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: Find the engine brake power.
Brake power comes from torque and speed: \( P_b = \dfrac{2\pi N T}{60} \), with \(T = 382\ \text{N.m}\) and \(N = 2000\ \text{rpm}\).
\( P_b = \dfrac{2 \times 3.14 \times 2000 \times 382}{60} = \dfrac{4797920}{60} = 79965.3\ \text{W} \)

Step 2: Find the power drawn by the cooling fan.
The fan takes 5% of the brake power to run.
\( P_{fan} = 0.05 \times 79965.3 = 3998.3\ \text{W} \)

Step 3: Find the useful (air) power delivered by the fan.
For an incompressible flow, useful fan power equals flow rate times static pressure rise.
\( P_{air} = Q \times \Delta p = 2.8 \times 900 = 2520\ \text{W} \)

Step 4: Find the fan efficiency.
Fan efficiency is the useful air power divided by the power the fan actually draws.
\( \eta_{fan} = \dfrac{P_{air}}{P_{fan}} \times 100 = \dfrac{2520}{3998.3} \times 100 = 63.0\% \)

Final Answer:
The fan efficiency works out close to 63%. \[ \boxed{\eta_{fan} \approx 63\%} \]