18-Env-A3 Geotechnical and Hydrogeological Engineering · May 2018
Question 3 of 6: Seepage and Tracer Travel Time Beneath an Earth Dam
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams May 2018 — 04-Env-A3, Geotechnical and Hydrogeological Engineering — 3 hours, open book. All SIX questions are answered below (the exam marks only the first five as submitted; all six are solved here as a complete study resource). Marking scheme: each question 20 marks, equal value.
Find. The daily seepage volume beneath the dam and the tracer travel time from point A (upstream toe) to point B (downstream toe).
Figure 2: earth dam and reservoir system, resting on a flat 10 m loam layer over impermeable shale.
Approach. The dam's own clay ($k=0.0001$ cm/s) is four orders of magnitude tighter than the loam foundation ($k=0.01$ cm/s), so essentially all seepage travels horizontally through the loam beneath the dam's impervious base. Because the pervious layer is flat and of uniform thickness, and both the headwater and tailwater flood the layer's top surface right up to the toes, this reduces to a single Method-of-Fragments flow tube of length $B$ (the dam base width) and cross-section $a\times 1$ (per metre of crest).
Convert $k$ and set up the flow tube. $$k = 0.01\text{ cm/s} = 1\times10^{-4}\text{ m/s} = 8.64\text{ m/day}$$ Head loss across the base: $\Delta h = H_1-H_2 = 18-8=10$ m, over a horizontal path length $B=100$ m (the geometry checks out: base $=$ crest $+2\times(\text{slope}\times\text{height}) = 10+2(2.5)(18)=100$ m, matching the reservoir depth given).
(a) Seepage rate. $$q' = \dfrac{k\,a\,\Delta h}{B} = \dfrac{8.64\times 10\times 10}{100} = 8.64\text{ m}^3/\text{day per metre of crest}$$ $$Q = q'L = 8.64\times 50 = \boxed{432\text{ m}^3/\text{day}}$$
(b) Tracer travel time. Flow through this single flow tube is uniform (same velocity everywhere along its length), so the seepage (average linear) velocity follows directly from the average gradient $i=\Delta h/B$ and an assumed effective porosity for the loam. Check: the loam's porosity is not given in the source; a typical value $n_e=0.40$ is assumed (loam, per Freeze & Cherry Table 2.4) — travel time scales directly with $n_e$. $$v_{Darcy} = k\,i = 8.64\times\dfrac{10}{100} = 0.864\text{ m/day}, \qquad v_{seep} = \dfrac{v_{Darcy}}{n_e} = \dfrac{0.864}{0.40} = 2.16\text{ m/day}$$ $$t = \dfrac{B}{v_{seep}} = \dfrac{100}{2.16} = \boxed{46.3\text{ days}}$$
