Seepage and phreatic line: horizontal filter at toe

For a homogeneous earth dam with a horizontal filter (drain) at the toe, Casagrande's method treats the top flow line (phreatic line) as a basic parabola with its focus at the starting point $F$ of the filter. The seepage follows Darcy's law with the Dupuit assumption.

Sketch and notation

        B     E
 FRL ~~~*~~~~~*.
        :    /  `.   phreatic line
   h    :   / u/s `-.
        :  /  face    `-._
 -------:-/---------------`|=========  filter
 heel    |<------- d ------>F
        (B is 0.3Δ upstream of E)

• $h$ = height of water above the dam base (head), $d$ = horizontal distance from point $B$ to the filter start $F$.
• $E$ is the point where the water surface meets the u/s face, and $\Delta$ is the horizontal projection of the wetted part of the u/s face. The basic parabola starts at $B$, on the water surface a distance $0.3\,\Delta$ horizontally upstream of $E$ (towards the reservoir).

Equation of the phreatic line

Put the origin at the focus $F$, with $x$ measured positive upstream. A parabola with its focus at the origin and focal length $y_0/2$ (vertex at $x=-y_0/2$) is

$$y^2 = 2y_0x + y_0^2$$

where $y_0$ is the ordinate of the parabola at the focus ($x=0$).

Seepage discharge (Dupuit)

At any section of the dam the flow per unit width is $q = k\,i\,A$, with the hydraulic gradient $i=dy/dx$ and the flow area $A=y\times1$:

$$q = k\,y\,\frac{dy}{dx}$$

Differentiating the parabola: $2y\,dy = 2y_0\,dx$, so $y\,\dfrac{dy}{dx} = y_0$. Therefore

$$q = k\,y_0$$

Finding $y_0$

The parabola passes through $B$, where $x = d$ and $y = h$:

$$h^2 = 2y_0d + y_0^2 \;\Rightarrow\; y_0^2 + 2dy_0 - h^2 = 0$$ $$y_0 = \sqrt{d^2+h^2} - d$$ $$\boxed{q = k\left(\sqrt{d^2+h^2}-d\right)}$$

Drawing the phreatic line

1. Mark $E$, then point $B$ at $0.3\Delta$ upstream of $E$, and measure $d$ and $h$.
2. Calculate $y_0$ and $q$.
3. Find points of the parabola from $y=\sqrt{2y_0x+y_0^2}$ for several values of $x$ and plot them.
4. Entry correction: start the actual phreatic line at $E$, perpendicular to the u/s face (an equipotential line), and join it smoothly to the basic parabola a little downstream.
5. Downstream, the parabola meets the horizontal filter vertically at its vertex, a distance $y_0/2$ downstream of $F$. With a horizontal filter no exit correction is needed.