Coupling free flow / porous-medium flow General idea free flow, Navier-Stokes wind 1 phase, 2 components, temperature sharp interface evaporation porous-medium / Darcy flow 2 phases, 2 component, temperature 2/14
Coupling free flow / porous-medium flow Applications CC BY-SA 3.0 Savant-fou Freeze-drying CC BY-NC-SA 2.0 Jesse Varner Agricultural Research Service, k4500-12 Potash evaporation ponds Salinization in California 3/14
Stokes / Darcy coupling Coupling of Stokes and Darcy [Mosthaf et al. 2011, Baber et al. 2012] 2 phases, 2 components, temp. / 1 phase, 2 components, temp. box scheme, implicit Euler assemble in single matrix, Newton, SuperLU oscillations with Navier-Stokes In collaboration with Dani Or, ETH Zurich 4/14
Navier-Stokes / Darcy coupling Schemes staggered grid conservative coupling cell-centered finite volume 5/14
Navier-Stokes / Darcy coupling Coupling Continuity of normal stresses p pm = p ff + n Γ ( ̺vv ⊺ − ̺ν ∇ v ) ff n Γ Continuity of normal mass fluxes v pm n Γ = v ff n Γ Beaver-Joseph-Saffman condition ff � t ⊺ Γ ,i Kt Γ ,i 0 = α BJ v + τn Γ · t Γ ,i ν̺ � ff � � = α BJ v + t ⊺ Γ ,i Kt Γ ,i ∇ vn Γ · t Γ ,i 6/14
Navier-Stokes / Darcy coupling Test case 1 v = 20 x 1 (1 − x 1 ) p = 0 No-flow Test case from [Kanschat, Rivi` ere 2010] Pressure Velocity 8/14
Current status Compare with analytic solution [Chidyagwai, Rivi` ere 2011] Physical laws for 2 components and temperature in Navier-Stokes Coupling for components and temperature SuperLU limits problem size Algebraic and k - ε turbulence models 9/14
Possible speed-up by adaptivity Different mesh sizes Coarser mesh for Darcy sufficient Couple with a mortar method Exists for Stokes / Darcy code [Baber unpublished] 10/14
Possible speed-up by adaptivity Multi time stepping Calculate free flow with smaller time steps ∆ t Calculate complete system with with ∆ T Unclear how conserve mass / energy Stokes / Darcy with 1 phase, linear [Rybak, Magiera 2014] ∆ t ∆ T 11/14
Possible speed-up by adaptivity Decouple linear systems Decouple linear systems using Schur complement [Discacciati, Quarteroni 2004] Ax + By = f Cx + Dy = g D − CA − 1 B y = g − CA − 1 f � � Ax = f − By Only invert systems A and D − CA − 1 B Coupling: free flow and porous-medium flow Navier-Stokes: mass and momentum balance equations Multiple equations: mass balance equations and temperature / components 12/14
Validation and Verification Wind tunnel experiments in collaboration with Kate Smits, Colorado School of Mines Direct numerical simulation courtesy of Wang et. al., iRMB TU Braunschweig 13/14
Outlook Unstructured grids, especially non-trivial topology for coupling interface [Ansanay-Alex et al. 2011, Verboven et al. 2006] Model adaptivity Reconsider Beavers-Joseph condition Really decouple domains, iterative coupling [Discacciati 2004] Use explicit method in Navier-Stokes Couple with different software tools like OpenFOAM Scale effects up 14/14
Backup Mass and momentum balance equations Navier-Stokes mass balance equation ∂ ∂t̺ + div( ̺v ) − q p = 0 Navier-Stokes momentum balance equation ∂ ∂t ( ̺v ) + div ( ̺vv ⊺ ) − div( ̺ν ∇ v ) + ∇ p − ̺g − q v = 0 Darcy flow equation � K � Φ ∂ ∂t ( ̺S ) − div ν ( ∇ p − ̺g ) − q pm = 0 15/14
Backup additional free flow equations energy balance equation ∂ ∂t ( ̺u ) + div ( ̺vh − λ ∇ T ) = q T transport equation ∂ ∂t ( ̺X ) + div ( ̺vX − D steam ̺ ∇ X ) = q steam 16/14
Backup additional Darcy flow equations energy balance equation Φ ∂ α S α ) + (1 − Φ) ∂ � ∂t ( ̺ α u κ ∂t ( ̺ pm c pm T ) α ∈{ g , l } � = q T + div ̺ α h α v α − λ pm ∇ T α ∈{ g , l } transport equation Φ ∂ � � ∂t ( ̺ α X κ � ̺ α v α X κ α − D κ α, pm ̺ α ∇ X κ � α S α ) + div α α ∈{ g , l } α ∈{ g , l } � q κ = α α ∈{ g , l } 17/14
Backup coupling conditions for thermal equilibrium temperature continuity, heat flux continuity ( T ) ff = ( T ) pm � � n · ( ̺vh − λ ∇ T ) ff = − n · ̺ g v g h g + ̺ l v l h l − λ pm ∇ T pm 18/14
Backup coupling conditions for chemical equilibrium continuity of mass fractions, continuity of the component fluxes ( X κ ) ff = � X κ � (1) g pm n · ( ̺vX κ − D̺ ∇ X κ ) ff = − n · � ( ̺ α v α X κ α − D α, pm ̺ α ∇ X κ α ) α ∈{ g , l } pm (2) 19/14
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