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Two-stroke motoring test: 1.5 kW motoring

Short answer

BP = 7.54 kW, IP = 9.04 kW, mechanical efficiency = 83.41%, indicated thermal efficiency = 32.30%.

A two stroke diesel engine was motored when the meter reading was 1.5 kW. Then the test on the engine was carried out for one hour and the following observations were recorded: Brake torque = 120 Nm; Speed = 600 rpm; Fuel used = 2.5 kg; calorific value of fuel = 40.3 MJ/kg; Determine: (a) Brake power, (b) Indicated power, (c) Mechanical efficiency and (d) Indicated thermal efficiency.

In a motoring test the engine is turned by the dynamometer without fuel; the power absorbed equals the friction power: FP=1.5FP = 1.5 kW.

Data: T=120T = 120 N m, N=600N = 600 rpm, fuel = 2.5 kg in 1 h, CV = 40.3 MJ/kg = 40 300 kJ/kg.

(a) Brake power

BP=2πNT60=2π×600×12060=7540 W=7.540 kWBP = \frac{2\pi N T}{60} = \frac{2\pi \times 600 \times 120}{60} = 7540\ \text{W} = 7.540\ \text{kW}

(b) Indicated power

IP=BP+FP=7.540+1.5=9.040 kWIP = BP + FP = 7.540 + 1.5 = 9.040\ \text{kW}

(c) Mechanical efficiency

ηm=BPIP=7.5409.040=0.8341=83.41%\eta_m = \frac{BP}{IP} = \frac{7.540}{9.040} = 0.8341 = 83.41\%

(d) Indicated thermal efficiency

Q˙=2.53600×40 300=27.99 kWηith=IPQ˙=9.04027.99=0.3230=32.30%\begin{aligned} \dot Q &= \frac{2.5}{3600}\times 40\,300 = 27.99\ \text{kW} \\ \eta_{ith} &= \frac{IP}{\dot Q} = \frac{9.040}{27.99} = 0.3230 = 32.30\% \end{aligned}

Answer: BP = 7.54 kW, IP = 9.04 kW, ηm\eta_m = 83.41 %, ηith\eta_{ith} = 32.30 % (brake thermal efficiency = 26.94 %).

✓ Answer checked by Claude Opus 5.5 (high effort)

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