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Split_00061 - 9.51(cont'd 1 mol s fuel gas 0.85 mol CH 4 s...

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9- 61 b. 1 mol / s fuel gas 0.85 mol CH / s 0.15 mol C H / s 4 2 6 , Theoretical oxygen 2 mol O 0.85 mol CH 1 mol CH s 3.5 mol O 0.15 mol C H 1 mol C H s mol O / s 2 4 4 2 2 6 2 6 2 = + = 2 225 . Assume 10% excess O O fed = 1.1 2.225 = 2.448 mol O / s 2 2 2 × C balance : mol CO / s 2 ± . . ± . n n 2 2 085 1 015 2 115 = + = b gb g b gb g H balance mol H O / s 2 : ± . . ± . 2 085 4 015 6 215 3 3 n n = + = b gb g b gb g 10% 01 2 225 0 223 4 excess O mol O / s mol O / s 2 2 2 = = ± . . . n b gb g Extents of reaction : ± ± . ± ± . ξ ξ 1 2 085 015 = = = = n n CH C H 4 2 6 mol / s, mol / s Reference states: CH g , C H g , N g , O g , H O l , CO (g) at 25 C 4 2 6 2 2 2 2 o b g b g b g b g b g (We will use the values of Δ ² H c o given in Table B.1, which are based on H O l 2 b g as a combustion product, and so must choose the liquid as a reference state for water) Substance mol kJ mol mol kJ mol CH C H O CO H O v in in out out 4 2 6 2 2 ± ² ± ² . . . . . . n H n H H 085 0 015 0 2 225 0 0 223 0 115 0 215 2 1 b g ² ² . H H 1 25 44 01 = = Δ v o C kJ / mol e j Energy Balance : ± ± ² ± ² ± ² ± ² Q n H n H n H n H i i i i = + + CH c o CH C H c o C H out in 4 4 2 6 2 6 Δ Δ e j e j mol / s CH kJ mol mol / s C H kJ mol mol / s H O kJ / mol
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