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:

To calculate the compressional wave velocity, ensure that all the moduli are in SI units (Pa), and the density is in kg/m³. Pay attention to unit conversions when needed.

Answer 1

The compressional wave velocity (\( V_p \)) can be calculated using the following formula:
\[ V_p = \sqrt{\frac{K + \frac{4}{3} \mu}{\rho}} \] where:
- \( K \) is the bulk modulus (36 GPa),
- \( \mu \) is the shear modulus (30 GPa),
- \( \rho \) is the density of the formation (calculated below).
First, we calculate the density of the formation using the formula:
\[ \rho = \phi \rho_f + (1 - \phi) \rho_m \] where:
- \( \phi \) is the porosity (0.15),
- \( \rho_f \) is the fluid density (1.0 g/cc),
- \( \rho_m \) is the matrix density (2.65 g/cc).
Substitute the values into the equation:
\[ \rho = (0.15)(1.0) + (1 - 0.15)(2.65) = 0.15 + 2.2525 = 2.4025 \, {g/cc} = 2402.5 \, {kg/m}^3 \] Now, convert the units of \( K \) and \( \mu \) to the SI unit of Pascals (Pa):
\[ K = 36 \times 10^9 \, {Pa}, \quad \mu = 30 \times 10^9 \, {Pa} \] Substitute the values of \( K \), \( \mu \), and \( \rho \) into the formula for \( V_p \):
\[ V_p = \sqrt{\frac{36 \times 10^9 + \frac{4}{3}(30 \times 10^9)}{2402.5}} = \sqrt{\frac{36 \times 10^9 + 40 \times 10^9}{2402.5}} \] \[ V_p = \sqrt{\frac{76 \times 10^9}{2402.5}} = \sqrt{31.61 \times 10^6} = 5622.7 \, {m/s} \] Thus, the compressional wave velocity in the formation is approximately \( 5.62 \times 10^3 \, {m/s} \), which lies between 5.50 and 5.70 x 10³ m/s.