Exercise 5 Solution

\displaystyle \textrm{horizontal }K=\frac{1000\;\frac{\textrm{m}}{\textrm{d}}(10\;\textrm{m})+1\;\frac{\textrm{m}}{\textrm{d}}(10\;\textrm{m})}{20\;\textrm{m}}=\frac{100010\;\frac{\textrm{m}^2}{\textrm{d}}}{20\;\textrm{m}}=5000.5\;\frac{\textrm{m}}{\textrm{d}}

\displaystyle \textrm{vertical }K=\frac{20\;\textrm{m}}{\frac{10\;\textrm{m}}{1000\;\frac{\textrm{m}}{\textrm{d}}}+\frac{10\;\textrm{m}}{1\;\frac{\textrm{m}}{\textrm{d}}}}=\frac{20\;\textrm{m}}{0.01\;\textrm{d}+10\;\textrm{d}}=1.98\;\frac{\textrm{m}}{\textrm{d}}

For a three-order of magnitude difference in hydraulic conductivity, horizontal flow is through the more permeable layer and the equivalent K is dominated by the more permeable layer, while in the vertical direction flow must pass through the low conductivity layer so the equivalent K is much lower. This is equivalent to the phenomenon of electrical current flowing through resistors in parallel and in series.

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Introduction to Karst Aquifers Copyright © 2022 by Eve L. Kuniansky, Charles J. Taylor, and Frederick Paillet. All Rights Reserved.