Seepage Beneath a Dam & Impervious Blanket: 18-Env-A3 Geotechnical and Hydrogeological Engineering, May 2014, Q3

Question 3: Seepage Beneath a Dam & Impervious Blanket (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.

Given data
Quantity	Symbol	Value
Dam crest elevation	—	135 m
Dam base / top of sand-gravel elevation	—	121 m
Upstream (reservoir) water elevation	—	133.5 m
Tailwater elevation	—	122.5 m
Upstream (and, by the figure, downstream) face slope	—	1V : 3H
Sand-gravel layer thickness	$a$	21 m
Sand-gravel hydraulic conductivity	$k$	0.2 cm/s
Dam length (transverse to section)	$B_w$	150 m
Downstream flooded sand-gravel strip (toe to layer end)	$L$	75 m

Figure 2 — the dam sits directly on the sand-gravel layer's top surface (elev. 121 m); both the reservoir and (since tailwater 122.5 m > 121 m) the downstream strip flood the layer's surface right up to the dam's toes, so the dam's own base acts as the only "impervious floor" forcing horizontal flow.

Find. (a) the seepage volume per day beneath the dam; (b) the upstream impervious-blanket length $X$ that halves that seepage.

Approach. The sand-gravel layer is a uniform-thickness ($a=21$ m) confined stratum, flat-topped and flat-bottomed (aquiclude), with the headwater and tailwater both flooding the layer's surface directly at the dam's own toes (no gap). This is the single-fragment case of the Method of Fragments: with no vertical cutoff, seepage beneath the dam reduces to horizontal, one-dimensional Darcy flow under an impervious "floor" of width equal to the dam's own base width $B$, over the full depth $a$.

Part (a) — dam base width and hydraulic conductivity in consistent units. The dam height is $135-121=14$ m; assuming the triangular section is symmetric (1V:3H on both faces, as drawn), each half-base runs $3\times14=42$ m, so $$B=2\times42=\boxed{84\ \text{m}}.$$ Converting $k$ to metres per day: $k=0.2\ \text{cm/s}=0.002\ \text{m/s}\times86{,}400\ \text{s/day}=\boxed{172.8\ \text{m/day}}.$
Apply Darcy's law across the dam's footprint. With head loss $\Delta h=133.5-122.5=11.0$ m over horizontal path length $B=84$ m through a layer of thickness $a=21$ m, the seepage per metre of dam length is $$q=\frac{k\,a\,\Delta h}{B}=\frac{172.8\times21\times11.0}{84}=\boxed{475.2\ \text{m}^3/\text{day per m}}.$$ Multiplying by the dam's transverse length $B_w=150$ m, $$Q=q\,B_w=475.2\times150=\boxed{71{,}280\ \text{m}^3/\text{day}}.$$
Part (b) — blanket length to halve the seepage. An upstream impervious blanket of length $X$ simply extends the impervious "floor" the water must travel under before it can enter the pervious layer, lengthening the flow path from $B$ to $B+X$ while $k$, $a$ and $\Delta h$ are unchanged: $$q'=\frac{k\,a\,\Delta h}{B+X}=\frac{1}{2}q=\frac{1}{2}\cdot\frac{k\,a\,\Delta h}{B}\ \Longrightarrow\ B+X=2B\ \Longrightarrow\ X=B=\boxed{84\ \text{m}}.$$ (Check: $q'=172.8\times21\times11.0/(84+84)=237.6\ \text{m}^3/\text{day per m}$, exactly half of $475.2$.)

Check: two assumptions worth flagging. (1) The figure dimensions only the upstream 1V:3H slope; a symmetric downstream face (also 1V:3H) is assumed from the drawing, giving $B=84$ m — an asymmetric dam would change $B$ and hence both answers proportionally. (2) This single-fragment Darcy estimate ignores the small extra head loss where flow turns from vertical to horizontal at the toes (the "entrance/exit" resistance terms of a full flow net or a rigorous Method-of-Fragments solution); it is offered as the APPROXIMATE estimate the question calls for in part (b), and the given downstream length $L=75$ m only confirms that the tailwater floods the sand-gravel surface all the way to the dam's toe (so the boundary condition is met right there, with no separate unflooded exit reach to add to the path).

Quantity	Value
Dam base width, $B$	84 m
$k$ (converted)	172.8 m/day
(a) Seepage volume, $Q$	71,280 m³/day
(b) Blanket length, $X$	≈ 84 m

Concept & Theory

Topic. Seepage beneath a dam on a permeable stratum of limited depth, with and without an upstream impervious blanket (Method of Fragments, single horizontal-flow region).

Key relations. Darcy's law $q=kiA$; for a confined layer of thickness $a$ under an impervious floor of length $L_f$, $q=k\,a\,\Delta h/L_f$ per unit width; an impervious blanket of length $X$ adds directly to $L_f$.

Why this works. When the pervious layer's top is sealed by an impervious floor (dam base, or blanket) over its whole width and the heads are applied right at that floor's two edges, the flow is essentially horizontal and uniform across the layer thickness — exactly the geometry Darcy's law was derived for, so no flow net is needed to get an engineering estimate.

Common pitfall. Forgetting to convert $k$ from cm/s to the m/day units the answer is requested in (a factor-of-500-ish error), and using the dam's CREST width instead of its BASE width for the flow-path length $B$.

Source. Das, Principles of Geotechnical Engineering, ch. 8 (seepage, flow nets, Method of Fragments).

Question 3 of 6: Seepage Beneath a Dam & Impervious Blanket

Question 3: Seepage Beneath a Dam & Impervious Blanket (20 marks)