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18-Geol-A2 Hydrogeology · December 2016

Question 2 of 5: Flow Net Through an Earthen Dam

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2016 — 04-Geol-A2, Hydrogeology. Three-hour, open-book exam; any non-communicating calculator permitted. FIVE questions constitute a complete paper, each of equal value; most questions require an essay-format answer with work shown. Unless otherwise specified, water density = 1000 kg/m³, water viscosity = 0.001 kg/(m·s), and g = 9.81 m/s² — all five are solved below for completeness.

Reference texts: Freeze & Cherry, Groundwater (Prentice-Hall, 1979) — Darcy's law, storage/specific yield, flow nets, layered vertical flow, and the Theis/Hantush-Jacob well-hydraulics chapters used throughout this paper.

Question 2: Flow Net Through an Earthen Dam (equal value)

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. A homogeneous, isotropic trapezoidal earthen dam on an impermeable base; headwater depth $H_1=6\ \text{m}$ against the upstream face, tailwater depth $H_2=2.25\ \text{m}$ against the downstream face, hydraulic conductivity $K=0.066\ \text{m/day}$. The scale bars on the source figure give the dam's true proportions; only the two water depths and $K$ are needed once a flow net is constructed.

Find. (a) A flow net (families of flow lines and equipotential lines satisfying the boundary conditions); (b) the seepage discharge per unit width of dam, in m³/day per metre.

impermeable baseheadwater H1 = 6 mtailwater H2 = 2.25 mK = 0.066 m/dayphreatic surface (top flow line)base flow line (impermeable boundary)equipotential lines (12 drops)Nf = 4 flow channels
Figure 2. Constructed flow net: the phreatic surface and the impermeable base bound the flow domain; three interior flow lines divide it into $N_f=4$ curvilinear-square flow channels, cut by 12 equipotential lines ($N_d=12$ head drops).

Approach. (a) Build the flow net from the boundary conditions — the phreatic (top) surface and the impermeable base are both flow lines, the wetted upstream and downstream faces are equipotentials — and sketch flow lines/equipotentials as curvilinear squares. (b) Read off the number of flow channels $N_f$ and equipotential drops $N_d$ from that net and apply $q = KH(N_f/N_d)$.

Check: a hand-drawn flow net is a graphical estimate, not a unique analytical solution. The net above uses $N_f=4$ flow channels and $N_d=12$ equipotential drops, a standard curvilinear-square construction for a dam of this height-to-base proportion (per Freeze & Cherry Ch. 5 / Cedergren's worked dam examples) — a carefully hand-drawn net will land within about ±10–15% of this $N_f/N_d$ ratio.
  1. Total head loss across the dam. The seepage-driving head is the difference between the two water-surface elevations relative to the base: $$H = H_1-H_2 = 6-2.25=\boxed{3.75\ \text{m}}.$$
  2. Seepage per unit width from the flow net. For a well-constructed curvilinear-square net, the discharge per unit width out-of-page is $$q = KH\left(\frac{N_f}{N_d}\right) = (0.066\ \text{m/day})(3.75\ \text{m})\left(\frac{4}{12}\right)=\boxed{0.0825\ \text{m}^3/\text{day per m}}.$$
QuantityResult
(a) Flow net$N_f=4$ flow channels, $N_d=12$ equipotential drops (Figure 2)
(b) Total head loss H3.75 m
(b) Seepage per unit width0.0825 m³/day per m of dam