Additional Material for Chapter 6

1PET 212E-Rock PropertiesAdditional Material on Relative

PermsSpring 2007

M. Onur

Introduction Primary Recovery

Production is obtained by only using intrinsic (or natural) energy of the reservoir as the driving force. (e.g., solution gas drive, gas cap drive, compaction, water drive)

Typically we recover 10 to 15% of original oil in place (OOIP) unless there is strong water drive mechanism

Enhanced Oil Recovery (EOR) Refers to production over and above that which

can be recovered by primary production (e.g., water injection, gas injection, CO2 injection, steam injection, polymer flooding, surfactant flooding)

Water Flooding

We inject water in one (or more) wells and recover oil from other wells. In doing this, we provide externally additional mechanical energy (e.g. pressure drop) to overcome viscous resistance to oil flow.

It is also known as Secondary Recovery because it is a second batch of oil after a field was depleted by primary production.

Water Flooding

Virtually all oil reservoirs are eventually water flooded.

Efficiency: 25% to 50% of oil recovery depending on:

Rock and fluid properties Degree of heterogeneity (spatial variation).

2Basic Definitions If the fluids, when mixed together, segregate into

two distinct phases, they are said to be immiscible (e.g., oil and water).

If the fluids, when mixed together, form a new single homogeneous phase, they are said to be miscible (CO2 and oil).

Oil and natural gas are always at least partially miscible because we can dissolve some gas in the oil.

Significance of Scale

10-10 10-7 10-6

m

10-5 10-4 10-3

Size of H2O molecules

1010

Earth to sun

107

Earth diameter

10-2 10-1 100 101102 103

mm cm m

deepest sediment

Block size In flow simulators

wellborediameter

pore sizes

vugs

core sample

Visible light

Microscopic scale(Navier stokes and PoiseuilleEquation)

Macroscopic scale(Darcys equation and continuum approach)

Microscopic vs. Macroscopic

Microscopic efficiency: denotes the ability of water to displace oil from the pore space where it is located. Limitations on microscopic efficiency are due to rock/fluid properties Interfacial tension, wettability, contact angle,

viscosities, etc. Thus, a residual oil always in the pore spaces,

trapped by the action of capillary forces that cannot be overcome by the viscous forces exerted by injection of water phase.

Microscopic vs. Macroscopic

Macroscopic efficiency: denotes the ability of flow system to displace the microscopically mobilized oil to production wells. Limitations on macroscopic efficiency are due to the macroscopic distribution of fluids in the reservoir (oil and water will prefer a flow path of least resistance).

3Microscopic vs. Macroscopic The macroscopic distribution of fluids will be

affected by Shape of the reservoir and the geometry of well

pattern. Variability of horizontal and vertical perms: High

contrast in perm causes injected fluid to follow the high perm regions.

Viscosity contrast of fluids: If the injected fluid has lower viscosity, it tends to preferably flow into already invaded regions (viscous fingering).

Density contrast of fluids. Effects of gravity cause lighter fluid to override, heavier fluid to under run.

Porous Media

Pore structure: Its geometry depends on Pore body (related to volume, fluid storage

capacity, ) Pore throat (related to conduction, i.e.,

function of permeability, k) Topology: Connectivity between pore

bodies (coordination number z)

Porous Media

Coordination number, z, is defined as the number of pore throats (or pore neck) emanating from a pore body

Pore body

Pore throat

z = 4

Porous Media

Connectivity is required to reach residual phases. As the coordination number, z, increases

displacement efficiency increases)

z is infinity for the bundle of capillary tubes model, i.e., zero oil residual saturation in the case of water displacement.

4Porous Media

Bundle of capillary tubes

For a real porous media, they report z to be 6 to 12.

Sor

Injectwater

After injection, remaining oil saturation will bezero

Fundamental Properties of Porous Media (PM)

Porosity

gpbb

p VVVVV +== ;

reseff

,res

,eff ;

>>==

b

derconnecteintp

b

connectedp

VV

VV

Fundamental Properties of PM

Saturation: Given a sample of rock where the pore volume contains a combination of oil, gas, and water, the saturation of phase m is defined as the fraction of the pore volume occupied by phase m, i.e.,

p

mm V

VmS ==Volume Pore

"" phase of Volume

Notes on Saturation

Saturation is dimensionless. Sum of saturations must be equal to unity,

i.e.,

Oil in place depends on oil saturation. Effective and relative permeabilities depend

on phase saturations.

1=++ gwo SSS

5Notes on Saturation Consider a core sample containing oil and

water only. If the dry mass and the 100% (i.e., Sw = 1) water saturated mass of the core are given, the water saturation in the core, when both mobile phases exist and the total mass is measured, can be computed from

( ) powpodryt

w VVmm

S

=

Notes on Saturation

In previous equation,

mt: the total mass of core when both phases exist.

mdry: the dry mass of coreo: the density of oil w: the density of waterVp: the pore volume of the core

Fundamental Properties of PM

Permeability Absolute permeability (if a single phase

with Sm = 1 flows in the pore, k)

Effective permeability (more than one phase is flowing inside the pore, km)

Relative Permeability

Effective permeability data are generally presented as relative permeability data. Defined as the ratio of the effective

permeability of a phase to a basepermeability, k

mrm

kk =k

6Relative Permeability

Three different base permeabilites can be used: The absolute air permeability,

The absolute water permeability,

The effective permeability to oil at residual water (in general, residual wetting phase saturation),

airkk =)1( == ww Skk

)( wiwo SSkk ==

Relative Permeability For most cases, the three base

permeabilities do not have the same value:

It is important to know which base permeability was used for a particular set of relative permeability data.

)()1( wiwowwair SSkSkk ==

Relative Permeability

Depends on area available to flow as well as Slippage between phases Rock wettability (which phase actually wets

the surface of the rock) Hysteresis (differences in path through

saturation)

Equations for Relative Permeability Curves

There are several simple equations of relative permeability curves that can be used for modeling purposes.

Straight line models (oil/water system).

iwor

worwro SS

SSSk =

11)( (47)

7Equations for Relative Permeability Curves

Straight line models (oil/water system).

In Eqs. 47 and 48, Sor is the irreducible oil saturation and Siw is irreducible water saturation. Note that Eqs. 47 and 48 define kro and krw as linear functions of water saturation Sw (see the figure in the next slide).

iwor

iwwwrw SS

SSSk =

1)( (48)


kro or krw

Sw

1.0

0.00.0 Siw 1-Sor

Curves intersect at ( )2

1 iworw

SSS +=1.0


Corys formula (for unconsolidated water-wet sand.)

and

where

( )4= wrw Sk (48)( ) ( )[ ]22 11 = wwro SSk (49)

iw

iwww S

SSS =

1(50)

Corys Relative Perm Curves

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

kro or krw

8Equations for Relative Permeability Curves

Straight line models with different end-point values (oil/water system).

Where and are the end-point values of water and oil relative permeability curves (see the figure in the next slide). Usually, .

=

iwor

iwwrwrw SS

SSkk1

0 (51)

=

iwor

worroro SS

SSkk110 (52)

0rwk

0rok

1/ 00

9Relative Perm Curves For Three-Phase Flow (oil/water/gas)

The assumptions are thatkrw depends only on the water saturation Sw and does not depend on the number of phases present; i.e., krw = krw(Sw).krg depends only on gas saturation Sg and does not depend on the number of phases present; i.e., krg = krg(Sg).kro is assumed to depend on both Sw and Sg; i.e., kro = kro(Sw ,Sg).

Relative Perm Curves For Three-Phase Flow (oil/water/gas)

A typical way to construct three-phase relative perms is first to measure effective oil perm (kow) and effective water perm (kw) with only oil and water flowing in the core, and obtain two-phase oil and water relative perm curves as

and)()(

iwow

wowrow Sk

Skk =

)()(

iwow

wwrw Sk

Skk =

(56)

(57)


Next, we measure two-phase effective oil and gas perms in the core in the presence of irreducible water saturation. Note that only oil and gas can flow under this condition. These two phase effective perms are a function of Sg and are denoted by kog and kg, respectively. Then, the oil and gas relative perm curves are constructed from

)0()(

== goggog

rog SkSk

k (58)

)0()(== goggg

rg SkSk

k (59)


Strones (second) model for three-phase oil relative perm is then given by

Does this reduce to correct two-phase oil relative permeability if Sg = 0 or Sw =Siw? Yes

( ) [ ][ ][ ])()(

)()()()(,

grgwrw

grggrogwrwwrowgwro

SkSkSkSkSkSkSSk

+++= (60)

10


Because two-phase oil-gas relative perm curves are measured in the presence of Siw, the two-phase oil/gas case with Sg = 0 is physically identical to the two phase oil/water system with Sw=Siw. That is

(61))0()( == gogiwow SkSk


Because relative perms are obtained by normalizing effective perms by kow(Siw) and kog(Sg = 0), then

We also know that

(62)1)0()( === grogiwrow SkSk

0)0()( === grgiwrw SkSk (63)


Using Eqs. 62 and 63, we can show that the three-phase formula of Eq. 60 reduces to the two-phase formulas under two-phase flow conditions, i.e.,

and(64))(),( groggiwro SkSSk =

)()0,( wrowgwro SkSSk == (65)


For example, if we evaluate Eq. 60 at Sw =Siw, we can show that Eq. 60 gives the oil relative permeability curve, relative oil permeability for the two-phase flow of oil and gas. For this case,

1)( =iwrow Sk

0)( =iwrw Sk

(66)

(67)

11


Using Eqs. 66 and 67 in Eq. 60

or

( ) [ ][ ][ ])()(

)()()()(,

grgwirw

grggrogwirwwirowgwiro

SkSk

SkSkSkSkSSk

+

++=1 0

0

( ) )()()()(, groggrggrggroggwiro SkSkSkSkSSk =+= (68)


Similarly, we can show that Eq. 60 reduces to oil relative perm curve in the case of oil/water flow; i.e., Sg in Eq. 60 is evaluated at 0 gas saturation.

or

( ) [ ][ ][ ])0()(

)0()0()()(0,

=+

=+=+==

grgwrw

grggrogwrwwrowgwro

SkSk

SkSkSkSkSSk1 0

0

( ) )()()()(0, wrowwrwwrwwrowgwro SkSkSkSkSSk =+== (69)


Now, lets consider a linear flow in the x-direction, we normally see Darcys law written as

which is correct provided that we define kkro equal to effective oil permeability denoted here by ko. If ko = kkro, how should we interpret k in Eq. 70?

(70)dx

dpkkcv oo

roox =,


Because all two-phase relative perms are normalized by kow(Siw)=kog(Sg=0), the three-phase oil relative permeabilitiesmust be be normalized the same way. That is, if ko(Sw,Sg)denotes the three-phase effective oil permeability, then

This means k in Eq. 70 is really kow(Siw).

(71)( ) ),()(, wgroiwowgwo SSkSkSSk =

12


If we want Eq. 70 be correct with k equal to the true absolute perm, we simply normalize all effective perms by k, e.g., we replace Eq. 56 by

And do the same changes in Eqs. 57, 58 and 59. However, we should modify Eq. 60 so that it will reduce to the correct two-phase relative perms.

(72)( )kSkSk wowwrow

)(=


One way to do this is to define a scaling constant given by

and replace Eq. 60 by

(73))0()( === grogiwrowD SkSkk

( )[ ]})()(

)()(

)()(,

grgwrw

grgD

grogwrw

D

wrowDgwro

SkSk

Skk

SkSk

kSkkSSk

+

+

+=

(74)


With this last definition, we can also show that Eqs.64 and 65 hold. For example, if we evaluate Eq. 74 at Sw = Siw, we obtain

or

(75)

( )[ ]})()(

)()(

)()(,

grgiwrw

grgD

grogiwrw

D

iwrowDgiwro

SkSk

Skk

SkSk

kSkkSSk

+

+

+=0

01

( ) )()()()(, groggrggrgD

grogDgiwro SkSkSkk

SkkSSk =

+=

Oil/water Relative PermeabilityFig. 3 - Oil/water rel perms

0

0.2

0.4

0.6

0.8

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Water saturation

Rel

. Per

m

13

Rules of Thumb

For a 2-phase oil-water system For water wet rock,

irreducible water saturation, 0.15 Siw 0.25 Curves intersect at Sm > 0.5 The value of krw at Sor is typically 0.3

For oil wet rock irreducible water saturation, 0.1 Siw 0.15 Curves intersect at Sm < 0.5 The value of krw at Sor is typically 0.5

Oil/water rel perms, water wet rock

0

0.2

0.4

0.6

0.8

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9Water Saturation

Rel

. Per

m.

Oil/water relative permeabilities, oil wet

0

0.2

0.4

0.6

0.8

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9Water Saturation

Rel

. Per

m.

Hysteresis

The curves obtained by displacing the wetting phase with the nonwettingphase are referred to as the drainage curves (i.e, increasing nw phase sat.)

If having displaced all of the wetting phase possible, we then let the wetting phase saturation increase, we obtain the imbibition curves.

14

Hysteresis Physical explanation for Hysteresis

Water always in small pores due to its smaller molecular size

Gas always in larger pores Oil is always in the intermediate pores

As a result, trapment (or discontunity) occurs during either wetting or non-wetting displacement.

Darcys Law

For multiphase flow, Darcys law can be written as

lp

BAk

lp

BAkkq

m

mm

m

m

mm

rmm

=

=

3

3

10127.1

or

10127.1

Water/Oil Ratio Consider an oil/water system and assume there is no

capillary pressure, so pressure in all phases is the same, we can derive the following important reservoir engineering results:

Water oil ratio

row

rwo

w

o

o

w

kk

BB

qqWOR

==

15

Interfacial Tension (Surface Tension), Capillary Pressure

Basic Concepts

A molecule I in the interior of a liquid is under attractive forces in all directions

The vector sum of these forces is zero. A molecule S at the surface of a liquid is acted on

by a net inward cohesive force that is perpendicular to the surface.

Results in a membrane that separates liquid from gas or one liquid from another. Immiscible fluids

Basic Concepts

Work is required to move molecules from the interior to the surface against this cohesive force.

The surface tension ( sigma) of a liquid is the work that must be done to bring enough molecules from inside the liquid to the surface to form one new unit area of that surface Also can be thought of as the force per unit length

acting on a curve or line in the surface

Interfacial tension (surface tension) is the force per unit length required to increase surface area by one unit.

Think of an elastic membrane stretched over a frame. IFT like a ``stretching force.

Basic Concepts

areaunitworklFl

lF === /2

16

Condition of Surface at Break Point

Interfacial surface tension is proportional to surface tension force

If we denote the interfacial tension by , and F is the minimum force required to raise the ring through the liquid surface,

( )RF= 2

Contact Angle

Suppose we have a dense liquid (Liquid 1) laying on a solid surface and surrounded by a light immiscible liquid, (Liquid 2)

The Contact angle () is defined as the angle between the solid and the liquid-liquid interface, measured through the dense fluid.

Liquid 1

Liquid 2

Contact Angle

Surface Tension Forces

Three distinct interfaces, and three sets of surface tension 1,2 is the interfacial tension between Fluid 1 and

Fluid 2. 1,s is the interfacial tension between Fluid 1 and

the solid. 2,s is the interfacial tension between Fluid 2 and

the solid.

Liquid 1

Liquid 2

1,2

2,S 1,S

Equilibrium Condition

If the system is in equilibrium, then the forces must be in balance.

Adhesion, AT

( )+= cos2,1,1,2 ss( )= cos2,1,1,2 ss

( )= cos2,1,1,2 ssTA

17

Effect of Increasing 2,s

Since 1,s and 1,2 have not changed, the only way to return to equilibrium is by decreasing , or increasing cos()

Liquid 1

Liquid 2

1,2

2,S 1,S

2,S increased - droplet smeared.

Effect of Decreasing 2,s

If we decrease 2,s, the contact angle would increase, and the denser liquid would tend to gather into a little droplet.

Liquid 1

Liquid 2

1,2

2,S 1,S

2,S decreased

Assume that Liquid 1 is Water and Liquid 2 is Oil

If the contact angle goes to zero, i.e., 0, water will spread out completely over the surface The water tends to adhere to the solid surface

more than the oil does.

Surface is water-wet, or water preferentially wets the surface

Note that AT > 0 implies that cos() > 0, and 0 < /2.

( ) 0cos,,,,1,2 >== owswsossTA

Oil-Water Systems

If = /2, the surface is said to be neutrally wet.

If > /2, the oil tends to spread under the water and make it into a spherical droplet; In this case, the surface is preferentially wet by oil,

and the surface is said to be oil-wet. Behavior of oil-gas or water gas systems is

analogous. Can you think of a liquid-gas system that is

preferentially wet by gas?

18

Reservoir Flow To facilitate oil flow, it is better to have water-

wet rock than oil-wet rock. When water sticks to the rock it spreads over the

rock surface and leaves the oil in channels surrounded by water.

The friction between flowing oil and the surrounding water cushion is much less than the friction between oil and rock, so oil flows easily.

Conversely, if the rock is preferentially oil-wet, it is much more difficult to displace oil from the reservoir. Residual oil saturation is higher.

Water Wet Rock

It is very difficult to measure the contact angle for field applications; to do so, we would need a clean even (flat) surface obtained from the reservoir rock.

Sand Grains

Water

Oil

Interfacial Tension

If we have a bubble of one fluid suspended in a second fluid, it is possible to show that there will be a difference inpressures inside and outside of the bubble. Which is greater? Think of a balloon.

Pin

Pout

Spherical bubble of one fluid suspended in another

Force Balance

Forces acting on one hemisphere A force due to a difference in pressure

between the inside and outside tending to blow the bubble apart

Surface tension force, tending to pull the halves of the bubble together

( )R

pp outin2=

19

Laplace-Young Equation

Non-spherical bubble

R1 and R2 are principal radii of curvature

( )

+=

21

11RR

pp outin

Note

We will apply the same basic equation to model the pressure difference across an interface between two immiscible fluids, e.g., an oil drop surrounded by water in a pore space. Unfortunately, it is not possible to measure the principal radii of curvature for this situation.

Initial Fluid Distribution

The common belief is that reservoirs are originally water wet. Oil migrates into the structure and displaces water. As oil is less dense than water, water is displaced downward due gravitational forces. Some water may be trapped, due to capillary effects. The water distribution at discovery is referred to as connate water saturation. Over time, reservoir may become oil wet.

Capillary Rise

Water wet,

in any absolute system of units. What if oil wet?

oil( ) ( ) ghr

ppp owwoc )(cos2 ===

20

Fluid Distribution-Capillary Tube Notes Capillary pressure must be

equal to gravitational forces if fluids are in equilibrium and not flowing

Capillary pressure is a function of adhesion tension (or IFT) and inversely proportional to the radius of the tube.

The greater the adhesion tension (or IFT), the greater the equilibrium height.

Effect of r

Effect of

Application to Porous Rocks If we take a sandstone core saturated with oil,

and place it in a jar of water (which preferentially wets the sand), what will happen? Since water preferentially wets sand, water will

begin entering the smallest pores where capillary pressure is greatest. (Imbibition).

Oil will be left to occupy the largest pores of the sandstone.

Imbibition will continue until the adhesion force is balanced by the gravity forces. (Capillary pressure force equals to adhesion force)

Oil Trappment in PM

At bottom of oil drop (B)

ghpp oHoBo +=ghpp wHwBw +=

B

woBwBo r

pp ,2=

21

Oil Trappment in PM Eliminating pressures at

the bottom of the droplet

At top of oil droplet

Droplet will only go through bottleneck if

( )B

woowHwHo r

ghpp ,2+=

H

woHwHo r

pp ,2=

H

woHwHo r

pp ,2>

Hysteresis in Capillary Pressure

Similar to relative permeability, capillary pressure exhibits hysteresis effects. The curve obtained by starting with a core completely saturated with the wetting phase and displacing the wetting phase with the nonwetting phase is referred to as the drainage curve. If having displaced all of the wetting phase possible, we then let the wetting phase saturation increase, we obtain the imbibition curve.

Drainage/Imbibition Capillary Pressure -Saturation

Relationship depends on: Size and distribution of pores The fluids and solids involved The history of the saturation process

22

Bundles of Capillaries

For cross-sectional flow area A cm2, equivalent permeability is found by equating flow rate from Poiseuilles law and Darcys Law:

( )=

=n

i

i

Lpprq

1

214

8( )

LppAkq 21

910869 =

.

darcies in . ==n

iirA

k1

4610712

Additional Material for Chapter 6

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