Question 2 of 7: Unconfined Aquifer on a Circular Island — Water-Table Sketches, Well Drawdown, and the Effect of Infiltration

Question 2: Unconfined Aquifer on a Circular Island — Water-Table Sketches, Well Drawdown, and the Effect of Infiltration (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.

a) Water-table profiles (sketch, no calculation). With no withdrawal and no infiltration (i) the aquifer is in equilibrium with the lake on every side, so the water table is simply flat at the lake level h₀ everywhere — no head gradient, no flow. Turning on withdrawal only (ii) draws a classic axisymmetric cone of depression centred on the well: the water table falls steeply close to the well (where the same discharge is squeezed through a small circumference) and flattens out approaching the lake edge, which acts as a constant-head boundary re-supplying the aquifer. Infiltration only (iii), with no well, has nowhere to leave the island except radially outward to the lake, so it builds a symmetric recharge mound that peaks at the centre of the island (r = 0) and falls smoothly to the lake level at the edge — the mirror image of (ii). With both effects together (iv) the two competing mechanisms produce a profile that starts near h₀ at the well (the local pumping drawdown is largely offset by recharge arriving from the whole island), rises to an interior mound where the outward-flowing recharge exceeds the water still being drawn toward the well, then falls back to h₀ at the lake edge — part (c) below locates that interior maximum quantitatively. Panel (iv) is therefore neither a pure cone nor a pure mound but a combination with one interior high point between the well and the shoreline.

Approach (b–d). Model the island as an unconfined, radially symmetric Dupuit aquifer bounded at r = R by a constant head h₀ (the lake); superpose a uniform areal recharge w with the well's point sink Q using the steady continuity balance that the net radial outflow crossing any circle of radius r equals the recharge collected inside that circle minus the well discharge.

1. (b) Thiem/Dupuit drawdown, no infiltration. For steady axisymmetric flow to a fully penetrating well in an unconfined aquifer with a fixed head h₀ at r = R, $$h_0^2 - h_w^2 = \frac{Q}{\pi K}\ln\!\left(\frac{R}{r_w}\right) = \frac{50}{\pi(10)}\ln\!\left(\frac{500}{0.25}\right) = 1.592\times 7.601 = 12.10\ \text{m}^2$$ so $h_w = \sqrt{25 - 12.10} = 3.592\ \text{m}$ and the drawdown at the well is $$s_w = h_0 - h_w = \boxed{1.41\ \text{m}}$$
2. (c) Locating the interior maximum with recharge. With uniform infiltration w and well discharge Q, the net radial flow crossing a circle of radius r (outward positive) is $Q_r(r) = w\pi r^2 - Q$: everything recharged inside r must either feed the well or flow on outward. The water table has zero slope, hence a local maximum, exactly where this net flow vanishes, independent of K or h₀: $$w\pi r_{max}^2 = Q \quad\Rightarrow\quad r_{max} = \sqrt{\frac{Q}{\pi w}} = \sqrt{\frac{50}{\pi(0.001)}} = \boxed{126\ \text{m from the well}}$$
3. (d) Drawdown at the well with recharge added. Integrating the Dupuit continuity balance $-2\pi r K h\,dh/dr = w\pi r^2 - Q$ from r out to R (h = h₀ at r = R) gives the combined profile $$h(r)^2 = h_0^2 + \frac{w}{2K}\left(R^2 - r^2\right) - \frac{Q}{\pi K}\ln\!\left(\frac{R}{r}\right)$$ (this correctly collapses to part (b) when w = 0, and to h = h₀ at r = R, the required boundary condition). Evaluating at the well radius, $$h_w^2 = 25 + \frac{0.001}{20}\left(500^2 - 0.25^2\right) - \frac{50}{\pi(10)}\ln\!\left(\frac{500}{0.25}\right) = 25 + 12.50 - 12.10 = 25.40\ \text{m}^2$$ $$h_w = 5.040\ \text{m} \quad\Rightarrow\quad s_w = h_0 - h_w = \boxed{-0.04\ \text{m (a slight mound, not a drawdown)}}$$ Physically, the recharge collected over the whole 500 m-radius island (12.5 m² of the $h^2$ budget) very nearly balances what the well removes (12.1 m²), so the water level at the well ends up almost exactly back at the undisturbed lake level, marginally higher.