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多層圓筒壁 = - = 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. 準(zhǔn)數(shù)符號(hào)及意義準(zhǔn)數(shù)符號(hào)及意義 準(zhǔn)數(shù)名稱準(zhǔn)數(shù)名稱符號(hào)符號(hào)意義意義 努塞爾特

14、準(zhǔn)數(shù)努塞爾特準(zhǔn)數(shù) （Nusselt） Nu=l/ 表示對(duì)流傳熱系數(shù)的準(zhǔn)數(shù)表示對(duì)流傳熱系數(shù)的準(zhǔn)數(shù) 雷諾準(zhǔn)數(shù)雷諾準(zhǔn)數(shù) （Reynolds） Re=lu/ 確定流動(dòng)狀態(tài)的準(zhǔn)數(shù)確定流動(dòng)狀態(tài)的準(zhǔn)數(shù) 普蘭特準(zhǔn)數(shù)普蘭特準(zhǔn)數(shù) （Prandtl） Pr=cp/ 表示物性影響的準(zhǔn)數(shù)表示物性影響的準(zhǔn)數(shù) 格拉斯霍夫準(zhǔn)數(shù)格拉斯霍夫準(zhǔn)數(shù) （Grashof） Gr=gtl32/2 表示自然對(duì)流影響的準(zhǔn)數(shù)表示自然對(duì)流影響的準(zhǔn)數(shù) s r n r m eu GPARN = FC in tube 低粘度液體 n reu PRN 8 . 0 023. 0= 14. 0 3 1 8 . 0 027. 0 = W reu PRN 高粘度

15、液體 np i i c ud d )()(023. 0 8 . 0 a= Nu=0.023Re0.8Prn 式中式中n值視熱流方向而定，當(dāng)流體被加熱時(shí)，值視熱流方向而定，當(dāng)流體被加熱時(shí)，n=0.4，被冷卻時(shí)，被冷卻時(shí)，n=0.3。 應(yīng)用范圍應(yīng)用范圍 ： Re10000，0.7Pr60。若。若 L/di1000010000，0.70.7Pr16700Pr6060。 特性尺寸特性尺寸 取管內(nèi)徑取管內(nèi)徑 定性溫度定性溫度 除除w w取壁溫外，均為流體進(jìn)、出口溫度的算取壁溫外，均為流體進(jìn)、出口溫度的算 術(shù)平均值。術(shù)平均值。 當(dāng)液體被加熱時(shí)當(dāng)液體被加熱時(shí)（/w）0.14=1.05 當(dāng)液體被冷卻時(shí)當(dāng)液體被

16、冷卻時(shí)（/w）0.14=0.95 對(duì)于氣體，不論加熱或冷卻皆取對(duì)于氣體，不論加熱或冷卻皆取1。 高粘度流體高粘度流體 FC around cylinder 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(guò)自然對(duì)流的散熱量。 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 食品工業(yè)中罐頭 殺菌，食品速凍 等許多過(guò)程都屬 于不穩(wěn)定傳熱。 傳熱計(jì)算主要有兩種類型：傳熱計(jì)

26、算主要有兩種類型： 設(shè)計(jì)計(jì)算設(shè)計(jì)計(jì)算 根據(jù)生產(chǎn)要求的熱負(fù)荷確定換熱器的傳熱面積。根據(jù)生產(chǎn)要求的熱負(fù)荷確定換熱器的傳熱面積。 校核計(jì)算校核計(jì)算 計(jì)算給定換熱器的傳熱量、流體的溫度或流量。計(jì)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(jì)算穩(wěn)定傳熱的計(jì)算 對(duì)間壁式換熱器作能量恒算，在忽略熱損失的情況下有對(duì)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。上式即為換熱器的熱量恒算式。 式中式中 Q換熱器的熱負(fù)荷，換熱器的熱負(fù)荷，kJ/h或或w W流體的質(zhì)量流量，流體的質(zhì)量流量，kg/h H單位質(zhì)量流體的焓，單位質(zhì)量流體的焓，kJ/kg 下標(biāo)下標(biāo)c、h分別表示冷流體和熱

27、流體，下標(biāo)分別表示冷流體和熱流體，下標(biāo)1和和2表示換熱器的進(jìn)口和出口。表示換熱器的進(jìn)口和出口。 Q=Wh(Hh1-Hh2)=Wc(Hc2-Hc1) 一、能量恒算一、能量恒算 若換熱器中兩流體若換熱器中兩流體無(wú)相變無(wú)相變時(shí)，且認(rèn)為流體的比熱不隨溫時(shí)，且認(rèn)為流體的比熱不隨溫 度而變，則有度而變，則有 式中式中 cp流體的平均比熱，流體的平均比熱，kJ/（kg ） t冷流體的溫度，冷流體的溫度， T熱流體的溫度，熱流體的溫度， Q=Whcph(T1-T2)=Wccpc(t2-t1) 若換熱器中的熱流體若換熱器中的熱流體有相變有相變，如，如飽和蒸汽冷凝飽和蒸汽冷凝時(shí)，則有時(shí)，則有 當(dāng)當(dāng)冷凝液的溫度低于

28、飽和溫度冷凝液的溫度低于飽和溫度時(shí)，則有時(shí)，則有 式式中中 Wh飽和蒸汽（熱流體）的冷凝速率，飽和蒸汽（熱流體）的冷凝速率，kg/h r飽和蒸汽的冷凝潛熱，飽和蒸汽的冷凝潛熱，kJ/kg Q=Whr=Wccpc(t2-t1) 注注：上式應(yīng)用條件是冷凝液在飽和溫度下離開(kāi)換熱器。：上式應(yīng)用條件是冷凝液在飽和溫度下離開(kāi)換熱器。 Q=Whr+cph(T1-T2)=Wccpc(t2-t1) 式中式中 cph冷凝液的比熱，冷凝液的比熱， kJ/（kg ） Ts冷凝液的飽和溫度，冷凝液的飽和溫度， 通過(guò)換熱器中任一微元面積通過(guò)換熱器中任一微元面積dS的間壁兩側(cè)流體的傳熱速率的間壁兩側(cè)流體的傳熱速率 方程（仿

29、對(duì)流傳熱速率方程）為方程（仿對(duì)流傳熱速率方程）為 dQ=K(T-t)dS=KtdS 式中式中 K局部總傳熱系數(shù)，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，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，換熱器的任一截面上冷流體的平均溫度， 上式稱為上式稱為總傳熱速率方程總傳熱速率方程。 二、總傳熱速率方程二、總傳熱速率方程 1 1 總傳熱速率微分方程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(duì)應(yīng)，選擇的傳總傳熱系數(shù)必須和所選擇的傳熱面積相對(duì)應(yīng)，選擇的傳 熱面積不同，總傳熱系數(shù)的數(shù)值也不同。熱面積不同，總傳熱系數(shù)的數(shù)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基于管內(nèi)表面積、外表面積、外表面平均面積 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)，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，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(zhǔn)。注：在工程大多以外表面積為基準(zhǔn)。 oomii dSdS b dS tT dQ aa 11 - = 對(duì)于管式換熱器，假定管內(nèi)作為加熱側(cè)，管外為冷卻側(cè)，對(duì)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(guò)任一微元面積則通過(guò)

31、任一微元面積dS的傳熱由三步過(guò)程構(gòu)成。的傳熱由三步過(guò)程構(gòu)成。 由熱流體傳給管壁由熱流體傳給管壁 dQ=i(T-Tw)dSi 由管壁傳給冷流體由管壁傳給冷流體 dQ=o(tw-t)dSo 通過(guò)管壁的熱傳導(dǎo)通過(guò)管壁的熱傳導(dǎo) dQ=(/b)(Tw-tw)dSm 由上三式可得由上三式可得 2 2 總傳熱系數(shù)總傳熱系數(shù) 2.1 2.1 總傳熱系數(shù)的計(jì)算式總傳熱系數(shù)的計(jì)算式 由于由于dQ及（及（T-t）兩者與選擇的基準(zhǔn)面積無(wú)關(guān)，則根據(jù)總）兩者與選擇的基準(zhǔn)面積無(wú)關(guān)，則根據(jù)總 傳熱速率微分方程，有傳熱速率微分方程，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 總傳熱系數(shù)（以外表面為基準(zhǔn)）為總傳熱系數(shù)（以外表面為基準(zhǔn)）為 om o ii o o d bd d d kaa 11 = 同理同理 總傳熱系數(shù)表示成熱阻形式為總傳熱系數(shù)表示成熱阻形式為 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 = 在計(jì)算總傳熱系數(shù)在計(jì)算總傳熱系數(shù)K時(shí)，污垢熱阻一般不能忽視，若管壁時(shí)，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內(nèi)、外側(cè)表面上的熱阻分別為Rsi及及Rso時(shí)，則有時(shí)，則有 當(dāng)傳熱面為平壁或薄管壁時(shí)，當(dāng)傳熱面為平壁或薄管壁時(shí)，di、do、dm近似相等，則有近似相等，則有 2.2 2.2 污垢熱阻污垢熱阻 當(dāng)管壁熱阻和污垢熱阻可忽略時(shí)，則可簡(jiǎn)化為當(dāng)管壁熱阻和污垢熱阻可忽略時(shí)，則可簡(jiǎn)化為 oio kaa 111 = o Ka 11 若若o i，則有，則有 總熱阻是由熱阻大的那一側(cè)的對(duì)流傳熱所控制，即當(dāng)兩個(gè)總熱阻是由熱阻大的那一側(cè)的對(duì)流傳熱所控制，即當(dāng)兩

34、個(gè) 對(duì)流傳熱系數(shù)相差不大時(shí)，欲提高對(duì)流傳熱系數(shù)相差不大時(shí)，欲提高K值，關(guān)鍵在于提高對(duì)流值，關(guān)鍵在于提高對(duì)流 傳熱系數(shù)較小一側(cè)的傳熱系數(shù)較小一側(cè)的。 若兩側(cè)的若兩側(cè)的相差不大時(shí)，則必須同時(shí)提高兩側(cè)的相差不大時(shí)，則必須同時(shí)提高兩側(cè)的，才能，才能 提高提高K值。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(shí)清除污垢。及時(shí)清除污垢。 由上可知：由上可知： 例例 一列管式換熱器，由一列管式換熱器，由252.5mm的鋼管組成。管的鋼管組成。管 內(nèi)為內(nèi)為CO2，流量為，流量為6000kg/h，由，由55冷卻到冷卻到30。管外。管外 為冷卻水，

35、流量為為冷卻水，流量為2700kg/h，進(jìn)口溫度為，進(jìn)口溫度為20。CO2與與 冷卻水呈逆流流動(dòng)。已知水側(cè)的對(duì)流傳熱系數(shù)為冷卻水呈逆流流動(dòng)。已知水側(cè)的對(duì)流傳熱系數(shù)為 3000W/m2K，CO2 側(cè)的對(duì)流傳熱系數(shù)為側(cè)的對(duì)流傳熱系數(shù)為40 W/m2K 。 試求總傳熱系數(shù)試求總傳熱系數(shù)K，分別用內(nèi)表面積，分別用內(nèi)表面積A1，外表面積，外表面積A2 表示。表示。 解：查鋼的導(dǎo)熱系數(shù)解：查鋼的導(dǎo)熱系數(shù)=45W/mK 取取CO2側(cè)污垢熱阻側(cè)污垢熱阻Ra1=0.5310-3m2K/W 取水側(cè)污垢熱阻取水側(cè)污垢熱阻Ra2=0.2110-3m2K/W 以內(nèi)、外表面計(jì)時(shí)，內(nèi)、外表面分別用下標(biāo)以內(nèi)、外表面計(jì)時(shí)，內(nèi)、

36、外表面分別用下標(biāo)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 換熱器 n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 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: 兩固體表面間的輻射傳熱 nF12反映物體 2可截獲物 體1輻射能 量的分?jǐn)?shù) 212121 FSFS= nF12稱為物 體1對(duì)物體 2的角系數(shù) 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。 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