Seepage discharge through dam body (k = 1×10⁻⁶ m/s)

Reading of the figure (assumptions): water depth $H=h=30$ m, free board 3 m so dam height $=33$ m; crest width 8 m; d/s slope 2.5:1; total base width $B=170$ m; drainage filter at the toe, so the focus of the parabola is at the d/s toe; impervious foundation; $k=1\times10^{-6}$ m/s.

Step 1: Geometry

$$\begin{aligned} \text{d/s projection} &= 2.5\times33=82.5\ \text{m}\\ \text{u/s projection} &= 170-8-82.5=79.5\ \text{m}\ \Rightarrow\ \text{u/s slope}=2.41:1\\ b &= 2.41\times30=72.3\ \text{m (wetted part)},\quad 0.3b=21.7\ \text{m}\\ d &= B-0.7b = 170-50.6=119.4\ \text{m} \end{aligned}$$

Step 2: Phreatic line (Casagrande)

$$y_0=\sqrt{h^2+d^2}-d=\sqrt{30^2+119.4^2}-119.4=3.71\ \text{m}$$ $$y=\sqrt{y_0^2+2y_0x}=\sqrt{13.77+7.42\,x}$$

The line is drawn from the corrected entry point (at $x=d$, $y=30$ m), made normal to the u/s face at the start, and ends at the drainage filter at the toe.

Step 3: Seepage discharge

Because $dy/dx=\dfrac{y_0}{y}$ and $q=k\,y\,\dfrac{dy}{dx}$ is constant along the line,

$$q=k\,y_0=1\times10^{-6}\times3.71=3.711 \times 10^{-6}\ \text{m}^3/\text{s per m}$$

Answer: $q\approx3.711 \times 10^{-6}$ m³/s per metre length of dam ($\approx0.321$ m³/day per m).