Question 6 of 6: Steady-State Well Discharge — Unconfined vs. Confined Aquifer

Question 6: Steady-State Well Discharge — Unconfined vs. Confined 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. Unconfined aquifer, medium sand, bedrock (impermeable base) at 10 m below ground. Monitoring well 1: $r_1=75$ m, water level $6$ m below ground $\Rightarrow$ saturated thickness $h_1=10-6=4$ m. Monitoring well 2: $r_2=96$ m, water level $5$ m below ground $\Rightarrow$ $h_2=10-5=5$ m. No hydraulic conductivity is given in the source; medium sand is assigned a representative $K=10\ \text{m/day}$ (mid-range of the textbook value for medium sand, see the check note below).

Find. (a) Steady-state well discharge $Q$. (b) The assumptions this estimate relies on. (c) $Q$ if the same drawdown pattern instead occurred in a 3 m thick confined (clay-capped) sand aquifer.

Approach. Apply the Thiem equation for steady radial flow to a fully-penetrating well between the two monitoring wells, using the unconfined form (which depends on $h^2$, since the saturated thickness itself varies with radius) for part (a), and the confined form (which depends on $h$ linearly, since the aquifer thickness $b$ is now fixed by the overlying clay) for part (c).

1. (a) Unconfined Thiem equation between the two monitoring wells. $$Q=\frac{\pi K\left(h_2^2-h_1^2\right)}{\ln(r_2/r_1)}=\frac{\pi(10)\left(5^2-4^2\right)}{\ln(96/75)}=\frac{\pi(10)(9)}{0.2469}=\boxed{1{,}145\ \text{m}^3/\text{day}\ (13.3\ \text{L/s})}$$
2. (c) Confined Thiem equation with fixed thickness $b=3$ m. The same drawdown pattern ($h_1=4$ m, $h_2=5$ m, measured now as piezometric heads above the aquifer base) but a confined aquifer only $b=3$ m thick: $$Q=\frac{2\pi K b\left(h_2-h_1\right)}{\ln(r_2/r_1)}=\frac{2\pi(10)(3)(5-4)}{0.2469}=\boxed{764\ \text{m}^3/\text{day}\ (8.84\ \text{L/s})}$$ Discussion: confining the aquifer to a fixed 3 m thickness reduces the discharge to $764/1{,}145=66.7\%$ of the unconfined value, exactly matching the purely geometric ratio $2b/(h_1+h_2)=2(3)/9=0.667$ — this ratio is independent of $K$, so the direction and size of the change would hold even if a different (equally defensible) conductivity assumption had been used.

(b) Assumptions required for part (a). The Thiem solution assumes: steady-state (equilibrium) radial flow has been reached, so drawdowns are no longer changing with time; the aquifer is homogeneous and isotropic with a constant hydraulic conductivity $K$ throughout the zone between the two monitoring wells; the pumping well fully penetrates the aquifer and flow is purely horizontal and radially symmetric about it (the Dupuit–Forchheimer assumption, which also neglects the small vertical flow components that really occur near a partially-penetrating or near-well zone); there is no recharge, leakage or areal pumping interference within the radius of influence; and the bedrock base is truly horizontal and impermeable, as stated.