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1、化工作业（Chemicaloperations）HeattransferinthefourchapterThefurnacewallof1.flatfireplaceconsistsofthreekindsofmaterials,anditsthicknessandheatconductivityseriesareinthetable:iftheinnersurfaceofrefractorybricklayerThetemperatureofT1is1150,thetemperatureoftheoutersurfaceofthes
2、teelplateis30T4,andtheheatlossthroughthefurnacearmis300W/mTwoTestmeterCalculatingheatfluxofheatconduction.Ifthecalculatedresultsdonotagreewiththeactualheatloss,thecauseandadditionalthermalresistancearecalculated.Materialthickness,mmthermalconductivity,w/(M.C)1(inner)ref
3、ractorybrick2001.072insulatingbrick1000.143steel645Solution:thedatagivenintheextractiontable:Firebrick=B1=0.2m,lambda1=1.07w/(M.C)Insulationlayer:B2=0.1M,2=0.14w/(lambdam+C)Steel:B3=0.006m,3=45w/(lambdam+C)AccordingtotheformulaofmultilayerflatwallheatconductionrateQ=(T1
4、-Tn)/sigma(bi/SlambdaI)andq=Q/SGetq=(T1-Tn)/sigma(bi/lambdaI)=1242w/mThreeThisisinlinewiththeactualheatlossofQ'=300w/mhasacertaingap,sotheremaybesomeadditionalThermalresistance,theadditionalthermalresistanceisR'Q'=(T1-Tn)/sigma(bi/Sxi)+R']=300R'=m+2.83C/WTheinnerlayerof
5、the2.combustionfurnaceis460mmthickrefractorybrick,andtheouterlayeris230mminsulatingbrick.TheinternalsurfacetemperatureT1ofthefurnaceis1400C,thesurfacetemperatureis100DEGT3,theinterfacetemperatureheatfluxtestforheatconductionbetweentwobricks.ThefurnacecontactiswellknownT
6、hethermalconductivityofrefractorybrickis1=0.9+0.0007t,andthethermalconductivityofinsulatingbrickis2=0.3+0.0003t.TscoreintwoformulaDon'ttaketheaveragetemperatureforeachlayerofmaterial,unitC,lambdaunitisW/(M+C).Solution:theinterfacetemperaturebetweenthetwobricksisT2,T1=14
7、00,T3=100A1=thermalconductivityofrefractorybricksof0.9+0.0007(T1+T2)/2=0.9+0.0007+(1400+T2)/2=1.39+0.00035T2Thethermalconductivityofadiabaticrotationislambda2=0.3+0.0003(T3+T2)/2=0.315+0.00015T2(T1-t2)/(b1/lambda1)=(T2-t3)/(b2/lambda2)*0.00065t2Two+1.5t2-2009=0Theinterf
8、acetemperatureisT2=949A1=thermalconductivityofeachlayer(1.722w/*m+C)A2=0.457w/(M+C)Accordingtothemultilayerfla
