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GATE-2017 (70)

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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(Tf) 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

 

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