A two-wheel drive tractor with rear wheel rolling radius of 600 mm develops a tractive force of 18.5 kN at a wheel slip of 12 %. The engine speed is 2400 rpm and the transmission ratio from the engine to the rear wheels is 120:1. If the tractor experiences a motion resistance of 1.75 kN, the drawbar power developed by the tractor will be ________ kW.
Official Answer Key:
18.00 to 21.00
Step-by-Step Solution & Explanation
Rear wheel speed N_w = 2400/120 = 20 rpm.
Theoretical speed V_t = 2πrN/60 = 2π(0.6)(20)/60 = 1.2566 m s⁻¹.
Actual speed V_a = V_t (1 − s) = 1.2566 × 0.88 = 1.1058 m s⁻¹.
Drawbar pull = tractive force − motion resistance = 18.5 − 1.75 = 16.75 kN.
DBP = 16.75 × 1.1058 = 18.52 kW — within the official range 18.00–21.00.
Common error: using the theoretical speed (gives 21.05 kW, the top of the accepted band, which is why the examiner set a wide range).
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