Question 4 of 5: Confined-Aquifer Drawdown — Theis, then a Leaky Aquitard (Hantush-Jacob)

Question 4: Confined-Aquifer Drawdown — Theis, then a Leaky Aquitard (Hantush-Jacob) (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 $Q=15$ L/s $=0.0150\ \text{m}^3/\text{s}$, $T=10^{-2}\ \text{m}^2/\text{s}$, $S=10^{-4}$; observation well $r=150$ m, pumping time $t=24$ h. Aquitard thickness $b'=5$ m, aquitard $K'=10^{-6}$ cm/s in part (b). (The aquifer's own hydraulic conductivity, $10^{-3}$ cm/s, is not needed for the drawdown calculation — $T$ and $S$ already fully characterize the confined response; it would only be relevant to back out the aquifer's thickness, $b=T/K\approx1000$ m, which the problem does not ask for.)

Find. (a) Theis drawdown with an impermeable aquitard. (b) Hantush-Jacob leaky-aquifer drawdown with the given aquitard conductivity. (c) whether a storative aquitard would increase or decrease the drawdown further.

Approach. (a) is the standard non-leaky Theis solution. (b) replaces the Theis well function $W(u)$ with the leaky Hantush-Jacob well function $W(u,r/B)$, where $B=\sqrt{Tb'/K'}$ is the leakage factor; both are evaluated here by direct numerical integration of their defining integrals (equivalent to, but more precise than, reading the printed $W(u)$ / $W(u,r/B)$ tables).

1. Part (a) — Theis drawdown, impermeable aquitard. With $t=24\ \text{h}=86{,}400$ s: $$u=\frac{r^2S}{4Tt}=\frac{(150)^2(10^{-4})}{4(10^{-2})(86{,}400)}=\boxed{6.51\times10^{-4}},$$ and from the exponential-integral well function $W(u)=6.760$: $$s=\frac{Q}{4\pi T}W(u)=\frac{0.0150}{4\pi(0.01)}(6.760)=\boxed{0.807\ \text{m}}.$$
2. Part (b) — leakage factor. Converting $K'=10^{-6}\ \text{cm/s}=10^{-8}\ \text{m/s}$: $$B=\sqrt{\frac{Tb'}{K'}}=\sqrt{\frac{(0.01)(5)}{10^{-8}}}=\boxed{2236\ \text{m}},\qquad \frac{r}{B}=\frac{150}{2236}=\boxed{0.0671}.$$
3. Leaky drawdown. $u$ is unchanged (it depends only on the aquifer's own $T$, $S$), so with $W(u,r/B)=5.572$: $$s'=\frac{Q}{4\pi T}W(u,r/B)=\frac{0.0150}{4\pi(0.01)}(5.572)=\boxed{0.665\ \text{m}}.$$ The leaky drawdown (0.665 m) is smaller than the impermeable-aquitard Theis drawdown (0.807 m), because leakage through the aquitard supplies part of the pumped water directly from the overlying source bed rather than drawing it all from the aquifer's own confined storage.
4. Part (c) — effect of a storative aquitard. If the aquitard itself had significant storativity, it would also release water from its own elastic storage as its head declined (in addition to steadily leaking water through at the rate the Hantush-Jacob solution already assumes). That extra, delayed source of water further reduces the load on the aquifer's own storage and transmissivity, so the drawdown would be smaller than the 0.665 m computed in part (b) (this is the physical basis of the more general Hantush "leaky aquifer with aquitard storage" / Neuman-Witherspoon solution, which always predicts less drawdown at a given time than the storage-free leaky solution used here).