Question 2 of 5: Layered Confined System — Effective Conductivities, Vertical Flow, and Regional Discharge

Question 2: Layered Confined System — Effective Conductivities, Vertical Flow, and Regional Discharge (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. Three horizontal layers of intrinsic permeability and thickness as below; the exam's default fluid properties (1000 kg/m³, 0.001 kg/m-sec, $g=9.81\ \text{m/s}^2$) convert permeability to hydraulic conductivity since no test temperature is stated for this question. (b) pressure head 20 m of water at the top of the system, 90 m of water at the base. (c) piezometers 200 m apart, water levels 260 m and 245 m, aquifer extent 200 m perpendicular to flow.

Find. (a) effective horizontal $K_h$ and vertical $K_v$ conductivity of the layered system. (b) vertical Darcy velocity through the stack. (c) volumetric discharge through the aquifer.

Approach. Convert each layer's permeability to hydraulic conductivity via $K=k\rho g/\mu$, then combine horizontally as a thickness-weighted arithmetic mean and vertically as a thickness-weighted harmonic mean. Part (b) uses the vertical $K_v$ with a total-head difference built from elevation head plus pressure head across the full 80 m stack. Part (c) uses the horizontal $K_h$ with the piezometric gradient over the full cross-sectional area.

1. Convert permeability to hydraulic conductivity. $K=k\rho g/\mu=k\times(1000)(9.81)/(0.001)=k\times9.81\times10^{6}$: $$K_1=3.2\times10^{-11}\times9.81\times10^6=3.139\times10^{-4}\ \text{m/s},\quad K_2=4.2183\times10^{-5}\ \text{m/s},\quad K_3=1.962\times10^{-6}\ \text{m/s}.$$
2. Effective horizontal conductivity (thickness-weighted mean). With $B=25+20+35=80$ m: $$K_h=\frac{K_1b_1+K_2b_2+K_3b_3}{B}=\frac{(3.139\times10^{-4})(25)+(4.2183\times10^{-5})(20)+(1.962\times10^{-6})(35)}{80}=\boxed{1.10\times10^{-4}\ \text{m/s}}.$$
3. Effective vertical conductivity (thickness-weighted harmonic mean). $$K_v=\frac{B}{b_1/K_1+b_2/K_2+b_3/K_3}=\frac{80}{25/(3.139\times10^{-4})+20/(4.2183\times10^{-5})+35/(1.962\times10^{-6})}=\boxed{4.35\times10^{-6}\ \text{m/s}}.$$ $K_h/K_v\approx25.2$ — strongly anisotropic, as expected when a very low-$K$ bottom layer dominates the series (vertical) path while barely affecting the parallel (horizontal) average.
4. Part (b) — total head at top and bottom. Taking the base of the system as datum ($z=0$): $h_{\text{top}}=B+\psi_{\text{top}}=80+20=\boxed{100\ \text{m}}$, $h_{\text{bot}}=0+\psi_{\text{bot}}=\boxed{90\ \text{m}}$. Since $h_{\text{top}}>h_{\text{bot}}$, flow is downward through the system.
5. Vertical Darcy velocity. $$v=K_v\frac{h_{\text{top}}-h_{\text{bot}}}{B}=(4.35\times10^{-6})\left(\frac{100-90}{80}\right)=\boxed{5.44\times10^{-7}\ \text{m/s downward}}.$$
6. Part (c) — horizontal gradient and discharge. Gradient $i=(260-245)/200=0.075$; cross-sectional area $A=B\times(\text{extent})=80\times200=16{,}000\ \text{m}^2$: $$Q=K_hiA=(1.095\times10^{-4})(0.075)(16{,}000)=\boxed{0.131\ \text{m}^3\text{/s}\ (131\ \text{L/s})}.$$