Seepage through dam with central core (33 m dam)

Data and assumptions: dam height 33 m, free board 3 m so reservoir depth $H=30$ m; top width 7 m; core width $t=3$ m with $k_c=4\times10^{-8}$ m/s; shells $k=4\times10^{-6}$ m/s; slopes 3:1 (u/s) and 2.5:1 (d/s); impervious foundation, no tail water. The u/s shell is much more pervious than the core, so the head on the core face is the full $H=30$ m.

Seepage length in the d/s shell

The d/s face has a horizontal projection of $2.5\times33=82.5$ m; half of the 4 m by which the crest exceeds the core width lies d/s of the core, so

$$L=82.5+\frac{7-3}{2}=84.5\ \text{m}$$

Continuity between core and d/s shell

Core: $q=\dfrac{k_c(H^2-h^2)}{2t}$; shell: $q=\dfrac{k\,h^2}{2L}$, where $h$ is the phreatic height at the d/s face of the core.

$$\begin{aligned} \frac{4\times10^{-8}(900-h^2)}{6}&=\frac{4\times10^{-6}h^2}{2\times 84.5}\\ 6.667\times10^{-9}(900-h^2)&=2.3669\times10^{-8}\,h^2\\ h^2&=197.8\ \text{m}^2,\quad h=14.06\ \text{m} \end{aligned}$$

Discharge

$$q=\frac{k\,h^2}{2L}=\frac{4\times10^{-6}\times197.8}{2\times84.5}=4.681 \times 10^{-6}\ \text{m}^3/\text{s per m}$$

Check using the core: $\dfrac{4\times10^{-8}(900-197.8)}{6}=4.681 \times 10^{-6}$ m³/s per m.

Answer: seepage discharge $=4.681 \times 10^{-6}$ m³/s per metre length of dam ($\approx0.404$ m³/day per m).