GATE-2018 (24) :Wyllie – Time average equation

 

70) The SP response of a thick, clean sandstone bed is -54mV. Given the mud filtrate resistivity to be 0.45 Ωm at a formation temperature(T_f) of 130⁰ F and the coefficient K=77.29, the formation water resistivity is _____Ωm.

 (Thanks to Chandrasekhar, ANU)

Solution:

The SP response can be defined as $SP =-K log(\frac{\rho_{mf}}{\rho_{w}})----(1)$

  Given that SP=-54mV

   Resistivity of mud filtrate$ (\rho_{mf}) $= 0.45 Ωm

   Co-efficient (K) =77.29

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

 By substituting the given value in the equation (1)

 

$SP =-K log(\frac{\rho_{mf}}{\rho_{w}})$

 

$-54 =-77.29 log(\frac{0.45}{\rho_{w}})$

 

$\frac{54}{77.29} = log(\frac{0.45}{\rho_{w}})$

 

$0.6986 = log(\frac{0.45}{\rho_{w}})$

 

$10^{0.6986}=\frac{0.45}{\rho_{w}}$

 

$4.996=\frac{0.45}{\rho_{w}}$

 

$\rho_{w}=\frac{0.45}{4.996}$

 

$\rho_{w}=0.09 $Ωm