Thermodynamics and Propulsion

15.2 Fuel-Air Ratio

The reaction for aeroengine fuel at stoichiometric conditions is

$$C_8H_{18} +12.5 O_2 +12.5 (3.76)N_2 \longrightarrow 8 CO_2 +9 H_2 O + 47.0 N_2.$$

On a molar basis, the ratio of fuel to air is $[1/(12.5+47.0)] = 1/59.5 = 0.0167$ .

To find the ratio on a mass flow basis, which is the way in which the aeroengine industry discusses it, we need to ``weight'' the molar proportions by the molecular weight of the components. The fuel molecular weight is $114\textrm{ g/mol}$ , the oxygen molecular weight is $32\textrm{ g/mol}$ and the nitrogen molecular weight is (approximately) $28\textrm{ g/mol}$ . The fuel/air ratio on a mass flow basis is thus

$$\textrm{Fuel-air ratio: }= \frac{1\textrm{ mol} \times 114\textrm... ...extrm{ g/mol} +12.5 \times 3.76\textrm{ mol} \times28\textrm{ g/mol}} = 0.0664.$$

If we used the actual constituents of air we would get 0.0667, a value about 0.5% different.

Muddy Points

Do we always assume 100% complete combustion? How good an approximation is this? (MP 15.3)