Well Hydraulics in an Unconfined Aquifer: 18-Env-A3 Geotechnical and Hydrogeological Engineering, May 2014, Q6

Question 6: 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.

Given data
Quantity	Symbol	Value
Well diameter	$2r_w$	30 cm
Bedrock depth below ground	$b$	20 m
Static water table depth below ground	—	2 m
Max. allowable drawdown at the well	$s_{max}$	2 m
Hydraulic conductivity	$K$	20 m/day
Porosity	$n$	0.35
Observation well radii	$r_1$, $r_2$	10 m, 100 m

Figure 4 — radial cone of depression to an unconfined pumping well, Dupuit–Thiem geometry.

Find. (a) maximum sustainable discharge $Q$; (b) travel time for a conservative tracer from $r_1$ to $r_2$.

Approach. Use the Thiem equation for steady radial flow to a well in an unconfined aquifer between the well itself and the far observation well (assumed, for lack of a stated radius of influence, to sit close enough to the undisturbed static head that $h(r_2)\approx h_0$) to get $Q$; then integrate the Darcy seepage velocity $v(r)=Q/(2\pi r\,h(r)\,n)$ along $r$ between the two observation wells for the travel time.

Part (a) — saturated thicknesses. The aquifer extends from the water table down to bedrock. Static (pre-pumping) saturated thickness: $$h_0=b-2=20-2=18\ \text{m}.$$ At the well, the head is drawn down by the allowable $s_{max}=2$ m: $$h_w=h_0-s_{max}=18-2=16\ \text{m},\qquad r_w=0.15\ \text{m}.$$
Apply the Thiem equation between the well and $r_2$. Taking $r_2=100$ m as (approximately) the point where drawdown has become negligible, $h(r_2)\approx h_0=18$ m: $$Q=\frac{\pi K\left(h_0^2-h_w^2\right)}{\ln(r_2/r_w)}=\frac{\pi\times20\times(18^2-16^2)}{\ln(100/0.15)}=\frac{\pi\times20\times68}{6.502}=\boxed{657\ \text{m}^3/\text{day}}.$$
Part (b) — head profile $h(r)$ between the observation wells. Re-arranging Thiem between $r_w$ and a general radius $r$, $$h(r)=\sqrt{h_w^2+\frac{Q}{\pi K}\ln(r/r_w)}\ \Rightarrow\ h(r_1{=}10)=17.32\ \text{m},\ \ h(r_2{=}100)=18.00\ \text{m (consistent)}.$$
Integrate the seepage velocity for travel time. By continuity the Darcy flux at radius $r$ is $q(r)=Q/(2\pi r\,h(r))$, and the interstitial (seepage) velocity is $v(r)=q(r)/n$. Travel time is $$t=\int_{r_1}^{r_2}\frac{dr}{v(r)}=\int_{r_1}^{r_2}\frac{2\pi n\,r\,h(r)}{Q}\,dr.$$ Evaluating this integral numerically with the $h(r)$ from Step 3 (Q, K, n as above): $$t=\boxed{296\ \text{days}\ (\approx0.81\ \text{yr})}.$$

Check: the problem does not state a radius of influence, so $r_2=100$ m is taken as the point of negligible drawdown ($h(r_2)\approx h_0$) for part (a) — a common simplification when no other outer boundary is given, and consistent with $r_2$ being the far observation well in the same figure. A materially larger true radius of influence would raise $Q$ slightly (the $\ln$ dependence makes the answer fairly insensitive to this choice).

Quantity	Value
$h_0$ (static saturated thickness)	18 m
$h_w$ (at well, max drawdown)	16 m
(a) Maximum discharge, $Q$	657 m³/day
$h(r_1)$	17.32 m
(b) Tracer travel time, $r_1\to r_2$	≈ 296 days (0.81 yr)

Concept & Theory

Topic. Steady radial flow to a well in an unconfined (water-table) aquifer — the Dupuit–Thiem equation — and seepage (interstitial) velocity for contaminant/tracer transport.

Key relations. $Q=\pi K(h_2^2-h_1^2)/\ln(r_2/r_1)$ (Thiem, unconfined); Darcy flux $q=Q/(2\pi r h)$; seepage velocity $v=q/n$ (n.b. NOT the Darcy flux itself — porosity slows the true pore-water/tracer velocity down).

Why this works. The Dupuit assumption (flow lines essentially horizontal, equipotentials vertical) reduces an inherently 2-D unconfined flow field to a 1-D radial ODE that integrates directly to the Thiem relation; continuity then fixes the local flux at any radius from the same constant $Q$, and dividing by porosity converts the bulk Darcy flux into the actual average speed of a water parcel (and hence a conservative tracer) moving through the pores.

Common pitfall. Using the Darcy flux $q$ directly as the tracer's travel speed instead of dividing by porosity $n$ (understating travel time by a factor of $1/n\approx2.9$ here), and forgetting that $h(r)$ — not a constant average thickness — must be used inside the flux at each radius for an unconfined aquifer.

Source. Freeze & Cherry, Groundwater (1979), ch. 2 (Darcy's law, seepage velocity) & ch. 8 (well hydraulics); Das, Principles of Geotechnical Engineering, ch. 7 (permeability).

Question 6 of 6: Well Hydraulics in an Unconfined Aquifer

Question 6: Well Hydraulics in an Unconfined Aquifer (20 marks)