Air flow and fuel flow of an SI engine

Given: $V_d = 1.6\times10^{-3}$ m³, $N = 3000$ rpm, $\eta_v = 0.85$, $\rho_a = 1.18$ kg/m³, $BP = 35$ kW.

(i) Air flow

A four-stroke engine inhales once every two revolutions.

$$\dot m_a = \eta_v \rho_a V_d \frac{N}{2\times 60} = 0.85 \times 1.18 \times 1.6\times10^{-3} \times \frac{3000}{120} = 0.0401\ \text{kg/s}$$

That is $0.0401 \times 3600 = 144.4$ kg/h.

(ii) Fuel flow at $AFR = 14.7$

$$\dot m_f = \frac{\dot m_a}{14.7} = \frac{144.4}{14.7} = 9.83\ \text{kg/h}$$

For a petrol density of 740 kg/m³, this is $9.83/0.74 = 13.3$ L/h.

(iii) Brake specific fuel consumption

$$bsfc = \frac{\dot m_f}{BP} = \frac{9.83 \times 1000}{35} = 281\ \text{g/kWh}$$

(iv) Full-load mixture

Taking the same air flow:

$$\dot m_f = \frac{144.4}{12.5} = 11.55\ \text{kg/h}, \qquad \lambda = \frac{AFR_{actual}}{AFR_{stoich}} = \frac{12.5}{14.7} = 0.85$$

The mixture is rich: it has 15 % less air than stoichiometric, or about 18 % excess fuel ($\phi = 1/\lambda = 1.18$).

Answer: $\dot m_a = 0.0401$ kg/s; $\dot m_f = 9.83$ kg/h at $\lambda = 1$; bsfc $= 281$ g/kWh; full-load fuel $= 11.55$ kg/h with $\lambda = 0.85$.