Munson 4 ed - Solution manual Fundamentos de Mecânica dos Fluidos 1

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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 of the chapters in the 
Fourth Edition of FUNDAMENTALS 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 of the 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=ae bx 
Determines the least squares fit for a function of the form 
y=bx 
Determines the least squares fit for a function of the form 
y=a+bx 
Determines the least squares fit for a function of the form 
y = do + d JX + d 2x2 + d 3x3 + ... 
Determines the least squares fit for a function of the form 
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_RA Y.BAS Calculates Fanno or Ray leigh 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 ne r.!~ 
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. 
= a 1'19 tI //1 r c/'spkce /?'J()~';' ..!. 
-time 
(.b) e he 1'"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 L T- 2 
e. n e rj tj ~ (M I- 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~) =. L If 
+-()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. 
a c c-e/e ro.: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 ...: I1L2 T- Z 
(a) volume Oen~f-h) 3.-:. L 3 
--
(e) Work - !=L 
/- '3 
I 
I 
I 
I 
i 
I 
I 
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 
d 3.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 fo llowing combinations are dimensionless: (a) 
vr", (b) VC/', (e) V'" (d) VIM 
(L T -'j(L)f1. zr) 
, 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 dlfnen sion!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 flu id mechanics. 
Make up five possible dimensionless parameters 
by using combinations of some of the quantities 
listed in Table 1.1. 
Some possible e"Qmple~ : 
(L r2)(T) u C( e Ie 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 
= - (1'7 zr: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 [p J 
, 
•• 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$ti/{,t/~H /.5 a. !J~n(lY~ I 
h~mo1enet!Jvs €lttA..-6I4;;' .fh{(.i: IS 1I11//c/ IH CiI1.!! ~!:f5Iem 
~f Un ,·f..:5. Yes. 
1.10 The pressure difference, Ap, across a 
partial blockage in an artery (called a stenosis) is 
approximated by the equation 
cosity (FL -~T), p the blood density (ML -3), D 
the artery diameter, Ao the area of the unob-
structed artery. and A I the area of the stenosis. 
Determine the dimensions of the constants K,. 
and K". Would this equation be valid in any sys-
tem of units? 
pV (All )2 1 
.1p = K! D + K" A I - 1 p V-
where V is the blood velocity, Jl the blood vis-
Since eac.h -terM mv.st h~lJe. the same dimensions; 
k'v Cll'ld Ku are dirnen5ionJe-:'5. Thu~.1 fhe efuafltJJI/ 
IS (;( ttener~1 h()f71~jel1eO"s e~ ua.l-;tJv, -tnCI'/- w{)uld be 
va/ic/ t'n Cfn!! C()tJ5isffnt sfjsl-em of U)1jf5. yes. 
/-7 
I. / / J I . II Assume that the speed of sound, c, in a fluid depends 
on an elastic modulus, Eu, with dimensions FL ~2, and the fluid 
density, p, in the form c = (Eu)"(p)h. If this is to be a dimen-
sionally homogeneous equation, what are the values for a and 
h? Is your result consistent with the standard formula for the 
speed of sound? (See Eq. 1.19.) 
0) 
FPr ~ d)J11eY1~/Of1tt/I'1 h(!)mt1ef1eDIJ5 -€$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) 
(.£. :!iJ 1-, ~ I-y ~Y1 dJ/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 v28 B (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/ledJ t(J eg Ua.,tIOJl -1'n¢,f WOf,{ /~ be. v t).. //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 /fA l' = ('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) l I. Cf Iff) )(. /0 W = 
~ 
~~ 
(d) 0.0320 
N-1'H1 (~, 0 j 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) (.°£,;</0 3 J.)= 7.g3X/~5J 
BTU 
C~) .2LjO int.' = (;'''10 tni ) (;'''Oq;(./(;.3 1"YY1,)::: 38iDX/oS" t?11 
I'n1L 
Cd) 71. / h p 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 == 0 X -Vol"rn~ 
d' =: jJ d = {0.2 ;'3 ) {!D- l ;g. )(r.8/ ;) = f. UU/iJ-;;J 
"lJ =- (I. '( (,,2 ;( JD -3 ;;', ) (10 1;m3) = /. '( ~2 X I D I, N 
= (I. "t,z X /D (. N ) (:1., 2tf8 x/D-1 -J& ) :::: ~, If! X JOS f h 
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 it 1..: (/ .ft'")f( a 301f.>') 2/1?1 ,,-] = 0, () q 29{) /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 If 3 - l' -It 3 ) 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..1 q 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 J 
1111 ~ "" \ ( T; (3. ZFO~)3 -f1:: 3 
- I 040 x /0- 3 S l u ~~ 
. 1 f-t3 
Th ~S.i m ul.f.i pJ'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. () g r i. I D 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' 
T h US) rn f.,( I t 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) m3/s. (b) 
liters/min. and (c) ft3/s? 
f./owrat e = 
-:2 
757 ;<. 10 /i'Y7.3 
.5 
(b) Since / Ii fer = / [) -3 t1"/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 ()- ~ if 3 ) (3 S 3 I X J 0 
-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.e i9h 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- de p~;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: () ft Is :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 r a 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 B 6 tI/lits / 
;0 /.25" 
In JI uni-t-s: 
V:: (to ft )(~. '3IJJfr ~):: 3.06 T 
S ft 
~;: 1',:g I ~ 
~ ::: (~+t:) (0. "3 04-g ~ ):: O. b I 0 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 I u 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 ft 3 
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 
-tf x 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 D IF b -Pi.] 
1p () I. y"j J ~:;:3 
Thus; the, I~crell~ In Vt) lumt: ,:s 
.3 
'I: 1/ J i 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.:= a rea :=. == 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/;/~. AI 0, 
,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'S5 oX 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 - '17 g'13.'3 · /.J :: f'/t. 2 ~ . 56 = 0. qqg 2. 
o J+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). 
A t 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 of oxygen 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.we r ,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<!3
3 In" 
" 
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 
( J 2 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- ~m JJIIYJ51J1} 
of -!7te '1q5 
6/ 1he.5e va/lies w/ fl, -tJ"e tlC/:t(o/ df!1>/~ 
/ rl the -ban i. / n ell 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.) 
.: J h. 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/03 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
eJl<) (flJf~t.A.) = '18'. 7 OSLo.. r2. 7; 0 ( 5z()-;e I . 
l-l7 
I. 3 i -'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 CD! 
--+--------+--------~~~--~----~----~ 
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 
':'1.3 l ) 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.) x 14LfJ;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. 
ABC D I 
Pressure, Temperature, Gas constant, Atm. Pressure, ' 
psi I of fi Ib/slug.oF psia 
o ! 59 1716 14.7 
E 
Density, 
slugs/ft3 
0.00238 
i 
i 
Row 12 
~---l-------+ ____ --+------il Formula: 
i=((A 12+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 = KR 4t 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./s 2.) (90 t) 5 ) 
-r-z/ 
(97 0 --k#~3) (7. If 'I x /0 -If fn1% ) = 
D. 727 ~ 
Im-S 
= u,727 
1-3/ 
I. ¥2 I 
J 042 The viscosity of a soft drink was determined by using 
a capillary tube viscometer similar to that shown in Fig. P 1.41 
and Vidl'O V 1.3. 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 
- t J 1< /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 I 03,,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 [3 x/oJ] [-¥.] 
5/~c~ each +rn11 ;'n +he egutL.f:lbJ1 !1?"fs-t hftlle -t-he stlme 
d / 1rJl'''''tM5 -tJte ~IJ 51-o",1-.s a..?petl r/n~ /rl 1J1e efllLa.:I:,clI 
.I ' . / -3 m u 51- have dllnen ~/pif..s l e. [ ] 
[)] :;; [TJ [11.>< JD 1-J ~ [.1:: ] D~ X I D3 J .:. -b-
7htl..5) w; 1h a. c.hol1"~ I;' Ut1/fs /he J/,,/tI~ "I 7h~ 
C(J)115-fz1l1i5 wPt// tI e..l1l1l1fe q ntl -t7J/~ I S 11~t:- tt j-enent / 
h orno jen e(J)tI.J. ..(2 g aa-i'£;;J . ;\1.0. 
(j;) Pr~m Ta /;Ie 8.2 /n A ppel1d1 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. C hll119E (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 o lf>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 a t ~ ~-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 air at 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 1 I 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 
/. 'i Z X 10 -6' z. 7sf X If) 8 
-(i ~. 963)L /0 8 I.f?;('JD 
/, Cj 7 ;( If) -tJ- 3.037 X 10 g 
-:,- l. ;; ~" X 1 () 8 J.o 7 ;( If) 
-5 3. 322;( It) 8 2../7.xJO 
T I~ Sh()Wn b<-/ow.' _3~ 1jP-
3. Si. J D 
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. re frt!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 V 1.1.) 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 of interest, 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 11 T. 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) - J 0 
3. 'III x If) -.3 -3 .10 ;'~3.1; /. (!)Ol x 10 
/.Ib '3 I 3. IG' .1.1f3 x1D-3 6:.. 5"29 ;( I~ -1(0 
~6 333./6 
-3 ~ ~'G'" X lo-if 3. 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 
. 
s hf){'()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 ftl s? 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 .f j{,lid I 
'1j" h (~t f_" (6.L1 ~)( 
J;;Op JUr 1TACt. 
iJl = 0.4 N • 51m 2 
N 20-
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';' -fl t(/d. 
A- p/oi of 7hf! cltt(:a. .£S .sh/)(,Qb bt!/ow. ;:PI" A. f/ewI:rUII~H 71 u,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 ii' I .1: I 
.-' 1 - - -If ._+-- -: '1'--1-1-f--, HI' 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.. 
a VI ~ b" ~ Tn j(: .. k: n t5S f)f t) i I I a. '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 n f) = 6. J2.5'1 
& = ,. Z 2. 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 r Ver " ... 1 =D 't~ 
nUS.) 1 
OW:. ~A tA ~ 
wkev! A = rrDi 
i 
aVId ,I)- (v e 1t'>C.:~) _ fr L= t- -( +ilm1hlc.l::lle.5s) 
.56 -th~t 
1»= (I'- t )(1TDj) 
~ T I'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 
~ 
I 
o 
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 V vx/ 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 C I 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't ue; 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 n d /-vrff lie. reg tI /re d to rtJ fa 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'W 
L 
~ 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 rev Imin. 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 !!.!. )(eillT 
milt 
~ )(1 mlh) = blT 
rev '0 s W= vad s 
.:l.7T (i£ft)3(-A ft:)(s,s)(/a- 3 !Jt.)(67T 0/) = 
( ~ -ft) 
120 
0, q If '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 
5.49 w (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~lIe J ~ J'J 
.e$Ua. /;/~11 ) 
Jl'e/a.bed .J-o -the tlnJU/dY (/(!/t:Jcil-!1.; UJ.) 
3 
:;, 7T R,: ) J t.U ~--
l!o - ftL' 
(see 
(1I'1d 
so/tJl-,{)~ ..J.t> Problem I. hI, ). Thus) /"y Ii
flx'ed ,et:Jmef,.J 
a. 'lIt/en /l1~CC).s;-f!:J J E~ ,/1) /05 ot 1h~ ~f'rn 
y=hx ( !1 rv'J Qni1 x' r.." W ) 
If. ~,,~tr,1'J i .fl/Utl / .fc 
j, = :z Tr ~.:1 ). != 
IS 
)2' 
(. 
WI ih LIAlfrF6 I. 
**************************************************~ 
** This program determines the least squares fit ** 
** for a function of the form y = b * x ** 
*************************************************** 
Number of points: 6 
Input X, Y 
? 1.0.13.1 
'? 2.0,26 .. 0 
~) 3.0,39.5 
'? 4.0.52.7 
? 5.0,640.9 
'? 6 .. 0,78.6 
b = +1.308E+Ol -Ft,Jb'5 
X 
+1.0000E+OO 
+2.0000E+OO 
+3.0000E+OO 
+4,.OOOOE+OO 
+5.0000E+OO 
+6.0000E+OO 
Y Y(predicted) 
+1.3100E+Ol +1.3082E+Ol 
+2.6000E+Ol +2.6165E+01 
+3.9500E+Ol +3.921±7E+01 
+5.2700E+Ol +5.2330E+Ol 
+6.4,900E+01 +6.51±12E+01 
+7.8600E+Ol +7.81±95E+Ol 
(C()I?'t ) 
/-5~ 
(/) 
().) 
(emit) 
! = (b) ( ~. - f?.: ') 
.27T ~.3,R 
find w/th -/he daf&" g/~tl1) 
r; g Of ft.f/;'$ ) (.? 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~ e 1 Uq / -1-" 
do;= ,. tdA 
~ ntrt d It.: 2.11" yo dr, Thel'5 J 
5;nce 
cJ C7J: "" T Z.".. r d '" 
"( = 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 25 psi. 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 D 0 
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 
, 
;'rll t, 4/ s fa i-e L ... t"'" - - -~. !j I- "V .fIn II / 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 )( J f) S h~f5 
-ft:'a 
, ~ ) 
-for IIJ I I/l. 
7f 
..... (7 c; 7ii: a. ) ( If tf -:t:;"A--
ft.R -('I. 25 ;I. / () - ~ S htf..s ) {3. O'If ;( j 0 3 ~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 tv a:te ". ( Tr,; bJ eo /, /, ) 
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-o Y 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 (! SWhSCh pi 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 = VIe R.r
7 
(EZ' I· 2o) 
t:;r 4< ::: lifO Clf1d If=- 17/t, 1i·1/, s/,,!! .t)~ 
C = '19. tJ Y T(~) 1 
Fr~m 1a6/t: C. / In 4pp end':x C at tI/1 (J / -fi .f.u d f! 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) 1 040 t---t---t---t--. .- ""-~--:-----1f---J!---L-...---l 
~1020 .t--r--+--+I_--+-~-~--+I---!--l 
~1 000 .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{"$Stlfe e~t",ls the 
(/t2for ,res'Sure. !=Or waiel" ai- /(p() df' (lj.~,." ?;bJl8,j ,~Apptw:l/J(B) 
Thus 
/ 
i = if. 7Lf pSI,' (IIbS) 
r 
1.79 Estimate the minimum absolute pressure (in pascals) 
that can be developed at the inlet of a pump to avoid cavitation 
if the fluid is carbon tetrachloride at 20 °c. 
Ca vi i-A.I/!)11 rnp'1 (pee tI r when fhe stlciion P;'(J$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 2.0IJC ...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;aol S -th~ va.~J' fYe.J5'tlre. ;:-r/)/7'I 74lie B. 2 I;' I+!'ff"c/J( fj' 
~ r wA,ter a t 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; (). J x J ~ 3 ::.. )(/. lj5"1; )( / J- ~ fl,.{ ) 
::: /0, 2 ps I.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,i J/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 e rel-
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 
e required 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 F 
V€r+t '4 I 
::.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 :: (v x. J enJ1n 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~ h jq de +0 "-J./Da.i It ~ -< T "SIn e. 
"StYlet.. rma)(Jf1'lUfn1 Value JoY" ~Ine IS \ J'+- follows 
I 
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, tj I 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 )( J 0 
.D ( I'n.) 
D -3 
f) /0 e y r" y = 3.7&f x J 0 x I D 0 
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' I 0 ;-"'l. . 
S/nc-e ~ ;st.andArd ~I:ee/ paptr c)lf hAS ~ 
d/~met;('r ~f ". ~3" il1') wh/ch IS Jess fr..a 11 
O.O'/'/- /n.) It sJ1()~J~ f/~4.t. A- ~/mpj~ e)l../Jtrirnmi 
iP f / I Vof v,' f ~ 111 I.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. g B 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, 7 g 7 
/-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 e for 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 e is 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 
*************************************************** 
** This program determines the least squares fit ** 
** 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 1hen 0--= 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, dul dy. 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 
(c On '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 
/- 9 I 
!. 'to ( c ~I") t ) 
Problem 1.90 
Weight, W, vs Angular Velocity, 0) 
for 
Silicone Oil 
Problem 1.90 
Weight, W, vs Angular Velocity, 0) 
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 
Weight, W, vs Angular Velocity, 0) 
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-