Capillary pressure and the transition zone

Quick refresher

$p_c$ is the difference between the pressures of the non-wetting and the wetting phase. In a cylindrical channel of radius $r$ it equals $2\sigma\cos\theta/r$ : a channel half as wide gives twice the pressure.
In an accumulation that has come to equilibrium, $p_c = (\rho_w - \rho_o)\,g\,h$ . Each height above the free water level (FWL) has its own capillary pressure, and through the core curve its own water saturation.
Oil starts to push water out of the rock only when the pressure reaches the entry pressure $p_d$ of the largest connected channels. So the oil–water contact (OWC) lies above the FWL by $h_d = p_d/(\Delta\rho\,g)$ .
The imbibition curve does not retrace the drainage curve (this is called hysteresis). Part of the oil is trapped, and at $p_c = 0$ the water saturation does not return to one.
To convert a laboratory curve to reservoir fluids, multiply it by the ratio of the adhesion tensions $\sigma\cos\theta$ . Mercury pressures, for an oil–water pair at typical values, become about 14 times smaller.
The Leverett J-function $J = p_c\sqrt{k/\phi}/(\sigma\cos\theta)$ brings the curves of samples of one rock type onto a single curve. In a less permeable rock there is more water at the same height, and the OWC is higher.

Key relationships

p_c = p_{\text{nw}} - p_{\text{w}} = \dfrac{2\sigma\cos\theta}{r}

capillary pressure in a cylindrical channel of radius $r$

p_c = (\rho_w - \rho_o)\,g\,h

capillary pressure at height $h$ above the free water level

p_{c,\text{res}} = p_{c,\text{lab}}\,\dfrac{(\sigma\cos\theta)_{\text{res}}}{(\sigma\cos\theta)_{\text{lab}}}

conversion of a laboratory curve to reservoir fluids

J(S_w) = \dfrac{p_c}{\sigma\cos\theta}\sqrt{\dfrac{k}{\phi}}

Leverett J-function, consistent units

Common pitfalls

Mixing up the OWC and the free water level. At the FWL the capillary pressure is zero. The OWC is above it, by a height that corresponds to the entry pressure.
Using the imbibition curve to build the initial saturation distribution, and the drainage curve for the watered-out part of the reservoir.
Converting a mercury curve with a fixed factor without checking $\sigma$ and $\theta$ , or taking the cosine of the angle measured through the mercury.
Applying one Leverett J-function to different rock types.
Expecting a single OWC in a layered reservoir. Each layer has its own contact and its own transition zone; only the free water level is the same for all.

After this lesson you can

show where the formula for capillary pressure in a cylindrical channel comes from, and explain how it depends on pore size, interfacial tension and wettability
read drainage and imbibition capillary pressure curves: the entry pressure, the irreducible water saturation, the residual oil and the hysteresis (the gap between the two curves)
compare laboratory measurement methods and convert a laboratory curve to reservoir fluids
build the water saturation profile with height above the free water level, and find the position of the oil–water contact and the thickness of the transition zone
carry a curve over to a rock with a different permeability and porosity using the Leverett J-function, and explain how saturation is distributed in a layered reservoir

Where capillary pressure comes from

The interface (the boundary) between two fluids that do not mix is stretched like a film (interfacial tension). If the interface is curved, the tension presses it towards its concave side (the inside of the curve). It can stay in equilibrium only if the pressure on that side is higher. That is why the pressure inside a drop or a bubble is always higher than outside. The pressure difference across a curved interface is called the capillary pressure.

In a pore, the oil–water interface is concave on the side of the non-wetting fluid, because the wetting fluid “creeps” along the walls. So the pressure in the non-wetting fluid is higher. For oil and water in a water-wet rock, $p_c = p_o - p_w > 0$ . In oil-wet pores the capillary pressure, written the same way, becomes negative.

Laplace equation for a curved interface

p_c = p_{\text{nw}} - p_{\text{w}} = \sigma\left(\frac{1}{R_1} + \frac{1}{R_2}\right)

where:

$p_{\text{nw}},\ p_{\text{w}}$ pressures of the non-wetting and the wetting phase at the interface, Pa
$\sigma$ interfacial tension, N/m
$R_1,\ R_2$ principal radii of curvature of the interface (how sharply it bends in two perpendicular directions), m

Units: SI

Holds when

The interface is stationary or moves very slowly (capillary equilibrium).
The interfacial tension is the same over the whole interface.

The more strongly the interface is curved (the smaller the radii of curvature), the greater the pressure difference. For a spherical interface $R_1 = R_2 = R$ and $p_c = 2\sigma/R$ .

In a cylindrical channel of radius $r$ , the interface is part of a sphere: a meniscus (the curved surface of a liquid in a narrow tube). Its radius is tied to the radius of the channel by the contact angle: $R = r/\cos\theta$ . Putting this into the Laplace equation gives the basic formula.

Capillary pressure in a cylindrical channel

p_c = \frac{2\sigma\cos\theta}{r}

where:

$\sigma$ interfacial tension, N/m
$\theta$ contact angle, measured through the wetting phase, degrees
$r$ radius of the channel; in a rock, the effective radius of the pore throat through which the meniscus passes, m
$p_c$ capillary pressure, Pa

Units: SI

Holds when

The channel has a circular cross-section, and the meniscus is part of a sphere.
The contact angle is measured through the wetting phase, so $\cos\theta > 0$ .
The system is in equilibrium: the meniscus is stationary.

Capillary pressure is inversely proportional to the size of the channel: it is large in small pores and small in large ones. It is proportional to the adhesion tension $\sigma\cos\theta$ , so it depends on both the fluids and the wettability.

In other unit systems (2)

Practical metric units

p_c\,[\text{kPa}] = \frac{2\,\sigma\,[\text{mN/m}]\,\cos\theta}{r\,[\text{$\mu$m}]}

$\sigma$ in mN/m, $r$ in µm, $p_c$ in kPa.

There is no constant: mN/m divided by µm gives kPa ( $10^{-3}/10^{-6} = 10^{3}$ Pa).

Oilfield units

p_c\,[\text{psi}] = \frac{0.290\,\sigma\,[\text{dyn/cm}]\,\cos\theta}{r\,[\text{$\mu$m}]}

$\sigma$ in dyn/cm (the same as mN/m), $r$ in µm, $p_c$ in psi.

$0.290 = 2/6.895$ : the factor 2 with kPa converted to psi (1 psi = 6.895 kPa).

Capillary rise. Lower a thin water-wet tube through the oil–water interface in a vessel. Water will rise in the tube above the level in the vessel. At the level of the interface in the vessel, the oil and water pressures are equal. At height $h$ , the pressure of each fluid is lower by the weight of its column: in water by $\rho_w g h$ , in oil by $\rho_o g h$ . Water is heavier, so at height $h$ the oil pressure is higher than the water pressure by $(\rho_w - \rho_o)\,g h$ . Water keeps rising until this difference equals the capillary pressure of the meniscus.

Height of capillary rise

(\rho_w - \rho_o)\,g\,h = \frac{2\sigma\cos\theta}{r}\quad\Rightarrow\quad h = \frac{2\sigma\cos\theta}{(\rho_w - \rho_o)\,g\,r}

where:

$\rho_w,\ \rho_o$ densities of water and oil, kg/m³
$g$ acceleration due to gravity, 9.81, m/s²
$h$ height of the meniscus above the level where the pressures of the two fluids are equal, m
$\sigma,\ \theta,\ r$ as in the formula for a channel, N/m, degrees, m

Units: SI

Holds when

Capillary and hydrostatic equilibrium.
The tube is water-wet, and water is the wetting phase.

The thinner the tube, the higher the water rises. A rock is a huge number of “tubes” of different sizes. So above the level of equal pressures, water stands higher in the small pores than in the large ones, and the water saturation decreases smoothly with height. This is how the transition zone arises.

Worked exampleCapillary pressure of pore channels and height above the contact

Given (assumed data)

Oil–water in the reservoir: $\sigma = 30$ mN/m, $\theta = 30^\circ$ ; water and oil densities 1040 and 800 kg/m³
Radii of the pore channels: 5 µm (a highly permeable sandstone) and 0.5 µm (a siltstone)
For comparison, gas–water in the reservoir: $\sigma = 50$ mN/m, $\theta = 0$ , gas density 250 kg/m³, channel 5 µm

Solution

Adhesion tension: $\sigma\cos\theta = 30\cdot 0.866 = 25.98$ mN/m.
Channel of 5 µm: $p_c = 2\cdot 25.98/5 = 10.39$ kPa. With each metre of height, the oil–water pressure difference grows by $(1040 - 800)\cdot 9.81 = 2354$ Pa. Oil will enter such a channel at a height $h = 10.39\cdot 10^3/2354 = 4.41$ m above the level of equal pressures.
Channel of 0.5 µm: the pressure is 10 times greater, 103.9 kPa, and the height is also 10 times greater, 44.1 m.
Gas–water, channel of 5 µm: $p_c = 2\cdot 50/5 = 20.0$ kPa. But the pressure grows faster with height, by $(1040 - 250)\cdot 9.81 = 7750$ Pa per metre, so the height is only 2.58 m.

Answer: Oil enters channels with a radius of 5 µm at a height of 4.41 m above the level of equal pressures, and channels of 0.5 µm only at a height of 44.1 m.

What it means. The size of the channels sets the scale of the transition zone: in a highly permeable sandstone it is metres thick, in a siltstone tens of metres. Gas has a greater interfacial tension, but it separates from water at a smaller height, because the density difference between gas and water is three times greater. The data in this example are assumed for illustration.

Capillary pressure curves: drainage and imbibition

A rock is a network of pores of different sizes, joined by narrow throats. Imagine a sample full of water, into which oil or gas is injected at a pressure raised step by step. This is drainage: the non-wetting fluid pushes out the wetting fluid.

How drainage proceeds

1
Up to the entry pressure
The non-wetting fluid does not enter the sample. Its menisci cannot get through even the largest throats on the surface of the sample.
2
Entry pressure
At the entry pressure $p_d$ (also called the threshold or displacement pressure) the non-wetting fluid penetrates the largest connected channels, and the water saturation begins to fall.
3
Filling of the pores
As the pressure rises, smaller and smaller pores are filled. If the channels are of similar size, the curve runs almost flat (a plateau). If their sizes vary a lot, the curve rises smoothly.
4
Irreducible water
At the irreducible water saturation $S_{wirr}$ the curve becomes almost vertical. The remaining water sits in pore corners, films and the smallest pores, and a further rise in pressure hardly pushes it out.

The reverse process is imbibition: the wetting fluid comes back into the rock and pushes out the non-wetting fluid. At first the wetting fluid is drawn in by itself, while the capillary pressure falls to zero. This is spontaneous imbibition. In a strongly water-wet rock the process almost ends there. If part of the pores is oil-wet, the rest of the oil can be pushed out only at a negative capillary pressure. This is forced imbibition. Even at a very large negative pressure, part of the oil stays trapped: the water saturation approaches $1 - S_{or}$ , not one.

SchematicThe capillary pressure cycle of a weakly water-wet rock. Primary drainage (the first drainage, starting from a rock full of water) starts at the entry pressure

p_d

and ends at the irreducible water saturation. Imbibition follows a different curve: first it is spontaneous, then it continues at negative capillary pressure, and it ends at

1 - S_{or}

Capillary pressure hysteresis explains why imbibition does not retrace drainage:

The advancing contact angle is larger than the receding angle, so for the same pore geometry the imbibition capillary pressure is lower.
Pore shape. In drainage the meniscus is held back by the narrow throats. In imbibition it is held back by the wide pore bodies, which fill at a lower pressure.
Trapping. The wetting fluid advances along corners and films and cuts off droplets of the non-wetting fluid in the large pores. The droplets stay there. This is residual oil or trapped gas.

If the process reverses halfway, the saturation follows intermediate curves between the main ones (scanning curves). Reservoir simulation models include a hysteresis calculation for this.

Which curve to use

Process: Oil or gas fills the trap
Curve: primary drainage
What it determines: the initial saturation and the transition zone

Process: Rise of the OWC, waterflooding of a water-wet reservoir
Curve: imbibition
What it determines: the saturation of the watered-out part and the residual oil

Process: Gas displaces oil or water
Curve: drainage: gas is the non-wetting phase
What it determines: the saturation in the gas cap and in the zone where gas comes out of solution

Process	Curve	What it determines
Oil or gas fills the trap	primary drainage	the initial saturation and the transition zone
Rise of the OWC, waterflooding of a water-wet reservoir	imbibition	the saturation of the watered-out part and the residual oil
Gas displaces oil or water	drainage: gas is the non-wetting phase	the saturation in the gas cap and in the zone where gas comes out of solution

In an oil-wet or mixed-wet rock, waterflooding includes parts with negative capillary pressure.

How capillary pressure is measured

On a core sample, the capillary pressure is set step by step, and when equilibrium has been reached the saturation is measured. The methods differ in how they create the pressure and in the fluids they use.

Laboratory methods

Method: Porous plate (semipermeable membrane)
Fluids: any, including reservoir fluids
Duration: weeks
Features: the reference method; its limit is breakthrough of the membrane (the non-wetting fluid passes through it)

Method: Centrifuge
Fluids: gas–water, oil–water
Duration: hours to days
Features: the data need a calculation to process

Method: Mercury injection (MICP)
Fluids: mercury–vacuum
Duration: 1–2 hours
Features: very high pressures; the sample is lost (it cannot be used again)

Method	Fluids	Duration	Features
Porous plate (semipermeable membrane)	any, including reservoir fluids	weeks	the reference method; its limit is breakthrough of the membrane (the non-wetting fluid passes through it)
Centrifuge	gas–water, oil–water	hours to days	the data need a calculation to process
Mercury injection (MICP)	mercury–vacuum	1–2 hours	very high pressures; the sample is lost (it cannot be used again)

There is also a dynamic method. Two fluids flow through the sample together at a steady rate, and the pressure of each is measured through semipermeable partitions. It is used mainly in research.

Porous plate (semipermeable membrane). The sample is placed on a ceramic plate full of the wetting fluid. The fine pores of the plate let water through but not oil or gas, as long as the pressure is below the breakthrough pressure of the plate. The pressure of the non-wetting fluid is raised in steps. At each step we wait until water stops flowing out of the sample. The method is direct, and it can use reservoir fluids, but one curve takes weeks.

Centrifuge. The sample is spun at several speeds. The centrifugal force creates a pressure difference between the fluids inside it. This difference is greatest at the inner end of the sample, nearest the axis of rotation, and zero at the outer end. We measure the volume of fluid pushed out. The capillary pressure and the saturation vary along the length of the sample, so the curve is reconstructed by calculation. The method is fast, and it gives both drainage and forced imbibition.

Mercury injection capillary pressure (MICP). A dried sample is put under vacuum (evacuated). Mercury, the non-wetting fluid, is then injected into it in pressure steps, up to hundreds of megapascals. The saturation is found from the volume of mercury that has entered. The first portions of mercury fill the roughness of the sample surface, so the entry pressure is taken where the saturation starts to grow noticeably. The method is fast, works on small samples and even on drill cuttings (small pieces of rock from the borehole), and gives the distribution of throats by size. But mercury is toxic, the sample cannot be used after the test, and the result has to be converted to reservoir fluids.

Converting the laboratory curve to the reservoir

In the laboratory we use convenient pairs of fluids: air and water, mercury and vacuum, laboratory oil and water. In the reservoir the fluids are reservoir oil, gas and water at reservoir temperature. If the pore structure is the same, the same channels are filled at the same saturation, and the capillary pressures are in the same ratio as the adhesion tensions.

Conversion of capillary pressure from laboratory to reservoir fluids

p_{c,\text{res}} = p_{c,\text{lab}}\,\frac{(\sigma\cos\theta)_{\text{res}}}{(\sigma\cos\theta)_{\text{lab}}}

where:

$p_{c,\text{res}}$ capillary pressure of the reservoir pair of fluids at the same saturation, Pa
$p_{c,\text{lab}}$ capillary pressure measured in the laboratory, Pa
$(\sigma\cos\theta)_{\text{res}},\ (\sigma\cos\theta)_{\text{lab}}$ adhesion tensions of the reservoir and the laboratory pairs, N/m

Units: both pressures in the same units; all tensions in the same units

Holds when

The pore structure is the same in the test and in the reservoir (shaly and poorly cemented rocks are tested under reservoir confining stress, the squeeze from the surrounding rock).
The contact angles are measured through the wetting fluid: for mercury this is the angle through the vacuum, about 40°, not 140°.
The wettability of the reservoir is known. This is the weakest link of the conversion.

Multiplying the laboratory curve by a constant factor turns it into the reservoir curve. Then the formula $h = p_c/(\Delta\rho\,g)$ turns it into the saturation profile with height above the FWL.

Approximate adhesion tensions

Pair of fluids: Air–water, laboratory
$\sigma$ , mN/m: 72
$\theta$: 0°
$\sigma\cos\theta$ , mN/m: 72

Pair of fluids: Mercury–vacuum
$\sigma$ , mN/m: 485
$\theta$: 140° through mercury
$\sigma\cos\theta$ , mN/m: 372

Pair of fluids: Oil–water, reservoir
$\sigma$ , mN/m: 30
$\theta$: 30°
$\sigma\cos\theta$ , mN/m: 26

Pair of fluids: Gas–water, reservoir
$\sigma$ , mN/m: 50
$\theta$: 0°
$\sigma\cos\theta$ , mN/m: 50

Pair of fluids	$\sigma$ , mN/m	$\theta$	$\sigma\cos\theta$ , mN/m
Air–water, laboratory	72	0°	72
Mercury–vacuum	485	140° through mercury	372
Oil–water, reservoir	30	30°	26
Gas–water, reservoir	50	0°	50

These values are often assumed when there are no measurements. The actual ones depend on the fluids, the temperature and the wettability. For mercury we use the absolute value of the cosine: 485 · cos 40° ≈ 372.

Worked exampleHeight above the FWL from a mercury curve

Given (assumed data)

Mercury curve of the sample: mercury starts to enter the sample noticeably at 0.10 MPa; at 0.30 MPa the mercury saturation is 0.40, and at 1.0 MPa it is 0.65
Reservoir fluids: oil–water: $\sigma = 30$ mN/m, $\theta = 30^\circ$ ; water and oil densities 1040 and 800 kg/m³
Mercury–vacuum: $\sigma = 485$ mN/m, $\theta = 140^\circ$ through mercury

Solution

Conversion factor: $(30\cos 30^\circ)/(485\cos 40^\circ) = 25.98/371.5 = 0.0699$ . The reservoir capillary pressures are 14.3 times smaller than the mercury ones.
Entry pressure in the reservoir: $0.10\cdot 10^6\cdot 0.0699 = 6993$ Pa. The capillary pressure grows by $(1040 - 800)\cdot 9.81 = 2354$ Pa with each metre of height. So the OWC lies above the FWL by $h_d = 6993/2354 = 2.97$ m.
The 0.30 MPa point: $p_c = 21.0$ kPa, $h = 8.9$ m. The water saturation is the fraction of the pores that mercury does not occupy: $1 - 0.40 = 0.60$ .
The 1.0 MPa point: $p_c = 69.9$ kPa, $h = 29.7$ m, and the water saturation is $1 - 0.65 = 0.35$ .
Radius of the largest connected channels, from the mercury entry pressure (0.10 MPa = 100 kPa) and the practical formula: $r = 2\cdot 371.5/100 = 7.43$ µm.

Answer: The OWC is 2.97 m above the free water level. At a height of 8.9 m the water saturation is 0.60, and at a height of 29.7 m it is 0.35.

What it means. The transition zone of this rock extends over tens of metres. Even 30 m above the FWL, water takes up a third of the pores, and the interval can produce water together with oil. An error in the contact angle goes straight into the height. If $\theta = 0$ is taken for the oil–water pair instead of 30°, all the heights become larger by a factor of $1/\cos 30^\circ = 1.15$ . The data in this example are assumed for illustration.

The Leverett J-function: one curve for many samples

Samples from the same formation differ in permeability and porosity, and their capillary pressure curves lie well apart: the curves of less permeable samples are higher. To describe the formation with a single relation, we bring the curves to a dimensionless form (one without units). We estimate a typical pore radius from permeability and porosity. For a bundle of capillary tubes, $k \propto \phi r^2$ , so $r \propto \sqrt{k/\phi}$ (the derivation is in the advanced section). Putting this radius into the formula for a channel gives the dimensionless Leverett J-function.

Leverett J-function

J(S_w) = \frac{p_c}{\sigma\cos\theta}\sqrt{\frac{k}{\phi}}

where:

$J$ dimensionless capillary pressure, —
$p_c$ capillary pressure at water saturation $S_w$ , Pa
$\sigma\cos\theta$ adhesion tension of the pair of fluids, N/m
$k$ permeability of the sample, m²
$\phi$ porosity of the sample, fraction

Units: SI (any consistent system)

Holds when

The samples belong to one rock type, with similar pore structure and tortuosity (how winding the flow paths are).
$k$ , $\phi$ and the $p_c$ curve are measured on the same sample.
If the contact angle is unknown, $\cos\theta$ is sometimes left out. Then it must be left out in the reverse conversion too.

For samples of one rock type, the points $J(S_w)$ fall on a single curve. Going the other way, this curve gives the capillary pressure curve for any $k$ , $\phi$ and fluids: $p_c = J\,\sigma\cos\theta/\sqrt{k/\phi}$ . At the same height (the same $p_c$ ), a less permeable rock has a smaller $J$ and a higher water saturation.

In other unit systems (2)

Metric oilfield units

J = 31.4\,\frac{p_c\,[\text{MPa}]}{\sigma\,[\text{mN/m}]\cos\theta}\sqrt{\frac{k\,[\text{mD}]}{\phi}}

$p_c$ in MPa, $\sigma$ in mN/m, $k$ in mD, $\phi$ as a fraction.

$31.4 = 10^{6}\cdot\sqrt{0.9869\cdot 10^{-15}}/10^{-3}$ : MPa → Pa, mD → m² under the root, mN/m → N/m.

Oilfield units

J = 0.2166\,\frac{p_c\,[\text{psi}]}{\sigma\,[\text{dyn/cm}]\cos\theta}\sqrt{\frac{k\,[\text{mD}]}{\phi}}

$p_c$ in psi, $\sigma$ in dyn/cm, $k$ in mD, $\phi$ as a fraction.

$0.2166 = 6895\cdot\sqrt{0.9869\cdot 10^{-15}}/10^{-3}$ (1 psi = 6895 Pa).

In practice the points scatter much less if $J$ is plotted not against the water saturation but against the normalised water saturation. This is the fraction of the interval from the irreducible water saturation up to one. The irreducible water itself reflects the pore structure of each sample. If the points of different samples still do not fall on one curve, the samples are of different rock types, and each type is described by its own function.

Normalised water saturation

S_w^{*} = \frac{S_w - S_{wirr}}{1 - S_{wirr}}

where:

$S_w^{*}$ normalised water saturation: 0 at the irreducible value, 1 at full water saturation, fraction
$S_{wirr}$ irreducible water saturation of the given sample or layer, fraction

Units: fraction

Holds when

$S_{wirr}$ is measured on the same sample, or for the same rock type, as the curve.

Worked exampleTwo layers with one Leverett J-function

Given (assumed data)

Leverett function of the rock type: $J = 0.15\,(S_w^{*})^{-1/2}$ ; when $J \le 0.15$ the rock is completely full of water
Layer A: $k = 200$ mD, $\phi = 0.22$ , $S_{wirr} = 0.15$
Layer B: $k = 10$ mD, $\phi = 0.16$ , $S_{wirr} = 0.30$
Fluids: $\sigma\cos\theta = 25.98$ mN/m (30 mN/m and 30°); water and oil densities 1040 and 800 kg/m³
Height above the FWL: 10 m

Solution

Typical pore size of layer A: $\sqrt{k/\phi} = 0.947$ µm, pressure scale $\sigma\cos\theta/\sqrt{k/\phi} = 27{,}429$ Pa. For layer B, 0.248 µm and 104,609 Pa.
The entry pressure is the scale multiplied by 0.15: 4114 Pa in layer A and 15,691 Pa in layer B. The OWC of layer A is 1.7 m above the FWL, and that of layer B is 6.7 m above it.
At a height of 10 m, $p_c = 10\cdot 2354 = 23{,}544$ Pa. Layer A: $J = 23{,}544/27{,}429 = 0.858$ , $S_w^{*} = (0.15/0.858)^2 = 0.0305$ , $S_w = 0.15 + (1 - 0.15)\cdot 0.0305 = 0.176$ .
Layer B: $J = 0.225$ , $S_w^{*} = 0.444$ , $S_w = 0.30 + (1 - 0.30)\cdot 0.444 = 0.611$ .
We take the top of the transition zone to be where the water saturation is 0.05 above the irreducible value: $S_w^{*} = 0.05/(1 - S_{wirr})$ and $J = 0.15/\sqrt{S_w^{*}}$ . For layer A this is 7.2 m above the FWL, for layer B 24.9 m.

Answer: At a height of 10 m, water fills 18% of the pores of layer A and 61% of the pores of layer B. The transition zone of layer A ends about 7.2 m above the FWL; that of layer B ends only at about 24.9 m.

What it means. The free water level is the same for both layers, but the contacts and transition zones differ. The tight layer B is completely full of water up to 6.7 m, while layer A already has oil above 1.7 m. If such layers alternate, a well may cross a water-filled tight bed above an oil-bearing permeable one. This is not a separate aquifer. It is a capillary effect. The data in this example are assumed for illustration.

CalculatedWater saturation profiles of layers A (200 mD) and B (10 mD) above the common free water level, calculated with the Leverett function from the example. The dots mark the water saturations at a height of 10 m.

Saturation profile, OWC and transition zone

In an accumulation that has come to equilibrium, the pressure of each fluid changes with depth according to the weight of a column of that fluid (hydrostatics). The water pressure line and the oil pressure line cross at the free water level (FWL). Here the two pressures are equal and $p_c = 0$ . Above the FWL the oil pressure is higher than the water pressure, and the difference grows with height.

Capillary pressure above the free water level

p_o(h) - p_w(h) = p_c(h) = (\rho_w - \rho_o)\,g\,h

where:

$p_o(h),\ p_w(h)$ oil and water pressures at height $h$ , Pa
$\rho_w,\ \rho_o$ densities of formation water and oil at reservoir conditions, kg/m³
$g$ acceleration due to gravity, 9.81, m/s²
$h$ height above the free water level, m

Units: SI

Holds when

Hydrostatic and capillary equilibrium: the accumulation is not in production, or has been shut in for a long time.
Both fluids are continuous over the whole height, and the formation water is not flowing (flow does not tilt the contact).
For a gas accumulation, use the gas density in place of the oil density. For a gas–oil contact, use the gas–oil pair of fluids.

The height above the FWL fixes the capillary pressure, and the core capillary pressure curve, converted to reservoir fluids, gives the water saturation at that pressure. This is how we build the saturation profile with height. It is the basis of the hydrocarbons-in-place calculation in reservoirs with a wide transition zone, and of the initial saturation in the reservoir simulation model.

In other unit systems (2)

Metric oilfield units

p_c\,[\text{MPa}] = 0.00981\,(\rho_w - \rho_o)\,[\text{g/cm}^3]\;h\,[\text{m}]

densities in g/cm³, height in m, pressure in MPa.

$0.00981 = 9.81\cdot 1000/10^6$ : g/cm³ → kg/m³ and Pa → MPa.

Oilfield units

p_c\,[\text{psi}] = 0.433\,(\gamma_w - \gamma_o)\;h\,[\text{ft}]

$\gamma_w,\ \gamma_o$ are the specific gravities of water and oil relative to fresh water, $h$ is in feet, $p_c$ in psi.

0.433 psi/ft is the pressure gradient of a column of fresh water.

From the bottom up: below the FWL is the water-bearing part. From the FWL up to a height $h_d = p_d/(\Delta\rho\,g)$ , the pores are still completely full of water, because the capillary pressure is below the entry pressure. The top of this interval is the oil–water contact (OWC), the one we see on logs. Above it lies the transition zone: the water saturation decreases with height, and both fluids are mobile. Higher still, the water saturation is close to irreducible, and the inflow has no water in it. For more on the zones see the topic on saturation.

Layered reservoir. Each rock type has its own capillary pressure curve, so each has its own OWC and its own transition zone. The free water level is the same for layers that are hydraulically connected (connected so that pressure can pass between them). So a tight bed may be completely full of water above an oil-bearing permeable one (see the example with the Leverett function), and the contact seen on logs “jumps” from layer to layer. The one level that is the same for the whole accumulation, the FWL, is found where the oil and water pressure lines cross. These lines are measured with a formation tester, a tool that takes pressure readings in the rock (Pressure gradients, fluid contacts and capillary equilibrium).

Terms and symbols

Capillary pressure $p_c$: The difference between the pressures of the non-wetting and the wetting phase at the curved interface between them. In a channel of radius $r$ it equals $2\sigma\cos\theta/r$ .
Entry pressure $p_d$: The capillary pressure at which the non-wetting phase begins to enter the largest connected channels of the rock. It is also called the threshold or displacement pressure.
Drainage: Displacement of the wetting phase by the non-wetting phase, for example of water by oil or gas in a water-wet rock.
Imbibition: Displacement of the non-wetting phase by the wetting phase. It may be spontaneous ( $p_c > 0$ ) or forced ( $p_c < 0$ ).
Free water level (FWL): The level where the oil and water pressures are equal and $p_c = 0$ . The OWC lies above it by $h_d = p_d/(\Delta\rho\,g)$ .
Leverett J-function $J$: The dimensionless capillary pressure $p_c\sqrt{k/\phi}/(\sigma\cos\theta)$ . It is the same for all samples of one rock type.
Capillary pressure hysteresis: The mismatch of the drainage and imbibition curves. It is caused by contact angle hysteresis, pore shape and trapping of the non-wetting phase.

Common mistakes and misconceptions

Wrong

“The OWC is the level where the capillary pressure is zero.”

In fact

That is the free water level. The OWC that we see on logs is above it, by a height that corresponds to the entry pressure. This is a fraction of a metre in coarse-grained rocks, and metres or tens of metres in tight ones.

Wrong

“There is a water-saturated bed above an oil-saturated one, so there is a separate aquifer there.”

In fact

A tight bed with a high entry pressure can be completely full of water above a permeable oil-bearing layer, with one common free water level. Check the capillary pressure curves and the pressure measurements along the section.

Wrong

“A mercury curve is converted to a reservoir one by dividing by a constant, 5 or 10.”

In fact

The factor is the ratio of the adhesion tensions of the reservoir pair and the laboratory pair, and it is calculated from the actual $\sigma$ and $\theta$ . A divisor of about 5 converts a mercury curve to an air–water pair, not to the reservoir. For an oil–water pair at 30 mN/m and 30°, the mercury pressures are divided by about 14. At a lower interfacial tension the divisor is even larger.

Wrong

“The capillary pressure curve is a property of the rock.”

In fact

It depends on the fluids (through $\sigma\cos\theta$ ), on the wettability, and on the direction of the process: drainage or imbibition.

Wrong

“One Leverett function is enough for the whole reservoir.”

In fact

The function brings together only samples with a similar pore structure. If the points $J(S_w)$ scatter widely, there are several rock types, and each needs its own relation.

In depth: advanced detail

The bundle-of-capillaries model and the Leverett function

Picture the rock as a bundle of $n$ identical winding tubes of radius $r$ . They cross a sample of length $L$ and cross-section $A$ . Each tube has length $\tau L$ , where $\tau \ge 1$ is the tortuosity (how winding the tubes are). By Poiseuille's law (the law of flow in a tube), the flow rate in one tube is $q_1 = \pi r^4\,\Delta p/(8\mu\,\tau L)$ , and the total flow rate is $q = n q_1$ .

We compare this with Darcy's law, $q = kA\,\Delta p/(\mu L)$ , and get $k = n\pi r^4/(8\tau A)$ . The porosity of the bundle is the volume of the tubes divided by the volume of the sample: $\phi = n\pi r^2\tau/A$ . If we eliminate $n$ (remove it from the two equations), we find $k = \phi r^2/(8\tau^2)$ , that is, $r = \tau\sqrt{8k/\phi}$ .

The entry capillary pressure of such tubes is $p_c = 2\sigma\cos\theta/r$ . The dimensionless combination $J = p_c\sqrt{k/\phi}/(\sigma\cos\theta)$ then turns out to equal $2/(\tau\sqrt{8})$ . It depends neither on the size of the tubes nor on the fluids. In a real rock the channels have different sizes, so $J$ changes with saturation. But if the samples have the same tortuosity and the same shape of channel-size distribution, their $J(S_w)$ curves coincide. The main assumption is equal tortuosity. That is why the function works only within one rock type.

If the contact rose: the imbibition profile and residual oil

The drainage curve describes an accumulation that formed when oil pushed water out. Later the water level may rise, because of structural changes, partial destruction of the accumulation, or production. Then the lower part of the accumulation goes through imbibition. There the saturation follows the imbibition curve: at the same height there is more water, and the oil left in the flushed interval is trapped and cannot move.

Such intervals are deceptive. On the resistivity log they may look partly oil-bearing, but when tested they give water. These are residual oil zones: traces of a former accumulation below the present contact. To build saturation profiles for the watered-out part of a reservoir, and for a model after waterflooding has started, we need imbibition curves, not drainage curves.

Pore throat size distribution from a mercury curve

Each mercury pressure step corresponds to the radius of the throats that the mercury passes through: $r = 2\sigma|\cos\theta|/p_c$ . At 485 mN/m and 140° this is $r\,[\mu\text{m}] \approx 0.74/p_c\,[\text{MPa}]$ . The rise in mercury saturation between two neighbouring steps is the fraction of the pore volume that can be reached through channels of this size. This is how a mercury curve becomes a distribution of pore channels by size. We use it to pick out rock types and to estimate permeability.

There are two cautions. First, a cavity behind a narrow throat is filled only at the pressure needed for that throat. So the mercury curve counts large pores that can be reached only through narrow channels as small pores. Second, a pressure of 400 MPa corresponds to channels with a radius of about 2 nm. Oil in the reservoir does not enter such pores, so we do not use the upper part of the mercury curve for the reservoir.

Connections to other topics

Fluid saturation — initial, irreducible and residual saturations and the zones above the contact.
Wettability and interfacial phenomena — interfacial tension, contact angle and the negative branch of capillary pressure.
Relative permeability — relative permeability curves with the same end points $S_{wirr}$ and $S_{or}$ .
Pressure gradients, fluid contacts and capillary equilibrium — how the free water level is found from pressure gradients.
Reservoir and seal — the entry pressure of the seal, and how tall an accumulation it can hold.
Volumetric estimation of hydrocarbons in place — how the saturation profile enters the hydrocarbons-in-place calculation.

Check yourself

1
The radius of the pore channels is halved, with the same fluids. How do the entry pressure and the height of the OWC above the free water level change?
1. A.They will not change
2. B.They will increase fourfold
3. C.They will double
4. D.They will halve
Show the answer
Correct answer: C. They will double
$p_c = 2\sigma\cos\theta/r$ is inversely proportional to the radius, and $h_d = p_d/(\Delta\rho\,g)$ is proportional to the entry pressure, so both quantities double.
2
The entry pressure of a rock at reservoir conditions is 10 kPa; the densities of water and oil are 1030 and 830 kg/m³. At what height above the free water level is the OWC?
Show the answer
$h_d = p_d/((\rho_w - \rho_o)\,g) = 10{,}000/(200\cdot 9.81) \approx 5.1$ m.
3
Which curve do we use to build the initial saturation distribution in an oil accumulation that formed when oil migrated into the trap?
1. A.Any of them: they coincide
2. B.Primary drainage
3. C.Forced imbibition
4. D.Spontaneous imbibition
Show the answer
Correct answer: B. Primary drainage
Oil filling the trap pushed out water, the wetting fluid. This is drainage. Imbibition curves describe watered-out intervals and waterflooding. Because of hysteresis, they give a different saturation at the same height.
4
On a mercury curve, the mercury saturation at 0.5 MPa is 0.55. What capillary pressure, and what height above the free water level, does this correspond to for oil and water? Take $\sigma\cos\theta = 26$ mN/m for oil and water, $\sigma|\cos\theta| = 372$ mN/m for mercury, and a water–oil density difference of 250 kg/m³.
Show the answer
$p_c = 0.5\cdot 26/372 \approx 0.035$ MPa, that is, 35 kPa; $h = 35{,}000/(250\cdot 9.81) \approx 14$ m. At this height the water saturation is $1 - 0.55 = 0.45$ .
5
How will the thickness of the transition zone change if in the same rock there is gas above the water instead of oil?
1. A.It will decrease
2. B.It will increase
3. C.It will not change
Show the answer
Correct answer: A. It will decrease
Capillary pressure builds up with the density difference. For the gas–water pair this difference is several times greater than for the oil–water pair, so the pressure needed is reached at a smaller height. The greater gas–water interfacial tension partly offsets this. Overall, though, the transition zone of gas accumulations is thinner.
6
Why can a tight bed in a well be completely full of water above an oil-bearing sandstone?
Show the answer
A tight rock has small pores and a high entry pressure, so its own OWC lies high above the common free water level. If the bed lies below that height, oil has not entered it, although oil has entered the permeable sandstone below it. This is a capillary effect, not a separate aquifer.

Updated: 19 October 2026