流体力学课后作业4

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1、1.Apressureof20psiismeasuredatadepthof20ft.Calculatethedensityoftheliquidifp=0onthesurface.解：1psi=6.895kPa而1ft=304.8mm根据P=ρgh得：ρ=P/gh=6.895×103/（9.8×0.3048）=2.308×103kg/m32Avacuumof31kPaismeasuredinanairflowatsealevel.Findtheabsolutepressurein:(a)kPa(b)mmH

2、g(c)psi(d)ftH20(e)in.Hg解：根据P=P’-Pa得：P’=P+PaP’=101.325-31=70.325kPa因为1mmHg=133.33Pa所以得：P’=70.325kPa=70325/133.33=527.45mmHg因为1psi=6.895kPa，所以70.325/6.895psi=10.20psi因为1mmH2o=9.807Pa所以P’=70.325/(304.8×9.807)=22.97ftH2O70kpa=20.77in.Hg3.Thewaterlevelissameint

3、heclosuretank,tube1,pipe3,and4.Pipe1canfluctuate,soastoadjustthewaterpressureinthebank.If(1)tube1declineinacertainheight,(2)tube1riseinacertainheight.Trytoillustratewhichisthehighestlevel,whichisthelowestlevel,andwhicharethesamelevelateachtime.解：在升降前有：P2=P

4、4=Pa在变化前后都有：h1=h3h2=h4筒一下降一定的高度，此时水箱中的水流向筒中，以保持h1=h3.有液面2.3.4均下降，而液面1下降的高度与液面3下降高度相同，但小于筒下降的高度。由于筒下降，水箱中的空气被扩充，则有：P2＜Pa=0又PA=P2+ρga得PA=0+ρgb=ρgb推出PA/ρg=P2/ρg+aPA/ρg=b因为P2＜0所以b＜ah3＜h2所以h2=h4＞h1=h3筒一上升一定的高度。同理可得：相反的b＞ah3＞h2有h1=h3＞h2=h44.Inordertoaccuratelyme

5、asurethepressuredifferencebetweenAandBintheliquidwithdensityof,amicromanometerwasdesignedspeciallyasshowninthefigure.Trytodeterminetherelationshipbetweenand’,andthepressuredifferencebetweenAandB.依题意有：从1-1面的静压相等可知PA-ρgH=PB-ρ’gH可得：PA-PB=ρgH-ρ’gH=gH(ρ-ρ’)由于1面

6、和3面中间是空气，则P1=P2由2-2平面的静压相等可得：PB-ρ’gb=PB-ρg(h1+h2)又b=a+h2+h1所以推出ρ’gb=ρg（b-a）进一步得：ρ’=ρ(b-a)/b将ρ’=ρ(b-a)/b代入PA-PB=ρgH-ρ’gH得：PA-PB=ρagH/b5.Therewasaflatrectangularmetalgatesupportedwith3beamsontheoppositesurfacetowater,theheightofgateandthedepthofwateraresamea

7、sh=3m.Calculatethelocationsofbeamswhichareuniformityinforce.解：作闸门的压强分布图如图右，将其面积分为Ⅰ、Ⅱ、Ⅲ三部分。各梁的位置分别为闸门各部分AE、EG、GB受力的压力中心D1、D2和D3A△ABC=1/2ρgh×h=1/2×103×9.807×32=44.13×103N/mA△AEF=1/3A△ABC=1/2ρgyE×yE=1/3×44.13×103=1/2×103×9.807y2E得：yE=1.73m由A△AGH=2/3A△ABC=1/2ρ

8、gyG×yG=1/2×103×9.807y2G得yG=2.449m△yGE=yG-yE=2.449-1.73=0.719m△yBG=yB-yG=3-2.449=0.551my1=2/3yE=2/3×1.73=1.15m对梯形EFHG分成矩形和三角形，根据合力距定理有：1/2△yGE（△yGE×ρgyE）+2/3△yGE(1/2△yGE×ρgyGE)=AEFHG(y2-yE)化简代入数据得：y2=2.11m同理解得