Water Logging and Drainage

Water logging is the condition in which the water table rises so close to the ground surface that the root zone of crops is saturated and the soil pores are filled with water, so air cannot enter. A field is generally said to be waterlogged when the water table is within about 1.5 m of the ground surface (within the capillary fringe of the root zone).

Causes

1. Seepage from unlined canals and distributaries, which recharges the groundwater continuously.
2. Over-irrigation and wasteful application of water by farmers, giving deep percolation beyond the crop need.
3. Inadequate natural drainage: flat topography, as in the Terai, with no outlet for excess water.
4. Obstruction of natural drains by canals, roads, railways and embankments built across the drainage lines.
5. Heavy rainfall and flooding with an impervious or clayey subsoil.
6. Impervious layer at a shallow depth, which holds the percolated water.
7. Absence or poor maintenance of drains, and lack of canal lining.
8. Rise of the water level in rivers/reservoirs, which raises the groundwater in adjacent land.

Effects

• Crop yield falls: roots lack oxygen, nutrient uptake and soil bacteria are reduced; crops die.
• Salinity and alkalinity: capillary rise brings salts to the surface, leaving white salt crusts after evaporation.
• Soil temperature falls and tillage becomes difficult; machines cannot work on wet soil.
• Reduced cultivable area and delayed sowing.
• Health problems: mosquitoes and water-borne diseases.
• Damage to foundations of buildings and roads; loss of land value.

Preventive and remedial (reclamation) measures

Of these, prevention (canal lining, water-use discipline) is cheaper than cure. Where the table has already risen, a combination of surface drains with tile drains or tube wells is used for reclamation.

The spacing of tile drains is obtained by applying Darcy's law to the steady flow of recharge water towards parallel drains laid over an impervious layer (Dupuit–Forchheimer assumptions: flow is horizontal, and the hydraulic gradient equals the slope of the water table).

Assumptions

• Soil homogeneous and isotropic, permeability $K$
• Drains parallel, equal spacing $S$, laid at height $a$ above the impervious layer
• Uniform recharge $q$ (m/s, per unit area) from rainfall/irrigation
• Steady state; the water table has height $b$ above the impervious layer midway between two drains, and $a$ at the drain

          ground level
 ------------------------------------------
        .--------. water table (b at middle)
     _.'           '._
    y|      q        |   <- recharge q
  ---o-----------------o---  drain level (height a)
 a   drain   S/2    drain
 //////////// impervious layer ///////////
        origin x = 0 at the drain

Derivation

Take $x$ from the drain, and $y$ the height of the water table above the impervious layer at $x$. Flow per unit length of drain through a vertical section at $x$ comes from the area between $x$ and the mid-point, a width $(S/2-x)$:

$$q_x = q\left(\frac{S}{2}-x\right)$$

By Darcy's law (area = $y\times1$, gradient $= dy/dx$):

$$q\left(\frac{S}{2}-x\right) = K\,y\,\frac{dy}{dx}$$

Integrating:

$$\frac{K y^2}{2} = q\left(\frac{S x}{2}-\frac{x^2}{2}\right) + C$$

At $x=0$, $y=a$, so $C = \dfrac{K a^2}{2}$. At the mid-point $x=S/2$, $y=b$:

$$\frac{K b^2}{2} = q\left(\frac{S^2}{4}-\frac{S^2}{8}\right)+\frac{K a^2}{2} = \frac{q S^2}{8}+\frac{K a^2}{2}$$ $$q S^2 = 4K\left(b^2-a^2\right)$$ $$\boxed{\,S = \sqrt{\frac{4K\,(b^2-a^2)}{q}}\,}$$

This is Donnan's formula. With $h=b-a$ (height of the water table above the drain at the mid-point) and $d=a$, it can be written as Hooghoudt's form:

$$q = \frac{8Kdh + 4Kh^2}{S^2}$$

Use

• $q$ = (fraction of rainfall to be drained) / (time), for example 1.5 % of annual rainfall in 24 h
• $b$ = depth of impervious layer − depth of highest water table below ground
• $a$ = depth of impervious layer − depth of drain below ground

Then $S$ is calculated. If the actual drain is on the impervious layer ($a=0$), the formula reduces to $S=\sqrt{4Kb^2/q}$.

Drainage is the removal of excess surface water and groundwater from land so that the crops get a good soil moisture and aeration condition. Irrigation without drainage raises the water table and leads to water logging and salinity.

Why drainage of irrigated land is necessary

• To prevent water logging: the water table must be kept below the root zone (at least 1.5 m below ground).
• To maintain soil aeration, so that roots get oxygen and the soil bacteria can work.
• To prevent salinity and alkalinity by leaching and by removing salts brought up by capillary action.
• To remove excess rainfall (floods) quickly, especially in the Terai paddy areas.
• To allow cultivation, machinery work and timely sowing, and to increase the soil temperature.
• To control mosquito-borne diseases and to protect roads, canals and buildings.

Methods of drainage

 Drainage
 |-- Surface drainage  (open drains)
 |-- Subsurface drainage (tile, mole, vertical)
 '-- Interceptor / relief drains

1. Surface drainage: removal of excess surface water through open drains. Field drains collect water from the field, collector drains gather it from several field drains, and main (outfall) drains carry it to a natural stream. Land levelling and grading help. Suitable for flat fields and heavy rainfall, as in Terai paddy fields.
2. Subsurface drainage: lowering of the water table by buried drains.
   • Tile (pipe) drains: perforated pipes or clay/concrete tiles laid at 1.5–2.5 m depth with laterals and collectors; spacing is from Donnan's/Hooghoudt's formula.
   • Mole drains: unlined cylindrical channels formed in clay soil by a mole plough, 0.5–1 m deep.
   • Open ditches (deep drains) for large water table control.
3. Vertical drainage: pumping from tube wells to lower the water table, where the aquifer is permeable. The pumped water can be used for irrigation.
4. Interceptor drains: deep drains laid along the foot of a canal or hill slope to intercept seepage before it reaches the field.
5. Natural drainage improvement: cleaning and straightening natural streams.

The choice depends on the soil permeability, depth of the impervious layer, rainfall, slope and cost: surface drains in flat, low-permeability, high-rain areas; tile or vertical drainage where the water table is the problem.

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. Rice in the Terai tolerates deep water only briefly.

Given data and assumptions

• Initial depth $h_0$ = 50 mm; design 3-day rainfall $R$ = 400 mm, falling uniformly, so the rain rate $r = R/3$ = 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 $D$ (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:

$$\begin{aligned} h_{peak} &= h_0 + R - 3D \\ t_1 &= \frac{h_e-h_0}{r-D} \quad (\text{time to reach } h_e) \\ t_2 &= \frac{h_{peak}-h_e}{D} \quad (\text{time to fall back to } h_e) \\ T_e &= (3-t_1)+t_2 \le 2 \end{aligned}$$

The duration condition above 250 mm governs. Trial values:

Solving $T_e$ = 2 days gives $D$ = 46.5 mm/day. Then:

• $h_{peak}$ = 50 + 400 − 3 × 46.5 = 311 mm (below 400 mm; the time above 400 mm is 0.00 day, less than 1 day, so this limit is satisfied)
• $t_1$ = 2.30 days, $t_2$ = 1.30 days, $T_e$ = (3 − 2.30) + 1.30 = 2.00 days

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

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

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

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. Plain (Terai) bunded rice fields.

Given data and assumptions

• Initial depth $h_0$ = 55 mm; design 3-day rainfall $R$ = 400 mm, falling uniformly, so the rain rate $r = R/3$ = 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 $D$ (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:

$$\begin{aligned} h_{peak} &= h_0 + R - 3D \\ t_1 &= \frac{h_e-h_0}{r-D} \quad (\text{time to reach } h_e) \\ t_2 &= \frac{h_{peak}-h_e}{D} \quad (\text{time to fall back to } h_e) \\ T_e &= (3-t_1)+t_2 \le 2 \end{aligned}$$

The duration condition above 250 mm governs. Trial values:

Solving $T_e$ = 2 days gives $D$ = 47.9 mm/day. Then:

• $h_{peak}$ = 55 + 400 − 3 × 47.9 = 311 mm (below 400 mm; the time above 400 mm is 0.00 day, less than 1 day, so this limit is satisfied)
• $t_1$ = 2.28 days, $t_2$ = 1.28 days, $T_e$ = (3 − 2.28) + 1.28 = 2.00 days

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

$$D = 47.9\times\frac{10}{86.4} = 5.54\ \text{l/s/ha}$$

Answer: drainage rate ≈ 47.9 mm/day (≈ 5.5 l/s/ha); adopt 48 mm/day.

Tile drain spacing is found from Donnan's (Dupuit–Forchheimer) formula for steady flow towards parallel drains over an impervious layer:

$$S=\sqrt{\frac{4K\,(b^2-a^2)}{q}}$$

where $S$ = spacing, $K$ = permeability, $q$ = drainage rate (discharge per unit area), $b$ = height of the water table above the impervious layer at the mid-point, $a$ = height of the drain above the impervious layer.

Data

• Depth of impervious layer = 9 m, depth of drain = 2 m, depth of highest water table = 1 m (all below ground)
• $K$ = 0.001 cm/s = $1\times10^{-5}$ m/s
• Rainfall 1600 mm; 1.5 % to be drained in 24 hours

Step 1: Drainage rate

$$q = \frac{1.5}{100}\times\frac{1600}{1000}\ \text{m} \Big/ (24\times3600\ \text{s}) = 2.778\times10^{-7}\ \text{m/s}$$

Step 2: Heights above the impervious layer

$$\begin{aligned} a &= 9-2 = 7\ \text{m} \\ b &= 9-1 = 8\ \text{m} \end{aligned}$$

Step 3: Spacing

$$\begin{aligned} S^2 &= \frac{4\times1\times10^{-5}\times(8^2-7^2)}{2.778\times10^{-7}} \\ &= \frac{4\times1\times10^{-5}\times15.00}{2.778\times10^{-7}} = 2160\ \text{m}^2 \\ S &= 46.5\ \text{m} \end{aligned}$$

Answer: spacing of the tile drains ≈ 46.5 m (say 46 m).

Drainage canals in paddy fields (Terai of Nepal) are designed to remove excess rain water within the time that 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 annual 3-day rainfall for each year; carry out 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 $R =$ mean annual maximum × growth factor.
2. Fix the crop tolerance: the initial water depth in the field $h_0$ (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: neglected (safe side) or deducted, if their values are known.
4. Draw the water-level history: the rain falls uniformly over 3 days, drainage $D$ removes water, and the level rises at $(R/3-D)$ per day; after the rain it falls at $D$ per day. $$h_{peak}=h_0+R-3D, \qquad T_e=\left(3-\frac{h_e-h_0}{R/3-D}\right)+\frac{h_{peak}-h_e}{D}$$
5. Find the drainage rate $D$ 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: $Q = D \times A$, where $A$ is the area served (ha), taking a reduction for large areas if justified.
7. Hydraulic design: choose side slope, bed slope (following the land slope), Manning's $n$ (0.025–0.035 for earthen drains), then solve $$Q = \frac{1}{n}AR^{2/3}S^{1/2}$$ for the bed width and depth. Check the velocity: 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 field drains 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 3-day design rainfall is 400 mm and the tolerance is 250 mm for 2 days, the rate $D$ comes out about 46.5 mm/day (5.4 l/s/ha), so a 10 ha block needs $Q = 5.4\times10 = 54$ l/s, and the drain section follows from step 7.

The drainage canal must remove the 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, growth factor 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 $d$ = 0.17 m
• $A=(b+d)d$ = 0.116 m², $P=b+2\sqrt{2}\,d$ = 0.987 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 checked by periodic cleaning)

Provide free board 0.15 m, so 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.

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).

Tile drain spacing is found from Donnan's (Dupuit–Forchheimer) formula for steady flow towards parallel drains over an impervious layer:

$$S=\sqrt{\frac{4K\,(b^2-a^2)}{q}}$$

where $S$ = spacing, $K$ = permeability, $q$ = drainage rate (discharge per unit area), $b$ = height of the water table above the impervious layer at the mid-point, $a$ = height of the drain above the impervious layer.

Data

• Depth of impervious layer = 10 m, depth of drain = 2.2 m, depth of highest water table = 1.2 m (all below ground)
• $K$ = 0.001 cm/s = $1\times10^{-5}$ m/s
• Rainfall 1700 mm; 1.5 % to be drained in 24 hours

Step 1: Drainage rate

$$q = \frac{1.5}{100}\times\frac{1700}{1000}\ \text{m} \Big/ (24\times3600\ \text{s}) = 2.951\times10^{-7}\ \text{m/s}$$

Step 2: Heights above the impervious layer

$$\begin{aligned} a &= 10-2.2 = 7.8\ \text{m} \\ b &= 10-1.2 = 8.8\ \text{m} \end{aligned}$$

Step 3: Spacing

$$\begin{aligned} S^2 &= \frac{4\times1\times10^{-5}\times(8.8^2-7.8^2)}{2.951\times10^{-7}} \\ &= \frac{4\times1\times10^{-5}\times16.60}{2.951\times10^{-7}} = 2250\ \text{m}^2 \\ S &= 47.4\ \text{m} \end{aligned}$$

Answer: spacing of the tile drains ≈ 47.4 m (say 47 m).

Tile drain spacing is found from Donnan's (Dupuit–Forchheimer) formula for steady flow towards parallel drains over an impervious layer:

$$S=\sqrt{\frac{4K\,(b^2-a^2)}{q}}$$

where $S$ = spacing, $K$ = permeability, $q$ = drainage rate (discharge per unit area), $b$ = height of the water table above the impervious layer at the mid-point, $a$ = height of the drain above the impervious layer.

Data

• Depth of impervious layer = 9 m, depth of drain = 2 m, depth of highest water table = 1 m (all below ground)
• $K$ = 0.001 cm/s = $1\times10^{-5}$ m/s
• Rainfall 1400 mm; 1 % to be drained in 24 hours

Step 1: Drainage rate

$$q = \frac{1}{100}\times\frac{1400}{1000}\ \text{m} \Big/ (24\times3600\ \text{s}) = 1.62\times10^{-7}\ \text{m/s}$$

Step 2: Heights above the impervious layer

$$\begin{aligned} a &= 9-2 = 7\ \text{m} \\ b &= 9-1 = 8\ \text{m} \end{aligned}$$

Step 3: Spacing

$$\begin{aligned} S^2 &= \frac{4\times1\times10^{-5}\times(8^2-7^2)}{1.62\times10^{-7}} \\ &= \frac{4\times1\times10^{-5}\times15.00}{1.62\times10^{-7}} = 3703\ \text{m}^2 \\ S &= 60.9\ \text{m} \end{aligned}$$

Answer: spacing of the tile drains ≈ 60.9 m (say 61 m).

Tile drain spacing is found from Donnan's (Dupuit–Forchheimer) formula for steady flow towards parallel drains over an impervious layer:

$$S=\sqrt{\frac{4K\,(b^2-a^2)}{q}}$$

where $S$ = spacing, $K$ = permeability, $q$ = drainage rate (discharge per unit area), $b$ = height of the water table above the impervious layer at the mid-point, $a$ = height of the drain above the impervious layer.

Data

• Depth of impervious layer = 10 m, depth of drain = 1.5 m, depth of highest water table = 1 m (all below ground)
• $K$ = 0.001 cm/s = $1\times10^{-5}$ m/s
• Rainfall 1000 mm; 1 % to be drained in 24 hours

Step 1: Drainage rate

$$q = \frac{1}{100}\times\frac{1000}{1000}\ \text{m} \Big/ (24\times3600\ \text{s}) = 1.157\times10^{-7}\ \text{m/s}$$

Step 2: Heights above the impervious layer

$$\begin{aligned} a &= 10-1.5 = 8.5\ \text{m} \\ b &= 10-1 = 9\ \text{m} \end{aligned}$$

Step 3: Spacing

$$\begin{aligned} S^2 &= \frac{4\times1\times10^{-5}\times(9^2-8.5^2)}{1.157\times10^{-7}} \\ &= \frac{4\times1\times10^{-5}\times8.75}{1.157\times10^{-7}} = 3024\ \text{m}^2 \\ S &= 55.0\ \text{m} \end{aligned}$$

Answer: spacing of the tile drains ≈ 55.0 m (say 55 m).

Tile drain spacing is found from Donnan's (Dupuit–Forchheimer) formula for steady flow towards parallel drains over an impervious layer:

$$S=\sqrt{\frac{4K\,(b^2-a^2)}{q}}$$

where $S$ = spacing, $K$ = permeability, $q$ = drainage rate (discharge per unit area), $b$ = height of the water table above the impervious layer at the mid-point, $a$ = height of the drain above the impervious layer.

Data

• Depth of impervious layer = 9 m, depth of drain = 1.5 m, depth of highest water table = 1 m (all below ground)
• $K$ = 0.001 cm/s = $1\times10^{-5}$ m/s
• Rainfall 800 mm; 1 % to be drained in 24 hours

Step 1: Drainage rate

$$q = \frac{1}{100}\times\frac{800}{1000}\ \text{m} \Big/ (24\times3600\ \text{s}) = 9.259\times10^{-8}\ \text{m/s}$$

Step 2: Heights above the impervious layer

$$\begin{aligned} a &= 9-1.5 = 7.5\ \text{m} \\ b &= 9-1 = 8\ \text{m} \end{aligned}$$

Step 3: Spacing

$$\begin{aligned} S^2 &= \frac{4\times1\times10^{-5}\times(8^2-7.5^2)}{9.259\times10^{-8}} \\ &= \frac{4\times1\times10^{-5}\times7.75}{9.259\times10^{-8}} = 3348\ \text{m}^2 \\ S &= 57.9\ \text{m} \end{aligned}$$

Rainfall 80 cm = 800 mm.

Answer: spacing of the tile drains ≈ 57.9 m (say 58 m).

The tile must carry the drainage discharge from the area served by the lateral, flowing full under gravity. Manning's formula for a full circular pipe is used.

Data

• Area drained by one lateral = 200 m × 50 m = 10 000 m² = 1 ha
• Drainage coefficient = 2 cm/day = 0.02 m/day; grade $S$ = 0.003
• Manning's $n$ = 0.011 for clay/concrete tile (assumed)

Discharge

$$Q = 0.02\times\frac{10000}{86400} = 0.00231\ \text{m}^3/\text{s} = 2.31\ \text{l/s}$$

Diameter

For a pipe flowing full, $A=\pi D^2/4$ and $R=D/4$:

$$Q=\frac{1}{n}\cdot\frac{\pi D^2}{4}\left(\frac{D}{4}\right)^{2/3}S^{1/2} \quad\Rightarrow\quad D^{8/3}=\frac{4^{5/3}\,nQ}{\pi\,S^{1/2}}$$ $$D=\left(\frac{4^{5/3}\times0.011\times0.00231}{\pi\times\sqrt{0.003}}\right)^{3/8} = 0.0871\ \text{m} = 87\ \text{mm}$$

The next standard tile size is 100 mm. Its full-flow capacity = 3.34 l/s (0.00334 m³/s) > 2.31 l/s, so it is safe (velocity 0.43 m/s).

Answer: provide 100 mm diameter tiles.

(i) Size of tiles for 2 cm/day

Each lateral drains 200 m × 50 m = 1 ha. The discharge is

$$Q = 0.02\times\frac{10000}{86400} = 0.00231\ \text{m}^3/\text{s} = 2.31\ \text{l/s}$$

For a pipe flowing full ($A=\pi D^2/4$, $R=D/4$, $n$ = 0.011, $S$ = 0.003):

$$D=\left(\frac{4^{5/3}\,nQ}{\pi\sqrt{S}}\right)^{3/8}=\left(\frac{4^{5/3}\times0.011\times0.00231}{\pi\times\sqrt{0.003}}\right)^{3/8} = 87\ \text{mm}$$

Adopt the next standard size, 100 mm. Full-flow capacity of the 100 mm tile:

$$Q_{full}=\frac{1}{0.011}\times\frac{\pi(0.1)^2}{4}\times\left(\frac{0.1}{4}\right)^{2/3}\times\sqrt{0.003} = 0.00334\ \text{m}^3/\text{s} = 3.34\ \text{l/s}$$

(ii) Spacing at 3 cm/day

The same 100 mm tile at the same grade can carry only 3.34 l/s. The area it can serve at a drainage coefficient of 0.03 m/day is

$$A = \frac{Q_{full}}{q} = \frac{0.00334}{0.03/86400} = 9630\ \text{m}^2$$

For a lateral of length 200 m:

$$\text{Spacing}=\frac{9630}{200} = 48.1\ \text{m}$$

So the laterals must be placed about 48 m apart (slightly closer than 50 m). If the 50 m spacing is to be kept, the tile size must be increased: at 3 cm/day the discharge is 3.47 l/s, which needs the next size, 125 mm.

Answer: (i) 100 mm tiles; (ii) spacing ≈ 48 m for 100 mm tiles.

Causes of water logging

• Seepage from unlined canals and over-irrigation raise the water table.
• Flat land and poor natural drainage in the Terai, with blocked or obstructed drains (roads, embankments).
• Heavy monsoon rain with flooding from the rivers, on an impervious or clayey subsoil.
• Shallow impervious layer; absence of proper drains and poor maintenance.

Remedial measures

1. Prevention: lining of canals and water courses, controlled irrigation and good water management, cropping pattern suited to the water availability.
2. Surface drainage: improving natural drains and constructing artificial open drains.
3. Subsurface drainage: tile or pipe drains, mole drains.
4. Vertical drainage: tube wells to lower the water table (Terai has a good aquifer) and to give irrigation water.
5. Interceptor drains near canals, land levelling and the use of deep-rooted crops.

Method and assumptions for surface drainage design in the Terai

The method is based on the drainage rate for bunded paddy fields:

1. Take the design 3-day rainfall for a return period of 5–10 years (mean annual maximum × growth factor).
2. Assume a bunded field with the initial water depth $h_0$ = 40–50 mm and no inflow from outside.
3. Rice tolerance: a maximum depth (300–400 mm) for 1 day and depth above a limit (200–250 mm) for 2–3 days.
4. ET and deep percolation are neglected (safe side); the rain falls uniformly over 3 days and drainage is constant.
5. Find the drainage rate $D$ (mm/day, then l/s/ha) satisfying the depth–duration limits by trial.
6. Design discharge $Q = D \times A$; the drains are sized by Manning's formula with $n$ = 0.025–0.035, non-silting and non-scouring velocities, and free board 0.15–0.3 m.
7. The drains follow the natural slope and discharge to a natural drain or river at a level lower than the field outlets.

Internal and external drainage

• Internal drainage is the system inside the command area or farm boundary: the field drains, lateral and collector drains (open or tile) that collect excess water from the fields and carry it up to the boundary of the project area.
• External drainage is the system outside the farm/project area: main drains, outfall channels and the improved natural streams that carry the collected water to the final outlet (river), together with flood protection works.

 field --> field drain --> collector --> main drain --> river
 |<--- internal drainage --->|<--- external drainage --->|

Causes of water logging

• Seepage from unlined canals and over-irrigation raise the water table.
• Flat land and poor natural drainage in the Terai, with blocked or obstructed drains (roads, embankments).
• Heavy monsoon rain with flooding from the rivers, on an impervious or clayey subsoil.
• Shallow impervious layer; absence of proper drains and poor maintenance.

Remedial measures

1. Prevention: lining of canals and water courses, controlled irrigation and good water management, cropping pattern suited to the water availability.
2. Surface drainage: improving natural drains and constructing artificial open drains.
3. Subsurface drainage: tile or pipe drains, mole drains.
4. Vertical drainage: tube wells to lower the water table (Terai has a good aquifer) and to give irrigation water.
5. Interceptor drains near canals, land levelling and the use of deep-rooted crops.

Method and assumptions for surface drainage design in the Terai

The method is based on the drainage rate for bunded paddy fields:

1. Take the design 3-day rainfall for a return period of 5–10 years (mean annual maximum × growth factor).
2. Assume a bunded field with the initial water depth $h_0$ = 40–50 mm and no inflow from outside.
3. Rice tolerance: a maximum depth (300–400 mm) for 1 day and depth above a limit (200–250 mm) for 2–3 days.
4. ET and deep percolation are neglected (safe side); the rain falls uniformly over 3 days and drainage is constant.
5. Find the drainage rate $D$ (mm/day, then l/s/ha) satisfying the depth–duration limits by trial.
6. Design discharge $Q = D \times A$; the drains are sized by Manning's formula with $n$ = 0.025–0.035, non-silting and non-scouring velocities, and free board 0.15–0.3 m.
7. The drains follow the natural slope and discharge to a natural drain or river at a level lower than the field outlets.

Anti-waterlogging measures are the steps taken to prevent the water table from rising, and to reclaim land that is already waterlogged.

1. Preventive measures

• Lining of canals and water courses to reduce seepage losses.
• Controlled irrigation: supply only the crop requirement, with a proper irrigation interval (avoid over-watering).
• Efficient water management: rotational supply (warabandi), farmer training, volumetric charges.
• Improving the natural drainage: clearing, deepening and straightening natural streams; providing cross-drainage works of adequate waterway so that roads, canals and embankments do not obstruct flow.
• Intensity of irrigation limited to the available drainage capacity; avoiding unlined canals on high-seepage ground; and a good cropping pattern.
• Interceptor drains along the canal.

2. Remedial (reclamation) measures

• Surface drains: open field, collector and main drains.
• Subsurface drains: tile or pipe drains and mole drains, to lower the water table to a safe depth (about 1.5 m or more).
• Vertical drainage: tube wells or pump wells; pumped water is reused for irrigation.
• Biological drainage: deep-rooted trees such as eucalyptus along canals and fields.
• Land levelling and construction of embankments against floods.

Prevention is cheaper than cure; a combination of lining and surface drains is generally adopted, with tile drains or tube wells for severely affected areas.

The layout of a surface drainage system is planned so that every field outlet discharges by gravity into a drain, and the water moves from small drains to larger ones until it reaches the natural outlet.

Components

1. Field drains (laterals): shallow drains at the lower side of the fields, which receive water from the bunded fields' outlets.
2. Collector drains: receive water from several field drains; run across or along the ground slope.
3. Main (outfall) drain: receives the flow of the collectors and delivers it to a natural drain/river.
4. Outlet: structure at the junction with the river with a flap gate or sill to prevent backflow.
5. Structures: culverts at crossings of roads/canals, drops where the slope is steep.

        irrigation canal ======================
        | field | field | field |  <- bunded fields
  ------+-------+-------+-------+----- field drains
        |       |       |       |
        v       v       v       v
  ===========================>  collector drain
                |
                v  main drain
              ~~~~~~~~~~~~~~~~~~~ natural river

Planning principles

• Drains follow the lowest lines of the land (natural depressions), and are kept parallel to the contours for the field drains and along the slope for collectors.
• The irrigation canal runs on the ridge, the drains in the valleys; drain and canal do not run side by side.
• The spacing of the field drains depends on the size of fields and the land slope (typically 100–300 m).
• Water level in each drain is at least 0.3 m below the outlets it receives.
• Minimum number of crossings and a short route to the outfall; the layout should avoid cutting through villages and roads.

The design of surface drains is based on the following criteria.

1. Design rainfall: the maximum 3-day rainfall of 5–10 years return period (taken as higher for main drains and outfalls, e.g. 10–25 years), based on frequency analysis and growth factors.
2. Crop tolerance: the permissible depth and duration of flooding of the crop. For rice: a maximum of 300–400 mm for 1 day and 200–250 mm for 2–3 days; the initial depth in the field is 40–50 mm.
3. Drainage rate (coefficient): found in mm/day or l/s/ha such that the depth–duration limits are not exceeded; ET and percolation are generally neglected.
4. Design discharge: $Q = D \times A$ for the area served.
5. Hydraulic design: Manning's formula $Q=\frac{1}{n}AR^{2/3}S^{1/2}$ with $n$ = 0.025–0.035 for earthen drains.
6. Velocity limits: non-silting (about 0.3 m/s) and non-scouring (0.6–0.9 m/s depending on the soil).
7. Section: trapezoidal; side slopes 1:1 to 2:1 as per the soil; economical bed width/depth ratio; free board 0.15–0.3 m.
8. Bed slope: following the natural ground slope (generally 1 in 500 to 1 in 2000 in the Terai).
9. Water level: the full-supply level of the drain is at least 0.3 m below the field level (or the outlets) so that the field drains freely.
10. Outlet: the lowest point of the system must be higher than the high flood level of the receiving river, or a flap gate is provided to prevent backflow.
11. Provision of structures (culverts, drops, check structures) and access for maintenance.

In bunded rice fields the design discharge of the surface drain is fixed from the drainage rate that satisfies the tolerance of the crop to depth and duration of flooding.

1. Design rainfall: find the maximum 3-day rainfall $R$ for the chosen return period (5–10 years), using the frequency analysis or the mean annual value × growth factor.
2. Initial depth: take the standing water $h_0$ in the bunded field (40–50 mm).
3. Crop tolerance: fix the maximum depth $h_{max}$ and its permissible duration, and the depth $h_e$ and the duration $T_e$ for which it may persist (for example 300–400 mm for 1 day, 200–250 mm for 2–3 days).
4. Water-level history: assuming the rain falls uniformly in 3 days, ET and percolation neglected, and drainage $D$ constant: $$h_{peak}=h_0+R-3D, \qquad T_e=\left(3-\frac{h_e-h_0}{R/3-D}\right)+\frac{h_{peak}-h_e}{D}$$ Find $D$ (mm/day) by trial so that the limits are just satisfied; convert: 1 mm/day = 0.1157 l/s/ha.
5. Design discharge: $$Q = D\,A$$ with $D$ in l/s/ha and $A$ in ha (gives l/s). For large areas, a reduction factor for the travel time is sometimes applied.

A drainage rate of 3 l/s/ha is a fixed removal rate. The field is inundated when the water level is above 200 mm, so the level history is traced for the design storm.

Data (assumed from the standard problem)

Initial depth $h_0$ = 40 mm; design 3-day rainfall $R$ = 300 mm (100 mm/day); ET and deep percolation neglected.

$$D = 3\ \text{l/s/ha} = 3\times\frac{86.4}{10} = 25.92\ \text{mm/day}$$

(1 l/s/ha = 86.4 m³/ha/day = 8.64 mm/day.)

Rise during rain

Net rise = 100 − 25.92 = 74.08 mm/day. Time to reach 200 mm:

$$t_1 = \frac{200-40}{74.08} = 2.16\ \text{days}$$

Peak at the end of day 3:

$$h_{peak} = 40 + 300 - 3\times25.92 = 262.2\ \text{mm}$$

Fall after rain

Time to fall from 262.2 mm to 200 mm:

$$t_2 = \frac{262.2-200}{25.92} = 2.40\ \text{days}$$

Duration above 200 mm = (3 − 2.16) + 2.40 = 3.24 days.

Will the depth exceed 300 mm?

No. The maximum depth is 262 mm, which is less than 300 mm, so the system is safe for both limits (the 300 mm limit is not reached, and the stay above 200 mm is about 3.2 days).

If the 10-year design rainfall of 450 mm (growth factor 1.5) were used, the peak would be 412 mm and the field would stay above 200 mm for 9.9 days, so 3 l/s/ha would then be inadequate.

Answer: about 3.2 days above 200 mm; maximum depth 262 mm, which does not exceed 300 mm.

Surface drainage of the 40 ha Terai field is designed from the design rainfall, the permissible ponding of the rice crop, and Manning's formula for the drain.

Step 1: Design rainfall and drainage rate

Design 3-day rainfall $R$ = 50 mm (16.7 mm/day). Assume the initial water depth in the bunded field $h_0$ = 50 mm (not given).

Water level at the end of the storm without drainage = 50 + 50 = 100 mm. This is below the permissible 300 mm and also below 200 mm, so the crop limits are met even before any drainage. To avoid accumulation for later storms, the 3-day rainfall is removed within the same 3 days:

$$D = \frac{R}{3} = \frac{50}{3} = 16.7\ \text{mm/day} = 16.7\times\frac{10}{86.4} = 1.93\ \text{l/s/ha}$$

Step 2: Design discharge

$$Q = D\,A = 1.93\times40 = 77.2\ \text{l/s} = 0.0772\ \text{m}^3/\text{s}$$

Step 3: Section of the drain

Assume an earthen trapezoidal drain, side slope 1:1, $n$ = 0.025 (given), slope $S$ = 1/400 (given), bed width $b$ = 0.5 m.

$$Q=\frac{1}{n}AR^{2/3}S^{1/2}$$

By trial: depth $d$ = 0.21 m, $A=(b+d)d$ = 0.147 m², $P=b+2\sqrt{2}d$ = 1.086 m, $R$ = 0.135 m, check $Q$ = (1/0.025) × 0.147 × 0.135^(2/3) × (1/400)^0.5 = 0.0772 m³/s.

$V$ = 0.0772/0.147 = 0.53 m/s, which is non-silting (> 0.3 m/s) and non-scouring (< 0.6 m/s for earth).

Provide free board 0.15 m: total depth = 0.21 + 0.15 = 0.36 m, adopt 0.40 m.

Step 4: Layout

Field drains (laterals) run along the slope into one collector drain along the lower edge of the 40 ha block; the collector discharges to the natural drain through an outlet structure with its sill 0.3 m below the lowest field level. The drain is made deep enough for the field outlets to discharge freely, with the drain water level at least 0.3 m below the field outlets.

Answer: D = 16.7 mm/day (1.93 l/s/ha), Q = 77.2 l/s; trapezoidal drain with bed width 0.5 m, flow depth 0.21 m (total 0.40 m), side slopes 1:1, slope 1 in 400.

Tile drain spacing is found from Donnan's (Dupuit–Forchheimer) formula for steady flow towards parallel drains over an impervious layer:

$$S=\sqrt{\frac{4K\,(b^2-a^2)}{q}}$$

where $S$ = spacing, $K$ = permeability, $q$ = drainage rate (discharge per unit area), $b$ = height of the water table above the impervious layer at the mid-point, $a$ = height of the drain above the impervious layer.

Data

• Depth of impervious layer = 12 m, depth of drain = 2 m, depth of highest water table = 1.5 m (all below ground)
• $K$ = 0.01 cm/s = 0.0001 m/s
• Rainfall 2000 mm; 2 % to be drained in 48 hours

Step 1: Drainage rate

$$q = \frac{2}{100}\times\frac{2000}{1000}\ \text{m} \Big/ (48\times3600\ \text{s}) = 2.315\times10^{-7}\ \text{m/s}$$

Step 2: Heights above the impervious layer

$$\begin{aligned} a &= 12-2 = 10\ \text{m} \\ b &= 12-1.5 = 10.5\ \text{m} \end{aligned}$$

Step 3: Spacing

$$\begin{aligned} S^2 &= \frac{4\times0.0001\times(10.5^2-10^2)}{2.315\times10^{-7}} \\ &= \frac{4\times0.0001\times10.25}{2.315\times10^{-7}} = 17712\ \text{m}^2 \\ S &= 133.1\ \text{m} \end{aligned}$$

Here $K$ = 1 × 10⁻⁴ m/s = 0.01 cm/s, and 2 % of 2000 mm = 40 mm is drained in 2 days = 48 h.

Answer: spacing of the tile drains ≈ 133.1 m (say 133 m).

Tile drain spacing follows from Donnan's formula for steady flow to parallel drains above an impervious layer:

$$S=\sqrt{\frac{4K\,(b^2-a^2)}{q}}$$

Here the drainage rate is given directly by the drainage coefficient, taken as 0.012 m/day (12 mm/day); the annual rainfall of 900 mm is therefore not needed.

Data

• $q$ = 0.012 m/day = 0.012/86400 = $1.389\times10^{-7}$ m/s
• $K$ = 0.012 cm/s = 0.00012 m/s
• $a$ = 9 − 1.6 = 7.4 m (drain above the impervious layer)
• $b$ = 9 − 1.2 = 7.8 m (highest water table above the impervious layer)

Calculation

$$\begin{aligned} S^2 &= \frac{4\times0.00012\times(7.8^2-7.4^2)}{1.389\times10^{-7}} \\ &= \frac{4\times0.00012\times6.08}{1.389\times10^{-7}} = 21012\ \text{m}^2 \\ S &= 145.0\ \text{m} \end{aligned}$$

Answer: spacing of the tile drains ≈ 145.0 m (say 145 m).

The drains must be placed so that the water table, midway between the drains, stays below the root zone plus the capillary rise. The depth of the drain then follows from Donnan's formula.

Data

• Root zone depth = 1.5 m, capillary rise = 0.3 m
• $K$ = 1.5 × 10⁻⁴ m/s; drain spacing $S$ = 200 m; depth of impervious layer = 10 m below ground
• Drainage capacity = 0.11 m³/s/km² = 0.11/10⁶ = $1.10\times10^{-7}$ m/s

Step 1: Highest permissible water table

Depth of the water table below ground at the mid-point = 1.5 + 0.3 = 1.8 m. Its height above the impervious layer:

$$b = 10 - 1.8 = 8.2\ \text{m}$$

Step 2: Height of the drain above the impervious layer

$$S^2=\frac{4K(b^2-a^2)}{q} \quad\Rightarrow\quad a^2 = b^2-\frac{qS^2}{4K}$$ $$\begin{aligned} a^2 &= 8.2^2-\frac{1.10\times10^{-7}\times200^2}{4\times1.5\times10^{-4}} \\ &= 67.24-7.333 = 59.907 \\ a &= 7.740\ \text{m} \end{aligned}$$

Step 3: Depth of drain below ground

$$\text{Depth} = 10 - 7.740 = 2.26\ \text{m}$$

Answer: lay the drains about 2.26 m (say 2.3 m) below the ground surface, i.e. 7.74 m above the impervious layer. The water table at the mid-point then stays 1.8 m below the surface, just below the root zone and capillary fringe.

Drainage of irrigated land is designed to keep the water table below the root zone and to remove excess surface water, so that water logging and salinity do not develop.

Design steps

1. Collect the data: rainfall, irrigation intensity and efficiency, soil permeability $K$, depth of the impervious layer, ground levels, the water table depth and the outlet level.
2. Find the drainage requirement (rate $q$): the part of the rainfall and irrigation water to be removed in a fixed time (for example 1–2 % of annual rainfall in 24 h), or a drainage coefficient in mm/day (or l/s/ha) for the paddy field from the crop tolerance.
3. Fix the water table level: the highest permissible water table is below the root zone plus capillary rise (about 1.5–1.8 m below ground).
4. Choose the system: surface drains (open field, collector and main drains) for excess rain water; subsurface (tile) drains for lowering the water table; tube wells where suitable.
5. Subsurface design: drain depth 1.5–2.5 m and spacing from Donnan's/Hooghoudt's formula $$S=\sqrt{\frac{4K(b^2-a^2)}{q}}$$ Size the lateral by Manning's formula for a pipe flowing full.
6. Surface drain design: discharge $Q=D\times A$; Manning's formula for the section; velocities within the non-silting and non-scouring limits; free board 0.15–0.3 m.
7. Layout and outlets: field, collector and main drains lead to a natural outlet higher than the high flood level, or provided with a flap gate; add structures and an operation and maintenance plan.