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 \(\underline{\hspace{1cm}} \times 10^3 \, m/s\) (rounded off to two decimal places).
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Solution and Explanation
Step 1: Effective density of formation.
Bulk density: \[ \rho = (1 - \phi)\rho_m + \phi \rho_f \] \[ \rho = (1 - 0.15)(2.65) + (0.15)(1.0) = 2.2525 \, g/cc \] \[ \rho = 2252.5 \, kg/m^3 \]
Step 2: P-wave velocity relation.
The compressional wave velocity is: \[ V_p = \sqrt{\frac{K + \frac{4}{3}G}{\rho}} \] where \(K = 36 \, GPa = 36 \times 10^9 \, Pa\), \(G = 30 \times 10^9 \, Pa\).
Step 3: Compute numerator.
\[ K + \frac{4}{3}G = 36 \times 10^9 + \frac{4}{3}(30 \times 10^9) \] \[ = 36 \times 10^9 + 40 \times 10^9 = 76 \times 10^9 \]
Step 4: Velocity calculation.
\[ V_p = \sqrt{\frac{76 \times 10^9}{2252.5}} = \sqrt{3.373 \times 10^7} \] \[ V_p \approx 5807 \, m/s = 5.81 \times 10^3 \, m/s \]
Final Answer: \[ \boxed{5.81 \times 10^3 \, m/s} \]
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