Toe-drain earthen dam: discharge and phreatic line

Given: u/s slope 2:1, d/s slope 3:1 (H:V), water depth $h=50$ m, free board 5 m so dam height $=55$ m, crest width 20 m, toe drain 12 m long, $k=2.5$ cm/hr $=\dfrac{2.5}{100\times3600}=6.944 \times 10^{-6}$ m/s.

   55 m   20 m
        ________
  2:1  /        \ 3:1
 50m  /  dam     \        toe drain 12 m
 ~~~ /            \_____[#######]
 ////////////////////////////////

Step 1: Geometry

$$\begin{aligned} B &= 2(55)+20+3(55)=295\ \text{m}\\ b &= 2\times50=100\ \text{m},\quad 0.3b=30\ \text{m}\\ d &= B-0.7b-12 = 295-70-12 = 213\ \text{m} \end{aligned}$$

($d$ is measured from the corrected entry point to the u/s end of the toe drain, which is the focus of the parabola.)

Step 2: (i) Specific discharge

$$y_0=\sqrt{h^2+d^2}-d=\sqrt{50^2+213^2}-213=5.790\ \text{m}$$ $$q=k\,y_0=6.944 \times 10^{-6}\times5.790=4.021 \times 10^{-5}\ \text{m}^3/\text{s per m}\quad(\approx3.47\ \text{m}^3/\text{day per m})$$

Step 3: (ii) Co-ordinates of the phreatic line

$$y=\sqrt{y_0^2+2y_0x}=\sqrt{33.52+11.58\,x}$$

with $x$ from the focus (u/s end of the toe drain) towards the u/s side:

At $x=213$ m the ordinate is $50$ m (corrected entry). The line is joined to the u/s face at right angles at the entry, and since the line enters the toe drain ($\alpha=180^\circ$) no exit correction is needed ($\Delta a=0$); it meets the drain at about $a=y_0/2=2.89$ m beyond the focus.

Answer: (i) $q=4.021 \times 10^{-5}$ m³/s per m; (ii) $y=\sqrt{33.52+11.58\,x}$ with $y_0=5.790$ m, as tabulated.