Two-stroke motoring test with specific fuel consumption

In a motoring test the engine is driven by the dynamometer without fuel; the motoring power equals the friction power: $FP = 4.5$ kW.

Data: $T = 250$ N m, $N = 1500$ rpm, fuel = 5 kg/h, CV = 40 MJ/kg = 40 000 kJ/kg.

Brake power and indicated power:

$$\begin{aligned} BP &= \frac{2\pi N T}{60} = \frac{2\pi \times 1500 \times 250}{60} = 39.27\ \text{kW} \\ IP &= BP + FP = 39.27 + 4.5 = 43.77\ \text{kW} \end{aligned}$$

Heat supplied:

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

i) Mechanical efficiency

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

ii) Indicated thermal efficiency

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

iii) Brake thermal efficiency

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

iv) Indicated specific fuel consumption

$$isfc = \frac{\dot m_f}{IP} = \frac{5}{43.77} = 0.1142\ \text{kg/kWh}$$

v) Brake specific fuel consumption

$$bsfc = \frac{\dot m_f}{BP} = \frac{5}{39.27} = 0.1273\ \text{kg/kWh}$$

Answer: $\eta_m$ = 89.72 %, $\eta_{ith}$ = 78.79 %, $\eta_{bth}$ = 70.69 %, isfc = 0.114 kg/kWh, bsfc = 0.127 kg/kWh.

(The efficiencies are unrealistically high for a real engine because the given fuel rate is low; the method is the point of the problem.)