Subtopic: Calculation of performance parameters – field capacity, efficiency, performance index, application rate and losses
Question Statement:
A 9×20 cm fluted roller type seed drill is operated at a forward speed of 3 km h⁻¹ in a field of size 120 m × 90 m. The effective ground wheel diameter of the seed drill is 0.5 m and the ratio of ground wheel rpm to the fluted roller rpm is 2. For one complete rotation of each fluted roller, 6.8 g seed is transferred from the seed box to the seed tube. The average time taken for each turn while operating length-wise is 50 s and the total time wasted in refilling the seed box for sowing the entire field is 40 min. The seed rate in kg ha⁻¹ will be
Options:
(A) 108.32
(B) 122.55
(C) 136.99
(D) 240.71
Official Answer Key:
A
Step-by-Step Solution & Explanation
Working width = 9 rows × 0.20 m = 1.8 m. Distance per ground-wheel revolution = πD = π×0.5 = 1.5708 m.
Ground-wheel:fluted-roller rpm ratio = 2:1, so each roller makes 0.5 rev per ground-wheel rev, discharging 6.8×0.5 = 3.4 g per roller per ground-wheel revolution; with 9 rollers: total seed = 9×3.4 = 30.6 g per ground-wheel revolution.
Area covered per ground-wheel revolution = width×distance = 1.8×1.5708 = 2.827 m².
Seed rate = (30.6 g/2.827 m²) × 10,000 m²/ha ÷ 1000 g/kg = 108.2 kg/ha ≈ 108.32. Answer: (A).
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