Question 4 of 5: Transient Drawdown Near a Straight-Line Boundary, and Unconfined Thiem Hydraulics

Question 4: Transient Drawdown Near a Straight-Line Boundary, and Unconfined Thiem Hydraulics (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)–(b) Confined aquifer $b=22$ m, $K=10^{-2}$ cm/s, $S_s=1.0\times10^{-4}\ \text{m}^{-1}$, a straight-line boundary 190 m due west of the pumping well, pumping rate $Q=6$ L/s, observation point 100 m due north of the well. (a) boundary is constant-head, drawdown wanted at $t=12$ h. (b) boundary is impermeable, pump runs 0–12 h then stops, drawdown wanted at $t=18$ h. (c) Unconfined aquifer, well diameter 32 cm, thickness 48 m, $Q=9.8$ L/s; water elevation 102 m.a.s.l. at the well, 106 m.a.s.l. at $r=145$ m.

Find. (a) drawdown 100 m north of the well after 12 h, with a constant-head boundary. (b) drawdown at $t=18$ h if the boundary is instead impermeable and the pump stops at $t=12$ h. (c) hydraulic conductivity of the unconfined aquifer.

Approach. Parts (a)–(b) both use the method of images: a straight-line boundary is replaced by an imaginary well of the same pumping rate, mirrored across the boundary at twice the well-to-boundary distance, so the boundary itself is a line 190 m west of the well and the image sits 380 m further west (390 m total from the real well). A constant-head boundary needs an injection (recharge) image well, whose drawdown is subtracted; an impermeable boundary needs a pumping image well, whose drawdown is added. Part (b) additionally needs step-off (Theis) superposition because the pump stops at $t=12$ h. Part (c) is the standard Thiem equation for a fully penetrating well in an unconfined aquifer.

1. Part (a) — aquifer properties and geometry. $T=Kb=(10^{-4}\ \text{m/s})(22)=2.2\times10^{-3}\ \text{m}^2/\text{s}$, $S=S_sb=(1.0\times10^{-4})(22)=2.2\times10^{-3}$. With the real well at the origin and the boundary a line 190 m west of it, the image (recharge) well sits at 380 m west, so: $$r_1=100\ \text{m (real well)},\qquad r_2=\sqrt{380^2+100^2}=\boxed{392.9\ \text{m (image well)}}.$$
2. Drawdown by superposition (constant-head boundary). $Q=0.006\ \text{m}^3/\text{s}$, $t=12\times3600=43{,}200$ s, so $u_1=r_1^2S/(4Tt)=0.0579$ and $u_2=r_2^2S/(4Tt)=0.894$; the injection image well's drawdown is SUBTRACTED: $$s=\frac{Q}{4\pi T}\left[W(u_1)-W(u_2)\right]=\frac{0.006}{4\pi(2.2\times10^{-3})}\left[2.329-0.263\right]=\boxed{0.448\ \text{m}}.$$ The constant-head boundary noticeably reduces the drawdown below what an infinite aquifer would give (an infinite aquifer would show $Q/4\pi T\cdot W(u_1)=0.506$ m) — the boundary is already "feeling" the recharge effect at 12 hours.
3. Part (b) — impermeable boundary with the pump stopping at 12 h. Now the image well is a second PUMPING well at the same location (380 m west), so both the real and image drawdowns ADD. To capture pumping stopping at $t_1=12$ h, apply Theis step-off superposition (a virtual injection well of the same rate switched on at $t_1$) to BOTH the real-well and image-well terms, evaluated at $t=18$ h ($t-t_1=6$ h $=21{,}600$ s): $$s=\underbrace{\frac{Q}{4\pi T}\left[W(u_1(t))-W(u_1(t-t_1))\right]}_{\text{real well}}+\underbrace{\frac{Q}{4\pi T}\left[W(u_2(t))-W(u_2(t-t_1))\right]}_{\text{image well}}.$$
4. Evaluate the step-off superposition. Computing each of the four Theis terms ($s(r,t)=\frac{Q}{4\pi T}W(u)$) and combining: $$s=\underbrace{\left[s(r_1,18\text{h})-s(r_1,6\text{h})\right]}_{0.589-0.367=0.222}+\underbrace{\left[s(r_2,18\text{h})-s(r_2,6\text{h})\right]}_{0.0995-0.0143=0.0852}=\boxed{0.307\ \text{m}}.$$ Continuous pumping to 12 h under this same impermeable-boundary geometry (real + adding image well, no shut-off) would have reached $s(r_1,12\text{h})+s(r_2,12\text{h})=0.563$ m; the observed 18 h value (0.307 m) is well below that because 6 hours of recovery after shut-off has already erased a large share of the drawdown, even though the impermeable boundary keeps reinforcing (not relieving) it.
5. Part (c) — Thiem equation for the unconfined well. With well radius $r_w=0.16$ m, $h_w=102$ m at the well and $h=106$ m at $r=145$ m: $$K=\frac{Q\ln(r/r_w)}{\pi(h^2-h_w^2)}=\frac{(0.0098)\ln(145/0.16)}{\pi(106^2-102^2)}=\frac{(0.0098)(6.809)}{\pi(832)}=\boxed{2.55\times10^{-5}\ \text{m/s}}.$$