Petrophysical input for simulation: relative permeability and capillary pressure

Quick refresher

The model needs the curves of the process that it calculates, for each rock type (a group of rocks with similar pore structure): imbibition curves for waterflooding of a water-wet reservoir, and drainage curves for the initial distribution of fluids and for gas invasion.
Laboratory curves are distorted by the capillary end effect: at the outlet of the sample the wetting phase is held back until the capillary pressure falls to zero. We check the experiments by changing the rate, by comparing methods, and by simulating the experiment itself.
The curves of samples are averaged in normalised coordinates, $S_n$ and $k_r/k_r'$ (each curve is stretched so that it runs from 0 to 1). The Corey exponents are averaged, and the end points are linked with permeability separately.
The capillary pressure is carried over to the cells by the Leverett J-function of the rock type, $p_c = J\,\sigma\cos\theta/(31.4\sqrt{k/\phi})$ . The same function sets the initial saturation by height above the free water level (FWL, the level where the capillary pressure is zero).
End-point scaling gives each cell its own irreducible water (water that cannot be displaced) and residual oil (oil left after flooding) without new tables. The Land relation sets the trapped oil and gas when the initial saturation is incomplete.
The table must be monotonic, and its end points must agree with each other and with the initial equilibrium. The rows are closer together where the curves are steeper.

Key relationships

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

normalising, and the Corey model

J = \dfrac{31.4\,p_c\sqrt{k/\phi}}{\sigma\cos\theta},\qquad J = J_d\,(S_w^{*})^{-b}

the Leverett function ( $p_c$ in MPa, $k$ in mD, $\sigma$ in mN/m) and its fit for a rock type

S_t = S_{wi,t} + (S_w - S_{wi})\,\dfrac{1 - S_{wi,t} - S_{or,t}}{1 - S_{wi} - S_{or}}

two-point scaling: the table saturation for the saturation of a cell

S_{or}^{*} = \dfrac{S_{oi}^{*}}{1 + C\,S_{oi}^{*}},\qquad S^{*} = \dfrac{S}{1 - S_{wi}}

the Land relation for trapped oil

\dfrac{k_{ro}(S_{w2})}{k_{ro}'} = \dfrac{f_{o2}}{\dd\bigl(1/(W_iI_r)\bigr)/\dd(1/W_i)}

the JBN method for an unsteady-state experiment

Common pitfalls

Taking drainage curves for waterflooding of a water-wet reservoir, or one curve for the whole reservoir.
Averaging the relative permeabilities of samples at the same water saturation. The end points blur, and the average curve looks like none of the samples.
Carrying over into the model the laboratory irreducible water and residual oil that were distorted by the end effect or by too small a pressure difference.
Leaving $k_{rw} = k_{rw}' < 1$ at $S_w = 1$ where there is a water zone: the water influx will be underestimated.
Setting the irreducible water differently in the relative permeability table and in the initial saturation: the model will contain mobile water that is not in the reservoir, or the reverse.

After this lesson you can

draw up a programme of special core analysis for the tasks of the model: which experiments, on which samples and under what conditions
recognise the capillary end effect and other distortions in the data, and process an unsteady-state experiment (water is injected at a constant rate while the saturations change) by the JBN method
normalise the relative permeability curves of samples, find the Corey exponents (the powers in the Corey formula that set the shape of a relative permeability curve) and average the curves of a rock type
build the Leverett J-function of a rock type (a function that brings the capillary pressure curves of rocks of one type to a single curve) from capillary pressure measurements, and calculate the capillary pressure and saturation for a cell of the model
scale the end points (the saturations and the relative permeabilities at the ends of the curves) cell by cell, account for trapping by the Land relation (a formula for the oil or gas that stays trapped in the pores), and know how three-phase relative permeabilities are set
assemble a table for the simulator and check it: monotonicity, consistency of the end points, initial equilibrium

What the model needs and where it comes from

For every cell the simulator needs the relative permeabilities and the capillary pressure as functions of saturation. They are set in tables by rock type, and the curves for the tables come from special core analysis (routine and special analysis). There are few experiments: they are long and expensive. So we plan the programme from the tasks of the model: which processes it will calculate, and which decisions depend on them.

What the model needs and which experiments give it

The model needs: Water–oil relative permeability, imbibition
Experiments: steady-state or unsteady-state, on restored-state core (core whose wettability has been restored)
What for: waterflooding, a rising oil–water contact

The model needs: Gas–oil relative permeability, drainage
Experiments: unsteady-state at irreducible water, centrifuge
What for: gas cap, solution gas, gas injection

The model needs: Capillary pressure, drainage
Experiments: porous plate (membrane), centrifuge, mercury with conversion
What for: initial saturation, transition zone

The model needs: Capillary pressure, imbibition
Experiments: centrifuge, porous plate; spontaneous and forced imbibition
What for: waterflooding, hysteresis

The model needs: Residual oil, end points
Experiments: long flooding, centrifuge
What for: $S_{or}$ , $k_{rw}'$ , recoverable reserves

The model needs: Wettability
Experiments: Amott, USBM (two standard wettability tests)
What for: choice of the curves and of the process

The model needs: Trapped gas
Experiments: imbibition at different initial gas saturations
What for: the Land coefficient

The model needs	Experiments	What for
Water–oil relative permeability, imbibition	steady-state or unsteady-state, on restored-state core (core whose wettability has been restored)	waterflooding, a rising oil–water contact
Gas–oil relative permeability, drainage	unsteady-state at irreducible water, centrifuge	gas cap, solution gas, gas injection
Capillary pressure, drainage	porous plate (membrane), centrifuge, mercury with conversion	initial saturation, transition zone
Capillary pressure, imbibition	centrifuge, porous plate; spontaneous and forced imbibition	waterflooding, hysteresis
Residual oil, end points	long flooding, centrifuge	$S_{or}$ , $k_{rw}'$ , recoverable reserves
Wettability	Amott, USBM (two standard wettability tests)	choice of the curves and of the process
Trapped gas	imbibition at different initial gas saturations	the Land coefficient

The methods of measuring relative permeability and capillary pressure are described in the lessons on relative permeability and on capillary pressure, and wettability in the lesson on wettability.

Samples. The curves are needed for rock types (rock types), so for each type we take several uniform samples (checked by tomography) that cover its range of permeability, and not only the best reservoir rock. The porosity and permeability of the samples are known beforehand from routine analysis, and the neighbouring pieces of core go to mercury injection.

A typical sequence of experiments on a sample

1
Preparation
Cleaning, saturation with reservoir (or model) water, and the permeability to water, which is the base for normalising the relative permeabilities.
2
Primary drainage
Oil displaces water down to the irreducible saturation. It is done on a semi-permeable membrane by pressure steps, or by flooding. The membrane steps give the drainage capillary pressure curve.
3
Restoring the wettability
Ageing the core in oil at reservoir temperature for weeks (Wettability and interfacial phenomena). A sample in its native state skips this step.
4
Imbibition and relative permeability
Water displaces oil by the steady-state or the unsteady-state method. At the end, long flooding or a centrifuge gives the residual oil.
5
Wettability and imbibition capillary pressure
Spontaneous and forced imbibition (Amott, USBM). Destructive experiments come at the very end.

Conditions. We use reservoir temperature and effective stress (the stress that squeezes the rock). We use reservoir oil, or oil with the same viscosity ratio, and water of reservoir composition. The rate is chosen so that capillary forces do not distort the result, but also not higher than Darcy's law and the strength of the rock particles allow (capillary length and the similarity criterion).

Checking the experiments: the end effect and the unsteady-state test

Beyond the outlet face of the sample the phases separate, and the capillary pressure there is zero. Inside a water-wet sample it is positive. So at the outlet the wetting phase, water, accumulates until its saturation reaches the value at which $p_c = 0$ . This is the capillary end effect. The zone of accumulation extends from the face over a length of about one capillary length, the distance over which capillary forces spread the saturation (capillary term), and in a short sample at a low rate it takes a noticeable part of the sample.

In drainage (oil displaces water), water is held at the outlet. The irreducible water saturation of the sample comes out too high, and the oil relative permeability at it too low.
In imbibition, water must accumulate at the outlet before it leaves the sample. Breakthrough is delayed, the average water saturation of the sample is higher than in its inner part, and the relative permeabilities assigned to the average saturation are distorted. In a rock of mixed wettability (some of the surface prefers water, and some prefers oil) the capillary pressure becomes zero while there is still a lot of oil in the pores, and extra oil stays at the outlet. So the residual oil saturation comes out too high.
In a steady-state experiment the saturation is not uniform along the length, and a relative permeability point refers not to the saturation that the volume balance gave.

How the end effect is recognised and reduced

1
Change the rate
Repeat the experiment, or a part of it, at another rate. If the result depends on the rate, it has a capillary component.
2
Lengthen the sample
Use composite samples made of several cylinders in good contact. In the steady-state method, use a mixing section at the inlet and the outlet, or semi-permeable membranes at both ends (Hassler's method).
3
Measure the saturation along the length
X-ray or tomographic scanning shows the saturation profile and the zone of accumulation at the outlet.
4
Simulate the experiment itself
Model the experiment on a fine grid with capillary pressure, and choose the relative permeabilities so that the production and the pressure drop agree in time. This is more reliable than analytical processing when capillary forces are noticeable.

Unsteady-state experiment. Water at a constant rate displaces oil from a sample at irreducible water saturation. During the experiment we measure the production of the phases and the pressure drop over time. How to find the water fraction and the saturation at the outlet by Welge's formulas is described in the lesson on Buckley–Leverett. To get the relative permeabilities themselves, we need a second relation, the one with the pressure drop. It is given by the JBN method (Johnson–Bossler–Naumann). The pressure drop is converted into the relative injectivity, the ratio of $q/\Delta p$ to its value at the start of the experiment, when only oil flowed through the sample.

The JBN method: relative permeability at the outlet face from an unsteady-state experiment

\begin{aligned} f_{o2} &= \frac{\dd N_{pD}}{\dd W_i},\qquad S_{w2} = \bar S_w - W_i\,f_{o2},\qquad I_r = \frac{q/\Delta p}{(q/\Delta p)_0},\\ \frac{k_{ro}(S_{w2})}{k_{ro}'} &= \frac{f_{o2}}{\dd\bigl(1/(W_iI_r)\bigr)/\dd(1/W_i)},\qquad k_{rw}(S_{w2}) = k_{ro}(S_{w2})\,\frac{1 - f_{o2}}{f_{o2}}\,\frac{\mu_w}{\mu_o} \end{aligned}

where:

$W_i,\ N_{pD}$ water injected and oil produced, in pore volumes of the sample ( $W_i$ is the same as $W_{iD}$ of the lesson Buckley–Leverett theory and the Welge method), —
$f_{o2}$ oil fraction in the flow at the outlet of the sample, $1 - f_{we}$ ; the index 2 means the outlet face, as is usual in this method, —
$\bar S_w,\ S_{w2}$ average water saturation of the sample ( $S_{wi} + N_{pD}$ ) and the saturation at the outlet face ( $S_{we}$ of the Buckley–Leverett lesson), —
$I_r$ relative injectivity. At a constant rate $I_r = \Delta p_0/\Delta p$ , where $\Delta p_0$ is the pressure drop at the start of the experiment (only oil flows, at $S_{wi}$ ), —
$k_{ro}'$ oil relative permeability at $S_{wi}$ : the method gives the relative permeabilities as fractions of this value, —
$\mu_w,\ \mu_o$ viscosities of water and oil at the conditions of the experiment, mPa·s

Units: dimensionless; viscosities in equal units

Holds when

The assumptions of Buckley–Leverett hold: a uniform sample, incompressible phases, negligible capillary pressure and end effect, and stable displacement.
The data are taken after water breakthrough. The derivatives are taken from smoothed dependences.
To get the relative permeabilities as fractions of the absolute permeability, the result is multiplied by $k_{ro}'$ .

Each measurement after breakthrough gives one point of the curves, at the saturation at the outlet. Before breakthrough the outlet sees only oil, so the experiment tells us nothing about the interval from $S_{wi}$ to the saturation at the front. With a sharp front ( $M' \le 1$ ) this is most of the range of mobile saturation.

Worked exampleProcessing an unsteady-state experiment by the JBN method

Given (assumed data)

Sample: a twin plug of plug 2 (the table in the next section): $S_{wi}$ = 0.227, $k_{ro}'$ = 0.81; oil 2.0 mPa·s, water 0.6 mPa·s; constant rate
Water breakthrough: after 0.422 pore volumes
Measurements after breakthrough: at $W_i$ = 0.5, 0.6 and 0.7: $N_{pD}$ = 0.4297, 0.4371 and 0.4430; $\Delta p/\Delta p_0$ = 1.9058, 1.8194 and 1.7529

Solution

The oil fraction at the outlet at $W_i$ = 0.6 is the derivative of production with respect to injection: $f_{o2} \approx (0.4430 - 0.4297)/(0.7 - 0.5) = 0.0665$ .
The average saturation is $\bar S_w = 0.227 + 0.4371 = 0.6641$ . At the outlet $S_{w2} = 0.6641 - 0.6\cdot0.0665 = 0.6242$ .
The quantity $1/(W_iI_r) = (\Delta p/\Delta p_0)/W_i$ is 3.8116, 3.0323 and 2.5041.
Its derivative with respect to $1/W_i$ is $(3.8116 - 2.5041)/(1/0.5 - 1/0.7) = 1.3075/0.5714 = 2.2882$ .
The relative permeabilities as fractions of $k_{ro}'$ : for oil $0.0665/2.2882 = 0.02906$ , and for water $0.02906\cdot(1 - 0.0665)/0.0665\cdot0.6/2.0 = 0.1224$ .
For comparison, the curves from which the experiment was built, at $S_w$ = 0.6242, give 0.02924 and 0.1230.

Answer: At

S_w

= 0.6242 the method gave

k_{ro}/k_{ro}'

= 0.02906 and

k_{rw}/k_{ro}'

= 0.1224, within 0.6% of the true values.

What it means. The measurements here are calculated from the Buckley–Leverett solution without capillary pressure, so the method returns the original curves. Only the finite-difference derivative gives an error. In a real experiment the end effect breaks this assumption, and the JBN relative permeabilities begin to depend on the rate. Note the range too. Breakthrough came at a water saturation at the front of 0.604, and the experiment tells us nothing about the whole interval from 0.227 up to it. We take this part from a steady-state experiment, or from a simulation of the experiment on a grid. The data in this example are assumed for illustration.

Capillary pressure measurements. On a semi-permeable membrane each step is held until water stops leaving the sample. Otherwise the point of the curve will be under-drained. A centrifuge gives the saturation averaged over the sample, while the capillary pressure and the saturation change along its length. The curve is recovered by calculation (see the advanced section). The mercury curve is converted to reservoir fluids (conversion to reservoir conditions), and its initial part, where mercury fills the roughness of the sample surface, is discarded.

Normalising and averaging the curves

Samples of one rock type give similar curves, but not equal ones. Each has its own irreducible water, its own residual oil and its own end-point relative permeabilities. If we average the curves at equal water saturation, the average curve starts at the smallest $S_{wi}$ , ends at the largest $1 - S_{or}$ , and is flatter than any of them. So we first normalise the curves, that is, we convert them to coordinates where all the end points are the same. Only then do we average them, and we describe the end points separately.

Normalised relative permeabilities and Corey exponents of a sample

S_n = \frac{S_w - S_{wi}}{1 - S_{wi} - S_{or}},\qquad k_{rw}^{*} = \frac{k_{rw}}{k_{rw}'},\quad k_{ro}^{*} = \frac{k_{ro}}{k_{ro}'},\qquad n_w = \frac{\sum \ln k_{rw}^{*}\,\ln S_n}{\sum (\ln S_n)^2},\quad n_o = \frac{\sum \ln k_{ro}^{*}\,\ln(1 - S_n)}{\sum \bigl(\ln(1 - S_n)\bigr)^2}

where:

$S_n$ normalised water saturation of the sample (Corey model), —
$k_{rw}^{*},\ k_{ro}^{*}$ relative permeabilities as fractions of the end-point values of the sample, —
$n_w,\ n_o$ Corey exponents: the slopes of straight lines through the origin, in logarithms, —
$S_{wi},\ S_{or},\ k_{rw}',\ k_{ro}'$ end points of the sample itself, —

Units: dimensionless; the sums are over the inner points of the sample

Holds when

In logarithms the Corey model is a straight line through the origin: $\ln k_r^{*} = n\ln S_n$ . The least-squares method for such a line gives the formulas above.
The points at the very ends (small $S_n$ or $1 - S_n$ ) are distorted more strongly by the error of the end points, so we check them separately.
If the Corey model does not describe the shape of the curves, we average the normalised relative permeabilities at equal $S_n$ , or we take a more flexible fit.

Normalising separates the shape of a curve (the exponents) from its ends. We average the exponents over the samples of the rock type, and link the end points with permeability or with height above the contact. Each cell of the model then gets its own ends with a common shape.

Samples of rock type 2: end points of the imbibition experiments

No.: 1
$k$ , mD; $\phi$: 22; 0.181
$S_{wi}$: 0.257
$S_{or}$: 0.262
$k_{ro}'$ ; $k_{rw}'$: 0.76; 0.20

No.: 2
$k$ , mD; $\phi$: 41; 0.207
$S_{wi}$: 0.227
$S_{or}$: 0.248
$k_{ro}'$ ; $k_{rw}'$: 0.81; 0.23

No.: 3
$k$ , mD; $\phi$: 66; 0.232
$S_{wi}$: 0.204
$S_{or}$: 0.236
$k_{ro}'$ ; $k_{rw}'$: 0.85; 0.26

No.	$k$ , mD; $\phi$	$S_{wi}$	$S_{or}$	$k_{ro}'$ ; $k_{rw}'$
1	22; 0.181	0.257	0.262	0.76; 0.20
2	41; 0.207	0.227	0.248	0.81; 0.23
3	66; 0.232	0.204	0.236	0.85; 0.26

Rock type 2 of the lesson Reservoir quality: geological controls on porosity and permeability: the FZI (flow zone indicator, a number that groups rocks by pore structure) of the samples is 1.57, 1.69, 1.75 µm. $S_{wi}$ is the saturation after the last step of the membrane. The relative permeabilities are as fractions of the absolute permeability. The data are assumed for illustration.

Steady-state imbibition experiments: inner points

No.: 1
$S_w$: 0.352
$k_{rw}$: 0.0009
$k_{ro}$: 0.462

No.: 1
$S_w$: 0.451
$k_{rw}$: 0.0093
$k_{ro}$: 0.243

No.: 1
$S_w$: 0.544
$k_{rw}$: 0.0342
$k_{ro}$: 0.103

No.: 1
$S_w$: 0.643
$k_{rw}$: 0.0951
$k_{ro}$: 0.0215

No.: 2
$S_w$: 0.334
$k_{rw}$: 0.0020
$k_{ro}$: 0.455

No.: 2
$S_w$: 0.436
$k_{rw}$: 0.0144
$k_{ro}$: 0.229

No.: 2
$S_w$: 0.543
$k_{rw}$: 0.0510
$k_{ro}$: 0.0800

No.: 2
$S_w$: 0.645
$k_{rw}$: 0.1150
$k_{ro}$: 0.0151

No.: 3
$S_w$: 0.317
$k_{rw}$: 0.0035
$k_{ro}$: 0.451

No.: 3
$S_w$: 0.427
$k_{rw}$: 0.0214
$k_{ro}$: 0.206

No.: 3
$S_w$: 0.541
$k_{rw}$: 0.0667
$k_{ro}$: 0.0640

No.: 3
$S_w$: 0.653
$k_{rw}$: 0.1450
$k_{ro}$: 0.0090

No.	$S_w$	$k_{rw}$	$k_{ro}$
1	0.352	0.0009	0.462
1	0.451	0.0093	0.243
1	0.544	0.0342	0.103
1	0.643	0.0951	0.0215
2	0.334	0.0020	0.455
2	0.436	0.0144	0.229
2	0.543	0.0510	0.0800
2	0.645	0.1150	0.0151
3	0.317	0.0035	0.451
3	0.427	0.0214	0.206
3	0.541	0.0667	0.0640
3	0.653	0.1450	0.0090

Reservoir fluids at reservoir temperature. The data are assumed for illustration.

Worked exampleCorey exponents of a rock type from three samples

Given (assumed data)

Samples and points: the two tables above
Task: common exponents $n_w$ and $n_o$ for rock type 2

Solution

Sample 2, the point $S_w$ = 0.436: $S_n = (0.436 - 0.227)/(1 - 0.227 - 0.248) = 0.209/0.525 = 0.3981$ ; $k_{rw}^{*} = 0.0144/0.23 = 0.0626$ ; $k_{ro}^{*} = 0.229/0.81 = 0.2827$ .
This single point would give $n_w = \ln 0.0626/\ln 0.3981 = 3.01$ and $n_o = \ln 0.2827/\ln(1 - 0.3981) = 2.49$ .
From all four points of each sample (the formulas above): $n_w$ = 3.349, 2.989 and 2.695; $n_o$ = 2.200, 2.505 and 2.809.
For the rock type: $n_w = (3.349 + 2.989 + 2.695)/3 = 3.011 \approx 3.01$ ; $n_o = 2.505 \approx 2.50$ .

Answer: For rock type 2 we adopt

n_w

= 3.01 and

n_o

= 2.50. The normalised points of all the samples lie near these curves (the figure below).

What it means. The scatter of the exponents is not random: for a less permeable sample the water curve is steeper ( $n_w$ is larger) and the oil curve is flatter. With three samples this is averaged. With more samples, the exponents, like the end points, are linked with permeability. How much the accuracy matters: at $S_n$ = 0.5 the exponents 3.349 and 2.695 give $k_{rw}^{*}$ = 0.098 and 0.154, which is 1.6 times more. The data in this example are assumed for illustration.

CalculatedNormalised relative permeability points of three samples of rock type 2, and Corey curves with the adopted exponents

n_w

= 3.01 and

n_o

= 2.50.

The end points are described separately, as functions of properties known in each cell. For the three samples of type 2, the irreducible water and the residual oil fall as permeability grows. The straight lines in coordinates $\lg k$ ( $\lg$ is the logarithm to base 10), by least squares, are: $S_{wi} = 0.4061 - 0.1111\lg k$ and $S_{or} = 0.3352 - 0.0544\lg k$ ( $k$ in mD). The end-point relative permeabilities also grow with permeability ( $k_{ro}'$ from 0.76 to 0.85, $k_{rw}'$ from 0.20 to 0.26). But in the example, for simplicity, they were averaged: 0.807 and 0.23. For tight cells this slightly overestimates the end-point relative permeabilities, and for permeable cells it underestimates them. The irreducible water of a cell can also be taken from the Leverett function at its height above the contact: the next section.

Rock types and the Leverett J-function

Each rock type of the model gets its own set: normalised relative permeabilities, links of the end points with the properties of a cell, and a Leverett J-function for the capillary pressure. The type of a cell is taken from the geological model (prediction of rock types). The samples are assigned to types by the same features as in the typing, for example by FZI. The capillary pressures of samples of one type are brought to one curve by the Leverett J-function (Leverett J-function). And back: from it we get the curve for any cell with its $k$ and $\phi$ .

The Leverett J-function of a rock type and the capillary pressure of a cell

J = \frac{31.4\,p_c\sqrt{k/\phi}}{\sigma\cos\theta},\qquad S_w^{*} = \frac{S_w - S_{wirr}}{1 - S_{wirr}},\qquad J = J_d\,(S_w^{*})^{-b},\qquad p_c(S_w) = \frac{J\,\sigma\cos\theta}{31.4\sqrt{k/\phi}}

where:

$p_c$ capillary pressure, MPa
$k,\ \phi$ permeability and porosity of the sample or of the cell, mD, fraction
$\sigma\cos\theta$ adhesion tension of the pair of fluids, mN/m
$S_{wirr}$ irreducible water saturation of the sample or of the cell, —
$J_d,\ b$ parameters of the rock type: the $J$ of the entry pressure and the exponent, —

Units: metric oilfield; $J$ is dimensionless

Holds when

$31.4 = 10^6\sqrt{0.9869\cdot10^{-15}}/10^{-3}$ is the constant of the Leverett function for MPa, mD and mN/m (the lesson on capillary pressure).
The power form is the same as in the example of the lesson on capillary pressure. Its parameters are chosen from the points by least squares in logarithms.
The drainage curve is for the initial distribution of fluids. For a waterflood calculation we need an imbibition curve of its own.

The smaller $\sqrt{k/\phi}$ of a cell, the higher its capillary pressure at the same saturation, and the thicker the transition zone (the zone above the free water level where the water saturation falls gradually). Dividing by $\sigma\cos\theta$ lets us go from laboratory fluids to reservoir ones.

Capillary pressure measurements of the samples of type 2: water saturation after each pressure step

$p_c$ , kPa: 4
Sample 1: 1.000
Sample 2: 0.872
Sample 3: 0.723

$p_c$ , kPa: 8
Sample 1: 0.588
Sample 2: 0.471
Sample 3: 0.395

$p_c$ , kPa: 16
Sample 1: 0.374
Sample 2: 0.314
Sample 3: 0.277

$p_c$ , kPa: 32
Sample 1: 0.300
Sample 2: 0.259
Sample 3: 0.227

$p_c$ , kPa: 64
Sample 1: 0.267
Sample 2: 0.236
Sample 3: 0.212

$p_c$ , kPa: 128
Sample 1: 0.257
Sample 2: 0.227
Sample 3: 0.204

$p_c$ , kPa	Sample 1	Sample 2	Sample 3
4	1.000	0.872	0.723
8	0.588	0.471	0.395
16	0.374	0.314	0.277
32	0.300	0.259	0.227
64	0.267	0.236	0.212
128	0.257	0.227	0.204

Primary drainage on a semi-permeable membrane, reservoir oil and water at reservoir temperature, $\sigma\cos\theta$ = 25.98 mN/m (30 mN/m and 30°). The last step leaves the sample at the $S_{wi}$ from which the imbibition experiment began. The data are assumed for illustration.

Worked exampleThe Leverett function of type 2 and the saturation of a cell

Given (assumed data)

Samples: the table above; $\sigma\cos\theta$ = 25.98 mN/m
Cell of the model: type 2, $k$ = 36 mD, $\phi$ = 0.20; the irreducible water from the trend is $S_{wi} = 0.4061 - 0.1111\lg 36 = 0.2332$
Fluids in the reservoir: densities of water and oil 1.04 and 0.80 g/cm³

Solution

$J$ of sample 2 at 8 kPa: $31.4\cdot0.008\cdot\sqrt{41/0.207}/25.98 = 31.4\cdot0.008\cdot14.074/25.98 = 0.1361$ . Its $S_{wirr}$ is the saturation at the last step, 0.227, so $S_w^{*} = (0.471 - 0.227)/(1 - 0.227) = 0.3157$ .
The same is done for all the steps. At the last step, 128 kPa, $S_w^{*} = 0$ by the very definition of $S_{wirr}$ , and these points do not enter the fit. We discard one more point, of sample 1 at 4 kPa (the pressure is below the entry pressure, $S_w$ = 1), and the points at 64 kPa, where $S_w^{*}$ is less than 0.02: there the normalised saturation is decided mostly by the error of $S_{wirr}$ . Of 18 points, 11 remain.
The line $\ln J = \ln J_d - b\ln S_w^{*}$ through these points gives $J_d$ = 0.0614 and $b$ = 0.680. The average scatter of the points is 2% in $J$ .
The pressure scale of the cell is $\sigma\cos\theta/(31.4\sqrt{k/\phi}) = 25.98/(31.4\cdot13.416) = 0.06167$ MPa. The entry pressure is $0.0614\cdot0.06167 = 0.00379$ MPa (3.79 kPa). The capillary pressure grows with height by $0.00981\cdot0.24 = 0.002354$ MPa/m, so the oil–water contact of the cell is $0.00379/0.002354 = 1.61$ m above the FWL.
At a height of 10 m: $p_c = 0.02354$ MPa, $J = 0.02354/0.06167 = 0.3817$ , $S_w^{*} = (0.3817/0.0614)^{-1/0.680} = 0.0681$ , and $S_w = 0.2332 + (1 - 0.2332)\cdot0.0681 = 0.285$ .
The last step of the membrane, 128 kPa, corresponds to a height of $0.128/0.002354 = 54$ m above the FWL.

Answer: The function of type 2 is

J = 0.0614\,(S_w^{*})^{-0.680}

. In the cell of 36 mD the oil–water contact lies 1.61 m above the FWL, and at a height of 10 m the water saturation is 0.285.

What it means. One function, with a scatter of a few per cent, described three samples with permeabilities from 22 to 66 mD (the figure below). The irreducible water of the samples is the saturation at the largest pressure of the experiment, that is, at a height of 54 m. If the accumulation is higher than that, the water saturation near the top comes from extrapolating the function, and we check it against well logs. The data in this example are assumed for illustration.

CalculatedThe Leverett function of rock type 2: the points of three samples (those used in the fit) and the power-law fit

J = 0.0614\,(S_w^{*})^{-0.680}

. The

J

scale is logarithmic.

Initial equilibrium and well logs. The simulator calculates the initial saturation of each cell from its height above the free water level and its capillary pressure curve (transition zone). We compare the resulting profile with the water saturation from well logs (saturation by height). If the difference is systematic, we look for the cause (the FWL level, the rock type, the wettability) and do not fit each cell. In some simulators the capillary pressure of a cell can be stretched so that it reproduces the log exactly. But then, in cells with strong stretching, the curve is no longer physical, and during development the water in them behaves differently from the water in the reservoir.

End-point scaling and trapping

Within one rock type the irreducible water and the residual oil change from cell to cell together with permeability. We do not need a table for each cell. The simulator takes the table of the rock type and scales it by the end points of the cell (end-point scaling). For the saturation of a cell it finds the table saturation with the same position between the end points. For the Corey model this is exactly the same normalised saturation $S_n$ . Then it takes the value at that table saturation.

Two-point scaling of a table by the end points of a cell

S_t = S_{wi,t} + (S_w - S_{wi})\,\frac{1 - S_{wi,t} - S_{or,t}}{1 - S_{wi} - S_{or}},\qquad k_{rw}(S_w) = k_{rw,t}(S_t)\,\frac{k_{rw}'}{k_{rw,t}'}

where:

$S_w,\ S_t$ water saturation of the cell and the corresponding saturation of the table, —
$S_{wi},\ S_{or}$ end-point saturations of the cell, —
$S_{wi,t},\ S_{or,t}$ end-point saturations of the table of the rock type, —
$k_{rw,t}(S_t)$ water relative permeability from the table (linear interpolation between rows), —
$k_{rw}'/k_{rw,t}'$ vertical scaling of the end-point value; for oil in the same way with $k_{ro}'$ , —

Units: dimensionless

Holds when

Horizontal scaling (along the saturation axis) is a linear mapping of the interval $[S_{wi}, 1 - S_{or}]$ of the cell onto the interval of the table. Three-point scaling adds an intermediate point (for example $1 - S_{or}$ between the critical and the largest water saturation) and maps the two parts separately.
The end points of a cell are set by maps, or by links with the properties of the cell. The capillary pressure is scaled consistently, through the Leverett function.
The shape of the curve (the exponents) is common to the rock type.

Scaling keeps the shape of the curve and moves its ends. Without it, cells with different irreducible water would get one table, and in the less permeable cells water would be mobile already at the initial saturation.

Trapping (oil left behind in the pores as isolated drops). The residual oil after waterflooding depends on how much oil there was in the rock at the start. In a transition zone the initial oil saturation is less than in the oil zone, and less oil is trapped. This link is described by the empirical Land relation. It was derived for trapped gas and is applied to oil as well. In simulators it is part of the hysteresis models (saturation history).

The Land relation: the trapped saturation of the non-wetting phase

S_{or}^{*} = \frac{S_{oi}^{*}}{1 + C\,S_{oi}^{*}},\qquad S^{*} = \frac{S}{1 - S_{wi}},\qquad C = \frac{1}{S_{or,\max}^{*}} - \frac{1}{S_{oi,\max}^{*}}

where:

$S_{oi}^{*},\ S_{or}^{*}$ initial and residual (trapped) oil saturation as fractions of the pore volume not occupied by irreducible water, —
$C$ Land coefficient: the larger it is, the less is trapped, —
$S_{or,\max}^{*}$ trapped oil at the largest initial oil saturation $S_{oi,\max}^{*}$ ; in the oil zone $S_{oi,\max}^{*} = 1$ , —

Units: fractions

Holds when

This is an empirical relation. $C$ is chosen from imbibition experiments at different initial saturations, or from one pair of end points.
For gas the same holds, with $S_{gi}^{*}$ and $S_{gr}^{*}$ : trapped gas when water invades a gas cap, in water-alternating-gas injection, and in underground gas storage.

At a small initial saturation almost all the non-wetting phase is trapped ( $S_{or}^{*} \approx S_{oi}^{*}$ ). As the initial saturation grows, the trapped amount grows more and more slowly, and at $S_{oi}^{*} = 1$ it equals $1/(1 + C) = S_{or,\max}^{*}$ .

Worked exampleEnd points of two cells and the transition zone

Given (assumed data)

Table of type 2: built for the cell of 36 mD: $S_{wi}$ = 0.2332, $S_{or}$ = 0.2505 (the table is in the last section)
A tighter cell: 15 mD, $\phi$ = 0.17: a little below the range of the samples, so this is an extrapolation of the trends
A cell of 36 mD in the transition zone: initial water saturation 0.45 (about 3.8 m above the FWL)

Solution

End points of the 15 mD cell: $S_{wi} = 0.4061 - 0.1111\lg 15 = 0.2754$ and $S_{or} = 0.3352 - 0.0544\lg 15 = 0.2712$ .
Its saturation of 0.45 in the table: $S_n = (0.45 - 0.2754)/(1 - 0.2754 - 0.2712) = 0.3851$ , and $S_t = 0.2332 + 0.3851\cdot(1 - 0.2332 - 0.2505) = 0.4320$ .
$k_{rw}$ from the table is between the rows 0.40 (0.0077) and 0.45 (0.0169): $0.0077 + (0.4320 - 0.40)/0.05\cdot(0.0169 - 0.0077) = 0.0136$ . By the Corey model with the end points of the cell it is 0.0130.
The capillary pressure of the cell at 0.45 is the same Leverett function with its $k$ and $\phi$ . The scale is $25.98/(31.4\sqrt{15/0.17}) = 0.08808$ MPa, against 0.06167 MPa for the 36 mD cell. $S_w^{*} = (0.45 - 0.2754)/(1 - 0.2754) = 0.2410$ , and $p_c = 0.08808\cdot0.0614\cdot0.2410^{-0.680} = 0.01423$ MPa, or 14.2 kPa. So a water saturation of 0.45 for this cell is at a height $0.01423/0.002354 = 6.0$ m above the FWL, and not at 3.8 m.
Trapping in the 36 mD cell: $C = (1 - 0.2332)/0.2505 - 1 = 2.0611$ ; $S_{oi}^{*} = 0.55/(1 - 0.2332) = 0.7173$ ; $S_{or}^{*} = 0.7173/(1 + 2.0611\cdot0.7173) = 0.2894$ ; and $S_{or} = 0.2894\cdot(1 - 0.2332) = 0.2219$ .

Answer: The 15 mD cell takes the values from the table at the saturation 0.4320 instead of 0.45. In the transition zone of the 36 mD cell, 0.2219 of oil will be left after waterflooding, and not 0.2505.

What it means. Two corrections to one table change the movable oil. In the transition zone the displaced oil is $0.55 - 0.2219 = 0.3281$ of the pore volume instead of $0.55 - 0.2505 = 0.2995$ , which is 10% more. Interpolation in the table gave a $k_{rw}$ higher than the exact one by 5%: rows 0.05 apart are a little coarse at small saturations. Where the curve is steeper, we put the rows closer. The data in this example are assumed for illustration.

Hysteresis in the model. If the saturation in a cell changes direction (water first rises and then retreats, or gas is injected and produced in turn), the cell follows intermediate, scanning curves (curves that run between the drainage curve and the imbibition curve). Simulators build them from the drainage and imbibition curves and the Land relation. The common models for the non-wetting phase carry the names of Killough and Carlson. For waterflooding of a water-wet reservoir with no changes of direction, hysteresis is usually not needed, and imbibition curves are enough. For gas it often decides the forecast.

CalculatedRelative permeability curves of the 36 mD cell (the table of type 2), and the same curves moved by scaling onto the 15 mD cell (dashed): the shape is common, and the end points are shifted.

Three phases: Stone's models

When water, oil and gas are all mobile in a cell (in the zone of gas coming out of solution under waterflooding, in gas injection after water, in alternating injection of water and gas), we need the oil relative permeability as a function of two saturations. It is almost impossible to measure. So the simulator calculates it from two two-phase tables: water–oil (with no gas) and gas–oil (at irreducible water). For a water-wet rock we assume that the water relative permeability depends only on $S_w$ , and the gas one only on $S_g$ . For oil we use interpolation models. The most common are Stone's models, the first and the second (three mobile phases).

Stone's second model for the oil relative permeability (normalised form)

k_{ro} = k_{rocw}\left[\left(\frac{k_{row}}{k_{rocw}} + k_{rw}\right)\left(\frac{k_{rog}}{k_{rocw}} + k_{rg}\right) - (k_{rw} + k_{rg})\right]

where:

$k_{row}(S_w),\ k_{rw}(S_w)$ oil and water relative permeabilities from the water–oil table, at the water saturation of the cell, —
$k_{rog}(S_g),\ k_{rg}(S_g)$ oil and gas relative permeabilities from the gas–oil table (at irreducible water), at the gas saturation of the cell, —
$k_{rocw}$ oil relative permeability at irreducible water and no gas: the common end point of both tables, —

Units: fractions

Holds when

A water-wet rock: water is in the small pores, gas in the large ones, and oil between them.
A negative result is set to zero. This is how the model sets the residual oil of three-phase displacement.
$k_{rocw}$ must be the same in both two-phase tables. Otherwise, at $S_g \to 0$ or $S_w \to S_{wi}$ the model does not return to the two-phase curves.

If there is no gas ( $k_{rog} = k_{rocw}$ , $k_{rg} = 0$ ), the formula gives $k_{row}$ . If the water is irreducible ( $k_{row} = k_{rocw}$ , $k_{rw} = 0$ ), it gives $k_{rog}$ . Stone's first model instead multiplies normalised factors, and we must set in it the smallest residual oil of three-phase displacement. The choice of the model noticeably changes the forecast of water-alternating-gas injection, so we check it against experiments or the production history.

The table for the simulator and its checks

The simulator gets the curves as tables. For the water–oil pair there are rows of water saturation with $k_{rw}$ , $k_{ro}$ and $p_c$ . For the gas–oil pair there are rows of gas saturation. Between rows the values are interpolated linearly, so we put the rows closer where the curves are steeper: near the irreducible water for the capillary pressure and the water relative permeability, and near the residual oil for the oil relative permeability. The end points of the table have a meaning of their own:

The first row, $S_{wc}$: The smallest water saturation in the table: the connate water to which the model brings the cells high above the contact. The capillary pressure in this row is set no lower than at the top of the accumulation. Then the whole height of the accumulation lies inside the table, and the saturation by height is set by its rows and not by cutting off at $S_{wc}$ .
The critical water saturation $S_{wcr}$: The last row where $k_{rw} = 0$ . Above it water is mobile. If $S_{wcr}$ is less than the initial saturation of the cells above the transition zone, water in them will be mobile from the first day.
$1 - S_{or}$: The row from which $k_{ro} = 0$ and $k_{rw} = k_{rw}'$ .
The row $S_w = 1$: This row is for the cells of the water zone, where there is no oil: by definition $k_{rw} = 1$ as a fraction of the absolute permeability, and $p_c$ is the entry pressure. Without it the water influx will go with $k_{rw}' < 1$ and will be underestimated.

Worked exampleA table for a cell of type 2

Given (assumed data)

Cell: type 2, $k$ = 36 mD, $\phi$ = 0.20
Rock type: $n_w$ = 3.01, $n_o$ = 2.50; $k_{ro}'$ = 0.807, $k_{rw}'$ = 0.23; the end-point trends and $J = 0.0614\,(S_w^{*})^{-0.680}$ are from the examples above
Accumulation: the top is 50 m above the FWL; the density difference of water and oil is 0.24 g/cm³

Solution

End points: $S_{wi} = 0.4061 - 0.1111\cdot1.5563 = 0.2332$ and $S_{or} = 0.3352 - 0.0544\cdot1.5563 = 0.2505$ . The range of mobile saturation is $1 - 0.2332 - 0.2505 = 0.5163$ .
The row $S_w$ = 0.45: $S_n = (0.45 - 0.2332)/0.5163 = 0.4199$ ; $k_{rw} = 0.23\cdot0.4199^{3.01} = 0.0169$ ; $k_{ro} = 0.807\cdot(1 - 0.4199)^{2.50} = 0.2068$ .
Its capillary pressure: $S_w^{*} = (0.45 - 0.2332)/(1 - 0.2332) = 0.2827$ ; $p_c = 0.06167\cdot0.0614\cdot0.2827^{-0.680} = 0.00894$ MPa, or 8.94 kPa.
The first row is $S_{wi}$ with the pressure at the top: $0.00981\cdot0.24\cdot50 = 0.1177$ MPa, or 117.7 kPa. The last ones are $1 - S_{or}$ = 0.7495 ( $k_{ro}$ = 0, $k_{rw}$ = 0.23) and $S_w$ = 1 ( $k_{rw}$ = 1, and $p_c$ is the entry pressure, 3.79 kPa).
Checks: the saturations grow, $k_{rw}$ does not fall, $k_{ro}$ and $p_c$ do not rise. 4 of 4 are met. At a height of 10 m the table gives, by interpolation between the rows 0.27 and 0.30, a water saturation of 0.289 against 0.285 from the Leverett function.

Answer: The table of 15 rows (below) is monotonic and consistent with the Leverett function: at a height of 10 m the initial water saturation from it differs from the exact one by 0.004.

What it means. The rows 0.24, 0.25 and 0.27 were added for the capillary pressure. Between 0.2332 and 0.30 it falls almost six times, and without these rows the upper part of the transition zone would get noticeably more water at initialisation: at a height of 20 m, 0.281 instead of 0.252. The residual oil 0.2505 and the end-point relative permeabilities in the table are the same ones that go into the scaling of other cells. Their consistency with the initial saturation is checked by a simple calculation with no wells: in a correctly initialised model, nothing should move in a year. The data in this example are assumed for illustration.

Water–oil table for a cell of type 2 (36 mD)

$S_w$: 0.2332
$k_{rw}$: 0
$k_{ro}$: 0.8070
$p_c$ , kPa: 117.7

$S_w$: 0.2400
$k_{rw}$: 0.000001
$k_{ro}$: 0.7807
$p_c$ , kPa: 94.1

$S_w$: 0.2500
$k_{rw}$: 0.000008
$k_{ro}$: 0.7429
$p_c$ , kPa: 50.9

$S_w$: 0.2700
$k_{rw}$: 0.000081
$k_{ro}$: 0.6708
$p_c$ , kPa: 29.9

$S_w$: 0.3000
$k_{rw}$: 0.000488
$k_{ro}$: 0.5707
$p_c$ , kPa: 19.9

$S_w$: 0.3500
$k_{rw}$: 0.0026
$k_{ro}$: 0.4250
$p_c$ , kPa: 13.6

$S_w$: 0.4000
$k_{rw}$: 0.0077
$k_{ro}$: 0.3043
$p_c$ , kPa: 10.7

$S_w$: 0.4500
$k_{rw}$: 0.0169
$k_{ro}$: 0.2068
$p_c$ , kPa: 8.9

$S_w$: 0.5000
$k_{rw}$: 0.0315
$k_{ro}$: 0.1310
$p_c$ , kPa: 7.8

$S_w$: 0.5500
$k_{rw}$: 0.0529
$k_{ro}$: 0.0749
$p_c$ , kPa: 6.9

$S_w$: 0.6000
$k_{rw}$: 0.0822
$k_{ro}$: 0.0364
$p_c$ , kPa: 6.3

$S_w$: 0.6500
$k_{rw}$: 0.1207
$k_{ro}$: 0.0132
$p_c$ , kPa: 5.7

$S_w$: 0.7000
$k_{rw}$: 0.1698
$k_{ro}$: 0.0023
$p_c$ , kPa: 5.3

$S_w$: 0.7495
$k_{rw}$: 0.2300
$k_{ro}$: 0
$p_c$ , kPa: 5.0

$S_w$: 1.0000
$k_{rw}$: 1.0000
$k_{ro}$: 0
$p_c$ , kPa: 3.8

$S_w$	$k_{rw}$	$k_{ro}$	$p_c$ , kPa
0.2332	0	0.8070	117.7
0.2400	0.000001	0.7807	94.1
0.2500	0.000008	0.7429	50.9
0.2700	0.000081	0.6708	29.9
0.3000	0.000488	0.5707	19.9
0.3500	0.0026	0.4250	13.6
0.4000	0.0077	0.3043	10.7
0.4500	0.0169	0.2068	8.9
0.5000	0.0315	0.1310	7.8
0.5500	0.0529	0.0749	6.9
0.6000	0.0822	0.0364	6.3
0.6500	0.1207	0.0132	5.7
0.7000	0.1698	0.0023	5.3
0.7495	0.2300	0	5.0
1.0000	1.0000	0	3.8

Imbibition relative permeabilities as fractions of the absolute permeability. The drainage capillary pressure is for the initial distribution. The first row is the pressure at the top of the accumulation (50 m above the FWL), and the last row is the water zone. The data are assumed for illustration.

Checks before the calculation

1
Monotonicity
Saturation grows along the rows, $k_{rw}$ does not fall, and $k_{ro}$ and $p_c$ do not rise. There are no repeated saturations.
2
End points
$k_{rw} = 0$ up to $S_{wcr}$ , and $k_{ro} = 0$ from $1 - S_{or}$ . $k_{ro}$ in the first row equals $k_{rocw}$ of the gas–oil table. There is a row $S_w = 1$ for the water zone.
3
Scaling
The end points of the cells lie within physical limits ( $S_{wi} + S_{or} < 1$ ), and they agree with the initial saturation and with each other on the maps.
4
Initial equilibrium
Compare the saturation profile by height with the well logs. A model with no wells must stay at rest: flows after initialisation mean that the tables, the contacts or the scaling are inconsistent.
5
Sensitivity
The exponents and the end points are the main “knobs” of uncertainty (Corey model). We set their variants from the scatter of the samples, and we do not choose them arbitrarily during history matching (History matching).

Terms and symbols

SCAL: Special core analysis: capillary pressure, relative permeability, wettability and other experiments on a few selected samples.
Capillary end effect: The accumulation of the wetting phase at the outlet face of a sample, where the capillary pressure is zero. It distorts the end points and the relative permeabilities.
JBN method: The processing of an unsteady-state experiment: relative permeabilities at the outlet face from the derivatives of production and of relative injectivity.
Normalising the curves: The conversion of the relative permeabilities to the coordinates $S_n$ and $k_r/k_r'$ , where the end points are the same for all samples.
End-point scaling: The transfer of the table of a rock type to a cell with other $S_{wi}$ , $S_{or}$ and $k_r'$ , with the same shape of the curves.
Land coefficient $C$: A parameter of the relation between the trapped and the initial saturation of the non-wetting phase.
$S_{wc}$ , $S_{wcr}$: The connate water saturation (the first row of the table) and the critical one, the largest at which water is still immobile.
Stone's models: Schemes for calculating the three-phase oil relative permeability from the two-phase tables water–oil and gas–oil.

Common mistakes and misconceptions

Wrong

“The average curve of a rock type is the average of the relative permeabilities of the samples at the same water saturation.”

In fact

This mixes curves with different end points: the average starts at the smallest $S_{wi}$ , ends at the largest $1 - S_{or}$ , and is flatter than any of the originals. We average in normalised coordinates, and we describe the end points separately.

Wrong

“The irreducible water of a sample is the irreducible water of the reservoir.”

In fact

It is the saturation at the largest pressure of the experiment. And if it was found by flooding with oil, it is also overestimated by the end effect. In the example of the lesson, the last step of the membrane, 128 kPa, corresponds to a height of about 54 m above the free water level. Above it the saturation is an extrapolation, and we check it against well logs.

Wrong

“An unsteady-state experiment gives the whole relative permeability curve.”

In fact

Only after breakthrough, from the saturation at the front to $1 - S_{or}$ . In the example of the lesson these are saturations above 0.604. The whole lower part is taken from a steady-state experiment, or from a simulation of the experiment on a grid.

Wrong

“If the simulator accepted the table, it is correct.”

In fact

The program checks the format, but not the physics: a non-monotonic row, $k_{rw}' < 1$ at $S_w = 1$ , inconsistent end points of the two tables, or $p_c$ of the first row below the pressure at the top. We check the table against a list, and by a calculation with no wells.

In depth: advanced detail

Where the JBN formulas come from

Behind the front each saturation moves at a speed proportional to $f_w'(S)$ (frontal advance equation). So in the dimensionless coordinate $x_D = x/L$ , after $W_i$ pore volumes have been injected, the saturation $S$ is at the point $x_D = W_if_w'(S)$ . At the outlet $f_w'(S_{w2}) = 1/W_i$ . At a constant rate, the pressure drop is the sum of the resistances along the length: $\Delta p = \dfrac{q\mu_oL}{kA}\int_0^1\dfrac{\dd x_D}{\lambda_r}$ , where $\lambda_r = k_{ro} + k_{rw}\mu_o/\mu_w$ is the total mobility in units of oil.

At the start of the experiment $\lambda_r = k_{ro}'$ , so $1/I_r = \Delta p/\Delta p_0 = k_{ro}'\int_0^1\dd x_D/\lambda_r$ . We change the variable: $u = x_D/W_i = f_w'(S)$ goes from 0 at the inlet to $1/W_i$ at the outlet, and $\dfrac{1}{W_iI_r} = k_{ro}'\int_0^{1/W_i}\dfrac{\dd u}{\lambda_r(u)}$ . The derivative with respect to the upper limit gives $\dd\bigl(1/(W_iI_r)\bigr)/\dd(1/W_i) = k_{ro}'/\lambda_r(S_{w2})$ . Finally the oil fraction at the outlet is $f_{o2} = k_{ro}/\lambda_r$ , from which $k_{ro}(S_{w2})/k_{ro}' = f_{o2}\big/\bigl[\dd(1/(W_iI_r))/\dd(1/W_i)\bigr]$ , the formula of the main section. The oil fraction and the saturation at the outlet come from Welge's formulas.

Capillary pressure from a centrifuge

In a sample that rotates with an angular velocity $\omega$ , the centrifugal force creates a pressure difference between the phases that grows towards the axis of rotation. At the outer face (radius $r_2$ ) the capillary pressure is taken as zero, since beyond it the phases separate freely. At the inner face (radius $r_1$ ) it is the largest:

Capillary pressure at the inner face and the saturation there (the Hassler–Brunner method)

p_{cL} = \tfrac12\,\Delta\rho\,\omega^2\left(r_2^2 - r_1^2\right),\qquad S_L = \frac{\dd\,(p_{cL}\,\bar S)}{\dd p_{cL}} = \bar S + p_{cL}\,\frac{\dd\bar S}{\dd p_{cL}}

where:

$p_{cL}$ capillary pressure at the inner face, Pa
$\Delta\rho$ density difference of the phases, kg/m³
$\omega$ angular velocity, $2\pi n/60$ at $n$ revolutions per minute, rad/s
$r_1,\ r_2$ distances from the axis to the inner and outer faces of the sample, m
$\bar S,\ S_L$ average saturation of the sample (from the volume produced) and the saturation at the inner face, —

Units: SI

Holds when

Equilibrium is reached at each rotation speed. At the outer face $p_c = 0$ .
The field of centrifugal force along the sample is taken as uniform. This is acceptable for short samples ( $r_1/r_2$ not less than about 0.7). Otherwise more accurate solutions are used.
The derivative is taken from the smoothed dependence $\bar S(p_{cL})$ built from the speed steps.

Each rotation speed gives a point $(p_{cL}, S_L)$ of the capillary pressure curve. For example, for a sample 2.5 cm long at 3000 rpm, with $r_2$ = 8.5 cm, $r_1$ = 6.0 cm ( $r_1/r_2 \approx 0.71$ ) and $\Delta\rho$ = 250 kg/m³: $\omega = 314.2$ rad/s and $p_{cL} = 0.5\cdot250\cdot314.2^2\cdot(0.085^2 - 0.060^2) \approx 45$ kPa.

Connections to other topics

Relative permeability — relative permeability curves, end points, the Corey model and the methods of measurement.
Capillary pressure and the transition zone — laboratory methods, conversion to the reservoir, the Leverett function and the transition zone.
Wettability and interfacial phenomena — restoring the wettability of core, and the Amott and USBM indexes.
Core analysis — sampling, preservation and preparation of core, routine and special analysis.
Buckley–Leverett theory and the Welge method — Welge's formulas, on which the processing of an unsteady-state experiment rests.
Reservoir quality: geological controls on porosity and permeability — rock types, to which the curves are assigned.
Fluid saturation — saturation by height and its comparison with well logs.
Two-phase upscaling: pseudo-functions — pseudo functions for coarse cells.
History matching — why relative permeabilities are not chosen arbitrarily during history matching.

Check yourself

1
What curves does the model of a water-wet reservoir need, if it is first initialised and then waterflooded?
1. A.Drainage curves for everything
2. B.Drainage capillary pressure for the initial saturation, and imbibition relative permeabilities for the waterflood
3. C.Imbibition curves for everything
4. D.Any curves, if they were measured on core of this reservoir
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Correct answer: B. Drainage capillary pressure for the initial saturation, and imbibition relative permeabilities for the waterflood
The initial distribution was formed when oil displaced water: this is drainage. The waterflood of a water-wet reservoir is imbibition: water displaces oil, and the oil relative permeability is then lower than in drainage.
2
A sample has $S_{wi}$ = 0.25, $S_{or}$ = 0.30 and $k_{rw}'$ = 0.30. At $S_w$ = 0.50 a $k_{rw}$ of 0.045 was measured. Estimate the Corey exponent $n_w$ from this point.
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$S_n = (0.50 - 0.25)/(1 - 0.25 - 0.30) = 0.25/0.45 \approx 0.556$ ; $k_{rw}^{*} = 0.045/0.30 = 0.15$ ; $n_w = \ln0.15/\ln0.556 = -1.897/(-0.587) \approx 3.2$ . From one point the estimate is rough. For a sample we use all the points, and for a rock type several samples.
3
A table is built for $S_{wi}$ = 0.20 and $S_{or}$ = 0.25. A cell with $S_{wi}$ = 0.30 and $S_{or}$ = 0.25 has a water saturation of 0.50. At what saturation will the simulator take the values from the table with two-point scaling?
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$S_n = (0.50 - 0.30)/(1 - 0.30 - 0.25) = 0.20/0.45 \approx 0.444$ ; $S_t = 0.20 + 0.444\cdot(1 - 0.20 - 0.25) = 0.20 + 0.444\cdot0.55 \approx 0.444$ . Without scaling the cell would get the values at 0.50, with a noticeably larger water relative permeability.
4
At $S_{wi}$ = 0.20 the largest residual oil is 0.30. How much oil will be trapped, by the Land relation, in a transition-zone cell with an initial oil saturation of 0.40?
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$C = (1 - 0.20)/0.30 - 1 \approx 1.667$ ; $S_{oi}^{*} = 0.40/0.80 = 0.5$ ; $S_{or}^{*} = 0.5/(1 + 1.667\cdot0.5) = 0.5/1.833 \approx 0.273$ ; $S_{or} = 0.273\cdot0.80 \approx 0.22$ . The oil that can be displaced is $0.40 - 0.22 = 0.18$ , while with a constant $S_{or}$ = 0.30 it would be only 0.10.
5
An imbibition experiment on a sample of mixed wettability was repeated at half the rate, and the residual oil saturation “grew” from 0.27 to 0.31, and water breakthrough came later. What is this most likely?
1. A.The end effect: at a lower rate its zone at the outlet is longer
2. B.The wettability has changed
3. C.An error in the viscosity measurement
4. D.Normal scatter, and the curves can be averaged
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Correct answer: A. The end effect: at a lower rate its zone at the outlet is longer
A dependence on the rate is the main sign of a capillary component. The zone at the outlet, where the capillary pressure tends to zero, grows as the rate falls. Breakthrough is delayed, and in a sample of mixed wettability more oil is left there. So the residual oil saturation, referred to the sample as a whole, is overestimated. Such experiments are processed with capillary pressure taken into account, or they are repeated on longer samples.
6
By Stone's second model, find the oil relative permeability if $k_{rocw}$ = 0.80, $k_{row}$ = 0.30, $k_{rw}$ = 0.05, $k_{rog}$ = 0.40 and $k_{rg}$ = 0.10.
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$k_{ro} = 0.80\,[(0.30/0.80 + 0.05)(0.40/0.80 + 0.10) - (0.05 + 0.10)] = 0.80\,[0.425\cdot0.60 - 0.15] = 0.80\cdot0.105 = 0.084$ . This is less than both two-phase oil relative permeabilities (0.30 and 0.40): the presence of both water and gas takes paths away from the oil.

Updated: 16 October 2026