Elementary Principles 106

# Elementary Principles 106 - 4.73 a. C3H8 +5 O2 3 CO2 + 4...

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Unformatted text preview: 4.73 a. C3H8 +5 O2 3 CO2 + 4 H2O, C4H10 + 13/2 O2 4 CO2 + 5 H2O Basis 100: mol product gas n1 (mol C3H8) n2 (mol C4H10) n3 (mol O2) 100 mol 0.474 mol H2O/mol x (mol CO2/mol) (0.526x) (mol O2/mol) x 69.4 = x = 0.365 mol CO 2 /mol 0.526 - x 30.6 Dry product gas contains 69.4% CO2 3 unknowns (n1, n2, n3) 3 balances (C, H, O) = 0 D.F. O balance: 2 n3 = 152.6 n3 = 76.3 mol O2 n1 = 7.1 mol C 3 H 8 C balance : 3 n1 + 4 n 2 = 36.5 65.1% C 3 H 8 , 34.9% C 4 H10 H balance : 8 n1 + 10 n 2 = 94.8 n 2 = 3.8 mol C 4 H10 b. nc=100 mol (0.365 mol CO2/mol)(1mol C/mol CO2) = 365 mol C nh = 100 mol (0.474 mol H2O/mol)(2mol H/mol H2O)=94.8 mol H 27.8%C, 72.2% H From a: 7.10 mol C 3 H 8 3.80 mol C 4 H10 4 mol C 3 mol C + mol C 3 H 8 mol C 4 H10 7.10 mol C 3 H 8 11 mol (C + H) 3.80 mol C 4 H10 14 mol (C + H) + mol C 3 H 8 mol C 4 H10 4.74 100% = 27.8% C Basis: 100 kg fuel oil Moles of C in fuel: 100 kg 0.85 kg C 1 kmol C = 7.08 kmol C kg 12.01 kg C 100 kg 0.12 kg H 1 kmol H = 12.0 kmol H kg 1 kg H 100 kg 0.017 kg S 1 kmol S = 0.053 kmol S kg 32.064 kg S Moles of H in fuel: Moles of S in fuel: 1.3 kg non-combustible materials (NC) 4-65 ...
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