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1.3AdifferentialmanometerasshowninFig.issometimesusedtomeasuresmallpressuredifference.Whenthereadingiszero,thelevelsintworeservoirsareequal.AssumethatfluidBismethane(甲烷),thatliquidCinthereservoirsiskerosene(specificgravity=0.815),andthatliquidAintheUtubeiswater.TheinsidediametersofthereservoirsandUtubeare51mmand6.5mm,respectively.Ifthereadingofthemanometeris145mm.,whatisthepressuredifferenceovertheinstrumentInmetersofwater,(a)whenthechangeinthelevelinthereservoirsisneglected,(b)whenthechangeinthelevelsinthereservoirsistakenintoaccount?Whatisthepercenterrorintheanswertothepart(a)?Solution:pa=1000kg/m3pc=815kg/m3pb=0.77kg/m3D/d=8R=0.145mWhenthepressuredifferencebetweentworeservoirsisincreased,thevolumetricchangesinthereservoirsandUtubesRdxD2244(1)soRDdx2(2)andhydrostaticequilibriumgivesfollowingrelationshipgRgxpgRpAcc21(3)sogRgxppcAc)(21(4)substitutingtheequation(2)forxintoequation(4)givesgRgRDdppcAc)(221(5)(a)whenthechangeinthelevelinthereservoirsisneglected,PagRgRgRDdppcAcAc26381.98151000145.0)()(221(b)whenthechangeinthelevelsinthereservoirsistakenintoaccountPagRgRDdgRgRDdppcAccAc8.28181.98151000145.081.9815145.0515.6)()(22221error=%=7.68.2812638.2811.4TherearetwoU-tubemanometersfixedonthefluidbedreactor,asshowninthefigure.ThereadingsoftwoU-tubemanometersareR1=400mm,R2=50mm,respectively.Theindicatingliquidismercury.Thetopofthemanometerisfilledwiththewatertopreventfromthemercuryvapordiffusingintotheair,andtheheightR3=50mm.TrytocalculatethepressureatpointAandB.Solution:ThereisagaseousmixtureintheU-tubemanometermeter.ThedensitiesoffluidsaredenotedbyHgOHg,,2,respectively.ThepressureatpointAisgivenbyhydrostaticequilibriumgRRgRgRpgHgOHA)(32232gissmallandnegligibleincomparisonwithHgandρH2O,equationabovecanbesimplifiedcApp=232gRgRHgOH=1000×9.81×0.05+13600×9.81×0.05=7161N/m²1gRpppHgADB=7161+13600×9.81×0.4=60527N/mFigureforproblem1.41.5Waterdischargesfromthereservoirthroughthedrainpipe,whichthethroatdiameterisd.TheratioofDtodequals1.25.TheverticaldistancehbetweenthetankAandaxisofthedrainpipeis2m.WhatheightHfromthecenterlineofthedrainpipetothewaterlevelinreservoirisrequiredfordrawingthewaterfromthetankAtothethroatofthepipe?Assumethatfluidflowisapotentialflow.Thereservoir,tankAandtheexitofdrainpipeareallopentoair.Solution:Bernoulliequationiswrittenbetweenstations1-1and2-2,withstation2-2beingreferenceplane:2222222111ugzpugzpWherep1=0,p2=0,andu1=0,simplificationoftheequation1Therelationshipbetweenthevelocityatoutletandvelocityuoatthroatcanbederivedbythecontinuityequation:22Dduuo22dDuuo2Bernoulliequationiswrittenbetweenthethroatandthestation2-23Combiningequation1,2,and3givesDdpapaHhAFigureforproblem1.5222uHg222200uupSolvingforHH=1.39m1.6Aliquidwithaconstantdensityρkg/m3isflowingatanunknownvelocityV1m/sthroughahorizontalpipeofcross-sectionalareaA1m2atapressurep1N/m2,andthenitpassestoasectionofthepipeinwhichtheareaisreducedgraduallytoA2m2andthepressureisp2.Assumingnofrictionlosses,calculatethevelocitiesV1andV2ifthepressuredifference(p1-p2)ismeasured.Solution:InFig1.6,theflowdiagramisshownwithpressuretapstomeasurep1andp2.Fromthemass-balancecontinuityequation,forconstantρwhereρ1=ρ2=ρ,2112AAVVFortheitemsintheBernoulliequation,forahorizontalpipe,z1=z2=0ThenBernoulliequationbecomes,aftersubstituting2112AAVVforV2,22121211212020pAAVpV===144.281.92100081.910002125.11112442ghdDuHgRearranging,2)1(21212121AAVpp12221211AAppV=PerformingthesamederivationbutintermsofV2,21221212AAppV=1.7AliquidwhosecoefficientofviscosityisµflowsbelowthecriticalvelocityforlaminarflowinacircularpipeofdiameterdandwithmeanvelocityV.ShowthatthepressurelossinalengthofpipeLpis232dV.Oilofviscosity0.05Pasflowsthroughapipeofdiameter0.1mwithaaveragevelocityof0.6m/s.Calculatethelossofpressureinalengthof120m.Solution:TheaveragevelocityVforacrosssectionisfoundbysummingupallthevelocitiesoverthecrosssectionanddividingbythecross-sectionalarea1Fromvelocityprofileequationforlaminarflow2substitutingequation2foruintoequation1andintegrating3rearrangingequation3givesRRrdruRudAAV02021122014RrRLppuL2032DLppVL1.8.Inaverticalpipecarryingwater,pressuregaugesareinsertedatpointsAandBwherethepipediametersare0.15mand0.075mrespectively.ThepointBis2.5mbelowAandwhentheflowratedownthepipeis0.02m3/s,thepressureatBis14715N/m2greaterthanthatatA.AssumingthelossesinthepipebetweenAandBcanbeexpressedasgVk22whereVisthevelocityatA,findthevalueofk.IfthegaugesatAandBarereplacedbytubesfilledwithwaterandconnectedtoaU-tubecontainingmercuryofrelativedensity13.6,giveasketchshowinghowthelevelsinthetwolimbsoftheU-tubedifferandcalculatethevalueofthisdifferenceinmetres.Solution:dA=0.15m;dB=0.075mzA-zB=l=2.5mQ=0.02m3/s,pB-pA=14715N/m2smdQVVdQAAAA/132.115.0785.002.044222smdQVVdQBBBB/529.4075.0785.002.044222Whenthefluidflowsdown,writingmechanicalbalanc
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