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Tile drain spacing: rainfall 1400 mm

Short answer

Donnan's formula S = sqrt[4K(b²-a²)/q]: K=1×10⁻⁵ m/s, q = 1% of 1400 mm in 24 h = 1.62×10⁻⁷ m/s, a=7 m, b=8 m. Spacing S ≈ 60.9 m (say 61 m).

Find the spacing of the tile drains for an area having average annual rainfall of 1400 mm, if 1% is to be drained in 24 hours. From ground level, depth of impervious stratum = 9 m; depth of drains = 2 m and depth of highest position of the water table = 1 m. Coefficient of permeability may be taken as 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 = 2 m, depth of highest water table = 1 m (all below ground)
  • KK = 0.001 cm/s = 1×10−51\times10^{-5} m/s
  • Rainfall 1400 mm; 1 % to be drained in 24 hours

Step 1: Drainage rate

q=1100×14001000 m/(24×3600 s)=1.62×10−7 m/sq = \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

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

Step 3: Spacing

S2=4×1×10−5×(82−72)1.62×10−7=4×1×10−5×15.001.62×10−7=3703 m2S=60.9 m\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).

✓ Answer checked by Claude Opus 5.5 (high effort)

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