Engineering - Farm Power and Machinery Calculations
Scope of the Engineering domain
GATE AG tests farm power and machinery here: engine and tractor performance, tillage and sowing implements, harvesting and threshing equipment, and the machinery-management arithmetic that converts these into field capacity, power demand and cost per hectare.
Power flow: engine to drawbar
Power reaches the soil through transmission and traction, losing energy at each stage. Drawbar power (kW) = draft (kN) × forward speed (m/s). Drawbar power = PTO power × tractive efficiency, and PTO power = engine power × transmission efficiency. Slip = (no-load travel distance − actual distance) ÷ no-load distance × 100, so actual speed = theoretical speed × (1 − slip); slip cuts travel speed, not draft.
Field capacity
Theoretical field capacity (ha/h) = working width (m) × speed (km/h) ÷ 10. Effective field capacity = theoretical × field efficiency, where field efficiency absorbs turns, filling, unloading and adjustment stops, so effective capacity is always the smaller figure.
Worked example
A 2.5 m cultivator works 12 cm deep at 6 km/h with a specific draft of 40 kN/m² of soil cross-section. Soil cross-section = 2.5 × 0.12 = 0.30 m². Draft = 40 × 0.30 = 12 kN. Speed = 6 km/h = 1.667 m/s, so drawbar power = 12 × 1.667 ≈ 20 kW. With tractive efficiency 0.70 and transmission efficiency 0.90, engine power required = 20 ÷ (0.70 × 0.90) ≈ 31.7 kW. Always divide by efficiencies when moving up the power chain, multiply when moving down.
Traps that cost marks
- Entering width in centimetres in the width × speed ÷ 10 formula, or dropping the factor 10.
- Returning theoretical field capacity when the question asks for effective capacity, total time, or cost per hectare.
- Applying slip as a correction to draft instead of to forward speed.
- Mixing specific draft per metre of width with specific draft per square metre of furrow cross-section.
- Quoting engine power when the question asks for drawbar or PTO power.