Question

The formation resistivity factor (\( F \)) is related to the formation porosity (\( \phi \)) in a water-bearing carbonate formation by the following correlation:
\[ F = 0.9 \phi^{-2} \] where \( \phi \) is in fraction. The resistivity of the invaded zone of the formation obtained by the Microspherically Focused Log (MSFL) is 4.5 \( \Omega \)m, and the resistivity of the mud-filtrate is 0.05 \( \Omega \)m. The formation porosity is .......... % (rounded off to one decimal place).

Hint:

In formation resistivity calculations, understanding the relationship between resistivity factors and porosity is key. Use the given formulas to solve for the unknown parameters.

Answer 1

In a water-bearing formation, the formation resistivity factor \( F \) is related to the porosity \( \phi \) by the equation:
\[ F = 0.9 \phi^{-2} \] The resistivity factor \( F \) can also be expressed using the formula:
\[ F = \frac{R_t}{R_0} \] where \( R_t \) is the resistivity of the formation and \( R_0 \) is the resistivity of the mud-filtrate. Given that \( R_t = 4.5 \, \Omega {m} \) and \( R_0 = 0.05 \, \Omega {m} \), we can calculate \( F \) as:
\[ F = \frac{4.5}{0.05} = 90 \] Now, using the first equation for \( F \), we substitute \( F = 90 \) and solve for \( \phi \):
\[ 90 = 0.9 \phi^{-2} \] Solving for \( \phi^{-2} \):
\[ \phi^{-2} = \frac{90}{0.9} = 100 \] Taking the reciprocal to find \( \phi \):
\[ \phi = \frac{1}{\sqrt{100}} = 0.1 \] Thus, the formation porosity \( \phi \) is 10.0%.