Question

For a given reservoir of porosity \( \phi = 0.1 \), the water saturation is equal to the irreducible water saturation. The BVI (Bulk Volume Irreducible) water saturation was estimated as 0.04 using NMR (Nuclear Magnetic Resonance) log.
The permeability, k in mD, is estimated using the relation \[ k = 10^4 \phi^4 \left(\frac{FFI}{\phi - FFI}\right)^2 \], where FFI is the Free Fluid Index.
Based on this information, the permeability (in mD) is __________. (Rounded off to two decimal places)

Hint:

First find FFI from phi minus BVI, then substitute directly into the given permeability formula.

Answer 1

Correct Answer: 2.25

Step 1: Find the Free Fluid Index (FFI):
The total porosity is split between the bulk volume irreducible water, BVI, and the free fluid index, FFI, so $\phi = FFI + BVI$. Rearranging gives $FFI = \phi - BVI = 0.1 - 0.04 = 0.06$.
Step 2: Compute the ratio inside the bracket:
The formula needs $\phi - FFI$, which is $0.1 - 0.06 = 0.04$. The ratio $\frac{FFI}{\phi - FFI}$ is then $\frac{0.06}{0.04} = 1.5$.
Step 3: Square the ratio:
Squaring 1.5 gives $1.5^2 = 2.25$.
Step 4: Compute $10^4 \phi^4$:
With $\phi = 0.1$, $\phi^4 = 0.1^4 = 0.0001$. Multiplying by $10^4$ gives $10^4 \times 0.0001 = 1$.
Step 5: Combine both parts:
The permeability is $k = 10^4 \phi^4 \times \left(\frac{FFI}{\phi-FFI}\right)^2 = 1 \times 2.25 = 2.25$ mD.
Final Answer:
\[ \boxed{2.25 \text{ mD}} \]