Chapter 9 Simultaneous Flow of Immiscible...

Chapter 9 – Simultaneous Flow of Immiscible Fluids

9.1

An important problem in petroleum engineering is the prediction of oil recovery

during displacement by water. Two common examples are a natural water drive and

secondary waterflood. The latter is displacement of oil by bottom or edge water, the

former is the injection of water to enhance production. In this chapter we will begin with

the development of equations of multiphase, immiscible flow, concluding with the frontal

advance and Buckley-Leverett equations. Next, we will discuss factors that control

displacement efficiency followed by limitations of immiscible displacement solutions.

9.1 Development of equations

The development of equations for describing multiphase flow in porous media

follows a similar derivation as given previously for single phase, i.e., combination of

continuity equation, momentum equation and equation of state. The mass balance of

each phase can be written as:

increment in time saccumulate

thatphase of mass

increment in time

leaving phase of mass

increment in time

entering phase of mass

Shown in Figure 9.1 is the differential element of porous media for oil.

uox│x uox│x+x

x

y z

Figure 9.1 Differential element in Cartesian coordinates. Only x-direction velocity is

shown.

As an example, the mass of oil entering and leaving the element is given by:

Entering: tAutAutAu zzozoyyoyoxxoxo (9.1)

Leaving: tAutAutAu zzzozoyyyoyoxxxoxo

(9.2)

Oil can accumulate by: (1). Change in saturation, (2). Variation of density with

temperature and pressure, and (3). Change in porosity due to a change in confining stress.

Thus we can write,

toottoo VSVS

(9.3)


9.2

Substitute Eqs. (9.1-9.3) into the conservation of mass expression, rearrange terms, and

take the derivative as t, x, y, z 0, then the phase dependent continuity equations

can be written as;

ooozooyooxo St

uz

uy

ux

(9.4)

wwwzwwywwxw St

uz

uy

ux

(9.5)

The oil and water continuity equations assume no dissolution of oil in the water phase.

That is, no mass transfer occurs between phases and thus flow is immiscible.

The next step is to apply Darcy’s Law to each phase, i. For example in the x-

direction,

x

ku i

i

iixix

(9.6)

where uix is the superficial velocity of phase i in the x-direction, kix, is the effective

permeability to phase i in the x-direction, and is the phase potential. Substitute Eq.

(9.6) into (9.4), apply Leibnitz rule of differentiation, and combine terms, results in,

ooo

o

o

ozoo

o

oyoo

o

oxo St

gz

pk

zy

pk

yx

pk

x

(9.7)

www

w

w

wzww

w

wyww

w

wxw St

gz

pk

zy

pk

yx

pk

x

(9.8)

Even though Eqs. (9.7) and (9.8) are written in Cartesian coordinates, they both can be

solved for a particular geometry. The solution will provide not only pressure and

saturation distributions, but also phase velocities at any point in the porous media.

To combine Eqs. (9.7) and (9.8) requires a relationship between phase pressures

and between phase saturations. The latter is easily understood from the definition of

saturations in Chapter 4, So + Sw = 1.0. The relationship between pressures was

developed in Chapter 5, and is known as capillary pressure.

wP

oPor

wP

nwP

cP (9.9)


9.3

9.2 Steady state, 1D solution

As a simple example, let’s consider the steady state solution to fluid flow in a

linear system as shown in Figure 9.2. This example is of primary interest in lab

experiments to determine relative permeabilities.

qo

qw L

poi

Pwi

poL

PwL

D

Figure 9.2 Steady state core flood of oil and water.

Oil and water are injected simultaneously, rates and pressures are measured, and core

saturation is determined gravimetrically. Permeability is unknown.

The steady state, incompressible fluid diffusivity equations are given by:

0

0

dx

dpk

dx

d

dx

dpk

dx

d

ww

oo

(9.10)

Integrating and combining with Darcy’s equations,

A

qc

dx

dpk

A

qc

dx

dpk

www

ww

ooo

oo

(9.11)

If water saturation is uniform throughout the core, then effective permeability is

independent of x. Therefore, for oil,

L

oo

o

p

p

o dxk

cdp

oL

oi

(9.12)

which upon integrating, becomes,

)( oLoi

ooo

ppA

Lqk

(9.13)

If kbase = ko at Swi is known, then it is possible to calculate relative permeability.


9.4

9.3 Capillary End Effect

During laboratory experiments, capillary equilibrium must be maintained; that is,

Pc = Po – Pw. Unfortunately, under certain conditions capillary end effects occur due to a

thin gap existing between the end of the core and the core holder. As shown in Figure

9.3, capillary pressure in this gap is zero.

gap

Pc=0

Figure 9.3 Schematic of gap between core and holder

The result is a rapid change in capillary pressure from a finite value immediately adjacent

to the outlet to zero in the gap. As a consequence, the saturation of the wetting phase

must increase to a value of Pc = 0. A generic core profile is shown in Figure 9.4 for both

pressures and saturation.

Sw

0 0 L L

Po

Pw

Pc=0+ Swc

Sor

P

Figure 9.4 Pressure and saturation profile through a core of length, L, with capillary end

effect.

Mathematically, we can describe this effect by investigating Darcy’s Law for the non-

wetting phase.

x

S

S

p

x

pAkq w

w

cw

nw

nw

nw

(9.14)


9.5

At the outlet, knw 0, but qnw ≠ 0; therefore,

x

S

Lx

wlim

(9.15)

Two plausible methods have been applied to avoid capillary end effect. The first

is to inject at a sufficiently high rate such that the saturation gradient is driven to a small

region at the end of the core. The second method is to attach a thin, (high porosity and

high permeability) Berea sandstone plug in series with the test core sample. The result is

to confine the saturation gradient in the Berea plug and thus have constant saturation in

the sample of interest.

A consequence of the saturation gradient in the core is that effective permeability

can no longer be considered constant from 0 < x < L. Subsequently, the convenient

steady state method of obtaining relative permeability outlined in Section 9.2 is not valid.

A solution to the saturation gradient can be obtained be combining the definition of

capillary pressure with the steady state, incompressible diffusivity equations. Begin with

defining the boundary conditions. Illustrated in Figure 9.5 is the capillary pressure –

saturation relationship in a core with end effects.

Sw

pc

0

inlet

outlet

Figure 9.5 Schematic representation of capillary pressure – saturation relationship in a

core sample with end effect.

From this figure we can deduce the following conditions,

Pc = 0 for both oil and water phases at x = L.

Sw = Swi at x = 0, thus Pc = Poi – Pwi

Sw = SwL at x = L, thus Pc = 0

From the definition of capillary pressure,


9.6

dx

dp

dx

dp

dx

dp woc (9.16)

Since pc = f(Sw),

dx

dS

S

p

dx

dp w

w

cc (9.17)

Substituting Eq. (9.17) into (9.16) for the capillary pressure gradient term, and Eq. (9.11)

into (9.16) for the oil and water gradient terms and rearranging, results in,

L

x

S

S

o

oo

w

ww

w

w

c

dx

Ak

q

Ak

q

dSS

pwL

w

(9.18)

Equation (9.18) can be solved either graphically or numerically for saturation gradient.

The result will be a calculated saturation profile similar to the one shown in the right-

hand side of Figure 9.4.

9.4 Frontal advance for unsteady 1D displacement

The unsteady-state displacement of oil by water is due to the change in Sw with

time. This can be visualized by looking at the schematics in Figure 9.6. These

schematics represent

Figure 9.6 Progression of water displacing oil for immiscible, 1D

Swi

Sor

Sw

A

Swi

Sor

Sw

C

0 1x/L

Swi

Sor

Sw

B

Swi

Sor

Sw

D

0 1x/L

Swi

Sor

Sw

A

Swi

Sor

Sw

A

Swi

Sor

Sw

C

0 1x/L

Swi

Sor

Sw

B

Swi

Sor

Sw

B

Swi

Sor

Sw

D

0 1x/L


9.7

snapshots in time of the frontal boundary as water is displacing oil. In sequence, A

depicts the initial state of the sample (or reservoir) where saturations are separated into

irreducible water, residual oil and mobile oil components. After a given time of

injection, the front advances to a position as shown in B. Ahead of the front water

saturation is at irreducible, but behind the front water saturation is increased. Continuing

in time, eventually the water will breakthrough the end of the core (reservoir) and both oil

and water will be produced simultaneously, C. Continued injection will increase the

displacing phase saturation in the core (reservoir), D.

Two methods to predict the displacement performance are 1) the analytical

solution by Buckley – Leverett (1941), and 2) applying numerical simulation. Only the

analytical solution will be described in this chapter.

9.4.1 Buckley – Leverett (1941)

The derivation begins from the 1D, multiphase continuity equations.

oooxo St

ux

(9.19)

wwwxw St

ux

(9.20)

In terms of volumetric flow rate,

oooo St

Aqx

(9.21)

wwww St

Aqx

(9.22)

Assume the fluids are incompressible and the porosity is constant. Eqs. (9.21) and (9.22)

simplify to,

t

SA

x

q oo

(9.23)

t

SA

x

q ww

(9.24)

Combining,

0

t

SSA

x

qq owow (9.25)

The result is qT = qo + qw = constant, the total flow rate is constant at each cross-section.


9.8

From the definition of fractional flow,

Two

Tww

qfq

qfq

)1(

(9.26)

Substitute into Darcy’s equation for each phase,

sin)1( gx

pAkqfq o

o

o

oTwo (9.27)

singx

pAkqfq w

w

w

wTww (9.28)

Rearranging Eqs. (9.27) and (9.28), we can substitute into Eq. (9.16) for the pressure

gradient terms. Solving the resulting equation for fractional flow of water, provides the

complete fractional flow equation.

ow

wo

c

To

o

ow

wow

k

k

gx

p

q

Ak

k

kf

1

sin

1

1 (9.29)

In the analytical solution it is difficult to analyze the derivative term (dpc/dx). If we

expand this derivative to,

x

S

S

p

x

p w

w

cc (9.30)

In linear displacement, dpc/dSw 0 at moderate to high water saturations as observed by

the capillary pressure curve such in Figure 9.7. As a result, dpc/dx 0.

Sw

Pc

0

w

c

S

p

Figure 9.7 Capillary pressure curve illustrating flat transition region at moderate to high

water saturations.


9.9

If the derivative term is negligible, and flow is in the horizontal direction such that no

gravity term is present, then the fractional flow equation reduces to,

ow

wow

k

kf

1

1 (9.31)

If we define mobility ratio as,

wo

ow

k

kM

(9.32)

then fw = 1/(1+1/M).

If we return to Eq. (9.24) and substitute for qw, we obtain,

t

S

q

A

x

f w

T

w

(9.33)

To develop a solution, Eq. (9.33) must be reduced to one dependent variable, either Sw or

fw. Observe, Sw = Sw(x,t) or,

dtt

Sdx

x

SdS

x

w

t

ww

(9.34)

Let dSw(x,t)/dt = 0, (Tracing a fixed saturation plane through the core) then

t

w

x

w

S

xS

tS

dt

dx

w

(9.35)

where the left-hand side is the velocity of the saturation front as it moves through the

porous media.

Observe fw = fw(Sw) only, then,

t

w

tw

w

t

w

x

S

S

f

x

f

(9.36)

Substitution of Eqs. (9.35) and (9.36) into Eq. (9.33), results in the frontal advance

equation.

tw

wT

S S

f

A

q

dt

dx

w

(9.37)


9.10

Equation (9.37) represents the velocity of the saturation front. Basic assumptions in the

derivation are incompressible fluid, fw(Sw) only and immiscible fluids. Furthermore, only

oil is displaced; i.e., the initial water saturation is immobile, and no initial free gas

saturation exists; i.e., not a depleted reservoir.

The location of the front can be determined by integrating the frontal advance

equation,

dtqS

f

Adx T

t

tw

w

x

S

wS

w

00

1

(9.38)

If injection rate is constant and if the dfw/dSw = f(Sw) only, then

w

w

Sw

wT

S S

f

A

tqx

(9.39)

We can evaluate the derivative from the fractional flow equation (Eq. 9.31), either

graphically or analytical. Figure 9.8 illustrates the graphical solution.

Swf Sw Swc

fw

fwf

Swbt

Figure 9.8 Fractional flow curve

The fractional flow of water at the front, fwf, is determined from the tangent line

originating at Swc. The corresponding water saturation at the front is Swf. The average

water saturation behind the front at breakthrough, Swbt, is given by the intersection at fw =

1. The location of the front is determined by Eq. (9.39), with the slope of the tangent to

the fractional curve used for the derivative function.


9.11

9.4.2 Displacement Performance (constant injection rate)

The displacement performance can be separated into two intervals, before and

after breakthrough. Until breakthrough, the volume of oil produced is equal to the

volume of water injected. After breakthrough, water saturation gradients exists, thus the

volume of water in the system can be defined as;

2

1

x

x

ww dxASV (9.40)

and the volume of oil displaced,

wiwo SxxAVV )( 12 (9.41)

Figure 9.9 illustrates the recovery of oil both before and after water breakthrough. Note

the 45 degree slope up to breakthrough and then the decrease in slope (reduced

performance) after breakthrough.

Np

Qi

breakthrough

Figure 9.9 Typical oil recovery performance plot for immiscible displacement

A solution for waterflood performance was developed by Welge in 1952. Define

the volumetric average water saturation as,

2

1

2

1

x

x

x

x

w

w

dxA

dxAS

S

(9.42)

For constant cross-sectional area (A) and porosity (), Equation (9.42) reduces to,

12

2

1

xx

dxS

S

x

x

w

w

(9.43)


9.12

The integrand can be expanded and the equation rearranged such that,

2

11212

1122 1w

www dSx

xxxx

SxSxS (9.44)

Substitute the frontal advance equation (Eq. 9.37) for the integral and solve,

12

2

1

2

1

wwT

w

Sw

wT

ffA

tq

dSS

f

A

tqdSwx

w

(9.45)

Thus the general Welge equation is,

12

12

12

1122

xx

ff

A

tq

xx

SxSxS wwTww

w

(9.46)

A useful simplification is to consider x1 = 0 at the inlet and x2 = L at the outlet end of the

core,

22 1 wT

ww fLA

tqSS

(9.47)

where fw1 is assumed to be one at the inlet.

Define the total volume injected, Wi, = qT*t, and the pore volume, Vp = AL. Combining

gives the number of pore volumes injected, Qi,

p

ii

V

WQ (9.48)

Thus we can write Eq. (9.47) in terms of Qi.

22 1 wiww fQSS (9.49)

The cumulative oil displaced, Np, can be expressed in terms of the difference in the

average water saturation and the exit end saturation, i.e.,

2wwpp SSVN (9.50)

Consider a special case immediately before breakthrough. In this case, Sw2 = Swi and fw2

= 0. Subsequently, Eq. (9.49) can be written as:

ibtwiwbt QSS (9.51)

and the cumulative oil displaced:

wiwbtpp SSVN (9.52)


9.13

9.4.3 Determination of relative permeability curves

Continuing from the previous section, the objective is to determine relative

permeability curves from linear displacement data. Writing Equation (9.49) in terms of

fractional flow of oil, fo,

022 fQSS iww (9.53)

Next, calculate ko/kw from the fractional flow equation.

1

1

22

ww

o

Sw

o

fk

k

w

(9.54)

In this example, both gravity and capillary pressure components are considered negligible

and thus are ignored. The average fractional flow of oil at the exit end for a given time

increment is given by,

i

WpWWp

ii

p

oW

NN

WdW

dNf iii

0

lim2 (9.55)

where dNp is cumulative oil produced and dWi is water injected during t. Alternative

expressions for dNp and dWi can be written as,

ii

wp

dQLAdW

SdLAdN

(9.56)

which results in a useful expression for fo2.

i

ww

i

wo

Q

SS

dQ

Sdf 2

2

(9.57)

The slope of a plot of average water saturation vs PVs of water injected provides an

estimate of fo2 (Figure 9.10).

Sw

Qi

breakthrough

0

Sw2

Figure 9.10 Plot of average water saturation vs Qi for determining fo2.


9.14

To estimate the permeability of each phase, begin with the Darcy multiphase flow

equation written in terms of pressure drop.

L

w

rw

o

rob

T

kkAk

dxqp

0

(9.58)

Pressure drop is measured across the core during the constant rate test. From the single

phase, steady state experiment we obtain,

b

bbb

p

LqAk

(9.59)

Define effective apparent viscosity,-1

, as:

1

1

w

rw

o

ro kk

(9.60)

Substitute Eqs. (9.59) and (9.60) into Eq. (9.58),

L

bb

bT dx

Lq

pqp

0

(9.61)

Define the average apparent viscosity as:

x

x

dx

dx

0

0

1

1

(9.62)

thus at the outlet end, x = L, Eq. (9.61) becomes,

bT

bb

pq

pq

1 (9.63)

Calculation of individual relative permeabilities requires values of -1

at known

saturations. For example, at the outlet end, where Sw2 is known,

1

2

2

1

2

2

wwrw

ooro

fk

fk

(9.64)

where the exit end apparent viscosity can be determined from the following relationship.


9.15

i

idQ

dQ

111

2

(9.65)

The derivative can be evaluated from the slope of a plot of the inverse of average

apparent viscosity vs. PVs water injected as shown in figure 9.11.

-1

Qi

breakthrough

Figure 9.11 Plot of inverse of average apparent viscosity vs Qi for determining fo2.

Example 9.1

An unsteady state test was performed at constant injection rate for the purpose of

determining the oil and water relative permeability curves. Table 1 lists the input

parameters for the test.

Swi = 0.35

Vp = 31.13 cc

w = 0.97 cp

o = 10.45 cp

q = 80 cc/hr

pb/qb= 0.1245 psi/cc/hr

Table 1 – Input Parameters


9.16

cumulative Cumulative

wtr injection oil produced p Qi Swave fo2 Sw2 fw2 kro/krw

Wi, (cc) Np, (cc) psi (PV)

0.00 0.00 138.6 0.000 0.350 1.000 0.350 0.000

3.11 3.11 120.4 0.100 0.450 1.000 0.350 0.000

7.00 7.00 97.5 0.225 0.575 0.585 0.443 0.415 15.166

11.20 7.84 91.9 0.360 0.602 0.154 0.546 0.846 1.963

16.28 8.43 87.9 0.523 0.621 0.083 0.577 0.917 0.980

24.27 8.93 83.7 0.780 0.637 0.038 0.607 0.962 0.425

39.20 9.30 78.5 1.259 0.649 0.019 0.625 0.981 0.208

62.30 9.65 74.2 2.001 0.660 0.009 0.641 0.991 0.103

108.90 9.96 70.0 3.498 0.670 0.005 0.653 0.995 0.053

155.60 10.11 68.1 4.998 0.675 0.002 0.666 0.998 0.018

311.30 10.30 65.4 10.000 0.681 0.001 0.669 0.999 0.013

Table 2. Calculations for relative permeability ratio

0.30

0.35

0.40

0.45

0.50

0.55

0.60

0.65

0.70

0.0 0.5 1.0 1.5 2.0

Qi

Sw

ave

Figure 1. Plot of average water saturation vs. pore volume water injected. Slope provides

exit end fractional flow of oil.

p fo2 fw2 Qi average m*

psi (PV) -1

2-1

Sw2 krw kro

138.6 1.000 0.000 0.000 13.50 13.50 0.350 0.000 0.774

120.4 1.000 0.000 0.100 11.73 -17.80 13.50 0.350 0.000 0.774

97.5 0.585 0.415 0.225 9.50 -10.68 11.90 0.443 0.034 0.514

91.9 0.154 0.846 0.360 8.95 -3.14 10.08 0.546 0.081 0.160

87.9 0.083 0.917 0.523 8.56 -1.90 9.56 0.577 0.093 0.091

83.7 0.038 0.962 0.780 8.15 -1.24 9.12 0.607 0.102 0.043

78.5 0.019 0.981 1.259 7.65 -0.76 8.60 0.625 0.111 0.023

74.2 0.009 0.991 2.001 7.23 -0.37 7.97 0.641 0.121 0.012

70.0 0.005 0.995 3.498 6.82 -0.20 7.51 0.653 0.129 0.007

68.1 0.002 0.998 4.998 6.63 -0.07 6.98 0.666 0.139 0.003

65.4 0.001 0.999 10.000 6.37 -0.05 6.90 0.669 0.141 0.002

Table 3. Calculations to determine relative permeabilites


9.17

0

2

4

6

8

10

12

14

16

0.0 5.0 10.0 15.0

Qi

-1

)av

e

Figure 2. Average apparent viscosity vs pore volume of water injected

Figure 3. Oil-water relative permeability curves


9.18

Step-by-step procedure

Measured data includes cumulative water injection (Wi), cumulative oil produced (Np)

and pressure drop (p) as shown in Table 2.

Step 1:

Calculate the cumulative pore volumes of water injected, Qi from Eq. 9.48.

p

ii

V

WQ

Step 2:

Calculate the average water saturation from Eq. 9.50.

p

p

wiwV

NSS

Step 3:

Calculate the exit end fractional flow of oil from the slope of Figure 1.

i

wo

Q

Sf

2

Step 4:

Calculate the exit end water saturation from Eq. 9.49.

22 oiww fQSS

Step 5:

Calculate exit end fractional flow of water by,

022 1 ffw

Step 6:

Calculate the relative permeability ratio as shown in Table 2 from Eq. 9.54.

1

1

22

ww

o

Sw

o

fk

k

w

Step 7:

Find the average apparent viscosity from Eq. 9.63.

bT

bb

pq

pq

1


9.19

Step 8:

Find the slope of the average apparent viscosity vs Qi plot, Figure 2.

iQm

1

*

Step 9:

Calculate the exitend apparent viscosity from Eq. 9.65. Results shown in Table 3.

*11

2 mQi

Step 10:

Calculate the individual relative permeabilities with respect to the outlet end,

where Sw2 is known,

1

2

2

1

2

2

wwrw

ooro

fk

fk

Results are shown in Table 3 and Figure 3, respectively.

9.4.4 Displacement Performance (constant pressure)

In some cases it is more advantageous to perform an experiment at a constant pressure

differential. As a result the injection rate is allowed to vary with time. One such

example, is the linear displacement of oil by gas. This technique of determining oil and

gas relative permeabilities by the unsteady state method is known as “gasflooding”.

The following data were obtained in a laboratory experiment to determine the relative gas

and oil permeability. Plot kro, krg vs. So.

Pinlet = 2.0 atm, abs

Poutlet = 1.0 atm, abs

o = 1.2 cp

g = 0.018 cp

Vp =180 cm3

qo = 0.40 cc/sec


9.20

Time (secs) Cumulative

gas injection

(cc)

Cumulative

oil produced

(cc)

0 0 0

104 50 42.5

134 75 49.0

199 150 56.0

238 200 58.5

276 250 60.3

381 400 63.7

447 500 65.5

518 600 66.3

577 700 67.4

635 800 68.1

693 900 69.0

750 1000 69.7

Solution

A laboratory experiment was run with a constant pressure drop between the inlet and

outlet. Measured were time, cumulative gas injected, and cumulative oil produced. Also

known are the oil and gas viscosities, pore volume of the sample and the single phase oil

rate prior to gas injection…saturate the core with oil, steady state process, at irreducible

water saturation.

Step 1: Plot cumulative oil production (Np) vs time. Determine oil flow rate by,

dt

pdN

oq

Step 2: Calculate the cumulative gas injected in terms of mean pressure and expressed in

pore volumes.

)

oP

iP(

iP2

pV

iG

)pv(i

Q

Step 3: Calculate the average gas saturation by,

giS

pV

pN

gS

Step 4: Determine the oil cut from the slope of a plot of average gas saturation vs Qi


9.21

idQ

gSd

of

Step 5: Determine the relative permeability ratio,

o

g

of

of1

rok

rgk

Step 6: Calculate the saturation at the outflow face

o

fi

Qg

Sg

S *2

Step 7: Determine kro by Darcy’s Law,

4.0

)t(o

q

oiq

)t(o

q

pA

Looi

q

pA

Lo

)t(o

q

k

ok

rok

Step 8: Determine the gas relative permeability

ro

k*

nrok

rgk

rgk

time

secs

Cumulative

Gas injection,

Gi, (cc)

Cumulative

oil produced

Np, (cc)

production

rate

qo (cc/sec)

Cumulative

Gas injection,

Qi, (pv)

Average gas

saturation Sg

oil cut

fo

0 0 0 0 0 0 1

104 50 42.5 0.366 0.370 0.236 0.490

134 75 49.0 0.142 0.556 0.272 0.101

199 150 56.0 0.091 1.111 0.311 0.057

238 200 58.5 0.056 1.481 0.325 0.032

276 250 60.3 0.036 1.852 0.335 0.020

381 400 63.7 0.030 2.963 0.354 0.016

447 500 65.5 0.019 3.704 0.364 0.010

518 600 66.3 0.015 4.444 0.368 0.007

577 700 67.4 0.015 5.185 0.374 0.007

635 800 68.1 0.014 5.926 0.378 0.006

693 900 69.0 0.014 6.667 0.383 0.006

750 1000 69.7 0.012 7.407 0.387 0.005


9.22

Figure 1. cumulative oil produced as a function of time

Figure 2. Average gas saturation vs pore volume of gas injected

Cumulative oil vs time

0

10

20

30

40

50

60

70

80

0 100 200 300 400 500 600 700 800

time, secs

Np

, cc

Average gas saturation vs PV gas injected

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0.45

0 1 2 3 4 5 6 7 8

Qi, pv

Sg

(av

e)


9.23

Figure 3. Oil and gas relative permeability curves as a function of exit end oil saturation

Cumulative

Gas injection,

Qi, (pv)

Average gas

saturation Sg krg/kro ratio

Exit end

saturation

Sg2

Exit end

saturation

So2 Kro Krg

0 0 0 0 1 1.000 0

0.37 0.236 0.016 0.055 0.945 0.914 0.014

0.56 0.272 0.133 0.216 0.784 0.355 0.047

1.11 0.311 0.248 0.248 0.752 0.228 0.057

1.48 0.325 0.450 0.277 0.723 0.140 0.063

1.85 0.335 0.754 0.299 0.701 0.091 0.069

2.96 0.354 0.947 0.308 0.692 0.076 0.072

3.70 0.364 1.523 0.328 0.672 0.047 0.072

4.44 0.368 2.090 0.337 0.663 0.037 0.076

5.19 0.374 2.207 0.339 0.661 0.038 0.085

5.93 0.378 2.485 0.343 0.657 0.034 0.086

6.67 0.383 2.485 0.343 0.657 0.035 0.086

7.41 0.387 2.842 0.348 0.652 0.031 0.087

Gas/oil relative permeability curves

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

So2

Kro

or

krg

krg

kro


9.24

9.5 Factors that control displacement efficiency

Before discussing the effect of various factors on displacement efficiency, we will

begin with an elaboration on “mobility ratio”. It is the most widely-used lumped

parameter used to estimate displacement performance. The general definition given by

Eq. (9.31) does not fully describe the terms. Consider a sharp front as illustrated in

Figure 9.11.

Sor

Swi

w)Sor

w)Swf o)Swi

Figure 9.11 Schematic of a sharp front in immiscible displacement including location of

mobility terms.

In this case, the mobility ratio is defined as the mobility of the displacing phase behind

the front to the mobility of the displaced phase ahead of the front.

d

DM

(9.66)

where,

wi

or

So

ood

Sw

wwD

k

k

(9.67)

Note both permeabilities are evaluated at the endpoints, Swi and Sor, respectively.

If no sharp front is evident, we define the “apparent mobility ratio”, Ms, as:

wiwf Sro

o

Sw

rws

k

kM

(9.68)

where the water mobility is evaluated at the average water saturation behind the front.

The apparent mobility ratio is a measure of the relative rate of oil movement ahead of the


9.25

front to the water movement behind the front, assuming the oil and water pressure

gradients are equal. Therefore, if

Ms < 1 oil rate > water rate….high displacement

Ms = 1 oil rate = water rate

Ms > 1 oil rate < water rate….poor displacement efficiency

The result is displacement efficiency (ED) decreases as apparent mobility increases as

shown schematically in Figure 9.12.

ED

Qi(pv)

Ms

3 1

5

Figure 9.12 Effect of apparent mobility ratio on displacement efficiency

In general form,

d

i

Sd

Si

M

(9.69)

where i is the mobility of the injected fluid evaluated at the average saturation of the

injected fluid at breakthrough, and d is the mobility of the displaced fluid evaluated at

the average saturation of the displaced fluid. Typical values are Ms of 0.2 to 10 for water

displacing oil, up to Ms of 1000 for gas displacing oil.

Wettability

The shape of the relative permeability curves are influenced by wettability, subsequently

impacting mobility ratio and fractional flow. Figure 9.13 illustrates the effect of

wettability on the fractional flow of water.


9.26

Sw

=47°

Slightly Water wet

fw

oil wet

=180°

Figure 9.13 Effect of wettability on fractional flow of water

A decrease in water wetness, results in an increase in krw and a corresponding decrease in

kro. The mobility term increases and thus becomes more unfavorable; i.e., poorer

displacement efficiency. Figure 9.14 illustrates the decrease in efficiency in terms of

pore volumes of oil produced vs pore volumes of water injected.

Np

(pv)

Qi(pv)

47°

2.5

0.4

0

0.3

180°

Incremental due

to wettability

Figure 9.14 Effect of apparent mobility ratio on displacement efficiency

Interfacial Tension

Recovery efficiency for displacement of oil by water is a weak function of the

interfacial tension. As shown in Figure 9.15, the difference in oil recovery is minimal for

a large range of interfacial tensions.


9.27

Np

(pv)

Qi(pv)

=0.5

0.4

0

=40

Figure 9.15 Effect of interfacial tension on displacement efficiency

Recall, capillary pressure is a function of interfacial tension. In displacement, the

capillary pressure gradient is included in the fractional flow (See Eq. 9.29). Therefore,

the weak function suggests capillary pressure is not a dominant component of

displacement.

Viscosity Ratio

As the viscosity of oil increases the mobility ratio will correspondingly increase,

resulting in an increase in the fractional flow of water. Figure 9.16 illustrates the effect

for three arbitrary viscosity ratios, 100, 10 and 1, respectively.

Sw

o/w= fw

100 10 1

Figure 9.16 Effect of viscosity ratio on fractional flow of water

In fact, according to Eq. 9.31, as M increases, then fw 1.0. The oil recovery for various

viscosity ratios can be seen in Figure 9.17.


9.28

Np

(pv)

Qi(pv)

o=1.8cp

0

Incremental due

to oil viscosity

o=151cp

Figure 9.17 Effect of viscosity ratio on displacement efficiency

The ultimate recovery is independent of the viscosity ratio; however, the time to recover

the oil is highly dependent on the ratio.

Gravity

The influence of gravity on displacement can be explained by observing the

reservoir configuration shown in Figure 9.18.

wtr

oil

Figure 9.18. Stratigraphic reservoir configuration

By definition, = o – w in the fractional flow equation for water (Eq. 9.29).

Assuming gravity is (+) upwards, then since o < w gravity will reduce the fractional

flow of water when the water is moving updip. Conversely, water injected at the crest of

the structure will move faster under the influence of gravity. The subsequent

displacement efficiency and oil recovery are less in this case.

9.6 Residual oil saturation

After displacement, there exists a remaining oil saturation known as the residual

oil saturation. It is dependent on wettability, pore size distribution, microscopic

heterogeneity and properties of the displacing fluids. As an example, consider for water


9.29

wet rock the oil is trapped as globules or ganglia. But for oil wet rock the oil is trapped

as a film on the grain surfaces. The importance of understanding the residual oil

saturation is it establishes a maximum efficiency for oil displaced by water on the

microscopic level. Furthermore, it is the initial oil saturation for the next possible phase

of development; i.e., EOR.

The measure of the effectiveness of the displacement process is defined by the

microscopic displacement efficiency, ED.

11 /

/1

by water contactedd/unit waterflooof beginningat OIP stock tank

by water contacted Vprecovered/ oil stock tank

oo

oo

D

BS

BS

E

(9.70)

Where, So1 is the volumetric average oil saturation at the beginning of the waterflood and

So is the volumetric average oil saturation at a particular point during the waterflood. The

oil displaced is given by;

1

1

o

opwDp

B

SVEN

w (9.71)

where Npw is the oil displaced by water and Vpw is the pore volume swept by water to the

volumetric average residual oil saturation.

The dependence of residual oil saturation on capillary and viscous forces was

verified by a series of experiments. Using dimensional analysis, define Capillary

Number, NCA, as the ratio of viscous to capillary forces.

cosow

wCA

vN (9.72)

Where v is the interstitial velocity (u/) and w is the viscosity of the displacing fluid.

Experimental data was obtained by measuring the oil saturation in cores when the first

water is detected at the outlet. Since the oil volume produced after breakthrough is small,

the results represent the trapping process. Figure 9.19 illustrates the general behavior of

the reduction in oil saturation in the core at breakthrough with increasing capillary

number.


9.30

0 NCA 10

-8 10

-3

50

So,%

pv

Figure 9.19 Reduction of oil saturation at breakthrough vs capillary number

At smaller capillary numbers, capillary forces dominate. With increasing capillary

number the viscous forces become more dominate.

A modification of the original definition of capillary number was developed for

waterfloods at constant injection rate.

4.0

cos)(

o

w

oworoi

wCAM

SS

vN

(9.73)

The following guidelines are suggested.

NCAM < 10-6

capillary forces dominate

10-4

< NCAM < 10-5

transition

NCAM > 10-4

viscous forces dominate

Correlating residual oil saturation with capillary and viscous forces has several

important implications for fluid flow in porous media. It demonstrates the independence

of Sor from flood velocity at reservoir rates. Furthermore, correlations illustrate that Sor

can be reduced below field waterflood residual in the laboratory corefloods if the lab

experiments are conducted at large NCAM.

In many field applications reservoir pressure has depleted to the point where

appreciable free gas saturation exists in the pores. Subsequently, prior to water injection

both a residual oil and gas saturation co-exist. If re-pressurization occurs during water

injection, the gas will dissolve back into the oil with little, if any, effect on the residual oil

saturation. However, if a trapped gas saturation is present at the time the residual oil is

trapped by water, a substantial reduction of residual oil saturation will occur. For

example, a correlation shown in Figure 9.20 illustrates the reduction in residual oil

saturation for a water-wet, consolidated rock. Implied in the figure is that an increase in


9.31

initial flowing gas saturation is proportional to an increase in trapped gas saturation and

thus a reduction in residual oil saturation.

Sgi,%

10

0 30

S

or,%

Figure 9.20 Reduction in Sor for increasing initial gas saturation

9.7 Limitations of the frontal advance solution

In development of the frontal advance solution several limitations are evident.

Based on the assumptions in deriving the solution, the fluids were considered immiscible

and incompressible. Furthermore, the porous media was assumed isotropic and

homogeneous, with uniform saturation distributions. And last, only one-dimensional,

linear flow was illustrated.

The frontal advance solution applies to a stabilized displacement process. In

other words, the displacement behavior is independent of injection rate and length of the

sample. Two parameters which must meet the criteria are the breakthrough saturation

and the recovery vs PV injected after breakthrough. An empirical correlation from

dimensional analysis was developed to determine if a flood was at stabilized conditions.

}/{75.762.0

}{1085.510835.0

2

99

daycpftto

NxtoxLu wT

(9.74)

The left-hand-side is known as the “critical scaling coefficient”. If the numerical value of

this coefficient exceeds the critical value, then stabilized flow will occur. Since rate and

length effects occur when capillary forces become important in the displacement process,

this scaling factor also indicates when capillary forces are minimized. In applications,

under field conditions the displacement process is almost always stable. Under lab

conditions, to compute relative permeabilities from linear displacement tests, it is


9.32

necessary to estimate the operating conditions to obtain stabilized flow. Two examples

from Willhite (1986) illustrate the point.

Example 1

A reservoir is 1000 ft long, and was flooded at an average frontal velocity of 1 ft/day.

The porosity of the reservoir is 19% and the displacing fluid viscosity is 0.7 cp. Estimate

the scaling coefficient and determine whether the displacement was stabilized.

Solution

In oilfield units, the value of uT = 0.19 ft/day (=1 ft/day*.19). Thus,

daycpftLu wT /133)7.0)(19.0)(1000( 2

This value is an order of magnitude greater than the critical values observed in lab

experiments, and therefore flow is stabilized.

Example 2

It is desired to conduct a laboratory waterflood experiment under stabilized conditions in

a core 2.54 cm in diameter and 5 cm long. The porosity of the core is 15% and the

viscosity is 1 cp [1 kPa-s]. Estimate the volumetric injection rate in cubic meters/second

if the critical scaling coefficient is 5.85 x 10-9

N.

Solution

TTwT uxsPaumLu 5105)001.0)()(05.0(

Substituting the critical value, results in uT = 1.17x10-4

m/s. Subsequently, the

volumetric rate becomes,

smxq

mxAuq T

/1093.5

)0254.0(4

1017.1

38

24

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Chapter 9 Simultaneous Flow of Immiscible...

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