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
Osubscostompa Karen Carter Yandex browser
ReplyDeleteSlack
Winamp Pro
dinoverwa