Flow net for a cutoff wall and seepage quantity: 16-Civ-A4 Geotechnical Materials and Analysis, December 2014, Q5

Question 5: Flow net for a cutoff wall and seepage quantity (Value: 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 single-line cutoff wall in a soil layer 6 m thick over an impermeable base. The wall penetrates 3 m below ground surface (its tip is 3 m above the impermeable layer). Upstream water stands 3 m above ground surface; downstream water is at ground surface. Permeability $k = 2.0\times10^{-5}$ m/s.

Given data
Quantity	Symbol	Value
Upstream head above ground	—	3 m
Downstream water level	—	at ground surface
Wall penetration below ground	$D$	3 m
Soil below wall tip to impermeable	—	3 m
Coefficient of permeability	$k$	$2.0\times10^{-5}$ m/s

Find. (a) a valid flow net drawn to the rules; (b) the steady seepage discharge $q$ per metre run of wall.

Figure Q5. Flow net for the cutoff wall: 4 flow channels (blue) passing under the wall tip and 8 equipotential drops (orange, dashed). The wall and the impermeable base are the bounding flow lines, the two ground surfaces are the bounding equipotentials, and the vertical line below the tip is the mid-drop (h = h_w/2) by symmetry. The lines are contours of a finite-difference Laplace solution, so the figures are true curvilinear squares.

Approach. The total head lost across the wall equals the difference in the two free-water levels; draw a curvilinear-square flow net around the wall tip to count the flow channels $N_f$ and equipotential drops $N_d$, then apply Darcy through the net: $q = k\,h_w\,N_f/N_d$.

Total head loss. Upstream water is 3 m above ground surface and downstream water is at ground surface, so $$h_w = 3.0 - 0 = 3.0\ \text{m}.$$ This is the head that is dissipated as water flows down the upstream face, under the wall tip and up the downstream side.
Flow-net rules. A valid net satisfies: flow lines and equipotential lines intersect at right angles; each figure is a “curvilinear square” (equal median width and length); the wall and the impermeable base are the boundary flow lines; the upstream and downstream soil surfaces are boundary equipotentials. Sketching to these rules around the wall tip (see figure) gives $N_f = 4$ flow channels and $N_d = 8$ equipotential drops, so the shape factor is $N_f/N_d = 0.5$. The net is symmetric: the wall reaches exactly half-way down the 6 m layer, so the line below the tip carries exactly half the head, and the exact conformal-mapping solution for this geometry ($d/T = 0.5$) also gives $N_f/N_d = 0.5$.
Head drop per potential. Each equipotential drop removes an equal slice of head, $$\Delta h = \frac{h_w}{N_d} = \frac{3.0}{8} = 0.375\ \text{m per drop}.$$
Seepage quantity per metre. For a unit width of wall, $$q = k\,h_w\,\frac{N_f}{N_d} = (2.0\times10^{-5})(3.0)\frac{4}{8}.$$ $$\boxed{\;q \approx 3.0\times10^{-5}\ \text{m}^3/\text{s per metre run}\;}$$ i.e. about $2.6\ \text{m}^3$ per day per metre of wall.

Seepage results
Quantity	Value
Total head loss $h_w$	3.0 m
Flow net (sketch)	$N_f = 4$, $N_d = 8$
Head per drop $\Delta h$	0.375 m
Seepage $q$ per metre	$3.0\times10^{-5}\ \text{m}^3/\text{s/m}$

Check: A flow net is a hand construction, so $N_f$ and $N_d$ each carry about ± one field; the ratio $N_f/N_d \approx 0.5$ (and hence $q$) is stable, and an equally valid coarser net (e.g. $N_f=3,\ N_d=6$) gives the same ratio. A finite-difference solution of the same section gives $q/(k\,h_w) = 0.49$. The reported $q$ should be read as $\approx 3\times10^{-5}\ \text{m}^3/\text{s/m}$.

Topic: Two-dimensional confined seepage solved graphically by a flow net.

Key relations: $q = k\,h_w\,N_f/N_d$ (per unit width); $\Delta h = h_w/N_d$; pressure head at any point $h_p = h - z$ from the net; exit gradient $i_e = \Delta h/\ell$.

Why this works: A flow net is a scaled solution of Laplace’s equation $\nabla^2 h = 0$. Curvilinear squares make the discharge through every flow tube equal ($k\,\Delta h$) and the head drop across every equipotential equal, so total discharge is simply channels × tube discharge.

Common pitfall: Taking the head loss as the wall penetration or the layer thickness instead of the difference in the two water surfaces; and drawing rectangles that are not squares, which corrupts the $N_f/N_d$ count.

Source: Das & Sobhan, Principles of Geotechnical Engineering, ch. 8 (seepage); Craig’s Soil Mechanics, ch. 2.

Question 5 of 6: Flow net for a cutoff wall and seepage quantity

Question 5: Flow net for a cutoff wall and seepage quantity (Value: 20 marks)