Question 2 of 6: Layered Aquitard Conductivities and a Density-Driven Flow Direction

Question 2: Layered Aquitard Conductivities and a Density-Driven Flow Direction (equal value)

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. (a) Upper layer $b_1=10$ m, $K_1=5\times10^{-6}$ cm/s; lower layer $b_2=15$ m, $K_2=2.5\times10^{-7}$ cm/s; head at the top of the aquitard 30 m, head at the bottom 56 m. (b) A 15 m aquitard between a fresh aquifer ($\rho_f=1000\ \text{kg/m}^3$) above and a saline aquifer ($\rho_s=1120\ \text{kg/m}^3$) below; the fresh-side well reads a 10 m water column, the saline-side well reads a 3 m water column.

Find. (a) the effective horizontal and vertical hydraulic conductivity of the two-layer aquitard, and the head at the layer interface. (b) whether flow across the aquitard is upward or downward.

Approach. Part (a) uses the standard layered-medium averages — a thickness-weighted arithmetic mean for horizontal (parallel) flow and a thickness-weighted harmonic mean for vertical (series) flow — then finds the interface head from continuity of the vertical Darcy flux across the two layers. Part (b) cannot compare the two piezometer readings directly because they were taken in fluids of different density; each pressure head must first be converted to an equivalent freshwater head at a common elevation datum before the heads can be compared.

1. Part (a) — effective horizontal conductivity. For flow parallel to the layering, the layers act in parallel and the effective $K$ is the thickness-weighted arithmetic mean: $$K_h=\frac{K_1b_1+K_2b_2}{b_1+b_2}=\frac{(5\times10^{-6})(10)+(2.5\times10^{-7})(15)}{25}=\boxed{2.15\times10^{-6}\ \text{cm/s}}.$$
2. Effective vertical conductivity. For flow across the layering the layers act in series and the effective $K$ is the thickness-weighted harmonic mean: $$K_v=\frac{b_1+b_2}{\dfrac{b_1}{K_1}+\dfrac{b_2}{K_2}}=\frac{25}{\dfrac{10}{5\times10^{-6}}+\dfrac{15}{2.5\times10^{-7}}}=\boxed{4.03\times10^{-7}\ \text{cm/s}}.$$ As expected for series flow, $K_v$ is dominated by (and close to) the less permeable lower layer.
3. Interface head. The Darcy flux is the same through both layers (continuity), so the head $h_m$ at the 10 m/15 m interface satisfies $K_1(h_m-h_{\text{top}})/b_1=K_2(h_{\text{bottom}}-h_m)/b_2$; solving the resulting linear equation for $h_m$: $$h_m=\frac{K_1b_2\,h_{\text{top}}+K_2b_1\,h_{\text{bottom}}}{K_1b_2+K_2b_1}=\boxed{30.84\ \text{m}}.$$ The interface head sits much closer to the top (30 m) than to the bottom (56 m) because almost the entire 26 m head loss is consumed crossing the far less permeable lower layer.
4. Part (b) — convert each reading to an equivalent freshwater head. With the aquitard base as datum ($z=0$) and its top at $z=15$ m, the freshwater-equivalent head at a point of elevation $z$ carrying a fluid of density $\rho$ and pressure head $\psi$ is $h_f=z+(\rho/\rho_f)\psi$. At the fresh-water well (already fresh, so no correction is needed): $$h_{f,\text{top}}=15+\left(\frac{1000}{1000}\right)(10)=\boxed{25.0\ \text{m}}.$$
5. Freshwater-equivalent head at the saline well. Applying the same conversion with the saline density: $$h_{f,\text{bottom}}=0+\left(\frac{1120}{1000}\right)(3)=\boxed{3.36\ \text{m}}.$$ Since $h_{f,\text{top}}$ (25.0 m) is far greater than $h_{f,\text{bottom}}$ (3.36 m), flow across the aquitard is downward — from the fresh aquifer, through the aquitard, into the saline aquifer.