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Thermodynamics HW Solutions 946

# Thermodynamics HW Solutions 946 - 48 R 1200 52 R Btu/h.ft...

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Chapter 12 Radiation Heat Transfer 12-56E A radiation shield is placed between two parallel disks which are maintained at uniform temperatures. The net rate of radiation heat transfer through the shields is to be determined. T 1 = 1200 R, ε 1 = 1 ε 3 = 0.15 T 2 = 700 R, ε 2 = 1 1 ft 1 ft T = 540 K ε 3 = 1 Assumptions 1 Steady operating conditions exist 2 The surfaces are black. 3 Convection heat transfer is not considered. Properties The emissivities of surfaces are given to be ε 1 = ε 2 = 1 and ε 3 = 0.15. Analysis From Fig. 12-44 we have 52 . 0 13 32 = = F F . Then . The disk in the middle is surrounded by black surfaces on both sides. Therefore, heat transfer between the top surface of the middle disk and its black surroundings can expressed as 48 . 0 52 . 0 1 34 = = F ]} ) K 540
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Unformatted text preview: ( [ 48 . ] ) R 1200 ( [( 52 . ){ R Btu/h.ft 10 1714 . ( ) ft 069 . 7 ( 15 . )] ( [ )] ( [ 4 4 3 4 4 3 4 2 8 2 4 2 4 3 32 3 4 1 4 3 31 3 3 − + − ⋅ × = − σ ε + − σ ε = − T T T T F A T T F A Q & Similarly, for the bottom surface of the middle disk, we have ]} ) K 540 ( [ 52 . ] ) R 700 ( [( 48 . ){ R Btu/h.ft 10 1714 . ( ) ft 069 . 7 ( 15 . )] ( [ )] ( [ 4 4 3 4 4 3 4 2 8 2 4 5 4 3 35 3 4 4 4 3 34 3 3 − + − ⋅ × = − σ ε + − σ ε = − − T T T T F A T T F A Q & Combining the equations above, the rate of heat transfer between the disks through the radiation shield (the middle disk) is determined to be and T Btu/h 866 = Q & 3 = 895 K 12-43...
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