Elementary Principles 412

# Elementary Principles 412 - 9.49(cont'd a x0(kg O 2 kg H H...

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Unformatted text preview: 9.49 (cont'd) a. x0 (kg O) 2 kg H H available for combustion = total H H in H 2 O ; latter is kg coal 16 kg O Eq. (9.6-3) HHV = 32,764C + 141,790 H - FG H O + 9261S 8 IJ K in water A This formula does not take into account the heats of formation of the chemical constituents of coal. b. C = 0. 758 , H = 0. 051, O = 0. 082 , S = 0. 016 HHV b g Dulong = 31,646 kJ kg coal 1 kg coal = 0. 0320 kg SO 2 kg coal 32.06 kg S burned 0.0320 kg SO 2 kg coal . = = 101 10 -6 kg SO 2 kJ 31,646 kJ kg coal 0.016 kg S 64.07 kg SO 2 formed c. Diluting the stack gas lowers the mole fraction of SO2, but does not reduce SO2 emission rates. The dilution does not affect the kg SO2/kJ ratio, so there is nothing to be gained by it. o CH 4 + 2O 2 CO 2 + 2H 2 O l , HHV = - Hc = 890.36 kJ mol Table B.1 9.50 bg b g 7 C 2 H 6 + O 2 2CO 2 + 3H 2 O l , HHV = 1559. 9 kJ mol 2 1 CO + O 2 CO 2 , HHV = 282. 99 kJ mol 2 2.000 L 273.2K 2323 mm Hg 1 mol Initial moles charged: = 0.25 mol 25 + 273.2 K 760 mm Hg 22.4 L STP (Assume ideal gas) bg a f a f Average mol. wt.: (4.929 g) (0.25 mol) = 19.72 g / mol Let x1 = mol CH 4 mol gas, x2 = mol C 2 H 6 mol gas 1 - x1 - x2 mol CO mol gas 1 4 2 1 2 g b gh MW = 19.72 x b16.04 g mol CH g + x b30.07g + b1 - x - x gb28.01g = 19.72 b1g HHV = 963.7 kJ mol x b890.36g + x b1559.9g + b1 - x - x gb282.99g = 963.7 b2g 1 2 1 2 c b Solving (1) & (2) simultaneously yields x1 = 0.725 mol CH 4 mol, x 2 = 0188 mol C 2 H 6 mol, 1 - x1 - x 2 = 0.087 mol CO mol . 9.51 a. Basis : 1mol/s fuel gas o CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(v), Hc = -890.36 kJ / mol 7 o C 2 H 6 (g) + O 2 (g) 2CO 2 (g) + 3H 2 O(v), Hc = -1559.9 kJ / mol 2 Excess O2, 25C n2 , mol CO 2 n3 , mol H 2 O n4 , mol O 2 25C 1 mol/s fuel gas, 25C 85% CH4 15% C2H6 9-60 ...
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## This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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