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

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

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

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

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

18. Bulk Undercooling We think of a solid changing to a liquid at a single equilibrium temperature T m (e.g., in ice-water T m = 273.2 K) Consider the liquid volume following the cooling curve below Time t T TmTm As volume temp Drops below T m we would expect it to change to solid 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 T u below T m is reached

19. Bulk Undercooling We think of a solid changing to a liquid at a single equilibrium temperature T m (e.g., in ice-water T m = 273.2 K) As volume temp Drops below T m we would expect it to change to solid Time t T TmTm TuTu 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 T u below T m is reached Consider the liquid volume following the cooling curve below

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

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

22. GLGL GSGS T TmTm Liquid at undercooled temperature is meta-stable –-given “encouragement” will transform to solid state. TuTu 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 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 can be undercooled below the equilibrium melting temperature 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

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

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

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

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

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

29. Molar Free Energy GLGL GSGS T TmTm TuTu Gibbs-Thomson-Surface Energy Undercooling Consider Bulk free energies in liquid and solid 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  Surface energy J/m 2

30. Gibbs-Thomson-Surface Energy Undercooling 1 2

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

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

33. How does solid grow T Initial State Small seed At bulk undercoling 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 SolidLiquid

34. What Controls Morphology Preferred growth direction in crystal structure Manifest in anisotropic surface tension, e.g., Four fold symmetry 0.25 1 since TBTB Will drive sold. Faster In preferred direction TBTB Due to the “pinching” In preferred dir. Growth Looks unstable TBTB But increase in  will reduce T 90 —can reach a Balance between Temp. grad. and  Steady tip vel; “Operating point”

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

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

37. 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) Diffusive interface model 1: Phase Field Smear out interface n Liquid fraction Advantage can be calculated ON A FIXED UNIFORM GRID

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

39. 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 Numerical Solution Very Simple—Calculations can be done on regular PC Typical grid Size 200x00 ¼ geometry Seed calc With small Solid part.

40. Produces nice answers BUT are they correct

41. Verification 1 Looks Right!! k = 0 (pure),  = 0.05, T 0 = -0.65,  x = 3.333d 0 Enthalpy Calculation Dimensionless time  = 0 (1000) 6000 k = 0 (pure),  = 0.05, T 0 = -0.55,  x = d 0 Level Set Kim, Goldenfeld and Dantzig Dimensionless time  = 37,600

42. Verification 2 Verify solution coupling by Comparing with one-d solidification of an under-cooled melt T 0 = -.5 Compare with Analytical Similarity Solution Carslaw and Jaeger Temperature at dimensionless time t =250 Front Movement

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 45 o Then Twist solution back Dimensionless time  = 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) Ease of CodingExcellent CPUVery Good (runs shown here took between 5 and 20 minutes on a regular PC) Convergence to known analytical sol. Excellent Convergence to known operating state Good (further study with finer grid indicates problem Grid AnisotropyReasonable (further study with finer grid required) CurvatureOK-But I have doubts --grid dependence ?

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