Question 2 of 5: Vertical Flow Through an Intact and a Fractured Clay Aquitard, and a Two-Layer Anisotropy Test

Question 2: Vertical Flow Through an Intact and a Fractured Clay Aquitard, and a Two-Layer Anisotropy Test (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) The total head and pore pressure at the base of the clay liner for the stated upward flux. (b) The vertical effective hydraulic conductivity once the clay is fractured. (c) The individual hydraulic conductivities $K_1$ and $K_2$ of the two soil layers.

Approach. Part (a) converts intrinsic permeability to hydraulic conductivity ($K=k\rho g/\mu$), then uses Darcy's law across the known thickness to back out the head difference needed for the target flux, and converts the resulting head at the base to a pressure. Part (b) applies the cubic (parallel-plate) law for a regularly fractured medium, $K_{\text{eff}}=\rho g b^3/(12\mu s)$. Part (c) recognizes the vertical test as flow perpendicular to layering (harmonic-mean $K_v$) and the horizontal test as flow parallel to layering (thickness-weighted arithmetic-mean $K_h$); the two means give two equations that are solved simultaneously for $K_1$ and $K_2$.

1. Part (a) — hydraulic conductivity of the intact clay. $$K=\frac{k\rho g}{\mu}=\frac{(1.1\times10^{-17})(1000)(9.81)}{0.001}=\boxed{1.079\times10^{-10}\ \text{m/s}}.$$
2. Head difference for the target flux. Convert the target velocity to SI: $v=1\times10^{-3}\ \text{m/yr}\div(3.156\times10^7\ \text{s/yr})=3.169\times10^{-11}\ \text{m/s}$. With the top of the clay as datum ($z_{\text{top}}=0$, so $z_{\text{bot}}=-2.5\ \text{m}$) and the piezometer reading giving $h_{\text{top}}=z_{\text{top}}+0.5=0.500\ \text{m}$, Darcy's law for upward flow ($h_{\text{bot}}>h_{\text{top}}$) gives: $$v=K\frac{h_{\text{bot}}-h_{\text{top}}}{L}\ \Rightarrow\ h_{\text{bot}}=h_{\text{top}}+\frac{vL}{K}=0.500+\frac{(3.169\times10^{-11})(2.5)}{1.079\times10^{-10}}=\boxed{1.234\ \text{m}}.$$
3. Pressure at the base. The pressure head there is $\psi_{\text{bot}}=h_{\text{bot}}-z_{\text{bot}}=1.234-(-2.5)=3.734\ \text{m}$, so: $$P_{\text{bot}}=\rho g\psi_{\text{bot}}=(1000)(9.81)(3.734)=\boxed{36{,}632\ \text{Pa}\ (36.6\ \text{kPa})}.$$
4. Part (b) — fractured aquitard, cubic law. For parallel fractures of aperture $b$ spaced $s$ apart in an otherwise impermeable matrix: $$K_{\text{eff}}=\frac{\rho g b^3}{12\mu s}=\frac{(1000)(9.81)(120\times10^{-6})^3}{12(0.001)(0.45)}=\boxed{3.14\times10^{-6}\ \text{m/s}}.$$ This is roughly four orders of magnitude larger than the intact-clay $K$ from part (a) — even widely spaced, hairline fractures dominate vertical flow through an otherwise tight aquitard.
5. Part (c) — vertical (series) test. Flow crosses both 0.5 m layers in series over the block's 1 m height: Darcy's law is driven by total head $h=z+\psi$, not pressure head alone. With the datum at the base of the block, the top has $h_{\text{top}}=1.0+0.10=1.10\ \text{m}$ and the atmospheric base has $h_{\text{bot}}=0+0=0$, so $\Delta h=1.10\ \text{m}$ over 1 m (gradient 1.10 — gravity supplies most of the driving head): $$K_{v,\text{eff}}=\frac{v_{\text{vert}}}{\Delta h/1\ \text{m}}=\frac{(0.1\ \text{mm/min})\div60}{1.10/1}=\boxed{1.515\times10^{-6}\ \text{m/s}},\qquad K_{v,\text{eff}}=\frac{1}{\dfrac{0.5}{K_1}+\dfrac{0.5}{K_2}}.$$
6. Horizontal (parallel) test. Flow runs along the 2 m width, parallel to the layering; both faces sit at the same elevation, so here the head difference is just the 0.10 m of applied pressure head: $$K_{h,\text{eff}}=\frac{v_{\text{horiz}}}{\Delta h/2\ \text{m}}=\frac{(1\ \text{mm/min})\div60}{0.10/2}=\boxed{3.333\times10^{-4}\ \text{m/s}},\qquad K_{h,\text{eff}}=\frac{0.5K_1+0.5K_2}{1}.$$
7. Solving the two-equation system. From $K_{h,\text{eff}}$: $K_1+K_2=6.667\times10^{-4}$. From $K_{v,\text{eff}}$: $\dfrac{1}{K_1}+\dfrac{1}{K_2}=\dfrac{2}{K_{v,\text{eff}}}=1.320\times10^6$, so $K_1K_2=\dfrac{K_1+K_2}{1.32\times10^6}=5.051\times10^{-10}$. These are the sum and product of the roots of $K^2-(6.667\times10^{-4})K+5.051\times10^{-10}=0$: $$K=\frac{6.667\times10^{-4}\pm\sqrt{(6.667\times10^{-4})^2-4(5.051\times10^{-10})}}{2}\ \Rightarrow\ \boxed{K_1=6.66\times10^{-4}\ \text{m/s}},\ \boxed{K_2=7.58\times10^{-7}\ \text{m/s}}.$$ (Check: $0.5K_1+0.5K_2=3.333\times10^{-4}$ and $1/(0.5/K_1+0.5/K_2)=1.515\times10^{-6}$ m/s, reproducing both tests.) The roughly three-order-of-magnitude contrast between the layers is why the parallel (horizontal) test is dominated by the more permeable layer while the series (vertical) test is dominated by the tighter one.