Question 2 of 6: Seepage Under a Dam — Flow Net and Flow Rate

Question 2: Seepage Under a Dam — Flow Net and Flow Rate (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) a flow net for seepage under the dam; (b) flow rate per unit width (per metre of dam length) beneath the dam.

Approach. Because the impervious base is sloped and the dam itself is embedded (not flush with the ground surface), a hand flow net is awkward to draw with any precision; solve the equivalent boundary-value problem with a masked finite-difference grid over the true (irregular) flow domain — Dirichlet heads at the reservoir/tailwater beds, no-flow at the sloped clay top and at the dam's own embedded base — and read the flow rate and the equivalent $N_f/N_d$ ratio straight off the converged head field. This is exactly a flow net, just refined numerically instead of by hand.

1. Set up the domain and boundary conditions. Take $x=0$ at the upstream edge of the figure ($x=78\ \text{m}$ total width). The top boundary is Dirichlet at $H_{up}=134.2\ \text{m}$ for $0\le x<33\ \text{m}$ and $H_{dn}=126.1\ \text{m}$ for $53
2. Solve Laplace's equation on the masked grid. With $\nabla^2 h=0$ inside the active (silty-sand) cells and the boundary conditions above, a 0.25–0.5 m finite-difference grid (5-point stencil, sparse direct solve) converges to under 1% change in flow rate on refinement. The flow per unit width follows from Darcy's law integrated down the mid-dam section: $$q=-\int k\,\frac{\partial h}{\partial x}\,dz \approx 5.30\times10^{-6}\ \text{m}^3/\text{s per m}.$$
3. Convert and cross-check against the flow-net ratio. In m³/day per metre of dam length, $$q=5.30\times10^{-6}\times86400=\boxed{0.459\ \text{m}^3/\text{day per m}}.$$ The equivalent flow-net ratio is $q=k\,\Delta H\,(N_f/N_d)$, so $$\frac{N_f}{N_d}=\frac{q}{k\,\Delta H}=\frac{5.30\times10^{-6}}{2.778\times10^{-6}\times8.1}=0.235,$$ consistent with a hand-drawn net of about 2 flow channels crossing roughly 8–9 equipotential drops (Fig. Q2) — the ratio is well below 1 because nearly all of the head is lost squeezing through the thin ($\approx8.3\ \text{m}$) downstream throat, where the clay rises to el. 116.0 m.