---
title: "Derive expression for tile drain spacing"
subject: "Irrigation and Drainage Engineering"
chapter: 9
key: "semester_6/irrigation_and_drainage_engineering:tile-drain-spacing-derivation"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/tile-drain-spacing-derivation
site: BCT Engineering
---

# Derive expression for tile drain spacing

Irrigation and Drainage Engineering · Chapter 9: Water Logging and Drainage

- Asked in: 2075 Baisakh, 5 marks; 2070 Chaitra (old course), 5 marks; 2064 Jestha (old course), 5 marks; 2064 Kartik (old course), 8 marks; 2078 Baisakh, 5 marks; 2079 Chaitra, 6 marks; 2077 Chaitra, 5 marks; 2071 Magh, 5 marks (asked 8 times)
- Video: [Unit:7 | Water logging & drainage | Derive Subsurface drain spacing | BE Civil | Prashant YT |](https://www.youtube.com/watch?v=lvH1OVIFutw) — Prashant YT
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/tile-drain-spacing-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-9

## Question

Derive the expression (equation) for the spacing of subsurface (tile) drains capable of lowering the water table.

## Short answer

Applying Darcy's law for horizontal flow between parallel drains over an impervious layer gives S = sqrt(4K(b² – a²)/q), where S is spacing, K permeability, b and a water table heights above impervious layer at midpoint and drain, and q recharge rate.

## Answer

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

```text
          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 $S=\sqrt{4Kb^2/q}$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which assumptions underlie the Dupuit-Forchheimer tile drain derivation?
   A. Flow is horizontal and the hydraulic gradient equals the slope of the water table
   B. Flow is vertical with a constant gradient
   C. The soil is non-homogeneous and the flow unsteady
   D. Recharge is zero
   Answer: A. Dupuit-Forchheimer assumes horizontal flow with gradient = dy/dx, allowing Darcy's law to be applied simply. (Key point: Assumptions of derivation)

2. What is the flow per unit drain length through a section at distance x from the drain?
   A. q_x = q x
   B. q_x = q (S/2 - x), from recharge on the strip between x and the mid-point
   C. q_x = q S
   D. q_x = K y
   Answer: B. Water passing section x is recharge falling on the width between x and mid-point S/2, i.e. S/2 - x. (Key point: Flow at distance x)

3. What boundary conditions fix the constant of integration and give the final result?
   A. y = 0 at x = 0 and y = a at x = S
   B. y = b at x = 0 and y = a at x = S/2
   C. y = a at x = 0 (drain) and y = b at x = S/2 (mid-point)
   D. dy/dx = 0 at the drain
   Answer: C. C = K a^2/2 from the drain; applying y = b at mid-point gives q S^2 = 4K(b^2 - a^2). (Key point: Boundary conditions)

4. What is the final spacing expression, and what does it reduce to for a drain on the impervious layer (a = 0)?
   A. S = sqrt(K(b - a)/q); reduces to zero
   B. S = 4K(b^2 - a^2)/q; reduces to S = 4K b^2/q
   C. S = sqrt(8K b/q); unchanged when a = 0
   D. S = sqrt(4K(b^2 - a^2)/q); reduces to S = sqrt(4K b^2/q)
   Answer: D. From q S^2 = 4K(b^2 - a^2), putting a = 0 gives S = sqrt(4K b^2/q). (Key point: Donnan's formula)

## Related in Irrigation and Drainage Engineering

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