Question 5 of 6: Well Hydraulics in an Unconfined Aquifer

Question 5: Well Hydraulics in an Unconfined Aquifer (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 $s_1$, $s_2$ at $r_1=50$ m and $r_2=500$ m; (b) travel time for a conservative tracer between $r_1$ and $r_2$.

Approach. Use the Thiem/Dupuit equation for steady unconfined radial flow between each observation point and the known boundary condition at $R$, then integrate the radial seepage (interstitial) velocity from the Thiem head field to get travel time.

1. Part (a) — drawdown via the unconfined Thiem equation. Between the boundary ($h=H$ at $r=R$) and any point $r$: $$h(r)=\sqrt{H^2-\frac{Q}{\pi k}\ln\frac{R}{r}}\ ,\qquad s(r)=H-h(r).$$ $$h(50)=\sqrt{36-\frac{500}{\pi(1728)}\ln\frac{2000}{50}}=\sqrt{36-0.3397}=5.9716\ \text{m}\ \Rightarrow\ \boxed{s_1=2.84\ \text{cm}}.$$ $$h(500)=\sqrt{36-\frac{500}{\pi(1728)}\ln\frac{2000}{500}}=\sqrt{36-0.1277}=5.9894\ \text{m}\ \Rightarrow\ \boxed{s_2=1.07\ \text{cm}}.$$
2. Part (b) — travel time. The radial seepage (interstitial) velocity at distance $r$ is $v(r)=\dfrac{Q}{2\pi r\,h(r)\,n_e}$, so the travel time between $r_1$ and $r_2$ is $$t=\int_{r_1}^{r_2}\frac{2\pi n_e\,r\,h(r)}{Q}\,dr.$$ Because the drawdown is tiny across this range ($h\approx H=6$ m throughout, varying by <0.03 m), this integrates essentially in closed form as $$t\approx\frac{\pi n_e H\left(r_2^2-r_1^2\right)}{Q}=\frac{\pi(0.30)(6)(500^2-50^2)}{500}=\boxed{2792\ \text{days}\ (7.65\ \text{yr})},$$ confirmed to within 0.3% by numerically integrating with the exact (non-constant) $h(r)$.