Question 1 of 5: Confined-Aquifer Storage, Three-Point Hydraulic Gradient, and Density-Driven Flow Across an Aquitard

Question 1: Confined-Aquifer Storage, Three-Point Hydraulic Gradient, and Density-Driven Flow Across an Aquitard (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 volume of water released by a 2 m piezometric decline. (b) The magnitude and direction of the hydraulic gradient from the three water levels. (c) The direction of flow across the aquitard under the two stated density assumptions.

Approach. Part (a) converts specific storativity to a dimensionless storativity ($S=S_sb$) and multiplies by plan area and head decline. Part (b) sets up a local $x$–$y$ (east–north) coordinate system at Well A and solves the two independent one-dimensional gradients (A→C for the east component, A→B for the north component) as a vector. Part (c) converts each well's pressure-head reading to a common freshwater-equivalent total head using $h_f=z+(\rho/\rho_f)\psi$, once with the saline water mis-assumed fresh (case i) and once with its true density (case ii), and compares the two heads to find the flow direction in each case.

1. Part (a) — storativity and released volume. $S=S_sb=(1.1\times10^{-6})(55)=6.05\times10^{-5}$ (dimensionless). Over the plan area $A=1500\times2400=3.6\times10^6\ \text{m}^2$: $$V=S\cdot A\cdot\Delta h=(6.05\times10^{-5})(3.6\times10^6)(2)=\boxed{435.6\ \text{m}^3}.$$
2. Part (b) — gradient components. Placing Well A at the origin, Well B is at $(0,250)$ (north) and Well C at $(250,0)$ (east). Each is a one-dimensional gradient along its own axis: $$\frac{\partial h}{\partial y}\bigg|_{\text{north}}=\frac{196-180}{250}=0.0640,\qquad \frac{\partial h}{\partial x}\bigg|_{\text{east}}=\frac{188-180}{250}=0.0320.$$
3. Magnitude and direction. The gradient vector's magnitude and bearing (measured clockwise from north) are: $$|\nabla h|=\sqrt{0.0640^2+0.0320^2}=\boxed{0.0716\ (7.16\%)},\qquad \theta=\tan^{-1}\!\left(\frac{0.0320}{0.0640}\right)=\boxed{N26.6^{\circ}E}\ \text{(direction of rising head)}.$$ Groundwater flows down-gradient, i.e. exactly opposite this bearing: $\boxed{S26.6^{\circ}W}$, from the Well-B corner of the triangle toward the Well-A/C side.
4. Part (c) — freshwater-equivalent heads. Take the top of the aquitard as datum ($z_1=0$) so the bottom sits at $z_2=-22\ \text{m}$. The fresh-water well head is unaffected by the density conversion (its own fluid is already the reference fluid): $$h_{f,1}=z_1+\psi_1=0+10=\boxed{10.00\ \text{m}}.$$ For the saline well, the freshwater-equivalent head is $h_f=z+(\rho/\rho_f)\psi$, applied twice with the two stated densities: $$\text{(i) }\rho=998:\ h_{f,2}=-22+\left(\frac{998}{998}\right)(30)=\boxed{8.00\ \text{m}}$$ $$\text{(ii) }\rho=1150:\ h_{f,2}=-22+\left(\frac{1150}{998}\right)(30)=\boxed{12.57\ \text{m}}.$$
5. Direction in each case. Case (i): $h_{f,1}=10.00\ \text{m} > h_{f,2}=8.00\ \text{m}$, so flow is downward, from the fresh aquifer into the saline aquifer. Case (ii): $h_{f,2}=12.57\ \text{m} > h_{f,1}=10.00\ \text{m}$ — the ranking flips, and flow is upward, from the saline aquifer into the fresh aquifer, even though the denser fluid sits below.