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Top flow line, dam with horizontal filter

Short answer

Casagrande's method: start point B on water surface 0.3Δ upstream of A, focus F at filter start. Compute d (B to F), y0 = √(d²+h²)−d, base parabola x = (y²−y0²)/(2y0). Plot parabola from B; apply entry correction at A (perpendicular to slope) merging into the parabola. No exit correction.

With a neat sketch, describe the method to find the top flow line for an earthen dam with horizontal filter.

The top flow line (phreatic line) of an earth dam with a horizontal filter is found by Casagrande's method, using the fact that the line is nearly a parabola (the "base parabola") whose focus is the start of the filter.

   B      A                 crest
   .  ▽  ___________________
    ·    /·               \
     ·  / ·  base parabola \
       / entry ··           \
      / correction ··  ·     \
     /___________________F=====\_ filter
   upstream toe        |<-d->|
   |<0.3Δ>|  (B is upstream of A)

Procedure

  1. Draw the dam section to scale. Mark the reservoir water level and point A, where it meets the upstream slope.
  2. Mark point B on the water surface at a horizontal distance 0.3 Δ0.3\,\Delta upstream of A (on the reservoir side, away from the dam), where Δ\Delta is the horizontal projection of the wetted upstream slope. B is the start of the base parabola.
  3. The focus F of the parabola is the upstream end of the horizontal filter. Let dd be the horizontal distance from B to F (so dd includes the 0.3Δ0.3\Delta) and hh the height of the water surface (point B) above the base.
  4. Find the parameter y0y_0 (distance from focus to directrix):
y0=d2+h2−dy_0 = \sqrt{d^2 + h^2} - d
  1. The parabola equation (origin at F, xx positive upstream) is
x=y2−y022y0x = \frac{y^2 - y_0^2}{2y_0}

Compute points for several values of yy and plot the base parabola from B to the filter. The vertex is at a distance y0/2y_0/2 downstream of F. 6. Apply the entry correction: draw the top flow line from A, starting perpendicular to the upstream slope (an equipotential), and merge it smoothly into the parabola by hand. 7. No exit correction is needed with a horizontal filter. The base parabola itself is the top flow line there: it passes vertically above F at height y0y_0 and meets the filter at right angles at the vertex.

Discharge

q=k y0q = k\,y_0

The top flow line is the upper boundary of the unconfined flow; once it is located, the flow net is completed below it.

✓ Answer checked by Claude Opus 5.5 (high effort)

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