Subtopic: Power efficiencies and measurement, engine performance curves
Question Statement:
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 a 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)
Options:
(A) 58
(B) 60
(C) 63
(D) 65
Official Answer Key:
C
Step-by-Step Solution & Explanation
Step 1 — engine brake power:
BP = 2πNT/60 = (2 × 3.14 × 2000 × 382)/60 = 79,965 W ≈ 80.0 kW
Step 2 — power supplied to the fan (input):
P_in = 5% of BP = 0.05 × 79,965 = 3998 W
Step 3 — useful air power (output). For incompressible flow the useful air power is the product of volumetric flow rate and static pressure rise:
P_air = Q × Δp = 2.8 × 0.9 = 2.52 kW = 2520 W
Step 4 — fan efficiency:
η = P_air/P_in = 2520/3998 = 0.6303 = 63 %
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