Elementary Principles 139 - 5.46 (cont'd) n7 = 0.9(0.440...

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Unformatted text preview: 5.46 (cont'd) n7 = 0.9(0.440 kmol C5 H 12 ) 6 kmol H 2 O = 2.38 kmol H 2 O(l ) / min min kmol C5 H 12 Condensate: VC5H12 = VH 2 O = 0.044 kmol 72.15 kg L = 5.04 L min min kmol 0.630 kg L 1 kg min kmol = 42.89 L min 2.38 kmol 18.02 kg Assume volume additivity (liquids are immiscible) Vliq = 5.04 + 42.89 = 47.9 L min b. C5 H 12 (l ) C 5 H 12 l H 2O l bg bg H 2O l bg 5.47 n air (kmol / min), 25 C, 1 atm 0.21 O 2 0.79 N 2 n 0 (kmol / min) 0.20 kmol H 2S / mol 0.80 kmol CO 2 / mol n 2 (kmol H 2 S / min) n1 (kmol H 2S / min) Furnace 3 H 2 S + 2 O 2 SO 2 + H 2 O 2H 2 S + SO 2 3S(g) + 2 H 2 O Reactor 10.0 m 3 / min @ 380 C, 205 kPa n exit (kmol / min) n 3 (kmol N 2 / min) n 4 (kmol H 2 O / min) n 5 (kmol CO 2 / min) n 6 (kmol S / min) n exit = PV 205 kPa 10.0 m3 / min = = 0.377 kmol / min m3 kPa RT 8.314 kmolK 653 K n1 = 0.20 n 0 / 3 = 0.0667n 0 ; n 2 = 2 n1 = 0.133n 0 b g 5-27 ...
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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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