16-Civ-A4 Geotechnical Materials and Analysis · December 2016

Question 2 of 5: Flow net and seepage under a dam with two cutoff walls

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

Question 2: Flow net and seepage under a dam with two cutoff walls (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. Concrete dam, base width 30 m, on an anisotropic pervious foundation 20 m deep to an impervious base; two cutoff walls each 6 m deep; upstream water 10 m, downstream tailwater 1.5 m; $k_x = 7\times10^{-5}\,$cm/s, $k_z = 1\times10^{-5}\,$cm/s.

Find. Construct the flow net and determine the seepage loss (m³/day per metre run of dam).

Approach. Because the soil is anisotropic, transform the section to an equivalent isotropic one, sketch a curvilinear-square net, count flow channels $N_f$ and potential drops $N_d$, and apply $q = k'\,H\,(N_f/N_d)$ with the equivalent permeability $k'=\sqrt{k_x k_z}$.

1. Transform the horizontal scale. For anisotropic flow the horizontal dimensions are scaled by $x_{tr}=x\sqrt{k_z/k_x}=x\sqrt{1/7}=0.378\,x$, while vertical dimensions are unchanged. The 30 m base becomes $30(0.378)=11.3\,$m; the 20 m depth and 6 m cutoffs stay as drawn. The net is then sketched as true curvilinear squares on this transformed section.
2. Equivalent isotropic permeability. $k' = \sqrt{k_x\,k_z} = \sqrt{(7\times10^{-5})(1\times10^{-5})} = \sqrt{7}\times10^{-5} = 2.65\times10^{-5}\ \text{cm/s} = 2.65\times10^{-7}\ \text{m/s}.$
3. Count the net. On the transformed section the layer (20 m) is almost twice as deep as the base is wide (11.3 m), so the flow dives well below the 6 m cutoffs. Curvilinear squares then need about three drops in the entry zone, three beneath the base and three in the exit zone: $\boxed{N_f = 4,\quad N_d = 9}$ (shape factor $N_f/N_d = 0.444$). A net sketched on the untransformed 30 m section looks square with fewer drops and overstates $q$.
4. Seepage per metre run. With the full head $H=8.5$ m acting between reservoir and tailwater, $$q = k'\,H\,\frac{N_f}{N_d} = (2.65\times10^{-7})(8.5)\left(\tfrac{4}{9}\right) = 1.00\times10^{-6}\ \text{m}^3/\text{s per m}.$$
5. Convert to m³/day. $q = 1.00\times10^{-6}\times86\,400 = \boxed{0.086\ \text{m}^3/\text{day per metre run of dam}}.$
6. Cross-check the count. A finite-difference solution of Laplace's equation on the transformed section gives the same flow entering and leaving, $q\approx0.086$ m³/day per m ($N_f/N_d\approx0.445$). Pavlovsky's method of fragments (entry cutoff + confined reach + exit cutoff, $\sum\Phi = 2.29$) gives $N_f/N_d = 0.436$ and $q = 0.085$ m³/day per m. All three agree within 3 %.