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The Stefan Problemand theExact Solution for the Two Phase

Stefan Problem

DT Project

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First Things First

We desire to formulate Two-Phase StefanModel of Melting and Freezing

Non linear model

Moving Boundary Problem

Contains an unknown which is the region to besolved..

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Why Stefan Problem???

The formulation of the

Stefan Problem is afoundation on which

more complex modelscan be built.

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First (contd)

Note: The phase changing process is governed

by the conservation of energy.

The unknowns are the temperature fieldand the location of the Interface.

Involves a phase change material(PCM)

with constant density( ), latent heat( ),melt temperature( ), Specific heats( ),and Thermal conductivities( ).

L

mT LS cc ,

LS kk ,

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Physical Assumptions

Conduction only

Constant latent heat (L)

Fixed melting temperature( ) which is

according to the phase change material(PCM)

Interface thickness is 0 and it is a sharp front;

it separates the phases

mT

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Assumptions (contd)

Thermophysical properties aredifferent for each phase

Conductivities ( )

Specific heats ( )

Density remains constant

L Sc c

L Sk k

L S

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Other Assumptions

Nucleation and supercooling are assumedto be not present

Surface tension and curvature isinsignificant

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Only Conduction

Conduction of Heat

Temperature

Heat(enthalphy)

Heat Flux

Characterizes phases

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Heat Equation

-Heat conduction equation (one space

dimension):

-well-posed

-Heat equation:

where

( )t x x

cT kT

t xxT T

kc

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The Two-Phase Stefan Problem

A slab, , initially solid attemperature , is melted byimposing a hot temperature at the

face and keeping the back face,insulated (all parameters constant).

lx 0

minit TT

mL TT

0x lx

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Lets Find a Solution

The solution of the

Stefan Problem is T(x,t)and X(t)!!!!

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Mathematical Model

PDE for

Interface(t>0)

Initial Condition

Boundary Condition

t L xx

t S xx

T T

T T

0 ( ), 0

( ) , 0

X t t

X t x t

( ( ), )

'( ) ( ( ) , ) ( ( ) , )

m

L x S x

T X t t T

LX t k T X t t k T X t t

lxTTxT

X

minit

0,)0,(

0)0(

0),(

0,),0(

tlTk

tTTtT

xS

mL

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Be Exact!

In order to explicitly solve the Two-PhaseProblem we need to assume the slab is semi-infinite.

Physical Problem: We want to melt a semi-infinite slab, ,

initially solid at a temperature , by imposingtemperature , on the face .

The alphas are different for each phase. All

parameters constant.

0 x

S mT T

L mT T 0x

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Two-Phase Neumann Solution

We derive the Neumann Solution

We use the similarity variable, ,and seek thesolution for both for the liquid and

for the solid.

Seek a solution for X(t) in the form:

t

x

( , ) ( )LT x t F

( , ) ( )ST x t F

( ) 2 LX t t

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Temperature

Temperature in the liquid region at t>0:

Temperature in the solid region at t>0:

0 ( )

( )

2( , ) ( ) LL L m

x X t

xerf

tT x t T T T erf

( )

( )2

( , ) ( )( / )

S

S m S

L S

x X t

xerfc

tT x t T T T

erfc

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Neumann similarity solution of the 2-phaseStefan Problem for the interface

( ) 2L

X t t

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Transcendental Equation

There is a different Transcendental Equations thatincorporates the TWO Stefan numbers (one foreach phase).

Trans. Equation:

Stefan Numbers and parameter v:( )

L L m

L

c T TSt

L

( )S m SS

c T TSt

L

L

S

v

)(exp()()exp(

222verfcvv

St

erf

St SL

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Newton!!!

Newton's method is an algorithm that findsthe root of a given function.

-Where F(x) = 0.

The fastest way to approximate a root.

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What now???

We can use the Newton algorithm tosolve the transcendental equation forthe Stefan problem

We do this in order to find a uniqueroot lambda and therefore a uniquesimilarity solution for each 0, 0, 0L SSt St v

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Newton Algorithm

1. Guess x0

2. Take a Newton Step

where

3. Terminate if

xxxnn

1

)(

)(

xf

xfx

10*

)(

TOLx

TOLxf

-(Want TOL to be very small soconvergence should be noticed)

-Stuck. Better guess

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Approximate the Root for Newton

For the 2-phase problem there is agood approximation which can be x0 inthe Newton program.

Approximation of Lambda:

2

1 22

S S

L

St St Stv v

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Glaubers Salt Input Values

-Tm = 32;

-CpL = 3.31;

-CpS = 1.76;

-kL = .59e-3;-kS = 2.16e-3;

-rho = 1460;

-Lat = 251.21;

-Tinit = 25;

-Tbdy = 90;

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Water Input Values

- Tm = 0;

- CpL = 4.1868;

- CpS = .5;- kL = .5664e-3;

- kS = 2.16e-3;

- rho = 1;- Lat = 333.4;

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Glaubers salt Example

maximum number of iterations to be performed,20tolerance for the residual,1.0e-7Iterations is 1xn = 5.170729e-001fx = -2.740474e-001

Iterations is 2xn = 5.207715e-001fx = 1.747621e-002

Iterations is 3xn = 5.207862e-001fx = 6.881053e-005

Done. Root is x=5.207862e-001, with Fx=1.068988e-009, Iterations is n=4

LAMBDA=

5.207861719133929e-001

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Plot Lambda vs StefanFor small Stefan numbers from 0:5 and M=51

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Portion of Exact Solution Code function lambda = neumann2p(CpL, CpS, kL, kS, rho, Tm, Lat, Tbdy,Tinit)

format long e %----------------------Input values--------------------------------------- CpL = input('Enter specific heat: '); CpS = input('Enter specific heat of solid: '); kL = input('Enter thermal conductivity of liquid: '); kS = input('Enter themal conductivity of solid: '); rho = input('Enter density which is constant: '); Tm = input('Enter the melting temperature of substance: '); Lat = input('Enter Latent heat: '); TL = input('Enter temperature at x=0: '); dat2p; %--------------------------Constants to derive----------------------------- alphaL = kL/(rho*CpL); alphaS = kS/ (rho *CpS);

dT = Tbdy - Tm; dTa = Tm - Tinit; v = sqrt(alphaL./alphaS);

StS = (CpS*dTa)/ Lat; StL = (CpL*dT)/Lat; %----------------------------Newton----------------------------------- x0 = 0.5 * (-StS/ (v*sqrt(pi)) + sqrt(2*StL + (StS/(v*sqrt(pi)))^2)); lambda = transnewton2p(x0,20,1.0e-7,StS,StL,v);

function Xt = XofT(lambda,alphaL,t) Xt = 2*lambda*sqrt(alphaL*t); function TofXTL = TofXTL(lambda,alphaL,Tbdy,dT,x,t) TofXTL = Tbdy-dT*(erf(x./(2*sqrt(alphaL*t)))./erf(lambda)); function TofXTS = TofXTS(lambda,alphaS,Tinit,dTa,x,t,v) TofXTS = Tinit +dTa*(erfc(x./(2*sqrt(alphaS*t)))./erfc(lambda*v))

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Exact Front for Glaubers Salt

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Exact Front for Water

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Exact Histories for Salt

-5,10,15 from top to bottom

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Exact Histories for Water

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Exact Profiles for Salt

-notice sharp turn at the melt temperature

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Exact Profiles for Water

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Enthalpy Method for

Stefan 2-Phase Problem

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Formulation/ Discretization of

Stefan 2-Phase Problem

Discretize Control Mesh

Discretize Heat Balance

Discretize Fluxes

Discretize Boundary Conditions

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Partition Control Volumes:

Subdivide the regions into M intervals, orcontrol volumes: with each

Subregion associate a node .

Volume of :

MVVV ,...,, 21

jx

MjxAV jj ,...,1, jV

jV

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Create the control mesh:

.

,,...,1,)1(

,0

/

2/1

2/1

2/1

lxMx

Mjxjx

x

Mlxx

M

j

j

steps)timediscretize(tntn

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Discretize Heat Balance

Integrating heat balance eqn over control

volume and over time interval

e

T

TTcdTTcE

qE

ref

ref

T

T

xt

ref

E

retemperatureferencewhere

][)(

Equation)Balance(Heat0

jV ],[ nnn ttt

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Continue

Dividing out the A and integrating the

derivatives yields:

Assuming E is uniform and is small:

dxdttxqAdtdxtxEAt

n

n

j

j

j

j

n

n

t

t

x

xx

x

x

t

t

),(),(1 2/1

2/1

2/1

2/1

1

1

1

2/1

2/1

)],(),([),( 2/12/1

n

n

n

n

j

j

t

t

jj

tt

tt

x

x

dttxqtxqdxtxE

j

x

x

j xtxEdxtxE

j

j

),(),(

2/1

2/1

jV

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Discretize Fluxes

Fouriers Law:

Approximate q discretely:

x

TkkTq x

Mjxx

TT

kqjj

jj

jj ,...,2,1

1

2/12/1

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Discretize Boundary Conditions

Imposing Temperature from left:

Impose exact temperature at back face.

The left boundary flux at x=0:

The right boundary flux at x=l:

,...2,1,0),(00 ntTT nn

1

12/1

2/1

012/1

2/1with,

k

xR

R

TTq

nnn

02/1 n

Mq

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Liquid Fraction & Mushy

is Solid

is liquid

is mushy

Liquid Fraction:

jV0jE

jVLEj

jVLEj 0

L

Ejj

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Energy & Temperature Relation

Solve for T:

(liquid)T,][(solid)T],[

mmL

mmS

TLTTcTTTcE

(liquid)E,

)(interfaceE0,

(solid)0E,

Lc

LET

LT

c

ET

T

L

m

m

S

m

G

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Energy vs. Temperature Graph

with Enthalpy Scheme:

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Explicit Time Scheme

Initial Temperature Known:

Initial Enthalpy

Set Resistance and Fluxes from InitialTemperature and Enthalpy

Update Enthalpies at

Update TemperatureUpdate Liquid Fraction

MjxTT jinitj ,...,2,1),(0

MjEj ,...,2,1,0

1njE 1n

t

i i

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Flux and Resistance Drive Heat

Flows

with2/1

1

2/1

j

n

j

n

jn

jR

TTq

SL k

x

k

xR

2

)1(

2

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Update Enthalpy at Next Time

Step(The Discretized Enthalpy)

Mjqqx

tEE

n

j

n

j

j

nn

j

n

j ,...,1],[ 2/12/11

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Update Liquid Fractions/Phases:

(liquid)if,1

(mushy)0if,

(solid)0if,0

n

j

nj

n

j

n

j

nj

EL

LEL

E

E

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Glaubers Salt Example

solidfortyconductivi10162

liquid)fority(conductiv10590

solid)forheat(specific761

liquid)forheat(specific313

heat)(latent21251re)temperatuimposing(90

re)temperatu(initial25

erature)(melt temp32

(density)1460

3

3

3

CkJ/ms.k

CkJ/ms.k

CkJ/kg.c

CkJ/kg.c

kJ/kg.LCT

CT

CT

kg/m

o

S

o

L

o

S

o

L

o

L

o

S

o

m

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Matlab Code Subroutines

Call INPUT: a data file contains all the

data needed for computing.

Call MESH: a function sets up controlvolume, the node and the face vectors.

Call START: a function initialize

temperature, enthalpy and liquid fractionat each control volume.

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Continue

Call FLUX: a function finds the fluxesfor each control volume at current time.

Call PDE: a function updates temperature,

enthalpy and liquid fraction at next timestep.

Call OUTPUT: a function outputs needed

and computed parameters. Call COMPARE: a function comparesthe exact and numerical solutions.

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Neumann Exact vs. Numerical

(Fronts, Histories and

Profiles) Plots for Varied M

Values

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Exact Front vs. Num. Front at M=32

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80

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E Hi N Hi M 32

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Exact Hist vs. Num.Hist at M=32

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Exact Hist vs. Num. Hist at M=60

i i

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E Hi N Hi M 160

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E Hi N Hi M 256

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Exact Profile vs. Num. Profile at M=32

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Exact Profile vs. Num.Profile at M=120

fil fil 160

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Summary on Plots

The numerical solution is getting closer

to closer to the exact solution as the

number of nodes M gets bigger and

bigger.

The numerical solution profile plots are

closer to the exact solution plots even forsmaller Ms.

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Stefan2p Errors (Front,

History and Profile) vs. M

Plots at hrst 50max

Melt Front Error vs M

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Melt Front Error vs. M

Melt Front Error vs M

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Melt Front Error vs. M

T Hi t E M

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Tem-History Error vs. M

T P fil E M

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Tem-Profile Error vs. M

Tem Profile Error vs M

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Tem-Profile Error vs. M

S h Pl

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Summary on the Plots

As the number of nodes M increases, the

errors for Stefan2p Front, History and

Profile plots appear decreasing trends.

These decreasing trends are even more so

for M equals binary numbers, i.e., 32, 64,

128, 256 and etc.

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Mushy2p:

An Alternative to theEnthalpy Scheme

Sherry Linn

E T h i h h l h

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E vs. T graph with enthalpy scheme

Wh h ?

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Why a new scheme?

Enthalpy schemes energy vs.

temperature curve not differentiable at

T = Tm!

Want a scheme based on a piecewise

differentiable energy vs. temperaturecurve.

To achieve piecewise

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p

differentiability

Impose a mushy zone of predetermined

length :

m

mm

m

TTLiquidTTTMushy

TTSolid

::

:

E vs. T graph with enthalpy scheme (solid) and

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E vs. T graph with enthalpy scheme (solid) and

mushy scheme (dashed)

I t d i h 2

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Introducing mushy2p

Explicit scheme

Independent from Stefan2p

Differs from Stefan 2p (PDE function)Imposed mushy zone affects

- Temperature

- Liquid fraction

T t

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Temperature

(solid)if

(mushy)0.0if

(liquid)0.0if

LEc

LE

T

LEL

ET

Ec

ET

T

n

jLp

n

j

m

n

j

n

jm

n

jS

p

n

j

m

n

j

D i i T t t M h Ph

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Deriving Temperature at Mushy Phase

Two points: (Tm,0),

(Tm+epsilon,rho*L)

Obtain equation ofline E in terms of T

Solve for T in terms

of E

Li id F ti i T f

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Liquid Fraction in Terms of

(liquid)if,1

(mushy)if,

(solid)if,0

n

jm

m

n

jm

m

n

j

m

n

j

n

j

TT

TTTTT

TT

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Temp History Error vs. Epsilon at

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p y p

x = .49 m, M = 64

Temp Profile Error vs. Epsilon at

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p p

t = 50hrs, M = 64

Melt Front Error vs. Epsilon at

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p

M = 128

Temp History Error vs. Epsilon at

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p y p

x = .49 m, M = 128

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Mushy vs. Enthalpy

How good is the mushy scheme?

Which is better, mushy or Stefan?

Recap: Glaubers salt

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solidfortyconductivi10162

liquid)fority(conductiv10590solid)forheat(specific761

liquid)forheat(specific313

heat)(latent21251

re)temperatuimposing(90

re)temperatu(initial25

erature)(melt temp32

(density)1460

3

3

3

CkJ/ms.k

CkJ/ms.kCkJ/kg.c

CkJ/kg.c

kJ/kg.L

CT

CT

CT

kg/m

o

S

o

L

o

S

o

L

o

L

o

S

o

m

Recap: Glauber s salt

Recap: Comparing Exact to

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p p gNumeric Solution

Requirements/Precautions: Input data for the explicitly solvable case

Impose exact temperature at back face

Error Analysis: L

-norm: err = max{ |Fapprox Fexact| }

Compare solution via three things:

1.Melt front X(t)2.Temperature T(x,t) history at fixed x

3.Temperature T(x,t) profile at fixed t

How good is mushy2p?

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1. Melt front X(t) location

2. Temperature T(x,t) history at fixed x

3. Temperature T(x,t) profile at fixed t

Melt Front X(t)M 32 1/32 03125 t 50 h

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M = 32, = 1/32 = .03125 = x, tmax = 50 hrs.

Max error 9.24 mm

Melt Front X(t), M = 128,

1/128 0078125 t 50 h

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= 1/128 = .0078125 = x, tmax = 50 hrs.

Max error 1.66 mm

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T(x,t) history at x 0.484 m

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M = 32, = 1/32 = .03125 = x, tmax = 50 hrs.

Max error 7.5510-2C

T(x,t) history at x 0.496 m, M = 128,

1/128 0078125 x tmax 50 hrs

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= 1/128 = .0078125 = x, tmax = 50 hrs.

Max error 1.9110-2C

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T(x,t) profile at t= tmax = 50 hrsM 32 1/32 03125 x

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M = 32, = 1/32 = .03125 = x

Max error 1.39C

T(x,t) profile at t= tmax = 50 hrsM = 128 = 1/32 = 03125 = x

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M = 128, = 1/32 = .03125 = x

Max error 0.639C

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Max errors for numeric schemes at various numbers of

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nodes: Temperature T(x,t) profile at t = 50 hrs.

M stefan2p Mushy2p ( = 1/M)

32 1.39721734740785 1.39111290103978

40 0.65724954768591 0.65923220965860

60 0.29444363002193 0.29550235005397

64 0.83534465188910 0.8296304512953580 0.14473018624196 0.14560810548880

120 0.22392243999277 0.22044558071511

128 0.64645276473467 0.63885783836963

160 0.40088509319875 0.39523873217349

240 0.19702938327533 0.19715111147884

256 0.12053866446253 0.11876976279872

Max errors for numeric schemes at various numbers of

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nodes: Temperature T(x,t) profile at t = 50 hrs.

M stefan2p Mushy2p ( = 1/M)

32 1.39721734740785 1.39111290103978

40 0.65724954768591 0.65923220965860

60 0.29444363002193 0.29550235005397

64 0.83534465188910 0.8296304512953580 0.14473018624196 0.14560810548880

120 0.22392243999277 0.22044558071511

128 0.64645276473467 0.63885783836963

160 0.40088509319875 0.39523873217349

240 0.19702938327533 0.19715111147884

256 0.12053866446253 0.11876976279872

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So which is better?

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So which is better?

Stefan2p Represents physical

reality

Jump in heat flux

Small error, dependingon number of nodes

Mushy2p Artificially-imposed

mushy zone

Energy E(T) is

continuous

Smaller error,depending on (withsame nodes)

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Is mushy2p a better scheme?

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Is mushy2p a better scheme?

Is it more efficient? Whats the optimal ?

Is the error different enough to be

significant?

Can we justify using a scheme that doesnt

seem to reflect reality?

Documents

DT Present (1)