Question

For a given fuel oil with composition C₁₁H₂₀N, calculate the stoichiometric amount of air required for complete combustion to products CO₂, H₂O and NO. Assume no thermal NO_x is formed. Present final answer as kg air/kg fuel.

Answer 1

Oxidation reaction:

C11H20N + (33/2)O2 ---> 11CO2 + 10H2O + NO

O required = 1*2 + 10*1 + 1 = 33

O2 required = 33/2

Basis: 1 kmole of fuel

Molecular weight of the fuel C11H20N = 12*11 + 10*1 + 14*1 = 156 kg/kmol

mass of fuel supplied = 1 kmol x 156 kg/kmol = 156 kg

O2 required stoichiometrically = (33/2) kmol = 16.5 kmol

Air contains 21 % by mole O2 and 79 % by mole N2

.: Air required theoretically = 16.5 kmol / 0.21 = 78.57142857 kmol

Molecular weight of O2 = 2*16 = 32 kg/kmol

Molecualr weight of N2 = 2*14 = 28 kg/kmol

Molecular weight of air = 0.21*32 + 0.79*28 = 28.84 kg/kmol

mass of air required = 78.57142857 kmol x 28.84 kg/kmol = 2266 kg

kg air / kg fuel = 2266 / 156 = 14.52564........................................................Answer