---
title: "Seepage through dam with central core (33 m dam)"
subject: "Hydropower Engineering"
chapter: 4
key: "semester_7/hydropower_engineering:earthen-dam-impervious-core-seepage-2070c"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/earthen-dam-impervious-core-seepage-2070c
site: BCT Engineering
---

# Seepage through dam with central core (33 m dam)

Hydropower Engineering · Chapter 4: Headworks of Storage Plants

- Asked in: 2070 Chaitra, 5 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-2070c
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4

## Question

Determine the seepage discharge for the earthen dam having 33 m total height with 3 m width impervious central core. Take top width of the dam as 7 m and freeboard 3 m. The coefficient of permeability of the dam material is $4 \times 10^{-6}$ m/sec and that of the impervious core is $4 \times 10^{-8}$ m/sec. The upstream and downstream slopes of the dam are 3:1 and 2.5:1 respectively.

## Short answer

Flow continuity gives phreatic height h = 14.06 m at the core exit. Seepage discharge q = 4.681×10⁻⁶ m³/s per m.

## Answer

**Data and assumptions**: dam height 33 m, free board 3 m so reservoir depth $H=30$ m; top width 7 m; core width $t=3$ m with $k_c=4\times10^{-8}$ m/s; shells $k=4\times10^{-6}$ m/s; slopes 3:1 (u/s) and 2.5:1 (d/s); impervious foundation, no tail water. The u/s shell is much more pervious than the core, so the head on the core face is the full $H=30$ m.

#### Seepage length in the d/s shell
The d/s face has a horizontal projection of $2.5\times33=82.5$ m; half of the 4 m by which the crest exceeds the core width lies d/s of the core, so
$$
L=82.5+\frac{7-3}{2}=84.5\ \text{m}
$$

#### Continuity between core and d/s shell
Core: $q=\dfrac{k_c(H^2-h^2)}{2t}$; shell: $q=\dfrac{k\,h^2}{2L}$, where $h$ is the phreatic height at the d/s face of the core.

$$
\begin{aligned}
\frac{4\times10^{-8}(900-h^2)}{6}&=\frac{4\times10^{-6}h^2}{2\times 84.5}\\
6.667\times10^{-9}(900-h^2)&=2.3669\times10^{-8}\,h^2\\
h^2&=197.8\ \text{m}^2,\quad h=14.06\ \text{m}
\end{aligned}
$$

#### Discharge
$$
q=\frac{k\,h^2}{2L}=\frac{4\times10^{-6}\times197.8}{2\times84.5}=4.681 \times 10^{-6}\ \text{m}^3/\text{s per m}
$$
Check using the core: $\dfrac{4\times10^{-8}(900-197.8)}{6}=4.681 \times 10^{-6}$ m³/s per m.

**Answer**: seepage discharge $=4.681 \times 10^{-6}$ m³/s per metre length of dam ($\approx0.404$ m³/day per m).

## Check yourself

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

1. With dam height 33 m and free board 3 m, what head H is applied on the core?
   A. 30 m
   B. 33 m
   C. 36 m
   D. 3 m
   Answer: A. Reservoir depth = dam height minus free board = 30 m. (Key point: Reservoir depth)

2. How is the d/s shell seepage length obtained?
   A. L = 2.5 x 33 + (7-3)/2 = 84.5 m
   B. L = 3 x 33 = 99 m
   C. L = 2.5 x 30 = 75 m
   D. L = 7 + 82.5 = 89.5 m
   Answer: A. The d/s face projects 82.5 m, plus half the crest excess over the core width (2 m) lying d/s of the core. (Key point: Seepage length L)

3. Which relation determines the phreatic height h at the core's d/s face?
   A. k_c(H^2-h^2)/(2t) = k h^2/(2L)
   B. k_c H = k h
   C. k_c h^2/(2t) = k H^2/(2L)
   D. H = h + t
   Answer: A. Continuity of flow between core and shell gives h^2 = 197.8 and q = 4.681e-6 m3/s per m. (Key point: Equating flows)

## Related in Hydropower Engineering

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- [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
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## Related in other subjects

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