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Rope dynamometer trial: brake and indicated power

Short answer

Effective radius is 0.325 m, net brake load 267.32 N, torque 86.88 Nm. Brake power = 5.91 kW, indicated power = 7.39 kW.

The following observations were recorded during a trial of a four stroke engine with rope dynamometer. Engine speed = 650 rpm, Dia. of brake drum = 600 mm, Dia. of rope = 50 mm, Dead load on the brake drum = 32 kg, spring balance reading = 4.75 kg, Mechanical efficiency = 80%. Calculate the brake power and indicated power.

Data: N=650N = 650 rpm, drum diameter D=0.6D = 0.6 m, rope diameter d=0.05d = 0.05 m, dead load W=32W = 32 kg, spring balance S=4.75S = 4.75 kg, ηm=0.80\eta_m = 0.80.

For a rope brake dynamometer the effective radius is measured to the rope centre:

Re=D+d2=0.6+0.052=0.325 mR_e = \frac{D + d}{2} = \frac{0.6 + 0.05}{2} = 0.325\ \text{m}

Net brake load:

(W−S)g=(32−4.75)×9.81=267.32 N(W - S)g = (32 - 4.75)\times 9.81 = 267.32\ \text{N}

Brake power

T=(W−S)g Re=267.32×0.325=86.88 N mBP=2πNT60=2π×650×86.8860=5914 W=5.914 kW\begin{aligned} T &= (W-S)g\,R_e = 267.32 \times 0.325 = 86.88\ \text{N m} \\ BP &= \frac{2\pi N T}{60} = \frac{2\pi \times 650 \times 86.88}{60} = 5914\ \text{W} = 5.914\ \text{kW} \end{aligned}

Indicated power

IP=BPηm=5.9140.80=7.392 kWIP = \frac{BP}{\eta_m} = \frac{5.914}{0.80} = 7.392\ \text{kW}

Answer: BP = 5.91 kW, IP = 7.39 kW (friction power = 1.48 kW).

✓ Answer checked by Claude Opus 5.5 (high effort)

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