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 field sprayer having a boom with 20 nozzles spaced 0.46 m apart is to be designed for a maximum application rate of 750 litre ha⁻¹ at 520 kPa pressure. The forward speed of travel is 6.5 km h⁻¹. Neglect field losses and assume that 10% of the pump output is bypassed. If mechanical agitation requires 375 W input power and the pump efficiency is 70%, the maximum power input required is
Options:
(A) 720 W
(B) 879 W
(C) 1095 W
(D) 1403 W
Official Answer Key:
D
Step-by-Step Solution & Explanation
Hydraulic power delivered by pump = Q_pump × pressure = (83.06/60,000 m³/s)×520,000 Pa = 719.8 W. Pump (shaft) input power = hydraulic power/efficiency = 719.8/0.70 = 1028.3 W. Total power input required = pump input + agitation input = 1028.3+375 = ≈1403 W.
Section 3: Farm Power
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