Heat Chap09094.docx
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Heat Chap09094.docx
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HeatChap09094
ReviewProblems
9-94EAsmallcylindricalresistormountedonthelowerpartofaverticalcircuitboard.Theapproximatesurfacetemperatureoftheresistoristobedetermined.
Assumptions1Steadyoperatingconditionsexist.2Airisanidealgaswithconstantproperties.3Thelocalatmosphericpressureis1atm.4Radiationeffectsarenegligible.5Heattransferthroughtheconnectingwiresisnegligible.
PropertiesThepropertiesofairat1atmandtheanticipatedfilmtemperatureof(Ts+T)/2=(220+120)/2=170Fare(TableA-15E)
AnalysisThesolutionofthisproblemrequiresatrial-and-errorapproachsincethedeterminationoftheRayleighnumberandthustheNusseltnumberdependsonthesurfacetemperaturewhichisunknown.Westartthesolutionprocessby“guessing”thesurfacetemperaturetobe220Ffortheevaluationofthepropertiesandh.Wewillchecktheaccuracyofthisguesslaterandrepeatthecalculationsifnecessary.Thecharacteristiclengthinthiscaseisthediameterofresistor,
Then,
and
whichissufficientlyclosetotheassumedtemperaturefortheevaluationofproperties.Therefore,thereisnoneedtorepeatcalculations.
9-95Anicechestfilledwithiceat0Cisexposedtoambientair.Thetimeitwilltakefortheiceinthechesttomeltcompletelyistobedeterminedfornaturalandforcedconvectioncases.
Assumptions1Steadyoperatingconditionsexist.2Airisanidealgaswithconstantproperties.3Heattransferfromthebaseoftheicechestisdisregarded.4Radiationeffectsarenegligible.5Heattransfercoefficientisthesameforallsurfacesconsidered.6Thelocalatmosphericpressureis1atm.
PropertiesThepropertiesofairat1atmandtheanticipatedfilmtemperatureof(Ts+T)/2=(15+20)/2=17.5Care(TableA-15)
AnalysisThesolutionofthisproblemrequiresatrial-and-errorapproachsincethedeterminationoftheRayleighnumberandthustheNusseltnumberdependsonthesurfacetemperaturewhichisunknown.Westartthesolutionprocessby“guessing”thesurfacetemperaturetobe15Cfortheevaluationofthepropertiesandh.Wewillchecktheaccuracyofthisguesslaterandrepeatthecalculationsifnecessary.Thecharacteristiclengthforthesidesurfacesistheheightofthechest,Lc=L=0.3mThen,
Theheattransfercoefficientatthetopsurfacecanbedeterminedsimilarly.However,thetopsurfaceconstitutesonlyaboutone-fourthoftheheattransferarea,andthuswecanusetheheattransfercoefficientforthesidesurfacesforthetopsurfacealsoforsimplicity.Theheattransfersurfaceareais
Thentherateofheattransferbecomes
TheoutersurfacetemperatureoftheicechestisdeterminedfromNewton’slawofcoolingtobe
whichisalmostidenticaltotheassumedvalueof15Cusedintheevaluationofpropertiesandh.Therefore,thereisnoneedtorepeatthecalculations.Thentherateatwhichtheicewillmeltbecomes
Therefore,themeltingoftheiceinthechestcompletelywilltake
(b)Thetemperaturedropacrossthestyrofoamwillbemuchgreaterinthiscasethanthatacrossthermalboundarylayeronthesurface.Thusweassumeoutersurfacetemperatureofthestyrofoamtobe19
.Radiationheattransferwillbeneglected.Thepropertiesofairat1atmandthefilmtemperatureof(Ts+T)/2=(19+20)/2=19.5Care(TableA-15)
Thecharacteristiclengthinthiscaseisthewidthofthechest,Lc=W=0.4m.Then,
whichislessthancriticalReynoldsnumber(
).Thereforetheflowislaminar,andtheNusseltnumberisdeterminedfrom
Thentherateofheattransferbecomes
TheoutersurfacetemperatureoftheicechestisdeterminedfromNewton’slawofcoolingtobe
whichisalmostidenticaltotheassumedvalueof19Cusedintheevaluationofpropertiesandh.Therefore,thereisnoneedtorepeatthecalculations.Thentherateatwhichtheicewillmeltbecomes
Therefore,themeltingoftheiceinthechestcompletelywilltake
9-96Anelectronicboxiscooledinternallybyafanblowingairintotheenclosure.Thefractionoftheheatlostfromtheoutersurfacesoftheelectronicboxistobedetermined.
Assumptions1Steadyoperatingconditionsexist.2Airisanidealgaswithconstantproperties.3Heattransferfromthebasesurfaceisdisregarded.4Thepressureofairinsidetheenclosureis1atm.
PropertiesThepropertiesofairat1atmandthefilmtemperatureof(Ts+T)/2=(32+15)/2=28.5Care(TableA-15)
AnalysisHeatlossfromthehorizontaltopsurface:
Thecharacteristiclengthinthiscaseis
.Then,
and
Heatlossfromverticalsidesurfaces:
ThecharacteristiclengthinthiscaseistheheightoftheboxLc=L=0.15m.Then,
and
Theradiationheatlossis
Thenthefractionoftheheatlossfromtheoutersurfacesoftheboxisdeterminedtobe
9-97Asphericaltankmadeofstainlesssteelisusedtostoreicedwater.Therateofheattransfertotheicedwaterandtheamountoficethatmeltsduringa24-hperiodaretobedetermined.
Assumptions1Steadyoperatingconditionsexist.2Airisanidealgaswithconstantproperties.3Thermalresistanceofthetankisnegligible.4Thelocalatmosphericpressureis1atm.
PropertiesThepropertiesofairat1atmandthefilmtemperatureof(Ts+T)/2=(0+20)/2=10Care(TableA-15)
Analysis(a)ThecharacteristiclengthinthiscaseisLc=Do=6.03m.Then,
and
Heattransferbyradiationandthetotalrateofheattransferare
(b)Thetotalamountofheattransferduringa24-hourperiodis
Thentheamountoficethatmeltsduringthisperiodbecomes
9-98Adouble-panewindowconsistingoftwolayersofglassseparatedbyanairspaceisconsidered.Therateofheattransferthroughthewindowandthetemperatureofitsinnersurfacearetobedetermined.
Assumptions1Steadyoperatingconditionsexist.2Airisanidealgaswithconstantproperties.3Radiationeffectsarenegligible.4Thepressureofairinsidetheenclosureis1atm.
PropertiesWeexpecttheaveragetemperatureoftheairgaptoberoughlytheaverageoftheindoorandoutdoortemperatures,andevaluateThepropertiesofairat1atmandtheaveragetemperatureof(T1+T2)/2=(20+0)/2=10Care(TableA-15)
AnalysisWe“guess”thetemperaturedifferenceacrosstheairgaptobe15C=15KforuseintheRarelation.Thecharacteristiclengthinthiscaseistheairgapthickness,Lc=L=0.03m.Then,
ThentheNusseltnumberandtheheattransfercoefficientaredeterminedtobe
Thentherateofheattransferthroughthisdoublepanewindowisdeterminedtobe
Check:
Thetemperaturedropacrosstheairgapisdeterminedfrom
whichisveryclosetotheassumedvalueof15CusedintheevaluationoftheRanumber.
9-99Anelectricresistancespaceheaterfilledwithoilisplacedagainstawall.Thepowerratingoftheheaterandthetimeitwilltakefortheheatertoreachsteadyoperationwhenitisfirstturnedonaretobedetermined.
Assumptions1Steadyoperatingconditionsexist.2Airisanidealgaswithconstantproperties.3Heattransferfromtheback,bottom,andtopsurfacesaredisregarded.4Thelocalatmosphericpressureis1atm.
PropertiesThepropertiesofairat1atmandthefilmtemperatureof(Ts+T)/2=(45+25)/2=35Care(TableA-15)
AnalysisHeattransferfromthetopandbottomsurfacesaresaidtobenegligible,andthustheheattransferareainthiscaseconsistsofthethreeexposedsidesurfaces.Thecharacteristiclengthistheheightofthebox,Lc=L=0.5m.Then,
and
Theradiationheatlossis
Thenthetotalrateofheattransfer,thusthepowerratingoftheheaterbecomes
Thespecificheatoftheoilattheaveragetemperatureoftheoilis1943J/kg.C.Thentheamountofheattransferneededtoraisethetemperatureoftheoiltothesteadyoperatingtemperatureandthetimeittakesbecome
whichisnotpractical.Therefore,thesurfacetemperatureoftheheatermustbeallowedtobehigherthan45C.
9-100Ahorizontalskylightmadeofasinglelayerofglassontheroofofahouseisconsidered.Therateofheatlossthroughtheskylightistobedeterminedfortwocases.
Assumptions1Steadyoperatingconditionsexist.2Airisanidealgaswithconstantproperties.3Thelocalatmosphericpressureis1atm.
PropertiesThepropertiesofairat1atmandtheanticipatedfilmtemperatureof(Ts+T)/2=(-4-10)/2=-7Care(TableA-15)
AnalysisWeassumeradiationheattransferinsidethehousetobenegligible.Westartthecalculationsby“guessing”theglasstemperaturetobe4Cfortheevaluationofthepropertiesandh.Wewillchecktheaccuracyofthisguesslaterandrepeatthecalculationsifnecessary.Thecharacteristiclengthinthiscaseisdeterminedfrom
.Then,
Usingtheassumedvalueofglasstemperature,theradiationheattransfercoefficientisdeterminedtobe
Thenthecombinedconvectionandradiationheattransfercoefficientoutsidebecomes
Againwetaketheglasstemperaturetobe-4Cfortheevaluationofthepropertiesandhfortheinnersurfaceoftheskylight.Thepropertiesofairat1atmandthefilmtemperatureofTf=(-4+20)/2=8Care(TableA-15)
Thecharacteristiclengthinthiscaseisalso0.357m.Then,
Usingthethermalresistancenetwork,therateofheatlossthroughtheskylightisdeterminedtobe
Usingthesameheattransfercoefficientsforsimplicity,therateofheatlossthroughtheroofinthecaseofR-5.34constructionisdeterminedtobe
Therefore,ahouseloses115/5.3621timesmoreheatthroughtheskylightsthanitdoesthroughaninsulatedwall
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