Drainage rate: 300 mm in 3 days, growth factor 1.5

Rice tolerates standing water only up to limits of depth and duration, so the drainage rate is chosen such that the field water level, after the design storm, stays within these limits. The 3-day rainfall of the 10-year return period is the mean annual maximum multiplied by the growth factor: $R = 1.5\times300 = 450$ mm.

Given data and assumptions

• Initial depth $h_0$ = 40 mm; design 3-day rainfall $R$ = 450 mm, falling uniformly, so the rain rate $r = R/3$ = 150.0 mm/day
• Maximum depth 300 mm may persist for 1 day; depth above 200 mm may persist for 3 days
• ET and deep percolation neglected (safe side); drainage $D$ (mm/day) starts with the rain and is constant

Water level history

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

$$\begin{aligned} h_{peak} &= h_0 + R - 3D \\ t_1 &= \frac{h_e-h_0}{r-D} \quad (\text{time to reach } h_e) \\ t_2 &= \frac{h_{peak}-h_e}{D} \quad (\text{time to fall back to } h_e) \\ T_e &= (3-t_1)+t_2 \le 3 \end{aligned}$$

The duration condition above 200 mm governs. Trial values:

Solving $T_e$ = 3 days gives $D$ = 60.6 mm/day. Then:

• $h_{peak}$ = 40 + 450 − 3 × 60.6 = 308 mm (slightly above 300 mm; the time above 300 mm is 0.23 day, less than 1 day, so this limit is satisfied)
• $t_1$ = 1.79 days, $t_2$ = 1.79 days, $T_e$ = (3 − 1.79) + 1.79 = 3.00 days

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

$$D = 60.6\times\frac{10}{86.4} = 7.01\ \text{l/s/ha}$$

Answer: drainage rate ≈ 60.6 mm/day ≈ 7.0 l/s/ha (adopt 7.0 l/s/ha).