Layered Confined Aquifer, Vertical Flow, and a Density-Driven Interface and Slug Test: 18-Geol-A2 Hydrogeology, May 2017, Q2

Question 2: Layered Confined Aquifer, Vertical Flow, and a Density-Driven Interface and Slug Test (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)–(b) Three confined layers — top $b_1=3.5$ m, $K_1=1.1\times10^{-3}$ cm/s; middle $b_2=2.5$ m, $K_2=4.2\times10^{-6}$ cm/s; bottom $b_3=2.8$ m, $K_3=1.3\times10^{-4}$ cm/s — with pressure $104$ kPa across the top of the whole aquifer and $184$ kPa across the bottom. (c) 11 m aquitard between a fresh aquifer above ($\rho_f=998\ \text{kg/m}^3$, 11.5 m water column) and a saline aquifer below ($\rho_s=1160\ \text{kg/m}^3$, 21 m water column). (d) Slug test: casing radius $r_c=5$ cm, screen radius $R=10$ cm, screen length $L_e=3.2$ m, aquifer thickness $b=15$ m; head ratio $y=0.52$ m at $t=0$ falling to $y=0.186$ m at $t=3$ s.

Layer	Thickness	Hydraulic conductivity
Top	3.5 m	$1.1\times10^{-3}$ cm/s
Middle	2.5 m	$4.2\times10^{-6}$ cm/s
Bottom	2.8 m	$1.3\times10^{-4}$ cm/s

Find. (a) effective horizontal $K_h$ and vertical $K_v$ conductivities of the 3-layer aquifer. (b) magnitude/direction of the Darcy velocity in each layer, and the pressure at the top/middle interface. (c) direction of flow across the aquitard. (d) transmissivity of the aquifer from the slug test.

Figure 1 — The three-layer confined aquifer of Q2(a)–(b). Total head is higher at the top (19.42 m) than the bottom (18.79 m), so the Darcy velocity is directed downward and, by continuity, has the same magnitude in every layer.

Approach. Part (a) applies the standard layered-medium averages: a thickness-weighted arithmetic mean for horizontal (parallel) flow, a thickness-weighted harmonic mean for vertical (series) flow. Part (b) first converts each boundary pressure to a total head (elevation + pressure head, base of the aquifer as datum), which shows flow is vertical and downward; continuity then forces the same Darcy velocity through all three layers, found from $K_v$ and the total head drop, and the interface pressure follows from a Darcy-law head loss across just the top layer. Part (c) converts each piezometer reading to a common freshwater-equivalent head before comparing them, since raw heads in fluids of different density cannot be compared directly. Part (d) applies the Hvorslev slug-test method for a partially penetrating well with $L_e/R\ge8$.

Part (a) — effective horizontal conductivity. For parallel (horizontal) flow the layers act side by side, so $K_h$ is the thickness-weighted arithmetic mean: $$K_h=\frac{K_1b_1+K_2b_2+K_3b_3}{b_1+b_2+b_3}=\frac{(1.1\times10^{-3})(3.5)+(4.2\times10^{-6})(2.5)+(1.3\times10^{-4})(2.8)}{8.8}=\boxed{4.80\times10^{-4}\ \text{cm/s}}.$$
Effective vertical conductivity. For series (vertical) flow the layers pass one common flux, so $K_v$ is the thickness-weighted harmonic mean: $$K_v=\frac{b_1+b_2+b_3}{\dfrac{b_1}{K_1}+\dfrac{b_2}{K_2}+\dfrac{b_3}{K_3}}=\frac{8.8}{\dfrac{3.5}{1.1\times10^{-3}}+\dfrac{2.5}{4.2\times10^{-6}}+\dfrac{2.8}{1.3\times10^{-4}}}=\boxed{1.42\times10^{-5}\ \text{cm/s}},$$ dominated almost entirely by the far less permeable middle layer, exactly as series flow should be.
Part (b) — total head at the top and bottom. Taking the base of the aquifer as datum ($z=0$, so the top is at $z=8.8$ m) and converting each pressure to a pressure head $\psi=P/(\rho_wg)$: $$h_{\text{top}}=8.8+\frac{104{,}000}{(998)(9.81)}=19.42\ \text{m},\qquad h_{\text{bot}}=0+\frac{184{,}000}{(998)(9.81)}=18.79\ \text{m}.$$ Since $h_{\text{top}}>h_{\text{bot}}$, flow is downward through the whole aquifer.
Darcy velocity (same in every layer, by continuity). With $K_v=1.42\times10^{-5}\ \text{cm/s}=1.42\times10^{-7}\ \text{m/s}$ applied over the full 8.8 m thickness: $$v=K_v\frac{h_{\text{top}}-h_{\text{bot}}}{B}=(1.42\times10^{-7})\frac{19.42-18.79}{8.8}=\boxed{1.01\times10^{-8}\ \text{m/s, downward, in all three layers}}.$$ A single steady vertical flux must pass through each layer in series, so the magnitude and direction are identical top to bottom even though each layer's own conductivity differs by orders of magnitude — only the head GRADIENT differs layer to layer.
Pressure at the top/middle interface. Applying Darcy's law across just the top layer ($K_1=1.1\times10^{-3}\ \text{cm/s}=1.1\times10^{-5}\ \text{m/s}$) to find the head lost crossing it, then converting the resulting head at $z=5.3$ m (top of the middle layer) back to a pressure: $$h_{\text{interface}}=h_{\text{top}}-\frac{vb_1}{K_1}=19.42-\frac{(1.01\times10^{-8})(3.5)}{1.1\times10^{-5}}=19.42\ \text{m (negligible loss)},$$ $$P_{\text{interface}}=\rho_wg\left[h_{\text{interface}}-z_{\text{interface}}\right]=(998)(9.81)(19.42-5.3)=\boxed{138.2\ \text{kPa}}.$$ Because the top layer is by far the most conductive, it carries almost no head loss for the tiny cross-layer flux, so the interface pressure is only slightly above what pure hydrostatics from the top boundary alone would give.
Part (c) — freshwater-equivalent head at each well. With the base of the aquitard as datum ($z=0$, top at $z=11$ m) and $h_f=z+(\rho/\rho_f)\psi$: $$h_{f,\text{fresh}}=11+\left(\frac{998}{998}\right)(11.5)=\boxed{22.5\ \text{m}},\qquad h_{f,\text{saline}}=0+\left(\frac{1160}{998}\right)(21)=\boxed{24.4\ \text{m}}.$$ Because $h_{f,\text{saline}}$ exceeds $h_{f,\text{fresh}}$, the corrected heads show flow is directed upward, from the saline aquifer through the aquitard into the fresh aquifer — the strong artesian head in the saline unit more than offsets its higher density, so this aquitard is a point of upward saline leakage rather than downward fresh-water recharge.
Part (d) — basic time lag and Hvorslev conductivity. The exponential head-decay model $y/y_0=e^{-t/T_0}$ gives the basic time lag directly from the one data point: $$T_0=\frac{t}{\ln(y_0/y_1)}=\frac{3}{\ln(0.52/0.186)}=2.92\ \text{s}.$$ With $L_e/R=3.2/0.10=32\gg8$, the simplified Hvorslev formula applies: $$K=\frac{r_c^2\ln(L_e/R)}{2L_eT_0}=\frac{(0.05)^2\ln(32)}{2(3.2)(2.92)}=\boxed{4.64\times10^{-4}\ \text{m/s}}.$$
Transmissivity. $$T=Kb=(4.64\times10^{-4})(15)=\boxed{6.96\times10^{-3}\ \text{m}^2/\text{s}}.$$ The very fast recovery (72% of the initial displacement dissipates in only 3 s) is itself a strong qualitative sign of a highly transmissive aquifer, consistent with the large computed $K$.

Quantity	Result
(a) Effective horizontal $K_h$	4.80×10⁻⁴ cm/s
(a) Effective vertical $K_v$	1.42×10⁻⁵ cm/s
(b) Darcy velocity (all layers)	1.01×10⁻⁸ m/s, downward
(b) Pressure at top/middle interface	138.2 kPa
(c) Flow direction across aquitard	Upward (saline → fresh)
(d) Transmissivity (Hvorslev)	6.96×10⁻³ m²/s

Concept & Theory

Topic. Layered-medium effective conductivity, vertical Darcy flow through a series system, freshwater-equivalent head across a density interface, and the Hvorslev slug-test method.

Key relations. $K_h=\sum K_ib_i/\sum b_i$; $K_v=\sum b_i/\sum(b_i/K_i)$; $h_f=z+(\rho/\rho_f)\psi$; $y/y_0=e^{-t/T_0}$; $K=r_c^2\ln(L_e/R)/(2L_eT_0)$.

Why this works. In series (vertical) flow, continuity forces one common Darcy flux through every layer, so the effective conductivity is dominated by the layer that resists that flux most (a harmonic average); the same continuity argument is why the Darcy velocity itself — not just the effective $K$ — is identical in every layer of Q2(b). Comparing raw piezometric heads across a density interface is invalid because a tall column of light fluid can balance a short column of dense fluid; converting every reading to what it would read as fresh water removes the density term and isolates the true driving potential, which is exactly how the counter-intuitive upward saline-to-fresh direction in part (c) is revealed. The Hvorslev method treats the well/aquifer system as a first-order linear reservoir, so the head ratio decays exponentially with the "basic time lag" $T_0$ playing the role of a time constant.

Common pitfall. Assuming denser fluid always sits below and flows (if any) go downward is not a substitute for the head calculation — part (c) is a direct counter-example, and the only way to get the right direction is the freshwater-equivalent-head comparison. In part (b), students often forget that vertical continuity requires the SAME Darcy velocity in every layer and instead (incorrectly) compute a different $v_i=K_i\,i_i$ per layer using the same overall gradient, which violates flux continuity entirely.

Source. Freeze & Cherry, Groundwater, Ch. 2.6 (layered-medium conductivity), Ch. 2.11 (density-dependent flow) and Ch. 8.4 (slug tests, Hvorslev method).

Question 2 of 5: Layered Confined Aquifer, Vertical Flow, and a Density-Driven Interface and Slug Test

Question 2: Layered Confined Aquifer, Vertical Flow, and a Density-Driven Interface and Slug Test (equal value)