18-Env-A3 Geotechnical and Hydrogeological Engineering · December 2019
Question 4 of 6: Steady Radial Flow to a Municipal Water Supply Well
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams December 2019 — 18-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.
Given. An unconfined aquifer pumped by a fully-penetrating well, with two observation wells at $r_1$ and $r_2$.
Given data
Quantity
Value
Well water level, $h_w$
5 m
Hydraulic conductivity, $k$
3 m/day
Well radius, $r_w$
0.05 m
Observation well 1 radius, $r_1$
30 m
Observation well 2 radius, $r_2$
50 m
Minimum allowable head at well 2, $h_2$
15 m
Find. (a) Maximum steady pumping rate $Q$. (b) The rise in $h_w$ if $Q$ is halved. (c) The 2-year time-of-travel WHPA radius.
Figure 2: unconfined aquifer pumping well with two observation wells (schematic, not to scale) — the drawdown (cone of depression) is deepest at the pumping well and shallows with distance.
Approach. Idealize the well itself as the innermost "observation point" (its own water level $h_w$ at $r=r_w$, ignoring well losses) and pair it with the Thiem/Dupuit equation for unconfined radial flow; use the $r_2$ constraint to size the maximum discharge, then use the head this implies at $r_1$ as the fixed outer reference for the reduced-discharge case in part (b), and a cylindrical mass balance for the capture zone in part (c).
(a) Maximum pumping rate. The Dupuit–Thiem equation for steady radial flow in an unconfined aquifer between the well ($r_w,h_w$) and the point where the head must not fall below 15 m ($r_2, h_2$) is $$Q = \dfrac{\pi k(h_2^2-h_w^2)}{\ln(r_2/r_w)} = \dfrac{\pi(3)(15^2-5^2)}{\ln(50/0.05)} = \dfrac{\pi(3)(200)}{6.908} = \boxed{273\text{ m}^3/\text{day}}$$
Head at the $r_1$ observation well (needed for part b). The same equation, now solved for the head at $r_1$ under this same $Q$: $$h_1 = \sqrt{h_w^2 + \dfrac{Q\ln(r_1/r_w)}{\pi k}} = \sqrt{25 + \dfrac{273\times\ln(600)}{\pi(3)}} = \boxed{14.50\text{ m}}$$ (sanity check: $h_1=14.50\text{ m} < h_2=15\text{ m}$ – the nearer well shows more drawdown, as it must.)
(b) Rise in $h_w$ when $Q$ is halved.Check: $r_1$ is taken as far enough from the pumping well that its head stays essentially unchanged when the discharge is reduced (the standard fixed-boundary simplification for a hand calculation); a full transient/superposition analysis would show a small additional change at $r_1$ too. Holding $h_1=14.50$ m fixed and solving the same equation for the new $h_w'$ at $Q'=Q/2=136.4$ m³/day: $$h_w' = \sqrt{h_1^2 - \dfrac{Q'\ln(r_1/r_w)}{\pi k}} = \sqrt{210.2 - \dfrac{136.4\times6.397}{\pi(3)}} = \boxed{10.84\text{ m}}$$ $$\text{Rise} = h_w' - h_w = 10.84-5.00 = \boxed{5.84\text{ m}}$$
(c) Wellhead Protection Area radius. Using the original design discharge $Q=273$ m³/day (the well's rated capacity from part a) and a cylindrical mass-balance capture zone (no regional gradient given, so radial inflow is assumed uniform): the volume pumped over 2 years must equal the pore volume swept within radius $R$, $$Qt = \pi(R^2-r_w^2)\,n_e\,b$$ Assuming an effective porosity $n_e=0.30$ (typical clean sand, Freeze & Cherry Table 2.4) and taking the local saturated thickness as the average of the two known heads near the well, $b=(h_w+h_1)/2=9.75$ m: $$R = \sqrt{r_w^2 + \dfrac{Qt}{\pi n_e b}} = \sqrt{(0.05)^2+\dfrac{273\times730}{\pi(0.30)(9.75)}} = \boxed{147\text{ m}}$$
