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CSIR-NET 2019 DEC

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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%



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