Application of the Enthalpy Method: From Crystal Growth to Sedimentary Basins

Grain Growth in Metal SolidificationFrom W.J. Boettinger

m

10km

“growth” of sediment delta into oceanGanges-Brahmaputra Delta

Vaughan VollerUniversity of Minnesota

volle001@umn.edu

The problem—simulate the growth of a crystal into an undercooled melt contained in an insulated cavity

How does solidificationproceed?

Why do we get a dendritic shape?

Solid-liquid interfacewith time

The Classic Stefan Problem-with curvature dependentPhase change temperature

seed

T0 < 0H = cT + fL

Initial liquid at a temperature below equilibrium solidification temperature T = 0seeded 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 = T0 < 0

Negative gradient into liquid removesresidual latent heat and drives solidification

If a solute is present the equilibrium tempand gradient slope will be lower—resultingin a slower advance for the solidification

How does solidification proceed?

Why do we get a dendritic shape?

Initial seed with radius

0*

om

m* T)CC(m

L

T)(TT

anisotropicsurface energy

liquidus slope

To < Tm

Surface of seed is under-cooled dueto curvature (Gibbs-Thompson) andsolute (Not Shown kinetic)

)C1(MC)4cos151(dT 0o Capillary length~10-9 for metal

With dimensionless numbers L

cmM,

C

CC,

c/L

mCTTT

0

*0f

*

Angle between normal and x-axis

)4cos151(

0.25

0.25 1

Anisotropic term makes under cooling less in preferred growth directions

As crystal growsthe sharper tempgrad at tip drives solharder BUT the increased tip curvature holds it backA steady tip operatingVelocity is reached

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, Wei Shyy

Interface Tracking

Juric Tryggvason

Zhao and HeinrichKim, Goldenfeld, DantzigAnd Chen, Merriman, Osher,andSmereka

Level Set

Solve for level setUsing speed functionfrom Stefan Cond.

Maintaindistance functionproperties by re-in

Solve heat con.Use level setto mod FD at interface

Enthalpy Method-First Proposed by Eyres et al 1947

For Crystal Growth by K H Tacke, 1988

)TK(t

H

fTH

A function of f if 0 < f < 1(f determines curvature)

In this work: use iterative sol. Include anisotropy and solute

0.60.70.3

0.0

1.0

liquid fraction0 in solid1 in liquid-a physical level set

Governing EquationsWith additional dimensional numbers

l

le2

*l

*

DL,

tt,

xx

Governing equations are

)TK(t

H

)VD(t

C

gTH

sl C)f1(fCC

k)k1(f

CV

l

s)f1(fK

l

s

D

D)f1(kf

Le

1D

)V1(MC

)4cos151(dT

0

o

Chemical potential

Continuous at interface

il

is kCC

concentration

If 0 < g < 1

2/32y

2x

yy2xxyyxxy

2y

)gg(

ggggg2gg

)

g

g(tan

y

x1

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 thenUpdate 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 solidintroduce very small solid seed in ALL neighboring cells.Required to advance the solidification

Numerical Solution

xUse square finite difference grid, set length scale to

)V1(MC)4cos151(dT 0oi

k)k1(f

CV

)fTH(ff i

fHT

tol)fTH( i

Very Simple—Calculations can be done on regular PC

Initial condition—Circle r = 2.5do

Typical gridSize 200x00¼ geometry

ON A FIXEDUNIFORM GRID

Verification 1 Looks Right!!

k = 0 (pure), = 0.05, T0 = -0.65, x = 3.333d0

Enthalpy Calculation

Dimensionless time = 0 (1000) 60002

odtk

k = 0 (pure), = 0.05, T0 = -0.55, x = d0

Level Set Kim, Goldenfeld and Dantzig

Dimensionless time = 37,600

Verification 2

Verify solution coupling by Comparing with one-d solidification of an under-cooled binary alloy

Constant Ti, Ci

k = 0.1, Mc = 0.1, T0 = -.5, Le = 1.0

Compare with Analytical Similarity Solution—Rubinstein Carslaw and Jaeger combo.

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

0 10 20 30 40x

C

T/T0k = 0.1

MC0 = 0.1

T0 = -0.5

Le = 1.0

Concentration and Temperatureat dimensionless time t =100

Symbol-numeric sol.

0

2

4

6

8

10

12

0 100 200 300

time

psoi

tion

Front Movement

Red-lineNumeric sol.Covers analytical

Verification 3

Compare calculated dimensionless tip velocity withSteady state operating state calculated from the microscopic solvability theory

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

0 1000 2000 3000 4000 5000 6000

time

velocity

5do

3.333do

2.5do

x

dvv ogridtip

Verification 4

Check for grid anisotropy

)V1(MC))4cos(151(dT 0oi

Solve with

4 fold symmetry twisted 45o

Then Twist solution back

x = .36do

x = .38do

Dimensionless time = 6000

0d5.2x

0Le

10Le

1Le

2.Le

k = 0.15, Mc = 0.1, T0 = -.65

= 0.05, x = 3.333d0

Result: Effect of Lewis Number

l

le DL

All predictions atDimensionless time =6000

0

0.5

1

1.5

2

2.5

3

3.5

4

0 50 100 150 200 250 300 350 400 450 500

distance

Concentration

k = 0.15, Mc = 0.1, T0 = -.55, Le = 20.0

= 0.02, x = 2.5d0

Dimensionless time = 30,000

Result: Prediction of Concentration

2

odtk

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

0 1000 2000 3000 4000 5000 6000

time

velocity

5do

3.333do

2.5do

Conclusion –Score card for Dendritic Growth Enthalpy Method (extension of original work byTacke)

Ease of Coding Excellent

CPU Very 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 Anisotropy Reasonable (further study with finer grid required)

Flexibility to add more physics

Excellent (adding solute only required 10% more lines)

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

0 10 20 30 40x

C

T/T0k = 0.1

MC0 = 0.1

T0 = -0.5

Le = 1.0

Fans Toes Shoreline

Two Sedimentary Moving Boundary Problems of Interest

Moving Boundaries in Sediment Transport

NCED’s purpose:to catalyze development of an integrated, predictive science of the processes shaping the surface of the Earth, in order to transform management of ecosystems, resources, and land use

The surface is the environment!

Research fields• Geomorphology• Hydrology• Sedimentary geology• Ecology• Civil engineering• Environmental economics• Biogeochemistry

Who we are: 19 Principal Investigators at 9 institutions across

the U.S.

Lead institution: University of Minnesota

1km

Examples of Sediment FansMoving Boundary

How does sediment-basement interfaceevolve

Badwater Deathvalley

Sediment mass balance gives

Sediment transported and deposited over fan surfaceby fluvial processes

xxt

From a momentum balance anddrag law it can be shown thatthe diffusion coefficient is a function of a drag coefficientand the bed shear stress

when flow is channelized = cont.

Convex shape

sediment

h(x,t)

x = u(t)

0q

bed-rock

ocean

x

shoreline

x = s(t)

land surface

A Sedimentary Ocean Basin

An Ocean Basin

Melting vs. Shoreline movement

pressurizedwater reservoir

to water supply

solenoidvalve

stainless steelcone

to gravel recycling

transport surface

gravel basement

rubber membrane

experimental deposit

Experimental validation of shoreline boundary model

~3m

Base level

Measured and Numerical results ( calculated from 1st principles)

1-D finite difference deforming grid vs. experiment

xxt+Shoreline balance

An Ocean Basin

Melting vs. Shoreline movement

Limit Conditions: A Fixed Slope Ocean

q=1

s(t)

similarity solution

q

2

)(erfe

)(erf1,t2s

22/1

21

21

2

21

21

0

5

10

15

20

25

0 100 200 300

Time

sh

ore

line

0if),x(LH

2

2

xt

H

Enthalpy Sol.

A Melting Problem driven by a fixed flux with SPACE DEPENDENT

Latent Heat L = s

dt

dss

x)t(sx0,

xt s2

2

s Depth at toe

)TK(t

H

L)(TH

h(x,y,t)

q

bed-rock

ocean

y

shoreline

x = s(t)

land surface

(x,y,t)

A 2-D Front -Limit of Cliff face Shorefront But Account of Subsidence and relative ocean level

0hif),t,y,x(LhH

)h(t

H

Enthalpy Sol.

xy

]L/H,[MINfrac 1

Solve on fixed gridin plan view

Track Boundary by calculating in each cell

0

5

10

15

20

25

30

0 100 200 300 400 500

time

shor

elie

n po

sitio

n

s(t)

0

5

10

15

20

25

30

0 500 1000 1500 2000

time

shor

elin

e po

sitio

n

numerical

steady state

s(t)

Hinged subsidence

q

2

)(erfe

)(erf1,t2s

22/1

21

21

2

21

21

A 2-D problem Sediment input into an oceanwith an evolving trench driven By hinged subsidence

With Trench

No Trench Trench

Plan view movement of fronts

Stratigraphy and Shoreline

-30-20-10

0102030405060

0 10 20 30

Models can predict stratigraphy“sand pockets” = OIL

The Po

Shoreline position is signature of channels

WHY Build a model

Shoreline Tracking Model has been

)t,y,x(LhH

)h(tH

Latent Heat a function of Space and time

Enthalpy methods can generate models for distinct moving boundary systems

)TK(tH

L)(TH

Crystal Growth in undercooled pure melt

Phase change temperature a function of interface