Diesel engine 75 kW: fuel consumption and efficiencies

Data: $\dot m_f = 5.5$ g/s $= 0.0055$ kg/s, $BP = 75$ kW, $\eta_m = 0.85$, LHV $= 44$ MJ/kg $= 44\,000$ kJ/kg.

Indicated power:

$$IP = \frac{BP}{\eta_m} = \frac{75}{0.85} = 88.24\ \text{kW}$$

Fuel used per hour: $0.0055 \times 3600 = 19.8$ kg/h.

a) Brake specific fuel consumption

$$bsfc = \frac{\dot m_f\ (\text{kg/h})}{BP} = \frac{19.8}{75} = 0.264\ \text{kg/kWh}$$

b) Indicated specific fuel consumption

$$isfc = \frac{19.8}{88.24} = 0.2244\ \text{kg/kWh}$$

c) Brake thermal efficiency

Heat supplied by fuel: $\dot m_f \times LHV = 0.0055 \times 44\,000 = 242$ kW.

$$\eta_{bth} = \frac{BP}{\dot m_f\,LHV} = \frac{75}{242} = 0.3099 = 30.99\%$$

d) Indicated thermal efficiency

$$\eta_{ith} = \frac{IP}{\dot m_f\,LHV} = \frac{88.24}{242} = 0.3646 = 36.46\%$$

Check: $\eta_{bth} = \eta_m \times \eta_{ith} = 0.85 \times 36.46 = 30.99\%$.

Answer: bsfc = 0.264 kg/kWh, isfc = 0.2244 kg/kWh, $\eta_{bth}$ = 30.99 %, $\eta_{ith}$ = 36.46 %.