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 rotary disc mower has 4 rotating discs, each cutting 60 cm width (no overlap). Specific power loss (air/stubble/gear friction) = 2.5 kW per m cutting width; specific cutting energy = 2.0 kJ/m². At forward speed 6 km/h, find the PTO power requirement (kW). (Rounded to 2 decimal places)
Official Answer Key:
13.90 to 14.10
Step-by-Step Solution & Explanation
Total cutting width = 4×0.6 = 2.4 m. Friction/air power loss = 2.5×2.4 = 6.0 kW. Area cut per second = width×speed = 2.4×(6000/3600) = 4.0 m²/s. Cutting power = 2.0 kJ/m² × 4.0 m²/s = 8.0 kW. Total PTO power = 6.0+8.0 = 14.0 kW.
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