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 hydraulic sprayer at 10 km/h and 420 kPa covers a width of 4.5 m; pump power = 0.75 kW, pump efficiency = 70%; 10% of discharge is bypassed for agitation. Working pressure is increased to 500 kPa. Assuming no change in coverage width, find the application rate (L/ha) at the new pressure. (Rounded to 1 decimal place)
Official Answer Key:
981.5 to 982.5
Step-by-Step Solution & Explanation
At 420 kPa: hydraulic power = 0.75×0.70 = 0.525 kW; total discharge Q₁ = 525/420000 = 1.25×10⁻³ m³/s = 1.25 L/s. Nozzle discharge varies with the square root of pressure (Q∝√P): Q₂(total) = 1.25×√(500/420) = 1.364 L/s. Effective spray discharge (after 10% bypass) = 0.9×1.364 = 1.227 L/s. Speed = 10 km/h = 2.778 m/s. Application rate = Q₂,eff/(width×speed)×10,000 = 1.227/(4.5×2.778)×10,000 ≈ 982.0 L/ha.
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