Skip to main content

Drainage rate (l/s/ha) for bunded rice fields

Short answer

By trial, the drainage rate that satisfies the constraints (peak water level ≤400 mm for ≤1 day, depth above 250 mm ≤2 days) for a 3-day 400 mm rainfall starting from 50 mm initial depth is 46.5 mm/day, which equals 5.38 l/s/ha, adopted as 5.4 l/s/ha.

Determine the drainage rate in l/s/ha required to meet the following conditions for healthy growth of rice paddies in bunded field in Terai of Nepal. Initial water level in field = 50 mm; Maximum water level is 400 mm which may persist for up to one day; Depth in excess of 250 mm may persist for up to 2 days; No rain follows the design rainfall for several days; Neglect ET and deep percolation losses; Design 3 day rainfall is 400 mm.

Rice tolerates standing water only within limits of depth and duration, so the drainage rate is chosen such that the field water level during and after the design storm stays within these limits.

Given data and assumptions

  • Initial depth h0h_0 = 50 mm; design 3-day rainfall RR = 400 mm, falling uniformly, so the rain rate r=R/3r = R/3 = 133.3 mm/day
  • Maximum depth 400 mm may persist for 1 day; depth above 250 mm may persist for 2 days
  • ET and deep percolation neglected (safe side); drainage DD (mm/day) starts with the rain and is constant

Water level history

During the storm the level rises at (r−D)(r-D) per day, and after it the level falls at DD per day:

hpeak=h0+R−3Dt1=he−h0r−D(time to reach he)t2=hpeak−heD(time to fall back to he)Te=(3−t1)+t2≤2\begin{aligned} h_{peak} &= h_0 + R - 3D \\ t_1 &= \frac{h_e-h_0}{r-D} \quad (\text{time to reach } h_e) \\ t_2 &= \frac{h_{peak}-h_e}{D} \quad (\text{time to fall back to } h_e) \\ T_e &= (3-t_1)+t_2 \le 2 \end{aligned}

The duration condition above 250 mm governs. Trial values:

D (mm/day)Peak level (mm)Time above limit (days)
403302.86
453152.18
46.53102.00
503001.60
552851.08

Solving TeT_e = 2 days gives DD = 46.5 mm/day. Then:

  • hpeakh_{peak} = 50 + 400 − 3 × 46.5 = 311 mm, which is below 400 mm, so the 400 mm level is never reached and the 1-day limit is satisfied
  • t1t_1 = 2.30 days, t2t_2 = 1.30 days, TeT_e = (3 − 2.30) + 1.30 = 2.00 days

Conversion: 1 mm/day = 10 m³/ha/day = 0.1157 l/s/ha, so

D=46.5×1086.4=5.38 l/s/haD = 46.5\times\frac{10}{86.4} = 5.38\ \text{l/s/ha}

Answer: drainage rate ≈ 46.5 mm/day ≈ 5.4 l/s/ha (adopt about 5.4 l/s/ha).

✓ Answer checked by Claude Opus 5.5 (high effort)

All Irrigation and Drainage Engineering chapter 9 questions →