Subtopic: Calculation of performance parameters – field capacity, efficiency, performance index, application rate and losses
Question Statement:
A multi-crop seed-cum-fertilizer drill is driven by a ground drive wheel of diameter $D = 70\text{ cm}$. The drive wheel drives the star-wheel fertilizer shaft through a sprocket ratio of $i = \frac{N_{\text{shaft}}}{N_{\text{wheel}}} = 0.5$. In $100$ revolutions of the ground wheel, the machine delivers $3.5\text{ kg}$ of urea. If the drill has $7$ furrow openers spaced $25\text{ cm}$ apart, the fertilizer application rate in $\text{kg/ha}$ is ________ (round off to 2 decimal places).
Answer Key:
90.95
Step-by-Step Solution & Explanation
```
Fertilizer Application Calculation:
Number of rows n = 7
Row spacing s = 0.25 m
Working width W = 7 * 0.25 = 1.75 m
Wheel diameter D = 0.70 m
```
**Step-by-step Calculation:**
1. Distance traveled in $100$ revolutions of ground wheel:
$S = 100 \times (\pi D) = 100 \times (\pi \times 0.70\text{ m}) = 70\pi = 219.911\text{ m}$
2. Effective width of drill ($W$):
$W = 7 \times 0.25\text{ m} = 1.75\text{ m}$
3. Total area covered during the run ($A$):
$A = W \times S = 1.75\text{ m} \times 219.911\text{ m} = 384.845\text{ m}^2$
4. Application rate ($R$ in $\text{kg/ha}$):
$R = \frac{\text{Fertilizer Mass (kg)}}{A\text{ (m}^2\text{)}} \times 10000 = \frac{3.50}{384.845} \times 10000 = 90.946\text{ kg/ha} \approx 90.95\text{ kg/ha}$