18-Env-A3 Geotechnical and Hydrogeological Engineering · May 2018
Question 6 of 6: Unconfined Aquifer Pumping Well — Discharge and Tracer Travel Time
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams May 2018 — 04-Env-A3, Geotechnical and Hydrogeological Engineering — 3 hours, open book. All SIX questions are answered below (the exam marks only the first five as submitted; all six are solved here as a complete study resource). Marking scheme: each question 20 marks, equal value.
Find. (a) the maximum well discharge; (b) the tracer travel time between the two observation wells.
Figure 4: pumping well in an unconfined aquifer, with the Dupuit drawdown curve and both observation wells.
Approach. Use the Dupuit–Thiem equation for steady radial flow to an unconfined well. The problem gives no explicit radius of influence, so the farther observation well ($r_2=50$ m) is taken as close enough to the edge of the cone of depression that the head there is effectively still the static level $H_0$ — this lets the well-to-$r_2$ pair define $Q$ directly from the allowable drawdown. Travel time between the two observation wells then follows from the seepage velocity implied by that same $Q$.
(a) Head at the well and maximum discharge. $$h_w = H_0-s_w = 10-2 = 8\text{ m}$$ Check: no radius of influence is stated; $r_2=50$ m (the farther observation well) is assumed to be at the edge of measurable drawdown, i.e. $h(r_2)\approx H_0$. $$Q = \dfrac{\pi k\left(H_0^2-h_w^2\right)}{\ln(r_2/r_w)} = \dfrac{\pi(5)(10^2-8^2)}{\ln(50/0.10)} = \dfrac{\pi(5)(36)}{\ln(500)} = \dfrac{565.5}{6.215} = \boxed{91.0\text{ m}^3/\text{day}}$$
Heads at the two observation wells. Using the same Thiem relation from the well: $$h_1 = \sqrt{h_w^2+\dfrac{Q}{\pi k}\ln\!\left(\dfrac{r_1}{r_w}\right)} = \sqrt{64+\dfrac{91.0}{\pi(5)}\ln(200)} = 9.73\text{ m}, \qquad h_2=\sqrt{64+\dfrac{91.0}{\pi(5)}\ln(500)}=10.0\text{ m}$$ ($h_2=10.0$ m recovers $H_0$ exactly, confirming the $r_2$-as-static-edge assumption is self-consistent.)
(b) Travel time between the observation wells. The seepage velocity varies with radius since both the flow area ($2\pi r h$) and $h$ itself change between $r_1$ and $r_2$; using the average saturated thickness $\bar h=(h_1+h_2)/2=9.87$ m over the annulus between them gives a clean hand estimate: $$t \approx \dfrac{\pi n_e \bar h\left(r_2^2-r_1^2\right)}{Q} = \dfrac{\pi(0.35)(9.87)(50^2-20^2)}{91.0} = \dfrac{\pi(0.35)(9.87)(2100)}{91.0} = \boxed{250\text{ days}}$$ Check: a full numerical integration of the variable-$h(r)$ seepage velocity over $r_1$–$r_2$ gives 251 days, within 0.5% of the average-head hand estimate above.
