Subtopic: Calculation of performance parameters – field capacity, efficiency, performance index, application rate and losses
Question Statement:
A mower with a cutter bar of $1.5\text{ m}$ length is cutting forage at a forward speed of $6\text{ km/h}$. If the total time lost in turning and unclogging is $20\%$ of operating time, calculate the time required in $\text{hours}$ to harvest a rectangular field of $3.6\text{ ha}$.
Answer Key:
5
Step-by-Step Solution & Explanation
Theoretical Field Capacity:
$TFC = \frac{W \times S}{10} = \frac{1.5 \times 6}{10} = 0.90\text{ ha/h}$
Time loss is $20\%$, meaning field efficiency $\eta = 80\% = 0.80$.
Effective Field Capacity:
$EFC = TFC \times 0.80 = 0.90 \times 0.80 = 0.72\text{ ha/h}$
Total time required to cut $3.6\text{ ha}$:
$t = \frac{\text{Area}}{EFC} = \frac{3.6}{0.72} = 5.0\text{ hours}$