Earth dam flow net and discharge

Given

Top width 15 m; slopes 2H:1V both sides; height 30 m; free board 5 m, so reservoir depth $h = 25$ m; horizontal toe drain 30 m long; $k = 40$ m/day.

Geometry (horizontal drain, Casagrande)

• Base width: $2(2\times30) + 15 = 135$ m.
• The wetted upstream slope has horizontal projection $\Delta = 2\times25 = 50$ m, so A (where the water surface meets the slope) is 50 m from the upstream toe.
• The base parabola starts at B, $0.3\Delta = 15$ m upstream of A, i.e. $50 - 15 = 35$ m from the upstream toe.
• The focus F (start of the drain) is at $135 - 30 = 105$ m from the upstream toe.
• Horizontal distance B to F: $d = 105 - 35 = 70$ m.

Theoretical (Casagrande) discharge

$$y_0 = \sqrt{d^2 + h^2} - d = \sqrt{70^2 + 25^2} - 70 = 74.33 - 70 = 4.33\ \text{m}$$ $$q = k\,y_0 = 40 \times 4.33 = 173.2\ \text{m}^3/\text{day per m}$$

Flow net

Base parabola ($x$ from F towards the upstream side): $x = \dfrac{y^2 - y_0^2}{2y_0}$

The vertex (where the line meets the drain) is $y_0/2 = 2.17$ m downstream of F.

  B    A            crest 15 m
  .  ▽\_____________
   ·   \·            \
       / ··  parabola  \  2:1
      / entry  ··.      \
     /________________F====\__ drain 30 m
   upstream toe

Plot these points, correct the entry from A (perpendicular to the upstream slope), and end the top flow line on the drain at right angles. Draw the flow lines between the top flow line and the base, and equipotentials at equal head drops so that the fields are curvilinear squares. A typical sketch gives $N_f = 3$ flow channels and $N_d \approx 17$ drops; the head drop per field is then $25/17 = 1.47$ m.

$$q = k\,h\,\frac{N_f}{N_d} = 40 \times 25 \times \frac{3}{17} \approx 176\ \text{m}^3/\text{day per m}$$

Comparison

The flow net value depends on how accurately the fields are drawn, but with $N_f/N_d \approx 0.17$ to $0.18$ it agrees with the theoretical value within a few percent.