---
title: "Seepage through dam with central core (45 m dam)"
subject: "Hydropower Engineering"
chapter: 4
key: "semester_7/hydropower_engineering:earthen-dam-impervious-core-seepage-2075c"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/earthen-dam-impervious-core-seepage-2075c
site: BCT Engineering
---

# Seepage through dam with central core (45 m dam)

Hydropower Engineering · Chapter 4: Headworks of Storage Plants

- Asked in: 2075 Chaitra, 4 marks
- Video: [Embankment Dams | Geotechnical Engineering Aspects](https://www.youtube.com/watch?v=G9fOUy6y08c) — GeoTechs With Clemency
- 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-impervious-core-seepage-2075c
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4

## Question

Find the seepage discharge through the homogeneous earthen embankment dam with 3 m width of central impervious core as shown in the figure. Given: (i) height of the dam = 45 m with free board as 3 m; (ii) upstream water level = 42 m, top width of the dam = 8 m; (iii) U/S and D/S side slope of the dam = 1V:3H; (iv) coefficient of permeability of the dam material = $4 \times 10^{-6}$ m/s and that of the impervious core = $5 \times 10^{-8}$ m/s.
[Figure: earthen dam with central impervious core 3 m wide, total base length L = 126 m, top width 8 m, 42 m water depth, 3 m free board, slopes 1V:3H.]

## Short answer

Equating flow through the core and downstream shell yields phreatic height h = 24.64 m at the core's downstream face. Seepage discharge q = 9.639×10⁻⁶ m³/s per m.

## Answer

**Assumptions** (from the figure data): the central core is 3 m wide with $k_c=5\times10^{-8}$ m/s; the shells have $k=4\times10^{-6}$ m/s; foundation impervious; no tail water. The u/s shell is far more pervious than the core, so the full reservoir head $H=42$ m acts on the u/s face of the core. The d/s shell carries the flow from the core to the toe over a horizontal length $L=126$ m (as given in the figure). Let $h$ be the height of the phreatic line at the d/s face of the core.

```text
        8 m
      _______
  1V:3H/|core|\ 1V:3H
 42m  / |3 m |  \
 ~~~ /  |    |   \
 ////////////////////
 |        L = 126 m |  (d/s shell)
```

#### Flow through the core (Dupuit)
$$
q=\frac{k_c\,(H^2-h^2)}{2t}
$$

#### Flow through the d/s shell (toe at ground level)
$$
q=\frac{k\,h^2}{2L}
$$

#### Equate (continuity)
$$
\begin{aligned}
\frac{5\times10^{-8}\,(42^2-h^2)}{2\times3}&=\frac{4\times10^{-6}\,h^2}{2\times126}\\
8.333\times10^{-9}(1764-h^2) &= 1.5873\times10^{-8}\,h^2\\
h^2 &= 607.3\ \text{m}^2 \Rightarrow h=24.64\ \text{m}
\end{aligned}
$$

#### Discharge
$$
q=\frac{k\,h^2}{2L}=\frac{4\times10^{-6}\times607.3}{2\times126}=9.639 \times 10^{-6}\ \text{m}^3/\text{s per m}
$$
Check with the core: $q=\dfrac{5\times10^{-8}(1764-607.3)}{6}=9.639 \times 10^{-6}$ m³/s per m.

**Answer**: seepage discharge $\approx9.639 \times 10^{-6}$ m³/s per metre run of dam ($\approx0.833$ m³/day per m); the phreatic line drops sharply across the core and stands $24.64$ m high at its d/s face.

## Check yourself

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

1. Why is the full head H = 42 m taken on the u/s face of the core?
   A. The u/s shell is far more pervious than the core, so it loses little head
   B. The core is thicker than the shell
   C. The shell is impervious
   D. Water level is 42 m only at the d/s
   Answer: A. The shell (k = 4e-6) is much more permeable than the core (5e-8), so nearly all head is lost across the core. (Key point: Head on the core)

2. How is h at the d/s face of the core found?
   A. Equate core flow k_c(H^2-h^2)/(2t) with shell flow k h^2/(2L)
   B. Set h equal to H/2
   C. Set the two permeabilities equal
   D. Use h = free board
   Answer: A. Flow through the core equals flow through the d/s shell, which gives h^2 = 607.3 m^2. (Key point: Continuity method)

3. What is the resulting seepage discharge?
   A. About 9.639 x 10^-6 m3/s per m
   B. About 9.639 x 10^-3 m3/s per m
   C. About 4 x 10^-6 x 42 = 1.68 x 10^-4 m3/s per m
   D. About 5 x 10^-8 m3/s per m
   Answer: A. q = k h^2/(2L) = 4e-6 x 607.3/252 = 9.639e-6 m3/s per m, matching the core check. (Key point: Final discharge)

## Related in Hydropower Engineering

- [Seepage through dam with central core (33 m dam)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aearthen-dam-impervious-core-seepage-2070c) — Hydropower Engineering, chapter 4
- [Seepage discharge through dam body (k = 1×10⁻⁶ m/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aearthen-dam-seepage-gradient-figure-2072k) — Hydropower Engineering, chapter 4
- [Seepage line for 22 m high earthen dam](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aearthen-dam-seepage-line-22m-45deg) — Hydropower Engineering, chapter 4
- [Toe-drain earthen dam: discharge and phreatic line](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aearthen-dam-toe-drain-2069) — Hydropower Engineering, chapter 4
- [Design criteria of earthen embankment dam](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aearthen-dam-design-criteria) — Hydropower Engineering, chapter 4
- [Seepage control measures in embankment dam](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aembankment-seepage-control-measures) — Hydropower Engineering, chapter 4

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