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# Answer to Question #169865 in Mechanical Engineering for Rachel

Question #169865

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

1
2021-04-01T01:03:39-0400

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

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