Well Drawdown in a Confined Aquifer — Fully Confined vs. Leaky: 18-Geol-A2 Hydrogeology, December 2016, Q4

Question 4: Well Drawdown in a Confined Aquifer — Fully Confined vs. Leaky (equal value)

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. Confined aquifer, transmissivity $T=248\ \text{m}^2/\text{day}$, storativity $S=0.0002$; new well pumped at $Q=2.8\ \text{m}^3/\text{min} = 4032\ \text{m}^3/\text{day}$, continuously, for $t=30\ \text{days}$. (b) The confining bed above the aquifer is $b'=3\ \text{m}$ thick with vertical hydraulic conductivity $K'=0.05\ \text{m/day}$ and negligible storage of its own (steady leakage, no aquitard storage release).

Quantity	Value
T	248 m²/day
S	0.0002
Q	2.8 m³/min = 4032 m³/day
t	30 days
Confining layer (b) b', K'	3 m, 0.05 m/day

Find. Drawdown $s$ at $r=50\ \text{m}$ and $r=150\ \text{m}$ after 30 days of continuous pumping, (a) for the fully confined aquifer and (b) for the same aquifer leaking through the overlying confining bed.

Figure 4. Radial cross-section: (a) fully confined aquifer drawdown cone (Theis, solid red) vs. (b) leaky-aquifer drawdown (Hantush–Jacob, dashed green) — vertical leakage down through the confining bed partly replenishes the pumped aquifer, so the leaky cone is shallower and narrower at the same $r$ and $t$.

Approach. (a) Apply the Theis nonequilibrium equation $s=\dfrac{Q}{4\pi T}W(u)$ with $u=r^2S/(4Tt)$, evaluating the exponential-integral well function $W(u)$. (b) Apply the Hantush–Jacob leaky-aquifer equation, which replaces $W(u)$ with the leaky well function $W(u,r/B)$, using the same $u$ but a leakage factor $B=\sqrt{Tb'/K'}$ built from the confining bed's properties.

Convert the pumping rate and set up $u$. $Q = 2.8\ \text{m}^3/\text{min}\times1440\ \text{min/day} = 4032\ \text{m}^3/\text{day}$. For $r=50\ \text{m}$: $$u_{50}=\frac{r^2S}{4Tt}=\frac{(50)^2(0.0002)}{4(248)(30)}=1.680\times10^{-5};\qquad u_{150}=\frac{(150)^2(0.0002)}{4(248)(30)}=1.512\times10^{-4}.$$
(a) Theis drawdown. $W(u)=-0.5772-\ln u + u - \cdots$ (exponential integral $E_1(u)$); for these small $u$, $W(u_{50})=10.417$ and $W(u_{150})=8.220$. With $Q/(4\pi T)=4032/(4\pi\cdot248)=1.293\ \text{m}\cdot\text{day}$: $$s_{50}=(1.293)(10.417)=\boxed{13.48\ \text{m}},\qquad s_{150}=(1.293)(8.220)=\boxed{10.63\ \text{m}}.$$
(b) Leakage factor for the aquitard. With no aquitard storage, the Hantush–Jacob leakage factor is $$B=\sqrt{\frac{Tb'}{K'}}=\sqrt{\frac{(248)(3)}{0.05}}=\boxed{121.98\ \text{m}}.$$ So $r/B$ is $50/121.98=0.410$ at 50 m and $150/121.98=1.230$ at 150 m.
Leaky well function and drawdown. $W(u,r/B)=\displaystyle\int_u^\infty \frac{1}{y}\exp\!\left[-y-\frac{(r/B)^2}{4y}\right]dy$, evaluated numerically: $W(u_{50},0.410)=2.186$ and $W(u_{150},1.230)=0.612$. Applying the same prefactor: $$s_{50}=(1.293)(2.186)=\boxed{2.83\ \text{m}},\qquad s_{150}=(1.293)(0.612)=\boxed{0.79\ \text{m}}.$$ Leakage down through the confining bed feeds the aquifer as it is pumped, so both drawdowns are far smaller than the fully-confined case — most dramatically at 150 m, where the cone has almost stabilized (0.79 m vs. 10.63 m).

Case	r = 50 m	r = 150 m
(a) Fully confined (Theis), $u$	$1.68\times10^{-5}$	$1.51\times10^{-4}$
(a) Fully confined, $W(u)$	10.417	8.220
(a) Drawdown, 30 days	13.48 m	10.63 m
(b) Leaky, $r/B$ (B = 121.98 m)	0.410	1.230
(b) Leaky, $W(u,r/B)$	2.186	0.612
(b) Drawdown, 30 days	2.83 m	0.79 m

Concept & Theory

Topic. Transient radial flow to a pumping well — Theis (non-leaky confined) vs. Hantush–Jacob (leaky confined) well hydraulics.

Key relations. Theis: $s=\dfrac{Q}{4\pi T}W(u)$, $u=\dfrac{r^2S}{4Tt}$, $W(u)=E_1(u)$. Hantush–Jacob (no aquitard storage): $s=\dfrac{Q}{4\pi T}W(u,r/B)$, leakage factor $B=\sqrt{Tb'/K'}$, and $W(u,r/B)\to W(u)$ as $B\to\infty$ (no leakage) while $W(u,r/B)$ approaches a finite steady-state value $2K_0(r/B)$ as $t\to\infty$ (Bessel function), so leaky drawdown stabilizes instead of growing indefinitely.

Why this works. Both solutions come from solving the same radial diffusion equation for $s(r,t)$, but the leaky case adds a source term equal to the vertical leakage rate through the aquitard, $K'(s/b')$ per unit area, driven by the drawdown itself. Physically, once the pumped aquifer is drawn down relative to the water table/overlying unit, water starts leaking down through the aquitard to replace what the well removes — the smaller $B$ is (thin, leaky aquitard), the sooner and more strongly this recharge suppresses the cone of depression, exactly as seen comparing $r/B=0.41$ vs. the non-leaky case here.

Common pitfall. Using the confined storativity $S$ for the aquitard in the leakage factor — $B$ uses only $T$, $b'$, and $K'$ because the problem states the aquitard has no storage of its own (an instantaneous, steady leakage assumption); mixing in an aquitard $S'$ would require the more general Hantush (with storage) solution instead. Also, forgetting to convert the pumping rate to consistent units (m³/min → m³/day) before combining with $T$ in m²/day.

Source. Freeze & Cherry, Groundwater, Ch. 8.2 (Theis nonequilibrium solution) and Ch. 8.4 (leaky aquifers, Hantush–Jacob method).

Question 4 of 5: Well Drawdown in a Confined Aquifer — Fully Confined vs. Leaky

Question 4: Well Drawdown in a Confined Aquifer — Fully Confined vs. Leaky (equal value)