=--=--b-------;---------'+f.J
Upsfream of rhe jvmp Fi;:: ( Vi)!?, :: 'f so thai,y,
v,::: (~Yi)~ ~ ::[(32,2~)(O.2ft)]J-i(If) =10.2 ~
Hence, frrJm ft{. (I):
10,2 ::: 1.'1-9 (o.2.)Zh S Y'2. .s - 0 0577
0.0/2 0, J or 01 - ='===
Downs/ream of +he jump Y2. == f[-)+JI f8Fr,2.']
or y,
Y2
=(i) (0.2 fn[-J+JI +8 (1f)2.' ] =I. 03~ ff
ThvsJ AI V, == 1)2 ~ or v,.=~ V, =(J.g3~~) (JO,2 ~) == 1.97l
so -Ihat £". 0) 9ives
1/17 =:,'o~~ (J.036{~ So~ J or Soz::: 0.00021./-0
IO-7q
(I)

10.8514- I
10.95* A rectangular channel of width b is to
carry water at flowrates from 30 :s Q :s 600 cfs.
The water depth upstream of the hydraulic jump
that occurs (if one does occur) is to remain 1.5
ft for all cases. Plot the power dissipated in the
jump as a function of ftowrate for channels of
width b = 10, 20, 30, and 40 ft.
t / ,
VI - Yi=1.5ff Y:L ~
774/ 717 777~7771
-po: rQhL J where hL " Yt [1- (~) + f (J - (f,J)] (J)
and
~ ::: ~[-J+~I+8 Fr~'] J provided F"r,?;O (2.)
Also, I
,- - Y, h I, - Q - Q fh .1
rr, - (9 >:)Y:& J were ,- II, - /.5 b so Q7
C = G-£J =O.OqSq Q LJ r £. 0)
10 [(32.2ft)(J.5fl)]~ b nenceJ rrom tt·
hL = (I.5J[J -(~) +(0. 00if60)(Z}'O-(f;.t)JffJ
where b"'flJ
Q,.,if (3)
and from E£(. (2.)
~ = i[-J+ (! +O.0736(~/")~J (IJ)
For the given values of plof -p from
p= 62.ifQhL
ff:s'lb for 30~ Q tr600 ¥ (5)
Nole; If Fr, </ fhere is noivRJp and p:::o. Frf)I')J above) J'h:::/
when Q= O. 0~5q = IO.lI- b (I)
Lei (f, == flowr"fe when Fij =/. from £'/- (6) we oblain
bJ f+ QJ J -¥3
10 lOll
2.0 208
30 3/2-
~o ¥16
Wilh h::: /~ 2.~:J~ or Jl.off calculate and plot -p from:
0) p::: 0 if ~<.QI
b) 'P:: 62.# QhL ~J where ohloin hI. from El(. (.3) wdh
~ from £,/-(,,) if Q/~Q ~ 600ij-3
The pro9ram and rcsulfs are 9iv81J he/ow. (See protjrdm P/o#85)
(conI/: )
/0-80

/0.85·' (conJi )
100 cis
110 print.# 1, H*** *** *** ******* ******* ** ****** **** *** *** *** *** H
120 print#l, "** This program calculates the power **"
130 print#l, !!** dissipated by a hydraulic jump for **"
1l,t0 print#l, H** various width channels and various **"
150 print#l, H** flowrat.es. **"
160 print#l, "***********************************************"
180 b = 0
190 for i = 1 to l,t
200 b = b + 10
220 print#l, " H
230 print#l, using "With b = ###.##H;b
2l,t0 01 = 10.l,t*b
250 print#l, using "If 0 < ###.# then P = 0 (no jump possible)";Ol
260 print#l, " Q, cfs P, ft. ibis"
270 Q = 50
280 for j = 1 to 11
300 0 = 0 + 50
310 if Q < 01 then goto 360
320 y2y1 = 0.5*(-1 + (1 + 0.0736*(0/b)~2)~0.5l
330 h = 1.5*(1- y2y1 + 0.00l,t60*(Q/bl-2*(1 - y2y1-(-2l 11
34-0 P = 62.l,t*Q*h
350 print#l, using H ####.## +#.###----";Q,P
3 60 nezt. j
370 next i
Sample OVlpVI!
***********************************************
** This program calculates the power **
** dissipated by a hydraulic jump for **
** various width channels and various **
** flowrates. **
***********************************************
With b = 10.00
If 0 < 10l,t.0 then P = 0 (no jump possible)
Q. cfs P. ft. ibis
150.00 +l,t.6l,tOE+02
200.00 +l,t.131E+03
250.00
300.00
350.00
l,t00.00
l,t50.00
500.00
550.00
600.00
+1.l,t32E+0i!
+3.i!27E+04.
+6. 72l,tE+0l,t
+1.165E+05
+1.852E+05
+2.766E+05
+3.939E+05
+5. l,t0 l,tE+ 0 5
With b = 20.00
If 0 ( 208.0 then P = 0 (no jump possible)
Q, cfs P, ft. ibis
250.00 +8.859E+01
300.00 +9.281E+02
350.00 +3.376E+03
l,tOO.OO +8.263E+03
l,t50.00 +1.6l,tlE+0l,t
500.00 +2.863E+0l,t
550.00 +l,t.57l,tE+04-
600 . 00 +6 . 855E+0l,t (con J/: )
10- 8/

/0, eSf. (con'/)
1.E+06
1.E+05
1.E+04
rJj
-.c
.:=
Q..
1.E+03
1.E+02
With b = 30.00
If Q < 312.0 then P = 0 (no jump possible:
Q, cfs P, ft.lb/s
350.00 +2.87~E+01
~OO.OO +3.628E+02
~50.00 +1.392E+03
500.00
550.00
GOO.OO
+3.~9~E+03
+7.039E+03
+1.239E+0~
With b = ~O.OO
If Q <: ~16.0 then P = 0 (no jump possible)
Q, cfs P. ft.lb/s
~50.00 +1.129E+Ol
500.00 +1.772E+02
550.00 +7.201E+02
600.00 +1.856E+03
PvsQ
--b=10ft
- - - b = 20 ft
..... b = 30 ft
1- . - .b =40 ft. t
1000
Q,ft/s
IO-~'}...

10.96
10.@6 Water flows in a rectangular channel at
a depth of y = 1 ft and a velocity of V = 20
ft/s. When a gate is suddenly placed across the
end of the channel, a wave (a moving hydraulic
jump) travels upstream with velocity Vw as is in-
dicated in Fig. PlO.66. Determine Vw' Note that
this is an unsteady problem for a stationary ob-
server. However, for an observer moving to the
left with velocity V•. , the flow appears as a steady
hydraulic jump.
v~ y
------:---:-:::---- --:----
v=o
FIGURE PIO.B6
FOr an ohserver moving 10 fhe lell with speed Vw the flolJl4ppe4rs as shown below.
Thlls) freQ/ fhe flow CIS q Jump wiih
H v, (20+ Vw )
r
,
::;(1 y,)Ji =[(32,:Z¥)(Jf+)] ~
or
Fr, = 0./76 (20+Vw) (I)
(:J.)
III A I IJ I Y2 _ v, _ 2. 0 +Vw
n so) I Y, = 2 V2 J or V - V - V.
and N 2 W
~ =l[-J+/JfBJ:;-/] which when combined with £,(.s.(J)Qna (2.) become.s
20~ v., = ~[-J+jJfB(O.I76i'(:lOfv.,l]
or ( k
2(ZO+V...,)+ ~ =Vw 1+(O.2'1-8)(ZO+Vwt)2.
or 2 2 [ 2
('f04- 3 Vw) = Vw I +(O.2.1f8)(Zo+Vw) ] J which CQn be wrilfen as
O.2Jf8 Vw
Lf
+9.'12 'lw3
+91.2 Yw2.- 2JfO Yw-1600 ::; 0 == f(VIAI
)
(3)
Trial and error solution of£f(.(3) fo/' f(Vw)::o:
Vw .Jf f(Vw )
50
'1-.20 -/87 f
~.2.S -1'30
0 VwJ.}.30 -72.2 #.2-
'1-.'35 -/2.9
¥.JfO '1-7. 6 -50
-100
Thus, Vw.:: If. 36 {t
/0-83

/0.87
10.87 When water flows over a triangular weir as shown in
Fig. PIO.87 and Vidl'O VIO.7, the cross-sectional shape of the
water stream is clearly triangular in the plane of the weir (Sec-
tion a-a). Farther downstream (Section b-b) the shape of the
water stream is definitely not triangular. Explain why this is so.
Hint: Consider the water velocity profile at Section' a-a.
a-a bOb
II FIGURE P10,87
a
......................
As d/sctJ.fJ'ed in Secfio/l /0. &. 2) fl;e speed ollhe wafer f/ow/1J9 overII
weir is a function of h (see the ff?Vre ):
U2. ::,j2?- (h +¥f)
ThlJ~ fvr () 7I'iaIJ9v/ar- weir
the wafer speed af (3) is 9reafel'
than fhlJ+ af (If),
The h"her -/he .speed) the faflflJefl "
.fhe wafer ''.shoots ()vtlJ
t)r 1114/1rIIfJtier
the tic/ion of11'tJlIltr The frtJjecfuf/e.r
of{he wtJJfer.l fhereforeJ al'e ~ .sh()wlI
be/owI
...Yt<V1t
...
(If) "-
"-
,
V.3
(~) .-- - "'
, " ....
'"
" '
The rBsv/f is fA dis/of lion ot fhe ori91/Ja/ rrilJllflli4r crfN..r-secf/()1J
of fhe lIJlJJier sil'eam 4S .shfJwl1 if) lite videlJ,
JO-8tf

10,88
IO.eq
10.8e Water flows over a 5-ft-wide, rectangular sharp-
crested weir that is P", = 4.5 ft tall. If the depth upstream is 5
ft, detennine the flowrate.
lO.8Q A rectangular sharp crested weir is used
to measure the flowrate in a channel of width 10
ft. It is desired to have the channel flow depth be
6 ft when the ftowrate is 50 cfs. Determine the
height, p... , of the weir plate:
Q= Cw,. f Vii h H
34
, where H-:::6ff - Pw and
CWr =: 0.61/ +0.075 -H;
Thus}
Q= (0.611 +-O.07.5(6~PJAl)) (t)(2~) (6-Pw)
or W
5oIf3== (0.6/J +0.075 (6-;:'",)) (})(6¥''f~ )~ (/0 f+) (6- Pw)
Hence}
[
(6-~)J ~B. J5 + P
w
(6 - Pw) - /2.5 ;:: 0 e F(p,#)
Trial and error so/vfion of £q. (J) for P(Pw) -::O!
3
Pw F(pw) F
'1-.5 3.08
if. 6 1150
if. 7 O.OOqq
If.e -/,116
o ~--~--~--~--~-
Thv.sJ
Pw = '1-.70 ff
-I
/0-85
(I)

IO.qo
10.q0 Water flows from a storage tank, over two trian-
gular weirs, and into two irrigation channels as shown in
Video VIO.7 and Fig. PIO.qO The head for each weir is
0.4 ft and the flowrate in the channel fed by the 90-degree
V-notch weir is to be twice the flowrate in the other chan-
nel. Determine the angle efor the second weir.
Q=Cwt k ton (!)1/i.j H
S
4.
where
• FIGURE P10.QO
e, ::qoo J HI:: H2o :: O.'fff J QlJd QI;::: 2 61:2-
ThvsJ
from Fit;. 10•25.1
C.~tl = o.sQO
From £'(s.U) an d (2-) J
C 8 (6, ) ~r::::-: H5/~ 8 1. (f},.)~r::-:. H. sA
'lit, Is tan -r Y2.j I == C__If2, 7$ TOil ~ y2-j ~ X 2
or
0,590 tan '1£" :: CW/ 2 iO/1 (~ )x2
or
(I)
(2)
C",t2. tan (~ ) -:: 0.295 (3)
Tr/a I and error so/vllon: 1I.r.svme &2::: 20°. From Fif. /tJ./61
CIII'I.z.:: 0,6u
ThIJsJ CWtz fa!} (~ ) = o. 626 tan (10') ::: 0./10 == O,ZPS. ThllsJ
e.2 wF 20°
Rel'Bo.fed trie..r result in fhe 9r.aphiJelol,ll froln which We GonC/(Jde thai
82, ==.53
0
o.1f
0.3
CWt,. fan(~)
0.2.
0.1
o
2,0 .30 e d '1-0
Jf:Kj
.50 6t)
10-86

/0.'1/ I
10." Water flows over a broad-crested weir that has a
width of 4 m and a height of Pw = 1.5 m. The free-surface well
upstream of the weir is at a height of 0.5 m above the surface
of the weir. Detennine the flowrate in the channel and the
minimum depth of the water Ilbove the weir block.
Q=Cwbbff(ff4H~J where
0.65 0.6£
Cwb :: (I t 1L)~ :;:: (1+ o.5m)V2. - 0,563
Pw l.S m
Thus,
~ (2 ):3/2. 3kQ= (O,S63)(q.rn)(q.8/~) 2. 3" (o.5m) 2==
Also,
Ymin::: Yc = 1-H == (t;)(o.sm) == 0.333 rn
/0-87
,

10,92. I
10.92 Determine the flowrate per unit width, q, over a
broad-crested weir that is 3.0 m tall if the head, H, is 0.60 m.
Q 0 6.5 ( 2)31. 3~ HQ =1) = ( .iL u 119 '3 H 2.
J
where :::O,6tn and Pw =3.0 m
o 1+ rw)'"
Thvs,
0.6.5 .I!! ~ (2-)~ ()~ rn2.Cf= (lt0.E.)~ (q.8J s~) :3 O,6m :::: O.Lf70 s
3.0
lO.Q3 Water flows under a sluice gate in a
channel of IO-ft width. If the upstream depth re-
mains constant at 5 ft, plot a graph of flowrate as
a function of the distance between the gate and
channel bottom as the gate is slowly opened. As-
sume free outflow.
~- f l..tQ
Y,=sfi ______
I +777/777]' 'f 7 7777/
Q::fb= b~aJ2.gy,'J where YJ::5{f, b=Joft,o/JdCd is fromhij.lO.2'1.
Thu~ k
Q::: ~ (loft) a [2(32.2ii~ )(5 rn] 2. = j7q ~ a 1/J where a"" (f
QJ rt ~ ~ Q, ¥
0 00 0.6 0
0.5 10 0.58 5/,9
1.0 .5 0.55 9B.5
1..5 3.33 0.53 IJf2
2..0 2.5 O•.5J 183
2.,5 2 Q.50 2.21f
250 --
200 -----.----~----.------------ -----
rn 150------------------- ------------ -------
",--
¢::
a 100
50 -------
O-¥=----r----,------,-------,.--------;
0.0 0.5 1.0 1.5 2.0 2.5
a, ft
10-88

}O·IILf I
lO.Q4 Water flows over the rectangular sharp
crested weir in a wide channel as shown in Fig.
PIO.94. If the channel is lined with unfinished
concrete with a bottom slope of 2 m/300 m, will
it be possible to produce a hydraulic jump in the
channel downstream of the weir? Explain.
~:~~::~--x~~ U";f"m flow J'm:'
~J;t;:J;2~~'~;!N//'YI So = 2/300
FIGURE PIO.Q4
Q;: CWI' t "(29' b H.3/2- J where Cwr = 0,611 + 0, 075 (!) wilh
H= 3m-2.2m =0.8117 and !i,=2,2/'n
Thus) I
Q=[0.611 +O.075(2~·~:)J (t)[2(Q.8J ~)J~ b (O,8m)~
or 3
Q== ',3ifQ b ~ J where b....m
HenceJ
V. - ~ -~ _ I. 3 'f9 b ~:::: J.3'f9
I - A, - b>'J b y, or J y,
For lJni forth flow
(I)
X :z.A ~ ' X ~ ~ h X / 2m
Q=n fJ Rh :3 So:2) or VI;:: 7f Rh/ So J W ere ~ J So::: 300m =0.00667
Also) for a wide chann8/ 1//= by! and R:2 y, f b .so that
R -..&. - by, - v 'f b» v
hl- P, -(2y,+b) -'I I /I
Thus with n= O.O/If (see 74ble /0.])
J ~ I
Q = /.3lfQ b== o.d'If- (by,) (y,):3 (0. 00667)~
or
Yt -::: 0, iflS 111
HenceJ
from £r· (/)
II __ 1,3Jf9 m. fh f Fi _ I, _ 3.25 fLVI =3,2S s So 4 r. - A.~ - r. /.L
I O.l/I.E I vljr, L(Q.8IIf)(o.JfISfb)]r.I.
or
Fr/;::: /. 6 I Since Fh >,if is possible +0 produce a jtJmp,
/o-gq

10. q.5 I
10._95 Water flows in a rectangular channel of
width b = 20 ft at a rate of 100 ft'/s_ The flowrate
is to be measured by using either a rectangular
weir of height Pw = 4 ft or a triangular (8 = 90~)
sharp crested weir. Determine the head, H, nec-
essary. If measurement of the head is accurate to
only ±0.04 ft, determine the accuracy of the mea-
sured flowrate expected for each of the weirs.
Which weir would be the most accurate? Explain.
(0) Recian9lJ/ar weir:
~= (0.611 +O.07S("*w))(1-)'YZib H
3
4.J
where Pw=:/fff
Thvs,
Q= [0.6/1 .,. 0.07.5t ~)1(})[2(32.2 ~2)J120ff) H
3
4
or
3A .tl3
Q= J07(0.6J1 +O.O/8BH)H 2 J where ~"'!f and H"'ff
With Qr:: /oo1jl this 9ive.s 0/135 =(0.6/1 +O.OI88H)H-%
or 3.
( 32.S + H)H~ - Iff/. 7 =0= F(H)
Trial and error so/ulion for F(H):::O:
H F(H)
,,20
/.25
1.30
Thus J
-.5.'1-0
-2,.53
O,~o
H::::I. 2ql/-(f
o
-2. -
-If -
-6
I
1.2.0
I
1.1.5
H=/,291-
V H1.30
(b) Trian9lJ/ar weir: ,
Q=GlUt ~ fan (t) V29 H.54
::: Cwt(~Xftm 1f5o)[2(32.2~)J~ H
S
/2
(I)
or ~ n3
Q= -¥o2e Cwt H:L I.f J where H,., ff and Cwf is from F,"9./0.25 (2.)
For Q:::: looff) assume Cw+::: 0.58 so ThQt
~.28 (O,EB) H~ J or H= ~39fl Noh: : The ass{)med Cw+:::O.S8
ohecks (see A'l. 10.25)
CaJclJJafe ~ forfPH,oo, ~()!o.o~J and #'00-0•0'1 from £f(s. (f) and (2):
(Rectangular) HJ H QJ cf.s (Trian9u/ar) HI ft GJ cfs
I. zStf 95.3 Lt. 35 QB.O
"'rio::: J. 2.9lf 100 H,oo =~. 39 100
J,33tf 10'1.9 ~.~3 }02.5
'v./iJh Hto.oifff if is seen thai frian9{)/fJr weir is more acClJrale
((.e. smQller variafion inQ).
IO-QO

/0.96
10.% Water flows over a triangular weir as shown in Fig.
1/
(, )
PIO.96a and VidCt) VJO.1. II is proposed that in order to in-
crease the flowrate. Q. for a given head. H. the triangular weir
should be changed to a trapezoidal weir as shown in Fig.
PIO.96b. (a) Derive ~,quation forthe flowrate as a function of
the head for the trapezoidal weir. Neglect the upstream veloc-
ity head and assume the weir coefficient is 0.60, independent
of H. (b) Use the equation obtained in part (a) to show that
when b « H the trapezoidal weir functions as if it were a
lriangular weir. Similarly. show that when b » H the trape-
zoidal weirfunctions as if it were a rectangular weir.
• FIGURE P 1 0 .96
h~H
(<IJ q=CJUZ i dh J where Uz c 1fi.jh
h'O
ofld
j = b +2(J/-h) =(:;.J;.J,)-2h
Thlls, H
Q;cJ1{ijJi[(2Hfb)-2h]dh
o 1/ 1/
",cfij (1.iI+b) S1[hJh-c"Yij.(Z)I h3
/"dh
o 3 0
= Cw6j (2Htb) ~ /I 'f -Cw ~. t /ISh.
or $
(J) Q= CwHV2ib/I~ +.J Yij.!I ~] J Where Cw ~o,D
(b)
Note: This e1v4/ilfl is s/!lf;(y Me SIIIII of q 101' aredo"9u1or weIr
all" Q{Of' t/ /ri41l9U/rJl' weiI', T/7alIs Q'Cl =' Qp ~ Q... .
(f,) From fr.(J)
Q='cwVij 1I~[ffb +J/I]
TIllIs, if h«II, Q'Z C", Yij. /I "''f.£ /I] '" clll 1fijIs /I'%. W/;/C/; Is
fhe e'1i1lliilfl for tI frI4f1~v/dr 41iel',
Si/llr1d~1~ il h»~ q~ C,..1(1.i hf II~ whidl IS f/;e e'1wI'hii/l for
areciafl'1v/tlr wei/'.
10-'I{

10.97
10.91 A water-level regulator (not shown) maintains a
depth of 2.0 m downstream from a 50-ft-wide drum gate as
shown in Fig. Pl O.97 . Plot a graph offlowrate. Q. as a function
of water depth upstream of the gate. Yl' for 2.0":;; Yl ..:;; 5.0 m.
,j·"·"'·',j'T""·'·"'·
y,
1
a= 1m
j~=2m
• FIG U REP 10.97
Q::: bq. -::: b Ca a V~1Yt I where a=/m and b=5011 (0, 301f8 ff) :::IS,2m
Thus,
Q= (ls,2fh) Cd, (/m) V2 (fl81!ffi) ('h m), == 67.3 Cd, yy; %3 where y, 'V m
Obfain Cd from Fiq. 10. zqw/lh ~ :: 2.
Y/J m Yt
Cd
Q J!!-?'
-a J S
2.0 2.0 0 0
2.5 2.5 0.'12 Jfl/.7
3.0 3.0 0,5.3 6/.8
3.5 .3.5 O.5~ 70.0
Q VSY1
If. 0 IA() O.5tf5 73./f 90
If, 5 -¥.s 0.55 78.S
5.0 5.0 0.S5 82.8 80
2.2 ~.2 0.25 25.0
7..lf ')..If 0.35 36.5
70
60
• Drowned outflow
2.6 2. tf O,~7 £1.0
2.B 2.9 o.S3 05'1,7 ...!D. 50
E
• Free outflow
d 40
30
20
10
0
2 3 4 5 6
/0-92

10.'18
10.98 Calibration of a Triangular Weir
Objective: The flowrate over a weir is a function of the weir head. The purpose of this
experiment is to use a device as shown in Fig. PlO.98 to calibrate a triangular weir and de-
termine the relationship between flowrate, Q, and weir head, H.
Equipment: Water channel (flume) with a pump and a flow control valve; triangular weir;
float; point gage; stop watch.
Experimental Procedure: Measure the width, b, of the channel, the distance, PW' be-
tween the channel bottom and the bottom of the V-notch in the weir plate, and the angle, e,
of the V-notch. Fasten the weir plate to the channel bottom, turn on the pump, and adjust the
control valve to produce the desired flowrate, Q, over the weir. Use the point gage to mea-
sure the weir head, H. Insert the float into the water well upstream from the weir and mea-
sure the time, t, it takes for the float to travel a known distance, L. Repeat the measurements
for various flowrates (i.e., various weir heads).
Calculations: For each set of data, determine the experimental flowrate as Q = VA, where
V = Lit is the velocity of the float (assumed to be equal to the average velocity of the water
upstream of the weir) and A = b(Pw + H) is the flow area upstream of the weir.
Graph: On log-log graph paper, plot flowrate, Q, as ordinates and weir head, H, as ab-
scissas. Draw the best-fit line with a slope of 5/2 through the data.
Results: Use the flowrate-weir head data to determine the triangular weir coefficient, CWI,
for this weir (see Eq. 10.32). For this experiment, assume that the weir coefficient is a con-
stant, independent of weir head.
Data: To proceed, print this page for reference when you work the problem and dick here
II FIGURE P10.98

/O.Q8 (con'/)
~
M
<
4:
ci
0.10
Solution for Problem 10.98: Calibration of a Triangluar Weir
e, deg b, in. Pw, in. L,ft
90 6.00 6.55 1.50
H,ft t, s V, ftls Q, ftA
3/s
0.231 8.2 0.183 0.0711
0.224 8.5 0.176 0.0679
0.211 10.7 0.140 0.0530
0.192 12.5 0.120 0.0443
0.176 16.5 0.091 0.0328
0.156 19.5 0.077 0.0270
0.136 27.1 0.055 0.0189
0.106 48.2 0.031 0.0101
0.091 62.9 0.024 0.0076
0.088 68.1 0.022 0.0070
Q =VA =V*b(Pw + H) where V =LIt
Q =Cwt (8/15) tan(8/2) (2g)1/2 H5/2 where from the graph
Q = 2.76 H2
.
5
Thus, Cwt = (15/8)*2.76/(2*32.2)1/2 = 0.645
Problem 10.98
Flowrate, Q, vs Head, H
------~-I
I
--I
-~ --~~I'---~-----i--~- . : +--- I
--+-I------i---!i-i---!_~i_~__=~
---------------:--------1
------.---------.- -~----- -II • Experimental
i '-------.:----=--
----------j------~
- I
I
I
1---+-----'-----,--1----'---'----.;
I
0.01 ~------------'------'
0.1 1
H,ft
jO-9Jf

10.9'1
10.99 Calibration of a Rectangular Weir
Objective: The flowrate over a weir is a function of the weir head. The purpose of this
experiment is to use a device as shown in Fig. PlO.99 to calibrate a rectangular weir and de-
termine the relationship between flowrate. Q, and weir head, H.
Equipment: Water channel (flume) with a pump and a flow control valve; rectangular
weir; float; point gage; stop watch.
Experimental Procedure: Measure the width, b. of the channel and the distance, PW'
between the channel bottom and the top of the weir plate. Fasten the weir plate to the chan-
nel bottom, turn on the pump, and adjust the control valve to produce the desired flowrate.
Q, over the weir. Use the point gage to measure the weir head. H. Insert the float into the
water well upstream from the weir and measure the time, t, it takes for the float to travel a
known distance, L. Repeat the measurements for various flowrates (i.e.• various weir heads).
Calculations: For each set of data, determine the experimental flowrate as Q == VA. where
V = Lit is the velocity of the float (assumed to be equal to the average velocity of the water
upstream of the weir) and A == b(Pw + H) is the flow area upstream of the weir.
Graph: On log-log graph paper. plot flowrate, Q. as ordinates and weir head. H. as ab-
scissas. Draw the best-fit line with a slope of 3/2 through the data.
Results: Use the flowrate-weir head data to determine the rectangular weir coefficient.
Cwn for this weir (see Eq. 10.30). For this experiment, assume that the weir coefficient is a
constant, independent of weir head.
Data: To proceed. print this page for reference when you work the problem and click here
III FIGURE P10.99
/0-95

10,99 (con'/J
1.00
.!eM
=0.10
ci
0.01
Solution for Problem 10.99: Calibration of a Rectangular Weir
b, in. Pw, in. L, ft
6.00 6.00 1.40
H,ft t, s V, fUs Q, ft"3/s
0.254 2.2 0.636 0.240
0.216 2.7 0.519 0.186
0.184 3.0 0.467 0.160
0.162 4.2 0.333 0.110
0.151 4.5 0.311 0.101
0.111 6.6 0.212 0.065
0.060 15.8 0.089 0.025
0.046 23.8 0.059 0.016
0.031 38.4 0.036 0.010
Q =VA =V*b(Pw + H) where V =LIt
Q = Cwr (2/3) (2g)112 H3/2
b where from the graph
Q = 1.79 H1
.
5
Thus, Cwr =(3/2)*1.79/(0.5*(2*32.2)1/2) =0.669
Problem 10.99
Flowrate, Q, vs Head, H
~---=-~-~~-~i~'~.~- ! _~-- -------~_ I ----------+-~--,__>_--~___ ------,_
---~-----------......--------+~~- ---- ~-~--
.----------------,-.-~~
'- --------,-- -----------.~----~-----.------.-
-- ---------->------_._-- ~~,~
-:,-----1
--------~--'-1
~_____t___:-__:::;:I
. -.~----.~
---=-~~~----------, ----<----+-----.-'-!~--=----==--:.-~.=J
! ; I
: i i
---~~~-,~~-~~=-==---~'==-=~==·=-=~:~-=-=I
--.--------- -·---..:.....1
, IQ = 1.~9H1.5
I • Experimental
0.01 0.10
H,ft
1.00
/0-96

/0./00
10.100 Hydraulic Jump Depth Ratio
Objective: Under certain conditions, if the flow in a channel is supercritical a hydraulic
jump will form. The purpose of this experiment is to use an apparatus as shown in Fig. PIO.IOO
to determine the depth ratio, YZ/YI> across the hydraulic jump as a function of the Froude
number upstream of the jump, Fri'
Equipment: Water channel (flume) with a pump and a flow control valve; sluice gate;
point gage; adjustable tail gate.
Experimental Procedure: Position the sluice gate so that the distance, a, between the
bottom of the gate and the bottom of the channel is approximately I inch. Adjust the flow
control valve to produce a flowrate that causes the water to back up to the desired depth, Yo,
upstream of the sluice gate. Carefully adjust the angle, 0, of the tail gate so that a hydraulic
jump forms at the desired location downstream from the sluice gate. Note that if 0 is too
small, the jump will be washed downstream and disappear. If 0 is too large, the jump will
migrate upstream and be swallowed by the sluice gate. With the jump in place, use the point
gage to determine the depth upstream from the sluice gate, Yo, the depth just upstream from
the jump, YI> and the depth downstream from the jump, Yz. Repeat the measurements for var-
ious flowrates (i.e., various Yo values).
Calculations: For each data set, use the Bernoulli and continuity equations between points
(0) and (1) to determine the velocity, VI> and Froude number, Fr = VI/(gYI)I/2,just upstream
from the jump (see Eq. 3.21). Also use the measured depths to determine the depth ratio,
YZ/YI' across the jump.
Graph: Plot the depth ratio, Yz/Yl> as ordinates and Froude number, Frl, as abscissas.
Results: On the same graph, plot the theoretical depth ratio as a function of Froude number
(see Eq. 10.24).
III FIGURE P10.100
(con'/)

/0./00 (con)')
Solution for Problem 10.100: Hydraulic Jump Depth Ratio
Yo, ft
0.855
0.759
0.691
0.578
0.492
0.414
0.289
0.248
hft
0.055
0.055
0.055
0.055
0.055
0.055
0.055
0.055
Y2, ft.
0.404
0.386
0.367
0.337
0.308
0.280
0.233
0.211
Experimental
V" ftls Fr, Y2/Y'
7.19 5.40 7.35
6.75 5.07 7.02
6.42 4.82 6.67
5.83 4.38 6.13
5.34 4.01 5.60
4.85 3.65 5.09
3.95 2.97 4.24
3.62 2.72 3.84
For flow under a sluice gate:
Theory:
V, =[2g*(yo - y,)/(1 - (Y,/YO)2)],12
Fr, =V,/(gY1)112
Problem 10.100
Depth Ratio, Y2JY1,
vs
Froude Number, Fr1
8 ------.- ------------- . _ - - - - -
7 - -- --------.--.----------- .---.---~------------~~ -.--.-----------
6 - ---. ----------------.-.-. -------P-+_
5
~
- ----~---.----:..--.______7"'L"L+___ - -
N
l
>- 4 ---.,--~---~- ._----_.-----j
3 ------1I
2 ---/-----~---,------------~
i
1 _ _ _ _---C--~---------- ___;
!,
1
o+-----~----~----~----r_----r_--~-----4
o 1 2 3 4 5 6 7
Theoretical
Fr, yh,
1 1.00
2 2.37
3 3.77
4 5.18
5 6.59
6 8.00
i
-+- Experimental I i
- - Theoretical

/0./01
10.101 Hydraulic Jump Head Loss
Objective: Under certain conditions, if the flow in a channel is supercritical a hydraulic
jump will form. The purpose of this experiment is to use an apparatus as shown in Fig. PlO.lOl
to determine the head loss ratio, hJYI> across the hydraulic jump as a function of the Froude
number upstream of the jump, FrI'
Equipment: Water channel (flume) with a pump and a flow control valve; sluice gate;
point gage; Pitot tubes; adjustable tail gate.
Experimental Procedure: Position the sluice gate so that the distance, a, between the
bottom of the gate and the bottom of the channel is approximately 1 inch. Adjust the flow
control valve to produce a flowrate that causes the water to back up to the desired depth, Yo,
upstream of the sluice gate. Carefully adjust the angle, e, of the tail gate so that a hydraulic
jump forms at the desired location downstream from the sluice gate. Note that if eis too
small, the jump will be washed downstream and disappear. If eis too large, the jump will
migrate upstream and be swallowed by the sluice gate. With the jump in place, use the point
gage to determine the depth upstream from the sluice gate, Yo, and the depth just upstream
from the jump, YI' Also measure the head loss, hb as the difference in the water elevations
in the piezometer tubes connected to the two Pitot tubes located upstream and downstream
of the jump. Repeat the measurements for various flowrates (i.e., various Yo values).
Calculations: For each data set, use the Bernoulli and continuity equations between points
(0) and (1) to determine the velocity, VI> and the Froude number, Fri ::::: VJi(gYl)IJ2, just up-
stream from the jump. Also calculate the dimensionless head loss, hJYl> for each data set.
Graph: Plot the dimensionless head loss across the jump, hJYI> as ordinates and the
Froude number, FrI, as abscissas.
Results: On the same graph, plot the theoretical dimensionless head loss as a function of
Froude number (see Eqs. 10.24 and 10.25).
Data: To proceed, print this page for reference when you work the problem and click ht're
Sluice gate Point gage
Pitot tube 1/1 FIGURE P10.101
( con)t)
/o-qq

IO./O} (con J
/)
Solution for Problem 10.101: Hydraulic Jump Head Loss
Experimental
Yo, ft Y1, ft Y2, ft. hL, ft V1, ftls Fr1 hL/Y1
0.855 0.055 OA04 0.364 7.19 5AO 6.62
0.759 0.055 0.386 0.313 6.75 5.07 5.69
0.691 0.055 0.367 0.271 6A2 4.82 4.93
0.578 0.055 0.337 0.201 5.83 4.38 3.65
0.492 0.055 0.308 0.152 5.34 4.01 2.76
0.414 0.055 0.280 0.117 4.85 3.65 2.13
0.289 0.055 0.233 0.058 3.95 2.97 1.05
0.248 0.055 0.211 0.042 3.62 2.72 0.76
For flow under a sluice gate:
Theory:
V - [ * ( ) 2 ]1/21 - 2g Yo - Y1 1(1 - (Y1 /yO) )
hL/y1= 1 - (Y2/y1) +Fr12[1 - (y1/Y2)2]/2
where
Y2/Y1 =[-1 + (1 + 8Fr/)112]/2
Problem 10.101
Dimensionless Head Loss, hL/Y1
vs
Froude Number, Fr1
10 - ---~----~-----.------------------ ---- ----
8 ---- -~---
i
!
1
--F--F------;-- ---I
I
1
i-----------------14 --~-----
i
2
o+---~--==~----+---~----_r----r_--~
o 1 2 3 4 5 6 7
10-/00
Theoretical
Fr1 Y2/Y1 hL/Y1
1 1.00 0.00
2 2.37 0.27
3 3.77 1.41
4 5.18 3.52
5 6.59 6.62
6 8.00 10.72
--+- Experimental I
I
- - Theoretical

11.1 As demonstrated in Vidt'll '11..1, fluid density differ-
ences in a flow may be seen with the help of a Schlieren opti-
cal system. Discuss what variables affect fluid density and how.
;:~Y (/11 ,tdelll/ ~4t :
e = ..e..Rr
,$tJ elf(I"I~J ,if elel'JJ'/Iy~f,w;11 a.ce(/n1pfJW?Y t:l,lln,~ i"
pressure) ?) ,tiS ~'pI)~jl/fJn ~ R" a.hd/rh' feMt~"'r~.J T.
'/tJ/Y;aniJt?r ,n ';/lti'd vslfJei'J Antipr J,(N"'h~ ILl?eI
c,f)(J/'n; yn~ reslll-l /" /Jresrul't! And t-ewtp!KIJIuI't:,
chan,es-. C/'4h'~ I;" JU ~PIJ.r/I,·tlJ? -/j,4t- Q/fer;f
fI, t, VIJ lue (J f ~e ,gr (,h1J'~I'JI.I 1</ will Y'eS'tll-/ In
C 1"ln,eJ ()I a'el'1f ily~ e.
II-I

I ,. 2 J
11.1 Describe briefly how a Schlieren optical visualization
system (Video" V11.1 and Vll.1, also Fig. 11.4) works.
A 5'eJ"/"ere,,, v'jIAA/'jlll,o~ syskl"l'1 de-lecl.r ~ve..,
$Yt'tllll 1/« yja-honI In -IAe iYJr/ex 01 re66ch~1'I of
'" {;anIpareHJ htlllrNiQ I. I", f/""/~ Wt~&J,a,,/~J~ rJis
ryflt!~ ir dr/en uJed.,fp JhJH~/iJe fh4t:i:.$ lind
~ fl"t!r VQ ria h(;nj Ih ~luld eI~nf/!J ahd' rhUf
V(J,ri~fo;;YI.$ I~ I;' elex of t'e.{y,c.,/on. VII)'J. i/'hJ
I~ -lite JeYl~'·-h; Cll'lei 'nt/ex or~~-6~ Ii, I) fluid
f/rAN "~fulf I;' J(JWle li!/'1 myJ he,;', t~'Jllt II
().y,"""d IAI'I ,h/e",I,'J'II./.I ',,1,'1 bllJ&~lr called />t..,
~ fIcn/te erJ,e II "t!$UI'f"'~ ,'", Vi,fH."j4fJ"n ~
LI~J1fify v4yi.. li()I'1.J· Ph' ¥t"l"yc. I,;~~....h(,n
,
/YI ellA. til"} A f'~ e~.lic, d/a1r/JNt;t fee
En,,;'~e-rlny FIlA;1'd ~ lltt1#1ic'(~ Iff W,/ey 6,<
6y hI,." f(dberJ ~n a""e/ C'7~ Crowe.
/1-2.

11·3 I
.113 Are the flows shown in Videos VIl.l and VII..:! com-
pressible? Do they involve high-speed flow velocities? Discuss.
Va Y'/tl h-()YlS l~ I'I~ iol d~nJ/& aye. ev,'den"f $"
(,...(/Ykl,yes.r/ .hIe IltlWS ~ /Y,VI) Iv~. All ()f
Ihe!-I!. f/8W$ OA'~ /fIlAI S''pe~eI ~c,¥f ~ th~
"o.lJ/e ex/-/- I'lfIW which if J,J't, Sfl'ttel.
JI· l./ I
J!A In cities where it can get very hot, airplanes are not
allowed to take off when the ambient temperature exceeds a cap
level. Does this make sense?
0", ~ hd-I d-t.l fit e d~n J,'':J 0 -I fhe ()ir If leI1
fJ. an q,., et, co/del' clDJ s'-nce ~:-':; ·
The liff :k,.t:..eJ e~er~rJ (JJII fJ.e.,. /,-{t/'n!l SUr"CfJ
~f IJII1 alYp/Ql'le Oy~ ye/4if!t:I -Iv deYJ>;1:J 7
2.
~: t f rJ ~C'-
So" #r fI.~ ,SQI'I'Je,. 4 /rpI4J?e ~iJJ.f~ () h&"~'"
t
~fJud V Ii ~f"'rfd (hJ A. h,f cl4J fnAh I}11 A..
cold d,. 7d fA ch"ell~ IhJof "Jl9J,~ $fe~""/ J?1t:#e
YUJIIW4y lel?Jfh If neeqed/·
11-3

II.> 1
11.5 Air flows steadily between two sections in a duct. At
section (1), the temperature and pressure are TI = 80°C,
PI = 301 kPa(abs), and at section (2), the temperature and pres-
sure are T2 = 180°C, P2 = 181 kPa(abs). Calculate the
(a) change in internal energy between sections (1) and (2),
(b) change in enthalpy between sections (1) and (2). (c) change
in density between sections (1) and (2), (d) change in entropy
between sections (1) and (2). How would you estimate the loss
of available energy between the two sections of this flow?
(b) C1,11.9 may be used fo eva. /u.~fe the cht:l"Je I;' en/ltaiI'Y.J
'" v
h:J. - hi '
..... v
h1.- 4I
Thus) r ~ plOl'/e~ /I.I(d)
:- c (7- T ) = I/OtJ'j j '~53}:: - "J5lK) = /O~IfDO z.."I' 1 I l I *!, j{)( , *j
(d) £~ . /1.22. /)'Jay be w~d ,f; evaIWA-fe- Me- cA0J11e. in t:I1lnpy.;
5. - s .2.. I
Thus;
s;,- 5, = CfJ In Tz - I< In ~ = (00'1 j y,{t:S3K
Y-r; ~ "'Il9. K (35~K)
(Z8',Q N. ... Y1t~l'dJ(JY
"k,. k (Jdl kl, )
;. - Sf 3'Jb :r= -
1I,.k
(con't )
//- 'I

5",;'(.(, the f/tMJ i"v"lwJ 0 fIJIt;h'OInf d,lWJf~ J~ tlenfl.,1 .r~~
$()Iu-h'~ Iv Pllyj (t.) ~b(}ve..l if if t:.lYr1fr'(.fJ;J,le (/lllul 11· s:loB
''''''-!If k ~ "" i!-IIII< I....-h·,.. fitL /"u ,-.. ,,1111ilatk ~hWJ'1 J,~n
~~c -h'fh/! (J) IV"tK (2). 50 ~ E;. '>'ltJt we ~I
~
loSS -:: u,.-Ii, .f- JP{-f) - ~he!
) f' /.,
a>?d -h> UIwoplek II"J JoI../,.... ~ ,,~,J >"If/rt. 1;,/rN",~.t,,;,. f. f"I<.
C(by. fl v 4. !I.t'"~ fI,e I i?Ie roYI / ~" CJ 6 h~f •
//·6 I
alt'tB/
in
11.6 Helium is compressed isothermally from
121 kPa (abs) to 301 kPa (abs). Determine the
entropy change associated with this process.
57- - S =- - /3 9() ..!:.-I
~.K
11-5

/1,7 I
11.7 Air at 14.7 psia and 70 of is compressed adiabatically
by a centrifugal compressor to a pressure of 100 psia. What is
the minimum temperature rise possible? Explain.
The YYllnI /'JIIUWi fempeY'~ful'e rise. wtJu /d (Jew" w / Ih
fY/c li(}n less process /,4/hlch inv/)Ive.r tf uns~nt en-/rt;I'Y /)y Ifen~ic
-Flow. ACc..lh'dlllj ft; -Ihe se.GQnd klw of fheyJt!fodYJ1tJl'JlliCJ) E9' S. 101 )
fhe enfyopi' Musf IhCret1se. or Y~1I11? CdI1.J'htn7 c:U.iYl,!! 4n atllablfle-
procesS'~ i'; Cdnnof d~ cYetl.fe. . The 7-.s d iOfYIIJII? f'Ke !rhed be/ow
illttsfvedef how fhe i5r3nfn:,pic p~sS' ~$ulfr I~ a. n-,I,,/mum 'kRl~K.
r~e. ~~
s
Far fhe isenfr-op/{..
T -(lut
mi"i,.,/.(""
pYocess J EI' J/.211- is- va/ld· 7iuAs I
&=1 . '-i:..!
~n (P...t J""= (£Jo'R) /fof)P$iA.) I.:1/7 "I(
f>,;" 7 (I¥. 7!114
lind
T T./Juf
In
() ()
_ '117 ~ - 530 ~
min;,..14""
11- 6

fl.g J
11.8 Methane is compressed adiabaticallyi
from 100 kPa (abs), 25°C to 200 kPa (abs). Whati
is the minimum compressor exit temperature pos-i
sible? Explain. i
me m/n/murn ctJmpressO'Y ex.t'f -fempeyt~lur-e wtJuld occ..u~ tv/fJ, tin
ad/al,~tlc Cll1d Ir)cf,(jnks..s /JYlJces$ whic4 invc/ves a unstzrnf
elr/rtJp}, bY i$el'rlrr;plc Flaw. A cceJYdI~ -10 /he.. '5ec..()~d law of
fherm()tilrnqm/c.J.; £9' 5. /0/) -Ihe eh1-rojJ,Y m",.sf /17c..yetlfe or
'Yemcl!'n ctJ;;;-/lttl1f dllr1n.J Qn adiabalic pr(Jce.5~ if- CQI/'1I'1Dt
clecrel1(s~. The T-s d/tl1Ytlm .riefcAed 6elM ;I/t{sfraks htJw fAL
iseh-iYPpic pn;cess resulf:> I;' D-- /()~r exi f fe?npeyt14"e, -/hQJ'l I1l7y
aciut:d ad/a6a{;c py()cess j;elwe.e~ .fhe. Sdl'J7e /1ressuyes.
~v.t
T
S
C/mc.lutte ~ ~ .//.2'/ Ihal
Or
= 3?/1<....
11-7

/I. q I
J1.9 Air expands adiabatically through a turbine from a
pressure and temperature of 180 psia. 1600 oR to a pressure of
14.7 psia. If the actual temperature change is 85% of the ideal
temperature change. determine the actual temperature of the ex-
panded air and the actual enthalpy and entropy differences
across the turbine.
To deret'm,;'e -th.e aclvtal fen;1pey~fwr'e. of I-l1e expal1atP/ air lJJ1tJ1 the
ac~1 ei/lfttall''1 t:iJ;Jd enn--ol'Y d;ffenHcef IICl'tJ$f Me fuy6/ne we nted
{ii'sf /D delent1lne -lite idf'tJl fetnperalu~e Chan!e (lcr(JSf lite fllr6/ne.
lhe ide,tI 7elNlperalul'e. cht:tJ19~ t:lCY()$> IAe. lurb/ne is tiss()c/ak, wi#' an
adialott-n'c- IifHfIJ fYiV'h4Y1/~SS t:lJlJd 11uI; iJ~}1 frt7pic turbine eYpt:l¥15Ipn.
The t:tchtetl Pl'lXefJ Ihwltl~s t:l Jmallw ~walwt'e ct,tlJ1!~ as ;Iklh-tkd
/Alilh ~ T- S c//&:~?r,1;a;, .r-kefrh 6e/gw.
£1. 11. 2 '/
~t(f
iel~~1
5ince
( ~lJ.t
1cA..&/
P,~
s
T. ) = O. 85 ( T - 7. )~ Duf ffl
it:l~« 1
11- g

11·10 I
11.10 An expression for the value of cp fori
carbon dioxide as a function of temperature is !
= 9210 _ 3.71 x 1()6 8.02 x 108
Cp T + P
where cp is in (ft . Ib)/(slug . OR) and where Tis
in oR. Compare the change in enthalpy of carbon
dioxide using the constant value of cp from Prob-
lem 11.2 with the change in enthalpy of carbon
dioxide using the expression above, for T2 - TJ
equal to (a) 10 oR; (b) 1000 oR and (c) 3000 OR.
Set TJ = 540 oR.
'" v
{f;v con5~1 c;" fhe cl1lflJ1,Je In e;r~~lpy.J h,- 1,,) ~y b~
eVdlfA~fepi will? Gf. I/. '}. rhus}
(h:-{):= C (T~ - ~ )
,,,,,sIMt ., v v
Fur VtlY)'lY:J '" So) the cha)1~ 'M enlJta/pYJ h'J,-4, J h1~ Ix
-e VtlluaW with 6'{ I/. 3. Thus"
v v I~ r~
Oy
h,-hi" c" dT '" J('210 - "i.7(XIO '+
7; r; T
~.()Z X/Of) d r
T"l. J
('h - h):::. 9210 (rl.- T,) - 3.71 XIO'k I~) -~.OlX IO~J.. _ ..!-)2. I ""YI't!, tT (7: T.
'it I 1- I
(tJ.JhrY r; = 5'/-0 (lK aAltd ~ = 55'0"R.
aYld
v If
(n,,-hi)VtMY/~J
c.,.
f~97 /'.;.. /6 ) (sSO 01<_ 5''M of() :::: if 9,{}()O -li-.lb
$/"" .1( $iu.J
51 000)
(emil)
fl. /1,
SI"

II. 10 I(con't)
and
(. v V)h -I-, .
2 J VtlYtf/~
C"
7
2 /0 11-. /6-::: . -, X 10
(V v
h- ~) .2. , Varytn.J
7
//.// I
11.1 1 Does sound travel faster in the winter or summer?
Why?
firy. ClIY" ~ ~ S"peL4 tJr .f4Jw1d ~ EI' /1.3, iJ
c -= Vf<Ti<"
Sf} I/IIhe~ the "Jtt.tln~~t -k""'r"lut'(!" 1.: if I~v.le,.. PS "vr/~ IJu
1/tI1;"~~ fI,.(! s-pef!d or JtJIU'UJ'.1 c~ If /~wel'" fJ,ar. c:lIAY/~, the
fla"l'?7eY'~ w/'el"l r p",p{ f1t.u.s C If j,~"),er:
//-10

1/·J2. T
Il.12 Estimate how fast sound travels at an altitude of
250,000 ft above the surface of the earth.
~ eg. 11.3b
C:; VRTIrI,
Af ~J;J ol-lit.,de ,,1 2S'O, ddtJ .f.J. ,) ~ Tlltb/~ C·I ~ ~r~~1I1e
~~ -Iv k - 81.77 6F 5D
C = If(?'6 fl. ~' ) ('160 -R_ ih)XI.l{-)
Y~ siCA,. 'R It Ih) :::
( SII#-7.!!..
s').
1/-11

11.13 J
11.13 Determine the Mach number of a car
moving in standard air at a speed of (a) 25 mph,
(b) 55 mph, and (c) 100 mph.
The M4ci1 nUY'1ber IS /he raflD of ~Cd/ ve/()cify -10 Jleed D/.rIJtmri.
711us
Ma..
_ V
- -c
h ~/t) J1daypl a /y
C :: VRT* =
or 06()O L )
_ (/117 If) hr - _ 76 I. 6 mph
C - sj(~2fO:~)
(a)Foy V = 25 WlI'~
MfA. .::
z5' I'1"PJ,
761.{, mph
(h)FI)Y V.::!i5 mf h
fVlA. = 5~ n?/h
761.£ mph
(C)Hrr V ~ 100 rnfh
Ma : lao mph
76/. b Y"ph
= O.03Z8
= 0.072 '2
O· 131
1/-/2..
0: /117 fl-
s

II. IS I
11.15 How would you estimate the distance between you
and an approaching storm front involving lightning and thun-
der?
(J¥'J,e. w()..! ft; e$-h~"ie, fhe d/f~J'lce ioe-lwf'en YblA
and f),ffYrJac,hu;'J Sffrm c./oud.$.J x) ;s -frJ C~UI!II
fhe 1?/..{l'Wbey tJ/ seco;ulJ f befwe.en 5eeJ;'q fj,e
/ / ./
/i!t.thJj tlJ'It/ hear;, !-""ouler. U fJ7 Pi'!
~'I'Yt?x"'ntf,(f(; value 01 lite, sl'eetl 01' .s1J"11~
IIf~ fj: (see Table B.3) ~ CdJi PjJjJrtJx;~/e
dJf/-r;.nct-.I}( ~
1/-/3

/ /. /6 J
11.16 If a high-performance aircraft is able
to cruise at a Mach number of 3.0 at an altitude
of 80,000 ft, how fast is this in: (a) mph; (b) ,
ftls; (c) m/s?
(6) Wi~ Etg. /1, 'It,
V == (Ma) c
ahd af eo, (Jot) 71- /n U· 5· siz:lYJdayo/ f?lrn",oS',Phere/ w~ ~vc
~ -fhe so lu fio 1'"1 0 f- p rob/.e;-n / /. / ~
C. ~ q7K frs
1hu.5
v ;:: C-~.o )('178 ~f ) == 2.'130 t!.
= 5
(cn lhel't
(C) Also
II ::::
1/-1 If
= 2000 mph
393 m
s

/ /.1'7 I
1I, l7 . Compare values of the speed of sound in m/s at 20°C
in the following gases: (a) air, (b) carbon dioxide, (c) helium,
(d) hydrogen, (e) methane. Give one example of why knowing
this may be important.
To CalUAlak -fhc 5flted. of sdund 111 tJfn /dea/ 91lS ~ can
lAse ~tj. //.J(,. Thus
C :. tfRT~
Wi!t1 va/ueJ of R aMti '* {Y7Jm TaJ:,le /.7 we o6h/n
(AJ f,;,y tl/r
c =
I()(JS In
-.s
-.5
(e) ~ me /lulYJ e
--------------------~---
__ (5iB.1.I. ) (7!13 K ) ('.31)
C -1J.9. K . 'f'f6
fJ I) (I ~~ )
l' }.}.Pt 11..9In
(con't)
1/-/5

//·17 I ("on 't )
ie$-I-f of Tuybtll'YJ« ch/Y)er A, ~Qf fAlAt
fA If) w~Y 'S"l'eer/ of flJLl he! if ftJmef,-WlCJ
fo hl!A - MA.ch -l?tlm6er t:JP~Y'~fJ'f:Jn
be-IIc.1,/'evet?! w"fh le.f'S" ,..falit)nil" I
-Sl'~ed anti t,e~e les{ sf~~.r;~J ()1 fhe
}'Y~chQI?/c" / PIY-1-S of fhe -ferf rj.
II. I tj I
JLib If a person inhales helium and then talks, his or her
voice sounds like "Donald Duck." Explain why this happens.
ThL 5r<-t~ of .s~j II? he/,',-<W1 ;!' "~(JY"; -/1,Yee
n1He s ftw. f p.t.t~ of .I/I1AA'Id I;' a I r f D I/J... pe-YfImJ VI)" u-
50u ndJ Ilk if- d~f when 5fea./c.'j fJ"YVVtJh he//un-"I.
11- /6

II. tq
11. Jq Explain how you could vary the Mach number but
not the Reynolds number in air flow past a sphere. For a con-
stant Reynolds number of 300,000, estimate how much the drag
coefficient will increase as the Mach number is increased from
0.3 to 1.0.
Cans"del"ln~ ~'r QS CAn ide"l ~~S I II(. can -f1<prtss +he Mac.k tL4t1~er
Mo., o.s
MAo= v
c
1he. Re.~r'l.o feh
v--
R~::: ~yd -== ?~d
_ }.A RT)It
(I)
Loo::.lt»q at ~A&.(.10tls. Q~ '2.. iJtC. v(a~n 1WJ..t IN'<. C4n v(J.'lj
Mo. lA)~i~ k.oO·l~ Re. cd)Sta.rf ~ va.v~{It'~ V A~d P OV~
v..J i +~ PV held c.Dlc;i'lllf.;' .
S-
Fro M +kc D)'(7Afh Ioetow wcz. C61c..lUGle +ko.-- a,1- Re -:: ~ ~ to)
the d.ra.9 c.tdf;"'e.n1 'I~C)C.<;' fro~ O,Lt1 +z> 0.15 C(+ Ma.
,ACr.eo.<;eS ~ o. ~ hI,D .
Ma = 1.2 2.0 3.0 ,,1.5
1.0~~~~~~~~:z~~~~~~4.5
0.9----.-------
1.1
O.B -.----.------ - - - - - - - ; - - - - 1
1.0
0.9
O.7 ~;:':;";;;~::;-'.:.;.;"::;...::::::::::::..--::;-:,.-:::;--=:;--;::==:;:::::;::::::::;1
0.6 r----1r----.--...E-
CD 0.51"'""'_ _
0.4
0.2---..
0.1 _. .--._--"'...__. . --------- FIGURE 11.2 The variation of
the drag coefficient of a sphere
o:---"--__________---l with Reynolds number and
2 3 4 5 6 7 B 9 Mach number. (Adapted from
Re x 10-5 Fig. 1.8 in Ref. 1 of Chapter 9)
11- 17

//·20 J
11.20 The flow of an ideal gas may be con-
sidered incompressible if the Mach number is less
than 0.3. Determine the velocity level in ft/s and
in mls for Ma = 0.3 in the following gases: (a)
standard air; (b) hydrogen at 68 oF.
Frowr bCO' //.46 we h.ave.
V= (Ma.) c
{;Jp,ic/r. when C/JrYt6;}u"d wifj" ~~. II. '36 / leads Iv
V = M~ VRT"k
v:: o. '3
A/so
v = (335 fJ) (0. 30'l-K ; ) ~ /02 !!!
=s
(I)
(b) For hydrogen af 68
D
F; R =-;.'I-")(I()~ ff· l/:; anel-lt=/''11 ~ Tt2lJe /-7.
A/so T:: S'21 d/(. lhus wilh E1' / s""J .~
II/so
1/-18
= 1'290 !!
=== J

11.21
11.21 At the seashore, you observe a high-
speed aircraft moving overhead at an elevation
of 10,000 ft. You hear the plane 8 s after it passes
directly overhead. Using a nominal air temper-
ature of 40 of, estimate the Mach number and
speed of the aircraft.
Mp. = 1 V- -SinK C
Also
hllltJr - r
-Vi
{om b;n/~ c1L /~d z. we
~/n ()( r 5'1'" tJ(
::.
{bStI( C t
Then
IX ::
aM-cI
tJ6~/~
=
1
Ma, ~
- - 2.0g• 0
SIlT 21.7
Fuy-/hw
V = (M~) c =(2.0!)(1096 ~f)
x=V./;
23.7
o
If2z7°= -s
(/ )
( 2.)
=/OCfh If
s

If. 22 I
11.22. A schlieren photo of a bullet moving through air at
1:1-.7 psia and 68 of indicates a Mach cone angle of 28°. How
fast was the bullet moving in: (a) mis, (b) ft/s, (c) mph?
W'-Ih &ps. I/. 31 tl/luJ //·36 we obk//1
V = ~ ::: VR-r-k
SIn 0("
For- tJ /r af
frcrm fable
(b) Thuf
/
v -=-
(CA.) or
(c) and
Sin C""
-:: 732.
//-2.0
-s
-: 2'-100 If
=s

/ J. 2.3
11.23 At a given instant of time, two of the
pressure waves, each moving at the speed ofsound,
emitted by a point source moving with constant
velocity in a fluid at rest are shown in Fig. P11.2:J
Determine the Mach number involved and indi-
cate with a sketch the instantaneous location of
the point source.
ihSto#tlneOUs loeo.fion
of lhe point
S()ur(e
1+----0.15 m--~
0.1 m
7he Mach number tlIs,flJc/ak4 w; fJ,
In V()/ve.r/ I~ -fJ,e skeM
as $ho~YI be,/fIW.
~J FIGURE Pll.2 j
Ine l'I1onlJYI DT f1Je.
p(),~f S~uyce
wi/I, E'I- 1/.3'1
Mti =
Prd»? ..;h e.
O. OJ ""
.R
we
(!). /1+7
(J.IS"1?1 +-L
p. dlm)(~./S"'" t- .£) ::: 0.1 ~) L
ab()l/(! i5 8/1$lIy tJ614111e4
I. :. (d. ()/o/ (f).ISm) .... Of 0/67 /?'I
(~. ()f!m)
0.0, J'J?
O. !7- 9'1
= I."7

IJ.2.if
11.2+ At a given instant of time, two of the
pressure waves. each moving at the speed of sound,
emitted by a point source moving with constant
velocity in a fluid at rest are shown in Fig. Pll.21t.
Determine the Mach number involved and indi-
cate with a sketch the instantaneous location of
the point source.
c t wlf,v-&
ifJslal/T4neIJUS I~on
~f fhe pllilf
Soulte 10
' sci, In.
2 in. :C(i-tW/M)
-i
~Vt .:;J
FIGURE PU.24-
(I)
J.I~, -f'roYJ,t fh e. Sk.e-k~ aloove. We hrv/~
C(t --C~lIe) 0:;: ~ In. -::: C t - c tvJ"~ = /0 In. - cfwa~
ThIA~,
C..J - / CJ In - 2 ,'". ...... 8 /n .I. wave - .
()J1d w,'ft" £1' I.
M~:::. SIn.
SIn.
A/so
vt vt ~.h2S"
Ma ::. -= -ct 10,n .
/lutf,
Vi /0-025 )(/1) ,;.. ) '::: 6.2.S iI'],
/ /-22.

11·25' J
11.2~ How much time in seconds will it take for the "bang"
of a firecracker exploding to be heard after the blast from 200
yards away on a standard day (see ',ilh'o VI L5)?
rlaYldarrf a/r.>
we.. CCln
///7 .f.J.
s )
We +he. s-peed (Jf sound t'n
fo fief the lime e.rh~afe SPtAJkfj ~ wilt,
t:: :s. -;: (2~()yJs ) ( 3ft ) -=
c
(1117 f!)
11·26 J
j 1.26 Sound waves are very small amplitude pressure
pulses that travel at the "speed of sound." Do very la:ge am-
plitude waves such as a blast wave caused by an explosIOn (see
Vidl'O - i 1.5) travel less than, equal to, or greater than the speed
of sound? Explain.
0·5 s
The ~peed ,,{ ,,$PUl1d IS fhe speed al wh/c.h a V'J
infinire.>/md'/ pl'eSSUye dl.!fuy/,ance -frfAlle/.r Ii,YdUft..
a f1u/J and it re?Ye.sent-..f fAe h'J/Y)/mtlPl .r;;-eecl
01 f-lJiJ di.rlurbO)'Jce. Fin; Ie pl'~t!tlye c//.rlurbl'u')cef
fral/~ I #.ft-e;- I-h~J-'J $t;tlnc/ waVeS' hectlu.le {he
/arr;ey f 1f.ffute diffe"ence. pcfs pf a 4y/veY" 61
fttf"/w m(}lIf:,»,)enf.
/1-23

/ 1.27 I
l1.27 Starting with the enthalpy fonn of the energy equa-
tion (Eq. 5.69) show that for isentropic flows. the stagnation
temperature remains constant.
we have
-I- J{e,.(1,..)]=- ~efof
,,,
W
s/,..ff
/tef i"
Fir/ ise.n-fr.6t;c., fiwv fhe en~J fl!.YYlIJ,'".r tI/YJJ-Jm"f
P. hd (Qnel-:: O. ShfYlI/lH#)., en ttIf2t,.J if rJeH1'Jed t7J
v v '2.
1,0 =- h -I- ~
z
~o / ~ 1?e,jtj/.b/e cknfe In ele va. --hdn (ol=~ .;;,JQ.re;)
a 17d ho .rA.tlft IN'()'(k...1 Vj.t,(.ff) fheh
//- 2'Jf

".2e I
11.28 Explain how fluid pressure varies with
cross section area change for the isentropic flow
of an ideal gas when the flow is (a) subsonic; (b)
supersonic.
With -the help "f £" II. l/ 7 (A.e CA.}') C~mmenf cJn haw 'Y~UYt. VPYlCS
Wi#' dYeA. chan~ In tin isen--h-o,o/c. f/(JW. FroJ'V1 13,. //.'/7 w~ t7b~/;"
dp ::: I' V'). c/ A ( t)
(l-tvla2) ft
raj Ftr; subson/c, fluwI £1' I sU1~.rIJ -fhaf- Ch4;'J9~S of P f/)/I~w
vJ"a~ges of A· If A 111 C f"ep..{es / p Incretl.Jef ~nd v/c~ verst{.
(b) FttY su.perftJXic fldW,J ~. I SlAjlfeIir thai cht1I?~J of p are "PfHf;ft
11.29 1
ro cAaJ?!es of A. I-/ A J;'crease.f / P clec~a.Je.J and VIC~ versa..
11.2.9 For any ideal gas, prove that the slope
of constant pressure lines on a temperature-en-
tropy diagram is positive and that higher pressure
lines are above lower pressure lines.
r:ri -
-r
tlYJd since .fw.. an iderl/ gel! ~. //.7 is' VA-It'dl
We hcwt.
v
d t, ::: C
f
d T
and -thus
dr = I. (I)
ds Cp
Wif~ E1'/ h'e CtJYIclude -fhdf the.. 5/ope of a Ct/J'Istanr- pye.rJu~ line 0)'1
a -femfJeva.lurt. - en-lrupt d/atjY4fi1 1$ p"sif/ve·
t-UYfl,eY, ~ fl' /1. ZLf we CUyzc!ut:!e -fhal
3= :: (ii (1,)
P, 1; )
fDy ~n-y iseh~/c PYlJce.ss aJ1cX rhus h1yher f'lesJtw(. lines t:J re abpve
lowe yo pr-e.JSU1'(. 1;J'?~s In fempeya:fu*_ - erd-Y17py c1;~!YAflfS.
/I-ZS

//·30 I
11.30 Determine the critical press~re and
temperature ratios for: (a) air; (b) carbon diox-
ide; (c) helium; (d) hydrogen; (e) methane; (f)
nitrogen; (g) oxygen.
1he- crifjc..a/ preSSUye rafio an.d The Crih'ca/ tempem/uye raf/o -iOYoY'}
ideal
90S a y-e I fyd'I¥! t::)s. /I.~ / and II· 63
p~
..iL
= (2- )* I
~ -Je.,.,
and
Ttl 2.
::- -~ 'k +I
(a) rdY air) 1<= /·I./-tJ fr.uw. Table /·7. Thus ,
,.'1<;
p*
~ (/~~I ) '..-1
6.5283=Pa
and
z 0.83'$1
~ ',lfo+1
(b) I7r; carbon d/()(/de I k = /·30 ftrm" 74b1c 1·7. Th(J~;
(c)
~ ~3o
f := /2- ),y;:; -:::L~ O· S'"ifS7
F:, /.'301"1
and
T¥- 0.3676
=
m hej,'u m I -k =- f. 66
/.(d,
p* -= I~)'~ -Po ( /·6t+!
()nd
O.l{-Kg I
T* _ 2-
~ - = 0.75'19
( con't)
11-2.6
(I)
(2)

//. '3 0 I (con If)
(I)
a~d
()nd
r* 2
- ==
r; PII+I
z
r; 1·11+/
:ft
T 2
- ::-
~ := /·'10+1
ThusJ
0.1658
fm;. Tdb/e /. 7.
0.£28'3
6. 83sj
(g) Fw tJxy'jen I k ~ /. 'kJ ~ 7J.l.le /. 7. JA.u~
~ ~
P (2 ',110-1
P.. :;:. )::: O· ~28J
tJ /.'forJ
and
T~ Z
O.KJJ'J=
//-2.7

11.3 , Air flows steadily and isentropically from
standard atmospheric conditions to a receiver pipe
through a converging duct. The cross section area
of the throat of the converging duct is 0.05 ft2.
Determine the mass flowrate through the duct if
the receiver pressure is (a) 10 psia; (b) 5 psia.
Sketch temperature-entropy diagrams for situa-
tions (a) and (b). Verify results obtained with
values from the appropriate qra.ph in Appendix D
with calculations involving ideal gas equations.
Th/s problem is .!";rn;/ar -10 Exam'ple II· 5"
Th e.- yy,t:l5 s fiOWVlff fe.. is' db fa IJ? e.d a f Ihe fhroAt w//h Gj. I;. Sit? 7JJ1I~
rn ": fth Afh ~h (t )
The ihr()af de.n!iiy Can b~ obhlJ1ePl with E'J' /l.btJ. 7hu~J
t1~ ~ /1 rI r (~) M42 ]i':i (2)
2. 1-J.
To de(e,-in,),e. the. thnsaf M(J~~ n /A"...w INt. use. E~. 1/·£"9. &JJ
The
.If
fhen
fhen +h~ f/()W IS
The l/eJocJty af fhe -fhYOtlf 15
~"hed tz, y/e/c(
~J., =} M4+h vr-R.-~-It(-
wheYe 4h is (J6/1/11;'etl wi~ ~f' /1·56'.
T =
7;.
~
/ 1- (-k.; ')Mo.+~
(~n'f )
//- 2. 8
Thus .I
(5)

I/, j J (con If )
(a]
F~ £.8 . 2. INe- ()bIP I;' ---.!-
~t, = (2.33 ~/O-J J'IU!)[ I ~/'VQ~
+tJ
I + (I'~ -') (0. 7618 ) "'J -
Fnm, Ef' $' Wt!- fief
T
ih
I -;- (I.¥~-I)(O.76Zg/"
tJl1d. wi It, Ef. 'I
/Iv; ftt E1' / we ObfaJ;'
:= &'06.2 ff
.5
,;, = (k07)(IO-7;~;J{(J.05ff1J{g/)6.Z {'J ~
AlfeYYlttf.ivd
J/ .u.Jlj Fig. D. I //Iliff,..
t;1, /0 f.>ia
:: -= 0.66'
P., 1'1. 7f1JiA.
The value of M"-h. j..f
Met := 0·7t
-fh
finr Maf~ -:. 0, 7~) we. rei ~
~ ::: (~. 9 ) -: :; (0. 9
'TheI"J Wi Ii. ?1' ~
~ = 0.76
11-2q
Pi'). P. I
)(s/r~) :: lI6 7 ~
=805f't-
T

1/·31 J (Contf)
fD.r Mtlft,. :: O.7h ~ 'Jei ~ F1' IJ. I
~ ~ tJ. 760lt £J :: (0.76 ) ( 2.'S.?Xlo-1 J/~) .::::~ ~ H2
-1
/. i XI() .r/tq
NtIW, wjn. B~./ w~ IJblaln
T =oft,
/ r (,'::'-j
OM""- witt, &j. LJ
V
In
WiHt Gj. / ~ ()Iob./~
Ii, ::: (t.5a9 XIO] flU!) (~.as- ff'2-)(1019
~f1
AlkYYlah've'J) ~ Fir- lJ./ fw Mf)..::
T :: (O.K? ) (Si,,,/() = Ifjl ~
+t,
OM.d
~ = (o.t,-/
Then wj tf". ~f' if
klh :: I(i716
(UJY/' t )
//-30
= /O/q f+
s
~) =
1.0
---f.f~

1/·31 (um'!)
0. 07'2 slllr
-S'
T
~ T / Po
Ttl T-7 0
(a)
~"'} ()..=Pre (b)
pi *1/ ~~b~J.
1* Tl'
/" '/ t:.e
s
11-31

1/·32 I
11.32 Helium at 68 OF and 14.7 psia in a large
tank flows steadily and isentropically through a
converging nozzle to a receiver pipe. The cross
section area of the throat of the converging pas-
sage is 0.05 ft2. Determine the mass flowrate
through the duct if the receiver pressure is (a) 10
psia; (b) 5 psia. Sketch temperature-entropy dia-
grams for situations (a) and (b).
]his p't?J6/eWJ is Jlm;/(lY h £X~mp/e 1/'S"Jexcepf helium Irli,vo/vfel.
The YY}t:lS.r H()wyx/-c is o6leune,( af- fAe #JYrlai w;1t, £,. //.'Id. Thtl~
I'h - f. A V (;)
fh -It.. Ita
The ffwo~t densify cal'! be I') b ItA. I i-Jed witl-t t-i. If. 10. ThMf,
..L..
t1.t, ~ r I j~-'=
/ f (*-') M~2 (2.)
2. :';'1,
To cle-bym,'n t fhe -Ihroa.-r Mpch nu."""/'-rr /Ne use £1'//' Sf. ~JI
i J dblruy,ed wilt,.
'·66
ft· 11·6/. TJu,S;
/ 2 )~=
( /:66 f I
7./7S1'5/(;..
If. fhe. yecei~ ptesS£.VIe.; r:.e) is qreafe" Tt,(;ll'l Q'I'"
-IheVI p - p.. ancJ.. ih e., II().A/ ,.!
-H, - rc
Y}of chok.ed.
f-l1en p - p.~ OVId fItL .fluw is
ofh - f",
chol:.e4. .
The veloc,ify eLf the -fhYOA.-f if
COntlo/l'!ed h y/e/el.
~l,:: Mq,h Vf< ~~
wheve ~ ;f oh-h I~ed wift-J
(s)
I-f (~-f) M~:
(U;1'1'f )
/1..32

II· 32 J (con If )
:: 0. 7~32.
We lise tie ideAl !ll1f ~wv/I'6'" of- sJtk (Gj.//./ ) ft> 6bhUn ~. 7ku~/
~ - to _
f( To
PY?Jn.. E5.~ W(.. Jef
528 -~
T = _ '15'3 DI(
+1, Ii- (1";- )(tJo 7tJI1.)"L
and Wi It,. {t. ~
v -= ((). 7oiz.. )
in
/ t-(;'b6 - ')
2-
(~n'-t )
/1-33
-II
= 3.2zlX/~ .fIU1
oft-]
I
-

//.32. (con'l)
With [1-/ iN£' ohla.t"n
$61 {i-
S
n. " (z.Of6 X/of ;::)( (7.05 HJ(ml 7) = tl. 03
T
~--""","---- t;, ....---~----~
(aJ
(1:»
I---~~---- T *
5
/1-31-

II· 33 I
11.33 What is the static pressure to stagna-
tion pressure ratio associated with the following
motion in standard air: (a) a runner moving at
the rate of 20 mph; (b) a cyclist moving at the
rate of 40 mph; (c) Acar moving at the rate of
65 mph; (d) an airplane moving at the rate of 500
mph.
M~
we. can
C}y
we use
ft!ta.,. -= 20 I'kl''''
16/. ~ I'kPh
r
E.. ::" LPo I-t-
(6) FOr V ~ 'f~ mph
-= ~.OZ62.
M tt; '10 mph ::: O. ()S 2 S
76I. h J11.tJJ.
til1d 3. t:;
.e- [ J 7 o. q98
~ -=- J+ ~17-(O.OS2.$/· =-
(C) ~ v:::: 6~ mflA
M~ -::; 6 ~ n?'ph =- 0, tJi"S'I
7t/. 6 Jnf)A "3 S
aJ1.d C ( ., 7 . --"" o. ?9'frt
Ii ::: I + ()."2-(0.08'5'+) J
(d) h v::: 5"OOtnph
/l1~ == 5ZJP rYlfJh -::- t11. 65"6
7('/.6 PIP/"
tin.d. 7·5
[ I+O'~
.l ::: ~'7,+9
.e. = (o.f/i6)'2. J~
/1-35
(/)
(11.5CJ)
:: 1/17 ff
.I

1/·34 T
(a)
11.34 The static pressure to stagnation pres-
sure ratio at a point in an ideal gas flow field is
measured with a Pitot-static probe as being equal
to 0.6. The stagnation temperature of the gas is
20°C. Determine the flow speed in m/s and Mach
number if the gas is (a) air; (b) carbon dioxide;
(c) hydrogen.
To defeY"'lne -fh~ flow .s~~d IJnd M4Clt number hpvl1! ke;r j/ven
rhe stan'- p;~S5UYe. fo s-htfntfJlitJn II'{JS$Uf'e yah~ 1') ami
~
S-1-4jI'Za/li(;1'7
the 9/ven
f~mfJen:llure / 7; <I -n,~ tI /1'"
volue of.e (JII'Jt:/ yea4
Fd
we. el1f~1'" f/9' j), I I wilh
C()I"~~sPtJ"""1Y vt/lue 0/ Mq. lkus wilt,
I/o/ue In
MIA = 0.89
--rdY" M~ =' t;·19" F;9' /). I fIves
I ::. 1).16
7;
p ~ tJI /'tf,u s
T =- (r) 1,; ~ (t).86
7;
Fig. iJ./
( h ) Fol' C4rbIn
Thu5 we halle
r.
-j(t)Ma =
1ff~ 'fr I ]e)- _ _/ Z
1·3-1 -
-)-11. ( d.6) -;:;- I. '3 - I
(}Y' I;
MA -::: 0.11 J (C,4nlf)-- //- 36

//.3 con'-f )
-
lhen
11-37
_ CJ.8t¥
~ 252.5k
:; 1070!!2
= oS

//·"35 I
11.35 The stagnation pressure and tempera-
ture of air flowing past a probe are 120 kPa (abs)
and 100°C, respectively. The air pressure is 80
kPa (abs). Determine the air speed and Mach
number considering the flow to be (a) incom-
pressible; (b) compressible.
(to A$SlAJII1ln!) inCtJrnpYtHiblc
-fo COJ1I'1e.ct ihe. slA.h'c
Iilw we ()~e 8eY'MIA/lis efuahiJn (Et.3.7)
anti sfz;fJ1(;o4bYl S'fttks t:iYld ,er
V =-/Z(Fo-P)
I"D
(I)
~ = 10
R7;
(2)
aJIld U/mb/YJl~ E"Is, / and 2 W~ Ob"IR/;'
2 [IZtJ 1tl'4(A6s)- 8(}~{~6F)](ze6,9 ~ V373~)
*'/. K /- -:: 267 ~
[i 20 fe~{"'b$)7 (/1!'i~," ) = f
number We need sa.
Ma = v v (~)
-c.
VRT1<
r W-t, fA fe Ih. t:.- ~u.a:f,~n of mat/on (€ij. 11.91) It; (}~/;'
V"(~-I) = ~73jt:. _ (2/7 f )'2.(t.4_1) (I.-!:!- )
~ ~r· !!!
2 R R. 2 (/."") (Z86.t1 N."" ) $"
'Ill k.
T = To
T :::. 337·S K..

1/.3£ Ct) fI'+ )
Wi~ Etj. 3 We o6f?J,;'
p
Nil ::: ()·78
=-
2D 7 !!!.
.s
_ 0.67
Also frlIWI F /9. f)·1 we ~4.c{
r ::= tJ. ali
~
() 11tl ftt.uJ
T ::: ( 0 .~1 ) (37~ f:-) - 132 Ie
V -::. Ma?l RT -k = (0.78J
- 0.725"

For
11.36 The stagnation pressure indicated by a
Pitot tube mounted on an airplane in flight is 45
kPa (abs). If the aircraft is cruising in standard
atmosphere at an altitude of 10,000 m, determine
the speed and Mach number involved.
/~ O()t) n7 sfandard tlfmtJsphere.
P = 26· 5"tJ kIll. (abs)
dYld
T - 223./k-
lhu.J
P =
zt. 5"tJ k~ (abs)
- 1).59
Pa '1-5 VIIt (tiJ,s )
and -frmn P/g. 0./ we reacf
v = {/vlA)C
/I-~O
vve ~ef -h-PYh Ta61e C·2

If. 39 I ':'11 .JK An ideal gas enters subsonicaJly and flows isen-
tropicaJly through a choked converging-diverging duct having
a circular cross-section area A that varies with axial distance
from the throat, x, according to the formula
A = 0.1 + xl
rhis is like Example //.8.
where A is in square feet and x is in feet. For this flow situa-
tion, sketch the side view of the duct and graph the variation
of Mach number, static temperature to stagnation temperature
ratio, TITo, and static pressure to stagnation pressure ratio, plpo,
through the duct from x = -0.6 ft to x = +0.6 ft. Also show
the possible fluid states at x = -0.6 ft, 0 ft, and +0.6 ft using
temperature-entropy coordinates. Consider the gas as being he-
lium (use 0.051 :5 Ma :5 5.193). Sketch on your pressure vari-
ation graph the nonisentropic paths that would occur with over-
and underexpanded duct exit flows (see Vilh:o V11 A) and ex-
plain when they wilJ occur. When will isentropic supersonic
duct exit flow occur?
S'il1c.e
and
A~
A = (J., 1" x"
r =
l.
0.1 +- X (n
1r _
valu.e5 cOffesponr)/nJ -Iv VrA/utJ f)fWitfa ~~. I we... UiYl detet'n1me r
X. The tAye Summol'J)ed In the 9r4.p""- aJ'Ld -ialole.$
r).uc,f is. cr.ol<.e.d,
If f:1 ~
A = 0./ T
and
A = I +- x'l..
A" 0·,
Wi ft.. &t. 2 we... CIA.J'1 delfHfYllne
(2..)
(.(N"ell'ol1d/~ tv value! of k.
The>e, .1 Values Clye
A*
!=rJy he"~YVI wt! en ~
MfA.. values w/fhiyr ~e rtlfl9(!"
CJ bl-a /;" values of d
Aif
pnJ~Y"~
s,ecih~ci I;" -ihe pY{)blem sfAkJ¥/~.".f Ci;za
(Gfj. fl· 71 )) X (£1' 2.) J T ('f.I/·~6) and
1;,
,E. (~1' 1/ Sq ) .
.(1,
The ~ e. VI) lues aft ~6ult4.k~ and f//"dpht.d on
p~,!es Ih~f ftJ/ldIJ/.
(()OfJ'f)
1/- H·I

//·38 COf'/'t )
A
Ma.- AI/!
0,0 Sf
0, tJ l'
0./2"3
0.223
(l.Lf' 0
1,00
1.95~
2.771
'l.l/17
J/.'f'f'J
?Iq~
~.O
/.j.{)
1·D
Ma..
2·0
J.G
/1.0'
7.'13
'1.61-
2·61
'·lIo
'.00
"l/a
2.'0
"1.,0
1. 'If)
/1.0
FrO}lYl
I
Pro9Y't:1-yt'l IS€NTI?OP w/ft. -I< =1·66
Ef· 2.
x(ffJ :r. 1To
Po
5ub5tJllic.. so((J. fJ'oY
!/.OO O.9'I(Ulf O. 9~71l{
t (J.g 0 O. ~ 'I gO" O. 'I9SZ2.
t 0.60 o,t19S03 fJ.'ji155
:!. D.I.fO 0.1138S' ~.~5fI9
! 0.2.0 0·93'173 0.91/36'
0 O.1f1g f ().tfWD8
. So{ufidn$"uper$OI1I c
0.20 O.~'/1'2.7 o. 1'I83J
oAo b.Z~lq5" 0.01./1'{1
~.60 O.(5»)' 0.0('1'1'
a.go 0./32.12- 0.00'2.'+
,.00 0.(0102- O.OO~(3
o~====~~~~----------~~==~~==~~
_1,0 -o.f -fJ., -0."1- -0., 0 D.l. (J·Y ~6 (J.g /,()
>( (II)
Va yit). fit)n of Ma...c"h, numoer ..fi,y hej,'uYYl-
(con't)
/f-lf2.
s-lr;re.
()./c
d

1/. '?8 (Con'-t )
I.Or--------'--~ .
o.~
O.b
r.
10
0.'1
o~--~--~--~----------------------------~-/.0 -0.1 - d. b _(J·Y -C.'J- () tJ."J. o.'{ ~, dot 1.0
x(ff)
Var/lili()n of slahc -kmfJemltwt:, 1D sm,l1alt'()1" fempeYfAtufe y():hp
lOY he1i't1.lJ1
/.0..---------__
().~
0.&
f
Po 0.'1
0.2
o~------------------------------~======--~
-/·0 -o.g -0.6 -tJ,1f -b.'2- 0 ()./.. ()·4 (7.6 ".f /.0
x(ff)
Vay j o.fjo h of ~fic ptesSUYC +0 5ftt1npt:f,-oY presSlAi( Y(;{1>'o
ftrv he lilA VV"I
(con'i)
~________________~_ _..._ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _---J
1/- J.f3

/1· ~8 I (COY/'t)
/f:~.q97'~
1·0
a,e Po
. ,/ I;: (J·f''1JL1
'(1
I
O.q
o.g
.l- 0. '18108
71:~.75'8gb
y'
1'00.7
1
1
0·6
i
I 10
1
0.5
:
j
[ o.~
0.3 ..~. .. - - .. . .
0., .- .....-.e. . .. ..=".06 ?/~
1 V"d I:: 0.10101.-f). I ... / .. Iq
0
$
--_. -- .- -- - ..- " ------
TerJ1pe r-Q.1u¥t, enirop'J dJa9yam Iw he//u~
(c.~n 'I; )
11- lflf

Over- and under-expanded duct exit flows will occur on
approximate paths sketched on the magnified pressure variation
graph below when the ambient pressure of the surroundings into
which the duct is discharging is respectively greater than and less
than the flowing fluid p-ressure at the duct exit.This illustrates how
the flow adjusts to these pressure differences through oblique
shock waves that involve irreversible and thus non-isentropic
flows. When these two pressures are equal, the flow is "ideally
expanded" and the flow into the immediate surroundings is nearly
isentropic.
1.0
><. (-ft)

II. 3'1
." 11.]9 An ideal gas enters supersonically and flows isen-
tropically through the choked converging-diverging duct de-
scribed in Problem 11.38. Graph the variation of Ma, TITo. and
plpo from the entrance to the exit sections of the duct for he-
lium (use 0.05 1 s: Ma:S: 5.193). Show the possible fluid states
at x = - 0.6 ft, 0 fI, and +0.6 ft using temperature-entropy
coordinates. Sketch on your pressure variation graph the non-
isentropic paths that would occur with over- and underexpanded
duct exit flows (see ilku V.l l...l) and explain when they will
occur. When will isentropic supersonic duel exit flow occur?
Thi! i! si,.,.i/aY' w E)(ample // 9 ·
Thi! f,-able Wl involve! the dud of fydl.kM /1'39. IId~vey fht.
flaw enf~Yf fv,,~ndY1;"'//y . We. CitY! usc value, fro"" the hb'es
"f p,.,bk... II·31 wi~ " li Hle yeRYYtlnq~mthf to (/~U>{JI1i fur Ik
;vpersd nic enf&¥IYJ9 ('low.
For heI"~.., Vie- hMe
FrtJyYI PnlfYa,n-, Ir€AJT1!df' wi~ ~. /. 66
M... ~
£1,2. oFII.J'!
r:X(ff) .!.. st.IG
A¥ T, Po
'S Upe Y'SIJn i"- SIIII{#,,'"
5 .IQ3 1/.0 -/.00 O,/D"7- 0.00J13 a.
~·wv 7.Q _ .:10 O. ln8~ O.DOiZV
3 .'V7 LJ.b - ' ., 0 0./0556 O.o/VV6
2.778 2.6 _0.40 0.211~S' 0 .0"1'(1
I.S~5 /.4 - 0· 1.0 O'~1817 0.14811
1·0 1·0 0 0.'>188 O·V1B08 b
1 . 1~5 1·4 O· 'l.D O.V611'-, O.I ~.1J
2·771 'H o· ~o 0.211115 O·041VI
J.6~7 'I.• 0.60 17.1 65>6 O· dNq,
4-441 7·4 O.So o.I~t.82 O. ~lV
5.113 /1. 0 f.00 0 . 10/07, O.003f'J C
5 lAh5bnic. Sol.hOn
OHiO 1·'/0 0:1.0 O. q1473 O. fYi/6
0.223 2.6J O·YO O.'8JB~ O .1f'8~
0·123 lU,?. O· 60 ~.'i'15'3 O·jr7'5
0.076 7.4J 0. 10 0."", 0·19522
0.051 /I. o. /.oa 0'''9/¥ o·,nlv d
(con't)

//,3q (con'l)
/. t>
oL---------------------______~~======~
-/.00 -0.8 -D.&. -08 -~.2 0 6.2 0.'1 o.b e,g /.0
/.0
D.E
0.'
I
0.4
0.2.
x (f+)
Von'afi4n of Mach lIJuMhe~ Iuv he//uJ'tl.
o ~------------------------------------------~
-1,00 -o.g -~.b -c.If - O. 2 0 a"2.. (). '+ 0.6 o.g /.0
x{ ff)
Va riafion of 'SftJ. h'c-- ternperaiufl- 11 '> M5V1ttHOvt femper4.lrJlt r~ H6
//- 1f7

11·3Q I (con't )
than the flowing fluid pressure at the duct exit.This illustrates how
isentropic.
x. (oft)
·0
II-tfi

1:
~
(con 'I)
/.0
o.~
0.1,
().Ij
0,1
o~======~~-------------------=~==~__~
-I.OD -0.8 -0.6 -0, IJ - 0.2 0 fJ,'2- O.Lf 0.6 0.1 ,.0
x(f+)
Varjllt./ioYl of 5~n'-c.- preSSf).re. -fr; r!-P.?I1.4t-h'oY p~SSU't yaNo
fr;y heliuW
D.q------ - --- -
~.y - -
0.'3
f: o.qt:/7K'I
P,
r:. =0.00313
Po
0,1 - - j L - - - I::; ~.IOIDZ
~
o l.--_ _ _ _ __
s
T - 5 d/aCf yaWl -{Py he //uW.
1/-.'19

1/. LfO I
'<1 lAO Helium enters supersonically and flows isentropi-
cally through the choked converging-diverging duct described
in Example 11.8. Compare the variation of Ma, TITo, and pipa
for helium with the variation ofthese parameters for air through-
out the duct. Use 0.163 ;s; Ma ;s; 3.221. Sketch on your pres-
sure variation graph the nonisentropic paths that would occur
with over- and underexpanded duct exit flows (see Viden V U A)
and explain when they will occur. When will isentropic super-
sonic duct exit flow occur?
.A ) T ~ 11t1 .t:. w ift-t fV/o... .fW he /itAm W~$
A1f To Po
o6-kJ./~-ed wi~
Mtte-L-t. numb.e.Y"! In
/iI1PtA.T. Values of
PYb9r4.",., .ISe N71<OF IN i It, It =/. 66. values of
tI1e Y'a.n9t:.. 0./, 3 ~ MfA. ~ '3 ·22 J w~e. user! ~5
)( Vtere ob-/QlitIJ/J. wi#,
)( • j($,) 0./ - O. J
whIch is £9. 5" of Example 1/.8
The VQ-yianah ()f Ma" T anol ?
~ GKtUY//'le 11.8. h-eli:w, Po
X(rn) I Ma 'I.. f
r; Po
- tJ.5
-0.4
- 0.3
-0.2
- O· I
o
O. I
0.2-
(!J,J
().'f
tJ. ~
(). (
(). 7..
0.3
0.4
0·5"
~, 2.21
2.17~
:;'.;2.0
'.85'9
,·lIO Z
/.0
J.'-I() 2
/·95'8
Z.~z. 0
2·778
3.221
(J.6~ 2-
0.%0
0.116
O.:2Z3
0./63
0.22' I O. 02~g
O.Z~2.0 0.01/14
(J·?602. 0.0767
(J.46t3 O.Iq.g)
O. ,06, O.21ltf4
0.7519 0·'188 I
O. ,06'
O.46g3
0.3'02-
O.29z0
O. '2.2., I
O.i66 't
O.'13 '11
"",(, B1
('j.'k31j
(?',9/3
0.2. "1'1
0.1"18 '3
0.07h'1
~.OLfI4
O.oz38
O·6't83
~. ~ If 39
O.~l.17
0.95 'C;
0/11B3
(can't)
/1-50
ye(J. Yrant'jeol. -Iv y/e Icl )( .
with. x !;;y al'; tJbtztIJ1eJ
M~
2.80
2.ljg
2. /L.f
/.76
'.37
/.0
/·37
1·76
1.14
1·48
2.~O
0.69
0.'-17
0.'32
0.23
O,f7
air
I
To
O.?;39Q
O.LI¥8,(
tJ.s2/9
O.617S'
0.72.7J
0. 83~3
0.7271
o.617S
tJ.~219
o.Lftfg'l
O·~SqLf
O,Ql31
Q.QS77
o.'77'1?
O.i8''15
0.991{3
.E..
Fa
(/,0369
(J.0{04
0.102.7
()./8tfii;
tJSZ78
~S]23
0.32.78
tJ.I'Jso
0./027
a.0604
d.0369
tJ.7Z7'I
().8~'
0·'13/)"
0.'1638
tJ.'l8ao

1/.40 (con If)
1.0
MtL
2.0
'.0
. /""
hellun.,~~
~ .
~ air
o~--------~----------------------------------~-D.S -o.'f -0.3 -cl'Z -0./ 0 0./ at..- 0.'3 D.4 a.S"
X(m)
/.0
o.i
I
7;,
0.(,
0.1./
O.2~--------------------------------------______~
-os -0.1.{ -03 -0..2-- -0.1 0 ~ I O. "2-- tJ. 3 (). l(
X(hl)
/·0
O.i
0.2
o ~--------------------------------------------~-o.!) _o.¥ -0,3 -a.2 -0. I 0 O. I a"2.-
x( rn )
( c,()n I~ )
//-51

/,. '1-0 I (c()n 't )
isentropic.
. ()
x(.ft)
//-52

) I· ifl
"1 JA j Helium enters subsonically and flows isentropically
through the converging-diverging duct of Example 11.8. Com-
pare the values of Ma, TITo, and plpo for helium with those for
air at several locations in the duct. Use 0.163 :s: Ma :s: 3.221.
Sketch on your pressure variation graph the nonisentropic paths
that would occur with over- and underexpanded duct exit flows
(see Vith:,1 " II.") and explain when they will occur. When will
isentropic supersonic duct exit flow occur?
The. vari(?fion 01 II ) T (J/Yld L wift., Ma ~ hel,'u1'Yl wtJS
~J(- ~ ~
06 fr:uned with PYOfyt:LVYI IJGNTROP tv,'ft-t k =/.6€. l4/ues of
Mach ntAYltbeys '" -the ytll'1re O. 163 ~ Mel. ~ ?'221 ~ye. used as
Inpuf. Values ~I' x weY'e r;6klnetJ witt..
x =/:") 0·1
O. I
/1·8 tetlYYtV1gec/ fz, YIeld x.which is £fj. 5 of &)((';l 11'1;;/<-
The., vayia.fjon tJf Md L
) T..
fhm, EXtffflp/e /I. 8 . I)
t1nd.E.. w/fit x fZ;y a/yo was ()6m/~ed
p.
x(WI)
-0-5
-0."1
-0. :3
- (). 2-
-0· 1
o
0.1
0.2-
o·?;
0·4
0·5
0·[
0·2
0·3
().L{
0.5
O./l> 3
O.z23
0·3/h
0.'1'60
(J .un
1.0
/.'102
1.658
~.'32.()
1.,. 778
3.'2'2.1
O.6~2
O.q.~()
0.3t6
0:2.2~
0.163
helium
!
TQ
(J.~9/3
O.'j83t:t
0.<)661
0·9'347
O. ~'6 q
0·751'1
P Mo
Pc
O.~7~3 0.l7
0.'5"9 0.2'3
0.92/7 0.37.-
O.gl(3Cf 0.47
o.('QI3 0.69
0.'1'68/ 1.0
supers~l7i c. sokrh'on
o. 60'~
0.4653
0·3~02
O·2g2o
0.2.26 ,
0.2'&44-
0.llf3;
(J.O]67
0.0'+1'1
O.C03~
/·37
,.71,
2.14
2·1ft
2.g0
5ubscnic... 50/r~:HoYl
0.866'/
o.tH~7
O.'J6H 1
o·~83cr
0.9913
1/-53
o.6Qs3
~. g-'f3,
o·97.17
O.9~¥J
O.978~
0.6'1
0·lf7
0·32
0.1.'3
0.17
T
T"
O.c(9Lf3
o.,)8'1S"
0.979'7
0.'/577
O.l]/~1
(J.~33
0.7271
(J.bOS
(J.~219
o.'ttftl(
0.'18'1'1
O. '7131
O·'jS77
o·9799
O.989~
O. &)1:1'f3
1:.
~
O.98{)O
O·C]6'3Z
(7:=]315
O. 'l5~b
0.727L/
0.5283
0.3278
('-Ig50
0.1027
O.06r)'/
0.0%1
O.727J.f
0.851"
f)'''31~
(}.~6 3g
o.ljgoo

//.4' , (Con'f)
I
-·--1-- _
j
i
, i
I
~.2~------------------------------~------------------~_os -0.1.{ -03 -0.).. -0. I 0 Q/
O.i
·~O.b
P
p.
·-·o.~
I :
- 0."2.
X (I"'n )
O.V fJ.3 0.5"
O·L---____________________________________________J
-t1,S'" -~~2. -0.' 0
Ix(rn')
0.1 a'2-
(ctJn It )
-0;'3I
1/-5"1-

11·'1-1 I (C()h't J
isentropic.
x.(.ft)
1/-55

/ /. LfZ
11.42* Helium flows subsonically and isen-
tropically through the converging-diverging duct
of Example 11.S. Graph the variation of Ma.
TlTo, and p/Po through the duct from x = -0.5
m to x = +0.5 m for p/Po = 0.99 at x = -0.5
m. Sketch the corresponding T - s diagram. Use
0.110 ::s Ma ::s 0.430.
Thi S' is like Extlt?1f/~ /1./0·
Foy he Hum we.. lA ~e
values of
x =
IJ T
.A~ ) To
+he.. yange.
/A~(:~)
pY09ratl1
().nol f
f!,
0·110 ~
o. I
ISEN7f(OP
coyyespondlnJ
fl.l(tA.. ~ o·'f3o .
wi ft.t
+0
We...
which is
chol<ed /
based On E'fj. z fJf exP.I'J'!fIe //·8.
Af=
A:I" F A
+J,n~t
(Ii )AAI -fl.l'o~f
Ie: /.66 +0 defc,ymJn~
va.lu.es of- M.a...
ulclAlo..k- x. wire...
(I )
S/nce the.. flow is har
wne-re (.R 1 is ofo.!-A,;'eo( wi~ P'f'(xi'{'·aw,. .lSfN1X~f fw. Ma.:: ().o/Jo.
Aol' '-fYI't~(}o..f
Thus I
"L
A*
O./m O. Of. 8'
,.
= ::: Yl1
/.'17
al'fcJ.. ~1' I heL~U
X == !O.~68 (fo) - (J·t (2)
wi ft1 f(r 2 W, ft... e.. ro,,¥a.~ IStNiROf wiftt It=- /.6'
I A I. f.
,
x{Itt) MA si&de
~ ~ ~
-0.5"0 0./10 15./6 O.CJ%O 0.<;900 a..
-0·1./0 O./l.fq 3.8'3 O.~'2.7 O.'J'lIV
-O.SO 0.206 2.~{ O.']g62. O.%§f::J
- 0.2. 0 0.28g 2.06 0.9734 O.93Y3
-0·( 0 0.3&1 "b2 0.95'13 a.ggqo
0 O.'f30 1·47 0.9425 0·81,16 b
0·/0 0'$81 1·6z. 0.95'13 O. tj~c;o
0.20- O·2gg Z.06 0.9734 0.9'11./3
0.30 0.206 'Z.~I 0·~g'6z (J.C}6?'
0.'+0 0.1'/-9 ;.~3 0·9927 o.~8Ig
0.1;0 O· (/O !;./ b 0.<]960 o.9~OO c.
(can 'f)
//-56

/1.1./2- (con 't )
0.5
O.'"i
Ma..
0·3
0.'2.
0./
- 0.5 -o.z.+ - 0.3 -O.l -0. / 0 0./ 0.7.. 0·3 0.4 0.5
X(Yn)
1.00·
O.9g
T
I~
0.%
0.9~
-D.5 - 0.4 -0.3 -0.2 -0. , 0 o. ! O. ?. Q3 0.4 0.5
X(M)
/.00
0.95
P
P-
o O.')fJ
l!.~S
0.80
-0.5 - O./.f -0.3 -0.1.- _0.1 0 O. { ~.'].; ~.3 O./f (j.S"
X(rn)
F :. 0.99
Po
,.00 7"
- =0.996
To
0.<)3
I
70
0.91.
f.. =tJ. <l?'1
Po
O.9~ I : (J.9lf25
7'0
o.n
/1-57

/I. i.f 3 I
11.43 An ideal gas flows subsonically and is-
entropically through the converging-<.liverging duct
described in Problem 11.39. Graph the variation
of Ma, T/To, and p/Po from the entrance to the
exit sections of the duct for (a) air; (b*) helium
(use 0.047 :s Ma :s 0.722). The value of p/Po is
0.6708 at x = 0 ft. Sketch important states on a
T - s diagram.
ThiS is like Example 1/.10·
Since .f..::: O.~7()g at- x'=' 0 is qtettfe.r Than
Po
e..>!'= = O.5"2gg fov air (sa..
F;.
fY'ohlenn 11·3(J(a,) stJ/u-h'f)n) and O.¥8fl fo,... hel/uYn (s(e ;:n>6/e""" I;'JO~)
50IlAh'oYl) -the. a/r aY'JP he/,itYVI I'ltJws fhvdlA9h the. c"ohver9Inj-d/veY9'ny
dud ()ye not choked. Fzrv values of t ()f dif!erenf values of x we
Ohfr:l/~ Coyve.>ptJndiY'1j vtllue s of Ma; ; And ?
• • d Po fll(o.) Foy aIr we entey- F=la., P.I wift.. values () -Iv 9e.f M~I T
./ II~ or
~ fiy.[).1
we gei
A = /,oS
n,r fhe
1
A = 1).1 If .
{IIJW vcr/tie of .E..
~
A ~ = 0./ f-f 'Z::: O.~'15 ff'1
/.05
A ~
x'l.-fc./
-::
~'J. t- O.t
A* II~ 0.,,'5
The, UJYyfspOYlcl/~ va /u~J of ~ Na.,; I
IJ~
)
~
Fij.D.1 a~ al.>o -fo6/111a. k PI 0 JI fhe Ylex f ra.ge.
~n't )
1/-58
If)
A (lIf x::: 0
(lJ
lOfA .e.
Ii
d.t7ol
)
Ir~

/1.43 (COtllt )
Witt. ~.I F 1""01'.. Fi9' D·I
X (f-+) A M",- T P S~teA-I'
To ~
-1,0 1/.6 0.05 O..,q O. 'j9 a"
-0. g 7.8 O.og O·Cf9 ~·19
-0·6 1I.g ~.IZ 0.'19 O.'J~
-0.4 2.7 0.2.2 O.tt'i 0.'j6'
-0·2 /·S ~.44 ()/jb 0.9.7
0 1.0 0.78 ~.H 0.&6 b
0.2- 1·5 t1·LfI.l 0·16 O·~7
0.4 2./ (J.2"1. 0.'19 O.':U·
O.h 4.2 0·/2 o..,., 0.4!f'i
C.g 7.8 0.08 0·19 o.'t~
I.D 11.6 a.o5 O'&]9 O.CJ'
c.
/.0
0.8
0.6
MtL
O.L1
O.L
0
-1.0 -0.'1 -D.b -0.'1' -D.'1. 0 P.2 (J.t( 0.6 Oog ,.0
x(ff)
Van'af/oY! 01 M~ch number ~ a;'y
/,00
O.~5
I
r;
o.lfo
~85
0·80
-/.0 -O.~ -4.~ -b.t..f -0.2- 0 p. '2, O·Lf 0.(:; 0.& ,.()
x{ff)
Vaviah'on of slztfic femptya/urt. -hJ sirA.'h~f,dn .f~J71p~Alure rdno
fDy 41r
(con't)
If-59

II. if 3 (Con/t)
/. D. r-----....'""'---
0.'1
E
Po O.g
a6 ~--------------------------------------------~
-/.0 -(J,g -D.6 -fJ.'{ -~.... 0 I.'], ().4 ~. 6 tJ.8
x(.f.f )
Vayj~fi"Y'I of s172h'c. pyes.ruye -hJ $-Ia,nlf,h'em pteS5ure y-af/o
/7Jy air
1·00
0.95
r.
TD
0.90
0.85
1: :: 0·'79
PD
! = 0·"''1
1D
4'<;0 ~------
1-5
,
aIr
I. ()
(b) h¥ hel/IA~, W~ lAse PYO'jyaYl>1 .zS€N77et1P wit4 ~ == /.6~ ~
defCvrnll17e va/u.~s /)f ~) L 1AJ1/1 P
A~ 10 Pe,
wi tlti,.". the ral?ye. 0·17 ~ MIA ~
U;yy~>I'Imdl'nj -fD VIJ/I.(~.1 "f
0.7'2.2. We. CAIc.~/~ k X wd!.
(I)X;; JA~ (~71) - 0./
Wf"ie"h is blJ.s~rJ. 0".. G'i' z. "f PI7>J,lell1i'L. ll·31j. 5 ince #ais lIowi5 no!
choked) A"" 1- A+hY'OtLf' ThlA.s we d.e~I'tI'I;V/e Il ~ w/J-{...
1/"* =- AtitnJAi
(r»n't)
1/- 60

//. If.3 (con't: )
(A14J.y()4.f
where
7iJusJ
flK- ::: 0./ ffl.::: 0.0'1115 if l
/.0135
and ~. I bec(}me.s
x ~ j""'(O-..(J-9-3-IS-)-{-:-:-~-)---O-.-I
Wi~ ~~. 2
X (if) M~
-1.0 O,O.ti7
- 0.8 0.071
- 0.6 0·/15
- f). if 0·207
- ()·2 ~.tfiq
0 D.722
0.2 {}·lfl't
(j.Lt 0.207
().~ O.{/~
D.<J 0.071
'·0 O.()47
o.g
0.6
Mcu
0.4
0.'2.
0
-/.0 -d.g -D.b -0·4
Va YIP" fioY'
I.00r---___~
0.95
-r-T., Of}o
Wit{" I>ENTP..Of
A rAll' 1'0
JJ.qq~lj o.m3
7.ttS'" ~."g3
4.'371 0."S7
'Z.7f1n 0.'86/
1.5IJlf'i o·'If52-
J·0735 O.gS]'}..
,·sol(q 0.'''52
2. ,Q73 0·9161
4.'t378 O· ,,'57
7· ;5'1' O.9H3
n'f1f3'i 0"~'3
-0,1.- () 0.1.- (J·lI
X( ff)
t7f MA.cV, nt..o-."bt:IY f&v
(2)
.f ~k
oJ~~:z. ().,
c;.9959
0.'719/
0·96£3
0.3' 7'1
(7. '708 b
0.'fJb7't
0.,65J
o·n'!'
O·~95g
O·lj'rn. C
(J.t, 0,£ ,. D
he//U;'YI
O'8$~-------------------------~--------------------------~
_I.~ -0.8 -0.6 -0.'1 -0.2. 0 0.2 '·4 (J.(, (J.g /.0
X (f+)
va.,/~.h·Qn of s~h'c. fem~(1.n.ye -fo slz<'jl'laf/t7Yl
(COI/1' t. )
//- 6/

II. 'f3 (UJI1'i)
f
0.1
0./,---------------------------'-1.0 -O.g -0.6 -0.'1 -fJ.J, 0 ().2. 0."1 (J., b.~ ,.0
)( (.fI..)
Va "/A.h'~n of smfle. pye$,H(y~ h, sf-a~/I'f'. h'o" pYe.JSuye ra.h·tJ '" he.I/~~
~.8S
f.. .. O.Ho~
Po
I .. 0.'1532-
10
0.80'--------
/1- 62

To
Foy
air )
11.4Jf An ideal gas contained in a large stor-
age container at a constant temperature and pres-
sure of 59 of and 25 psia is to be expanded
isentropically through a duct to standard atmo-
spheric discharge conditions. Describe in general
terms the kind of duct required and determine
the duct exit cross section area if the discharge
mass fiowrate required is 1.0 lbm/s and the gas
is (a) air; (b) carbon dioxide; (c) helium.
de leYJIYII;' e -fhe duc.f ex;! flow aveC(
.
Aex.i-!-
= m
t£jr ~)(it
fhe ex; f- flokl ctensify I /eXi/- ) we
~;9. O. / . Thu$ I
]~~I)
~)(if
::
/;/ /
(i!- ') Ma 2. .I t-
2. e'Xd
vve.
US(!
where.
~ (jlr
~xit = ~
RJ;
( ;::I). D· / valu.. of ~'I)
J;
us~
(j)
Er //·60 or loy
We.. obrn,;'
Fig- D. /.
Mq -frdWI
exit
?exit uSlny €-1- II- S1/ tJY ~ t:IIY/
Ma.
exit-
/VI4 e)(i f-
lo o6-h l'n
7htlf
~----------------------------
-r~1i;.1) - J(t;) (#-)
F;j. D. / valuf:- fA S fA /uhc.-//()Yl of
,t;
(5)
lI~Xif :: (M4eXi f-) C = (M~jt) f//~ ~)f *
(con't)
'------~---~~------~----------..-.---------~
/1-63

/J./fLf (COIJ't- )
where fr.trm E~. /I. S"6 I OY IPv clIr" ~ Fiq· P. I
TeX.it - __7;____
I t- 1~-I)Ma 'l .
( , 2. exJf
(JY ~ air
7: - ~ ( r.TeXi!' Vel lite ~ F'1' D,I .fW A1t;.~w/'f)~x.i f - """.
"
anA thu!
Vexif = (M4eXl'l- )
(tlJ Fov air
~x.if ILl· 7,os/tL Q
= tJ.5 I
The
-t;; 2>",PS/A
covyesjJona'IYIJ val",es
lex.ij- _ tJ.86
- -
To
~xil- = 0.63
~
Now win, £'1. 3
::: (25"(J$/a ) (1'I't ;.,~'l. ) (0. '3
(17/6 f'f· 16 )( SI5 -IZ-)
S"IUf. ";2.
(UlYJ'+)
IJ~ 6f.j-
)
(5)
-J
2 ·7S' )(ItJ .1'114.J
.ff)

//.'f4 (C()J'7 'f)
'P) = (). tJ / z If 2.
(2.1, XIP-l sl,; ) (q32- ~I- )(nZ~)
H $//'(9
11 Ct;n l/f!/rJIYlJ dwc,./- would. 5ufh'ce $/;'ce ..fk eKil- fttJw i.s
S lIb?Sd>1 i C .
~Xif =
-3
if. ()J' X 10 or !U'j
Wi Itt ~1' $" we fel-
~lC/f = (0.1123)
-(./3
(eon'c)
/1-- 6S
I
-/·3 - /
/
-:. 76s: 7 /1-
-.J
S'u.b.s())1/c . !

II. 'fLf
(C) hJr he/,ulYJ ; Gj. 'I yi~/d5
r-----------------------
=0.3'1'1
Mae)(,j f
/
-/·66-1
-¥
~.05'6 x/a f/U!
rf3
WI/-h ~1' t w€- gef
1/ = (0.81.1£1)veXjr
Wi ff,. 1:-). I W'-
(I ~)
5
--------------------------------:
(t.OS"6 )(10-ii SI/,(~)(2lff'f fr13 Z. 2 16no)
~3 T~1
O. 030? {flo
Aexif
A C()l1VerjlrJj I?tJl}/C w;J/ d{) 51V)ce -Ik exit f/()w If f"b.xmic.

1/. L/-
11.45 An ideal gas is to flow isentropically
from a large tank where the air is maintained at
a temperature and pressure of 59 of and 80 psia
to standard atmospheric discharge conditions.
Describe in general terms the kind of duct in-
volved and determine the duct exit Mach number
and velocity in ftls if the gas is (a) air; (b) meth-
ane; (c) helium.
-10 de/eywz/ne the. dlAcJ exit Mach hlAm6ev Ma~'f/ XI
E~.I/·s"'j 0,,-
Mer 'r =ex,
Ov f,v IJIlr
/lA~
ex.it
VeXif
wheye
1exit
1C~if
(a.) Ft;y CUr
Pexit
-~
::
==
.fv.,. Ildr J Fi9' 0./ . TAu>1
/ ](~~)
Fi9- f).1 VA/tfc. 'if ~ Ibnc b-on (7/ ~)(.il-
~
(]x/f ve/pci.j,A v,c: 'f W~ IAf~
'7" XI)
1-14,,,.if. YR TcXif- -k
:; ()./f38
9" /sid.-
) we lA~e
(I )
(2 )
{"3J
(5)
v.JA.d f1,u,> kvw. F /9. 0 '/1 ~ c.o-rr€$fol1d!n5 Vt7""~S are,
M4
'
)1.i/
:' /·Z
ttJt.d
"Tex/f" = tJ.b2
7;
/1- '61

1/. £/5 Con If)
!he ~ wiftt. ftj. 5 we ohfa.tJ1.
/.. ::: (S?'1DI<)(~.61.. ) -= 322.
rJ
p.,.
ex-if
CIP'II:A w;fh Erg. 3 we C-vn.ctude. fha f-
Vexit -: (t."l) (17/6 -FI-./I; )(~22 ~1!.)(I.Lf) = ISSO ti
SI"1' 0.e ( I ~ ) = J
• ~/Lilj. !!:
fJ C()nJleY7/~j - clive,Yf'J1j n oJ)/e ; r yei:/~c1 6ectUUe M..t. ~xif
:f/()W i; 5'u(J(!'ySdn/c-.
(b) ~ Me..tiuJne we. dblrA.,~ ~ €'1' I
M()exit ::: ~ / ,,')1-1
(!~. 7~s;a) Tj(
gO~r/~
Then wi Ih clJ- 'f w~ gef
~x;t =
5'19 III?
1+- (1.31 - )(1-7<fJ~
3 iNe. l"fr,u;"
A c(/m)~/~ - dt'~71"'J
fI(fW if .5(Al'ey~()nic-.
(C) Fw he//ui'YI we cJ61a,;'" ~ Gj. I
- /
They, with
- 26 'f. I "/?
(VOy/lt )
//- 6e
= /.78
e-tiJ
- /.71

1/. tf5 Con't)
Ve'{ir = (1.71)
= 577~ if_ s
/J C!Mllw,t'nJ - d/.)t:IY'iJ~J t1/)31/~ I) ve5",yed Slnc~ the ex.jl-
fft!W I r 5'lA(JeySdY1;e, .

II. '16
ro
0)'
10
0"'-
To
Oy
11.46 An ideal gas flows isentropically through
a converging-diverging nozzle. At a section in the
converging portion of the nozzle, AI = 0.1 m',
PI = 600 kPa (abs) , TI = 20°C, and Mal = 0.6.
For section (2) in the diverging part of the nozzle
determine A,. p,. and T~ if Ma, = 3.0 and the
gas is (a) air; (b) helium.
de-feYIfII;'e Alo we use Ecp. 1/.71 0'1' rvY' t;l!r.) PIt]. PI/
-It 1- I
(~~ ) j;~ [
11- (~, ) M4; j",(*-<l
A:z. II, A, f r{1I..~I 2::: -- =
(:;)
~+I
.L
I- I t (~) Mtt,2J:l.f*-I)
fo1q, , ~ (!~ f)
lIN .a,y
A -:; A, !(F;').;..
( Fi 9,
dekH-rn,~ e. { we
~
(?-)= p~ = ~
I (P.)
~
frrr airI
p. ::::. i(f;:'9' 0· I V6/U' o.f P,. f.... M<i ~ l]~
p,
2-
(F/9 f),,/ vdlue. of..& fw M.&o,)
defe.,m/,,-e T~ I':, Iw' (:'9. 0./.w~ (A.s-e. G'-g. 11.56 or ti ';"/
(~) [, + (~~')M~,~])7:2- ~= T 7; =, --
fT.) [ I r-(~-I) Mtj/" J7;
.frH 2-
a,Y"
T ! (fiq. f). I Va llA.!. of T~ f,.. 11,,,, )1T~ =- To
, (Flq D.I Va. I",,! or 5 !w M~)
'"(COn' f )
11- 70
(/)
(2)
(3)
('f)
lhtls,
(5)
(~)

(~) p.()y a/,. .. £'1- 2. 1w.S' -Iv
A2. : (0.//11') ('1·3 :J
(I. 2. )
£9' 1./ yt'elds
~ : [ 6o(J 1t.1i{~"r)7
aYl~ ~f" 9/lle s
r :: (293 fC..) (0. ~6 )
:L {d.q~}
= 1/3 I<.
:= O.2.S7n-?L.
11-71

//.!.f7
11.47 Upstream of the throat of an isentropic
converging-diverging nozzle at section (1), VI =
150 mis, PI = 100 kPa (abs), and TI = 20°C. If
the discharge flow is supersonic and the throat
area is 0.1 m~, determine the mass ftowrate in,
kg/s for the flow of (a) air; (b) methane; (c) he-.
lium.
we. de feYJI'JIIIYJe.. fhe Mach num6ey- at sec//on. (/) w/It,
(;)
ror a/I of Ih"e .1tl~e.s
he cause ~ I slow.
fAe. ~ YI/wln3 flow is
Ma~s Illlwyak. We use
//JvoIVu:/ i-l- is /;/::.(!./y fhcd MQ, is /t!ss 'lhdh /.~
TJruS'.. t1, t.. .flow ~f Ih e.. -f),Ylx~.f /J rJ?1l/.:.ed 11;"~
~tl6s(')n/c. tl¥Ja. the. leaj//~ HIl/AI is sVl'erstJnic. my
6fj . 1/. '1tJ -fo 0 61rJl'n
,;, =-,..<J ~A-¥ V of (2)
FoY' fhY()tlf ve/~ci fy / V~ we, lAse
V*~ VRT~-k (3)
To o!om/;' T ~ W~ use ff· I/, 63. ThusJ
(tf)
~r .fur cur)
T~ -= ~ (value of' T ~ 1=,'9. O. J fiJv /I1a -=- /.0 )
7"
(5)
To de/e..ym,ne TD
wt:.. use €~ . 11-S6. Thus"
~ =: r; [/ r (: I ) ~/2 ]
~y 4/y
,
T,
(~nll::)
//-72

//,47 ~OP'l't )
(q)
or ..foy tllY)
p¥
=Po (vAlue of j?'" ~ ':::;g_ P. / ~ /Vtt:(. =' I- 0 )
Po
(I tJ)
FaJ' Po we. use E'I_ 1/-$'9 - ThMc;~
frI.
~
::::
P, [ I f- (~; I) /v1Q,L ]r-i (II)
oy fw t[ ,;,.
~ ::.
?,
(va.lue (If f: .;;~ ;;'9. 0, 1 ,tOy Ma,)P,.
(a) '::w aIr we use G$_ / h LJ !,-hi,;"
Mil, (). tt 372--
2.'13 k...
:: ]05 K..
( d_ '16 )
we.. tJ/:'hr I;'
/1- 73

Po -
we... ololA,;'
/00 iPtl (ab$ )
~.g7
w;tt.. f~e ~e}f of Fi~. a.1
= /I 5 1t..1't:i (4f?S )
6f. It)
pK ;;: [1/5 /tIjJI{.(llI;s)] (0. 52 g2 8) = 6~. 8 kJ'~{a"$)
7heJ'l wi fA ~f- 3
;;>1 ':::
(~t)B >(10
1
IJ )
/'I?'l- =. ".8'3 ~f
(236.9 N·n? )(25 if kJ ,.,.,3
H"na.t'J- I t#itt.
~l. I<. •
fi' z W"1! obI-.U'l
.
fJ· BJ 3)(0./ Yh2.)(:1lttM = ;) 2t.5 'k}:::;.
-.J
(6) F~,.. meihanG we- lASe.. Gj. I -fr; db~;"
050 ':) 03363
'(~/I· 3 N. YI? )(2.'i3Ie) (,.'11 )
~.I< (I L )
a,.l!!.
Ilow is c-tuk..ed ",f- f3.. +J,e
-::: tJ. 336']
~ :=. (213 ~) [I +- (t.1~-1)(0.J163)).] - Z~!·II<
and.. i#; ~ cf Lf we. ohm 4;"
T ~ ::: (2.-93·/I<)C 2.. )
/.31 + I
-= 25"S./ K.
Th.<l!
v,f. - (Slg.) '!:..!!!.) (2.~C.II<){1.31)
1c.,. I:. (I ~
. *7. !!!)F-~ £1'// we.. ob!.,,.. f~ ~
~ ;:::. [/()d kl(((4ks)] rI + (t.1~-') (tJ.3361/] '.11-'
()ltd /AI iIt, Elj. 'i I/X- 9ef 1.11
D.JI ::: [i07. ~ M(4bs)](~)/.7/-1 -::. 5'3.5? k
r t'.31-t-1
1"heJ/7 WI ~ E-1· 3 we. ho,lIC
;y (5-g. J x 10 '3 !,.. )
1--_ _ _ _;O__-.::_~~~31:~ )(253./~)
~J
= o. ~3S ;;;j
(con/f)
---------------------------~
/1- 7Jf

/1. If 7 (con l-t )
r/yzally, witt, £~. 2- we ohm.,;"
vi? == (0.tf3f ~f)((J.I hI·~)(~/!.6 ;')
",,3
it.~_ /8·1 _
~~ .s
_ 0./'/92..
7; = (2-q31<) [ / + (1.~-I)(O.14.ti2)2] = 2.9>.21<..
aMrA vJ;tt, C1. If Wf!. ()6 m.,"n
T.r = (295".2. k / _2_) -= 222. k.
"ZI. 66 1-1
~ :: [ioo 1efl{{aI,~)][
and.. wiHt fq. 'i 1
o .f, 6
p.r::: /()1.9 fq, I'f( (4'S) - :::[ J( 2 )'.~-'
/.6( tf
Thc¥I wi~ £,. B'
3
f1 if ~ ('If 7'1 X 10 ~)
- 0.107'7
(2077 ~ ) (22J../<. )
ai·1< .
r:,·Y)().lly wilt. €~.].. we 6/;'-IA II'?
I
/J.1
-
1/-75 .
-s

11.Lf8 The flow blockage associated with the
use of an intrusive probe can be important. De-
termine the percentage increase in section veloc-
ity corresponding to a 0.5% reduction in flow area
due to probe blockage for air flow if the section
area is 1.0 m:', Tll = 20°C, and the unblocked
flow Mach numbers are (a) Ma = 0.2; (b) Ma =
0.8: (c) Ma = 1.5: (d) Ma = 3.0.
We won-l Iv asceyll1/~
~l!JblD"J::.LC'
To de'lwvn/;'e fht. Ul"lb/f)c):,ed aveA.. ve,t"c;1y / '{~bIDc..b') we U5e
V ::: M~ ,/R. T k..
U 11hi()c/c.e.d.. M" "IfI ,,/cui V U~bIDtJcc.J
H;y ~Yl6I()C.~~ we lAse
{lJ
T ~ 7;; ( I ~ Ma. rn.,.,. E~. /I.?G ~ Ma J (z)
(mb/ockeol 7; Ullh/~d:£r/ Ii"VN
II) dc-~JII'1J;"e. I1u 6lodu..d tu'&. veLou'",~ ~/()c.kuI) ~ £A4..t.
~/()~J = ~tloc-lael ~R~,tk,fk.
Pw M~ I'Ve.. we. AJlochcI t)Jt1d dek¥~Jneb/d(.k..v/
A~
£f. / /.7/.
,!1bldc,h.t/ ;'"~'jfA '''t!r nr/p/ 4Atd ernrr·
A~
11-76
(5)

Iliff ~()nlt)
(a) Fiff M~ ~ tJ. l we. oh,*,;' vV il't, E1S. 2 CI..,d J).S",
tfI'/6/~d~
T II. = (ZQ3 t:-) (~. 9rz()() '= 2.tf()· 7 k.U ~f:I ~ c.k.ec/
Then w itt... E~· I tv!- ha~
r-----------------------
~h61"t4.I -= (0.2..) (2gb.] ~ f~qO.71<)(I.'-IJ
~.'" (' "',..)
We us-e e1f . ~ (,fn~ 11.7/ +v ret 1f .;=
46/ocJ:.e4
- =(o.19SJ(2.'163S) = 2.9'1'7
/1-;('
/JJ1tX w;/4 1d'$./l71 we olht/~
MI( ,~J =~. 2 ()/
blluc.u(
W/I"tt £~~ s ~ ~f
z,f"j:; K -= Z O/()., k
(b) ~ M~ == o.g V-/C obh,;' IN;/(, £,s:.2 D-ttP( 11·5"6
~nb/~tk.wI = (zQ31c.)(~.8g65Z.) =- 2.>"'13 ;:
men tN)itt £9' I we jef
~-------------------
Vlutb1oU:uJ:::: O. 8 (2gt.'! f./.I?' ) (zS9.!I<)(I.'1)
~j.1<. (/ IV )
We. USc.. ~'i.(.¥ CYltJ/. 1/.7/ !J.?- 1'>'1
A s~
r;~/8c1c.e4 -c: (0. 91S'")(/·03Izs) - 1·03]
lit¥:
aJIId.. w; It, £,,- /I, 71 we 0 hho;'
M~;/ockuJ ;;: ()·313
Wilt, E~. S we re"f
7i1"~ = 293 k.. = 2S"3.8 K..
I t-(t./f- /)(c1,g/~l
"2. (COn Ii )
/1- 77
= "·l/i] %
-

w; fh £,. '3 we have
V, -;:: (0.113)
j,Il}t/<Jl.r;/
tlnd
(~/I)cJ:e4 - ~11h1(J(,~) x 1110 ";>.. (2.b2.1 ~ - 258· 'I ;' ) (/tJtJ) =l.lf3 %s ~
VLtnblD~ (?t"i.lf ;')
{Cj hr MA... =- ,.S- M tJ);,Itl/;',,-
wilt, Gjs.2 ew.'{ 1/.5'6
r;~b/"~ :: (2.93K) (~,6g9ts) -= 2()2. / k
l),~n wi It, cf. / We. gef .
~J'fblodu.(/ == (I' S-) ;-(234-b.-9-~-·-:-)-t-2-0).-.-,-)-(1-"-tj)
V I F<7. ,.... (I /1/ )
We. use.. EjS.lf and /1·7, -h -I- Iz!. -;
7....' .f"2-
A bl"cJ.erJ. =' (~.qr;S)(1.17, "2-) -= I. /7
A-t
~Mci w/#t 1::7' 1/. 71 w~ tJ"-k/~
ana
(~/othd - II "/orkd)Urt X /00
(d) FoY ~ =- 3·0 we.. ~~;"I;' "",lit, Ets'2 e(hol II.S6
Gn/:'/t/du-t! '=- (2-1J 1<) (0.15 71lf) = /(JI/.6 I:.
Tht Yl w;~ Gf. 1 W~ gef
(ct7I1'f)
/1... 78

we.. {,(se E1f. if (J nd II. 7 /
A61od<.t..4 -:: (tJ 0 Cf9S ) (Lf.23lf6) - If. '2.1 3
-A~
Pl'7d Wift, £'g.11.71 we. ~bhl;'
Mil/ -:: 2. '19)
blocicul
With. E~. 5" we. 'let-
Ond
T
b/()ckuI
2'13 k..
I +- roll -I ) (2.ct9S)),
2.
w!. have.
('{)()d~~ - ~1'J61()Ci~.d) X I () 0 -G (6fL!. 't If- - b/,-/. 9 -;- )(i()b) =_ f1, bit] %
~M61odu/ ( 6Ii(. 'j ;J
-

1/.50
11.50 For Fanno flow, prove that
dV jk(Ma1/2)(dx/ D)
-=
V 1 - Ma~
and in so doing show that when the flow is sub-
sonic, friction accelerates the fluid, and when the
flow is supersonic. friction decelerates the fluid.
5f~yft~ wifh E'j. /1. qs we. h~ve
.L (/ t- '*M( 2) d(v'") d (M4).)
2 V~ Mal
FrtJm €$. 11~1~ we j1lfve
(I)
d (/.111.
2
) = d(l/2.jrI -f (1c- ,) Mal.] (2)
MAl vz. 2
COMb/niYIJ G~,), , And Z we tJbfa/Y1
~ ( /+ 1i M/./'") d (v~) _ [It- (i!-I) M4"] d~ / f :fA ft11l~)(:: t1 (3)
1/ 2 :2. v" 2. /)
.L ( 2 ) d{/.I1.j.2 Ma - , -= -
and
d (II l) -=- M ,/. f -k dx
V 2.. ( M4: 1._ I ) f)
t-/oW(. II'tW'
d(I/2.) :: :z.VdV
~s C(/Wl41/n,""..J 1:11." 4Hd S we ,ef
rf.!! = ,ck (~"l)( ~}
II"
/111(''2/-
When -fJ,~
Ft""" is .s~6.s1J11;c (M({ < I. ())
~
~/c,f/ol'l IA.C( e/eya.lcs 'lite -/114#1:1. On I1tL
i~ .f'vpeYJIJJlic (Ma> 1.0)., €f· 6 leads n,
n,iG -//on de ce(eY4!u #,e -/i",,/el.
//-80
(5)
(,,)
1£,. ~ Iet.ttis -h> dV = "/- tI"d flutJ
V
dtt.e.y ha.Hd ,..,4elo1 fhe. H~k/
d v = _ Q neJ/ In jl,ir Cl!ue
v

11.5/ Standard atmospheric air (To = 59 of,
Pu = 14.7 psia) is drawn steadily through a fric-
tionless and adiabatic converging nozzle into an
adiabatic, constant cross section area duct. The
duct is 10 ft long and has an inside diameter of
0.5 ft. The average friction factor for the duct
may be estimated as being equal to 0.03. What
is the maximum mass flowrate in slugs/s through
the duct? For this maximum flowrate determine
the values of static temperature, static pressure,
stagnation temperature, stagnation pressure. and
velocity at the inlet [section (1)] and exit [section
(2)] of the constant area duct. Sketch a temper-
ature-entropy diagram for this flow.
7J,/s is 51",aay 7z, E:xaYH/J!e //- 12. As (..x-pllJ.';'uI II? £-xa""'I/~ 1/· /2/
tlte I'YttlX/""uJ'V/ {'low ~ak thYfJH7h fIt(.. due-I will C1C.c"u Y wl,~n I/,~
(..tTn{/..""f af'eJ ducf choi:t!$ and f1tt Math nUWlber ar thl. c4tc.f
ex./f ['see-f/on (2)] is 1.1). lJr.e l'I'lax/munr, liowrak CAn be (}b/7l/~eP{
wi111
(f)
We VIOle..
ft,e. {low
the! 7; is
is a.4/alHlh·c -
fhl'ol<f1hIJI.l f T1te ~m{,~ hIJw J/"u
::: 1,,,1. =5/qD~_ .A/s01 Po is
CUn4l4nf
COlliS Ittn f
II? -the C.(}n ~rf i':f
Arelflf duc.f belAIlIe
nO.Jl/e.. bu.f
01 lYi~li()n.
ple&r~RSe.J ft,'(ll,,(fi. IA~
TAt/5.1 ~ =1'1. 7l'fia. .~,
kv c.h.ok.ed flow
f" ( "-1. - £, ) (~.()3) (It; Ff)
0.6 f" (,,(~-.f). "=' -:: =
J)
_ /
t). S -1+ ()
a~d from Fifj'p, :2
~ CAn re.tA.()( va'/ues o~ Mill) 1; v,
" 4~t(:( 7he n
--~
];J vii' ) ~,
i:= 7; Ct/n b~ ()btR.Ine,f witt, €,. //.03
p~
ptl- .
lJu,/S/hCe. 7; ;.s CtmJ/It'"l-
T"" ~ ~2:-)7
./
= (;.:+1 /51'1'12-J = ~32D,e :-~+ " ~and " I
V =V2 COn 6~ ple/(""""/I? t:d wi""
(t 716
so /020 If" = L(
=.r <6
(con't)
//- 8/

/ /.51 I (cOJ1't)
*'f (1-,(, ) ':S
b
r1t/ -::: ~. S7
I
~ :: 1·13
T'f-
tI
I :;: 0, 6
v~
; . 0; 1.86
Po,-L- -= /. ').2.
pr¥
4
fZrom C1' 2. We.. 9e.f
T :: (I.13 )('13)" f)12.) _ 'fRYif)It!.
I " -===
Witt. £,.1 IN'" dbh.Jn
V, .=. (0.6 )(1172.0 {f)::: 6/2 Ii
J - .J
:: ~II
- Itt· 7 pii..
,.22
= '2-I'.s/a,
/.21.
" :::; (0.8
= If. i' I's/~
/.16-
:-
//- 82.
=
(~)

11-5'2.1
11.52 The upstream pressure of a Fanno flow venting to
the atmosphere is increased until the flow chokes. What will
happen to the flowrate when the upstream pressure is further
increased?
.-E f./}aIr;;;:;; :: (AJ)1S~/It fJ'fT V"" 1 I'
fV -
Alf() /!4f- 4/,
"YI"e QX iA/ ItJCAJ'/4Y1 ,;, -ft,e- -h~ ~ f3' II.SG
T -- --
/
5"0 -4.... ani fJne axiAl IDUtti~ ()f Ike hlW w~eYe tIu
Ma level Ij' flu. Ja-me 7; IS At's/) tJ,e .}al'lre, blA-f
}
f if ",jher 7kuJ f7V If ~Lrt) !-,14,Y" 41'1r1 we
UMcllA.~
-flbW e., Iv
hILVwJ~
,10 + a
Jn~~tI
ft,/Lf
~
-I1u. ,'r,/e-I !YrSJ ~ 1')/ a cJ,okd J#nno1;'u-etJ-f ':J
fJv ~-In-t ,.!, her'C. vetil/it In
~ p~ df EX~/e
s-e". iCJ dj Fa;'I?o IJ'ne./'
I ~ I~f fW!,{".{'lI foe ~
//- 83
tlfJ-a /nCff-Ue (1/ .fIWlY4k Alto.
1/.11 ~ ~
.;;:., d,'{'~1 v,l/ueJ df

11.53 The duct in Problem 11.51 is shortened
by 50%. The duct discharge pressure is main-
tained at the choked flow value determined in
Problem 11.5/. Determine the change in mass
flowrate through the duct associated with the 50%
reduction in length. The average friction factor
remains constant at a value of 0.03.
Jh/s is like Example /1.13. We 9uess 1hN"- fhe, s-horknt&/ dud
will sf/II c-htJ/(e QHd chee-K. OUr as~whlfJf,'on by CdH1f?""~ fd
wi th P ~ .Lf p,; ~ p:tt" fire .flow is vhoI:.ed. .If noll
t:11'1()ff1ey ass umpf,'oY1 mus-I- b~ Y'n.llde. For choked flow ~
ca./cu./,:d-e t-ht?- vntlSf HOWYA!e &IS w-e- 01/&1 /n 6xtl~/~ //./2
Oy In iltfZ st)/uh'on 01- pyo~/eYVI II. S /. F()y Gli1clulced flow,.
we.. mus -f de v/f'e al1t7/11ey ~fya,k9!/.
Fey choked .flow'
P (d.S/'I)
FrO>?? F;9' 0.2. we ~etd
M~I .:
o· 6"
T,
1./
,If ::.
~ (J,7-If -=-
V
PI
/.G
~
:-
~I
Thus
;:: :::(0.75" )(N. 7psilA.) ~ I I ;os1A.,
(con If)
//- at(-
:f' fI. -.,( )
D
(I)
(2)

//.~3' J (con't)
;.0
5'ince
~ ; •. 3{J I'I/tJ.. > ~ ; t. 3'1 1'~/4..
Itte fl(}w i.t che1k.e4 4. f" tlJ.U.lnH!d.
loA' -= r;. Cdn be o6ht,J,ed wil4 G1.1/.63 SInce ~ 't CtJ)1.f/rnd. 7h1l$...
_.r
I
lAIiftt. E1' I 1A/e. have
~
~ A:l, ~/I'M
- ':
~
,
~.Z.68 flit!1'}1 = -.f
The. vhtU1lj e. ,n h1,uj !lflWYtItIt: I"J
?~':- M,09x /00
'() If
The.. 1'Y'·~Hs I'lowr..!c. ;;'cvt4l.ted "r 9. ~)f) i;/lrel<'l. f1....t fub~ W4J
J4/~ i:J7 sa)..
1/-.85

11·5't I
l1.slf If the same mass ftowrate of air ob-
tained in Problem 11.51 is desired through the
shortened duct of Problem 11.53, determine the
back pressure, P2, required. Assume f remains
constant at a value of 0.03.
rhis i1 .r/"",;'/t:{y -fo EKe:tht'pIB /1.1</. Since the stJ~e IMIlS$
fl(}w~ale ach/el/ed m Prt.J6IeJ1l-l 11.51 is desil"l!c/ wi!A n,~ ~h/),.ft",-t(l/
due-I- of Pmhle., 11·~3", /Me. I?eu .;." Acl.iel/-e, fl,e. val'te ,,/- Mit,
C;'felIHed II't P1"llb/~""" 11.,1 _ Tltu$,.frJy '/17e. Stlm(. vcdue. of M4.J.. t;tf~ I;'
f'yz,bl e,-. lIS'! we. ~ ve.
D
oy
./:)
F(J., -l,)
£)
".~ ff
W/fh f(.lY-~'l.) -= ().3 wt.. enfeJr FiJ. Po Z and r~~d
lJ
= 1.6 0)
1h.e Vlt/ue of p '" ~bhJ;'ed /~ PYDJ,/e-, 1/.53 iI
f ~ == 6 ·ttl psit;..;.
(A.II.;" wi/1t 1:1_ I we ,e!
P." =- ( ,.6" )(~. ~fi psi"-) ~ (l-. ~$;a.
-
11- i6
.rh'll va.I/Ii Sf)
I

//.55 I
11.55 If the average friction factor of the duct
of Example 11.12 is changed to (a) 0.01 or (b)
0.03. determine the maximum mass flowrate of
air through the duct associated with each new
friction factor and compare with the maximum
mass flowrate value of Example 11.12.
(a) For f' = ().Ol we have.
and
F (~ 1'_.l., ) _ (1/.01) (7.", )
/) (()./ rn)
fig. P.2 we ret2d
fv1~1 ~.7
T, = /. I
rtf
VI ()- ~ ·73
v~
FnIn-t 6 xantple /1·/2
7,f :: 2 'f0 J<.
and
vAl = "3/0 ~
s
7hU5) wifh £,.I we 'jei
T =(I. I ) (2.'I() /<-) = 2& Lf k.I
aYld wif1,.. f'j.2 we %fa,;'"
c 0.2
V : (0,73 ) (310 ~) = 22.6 c:!
, $ S
= 0.72
f>J == (().72
1/- 81
0)
(2.. )
= /.7 1l"
S

Poy f;; 0.03 we have
r(~ tl_l/) = ((J.03) (2.,.,.. ) ::. (}.6
/) ((J./ ~ )
Ol1d on Fig. P.2. we- Yedd
/V1 ~ I
?;
rlf-
V,
V.f'
1Au5J
-= 0.57
-= 1.13
= ().6
1; ~ (J./.3 )(z'ftJ t-) == 27 /1<
V = (0.6 )(3/0 '!?) -= /g6 ft'1
I S S
Fro;'Yl Fit]. [). I We read ~V" Mtf, == ().S7
1, :: O. <;
P..
~J
1h1.A5I
R= (0.2
10 de-k¥IM/ne
The max./mul"l? (ch()lc~d
. 1·70
-Ii,
WI - -f"" O.t:J1
f
IY1 :: 1.65 k,
{ -= 0.02- -J
In ::: 1.5"'2. ~
f=(J.03 ,.)
/1- g8
/.52- -k.,
= 7

II. 50 I
11.5'6 If the length of the constant area duct
of Example 11.12 is changed to (a) 1 m or (b) 3
m, and all other specifications in the problem
statement remain the same, determine the max-
imum mass flowrate of air through the duct as-
sociated with each new length and compare with
the maximum mass flowrate of Example 11.12.
FoY' maXI"nUrn I'Iowrak- the duct IS choked.
raJ rOY 12. - /', =- I"" we. have
Fyc'f")-. F .q [J", ..I' • .-
Ma, .:: ~. 7
T, I
- =I.
T"
V, n._ ': v.73
II'"
Frtn'YI Ex.~p/€ /1./2
tina
T~':: J. '10 k
1I:t< = 310 ~
~
.J) tJ. / ".,
1hus I wif'h £'1' J we, tJb h ,;"
~ ~ (I. I )(:J.'10I<): 26iJK.
and wili, £1, 2
ltj::; (0.73
we 'Jef
) ('310 rn) = 2. ,,6 !!!.
s S'
70 deieYm;¥Je f, we €nJw. F i9. O·J
P, =o./l.
~,,
nus
~ - (0. 72-(I -
(C On't )
/1- 8'1
(I)
(20)
wi#' Ma =0.70 CLw;( readJ

1/.5"6 I (con'f)
f (1 ~ 1,) _ f ( 1.2 - £, )
- --J) /)
~ Fig. [J.2.
NlfA ::: O. ~ 7
J
T, ::. I. /3
rlf
V, 0.6
II'" -
Thus
-r; :;(/.13 J(2¥OJ<) : 27/F:-
(O.tJ2)(3 W1 )
(d.1;If)
V -= (tJ.t ) (3/tJ !!!) = /3' !!!, $ S
Fr~m h1- D· / we. Yead f,.". MIA, =0.57
m == 1·70 Ie,
-.)-1=1"" .$
l I
.
~hi
= 1.65
le.(,: ;<,..., .J'
, :: 1.~'l1t9J'YI
.l")-~ =3..... J
//-.'10
_ 1J.6

1/.57 I
11.51 The duct of Example 11.12 is lengthened
by 50%. If the duct discharge pressure is set at a
value of Pd = 46.2 kPa (abs), determine the mass
ftowrate of air through the lengthened duct. The
average friction factor for the duct remains con-
stant at a value of 0.02.
:f (12. -.(, )
=
[)
Or
(O.tJ2 2(?n,2 = d. , :::
,cu':.~) _ F(L'--t1 )
- -- ti)
d'/hl
f) £)
We 9lAfSS a. va(~(! of Mt:i2. Ilhrl ger a. CdYl'tSI'OrJlil,y Vttltte of fa!~)
L - • ,..,..., H. I of £)
rr~ r'9' v, II- • Wi"", £,_ I we t/fel-"/. c"lc.ei/"fe ~ value
r !..et'...l,) and wif;' MiS vnlue of f/.(J!..~) w~ 9f!f ~ r/~. p.2-
o p
a. V()/tle of MQ,. Witt-. M~J ()'ltd MtAl. we.. t76Iai" Ii-rm, ':iys: P./
Then with
(2)
Pl. thai we COn (b)¥tftlTt fAlift" 11. = '11,2 kla(dhJ)~
of .M~2.. =O.9S. CtJyye-'l'0nt::t/~ -h Ma2.=~''1S'
we fie +- &f VA/{,fe of
We 91Ae5f ~ value
w(. rea.d on F ,'3. D.2.
J)
J),en with 6'1' I we Oblal~
Fre'-L, ) = (). 6 + ~.OO128 - tJ. (;
D
fw fl.Jhie.h ~ FIJ- 0.2-
Mttj
"" tJ· 57
( COn 't)
//- 9/

/1.57 I (con 't )
Now wi ft" [1- 2 Wt 'Ie f
R ::: [/0 I ~,o~ (tl6s-)] (0, BI
2.
)(- I _I (l.tJ6 ) = 'I6,61t.At(ab.s)
1.96' /
which is (lose... el10UYh fo flu JI'v~11 c//5Ct,(1Yjt.. fll'tr$UYe k t.(I h ALte,.;.
fhe aJfll""I'-h't1r& cl tv1fA~:: ().f)5".
7lt1l1 P = /1:;) P :: (tJ,81) [/tJr~Ik(abs)]::: '31.kIQ{4bJ)
' J (~ °1 ,
~I
aHcI
T =I r. ) 7;, ::: (0.1 'f ) (233J::-) :; 2-7 I I<.
J ( 7: I
1)/ J
IJIso -from F 19. P.2 fur Ma, = t7. s; 7
V, = 0.'/
V~
anti f/1u5
V. :. (t). (JI
J
• -119111 - /.55 _
s
1/- 92
(s~~f,.)7r(O'/h/)2.(;S1 f!)
(286.? N.fJ'1 )(27/;:)('1)
*,.1::.

/1.58 1
11. 58 An ideal gas flows adiabatically with
friction through a long, constant cross section area
pipe. At upstream section (1), p, = 60 kPa (abs),
T, = 60°C and V, = 200 m/s. At downstream
section (2), T2 = 30°C. Determine P2, V2, and
the stagnation pressure ratio Po)po.' if the gas is
(a) air; (b) helium.
Ma =1
:::
F Ij. p.2 'Kif?.
T, ~ /./ 'f
r""
~ ::: 19
?if
~II ; /.25
p~
D
rrom the ftmper"fzo'e.- raf/()
r:lf::. T, ~ (~J3/C.)
/.1 Lf (/. 11f)
,if (z97. Jt:.)
Fiy. p.?. (,.vi H, 7'l. ::; /.01.{ (iifhd rtl/fd
&. :: /./
p~
!J:i.,. /
-r
(con/i)
=0.55
(I)
('~...)
(3)
(4)

1/. sa I (contf)
WiN,
~:: ~ ( P )F)(P,. )
P, pi(
(}nd E.<js. I and '3 we obl.,;'
Ii:: [60 f<Pa.(abs)] ( I -(I/) == 3'1. 7 1zf" (4bs)
I.e; I
AIs() wllhI
~t.I
Pc,I
~/~ :; (;
~II
) /_1_1 = (J.g
{/.25' -;
..-- .
r ',-tally
V. ::'J..
(216.9 N.I?/ )(]()]fc.)(f.tf)
"kf.k (I N) ;: ?I'f ;'
(6) Hr/ heiJ~m) It? =/. 6{' Itnd R = "2.077 N.m
-/<!J.K
The Mach. nUMbey of-
::.
T~
Or
sec-han (/) is
(2.00 ': )
(con'+ )
/I-9/f
ky.e
s,.
.f~ Ta61e /. B.
0;:. 0./17
(5)
(7)

II. 58 I (Ct)f'J If)
Wit? ftp - 5 we obiaJ'n
T ~::: (331j() :;: 2!i"3.2 K.
(I-j/~)
Thu)
T-z. - (3()3 ~) ~
- 1.ICf7
r~ (Z~]. 2/:..)
Wi~ T'j. -::: t. /q7 we...
T.I(-
M~2. = if(R-fL ) - I ] (ft~,) ; fr¥) _I 7(f-)::: O. fig~3
I (!!:) IL (/.19 7) J(/.66-/
i*
!l : J- f (1<f) . l =,1-' ) [ (i~) J: IffS
p.r Mal. / r(*;')M&t{] (().SED3 1+(!-'~-I)(~.5"g03/J
~ ::: .l. flZ. )(1+~Ma;)1';::':/--'.-)[;2-iT;fe'!!:}jo.)gDl)"?;:.:~';.2
;:ft MOl {(ir' '2. Jj (O.>80~ ll.,,+,)L I 2. 1Jj
W;ff,.. ~
!{ = ~ (: )( f$ )
alid ~$_ ~ tl'Yt4 8 /,Ale.. ha lie
f?;; [6o*Pq(~S)J L!-)(/.81)) _ 18.¥ W'i(4h5)
~ (',/3 -
Wi~
F/Y'J t:iII'I ~
~::: M"'l..VR~'k.
5<;3 !:!:!.
- 5
--
r077 1'0 )f303K)(t.66)
1J..J.1< (I ~)
1<..9- ~
.r~
1/- CIS
(8)

/I. 5"9 I
11.5"q For the air flow of Problem 11.58, de-
termine T, p, and V for the section halfway be-
If secf/on (A) is p/ac(ld halfway belween sec/70/'u (/) tlnd[2.) we ha~
f (). ~J..,) _ -f ('(~-.l2)
[) D
(I)
D
and wifh f/!~.fll)
j)
values
we Pi,., P~2. (',IYld Ye~P/ (O"l'"l"es,Pone/)Y:J
(.2)
where. fl- (pas ob-/Q;"Y1I!'/ Ii? t'lte so/u';'/()r} ()f PY7Jblem II. 58'
pJf .
Ft'Ylctll'l) l1f IS" () b..frL/rl ed witt,
I/. := Ma.
" A
VR~ I(
F(,R~.I,) and
--70 deferrn,;'~
D
Ma.l = a. S~ aVId.
and vead
~O;.. :. 0.9/
_ 0·7
-= 0.01
D
TheY!,. wi ~ ft. {
:f (./.1_1.,,):; ~_~_1_ _-_0_._0/~)_
D 2...
(C OI"1't)
/1-96
we enk,. 1='7- 1J·2 with
So /14 '17'0.., of Prob/el'J? II. S!
+ 0.01

1/.51 1 (CO;1'+)
Wifh f(j/-.l.A) = ().36
f)
Ma = (),61
"
TA ;;; /./
r*
~ = /·7
Now wi'M f" 2 we obtai;'
r.. -= (j./ ' )(2'1'1;:') ; 321 K.~
P-. :: (/. 7
A
)f-. / 5 if '1(~ (lIbs)
/. ?
£1- if we have
v. ;" (tJ, 63)
II
(Z~6J; /v."" )(32..3(1::.)(/''1-)
'kJ. I< (I 2:!.- )
;t.,. ~
.f~
_ 2'27 ~
- - $

11.60 I
11. 60 An ideal gas flows adiabatically be-
tween two sections in a constant cross section area
pipe. At upstream section (1), PO.I = 100 psia,
To. I = 600 OR, and Mal = 0.5. At downstream
section (2), the flow is choked. Determine the
magnitude of the force per unit cross section area
exerted by the inside wall of the pipe on the fluid
between sections (1) and (2) if the gas is (a) air;
(b) helium.
r
I
p,A ~l
I
I
(I)
/" CMSItI,,f 0_ pipe
~ / cc,,~1 voJuWte
(2)
flow fVfh'lIdJ) -f7, (2.)
The. (,011 fr-o/ V" lume s-/:;efch eel ~h~lIe is IAse4. Afl'lr/~ 1I1t. A'(./a,/
cam.,()t1enT t7f fhe //net1y l'JI}()menful'Yl ~u.ah'oh (€f' 5. 22.) -Iv i11e
col1Tel1fJ of fhiJ (;1);11n;/ VOIUW1~ w~ ~f -frff tlte -ffm:;(.... exe;led by lite.
pipe wall "11 the flu/ell 1<;( )
Rx == p,A-~A -t rn(~-v2.)
I<x :::
A
~ - p.,. + ;0, ~ (~-- V1. )
Tt,us we JIleetA ~) Pl.; fJl) ~ aYJd ~.
we.. enter ,:;J' D. I with(a.) Po;" lAIr
tJlJd
TAus
and
T, ::- CJ Or,7-;)
TolJ
f, :: OJIJ.f
F:,.I
T = ((),95
J
/1- '18
(I)
535 t!.f

1/.60 I (Con 'f )
Af sechi;n (2.) the flow is c/1bked. 7J1.M<; we. use fhe '" s!z:.k. p/ fAe
F-anno f//Jw/ Fig. 0·2 lilY s(.cH0f1(2). 6nk¥J~J Fif· /)·2 wi/1, Ma,=IJ.S
we ,eAOl
and
v, = (J,5'f = It;
V*" I,V,.
'fl1u.~
~: ~ ; £~i6 if2 - logo f!
''.>If (0.5 J.f) S
f'/I)w w/It, £". 1 I#l. ~1Ie.
R.AJ( =('1if ps/()..)(i'f'f Ij,.~) -(3't''ffJ.ri,,-)(I'Ilf 1;,.2)
-fi1- if"
(b) For" hell~WI (R= /2/120 f-t·lb an(/{ 1<. =1·66 .fi.trw, fable /,7) we. use
5/£,(,/. PI(
£'J~' /1.~6 q""d //,Sf /IV;J-!j.. Mal -= ~·S QJlld o6trAin
7; 6~tJ d~
/f(~t)Mq,' /+ f·66 -')d.~l
2 1.6'
and
P, =' f. f-I f- (~Y'W" 1t~(IM,Si·i, f-{l;"}(M/1 ,.":'MIS;...
aJ1t:A
1/- qq

/ /. b0 I (CO/1' t )
AI- secfioYl (2) fhe fl()w IS ch()ked. TJuA~/ we use. fhe '" ~-lA1e
of Fanno f/tJW fuy secfidYl (2). With EfjS. //./07 a.JIICi /1./01 tfnd
MA -= 0.5 we qef-I
~/. q pr/a.
= .L
~!(~) 12(P.5) _-=Z-=--__
'+{I·;:-0(~·sl
1{ 3 . /?
P = 6· q PsI~ = (~
:::; ~05() ff = V
.J 2.
N()w w;tt.. G;. / we hdll~
11- 100

11.6 I An ideal gas enters [section (1)] a fric-
tionless, constant flow cross section area duct with
the following properties:
To = 293 K
po = 101 kPa (abs)
Mal = 0.2
For Rayleigh flow, determine corresponding val-
ues of fluid temperature and entropy change for
various levels of pressure and plot the Rayleigh
line if the gas is helium.
ThiS if 5/11'1;lar fr; lXt:lmple. 1/. IS.
10 pJDf ihe. Rayle/9h //ne ~s*uJ for we. uJe E,. 1/.111
2-
1-(,IlV) RT
::: ('onsfant == p + (~V,) RT,
I
p +
and Ell' //.7b
S-~ :::
P
,r R In p
Cp In - -
r; r:
P,
II)
(2)
Iv ('OI'l.J1ru"t a fable of vtllue~ of femperaturG (lnt:l/ t!nlnJl')' t.hllnJe
C() ('I'esf1(h?d/~
f( aylei?h flow.
Ir> d;ffe.,e""f levels of press"r~ dow1JJfy~,~ t?-
10 de feKjNJ./;' e
f() IJh /Q/YI P,
,P"I
1() de fe?'~"";'e. 0
P, we. use M4, aJ1P- E,. 11·59
lAnd flte 17 ~ -/rdm
we. use M~l ani £1;; /1·56
-Iv 1)6 -m,'n r, tJlI7t7 fhen 7; -h-6W1
74,I
~ ==(}) r..
7;, ~I
I
We. ob1rJ.,'n ~~~"oV- C4l'1rreu'-/ ~
(C()n't)
//- /0/
(~)
(5)

1/.61 I (C()n't)
With ffj. 5 we. ~Illlt.
('17.72)(11) 1flo}(O.}.)
~ v, =
281.2. I::.
(~)
/1- /02

11.61
1100
leDO
T(t.)
%0
7DO
;00
(('0;'1 't )
P[~'IJJp.bs)]
90
KO
7(J
60
~O
if5
'f0
35
25
I~
T(/(:.) S"-~~L)
583
11.,. /(.
3g33
8B3.2 61tfB
IOq'2. 763 '-I
J2J2 '84- '10
/236 fflZO
12/5 q,oCf
, /72 qlh 5
/105 1135
903.3 g7g /
€Jlo.'1 77CJ5
~o~------------------------------------------------
~ooo 7N)O 'l()()O roo0
11- }03

1/.67- I
11.6 Z- Standard atmospheric air [To = 288 K,
Po = 101 kPa (abs)] is drawn steadily through an
isentropic converging nozzle into a frictionless and
diabatic (q = 500 kJIkg) constant cross section
area duct. For maximum flow determine the val-
ues of static temperature, static pressure, stag-
nation temperature, stagnation pressure, and flow
velocity at the inlet [section (1)] and exit [section
(2)] of the constant area duct. Sketch a temper-
ature-entropy diagram for this flow.
For yYI(JX /rnlAYYI fiow/ f-he Ray/e/9A fl()w is c.hok.ed. fbY' I-he. isenfrup/~
nOJ3/e.
1. ~ T = 288;:'
~I 0
P., -:;. p ;:. /0 I itflfL (A~~)
~I "
10 defeyYJ7/-"e fhe ~-/r;l.fic.- 5/-al-e af f/1~ nOJ}/e exif Rayleigh /;tJw I~/ef
.I /
we need the value of Ma" 70 deJcyYH'YJe M"t we... use
hfJ 2. - ~(J - ~ -= c,o (7: - ~, )/ )/ ~2
Or--
(5"O~ (JIM ~)
T :: !l -r r = 4LJ -I- 29$1::Gp. ~
~I == 73£ k.
rIM'! ~)
P.f.1<
Clnd nof/nJ fhat ~ c.h(Jked f/dw T :
~a we ~f)
~2 J
~I 7;, ZSs k.
tJ."37" ::. I :: ==
7;7. T 7g6 I<I
~4
With ~I 0,37 we.- eYlIe¥ -F;~. lJ.3 and rene!I ";;
-7,;4,
~
= 2./ (I)
~
T, : &.S'- .2.... (2)
~
/1- 104

/1.,,2- I {Coni!}
v, = (}.2
Va.
~ = 1.lq
forAI
Wi#! E~. '-I we olofa,'n
~().
::: p'" I
100/effA(abs) 8q.· q fefa.. (().~s):;; 1:.-= ::
", "2-
w;th
I
I./q
1Vl", ~ 0.31I
P, = 0.94
~/I
'£. ::: 0.93
7;"J
/.1'1
we 'f'e.a. d fYtrM r::ig.O./
Witt, frs. ~ and 6 we, get
~ ::= (O.9Lf .)[IOI1e..P~{~S)] :: C/'5 f<Pa.{ltb5)
~ :. (0. rSfll.f ) (Z.SB t::..) .:= ZBZ IZ
1huJ
V ::. MfA, rR T, f< = (().JI)
J
?,
2./ .
II: = v,({
~. '2
{}.2
(ctJ1'1 't )
1/- /05
(4 )
(6)

1/.62 (un't)
5 -s ':.2. )
c ~ T. - f In ?~
til -
(,/ P.
7(1<)
$2- - ~J': /6tJO .IJ. n-.
~!. k
60()
"/
1100 p.; q5 kf4.(";$)
I •
~;'=2F£Xz~o
'"/
I
/1- 106
S-s ~.). )
/

/1.6 '3
11.63 An ideal gas enters a 0.5-ft inside di-
ameter duct with PI = 20 psia, TI = 80 of, and
VI = 200 ft/s. What frictionless heat addition rate
in Btu/s is necessary for an exit gas temperature,
T2 = 1500 OF? Determine P2, V2, and Ma~ also.
The gas is (a) air; (b) helium.
10 defel'm/ne the f,eaf -/ran.Jfer yak we Use fhe eneY9J qUaftbn
(fer 5'.69) -fo gef
•
Clhef =
Fo.". n1a$$ .flow'ra fe we. use.
<7Y
To
.m
T ~
To
-!r;y air
/
= P,
Rr;
2-
1rJ) V
- I J
Lf
-r rAY1d r we. U~e £0. /1.56. 7hu~,-
~'- ~J P
/ +(~2.-I) /vlt{2.
-r _ {(Mo.) In F/:J. P. I- -7;
def.eYm,ne If we. use.
!?:: P, (fa.)(J )p, ItA
whel'e wi fh 1:$_ //. /23 fp, /?~y/e;j), flow
p It It.
- -:::
~ If It!vf().1.
or -fiN. (1 I~
P :: f (Mill)
flo.
111
Foy eXir ve/()cify I ~ I we use
'{::. Mtl2. VRT"l. k
We. dekYm,'rJe Ma wift..I
/'vfa. ::. V, :::
I C,
V.
(C()n , f )
/1- /07
(I)
(2.)
(3)
(L/j
(5)
(1)
(q)

II. 63 I(conI! )
Gl1d we determIne Mal. wifh
r..
(~~)(; )- =
7;,. ~
elY/of f~ . /I. lit -fot' /(ayle:Jh How/ nlllHe/y
T -:: friT k ) M.
~ / t- k M~"
l'
(a) Foy 0/1' we defeYml;'e Mal wilt, £(r q. 1}1I'~
T, -:: f). 19
7;,I
T :::
~I
0.99"
(ZotJ tf)
= Sr5~
Wdl, M", = O.lt we read QrI
and
T. = 0./7
? :: 2·3
PP...
TAu> will,. G~. It) we obm/~
f..;. -::;; (Iy60 'r< ) (0. 17 ):=.~. 6 2-
7;.. ~'f() til<
(C&n /t)
11- 108
= o·/g
(JO)
(II)
02)

/I.G? I (COf'l't)
F~ h. ~ 0.62 we gef ~ Fig. D· 3
~
Ma ::: ().1/-0
2-
GVld
~ = 1."6
Pa.
Wirh MaJ.. ::: 0.'10 we. Yed.ci On PIg. £) ·1
1?- 0.97=
7;;2-J
Thus
/
1'1' (J
~"]., ~ - Z020 DR.
-J
0.97
Then. wiM q.§" we have
R. ;: (2CJfJ/IA.)!~ ) (1.'16 ) = 17 pfJ&t.
:1. (2.3
Wi th 8t.? we have
t( := (0. 'fp) (t7)t ff. /6 ) (;q6~ tJlZ)(f.l-f)
51u.9· '~ (I .1!z.. ::: ~&I :'
s~.fl)
Wi I?t Gj~ 2 we ,ef ;;.
. (zOPJ/a. )(1'1'1 In.'l) 7T (0,5 ft )).(ictJ ft)111 = F/). :: 0./22 ~
(1716 1'1. /6)(5~ 'I?-)
.slll~ !I?
11110< (NiH, Gj.( t,Je 0.6*',;'
(tf)
f2~ef : (tJ,/22. J'Iu,)(to()6 -flit) (Z02cJ~ - ~Lf5~~) :: 13'M 8~
'1 ..r 54) tJ/?
I;, . (778 fl.I") J"
Bn..
( ConI f )
/ /- loq

//.63 I (c()/I'f)
(hJ!iJ-y- hellu'm J f<.:;;. /,66 And If == /;;'10/20 H.16 fY-~ Ta6/~ j. 7.
W;Ht ~. '1 we h(lve $/~. d"f
(Z()tJ (t)Mt{) :: ., - a os97
Witt.
~I :: "': f f/!-/)M~1;;(S'IIJ ~)~t-(I. 6~- (J'tJ1 ()S99))-= ~'I~.' ()I!
FtJy 1v14, =~ .()Sf( i.> EI5. 6 (/f;'HO{ / / ,y/e/d
lind
or
(bId
If- /.66
-------------------: 2,6o/~
/ I- (1.61)(d. d5''1'i) l
Mq :: d.1/6 bY s-, /t;
2 -
We use !he JuhJ(h''/c f~/uh~/ Mtil.. ... O.J/b >/},c.e h~4.f is be,:,
added If"l1d /IV:Ih h~f dddil,dn we- cal?'?of a.cce/e Yt:<~ ~
su,tH7rSd11/( I<aI/e&~ .flow ~ a.... .>'ubfdn/c ctJV1dt'I/OJ.-a vf/JiYrta"".
(Con If: )
/1- 110

/1.63 I (Con 'i)
Wifh J.1al
= 0.//6 aJ1ci wif{" bj.] We obi7A./~
T. :: (1960 4~) [ I f-(i..6'-')(o.1I6)2] ;; /fI~7 DR.
~~ 2
Wi~ £1. 6 a-"".rJ M.a.).. =O. //6 we have
_1_r_/..-:'6:..-"___ ; z.60 Z
1+ (1.66)(tJ.U6/
IN iIt,. Ef . 5" we
P.. :: (Z() f.Jif). )/_1_ ) (2.602) -= 19· 7 (ISla..
l. ( 2.6tfll
w;ty, Gtj. g we Yv; V~
----------------
~ = (6.116) (/),,'f-?,O Ff.11; )(/q6~·R)(1.66)
5/"9. DIl. ( II,
!Alif?, {'J. 2 we- ')ef / 5IUff!)
-=- 737 !!
=s
m = (20jJS/a.)(I'I'I f:.~~) 7r (()·'5)#-(ZOO jt) -:::
('1)
Ql1et
In
(1- /1/
-.f

11.64- Air enters a length of constant cross
section area pipe with PI = 200 kPa (abs). T, =
500 K, and VI = 400 m/s. If 500 kJ/kg of energy
is removed from the air by frictionless heat trans-
fer between sections (1) and (2). determine P:-
T2• and V2• Sketch a temperature-entropy dia-
gram for the flow between sections (1) and (2).
To defeYrnine fhe slrAfe of the aIr Cff rech'on(z} we.- use fhe
eheyCit, e~n{(ll/on (E~.5".61) Iv c.a/culak ihe value of T . ~5
iI/ {) C12. ~
~Ylet
:=
in
or-
~ ::
~2.
We. ()b~/~
a v~/ue
111et-
+ 70, ~nef
7;;, (I)I., - -) ~ +
c" c,P I
To, ~ T, w/'ic.r. M ve~tt fY.~ FI·9· /). / w/tt.
J
7;1 1
of /V(().., • we de fe.Ym/;1e Ma IV"~I
M~ =I
~ ~:=
('/
{R7;~ (2)
we abo enfey Pi9- P. 3 t)11d. tea.c/ volu es of
p, wif"
p~
(~)
W//-h thi! value or
de feYYYllhe ~ I 7;,. Ill1d v wi~
2
P-, :; (1/it') f~ ~ P, , ('1)
~ =: (~2-)(~) '0
<iii Ti
(5')
(oul
~ =(~~)(:) v, (6)
(con/t)
//- //2

u6'+1 (con/f)
We.. use
M~,
Gtj. 7, ~ gef
_ (lfo() 1)1)
(z?{.] Mffl )(S~OI<)(i.lf2
1;.9- fr. 7 (I .!:!... )
t?,. l!!
S?.
aVId
r, -= O. It
~,
I
p, '= t·/if
~
T, =/.02
~
V, = o.'1
1I't:t
~,
, =- ~ 99
T.
~""
fI~w wIft.t 7;, I = b. qq
~&(
((Of/It)
//-1/3

aka
M~ ;; f). IS
2-
.5. :: 2.:3
~
7;. ;. d.1l
7;
V2,. = IJ. () 7
lie..
With 1-hese. yah'os antii fhose ral/os CPfYefl()ndJYy fo Mq, =P.f7 We lAse
Gjs-. LfJ
~ awi 6 -1-0 ,6ttun
~ =: ( 2 . 3 )r;/~ }[lao l/Uai¥)] =.t!!..! ItJ/;. ("/;$)
~ ::: (C'!7
and
)~ 1
( I. f) Z
)~'f 1
Nok That acetfh;/I ~J
Thi 1 if no! Cf7n"f!cf
({sstJc/a/ed tvjlt, UJ/ny
_ <63 k..
) ('ftJ()!!2.) :: 31 1?1
s - s
resull-
To~ ~ U~/I1'.J If. 11.13/- I1rs7 Juwe-W/Y;, we..- deJ~WJ/~-e
~'"
!vftl
J
w/M E1' 1/./3/. ThuJ,
~ I :: ;2.(iu-/) Mq/ ( / t- ~ Mq/) =
~~ (I t- kMq/)1.
ov
( C1JYl 'f)
/1- litt-
2 (I." rl)(0. m;1ir(.{:!t:.if?tj
[/ f- (I. if) (0. !13/]

II. GLf I (cOn '+)
Now
T, I
7;"
I
r;, = ~ - (50PJt:-) ::::
J O. !62'1 (J.!6 21.f
Now (IIJ; f4 E1- I we J"ave
r;1 , ; _(~(}~m t ))
f()iI L )
f<r. jc..
Wi it... {'t. ~ We ,,6~/h
Tc A
I
/v1(f... = ~./776
I
T,
10,.I
------------- ~ aS62¥
I+(t,~-/) (tJ,F'i3)"l.
::: el· 7'1 !<
7hert w;ftJ 6$. I/. /U a.J-td Mil, = ().gq] Clnd Ma.z.. =()./776 /1.1(. 1~f
":: /- ==- /. 0z,r; r(u-- k-) Ma ,]Z, ::: [ut- ,·if)(~,rIj3) i2-
~ :--. / f- -k ~4,"Z- / + {to If)(0. ~9'3l
1/-1/5

1/.{,4 J (~f1(f::)
NOlA! witt., E~ . ~ we h.ave
l;::: (1)./666 J(--.!-)(5t;/)/C.) == 8'1. /'1 k.
/.026
T;, :; &'/./1 K. < T:., = gr 7'11<
~~
(is if si2dv.-Ld ~.
Fov OlAV T- S" ~1=.L1vL We u.se £7' //.7t -fl? CA/t".J4 fc- ~-~. TJrllJ,
S-)'.
l J
200
100
:= - 7,030 2
1<.9.1<.
o ~-----------------------------------------
1/- //6

11.65 -'
11. 6'5 Air flows through a constant cross sec-
tion area pipe. At an upstream section (1), PI =
15 psia, TI = 530 oR, and VI = 200 ft/s. Down-
stream at section (2), P2 = 10 psia and T~ =
1760 oR. For this flow, determine the stagnation
temperature and pressure ratios, To.21To.l and
Po.]1PO.I' and the heat transfer per unit mass of air
flowing between sections (1) and (2). Is the flow
between sections (1) and (2) frictionless? Explain.
To deferWI/;"e fhe. :da9na.noYl -femp€rafuYe. Cll1d the s1zl:JYlahon pyeJ.fuYe
ro.i-io) we Lae.
7;;2
t~'~(T, )(!i), :.
To,I Tz ~, ;' T"
OJ
one(
= (r.'·)(f )(?)fo-z.I
~,
I. p~ f;., P,
/
(2)
whe.,e
T f(~,,)
.
-:: 111 Fig. O. I
7;
(;/1'101
f'
= r(?1a) 117
~
70 defeYWI,;,e the. Mach. J-1u;Hbey- ed eoch. st-ch"OI'7 we. (.,fse
Ma :: v :: v
C
For fhe Ve.IDC,' ~ at- s-ecfion (2)" ~ J t,v~ (lS!- the. ~~h.Jenlafi;'Y) of
mass PY-/yZC.,f;a.1 -Iv ob~/;'
~~ = ~ ~
dy
~:. ~ ~ :: IP, )(!2-) ~
,.o~ (~ / r;
Fo". Me he~t /-YOI1J/e¥ pev UYI/t I17tl5S of dir /hWI-., 6elw~eY1
S-f.cllons (I) aMd {Z} we l1Se. IAe. el1eY'!i:; e$ U~..J1(,n ( E$. 5".6,) iv
tJb~/n
CJlJel- = h0 2. - ~~/I -=
111
We.. ~~fz:t/~
(COy/It)
1/-117

/I. C-t;" J (con 'f )
T~ asce.r-/Q/n wl1(~tI,~y or" no! f;,e air flow belwteYl secn~y)J (/) dht:/
(2.) i r {Y/cndn/eSJ ~ lA>~ fhe ax/a.I COm,bonel)f or I/,e //neerr
M~m~YlIun-, ep-tmt;n (£~. S72.2.) fz> gef
Rx = A(~ - f2) of yj" (v, - ~)
~x == ~ - f2. 1- A ~ /v -~) ~ R-P1. + Ii V; (~-~)n { "I R7j
Fiy}t we U5~ €tj." -h detey"",,;'e ~. 7hu~
{ = (ISP.J/~ f76~ Ote)(2~~ j"") : 9'16 (I-
/tlI'J/~ 5stJ ",( S
N(fvV witt.. ECi. 3 we CAlculale Mt/ turd Mti . We. Jv:w~p J Z
Ma. =I
With
(zoo §t )
M4 ::. ~.I f
J )
7; -= 0,19
~I
e... -:::. ~.'97
~I
(AYld -few.
Mf{"2:: 0, ,/-f .)
-("). = 0.96
~J2.
&. =0.85
Po) "J.,
:= ~. I B
= ().'f8
(con 't )
/1-118
(7)

J/.65
Thesf:. ~perafu'Ie tH1d fye5~tJr<e ratio! tr-ye UJed wiItt &;1. I &tI1A 2.
+v oh~/'y,
--
61' 5" we. d bf7../~
In
(6(){){ -F1.16)(~~S'<) ( 3.112-/) - 7.37XIf)' ff. /~
51U.7. D/( s(1tf
al7d wi~ €J. 7 w~ hA.ve
Rx = (I~I'.J;t< _If)f1.JI'a.-)f'f'f I;', 1.)
A ~~
/<)1, ;= 3'12 16
A f.j;'
-f (l5"fJI'~)(Nt! ./fi')(2()O ttA 2J)O fi- r'it f~ (~A IbIf)'1--
I 71" -(f. Ih ) (£f~ ~I<) J2.
.>It<j. 6,R
and vv!.- (.Qytdude. fkq/ Me- flow JJ l10t fr,'c{;'t/hlesJ.
/I-/lq

1/.66
W-e
11. 66 The Mach number and stagnation pres-
sure of an ideal gas are 2.0 and 200 kPa (abs) just
upstream of a normal shock. Determine the stag-
nation pressure loss across the shock for the fol-
lowing gases: (a) air; (b) helium. Comment on
the effect of specific heat ratio. k, on shock loss.
~hoc*.. I (/V"
= F:. (1-c:); x ~)1;)(I
To defe YW//ne.
ThusI
fh~ J'1a~Y1a 1/~J1 PJ-eSJ(jy€ yan6
1... 1
Po) y
PoxI
aIr"
[(¥)Na:]~-{I r(~)~d:}I-~
((~)~:- I~)I 7{"!-;
L ~+I (1<11 J
(I)
[j. 1/. /S"b'
(2)
(a) FtIVtI!~(i;'~/.'I) we ha ~ J'v/P :: 2.()
x ./
~y
~x"I
:: (J.72
TlnlS / wi tI1 E'j. I We t) 6';"I h
~x - t;, = (fd4 k.~{tf6s) ](1- IJ.72 ) = §6" ~,att
I JY =:=::..=.
(b) Ft¥ he//uYVt{~::1.66)we At:We wi~ &J.2.
(!.t6) f~ )~ :: [{I¥')(2.0)"2.]~·6W'[1 t(t~)(2.ot I-I." .. = 0.7621
~/JC J[:l (/.~I)](2.~~2..- /i.66-')1'~-1)
2l 1.6t f-I ( Util
ClYl d with Et· I we 9e-1
R,x
- 1:1
,/ -= ["20"VP-.{aJ,.sJ] (1- ().7'7..1) == J2i.. I<IC(
~ pcuff (a) OA-Id (b) we UMctYrda M4 Ittt los> cJ( kkl !i'a.5UI'e
OC¥bSJ {)t... ntNMA../ shtJck... decrea...J't:...J wilt,. aJil iV/c~e In 1<..
1/-/20

1/.67
11. 67 The stagnation pressure ratio across a nonna! shock
in an ideal gas flow is 0.8. Detennine the Mach number of the
flow entering the shock if the gas is air.
-To d eferrn I;'e ClI i" flDW
eMteY/~
pY..(!$st.u'e
>ltoU:...1 Max.l ,,/ve.Y1 fl,e 5 ~nq;f;;Y!
fbl )( we ~Mkr Flo. f). 'I w/ft,
- .I .7
Pc;y
~x 0·8_J_
=
~y
{,{vtd read On F/g. D.Lf
/.83
1/-/7../

11. bB Just upstream of a normal shock in an
ideal gas flow, Ma = 3.0, T = 600 oR, and p =
30 psia. Determine values of Ma, To, T, Po- p_
and V downstream of the shock if the gas is (a)
air; (b) helium.
Ma
y
(2~ )1v1o.:_I
(I )
~-I •
oy .f;y ~/Y' W£ U5e Fit}. D.l.} fi.,. fvl4; tlS 4 funchfM of A14.J('
To
or
rClY
de fe n''''~lIrle T we.. use E~. /1.56. ThlA.J~
CJY
~y = 7j, [ Ii- f*;.I) M4/ ]
we tH"~ ':,'9' lJ.1 ~
1/- 122
(z)
("3)
(5)
!} as a: Fi<hch~ of" /vIqx'
p)f.

I /. ~~ I (CO/1'-f)
(to FaY tllY we reQd ~ Fig. p.Lj -mv Ma)(:: 3,0
Mit = 0.4-75
'}
R {n.1 = 10."3
PJ<
-= 12
'r :: fJ.96"
T
~y
~ £~.I w(: qef
~ : (7...7 ) (loolJf<)
and t/1us with. E'I./tJ
7. :: Iy ==
~ y 6.9~
Wilt, 6,. 7 we (J j, .frt ;"
= 16w of!!-
~ =VO.3 J{3tJl'flP.) = 3tJ11/'s/o..
aYlc{ E'j. 9 y/e(p(.J
p.. = (12 )(3tJJ?.fJf:,..) :: '3~ j'75JCt..
~y -=
TheY1 wi#, Gi. b we 0 b IR,;'
~ :: (0. q,75 )
(CDn'-t)
1/- /23
(10 )
-= 937 fl-
-== ..5

II. C8 (con If)
~ = (J.t;z/ 35"00 {f
= J
/1- 12'1

II. 69 I
Ii):.e +he one sht;wn ,..., Vicleo V. a.LI
11. ((/7 A total pressure probe"is inserted into a supersonic
air flow. A shock wave fonns just upstream of the impact hole.
The probe measures a total pressure of 500 kPa(abs). The stag-
nation temperature at the probe head is 500 K. The static pres-
sure upstream of the shock is measured with a wall tap to be
100 kPa(abs). From these data, detennine the Mach number and
velocity of the flow.
/hiJ iJ like GXdJ"J1p/e. 11.19·
We eh lev r::-;~. J). 'I w iff"
fo,y ::
f'x
llJ1d r-etJ..d
5 O~ -kf~ (4(,,r)
IO~ -lu'a..{t

Solution manual fundamentals of fluid mechanics (4th edition)

Solution manual fundamentals of fluid mechanics (4th edition)

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