Regional Confined Discharge and Dupuit-Forchheimer Flow with Recharge: 18-Geol-A2 Hydrogeology, May 2017, Q3

Question 3: Regional Confined Discharge and Dupuit-Forchheimer Flow with 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. (a) Confined aquifer $b=5.0$ m, $K=7\times10^{-3}$ cm/s, effective porosity $n_e=0.25$, piezometric drop $\Delta h=0.2$ m over 240 m. (b) Unconfined aquifer between two water bodies, $K=14.3$ m/day, $h_1=17.7$ m, $h_2=13.2$ m, $L=525$ m, uniform areal recharge $w=0.007$ m/day.

Find. (a) discharge through a 10 m wide strip and the average linear (seepage) velocity. (b)(i) discharge per unit width at $x=480$ m; (ii) whether a water-table divide exists and where; (iii) the maximum water-table height.

Figure 2 — Unconfined aquifer between two water bodies with uniform areal recharge. Because the head drop from $h_1$ to $h_2$ dominates the small recharge, the water table (dashed) declines monotonically from left to right — there is no interior mound/divide.

Approach. Part (a) is a straight application of Darcy's law: the piezometric gradient gives the Darcy velocity, and discharge follows from the cross-sectional area, while the average linear velocity divides the Darcy velocity by the effective porosity. Part (b) uses the Dupuit-Forchheimer solution for unconfined flow with uniform areal recharge, $h^2(x)=h_1^2-\dfrac{h_1^2-h_2^2}{L}x+\dfrac{w}{K}x(L-x)$; differentiating gives the discharge per unit width $q(x)=-\dfrac{K}{2}\dfrac{d(h^2)}{dx}$, and a water-table divide exists wherever $q(x)=0$ inside $(0,L)$.

Part (a) — Darcy velocity and discharge. $K=7\times10^{-3}\ \text{cm/s}=7\times10^{-5}\ \text{m/s}$; gradient $i=0.2/240=8.33\times10^{-4}$: $$v=Ki=(7\times10^{-5})(8.33\times10^{-4})=5.83\times10^{-8}\ \text{m/s},$$ $$Q=v\,(b\times W)=(5.83\times10^{-8})(5.0\times10.0)=\boxed{2.92\times10^{-6}\ \text{m}^3/\text{s}\ (0.252\ \text{m}^3/\text{day})}.$$
Average linear velocity. The seepage (average linear) velocity is the Darcy velocity divided by the effective porosity, since only the pore space carries flow: $$v_s=\frac{v}{n_e}=\frac{5.83\times10^{-8}}{0.25}=\boxed{2.33\times10^{-7}\ \text{m/s}}.$$
Part (b)(i) — discharge per unit width at x=480 m. Differentiating the Dupuit-Forchheimer head-squared profile gives $q(x)=\dfrac{K(h_1^2-h_2^2)}{2L}-\dfrac{wL}{2}+wx$; with $h_1^2-h_2^2=17.7^2-13.2^2=139.05\ \text{m}^2$: $$q(480)=\frac{(14.3)(139.05)}{2(525)}-\frac{(0.007)(525)}{2}+(0.007)(480)=\boxed{3.42\ \text{m}^2/\text{day}}\ \text{(toward the } h_2\text{ side)}.$$
Part (b)(ii) — test for a water-table divide. A divide exists only where $q(x)=0$ for some $x$ strictly inside $(0,L)$. Evaluating at the left boundary: $$q(0)=\frac{K(h_1^2-h_2^2)}{2L}-\frac{wL}{2}=1.894-1.838=\boxed{+0.056\ \text{m}^2/\text{day}}\ (>0).$$ Because $q(x)$ increases monotonically with $x$ (its slope is $w>0$) and is already positive at $x=0$, $q(x)>0$ for the entire domain $0\le x\le L$ — the flow is directed toward the $h_2$ side everywhere, augmented by recharge along the way. No interior water-table divide exists for this combination of head difference, length and recharge rate: the recharge here is too small, relative to the $h_1$-to-$h_2$ head drop over 525 m, to reverse or stall the throughflow anywhere between the two boundaries.
Part (b)(iii) — maximum water-table height. Since $q(x)>0$ everywhere, $d(h^2)/dx=-2q(x)/K<0$ throughout, so $h(x)$ decreases monotonically from $x=0$ to $x=L$ with no interior maximum. The maximum height is therefore simply the higher boundary value: $$\boxed{h_{\max}=h_1=17.7\ \text{m, occurring at }x=0}.$$

All three parts of (b) tell one consistent story: the recharge is real (it does add to the discharge, raising $q$ from 1.894–1.838≈0.056 m²/day at $x=0$ to 3.42 m²/day at $x=480$ m) but it is not large enough to overcome the underlying head gradient from $h_1$ to $h_2$, so the water table never mounds above $h_1$ and the discharge never reverses direction.

Quantity	Result
(a) Discharge $Q$ (10 m strip)	2.92×10⁻⁶ m³/s (0.252 m³/day)
(a) Average linear velocity	2.33×10⁻⁷ m/s
(b)(i) $q$ at $x=480$ m	3.42 m²/day
(b)(ii) Water-table divide	None — $q(x)>0$ for all $0\le x\le L$
(b)(iii) Maximum water-table height	17.7 m, at $x=0$

Concept & Theory

Topic. Regional Darcy discharge and seepage velocity in a confined strip, and the Dupuit-Forchheimer solution for unconfined flow between two fixed-head boundaries with uniform areal recharge.

Key relations. $Q=Kib A$; $v_s=v/n_e$; $h^2(x)=h_1^2-\dfrac{h_1^2-h_2^2}{L}x+\dfrac{w}{K}x(L-x)$; $q(x)=-\dfrac{K}{2}\dfrac{d(h^2)}{dx}$; divide exists iff $q(x)=0$ for some $x\in(0,L)$.

Why this works. The Dupuit assumption treats flow as horizontal at every $x$, so the discharge per unit width is exactly proportional to $-d(h^2)/dx$ from the 1-D form of Darcy's law integrated over the local saturated thickness $h(x)$; recharge is then just mass added along the flow path, integrated into $q(x)$ as a linear term in $x$. Whether a mound/divide forms is a genuine competition between the two boundary heads (which drive throughflow one way) and the total recharge volume (which pushes water outward from the middle) — only when recharge is large enough relative to $K(h_1^2-h_2^2)/L^2$ does the interior term win and create an interior maximum.

Common pitfall. Assuming a divide always exists whenever recharge is present, and searching for one algebraically without first checking that $q(0)$ and $q(L)$ actually bracket zero — here $q(0)>0$ already, so the formula for $x_d$ would return a value outside $[0,L]$, which is the tell that no physical divide exists rather than a computation error.

Source. Freeze & Cherry, Groundwater, Ch. 2.11 (Darcy's law, seepage velocity) and Ch. 5.3 (Dupuit-Forchheimer theory, areal recharge between fixed boundaries).

Part (c) — Assumptions Behind the Dupuit-Forchheimer Analytical Solution

The closed-form solution used in part (b) rests on the classical Dupuit-Forchheimer assumptions for unconfined flow, together with the standard idealizations needed to superimpose a uniform recharge term onto it:

Dupuit assumption (essentially horizontal flow). Equipotential lines are treated as vertical and flow lines as horizontal, so the hydraulic gradient at any $x$ is approximated by the slope of the water table, $dh/dx$, rather than the true (slightly curved) gradient of the flow net. This is acceptable only where the water table slope is small and $L\gg h$.
Homogeneous, isotropic aquifer. $K$ is constant in space and direction, so it can be pulled outside the integral that produces the $h^2(x)$ parabola.
Impermeable, horizontal base. The datum ($z=0$) is a true no-flow boundary with no leakage out the bottom of the aquifer.
Steady state. $h(x)$ and $q(x)$ do not change with time; recharge $w$ is applied at a constant, uniform rate over the entire domain, and the two boundary heads $h_1$, $h_2$ are held fixed.
Two-dimensional, vertical-plane flow. The aquifer is treated as uniform (and effectively infinite) in the direction perpendicular to the page, so the problem reduces to a single horizontal coordinate $x$; there is no lateral (out-of-plane) flow component.
Fully saturated, unconfined aquifer with a free water table. The saturated thickness at any $x$ is $h(x)$ itself (not a fixed confined thickness), which is why the governing equation is written in terms of $h^2$ rather than $h$ — the Dupuit-Forchheimer form of the Boussinesq equation.

Each assumption is a genuine simplification of the real (three-dimensional, generally transient, anisotropic) flow field; the solution's accuracy degrades most visibly near the two boundary water bodies, where the true flow has a real vertical component that the Dupuit approximation discards by construction.

Concept & Theory

Topic. The physical assumptions underlying the Dupuit-Forchheimer approximation for unconfined groundwater flow.

Key relations. Dupuit assumption $i\approx dh/dx$ (instead of the true 2-D gradient); Boussinesq/Dupuit-Forchheimer governing equation $\dfrac{d}{dx}\!\left(h\dfrac{dh}{dx}\right)=-\dfrac{w}{K}$.

Why this works. Replacing the true curved equipotentials with vertical ones is only valid when the aquifer is long and thin relative to the head variation across it ($L\gg h_1,h_2$), which is exactly the geometry of this problem (a 525 m length against water levels of order 15–18 m).

Common pitfall. Listing only "Darcy's law applies" is not sufficient — Darcy's law alone does not produce the $h^2(x)$ parabola; it is specifically the Dupuit horizontal-flow assumption applied to Darcy's law, integrated over the locally varying saturated thickness, that yields the closed-form solution.

Source. Freeze & Cherry, Groundwater, Ch. 5.3 (Dupuit-Forchheimer theory and its assumptions).

Question 3 of 5: Regional Confined Discharge and Dupuit-Forchheimer Flow with Recharge

Question 3: Regional Confined Discharge and Dupuit-Forchheimer Flow with Recharge (equal value)

Part (c) — Assumptions Behind the Dupuit-Forchheimer Analytical Solution