Simulation of Subsurface Two-Phase Flow in an Oil

Reservoir

Carsten Volcker, John Bagterp Jørgensen, Per Grove Thomsen andErling Halfdan Stenby

Technical University of Denmark (DTU)Department of Informatics and Mathematical Modeling (IMM)

January 28, 2009

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Outline

1 Introduction

2 Two-Phase Isothermal Immiscible FlowConservation EquationsConstitutive Equations

3 Numerical IntegrationODE ModelRunge-KuttaESDIRK Integration

4 Solving The Linear EquationsModified NewtonLinear Solvers

5 Error Control

6 Test Case

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Introduction

2D Reservoir

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Introduction

Optimizing Production

Maximizing net present value (NPV):

Closed loop optimizer:

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Introduction

Water Flooding without/with Optimal Control

Permeability field with two streaks:

Without optimal control:

With optimal control:

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Two-Phase Isothermal Immiscible Flow Conservation Equations

Conservation Equations

Mass conservation of water and oil:

∂

∂tCw(Pw, Sw) = −∇Nw(Pw, Sw) +Qw

∂

∂tCo(Po, So) = −∇No(Po, So) +Qo

No flow potential due to gravitation.

Homogenous permeability field.

Capillary pressure neglected.

Incompressible rock.

Mass concentrations:

Cw = φρw(Pw)Sw

Co = φρo(Po)So

Fluxes through the porous medium:

Nw = ρw(Pw)uw(Pw, Sw)

No = ρo(Po)uo(Po, So)6 / 21

Two-Phase Isothermal Immiscible Flow Conservation Equations

Darcy Velocity and Boundary Conditions

Pressure driven flow, Darcy’s law:

uα = −kkrα(Sα)

µα∇Pα α = o,w

No flow across boundaries:

uα · n = 0 and uα · n = 0

Internal sources/sinks due to wells:

Water is injected to maintain pressure and replace the oil.

Oil and water are produced.

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Two-Phase Isothermal Immiscible Flow Conservation Equations

Reduction Of State Variables:

Water saturation (volume fraction):

Sw + So = 1

Pressure difference due to capillary pressure:

Pcow = Po − Pw

Reduction of variables:

Sw = 1 − So ⇒ S = Sw = 1 − So

Pcow = 0 ⇒ P = Pw = Po

State variables (S,P ) = (Sw, Po):

S = S(t, s)

P = P (t, s)

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Two-Phase Isothermal Immiscible Flow Constitutive Equations

Density and compressibility

Compressible fluids:

ρw = ρw0eP−Pw0

ρo = ρo0eP−Po0

0 100 200 300 400 500 600 700 800 900 1000750

800

850

900

950

1000

1050Densities of water and oil [kg/m3]

P [bar]

ρ

w

ρo

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Two-Phase Isothermal Immiscible Flow Constitutive Equations

Relative Permeabilities by The Corey Relations

Relative permeabilities:

krw = krw0snw

kro = kro0(1 − s)nw

Reduced water saturation:

s =S − Swc

1 − Swc − Sor

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

← Swc

← 1−Sor

Rel. perm. of water and oil by the Corey relations

S

k

rw

kro

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Numerical Integration ODE Model

Different Formulation

Partial differential equation (PDE) model:

∂

∂tCw(Pw, Sw) = −∇Nw(Pw, Sw) +Qw

∂

∂tCo(Po, So) = −∇No(Po, So) +Qo

Different formulation of an ordinary differential equation (ODE) modelafter discretizing spatially:

d

dtg(x(t)) = f(t, x(t)) x(t0) = x0

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Numerical Integration Runge-Kutta

Runge-Kutta Methods

Tailored formulation of an s-stage Runge-Kutta method:

Ti = tn + cihn i = 1, 2, . . . , s

g(Xi) = g(xn) + hn

s∑

j=1

aijf(Tj,Xj) i = 1, 2, . . . , s

g(xn+1) = g(xn) + hn

s∑

j=1

bjf(Tj,Xj)

g(xn+1) = g(xn) + hn

s∑

j=1

bjf(Tj,Xj)

en+1 = g(xn+1) − g(xn+1) = hn

s∑

j=1

djf(Tj,Xj) dj = bj − bj

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Numerical Integration ESDIRK Integration

Fast Integration by ESDIRK

Only s− 1 implicit stages:

0 0c2 a21 γ

c3 a31 a32 γ...

.... . .

1 b1 b2 b3 · · · γ

b1 b2 b3 · · · γ

b1 b2 b3 · · · bsd1 d2 d3 · · · ds

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Solving The Linear Equations Modified Newton

Modified Newton Step

The state values Xi are obtained by sequential solution of the residual:

R(Xi) = g(Xi) − hnγf(Ti,Xi) − ψi = 0 i = 2, 3, . . . , s

ψi = g(xn) + hn

i−1∑

j=1

aijf(Tj ,Xj) i = 2, 3, . . . , s

The Jacobian of the residual R(Xi):

J(Xi) =∂R

∂Xi(Xi) =

∂g

∂x(Xi) − hnγ

∂f

∂x(Ti,Xi)

≈∂g

∂x(xm) − hmγ

∂f

∂x(tm, xm)

= J(xm) = LU

Only updating the Jacobian by slow convergence or divergence:

LU∆Xi = R(Xi)

Xi := Xi − ∆Xi

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Solving The Linear Equations Linear Solvers

Jacobian Structure

1D: 3 Non-zeros: 2D: 5 Non-zeros: 3D: 7 Non-zeros:

Solving the linear equations:

Sparse direct solver: LU factorization and back sustitution.

Iterative solver: GMRES.

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Error Control

Adaptive Time Stepping

In most commercial simulators:

Simple heuritics implemented e.g. maximum variation of saturations.

ESDIRK, embedded error estimator:

en+1 = g(xn+1) − g(xn+1) = hn

s∑

i=1

dif(Ti,Xi)

Measures of the error may be controlled adjusting the time step accordingto

hn+1 =hn

hn−1

(

ε

rn+1

)k2/k (

rn

rn+1

)k1/k

hn

en+1 is an error estimate of the conserved quantities g(xn+1).

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Error Control

ESDIRK Performance

ESDIRK23 performance and statistics, 45 × 45 grid blocks:

0 20 40 60

10−6

10−4

10−2

Ste

psiz

e

0 20 40 60

10−2

100

102

Time

Err

/Tol

0 20 40 600

10

20

Sta

ge 2

0 20 40 600

10

20

Time

Sta

ge 3

Method : ESDIRK23absTol : 4050x1relTol : 4050x1

nIter 2 : 4734nIter 3 : 5506

nStep : 1289nFail : 129nDiv : 10nSlow : 1

nFun : 12815nJac : 1279nLU : 1290nBack : 10240

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Error Control

Performance by Significant Digits

ESDIRK performance on 1D case, 1000 grid blocks:

ESDIRK12 = red, ESDIRK23 = green, ESDIRK34 = blue.18 / 21

Test Case

2D Test Case, 45 × 45 Grid Blocks

Permeability field with two streaks:

Oil saturation after 31 days:

5 10 15 20 25 30 35 40 45

5

10

15

20

25

30

35

40

45 0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.55

0.6

0.65

0.7

Oil saturation after 62 days:

5 10 15 20 25 30 35 40 45

5

10

15

20

25

30

35

40

45 0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.55

0.6

0.65

0.7

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Test Case

Butcher Tableau’s of Runge-Kutta Methods

The explicit Runge-Kutta (ERK) method:

0 0c2 a21 0c3 a31 a32 0...

.... . .

cs as1 as2 as3 · · · 0

b1 b2 b3 · · · bs

b1 b2 b3 · · · bsd1 d2 d3 · · · ds

The A-matrix in Runge-Kutta methods:

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Test Case

Butcher Tableau’s of Runge-Kutta Methods

The diagonally implicit Runge-Kutta (DIRK) method:

c1 a11

c2 a21 a22

c3 a31 a32 a33

......

. . .

cs as1 as2 as3 · · · ass

b1 b2 b3 · · · bs

b1 b2 b3 · · · bsd1 d2 d3 · · · ds

The A-matrix in Runge-Kutta methods:

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Test Case

Butcher Tableau’s of Runge-Kutta Methods

The singly diagonally implicit Runge-Kutta (SDIRK) method:

c1 γ

c2 a21 γ

c3 a31 a32 γ...

.... . .

cs as1 as2 as3 · · · γ

b1 b2 b3 · · · bs

b1 b2 b3 · · · bsd1 d2 d3 · · · ds

The A-matrix in Runge-Kutta methods:

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Test Case

Butcher Tableau’s of Runge-Kutta Methods

The explicit singly diagonally implicit Runge-Kutta (ESDIRK) method:

0 0c2 a21 γ

c3 a31 a32 γ...

.... . .

cs as1 as2 as2 · · · γ

b1 b2 b3 · · · bs

b1 b2 b3 · · · bsd1 d2 d3 · · · ds

The A-matrix in Runge-Kutta methods:

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Test Case

Butcher Tableau’s of Runge-Kutta Methods

The fully implicit Runge-Kutta (FIRK) method:

c1 a11 a12 a13 · · · a1s

c2 a21 a22 a23 a2s

c3 a31 a32 a33 a3s...

.... . .

...cs as1 as2 as3 · · · ass

b1 b2 b3 · · · bs

b1 b2 b3 · · · bsd1 d2 d3 · · · ds

The A-matrix in Runge-Kutta methods:

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Test Case

Fast Integration by ESDIRK

Only s− 1 implicit stages:

0 0c2 a21 γ

c3 a31 a32 γ...

.... . .

1 b1 b2 b3 · · · γ

b1 b2 b3 · · · γ

b1 b2 b3 · · · bsd1 d2 d3 · · · ds

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Test Case

Fast Integration by ESDIRK

Only s− 1 implicit stages:

0 0c2 a21 γ

c3 a31 a32 γ...

.... . .

1 b1 b2 b3 · · · γ

b1 b2 b3 · · · γ

b1 b2 b3 · · · bsd1 d2 d3 · · · ds

The first stage is explicit, which implies that:

X1 = xn

xn+1 = Xs

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Test Case

Fast Integration by ESDIRK

Only s− 1 implicit stages:

0 0c2 a21 γ

c3 a31 a32 γ...

.... . .

1 b1 b2 b3 · · · γ

b1 b2 b3 · · · γ

b1 b2 b3 · · · bsd1 d2 d3 · · · ds

The Butcher tableau is constructed such that:

X1 = xn

xn+1 = Xs

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Test Case

Fast Integration by ESDIRK

Only s− 1 implicit stages:

0 0c2 a21 γ

c3 a31 a32 γ...

.... . .

1 b1 b2 b3 · · · γ

b1 b2 b3 · · · γ

b1 b2 b3 · · · bsd1 d2 d3 · · · ds

The state values Xi are obtained by sequential solution of the residual:

R(Xi) = g(Xi) − hnγf(Ti,Xi) − ψi = 0 i = 2, 3, . . . , s

ψi = g(xn) + hn

i−1∑

j=1

aijf(Tj ,Xj) i = 2, 3, . . . , s

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