Answer to Question #169865 in Mechanical Engineering for Rachel

Question

Four horizontal, homogeneous, isotropic geologic formations, each 5m thick over lie one another. If the hydraulic conductivities are 10-4,10-6,10-4 and 10-6 respectively,, calculate the horizontal and vertical components of hydraulic conductivity for the equivalent homogeneous but anisotropic formation

Answer 1

Here different conductivity is given as

$K_1= 10^{-4},K_2= 10^{-6},K_3=10^{-4} and K_4=10^{-6}$

here $z_1=z_2=z_3=z_4=5$

Now, horizantal component conductivity is given by following equation

$K_x= \frac{K_1z_1+K_2z_2+K_3z_3+K_4z_4}{z_1+z_2+z_3+z_4}$

$K_x= \frac{10^{-4}\times 5+10^{-6}\times 5+10^{-4}\times 5+10^{-6}\times 5}{5+5+5+5+5}=\frac{10\times 10^{-4}+10\times10^{-6}}{20}$

$K_x=5.05\times 10^{-5} \frac{m}{s}$

And vertical component hydraulic conductivity is given by following equation as

$K_z=\frac{ z_1+z_2+z_3+z_4}{\frac{z_1}{K_1}+\frac{z_2}{K_2}+\frac{z_3}{K_3}+\frac{z_4}{K_4}}$

$K_z=\frac{ 20}{\frac{5}{10^{-4}}+\frac{5}{10^{-6}}+\frac{5}{10^{-6}}+\frac{5}{10^{-4}}}$

$K_z= 1.98 \times 10^{-6}$ m/s