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Tile drain spacing: rainfall 1000 mm

Short answer

Donnan's formula: K=1×10⁻⁵ m/s, q = 1% of 1000 mm in 24 h = 1.157×10⁻⁷ m/s, a=10−1.5=8.5 m, b=10−1=9 m. Spacing S = 55.0 m (adopt 55 m).

In a tile drainage system, the drains are laid with their centers 1.5 m below the ground level. The impervious layer is 10 m below the ground level and the average annual rainfall in the area is 1000 mm. If 1% of the annual rainfall is to be drained in 24 hours to keep the highest position of the water table to 1 m below the ground level, determine the spacing of the drain pipes. Take K = 0.001 cm/sec.

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

S=4K (b2−a2)qS=\sqrt{\frac{4K\,(b^2-a^2)}{q}}

where SS = spacing, KK = permeability, qq = drainage rate (discharge per unit area), bb = height of the water table above the impervious layer at the mid-point, aa = 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)
  • KK = 0.001 cm/s = 1×10−51\times10^{-5} m/s
  • Rainfall 1000 mm; 1 % to be drained in 24 hours

Step 1: Drainage rate

q=1100×10001000 m/(24×3600 s)=1.157×10−7 m/sq = \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

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

Step 3: Spacing

S2=4×1×10−5×(92−8.52)1.157×10−7=4×1×10−5×8.751.157×10−7=3024 m2S=55.0 m\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).

✓ Answer checked by Claude Opus 5.5 (high effort)

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