Thermodynamics HW Solutions 953

# Thermodynamics HW Solutions 953 - bands The emissivity...

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Chapter 12 Radiation Heat Transfer Analysis The volumetric analysis of a gas mixture gives the mole fractions y i of the components, which are equivalent to pressure fractions for an ideal gas mixture. Therefore, the partial pressures of CO 2 and H 2 O are atm 09 . 0 atm) 1 ( 09 . 0 atm 10 . 0 atm) 1 ( 10 . 0 2 2 H CO = = = = = = P y P P y P O w c 6 m Combustion gases 1000 K The mean beam length for a cube of side length 6 m for radiation emitted to all surfaces is, from Table 12-4, L = 0.66(6 m) = 3.96 m Then, atm ft 1.57 atm m 48 . 0 m) atm)(3.96 09 . 0 ( atm ft .30 1 atm m 396 . 0 m) atm)(3.96 10 . 0 ( = = = = = = L P L P w c The emissivities of CO 2 and H 2 O corresponding to these values at the gas temperature of T g = 1000 K and 1atm are, from Fig. 12-36, 17 . 0 atm 1 , = ε c and 26 . 0 atm 1 , = ε w Both CO 2 and H 2 O are present in the same mixture, and we need to correct for the overlap of emission
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Unformatted text preview: bands. The emissivity correction factor at T = T g = 1000 K is, from Fig. 12-38, 039 . 474 . 10 . 09 . 09 . 87 . 2 57 . 1 30 . 1 = ε Δ ⎪ ⎭ ⎪ ⎬ ⎫ = + = + = + = + c w w w c P P P L P L P Note that we obtained the average of the emissivity correction factors from the two figures for 800 K and 1200 K. Then the effective emissivity of the combustion gases becomes 0.391 = − × + × = ε Δ − ε + ε = ε 039 . 26 . 1 17 . 1 atm 1 , atm 1 , 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. 12-50...
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