1-10.The conical bearing is placed in a lubricatingNewtonian fluid having a viscosity μ.Determine thetorque T required to rotate the bearing with a constantangular velocity of o.Assume the velocity profile along thethicknesst of thefluid is linear.SOLUTIONSince the velocity distribution is linear, the velocity gradient will be constant.Thevelocityoftheoilincontact withtheshaft at an arbitrarypoint isU=wor.Thus.du_μuor=PdytFrom thegeometry shown inFig.a,drT(1)dz =Z=tanetaneAlso,fromthegeometryshown inFig.b,(2)dz = ds cos oEquatingEqs.(1)and (2),drdr=dscosads =tanesindThe area of the surface of the differential element shown shaded in Fig.a is2㎡dA=2rds=-rdr.Thus,the shearforcetheoil exertson this area issin2m2mμo(or)dF= rdA =r-drrdrsinetsing1Considering themomentequilibrium of the shaft,Fig.a,TrdF=0EM,=0;2mμ0rdF=T=rdrtsineJ2mu0(r4tsin4THOR+Ans.2tsine12
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1-11.For water falling out of the tube, there is a differencein pressureApbetweenapoint located justinsideandapointjust outside of the stream due to the effect of surface tension.Determinethe diameterdofthe stream at thislocation.SOLUTIONConsider a length L of the water column.The free-body diagram of halfcolumn is shown in Fig.a.Considertheforce equilibrium along the y-axis,ZF,=0,20L+po[d(L)] -p[d(L)] =02o=(p-po)dHowever,pi-Po=Ap.Then20d=Ap(a)13
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Chapter 2Fluid Statics2-1.If the piezometer measures a gage pressure of 10 psiat point A, determine the height h of the water in thetube. Compare this height with that using mercury.TakePw= 1.94 slug/ff and pHg = 26.3 slug/ff.05ASOLUTIONHere, the absolute pressure to be measured isIb)(12in.)21+ Patm= (1440 + Patm)10p=Pg+Patmin八1ftftIpIp(a)(b)For the water piezometer, Fig. a,lb= patm+ (1.94 slug/ft)(32.2 ft/s)(h + 0.5ft)(1440 + Patm)p=Patm+pwsft3h=22.55ft=22.6ftAns.Forthemercurypiezometer,Fig.b,1b= Patm+ (26.3 slug/ft)(32.2ft/s-)(h+0.5ft)P= Pam Pui (144 + Pam),hHg=1.20ftAns.14
14 Chapter 2 Fluid Statics 2-1
2-2.The field storage tank is filled with oil.The standpipeisconnected tothetank at C,and the system is open to theatmosphereatBandE.Determinethemaximumpressureinthetankinpsi if the oil reaches a levelof Fin thepipe.Also, at whatlevelshouldtheoilbeinthetank,sothatthemaximumpressureoccurs in thetank? Whatis thisvalue? Take p。=1.78 slug/ff3.0SOLUTIONSince thetop ofthe tank is open to the atmosphere,the free surfaceof the oil in thetank will be the same height as that of pointF.Thus,the maximum pressurewhichoccursatthebaseofthetank(levelA)is(PA)g=yh= (1.78 slug/ft)(32.2 ft/s2)(4ft)Ib(1ft)2=229.26= 1.59 psiAns.ftz（12in.）Absolutemaximumpressureoccurs atthebaseofthetank(level A)whentheoilreacheslevel B.(PA) abs =yhmax= (1.78 slug/ff)(32.2ft/s2)(10 ft)121ft=573.16lb/ft3.98psi Ans.=(12in.）15
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2-3.Determinethe level h'of water in thetube if thedepths of oil and water in the tank are 0.6m and 0.8mrespectively,and the height of mercury in the tube ish=0.08m.Takep。=900kg/m.pw=1000kg/m2,andPHg=13550kg/m089T100mtSOLUTIONReferringtoFig.a,hAB=0.6m,hc=0.8-h'andhcp=h=0.08m.Then themanometerrulegivesPA+PoghAB+PughBC-PHghcD=PDHere,Pa=Pp=O,sincepoints AandDareexposedto the atmosphere.0+（900kg/m）g）(0.6m)+(1000kg/m）（g）(0.8m-h)-(13550kg/m)g)(0.08m)= 0hl=0.256m=256mmAns,Note:Since0.1m<h'<0.8m,thesolution isOK!hAB=0.6mB-D=0.08mChBc=0.8m-h(a)16
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