Thermodynamics HW Solutions 960

Thermodynamics HW Solutions 960 - . 073 . 084 . = + = + = w...

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Chapter 12 Radiation Heat Transfer 073 . 0 ) 027 . 0 ( K 600 K 1500 ) 8 . 1 ( 084 . 0 ) 031 . 0 ( K 600 K 1500 ) 5 . 1 ( 45 . 0 atm 1 , 45 . 0 65 . 0 atm 1 , 65 . 0 = = ε = α = = ε = α w s g w w c s g c c T T C T T C Also Δα = Δε , but the emissivity correction factor is to be evaluated from Fig. 12-38 at T = T s = 600 K instead of T g = 1500 K. There is no chart for 600 K in the figure, but we can read Δε values at 400 K and 800 K, and take their average. At P w /( P w + P c ) = 0.6 and P c L + P w L = 0.07 we read Δε = 0.0. Then the absorptivity of the combustion gases becomes 0.157
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Unformatted text preview: . 073 . 084 . = + = + = w c g The surface area of the pipe per m length of tube is 2 m 4712 . m) 1 ( m) 15 . ( = = = DL A s Then the net rate of radiation heat transfer from the combustion gases to the walls of the furnace becomes W 10,276 = = = ] ) K 600 ( 157 . ) K 1500 ( 08 . )[ K W/m 10 67 . 5 )( m 4712 . ( ) ( 4 4 4 2 8 2 4 4 net s g g g s T T A Q & 12-57...
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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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