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Earth dam with filter, seepage quantity

Short answer

Using Casagrande's method: d = 92 m, h = 30 m, y0 = √(92²+30²)−92 = 4.77 m. Seepage q = k y0 = 3×10⁻⁶ m/s × 4.77 m = 1.43×10⁻⁵ m³/s per metre (≈1.24 m³/day per m).

An earth dam of homogeneous section with a horizontal filter is shown in the figure below. If the coefficient of permeability of the soil is 3×10−33 \times 10^{-3} mm/s, find the quantity of seepage per unit length of the dam. [Figure: homogeneous earth dam, upstream and downstream slopes 3:1, reservoir level E at 30 m above the base, crest width 8 m, horizontal distance 27 m marked from A to B on the upstream slope, 32 m marked between B and B', 90 m marked at the upstream base, total base length 200 m, horizontal filter of length S = 45 m at the downstream toe, directrix of base parabola marked. Some dimension labels are hard to read.]

Method

Casagrande's base parabola for a dam with a horizontal filter. Seepage per unit length: q=k y0q = k\,y_0 where y0=d2+h2−dy_0 = \sqrt{d^2 + h^2} - d.

Geometry from the figure (some labels were unclear, so values are deduced)

  • Slopes 3H:1V. Water height h=30h = 30 m, so the wetted upstream slope has horizontal projection Δ=3×30=90\Delta = 3\times30 = 90 m. The water surface meets the upstream slope 90 m from the upstream toe.
  • Base width 200 m = 3Hd+8+3Hd3H_d + 8 + 3H_d gives dam height Hd=32H_d = 32 m.
  • The base parabola starts at a point on the water surface 0.3Δ=0.3×90=270.3\Delta = 0.3\times90 = 27 m upstream of where the water meets the slope (this is the 27 m in the figure). So it starts 90−27=6390 - 27 = 63 m from the upstream toe.
  • Filter length 45 m at the downstream toe, so the focus F is at 200−45=155200 - 45 = 155 m from the upstream toe.
d=155−63=92 md = 155 - 63 = 92\ \text{m}

Base parabola parameter

y0=922+302−92=96.77−92=4.77 my_0 = \sqrt{92^2 + 30^2} - 92 = 96.77 - 92 = 4.77\ \text{m}

Seepage

k=3×10−3k = 3\times10^{-3} mm/s =3×10−6= 3\times10^{-6} m/s

q=k y0=3×10−6×4.77=1.43×10−5 m3/s per mq = k\,y_0 = 3\times10^{-6} \times 4.77 = 1.43\times10^{-5}\ \text{m}^3/\text{s per m} q=1.43×10−5×86400=1.24 m3/day per metre lengthq = 1.43\times10^{-5} \times 86400 = 1.24\ \text{m}^3/\text{day per metre length}

Answer: q≈1.43×10−5q \approx 1.43\times10^{-5} m³/s per metre (≈1.24\approx 1.24 m³/day per m).

✓ Answer checked by Claude Opus 5.5 (high effort)

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