Irrigation planning: A tube well delivers 4 m^3/hour and operates 20 hours per day during the season. If the irrigation interval is 20 days and the required depth per irrigation is 7 cm, what is the maximum command area (in m^2)?

Introduction / Context:
Command area calculations translate discharge and operating schedule into the land area that can be adequately irrigated at a specified interval and depth. This is essential for sizing well fields and planning water distribution.

Given Data / Assumptions:

• Discharge Q = 4 m^3/hour.
• Operating time = 20 hours/day.
• Irrigation interval = 20 days.
• Depth per irrigation = 7 cm = 0.07 m.

Concept / Approach:

Total volume available per interval equals discharge × operating time over the interval. Command area A = Volume / required depth. Uniform application and negligible losses are assumed.

Step-by-Step Solution:

Verification / Alternative check:

Units check: m^3 / m = m^2 → correct. Result matches option D.

Why Other Options Are Wrong:

• Values smaller or larger than 2.29 × 10^4 m^2 result from arithmetic errors (wrong interval volume or depth conversion).

Common Pitfalls:

• Forgetting to convert 7 cm to 0.07 m.
• Multiplying instead of dividing by depth when computing area.

Final Answer:

2.29 × 10^4 m^2