Unconfined Dupuit-Forchheimer Flow With and Without Areal Recharge, and a Confined-Aquifer Slug Test: 18-Geol-A2 Hydrogeology, December 2017, Q5

Question 5: Unconfined Dupuit-Forchheimer Flow With and Without Areal Recharge, and a Confined-Aquifer Slug Test (20 marks)

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.

Part	Symbol	Value
(a)	Width, $K$, $n$, $h_A$, $h_B$, spacing $L$	600 m, $1.2\times10^{-5}$ m/s, 30%, 44 m, 33 m, 160 m
(b)	Recharge $R$; location	0.20 m/yr; 100 m from A / 60 m from B
(c)	$r_c$, $r_w$, $L_e$, $H_0$, $H(t)$, $t$	6 cm, 10 cm, 3 m, 0.6 m, 0.3 m, 5 s

Find. (a) Total flow through the aquifer with negligible recharge. (b) Pore-water (seepage) velocity at the stated location once areal recharge is included. (c) Hydraulic conductivity from the Hvorslev slug-test analysis.

Approach. Part (a) is a straight Dupuit-Forchheimer discharge calculation between two saturated-thickness readings. Part (b) uses the Dupuit equation with uniform areal recharge, $h^2(x)=h_A^2+(h_B^2-h_A^2)(x/L)+(R/K)x(L-x)$, differentiates to get the local Darcy velocity $v_x=-\dfrac{K}{2h(x)}\dfrac{d(h^2)}{dx}$, and divides by porosity for the pore velocity. Part (c) fits a single-exponential decay to the two slug-test readings to get the basic time lag $T_0$, then applies the Hvorslev piezometer formula.

Part (a) — Dupuit discharge, no recharge. $$Q=\frac{K\cdot W\cdot(h_A^2-h_B^2)}{2L}=\frac{(1.2\times10^{-5})(600)(44^2-33^2)}{2(160)}=\frac{(1.2\times10^{-5})(600)(847)}{320}=\boxed{0.01906\ \text{m}^3/\text{s}\ (19.06\ \text{L/s},\ 1647\ \text{m}^3/\text{day})}.$$
Part (b) — saturated thickness with recharge, at $x=100$ m. $R=0.20\ \text{m/yr}=6.338\times10^{-9}\ \text{m/s}$: $$h^2(100)=44^2+(33^2-44^2)\frac{100}{160}+\frac{R}{K}(100)(60)=1936-529.4+3.17=1409.8\ \Rightarrow\ h(100)=\boxed{37.55\ \text{m}}.$$
Local head-squared gradient and Darcy velocity. $$\frac{d(h^2)}{dx}\bigg|_{100}=\frac{33^2-44^2}{160}+\frac{R}{K}(160-200)=-5.294-0.0211=-5.315\ \text{m},$$ $$v_x=-\frac{K}{2h(100)}\frac{d(h^2)}{dx}=-\frac{1.2\times10^{-5}}{2(37.55)}(-5.315)=\boxed{8.49\times10^{-7}\ \text{m/s}}\ \text{(directed from A toward B)}.$$
Pore-water velocity. $$v_{\text{pore}}=\frac{v_x}{n}=\frac{8.49\times10^{-7}}{0.30}=\boxed{2.83\times10^{-6}\ \text{m/s}\ (0.245\ \text{m/day})}.$$
Part (c) — basic time lag from the single slug-test reading. Assuming exponential recovery $H(t)/H_0=e^{-t/T_0}$: $$T_0=\frac{t}{\ln(H_0/H)}=\frac{5}{\ln(0.6/0.3)}=\boxed{7.21\ \text{s}}.$$
Hvorslev hydraulic conductivity. With $L_e/r_w=3/0.10=30>8$ (satisfying the Hvorslev line-source condition): $$K=\frac{r_c^2\ln(L_e/r_w)}{2L_eT_0}=\frac{(0.06)^2\ln(30)}{2(3)(7.21)}=\boxed{2.83\times10^{-4}\ \text{m/s}}.$$

Figure 6 — Unconfined aquifer between Wells A and B. The dashed curve (part b) sits slightly above the no-recharge line (solid, part a) because areal recharge adds water to the flow system as it moves downgradient.

Quantity	Result
(a) Total flow $Q$ (no recharge)	0.01906 m³/s (1647 m³/day)
(b) Saturated thickness at x = 100 m	37.55 m
(b) Darcy velocity at x = 100 m	8.49×10⁻⁷ m/s
(b) Pore-water velocity at x = 100 m	2.83×10⁻⁶ m/s (0.245 m/day)
(c) Basic time lag $T_0$	7.21 s
(c) Hydraulic conductivity $K$	2.83×10⁻⁴ m/s

Concept & Theory

Topic. Dupuit-Forchheimer unconfined flow with and without areal recharge, and the Hvorslev slug-test method.

Key relations. No recharge: $Q=\dfrac{KW(h_A^2-h_B^2)}{2L}$; with recharge: $h^2(x)=h_A^2+(h_B^2-h_A^2)\dfrac{x}{L}+\dfrac{R}{K}x(L-x)$, $v_x=-\dfrac{K}{2h}\dfrac{d(h^2)}{dx}$, $v_{\text{pore}}=v_x/n$; Hvorslev $K=\dfrac{r_c^2\ln(L_e/r_w)}{2L_eT_0}$, $T_0=t/\ln(H_0/H)$.

Why this works. The Dupuit assumption treats flow as horizontal and uniform over the local saturated depth, which is what lets the governing equation be written directly in terms of $h^2$; adding uniform areal recharge turns the otherwise-linear $h^2(x)$ profile into a parabola, and because $Q=-\dfrac{K}{2}\dfrac{d(h^2)}{dx}$ integrates the recharge added from $x=0$ up to that point, the local discharge (and hence Darcy velocity) is no longer constant along the flow path the way it is without recharge. The Hvorslev method treats the well-plus-formation system as a single first-order (exponential) decay toward equilibrium, so a single measured $(t,H)$ pair fully determines the time constant $T_0$ without needing the full recovery curve.

Common pitfall. Reporting the total discharge $Q(x)$ (which is not constant once recharge is present) when the question actually asks for the local pore/seepage velocity $v_x/n$ at a specific point; and forgetting to divide the Darcy (specific-discharge) velocity by porosity to obtain the actual average linear (pore) velocity that a tracer would experience.

Source. Freeze & Cherry, Groundwater, Ch. 5.3–5.4 (Dupuit-Forchheimer flow with recharge, water-table divides); Kruseman & de Ridder, Analysis and Evaluation of Pumping Test Data, Ch. 12 (Hvorslev slug test).

Question 5 of 5: Unconfined Dupuit-Forchheimer Flow With and Without Areal Recharge, and a Confined-Aquifer Slug Test

Question 5: Unconfined Dupuit-Forchheimer Flow With and Without Areal Recharge, and a Confined-Aquifer Slug Test (20 marks)