Engine test: four-cylinder at 2000 rpm

Data: 4 cylinders, four-stroke, $D = 0.12$ m, $L = 0.14$ m, $N = 2000$ rpm, $\dot m_f = 15$ kg/h, CV = 42 000 kJ/kg, net tension $= 40$ kgf, brake circumference $= 3.0$ m, $\eta_m = 0.80$.

Brake drum radius: $R = \dfrac{3.0}{2\pi} = 0.4775$ m (no rope size given, so this is taken as effective radius). Net force: $F = 40 \times 9.81 = 392.4$ N.

Brake power:

$$BP = \frac{2\pi N F R}{60} = F \times \frac{\text{circumference} \times N}{60} = 392.4 \times \frac{3.0 \times 2000}{60} = 39\,240\ \text{W} = 39.24\ \text{kW}$$

i) Indicated horse power

$$IP = \frac{BP}{\eta_m} = \frac{39.24}{0.80} = 49.05\ \text{kW}$$

With 1 hp = 0.746 kW: $IHP = 49.05/0.746 = 65.75$ hp (66.69 metric hp if 1 hp = 735.5 W).

ii) Brake thermal efficiency

$$\begin{aligned} \dot Q &= \frac{15}{3600}\times 42\,000 = 175\ \text{kW} \\ \eta_{bth} &= \frac{39.24}{175} = 0.2242 = 22.42\% \end{aligned}$$

iii) Indicated mean effective pressure

$$\begin{aligned} A &= \frac{\pi}{4}(0.12)^2 = 0.011310\ \text{m}^2,\quad n = \frac{N}{2} = 1000,\ k = 4 \\ p_m &= \frac{60\,IP}{L A n k} = \frac{60 \times 49\,050}{0.14 \times 0.011310 \times 1000 \times 4} \\ &= 4.647\times10^5\ \text{N/m}^2 = 4.65\ \text{bar} \end{aligned}$$

iv) Brake specific fuel consumption

$$bsfc = \frac{15}{39.24} = 0.3823\ \text{kg/kWh}$$

Answer: IHP = 49.05 kW (65.75 hp), $\eta_{bth}$ = 22.42 %, imep = 4.65 bar, bsfc = 0.382 kg/kWh.