Discussion
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 m3.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 π = 3.14)
Answer: C
The fan efficiency is about 63%.
Brake power of the engine:
\[ \begin{aligned} P_{b} &= \frac{2\pi N T}{60} = \frac{2(3.14)(2000)(382)}{60} \\ &\approx 79\,965\ \text{W} \end{aligned} \]
The fan takes 5% of this, about 3998 W. The useful (air) power it delivers is the volume flow times the pressure rise:
\[ \begin{aligned} P_{air} &= Q\,\Delta p = 2.8 \times 900 = 2520\ \text{W} \\ \eta &= \frac{2520}{3998} \approx 0.63 \end{aligned} \]
So the efficiency is nearest to 63%. Rounding the brake power to 80 kW (fan input 4000 W) still gives 63%. The distractors would need different fan inputs: 65% about 3880 W, 60% 4200 W and 58% about 4350 W, none of which is 5% of this engine's brake power.
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