Seepage line for 22 m high earthen dam

Given: dam height 22 m, top width 6 m, water depth $h=20$ m, both slopes $45^\circ$ (1:1), length 3 km, $k=3\times10^{-3}$ mm/s $=3\times10^{-6}$ m/s. Assume impervious foundation, no drain and no tail water; the line is drawn by Casagrande's method.

  22 m _6 m_
      /|     |\
 45° / |     | \ 45°
    /  | 20m |  \
   ~~~~|     |   \
  /////////////////
  |<-- 50 m ------>|

Step 1: Geometry

$$\begin{aligned} B &= 22+6+22=50\ \text{m}\\ b &= 20\ \text{m (1:1 slope)},\quad 0.3b=6\ \text{m}\\ d &= B-0.7b = 50-14 = 36\ \text{m} \end{aligned}$$

Step 2: Basic parabola (focus at the d/s toe)

$$y_0=\sqrt{h^2+d^2}-d=\sqrt{400+1296}-36=5.18\ \text{m}$$ $$y=\sqrt{y_0^2+2y_0x}=\sqrt{26.86+10.37\,x}$$

At $x=d=36$ m, $y=20$ m (corrected entry point).

Step 3: Exit correction (d/s slope $\alpha=45^\circ$ > $30^\circ$)

$$a+\Delta a=\frac{y_0}{1-\cos45^\circ}=\frac{5.18}{0.2929}=17.69\ \text{m}$$

From Casagrande's table, $\Delta a/(a+\Delta a)\approx0.34$ at $45^\circ$ (0.36 at $30^\circ$, 0.32 at $60^\circ$), so

$$\Delta a=6.02\ \text{m},\qquad a=11.68\ \text{m}$$

The seepage line leaves the d/s face about 11.68 m (along the slope, about 8.26 m vertically) above the toe. The part between the parabola and this point is drawn tangent to the parabola.

Step 4: Seepage discharge

$$\begin{aligned} q &= k\,y_0 = 3\times10^{-6}\times 5.18 = 1.555 \times 10^{-5}\ \text{m}^3/\text{s per m}\\ Q &= q\times 3000 = 4.664 \times 10^{-2}\ \text{m}^3/\text{s} \end{aligned}$$

Answer: seepage line $y=\sqrt{26.86+10.37\,x}$ with $y_0=5.18$ m, leaving the d/s face at $a=11.68$ m along the slope above the toe; total seepage for 3 km $=4.664 \times 10^{-2}$ m³/s ($\approx4030$ m³/day).