Solidification / Melting Moving Boundary Problems: A finite difference method

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Solidification / Melting Moving Boundary Problems: A finite difference method. Final Project MANE 6640 – Fall 2009 Wilson Braz. Background. Solidification has obvious application to engineering problems such as: Casting, metallurgy, soil mechanics, freezing of food, etc. - PowerPoint PPT Presentation

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Solidification / MeltingMoving Boundary Problems:A finite difference method

Final ProjectMANE 6640 – Fall 2009 Wilson Braz

Background Solidification has obvious application to

engineering problems such as: Casting, metallurgy, soil mechanics, freezing of

food, etc. Solidification may be modeled using a

moving boundary. Several techniques for solving the moving

boundary problem. Isotherm Migration Method (IMM), method of

lines, finite element, finite difference, enthalpy method, and others.

The problem: Use 2-D Finite Difference method to

analyze the solidification of square plate

CL

CL

2

2

Tw < Freezing

Tw < Freezing

Symmetry B.C. Tinitial > Freezing0

y

H

x

H

y

u

x

u

The Problem - Continued.Comparison of Results

Results from 2 different sources are in disagreement

Method was coded in MATLAB Results were compared with those

given in sources

Approach - General Enthalpy method

Use an explicit, finite difference routine to numerically solve Develop numeric approximation equations,

discretize domain, set initial conditions, set boundary conditions, march through domain, step through time.

y

u

x

uk

yxt

H tyxu ,,Find: Such that:

rp uuCH

Approach Technique

Material properties vary depending on state (liquid or solid) Conditional statements test for material

state using temperature. Apply appropriate values for material

properties depending on state. Calculate ‘H(x,y)’ using finite differencing Find ‘u(x,y)’ given using new ‘H(x,y)’

Non-Real material properties, initial and boundary conditions: To simplify calculation, and to compare directly with

published results, the following material properties were used:

Mesh size varied Time increment set to satisfy CFL condition

0001.1

561.1

0001.1

9999.0

0.1__

init

solfusion

melt

freeze

liqpsolpsolliqsol

T

H

T

T

CCkk

22

22

max 4

1

yx

yx

k

Ct p

Determining solid/liquid interface

Tem

p

Y-coord @ x=0

Solid Liquid Interface Temp < Freezing Temp

Results T(x,y,t) #grid pts. = 11x11, time = 0.0001

Results table: 11x11 mesh

x distance

time 0 0.1 0.2 0.3 0.4 0.5 0.60.05 0.7536 0.7536 0.7536 0.7536 0.7536 0.7535 0.75340.10 0.6590 0.6590 0.6590 0.6589 0.6586 0.6577 0.65010.15 0.5963 0.5963 0.5963 0.5949 0.5911 0.5827 0.50040.20 0.5001 0.5001 0.5001 0.5001 0.5001 0.4999 0.39990.25 0.4649 0.4649 0.4649 0.4600 0.4001 0.3999 0.30 0.4000 0.4000 0.4000 0.3999 0.3669 0.2358 0.35 0.3525 0.3525 0.3525 0.3001 0.2801   0.40 0.2999 0.2999 0.2999 0.2611 0.1203   0.45 0.2000 0.2000 0.2000 0.1776     0.50 0.1558 0.1558 0.1558       

x distance

time 0 0.1 0.2 0.3 0.4 0.5 0.60.05 0.8125 0.8106 0.8048 0.7940 0.7764 0.7476 0.69040.10 0.6979 0.6965 0.6921 0.6836 0.6683 0.6392 0.56060.15 0.6157 0.6141 0.6095 0.6000 0.5810 0.5201 0.20 0.5473 0.5453 0.5394 0.5268 0.4789   0.25 0.4865 0.4838 0.4755 0.4567 0.3894   0.30 0.4302 0.4263 0.4146 0.3654     0.35 0.3766 0.3708 0.3534 0.2859     0.40 0.3337 0.3158 0.2623       0.45 0.2816 0.2585 0.1893       0.50 0.2376 0.2056 0.1097       

Values of the y-coordinate on the solid-liquid interface for fixed values of x at various times

Values using method coded in MATLAB Values taken from John Crank

time 0 0.1 0.2 0.3 0.4 0.5 0.60.05 -7% -7% -6% -5% -3% 1% 9%0.10 -6% -5% -5% -4% -1% 3% 16%0.15 -3% -3% -2% -1% 2% 12% 0.20 -9% -8% -7% -5% 4%   0.25 -4% -4% -2% 1% 3%   0.30 -7% -6% -4% 9%     0.35 -6% -5% 0% 5%     0.40 -10% -5% 14%       0.45 -29% -23% 6%       0.50 -34% -24% 42%       

Comparison

NOTE: Values of x-coordinate shown in left table were found by liniearly interpolating location where T = 1.0000. Method used on right table is unknown

Results – Non real solidPlot of Enthalpy: Red Solid – Blue Liquid#grid pts. = 11x11, time = 0.0001

Results Table showing solid-liquid interface #grid pts. = 41x41, dt = 0.00005

x-coord.

time (sec) 0 0.1 0.2 0.3 0.4 0.5 0.6

0.0500 0.7664 0.7664 0.7664 0.7664 0.7664 0.7662 0.7653

0.1000 0.6719 0.6719 0.6718 0.6713 0.6699 0.6662 0.6500

0.1500 0.5991 0.5989 0.5980 0.5955 0.5905 0.5750 0.4819

0.2000 0.5251 0.5251 0.5251 0.5250 0.5132 0.4625 0.0000

0.2500 0.4749 0.4749 0.4749 0.4500 0.4250 0.1744 0.0000

0.3000 0.4244 0.4227 0.4141 0.3929 0.3176 0.0000 0.0000

0.3500 0.3694 0.3675 0.3500 0.3140 0.0000 0.0000 0.0000

0.4000 0.3157 0.3126 0.2902 0.2121 0.0000 0.0000 0.0000

0.4500 0.2500 0.2500 0.2179 0.0000 0.0000 0.0000 0.0000

0.5000 0.2000 0.1921 0.1219 0.0000 0.0000 0.0000 0.0000

0.5500 0.1250 0.1177 0.0000 0.0000 0.0000 0.0000 0.0000

0.6000 0.0417 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

ResultsPlot of Temperature: #grid pts. = 41x41, dt = 0.00005

ResultsPlot of Enthalpy: Red Solid – Blue Liquid#grid pts. = 41x41, dt = 0.00005

Results – Non real solidPlot of Enthalpy: Red Solid – Blue Liquid#grid pts. = 61x61, dt = 0.000025

Difficulties & Limitations with this approach

Trouble matching results presented by John Crank, and Ernesto Gutierrez-Miravete

Suspect an issue with initial calculation of H(x,y,0), or u(x,y,0) 1st time step shows temperature jump up to ~2