Thermodynamics HW Solutions 947

Thermodynamics HW Solutions 947 - + = T T Q & The...

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Chapter 12 Radiation Heat Transfer 12-57 A radiation shield is placed between two large parallel plates which are maintained at uniform temperatures. The emissivity of the radiation shield is to be determined if the radiation heat transfer between the plates is reduced to 15% of that without the radiation shield. 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 and ε 2 = 0.9. Analysis First, the net rate of radiation heat transfer between the two large parallel plates per unit area without a shield is 2 4 4 4 2 8 2 1 4 2 4 1 shield no , 12 W/m 4877 1 9 . 0 1 6 . 0 1 ] ) K 400 ( ) K 650 )[( K W/m 10 67 . 5 ( 1 1 1 ) ( = + × =
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Unformatted text preview: + = T T Q & The radiation heat transfer in the case of one shield is 2 2 shield no , 12 shield one , 12 W/m 6 . 731 W/m 4877 15 . 15 . = = = Q Q & & Then the emissivity of the radiation shield becomes + + = + + + = 1 2 1 9 . 1 6 . 1 ] ) K 400 ( ) K 650 )[( K W/m 10 67 . 5 ( W/m 731.6 1 1 1 1 1 1 ) ( 3 4 4 4 2 8 2 2 , 3 1 , 3 2 1 4 2 4 1 shield one , 12 T T Q & T 2 = 400 K 2 = 0.9 T 1 = 650 K 1 = 0.6 Radiation shield 3 which gives 0.18 = 3 12-44...
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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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