Seepage through dam with central core (45 m dam)

Assumptions (from the figure data): the central core is 3 m wide with $k_c=5\times10^{-8}$ m/s; the shells have $k=4\times10^{-6}$ m/s; foundation impervious; no tail water. The u/s shell is far more pervious than the core, so the full reservoir head $H=42$ m acts on the u/s face of the core. The d/s shell carries the flow from the core to the toe over a horizontal length $L=126$ m (as given in the figure). Let $h$ be the height of the phreatic line at the d/s face of the core.

        8 m
      _______
  1V:3H/|core|\ 1V:3H
 42m  / |3 m |  \
 ~~~ /  |    |   \
 ////////////////////
 |        L = 126 m |  (d/s shell)

Flow through the core (Dupuit)

$$q=\frac{k_c\,(H^2-h^2)}{2t}$$

Flow through the d/s shell (toe at ground level)

$$q=\frac{k\,h^2}{2L}$$

Equate (continuity)

$$\begin{aligned} \frac{5\times10^{-8}\,(42^2-h^2)}{2\times3}&=\frac{4\times10^{-6}\,h^2}{2\times126}\\ 8.333\times10^{-9}(1764-h^2) &= 1.5873\times10^{-8}\,h^2\\ h^2 &= 607.3\ \text{m}^2 \Rightarrow h=24.64\ \text{m} \end{aligned}$$

Discharge

$$q=\frac{k\,h^2}{2L}=\frac{4\times10^{-6}\times607.3}{2\times126}=9.639 \times 10^{-6}\ \text{m}^3/\text{s per m}$$

Check with the core: $q=\dfrac{5\times10^{-8}(1764-607.3)}{6}=9.639 \times 10^{-6}$ m³/s per m.

Answer: seepage discharge $\approx9.639 \times 10^{-6}$ m³/s per metre run of dam ($\approx0.833$ m³/day per m); the phreatic line drops sharply across the core and stands $24.64$ m high at its d/s face.