Robust Monolithic - Multigrid FEM Solver for Three Fields Formulation Rising from non-newtonian Flow Problems

Transcription

1 Robust Monolithic - Multigrid FEM Solver for Three Fields Formulation Rising from non-newtonian Flow Problems M. Aaqib Afaq Institute for Applied Mathematics and Numerics (LSIII) TU Dortmund 13 July 2017

2 Contents 1 Motivation 2 Governing Equations 3 Variational Formulation 4 Finite Element Approximation 5 Numerical Results 6 Summary

3 Contents 1 Motivation 2 Governing Equations 3 Variational Formulation 4 Finite Element Approximation 5 Numerical Results 6 Summary

4 Three Fields Formulation Formulation involving stress Bingham/Viscoplastic flows Viscoelastic flows Multiphase flows Saddle point problem Requires two inf-sup conditions Velocity-Pressure Stress-Velocity

5 Bingham Model with additional symmetric viscoplastic stress tensor D(u) ɛ W D(u) = 0 (2ηD(u) + τ s W) + p = 0 u = 0 u = g D in Ω in Ω in Ω on Γ D W = D(u) D(u) ɛ W = new stress tensor τ s = yield stress D = strain rate tensor

6 Viscoelastic Model with additional elastic stress tensor Oldroyd-B (γ = 0) Giesukus (γ = 1) u t + (u )u = p + η s u + η p eψ u = 0 ψ t + (u )ψ (Ωψ ψω) 2B = 1 (e ψ I) γe ψ (e ψ I) 2 ψ = log conformation tensor D = strain rate tensor η s = solvent viscosity η p = polymer viscosity

7 Multiphase Model with additional multiphase stress tensor τ = τ s + τ m ( ) u ρ(ψ) t + u u τ + p = 0 u = 0 ( ) ψ ψ τ m = σ, τ s = 2µ(ψ)D(u) ψ ϕ t + u ϕ γ nd (( ϕ ϕ ϕ 1) ϕ ϕ ) = 0 ψ τ + (γ ncψ(1 ψ) ϕ) (γ nd ( ψ ϕ) ϕ) = 0 τ m = multiphase stress τ s = viscous stress

8 Jump Discontinuity Rising bubble in non-newtonian fluids In viscoelastic liquid Volume exceeds a critical value Rise velocity jump discontinuity shape changes: convex to tear drop

9 Jump Discontinuity Prediction: Boundary condition changes from rigid to free Jump occurs due to negative wake Jump occurs due to shear dependence of the viscosity

10 Industrial Applications Bubble Phenomena Pipeline transport application Bio reactors Combustion engines Underwater explosions Medicine: Angioplasty Food industry Fermentation process Bread and yogurt Nature: Bubble fence

11 Industrial Applications Bubble Phenomena Chemical separator Dispersion of gas bubble Efficient mass transfer Polymer and sludge processes

12 Contents 1 Motivation 2 Governing Equations 3 Variational Formulation 4 Finite Element Approximation 5 Numerical Results 6 Summary

13 Goals to Achieve Efficient three fields solver Solver: Monolithic multigrid approach Constraints free for the choice of finite element spaces

14 Navier Stokes Equation Primitive Formulation (u )u 2ηD(u) + p = 0 in Ω u = 0 in Ω (1) u = g D on Γ D Three Field Formulation σ D(u) = 0 in Ω ( ) (u )u 2η(1 α)d(u) + 2ηασ + p = 0 in Ω u = 0 u = g D in Ω on Γ D (2)

15 Three Fields Formulation Stokes System σ D(u) = 0 in Ω ( ) 2η(1 α)d(u) + 2ηασ + p = 0 in Ω u = 0 u = g D in Ω on Γ D (3) σ = extra stress tensor α = parameter η = viscosity

16 Contents 1 Motivation 2 Governing Equations 3 Variational Formulation 4 Finite Element Approximation 5 Numerical Results 6 Summary

17 Variational Formulation V = H 1 0(Ω) := ( H 1 0 (Ω)) 2 velocity and dual space V Q = L 2 0 (Ω) pressure and dual space Q T = ( L 2 (Ω) ) 4 stress and dual space T sym A u defined on V V A u u, v := 2η(1 α) A σ defined on T T A σ σ, τ = 2ηα The associated bilinear forms Ω Ω D(u) : D(v) dx (4) σ : τ dx (5) a u (u, v) = A u u, v, a σ (σ, τ ) = A σ σ, τ (6)

18 Variational Formulation B and C defined on V Q respectively, V T Bv, q := v q dx (7) Ω Cv, τ :=2ηα τ : D(v) dx (8) Associated bilinear forms b(, ) and c(, ) defined on V Q R and V T R respectively Ω b(v, q) := Bv, q (9) c(v, τ ) := Cv, τ (10)

19 Two-Folds Saddle Point Problem A u C T B T u RHS σ C A σ 0 σ = RHS u (11) B 0 0 p RHS p The system (3) has the following weak formulation a σ (σ, τ ) + c(u, τ ) = RHS σ, τ τ T a u (u, v) + c(v, σ) + b(v, p) = RHS u, v v V b(u, q) = RHS p, q q Q (12)

20 Variational Formulation Introduce the null spaces Ker B Ker B = {v V; b(v, q) = 0 q Q} Let X := Ker B T A(u, σ), (v, τ ) = A u u, v + Cv, σ + A σ σ, τ Cu, τ (13) a(u, V) = a u (u, v) + c(v, σ) + a σ (σ, τ ) c(u, τ ) (14)

21 Solvability of Problem Find U X such that: a(u, V) = f, V V X (15) Theorem Let X be a Hilbert space and f X, topological dual space of X, and let a(.,.) be a bilinear form on X satisfying the following hypothesis: 1). a(, ) is continuous: there exists a constant C > 0 such that: a(u, V) C U V U, V X (16)

22 Solvability of Problem Theorem (cont...) 2). a(, ) is X-elliptic: there exists a constant β > 0 such that: a(v, V) β V 2 V X (17) then problem has a unique solution U X. Remark Besides from theory of saddle point problem it is easy to show that there exists a unique p Q such that (σ, u, p) is the unique solution of problem (12)

23 Solvability of the Problem Illustration for α 1 a(v, V) = a u (v, v) + a σ (τ, τ ) = 2η(1 α) D(v) 2 dx + 2ηα τ 2 dx Ω Ω C2η(1 α) v 2 V + 2ηα τ 2 T C (v, τ ) 2 X Remark When α = 1 we can show existence and uniqueness using saddle point theory

24 Contents 1 Motivation 2 Governing Equations 3 Variational Formulation 4 Finite Element Approximation 5 Numerical Results 6 Summary

25 Finite Element Approximation Domain Ω R d Quadrilaterals elements K, Ω = int ( K T h K) Finite element spaces approaximations T h = { τ h M, σ h K Q 2 (K) } V h = { v h V, v h K Q 2 (K) } { } Q h = q h Q, q h K P1 disc (K) (18)

26 Finite Element Approximation Approximation of Stokes problem for finite element spaces T h T, V h V, Q h Q a σ (σ h, τ h ) + c(u h, τ h ) = RHS σ, τ h τ h T h a u (u h, v h ) + c(v h, σ h ) + b(v h, p h ) = RHS u, v h v h V h b(u h, q h ) = RHS p, q h q h Q h (19) Two LBB conditions required Firstly, between velocity and stress Secondly, between velocity and pressure

27 Finite Element Approximation Approximation of Stokes problem for finite element spaces T h T, V h V, Q h Q a σ (σ h, τ h ) + c(u h, τ h ) = RHS σ, τ h τ h T h a u (u h, v h ) + c(v h, σ h ) + b(v h, p h ) = RHS u, v h v h V h b(u h, q h ) = RHS p, q h q h Q h Two LBB conditions required Firstly, between velocity and stress Secondly, between velocity and pressure

28 Stable Element Choice 1 Fortin and Fortin element (u, p) spaces LBB compatible σ and D(u) same FEM discontinuous space

29 Stable Element Choice 2 Marchal-Crochet element Subcell discretization to enrich the local d.o.f for σ n σ > n u condition satisfied Computational cost is increased due to more d.o.f

30 EO-FEM Stabilization Remedy: Jump term addition to ensure element pair is stable Penalizing the jump of the solution gradient over E J u (u h, v h ) = γ u 2ηαh [ u h ] : [ v h ] dω (20) e E e h A u + J u C B T u RHS u C T A σ 0 σ = RHS σ (21) B 0 0 p RHS p

31 Discrete Problem Theorem Find U h X h such that: a(u h, V h ) + j(u h, V h ) = f h, V h V h X h (22) V h 2 = V h 2 + j(v h, V h ) (23) Let X h be a Hilbert space and f h X h, topological dual space of X, and let a(.,.) be a bilinear form on X h satisfying the following hypothesis: 1). a(, ) is continuous: there exists a constant C h > 0 such that: a(u h, V h ) C h U h V h U h, V h X h (24)

32 Solvability of Problem Theorem (cont...) 2). There exists a constant β h > 0 such that : a(u h, V h ) sup β h U h U h X h (25) V h X h V h then problem has a unique solution U h X h. Remark Besides from theory of saddle point problem it is easy to show that there exists a unique p h Q h such that (σ h, u h, p h ) is the unique solution of problem (19) No constraints on the choice of discrete finite space for stress

33 Contents 1 Motivation 2 Governing Equations 3 Variational Formulation 4 Finite Element Approximation 5 Numerical Results 6 Summary

34 Benchmark Configuration Inlet: Dirichlet parabolic profile u(0, y) = 4 Uy(0.41 y) (0.41) 2 No-slip at upper and lower walls Γ 1 and Γ 3 Outlet: Neumann boundary Γ 2 Kinematic viscosity η = 10 3 Characteristic length of cylinder l c = 0.1

35 Numerical Results

36 Monolithic Multigrid Solver Three fields Stokes vs Stokes solver in primitive variables Level Lift Drag NL/LL Lift 1 Drag NL/LL / / / / / / / / /2 Three field solver performance efficient as primitive Stokes solver Check robustness and consistency! 1 Damanik. H FEM Simulation of Non-isothermal Viscoelastic fluids, PhD Thesis

37 EO-FEM : Consistency Consistency for case α = 0 Level α Lift Drag NL/LL Lift Drag NL/LL No Stab. Stab / / / / / / / /3 Edge Oriented FEM is consistent Side effect neither on solution nor on the solver

38 EO-FEM :Robustness Two extreme cases α=0 viscous contribution α=1 no viscous contribution Level α Lift Drag NL/LL Lift Drag NL/LL No Stab. Stab / / / / / / / / / / / /3 Robustness w.r.t. problem! Consistent and grid independent solver

39 Contents 1 Motivation 2 Governing Equations 3 Variational Formulation 4 Finite Element Approximation 5 Numerical Results 6 Summary

40 Summary Robust monolithic multigrid solver for three fields Stokes formulation using EO-FEM: Advantages Taking away the 2nd inf-sup condition (no constraints on the choice of the σ space) Allowing large class of discretiztation for the stress Taking the efficiency of the Stokes solver in primitive variables to Stokes solver in three fields formulation