Two-stroke motoring test: 4.5 kW motoring

In a motoring test, the engine (without fuel) is driven by the dynamometer acting as a motor. The power needed to turn it at the test speed equals the friction power of the engine. So $FP = 4.5$ kW.

Data: $T = 250$ N m, $N = 1500$ rpm, fuel = 5 kg in 1 h, CV = 40 MJ/kg.

Brake power:

$$BP = \frac{2\pi N T}{60} = \frac{2\pi \times 1500 \times 250}{60} = 39\,270\ \text{W} = 39.27\ \text{kW}$$

Indicated power:

$$IP = BP + FP = 39.27 + 4.5 = 43.77\ \text{kW}$$

Heat supplied by fuel:

$$\dot Q = \frac{5}{3600} \times 40\,000 = 55.56\ \text{kW}$$

a) Mechanical efficiency

$$\eta_m = \frac{BP}{IP} = \frac{39.27}{43.77} = 0.8972 = 89.72\%$$

b) Indicated thermal efficiency

$$\eta_{ith} = \frac{IP}{\dot Q} = \frac{43.77}{55.56} = 0.7879 = 78.79\%$$

c) Brake thermal efficiency

$$\eta_{bth} = \frac{BP}{\dot Q} = \frac{39.27}{55.56} = 0.7069 = 70.69\%$$

Answer: $\eta_m$ = 89.72 %, $\eta_{ith}$ = 78.79 %, $\eta_{bth}$ = 70.69 %.

(These thermal efficiencies are far higher than any real diesel engine, about 30–40 %, because the given fuel consumption is low; the method is what matters.)