Coupling free and porous-media flows: modeling, analysis and - PowerPoint PPT Presentation

Coupling free and porous-media flows: modeling, analysis and numerical approximation Marco Discacciati Special Semester on Multiscale Simulation & Analysis in Energy and the Environment RICAM, Linz , October 5, 2011 Partial support of the Marie Curie Career Integration Grant 2011-294229

MOTIVATION Modeling free and porous media flows requires to consider coupled differential models featuring Navier-Stokes equations in the fluid domain and a filtration model in the porous domain, like the Darcy equation . ⇒ Global coupled heterogeneous differential problem.

Environmental application Blood flow simulations 115.1 109.96 109.9 110 109.84 109.78 109.72 109.66 109.6 100 109.54 109.48 109.42 109.36 109.3 109.24 90 109.18 109.12 109.06 83.22 109 63.75 70 80 90 98.44 Industrial applications: filters, porous foams, fuel cells...

PROBLEM SETTING Fluid flow: Navier-Stokes equations − div T( u f , p f ) + ( u f · ∇ ) u f = f in Ω f div u f = 0 where T( u f , p f ) = ν ( ∇ u f + ∇ T u f ) − p f I is the Cauchy stress tensor. Fluid through porous media: Darcy’s equations K − 1 u p + ∇ p p = 0 in Ω p ⇔ − div (K ∇ p p ) = 0 in Ω p div u p = 0 Ω f Free-fluid domain Γ Porous media domain Ω p

COUPLING (INTERFACE) CONDITIONS The solution must satisfy three regularity conditions across Γ: the continuity of the normal velocities u f · n = u p · n ⇔ u f · n = − K ∇ p p · n a consequence of the incompressibility; the continuity of the normal stresses − n · T( u f , p f ) · n = p p (pressures can be discontinuous across Γ); a condition on the tangential component of the normal stress: Beavers–Joseph–Saffman equation − τ · T( u f , p f ) · n = α u f · τ ( − α u p · τ ) [Miglio, Discacciati, Quarteroni (2002); Layton, Schieweck, Yotov (2003)]

COUPLING (INTERFACE) CONDITIONS Experimental approach: - 1967: Beavers and Joseph; - 1971: corrected by Saffman (removed u p in the tangential part). Mathematical approach: - 1996, 2000, 2001: justification by J¨ ager and Mikeli´ c via homogenization theory.

LITERATURE A far-from-complete list of names: Arbogast et al.; Badia, Codina; Becker et al; Bernardi et al.; Burman, Hansbo; Correa, Loula; D’Angelo, Zunino; Discacciati, Quarteroni; Iliev, Laptev; Kanschat; Layton; Fuhrmann et al.; Galvis, Sarkis; Gatica, Oyarzua; Girault; Gunzburger, Hua; J¨ ager, Mikeli´ c, Neuss; Mu, Xu, Zhu; Nassehi et al; Rivi` ere; Urquiza et al.; Wolmuth, Helmig; Yotov; ...

MATHEMATICAL ANALYSIS OF THE COUPLED PROBLEM

WEAK FORM OF THE COUPLED DARCY – NAVIER-STOKES PROBLEM (I) Find u f ∈ H 1 (Ω f ), p f ∈ L 2 (Ω f ), p p ∈ H 1 (Ω p ): � � � ν ∇ u f · ∇ v + α ( u f · τ )( v · τ ) + [( u f · ∇ ) u f ] v Ω f Γ Ω f � � � − p p ( v · n ) = f · v p f div v + Ω f Γ Ω f � q div u f = 0 Ω f � � K ∇ p p · ∇ ψ − ψ ( u f · n ) = 0 Ω p Γ

WEAK FORM OF THE COUPLED DARCY–NS PROBLEM (II) Find u f ∈ H 1 (Ω f ), p f ∈ L 2 (Ω f ), u p ∈ L 2 (Ω p ), p p ∈ H 1 (Ω p ): � � � ν ∇ u f · ∇ v + α ( u f · τ )( v · τ ) + [( u f · ∇ ) u f ] v Ω f Γ Ω f � � � − p f div v + p p ( v · n ) = f · v Ω f Γ Ω f � q div u f = 0 Ω f � � K − 1 u p · w + w · ∇ p p = 0 Ω p Ω p � � u p · ∇ ψ + ( u f · n ) ψ = 0 Ω p Γ [Urquiza et al. (2008); Masud, Hughes (2005)]

ON THE WELL-POSEDNESS OF THE COUPLED PROBLEMS The Darcy-Stokes case: well-posedness can be easily proved using the theory by Brezzi for saddle point problems. [Miglio et al. (2002); Layton et al. (2003); Urquiza et al. (2008)] The Darcy-Navier–Stokes case: if there holds � f � L 2 (Ω f ) ≤ C ν 2 the Darcy-Navier–Stokes problem has a solution which is unique if the normal velocity across the interface is ‘small enough’ : u f · n ∈ S r m = { η ∈ H 1 / 2 00 (Γ) ≤ r m } ⊂ H 1 / 2 00 (Γ) : � η � H 1 / 2 00 (Γ) for a suitably defined radius r m . [Girault, Rivi` ere (2009); Badea, Discacciati, Quarteroni (2010)]

FINITE ELEMENT APPROXIMATION A conforming FE approximation of this problem would lead to solve a global nonlinear system, generally large, sparse and ill-conditioned. “Laplace” – Navier-Stokes problem: D T  A ff ( u f ) A f Γ ( u f ) 0 0    u f f D f 0 D f Γ 0 0 p f      D T A f  A Γ f ( u f ) ΓΓ ( u f ) 0 M ΓΓ  u Γ  = F   f Γ  f      p p  A T  0 0 0 A pp    Γ p  p Γ A p − M T 0 0 A Γ p p ΓΓ ΓΓ with u Γ f → nodal values of u h f · n on Γ p Γ p → nodal values of p h p on Γ

Darcy – Navier-Stokes problem: D T  A ff ( u f ) A f Γ ( u f ) 0 0 0   u f  f D f 0 D f Γ 0 0 0 p f      D T A f  A Γ f ( u f ) ΓΓ ( u f ) 0 0 M ΓΓ  u Γ    f Γ  f  = F     u p G T G T   0 0 0 A p    p Γ p    p p   0 0 0 G p S pp S p Γ     p Γ S p M T 0 0 G Γ p S Γ p p ΓΓ ΓΓ

NUMERICAL ALGORITHMS

THE DARCY – NAVIER-STOKES CASE Fixed-point (Picard) method Newton method Convergence result: if • � f � L 2 (Ω f ) ≤ ˜ C ν 2 then • the Navier-Stokes/Darcy problem has a unique solution • the Newton method converges to this solution provided the initial normal velocity u 0 f · n on Γ is chosen ‘close enough’ to the solution. We have to solve a linearized coupled problem at each iteration. [Badea, Discacciati, Quarteroni (2010)]

NUMERICAL RESULTS We take Ω f = (0 , 1) × (1 , 2) and Ω p = (0 , 1) × (0 , 1). We use Taylor-Hood elements for the Navier-Stokes equations and quadratic Lagrangian elements for the Darcy equation. √ The exact solution is u f = (( y − 1) 2 + ( y − 1) + K x ( x − 1)), p f = 2 ν ( x + y − 1), ϕ = K − 1 ( x (1 − x )( y − 1) + ( y − 1) 3 / 3) + 2 ν x . Number of iterations with respect to the parameters ν and K: ν K h = 0 . 1429 h = 0 . 0714 h = 0 . 0357 FP N FP N FP N 1 1 7 4 7 4 7 4 10 − 4 1 5 4 5 4 5 4 10 − 1 10 − 1 10 5 10 5 10 5 10 − 2 10 − 1 15 6 15 6 15 6 10 − 2 10 − 3 13 6 13 6 13 6

AN APPLICATION: internal ventilation of motorcycle helmets (collaboration with F. Cimolin, Politecnico di Torino) The ventilation system is realized by means of a series of channels crossing the helmet. The air enters the channels from the air intakes, and the objective is to extract as much heat as possible. [Cimolin, Discacciati (2010)]

This simplified scheme shows how the heat is extracted by the fresh air, which flows above and through the porous comfort layer surrounding the head: (Schematic frontal cross section of the helmet) (Schematic longitudinal cross section of an air channel)

NUMERICAL RESULTS (I) 2D problem discretized using about 200,000 elements. Navier-Stokes/Darcy problem solved by the Newton method. Velocity field

A “DECOUPLED” STRATEGY We would like (in this case for software availability reasons) to solve the coupled problem exploiting the “intrinsic” decoupled structure of our physical problem ⇒ alternate the solution of the Navier-Stokes problem in Ω f and of Darcy equations in Ω p : Given a normal stress on the interface, solve the Navier-Stokes 1 equations in Ω f and recover the corresponding normal velocity across the interface; Use the computed normal velocity across the interface as 2 boundary condition for the Darcy equations in Ω p , solve them and recover the corresponding normal stress on the interface; Iterate using a suitable convergence criterion and a relaxation 3 procedure to enhance convergence (if necessary). Use a domain decomposition approach ⇒ write the global problem as an interface problem, choosing suitable interface variables.

NUMERICAL RESULTS (II) Steady-state flow field computed using a Navier-Stokes/ Forchheimer model: The normal component of the velocity through the interface and velocity profile at the outlet: y u y 4 0.2 3 2 0.1 1 u x 0.5 1.0 1.5 2.0 2.5 3.0 50 x 0.0 10 20 30 40 � 1 � 2 � 0.1 � 3

THE DOMAIN DECOMPOSITION FRAMEWORK

CHOICE OF THE INTERFACE VARIABLE There are two possible strategies to choose the interface variable: λ = u f · n on Γ; in that case we aim at satisfying − K ∇ p p · n = λ on Γ σ = p p on Γ; here, we aim at satisfying − n · T( u f , p f ) · n = σ on Γ Both choices are suitable from a mathematical standpoint since they yield well-posed subproblems in the fluid and the porous part.

INTERFACE EQUATION FOR DARCY–STOKES We can equivalently express the Darcy-Stokes problem in terms of the solution λ (normal velocity across Γ) of the interface problem (1) S s λ + S d λ = χ on Γ S s continuous and coercive fluid operator: solve S s : λ (normal velocities on Γ) Stokes ξ = − n · T( u f , p f ) · n (normal stresses on Γ). − − − → S d continuous, positive porous media operator: solve S d : λ (fluxes of p p on Γ) − Darcy ξ = p p | Γ . − − → S s is spectrally equivalent to S s + S d : there exist two positive constants k 1 and k 2 (independent of η ) such that ∀ η ∈ Λ 0 ⊂ H 1 / 2 k 1 � S s η, η � ≤ � ( S s + S d ) η, η � ≤ k 2 � S s η, η � 00 (Γ) . There exists a unique solution λ ∈ H 1 / 2 00 (Γ) for (1).