Discharge and Tracer Travel Time, Unconfined Well: 18-Env-A3 Geotechnical and Hydrogeological Engineering, December 2018, Q4

Question 4: Discharge and Tracer Travel Time, Unconfined Well (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.

Fig. Q4 — radial flow to an unconfined well (schematic, not to scale): static water table 5 m below grade, impermeable shale at 10 m below grade, observation well at r₁ = 20 m, radius of influence R ≈ 500 m.

Given.

Given data
Quantity	Symbol	Value
Well diameter	—	20 cm ($r_w=0.10$ m)
Depth to aquifer base (impermeable shale)	—	10 m below grade
Static water-table depth	—	5 m below grade
Saturated thickness (static)	$H_0$	5 m
Observation-well distance	$r_1$	20 m
Porosity	$n$	0.30
Hydraulic conductivity	$K$	10 m/day
Max allowable drawdown at well	$s_{max}$	1 m
Radius of influence	$R$	500 m

Find. (a) maximum discharge $Q$ (L/s); (b) travel time for a conservative tracer from the observation well to the pumping well.

Approach. Use the unconfined (Dupuit–Thiem) equation with the well's own radius and the given radius of influence to get $Q$, then integrate the radially-varying seepage velocity from $r_1$ down to $r_w$ to get the travel time.

Part (a) — water level in the well. The 1 m allowable drawdown gives $h_w=H_0-s_{max}=5-1=4.00\ \text{m}$.
Thiem discharge. $$Q=\frac{\pi K\left(H_0^2-h_w^2\right)}{\ln(R/r_w)}=\frac{\pi\times10\times(5^2-4^2)}{\ln(500/0.10)}=\frac{282.7}{8.517}=33.19\ \text{m}^3/\text{day}.$$ Converting: $$Q=33.19\times\frac{1000}{86{,}400}=\boxed{0.384\ \text{L/s}}.$$
Part (b) — drawdown profile. The Thiem profile gives the saturated thickness at any radius: $h(r)^2=h_w^2+\dfrac{Q}{\pi K}\ln(r/r_w)$. At the observation well, $h(20)=4.65$ m (drawdown of only 0.35 m there, versus 1.00 m at the well — the cone flattens quickly with distance).
Travel time by direct integration. The tracer's local seepage velocity is $v(r)=\dfrac{Q}{n\cdot2\pi r\,h(r)}$, so the travel time from $r_1$ to $r_w$ is $$t=\int_{r_w}^{r_1}\frac{n\cdot2\pi r\,h(r)}{Q}\,dr,$$ evaluated numerically (Simpson's rule) using the $h(r)$ profile above: $$t=\boxed{52.1\ \text{days}}.$$

Check: a quick cross-check using the average saturated thickness $\bar h=(h_w+h(r_1))/2=4.32$ m in the closed-form $t\approx\pi n\bar h(r_1^2-r_w^2)/Q$ gives 49.1 days — about 6% below the full numerical integral. This is a wider gap than usual for the $\bar h$-approximation because the drawdown here (20% of $H_0$ at the well) is larger than in cases where that shortcut has previously checked to <1%; the numerically-integrated 52.1 days is reported as the primary answer.

Quantity	Value
(a) Maximum discharge, $Q$	33.19 m³/day = 0.384 L/s
(a) Water level in well, $h_w$	4.00 m
(b) Saturated thickness at r₁	4.65 m
(b) Tracer travel time	52.1 days

Concept & Theory

Topic. Steady radial flow to a well in an unconfined aquifer (Dupuit–Thiem).

Key relations. $Q=\pi K(H_0^2-h_w^2)/\ln(R/r_w)$; $h(r)^2=h_w^2+(Q/\pi K)\ln(r/r_w)$; seepage velocity $v(r)=Q/(n\cdot2\pi r\,h(r))$; travel time $t=\int n\cdot2\pi r\,h(r)/Q\,dr$.

Why this works. Dupuit's assumption (horizontal flow, vertical equipotentials) turns the unconfined radial-flow problem into a 1-D ODE in $r$ that integrates directly to the Thiem equation once boundary heads at $r_w$ and $R$ are fixed. The travel time integral is just "distance over local velocity," but the velocity itself varies with $r$ because both the flow area ($2\pi r\,h(r)$) and the saturated thickness change along the path — a uniform-velocity shortcut is only accurate when drawdown is small relative to $H_0$.

Common pitfall. Using $R$ (radius of influence, the outer no-drawdown boundary) instead of the actual pumping distance in the Thiem formula, and confusing the aquifer's Darcy (superficial) velocity with the tracer's true (interstitial) seepage velocity, which is faster by a factor of $1/n$.

Source. Freeze & Cherry, Groundwater (1979), Ch. 8 (well hydraulics, the Thiem equation, seepage velocity).

Question 4 of 6: Discharge and Tracer Travel Time, Unconfined Well

Question 4: Discharge and Tracer Travel Time, Unconfined Well (20 marks)