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Unformatted text preview: 133 A man is considering buying a IE—oz steak for 53.15.. or a EEGg steak for $1.8G. The steak that is a
better hn}r is to he determined. Assmuptt'eus The steaks are of identical quality. Arralrsl's To make a comparison possible: we need to express the cost ofeach steak on a common hasis. Let
us choose 1 kg as the basis for comparison. Using proper conversion factors: the unit cost of each steak is
determined to he \ t's3.15“’1ssz'~.” llhin .52 ounce steak: Unit Cost = II '._12 oz ,.H1]hn13"._D_15339kg£
320 gram steak:
UnitCost= $13”  190% =$B.T5tkg
.NEEGg/H 1kg {, Therefore: the steak at the international market is a better bu}: 131 The thrust developed by the jet engine ofa Boeing 77? is given to he 359300 pounds. This thrust is to
be expressed in N and kgf. Arralrsl's Noting that l lbf = 4.448 N and 1 kgf = 9.81 N, the thrust developed can he expressed in two
other units as [4.443 N" Thrust in I: Thrust = [83,030 lhf}: = 3.“! K105 H """l' r , .
TIJIuHinksf: 11111151431311? N}; “‘3‘ f
teem; 4355:1114 kgf a a .r 1? Sin automobile tire is inﬂated with air. The pressure rise of air in the tire when the tire is heated and the
amount of air that must he bled oﬂ'to reduce the temperature to the original s‘alue are to he determined Assmrrpn'ous 1 Est speciﬁed conditions. air behaves as an ideal gas. I The volume of the tire remains
constant. Properties The gas constant of air is R = D2 37' kPaml‘kgK.
Analysis Initially. the absolute pressure in the tire is
Pi = P3+Em = 210+1CC=313kPa Treating air as an ideal gas and assuming the volume of the tire to
remain constant. the final pressure in the tire can he determined ﬁ'om Epl Ply: _ T3 _ 323K _ __, _
T; = T: _}P: _ T1 P1 _ 293K {:3 lePa] — sdePa 3:1:
23am Thus the pressure rise is
nP=P3—P1 =33ﬁ—31D=2l§ ItPa The amount of air that needs to he hled oﬂ'to restore pressure to its original salue is 1 1; 3
m. =I—§_{J= {JlCIkPaHll'ﬂssm } =0Dgﬂﬁkg
' RT: {DIETkPam3:'l:gK:II:293IQ H
. ‘ 2s 3
a _ P‘tr' {leIltPaKﬂﬂ m } = mmkg ‘ ' ar2 ={a_237kPa.m1..kg.1t){323to
am = m; —m1 moanstosses =a.ooto kg llTE The inininiunipressure in a punip is given. It is to be determined ifthere is a danger (ifcavitation
Properties The vapor pressure of water at TWP is 0.3 I532 psia. Analysis To amid cavitation, the pressure everywhere in die flew should remain abm‘e me vapor {er
saturation ] pressure at the given temperature, 1which is P» = Puntman ='i3'3¢533PEiﬁ
The niininnnn pressure in the pump is [11 psia. which is less than the L'ﬁpDI pressure. Therefore: there is
danger of sanitation in the pump. Discussant Hate that the vapor pressure increases with increasing temperature: and the danger of eatitatinn
increases at higher ﬂuid temperatures. 130 The percent increase in the density of an ideal gas is given for a moderate pressure. The percent
increase in density ofthe gas when compressed at a higherpressure is to be determined. .«lssmrgln'ous The gas behat‘es an ideal gas.
Analysis For an ideal gas. P = cm" and (EFF 5,0). = R1" = P ,o . and thus 1: :iulgal =P . Therefore. the coeﬁicienc of compressibility of an ideal gas Is equal to its absolute pressure. and the coefﬁcient of
compressibility of the gas increases 1with increasing pressure. Substituting x= P into the definition of the coefficient of compressibility a' E — a cw ' u .o
and rearranging gives
*3 _ E
o P Therefore. we percent increase of density of an ideal gas isothermal compression is equal to the
percent increase inpressure. _ no a? 11—1[2l n.
A 10 . _=_=_=1.:.....
I dllIl. F P op 3P lCll—lﬂl:I _
a 1m ten: —=—= =15.)
t a .o P mu Therefore. a pressure change of 1 arm causes a density change of 1091: at 1D arm and a density change of
1".’.: at IUD atm. 130 The percent increase in the density of an ideal gas is given for a moderate pressure. The percent
increase in density ofthe gas when compressed at a higherpressure is to be determined. .«lssmrgln'ous The gas behat‘es an ideal gas.
Analysis For an ideal gas. P = cm" and (EFF 5,0). = R1" = P ,o . and thus 1: :iulgal =P . Therefore. the coeﬁicienc of compressibility of an ideal gas Is equal to its absolute pressure. and the coefﬁcient of
compressibility of the gas increases 1with increasing pressure. Substituting x= P into the definition of the coefficient of compressibility a' E — a cw ' u .o
and rearranging gives
*3 _ E
o P Therefore. we percent increase of density of an ideal gas isothermal compression is equal to the
percent increase inpressure. _ no a? 11—1[2l n.
A 10 . _=_=_=1.:.....
I dllIl. F P op 3P lCll—lﬂl:I _
a 1m ten: —=—= =15.)
t a .o P mu Therefore. a pressure change of 1 arm causes a density change of 1091: at 1D arm and a density change of
1".’.: at IUD atm. 1. T The density of seawater at the free surface and the bull: modulus ofelasticity are given. The density
and pressure at a depm of 2500 in are to be determined. Assumptions 1 The temperature and the hull: modulus of elasticity of seawater is constant. 1 The
gravitational acceleration remains constant. Pr'apai'n'as The density of seawater at free sttrface where the pressure is given to be 1030 kg‘nid: and the
bulls modulus ofelastjcity of seawater is given to he 2.34t10g Mm]. Analysis The coefﬁcient of compressibility or the bull: modulus ofelasticity ofﬂuids is expressed as I E: El
1 or 1:: pg {at constant T)
.__cp .'1. tip The differential pressure change across a diﬁ'erential ﬂuid height ofa‘: is given as cl” = Jogpt': 2500 III.
Combining the two relations above and rearranging:
Fpﬁﬂori _;. de=si=
do do p‘ 1: Integrating from 3 = 0 where ,o = ,oﬂ =1030 ls:g'n.t'1 to z = z where ,o = ,0 gives .=' aﬂo_£{5dT 1 1 g?
  E, o. _> —_—=—
a p‘ r Po :0 s Solving for ,0 gives the variation ofdensity with depth as
1 1.,o.;.—g:. Pt' Substituting into the pressure change relation :1? = pga‘: and integrating ﬁom s = 0 where
P = P: = QEkPa to:=:whereP =Pgives .1“ JP: r 3:31: .o 1"pU—gE.'r I _v P = P: +a'l.t:t 1 l
w x1 — Po 33 1' , which is the desired relation for the variation of pressure in seawater with depth.
At a = 2500 m. the values ofdensity and pressure are determined by substitution to be 1 = =1041 It rml
1:'[1030]tg:'m’}—[9.31m‘s'JCi00 m):'[2.34a10p tomJ g ,o NI P=tsaootrPaJ—{2.34x109u.'m2)ml+3+t;rt—l
txl—{10301cgm )[9.Slm‘s']{2300mj.'{2.34><10 armJ;
=2.sjoamlsa
=2s.5o MPa since 1 Pa = 1‘_‘~I.'n1s = 1]tg:'ms2 and lltPa =1000 Pa. Discussion Note that if we assumed o = oc. = constant at 1030 kgs'm3, The pressure at 2500 m would be
P = P: hog": = 0.093  23.25 = 25.36 MPa. Then the density at 2300 n1 can be estimated to he as = pastP= (1030}[2340 ItfE‘aTIQilo hIPa] =11.1]tg.'m‘:' and thus p = 1041 lsg"nii‘ 24? A clutch system is used to transmit torque through an oil film between two identical disks. For
speciﬁed rotational speeds: the torque transmitted is to be determined. Assmuplt'ons l The thiclcness of the oil film is uniform. 2 The rotational speeds ofdisks remains constant. Properties The absolute viscosity of oil is given to be p. = 0.38 Ns.'I:n2. Arralrsr's The distcs are rotting in the same direction at different angular speeds of cal and of mg . Therefore.
we can assume one ofthe disks to be stationary and the other to be rotating at an angular speed of co; — co: . The velocity gradient anywhere in the oil of ﬁlm thickness is is l’sil where i” = (to; — cs12 }r is the tangential
velocity. Then the wall shear stress anywhere on the surﬁce of the faster disk at a distance r ﬂ'om the axis
of rotation can be expressed as do i” [all — or: jr' 1'“. = Ir:—= y— = Jr: o‘r l.‘ ll .
Then the shear force acting on a differential area obi on the surface n
and the torque generation associated with it can be expressed as 0,]
[so] — £12 }J'
o‘F = two‘s! = ,r: _— [Emira‘r 011
n
— 'l 2 :2 . — 'u *
dT = rdF = ,u in" f 3" {Em'jdr = —"“E“;1 “rdj r’o‘r
. I
integrating,
. ,I 4 D '2 L
T = 2x;r(col —co]: I'D Elgar = mrrEro] —o:2} r = statics; —oo3 JD
l':  rD F: —l 32!?!
hi]
Noting that co = Earl , the relative angular speed is
. . , t _ 'r l min";
so; — co] = lrrﬁrrl — in) = (Est rad'revj[~145£ —1393,tre1.'.'min] 6'3] P: 5.445 rad's ._
s 5 . Substituting. the torque transmitted is determined to be _ gross N  3me 315.445 5993.30 m}l
310.003 1.11} T =[l.55Nm Discussion Hote that the torque transmitted is proportional to the fourth power of disk diameter. and is
inversely proportional to the thickness ofthe oil ﬁlm. 249 _'‘L muEtidislc Electrorheological “ER” clutch 1arith a ﬂuid in which shear stress is expressed as
r = r3. —,1e:dul.l"a‘v1'} is considered. A relationship for the torque transmitted by the clutch is to be obtained. and the numerical T.‘alue ofthe torque is to be calculated. Aismuplfom l The thickness of the oil layer between the disks is constant. I The Eingham plastic model
for shear stress expressed as r = r1. + Irtﬂdufldjr} is valid. Properties The constants in shear stress relation are given to be .u = 13.1 Pas and T:F2.5 era. F1=l.1n1m "T.— Clutput shaft
Input shaft Plates mounted on shell
Plates mounted on input shaﬁ T.'ariable magnetic ﬁeld Analyst's (a) The T.‘elocity gradient anywhere in the oil of film thickness to is 1’ it": 1a‘here I’= on?“ is the
tangential velocity relative to plates mounted on the shell. Then the wall shear stress anywhere on the
surface of a plate mounted on the input shaﬂ at a distance 1' ﬁ'om the axis ofrotation can be expiessed as du F" for T=r+If—=T+I—=T.'Lt'—
” ‘ 'a'r t “is J is Then the shear force acting on a differential area all on the surface of a
disk and the torque generation associated with it can be expressed as a? = r.“ at =f «.3 +113 IlEZm'jwtr I_ if I.
.« s r' x .3'
d'T =m‘F=r‘i 1'1. +1:ﬂ [2arjdr=2:'r .71.?" +111 id:
x ‘ ii 2 .l ' n t
lutegrating.
3 'z (or! I" r3 alt1TH: T   a}  1
T=23J' irrJwa— iii=2: 1'}. —+’“ 2 =2: its? —Rfj+’“—[ag 424}
rR k ' in _. 3 4h Hi“J12 3 ' 4h ' g This is the torque transmitted by one surface of a plate mounted on the input shaft. Then the torque
transmitted by both surfaces of'lkT plates attached to input shaft in the clutch becomes
l—T . T = test "
L 3 {b} Noting that at = 2n] = Jrﬁﬂﬂﬂ less"mind = 15.030 rad"min = 251.3 rad"s and substituting: 3 3 #53 4 L
Rs—R —— Rs—R
{ _ L] “if _ 1:1] {UlKs"mjjtlﬁljs'sj
trust] 12 m} l [[uromf —[o.oim}3]+ T = {Hm 1;][25410 :‘s'fm  action:ij4 {0.01 inf] = anal] Nm 251 The torque and the rpm of a double cylinder viscometer are given. The viscosity of the ﬂuid is to be
determined. Assmuplfaiis l The inner cylinder is completely submerged in oi]. I The viscous effects on the two ends of
the inner cylinder are negligible. 3 The ﬂuid is Newtonian. Analyst's Substituting the given values, the viscosity of the ﬂuid is determined to be If {U.ENmj{D.D012 m} Dl'sr'nrsi'sn This is the viscosity value at the temperature that existed
during the experiment. T.'iscosity is a strong function of temperature,
and the values can be significantly different at different temperatures. 261 A. glass tube is inserted into a liquid, and the capillary iise is measured. The sulface tension ofthe
liquid is to be deteimtned. Assumptions 1 There are no impua'ities in the liquid. and no containination on the surfaces ofthe glass tube.
2 The liquid is open to the atmospheric air. Properties The denaitjf' of The liquid is given to he 96E] kg‘n13. The contact angle is given to be 15°. .lirt:r{1'si5 Substituting the ﬂ'llﬂlE'I'lC al values: the sniface tension is determined ﬁ'onzt the capillar}r Iise relation
to he 0",: o . 3 * . 3 ". ' 'I ' I'
pgiil ={96EI kgm 3.3311113 Hilﬁﬂlg. mj{D.DICI)n:L} 11s 1 Lungazmm
Ecos; licoslia} l 1kEm.'5‘ _.' Discussion The stuface tension depends on tempeiature. Therefore: the
value determined is valid at the temperature of the liquid. 265 A steel ball ﬂo ats on 1a‘ater due to the surface tension eﬁ‘ect. The maximum diameter of the hall is to
be determined: and the calculations are to be repeated for aluminum.
Assmuplt'ans 1 The svater is pure, and its temperature is constant. 2 The ball is dropped on water slots'13; so that the inertial effects are negligible. 3 The contact angle is taken to be '3'” for maximum diameter. Properties The surface tension ofsvater at 20°C is o; = CLUB N"m. The contact angle is_ talcen to he D“. The
densities ofsteel and aluminum are given to be m = 73GB Ichm‘1 and log] = ETGU leg"ml. Anaijsis The surface tension force and the weight of
the hall can be expressed as F: =ail‘3hrrj and H’=mg =,ogV=pg.ID3 ."I5 When the ball ﬂoats. the net force acting on the hall in the vertical direction is zero. Therefore, setting
F__. = H" and solving for diameter D gives
'—
1}: IE
1' as
Substititing the Imown quantities: the maximum diameters for the steel and aluminum halls become
,— .— _. K 2\'
‘50.  sun33. 1c . .
‘ as t {TEDt}kg"m’}l:9_31m.g:lk 1N /_
c  " . 2 a
ﬁ   ans33.. 1b . ‘
D—‘' = ii: Iﬁs—W.‘ g ms =4.1x10_’ m=4.1mm
1 PE ‘t (limkgimJJEQSIm‘sja 1N {J Discussion Note that the ball diameter is inversely proportional to the square root of density, and thus for a
given material. the smaller halls are more likelyr to ﬂoat. lTDI. The minimum pressure on the suction side of a 1a‘ater pump is given. The maximum water
temperature to avoid the danger of cavitation is to he determined. Properties The saturation temperature ofwater at 13.95 psia is llﬂUDF. A:ra{1.'st's To avoid cavitation at a specified pressm'e: the ﬂuid temperature everywhere in the ﬂow should
remain below the saturation temperature at the git'eu pressure, 1which is Total: = T'arui IZISSpJfa = 1ﬂﬂoF a Themfore. the 1'" <1 IGGDF to avoid cat‘itatiou. Discussion Kore that saturation temperature increases with pressure. and thus cavitation may occur at
higher pressure locations at higher ﬂuid temperatures. 2?5 A circular disk immersed in oil is used as a damper. as shown in the ﬁgure. It is to be shown that the damphrg torque is Tmplg = Cm 1.t'here C = ﬁ.5;‘ry[:lfrt+ lift: :LFE'l . Assumptions 1 The thickness of the oil layer on each side remains constant. I The velocity proﬁles are
linear on both sides of the disk. 3 The tip effects are negligible. 4 The effect ofthe shaﬁ is negligible. Damping oil Arrafi'sis The L'elo cit3' gradient anywhere in the oil ofﬁlm thickness :1 is I’.'a u‘here I’ = car is the tangential
T.‘elocitp. Then the wall shear stress anywhere on the upper surface of the disk at a distanee r from the axis
ofrotation can be expressed as ah: I" (or
I“. = Irr—= ,u—= ,u—
a‘i' a n Then the shear force acting on a differential area rid on the surface and the torque it generates can be
expressed as are = was =r1£dd
{I 1
n m. a"T=ra‘F=,u titl a!
Noting that it = 211113?“ and integrating: the torque on the top surface is determined to be H
4 t =3 1 a I 1   are"
Twp =ﬂ [was =ﬂ r' [ea31m = "Wall Has = “Tim’— = '”
at at at  rIZI a 4: a 4 “U in
The torque on the bottom surﬁce is obtained by replaping a! by b,
sauna? 4
Thaibunt = 2b
The total torque acting on the disk is the sum ofthe torques acting on the top and bottom surfaces.
.1 3
mmﬂ 1 l
Tdampsng.tatal = Tatiana: +Ttop = I 2 _\;+;Jl
or.
T —c 11 c—E‘ﬂﬂl f1 1‘
ﬂlpmgltmal — 6:} “ EIE — 2 IHE+ This completes the proof. Discussion Note that the damping torque [and thus damping power} is inversely proportional to the
thickness of oil ﬁlms on either side, and it is propoitional to the 4111 power o‘fthe radius ofthe damper disk. 2HT. _''1 glass Tube is inserted 111m meremjs'. The capillary drop of mercury in The Tube is To be determined. slismupn'am 1 There are no impunties in mercury: and no eonTaannaTion on The siufaees ofThe glass mhe.
2 The LuereLTTjr' is open To The atmospheiio air. Properties The surface Tension ofmerouryglass ir atmospherie air at 63”? [20”le 1s :3; = 0.44ﬂx0ﬂ633s =
{$6015 lbffﬁ. The derisiT}r of LDEI'C'lTl'jr' is p = 34'." lbm."ft’ at FTP. but we can also use This s‘alue at 63°F.
The contact angle is given To be 140‘“. .lnn!_1'51'5 3111:: sTiTuTiJJg The meerioal Values: The cap:l:illa:r1r drop 1s deTermined To be ;. 35; €05 .d' 2mm 15 Tbsﬁllings 140°) n" 32.2 1!:I111fts's2 "
I = — = —_ 1 I _
LE3 [34? 1h111‘frj j:32.2 ﬁ"s‘ ]I:I:I.Ti L‘ frj '.\ 1 lbf j
=—;].CDl:1:lﬂ m:
= —C.C 175 in
Discussion The uegaTis'e sign indicates capillar; drop i11sTead ofrise. The MEIcm}. .—. ” drop is very small 111 This case heoause of The large diameter of The Tube. ...
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 Fall '08
 SMITH
 Shear Stress, Fundamental physics concepts

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