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Designing drainage canals in paddy fields

Short answer

First, find design 3-day rainfall for the return period. With initial water depth and crop tolerance (maximum depth and duration limits), determine drainage rate D by trial so that field water levels stay within limits. Then design discharge Q = D × area, and size the drain cross-section using Manning’s equation, ensuring non-silting, non-scouring velocities and freeboard.

Explain in details the procedures (all steps required to arrive at the design discharge) of designing drainage canals in irrigated paddy fields.

Drainage canals in paddy fields (Terai of Nepal) are designed to remove excess rainwater within the time for which rice can tolerate flooding. The design discharge is obtained from the drainage rate and the area, and the canal is sized by Manning's formula.

Procedure

  1. Collect rainfall data and find the maximum 3-day rainfall of each year; carry out a frequency analysis (Gumbel or growth factors) to get the 3-day rainfall for the selected return period (5–10 years for field drains; larger for main drains). Design rainfall R=R = mean annual maximum × growth factor.
  2. Fix the crop tolerance: the initial water depth in the field h0h_0 (about 40–50 mm) and the maximum depth and duration the rice can bear, for example 300 mm for 1 day and 200 mm for 3 days (bunded fields).
  3. Fix ET and percolation: neglect them (safe side) or deduct them if their values are known.
  4. Draw the water-level history: the rain falls uniformly over 3 days and the drainage DD removes water, so the level rises at (R/3−D)(R/3-D) per day; after the rain it falls at DD per day. hpeak=h0+R−3D,Te=(3−he−h0R/3−D)+hpeak−heDh_{peak}=h_0+R-3D, \qquad T_e=\left(3-\frac{h_e-h_0}{R/3-D}\right)+\frac{h_{peak}-h_e}{D}
  5. Find the drainage rate DD such that the peak depth and the duration above the limit depths do not exceed the tolerance (solve by trial). Convert: 1 mm/day = 0.1157 l/s/ha.
  6. Design discharge of the drain: Q=D×AQ = D \times A, where AA is the area served (ha), with a reduction for large areas if justified.
  7. Hydraulic design: choose the side slope, bed slope (following the land slope) and Manning's nn (0.025–0.035 for earthen drains), then solve Q=1nAR2/3S1/2Q = \frac{1}{n}AR^{2/3}S^{1/2} for the bed width and depth. Check that the velocity lies between the non-silting limit (about 0.3 m/s) and the non-scouring limit (about 0.6–0.9 m/s for earth).
  8. Add free board (about 0.15–0.3 m) and fix the water level of the drain at least 0.3 m below the field outlets so that the fields drain freely.
  9. Layout and structures: field drains, collector drains, main drain and outlet; provide culverts, drops and check structures where necessary.

Worked outline

If the initial depth is 50 mm, the 3-day design rainfall is 400 mm and the tolerance is 250 mm for 2 days, the rate DD comes out at about 46.5 mm/day (5.4 l/s/ha), so a 10 ha block needs Q=5.4×10=54Q = 5.4\times10 = 54 l/s, and the drain section follows from step 7.

✓ Answer checked by Claude Opus 5.5 (high effort)

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