Question

A stratified soil deposit has three layers of thicknesses \(z_1=4,\;z_2=1,\;z_3=2\) units and the corresponding permeabilities of \(k_1=2,\;k_2=1\) and \(k_3=4\) units, respectively. The average permeability perpendicular to the bedding planes will be

• A. \(4\)
• B. \(2\)
• C. \(8\)
• D. \(16\)

Hint:

For layered soils: Parallel to bedding: \[ k_h=\frac{\sum k_i z_i}{\sum z_i} \] Perpendicular to bedding: \[ k_v=\frac{\sum z_i}{\sum\left(\frac{z_i}{k_i}\right)} \] Use the harmonic mean for flow perpendicular to bedding.

Answer 1

The Correct Option is B

Concept: For a stratified soil deposit, the equivalent permeability perpendicular to the bedding planes is given by the harmonic mean, \[ \boxed{k_v=\frac{\sum z_i}{\sum \left(\frac{z_i}{k_i}\right)}} \] where \[ z_i=\text{Thickness of }i^{th}\text{ layer}, \] and \[ k_i=\text{Permeability of }i^{th}\text{ layer}. \]

Step 1: Write the given data. \[ z_1=4,\qquad z_2=1,\qquad z_3=2 \] \[ k_1=2,\qquad k_2=1,\qquad k_3=4 \]

Step 2: Calculate the total thickness. \[ \sum z_i=4+1+2=7 \]

Step 3: Calculate the denominator. \[ \sum\frac{z_i}{k_i} = \frac42+\frac11+\frac24 = 2+1+\frac12 = 3.5 \]

Step 4: Compute the average permeability. \[ k_v = \frac{7}{3.5} = 2 \] Hence, \[ \boxed{k_v=2} \] Therefore, the correct option is \[ \boxed{(B)\;2.} \]