Thermodynamics HW Solutions 988

# Thermodynamics HW Solutions 988 - 0.245 1555 0896 = − =...

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Chapter 12 Radiation Heat Transfer Then the effective emissivity of the combustion gases becomes 0.194 0 . 0 062 . 0 8 . 1 055 . 0 5 . 1 atm 1 , atm 1 , = × + × = ε Δ ε + ε = ε w w c c g C C 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 1555 . 0 ) 062 . 0 ( K 378 K 790 ) 8 . 1 ( 0896 . 0 ) 037 . 0 ( K 378 K 790 ) 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 = 378 K instead of T g = 790 K. We use the chart for 400 K. At P w /( P w + P c ) = 0.67 and P c L + P w L = 0.112 we read Δε = 0.0. Then the absorptivity of the combustion gases becomes
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Unformatted text preview: 0.245 . 1555 . 0896 . = − + = α Δ − α + α = α w c g The emissivity of the inner surfaces of the tubes is 0.9. Then the net rate of radiation heat transfer from the combustion gases to the walls of the tube becomes W 10,745 = − ⋅ × + = α − ε σ + ε = − ] ) K 378 ( 245 . ) K 790 ( 194 . )[ K W/m 10 67 . 5 )( m 827 . 2 ( 2 1 9 . ) ( 2 1 4 4 4 2 8 2 4 4 rad s g g g s s T T A Q & ( b ) The heat of vaporization of water at 1 atm is 2257 kJ/kg (Table A-9). Then rate of evaporation of water becomes kg/s 0.0773 = × + = + = ⎯→ ⎯ = + J/kg 10 7 . 333 W ) 745 , 10 050 , 15 ( 3 rad conv evap evap rad conv fg fg h Q Q m h m Q Q & & & & & & 12-104 …. . 12-106 Design and Essay Problems KJ 12-85...
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