# Porosities, velocities, and densities of rocks

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 5.3a

Assume that sandstone is composed only of grains of quartz, limestone only of grains of calcite, and shale of equal quantities of kaolinite and muscovite. For sandstone, limestone, and shale saturated with salt water ${\displaystyle (\rho =1.03\ {\rm {g/cm}}^{3})}$, what porosities are implied by the densities shown in Figure 5.3a? (Mineral densities are: ${\displaystyle \rho _{\rm {quartz}}=2.68}$; ${\displaystyle \rho _{\rm {calcite}}=2.71}$; ${\displaystyle \rho _{\rm {kaolinite}}=2.60}$; ${\displaystyle \rho _{\rm {muscovite}}=2.83}$, all in g/cm${\displaystyle ^{3}}$.)

### Background

Gardner et al. (1974) plotted the log of velocity against the log of density for sedimentary rocks and obtained the empirical relation known as Gardner’s rule:

 {\displaystyle {\begin{aligned}\rho =aV^{1/4}\end{aligned}}} (5.3a)
Table 5.3a. Rock densities and porosities.
Porosity
Rock Density range Density av. Mineral density Max Av. Min
Ss 2.00–2.60 g/cm${\displaystyle ^{3}}$ 2.35 g/cm${\displaystyle ^{3}}$ 2.68 g/cm${\displaystyle ^{3}}$ 41% 20% 5%
Ls 2.20–2.75 2.55 2.71 30 10 0
Sh 1.90–2.70 2.40 2.72 48 19 0

${\displaystyle \rho }$ being in g/cm${\displaystyle ^{3}}$ and ${\displaystyle a=0.31}$ or 0.23 when ${\displaystyle V}$ is in m/s or ft/s, respectively. The rule is valid for the major sedimentary rock types, but not for evaporites or carbonaceous rocks (coal, lignite).

When a porous rock is saturated with a fluid, its density ${\displaystyle \rho _{s}}$ is given by the equation

 {\displaystyle {\begin{aligned}\rho _{s}=\phi \rho _{f}+\left(1-\phi \right)\rho _{m},\end{aligned}}} (5.3b)

${\displaystyle \phi }$ being the porosity, ${\displaystyle \rho _{f}}$ and ${\displaystyle \rho _{m}}$ the densities of the fluid and rock matrix, respectively.

### Solution

The density ranges in Table 5.3a were obtained from Figure 5.3a. The mineral densities are for ${\displaystyle \phi =0}$, the values for shale being averages for kaolinite and muscovite.

We solve equation (5.3b) for ${\displaystyle \phi }$, obtaining

 {\displaystyle {\begin{aligned}\phi ={\frac {\rho _{m}-\rho _{s}}{\rho _{m}-\rho _{f}}}.\end{aligned}}} (5.3c)
Figure 5.3a.  Histogram of rock densities (from Grant and West, 1965).

The histogram in Figure 5.3a does not encompass the complete range of samples and the range limits have been picked somewhat arbitrarily. Porosity in rocks ranges from about 50% to 0%. The upper limits of the density range sometimes exceed the mineral densities, hence heavier minerals must be present in the rocks; in these cases we assume that ${\displaystyle \phi =0}$. We take ${\displaystyle \rho _{f}=1.03\ {\rm {g/cm}}^{3}}$ as the fluid density.

## Problem 5.3b

What velocities would be expected for the density values in Table 5.3a according to Gardner’s rule? Plot these on Figure 5.3b.

### Solution

We solve equation (5.3a) for the velocity ${\displaystyle V}$, obtaining

{\displaystyle {\begin{aligned}V=(\rho /a)^{4}=(\rho /0.31)^{4}\times 10^{-3}=0.11\rho ^{4}\ {\rm {km/s}}.\end{aligned}}}

The velocities in Table 5.3b are plotted as triangles on Figure 5.3b.

Table 5.3b. Velocities obtained from densities in Table 5.3a.
Rock ${\displaystyle V_{\hbox{min}}\left({\hbox{km/s}}\right)}$ ${\displaystyle V_{\hbox{av}}\left({\hbox{km/s}}\right)}$ ${\displaystyle V_{\hbox{max}}\left({\hbox{km/s}}\right)}$
Ss 1.8 (41%) 3.4 (20%) 6.8 (0%)
Ls 2.6 (30%) 4.6 (10%) 6.8 (0%)
Sh 1.4 (48%) 3.4 (19%) 6.8 (0%)
 Note: The values in parentheses are the porosities.
Figure 5.3b.  P-wave velocities from different sources. (1) Grant and West, 1965; (2) Kearey and Brooks, 1984; (3) Lindseth, 1979; (4) Mares, 1984; (5) Sharma, 1997; (6) Sheriff and Geldart, 1995; (7) Waters, 1987.

## Problem 5.3c

From Figure 5.3c, what densities would you expect at 7500 ft and how do these compare with Figures 5.3d and 5.3e from offshore Louisiana?

Figure 5.3c.  Porosity versus depth (from Atkins and McBride, 1992).
Figure 5.3d.  Average sand density between 7000 and 8000 ft (courtesy Geophy. Develop. Corp.).
Figure 5.3e.  Average shale density between 7000 and 8000 ft (courtesy Geophy. Develop. Corp.).

### Solution

Using ${\displaystyle \rho _{\rm {quartz}}=2.68\ {\rm {g/cm}}^{3}}$ and ${\displaystyle \phi =31\%}$ from Figure 5.3c, equation (5.3c) gives ${\displaystyle \rho =2.19\ {\rm {g/cm}}^{3}}$, which is in accord with Figure 5.3d. Using ${\displaystyle \rho _{\rm {shale\ minerals}}=2.71}$, equation (5.3c) gives ${\displaystyle \rho =2.22\ {\rm {g/cm}}^{3}}$, which is slightly lower than most values in Figure 5.3e.