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Two-stroke motoring test: 2.2 kW motoring

Short answer

Motoring test gives FP=2.2 kW. BP=2π(800)(150)/60=12.57 kW, IP=14.77 kW, fuel heat=27.78 kW. Mechanical efficiency=85.10%; indicated thermal efficiency=53.16%.

A two stroke diesel engine was motored when the meter reading was 2.2 kW. Then the test on the engine was carried out for one hour and the following observations were recorded: Brake torque = 150 Nm; Speed = 800 rpm; Fuel used = 2.5 kg; calorific value of fuel = 40 MJ/kg; Determine: (a) Brake power (b) Indicated power, (c) Mechanical efficiency and (d) Indicated thermal efficiency.

In the motoring test the engine is turned by the dynamometer without fuel; the power required equals the friction power, so FP=2.2FP = 2.2 kW.

Data: T=150T = 150 N m, N=800N = 800 rpm, fuel = 2.5 kg in 1 h, CV = 40 MJ/kg = 40 000 kJ/kg.

(a) Brake power

BP=2πNT60=2π×800×15060=12 566 W=12.566 kWBP = \frac{2\pi N T}{60} = \frac{2\pi \times 800 \times 150}{60} = 12\,566\ \text{W} = 12.566\ \text{kW}

(b) Indicated power

IP=BP+FP=12.566+2.2=14.766 kWIP = BP + FP = 12.566 + 2.2 = 14.766\ \text{kW}

(c) Mechanical efficiency

ηm=BPIP=12.56614.766=0.8510=85.10%\eta_m = \frac{BP}{IP} = \frac{12.566}{14.766} = 0.8510 = 85.10\%

(d) Indicated thermal efficiency

Heat supplied:

Q˙=2.53600×40 000=27.78 kW\dot Q = \frac{2.5}{3600}\times 40\,000 = 27.78\ \text{kW} ηith=IPQ˙=14.76627.78=0.5316=53.16%\eta_{ith} = \frac{IP}{\dot Q} = \frac{14.766}{27.78} = 0.5316 = 53.16\%

Answer: BP = 12.57 kW, IP = 14.77 kW, ηm\eta_m = 85.10 %, ηith\eta_{ith} = 53.16 % (brake thermal efficiency would be 45.24 %).

✓ Answer checked by Claude Opus 5.5 (high effort)

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