Heat Chap12088.docx

Heat Chap12088.docx

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HeatChap12088

ReviewProblems

12-88Thetemperatureofairinaductismeasuredbyathermocouple.Theradiationeffectonthetemperaturemeasurementistobequantified,andtheactualairtemperatureistobedetermined.

AssumptionsThesurfacesareopaque,diffuse,andgray.

PropertiesTheemissivityofthermocoupleisgiventobe=0.6.

AnalysisTheactualtemperatureoftheaircanbedeterminedfrom

12-89Thetemperatureofhotgasesinaductismeasuredbyathermocouple.Theactualtemperatureofthegasistobedetermined,andcomparedwiththatwithoutaradiationshield.

AssumptionsThesurfacesareopaque,diffuse,andgray.

PropertiesTheemissivityofthethermocoupleisgiventobe=0.7.

AnalysisAssumingtheareaoftheshieldtobeveryclosetothesensorofthethermometer,theradiationheattransferfromthesensorisdeterminedfrom

Thentheactualtemperatureofthegascanbedeterminedfromaheattransferbalancetobe

Withouttheshieldthetemperatureofthegaswouldbe

12-90EAsealedelectronicboxisplacedinavacuumchamber.Thehighesttemperatureatwhichthesurroundingsurfacesmustbekeptifthisboxiscooledbyradiationaloneistobedetermined.

Assumptions1Steadyoperatingconditionsexist2Thesurfacesareopaque,diffuse,andgray.3Convectionheattransferisnotconsidered.4Heattransferfromthebottomsurfaceoftheboxisnegligible.

PropertiesTheemissivityoftheoutersurfaceoftheboxis=0.95.

AnalysisThetotalsurfaceareais

Thenthetemperatureofthesurroundingsurfacesisdeterminedtobe

12-91Adouble-walledsphericaltankisusedtostoreicedwater.Theairspacebetweenthetwowallsisevacuated.Therateofheattransfertotheicedwaterandtheamountoficethatmeltsa24-hperiodaretobedetermined.

Assumptions1Steadyoperatingconditionsexist2Thesurfacesareopaque,diffuse,andgray.

PropertiesTheemissivitiesofbothsurfacesaregiventobe1=2=0.15.

Analysis（a）Assumingtheconductionresistancesofthewallstobenegligible,therateofheattransfertotheicedwaterinthetankisdeterminedtobe

（b）Theamountofheattransferduringa24-hourperiodis

Theamountoficethatmeltsduringthisperiodthenbecomes

12-92Twoconcentricsphereswhicharemaintainedatuniformtemperaturesareseparatedbyairat1atmpressure.Therateofheattransferbetweenthetwospheresbynaturalconvectionandradiationistobedetermined.

Assumptions1Steadyoperatingconditionsexist2Thesurfacesareopaque,diffuse,andgray.3Airisanidealgaswithconstantproperties.

PropertiesTheemissivitiesofthesurfacesaregiventobe1=2=0.5.Thepropertiesofairat1atmandtheaveragetemperatureof（T1+T2）/2=（350+275）/2=312.5K=39.5Care（TableA-15）

Analysis（a）NotingthatDi=D1andDo=D2,thecharacteristiclengthis

Then

Theeffectivethermalconductivityis

Thentherateofheattransferbetweenthespheresbecomes

（b）Therateofheattransferbyradiationisdeterminedfrom

12-93Asolarcollectorisconsidered.Theabsorberplateandtheglasscoveraremaintainedatuniformtemperatures,andareseparatedbyair.Therateofheatlossfromtheabsorberplatebynaturalconvectionandradiationistobedetermined.

Assumptions1Steadyoperatingconditionsexist2Thesurfacesareopaque,diffuse,andgray.3Airisanidealgaswithconstantproperties.

PropertiesTheemissivitiesofsurfacesaregiventobe1=0.9forglassand2=0.8fortheabsorberplate.Thepropertiesofairat1atmandtheaveragetemperatureof（T1+T2）/2=（80+32）/2=56Care（TableA-15）

AnalysisFor

wehavehorizontalrectangularenclosure.ThecharacteristiclengthinthiscaseisthedistancebetweenthetwoglassesLc=L=0.03mThen,

Neglectingtheendeffects,therateofheattransferbyradiationisdeterminedfrom

DiscussionTheratesofheatlossbynaturalconvectionforthehorizontalandverticalcaseswouldbeasfollows（NotethattheRanumberremainsthesame）:

Horizontal:

Vertical:

12-94EThecirculatingpumpofasolarcollectorthatconsistsofahorizontaltubeanditsglasscoverfails.Theequilibriumtemperatureofthetubeistobedetermined.

Assumptions1Steadyoperatingconditionsexist.2Thetubeanditscoverareisothermal.3Airisanidealgas.4Thesurfacesareopaque,diffuse,andgrayforinfraredradiation.5Theglasscoveristransparenttosolarradiation.

PropertiesThepropertiesofairshouldbeevaluatedattheaveragetemperature.Butwedonotknowtheexittemperatureoftheairintheduct,andthuswecannotdeterminethebulkfluidandglasscovertemperaturesatthispoint,andthuswecannotevaluatetheaveragetemperatures.Therefore,wewillassumetheglasstemperaturetobe85F,andusepropertiesatananticipatedaveragetemperatureof（75+85）/2=80°F（TableA-15E）,

AnalysisWehaveahorizontalcylindricalenclosurefilledwithairat0.5atmpressure.Theprobleminvolvesheattransferfromthealuminumtubetotheglasscoverandfromtheoutersurfaceoftheglasscovertothesurroundingambientair.Whensteadyoperationisreached,thesetwoheattransferratesmustequaltherateofheatgain.Thatis,

（perfootoftube）

Theheattransfersurfaceareaoftheglasscoveris

（perfootoftube）

TodeterminetheRayleighnumber,weneedtoknowthesurfacetemperatureoftheglass,whichisnotavailable.Therefore,solutionwillrequireatrial-and-errorapproach.Assumingtheglasscovertemperaturetobe85°F,theRayleighnumber,theNusseltnumber,theconvectionheattransfercoefficient,andtherateofnaturalconvectionheattransferfromtheglasscovertotheambientairaredeterminedtobe

Also,

Thenthetotalrateofheatlossfromtheglasscoverbecomes

whichismorethan30Btu/h.Therefore,theassumedtemperatureof85°Ffortheglasscoverishigh.Repeatingthecalculationswithlowertemperatures（includingtheevaluationofproperties）,theglasscovertemperaturecorrespondingto30Btu/hisdeterminedtobe81.5°F.

Thetemperatureofthealuminumtubeisdeterminedinasimilarmannerusingthenaturalconvectionandradiationrelationsfortwohorizontalconcentriccylinders.Thecharacteristiclengthinthiscaseisthedistancebetweenthetwocylinders,whichis

Also,

（perfootoftube）

Westartthecalculationsbyassumingthetubetemperaturetobe118.5°F,andthusanaveragetemperatureof（81.5+118.5）/2=100F=640R.Usingpropertiesat100F,

Theeffectivethermalconductivityis

Thentherateofheattransferbetweenthecylindersbyconvectionbecomes

Also,

Thenthetotalrateofheatlossfromtheglasscoverbecomes

whichismorethan30Btu/h.Therefore,theassumedtemperatureof118.5°Fforthetubeishigh.Bytryingothervalues,thetubetemperaturecorrespondingto30Btu/hisdeterminedtobe113.2°F.Therefore,thetubewillreachanequilibriumtemperatureof113.2°Fwhenthepumpfails.

12-95Adouble-panewindowconsistsoftwosheetsofglassseparatedbyanairspace.Theratesofheattransferthroughthewindowbynaturalconvectionandradiationaretobedetermined.

Assumptions1Steadyoperatingconditionsexist2Thesurfacesareopaque,diffuse,andgray.3Airisanidealgaswithconstantspecificheats.4Heattransferthroughthewindowisone-dimensionalandtheedgeeffectsarenegligible.

PropertiesTheemissivitiesofglasssurfacesaregiventobe1=2=0.9.Thepropertiesofairat0.3atmandtheaveragetemperatureof（T1+T2）/2=（15+5）/2=10Care（TableA-15）

AnalysisThecharacteristiclengthinthiscaseisthedistancebetweentheglasses,

Thentherateofheattransferbynaturalconvectionbecomes

Therateofheattransferbyradiationisdeterminedfrom

Thentherateoftotalheattransferbecomes

DiscussionNotethatheattransferthroughthewindowismostlybyradiation.

12-96Asimplesolarcollectorisbuiltbyplacingaclearplastictubearoundagardenhose.Therateofheatlossfromthewaterinthehosebynaturalconvectionandradiationistobedetermined.

Assumptions1Steadyoperatingconditionsexist2Thesurfacesareopaque,diffuse,andgray.3Airisanidealgaswithconstantspecificheats.

PropertiesTheemissivitiesofsurfacesaregiventobe1=2=0.9.Thepropertiesofairareat1atmandthefilmtemperatureof（Ts+T）/2=（40+25）/2=32.5Care（TableA-15）

AnalysisUndersteadyconditions,theheattransferratefromthewaterinthehoseequalstotherateofheatlossfromtheclearplastictubetothesurroundingsbynaturalconvectionandradiation.Thecharacteristiclengthinthiscaseisthediameteroftheplastictube,

.

Thentherateofheattransferfromtheoutersurfacebynaturalconvectionbecomes

Therateofheattransferbyradiationfromtheoutersurfaceisdeterminedfrom

Finally,

DiscussionNotethatheattransferismostlybyradiation.

12-97Asolarcollectorconsistsofahorizontalcoppertubeenclosedinaconcentricthinglasstube.Theannularspacebetweenthecopperandtheglasstubesisfilledwithairat1atm.Therateofheatlossfromthecollectorbynaturalconvectionandradiationistobedetermined.

Assumptions1Steadyoperatingconditionsexist2Thesurfacesareopaque,diffuse,andgray.3Airisanidealgaswithconstantspecificheats.

PropertiesTheemissivitiesofsurfacesaregiventobe1=0.85forthetubesurfaceand2=0.9forglasscover.Thepropertiesofairat1atmandtheaveragetemperatureof（T1+T2）/2=（60+40）/2=50Care（TableA-15）