# Effect of burial history on velocity

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 5.6a

Assume a subsiding area where there has been no uplift and a shale that is normally pressured until it reaches a depth of burial of 1.40 km, at which point fluid communication is cut off, that is, interstitial fluid can no longer escape. If it is at a depth of 2.00 km, what velocity and fluid pressure would you expect? If at a depth of 3.00 km?

### Background

The normal fluid pressure at depth ${\displaystyle z}$ is ${\displaystyle z\rho _{f}}$, where ${\displaystyle \rho _{f}}$ is the fluid density. The fluid pressure gradient is ${\displaystyle \partial {\mathcal {P}}_{f}/\partial z=\rho _{f}\approx 10}$ MPa/km, whereas the pressure gradient due to the rock overburden is about 22.5 MPa/km. The differential pressure gradient is therefore approximately 12 MPa/km.

### Solution

At a depth of 1.40 km, the velocity is about 2.5 km/s (see Figure 5.6a) and the fluid pressure ${\displaystyle {\mathcal {P}}_{f}}$ is about 10 MPa/km ${\displaystyle \times }$ 1.4 km = 14 MPa, while the pressure on the matrix ${\displaystyle {\mathcal {P}}_{m}\approx 22.5\times 1.4=32\ {\rm {MPa}}}$, the differential pressure being 18 MPa. The porosity will be about 30% (see Figure 5.1a). If the shale is cut off at a depth of 1.4 km and the depth is then increased to 2.0 km, the differential pressure and velocity will not change greatly.

At 2.0 km, the overburden pressure ${\displaystyle {\mathcal {P}}_{m}=22.5\times 2.0=45\ {\rm {MPa}}}$, the fluid pressure will be about ${\displaystyle {\mathcal {P}}_{f}\approx 45-18=27\ {\rm {MPa}}}$. At 3.0 km, ${\displaystyle {\mathcal {P}}_{m}\approx 22.5\times 3.0=68}$ MPa and ${\displaystyle {\mathcal {P}}_{f}\approx 68-18=50\ {\rm {Mpa}}}$.

Figure 5.6a.  Velocity/depth relations (Venezuela data from Gregory, 1977). Triangles show data from problem 5.8, squares are from problem 5.9b.

## Problem 5.6b

Assume a shale buried to 3.00 km and then uplifted to 2.00 km, being normally pressured all the time. What velocity and fluid pressure would you expect?

### Solution

The velocity at 3.00 km is about 3.2 km/s from Figure 5.6a and the fluid pressure about 30 MPa. Burial to this depth will have reduced the porosity to about 25% (see Figure 5.1a) and very little of the porosity will be recovered upon uplift. Hence the velocity will be larger than for a situation where the rock had not been buried deeper.

## Problem 5.6c

Assume the shale in part (a) is buried to 3.00 km and then uplifted to 2.00 km without fluid communication being established. What velocity and fluid pressure would you expect? What if the shale is uplifted to 1.00 km?

### Solution

Since the differential pressure was not changed, we expect about the same result at 2.00 km as in part (a). At 2.00 km, ${\displaystyle {\mathcal {P}}_{f}\approx 20\ {\rm {MPa}}}$. When the shale is then uplifted to 1.40 km, its pressure will again be normal. If further uplifted to 1.00 km without fluid communication being established, the porosity will remain about 32%. The velocity and differential pressure will be about the same as they were at 1.40 km, but ${\displaystyle {\mathcal {P}}_{f}}$ will now be ${\displaystyle 22.5\times 1.00-18=4.5\ {\rm {MPa}}}$ (instead of the usual value of 10 Mpa); the shale will be under-pressured and will have approximately the same velocity as at 1.40 km.