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Tile drain spacing: rainfall 80 cm

Short answer

Donnan's formula: with 80 cm = 800 mm, 1% in 24 h gives q=9.259×10⁻⁸ m/s, a=9−1.5=7.5 m, b=9−1=8 m, K=1×10⁻⁵ m/s. Spacing S ≈ 57.9 m (say 58 m).

In a tile drainage system, the drains are laid with their centers 1.5 m below the ground level. The impervious layer is 9.0 m below the ground level and the average annual rainfall in the area is 80 cm. 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 ground level, determine the spacing of the drain pipes. The coefficient of permeability is 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 = 9 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 800 mm; 1 % to be drained in 24 hours

Step 1: Drainage rate

q=1100×8001000 m/(24×3600 s)=9.259×10−8 m/sq = \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

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

Step 3: Spacing

S2=4×1×10−5×(82−7.52)9.259×10−8=4×1×10−5×7.759.259×10−8=3348 m2S=57.9 m\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).

✓ Answer checked by Claude Opus 5.5 (high effort)

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