Consider the combustion of methanol (CH3OH) in air at an equivalence ratio of 0.

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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 this reaction?

Nov 30th, 2014

Composition of air:

N2: 78.08%

O2: 20.95%

Argon: 0.93%

CO2: 0.03%

Others: 0.01%

Water vapor: variable

For simplicity, air is considered as a mixture that contains 79% N2 and 21% O2 only. Other components are neglected.

0.21O2 + 0.79N2 = 0.21(O2 + 3.76N2)

If oxygen 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.

CH3OH + 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

Nov 30th, 2014

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