食品工程原理：chapter4 Heat transfer

1、Chapter4 Heat transfer Heat transfer nHeat transfer is the movement of energy from one point to another by virtue of a difference in temperature. Heat transfer nConduction: Heat will be transferred between adjacent molecules. nConvection: Heat is transferred when molecules move from one point to ano

2、ther and exchanges energy with another molecule in the other location. nRadiation: the phenomenon of heat transfer by electromagnetic waves. Heat transfer by conduction nFouriers First Law of Heat Transfer Q is the rate of heat flow, A is the area through which heat is transferred. The expression q/

3、A, the rate of heat transfer per unit areas, is called the heat flux.q. The derivative dT/dx is the temperature gradient. K is thermal conductivity. Q A =-k dT dx Estimation of thermal conductivity of food products nChoi and Okos (1987) nK is calculated from the thermal conductivity of the pure comp

4、onent ki and the volume fraction of each component, Xvi: =)Xk(k vii 26 a 26 fi 26 c 27 f 26 p 24 ic 26 w T109069. 2T001401. 03296. 0k T101683. 3T0012497. 018331. 0k T103312. 4T0013874. 02014. 0k T107749. 1T0027604. 01807. 0k T107178. 2T0011958. 01788. 0k T100154. 1T0062489. 02196. 2k T107306. 6T0017

5、625. 057109. 0k - - - - - - - -= -= -= -= -= -= -= Protein:Cpp= 2008.21208.910-3T-1312.910-6T 2 T28063. 08 .2423 T36589. 05 .1311 T31046. 01 .1599 T41757. 059.925 T51814. 09 .1329 T13071. 089.916 T0037574. 0T0031439. 018.997 a fi c f p ic 2 w -= -= -= -= -= -= -= Xi, mass fraction = 2 2 2 2 2 2 p z

6、T y T x T C k t T Heat transfer by conduction nHeat transfer though a slab 11 T)xx( x T T- -= Example 7.3. Thermocouples embedded at two points within a steel bar, 1 and 2 mm from the surface, indicate temperatures of 100C and 98C, respectively. Assuming no heat transfer occurring from the sides, ca

7、lculate the surface temperature. Solution: T2 = 98, T1 = 100, x2 = 2 mm and x1 = 1 mm. The temperature gradient T/ x = (T2 T1)/(x2 x1) = (98 100)/0.001(2 1) = 2000. T = (2000)(x1 x) + T1 At the surface, x = 0, and at point x1 = 0.001, T1 = 100. Thus: T = 2000(0.001) + 100 = 102 C Example 7.4. A cyli

8、ndrical sample of beef 5 cm thick and 3.75 cm in diameter is positioned between two 5-cm- thick acrylic cylinders of exactly the same diameter as the meat sample. The assembly is positioned inside an insulated container such that the bottom of the lower acrylic cylinder contacts a heated surface mai

9、ntained at 50C, and the top of the upper cylinder contacts a cool plate maintained at 0C. Two thermocouples each are embedded in the acrylic cylinders, positioned 0.5 cm and 1.5 cm from the sample-acrylic interface. If the acrylic has a thermal conductivity of 1.5 W/(m K), and the temperatures recor

10、ded at steady state are, respectively, 45 C, 43 C, 15C, and 13C, calculate the thermal conductivity of the meat sample. Conduction heat transfer through walls of a cylinder dr dT )rL2(kq-= 1 2 21 ln 2 r r TT Lkq - = n多层圆筒壁 = - = n i i i i n r r k TTL q 1 1 11 ln 1 2 If the wall of the cylinder consi

11、sts of layers having different thermal conductivities. T1 and T2 must transect a layer bounded by r1 and r2, which has a uniform thermal conductivity k1. Similarly, the layer bounded by r and r where the temperatures are T and T must also have a uniform thermal conductivity, k2. Heat transfer by con

12、vection n is the heat transfer coefficient, A is the area of the fluid solid interface where heat is being transferred, and T, the driving force for heat transfer, is the difference in fluid temperature and the solid surface temperature Q=aA(T m -T s)= aAT Natural and forced Convection nNC depends o

13、n gravity and density and viscosity changes associated with temperature differences in the fluid to induce convective currents. nHeat transfer coefficients of FC depends on the velocity of the fluid, its thermophysical properties, and the geometry of the surface. 准数符号及意义准数符号及意义 准数名称准数名称符号符号意义意义 努塞尔特

14、准数努塞尔特准数 （Nusselt） Nu=l/ 表示对流传热系数的准数表示对流传热系数的准数 雷诺准数雷诺准数 （Reynolds） Re=lu/ 确定流动状态的准数确定流动状态的准数 普兰特准数普兰特准数 （Prandtl） Pr=cp/ 表示物性影响的准数表示物性影响的准数 格拉斯霍夫准数格拉斯霍夫准数 （Grashof） Gr=gtl32/2 表示自然对流影响的准数表示自然对流影响的准数 s r n r m eu GPARN = FC in tube 低粘度液体 n reu PRN 8 . 0 023. 0= 14. 0 3 1 8 . 0 027. 0 = W reu PRN 高粘度

15、液体 np i i c ud d )()(023. 0 8 . 0 a= Nu=0.023Re0.8Prn 式中式中n值视热流方向而定，当流体被加热时，值视热流方向而定，当流体被加热时，n=0.4，被冷却时，被冷却时，n=0.3。 应用范围应用范围 ： Re10000，0.7Pr60。若。若 L/di1000010000，0.70.7Pr16700Pr6060。 特性尺寸特性尺寸 取管内径取管内径 定性温度定性温度 除除w w取壁温外，均为流体进、出口温度的算取壁温外，均为流体进、出口温度的算 术平均值。术平均值。 当液体被加热时当液体被加热时（/w）0.14=1.05 当液体被冷却时当液体被

16、冷却时（/w）0.14=0.95 对于气体，不论加热或冷却皆取对于气体，不论加热或冷却皆取1。 高粘度流体高粘度流体 FC around cylinder 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。 nThe problem of heat transfer through multiple layers can be analyzed as a problem involving a series of resistance to heat transfer. nThe transfe

17、r of heat can be considered as analogous to the transfer of electrical energy through a conductor. nT is the driving force equivalent to the voltage E in electrical circuits. The heat flux q is equivalent to the current, I. STEADY-STATE HEAT TRANSFER The Concept of Resistance to Heat Transfer Overal

18、l resistance to heat transfer is the sum of the individual resistance in series: R = R1 + R2 + R3 + . . . . . . Rn; Thus, nFor heat transfer through a cylinder, For convection heat transfer: Combined Convection and Conduction: The Overall Heat Transfer Coefficient The temperatures of fluids on both

19、sides of a solid are known and the rate of heat transfer across the solid is to be determined. Heat transfer involves convective heat transfer between a fluid on one surface, conductive heat transfer through the solid and convective heat transfer again at the opposite surface to the other fluid. Rat

20、e of heat transfer: U nExample 7.11. Calculate the rate of heat transfer across a glass pane that consists of two 1.6-mm thick glass separated by 0.8-mm layer of air. The heat transfer coefficient on one side that is at 21C is 2.84 W/(m2K) and on the opposite side that is at 15C is 11.4 W/(m2K). The

21、 thermal conductivity of glass is 0.52 W/(mK) and that of air is 0.031 W/(mK). The Logarithmic Mean Temperature Difference nExample 7.13. Applesauce is being cooled from 80C to 20C in a swept surface heat exchanger. The overall coefficient of heat transfer based on the inside surface area is 568 W/m

22、2K. The applesauce has a specific heat of 3187 J/kgK and is being cooled at the rate of 50 kg/h. Cooling water enters in countercurrent flow at 10C and leaves the heat exchanger at 17C. Calculate: (a) the quantity of cooling water required; (b) the required heat transfer surface area for the heat ex

23、changer. UNSTEADY-STATE HEAT TRANSFER nHeating of Solids Having Infinite Thermal Conductivity Example 7.17. A steam-jacketed kettle consists of a hemispherical bottom having a diameter of 69 cm and cylindrical side 30 cm high. The steam jacket of the kettle is over the hemispherical bottom only. The

24、 kettle is filled with a food product that has a density of 1008 kg/m3 to a point 10 cm from the rim of the kettle. If the overall heat transfer coefficient between steam and the food in the jacketed part of the kettle is 1000 W/(m2K), and steam at 120C is used for heating in the jacket, calculate t

25、he time for the food product to heat from 20C to 98C. The specific heat of the food is 3100 J/(kgK). 3 2323 16075. 0 )10. 030. 0()69. 0( 2 1 )69. 0( 12 1 2 1 6 1 2 1 m HddVolume = -= nSolids with Finite Thermal Conductivity UNSTEADY-STATE HEAT TRANSFER 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：传热计

26、算主要有两种类型： 设计计算设计计算 根据生产要求的热负荷确定换热器的传热面积。根据生产要求的热负荷确定换热器的传热面积。 校核计算校核计算 计算给定换热器的传热量、流体的温度或流量。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。上式即为换热器的热量恒算式。 式中式中 Q换热器的热负荷，换热器的热负荷，kJ/h或或w W流体的质量流量，流体的质量流量，kg/h H单位质量流体的焓，单位质量流体的焓，kJ/kg 下标下标c、h分别表示冷流体和热

27、流体，下标分别表示冷流体和热流体，下标1和和2表示换热器的进口和出口。表示换热器的进口和出口。 Q=Wh(Hh1-Hh2)=Wc(Hc2-Hc1) 一、能量恒算一、能量恒算 若换热器中两流体若换热器中两流体无相变无相变时，且认为流体的比热不随温时，且认为流体的比热不随温 度而变，则有度而变，则有 式中式中 cp流体的平均比热，流体的平均比热，kJ/（kg ） t冷流体的温度，冷流体的温度， T热流体的温度，热流体的温度， Q=Whcph(T1-T2)=Wccpc(t2-t1) 若换热器中的热流体若换热器中的热流体有相变有相变，如，如饱和蒸汽冷凝饱和蒸汽冷凝时，则有时，则有 当当冷凝液的温度低于

28、饱和温度冷凝液的温度低于饱和温度时，则有时，则有 式式中中 Wh饱和蒸汽（热流体）的冷凝速率，饱和蒸汽（热流体）的冷凝速率，kg/h r饱和蒸汽的冷凝潜热，饱和蒸汽的冷凝潜热，kJ/kg Q=Whr=Wccpc(t2-t1) 注注：上式应用条件是冷凝液在饱和温度下离开换热器。：上式应用条件是冷凝液在饱和温度下离开换热器。 Q=Whr+cph(T1-T2)=Wccpc(t2-t1) 式中式中 cph冷凝液的比热，冷凝液的比热， kJ/（kg ） Ts冷凝液的饱和温度，冷凝液的饱和温度， 通过换热器中任一微元面积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的间壁两侧流体的传热速率 方程（仿

29、对流传热速率方程）为方程（仿对流传热速率方程）为 dQ=K(T-t)dS=KtdS 式中式中 K局部总传热系数，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，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，换热器的任一截面上冷流体的平均温度， 上式称为上式称为总传热速率方程总传热速率方程。 二、总传热速率方程二、总传热速率方程 1 1 总传热速率微分方程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。热面积不同，总传热系数的数值也不同。

30、 dQ=Ki(T-t)dSi=Ko(T-t)dSo=Km(T-t)dSm 式中式中 Ki、 、Ko 、Km基于管内表面积、外表面积、外表面平均面积 基于管内表面积、外表面积、外表面平均面积 的总传热系数，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，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。注：在工程大多以外表面积为基准。 oomii dSdS b dS tT dQ aa 11 - = 对于管式换热器，假定管内作为加热侧，管外为冷却侧，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则通过

31、任一微元面积dS的传热由三步过程构成。的传热由三步过程构成。 由热流体传给管壁由热流体传给管壁 dQ=i(T-Tw)dSi 由管壁传给冷流体由管壁传给冷流体 dQ=o(tw-t)dSo 通过管壁的热传导通过管壁的热传导 dQ=(/b)(Tw-tw)dSm 由上三式可得由上三式可得 2 2 总传热系数总传热系数 2.1 2.1 总传热系数的计算式总传热系数的计算式 由于由于dQ及（及（T-t）两者与选择的基准面积无关，则根据总）两者与选择的基准面积无关，则根据总 传热速率微分方程，有传热速率微分方程，有 o i o i i o d d dS dS k k = o m o m m o d d dS

32、 dS k k = om o ii o o d bd d d tT dS dQ aa 1 - = 所以所以 om o ii o o d bd d d K aa 1 1 = oo i m i i i d d d bd K aa = 1 1 oo m ii m m d db d d K aa = 1 总传热系数（以外表面为基准）为总传热系数（以外表面为基准）为 om o ii o o d bd d d kaa 11 = 同理同理 总传热系数表示成热阻形式为总传热系数表示成热阻形式为 o so m o i o si ii o o R d bd d d R d d kaa 11 = o sosi io

33、 R b R kaa 111 = 在计算总传热系数在计算总传热系数K时，污垢热阻一般不能忽视，若管壁时，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内、外侧表面上的热阻分别为Rsi及及Rso时，则有时，则有 当传热面为平壁或薄管壁时，当传热面为平壁或薄管壁时，di、do、dm近似相等，则有近似相等，则有 2.2 2.2 污垢热阻污垢热阻 当管壁热阻和污垢热阻可忽略时，则可简化为当管壁热阻和污垢热阻可忽略时，则可简化为 oio kaa 111 = o Ka 11 若若o i，则有，则有 总热阻是由热阻大的那一侧的对流传热所控制，即当两个总热阻是由热阻大的那一侧的对流传热所控制，即当两

34、个 对流传热系数相差不大时，欲提高对流传热系数相差不大时，欲提高K值，关键在于提高对流值，关键在于提高对流 传热系数较小一侧的传热系数较小一侧的。 若两侧的若两侧的相差不大时，则必须同时提高两侧的相差不大时，则必须同时提高两侧的，才能，才能 提高提高K值。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。及时清除污垢。 由上可知：由上可知： 例例 一列管式换热器，由一列管式换热器，由252.5mm的钢管组成。管的钢管组成。管 内为内为CO2，流量为，流量为6000kg/h，由，由55冷却到冷却到30。管外。管外 为冷却水，

35、流量为为冷却水，流量为2700kg/h，进口温度为，进口温度为20。CO2与与 冷却水呈逆流流动。已知水侧的对流传热系数为冷却水呈逆流流动。已知水侧的对流传热系数为 3000W/m2K，CO2 侧的对流传热系数为侧的对流传热系数为40 W/m2K 。 试求总传热系数试求总传热系数K，分别用内表面积，分别用内表面积A1，外表面积，外表面积A2 表示。表示。 解：查钢的导热系数解：查钢的导热系数=45W/mK 取取CO2侧污垢热阻侧污垢热阻Ra1=0.5310-3m2K/W 取水侧污垢热阻取水侧污垢热阻Ra2=0.2110-3m2K/W 以内、外表面计时，内、外表面分别用下标以内、外表面计时，内、

36、外表面分别用下标1、2表示。表示。 kmw RR d d d db K m = = = 2 21 2 1 2 1 1 1 /5 .38 00021. 000053. 0 025. 0 02. 0 3000 1 0225. 0 02. 0 45 0025. 0 40 1 1 11 1 aa aa kmw RR d db d d K m = = = 2 21 2 2 1 2 1 2 /3 .31 00021. 000053. 0 3000 1 0225. 0 025. 0 45 0025. 0 02. 0 025. 0 40 1 1 11 1 aa aa Heat Exchange Equipme

37、nt Swept surface heat exchanger To heat, cool or provide heat to concentrate viscous food products Heat Exchange Equipment Double pipe heat exchanger A major disadvantage is the relatively large space it occupies for the quantity of heat exchanged Heat Exchange Equipment Shell and tube heat exchange

38、r Heat Exchange Equipment plate heat exchanger 换热器 n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 Heat transfer by radiation nspectral Irradiation DRA QQQQ= Radiosity, absorptivity, reflectivity, transmissivity nBlack body is one that absorbs all incident radiation. nEmissivity () is a property that is th

39、e fraction of radiation emitted or absorbed by a black body at a given temperature that is actually emitted or absorbed by a surface at the same temperature. nBlack bodies have = 1, q/A = T4. nGray bodies have 1, q/A = T4. , Stephan-Boltzman constant, 5.6732 108 W/(m2 K4). Stephan-Boltzman Law Theen

40、ergyfluxfromaBlacksurfaceatanabsolute temperatureT,asafunctionofthewavelengthis: 两固体表面间的辐射传热 nF12反映物体 2可截获物 体1辐射能 量的分数 212121 FSFS= nF12称为物 体1对物体 2的角系数 Calculation of radiation heat transfer nFor slabs of S1=S2 1 11 21 4 2 4 110 12 - - = TTS q nFor S1S2 4 2 4 110112 TTSq-= nTwo parallel finite surfa

41、ces 4 2 4 112102112 TTFSq-= ) 1 1 ( 1 1 22 1 1 - = S S n 4 2 4 11012 TTSq n -= Homework 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。 nElectromagneticspectrumbetween300MHzand300GHz; nMwmaybereflectedorabsorbedbymaterialsortransmit through

42、materialswithoutanyabsorption,dependingonthe dielectricpropertiesofamaterial. nMwpenetrateafood,andheatfoodwithintheentirefoodmore rapid. nMwisnonionizingradiation,itgeneratesheatbyinteraction withfood. Microwave and Dielectric Heating Microwave Heating q/V = energy absorbed, W/cm3; f = frequency, H

43、z; e = dielectric constant, an index of the rate at which energy penetrates a solid; tan() = dielectric loss factor, an index of the extent to which energy entering the solid is converted to heat; E = field strength in volts/cm2, set by the type of microwave generator used. e and tan() properties of

44、 the material and are functions of composition and temperature Example 7.10. The dielectric constant of beef at 23C and 2450 MHz is 28 and the loss tangent is 0.2. The density is 1004 kg/m3 and the specific heat is 3250 J/(kgK). Potato at 23C and 2450 MHz has a dielectric constant of 38 and a loss tangent of 0.3. The density is 1010 kg/m3 and the specific heat is 3720 J/(kgK). n(a) A microwave oven has a rated output of 600 W. When 0.25 kg of potatoes were placed inside the oven, the temperature rise af