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Friction power as percentage of indicated power

Short answer

BP=2π(1800)(8)/60=1.508 kW; FP=IP-BP=1.8-1.508=0.292 kW. Friction loss is 0.292/1.8×100=16.22% of indicated power.

A single cylinder engine running at 1800 rpm develops a torque of 8 Nm. The indicated power of the engine 1.8 kW. Find the loss due to friction power as the percentage of indicated power.

Data: N=1800N = 1800 rpm, T=8T = 8 N m, IP=1.8IP = 1.8 kW.

Brake power:

BP=2πNT60=2π×1800×860=1508 W=1.508 kWBP = \frac{2\pi N T}{60} = \frac{2\pi \times 1800 \times 8}{60} = 1508\ \text{W} = 1.508\ \text{kW}

Friction power:

FP=IP−BP=1.8−1.508=0.292 kWFP = IP - BP = 1.8 - 1.508 = 0.292\ \text{kW}

Friction loss as a percentage of IP:

FPIP×100=0.2921.8×100=16.22%\frac{FP}{IP}\times 100 = \frac{0.292}{1.8}\times100 = 16.22\%

Answer: Friction power = 0.292 kW, which is 16.22 % of the indicated power (mechanical efficiency = 83.78 %).

✓ Answer checked by Claude Opus 5.5 (high effort)

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