Chapter 12 Radiation Heat TransferPropertiesThe emissivities of all surfaces are ε= 1 since they are black. AnalysisBoth disks possess same properties and they are black. Noting that environment can also be considered to be blackbody, we can treat this geometry as a three surface enclosure. We consider the two disks to be surfaces 1 and 2 and the environment to be surface 3. Then from Figure 12-7, we read FFF1221130261026 074===−=...(summation rule)The net rate of radiation heat transfer from the disks into the environment then becomes ()W5505=−⋅×=−==+=−]K300K700)[KW/m1067.5]()m3.0()[74.0(2)(22444282434111331323133πσTTAFQQQQQ&&&&&Disk 1, T1= 700 K, ε1= 1 Disk 2, T2= 700 K, ε2= 1 0.40 mEnvironment T3=300 K ε1= 1 D = 0.6 m12-34 A furnace shaped like a long equilateral-triangular duct is considered. The temperature of the base surface is to be determined. Assumptions1Steady operating conditions exist 2The surfaces are opaque, diffuse, and gray.
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This note was uploaded on 01/24/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.