---
title: "Steady 2D model: seepage under a dam"
subject: "Computational Techniques in Civil Engineering"
chapter: 7
key: "semester_8/computational_techniques_in_civil_engineering:seepage-under-dam-2d-model"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/seepage-under-dam-2d-model
site: BCT Engineering
---

# Steady 2D model: seepage under a dam

Computational Techniques in Civil Engineering · Chapter 7: Simulation of Ground water flow

- Asked in: 2079 Shrawan, 8 marks; 2075 Bhadra, 5+3 marks; 2074 Bhadra, 6 marks (asked 3 times)
- Video: [Seepage Analysis & Slope Stability Analysis of Earthen Dam](https://www.youtube.com/watch?v=TXE9RN1oz74) — CIVIL Knowledge Centre
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/seepage-under-dam-2d-model
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-7

## Question

Develop a steady state 2D model (suitable finite difference expression for homogeneous and isotropic aquifer) for the simulation of seepage under a dam. Also describe the iterative procedure for computing potential at each grid and seepage rate.

## Short answer

Laplace equation governs steady seepage. Interior node: φ_{i,j}=¼(φ_{i+1,j}+φ_{i-1,j}+φ_{i,j+1}+φ_{i,j-1}). Fixed heads on upstream/downstream beds; no-flow boundaries use image points (φ_{i,j-1}=φ_{i,j+1}). Gauss–Seidel iteration solves potentials. Seepage per unit width: Q=K∑(φ_{i,j}−φ_{i+1,j}) across a vertical section.

## Answer

#### Physical situation
```text
   h1 (upstream)      dam        h2 (downstream)
 ~~~~~~~~~~~~~~~~ [=========] ~~~~~~~~~~~~~~
 phi=h1  . . . . . no flow . . . . phi=h2
 . . . . . . . . . . . . . . . . . . . . .
 . . . . . . . . . . . . . . . . . . . . .
 ///////// impervious layer (no flow) /////
```
Water seeps under the dam through the foundation soil from the upstream bed (head $h_1$) to the downstream bed (head $h_2$). The flow is steady and two-dimensional in a vertical section.

#### Governing equation
Continuity for steady flow, $\dfrac{\partial q_x}{\partial x}+\dfrac{\partial q_y}{\partial y}=0$, with Darcy's law $q_x=-K_x\dfrac{\partial\phi}{\partial x}$, $q_y=-K_y\dfrac{\partial\phi}{\partial y}$ ($\phi$ = potential or hydraulic head). For a homogeneous, isotropic aquifer ($K_x=K_y=K$):
$$
\frac{\partial^2\phi}{\partial x^2}+\frac{\partial^2\phi}{\partial y^2}=0\qquad\text{(Laplace equation)}
$$

#### Finite difference expression
Cover the section by a square grid ($\Delta x=\Delta y$). Central differences from Taylor's series:
$$
\frac{\partial^2\phi}{\partial x^2}\approx\frac{\phi_{i+1,j}-2\phi_{i,j}+\phi_{i-1,j}}{\Delta x^2},\qquad
\frac{\partial^2\phi}{\partial y^2}\approx\frac{\phi_{i,j+1}-2\phi_{i,j}+\phi_{i,j-1}}{\Delta y^2}
$$
Adding and setting $\Delta x=\Delta y$:
$$
\boxed{\ \phi_{i,j}=\frac14\left(\phi_{i+1,j}+\phi_{i-1,j}+\phi_{i,j+1}+\phi_{i,j-1}\right)\ }
$$
The potential at each grid point is the average of its four neighbours.

#### Boundary conditions
- Upstream bed: $\phi=h_1$; downstream bed: $\phi=h_2$ (specified head).
- Base of the dam and the impervious layer: no flow, $\partial\phi/\partial n=0$. Use an image (ghost) point: for a point on a horizontal no-flow boundary, $\phi_{i,j-1}=\phi_{i,j+1}$ so
$$
\phi_{i,j}=\frac14\left(\phi_{i+1,j}+\phi_{i-1,j}+2\phi_{i,j+1}\right)
$$
- Far left and right ends: assumed no-flow (or fixed heads equal to $h_1$, $h_2$), placed far enough from the dam.

#### Iterative procedure (Gauss-Seidel / Liebmann)
1. Set up the grid and number the nodes; give the boundary nodes their values.
2. **Guess** the potential at all interior nodes (for example, a linear variation from $h_1$ to $h_2$).
3. **Sweep** the grid row by row. At each interior node compute $\phi_{i,j}^{new}$ with the equation above, using the latest available neighbour values. Optionally over-relax: $\phi^{new}=\phi^{old}+\omega(\phi^{GS}-\phi^{old})$, $1<\omega<2$.
4. Find the largest change $\max|\phi^{new}-\phi^{old}|$ over the grid.
5. If it is greater than the tolerance $\varepsilon$ (e.g. 0.001 m), repeat from step 3; otherwise stop.
6. **Seepage rate.** Between two adjacent nodes the flow through one cell face (per unit length of dam) is
$$
q=K\,\frac{\phi_a-\phi_b}{\Delta x}\,\Delta y=K(\phi_a-\phi_b)\quad(\Delta x=\Delta y)
$$
   Choose a vertical section between two columns $i$ and $i+1$ below the dam; the total seepage is the sum over all rows:
$$
Q=K\sum_{j}\left(\phi_{i,j}-\phi_{i+1,j}\right)
$$
   (check with a different section; the totals must agree if the solution has converged, and also equal the flow entering at the upstream bed).
7. Print the potentials (to draw equipotential lines and the flow net) and the seepage $Q$ in m$^3$/s per metre length of dam.

## Check yourself

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

1. For steady 2D seepage in a homogeneous isotropic aquifer, what is the governing equation?
   A. d(phi)/dt = K d2(phi)/dx2
   B. d2(phi)/dx2 + d2(phi)/dy2 = 0 (Laplace equation)
   C. d2(phi)/dx2 = S d(phi)/dt
   D. d(phi)/dx + d(phi)/dy = 0
   Answer: B. Continuity with Darcy's law and Kx = Ky gives the Laplace equation. (Key point: Governing equation)

2. With dx = dy, the finite difference form gives phi(i,j) as what?
   A. Sum of the four neighbours
   B. Average of the four neighbouring potentials
   C. Average of the two x-neighbours only
   D. Difference of opposite neighbours
   Answer: B. phi(i,j) = (1/4)(phi(i+1,j)+phi(i-1,j)+phi(i,j+1)+phi(i,j-1)). (Key point: Finite difference expression)

3. How is a horizontal no-flow boundary node handled?
   A. Set phi = 0
   B. Set phi = h1
   C. Ignore the node
   D. Use an image point phi(i,j-1) = phi(i,j+1), so phi = (1/4)(phi(i+1,j)+phi(i-1,j)+2phi(i,j+1))
   Answer: D. Zero gradient across the boundary makes the ghost value equal to the opposite neighbour, doubling that term. (Key point: No-flow boundary)

4. How is the total seepage computed once the iteration converges?
   A. Q = K sum over rows of (phi(i,j) - phi(i+1,j)) across a vertical section
   B. Q = K times the largest phi
   C. Q = (h1 + h2)/2
   D. Q = K times the tolerance
   Answer: A. Each cell face carries K(phi_a - phi_b) when dx = dy; summing across a section gives Q per metre of dam. (Key point: Seepage rate)

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