Flow Net and Seepage Beneath a Dam with a Cut-off Wall: 07-Str-A3, May 2014, Q5

Question 5: Flow Net and Seepage Beneath a Dam with a Cut-off Wall (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. A confined seepage problem: a permeable stratum of finite thickness, bounded below by an impervious clay, with an impervious dam base and cut-off wall imposed on it.

Given data
Quantity	Symbol	Value
Head of water retained (upstream level above tailwater)	$h_w$	10 m
Thickness of the silty sand stratum	—	20 m
Penetration of the cut-off below the dam base	—	7 m
Depth of the dam base below original ground	—	2.5 m
Length of the impervious dam base, cut-off to toe	—	25 m
Coefficient of permeability of the silty sand	$k$	$2.0\times10^{-4}$ cm/s = $2.0\times10^{-6}$ m/s
Underlying stratum	—	thick clay, treated as impervious

Find. (i) a flow net for the confined flow beneath the structure, stating the number of flow channels $N_f$ and equipotential drops $N_d$, and (ii) the seepage quantity $q$ per metre run of the wall.

Approach. Sketch the flow net by fixing the four boundary conditions first — two equipotentials at the entry and exit surfaces, two flow lines along the impervious base of the structure and along the top of the clay — then fill the interior with curvilinear squares, and finally read $N_f$ and $N_d$ off the completed net and substitute into the flow-net discharge equation printed on the formula sheet.

(i) Constructing the flow net

Flow beneath the dam is confined: it has no free surface, so the region is fully bounded and the net is determined entirely by geometry. Four boundaries fix it:

The upstream ground surface, everywhere upstream of the dam, is an equipotential at total head $h = 10\ \text{m}$ (taking the tailwater surface as datum). Water enters the soil here.
The downstream ground surface beyond the toe of the dam is an equipotential at $h = 0$. Water leaves here.
The underside of the dam together with the cut-off wall forms the uppermost flow line. It is a single continuous impervious surface, running down the upstream face of the dam, down one side of the sheet pile, around its toe, back up the other side, along the base to the downstream corner D, and finally up the 2.5 m embedded downstream face of the dam to the ground surface.
The top of the clay, 20 m below original ground, is the lowest flow line, since no water crosses into the clay.

Between these boundaries the net is drawn so that flow lines and equipotentials meet at right angles and every figure is a curvilinear square. Working outwards from the cut-off — where the elements are smallest, because the flow is squeezed around the toe — and adjusting until the squares close gives a net with

$$N_f = 4 \ \text{flow channels}, \qquad N_d = 12 \ \text{equipotential drops}$$

Figure 5.1 — The completed flow net. Flow lines (solid) start on the upstream ground surface, pass beneath the cut-off and re-emerge downstream; equipotentials (dashed) cross them at right angles. The elements are smallest at the toe of the cut-off, where the hydraulic gradient is highest.

The head lost across each equipotential drop is the same:

$$\Delta h = \frac{h_w}{N_d} = \frac{10}{12} = 0.833\ \text{m per drop}$$

so the total head at any point in the net is $h = h_w - n_d\,\Delta h$, where $n_d$ is the number of drops already traversed. At the toe of the cut-off only about 3.6 of the 12 drops have been used, so the total head there is roughly $10 - 3.6(0.833) \approx 7.0\ \text{m}$ (the finite-difference solution in step 3 below gives 7.2 m on the upstream face and 6.9 m on the downstream face at tip level). A cut-off at the upstream edge of a long base is not a free-standing sheet pile, whose tip would sit on the middle equipotential: most of the remaining head is dissipated along the 25 m impervious base, so most of the drops lie downstream of the wall.

(ii) Seepage quantity

Convert the permeability to SI units. Flow-net work is done in metres and seconds, so $$k = 2.0\times10^{-4}\ \frac{\text{cm}}{\text{s}} \times \frac{1\ \text{m}}{100\ \text{cm}} = 2.0\times10^{-6}\ \text{m/s}$$
Apply the flow-net discharge equation. The formula sheet gives, per unit width perpendicular to the section, $$q = k\,h_w\,\frac{N_f}{N_d}$$ Every quantity in it has already been established: $k$ from the site investigation, $h_w$ from the reservoir level, and the ratio $N_f/N_d$ — the shape factor — from the net. Substituting, $$q = \left(2.0\times10^{-6}\right)(10)\left(\frac{4}{12}\right) = \left(2.0\times10^{-5}\right)(0.3333)$$ $$\boxed{\,q = 6.67\times10^{-6}\ \text{m}^3/\text{s per metre run}\,}$$
Express the result in units an engineer can use. A seepage rate quoted in cubic metres per second per metre is hard to judge, so convert: $$q = 6.67\times10^{-6} \times 86\,400 = 0.576\ \text{m}^3/\text{day per metre} = 576\ \text{litres/day per metre}$$ which is $210\ \text{m}^3$ per year for every metre of dam length. For a 100 m long structure that is roughly $2.1\times10^{4}\ \text{m}^3$ per year — a loss that is significant for reservoir yield but entirely manageable, and one that the cut-off has already reduced.
Confirm the shape factor independently. A hand-drawn net always carries some judgement in it, so the shape factor was checked by solving Laplace’s equation $\partial^2h/\partial x^2 + \partial^2h/\partial y^2 = 0$ directly on the drawn geometry by finite differences, with no-flow conditions on the dam base, both embedded faces of the dam, the cut-off and the top of the clay, and fixed heads on the two ground surfaces. That solution returns $$\left(\frac{N_f}{N_d}\right)_{\text{exact}} = 0.317 \quad \text{against} \quad \frac{4}{12} = 0.333$$ a difference of about 5 %, which is within the accuracy normally claimed for a hand-drawn flow net and confirms that four channels and twelve drops is the right net rather than, say, five and fourteen (0.357, 13 % high). Carrying the exact value instead would give $q = 6.34\times10^{-6}\ \text{m}^3$/s per metre (0.548 m$^3$/day).
Check the exit gradient, which is what actually governs safety. The last equipotential drop occurs over the shortest element in the net, immediately downstream of the toe of the dam, where the flow rises to the surface. With that element measuring roughly 2.5 m in the completed net, $$i_{\text{exit}} \approx \frac{\Delta h}{\ell} = \frac{0.833}{2.5} = 0.33$$ Against a critical gradient of $i_c = (G_s - 1)/(1 + e) \approx 1.0$ for a silty sand at $e \approx 0.65$, this gives a factor of safety against piping of about 3; the finite-difference solution confirms the estimate, giving $i \approx 0.32$ in the soil immediately beyond the toe. That is acceptable but not generous — a downstream filter blanket would normally be specified. This is the number a designer cares about far more than the seepage quantity itself.

Results
Quantity	Symbol	Value
Permeability in SI units	$k$	$2.0\times10^{-6}$ m/s
Flow channels in the net	$N_f$	4
Equipotential drops in the net	$N_d$	12
Head loss per drop	$\Delta h$	0.833 m
Total head at the toe of the cut-off (FD)	$h$	≈ 7.0 m (about 3.6 drops used)
Seepage per metre run	$q$	$6.67\times10^{-6}$ m$^3$/s = 0.576 m$^3$/day
Annual seepage per metre run	—	210 m$^3$/yr
Shape factor, finite-difference check	$N_f/N_d$	0.317 against 0.333 drawn (≈ 5 %)
Approximate exit gradient	$i_{\text{exit}}$	0.33 (FD: 0.32), giving FS ≈ 3 against piping

Topic: Confined seepage beneath a hydraulic structure — flow-net construction, the shape factor, and the seepage and exit-gradient quantities it delivers.

Key relations: Laplace’s equation for an isotropic homogeneous soil; $q = k h_w N_f/N_d$ per unit width; $\Delta h = h_w/N_d$ and $h_p = (n_d/N_d)h_w$ from the printed formula sheet; $v = ki$ with $i = h/L$; and $i_c = (G_s-1)/(1+e)$ for the critical gradient against piping.

Why this works: For steady flow through a saturated incompressible soil obeying Darcy’s law, continuity plus $v = -k\,\nabla h$ gives Laplace’s equation, whose solution is a pair of orthogonal families of curves — equipotentials and flow lines. Drawing them as curvilinear squares forces the same discharge through every channel and the same head drop across every drop, so the total discharge collapses to the product of $k$, the total head, and the purely geometric ratio $N_f/N_d$. Nothing about the soil except $k$ enters, which is why one flow net serves for any permeability, and why the net is worth drawing carefully once.

Common pitfall: Using the wrong head. $h_w$ is the difference between upstream and downstream water levels, not the depth of water upstream and not the depth of the stratum; here both happen to be numerically 10 m only because the tailwater stands exactly at ground level. A second pitfall is counting the boundary flow lines as channels: four flow lines drawn between the two boundary flow lines give five channels, not four, so the count must be of the spaces, not the lines. The third is stopping at the seepage quantity — the exit gradient, not the discharge, is what causes dam failures.

Source: Das, Principles of Geotechnical Engineering, 9th ed., Ch. 8 — seepage, flow-net construction rules, seepage calculation from a flow net, and uplift/piping checks beneath hydraulic structures; Knappett & Craig, Craig’s Soil Mechanics, 8th ed., Ch. 2, for the boundary-condition method used above; Canadian Dam Association, Dam Safety Guidelines, for the Canadian requirements on seepage control and filter design.

Question 5 of 6: Flow Net and Seepage Beneath a Dam with a Cut-off Wall

Question 5: Flow Net and Seepage Beneath a Dam with a Cut-off Wall (20 marks)

(i) Constructing the flow net

(ii) Seepage quantity