Example:

You have just pressed a pellet (disk) of kaolinite to be used for a flow through experiment to study fluid-rock interactions. You need to know the porosity of the pellet. Here's one way to proceed.

 

Given:

Io = 10,000 counts per second (CPS). This is the intensity measurement with no sample in place.

I = 3,500 CPS Intensity of transmitted beam with sample in place.

Kaolinite r = 2.6 g/cm3

Kaolintite µ* = 30.8 cm2/g

x = 0.02 cm (thickness of pellet)

 

Assuming that air does not influence the absorption is not totally true. However, given the small value of r for air ~ 0.0 g/cm3, we can simplify the calculation. The pellet is a mixture of air and kaolinite.

I = Io exp(-µ*r x) mixture

3500 = 10000 exp(-µ*r 0.02)

0.35 = exp(-µ*r 0.02)

Find natural log of each side.

-1.0498 = - (µ* r) 0.02

52.49 = (µ* r) mixture

 

The value of 52.49 represents kaolinite's absorption contribution to the thickness of the pellet.

If the pellet were 100% kaolinite, then (µ* r) kaolinite = 80.08

52.49 = (30.8) (2.6) (fraction of thickness)kaolinite

Thicknesskaolinite = 0.66

Therefore, the porosity = 34%