Subtopic: Power transmission systems – gear trains, differential, final drives and power take-off
Question Statement:
A two-wheel drive tractor has a PTO speed of 540 rpm and it produces 35 kW net engine power. Corresponding torque available at PTO in N m will be
Options:
(A) 435–457
(B) 465–485
(C) 495–505
(D) 535–558
Official Answer Key:
D
Step-by-Step Solution & Explanation
T = 9550·P/N. The 35 kW figure is the *engine's* net power; it must first pass through the transmission/PTO clutch and gear train before reaching the PTO shaft. With a typical PTO transmission efficiency of ~88–90% (standard textbook assumption), PTO power ≈ 35×0.88 ≈ 30.8 kW, so T = 9550×30.8/540 ≈ 545 N·m, which falls in range (D). Using the full 35 kW directly (T = 9550×35/540 ≈ 619 N·m) overshoots every option, confirming that a transmission-loss correction is intended.
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