How Is the Effective Conductivity of Sediment Layers Calculated?

[KEØM]

Homework Statement

We consider the following configuration. A pile of N layers of sediments. Each layer has a conductivity \sigma_{i} (i from 1 to N) and a thickness m_{i}. The length of the system is L. In this case, the electrical conductivity is not a scalar but a second order symmetric tensor.

Using Ohm's Law and the idea of resistors working in parallel, demonstrate that the effective conductivity of the block along the x-direction is:

\sigma_{x} = \frac{\Sigma_{i=1}^{N} m_{i}\sigma_{i}}{\Sigma_{i=1}^{N} m_{i}}<br />

Homework Equations

\vec{j} = -\underline{\sigma} \vec{\nabla} V, <br /> <br /> \underline{\sigma} = \begin{bmatrix}<br /> \sigma_{x} &amp; 0 \\<br /> 0 &amp; \sigma_{z} <br /> \end{bmatrix}} <br /> <br />

The Attempt at a Solution

I am not exactly sure if this qualifies as advanced physics. This is a geophysics course and I have never seen this before (not even in the course).

The first thing I did was I used the product over sum formula for two parallel resistors.

R_{eq} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}

Now using the formula R = \frac{\rho}{g}

where \rho is the resistivity of the specific layer and g is the geometrical factor (I don't know what this is in this case and I don't know how to use it).

Substituting this in for R and using \rho = \frac{1}{\sigma} this parallel equation, after a lot of algebra, becomes:

R_{eq} = \frac{1}{g_{1} \sigma_{1} + g_{2}\sigma_{2}}

Now, can I set R_{eq} = \frac{1}{\sigma_{x} g_{eq}} , g_{eq} = g_{1} + g_{2}?

And then can I go as far as setting g equal to m?

If that is the case then I will have the correct answer because,

\frac{1}{\sigma_{x}(m_{1} + m_{2})} = \frac{1}{m_{1} \sigma_{1} + m_{2}\sigma_{2}}<br /> <br /> \Rightarrow \sigma_{x} = \frac{m_{1} \sigma_{1} + m_{2}\sigma_{2}}{m_{1} + m_{2}}.<br /> <br />

Which is the formula for the specific case of N =2.

Could someone please verify my approach or let me know how horribly wrong I am?

Thanks in advance,

KEØM

 

[KEØM]

I did some thinking and I think I got it. If I manipulate \vec{j} = -\underline{\sigma} \vec{\nabla}V into V = IR by letting \nabla V_{x} = \frac{\Delta V}{\Delta x} and

\nabla V_{z} = \frac{\Delta V}{\Delta z}

then the magnitude of my first equation will become

j = \sigma_{x}\frac{\Delta V}{\Delta x} in the x-direction

and

j = \sigma_{z}\frac{\Delta V}{\Delta z} in the z-direction.

Then in the x-direction:\frac{I}{A}= \sigma_{x}\frac{\Delta V}{\Delta x} <br /> <br /> \Rightarrow \Delta V = \left( \frac{\Delta x}{A \sigma_{x}} \right)I.<br /> <br />

But as is apparent in the diagram (I will post this in a bit)\Delta x = L and A = \Delta x m = Lm

The above equation then becomes:

\Delta V = \left( \frac{\rho_{x}L}{mL} \right)I

The L's cancel and this gives me V = IR where

R = \frac{\rho_{x}}{m}}.

Hence m is my geometrical factor.

I do this same process for the z-direction (with \Delta z = m) and I get my R to be:

R = \frac{\rho_{z}}{L}}.

So then L must be my geometrical factor for the z-direction? A problem further into my worksheet asks for me to give the expression for \sigma_{z} similar to that of \sigma_{x} in the first problem. But using L as my geometrical factor I do not get the correct result.

Can someone please verify my answers here and help me with the geometrical factor in the z-direction?

Thanks in advance,

KEØM

 

[KEØM]

Here is the link to the file which contains my worksheet. The picture that describes the configuration of the layers is at the bottom of the page.

https://docs.google.com/fileview?id...QtNTg1ZC00N2M3LTliODEtOTBjZGYzZmMzZDE4&hl=en"

Thanks again,

KEØM