Unit Operations of Chemical Engineering(化工单元操作)

1.1Whatwillbethe(a)thegaugepressureand(b)theabsolutepressureofwateratdepth12mbelowthesurface?ρwater=1000kg/m3,andPatmosphere=101kN/m2.Solution:Rearrangingtheequation1.1-4ghppabSetthepressureofatmospheretobezero,thenthegaugepressureatdepth12mbelowthesurfaceiskPaghppab72.1171281.910000Absolutepressureofwateratdepth12mkPaPaghppab72.2182187201281.91000101000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)andhydrostaticequilibriumgivesfollowingrelationshipgRgxpgRpAcc21(3)sogRgxppcAc)(21(4)substitutingtheequation(2)forxintoequation(4)givesgRgRDdppcAc)(221(5)（a）whenthechangeinthelevelinthereservoirsisneglected,PagRgRgRDdppcAcAc26381.98151000145.0)()(221（b）whenthechangeinthelevelsinthereservoirsistakenintoaccountPagRgRDdgRgRDdppcAccAc8.28181.98151000145.081.9815145.0515.6)()(22221error=％＝7.68.2812638.2811.4TherearetwoU-tubemanometersfixedonthefluidbedreactor,asshowninthefigure.ThereadingsoftwoU-tubemanometersareR1=400mm，R2=50mm,respectively.Theindicatingliquidismercury.Thetopofthemanometerisfilledwiththewatertopreventfromthemercuryvapordiffusingintotheair,andtheheightR3=50mm.TrytocalculatethepressureatpointAandB.Solution:ThereisagaseousmixtureintheU-tubemanometermeter.ThedensitiesoffluidsaredenotedbyHgOHg,,2,respectively.ThepressureatpointAisgivenbyhydrostaticequilibriumFigureforproblem1.4gRRgRgRpgHgOHA)(32232gissmallandnegligibleincomparisonwithHgandρH2O,equationabovecanbesimplifiedcApp=232gRgRHgOH=1000×9.81×0.05+13600×9.81×0.05=7161N/m²1gRpppHgADB=7161+13600×9.81×0.4=60527N/m1.5Waterdischargesfromthereservoirthroughthedrainpipe,whichthethroatdiameterisd.TheratioofDtodequals1.25.TheverticaldistancehbetweenthetankAandaxisofthedrainpipeis2m.WhatheightHfromthecenterlineofthedrainpipetothewaterlevelinreservoirisrequiredfordrawingthewaterfromthetankAtothethroatofthepipe?Assumethatfluidflowisapotentialflow.Thereservoir,tankAandtheexitofdrainpipeareallopentoair.Solution:Bernoulliequationiswrittenbetweenstations1-1and2-2,withstation2-2beingreferenceplane:2222222111ugzpugzpWherep1=0,p2=0,andu1=0,simplificationoftheequationDdpapaHhAFigureforproblem1.5222uHg1Therelationshipbetweenthevelocityatoutletandvelocityuoatthroatcanbederivedbythecontinuityequation:22Dduuo22dDuuo2Bernoulliequationiswrittenbetweenthethroatandthestation2-23Combiningequation1,2,and3givesSolvingforHH=1.39m1.6Aliquidwithaconstantdensityρkg/m3isflowingatanunknownvelocityV1m/sthroughahorizontalpipeofcross-sectionalareaA1m2atapressurep1N/m2,andthenitpassestoasectionofthepipeinwhichtheareaisreducedgraduallytoA2m2andthepressureisp2.Assumingnofrictionlosses,calculatethevelocitiesV1andV2ifthepressuredifference(p1-p2)ismeasured.Solution:InFig1.6,theflowdiagramisshownwithpressuretapstomeasurep1andp2.Fromthemass-balancecontinuityequation,forconstantρwhereρ1=ρ2=ρ,222200uup＝＝＝144.281.92100081.910002125.11112442ghdDuHg2112AAVVFortheitemsintheBernoulliequation,forahorizontalpipe,z1=z2=0ThenBernoulliequationbecomes,aftersubstituting2112AAVVforV2,22121211212020pAAVpVRearranging,2)1(21212121AAVpp12221211AAppV＝PerformingthesamederivationbutintermsofV2,21221212AAppV＝1.7AliquidwhosecoefficientofviscosityisµflowsbelowthecriticalvelocityforlaminarflowinacircularpipeofdiameterdandwithmeanvelocityV.ShowthatthepressurelossinalengthofpipeLpis232dV.Oilofviscosity0.05Pasflowsthroughapipeofdiameter0.1mwithaaveragevelocityof0.6m/s.Calculatethelossofpressureinalengthof120m.Solution:TheaveragevelocityVforacrosssectionisfoundbysummingupallthevelocitiesoverthecrosssectionanddividingbythecross-sectionalarea1Fromvelocityprofileequationforlaminarflow2substitutingequation2foruintoequation1andintegrating3rearrangingequation3gives1.8.Inaverticalpipecarryingwater,pressuregaugesareinsertedatpointsAandBwherethepipediametersare0.15mand0.075mrespectively.ThepointBis2.5mbelowAandwhentheflowratedownthepipeis0.02m3/s,thepressureatBis14715N/m2greaterthanthatatA.AssumingthelossesinthepipebetweenAandBcanbeexpressedasgVk22whereVisthevelocityatA,findthevalueofk.IfthegaugesatAandBarereplacedbytubesfilledwithwaterandconnectedtoaU-tubecontainingmercuryofrelativedensit