Question

A rock is fully saturated with water of resistivity \(\rho_w=0.25\ \Omega\text{m}\). The bulk resistivity is \(\rho=60\ \Omega\text{m}\). Using Archie's law with \(a=1\) (tortuosity factor) and \(m=2\) (cementation exponent), determine the porosity \(\phi\) in percent. \([\,\text{round off to 2 decimals}\,]\)

Hint:

Mnemonic for \(a=1,m=2\): \(\phi(%)\approx 100\sqrt{\rho_w/\rho}\). Ensure units of \(\rho\) and \(\rho_w\) match.

Answer 1

Step 1: Write Archie's law for fully saturated rock.
For brine-saturated clean formations: \[ \rho=a\,\rho_w\,\phi^{-m} \quad \Rightarrow \quad \frac{\rho}{\rho_w}=a\,\phi^{-m}. \] Given \(a=1,\ m=2\), \[ \phi^{-2}=\frac{\rho}{\rho_w}\quad\Rightarrow\quad \phi=\left(\frac{\rho_w}{\rho}\right)^{1/2}. \]

Step 2: Insert numbers and keep significant figures.
\[ \phi=\sqrt{\frac{0.25}{60}} =\sqrt{0.0041666667} =0.064549722\ldots \] Convert to percentage: \[ \phi(%)=0.064549722\times100=6.4549722%. \]

Step 3: Round and sanity-check.
- Rounded to two decimals: \(6.45%\). - Check: Higher rock resistivity relative to water implies low porosity (current path restricted) — value \(6.45%\) is physically reasonable for a tight rock.

Final Answer:\ \(\boxed{6.45%}\)