Question 2 of 5: Layered-Aquitard Darcy Velocity Under Fresh Water and Brine

Question 2: Layered-Aquitard Darcy Velocity Under Fresh Water and Brine (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. Two-layer aquitard, each layer $b=5$ m thick: top layer $K_1=10^{-8}$ cm/s, bottom layer $K_2=10^{-6}$ cm/s (values stated for the freshwater case). Pressure at the top $P_{\text{top}}=105$ kPa, at the bottom $P_{\text{bottom}}=210$ kPa. Part (b) repeats with brine of density $\rho_{br}=1150\ \text{kg/m}^3$ filling the same aquitard (same measured pressures, same viscosity).

Find. The magnitude and direction of the Darcy (specific-discharge) velocity across the aquitard, for (a) fresh water and (b) brine.

Approach. Convert each layer's conductivity into an effective vertical (series/harmonic) conductivity for the two-layer aquitard, convert each pressure to a total head using the appropriate fluid density, then apply Darcy's law over the full 10 m thickness. For part (b), because the layer $K$-values given in the statement are intrinsic-permeability-controlled but only quoted for the freshwater case, the same soils under brine have a proportionally larger hydraulic conductivity, $K_{br}=K(\rho_{br}/\rho_w)$, since $K=k\rho g/\mu$ and only $\rho$ changes (the stated viscosity is unchanged).

1. Part (a) — effective vertical conductivity (series/harmonic mean). Converting to SI, $K_1=10^{-10}$ m/s, $K_2=10^{-8}$ m/s, and for two layers in series (flow crosses both): $$K_v=\frac{b_1+b_2}{\dfrac{b_1}{K_1}+\dfrac{b_2}{K_2}}=\frac{10}{\dfrac{5}{10^{-10}}+\dfrac{5}{10^{-8}}}=\boxed{1.980\times10^{-10}\ \text{m/s}}.$$ As expected for series flow, $K_v$ is dominated by (close to double) the less permeable top layer's contribution.
2. Convert pressures to total heads (fresh water). Taking the aquitard base as datum ($z=0$, top at $z=10$ m), $h=z+P/(\rho g)$: $$h_{\text{top}}=10+\frac{105{,}000}{(1000)(9.81)}=10+10.70=\boxed{20.70\ \text{m}},\qquad h_{\text{bottom}}=0+\frac{210{,}000}{(1000)(9.81)}=\boxed{21.41\ \text{m}}.$$
3. Darcy velocity (fresh water). $\Delta h=h_{\text{bottom}}-h_{\text{top}}=21.41-20.70=0.703$ m; since this is positive, head is higher at the base, so flow is upward, and $$v=K_v\frac{|\Delta h|}{L}=(1.980\times10^{-10})\frac{0.703}{10}=\boxed{1.39\times10^{-11}\ \text{m/s, upward}}.$$
4. Part (b) — scale the conductivities to brine. With the same intrinsic permeability and the same viscosity, $K\propto\rho$, so $$K_{1,br}=K_1\frac{1150}{1000}=1.15\times10^{-10}\ \text{m/s},\qquad K_{2,br}=K_2\frac{1150}{1000}=1.15\times10^{-8}\ \text{m/s},$$ $$K_{v,br}=\frac{10}{\dfrac{5}{1.15\times10^{-10}}+\dfrac{5}{1.15\times10^{-8}}}=\boxed{2.277\times10^{-10}\ \text{m/s}}.$$
5. Convert pressures to heads with brine. The same measured pressures now correspond to a smaller pressure head because brine is denser ($\psi=P/(\rho_{br}g)$): $$h_{\text{top}}=10+\frac{105{,}000}{(1150)(9.81)}=\boxed{19.31\ \text{m}},\qquad h_{\text{bottom}}=0+\frac{210{,}000}{(1150)(9.81)}=\boxed{18.61\ \text{m}}.$$
6. Darcy velocity (brine) — the direction reverses. $\Delta h=h_{\text{bottom}}-h_{\text{top}}=18.61-19.31=-0.693$ m: now the top has the higher head, so $$v=K_{v,br}\frac{|\Delta h|}{L}=(2.277\times10^{-10})\frac{0.693}{10}=\boxed{1.58\times10^{-11}\ \text{m/s, downward}}.$$ The flow direction flips between (a) and (b) even though the raw pressures are unchanged, because converting pressure to head is fluid-density-dependent: a denser fluid needs a taller column to exert the same pressure, so the same 210 kPa at the base produces a smaller head increment under brine than under fresh water, and the balance that favoured upward flow in (a) reverses.