食品工程-熱量轉(zhuǎn)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多層圓筒壁Ifthewallofthecylinderconsistsoflayershavingdifferentthermalconductivities.

T1andT2musttransectalayerboundedbyr1andr2,whichhasauniformthermalconductivityk1.Similarly,thelayerboundedbyrandrwherethetemperaturesareTandTmustalsohaveauniformthermalconductivity,k2.Heattransferbyconvectionαistheheattransfercoefficient,Aistheareaofthefluidsolidinterfacewhereheatisbeingtransferred,and?T,thedrivingforceforheattransfer,isthedifferenceinfluidtemperatureandthesolidsurfacetemperatureNaturalandforcedConvectionNCdependsongravityanddensityandviscositychangesassociatedwithtemperaturedifferencesinthefluidtoinduceconvectivecurrents.HeattransfercoefficientsofFCdependsonthevelocityofthefluid,itsthermophysicalproperties,andthegeometryofthesurface.準數(shù)符號及意義準數(shù)名稱符號意義努塞爾特準數(shù)（Nusselt）Nu=αl/λ

表示對流傳熱系數(shù)的準數(shù)雷諾準數(shù)（Reynolds）Re=luρ/μ

確定流動狀態(tài)的準數(shù)普蘭特準數(shù)（Prandtl）Pr=cpμ/λ

表示物性影響的準數(shù)格拉斯霍夫準數(shù)（Grashof）Gr=βgΔtl3ρ2/μ2

表示自然對流影響的準數(shù)FCintube低粘度液體高粘度液體

Nu=0.023Re0.8Prn

式中n值視熱流方向而定，當流體被加熱時，n=0.4，被冷卻時，n=0.3。應(yīng)用范圍

：Re>10000，0.7<Pr<120，管長與管徑比L/di>60。若

L/di<60時，α須乘以（1+（di/L）0.7）進行校正。特性尺寸

：

取管內(nèi)徑，

定性溫度：

流體進、出口溫度的算術(shù)平均值。

低粘度流體

Nu=0.023Re0.8Pr1/3（μ/μw）0.14應(yīng)用范圍

Re>10000，0.7<Pr<16700，L/di>60。特性尺寸

取管內(nèi)徑定性溫度

除μw取壁溫外，均為流體進、出口溫度的算

術(shù)平均值。當液體被加熱時（μ/μw）0.14=1.05當液體被冷卻時（μ/μw）0.14=0.95

對于氣體，不論加熱或冷卻皆取1。高粘度流體

FCaroundcylinder繞方形物體繞柱形物體例題：水平放置的蒸氣管道，外徑為100mm，若管外壁溫度為100℃，周圍大氣溫度為20℃，試求每米管道通過自然對流的散熱量。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食品工業(yè)中罐頭殺菌，食品速凍等許多過程都屬于不穩(wěn)定傳熱。傳熱計算主要有兩種類型：

設(shè)計計算

根據(jù)生產(chǎn)要求的熱負荷確定換熱器的傳熱面積。

校核計算

計算給定換熱器的傳熱量、流體的溫度或流量。穩(wěn)定傳熱的計算對間壁式換熱器作能量恒算，在忽略熱損失的情況下有上式即為換熱器的熱量恒算式。式中

Q——換熱器的熱負荷，kJ/h或w

W——流體的質(zhì)量流量，kg/h

H——單位質(zhì)量流體的焓，kJ/kg

下標c、h分別表示冷流體和熱流體，下標1和2表示換熱器的進口和出口。Q=Wh(Hh1-Hh2)=Wc(Hc2-Hc1)一、能量恒算

若換熱器中兩流體無相變時，且認為流體的比熱不隨溫度而變，則有式中

cp——流體的平均比熱，kJ/（kg·℃

）t——冷流體的溫度，℃

T——熱流體的溫度，℃Q=Whcph(T1-T2)=Wccpc(t2-t1)若換熱器中的熱流體有相變，如飽和蒸汽冷凝時，則有

當冷凝液的溫度低于飽和溫度時，則有

式中Wh——飽和蒸汽（熱流體）的冷凝速率，kg/h

r——飽和蒸汽的冷凝潛熱，kJ/kgQ=Whr=Wccpc(t2-t1)注：上式應(yīng)用條件是冷凝液在飽和溫度下離開換熱器。Q=Wh[r+cph(T1-T2)]=Wccpc(t2-t1)式中

cph——冷凝液的比熱，kJ/（kg·℃

）

Ts——冷凝液的飽和溫度，

℃

通過換熱器中任一微元面積dS的間壁兩側(cè)流體的傳熱速率方程（仿對流傳熱速率方程）為

dQ=K(T-t)dS=KΔtdS式中

K——局部總傳熱系數(shù)，w/（m2·℃

）

T——換熱器的任一截面上熱流體的平均溫度，

℃t——換熱器的任一截面上冷流體的平均溫度，

℃上式稱為總傳熱速率方程。二、總傳熱速率方程

1總傳熱速率微分方程

總傳熱系數(shù)必須和所選擇的傳熱面積相對應(yīng)，選擇的傳熱面積不同，總傳熱系數(shù)的數(shù)值也不同。dQ=Ki(T-t)dSi=Ko(T-t)dSo=Km(T-t)dSm式中

Ki、Ko、Km——基于管內(nèi)表面積、外表面積、外表面平均面積

的總傳熱系數(shù)，w/（m2·℃

）Si、So、Sm——換熱器內(nèi)表面積、外表面積、外表面平均面積，

m2

注：在工程大多以外表面積為基準。

對于管式換熱器，假定管內(nèi)作為加熱側(cè)，管外為冷卻側(cè)，則通過任一微元面積dS的傳熱由三步過程構(gòu)成。由熱流體傳給管壁

dQ=αi(T-Tw)dSi由管壁傳給冷流體

dQ=αo(tw-t)dSo通過管壁的熱傳導

dQ=(λ/b)·(Tw-tw)dSm由上三式可得2總傳熱系數(shù)

2.1總傳熱系數(shù)的計算式

由于dQ及（T-t）兩者與選擇的基準面積無關(guān)，則根據(jù)總傳熱速率微分方程，有所以總傳熱系數(shù)（以外表面為基準）為同理總傳熱系數(shù)表示成熱阻形式為

在計算總傳熱系數(shù)K時，污垢熱阻一般不能忽視，若管壁內(nèi)、外側(cè)表面上的熱阻分別為Rsi及Rso時，則有當傳熱面為平壁或薄管壁時，di、do、dm近似相等，則有2.2污垢熱阻當管壁熱阻和污垢熱阻可忽略時，則可簡化為若αo<<

αi，則有總熱阻是由熱阻大的那一側(cè)的對流傳熱所控制，即當兩個對流傳熱系數(shù)相差不大時，欲提高K值，關(guān)鍵在于提高對流傳熱系數(shù)較小一側(cè)的α。若兩側(cè)的α相差不大時，則必須同時提高兩側(cè)的α，才能提高K值。若污垢熱阻為控制因素，則必須設(shè)法減慢污垢形成速率或及時清除污垢。由上可知：例

一列管式換熱器，由?25×2.5mm的鋼管組成。管內(nèi)為CO2，流量為6000kg/h，由55℃冷卻到30℃。管外為冷卻水，流量為2700kg/h，進口溫度為20℃。CO2與冷卻水呈逆流流動。已知水側(cè)的對流傳熱系數(shù)為3000W/m2·K，CO2

側(cè)的對流傳熱系數(shù)為40W/m2·K。試求總傳熱系數(shù)K，分別用內(nèi)表面積A1，外表面積A2表示。

解：查鋼的導熱系數(shù)λ=45W/m·K

取CO2側(cè)污垢熱阻Ra1=0.53×10-3m2·K/W

取水側(cè)污垢熱阻Ra2=0.21×10-3m2·K/W以內(nèi)、外表面計時，內(nèi)、外表面分別用下標1、2表示。

HeatExchangeEquipmentSweptsurfaceheatexchangerToheat,coolorprovideheattoconcentrateviscousfoodproductsHeatExchangeEquipmentDoublepipeheatexchangerAmajordisadvantageistherelativelylargespaceitoccupiesforthequantityofheatexchangedHeatExchangeEquipmentShellandtubeheatexchangerHeatExchangeEquipmentplateheatexchanger換熱器板式換熱器單程列管式換熱器噴淋式換熱器螺旋管式換熱器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:兩固體表面間的輻射傳熱F12反映物體2可截獲物體1輻射能量的分數(shù)F12稱為物體1對物體2的角系數(shù)CalculationofradiationheattransferForslabsofS1=S2ForS1<<S2TwoparallelfinitesurfacesHomework有一表面積為0.1m2的面包塊在烤爐內(nèi)烘烤，爐內(nèi)壁輻射換熱面積為1m2，壁面溫度為250℃，面包溫度為100℃，假設(shè)爐壁和面包之間為封閉空間，求面包得到的輻射熱量。面包黑度取0.5，爐壁黑度取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