Irrigation outlet discharge to meet evapotranspiration An outlet serves 20 hectares of cropped area. The evapotranspiration requirement is 20 mm, occurring uniformly over 20 days. Neglecting other field losses, determine the continuous discharge required at the outlet in L/s.

Introduction:
Irrigation flow sizing often reduces to converting a depth of water over an area and time into a steady discharge. This problem illustrates the straightforward volumetric approach for evapotranspiration (ET) demands when losses are neglected and application is uniform over the design period.

Given Data / Assumptions:

• Area A = 20 ha = 200,000 m^2.
• Total ET depth over period = 20 mm = 0.020 m.
• Duration T = 20 days; assume continuous supply.
• Losses like conveyance and field application are neglected.

Concept / Approach:

Compute required daily (or total) volume and divide by time to obtain discharge. Since ET occurs uniformly over 20 days, the daily depth is 1 mm (0.001 m). Multiply by area to get daily volume; convert to L/s using 1 m^3 = 1000 L and 1 day = 86400 s.

Step-by-Step Solution:

Verification / Alternative check:

Alternatively, compute total volume (A * 0.020 m = 4000 m^3) and divide by 20 days to obtain the same daily rate; conversion yields 2.31 L/s again.

Why Other Options Are Wrong:

2.52 and 2.01 result from rounding or area mis-conversions; 1.52 assumes a smaller daily depth; 3.10 is too high for the given ET.

Common Pitfalls:

Forgetting that 1 ha = 10,000 m^2; confusing mm/day with cm/day; neglecting to convert days to seconds correctly when expressing L/s.

Final Answer:

2.31