Drawdown-Curve Diagnostics, Cooper-Jacob Analysis, and Superposition for Two Wells with Different Pumping Schedules: 18-Geol-A2 Hydrogeology, December 2017, Q3

Question 3: Drawdown-Curve Diagnostics, Cooper-Jacob Analysis, and Superposition for Two Wells with Different Pumping Schedules (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.

Part	Symbol	Value
(b)	$Q$, $r$, $s_1$ at $t_1$, $s_2$ at $t_2$	3550 L/min, 200 m, 3.2 m @ 4 h, 5.2 m @ 12 h
(c)	$r_A$, $Q_A$, on-time A; $r_B$, $Q_B$, on-time B	50 m, 20 m³/hr, 0–24 h; 90 m, 10 m³/hr, 0–36 h
(c)	$b$, $K$, $S_s$; target time	22 m, $1\times10^{-5}$ m/s, $1\times10^{-5}$ m⁻¹; $t=48$ h

Find. (a) A qualitative log-log drawdown-vs-time sketch for the four aquifer/boundary types. (b) Storativity $S$ and transmissivity $T$ from the two-point drawdown record. (c) Total drawdown at the observation well 48 hours after both wells started, given each well's finite pumping duration.

Approach. Part (a) is qualitative: all four curves share the same early-time (small-$r^2S/4Tt$) Theis trend before diverging once the aquifer "feels" the leaky, barrier, or recharge boundary. Part (b) applies the Cooper-Jacob straight-line method between the two given (t, s) points to get the slope per log cycle, from which $T$ follows directly and $S$ follows from the time-axis intercept. Part (c) uses Theis superposition: each well that has been pumping and is later shut off (or, for Well B, simply has not yet reached 48 h of continuous pumping) is represented as a pumping well switched on at $t=0$ plus an equal-rate injection ("image-in-time") well switched on at the moment pumping actually stopped.

Part (a) — qualitative curve shapes. See Figure 4. All four curves coincide at early time (Theis behaviour, before the boundary/leakage is felt). (i) The ideal non-leaky confined aquifer keeps following the Theis type curve indefinitely: drawdown keeps rising, with a log-log slope that slowly decreases as the curve approaches its semi-log (Cooper-Jacob) straight line. (ii) A leaky aquifer's curve bends away from the Theis line and flattens as steady vertical leakage begins supplying the pumped water. (iii) A low-permeability barrier boundary steepens the curve above the Theis line (drawdown accelerates, doubling the late-time slope) because no water can cross the barrier. (iv) A constant-head recharge boundary flattens the curve toward a fully horizontal asymptote as the boundary supplies water at no further drawdown cost.
Part (b) — Cooper-Jacob slope. $Q=3550\ \text{L/min}=0.05917\ \text{m}^3/\text{s}$. Slope per log cycle between the two points: $$\Delta s=\frac{2.30Q}{4\pi T}\log_{10}\!\left(\frac{t_2}{t_1}\right)\ \Rightarrow\ \text{slope}=\frac{5.2-3.2}{\log_{10}(12/4)}=\frac{2.0}{0.4771}=4.192\ \text{m/log-cycle}.$$
Transmissivity. $$T=\frac{2.30Q}{4\pi\,(\text{slope})}=\frac{2.30(0.05917)}{4\pi(4.192)}=\boxed{2.58\times10^{-3}\ \text{m}^2/\text{s}}.$$
Storativity from the time intercept. Extrapolating the straight line to $s=0$ at $t_0$: using $s_1=\text{slope}\cdot\log_{10}(t_1/t_0)$, $t_0=t_1/10^{s_1/\text{slope}}=4/10^{(3.2/4.192)}=0.690\ \text{h}=2483\ \text{s}$. Then: $$S=\frac{2.25Tt_0}{r^2}=\frac{2.25(2.58\times10^{-3})(2483)}{200^2}=\boxed{3.61\times10^{-4}}.$$
Part (c) — aquifer properties and superposition set-up. $T=Kb=(1\times10^{-5})(22)=2.2\times10^{-4}\ \text{m}^2/\text{s}$, $S=S_sb=(1\times10^{-5})(22)=2.2\times10^{-4}$. Both wells start at $t=0$; the observation-well drawdown 48 h later is the sum of each well's contribution, each built from a pumping well plus a compensating equal-rate injection well starting when that pump actually stopped: Well A off at 24 h ($48-24=24\ \text{h}$ of "recovery" running time), Well B off at 36 h ($48-36=12\ \text{h}$ of "recovery" running time): $$s=\frac{Q}{4\pi T}\Big[W(u_{\text{on}})-W(u_{\text{off}})\Big],\qquad u=\frac{r^2S}{4Tt}.$$
Well A contribution ($r_A=50\ \text{m}$, $Q_A=20/3600=5.556\times10^{-3}\ \text{m}^3/\text{s}$): $u(t=48\text{h})=3.62\times10^{-3}$, $u(t=24\text{h})=7.23\times10^{-3}$, giving $W(u)=5.048$ and $4.359$ respectively: $$s_A=\frac{5.556\times10^{-3}}{4\pi(2.2\times10^{-4})}\big(5.048-4.359\big)=\boxed{1.39\ \text{m}}.$$
Well B contribution ($r_B=90\ \text{m}$, $Q_B=10/3600=2.778\times10^{-3}\ \text{m}^3/\text{s}$): $u(t=48\text{h})=1.17\times10^{-2}$, $u(t=12\text{h})=4.69\times10^{-2}$, giving $W(u)=3.881$ and $2.529$ respectively: $$s_B=\frac{2.778\times10^{-3}}{4\pi(2.2\times10^{-4})}\big(3.881-2.529\big)=\boxed{1.36\ \text{m}}.$$
Total drawdown. $$s_{\text{total}}=s_A+s_B=1.39+1.36=\boxed{2.74\ \text{m}}.$$

Figure 4 — Qualitative log-log drawdown signatures (Q3a). All four coincide early, then diverge once the leaky aquitard, barrier, or recharge boundary is "felt": leakage (ii) and recharge (iv) both flatten the curve — (iv) toward a fully horizontal steady-state asymptote — while an impermeable barrier (iii) steepens it, roughly doubling the late-time slope.

Quantity	Result
(b) Transmissivity $T$	2.58×10⁻³ m²/s
(b) Storativity $S$	3.61×10⁻⁴
(c) Drawdown from Well A	1.39 m
(c) Drawdown from Well B	1.36 m
(c) Total drawdown at t = 48 h	2.74 m

Concept & Theory

Topic. Diagnostic drawdown-curve shapes, the Cooper-Jacob straight-line simplification of the Theis solution, and superposition for wells with a finite pumping duration.

Key relations. Theis $s=\dfrac{Q}{4\pi T}W(u)$, $u=\dfrac{r^2S}{4Tt}$; Cooper-Jacob slope $\Delta s=\dfrac{2.30Q}{4\pi T}$ per log cycle of $t$, $S=\dfrac{2.25Tt_0}{r^2}$; superposition for a well shut off at $t_{\text{off}}$: add an equal-rate injection well "born" at $t_{\text{off}}$.

Why this works. The Cooper-Jacob approximation replaces the full Theis well function with its late-time logarithmic asymptote, which is valid once $u$ is small (commonly $u<0.05$). Here the fitted values give $u=0.097$ at 4 h and $u=0.032$ at 12 h, so the 4-hour point sits at the edge of that range: an exact two-point Theis fit to the same data gives $T=2.40\times10^{-3}\ \text{m}^2/\text{s}$ and $S=4.28\times10^{-4}$, i.e. the Cooper-Jacob answer the question calls for is about 7% high on $T$ and 16% low on $S$ — two points on that straight line then fully determine both unknowns ($T$ from the slope, $S$ from where the line crosses zero drawdown). Superposition works because the governing groundwater-flow equation is linear: a finite-duration pumping period is exactly reproduced by an infinite pumping well plus an infinite injection well of the same rate starting at the shut-off time, since their effects after that time cancel out the "continued pumping" that never really happened.

Common pitfall. Using the raw time since the well was switched ON (rather than since the shut-off/switch event) inside the second, subtracted well-function term; and forgetting that "pumped for 36 hours" means Well B has already stopped by the 48-hour observation time, so its own recovery period must also be superposed in.

Source. Kruseman & de Ridder, Analysis and Evaluation of Pumping Test Data, Ch. 3 (Cooper-Jacob method) and Ch. 2 (superposition, image wells in time); Freeze & Cherry, Groundwater, Ch. 8.1–8.2.

Question 3 of 5: Drawdown-Curve Diagnostics, Cooper-Jacob Analysis, and Superposition for Two Wells with Different Pumping Schedules

Question 3: Drawdown-Curve Diagnostics, Cooper-Jacob Analysis, and Superposition for Two Wells with Different Pumping Schedules (20 marks)