solution fm manual uploaded by

/.1 I
1.1 The force. F. of the wind blowing against a building is given by
F = CDPV'A/2. where Vis the wind speed. P the density of the air.
A the cross-sectional area of the building. and CDis a constant tenned
the drag coefficient. Detennine the dimensions of the drag coefficient.
Qr
Thus)
F i: ML T-'
e,;, (VlL-3
V.= L 7-1
I/�L2.
C/) -J; (!YIL T �2)/UML"l)(L T -,)2- (L")] =: M 0 LO TO
HenceJ
CD is dimensionless.
I-I

/.2. I
(6 )
L2 YeJify the dimensions, in both the FLTand MLTsystems,
of the following quantities which appear in Table 1.1: (a) vol·
ume, (b) acceleration, (e) mass, (d) moment of inertia (area),
and (e) work.
Volume L3
t-ITn! r�'� of- c.hl/n1� � � yt/o,/';'y
L T -, -=- L T -.2.
T
(�) 1ir1.a.sS M
or- w;tn F"':" M L 7-2
rm;l5S= !=L-'TZ
(vi) mOI7lt!I'Jt- 11- ,nu!:,p.. (Art'.c.) :: se�l1d l7111mt'"f "f art!£.
� (L2. ) (L 2) . L If
!=L
WtN"k:
/-1..

1.3 I
1.3 Determme the dimensions, in both the
FLT system an<Lthe MLT system, for <a) the
product of force times acceleration, (b) the prod-
uct of force times velocity divided by area. and
<c) momentum divided by volume.
(a) f�Y(.e I( cccce/eraf.11!"/ .:... (P)(L T-;J.) - FL 7 -,2
(b)
(C)
Since
..J.crc.e x ve/c>cd'j
a reA.
rn �meni-llm
volume
(P)(LT-') �
PL-' r-I
L2-
. (I1f.T-1(LT-').=.. I1T-3
rnp�s " Ve/oci.f�
V&)/Unt�
� (FT2L-t)(L 7-1)
L3
/-3

1.4- I
(a.)
(b)
(c)
VI- Verify.lhe dimensions. in both the FLT
system and the MLT system. of the following
quantities whicA-.appear in Table 1.1: (a) fre
quency. (b) stress. (e) strain. (d) torque, and (e)
work.
.51-r�ss
Since.
�rce --
Ol"t!'tt
F-'- M L rl.- )
.5.J-r�5s ;"
F :;-
L2.
ML r-2
L"-
C)"'1.n1t!' I", /e/lji"h ...:
It'l1fll1
FL-2
- 1'1L -I T-2
L .
L
FL
=
(e) work = ..force. � dis/t,nu!. - FL
I-lf

/.5 I
(aJ
(h)
(C)
1.5 Ifu is a velocity, x a length, and t a time,
what are the dimensions (in the MLTsystem) of
(a) aulat, (b) a'ulaxat, and (c) I (aulat) dx?
PtA.. L T-1
L T-Z-
- -
dt:- T
�zu. L T-1
.
T-2-
-
h(}t {L){T}
fJu. dx
.
(iT-I) fL)-
-
dt; T
/- 5
L 2 T-.2

/. 6 I
1.6 If p is .8 pressure, Va velocity, and p a fluid density
what are the dimensions (in the MLT system) of (a) pip (b)
pVp, and (c) p/pV2?
'
(a. )
1- 6

1.7 I
(a.J
(h)
(C: )
1.7 If V is a velocity, e a length, and v a fluid properly (the kine
matic viscosity) having dimensions of L2T-', which of the fol
lowing combinations are dimensionless: (a) vev, (b) velv, (c) vtv,
(d) Vlev?
VJ.-z/ - (L T-,)(L)fL2.r� - L �T-l
fn"f tltmen�iot1less)
vj (L7-')(L) . LOTfj (dim�nsicm less)- -
-V
(L '2. r-I)
V
2
7/
- (L T-�'2.(L2. r-'j � L"T-3
(nof dimfns/on/(ss)
(d) V . (LT-1) , -2..
(no! dlfnension/e:'j)L
).7/ (L )(e r')
-
/-7

1.8 T
1.8 If V is a velocity, detennine the dimensions of Z, a, and G,
which appear in the dimensionally homogeneous equation
V = Z(a - I) + G
v == r (o(-I) -t- G
[LT-J == [r][0<-iJ -t- [6]
Since. eltU, ierm I'; 1Jt�
t:he Stll11e til;''It'n5Ipn.s It:I
E...:: L T-1
etglllb!:,dJ"" musi
;;//t!JtVS i:h,. i
01. - P"j,() TO (dIml'l'J5/�11/t'ss �;nu 6!w111117r..I.
tuiih ., nllm)xr)

1,9 I I.q The volume rate of flow, Q, through a pipe containing a
slowly moving liquid is given by the equation
7rR'
t:.p
Q= --
81-'(
where R is the pipe radius, t:.p the pressure drop along the pipe,
i.L a fluid property called viscosity (FL-27), and ( the length of
pipe. What are the dimensions of the constant 7r/8?Would you
classify this equation as a general homogeneous equation?
Explain.
The. c.�() s ttl11'i:: IT/? Is d i 1"1 e r'l S, 01 /,'s S-
)
-il1e e�L(,J./�1'I I" "'- �e..,ert<1 hom"9£11 eo 145
and
€ bu.".:hD I! fl.,o..t- IS VA It'd I� ClVl,=, �OI1 s lOS �elt
l..! � ,'t- j �.s.J..eIt) , 'Ie .5.

/.10 T
I. /1 I
1.10 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 fonnula
h = (0.04 to 0.09)(D/d)'V2/2g
where Ii 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 f).f)'!) (.!})If2.�l.
[f;]:: [O.O�J-o o.o�[:J&J[�:J[t]
[L J = [0.O/f .f-o 0.01] [LJ
Since each -fa,." 111 -the ezu.a-t,;n rnusl: nt/lie fhe
Sqrne. dt'me":$It9h.5 the C&If:iIrJHi -ferm (t), ()'f h o. ()q) rnusf
/
bt ciirnfns/clI/(,ss. Thus1 1J,� e$aA..iltiH /.5 a. Jfnt'Y4/
h19mo1('net:Jvs eltut.!:u;" -Inaf /.5 I/ti//d /ff 4n!f sfj5km
�f unl·i-�. Yes.
1.11 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-' ). D
the artery diameter. Ao the area of the unob
structed artery, and AI 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?
-
/IV (Ao )'_.Ap = K,. Ii + K.
AI - 1 pV'
where V is the blood velocity. /I the blood vis-
Sinc.e eac.h +errn must have. The S4me d/rnens/ol1s�
k'v CtVld KIA are dirnif1sjonJe�s. Thus1 fhe etuaHM
is (( f)ener1i/ hOJ?7()Jeneotis e� J).().J-itJ"7 -fhcd- woulc/ be
valid In Cftl!;l tOl1sisi-fnt 5lfsl-em 0';' unih. yes.
/-/0

/. /1- I 1.12. 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 fonn c = (Eu)"(p)
b
. If this is to be a dimen
sionally homogeneous equation, what are the values for a and
b? Is your result consistent with the standard formula for the
speed of sound? (See Eq. 1.19.)
c= (£vt(f/
51nte C::: 1.7-1 Ey :: Fe' f:;- FC"f 2
[�J� [f:4][:b_:2.b) (/)
.k:Pl
I, I)
{;PI' ;(. dllnfrmcmp//'f h(!)moqeneOIJ5..(!.!tl.dlc/1 ea.ch .f-�rm
/11 the e!ua.tlo/1 /nUS';' have ih� 511/?1� dlmel1,)J()�.5. Thtl5,
-tne. YI1M hand s/4e ()+ E<g. (/) tnt/sf }?a� 1h� dJmel1sliJAs
of- LT-I. Therei'tJre)
a-tb==o
2..b,,-1
.ta. -t 'I b = -I
(-i:" e //m111(J. te F)
(-Co Sa f, •.f.'1 C4Hd, f/(2� "n T)
(-b :!ill-I� Iy tB/'14;+';" " " L)
a. ::: 1.. ClMI h = -J.2. 2.
So
C :: rEI'/
Thb rt'su'+ (s Csnsisff>J1(/Aldh
::5peed (},f- So/,{fld. 'Ies .
/-1/

1,13 I
1.13 A formula to estimate the volume rate
of flow, Q, flowing over a dam of length, B, is
given by the equation
of the dam (called the head). This formula gives
-Q in ft'ls when Band H are in feet. Is the con
stant, 3.09, dimensionless? Would this equation
be valid if units other than feet and seconds were
used?
Q = 3.09BH'"
where H is the depth of the water above the top
Q= 3,O? B f..I%
3/,
[L3r-J � @,O'lJ[LJ[L] Z
[L3T-J::: [3.M] [LJS"12
Since epch .f.errn t"n +-he e�ual-if)n mUJf. htCjI� fhe saine
dimen�ion.5 -the {of/S/tlnf 3. oq rnus! htlve dlmemiollS
t)f L'/'1..r-1 Qnel is fhcY'�-k>re net dlmrnSIOI//e5s. Nt?
5ince fh� �,,:,.f(ull hils dlm(I/SIOI1S ifs va/tie wil/ chll"f€.
wi fh A chtl-ne;e in uni.fs. &:
/-/2

' I.;-;:::--r----------------------------,/. /5 I 1.15 Make use of Table 1.3 to express the
following quantities in SI units: (a) 10.2 in.lmin,
(b) 4.81 slugs, (c) 3.02Ib, (d) 73.1 ft/s', (e) 0.0234
lb·s/ft'.
(c:t.) /0, 2 ::;"
(h) 'f.31 s/u1s= (t.?/ s/�f5.) 01l5f')c/O s1u�)= 70,2 J.j
(t.) 3.tJ2 /b: (3.1)2./6 ) (If. 'Ilff f/;) � /3. if AI
(d) 73. / Ifi :
ce) tY,�231f
Ib'5 (0. /J231f
lb' $ ) ('I. Uti10
/11. ..1
)-- � -;:;;;'i'
If'- f.e Ih. s
.ft>-
I. /'2
N·o5
-
MI'2..
/-/3

( ,
/
,!) ....._----.----------- ------------------,
/. 16 I
(a.)
1.16 Make use of Table 1.4 to express the
following quantities in BG units: (8) 14.2 km,
(b) 8.14 N/m', (c) 1.61 kg/m" (d) 0.0320 N·m/s,
(e) 5.67 mm/hr.
I'!. 2 kl1?1 -
(i'l:2)(I03an) (3.2.4'/ �)=-
'1./'6;< lOft- If
(b) 8'/1{- N
::: (g.J 'f � ) �. 3U ;( /0.3 :;3)=
-2 Ib
.1)'>1 3 5',/gxlo
H
3
..!Y.
(d)
N·rm
(!),t)320 --
s
(e) s'b7
11>'13
-- (t!). ())Zo N�tm ) (7.37iD.x /0
-:1.
- 2. 3 b X JD
_L
- 5,/7)C/o
J-Ilf
,-"
....,.'
-f.f./J,
s
f.f
- I
- 3
3, /;. ;( 10 Slugs
.{-tl
/f·/b
)--:s-
N·rM
5

/17' I
1.17 Express the following quantities in SI units: (a) 160 acre,
(b) 15 gallons (U.S.), (e) 240 miles, (d) 79.1 hp, (e) 60.3 of.
(a.) /�O
(b)
C�)
(d..)
c e)
.'-.Jfo rni = ()"40 rni)(s"28Ll!�)(-3.b'18)(lb-I;:)= 3.2�)(IOS-rm
''t.} hp::(7,!.lhp)(SS'D4�1j, )(I.3S"lP� )= 5".qbJ<ID�.J
hP �,ij 5
"'n d 1.,J ;: W :So -I).,..-:
5
�
/9. hp =- 5, qD '/. ID W
Tc. :: §.. (bfJJ'f:. -32) = 15,7·C
cr
= 15'.7°( -1-273 = �8'l1<'
1-/5

/./8 I 1.1S 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; 1 lb = 4.4482 N; I
and 1 slug = 14.594 kg.
(a) / ft1..: (/ .fe-i((), 30n;) 2nn 7..]= O. 0929tJ "m-<-
L -/--e-
Thus) muffiil:; It2. by 9. 290 E- 2 -1-0 �tMtJty-t
fo 1':7 2..
fb) /
(t:. )
(d)
Thus) rnullipJ'j slugs/Ie b!J s: /5"'1 E-r 2. .fo conm-I
/:0 -Ie? //m �
/ .ff =
5"
Thus) mull,ll!) -Ills b'J
1:0 /!n./s.
3.0H E -/ h cont/�rt
I II; _ (I jJ:. ).rtf.INt2 IV)[ / II:3 l
If:!> - l' .ft3 l�' Ib {O,3{;lff)3M13 J
IV
-= IS7. / ,nn�
TnuS; -rn u /fipl:; IblIi � b!:J /. 57/ £Of2 -fo t'c>nJlfyt
-/:0 ;V/--m 3,
/-/6
o

(a)
(b)
- -- .
o
o
1.1q For Table 1.4 verify the conversion re-
lationships for: (a) acceleration, (b) density,
(e) pressure, and (d) volume flowrate. Use the
basic conversion relationships: 1 m = 3.2808 ft;
1 N = 0.22481 lb; and 1 kg = 0.068521 slug.
3.2;/ 1-1:
5"
Thf.;ls; me.c/h'pllj trn/s' b:f 3.;28/ 1:0 �e'nileyi
-to -!-t/.s '-.
_
I °40 ;< 10-3 S//.{5S. 1
.f-t3
Thl.4s) mul+ipllj --'<.j/rm3 h,!1 I.q/fo E-3 -to �l1l1erl..
-to s/lA�h 3.
(C) IJi =(1 N '(O.22481�)r 11M/. l
/Yv> 'l. f»1 4 ) IV l U' :tfO,g)4 -FeJ
-2 Ib
-=- /..,Ogr :I- 10 .ft'-
ThU5) mu/Lpilj N/rrnl b� ;;',081 E'-1 fo ��nVfYI:
1::-0 /h /It 2-.
ftJ35".3/ _
s
Thus.) rnultip/':J (}113/s h� 3. !)3! £+1 to (ePl//Hyt
-t-o ft3/s.
/- '7
o

/.2.01
("'- )
1.10 Water flows from a large drainage pipe at a rate of
1260 gal/min. What is this volume rate of flow in (a) m3/s. (b)
liters/min. and (e) ft3/s?
.f./owr().i� = (/200 gal
tnin
-2
757 ;<.. 10
1J?1.3
.s
(J:,) 51ilCe / /il:er = //)-3IJ?1 �
(n
J/owra../e= (757 ;'/0-2 �3)(/031;kf'.5)((Po.s)5 h>1 :3 hn/"
f/() II,) rQ. l (. = (757 X } D- 1 iFJ )(3. S3 ( X ) D
-f'tJ
- 2. �7 s
1-18
c
)

o
o
1.2 I An important dimensionless parameter
in certain types offluidflowproblems is the Froude
number defined as VIVge, where V is a velocity,
g the acceleration of gravity, and e a length. De
termine the value of the Froude number for V =
10 ftls, g = 32.2 ftls', and e = 2 ft. Recalculate
the Froude number using SI units for V, g, and
e. Explain the significance of the results of these
calculations.
In 86 1N1iis /
v /0 .li:-
s
/.25"::-
In 3I unil-5 :
V = (r" It- ) (iJ. 3tJ'ff 2!!-)= 3. ()� T
S
ft
Cj"" cr,:8/ �
.£ = (�+1:)(O.304-g m1)= O.bIOm,
IT
v
Yi�
=
The va/tie (>1 a.
il1 de'pend�,d t!)f
3 - IY>1, os S
d/rnMsion/ess
-the. uni t
/-1 'I
c ')
/.25
partlrne-t ey
:Jfjs-fem.
IS

1.2-3
�.23 A ta
.
nk contains 500 kg of a liquid whose specific gravity is
. DetennlOe the volume of the liquid in the tank.
m::: �V := SG r/hO V
Thus,
V:= m/(SG PI/1.0 ) = SOO K9j((2)(qqq � ))
= 0.2';0 m3
J.2Jf
, I
I.Z'f- Clouds can weigh thousands of pounds due to their
liquid water content. Often this content is measured in grams
J + . ,
� per cubic meter (glm3). Assume that a cumulus cloud occupies -
��++t
a volume of one cubic kilometer. and its liquid water content �� e
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
t
pounds?
, ,
. i �.'.' I
r = 3]1; u�l 1+
+
��ti
r-
I
+
I
+
+
��,.-Yot,LL�L .' (;01,1114(:-5'.i8'1 �).J
�
,
.
I
� (s.Z3b IIJJ 'ft) +--1----,--1-�
+
.. 1:-p. 2 1·3
/-20

o
o
"
'
:,,v
r-��-----------------------------------------,
I.;Z .5 I 1.25 A tank of oil has a mass of ZS- 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 10-
.. cated on the moon's surface where the gravita
tional attraction is approximately one-sixth that
at the earth's surface?
( It) w.e i:Jh f ;: t»f as.s x. d
- (25 5/uqS) (32,.2 �)= _80S Ib
- (ZSS/U9S)(It;,Sf � ) (l.g/ ;,.)=- 3S"8o.lI/
(b) t???4S,$ = .25' :;/'.((lS (tyr.ASS clots j'J/)t depfnd �II
JY'�vikfiIJHQ/ a H-rad-it)f1 )
w.eijlJi = (25"' S/U'j5 ) (32.:�) :::: /3Lf 11,
�/�.��6-'-1------------------------�------�------�1.2,6 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.
l?'yta,55 -
--
fo fi·/s2
)
- (30,r;, Jj. ) (If, 0 r;;) (0. 301ft ;; )
N
/- 2.1

1,2.7 I
1.27 The density of a certain type of jet fuel
is 775 kg/m'. Determine its specific gravity and
specific weight.
0,775"
- (775 !::1.) (q; �/ �)= 7. fa() Y3/?n 3 S � /111

1,1-1 I
1.28 A hydrometer is used to measure the specific gravity of liq
uids. (See Video V2.8.) For a certain liquid a hydrometer read
ing indicates a specific gravity of 1.15. What is the liquid's den
sity and specific weight? Express your answer in SI units.
5G -
,
-
t-
t
t-
/-2.3
I
r

1,1).
j, Z 9 J
1.'2..q 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.
InllsS of. WtJ.-/;ey =- -¥Xf
tJheve -If Is jne lIo/l1me and;; 171e derlsrf<:J. Jln�e., 1h�
!i7It$;' m«$i- YetrJllln �rlsfal1i t?5 -the -temjlRra.-tuYe ehtll1qe..5
-r-x /J - -tf- J(;J
'jc"I,/(), f�"(ftJ°
0)
fitJt;1 Ta6)e S, / J} =- /. rtf" 5/1(£5III"O�'Io'F �
£J - /. f / .5iH�
/1120 rtp rpOF - , 3 H3
Tltef'e'(;'Yf'; /Y1J1I1 �tJj)
/ ';/'(f.j)-if" = (If.f1t.3)( j, 9'11) .f.r3
If. DIf/' .fiJ
1po J.t;3j �3
Thus/ 1h� I/iCyet1� in /10 /wm� /:S
i. OJ,?L - f. 000 = 0, OJ fi, #
3
The e-hQl1fe )n uN.kY elelI/ltl 41) id 1t1�J P
.1-11- �. 01 n .ft3 -3
ji= are. :=. =5.r2x/v+-t,=O,0710in.I:t 71
rift) 2-If
Th,:s 5/nA// e-h1(11ge In def1h WtJ/JIJ n�j. he VfYfj
110frce�ble, iVt',
!J S/;1hf/fj d'�(lIi vO/lle h" ,6} /.JI;" b... cbiajl1(',(
If sjncif,i. i.Jc'1hf cfwIII,,, Ir IU'" rIt1li{'Y' 1M" 4frlsii-!J'
TIII� I'.; dUll 10 1M. /Q(.c 1hof iltele i.s s&>/}/( IIHCfmJit-/;J
111- -the li>1(r/h SI;"J7c4111 11911Y'1! of t;,t'� -h<.J� iJ4/t.(t'S" 0111'/
-Ih.f' �(>/I<'f-;o" ,:. s{'t1S,f'';f! -ft> '1},,} unC.(,..hin�.

1.31 I
1.31 A mountain climber's oxygen tank contains 1 lb of oxygen
when he begins his trip at sea level where the acceleration of grav
ity is 32.174 ftls'. What is the weight of the oxygen in the tank
when he reaches to top of Mt. Everest where the acceleration of
gravity is 32.082 ftls'? Assume that no oxygen has been removed
from the tank; it will be used on the descent portion of the climb.
vv == rn�
Lei ( )sl denote se� level cmd ( )MIE denofefhe fop ofMf. Everesl
1hvsJ
Ws, � JIb :: m�1 �sl and
�IE ;: ff1/HI£ �/tJf£
However msl:: )rIllllf so -Ih4f S/flce In :: � j
�Im ,,; - -
sl psi
or
�/E =
Jib 32..082 ft/s1.. :::; 0 qq711b
32..17Ifff/sl. ='===

I ,�q
r7�'-----------------------------------------�
/,4.2 I
1.32. The infonnation 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. Detennine 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.
t./�/Jhf of. f/tf,d
{/t!/"'m� �I /Iu/c/
�k/ we/fltf = maSS x f} =
we/11J1 "I C/Jt7 =- &. /53111
r' /
-3 ) / -3 /n13) -� 3
vj}lt(tn� ",/ rltl'il:' U's5x /tJ L (/() L :::. :1'55'x/() 1m
ThiI� i'r()111 E{ (I)
(;'=
•
3. "2 t/ -
�. /5'3 N
tt770 N
.1)1.3
F&Y' wafer d 20'C (see -r;.,J/e B. z /� Itp!fSdl'x J])
(f. :: r7gr 3.3 ' /.J :: 1ft. 2 � . 56 = CJ. 1''fn.
Uz. 0 /h1 ) (11-1..1) . ,,",3 )
•
IJ etPml)tm�11 ,,/ �e5(, Jla/lln .for /()1I.&fV b,Ji1h 1h()S�
fz.r 171e. po; s/7()W..5 171",1 -!J,t:. �c;ltc tJ! Ijht�
dtl1sa-y; I1l1d �eCi;:'c' ji'rtlv,f:t 0;:. 7'7te ;op czre all
s/'j}JfI� /Pl()eY '/},Ifn 1ne Cc>rre$f"l1d;� 1/(}/tlfS -hr wa1er.
1-'16

,"1.33 I
'1.33 The variation in the density of water, p, with tem
perature, T. in the range 20°C,;; T,;; 50°C. is given in the
following table.
Density (kg/m') I 998.2 I 997.1 I 995.7 I 994.1 I 992.2 I 990.2 I 988.1
Temperature (0C)I 20 I 25 I 30 I 35 I 40 I 45 I 50
Use these data to determine an empirical equation of the fonn
p = Ct + c,T + c,T' 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?
d",l« .£e- tJ :S�t:t?;tp PYlI{'r p,JynDItl;1¥I
tiSJn� a �6?"ufq,.d' C�rvl'- F, :;ftrl� 'pl"Djn?m $vch
115 �V",o' In EXCeL. Thu.s,.,
f = /p�/ - O. 0S".3:l r - o. 01) �I r2. (J)
II? tht!.. -b;J� below/ f (Pt'f'dlcfr.tI)
IS I;' fmlf t1?'�e/)/- ",/fJ.,
I
('J/�II).
T.OC P. kglm"3 p. Predicted
20 998.2 998.3
25 997.1 997.1
30 995.7 995.7
35 994.1 994.1
40 992.2 992.3
45 990.2 990.3
50 988.1 988.1
T= If2.! "c
/eJo/ _ 0.0533 (if2./-C)- ().()ol{1 frl.!.c)Z-_q_q'_I._S ��
/-2. 7

1,3,+ I
l.3'f- If 1 cup of cream having a density of 1005 kg/m
'
is turned
into 3 cups of whipped cream, determine the specific gravity
and specific weight of the whipped cream.
11A�s ,,/ crellm) m1 = (/ 00 f,- �1)J< (-V�p )
who'( -V "'- Volume.
Since
:::
=
= D, 335"
'Ow�ipp.,� ::
fwkip.,.... )( #
;;-
(3i5S !!o)(ergI ;.)tt"t:d M
Cy",_ ",
=3 :ZQO
N
- rm'3
/-18

/.36 I
1.36 Detennine the mass of air in a 2 m' tank if the air is at room
temperature, 20 ·C, and the absolute pressure within the tank is
200 kPa (abs).
e = f/RT
Th'.IS,
wifh T::J.oQc ::(20f273)K::H3K
anJ. fJ:: 2ookPa:;: 200X/0
3
.g.,.
� :;: (200X I03-ffi)/[(2,g69x/l �;1)(2q3K)j
z:: 2.38 �
HenceJ
m::pV�2.38�(2mJ)& 'l:76/<'1

1.37 I
1.38 I
1.31 Nitrogen is compressed to a density of
4 kg/m' under an absolute pressure of 400 kPa.
Determine the temperature in degrees Celsius.
-PT=-=
jJR.
337 K - ;}.73
1.3& The temperature and pressure at the surface of Mars
during a Martian spring day were detennined to be -50 ·C and
900 Pa, respectively. (a) Determine the density of the Martian
atmosphere for these conditions if the gas constant for the
Martian atmosphere is assumed to be equivalent to that of
carbon dioxide. (b) Compare the answer from part (a) with the
density of the ean,h's atmosphere dUling a spring day when the
temperature is 18 ·C and the pressure 101.6 kPa (abs).
(0-) - -t ::-
q00 if;....- -
337 I<.
fMI1YS RT
(In.�-k-?<)[(-5'bo C + 2.13)3
-P
3 N
Cb) F. I D I. iJ> >< 10 n;:;-�-= =
e..r'flJ RT (2.AI..'1 .J' )[(18'C i'Z13)k]4;>3-'1<
'Th",s)
fll'llllr$ 0.0214- !�=
O. D 1'15-
fe..
,.-r...
/.2 Z �
/Il"2
1-30
Je,- 0.D2.14 """3I'm
::
I. 2. '2 -h
/l!'13
- 1. t'J 5 010

,,",,,
v
1.3 q A closed tank having a volume of 2 ft3 is filled with
0.30 lb of a gas. A pressure gage attached to the tank reads 12
psi when the gas temperalUre 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. ' .
id�/j� t =
tJ. go IJ,
1= #x. volwme (.J2z-Pj.. )6.. fe)
Since.
( ;: -trno.sther,c
,;{h� wftn
jOYe.JStlre as.,Sumed -h,
T:::. (J'IJor + If.�iJ) OR
(2/,,7
ty:::
-3 /
1.t, � )( /0 oS wl/S
H3
(121'" /'1':7) tS/�
h( � II/-: 7 ,..r/a )
ti .ft, /fellj 1:hll b
3
Ttt/,Ic I. 7 R :: / 5"SIf)( If) /OY
A
of
;€ = /. 2 t;.Z )( /0 If Ii, /), ./-0;-
Sf/.(9 ' Dj<.
£� .f!) I.f. -!he
/J
7,/2
( -
! 5"5if .(/� 3
he /uim
- 'f ,/
./J - 7. /2-
(- -
/.ZIf2.XIO '+
.x I b
-U..!4..S
- 5. 73
ft-3
6/ 1l1e$e (/Il/t(es WI fl, -the
1It the -btll11. /;'dl C.1lh!,S
/-31
aC/:t(a/ detl5/-i:t
'1h4 t 11t e..
(I)

t.'+0 I
1.40 A compressed air tank contains 5 kg of
air at a temperature of80 0c. A gage on the tank
reads 300 kPa. Determine thevolumeofthe tank.
V(!)/tlme -
1=
1.'-1-/ 1
/ff14�S
, ,
J.;U tnI 3
I.!H 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 I0 OF?
/-32.

1.42 The helium-filled blimp shown in Fig. P1.42 is used at var
ious athletic events. Detennine the number of pounds of helium
within it if its volume is 68,000 ft' and the temperature and pres
sure are 80 'F and 14.2 psia, respectively.
• FIG U R E P1.42
W := 0" V where '1/,;: 6 � 000 (1.3 and 0'-:: ft :: (f/RT),
ThVJ, . ,..
ff �t = [111.2 *� (IJflf�)/«/,J-'fJ.XIOIf�I"/'R)(80-+'f60)'R)J (32.2 ft>
::: 9. 81X/O-a..J!V1 (Jlbj(s/vJfl/S2)) = Q.82X10-;Wff . s"
Hence,
'W == 9.f2XIO-;�(68,.ooofP) ,;: 668 /b
/-33

J--
Master Typing Sheet
I00( Reduct{)I
8 1/:2 x I I tnm size
-
I
*1.43 Develop a computer program for calculating the density
of an ideal gas when the gas pressure in pascals (abs), the tem
perature in degrees Celsius, and the gas constant in J/kg' K are
specified. Plot the density of helium as a function of temperature
from 0 ·C to 200 ·C and pressures of 50, 100, ISO, and 200 kPa
(abs).
I
I
--...., .....- . �
I I
T
I
I
I
(')htl'< -r iJ IIl.j,,/"j.� ;oY'usJ� kl1h� 9AS aJ..,s J.�11 t anA T
i�-(lbs()/",,-k -iempua-blre. Ti-t-'ws 1-1 1ht! i:�mJ� ��cuye
Is 111 "C I --tn1el1 �- 1 ,1 I
�-+--1-+1 --<---T=�6--rt-Z::,7J', J� -
i Sfreadshl'et reX(,fLJ 1rbjYlltrf -hI" etdeu/ailllj fJ /oJ/oU)s,
I
� �T7h� is�p�ro�gl�Tm�ca�lc�u�l a�re� s�t�he�d�e� n�si�w o�f�an�id� ea�l�g=as��
____�____�
I
when the absolute pressure in Pascals, the temperature,
in degrees C, and thegas constant in J/kgoK are specified.
t To use, replace current values with desired values of
temperature, pressure, and gas constant.
�--
�
---+-- ��A
--�--�B�--�--�C�--��D��------�----�
t
Pressure, Temperature, Gas constant, Density,
.-----...-..!
Pa °c J/kg'K kg/m3
1.01E+05 15 286.9 1.23
. Jf
Row 10
Formula:
�----�--------�-------4=A10/((B10+273.15)*C10)
-
T ,
��
f---,-" _+-�i _+�5.trrnl!/e.,' J:a/c.u.LAre ;L Ale p:: Z(J)o"k pf).) l:�mttrl.l:l4re =
c I, I J.eJ·C) (Mil R. � L87 .J/�. k.
t-----
+
,
.,...--- -----t-...... -
�.��__ ��T _J_ _��,. A
' -
B C D
Pressure, Temperature, Gas constant,
Pa °c J/kg-K
2.00E+05 20 287
I (Gon'f)I
-/-3'/'
Density,
kg/m3
2.38 Row 10
"...............- -

"".43 I (con't)
fhe dens/fy of he/ivrn is ploifed in/he 9faph be/ow.
Density of Helium
0.4 ,---------.-----,----,-------,
0.35 +-�c------I-------jc-----+-------j
0.3
� ,0.25 +'--=:.,'---;;;::-----j---------1"'-"""'""-:::---+----1
-,'- -- ----r--------0.2 -I------j---=::-�-_= +------+--�
-r::-___
-...... . ----0.15 -1---=-=---'-----+--0_-:-__---+---- +---=-=---=----- -- --- -...
0.1 +------+----+----+------- -:.-=--=---=..,..- ---
1-- ----- - ------
-
- 1-- __
-. ---0.05 +------+----I----=-:F-=��-
0-1------1----+----+-----
o 50 100 150 200
f = 2.00 !<Pa Cds)
'" ISo
;# 100

I. Lf£ I
I. Lfb I
1.45 Por flowing water. what is the magnitude of the velocity gra
dient needed to produce a shear stress of 1.0 N/m
2
?
Thvs)
rJl). - 'r
rJy
- =
fo
h
�3 N·s J II
w ere .Ll:: /./2 x/o --;:- ano '( =I.o-�r- m m
/.0 �
893
I
- -
1.12 X 10�3 If·s..
-
s
m�
1.46 Make use of the data in Appendix B to determine the
dynamic viscosity of glycerin at 85 0p. Express your answer in
both SI and BG units.
Te, :: � (IF -3.7.)= � (1'$";:-32):: ;2'U·c
Frt>m Fi9. B. / /" Afpf'nd/,t 8:
/ (yJ'II!e"/11 �I ,f5°,!!- (,2<;.'1-#(.)):::. o. 10 �2. (SI uni+S)
(tJ � NoS) (
-2 Ibd
j< � . ;t>fl. .2.05''I,c/O h">. );::: 1.3x/O-Z'��' (8G""�
/-36
iJ.5 Tl;
"" ...

/. /f7 I
1.41 One type of capillary-tube viscometer is shown in
Video V 1.5 and in Fig. Pl.41 . 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 m'ts
is then obtained from the equation v = KR4t 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 20· C 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 kglml is
tested in the same viscometer the drain time is 900 s. What
is the dynamic viscosity of this liquid?
Glass
strengthening -
bridge
-."1:>Etched lines
Capillary --'<+' ....
tube ;�
)
• FIGURE P1.41
• •
�r '}lfj�er;')1 @
/. /r )(/()-j)hn � Is ::
20D{ f}= I.JCfx/P -�1s
(;CR�)(�tfgO s)
k. R'f=
8,32 X/0-7 /'I'1121s2.
v=
=
/Iiu/d wdn t=- 100s
(g. $.l j, /()-7/'t11 "-Is2.) (C,00 S )
=�-v
(970 -k#�3)(7.If 9 X/0-If tm1s)=
6_727 �(In.J
= =
(),717
/-37

Master Typing "hee'
I OCk ReductinJ1
8 112 x I I trim SIlC
I. '18 I
--"':
1.+8 The viscosity of a soft drink was determined by using
a capillary tube viscometer similar to that shown in Fig. PI.47
and Video V..5. 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;" = kr:-The fcrowing 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, Ii, of regular pop greater than that of diet
pop?
I(S)
SG
Regular pop
377.8
1.044
Diet pop
300.3
1.003
= [,(+,J(16 )1'"1'-' �1 I I<
(.f I< 5j6 Ja'Jt -!
LoO
,
r
II(3010.35)(1.005) -J x J OD
II I
-- 1
I I
� 1 .3 II 0 ()t,
; I
I
I---�-->-'--�-t· , --.-+- -+- - + -
I �
I �
f-------+------.---+---l-' --I---��...........
-
-
I
I II ,
I
t
t
.. ..._.
+ -I-
I . I I
_ /-38
1
I
i
t
t
,
I
J
I..�--l.�
�r
1
-1
j
,
, II I
I
i}IOO
1.� -1
,

/JH I
1.4.Q Determine the ratio of the dynamic vis
cosity of water to air at a temperature of 60 "C.
Compare this value with the corresponding ratio'
of kinematic viscosities. Assume the air is at stan
dard atmospheric pressure.
Fre>rn Tt, 6/� 8,2. /n Ap�ndiJl B :
(.f.r/r /Vafe,. ai fpo'�) jA- = It. ;""5 )LIO-If�� j
Thus.) -'I-
/'H..() If. r..r..5 J{. /0 =
�3,7=
�/II;' I. fIT ;( I()-�
�L()
-7
if. 71f!i )I. 10 -2
:: = :l.!J5")I.J0
"Y.tti,. /. ek ;(/0-5"

Master Typing: Sheet
lOCk Reduction
8 1/2 x I I trilll size
/,:50 I I
��I==�J====;!========�'========�I========�==�- ----
I--
r-
-
+
-
--
/,IsI
1.50 The viscosity of a certain fluid is 5 x
10-' poise. Determine its viscosity in both SI and
BG units.
t:lne/ Frp/?f TctbJ� 1.1;-
/'<- :: (5;t.. /D-�- :51. ) (�. O?f X/o- 2
j'l
1.·SI The kinematic viscosityofoxygenat20 °C
. and a pressure of 150 kPa (abs) is 0.104 stokes.
Detennine the dynamic viscosityofoxygen at this
temperature and pressure.
-v = 6). /tJlf sl-okes - .s
= (6).!()'f �2.) (/0-'/- mt" ) (I. '17S
tJm;J.
{
--"" = .1. tJ5 .K I()- b-
J
l-lfO
-
-]
I
N·s
-_.1

Master Typing Sheet
1 0% Reduction
8 1/2 x I I trim size
"/ 5 2
I-
J
I
�
I
I
f-+- -
Rate of
I
. '•.. 52. Auids for which the shearing stress, -r, is not linearly
related to the rate of shearing strain, 'Y, are designated as non
Newtonian fluids. Such fluids are commonplace and can exhibit
unusual behavior as shown in Vicltll V 1 .6. Some experimental
data obtained for a particular non-Newtonian fluid at 80 'F are
shown below.
-rOb/n2)
I
0
I 2.11
I 7.82 I 18.5 I 31.7 I'Y (s-') 0 50 100 150 200
Plotthese 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
2.11
7.82
18.5
31.7
; 40 .,-_,-_+-'-!OLjIOollLlj'OIIRa...1v/-'2<-"+'-'1-'10=>.0,'<-+,"
..
a 30 +--+--+---f----,,,,.----I50
100
150
200
ui
./� 20 +---+---+-/--J.tI''----+---i
.� 10 +--+-�-----:�-+-t----I
:l 0 �-..,..- -
J:
Ul 0 50 100 150 200
Rate of shearing strain, 1/s
250
�
I
t
I
,
where.
1'ht ra.,te

Master Typing Sheet
1 0% Reduction
8 112 x I I trim size
�/,�5=3==�= 1.5.3 Water flows near a flat surface and some measure-
1 ments of the water velocity, u, parallel to the surface. at different
f-c- heights. y. above the surface are obtained. At the surface y = O.
- -+-+t..; 1 After an a?alysis
.
of the data. the lab technician repons that the
I j I
t '
-I velocity dlstnbutton In the range 0 < Y < 0. 1 ft is given by
, ++ 1 - the equation
.....++ .
d+l I u = 0.81 + 9.2y + 4.1 X 1O'y'
t
+ +•
i . • '----r-- I
·1.1 . 1 I
-,----T
I
I ++
i I
1
+
-I---
+
I
-I--
I

Master Typing Sheet
1 0% Reduction
8 1/2 x I I trim size
/.Stf I
-I • _ � 1
1•.5lf Calculate the Reynolds numbers for the flow of water
, and for air through a 4-mm-diameter tube. if the mean velocity
. is 3 mls and the temperature is 30 ·C in both cases (see Example
1 .4). Assume the air is at standard atmospheric pressure.
(horn Ta "Ie B. l Ih Irppetld/x B) :
)£ _ 7. '? 7'.i" ;<. 10- If
N.�
/ ,-m "Z.
Re
_ r?. V D
/"
_ (rt?S:7 1!3) (3 if) ( O. tPOlf tm) = /� ooo
-
-
I.
I
- t-
7. 9 75" x IO- if N. s
/l?'1 l.
Cit iO' e ( +r()m r;. Ue B. If if') 4pPfndi)( B) :
jj = I. / �£ � $.
{ -;:;;;-�
/(e =
----
- -
,
tI
l-
t-
, ,
1--
T I�
r
= (;. /�5 :J!3) ( � 0/) ( �. ()OLf an)=
/. 8" ,( 10- �- N. 5
"",, :I.
1
t
I
---- -
T
I
� -
-
L
I ,
t
I
752
-
II- -
t-
-t
-
t
--i
� + .- +-
, I ii I-
--;++" t+ 1- -�+�--t--
-, f ; - � t l-
t t I I .,. .. t
+ -
1--1-_'_'..;-'_+_'_--1--_+-_--"--' _.-11--_+-___
f--+ -
+
- -
1 -
, I
+1
---1 -
I
i-- +
t
I
,
-II
I
f I
-
I
,
I
rI
,
t

/, 5 5 I � .. _ . a ", , ._. ... •
o
o
1.·5 5 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
10-6 kg/(m·s·KI ') and S = 1 10.4 K. Use these'
values to predict the viscosity of air at 10 °C and
90 °C and compare with values given in Table B.4
in Appendix B.
'3
(. T T.
=
T -t 5
T r / l 0. /f 1<
FtP'r :2 6'�. 15 k )
1- =
From Ta /'/e
'3 (P �. IS 1<
)
//0. 'f
From Ta. ble B. If) /"
o

I,.
I,)D
1.,5"6 '* I
o
o
1.56* 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 1.55. (Hint: Rewrite the equation in
the form
- -
T3/2
=
(!)T +
S
J1 -C C
and plot TJl2/J1 versus T. From the slope and in
tercept of this curve C and S can be obtained_)
C!/(.. tl"� I. /0 (IU, be "''', /fen
.
1he (r:;1"1>'1I "
3/...
S
� = (� ) TT- C (J)
�
qm( WI'flt 17te rJA-bt. Ir�m -r;ble 13. J.f
T(od T (Ie ) I- (N.s/hI1l.) T�� [J///(lJ�'5)]
0 J73. I:; 1. 71 :£ IO·S .2. 6tf� ;c10 �
,;lo J-Q3. lb /. �2 x /o ·S ,Z. tsf x 10 8
- 5'
;2 . '163 :L 10 8
tj.D 313. lfi 1. g ? .x JD
3 . 037 x: 10 g
6 0 333. I£' I. 9 7 ,(/0-i'
.3 53. /�
- �- 1. .2 � 6 X 10 8
8 0 '). 0 7 ;( /0
lOt) 3 73. 1;
-5
3 . 322. )( It) 82. / 7 .x 10
A- plot 0 / T� V,s . T I:' shown tHolo w .'
s.S/.JO!
·Uo

(C()n't )
5/�,e tJ,� d(.t6.. P/Dt as Qff QfProx'WI.ie Sf"419Jd /Jff'e,;
RS' (/ ) Gil! b£ ref�l1kd "'1 tll1 t'S'ILi:tO# of 11ft!.
Icrm
.b ...... lie I
tHl tA a. N J/C ,
Ft'!- in! daAa.. -/-0 a. ftn �M €,!UaiIO;1 USl n � a...
..5i:t'(ndtii'd (tuY"l/� - /' fhnJ ,roJYllm s�clt AS l'oul1 d
I n ex CeL . Th u s)
a'n d
fj = �, 9b 9x /0 5"x. -+ 7, Lj-If/ oX J b7
7
=
7. If./f/ X ID
s = /D7 I<
77wt! /ltt/ues �y C fll1l1 S tll"e //1 �iJM 4'lN'emftlt
fA;/tn Iltdll�S rivf/'I il1 Problem /. 55 ,
l - if6

MaSler Typing Sheet
H Y1r Reuuction
8 I/� x I I trm ,iz'
/. S 7
-
I--
1.5 7 The viscosity of a fluid plays a very imponant role in
determining how � fluid flows. (See Video V1..3) 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 = KIfL. In this equation K
is a constant for a given tube and pressure, and " is the dynamic
viscosity. For a panicular liquid of interest, the viscosity is given
by Andrade's equation (Eq. 1 . 1 1 ) with D = 5 X 10-7 1b ' sift'
andB = 4000 oR. By what percentage will the velocity increase
as the liquid temperature is increased from 40 OF to 1 00 °F?
I
_
A1ssume all other f�1ctors remain cor:nt.
Ij V�6° = fA'foc
I,
-+
-. + . . i . . ... . , ._--- l-
tI
I
T
[V(OI)o _
�,:1 00
"'to" IJ
I
I
t
,
It
J
t
r -- -- �
, I I I
Ii
I-
.--- . .-�
.. .--.. - �
,
I I II -
/- .'f1
( 11
( 2. )
(3 )
-
�

"/'.5 8
o
o
"1.5B Use the value of the viscosity of water
given in Table B.2 at temperatures of 0, 20, 40,
60, 80, and 100 ·C to determine the constants D
and 8 which appear in Andrade's equation (Eq.
1.11). Calculate the value ofthe viscosity at 50 ·C
and compare with the value given in Table B.2.
(Hint: Rewrite the equation in the form
1
In J1 = (8)
T
+ In D
and plot In J1 versus l i T. From the slope and
intercept of this curve 8 and D can be obtained.
If a nonlinear curve fitting program is available
the constants can be obtained directly from Eq.
1.11 without rewritjng the equation,)
E8"4 ,tJOI1 1./1 urn k t.Jrdlf'I1 NI 1h� /:,rm
In;" =- (13) -f t In j)
IIlI(t w/f;, 1lte d-t6t. +n>m 74b/� 8. 2 .'
T roC) T(I<.) I/T(K) I- (/./.sk,�)
:1 73, }S 3. Ii,,/ ;L/o
- 3
/. 7K 7 ;(}o- J0
3. iii/ .1/0
-.1 - 3
.J. o ,;1., '13, Ii /. (')0.1. .( /0
40 3 ) 3. /� 3. /0 x/o-] �. S'.2'f ;( fa- If
�o 3 33. /5
-3
-1'. U.�;(lo-1f
3, t)oZ xIO
-3
3. S"1.f. 7 ;(/O- If
90 3 5'3, /5 .<. f52 ..( /0
-J - If
100 3 7 3. /S- �, (.S'� XIf) :l.KI/?.x 10
A- plot of IfJ/' VS, Ilr 1.$ ShOUln 6e/f)ltJ :
- r, o
- 8. 0
11'1 j<
- t. '3Z 7
- {. . 'lo t,
- 7. �3'f
- 7 . �70
- 7. "''It;
- 8, i 7'f
_ 3
'f.Ox/0
rCo/J� )
( I )

-'/, 5'8 I
::»;,c<. 1J,e d�/;� plot �$ �)1 o./JpYO:l.im 4+e. S.t-r4;9hi.
1,'rJ-e E�. 0 ) ClJI1 b� Jlst'ej -Iv re..pY't'�1t t: these d#....f:a.
I
To " bt:tu� B tin A. D/ //'+ fh (. cI"".f:tL &0 a n
b><
eJ<-?�nM h a / � 0 (I. ec.�/�1'3 � .{ th e .form f:J = Cl. e
:5uV! Ifs fblAI1 Ci I n H)<' CP'L .
Th eA s )
Qnd 3
B = .b : /, ,f70 ;( /0 I<.
-6
)< = / 71.. 7 ;t ID e
Ai:- 50 '( (323. /!iJdl
/- =
- I.
/. 7�7 .x./D e
- ¥-
= S. 7/,;, X /0 N.S/I1>1�
7
- �
/
From It:lb/� B. 2 ) )t = S: �!;( IO /J. s/,.,,/-.

1.59 For a parallel plate arrangement of the
type shown in Fig. I. 5 it is found that when the
distance between plates is 2 mm, a shearing stress
of 150 Pa develops at the upper plate when it is
pulled at a velocity of I m/s. Determine the vis
cosity of the fluid between the plates. Express
your answer in SI units.
T =)A ��
dlA = JL
d!:f h
r
/-
=
(l/ )b
-
/ 00 A::
""' >-
( r ; )4 o();l.'1>?
/- S O
=

1. 60 I
, " .. ' . , ; " ��', •
I 21'
I
b
1.60 Two flal plales are oriented parallel above a fixed lower plale
as shown in Fig, PL60, The lap plale, localed a dislance b above
the fixed piale, is pulled along wilh speed V. The olher thin plate
is localed a dislance cb. where 0 < c < I , above the fixed plale.
This plate moves wilh speed V" which is delermined by lhe vis
cous shear forces imposed on it by the fluids on its top and bOl
tom. The fluid on the lOp is lwice as viscous as lhal on the bOl
lOrn. Pial lhe ralio V,iV as a funclion of c for 0 < c < L
1 16
, "[
I'
--:1
" i-V,'�'".,.'
�,, .j
IiII F I G U R E P1.60
For cons/an! speed) l-L of the mldcile flcde, fhe net force
on the pIC/Ie is O. Hence) �'P ::: rbollom J where r::: "(II.
Thvsj the sheaf slress on the lup af)oI boi/om o f fhe p/ale
m tis f be equal,
h ?" au (I)
?;,p '" ?boNum !AI ere "')A cry
V
For {he boTfom {/vid. rl#- ::: -¥- while fut +he 10p flilicidd!i :;: (V-v,t
ely C D ' Y b-d
Hence) from Eqn. ({),
( V - V,) V,
(21L) =: (11.) -
J
wh'c.h can he wnHen as :
r b ( I-c.) r c- b
1-eV - zeV, ::: V, - dl,
or
V, .::
1- C
-
V c f I
t/ofe: If c " 0 v, ::: 0
J V
If c$' � � == J:..
' V .3
If c ::: / V, = I
) V
�-
V
1
0.8 .
0.6 .
/0.4
/
/0.2
/0
o 0.2 0.4 0.6
c
/- S I
0.8 1

1 . 6/ T
1.61 There are many fluids that exhibit non-Newtonian behavior
(see, for example, Video VI.6). 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. As
sume that the viscosity of blood is to be determined by measure·
ments of shear stress, T. and rate of shearing strain. duldy, ob·
tained from a small blood sample tested in a suitable viscometer.
Based on the data given below determine if the blood is a New·
tonian or non·Newtonian fluid. Explain how you arrived at your
answer.
T(N/m
'
) I 0.04 1 0.0610.12 10.1810.3010.5211.121 2.10
duldy (5-1 ) 1 2.2514.50 1l.25 22.5 45.0 90.0 225 450
For "- Net.Jl:tJ",QI1 .Pluid -th( ra.l:/o of 1: n du/d!} IS �
C(!)ftS-UMt . F;py -jJt( dAia.. 9'/J�n
1-
dt</I'J
Thl! ra t.io I:; noi. "- �"S-&l1t ",,-I: decr(J2s�s tiS the ra-tl! ",f Shf't/r/II�
stt'll/n /ncrN��!. Th�J., 1hls -FlUid (b/{)()d) I"; .:t /JM- New/;oJ1/trn -f//.(,'<1.
,if p/oi of i1t1! ci4f;a .d .show" b(!./olV. !T>r A lIe.Jb:MIIH ;t/wid 1n�
Ctlrv< w",,,I'( b< .. si:nll;l!i //11( 4)/'17, I( 5/bPf!: "f / 1-0 / .
v V
v� / 1
0.1
��II��!I�llllrNewlonian{Ivia
0.01
1 1 0
du J
-ry ) "5
I I
1 00 1 000
where a :::/ for (.i IVewfonian flvid.

1. 6 2.
1.62. The sled shown in Fig. PI.62. slides along on a thin
horizontal layer of water berween the ice and the runners. The
horizontal force that the water puts on the runners is equal to
1.2 lb when the sled's speed is 50 ftls. The total area of both
runners in contact with the water is 0.08 ft2, and the viscosity
of the water is 3.5 X lO-l lb s/ft2. Determine the thickness of
the water layer under the runners. Assume a linear velocity
distribution in the water layer.
F rl'one) = l' A
'[= f ��= I'
Th H S)
V
d
• F I G U R E P1.62.
d
= (�!)-<1v-sJ)f:)(5D Pf)(a, !J8.fl-)
- if (l l
1 1.1 )( I D +t
) - £3
1 , 2.. I",

1. 63
1•. 63 A 25-mm-<liametershaftis pulled through
a cylindrical bearing as shown in Fig. Pl.63 The
lubricant that fills the O.3-mm gap between the
shaft and bearing is an oil having a kinematic
viscosity of 8.0 x 10-' m'/s and a specific gravity
of 0.91. Determine the force P required to pull
the shaft at a velocity of 3 m/s. Assume the ve
locity distribution in the gap is linear.
1---0.5 m---..,
FIGURE Pl.i!>3
2 F;. = o
J
p= 7A
"""- �
:D� �
TA
� ""-
�..... �
A = rr D ... (sha.ft /en91h I� bRPi'"/'r>, ) 2 TT D.1.
.so -/h, d
(I/elt>,if'j Df shaH._)
= j< �( :lap widtn ) / ./,
p= 0 r)(lTD.1 )
p
;>
?=
(3. () ,( JD-,7�)(tJ.qI )G /t/1�)/3f')(Tr)fCJ, 02b",,)(0.5""" )
( 0. 000 3tm )
/- 5'f
p
-- x

1. 64-
1.6.'+ A IO-kg block slides down a smooth in
clined surface as shown in Fig. P1.6'/'. Detennine
the tenninal velocity of the block if the O.I-mm
gap between the block and the surface contains
SAE 30 oil at 60 OF. Assume the velocity distri
bution in the gap is linear, and the area of the
block in contact with the oil is 0.1 m'.
'2 F,<. = o
Th u s)
Sin ce
W · 2 ° = -"-L ASI n 0
W �/n 2 0 ° = )' � A
Th U 5
J (/#,1'h w= n>'> 9 )
v =
b '4 51;" 20· =
0.1 mm gap
FIGURE p1.6'1-
(0.000I /)n HIOJ.;)/f.JI�..Ys/� lDo)
( O. H N.S )({). l tm2)
/1'>1 '"
J - 55

1. 6 5
1.65 A layer of water flows down an inclined
fixed surface with the velocity profile shown in
Fig. P1.6S. Detennine the magnitude and direc
tion of the shearing stress that the water exerts
on the fixed surface for U = 2 mls and h =
0.1 m.
Tht/' ;, a.1: -th�
C��)Fo
So
fixe,! sur/dee
= dl O
h
/ - 5 6
FIGURE PI-55

·1.6b Standard air flows past a flat surface and
velocity measurements near the surface indicate
the following distribution:
y (ft) 10.005 10.01 10.02 10.04 1 0.06 1 0.08
u (ft/s) 0.74 1.51 3.03 6.37 10.21 14.43
The cuordinate 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 = C,y + C,y'
and use a standard curve-fitting technique to de
termine the constants C, and C,. (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.
( It) Use I1M/;;'ertr rejr�SSI';" proghlm
(6)
-fo OUI:tIh &JefHC/Mi::s (I tmti C.z. . The prt:J/Jrr:lm PNJrluc&s
Iells t slp'tlrt's est/males "I /he ptlY'lIl11ei:frs of tl ntJl1itll/?4r
rt/1Pt/e/. FbI'" -the dlt/;a. J/l/fJ1)
C = 153 5 - '
1
SIYl C e., du
t=r d !1
;f .k1/()UlS 1h,rt
t =j' (� t 3 C� �L )
Thus, at- the wil li ('1 = 0)
Ifi
/- 5 7
- 5' II:,
= 5: 7:J. x IO it..

t, 67
1.67 A new computer drive is proposed t o have a disc, as shown
in Fig. P1.67. The disc is to rotate at 10,000 rpm, and the reader
head is to be positioned 0.0005 in. above the surface of the disc.
Estimate the shearing force on the reader head as result of the air
between the disc and the head.
IIll F I G U R E P1.67
Stationary reader head
Rotating disc
Q,2·in,dia.
-j f--
0.0005 in.
F '" ..rheal' force on head. � CA ) whenJ
if Ihe ve/oc/ly prolJle
if} fhe qap he/ween fhe Jire anJ head ir /iYJfrir a/ld {/11I"f{Jl'm
tI (jIGSs fhe hetlrJ.J 1hen
au U
(:= r Ty =)A- b ) where
rr'" (,() R :: 10 aDo!!!.. ( !min)(21l''A!) (.!- 1-1) :: / 75" 1t,/ min 60-I rev 1'2- s
Thv�
#
r "'(3.7IfX/O-7�) / 75 -:s
W ( O',o:.s;0)
$0 fh",f
F :: ,...,1 = (1,5"7#-.. )* (W=-Nt ::
3. '13 x IO-If16

1 . 68
1.68 The space between two 6-in.-Iong 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 1 80 rev/min. The outer cylinder is fixed. As
sume the velocity distribution in the gap to be linear.
Tortuf; d � du� -/-0 �11eIIr;"j smss
on /nnt'r c:;/in c/o- ,j qual .f"
d 'T: �. i dA
wh-er� cjA .= (1::. de) -f .c
d 'J ::
:z.
�. J T dB
and Ivr3H! regu;"rrd to
I n ne r 01f/tlder is 2 Tr
CO? view
Fixed
outer
cylinder
1I
1
):: !</J ride() (J. ... CfjllfJdrr lenqt"h )
ve /Oc./.j.� disfribw.f../ol1 / n fhe 3fA. />
S o -fhtl. !:
R.. W
R() - R.�.
.27i R/}.f cu
Re> - R,'
(/80 r..� )(e3lTm lJ,
w = rAg )(/ n1l;'):. 6 rr
re v t-, o s
Yad-
s
.<.7T (-!£.ft/(!i ft)(8.S;</D-3!/f!..)(67T 0/) =
( � -ft )
1 2.
power =
Ie/lows
fower = (C f'f'f ft.fJ,)(67T r;d'J =: 17. 8 �.I.b
/-5 9

1.69 A pivot bearing used on the shaft of an electrical instrument
is shown in Fig. P1.69. An oil with a viscosity ofJ.L = 0.010 Ib·S/ft'
fills the O.OOI-in. gap between the rotating shaft and the station
ary base. Determine the flictional torque on the shaft when it ro
tates at 5,000 rpm.
0.001 in.
<--J...� 5,000 rpm
0.2 in.
!l = 0.010 Ib . sift'
m F I G U R E P1.S9
LeI dfl::: fOf'Cfve on area elemenf dill
where d/} :: '-71r dJ = 2 71r dr/s;n(}
Tht/s, du ('(}r
dr::: r dF :: r ' dfl where (' �,P dj ==j-l T
so -Ihtl�
dlT:: r (I'- T) (27trdr/sin e)
= 21l')JfJJ
b sjhfJ
Hence,
u� fd;r :::
r �o
/- 6 0
17')N W
z. b sin 8
(/)

t. 70
1.70 The viscosity of liquids can be measured through the use of a
rotating cylinder viscometer of the type illustrated in Fig. Pl.70.In
this device the outer cylinder is fIXed and the inner cylinder is rotated
with an angular velocity. w. The torque � required to develop w is
measured and the viscosity iscalculated fromthese two measurements.
(a) Develop an equation relating J.'. w. �. e. RD. and Ri• Neglect
end effects and assume the velocity distribution in the gap is lin
ear. (b) The following torque-angular velocity data were obtained
with a rotating cylinder viscometer of the type discussed in part (a).
Torque (ft · Ib) 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 R, = 2.50in.• Ri = 2.45in. • and e = 5.00in.
Make use ofthese data and a standard curve-fitting program to de
termine the viscosity of the liquid contained in the viscometer.
(Q.)7orzut'; d J: du� -Iv :shetll'll1J sms.5
on /nnf'r C:Jh l1 dO /,j elI/til p,
d 'T=- R.· T dA
wh-fr� clJJ .: �. di9) -f .
d J=
:z.
�. J T dB
{Jlld /vr311t. regtl/rrd to
' n ne " c'I/lncler is
2 TT
J:: R/J ride(>
..
';UT R. . j l'L -
disfribl.{/-/on 1 11 f,4e 3a p
T=/-
J::
,�)
(lI1d a. (}111M
Wh�Y"e h I �
R · wL
R. - R ·c) ,
:;.;R/};
S o inti. -I:
cU
R - ,R .
o L
III�c()s;f!:JI
y = h J(
Ef , tl)
Thu5;
IS a t
( '1 rv�
If. �".sb,11 i .R�tltlJ -10
(con'�)
/- 6 1
;;'y "
the
(/11 "
liquid
Fixed
outer
cyli
1l
J
• F I G U R E P1.70
, ,
top VIew
(.1 A. C'j/ll')drr /en� /i, )
!';;ed Jt't!Jmeir.!l
/tJl"m
x � w )

1.10 I (con'�)
"3
-
j, =
:lIT"�' ).)A-
Pt) - .t2,:
r; oHal � h .ft:1:. -the dC(../:-&<.- h a I in e�v ..e�/.A.o...Jl�n
� .f fhe -foyl'Yl !J !:. hx USIi?1 a... S/:I/I'1 dQi'd r:.UI'"i1� - I, H-U?j
proJfll1r/ S ll c-h '1 .5 .f!:,und , n J::X C�L .
. Th us, from £g . ( 2)
;-=
(6 ) ( f?o - f2c: ')
.z 7T l? 3.-fL
Qn cl w,'1H iJ,e da.f.t<... gl;tJ1� b = 15. D8 -fHb'5 ) 50 1nd
�/3.Of .ft'//" 5)(.1.5"0 - 2.. LJ.5"
It:)
. / 2
.JIT" (lJf.� ft).3 (5.�o it)/2-
1 2..
.1. Jf..5 //" 5

I. 7 J
1 .71 A l2·in.-diamcter 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.71, Detennine 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.
kY'?Ui; day I
pn p/,.+� r.i efU4/
d 7= r 1: ciA
dA- : 2.11" r- dr, Th&l�J
c/ ClJ = y T 271" Y d ".
or::.urfRr>. r ell'
o
• FIGURE P1.71
7= )A- �� I 11M' "r «
ve/(),,'ry J,sty; b ... .f"';H (SRefij"N:)
::: () . 0 772 1-1:. . I/,
/- 6 3
Rotating plate
Torque
0.1 in. gap

/,13 I
1.73 Some measurements on a blood sample at 37 °C (98.6 OF)
indicate a shearing stress of 0.52 N/m' for a corresponding rate
of shearing strain of 200 s-'. Determine the apparent viscosity
of the blood and compare it with the viscosity of water at the
same temperature.
du.
•
�=)A = t 't-
d �
'L
N
)J-bJODc!
�
= iJ. 5"2. ;;;...
)( I 0- 'I !:!.:..:.,
2 �. 0
-;
-
I=roln Ta j,J�
@) 30°C
@ 40°C
Th us, w,'th I l�f/l.r
a n 01.
=
j,."iJ .L
$ _
I. Q75' ;;< 1 0- if N·.!
•
/1)1 •
10. 5' 2./f X I O-V N· s
m, '"
1'I�rpD J..hon ) Ai (noc)
/ 'Hz.o
- � N . �
Z �. 0 X Ib (i;;"i
10. q� ;( I D-If�
"" ..

/.75 T
1.75 A sound wave is observed to travel through a liquid with a
speed of 1500 mls. The specific gravity of the liquid is 1.5. De
termine the bulk modulus for this fluid.
where 0 "" $G 0 and SG :=/. S, H�o
Thlls,
[", = cz-p :: CZ 5 G P)lz.q
::(Isoo!ft(I.S)(nq�)
== 3. 37x /0
9
'59'111 ,;s2. m
Of'
9 N
E/V' � 3.37 x IO fiii
/ - 65

1. 7 6 J
1. 17 J
1.16 Estimate the increase in pressure (in psi)
required to decrease a unit volume of mercury
by 0.1%.
E
v
= - dp
4�/f,J
ThH5J
( Ft. 1, /;). )
tJ -p
..,.
- £y LH' _
- (If:/If XJO' i��)f- tJ, �()/)
'""
¥
or
.t1 f
�
-
3 .
tf.. I 'f X /0 fSL
1 .17 A I-m' volume of water is contained in
a rigid container. Estimate the change in the vol
ume of the water when a piston applies a pressure
of 35 MPa.
E." =
-
Li¥ "..- -
4f
fit = -
( J 1))1 3 )(35)(ID'::'� )�
£V .2. /5x. IlJ 9�
...... �
decreAse In 1-''' It/me ::t O. 0/63 /11'1 3
/ - 6 6
= - tJ. 0/63 n>1
3

1 .7 8 I
J . 78 Determine the speed of sound at 20 ·C
in (a) air, (b) helium. and (c) natural gas. Express
your answer in m/s.
c = VI€ICT
Wdh T = ,j/)'C + .l. 73 = .;2.93 k :
(Er,. /. �() )
(Po) �y "It c =V(i.IfO)(.2.n. . 1 J )6f�k) ::
kJ'1<.)
1 - 67

,
1.7q 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.
-P�'- :: 1:£
/'1.' I'f.
ThU$)
-P - � ? ,
f -
�' (.
:5l'n c.e
m4 S S
f :: I/t)/ume..
-therefore
)
(.Jj,ere l ' I'V t'ni1-,;'/ st;c d.Co (i""I
f "" f;'nA/ .:Ji�ie..
1;:. = (3J{(o15of 1'1, 7) pH' c46s)j=: I I cr 1>5(.' (ds )
� (p<.je ) = (i/ '1- /Lf. ��i .:: / 0 if 1'S� (3Ilje)
/ - 6 8

/. 80 I
1.80 Repeat Problem 1.7'! ifthe compression
process takes place without friction and without
heat transfer (isentro�c process).
ThusI
tJ;'e� i ....... /"j-t/p/ sill.te
.J 'V /-/n41 :,-ta;ce.
1.-
� =(�) �.
/m 4 S S
t = v(J/ume )
/ - 6q
so 7J,ll.i
(:/IU/

/. 8 1 I
1. BI Carbon dioxide at 30 'C and 300 kPa absolute pressure
expands isothermally to an absolute pressure of 165 kPa. Deter·
mine the final density of the gas.
R T·L
uJhere " '",, ; 'rlliJa/ sIttIe tlh H
-!- 'V -fInIf / s·h.te .
/ - 70

I J
�
/. 8 2 I
For
1 .9Z- Natural gas at 70 OF and standard atmospheric pressure
of 14.7 psi Cabs) is compressed isentropically to a new absolute
pressure of 70 psi. Detelmine the final density and temperature
of the gas.
/sen tropIC. ci!/m;;ress/i!/11 ,
-1> =:
Cons t:Ql1t
{'-Ie
-P, . ,pF t' rv In;hAJ .rhte
� �
where tind
p.� 1, * f. � hr1tJ/ s k-ce .L .f-
So -inA}
-41050 i
!} _
=
C(nc/
1-
-
--e> r
7f :
-fr -
frR
7 (' 5 �R
( .it )(Ilflf i!2.t )7C> 111. ). +-t:...
('1.25" )f./D-B S/lIfJ )(3.offx 10 $ -1& 11.
)h3 sJuf . D�
7(P 5 oR - /.jjpCJ = 305 of
} - 1 1

1.8 3 I
1.B3 Compare the isentropic bulk modulus of
air at 101 kPa (abs) with that ofwater at the same
pressure.
�r ti lt (E''g. /./7 ))
Fy � -k f = (I.Jfo ) (10/ x /,:?�t. ) = 1. 11/ x /� 5"�
I='bl- Ultl-ier ( TaI.J� I.J, )
13,., = ,1. /5" ,>( 10 1 ,q
Thus)
Ev (Wit--ter )
£", (cur )
,;(. /5" ;< /0 'iA.
I. If/ ,I( 1�5"1}
/ -72.
- 1. 5";l ,x/o i'

4/. 8if I " 1.elf D�-;e�Pacomputer 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 1.7'1.
? = (!tMs-toni.
fA
wheye -h= / trw isoth�N11Q/ PI''' cess ) and k = .:Jj'ecific. heLl 1-"4. 1:10
lOr 1.st'nfrtJp'c proc�ss. THus J
-I>�. = 1}.
;:..,. ;;..p,
where £ '" In/flll/ �kc.t�.1 .f'""' /ihq/ .5hik1 So th,;i;
1; = ft.)"!:..
fhen � = 'It.
fL' fj.
w hel"e �.) � ) tire
Thus) Ir�m 131 /1)
�3 r 1;.ttn ;:
Cal1 be. U./"i ffen a s
if, = (�)-k (�j T tCIh<) - 1;.1:1»1
A .5preadsj,�c (P)({EL ) pro'1rt1'm ';;1' {'a /�,, /c.J-7;';3
-the And / Q d9e.. p rt!SStlYe -to/lo ws .
/- 73
{/ J
{ .3 }

�/. a ll- I
This program calculates the final gage pressure of an ideal gas when the
initialgagepressure in psi, the initial volume, the final volume, the
atmospheric pressure in psia, and the type of process(isothermal or
isentropic) is specified. To use, replace current values and let k - 1 for isothermal
Iprocess or k - specific heat for isentropicprocess.
A 9 C D E F
nitial gage Initial Final Atmospheric Final gage
pressure volume volume pressure pressure
pig(psi) V, Patm(psia) k Prg(psi)
25 1 0.3333 14.7 1 04.4 Row 1 0
i-__-t___
-+-I Formula: v-
((910lC1 0)AE1 O)*(A1 0+D1 0)-D1 0
D 1 I' 11 I ) J 7 q ti l'( IM chaie'" / 11 % <
a. t.. 1'rpm rob �m ,
�.bi",/� b.bI� I
7/�"i1 &t !';'ul J'-j< pnSsrp'e "I /� If. !.f psi
/-7/f

/. 85 I
I.as An imponant 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?
111 (lCh n '" 111 hey
To.'}'Je 6. -3 I n
�"r � 50 ·F
A-ppend/� 8
+t
= I I D b
.s
(goo /Itlph){n8 D tt)(�)
( l o b .f4:-
5
1. 0 (.,

1. 8 6 I
I. e 6 Jet airliners typically fly at altitudes between approx
imately 0 to 40,000 fl. Make use of the data in Appendix C to
show on a graph how the speed of sound varies over this range.
c = VleR.r1 (EZ, j. 2cJ)
Frt?hl Ia6k C. / /� 4pP{';1(i/'( C' a-l- (//1 aJ./-i-J-ur/(! D-t 0 +-t
T= S'/.OO + f60 ::: 5J crf( .5", 1h£--t
C-- "ff, 0 YSf'1,z ::' /11t. .f+
s
elf/c.,,/ait,plfs C411 /;e Millie
reslI/f;ft, 1t-a,Ph is �hpwlI
Altitude, It Temp.,· F Temp.," R c, fUs
0 59 519 1 1 16
5000 41 .17 501 .17 1 097
1 0000 23.36 483.36 1 077
1 5000 5.55 465.55 1 057
20000 -12.26 447.74 1 037
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--,---r-,.----r---,---,----;---,
1100 ""
�1080 i---+�-�r_.......;:---I--+--+--+-+---I
'g1060 1---+---ii-""�k--t--+---+-+-_I
�1040 1--+--+-+�--.:"'!:_-__+-_+-__I-__I
�1020 -1--t---t--+--1�f-�+_-+_-+_-_I
[1000 -r--+--+-+-+--+"'��___I-__ICI)
980 -r--t-;-. .-t---j---+--+--+�--=-:'-..d---J
i'---960 -!-,-�----'--�-L---+_-...L--'---I
o 5000 10000' 15000 20000 25000 30000 35000 40000
Altitude, ft
/- 7&

/ . 87 I
10 8 8 I
1.87 (See Fluids in the News article titled "This water jet is a
hlast." Section 1.7.1) By what percent is the volume of water de
creased if its pressure
i
is increased to an equivalent to 3000 at
mospheres (44.100 psi)?
lJ¥- _
- -
-V-
= -
(�'b. ). ;1. )
4 Lf, l oa ��l·a.. - 1 4.1 p�L"--
= _ O . 1 + I
3. 1 � x: 1 0 '5" 1m: 0..
1.88 During a mountain climbing trip it is observed that the wa
ter used to cook a meal boils at 90 'C rather than the standard 1 00
'C at sea level. At what altitude are the climbers preparing their
meal? (See Tables B.2 and C.2 for data needed to solve this prob
lem.)
When fhe wafer boils,
Iioil = fJtV J
where from TaMe B.2j al 1:: f/ofJc
fAr :: 7. 0 I x /0'1 1-2. (abs)
Also, from 14 b/e C. k, for a sfandarcA afmo.sphere
fJ :: 7.o/'Xlo'f�(abs) af an qlfiluole o f � ooom
/- 77

1. 8 Q I
1 . Sq 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.
Ca pihl-/on mdlj pcolr whfn Ihe /�C(d pr{'ssure e(gtlll/s the
va'ptPi'" jJrt'ssure . ;:;"r wR iel- a i- 1(P() �f' (#�"" Tt.6/i: B, ! ,�lttfJfJl(/,itB)
/. 90 I
ThusI
-?. = 'I. 7Lf PSI.: (1/1,5 )
r
J .qO Estimate the minimum absolute pressure (in pascals)
that can be developed at the inlet of a pump to avoid cavilation
if the fluid is carbon tetrachloride at 20°C.
Cavi .fa 1-;()11 rna,! N'CtH when !he sue-lion pf"f'55Uf/e
(i t- -the ptlmj> in )� t E''$ua/s the va-I'0'" pressure .
/;;- car};on td....ac.�/oy;d� ai :LIfe � :: I] �R.. (,ds )
1hu�) 13 Ie PI<. (4j,S )
/- 7 8

/ , qI I
1.91 When water at 70 °C flows through a converging section of
pipe, the pressure decreases in the direction of flow. Estimate the
minimum absolute pressure that can develop without causing cav
itation. Express your answer in both BO and SI units.
("tll/ifillJPII nut,! OCC(lr 'n t/f� � vt'r91/1� sec.-tlDI1 b/" p'l'e WhM
t"ht:. ?r�sS"l'e eZlLols -th� va�? pYeJ�"re . t"YtJt?1 746h B. l In Itff''''''".r Ii
./r;y- wd�r a.t 70 ·(./ 1;, : 31. 2. -h Pa.. ttl"" ) , Thu�
86 Hn if-...s
'pr(S�Jtre ::- (31.Zx Jb3:1)(/.1f5'bx/(1-!:t.'),..... �
I?1lnlmum
- Ll. S2 fda.
1.q 2. At what atmospheric pressure will water
boil at 35 °C? Expressyour answer in both 51 and
BG units.
The VIJ..j>0Y fY�s. s,.trc o/- wilier 0 6 3S'C 1.5
S. gl ....t� (;,5$) (-/I'4n7 To!'//! 8. 2 ,,, /ip?"d"� j3
uSI"7 1/",,"; l;,krfo/lJi;(f)� J. Titus, ;,c &V�ie,. bo,"/s
at 1l1is femp�r� dl(re f11e 4tm".spHPrtC p,.'.I5ure. !nus!:
.he eftlDI i� S. i/ .lJ.f'a.. (�/,s ) I;' :5I t(l'1ih. In f.J6 l/'1i fs)
(S.J'/.xI03::"Z.)(/''15"PxJt)-'f-:� ) = O. !If.;L p�i (o"s)
7:�
/- 79

/ , q'f I
1.94 When a 2-mm-diameter tube is inserted into a liquid in an
open tank. the liquid is observed to rise 10 mm above the free sur
face of the liquid. the contact angle between the liquid and the tube
is zero. and the specific weight of the liquid is 1 .2 X 10' N/m'.
Determine the value of the surface tension for this liquid.
ThvsJ
2 rJ c os e
O' R
- 0.0 6 0 %
J
where e � 0
'f N ( -3 ) -3
1. 2 XIO ;;j3 1 0 '1- 1 0 m (2X/V m12 )
2- eN 0

/. qS I
Lq 5 Small droplets ofcarbon tetrachloride at
68 OF are formed with a spray nozzle. If the av
erage diameter of the droplets is 200 I,m what is
t,he difference in pressure between the inside and
outside of the droplets?
5ince
p=
- :z.
rr = .J. b '? A IO it.....
==
j- 8 1
(£t 1. :2.1 )

I. q 6 I
1 .9 6 A l2-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.
'For e�lIi//6riUII1 fsf'� !/jtlt"'e ),;
t(z�U):: rr(z tl.)
SD 1/Ia t
1> ==
=
12 '" Jb- .3 Mt
-z
= 12. 2 Pa.

/· 9 7 I
1.97 As shown in Video VI.9, surface tension forcescan be strong
enough to allow a double-edge steel razor blade to "float" on wa-
ter, but a single-edge blade will sink. Assume that the surface ten
sion forces act at an angle 8 relative to the water surface as shown
in Fig. P1.97. (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. De
termine the value of 8 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. Sup
port your answer with the necessary calculations.
I
(a. ) L F =- 0
V€r+1 Ca I ,
tow
�
• •
'1;J = T S in B
lu heve ttJ ::: (YyI x
� cHI d T = CT x. iel'lfhl of. 5 1 cie5_
blode.
(o. lP� )( ID-3--k� ) (�. � I fYYI/sl.): 0.n I.)D-2..!j;,) ({J. 20� !)n) 5'1 ,, 8
:5 111 &- =- D . Lf-15
[;) = :2.. 4-. 5 °
( b) For slnrle-ed<Je blade
'2U = /fr1Mad� X /} :: (2. /,,! )(IO-3 -k�) (1.�J !)n/�l.)
::: () . O Z 5'� tV
Ci YJ d
,. :s i f] � ::: (cr;;<. len,57n of. blade) :5ln tJ
:- (7.3J.1. X /o-z Nb) ( O./5lf/Yrl) '5 /11 fJ
� 0 . 0 1 1 3 5/� e
! /1 or-ael" +Or- h l4 de +0 " J.loa.i ' .::zu -< T Sln B .
; >J n a lY'r'I a X: / m tarn Va. l u e +o y- � l n e I S I ,'+ followsI
fi1cd CZU :> T SIr, e Cl n d �jn:J(� -ec15e hlade w ; I I sill k .

I. 98 I
( � )
1 . q S To measure the water depth in a large open tank with
opaque walls, an open vertical glass tube is anached 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 Eq. 1 .22 (with
e = 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 errono be less than 1 0/0, what is the smallest
tube diameter allowed?
Th� eX CeSS h�ljh t I
h) caused 6(
h = 2 cr� G
'oR
1J,e, surfr" c. I:-enf,/�n /s
(EZ . j, '2z. )
f;?r G-::: 0" tv; th b = 2. R.
h = if 0-
rD
Prt9r"n TAJ,/, /3, ( / 11 A-ppendl X 8 /Dr tVo.1:ev a.,i.
fT= Jf. 9/ xIf)- 3 fb/Pi (;111&/ r = ' Z, 2. z lob/ft:".
Th�s .fnm Pq. ( J )I P -J
I
I. )
h (F-1:) :: If (if.fit x: 10 -:;:L
(�z lZ. l.!!. ) D (,'rI.)
. k ' ' 2. I�.f.f.c
- 3
::l.l q )( 1 0
.D ( (n.)
0 )
%O"F
010 f'rror =- h t+t J )( 1 0 0
3 ft
(toI 7lt -The frue. c.1eptn
= 3 ft-)
('r toJJ0 Ws frc>WI E� l �) -tJ.../lf:
_ 3
�/o e y rt'Y =
3, 7tf X If>
:;( 10 0
3 D (ln. )
A- plot of io tvr,pr vusus t:Wbe dlt1tt1e.fer IS
S/1()W/1 "n 17" I1ft-t pa 'l e .
/ - 8 1f
c � )

I. 9 8 I (Ci:JI'/t )
Diameter % Error
of tube, in.
0.1 1 .26
0 . 1 5 0.84
0.2 0.63
0.3 0.42
0.4 0.32
0.5 0.25
0.6 0.21
0.7 0. 1 8
0.8 0. 1 6
0.9 0.14
1 0 . 1 3
"
Values obtained
from Eq. (3)
(b) For Jio
1
1 .50
�
1 .000

�
�
w
�0
0.50
�
0.00
0 0.2 0.4 0.6 0.8 1 1 .2
Tube diameter, in.
j
eYYtJY ;;'PI11 £g . O)
) =
6. /2b
tx/".J
D -= O. / 2 "" in .
/ - 85

/. (/q I
1.q q Under the right conditions. it is possible. due to surface
tension. to have metal objects float on water. (See " ide" Vl .q.)
Consider placinr, a short length of a small diameter steel (sp.
wI. = 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.
NOle: 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?
In "cIty Ivr yo� .J-.o -float (s�e (',?ille)
I i .ft>/Iows Pta t
1. 0-1 ::: CW .:: r-�)(n4)� �tt'eI
Thus ) #'r 1J1( /,m;hnJ ca.se
:l :z.. rr.l-
DmRI. -
[) mao(,
.
0 . 0 "" '+ I n .
g cr
rrl 'V1.
-3 u..
5 l l x l b t'l•
St'nct. a. �t.tlf1t1lfYd �ieel p"-pty dip hils 4
dl4mdrr D f tJ. � 3/' in. ) wh,'c), I So /(55 fr..a.J1
O. D'-/f/. ;'n.) It 5nol{M f/o tL /;- . ,4 �/mp/( e;t..painJflti
f.,t) t // V-ev;.f� Tht.!. . Yes .
/- 86

/. I D O I
I. 101 I
1. 1 0 0 An open, clean glass tube. having a diameter of 3 mm,
is insened venically into a dish of mercury at 20 'C. How far
will the column of mercury in the tube be depressed?
� =
'2 cr (!i95 f)
?r R
WI // be del'resseel
1. 101 An open, clean glass tube (8 =
0') is inserted veltically
into a pan of water. What tube diameter is needed if the water
level in the tube is to rise one tube diameter (due to surface
tension)?
For
---h =
-It = ZR.
J.. R :
'2 <T CDS e
� R
an d B = 0 °
Z r:r ( , )
'(fR.
_ 3 I j,
57 /)3 x JD :r:e-
I 1.1..
f<> :t. Lf
.tt J
- 3
R :: �. 9'� X ) D +1:.
d Ia rn e be j.- - .2.. R ::- l. g o x ) 0- 2 II:-
(�g. I. 22 )
- 3
3. 00 x /D tm
3. 0 0 ,.,.,., m?

1 . 102.. I
1 ./01. Determine the height water at 60 'F will
rise due to capillary action in a clean. !·in.-di·
ameter tube. What will be the height if the di
ameter is reduced to 0.01 in.o
2 rT cos G
'!' R
TJ,,,� with (9 : 0 J
/
S/rn//tl rl'J
)
( lor R =- 19. W!i" In. )
I ' )(C>,J:Z.5"/�. )r tJ. /Kb lh. . =
o, oO S' In.

1. /03
1.1113 (See Fluids in the News article titled "Walking on water,"
Section l.9.) (a) The water strider bug shown in Fig. Pl.l03 is
supported on the surface of a pond by surface tension acting along
the interface between the water and the bug's legs. Determine the
minimum length of this interface needed to support the bug. As
sume the bug weighs 10-' N and the surface tension force acts
vertically upwards. (b) Repeat part (a) if surface tension were to
support a person weighing 750 N.
(0..)
-e�III L bl'l;"1r J
f7UJ :: rrj.
C1lv
- 3
1 . 3 � )( 1D fWI
11 F I G U A IE P1 .103
4.v r"V W-e'18 ht-
a- rv SII....tac� .J..r.,SIOI1
f I'J It'nj"tn 61 "f)m.�c!...
(,=1I. )(IO-3tyl) (ID3�) ;;
(.1) If-
. 0 2 )( I D f{YI

/. /OIJ
1.lO't 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. PJ.lO'lj; 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, I, 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 = Nil.
Graph: For each fluid lested, 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. 1,7).
Data: To proceed, print this page for reference when you work the problem and c/i" k I,,'rl'
to bring up an EXCEL page with the data for this problem.
,·�4f---Re,tating Inner cylinder
.�·:I'I--Ol.ter cylinder
Drive weIght
• F I G U R E P l .IOlf

/. / OIf I
Problem 1 . 104-
Weight, W, vs Angular Velocity, (0
for
Silicone Oil
4.50 ,----,------,---..,-----,
4.00 +----+---+----l.-----I
/3.50 +---+---1--/
3.00
Vz 2.50 +----f---�'-----+--___1
;: 2.00
/ W=2.5 '"
./1 .50 +---��-_l---+_---I
1 .00 ./
0.50
I
......
0.00 �---+---_I---+----I
Problem 1 . JOy.
Weight, W, vs Angular Velocity, <0
for
Corn Syrup
4.50 ,----r----,c---�
4.00 +----+---+v= -3.50 -
I3.00
t --
z 2.50
./
-
;: 2.00
7
-I W = 1 2 S ",
1.50
i
j1 .00 ../
0.50 +--.<'--+----+
-
0.00 -I-----I----I---�----1
0.00 0.50 1 .00 1.50 2.00 0.00 0.10 0.20 0 30 OAO
<0, revls
1 .20
1 .00
0.80
z
�.
0.60
0.40
0.20
0.00
0.00
Problem 1.10'#0
Weight, W, vs Angular Velocity, (j)
for
Latex Paint
I
,
W = 1 .466
(flO 707
0.20 0.40 0.60
(fl rev/s
00, revls
I
0.80
I
I

I. 1 0 5
1. 105 Capillary Thbe Viscometer
Objective: The f10wrate 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. PI.IOS, the kine
matic viscosity, v, is inversely proportional to the f1owrate, 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 15.6'C and determine the
f10wrate 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
f10wrate 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: Foreach temperature tested, determine the f1owrate. Q = Vlt. Use the data
for the 1 5.6'C water to determine the calibration constant, K, for this device. That is, K = vQ,
where the kinematic viscosity for 1 5.6'C water is given in Table 1 .5 and Q is the measured
f10wrate 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 h" !'f'
to bring up an EXCEL page with the data for this problem.
Water
Capillary tube
Q
Graduated cylinder
• F I G U R E P 1 . 1 05

I. IQ5 I ( Cu,'t )
Solution for Problem 1.105 Capillary Tube Viscometer
V, ml t, s T, deg C Q, mils v, mA2/s
9.2 19.8 15.6 0465 1.12E-06
9.7 15.8 26.3 0.614 849E-07
9.2 16.8 21.3 0.548 9.51E-07
9.1 21.3 12.3 0427 1.22E-06
9.2 13.1 34.3 0.702 742E-07
94 10.1 504 0.931 5.60E-07
9.1 8.9 58.1 1 .022 5.10E-07
v = KlQ K, mA2 mllsA2 v (at 15.6 deg C), mA2/s
5.21E-07 1.12E-06
K = v Q = 1.12E-6 mA2/s ' 0.465 mils = 5.21E-7 mA2 mllsA2
..!!!
N
<
E
Problem 1.105
Viscosity, v, vs Temperature, T
1.5E-06 ,----,.--------,-----,------,
1 .0E-06 -1------.:'k----+------1--.-
��5.0E-07 -I----+----+---=::,�I-----/
O.OE+OO -I-----r----+-'-----.-----l
o 20 40
T, deg C
60 80
From Table B.2
T, deg C v, mA2/s
10 1.31E-06
20 1.00E-06
30 8.01 E-07
40 6.58E-07
50 5.53E-07
60 4.75E-07
• Experimental .
I-- From Table B.21

2.,2 I
2.2 A closed, 5·m·tall tank is filled with water to a depth of 4 m.
The top portion of the tank is fllled with air which, as indicated by
a pressure gage at the top of the tank, is at a pressure of 20 lcPa.
Determine the pressure that the water exerts on the bouom ofthe tank.
3";
7..0X/0 -;.
m
:J/V. 3N
== :z.OX/O � + tf. 80x/O n? (ifm)
= .5'9,:;' X /0.1!!..... -::: 5P,2 kPa
m
(f)
j ..:-., ,':<:-' tfJrfI:h' • • • ' t
' . . . . .
.
.
� .
. : .(2) ,

2.3 I
2.3 A closed lank is partially filled with glycerin. If the air
pressure in the tank is 6 Ib/in.2 and the depth of glycerin is 10
ft. what is the pressure in Ib/ft2 at the bottom of the tank?
= in!. 1173 Yo ft) + (t:, /1,... )(I'f'l /n....).f1:
;11. .ft "-
= ItOS-O
/"
I..e..
2...tt I
(a)
(/,)
2.1/0 Blood pressure is usually given as a ratio of the
maximum pressure (systolic pressure) to the minimum
pressure (diastolic pressure). As shown in Video V2..2 such
pressures arc commonly measured with a mercury mano
meter. A typical value for this ratio for a human would be
120170. where the pressures arc in mm Hg. (a) What would
these pressures be in pascals? (IJ) If your car tire was
innatcd to 120 mm Hg. would it be sumcient for normal
driving?
f::: 7th
For /2.0 t?n-m
}...J..J
;:::Or 70 an /1'11 +Id-
..
o
0
p= (/:3.1 X /Oi� )((), !:t0 1trI)= /�oO-kPa..
'1:J = (;33 X /03£)(0. 070/)71)= r.3/ -iT?.
/ . 11'113
FOl- 120 mt-m H:; tJ :: (it.. 0 X 1�3..it)f;'1S'{)X//)
-
'f
1b/i11�)I /1n'2. il IV/hit1.
5/�ce
IS n"i:
� :2" g2.. 1:Si.
-t'jPICa I ..hr� pVRSStlye IJ
:5U P-hc.J erJ t lOr /Jt:Jrputi
3tJ-gs -p5"� J 20"""".fIJ
dl'lv,,1j.

2·5 I
2.5 An unknown immiscible liquid seeps into the bonom of an
open oil tank. Some measurements indicate that the depth of
the unknown liquid is 1.5 m and the depth of the oil (specific
weight = 8.5 kN/m'l floating on top is 5.0 m. A pressure gage
connected to the bottom of the tank reads 65 !cPa. What is the
specific gravity of the unknown liquid?
==
(�j I)(5"rm) ,. (�(,.)(ISffl �
-1={..lf."" - (ls'oi,)(S-tm)
15 tVY
15" }(ltl!!..
1111"
56 =
ill. =
'). .6 I
2.11 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/m'?
Express your answer in pascals and psi.
P=-J'J. +�
.4f -the Sur/"ce. to =0 �t> tn�'
1= (;lJ./j(ID"3:!;J{S'XI03hr» =50.5
,4./$0I
t=(50.5
2-3
= 50.5" 1'1 p...

;J.. 7 I
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
(a..)
Thus)
� '" -6' =-f3
'!.:1: ::
- i dr
f'
1ft's a. fttnd'/lP/1 0 f p) W(,
111 fe'lI't1f-ln? lit (I ). :s Inee)
p-
!if.
df/f (J
=
fv /7
- E 1;,v
t:
of part (a) to determine the pressure at a depth
of 6 km assuming seawater has a bulk modulus
of 2.3 x 10' Pa, and a density of 1030 kg/m' at
the surface. Compare this result with that ob
tained by assuming a constant density of 1030
kg/m'.
( £g. Vf)
(r)
(Eg. 1./3)
a.t p=o
t.Jhere

2.7 I (can'l)
(c) h:,,, UJI?S /:anI: densi�
P=?fJ.. =;;3- J.. =(t.fJ3�JD3�)(?,f/;..)({,�/1>3rm)
=
fa tJ. t. f.1R..

2.8 I
2.8 Sometimes when riding an elevator or driving up or down a
hilly road a person's ears "pop" as the pressure difference between
the inside and outside of the ear is equalized. Determine the
pressure difference (in psi) associated with this phenomenon if it
occurs during alSO ft elevation change.
bf ::: tDh ::: 0.0765 }�J (IS�FIJ
Ib I H _)
'"' II,,05
1P' (I'H in.'-
� 0.0797 pSI
.
2.q 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 = _�
dr
Let -p..::];o - Z
So -mat d-P. : -eli:
and
1 ::
v
l"
II 7 / � I

"'2.10 J
;j<2. JO In a certain liquid at rest, measurements of the spe
cific weight at various depths show the following variation:
h (ft) 'Y (Ib/rt')
0 70
(CDI'! t. )
60
70
80
90
100
107
110
112
114
115
10
20
30
40
50
76
84
91
97
102
-
The depth h = 0 cor:responds to a free surface at atmo
spheric pressure. Determine, through numerical 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).
Le.J:: -i� � -� (.rt"� .;; (jurt! J s ..
<>
1J1�i cJi.=-dh lind tht'y� Ioy�
r
- - =-
{
dp == -}- ell =:: � dh
Thus,
[Ii-
J,..
f�'dhdp ==
() 0
t
t
I r f / I (' / / / //
01"
[I)
where '/;.. IS --!he ?l'eSSL(Ye a.:1: depth -A,:.
E'!U/l,b/on (J) CIIII � I� -ferntl-t>a' n tim en c q Jf'J lASI"J
--the {rapt" ,?o,'da / ru Je J
(. : e.)
"" - I
I .: { L(':1,- + ':1t:.f'/)(�' - >t. )£=-1 £+/ L
/JheY'e !J � d' ) >< IV h) tlntl n == hum brY" �.f. c1a-+1{
po/n7-J
(CtJl'/t)
2-7
sz:
-
--

ft2./o I
7he -1:0J,U)4.f.e,( ri'suJf� a i"'C 'lIven be/�� a 11M �
WI'!'h "/he e�rri'.s.l'M(ItA� pl(Ji "I prr!'ssUY.e vS. depth.
heft) Y. Ib/ftA3 Pressure.psf
0 70 0
10 76 730
20 84 1530
30 91 2405
40 97 3345
50 102 4340
60 107 5385
70 110 6470
80 112 7580
90 114 8710
100 115 9855
12000
10000
/c;::-
8000In
Q.
/Q.
.; 6000�
V:J
/
..
..
..
4000�
D..
V
/'2000
/
v-
0
0 20 40 60 80 100
Depth. h (ft)
:z-z

11>2.12 I
*2.12. Under nonnal conditions the temperature of the
atmosphere decreases with increasing elevation. In some
situations
:
however, a temperature inversion may exist so
that the aIr temperature increases with elevation. A series
of temperature probes on a mountain give the elevation
-temperature data shown in Table P2.12. 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
lhe mountain.
Elevation (ft)
5000
5500
6000
6400
7100
7400
8200
8600
9200
9900
TABLE P2.12.
I-vtPm I"r· ).q)
i
l�
1M
-f.%, � d�
::' -
-
-
PI R -2- T
I
In "the. 1-"hie 6c/�w -/;,« t.<mp.tYA iu>'< In OJ'( IS
.
I/r("�) t4bl(/4 J..,.,/.
and -n,� '" J, ,yo" If d
-
Elevation, ft T, of T, oR 11 T(oR)
5000 50.1 509.8 0.001962
5500 55.2 514.9 0.001942
6000 60.3 520.0 0.001923
6400 62.6 522.3 0.001915
7100 67.0 526.7 0.001899
7400 68.4 528.1 0.001894
8200 70.0 529.7 0.001888
8600 69.5 529.2 0.00189
9200 68.0 527.7 0.001895
9900 67.1 526.8 0.001898
Temperature (OF)
,.11'11
50.1 (base)
55.2
60.3
62.6
67.0
68.4
70.0
69.5
68.0
67.1 (lOp)

.rI- ;;//e>I<JS ;;'''111 £"� , (J) kJ J 'tH 'A = 12, J P':JL'a. "/J14f
- 0./753,
� = (IZ, / ,s/a) e = 10,2 pSi'a.
(!Vote,' Slna 1h� -k",ptnl-k� fltll',,;hl!iH IS not 1If'''� j,mle
If w"v!t:l br -eXfeC.ffd 7hflc 111(! Ifss/.{/1Ipho)f of If t4JI1skl//
kmten!f"r� k/IPII/4 f/l.l(! fP$tI I'e.slllf.s, I� "the fflltpti'4f11J"C
I� 4S511/11�d .;., b< &Plfshl;t.i "i- the bllS€ ff>111'ppYl!d::ilye
(So.! of); �: 10, / ps/4.. / whiCh /s ()11/1 sl''lhfl;;
dllla�,d: irlP/?? 1Jte l'es{'(li '1'(;01 I!IbollfC , )

2.14 (See Fluids in the News article titled "Giraffe's blood pres
sure," Section 2.3.1.) (a) Detennine the change in hydrostatic pres
sure in a giraffe's head as it lowers its head from eating leaves 6 m
above the ground to getting a drink of water at ground level as
shown in Fig. P2.14. Assume the specific gravity of blood is
SG = I. (b) Compare the pressure change calculated in part (a) to
the normal 120 mm of mercury pressure in a human's hean.
(().J h,y h'1dn�s�lt(;. frezuy( c-traYl�f!....1
4 p == 'If-J,. = (Cf:K() �)(l>1m) = S9. g::'--= S� fl-kFA.
£b) To etJIml(J"� JCl/� pY�JSUYc. 1, hl/rfJlYt heart.
Co/1 Vt:yt;- Pr'�S.sIlI'<' 'n Pity/:- (4) 10 ttn-1rt 1+,5
53. .8 -leN = 'OIJ. -fz. = (�3�')){H/l'n .... 'J- J.l.J- rm 3--
--h. = (0.4 if l /)T)(u/�) = Lf'f2 MlMC Itj
"I
�
Thus '11rc. pYeJ.Sur'c. �ha"1e
I
I� I.fy.z /1H"" Jlj e�mplt'e;{ I ..
-ht� huma'l heayt-.

2.15 I
2.15 Assume that a person skiing high in the mountains at an
altitude of 15,000 ft takes in the same volume of air with each
breath as she does while walking at sea level. Determine the ratio
of the mass of oxygen inhaled for each breath at this high altitude
compared to that at sea level.
Lef ( )0 denote seQ level and ( ),.5 denofe /5,000 It ()/fiftJ4e.
Thvsj since m ,;: mC4.f.S -::: (? VJ where 'V:: VfJjV/IIB,
m9 :: eo Va t/lnd �s =: P/� t{sJ where 11; � t{s.
Hence,
m,s (J,sUs = �s
nf; == po 1.10 fa
If if is 4.f.Sv/)fU4 fhqf ffoe ail' c()m,o.r/l'(lfJ (el') 0;; fJ/
tAil' fpPif I� QXY9frJ) is fhe- .f4me at sep,/evel as' /1 /.r4f/�O()df�
fhen we cafl we fhe r V�/I/es f!'o/i1 Ttlb/e c, /:
eo::: 2..377X/O-3 .r:�.s. anrA Bs -;/.Jf?lx/o-.l �:�'f So thaI
- .::: 0.6').. 9 --; 62.910
2-/2

2.16 I
(a)
cc. )
2.J 6 Pikes Peak near Denver, Colorado has
an elevation of 14,110 ft. (a) Determine the pres
sure at this elevation, based on Eg. 2.12. (b) If
the air is assumed to have a constant specific
weight of 0.07647Ib/ft', what would the pressure
be at this altitude? (c) If the air 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).
-fhtn
t=
(/711. H:, ·/6 )�dP3J;7£)s ''t� /o. TL
I Z If D
II:,
-Ft.'-
P -= f:.. - '(r-lr.
Cabs)
(Er 2. JZ)
q: 3z.I7'f ft:
iT 5-" )
� J 1/6,. z �'- - ((). 671..'1-7 ;;J{t� 110 I-d
/ () I.f 0 l:k.. (o.f:, s )
fe-
1270 Jk ( q;l:"s )
f/;'-

2.17
I I 2.'7 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
Le t
T/lI-1S)
��bf.
le c�in Appendix C forr an;�evation of 5
d f g. dl:- - - - -
? R T
�, t-,
t�t Io�
,p
f: =- -
�
.z. =. 0
I ) � -v p
.1
dl:
I
t
(?. Ta.. 7 z
<l
j.".... i2:. � C('" d T= Ta.-[Sf.J
1:
0>1'"
1:.
" [-fo 1M (Y·-r'·1 � [� (Tqn) -� Ttt];... :- ::
�
=- J; h ( 1- @.r)
4114 �kil11
R� T...
/o5ariinm 0/ b"th SlcJeJ of.
....1
eg u. 12.:1::,()'H ':! H.ld.s
.p:= � (I - �)�
1-" 5"'�1'm ()) ' fh Ii =101. 33-4?a.
().t)C)��O &. R,= �K7...:!..M1 )
.#.#./<.
)
(£Z.:l·IZ)
Ta. :: :In. lG'k r= 'un �, / (J S )

2./6 I
2./8 As shown in Fig. 2.6 for the U.S. stan
dard atmosphere, the troposphere extends to an
altitude of 1 1 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 mis'
in your calculations. Compare your results with
those given in Table C.2 in Appendix C.
FlPr I��ih-trma/ C4>n(hft�H5)
Lei:.
one!
-1 ('i!-z--�')
1:: :: 1'. e
R To
2. I
p, = 22, ,�p,," )
-r;:: - 5 t.. 5·C T ;J. 73.6 =: 2/1"," !) k .
I;).. J -k Po-
_ �77 7...)(/.f')i10'''''- II )lIb
3"", )J
l (:l??.if/( ) (:1I�. /'5'/<')
M.(
'1.77 Si )
41�" ) :3 N
O./9b -k91> 1;2./x.ltJ ,.,.. ....
tf=
= -
RT
(t'r�m Table
t1. ::
O. /9t.fg
(.2 /11
h . )1m3
(J-8'7 ./ )(:2/�. /'f7k) 11113
'1.1<
4.ppef1d/x. C
) �
:: I.?. II ! Pa.. Clnd

2..1 If I
2.19 (See Huids in the News article tilled "Weather, barometers,
and bars," Section 2.5.) The record low sea-level barometric pres
sure ever recorded is 25.8 in. of mercury. At what altitude in the
standard atmosphere is the pressure equal to this value?
2-16

2.2.0 I
2.241 On a given day, a barometer at the base of the Washington
Monument reads 29.97 in. of mercury. What would the barometer
reading be when you carry it up to the observation deck 500 ft
above the base of the monument?
Lef ( )b and ( �d corre�pofJd. fo fhe b4se and ohservalion
deckJ respectively.
-rhl)s) wifh H:: hei9hf of fhe m onvmenI)
£)b - i7d - 0... H - l.tsx/o-11 (soON) == 38,S !E.r' 1'0 - air - . fI If'-
But
n - cr: h where d':lIa == 81f7 -J-fb3 and h :: haromeier reqd'1J9.7r - II!] ) ./ r
Thvs)
� (2.9.91 ff) -I. h �38 slk., 11.. 119 or/. • ff2
or
hod ==(��:7ff) -
or
hJ = 29.'13 /n.
2-/7
==[(::PfI) -o.olfs'Sflk,ZfH
= (1-9.97 -O.�'fS)"fI.

2.1../
2..21 Bourdon gages (see Video V2.3 and Fig. 2.13) are
commonly used to measure pressure. When such a gage is
attached to the closed water tank of Fig. P2.lo' the gage reads
5 psi. What is the absolute air pressure in the tank? Assume
standard atmospheric pressure of 14.7 psi.
f=rh + Po
i?,�- M H)�" = fui-
�� = (5/�7. -t 1<1.7 ::1)-
(j .pt)(�1.if �3)
/'fif In. 2-
.ftl-
Air
T12 in.
Bourdon gage
6 in.
• FI G U R E P2.'-1

2.2.2 I
2.'l.J.f I
2.22 Onthe suction side ofa pump a Bourdon
pressure gage reads 4O-kPa vacuum. What is the
corresponding absolute pressure if the local at
mospheric pressure is 100 kPa (abs)?
= =
2.24 A water-filled U-tube manometer is used to measure the pressure
inside a tank that contains air. The water level in the U-tube on the side
that connects to the tank is 5 ft above the base of the tank. The water
level in the other side of the U-tube (which is open to the almosphere)
is 2 ft above the base. Determine the pressure within the tank.
fai" +0" (.6-m -K, (2(1) =0
11..0 11).0
or
fa,r :: - (3 H) �t.d - -(3ff)(62,1f�)
/h
::: -/871(i
2-/9
�T -i
�H t �. _
�
iff
I
.
L------'�
T ;�

2.25 I
2.�5 A barometric pressure of 29.4 in. Hg
corresponds to what value of atmospheric pres
sure in psia. and in pascals?
2-20

�,�6 I 2.2.6 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?
(]11G /I< tim, varY' fY'�Hilre.)
r(A.6.n) : -y-R. + -Pv-
wher'e ? 'V vAf()r frtssllr'e
3 IV
-I N
(A) For meY'cur':} :
to/ x/D -. - I. �;(IO -..
"" - ",.,
133 x /0 3JL
".,3
(j,) /;Y' /VA.-t.eY' :
ct) For efA,/
a/coh()1 :
:: 0.751 I'rY1
3 tV J
�"
1(1)( ID -1-
-
t. 77i 10 .l!!.
,.,.. "" ,
7. 80 X 10J J!.
",,3
= /O. J fWI
�
3# 3N
JtJ/X/() -;, -s'9x/o -...
"'" ,.,..
;:'
7.7'fXIO'.!:L
/»13
-
l.;l, 3 I'n1-
(Wdhout VfJ.for f¥'wL/re )
f (tt.1:"" ) � d'h
--R.
�
3 N
10/ ;(.ID ;;;1.
1'33)(. 10'3.J!..
"".3
- 6.751 I'l71
3 II
101 '/../D �..=
r.rOXI03.!£.",,3
= /0.3 rm
)I.. IbJ AI
101
::
,.", ...
3 N
7.7't;< 10 -3
11M
-
/3.() m,
Ye s. F;r rnercut'!J btlY'tlrneters -fJ,e effect 01 IIdpn' fWSfllre
/s n f't'jli'qibfe
I
t(nd t:ht! ye$I/lred he/'jhi of- the merCul'1I
colum n /s Y'uIs,pn<?ble,

2..27 I
2.27 A mercury manometer is connected to a large reservoir of
water as shown in Fig. P2.27. Determine the ratio. h.lhm• of the
distances It. and Itm indicated in the figure.
f, ::: �hw + owhm
bur fl � f,; ::. Om (zh6l)
ThIJsJ
Ow hw + Owhn ::; 2 Om hm
or
(tw) hw :::(20'",- 'tw)hm
.s 0 That
. -,' '--
-
'''-:�::�'
:-
'
-
-
'
-,',-
'
-
-
-
'
-,'�
IWater
,---- h",
I t.Cll
. 1fT(I) • (3)
.1
o'm !:m" �
'-- Mercury
III FIG U R E P2.27
hw == (2 «m-tl"w) ::2 SGm -/ J
where SGm ==* :::/3..56
h", 0"111
rhf)SJ
t; =2(/3..5'6)-/ == 26,{
2-22

.t. :2. 8 I
2.2$ A U·tube manometer is connected to a closed tank
containing air and water as shown in Fig. P2.2B. 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.
- Closed valve
';;'
f---Air pressure = 16 pSia
(I)
Gage fluid
(y = 90 Ib/ft3)
Water
• FI G U R E P2.28
T2 ft
�1
Pressure
gage

z. Zq T
�. 2 q A closed cylindrical tank filled with water has a hem
ispherical dome and is connected to an invened piping system
as shown in Fig. P2.2'. The liquid in the lOp pan of the piping
system has a specific gravity of 0.8, and the remaining parts of
the system are filled with water. If the pressure gage reading at
A is 60 kPa, detennine: (a) the pressure in pipe B, and (b) the
pressure head, in millimeters ofmercury, at the top of the dome
(point C).
Hemispherical dome
c
Water
-'(' If E P2..:z.q
(aJ 14 + LS(,){�2-tl) (d'/YY1) -1- �ZO (2/1'>1) =-fB
i4m
3m
2m
VSG=o.a
�Q
waC: �
Fe = c,o;'({ + (o.e)(r.8/.(//:.)(3"",)+ (U�X/D3�:J)(ztM)
:= /03 -kPa.
(b)
� =
-p - � (31M).4 zO
:: �o.kPa. - ('I.�OXID3�) )[1M" )
5�. �
3 N
- X JD -
/)71).
..pc
3 N
-I,.
30.lz.x)! --= -- :: ",, '-
tJ, 230,.,.,.,
d'4
-=
/33)( II) 3 J:L
:J ",,3
= �. Z 30 ".,.,
(/�
3
",,� ) = 230M1M1
.......

2,30 J
2.30 Two pipes are connected by a manometer as shown in Fig.
P2.30. Determine the pressure difference, PA - PB, between the pipes.
Gage�id(SG = 2.6) Water
II FIG U R E P2.30
13 m
� T d;;"o ((.IS"", rO,(,,..) - �f (tJ,I._)..,. d'�o (/.3",,-0'5"... )=.fa
Thus)
Pff--PB = �f (o.{,,..,) -�" (OS''''' 1'0,"_ t-I.]"" -0" ".,)
:= - 3,32 -AI?.
2-25

2.3 /
2•.3/ A U-tube manometer is connected to a
closed tank as shown in Fig. P2.3/. The air pres
sure in the tank is 0_50 psi and the liquid in the
tank is oil (y = 54.0 Iblft'). The pressure at point
A is 2.00 psi. Determine: (a) the depth of oil, Z,
and (b) the differential reading, fr. on the ma
nometer.
(a.)
(b)
Thus
(.;.
)
-PA T (rDii (1.·Ft)
(S6) ( (l'�1D)
Air
FIGURE P2.31
51f. 0 �3
=
=
(3'.05)(�2.1f ��)
2, Og .{-t
Open
= 'f;O()Jt

.2.32
2.32 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 of2.6. What is the pressure in pipe B corresponding to
the di�erential reading shown? .-
3 in.
I
SG = 2.6
FIGURE P2.32
-PI} -t 6'#:;.0 (/�ft) - �f (t ft) si;,Jo· - 6'1Izt) (,: ft) = fB
(whe>'(! 'ON /.s f1f� �,ec/h'c w�lj;, t "i tile JqJe f-Iu'-d)
PI:> :: P,4 - ?r#f. C� ft) oS,., 30"
Water
= (0,6 �.)(N4-�:) - �.;")(�2,'fI-��)�-!.ft)(O,fi) = :3?�
:: 32.3Ibl,f-t2-/II(t{- fit.·IPt,'- := 0.22'-f j='Sf..:
2.33 A flowrate measuring device is installed
in a horizontal pipe through which water is flow
ing. A U-tube manometer is connected to the
pipe through pressure taps located 3 in. on either
side of the device. The gage fluid in the manom
eter has a specific weight of112 Ib/ft'. Determine
the differential reading of the manometer corre
sponding to a pressure drop between the taps of
O.S lb/in'.
Le C P, tlnd 1'2- bl fl"tsstll't's III: !YI!H/JI'F! -lats.
WYlie rnlll1pmei;Py -eglll-btO'I betwun p, lin&{ 1';z. Thus,
1't r �z./-P., + -R. )
't#f.{ - (fJl.:t.0 -tI ;0 -f'2.
=
I. 'l-S- H
2-27

"
J.31.f I
f$
h
2.31/- Small differences in gas pressures are
commonly measured with a micromanometer of
the type illustrated in Fig. P2.3f. Thisdevice con
sists of two large reservoirs each having a cross
sectional area, An which are filled with a liquid
having a specific weight, y" and connected by a
U-tube of cross-sectional area, A" containing a
liquid of specific weight, y,. When a differential
gas pressure, p, - p" is applied a differential
reading, h, develops. It is desired to have this
reading sufficiently large (so tliat it can be easily
read) for small pressure differentials. Determine
the relationship between handp, - p, when the
area ratio A,IA, is small , and show that the dif
ferential reading, h, can be magnified by making
the difference in specific weights, y, - y" small.
Assume that initially (withp, = p,) the fluidlevels
____--.:Ic... t---�2
FIGURE P2.3i!-
in the two reservoirs are equal.
'�/h�1 /�f/el
for gajf .f!",'c/
-
,.
,
{�
ff
whrn a d,'{-ffYfn-t'o J preSSUre I:- f>.. /� a.f>pl/elf we �S"�/,I(
re�r/lpi, dytPf>' b'1 ;' c//SHlI:C(',1 t:.1t/ Qllt! ri'jlti /-l've/ ",is"s
11J� nUI/? "m ei:n e(POlt.iCJJ1 lH!cl>me.s
fltl/; Iev� I 0,
b!l An. ThIlS.I
1; -r 0; (1, -t --A - i -1..) - 32 � - d', (/., .,. J. h) :: 1'2.
tJ - 1;. = d;. -i. - 61 -t. -r � {2. . .{ )
SInce -the /liw1d.s
LJl A =- ..p,r .:z.
I" the rntlnfJmeter Qre InComIr't'5Sih)e;,
4i 01" :z iJ h A-t
-I... ::- Ar
le-H
0)
A.t:- 1$ sm,,)1 7hfn
Ar
;lilA « � (Utd EZ.tI)
be ne'1JeG1;e A. Thus,
-t; - 12 = (� - rl ) I..
� -,p...
C(nt/ I"r1e 1/,';/1"5 pi h
dd�·feY't'ntl4J.s if 0;2..- Crl ,�
2-28

2.3.5" I
2.35 The cyclindricaJ tank with hemispherical ends shown
in Fig. P2.35 contains a volatile liquid and its vapor. The liq
uid density is 800 kg/m', and its vapor density is negligible.
The pressure in the vapor is 120 kPa (abs), and the atmospheric
pressure is !OI kPa (abs). Determine: (a)the gage pressure read
ing on the pressure gage; and (b) the height, h, of the mercury
manometer.
Vapor A"/1m'
1m
.Ll__ f-________
!ill FIG U R E P2.35
Open
�MerCUry
cc:v Le-f; dt= �p.wh. bF liU1'e( = Ci'oo :3)(1,81;,,)= 185D;3
QVliA
i? (�Ct9e.):: I 20.kPo. Co.!.s) - lol.k Po. (ahs);: ) q -It RVlJ.pDr
nus)
�A5c.:: -PVI<Pdr -t '0,( (ll'M)
= 'l)(c3;;-" -t hllSD�!»(IWI)
-:
Z�, � -k pQ.
{=(),1021'oM
2-29

2.3ta
2.36 Determine the elevation difference. tJ.h, between the
water levels in the two open tanks shown in Fig. P2.36.
SG = 0.90
71'
�
0.4 m
--.---
T
�-----.I ---l--
.�iTS7-
1m
• FIGURE P2.36
£). .{ .:: C>. If/YI1 - ((J, q ) ((). 4-""" )
2.37
2.31 For the configuration shown in Fig.
P2.31 what must be the value of the specific
weight of the unknown ftuid? Express your an
swer in Ib/ft'.
rI'.L---water
5.5 in.
FIGURE P2.37
Ld..- r he $�cihc. UJ�/Jht rJ/ tlnKMUln l-IuJd. the",
in.
(( [(5'.5'-1.4)-ftl_ �r(3.3-1'If)h]_;r. [(If.C;-3'S)ft1=o#;)'0 I:z. :J l 'z. if2.0 17- J
Y.. [(s.5'-1. 'I) -('f.f-3.3 )]t�.W�O =
(3.�- 1.q..) in..
- 2;;'./ ita

2.38 An air-tilled, hemispherical shell is at
tached to the ocean floor at a depth of 10 m as,
shown in Fig. P2.3.1'. A mercury barometer 10-'
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?
73
Ocean surface
Mercury
Shell
FIGURE 1'2.38
-Fa. '" UP1c/llte (IIi- p"'�ssure I;'s/d� shell = �3- (0.7��M»
050 1hat
�I:-tm -v 5 uy/ace aill?ltlsphel'l(" fV't'5suY'e
�w- N specilic- Wflf"t of selu<JI_:beV'
t - -t � (o.ns--.)
;;
- (;0.1 g)(;0. 3b"..) -t (J33 �.Xo.n.....)

·2.j'7 Both ends of the U-tube mercury ma-
nometer of Fig. P2.3'f 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 h,. 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, P,. Show on
a plot how the differential reading varies with P,
for hi = 25, 50, 75, and 100 mm over the range
a :5 P, :5 300 kPa. Assume that the temperature
of the trapped air remains constant.
p.
-F. dJ;JValve
41.._"1
�
h,
T
1""" l>�
L
Mer,ury
FIGURE P2.��
W/·f/, 1h( vtll(I( cJ"sed. Q"� a.. fyeSSllre; 1;1 o.ll'll�""1
.-f-.. _ 0' .lJ1:- .h
'I J/3
r"'-
1J-1i
an" f:.. tire
cf tra.pted (/II"
OJ/I
.JIfJt? ty�ss"r-rs. !PI' IS()1h�rm4J
::£ = eM.! It1l1i
L t .("y �/'IJItIl,t til r mass
1:,. Pi = 1 -v;
-If Is lilt- 11"/""'(" I
? Ij 4b�j,de !V�5Sllyel fJlul
fo /n"It;,/ 411d 1711·"I s I-rt ffS/ rf.!.f�C.f,vdfj. Tn U5 )
1: if -
a.1:"" t' •
(-I;., 1- �t,., ) *f-
for t1lr frAlPul t"n rlyH leJ -V::: -l (AN'a. �f /"h< )t L)
£tj.fZj eq" he IV/,;/fen 4.5
[ �L'
- I]� :: -/;.itm �. _ A h
I.. ;2...
Substr/llt!! t=p .t3) /"fr, Eg. (J) -fr, () btill�
, ,,
(I )
( 2.)
1J,.. i
( 3)
�h � �1 [PI r 1;.J� (J - {:-.
.
� )] (If)
(CtJn't )

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