Question

The porosity of a formation with matrix density of 2.65 g/cc and fluid density of 1.0 g/cc is 0.15. The formation has shear modulus of 30 GPa and bulk modulus of 36 GPa. The compressional wave velocity in the formation is _________ $\times 10^3$ m/s (rounded off to two decimal places).

Hint:

For isotropic elastic media: \[ V_p = \sqrt{\frac{K+\frac{4}{3}G}{\rho}} \] Always convert: g/cc $\rightarrow$ kg/m$^3$ (multiply by 1000), and GPa $\rightarrow$ Pa (multiply by $10^9$).

Answer 1

Correct Answer: 5.62

Step 1: Find the bulk density of the fully saturated rock. 
The bulk density of a porous, fluid-filled formation is calculated using:

\[ \rho = (1-\phi)\rho_m + \phi \rho_f \]

Here, the porosity is $\phi = 0.15$, the matrix density is $\rho_m = 2.65$ g/cc, and the fluid density is $\rho_f = 1.0$ g/cc.

Substituting the values:

\[ \rho = 0.85(2.65) + 0.15(1.0) = 2.2525 + 0.15 = 2.4025 \text{ g/cc} \]

Converting this to SI units:

\[ 2.4025 \text{ g/cc} = 2402.5 \text{ kg/m}^3 \]

Step 2: Apply the P-wave velocity formula.
The compressional wave velocity is given by:

\[ V_p = \sqrt{\frac{K + \frac{4}{3}G}{\rho}} \]

The bulk modulus is $K = 36$ GPa and the shear modulus is $G = 30$ GPa.

First, evaluate the elastic term:

\[ K + \frac{4}{3}G = 36 + \frac{4}{3}(30) = 36 + 40 = 76 \text{ GPa} \]

Converting gigapascals to pascals:

\[ 76 \text{ GPa} = 76 \times 10^9 \text{ Pa} \]

Step 3: Calculate the P-wave velocity.

\[ V_p = \sqrt{\frac{76 \times 10^9}{2402.5}} = \sqrt{31.634 \times 10^6} \approx 5624.39 \text{ m/s} \]

Step 4: Write the final value in standard form.
 

\[ 5624.39 \text{ m/s} \approx \boxed{5.62 \times 10^3 \text{ m/s}} \]