Question

Irrigation requirement of 90 days duration maize variety is 50 cm. How much area can be irrigated with a flow rate of 20 litre per second for 8 hours in a day.

• A. 2 ha
• B. 5 ha
• C. 12 ha
• D. 4 ha

Hint:

Remember the standard conversion factor:
A flow rate of \( 1 \text{ lps} \) operating continuously for \( 24 \text{ hours} \) supplies approximately \( 64 \text{ ha-cm} \) of water.
Adjusting this for \( 8 \text{ hours} \) of operation yields \( 88 \text{ ha-cm} \) of water per day.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The total area that can be irrigated by a water source depends on the volumetric flow rate, the daily operation time, the total crop growth duration, and the net depth of irrigation required by the crop.
Key Formula or Approach:
Volume of water supplied per second:
\[ q = 20 \text{ litres/second} = 0.02 \text{ m}^3/\text{second} \]
Volume of water supplied in one day (\( V_d \)):
\[ V_d = q \times t_{\text{working}} \times 3600 \]
Total volume of water supplied over the crop duration (\( V_{\text{total}} \)):
\[ V_{\text{total}} = V_d \times \text{Crop duration (days)} \]
Commanded area (\( A \)) in square meters:
\[ A = \frac{V_{\text{total}}}{\text{Irrigation requirement (depth, } d)} \]

Step 2: Detailed Explanation:
Let us perform the calculations step-by-step:
Identify the given parameters:
- Flow rate, \( q = 20 \text{ lps} = 0.02 \text{ m}^3/\text{s} \).
- Daily working hours, \( t_{\text{working}} = 8 \text{ hours} \).
- Crop duration, \( T = 90 \text{ days} \).
- Total irrigation depth required, \( d = 50 \text{ cm} = 0.5 \text{ m} \).
Calculate the volume of water supplied in one working day:
\[ V_d = 20 \text{ l/s} \times (8 \text{ hours/day} \times 3600 \text{ seconds/hour}) \]
\[ V_d = 20 \times 28,800 = 576,000 \text{ litres/day} = 576 \text{ m}^3/\text{day} \]
Calculate the total volume of water supplied over the 90-day crop period:
\[ V_{\text{total}} = 576 \text{ m}^3/\text{day} \times 90 \text{ days} = 51,840 \text{ m}^3 \]
Calculate the total area that can be irrigated with this water volume:
\[ A = \frac{V_{\text{total}}}{d} = \frac{51,840 \text{ m}^3}{0.5 \text{ m}} = 103,680 \text{ m}^2 \]
Convert the area from square meters to hectares:
\[ A_{\text{ha}} = \frac{103,680 \text{ m}^2}{10,000 \text{ m}^2/\text{ha}} = 368 \text{ ha} \approx 4 \text{ ha} \]
Therefore, the total area of maize that can be successfully irrigated is \( 4 \text{ ha} \).

Step 3: Final Answer:
The total area that can be irrigated is 4 ha.