Elementary Principles 285

# Elementary - 8.20 n = 1000 m 3 1 min 273 K min 60 s 1 kmol 303 K 22.4 m 3 STP Table B.8 for H b g = 0.6704 kmol s = 670.4 mol s 670.4 mol 0.73 kJ s

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8- 6 8.20 n == = 1000 m 1 min 273 K 1 kmol min 60 s 303 K 22.4 m STP kmol s mol / s 3 3 bg 06704 6704 .. Energy balance on air: QH n H Q H = = ΔΔ Δ 670.4 mol 0.73 kJ 1 kW s mol 1 kJ s kW Table B.8 for ± . 489 4 Solar energy required 489.4 kW heating 1 kW solar energy 0.3 kW heating kW 1631 Area required 2 2 1631 kW 1000 W 1 m 1813 m 1 kW 900 W 8.21 CH 5O 3CO 4HO 38 2 2 2 +→ + ² . ² . . n n fuel 3 air 2 2 SCFH h lb - mol 359 ft lb - mol h lb mol h lb - mol O 1b - mol C H 1 lb - mol air 0.211b - mol O lb mol h = × = = 135 10 1 376 376 5 115 103 10 5 4 2 1 302 45 8 2 1 2 3 0 4- 1 7 lb mol = 1.03 10 [0.02894 0.4147 10 0.3191 10 1.965 10 ] h 1.03 10 lb-mol 8.954 kJ 453.593 mol 9.486 10 Btu = =3.97 10 Btu/h h mol lb-mol kJ T p T QHnC d T TT T d T −−− Δ⋅ ⎛⎞ ×⋅ + × + × × ⎜⎟ ⎝⎠ ×× × ² 8.22 a. Basis : 100 mol feed (95 mol CH 4 and 5 mol C 2 H 6 ) CH O CO 2H O C H 7 2 O2 C O 3 H O 42 22 2 6 2 + + + 2 n O 4 26 2 2 2 mol CH mol O 1 mol CH mol C H mol O 1 mol C H mol O =⋅ + L N M M O Q P P = 125 95 2 5 35 259 4 . . . Product Gas : 2 2 CO : 95(1)+5(2)=105 mol CO H O: 95(2)+5(3)=205 mol H O O : 259.4-95(2)-5(3.5)=51.9 mol O N : 3.76(259.4)=975 mol N
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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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