Subtopic: Calculation of performance parameters – field capacity, efficiency, performance index, application rate and losses
Question Statement:
A $9$-row tractor-drawn seed drill with row-to-row spacing of $20\text{ cm}$ travels at a forward speed of $4.5\text{ km/h}$. The field efficiency of the operation is $75\%$. The time required to plant a $4.86\text{ ha}$ field is ________ hours (answer in integer).
Answer Key:
8
Step-by-Step Solution & Explanation
1. Effective working width of the seed drill:
$W = 9 \text{ rows} \times 0.20\text{ m} = 1.80\text{ m}$
2. Effective Field Capacity (EFC):
$\text{EFC} = \frac{S \times W \times \eta}{10} = \frac{4.5\text{ km/h} \times 1.80\text{ m} \times 0.75}{10} = \frac{6.075}{10} = 0.6075\text{ ha/h}$
3. Time required to complete $4.86\text{ ha}$:
$t = \frac{\text{Total Area}}{\text{EFC}} = \frac{4.86\text{ ha}}{0.6075\text{ ha/h}} = 8.0\text{ hours}$