IP, BP and efficiencies of a petrol engine

Given: $D = 0.08$ m, $L = 0.1$ m, $n = 4$, $N = 3000$ rpm, $p_{mi} = 9 \times 10^5$ Pa, $W = 24 \times 9.81 = 235.4$ N, $R = 0.5$ m, $\dot{m}_f = 11.2$ kg/h, $CV = 44\,000$ kJ/kg.

Indicated power

For a four-stroke engine the number of power strokes per cylinder per second is $N/2 = 25$.

$$\begin{aligned} A &= \frac{\pi}{4}(0.08)^2 = 5.027 \times 10^{-3}\ \text{m}^2 \\ IP &= \frac{p_{mi} L A \cdot (N/2) \cdot n}{1000} = \frac{9 \times 10^5 \times 0.1 \times 5.027 \times 10^{-3} \times 25 \times 4}{1000} = 45.24\ \text{kW} \end{aligned}$$

Brake power

$$\begin{aligned} T &= W R = 235.44 \times 0.5 = 117.7\ \text{N m} \\ BP &= \frac{2\pi N T}{60\,000} = \frac{2\pi \times 3000 \times 117.7}{60\,000} = 36.98\ \text{kW} \end{aligned}$$

Friction power and mechanical efficiency

$$\begin{aligned} FP &= IP - BP = 45.24 - 36.98 = 8.26\ \text{kW} \\ \eta_m &= \frac{BP}{IP} = \frac{36.98}{45.24} = 0.8175 \end{aligned}$$

Thermal efficiencies

$$\begin{aligned} \text{Fuel energy} &= \frac{11.2}{3600} \times 44\,000 = 136.9\ \text{kW} \\ \eta_{bth} &= \frac{36.98}{136.9} = 0.270 \\ \eta_{ith} &= \frac{45.24}{136.9} = 0.330 \end{aligned}$$

Check: $\eta_m \eta_{ith} = 0.8175 \times 0.330 = 0.270$.

Specific fuel consumption

$$bsfc = \frac{11.2}{36.98} = 0.303\ \text{kg/kWh}, \qquad isfc = \frac{11.2}{45.24} = 0.248\ \text{kg/kWh}$$

Answer: IP = 45.2 kW, BP = 37.0 kW, FP = 8.26 kW, $\eta_m$ = 81.8%, $\eta_{ith}$ = 33.0%, $\eta_{bth}$ = 27.0%, bsfc = 0.303 kg/kWh, isfc = 0.248 kg/kWh.