Question 5 of 6: Transient Drawdown — Stepped Pumping, a River Boundary and a Leaky Aquitard
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2014 — 04-Geol-A2 Hydrogeology. Three-hour, open-book exam; any non-communicating calculator permitted. Five questions constitute a complete paper and all five are of equal value; most call for an essay-format answer with clarity and organization counted. Unless stated otherwise, water density is taken as 1000 kg/m³, water viscosity as 0.001 kg/m-sec, and g as 9.81 m/s². All six printed questions are solved below for completeness.
Reference texts: Freeze & Cherry, Groundwater (Prentice-Hall, 1979) — Darcy's law, the Theis and Thiem well equations, leaky-aquifer (Hantush-Jacob) theory, image-well boundary methods, and density-dependent flow; Todd & Mays, Groundwater Hydrology — unconfined-well hydraulics and dam-seepage (Dupuit-Forchheimer) solutions; EGBC Geoscience Professional Practice Guidelines for assumption-disclosure conventions on open-book calculations.
Question 5: Transient Drawdown — Stepped Pumping, a River Boundary and a Leaky Aquitard (equal value)
Given. Confined aquifer thickness $b=60$ m, $K=10^{-4}$ cm/s, specific storativity $S_s=1\times10^{-5}\ \text{m}^{-1}$, bounding aquitard 10 m thick. (a) Pumping schedule: $Q_1=5$ L/s for the first 12 h, then $Q_2=10$ L/s for the next 12 h, then shut-off; observation point $r=200$ m. (b) A river (constant-head boundary) 100 m north of the well; observation well 150 m east, at $Q_1=5$ L/s, $t=12$ h. (c) Leaky aquitard with $K'=10^{-9}$ m/s (negligible aquitard storage), same 150 m observation well, $t=12$ h, no river.
Symbol
Value
Transmissivity $T=Kb$
$6.00\times10^{-5}\ \text{m}^2/\text{s}$
Storativity $S=S_sb$
$6.00\times10^{-4}$
Find. (a) drawdown at $r=200$ m at $t=24$ h and $t=36$ h. (b) drawdown at the river-bounded observation well at $t=12$ h. (c) drawdown at the same well at $t=12$ h with the leaky aquitard instead of the river.
Approach. The pumping/shut-off schedule is handled by superposing Theis solutions of the well function $W(u)$, one term per rate change, each "clock" starting at the moment that change occurs. The river boundary is handled by the standard image-well method (a mirrored recharge well of equal rate on the far side of the boundary). The leaky aquitard uses the Hantush-Jacob leaky well function $W(u,r/B)$ with leakage factor $B=\sqrt{Tb'/K'}$, evaluated here by direct numerical integration of its defining integral (equivalent to reading the exam's supplied Table 5.2, which the same $u$ and $r/B$ locate).
Part (a) — superposition terms. The stepped/shut-off schedule is three superposed constant-rate wells at the same location: $+Q_1$ from $t=0$, $+(Q_2-Q_1)=+5$ L/s starting at $t=12$ h, and $-Q_2=-10$ L/s starting at $t=24$ h (the shut-off). Each term's drawdown uses $u=r^2S/(4Tt')$ and $s=\dfrac{Q}{4\pi T}W(u)$, with $t'$ the time elapsed since that term started.
Drawdown at $t=24$ h. Only the first two terms have started; with $r=200$ m, $u(24\,\text{h})=1.157\Rightarrow W=0.1695$ and $u(12\,\text{h})=2.315\Rightarrow W=0.0319$:
$$s(24\,\text{h})=\frac{Q_1}{4\pi T}W(1.157)+\frac{Q_2-Q_1}{4\pi T}W(2.315)=(6.63)(0.1695)+(6.63)(0.0319)=\boxed{1.34\ \text{m}}.$$
Drawdown at $t=36$ h. All three terms are now active, evaluated at their own elapsed times (36 h, 24 h and 12 h respectively; $u=0.772,1.157,2.315$ giving $W=0.3271,0.1695,0.0319$):
$$s(36\,\text{h})=(6.63)(0.3271)+(6.63)(0.1695)+(13.26)(-0.0319)=2.17+1.12-0.42=\boxed{2.87\ \text{m}}.$$
Drawdown is still rising 12 hours after shut-off, not yet recovering — at $r=200$ m the aquifer's characteristic response time $r^2S/4T\approx28$ h is longer than the 12 h since shut-off, so the pulse from the higher-rate second stage is still propagating outward and dominates the (much smaller) recovery contribution.
Part (b) — image well for the river. With the pumping well at the origin and the river along $y=100$ m, an image recharge well of the same rate $Q_1$ sits at the mirror point $(0,200)$. The 150 m-east observation well is $r_1=150$ m from the real well and $r_2=\sqrt{150^2+200^2}=250$ m from the image well; at $t=12$ h, $u_1=1.302\Rightarrow W=0.1350$ and $u_2=3.617\Rightarrow W=0.0060$:
$$s=\frac{Q_1}{4\pi T}\left[W(u_1)-W(u_2)\right]=(6.63)(0.1350-0.0060)=\boxed{0.855\ \text{m}}.$$
The image (recharge) well's drawdown is subtracted because the constant-head river boundary is simulated by injection, which partially offsets the real well's drawdown — the river-bounded drawdown (0.855 m) is smaller than it would be in an infinite aquifer.
Part (c) — leakage factor. Removing the river and instead allowing leakage through the aquitard ($b'=10$ m, $K'=10^{-9}$ m/s) introduces the leakage factor
$$\begin{aligned} B &=\sqrt{\frac{Tb'}{K'}}=\sqrt{\frac{(6.00\times10^{-5})(10)}{10^{-9}}}=\boxed{775\ \text{m}} \\ \frac{r}{B} &=\frac{150}{775}=0.194 \end{aligned}$$
Leaky drawdown at $t=12$ h. At the same $r=150$ m and $t=12$ h as before, $u=1.302$ (unchanged by the aquitard) and, from the Hantush-Jacob leaky well function at $u=1.302$, $r/B=0.194$: $W(u,r/B)=0.134$:
$$s=\frac{Q_1}{4\pi T}W(u,r/B)=(6.63)(0.134)=\boxed{0.891\ \text{m}}.$$
This is barely below the non-leaky Theis value at the same $u$ (0.895 m, part-b's real-well-only term) — at $u>1$ (i.e. this early in the test relative to the aquifer's response time) the aquitard has not yet had time to deliver meaningful leakage, so the leaky and non-leaky solutions have not yet diverged; leakage would suppress drawdown far more strongly at later times (smaller $u$).
