流体力学课后作业1

《流体力学课后作业1》由会员上传分享，免费在线阅读，更多相关内容在教育资源-天天文库。

1、1.14Fortwo0.2-m-longrotatingconcentriccylinders,thevelocitydistributionisgivenbyu(r)=0.4/r－l000rm/s.Ifthediametersofthecylindersare2cmand4cm,respectively,calculatethefluidviscosityifthetorqueontheinnercylinderismeasuredtobe0.0026N·m.解：已知：u(r)=0.4/r－l000rT=0.0026N·ml=0.2m又T=F×R=τ·πdl·r对于极坐标

2、而言，切应力表达式应为：=μ（-0.4/r2-1000-0.4/r2+1000）=-0.8μ/r2，方向与速度方向相反。推出：μ=Tr2/（0.8πdl·r）=0.2586Pa·s1.20Aflowmightreasonablybeapproximatedasincompressibleifthechangeofdensitywithpressureatconstanttemperatureislessthan0.5%.a.Whatisthemaximumpressurechangepossibleforwatertobeconsideredincompressibleac

3、cordingtothiscriterion?b.Whatisthemaximumpressurechangepossibleforair(initiallyatstandardsea-levelconditions)tobeconsideredincompressibleaccordingtothiscriterion?解：a.已知△ρ/ρ≤0.5%求△P=?αp·△P=△ρ/ραp外插至标准大气压（101325Pa）下，得αp=0.54×10-9有：△P=0.5%/（0.54×10-9）≈9.26MPab．P=ρRT当T不变时P/ρ=△P/△ρ得.△P=P·△ρ/ρ=1

4、01325×0.5%=507Pa