multi phase flow

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Shiraz University

Multiphase Flow in Porous MediaShahab Ayatollahi

([email protected])



Table of Contents

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TABLE OF CONTENTS 2

1. MICROSCOPIC ORIGIN OF PETROPHYSICAL PROPERTIES 4

1.1 Representative Elemental Volume (REV) 4

1.2 Permeability (Carman-Kozeny Equation) 5

1.3 Pores, Throats and Crevices 7

2. INTERFACIAL TENSION AND CAPILLARITY 9

2.1 Interfacial Tension 9

2.2 Mechanical Equilibrium of a Surface – Laplace Equation 9

2.3 Equilibrium at a Line of Contact – Young’s Equation 11

2.4 Capillary Pressure 13 2.4.1 Fundamentals 13

2.4.2 Rise in a Capillary Tube 18

2.4.3 Capillary Transition Zone 192.4.4 Measuring Capillary Pressure 20

2.4.5 Capillary Pressure Models 222.4.6 Capillary Number and Bond Number 242.4.7 Leverett J-Function 25

2.5 Residual Saturations 27 2.5.1 Residual Nonwetting Phase Saturation 28

2.5.2 Trapping Mechanisms 32

2.5.3 Immobile Wetting Phase Saturation 36

2.5.4 CDC Curves 36

2.6 Wettability 37



3. MULTIPHASE FLOW IN POROUS MEDIA 42

3.1 Relative Permeability 42

3.1.1 Two-Phase Flow 423.1.2 Drainage Relative Permeability – Burdine’s Theory 44

3.1.3 Imbibition Relative Permeability 473.1.4 Relative Permeability Correlations 48

3.1.5 Capillary End Effect 49

3.1.6 Measurement of Relative Permeability 52

3.1.7 Three-Phase Flow 553.1.8 Three-Phase Relative Permeability 55

4. MULTIPHASE DISPLACEMENT 56

4.1 Immiscible Displacement Equations of Motion 56 4.1.1 Conservation of Mass and Momentum 564.1.2 Fractional Flow 58

4.2 Buckley-Leverett Theory – One-Dimensional Flow 60 4.2.1 Equations of Motion 60

4.2.2 Capillary Pressure Gradient Terms 62

4.2.3 Fractional Flow with Gravity Only 644.2.4 The Buckley-Leverett Solution 65

4.2.5 Method of Characteristics 67

4.2.6 Shock Velocities 70

4.2.7 Welge Construction 714.2.8 Classification of Waves 73

4.2.9 Average Saturations 754.2.10 Oil Recovery Calculations 76

4.2.11 Effect of Mobility on Recovery 79

4.2.12 Effect of Gravity on Recovery 82

4.3 Derivation of Relative Permeability from Displacements 85 4.3.1 JBN Method 854.3.2 Jones and Roszelle Method 88

4.4 Gravity Segregation 88 4.4.1 Vertical Segregation 88

4.4.2 Displacement Under Segregated Flow Conditions 97



1. Microscopic Origin of Petrophysical Properties

When we study the flow of fluids in channels and open spaces, we make a continuumassumption. However porous materials have complex internal structure – in principle we

could compute the Navier-Stokes flow behavior within the pore spaces but in practice the

mathematics would be intractable. Instead we attempt to describe the properties of themedium in terms of some kind of equivalent continuum.

Figure 1.1: Conceptualization of flow in a porous medium

1.1 Representative Elemental Volume (REV)

The usual approach is to associate point properties with an equivalent continuum whichhas properties and behavior based on the local properties averaged over a “representative

elemental volume” (REV). The REV is taken to be large in comparison with a pore, yet

small in comparison with regional variations in properties of the medium. See Figure1.2and Figure 1.3

0

10

20

30

40

φ

Volume

Figure 1.2: Computed porosities as a function of volume considered.



0

10

20

30

40

φ

Volume

Figure 1.3: Computed porosities as a function of volume considered.

1.2 Permeability (Carman-Kozeny Equation)

[Dullien, p. 254-259] [Lake, p. 44-48]

Darcy’s Law: L

p A

k q

∆µ

=(1.1)

Where ∆ p is the pressure drop, q is the volumetric flow rate (L3/t). The dependence of

the macroscopic property permeability, k , can be determined in terms of the microscopic

properties of the pore space. One way to do this is to treat the pore space as a bundle of

capillary tubes (Figure 1.4).

Actually, tubes are not straight

Figure 1.4: Conceptual model of porous medium -- capillary tube bundle.

The Hagen-Poiseuille equation can be derived for flow of a Newtonian fluid in a single

tube or radius R and length Lt :

t

t L

p Rq

∆µ

π=

8

4

(1.2)



In steady laminar flow, the velocity distribution across the tube is parabolic. The average

velocity is given by:

t

t

L

p R

R

qv

∆µ

=π

=8

2

2

(1.3)

Since we know that in a true porous medium the tubes are not straight, we need to define

a representative volume (REV) over which we can define the representative length andflow velocity. The time taken for the fluid to pass through the tortuous path will be the

same as the time to pass through the REV:

φ=

=

=

A

qvwhere

v

L

v

Lt

REV tube

t ;(1.4)

Notice the distinction between the flux velocity, u = q/A, also known as the Darcy

velocity or the superficial velocity, and the interstitial velocity, v = u/ φ, which is thespeed at which the fluid actually moves in the pore space. The interstitial velocity is the

speed at which a tracer front would move in the medium.

Combining with Darcy’s Law:

pk

L

v

L

p R

L

v

L

REV

t

tube

t

∆φµ

=

=

∆µ

=

2

2

28

(1.5)

22

2

22

,88

=τ

τφ

=φ

= L

Lwhere

R

L

L Rk t

t (1.6)

The tortuousity, τ, is a variable that defines the “straightness” of the flow paths. Astraight tube has a tortuousity of 1, whereas common porous materials have tortuousity

values between 2 and 5. Tortuousity can be determined experimentally from resistivity

measurements.

In an actual porous medium, the radius of the “tubes” is not going to be uniform. We can

define the hydraulic radius for a noncircular tube as:

perimeter Wetted

flowtoopenareasectionalCross2

−×=

h R(1.7)

This definition differs from that in Lake, who leaves out the 2. With the definition of Eq.

(1.7), the hydraulic radius of a cylindrical tube would be R, the radius of the tube – which

makes more sense.

Since a porous medium is not really made up of actual tubes, we can define the hydraulic

radius instead as:



( )φ−φ

=×=1

2

areasurfaceWetted

flowtoopenVolume2

v

ha

R(1.8)

where av is the specific surface area, the surface area of the pore space per unit volume of

solid. Specific surface area is an intrinsic property of the porous medium.

Substituting in Eq. (1.6):

( ) 22

3

12 vak

φ−τ

φ=

(1.9)

For a porous medium made up of uniform spheres of radius R (diameter D p):

p

v D R R

Ra

6343

34

2

==ππ

=(1.10)

Hence: ( )2

23

172

1

φ−

φ

τ= p D

k

(1.11)

This is what is know as the Carman-Kozeny equation and defines the Carman-Kozeny

permeability. The Carman-Kozeny equation provides good estimates of the permeability

of packs of uniform spheres, for real porous media it reveals the dependence on pore size

( D p), tortuosity (τ) and of packing (through porosity φ).

1.3 Pores, Throats and Crevices

[Dullien, Section 1.2]

When we talk about flow through porous media (as we will throughout this course) we

often make use of conceptual models of the pore structure in terms of pores and throats,as in Figure 1.5.

Pore

Throat

Pore

Throat

Figure 1.5: Pores and throats in a conceptual two-dimensional model.

Such pore-and-throat models are useful to represent many phenomena observed in real

porous flow, even though the actual structure of the connections may be three-



dimensional. However there are other important instances where the three-dimensional

structure of the throats and pores themselves may significantly govern the flow behavior.For example, if the cross-section of a throat is square or triangular, there are crevices in

the corners that may trap fluids, Figure 1.6.

CreviceCrevice

Figure 1.6: Crevices in the corners of throats or pores.



2. Interfacial Tension and Capillarity

[Dullien, p. 118-139] [Lake, p. 48-58]

2.1 Interfacial Tension

When fluids are immiscible it means that when you bring them in contact they do not mix

but remain separated by an interface. The molecules do not mix because they have a

greater affinity for molecules of their own kind. Molecules near the surface are morestrongly attracted; this leads to an apparent mechanical force at the surface called

interfacial tension. The force arises from the need to expend work to create new surface

area at the contact.

Interfacial tension σ has units of force/length, and can be visualized in terms of the force

that acts along any line in the plane of the surface, as in Figure 2.1.

dl σdldldl σdlσdlσdldl

Figure 2.1: Interfacial tension force at an interface between two fluids.

Consider an experiment where we take a loop of cotton placed in the surface of a soap

bubble. If we pop the soap film inside the loop of cotton, what will happen?

Soap film

Thread

Pop

Figure 2.2: What happens when the film within the thread is popped?

2.2 Mechanical Equilibrium of a Surface – Laplace Equation

Consider the equilibrium of the spherical cap of an interface between two fluids, as

shown in Figure 2.3.



σσ

p2

p1

σσ σσ

p2

p1

θ

σR r

θ

σR r

Figure 2.3: Spherical cap of an interface between two fluids.

If we take a force balance on the cap, balancing the differences in pressure on the inside

(fluid 1) and outside (fluid 2) with the force along the edge of the cap due to interfacialtension, then:

( ) r r p p π×θσ=π− 2sin221 (2.1)

( )r

p pθσ

=−sin2

21(2.2)

( ) R

p pσ

=−2

21for a spherical cap since θ= sin Rr (2.3)

More generally, we may consider a nonspherical cap with two differing radii of curvature

R1 and R2 as in Figure 2.4.

Figure 2.4: Nonspherical cap (Figure 2.3 from Dullien).

The force balance then gives us:

( )

+σ=−

21

21

11

R R p p

(2.4)

It can be shown that for orthogonal lines such as AB and CD:



2121

1111

r r R R+=+

(2.5)

where r 1 and r 2 are the principal radii of curvature. Hence the quantity 1/R1 + 1/R2 is an

invariant for any pair of orthogonal lines in the surface.

If we define the mean radius of curvature r m as:

+=

21

21

111

R Rr m (2.6)

then

( )mr

p pσ

=−2

21

(2.7)

This is a generalization of the equation for the spherical surface and is known as the

Laplace equation. Note that p1 – p2 is the capillary pressure which will be discussedfurther in Section 2.4.

2.3 Equilibrium at a Line of Contact – Young’s Equation

In porous media we are often looking at two immiscible fluids (e.g. water and oil) incontact with a solid phase (rock grains). Consider a spherical interface between two

fluids in a cylindrical tube, as in Figure 2.5.

r

R

θ σ

p2

p1

r

R

θ σ

p2

p1

Figure 2.5: Spherical interface between two fluids in a cylindrical tube.

We can write the force balance much as we did previously for the spherical cap:

( )r

p pθσ

=−cos2

21(2.8)

Here the interfacial tension σ is a function of the two fluids, and the contact angle θ is a

function of the solid and the two fluids. Note that if (p1 – p2 ) is not equal to 2 σ cos θ / r then the interface will move in the tube.



The contact angle can be understood by considering a drop of liquid on a solid surface, as

in Figure 2.6.

Figure 2.6: Surface forces at a line of contact (Figure 2.5 from Dullien).

For this system to be in static equilibrium:

ls sg σ−σ=θσ coslg (2.9)

This is known as Young’s Equation. σlg, σsg, and σls are properties of the fluids and the

solid material. We cannot measure σls or σsg, but experiments confirm that such surfaceforces do exist. For rough surfaces the actual area of contact may differ from the

apparent area (see Figure 2.6 in Dullien), so a general form of Young’s equation can bewritten:

)ls sg aa r σ−σ=θσ coslg (2.10)

where θa is the apparent contact angle and r a is the ratio of the actual surface area to theapparent surface area.

By convention, contact angle is always measured though the denser phase (usually water for our problems).

Young’s equation can also be written (using o for oil and w for water, this being the most

common case of interest):

ow

sw so

σσ−σ=θcos

(2.11)

For a water-wet surface (a polar surface), σso > σsw and therefore θ < 90º. For an oil-wet

(a nonpolar surface) σso < σsw and therefore θ < 90º.

Note that it is physically possible for σso - σsw > σow in which case cos θ > 1 and the

water would spontaneously spread across the surface.



Most clean rock surfaces are polar and therefore water-wet, but many reservoir rocks are

not completely water-wet. This is due to the presence of surfactants in the oil that

contain polar molecules that alter the surface attraction forces, as in Figure 2.7.

water

oil

water oil

water-wet oil-wet

Figure 2.7: Surfactants present in the oil can change the surface to be oil-wet.

Contact angle can also be a function of the direction of movement of the interface, with

advancing and receding contact angles being different. This is known as contact anglehysteresis (Figure 2.8, see also Figures 2.6 and 2.9 in Dullien).

Figure 2.8: Contact angle hysteresis between advancing and receding contacts. Figure 3-

8 from Lake).

2.4 Capillary Pressure

2.4.1 Fundamentals

As seen earlier, a pressure difference occurs at an interface between two fluids. Consider a tapered capillary tube filled with wetting fluid that is in contact with a nonwetting fluid,

as in Figure 2.9.



Figure 2.9: Fluid interface in a tapered capillary tube (Figure 2.11 from Dullien).

In order for the nonwetting fluid to enter the tube, the pressure has to be raised above that

of the wetting fluid. The pressure difference between the nonwetting and the wettingfluids is the capillary pressure. The pressure is higher in the nonwetting fluid, on theconcave side of the interface. By definition, the capillary pressure is the nonwetting fluid

pressure minus the wetting fluid pressure, hence the capillary pressure is (almost always)

a positive number.

( )ϕ+θσ

=−= cos2

R p p P wnwc

(2.12)

For a diverging capillary tube, as in Figure 2.10, the capillary pressure is given by:

( )ϕ−θσ=−= cos2

R p p P wnwc

(2.13)

ϕ

θ

ϕ

θ

Figure 2.10: Fluid interface in a diverging capillary tube.

You will note that the interface in the diverging tube is unstable, since as R increases P c

decreases and the interface will therefore move towards the larger radius end of the tube.

If we imagine a capillary tube with tapered constrictions, the interface will move steadily

into the converging part of the tube as we increase the nonwetting phase pressure, will

hang up at the throat until the pressure reaches 2 σ/ R cos (θ – ϕ), since 2 σ/ R cos (θ – ϕ)

is greater than 2 σ/ R cos (θ + ϕ), then will jump suddenly through the diverging part of



the tube (Figure 2.11). This is known as a Haines jump (see Dullien, P. 163-166 for more

description).

jump jump

Figure 2.11: Fluid interface in a converging/diverging capillary tube.

In a smooth throat, as in Figure 2.12, the interface will not hang up, but the Haines jump

will still occur. Note that the capillary pressure variation is not symmetric, and has larger values in the divergent part of the tube.

P c

Figure 2.12: Fluid interface in a smooth converging/diverging capillary tube.

Notice that it requires a certain threshold pressure or entry pressure for the nonwetting

fluid to enter a porous material. This entry pressure corresponds to the capillary pressureof the largest pore throat across which a nonwetting/wetting phase interface occurs.

Figure 2.13: Nonwetting phase enters the largest pore first, overcoming the entry

pressure to do so.

As more nonwetting fluid enters the porous material, the wetting phase is displaced from

successively smaller pores. Hence the capillary pressure rises as the nonwetting phasesaturation increases (in Figure 2.14, mercury is the nonwetting phase).



Entry

pressure

Figure 2.14: Capillary pressure curve (Figure 3-4 in Lake).

Saturation is defined as the volume of fluid in the pore space relative to the total pore

volume. Saturation can of course be defined over any volume of the medium.

The sudden reduction in saturation once the entry pressure is reached is due to invasion percolation, which is the Haines jump process shown earlier in Figure 2.11. Once a pore

throat has filled then the adjacent pore will fill immediately, but a throat cannot fill until

the adjacent (upstream) pore has filled.

The shape of the capillary pressure curve depends on the pore size distribution. A moreuniform distribution of pore sizes gives rise to a flatter capillary pressure curve (Figure

2.15).

The capillary pressure curve is also a function of whether the nonwetting phase isdisplacing the wetting phase (drainage – line 1 in Figure 2.16) or the wetting phase is

displacing the nonwetting phase (imbibition – line 2 in Figure 2.16). Drainage and

imbibition always refer to the wetting phase. The difference between drainage and

imbibition is due to a number of phenomena, most prominently trapping , which isdiscussed in the next section.



P c

S w 1.0

very nonuniform

pore distribution

more uniform

pore distribution

Figure 2.15: Capillary pressure curve as a function of pore size distribution.

Figure 2.16: Drainage (1) and imbibition (2) capillary pressure curves (Figure 2.18 from

Dullien). Water-wet rock.

Capillary pressure is also a function of wettability – an oil-wet material will have

capillary pressure curves that are unlike water-wet materials, see Figure 2.17. Incomparison with the water-wet material in Figure 2.16, the imbibition curve crosses the

zero capillary pressure axis at a much larger nonwetting phase saturation.



Figure 2.17: Drainage (1) and imbibition (2) capillary pressure curves (Figure 2.19 from

Dullien). Oil-wet rock.

2.4.2 Rise in a Capillary Tube

A consequence of the force imbalance at an interface between two fluids in a capillarytube is that the meniscus will rise up the tube, Figure 2.18.

θ σ

p2

p1

r r

R θ σ

p2

p1

h

Figure 2.18: Capillary rise in a thin tube.

Considering the force balance in the vertical direction:

)(cos2 122 p pr r −π=θσπ (2.14)

r p p P c

θσ=−=

cos212

(2.15)

And since p2 = patm and p1 = patm – ρ gh, then:

ghr

p p P c ρ=θσ

=−=cos2

12(2.16)



Hence the height to which the wetting phase rises in the tube is a function of the tube

diameter as well as the interfacial tension and the contact angle.

2.4.3 Capillary Transition Zone

In a reservoir, a similar process causes the wetting phase (commonly water) to rise upinto the nonwetting phase (typically oil, then gas) above the apparent “water-oil contact”

(WOC) which clearly represents a transition zone rather than a sharp interface. This is

often referred to as the capillary fringe or the capillary zone. The height depends on thecapillary pressure, which in turn depends on the saturation. So the saturation becomes a

function of height, the same in shape as the capillary pressure curve.

In the oil: p gH p oo =ρ+

In the water: p gH p ww =ρ+

So g

S P H or gH p p P wc

woc ρ∆=ρ∆=−=

)(,

H

P c

po

pw

p

Figure 2.19: Capillary pressure difference in a reservoir.

S w1.0

H=

P c /∆ρ g

transitionzone

S w1.0

H=

P c /∆ρ g

transitionzone H=

P c /∆ρ g

transitionzone

Figure 2.20: Capillary transition zone in a reservoir.



2.4.4 Measuring Capillary Pressure

Capillary pressure is often measured using the “porous plate method” as shown in Figure

2.21. The purpose of the porous plate (sometimes also referred to as a semipermeablediaphragm) is to allow passage of water but not oil. The diaphragm is strongly water wet

and has a high entry pressure for oil. Depending on the pressure of the experiment, thegraduated tube is used to measure both the capillary pressure and the change in

saturation. In some forms of this experiment, the U-tube is flexible and the graduated

tube is raised and lowered to change the pressure difference.

Figure 2.21: Porous plate method to measure capillary pressure curves (Figure 2.16

from Dullien).



Figure 2.22: Porous plate capillary pressure apparatus in Stanford’s geothermal lab.

Figure 2.23: Porous plate capillary pressure apparatus in Stanford’s geothermal lab.



Figure 2.22 and Figure 2.23 show a dual-chamber capillary pressure apparatus made by

Ruska instruments. The right chamber has a core in place under test, and the left

chamber has been opened to reveal the porous diaphragm. This instrument is usuallyused to measure nitrogen-water capillary pressure.

These experiments are typically very time-consuming since it takes a long time for fluids

to equilibrate between the displacement steps.

Other methods to measure capillary pressure include mercury injection and centrifuge

methods. In the centrifuge method, the core is spun at high speed as in Figure 2.24.

r 1

r 2

ω

Figure 2.24: Centrifuge method to measure capillary pressure.

The pressures in the fluids due to the rotation of the sample are given by:

22

21 r p ρω= (2.17)

At the outer radius the fluids have the same pressure ( P c = 0) because they are in contactoutside the core, but at the inner radius the density difference between them causes a

pressure difference. The resulting capillary pressure causes the nonwetting phase

(usually oil) to displace the wetting phase (usually water) out through the outer end of the

core. After spinning at a fixed angular velocity ω, the core is taken out and weighed todetermine the average saturation. The core is then returned to the centrifuge and spun at

a higher speed. The resulting capillary pressure at the top of the core is:

2

1

2

2

2

21 r r P c −ρω∆= (2.18)

This kind of experiment is able to achieve very high values of the capillary pressure, butis only useful for drainage. There is always a pressure gradient along the core due to the

differing radii of rotation, which can cause difficulties in some cases.

2.4.5 Capillary Pressure Models

Two commonly used empirical models for the capillary pressure curves are the Brooks-

Corey and van Genuchten models. The Brooks-Corey model for the drainage capillary pressure curve can be written:



λ−

−−

= /1)1

(wr

wr ec

S

S S p P

(2.19)

where pe is the entry pressure. The imbibition curve is:

)1(/1 −= λ−

eff ec S p P (2.20)

where the “effective” saturation is:

nwr wr

wr weff

S S

S S S

−−−

=1 (2.21)

λ is a parameter that depends on pore size distribution:• λ = 2 for a wide range of pore sizes.

• λ = 4 for a medium range of pore sizes.

• λ = ∞ for a single uniform pore size.

An example of a Brooks-Corey curve is shown in Figure 2.25. Note that a consequenceof the Brooks-Corel model is that log of capillary pressure will be a linear function of log

of effective saturation (for drainage):

eff ec S p P ln1

lnlnλ

−=(2.22)

0

1

2

3

4

5

6

7

8

9

10

0 0.2 0.4 0.6 0.8 1 S w

P c

Pc (drainage)

Pc (imbibition)

Figure 2.25: Brooks-Corey capillary pressure curves with λ=2, S wi=0.24, S nwr =0.19,

pe=2.

The van Genuchten model is more commonly used in hydrology, and may be written:

( )[ ] nn

n

eff c S P 1

1 11

−α

= −

(2.23)



2.4.6 Capillary Number and Bond Number

We can define the capillary number as a dimensionless number that relates the

magnitudes of the capillary and viscous forces. The capillary number allows us tounderstand whether capillary forces dominate at the pore scale.

The pressure difference within a pore of radius r due to capillarity is given by:

r P c

σ=

2

(2.24)

The viscous pressure drop across the same pore, assuming it has a length L and

permeability k is:

k Lu p so

L pk u visc

visc µ=∆∆µ−= ,(2.25)

Hence the ratio of viscous pressure drop to capillary pressure is of order:

σµ

≈2

r

k

Lu

pressureCapillary

drop pressureViscous

(2.26)

Recalling the Carman-Kozeny relationship (Eq. (1.11) between permeability and pore

size:

( )23

2

23

10172

1r

Dk

p −≈φ−

φ

τ=

(2.27)

For a pack of spheres, typically L/r is around 10, so we can write:

c N u

r

r ur

k

Lu

pressureCapillary

drop pressureViscous 22

23

2

101010

10

2=

σµ

=σµ

=σ

µ≈ −

(2.28)

Here we have defined the capillary number uµ/σ. For a more realistic porous material:

c N pressureCapillary

drop pressureViscous 310≈(2.29)

For a capillary number of around 10-3

, the capillary and viscous forces would be about

the same.

Considering typical values of the porous medium parameters in common types of

problems, for groundwater flow (where air and water are the nonwetting and wetting

phases), u is of order 1 m/day to 1 m/year (~10-5

to 10-8

m/sec), µ is 1 cp = 10-3

Pa.s, and



σ is around 60×10-3

N/m, hence N c is around 10-6

to 10-9

. In this case capillary forcesdominate. In infiltration into soil, u may be much larger, for example 1 m/hr, so N c may

be as large as around 10-4

in which case viscous forces may become equally important.

For oil reservoirs, u is of order 10-5

m/sec at most (except very close to gas wells), µ is of

order 1 cp = 10-3

Pa.s, and σ is around 50×10-3

N/m, hence N c is around 10-7

. Again,capillary forces dominate.

In some cases we may also find it useful to relate the magnitudes of capillary and

buoyancy forces to understand the effects of gravity. For this we define the Bond number . The buoyancy force due to a density change over length L can be written:

gL p grav ρ∆=∆(2.30)

B N gL gLr

pressureCapillary

pressure Buoyancy 12

1 10102

−− =σ

ρ∆≈

σρ∆

≈(2.31)

The Bond number is ∆ρ gL2 / σ. If the Bond number were to be around 10, then the two

forces would be of similar significance.

Typical values of the Bond number for hydrology problems, in which the density

difference between air and water gives rise to ∆ρ g values around 104

N/m2

and L isaround 10

-3m (grain size) would be around 0.2. This means the ratio of buoyancy to

capillary forces is around 0.02. In oil reservoirs, the density difference is smaller, and the

pore size is also smaller, so Bond number is of order 4×10-4

. In both cases, capillaryforces dominate over buoyancy forces.

2.4.7 Leverett J-Function

[Dullien 141-142, Lake 53-54]

Since capillary pressure can be measured more easily for laboratory fluids (e.g. air-water

or air-mercury) than oilfield fluids, it is useful to develop a dimensionless group in anattempt to correlate capillary pressure curves. To take account of the different interfacial

tension and contact angle between different fluid pairs, we can scale the capillary

pressures by these variables:

lablab

field field

labc

field c

P

P

θσ

θσ=

cos

cos

(2.32)

Given that reservoir rocks with similar lithology are likely to have similar pore size

distributions (albeit not necessarily at the same size scale), Leverett in 1941 developed adimensionless variable (since known as the Leverett J-function) which interrelates the

rock properties and fluid properties. The pore size distribution is itself interrelated with

the capillary pressure, which means the Leverett J-function can be written in terms of the



capillary pressure (easy to measure) rather than the pore size distribution (not so easy to

measure).

• The capillary pressure at a given saturation is a measure of the smallest poresentered by the continuous phase.

• The shape of the capillary pressure curve is a function of the pore sizedistribution.

• The magnitude of the capillary pressure depends on the mean pore size.

• The J-function combines these dependencies to remove the mean pore size andtherefore provide a single defining function for similar rocks. J is a function of

the nonwetting phase saturation, and is used to replace the pore size with themean pore size Rm:

)('cos2

)(

cos2)( nw

mnw

nwc S J RS R

S P θσ

=θσ

=(2.33)

From the Carman-Kozeny derivation (Eq. (1.6)) we can write:

φτ=

τφ

=k

R so R

k m 8,8

2

(2.34)

φτ

θσ=

θσ=

k P R P S J cmc

nw 8cos2cos2

)('(2.35)

The final definition of the J-function is then:

φθσ=

k P S J c

nwcos

)((2.36)

Where)('

8

2)( nwnw S J S J

τ=

(2.37)

The J-function is strictly only applicable to primary drainage situations, however the

form of the equation emphasizes some general truths. Since P c depends on Rm and k alsodepends on Rm, the interrelation between P c and k holds in other cases too. It is generallythe case that tighter formations (low k ) will have higher capillary pressure and will

imbibe water faster than higher permeability rocks. This leads to rate-dependent

recoveries in heterogeneous formations, and capillary cross-flow of oil from low to high permeability rocks. Figure 2.28 from Corbett, Ringrose, Jensen and Sorbie, SPE 24699

“Laminated Clastic Reservoirs: The Interplay of Capillary Pressure and Sedimentary

Architecture” (1992 SPE Fall Meeting) shows observations of this effect.



Water in Oil

Oil

Oil high k

low k

Figure 2.26: High rate case – oil is expelled into nearly residual oil saturation part of

reservoir and is therefore hardly mobile.

Water in

Oil

Oil

Oil high k

low k Water

Figure 2.27: Low rate case – oil is expelled into high oil saturation part of reservoir with

good mobility.

Figure 2.28: Recovery as a function of rate from a laminated reservoir (Figure 5 from

Corbett, Ringrose, Jensen and Sorbie, 1992).

2.5 Residual Saturations

There is a residual saturation at which the capillary pressure appears to head towardsinfinity or zero (Figure 2.29). In the case of the wetting phase we often refer to this asthe immobile saturation. No matter how much pressure we apply, we cannot reduce the

wetting phase saturation any further (actually this is an oversimplification, since what we

mean is that we cannot reduce it in any reasonable amount of time).



Residual nonwetting

phase saturation

immobile wetting

phase saturation

Figure 2.29: Capillary pressure curve (Figure 3-4b in Lake).

Figure 2.30: Invasion sequence1-6 giving rise to capillary pressures in Figure 2.29

(Figure 3-4a in Lake).

2.5.1 Residual Nonwetting Phase SaturationThe residual nonwetting phase saturation occurs because small blobs of nonwetting phase

become trapped in the pores, and once disconnected from each other they can no longer

flow. In a sequence of injections of nonwetting followed by wetting phase fluids (Figure

2.30) the capillary pressure is governed by the connected nonwetting phase saturation.

For each initial saturation of nonwetting phase, there is a certain residual saturation thatwould remain after flooding with the wetting phase. For example, in Figure 2.31, if a

rock had the initial nonwetting phase saturation represented by point A, then were it to be

saturated with wetting phase it would have the residual saturation represented by A’. Thesaturation at A’ can be determined from the saturations at points A, B and C. This means



that the relationship between initial and residual nonwetting phase saturations can be

determined from the drainage and imbibition capillary pressure curves.

A’ A’

Figure 2.31: Initial and residual nonwetting phase saturation (based on Figure 3-6 in

Lake).

In Figure 2.31, the initial and residual nonwetting phase saturations are:

Anwi S S = (2.38)

)( AC Bnwr S S S S −−= (2.39)

Note that the capillary pressure curve in Figure 2.31 is plotted as a function of wetting phase saturation, so the nonwetting phase saturation is measured from right to left.

The relationship between the initial and residual nonwetting phase saturation is of great

importance in determining the effectiveness of oil recovery during waterflooding. Wecan plot initial vs. residual saturation in an IR curve, such as Figure 2.32. The curve can

be characterized using the Land trapping coefficient , C , defined as:

C S S nwinwr

=−**

11

(2.40)

Where the normalized saturations S* are defined:

wi

nwr nwr

wi

nwinwi

S

S S

S

S S

−=

−=

1;

1

**

(2.41)

We can also write:



*

**

1 nwi

nwinwr

CS

S S

+=

(2.42)

C varies from zero (complete trapping) to infinity (no trapping).

untrapped oil

Figure 2.32: Initial vs. residual nonwetting phase saturation [IR curve] (Figure 3-5 in

Lake).

The maximum amount of trapped nonwetting phase is given by:

( )1111

max

***−==−

nwr nwinwr S C

S S (2.43)

This provides a useful experimental procedure to determine C as follows:

1. Saturate the medium 100% with wetting phase.

2. Conduct primary drainage to S w = S wi (S *

w = 0, S *nw = 1).

3. Conduct imbibition to max

*

nwr S .4. Calculate C .

In real reservoirs, the starting condition is usually S w = S wi, that is the maximum S nw,

which leads to the maximum S nwr . The phase permeability for the nonwetting phase isdue only to the “free” saturation (connected saturation) S nwf and not the trapped saturation

S nwt .



free trapped

Figure 2.33: Free and trapped nonwetting phase saturation (from Figure 3-6 in Lake).

At intermediate saturations:

nwt nwf nw S S S +=;

***

nwt nwf nw S S S += (2.44)

We can calculate*

nwt S from Land’s relation and the knowledge that*

nwt S equals max

*

nwr S

less the amount of *

nwf S that will be trapped as the imbibition proceeds to ( )max

*

nwr S ,namely:

***

max

**

nwt nwr nwt nwr nwt S S S S S ∆−=∆−= (2.45)

This can be obtained from Land’s relation starting with

*

nwf S =

*

nwiS :

C S S nwf nwt

=−∆ **

11

(2.46)

*

*

*

1 nwf

nwt

nwf CS

S

S =−

∆ ;*

*

*

1 nwf

nwf

nwt CS

S S

+=∆

(2.47)

So*

*

****

1 nwf

nwf

nwr nwt nwr nwt CS

S S S S S

+−=∆−=

(2.48)

The free saturation that contributes to flow is:

*

*

*****

1 nwf

nwf

nwr nwnwt nwnwf CS

S S S S S S

++−=−=

(2.49)

*******2** )( nwf nwf nwr nwr nwf nwnwnwf nwf S S CS S S CS S S C S +−−+=+ (2.50)



)()()( *****2*

nwr nwnwr nwnwf nwf S S CS CS S S C −−+−+ (2.51)

This is a quadratic of the form ax2

+bx+c=0, where:

*

nwf S x = ; C a = ; )( **

nwr nw S S C b −−= ; )( **

nwr nw S S c −−=

Hence

−+−+−= )(

4)()( **2****

21*

nwr nwnwr nwnwr nwnwf S S C

S S S S S

(2.52)

2.5.2 Trapping Mechanisms

We are fundamentally interested in trapping of the nonwetting phase. In petroleum

recovery the nonwetting phase is usually the oil, which we would earnestly like not to be

trapped in the ground. In hydrology the nonwetting phase is often the contaminant,which again we would like not to be left in the ground.

There are a number of trapping mechanisms, of which we will discuss two. The first is

dependent on pore network topology, the second describes trapping within an individual pore.

(a) Pore Doublet Model

[Dullien, p. 426-429] [Lake, p. 63-67]Consider two adjacent pores of different sizes initially filled with nonwetting fluid, as in

Figure 2.34. Wetting fluid is introduced to displace the nonwetting fluid (imbibition).

Figure 2.34: Pore doublet model (Figure 3-14 in Lake).

If we assume that: (a) Poiseuille flow describes the behavior of the fluids in the tubes (noeffect due to the interface), and (b) that viscosities are equal, then:



)(8

2

4

21

4

121 p R p R L

qqq ∆+∆µπ

=+=(2.53)

The driving force for the two channels must be the same:

2211 cc P p P p p −∆=−∆=∆ (2.54)

While the interfaces are in the tubes:

−θσ=−

21

21

11cos2

R R P P cc

(2.55)

We can combine Eqs. 2.54 and 2.55 to eliminate ∆ p1, then eliminate ∆ p2 in favor of q:

4

1

2

21

4

2

1

1

11

4

cos

+

−

µθσπ

−

=

R

R

R R L

Rq

q

(2.56)

4

1

2

21

4

2

4

1

2

2

1

11

4

cos

+

−

µθσπ

+

=

R

R

R R L

R

R

Rq

q

(2.57)

To investigate the trapping behavior, we can investigate the ratio of the average velocities

(v = q/πr 2) for each tube:

−

ββ−

β

−

β+

=

114

11

4

2

2

1

2

cap

cap

N

N

v

v

(2.58)

where β = R2 /R1 is a heterogeneity factor , and the local capillary number Ncap is:

θσπµ

=cos3

1 R

Lq N cap

(2.59)

When q is large, capillary forces are negligible (capillary number approaches infinity)

and the velocity ratio is:



2

1

22

1

2

=β≈

R

R

v

v

(2.60)

In this case the interface advances faster in the larger tube and traps nonwetting fluid inthe small tube.

When q is small and capillary forces dominate, then v2 /v1 < 1 and the interface moves

faster in the small tube and traps nonwetting phase in the large tube, as in Figure 2.34(b).This condition is typical for realistic values of q, R1 and R2.

Note that:

• Nonwetting fluid is trapped in the large pores and the wetting phase flows past it

in the small pores.

• Lowering capillary forces (increasing capillary number) decreases the trapping phenomenon.

• There is no trapping without local heterogeneity.

In general this model overestimates the amount of trapping in real porous media.

(b) Snap-Off Model

[Dullien, p. 429-436] [Lake, p. 67-68]The snap-off model is a better description of the trapping phenomenon in real porous

media, and can account for 80% of the trapped nonwetting phase. In this model, the pore

is envisaged of a tube of varying cross section. With a low aspect ratio of area variation,the wetting phase can effectively displace the oil in a piston-like fashion as in Figure

2.35(a). For a higher aspect-ratio channel, there is a higher gradient of capillary pressure

in the nonwetting phase than in the (continuous) wetting phase, so the nonwetting phase

wants to flow backwards locally and the collar of wetting phase “snaps off” as in Figure2.35(b).

Water

Oil

Water

Oil

Trapped oil Collar of water

Water

Trapped oil Collar of water

Water

(a)

(b)

Figure 2.35: (a) Piston-like flow in low aspect-ratio channel; (b) Snap-off in high aspect-ratio channel.



Rt R f

2Rth

2Rb

∆ pw

Rt R f

2Rth

2Rb

∆ pw

Figure 2.36: Pressures around a trapped ganglion of nonwetting phase.

Looking at the pressures at the upstream and downstream ends of the trapped ganglion(Figure 2.36), if p is the pressure in the wetting phase at the upstream point, then the

nonwetting phase pressure is:

t R p p

σ+=

21

(2.61)

The pressure in the ganglion at the downstream end is:

f

w R

p p pσ

+∆−=2

2

(2.62)

The ganglion can only move downstream if p1 > p2, hence we can say that the ganglion

becomes trapped unless:

−σ≈

−σ>∆

btht f

w R R R R

p11

211

2

(2.63)

This explains why the trapping occurs in high aspect-ratio pores, and also trapping is a

rate-dependent phenomenon. When the flow is slow, the displacement is more likely to be piston-like (why?).

It is also interesting and important to understand how the collar of wetting phase can be present in the pore throat, apparently “ahead” of the nonwetting phase interface. To see

this, we have to consider the three-dimensional nature of the pore, with crevices along the

side as in Figure 2.37.



CreviceCrevice

Figure 2.37: Capillary pressure allows fluid to travel along the crevices in pore corners.

The wetting phase can move along the crevices in pore corners to the pore throat before

the main interface reaches the throat. Figure 5.59 in Dullien shows this in a moreskillfully drafted picture.

2.5.3 Immobile Wetting Phase Saturation

In a drainage process the invading nonwetting phase flows through the center of the pores

while the wetting phase forms a film on the walls.

In the formation of most oil reservoirs, oil migrates into sedimentary rocks that are

initially filled with water as the wetting phase. As oil saturation increases, the water is

forced out of the largest pores first and occupies smaller and smaller spaces in the rock.When the water saturation becomes low enough, the water becomes disconnected and

forms pendular rings. Once disconnected, the wetting phase can no longer flow and

hence the remaining fluid is at immobile wetting phase saturation.

Water

Oil

θ

Water

Oil

θ

Figure 2.38: Immobile wetting phase saturation in the form of pendular rings.

The capillary pressure at wetting phase saturation is a function of the two principal radiiof the donut-shaped ring (the “waist” radius and the “groove” radius):

+θσ=−=

21

11cos

R R p p P wnwc

(2.64)

2.5.4 CDC Curves

[Lake 68-77, Dullien 443-458]

Now that we have seen that the residual saturations are a function of the trapping

mechanisms and that trapping mechanisms are a function capillarity, we can examine the

capillary desaturation curve (CDC), Figure 2.39. The CDC relates the amount of trapped



nonwetting or wetting phase as a function of Capillary number (=µu/σ, Eq. 2.28). Recallthat the Capillary number is a ratio between viscous and capillary forces.

Figure 2.39: Capillary desaturation curve (CDC). (Figure 3-17 from Lake).

In most cases the nonwetting phase has a higher residual (more trapping) than the wetting

phase. Both phases tend to have a critical Capillary number at which the trapped phase

begins to mobilize. The critical Capillary number for wetting phase is often higher thanthat for nonwetting phase, hence the target for enhanced oil recovery is to modify the

Capillary number to lie between the two critical values (why is that a good thing?). In

practice it is difficult to raise either the viscosity or velocity, so the most accessible way

to increase Capillary number is to reduce the interfacial tension σ, for example by addingsurfactant.

Lake, in pages 73-77 outlines a method due to Stegemeier in which the CDC curve can

be estimated from the IR curve by imposition of the snap-off model (Eq. 2.63).

2.6 Wettability

Clean rocks, sandy aquifers and surface soils with a low organic content are usually

water-wet . Reservoir rocks and surface soils with high organic content are often mixed-wet rather than completely water-wet, as typically some of the pores are water-wet andothers are oil-wet. This is not the same as a neutral-wet medium in which the contact

angle is zero everywhere. The reason for mixed-wetness is that pores in contact with

crude oil become oil-wet, as we saw earlier in Figure 2.7, whereas the small throats andcrevices remain water-wet since crude oil never reaches them.



(a)

P c 0

S w0 1.0

P c 0

S w0 1.0

(b)

(c)

P c 0

S w0 1.0

P c 0

S w0 1.0

(d) Figure 2.40: (a) Water wet, (b) oil wet, (c) mixed wet 1, (d) mixed wet 2.

We refer to imbibition as the increase in the wetting phase saturation. Spontaneous imbibition occurs when the capillary pressure is positive, forced imbibition occurs when

the capillary pressure is negative. The only way to have forced imbibition of water (for

example) is to have a connected network of oil-wet pores so that the water can displace

the oil.



P c 0

+ve

-ve

S w0 1.0

A1

A2S wi ∆S ws S or

∆S os

Primary drainage (forced)

Secondary drainage (forced)

Secondary drainage

(spontaneous)Imbibition(forced)

Imbibition

(spontaneous)

Figure 2.41: Capillary pressure diagram used to characterize wettability.

We have seen that a porous material can defined as water-wet , oil-wet or mixed-wet . Thedegree to which a reservoir is one or another of these can be determined by considering

the capillary pressure curve, or by characterizing it in terms of wettability indices. Thereare a number of different indices in common usage.

Amott Indices:

Referring to Figure 2.41, we can define the Amott indices as:

or wi

wsw

or wi

oso

S S

S I

S S

S I

−−∆

=−−

∆=

1;

1 (2.65)

If the material is completely water-wet, then I o = 0 and I w = 1. If the material is stronglyoil-wet then I o = 1 and I w = 0. I o is the displacement-by-oil ratio, and is the water volume

displaced by spontaneous oil imbibition, relative to the total water volume displaced byoil imbibition (spontaneous and forced). I w is the displacement-by-water ratio, and is the

oil volume displaced by spontaneous water imbibition, relative to the total oil volume

displaced by water imbibition (spontaneous and forced). If we have connected pathwaysof both oil and water then both indices can be greater than zero.



Amott-Harvey Index:

or wi

oswsow AH

S S

S S I I I −−

∆−∆=−= 1 (2.66)

The Amott-Harvey index ranges from +1 for a completely water-wet medium to –1 for a

completely oil-wet medium.

USBM Wettability Index:

This index is based on the ratio of the two areas representing forced imbibition in Figure2.41:

=

2

1log A

A N

w(2.67)

The range is from +∞ for a completely water-wet material to -∞ for a completely oil-wet

material. Typical values are in the range –1.5 to +1.0. In general this index is not usedvery much.

Let us consider the effects of wettability on the residual saturations. Looking first at the

residual oil saturation, we can see that:1. If the contact angle becomes greater, then we have less snap-off and S or will be

less.

2. If there are unconnected networks of oil-wet pores, S or will be higher.3. In strongly oil-wet systems, oil can flow in layers, so S or will be less.

In general, the residual oil saturation becomes smaller as the medium becomes more oil-wet.

Looking at the immobile water saturation, S wi becomes higher in oil-wet materials since

water can be trapped in big oil-wet pores.

Leverett J scaling does not work for mixed-wet rocks, because it is defined for water-wet

materials.



Water-wet Oil-wet Mixed-wet 1 Mixed wet 2

θ<30º

wo

θ

θ>150º

woθ

Patches of oil-wet

and water-wet

30º<θ<150º

wo

θ

P c=po - pw>0 P c=po - pw<0

S wi < S or

Because water remains connected

but oil does not.

S wi > S or Because oil remainsconnected but water

does not.

S or a bit less than

when θ<30º and S wi a bit more.

Water in crevices andsmall pores.

Oil in crevices andsmall pores.

No oil or water increvices.

Water remains

connected.

Oil remains

connected.

Both water and oil

can get trapped.

P c 0

S w0 1.0

P c 0

S w0 1.0

P c 0

S w0 1.0

P c 0

S w0 1.0

Spontaneous

imbibition of water.

Spontaneous

imbibition of oil.

No spontaneous

imbibition. Only

piston-like

displacement, no

snap-off. I w = 1

I o = 0

I w = 0

I o = 1

I w > 0

I o > 0

I w = 0

I o = 0



3. Multiphase Flow in Porous Media

3.1 Relative Permeability

[Dullien 338-380, Lake 58-62]

3.1.1 Two-Phase Flow

So far we have considered the capillary behavior of two fluids in a porous medium,

essentially as a static phenomenon. However capillary effects also make a difference tothe way fluids flow together through the porous medium.

In single-phase flow, the absolute permeability, k , is defined by Darcy’s Law and is

imagined to be a physical property of the rock (although in some cases this may not be

so).

x

p A

k q x ∂

∂µ

=(3.1)

Although the interactions between two flowing phases can be due to a variety of different

effects (see Figure 5.1 in Dullien for example), it is easy to see that even in a plain

capillary tube it is harder to flow two fluids than it is to flow one.

L

q q

∆ p ∆ p+P c

Figure 3.1: Apparent reduction of permeability when an interface separates two flowing

fluids.

In Figure 3.1, in single-phase flow (left diagram) the apparent permeability is:

p

L

A

qk

∆

µ=

(3.2)

whereas the apparent permeability in two-phase flow (right diagram) has the lesser value:

c P p

L

A

qk

+∆µ

=(3.3)

In real porous media, the individual phase permeabilities depend on how difficult it is to

propagate an interface through the porous medium – this means that the phase



permeability depends on saturation (as capillary pressure depends on saturation), as in

Figure 3.2.

S w

k o k w

k abs

S wi 1-S or

Figure 3.2: Phase permeabilities as a function of saturation.

Rather than refer to individual phase permeabilities, we usually make use of the term

relative permeability which refers to the phase permeability relative to the absolute

permeability k :

k

S k S k

k

S k S k ww

wrwwo

wro

)()(;

)()( ==

(3.4)

S w

k ro k rw

1

S wi 1-S or

Figure 3.3: Relative permeability.

Notice in Figure 3.3 that k ro(S wi ) is a little less than 1 – any incremental increase in oil

saturation will not increase the connected oil network much, so oil relative permeabilitydoes not change significantly. Sometimes relative permeability is scaled so that the value



is normalized to k o(S wi ) so that the maximum value of k ro is 1.0 – be sure to check

whether values of relative permeability you are using have been so normalized.

Notice that oil interferes with water more than water interferes with oil (in a water-wetrock). This is because the oil occupies the largest pores. The basic nature of relative

permeability effects is due to the flowing of the fluids in different channels.

As for capillary pressure curves, we see different relative permeability curves for drainage and imbibition (for pretty much the same reasons).

Drainage curves are important for:

• solution gas drive (since oil and water are generally wetting relative to gas),

• for gravity drainage (gas displaces drained oil),

• gas injection processes,

• oil or gas displacing water (in tertiary recovery processes).

Imbibition curves are relevant to:

• waterflood calculations,

• water influx,

• oil displacing gas (e.g. oil moving into a gas cap).

3.1.2 Drainage Relative Permeability – Burdine’s Theory

Burdine’s theory is a way to derive the relationship between relative permeability and

capillary pressure, based on hydraulic radius concepts based on a capillary tube bundle

model. The same result in derived in Dullien, Section 5.2.4.2, page 374, based on

statistical concepts in a cut-and-random-rejoin model of the capillary tube bundle. Sincethe statistical model is described in Dullien, we will develop the original Burdine theory

here using the hydraulic radius approach.

Using capillary tubes as a model for the pores, the Carman-Kozeny theory (Section 1.2)

gives us, for each tube:

i si

iii

i

ii

l

A p Rq

l

p Ru

µα∆

=µ∆

=8

;8

22

(3.5)

where Ri is a characteristic radius of a (noncircular) pore or cross-sectional area Ai and

length Li. α si is a shape factor to account for the noncircularity of the pore.

For the drainage process, the nonwetting phase enters the largest pores first. If the

nonwetting phase invades the dni next largest pores, the change in volume of the wetting

phase is:

ii

wiiiiw

Al

dV dndn Al dV −=−= ;

(3.6)

which decreases the water flow rate by:



w

i si

ii

i si

iiw dV

l

p Rdn

l

A p Rdq

2

22

88 µα∆

+=µα

∆−=

(3.7)

Darcy’s law for two-phase flow at the macroscopic scale gives:

wwww dk L

p Adqor

L

p Ak q

µ∆

=µ∆

= ;(3.8)

Equating the incremental flow rates at microscopic and macroscopic scales:

w

i si

iw dV

Al

L Rdk

2

2

8α=

(3.9)

However, since dV w = φ ALdS w, at the macroscopic scale, then:

w

i si

iw dS

l

L Rdk

2

22

8αφ

=(3.10)

∫ αφ

=w

wi

S

S

w

i si

iw dS

l

L Rk

2

22

8(3.11)

The lower limit of the integral is S wi since none of the saturation below immobile

saturation contributes to the flow. To evaluate the integral we need to know how thegeometric parameters Ri, α si and l i behave as a function of saturation S w. The hydraulicradius Ri can be related to the capillary pressure, since capillary pressure is related to thelargest pore size occupied by the wetting phase. A balance of forces across a pore gives:

pc A P ωθσ= cos(3.12)

where ω p is the wetted perimeter. The characteristic radius is given by:

i

i

i

pi

i R

R

R A21

2

2

=π

π=

ω (3.13)

So, )(

cos2

wc

iS P

Rθσ

=(3.14)

Substituting in Eq. (3.11):

∫ αφθσ

=w

wi

S

S

w

wcww s

w dS S P S l S

Lk

)()()(2

)cos(22

22

(3.15)



If we define the effective saturation as:

wi

wiww

S

S S S −

−= 1

*

(3.16)

Then we can simplify Eq. (3.15) by expressing:

Average tortuosity2

*2 )1(

L

S l w ==τ

(3.17)

So,2**2

*2

2

2

)()1(

)()()(

w

s

w

w s

ww s

S S l

S l

L

S l S τα=

=τα≈

α

(3.18)

Notice that l approaches infinity as S w approaches S wi and the wetting phase becomes

disconnected.

Substituting,∫τα

φθσ=

w

wi

S

S

w

wc s

ww dS

S P

S k

)(

1)()cos(2

2*2

(3.19)

Them making use of the definition of relative permeability, k rw=k w(S w )/k w(S w=1):

∫

∫=

1

2

2

2*

)(

1

)(1

)(

wi

w

wi

S

w

wc

S

S

w

wc

wrw

dS S P

dS S P

S k

(3.20)

By a similar argument, the nonwetting phase drainage relative permeability is:

∫

∫=

1

2

1

2

2*

)(1

)(

1

)).((

wi

w

S

w

wc

S

w

wc

owiroro

dS S P

dS S P

S S k k

(3.21)

where wioc

ocoo

S S

S S S

−−−

=1

*

(3.22)

Notice that S oc is not the same as S or because we are on the primary drainage curve. S oc is

related to percolation.

If we make use of the Brooks-Corey relation for capillary pressure, Eq. (2.19), then:



λ=/1*

)( w

ec

S

p P

(3.23)

Then

b

a

w

wie

b

a

ww

b

a wiec

S

S pdS S

S pdS

S P

+λ−

=−

=+λ

λ∫∫12

)(

)1(

1)(

)1(

1

)(

112

*

2

*/2*

22

(3.24)

After which we can evaluate:

32* )(

+λ= wrw S k (3.25)

and

−=

+λ 12*2*

)(1)).(( wowiroro S S S k k (3.26)

Normally we assume that S oc is roughly zero, so:

** 11

)(1

1

1

1w

wi

wiwwi

wi

w

wi

oo S

S

S S S

S

S

S

S S −=

−−−−

=−−

=−

=(3.27)

So

−−=

+λ 12*2* )(1)1).(( wwwiroro S S S k k

(3.28)

These are known as the Brooks-Corey relations for relative permeability.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

Sw

k r

krw

kro

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

Sw

k r

krw

kro

Figure 3.4: Brooks-Corey relative permeability curves for λ=2 (left) and λ=4 (right).

3.1.3 Imbibition Relative Permeability

During imbibition we have to worry about the trapped nonwetting phase, which does not

contribute to flow. Rewriting Eq. (3.26) in terms of nonwetting phase saturations:



−−=

+λ 12*2* )1(1)).(( nwnwwirnwrnw S S S k k

(3.29)

Then we need to recognize that of the nonwetting phase saturation*

nwS , it is only theflowing part of the saturation that contributes to the relative permeability. Hence we can

write:

−−=

+λ 12*2* )1(1)).(( nwf nwf wirnwrnw S S S k k

(3.30)

Where the flowing nonwetting phase saturation*

nwf S was found earlier from Eq. (2.52),namely:

−+−+−= )(

4)()( **2****

21*

nwr nwnwr nwnwr nwnwf S S C

S S S S S

(3.31)

The wetting phase flows in pores of radius characterized by the saturation:

nwt wnwf S S S S +=−=1(3.32)

If we use the drainage capillary pressure curve as a substitute for radius (as we did in the

last section) then we need to include the effect of pores from S wi to pores now containing

S w + S nwt , hence:

∫

∫

+

+=

1

2

2

2*

)(

1

)(

1

)(

wi

w

wi

S

w

nwt wc

S

S

w

nwt wc

wrw

dS S S P

dS S S P

S k

(3.33)

For a constant pore size index, λ, and Lands trapping coefficient, C , this can be integrated

to give:

( )[ ]

−+

+

λ−+= ∫

+ λ+λ

**

0

2

2

12**2*

)1(11

2)(

nwt w S S

nwt wwrw dS S C

S S S S k

(3.34)

3.1.4 Relative Permeability Correlations

There a number of standard correlations in common usage. We have already seen the

Brooks-Corey correlation, Eqs. (3.25) and (3.28):

32* )(

+λ= wrw S k (3.35)



−−=

+λ 12*2* )(1)1).(( wwwiroro S S S k k

(3.36)

A common special case is when λ=2, giving rise to a correlation known as the Coreycurves:

4* )( wrw S k = (3.37)

2*2* )(1)1).(( wwwiroro S S S k k −−= (3.38)

The Corey and Brooks-Corey models have some theoretical basis, but there are also a

number of empirical relations in exponential form. One of the most useful is due to

Honarpour et al. (1982):

w

nw

n

nwr wr

wr wrwrw

n

nwr wr

nwr wrnwrnw

S S

S S k k

S S

S S k k

−−−

=

−−−−

=

1

1

1

0

0

(3.39)

Another exponential form is due to Chierici:

−−−

−=

−− −−=

α

α

w

or wr

wiwwrwirw

or wr

wiworoiro

S S

S S Ak k

S S S S Ak k

o

1exp

1exp

(3.40)

The Chierici relations are less convenient because they have four parameters, however this also allows them to describe a wider variety of types of relative permeability curves.

A form similar to the Corey model is used in the hydrology field and is due to van

Genuchten:

[ ]k S S rl ek ek

m m= − −1 2 12

1 1/ /( ( )

[ ]k S S rg ek ek

mm

= − −( ) / /1 11 3 12

(3.41)

3.1.5 Capillary End Effect

Before discussing ways of measuring relative permeability in the laboratory, it is

necessary to understand the issue of capillary end effect . This is a phenomenon that



occurs often in laboratory experiments and makes it difficult to know exactly what the

saturation may be inside a core sample.

If we consider a steady-state flow of oil and water through a (water-wet) core that was

filled originally with water, as in Figure 3.5, then we can see that if the outlet end of the

core is at atmospheric pressure then both phases must have the same pressure(atmospheric) and the capillary pressure will therefore be zero. Given that the capillary

pressure has a specific relationship with saturation, this in turn means that the saturation

at the outlet face must have a specific value.

qw

qo

qw

qo

A

P c=0

po=pw=patm

qw

qo

qw

qo

Figure 3.5: Simultaneous flow of oil and water in a steady-state experiment.

Based on the capillary pressure diagram (Figure 3.6), if the capillary pressure is zero then

the water saturation must be at 1-S or . Experimental observations have confirmed this to

be the case. How can the oil be flowing out of the core if it is at residual (and thereforeimmobile) saturation? Remember that the capillary pressure diagram represents a static

condition whereas we are now talking about flow – hence the diagram is not fully

applicable.

P c

S w0 1-S or 1.0

P c=0

Figure 3.6: Capillary pressure must be zero at the outlet end.



S w

1-S or

P c

0

Figure 3.7: Capillary pressure end effect gives rise to a changing saturation towards the

outlet.

The result of the strong variation in capillary pressure at the outlet, and the correspondingchange in saturation (Figure 3.7), is known as the capillary end effect . This effect can be

a major difficulty when making experimental measurements that require some kind of

inference of the saturation in the core or which assume the saturation to be constant.The capillary end effect can be analyzed based on the flow equations:

∂∂

+∂

∂µ

−=∂∂

µ−=

x

P

x

p A

k k

x

p A

k k q cw

o

roo

o

roo

(3.42)

Ak k

q

x

p

x

p A

k k q

rw

wwww

w

rww

µ−=

∂

∂⇒

∂

∂

µ

−=

(3.43)

x

P

Ak k

q

Ak k

q c

rw

ww

ro

oo

∂∂

−µ

=µ

(3.44)

So Ak k

q

k

q

x

P

ro

oo

rw

wwc 1

−=

∂∂ µ µ

(3.45)

or w

cro

oo

rw

www

S P k

q

k

q

Ak x

S

∂∂

−=

∂

∂ 11 µ µ

(3.46)

At the outlet face S w = 1-S or , and Eq. (3.46) can be used to compute the saturation at

decreasing values of x proceeding upstream of the outlet face, as in Figure 3.8.



0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0 0.02 0.04 0.06 0.08

x

S w

Figure 3.8: Capillary end effect as a function of distance from outlet.

The capillary end effect depends on the flow rates, the wetting/nonwetting flow ratio, the

viscosities, the core size and permeability, and the capillary pressure and relative permeability curves.

Depending on the configuration of the experiment, there is often also a capillary endeffect at the inlet end of the core.

3.1.6 Measurement of Relative Permeability

[Dullien 367-373]

There are a variety of different methods to measure relative permeability in thelaboratory. In general these fall into the categories of steady-state or unsteady-statemethods.

In a steady-state method, the experiment proceeds as follows:(1) Begin with the core saturated with one of the two phases (usually the wetting

phase).

(2) Inject both phase simultaneously, starting with a large ratio of qw /qnw, as in Figure

3.5.(3) Continue to inject (for a long time) until the output flow rates of each phase

stabilize to equal the input flow rates.

(4) Measure the pressure drop ∆ p across the core. Usually we assume that the pressure drop in the wetting and nonwetting phases are the same, since it is

implied that the saturation is constant except within the capillary end effect

(therefore the capillary pressure would be constant and the differences betweenthe phase pressures would also be constant).

(5) Measure the saturation in the core. Note that the saturation is not equal to

qw /qw+qnw, hence it necessary to measure the value independently, either by

weighing or using X-ray CT methods.(6) Go back to step (2) and decrease the ratio of qw /qnw. This therefore represents a

drainage relative permeability experiment. It may be useful to adjust the values



of qw and qnw while changing the ratio of qw /qnw in order keep the pressure drops

more or less constant – this would be desirable if one of the phases is gas, sincethe viscosity and Klinkenberg effect may change with pressure.

Following a sequence of decreases in qw /qnw until only nonwetting phase is flowing, we

can then perform another sequence of steps increasing qw /qnw to measure an imbibition relative permeability curve.

The steady-state approach can be extremely time-consuming, especially for low

permeability materials. Also, the capillary end effects can be problematic. The capillaryend effect can be accommodated be only measuring the pressure drop in the central

region of the core. This approach is not convenient for routine measurement however, so

often a “batch” type of apparatus makes use of some kind of capillary barrier to confine

the capillary end zone inside the barrier so that it does not exist in the core itself. One

common configuration is the so-called “Penn State” method, which uses short cores placed ahead and behind the tested core, as in Figure 3.9. In the Penn State method the

central test section can be quickly removed, weighed and replaced at each flow step.

Measuring relative permeability in fluids in which there is a phase change between the

two phases is particularly difficult. This is because the individual phase flow rates maydiffer from the injected rates as phase interchange occurs due to evaporation and

condensation inside the core.

Figure 3.9: Penn State method (Figure 5.19 from Dullien).

Unsteady-state methods depend on computing the relative permeability values based on a

flow model of the dynamic behavior taking place inside the core. These generally make

use of models such as Buckley-Leverett, that we will consider later in the course.

Common variants of unsteady-state methods are the Welge method and the similar JBN method (which stands for Johnson, Bossler and Naumann, who published a description of

it in 1959). Another common method is due to Jones and Roszelle in 1978. All these

approaches will be described after we have talked about the Buckley-Leverett theory.



Unsteady-state methods are much faster than steady-state methods but suffer from a

number of difficulties. If the two phases are similar in mobility the displacement may bemore-or-less pistonlike and only two values of the relative permeability will be revealed

(at the two residual saturation values). Capillary end effect is also still a difficulty.

In modern experiments it is possible and useful to measure the pressure and saturation atmultiple points along the core using X-ray CT methods, either in steady-state or

unsteady-state experiments. This allows for careful monitoring of the capillary end

effect, which can then be avoided by appropriate selection of the section of the core over

which pressure differences can be measured. Several different examples of this type of experiment have been performed in the last several years at Stanford (Figure 3.10 and

Figure 3.11). The apparatus shown in Figure 3.10 was designed in advance to minimize

capillary end effect – the measured saturations in Figure 3.11 show that the effect is

minimal and that the saturations are more-or-less uniform (except in one case [step 3]

where the X-ray CT scan captured the image of a displacement front as it passed alongthe core).

Steam

Generator

FlexibleHeaters

Insulation

P and T

Transducers

Data Acquisition Computer

LabViewWater Pump

w S t e a m & w a t e r

W a t e r

Core

Heat Flux

SensorsWater Pump

Heater for

Hot Water

Epoxy

Figure 3.10: Steady-state steam-water relative permeability experiment used in various forms by Ambusso, Tovar, Satik, Mahiya, O’Connor and Li at Stanford, 1995-2001.



0.00

0.20

0.40

0.60

0.80

1.00

0 10 20 30 40

Distance, cm.

S t e a m s

a t u r a t i o n

Step 1

Step 3

Step 4

Step 5

Step 6

Step 7

Step 8

Step 9

Step 10

Step 11

Step 12

Figure 3.11: Saturations measured by X-ray CT methods by Mahiya in 1999.

3.1.7 Three-Phase Flow

3.1.8 Three-Phase Relative Permeability



4. Multiphase Displacement

4.1 Immiscible Displacement Equations of Motion

[Dullien, Section 5.3; Lake, Section 5-2; Dake, Chapter 10]

4.1.1 Conservation of Mass and Momentum

We start with Darcy’s law for multiphase flow:

)( gh pk k k k

u ww

w

rww

w

rww ρ+∇

µ−=Φ∇

µ−=

r

(4.1)

)( gh pk k k k

u oo

o

roo

o

roo ρ+∇µ−=Φ∇µ−=r

(4.2)

Considering capillary pressure:

)( wcwoc S P p p P =−= ; cow P p p −= (4.3)

So)( gh P p

k k u wco

w

rww ρ+−∇

µ−=

r

(4.4)

Next, consider conservation of mass on a control volume (Figure 3.10):

[Rate of change of mass of water in ∆ x.∆ y] = [Net influx of mass of water]

∆ x

∆ y y xS ww ∆∆φρ

yuwxw ∆ρ

xuwyw ∆ρ

( ) y xu x

u wxwwxw ∆

∆ρ

∂∂

+ρ

( ) x yu y

u wywwyw ∆

∆ρ

∂∂

+ρ

Figure 4.1: Conservation of water mass over a control volume.



[ ]

( ) ( ) x yu y

u y xu x

u

xu yu y xS t

wywwywwxwwxw

wywwxwww

∆

∆ρ

∂∂+ρ−∆

∆ρ

∂∂+ρ−

∆ρ+∆ρ=∆∆φρ∂∂

(4.5)

[ ] ( ) ( ) x yu y

y xu x

y xS t

wywwxwww ∆∆ρ∂∂

−∆∆ρ∂∂

−=∆∆ρ∂∂

φ(4.6)

So,[ ] ( ) ( )wywwxwww u

yu

xS

t ρ

∂∂

−ρ∂∂

−=ρ∂∂

φ(4.7)

[ ] ( ) ( ) 0=ρ∂

∂

+ρ∂

∂

+ρ∂

∂

φ wywwxwww u yu xS t (4.8)

[ ] ( ) 0. =ρ∇+ρ∂∂

φwwww uS

t

r

(4.9)

For constant water density:

( ) 0. =∇+∂

∂φ

ww ut

S r

(4.10)

Similarly, for constant oil density:

( ) 0. =∇+∂

∂φ

oo ut

S r

(4.11)

Now, since S o + S w = 1, we can write:

( ) 0. =∇+∂

∂φ− o

w ut

S r

(4.12)

Then, adding the two equations together:

( ) 0. =+∇ wo uurr

(4.13)

If we multiply by area to convert the flux velocity to volumetric flow rate q, then:

( ) 0. =∇ T qr

(4.14)

To obtain an equation in terms of pressure, substitute Eqs. (4.1) and (4.2):



)()( gh pk k

gh P pk k

uu oo

o

rowco

w

rwwo ρ+∇

µ−ρ+−∇

µ−=+

rr

(4.15)

Or, h g P puu Aq wwoocwoT woT ∇ρλ+ρλ−∇λ+∇λ−=+= )(/rrr

(4.16)

Where w

rww

k k

µ=λ

, o

roo

k k

µ=λ

, woT λ+λ=λ (4.17)

Substituting:

( ) 0][. =∇ρλ+ρλ+∇λ−∇λ∇ h g P p wwoocwoT (4.18)

For negligible gravity and capillary forces:

( ) 0. =∇λ∇ oT p (4.19)

• This is a nonuniform Laplace equation for the pressure.

• Laplace equations are elliptic and exhibit smooth “diffusive” kinds of solutionwith no internal maxima and minima.

In reservoir simulation, Eq. (4.9) is the saturation equation and Eq. (4.18) is the pressureequation. Usually we solve these two equations by finite difference, for example using

the IMPES approach that solves the pressure equation implicitly and the saturation

equation explicitly.

4.1.2 Fractional Flow

It is convenient to reformulate the saturation equation in terms of a function we call the

fractional fl ow, defined as:

T

w

ow

ww

q

q

qq

q f =

+=

(4.20)

We an rearrange Eq. (4.16) as follows:

∇ρλ+ρλ+∇λ−

λ=∇− h g P

A

q p wwoocw

T

T

o )(1

r

(4.21)

Substituting into the water flow equation, Eq. (4.1):

h g P p gh P pu wwcwowwcoww ∇ρλ−∇λ+∇λ−=ρ+−∇λ−= )(r

(4.22)



h g P h g P A

qu wwcwwwoocw

T

T

ww ∇ρλ−∇λ+

∇ρλ+ρλ+∇λ−λλ

= )(

r

r

(4.23)

h g h g P A

qu oo

T

www

T

wcw

T

wT

T

ww ∇ρλ

λλ

+∇ρλ

−

λλ

+∇λ

−

λλ

−

λλ

= 11

r

r

(4.24)

Now T

o

T

oww

T

w

λλ

−=λ

λ−λ−λ=

−

λλ

1

(4.25)

Soh g h g P

A

qu oo

T

www

T

ocw

T

oT

T

ww ∇ρλ

λλ

+∇ρλλλ

−∇λλλ

+

λλ

=r

r

(4.26)

∇ρ−ρλ−∇λ+

λλ

= h g P A

qu owoco

T

T

ww )(

r

r

(4.27)

So

( )

∇ρ∆−∇

λ+

λλ

=

== h g P q

A

Aq

u

q

q f c

T

o

T

w

T

w

T

ww rr

r

1

(4.28)

In a somewhat cavalier fashion we are mixing vector and scalar designations in this lastequation. We could define fractional flow as a vector, since it represents flow and

therefore has an associated direction, however unless relative permeability is direction-dependent then f w will not be. Most often we will use the concept here to describe one-

dimensional flow and therefore we will never need the (improbable) vector notation.

Notice that in the absence of capillary and gravity effects, the fractional flow is simply

equal to the mobility of water relative to the total mobility. In this case, fractional flow

would be invariant with direction and therefore truly can be defined as a scalar quantity.

Now let’s look again at the equation describing conservation of mass of water, Eq. (4.10),

and substitute the fractional flow:

( ) 0. =∇+∂

∂φw

w ut

S r

(4.29)

0. =

∇+

∂∂

φ A

q f

t

S T w

w

r

(4.30)

Recall from Eq. (4.14) that ( ) 0. =∇ T qr

, so:

( ) wT wT f q f q ∇=∇ ..rr

(4.31)



Thus0. =∇

φ

+∂

∂w

T w f A

q

t

S r

(4.32)

Remembering the interstitial velocity v, where:

A

qv T

φ=

r

r

(4.33)

then0. =∇+

∂∂

ww f vt

S r

(4.34)

This important equation is a rewrite of the saturation equation in terms of fractional flow.

4.2 Buckley-Leverett Theory – One-Dimensional Flow

4.2.1 Equations of Motion

Consider linear one-dimensional flow in reservoir with constant dip angle, as in Figure

4.2. Approximately linear flow occurs in reservoirs with line-drive development patterns,

as in Figure 4.3.

q T

A, k, k r o

, k r w

x q T

A, k, k r o

, k r w

x

hθ

Figure 4.2: Linear flow in a reservoir with constant dip angle θ.



L L

Figure 4.3: Approximately linear flow occurs in reservoirs with line-drive development

patterns.

In this one-dimensional flow the pressure equation, Eq. (4.18), becomes:

0=

∂Φ∂

λ∂∂

x xT

(4.35)

Where Φ is the potential, for example:

gh p ooo ρ+=Φ (4.36)

Solving the pressure equation gives us the velocity as a function of x. Actually we do not

need to show how to calculate the pressure solution here, however you can easily see that

for single-phase flow we would have:

02

2

=∂

Φ∂ x , so, L

x∆Φ−Φ=Φ

0(4.37)

Clearly, for a two-phase problem λT will not be constant so the pressure distribution will

change with both position and time.

If we turn our attention now to the saturation equation, Eq. (4.34), for this one-dimensional problem the equation becomes:

0=∂∂

+∂

∂ x

f v

t

S ww

(4.38)

where the fractional flow, f w, is now given by:



θρ∆−∂∂λ

+λλ

= sin1 g x

P

q

A f c

T

o

T

ww

(4.39)

since dh/dx = sin θ.

Clearly the fractional flow, f w, is a function of saturation and dip angle:

( )θ= ,www S f f (4.40)

• For flow updip (θ>0) gravity retards the water flow.

• For flow downdip (θ<0) gravity increases the water flow.

• If qT /A is large, then the flow is viscous-dominated and capillary pressure and

gravity have less influence.

4.2.2 Capillary Pressure Gradient Terms

Consider a water/oil displacement front, as in Figure 4.4. Anticipating the solution a bit,

the saturation gradient is small everywhere except in the vicinity of the front.

S w

x

S wi

∆S wWater Oil

Lc

Figure 4.4: One-dimensional displacement of oil by water.

The capillary pressure gradient can be considered as:

x

S

S

P

x

P w

w

cc

∂∂

∂∂

=∂∂

(4.41)

Hence the capillary pressure gradient is only of importance in the vicinity of the frontalso. In the vicinity of a stable front, the shape will be determined by a balance between

the viscous and capillary forces:

capillaryviscous p p ∆∆ ~and c

viscous

L

p A

k q

∆µ

=

(4.42)

Hencecapillary

c pkA

Lq∆

µ~

or µ

∆

q

pkA L

capillary

c ~(4.43)



We can look at a local capillary-to-viscous number as the relative size of the region

where capillary forces are important:

Lq

pkA

L

L N

capillaryccv µ

∆==(4.44)

Notice that this is different from the capillary number we defined in Eq. 2.28 which is a

viscous-to-capillary number. When N cv << 1, we can neglect the small region where

capillary pressure gradient is important.

Example:

Consider a typical case in which:

q/A 1 ft/day

k 100 mdµ 1 cp = 6.72×10

-4lbm/ft-sec

∆ pc 3 psi

day ft

in

lbf

ft lbm

lbm

ft cp

md

ft

ft Lcpday ft

inlbf md N cv

sec360024144

sec

2.32

1072.6

sec110

))(1)(/1(

)/3)(100(2

2

24

2142 ×−×

−= −

−

(4.45)

L N cv

9.1=

(4.46)

In the laboratory, L is of order 1 foot so N cv is not small compared to 1 when q/A is of

order 1 ft/day. We would need to flow at 20 ft/day to make the capillary region small.

For core plug measurements, L is of order 0.1 ft so we would have to flow at 200 ft/day.

In the field L is of order 100 to 1000 ft so N cv is small when q/A is of 1 ft/day.

This is the justification for ignoring capillary pressure gradients in most applications of Buckley-Leverett calculations at the field scale, but we must remember that we cannot

ignore them at the lab scale.

A broader discussion of this point can be found in Lake, Section 5-3, which alsoconsiders the effects of capillary dissipation. In particular you should know about the

Rapoport and Leas number, which is similar in function to the N cv defined here (although

inverse in its ratio of capillary to viscous effects). The Rapoport and Leas number isdefined as:

θφσµ

φ=

cos12

0

1

1

2/1

r

RLk

uL

k N

(4.47)

where the subscript 1 refers to the displacing phase (water in a waterflood).



4.2.3 Fractional Flow with Gravity Only

Having decided that we can usually ignore the gradient of capillary pressure (recognizing

that this will be wrong in a small region around the displacement front), we can write:

θρ∆

λ+

λλ

=θ sin1),( g q

AS f

T

o

T

www

(4.48)

We can define the gravity number , N g , as a ratio of gravity vs. viscous forces:

T o

g q

g kA N

µρ∆

=(4.49)

We note that:

orw

wro

w

oT

w

k

k

µµ

+=

λλ

+=

λλ

1

1

1

1

(4.50)

Then o

w

rw

ro

ro g

ww

k

k

k N S f

µµ

+

θ−=θ

1

sin1),(

(4.51)

If we make use of Honarpour’s relation for relative permeability, Eq. (3.39), we canwrite:

w

o

o

n

w

n

w

n

wro g

ww

S M

S

S k N S f

)(

)1(1

sin)1(1),(

*0

*

*0

−+

θ−−=θ

(4.52)

where wior

wiww

S S

S S S

−−−

=1

*

(4.53)

and w

o

ro

rw

k

k M

µµ=

0

00

(4.54)

is the water/oil end-point mobility ratio.

Typical fractional flow curves for no = nw = 2 and S wi = S or = 0.2 are shown in Figure 4.5for horizontal flow at different mobility ratios, and in Figure 4.6 for updip and downdip

flow for mobility ratio M = 1.



0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

S w

w

1

0.1

10

M=

Figure 4.5: Fractional flow curve for θsin0

ro g k N =0 (horizontal flow).

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0 0.2 0.4 0.6 0.8 1

S w

f w0

3

-3

N g .k ro .sinθdowndip

updip

Figure 4.6: Fractional flow curve for 0=1 (unit mobility, inclined flow).

Notice from Figure 4.6 that it is entirely possible for f w to be greater than one or less than

zero. When these two situations occur, there is countercurrent flow of oil and water.

4.2.4 The Buckley-Leverett Solution

As we saw in Eq. (4.34),

0=∂∂

+∂

∂ x

f v

t

S ww

, where the interstitial velocity A

qv T

φ=

(4.55)

For constant dip angle θ, we can write this as:



0=∂

∂+

∂∂

x

S

dS

df v

t

S w

w

ww

(4.56)

If the medium is initially at saturation S wI at time t = 0, we start injecting water at x = 0 to

raise the saturation to S w0 (at x = 0) – this gives us the initial and boundary conditions:

Initial condition: wI w S xS =)0,( (4.57)

Boundary condition: 0),0( ww S t S = (4.58)

In practice, we control f w(0, t ) not S w(0, t ), so the boundary condition is:

[ ] [ ]00

),0(),0(wwwwww

S S f t S f t f ⇒==(4.59)

The practical problem of most interest is where the initial water saturation is S wi and we

inject 100% water:

or ww S S t f −=⇒= 11),0( 0 (4.60)

Initial condition: wiw S xS =)0,( (4.61)

Boundary condition: or w S t S −= 1),0( (4.62)

Note that since the equation is first order in distance, only one boundary condition isrequired and the solution is not dependent on system length, L.

It is convenient to introduce dimensionless variables as follows:

L

x x D =

(4.63)

∫∫∫ =

φ

==t

T

p

t

T

t

D dt qV

dt AL

q

L

dt vt

000

1

= pore volumes injected (4.64)

Since L was introduced artificially into the dimensionless variables, its influence will

always cancel out.

With this change of variables, the saturation equation becomes:

0=∂∂

+∂∂

D

w

w

w

D

w

x

S

dS

df

t

S

(4.65)



4.2.5 Method of Characteristics

The Buckley-Leverett equation, Eq. (4.65), is a first-order, hyperbolic equation. We can

solve hyperbolic equations conveniently by the Method of Characteristics.

Define a characteristic curve as a line in the x-t plane along which the value of S w may bedescribed by an ordinary (total) differential equation.

The total derivative of S w is:

D

D

t D

w

D

w

D

w

t

x

x

S

t

S

dt

dS

∂∂

∂∂

+∂∂

=(4.66)

We can note that the Buckley-Leverett equation, Eq. (4.65), is in the form of an advection

equation in which df w /dS w represents the velocity of propagation of a saturation front. If we define this velocity as v sw, then along lines for which:

w

w sw

D

D

dS

df v

t

x==

∂∂

(4.67)

the combination of Eqs. (4.65) and (4.66) shows that the saturation is described by:

0= D

w

dt

dS

(4.68)

Which is to say S w is constant. This means that constant values of S w propagate withconstant velocity and that we can define our characteristic lines in x D – t D space from Eq.

(4.67).

Let us now talk about characteristic diagrams, which for the problem of interest are

contours of constant S w plotted on an x D – t D plot.

In order to understand the plot itself first, consider the easier problem of mainly oilinjection into a reservoir that is filled 50% with water. The fractional flow and its

gradient are shown in Figure 4.7.



0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

S w

f w

0

0.4

0.8

1.2

1.6

2

2.4

2.8

3.2

3.6

4

d f w / d S w

Figure 4.7: Fractional flow curve for example problem, including gradient

At the initial condition, S w is 0.5, so df w /dS w is 3.2. At the inlet boundary, largely oil isinjected such that f w is 0.008, so S w is 0.3, and df w /dS w is 0.8. Therefore on the x D – t D

plot the slope of the lines dx D /dt D are 3.2 on the t D=0 line and 0.1 on the x D=0 line, as in

the characteristic diagram shown in Figure 4.8. Usually we draw the characteristic

diagram with time on the vertical axis (opposite to Figure 4.8), it is drawn with time onthe horizontal axis here to emphasize the correspondence between velocity and slope of

the x D vs. t D lines.

We can then construct the spatial solution for a specific value of t D, or the time solutionfor a specific value of x D, as in Figure 4.9, by reading off the saturation values

represented by the characteristic lines crossed.



0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.05 0.1 0.15 0.2

t D

x D

V sw =df w /dS w0.5

0.5

0.5

0.30.30.3

0.5

0.4

0.35

Figure 4.8: Characteristic diagram for oil displacing water

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.05 0.1 0.15 0.2

t D

x D

V sw =df w /dS w0.5

0.5

0.5

0.30.30.3

0.5

0.4

0.35

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.55

0 0.05 0.1 0.15 0.2

t D

S

w

xD=0.01

xD=0.1

tD=0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6

Sw

x D

Figure 4.9: Construction of solution from characteristic diagram.



In this first example, the velocities of successive saturation values decreased with time

(for a specific value of x D). This results in a spreading wave. If the velocities increase

with time then the characteristic lines run into each other and form a shock , as in Figure4.10. Along the shock characteristic, the saturation is multivalued as shown in Figure

4.11. This is commonly the situation when water displaces oil, because of the shape of

the fractional flow curve.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.05 0.1 0.15 0.2

t D

x D

Shock

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.05 0.1 0.15 0.2

t D

x D

Shock

Figure 4.10: Characteristic diagram for water displacing oil.

S w

x

S w

x

Figure 4.11: Higher velocities for upstream values steepen the front and create a shock.

4.2.6 Shock VelocitiesConsider a small element around a shock propagating at constant velocity v sh, as in Figure4.12.

v sh

2 1

Figure 4.12: Control volume in the vicinity of a propagating shock.



We will write the conservation equations in a reference frame moving with the shock at

speed v sh. This changes the transient problem into a steady state problem. In the steady

state we can write, flow rate in = flow rate out :

1122 w shwT w shwT AS v f q AS v f q φ−=φ− (4.69)

( ) ( )1212 ww shwwT S S Av f f q −φ=− (4.70)

( )( ) w

w

w

wT

ww

wwT sh

S

f v

S

f

A

q

S S

f f

A

qv

∆∆

=∆∆

φ

=−−

φ

=12

12

(4.71)

We can define the dimensionless shock velocity as:

v

v

S

f v sh

w

w shD =

∆∆

=(4.72)

The chord connecting two points on the fractional flow curve gives this velocity, as in

Figure 4.13.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

S w

f w

∆ f w

∆S w0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

S w

f w

∆ f w

∆S w

Figure 4.13: Dimensionless shock velocity is a function of upstream and downstreamconditions.

4.2.7 Welge Construction

This method provides a way to construct the Buckley-Leverett solution based on the

shock velocities derived from the fractional flow curve. This is useful as a way tovisualize how the solution behaves and what causes the shocks to appear the way they do.

Let us consider the more interesting problem of water displacing oil, in which we have areservoir initially filled with oil such that S w = S wi and f w = 0. We displace with water

such that S w = 1 – S or and f w = 1. Consider changing saturation in increments as in Figure

4.14. Saturations less that S wf move faster than the front and therefore catch up to it,



thereby making the jump in saturation bigger. Saturations greater than S wf move slower

than the front and therefore get left behind.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

S w

f w

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

S w

f w

wf S w

w

w

w

S

f

dS

df

∆∆

=

S wf 1-S or S wi

wf S w

w

w

w

S

f

dS

df

∆∆<

wf S w

w

w

w

S

f

dS

df

∆∆

> so each new S w

catches up withfront

Velocity

increases

with S w

Figure 4.14: Velocity increases with saturation up to S wf , then decreases.

The last saturation that can catch up to the shock front is S wf , the velocity of which is

represented by the tangent of the fractional flow curve.

The saturation distribution can be computed both at and behind the shock using the same

Method of Characteristics approach, as in Figure 4.15. Notice that here the characteristicdiagram is plotted in the more conventional manner with x D on the horizontal axis.

Breakthrough occurs when:

10 =∆∆

D

w

w t S

f

(i.e. when x = L, x D = 1) (4.73)

or, wiwf

wiwf

w

w D

f f

S S

f

S t

−

−=

∆∆

=0

(4.74)



tD=0.25

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

x D

S w

tD=0.25

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

x D

S w

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0 0.2 0.4 0.6 0.8 1

x D

t D

Figure 4.15: Buckley-Leverett solution for water injection into an oil-filled medium.

4.2.8 Classification of Waves

Changes of saturation and fractional flow propagate as waves. The wave velocity is not

the same as the particle velocity. The particle velocity is given by:

o

oT o p

AS

f qv

φ=

,

(4.75)

w

w

o

o

T

o p

oD pS

f

S

f

v

vv

−−

===1

1,

,

(4.76)

Similarly, w

wwD p

S

f v =,

(4.77)



These can be shown graphically by drawing lines from the origins (0, 0) and (1, 1), as in

Figure 4.16. We will come back to consider the interrelationship between particlevelocities and saturation curves later.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

S w

f w

v po,D

v pw,D

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

S w

f w

v po,D

v pw,D

Figure 4.16: Particle velocities for oil and water.

Given that wave velocities are not the same as particle velocities, we can characterize

waves in terms of the nature of their (wave) velocities:

1. A wave that becomes more diffuse (a sequence of decreasing velocities behind the

wave front) is called a spreading wave, or a rarefaction.2. A wave that becomes less diffuse (a sequence of increasing velocities behind the

wave front) is called a sharpening wave. These will eventually become shocks.

3. A wave that has both spreading and sharpening character is called mixed . TheBuckley-Leverett solution as in Figure 4.15 is a good example.

4. A wave that neither spreads nor sharpens on propagation is known as indifferent . In

the absence of dispersion such waves propagate as shocks. With dispersion, the wave

spreads although not as rapidly as a spreading wave.

x D

S w

x D1 x D

S w

x D1

Figure 4.17: Average saturation calculation.



4.2.9 Average Saturations

For any position x D1 behind the Buckley-Leverett shock front (i.e. on a continuous

spreading wave), Figure 4.17, the average saturation behind at given time t D is:

∫=1

01

1)(ˆ

D x

Dw

D

Dw dxS x

t S

(4.78)

Integrating by parts:

−= ∫

−

1

1

1

01

1)(ˆ

w

or

D

S

S

w D

x

w D

D

Dw dS xS x x

t S

(4.79)

Since x D1 is in the spreading part of the wave, we can write:

Dw D t f x '= (4.80)

So,∫

−

−=1

11

1 '1

)(ˆw

or

S

S

ww D

D

w Dw dS f t x

S t S

(4.81)

∫−

−=1

11

1

w

or

S

S

w

w

w

D

Dw dS

dS

df

x

t S

(integration at constant t D) (4.82)

)1()(ˆ1

1

1−−=

w

D

Dw Dw f

x

t S t S

(4.83)

Or, 1

11

'

)1()(ˆ

w

ww Dw

f

f S t S

−−=

(4.84)

Rearranging,1

1

1 ')(ˆ

1w

w Dw

w f S t S

f =

−

−

(4.85)

This shows that the average saturation is given by the intersection of the tangent to thefractional flow curve to the axis S w = 1, as in Figure 4.18.

Note that the average saturation upstream of a given value of S w1 does not change in time,even though the location x D1 does change. The entire saturation profile grows linearly, so

the fraction of the profile that a given saturation occupies remains the same.



0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

S w

f w

S wf S w1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

S w

f w

S wf S w1

f w1

S w1

1- f w1

1)(ˆw Dw S t S −

)(ˆ Dw t S

f’ w1

1

Figure 4.18: Average saturation determined from intercept with f w = 1.

4.2.10 Oil Recovery Calculations

In common core analysis in the laboratory, we measure:

• Cumulative oil recovery, N p(t D)

• Producing water cut, f we(t D)

We can use this information to construct the fractional flow curve as a function of saturation, as shown in the following.

We can measure the pore volume, V p = φ AL.

We measure the original oil in place (OOIP), S oI = V p (1 – S wi), hence we can obtain S wi.

From measurements of N p we can obtain the average water saturation:

p

p

pD pDwiwV

N N where N S S =+= ,

(4.86)

If we consider x D1 = 1 (i.e. x = L, the outlet end of the core), then S w1 = S we.

We know that after breakthrough ( x D1 < x Df ):

)1('

1we Dwe

we

wewew f t S

f

f S S −+=

−+=

(4.87)

So, )1( we Dwwe f t S S −−= (4.88)



The terms on the right hand side are obtainable directly from the measurements of

N pD(t D) and f we(t D). In particular, the average water saturation behind the shock front is

given by:

wf

wf w

wf f

S S

f '

1=

−

−

(4.89)

As seen in Figure 4.19, this slope is also equal to:

wf

wiw

f S S

'1

=− (4.90)

In other words, the extension of the tangent to f w = 1 gives the average saturation behind

the front and the average saturation at breakthrough.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

S w

f w

S wf

S wi

1- f wf

wf w S S −

wS

Figure 4.19: Average saturation determined from intercept with f w = 1.

Prior to breakthrough the average saturation is given by:

AL

t q ALS S T wi

w φ+φ

=)(

(4.91)

So, Dwiw t S S += (4.92)

Since all the oil produced has been replaced by water, the average saturation can be

related to the cumulative oil production, N p:



p

p

wiwV

N S S +=

(4.93)

At breakthrough:

wf

BT D f

t '

1, =

(4.94)

After breakthrough the oil recovery can be related to the saturation at the producing end,S we, as in Figure 4.20.

x D

S w

1

S wf

S we t D>t D,BT

Figure 4.20: Saturation determined at outlet, x D = 1.

Previously, Eq. (4.85), we showed that the average saturation upstream any saturationthat lies behind the shock front is:

1

1

1 'ˆ

1w

ww

w f S S

f =

−

−

(4.95)

Letting S w1 = S we, then:

we

wewew

f

f S S

'

1−+=

(4.96)

and wiw pD S S N −= (4.97)

Remembering also that when x D = 1, then t D = 1 / f’ w, we can then construct the entire

solution graphically, determining the S we vs. t D history and N pD vs. t D history, as in Figure4.21. We can also do the reverse and calculate f w vs. S w once we measure S we and N p.



N pD

t Dt D,BT =1/ f’ wf

1-S or -S wi

BTwiw S S −= wS from extension of slope

from S we

t Dt D,BT =1/ f’ wf

S wi

1-S or

S wf S we

22 '/1 w D f t =

Figure 4.21: Saturation history determined at outlet, x D = 1, and cumulative production

history.

4.2.11 Effect of Mobility on RecoveryAs we saw earlier, the mobility ratio affects the fractional flow curve – hence the ultimate

oil recovery is also a function of the mobility ratio.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

S w

w

1

0.1

10

M=

Figure 4.22: Fractional flow curves for horizontal displacement at different mobility

ratios (same as Figure 4.5).



High (Unfavorable) Mobility Ratio

In this case the breakthrough is early and a lot of oil is left behind the front (in fact we

may not even have a front in some cases). For example, when M =10 in Figure 4.22, theshock velocity is around 3.6 – this places the shock at a value of x D/t D = 3.6 (notice thatwhen t D is 0.25, the front is at x D=0.9 in Figure 4.23), and the breakthrough at t D = 1/3.6 =

0.28 as seen in Figure 4.24.

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0 0.2 0.4 0.6 0.8 1

x D

t D

tD=0.25

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

x D

w

Figure 4.23: Characteristic diagram and saturation distribution for unfavorablemobility ratio (M=10 from Figure 4.22).

N pD

t D0.28

1-S or -S wi

BT

Figure 4.24: Recovery curve for unfavorable mobility ratio (M=10 from Figure 4.22).

Modest Mobility RatioWith a unit mobility ratio, the breakthrough is delayed and the sweep efficiency is

greater. For the mobility ratio 1 in Figure 4.22, the shock velocity is 2, – this places the

shock at a value of x D/t D = 2 (notice that when t D is 0.25, the front is at x D=0.5 as shown

earlier in Figure 4.15), and the breakthrough at t D = 1/2 = 0.5 as seen in Figure 4.25.



N pD

t D0.28

1-S or -S wi

BT

0.5

Figure 4.25: Recovery curve for unit mobility ratio (M=1 from Figure 4.22).

Low (Favorable) Mobility Ratio

With low mobility ratio we have a late breakthrough and the oil is swept almost

completely. For example, when M =0.1 in Figure 4.22, the shock velocity is around 1.66

– this places the shock at a value of x D/t D = 1.66 (notice that when t D is 0.25, the front isat x D=0.416 in Figure 4.26), and the breakthrough at t D = 1/1.66 = 0.6 as seen in Figure

4.27. Note that the maximum possible shock velocity for this example (in which S or = S wi

= 0.2) is given by:

6667.12.02.01

1

1

1=

−−=

−−=

wior

shS S

v(4.98)

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0 0.2 0.4 0.6 0.8 1

x D

t D

tD=0.25

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

x D

w

2

Figure 4.26: Characteristic diagram and saturation distribution for unfavorable

mobility ratio (M=0.1 from Figure 4.22).



N pD

t D

1-S or -S wi

BT

0.6

Figure 4.27: Recovery curve for unit mobility ratio (M=0.1 from Figure 4.22).

Note that the entire saturation change at the outlet occurs at once, and there is no

fractional flow information obtainable from the production history. If we wanted to

obtain relative permeability information by matching the recovery as a function of time,we would obtain only two values (the end-points) and nothing in between.

4.2.12 Effect of Gravity on Recovery

As we saw in Section 4.2.3, gravity affects the fractional flow relationship, through Eq.

(4.51), repeated here:

o

w

rw

ro

ro g

ww

k

k

k N S f

µ

µ+

θ−=θ

1

sin1),(

(4.99)

The consequence of this is that oil recovery will be different for different inclination

angle θ.

Flooding Updip ( N g sinθ > 0)

As can be seen in Figure 4.28, the effect of flooding updip is to increase the breakthrough

water saturation S BT , and to reduce the velocity of the front v shD. This results in more oil being displaced at breakthrough, although breakthrough occurs later (Figure 4.29). In

many cases this would be an attractive process, since displacement efficiency is poor

after breakthrough.



0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0 0.2 0.4 0.6 0.8 1

S w

f w0

1.5

N g .k ro .sinθ

updip

Figure 4.28: Fractional flow curves for horizontal and updip displacement (M=1).

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

x D

S

Horizontal

Updip

Figure 4.29: Saturation distribution for horizontal and updip displacement (M=1).

Flooding Downdip ( N g sinθ < 0)

As can be seen in Figure 4.30, the effect of flooding downdip is to decrease the

breakthrough water saturation S BT , and to increase the velocity of the front v shD. Thisresults in less oil being displaced at breakthrough, although breakthrough occurs sooner

(Figure 4.31). In many cases this would not be an attractive process, since displacement

efficiency is poor after breakthrough.



0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0 0.2 0.4 0.6 0.8 1

S w

f w0

-1.5

N g .k ro .sinθdowndip

Figure 4.30: Fractional flow curves for horizontal and downdip displacement (M=1).

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

x D

S w

Horizontal

Downdip

Figure 4.31: Saturation distribution for horizontal and downdip displacement (M=1).

In a steeply downdipping case in which N g sinθ << 0, the fractional flow curve might

look as in Figure 4.32. In this case the fractional flow exceeds one, so that:

0,;1)( <>+

= o

wo

www u so

uu

uS f

(4.100)

This means that the oil flows updip while the water is flowing downdip, in a process

called countercurrent flow. There is no way to push the oil ahead of water for S w > S wm in Figure 4.32. Hence the overall oil recovery is limited.



0

0.2

0.4

0.6

0.8

1

1.2

0 0.2 0.4 0.6 0.8 1

S w

f w0

-3

N g .k ro .sin◊downdip

horizontal

S wm

Figure 4.32: Fractional flow curves for strongly downdip displacement (M=1).

4.3 Derivation of Relative Permeability from Displacements

In Section 3.1.6 we talked about steady-state methods of measuring relative permeability

in the laboratory. Now that we have learned about Buckley-Leverett displacements, weare ready to talk about unsteady-state experiments.

The main reason to consider unsteady experiments is that steady-state experiments can beextremely time-consuming. This is because achievement of a stable and uniform

saturation distribution takes a long time, especially in a low permeability core. If youthink about the Buckley-Leverett displacement process, you will realize that you would

need to flow several pore volumes to achieve a steady saturation distribution, such asthose shown earlier in Figure 3.11. In practical cases, this might require days, weeks or

even months. Hence the interest in unsteady state methods.

We have seen in Section 4.2.11 that the output oil (and water) production characteristics

are a function of the fractional flow curve f w vs. S w, and hence are also a function of therelative permeability curves. The essence of unsteady-state methods is to estimate therelative permeability curves based on the measured oil and water production histories.

4.3.1 JBN Method This method is due to Johnson, E.F., Bossler, D.P., and Naumann, V.O.: “Calculation of

Relative Permeability from Displacement Experiments”, Transactions AIME (1959), 216,

107-116.

In the laboratory, we measure the output water and oil flow rates to determine:

orw

wrowo

ww

k

k uu

u f

µ

µ +

=+

=1

1

(4.101)



So,1=+

orw

wroww

k

k f f

µ and we

oe

we

we

o

w

rw

ro

f

f

f

f

k

k =

−=

1

µ

µ

(4.102)

This would give us a way to estimate k ro/k rw. We then need a way to determine one of the

two relative permeabilities independently. We note that during the displacement the pressure drop will be a function of the mobility along the core, which we can determine

by making some kind of assumption about the displacement process (for example, that it

is governed by Buckley-Leverett theory).

At any instant during the displacement:

x

p A

k k f qq

o

rooT o ∂

∂−==

µ

(4.103)

∫ −=∆+=∆ L

oro

ooT L p p pwheredxk

f

kA

q p )()0(;

(4.104)

So,∫ ==∆

1

/;o

D D

ro

ooT L x xwheredxk

f

kA

Lq p

(4.105)

Now at time t D, from Buckley-Leverett theory, Eq. (4.67), we can write:

we

w

Dwe

Dw D f

f

t f

t f

L

x x

'

'

'

'===

andw

we

D df f

dx ''

1=

(4.106)

So,∫=∆

we

w

f

f w

ro

o

we

oT df k

f

kAf

Lq p

'

''

' (4.107)

Instead of making detailed record of all the variables, we can just denote the “intake

capacity” as qT /∆ p.

At the start of the displacement:

p L

Ak qq

o

oT ∆==µ if 1)( =wiro S k (4.108)

Defining a “relative injectivity” as:

( )( ) Ak p

Lq

pq

pq I oT

sT

T R ∆

=∆∆

=/

/

(4.109)



Then, R

we f

f w

ro

o

I

f df

k

f we

w

''

'

'=∫

(4.110)

Now from basic calculus we know that:

)()( q f dx x f dq

d q

p=∫

for p constant (4.111)

So,

ro

o

D

R D

R

we

we f ro

o

k

f

t d

I t d

I

f

df

d

k

f

we

=

=

=

1

1

'

''

at the outlet end (4.112)

Thus for a given t D, the slope of the line 1/t D I R vs. 1/tD gives f o/k ro at the outlet.

The saturation at the outlet is given by Eq. (4.88), or:

oe Dwwe f t S S −= (4.113)

The overall “recipe” for the JBN method is therefore as follows:

1. Measure:

( )( )

sT

T R

pq pq I

∆∆=

//

relative injectivity

L A

t qt T D

φ =

pore volumes injected

)(1)( Dwe Doe t f t f −= fractional flow of oil in effluent

p

p

V

N

pore volumes of oil produced2. Calculate:

p

pwiw

V

N S S +=

from material balance

oe Dwwe f t S S −= from Welge construction

=

D

R D

ero

o

t d

I t d

k

f

1

1

from plot of 1/t D I R vs. 1/tD



This gives )(

)(

wio

woro

S k

S k k =

, so implies that 1)( =wio S k .

Knowing k ro, we can use Eq. (4.101) to find k ro/k rw and hence k rw:

=

−

=o

wro

o

w

w

wro

o

wrw

f

f k

f

f k k

µ

µ

µ

µ

1

Notice the implicit assumption that 1)( =wio S k . If this is not the case, we could measure

the oil permeability at immobile water saturation during core preparation.

The JBN method is quite commonly applied, but it is important to note that there are

some important limitations on its use. The calculated relative permeability numbers are

only available at saturation values between the breakthrough saturation (the shock frontsaturation) and the water saturation at residual oil. This could be a very small range in

many cases, especially in the case of unit mobility or favorable mobility ratio (see Figure4.5). The method also assumes that the exit saturation is in fact given by Buckley-Leverett theory and therefore ignores the capillary end effect (Section 3.1.5). Finally, the

method assumes a sharp-fronted displacement, and therefore would not work well when

capillary pressure smears the front.

Hence the JBN method works best when:1. Capillary effects are negligible (rapid flow).2. Mobility ratio is unfavorable.

4.3.2 Jones and Roszelle Method

Another variant of the unsteady-state displacement approach to relative permeabilitymeasurement is due to Jones S.C., and Roszelle, W.O., “Graphical Techniques for

Determining Relative Permeability from Displacement Experiments,” J. Pet. Tech., (May1978), 807-817. The calculation procedure is different, but still depends on the Buckley-

Leverett solutions, and therefore works under the same conditions and constraints as the

JBN method. We will not go into more detail in this course.

4.4 Gravity Segregation

Another important flow problem of practical interest is gravity segregation. Having

understood Buckley-Leverett displacement theory, we can apply it to these new situations

as well. We will look at two cases. The first one is considered as a one-dimensional(vertical) flow problem in which an unstable oil-water column rearranges itself due to

gravity. In the second problem we will look more closely at the stability of a displacing

front in two-dimensions.

4.4.1 Vertical Segregation

Consider the case we talked about earlier in Section 2.4.7 in which rapid displacement

brings an unstable vertical arrangement of fluids with water overlying oil in adjacent(communicating) layers, as shown in again in Figure 4.33. Previously we were interested

in the phenomenon of capillary cross-flow, which would occur whether the oil layer was

on the top or underneath. Here we are interested in a different phenomenon, in which the



oil wants to be in the upper layer because of its lower density. Due to the effects of

gravity alone, the state in which water fills the upper layer is not in vertical equilibrium

(VE).

Water inOil

Oil

Oil

Figure 4.33: Unstable vertical segregation due to faster flow in the upper layer.

This problem is truly two-dimensional, however we can learn much about its behavior bylooking at a simplified one-dimensional problem, as in Figure 4.34.

Water at 1 - S or

Oil at S wc

S w

z

+H

-H

0

S wc 1-S or

Figure 4.34: Gravity segregation problem simplified to one dimension.

With no flow in the horizontal direction, we can write the vertical velocities, v, of oil and

water from Darcy’s law and relative permeability as:

+

∂

∂−= g

z

pk k v w

w

w

rww ρ

µ (4.114)

+

∂∂

−= g z

pk k v o

o

o

roo ρ

µ (4.115)

Conservation of mass requires that:

0=∂

∂+

∂∂

z

v

t

S wwφ (4.116)



0=∂∂

+∂

∂ z

v

t

S ooφ (4.117)

and 1=+ ow S S (4.118)

For thick layers we can ignore the effects of capillary pressure (notice that we are thereby

discounting the capillary cross-flow phenomenon considered in Section 2.4.7), hence:

( ) 0=−∂∂

=∂∂

woc p p

z z

P

(4.119)

Combining the continuity equations, Eqs. (4.116), (4.117) and (4.118):

( ) ( ) const vvhence z

vv

t

S S ow

owow =+=∂+∂+

∂+∂ ;0φ

(4.120)

Our boundary conditions are that vo = vw = 0 at top and bottom boundaries, so in Eq.(4.120) we can infer that the constant must be zero, and hence:

woow vvand vv −==+ ;0 (4.121)

Rearranging the pressure equations:

g k k v

z p w

rw

www ρ +=∂∂−(4.122)

g k k

v

z

po

ro

ooo ρ +=∂∂

−(4.123)

So,

( )( ) 0=−+−=

∂−∂

ow

ro

oo

rw

wwwo g k k

v

k k

v

z

p p ρ ρ

, since0=

∂∂ z

P c

(4.124)

Now since that vo = - vw :

( )ow

ro

o

rw

wo g

k k k k v ρ ρ

µ µ −=

+

(4.125)

( ) ( )

+

×−

=

+

−=

o

w

rw

ro

ro

o

ow

ro

o

rw

w

owo

k

k

k k g

k k k k

g v

µ

µ µ

ρ ρ

µ µ

ρ ρ

1

(4.126)



( )

+

=−

=

o

w

rw

ro

rooo

o

owo

k

k

k S F S F

k g v

µ

µ µ

ρ ρ

1

)(),(

(4.127)

Hence, from Eq. (4.117):

( )0=

∂∂

∂∂−

+∂

∂=

∂∂

+∂

∂ z

S

S

F k g

t

S

z

v

t

S o

oo

owooo

µ

ρ ρ φ φ

(4.128)

( )0)(' =

∂∂−

+∂

∂ z

S S F

k g

t

S oo

o

owo

φµ

ρ ρ

(4.129)

Introducing the dimensionless time and distance:

( )τ φµ

ρ ρ t

H

k g t t

o

ow D

=−

=(4.130)

H

z z D =

(4.131)

where the variable τ is a characteristic time scale for gravity segregation:

( )k g H

ow

o

ρ ρ φ τ

−=

(4.132)

Then finally we can write the governing equation as:

0)(' =∂∂

+∂∂

D

oo

D

o

z

S S F

t

S

(4.133)

For typical relative permeabilities, the F(So) curve might look as in Figure 4.35.



0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

0 0.2 0.4 0.6 0.8 1

S o

F ( S

1

A

C

B

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

0 0.2 0.4 0.6 0.8 1

S o

F ( S

1

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

0 0.2 0.4 0.6 0.8 1

S o

F ( S

1

A

C

B

Figure 4.35: Example of F(S) function (Honarpour relative permeability curves with no =nw = 2, and S wi = S or = 0.2, end-points 1 [oil] and 0.6 [water]).

In Figure 4.35, saturations to the right of point C will move positive velocity, and those tothe left will move with negative velocity. Saturations at point C will not move. Given

the initial saturation distribution as in Figure 4.36, we see that one saturation at theinterface will remain constant and that two shocks will propagate, one upwards with

shock-front saturation S oA and one moving downwards with shock-front velocity S oB.This will give rise to a characteristic diagram as in Figure 4.37. This is much like theBuckley-Leverett problem, only a little more complex.

S o

z D

+1

-1

0

1-S wcS or

C

S oC S o

z D

+1

-1

0

1-S wcS or

C

S oC

Figure 4.36: Oil saturation as a function of height, with shock velocities up and down.



t D

z D

+1

-1

0S oC

S oA

S oB

t D1 t D

z D

+1

-1

0S oC

S oA

S oB

t D1

Figure 4.37: Characteristic diagram for the gravity segregation problem.

Considering a time t D1, which precedes the arrival of either shock at the boundaries, we

can use the characteristic diagram to construct the saturation profile, as in Figure 4.38.

The oil shock moves upwards and the water shock moves downwards. Notice that the

saturation S oC does not move.

S o

z D

+1

-1

0

1-S wcS or S oC

S oA S oB

F’ (S o)=0

S o

z D

+1

-1

0

1-S wcS or S oC

S oA S oB

S o

z D

+1

-1

0

1-S wcS or S oC

S oA S oB

F’ (S o)=0

Figure 4.38: Oil saturation as a function of height, at time t D1 , from intersections with the

characteristics in Figure 4.37.

Once the shocks arrive at top and bottom of the region, an interesting situation arises.For example, as soon as S oB arrives at z D = -1, the oil saturation must go to S or in order to

satisfy vo = 0. Similarly, as soon as S oA arrives at z D = +1, the oil saturation must go to 1-

S wc in order to satisfy vw = 0. This creates a new discontinuity in saturation at the boundaries, and sets of shocks moving in the opposite direction. The tangent

construction, as shown in Figure 4.40, now goes from S oB ahead of the shock to S oD , with

S oD - S or traveling as a spreading wave, as in Figure 4.41.



t D

z D

+1

-1

0S oC

S oA

S oB

t D1

S oE

S oD

t D

z D

+1

-1

0S oC

S oA

S oB

t D1

S oE

S oD

Figure 4.39: Characteristics following arrival of the shocks at the top and bottom. S oD and S oE defined as in Figure 4.40.

0

0.01

0.02

0.030.04

0.05

0.06

0.07

0.08

0.09

0.1

0 0.2 0.4 0.6 0.8 1

S o

F ( S

1

0

0.01

0.02

0.030.04

0.05

0.06

0.07

0.08

0.09

0.1

0 0.2 0.4 0.6 0.8 1

S o

F ( S

1

A

C

B

D E

Figure 4.40: Shock construction on the F(S) diagram after shock arrival at top and

bottom.



S o

z D

+1

-1

0

1-S wcS or S oC

S oA S oB

S oD S oE

oil shock

water shock

S o

z D

+1

-1

0

1-S wcS or S oC

S oA S oB

S oD S oE

oil shock

water shock

Figure 4.41: Reverse shock directions and resulting saturation profiles.

As S oD and S oE propagate into the nonuniform saturation profile left behind the original

shocks, the tangent construction changes and the shocks speed up, as in Figure 4.42. Thiscauses the characteristic lines to become curves, as in Figure 4.43.

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

0 0.2 0.4 0.6 0.8 1

S o

F ( S

1

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

0 0.2 0.4 0.6 0.8 1

S o

F ( S

1

A

C

B

D E

Figure 4.42: As SoD and SoE propagate into the nonuniform saturation profile left behind

the original shocks, the tangent construction changes and the shocks speed up.



t D

z D

+1

-1

0S oC

S oA

S oB

t D1

S oE

S oD

t D

z D

+1

-1

0S oC

S oA

S oB

t D1

S oE

S oD

Figure 4.43: Characteristic lines curve as the shocks speed up.

The final state is complete segregation of the oil and water, as in Figure 4.44.

Note that the tangent constructions all gave velocities in terms of a characteristic velocity

group v g :

( ))(')(' o g o

o

owS S F vS F

k g v

o=

−=

φµ

ρ ρ

(4.134)

The time for gravity segregation to occur is of order:

( )k g

H

v

H

ow

o

g

g ρ ρ

φ τ

−=~

(4.135)

S o

z D

+1

-1

0

1-S wcS or

oil

water

S o

z D

+1

-1

0

1-S wcS or

oil

water

Figure 4.44: Final saturation as a function of height, with oil and water fully segregated.

If we superimpose a mean flow in the horizontal direction, the saturations will propagate

with velocity vT = qT / φ A. The relative importance of gravity segregation in a flow



depends on the viscous/gravity ratio, which compares the convective flow time to the

gravity segregation time.

( ) ( )

−=

−==

L

H

k g

v

v

L

k g

H N

ow

oT

T ow

o

c

g

GV ρ ρ

φµ

ρ ρ

φµ

τ

τ /

(4.136)

When N V/G >> 1 there is little gravity segregation (the gravity effect is slow) and when

N V/G << 1 the flow will segregate before it reaches x = L. Note that this ratio is different

than one based on forces:

∆=

∆=

∆∆

= H

L

k g

v

H g

k Lv

p

p N oT oT

g

vGV

ρ

µ

ρ

µ //

(4.137)

4.4.2 Displacement Under Segregated Flow Conditions

[Dake 372-383, Lake 214-216]

Up to this point we have considered only one-dimensional displacements. However it is

clear in real flow situations that the effects of multidimensional flow may be important.So in this section we will look at the problem of water-oil displacement under segregated

flow conditions, taking into account that the flow may not be just one-dimensional. We

can consider a displacement such as in Figure 4.45.

x θ

H

h

o i l

S w = S w i

S o = 1 - S w i

W a t e r

S w = 1 -

S o r

S o = S o r z

Figure 4.45: Water-oil displacement under segregated flow conditions.

For simplicity, we will assume that only a single phase flows in each region, with a shock from S wi to 1 – S or . At the shock front ∆S w = 1 – S wi - S or .

The relative permeabilities ahead of the front are:

)(0

wiroro S k k = and 0=rwk

The relative permeabilities behind the front are:

)1(0

or rwrw S k k −= and 0=rok

The zero superscript emphasizes the end-point value of the relative permeability.



Depending on the flow rate, mobility ratio and density difference, the flow may be either

stable or unstable. The condition for stability is that the gravity and viscous forces are in

balance. The interface is stable when the interface slope is constant, as in Figure 4.46.

=−== β tandx

dh

dx

dz

constant (4.138)

x θ

o i l

w a t e r

β

x θ

o i l

w a t e r

β

Figure 4.46: Stable displacements: (left) tan β < tan θ ; (right) tan β > tan θ .

Instability occurs when the water tongue underruns the oil and never establishes a stable

interface, as in Figure 4.47.

xθ

o i l

w a t e r

Figure 4.47: Unstable displacements as tan β Æ 0.

If the interface is stable (β is constant) and if only oil flows on one side and only water flows on the other, then at the interface:

t wo uuu == (4.139)

For a point on the interface:

+

∂∂

−=

+

∂∂

−== θ ρ λ θ ρ µ

sinsin0

g x

p g

x

pk k uu o

ooo

o

o

rot o

(4.140)

+

∂∂

−=

+

∂∂

−== θ ρ λ θ ρ µ

sinsin0

g x

p g

x

pk k uu w

www

w

w

rwt w

(4.141)

The assumption of vertical equilibrium requires that:



( )n g pnn

www ρ +

∂∂

=∂Φ∂

(4.142)

where n is the vertical direction (pointing upwards): n = z cos θ.

Hence:

1C n g p ww =+ ρ (4.143)

Denoting pw( x) as the water pressure at z = 0 (also equal to C 1), we can therefore write:

θ ρ cos)(),( z g x p z x p www −= (4.144)

Similarly we can write:

θ ρ ρ cos),(2 z g z x pC n g p oooo +==+ (4.145)

At the oil/water interface we can evaluate C 2 as:

θ ρ cos),(2 h g h x pC oo += (4.146)

We can evaluate po( x, h) knowing the capillary pressure P c for a given saturation

difference ∆S w:

cwwcwo P gh x p P h x ph x p +−=+= θ ρ cos)(),(),( (4.147)

θ ρ θ ρ coscos)(2 gh P gh x pC ocww ++−= (4.148)

[ ] θ ρ θ ρ ρ coscos)(),( gz gh P x p z x p oowcwo −−−+= (4.149)

Comparing Eqs. (4.144) and (4.149), we can see that the phase pressure differences can

be represented as a pseudocapillary pressure function of the form:

θ ρ cos gh P P cc

∆−=(4.150)

[Note that Dake describes a similar pseudocapillary pressure, but leaves out the real

capillary pressure, P c.]

Along the direction of the bed, x, we can write:

{

0

cos)(),(

→

−=∂

∂dx

dz g x

dx

dp

x

z x pw

ww θ ρ

(4.151)



So, dx

dp

x

z x p ww =∂

∂ ),(

(4.152)

and{

[ ]{

00

coscos),(

→→

−−−∂∂

+=∂

∂dx

dz g

dx

dh g

x

P

dx

dp

x

z x poow

cwo θ ρ θ ρ ρ

(4.153)

dx

dh g

dx

dp

x

z x p wo θ ρ cos),(

∆−=∂

∂

(4.154)

Along the interface:

+∂

∂

−= θ ρ λ sin g x

p

u o

o

oo(4.155)

+∆−−= θ ρ θ ρ λ sincos g

dx

dh g

dx

dpu o

woo

(4.156)

+

∂∂

−= θ ρ λ sin g x

pu w

www

(4.157)

Subtracting these last two equations, and replacing uo and uo by ut :

θ ρ θ ρ λ λ

sincos11

g dx

dh g u

wo

t ∆+∆=

−

(4.158)

+∆=

− θ θ ρ λ

λ

λ sincos1

dx

dh g u w

o

wt

(4.159)

+

∆=

− 1

tan

1sin1

θ

θ ρ λ

λ

λ

dx

dh

u

g

t

w

o

w

(4.160)

or,( )

+=− 1

tan

11

θ dx

dhG M

(4.161)

where, wro

orw

o

w

k

k M

µ

µ

λ

λ 0

0

==(4.162)

and, t

w

q

gAG

θ ρ λ sin∆=

(4.163)



Solving for the interface slope:

θ β tan11

tan

−−

=−= G

M

dx

dh

(4.164)

or,θ β tan

1tan

−−

=−=G

G M

dx

dh

(4.165)

In a given displacement, M and G are constants, so dh/dx = tan β is also constant. For stability (of either type), we require that dh/dx < 0, that is the interface should approach h

= 0 as x increases (see Figure 4.46).

For this to happen, we require that:

1−> M G (4.166)

and thus the stability limit is when:

1−= M G (4.167)

This stability limit defines a critical flow rate that must not be exceeded, otherwise water

underrun will occur:

crtical t

w

q

gA

M G ,

sin

1

θ ρ λ ∆

=−= (4.168)

or, ( )1

sin

1

sin 0

, −∆

=−

∆=

M

gAk k

M

gAq

w

rwwcrtical t

µ

θ ρ θ ρ λ

(4.169)

For flow rates less than this critical value, gravity will stabilize the water tongue. Noticethat the mobility ratio M is also important in defining the behavior of the interface.

G

M G

G

G M )1(1

tan

tan −−=

−−−=

θ

β

(4.170)

• If M > 1 the interface is stable if G > ( M – 1) and β < θ. Unstable if G < ( M – 1).This is the left diagram in Figure 4.46.

• If M = 1 the interface is unconditionally stable and β = θ. The interface will behorizontal.

• If M < 1 the interface is unconditionally stable and β > θ. This is the rightdiagram in Figure 4.46.

The concepts used in this section can also be used to consider the downdip displacementof oil by gas.



xθ

H

h

g a s

S o = S o r

S g = 1 - S o r

o i l

S g = S g c

S o = 1 -

S g c z

Figure 4.48: Gas-oil displacement under segregated flow conditions.

multi phase flow

Documents

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phase relative permeability

twophase flow

flow of fluids

conceptualization of

capillary number

capillary tube

measuring capillary