LL4 ENPE470-2013 Instability

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February 12, 2013 ENPE 470 RESERVOIR MODELING Dr. Ezeddin Shirif

INTRODUCTION TO RESERVOIR SIMULATION

Instability

Nomen References AuthorInfo

Derivation of the Flow Equations

REFERENCES ABOUT EXITHELPNomenclatureHome

ENPE 470 RESERVOIR MODELING2

Flow between

block

Transmissibility

Conservation of Mass

Constitutive Equations

Questions

Flow Equation

Single-phase

Flow

Non-horizontal

FlowMutlidimensional

FlowCoordinate Systems

Nomenclature

Introduction

Home

Handouts(pdf file)

Contents

Neumanns Method

Karpluss Method

Implicit Scheme

Matrix Method

Instability

Explicit Scheme

Crank Nickolson

Fourier series Method

Explicit Method

Implicit Method

Crank Nickolson Method

Questions

Introduction





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability

An unstable solution scheme will accumulate round-off error until the solution becomes meaning-less

Single precision, the truncation error is about 10-8

Double precision, the truncation error is about 10-16

Determination of stable time





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Determination of stable t:

1. Karpluss method 99 % of time gives correct results. This method does not include BCS

2. Fourier series (Newmanns method) 100 % of time gives correct results. This method does not include BCS

3. matrix method 100 % of time gives correct results. This method includes BCS and ICS

Determination of stable time

Instability





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Express the flow equation in the following form:

Neglect Gi and qi , then:

1. If all coefficients (a, b, c, ) are positive, then scheme is stable.

2. If one or more of the coefficients are negative, then for stability:

Karpluss method

Instability

0......)()()()( 11111 =++++ ++++ nininininininini ppdppcppbppa





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

For stability:

A + b + c + d + .. 0

Karpluss method

Instability

0......)()()()( 11111 =++++ ++++ nininininininini ppdppcppbppa





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

in

in

ibii

in

in

iin

in

ii GPPtcvqppTppT +

=

+++ )()()( 1'12/112/1

Karpluss method for explicit scheme

Rewrite the above equation and ignore q and G:

0)()()( 112/112/1 =+

++n

in

ibiin

in

iin

in

ii PPtcvppTppT

Instability





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

02/12/1 +

+ t

cvTT biiii

Karpluss method for explicit scheme

For stability:

or:

2/12/1 + +

ii

bii

TTcv

t

Calculate t for every i, then choose the smallest value

Instability





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction Karpluss method for explicit scheme

2-D:

2/1,2/1,,2/1,2/1

,,

++ +++

JiJiJiJi

jbijiTTTT

cvt

Instability





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

in

in

ibii

in

in

iin

in

ii GPPtcvqppTppT +

=

++

+

++++ )()()( 1'1112/11112/1

Karpluss method for implicit scheme


0)()()( 11112/11112/1 =++

+

++++

n

in

ibiin

in

iin

in

ii PPtcvppTppT

Instability





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction Karpluss method for implicit scheme

Rewrite the above equation

0)()()(

)()(11

12/11

2/1

12/1

112/1

=

+

++

+

++

+++

n

in

ibiin

in

iin

in

ii

n

in

iin

in

ii

PPt

cvppTppT

ppTppT

0)()(

)(11

11

2/1

1112/1

=

+

+

++

+

++++

n

in

ibiin

in

in

in

ii

n

in

in

in

ii

PPt

cvppppT

ppppT

Instability





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

02/12/12/12/1 +

++ t

cvTTTT biiiiii

Karpluss method for implicit scheme

For stability:

or:

t allfor true is this t

cvbii

0

This means, that the scheme is absolutely stable

Instability





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

in

in

ibii

in

in

iin

ii

n

in

iin

in

ii

GPPt

cv

qppTppT

ppTppT+

=

++

++

+

+

++++

)()]()([

21

)]()([21

1

'

12/112/1

11

12/1

1112/1

Karpluss method for Crank-Nickolson scheme


0)(2)()(

)()( 112/112/1

11

12/1

1112/1

=

+ +

++

+

+

++++ n

in

ibii

n

in

iin

in

ii

n

in

iin

in

ii PPt

cv

ppTppT

ppTppT

Instability





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction Karpluss method for Crank-Nickolson scheme

Rewrite the above equation

Instability

0)(2)()(

)()(1

12/112/1

11

12/1

1112/1

=

+++

+++

+

+

++++

n

in

ibiin

in

iin

in

ii

n

in

in

in

iin

in

in

in

ii

PPt

cvppTppT

ppppTppppT

0)(2)()(

)()()()(1

12/112/1

112/1

12/1

12/1

112/1

=

++

+++

+

+

++

+++

n

in

ibiin

in

iin

in

ii

n

in

iin

in

iin

in

iin

in

ii

PPt

cvppTppT

ppTppTppTppT





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

022/12/1 ++++ ++++ tcvTT biiii

Karpluss method for Crank-Nickolson scheme

For stability:

or:

TTcv

tii

bii

2/12/1

2++++ ++++

This means, that the scheme is conditionally stable

Karpluss criteria gives us a conservative answer

Instability





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Continue

)]()([(111,1

)(11,

])([(,1

)]([1,

])([,1

)(,

21

21

21

21

21

21

yyxxinnji

yxinnji

yxxinnji

yyxinnji

yxxinnji

yxinnji

eP

eP

eP

eP

eP

eP

+++++

+++

+

+++

+++

+

=

=

=

=

=

=

Instability

Stability analysis by Fourier Series method

The finite difference solution discrete value can be decomposed into a product of space and time dependent terms as:





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Continue

1max

1

max

=

+

n

n

AF

Instability

Stability analysis by Fourier Series method

The Fourier Series states that a scheme is stable as long as the amplification factor, AFmax, is less than one.

the amplification factor, AFmax describes how an error grows with time





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability

Fourier Series method for Explicit scheme

Consider the following flow equation

)()()(1

211

2

2

t

PPk

c

x

pppp

t

Pk

c

x

P

n

in

in

in

in

in

i

=

=

++





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability



n

in

in

in

in

i PPpppr

xc

tkrlet

=+

=

++

111

2

)2(





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability


Using Fourier Series definitions:

][][1)]([

][)]([11

1

11 2xinxin

xxin

xinxxin

eee

eer

=

+

+

+

Dividing by and rearranging:xi

e 1





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability



nnxinnxin eer =+ + 1]2[ 11

Rearranging:

1]12)([ 11 + =++ nxixin reer





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability


Eulers identity:

sincos ie i =

Now, applying Eulers identity:

11111 ]12)sin(cos)sin(cos[ +=+++ nn rxixrxixr





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability


Rearrange:

11 ]12cos2[ +=+ nn rxr

11 )]cos1(21[ += nn xr





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability


Dividing both sides by n and rearranging:

)cos1(21 11

max xrAF nn

==+

For stability, AF max1





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability


1)cos1(21 1 xr

Consider the following three situation for the above argument:

pi

pi

=

=

=

x

x

x

1

1

1

)32

)2

0)1





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability


111)11(211)cos1(21

10cos0

1

1

==

r

xr

x

The above is always true but it provides no useful information.

Situation #1





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability


Lets wait for our final decision.

Situation #2

121,1)01(21,1)cos1(21

,02

cos,2

1

1

==

rr

xr

x

pipi





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability


Lets wait for our final decision.

Situation #3

141,1)11(21,1)cos1(21

,1cos,

1

1

+

==

rr

xr

x

pipi





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability


Situation #3

21

,24,1)41(,

0,04,1)41(,141

+

rrr

ORrrrr

2xc

tkr Recall

=





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability

210 r


Situation #3

kxc

t2

21

21

2

xc

tk

This means, this scheme is conditionally stable





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability

Fourier Series method for Implicit scheme


)()()(1

2

11

1111

2

2

t

PPk

c

x

pppp

t

Pk

c

x

P

n

in

in

in

in

in

i

=

=

++

++++





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability



n

in

in

in

in

i PPpppr

xc

tkrlet

=+

=

++

+++

111

111

2

)2(





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability



][][1)]([1

][1)]([111

1

11 2xinxin

xxin

xinxxin

eee

eer

=

+

+

+

+++


e 1





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability



nnxinnxin eer =+ ++++ 1111 ]2[ 11

Rearranging:

nxixinreer =+ + ]12)([ 111





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability


Eulers identity:

sincos ie i =


nnrxixrxixr =+++ ]12)sin(cos)sin(cos[ 11111





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability


Multiply by -1 and rearrange:

nnrxr =+ ]12cos2[ 11

nn xr =++ ]1)cos1(2[ 11





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability



1)cos1(21

1

1

max +==

+

xrAF

n

n






Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability


11)cos1(2

11

+ xr

11)cos1(2 1 + xr

0)cos1(2 1 xr

11)cos1(2 1 + xr





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability


0)cos1(2 1 xr

0cos1 1 x

1cos 1 x

This is true for all values of , regardless of the value of r.

The scheme is unconditionally stable.

x1





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability

Fourier Series method for Crank-Nickolson scheme


)(2)()(

)()(1

211

2

11

1111

t

PPk

c

x

ppppx

ppppn

in

in

in

in

in

i

n

in

in

in

i

=

+

+

+

+

++++

t

Pk

c

x

P

=

2

2





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability



n

in

in

in

in

in

in

in

i PPppppppr

xc

tkrlet

=+++

=

++

+

+++

111

11

111

2

)22(2

2





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability



][][1

)]([][

)]([)]([1

][1)]([1

11

11

11

11

2

22 xinxin

xxinxin

xxinxxin

xinxxin

ee

ee

ee

ee

r

=

+

++

+

++

+++


e 1





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability



nn

xinnxin

xinnxin

ee

eer

=

+

++ +

+++1

111

11

11

22

2

Rearranging:

++

=+

+

]14)(2[]14)(2[

11

111

reer

reer

xixin

xixin





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability


Eulers identity:

sincos ie i =


+++=+++

]14)sin(cos2)sin(cos2[]14)sin(cos2)sin(cos2[

1111

11111

rxixrxixrrxixrxixr

n

n





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability

]14cos4[]14cos4[ 111 +=+ rxrrxr nn


Multiply by -1 and rearrange:

]1)cos1(4[]1)cos1(4[ 111 =++ xrxr nn





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability



1)cos1(41)cos1(4

1

11

max ++

==

+

xr

xrAFn

n






Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability

11)cos1(41)cos1(4

1

1 ++

xr

xr


1)cos1(41)cos1(4 11 ++ xrxr

1)cos 1 xThis is true for all values of , regardless of the value of r.


x1





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Instability

11)cos1(41)cos1(4

1

1 ++

xr

xr


1)cos1(41)cos1(4 11 + xrxr

11

This is true for all values of , regardless of the value of r.


x1





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

Incompressible fluid flow in 1-D

x3=150mx2=190m

Continue

Calculate pressure distribution as follows:

1. Explicit scheme at t1=t, t2=2t, and t100=100t

2. Implicit scheme at t2=2t, and t100=100t

3. Also calculate the MB

A2 A4

P3P2

K2=1m2 K4=0.1

P1

K1=.5A1=5000m

2

2=.25 3=.151=.2

x1=200m

T1/2=0

T7/2=0

Example: No flow boundary

All blocks have the same depth and area

qi=100m3/D

=50mPa.s, c=10-6kPa-1, Di=0, and initial conditions, Pio=10MPa





Flow between

block

Transmissibility



Questions

Flow Equation

Single-phase

Flow

Non-horizontal



Nomenclature

Introduction

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LL4 ENPE470-2013 Instability