# Velocities in limestone and sandstone

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 5 141 - 180 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 5.4

Assume that the velocity in calcite is 6.86 km/s and in quartz 5.85 km/s. What velocities should be expected for 10, 20, and 30% porosities in (a) limestone composed only of calcite; (b) sandstone composed only of quartz? Where do these values plot on lithology versus velocity curves (Figure 5.4a) and porosity versus velocity curves (Figure 5.4b)? Assume that the pore fluid is water with velocity 1.55 km/s.

### Background

The velocity in a porous rock depends primarily on the matrix velocity, the porosity, and the nature of and velocity in the pore fluid. The empirical time-average equation [of the same form as equation (5.3b)] relates the specific transit time (reciprocal of the velocity) or slowness to the volume fraction of pore space $\phi$ and the remaining volume $\left(1-\phi \right)$ :

 {\begin{aligned}1/V=\phi /V_{f}+\left(1-\phi \right)/V_{m},\end{aligned}} (5.4a)

where $V$ , $V_{f}$ , and $V_{m}$ are the velocities in the saturated rock, the fluid, and the rock matrix, respectively.

### Solution

Substituting $V_{f}=1.55\ {\rm {km/s}}$ in equation (5.4a) yields $V={\frac {1.55}{\phi +(1-\phi )(1.55/V_{m})}}$ . For $\phi$ values of 10, 20, and 30%, we obtain the velocities in Table 5.4a. Figure 5.4b.  Velocity versus porosity (after Wyllie et al., 1958). The limestone values (plotted as crosses on both Figures 5.4a and 5.4b) are slightly high and the sandstone values (plotted as circles) are reasonable. The line in Figure 5.4b is the time-average equation.

Table 5.4a. Velocities calculated from porosities.
Rock $V_{m}$ $\phi =10\%$ $\phi =20\%$ $\phi =30\%$ Limestone 6.86 km/s 5.11 km/s 4.07 km/s 3.38 km/s
Sandstone 5.85 4.58 3.76 3.19