Subtopic: Calculation of performance parameters – field capacity, efficiency, performance index, application rate and losses
Question Statement:
A fertilizer drill with a row to row spacing of 40 cm, discharges 38 g of fertilizer per row per revolution of the metering wheel. The metering wheel is driven through a chain transmission system by ground wheel having 60 cm diameter. Neglecting skid of the ground wheel, for an application rate of 200 kg·ha⁻¹, the speed ratio of ground wheel to metering wheel will be (Take π = 3.14)
Options:
(A) 1.40 : 1
(B) 2.52 : 1
(C) 3.64 : 1
(D) 4.76 : 1
Official Answer Key:
B
Step-by-Step Solution & Explanation
Area covered per row per revolution of the ground wheel = row spacing × ground-wheel circumference = 0.40 × (π×0.60) = 0.40×1.884 = 0.7536 m². Fertilizer needed for that area at 200 kg/ha (= 0.02 kg/m²): 0.7536×0.02 = 0.015072 kg = 15.07 g. The metering wheel discharges 38 g per its own revolution, so metering-wheel revolutions needed per ground-wheel revolution = 15.07/38 = 0.3966. Speed ratio (ground : metering) = 1 : 0.3966 = 2.52 : 1, matching (B).
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