第三章 管式反應器

第三章管式反應器3.1平推流反應器thePlugFlowReactor平推流反應器（PFR）：反應器中的流動狀態(tài)是人們設想的一種理想流動，即在反應器內(nèi)具有嚴格均勻的速度分布，且軸向沒有任何混合。PFRischaracterizedbythefactthattheflowoffluidthroughthereactorisorderlywithnoelementoffluidovertakingormixingwithanyotherelementaheadorbehind.Actually,theremaybelateralmixingoffluidinaPFR;however,theremustbenomixingordiffusionalongtheflowpath.Thenecessaryandsufficientconditionforplugflowisfortheresidencetimeinthereactortobethesameforallelementsoffluid.平推流反應器特點：（1）在正常情況下，它是連續(xù)定態(tài)操作，故在反應器的各個截面上，過程參數(shù)（濃度、溫度等）不隨時間而變化；（2）反應器內(nèi)濃度、溫度等參數(shù)隨軸向位置變化，故反應速率隨軸向位置變化。（3）由于徑向具有嚴格均勻的速度分布，也就是在徑向不存在濃度分布。

PFR的基礎設計方程對PFR建立物料衡算式，就可以得到PFR的基礎設計方程式。在PFR中進行平推流動時，物料衡算式有如下特點：（1）由于流動處于穩(wěn)定狀態(tài)，各點濃度、溫度和反應速度均不隨時間而變化，故單元時間上t可任取；（2）

由于沿流動方向濃度、溫度和（－rA）都在改變，故應取單元體積△V＝dV；（3）穩(wěn)定狀態(tài)下，單元時間、單元體積內(nèi)反應物的積累量為零。Steady-statePlugFlowReactorInaplugflowreactorthecompositionofthefluidvariesfrompointtopointalongflowpath;consequently,thematerialbalanceforareactioncomponentmustbemadeforadifferentialelementofvolumedV.ThusforreactantA,thematerialbalancebecomesinput=output+disappearancebyreaction+accumulation=0ReferringtoFigleft,weseeforvolumedVthat:InputofA,moles/time=FAOutputofA,moles/time=FA+dFADisappearanceofAbyreaction,moles/time=(-rA)dVFig5.5onpage101LevIntroducingthesethreetermsinthematerialbalanceequationweobtainFA=(FA+dFA)+(-rA)dVNotingthatdFA=d[FA0(1-xA)]=-FA0dxAWeobtainonreplacementFA0dxA=(-rA)dVThis,then,istheequationwhichaccountsforAinthedifferentialsectionofreactorofvolumedV.Forthereactorasawholetheexpressionmustbeintegrated.NowFA0,thefeedrate,isconstant,but(-rA)iscertainlydependentontheconcentrationorconversionofmaterials.Groupingthetermsaccordingly,weobtainEquation3.2-5allowsthedeterminationofreactorsizeforagivenfeedrateandrequiredconversion.Asamoregeneralexpressionforplugreactors.Ifthefeedonwhichconversionisbased,subscript0,entersthepartiallyconverted,subscripti,andleavesataconversiondesignatedbysubscriptf,wehaveForthespecialcaseofconstant-densitysystemsxA=1-CA/CA0ordxA=-dCA/CA0Inwhichcasetheperformanceequationcanbeexpressedintermsofconcentrations,orTheseperformanceequationscanbewritteneitherintermsofconcentrationorconversion.Whateveritsform,theperformanceequationsinterrelatetherateofreaction,theextentofreaction,thereactorvolume,andthefeedrate,andifanyoneofthesequantitiesisunknownitcanbefoundfromtheotherthree.Fig.Belowdisplaystheseperformanceequationsandshowsthatthespace-timeneededforanyparticulardutycanalwaysbefoundbynumericalorgraphicalintegration.However,forcertainsimplekineticformsanalyticintegrationispossible–andconvenient.Someofthesimplerintegratedformsforplugflowareastable3.2-1.Fig5.6onpage103Bycomparingthebatchexpressionswiththeseplugflowexpressionswefind:Forsystemsofconstantdensity(constant-volumebatchandconstant-densityplugflow)theperformanceequationsareidentical,τforplugflowisequivalenttotforthebatchreactor,andtheequationscanbeusedinterchangeable.Forsystemsofchangingdensitythereisnodirectcorrespondencebetweenthebatchandplugflowequationsandthecorrectequationmustbeusedforeachparticularsituation.Inthiscasetheperformanceequationscannotbeusedinterchangeable.Example3.2-2PlugFlowReactorperformanceAhomogenousgasreactionA3Phasareportedrateat215℃,(-rA)=0.01CA1/2,(mol/liter·sec).Findthespace-timeneededfor80%conversionofa50%A-50%inertfeedtoaplugflowreactoroperatingat215℃and5atm(CA0=0.0625mol/liter).SolutionForthisstoichiometryandwithinerts,InwhichcasetheplugflowperformanceequationbecomesTheintegralcanbeevaluatedinanyoneofthreeways:graphically,numerically,oranalytically.LetusillustratethesemethodsGraphicalIntegration.Firstevaluatethefunctiontointegratedatselectedvalues(seetable)andplotthisfunction(seeFig).CountingsquaresorestimatingbyeyewefindNumericalIntegration.UsingSimpson’srule,applicabletoanevennumberofuniformlyspacedintervalsonthexAaxis,wefindAnalyticalIntegration.FromatableintegralsThemethodofintegrationrecommendeddependsonthesituation.Inthisproblemprobablythenumericalmethodisthequickestandsimplestandgivesagoodenoughanswerformostpurposes.Sowiththeintegralevaluated,Eq.BecomesThehomogenousgasdecompositionofphosphine4PH3(g)→P4+6H2Proceedsat649℃withthefirst-orderrate(-rPH3)=(10/hr)CPH3

Whatsizeofplugflowreactoroperatingat649℃and460kPacanproduce80%conversionoffeedconsistingof40molofpurephosphineperhour?Fig5.5onpage106Solution:

letA=PH3,R=P4,S=H2,thenthereactionbecomes4A→R+6Swith(-rA)=(10/hr)(CA)Thevolumeofplugflowreactorisgivenby3.2-5

EvaluatingtheindividualtermsinthisexpressiongivesFA0=40mol/hrCA0=PA0/RT=460000/(8.314×922)=60mol/m3

εA=(7-4)/4*1=0.75xA=0.8hencethevolumeofreactor3.2變溫平推流反應器

平推流反應器能量方程的推導，與間歇反應器相似。對于定常流動的PFR，系對微元反應體積作熱量衡算，而不是對微元反應時間。1）

基礎設計式：作PFR中微元段dl的物料衡算FtFt+dFtdlTmLFA0xAxA+dxATT+dT0(1)作熱量衡算FtCptT=(Ft+dFt)Cpt(T+dT)±（-rA）(-△HA)T0S·dl+UA（T-Tm）dl+0FtCpt≈(Ft+dFt)Cpt（2）（1）代入（2）式，得聯(lián)立（1）、（3）式得：這是非線性常微分方程組，可以用數(shù)值法求解，如改進歐拉法等。（3）0LTT0xA2）絕熱PFR設計式絕熱平推流反應器得熱量衡算式中，對外傳熱項為零，故（2）式變?yōu)椋菏街笑私茷槌?shù)，稱為變溫絕熱PFR的絕熱溫升于是：T-T0=λ(xA-xA0)，從PFR進口處狀態(tài)T0，xA開始，由各點xA可得相應的反應溫度T，結(jié)合起來計得（-rA）之值，再代入3）作圖法求解變溫PFR：xA1.00T0,xA0T0xAfVP/FA0(-rA)斜率=1/λ3.3組合反應器

Multiple-ReactorSystem

3.3.1(a)PlugFlowReactorsinSeriesand/orinParallelConsiderNplugflowreactorconnectedinseries,andletx1,x2,…,xNbethefractionalconversionofcomponentAleavingreactor1,2,…,N.BasingthematerialbalanceonthefeedrateofAtothefirstreactor,wefindfortheithreactorfromEq.3.2-6

OrfortheNreactorsinseriesHence,NplugflowreactorsinserieswithatotalvolumeVgivesthesameconversionasasingleplugflowreactorofvolumeFortheoptimumhookup(連接)ofplugflowreactorsconnectedinparallelorinanyparallel-seriescombination,wecantreatthewholesystemasasingleplugflowreactorofvolumeequaltothetotalvolumeoftheindividualunitsifthefeedisdistributedinsuchamannerthatfluidstreamsthatmeethavethesamecomposition.Thus,forreactorsinparallelV/Forτmustbethesameforeachparallelline.Anyotherwayoffeedingislessefficient.Example:ThereactorsetupshowninFig.Consistsofthreeplugflowreactorsintwoparallelbranches.BranchDhasareactorofvolume50litersfollowedbyareactorofvolume30liters.BranchEhasareactorofvolume40liters.WhatfractionofthefeedshouldgotobranchD?Solution:BranchDconsistsoftworeactorsinseries;hence,itmaybeconsideredtobeasinglereactorofvolumeVD=50+30=80litersNowforreactorsinparallelV/Fmustbeidenticaliftheconversionistobethesameineachbranch.Therefore,Therefore,two-thirdsofthefeedmustbefedtobranchD.3.4循環(huán)反應器RecycleReactorIncertainsituationsitisfoundtobeadvantageoustodividetheproductstreamfromaplugflowreactorandreturnaportionofittotheentranceofthereactor.LettherecycleratioβbedefinedasThisrecycleratiocanbemadetovaryfromzerotoinfinity.Reflectionsuggeststhatastherecycleratioisraised