Question 4 of 5: Steady Radial Flow to a Well — Confined and Unconfined Thiem Analysis, and a Leaky-Aquifer Drawdown

Question 4: Steady Radial Flow to a Well — Confined and Unconfined Thiem Analysis, and a Leaky-Aquifer Drawdown (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.

Find. (a) Hydraulic conductivity from the confined-aquifer Thiem equation. (b) Hydraulic conductivity from the unconfined-aquifer Thiem equation. (c) Leaky-aquifer drawdown, and whether including aquitard storage would make it larger or smaller.

Approach. Parts (a) and (b) are steady, two-point Thiem problems: confined flow is linear in $h$, unconfined flow is linear in $h^2$ (Dupuit). Part (c) computes the leakage factor $B=\sqrt{Tb'/K'}$ and evaluates the Hantush-Jacob leaky well function $W(u,r/B)$ by direct numerical integration of its defining integral (the exam's own printed $W(u,r/B)$ table does not reconcile with the closed-form steady-state asymptote $W(0,r/B)=2K_0(r/B)$, so the integral is evaluated directly rather than read off the table; see the check note below).

1. Part (a) — confined Thiem equation. $Q=16\ \text{L/s}=0.016\ \text{m}^3/\text{s}$, $r_w=0.12\ \text{m}$, $r_2=150\ \text{m}$, $b=60\ \text{m}$: $$K=\frac{Q\ln(r_2/r_w)}{2\pi b(h_2-h_1)}=\frac{(0.016)\ln(150/0.12)}{2\pi(60)(110-105)}=\boxed{6.05\times10^{-5}\ \text{m/s}}.$$
2. Part (b) — unconfined Thiem equation. $Q=12\ \text{L/s}=0.012\ \text{m}^3/\text{s}$, $r_w=0.125\ \text{m}$, $r_2=150\ \text{m}$, The unconfined Thiem equation needs saturated thicknesses measured from the aquifer base, not elevations above sea level (102 m and 105 m cannot be thicknesses of a 50 m aquifer). The question gives no base elevation, so assume the aquifer is fully saturated (50 m) at the 150 m observation point: base $=105-50=55$ m.a.s.l., giving $h_2=50\ \text{m}$ and $h_1=102-55=47\ \text{m}$: $$K=\frac{Q\ln(r_2/r_w)}{\pi(h_2^2-h_1^2)}=\frac{(0.012)\ln(150/0.125)}{\pi(50^2-47^2)}=\frac{(0.012)(7.090)}{\pi(291)}=\boxed{9.31\times10^{-5}\ \text{m/s}}.$$ (Using the sea-level elevations directly as if they were thicknesses would give $4.36\times10^{-5}$ m/s, about half the correct value.)
3. Part (c) — leakage factor. $T=10^{-2}\ \text{m}^2/\text{s}$, $b'=5.5\ \text{m}$, $K'=10^{-6}\ \text{cm/s}=10^{-8}\ \text{m/s}$: $$B=\sqrt{\frac{Tb'}{K'}}=\sqrt{\frac{(10^{-2})(5.5)}{10^{-8}}}=\boxed{2345\ \text{m}},\qquad \frac{r}{B}=\frac{150}{2345}=0.0640.$$
4. Dimensionless time parameter. $S=10^{-4}$, $t=24\ \text{h}=86{,}400\ \text{s}$: $$u=\frac{r^2S}{4Tt}=\frac{150^2(10^{-4})}{4(10^{-2})(86{,}400)}=\boxed{6.51\times10^{-4}}.$$
5. Leaky well function and drawdown. Numerically integrating $W(u,r/B)=\displaystyle\int_u^\infty\frac{1}{y}\exp\!\left[-y-\frac{(r/B)^2}{4y}\right]dy$ gives $W(6.51\times10^{-4},\,0.0640)=5.649$, so: $$s=\frac{Q}{4\pi T}W(u,r/B)=\frac{0.018}{4\pi(10^{-2})}(5.649)=\boxed{0.809\ \text{m}}.$$ (For reference, treating the same aquifer as non-leaky — Theis with the same $u$ — would give $s_{\text{Theis}}=0.968\ \text{m}$; leakage measurably reduces the drawdown, as expected.)
6. Effect of aquitard storage. The calculation above uses the Hantush-Jacob assumption that the aquitard stores no water, so leakage into the aquifer is simply $K'(s/b')$ and comes only from the overlying source bed. If the aquitard's own storativity is significant, it becomes an additional source: as the aquifer head falls, water is released elastically from storage in the aquitard, and while the drawdown front is still diffusing up through the aquitard the head gradient at its base (and hence the flux into the aquifer) is steeper than the linear $s/b'$ profile assumed above. More water supplied at the same pumping rate means less water taken from aquifer storage, so the drawdown would be smaller than the 0.809 m calculated here (Hantush's 1960 modified leaky theory; Neuman & Witherspoon).