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1、1.3AdifferentialmanometerasshowninFig.issometimesusedtomeasuresmallpressuredifference.Whenthereadingiszero,thelevelsintworeservoirsareequal.AssumethatfluidBismethane（甲烷）,thatliquidCinthereservoirsiskerosene(specificgravity=0.815),andthatliquidAintheUtu
2、beiswater.TheinsidediametersofthereservoirsandUtubeare51mmand6.5mm,respectively.Ifthereadingofthemanometeris145mm.,whatisthepressuredifferenceovertheinstrumentInmetersofwater,(a)whenthechangeinthelevelinthereservoirsisneglected,(b)whenthechangeinthelev
3、elsinthereservoirsistakenintoaccount?Whatisthepercenterrorintheanswertothepart(a)?Solution：pa=1000kg/m3pc=815kg/m3pb=0.77kg/m3D/d=8R=0.145mWhenthepressuredifferencebetweentworeservoirsisincreased,thevolumetricchangesinthereservoirsandUtubes(1)so(2)andh
4、ydrostaticequilibriumgivesfollowingrelationship(3)so(4)substitutingtheequation(2)forxintoequation(4)gives(5)（a）whenthechangeinthelevelinthereservoirsisneglected,（b）whenthechangeinthelevelsinthereservoirsistakenintoaccounterror=1.4TherearetwoU-tubemanom
5、etersfixedonthefluidbedreactor,asshowninthefigure.ThereadingsoftwoU-tubemanometersareR1=400mm，R2=50mm,respectively.Theindicatingliquidismercury.Thetopofthemanometerisfilledwiththewatertopreventfromthemercuryvapordiffusingintotheair,andtheheightR3=50mm.
6、TrytocalculatethepressureatpointAandB.Figureforproblem1.4Solution:ThereisagaseousmixtureintheU-tubemanometermeter.Thedensitiesoffluidsaredenotedby,respectively.ThepressureatpointAisgivenbyhydrostaticequilibriumissmallandnegligibleincomparisonwithandρH2
7、O,equationabovecanbesimplified==1000×9.81×0.05+13600×9.81×0.05=7161N/m²=7161+13600×9.81×0.4=60527N/mDdpapaHhAFigureforproblem1.51.5Waterdischargesfromthereservoirthroughthedrainpipe,whichthethroatdiameterisd.TheratioofDtodequals1.25.Theverticaldistance
8、hbetweenthetankAandaxisofthedrainpipeis2m.WhatheightHfromthecenterlineofthedrainpipetothewaterlevelinreservoirisrequiredfordrawingthewaterfromthetankAtothethroatofthepipe?Assumethatfluidflowisapotentialflow.Thereservoir,tankAandtheexito
