9
$\begingroup$

Why is the geostatic (or is it the lithostatic?) gradient 1 psi/ft? How is that derived from g/cc? Thanks.

$\endgroup$
9
$\begingroup$

Lithostatic pressure might be roughly 23 kPa/m (about 1 psi/ft) in young basins dominated by siliciclastic rocks, but it varies substantially. Pressure has a linear relationship to density. Let's explore the relationship, and the variance.

First some definitions:

  • Lithostatic pressure — vertical pressure due to the weight of rock and fluid.
  • Hydrostatic pressure — vertical pressure due to the weight of fluid only.
  • Fracture pressure — the pressure at which the rock fractures.
  • Pore pressure — the pressure in the pores of a rock; usually close to hydrostatic.

There's a lot of jargon in this field — lithostatic pressure is sometimes called confining pressure, overburden pressure, vertical stress, total stress, and various other things. I hadn't heard of geostatic pressure before, and a quick bit of research revealed that some people call the pressure due to rock and fluid 'lithostatic' and some call it 'geostatic', reserving 'lithostatic' for the pressure due to the dry rock only. In my experience, it doesn't pay to be pedantic about these inconsistencies, all you can do is spell out what you mean.

We'll do a quick derivation (there's a more comprehensive one here). Pressure $P$ is force $F$ per unit area, and Newton told us that $F = ma$, where $m$ is mass and $a$ is acceleration. Since we're interested in vertical pressure, acceleration is due to gravity so we can use $g$. We know the mass of a rock normalized to volume as density $\rho$, and we're interested in a depth $d$, so:

$$P = \rho d g$$

The units are $[\mathrm{kg/m}^3] \times [\mathrm{m}] \times [\mathrm{m/s}^2]$ or newtons per square metre, N/m$^2$, which we call pascals, Pa, for convenience (10 kPa is 1.45 psi). Note that in the real world density varies with depth, so you'll need to integrate.

Because the pressure increases roughly linearly with depth, it's often convenient to talk about the pressure gradient, which is usually expressed in units of kPa/m, and of course is just the gradient of a line on a pressure–depth plot, like in the example below from a fictional offshore well.

So where does 23 kPa/m come from? If we take an approximate average density of sedimentary rock as 2300 kg/m$^3$, then $P = 2300 \times g$ Pa/m, or about 23 kPa (we often round $g$ to 10). Different mineralogies, different porosities, different fluids — all these will change the lithostatic pressure gradient.

Pressure in an offshore well

$\endgroup$
  • $\begingroup$ Using the equation found in this link what are the assumed porosity and water density required to get the approximate avg. density of sedimentary rock as 2.3 g/cc or 2300 kg/m3? $\endgroup$ – Armadillo Dec 9 '14 at 15:00
  • $\begingroup$ I still don't see how the gradient is found. I see how you got the pressure: $$2300 \frac{kg}{m^3} \cdot 10 \frac{m}{s^2}=23,000 \frac{kg \ m }{m^3 \ s^2} = 23,000 Pa$$ $$23,000 Pa \cdot \frac{1.45 psi}{10,000 Pa}=3.335 psi$$ how do we get to 1 psi/ft? $\endgroup$ – Armadillo Dec 9 '14 at 15:23
  • $\begingroup$ We're just considering a depth of 1 m to make the math simpler (since our model has one rock it's a constant gradient). We get 23 kPa for that metre, and as you show that is 3.335 psi/m, or 1.02 psi/ft (note the change in depth units). $\endgroup$ – kwinkunks Dec 9 '14 at 17:40
  • $\begingroup$ I totally agree with you about spelling out what you intend to mean. $\endgroup$ – Armadillo Dec 9 '14 at 18:04
5
$\begingroup$

Overburden/Lithostatic/Geostatic pressure (or stress) at a given depth is the vertical pressure (or stress) due to the weight of a column of rock and the fluids contained in the rock above that depth. The pressure applied by a resting rock mass (this includes the fluids within the rock's pore space) under the acceleration of gravity is related to the rock's mass density by the formula,

$$\tag{1} p(z) = p_o+\int_0^z\rho(z) g \ dz$$

where $p(z)$ is pressure at depth $z$, $p_o$ is the datum pressure, $\rho (z)$ is the bulk density for the rock mass (rock matrix plus the mass in its pore space) as a function of depth, and $g$ is the acceleration due to gravity.

See here for a more comprehensive derivation for the pressure-density-depth relationship shown in Equation 1. This explanation assumes the International System of units, and provides a gradient in units Pa/m. See my explanation here for how to convert SI units to the American Engineering System of units, which uses psi/ft.

See here for a discussion on why overburden pressure is assumed to be equal to overburden stress.

For the assumption of 1 psi/ft, the bulk density of the rock will be assumed a constant. Gauge pressure will be assumed, therefore $p_o$ will be 0 psig since the datum will be selected to be at the surface of the earth. A constant bulk density can be assumed if the grain density of the rock matrix, liquid water density in the pore space of the rock, and porosity value of the rock are assumed. The bulk density can be estimated as,

$$\rho_t = [\rho_m \times (1-\phi)]+[\rho_w \times \phi]$$ where $\rho_m$ is the weighted average of grain (mineral) density (sandstone and shale = 2.65 g/cc), $\rho_w$ is the weighted average of pore-water density (g/cc), and $\phi$ is the weighted average of the rock porosity.

To get a value of 1 psi/ft, we will assume $\rho_m = 2.65$ g/cc, $\rho_w = 1.05$ g/cc, and $\phi \approx 21.457$%. This will give a bulk density of $\rho_t \approx 2.306682$ g/cc.

There is 0.0022046 lb-mass per gram and 28316.85 cubic centimeters per cubic foot. Therfore $\rho_t \approx 2.306682$ g/cc $= 144$ lbm/ft^3.

Substituting this density value into the equation presented in the link for using American Engineering System of units, assuming $D=1$ft:

$$\frac{p}{D}=\nabla p=\frac{(\rho g_c)}{144g}$$ where $g_c$ is the gravitational constant of acceleration (32.2 ft/sec^2) and $g$ is the units conversion factor to lbf (32.2 $\frac{lbm \ ft/sec^2}{lbf}$), you should get 1 psi/ft.

$\endgroup$
  • 1
    $\begingroup$ Nice example (apart from the Imperial units :) but note that a formation fluid density of about 1030–1070 $\mathrm{kg/m}^3$ would be more realistic in many basins. $\endgroup$ – kwinkunks Dec 9 '14 at 18:01
  • $\begingroup$ Very nice to know. Obviously to get the math to work out to 1 psi/ft there is no unique solution. It makes sense for the water density to be greater than 1 since there are dissolved solids (salinity). I will edit my post using the mean value of 1.05 g/cc $\endgroup$ – Armadillo Dec 9 '14 at 18:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.