18-Env-A3 Geotechnical and Hydrogeological Engineering · December 2019

Question 3 of 6: Seepage Through a Sand Seam Beneath a Levee

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

Notes on this paper

National Exams December 2019 — 18-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.

Reference texts: Das, Principles of Geotechnical Engineering, 9th ed.; Craig & Knappett, Craig's Soil Mechanics, 8th ed.; Freeze & Cherry, Groundwater (1979).

Question 3: Seepage Through a Sand Seam Beneath a Levee (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. Reservoir water surface at Elev. 38 m; sand seam daylights at Elev. 30 m on the reservoir side and Elev. 28 m on the ditch side (matching the ditch water surface); 200 m horizontal run; seam inclined at 10° to the horizontal; seam thickness 1 m over the full 500 m levee length; $k=2.6\times10^{-3}$ cm/s.

Find. (a) The daily seepage volume into the ditch. (b) The piezometric (total) head at Point A, located on the seam roughly 69% of the way from the reservoir side toward the ditch (read off the figure).

Figure 1: levee cross-section (schematic, not to scale) — reservoir and ditch connected by an inclined sand seam through the clay levee; Point A sits about 69% of the way along the seam toward the ditch.

Approach. Treat the seam as a confined conduit open directly to the reservoir at one end and the ditch at the other, so the total head loss along it equals the difference between the two free-surface elevations; get the seam's actual (inclined) length from the 10° slope, apply Darcy's law for the flow rate, then interpolate linearly along the seam for the head at Point A.

Hydraulic conductivity in consistent units. $$k = 2.6\times10^{-3}\text{ cm/s} \times \dfrac{1\text{ m}}{100\text{ cm}}\times 86{,}400\text{ s/day} = 2.246\text{ m/day}$$
Seam length and gradient. The seam's horizontal run is 200 m at 10° to the horizontal, so its true (slant) length is $$L = \dfrac{200}{\cos10^\circ} = 203.1\text{ m}$$ Both seam ends are directly exposed to open water (reservoir at Elev. 38 m; ditch at Elev. 28 m), so the total head lost along the seam is simply the difference between those two free surfaces, $\Delta H = 38-28=10$ m (the seam's own end-point elevation labels, 30 m and 28 m, describe where it daylights – not the driving head, which is set by the two reservoirs). $$i = \dfrac{\Delta H}{L} = \dfrac{10}{203.1} = 0.0492$$
(a) Seepage discharge. Flow cross-section = seam thickness (1 m) × levee length (500 m, the "into the page" dimension): $$A = 1\times500 = 500\text{ m}^2$$ $$Q = kiA = 2.246\times0.0492\times500 = \boxed{55.3\text{ m}^3/\text{day}}$$
(b) Piezometric head at Point A. With constant $k$ and a constant cross-section, head drops linearly with distance along the seam. Point A sits at $\approx$69% of the horizontal run from the reservoir side (read off the figure, which is explicitly marked "not to scale"): $$h_A = 38 - 0.687\times10 = \boxed{31.1\text{ m}}$$ (i.e. a piezometer standpipe at A would show water rising to elevation 31.1 m – roughly a third of the way down from the reservoir toward the ditch, consistent with A sitting closer to the ditch end of the seam).

Final results — Question 3
Quantity	Value
Seam length, $L$	203.1 m
Hydraulic gradient, $i$	0.0492
(a) Seepage into ditch, $Q$	55.3 m³/day
(b) Piezometric head at A, $h_A$	≈31.1 m

Topic: Confined seepage through an inclined permeable seam (Darcy's law, linear head loss).

Key relations: $Q=kiA$; $i=\Delta H/L$; total head $h=$ elevation head $+$ pressure head, linear along a uniform conduit.

Why this works: Once both ends of the seam are directly open to still water, the total head at each end equals that reservoir's own free-surface elevation (no separate pressure-head bookkeeping is needed); with constant $k$ and constant cross-section along the seam, Laplace's equation reduces to a straight-line head profile, so any interior point's head is a simple linear interpolation by distance along the flow path.

Common pitfall: Using the seam's own labelled end elevations (30 m, 28 m – only a 2 m difference) as the driving head instead of the true 10 m difference between the reservoir and ditch free surfaces; or using the 200 m horizontal distance directly as the flow length instead of correcting for the 10° incline.

Source: Das, Principles of Geotechnical Engineering, Ch. 7 (Permeability); Craig & Knappett, Soil Mechanics, Ch. 2.

← Question 2 Next: Question 4 →