Single cylinder engine: 20-minute trial

Data: $D = 0.2$ m, $L = 0.28$ m, $p_m = 3.1$ bar, $N = 350$ rpm, net brake load $W = 640$ N, drum diameter 1 m (no rope diameter given, so effective radius $R = 0.5$ m), fuel 1.52 kg in 20 min, CV = 43 900 kJ/kg.

For a four-stroke engine the number of power strokes per minute is $n = N/2 = 175$.

(i) Indicated power

$$\begin{aligned} A &= \frac{\pi}{4}D^2 = \frac{\pi}{4}(0.2)^2 = 0.031416\ \text{m}^2 \\ IP &= \frac{p_m L A n}{60} = \frac{3.1\times10^5 \times 0.28 \times 0.031416 \times 175}{60} \\ &= 7953\ \text{W} = 7.953\ \text{kW} \end{aligned}$$

(ii) Brake power

$$\begin{aligned} T &= W R = 640 \times 0.5 = 320\ \text{N m} \\ BP &= \frac{2\pi N T}{60} = \frac{2\pi \times 350 \times 320}{60} = 11\,729\ \text{W} = 11.729\ \text{kW} \end{aligned}$$

(iii) Mechanical efficiency

$$\eta_m = \frac{BP}{IP} = \frac{11.729}{7.953} = 1.475 = 147.5\%$$

(iv) Indicated thermal efficiency

$$\begin{aligned} \dot m_f &= \frac{1.52}{20\times60} = 0.0012667\ \text{kg/s}\ (4.56\ \text{kg/h}) \\ \text{Heat supplied} &= \dot m_f \times CV = 0.0012667 \times 43\,900 = 55.61\ \text{kW} \\ \eta_{ith} &= \frac{IP}{\dot m_f\,CV} = \frac{7.953}{55.61} = 0.1430 = 14.30\% \end{aligned}$$

Answer: IP = 7.953 kW, BP = 11.729 kW, $\eta_m$ = 147.5 %, $\eta_{ith}$ = 14.30 %.

Note on the data: a mechanical efficiency above 100 % is impossible (BP can never exceed IP), so the figures as printed are inconsistent. If the engine is taken as two-stroke ($n = N = 350$), IP doubles to 15.906 kW, giving $\eta_m = 11.729/15.906 = 73.7\%$ and $\eta_{ith} = 15.906/55.61 = 28.6\%$, which are realistic values. In the exam, show the working as above and point out the inconsistency.