Farm Power · Farm Power · 1 mark · Multiple choice
A tractor engine delivers brake torque 382 N·m at 2000 rpm. Its radiator fan draws 5% of brake power and supplies $2.8\,\mathrm{m^3/s}$ of incompressible air against 0.9 kPa static pressure rise. Ignore other losses and take $\pi=3.14$. Find the nearest fan efficiency in percent.
Engine brake power is $P_b=2\pi NT/60=2(3.14)(2000)(382)/60=79965.3$ W. Fan input is $0.05P_b=3998.27$ W. Useful air power is $Q\Delta p=2.8(900)=2520$ W. Therefore $\eta=2520/3998.27=0.6303$,or **63%, C**.
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