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 actual field capacity of the machine in ha h⁻¹is
Options:
(A) 0.30
(B) 0.32
(C) 0.36
(D) 0.40
Official Answer Key:
B
Step-by-Step Solution & Explanation
Field area = 120×90 = 10,800 m² = 1.08 ha. Working width = 1.8 m, so number of length-wise passes ≈ 90/1.8 = 50 passes, each of length 120 m; total travel distance = 50×120 = 6000 m.
Travel time = 6000 m ÷ (3 km/h = 833.3 m/min... using 3000 m/h) = 6000/3000 = 2.0 h.
Turn time = 49 turns (between 50 passes) × 50 s = 2450 s = 0.6806 h.
Refill time = 40 min = 0.6667 h.
Total effective time = 2.0+0.6806+0.6667 = 3.347 h.
Actual field capacity = Area/Time = 1.08/3.347 = 0.323 ≈ 0.32 ha/h. Answer: (B).
Section 3 — Farm Power
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