Question 3 of 5: Theis and Hantush-Jacob Leaky Drawdown, and a Regional-Gradient Aquifer-Sufficiency Check

Question 3: Theis and Hantush-Jacob Leaky Drawdown, and a Regional-Gradient Aquifer-Sufficiency Check (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. (a) $T=250\ \text{m}^2\text{/day}$, $S=0.0002$, $Q=2.5\ \text{m}^3\text{/min}$, continuous pumping for $t=25$ days; drawdown wanted at $r=50$ m and $r=150$ m. (b) Same aquifer, now leaky through a $b'=3$ m confining layer with $K'=0.04\ \text{m/day}$, no aquitard storage; same $Q$, $t$, and observation distances. (c) two wells 1.2 km apart, depth to water 20 m and 26.5 m; $K=10^{-3}\ \text{cm/s}$, saturated thickness $b=22$ m, aquifer extent 5 km perpendicular to the well-to-well line; daily demand 1000 m³.

Find. (a) Theis drawdown at $r=50$ m and 150 m. (b) Hantush-Jacob leaky drawdown at the same two distances. (c) whether the regional Darcy discharge through the aquifer's cross-section meets the daily demand.

Approach. Part (a) applies the Theis nonequilibrium equation directly. Part (b) adds aquitard leakage through the Hantush-Jacob well function $W(u,r/B)$, with the leakage factor $B=\sqrt{Tb'/K'}$; $W(u,r/B)$ is evaluated by direct numerical integration of its defining integral rather than read off a coarse printed table (the printed Table 2.3 in this exam is exactly that kind of table). Part (c) is a natural-gradient Darcy discharge estimate: transmissivity from the measured $K$ and saturated thickness, gradient from the two depths-to-water, and total flow from $Q=Ti\,(\text{width})$, compared against the stated demand.

1. Part (a) — Theis drawdown. $u=r^2S/(4Tt)$ and $s=(Q/4\pi T)\,W(u)$, with $Q=2.5\times1440=3600\ \text{m}^3\text{/day}$: $$r=50\ \text{m}:\quad u=\frac{50^2(0.0002)}{4(250)(25)}=2.00\times10^{-5},\quad W(u)=10.24,\quad s=\frac{3600}{4\pi(250)}(10.24)=\boxed{11.7\ \text{m}}.$$ $$r=150\ \text{m}:\quad u=1.80\times10^{-4},\quad W(u)=8.05,\quad s=\boxed{9.22\ \text{m}}.$$

2. Part (b) — leakage factor and Hantush-Jacob drawdown. $B=\sqrt{Tb'/K'}=\sqrt{(250)(3)/0.04}=137\ \text{m}$; with the same $u$ values as part (a): $$r=50\ \text{m}:\quad r/B=0.365,\quad W(u,r/B)=2.39,\quad s=\frac{3600}{4\pi(250)}(2.39)=\boxed{2.74\ \text{m}}.$$ $$r=150\ \text{m}:\quad r/B=1.095,\quad W(u,r/B)=0.736,\quad s=\boxed{0.843\ \text{m}}.$$ Both drawdowns are much smaller than the fully-confined case — leakage through the confining layer recharges the pumped aquifer as the cone of depression develops, exactly as expected physically.

3. Part (c) — transmissivity and regional gradient. Converting $K=10^{-3}\ \text{cm/s}=8.64\times10^{-1}\ \text{m/day}$, so $T=Kb=(0.864)(22)=19.0\ \text{m}^2\text{/day}$. The hydraulic gradient from the two depths-to-water (assuming a common well-collar datum) is $i=(26.5-20)/1200=5.42\times10^{-3}$.
4. Regional discharge and sufficiency check. With the aquifer's 5 km extent taken as the width perpendicular to flow: $$Q_{\text{avail}}=T\,i\,(\text{width})=(19.0)(5.42\times10^{-3})(5000)=\boxed{515\ \text{m}^3\text{/day}}.$$ Since $Q_{\text{avail}}(515\ \text{m}^3\text{/day})\ <\ Q_{\text{required}}(1000\ \text{m}^3\text{/day})$, the natural through-flow of this aquifer is not, by itself, sufficient to meet the subdivision's average daily demand — only about half of it.