Subtopic: Functional requirements, principles of working, construction and operation of manual, animal, tractor and renewable energy operated equipment for tillage, sowing, planting, fertilizer application, inter-cultivation, spraying, mowing, chaff cutting, harvesting and threshing
Question Statement:
A tractor PTO operated 4-disc rotary mower is harvesting with a forward speed of 3.5 km h⁻¹. The cutting circle diameter and rpm of each disc are 60 cm and 1400, respectively. The peak cutting force experienced by each rotary disc is 110 N. The peak overall motion resistance is found to be 3.6 kN. If the overall power conversion efficiency of the tractor is 82%, required peak engine brake power in kW will be ______
Official Answer Key:
27.0 to 28.5
Step-by-Step Solution & Explanation
Power to overcome motion resistance: P_resist = F×v = 3600×0.9722 = 3500 W (v = 3.5 km/h = 0.9722 m/s). Disc tip speed: ω = 2π×1400/60 = 146.6 rad/s, V_tip = ω×D/2 = 146.6×0.3 = 43.98 m/s. Power per disc for cutting = F_cut×V_tip = 110×43.98 = 4838 W; for 4 discs, P_cut = 19,352 W. Total output power = P_resist + P_cut = 3500+19,352 = 22,852 W. Accounting for 82% conversion efficiency, required engine brake power = 22,852/0.82 = 27,868 W ≈ 27.9 kW, within the accepted band 27.0–28.5.
Section 3 — Farm Power
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