Question

The indicated power of the single-cylinder engine is 1.9 kW. It develops torque of 10 Nm at 1600 rpm. What is the friction power as the percentage of brake power?

(a) 12.43%

(b) 13.43%

(c) 14.43%

(d) 15.43%

Answer 1

The correct choice is (b) 13.43%

Easiest explanation: Brake power = bp, Indicated power = ip = 1.9kW, friction power = fp, speed = N = 1600 rpm, and torque = T = 10 Nm. Brake power = bp = 2πNT / 60000 = 2π * 1600 * 10 / 60000 = 1.675 kW. Friction power = fp = ip – bp = 1.9 – 1.675 = 0.225 kW. Percentage loss = (fp / bp)*100 = (0.225 / 1.675)*100 = 13.43%.