Single cylinder engine: 60-minute trial

Data: single cylinder, four-stroke, $D = 0.3$ m, $L = 0.45$ m, fuel 11.4 kg in 60 min, CV = 42 000 kJ/kg, $p_m = 6.0$ bar, net brake load $W = 1500$ N, $N = 300$ rpm, drum diameter 1.8 m, rope diameter 20 mm.

Power strokes per minute: $n = N/2 = 150$.

(a) Indicated power

$$\begin{aligned} A &= \frac{\pi}{4}(0.3)^2 = 0.070686\ \text{m}^2 \\ IP &= \frac{p_m L A n}{60} = \frac{6\times10^5 \times 0.45 \times 0.070686 \times 150}{60} \\ &= 47\,713\ \text{W} = 47.71\ \text{kW} \end{aligned}$$

(b) Brake power

Effective radius $R_e = \dfrac{1.8 + 0.02}{2} = 0.91$ m.

$$\begin{aligned} T &= W R_e = 1500 \times 0.91 = 1365\ \text{N m} \\ BP &= \frac{2\pi N T}{60} = \frac{2\pi \times 300 \times 1365}{60} = 42\,883\ \text{W} = 42.88\ \text{kW} \end{aligned}$$

(c) Mechanical efficiency

$$\eta_m = \frac{BP}{IP} = \frac{42.88}{47.71} = 0.8988 = 89.88\%$$

(d) Indicated thermal efficiency

$$\begin{aligned} \dot Q &= \frac{11.4}{3600}\times 42\,000 = 133.0\ \text{kW} \\ \eta_{ith} &= \frac{IP}{\dot Q} = \frac{47.71}{133.0} = 0.3587 = 35.87\% \end{aligned}$$

Answer: IP = 47.71 kW, BP = 42.88 kW, $\eta_m$ = 89.88 %, $\eta_{ith}$ = 35.87 %.