Calculate the Stoichiometric air/fuel ratio for the complete combustion of Octane (CH8H18)
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Expert's answer
2011-01-13T12:18:12-0500
2CH8H18 + 15O2 = 2CO2 + 26H2O
a molecular weight of Octane is 12 + 26 = 38 a molecular weight of oxygen molecula is 16*2 = 32 Thus the oxygen-fuel mass ratio is OFR = 15*32/2*38 = 6.316 There is 23.2 mass-percent of oxygen in the air, then AFR = M(air)/M(fuel) = OFR / 0.232 = 6.316/0.232 = 27.22
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