Subtopic: Calculation of performance parameters – field capacity, efficiency, performance index, application rate and losses
Question Statement:
A cultivator with a working width of 1.2 m utilizes 95% of its width due to overlapping while operating at a forward speed of 2 km h⁻¹. If the time lost in turning and other interruptions is 50 minutes per hectare, the field efficiency of the cultivator in percent is
Options:
(A) 79.84
(B) 83.34
(C) 86.97
(D) 90.20
Official Answer Key:
A
Step-by-Step Solution & Explanation
Theoretical field capacity (rated width): C_t = W·S/10 = (1.2×2)/10 = 0.24 ha/h → theoretical time for 1 ha = 1/0.24 = 4.1667 h = 250.0 min.
Effective (utilised) width = 0.95×1.2 = 1.14 m → effective capacity = (1.14×2)/10 = 0.228 ha/h → operating time for 1 ha at effective width = 1/0.228 = 4.386 h = 263.16 min.
Adding the 50 min/ha lost in turning: actual total time for 1 ha = 263.16 + 50 = 313.16 min.
Field efficiency = (theoretical time / actual time) × 100 = (250.0/313.16) × 100 = 79.84%. Answer: (A).
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