Step-Rate Pump Test Superposition, Two-Well Unconfined Flow, and Areal Recharge: 18-Geol-A2 Hydrogeology, May 2018, Q4

Question 4: Step-Rate Pump Test Superposition, Two-Well Unconfined Flow, and Areal Recharge (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) Confined aquifer $b=40$ m, $K=1.3\times10^{-3}$ cm/sec, $S_s=1.2\times10^{-5}\ \text{m}^{-1}$; pump rate 8 L/s for the first 12 h, then 5 L/s for the next 12 h, then shut off; drawdown wanted at $r=100$ m, $t=36$ h. (b) Unconfined aquifer $K=3\times10^{-3}$ cm/sec, width 2400 m, porosity $n_e=0.33$, base 30 m below ground; two wells 150 m apart, water levels 8 m and 11 m below ground; no recharge. (c) vertical recharge 0.25 m/yr added, otherwise unchanged.

Find. (a) drawdown at $r=100$ m, 36 h after the start of the multi-rate test. (b) (i) total flow through the aquifer, (ii) head halfway between the wells, (iii) pore (linear) velocity halfway between the wells. (c) head halfway between the wells once areal recharge is included.

Approach. Part (a) superposes three Theis solutions — the initial rate starting at $t=0$, plus a rate CHANGE of $-3$ L/s starting at $t=12$ h, plus a further change of $-5$ L/s (shut-off) starting at $t=24$ h — each evaluated at its own elapsed time since it began. Part (b) applies the Dupuit-Forchheimer approximation for steady unconfined flow between two fixed heads with no recharge. Part (c) reuses part (b)'s two-well geometry with an added areal-recharge term in the Dupuit head-squared profile.

Check: part (c) as printed says "the aquifer in Question 4(a)," but asks for the head "halfway between the wells" — that phrase only has meaning for the two-well system of part (b) (part (a) has a single pumping well and one observation point, not two wells). This is treated as a reference slip for "Question 4(b)," and part (c) is solved as part (b)'s system with recharge added, consistent with the "answer what the data actually supports" convention for internally inconsistent exam wording.

Part (a) — aquifer properties. $T=Kb=(1.3\times10^{-3}/100)(40)=\boxed{5.20\times10^{-4}\ \text{m}^2\text{/s}}$, $S=S_sb=(1.2\times10^{-5})(40)=\boxed{4.80\times10^{-4}}$.
Step-rate + shutoff superposition. At $t=36\ \text{h}=129{,}600$ s and $r=100$ m, three Theis terms apply: $Q_1=8$ L/s since $t=0$ (elapsed 36 h), $\Delta Q_2=5-8=-3$ L/s since $t=12$ h (elapsed 24 h), $\Delta Q_3=0-5=-5$ L/s since $t=24$ h (elapsed 12 h): $$s=\frac{Q_1}{4\pi T}W(u_{36})+\frac{\Delta Q_2}{4\pi T}W(u_{24})+\frac{\Delta Q_3}{4\pi T}W(u_{12})=\boxed{0.996\ \text{m}}.$$
Part (b)(i) — heads above the aquifer base. With the base as datum, $h_1=30-8=22$ m, $h_2=30-11=19$ m over $L=150$ m. Dupuit-Forchheimer unit-width flow: $$q=\frac{K(h_1^2-h_2^2)}{2L}=\frac{(3\times10^{-5})(22^2-19^2)}{2(150)}=1.23\times10^{-5}\ \text{m}^2\text{/s},$$ $$Q_{\text{tot}}=q\times(\text{width})=(1.23\times10^{-5})(2400)=\boxed{0.0295\ \text{m}^3\text{/s}\ (29.5\ \text{L/s})}.$$
Part (b)(ii) — head at the midpoint. $$h^2(x)=h_1^2-\frac{h_1^2-h_2^2}{L}x\ \Rightarrow\ h(75)=\sqrt{22^2-\frac{22^2-19^2}{150}(75)}=\boxed{20.55\ \text{m above the base}}$$ (depth below ground $\approx30-20.55=9.45$ m).
Part (b)(iii) — pore (linear) velocity at the midpoint. With no recharge, unit-width discharge $q$ is constant along $x$, so the local Darcy flux is $q/h(x)$: $$v_{\text{Darcy}}=\frac{q}{h(75)}=\frac{1.23\times10^{-5}}{20.55}=5.98\times10^{-7}\ \text{m/s},\qquad v_{\text{linear}}=\frac{v_{\text{Darcy}}}{n_e}=\frac{5.98\times10^{-7}}{0.33}=\boxed{1.81\times10^{-6}\ \text{m/s}}.$$
Part (c) — head at the midpoint with areal recharge. Recharge $W=0.25\ \text{m/yr}=7.92\times10^{-9}\ \text{m/s}$; the Dupuit head-squared profile gains a mounding term: $$h^2(x)=h_1^2-\frac{h_1^2-h_2^2}{L}x+\frac{W}{K}x(L-x)\ \Rightarrow\ h(75)=\sqrt{422.5+\left(\frac{7.92\times10^{-9}}{3\times10^{-5}}\right)(75)(75)}=\boxed{20.59\ \text{m above the base}}$$ (depth below ground $\approx30-20.59=9.41$ m) — slightly higher than the no-recharge case, as the added infiltration mounds the water table.

Quantity	Result
(a) $T$, $S$	5.20×10⁻⁴ m²/s, 4.80×10⁻⁴
(a) Drawdown at $r$=100 m, $t$=36 h	0.996 m
(b)(i) Total flow $Q_{\text{tot}}$	0.0295 m³/s (29.5 L/s)
(b)(ii) Head at midpoint (no recharge)	20.55 m above base (≈9.45 m depth)
(b)(iii) Pore (linear) velocity at midpoint	1.81×10⁻⁶ m/s
(c) Head at midpoint (with recharge)	20.59 m above base (≈9.41 m depth)

Concept & Theory

Topic. Superposition of Theis solutions for a stepped pumping-rate history, and the Dupuit-Forchheimer approximation for unconfined flow with and without areal recharge.

Key relations. Step-rate superposition $s(t)=\sum_i\dfrac{\Delta Q_i}{4\pi T}W(u_i)$, each term using its OWN elapsed time since that rate began; Dupuit $h^2(x)=h_1^2-\dfrac{h_1^2-h_2^2}{L}x+\dfrac{W}{K}x(L-x)$.

Why this works. The confined-flow (Theis) equation is linear in $Q$, so any piecewise-constant pumping history can be built by superposing a separate Theis response for every rate CHANGE, each one "starting its own clock" at the moment that change occurred and continuing indefinitely afterward (including through the shut-off, treated as one more rate change back to zero). The Dupuit approximation assumes essentially horizontal flow lines and integrates the 1-D continuity equation with a distributed source (recharge) term, producing a parabolic rather than linear $h^2(x)$ profile whenever $W\ne0$.

Common pitfall. Using the SAME elapsed time (36 h) for every superposed term instead of each term's own time-since-it-began collapses the calculation to a single Theis term and gives a materially wrong answer; and applying $q=$const along $x$ once recharge is present (part c) — with recharge, $q(x)$ is no longer constant, only the RECHARGE-FREE case (part b) has that shortcut.

Source. Freeze & Cherry, Groundwater, Ch. 8.3 (Theis solution, superposition) and Ch. 5.3 (Dupuit-Forchheimer theory, areal recharge between fixed boundaries).

Question 4 of 5: Step-Rate Pump Test Superposition, Two-Well Unconfined Flow, and Areal Recharge

Question 4: Step-Rate Pump Test Superposition, Two-Well Unconfined Flow, and Areal Recharge (equal value)