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%
$=0.2$
$=0.2$
$\rho_{w}=0.005$
$\rho_{t}=50$
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,
$S_{h}+S_{w}=1$
$S_{h}=1-S_{w}$
$S_{h}=1-0.05$
$S_{h}=0.95$
Excellent, Both did great job
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