Relative permeability

Quick refresher

The effective permeability $k_x$ is the permeability for a phase when the phases flow together. The relative permeability is $k_{rx} = k_x/k$ . The base of normalisation (the permeability that we divide by) can differ: the absolute permeability to gas or water, or the permeability to oil at irreducible water saturation (the water that cannot be pushed out of the pores). Always state it.
Each phase flows under the gradient of its own pressure (how fast its pressure changes along the path), with its own mobility $k\,k_{rx}/\mu_x$ (how easily the phase moves). The phases hinder each other, so $k_{ro} + k_{rw} < 1$ .
The curves are bounded by the end points (where one phase stops moving). At $S_{wi}$ , the irreducible water saturation, the water is immobile and $k_{ro}$ is at its maximum. At $1 - S_{or}$ , where $S_{or}$ is the residual oil saturation, the oil is immobile and $k_{rw}$ is at its maximum. The range of mobile saturation is $1 - S_{wi} - S_{or}$ .
The Corey model, a common formula for the curves, describes them with power functions of the normalised water saturation (the water saturation scaled to run from 0 to 1 between the end points), with the end points and two exponents.
The fractional flow of water is set by the ratio of the phase mobilities: the more viscous the oil, the earlier and the faster the water cut grows. A mobility ratio at the end points $M' > 1$ means unstable displacement (the water runs ahead of the oil) and early water breakthrough (water reaching the well).
In a water-wet rock (one whose surface prefers water) the curves cross at $S_w > 0.5$ , and $k_{rw}$ at the end of displacement is small. In an oil-wet rock it is the other way round. The drainage curves (oil pushes water out) and the imbibition curves (water pushes oil out) differ, and for waterflooding of a water-wet reservoir we use the imbibition curves.

Key relationships

k_{rx} = \dfrac{k_x}{k}

relative permeability of phase $x$ (oil, water, gas)

q_x = \dfrac{k\,k_{rx}A}{\mu_x}\cdot\dfrac{\Delta p_x}{L}

Darcy's law for a phase in multiphase flow

k_{rw} = k_{rw}'\,S_n^{\,n_w},\qquad k_{ro} = k_{ro}'\,(1 - S_n)^{n_o}

the Corey model; $S_n$ is the normalised water saturation

f_w = \dfrac{1}{1 + \dfrac{k_{ro}}{k_{rw}}\cdot\dfrac{\mu_w}{\mu_o}}

fractional flow of water: a horizontal reservoir, without capillary pressure

Common pitfalls

Comparing curves from different reports without checking the base of normalisation: $k_{ro}(S_{wi}) = 1$ only means that the curves are referred to the permeability to oil.
Mixing up relative permeability with the share of a phase in the flow: at equal relative permeabilities, tens of times more gas flows than oil, because gas is much less viscous.
Taking drainage curves to calculate the waterflooding of a water-wet reservoir.
Transferring laboratory relative permeabilities to a model without regard to the test conditions: the rate, the fluids, the wettability of the core, the capillary end effect (capillary forces hold the wetting phase back at the outlet end of the core).

After this lesson you can

distinguish absolute, effective and relative permeability, and write Darcy's law for two phases
read relative permeability curves for the oil–water and gas–oil pairs: the end points, the residual saturations, the crossover point of the curves
describe the curves with the Corey model and calculate relative permeability at a given saturation
estimate the fractional flow of water (the share of water in the flow), the water cut of the production and the mobility ratio (how easily the water moves compared with the oil) from relative permeabilities and viscosities
explain the effect of wettability and of the saturation history on the curves, and know how relative permeability is measured in the laboratory and why laboratory curves do not always represent the reservoir

Effective and relative permeability

Darcy's law describes the flow of one fluid that fills all the pores. The permeability in it is a property of the rock, and it is called the absolute permeability. When the pores hold oil and water together, or oil and gas, each phase moves through its own part of the pore space, and the phases hinder each other. The ability of the rock to pass one phase at a given saturation is called the effective permeability ( $k_o$ , $k_w$ , $k_g$ ), and its ratio to a base permeability is the relative permeability.

Darcy's law for each phase in multiphase flow

q_o = \frac{k\,k_{ro}\,A}{\mu_o}\cdot\frac{\Delta p_o}{L},\qquad q_w = \frac{k\,k_{rw}\,A}{\mu_w}\cdot\frac{\Delta p_w}{L}

where:

$q_o,\ q_w$ flow rates of oil and water at reservoir conditions, m³/s
$k$ absolute (base) permeability, m²
$k_{ro},\ k_{rw}$ relative permeabilities to oil and to water at the current saturation, —
$\mu_o,\ \mu_w$ viscosities of oil and water, Pa·s
$\Delta p_o,\ \Delta p_w$ pressure differences in the oil and in the water over the length $L$ , Pa
$A,\ L$ cross-section area and length, m², m

Units: SI (any consistent system; in Darcy units, as for single-phase flow)

Holds when

Steady-state flow at constant saturation.
Horizontal flow. For inclined flow the weight of the column of the phase is added.
The pressure differences in the phases differ by the change of capillary pressure along the length. In tests we try to make this difference negligibly small.

This is an empirical extension of Darcy's law (found from experiments): the effect of the second phase is reduced to a factor $k_{rx}$ that depends on saturation. The product $k\,k_{rx}$ is the effective permeability, and $k\,k_{rx}/\mu_x$ is the mobility of the phase, which decides which phase moves more easily.

Relative permeability

k_{rx}(S) = \frac{k_x(S)}{k_{\text{base}}}

where:

$k_x(S)$ effective permeability for phase $x$ (oil, water, gas) at saturation $S$ , mD
$k_{\text{base}}$ base permeability, mD
$k_{rx}$ relative permeability of the phase, fraction

Units: identical units of permeability

Holds when

The base of normalisation is stated explicitly: the absolute permeability (to gas with the Klinkenberg correction, or to water), or the effective permeability to oil at irreducible water saturation $k_o(S_{wi})$ .
To calculate a rate, we multiply the relative permeabilities by the same permeability that they are referred to.

Relative permeability makes it possible to compare samples of different permeability: their curves are brought to one scale from 0 to 1. When the curves are normalised to $k_o(S_{wi})$ , the oil end point is equal to one by definition.

The pressures of oil and water at one point of the reservoir are not equal: their difference is the capillary pressure (the pressure jump across the boundary between the two fluids in a pore). So in the general form we write Darcy's law for each phase with its own pressure gradient (see the advanced section). In a laboratory test the pressure differences in the phases are usually close, and we calculate the relative permeabilities from one measured pressure difference.

Worked exampleRelative permeability from a steady-state core test

Given (assumed data)

Sample: length 7.0 cm, diameter 3.8 cm, absolute permeability to water 240 mD
Steady state at $S_w = 0.50$: water 0.032 cm³/s, oil 0.060 cm³/s, pressure difference 1.00 atm in both phases
Viscosities at the test conditions: water 1.0 cP, oil 2.0 cP

Solution

Cross-section area: $A = \pi\cdot 3.8^2/4 = 11.34$ cm².
Effective permeability in Darcy units: $k_x = q_x\mu_x L/(A\,\Delta p)$ . For water: $0.032\cdot 1.0\cdot 7.0/11.34 = 0.0198$ D, that is 19.8 mD.
For oil: $0.060\cdot 2.0\cdot 7.0/11.34 = 0.0741$ D, that is 74.1 mD.
Relative permeabilities: $k_{rw} = 19.8/240 = 0.082$ , $k_{ro} = 74.1/240 = 0.309$ ; their sum is 0.391.

Answer: At a water saturation of 0.50,

k_{rw} \approx 0.082

and

k_{ro} \approx 0.309

What it means. Together the two phases use only about 39% of the flow capacity of the rock: the boundaries between the phases, trapped drops and winding paths noticeably raise the resistance. Water occupies half of the pores, but its share in the flow is 35%: water flows through small pores and films, and oil through large pores (the rock is water-wet). One point gives one pair of values. We get a full curve from a series of runs with different ratios of flow rates. The data in this example are assumed for illustration.

Relative permeability curves for oil and water

A series of such tests at different saturations gives the curves $k_{ro}(S_w)$ and $k_{rw}(S_w)$ . Their shape is explained by where the phases sit. Below are the curves of a water-wet rock, in which the water occupies small pores and films on the grains, and the oil occupies large pores.

CalculatedRelative permeability curves of a water-wet rock, calculated with the Corey model with the parameters of the example below. The dashed lines are the irreducible water saturation

S_{wi}

and the saturation

1 - S_{or}

, beyond which only one phase flows; the curves cross at

S_w \approx 0.56

At $S_w = S_{wi}$: the water is in films and small pores and does not form continuous paths: $k_{rw} = 0$ . The oil in the large pores flows almost as a single phase: $k_{ro}$ equals the end point $k_{ro}'$ , usually 0.5–1 of the absolute permeability.
As $S_w$ grows: the water fills larger and larger pores and starts to move, and the oil loses its best paths: $k_{ro}$ falls and $k_{rw}$ rises. The sum of the relative permeabilities is noticeably less than one: the phases hinder each other.
At $S_w = 1 - S_{or}$: the oil is broken into immobile drops: $k_{ro} = 0$ . The water flows, but the drops sit in the centres of large pores, so the water end point $k_{rw}'$ is noticeably less than one, usually 0.1–0.3 in a water-wet rock.
The crossover point: the saturation at which $k_{ro} = k_{rw}$ . In a water-wet rock it lies to the right of 0.5, and in an oil-wet rock to the left.

The end saturations $S_{wi}$ and $S_{or}$ are the same irreducible water saturation (the water that stays in the pores) and residual oil saturation (the oil left behind after flooding). They determine oil in place (the amount of oil in the reservoir) and the displacement efficiency, the share of the oil that water pushes out of the swept rock (end-point saturations). Between them lies the mobile interval $1 - S_{wi} - S_{or}$ , the share of the pores from which oil can in principle be displaced by water. In the lesson on saturation the irreducible water saturation is written $S_{wirr}$ , and the letters $S_{wi}$ stand for the initial one. The end point of the curves is written $S_{wi}$ here and in the lessons on displacement: in the main part of a reservoir, above the transition zone (the zone above the oil–water contact where the water saturation changes with height), the initial water saturation is equal to the irreducible one.

The Corey model

We describe the laboratory points with smooth functions: for interpolation (filling in between the measured points), for comparing samples and as input to a model. The most widespread approximation is the power-law Corey model.

The Corey model for the oil–water pair

S_n = \frac{S_w - S_{wi}}{1 - S_{wi} - S_{or}},\qquad k_{rw} = k_{rw}'\,S_n^{\,n_w},\qquad k_{ro} = k_{ro}'\,(1 - S_n)^{n_o}

where:

$S_n$ normalised water saturation: 0 at $S_{wi}$ , 1 at $1 - S_{or}$ , —
$S_{wi},\ S_{or}$ irreducible water saturation and residual oil saturation, fraction
$k_{rw}',\ k_{ro}'$ the end points: the relative permeability of water at residual oil, and of oil at irreducible water, —
$n_w,\ n_o$ Corey exponents for water and oil, —

Units: saturations and relative permeabilities as fractions

Holds when

An empirical approximation: the parameters are fitted to the laboratory points.
Outside the interval from $S_{wi}$ to $1 - S_{or}$ a phase is immobile: $k_{rw} = 0$ at $S_w \le S_{wi}$ and $k_{ro} = 0$ at $S_w \ge 1 - S_{or}$ .
One set of parameters describes one process, drainage or imbibition.

The exponents set the curvature: the larger $n_w$ , the longer the water relative permeability stays small after the water starts to move; the larger $n_o$ , the faster the oil relative permeability falls. The exponents usually lie between 1.5 and 6. The end points and the exponents are the settings that are easiest to adjust in sensitivity analysis (checking how much the results change when an input changes) and in history matching (adjusting a model until it reproduces the production history).

What the relative permeability curves mean for the water cut

Relative permeability is not yet the share of a phase in the flow. The flow rate of each phase is proportional to its mobility $k\,k_{rx}/\mu_x$ , so for the same pressure gradient the share of water depends on both the relative permeabilities and the viscosities. For a horizontal reservoir, if the gradient of capillary pressure is neglected, the share of water in the reservoir flow is:

Fractional flow of water

f_w = \frac{q_w}{q_w + q_o} = \frac{1}{1 + \dfrac{k_{ro}}{k_{rw}}\cdot\dfrac{\mu_w}{\mu_o}}

where:

$f_w$ share of water in the total flow at reservoir conditions, —
$k_{ro},\ k_{rw}$ relative permeabilities of oil and water at the current water saturation, —
$\mu_w,\ \mu_o$ viscosities of water and oil at reservoir conditions, mPa·s

Units: dimensionless; viscosities in identical units

Holds when

Horizontal flow: gravity is not taken into account.
The gradient of capillary pressure is neglected.
The flow rates are at reservoir conditions, and there is no free gas.

The share of water grows with the water saturation from 0 at $S_{wi}$ to 1 at $1 - S_{or}$ . The more viscous the oil, the higher $f_w$ is at the same saturation: in a reservoir with viscous oil, most of the oil is produced at a high water cut. The full theory, with gravity, capillary forces and the movement of the front, is in the lessons Fractional flow function and mobility ratio and Buckley–Leverett theory and the Welge method.

Water cut of the production at the surface

\text{WC} = \frac{f_w/B_w}{f_w/B_w + (1 - f_w)/B_o}

where:

$\text{WC}$ water cut: the share of water in the produced liquid at standard conditions, —
$B_w,\ B_o$ formation volume factors of water and of oil, m³/m³

Units: dimensionless

Holds when

Only oil and water flow in the reservoir.
We convert the reservoir flow rates to standard conditions with the formation volume factors.

Oil in the reservoir takes more volume than at the surface, so the water cut at the surface is slightly higher than the share of water in the reservoir flow.

The mobility ratio at the end points

M' = \frac{k_{rw}'/\mu_w}{k_{ro}'/\mu_o}

where:

$M'$ ratio of the mobility of the water behind the displacement front (the moving boundary between the injected water and the oil) to the mobility of the oil ahead of it, —
$k_{rw}',\ k_{ro}'$ end points of the relative permeability of water and of oil, —
$\mu_w,\ \mu_o$ viscosities of water and of oil at reservoir conditions, mPa·s

Units: dimensionless

Holds when

We take the end points from the curves of the same process, usually imbibition.
The viscosities are at reservoir conditions.

At $M' \le 1$ the water moves no more easily than the oil, and the displacement front is stable. At $M' > 1$ the water runs ahead of the oil and tends to break through in tongues (fingers), and breakthrough comes earlier (mobility ratio).

Worked exampleRelative permeability, fractional flow and water cut at a given saturation

Given (assumed data)

Corey parameters (water-wet rock): $S_{wi} = 0.22$ , $S_{or} = 0.25$ , $k_{ro}' = 0.85$ , $k_{rw}' = 0.25$ , $n_o = 2.5$ , $n_w = 3$
Water saturation of the reservoir at the well: $S_w = 0.55$
Viscosities in the reservoir: water 0.5 mPa·s; oil 3.0 mPa·s (for comparison, 30 mPa·s)
Formation volume factors: oil 1.25 m³/m³, water 1.02 m³/m³

Solution

Normalised saturation: $S_n = (0.55 - 0.22)/(1 - 0.22 - 0.25) = 0.623$ .
Relative permeabilities: $k_{rw} = 0.25\cdot 0.623^3 = 0.0603$ ; $k_{ro} = 0.85\cdot(1 - 0.623)^{2.5} = 0.0744$ .
Ratio of the mobilities of oil and water: $(0.0744/3.0)/(0.0603/0.5) = 0.205$ , and the share of water in the reservoir is $f_w = 1/(1 + 0.205) = 0.830$ .
At the surface: water $0.830/1.02 = 0.813$ , oil $0.170/1.25 = 0.136$ (as fractions of the reservoir flow); the water cut is $0.813/(0.813 + 0.136) = 0.856$ .
With an oil viscosity of 30 mPa·s the ratio of the mobilities is 10 times smaller, 0.0205, and $f_w = 0.980$ .
The mobility ratio at the end points: $M' = (0.25/0.5)/(0.85/3.0) = 1.76$ ; for oil of 30 mPa·s, 17.6.

Answer: At

S_w = 0.55

the relative permeability of water is 0.0603 and that of oil is 0.0744. The reservoir flow contains 83% water, and the water cut of the production is 86%. With oil of viscosity 30 mPa·s, the share of water in the flow would be 98%.

What it means. Water fills a little more than half of the pores, but it gives more than four fifths of the flow: at comparable relative permeabilities it is six times less viscous and flows much more easily than oil. $M' \approx 1.76 > 1$ , so the displacement is unstable, and water will break through earlier than in piston-like displacement (a flat front that pushes all the oil ahead of it). With viscous oil ( $M'$ about 17.6) the effect is much stronger. This is why reservoirs with viscous oil are developed mostly at a high water cut. The data in this example are assumed for illustration.

CalculatedFractional flow of water for the relative permeability curves of the example, with a water viscosity of 0.5 mPa·s and two oil viscosities. With viscous oil the share of water exceeds 90% already at

S_w \approx 0.47

, and stays above 90% over more than half of the range of mobile saturation.

Gas and oil

When the pressure in the reservoir falls below the bubble point pressure (the pressure at which gas starts to come out of the oil), or gas is injected into the reservoir, gas and oil flow together. Gas is the non-wetting phase (the phase that the rock surface does not prefer) relative to both water and oil, so it first occupies the largest pores. The irreducible water is immobile in this case, and the relative permeabilities of gas and oil are plotted against the gas saturation, or against the total liquid saturation $S_L = S_o + S_{wi}$ .

While the gas saturation is below the critical one, $S_{gc}$ , usually a few per cent, the gas bubbles are disconnected and immobile, but they already block part of the channels and lower $k_{ro}$ (critical gas saturation).
Above $S_{gc}$ the gas forms continuous channels in the large pores, and $k_{ro}$ falls faster than when oil is displaced by water in a water-wet rock: the gas takes away from the oil exactly its best paths.
The viscosity of gas in the reservoir is tens of times lower than that of oil, so even a small $k_{rg}$ gives a large flow of gas. For example, at $k_{rg} = 0.06$ , $k_{ro} = 0.25$ , $\mu_g = 0.02$ mPa·s and $\mu_o = 1.0$ mPa·s, the ratio of the mobilities of gas and oil is $(0.06/0.02)/(0.25/1.0) = 12$ : for each volume of oil the reservoir gives 12 volumes of gas at reservoir conditions.

That is why, after the critical gas saturation is reached, the gas–oil ratio of the wells (the gas produced per unit of oil) grows quickly, and oil production falls. This is the typical picture of a solution gas drive (a drive in which the gas that comes out of the oil pushes the oil toward the wells) (Performance prediction for solution gas drive reservoirs; for the effect of free gas on inflow, see Inflow performance relationship (IPR)). The residual oil saturation in displacement by gas, $S_{org}$ , differs from the residual in displacement by water, and we find it in separate tests.

Wettability and the history of saturation

The shape of the relative permeability curves reflects where the phases are, and so it reflects the wettability, the preference of the rock surface for water or for oil. In a water-wet rock the water occupies small pores and films: the relative permeability of water stays low, and that of oil falls slowly. In an oil-wet rock the water flows through large pores: the relative permeability of water is high, and that of oil falls fast. These differences are summarised in the empirical rules of Craig.

Signs of wettability on relative permeability curves (Craig's rules)

Sign: Irreducible water saturation
Water-wet: usually more than 0.20–0.25
Oil-wet: usually less than 0.15, often less than 0.10

Sign: $S_w$ at the point where $k_{ro} = k_{rw}$
Water-wet: more than 0.5
Oil-wet: less than 0.5

Sign: $k_{rw}$ at $S_{or}$ , as a fraction of $k_{ro}$ at $S_{wi}$
Water-wet: usually less than 0.3
Oil-wet: more than 0.5, up to 1

Sign	Water-wet	Oil-wet
Irreducible water saturation	usually more than 0.20–0.25	usually less than 0.15, often less than 0.10
$S_w$ at the point where $k_{ro} = k_{rw}$	more than 0.5	less than 0.5
$k_{rw}$ at $S_{or}$ , as a fraction of $k_{ro}$ at $S_{wi}$	usually less than 0.3	more than 0.5, up to 1

The rules are empirical, and there are exceptions: the structure of the pores and their connectivity change the picture. Wettability is judged by the whole set of signs and by direct measurements.

Relative permeabilities, like capillary pressure, depend on the history of saturation. The curves of drainage (oil displaces water, as when a reservoir is formed) and of imbibition (water displaces oil, as in the waterflooding of a water-wet reservoir) are different. In a water-wet rock the water curve is almost the same for both processes, but the oil relative permeability is lower in imbibition: the water cuts off and traps drops of oil. So to calculate the waterflooding of a water-wet reservoir, we need the imbibition curves.

If the saturation changes direction half-way, the reservoir follows intermediate curves, and simulators include hysteresis (the dependence of the curves on the path the saturation has taken) for this. It is especially important for gas: when water invades a gas reservoir, or when water and gas are injected in turn, part of the gas is trapped, and its relative permeability falls sharply.

How relative permeability is measured and how far to trust it

Relative permeability is measured on core samples. There are three main ways.

Steady-state method: we inject both phases at the same time at a constant ratio of the flow rates, wait for the steady state, and repeat the test at other ratios. Each point comes directly from Darcy's law, but a curve takes days and weeks.
Unsteady-state method: one phase displaces the other at a constant flow rate. We measure the production of the phases and the pressure difference over time, and work out the relative permeabilities by calculation. The method is fast and gives the end points well, but it is sensitive to the capillary end effect (explained below) and to instability of the front.
Centrifuge: displacement under centrifugal force; it gives the relative permeability of the displaced phase near the residual saturation.

For laboratory curves to describe the reservoir, you need:

Core with reservoir wettability: core in the native state (its wettability not changed by the mud or by storage), or core with restored wettability (cleaned, then aged in reservoir fluids) (Wettability and interfacial phenomena).
Reservoir fluids, or fluids close to them, and reservoir temperature and load on the sample.
Flow rates at which the ratio of the capillary forces to the viscous forces is close to that in the reservoir. At the outlet end of the sample, capillary forces hold back the wetting phase. This is the capillary end effect, and the saturation there differs from the mean. We reduce it with high flow rates and long composite samples (several plugs placed end to end), or allow for it by calculation.
The right process: imbibition for the waterflooding of a water-wet reservoir, and drainage for gas invasion.

Even good laboratory relative permeabilities refer to a sample a few centimetres long. In a model with cells tens of metres across, the non-uniformity of the reservoir changes the effective curves, so we refine them during history matching (What a reservoir simulation model is and why it is needed, Petrophysical input for simulation: relative permeability and capillary pressure).

Terms and symbols

Absolute permeability: The permeability of the rock when one fluid fills all the pores.
Effective permeability $k_o$ , $k_w$ , $k_g$: The permeability for one phase when several flow together; it is always less than the absolute permeability.
Relative permeability $k_{rx}$: The ratio of the effective permeability to the base permeability; a function of saturation from 0 to 1.
End points $k_{ro}'$ , $k_{rw}'$: The relative permeability of oil at irreducible water saturation, and of water at residual oil saturation.
Mobility $k\,k_{rx}/\mu_x$: How easily a phase flows through the rock, with its viscosity taken into account. The mobility ratio decides whether the displacement is stable.
Fractional flow of water $f_w$: The share of water in the total reservoir flow rate at a given saturation; together with the formation volume factors it determines the water cut of the production.
Capillary end effect: The holding back of the wetting phase at the outlet end of the sample by capillary forces, which distorts the laboratory relative permeabilities.

Common mistakes and misconceptions

Wrong

“ $k_{ro} + k_{rw} = 1$ : what the oil does not get, the water gets.”

In fact

The phases hinder each other (the boundaries between the phases, trapped drops, winding paths), so the sum of the relative permeabilities is noticeably less than one, and in the middle of the saturation interval it can be several times less.

Wrong

“The relative permeabilities of gas and oil are equal, so the phases flow in equal amounts.”

In fact

The flow rate is set by the mobility $k_r/\mu$ . Gas in the reservoir is tens of times less viscous, so at equal relative permeabilities tens of times more of it flows.

Wrong

“The water cut of the production is about equal to the water saturation of the reservoir.”

In fact

The water cut is set by the share of water in the flow, and that depends on the relative permeabilities and the viscosities. In the lesson's example at $S_w = 0.55$ the reservoir flow has about 83% water, and with viscous oil, 98%.

Wrong

“Any relative permeability curves from the core of this reservoir will do for a waterflood calculation.”

In fact

You need imbibition curves, measured on core with reservoir wettability under conditions that represent the reservoir. Drainage curves, and data on cleaned core, give a wrong forecast of the water cut.

Wrong

“Residual oil saturation is a constant for a rock.”

In fact

It depends on the wettability, on the displacing agent (water or gas), on the interfacial tension, on the displacement rate and on the volume pumped through, so we measure it for the particular process.

In depth: advanced detail

Two-phase Darcy's law with gravity and capillary pressure

In general each phase moves under the gradient of its own pressure and under its own weight. For one-dimensional flow along an axis $x$ inclined to the horizontal at an angle $\alpha$ :

Two-phase Darcy's law in differential form

u_o = -\frac{k\,k_{ro}}{\mu_o}\left(\frac{\partial p_o}{\partial x} + \rho_o g\sin\alpha\right),\qquad u_w = -\frac{k\,k_{rw}}{\mu_w}\left(\frac{\partial p_w}{\partial x} + \rho_w g\sin\alpha\right),\qquad p_o - p_w = p_c(S_w)

where:

$u_o,\ u_w$ flow velocities of oil and water: the flow rate per unit area, m/s
$p_o,\ p_w$ pressures in the oil and in the water, Pa
$\rho_o,\ \rho_w$ densities of oil and water, kg/m³
$\alpha$ angle of the axis $x$ to the horizontal, positive if the axis points up the dip, rad
$p_c(S_w)$ capillary pressure at the current water saturation, Pa

Units: SI

Holds when

Immiscible phases, laminar flow.
The relative permeabilities and the capillary pressure depend only on saturation and on the direction of its change.
At $\alpha > 0$ the weight of a phase holds back its movement up the dip.

The phases are linked by the capillary pressure: from the pressure of one phase and the saturation, we can find the pressure of the other. In this form a reservoir simulator solves the equations. The fractional flow of the main section follows from them if we drop the inclination and the gradient of capillary pressure.

Three mobile phases

When water, oil and gas move in the reservoir at the same time (in the waterflooding of a reservoir with a gas cap, a layer of free gas above the oil; in gas injection after water; or in alternate injection of water and gas), we need three-phase relative permeabilities. They are hard and slow to measure, so we usually calculate them from two sets of two-phase curves: water–oil and gas–oil.

In a water-wet rock it is assumed that the relative permeability of water depends only on the water saturation, that of gas only on the gas saturation, and that of oil, which occupies the pores of medium size, on both. For it, engineers use empirical interpolation schemes (rules for filling in between the two sets of curves), and the best known are Stone's models. The choice of scheme noticeably affects the forecast, especially in alternate injection of water and gas, so we check it against test data or the production history.

Connections to other topics

Rock permeability — absolute permeability and Darcy's law for one phase.
Fluid saturation — the end saturations, which are the end points of the relative permeability curves.
Wettability and interfacial phenomena — why the shape of the curves depends on wettability, and how to preserve it in core.
Capillary pressure and the transition zone — capillary pressure links the pressures of the phases and sets the initial saturation.
Fractional flow function and mobility ratio — the fractional flow of water with gravity and capillary forces taken into account.
Buckley–Leverett theory and the Welge method — how the displacement front and the water breakthrough time are calculated from the relative permeability curves.
Petrophysical input for simulation: relative permeability and capillary pressure — preparing relative permeability tables for a reservoir simulation model.

Check yourself

1
The relative permeabilities are referred to the permeability to oil at irreducible water saturation. What is $k_{ro}$ at $S_{wi}$ ?
1. A.Zero
2. B.The ratio of the absolute to the effective permeability
3. C.It depends on the oil viscosity
4. D.One, by definition
Show the answer
Correct answer: D. One, by definition
With this normalisation the oil end point is the base itself, so $k_{ro}(S_{wi}) = 1$ . To go over to normalisation to the absolute permeability, all the values are multiplied by $k_o(S_{wi})/k$ .
2
In a steady-state test, 0.050 cm³/s of oil with a viscosity of 4.0 cP flows through a sample 5.0 cm long with a cross-section of 10.0 cm² at a pressure difference of 2.0 atm. The absolute permeability is 300 mD. Find $k_{ro}$ .
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$k_o = q\mu L/(A\,\Delta p) = 0.050\cdot 4.0\cdot 5.0/(10.0\cdot 2.0) = 0.050$ D, that is 50 mD; $k_{ro} = 50/300 \approx 0.17$ .
3
Which set of signs points to an oil-wet rock?
1. A. $S_{wi} = 0.30$ , the curves cross at $S_w = 0.60$ , $k_{rw}' = 0.15$
2. B. $S_{wi} = 0.10$ , the curves cross at $S_w = 0.40$ , $k_{rw}' = 0.70$
3. C. $S_{wi} = 0.25$ , the curves cross at $S_w = 0.55$ , $k_{rw}' = 0.20$
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Correct answer: B. $S_{wi} = 0.10$ , the curves cross at $S_w = 0.40$ , $k_{rw}' = 0.70$
By Craig's rules, an oil-wet rock has a low irreducible water saturation, the curves cross at a water saturation below 0.5, and the relative permeability of water at the end of displacement is high. The first and the third sets are typical of a water-wet rock.
4
At some saturation $k_{ro} = 0.20$ and $k_{rw} = 0.10$ ; the oil viscosity is 2.0 mPa·s and the water viscosity is 0.5 mPa·s. What is the share of water in the reservoir flow?
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The ratio of the mobilities of oil and water is $(0.20/2.0)/(0.10/0.5) = 0.5$ , so $f_w = 1/(1 + 0.5) \approx 0.67$ : two thirds of the flow is water, although its relative permeability is half that of oil.
5
The mobility ratio at the end points is $M' = 0.6$ . What should you expect in waterflooding?
1. A.Unstable displacement with early water breakthrough
2. B.The water will not be able to displace the oil
3. C.A stable front: most of the mobile oil is produced before water breakthrough
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Correct answer: C. A stable front: most of the mobile oil is produced before water breakthrough
At $M' < 1$ the water behind the front moves no more easily than the oil ahead of it, disturbances of the front die out, the front stays sharp, and most of the mobile oil is produced before water breakthrough. The displacement approaches piston-like when $M'$ is much smaller than one. At $M' > 1$ the water runs ahead of the oil and breaks through in tongues.
6
Why does the gas–oil ratio of a well grow so fast after the gas saturation exceeds the critical one?
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Gas is the non-wetting phase and occupies the large pores, so once it becomes continuous it quickly gets good paths, and the oil relative permeability falls sharply. The viscosity of gas is tens of times lower than that of oil, so even a small gas relative permeability gives a large flow rate: the ratio of the gas and oil flow rates in the reservoir easily reaches ten or more.

Updated: 19 October 2026