UnitOperationsofChemicalEngineering化工单元操作.docx
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UnitOperationsofChemicalEngineering化工单元操作.docx
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UnitOperationsofChemicalEngineering化工单元操作
1.1Whatwillbethe(a)thegaugepressureand(b)theabsolutepressureofwateratdepth12mbelowthesurface?
ρwater=1000kg/m3,andPatmosphere=101kN/m2.
Solution:
Rearrangingtheequation1.1-4
Setthepressureofatmospheretobezero,thenthegaugepressureatdepth12mbelowthesurfaceis
Absolutepressureofwateratdepth12m
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=0.145m
Whenthepressuredifferencebetweentworeservoirsisincreased,thevolumetricchangesinthereservoirsandUtubes
(1)
so
(2)
andhydrostaticequilibriumgivesfollowingrelationship
(3)
so
(4)
substitutingtheequation
(2)forxintoequation(4)gives
(5)
(a)whenthechangeinthelevelinthereservoirsisneglected,
(b)whenthechangeinthelevelsinthereservoirsistakenintoaccount
error=
1.4TherearetwoU-tubemanometersfixedonthefluidbedreactor,asshowninthefigure.ThereadingsoftwoU-tubemanometersareR1=400mm,R2=50mm,respectively.Theindicatingliquidismercury.Thetopofthemanometerisfilledwiththewatertopreventfromthemercuryvapordiffusingintotheair,andtheheightR3=50mm.TrytocalculatethepressureatpointAandB.
Figureforproblem1.4
Solution:
ThereisagaseousmixtureintheU-tubemanometermeter.Thedensitiesoffluidsaredenotedby,respectively.ThepressureatpointAisgivenbyhydrostaticequilibrium
issmallandnegligibleincomparisonwithandρH2O,equationabovecanbesimplified
=
=1000×9.81×0.05+13600×9.81×0.05
=7161N/m²
=7161+13600×9.81×0.4=60527N/m
1.5Waterdischargesfromthereservoirthroughthedrainpipe,whichthethroatdiameterisd.TheratioofDtodequals1.25.TheverticaldistancehbetweenthetankAandaxisofthedrainpipeis2m.WhatheightHfromthecenterlineofthedrainpipetothewaterlevelinreservoirisrequiredfordrawingthewaterfromthetankAtothethroatofthepipe?
Assumethatfluidflowisapotentialflow.Thereservoir,tankAandtheexitofdrainpipeareallopentoair.
Solution:
Bernoulliequationiswrittenbetweenstations1-1and2-2,withstation2-2beingreferenceplane:
Wherep1=0,p2=0,andu1=0,simplificationoftheequation
1
Therelationshipbetweenthevelocityatoutletandvelocityuoatthroatcanbederivedbythecontinuityequation:
2
Bernoulliequationiswrittenbetweenthethroatandthestation2-2
3
Combiningequation1,2,and3gives
SolvingforH
H=1.39m
1.6Aliquidwithaconstantdensityρkg/m3isflowingatanunknownvelocityV1m/sthroughahorizontalpipeofcross-sectionalareaA1m2atapressurep1N/m2,andthenitpassestoasectionofthepipeinwhichtheareaisreducedgraduallytoA2m2andthepressureisp2.Assumingnofrictionlosses,calculatethevelocitiesV1andV2ifthepressuredifference(p1-p2)ismeasured.
Solution:
InFig1.6,theflowdiagramisshownwithpressuretapstomeasurep1andp2.Fromthemass-balancecontinuityequation,forconstantρwhereρ1=ρ2=ρ,
FortheitemsintheBernoulliequation,forahorizontalpipe,
z1=z2=0
ThenBernoulliequationbecomes,aftersubstitutingforV2,
Rearranging,
PerformingthesamederivationbutintermsofV2,
1.7AliquidwhosecoefficientofviscosityisµflowsbelowthecriticalvelocityforlaminarflowinacircularpipeofdiameterdandwithmeanvelocityV.Showthatthepressurelossinalengthofpipeis.
Oilofviscosity0.05Pasflowsthroughapipeofdiameter0.1mwithaaveragevelocityof0.6m/s.Calculatethelossofpressureinalengthof120m.
Solution:
TheaveragevelocityVforacrosssectionisfoundbysummingupallthevelocitiesoverthecrosssectionanddividingbythecross-sectionalarea
1
Fromvelocityprofileequationforlaminarflow
2
substitutingequation2foruintoequation1andintegrating
3
rearrangingequation3gives
1.8.Inaverticalpipecarryingwater,pressuregaugesareinsertedatpointsAandBwherethepipediametersare0.15mand0.075mrespectively.ThepointBis2.5mbelowAandwhentheflowratedownthepipeis0.02
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