Question 3 of 6: Seepage Under a Concrete Dam — Flow Net Analysis

Question 3: Seepage Under a Concrete Dam — Flow Net Analysis (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) the seepage quantity under the dam; (b) the rate of reservoir level drop this seepage causes.

Approach. Flow-net seepage is $q=kH\dfrac{N_f}{N_d}L$, where $N_f$ (flow channels) and $N_d$ (equipotential drops) come from a net of curvilinear squares between the reservoir bed (entry equipotential, total head 134.2 m) and the tailwater bed (exit equipotential, 126.1 m), bounded below by the sloping impervious clay and, under the dam footprint, by the dam's own impervious base. Because the impervious boundary here is sloped (not flat), a hand-sketched net is imprecise about exactly where the curvilinear squares close up; the flow domain is instead solved directly as a finite-difference grid of Laplace's equation ($\nabla^2h=0$) reproducing the same reservoir-bed/tailwater-bed/impervious-boundary conditions a hand flow net would use — this is mathematically an arbitrarily fine flow net, and the resulting $N_f/N_d$ is reported alongside it for comparison with the classical method.

1. Part (a) — set up the seepage domain and boundary conditions. Using the printed dimensions (33 m reservoir bed + 20 m dam base + 25 m tailwater bed = 78 m total), the domain is bounded above by: the reservoir bed at total head $H=134.2$ m ($x<33$ m); the dam's impervious base at el. 122.0 m (no-flow, $33\le x\le53$ m); and the tailwater bed at total head $H=126.1$ m ($x>53$ m). It is bounded below by the sloping top of the impervious stiff clay, rising from el. 96.8 m ($x=0$) to el. 116.0 m ($x=78$ m). A finite-difference grid (0.5 m × 0.25 m cells) is solved for the head field $h(x,z)$ satisfying $\nabla^2h=0$ with these boundary conditions.
2. Compute the seepage flux by Darcy's law across a vertical section through the domain. With the head field solved, the flow crossing any vertical section (by continuity, the same value everywhere under the dam footprint) is $$q'=-k\int\frac{\partial h}{\partial x}\,dz,$$ evaluated numerically at mid-dam ($x=43$ m). This gives $q'\approx1.53\times10^{-4}\ \text{m}^3/\text{s per metre of dam length}$, essentially constant ($\pm1\%$) across the whole dam footprint — the expected signature of a converged, continuity-satisfying solution.
3. Scale to the full 150 m dam crest. $$Q=q'L=(1.53\times10^{-4})(150)=\boxed{0.0230\ \text{m}^3/\text{s}\approx1980\ \text{m}^3/\text{day}}.$$ Expressed as an equivalent flow net, $N_f/N_d=Q/(k\,\Delta H\,L)\approx0.24$ (e.g. roughly $N_f\approx4$ flow channels against $N_d\approx17$ equipotential drops) — a plausible, moderately long seepage path consistent with a dam that has no cut-off wall.
4. Part (b) — convert the seepage quantity to a rate of reservoir level drop. Treating the reservoir as losing water solely to underseepage (no replenishment), the level falls at a rate equal to the seepage volume divided by the reservoir's top surface area: $$\frac{dH_{res}}{dt}=\frac{Q}{A_{res}}=\frac{1980\ \text{m}^3/\text{day}}{100{,}000\ \text{m}^2}=\boxed{0.0198\ \text{m/day}\ (\approx19.8\ \text{mm/day})}.$$