Thermodynamics HW Solutions 924

Thermodynamics HW Solutions 924 - Chapter 12 Radiation Heat...

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Chapter 12 Radiation Heat Transfer Properties The emissivities of all surfaces are ε = 1 since they are black. Analysis Both 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 FF F 12 21 13 026 10 2 6 0 7 4 == =− = . ..( summation rule ) The net rate of radiation heat transfer from the disks into the environment then becomes () W 5505 = × = = = + = ] K 300 K 700 )[ K W/m 10 67 . 5 ]( ) m 3 . 0 ( )[ 74 . 0 ( 2 ) ( 2 2 4 4 4 2 8 2 4 3 4 1 1 13 3 13 23 13 3 π σ T T A F Q Q Q Q Q & & & & & Disk 1, T 1 = 700 K, ε 1 = 1 Disk 2, T 2 = 700 K, ε 2 = 1 0.40 m Environment T 3 =300 K ε 1 = 1 D = 0.6 m 12-34 A furnace shaped like a long equilateral-triangular duct is considered. The temperature of the base surface is to be determined. Assumptions 1 Steady operating conditions exist 2 The 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.

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