Two-stroke motoring test: 2 kW motoring

In a motoring test the engine is driven by the dynamometer without fuel; the motoring power equals the friction power: $FP = 2$ kW.

Data: $T = 150$ N m, $N = 900$ rpm, fuel = 3 kg in 1 h, CV = 40 MJ/kg = 40 000 kJ/kg.

(a) Brake power

$$BP = \frac{2\pi N T}{60} = \frac{2\pi \times 900 \times 150}{60} = 14\,137\ \text{W} = 14.137\ \text{kW}$$

(b) Indicated power

$$IP = BP + FP = 14.137 + 2 = 16.137\ \text{kW}$$

(c) Mechanical efficiency

$$\eta_m = \frac{BP}{IP} = \frac{14.137}{16.137} = 0.8761 = 87.61\%$$

(d) Indicated thermal efficiency

$$\begin{aligned} \dot Q &= \frac{3}{3600}\times 40\,000 = 33.33\ \text{kW} \\ \eta_{ith} &= \frac{IP}{\dot Q} = \frac{16.137}{33.33} = 0.4841 = 48.41\% \end{aligned}$$

Answer: BP = 14.14 kW, IP = 16.14 kW, $\eta_m$ = 87.61 %, $\eta_{ith}$ = 48.41 % (brake thermal efficiency = 42.41 %).