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 ,
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