Skip to main content

Tile drain spacing: rainfall 1700 mm

Short answer

Use Donnan's formula S = sqrt[4K(b²-a²)/q] with K=0.001 cm/s, q from 1.5% of 1700 mm in 24 h = 2.951×10⁻⁷ m/s, a=7.8 m, b=8.8 m. Spacing S ≈ 47.4 m (adopt 47 m).

Calculate the spacing of the tile drains for an area having rainfall of 1700 mm, if 1.5% is to be drained in 24 hrs. From ground level, depth of impervious stratum is 10 m, depth of drain is 2.2 m and depth of highest position of water table is 1.2 m, coefficient of permeability = 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 = 2.2 m, depth of highest water table = 1.2 m (all below ground)
  • KK = 0.001 cm/s = 1×10−51\times10^{-5} m/s
  • Rainfall 1700 mm; 1.5 % to be drained in 24 hours

Step 1: Drainage rate

q=1.5100×17001000 m/(24×3600 s)=2.951×10−7 m/sq = \frac{1.5}{100}\times\frac{1700}{1000}\ \text{m} \Big/ (24\times3600\ \text{s}) = 2.951\times10^{-7}\ \text{m/s}

Step 2: Heights above the impervious layer

a=10−2.2=7.8 mb=10−1.2=8.8 m\begin{aligned} a &= 10-2.2 = 7.8\ \text{m} \\ b &= 10-1.2 = 8.8\ \text{m} \end{aligned}

Step 3: Spacing

S2=4×1×10−5×(8.82−7.82)2.951×10−7=4×1×10−5×16.602.951×10−7=2250 m2S=47.4 m\begin{aligned} S^2 &= \frac{4\times1\times10^{-5}\times(8.8^2-7.8^2)}{2.951\times10^{-7}} \\ &= \frac{4\times1\times10^{-5}\times16.60}{2.951\times10^{-7}} = 2250\ \text{m}^2 \\ S &= 47.4\ \text{m} \end{aligned}

Answer: spacing of the tile drains ≈ 47.4 m (say 47 m).

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Irrigation and Drainage Engineering chapter 9 questions →