Engine test: four-cylinder with rope radius

Data: 4 cylinders, four-stroke, $D = 0.1$ m, $L = 0.12$ m, $N = 1800$ rpm, $\dot m_f = 0.2$ kg/min, CV = 44 000 kJ/kg, net rope tension $(W - S) = 40$ kgf, brake circumference = 3.0 m, rope radius = 25 mm, $\eta_m = 0.90$.

Brake power

Drum radius from circumference: $R_d = \dfrac{3.0}{2\pi} = 0.4775$ m. Effective radius (to rope centre): $R_e = 0.4775 + 0.025 = 0.5025$ m. Net force: $F = 40 \times 9.81 = 392.4$ N.

$$\begin{aligned} T &= F R_e = 392.4 \times 0.5025 = 197.17\ \text{N m} \\ BP &= \frac{2\pi N T}{60} = \frac{2\pi \times 1800 \times 197.17}{60} = 37.165\ \text{kW} \\ IP &= \frac{BP}{\eta_m} = \frac{37.165}{0.9} = 41.295\ \text{kW} \end{aligned}$$

Heat supplied: $\dot m_f\, CV = \dfrac{0.2}{60} \times 44\,000 = 146.67$ kW.

(i) Brake thermal efficiency

$$\eta_{bth} = \frac{37.165}{146.67} = 0.2534 = 25.34\%$$

(ii) Indicated thermal efficiency

$$\eta_{ith} = \frac{41.295}{146.67} = 0.2816 = 28.16\%$$

(iii) Indicated mean effective pressure

For a four-stroke, $k$-cylinder engine: $IP = \dfrac{p_m L A (N/2) k}{60}$.

$$\begin{aligned} A &= \frac{\pi}{4}(0.1)^2 = 7.854\times10^{-3}\ \text{m}^2 \\ p_m &= \frac{60\, IP}{L A (N/2) k} = \frac{60 \times 41\,295}{0.12 \times 7.854\times10^{-3} \times 900 \times 4} \\ &= 7.302\times10^5\ \text{N/m}^2 = 7.30\ \text{bar} \end{aligned}$$

(iv) Brake specific fuel consumption

$$bsfc = \frac{0.2 \times 60}{37.165} = \frac{12}{37.165} = 0.3229\ \text{kg/kWh}$$

Answer: $\eta_{bth}$ = 25.34 %, $\eta_{ith}$ = 28.16 %, imep = 7.30 bar, bsfc = 0.323 kg/kWh (BP = 37.17 kW, IP = 41.30 kW).