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: Compute bulk density of the saturated formation.
Bulk density is: \[ \rho = (1-\phi)\rho_m + \phi \rho_f \] Given $\phi = 0.15$, $\rho_m = 2.65$ g/cc, $\rho_f = 1.0$ g/cc: \[ \rho = 0.85(2.65) + 0.15(1.0) = 2.2525 + 0.15 = 2.4025 \text{ g/cc} \] Convert to SI units: \[ 2.4025 \text{ g/cc} = 2402.5 \text{ kg/m}^3 \] Step 2: Use P-wave velocity relation.
Compressional (P-wave) velocity is: \[ V_p = \sqrt{\frac{K + \frac{4}{3}G}{\rho}} \] Given $K = 36$ GPa and $G = 30$ GPa: \[ K + \frac{4}{3}G = 36 + \frac{4}{3}(30) = 36 + 40 = 76 \text{ GPa} \] \[ 76 \text{ GPa} = 76 \times 10^9 \text{ Pa} \] Step 3: Calculate $V_p$.
\[ V_p = \sqrt{\frac{76 \times 10^9}{2402.5}} = \sqrt{31.634 \times 10^6} \approx 5624.39 \text{ m/s} \] Step 4: Express in the required form.
\[ 5624.39 \text{ m/s} = 5.62439 \times 10^3 \text{ m/s} \approx \boxed{5.62 \times 10^3 \text{ m/s}} \]