Elementary - 4.70 a C5H12 8 O2 5 CO2 6 H2O Basis 100 moles dry product gas n1(mol C5H12 Excess air n2(mol O2 3.76n2(mol N2 100 mol dry product

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Unformatted text preview: 4.70 a. C5H12 + 8 O2 5 CO2 + 6 H2O Basis: 100 moles dry product gas n1 (mol C5H12) Excess air n2 (mol O2) 3.76n2 (mol N2) 100 mol dry product gas (DPG) 0.0027 mol C5H12/mol DPG 0.053 mol O2/mol DPG 0.091 mol CO2/mol DPG 0.853 mol N2/mol DPG n3 (mol H2O) 3 unknowns (n1, n2, n3) -3 atomic balances (O, C, H) -1 N2 balance -1 D.F. Problem is overspecified b. N2 balance: 3.76 n2 = 0.8533 (100) n2 = 22.69 mol O2 C balance: 5 n1 = 5(0.0027)(100) + (0.091)(100) n1 = 2.09 mol C5H12 H balance: 12 n1 = 12(0.0027)(100) + 2n3 n3 = 10.92 mol H2O O balance: 2n2 = 100[(0.053)(2) + (0.091)(2)] + n3 45.38 mol O = 39.72 mol O Since the 4th balance does not close, the given data cannot be correct. c. n1 (mol C5H12) Excess air n2 (mol O2) 3.76n2 (mol N2) 100 mol dry product gas (DPG) 0.00304 mol C5H12/mol DPG 0.059 mol O2/mol DPG 0.102 mol CO2/mol DPG 0.836 mol N2/mol DPG n3 (mol H2O) N2 balance: 3.76 n2 = 0.836 (100) n2 = 22.2 mol O2 C balance: 5 n1 = 100 (5*0.00304 + 0.102) n1 = 2.34 mol C5H12 H balance: 12 n1 = 12(0.00304)(100) + 2n3 n3 = 12.2 mol H2O O balance: 2n2 = 100[(0.0590)(2) + (0.102)(2)] + n3 44.4 mol O = 44.4 mol O Fractional conversion of C5H12: 2.344 - 100 0.00304 = 0.870 mol react/mol fed 2.344 Theoretical O2 required: 2.344 mol C5H12 (8 mol O2/mol C5H12) = 18.75 mol O2 % excess air: 22.23 mol O 2 fed - 18.75 mol O 2 required 100% = 18.6% excess air 18.75 mol O 2 required 4-62 ...
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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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