---
title: "Seepage discharge through dam body (k = 1×10⁻⁶ m/s)"
subject: "Hydropower Engineering"
chapter: 4
key: "semester_7/hydropower_engineering:earthen-dam-seepage-gradient-figure-2072k"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/earthen-dam-seepage-gradient-figure-2072k
site: BCT Engineering
---

# Seepage discharge through dam body (k = 1×10⁻⁶ m/s)

Hydropower Engineering · Chapter 4: Headworks of Storage Plants

- Asked in: 2072 Kartik, 8 marks
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/earthen-dam-seepage-gradient-figure-2072k
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4

## Question

The following figure shows the cross-section of an earthen dam having coefficient of permeability $1 \times 10^{-6}$ m/s. Calculate the seepage discharge through the body of the dam with the help of the phreatic line.
[Figure: earthen dam with crest width 8 m, 3 m dimension at the top (upstream free-board/water level mark), H = 30 m, downstream slope 2.5 : 1, drainage filter at the toe, horizontal dimensions 30 m and 170 m at the base.]

## Short answer

Casagrande's parabola gives y0 = 3.71 m and phreatic line y = sqrt(13.77+7.42x). Seepage discharge q = k·y0 = 3.711×10⁻⁶ m³/s per m.

## Answer

**Reading of the figure (assumptions)**: water depth $H=h=30$ m, free board 3 m so dam height $=33$ m; crest width 8 m; d/s slope 2.5:1; total base width $B=170$ m; drainage filter at the toe, so the focus of the parabola is at the d/s toe; impervious foundation; $k=1\times10^{-6}$ m/s.

#### Step 1: Geometry
$$
\begin{aligned}
\text{d/s projection} &= 2.5\times33=82.5\ \text{m}\\
\text{u/s projection} &= 170-8-82.5=79.5\ \text{m}\ \Rightarrow\ \text{u/s slope}=2.41:1\\
b &= 2.41\times30=72.3\ \text{m (wetted part)},\quad 0.3b=21.7\ \text{m}\\
d &= B-0.7b = 170-50.6=119.4\ \text{m}
\end{aligned}
$$

#### Step 2: Phreatic line (Casagrande)
$$
y_0=\sqrt{h^2+d^2}-d=\sqrt{30^2+119.4^2}-119.4=3.71\ \text{m}
$$
$$
y=\sqrt{y_0^2+2y_0x}=\sqrt{13.77+7.42\,x}
$$

| x from focus (m) | y (m) |
|---|---|
| 0.0 | 3.71 |
| 10.0 | 9.38 |
| 25.0 | 14.12 |
| 50.0 | 19.62 |
| 75.0 | 23.88 |
| 100.0 | 27.49 |
| 119.4 | 30.00 |

The line is drawn from the corrected entry point (at $x=d$, $y=30$ m), made normal to the u/s face at the start, and ends at the drainage filter at the toe.

#### Step 3: Seepage discharge
Because $dy/dx=\dfrac{y_0}{y}$ and $q=k\,y\,\dfrac{dy}{dx}$ is constant along the line,
$$
q=k\,y_0=1\times10^{-6}\times3.71=3.711 \times 10^{-6}\ \text{m}^3/\text{s per m}
$$

**Answer**: $q\approx3.711 \times 10^{-6}$ m³/s per metre length of dam ($\approx0.321$ m³/day per m).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is q = k y0 valid anywhere along the phreatic line?
   A. Because dy/dx = y0/y, so q = k y (dy/dx) = k y0 is constant
   B. Because the line is horizontal
   C. Because the gradient equals 1 everywhere
   D. Because k varies with y
   Answer: A. For the parabola y dy/dx = y0, so the flow k y dy/dx is constant along the line and equal to k y0. (Key point: Why q = k y0)

2. Where is the focus of the parabola in this dam?
   A. At the d/s toe, where the drainage filter is
   B. At the dam crest
   C. At the u/s toe
   D. At the mid-base
   Answer: A. A drainage filter at the toe makes the d/s toe the focus. (Key point: Focus location)

3. Using h = 30 m and d = 119.4 m, what is y0 and q?
   A. y0 = 3.71 m, q = 3.711 x 10^-6 m3/s per m
   B. y0 = 30 m, q = 3 x 10^-5 m3/s per m
   C. y0 = 119.4 m, q = 1.19 x 10^-4 m3/s per m
   D. y0 = 7.42 m, q = 7.4 x 10^-6 m3/s per m
   Answer: A. y0 = sqrt(900 + 14256) - 119.4 = 3.71 m, then q = 1e-6 x 3.71. (Key point: Computing y0)

## Related in Hydropower Engineering

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