Efficiency ratio and specific fuel consumption

Fuel energy

$$Q_f = \dot{m}_f \times CV = \frac{6.2}{3600} \times 44\,000 = 75.78\ \text{kW}$$

Efficiencies

$$\begin{aligned} \eta_m &= \frac{BP}{IP} = \frac{18}{22} = 0.818 \\ \eta_{ith} &= \frac{IP}{Q_f} = \frac{22}{75.78} = 0.290 \\ \eta_{bth} &= \frac{BP}{Q_f} = \frac{18}{75.78} = 0.2375 \end{aligned}$$

Air-standard efficiency and relative efficiency

$$\begin{aligned} \eta_{Otto} &= 1 - \frac{1}{r^{\gamma-1}} = 1 - 7^{-0.4} = 0.541 \\ \eta_{rel} &= \frac{\eta_{ith}}{\eta_{Otto}} = \frac{0.290}{0.541} = 0.537 \end{aligned}$$

The relative efficiency of 53.7% shows how well the real engine approaches the ideal cycle: the rest is lost to the real-gas properties, heat loss, finite burning time and incomplete combustion. Typical values are 50-60% for SI engines.

Specific fuel consumption

$$bsfc = \frac{6.2}{18} = 0.344\ \text{kg/kWh}, \qquad isfc = \frac{6.2}{22} = 0.282\ \text{kg/kWh}$$

Answer: $\eta_m$ = 81.8%, $\eta_{ith}$ = 29.0%, $\eta_{bth}$ = 23.8%, $\eta_{rel}$ = 53.7%, bsfc = 0.344 kg/kWh, isfc = 0.282 kg/kWh.