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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