Step 1: Understanding the Concept:
Fertilizer application through a sprinkler irrigation system (fertigation) requires calculating the exact land surface area covered during one single operation (setting) of the sprinkler lateral.
Step 2: Key Formula or Approach:
1. The area covered per setting (\(A_s\)) is given by:
\[ A_s = N \times S_e \times S_m \]
where:
\(N\) = number of sprinklers on the lateral
\(S_e\) = spacing between individual sprinklers along the lateral
\(S_m\) = spacing between laterals along the main line
2. Convert this area from square meters (\(\text{m}^2\)) to hectares (ha).
3. Calculate the weight of fertilizer \(W\) required for this area:
\[ W = \text{Recommended Dose } (\text{kg/ha}) \times A_s \text{ (ha)} \]
Step 3: Detailed Explanation:
Given values:
- Number of sprinklers (\(N\)) = \(12\)
- Sprinkler spacing (\(S_e\)) = \(14\text{ m}\)
- Lateral spacing (\(S_m\)) = \(20\text{ m}\)
- Recommended dose = \(80\text{ kg/ha}\)
First, calculate the area \(A_s\) covered during one setting of the lateral:
\[ A_s = 12 \times 14 \text{ m} \times 20 \text{ m} \]
\[ A_s = 168 \text{ m} \times 20 \text{ m} = 3360 \text{ m}^2 \]
Convert the calculated area to hectares:
\[ A_s = \frac{3360}{10000} = 0.336 \text{ ha} \]
Now, calculate the quantity of fertilizer to be added for this setting:
\[ W = 80 \text{ kg/ha} \times 0.336 \text{ ha} \]
\[ W = 26.88 \text{ kg} \approx 27 \text{ kg} \]
Step 4: Final Answer:
The correct option is 3, which corresponds to 27 kg.