Subtopic: Calculation of performance parameters – field capacity, efficiency, performance index, application rate and losses
Question Statement:
A 9-row tractor-drawn seed-cum-fertilizer drill with a row-to-row spacing of $20\text{ cm}$ has a ground wheel diameter of $0.8\text{ m}$. During a stationary calibration test, the ground wheel is rotated by $50$ revolutions. The total seed collected from all $9$ tubes is $1.80\text{ kg}$. The calibrated seed sowing rate of the drill in $\text{kg/ha}$ is ________ (round off to 2 decimal places).
Answer Key:
79.58
Step-by-Step Solution & Explanation
```
Seed Drill Calibration Test:
Number of furrows: n = 9
Row spacing: s = 0.20 m
Drill width: W = 9 * 0.20 = 1.80 m
Wheel diameter: D = 0.80 m
```
**Method 1: Circumference & Sown Area Formula**
1. Effective working width ($W$):
$W = n \times s = 9 \times 0.20\text{ m} = 1.80\text{ m}$
2. Distance traveled in $N = 50$ wheel revolutions ($S$):
$S = N \times (\pi D) = 50 \times (\pi \times 0.80\text{ m}) = 40\pi = 125.6637\text{ m}$
3. Area covered during the calibration test ($A$):
$A = W \times S = 1.80\text{ m} \times 125.6637\text{ m} = 226.1947\text{ m}^2$
4. Calibrated seed rate ($R$ in $\text{kg/ha}$):
$R = \frac{\text{Seed Collected (kg)}}{A\text{ (m}^2\text{)}} \times 10000\text{ m}^2\text{/ha}$
$R = \frac{1.80\text{ kg}}{226.1947\text{ m}^2} \times 10000 = 79.577\text{ kg/ha} \approx 79.58\text{ kg/ha}$
**Method 2: Direct Calibration Constant Shortcut**
$\text{Area} = 50 \times \pi \times 0.8 \times 1.8 = 226.195\text{ m}^2$
$\text{Rate} = \frac{1.80 \times 10000}{226.195} = 79.58\text{ kg/ha}$