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2、1.1Whatwillbethe(a)thegaugepressureand(b)theabsolutepressureofwateratdepth12mbelowthesurface?ρwater=1000kg/m3,andPatmosphere=101kN/m2.Solution:Rearrangingtheequation1.1-4Setthepressureofatmospheretobezero,thenthegaugepressureatdepth12mbelowthesurfaceisAbsolutepressureofwateratdep
3、th12m1.3AdifferentialmanometerasshowninFig.issometimesusedtomeasuresmallpressuredifference.Whenthereadingiszero,thelevelsintworeservoirsareequal.AssumethatfluidBismethane（甲烷）,thatliquidCinthereservoirsiskerosene(specificgravity=0.815),andthatliquidAintheUtubeiswater.Theinsidediame
4、tersofthereservoirsandUtubeare51mmand6.5mm,respectively.Ifthereadingofthemanometeris145mm.,whatisthepressuredifferenceovertheinstrumentInmetersofwater,(a)whenthechangeinthelevelinthereservoirsisneglected,(b)whenthechangeinthelevelsinthereservoirsistakenintoaccount?Whatisthepercent
5、errorintheanswertothepart(a)?Solution：pa=1000kg/m3pc=815kg/m3pb=0.77kg/m3D/d=8R=0.145mWhenthepressuredifferencebetweentworeservoirsisincreased,thevolumetricchangesinthereservoirsandUtubes(1)so(2)andhydrostaticequilibriumgivesfollowingrelationship(3)so(4)substitutingtheequation(2)f
6、orxintoequation(4)gives(5)（a）whenthechangeinthelevelinthereservoirsisneglected,
7、（b）whenthechangeinthelevelsinthereservoirsistakenintoaccounterror=1.4TherearetwoU-tubemanometersfixedonthefluidbedreactor,asshowninthefigure.ThereadingsoftwoU-tubemanometersareR1=400mm，R2=50mm,respecti
8、vely.Theindicatingliquidismercury.Thetopofthemanometerisfilledwiththewatertopreventfromthemercuryvapordiffusingintotheair,andtheheightR3=50mm.TrytocalculatethepressureatpointAandB.Figureforproblem1.4Solution:ThereisagaseousmixtureintheU-tubemanometermeter.Thedensitiesoffluidsarede
9、notedby,respectively.ThepressureatpointAisgivenbyhydrostaticequilibrium
10、issmallandnegligibleincomparisonwithandρH2O,equationabovecanbesimplified==1000×9.81×0.05+13600×9.81×0.05=7161N/m²=7161+13600×9.81×0.4=60527N/mDdpapaHhAFigureforproblem1.51.5Waterdischargesfromthereservoirthrou
11、ghthedrainpipe,whichthethroatdiam
