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).
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).
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Solution and Explanation
Step 1: Recall definition of resistivity factor.
\[ F = \frac{R_t}{R_w} \] where \(R_t\) is true formation resistivity (from MSFL invaded zone log), and \(R_w\) is mud-filtrate resistivity.
Step 2: Substitute given values.
\[ F = \frac{4.5}{0.05} = 90 \]
Step 3: Apply correlation.
\[ F = 0.9 \phi^{-2} \] \[ 90 = 0.9 \phi^{-2} \] \[ \phi^{-2} = \frac{90}{0.9} = 100 \] \[ \phi = \frac{1}{\sqrt{100}} = 0.1 \]
Step 4: Convert to percentage.
\[ \phi = 0.1 \Rightarrow 10.0 \% \]
Final Answer: \[ \boxed{10.0 \%} \]
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