Agricultural Engineering: Farm Machinery, Irrigation, and Post-Harvest Technology
Agricultural Engineering tests machinery selection, irrigation efficiency and post-harvest handling. Numerical questions often involve field capacity, power, moisture and storage losses.
Core rules
- Theoretical field capacity = width in m × speed in km/h ÷ 10, in ha/h. Effective field capacity includes turning, loading and other lost time.
- Irrigation efficiency compares water beneficially used with water delivered. Drip and sprinkler can raise application efficiency but need filtration and pressure management.
- Moisture content on a dry basis = (wet mass − dry mass) ÷ dry mass × 100. Wet-basis moisture uses wet mass as the denominator.
- Post-harvest losses fall with timely harvest, clean threshing, safe moisture, proper bags or bins, and pest-proof storage.
Worked example
A 3 m combine operates at 5 km/h. Theoretical field capacity = 3 × 5 ÷ 10 = 1.5 ha/h. At 80% field efficiency, effective capacity = 1.5 × 0.80 = 1.2 ha/h. Time for 6 ha = 6 ÷ 1.2 = 5 hours. Reasoning: the formula gives capacity with no lost time; efficiency corrects for real field conditions. Keep width in metres and speed in km/h to get ha/h.
Traps that cost marks
Do not use effective capacity when the question says theoretical. Do not mix horsepower and kilowatts; 1 hp is about 0.746 kW. Do not use wet-basis moisture in a dry-basis formula. In irrigation, distinguish conveyance loss from application loss. In storage, check whether the question asks about quantity loss or quality loss.