---
title: "Tile drain spacing, Biratnagar rainfall"
subject: "Irrigation and Drainage Engineering"
chapter: 9
key: "semester_6/irrigation_and_drainage_engineering:tile-drain-spacing-biratnagar-2000"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/tile-drain-spacing-biratnagar-2000
site: BCT Engineering
---

# Tile drain spacing, Biratnagar rainfall

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

- Asked in: 2066 Bhadra (old course), 6 marks
- 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-biratnagar-2000
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-9

## Question

The annual rainfall in Biratnagar is 2000 mm. Find the spacing of sub-surface drains if 2% of average annual rainfall is to be drained in 2 days. Given: Depth of impervious stratum from the top of soil surface = 12 m; Position of drain is 2 m below the top soil surface and the depth of highest position of water table below the top soil surface = 1.5 m; Permeability, K = 1 × 10⁻⁴ m/s.

## Short answer

Donnan's formula: K=1×10⁻⁴ m/s, q=2% of 2000 mm in 48 h =2.315×10⁻⁷ m/s, a=12−2=10 m, b=12−1.5=10.5 m. Spacing S ≈133.1 m (say 133 m).

## Answer

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).**

## Check yourself

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

1. K = 1 x 10^-4 m/s. How does this compare with the earlier problems?
   A. It is already in m/s (0.01 cm/s) and is used directly as 0.0001
   B. It must be divided by 100 again
   C. It equals 1 x 10^-4 cm/s
   D. It equals 1 x 10^-6 m/s
   Answer: A. Using the unit-correct 1e-4 m/s in the formula avoids a 100x error. (Key point: Permeability unit)

2. How is q found for 2% of 2000 mm drained in 2 days?
   A. 40 mm in 24 hours, q = 4.63 x 10^-7 m/s
   B. 40 mm in 48 hours, q = 2.315 x 10^-7 m/s
   C. 2000 mm in 48 hours, q = 1.16 x 10^-5 m/s
   D. 4 mm in 48 hours
   Answer: B. 2% of 2000 mm = 40 mm; divide by 48 x 3600 s. (Key point: Drainage rate)

3. What are a and b (impervious 12 m, drain 2 m, water table 1.5 m below top)?
   A. a = 2 m, b = 1.5 m
   B. a = 10.5 m, b = 10 m
   C. a = 10 m, b = 10.5 m
   D. a = 12 m, b = 10.5 m
   Answer: C. a = 12 - 2 = 10 m; b = 12 - 1.5 = 10.5 m. (Key point: Heights)

4. What spacing results?
   A. About 13 m
   B. About 42 m
   C. About 17712 m
   D. About 133 m
   Answer: D. S^2 = 4 x 1e-4 x 10.25/2.315e-7 = 17712, so S = 133.1 m. (Key point: Spacing)

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