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1、ElevenChapterRadiationExchangeBetweenSurfaces13.1TheViewFactorFigure11.1ViewfactorassociatedwithradiationexchangebetweenelementalsurfaceofareadAianddAj.1.TheViewFactorIntegralViewfactorFijisdefinedasthefractionoftheradiationleavingsurfacei,whichisinterceptedbysurfacej.Assumingthat
2、surfaceiemitsandreflectsdiffusely,Thetotalrateatwhichradiationleavessurfaceiandisinterceptedbyjmaythenbeobtainedbyintegratingoverthetwosurface.ItisassumedthatradiosityJjisuniformoverthesurfaceAi.ThatisTheviewfactorasthefractionoftheradiationthatleavesAiandisinterceptedbyAj,Similar
3、y,theviewfactorFjiisdefinedasthefractionoftheradiationthatleavesAjandisinterceptedbyAi.Thetwoequationsmaybeusedtodeterminetheviewfactorassociatedwithanytwosurfacesthatarediffuseemittersandrefletorsandhaveuniformradiosity.2.ViewFactorRelationsReciprocityrelation:Itisuesfulindetermi
4、ningoneviewfactorfromknowledgeoftheother.Summationruleenclosure:Viewfactorrelationpertainstothesurfaceofanenclosure.Ifthesurfaceisconcave,itseesitselfandFijisnonzero.However,foraplaneorconvexsurface,Fij=0.A1Figure11.4Radiationtransferbetweentwosurfacethatmaybeapproximatedasblackbo
5、diesSinceradiosityequalsemissivepowerforablacksurfaceThenetradiativeexchangebetweenthetwosurfacesmaythenbedefinedas11.2BlackbodyRadiationExchangeExampleAfurnacecavity,whichisintheformofacylinderof75mmdiameterand150mmlength,isopenatoneendtosurroundingsthatareat27℃.thesidesandbottom
6、maybeapproximatedasblackbodies,areheatedelectrically,arewellinsulated,andaremaintainedattemperaturesof1350℃and1650℃,respectively.Side,T1InsulationBottom,T2HeaterwireLDHowmuchpowerisrequiredtomaintainthefurnaceconditions?SolutionKnown:surfacetemperaturesofcylindricalfurnace.Find:Po
7、werrequiredtomaintainprescribedtemperatures.Schematic:A1,T1=1350℃A2,T2=1650℃L=0.15mD=0.057mA3,T3=TsurTsur=300KAssumptions:1.Interiorsurfacesbehaveasblackbodies.2.Heattransferbyconvectionisnegligible.3.Outersurfaceoffurnaceisadiabatic.4.Thesurroundingmaybetreatedbyapproximatingthes
8、urfaceasablackbody.Analysis:Subje
