Calculation of true thickness of the bed

Q) A sandstone formation has porosities varying from 10-30%. The 100% water saturated sand with 10% porosity has a formation resistivity of 0.5 Ωm. what is the hydrocarbon saturation (%) in the sand formation with 20% porosity and formation resistivity of 50 Ωm( assume a=1,n=2)

(Thanks to Chandrasekhar, ANU)

Solution:

Given that porosity varies $\phi$=10-30%

Formation resistivity ($\rho_ {0}$) = 0.5 Ωm when porosity ($\phi$) is 10%

Then resistivity of water ($\rho_ {w}$) =?

Formation factor $F=\frac{\rho_{0}}{\rho_{w}}$

$\frac{1}{\phi^{2}}=\frac{\rho_{0}}{\rho_{w}}$

$\rho_{w}=\phi^{2}\times\rho_{0}$

$\rho_{w}=(0.1)^2\times (0.5)$

$\rho_{w}=0.005$

For the calculation of hydrocarbon saturation we need to calculate the saturation of water

$S_{h}+S_{w}=1$

$S_{w}=\sqrt{\frac{F \rho_{w}}{\rho_{t}}}$

$S_{w}=\sqrt{\frac{\rho_{w}}{\phi^{2}\times\rho_{t}}}$

Here

 

$\phi$= 20%

      $=0.2$

$\rho_{w}=0.005$

$\rho_{t}=50$

Substitute these values in the above equation

We get,

$S_{w}=\sqrt{\frac{\rho_{w}}{\phi^{2}\times\rho_{t}}}$

$S_{w}=\sqrt{\frac{0.005}{(0.2)^{2}\times(50)}}$

$S_{w}=\sqrt{\frac{0.005}{2}}$

$S_{w}=0.05$

The relation between water saturation and hydrocarbon saturation is given that,

$S_{h}+S_{w}=1$

$S_{h}=1-S_{w}$

$S_{h}=1-0.05$

$S_{h}=0.95$

$S_{h}$=95%