1Multiphase Flow

Darcys Law

When two or three fluids flow through a core, the fluids must share the pore space Each fluid has less cross-sectional area available

for flow. Multiphase Flow Darcys Law

m refers to the phase, o oil, g - gas, w - water

dldp

BAkq m

mm

mm =

310127.1

Phase Permeability

The new term in Darcys law is the (effective) phase permeability, km

Effective permeability is a function of saturation. If Sm = 1, so only phase m is in the core, km = k. If phase m is not flowing (i.e., it is discontinuous

and the other phase are flowing around it), then km = 0.

Limits: 0 km k

2Phase Permeability

The reduction in phase permeability below absolute permeability reflects the fact that only a fraction of the cross-sectional area is occupied by phase m.

If Am is the fraction of the area occupied by phase m Note Am = SmA We might expect

dldp

BkAq m

mm

mm =

310127.1

Phase Permeability

Equivalently kmA = kAm = kSmA Or km = kSm

This cannot be true If we take a dry core and add a few drops of water,

the water will enter the pores, and Sw > 0. If we now flow air through the core, only air will

flow, even though there is a water saturation. So kw kSw We can find similar cases with any phases

occupying the rock.

Critical Saturation

In fact, if we take a dry core sample and gradually add a wetting phase (e.g., water or oil), we will find that there is a minimum saturation that we must attain before that phase starts to flow. This saturation is called the critical saturation for

phase m, Sm,c Taking critical saturation into account, we

might guess that

k

SSS

kcm

cmmm

=

,

,

1

3Relative Permeability We define relative permeability to phase m by

Based on our guess for phase permeability, we can infer

Note krm = 1 when Sm = 1 krm = 0 when Sm Sm,c

kkk mrm =

=

cm

cmmrm S

SSk

,

,

1

Relative Permeability

This is not a bad guess, however, relative permeability is rarely a linear function of saturation

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

surface of the rock)

Oil-Water Relative Permeability

Fig. 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

4Hysteresis The relative permeability curves we obtain

depend on the saturation history of the displacement process of the core sample. For example, we can start with a core completed saturated with water and displace water by injecting oil until we reach irreducible (or critical) water saturation. Then we can reverse the process by injecting water until we reach residual oil saturation (under water injection). Hysteresis refers to the fact that the two sets of relative permeability curves are different.

Hysteresis

The curves obtained by displacing the wetting phase with the nonwettingphase are referred to as the drainage curves. If having displaced all of the wetting phase possible, we then let the wetting phase saturation increase, we obtain the imbibition curves.

Hysteresis

The curves obtained by displacing the wetting phase with the nonwetting phase 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 imbibitioncurves.

5Hysteresis

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.

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 Sw > 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 Sw < 0.5 The value of krw at Sor is typically 0.5

6---------------------------------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.

kro

krw

Oil/Water Rel Perm CurvesOil/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.

Darcys Law

In terms of relative permeability, Darcys law is written as

dldp

BAkkq mmm

rmm =

310127.1

7Some Results

If 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

Producing Gas-Oil Ratio

row

rwo

w

owo k

kBBF

=

srog

rgo

g

op Rk

kBBGORR +

==

Results Note that for gas flow, some of the gas that is being

produced is coming out of solution, while some is due to free flow of gas in the reservoir.

Total In situ Flow Rate (at reservoir conditions)

Note that pressure gradient required to produce a fixed rate of fluid changes as mobility changes.

g

rg

w

rw

o

rot

mtwgot

kkkdl

dpAkqqqq

++=

=++= 310127.1

Interfacial Tension (Surface Tension), Capillary Pressure

8Motivation Typically, more than one fluid is flowing at any given

reservoir location at any given time. This is particularly true when one fluid is injected to

displace another; e.g., during water flooding, when water is injected to displace oil.

Two fluids are said to be miscible if they can be mixed together in any proportion to form a single homogeneous mixture. For example, ethanol (grain alcohol) and water are

completely miscible fluids. If two fluids are not miscible, they are said to be

immiscible.

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

9Basic Concepts

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.

areaunitworklFl

lF === /2

Lab Experiment

Measurement of surface tension Immerse platinum

ring beneath surface of liquid and pull up.

A force is required to pull the ring through the surface.

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

10

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 Liquid 1 and Liquid 2. 1,s is the interfacial tension between Liquid 1 and the solid. 2,s is the interfacial tension between Liquid 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

11

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.

( )= cos2,1,1,2 ssTA

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

12

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?

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

Sand Grains

Water

Oil

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.

13

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 in pressures 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

Laplace-Young Equation Will do a force balance in

the vertical direction. Make similar triangles. Then component of IFT in vertical direction can be found as follows:

R)v

= sin( v=)sin(

RF )(2=

14

Laplace-Young Equation In the vertical direction

Pressure difference

Non-spherical bubble

R1 and R2 are principal radii of curvature

( )R

pp outin 22 =

( )RR

pp outin

22

2 ==

( )

+=

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.

Note

Interaction between the surface of the reservoir rock and the fluid phases confined in the pore space influences fluid distribution in rocks as well as flow properties.

15

Fluid Saturation and Capillary Properties

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.

Reservoir Fluids

Reservoir rocks may contain the following fluids Liquid hydrocarbons (light oil to asphalt) Gaseous hydrocarbons Water

Distribution in the reservoir is different from cores brought to surface Invasion of drilling mud filtrate Gas evolution and expansion with reduced

pressure Handling errors (washing or improper drying)

16

Interstitial Water

Interstitial or connate water is present in all reservoirs Surrounds grains

and fills small pores If water wet,

hydrocarbons occupy the center of large pores and cracks

Interstitial Water

Water saturation depends on Pore size distribution (capillary pressure), densities of reservoir fluids Sat varies with distance above oil-water contact.

Coarse-grained rocks with large pores have low connate water saturation

Fine grained rocks have smaller pore throats and high irreducible water saturation.

Water saturation determined by capillary effects.

Capillary Effects

Place a glass capillary tube in water Water adheres to glass -

rises in tube Stops rising when

adhesion force is balanced by the weight of liquid in the supported column

Force Balance

( ) ( )2cos2 rghr w =

17

Capillary Effects Force balance in terms

of pressure Pressure at B = pa Pressure at A = pa Pressure at B = pa gh =

pw Equating pressures

Capillary pressure = pc= pnon-wetting pwetting

( ) ( ) ( )+= cos222 rrprp wa

( ) ( )r

ppp wac== cos2

Capillary Effects If air is replaced by

oil,

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

( ) ( ) ghr

ppp owwoc )(cos2 === oil

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

18

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 weight of imbibed water. (Capillary pressure force equal to adhesion force).

Oil Migration

Modern oil reservoirs were formed by migration of hydrocarbons from source rock into previously water-saturated reservoir rock.

Effect of capillary pressure on migration?

Migration through Capillaries Droplet of oil migrating

through a series of narrow and wide channels

Oil less dense than water Buoyant force Capillary pressure forces

At bottom of oil drop (B)ghpp oHoBo +=ghpp wHwBw +=

B

woBwBo r

pp ,2=

19

Migration through Capillaries Eliminating pressures at

the bottom of the droplet

At top of oil droplet

Droplet will only go through bottleneck if

H

woHwHo r

pp ,2=

( )B

woowHwHo r

ghpp ,2+=

H

woHwHo r

pp ,2>

Hysteresis 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.

Hysterisis Same capillary pressure is

obtained for two different saturations Relationship between

wetting phase Saturation and capillary pressure depends on saturation process.

For a given capillary pressure, higher saturation is obtained with drainage than imbibition.

Hysterisis process is not reversible.

20

Hysterisis

Drainage/Imbibition

Relative Permeability Curves

21

Imbibition Rel Perm Curves

Capillary Pressure -Saturation

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

Initial Saturation Distribution in the Reservoir

Free water level is location at which capillary pressure is zero below this level water saturation is equal to 1.

Initial oil water contact (IWOC) Lowest point where oil is present below this level, water saturation is one.

In lab experiment, water saturation is one until (cap pressure) increases to threshold pressure,

cpthp

22

Saturation Distribution

Swi (Swc)

Initial Saturation Distribution

Similar thing occurs in the reservoir, from the free water level up to the IWOC where

Solve for the height of the IWOC above the free water level.

c

thowthc g

ghpp144

)( ==

1=wS

thh

Initial Saturation Distribution

Object: find saturation distribution as a function of h where h is the height above the free water level.

Either the free water level from gradient surveys or the IWOC from logs will be known.

If IWOC is given at depth D, solve

for and set c

thowthc g

ghpp144

)( ==thh thf hDD +=

23

Initial Saturation Distribution

Find h at the top of the transition zone, the zone that contains both oil and mobile water. At the top, water saturation is equal to irreducible water saturation and capillary pressure takes on its maximum value. Solve for

c

topowc g

ghp

144)(

max, =

toph

Initial Saturation Distribution, Method 1

Specify values of h from to , then compute the value of capillary pressure for each h from

Compute the value of from the capillary pressure versus water saturation curve or table.

Graph water saturation versus h or vice versa.

c

owc g

ghp144

)( =

thh toph

wS

Initial Saturation Distribution, Method 2

Specify values of from 1 to irreducible water saturation. Compute the corresponding value of capillary pressure from the versus

curve. To find the corresponding value of h solve the following equation for h:

Plot water saturation versus h or vice versa.

c

owc g

ghp144

)( =

wS

cpwS

24

Water Saturation Versus height Above Contact

S_w vs. h

0

0.2

0.4

0.6

0.8

1

1.2

0 50 100 150 200

h, height above free water level, ft

Wat

er S

atur

atio

n, S

_w

N or OIIP Oil Initially in Place

Once we have water saturation as a function of h, we can estimate the oil initially in place. We do this in two steps. First we estimate the oil in the transition zone, then we estimate the oil in the oil zone. The oil zone contains no free gas, but we could have a gas cap on top of the oil zone.

Oil in Transition Zone

Divide the transition zone into layers of thickness

Find the area, average porosity and average oil saturation at the center of each layer

Sum the to get the oil in STB in the transition zone. ( denotes oil FVF at initial reservoir pressure.)

njhhh jjj ,,1,1 L==

)615.5/()( 0 oijjjjj BhASN = jN

oiB

25

Initial Oil in Place

Transition zone

Oil zone: Same procedure exceptthroughout the oil zone, thus oil initially in place in oil zone is

==N

i itzNN

1

iwo SS = 1

)615.5/()()1(1 oim

k

m mmwiozBhASN = =

Total Oil Initially In Place

We could also let oil FVF factor vary with depth in the equations for computing oil in place. This would require we compute oil pressure at each depth of interest. We need one oil pressure measurement to do this.

ztz NNN 0+=

FWL from Pressure Gradients

Suppose we measure oil pressure at a point in the oil zone which is at a depth

Then the oil pressure at any point down to the IWOC is given by

Also apply this down to the depth of the FWL even though no oil exists below the IWOC.

0d

c

ooo g

ddgdpdp144

)()()( 00+=

26

FWL from Pressure Gradients

If water pressure is measured at some depth in the water zone then at any depth d,

c

ooo g

ddgdpdp144

)()()( 00+=

wd

c

wwwww g

ddgdpdp144

)()()( +=

FWL from Pressure Gradients

Both phase pressure are linear functions of depth. At the point where these two lines intersect, the phase pressures are equal so the capillary pressure is equal to zero. Thus, the value of d at the point of intersection is the depth of the free water level.

c

ooo g

ddgdpdp144

)()()( 00+=

c

wwwww g

ddgdpdp144

)()()( +=

Free Water Level From Pressure Gradients

If we use RFTs to take measurements of oil pressure in the oil zone and water pressure in the water zone, then the intersection of the gradients gives the location of the FWL.

Gradients

2380240024202440246024802500252025402560

5200 5300 5400 5500 5600 5700

depth

p_o

or p

_w, p

sia

p_o p_w

27

Gradients

2380240024202440246024802500252025402560

5200 5300 5400 5500 5600 5700

depth

p_o

or p

_w, p

sia

p_o p_w

Comments on Threshold Pressure

At any height , we havethhh

))(()()()()( thowthwthowo hhhphphphp += )()()( thmthmm hhhphp =

hhhphphp owthowthwthoc )()()()()( +=hhhp owthowthc )()()( +=

Threshold Pressure

If we go ahead and evaluate

at and use the fact that capillary pressure is zero at , we obtain

which effectively defines

0=hhhhphphp owthowthwthoc )()()()()( +=

thowthc hhp )()(0 =0=h

thowthcthwtho hhphphp )()()()( ==thh

28

Threshold Pressure

and

This implies that we can compute a superficial oil pressure for so

thwothwwthw hphphp == )0()0()(

thowthwtho hhphp )()()( =

thothothwthwo hhphhpp +=+= )()()0(

thhh 0

Comments on Threshold Pressure

at any height , we have

for both phases as long as h corresponds to a point between the free water level and the top of the transition zone. Similarly at any depth dcorresponding to such a point

0hhhphp mthmm = )()(

hdpdp mtopmm += )()(

Leverett J Function

In some cases, we can estimate capillary pressure from the Leverett J function constructed from core data on an analogue well or reservoir. You should use this with caution. One condition for it to be valid is that irreducible water saturation is identical for all cases for which the J-function is used.

29

Leverett J Function

This is a mixed unit system. J is dimensionless, capillary pressure is in psi, interfacial tension is in dynes/sq cm, permeability is in millidarcies. Given data from one set of cores, including capillary pressure, we can compute J.

kpJ cSw )cos(

219.0=

Leverett J Function

Now if we have another system where we have made measurements of permeability and porosity, we can compute capillary pressure from the above equation.

kpJ cSw )cos(

219.0=

Lab Data to Field Data

Often use non-reservoir fluids in the lab to measure capillary pressure. Can convert as

The justification is as follows:

lab

fieldlabcfieldc pp ))cos((

))cos((,,

=

30

Lab Data to Field Data

Solve the first equality for to obtain the equation on the preceding slide.

rpp

field

fieldc

lab

labc 2))cos(())cos((

,, ==

labcp ,