PROBLEM1.1 KNOWN:Thermalconductivity,thicknessandtemperaturedifferenceacrossasheetofrigid extrudedinsulation.
FIND:(a)Theheatfluxthrougha2m×2msheetoftheinsulation,and(b)Theheatrate throughthesheet.
SCHEMATIC:
k=0.029 ⋅ W mK x L=20mm
ASSUMPTIONS:(1)One-dimensionalconductioninthex-direction,(2)Steady-state conditions,(3)Constantproperties.
ANALYSIS:FromEquation1.2theheatfluxis
12 x qdTT-T =-k=kdxL
Solving, " x W10K q=0.029×mK0.02m ⋅ x2 W q=14.5 m ′′ <
Theheatrateis 2 xx2 W q=qA=14.5×4m=58W m
<
COMMENTS:(1)Besuretokeepinmindtheimportantdistinctionbetweentheheatflux (W/m2)andtheheatrate(W).(2)Thedirectionofheatflowisfromhottocold.(3)Notethat atemperaturedifferencemaybeexpressedinkelvinsordegreesCelsius.
PROBLEM1.2 KNOWN:Thicknessandthermalconductivityofawall.Heatfluxappliedtoonefaceand temperaturesofbothsurfaces.
FIND:Whethersteady-stateconditionsexist.
SCHEMATIC:
L=10mm
q”=20W/m2
k=12W/m·K T1=50°C T2=30°C
ASSUMPTIONS:(1)One-dimensionalconduction,(2)Constantproperties,(3)Nointernalenergy generation.
ANALYSIS:Understeady-stateconditionsanenergybalanceonthecontrolvolumeshownis
2inoutcond12()/12W/mK(50C30C)/0.01m24,000W/m qqqkTTL ′′′′′′===−=⋅°−°=
Sincetheheatfluxinattheleftfaceisonly20W/m2,theconditionsarenotsteadystate.<
COMMENTS:Ifthesameheatfluxismaintaineduntilsteady-stateconditionsarereached,the steady-statetemperaturedifferenceacrossthewallwillbe
ΔT=/202W/m0.01m/12W/mK0.0167KqLk′′=×⋅= whichismuchsmallerthanthespecifiedtemperaturedifferenceof20°C.
PROBLEM1.3 KNOWN:Innersurfacetemperatureandthermalconductivityofaconcretewall.
FIND:Heatlossbyconductionthroughthewallasafunctionofoutersurfacetemperaturesrangingfrom -15to38°C.
SCHEMATIC:
ASSUMPTIONS:(1)One-dimensionalconductioninthex-direction,(2)Steady-stateconditions,(3) Constantproperties.
ANALYSIS:FromFourier’slaw,ifx q ′′andkareeachconstantitisevidentthatthegradient, x dTdxqk ′′ =−,isaconstant,andhencethetemperaturedistributionislinear.Theheatfluxmustbe constantunderone-dimensional,steady-stateconditions;andkisapproximatelyconstantifitdepends onlyweaklyontemperature.TheheatfluxandheatratewhentheoutsidewalltemperatureisT2=-15°C are
CombiningEqs.(1)and(2),theheatrateqxcanbedeterminedfortherangeofoutersurfacetemperature, -15≤T2≤38°C,withdifferentwallthermalconductivities,k.
Fortheconcretewall,k=1W/mK,theheatlossvarieslinearlyfrom+2667Wto-867Wandiszero whentheinsideandoutersurfacetemperaturesarethesame.Themagnitudeoftheheatrateincreases withincreasingthermalconductivity.
COMMENTS:Withoutsteady-stateconditionsandconstantk,thetemperaturedistributioninaplane wallwouldnotbelinear.
PROBLEM1.4 KNOWN:Dimensions,thermalconductivityandsurfacetemperaturesofaconcreteslab.Efficiency ofgasfurnaceandcostofnaturalgas.
FIND:Dailycostofheatloss.
SCHEMATIC:
ASSUMPTIONS:(1)Steadystate,(2)One-dimensionalconduction,(3)Constantproperties.
ANALYSIS:Therateofheatlossbyconductionthroughtheslabis
Thedailycostofnaturalgasthatmustbecombustedtocompensatefortheheatlossis
COMMENTS:Thelosscouldbereducedbyinstallingafloorcoveringwithalayerofinsulation betweenitandtheconcrete.
PROBLEM1.5 KNOWN:Thermalconductivityandthicknessofawall.Heatfluxthroughwall.Steady-state conditions.
FIND:ValueoftemperaturegradientinK/mandin°C/m.
SCHEMATIC: L=20mm k=2.3W/m·K q ” x=10W/m2
ASSUMPTIONS:(1)One-dimensionalconduction,(2)Constantproperties.
ANALYSIS:Understeady-stateconditions,
SincetheKunitshererepresentatemperaturedifference,andsincethetemperaturedifferenceisthe sameinKand°Cunits,thetemperaturegradientvalueisthesameineitherunits.
COMMENTS:Anegativevalueoftemperaturegradientmeansthattemperatureisdecreasingwith increasingx,correspondingtoapositiveheatfluxinthex-direction.
PROBLEM1.6 KNOWN:Heatfluxandsurfacetemperaturesassociatedwithawoodslabofprescribed thickness.
FIND:Thermalconductivity,k,ofthewood.
SCHEMATIC:
ASSUMPTIONS:(1)One-dimensionalconductioninthex-direction,(2)Steady-state conditions,(3)Constantproperties.
ANALYSIS:Subjecttotheforegoingassumptions,thethermalconductivitymaybe determinedfromFourier’slaw,Eq.1.2.Rearranging,
COMMENTS:Notethatthe°CorKtemperatureunitsmaybeusedinterchangeablywhen evaluatingatemperaturedifference.
PROBLEM1.7 KNOWN:Innerandoutersurfacetemperaturesofaglasswindowofprescribeddimensions.
FIND:Heatlossthroughwindow.
SCHEMATIC:
ASSUMPTIONS:(1)One-dimensionalconductioninthex-direction,(2)Steady-state conditions,(3)Constantproperties.
ANALYSIS:Subjecttotheforegoingconditionstheheatfluxmaybecomputedfrom Fourier’slaw,Eq.1.2.
q2800W/m. 12 x x 2 x ′′ = ′′ = ⋅ ′′ = o
W15-5C
q1.4mK0.005m
Sincetheheatfluxisuniformoverthesurface,theheatloss(rate)is
q=qxA
q=2800W/m23m2 ′′× ×
q=8400W. <
COMMENTS:Alineartemperaturedistributionexistsintheglassfortheprescribed conditions.
PROBLEM1.8 KNOWN:Netpoweroutput,averagecompressorandturbinetemperatures,shaftdimensionsand thermalconductivity.
FIND:(a)Comparisonoftheconductionratethroughtheshafttothepredictednetpoweroutputof thedevice,(b)Plotoftheratiooftheshaftconductionheatratetotheanticipatednetpoweroutputof thedeviceovertherange0.005m≤L≤1mandfeasibilityofaL=0.005mdevice.
SCHEMATIC:
ASSUMPTIONS:(1)Steady-stateconditions,(2)Constantproperties,(3)Netpoweroutputis proportionaltothevolumeofthegasturbine.
PROPERTIES:Shaft(given):k=40W/mK.
ANALYSIS:(a)TheconductionthroughtheshaftmaybeevaluatedusingFourier’slaw,yielding
Theratiooftheconductionheatratetothenetpoweroutputis
(b)ThevolumeoftheturbineisproportionaltoL3.DesignatingLa=1m,da=70mmandPaasthe shaftlength,shaftdiameter,andnetpoweroutput,respectively,inpart(a),
andtheratiooftheconductionheatratetothenetpoweroutputis
PROBLEM1.8(Cont.) Theratiooftheshaftconductiontonetpowerisshownbelow.AtL=0.005m=5mm,theshaft conductiontonetpoweroutputratiois0.74.Theconceptoftheverysmallturbineisnotfeasiblesince itwillbeunlikelythatthelargetemperaturedifferencebetweenthecompressorandturbinecanbe maintained. <
Ratioofshaftconductiontonetpower
COMMENTS:(1)Thethermodynamicsanalysisdoesnotaccountforheattransfereffectsandis thereforemeaningfulonlywhenheattransfercanbesafelyignored,asisthecasefortheshaftinpart (a).(2)Successfulminiaturizationofthermaldevicesisoftenhinderedbyheattransfereffectsthat mustbeovercomewithinnovativedesign.
PROBLEM1.9 KNOWN:Width,height,thicknessandthermalconductivityofasinglepanewindowand theairspaceofadoublepanewindow.Representativewintersurfacetemperaturesofsingle paneandairspace.
FIND:Heatlossthroughsingleanddoublepanewindows.
SCHEMATIC:
ASSUMPTIONS:(1)One-dimensionalconductionthroughglassorair,(2)Steady-state conditions,(3)Enclosedairofdoublepanewindowisstagnant(negligiblebuoyancyinduced motion).
ANALYSIS:FromFourier’slaw,theheatlossesare
COMMENTS:Lossesassociatedwithasinglepaneareunacceptableandwouldremain excessive,evenifthethicknessoftheglassweredoubledtomatchthatoftheairspace.The principaladvantageofthedoublepaneconstructionresideswiththelowthermalconductivity ofair(~60timessmallerthanthatofglass).Forafixedambientoutsideairtemperature,use ofthedoublepaneconstructionwouldalsoincreasethesurfacetemperatureoftheglass exposedtotheroom(inside)air.
PROBLEM1.10 KNOWN:Dimensionsoffreezercompartment.Innerandoutersurfacetemperatures.
FIND:Thicknessofstyrofoaminsulationneededtomaintainheatloadbelowprescribed value.
SCHEMATIC:
ASSUMPTIONS:(1)Perfectlyinsulatedbottom,(2)One-dimensionalconductionthrough5 wallsofareaA=4m2,(3)Steady-stateconditions,(4)Constantproperties.
ANALYSIS:UsingFourier’slaw,Eq.1.2,theheatrateis
q=qA=k T LAtotal ′′ ⋅ Δ
SolvingforLandrecognizingthatAtotal=5×W2,find
L= 5kTW q 2Δ
L=0.054m=54mm. <
COMMENTS:Thecornerswillcauselocaldeparturesfromone-dimensionalconduction andaslightlylargerheatloss.
PROBLEM1.11 KNOWN:Heatfluxatonefaceandairtemperatureandconvectioncoefficientatotherfaceofplane wall.Temperatureofsurfaceexposedtoconvection.
FIND:Ifsteady-stateconditionsexist.Ifnot,whetherthetemperatureisincreasingordecreasing.
SCHEMATIC:
h=20W/m2·K T∞=30°C Air q” conv
Ts=50°C
ASSUMPTIONS:(1)One-dimensionalconduction,(2)Nointernalenergygeneration.
ANALYSIS:Conservationofenergyforacontrolvolumearoundthewallgives
SincedEst/dt≠0,thesystemisnotatsteady-state. <
SincedEst/dt<0,thestoredenergyisdecreasing,thereforethewalltemperatureisdecreasing.<
COMMENTS:Whenthesurfacetemperatureofthefaceexposedtoconvectioncoolsto31°C,qin= qoutanddEst/dt=0andthewallwillhavereachedsteady-stateconditions. q”=20W/m2
PROBLEM1.12 KNOWN:Dimensionsandthermalconductivityoffood/beveragecontainer.Innerandouter surfacetemperatures.
FIND:Heatfluxthroughcontainerwallandtotalheatload.
SCHEMATIC:
ASSUMPTIONS:(1)Steady-stateconditions,(2)Negligibleheattransferthroughbottom wall,(3)Uniformsurfacetemperaturesandone-dimensionalconductionthroughremaining walls.
ANALYSIS:FromFourier’slaw,Eq.1.2,theheatfluxis
Sincethefluxisuniformovereachofthefivewallsthroughwhichheatistransferred,the heatloadis
q16.62W/m0.6m1.6m1.2m0.8m0.6m35.9W ⎡⎤ =++×= ⎣⎦
COMMENTS:Thecornersandedgesofthecontainercreatelocaldeparturesfromonedimensionalconduction,whichincreasetheheatload.However,forH,W1,W2>>L,the effectisnegligible.
PROBLEM1.13 KNOWN:Masonrywallofknownthermalconductivityhasaheatratewhichis80%ofthat throughacompositewallofprescribedthermalconductivityandthickness.
FIND:Thicknessofmasonrywall.
SCHEMATIC:
ASSUMPTIONS:(1)Bothwallssubjectedtosamesurfacetemperatures,(2)Onedimensionalconduction,(3)Steady-stateconditions,(4)Constantproperties.
ANALYSIS:Forsteady-stateconditions,theconductionheatfluxthroughaonedimensionalwallfollowsfromFourier’slaw,Eq.1.2,
′′ q=k T L Δ
whereΔTrepresentsthedifferenceinsurfacetemperatures.SinceΔTisthesameforboth walls,itfollowsthat
Withtheheatfluxesrelatedas
COMMENTS:Notknowingthetemperaturedifferenceacrossthewalls,wecannotfindthe valueoftheheatrate.
PROBLEM1.14 KNOWN:Expressionforvariablethermalconductivityofawall.Constantheatflux. Temperatureatx=0.
FIND:Expressionfortemperaturegradientandtemperaturedistribution.
SCHEMATIC:
ASSUMPTIONS:(1)One-dimensionalconduction.
ANALYSIS:TheheatfluxisgivenbyFourier’slaw,andisknowntobeconstant,therefore
Solvingforthetemperaturegradientandsubstitutingtheexpressionforkyields
Thisexpressioncanbeintegratedtofindthetemperaturedistribution,asfollows:
Since x qconstant ′′ = ,wecanintegratetherighthandsidetofind
wherecisaconstantofintegration.ApplyingtheknownconditionthatT=T1atx=0, wecansolveforc.
PROBLEM1.14(Cont.) Therefore,thetemperaturedistributionisgivenby
COMMENTS:Temperaturedistributionsarenotlinearinmanysituations,suchaswhenthe thermalconductivityvariesspatiallyorisafunctionoftemperature.Non-lineartemperature distributionsmayalsoevolveifinternalenergygenerationoccursornon-steadyconditionsexist.
PROBLEM1.15 KNOWN:Thickness,diameterandinnersurfacetemperatureofbottomofpanusedtoboil water.Rateofheattransfertothepan.
FIND:Outersurfacetemperatureofpanforanaluminumandacopperbottom.
SCHEMATIC:
ASSUMPTIONS:(1)One-dimensional,steady-stateconductionthroughbottomofpan.
ANALYSIS:FromFourier’slaw,therateofheattransferbyconductionthroughthebottom ofthepanis
COMMENTS:Althoughthetemperaturedropacrossthebottomisslightlylargerfor aluminum(duetoitssmallerthermalconductivity),itissufficientlysmalltobenegligiblefor bothmaterials.Toagoodapproximation,thebottommaybeconsideredisothermalatT≈ 110°C,whichisadesirablefeatureofpotsandpans.
PROBLEM1.16 KNOWN:Dimensionsandthermalconductivityofachip.Powerdissipatedononesurface.
FIND:Temperaturedropacrossthechip.
SCHEMATIC:
ASSUMPTIONS:(1)Steady-stateconditions,(2)Constantproperties,(3)Uniformheat dissipation,(4)Negligibleheatlossfrombackandsides,(5)One-dimensionalconductionin chip.
ANALYSIS:Alloftheelectricalpowerdissipatedatthebacksurfaceofthechipis transferredbyconductionthroughthechip.Hence,fromFourier’slaw,
or
kW150W/mK0.005m 22
COMMENTS:ForfixedP,thetemperaturedropacrossthechipdecreaseswithincreasingk andW,aswellaswithdecreasingt.
PROBLEM1.17 KNOWN:Heatfluxandconvectionheattransfercoefficientforboilingwater.Saturation temperatureandconvectionheattransfercoefficientforboilingdielectricfluid.
FIND:Uppersurfacetemperatureofplatewhenwaterisboiling.Whetherplanforminimizing surfacetemperaturebyusingdielectricfluidwillwork.
SCHEMATIC:
hw=20,000W/m2K
hd=3,000W/m2K
ASSUMPTIONS:Steady-stateconditions.
PROPERTIES:Tsat,w=100°Catp=1atm.
ANALYSIS:Accordingtotheproblemstatement,Newton’slawofcoolingcanbeexpressedfora boilingprocessas
Thus,
Whenthefluidiswater,
,sat, 32 2010W/m /100C200C 2010W/mK swww
Whenthedielectricfluidisused,
,sat,
/52C719C 310W/mK
Thus,thetechnician’sproposedapproachwillnotreducethesurfacetemperature. <
COMMENTS:(1)Eventhoughthedielectricfluidhasalowersaturationtemperature,thisismore thanoffsetbythelowerheattransfercoefficientassociatedwiththedielectricfluid.Thesurface temperaturewiththedielectriccoolantexceedsthemeltingtemperatureofmanymetalssuchas aluminumandaluminumalloys.(2)Dielectricfluidsare,however,employedinapplicationssuchas immersioncoolingofelectroniccomponents,whereanelectrically-conductingfluidsuchaswater couldnotbeused.
PROBLEM1.18 KNOWN:Handexperiencingconvectionheattransferwithmovingairandwater.
FIND:Determinewhichconditionfeelscolder.Contrasttheseresultswithaheatlossof30W/m2under normalroomconditions.
SCHEMATIC:
ASSUMPTIONS:(1)Temperatureisuniformoverthehand’ssurface,(2)Convectioncoefficientis uniformoverthehand,and(3)Negligibleradiationexchangebetweenhandandsurroundingsinthecase ofairflow.
ANALYSIS:Thehandwillfeelcolderfortheconditionwhichresultsinthelargerheatloss.Theheat losscanbedeterminedfromNewton’slawofcooling,Eq.1.3a,writtenas
Fortheairstream:
Forthewaterstream:
COMMENTS:Theheatlossforthehandinthewaterstreamisanorderofmagnitudelargerthanwhen intheairstreamforthegiventemperatureandconvectioncoefficientconditions.Incontrast,theheat lossinanormalroomenvironmentisonly30W/m2whichisafactorof400timeslessthanthelossinthe airstream.Intheroomenvironment,thehandwouldfeelcomfortable;intheairandwaterstreams,as youprobablyknowfromexperience,thehandwouldfeeluncomfortablycoldsincetheheatlossis excessivelyhigh.
