Thermodynamics HW Solutions 986

# Thermodynamics HW Solutions 986 - Chapter 12 Radiation Heat...

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Chapter 12 Radiation Heat Transfer 0 . 0 67 . 0 08 . 0 16 . 0 16 . 0 112 . 0 075 . 0 037 . 0 = ε Δ = + = + = + = + c w w w c P P P L P L P Then the effective emissivity of the combustion gases becomes 0.117 0 . 0 062 . 0 1 055 . 0 1 atm 1 , atm 1 , = × + × = ε Δ ε + ε = ε w w c c g C C Note that the pressure correction factor is 1 for both gases since the total pressure is 1 atm. For a source temperature of T s = 105 ° C = 378 K, the absorptivity of the gas is again determined using the emissivity charts as follows: atm ft 0.036 atm m 0109 . 0 K 790 K 378 m) 5 atm)(0.142 16 . 0 ( atm ft 0.018 atm m 00545 . 0 K 790 K 378 m) 5 atm)(0.142 08 . 0 ( = = = = = = g s w g s c T T L P T T L P The emissivities of CO 2 and H 2 O corresponding to these values at a temperature of T s = 378 K and 1atm are, from Fig. 12-36, and 037 . 0 atm 1 , = ε c 062 . 0 atm 1 , = ε w Then the absorptivities of CO 2 and H 2 O become 0864 . 0 ) 062 . 0 ( K 378 K 790 ) 1 ( 0597 . 0 ) 037 . 0 ( K 378 K 790 ) 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 = 378 K instead of T g = 790 K. We use the chart for 400 K. At P w /( P w + P
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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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