Consider the combustion of methanol (CH3OH) in air at
an equivalence ratio of 0.85.
a) What is the fuel to air ratio on a volume basis for
simplicity, air is considered as a mixture that contains 79% N2 and 21% O2
only. Other components are neglected.
0.79N2 = 0.21(O2 + 3.76N2)
is contained in the fuel, the amount of air required to oxide the fuel is
reduced due to the presence of oxygen in the air fuel mixture.
1.5(O2 + 3.76N2) → CO2 + 2H2O + 5.64N2
On a molar
basis, the ratio of fuel to air is
[1/(3.76+5.64)] = 1/59.5 = 0.1067.
Please chose it as a best answer. Thx
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