Simultaneous Flow of Immiscible Fluids

Simultaneous Flow of Immiscible Fluids

• Purpose: predict the displacement of oil by water

• Organization: – development of equations of multiphase, immiscible flow

• concluding with the frontal advance and Buckley-Leverett equations.

– factors that control displacement efficiency

– limitations of immiscible displacement solutions


(1). Change in saturation

(2). Variation of density with temperature and pressure

(3). Change in porosity due to a change in confining stress.

increment in time saccumulate

thatphase of mass

increment in time

leaving phase of mass

increment in time

entering phase of mass

uox│x uox│x+x

x

y z

Development of Equations

tAutAutAu zzozoyyoyoxxoxo

tAutAutAu zzzozoyyyoyoxxxoxo

toottoo VSVS


• Phase dependent continuity equations

• Apply Darcy’s Law


ooozooyooxo St

uz

uy

ux

wwwzwwywwxw St

uz

uy

ux

x

ku i

i

iixix

ooo

o

o

ozoo

o

oyoo

o

oxo St

gz

pk

zy

pk

yx

pk

x

www

w

w

wzww

w

wyww

w

wxw St

gz

pk

zy

pk

yx

pk

x


• To combine requires:

So + Sw = 1.0

And


wP

oPor

wP

nwP

cP


• Oil and water are injected simultaneously

• rates and pressures are measured

• core saturation is determined gravimetrically.

• Permeability is unknown.

• Steady state, incompressible diffusivity Eqs.

• Assume water saturation is uniform throughout the core

Steady state, linear solution

qo

qw L

poi

Pwi

poL

PwL

D

0

0

dx

dpk

dx

d

dx

dpk

dx

d

ww

oo

)( oLoi

ooo

ppA

Lqk


Methods to avoid 1. inject at a sufficiently high rate 2. The second method is to attach a thin, (high porosity and high permeability) Berea sandstone

plug in series

Capillary End Effects

gap

Pc=0

Sw

0 0 L L

Po

Pw

Pc=0+ Swc

Sor

P

Simultaneous Flow of Immiscible Fluids Frontal advance – USS, 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

Progression of water displacing oil for immiscible, 1D


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

• In terms of volumetric flow rate,

• Assume the fluids are incompressible and the porosity is constant.

• Combining,

Buckley - Leverett

oooxo St

ux

wwwxw S

tu

x

t

SA

x

q oo

t

SA

x

q ww

oooo St

Aqx

wwww St

Aqx

0

t

SSA

x

qq owow qT = qo + qw = constant


• From the definition of fractional flow,

• Substitute into Darcy’s equation for each phase,

• complete fractional flow equation.

Buckley - Leverett

Two

Tww

qfq

qfq

)1(

sin)1( gx

pAkqfq o

o

o

oTwo

singx

pAkqfq w

w

w

wTww

ow

wo

c

To

o

ow

wow

k

k

gx

p

q

Ak

k

kf

1

sin

1

1


• fractional flow equation reduces to,

• If we define mobility ratio as,

• then fw = 1/(1+1/M)

Buckley - Leverett

ow

wo

c

To

o

ow

wow

k

k

gx

p

q

Ak

k

kf

1

sin

1

1

x

S

S

p

x

p w

w

cc

Sw

Pc

0

w

c

S

p

ow

wow

k

kf

1

1

wo

ow

k

kM


• From

• substitute for qw,

• reduce to one dependent variable. Observe, Sw = Sw(x,t) or,

• Let dSw(x,t)/dt = 0, then the velocity of the saturation front is given by

Frontal Advance

t

SA

x

q ww

t

S

q

A

x

f w

T

w

dtt

Sdx

x

SdS

t

w

t

ww

t

w

x

w

S

xS

tS

dt

dx

w


• Observe fw = fw(Sw) only, then,

• Substitution results in the frontal advance equation

Frontal Advance

t

w

tw

w

t

w

x

S

S

f

x

f

tw

wT

S S

f

A

q

dt

dx

w

Represents the velocity of the saturation front. Basic assumptions in the derivation are : 1. incompressible fluid, fw(Sw) only 2. immiscible fluids. 3. only oil is displaced; i.e., the initial water

saturation is immobile, and 4. no initial free gas saturation exists; i.e., not

a depleted reservoir.


• injection rate is constant and if the dfw/dSw = f(Sw) only, then the location of the front is given by:

Frontal Advance

Swf Sw Swc

fw

fwf

Swbt w

w

Sw

wT

S S

f

A

tqx

fractional flow of water at the front

water saturation at the front

average water saturation behind the front at breakthrough


Prior to breakthrough

Volume of oil produced (Np) =

Volume of water injected (Wi)

Displacement performance Constant injection rate

Np

Qi

breakthrough

after breakthrough

• Water saturation gradients exists

• Thus the rate of oil recovery decreases

• Apply Welge’s solution to predict waterflood performance


• Average water saturation

where fw1 is assumed to be one at the inlet.

• pore volumes injected, Qi,

• Thus in terms of Qi,

• 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.,


22 1 wT

ww fLA

tqSS

p

ii

V

WQ

22 1 wiww fQSS

2wwpp SSVN

Total volume injected, Wi

Pore volume, Vp


• Consider a special case immediately before breakthrough. In this case, Sw2 = Swi and fw2 = 0. Subsequently:

• and the cumulative oil displaced:


ibtwiwbt QSS

wiwbtpp SSVN


Example: An unsteady state test was performed at constant injection rate for the purpose of determining the oil and water relative permeability curves.

Determination of Relative permeability curves

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

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

Input data


Step 1: Calculate the cumulative pore volumes of water injected,





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

p

ii

V

WQ

p

p

wiwV

NSS

Step 2: Calculate the average water saturation


Step 3: Calculate the exit end fractional flow of oil from the slope





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

i

wo

Q

Sf

2

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


Step 4: Calculate the exit end water saturation





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

Step 5: Calculate exit end fractional flow of water

22 oiww fQSS 022 1 ffw


Step 6: Calculate the relative permeability ratio





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

1

1

22

ww

o

Sw

o

fk

k

w

Simultaneous Flow of Immiscible Fluids Determination of Relative permeability curves

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

bT

bb

pq

pq

1

Step 7: Find the average apparent viscosity


Step 8: Find the slope of the average apparent viscosity vs Qi plot


iQm

1

*


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

0

2

4

6

8

10

12

14

16

0.0 5.0 10.0 15.0

Qi

-1

)av

e


Step 9: Calculate the exitend apparent viscosity



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

Step 10: Calculate the individual relative permeabilities with respect to the outlet end, where Sw2 is known

*11

2 mQi 1

2

2

1

2

2

wwrw

ooro

fk

fk



Example: An unsteady state test was performed at constant pressure differential for the purpose of determining the oil and gas relative permeability curves.

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


Input data

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


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


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

dt

pdN

oq

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


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


Step 3: Calculate the average gas saturation

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

)o

Pi

P(

iP2

pV

iG

)pv(i

Q

gi

S

pV

pN

gS


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


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

idQ

gSd

of

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)


Step 5: Determine the relative permeability ratio


Step 6: Calculate the saturation at the outflow face

o

g

of

of1

rok

rgk

of

iQ

gS

gS *

2

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


Step 7: Determine kro by Darcy’s Law


Step 8: Determine the gas relative permeability

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

4.0

)t(o

q

oiq

)t(o

q

pA

Looi

q

pA

Lo

)t(o

q

k

ok

rok

ro

k*

nrok

rgk

rgk


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


• 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.

• Endpoint mobility ratio

• Apparent mobility ratio, Ms

Controlling factors of Displacement efficiency

Sor

Swi

w)Sor

w)Swf o)Swi

d

DM

wi

or

So

ood

Sw

wwD

k

k

wiwf Sro

o

Sw

rws

k

kM


• Apparent mobility ratio, Ms - measure of the relative rate of oil movement ahead of the front to the water movement behind the front

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

Ms = 1 oil rate = water rate

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


ED

Qi(pv)

Ms

3 1

5

Effect of apparent mobility ratio on displacement efficiency


A decrease in water wetness

Increase/decrease in krw ?

Increase/decrease in kro?

mobility term becomes more/less unfavorable ?

Poorer or better displacement efficiency?


Sw

=47°

Slightly Water wet

fw

oil wet

=180°

Np

(pv)

Qi(pv)

47°

2.5

0.4

0

0.3

180°

Incremental due

to wettability

Wettability – contact angle


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

• suggests capillary pressure is not a dominant component of displacement.


Wettability – interfacial tension

Np

(pv)

Qi(pv)

=0.5

0.4

0

=40

ow

wo

c

To

o

ow

wow

k

k

gx

p

q

Ak

k

kf

1

sin

1

1

rPc

cos2

Simultaneous Flow of Immiscible Fluids Controlling factors of Displacement efficiency

Sw

o/w= fw

100 10 1

Viscosity ratio

Np

(pv)

Qi(pv)

o=1.8cp

0

Incremental due

to oil viscosity

o=151cp

wo

ow

k

kM

As viscosity of oil increases:

Mobility ratio will Increase or decrease?

fw will Increase or decrease?

Poorer or better displacement efficiency?

ow

wow

k

kf

1

1


• fractional flow equation for water

• where = o – w

• Water moving updip will reduce the fractional flow of water

• 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.


Gravity effect

wtr

oil

owk

wok

1

singx

cp

Tq

o

Ao

k

owk

wok

1

1w

f

+

_


• ROS is dependent upon: – Wettability

– Pore size distribution

– Hetergeneity

– Properties of displacing fluid

• Importance of ROS: – Establishes the maximum efficiency for the displacement of oil by water on a

microscopic level

– It is the initial saturation for EOR processes in regions of a reservoir previously swept by a waterflood


Residual Oil Saturation


• measure of the effectiveness of the displacement process is defined by the microscopic displacement efficiency, ED.

• 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.

• Maximum displacement efficiency,

• Oil displaced is given by;



11 /

/1

by water contactedd/unit waterflooof beginningat OIP stock tank

by water contacted Vprecovered/ oil stock tank

oo

oo

D

BS

BS

E

1

1max

o

opwDp

B

SVEN

w

11

max/

/1

oo

oorD

BS

BSE

Npw is the oil displaced by water Vpw is the pore volume swept by water to the volumetric average residual oil saturation.


• define Capillary Number, NCA, as the ratio of viscous to capillary forces

• At small capillary numbers,

capillary forces dominate.

• With increasing capillary number

the viscous forces become more

dominate

• Importance: guide to ensure ROS from lab tests on small cores at high rates are representative of the ROS in the field



cosow

wCA

vN

0 NCA 10

-8 10

-3

50

So,%

pv

Reduction of oil saturation at breakthrough vs capillary number


• modification of the original definition of capillary number was developed for waterfloods at constant injection rate

• NCAM < 10-6 capillary forces dominate

• 10-4 < NCAM < 10-5 transition

• NCAM > 10-4 viscous forces dominate



4.0

cos)(

o

w

oworoi

wCAM

SS

vN

0 NCA 10

-8 10

-3

50

So,%

pv

Capillary forces

dominate

viscous forces

dominate

transition


• 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.


Free Gas Saturation

Invading Water bank

Connate water

Initial free gas

Initial oil saturation

oil bank


• However, a significant decrease in displacement performance occurs


Free Gas Saturation

0

10000

20000

30000

40000

50000

60000

70000

0.0 0.5 1.0 1.5 2.0

oil

dis

pla

ced

, b

bls

Qi, PV injected

Sgi = 15%

Sgi = 0

Decrease in oil recovery

increase in fillup time

Initial gas saturation = 15% Mobile gas saturation = 10% Trapped gas redissolves in oil


• 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.


Free Gas Saturation

Sgi,%

10

0 30

S

or,%

Reduction in Sor for increasing initial gas saturation (increasing trapped gas saturation)

Invading Water bank

Connate water

Initial free gas

Initial oil saturation

oil bank

Trapped gas


• fluids were considered immiscible and incompressible.

• The porous media was assumed isotropic and homogeneous, with uniform saturation distributions.

• Only one-dimensional, linear flow was illustrated.

• Stabilized displacement process displacement behavior is independent of injection rate and length of the sample

• Critical scaling factor - empirical correlation from dimensional analysis was developed to determine if a flood was at stabilized conditions

• Applications, – under field conditions the displacement process is almost always stable.

– Under lab conditions, to compute relative permeabilities from linear displacement tests, it is necessary to estimate the operating conditions to obtain stabilized flow

Limitations of the Frontal advance solution

}/{75.762.0

}{1085.510835.0

2

99

daycpftto

NxtoxLu wT


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).

• This value is an order of magnitude greater than the critical values observed in lab experiments, and therefore flow is stabilized


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


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

• Substituting the critical value, results in uT = 1.17x10-4 m/s. Subsequently, the volumetric rate becomes,


TTwT uxsPaumLu 5105)001.0)()(05.0(

smxq

mxAuq T

/1093.5

)0254.0(4

1017.1

38

24

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Simultaneous Flow of Immiscible Fluids

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