A non-homogenous soil deposit consists of a slit layer sandwiched between the fine-sand layer at top and clay layer below. Permeability of the silt layer is 10 times the permeability of the clay layer and one-tenth of the permeability of the sand layer. Thickness of the silt layer is 2 times the thickness of the sand layer and two-third of the thickness of the clay layer. The ratio of equivalent horizontal and equivalent vertical permeability of the deposit is.
10.9684

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Solution

The correct option is A 10.9684
$K_{2} = 10 K_{3} o r K_{3} = \frac{K_{2}}{10};$

$K_{2} = \frac{1}{10} K_{1} o r K_{1} = 10 K_{2};$

$Z_{2} = 2 Z_{1} o r Z_{1} = \frac{Z_{2}}{2};$

$Z_{2} = \frac{2}{3} Z_{3} o r Z_{3} = \frac{3}{2} Z_{2}$

$\Rightarrow K_{x} = \frac{K_{1} Z_{1} + K_{2} Z_{2} + K_{3} Z_{3}}{Z_{1} + Z_{2} + Z_{3}}$

$= \frac{\frac{10 K_{2} Z_{2}}{2} + K_{2} Z_{2} + (\frac{K_{2}}{10} \times \frac{3}{2} Z_{2})}{\frac{Z_{2}}{2} + Z_{2} + \frac{3}{2} Z_{2}}$

$= \frac{5 K_{2} Z_{2} + K_{2} Z_{2} + 0.15 K_{2} Z_{2}}{3 Z_{2}}$

$= \frac{6.15}{3} \times K_{2} = 2.05 K_{2}$

$K_{z} = \frac{\frac{Z_{2}}{2} + Z_{2} + \frac{3}{2} Z_{2}}{\frac{Z_{2}}{2 \times 10 K_{2}} + \frac{Z_{2}}{K_{2}} + \frac{1.5 Z_{2} \times 10}{K_{2}}}$

$= \frac{3 Z_{2}}{\frac{Z_{2}}{K_{2}} (\frac{1}{20} + 1 + 15)} = 0.1869 K_{2}$

$= \frac{K_{x}}{K_{z}} = \frac{2.05 K_{2}}{0.1869 K_{2}} = 10.9684$

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