Thermodynamics HW Solutions 984

Thermodynamics HW Solutions 984 - + = 1 15 . 1 15 . 1 1 10...

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Chapter 12 Radiation Heat Transfer 12-101 Two thin radiation shields are placed between two large parallel plates that are maintained at uniform temperatures. The net rate of radiation heat transfer between the plates with and without the shields, and the temperatures of radiation shields are to be determined. Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered. Properties The emissivities of surfaces are given to be ε 1 = 0.6, 2 = 0.7, 3 = 0.10, and 4 = 0.15. Analysis The net rate of radiation heat transfer without the shields per unit area of the plates is 2 W/m 3288 = + × = ε + ε σ = 1 7 . 0 1 6 . 0 1 ] ) K 300 ( ) K 600 )[( K W/m 10 67 . 5 ( 1 1 1 ) ( 4 4 4 2 8 2 1 4 2 4 1 shield no 12, T T Q The net rate of radiation heat transfer with two thin radiation shields per unit area of the plates is 2 W/m 206 = + + + + + × = ε + ε + ε + ε +
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Unformatted text preview: + = 1 15 . 1 15 . 1 1 10 . 1 10 . 1 1 7 . 1 6 . 1 ] ) K 300 ( ) K 600 )[( K W/m 10 67 . 5 ( 1 1 1 1 1 1 1 1 1 ) ( 4 4 4 2 8 4 4 3 3 2 1 4 2 4 1 shields two 12, T T Q & T 2 = 300 K 2 = 0.7 T 1 = 600 K 1 = 0.6 3 = 0.10 4 = 0.15 The equilibrium temperatures of the radiation shields are determined from K 549 = + = + = 3 4 3 4 4 2 8 2 3 1 4 3 4 1 13 1 10 . 1 6 . 1 ] ) K 600 )[( K W/m 10 67 . 5 ( W/m 206 1 1 1 ) ( T T T T Q & K 429 = + = + = 4 4 4 4 4 2 8 2 2 4 4 2 4 4 42 1 7 . 1 15 . 1 ] ) K 300 ( )[ K W/m 10 67 . 5 ( W/m 206 1 1 1 ) ( T T T T Q & 12-81...
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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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