Subtopic: Calculation of performance parameters – field capacity, efficiency, performance index, application rate and losses
Question Statement:
A tractor-mounted cultivator with an effective operating width of $W = 2.4\text{ m}$ operates at a speed of $v = 6.0\text{ km/h}$ in a rectangular field of length $L = 300\text{ m}$. The average turning time at each headland is $t_t = 20\text{ s}$. Assuming turns are made at each end of the $300\text{ m}$ run and neglecting all other non-operating time interruptions, the effective field efficiency of the cultivator is ________ $\%$ (round off to 1 decimal place).
Answer Key:
90.0
Step-by-Step Solution & Explanation
```
Field Bout Timing:
Run length L = 300 m
Travel speed v = 6.0 km/h = 1.667 m/s
Run time t_r = 300 / 1.667 = 180 s
Turn time t_t = 20 s
```
**Step-by-step Calculation:**
1. Forward speed in $\text{m/s}$:
$v = \frac{6.0 \times 1000}{3600} = 1.6667\text{ m/s}$
2. Time taken to cultivate one full run of length $L = 300\text{ m}$ ($t_{\text{operating}}$):
$t_{\text{op}} = \frac{L}{v} = \frac{300\text{ m}}{1.6667\text{ m/s}} = 180\text{ s}$
3. Total cycle time per run including headland turn:
$t_{\text{total}} = t_{\text{op}} + t_t = 180\text{ s} + 20\text{ s} = 200\text{ s}$
4. Field efficiency ($\eta_f$):
$\eta_f = \frac{t_{\text{op}}}{t_{\text{total}}} \times 100 = \frac{180}{200} \times 100 = 90.0\%$