流体力学课后作业VIP免费

1.1Apressureof2106N/m2isappliedtoamassofwaterthatinitiallyfilleda1,000cm3volume.Estimateitsvolumeafterthepressureisapplied.将2106N/m2的压强施加于初始体积为1,000cm3的水上，计算加压后水的体积。(999.1cm3)1.2AsshowninFig.1-9,inaheatingsystemthereisadilatationwatertank.Thewholevolumeofthewaterinthesystemis8m3.Thelargesttemperatureriseis500Candthecoefficientofvolumeexpansionisv=0.00051/K,whatisthesmallestcubageofthewaterbank?如图1-10所示，一采暖系统在顶部设一膨胀水箱，系统内的水总体积为8m3，最大温升500C，膨胀系数v=0.0051/K，求该水箱的最小容积？(0.2m3)Fig.1-9Problem1.21.3Whentheincrementofpressureis50kPa,thedensityofacertainliquidis0.02%.Findthebulkmodulusoftheliquid.当压强增量为50kPa时，某种液体的密度增加0.02%。求该液体的体积模量。(2.5108Pa)1.4Fig.1-10showsthecross-sectionofanoiltank,itsdimensionsarelengtha=0.6m,widthb=0.4m,heightH=0.5m.Thediameterofnozzleisd=0.05m,heighth=0.08m.Oilfillstotheupperedgeofthetank,find:(1)Ifonlythethermalexpansioncoefficientv=6.510-41/Koftheoiltankisconsidered,whatisthevolumeFig.1-10Problem1.4ofoilspilledfromthetankwhenthetemperatureofoilincreasesfromt1=-200Ctot2=200C?(2)Ifthelinearexpansioncoefficientl=1.210-51/Koftheoiltankisconsidered,whatistheresultinthiscase?图1-10为一油箱横截面，其尺寸为长a=0.6m、宽b=0.4m、高H=0.5m，油嘴直径d=0.05m，高h=0.08m。由装到齐油箱的上壁，求：(1)如果只考虑油液的热膨胀系数v=6.510-41/K时，油液从t1=-200C上升到t2=200C时，油箱中有多少体积的油溢出？(2)如果还考虑油箱的线膨胀系数l=1.210-51/K，这时的情况如何？((1)2.49210-3m3(2)2.3210-3m3)1.5Ametallicsleeveglidesdownbyselfweight,asshowninFig.1-11.Oilof=310-5m2/sand=850kg/m3fillsbetweenthesleeveandspindle.TheinnerdiameterofthesleeveisD=102mm,theouterdiameterofthespindleisd=100mm,sleevelengthisL=250mm,itsweightis100N.Findthemaximumvelocitywhenthesleeveglidesdownfreely(neglectairresistance).Fig.1-11Problem1.5有一金属套由于自重沿垂直轴下滑，如图1-11所示。轴与套间充满了=310-5m2/s、=850kg/m3的油液。套的内径D=102mm，轴的外径d=100mm，套长L=250mm，套重100N。试求套筒自由下滑时的最大速度为多少(不计空气阻力)。(50m/s)1.6Thevelocitydistributionforflowofkeroseneat200C(=410-3Ns/m2)betweentwowallsisgivenbyu=1000y(0.01-y)m/s,whereyismeasuredinmetersandthespacingbetweenthewallsis1cm.Plotthevelocitydistributionanddeterminetheshearstressatthewalls.在200C时，煤油(=410-3Ns/m2)在两壁面间流动的速度分布由u=1000y(0.01-y)m/s确定，式中y的单位为m，壁面间距为1cm。画出速度分布图，并确定壁面上的剪应力。(410-2Pa)1.7AsshowninFig.1-12,thevelocitydistributionforviscousflowbetweenstationaryplatesisgivenasfollows:21dpV=(Byy)2dxFig.1-12Problem1.7Ifglycerinisflowing(T=200C)andthepressuregradientdp/dxis1.6kN/m3,whatisthevelocityandshearstressatadistanceof12mmfromthewallifthespacingByis5.0cm?Whataretheshearstressandvelocityatthewall?如图1-12所示，两固定平板间粘性流动的速度分布由21dpV=(Byy)2dx给出。如果流体为甘油(T=200C)且压强梯度dp/dx为1.6kN/m3，间距By为5.0cm，距平板12mm处的速度与剪应力为多少？平板处的剪应力与速度为多少？(u12=0.59m/s；τ12=20.8N/m2；u0=0；τ0=40.4N/m2)1.8WhatistheratioofthedynamicviscosityofairtothatofwateratstandardpressureandT=200C?Whatistheratioofthekinematicviscosityofairtowaterforthesameconditions?在标准大气压、T=200C时，空气与水的动力粘度之比为多少？同样条件下它们的运动粘度之比又为多少？(A/W=0.0018；A/W=15.1)1.9Thedeviceshow...