Question

A homogeneous earthen embankment has an upstream water level of \(12\) m. The top width of the embankment is \(3\) m, and its upstream and downstream slopes (horizontal:vertical) are \(2:1\) and \(2.5:1\), respectively. A freeboard of \(2\) m is provided. The coefficient of permeability of the fill material is \(2.5 \times 10^{-5}\) m/s. If a \(20\) m long horizontal filter is placed inward from the downstream toe, the seepage discharge per unit length of the embankment is \(n \times 10^{-5}\) m\(^3\)/s per m. The value of \(n\) is ________. (Rounded off to two decimal places)

Hint:

Use the base parabola method with the corrected entry point for seepage into a horizontal filter.

Answer 1

Correct Answer: 5.92

Step 1: Work out the dam geometry.
Total dam height \(H\) = water level + freeboard \(= 12 + 2 = 14\) m.
On the \(2:1\) upstream slope, the water surface meets the face at a horizontal distance \(2 \times 12 = 24\) m from the upstream heel; call this point \(A\).
The full upstream horizontal run (for height \(14\) m) is \(2 \times 14 = 28\) m, and the full downstream run (slope \(2.5:1\)) is \(2.5 \times 14 = 35\) m.
Base width \(= 28 + 3 + 35 = 66\) m, so the filter, being \(20\) m long from the toe, has its inner edge at \(66 - 20 = 46\) m from the heel.

Step 2: Shift the starting point of the seepage line.
The actual phreatic line curves near the water surface, so the equivalent base parabola is taken to start not at \(A\) but at a point \(A'\), moved back by \(0.3\) times the distance \(GA = 24\) m, i.e. by \(0.3 \times 24 = 7.2\) m.
So \(A'\) sits at \(24 - 7.2 = 16.8\) m from the heel, still at the full head \(h = 12\) m.

Step 3: Get the horizontal distance to the filter.
\(L = 46 - 16.8 = 29.2\) m, measured from \(A'\) to the point where the filter begins (the focus of the base parabola).

Step 4: Apply the base-parabola discharge formula for a horizontal filter.
For seepage draining into a horizontal filter, the discharge per unit length of dam is \[ q = k\left(\sqrt{L^2+h^2}-L\right) \] \[ q = 2.5\times10^{-5}\left(\sqrt{29.2^2+12^2}-29.2\right) = 2.5\times10^{-5}\left(\sqrt{996.64}-29.2\right) \] \[ q = 2.5\times10^{-5}\left(31.57-29.2\right) = 2.5\times10^{-5}\times 2.37 = 5.92\times10^{-5}\ \text{m}^3/\text{s per m} \]
Final Answer:
The seepage discharge works out close to \(5.92\times10^{-5}\) m\(^3\)/s per m of embankment. \[ \boxed{n \approx 5.92} \]