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Sheet pile seepage quantity numerical

Short answer

For sheet pile with k=4×10⁻⁴ m/s, H=3 m, penetration ratio 0.7, flow net gives Nf=3, Nd=8. Seepage q = kH(Nf/Nd) = 4.5×10⁻⁴ m³/s per m (≈39 m³/day per m).

A deposit of cohesionless soil with a permeability of 4×10−24 \times 10^{-2} cm/s has a depth of 10 m with an impervious ledge below. A sheet pile wall is driven into this deposit to a depth of 7 m. The wall extends above the surface of the soil and a 3 m depth of water acts on one side. Sketch the flow net and determine the seepage quantity per metre length of the wall.

Given

k=4×10−2k = 4\times10^{-2} cm/s =4×10−4= 4\times10^{-4} m/s. Permeable layer depth T=10T = 10 m, sheet pile penetration s=7s = 7 m (so s/T=0.7s/T = 0.7), net head H=3H = 3 m.

Flow net

  H=3m ▽       |P|
  ~~~~~~~~~~~~~|P|~~~~~~ (downstream ground)
   ground      |P|   7 m
   3 flow channels, 8 equipotential drops
   ______________|__|________
   ======== impervious, 10 m deep ========

Draw the flow net with the pile tip 3 m above the impervious layer. Sketching it gives about

  • Nf=3N_f = 3 flow channels
  • Nd=8N_d = 8 equipotential drops

Seepage quantity

q=k H NfNd=4×10−4×3×38=4.5×10−4 m3/s per mq = k\,H\,\frac{N_f}{N_d} = 4\times10^{-4} \times 3 \times \frac{3}{8} = 4.5\times10^{-4}\ \text{m}^3/\text{s per m} q=4.5×10−4×86400=38.9 m3/day per metre lengthq = 4.5\times10^{-4} \times 86400 = 38.9\ \text{m}^3/\text{day per metre length}

A check with the exact solution for a single sheet pile of penetration ratio s/T=0.7s/T = 0.7 gives q/(kH)=0.371q/(kH) = 0.371, i.e. q=4.45×10−4q = 4.45\times10^{-4} m³/s per m, so the flow net values agree.

Answer: q≈4.5×10−4q \approx 4.5\times10^{-4} m³/s per metre length (≈39\approx 39 m³/day per m).

✓ Answer checked by Claude Opus 5.5 (high effort)

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