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1、Chapter TwelveMultimode Heat And Mass TransferThat most technically important problems involve multimode heat transfer effects.qcondqradqconvqradqcondqconvTo solve such problems,it may be necessary to simultaneously work with heat equation,surface energy balance,overall energy balances,and approprit
2、e rate rate equations for conduction,convection and radiation heat transfer.Example 12.1The 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-state conditions for which the interior su
3、rface 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 heating rate needed maintain the interior window su
4、rface at Ts,i=15.solutionKnown: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=240KGlassAssumptions:1.steady-state,one-dimensional conditions.2.Constant prope
5、rties.3.Uniform volumetric heating in window.4.Convection heat transfer from interior surface of window to interior air may be approximated as free convection from a vertical plate.Radiation exchange with automobile interior is negligible.5.Heat transfer from outer surface is due to forced convectio
6、n over a flat plate and radiation exchange with the sky.6.Boundary layer separation on the outer surface in negligible.7.Outer surface is diffuse and gray.8.Properties:9.Glass;Air;k,v,Pr,(look at tables)Analysis:The temperature distribution in the glass is governed by the appropriate form of the hea
7、t equation.Boundary conditionsApplying an energy balance to the inner surfaceHenceApplying an energy balance to the outer surface.(1)(2)(3)The inside convection coefficient may be obtained from Equation The outside convection coefficient may be obtained by first evaluating the Reynolds number.Mixed
8、boundary layer conditions(4)(5)Equation 5 may be expressed as(6)Substituting into equation 4Example 12.2 The composite insulation 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 therma
9、l resistance.Schematic:because is no lateral heat from cell to cell,and it suffices to consider the heat transfer across a single cell.Ts,oTs,iSurface slabsCellular air spacesCell cross section A-AAir spaceT1Inner slab,k1Side wall,k2T2Outer slab,k1qAAAssumptions:1.One-dimensional,steady-state condit
10、ions.2.Equivalent conditions for each cell.3.Constant properties.4.Diffuse,gray surface behavior.Properties:Analysis:The total resistance of the composite is determined by conduction,convection and radiation processes occurring within the honeycomb and by conduction across the inner and outer slabs.
11、The corresponding thermal circuit is shown.The total resistance of the composite isThe equivalent resistance for the honecomb isThe cell forms an enclosure that may be classified as a horizontal cavity heated from below,and the appropriate form of the Rayleigh number isThe resistance to heat transfe
12、r by radiation may be obtained by first noting that the cell forms a three-surface enclosure for which the sidewalls are reradiating.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 i
13、nside surface is at known temperature Ts which less then T.Known:Dry-bulb temperature associated with air flow through a large diameter duct of prescribed surface temperature.Find:Temperature of airflow.Schematic:Dry-bulb thermometer,AirAssumption:1.Steady-state conditions exit.2.Heat transfer by co
14、nduction along the thermometers is negligible.3.Duct wall forms a large enclosure about the thermometer.4.Thermometer surface are diffuse gray.5.Properties:(air)v,k,Pr.Analysis:Since TdbTs,there is net radiation transfer from the surface of the dry-bulb thermometer to the duct wall.Hence to maintain
15、 steady-state conditions,the thermometer temperature must be less than that of the air to allow for convection heat transfer from the air.From application of a surface energy balance to the thermometerThe air temperature is then(1)For the prescribed conditions the Reynolds number associated with the dry-bulb thermometer isComments:The effect of radiation exchange between the duct wall and the thermometer is small.For this reason .