---
title: "Seepage line for 22 m high earthen dam"
subject: "Hydropower Engineering"
chapter: 4
key: "semester_7/hydropower_engineering:earthen-dam-seepage-line-22m-45deg"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/earthen-dam-seepage-line-22m-45deg
site: BCT Engineering
---

# Seepage line for 22 m high earthen dam

Hydropower Engineering · Chapter 4: Headworks of Storage Plants

- Asked in: 2080 Bhadra, 10 marks
- Video: [[HYDROPOWER #3] Earthen Gravity Dam | Phreatic Line Equation| Numerical Problem | Failure of Dam](https://www.youtube.com/watch?v=TN1xkNzh7fo) — Just Learn Civil
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/earthen-dam-seepage-line-22m-45deg
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4

## Question

Determine the seepage line for a homogeneous earthen dam of height 22 m and top width 6 m retaining 20 m depth of water in the reservoir. The slope of the upstream and downstream faces of the dam is 45°. Also determine the seepage discharge if the length of the dam is 3 km and the value of the coefficient of permeability of the dam material is $3 \times 10^{-3}$ mm/s.

## Short answer

Casagrande's method yields y0 = 5.18 m, phreatic line y = sqrt(26.86+10.37x), and exit point a = 11.68 m along the downstream face. Specific discharge q = k·y0 = 1.555×10⁻⁵ m³/s per m; total for 3 km is 4.664×10⁻² m³/s.

## Answer

**Given**: dam height 22 m, top width 6 m, water depth $h=20$ m, both slopes $45^\circ$ (1:1), length 3 km, $k=3\times10^{-3}$ mm/s $=3\times10^{-6}$ m/s. Assume impervious foundation, no drain and no tail water; the line is drawn by Casagrande's method.

```text
  22 m _6 m_
      /|     |\
 45° / |     | \ 45°
    /  | 20m |  \
   ~~~~|     |   \
  /////////////////
  |<-- 50 m ------>|
```

#### Step 1: Geometry
$$
\begin{aligned}
B &= 22+6+22=50\ \text{m}\\
b &= 20\ \text{m (1:1 slope)},\quad 0.3b=6\ \text{m}\\
d &= B-0.7b = 50-14 = 36\ \text{m}
\end{aligned}
$$

#### Step 2: Basic parabola (focus at the d/s toe)
$$
y_0=\sqrt{h^2+d^2}-d=\sqrt{400+1296}-36=5.18\ \text{m}
$$
$$
y=\sqrt{y_0^2+2y_0x}=\sqrt{26.86+10.37\,x}
$$

| x from focus (m) | y (m) |
|---|---|
| 0.0 | 5.18 |
| 2.0 | 6.90 |
| 5.0 | 8.87 |
| 10.0 | 11.42 |
| 15.0 | 13.50 |
| 20.0 | 15.30 |
| 25.0 | 16.91 |
| 30.0 | 18.38 |
| 36.0 | 20.00 |

At $x=d=36$ m, $y=20$ m (corrected entry point).

#### Step 3: Exit correction (d/s slope $\alpha=45^\circ$ > $30^\circ$)
$$
a+\Delta a=\frac{y_0}{1-\cos45^\circ}=\frac{5.18}{0.2929}=17.69\ \text{m}
$$
From Casagrande's table, $\Delta a/(a+\Delta a)\approx0.34$ at $45^\circ$ (0.36 at $30^\circ$, 0.32 at $60^\circ$), so
$$
\Delta a=6.02\ \text{m},\qquad a=11.68\ \text{m}
$$
The seepage line leaves the d/s face about 11.68 m (along the slope, about 8.26 m vertically) above the toe. The part between the parabola and this point is drawn tangent to the parabola.

#### Step 4: Seepage discharge
$$
\begin{aligned}
q &= k\,y_0 = 3\times10^{-6}\times 5.18 = 1.555 \times 10^{-5}\ \text{m}^3/\text{s per m}\\
Q &= q\times 3000 = 4.664 \times 10^{-2}\ \text{m}^3/\text{s}
\end{aligned}
$$

**Answer**: seepage line $y=\sqrt{26.86+10.37\,x}$ with $y_0=5.18$ m, leaving the d/s face at $a=11.68$ m along the slope above the toe; total seepage for 3 km $=4.664 \times 10^{-2}$ m³/s ($\approx4030$ m³/day).

## Check yourself

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

1. For the 22 m dam with 1:1 slopes and 20 m water depth, how is d computed in Casagrande's method?
   A. d = B - 0.7b = 50 - 14 = 36 m
   B. d = B + 0.3b = 50 + 6 = 56 m
   C. d = B - b = 50 - 20 = 30 m
   D. d = B/2 = 25 m
   Answer: A. The entry point is corrected by 0.3b from the water edge, so the horizontal distance to the focus is B - 0.7b = 36 m. (Key point: Corrected entry point)

2. Which formula gives y0 for the basic parabola with focus at the d/s toe?
   A. y0 = sqrt(h^2 + d^2) - d
   B. y0 = sqrt(h^2 - d^2) + d
   C. y0 = (h + d)/2
   D. y0 = h^2/d
   Answer: A. Casagrande's basic parabola has y0 = sqrt(h^2 + d^2) - d, giving 5.18 m here. (Key point: Parabola constant y0)

3. Why is an exit correction (Delta a) applied for this dam?
   A. The d/s slope of 45 degrees is greater than 30 degrees, so the parabola must be corrected near the exit
   B. The foundation is pervious
   C. The tail water is high
   D. The upstream slope is flatter than 45 degrees
   Answer: A. For d/s slopes steeper than 30 degrees the line leaves the face above the parabola's intersection, so Casagrande's correction is used; a = 11.68 m along the slope. (Key point: Exit correction)

4. How is the seepage discharge per metre obtained, and what is its value here?
   A. q = k y0 = 1.555 x 10^-5 m3/s per m, then multiplied by 3000 m
   B. q = k h = 6 x 10^-5 m3/s per m
   C. q = k d = 1.08 x 10^-4 m3/s per m
   D. q = k y0 / L, divided by 3000 m
   Answer: A. From the parabola q = k y0; multiplying by the 3 km length gives about 4.66 x 10^-2 m3/s. (Key point: Seepage discharge)

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