Question 3 of 5: Unconfined Flow Between Two Wells — Dupuit-Forchheimer With and Without Recharge

Question 3: Unconfined Flow Between Two Wells — Dupuit-Forchheimer With and Without 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. Unconfined aquifer, width $W=2500$ m, $K=3\times10^{-3}$ cm/s, wells $L=150$ m apart. Aquifer base 30 m below ground; saturated thickness (measured from the base) $h_1=30-8=22$ m at the upstream well, $h_2=30-10=20$ m at the downstream well. Porosity $n=0.33$. Part (d): vertical recharge $w=0.25$ m/yr.

Find. (a) total flow $Q$ through the aquifer. (b) head (saturated thickness) at the midpoint, no recharge. (c) linear (pore) velocity at the midpoint. (d) head at the midpoint with recharge included.

Approach. With negligible recharge, one-dimensional unconfined (Dupuit) flow between the two wells gives $Q$ directly from the Dupuit discharge formula, and the water-table shape follows the parabola implied by a constant $Q$; the midpoint head follows by substitution. The linear velocity divides the local Darcy flux (from $Q$ and the local saturated thickness) by porosity. With recharge, the governing ODE gains a source term whose closed-form solution is the standard Dupuit-Forchheimer parabola-plus-recharge profile.

1. Part (a) — total flow (Dupuit discharge, no recharge). Converting $K=3\times10^{-3}\ \text{cm/s}=3\times10^{-5}\ \text{m/s}$: $$Q=\frac{K\left(h_1^2-h_2^2\right)W}{2L}=\frac{(3\times10^{-5})(22^2-20^2)(2500)}{2(150)}=\boxed{0.0210\ \text{m}^3/\text{s}\ (1814\ \text{m}^3/\text{day})}.$$
2. Part (b) — head at the midpoint. With no recharge, $Q$ (and hence $Kh\,dh/dx$) is constant along the flow path, giving the parabolic profile $h(x)^2=h_1^2-(h_1^2-h_2^2)x/L$; at $x=L/2=75$ m: $$h_{\text{mid}}^2=22^2-(22^2-20^2)\frac{75}{150}=484-42=442\ \text{m}^2\ \Rightarrow\ h_{\text{mid}}=\boxed{21.02\ \text{m}}$$ (the water table sits $30-21.02=8.98$ m below ground at the midpoint).
3. Part (c) — pore-water (linear) velocity at the midpoint. Because $Q$ is constant along the flow path, the local Darcy (specific-discharge) velocity is $Q$ divided by the local cross-sectional area $W\,h_{\text{mid}}$: $$v_{\text{Darcy}}=\frac{Q}{W\,h_{\text{mid}}}=\frac{0.0210}{(2500)(21.02)}=4.00\times10^{-7}\ \text{m/s},$$ and the linear (average interstitial) velocity divides by the effective porosity: $$v_{\text{linear}}=\frac{v_{\text{Darcy}}}{n}=\frac{4.00\times10^{-7}}{0.33}=\boxed{1.21\times10^{-6}\ \text{m/s}}.$$
4. Part (d) — midpoint head with recharge. With uniform vertical recharge $w$ added, the steady 1-D Dupuit-Forchheimer solution adds a parabolic mound term that vanishes at both wells ($x=0$ and $x=L$): $$h(x)^2=h_1^2-\left(h_1^2-h_2^2\right)\frac{x}{L}+\frac{w}{K}\,x(L-x).$$ Converting $w=0.25\ \text{m/yr}=7.92\times10^{-9}\ \text{m/s}$ and evaluating at $x=75$ m: $$h_{\text{mid}}^2=442+\frac{7.92\times10^{-9}}{3\times10^{-5}}(75)(75)=442+1.49=443.5\ \text{m}^2\ \Rightarrow\ h_{\text{mid}}=\boxed{21.06\ \text{m}}.$$ The recharge raises the midpoint head only slightly (442 → 443.5 m² under the square root) because $0.25$ m/yr is a very small flux compared with the lateral throughflow driven by the 2 m head difference between the wells over 150 m.