Seepage with impervious foundation: derivation

Dupuit's assumptions: (i) the hydraulic gradient at any point on the phreatic line equals the slope of the line, $i=\dfrac{dy}{dx}$; (ii) the gradient is constant over the whole vertical section; (iii) flow is horizontal and Darcy's law holds.

   h1 ___
      \  \ phreatic line         y
       \  \____                  |
        \       \____ h2 ........|  tail water
  ______ \  dam       \_____     |
 /////////// impervious base ////|///
  |<--------- L ---------------->|

Derivation

Take $x$ along the flow from the u/s end, and $y$ as the height of the phreatic line above the impervious base at distance $x$. The line falls in the flow direction, so $dy/dx$ is negative. Per unit length of dam, the discharge through the vertical section of height $y$ is

$$q = k\, i\, A = -k\,\frac{dy}{dx}\,(y\times 1)$$

Hence

$$q\,dx = -k\,y\,dy$$

Integrate from the u/s end ($x=0$, $y=h_1$) to the exit at tail water ($x=L$, $y=h_2$), where $h_1$ and $h_2$ are the water depths above the base on the u/s and d/s side, and $L$ is the horizontal seepage length:

$$\begin{aligned} q\int_0^L dx &= -k\int_{h_1}^{h_2} y\,dy \\ q\,L &= \frac{k}{2}\left(h_1^2-h_2^2\right) \end{aligned}$$ $$\boxed{\,q=\frac{k\left(h_1^{2}-h_2^{2}\right)}{2L}\,}$$

The phreatic line is a parabola: integrating only up to the section $x$ gives $q\,x = \frac{k}{2}(h_1^2-y^2)$, so

$$y=\sqrt{h_1^{2}-\frac{(h_1^{2}-h_2^{2})}{L}\,x}$$

Here $L$ is measured from the corrected entry point (0.3 times the horizontal projection of the wetted u/s slope inside the water line) to the point where the line meets the tail water. If the line emerges above tail water on the d/s face, the exit zone is treated by Schaffernak's or Casagrande's correction. For $h_2=0$ this becomes $q=\dfrac{k h_1^2}{2L}$.