---
title: "Impervious dam seepage discharge"
subject: "Soil Mechanics"
chapter: 7
key: "semester_4/soil_mechanics:impervious-dam-seepage-discharge-numerical"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/impervious-dam-seepage-discharge-numerical
site: BCT Engineering
---

# Impervious dam seepage discharge

Soil Mechanics · Chapter 7: Seepage through Soils

- Asked in: 2073 Bhadra, 4 marks
- Video: [Seepage though Earthen Dam || Soil Mechanics | Chapter 7 | Most important theory](https://www.youtube.com/watch?v=wmVYDaAoP7s) — Sudip Dhakal
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/impervious-dam-seepage-discharge-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-7

## Question

If the upstream and downstream heads of an impervious dam are 8 m and 1 m respectively, then find the seepage discharge when seepage of water takes place from upstream to downstream via the isotropic soil lying below the impervious dam. The total number of flow channels and equipotential drops are 9 and 12, respectively. Also, take the coefficient of permeability of the soil layer, k = $3 \times 10^{-4}$ cm/s.

## Short answer

Head difference H = 7 m. Seepage q = k H (Nf/Nd) = 3×10⁻⁶ m/s × 7 m × (9/12) = 1.575×10⁻⁵ m³/s per metre length (≈1.36 m³/day per m).

## Answer

#### Given

Upstream head 8 m, downstream head 1 m, so net head $H = 8 - 1 = 7$ m. $N_f = 9$, $N_d = 12$. $k = 3\times10^{-4}$ cm/s $= 3\times10^{-6}$ m/s.

#### Calculation

$$
q = k\,H\,\frac{N_f}{N_d} = 3\times10^{-6} \times 7 \times \frac{9}{12}
$$

$$
q = 1.575\times10^{-5}\ \text{m}^3/\text{s per metre length of dam}
$$

$$
q = 1.575\times10^{-5} \times 86400 = 1.36\ \text{m}^3/\text{day per metre}
$$

**Answer: $q = 1.58\times10^{-5}$ m³/s per m ($\approx 1.36$ m³/day per m).**

## Check yourself

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

1. With heads of 8 m and 1 m, H is:
   A. 9 m
   B. 8 m
   C. 7 m
   D. 1 m
   Answer: C. H = 8 - 1 = 7 m. (Key point: Net head)

2. 3 x 10^-4 cm/s equals:
   A. 3 x 10^-6 m/s
   B. 3 x 10^-2 m/s
   C. 3 x 10^-4 m/s
   D. 3 x 10^-8 m/s
   Answer: A. Divide by 100. (Key point: Unit conversion)

3. Which formula is applied?
   A. q = k H N_f/N_d
   B. q = k H N_d/N_f
   C. q = k H (N_f + N_d)
   D. q = k N_f N_d
   Answer: A. Standard flow net discharge. (Key point: Formula)

4. What is q?
   A. 1.575 x 10^-5 m3/s per m
   B. 2.1 x 10^-5 m3/s per m
   C. 1.575 x 10^-3 m3/s per m
   D. 2.8 x 10^-6 m3/s per m
   Answer: A. 3e-6 x 7 x 9/12 = 1.575e-5. (Key point: Result)

## Related in Soil Mechanics

- [Inclined specimen, total heads, gradient](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Ainclined-specimen-total-head-gradient) — Soil Mechanics, chapter 6
- [Earth dam flow net and discharge](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Aearth-dam-flow-net-compare-discharge) — Soil Mechanics, chapter 7
- [Earth dam with filter, seepage quantity](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Aearth-dam-horizontal-filter-seepage) — Soil Mechanics, chapter 7
- [Flow net discharge, 200 cc/day](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Aflow-net-discharge-numerical-200cc) — Soil Mechanics, chapter 7
- [Why a filter on dam downstream](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Afilter-downstream-earth-dam) — Soil Mechanics, chapter 7
- [Discharge velocity vs seepage velocity](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Adischarge-vs-seepage-velocity) — Soil Mechanics, chapter 6, asked 2×

## Related in other subjects

- [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 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
- [Jump upstream of a low dam](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ajump-low-dam-uniform-depth-0-5m) — Hydraulics, chapter 9
- [Seepage through dam with central core (45 m dam)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aearthen-dam-impervious-core-seepage-2075c) — Hydropower Engineering, chapter 4