reservoir simulation

FUNDAMENTALS OF RESERVOIR

SIMULATION

Dr. Mai Cao Lan,

GEOPET, HCMUT, Vietnam

Jan, 2014

ABOUT THE COURSE

16-Jan-2014 Dr. Mai Cao Lan, Faculty of Geology & Petroleum Engineering, HCMUT, Vietnam 2

COURSE OBJECTIVE

COURSE OUTLINE

REFERENCES

Course Objective

• To review the background of petroleum reservoir

simulation with an intensive focus on what and how

things are done in reservoir simulations

• To provide guidelines for hands-on practices with

Microsoft Excel


INTRODUCTION

FLOW EQUATIONS FOR PETROLEUM RESERVOIRS

FINITE DIFFERENCE METHOD & NUMERICAL SOLUTION FOR

FLOW EQUATIONS

SINGLE-PHASE FLOW SIMULATION

MULTIPHASE FLOW SIMULATION

COURSE OUTLINE

16-Jan-2014 5Mai Cao Lân – Faculty of Geology & Petroleum Engineering - HCMUT

T. Eterkin et al., 2001. Basic Applied Reservoir Simulation,

SPE, Texas

J.H. Abou-Kassem et al., 2005. Petroleum Reservoir

Simulation – A Basic Approach, Gulf Publishing Company,

Houston, Texas.

C.Mattax & R. Dalton, 1990. Reservoir Simulation, SPE,

Texas.

References

INTRODUCTION

NUMERICAL SIMULATION – AN OVERVIEW

COMPONENTS OF A RESERVOIR SIMULATOR

RESERVOIR SIMULATION BASICS


Numerical Simulation – An Overview


Mathematical Formulation


Numerical Methods for PDEs


Numerical Methods for Linear Equations


Mathematical Model

Physical Model

Numerical Model

Computer Code

Reservoir Simulator

Components of a Reservoir Simulator


• A powerful tool for evaluating reservoir performance

with the purpose of establishing a sound field

development plan

• A helpful tool for investigating problems associated with

the petroleum recovery process and searching for

appropriate solutions to the problems

What is Reservoir Simulation?


Reservoir Simulation Basics

• The reservoir is divided into a number of cells

• Basic data is provided for each cell

• Wells are positioned within the cells

• The required well production rates are specified as a

function of time

• The equations are solved to give the pressure and

saturations for each block as well as the production of

each phase from each well.


Simulating Flow in Reservoirs

• Flow from one grid block to the next

• Flow from a grid block to the well completion

• Flow within the wells (and surface networks)

Flow = Transmissibility * Mobility * Potential Difference

Geometry & Properties

Fluid Properties

Well Production


SINGLE-PHASE FLOW

EQUATIONS

ESSENTIAL PHYSICS

CONTINUITY EQUATION

MOMENTUM EQUATION

CONSTITUTIVE EQUATION

GENERAL 3D SINGLE-PHASE FLOW EQUATION

BOUNDARY & INITIAL CONDITIONS


Essential Physics


The basic differential equations are derived from the

following essential laws:

Mass conservation law

Momentum conservation law

Material behavior principles

Conservation of Mass


Mass conservation may be formulated across a control element with one fluid

of density r, flowing through it at a velocity u:

Dx

ur

element theinside

mass of change of Rate

Dx+at xelement

theofout Mass

at xelement

theinto Mass

Continuity Equation


Based on the mass conservation law, the continuity equation can be

expressed as follow:

A u Ax t

r r

ux tr r

For constant cross section area, one has:

Conservation of Momentum


Conservation of momentum for fluid flow in porous materials

is governed by the semi-empirical Darcy's equation, which for

one dimensional, horizontal flow is:

x

Pku

Equation Governing Material Behaviors


The behaviors of rock and fluid during the production

phase of a reservoir are governed by the constitutive

equations or also known as the equations of state.

In general, these equations express the relationships

between rock & fluid properties with respect to the

reservoir pressure.

Constitutive Equation of Rock


The behavior of reservoir rock corresponding to the

pressure declines can be expressed by the definition of the

formation compaction

1f

T

cP

For isothermal processes, the constitutive equation of rock becomes

f

dc

dP

Constitutive Equation of Fluids


The behavior of reservoir fluids corresponding to the

pressure declines can be expressed by the definition of fluid

compressibility (for liquid)

1, , ,l

T

Vc l o w g

V P

For natural gas, the well-known equation of state is used:

PV nZRT

Single-Phase Fluid System


Normally, in single-phase reservoir simulation, we would

deal with one of the following fluids:

One Phase Gas One Phase Water One Phase Oil

Fluid System

Single-Phase Gas


The gas must be single phase in the reservoir, which means

that crossing of the dew point line is not permitted in order

to avoid condensate fall-out in the pores. Gas behavior is

governed by:

rg rgs

Bg

constant

Bg

Single-Phase Water


One phase water, which strictly speaking means that the

reservoir pressure is higher than the saturation pressure of

the water in case gas is dissolved in it, has a density

described by:

rw rwsBw

constant

Bw

Single-Phase Oil


In order for the oil to be single phase in the reservoir, it

must be undersaturated, which means that the reservoir

pressure is higher than the bubble point pressure. In the

Black Oil fluid model, oil density is described by:

ro roS rgSRso

Bo

Single-Phase Fluid Model


For all three fluid systems, the one phase density or

constitutive equation can be expressed as:

r

constant

B

Single-Phase Flow Equation


The continuity equation for a one phase, one-dimensional system of

constant cross-sectional area is:

rrt

ux

The conservation ofmomentum for 1D,horizontal flow is: x

Pku

The fluid model:

r

constant

B

Substituting the momentum equation and the fluid model into the

continuity equation, and including a source/sink term, we obtain the

single phase flow in a 1D porous medium:

sc

b

qk P

x B x V t B

(1/ ), , ,l

d Bc B l o g w

dP

sc tf l

b

q ck P P Pc c

x B x V B t B t

Based on the fluid model, compressibility can now be defined in terms of the formation volume factor as:

Then, an alternative form of the flow equation is:

(1/ )fsc

b

cqk P d B P

x B x V B dP t

Single-Phase Flow Equation for Slightly Compressible Fluids


Single-Phase Flow Equation for Compressible Fluids


sc

b

qk P

x B x V t B

Boundary Conditions (BCs)


Mathematically, there are two types of boundary conditions:

• Dirichlet BCs: Values of the unknown at the boundaries

are specified or given.

• Neumann BCs: The values of the first derivative of the

unknown are specified or given.

Boundary Conditions (BCs)


From the reservoir engineering point of view:

Dirichlet BCs: Pressure values at the boundaries are

specified as known constraints.

Neumann BCs: The flow rates are specified as the known

constraints.

Dirichlet Boundary Conditions


For the one-dimension single phase flow, the Dirichlet boundary

conditions are the pressure the pressures at the reservoir boundaries,

such as follows:

R

L

PtLxP

PtxP

0,

0,0

A pressure condition will normally be specified as a bottom-hole

pressure of a production or injection well, at some position of the

reservoir.

Newmann Boundary Conditions


In Neumann boundary conditions, the flow rates at the end faces of the

system are specified. Using Darcy's equation, the conditions become:

For reservoir flow, a rate condition may be specified as a production or

injection rate of a well, at some position of the reservoir, or it is

specified as a zero-rate across a sealed boundary or fault, or between

non-communicating layers.

0

0x

kA PQ

x

Lx

Lx

PkAQ

General 3D Single-Phase Flow Equations


The general equation for 3D single-phase flow in field units (customary

units) is as follows:

c

p Z

g

r

Z: Elevation, positive in downward direction

c, c, c: Unit conversion factors

y yx xc c

bz zc sc

c

A kA kx y

x B x y B y

VA kz q

z B z t B

D D

D

3D Single-Phase Flow Equations for Horizontal Reservoirs


The equation for 3D single-phase flow in field units for horizontal

reservoir is as follow:

y yx xc c

bz zc sc

c

A kA k p px y

x B x y B y

VA k pz q

z B z t B

D D

D

xx

Z

B

kA

x

Bt

Vqx

x

p

B

kA

x

xxc

c

bsc

xxc

D

D


1D Single-Phase Flow Equation with Depth Gradient

Quantities in Flow Equations


FINITE DIFFERENCE METHOD & NUMERICAL SOLUTION OF SINGLE-PHASE

FLOW EQUATIONS

FUNDAMENTALS OF FINITE DIFFERENCE METHOD

FDM SOLUTION OF THE SINGLE-PHASE FLOW EQUATIONS


Numerical Solution of Flow Equations


The equations describing flui flows in reservoirs are of

partial differential equations (PDEs)

Finite difference method (FDM) is traditionally used for

the numerical solution of the flow equations

Fundamentals of FDM

In FDM, derivatives are replaced by a proper difference formula based on the Taylor series expansions of a function:

1 2 2 3 3 4 4

2 3 4

( ) ( ) ( ) ( )( ) ( )

1! 2! 3! 4!x x x x

x f x f x f x ff x x f x

x x x x

D D D D D

2 2 3

2 3

( ) ( ) ( )

2! 3!x x x

f f x x f x x f x f

x x x x

D D D

D

The first derivative can be written by re-arranging the terms:

( ) ( )( )

x

f f x x f xO x

x x

D D

D

Denoting all except the first terms by O (Dx) yields

The difference formula above is of order 1 with the truncation error being proportional to Dx


Fundamentals of FDM (cont.)

To obtain higher order difference formula for the first derivative, Taylor series expansion of the function is used from both side of x

2 3

3

( ) ( ) ( )

2 3!x x

f f x x f x x x f

x x x

D D D

D

Subtracting the second from the first equation yields

2( ) ( )( )

2x

f f x x f x xO x

x x

D D D

D

The difference formula above is of order 2 with the truncation error being proportional to (Dx)2

1 2 2 3 3 4 4

2 3 4

( ) ( ) ( ) ( )( ) ( )

1! 2! 3! 4!x x x x


x x x x

D D D D D

1 2 2 3 3 4 4

2 3 4

( ) ( ) ( ) ( )( ) ( )

1! 2! 3! 4!x x x x


x x x x

D D D D D


Typical Difference Formulas

Forward difference for first derivatives (1D)

( ) ( )( )

x

f f x x f xO x

x x

D D

D

1 ( )i i

i

f ffO x

x x

D

D

or in space index form


i-1 i i+1

Dx


Backward difference for first derivatives (1D)

( ) ( )( )

x

f f x f x xO x

x x

D D

D

1 ( )i i

i

f ffO x

x x

D

D



i-1 i i+1

Dx


Centered difference for first derivatives (1D)

2( ) ( )( )

2x

f f x x f x xO x

x x

D D D

D

21 1 ( )2

i i

i

f ffO x

x x

D

D



i-1 i i+1

Dx


Centered difference for second derivatives (1D)

22

2 2

( ) 2 ( ) ( )( )

x

f f x x f x f x xO x

x x

D D D

D

221 1

2 2

2( )i i i

i

f f ffO x

x x

D

D



i-1 i i+1

Dx


Forward difference for first derivatives (2D)

( , )

( , ) ( , )( )

x y

f f x y y f x yO y

y y

D D

D

, 1 ,

( , )

( )i j i j

i j

f ffO y

y y

D

D


i-1,j i,j i+1,j

i,j+1

i,j-1



Backward difference for first derivatives (2D)

( , )

( , ) ( , )( )

x y

f f x y f x y yO y

y y

D D

D

, , 1

( , )

( )i j i j

i j

f ffO y

y y

D

D



i-1,j i,j i+1,j

i,j+1

i,j-1


Centered difference for first derivatives (2D)

2

( , )

( , ) ( , )( )

2x y

f f x y y f x y yO y

y y

D D D

D

, 1 , 1 2

( , )

( )2

i j i j

i j

f ffO y

y y

D

D



i-1,j i,j i+1,j

i,j+1

i,j-1


Centered difference for second derivatives (2D)

22

2 2

( , )

( , ) 2 ( , ) ( , )( )

x y

f f x y y f x y f x y yO y

y y

D D D

D

2, 1 , , 1 2

2 2

( , )

2( )

i j i j i j

i j

f f ffO y

y y

D

D



i-1,j i,j i+1,j

i,j+1

i,j-1

Solving time-independent PDEs

Divide the computational domain into subdomains

Derive the difference formulation for the given PDE by replacing all

derivatives with corresponding difference formulas

Apply boundary conditions to the points on the domain boundaries

Apply the difference formulation to every inner points of the

computational domain

Solve the resulting algebraic system of equations


Exercise 1

Solve the following Poisson equation:

22

216 sin(4 )

px

x

subject to the boundary conditions:

p=2 at x=0 and x=1

10 x


Exercise 2


2 sin( )sin( )

0 1,0 1

u x y

x y


0 along the boundaries 0, 1, 0, 1u x x y y



Boundary Condition Implementation

b

pC

x

Newmann BCs:

1 0

1 1/2 1 0

0 1 1

p ppC

x x x

p p C x

D

1

1/2 1

1

x x

x x x

x x x

n n

n n n

n n n

p ppC

x x x

p p C x

D


Boundary Condition Implementation

Dirichlet BCs:

bp C

1 2

1

1 2

1 p p C

x

x x

D

D D

1

1

1x x

x

x x

n n

n

n n

p p C

x

x x

D

D D

Exercise 3


2 2 2( )exp( )

0 1,0 1, 2, 3

u x y

x y


exp( ); 0, 1u x y y y

exp( ); 0, 1u

x y x xx


Solving time-dependent PDEs

Divide the computational domain into subdomains

Derive the difference formulation for the given PDE by replacing all

derivatives with corresponding difference formulas in both space

and time dimensions

Apply the initial condition

Apply boundary conditions to the points on the domain boundaries

Apply the difference formulation to every inner points of the

computational domain

Solve the resulting algebraic system of equations


Exercise 4

Solve the following diffusion equation:

2

2,0 1.0, 0

u ux t

t x

subject to the following initial and boundary conditions:

( 0, ) ( 1, ) 0, 0u x t u x t t

( , 0) sin( ),0 1u x t x x

Hints: Use explicit scheme for time discretization


Explicit Scheme

The difference formulation of the original PDE in Exercise 4 is:

1

1 1

2

2

( )

n n n n n

i i i i iu u u u u

t x

D D

where

n=0,NT: Time step

i =1,NX: Grid point index


Implicit Scheme

The difference formulation for the original PDE in Exercise 4

1 1 1 1

1 1

2

2

( )

n n n n n

i i i i iu u u u u

t x

D D

where

n=0,NT: Time step



Semi-Implicit Scheme

Semi-Implicit Scheme for the Diffusion Equation in Exercise 4 is

1 1 1 1

1 1 1 1

2 2

2 2(1 )

( ) ( )

n n n n n n n n

i i i i i i i iu u u u u u u u

t x x

D D D

where

0 ≤ ≤ 1

n=0,NT: Time step


When =0.5, we have Crank-Nicolson scheme


Discretization in Conservative Form

21/2 1/2

( ) ( )

( ) i i

i i

P Pf x f x

P x xf x O x

x x x

D D

1

11/2 12

( )( )

i i

i i i

P PPO x

x x x

D

D D

1

11/2 12

( )( )

i i

i i i

P PPO x

x x x

D

D D

1 11/2 1/2

1 1

( ) ( )2 ( ) 2 ( )

( ) ( )( ) ( )

i i i ii i

i i i i

i i

P P P Pf x f x

x x x xPf x O x

x x x

D D D D D D

( )P

f xx x


i-1 i i+1

Dx

FDM for Flow Equations

FD Spatial Discretization

FD Temporal Discretization



For slightly compressible fluids (Oil)

x x b tc sc

c

A k V cp px q

x B x B t

D

For compressible fluids (Gas)

x x bc sc

c

A k Vpx q

x B x t B

D

Single-Phase Flow Equations

FDM for Slightly Compressible Fluid Flow Equations




Discretization of the left side term

The discretization of the left side term is then

1 12 21 1

2 2

( ) ( )

( ) ( )

i i

i i

i i

P Pf x f x

x xPf x O x

x x x

D D

where ( ) x xc

A kf x

B

1

11

2

( )

( ) / 2

i i

i i i

P PP

x x x

D D

1

11

2

( )

( ) / 2

i i

i i i

P PP

x x x

D D

FD Spatial Discretization of the LHS


1 12 2

1 1( ) ( )x x x x x xc i c i i c i i

i i i

A k A k A kpx P P P P

x B x B x B x

D

D D

Define transmissibility as the coefficient in front of thepressure difference:

2

1

2

1

1

21

D

ii

xxcx

Bx

kAT

i


Transmissibility



The left side term of the 1D single-phase flow equation is now discritized as follow:

1 12 2

1 1( ) ( )x xc i i i i ii i

i

A k Px Tx P P Tx P P

x B x

D

12 1 1

2 2

1

i

x xx c

i i

A kT

x B

D

Transmissibility



1

11 1

2

2x x x xx x i i

c c

i x x i x x ii i

A k A kA k

x A k x A k x

D D D

or

1 1 1

11

2

1

2

x x x x x xc c c

i i i

A k A k A k

x x x

D D D

Transmissibility (cont’d)

ii

iiii

ixx

xx

DD

DD

1

11

21

ii

iiii

ixx

xx

DD

DD

1

11

21

B

1


Weighted Average of Mobility

2

1

2

1

1

21

D

ii

xxcx

Bx

kAT

i

D

D

DD

DD

i

i

i

i

ii

iixxiixx

ixxixx

cx

Bx

Bx

xx

xkAxkA

kAkAT

i

111

2

1

1

1

11

1

2

1


Discretized Transmissibility



Explicit Method

1/2 1/2

1

1 1i i i

n n

i in n n n n n b tx i i x i i sc

c i

p pV cT p p T p p q

B t

D Implicit Method

1/2 1/2

1

1 1 1 1 1 1

1 1i i i

n n

i in n n n n n b tx i i x i i sc

c i

p pV cT p p T p p q

B t

D

Semi-implicit Method

1/2 1/2

1/2 1/2

1 1 1 1 1 1

1 1

1

1 11

i i i

i i

n n n n n n

sc x i i x i i

n n

i in n n n n n b tx i i x i i

c i

q T p p T p p

p pV cT p p T p p

B t

D

0 1

For the 1D, block-centered grid shown on the screen,

determine the pressure distribution during the first year of

production. The initial reservoir pressure is 6000 psia. The

rock and fluid properties for this problem are:

6 -1

t

1000ft; 1000ft; 75ft

1RB/STB; =10cp;

k =15md; =0.18; c =3.5 10 psi ;

Use time step sizes of =10, 15, and 30 days.

Assume B is unchanged within the pressure range

of interest.

x

x y z

B

D D D

Exercise 5


1 2 3 4 5

0p

x

0p

x

150 STB/Dscq

1000 ft

75 ft

1000 ft

Exercise 5 (cont’d)






6 -1

t

1000ft; 1000ft; 75ft

1RB/STB; =10cp;

k =15md; =0.18; c =3.5 10 psi ;

Use time step sizes of =10, 15, and 30 days.

Assume B is unchanged within the pressure range

of interest.

x

x y z

B

D D D

Exercise 6


1 2 3 4 5

0p

x

6000psiap

150 STB/Dscq

1000 ft

75 ft

1000 ft



FDM for Slightly Compressible Fluid Flow Equations




1 12 2

1 1( ) ( )x xc i i i i ii i

i

A k px Tx p p Tx p p

x B x

D

FD Spatial Discretization of the LHS for Compressible Fluids

Same as that for slightly compressible fluids


2

1

2

1

1

21

D

ii

xxcx

Bx

kAT

i


Transmissibility

12

1 1

1

if

if

i i i

i

i i i

p p

p p

1

B

Upstream Average of Mobility


1n n

b b

c ci i

V V

t B t B B

D

1ref ref

fc p p

FD Spatial Discretization of the RHS for Compressible Fluids







6 -1

t

1000ft; 1000ft; 75ft

k =15md; =0.18; c =3.5 10 psi

Use time step sizes of =10 days.

x

x y z

D D D

Exercise 7


PVT data table:p (psia) (cp) B (bbl/STB)

5000 0.675 1.292

4500 0.656 1.299

4000 0.637 1.306

3500 0.619 1.313

3000 0.600 1.321

2500 0.581 1.330

2200 0.570 1.335

2100 0.567 1.337

2000 0.563 1.339

1900 0.560 1.341

1800 0.557 1.343



1 2 3 4 5

0p

x

0p

x

150 STB/Dscq

1000 ft

75 ft

1000 ft


MULTIPHASE FLOW SIMULATION

MULTIPHASE FLOW EQUATIONS

FINITE DIFFERENCE APPROXIMATION TO MULTIPHASE FLOW EQUATIONS

NUMERICAL SOLUTION OF THE MULTIPHASE FLOW EQUATIONS


Continuity equation for each fluid flowing phase:

llll St

AuAx

rr

x

Pkku l

l

rll

gwol ,,

wocow PPP

ogcog PPP

Sll o,w, g

1gwol ,,

Momentum equation for each fluid flowing phase:

16-Jan-2014 89Dr. Mai Cao Lan, Faculty of Geology & Petroleum Engineering, HCMUT, Vietnam

Multiphase Flow Equations

• Considering the fluid phases of oil and water only, the flow equations for the two phases are as follows:

scw

w

w

c

bw

w

ww

rwxxc q

B

S

t

Vx

x

Z

x

P

B

kAk

x

D

sco

o

o

c

bo

o

oo

roxxc q

B

S

t

Vx

x

Z

x

P

B

kAk

x

D

cowow PPP 1 wo SS


Oil-Water Flow Equations

scw

w

w

c

bw

cowo

ww

rwxxc q

B

S

t

Vx

x

Z

x

P

x

P

B

kAk

x

D

sco

o

w

c

bo

o

oo

roxxc q

B

S

t

Vx

x

Z

x

P

B

kAk

x

D

1


Oil-Water Flow Equations

)()(11

21

21 ioioixoioioixo

i

i

oo

oo

roxxc

PPTPPT

xx

Z

x

P

B

kAk

x

D

)()(11

21

21 ioioixwioioixw

i

i

wcowo

ww

rwxxc

PPTPPT

xx

Z

x

P

x

P

B

kAk

x

D

Left side flow terms


Discretization of the Flow Equation

o krooBo

ww

rww

B

k


Phase Mobility

1 2

ii oo

21

1

11

21

DD

DD

ii

ioiioi

ioxx

xx

Upstream: weighted average:

x

Swir

Sw

1-Swir

Qw

average

upstream

exact

OIL


Averaging of Phase Mobility

ioioio

ioioio

io

PPif

PPif

1

11

21

iwiwiw

iwiwiw

iw

PPif

PPif

1

11

21


Upstream Average of Mobility

1 1 1 12 2

( ) ( )i i i ii i

ro oc x x o i

o o i

xo o o xo o o

k P Zk A x

x B x x

T P P T P P

D

1 1 1 12 2

( ) ( )i i i ii i

rw o cowc x x w i

w w i

xw o o xw o o

k P P Zk A x

x B x x x

T P P T P P

D

Left side flow terms


Discretization of Multiphase Flow Equation

Right side flow terms

o

oo

oo

o

BtS

t

S

BB

S

t

)()/1( 1 n

ion

o

io

o

o

roi

io

o PPdP

Bd

B

c

t

S

BtS

i

D

The second term:

)( 1

1

n

iwn

w

iio

i

n

i

o

o

SStBt

S

B i

D

wo SS 1

The first term:


Discretization of the Oil-Phase Equation

and

)()( 11 n

iwwiswon

ion

oipoo

io

o SSCPPCB

S

t

n

ii

io

o

o

riwiipoo

dP

Bd

B

c

t

SC

D

)/1()1(Where:

iio

iiswo

tBC

D


Discretization of Oil-phase RHS

Right side flow terms

w

ww

ww

w

BtS

t

S

BB

S

t

t

P

t

P

BPt

P

BPBt

cowo

ww

w

www

t

w

w

cow

t

cow S

dS

dPP


Discretization of Water-Phase Equation

and

Where:

)()( 11 n

iwwiswwn

ion

oipow

iw

w SSCPPCB

S

t

n

ii

iw

w

w

riwiipow

dP

Bd

B

c

t

SC

D

)/1(

ipow

iw

cow

iwi

iisww C

dS

dP

tBC

D


Discretization of Water-phase RHS

Ni ,...,1

1 11 12 2

1 1 1

1

i i i i i

i i

n n n n n nxo xo poo oo o o o i o ii i

n nswo wi w i osc

T P P T P P C P P

C S S q

1 11 1 1 12 2

1 1

1 1

i i i i i i i i

i i i

n n n n n n n nxw xwo o cow cow o o cow cowi i

n n n no sww wpowi o i i w i wsc

T P P P P T P P P P

C P P C S S q


Fully Discrete Oil-Water Flow Equations

n

iswo

n

isww

iC

C

iii

iiii

iiii

wsciosc

n

ion

o

n

iswoi

n

ipoo

n

cow

n

cow

n

ixwi

n

cow

n

cow

n

ixwi

n

o

n

o

n

ixwi

n

ixo

n

o

n

o

n

ixwi

n

ixo

qqPPCC

PPTPPT

PPTTPPTT

1

1111

121

121

121

21

121

21

First, the pressure is found by solving the following equation:


IMPES Solution of Oil-Water Flow Equations

11

1

1111

1

1

nn

i

nn

i

nn

i

n

iiiigPEPCPW ooo


n

ixwi

n

ixo

n TTWi 2

121

1

)()(

)(

11

1

21

21

n

icown

icown

ixwi

n

icown

icown

ixwi

wsciosc

n

ion

ipowi

n

ipoon

PPTPPT

qqPCCgiii

n

iswo

n

isww

iC

C

n

ixwi

n

ixo

n TTEi 2

121

1

1 12 2

1 12 2

1

i

n n n nxo xo pooii i

n n nxw xw powi ii i

C T T C

T T C

IMPES Pressure Solution


Once the oil pressures have been found, water saturationscan be obtained by either the oil-phase equation or thewater-phase equation.

n

ion

o

n

ipooosc

n

o

n

o

n

ixo

n

o

n

o

n

ixo

n

iswo

n

iwn

w

PPCq

PPTPPT

CSS

ii

iiii

i 1

1111

1 121

1211

Ni ,...,1

IMPES Water Saturation


A homogeneous, 1D horizontal oil reservoir is 1,000 ft long

with a cross-sectional area of 10,000 ft2. It is discretized into

four equal gridblocks. The initial water saturation is 0.160

and the initial reservoir pressure is 5,000 psi everywhere.

Water is injected at the center of cell 1 at a rate of 75 STB/d

and oil is produced at the center of cell 4 at the same rate.

Rock compressibility cr=3.5E-6 psi-1 . The viscosity and

formation volume factor of water are given as w=0.8cp and

Bw=1.02 bbl/STB.

Exercise 8


The gridblock dimensions and properties are: Dx=250ft,

Dy=250ft, Dz=40ft, kx=300md, =0.20. PVT data

including formation volume factor and viscosity of oil is

given as in Table 1 as the functions of pressure. The

saturation functions including relative permeabilities and

capillary pressure.

Using the IMPES solution method with Dt=1 day, find the

pressure and saturation distribution after 100 days of

production.


1 2 3 4

0p

x

250 ft

Ax=10,000 ft2

0p

x

Qo=-75 STB/dQw=75 STB/d




The relative permeability data:

Sw Krw Kro

0.16 0 1

0.2 0.01 0.7

0.3 0.035 0.325

0.4 0.06 0.15

0.5 0.11 0.045

0.6 0.16 0.031

0.7 0.24 0.015

0.8 0.42 0


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