Question 6 of 6: Unconfined Aquifer — Well Drawdown & Tracer Travel Time

Question 6: Unconfined Aquifer — Well Drawdown & Tracer Travel Time (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) drawdown at $r_1$ and $r_2$; (b) travel time for a conservative tracer from $r_1$ to $r_2$ (or $r_2$ to $r_1$, the direction of flow) at this pumping rate.

Approach. Apply the Dupuit–Forchheimer steady, unconfined, radial form of Darcy's law between the far-field radius of influence and each observation well to get the saturated thickness (hence drawdown) at $r_1$ and $r_2$, then integrate the porosity-corrected (actual) seepage velocity along the radial flow path between the two wells to get the travel time.

1. Governing equation. Steady radial (Dupuit) unconfined flow between the far-field boundary and radius $r$ gives $$Q=\frac{\pi K\left(h_0^2-h(r)^2\right)}{\ln(R/r)}\ \Rightarrow\ h(r)=\sqrt{h_0^2-\frac{Q}{\pi K}\ln\!\left(\frac{R}{r}\right)}.$$
2. Part (a) — drawdown at $r_1$ and $r_2$. With $K=4320\ \text{m/day}$, $h_0=7\ \text{m}$, $R=4000\ \text{m}$, $Q=100\ \text{m}^3/\text{day}$: $$h(10)=6.99685\ \text{m}\ \Rightarrow\ s_1=h_0-h(10)=\boxed{3.15\ \text{mm}},$$ $$h(50)=6.99769\ \text{m}\ \Rightarrow\ s_2=h_0-h(50)=\boxed{2.31\ \text{mm}}.$$
3. Part (b) — tracer travel time. The actual (seepage) velocity at radius $r$ is the Darcy (superficial) velocity divided by porosity, $v(r)=\dfrac{Q}{2\pi r\,h(r)\,n}$, so the travel time between $r_1$ and $r_2$ is $$t=\int_{r_1}^{r_2}\frac{2\pi n\,r\,h(r)}{Q}\,dr.$$ Evaluating this integral numerically (with $h(r)$ from Step 1) gives $$t=\boxed{184.7\ \text{days}}\ (\approx6.1\ \text{months}).$$