Question

In a sandstone reservoir, the density log reads 2.11 g/cc and sonic log reads 90 μs/ft. The other parameters are given below: Matrix density (\(\rho_{ma}\)) = 2.68 g/cc Fluid density (\(\rho_{fl}\)) = 1.0 g/cc Compressional wave travel time in matrix (\(\Delta t_{ma}\)) = 54 μs/ft Compressional wave travel time in fluid (\(\Delta t_{fl}\)) = 189 μs/ft The calculated secondary porosity of the reservoir is ________________ % (rounded off to the nearest integer).

Hint:

Secondary porosity is estimated by subtracting sonic porosity from density porosity when both logs are available.

Answer 1

Correct Answer: 6

Density porosity: \[ \phi_d = \frac{\rho_{ma} - \rho_b}{\rho_{ma} - \rho_{fl}} = \frac{2.68 - 2.11}{2.68 - 1.0} = \frac{0.57}{1.68} \approx 0.339 \] Sonic porosity using Wyllie time-average: \[ \phi_s = \frac{\Delta t - \Delta t_{ma}}{\Delta t_{fl} - \Delta t_{ma}} = \frac{90 - 54}{189 - 54} = \frac{36}{135} \approx 0.266 \] Secondary porosity: \[ \phi_{sec} = \phi_d - \phi_s = 0.339 - 0.266 = 0.073 \approx 7% \] Thus the secondary porosity lies between 6% and 8%. Final Answer: 6–8%