Phreatic line and discharge (k = 5.1×10⁻⁴ cm/s)

Given data and assumptions: homogeneous dam without any drain; foundation impervious; tail water ignored. Casagrande's method (basic parabola with focus at the d/s toe) is used.

• Dam height $= 290-235 = 55$ m; water head $h = 285-235 = 50$ m; free board $= 5$ m.
• $k = 5.1\times10^{-4}$ cm/s $= 5.1\times10^{-6}$ m/s.

 290 ______5 m______
     /|              \
  3:1/ |   homogeneous \ 2:1
    /  |    dam        \
 285 ~~|                 \
    /  |                  \
 235 ///////////////////////////
     |<--------- 280 m ------->|

Step 1: Geometry

$$\begin{aligned} \text{Base width } B &= 3(55)+5+2(55) = 280\ \text{m} \\ \text{Wetted horizontal projection } b &= 3\times 50 = 150\ \text{m} \\ \text{Entry correction } 0.3b &= 45.0\ \text{m} \\ d &= B-0.7b = 280-105.0 = 175.0\ \text{m} \end{aligned}$$

Here $d$ is the horizontal distance from the corrected entry point to the d/s toe (the focus $F$).

Step 2: Basic parabola

$$y_0=\sqrt{h^2+d^2}-d=\sqrt{50^2+175.0^2}-175.0=7.00\ \text{m}$$ $$y=\sqrt{y_0^2+2y_0x}$$

with $x$ measured from the focus (d/s toe) towards the u/s side:

At $x=d=175.0$ m the parabola reaches $y=h=50$ m, which is the corrected entry point. The u/s end of the actual line is drawn to meet the u/s face at right angles.

Step 3: Exit correction (d/s face)

The d/s slope angle is $\alpha=\tan^{-1}(1/2)=26.57^\circ$. The parabola meets the d/s face at a distance

$$a+\Delta a=\frac{y_0}{1-\cos\alpha}=\frac{7.00}{1-\cos 26.57^\circ}=66.3\ \text{m}$$

(measured along the slope from the toe). Casagrande's chart covers $30^\circ\le\alpha\le180^\circ$, so for $\alpha<30^\circ$ the exit point is found from Schaffernak's formula:

$$a=\frac{d}{\cos\alpha}-\sqrt{\frac{d^2}{\cos^2\alpha}-\frac{h^2}{\sin^2\alpha}}=195.66-\sqrt{38,281-12,500}=35.1\ \text{m}$$

So the line leaves the d/s face about 35.1 m along the slope from the toe (about 15.7 m above the base). (Extrapolating Casagrande's factor 0.36 would give about 42.5 m.) The last part of the line is drawn tangent to the parabola and blended into this exit point.

Step 4: Discharge

$$q=k\,y_0=5.1\times10^{-6}\times 7.00=3.571 \times 10^{-5}\ \text{m}^3/\text{s per m}$$

(Schaffernak's $q=k\,a\sin\alpha\tan\alpha=5.1\times10^{-6}\times35.1\times0.447\times0.5=4.0\times10^{-5}$ m³/s per m gives a similar order of value.)

Answer: the phreatic line is the parabola $y=\sqrt{49.04+14.01\,x}$ (x from the d/s toe, with entry and exit corrections), $y_0=7.00$ m, exit point about 35 m up the d/s face; seepage $q=3.571 \times 10^{-5}\ \text{m}^3/\text{s}$ per metre length of dam ($\approx 3.09$ m³/day per m).