食品工程-热量转移

Chapter4HeattransferHeattransferHeattransferisthemovementofenergyfromonepointtoanotherbyvirtueofadifferenceintemperature.HeattransferConduction:Heatwillbetransferredbetweenadjacentmolecules.Convection:Heatistransferredwhenmoleculesmovefromonepointtoanotherandexchangesenergywithanothermoleculeintheotherlocation.Radiation:thephenomenonofheattransferbyelectromagneticwaves.HeattransferbyconductionFourier’sFirstLawofHeatTransferQistherateofheatflow,Aistheareathroughwhichheatistransferred.Theexpressionq/A,therateofheattransferperunitareas,iscalledtheheatflux.q.ThederivativedT/dxisthetemperaturegradient.Kisthermalconductivity.EstimationofthermalconductivityoffoodproductsChoiandOkos(1987)Kiscalculatedfromthethermalconductivityofthepurecomponentkiandthevolumefractionofeachcomponent,Xvi:kiobtainedfromChoiandOkos(1987):Purewater,protein,fat,carbohydrate,fiber,andashXviobtainedfromChoiandOkos(1987):Purewater,protein,fat,carbohydrate,fiber,andashXi,massfractionFourier’ssecondlawofheattransferTheratiok/(Cp)ish,thethermaldiffusivityHeattransferbyconductionHeattransferthoughaslabExample7.3.Thermocouplesembeddedattwopointswithinasteelbar,1and2mmfromthesurface,indicatetemperaturesof100Cand98C,respectively.Assumingnoheattransferoccurringfromthesides,calculatethesurfacetemperature.Solution:

T2=98,T1=100,x2=2mmandx1=1mm.ThetemperaturegradientT/x=(T2−T1)/(x2−x1)=(98−100)/0.001(2−1)=−2000.T=−(−2000)(x1−x)+T1Atthesurface,x=0,andatpointx1=0.001,T1=100.Thus:T=2000(0.001)+100=102CExample7.4.Acylindricalsampleofbeef5cmthickand3.75cmindiameterispositionedbetweentwo5-cm-thickacryliccylindersofexactlythesamediameterasthemeatsample.Theassemblyispositionedinsideaninsulatedcontainersuchthatthebottomoftheloweracryliccylindercontactsaheatedsurfacemaintainedat50C,andthetopoftheuppercylindercontactsacoolplatemaintainedat0C.Twothermocoupleseachareembeddedintheacryliccylinders,positioned0.5cmand1.5cmfromthesample-acrylicinterface.Iftheacrylichasathermalconductivityof1.5W/(m·K),andthetemperaturesrecordedatsteadystateare,respectively,45◦C,43◦C,15C,and13C,calculatethethermalconductivityofthemeatsample.Conductionheattransferthroughwallsofacylinder多层圆筒壁Ifthewallofthecylinderconsistsoflayershavingdifferentthermalconductivities.

T1andT2musttransectalayerboundedbyr1andr2,whichhasauniformthermalconductivityk1.Similarly,thelayerboundedbyrandrwherethetemperaturesareTandTmustalsohaveauniformthermalconductivity,k2.Heattransferbyconvectionαistheheattransfercoefficient,Aistheareaofthefluidsolidinterfacewhereheatisbeingtransferred,and∆T,thedrivingforceforheattransfer,isthedifferenceinfluidtemperatureandthesolidsurfacetemperatureNaturalandforcedConvectionNCdependsongravityanddensityandviscositychangesassociatedwithtemperaturedifferencesinthefluidtoinduceconvectivecurrents.HeattransfercoefficientsofFCdependsonthevelocityofthefluid,itsthermophysicalproperties,andthegeometryofthesurface.准数符号及意义准数名称符号意义努塞尔特准数（Nusselt）Nu=αl/λ

表示对流传热系数的准数雷诺准数（Reynolds）Re=luρ/μ

确定流动状态的准数普兰特准数（Prandtl）Pr=cpμ/λ

表示物性影响的准数格拉斯霍夫准数（Grashof）Gr=βgΔtl3ρ2/μ2

表示自然对流影响的准数FCintube低粘度液体高粘度液体

Nu=0.023Re0.8Prn

式中n值视热流方向而定，当流体被加热时，n=0.4，被冷却时，n=0.3。应用范围

：Re>10000，0.7<Pr<120，管长与管径比L/di>60。若

L/di<60时，α须乘以（1+（di/L）0.7）进行校正。特性尺寸

：

取管内径，

定性温度：

流体进、出口温度的算术平均值。

低粘度流体

Nu=0.023Re0.8Pr1/3（μ/μw）0.14应用范围

Re>10000，0.7<Pr<16700，L/di>60。特性尺寸

取管内径定性温度

除μw取壁温外，均为流体进、出口温度的算

术平均值。当液体被加热时（μ/μw）0.14=1.05当液体被冷却时（μ/μw）0.14=0.95

对于气体，不论加热或冷却皆取1。高粘度流体

FCaroundcylinder绕方形物体绕柱形物体例题：水平放置的蒸气管道，外径为100mm，若管外壁温度为100℃，周围大气温度为20℃，试求每米管道通过自然对流的散热量。Theproblemofheattransferthroughmultiplelayerscanbeanalyzedasaprobleminvolvingaseriesofresistancetoheattransfer.Thetransferofheatcanbeconsideredasanalogoustothetransferofelectricalenergythroughaconductor.TisthedrivingforceequivalenttothevoltageEinelectricalcircuits.Theheatfluxqisequivalenttothecurrent,I.STEADY-STATEHEATTRANSFERTheConceptofResistancetoHeatTransferOverallresistancetoheattransferisthesumoftheindividualresistanceinseries:R=R1+R2+R3+......Rn;Thus,Forheattransferthroughacylinder,Forconvectionheattransfer:CombinedConvectionandConduction:TheOverallHeatTransferCoefficientThetemperaturesoffluidsonbothsidesofasolidareknownandtherateofheattransfer

acrossthesolidistobedetermined.

Heattransferinvolvesconvectiveheattransferbetweenafluid

ononesurface,conductiveheattransferthroughthesolidandconvectiveheattransferagainatthe

oppositesurfacetotheotherfluid.Rateofheattransfer:UExample7.11.Calculatetherateofheattransferacrossaglasspanethatconsistsoftwo1.6-mmthickglassseparatedby0.8-mmlayerofair.Theheattransfercoefficientononesidethatisat21Cis2.84W/(m2·K)andontheoppositesidethatisat−15Cis11.4W/(m2·K).Thethermalconductivityofglassis0.52W/(m·K)andthatofairis0.031W/(m·K).TheLogarithmicMeanTemperatureDifferenceExample7.13.Applesauceisbeingcooledfrom80Cto20Cinasweptsurfaceheatexchanger.Theoverallcoefficientofheattransferbasedontheinsidesurfaceareais568W/m2·K.Theapplesaucehasaspecificheatof3187J/kg·Kandisbeingcooledattherateof50kg/h.Coolingwaterentersincountercurrentflowat10Candleavestheheatexchangerat17C.Calculate:(a)thequantityofcoolingwaterrequired;(b)therequiredheattransfersurfaceareafortheheatexchanger.UNSTEADY-STATEHEATTRANSFERHeatingofSolidsHavingInfiniteThermalConductivityExample7.17.Asteam-jacketedkettleconsistsofahemisphericalbottomhavingadiameterof69cmandcylindricalside30cmhigh.Thesteamjacketofthekettleisoverthehemisphericalbottomonly.Thekettleisfilledwithafoodproductthathasadensityof1008kg/m3toapoint10cmfromtherimofthekettle.Iftheoverallheattransfercoefficientbetweensteamandthefoodinthejacketedpartofthekettleis1000W/(m2·K),andsteamat120Cisusedforheatinginthejacket,calculatethetimeforthefoodproducttoheatfrom20Cto98C.Thespecificheatofthefoodis3100J/(kg·K).SolidswithFiniteThermalConductivityUNSTEADY-STATEHEATTRANSFER食品工业中罐头杀菌，食品速冻等许多过程都属于不稳定传热。传热计算主要有两种类型：

设计计算

根据生产要求的热负荷确定换热器的传热面积。

校核计算

计算给定换热器的传热量、流体的温度或流量。稳定传热的计算对间壁式换热器作能量恒算，在忽略热损失的情况下有上式即为换热器的热量恒算式。式中

Q——换热器的热负荷，kJ/h或w

W——流体的质量流量，kg/h

H——单位质量流体的焓，kJ/kg

下标c、h分别表示冷流体和热流体，下标1和2表示换热器的进口和出口。Q=Wh(Hh1-Hh2)=Wc(Hc2-Hc1)一、能量恒算

若换热器中两流体无相变时，且认为流体的比热不随温度而变，则有式中

cp——流体的平均比热，kJ/（kg·℃

）t——冷流体的温度，℃

T——热流体的温度，℃Q=Whcph(T1-T2)=Wccpc(t2-t1)若换热器中的热流体有相变，如饱和蒸汽冷凝时，则有

当冷凝液的温度低于饱和温度时，则有

式中Wh——饱和蒸汽（热流体）的冷凝速率，kg/h

r——饱和蒸汽的冷凝潜热，kJ/kgQ=Whr=Wccpc(t2-t1)注：上式应用条件是冷凝液在饱和温度下离开换热器。Q=Wh[r+cph(T1-T2)]=Wccpc(t2-t1)式中

cph——冷凝液的比热，kJ/（kg·℃

）

Ts——冷凝液的饱和温度，

℃

通过换热器中任一微元面积dS的间壁两侧流体的传热速率方程（仿对流传热速率方程）为

dQ=K(T-t)dS=KΔtdS式中

K——局部总传热系数，w/（m2·℃

）

T——换热器的任一截面上热流体的平均温度，

℃t——换热器的任一截面上冷流体的平均温度，

℃上式称为总传热速率方程。二、总传热速率方程

1总传热速率微分方程

总传热系数必须和所选择的传热面积相对应，选择的传热面积不同，总传热系数的数值也不同。dQ=Ki(T-t)dSi=Ko(T-t)dSo=Km(T-t)dSm式中

Ki、Ko、Km——基于管内表面积、外表面积、外表面平均面积

的总传热系数，w/（m2·℃

）Si、So、Sm——换热器内表面积、外表面积、外表面平均面积，

m2

注：在工程大多以外表面积为基准。

对于管式换热器，假定管内作为加热侧，管外为冷却侧，则通过任一微元面积dS的传热由三步过程构成。由热流体传给管壁

dQ=αi(T-Tw)dSi由管壁传给冷流体

dQ=αo(tw-t)dSo通过管壁的热传导

dQ=(λ/b)·(Tw-tw)dSm由上三式可得2总传热系数

2.1总传热系数的计算式

由于dQ及（T-t）两者与选择的基准面积无关，则根据总传热速率微分方程，有所以总传热系数（以外表面为基准）为同理总传热系数表示成热阻形式为

在计算总传热系数K时，污垢热阻一般不能忽视，若管壁内、外侧表面上的热阻分别为Rsi及Rso时，则有当传热面为平壁或薄管壁时，di、do、dm近似相等，则有2.2污垢热阻当管壁热阻和污垢热阻可忽略时，则可简化为若αo<<

αi，则有总热阻是由热阻大的那一侧的对流传热所控制，即当两个对流传热系数相差不大时，欲提高K值，关键在于提高对流传热系数较小一侧的α。若两侧的α相差不大时，则必须同时提高两侧的α，才能提高K值。若污垢热阻为控制因素，则必须设法减慢污垢形成速率或及时清除污垢。由上可知：例

一列管式换热器，由Ø25×2.5mm的钢管组成。管内为CO2，流量为6000kg/h，由55℃冷却到30℃。管外为冷却水，流量为2700kg/h，进口温度为20℃。CO2与冷却水呈逆流流动。已知水侧的对流传热系数为3000W/m2·K，CO2

侧的对流传热系数为40W/m2·K。试求总传热系数K，分别用内表面积A1，外表面积A2表示。

解：查钢的导热系数λ=45W/m·K

取CO2侧污垢热阻Ra1=0.53×10-3m2·K/W

取水侧污垢热阻Ra2=0.21×10-3m2·K/W以内、外表面计时，内、外表面分别用下标1、2表示。

HeatExchangeEquipmentSweptsurfaceheatexchangerToheat,coolorprovideheattoconcentrateviscousfoodproductsHeatExchangeEquipmentDoublepipeheatexchangerAmajordisadvantageistherelativelylargespaceitoccupiesforthequantityofheatexchangedHeatExchangeEquipmentShellandtubeheatexchangerHeatExchangeEquipmentplateheatexchanger换热器板式换热器单程列管式换热器喷淋式换热器螺旋管式换热器HeattransferbyradiationspectralIrradiationRadiosity,absorptivity,reflectivity,transmissivityBlackbodyisonethatabsorbsallincidentradiation.Emissivity()isapropertythatisthefractionofradiationemittedorabsorbedbyablackbodyatagiventemperaturethatisactuallyemittedorabsorbedbyasurfaceatthesametemperature.Blackbodieshave=1,q/A=T4.Graybodieshave<1,q/A=T4.,Stephan-Boltzmanconstant,5.6732×10−8W/(m2·K4).Stephan-BoltzmanLawTheenergyfluxfromaBlacksurfaceatanabsolutetemperatureT,asafunctionofthewavelengthis:两固体表面间的辐射传热F12反映物体2可截获物体1辐射能量的分数F12称为物体1对物体2的角系数CalculationofradiationheattransferForslabsofS1=S2ForS1<<S2TwoparallelfinitesurfacesHomework有一表面积为0.1m2的面包块在烤炉内烘烤，炉内壁辐射换热面积为1m2，壁面温度为250℃，面包温度为100℃，假设炉壁和面包之间为封闭空间，求面包得到的辐射热量。面包黑度取0.5，炉壁黑度取0.8。Electromagneticspectrumbetween300MHzand300GHz;Mwmaybereflectedorabsorbedbymaterialsortransmitthroughmaterialswithoutanyabsorption,dependingonthedielectricpropertiesofamaterial.Mwpenetrateafood,andheatfoodwithintheentirefoodmorerapid.Mwisnonionizingradiation,itgeneratesheatbyinteractionwithfood.MicrowaveandDielectricHeatingMicrowaveHeatingq/V=energyabsorbed,W/cm3;f=frequency,Hz;

e=dielectricconstant,anindexoftherateatwhichenergypenetratesasolid;tan()=dielectriclossfactor,anindexoftheextenttowhichenergyenteringthesolidisconvertedtoheat;E=fieldstrengthinvolts/cm2,setbythetypeofmicrowavegeneratorused.eandtan()propertiesofthematerialandarefunctionsofcompositionandtemperature

Example7.10.Thedielectricconstantofbeefat23Cand2450MHzis28andthelosstangentis0.2.Thedensityis1004kg/m3andthespecificheatis3250J/(kg·K).Potatoat23Cand2450MHzhasadielectricconstantof38andalosstangentof0.3.Thedensityis1010kg/m3andthespecificheatis3720J/(kg·K).(a)Amicrowaveovenhasaratedoutputof600W.When0.25kgofpotatoeswereplacedinsidetheoven,thetemperatureriseafter1minuteofheatingwas38.5C.When60gofpotatowasheatedintheoven,atemperatureriseof40Cwasobse