2Phase Stefan problem.ppt

transcript

The Stefan Problemand the

Exact Solution for the Two Phase Stefan Problem

DT Project

First Things First

• We desire to formulate Two-Phase Stefan Model of Melting and Freezing– Non linear model– “Moving Boundary Problem”

• Contains an unknown which is the region to be solved…..

Why Stefan Problem???

•The formulation of the Stefan Problem is a foundation on which more complex models can be built.

First (cont’d)

• Note:– The phase changing process is governed

by the conservation of energy.

– The unknowns are the temperature field and the location of the Interface.

– Involves a phase change material(PCM) with constant density( ), latent heat( ), melt temperature( ), Specific heats( ), and Thermal conductivities( ).

mT LS cc ,

LS kk ,

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

Assumptions (cont’d)

• Thermophysical properties are different for each phase– Conductivities ( )– Specific heats ( )

• Density remains constant

L Sc c

L Sk k

Other Assumptions

• Nucleation and supercooling are assumed to be not present

• Surface tension and curvature is insignificant

Only Conduction

• Conduction of Heat– Temperature– Heat(enthalphy)– Heat Flux

• Characterizes phases

Heat Equation

-Heat conduction equation (one space dimension):

-well-posed

-Heat equation:

( )t x xcT kT

t xxT Tk

The Two-Phase Stefan Problem

• A slab, , initially solid at temperature , is melted by imposing a hot temperature at the face and keeping the back face, insulated (all parameters constant).

lx 0minit TT

Let’s Find a Solution

• The solution of the Stefan Problem is T(x,t) and X(t)!!!!

Mathematical Model

• PDE for

• Interface(t>0)

• Initial Condition

• Boundary Condition

t L xx

t S xx

0 ( ), 0

( ) , 0

X t x t

( ( ), )

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

L x S x

T X t t T

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

lxTTxT

0,)0,(

0,),0(

Be Exact!

• In order to explicitly solve the Two-Phase Problem 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 imposing temperature , on the face .

– The “alphas” are different for each phase. All parameters constant.

0 x S mT T

L mT T 0x

Mathematical Model

• Heat Equations for

• Interface(t>0)– Stefan condtion:

• Initial Condition

• Boundary Condition

t L xx

t S xx

0 ( ), 0

( ) , 0

X t x t

( ( ), )

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

L x S x

T X t t T

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

0,)0,(

),(lim

0,),0(

Two-Phase Neumann Solution

• We derive the Neumann Solution

– We use the similarity variable, ,and seek the solution for both for the liquid and for the solid.

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

( , ) ( )LT x t F

( , ) ( )ST x t F

( ) 2 LX t t

Temperature

• Temperature in the liquid region at t>0:

• Temperature in the solid region at t>0:

( , ) ( ) LL L m

tT x t T T T

( , ) ( )( / )

SS m S

tT x t T T T

• Neumann similarity solution of the 2-phase Stefan Problem for the interface

( ) 2 LX t t

Transcendental Equation

• There is a different Transcendental Equations that incorporates the TWO Stefan numbers (one for each phase).– Trans. Equation:

– Stefan Numbers and parameter “v”:( )L L m

c T TSt

( )S m S

c T TSt

)(exp()()exp( 222 verfcvv

Newton!!!

• Newton's method is an algorithm that finds the root of a given function.-Where F(x) = 0.

• The fastest way to approximate a root.

What now???

• We can use the Newton algorithm to solve the transcendental equation for the Stefan problem

• We do this in order to find a unique root lambda and therefore a unique similarity solution for each 0, 0, 0L SSt St v

Newton Algorithm

1. Guess x0

2. Take a Newton Step

3. Terminate if

xxx nn 1

-(Want TOL to be very small so convergence should be noticed)

-Stuck. Better guess

Approximate the Root for Newton

• For the 2-phase problem there is a good approximation which can be x0 in the Newton program.

Approximation of Lambda:

St StSt

Glauber’s 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;

Water Input Values

- Tm = 0;

- CpL = 4.1868;

- CpS = .5;

- kL = .5664e-3;

- kS = 2.16e-3;

- rho = 1;

- Lat = 333.4;

Glauber’s salt Examplemaximum 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

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

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))

Exact Front for Glauber’s Salt

Exact Front for Water

Exact Histories for Salt

-5,10,15 from top to bottom

Exact Histories for Water

Exact Profiles for Salt

-notice sharp turn at the melt temperature

Exact Profiles for Water

Enthalpy Method for Stefan 2-Phase Problem

Formulation/ Discretization of Stefan 2-Phase Problem

• Discretize Control Mesh

• Discretize Heat Balance

• Discretize Fluxes

• Discretize Boundary Conditions

Partition Control Volumes:

Subdivide the regions into M intervals, or

control volumes: with each

Subregion associate a node .

Volume of :

MVVV ,...,, 21

jxMjxAV jj ,...,1 ,

Create the control mesh:

,,...,1 ,)1(

steps) timediscretize(tntn

Discretize Heat Balance

Integrating heat balance eq’n over control

volume and over time interval

TTcdTTcE

re temperatureference where

Equation) Balance(Heat 0

jV ],[ nnn ttt

Continue…

Dividing out the A and integrating the derivatives yields:

Assuming E is uniform and is small:

dxdttxqAdtdxtxEAt

),(),(1 2/1

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

jjtttt

dttxqtxqdxtxE

j xtxEdxtxEj

),(),(2/1

Discretized Enthalpy

From Last Slide:

nj ,...,1 ],[ 2/12/1

)],(),([

2/12/1

dttxqtxq

xtxEtxE

Discretize Fluxes

Fourier’s Law:

Approximate q discretely:

TkkTq x

jjjj ,...,2 ,

12/12/1

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

2/1 with ,

02/1 nMq

Liquid Fraction & “Mushy”

• is Solid

• is liquid

• is “mushy”

Liquid Fraction:

jVLE j

jVLE j 0

Energy & Temperature Relation

Solve for T:

(liquid) T ,][

(solid) T ],[

(liquid) E ,

)(interface E0 ,

(solid) 0E ,

Energy vs. Temperature Graph with Enthalpy Scheme:

Explicit Time SchemeInitial Temperature Known: Initial Enthalpy Set Resistance and Fluxes from Initial Temperature and Enthalpy Update Enthalpies at Update Temperature Update Liquid Fraction

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

1njE 1nt

Flux and Resistance Drive Heat Flows

with2/1

Update Enthalpy at Next Time Step

(The Discretized Enthalpy)

nj ,...,1 ],[ 2/12/1

Update Temperature at Next

Time Step:

(liquid) ,

)(interface 0 ,

(solid) 0 ,

Update Liquid Fractions/Phases:

(liquid) if ,1

(mushy) 0 if ,

(solid) 0 if ,0

Glauber’s Salt Example

solidfor ty conductivi10162

liquid)for ity (conductiv10590

solid)for heat (specific761

liquid)for heat (specific313

heat)(latent 21251

re) temperatuimposing (90

re) temperatu(initial25

erature)(melt temp32

(density)1460

CkJ/ms.k

CkJ/kg.c

kJ/kg.L

kg/mρ

Matlab Code Subroutines

• Call INPUT: a data file contains all the data needed for computing.

• Call MESH: a function sets up control volume, the node and the face vectors.

• Call START: a function initialize temperature, enthalpy and liquid fraction at each control volume.

Continue…• Call FLUX: a function finds the fluxes

for each control volume at current time.• Call PDE: a function updates

temperature, enthalpy and liquid fraction at next time step.

• Call OUTPUT: a function outputs needed and computed parameters.

• Call COMPARE: a function compares the exact and numerical solutions.

Defining Errors• Front error at time:

• T(xout,time) error:

• History Error at :

• Profile Error at :

exioutXoutXout TioutTerrorTerrorT )(,max

exactXfronterrorXerrorX ,max

exacthistiXoutXout TiTerrorTerrorT )(,max

exactprofitt TiTerrorTerrorT )(,maxmaxmax

Neumann Exact vs. Numerical (Fronts, Histories and

Profiles) Plots for Varied M Values

Exact Front vs. Num. Front at M=32

Exact Hist vs. Num.Hist at M=32

Exact Hist vs. Num. Hist at M=60

Exact Profile vs. Num. Profile at M=32

Exact Profile vs. Num.Profile at M=120

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 for smaller M’s.

Stefan2p Errors (Front, History and Profile) vs. M Plots at

hrst 50max

Melt Front Error vs. M

Tem-History Error vs. M

Tem-Profile Error vs. M

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.

Mushy2p:

An Alternative to the Enthalpy Scheme

Sherry Linn

E vs. T graph with enthalpy scheme

Why a new scheme?

Enthalpy scheme’s energy vs. temperature curve not differentiable at T = Tm!

Want a scheme based on a piecewise differentiable energy vs. temperature curve.

To achieve piecewise differentiability…

Impose a mushy zone of predetermined length ε:

TTLiquid

TTTMushy

TTSolid

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

Introducing mushy2p

Explicit schemeIndependent from Stefan2pDiffers from Stefan 2p (PDE function)Imposed mushy zone affects

- Temperature

- Liquid fraction

Temperature

(solid) if

(mushy) 0.0 if

(liquid) 0.0 if

Deriving Temperature at Mushy Phase

Two points: (Tm,0), (Tm+epsilon,rho*L)

Obtain equation of line E in terms of T

Solve for T in terms of E

Liquid Fraction in Terms of ε

(liquid) if ,1

(mushy) if ,

(solid) if ,0

ε TTTTT

Melt Front Error vs. Epsilon at M = 64

Temp History Error vs. Epsilon at x = .49 m, M = 64

Temp Profile Error vs. Epsilon at t = 50hrs, M = 64

Melt Front Error vs. Epsilon at M = 128

Temp History Error vs. Epsilon at x = .49 m, M = 128

Temp Profile Error vs. Epsilon at t = 50hrs, M = 64

An Optimal Epsilon

Error decreases as epsilon increasesIs there a larger epsilon that causes error to

increase?An optimal epsilon (topic for further research)

Mushy vs. Enthalpy

How good is the mushy scheme?

Which is better, mushy or Stefan?

solidfor ty conductivi10162

liquid)for ity (conductiv10590

solid)for heat (specific761

liquid)for heat (specific313

heat)(latent 21251

re) temperatuimposing (90

re) temperatu(initial25

erature)(melt temp32

(density)1460

CkJ/ms.k

CkJ/kg.c

kJ/kg.L

kg/mρ

Recap: Glauber’s salt

Recap: Comparing Exact to Numeric 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 x3.Temperature T(x,t) profile at fixed t

How good is mushy2p?

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 = x, tmax = 50 hrs.

Max error 9.24 mm

Melt Front X(t), M = 128, = 1/128 = .0078125 = x, tmax = 50 hrs.

Max error 1.66 mm

T(x,t) history at x 0.484 mM = 32, = 1/32 = .03125 = x, tmax = 50 hrs.

Max error 7.55×10-2 °C

T(x,t) history at x 0.496 m, M = 128, = 1/128 = .0078125 = x, tmax = 50 hrs.

Max error 1.91×10-2 °C

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

Max error 1.39 °C

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

Max error 0.639 °C

Max errors for numeric schemes at various numbers of 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 0.82963045129535

80 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 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 0.82963045129535

80 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

So which is better?

• Stefan2p– Represents physical

reality– Jump in heat flux– Small error, depending

on number of nodes

• Mushy2p– Artificially-imposed

mushy zone– Energy E(T) is

continuous– Smaller error,

depending on (with same nodes)

Is mushy2p a better scheme?

• Is it more efficient?

• What’s the optimal ?• Is the error different enough to be

significant?

• Can we justify using a scheme that doesn’t seem to reflect reality?

2Phase Stefan problem.ppt

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