Numerical Approximation of Filtration Processes through Porous Media

Numerical Approximation of Filtration Processes throughPorous Medium

Master Thesis Presentation

Raheel Ahmed

Supervisor: Marco Discacciati

Universitat Politecnica de Catalunya - Barcelona Tech

CIMNE

June 25, 2012

Raheel Ahmed (UPC) Stokes - Darcy Coupling June 25, 2012 1 / 1

Introduction

Free fluid flow coupled with porous medium flow.

Importance in industrial and natural processes

Membrane filtration processesAir or oil filtersBlood flow through body tissuesForward osmosis processes

Numerical analysis of the coupled problem

[Campbell, Biology 2009]


Objectives

Mixed finite element discretization for both fluid regions.

Investigation of the optimum solution method based on preconditioning theSchur complement system.

“...at this very moment the search is on - every numerical analyst has a favoritepreconditioner, and you have a perfect chance to find a better one.”

-Gil Strang (1986)


Problem Statement

nd Γ

Ωs

Ωd

∂Ωs,N

∂Ωs,N

∂Ωd,D

∂Ωd,N

∂Ωs,D

∂Ωd,N

nn

Stokes Equation

−νus +∇ps = f inΩs

∇ · us = 0 inΩs

Darcy Equation

ud = −K∇pd inΩd

∇ · ud = 0 inΩd


Problem StatementInterface Conditions

nd Γ

Ωs

Ωd

∂Ωs,N

∂Ωs,N

∂Ωd,D

∂Ωd,N

∂Ωs,D

∂Ωd,N

nn

Conservation of mass across the interface:

us · n = ud · n, on Γ

Balance of normal forces across the interface:

−νn ·∂us

∂n+ ps = gpd on Γ

Beavers-Joseph-Saffman condition:

−ντ j ·∂us

∂n=

ν

ǫus · τ j (j = 1, 2 for 2D) on Γ


Steady Stokes Darcy ProblemWeak Formulation

Stokes Equations:

∫Ωs

ν∇u0s · ∇vs −

∫Ωs

ps ∇ · vs +

∫Γ

gp0d (vs · n) +

∫Γ

n−1∑j=1

ν

ǫ(u0

s · τ j )(vs · τ j ) =

∫Ωs

f · vs

−

∫Ωs

∇ · u0s qs = 0

Darcy Equations:

Primal-mixed formulation

Addition of stability terms proposed by [Masud. 2002].

1

2K−1g

∫Ωd

ud · vd +1

2g

∫Ωd

∇p0d · vd = 0

1

2

∫Ωd

gud · ∇qd +

∫Γ

g(u0s · n)qd −

1

2

∫Ωd

g(K∇p0d · ∇qd) = 0


Steady Stokes Darcy ProblemAlgebraic Formulation

By the finite element discretization, we get algebraic system as follows.

Stokes

P1 − P1 P2 − P2

Darcy

A1,ii A1,iΓ B1i 0 0 0A1,Γi A1,ΓΓ B1Γ 0 0 PΓ

BT1i BT

1Γ 0 0 0 00 0 0 A2 B2i B2Γ

0 0 0 BT2i Sii SiΓ

0 PTΓ 0 BT

2Γ SΓi SΓΓ

uis

uΓ

psud

pid

pΓ

=

F1i

F1Γ

F12

F21

F2i

F2Γ

Large

Ill-Conditioned

Symmetric

Indefinite


Steady Stokes Darcy ProblemAlgebraic Formulation

A1,ii A1,iΓ B1i 0 0 0A1,Γi A1,ΓΓ B1Γ 0 0 PΓ

BT1i BT

1Γ 0 0 0 00 0 0 A2 B2i B2Γ

0 0 0 BT2i Sii SiΓ

0 PTΓ 0 BT

2Γ SΓi SΓΓ

uis

uΓ

psud

pid

pΓ

=

F1i

F1Γ

F12

F21

F2i

F2Γ

As BsΓ 0 0BTsΓ A1,ΓΓ 0 PΓ

0 0 Ad BdΓ

0 PTΓ BT

dΓ SΓΓ

us

uΓ

ud

pΓ

=

Fs

F1Γ

Fd

F2Γ

Schur complement system:

(Σs + Σd ) uΓ = f1Γ − PΓΣ−1c f2Γ

(Σc + Σf ) pΓ = f2Γ − PTΓ Σ−1

s f1Γ

Properties of Schur Complement

Smaller than original system

Better conditioned than the original system (O(h−1)).


Solution Method

(Σs + Σd)uΓ = f1Γ − PΓΣ−1c f2Γ

Solution by Krylov iterative methods


Numerical Tests

(Σs + Σd)uΓ = f1Γ − PΓΣ−1c f2Γ

(0, 1) × (0, 2)

Γ

∂Ωs,D

∂Ωd,N∂Ωd,N

∂Ωd,D

Ωs

∂Ωs,D

Ωd

∂Ωs,N

Structured mesh;Elements: MINI &P1 − P1

Residual tolerance forsolution = 1.e − 9

Conjugate Gradient without preconditioner

Number of iterationsGrid ν = 10−4, ν = 10−6, ν = 10−6, ν = 100,

K = 10−3 K = 10−5 K = 10−8 K = 100

1 4 4 4 42 9 9 10 83 20 20 24 164 33 34 39 28

Iterations increase with decrease in mesh size.


Iteration tests with Preconditioner

Dirichlet-Neumann Preconditioner

P−1 = Σ−1

s

Grid ν = 10−4, ν = 10−6, ν = 10−6, ν = 100,

K = 10−3 K = 10−5 K = 10−8 K = 100

1 4 4 4 42 10 10 10 53 24 24 30 54 50 53 64 5



GHSS Preconditioner (work by [Benzi, 2009])

P−1 = 2α(Σd + αI )−1(Σs + αI )−1

Number of iterationsGrid ν = 10−4,K = 10−3 ν = 10−6,K = 10−5 ν = 10−6,K = 10−8 ν = 100,K = 100

1 4 (α = 1.e − 2) 2 (α = 1.e − 1) 1 (α = 1.e − 1) 4 (α = 1.e − 2)2 5 (α = 1.e − 2) 2 (α = 1.e − 1) 1 (α = 1.e − 1) 8 (α = 1.e − 2)3 5 (α = 1.e − 2) 2 (α = 1.e − 1) 1 (α = 1.e − 1) 11 (α = 1.e − 2)4 7 (α = 1.e − 2) 3 (α = 1.e − 1) 1 (α = 1.e − 1) 10 (α = 1.e − 2)

−8 −7 −6 −5 −4 −3 −2−4

−3

−2

−1

0

1

2ν = 10−4

log10

K

log 10

α

y = − 0.05*x3 − 0.74*x2 − 4*x − 8.4



GHSS-Variant (1)

P−1 = 2αd(Σd + αd I )

−1(Σs + αs I )−1

Optimum results for αs > αd when ν,K < 1

GHSS-Variant (2)P

−1 = 2αd(Σd + αd I )−1

→ αd = 1.e − 3Grid ν = 10−4

, ν = 10−6, ν = 10−6

,

K = 10−3 K = 10−5 K = 10−8

1 2 2 12 3 2 13 3 2 14 3 2 1



Neumann-Neumann Preconditioner

P−1 = θs

2(Σs)−1 + θd

2(Σd)−1

where, θs =νK

νK+hand θd = h

νK+hwith h being the mesh size.

Grid ν = 10−4, ν = 10−6, ν = 10−6,K = 10−3 K = 10−5 K = 10−8

1 2 1 12 2 1 13 2 1 24 3 1 1


Unsteady Stokes-Darcy Problem

∂us

∂t− νus + ∇ps = f inΩs

∇ · us = 0 inΩs

ud = −K∇pd inΩd

So

∂pd

∂t+ ∇ · ud = 0 inΩd

us · n = ud · n, on Γ

−νn ·∂us

∂n+ ps = gpd on Γ

−ντ j ·∂us

∂n=

ν

ǫus · τ j (j = 1, . . . , n − 1) on Γ.

nd Γ

Ωs

Ωd

∂Ωs,N

∂Ωs,N

∂Ωd,D

∂Ωd,N

∂Ωs,D

∂Ωd,N

nn


Unsteady Stokes-Darcy Problem

Weak Formulation

Mixed finite element discretization

Time discretisation: Backward Euler Difference Scheme

Interface Systems

(Σs + Σd) um+1Γ = f1Γ − PΓΣ

−1c f2Γ

(Σc + Σf ) pm+1Γ = f2Γ − PT

Γ Σ−1s f1Γ

for every time tm, m = 0, . . . ,N where N is number of time intervals


Unsteady Stokes-Darcy ProblemPreconditioners

Preconditioner Properties

Dirichlet Neumann CG solverΣ−1

s K , ν ≥ 1

GHSS GMRES solver2α(Σd + αI )−1(Σs + αI )−1 α is not fixed

Multiplicative

GHSS variant(1) → GHSS variant (2) CG solver2αd (Σd + αd I )

−1 K , ν ≤ 1

Neumann-Neumann CG solverθs

2(Σs)−1 + θd

2(Σd )−1 K , ν ≤ 1

θs , θd can be controlled.

For Unsteady, identical behaviour as presented for Steady problem.


Cross-Flow Filtration Problem

[novasep]

ν = 0.08247m2/s

by [Hanspal et al., 2009]


Cross-Flow Filtration Problem (Steady) I

K =

1.1882× 10−4m/s

K =

1.1882× 10−10m/s

0 1 2 3 4 5 6 7 8−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

3Velocity vectors

0.0E+000

1.6E−002

3.2E−002

4.9E−002

6.5E−002

8.1E−002

9.7E−002

1.1E−001

1.3E−001

1.5E−001

0 1 2 3 4 5 6 7 8 9−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

3Velocity Vectors

0.0E+000

6.6E−002

1.3E−001

2.0E−001

2.7E−001

3.3E−001

4.0E−001

4.6E−001

5.3E−001

6.0E−001


Cross-Flow Filtration Problem (Steady) II

Number of elements Number of iterations for solutionStokes Darcy Non-Preconditioned system Neumann-Neumann Preconditioned system

K = 1.1882× 10−4

432 104 10 51728 416 17 56912 1664 25 5

K = 1.1882× 10−10

432 104 10 31728 416 25 36912 1664 43 3


Cross-Flow Filtration Problem (Unsteady) I

Variation of Hyd. Conductivity near interface, with time.

Time Hyd. Conductivity Number of iterationst(s) K(m/s) Non-Preconditioned Neumann-Neumann Preconditioner1 1.1883 17 210 0.11883 17 220 0.00297 17 330 4.4009e-5 16 540 4.641e-7 12 1250 3.802e-9 22 8

Simulation


Unsteady Stokes-Darcy ProblemDecoupled Method

Work by [Shan et al. 2011] extended to mixed discretisation in porous region.

Different time steps for different sub-domains

Stokes Darcy∆s = n∆t

SΓ

∆t

Less number of time intervals for Darcy

Accuracy is compromised in Darcy domain.

Require refined mesh and small time intervals for better accuracy.


Conclusion

Optimum solution methods have been presented for coupled problem.

Can be employed for the numerical analysis of large practical problems.

Can be implemented into already available solvers.

Thank You!


Technology

Numerical Approximation of Filtration Processes through Porous Media