Drainage canal for 5 ha paddy field

The drainage canal must remove water from the paddy field fast enough that the level stays within the permissible depth and duration. The design discharge is the drainage rate multiplied by the area, and the canal is then sized by Manning's formula.

Step 1: Design rainfall

Mean annual 3-day maximum = 300 mm; for the 10-year return period the growth factor is 1.5, so $R = 1.5 \times 300 = 450$ mm in 3 days, i.e. 150 mm/day.

Step 2: Drainage rate

Initial level $h_0$ = 40 mm. Permissible: 300 mm for 1 day, and above 200 mm for up to 3 days. With rain falling uniformly and drainage $D$ starting with the rain:

$$\begin{aligned} h_{peak} &= 40 + 450 - 3D \\ T_{200} &= \left(3-\frac{200-40}{150-D}\right)+\frac{h_{peak}-200}{D} \le 3 \end{aligned}$$

By trial, $T_{200}$ = 3.00 days at $D$ = 60.6 mm/day, giving $h_{peak}$ = 308 mm; the time above 300 mm is only 0.23 day (< 1 day). So

$$D = 60.6\ \text{mm/day} = 7.01\ \text{l/s/ha}$$

Step 3: Design discharge

$$Q = D\,A = 7.01 \times 5 = 35.0\ \text{l/s} = 0.0350\ \text{m}^3/\text{s}$$

Step 4: Section of the drain

Assumptions: earthen canal, side slope 1:1, Manning's $n$ = 0.025, bed slope $S$ = 1 in 1000, bed width $b$ = 0.5 m. Solving $Q=\frac{1}{n}AR^{2/3}S^{1/2}$ by trial:

• Depth of flow $d$ = 0.172 m (say 0.17 m)
• $A=(b+d)d$ = 0.116 m², $P=b+2\sqrt{2}\,d$ = 0.986 m, $R$ = 0.117 m
• $V$ = 0.0350/0.116 = 0.30 m/s (about 0.3 m/s: low but non-eroding; silting is controlled by periodic cleaning)

Provide a free board of 0.15 m, so the total depth = 0.17 + 0.15 ≈ 0.32 m.

Answer: D = 60.6 mm/day (7.01 l/s/ha), Q = 35.0 l/s; canal with bed width 0.5 m, flow depth 0.17 m (total depth 0.32 m), side slopes 1:1, slope 1 in 1000.