Question 2 of 5: Layered Vertical-Flow Pressures, Density-Dependent Flow Direction Across an Aquitard, and Confined Storage Volume

Question 2: Layered Vertical-Flow Pressures, Density-Dependent Flow Direction Across an Aquitard, and Confined Storage Volume (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) Three horizontal formations, each 25 m thick, as below; total head 180 m at the top of the stack, 80 m at the base. (b) 20 m thick aquitard between a freshwater aquifer above ($\rho=1000\ \text{kg/m}^3$, 12 m of water in the top-screened well) and a saline aquifer below ($\rho=1160\ \text{kg/m}^3$, 30 m of water in the bottom-screened well). (c) $S_s=2.5\times10^{-6}\ \text{m}^{-1}$, $b=35$ m, plan area $2200\ \text{m}\times2000\ \text{m}$, target head decline $\Delta H=1.2$ m.

Find. (a) water pressure at the two internal boundaries. (b) the direction of flow across the aquitard under each density assumption. (c) the volume of water pumped.

Approach. Part (a) uses continuity of specific discharge through the three series layers to apportion the 100 m total head loss, then converts each boundary's total head to a pressure head using its elevation. Part (b) compares total head at the two screened points; crossing a density interface requires converting the saline-side reading to a freshwater-equivalent head before the comparison is valid. Part (c) combines specific storativity with aquifer thickness to get storativity, then applies $\Delta V=S\,A\,\Delta H$.

1. Part (a) — specific discharge through the series stack. Steady vertical flow through layers in series carries the same $q$ through each, so the 100 m head loss apportions as $\sum L_i/K_i$: $$q=\frac{\Delta H}{\sum L_i/K_i}=\frac{180-80}{25/0.001+25/0.005+25/0.001}=\frac{100}{55{,}000}=\boxed{1.818\times10^{-3}\ \text{m/s}}.$$
2. Head at each internal boundary. Head loss in a layer is $q\,L_i/K_i$: top layer loses $1.818\times10^{-3}\times25{,}000=45.45$ m, giving $h_{1/2}=180-45.45=\boxed{134.5\ \text{m}}$ at the top/middle boundary; the middle layer then loses $1.818\times10^{-3}\times5{,}000=9.09$ m, giving $h_{2/3}=134.5-9.09=\boxed{125.5\ \text{m}}$ at the middle/bottom boundary (check: $125.5-45.45=80.0$ m at the base, matching the given bottom head).
3. Pressure at the boundaries. With the base of the stack as datum ($z=0$), the top/middle boundary sits at $z=50$ m and the middle/bottom boundary at $z=25$ m, so the pressure head is $\psi=h-z$: $$p_{1/2}=\rho_wg(h_{1/2}-50)=1000(9.81)(134.5-50)=\boxed{829\ \text{kPa}},$$ $$p_{2/3}=\rho_wg(h_{2/3}-25)=1000(9.81)(125.5-25)=\boxed{985\ \text{kPa}}.$$

4. Part (b) — total head at the freshwater well. Taking the bottom of the aquitard as datum ($z=0$, top at $z=20$ m), the freshwater well is screened at the top with a 12 m column: $h_1=z_{\text{top}}+\psi_1=20+12=\boxed{32.0\ \text{m}}$.
5. Case (i) — saline density assumed 1000 kg/m³. Treating the 30 m saline column as if it were fresh water, $h_2=z_{\text{bot}}+\psi_2=0+30=\boxed{30.0\ \text{m}}$. Since $h_1(32.0)\ >\ h_2(30.0)$, water would appear to flow downward, from the fresh aquifer into the saline aquifer.
6. Case (ii) — true saline density used. Comparing heads across a density interface requires the freshwater-EQUIVALENT head at the saline point, $h_f=z+(\rho_s/\rho_f)\psi$: $$h_{f,2}=0+\left(\frac{1160}{1000}\right)(30)=\boxed{34.8\ \text{m}}.$$ Now $h_{f,2}(34.8)\ >\ h_1(32.0)$, so the true flow direction is upward — saline water rising into the freshwater aquifer — the OPPOSITE of the naive answer in case (i).

7. Part (c) — storativity and pumped volume. $S=S_sb=(2.5\times10^{-6})(35)=8.75\times10^{-5}$; plan area $A=2200\times2000=4.40\times10^6\ \text{m}^2$: $$\Delta V=S\,A\,\Delta H=(8.75\times10^{-5})(4.40\times10^6)(1.2)=\boxed{462\ \text{m}^3}.$$