Question 4 of 6: Unconfined Aquifer Well Yield & Source-Water Protection Area

Question 4: Unconfined Aquifer Well Yield & Source-Water Protection Area (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.

[Figure not reproduced: Figure 3 — unconfined aquifer with a single pumping well (schematic, redrawn from the source figure's setup; not to numeric scale). The underlying clay till (k ≈ 8.6×10 -4 m/day, five orders of magnitude below the sand) acts as the aquifer's effectively impermeable base. See the official exam paper.]

Find. (a) maximum sustainable discharge $Q$; (b) area protected by a 1-year time-of-travel capture zone around the well, at that discharge.

Approach. Use the Dupuit–Thiem equation for steady radial flow to a well in an unconfined aquifer, $Q=\pi K(h_0^2-h_w^2)/\ln(R/r_w)$. The source doesn't give an observation-well distance, so (per Note 1 on the exam, "state your assumptions") the radius of influence $R$ is estimated with Sichardt's empirical formula, standard practice for a single-well test with no observation data. Part (b) then uses a simple cylindrical-flow mass balance: the water pumped in one year must be supplied from the pore volume of a circular capture zone of the undisturbed aquifer thickness.

1. Part (a) — radius of influence (Sichardt's formula) and drawdown geometry. With $K=10\ \text{m/day}=1.157\times10^{-4}\ \text{m/s}$ and $s_w=7.5$ m, $$R=3000\,s_w\sqrt{K_{[\text{m/s}]}}=3000\times7.5\times\sqrt{1.157\times10^{-4}}=\boxed{242\ \text{m}}.$$ The saturated thickness at the well is $h_w=h_0-s_w=15-7.5=\boxed{7.5\ \text{m}}$.
2. Apply the Dupuit–Thiem equation. $$Q=\frac{\pi K\left(h_0^2-h_w^2\right)}{\ln(R/r_w)}=\frac{\pi\times10\times(15^2-7.5^2)}{\ln(242/0.10)}=\frac{\pi\times10\times168.75}{7.79}=\boxed{680\ \text{m}^3/\text{day}}.$$
3. Part (b) — 1-year capture-zone volume. The volume pumped in one year is $$V=Q\,t=680\times365=\boxed{248{,}400\ \text{m}^3}.$$ Idealizing the capture zone as a cylinder of the undisturbed aquifer thickness $h_0$ and porosity $n$ (radial flow, mass balance: pumped volume = pore volume drained from the zone), $$\text{Area}=\frac{V}{n\,h_0}=\frac{248{,}400}{0.25\times15}=\boxed{66{,}200\ \text{m}^2\ (\approx6.6\ \text{ha})},\qquad r=\sqrt{\frac{\text{Area}}{\pi}}=\boxed{145\ \text{m}}.$$ This radius (145 m) is comfortably inside the assumed 242 m radius of influence, as it must be — a useful check that the assumption in Step 1 is at least self-consistent.