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Split_00060

# Split_00060 - 9.49(cont'd a x0(kg O 2 kg H H available for...

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9- 60 a. b. c. H available for combustion = total H – H in H O 2 ; latter is x 0 16 (kg O) 2 kg H kg coal kg O in water A Eq. (9.6-3) HHV = + F H G I K J + 32 764 141 790 8 9261 , , C H O S This formula does not take into account the heats of formation of the chemical constituents of coal. C = 0 758 . , H = 0 051 . , O = 0 082 . , S = 0 016 . = HHV b g Dulong kJ kg coal 31 646 , 1 kg coal 0.016 kg S 64.07 kg SO formed 32.06 kg S burned kg SO kg coal 2 2 = 0 0320 . φ = = × 0 0320 101 10 6 . . kg SO kg coal 31,646 kJ kg coal kg SO kJ 2 2 Diluting the stack gas lowers the mole fraction of SO 2 , but does not reduce SO 2 emission rates. The dilution does not affect the kg SO 2 /kJ ratio, so there is nothing to be gained by it. 9.50 CH + 2O CO 2H O l 4 2 2 2 + b g , HHV H = − = Δ ± . c o kJ mol Table B.1 890 36 b g C H + 7 2 O CO 3H O l 2 6 2 2 2 + 2 b g , HHV = 1559 9 . kJ mol CO + 1 2 O CO 2 2 , HHV = 282 99 . kJ mol (Assume ideal gas) Initial moles charged: 2.000 L 273.2K 2323 mm Hg 1 mol 25+ 273.2 K 760 mm Hg L STP mol a f a f 22 0 25 .4 . = Average mol. wt.: ( . ( . 4 929 0 25 g) mol) = 19.72 g / mol Let x 1 = mol CH
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