Question 3 of 6: Seepage Through a Zoned Earth Dam

Question 3: Seepage Through a Zoned Earth Dam (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.

Find. (a) seepage volume through the dam per day; (b) maximum seepage flow velocity through the dam.

Approach. The dashed lines in the figure show the core is a uniform, full-height (11 m) vertical band, 10 m thick, flanked by two tapering silty-loam shoulders. Because $k_{shoulder}/k_{core}=240$, the low-permeability core controls the seepage: treat it as the dominant resistance in the flow path (standard composite/zoned-dam idealisation) and apply Darcy's law across it directly, then confirm with a full series-resistance solve of all three zones.

1. Core geometry. The core spans the full dam height uniformly (it does not taper like the shoulders): flow area $A_{core}=H\times\ell=11.00\times50=550\ \text{m}^2$, path length $L_{core}=10\ \text{m}$.
2. Part (a) — core-controlled seepage. With $k_{shoulder}$ 240× larger than $k_{core}$, essentially the whole 11 m head is lost crossing the core, so Darcy's law applies directly across it: $$i_{core}=\frac{H}{L_{core}}=\frac{11.00}{10}=1.10,$$ $$Q=k_{core}\,i_{core}\,A_{core}=0.001\times1.10\times550=\boxed{0.605\ \text{m}^3/\text{day}}.$$
3. Rigour check — full series-resistance solve. Modelling the upstream wedge (24 m run + 4 m flat crest strip), the core, and the downstream wedge (14 m run) as three resistances in series, $R=L/(kA)$ with each zone's average cross-section, gives $R_{shoulder,total}\approx0.61$ day/m² against $R_{core}=18.18$ day/m² — the core alone carries 97% of the total resistance. Solving $Q=H/R_{total}$ over all three zones gives $Q=0.585\ \text{m}^3/\text{day}$, within 3.4% of the core-only value, confirming the shoulders' resistance is a small, safely-neglected correction.
4. Part (b) — maximum seepage velocity. By continuity the same $Q$ crosses every section along the flow path, so velocity is highest where the flow area is smallest and non-tapering — that is the core (the shoulders' average area is comparable but they pinch to zero only at the very heel/toe, a non-physical edge singularity, not a meaningful "maximum"). The Darcy (superficial) velocity through the core is $$v=\frac{Q}{A_{core}}=\frac{0.605}{550}=1.10\times10^{-3}\ \text{m/day}.$$ Converting to true (interstitial) seepage velocity with an assumed clay porosity $n_{core}\approx0.45$ (typical for compacted clay, Freeze & Cherry Table 2.4): $$v_s=\frac{v}{n_{core}}=\frac{1.10\times10^{-3}}{0.45}=\boxed{2.44\times10^{-3}\ \text{m/day}}.$$