传热学综合传热PPT课件

1、传热学综合传热 Chapter Twelve Multimode Heat And Mass Transfer 传热学综合传热 That most technically important problems involve multimode heat transfer effects. qcond qrad qconv qrad qcond qconv To solve such problems, it may be necessary to simultaneously work with heat equation, surface energy balance, overall e

2、nergy balances, and approprite rate rate equations for conduction, convection and radiation heat transfer. 传热学综合传热 Example 12.1 The rear window of an automobile is of thickness L=8mm and height H=0.5m and contains fine-meshed heating wires that can induce nearly volumetric heating. Consider steady-s

3、tate conditions for which the interior surface of the window is exposed to air at 10 , while the exterior surface is exposed to air at -10 and a very cold sky with an effective temperature of -33 . The air moves in parallel flow over the surface with a velocity of 20m/s. Determine the volumetric hea

4、ting rate needed maintain the interior window surface at Ts,i=15. solution Known: Boundary conditions associated with a rear window experiencing uniform volumetric heating. Find: Volumetric heating rate needed to maintain inner surface temperature at Ts,i=15. Schematic: Tsky=240K os T , K288 , is T

5、Glass m/s20u K263 , o T K283 Air , i T ,oconv q rad q cond q ,iconv q cond q 传热学综合传热 Assumptions: steady-state, one-dimensional conditions. Constant properties. Uniform volumetric heating in window. Convection heat transfer from interior surface of window to interior air may be approximated as free

6、convection from a vertical plate. Radiation exchange with automobile interior is negligible. Heat transfer from outer surface is due to forced convection over a flat plate and radiation exchange with the sky. Boundary layer separation on the outer surface in negligible. Outer surface is diffuse and

7、gray. Properties: 1.Glass; Air; k, v, Pr, , , ( look at tables) Analysis: The temperature distribution in the glass is governed by the appropriate form of the heat equation. 21 2 2 CxCx k q xT 传热学综合传热 Boundary conditions isis TCTTx ,2, 0, 0 Applying an energy balance to the inner surface isii x icon

8、vcond TTh dx dT kqq , 0 , or isii x TThCx k q k , 0 1 isi i TT k h C ,1 Hence is isii Tx k TTh x k q xT , ,2 2 Applying an energy balance to the outer surface. 44 , , or skyosioso Lx radoconvcond TTTTh dx dT qqq isiiisii Lx TThLqTTh k Lq k dx dT , (1) (2) (3) 传热学综合传热 44 ,skyosiisiooso TTTThTThLq is

9、isii os TL k TTh k qL T , , 2 , 2 The inside convection coefficient may be obtained from Equation 2 278 61 169 Pr492. 01 387. 0 825. 0 H H Ra Nu 7 3 ,1 , 10137. 7 HTTg Ra is H 56 HNu KmW81. 2 2 H k Nuh H i The outside convection coefficient may be obtained by first evaluating the Reynolds number. 5

10、10413. 7Re Hu H Mixed boundary layer conditions 864Pr871Re037. 0 31 54 H HNu KmW6 .41 2 H k Nuh H o (4) (5) 传热学综合传热 Equation 5 may be expressed as K1 .28810286. 2 5 , qT os(6) Substituting into equation 4 44 ,skyosiisiooso TTTThTThLq K285 , os T 3 mKW136 q 传热学综合传热 Example 12.2 The composite insulati

11、on shown is being considered as a roofing material. Known: Dimensions of a composite insulation consisting of a honeycomb core sandwiched between solid slabs. Find: Total thermal resistance. Schematic: because is no lateral heat from cell to cell, and it suffices to consider the heat transfer across

12、 a single cell. Ts,o Ts,i Surface slabs Cellular air spaces osis TT , 1 L 3 L 2 L q W t 传热学综合传热 mm10W mm1 2 t Cell cross section A-A Air space T1 Inner slab, k1 Side wall, k2 T2 C25 0 , is T C10 0 , os T mm5 .12 3 L mm0 .50 2 L mm5 .12 1 L Outer slab, k1 q AA Assumptions: 1. One-dimensional, steady-

13、state conditions. 2. Equivalent conditions for each cell. 3. Constant properties. 4. Diffuse, gray surface behavior. Properties: KmW078. 0 1 kKmW170. 0 2 k85. 0 传热学综合传热 Analysis: The total resistance of the composite is determined by conduction, convection and radiation processes occurring within th

14、e honeycomb and by conduction across the inner and outer slabs. The corresponding thermal circuit is shown. icond R ,ocond R , 2 T 1 T hcrad R , hccond R , hcconv R , is T ,os T , The total resistance of the composite is ocondeqicond RRRR , The equivalent resistance for the honecomb is hc radconvcon

15、deq RRRR 1111 WK1603 2 1 1 , Wk L RR ocondicond WK8170 2 2 2 2 , twWk L R hccond 2 , 1 tWh R hcconv The cell forms an enclosure that may be classified as a horizontal cavity heated from below, and the appropriate form of the Rayleigh number is 5 3 221 1011. 3 LTTg RaL KmW25. 2Pr069. 0 2074. 031 2 L

16、Ra L k h WK6944 1 2 , tWh R hcconv 传热学综合传热 The resistance to heat transfer by radiation may be obtained by first noting that the cell forms a three-surface enclosure for which the sidewalls are reradiating. 21 2 21 tWAA 1 212112 4 2 4 1 2 1 1 1 2 RRRR rad FFFFF TTtW q WK7471 1 2 1 1 2 1 2 2 2 1 2 12

17、21 , TTTTtW F q TT R rad hcrad 1221 1FFF RR 传热学综合传热 Example 12.3 Cylindrical dry-bulb thermometer is installed in a large diameter duct to obtain the temperature T air flowing through the duct at a velocity V. the duct inside surface is at known temperature Ts which less then T. Known: Dry-bulb temp

18、erature associated with air flow through a large diameter duct of prescribed surface temperature. Find: Temperature of airflow. Schematic: Dry-bulb thermometer, 95. 0,mm3,C45 0 gdbdb DT sm5, VT Air 传热学综合传热 Assumption: Steady-state conditions exit. Heat transfer by conduction along the thermometers is negligible. Duct wall forms a large enclosure about the thermometer. Thermometer surface are diffuse gray. 1.Properties: (air) v, k, Pr. Analysis: Since TdbTs, there is net radiation transfer from the surface of the dry-bulb thermometer to the duct w