《流体力学》课程教学资源（英文复习）Review Problems for Fluid Mechanics

Review Problemsfor Fluid MechanicsEdited byThe Department of Mechanics2019.042

2 Review Problems for Fluid Mechanics Edited by The Department of Mechanics 2019.04

Chapter 1Introductionand Properties ofFluids1-1,Determine the change in the density of oxygen whenthe absolute pressure changes from 345kPa to286kPa,while thetemperatureremains constant at 25°C.This is calledanisothermalprocess.SOLUTIONApplying the ideal gas law with T,= (25C +273)K= 298 K,pi=345kPa andR=259.8J/kg·Kforoxygen (table inAppendixA),345(103)N/m2= pi(259.8J/kg·K)(298 K)PI=p/RTI;pi=4.4562kg/mFor p2=286kPa andT2=T,=298K,286(10)N/m2= p2(259.8J/kg·K)(298 K)P2=p2RT2p2=3.6941kg/m2Thus, the change in density isAp=p2-pl=3.6941kg/m2-4.4562kg/mAns.=-0.7621kg/m2m-0.762kg/m2The negative sign indicates a decrease in densityn

3 Chapter 1 Introduction and Properties of Fluids 1-1. Thus, the change in density is

1-2.Anexperimental testusinghumanbloodatT=30°Cindicates that it exerts a shear stress of =0.15 N/monsurface A,where the measured velocity gradient is 16.8 s-Since blood is a non-Newtonian fluid,determine its apparentviscosity at A.SOLUTIONduHere,= 16.8 s-" and ↑ = 0.15 N/m2. Thus,dydu0.15N/m2=μa(16.8s-)T=podyMa= 8.93(10-3)N ·s/m2Ans.Realize thatblood is a non-Newtonian fluid.For this reason,wearecalculating theapparent viscosity.4

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1-3.When theforcePis appliedtotheplate,thevelocityprofile for a Newtonian fluid that is confined under the plateis approximated by u= (4.23y/3)mm/s, where y is in mm.Determinetheshear stresswithinthefluid aty=5mm.Take μ= 0.630(10~3) N·s/m2SOLUTIONSince the velocity distribution is not linear,the velocity gradient varies with y.u = (4.23yl/3) mm/s崇 [6(423)-23]-1+dyAty=5mm,du1.41=0.4822 s-152/3dyThe shear stress isdu= [0.630(103) N s/m2] 0.4822 s-1T=μdy=0.3038(10-3) N/m2=0.304mPaduNote:When y=O,00andsoT→00.dyat this point.Hence,Theequation cannotbeapplied at this point.5

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1-4.Ifthekinematicviscosityofglycerinisv=1.15(10-)m/s,determineitsviscosityinFPSunits.Atthe temperature considered,glycerin has a specific gravityofS,=1.26.SOLUTIONThe density of glycerin isPg=SgPw=1.26(1000kg/m)=1260kg/m3Then,ugHly1.15(10-3)m2/s=Vg=1260kg/mPg ()())/=0.03026b·s/ft=0.03031bs/ftAns.6

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1-5.IfaforceofP=2Ncausesthe30-mm-diameter50mmshaft to slide along the lubricated bearing with a constantspeed of 0.5 m/s, determine the viscosity of the lubricant0.5m/sand the constant speed of the shaft when P=8N.AssumePthe lubricant is a Newtonian fluid and the velocity profilebetween the shaftand thebearing is linear.The gap betweenthebearing andthe shaftis 1mm.SOLUTIONSince the velocity distribution is linear,the velocitygradient willbe constant.duT=Hdy2N(0.5m/s)=E[2m(0.015 m)](0.05 m)(0.001m）μ= 0.8498 N·s/mAns.Thus,8N= (0.8488N·s/m2)[2(0.015m)(0.05m)（0.001m）Ans.w=2.00m/sAlso,byproportion,(0.5m/s）(2N)At(8N)?MAt+m/s=2.00m/sAns.V=7

7 1-5. SOLUTION

AP1-6.8mm)Jic6mmHY5Aplasticstriphavingawidthof0.2mandamassof150gpasses between twolayersAandBofpainthavinga viscosity of5.24N·s/m?.Determine theforcePrequiredto overcome theviscous frictionon each sideifthe strip0.30 mmoves upwards at aconstantspeedof 4mm/s.Neglectany福friction at the top and bottom openings,and assume the7W = 0.15(9.81)Nvelocity profile through each layer is linear.EVFB(a)SOLUTIONSince the velocity distribution is assumed to be linear,the velocity gradient willbeconstant.For layers Aand B,4mm/s4mm/s(du)du=0.5s-=0.66675s18mmdy)6mm(dy/The shear stresses acting on the surfaces in contact with layers A and B aredu= (5.24Ns/m2)(0.5 s-l)= 2.62N/m2TA=Adydu= (5.24N·s/m)(0.6667s-l)= 3.4933N/mTB=ALdyThus, the shearforces acting on the contact surfaces areFA=TAA=(2.62N/m)[(0.2m)(0.3m)l=0.1572NF= TgA= (3.4933N/m)[(0.2m)(0.3 m) =0.2096 NConsidertheforceequilibriumalongyaxis fortheFBDofthestrip,Figa+1EF=0:P-0.15(9.81)N-0.1572N-0.2096N=0P=1.8383N=1.84NAns.00

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1-7.The read-write head for a hand-heldmusic playerhasasurfaceareaof 0.04mm.Thehead isheld0.04μmabovethedisk, whichisrotating at a constant rate of1800 rpm.Determine the torqueT that must be applied to8mmthe disk to overcome the frictional shear resistance of theair between the head and the disk.The surrounding airis at standard atmospheric pressure and a temperature ofT20°C.Assume thevelocityprofileis linearSOLUTIONSince thevelocity distribution is assumed to be linear,the velocitygradient will beconstant.For layers Aand B,.4mm/s4mm/sdu= 0.5 s- (芸)s = 0.6675 s-8mm6mmdy)A(dy/BTheshearstressesactingonthesurfaces incontact withlayersAand Baredu= (5.24 N·s/m2)(0.5 s-1) =2.62N/m2TA=u(dy/Adu= (5.24N·s/m2)(0.6667s-l) = 3.4933N/m2TB=μdyThus,theshearforces acting onthecontact surfaces areFA= TAA = (2.62N/m)[(0.2 m)(0.3 m)I = 0.1572 NF=TA=(3.4933N/m)[(0.2m)(0.3m))=0.2096NConsidertheforceequilibrium alongyaxisfortheFBDofthestripFig.a.P-0.15(9.81)N-0.1572N-0.2096N=0+1F,=0;P=1.8383N=1.84NAns.9

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1-8.The verythin tube A of mean radius rand length L isplacedwithinthefixed circularcavityas shown.Ifthecavityhasasmallgapof thicknesstoneach sideofthetube,and isfilled with a Newtonian liquidhavinga viscosityμ,determinethe torque T required to overcome the fluid resistance androtatethetubewithaconstantangularvelocityofw.Assumethevelocityprofile within the liquid is linear.SOLUTIONSince the velocity distribution is assumed to be linear,the velocity gradient will beconstant.F=2muorLdu.T=HMdy(or)=μtOConsidering the moment equilibrium of the tube,Fig.a,EM=0;T-2Ar=0(a)(or)T = 2(μ)(2mrL)-4murLT=Ans.t10

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1-9.4.5rad/s40mrThetuberests ona1.5-mmthinfilm of oil havingaviscosityofμ=0.0586N·s/m.Ifthetubeisrotatingataconstantangularvelocityofo=4.5rad/s,determinethe801torqueTthat must be appliedtothetubetomaintainthemotion.Assume the velocityprofile within the oil is linearSOLUTIONOil is a Newtonian fluid.Sincethevelocity distribution is linear,the velocitygradientwill be constant.The velocity of the oil in contact with the shaft at an arbitrary pointis U=or.Thus.dy1du_MorT=Hdy1Thus, the shear force the oil exerts on the differential clementof area dA=2r drshownshadedinFig-ais2mue(2ardr)r-drdF-TdA--Considering the moment equilibrium of the tube about point D,FardF-T=0C+EMo=0;2auoT-rdF:rd2uoTuO(ra-)21Substitutingthenumerical values,(0.0586N·s/m)(4.5rad/s)(0.08%-0.04)T=2[1.5 (10-)m]=0.01060N·m=0.0106N·mAns.11

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