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Voller, University of Minnesota volle001@umn,edu Application of an Enthalpy Method for Dendritic Solidification Vaughan R. Voller and N. Murfield
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Voller, University of Minnesota volle001@umn,edu The problem—simulate the growth of a crystal into an undercooled melt contained in an insulated cavity How does solidification proceed? Why do we get a dendritic shape?
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Voller, University of Minnesota volle001@umn,edu seed T 0 < 0 H = cT + fL Initial liquid at a temperature below equilibrium solidification temperature T = 0 seeded with solid at solidification temperature H = 0 + fL, 0 < f < 1 Liquid layer adjacent to seed uses latent heat to heat up to T = 0 T = T 0 < 0 Negative gradient into liquid removes residual latent heat and drives solidification If a solute is present the equilibrium temp and gradient slope will be lower—resulting in a slower advance for the solidification How does solidification proceed?
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Voller, University of Minnesota volle001@umn,edu Why do we get a dendritic shape? Initial seed with radius anisotropic surface energy liquidus slope T o < T m Surface of seed is under-cooled due to curvature (Gibbs-Thompson) and solute (Not Shown kinetic) Capillary length With dimensionless numbers Angle between normal and x-axis 0.25 1 Anisotropic term makes under cooling less in preferred growth directions As crystal grows the sharper temp grad at tip drives sol harder BUT the increased tip curvature holds it back A steady tip operating Velocity is reached
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Voller, University of Minnesota volle001@umn,edu Alain Karma and Wouter-Jan Rappel Phase Filed Current Approaches (Pure Melt) Thermodynamic equation –minimizing free energy across a diffusive interface -1 < phase marker < 1 Heat equation with source H. S. Udaykumar, R. Mittal,y and Wei Shyy Interface Tracking Also see Juric Tryggvason and Zhao and Heinrich Kim, Gldenfeld, Dantzig And Chen, Merriman, Osher,andSmereka Level Set Solve for level set Using speed function from Stefan Cond. Maintain distance function properties by re-in Solve heat con. Use level set to mod FD at interface Enthalpy Method-Proposed by K H Tacke A function of f if 0 < f < 1 (f determines curvature) In this work: use iterative sol. Include anisotropy and solute
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Voller, University of Minnesota volle001@umn,edu Governing Equations With additional dimensional numbers Governing equations are Chemical potential Continuous at interface concentration If 0 < g < 1
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Voller, University of Minnesota volle001@umn,edu In a time step Solve for H and C (explicit time integration) Calculate curvature and orientation from current nodal g field Calculate interface undercooling If 0 < g < 1 then 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 Use square finite difference grid, set length scale to Very Simple—Calculations can be done on regular PC Initial condition— Circle r = 15do Typical grid Size 200x200 ¼ geometry ON A FIXED UNIFORM GRID
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Voller, University of Minnesota volle001@umn,edu 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
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Voller, University of Minnesota volle001@umn,edu Verification 2 Verify solution coupling by Comparing with one-d solidification of an under-cooled binary alloy Constant T i, C i k = 0.1, Mc = 0.1, T 0 = -.5, Le = 1.0 Compare with Analytical Similarity Solution—Rubinstein Carslaw and Jaeger combo. Concentration and Temperature at dimensionless time t =100 Symbol-numeric sol. Front Movement Red-line Numeric sol. Covers analytical
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Voller, University of Minnesota volle001@umn,edu Verification 3 Compare calculated dimensionless tip velocity with Steady state operating sate calculated from the microscopic solvability theory
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Voller, University of Minnesota volle001@umn,edu Verification 4 Check for grid anisotropy Solve with 4 fold symmetry twisted 45 o Then Twist solution back x =.36do x =.38do Dimensionless time = 6000
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Voller, University of Minnesota volle001@umn,edu k = 0.15, Mc = 0.1, T 0 = -.65 = 0.05, x = 3.333d 0 Result: Effect of Lewis Number All predictions at Dimensionless time =6000
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Voller, University of Minnesota volle001@umn,edu k = 0.15, Mc = 0.1, T 0 = -.55, Le = 20.0 = 0.02, x = 2.5d 0 Dimensionless time = 30,000 Result: Prediction of Concentration
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Voller, University of Minnesota volle001@umn,edu Conclusion –Score card for Dendritic Growth Enthalpy Method (extension of original work byTacke) Ease of CodingExcellent CPUVery Good (runs shown here took between 30 minutes and 2 hours on a regular PC) Convergence to known analytical sol. Excellent Convergence to known operating state Good (further study with finer grid required) Grid AnisotropyReasonable (further study with finer grid required) Flexibility to add more physics Excellent (adding solute only required 10% more lines)
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