module 5 - fractional flow theory

PETE 609 - Module 5

Fractional Flow Theory

Class Notes for PETE 609 – Module 5 /65

Author: Dr. Maria Antonieta Barrufet –Fall 2001

This module is protected by copyright. No part of any of these

pages may be reproduced in any form or by any means,

electronic or otherwise, without written permission from the

copyright owner

© Maria Barrufet – 979-8450314 / [email protected]

Learning Objectives

After completion of this module, you will be able to:

Estimate oil recoveries using fractional flow theory and Buckley-Leverett

1-D displacement

Module 5 – Fractional Flow Theory

Estimated Duration: 1 week

Fractional Flow Theory.

Homogeneous Reservoirs.

Application of Fractional Theory in Oil Recovery Calculations: Homogeneous

and Stratified Reservoirs.

Dissipation in Immiscible Displacements.

Suggested reading: MAB, R8, R18, S

mailto:[email protected]

PETE 609 - Module 5




Evaluate the effect of relative permeabilities and fluid viscosities in the

fractional flow equation

Estimate oil recoveries using Buckley-Leverett method

Estimate oil recoveries using Styles method

Estimate oil/water production ratios

PETE 609 - Module 5





To derive the fractional flow equation oil displacement will be assumed to take

place under the so-called diffusive flow condition. The constraints are that fluid

saturations at any point in the linear displacement path are uniformly distributed

with respect to thickness, This allows the displacement to be described

mathematically in one dimension.

The simultaneous flow of oil and water can be modeled using thickness averaged

relative permeabilities, along the centerline of the reservoir.

The condition for fluid potential equilibrium is simply that of hydrostatic

equilibrium for which the saturation distribution can be determined as a function

of capillary pressure and therefore, height. That is the fluids are distributed in

accordance with capillary-gravity equilibrium.

The condition of vertical equilibrium will be favored by

1. a large vertical permeability kv

2. small reservoir thickness (h)

3. large density difference between the fluids

4. high capillary forces meaning large capillary transition zone (H)

5. low fluid viscosities

6. low injection rates

The diffuse flow condition occurs when:

1. The displacement occurs at very high injection rates so that the effects

of capillary and gravity forces are negligible. The vertical equilibrium

condition is not satisfied.

2. The displacement is at low injection rates in reservoirs for which the

measured capillary transition zone greatly exceeds the reservoir

thickness and the vertical equilibrium condition applies.

PETE 609 - Module 5




The second condition can be can be visualized by observing ‎Figure 1. Since the

transition zone (H) is much larger than the reservoir thickness. The water

saturation can be considered uniformly distributed with respect to the reservoir

thickness.

Reservoir

Thickness

Swc 1-Sor

PcH

h Reservoir

Thickness

Swc 1-Sor

PcH

Swc 1-Sor

PcH

h

Figure 1 - Approximation to the diffuse flow condition for H >>h.

As a reference ‎Figure 2 indicates a small transition curve.

PETE 609 - Module 5




Pc

Swc 1-Sor

H

Small transition zone

Pc

Swc 1-Sor

H

Small transition zone

Figure 2 - An example of a capillary curve with a small transition zone.

‎Figure 3 shows a schematics of the top view of a linear reservoir which has

uniform cross sectional area A. Displacement will be considered in this prototype

reservoir model which can be tilted as indicated in ‎Figure 4.

PETE 609 - Module 5




Figure 3 - Linear prototype reservoir model. 1-D displacement. Top view.

L

Production

Injection

w

PETE 609 - Module 5




Figure 4 - Linear prototype reservoir mode. 1-D displacement. Cross section.

Both injection and production wells are considered to be perforated across the

entire, formation thickness, in the dip-normal direction.

The objective in this module is to describe the fluid saturation distributions in the

y-direction as the fluid moves through the x-direction.

Considering oil displacement in the tilted reservoir of ‎Figure 4, we apply Darcy’s

equation, for linear flow. The 1-D equations for the simultaneous flow of oil and

water are

L

Production

Injection

h

xy

z

qt

qi

sin

sin

dx

dz

xzL

Production

Injection

h

xy

z

qt

qi

L

Production

Injection

h

xy

z

qt

qi

LL

Production

Injection

h

xy

z

qt

qi

sin

sin

dx

dz

xz

sin

sin

dx

dz

xz

PETE 609 - Module 5




ro o oo

o

k k Aq

x (1)

rw w ww

w

k k Aq

x (2)

where the potential is defined as

P Pgz gx sin

(3)

thus, the flow rate for oil is

ro o oo

o

k k P g sinq

. x

61 0133 10x

(4)

and the flow rate of water is

rw w ww

w

k k P g sinq

. x

61 0133 10x

(5)

Subtract Equation ‎(4) from Equation ‎(5) and recall

PETE 609 - Module 5




c o w

t o w

o wo o cw w

ro rw

P P P capillary pressure

q q q total flow rate

g sinq Pq

k k A k k A . x

61 0133 10x

(6)

The number in Equation ‎(6) is a conversion factor when the units in this equation

are expressed as:

P atm

cp

mass gr

3gr/cm

k darcy

3q cm / sec

2A cm

If in Equation ‎(6) we substitute the oil flow rate in terms of water and total flow

rate we obtain,

o t o cww

rw ro ro

q P g sinq A

k k k k k k x . x

61 0133 10

(7)

PETE 609 - Module 5




Define fractional water flow wf in the reservoir as:

w ww

w o t

q qf

q q q

(8)

Substitute Equation ‎(8) in Equation ‎(7),

ro c

t ow

w ro

o rw

k k A P g sin

q x . x f

k

k

61

1 0133 10

1

(9)

or in field units,

ro c

t ow

w ro

o rw

k k A P. x . sin

q xf

k

k

31 1 127 10 0 4335

1

(10)

Recall that when field units are used the following properties are defined in the

following units

L ft

velocity ft/sec

time hr

liquid rate stb/day

PETE 609 - Module 5




gas rate MSCF/day

mass lbm

Viscosity and permeability have the same units as in the SI system.

General Shape of the Fractional Flow Equation

Since the viscosities and densities are assumed constant, the fractional flow

equation depends only upon saturation through the relative permeabilities. ‎Figure

5 shows a general shape of the fractional flow equation.

PETE 609 - Module 5




fw= 0

fw= 1

Swc

fw

Sw

1-Sor

Swf

Swbt

Figure 5 - Typical fractional flow curve as a function of water saturation.

For horizontal flow and neglecting the capillary pressure gradient we have

w hro o

rw w

fk

k

1

1

(11)

Then we can express Equation ‎(10) as

PETE 609 - Module 5




o cw w h

o t o t

k . x A P . x A sinf f

q x q

3 41 127 10 4 8855 101

(12)

or

w w c ghf f N N 1 (13)

cN = Capillary number, dimensionless

gN = Gravity number, dimensionless

In a previous exercise we have already found typical ranges for these

dimensionless numbers.

Effects of Terms in fw

In this module we have adopted Dake’s (1988) convention for the angle which

is measured from the horizontal to the line indicating the direction of flow. Thus

the gravity term . sin 0 4335 will be positive for oil displacement in the

up-dip direction 0<< (‎Figure 4), and negative for displacement in the down-dip

direction (<<2). Therefore gravity tends to suppress the flow of water.

The effect of the capillary pressure gradient can be qualitative understood by

expressing the gradient following the chain rule of differentiation.

PETE 609 - Module 5




c c w

w

P P S

x S x

(14)

‎Figure 6 indicates the first term of the capillary gradient. The slope of the

capillary pressure curve versus saturation is always negative. That is to decrease

the water saturation the capillary pressure must increase.

Sw

Pc

Swc 1-Sor

+ dSw

-dPc

Figure 6 - Capillary pressure as a function of water saturation.

The second part of the capillary gradient is indicated in ‎Figure 7 where it can also

be seen that the slope is always negative as well. Therefore cP / x is always

PETE 609 - Module 5




positive and consequently its effect would be to increase the fractional flow of

water.

x

Sw

Swc

1-Sor

front+ dx

-dSw

Saturation profile at a given time

Swf

Figure 7 - Water saturation distribution as a function of distance in the

displacement path.

The capillary gradient will increase fractional flow of water, but it is normally

ignored. Since it would involve an iterative procedure because wS / x is not

known as that is what we are trying to find out as a result of the displacement

calculations.

The water saturation distribution sketched in ‎Figure 7 corresponds to the

situation of after injecting a given volume of water. The diagram shows that there

is a distinct flood front ( a shock front), at which point there is a discontinuity in

PETE 609 - Module 5




the water saturation which increases abruptly from wcS to wfS , the front flood

saturation.

Behind the front there is a gradual increase of saturation from wfS to the

maximum value orS1 .

The fractional flow equation is used to calculate the fraction of the total flow

which is water, at any point in the reservoir, assuming the water saturation at that

point is known. To determine when a given water saturation plane reaches a

particular point in the linear system requires the use of the displacement theory

of Buckley-Leverett.

Homogeneous Reservoirs

Buckley-Leverett 1-D Displacement

For simplicity we will consider the displacement in a horizontal reservoir

(sin ) 0 , and we will neglect the capillary pressure gradient.

The conservation of mass of water flowing through a volume element A dx is

sketched in ‎Figure 8.

PETE 609 - Module 5




qw wx qw wx+dx

x x+dx

dx

m=v

m=q

Figure 8- Mass flow rate of water through a linear volume elements A dx .

The equation of conservation of mass of water flowing through a volume element

A dx is

w w w w w w w wx xq q q dx A dx S

x t

(15)

which can be reduced to

w w w wq A Sx t

(16)

and for the assumption of incompressible displacement w constant

PETE 609 - Module 5




w w

t x

q SA

x t

(17)

These derivatives followed with a bar with a subscript such as t or x indicate that

the derivative is evaluated at a given time and at a given distance.

Saturation Waves

Changes in saturation o wS ,S with time and position are called saturation

waves.

oS vs t @ fixed x saturation history

oS vs x @ fixed t saturation profile

Recall the dimensionless variables

D

xx

L (18)

t t

D

po o

udt qdtt

L V

(19)

Note: D Dx ,t plots show profiles with histories and transition or mixing

zones. The length of the dimensionless mixing zone is arbitrarily defined as

PETE 609 - Module 5




. .

D

D D S D St

x x x 0 1 0 9

(20)

. wJ wI wI

. wJ wI wI

S . S S S

S . S S S

0 1

0 9

0 1

0 9 (21)

Where

I = initial conditions

J = injected conditions

The velocity of a plane of constant saturation wdS 0 is given by the

derivative of the dimensionless distance with the dimensionless time, which can

be expressed as

D

w

w D

w D xDS

D w DS t

S / tdxv

dt S / x

(22)

‎Figure 9 shows an example of a plot of vsD Dx t where we can visualize

saturation profiles and saturation histories. For example at Dx 1 we have the

effluent history, and for a fixed dimensionless time such as Dt . 1 2 we can

observe the saturation profile.

PETE 609 - Module 5




0

0 .2

0 .4

0 .6

0 .8

1

0 0 .2 0 .4 0 .6 0 .8 1 1 .2 1 .4 1 .6 1 .8 2

xD

tD

S1=S1*

S1=0.6

S1=0.7

S1=0.76

S1=S1J

Saturation history

at xD=0.6

Saturation profile

at tD=1.2

Effluent history

Constant

saturation

S1I

Spreading

wave region

Constant

saturation S1J

Mixing zone

xD at tD=0.14

Figure 9 - Time-distance diagram for displacement

PETE 609 - Module 5




Classification of Waves

Waves can be classified according to the behavior of the mixing zone with the

dimensionless time. ‎Table 1 summarizes this classification.

1. Spreading

becomes more diffuse on propagation

Dx increases with Dt

2. Sharpening

becomes less diffuse on propagation, if no

dissipation present, and will become shocks

Dx decreases with Dt

3. Mixed may be spreading or sharpening; depends upon time

and limits of saturation

4. Indifferent Dx constant; typical of miscible waves

Table 1 - Classification of waves.

If we take the total differential of the water saturation using time and space

coordinates (not dimensionless) w wS S x,t

w ww

t x

S SdS dx dt

x t

(23)

and since it is the intention to study the movement of a plane of constant water

saturation, that is, wdS 0 , then

w

w w

x t S

S S dx

t x dt

(24)

PETE 609 - Module 5




The subscript below the parenthesis indicates that the derivative is evaluated

holding constant that property (i.e., space, time, or water saturation).

Furthermore, using the chain rule we can express

w w w

t w t

q q S

x S x

(25)

Combine Equation ‎(17), ‎(24) and ‎(25),

w

w w w

t t Sw t

q S S dxA

S x x dt

(26)

Recall t w w tq constant; q f q

w

w

Sw t

q dxA

S dt

(27)

And the velocity of a plane of constant saturation is

w

w

tw wS

S w wt t

qdx q fv

dt A S A S

1 (28)

ww

t w

S w t S

qdx df

dt A dS

(29)

PETE 609 - Module 5




Note that we could drop the subscript t in w

w t

f

S

because wf does not

depend upon time, and also it can be written as a total derivative.

Equation ‎(29) is the Buckley-Leverett Equation which implies that for a constant

rate of water injection t iq q , the velocity of a plane of constant water

saturation is directly proportional to the derivative of the fractional flow equation

evaluated for that saturation.

For a constant rate of water injection let

i tW q t = cumulative water injected

Integrate Equation‎(29)

Sw

w

w

w

x tw

to o

w S

w iS

w S

dfdx q dt

dS A

df Wx

dS A

1

(30)

Therefore at any time after the start of the injection the position of the different

water saturation planes can be plotted using Equation ‎(30), just by determining

the slope of the fractional flow curve at the particular value of each saturation.

The following procedure outlines the major steps taken to solve this problem.

PETE 609 - Module 5




Procedure to Evaluate the Buckley-Leverett Equation

Construct fractional flow equation given ro rwk , k as a function of

w o w o wS , , , , ...etc.

Choose time i tt t ; W q t 1 1 .

Cover wS range from wcS to orS1 with wS . 0 1

Evaluate w

w

df

dSat each wS and evaluate Equation ‎(30)

A typical figure obtained through this procedure is drawn in ‎Figure 5, which is

repeated below.

PETE 609 - Module 5




fw= 0

fw= 1

Swc

fw

Sw

1-Sor

Swf

Swbt

‎Figure 5 - Typical fractional flow curve as a function of water saturation.

However, there is a mathematical difficulty when using this technique. Since

frequently there is an inflexion point in the fractional flow curve then the plot of its

derivative will exhibit a maximum and will look like the one sketched in ‎Figure 10,

which can be appreciated by looking at the derivative of the fractional flow curve.

PETE 609 - Module 5




Sw

Swc 1-Sor

w

wS

dS

dfv

W

Figure 10 - Saturation derivative of a typical fractional flow curve.

If we use these derivatives to plot the saturation distribution at a particular time

the result would be the red line in ‎Figure 11. This profile is physically impossible

since it would indicate that three water saturations could co-exist at a given point

in a reservoir.

PETE 609 - Module 5




Swc

1-Sor

Sw

B

A

Swf

fixed time

x

Figure 11 - Saturation distribution at a particular time using Buckley-Leverett

Equation.

What happens is that the values of saturation that correspond to the maximum

velocity will initially tend to overtake the lower saturations resulting in the

formation of a saturation discontinuity or a shock front. Due to this discontinuity,

since Buckley-Leverett theory assumed that the saturation was continuous and

differentiable, the theory cannot describe the situation at the front. This, however,

can be applied behind the front, in the saturation range.

wf w orS S S 1

To draw the correct saturation profile requires to determine the vertical location

that separates the two shaded areas A and B. of ‎Figure 11. These areas must

cancel and the dividing line represents the saturation at the shock front wfS .

PETE 609 - Module 5




Determination of Average Water Saturation behind the

Shock Front

We start with a material balance for the injected water. ‎Figure 12 shows a

saturation profile before breakthrough. Water has been injected for a certain time

and at the position x1 the water saturation of the plane takes its maximum value

while at x2 the water saturation is the shock front saturation. We want to

determine both, the location and the value of this saturation. We also need to

know the average water saturation behind the front.

0 Lx1 x2

Swc

1-Sor

Swf

x

Sw

Sw

Saturation profile

at t < tbt

0 Lx1 x2

Swc

1-Sor

Swf

x

Sw

Sw

Saturation profile

at t < tbt

Figure 12 - Saturation profile before breakthrough indicating the shock front

saturation.

A material balance for the injected water gives

iW volume swept

average water saturation - connate water saturation

(31)

PETE 609 - Module 5




wf

i wi w wc w wc

w S

W dfW A x S S A S S

A dS

2 (32)

where we have replaced x2 using Buckley-Leverett. After cancellation of terms

we obtain

wf

w wc

w

w S

S Sdf

dS

1

(33)

Another expression for the average saturation behind the front can be obtained

from integrating saturation profiles.

By using the mean value theorem, the average water saturation from the injector

x 0 to the front x x 2 is given by

x xx

w ww o xow x

o

S dx S dxS dxS

xdx

1 22

1

2

2

(34)

x

or wx

w

S x S dxS

x

2

11

2

1 (35)

Replace x1, x2 and dx using Buckley-Leverett evaluated at the corresponding

saturations,

PETE 609 - Module 5




w

w

i wS

w S

W dfB L x

A dS

(36)

wf

or

or

wf

Sw w

or wS

w wS

w

w

w S

df dfS S d

dS dSS

df

dS

11

1

(37)

Evaluate wf

or

S

S1 by parts,

wf

wf wf

or or

or

SS S

w w ww w w

S Sw w wS

df df dfS d S dS

dS dS dS

1 1

1

(38)

wf

wf

or

or

S

Sww w S

w S

dfS f

dS

1

1

(39)

wf

or

S

ww wf

w S

dfS f

dS

1

1 (40)

Substitute Equation ‎(40) into Equation ‎(37) and simplify,

PETE 609 - Module 5




or wf

wf

or

wf

w wor wf

w wS S

w

w

w S

wor wf

w S

w

w S

df dfS S

dS dSS .

df

dS

dfS f

dS .

df

dS

1

1

1

1 1

(41)

wf

wfw wf

w

w S

fS S

df

dS

1 (42)

Comparing Equation ‎(33) and Equation ‎(42)

wf

w wc

w

w S

S Sdf

dS

1 ‎(33)

wf

wfw wf

w

w S

fS S

df

dS

1 ‎(42)

we obtain,

PETE 609 - Module 5




wf wf

wfwc wf

w w

w wS S

fS S

df df

dS dS

1 1 (43)

wf

w wf

w w wf w wcS

df f

dS S S S S

1 1 0 (44)

This is a key result!

The tangent to the fractional flow curve from the point w wc wS S ,f 0 must

have a point of tangency with coordinates wf

w w wfSf f f ; and the

extrapolated tangent must intercept the line wf 1 at the point

w w wS S ; f 1.

Need to plot w wf vs S and obtain derivative!

‎Figure 13 indicates the point of convergence of the two slopes at the shock front.

PETE 609 - Module 5




wff1

w wfS S

1 0

w wcS S

wSwcS wfS

w wff f

wf 1

wf 0

wff1

w wfS S

1 0

w wcS S

wSwcS wfS

w wff f

wf 1

wf 0

Figure 13 - Slope of the fractional flow curve.

Application of Fractional Flow Theory in Oil

Recovery Calculations

There are different methods for calculating the oil recovery depending on the

type of reservoir, either homogeneous or layered.

For Homogeneous Reservoirs:

1. Buckley-Leverett Method

PETE 609 - Module 5




For Layered or Stratified Reservoirs:

1. Stiles Method

2. Dykstra-Parsons Method

3. Johnson Method

Homogeneous Reservoirs - Buckley-Leverett

Method

Before water breakthrough we can easily obtain the saturation profiles and the oil

recovery is equal to the water injected (a trivial result). We need to evaluate the

oil recovery after breakthrough as well.

After breakthrough at producing well x =L2 .

Let i

id

WW

LA

= dimensionless number of pore volumes of injected water

PV LA 1 .

‎Figure 14 shows water saturation distributions at two different times one is at

breakthrough and the other at a later time in a linear waterflood.

PETE 609 - Module 5




x

Sw

Swc

1-Sor

Saturation profile atbreakthrough, tb

Swbt=Swf

L

Sw

0

Saturation profileat t > tb

SweSwbt

Figure 14 - Water saturation distributions at breakthrough and subsequently, in a

linear waterflood.

At breakthrough, wbtS = water saturation at breakthrough wfS front reaches

production well. And the reservoir water production increases suddenly from

zero to wbtf . This confirms existence of shock.

id iq q LA (45)

Dimensionless oil production at breakthrough

bt bt

wbt

pd id id bt w wcbt

w

w S

N W q t S Sdf

dS

1 (46)

PETE 609 - Module 5




Using Equation ‎(32)

btid

bt

id

Wt

q (47)

After breakthrough, both oil and water will be produced.

we

iid

w

w S

WW

LA df

dS

1 (48)

At this stage to evaluate oil recoveries,

we

w we we

w

w S

S S fdf

dS

11 (49)

or

w we we idS S f W 1 (50)

Subtract wcS from both sides of the equation

PETE 609 - Module 5




pd w wc we wc we idN S S S S f W 1 (51)

Exercise # 1 - Fractional Flow

Oil is being displaced by water in a horizontal, direct line drive under the diffuse flow

condition. The rock relative permeability functions for water and oil are listed in

‎Table 2,

wS rwk rok wS rwk rok

.20 0 .800 .50 .075 .163

.25 .002 .610 .55 .100 .120

.30 .009 .470 .60 .132 .081

.35 .020 .370 .65 .170 .050

.40 .033 .285 .70 .208 .027

.45 .051 .220 .75 .251 .010

.80 .300 0

Table 2 - Relative permeability saturation data for exercise # 1 (from Dake, 1988).

Pressure is being maintained at its initial value for which,

o wB . rb/stb and B . rb/stb 1 3 1 0

Compare the values of the producing water-cut (at surface conditions) and the

cumulative oil recovery at breakthrough for the following fluid combinations.

PETE 609 - Module 5




Case Oil viscosity Water

viscosity

1 50 cp .5 cp

2 5 cp .5 cp

3 .4 cp 1.0 cp

Table 3 - Cases to analyze the different fractional flow results for exercise #1.

Assume that the relative permeability and PVT data are relevant for all three cases.

Solution to Exercise #1 - Fractional Flow

1) For horizontal flow the fractional flow in the reservoir is

wrow

rw o

fk

k

1

1

(52)

while the producing water-cut at the surface, wsf , is

w wws

w w o o

q Bf

q B q B

(53)

where the rates are expressed in rb/d , and where w oB ,B take into account

compressibility effects B stb/rb

Combining the above two equations leads to an expression for the surface water-cut as

PETE 609 - Module 5




ws

w

o w

fB

B f

1

11 1

(54)

The fractional flow in the reservoir for the three cases can be calculated as follows:

Case 1 is w

o

.

01

Case 2 is w

o

.

1

Case 3 is w

o

.

2 5

PETE 609 - Module 5




wS rwk rok ro rwk k Fractional Flow wf

Case 1 Case 2 Case 3

.20 0 .800 0 0 0

.25 .002 .610 305.000 .247 .032 .001

.30 .009 .470 52.222 .657 .161 .008

.35 .020 .370 18.500 .844 .354 .021

.40 .033 .285 8.636 .921 .537 .044

.45 .051 .220 4.314 .959 .699 .085

.50 .075 .163 2.173 .979 .821 .155

.55 .100 .120 1.200 .988 .893 .250

.60 .132 .081 .614 .994 .942 .394

.65 .170 .050 .294 .997 .971 .576

.70 .208 .027 .130 .999 .987 .755

.75 .251 .010 .040 .999 .996 .909

.80 .300 0 0 1.000 1.000 1.000

Table 4 - Evaluation of fractional flow equation for cases 1 to 3.

Fractional flow plots for the three cases are shown in ‎Figure 15 and the results

obtained by applying Welge's graphical technique, at breakthrough, are listed below:

PETE 609 - Module 5




Case btwS

btwf

(reservoir)

btwsf

(surface) btwS

btpdN

(PV)

1 .28 .55 .61 .34 .14

2 .45 .70 .75 .55 .35

3 .80 1.00 1.00 .80 .60

Table 5 - Oil recoveries and saturation at breakthrough for exercise #1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

Case 1

w/ o=0.01

Case 2

w/ o =0.1

Case 3

w/ o =2.5

f w [

rb/r

b]

Sw

Figure 15 - Fractional flow plots for different oil-water viscosity ratios.

PETE 609 - Module 5




An important parameter in determining the effectiveness of a waterflood is the end

point mobility ratio defined as

rw w

ro o

kM

k

(55)

And, for horizontal flow, stable, piston-like displacement will occur for M 1. An

even more significant parameter for characterizing the stability of Buckley-Leverett

displacement is the shock front mobility ratio, sM , defined as

ro wf o rw wf w

s

ro o

k S k SM

k

(56)

In which the relative permeabilities in the numerator are evaluated for the shock front

water saturation, wfS . Hagoort has shown, using a theoretical argument backed by

experiment, that Buckley-Leverett displacement could be regarded as stable for the less

restrictive condition that sM 1. If this condition is not satisfied, there will be

severe viscous channeling of water through the oil and breakthrough will occur even

earlier than predicted using the Welge technique. Values of M and sM for the three

cases defined in the previous exercise are listed in the following table.

PETE 609 - Module 5




Case No. o

w

wfS rw wfk S ro wfk S sM M

1 100 .28 .006 .520 1.40 37.50

2 10 .45 .051 .220 .91 3.75

3 .4 .80 .300 0 .15 0.15

Figure 16 - Values of the shock front and end point relative permeabilities calculated

using the data of exercise #1 (Fractional Flow).

Using these data the results of the previous exercise can be analyzed as follows:

Case 1 - this displacement is unstable due to the very high value of the oil/water

viscosity ratio. This results in the by-passing of oil and consequently the

premature breakthrough of water. The oil recovery at breakthrough is very

small and many pore volumes of water will have to be injected to recover all

the movable oil. Under these circumstances oil recovery by water injection

is hardly feasible and consideration should be given to the application of

thermal recovery methods with the aim of reducing the viscosity ratio.

Case 2 - the oil/water viscosity ratio is an order of magnitude lower than in Case 1,

which leads to a stable and much more favorable type of displacement

sM 1 . This case will be analyzed in greater detail in the next

exercise (Exercise Oil Recovery Prediction for a Waterflood), in which the oil

recovery after breakthrough is determined as a function of the cumulative

water injected and time.

Case 3 - for the displacement of this very low viscosity oil o . cp 4 both the

end point and shock front mobility ratios are less than unity and piston-

like displacement occurs The tangent to the fractional flow curve, from

PETE 609 - Module 5




w wcS S , wf 0 , meets the curve at the point btw orS S 1 ,

btwf 1 and therefore btbt

ww orS S S . The total oil recovery at

breakthrough is btw wc or wcS S S S 1 , which is the total

movable oil volume.

Exercise # 2 - Oil Recovery Prediction for a

Waterflood

Water is being injected at a constant rate of 1000 b/d/well in a direct line drive in a

reservoir that has the following rock and fluid properties.

. 0 18

wcS . 0 20

orS . 0 20

o cp 5

w . cp 0 5

The relative permeabilities for oil and water are presented in ‎Table 2 and the flood

pattern geometry is as follows:

Dip angle = 0°

Reservoir thickness = 40 ft

Distance between injection wells = 625 ft

PETE 609 - Module 5




Distance between injectors and producers = 2000 ft

Figure 17 - Schematics of the direct line drive.

Assuming that diffuse flow conditions prevail and that the injection project starts

simultaneously with oil production from the reservoir

1) Determine the time when breakthrough occurs.

2) Determine the cumulative oil production as a function of both the cumulative water

injected and the time.

Solution to Exercise # 2 - Oil Recovery Prediction for a Waterflood

The relative permeabilities and viscosities of the oil and water are identical with those

of Case 2 in exercise #1 (Fractional Flow). Therefore, the fractional flow curve is the

same as drawn in ‎Figure 15, for which the breakthrough occurs when,

btwS . 0 45

btwf . 0 70

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and bt btid pdW N . 0 35

1) Calculation of the Breakthrough Time

For a constant rate of water injection the time is related to the dimensionless water

influx by the general expression

id

id

W (one pore volume ) (cu. ft )t

q . (cu. ft/year )

5 615 365 (57)

idW .t ( years )

.

625 40 2000 18

1000 5 615 365

idt . W ( years ) 4 39 (58)

Therefore breakthrough will occur after a time

btt . . . years 4 39 0 35 1 54

2) Cumulative Oil Recovery

The oil recovery after breakthrough, expressed in pore volumes, can be calculated using

the equation

pd we wc we idN S S f W 1 (59)

where

PETE 609 - Module 5




we

id

w

w S

Wdf

dS

1

(60)

Allowing weS , the water saturation at the producing end of the block, to rise in

increments of 5% btwe wfor S S the corresponding values of idW are

calculated ‎Table 6 using the data listed in ‎Table 4 for Case 2.

PETE 609 - Module 5




weS wef weS wef we wef S weS

idW

.45 (bt) .699

.05 .122 2.440 .475 .410

.50 .821

.05 .072 1.440 .525 .694

.55 .893

.05 .049 .980 .575 1.020

.60 .942

.05 .029 .580 .625 1.724

.65 .971

.05 .016 .320 .675 3.125

.70 .987

.05 .009 .180 .725 5.556

.75 .996

.05 .004 .080 .775 12.500

.80 1.000

Table 6 - Results for exercise #2 (Oil recovery).

In this table, values of we wef S have been calculated rather than determined

graphically. The values of weS

in Column 6 are the mid points of each saturation

increment, at which discrete values of idW have been calculated using Equation

‎(60). The oil recovery as a function of both idW and time can now be determined

using Equation ‎(59) as listed in ‎Table 7.

PETE 609 - Module 5




weS

we wcS S wef

wef

1

idW

(PV)

pdN

(PV)

Time

(yrs)

.475 .275 .765 .235 .410 .371 1.80

.525 .325 .870 .130 .694 .415 3.05

.575 .375 .925 .075 1.020 .452 4.48

.625 .425 .962 .038 1.724 .491 7.57

.675 .475 .982 .018 3.125 .531 13.72

.725 .525 .993 .007 5.556 .564 24.39

Table 7 - Oil recovery as a function of time and water injected

The values of wef

in Column 3 of ‎Table 7 have been obtained from ‎Figure 15 (Case

2), for the corresponding values of weS

. The oil recovery, in reservoir pore volumes is

plotted as a function of idW and time in ‎Figure 18. The maximum possible recovery

is one movable oil volume, i.e., wc orS S . PV 1 0 6 .

PETE 609 - Module 5




0

0 .1

0 .2

0 .3

0 .4

0 .5

0 .6

0 1 2 3 4 5 6

qi=1,000 rb/dNp

d (

PV

)

Wid(PV)

5 155 10 2520time (years)

Figure 18 - Dimensionless oil recovery (PV) as a function of dimensionless water

injected (PV), and time.

In the general case in which the displacement takes place at a fixed pressure, which is

above the bubble point pressure, then

p o

pd wc

oi

N Boil production (rb)N S

one pore volume (rb) N B 1 (61)

and the conventional expression for the oil recovery is

p pdoi

o wc

N NB (stb.oil)

N B S STOIIP (stb)

1 (62)

PETE 609 - Module 5




In the last exercise, o oiB B , since displacement occurs at the initial reservoir

pressure, p pd wcN N N S 1 .

When the mobility ratio is unfavorable (greater than 10) the Buckley-Leverett method

no longer applies and the viscous fingering method can be used instead, this material

can be seen in more detail in another course (Waterflooding – Pete 623).

Stratified Reservoirs - Stiles Method

This method applies when the mobility ratio is close to unity. Stiles makes

calculations for a layered reservoir using the following assumptions.

(a) The formation is made up of a number of layers of constant thickness

(b) There is no fluid segregation within a bed or communication (i.e. no cross

flow)

(c) The displacement is piston like which means the length of the transition

zone is zero.

(d) The system is linear

(e) The position of the front in each layer is directly proportional to the

absolute permeability of the bed

(f) The fractional flow of produced water depends upon the product of Kihi of

the layers in which water has already broken through compared to the

total Kh of the system.

(g) The layers may have different thickness and absolute permeability.

‎Figure 19 shows a stratified 6 layered reservoir. For convenience the natural

layering of the reservoir is ordered into a sequence of layers with decreasing

permeability, as required by the Stiles method.

PETE 609 - Module 5




Natural layering Re-ordered layersNatural layering Re-ordered layers

Figure 19-Stratified reservoir arranged to use Stiles method.

The left part ‎Figure 19 shows the highest permeability layer on top and the

lowest on the bottom.

We number the layers from the highest permeability, which breaks through first,

to the lowest.

That is for n layers the permeabilities are: K1 (highest), K2,…..Kn (lowest).

The thickness of the n layers are

h1, h2,….. hn

Total physically recoverable oil expressed in standard barrels is

w orc

pt

o

WHL S SN STB

B

1

7758

(63)

Where:

W =reservoir width-ft

=porosity, pore volume/bulk volume

H =total reservoir thickness, ft

L =reservoir length,ft

PETE 609 - Module 5




Bo =oil formation volume factor, reservoir volume/surface volume

The following example shows the calculation procedure using Stiles method for a

seven-layered reservoir of ‎Figure 20.

Absolute k-md Thickness-ft

210 20

190 12

70 5

50 7

30 15

10 30

3 18


210 20

190 12

70 5

50 7

30 15

10 30

3 18


210 20

190 12

70 5

50 7

30 15

10 30

3 18

Figure 20 - Permeability and thickness for a seven layered reservoir

Mathematical development

At the time, tj, that the jth layer has broken through, all of the physically

recoverable oil will have been recovered for that layer and from all layers having

higher permeability.

Since the velocities of the flood fronts in each layer are proportional to the

absolute permeabilities in the layers, the fractional recovery at tj in the j+1 layer

will be

j

j

K

K

1 (64)

In the above example, the fractional recovery in layer 2 at the time layer 1 has

broken through (t1) will be

PETE 609 - Module 5




K.

K 2

1

1900 905

210 (65)

That is, over 90% of layer 2 will be flooded out.

The recovery at time tj is given by

j ni

i i

i i j j

j j n

i

i

Kh h

KR R t

h

1 1

1

(66)

The first term in the numerator indicates the layers that have already been

flooded, while the second part gives the partially flooded portion.

jn

i j i

i i

H h , h h

1 1

(67)

Using the definitions of the above equation we can write

n

j i i

i jj

j j

h h Kk

R R tH

1

1

(68)

We can also write Equation ‎(68) as

PETE 609 - Module 5




t jj

j

j

C ChR

H K H

(69)

Where

jn

t i i j j i

i i

C K h , C K h

1 1

(70)

We can define the fractional flow of water at bottomhole conditions and at

surface at breakthrough for layer j. as.

j

w j

j t j

MCf t

MC C C

(71)

j

ws j

j t j

ACf t

AC C C

(72)

The fractional flow of water at surface is evaluated considering the formation

volume factors of oil and water.

Where

rw o

ro w

KM

K

(73)

and

PETE 609 - Module 5




rw o o

ro w w

K BA

K B

(74)

The proof of Equations‎(71) and ‎(72) is left as an exercise.

Recall that

wws j

w o surface

qf t

q q

(75)

‎Table 8 summarizes Stiles results for the example provided. Oil and fluid

properties are included in the table. Revise the calculations.

Exercise using Stiles methodBw = 1.02 krw = 0.35

Bo = 1.37 kro = 0.93

w = 0.6 cp

o = 0.83 cp

Recoverable oil= 100,000 STB

A= 0.699249772

Layer absolute k-md h h kh kh R at bt Np-STB fw

1 210 20 20 4200 4200 0.3553 35532 0.4370

2 190 12 32 2280 6480 0.3730 37304 0.7508

3 70 5 37 350 6830 0.4999 49987 0.8054

4 50 7 44 350 7180 0.5615 56150 0.8620

5 30 15 59 450 7630 0.6617 66168 0.9378

6 10 30 89 300 7930 0.8822 88224 0.9904

7 3 18 107 54 7984 1.0000 100000 1.0000

Table 8 - Stiles results for the example provided.

‎Figure 21 presents the fractional flow of water versus cumulative oil recovered

using Stiles method.

PETE 609 - Module 5




Stiles Method

0.40

0.50

0.60

0.70

0.80

0.90

1.00

0 20000 40000 60000 80000 100000

Np

fw

Figure 21 Recovery at breakthrough for all the layers

Note that the final 80,000 STB recovered at water cuts >90% may be

uneconomical!

Stratified Reservoirs - Dykstra-Parsons and

Johnson Methods

Stiles’ calculations do not take into account the continual variation of the

injectivity of each layer according to the advance of the water front. However, the

Dykstra-Parsons method overcomes this problem, and is valid for a wide variety

of mobility ratios. It is based also in the theory of piston like displacement.

Johnson’s has presented a simplified graphical approach to the Dykstra-Parsons

method for the case with a log-normal or “gaussian” permeability distribution,

characterized by its variance. ‎Figure 22 to ‎Figure 25 show the correlations

obtained between the vertical variation of permeability V, the initial water

PETE 609 - Module 5




saturation Sw, the mobility ratio and the oil recovery fraction of the initial oil in

place (R), for different producing water oil ratios (WOR).

To use this method we require to know V which is calculated from the statistical

analysis of the permeability distribution by plotting the permeability values on a

log probability paper and choosing the best straight line through the points.

If K 84.5 is the permeability read from the line such that 84.1% of the permeability

values are greater than K 84.5, then V is defined as follows.

.K KV

K

50 84 1

50

(76)

The following figures have been taken from Latil (1980). To use these

correlations one requires values of water saturation, mobility ratio, and V.

PETE 609 - Module 5




wR S .

.R .

.

1 0 15

0 150 27

1 0 45 w

V .

M .

S .

0 54

1 8

0 45

wR S .

.R .

.

1 0 15

0 150 27

1 0 45 w

V .

M .

S .

0 54

1 8

0 45

wR S .

.R .

.

1 0 15

0 150 27

1 0 45 w

V .

M .

S .

0 54

1 8

0 45

Figure 22 - Johnson's correlation for a producing water oil ratio (WOR) of 1.

PETE 609 - Module 5





PETE 609 - Module 5





Johnson's method gives good results when the initial oil saturation is more than

45%.

Dissipation in Immiscible Displacements

Two common dissipative effects in one-dimensional flows are capillary pressure

and fluid compressibility. Both phenomena are dissipative in the sense that they

cause mixing zones to grow faster. Capillary pressure and fluid compressibility

also bring additional effects.

PETE 609 - Module 5




Capillary Pressure

The effect of capillary pressure on a one-dimensional displacement is to spread

out the water saturation wave, particularly around shocks.

With capillary pressure

Without capillary pressure

Sw

x (distance)

With capillary pressure

Without capillary pressure

Sw

x (distance)

Figure 26 - Water saturation profiles with and without capillary pressure (Lake

1992)

PETE 609 - Module 5




With capillary spreading

Without capillary spreading

P

x (distance)

Oil pressure

Water Pressure

Pc

With capillary spreading

Without capillary spreading

P

x (distance)

Oil pressure

Water Pressure

Pc

Figure 27 - Water and oil phase pressure profiles with and without capillary

pressure (Lake 1992).

Fluid Compressibility

A second dissipative effect is fluid compressibility. Shows water saturation

profiles for two waterfloods having compressible oil and incompressible water.

Shows water saturation profiles for two waterfloods having incompressible oil and

compressible water

The effect of either oil or water compressibility is to spread out the Buckley-

Leverett shock front in addition to the spreading caused by numerical dispersion.

Additional references for this section can be found in Chapter 5 of Lake (1992).

PETE 609 - Module 5




Figure 28 - Water saturation profile for 1-D water-displacing-oil floods at t=200

days, considering compressible water and incompressible oil (adapted from

Samizo, 1982).

Figure 29 - Water saturation profile for 1-D water-displacing-oil floods at t=200

days, considering incompressible water and compressible oil (adapted from

Samizo, 1982).

module 5 - fractional flow theory

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