Question

A tractor engine delivers 382 N.m brake torque at a rated speed of 2000 rpm. The radiator cooling fan draws 5% of the engine brake power. The fan pushes \(2.8\ \text{m}^3\text{s}^{-1}\) of air against 0.9 kPa static pressure rise. Assuming air as incompressible, and ignoring other losses, the fan efficiency, in %, is nearest to (Take \(\pi = 3.14\))

• A. 58
• B. 60
• C. 63
• D. 65

Hint:

Find brake power from torque and speed, then compare the fan's flow-pressure power to the power it actually draws.

Answer 1

The Correct Option is C

Step 1: Convert torque and speed into shaft power in kW.
Angular speed $\omega = \dfrac{2\pi N}{60} = \dfrac{2 \times 3.14 \times 2000}{60} = 209.33\ \text{rad/s}$.
Brake power $P_b = T \omega = 382 \times 209.33 = 79963\ \text{W} \approx 79.96\ \text{kW}$.

Step 2: Work out how much of that power goes to the fan.
Only 5% of the engine power drives the fan, so $P_{fan} = 0.05 \times 79.96 = 3.998\ \text{kW}$.

Step 3: Work out the power actually used to move the air.
Flow work rate equals volume flow times pressure rise: $P_{air} = 2.8\ \text{m}^3\text{s}^{-1} \times 0.9\ \text{kPa} = 2.52\ \text{kW}$.

Step 4: Divide output by input to get efficiency.
$\eta = \dfrac{2.52}{3.998} \times 100 = 63.0\%$, so among the choices 63% is the closest match.

Final Answer:
The cooling fan runs at about 63% efficiency. \[ \boxed{\eta_{fan} \approx 63\%} \]