Growing Numerical Crystals Vaughan Voller, SAFL Feb 1 2006

1. Growing Numerical Crystals Vaughan Voller, SAFL Feb 1 2006

2. Germs snow-flakes-ice crystals Dendrite grains in metal systems science.nasa.gov (IACS), EPFL Growing Numerical Crystals Objective: Simulate the growth (solidification) of crystals from a solid seed placed in an under-cooled liquid melt Some Physical Examples

3. Simulation can be achieved using modest models and computer power Solved in ¼ Domain with A 200x200 grid Growth of solid seed in a liquid melt Initial dimensionless undercooling T = -0.8 Resulting crystal has an 8 fold symmetry

16. By changing conditions can generate any number of realistic shapes in modest times PC CPU ~5mins Complete garbage BUT—WHY do we get these shapes—WHAT is Physical Bases for Model HOW does the Numerical solution work, IS the solution “correct”

17. A Physical Basis What do I mean when I say that the bulk liquid is undercooled

18. Bulk Undercooling T As volume temp Drops below Tm we would expect it to change to solid Tm Time t We think of a solid changing to a liquid at a single equilibrium temperature Tm (e.g., in ice-water Tm = 273.2 K) Consider the liquid volume following the cooling curve below But due to the thermodynamics of the phase change it is more Than likely that the liquid will become undercooled i.e. remain in the liquid state until a temperture Tu below Tm is reached

19. Bulk Undercooling T As volume temp Drops below Tm we would expect it to change to solid Tm Tu Time t We think of a solid changing to a liquid at a single equilibrium temperature Tm (e.g., in ice-water Tm = 273.2 K) Consider the liquid volume following the cooling curve below But due to the thermodynamics of the phase change it is more Than likely that the liquid will become undercooled i.e. remain in the liquid state until a temperture Tu below Tm is reached

20. Entropy-measure of chaos-larger in liquid (less atomic structure) T Enthalpy (total heat) larger in liquid due to latent heat Tm Tu The undercooling behavior can be quantified By considering the thermodynamic state of a system given by the the Gibbs free energy (free enthalpy) G

21. GS Molar Free Energy GL T Tm Liquid here is meta-stable –given “encouragement” will transform to solid state. LIQUID SOLID A plot of the molar Gibbs free energies of the liquid and solid states looks like A substance will always tend to find a state that minimizes G

22. GS Molar Free Energy The process can be facilitated by the introduction of a small solid particle (nuclide) So called —heterogeneous nucleation— resulting in the growth of a solid with a crystalline morphology GL Liquid at undercooled temperature is meta-stable –-given “encouragement” will transform to solid state. T Tu Tm Note columnar crystals form when we have another type of undercooling- Constitutional under-cooling

23. For a given undercooling Not just any particle will be able to nucleate the solidification In order to work (grow solid) the solid seed particle needs to be able to establish an equilibrium with the under-cooled bulk liquid. To see how this works we need to understand how a solid-liquid interface can be undercooled below the equilibrium melting temperature

24. Interface undercooling A solid changes to a liquid at a single equilibrium temperature Tm (in ice Tm = 273.2 K) Conditions can exist, however, where the temperature at the solid-liquid interface, Ti, is below the equilibrium melting temperature solute Curvature-surface energy Gibbs -Thomson Kinetic

25. Solute Undercooling The distribution Of impurities (solutes) Will change the melting temperature Tm liquidus C –solute concentration

26. Interface undercooling A solid changes to a liquid at a single equilibrium temperature Tm (in ice Tm = 273.2 K) Conditions can exist, however, where the temperature at the solid-liquid interface, Ti, is below the equilibrium melting temperature Solute-presence of impurities lower melt Temp. Curvature-surface energy Gibbs -Thomson Kinetic

27. Kinetic undercooling Attachment of atoms to the solid Could be “sluggish”, if front advance is rapid Position of front will lag Tm isotherm

28. Interface undercooling A solid changes to a liquid at a single equilibrium temperature Tm (in ice Tm = 273.2 K) Conditions can exist, however, where the temperature at the solid-liquid interface, Ti, is below the equilibrium melting temperature Solute-presence of impurities lower melt Temp. Curvature-surface energy Gibbs -Thomson Kinetic sluggish attachment of atoms

29. Then consider spherical particle in liquid The particle will have an additional free energy due surface energy induced pressure difference between liquid and solid Equating g Surface energy J/m2 Gibbs-Thomson-Surface Energy Undercooling Consider Bulk free energies in liquid and solid Molar Free Energy GS GL T Tu Tm

30. Gibbs-Thomson-Surface Energy Undercooling 1 2

31. Interface undercooling A solid changes to a liquid at a single equilibrium temperature Tm (in ice Tm = 273.2 K) Conditions can exist, however, where the temperature at the solid-liquid interface, Ti, is below the equilibrium melting temperature Solute-presence of impurities lower melt Temp. Curvature-surface energy Gibbs -Thomson Kinetic sluggish attachment of atoms

32. BUT why does solid grow AND how is Morphology Controlled Explained Undercooling 1 2

33. Early Stage Liquid around seed is encouraged to change state Increase in temperature in surrounding region due to latent heat release Later Stage Temperature gradient removes Latent heat and heats bulk T How does solid grow Initial State Small seed At bulk undercoling Solid Liquid

34. What Controls Morphology since Four fold symmetry 0.25 1 0.25 TB TB Due to the “pinching” In preferred dir. Growth Looks unstable Will drive sold. Faster In preferred direction Preferred growth direction in crystal structure Manifest in anisotropic surface tension, e.g., Steady tip vel; “Operating point” TB But increase in k will reduce T90—can reach a Balance between Temp. grad. and k

35. On interface n vn Capillary length ~10-9 for metal Angle between normal and x-axis Can describe process with the Sharp Interface Model With dimensionless numbers Assumed constant properties Insulated domain Initiated with small solid seed

36. Interface Tracking-reconstruction Level Set H. S. Udaykumar, R. Mittal, Wei Shyy Juric Tryggvason Zhao and Heinrich Kim, Goldenfeld, Dantzig And Chen, Merriman, Osher,andSmereka Solutions Based on Sharp Interface Model

37. Advantage can be calculated ON A FIXED UNIFORM GRID Diffusive interface model 1: Phase Field Liquid fraction Smear out interface n Applies throughout domain obtain an evolution equation for that satisfies curvature and kinetic undercooling Can be done by Minimize a free energy functional OR Direct geometric modeling (Beckermann)

38. Diffusive interface model 2: Enthalpy (extension of Tacke) Liquid fraction Smear out interface f=1 f=0 n , where enthalpy Also Applies throughout domain If we only consider curvature undercooling with undercooling curvature and orientation A non-linear system for evolution of f

39. Numerical Solution Seed calc With small Solid part. Very Simple—Calculations can be done on regular PC In a time step Solve for H (explicit time integration on FIXED grid) If 0 < f < 1 then Calculate curvature and orientation from current nodal f field Calculate interface undercooling Update f from enthalpy as Check that calculated liquid fraction is in [0,1] Update Iterate until At end of time step—in cells that have just become all solid introduce very small solid seed in ALL neighboring cells. Required to advance the solidification Typical grid Size 200x00 ¼ geometry

40. Produces nice answers BUT are they correct

41. Level Set Kim, Goldenfeld and Dantzig k = 0 (pure), e = 0.05, T0 = -0.55, Dx = d0 Dimensionless time t = 37,600 Looks Right!! Verification 1 Enthalpy Calculation k = 0 (pure), e = 0.05, T0 = -0.65, Dx = 3.333d0 Dimensionless time t = 0 (1000) 6000

42. Verification 2 Verify solution coupling by Comparing with one-d solidification of an under-cooled melt T0 = -.5 Compare with Analytical Similarity Solution Carslaw and Jaeger Front Movement Temperature at dimensionless time t =250

43. Verification 3 Compare calculated dimensionless tip velocity with Steady state operating state calculated from the microscopic solvability theory

44. Verification 4 Check for grid anisotropy Solve with 4 fold symmetry twisted 45o Then Twist solution back Dimensionless time t = 6000

45. Verification 5: Curvature Calculation Operates over narrow band

46. Conclusion –Score card for Dendritic Growth Enthalpy Method (extension of original work by Tacke)

47. Alloy— Solve for concentration Interface Tracking-reconstruction ~10km Fotis H. S. Udaykumar, R. Mittal, Wei Shyy Juric Tryggvason Zhao and Heinrich Does any of this this have any relevance to shorelines Matt Extensions