Rock compressibility and the effect of stress on core properties

Quick refresher

The weight of the overlying rocks is shared between the fluid in the pores and the rock skeleton. The skeleton carries the effective stress: the overburden pressure (the weight of the rocks above, per unit of area) minus the pore pressure (the pressure of the fluid in the pores). During development the overburden pressure does not change, so each megapascal of fall in the reservoir pressure adds a megapascal of load on the skeleton.
There are three compressibilities: of the grains (very small), of the rock as a whole (important for subsidence, the sinking of the ground), and of the pore volume $c_f$ (important for oil in place and for the energy of the reservoir). If the grains are taken as incompressible, $c_f \approx c_b/\phi$ .
The pore compressibility of cemented rocks is usually $(0.3\text{–}1.5)\cdot10^{-3}$ MPa⁻¹, and for weak and loose rocks it is several times more. This is as much as for water and for undersaturated oil, so in the compressibility drive (the drive in which the expansion of the fluids and the rock pushes the oil out) we cannot neglect it.
When the pressure falls by $\Delta p$ , the pore volume decreases by $c_fV_p\Delta p$ . We measure the porosity and permeability of core under a confining pressure (a pressure that squeezes the sample from all sides) that corresponds to the effective stress in the reservoir.
In the laboratory a sample is usually squeezed equally from all sides, but in the reservoir the rock is squeezed only vertically. So we multiply the laboratory compressibility by $(1 + \nu)/(3(1 - \nu))$ , about 0.5–0.6, where $\nu$ is Poisson's ratio (how much a rock bulges sideways when it is squeezed lengthwise). We also take the confining pressure equal to the mean effective stress, not the vertical one.
Permeability is more sensitive to load than porosity. In weak rocks, beyond the elastic limit (the load above which the rock no longer springs back) the pores collapse irreversibly: the reservoir is compacted, and the surface subsides.

Key relationships

\sigma'_v = \sigma_v - p

vertical effective stress (the Biot coefficient, which shows how much of the pore pressure acts against the load, is taken as 1)

\Delta V_p = c_f\,V_p\,\Delta p

decrease of the pore volume when the reservoir pressure falls by $\Delta p$

c_f \approx \dfrac{c_b}{\phi}

pore compressibility from the bulk compressibility of the rock (the grains are incompressible)

c_f = c_f^{\text{lab}}\,\dfrac{1 + \nu}{3\,(1 - \nu)}

from the all-round compression in the laboratory to uniaxial strain (squeezing in one direction only) in the reservoir

\Delta h \approx \phi\,c_f\,h\,\Delta p

compaction of a reservoir of thickness $h$

Common pitfalls

Mixing up pore compressibility (per unit of pore volume) and the compressibility of the rock as a whole (per unit of rock volume): they differ by a factor of $1/\phi$ .
Taking the porosity and permeability measured without confining pressure as reservoir values.
Squeezing a sample with an all-round pressure equal to the overburden pressure or to the vertical effective stress: the rock will be over-compressed.
Using the laboratory compressibility in calculations without converting it to reservoir conditions.
Taking the compressibility of weak rocks as constant and reversible. Beyond the elastic limit it increases several times, and when the pressure recovers the pores do not reopen.

After this lesson you can

explain which stresses act on the rock in the reservoir, and what changes when the reservoir pressure falls and when core is brought to the surface
distinguish grain compressibility, bulk compressibility and pore volume compressibility, and convert one into another
calculate the change of pore volume and porosity when the pressure falls
describe the test that measures pore compressibility, choose the confining pressure and bring the laboratory value to reservoir conditions
estimate the fall of permeability as the effective stress grows, and the compaction of the reservoir

What presses on the rock in the reservoir

At depth the rock is loaded by the weight of all the overlying layers. This is the overburden pressure $\sigma_v$ , on average about 22–23 kPa for every metre of depth. From the sides the rock is squeezed by the neighbouring rocks: these are two horizontal stresses, usually smaller than the vertical one. And from inside the grains are held apart by the fluid, compressed in the pores to the reservoir pressure $p$ . Only the difference falls on the rock skeleton: the effective stress $\sigma'_v = \sigma_v - p$ . It is this that compresses the pores (Pore pressure and overburden pressure).

During development the weight of the overlying rocks does not change. The reservoir pressure changes: it falls with production, and the effective stress grows by the same amount. The rock then behaves like a very stiff sponge: the pores are slightly compressed and push out fluid, and the channels between the pores narrow.

CalculatedWho carries the weight of the overlying rocks in the example below: before development, and after the reservoir pressure falls from 30 to 18 MPa. The overburden pressure has not changed, but the load on the skeleton has grown by 12 MPa.

Worked exampleThe load on the skeleton before and after the pressure fall

Given (assumed data)

Depth of the reservoir: 2800 m
Mean overburden pressure gradient: 22.6 kPa/m
Reservoir pressure: initially 30 MPa; after production 18 MPa

Solution

Overburden pressure: $\sigma_v = 22.6\cdot2800/1000 = 63.3$ MPa.
Effective stress before development: $63.3 - 30 = 33.3$ MPa.
After the pressure fall: $63.3 - 18 = 45.3$ MPa, which is 36% more.

Answer: The load on the rock skeleton grew from 33.3 to 45.3 MPa.

What it means. The reservoir pressure fell by 40%, and the load on the skeleton grew by more than a third. For a strong rock this means a small elastic compression of the pores. For a weak rock it can mean the breaking of the bonds between the grains. Here we take the Biot coefficient as equal to one. For tight rocks it is smaller, and the load grows more slowly (see the advanced section). The data in this example are assumed for illustration.

Three compressibilities: of the grains, the rock and the pores

Compressibility is the relative change of volume when the pressure changes by one unit, at constant temperature.

Isothermal compressibility

c = -\frac{1}{V}\,\frac{\dd V}{\dd p}

where:

$c$ compressibility, 1/MPa
$V$ the volume considered, m³
$p$ pressure applied to this volume, MPa

Units: 1/MPa or 1/psi: $1\ \text{MPa}^{-1} = 6.895\cdot10^{-3}\ \text{psi}^{-1}$ ; $10^{-6}\ \text{psi}^{-1} = 0.145\cdot10^{-3}\ \text{MPa}^{-1}$

Holds when

The temperature is constant.
The minus sign makes the compressibility positive: as the applied pressure grows, the volume decreases.
For a rock you must say which volume you mean and which pressure you change, the external or the pore pressure.

For a liquid or gas everything is clear: there is one volume and one pressure. A porous rock has three volumes (of the grains, of the pores and of the rock as a whole) and two pressures, so there are several compressibilities.

Grain compressibility $c_r$: the change in the volume of the mineral grains themselves. It is very small: for quartz about $0.03\cdot10^{-3}$ MPa⁻¹. In most calculations the grains are taken as incompressible.
Bulk compressibility $c_b$: the change in the external volume of the rock together with its pores. Another name is volumetric compressibility. It controls the compaction of the reservoir, the subsidence of the surface, movements on faults and the loads on casing strings.
Pore volume compressibility $c_f$: the change in the pore volume; for short, pore compressibility. Porosity, oil in place and the elastic energy of the reservoir depend on it. It is this compressibility that enters the total compressibility $c_t$ and the material balance.

Pore volume compressibility

c_f = \frac{1}{V_p}\left(\frac{\partial V_p}{\partial p}\right)_{\sigma_v} = -\frac{1}{V_p}\,\frac{\partial V_p}{\partial \sigma'}

where:

$c_f$ pore volume compressibility, 1/MPa
$V_p$ pore volume, m³
$p$ pore (reservoir) pressure, MPa
$\sigma_v$ overburden pressure, which stays constant, MPa
$\sigma'$ effective stress, MPa

Units: consistent units of pressure and compressibility

Holds when

The overburden pressure is constant, and only the pore pressure changes.
The effective stress equals the difference between the overburden and pore pressures, so a growth of the stress by 1 MPa acts on the pores in the same way as a fall of the pore pressure by 1 MPa.
The value refers to a certain level of stress: as the load grows, $c_f$ usually decreases.

The quantity is positive: the pore volume decreases when the pore pressure falls, that is, when the load on the skeleton grows. We use this compressibility in flow calculations and in the material balance (Flow regimes, General material balance equation for oil reservoirs).

The link between pore compressibility and bulk compressibility

c_f \approx \frac{c_b}{\phi}

where:

$c_b$ bulk compressibility of the rock (per unit of rock volume), 1/MPa
$\phi$ porosity, fraction
$c_f$ pore compressibility (per unit of pore volume), 1/MPa

Units: any identical units of compressibility

Holds when

The grains are incompressible: the whole volume by which the rock has shrunk is lost by the pores.
Both compressibilities refer to the same kind of loading.

The same decrease of volume, referred to the pore volume, is $1/\phi$ times greater than when it is referred to the rock volume: for a porosity of 20%, five times. In reference tables both quantities are found under similar names (“rock compressibility”, “formation compressibility”), and mixing them up means an error of several times.

Approximate compressibilities

What is compressed: Quartz grains, $c_r$
10⁻³ MPa⁻¹: about 0.03
10⁻⁶ psi⁻¹: about 0.2

What is compressed: Cemented sandstone as a whole, $c_b$
10⁻³ MPa⁻¹: 0.05–0.3
10⁻⁶ psi⁻¹: 0.3–2

What is compressed: Pores of cemented sandstones and limestones, $c_f$
10⁻³ MPa⁻¹: 0.3–1.5
10⁻⁶ psi⁻¹: 2–10

What is compressed: Pores of weakly cemented and loose rocks, $c_f$
10⁻³ MPa⁻¹: 1.5–4.5 and more
10⁻⁶ psi⁻¹: 10–30 and more

What is compressed: For comparison: reservoir water
10⁻³ MPa⁻¹: about 0.45
10⁻⁶ psi⁻¹: about 3

What is compressed: For comparison: undersaturated oil
10⁻³ MPa⁻¹: 0.7–3
10⁻⁶ psi⁻¹: 5–20

What is compressed	10⁻³ MPa⁻¹	10⁻⁶ psi⁻¹
Quartz grains, $c_r$	about 0.03	about 0.2
Cemented sandstone as a whole, $c_b$	0.05–0.3	0.3–2
Pores of cemented sandstones and limestones, $c_f$	0.3–1.5	2–10
Pores of weakly cemented and loose rocks, $c_f$	1.5–4.5 and more	10–30 and more
For comparison: reservoir water	about 0.45	about 3
For comparison: undersaturated oil	0.7–3	5–20

Orientation values for estimates. For cemented rocks the pore compressibility is usually greater the lower the porosity, but the scatter is large; a reliable value comes only from a test on core from your own reservoir.

Pore volume and porosity when the pressure falls

Change of the pore volume when the reservoir pressure falls

\Delta V_p = c_f\,V_{pi}\,\Delta p,\qquad V_p = V_{pi}\,(1 - c_f\,\Delta p)

where:

$\Delta V_p$ decrease of the pore volume, m³
$V_{pi},\ V_p$ pore volume at the initial and at the current pressure, m³
$\Delta p$ fall of the reservoir pressure, $p_i - p$ , MPa
$c_f$ pore volume compressibility at reservoir conditions, 1/MPa

Units: consistent units of pressure and compressibility: MPa and 1/MPa, or psi and 1/psi

Holds when

$c_f$ is constant over the range of pressure considered. For large changes, take the dependence of the pore volume on stress directly from a test.
$c_f\Delta p \ll 1$ . Otherwise use the exact form $V_p = V_{pi}\,\e^{-c_f\Delta p}$ .
The pressure has fallen in the whole volume considered.

The volume by which the pores have shrunk pushes fluid towards the wells. In the material balance this term stands next to the expansion of oil and water. In the diffusivity equation (which describes how a pressure change spreads through the reservoir), $c_f$ is added to the compressibilities of the fluids to give the total compressibility $c_t$ .

Porosity changes a little less than the pore volume, because with the pores the volume of the rock itself also decreases. If the grains are incompressible, the relative change of porosity is $(1 - \phi)\,c_f\,\Delta p$ . Reservoir simulation models do it more simply: they keep the volume of a cell unchanged and multiply its pore volume by $1 - c_f\Delta p$ . For the pore volume the result is the same.

Worked exampleHow much fluid the compressing pores push out

Given (assumed data)

Pore volume of the oil part of the reservoir: 16.0 million m³; porosity 20%
Pore compressibility $c_f$: $0.8\cdot10^{-3}$ MPa⁻¹
Reservoir pressure: falls from 30 to 18 MPa
Expansion of the oil and of the connate water (the water that stays in the pores from the start) for the same pressure fall: 235.6 and 21.6 thousand m³ (the example of the lesson Reservoir drive mechanisms)

Solution

Relative decrease of the pore volume: $c_f\Delta p = 0.8\cdot10^{-3}\cdot12 = 0.96$ %.
Decrease of the pore volume: $\Delta V_p = 16.0\cdot10^6\cdot0.96/100 = 153.6$ thousand m³.
Porosity: $20\cdot\big[1 - (1 - 0.20)\cdot0.96/100\big] = 19.85$ %.
Share of the pores in the elastic expansion of the reservoir: $153.6/(235.6 + 21.6 + 153.6) = 37$ %.
If the pore compressibility were half as large, the elastic expansion, and with it the production down to the bubble point pressure, would be smaller by 19%.

Answer: The pores shrank by 153.6 thousand m³, which is less than 1%, but this is 37% of all the elastic energy of the reservoir.

What it means. The porosity hardly changed, from 20.0 to 19.85%. For oil in place this is unimportant. But for the energy balance the compression of the pores is comparable with the expansion of the oil. While there is no free gas in the reservoir, an error of a factor of two in $c_f$ changes the calculated production in the compressibility drive by almost a fifth. It distorts in the same way the oil in place found from the pressure decline (General material balance equation for oil reservoirs). The data in this example are assumed for illustration.

Measurements under load

Core at the surface is unloaded: there is neither overburden pressure nor pore pressure (Core analysis). To return the rock to its reservoir state, the laboratory puts the sample in a core holder (a cell that grips the sample), where a liquid squeezes it through a rubber sleeve. The pore pressure is usually left atmospheric, so the confining pressure is the effective stress applied to the sample: the net confining stress. We measure porosity and permeability under this load. Ordinary core holders are built for a confining pressure of up to a few tens of megapascals, usually about 70 MPa (10,000 psi).

To measure pore compressibility, the laboratory fills the sample with water and connects its pores to a thin graduated tube or to a vessel on a balance. It raises the confining pressure in steps and waits at each step until water stops coming out of the sample. The volume of water pushed out equals the decrease of the pore volume.

Pore compressibility from a test with stepwise confining pressure

c_f^{\text{lab}} = \frac{1}{V_p}\cdot\frac{\Delta V_w}{\Delta\sigma_c}

where:

$c_f^{\text{lab}}$ pore compressibility under all-round compression, 1/MPa
$\Delta V_w$ volume of water pushed out of the pores between two steps, cm³
$\Delta\sigma_c$ increase of the confining pressure between the steps, MPa
$V_p$ pore volume at the beginning of the interval, cm³

Units: volumes in identical units; pressure in MPa or psi

Holds when

The pore pressure is constant (atmospheric), and only the confining pressure changes.
A growth of the confining pressure acts on the pores in the same way as an equal fall of the pore pressure. This is true while the grains can be considered incompressible. For tight low-porosity rocks the test is run at the real overburden and pore pressures.
The sample is squeezed equally from all sides.

The compressibility is the slope of the curve “pore volume against confining pressure”. The curve is not a straight line. At low loads the microfractures that appeared when the core was unloaded close, and the compressibility is several times higher than at reservoir stresses. So we take the value in the range of stresses in which the reservoir works.

A test on a sample: volume 80.0 cm³, pore volume without load 16.00 cm³

Confining pressure: 0 MPa
Water expelled, cm³: 0
Pore volume, cm³: 16.000
Porosity, %: 20.00

Confining pressure: 2.5 MPa
Water expelled, cm³: 0.174
Pore volume, cm³: 15.826
Porosity, %: 19.83

Confining pressure: 5 MPa
Water expelled, cm³: 0.298
Pore volume, cm³: 15.702
Porosity, %: 19.70

Confining pressure: 10 MPa
Water expelled, cm³: 0.469
Pore volume, cm³: 15.531
Porosity, %: 19.53

Confining pressure: 15 MPa
Water expelled, cm³: 0.592
Pore volume, cm³: 15.408
Porosity, %: 19.40

Confining pressure: 20 MPa
Water expelled, cm³: 0.698
Pore volume, cm³: 15.302
Porosity, %: 19.30

Confining pressure: 30 MPa
Water expelled, cm³: 0.895
Pore volume, cm³: 15.105
Porosity, %: 19.09

Confining pressure: 40 MPa
Water expelled, cm³: 1.088
Pore volume, cm³: 14.912
Porosity, %: 18.90

Confining pressure	Water expelled, cm³	Pore volume, cm³	Porosity, %
0 MPa	0	16.000	20.00
2.5 MPa	0.174	15.826	19.83
5 MPa	0.298	15.702	19.70
10 MPa	0.469	15.531	19.53
15 MPa	0.592	15.408	19.40
20 MPa	0.698	15.302	19.30
30 MPa	0.895	15.105	19.09
40 MPa	1.088	14.912	18.90

The data are assumed for illustration. We calculate the porosity under load on the assumption that the grains are incompressible: the volume of the sample decreases by the volume of the water pushed out.

CalculatedThe pore volume of the sample from the table against the confining pressure. The steep first part is the closing of microfractures; after it the curve is almost a straight line. The mean effective stresses in the reservoir are marked, at the initial pressure of 30 MPa and after it falls to 18 MPa.

Which confining pressure to apply. In an ordinary core holder the sample is squeezed equally from all sides, which is all-round, or hydrostatic, compression. In the reservoir it is different. The vertical stress is usually greater than the horizontal ones, and when the reservoir pressure falls, the rock is compressed only vertically, because the neighbouring rocks hold it at the sides. This state is called uniaxial strain (the rock deforms in one direction only). Two rules follow.

Set the confining pressure equal to the mean effective stress in the reservoir, $(\sigma'_v + 2\sigma'_h)/3$ , not to the vertical one. A sample squeezed from all sides with a pressure equal to the vertical effective stress will be over-compressed, and its porosity will come out too low.
Convert the laboratory compressibility to reservoir conditions. A fall of the reservoir pressure by 1 MPa raises the vertical effective stress by 1 MPa, but the mean stress by noticeably less.

SchematicLeft: all-round compression in a core holder. The stresses are equal on all sides, and the sample shrinks in all directions. Right: uniaxial strain in the reservoir. The rock is compressed only vertically, the lateral dimensions do not change, and the horizontal stresses are smaller than the vertical one (a schematic, the strains are greatly exaggerated).

Converting the pore compressibility from all-round compression to uniaxial strain

c_f = c_f^{\text{lab}}\;\frac{1 + \nu}{3\,(1 - \nu)}

where:

$c_f$ pore compressibility at reservoir conditions, 1/MPa
$c_f^{\text{lab}}$ pore compressibility measured under all-round compression, 1/MPa
$\nu$ Poisson's ratio of the rock (how much a rock bulges sideways when it is squeezed lengthwise), —

Units: any identical units of compressibility

Holds when

The rock is elastic and the same in all directions, and the grains are incompressible.
The overburden pressure is constant and there are no lateral strains: the reservoir is wide compared with its thickness.
$\nu$ was measured on the same rock under slow loading, when the fluid has time to leave the pores.

At $\nu$ from 0.2 to 0.3 the factor is 0.50–0.62: in the reservoir the pores are compressed one and a half to two times less than in a hydrostatic test for the same change of pressure. Some laboratories run the test in uniaxial strain from the start, selecting the lateral pressure so that the diameter of the sample does not change. Then no conversion is needed. The derivation of the factor is in the advanced section.

Worked exampleFrom a test on a sample to the reservoir porosity and pore compressibility

Given (assumed data)

Test: the table above
Vertical effective stress in the reservoir: 33.3 MPa before development; 45.3 MPa after the pressure falls to 18 MPa
Poisson's ratio $\nu$: 0.30

Solution

The horizontal effective stress in uniaxial strain is $\nu/(1 - \nu) = 0.43$ of the vertical one (Fracture gradient and leak-off testing): $0.43\cdot33.3 = 14.3$ MPa. The mean is $(33.3 + 2\cdot14.3)/3 = 20.6$ MPa; after the pressure fall it is 28.0 MPa. The working range of the test is the steps of 20 and 30 MPa.
Porosity at 20 MPa: $\dfrac{16.00 - 0.698}{80.0 - 0.698}\cdot100 = 19.3$ %, which is 0.965 of the value without load.
Compressibility in the test between 20 and 30 MPa: $\dfrac{0.895 - 0.698}{15.302\cdot10} = 1.29\cdot10^{-3}$ MPa⁻¹.
For comparison, between 2.5 and 5 MPa: $3.13\cdot10^{-3}$ MPa⁻¹, which is 2.4 times more.
To reservoir conditions: the factor is $(1 + 0.30)/(3\cdot0.70) = 0.619$ ; $c_f = 1.29\cdot0.619 = 0.80\cdot10^{-3}$ MPa⁻¹, or $5.5\cdot10^{-6}$ psi⁻¹.

Answer: Porosity at reservoir conditions is 19.3% instead of 20.0%; the pore compressibility is about

0.80\cdot10^{-3}

MPa⁻¹.

What it means. This value, $0.8\cdot10^{-3}$ MPa⁻¹, is used in the examples of the lessons Flow regimes, Reservoir drive mechanisms and General material balance equation for oil reservoirs. Three typical errors would give other numbers. Porosity without confining pressure would overstate the pore volume, and with it the oil in place, by 3.6%. A compressibility taken from the initial part of the curve would overstate the elastic energy of the pores 2.4 times. The laboratory value without conversion to uniaxial strain would be wrong by a factor of 1.6, and the two errors together by a factor of 3.9. The data in this example are assumed for illustration.

Permeability under load

Permeability is determined not by the pores themselves but by the narrowest places between them, the pore throats (Rock permeability). Under load it is they that narrow first, and in addition microfractures close. So permeability reacts to a growth of the effective stress more strongly than porosity: a loss of a few per cent of the pore volume can cost tens of per cent of the permeability.

How sensitive permeability is to load

Rock: Strong cemented sandstone
Loss of permeability under reservoir load: a few per cent
Reason: a stiff skeleton; the channels hardly change

Rock: Weakly cemented sandstone
Loss of permeability under reservoir load: tens of per cent
Reason: the grains move closer and turn

Rock: Loose sand
Loss of permeability under reservoir load: several times
Reason: the packing of the grains is rearranged

Rock: Tight clay-rich rock
Loss of permeability under reservoir load: from tens of per cent to several times
Reason: thin channels and microfractures close

Rock: Fractured rock
Loss of permeability under reservoir load: depends on the direction of the stresses
Reason: a fracture closes if it is compressed across

Rock	Loss of permeability under reservoir load	Reason
Strong cemented sandstone	a few per cent	a stiff skeleton; the channels hardly change
Weakly cemented sandstone	tens of per cent	the grains move closer and turn
Loose sand	several times	the packing of the grains is rearranged
Tight clay-rich rock	from tens of per cent to several times	thin channels and microfractures close
Fractured rock	depends on the direction of the stresses	a fracture closes if it is compressed across

Qualitative guidelines from tests on samples of different rocks; for your own reservoir the dependence is measured. The lower the original permeability and the more clay in the rock, the stronger the relative decrease usually is.

For fractures, direction decides everything. If the greatest horizontal stress acts across a fracture, it compresses it, and the permeability falls. If it acts along the fracture, the fracture stays open, and sometimes opens wider. With unequal stresses, even an intact sample can expand in one direction, and the permeability along it grows. So for fractured and weak rocks an all-round confining pressure alone is not enough: the tests run in rigs where we set the three stresses independently.

For calculations the dependence of permeability on the reservoir pressure in a limited range is often approximated by an exponential.

Exponential dependence of permeability on the reservoir pressure

k = k_i\,\e^{-\gamma\,(p_i - p)}

where:

$k_i,\ k$ permeability at the initial and at the current reservoir pressure, mD
$\gamma$ coefficient of the sensitivity of permeability to pressure (permeability modulus), 1/MPa
$p_i - p$ fall of the pore pressure; at constant overburden pressure it equals the growth of the effective stress, MPa

Units: $\gamma$ and pressure in mutually inverse units

Holds when

An empirical approximation for a limited range of pressures. $\gamma$ is found from tests on core under load, or from the history of well productivity.
The overburden pressure is constant.
We take the change as reversible; for weak rocks this is not so.

While the product $\gamma\Delta p$ is small, the permeability falls by a share equal to it: at $\gamma = 0.01$ MPa⁻¹ every 10 MPa of pressure fall takes away about 10%. For strong rocks $\gamma$ is thousandths per megapascal, and for weak and fractured ones it is hundredths.

CalculatedThe permeability of the two rocks of the example when the reservoir pressure falls from 30 MPa, as a fraction of the initial value: cemented sandstone (

\gamma = 0.004

MPa⁻¹) and weak sand (

\gamma = 0.03

MPa⁻¹). Calculated with the exponential formula.

Worked examplePermeability after the pressure fall, and at the bottom of the well

Given (assumed data)

Cemented sandstone: $k_i = 50$ mD; $\gamma = 0.004$ MPa⁻¹
Weak sand: $k_i = 800$ mD; $\gamma = 0.03$ MPa⁻¹
Reservoir pressure: falls from 30 to 18 MPa
Bottomhole pressure of the producing well: 12 MPa

Solution

Sandstone at 18 MPa: $50\cdot\e^{-0.004\cdot12} = 47.7$ mD, a loss of 5%.
Weak sand at 18 MPa: $800\cdot\e^{-0.03\cdot12} = 558$ mD, a loss of 30%.
At the bottom of the well the pressure is 18 MPa below the initial one: the sandstone has 46.5 mD, and the weak sand 466 mD, that is 58% of the initial permeability.

Answer: After the pressure fall the sandstone loses about 5% of its permeability, the weak sand about 30%, and at the bottom of the well about 42%.

What it means. For a strong rock the correction is comparable with the error of the initial data. For a weak one it changes the forecast. The productivity of the wells falls together with the reservoir pressure. A larger drawdown (the pressure difference between the reservoir and the well) gives a smaller and smaller gain in rate, because the rock is most squeezed right at the bottom of the well, where most of the pressure difference is spent (Skin factor). Pressure maintenance (injecting water or gas to keep the pressure up) in such reservoirs preserves not only the energy but also the permeability. The data in this example are assumed for illustration.

Compaction of the reservoir and subsidence of the surface

In strong rocks the decrease of volume during development is negligible. In weak ones, such as loose sands and highly porous chalk, it is counted in per cents of the pore volume. While the load is small, the rock deforms elastically (it springs back when the load is removed). When the effective stress exceeds the elastic limit, the bonds between the grains break, the pores collapse, and the compressibility grows several times. This deformation is irreversible: if the pressure is raised afterwards, the pores do not reopen.

Compaction of the reservoir when the pressure falls

\Delta h \approx \phi\,c_f\,h\,\Delta p

where:

$\Delta h$ decrease of the thickness of the reservoir, m
$h$ thickness of the reservoir over which the pressure has fallen, m
$\phi$ porosity, fraction
$c_f$ pore compressibility at reservoir conditions, mean for the range of pressures, 1/MPa
$\Delta p$ fall of the reservoir pressure, MPa

Units: consistent units of pressure and compressibility; $\Delta h$ in the same units as $h$

Holds when

The reservoir is wide compared with its thickness and is compressed only vertically.
The grains are incompressible, so the volume of the rock decreases exactly by as much as the volume of the pores decreased.
The pressure has fallen over the whole thickness $h$ .

The product $\phi c_f$ is the compressibility of the rock as a whole in uniaxial strain; it is called the compaction coefficient (the compaction per unit of thickness and per unit of pressure fall). The subsidence of the surface is no more than the compaction of the reservoir: above a wide shallow reservoir it is almost the same, and above a narrow deep one it is noticeably smaller, because the overlying rocks work like an arch.

Worked exampleA strong sandstone and a weak highly porous rock

Given (assumed data)

Sandstone: thickness 40 m; porosity 20%; $c_f = 0.8\cdot10^{-3}$ MPa⁻¹; pressure fall 12 MPa
Weak highly porous rock: thickness 100 m; porosity 38%; $c_f = 4\cdot10^{-3}$ MPa⁻¹; pressure fall 15 MPa

Solution

Sandstone: $\Delta h = 0.20\cdot0.8\cdot10^{-3}\cdot40\cdot12 = 0.077$ m, that is 8 cm.
Weak rock: $\Delta h = 0.38\cdot4\cdot10^{-3}\cdot100\cdot15 = 2.28$ m. The pore volume has decreased by $c_f\Delta p = 6$ %.

Answer: The sandstone reservoir will be compacted by about 8 cm, and the weak rock reservoir by more than 2 m.

What it means. Eight centimetres at a depth of kilometres cannot be noticed at the surface. Two metres is subsidence that damages wells and structures, but it is also an enormous source of energy: the compressing pores push out 6% of their volume, whereas water expands by less than 1% for the same fall of pressure. For such a rock we cannot take the compressibility as constant. We give it as a table “pore volume against pressure”, from tests carried to the failure of the structure. The data in this example are assumed for illustration.

A source of energy. The fluid pushed out of the pores supports the pressure: this is the compaction drive (Reservoir drive mechanisms).
Subsidence. The compaction is passed upwards: the ground surface or the sea floor goes down.
Casing strings. In a compacting reservoir the strings are compressed along the axis and lose stability, and at its boundaries the moving rock shears them off.
Faults. Non-uniform compaction can move rocks along a fault and change how easily fluid crosses it.
Productivity. Together with the pores, the channels between them are compressed, and permeability falls.

Terms and symbols

Effective stress $\sigma'$: The part of the overburden pressure that the rock skeleton carries: the overburden pressure minus the pore pressure.
Pore volume compressibility $c_f$: The relative change of the pore volume when the pore pressure changes by one unit at constant overburden pressure.
Bulk compressibility $c_b$: The relative change of the external volume of the rock; for incompressible grains $c_b = \phi c_f$ .
Grain compressibility $c_r$: The relative change of the volume of the mineral matter; usually negligibly small.
Confining pressure: The pressure with which the sample is squeezed in a core holder; with the pore pressure subtracted, it is the net confining stress.
All-round (hydrostatic) compression: Loading with the same stress from all sides.
Uniaxial strain: Compression only vertically, with no change of the lateral dimensions: the condition of a wide reservoir when the pressure falls.
Poisson's ratio $\nu$: The ratio of the lateral expansion to the longitudinal compression of an elastic body; for reservoir rocks it is usually 0.15–0.35.

Common mistakes and misconceptions

Wrong

“The rock is practically incompressible, so pore compressibility can be ignored.”

In fact

The grains are incompressible, not the pores. The compressibility of the pore volume is of the same order as that of water and of undersaturated oil. In the lesson's example the compression of the pores gave 37% of the elastic energy of the reservoir.

Wrong

“The rock compressibility from a reference book can be used directly as $c_f$ .”

In fact

First find out to which volume it refers. A compressibility per unit of rock volume has to be divided by the porosity; at 20% the difference is a factor of five.

Wrong

“To reproduce the reservoir, the sample must be squeezed with a pressure equal to the overburden pressure.”

In fact

The skeleton is acted on not by the overburden pressure but by the effective stress, and the lateral stresses are usually smaller than the vertical one. In all-round loading, apply the mean effective stress, usually 0.5–0.8 of the vertical one.

Wrong

“If the reservoir pressure is raised by injection, the porosity and permeability will come back.”

In fact

The elastic part of the deformation will come back, but the microfractures, the shift of the grains and the collapse of the pores are irreversible. It is better not to let the pressure fall than to restore it.

In depth: advanced detail

Four compressibilities of a porous rock and the Biot coefficient

A porous rock has two volumes, of the rock as a whole $V_b$ and of the pores $V_p$ , and two pressures: the external $\sigma_c$ and the pore pressure $p$ . So, strictly speaking, there are four compressibilities. The first index shows which volume changes, and the second shows which pressure is changed while the other stays constant.

Four compressibilities of a porous rock

\begin{aligned} c_{bc} &= -\frac{1}{V_b}\frac{\partial V_b}{\partial \sigma_c}, & c_{bp} &= \frac{1}{V_b}\frac{\partial V_b}{\partial p},\\[4pt] c_{pc} &= -\frac{1}{V_p}\frac{\partial V_p}{\partial \sigma_c}, & c_{pp} &= \frac{1}{V_p}\frac{\partial V_p}{\partial p}\end{aligned}

where:

$c_{bc},\ c_{bp}$ compressibility of the rock as a whole when the external and the pore pressure change, 1/MPa
$c_{pc},\ c_{pp}$ compressibility of the pore volume when the external and the pore pressure change, 1/MPa
$V_b,\ V_p$ volume of the rock and volume of the pores, m³
$\sigma_c,\ p$ external (confining) pressure and pore pressure, MPa

Units: consistent units of pressure and compressibility

Holds when

The rock is elastic and uniform, and the grains consist of one mineral with compressibility $c_r$ .
When one pressure changes, the other is constant.
The loading is all-round.

In a test with constant pore pressure, $c_{pc}$ is measured, while $c_{pp}$ enters the reservoir calculations. There are exact relations between the four quantities, and it is enough to know one of them, the porosity and the grain compressibility.

Relations between the compressibilities

c_{bp} = c_{bc} - c_r,\qquad c_{pc} = \frac{c_{bc} - c_r}{\phi},\qquad c_{pp} = c_{pc} - c_r

where:

$c_r$ grain compressibility, 1/MPa
$\phi$ porosity, fraction
$c_{bc},\ c_{bp},\ c_{pc},\ c_{pp}$ as in the previous formula, 1/MPa

Units: any identical units of compressibility

Holds when

The same as for the definitions of the four compressibilities.

For incompressible grains ( $c_r = 0$ ) what remains is $c_{pp} = c_{pc} = c_{bc}/\phi$ , the formula of the main section. The ratio $c_{bp}/c_{bc} = 1 - c_r/c_{bc}$ is the Biot coefficient $\alpha$ from the effective stress law $\sigma' = \sigma - \alpha p$ (Pore pressure and overburden pressure). It shows what share of the pore pressure relieves the skeleton. For loose rocks $c_{bc} \gg c_r$ and $\alpha \approx 1$ ; for tight stiff rocks $\alpha$ is noticeably less than one.

Worked exampleA cemented sandstone: how much the grains matter

Given (assumed data)

Compressibility of the rock under external loading $c_{bc}$: $0.16\cdot10^{-3}$ MPa⁻¹
Compressibility of the quartz grains $c_r$: $0.027\cdot10^{-3}$ MPa⁻¹
Porosity: 20%

Solution

$c_{pc} = (0.16 - 0.027)/0.20 = 0.665$ , in units of $10^{-3}$ MPa⁻¹.
$c_{pp} = 0.665 - 0.027 = 0.638$ .
Biot coefficient: $\alpha = 1 - 0.027/0.16 = 0.83$ .
A simplified estimate that ignores the grains: $c_{bc}/\phi = 0.80$ , which is 25% more.

Answer: The compressibility of the pores with respect to the pore pressure is about

0.638\cdot10^{-3}

MPa⁻¹; the Biot coefficient is 0.83.

What it means. For a stiff cemented rock the grain compressibility is no longer negligible: the formula $c_b/\phi$ overstates the result by a quarter, and the effective stress, as the pressure falls, grows not by 1 but by 0.83 MPa for each megapascal. For weak rocks, where $c_{bc}$ is tens of times larger than $c_r$ , both corrections disappear. The data in this example are assumed for illustration.

Where the factor for uniaxial strain comes from

Let the rock be elastic and the same in all directions, and let the grains be incompressible. Then the volume of the rock changes in proportion to the mean effective stress $\sigma'_m = (\sigma'_v + 2\sigma'_h)/3$ , and the pore volume decreases by exactly as much as the volume of the rock.

In a hydrostatic test all three stresses equal the confining pressure, and the mean stress grows by $\Delta\sigma_c$ . In the reservoir, when the pressure falls by $\Delta p$ , the vertical effective stress grows by $\Delta p$ , because the overburden pressure is constant. The horizontal strains are forbidden, and Hooke's law gives that the horizontal effective stresses grow only by $\dfrac{\nu}{1 - \nu}\,\Delta p$ (Fracture gradient and leak-off testing).

Growth of the mean effective stress in uniaxial strain

\Delta\sigma'_m = \frac{1}{3}\left(1 + \frac{2\nu}{1 - \nu}\right)\Delta p = \frac{1 + \nu}{3\,(1 - \nu)}\,\Delta p

where:

$\Delta\sigma'_m$ growth of the mean effective stress, MPa
$\Delta p$ fall of the pore pressure, MPa
$\nu$ Poisson's ratio, —

Units: any identical units of pressure

Holds when

An elastic isotropic rock and incompressible grains (the Biot coefficient is 1).
A constant overburden pressure and zero lateral strains.

The pore volume responds to the mean stress in the same way in the test and in the reservoir. But in the reservoir each megapascal of pressure fall adds to the mean stress only $(1 + \nu)/(3(1 - \nu))$ megapascals. That is where the factor comes from: 0.50 at $\nu = 0.2$ , 0.56 at 0.25 and 0.62 at 0.3. The same ratio links the levels of the stresses themselves: the mean effective stress is $(1 + 2K)/3$ of the vertical one, where $K = \sigma'_h/\sigma'_v$ is the coefficient of lateral stress (the ratio of the horizontal to the vertical stress).

A real reservoir is only approximately elastic, the horizontal stresses in it depend on tectonics, and at the edges of the reservoir the condition “no lateral strain” is broken. So for reservoirs where compaction is important, engineers use tests in the uniaxial-strain mode and calculations of the stresses in the reservoir and in the surrounding rocks instead of the factor.

Estimating pore compressibility without core

When there are no tests on core, the pore compressibility of cemented rocks is estimated from the porosity. One of the oldest correlations is the Hall correlation, proposed in the 1950s from tests on sandstones and limestones.

The Hall correlation for the pore compressibility of cemented rocks

c_f \approx \frac{0.2585}{\phi^{0.438}}\cdot10^{-3}\ \text{MPa}^{-1}

where:

$c_f$ pore volume compressibility, 1/MPa
$\phi$ porosity, fraction

Units: empirical: $\phi$ as a fraction, $c_f$ in 1/MPa

Holds when

Cemented sandstones and limestones with a porosity of roughly 2 to 25%.
Not applicable to weakly cemented and loose rocks, to chalk or to fractured reservoirs.
Later tests showed a scatter of several times around this relation, so it is suitable only for a preliminary estimate.

The lower the porosity, the higher the pore compressibility: at a porosity of 20% the formula gives $0.52\cdot10^{-3}$ MPa⁻¹, and at 5% it gives $0.96\cdot10^{-3}$ MPa⁻¹. The pores of a tight rock are compressed relatively more, because their volume is small and the skeleton has many microfractures and slit-like pores.

In other unit systems (1)

Oilfield units

c_f \approx \frac{1.782}{\phi^{0.438}}\cdot10^{-6}\ \text{psi}^{-1}

$\phi$ as a fraction, $c_f$ in 1/psi.

$0.2585 = 1.782\cdot0.14504$ : the conversion of $10^{-6}$ psi⁻¹ to $10^{-3}$ MPa⁻¹ (1 MPa = 145.04 psi).

Connections to other topics

Pore pressure and overburden pressure — overburden and pore pressure, effective stress.
Porosity — what the pore volume is and how it is measured.
Rock permeability — why permeability is determined by the pore channels.
Core analysis — how samples are taken and prepared for tests under load.
Flow regimes — pore compressibility in the total compressibility and in the diffusivity equation.
Reservoir drive mechanisms — the compressibility drive and the compaction drive.
General material balance equation for oil reservoirs — the term with the compressibility of the water and the pores in the balance equation.
Fracture gradient and leak-off testing — horizontal stresses and the coefficient of lateral stress.
Abnormal formation pressures: origin, prediction and recognition — compaction of rocks with burial, and abnormal pressures.

Check yourself

1
The pressure in a reservoir has fallen by 8 MPa. How did the vertical effective stress change, if the Biot coefficient is taken as 1?
1. A.It fell by 8 MPa
2. B.It grew by 8 MPa
3. C.It grew by $8/\phi$ MPa
4. D.It did not change: the weight of the overlying rocks is the same
Show the answer
Correct answer: B. It grew by 8 MPa
The overburden pressure is constant, and the effective stress is its difference with the pore pressure. The fluid now props the grains up by 8 MPa less, and the load on the skeleton has grown by the same amount.
2
A report gives “rock compressibility $0.12\cdot10^{-3}$ MPa⁻¹ per unit of rock volume” and a porosity of 15%. What value of $c_f$ should go into the total compressibility $c_t$ ?
Show the answer
The total compressibility takes the compressibility per unit of pore volume: $c_f \approx c_b/\phi = 0.12\cdot10^{-3}/0.15 = 0.8\cdot10^{-3}$ MPa⁻¹. If you put in $0.12\cdot10^{-3}$ , you would understate the contribution of the pores almost seven times.
3
The pore volume of a reservoir is 50 million m³, and the pore compressibility is $1.0\cdot10^{-3}$ MPa⁻¹. How much fluid will the compression of the pores push out when the pressure falls by 6 MPa?
Show the answer
$\Delta V_p = c_fV_p\Delta p = 1.0\cdot10^{-3}\cdot50\cdot10^6\cdot6 = 300$ thousand m³ at reservoir conditions, which is 0.6% of the pore volume.
4
A hydrostatic test gave a pore compressibility of $1.5\cdot10^{-3}$ MPa⁻¹, and the Poisson's ratio of the rock is 0.25. Which value should you use for the reservoir?
1. A.About $0.5\cdot10^{-3}$ MPa⁻¹
2. B.About $0.83\cdot10^{-3}$ MPa⁻¹
3. C.About $2.7\cdot10^{-3}$ MPa⁻¹
4. D. $1.5\cdot10^{-3}$ MPa⁻¹
Show the answer
Correct answer: B. About $0.83\cdot10^{-3}$ MPa⁻¹
In the reservoir the rock is compressed without lateral strain, and the mean stress grows more slowly than the pore pressure falls: the factor is $(1 + 0.25)/(3\cdot0.75) = 0.556$ . $1.5\cdot0.556 \approx 0.83\cdot10^{-3}$ MPa⁻¹.
5
Why does the porosity of a sample squeezed from all sides with a pressure equal to the vertical effective stress in the reservoir come out lower than the reservoir porosity?
Show the answer
In the reservoir the horizontal effective stresses are smaller than the vertical one, and the mean stress, on which the pore volume depends, is only 0.5–0.8 of the vertical one. All-round loading with a pressure equal to the vertical stress creates a mean stress higher than the reservoir one: the rock is over-compressed. The right way is to apply the mean effective stress, or to run the test in uniaxial strain.
6
The permeability of a weakly cemented sandstone at the initial pressure is 200 mD, and $\gamma = 0.02$ MPa⁻¹. Estimate the permeability after the reservoir pressure falls by 10 MPa. Will it come back if the pressure is restored by injection?
Show the answer
$k = 200\cdot\e^{-0.02\cdot10} = 200\cdot0.819 \approx 164$ mD, a loss of about 18%. Most likely it will not come back fully: the elastic part of the deformation is reversible, but the shift of the grains and the breaking of the cement are not. So in stress-sensitive rocks engineers try to keep the pressure up from the very beginning.

Updated: 2 October 2026