Subtopic: Calculation of performance parameters – field capacity, efficiency, performance index, application rate and losses
Question Statement:
A 4-bottom tractor-mounted moldboard plow has a total cutting width of $1.4\text{ m}$. It operates at an average forward speed of $5.4\text{ km/h}$. If the total time lost in turning at headlands and unclogging represents $25\%$ of the total field time, the effective field capacity of the plow is ________ $\text{ha/h}$ (round off to two decimal places).
Answer Key:
0.57
Step-by-Step Solution & Explanation
1. Theoretical field capacity ($TFC$):
$TFC = \frac{W \times S}{10} = \frac{1.4 \times 5.4}{10} = 0.756\text{ ha/h}$
2. Field efficiency ($\eta$):
Since time lost is $25\%$, the effective operating fraction is $100\% - 25\% = 75\%$, so $\eta = 0.75$.
3. Effective field capacity ($EFC$):
$EFC = TFC \times \eta = 0.756 \times 0.75 = 0.567\text{ ha/h} \approx 0.57\text{ ha/h}$.