18-Env-A3 Geotechnical and Hydrogeological Engineering · Undated paper
Question 4 of 6: Well in a Confined Aquifer
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — 18-Env-A3, Geotechnical and Hydrogeological Engineering — 3 hours, open book. Six questions, each 20 marks, equal value; all six are solved below.
Find. (a) Steady-state discharge $Q$. (b) Rise in well water level if $Q$ is halved. (c) Radius of the 2-year time-of-travel WHPA.
Fig. 2 — Fully penetrating well in a confined aquifer, drawn down from $h_0$ to $h_w$ at the well face.
Approach. Confined radial flow follows the Thiem equation, $Q = 2\pi kH(h_0-h_w)/\ln(R/r_w)$. The radius of influence $R$ isn't given, so it is estimated with Sichardt's empirical formula — $Q$ is only logarithmically sensitive to $R$, so this is a low-risk, standard assumption. Part (b) exploits the fact that confined discharge is linear in drawdown; part (c) uses a cylindrical continuity/travel-time balance.
(a) Steady-state discharge. Drawdown at the well, $s_w = h_0-h_w = 8\text{ m}$. Sichardt's formula, $R=3000\,s_w\sqrt{K}$ ($K$ in m/s): $K = 8/86400 = 9.259\times10^{-5}\text{ m/s}$, so $$R = 3000\times8\times\sqrt{9.259\times10^{-5}} = 230.9\text{ m}$$ Then $$Q = \dfrac{2\pi kH(h_0-h_w)}{\ln(R/r_w)} = \dfrac{2\pi\times8\times10\times8}{\ln(230.9/0.25)} = \dfrac{4021.2}{6.828} = \boxed{589\text{ m}^3/\text{day}}$$
(b) Rise in well level at half discharge. With $k$, $H$, $R$ and $r_w$ unchanged, $Q$ is exactly proportional to drawdown $(h_0-h_w)$ in the Thiem equation, so halving $Q$ exactly halves the drawdown: $$s_w' = \dfrac{s_w}{2} = 4\text{ m}\ \Rightarrow\ h_w' = 20 - 4 = 16\text{ m}$$ $$\text{rise} = h_w' - h_w = 16-12 = \boxed{4.0\text{ m}}$$
(c) 2-year WHPA radius. Effective porosity isn't given; take $n_e=0.30$ (typical clean sand). For steady radial flow, the travel time from radius $r$ to the well is $t = n_e\pi H(r^2-r_w^2)/Q$ (integrating $dt = n_e\,dr/(v_s)$ with $v_s=Q/(2\pi r H n_e)$). Solving for $r$ at $t=2\text{ yr}=730$ days, using the discharge from part (a): $$r^2 = \dfrac{Qt}{n_e\pi H} + r_w^2 = \dfrac{589\times730}{0.30\times\pi\times10}+0.25^2 = 45{,}619$$ $$r = \boxed{214\text{ m}}$$
Check: (i) $R$ (radius of influence, part a) is estimated with Sichardt's empirical formula since the source gives no observation-well data; $Q$ varies only as $1/\ln(R)$, so a ±50% error in $R$ moves $Q$ by roughly ±7%. (ii) $n_e=0.30$ (part c) is assumed (typical clean sand, Freeze & Cherry Table 2.4) since the source doesn't give an effective porosity; the WHPA radius scales as $\sqrt{1/n_e}$, so this is a moderate-sensitivity assumption, flagged rather than hidden.
