Elementary Principles 138

# Elementary Principles 138 - 5.46 Basis 50.4 liters C 5 H 12...

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Unformatted text preview: 5.46 Basis: 50.4 liters C 5 H 12 l min bg 50.4 L C5 H12 (l ) / min n1 (kmol C5 H12 / min) heater n1 , n 2 Combustion chamber 15% excess air, Vair (L / min) n 2 kmol air 0.21 O 2 0.79 N 2 336 K, 208.6 kPa (gauge) n 3 (kmol C 5 H 12 / min) n 4 (kmol O 2 / min) n 5 (kmol N 2 / min) n 6 (kmol CO 2 / min) n 7 (kmol H 2 O / min) Condenser Vgas (L/min), 275 K, 1 atm n 4 (kmol O 2 / min) n 5 (kmol N 2 / min) n 6 (kmol CO 2 / min) C5 H 12 + 802 5CO 2 + 6H 2 O a. Vliq (L/min) m=3.175 kg C5 O12 / min n 3 (kmol C5 O12 / min) n 7 (kmol H 2 O(l ) / min) n1 = n3 = 50.4 L 0.630 kg 1 kmol = 0.440 kmol min min L 72.15 kg 3175 kg 1 kmol . = 0.044 kmol / min min 72.15 kg 0.440 - 0.044 kmol 100 = 90% C5 H 12 converted 0.440 frac. convert = n2 = 0.440 kmol C5 H12 1.15 ( 8 kmol O 2 ) 1 mol air = 19.28 kmol air min min kmol C5 H12 0.21 mol O 2 19.28 kmol 22.4 L STP min mol Vair = b g 336K 101 kPa 1000 mol = 173000 L min 273K 309.6 kPa kmol n 4 = [(0.21)(19.28) - (0.90)(0.440)(8)] n5 = n6 = kmol O 2 = 0.882 kmol O 2 / min min 19.28 kmol air 0.79 kmol N 2 = 15.23 kmol N 2 / min min kmol air 0.90(0.440 kmol C5 H 12 ) 5 kmol CO 2 . = 198 kmol CO 2 / min min kmol C5 H 12 Vgas = 0.882+15.23+1.98 kmol 22.4 L(STP) 275 K 1000 mol min mol 273 K kmol = 4.08 105 L/min 5-26 ...
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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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