Calculation of true thickness of the bed

111. The formation resistivity factors of two sedimentary rocks of porosity 10% and 20% are in the ratio 4:1. If the porosities of these are 12% and 24%, respectively, their formation resistivity factors are in the ratio of

1) 5:1                          

2) 4:1

3) 3.2:1                         

4) 2:1

Solution:

Given Porosities are,

$\phi_1$=10%=0.1

$\phi_2$=20%=0.2

The relation between formation factor and porosity is given form Archie's law.

$F =\frac{1}{\phi^2}$

$\frac{1}{\phi_1^2}=\frac{1}{(0.1)^2}$

$\frac{1}{\phi_2^2}=\frac{1}{(0.2)^2}$

$\frac{F_1}{F_2}=\frac{4}{1}$

$\frac{\frac{1}{\phi_1^2}}{\frac{1}{\phi_2^2}}=\frac{4}{1}$

$\frac{{\phi_2^2}}{{\phi_1^2}}=\frac{4}{1}$

$\frac{(0.2)^2}{{(0.1)^2}}=\frac{4}{1}$

$\frac{(0.04)}{{(0.01)}}=\frac{4}{1}$



$\frac{(4)}{{(1)}}=\frac{4}{1}$


Therefore, above relation is correct.

If porosities are 12% and 24% then what is the ratio of formation factor





$\phi_1$=12%=0.12


$\phi_2$=24%=0.24


$\frac{F_1}{F_2}$=?


$\frac{\frac{1}{\phi_1^2}}{\frac{1}{\phi_2^2}}$=?


$\frac{\phi_2^2}{\phi_1^2}$=?


=$\frac{(0.24)^2}{(0.12)^2}$


=$\frac{0.24*0.24}{0.12*0.12}$


=$\frac{4}{1}$