Elementary Principles 92

# Elementary Principles 92 - 4.59 a. Basis: 100 mol fed to...

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Unformatted text preview: 4.59 a. Basis: 100 mol fed to reactor/h 25 mol O2/h, 75 mol C2H4/h n1 (mol C 2H 4 //h) n2 (mol O 2 /h) n3 (mol C 2H 4O /h) reactor nC2H4 ( mol C 2H 4 /h) nO2 (mol O 2 /h) 75 mol C 2H 4 //h 25 mol O 2 /h n1 (mol C 2H 4 //h) n2 (mol O 2 /h) n3 (mol C 2H 4O /h) n4 (mol CO 2 /h) n5 (mol H 2O /h) Seperator separator n4 (mol CO 2 /h) n5 (mol H 2O /h) Reactor 5 unknowns (n1 - n5) -3 atomic balances -1 - % yield -1 - % conversion 0 D.F. Strategy: 1. Solve balances around reactor to find n1- n5 2. Solve balances around mixing point to find nO2, nC2H4 (1) % Conversion n1 = .800 * 75 (2) % yield: (.200)(75) mol C 2 H 4 90 mol C 2 H 4 O = n 3 (production rate of C 2 H 4 O) 100 mol C 2 H 4 (3) C balance (reactor): 150 = 2 n1 + 2 n3 + n4 (4) H balance (reactor): 300 = 4 n1 + 4 n3 + 2 n5 (5) O balance (reactor): 50 = 2 n2 + n3 + 2 n4 + n5 (6) O2 balance (mix pt): nO2 = 25 n2 (7) C2H4 balance (mix pt): nC2H4 = 75 n1 Overall conversion of C2H4: 100% b. n1 = 60.0 mol C2H4/h n2 = 13.75 mol O2 /h n3 = 13.5 mol C2H4O/h n4 = 3.00 mol CO2/h n5 = 3.00 mol H2O/h nO2 = 11.25 mol O2/h nC2H4 = 15.0 mol C2H4/h 100% conversion of C2H4 c. Scale factor = 2000 lbm C 2 H 4 O 1 lb - mole C 2 H 4 O h lb - mol / h = 3.363 h 44.05 lbm C 2 H 4 O 13.5 mol C 2 H 4 O mol / h nC2H4 = (3.363)(15.0) = 50.4 lb-mol C2H4/h nO2 = (3.363)(11.25) = 37.8 lb-mol O2/h 4-51 ...
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