Citation
Source-and-sink method of solution of moving boundary problems

Material Information

Title:
Source-and-sink method of solution of moving boundary problems
Creator:
Akbari, Mehdi, 1954-
Publication Date:
Language:
English
Physical Description:
xii, 229 leaves : ill. ; 29 cm.

Subjects

Subjects / Keywords:
Boundary conditions ( jstor )
Heat ( jstor )
Heat flux ( jstor )
Heat transfer ( jstor )
Liquids ( jstor )
Melting ( jstor )
Solids ( jstor )
Subroutines ( jstor )
Surface temperature ( jstor )
Temperature distribution ( jstor )
Dissertations, Academic -- Mechanical Engineering -- UF
Mechanical Engineering thesis Ph. D
Genre:
bibliography ( marcgt )
non-fiction ( marcgt )

Notes

Thesis:
Thesis (Ph. D.)--University of Florida, 1993.
Bibliography:
Includes bibliographical references (leaves 223-228).
General Note:
Typescript.
General Note:
Vita.
Statement of Responsibility:
by Mehdi Akbari.

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University of Florida
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Resource Identifier:
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29713762 ( OCLC )

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SOURCE-AND-SINK METHOD
MOVING BOUNDARY


OF SOLUTION OF PROBLEMS


BY

MEHDI AKBARI


A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT
OF THE REQUIREMENTS FOR THE DEGREE OF
DOCTOR OF PHILOSOPHY

UNIVERSITY OF FLORIDA


1993



























TO MY WIFE VAHIDEH, AND DAUGHTERS SARA AND MONA
















ACKNOWLEDGEMENTS


committee,


study


am most grateful Dr. C. K. Hsieh,


as well


as many


hours


to the chairman


who provided of guidance


of


my supervisory


the inspiration


for this


and encouragement.


His


enthusiasm


has been contagious,


and I hope


to emulate


his attitude


in the future.


I wish to


express


my


sincere


thanks


to Drs. R. A.


Gater,


G. Emch,


counsel


R. Abbaschian,


and understanding


and H. A. Ingley have been essential


whose


unselfish


in this study.


Last but by


Vahideh,


no means


least,


for her love and spiritual


I would support


like to thank


during


my wife,


this period


of


time of work


on the dissertation.


iii


G �


good
















TABLE OF CONTENTS


Page


ACKNOWLEDGEMENT LIST OF TABLES LIST OF FIGURES NOMENCLATURE ...


ABSTRACT CHAPTERS


* . . 0 0 0 0 0 * 0 * 0 0 a 0 0 0 0 0 0 0 * 0 0 0 * * * 0 0 * t 0 6 0 0 0

* . 0 *0 0 0 0 a 0 * *60 0 0 0 0 0 0 0 0 0 0 0 0 * 0 * 0 0 9 0 * * 0 * 0 0 * 0 0 0

* 0 0 0 0 0 6 S S S 0 * 0 * 00 0 0 0 0 0 0* S 0 0 0 S 6 * * 0 0S 0 0 0 0*

* . 0 0 0 0 0 0 0 0 0 0 0 0 9 0 0 0 0 0 * 0 0 0 0 * * * 0 * *0*


0 0 0 0 0 0& 0 * 0& 000 0 * * 0 0 0 0 0 0 * 0 0 & 0 O* 0 *0 *0*a 00 0 *00a 0 0 0 0* 0*0O* * 0 0 000 0*


INTRODUCTION .............

LITERATURE REVIEW


0 0 0 0 0 0 0 9 0 & 0 0 0 0 *0

0 & 0 050* 0 0 0 0 **


*0 0 0 0 0 0 *00 * 0 0 0 18
.............1


SOLUTION OF INVERSE STEFAN PROBLEMS BY SOURCE-AND-SINK METHOD ...............


Motivation ...........
General Analysis .....
Pre-melt Stage .....
Melting Stage ...... Example Problems ..... Numerical Solution ... Critique of The Method Results and Discussion


0 0 * 00 00 0 0 * 0 0 0 0


0 0* 0 * 05*0 0 0 0 0 0 *00*00 0 00 0*S0


3.7 Extension and Concluding Remarks


* 0 0 0 0 9 0 0 0*0 0 0 0


SOLUTION OF ABLATION PROBLEMS WITH ONE MOVING BOUNDARY BY A SOURCE-AND-SINK METHOD . . . . . 0 . a * 0 * * * 0 0 . 0 0 . 0 0 * 0 0 . 0 0 0 . . . .


4.1 Solution Methodology ........
Pre-ablation Stage ........
Ablation Stage ............
Equivalent Problem ........ 4.2 Illustrative Examples ....... 4.3 Numerical Solution of Ablated 4.4 Results and Discussion ......


0 0 0 0~


a 0 0 0 0
Front
S0 0 0 0


SOLUTION OF ABLATION PROBLEMS WITH TWO MOVING BOUNDARIES BY A SOURCE-AND-SINK METHOD . . . ... ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90


iv


. iii

. .vi

.vii

* .ix

* .xi


I II


III


3.1 3.2


3.3 3.4 3.5 3.6


IV


.21

.21 .25 .28 .28 .31
.33 .39 .40 .56


0 9 0O9 0O 0 0 0 0 0 0 0O0


.58

.58 .59 .62 .63 .65 .67 .69


0 ~ . . 0
0 0 ~ . S ~ 0 0 ~ . . S
000

























VI

APPENDICES

A


5.1 General Analysis 0...
Pre-melt Stage 0600 Melting Stage .....
Ablation Stage
Equivalent Problem 5.2 Examples ............
5.3 Numerical Solution of
Problem ............
5.4 Numerical Solution of
and Energy Storage 5.5 Numerical Examples 00

DISCUSSION AND RESULTS 006



EXTENSION OF THE SSM .....


0~~~~ 0a 0 0


O 0 0 0 0 000 0 0 0


The Ablation Temperature
0 0 0 6 0 0 0 0 0 0 0 0 000000000000


0 0 0 0 0 0 0 6. 0 � 0


0 0 0 0.* 0 0 6�� 0 0 0 0 0 0�* 0 0 0 0 0 0 00


0 0 0 0 60 0 00 00 0 0


Page

S..91
S.�91 .��94 �..95
..98
..101

..104

..109
..110

..130


....60... 0.135


B SSM FORTRAN
PROBLEM .

C SSM FORTRAN
PROBLEM ..

D SSM FORTRAN
PROBLEM ..
E SSM FORTRAN
PROBLEM ..

F SSM FORTRAN G SSM FORTRAN REFERENCES .............

BIOGRAPHICAL SKETCH ....


PROGRAM

PROGRAM
PROGRAM


PROGRAM
000 0�0 0

PROGRAM


PROGRAM PROGRAM


FOR FOR


FOR o 0 0 6 FOR FOR

FOR


INVERSE
INVERSE"

INVERSE
0 00 0 0 0

INVERSE


600*00*0
INVERSE
. . . ��0�0 ' "0


STEFAN
0����FAN

STEFAN
a 0 0 0 0 0 0 0

STEFAN


STEFAN
0 0 . . �.0 0 �0


0 � 0 0 0 0 0 6 � 0 6 0 .136 �*.000000 000147


0 0 . 0 0 0 0 0 0 * .154


0 0 0 0 0 0 � 0 0 0 "


ABLATION PROBLEM ... COMBINATION PROBLEM


0 .0 0 0 0 0 00 0 0 0 0 60 0 0


0 0 0 00 0 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 06 0 0 0 0 0 0 60 0 0 0 0 0 0 * 0 0 0 0 0 0 0 0 0 0 a 0 0 0 0 0 0 0 * 0 0 0


.165 .172 .186 .223 .229
















LIST OF TABLES


Page


Table


Expressions to account for


Properties Conditions Comparison for first


effects


of aluminum ...............

tested in eight examples .. between true and retrieved Stefan-Neumann example solve


of boundary


0 0 *0 0 0 0* 0 0 0000 0� 0


0 0 * 0 0 ~ ~~ 0 * 0 ~0 0 0 0 00

conditions d for boundary


Comparison between true and retrieved


conditions


for second


Stefan-Neumann example solved for


boundary


flux then temperature ............... . 0 0 � 0 � � . . . . . . . . 46


Comparison between true and for third Stefan-Neumann ex


retrieved ample solve


conditions d for boundary


Comparison for fourth


between true and


retrieved


conditions


Stefan-Neumann example solved for


boundary


flux then temperature ............................


Comparison between true and retrieved for fifth inverse example problem sol


conditions ved for


Comparison between true and for sixth inverse example pi


retrieved


conditions


roblem solved for


� . . . 0 0 � � � .51


Comparison between true and


retrieved


conditions


for seventh


inverse example


problem solved


temperaturet....................ween.true�and�ret
Comparison between true and retrieved conditions


for eight


inverse example


problem solved for


S 0 0 0 0 & 0 0 0 0


Conditions Conditions


tested tested


in four examples ..... .......


in three examples . . ..


0 0& 0 0 0 0 0 *


* 0 0 0 0 0 0 500* 0 0 0


vi


3.1


3.2 3.3 3.4


3.5


.23 .41 .42


3.6


3.7


3.8


3.9


3.10


3.11


for


4.1 5.1


.53

.70 111


conditions ........................


temperature ....................�.� ..�.� ..�.� ..�.�..�.�..�.�..��.. 43


temperature ...............................................� �� �� �� �48


� . � � � 49


�.........5


temperature .............................. �� ......


heat flux .......................................


� . . �� � � 52


heat flux ......................................
















LIST OF FIGURES


Page


Figure


Linearization of solid-liquid interface position curves for the numerical solution ............


* . o o35


System analyzed ............................................61

Trend of ablated surface position for a medium initially at phase-change temperature imposed with a constant heat-flux condition ............73

Accuracy, stability and convergency test of the SSM in the solution of ablation for a medium initially at phase-change temperature imposed with a constant heat-flux condition ...............75


Stabili in the medium conditi


ty and convergency test of the SSM solution of ablation for a subcooled imposed with a constant heat-flux o n . . . . . . . . . . . . . . . � � � � � � � � � � � � � � � � � � �.. .7 8


Comparison of ablated surface position with two methods in the literature for a subcooled medium imposed with a constant heat-flux condition ................................................. 80

Comparison of ablated surface position with the moment method in the literature for a subcooled medium imposed with a linear heat
-flux condition ........................................... 83

Stability and convergency test of the SSM in the solution of ablation for a subcooled medium imposed with a linear heat-flux condition ................................................. 85

Comparison of ablated surface position with the moment method in the literature for a subcooled medium imposed with a quadratic heat-flux condition ....................................... 87

Stability and convergency test of the SSM in the solution of ablation for a subcooled medium imposed with a quadratic heat-flux condition .................................................89
vii


3.1

3.2 4.1 4.2


4.3




4.4 4.5 4.6




4.7 4.8


4.9


System analyzed .. . ........... ������������������������2









Figure


5.1 5.2 5.3


....*. ** . . . .106




.............114


Trends of solid-liquid interface and ablated surface positions for a combination problem of subcooled medium imposed with a quadratic heat-flux condition ..... 0................. .


Temperature distributions in the medium at different times during ablation for a combination problem of subcooled medium imposed with a constant heat-flux condition .............................120

Stability and convergency test of the SSM in the solution of ablation for a combination problem of subcooled medium imposed with a constant heat-flux condition .............................122

Stability and convergency test of the SSM in the solution of ablation for a combination problem of subcooled medium imposed with a linear heat-flux condition ................................124

Stability and convergency test of the SSM in the solution of ablation for a combination problem of subcooled medium imposed with a quadratic heat-flux condition ............................126


5.5 5.6 5.7 5.8 5.9





5.10


viii


Linearization of solid-liquid interface and ablated surface position curves for the numerical solution ..........................

Trends of solid-liquid interface and ablated surface positions for a combination problem of subcooled medium imposed with a constant heat-flux condition .........................

Trends of solid-liquid interface and ablated surface positions for a combination problem of subcooled medium imposed with a linear heat-flux condition .........................


5.4


Overall accuracy test of the SSM in the solution of the combination problem of subcooled medium imposed with a constant heat-flux condition . . .............................. ���������129


Page


System analyzed ... . . . .............................. ��������������93


.��.o.� ... 116


�.............00 09 *118
















NOMENCLATURE


Specific Equation


heat


Temperature


function


used for boundary condition


Flux function


used for boundary


condition,


Green's


function, Heat flux


elements


in coefficient


matrix


Thermal conductivity Latent heat of fusion Latent heat of vaporization Normal unit vector


Distance


from sink or


source


to sense


point


Interface


Stefan


positions


number


Dummy


variable


for integration,


interface


speed


Temperature,


temperature


vector


Time


velocity


of the interface


Position,


source


point


Thermal


diffusivity


Dirac


delta


function,


Kronecker


delta


ix


Lf


Ste


R1, R2









Density


Dummy


variable


for time


Subscripts ai m,v Phase


nd Superscripts change


Upper


time limit


index


Time index Transition


Melt and


vapor


fronts


Initial


1.2


N1, N2


time
















Dissertation Presented


of the Universit
Requirements


y of Florida for the Degre


in Partial me of Docto]


Fulfillment


of the


of Philosophy


SOURCE-AND-SINK METHOD OF SOLUTION OF
MOVING BOUNDARY PROBLEMS By


Mehdi

May


Chairman: Dr. Chung K. Hsieh Major Department: Mechanical


Akbari 1993


Engineering


Source-and-sink


methods


have been developed


for the solution


of inverse of ablation


diffusion and Stefar


problems, i problems.


ablation

Green ' s


problems, functions


and combination


have


been used


and integrodifferential


positions equations


are solved


equations


by using


are derived


for the interface


in the phase-change medium.


local


linearization


These


position


and the boundary


heat flux if it


were treated as unknown.


The results


have shown


to be accurate,


convergent,


and stable.


The methods


developed


for the solution


of the inverse


diffusion


problems


have been used


to find the boundary


conditions


for the i approaches: approach. conditions


inverse


without


solution


approach


They


and


to be useful


are


solved


by


two


a time incremental


to find the boundary


to be supplied


at both sides


efficient


of the phase


in that they require


change

fewer


interface. equations to


The methods


be solved


are


for the


xi


and the temperature


of the interface


Stefan


problems.


a series


Both have shown


reliance


on the flux information


Abstract of


to the Graduate School









unknowns. solution


can also be easily


of the


problems.


In the


ablation


use of the source-and-sink


and combination


problems,


method


heat transfer


to analyze

is solved


fixed


imposed


domain


so that the original


on the moving


boundary is


taken


heat-flux


condition


to be the condition


that is imposed


on the solid-vapor


is then used


or liquid-vapor


together


interface.


with the temperature


This flux condition


at the interface


to


solve


for the interface


position as well as the


condition


imposed


on the fixed


used


boundary.


in the temperature


Finally,


equation


these


positions


to complete


and conditions


the solution.


method


has been used successfully


in solving


examples


encompassing


one-,


two-,


and three-phase


ablation


with the boundary


of the


medium


imposed


with


constant,


linear,


and


quadratic


conditions.


Numerical


solutions


for these


examples


as well


as the


trends


of the moving


boundary


positions


are discussed


in detail.


xii


the


in


are The


flux


developed


f or the


The algorithms
















CHAPTER I INTRODUCTION


Phase changes occurs when


a body


is exposed to


a large


heat flux


so that it melts


or vaporizes


at the surface.


With continuous


heating,


the melting


or vaporization


front


moves


inward


with time.


Since priori,


boundary problems treatment


the position


of the phase-change


this type of problem


problem


is commonl


in the heat transfer


find applications and processing, th


in thermal


iermal


protection


interface v known


literature.


energy


is unknown


as the moving


Phase


storage,


of re-entry


=change material


vehicles


in


space


technology,


laser


drilling


and cutting


in manufacturing,


freezing


and thawing


in soils


and foodstuff,


and photocoagulation


opthalmological


procedures,


among


others.


It is difficult to


solve


a phase-change


problem


because


of the


presence


such


of the nonlinear


a problem,


the number


condition


of phases


at the interface.


encountered


Usually


for


in the medium


depends appears triple place. solid, occurs


on the ambient


in the medium.


point


pressure


Thus for


of the substance,


On the other


liquid,


for


and


a medium


hand,


gas


and the temperature


a medium


only


for phase


states


heated


sublimation


change


may appear


of low thermal


above


range


or cooled below


and fusion


the triple


simultaneously,


conductivity


that


the


may take


point,


and this


and exposed


to


In practice,


in


large


heat flux.


the changed


phase


may b e


removed


as









in the re-entry where the


substance


is removed


by


a fluid


dynamic


force.


This material


removal


may


also be effected


naturally


such


as in sublimation,


where


the sublimed


In either


vapor


event,


is dissipated


the imposed heat


by diffusion flux follows


to the surroundings.


the boundary


motion,


and such problems the literature.


A different


are commonly


state


referred


of affairs


to as ablation


is encountered


problems


in


when the changed


phase


remains


stationary


and continues


to occupy


the


space


taken


its previous


state.


melt is not removed


Thus for example


by


an external


in melting


force,


a solid,


it will adhere


if the to the


solid


from which


it changes


phase.


Then in the absence


convection


in the melt,


the heat transfer


in the medium


can be


analyzed first D


by solving


resented


a Stefan


a formal


problem in


solution


recognition


of the problem


of Stefan in the


literature


Stefan that,


(see


problem


next chapter is basically


in the former,


the boundary whereas, in


condition the latter,


for


a survey


different


he problem is is always the imposed


of the literature).


from the ablation


solved ir imposed o condition


a fixed


i the fixed follows th(


problem domain,


The


in


and


boundary e boundary


motion. diminishes

the recedi


have been


Thus in the ablation in size with time, an(


ng


boundary


numerous


problem,


d the impos,


due to the shrinkage


efforts


in the literature


the domain ed condition


for analysis


is acting


of the medium. for the solution


on


There

of the


Stefan


problems.


Much less research,


however,


has been devoted


the solution


of the ablation


problems.


Also for those


ablation


that have been solved and documented


by


of


who

open


to


melted


This


soon as it is formed.


occurs


[1


problems


in the literature,









most


moving study found


are for the analysis


boundary


appears


for problems


of generic


in the medium.


involving


at the solid-liquid


two moving


and liquid-vapor


ablation


There


in which has been


boundaries


such


interfaces.


only


a lack of

as those


They


are


encountered


when


the


solid,


liquid,


and


gas


states


appear


simultaneously


in the medium.


Three


dissertation.


phase-change In the first


problems prob1 em,


will


be


an inverse


solved


Stefan


in


problem


be solved.


This problem


differs


from


a conventional


Stefan


problem


in the imposed


sense

on the


that,


boundary


for the conventional


is fully


specified.


problem,


the condition


This condition


is then


used together


with others


to solve


for interface


motion


as well


the temperature


problem,


however,


distribution the interface


in the medium.


motion


is given,


For the inverse


and this motion


is


as a part of the input


to solve


for the boundary


condition


is


necessary


found


to provide


in material


for this interface


treatment


and processing,


motion.


where


This problem the property


the material


may


be


a function


of the phase-change


rate at the


interface.


Solution


of the problem


thus enables


a better


control


the property


of the material.


It should


above


be noted


has recently


literature. dissertation


However


is


that the inverse


been solved , the method


more efficient


by


Stefan


Zabaras


developed


because


problem


et al. [I1 for solution


it requires


described

in the in this


no information


for the heat flux at the interface.


As will be shown,


two methods


are developed


for the solution


of the problem:


one expands


and the other uses


this will


used


as


that


is of


of


the


one


condition in


polynomial


or an inf inite


series,









the interface


a time incremental


position


information


at discrete


times


to solve


for the coefficients


in the polynomial


or series


expansion,


while


the latter


uses the


interface


position


at consecutive


times


to solve for


the conditions


incrementally.


Both can be used to solve


for the conditions


accurately


as supported


by eight


examples


encompassing


the search


for constant and


boundary.


time-variant


The methods


temperature and


are also well adapted to


flux conditions


a numeral


at the


solution.


The second


phase-change


problem


solved


in this work is


ablation


problem


that contains


only


one moving


boundary.


problem is solved fixed domain. In


by


a new method,


1 this method,


which


the ablated


treats phase


the problem


in


is considered


remain moving


in the


space


boundary


can


that is ablated


be taken


in the


to be


process.


an interior


As such,


the


phase-change


interface,


and the heat flux that is derived


at this interface


be used to match


the condition


imposed


at the moving


boundary.


with the additional at this interface,


the position


prescription


the flux condition


of the interface


of the temperature


there


for phase


change


can be used to determine


as well as the hypothetical


condition


that is imposed


conditions solution.


at the fixed


boundary.


can be used in the temperature


Finally


these


equation


positions


to complete


Solution


of the ablation


problems


has rarely


been attempted


a fixed


domain.


The method


presented


in this work thus represents


new endeavor,


which


will be shown


to be simpler


than most methods


described


in the literature.


To validate


the method


used in the


and


solution of the ablation


an


This


a to


can


Then


and

the


in


The former


utilIizes


method.


are provided


four examplIes


problems,









they


include


with constant


the analysis


and


of


time-variant


one- and two-phase


flux conditions


ablation


at


the


imposed moving


boundary.


Of those


four,


the


ones imposed


with constant


heat flux


can be checked is thus selecte


with an exact :d for checking


solution.

the soluti


A one-phase on over the


ablation entire p


problem


eriod


of


the ablation,


testing


while


the solution


a two-phase


in


ablation


a quasi-steady


problem


state.


is selected


Both yield


for


accurate


results stabi lity


as shoi

4 tests


later


in this dissertation.


have also been made rendering


Convergence


further


assurance


success


of the method.


above


The good provide


addresses combination


results


impetus


ablation


obtained


in the second


for the solution


with


of the ablation


two


moving


and Stefan


problem


of the third boundaries.


problems,


described


problem


This


an area that has


rarely


been studied


in the literature


yet


is important


in practice


when solid,


medium. moving


liquid,


Again


boundary


and


gas


the problem is treated


states


is solved


appear


in


as an interior


simultaneously


a fixed


domain,


phase-change


in the and the


interface.


flux condition


at


this


boundary


is


again


used with


temperatures


at the interfaces


to determine


the heat flux at the


fixed


boundary


solution.


as well as the interface


Unlike


its predecessor


positions


for which


to complete


the


two simultaneous


integrodifferential


time step,


equations


for the combination


are solved


problem,


simultaneously


three


equations


for each will be


solved three


simultaneously.


examples


Again


including


the results


imposition


of


are good as constant,


evidenced


linear,


flux conditions.


the


and for


that


is


The


the


by


and


wn


quadrati c









of all


will be used for the solution


the problems to a through


in this work. development


This method


[2-5],


where


has recently


the method has


been subjected


shown


to be


particularly addressed in


suited


i this wo


for the solution ,rk. In this meth


of the phase-change od, the problem is


problems


solved


using


one


set


of


governing


equation,


initial,


and


boundary


conditions, literature, equations,


region.


whereas


in the conventional


the problem


is solved


with each set focused


The conventional


methods


methods


by using


; developed different


on the solution


thus cannot


of


compete


in the sets of


one phase with the


source-and-sink


sink method,


only


it is in the solid


assigned


conventional


methods,


for simplicity.


one temperature


or liquid


region


equation


depends


in which


in the source-and-


will be derived;


on the position


This is in sharp


several


contrast equations


whether that is to the must be


derived,


again


one for each phase


region.


The time and effort


saved


in the source-and-sink


method


can thus be readily


appreciated.


With the inclusion


of the three


separate


problems


in this


dissertation, approach. Thi each problem;


coverage


of


methodology, felt that,


the presentation of


us one chapter


the


statement


analysis, e

only through


of


materials


follows


will be self-complete


the


problem,


.xamples, and results


such


an organizatior


a nontraditional


of


with the solution


It is

can be


general


and discussion. 1., the material


presented


in


a coherent


manner


without


severe


fragmentation.


what follows


in the next chapter, a literature


review


will


be given


It is then followed by


by


method


Moreover,


in the equation.


temperature


will be devoted


and the chapters


for the presentation


In


A source-and-sink


method


the presentation


of the three


f irst.









problems. conclude


to


and recommendations


this dissertation.


As usual,


comp ilI ed


all computer


in the appendix.


this research the future.


will be useful


programs


It is hoped


developed


in this work


that the insight


in the analysis


of the phase


gained


change


are from


in


will be


given


conclusions


Finally,

















CHAPTER II
LITERATURE REVIEW


Heat conduction


boundary


problems.


These


change


problems


constitutes


a large class


of

to


as Stefan

Clapeyron


problems


[6],


and


were first


solved


the thickness


in 1831 by of the solid


Lame and


crust


by


freezing


of water


under


a constant


temperature condi t ion.


able to find the thickness


to be proportional


to the


square


root of


time but did not find the constant


of proportionality.


Some thirty


years


later,


Neumannn


presented


an exact


solution


to the problem


his unpublished


the Stefan

solution w


lecture


problem


notes,


was solved


and for the first in its entirety


change


in


time in history,


[7-8].


space


Neumann's


imposed


a constant


temperature


condition.


His work is important


because, irrespective search for additional


solved Neumann


exactly


today


of the exact remain


some one hundred


numerous

solutions


those


and thirty


efforts . Stefan


that


years


made by


many


problems


were originally


others


that


can be


analyzed


ago.


While


obtained


the exact


by Neumann,


solution

the problem


to the Stefan p n has been named


)roblem

after


was first


Stefan


in


recognition of in 1889 [6-9]. who expressed 1


his published Practically the temperature


work appearing


the


same problem


of the phase-chan


in the


open


was solved ge medium i


literature by Stefan,


n terms


of


moving


with phase


are commonly referred


who determined


They


were


with


as for the phase


in


a semi-infinite


in


by










a similarity


variable,


to be proportional


and the position


to the


square


of


root of time.


the moving


Thus,


in


boundary essence,


the


moving


boundary


transformation.


general Stefan


problems,


problem


Since it has


that


was


solved


by


the transformation


now been firmly


can be solved


means


of


a similarity


can not be used for


established


exactly,


that,


it must be


for


in an


unbounded


domain,


of medium


of constant


properties,


and imposed


a constant


temperature


condition.


severe


limitation


imposed


by


the


similarity


transformation,


it is not surprising


to


see a wide variety


approximate monographs


what


solutions


developed


have been prepared


follows,


reviewed;


some popular


interested


readers


in


the literature.


for presentation


solution


are referred


of these


methods to 10-14


will


Books methods.


be


briefly


for details.


One of the early


methods


developed


for the solution


of the


Stefan


problems


imposed


with heat flux and convection


conditions


the power-series


position


expansion


and the temperature


method.


In this method,


in the medium


are expanded


the interface


in terms


of


power series


series


[10] or complimentary


are then substituted


conditions


to determine


error


functions


into the governing


the coefficients


equation


in the series


[11-14].


These


n and boundary expansion.


Power


series


method


works well


for the short-time


solutions.


At that time, the series ai


the series

re sufficient


converge r4 t to yield


apidly, ai accurate


ad only a results.


few terms


of


The method


also works


for the solution


in the neighborhood


of singularities


occur


as a phase


degenerates.


For large


time,


more terms


are needed in the series,


With


the


with


of


and In


is


that


may


X14-t,


and the method


rapidly


loses its appeal.







10


Stefan


problems


integrodifferential


can also be solved


equations.


by reducing


Lightfoot


[15]


solidification


as a moving


heat


source


front.


With the


use of


Green' s equations


function,


he


was able to derive


one for the temperature


two integrodifferential


and the other


for the interface


position.


Then by assuming


the position


a function


of the


square


root of time enabled


him to retrieve


the Stefan-Nuemann


solution.


Lightfoot' s dimensional


method Stefan


has been extended


problems


involving


to the solution


phase


change


in


of


two


an infinite


wedge


[16].


interface correction


To avoid


position


term


a tedious


was taken


was introduced


iteration


to be


a hype


into the solution


in two dimensions, lrbolic function a


to account


the


nd


for the


property


variation


in different


phase


regions.


Indeed,


Lightfoot' s


method


has been limited


to phase


change


of equal


properties.


limitation


has been lifted


by


Kolodner


[17] by using


double


source


and sink at the phase-change


interface.


Integrodifferential


equations


of Volterra


or Fredholm


also be derived


in the solution


of the Stefan


problems


by


a Fourier


transform inverted is useful


[18-19].


in the solution.


in providing


This occurs when


In general,


an exact


solution


the transformed


the integral


in integral


equations


equation


form.


are


method


However,


the integral


equations


must still


be solved


numerically


solution.


One of the unique


occurrence continuously


of


multiple


with time.


features


phase


The boundary


of the Stefan


regions


position


whose


problems


domain


of the domain


is the changes is also


of the solution. Boley


them took


to


the


This


type


can


for


introduced an


sought


as a part


[20]







11


embedding a large


method,


fixed


in which


domain


the time-variant


for the solution


domains


are embedded


of the problem.


This


introduces an unknown


must be solved


heat flux at the fixed


together


with the moving


boundary, boundary


and this flux


position


to


complete


the solution.


Boley's


embedding


method


has been extended


to the solution


of multidimensional


problems


without


internal


generation


problems however,


convenient


[21].


solved


It should


be noted


in this dissertation


a source-and-sink


method


to use in the solution


that all the moving


also work in


will be employed of the combination


a fixed


which


boundary domain;


is


of Stefan


more


and


ablation


problems.


Stefan [22-29]. I


problems


can also be solved


n this method,


by


a quasi-steady


an asymptotic


solution


expansion


is derived


by


dropping to find derived


the unsteady a long-time by dropping


y temperature solution. A


the interface


terms


in the governing


quasi-stationary


velocity


terms


solution


in


equations is also


one of the


governing short-time


equations solution.


and the interface

Asymptotic expa


flux condition


mansions


to establish


of the temperature


and


the moving


boundary


position


are then constructed


by using


these


solutions


for limits


Asymptotic


expansions


problems with nonlinear melt [22]. Basically


work best in the solution


r conditions a a perturbation


ind convective motions technique, it offers


of Stefan


of the insight


into the physics tedious mathemati


of the problem.


cally;


However,


even the determination


the method


is


very


of the first-order


terms


in the temperature


expansion


has


proven


to be


an intractable


task [26].


in


heat


[26].







12


Higher-order


distinct


terms


drawback


Coordinate


become


of the method. transformation


increasingly


offers


difficult


another


to obtain,


approach


to


solution embedding


of the Stefan


method,


problems


the transformation


[30]. n works


Somewhat


in


a fixed


similar domain


to the without


the need of introduction


of


any


new unknowns


such


as the boundary


heat flux in the embedding


method.


In fact,


the time-variant


domain


is mapped


into


an invariant


domain


in the transformation


method;


moving


boundary


is thus immobilized


in the solution.


solve


It is noted the Stefan


that


use of the transformation


problem.


It only provides


does not by


the


itself


convenience


of


working


in


a fixed


domain.


The coordinate


transformation


has thus


been used together


accuracy solution


with other


methods


and also in the numerical


for the improvement


methods


for facilitation


[31-37].


Like the Stefan


problems problems.


Stefan fixed


also fall

However,


problem domain,


problems


into the general the ablation p


because whereas


briefly


category


)roblem


in the latter in the forme


is


reviewed


above,


of the moving more complex


the problem c< r the domain


ablation boundary than the


an be solved


in


is continuously


changing


with time.


There


have been


numerous


efforts


developed


the solution however, has bE


of the Stefan een devoted to


problems;


the solution


only


very


1 i ini ted


of the ablation


work,


problems,


which


will now be reviewed


as follows.


Solution


of the ablation


problems


dates


back to Landau


who first


used the coordinate


transformation


to change


the variable


a fixed domain. An


the


the


of of


for


[38]


domain encountered


in the ablation


problem to







13


exact, medium in his medium


quasi-steady state solution


imposed paper.


after


with the constc Also studied


dropping


for the


case of


heat flux condition


is the ablation


a semi-infinite has been given


for the semi-infinite


assumption.


Using


Laplace


transformation, a time-variant


the ablation


Landau


was able to derive


heat flux imposed


solution


was only


on the fixed


boundary.


solution


for


However,

heat flux


condition.


include


Specifically,


(i)


Stefan


number


two limiting


approaching


cases


zero


were examined

(negligible


capacity)


and (ii)


Stefan


number


approaching


infinity


(negligible


latent un i ty,


heat). Landau


For the employed


case of the Stefan


a finite


difference


number


method


of the order


to solve


ablation


problem with results


method


presented


graphically.


has also been used by Rogerson


and Chayt


to find the


exact


melt-through


time for ablation


of


a slab imposed


a constant heat


flux condition


on one side and


an insulated


condition conduction thermal pr


on


the


equation operties


other


side.


1 and showed th

of the material


Rogerson


e results engaged


integrated


to be independent


the heat


of the


change.


The integral


approach


has been used in the solution


of the


phase-change

residuals,


problems


the method


[9,40,41].


was first


Essentially

introduced


a method


by


of weighted


von Karman


Pohlhausen


in the approximate


solution


of boundary


layer


equations.


The heat integral


reasonably a detailed


accurate


analysis


approach results.


is simple


However,


to use; it also provides


handicapped


Specifically,


the form of the profile


the steady-state


the pre-melt


derived


for the constant


that

heat


Landau's


of


with


the


[39]


in phase


and


of the temperature


it is somewhat


field.


in


the


is limited by


of the temperature


accuracy







14


initially


chosen


for analysis.


Also the approximation


can not be


systematically


improved


for accuracy.


The moment


method


proposed


by Zien [41]


carries


promise


improvement


method


over the classical


has been used to solve


time-dependent


heat-flux


L heat integral one-dimensional


conditions.


method. ablation


Both pre-melt


The moment


imposed


with


and ablation


and the integral


multiplying profile was


Again,


of the original


the integrand


s chosen


by


the heat balance


heat equation


powers


and substituted


integral


was carried


of temperature. A into the integrated


was used,

out after


temperature


version


of


the heat conduction


equation.


The heat balance


integral


based


the approximate for the boundary


temperature heat flux.


profile


Although


was then used


the method


as the expression


appears


to work for


the general


case of the time-variant


heat flux condition,


it is


expected


that the choice


of


an approximate


temperature


profile


works


for the nonmonotonic


heat flux


may not be applicable


to the


case of


more general


time-variant


conditions.


The ablation


problems


can also be solved


by


a variational


approach. Lagrangian


Biot and Agrawal thermodynamics to


applied


the solut-


the variational ion of ablation


analysis problems


variable


thermal


properties


[42].


They


considered


one-dimensional


heat transfer


in


a semi-infinite


cylinder


imposed


with


a constant


heat flux condition.


solved.


In the procedure,


Both pre-ablation


the governing


and ablation equations were


stages


were


transformed


and the Lagrangian


heat-flow


equation


was derived


which


provided


relationship


between


the surface


velocity


and the heat penetration


relation between these


of


solutions


were derived.


on


that


and

with


was also


two quantities


depth.


Another







15


found solved


by using


the


simultaneously


energy


for


equation.


These


equations


the heat penetration


depth


were then


and


the


velocity


of the ablated


surface.


The variational


method


has


been applied


for approximate


analysis of a and radiative


blation


of


heating


a semi-infinite


[43,44].


solid


Solutions


subject


were obtair


to convective ied in closed


form for both the pre-melt


and melt-removal


heating


regimes.


these


studies,


a cubic


temperature


profile


was taken.


The surface


temperature, melt removal


the thermal


were treated


penetration


depth,


as unknown


and


and the depth


of the


were determined


as


functions


of time.


The ablation Blackwell [45] has


solve


an ablation


a moving


coordinate


problem employed


problem


has been solved the exponential


in one-dimension.


system


which


numerically. differencing He proposed


was attached


Recently method to


the


use of


to the ablated


surface. T approach, th convection oi all elements


hen,


by invoking


ie element n f the moving and control


with the exception


The node point


the


matrices


grid,


use of


could


and


energy


volumes were


of the last ablating


at the moving


interface


a finite


be defined


storage.


moving


element


was fixed


control-volume for conduction, In this method,


at a uniform


and control


in


space.


velocity volume. As such,


the last ablating


boundary.


The method


e 1 ement


had one moving


has been applied to


boundary


and


the solution


of


one fixed a steady-


state


ablation


problem


for which an


exact solution


was available.


Comparisons conduction


were also made with central


terms


and upwind


differencing


differencing


for


for the convective


the


terms;


appear to be better


In


n umer ical1l1y.


the exponential


schemes







16


Ablation


problems


have also been solved by


a finite


element


method


with deforming


spatial


grids


[46].


In this method,


classical finite-element the continuous deformation the ablated surface. Th


equations ar

of the grid


e transformed


for


lis is done at t


a precise I he expense


to account for localization of of chores of


construction


of


an additional


convective


matrix.


Recently, as laser 1


ablation


beam.


has been applied


Masters


[47]


to model


used the finite


intense heating difference method


to solve


the ablation


of


a one dimensional


slab imposed


intense


uniform


heat flux and analyzed


the effect


of melt


on the


temperature solutions w


distribution ere found foi


the temperature studied the abl


history


ation


during


the heat pulse.


r the velocity


duri ng


due to


ablation.


of the surface


Abak i an

e laser


a continuous-wav


The steady-state


recession


and Modest


and [48]


beam irradiating


on a moving


semi-infinite


and semitransparent


solid.


Using


integral


method,


they were able to


derive


a set of nonlinear


partial


differential


equations


wh ich


were solved


numerically


for the


groove


depth


and shape


due to ablation.


They


also considered


the ablation


a moving


slab caused


by


irradiation


from continuous-wave


pulsed


laser


distribution processing a vaporization


beams


[49].


and derived The laser


Ls investigated occurred at


a solution


for the temperature


has also been used in material


by Dabby ar the surface;


ad Paek [50].


however,


In


below


their work, the surface,


the material


was heated


by absorption


of the


laser


radiation,


which


might


reach


Explosion


a temperature


may thus take place,


higher


which


than


that


provides a


for


means


vaporization. for material


in the drilling


the


such


with


an


of


an


and


removal


process.







17


The review


heat conduction


given


above


problems.


is strictly


In these


problems,


of regular


geometry,


the system


governing specified,


equation,


initial


and the problems


and boundary


are solved


mainly


conditions


are fully


for the determination


of the temperature


field.


In the


inverse


problems,


however,


roles


of known


and unknown


quantities


are exchanged.


There


been abundant of the regular the solution


studies doc problems; of inverse


cumented


in the literature


not much work, problems, an


however, has d for those


for the solution


been devoted


to


that have been


solved


and documented


in the literature,


nearly


all of them


are for


heat conduction


study


for inverse


without


problems


phase


change


[51-60].

change.


There


is


a lack of


Most inverse


heat conduction


problems


deal with


a situation


an extra


and this temperature


temperature is available


is used together


at one point


with others


in the domain, to find the


condition


imposed


on the boundary.


Stolz


[51]


first


solved


problem


formulated conduction principle. condition


found


a small


numerically.


as though


problem


An integral


and


it


time step


[52] was able to improve


In


his


work,


were a direct


he


equation


by


the


problem.


was able to


was derived


numerical


problem a linear


use the superposition


The method


inversion.


used


were too small.


an accurate


by using


solution.


a procedure


involved


minimization


of the


sum of the squared


difference


between


the actual


and the calculated


temperatures


at the location [53] developed


where


the


temperature data


for the solution


the


have


with phase


where


such


was solved,


inverse


Since


was


heat


was solved


to be inefficient


for the unknown


surface


if the time steps


must still


be used for


this method


was Yet, Beck that


were given.


Burggraf


an exact







18


series


solution


capacitance temperature


point.


data


to a one-dimensional


approximation


serving


and heat flux histories


Approximate


were used for input


inverse


as


results were found


in solution.


the


problem leading


were provided


if discrete

Beck [54]


with the lump


term.


The


at an interior or experimental


moved


one step


further


by solving


an inverse


problem


in which


the material


properties


were treated


as functions


of temperature;


problem element


solved

solver


becomes


nonlinear.


has also been reported


A two-dimensional


inverse


in the literature


finite


[55].


can be used for the solution


of the heat flux imposed


on the surface


of nuclear measurements


fuel rod with the


use of the interior


temperature


for input.


a handful


of studies


are found


for the solution


of inverse


heat transfer


with phase


change.


Macqueene


et al.


[61] proposed


inverse welding


finite


element


process.


method


In their


to determine


method,


the efficiency


the latent


of


an arc


heat of fusion


is


taken


into consideration


by


the variation


of the elemental


specific


heat when the


point.


average


Conduction


temperature


of the element


in both the solid


and


reaches liquid


the melting


regions


was


accounted


for.


Katz and Rubinsky


[62] proposed


a front-tracking


finite-element


method


for the solution


of one-dimensional


inverse


Stefan


problems.


His method


was applied


for the determination


the position


of the solid-liquid


interface


and the transient


temperature


distribution


in the solid


region


during


stationary


welding.


An inverse


method


was also used by


Landram


[63]


for the


analysis of the


thus


the


It


Only


an


of


arc


interface


pos i tion


and


the


energy transport







19


mechanisms


during


to be important


welding.


during


The vaporization energy


the motion


loss


of the solid-liquid


was found interface.


The interface


shape,


being


close


to hemispherical,


gives


clear


indication transfer.


of


a nearly


An analytic


one-dimensional


solution


to inverse


radial


Stefan


symmetry


problems


for the heat


in Cartesian


and spherical


geometries


was provided


by Rubinsky


and Shitzer


In their


analysis,


the inverse


Stefan


problem


was characterized


two boundary


conditions


at the moving


front.


This is in sharp


contrast


to the technique


developed


in this dissertation


Chapter


3)


which


needs


only


one condition


at the interface


to solve


for the boundary at the phase cl


derived

solved guessed problem.


condition.


hange


by integrating


by


the method


solution Series


In their


temperature. the governing of analytic


was taken solution w;


work,


the medium


An integral equation, iteration


to be the long-ti


as then developed


by


I


was initially


equation


which in whi me sol

induct


was then


was, in turn, ch the first lution to the tion following


a number


of iterations.


A boundary


element


analysis


with constant


elements


has been


developed


inverse analysis


by


Zabaras


solidification


developed


by


et al.


problem


for the solution


[1.


Beck [52-54]


They


and Burggraf


of


a one-dimensional


used the sensitivity


[53]


for inverse


heat transfer able to solve


region given


solution.


two separate


and the other


by Rubinsky


Using


inverse


in the liquid


and Shi tzer,


an integral


Stefan region.


two conditions


formulation,


problems,


they


were


one in the solid


Thus similar


to the


ones


must be provided


at


front, also an inefficient method.


E64].


by


(see


the freezing







20


It should


be mentioned


that


a source-and-sink


method


recently


been developed


in the heat transfer


literature,


which


particularly


suited


for the solution


of regular


and inverse


Stefan


problems.

to apply


In


a series


this method


of


papers,


Hsieh


to the solution


and his associates


of regular


were able


and inverse one-


and


two-phase


without


melting


subcooling


and solidification and superheating


problems


and imposed


for medium


with and


with constant


and


monotonic


temperature


has been applied


and heat flux conditions


to the solution


of phase


change


[2,3]. imposed


The method with cyclic


conditions developed


[4,5,65].


Two inverse


with the problem


solution


formulated


with


techniques


have also been


a source-and-sink


method


as shown


in reference


to the boundary


3


element


The method


method


has shown


as reported


by


to be closely


Hsieh


et al.


related [66,67].


a further extension of these


has


is


study


is


works.


The present
















CHAPTER III
SOLUTION OF INVERSE STEFAN PROBLEMS
BY A SOURCE-AND-SINK METHOD


This chapter


presents


the development


of


a source-and-sink


method


to solve


inverse


Stefan


problems.


In these


problems,


conditions boundary. the phases


are specified


Typically,


is given


at the moving


the temporal


location


and this location


rather


than


of the interface


is used together


the fixed


between


with others


to determine the boundary


temperature


or heat flux that is required


to provide


for this interface


motion.


In what


will be given


designed


follows

first.


for the solution


in this chapter, It is followed


the motivation


by


of the inverse


a general MStefan


for this study


analysis problems


that i s in three


dimensions. dimensional


examples extensions


This analysis


inverse


are


problem


provided


of the method


and


is then used in the solution


examples.


discussed


are included


Finally,


in


results


details


to conclude


and


of


one-


for these possible


this chapter.


3.1 Motivation


Heat diffusion


in


a medium


with constant


properties


governed


by


the partial


differential


equation


V 2T(ft) + k -


aT(ft)
Ot


the


is


YER t>0


21


(3.1)







22


where


all notations


have their


usual


meaning.


For this medium,


conditions


imposed


on the boundary are usually


one of the following


types


T(f,,t) = Fi(Yi,t), ri E Bi


T(fi,t) = Gi(iBt) Oni ki


OT(f,t)
o9ni


T(F,t) -


1 Hi(fit),


(3.4)


which


represent


conditions,


the


respectively.


familiar I

In (3.3)


Dirichlet, and (3.4),


Neumann,


denotes


and


Robin


an outward


drawn


normal.


Then,


with the additional


initial


condition


given


T(FO) = Ti(i)


the temperature


solution


can be expressed


by


means


of Green's


function


as [68]


+0


f G(fit I r',O)Ti(r')dV' R/


J f G(ft I r',r)u"'(r',r)dV'dr + OR'


Here,


the braced


conditions


given


term is used to account


earlier.


Their


expressions


for the three


are 1 i sted


boundary in Table


3.1.


A distinct


feature


is found


in the Green's


function


method


above--the


effects


of the initial


condition,


heat generation


conditions are embodied


the


(3.2)


(3.3)


as


(3-5)


T(ft)


(3.6)


1


(or


Yi E Bi


destruct ion),


respectively in


and boundary







23


Table


3.1


Expressions conditions


to account


for effects


of boundary


U


BOUNDARY CONDITION


4


t)F1 j


t)


t)


-G1 �l,


t)


i


t)-=Hi


(T~i


t)


I


{ EXPRESSION


0 a G (r,
osl C


tIT'~, V


0/


t
fG(T tI , os/


,7) dS;dtc


dn.


�T (T/i, t) h
adnrI) +
dn1


dS1 dt ,k


i


w


(7-4j I


Fj (74i


)n f


D (-I jf


' ki '...







24


the first,


second,


and third


terms on


the right


of equation


(3.6).


Then,


in the


case of regular


problems,


once these


conditions


fully specified, t effort, the Green's


;he temperature


function


can be easily


can be obtained


found.


by using


In this


the concept


of


point


charges


[68] or by solving


auxiliary


problems [69].


The format


the solution


of (3.6)


of the inverse


turns


out to be particularly


problems.


For the inverse


suited


problems,


left hand side of this equation


can be used to represent


the extra


temperature points. Th: information,


that is provided


en this temperature


such


as the init


either


at the boundary


can be used to determine ial condition, heat ger


or at interior


the missing =ration, or


boundary


conditions.


In these


efforts,


the missing


quantities


be expanded

substituted


either


in


into their


a polynomial


respective


or an infinite


integrals


in (3.6).


series,


which


is


The resulting


equation


series initial


is then solved


to complete condition


numerically


the solution.


for the coefficients


This method


and the heat generation.


clearly


works


in the for the


for


the


boundary condition, assume a particular


since it is unknown 'type' of condition


a priori,


one must first


on the


boundary.


For convenience,


one could


use equation


(1.2)


or (1.3)


for this condition.


Green'I s


function


is then found


with this


assumed


exchanged


condition.


Notice


in by


Hsieh


condition


[70].


can be That i s


a Dirichlet


condition


can be accomplished


by


means


Neumann


condition


and vice versa.


The search


for the condition


thus not restricted yet, an incremental


by


the type


of the conditions


assumed.


Better


approach can also


are


for

the


can


However,


that is imposed


to


if necessary as shown


say,


that the boundary


and Shang


of


is


-- - - - - - - -- 7


lie


be developed


to track


solution







25


the boundary


conditions


accurately


as will be shown


later.


concept problems


above


will now be applied


in which


the interface


for the solution


positions


of inverse


are given,


Stefan


and these


positions


are used to find the boundary


conditions


that must be


imposed


to


cause


the interface


motions.


3.2 General


Analysis


For the sake of illustration


in what follows


the Stefan


problems


consist


of two stages:


a pre-melt


stage,


when heat is added


to the surface


of


a subcooled


med i um


to raise


its temperature


to the


phase-change


temperature;


and


a melting


stage,


when the medium


changes


phase


and


the melting


starts


at the surface


interface


moves


inward


with time.


It is assumed


that the properties


for different


has


a distinct


medium.


will


be


phases are constant melting temperature


Convection developed


is negligible.


for


the


and of equal value.

; that is, no mushy


For generality,


solution


of


both


The medium zone in the the analysis


melting


and


solidification


problems.


The analysis


can also be extended


for the


solution unequal moment,


medium


of Stefan


phase


problems


properties


the simplified


shown


in Figure


in multiple


phases


and in


as will be discussed


problems


3.1.


will be solved The formulation


a medium


later.


with


For the


by considering


of these


the


problems


follows below.


The


and


the

























Figure


3.1


System analyzed












Pre-melt


Stage


Governing


equation--


FE (L US)


V 2To(f,t) =


1T0(rt)
a t


, (3.7)


to> t>


Initial


To(F,0) = Ti(f)


Melting


condition--


(3.8)


Stage


Liquid


Region:


Governing





V 2TL(f,t) =


equation--


1


aTL(ft)
at


(3.9)


t>to


Solid


Region:


Governing


V 2Ts(f,t)


equation--


-1


OTs(ft)
at


28


t>to


(3.10)









Initial


condition--


Ts(fto) = To(ft0)


Interface



TL(Fft) =TV


Conditions:



Ts(rf t)


OTs(rf,t)
On


OTL(rf,t)
On


Svn(t)


vn(t) =V.n


Here F 1 interface


denotes motion.


the interface


position


For the inverse


and


Stefan


v,, the history


problems


of the


of interest


in


this study,


this history


is used for the determination


of the


missing


boundary conditions.


The problems


as posed


can be solved


by


use of the Green's


function


method


described


in the preceding


section.


However,


direct


use of this method


would


require


(3.6)


to be applied


to two


separate


regions,


liquid


and solid,


and the solution


so obtained


not be


as efficient


thus used [2,4,15].


as one desires. In this method,


A source-and-sink


the melting


interface


method


is


is taken


a moving


be a moving conventional


represent


heat-sink


heat-source methods in


the temperatures


front


and


front.


which


a freezing


Then,


different


in different


interface


in sharp equations


regions,


only


i s taken contrast


to to


are used to one equation


it is in the solid


29


(3.11)


(3.12)


(3.13)


(3.14)


to be


may


will be derived.


Whether


or liquid


region


is







30


determined equation. simplified


by


the position


The solution with this method


that is assigned


of the inver

1. Following


in the temperature


problems


this approach


can then be the melting


stage


is solved by


considering an


equivalent


problem


as follows:


Governing


equation


for the equivalent


problem:


FE (LUS)


V 2T(ft)


PL


vn(t)b(f


- f)


1


0T(ft)
(9t


Initial


condition


for the equivalent


problem:


T(fit0) = T0(ft0)


Interface


conditions


for the equivalent


problem:


T(?1,t) = Tm,


vn(t) = Vii


(3.17a, b)


where


denotes


a Dirac


delta


function.


The signs


preceding


this function


are used for freezing


(+)


and melting (-).


can be shown


readily


that (3.15)


reduces


to (3.9)


(3.10).


Furthermore,


by integrating


(3.15)


across


the interface


rf-e


to Ff+e


and forcing


e to be


zero in


a limiting


process,


equation


pill


(3.15)


reduces


box at the interface


to (3.13).


This


in Figure 3.1.


can be proved by using


Other equivalences


are apparent.


t>to


(3.15)


(3.16)


It


and


from


the for


b(lz- Ff)


this stage







31


The equivalent


which


problem


the heat generation


m can be solved term is changed


by referring


to (3.6)


to the interface


in


motion


term as


t
L SG(ft t0LuS


LI
LuS


- ff)dV dr +


(3.18)


where


the plus sign


is used for freezing


and minus


sign


for melting.


Finally, equation


the boundary


conditions


to the interface


position,


can be found


rf,


and T(FI


by setting


t)


F in this


to the melting


temperature,


Tm 9 as


L


G(Ff,t I f',t0)Ti(f')dV'


t4'


toLuS


G(?ft I F',r)v


- r1)dV'dr + .j{


The missing implicitly.


boundary


conditions


In this effort,


can then be found


the time when melting


by solving


starts


(to)


them


can be


determined be taken generation


by solving


the pre-melt


to be the special


term is


zero.


Solution


problem, whose


case of (3.6)


in this stage


solution


in which


can again the heat


is thus elementary.


3.3 Example


Problems


The analysis


above


is


now used to


solve


example


problems


which


are in semi-infinite


domain


in which


the interface


motion


is given.


For the sake of generality,


the analysis


will be developed


determining


either


the Dirichlet


condition


or the Neumann


condition


the interface motion may


Tm


=I
LuS


(3.19)


for


G(Fjt I f',t0)Ti(f')dV'


T(fit)=


1 71(7)6(f/


that is imposed


on the boundary


at x=O, and







32


be the result


of either


melting


or solidification


in


a one- and two-


phase


medium.


Also for illustration


purposes,


the heat flow is


dimensional,


the initial


temperature


being


un i form.


Extensions


more dimensions


are provided


in Appendix


For the problem


given,


Green'Is


function


can be found


to be


G(x,t I x',r)= 1 [exp(
2 i-ra(t - -r)


(X - x)2 4a(t -- ))


where


the plus


and minus


signs


are to be used when the flux and


temperature respectively.


condition


is assumed


to


appear


at


can then be obtained


the


by using


boundary,


(3.18),


which


is recast


in


a general


format


To(x,t) Tm


^* t-to H(t - to)
Ste f


dR~r+t0)
d- G(xt I R(


r+to),r)dr


where


all


temperatures,


including


Tm , are


measured


in


excess


of the


initial


temperature,


To(x,t)


t


E(s) =


Here F(s)


x F(s)
2a t-s G(s)


and G(s)


denote


the assumed temperature and


heat flux


c respectively.


one


A


to


(3.20)


The temperature


as


T(x,t)
Tm


(3.21)


and


x2
exp[ - 4( )


]ds


in which


(3.22)


(3.23a,b)


condition


Also











H(t-t�) - 0


for


t>to t

CTM
Ste -L


where


Ste is known


as the Stefan


number.


With the


use of the


circumflexed


Heaviside


function


given


by (3.24),


equation


(3.21)


holds


for all time and for both one- and two-phase


problems.


Equation


(3.21)


can be used


to determine


the unknown


boundary


condition


by invoking


T0(R(t)
a=1- t)


use of the condition


t
H(t - to)
- Ste


-t

0


dR(r+t) G(R(t),t
d7-


t the interface





to IR(r+to),r)dr


where


the plus


and minus


signs


on the left hand side


are to be used


for freezing


and melting,


respectively.


3.4 Numerical


Solution


A local


In this effort,


can be used


time


to solve


range


(3.26)


is divided


numerically. into small


increments see Figure


in which


3.2.


each increment,


the interface


position


can be taken


integral


is taken


to be linear;


out of the integral


for


written as a summation


as


33


(3.24)


(3.25)


as


(3.26)


linearization


the entire


Then the dR/dt


and the convolution



































Figure


3.2


Linearization


position


of solid-liquid


curves for


interface


the numerical


solution






35


R(t) R(tN) R(t N-1)

R(tn)
R(tn-I)



R(t2>

R(t 1)


R (t) R (tn -1) -I+n(A-1t
(A 0


(AR)n


(A t)I


to* tI t2 tn 1 tn


tN1 tN t


I I I I I I I


0 -to


T -to







36


TO(R(tN)'tN) =l H(tN- to,
Tm Ste
m n=l


dR(t-)
dt


t to


G(R(tN),tN


- to I R(-+to),r)dr]


tn_1 - to


Here the signs


are to be selected


following


the statement


(3.27)

below


(3.26). evaluated


For the invers with the input


problems,


the right


of the interface


motion


hand side data. As


can be for the


left hand side,


if the boundary


conditions


are represented


power


series,


the number


then the number


of the terms


that


of terms are taken


on this side must be equal


in the


series.


to


In practice,


equation


(3.27)


can be written


by


means


of matrix


elements


melting


problem


for the series


method


N
Sbmnan = Cm n=l


N
+ dmnen
n=l


where


m=N;






10


tN > to; for


N=I ,2, .. m

bmn =


R(tm)
2a


sn-1
(tin -S3/


exp[


R(t7n)2
4a(tin - s)"ds


for in> n;


N
F(s) = L ans"-1
n-1


tm
1f sn-i
k J (tm- S)1/2
0


R(tm)2
exp[ - 4a(:tm-s )]ds


for m n;


N
G(s) = Lann-1
n=l


(3.29a,b,c)


(3.30)


cm - iaTm


by


as


for


(3.28)


and














m

tn - to Lif


G(R(tm),tm - toR(r + t0),r)dr


for m> n


tn_1 - to


dR(t-)
dt


Notice closed series


that the integrals


form if the missing


as shown


coefficient


in (3.29)


matrices


Band D


in these


conditions or in terms


are of


of


a lower


can be performed


in the series.


a Fourier triangular


in


power


The


structure,


characteristic


permitting


the solution


of (3.28)


to be carried


simply


by using


forward


substitution,


a simple


numerical


procedure.


The boundary


conditions


can also be determined


by


use of


incremental


tN tN-1


approach.


In this


E(s) exp (tN- S)1/2


method


, (3.27)


is recast


]ds =


4 !Tm{� 1


H(tN - to)N
Ste n-1


ti - to [dR(t-) I


G(R(tN),tN


- t0 I R(r+to),r)dr]} -


tn_1 - to


N-1
{�zF
n=1


where to 1


tn

f
tn__


E(s)
(tN _ s)1/2 exp[-


]ds}


(3.33)


to zero all the time in the integral


dmn


=1


(3.3 1a,b)


(3.32)


equations


are expressed


out


an


as


0 for


on the left and


s set







38


the last integral


on the right.


In practice,


equation


(3.33)


can be


recast


F(tN) G(tN)


PNG


N
Tm + EdNnen
n=l


N-1
-E
n=l


F(tn)Pn,F G(tn)Pn,G


(3.34 ab)


where


tN> to


;N=I ,2, � ,


dNn


can be obtained


by referring


(3.31);


tn
R(tN)f
2a f tn -


tn

I
tn-1


1
(tN s


1
(tN-S)/


exp[ - R(tN)2
4a(tN -s)


R(tN)2
exp[ 4a(tN-s)


(3.35 a,b)


In (3.34),


the summation


vanishes


when N-1=0.


Then starting


so on,


the conditions


can be evaluated


incrementally.


this effort, are used as


the F and G values


the input


found


in the computati


from the previous time steps on of the right hand side of


(3.34),


which


immediately


gives


the F and G values


at the succeeding


new time step. The algorithms


This continues


until


can be developed


the desired


for


a rapid


time


solution


i s reached.


of the


conditions.


Computer


programs


developed


for the solution


of inverse


Stefan


problems


given


in this chapter


are provided


in Appendices


through E.


as


and


to


PnF PnG


]ds


N=I,


and


from


In







39


3.5 Critique


of the Method


Using


Green's


functions,


the present


method


is limited


to the


solution


of


problems


whose


Green'Is


function


can


be obtained


analytically.


This


excludes


those


problems


whose


boundary


conditions nonlinearity Kirchhoff tr


and


governing


equations


can not be resolved


'ansformation).


are


by using


nonlinear


and


transformation


use of Green's


function,


projects


an impression


of being


related


to the boundary


element


method that [64]. Yet method, th the liquid


t has been undergoing


t, they are e boundary and solid


. functionally


element regions,


through


development


different.


equations whereas


are written s in the present


in recen"


;eparatf metho(


t years element ely for d, only


one equation


solved


(3.27)


in the prese


is derived. nt method


The number of is thus reduced


the equations


by half,


which


particularly


true in the solution


by


the incremental


approach


described


earlier.


More important,


in the boundary


element


method,


the problems


cannot


be solved


without


the information


for the heat fluxes


appear


on both sides


of the interface


[64,1].


Such fluxes,


however,


are unnecessary


in the present


work.


Instead,


equation


(3.26)


embodies


conditions


(3.12)


and (3.13)


in the form of


a single


integrodifferential


equation.


There


is


no need for the satisfaction


of the flux conditions;


as a result,


the present


method


effective


because


it requires less information


in the next section,


for input.

the present


method


also accurate.


In fact, is exact.


such


accuracy


is not unexpected


because


The only approximation


Yet with the


such


(e.g.,


it


In boundary


to be


is


that


As will be shown


is


more


is


(3.26)


in the analysis is


equation







40


the local


linearization


which


has been applied


to the interface


motion


and the boundary


in the present C constant elements.


analysis,


condition;


they


The analysis


see (3.33)


and (3.35).


have been approximated


can thus be improved


for


In fact, by using accuracy


by using


elements


higher


order


as in references


elements [59,71].


such


as linear


and quadratic


3.6 Results


and Discussion


For the numerical


experiments


performed


in this study,


interface


motion


data


are taken


from reference


5.


Aluminum


will be


used for tests;


its properties


are given


in Table


3.2.


The inverse


solution


techniques


developed


in this chapter


are used to solve


eight


example


problems


of which


four having


exact


solutions.


retrieved the exact interfaces interface conditions


conditions solutions


for these


for


error.


move as a function


motion


that


Stefan-Neumann


data


examples


In


square


of


are used to retrieve


appear on the surface problem, and Table


can thus be compared


these


four examples,


root of time, the constant t(


This problem


3.2 provides


with


the


and the emperature


i s known


a summary


as the of the


conditions


tested


The first


in the example


examples.


deals


with


a general


two-phase


Stefan-


Neumann


problem;


the medium


is initially


subcoo led


to 300 K,


which


is lower


than the melting


temperature


(see


Table


3.3).


For this


example, the temperature temperature is thus assun


at the boundary ned to appear a


is unknown;


t the boundary


an unknown S, and the


interface


motion


data


are used to find this temperature.


two sets of results are


the


The


The


results


are listed


in Table 3.4, where






41


Table 3.2


Properties of aluminum


Melting temperature, T. (K) 932 Vaporization temperature, T, (K) 2,543 Heat of fusion, Lf (J/kg) 389,600 Heat of vaporization, L, (J/kg) 9,462,000 Thermal conductivity, k (W/m K) 200 Density, p (kg/rn3) 2,710
Specific heat, c (J/kg K) 1200







42


Table 3.3


Conditions tested in


eight examples


PROBLEM INPUT TRUE CONDITION TYPE OF BOUNDARY DESCRIPTION DATA F(t):K; G(t):W/m2 CONDITION

ASSUMED SOLVED

1 two-phase R(t)-4t F(t)=1000 tempeatur temperature
Tj=300K 2 two-phase R(t)--dt F(t)= 1000 heat flux temperature
TpS300K Ti=932K=T

4 one-phase R(t)-4t F(t)= 1000 heat flux temp ure
Ti=932K=T

5 one-phase reference F(t)= 1000+5t temperature temperature
Ti=932K=T, [5]

6 two-phase reference G(t)=6.388034x 106 heat flux heat flux
T300K
7 two-phase reference F(t)f932+40t temperature temperature
Tr=300K
8 one-phase reference G(t)=3xl06+5xl0'ta heat flux heat flux
T1932K=Tm [5]
LIII I I







43


Table 3.4


Comparison between true and retrieved conditions for first Stefan-Neumann example solved for boundary temperature


________Boundary Conditions, T (K)


Imposed True Condition


Retrieved_____


Series Method


N=3


ii' i


1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00


1003.22 1002.97 1002.72 1002A8 1002.25 1002.03 1001.82 1001.61
1001.42 100123 1001.06 1000.89 1000.73 1000.58 1000A3 1000.30 1000.17 1000.06 999.95 999.85


II I aI


N=4


1003.17 1002.87 1002.58 1002.30
1002.04 1001.79 100136
100134 1001.14 1000.95 1000.78 1000.63 1000.50 1000.38
100028 100020 1000.14 1000.09 1000.07 1000.07


Incremental Method


1003.49 1000.41
100020 1000.11 1000.03 1000.05
1000.04 1000.02 1000.68 999.88 999.97 999.99 999.99 999.99 999.99 999.98 999.98 999.98 999.90
999.94


F (S) =� ansn-1
nul


For N=3, a1=1003.490405
a = -0.268732
a%= 4.350448xl03 For N-4, a1=1003.490405
-0.322478
6.264645x10"3 6.552981x10"s


Time (sec)


1
2
3
4
5
6
7
8
9 10 11
12 13
14 15
16 17
18 19
20







44


given--one incremental


for the series


solution


In the series


method. solution


solution


method,


method


a power


and the other


series


of degree


for the


N-1 is


used to represent


the temperature


(see (3.29b)),


and two values


are tested. I time intervals


In this method,


the interface


are used for input.


Thus,


position da for example,


ta at equal for N=3, R


data at times


determine


equal


the series,


to 0, which


6, 12, and 18 seconds


is,


in turn,


are used to


used to generate


the


temperature v the retrieved


alues listed 1 temperatures


for 20 time steps at the boundary


in the table.


by


both series


Comparing with the


true condition


(exact


temperature)


listed


to the left indicates


are in good agreement.


these


series


are listed


In this


case,


at the bottom


the coefficients of the table.


found


for


From the


temperature values


tabulated,


the results for


the low


power


series


appear


trend


to be


persists


10 (results temperatures


as good


as those


of the high


even with the test of


not shown).


a higher


This gives


have been converged.


power


power


series, series


the indication


As for the incremental


and such of degree that the solution


method,


the results


are also good; errors


are of the order


of 10-3


at large


time.


For the incremental


method,


the boundary


conditions


are foun given.


d at exactly The time step


the


same times


for the solution


en the interface positions are is thus identical to that for


the interface


data input.


Also notice


that the


convergence


stability difference solution o


that


are normally encountered


methods


)f integral


are nonexistent


equations.


Also


in


in the conventional


the present


as in the


finite


incremental


case of the series


the incremental solution


of N


they


and


are generally


results


solution method,







45


better


at large


time than small


time,


which


will


be further


discussed later.


In the example


above,


a temperature


condition


is imposed,


same 'type'


of condition


is assumed


in the


process


of the


inverse


solution.


Since


the type


of the condition


that is imposed


the boundary


is unknown


a priori,


it


may


well be the heat flux


condition


that


one assumes,


and the question


to be addressed


now is


whether


it is still


possible


to retrieve


the temperature condition


via the assumed


flux condition.


Use will now


be made of equation


(3.21)


to determine


flux condition


is fou


this temperature co ind, and the results


ndition


once the boundary


are listed


in Table


As shown good. but


in the table,


the series


the incremental


solution


results


solution


results


are not as accurate


are still as those


i sted


in Table


3.4.


This is certainly a result


of the


errors


being


accumulated evaluation


flux.


first


in the evaluation


of the temperature


While


this example


using serves


of the heat flux next in the


the previously


well


determined


to illustrate


heat


that the


conditions condition


are still


may


exchangeable,


lead to large


such


errors,


a two-step particularly


solution


of the


in the series


solution according approach compared


method,


and should


to experience, of solving fl


thus be avoided


the computer


Lux then


with that in the separate,


in practice.


time saved


temperature


one-step


is


In fact,


in this two-step


insignificant


approach


as


of direct


evaluation


of the flux and temperature.


A slight


modification


is made in the next two examples:


time the medium


is not subcooled,


the initial


temperature


temperature of the medium


being and 4


the


and


on


3.5.


this


(see examples 3


equal


to the melting







46


Table 3.5


Comparison between true and retrieved conditions for second Stefan-Neumann example solved for boundary flux then temperature


Retrieved


Series Method


N-4


N=5


i i i i


1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00


585.89 715.58 799.99 862.06 909.11 945.07 972.34 992.57
1006.96
1016.46 1021.85 1023.76 1022.76
1019.34 1013.95 1007.01 998.89 989.05 980.54 971.00


620.85
758.46 844.90 905.50 948.70 979.11 999.67
1012.44 1019.03 1020.76 1018.76
1014.02 1007.48 999.99
992.40 985.51 980.13 977.04 977.04 980.93


__ _ __II Na


Incremental Method


1023.06 1007.71 1003.60 1000.88
1000.04 999.68 999.52
999.44 998.11 998.30 998.37
998.44 998.51
998.59
998.66
999.45
999.16 999.11 999.07 997.15


G(S) =� ansn-1
nsl


For N--4,


For N=5,


a,=8001187.956002 a2= -387201.858263 a3= 1581.702473 a,= 148.390536 al=8945600.085189 a2= -541131.047558
2763.120508
324.034498 a,= 3.817008


[________Boundary Conditions, T (K)


Time (sec)


Imposed
True
Condition


1
2
3
4
5
6
7
8
9 10
11 12 13 14 15 16 17 18 19
20







47


in Table one-phase


3.3).


problem.


The Stefan-Neumann


It will


be


prob1 ems


tested


solved


that


thus become


the results


unaffected


series


by the change


of tests


to the one-phase


are made and the results


problem. A are listed


gain,


the


in Tables


and 3.7. Again, The two- step


reinforcing


the one-step solution


solution


results


the recommendation


results


are poorer


made earlier


are good (Table


(Table


3.7),


in the testing


3.6).


further of the


two-phase


problems.


Having problems, at


satisfactorily


ttention


is


completed


now directed


testing


of the Stefan-Neumann


to the solution


of inverse


Stefan


problems


whose


interface


motions


must be met by imposing


time-variant


temperature and


flux conditions.


There are


no exact


solutions


for these


problems,


and the interface


motion


taken


from Choi [5]


who have solved


the regular


(forward)


version


the problems


with great


accuracy.


The interface


position


then used to retrieve


listed results


in Tables


the boundary


3.8 through


for the direct


retrieval


3.11.


conditions


Tables


of the linear


and the results


3.8 and 3.9 give


temperature and


flux conditions



F(t) = 1000 + 5t



G(t) - 6.388034 x 106 + 2.82752 x 105t


while


Tables


3.10 and 3.11 give


the results for


the direct


retrieval


of the linear temperature


are


same 3.6


the


data


are


of


data


are


are the heat


(3.36)


(3.37)


heat flux condi ti ons


and quadratic







48


Table 3.6


Comparison between true and retr for third Stefan-Neumann example temperature


ieved conditions solved for boun


[I ___ Boundary Conditions, T (K)


unposed
True
Condition


Retrieved
.. ... I '


Series Method


N=3


N=4


Incremental Method


a a a a


1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00


1033.83 1030.31 1026.931023.69 1020.59 1017.63
1014.81 1012.13 1009.59 1007.19
1004.93 1002.81 1000.82 998.98 99728 995.71
994.29 993.00 991.86 990.85


1033.12 1028.96 1025.01 1021.29 1017.80
1014.55 1011.55 1008.81 1006.33
1004.13 1002.21 1000.57 999.23 998.20 997.47 997.07 997.00 997.26 997.87 998.82


1037.49 99822 998.58 998.85 999.05 999.20
999.24 999.38
999.44 999.49 999.53 999.59 999.60 999.65 999.65 999.69 999.69 999.73 999.71 999.78


a a a a


N
F(s) = asn-1
nul


For N=3, a1=1037A96339
a27= -3.728416
a*= 6.982696xl02 For N-4, aj=1037A96324
ai -4.474102
a3= 1.005513x10' ak,= 1.324365x103


dary


Time (sec)


1
2
3
4
5
6
7
8
9 10 11
12 13 14 15 16 17 18 19
20


l lNO







49


Table 3.7


Compar i son for fourth flux then t


between true and retrieved conditions Stefan-Neumann example solved for bou ;emperature


ndary


I______BoundaryConditions,_T_(K)


Imposed
True
Condition


Retrieved


Series Method


N-4


ii i i


1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00


906.32 953.87 980.48 997.67 1008.95 1016.02 1019.89 1021.28 1020.73 1018.66
1015.44 101139 1006.79 1001.89 996.94 992.16
987.77 983.98 980.99 979.00


N=5


917.43 96625 991.90
1006.84 1015.10 1018.67 1018.86 1016.61 1012.66 1007.67 1002.18 996.73 991.80 987.87 985.38 984.80 986.56 991.12 998.92
1010.44


J I I a


Incremental Method


1056.39 996.60 1002.26 999.33 999.70 999.45 999.16 999.22 999.09 999.23 999.05 999.32 998.89 999.72 998.27 1001.19
994.04 1014.19 951.97
1162.46


N
na sn-1 nul


For N=4,


For N=5,


a,=2805963.114624 a2= -204888.890714 a3. 2268.882336 4a-= 99.892255
a,=3137161.977178 a2= -286340.839757 a3= 3963.574559 a4= 218.130426
a,-- 1.467676


Time (sec)


1
2
3
4
5
6
7
8
9 10
11 12 13 14 15 16 17
18
19 20


G(s)


mummi







50


Table 3.8


Comparison between true and retrieved conditions for fifth inverse example problem solved for temperature


I____ Boundary Conditions, T (K)


Imposed
True
Condition


iRetrieved


Series Method


N=3


Error


N-4


Err


Incremental Method


I I 'U I I


1005.00 1010.00 1015.00
1020.00 1025.00 1030.00 1035.00
1040.00 1045.00 1050.00 1055.00 1060.00 1065.00 1070.00 1075.00 1080.00 1085.00
1090.00 1095.00 1100.00


1073.05
1064.27 1056.80
1050.64 1045.81 104229 1040.09 1039.21
1039.64 1041.40 1044.47 1048.86 1054.56 1061.59 1069.93 1079.58 1090.56 1102.85 1116.47 1131.39


6.77
5.37 4.11 3.00 2.03
1.19
0.49
-0.07
-0.51
-0.81
-0.99
-1.05
-0.97
-0.78
-0.47
-0.03
0.51 1.17 1.96 2.85


1078.13 1061.16
1048.37 1039.36 1033.71
1031.04 1030.94 1033.01 1036.86
1042.07 1048.26 1055.02 1061.95 1068.65 1074.72 1079.77 1083.38 1085.17
1084.72 1081.65


7.28 5.07
329 1.90 0.85 0.10
-0.39
-0.67
-0.78
-0.75
-0.64
-0.47
-0.29
-0.13
-0.02
-0.02
-0.15
-0.44
-0.94
-1.67


1003.61 1007.83 1012.50 1017.33 1022.27 1027.17
1032.24 1037.26
1042.29 1047.31 1052.35 1057.36 1062.38 1067.38 1072.39 1077.40 1082.37 1087.50 1092.06 1098.80


II Ii ,l II I a a


= ansn-I nul


For N=3, For N-4,


a1=1083.1636095 a2= -10.7640380 a3= 0.6587919
a,=1099.6680063 a2= -23.9563496
2.4866623 a4= -0.0666941


Time (sec)


1
2
3
4
5
6
7
8
9 10 11
12 13
14 15
16 17
18
19 20


Error


-0.13
-0.21
-0.24
-0.26
-0.26
-0.27
-0.26
-0.26
-0.25
-0.25
-0.25
-0.24
-0.24
-0.24
-0.24
-0.23
-0.24
-0.22
-0.26
-0.10


F(s)







51


Table


3.9


Comparison between true and retrieved conditions for sixth inverse example problem solved for heat flux


Boundary Conditions, q (Wlm)


Imposed
True
Condition


Retrieved


Series Method


N=4


Error


N=5


Error


Incremental Method


1 i I I I i


7589730.94 7660418.94 7731106.94 7801794.94 7872482.94 7943170.94 8013858.94 8084546.94 8155234.94 8225922.94
8296610.94 8367298.94 8437986.94 8508674.94 8579362.94 8650050.94 8720738.94 8791426.94 8862114.94 8932802.94


8076200.72 8147001.48
8206716.05 8257180.74
8300231.83 8337705.60 8371438.35 8403266.35 8435025.90 8468553.29 8505684.79
8548256.71 8598105.32 8657066.91 8726977.77 8809674.19 8906992.45 9020768.84 9152839.65 9305041.17


6.41 6.35 6.15 5.83
5.43 4.96
4.46 3.94 3.43 2.94 2.52 2.16 1.89
1.74 1.72
1.84 2.13 2.60 3.28
4.16


7558675.50 7694176.58 7815212.59 7920261.67 8008672.53 8080664.49 8137327.48 8180622.00 8213379.15 8239300.64 8262958.75 8289796.37 8326126.99 8379134.69
8456874.13 8568270.59
8723119.93 8932088.60 9206713.65 9559402.73


-0.40
0.44 1.08 1.52 1.73 1.73
1.54 1.19 0.71 0.16
-0.40
-0.93
-1.33
-0.61
-1.42
-0.94
0.02 1.60 3.88 7.01


7184872.69 7371675.43 7493717.81
7596454.24 7689904.72
7775533.53
7859641.15 7940899.93
8020420.49 8098888.64
8176244.61 8252784.49 8329871.72
8402584.80 8480061.68 8553180.08 8628203.50
8701314.04 8775467.98 8847997.32


a a a a a i m


For N-4,


G (S) =f anbs-'
n s


For N=5,


a,=3397040.84 a2=2247925.15 %=-353114.747 a4= 19587.0623 a,=8734725.30 a2=-3092029.40 a3= 1293821.95 a4=-188030.644 as= 9286.34410


Time (sec)


4.25 4.50 4.75 5.00 5.25 5.50
5.75 6.00 6.25 6.50 6.75 7.00 7.25 7.50 7.75 8.00 8.25 8.50 8.75 9.00


Error


-5.33
-3.77
-3.07
-2.63
-2.31
-2.11
-1.92
-1.77
-1.65
-1.54
-1.45
-1.36
-1.28
-1.25
-1.15
-1.12
-1.06
-1.02
-0.98
-0.95







52


Table


3.10 Comparison between true and retrieved conditions
for seventh inverse example problem solved for
temperature


[IBoundary Conditions, T (K)


Imposed
True
Condition


Retrieved


Series Method


N=3


Error


N-4


Error


Incremental Method


a i i i a i U


972.00 1012.00 1052.00 1092.00 1132.00 1172.00 1212.00 1252.00 1292.00 1332.00 1372.00
1412.00 1452.00 1492.00 1532.00 1572.00 1612.00 1652.00 1692.00 1732.00


976.35 1014.88
1053.63 1092.58
1131.74 1171.11 1210.69 1250.48 1290.48 1330.70 1371.12
1411.75 1452.59 1493.64 1534.91 1576.38 1618.06 1659.95 1702.06
1744.37


0.44 028 0.15 0.05
-0.02
-0.07
-0.10
-0.12
-0.11
-0.09
-0.06
-0.01
0.04 0.11 0.19 0.27 0.37 0.48 0.59 0.71


1020.27 1044.41 1071.70 1101.84 1134.51 1169.40 120612 1244.64 1284.37 1325.09 1366.49 1408.27
1450.12 1491.73 1532.80 1573.01 1612.05 1649.63 1685.42 1719.12


4.97 3.20 1.87 0.90
0.22
-022
-0A7
-0.58
-0.59
-0.51
-0A0
-0.26
-0.12
-0.01
0.05 0.06
0.003
-0.14
-0.39
-0.74


969.93
1006.51 1042.85 1079.43
1116.09 1153.87 1192.15 1230.89 1270.11 1310.73
1348.84 1389.93 1429.95 1470.27 1510.54 1550.77 1590.93 1631.04 1671.39 1710.13


a a a I I a *


F(s) =n ansnnal


For N=3, a,=938.0379452
a2= 38.2157864
a3= 0.1050509 For N--4, a,=999.5873340
a27 18.8517864 a3= 1.8834608 a4- -0.0513597


Error


Time (sec)


1
2
3
4
5
6
7
8
9 10 11
12 13 14 15
16 17
18
19 20


-0.21
-0.54
-0.87
-1.15
-1.40
-1.55
-1.64
-1.69
-1.69
-1.60
-1.69
-1.56
-1.52
-1.46
-1.40
-135
-1.31
-1.27
-1.22
-1.26


WM.j







53


Table


3.11 Comparison between true and retrieved conditions
for eight inverse example problem solved for
heat flux


[______ Boundary Conditions, q (W/m2)


Imposed
True
Condition


Retrieved


Series Method


N-4


Error


N=5


Error


Incremental Method


� I I I I I


3003125.00 3012500.00 3028125.00 3050000.00 3078125.00 3112500.00
3153125.00 3200000.00 3253125.00 3312500.00 3378125.00
3450000.00 3528125.00 3612500.00 3703125.00 3800000.00 3903125.00
4012500.00 4128125.00 4250000.00


3007778.76 3021875.02 3040169.58 3062934.56
3090442.06 3122964.20 3160773.08
3204140.84 3253339.57 3308641.38 3370318.40
3438642.74 3513886.50 3596321.81 3686220.76 3783855.48 3889498.09
4003420.68 4125895.38
4257194.29


0.15 0.31
0.40 0.42
0.40 0.34 0.24 0.13 0.01
-0.12
-0.23
-0.33
-0.40
-0.45
-0.46
-0.42
-0.35
-0.22
-0.05
0.16


3033589.06 3035992.94 3043865.79 3057990.24 3078999.09
3107375.31 3143452.06 3187412.63 3239290.49 3298969.30 3366182.86
3440515.14 3521400.30 3608122.64 3699816.64
3795466.96 3893908.40 3993825.95 4093754.75 4192080.13


1.01 0.78 052 0.26
0.02
-0.16
-0.31
-0.39
-0.43
-0.41
-0.35
-0.27
-0.19
-0.12
-0.08
-0.12
-0.23
-0.46
-0.83
-1.36


II 1-. L - I II


2995616.25
3004320.50 3019590.53 3036905.38 3061522.20 3092582.42
3130006.55 3173845.90
3223966.45 3280396.20
3343088.04 3411906.45 3487162.49 3568166.70 3656322.04 3748732.19 3851068.35 3951711.24 4084808.11 4096803.88


For N-4, a,=2997608.684
a2=33190.75084 a3=29232.59062 a4=2902.536744 For N=5, a,=3035721.67
a2=- 162102870 a3= 27733.4034 a4=12343.6674 a,=-1598.21379


Time (sec)


0.25
0.50 0.75
1.00 1.25 1.50
1.75 2.00
2.25 2.50 2.75
3.00 3.25
3.50 3.75 4.00
425
4.50 4.75 5.00


Error


-0.25
-0.27
-0.28
-0.43
-0.54
-0.64
-0.73
-0.82
-0.90
-0.97
-1.04
-1.10
-1.16
-1.23
-126
-1.35
-1.33
-1.51
-1.05
-3.60


nl


antsn-1









F(t) = 932 + 40t


G(t) = 3 x 106


+5 x 104t2


In these following


tables,


the computational


errors


are calculated


using


definition:


Error=1 q


where


pand


q represent


retrieved


and imposed


true conditions,


respectively.


As described


in Table


3.3,


those


in Tables


3.8 and 3.11


for the medium


initially


at phase


change


temperature,


while


those


Tables


3.9 and 3.10


are for the medium


initially


subcooled


at 300 K.


The former


are thus


one phase


problems


while


the latter


are two-


phase


problems.


Again


both series


and incremental


solution


methods


are used for solution


Table


3.8,


N that is


the series


as low


and their


solution


as 3,


whereas


results


results


are good.


converge


in seeking


For example


even with


a value


the flux condition


in of in


Table degree


3.9,


the series


results


not shown).


converges


rapidly


In Tables


from N=4 to N=5 (higher


3.8 and 3.11 the medium


melts


as soon as the boundary


conditions


are imposed,


whereas


in Tables


3.9 the medium


all


cases,


the


starts


accuracy


to melt at time greater


of the results


appears


than 4 seconds. to be unaffected


the time when melting


3.11 follow


the


takes


same trends


place.

as the


The results ones in Table


in Tables 3.10 and 3.8 and 3.9. Tests


conditions are thus successful.


54


(3.38)


(3.39)


the


(3.40)


are


in


In by


for the time variant







55


The inverse


solution


techniques


developed


in this study


expected


to be accurate


as mentioned


earlier


that is close


to the


end of the previous


this work, than small


the


time.


section.


accuracy


This


Yet for the Stefan


of the techniques can be attributed


problems


is better


solve


at large


to the curvature


d in time


of the


interface


position


curve,


which


is always


large


at small


time [2];


see Figure


will be position


3.2.


a slight at small


Then, error time.


in the numerical


associated


At large


solution


of (3.27),


with the linearization


time, however,


the position


there


of the


curve


tends


fact, as (3.27) r boundary present


to be linear;


time apidly


the linearization error will be


progresses, outnumber


conditions


method


can


at large


the accurate terms


under


the inaccurate terms to always be evaluated


time;


see the results


diminished. the summation


the effect accurately


in Tables


In in


that the

with the


3.4 through


3.11.

marching


This is schemes


a distinct reported


departure


from the trends


in the literature


in which


of other


the


errors


with time.


It should


also be pointed


out that,


for the


Stefan-Neumann


problem


chosen


for compar i so'n


in the present


study,


there


is


a singularity


of the temperature


at zero time.


This also


contributes


to the large


discrepancy


of the results


at small


time,


which


must not be overlooked.


It has been firmly


established


that,


in the solution


of the


Stefan


problems,


only


the Stefan-Neumann


problems


can be solved


exactly.


Yet,


it is also possible


to develop


an exact solution


an exponential


condition


imposed


on the boundary;


such condition,


however,


has


been


considered


as physically


untenable


yet, such a condition


[7].


are


to


grow


time tend


for


in


the


Worse


literature


- - - - %F


gives rise to a







56


constant


velocity


linearization


scheme


of the interface,


exact


a situation


in the solution


making


of the inver


the present se problems


Perfect


results


will be obtained,


rendering


the test of the


exponential


conditions


meaningless.


3.7 Extension


and Concluding


Remarks


The analysis developed for the solution of inverse


problems,


times


for re-melt


in this chapter


problems


can be


with multiple


and re-freeze


readily phases.


of the medium


extended For such must be


closely adapted


accounted


for,


for the development


and the analyses


[4,65]


of the inverse


can be readily


solution


techniques


presented


in this study.


for the solution


unequal double


of inver


for different


source


of the regular


The present

-se problems


phases


and sink fronts


problems


analysis in which


of the medium.


must be used


in reference


17.


can also be extended


the properties


are


For such problems,


as given Finally,


in the solution


it is noted


that


although


problems


in one-dimensional,


semi-infinite


doma in


have been


solved plane


for examples


wall)


in this work,


can also be solved


prob 1 ems


in finite


with the present


domains


methods.


(e.g.,


For these


problems, imposed. requires


there


are two boundaries


Two unknowns


the input


of


and two boundary


are thus sought


one additional


conditions


simultaneously,


condi ti on


are


and this


in the form of either


temperature or heat


flux at any


interior


point


close


to the boundary


where


no phase


with two phase simultaneously


change


change


takes


place.


interfaces


from both sides


On the caused


other hand, by separate


of the boundaries,


motion data for the second


for problems heat input


the interface this additional


[5].


interface will serve as







57


condition.


sides


In


any event,


of the interfaces.


be applied


to the soluti4


no flux information


Furthermore, th on of problems


i s needed


e present method in multiple di


at both can also


dimensions.


Again


the inverse solution


for these problems


can be developed on


the basis of the solution


problems.


of regular versions


of these
















CHAPTER IV
SOLUTION OF ABLATION PROBLEMS WITH ONE MOVING BOUNDARY
BY A SOURCE-AND-SINK METHOD


It is the


method boundary


purpose


for the soluti4

As will be sh


of this chapter to on of ablation p own, the essential


,present )roblems feature


a source-and-sink


with


one moving


of this method


is


that the solution


will be sought


in


a fixed


domain


which


does not


change


with time.


In fact,


the ablated


region


is to be treated


fictitious boundary boundary.


domain

will match


Then,


in which


the flux condition


that of the imposed


with the additional


conditior


vaporization


at the ablated

i at the moving


temperature


given


for the moving


boundary,


the conditions


at this boundary


overspecified


for the fictitious


domain.


The problem


can thus be


taken fixed


as an inverse


boundary


problem


is sought


in the


during


sense


the ablation


that the condition


period.


on the


This condition


will provide the solution


for the


necessary


conditions


at the moving


boundary


of the problem.


4.1 Solution


Methodology


a subcooled


medium,


the problem


can be divided


into two


stages;


namely,


pre-ablation


stage


and ablation


stage.


During


pre-ablation


order


stage,


to raise


heat


is added


its temperature


to the surface of to the phase-change


the medium


in


temperature.


Then with continuous


heating,


ablation


takes


place.


The ablated


58


as a


are


for


For


the







59


surface


moves


this boundary


As shown homogeneous an


inward motion.


with time and the imposed


in Figure


4.1,


nd isotropic.


it is assumed The thermophysic


heat flux follows


that the medium cal properties


constant.


For


the ablation


problem,


the ablated


region


immediately


surface


removed


phenomenon.


upon


formation.


Radiation


It is also assumed


is taken


that the medium


to be a changes


phase medium.


at a distinct

Moreover, I


temperature;


there


is


that is,


no volumetric


no mushy zone in the heat generation in the


medium.


Under


these


assumptions,


the ablation


problem


can be


formulated Pre-ablatic


as follows:



)n Stage


Governing


equation--


FE (AUS)


V 2T0(Ft) -1


dT0(ft)
0t


t0>t>0


Initial


T0(?,O) =Ti(f)


Boundary



G(t)= -k


condition--


(4.2)


condition-0T0(f0 ,t)
Oni'


(4.3)


is are


is


(4.1)

























Figure 4.1


System analyzed








61


Vn


m
r


ni


r


oa SI
Si

SI to
S









Ablation


Stage


Solid


Region:


Governing





V 2Ts(ft) -


equation--


1ES


~1


OTs(ft)
at '


(4.4)


t>to


Initial condition-Ts(fto) =T0(ft0) Boundary conditions--


Ts(fs,t) =TV


G(t) -- kTs(is,t)
G-t)=n


+pL V.


where


all notations


have their


usual


meaning;


see nomenclature.


rS represents


the position


for the ablated


surface,


which


is unknown


for the ablation


TS(F,to) is of ablation.


problem;


to


is the time when ablation


the temperature distribution


Both to and To(Fto)


in the medium


are to be found


starts;


and


at the onset


from the solution


of pre-ablation


equation


temperature


(4.7),


stage. which


gradient,


Notice relates


is the only


that the boundary


the


ablation


source


flux condition,


velocity


of nonlinearity


to


in


ablation


62


(4.5)


(4.6)


(4.7)


Here


the the


problem.







63


The ablation


problem


above


is solved


by


using


a source-and-


method.


In this method,


the domain


for investigation


extended


to


cover


AuS;


the ablation


front


becomes


an interior


phase-


change which


interface. the ablated


Solution


region


is thus rendered


is fictitious


in


in the


a fixed


sense


domain,


in


that it is


physically


nonexistent;


however,


the unablated


(solid)


region


real.


The conditions


imposed


at the


moving


boundary


in the original


ablation


problem


phase-change


are taken


interface.


to be those


Thus,


for the


imposed on new problem


* the interior in the fixed


domain,


the interface


position


and the fictitious


condition


on the


fixed found,


boundary


they


are the unknowns


can be used for input


to be determined.


Once they


in the determination


are


of the


temperature totally. A


in the solid


,n equivalent


region.


problem


The problem


is formulated


with


is thus solved a source-and-sink


method


as follows:


Equivalent


Problem


Governing


equation--


FE (A US)


V 2T(ft) -


Initial


OT(ft)
at


pL
Fk~,"~


condition--


(4.9)


T(fr t0) = T0(ft0)


Interface conditions--


sink


is


is


t>to


(4.8)


-- S) 1











T(?s,t) =TV


G(t) k - OT(rs't)
On


+pLV.i


where


denotes


a Dirac


delta


function.


As has been shown


Chapter across


3, equation (4.8) r the interface at FS


'educes t enables


o (4.4). the G(t)


Also integrating


to represent


(4.8)


the heat


flux imposed


Equation


on the interface


(4.8)


at the fictitious


can be solved


by using


side. (3.18)


in which


domains


for integration


for the first


and second terms on


the right


are changed


to AUS.


Also minus


sign


is taken


for the Dirac


delta


function


term in which


rf


is changed


to FS


G(ft I ?',r)Vii6(?' - Fs)dV'dr +


This


temperature


temperature


equation


equation


(4.10)


is forced by setting


to satisfy


the


F to the ablated


interface


surface


position,


rs, and T(Fs,t)


to the phase-change


temperature,


G(Ys,t I f',t0)Ti(?')dV'


AuS


t
Lff


G(fs,t I ',r)V. 6(?'


-Ys)dV'dr + { }


toAuS


is also differentiated to satisfy


64


(4.10)


(4.11)


in


the


T(ft)


fS AuS


as


toAuS


(4.12)


as


(4.13)


6 (-F - -FS)


G(fLt I f',t0)Ti(f')dV'


TV-=


(4.12)


Equation


as


t
Lff







65


- ______I __ - J aG (F?t 1 F117)
G(t)= - k ArfStan �Ti(')dV'-TL S an V.n(


ai4


- Ys)dV'dr


} +pLvii


r=rS


where


differentiation


is to be effected


at the solid


side of the


interface. summation


Notice


term,


that the boundary


which,


according


condition


to Table


is embodied


3.1, should


in the


be expressed


in the form


a G( ,t i /T)OT(rt)
f On
0o


where


the following


condition


i s taken


at the fixed


boundary:


aT(fot)
Oni


- G(?ft)
k


for t

and


- (f ,t)
k


for t > to


where there


g(Fot)


represents


are two unknowns


the imposed


in (4.13)


fictitious


and (4.14):


heat flux.


g(o ,t)


and V;


Clearly,


there


are


two equations


be found


to solve


them.


from the solution


of


In this effort, a pre-ablation


and TO(-Ft0)


problem


are to


as described


previously.


4.2 Illustrative


Examples


Use is


now made of the solution


methodology


given


in the


preceding examples


section


to solve


will be provided


examples

and they


in semi-infinite


include


medium


med i um.


with


Four


or without


subcooling


with the boundary


imposed


with constant or variable


flux conditions.


(4.14)


dfodr


(4.15)


(4.16)


heat







66


For


been given


a subcooled in Chapter


medi un,


3�


the pre-ablation


The pre-ablation


stage


temperature


solution


has


can be taken


from (3.22) as


T0(x,t) =


T-if
0


G(s) exp[
- s)1/2xp


x2
-4-xs]ds
4a(t-S)


Thus the time when ablation


starts


(to)


can be found


by solving


following


equation


to


0


implicitly


ds


Also at the moment


when ablation


starts,


the temperature


in the


solid


region


is given


by


the relation


T0(x,t0) =Ti


(4.19)


exp[- x ]ds
4a(to - S)]d


which


serves


as the initial


condition


for the ablation


stage.


For the problem


at hand,


(4.12)


can be used to derive


temperature


T(x,t) = Ti +


to fa


t

kr t0


exp[ -


x ]dr
4a(t-r)


exp[- 4a(t - ]dr


L






Also (4.13)


t

"--to


dRl(7)


1 {exp( - (x - R(0))2)+ exp( (x+R(r))2 4ra(t- r) p- 4oat(t -t)- r) (4.20)


is used to derive


(4.17)


the


(4.18)


as


TV = Ti


k 4 2'r









R 2(t) exp[ 1)
4a(t - 7


67
t

44t0


exp- R-(t)]d4a(t- r)]d


dRl(r)
dr


1
4ira(t - r){exp( -


(Rl(t) - R,(r))2


4a(t --r)


) + exp( -


(Rl(t)+RI(r)


2


4a(t- ,r)


Equation


(4.14)


follows


_ Rl(t) G(t)- 2---


R1(t)
exp[- 4( )


Rl(t) ]dr + RFf( j
r=ta


g(r) exp[
(t-r)3/2
)


R2(t) 4a(t-- r)


dR(-)
d7r


(t-)312(R


R1(r))exp( -


(Rl(t) -


R1(.r))2


+ (Rl(t)+Rl(r))exp(-


(Rl(t)+l(r))2 1
4a(t -r)


dRl(t)
+ pL dt


Solution


of (4.21) and (4.22)


subject


to the initial


condition,


to) = 0,


yields


the results


for two unknown


functions


Rl(t)


g(t).


4.3 Numerical


Solution


of Ablated


Front


Equations (

integrodi fferential


4.21)


and


equations,


(4.22)


which


are


coupled


nonlinear


this effort,


a local


linearization


scheme


is employed as described


in Section


3.4.


For the present


problem,


the entire


time


range


divided treated


into small


increments


as constants.


They


in which


both


dRl(t)
dt


can thus be taken


and g(t)


are


out of their


respective


integrals,


and


the


convolution


integrals


in


these


are changed to summations


G(r)


27r


t

-=t


I


as


(4.21)


t

:ft


]d'r


4a(t - r)


(4.22)


and


will be solved


numerically.


In


is


TV = Ti+


Lp
4-fr


equations


as







68


G(7)


tn
g(tn) J
r=tn_1


exp[- R2(tN) ]dr
4a(tN - 7)


exp[ -


dR1 (tn)
dt


1
NJ4wa(tN - 'r)


exp( -


(RI(tN)


- Rl(r))2


4a(tN -7-)


) +exp(


_((tNR-Rr ))2)' '
4a(tg 7


(4.23)


and


to
Rl(tN) G(r) 2" ra JIo( tN -T)3/2


4a(tN-) ]dr


N
+E g(tn) 11=1


Lp N1


tn
dR(tn)
dt
T=tn-1


1
(tN - 7)3/2


{(RI (tN)


- R1(r))exp( -


(Rl(tN)


-2


4a(tN--)


+ (Rl(tN)+Rl(r))exp(-


(Rl(tN)+Rlr))2L dRl(tN)
4a(tN - 7) )jr + pL dt


Notice


that the removal


of


dRl(t)
dt


and g(t)


will


cause


a slight


error


in the numerical


is


a gradual


L solution function o


of the ablated


)f t,


and


front


posi tion.


so is g(t) during


However,


the early


of ablation.


expected


The


to be small.


error


associated


In practice,


with the linearization


such


an error


can always


reduced by taking small


Tv = Ti + a


N
n=l


L


N
n=l


]dr


G(tN)


exp[-


]dr}


(4.24)


Ri


stage


is be


+1
k4ir


time increments.







69


It is also noted


that the. ablated


position


Rl(t)


in the


Green'Is f

velocity,


Function can
dRl(tn-1)
dt


time t should


and g(t)


be relate as shown


thus be discounted


in these


equations


to R(tn-1) by


in Figure


3.2.


as unknown.


will be solved


means of the The position Numerically,


incrementally


interface


R1


at any


the Rl(t) starting


from to


until


the final


time tN


is reached.


This corresponds


time marching developed for


scheme


in the numerical


the solution


solution.


of the ablation


Computer


problems


given


programs in this


chapter


are compiled


in Appendix


4.4 Results


and Discussion


Four examples


are tested


and they


include


semi-infinite


med i um


with and without


constant


subcooling


and time-variant


with the moving


conditions


(see


boundary


Table


imposed


with


4.1 for summary).


Notice


that,


in this table,


under


the column


of problem


description,


the ablated


medium in heat-flux


region


itially


condition,


is always


at phase-change


such


counted


as one phase.


temperature


as Example


1, there


imposed


Thus,


with


for


a constant


is one ablated


phase.


It is thus called


one-phase


prob 1 em


in the table.


Example


1 deals


with


an ablation


prob 1 em


that


can be solved


exactly medium


As shown


ablated


with


in Carslaw a constant


and Jaeger


velocity


[71,


for


a semi-infinite


, the heat flux imposed


on


the boundary


is given


by


the relation


G-[L + c(Tv -Ti) ]pU


where


is constant


and the velocity


is related


to the ablation


position


to a


f


(4.25)


as







70


Table 4.1


Conditions tested in four examples


Problem Material Heat Flux Remarks
Description Condition Imposed G(t)_(W/m2);t_(S)

1. One-phase Aluminum G(t)=5xl0 Exact solution is available for this
Ti=932 K=Tn-=Tv case.

2. Two-phase Aluminum G(t)=2x106 No exact solution for this case;
Ti=882 K literature.

3. Two-phase Aluminum G(t)=8xl0O t No exact solution for this case;
Tj=882 K
4. Two-phase Aluminum G(t)=2xl0 t2 No exact solution for this case;
Ti=882 K






71


Rl(t)
t


It is expected


that,


for


a medium


initially


at the phase-change


temperature,


C is equal


to pLU product.


It is also noted


equation


state.

given


as


(4.25)


is strictly


At that time,


valid


the temperature


for the medium


in


in the solid


a quasi-steady is stabilized


u
W(X - Ut)


(4.27)


which time.


occurs


The first


for the medium


two examples


having


are


ablated


thus chosen


for


a long period


to substantiate


of


points.


For the first


example,


the ablation


is due to melting.


med i um

ablates surface


is initially


at the melting


as soon as the heat is applied.


is shown


in Figure


4.2,


where


temperature


so that the surface


The position


the


curve


of the ablated


appears


linear error


Using


curves


the exact


shown


solution


4.3.


(4.25) for comparison Clearly, the results


yields the are stable


converge


satisfactorily,


errors


being


less than


one tenth


percent slightly


for all the time increments


larger


at


smallI


time,


tested.


a result


Also the of the


errors


are


singularity


associated


with the heat flux that is abruptly


applied


on the


surface


at time


zero.


In what


follows


in the numerical


computation


of all the examples


to be presented


in this chapter,


the time


increment


i s taken


as 0.5


sec,


unless


otherwise


noted.


for the algorithms


(4.26)


that


T(x,t)


= Tve


these


The


and


in Figure


to be


of


one


an excellent test


Example 1 provides

























Figure 4.2


Trend o med i um imposed


f
i


ablated surface position nitially at phase-change with a constant heat-flux


for a temperature condition








(s) I sewi


OL


GL,


0


000


CA
rm




0 c
C,)
0

"3
0
rme ~0

0
3,


two 9*0 9"0


O'L


oz























Figure 4.3


Accuracy, stability and convergency test of the SSM in the solution of ablation for a medium initially at phase-change temperature imposed with a constant heat-flux condition







(s) ' *wl.L


0


LO'O


OVO 003L


9L


01


9


0


Oz







76


developed


for


the solution


of the ablation


problems


dissertation.


In fact,


the good


results


obtained


in the first


examplI e


are somewhat


expected


because


of the linearization


used in


the solution


of the integrodifferential


equations


for which


ablation


rate itself is


linear.


ExamplI e


initially solution


2 is different


subcooled


for checking


(see


Table


because


4.1).


the results,


it deals


with


This time,


which


a medium there is


which


is


no exact


have also been tested


for


convergence


in Figure


4.4.


Of interest


in this figure


is the


slightly


smaller


ablation


rate at small


time,


which


attributed


to the large


side of the moving


slope


boundary


of the temperature curve at


(not


shown


in dissertation).


the solid At large


time,


this slope


diminishes,


and finally


it


is stabilized


evidenced


by


the linearity


of the position


curve


at large


time.


Using yields


the numerical


an ablation


data


over the last time step in Figure


rate of 0.00161


m/s,


which


4.4


is in good agreement


with that computed


This


gives


using


a good


equation indication


(4.25),


error


being


of the ablation


less than 2.5


approaching


the


quasi-steady


state.


An effort


is also made to check


the results


of Example


2 with


the analytical


solutions


reported


in the literature.


As shown


Figure


4.5,


both the heat integral


method


(HIM)


of Vallerani


and the moment


method


(MOM)


of Zien [41] yield


results


that


are in


close agr results a the plot.


reement


with the present


Llso fall right


Example


on the


2 has thus been


source-and-sink


curve


method.


and have thus been omitted


tested


successfully.


with


3 and 4 deal with a


in


this


the


can


be


as


in


[40]


in


subcool1ed med ium


imposed


Examples
























Figure 4.4


Stability and convergency test of the SSM in the solution of ablation for a subcooled medium imposed with a constant heat-flux condition







Ablated Surface Position, R1 (cm) 0 0 CA o - I0cn b



001
0o oo Fob0





0























Figure 4.5


Comparison of two methods in medium imposed condition


ablated surface the literature with a constant


position with for a subcooled heat-flux








Ablated Surface Position, R, (cm)

0. 0- - I 3 Ii 0 CA 0 Ci 0 CA0 00 >


X i



. 0 "/)
ic



















0 I~CA









UC


08







81


time-variant heat-flux


Example


3,


Like Example


phases


while


2,


in the medium


conditions.


a quadratic


the medium


A linear


heat-flux


is subcooled


for both examples.


heat-flux


is imposed initially; Comparison


is imposed in Example


there


in

4.


are two


of the present


solution

is given


with the moment


in Figure


4.6;


method


for the linear


convergence test based


heat-flux


on the change


condition


of time


increments


for this condition


in given


in Figure


4.7.


As shown


both figures,


the results


for the source-and-sink


method


are in


close


agreement


with the moment


method,


and the results


converged with a seen in Figures


time step


as large


as 0.5


sec.


4.8 and 4.9 for the case of


Similar results are


quadratic


heat-flux.


The tests


for ablation problems with one moving


boundary have thus


satisfactorily.


in


have


been completed























Figure 4.6


Comparison of ablated surface position with the moment method in the literature for a subcooled medium imposed with a linear heat
-flux condition








Ablated Surface Position, R1 (cm)

o 0 0 0 _b



0t>
xrcn
0 C








CA, Ag
3



00
i'
040






0" cc
%MOO









-
























Figure 4.7


Stability and convergency test of the SSM in the solution of ablation for a subcooled medium imposed with a linear heat-flux condition







(s) 1 lBWI.L


OL


0"0


g9


CA
0
0

0
-9
0


9'0 6"0 ZI g'l, 9"3 V':


C)


9z oz























Figure 4.8


Comparison of ablated surface position with the moment method in the literature for a subcooled medium imposed with a quadratic heat-flux condition








Ablated Surface Position, R, (cm)


o 0A 0C O h3 CA I0i





0 "







301






,



0 CA'























Figure 4.9


Stability and convergency test of the SSM in the solution of ablation for a subcooled medium imposed with a quadratic heat-flux condition




Full Text

PAGE 1

SOURCE-AND-SINK METHOD OF SOLUTION OF MOVING BOUNDARY PROBLEMS BY MEHDI AKBARI A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA 1993

PAGE 2

TO MY WIFE VAHIDEH, AND DAUGHTERS SARA AND MONA

PAGE 3

ACKNOWLEDGEMENTS I am most grateful to the chairman of my supervisory committee, Dr. C. K. Hsieh, who provided the inspiration for this study as well as many hours of guidance and encouragement. His enthusiasm has been contagious, and I hope to emulate his attitude in the future. I wish to express my sincere thanks to Drs . R. A. Gater, G. G. Emch , R. Abbaschian, and H. A. Ingley whose unselfish good counsel and understanding have been essential in this study. Last but by no means least, I would like to thank my wife, Vahideh, for her love and spiritual support during this period of time of work on the dissertation. m

PAGE 4

TABLE OF CONTENTS Page ACKNOWLEDGEMENT iii LIST OF TABLES vi LIST OF FIGURES • • VI 1 NOMENCLATURE ix ABSTRACT xi CHAPTERS I INTRODUCTION 1 II LITERATURE REVIEW 8 III SOLUTION OF INVERSE STEFAN PROBLEMS BY A SOURCE-AND-SINK METHOD 21 3.1 Motivation 21 3.2 General Analysis 25 Pre-melt Stage 28 Melting Stage 28 3.3 Example Problems 31 3.4 Numerical Solution 33 3.5 Critique of The Method 39 3.6 Results and Discussion 40 3.7 Extension and Concluding Remarks 56 IV SOLUTION OF ABLATION PROBLEMS WITH ONE MOVING BOUNDARY BY A SOURCE-AND-SINK METHOD 58 4.1 Solution Methodology 58 Pre-ablation Stage 59 Ablation Stage 62 Equivalent Problem 63 4.2 Illustrative Examples 65 4.3 Numerical Solution of Ablated Front 67 4.4 Results and Discussion 69 V SOLUTION OF ABLATION PROBLEMS WITH TWO MOVING BOUNDARIES BY A SOURCE-AND-SINK METHOD 90 IV

PAGE 5

Page 5.1 General Analysis 91 Pre-melt Stage 91 Melting Stage 94 Ablation Stage 95 Equivalent Problem 98 5.2 Examples 101 5.3 Numerical Solution of The Ablation Problem 104 5.4 Numerical Solution of Temperature and Energy Storage 109 5.5 Numerical Examples 110 VI DISCUSSION AND RESULTS 130 APPENDICES A EXTENSION OF THE SSM 135 B SSM FORTRAN PROGRAM FOR INVERSE STEFAN PROBLEM 136 C SSM FORTRAN PROGRAM FOR INVERSE STEFAN PROBLEM 147 D SSM FORTRAN PROGRAM FOR INVERSE STEFAN PROBLEM 154 E SSM FORTRAN PROGRAM FOR INVERSE STEFAN PROBLEM 165 F SSM FORTRAN PROGRAM FOR ABLATION PROBLEM 172 G SSM FORTRAN PROGRAM FOR COMBINATION PROBLEM 186 REFERENCES 223 BIOGRAPHICAL SKETCH 229 V

PAGE 6

LIST OF TABLES Table Page 3.1 Expressions to account for effects of boundary conditions 23 3.2 Properties of aluminum 41 3.3 Conditions tested in eight examples 42 3.4 Comparison between true and retrieved conditions for first Stefan-Neumann example solved for boundary temperature 43 3.5 Comparison between true and retrieved conditions for second Stefan-Neumann example solved for boundary flux then temperature 46 3.6 Comparison between true and retrieved conditions for third Stefan-Neumann example solved for boundary temperature 48 3.7 Comparison between true and retrieved conditions for fourth Stefan-Neumann example solved for boundary flux then temperature 49 3.8 Comparison between true and retrieved conditions for fifth inverse example problem solved for temperature 50 3.9 Comparison between true and retrieved conditions for sixth inverse example problem solved for heat flux 51 3.10 Comparison between true and retrieved conditions for seventh inverse example problem solved for temperature 52 3.11 Comparison between true and retrieved conditions for eight inverse example problem solved for heat flux 53 4.1 Conditions tested in four examples 70 5.1 Conditions tested in three examples Ill vi

PAGE 7

LIST OF FIGURES Figure P&ge 3.1 System analyzed 27 3.2 Linearization of solid-liquid interface position curves for the numerical solution 35 4.1 System analyzed 61 4.2 Trend of ablated surface position for a medium initially at phase-change temperature imposed with a constant heat-flux condition 73 4.3 Accuracy, stability and convergency test of the SSM in the solution of ablation for a medium initially at phase-change temperature imposed with a constant heat-flux condition 75 4.4 Stability and convergency test of the SSM in the solution of ablation for a subcooled medium imposed with a constant heat-flux condition 78 4.5 Comparison of ablated surface position with two methods in the literature for a subcooled medium imposed with a constant heat-flux condition 80 4.6 Comparison of ablated surface position with the moment method in the literature for a subcooled medium imposed with a linear heat -flux condition 83 4.7 Stability and convergency test of the SSM in the solution of ablation for a subcooled medium imposed with a linear heat-flux condition 85 4.8 Comparison of ablated surface position with the moment method in the literature for a subcooled medium imposed with a quadratic heat-flux condition 87 4.9 Stability and convergency test of the SSM in the solution of ablation for a subcooled medium imposed with a quadratic heat-flux condition 89 VI 1

PAGE 8

Figure Page 5.1 System analyzed 93 5.2 Linearization of solid-liquid interface and ablated surface position curves for the numerical solution 106 5.3 Trends of solid-liquid interface and ablated surface positions for a combination problem of subcooled medium imposed with a constant heat-flux condition 114 5.4 Trends of solid-liquid interface and ablated surface positions for a combination problem of subcooled medium imposed with a linear heat-flux condition 116 5.5 Trends of solid-liquid interface and ablated surface positions for a combination problem of subcooled medium imposed with a quadratic heat-flux condition 118 5.6 Temperature distributions in the medium at different times during ablation for a combination problem of subcooled medium imposed with a constant heat-flux condition 120 5.7 Stability and convergency test of the SSM in the solution of ablation for a combination problem of subcooled medium imposed with a constant heat-flux condition 122 5.8 Stability and convergency test of the SSM in the solution of ablation for a combination problem of subcooled medium imposed with a linear heat-flux condition 124 5.9 Stability and convergency test of the SSM in the solution of ablation for a combination problem of subcooled medium imposed with a quadratic heat-flux condition 126 5.10 Overall accuracy test of the SSM in the solution of the combination problem of subcooled medium imposed with a constant heat-flux condition 129 VI 1 1

PAGE 9

NOMENCLATURE c E F G r Ste s T t v X a 6 Specific heat Equation Temperature function used for boundary condition Flux function used for boundary condition, GreenÂ’s function, elements in coefficient matrix Heat flux Thermal conductivity Latent heat of fusion Latent heat of vaporization Normal unit vector Distance from sink or source to sense point Interface positions Stefan number Dummy variable for integration, interface speed Temperature, temperature vector Time velocity of the interface Position, source point Thermal diffusivity Dirac delta function, Kronecker delta IX

PAGE 10

p Density r Dummy variable for time Subscripts and Superscripts m,v Phase change N 19 N 2 Upper time limit index n Time index t Transition 1,2 Melt and vapor fronts 0 Initial time x

PAGE 11

Abstract of Dissertation Presented to the Graduate School of the University of Florida in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy SOURCE-AND-SINK METHOD OF SOLUTION OF MOVING BOUNDARY PROBLEMS By Mehdi Akbari May 1993 Chairman: Dr. Chung K. Hsieh Major Department: Mechanical Engineering Source-and-s ink methods have been developed for the solution of inverse diffusion problems, ablation problems, and combination of ablation and Stefan problems. GreenÂ’s functions have been used and integrodif f erential equations are derived for the interface positions and the temperature in the phase-change medium. These equations are solved by using local linearization of the interface position and the boundary heat flux if it were treated as unknown. The results have shown to be accurate, convergent, and stable. The methods developed for the solution of the inverse diffusion problems have been used to find the boundary conditions for the inverse Stefan problems. They are solved by two approaches: a series solution approach and a time incremental approach. Both have shown to be useful to find the boundary conditions without reliance on the flux information to be supplied at both sides of the phase change interface. The methods are efficient in that they require fewer equations to be solved for the xi

PAGE 12

unknowns. The algorithms can also be easily developed for the solution of the problems. In the use of the source-and-sink method to analyze the ablation and combination problems, heat transfer is solved in a fixed domain so that the original heat-flux condition that is imposed on the moving boundary is taken to be the condition imposed on the solid-vapor or liquid-vapor interface. This flux condition is then used together with the temperature at the interface to solve for the interface position as well as the condition imposed on the fixed boundary. Finally, these positions and conditions are used in the temperature equation to complete the solution. The method has been used successfully in solving examples encompassing one-, two-, and three-phase ablation with the boundary of the medium imposed with constant, linear, and quadratic flux conditions. Numerical solutions for these examples as well as the trends of the moving boundary positions are discussed in detail. xi 1

PAGE 13

CHAPTER I INTRODUCTION Phase changes occurs when a body is exposed to a large heat flux so that it melts or vaporizes at the surface. With continuous heating, the melting or vaporization front moves inward with time. Since the position of the phase-change interface is unknown a priori, this type of problem is commonly known as the moving boundary problem in the heat transfer literature. Phase change problems find applications in thermal energy storage, material treatment and processing, thermal protection of re-entry vehicles in space technology, laser drilling and cutting in manufacturing, freezing and thawing in soils and foodstuff, and photocoagulation in opthalmological procedures, among others. It is difficult to solve a phase-change problem because of the presence of the nonlinear condition at the interface. Usually for such a problem, the number of phases encountered in the medium depends on the ambient pressure and the temperature range that appears in the medium. Thus for a medium heated or cooled below the triple point of the substance, only sublimation and fusion may take place. On the other hand, for phase change above the triple point, solid, liquid, and gas states may appear simultaneously, and this occurs for a medium of low thermal conductivity and exposed to a large heat flux. In practice, the changed phase may be removed as 1

PAGE 14

2 soon as it is formed. This occurs in the re-entry where the melted substance is removed by a fluid dynamic force. This material removal may also be effected naturally such as in sublimation, where the sublimed vapor is dissipated by diffusion to the surroundings. In either event, the imposed heat flux follows the boundary motion, and such problems are commonly referred to as ablation problems in the literature. A different state of affairs is encountered when the changed phase remains stationary and continues to occupy the space taken by its previous state. Thus for example in melting a solid, if the melt is not removed by an external force, it will adhere to the solid from which it changes phase. Then in the absence of convection in the melt, the heat transfer in the medium can be analyzed by solving a Stefan problem in recognition of Stefan who first presented a formal solution of the problem in the open literature (see next chapter for a survey of the literature). The Stefan problem is basically different from the ablation problem in that, in the former, the problem is solved in a fixed domain, and the boundary condition is always imposed on the fixed boundary whereas, in the latter, the imposed condition follows the boundary motion. Thus in the ablation problem, the domain for analysis diminishes in size with time, and the imposed condition is acting on the receding boundary due to the shrinkage of the medium. There have been numerous efforts in the literature for the solution of the Stefan problems. Much less research, however, has been devoted to the solution of the ablation problems. Also for those ablation problems that have been solved and documented in the literature,

PAGE 15

3 most are for the analysis of generic ablation in which only one moving boundary appears in the medium. There has been a lack of study for problems involving two moving boundaries such as those found at the solid-liquid and liquid-vapor interfaces. They are encountered when the solid, liquid, and gas states appear simultaneously in the medium. Three phase-change problems will be solved in this dissertation. In the first problem, an inverse Stefan problem will be solved. This problem differs from a conventional Stefan problem in the sense that, for the conventional problem, the condition imposed on the boundary is fully specified. This condition is then used together with others to solve for interface motion as well as the temperature distribution in the medium. For the inverse problem, however, the interface motion is given, and this motion is used as a part of the input to solve for the boundary condition that is necessary to provide for this interface motion. This problem is found in material treatment and processing, where the property of the material may be a function of the phase-change rate at the interface. Solution of the problem thus enables a better control of the property of the material . It should be noted that the inverse Stefan problem described above has recently been solved by Zabaras et al . [1] in the literature. However, the method developed for solution in this dissertation is more efficient because it requires no information for the heat flux at the interface. As will be shown, two methods are developed for the solution of the problem: one expands the condition in a polynomial or an infinite series, and the other uses

PAGE 16

4 a time incremental method. The former utilizes the interface position information at discrete times to solve for the coefficients in the polynomial or series expansion, while the latter uses the interface position at consecutive times to solve for the conditions incrementally. Both can be used to solve for the conditions accurately as supported by eight examples encompassing the search for constant and time-variant temperature and flux conditions at the boundary. The methods are also well adapted to a numeral solution. The second phase-change problem solved in this work is an ablation problem that contains only one moving boundary. This problem is solved by a new method, which treats the problem in a fixed domain. In this method, the ablated phase is considered to remain in the space that is ablated in the process. As such, the moving boundary can be taken to be an interior phase-change interface, and the heat flux that is derived at this interface can be used to match the condition imposed at the moving boundary. Then with the additional prescription of the temperature for phase change at this interface, the flux condition there can be used to determine the position of the interface as well as the hypothetical condition that is imposed at the fixed boundary. Finally these positions and conditions can be used in the temperature equation to complete the solution . Solution of the ablation problems has rarely been attempted in a fixed domain. The method presented in this work thus represents a new endeavor, which will be shown to be simpler than most methods described in the literature. To validate the method used in the solution of the ablation problems, four examples are provided and

PAGE 17

5 they include the analysis of oneand two-phase ablation imposed with constant and time-variant flux conditions at the moving boundary. Of those four, the ones imposed with constant heat flux can be checked with an exact solution. A one-phase ablation problem is thus selected for checking the solution over the entire period of the ablation, while a two-phase ablation problem is selected for testing the solution in a quasi-steady state. Both yield accurate results as shown later in this dissertation. Convergence and stability tests have also been made rendering further assurance for the success of the method. The good results obtained in the second problem described above provide impetus for the solution of the third problem that addresses ablation with two moving boundaries. This is a combination of the ablation and Stefan problems, an area that has rarely been studied in the literature yet is important in practice when solid, liquid, and gas states appear simultaneously in the medium. Again the problem is solved in a fixed domain, and the moving boundary is treated as an interior phase-change interface. The flux condition at this boundary is again used with the temperatures at the interfaces to determine the heat flux at the fixed boundary as well as the interface positions to complete the solution. Unlike its predecessor for which two simultaneous integrodif f erent ial equations are solved simultaneously for each time step, for the combination problem, three equations will be solved simultaneously. Again the results are good as evidenced by three examples including imposition of constant, linear, and quadratic flux conditions.

PAGE 18

6 A source-and-sink method will be used for the solution of all the problems in this work. This method has recently been subjected to a through development [2-5], where the method has shown to be particularly suited for the solution of the phase-change problems addressed in this work. In this method, the problem is solved by using one set of governing equation, initial, and boundary conditions, whereas in the conventional methods developed in the literature, the problem is solved by using different sets of equations, with each set focused on the solution of one phase region. The conventional methods thus cannot compete with the source-and-sink method for simplicity. Moreover, in the source-andsink method, only one temperature equation will be derived; whether it is in the solid or liquid region depends on the position that is assigned in the equation. This is in sharp contrast to the conventional methods, in which several temperature equations must be derived, again one for each phase region. The time and effort saved in the source-and-sink method can thus be readily appreciated. With the inclusion of the three separate problems in this dissertation, the presentation of materials follows a nontradi tional approach. Thus one chapter will be devoted for the presentation of each problem; and the chapters will be self -complete with the coverage of the statement of the problem, general solution methodology, analysis, examples, and results and discussion. It is felt that, only through such an organization, the material can be presented in a coherent manner without severe fragmentation. In what follows in the next chapter, a literature review will be given first. It is then followed by the presentation of the three

PAGE 19

7 problems. Finally, conclusions and recommendations will be given to conclude this dissertation. As usual, all computer programs developed in this work are compiled in the appendix. It is hoped that the insight gained from this research will be useful in the analysis of the phase change in the future.

PAGE 20

CHAPTER II LITERATURE REVIEW Heat conduction with phase change constitutes a large class of moving boundary problems. These problems are commonly referred to as Stefan problems and were first solved in 1831 by Lame and Clapeyron [6], who determined the thickness of the solid crust by freezing of water under a constant temperature condition. They were able to find the thickness to be proportional to the square root of time but did not find the constant of proportionality. Some thirty years later, Neumannn presented an exact solution to the problem in his unpublished lecture notes, and for the first time in history, the Stefan problem was solved in its entirety [7-8]. NeumannÂ’s solution was for the phase change in a semi inf ini te space imposed with a constant temperature condition. His work is important because, irrespective of the numerous efforts made by many others in search for additional exact solutions, Stefan problems that can be solved exactly today remain those that were originally analyzed by Neumann some one hundred and thirty years ago. While the exact solution to the Stefan problem was first obtained by Neumann, the problem has been named after Stefan in recognition of his published work appearing in the open literature in 1889 [6-9]. Practically the same problem was solved by Stefan, who expressed the temperature of the phase-change medium in terms of 8

PAGE 21

9 a similarity variable, x/^ ? and the position of the moving boundary to be proportional to the square root of time. Thus, in essence, the moving boundary was solved by means of a similarity transformation. Since the transformation can not be used for general problems, it has now been firmly established that, for a Stefan problem that can be solved exactly, it must be in an unbounded domain, of medium of constant properties, and imposed with a constant temperature condition. With the severe limitation imposed by the similarity transformation, it is not surprising to see a wide variety of approximate solutions developed in the literature. Books and monographs have been prepared for presentation of these methods. In what follows, some popular solution methods will be briefly reviewed; interested readers are referred to 10-14 for details. One of the early methods developed for the solution of the Stefan problems imposed with heat flux and convection conditions is the power-series expansion method. In this method, the interface position and the temperature in the medium are expanded in terms of power series [10] or complimentary error functions [11-14]. These series are then substituted into the governing equation and boundary conditions to determine the coefficients in the series expansion. Power series method works well for the short-time solutions. At that time, the series converge rapidly, and only a few terms of the series are sufficient to yield accurate results. The method also works for the solution in the neighborhood of singularities that may occur as a phase degenerates. For large time, more terms are needed in the series, and the method rapidly loses its appeal.

PAGE 22

10 Stefan problems can also be solved by reducing them to integrodif f erential equations . Lightf oot [15] took the solidification as a moving heat source front. With the use of GreenÂ’s function, he was able to derive two integrodif f erential equations, one for the temperature and the other for the interface position. Then by assuming the position a function of the square root of time enabled him to retrieve the Stefan-Nuemann solution. LightfootÂ’s method has been extended to the solution of two dimensional Stefan problems involving phase change in an infinite wedge [16]. To avoid a tedious iteration in two dimensions, the interface position was taken to be a hyperbolic function and a correction term was introduced into the solution to account for the property variation in different phase regions. Indeed, LightfootÂ’s method has been limited to phase change of equal properties. This limitation has been lifted by Kolodner [17] by using double source and sink at the phase-change interface. Integrodif f erential equations of Volterra or Fredholm type can also be derived in the solution of the Stefan problems by a Fourier transform [18-19]. This occurs when the transformed equations are inverted in the solution. In general, the integral equation method is useful in providing an exact solution in integral form. However, the integral equations must still be solved numerically for a solution . One of the unique features of the Stefan problems is the occurrence of multiple phase regions whose domain changes continuously with time. The boundary position of the domain is also sought as a part of the solution. Boley [20] introduced an

PAGE 23

11 embedding method, in which the time-variant domains are embedded in a large fixed domain for the solution of the problem. This introduces an unknown heat flux at the fixed boundary, and this flux must be solved together with the moving boundary position to complete the solution. BoleyÂ’s embedding method has been extended to the solution of multidimensional problems without internal heat generation [21]. It should be noted that all the moving boundary problems solved in this dissertation also work in a fixed domain; however, a source-and-sink method will be employed which is more convenient to use in the solution of the combination of Stefan and ablation problems. Stefan problems can also be solved by an asymptotic expansion [22-29]. In this method, a quasi-steady solution is derived by dropping the unsteady temperature terms in the governing equations to find a long-time solution. A quasi -stationary solution is also derived by dropping the interface velocity terms in one of the governing equations and the interface flux condition to establish a short-time solution. Asymptotic expansions of the temperature and the moving boundary position are then constructed by using these solutions for limits [26]. Asymptotic expansions work best in the solution of Stefan problems with nonlinear conditions and convective motions of the melt [22]. Basically a perturbation technique, it offers insight into the physics of the problem. However, the method is very tedious mathematically; even the determination of the first-order terms in the temperature expansion has proven to be an intractable task [26].

PAGE 24

12 Higher-order terms become increasingly difficult to obtain, a distinct drawback of the method. Coordinate transformation offers another approach to the solution of the Stefan problems [30]. Somewhat similar to the embedding method, the transformation works in a fixed domain without the need of introduction of any new unknowns such as the boundary heat flux in the embedding method. In fact, the time-variant domain is mapped into an invariant domain in the transformation method; the moving boundary is thus immobilized in the solution. It is noted that use of the transformation does not by itself solve the Stefan problem. It only provides the convenience of working in a fixed domain. The coordinate transformation has thus been used together with other methods for the improvement of accuracy and also in the numerical methods for facilitation of solution [31-37]. Like the Stefan problems briefly reviewed above, ablation problems also fall into the general category of the moving boundary problems. However, the ablation problem is more complex than the Stefan problem because in the latter the problem can be solved in a fixed domain, whereas in the former the domain is continuously changing with time. There have been numerous efforts developed for the solution of the Stefan problems; only very limited work, however, has been devoted to the solution of the ablation problems, which will now be reviewed as follows. Solution of the ablation problems dates back to Landau [38] who first used the coordinate transformation to change the variable domain encountered in the ablation problem to a fixed domain. An

PAGE 25

13 exact, quasi-steady state solution for the case of a semi inf ini te medium imposed with the constant heat flux condition has been given in his paper. Also studied is the ablation for the semi inf ini te medium after dropping the steady-state assumption. Using Laplace transformation, Landau was able to derive the pre-melt solution for a time-variant heat flux imposed on the fixed boundary. However, the ablation solution was only derived for the constant heat flux condition. Specifically, two limiting cases were examined that include (i) Stefan number approaching zero (negligible heat capacity) and (ii) Stefan number approaching infinity (negligible latent heat). For the case of the Stefan number of the order of unity, Landau employed a finite difference method to solve the ablation problem with results presented graphically. LandauÂ’s method has also been used by Rogerson and Chayt [39] to find the exact melt-through time for ablation of a slab imposed with a constant heat flux condition on one side and an insulated condition on the other side. Rogerson integrated the heat conduction equation and showed the results to be independent of the thermal properties of the material engaged in phase change. The integral approach has been used in the solution of the phase-change problems [9,40,41]. Essentially a method of weighted residuals, the method was first introduced by von Karman and Pohlhausen in the approximate solution of boundary layer equations. The heat integral approach is simple to use; it also provides reasonably accurate results. However, it is somewhat handicapped in a detailed analysis of the temperature field. Specifically, the accuracy of the temperature is limited by the form of the profile

PAGE 26

14 initially chosen for analysis. Also the approximation can not be systematically improved for accuracy. The moment method proposed by Zien [41] carries promise of improvement over the classical heat integral method. The moment method has been used to solve one-dimensional ablation imposed with time-dependent heat-flux conditions. Both pre-melt and ablation solutions were derived. Again, the heat balance integral was used, and the integral of the original heat equation was carried out after multiplying the integrand by powers of temperature. A temperature profile was chosen and substituted into the integrated version of the heat conduction equation. The heat balance integral based on the approximate temperature profile was then used as the expression for the boundary heat flux. Although the method appears to work for the general case of the time-variant heat flux condition, it is expected that the choice of an approximate temperature profile that works for the nonmonotonic heat flux may not be applicable to the case of more general time-variant conditions. The ablation problems can also be solved by a variational approach. Biot and Agrawal applied the variational analysis and Lagrangian thermodynamics to the solution of ablation problems with variable thermal properties [42]. They considered one-dimensional heat transfer in a semi-infinite cylinder imposed with a constant heat flux condition. Both pre-ablation and ablation stages were solved. In the procedure, the governing equations were transformed and the Lagrangian heat-flow equation was derived which provided a relationship between the surface velocity and the heat penetration depth. Another relation between these two quantities was also

PAGE 27

15 found by using the energy equation. These equations were then solved simultaneously for the heat penetration depth and the velocity of the ablated surface. The variational method has been applied for approximate analysis of ablation of a semi inf ini te solid subject to convective and radiative heating [43,44]. Solutions were obtained in closed form for both the pre-melt and melt-removal heating regimes. In these studies, a cubic temperature profile was taken. The surface temperature, the thermal penetration depth, and the depth of the melt removal were treated as unknown and were determined as functions of time. The ablation problem has been solved numerically. Recently Blackwell [45] has employed the exponential differencing method to solve an ablation problem in one-dimension. He proposed the use of a moving coordinate system which was attached to the ablated surface. Then, by invoking the use of a finite control-volume approach, the element matrices could be defined for conduction, convection of the moving grid, and energy storage. In this method, all elements and control volumes were moving at a uniform velocity with the exception of the last ablating element and control volume. The node point at the moving interface was fixed in space. As such, the last ablating element had one moving boundary and one fixed boundary. The method has been applied to the solution of a steadystate ablation problem for which an exact solution was available. Comparisons were also made with central differencing for the conduction terms and upwind differencing for the convective terms; the exponential schemes appear to be better numerically.

PAGE 28

16 Ablation problems have also been solved by a finite element method with deforming spatial grids [46]. In this method, the classical finite-element equations are transformed to account for the continuous deformation of the grid for a precise localization of the ablated surface. This is done at the expense of chores of construction of an additional convective matrix. Recently, ablation has been applied to model intense heating such as laser beam. Masters [47] used the finite difference method to solve the ablation of a one dimensional slab imposed with an intense uniform heat flux and analyzed the effect of melt on the temperature distribution during the heat pulse. The steady-state solutions were found for the velocity of the surface recession and the temperature history during ablation. Abakian and Modest [48] studied the ablation due to a continuous-wave laser beam irradiating on a moving semi inf ini te and semitransparent solid. Using an integral method, they were able to derive a set of nonlinear partial differential equations which were solved numerically for the groove depth and shape due to ablation. They also considered the ablation of a moving slab caused by irradiation from continuous-wave and pulsed laser beams and derived a solution for the temperature distribution [49]. The laser has also been used in material processing as investigated by Dabby and Paek [50]. In their work, vaporization occurred at the surface; however, below the surface, the material was heated by absorption of the laser radiation, which might reach a temperature higher than that for vaporization. Explosion may thus take place, which provides a means for material removal in the drilling process.

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17 The review given above is strictly for the solution of regular heat conduction problems. In these problems, the system geometry, governing equation, initial and boundary conditions are fully specified, and the problems are solved mainly for the determination of the temperature field. In the inverse problems, however, the roles of known and unknown quantities are exchanged. There have been abundant studies documented in the literature for the solution of the regular problems; not much work, however, has been devoted to the solution of inverse problems, and for those that have been solved and documented in the literature, nearly all of them are for heat conduction without phase change [51-60]. There is a lack of study for inverse problems with phase change. Most inverse heat conduction problems deal with a situation where an extra temperature is available at one point in the domain, and this temperature is used together with others to find the condition imposed on the boundary. Stolz [51] first solved such a problem numerically. In his work, the inverse problem was formulated as though it were a direct problem. Since a linear heat conduction problem was solved, he was able to use the superposition principle. An integral equation was derived for the unknown surface condition and was solved by numerical inversion. The method was found to be inefficient if the time steps used were too small. Yet, a small time step must still be used for an accurate solution. Beck [52] was able to improve this method by using a procedure that involved minimization of the sum of the squared difference between the actual and the calculated temperatures at the location where the temperature data were given. Burggraf [53] developed an exact

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18 series solution to a one-dimensional inverse problem with the lump capacitance approximation serving as the leading term. The temperature and heat flux histories were provided at an interior point. Approximate results were found if discrete or experimental data were used for input in solution. Beck [54] moved one step further by solving an inverse problem in which the material properties were treated as functions of temperature; thus the problem solved becomes nonlinear. A two-dimensional inverse finite element solver has also been reported in the literature [55]. It can be used for the solution of the heat flux imposed on the surface of nuclear fuel rod with the use of the interior temperature measurements for input. Only a handful of studies are found for the solution of inverse heat transfer with phase change. Macqueene et al . [61] proposed an inverse finite element method to determine the efficiency of an arc welding process. In their method, the latent heat of fusion is taken into consideration by the variation of the elemental specific heat when the average temperature of the element reaches the melting point. Conduction in both the solid and liquid regions was accounted for. Katz and Rubinsky [62] proposed a front-tracking finite-element method for the solution of one-dimensional inverse Stefan problems. His method was applied for the determination of the position of the solid-liquid interface and the transient temperature distribution in the solid region during stationary arc welding . An inverse method was also used by Landram [63] for the analysis of the interface position and the energy transport

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19 mechanisms during welding. The vaporization energy loss was found to be important during the motion of the solid-liquid interface. The interface shape, being close to hemispherical, gives clear indication of a nearly one-dimensional radial symmetry for the heat transfer . An analytic solution to inverse Stefan problems in Cartesian and spherical geometries was provided by Rubinsky and Shitzer [64]. In their analysis, the inverse Stefan problem was characterized by two boundary conditions at the moving front. This is in sharp contrast to the technique developed in this dissertation (see Chapter 3) which needs only one condition at the interface to solve for the boundary condition. In their work, the medium was initially at the phase change temperature. An integral equation was then derived by integrating the governing equation, which was, in turn, solved by the method of analytic iteration in which the first guessed solution was taken to be the long-time solution to the problem. Series solution was then developed by induction following a number of iterations. A boundary element analysis with constant elements has been developed by Zabaras et al . for the solution of a one-dimensional inverse solidification problem [1]. They used the sensitivity analysis developed by Beck [52-54] and Burggraf [53] for inverse heat transfer solution. Using an integral formulation, they were able to solve two separate inverse Stefan problems, one in the solid region and the other in the liquid region. Thus similar to the ones given by Rubinsky and Shitzer, two conditions must be provided at the freezing front, also an inefficient method.

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20 It should be mentioned that a source-and-s ink method has recently been developed in the heat transfer literature, which is particularly suited for the solution of regular and inverse Stefan problems. In a series of papers, Hsieh and his associates were able to apply this method to the solution of regular and inverse oneand two-phase melting and solidification problems for medium with and without subcooling and superheating and imposed with constant and monotonic temperature and heat flux conditions [2,3]. The method has been applied to the solution of phase change imposed with cyclic conditions [4,5,65]. Two inverse solution techniques have also been developed with the problem formulated with a source-and-s ink method as shown in reference 3. The method has shown to be closely related to the boundary element method as reported by Hsieh et al . [66,67]. The present study is a further extension of these works.

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CHAPTER III SOLUTION OF INVERSE STEFAN PROBLEMS BY A SOURCE-AND-SINK METHOD This chapter presents the development of a source-and-sink method to solve inverse Stefan problems. In these problems, the conditions are specified at the moving rather than the fixed boundary. Typically, the temporal location of the interface between the phases is given and this location is used together with others to determine the boundary temperature or heat flux that is required to provide for this interface motion. In what follows in this chapter, the motivation for this study will be given first. It is followed by a general analysis that is designed for the solution of the inverse Stefan problems in three dimensions. This analysis is then used in the solution of onedimensional inverse problem examples. Finally, results for these examples are provided and discussed in detail and possible extensions of the method are included to conclude this chapter. 3 . 1 Motivation Heat diffusion in a medium with constant properties is governed by the partial differential equation V 2 T(f,t) + _ 1 5T(r,t) k “ 01 dt ’ r G R t > 0 ( 3 . 1 ) 21

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22 where all notations have their usual meaning. For this medium, the conditions imposed on the boundary are usually one of the following types T(r,-,t) = F,(r,-,t), r,G BdT(f,,t) G,.(f ,,t) ~s^r — r ' €B ' GB, (3-2) (3.3) (3.4) which represent the familiar Dirichlet, Neumann, and Robin conditions, respectively. In (3.3) and (3.4), n t denotes an outward drawn normal. Then, with the additional initial condition given as T(r,0) = T-(r) (3.5) the temperature solution can be expressed by means of Green’s function as [68] t T(f,t) = G(f,t | f',0)T.(f / )dV / + f [ G(f,t | f',r )dV , dr + E{ } (3.6) R' Or' Here, the braced term is used to account for the three boundary conditions given earlier. Their expressions are listed in Table 3.1. A distinct feature is found in the Green’s function method above--the effects of the initial condition, heat generation (or destruction), and boundary conditions are embodied respectively in

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23 Table 3.1 Expressions to account for effects of boundary conditions BOUNDARY CONDITION { } EXPRESSION T(T' / t) -F i (T' , t) -aff 3G ds'i dx J 0 i dn i am', t) = 1) 377: Y i -affc (T, 1 1 t' , X ) Gi ( ^' , T ] ds'i dx °s' i dT(r it t) + h s T( y _ U (-y /-) r r i H. cr' i , x ) , / , ajj G(r, t|T^, x ) dsidx °s' 1

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24 the first, second, and third terms on the right of equation (3.6). Then, in the case of regular problems, once these conditions are fully specified, the temperature can be easily found. In this effort, the GreenÂ’s function can be obtained by using the concept of point charges [68] or by solving auxiliary problems [69]. The format of (3.6) turns out to be particularly suited for the solution of the inverse problems. For the inverse problems, the left hand side of this equation can be used to represent the extra temperature that is provided either at the boundary or at interior points. Then this temperature can be used to determine the missing information, such as the initial condition, heat generation, or boundary conditions. In these efforts, the missing quantities can be expanded either in a polynomial or an infinite series, which is substituted into their respective integrals in (3.6). The resulting equation is then solved numerically for the coefficients in the series to complete the solution. This method clearly works for the initial condition and the heat generation. However, for the boundary condition, since it is unknown a priori, one must first assume a particular 4 type Â’ of condition that is imposed on the boundary. For convenience, one could use equation (1.2) or (1.3) for this condition. GreenÂ’s function is then found with this assumed condition. Notice that the boundary condition can be exchanged if necessary as shown in by Hsieh and Shang [70]. That is to say, a Dirichlet condition can be accomplished by means of a Neumann condition and vice versa. The search for the condition is thus not restricted by the type of the conditions assumed. Better yet, an incremental solution approach can also be developed to track

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25 the boundary conditions accurately as will be shown later. The concept above will now be applied for the solution of inverse Stefan problems in which the interface positions are given, and these positions are used to find the boundary conditions that must be imposed to cause the interface motions. 3.2 General Analysis For the sake of illustration in what follows, the Stefan problems consist of two stages: a pre-melt stage, when heat is added to the surface of a subcooled medium to raise its temperature to the phase-change temperature; and a melting stage, when the medium changes phase and the melting starts at the surface and the interface moves inward with time. It is assumed that the properties for different phases are constant and of equal value. The medium has a distinct melting temperature; that is, no mushy zone in the medium. Convection is negligible. For generality, the analysis will be developed for the solution of both melting and solidification problems. The analysis can also be extended for the solution of Stefan problems in multiple phases and in a medium with unequal phase properties as will be discussed later. For the moment, the simplified problems will be solved by considering the medium shown in Figure 3.1. The formulation of these problems follows below.

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Figure 3.1 System analyzed

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27

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28 Pre-melt Stage Governing equation-V *r 0 (r,t) = i 1 ST 0 (r,t) dt r G (LUS) t 0 > t > 0 Initial condition-T o (r,0) = T-(r) Melting Stage Liquid Region: Governing equation-V *r L (r,t) = i 1 0T L (r,t) dt r G L t > t Q Solid Region: Governing equation-V 2 T 5 (i,t) 1 9T s (r,t) a dt f€S t > t 0 (3.7) (3.8) (3.9) (3.10)

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29 Initial condition-Ts( r >t 0 ) — T 0 (r,t 0 ) (3.11) Interface Conditions: T L (T f ,t) = T m = T 5 (r /)t ) (3.12) dT 5 (f/,t) dT L (t f ,t) _ pL dn On ~ k 6} v „(t)=V.n (3-14) Here ry denotes the interface position and v n the history of the interface motion. For the inverse Stefan problems of interest in this study, this history is used for the determination of the missing boundary conditions. The problems as posed can be solved by use of the Green’s function method described in the preceding section. However, a direct use of this method would require (3.6) to be applied to two separate regions, liquid and solid, and the solution so obtained may not be as efficient as one desires. A source-and-s ink method is thus used [2,4,15]. In this method, the melting interface is taken to be a moving heat-sink front and a freezing interface is taken to be a moving heat-source front. Then, in sharp contrast to conventional methods in which different equations are used to represent the temperatures in different regions, only one equation will be derived. Whether it is in the solid or liquid region is

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30 determined by the position that is assigned in the temperature equation. The solution of the inverse problems can then be simplified with this method. Following this approach, the melting stage is solved by considering an equivalent problem as follows: Governing equation for the equivalent problem: V 2 T(r,t)+^ v n (t)5(r-r / ) = £ P L _ _ , i <9T(r,t) dt f G (LUS) (3.15) t > t 0 Initial condition for the equivalent problem: T(f,t 0 ) = T 0 (r,t 0 ) Interface conditions for the equivalent problem: (3.16) T(f / ,t) = T m , v„(t) = V.n (3.17a, b) where S( r— Fy) denotes a Dirac delta function. The signs preceding this function are used for freezing (+) and melting (-). It can be shown readily that (3.15) reduces to (3.9) and (3.10). Furthermore, by integrating (3.15) across the interface from Fy — e to Fy + e and forcing e to be zero in a limiting process, equation (3.15) reduces to (3.13). This can be proved by using the pill box at the interface in Figure 3.1. Other equivalences for this stage are apparent.

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31 The equivalent problem can be solved by referring to (3.6) in which the heat generation term is changed to the interface motion term as T(r,t)= G(r,t | r , > < 0 )T t (f / )dV / LuS G(r,t |r , ,r)v n (r)i(r'-r / )dV , dr+ E{ } t Q L U S (3.18) where the plus sign is used for freezing and minus sign for melting. Finally, the boundary conditions can be found by setting r in this equation to the interface position, Fy, and T(ry,t) to the melting temperature, T m , as T = m G(r / ,t|f',< 0 )T t .(f / )dV , ±^[ [ G(f / ,t|r / ,r)v n (r)5(r'-r / )dV / dr+ E{ } LuS t Q Lu5 (3.19) The missing boundary conditions can then be found by solving them implicitly. In this effort, the time when melting starts (t 0 ) can be determined by solving the pre-melt problem, whose solution can again be taken to be the special case of (3.6) in which the heat generation term is zero. Solution in this stage is thus elementary. 3.3 Example Problems The analysis above is now used to solve example problems which are in semi inf ini te domain in which the interface motion is given. For the sake of generality, the analysis will be developed for determining either the Dirichlet condition or the Neumann condition that is imposed on the boundary at x=0 , and the interface motion may

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32 be the result of either melting or solidification in a oneand twophase medium. Also for illustration purposes, the heat flow is one dimensional, the initial temperature being uniform. Extensions to more dimensions are provided in Appendix A. For the problem given, Green’s function can be found to be G(x,t | x',r)= 1 2^7ro(t — r) [ ex P( “ (x-x') 2 N ^ , (x + x') 2 3S(t^) )±exp< -4^(t3T) )] (3.20) where the plus and minus signs are to be used when the flux and temperature condition is assumed to appear at the boundary, respectively. The temperature can then be obtained by using (3.18), which is recast in a general format as G(x,t | R(r-Ft 0 ),r)dr (3.21) where all temperatures, including T m , are measured in excess of the initial temperature, and T 0 ( X ,t) — > 7T t 0 E ( s ) r x 2 ex p[ — -jTu ~ ]ds -si 1 / 2 4a(t — s) (t-s) ( 3 . 22 ) in which jlZM 2a t — s E(s) = ^ (3.23a,b) r G ( s ) Here F(s) and G(s) denote the assumed temperature and heat flux condition, respectively. Also

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33 1 t > t 0 for 0 t < t 0 (3.24) (3.25) where Ste is known as the Stefan number. With the use of the circumflexed Heaviside function given by (3.24), equation (3.21) holds for all time and for both oneand two-phase problems. Equation (3.21) can be used to determine the unknown boundary condition by invoking use of the condition at the interface as t-t ±[iT ° ( y t),t) ] 0 H(t 1 0 ) f dR(r+t 0 ) m Ste dr G(R(t),t t 0 | R(r-ft 0 ),r)dr (3.26) 0 where the plus and minus signs on the left hand side are to be used for freezing and melting, respectively. 3 . 4 Numerical Solution A local linearization can be used to solve (3.26) numerically. In this effort, the entire time range is divided into small increments in which the interface position is taken to be linear; see Figure 3.2. Then the dR/dt can be taken out of the integral for each increment, and the convolution integral written as a summation as

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Figure 3.2 Linearization of solid-liquid interface position curves for the numerical solution

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35 0 T

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36 i T 0 (R(t Ar ),t 7V -) l H (t N t 0 ) N, dR(t n ) T m ~ ±1 Ste *»» t 0 dt G(R(t N ),t N t 0 | R(r+t 0 ),r)dr] ^n-l “ ^0 (3.27) Here the signs are to be selected following the statement below (3.26). For the inverse problems, the right hand side can be evaluated with the input of the interface motion data. As for the left hand side, if the boundary conditions are represented by a power series, then the number of terms on this side must be equal to the number of the terms that are taken in the series. In practice, equation (3.27) can be written by means of matrix elements for a melting problem for the series method as N N 53 ^mn a n C ra 53 ^mn e n n=l n=l (3.28) where m=N ; t^>t 0 ; N=1,2,...,N; and * 0 for m < n N for m>n; F(s) = ^ a n 5 n n=l exp[ R(U 2 Mt ra -s) for m > n ; G(s) = 5Z a n s " 1 n=l c m ~ N 7 T rp a 1 m (3.29a,b,c) (3.30)

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37 r 0 for m < n = < mn r 7 T t -t n-1 G(R(t m ),t m -t 0 | R(r + t 0 ),r)dr for m>n 0 (3.31a,b) dR (.S) (3.32) n dt v ' Notice that the integrals in these equations can be performed in closed form if the missing conditions are expressed in the power series as shown in (3.29) or in terms of a Fourier series. The coefficient matrices B and D are of a lower triangular structure, a characteristic permitting the solution of (3.28) to be carried out simply by using forward substitution, a simple numerical procedure. The boundary conditions can also be determined by use of an incremental approach. In this method, (3.27) is recast as t N t N-1 2L T / 41 _ Ip) yv r A m\ Ste n 4=l dR(f n ) “ t 0 dt G(R(t yv ),t N -t 0 | R(r+t 0 ),r)dr]jt N-1 n { jv — j ±£ n=l E(s) _.^/2 ^n-l “ ^0 R{t N y \ T72“ p1 -i»( 1 „-s) 1< ' s ) n — 1 (t N — s) 4a(t N -s) (3.33) where t 0 is set to zero all the time in the integral on the left and

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38 the last integral on the right. In practice, equation (3.33) can be recast as F ( t Af) / P N,F P 7V,G -1 7T rr\ >Ja 1 m N + E d 11=1 Nn c n e„ — N -1 E n=l F ( t n) P n,F 1 G(t n )P n , G J (3.34 a,b) where t N > t 0 ; N=1,2,...,N; d Nn can be obtained by referring to (3.31); and (3.35 a,b) In (3.34), the summation vanishes when N-1=0. Then starting from N=1 , 2, and so on, the conditions can be evaluated incrementally. In this effort, the F and G values found from the previous time steps are used as the input in the computation of the right hand side of (3.34), which immediately gives the F and G values at the succeeding new time step. This continues until the desired time is reached. The algorithms can be developed for a rapid solution of the conditions. Computer programs developed for the solution of inverse Stefan problems given in this chapter are provided in Appendices B through E .

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39 3.5 Critique of the Method Using GreenÂ’s functions, the present method is limited to the solution of problems whose GreenÂ’s function can be obtained analytically. This excludes those problems whose boundary conditions and governing equations are nonlinear and such nonlinearity can not be resolved by using transformation (e.g., Kirchhoff transformation). Yet with the use of GreenÂ’s function, it projects an impression of being related to the boundary element method that has been undergoing through development in recent years [64]. Yet, they are functionally different. In boundary element method, the boundary element equations are written separately for the liquid and solid regions, whereas in the present method, only one equation (3.27) is derived. The number of the equations to be solved in the present method is thus reduced by half, which is particularly true in the solution by the incremental approach described earlier. More important, in the boundary element method, the problems cannot be solved without the information for the heat fluxes that appear on both sides of the interface [64,1]. Such fluxes, however, are unnecessary in the present work. Instead, equation (3.26) embodies conditions (3.12) and (3.13) in the form of a single integrodif f erent ial equation. There is no need for the satisfaction of the flux conditions; as a result, the present method is more effective because it requires less information for input. As will be shown in the next section, the present method is also accurate. In fact, such accuracy is not unexpected because equation (3.26) is exact. The only approximation in the analysis is

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40 the local linearization which has been applied to the interface motion and the boundary condition; see (3.33) and (3.35). In fact, in the present analysis, they have been approximated by using constant elements. The analysis can thus be improved for accuracy by using higher order elements, such as linear and quadratic elements as in references [59,71]. 3 . 6 Results and Discussion For the numerical experiments performed in this study, the interface motion data are taken from reference 5. Aluminum will be used for tests; its properties are given in Table 3.2. The inverse solution techniques developed in this chapter are used to solve eight example problems of which four having exact solutions. The retrieved conditions for these examples can thus be compared with the exact solutions for error. In these four examples, the interfaces move as a function of square root of time, and the interface motion data are used to retrieve the constant temperature conditions that appear on the surface. This problem is known as the Stefan-Neumann problem, and Table 3.2 provides a summary of the conditions tested in the examples. The first example deals with a general two-phase StefanNeumann problem; the medium is initially subcooled to 300 K, which is lower than the melting temperature (see Table 3.3). For this example, the temperature at the boundary is unknown; an unknown temperature is thus assumed to appear at the boundary, and the interface motion data are used to find this temperature. The results are listed in Table 3.4, where two sets of results are

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41 Table 3.2 Properties of aluminum Melting temperature, T m (K) 932 Vaporization temperature, T v (K) 2,543 Heat of fusion, L f (J/kg) 389,600 Heat of vaporization, L* (J/kg) 9,462,000 Thermal conductivity, k (W/m K) 200 Density, p (kg/m 3 ) 2,710 Specific heat, c (J/kg K) 1200

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42 Table 3.3 Conditions tested in eight examples PROBLEM DESCRIPTION INPUT DATA TRUE CONDITION F(t):K; G(t):W/m 2 TYPE OF BOUNDARY CONDITION ASSUMED SOLVED 1 two-phase Tj=300K
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43 Table 3. 4 Comparison between true and retrieved conditions for first Stefan-Neumann example solved for boundary temperature Boundary Conditions, T (K) Time (sec) Imposed True Condition Retrieved Series Method N=3 N=4 Incremental Method 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1003.22 1002.97 1002.72 1002.48 1002.25 1002.03 1001.82 1001.61 1001.42 100123 1001.06 1000.89 1000.73 1000.58 1000.43 1000.30 1000.17 1000.06 999.95 999.85 1003.17 1002.87 1002.58 1002.30 1002.04 1001.79 1001.56 1001.34 1001.14 1000.95 1000.78 1000.63 1000.50 1000.38 100028 100020 1000.14 1000.09 1000.07 1000.07 1003.49 1000.41 100020 1000.11 1000.03 1000.05 1000.04 1000.02 1000.68 999.88 999.97 999.99 999.99 999.99 999.99 999.98 999.98 999.98 999.90 999.94 N F(S) a n Sn ' 1 n m 1 For N=3, a,=1003.490405 a 2 = -0.268732 a 3 = 4.350448xl(T 3 For N=4, a, =1003.490405 aj= -0.322478 a 3 = 6.264645xia 3 a«= 6.552981xia s

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44 given--one for the series solution method and the other for the incremental solution method. In the series solution method, a power series of degree N-l is used to represent the temperature (see (3.29b)), and two values of N are tested. In this method, the interface position data at equal time intervals are used for input. Thus, for example, for N=3 , R data at times equal to 0, 6, 12, and 18 seconds are used to determine the series, which is, in turn, used to generate the temperature values listed for 20 time steps in the table. Comparing the retrieved temperatures at the boundary by both series with the true condition (exact temperature) listed to the left indicates they are in good agreement . In this case , the coefficients found for these series are listed at the bottom of the table. From the temperature values tabulated, the results for the low power series appear to be as good as those of the high power series, and such trend persists even with the test of a higher power series of degree 10 (results not shown). This gives the indication that the temperatures have been converged. As for the incremental solution method, the results are also good; errors are of the order of 10 % at large time. For the incremental method, the boundary conditions are found at exactly the same times when the interface positions are given. The time step for the solution is thus identical to that for the interface data input. Also notice that the convergence and stability that are normally encountered in the conventional finite difference methods are nonexistent in the present incremental solution of integral equations. Also as in the case of the series solution method, the incremental solution results are generally

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45 better at large time than small time, which will be further discussed later. In the example above, a temperature condition is imposed, and the same ‘type’ of condition is assumed in the process of the inverse solution. Since the type of the condition that is imposed on the boundary is unknown a priori, it may well be the heat flux condition that one assumes, and the question to be addressed now is whether it is still possible to retrieve the temperature condition via the assumed flux condition. Use will now be made of equation (3.21) to determine this temperature condition once the boundary flux condition is found, and the results are listed in Table 3.5. As shown in the table, the incremental solution results are still good but the series solution results are not as accurate as those listed in Table 3.4. This is certainly a result of the errors being accumulated first in the evaluation of the heat flux next in the evaluation of the temperature using the previously determined heat flux. While this example serves we ii to illustrate that the conditions are sti 1 1 exchangeable, such a twostep solution of the condition may lead to large errors, part i cu larly in the series solution method, and should thus be avoided in practice. In fact, according to experience, the computer time saved in this two-step approach of solving flux then temperature is insignificant as compared with that in the separate, one-step approach of direct evaluation of the flux and temperature. A slight modification is made in the next two examples: this time the medium is not subcooled, the initial temperature being equal to the melting temperature of the medium (see examples 3 and 4

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46 Table 3. 5 Comparison between true and retrieved conditions for second Stefan-Neumann example solved for boundary flux then temperature Boundary Conditions, T (K) Retrieved Time (sec) Imposed True Series Method Incremental Condition N=4 N=5 Method i 1000.00 585.89 620.85 1023.06 2 1000.00 715.58 758.46 1007.71 3 1000.00 799.99 844.90 1003.60 4 1000.00 862.06 905.50 1000.88 5 1000.00 909.11 948.70 1000.04 6 1000.00 945.07 979.11 999.68 7 1000.00 972.34 999.67 999.52 8 1000.00 992.57 1012.44 999.44 9 1000.00 1006.96 1019.03 998.1 1 10 1000.00 1016.46 1020.76 998.30 11 1000.00 1021.85 1018.76 998.37 12 1000.00 1023.76 1014.02 998.44 13 1000.00 1022.76 1007.48 998.51 14 1000.00 1019.34 999.99 998.59 15 1000.00 1013.95 992.40 998.66 16 1000.00 1007.01 985.51 999.45 17 1000.00 998.89 980.13 999.16 18 1000.00 989.05 977.04 999.11 19 1000.00 980.54 977.04 999.07 20 1000.00 971.00 980.93 997.15 N G(s)=£a n s J7-1 71*1 For N=4, a,=8001 187.956002 & 2 = -387201.858263 a 3 = 1581.702473 a 4 = 148.390536 For N=5, a,=8945600.085189 aj= -541131.047558 a 3 = 2763.120508 a 4 = 324.034498 aj= 3.817008

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47 in Table 3.3). The Stefan-Neumann problems solved thus become a one-phase problem. It will be tested that the results are unaffected by the change to the one-phase problem. Again, the same series of tests are made and the results are listed in Tables 3.6 and 3.7. Again, the one-step solution results are good (Table 3.6). The twostep solution results are poorer (Table 3.7), further reinforcing the recommendation made earlier in the testing of the two-phase problems. Having satisfactorily completed testing of the Stefan-Neumann problems, attention is now directed to the solution of inverse Stefan problems whose interface motions must be met by imposing the time-variant temperature and flux conditions. There are no exact solutions for these problems, and the interface motion data are taken from Choi [5] who have solved the regular (forward) version of the problems with great accuracy. The interface position data are then used to retrieve the boundary conditions and the results are listed in Tables 3.8 through 3.11. Tables 3.8 and 3.9 give the results for the direct retrieval of the linear temperature and heat flux conditions F(t) = 1000 + 5t (3.36) G(t) = 6.388034 x 10 6 + 2.82752 x 10 5 t (3.37) while Tables 3.10 and 3.11 give the results for the direct retrieval of the linear temperature and quadratic heat flux conditions

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48 Table 3.6 Comparison between true and retrieved conditions for third Stefan-Neumann example solved for boundary temperature Boundary Conditions, T (K) Retrieved Time (sec) Imposed True Series Method Incremental Condition N=3 N=4 Method 1 1000.00 1033.83 1033.12 1037.49 2 1000.00 1030.31 1028.96 998.22 3 1000.00 1026.93 1025.01 998.58 4 1000.00 1023.69 1021.29 998.85 5 1000.00 1020.59 1017.80 999.05 6 1000.00 1017.63 1014.55 999.20 7 1000.00 1014.81 1011.55 999.24 8 1000.00 1012.13 1008.81 999.38 9 1000.00 1009.59 1006.33 999.44 10 1000.00 1007.19 1004.13 999.49 11 1000.00 1004.93 1002.21 999.53 12 1000.00 1002.81 1000.57 999.59 13 1000.00 1000.82 999.23 999.60 14 1000.00 998.98 998.20 999.65 15 1000.00 997.28 997.47 999.65 16 1000.00 995.71 997.07 999.69 17 1000.00 994.29 997.00 999.69 18 1000.00 993.00 997.26 999.73 19 1000.00 991.86 997.87 999.71 20 1000.00 990.85 998.82 999.78 For N=3, a,=1037 .496339 N F ( s) =52 a n s n ' 1 i n ii 3.728416 6.982696xl0" 2 For N=4, a,=1037 .496324 a,= -4.474102 a 3 = 1.005513x1(7' a<= 1.324365xia 3

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49 Table 3 .7 Comparison between true and retrieved conditions for fourth Stef an-Neumann example solved for boundary flux then temperature Boundary Conditions, T (K) Retrieved Time (sec) Imposed True Series Method Incremental Condition N=4 N=5 Method 1 1000.00 906.32 917.43 1056.39 2 1000.00 953.87 966.25 996.60 3 1000.00 980.48 991.90 1002.26 4 1000.00 997.67 1006.84 999.33 5 1000.00 1008.95 1015.10 999.70 6 1000.00 1016.02 1018.67 999.45 7 1000.00 1019.89 1018.86 999.16 8 1000.00 1021.28 1016.61 999.22 9 1000.00 1020.73 1012.66 999.09 10 1000.00 1018.66 1007.67 999.23 11 1000.00 1015.44 1002.18 999.05 12 1000.00 1011.39 996.73 999.32 13 1000.00 1006.79 991.80 998.89 14 1000.00 1001.89 987.87 999.72 15 1000.00 996.94 985.38 998.27 16 1000.00 992.16 984.80 1001.19 17 1000.00 987.77 986.56 994.04 18 1000.00 983.98 991.12 1014.19 19 1000.00 980.99 998.92 951.97 20 1000.00 979.00 1010.44 1162.46 N G(s)=£a n s n-l J7*l For N=4, a,=2805963.1 14624 aj= -204888.890714 a,= 2268.882336 99.892255 For N=5, a,=3 137 161 .977 178 a,= -286340.839757 a 3 = 3963.574559 a 4 = 218.130426 a,= 1.467676

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50 Table 3.8 Comparison between true and retrieved conditions for fifth inverse example problem solved for temperature Boundary Conditions, T (K) Retrieved Time Imposed Series Method Incremental Error (sec) true Condition N=3 Error N=4 Error Method % % % i 1005.00 1073.05 6.77 1078.13 7.28 1003.61 -0.13 2 1010.00 1064.27 5.37 1061.16 5.07 1007.83 -0.21 3 1015.00 1056.80 4.11 1048.37 3.29 1012.50 -0.24 4 1020.00 1050.64 3.00 1039.36 1.90 1017.33 -0.26 5 1025.00 1045.81 2.03 1033.71 0.85 1022.27 -0.26 6 1030.00 1042.29 1.19 1031.04 0.10 1027.17 -0.27 7 1035.00 1040.09 0.49 1030.94 -0.39 1032.24 -0.26 8 1040.00 1039.21 -0.07 1033.01 -0.67 1037.26 -0.26 9 1045.00 1039.64 -0.51 1036.86 -0.78 1042.29 -0.25 10 1050.00 1041.40 -0.81 1042.07 -0.75 1047.31 -0.25 11 1055.00 1044.47 -0.99 1048.26 -0.64 1052.35 -0.25 12 1060.00 1048.86 -1.05 1055.02 -0.47 1057.36 -0.24 13 1065.00 1054.56 -0.97 1061.95 -0.29 1062.38 -0.24 14 1070.00 1061.59 -0.78 1068.65 -0.13 1067.38 -0.24 15 1075.00 1069.93 -0.47 1074.72 -0.02 1072.39 -0.24 16 1080.00 1079.58 -0.03 1079.77 -0.02 1077.40 -0.23 17 1085.00 1090.56 0.51 1083.38 -0.15 1082.37 -0.24 18 1090.00 1 102.85 1.17 1085.17 -0.44 1087.50 -0.22 19 1095.00 1116.47 1.96 1084.72 -0.94 1092.06 -0.26 20 1100.00 1131.39 2.85 1081.65 -1.67 1098.80 -0.10 a,=1083. 1636095 %= -10.7640380 a 3 = 0.6587919 a,=1099 .6680063 aj= -23.9563496 a 3 = 2.4866623 34 = -0.0666941 For N=3, F(S) =52 a n sn ‘ 1 n 1 For N=4,

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51 Table 3.9 Comparison between true and retrieved conditions for sixth inverse example problem solved for heat flux Boundary Conditions, q (W/m 2 ) || Retrieved Time Imposed Series Method Incremental Error (sec) irue Method % Condition N=4 Error N=5 Error % % 4.25 7589730.94 8076200.72 6.41 7558675.50 -0.40 7184872.69 -5.33 II 4.50 7660418.94 8147001.48 6.35 7694176.58 0.44 7371675.43 -3.77 4.75 7731106.94 8206716.05 6.15 7815212.59 1.08 7493717.81 -3.07 5.00 7801794.94 8257180.74 5.83 7920261.67 1.52 759645424 -2.63 5.25 7872482.94 8300231.83 5.43 8008672.53 1.73 7689904.72 -2.3 i 5.50 7943170.94 8337705.60 4.96 8080664.49 1.73 7775533.53 2.11 5.75 8013858.94 8371438.35 4.46 8137327.48 1.54 7859641.15 -L92 6.00 8084546.94 8403266.35 3.94 8180622.00 1.19 7940899.93 -1.77 6.25 8155234.94 8435025.90 3.43 8213379.15 0.71 8020420.49 -1.65 6.50 8225922.94 8468553.29 2.94 8239300.64 0.16 8098888.64 -1.54 6.75 8296610.94 8505684.79 2.52 8262958.75 -0.40 8176244.61 -i.45 7.00 8367298.94 8548256.71 2.16 8289796.37 -0.93 8252784.49 -1.36 7.25 8437986.94 8598105.32 1.89 8326126.99 -1.33 8329871.72 -L28 7.50 8508674.94 8657066.91 1.74 8379134.69 -0.61 8402584.80 -1.25 7.75 8579362.94 8726977.77 1.72 8456874.13 -1.42 8480061.68 -1.15 8.00 8650050.94 8809674.19 1.84 8568270.59 -0.94 8553180.08 1.12 8.25 8720738.94 8906992.45 2.13 8723119.93 0.02 8628203.50 -1.06 8.50 8791426.94 9020768.84 2.60 8932088.60 1.60 8701314.04 1.02 8.75 8862114.94 9152839.65 3.28 9206713.65 3.88 8775467.98 -0.98 9.00 8932802.94 9305041.17 4.16 9559402.73 7.01 8847997.32 -0.95 For N=4, a,=3397040.84 N ^=2247925.15 G(S) =^^n S a3=-353 114.747 n 1 a 4 =19587.0623 . For N= =5. a,=8734725.30 a 2 =-3092029.40 83 = 1293821.95 a^-188030.644 aj= 9286.34410 1

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52 Table 3.10 Comparison between true and retrieved conditions for seventh inverse example problem solved for temperature Boundary Conditions, T (K) Retrieved Time Imposed Series Method Incremental Error (sec) true Method % Condition N=3 Error N=4 Error % % 1 972.00 976.35 0.44 1020.27 4.97 969.93 -0.21 2 1012.00 1014.88 028 1044.41 3.20 1006.51 -0.54 3 1052.00 1053.63 0.15 1071.70 1.87 1042.85 -0.87 4 1092.00 1092.58 0.05 1101.84 0.90 1079.43 -1.15 5 1132.00 1131.74 -0.02 1134.51 0.22 1116.09 -1.40 6 1172.00 1171.11 -0.07 1169.40 -0.22 1153.87 -1.55 7 1212.00 1210.69 -0.10 1206.22 -0.47 1192.15 -1.64 8 1252.00 1250.48 -0.12 1244.64 -0.58 1230.89 -1.69 9 1292.00 1290.48 -0.11 1284.37 -0.59 1270.11 -1.69 10 1332.00 1330.70 -0.09 1325.09 -0.51 1310.73 -1.60 11 1372.00 1371.12 -0.06 1366.49 -0.40 1348.84 -1.69 12 1412.00 1411.75 -0.01 1408.27 -0.26 1389.93 -1.56 13 1452.00 1452.59 0.04 1450.12 -0.12 1429.95 -1.52 14 1492.00 1493.64 0.11 1491.73 -0.01 1470.27 -1.46 15 1532.00 1534.91 0.19 1532.80 0.05 1510.54 -1.40 16 1572.00 1576.38 0.27 1573.01 0.06 1550.77 -1.35 17 1612.00 1618.06 0.37 1612.05 0.003 1590.93 -1.31 18 1652.00 1659.95 0.48 1649.63 -0.14 1631.04 -1.27 19 1692.00 1702.06 0.59 1685.42 -0.39 1671.39 -1.22 20 1732.00 1744.37 0.71 1719.12 -0.74 1710.13 -1.26 For N=3, a, =938.0379452 N a?= 38.2157864 F(s)=^a n s n 1 a 3= 0.1050509 /j* 1 For N=4, a,=999.5873340 %= 18.8517864 a 3 = 1.8834608 a 4 = -0.0513597

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53 Table 3.11 Comparison between true and retrieved conditions for eight inverse example problem solved for heat flux Boundary Conditions, q (W/m 2 ) Retrieved Time Imposed Series Method Incremental Error (sec) i rue Condition N=4 Error N=5 Error Method % % % 0.25 3003125.00 3007778.76 0.15 3033589.06 1.01 2995616.25 -0.25 0.50 3012500.00 3021875.02 0.31 3035992.94 0.78 3004320.50 -0.27 0.75 3028125.00 3040169.58 0.40 3043865.79 0.52 3019590.53 -0.28 1.00 3050000.00 3062934.56 0.42 3057990.24 0.26 3036905.38 -0.43 1.25 3078125.00 3090442.06 0.40 3078999.09 0.02 3061522.20 -0.54 1.50 3112500.00 3122964.20 0.34 3107375.31 -0.16 3092582.42 -0.64 1.75 3153125.00 3160773.08 0.24 3143452.06 -0.31 3130006.55 -0.73 2.00 3200000.00 3204140.84 0.13 3187412.63 -0.39 3173845.90 -0.82 2.25 3253125.00 3253339.57 0.01 3239290.49 -0.43 3223966.45 -0.90 2.50 3312500.00 3308641.38 0.12 3298969.30 -0.41 3280396.20 -0.97 2.75 3378125.00 3370318.40 -0.23 3366182.86 -0.35 3343088.04 -1.04 3.00 3450000.00 3438642.74 -0.33 3440515.14 -0.27 3411906.45 1.10 3.25 3528125.00 3513886.50 -0.40 3521400.30 -0.19 3487162.49 -1.16 3.50 3612500.00 3596321.81 -0.45 3608122.64 0.12 3568166.70 -1.23 3.75 3703125.00 3686220.76 -0.46 3699816.64 -0.08 3656322.04 -1.26 4.00 3800000.00 3783855.48 -0.42 3795466.96 0.12 3748732.19 -1.35 4.25 3903125.00 3889498.09 -0.35 3893908.40 -0.23 3851068.35 -1.33 4.50 4012500.00 4003420.68 0.22 3993825.95 -0.46 3951711.24 -1.51 4.75 4128125.00 4125895.38 -0.05 4093754.75 -0.83 4084808.11 -1.05 5.00 4250000.00 4257194.29 0.16 4192080.13 -1.36 4096803.88 -3.60 N G(s)=£a„s n-1 71*1 For N=4, a, =2997608.684 aj=33 190.75084 a 3 =29232.59062 a 4 =2902.536744 For N=5, a, =303572 1.67 a^-16210.2870 aj= 27733.4034 a 4 = 12343.6674 35 =1598.2 1379

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54 F(t) = 932 + 40t (3.38) G(t) = 3xl0 6 + 5xl0 4 < 2 (3.39) In these tables, the computational errors are calculated using the following definition: respectively . As described in Table 3.3, those in Tables 3.8 and 3.11 are for the medium initially at phase change temperature, while those in Tables 3.9 and 3.10 are for the medium initially subcooled at 300 K . The former are thus one phase problems while the latter are twophase problems. Again both series and incremental solution methods are used for solution and their results are good. For example in Table 3.8, the series solution results converge even with a value of N that is as low as 3, whereas in seeking the flux condition in Table 3.9, the series converges rapidly from N=4 to N=5 (higher degree results not shown). In Tables 3.8 and 3.11 the medium melts as soon as the boundary conditions are imposed, whereas in Tables 3.9 the medium starts to melt at time greater than 4 seconds. In all cases, the accuracy of the results appears to be unaffected by the time when melting takes place. The results in Tables 3.10 and 3.11 follow the same trends as the ones in Table 3.8 and 3.9. Tests Error=l —2 (3.40) where p and q represent retrieved and imposed true conditions, for the time variant conditions are thus successful.

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55 The inverse solution techniques developed in this study are expected to be accurate as mentioned earlier that is close to the end of the previous section. Yet for the Stefan problems solved in this work, the accuracy of the techniques is better at large time than small time. This can be attributed to the curvature of the interface position curve, which is always large at small time [2]; see Figure 3.2. Then, in the numerical solution of (3.27), there will be a slight error associated with the linearization of the position at small time. At large time, however, the position curve tends to be linear; the linearization error will be diminished. In fact, as time progresses, the accurate terms under the summation in (3.27) rapidly outnumber the inaccurate terms to the effect that the boundary conditions can always be evaluated accurately with the present method at large time; see the results in Tables 3.4 through 3.11. This is a distinct departure from the trends of other time marching schemes reported in the literature in which the errors tend to grow with time. It should also be pointed out that, for the Stef an-Neumann problem chosen for comparison in the present study, there is a singularity of the temperature at zero time. This also contributes to the large discrepancy of the results at small time, which must not be overlooked. It has been firmly established that, in the solution of the Stefan problems, only the Stefan-Neumann problems can be solved exactly. Yet, it is also possible to develop an exact solution for an exponential condition imposed on the boundary; such condition, however, has been considered as physically untenable in the literature [7]. Worse yet, such a condition gives rise to a

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56 constant velocity of the interface, a situation making the present linearization scheme exact in the solution of the inverse problems [5]. Perfect results will be obtained, rendering the test of the exponential conditions meaningless. 3 . 7 Extension and Concluding Remarks The analysis developed in this chapter can be readily extended for the solution of inverse problems with multiple phases. For such problems, times for re-melt and re-freeze of the medium must be closely accounted for, and the analyses [4,65] can be readily adapted for the development of the inverse solution techniques presented in this study. The present analysis can also be extended for the solution of inverse problems in which the properties are unequal for different phases of the medium. For such problems, double source and sink fronts must be used as given in the solution of the regular problems in reference 17. Finally, it is noted that although problems in one-dimensional, semi-infinite domain have been solved for examples in this work, problems in finite domains (e.g., plane wall) can also be solved with the present methods. For these problems, there are two boundaries and two boundary conditions are imposed. Two unknowns are thus sought simultaneously, and this requires the input of one additional condition in the form of either temperature or heat flux at any interior point close to the boundary where no phase change takes place. On the other hand, for problems with two phase change interfaces caused by separate heat input simultaneously from both sides of the boundaries, the interface motion data for the second interface will serve as this additional

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57 condition . In any event, no flux information is needed at both sides of the interfaces. Furthermore, the present method can also be applied to the solution of problems in multiple dimensions. Again the inverse solution for these problems can be developed on the basis of the solution of regular versions of these problems.

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CHAPTER IV SOLUTION OF ABLATION PROBLEMS WITH ONE MOVING BOUNDARY BY A SOURCE-AND-SINK METHOD It is the purpose of this chapter to present a source-and-s ink method for the solution of ablation problems with one moving boundary. As will be shown, the essential feature of this method is that the solution will be sought in a fixed domain which does not change with time. In fact, the ablated region is to be treated as a fictitious domain in which the flux condition at the ablated boundary will match that of the imposed condition at the moving boundary. Then, with the additional vaporization temperature given for the moving boundary, the conditions at this boundary are overspecified for the fictitious domain. The problem can thus be taken as an inverse problem in the sense that the condition on the fixed boundary is sought during the ablation period. This condition will provide for the necessary conditions at the moving boundary for the solution of the problem. 4.1 Solution Methodology For a subcooled medium, the problem can be divided into two stages; namely, pre-ablation stage and ablation stage. During the pre-ablation stage, heat is added to the surface of the medium in order to raise its temperature to the phase-change temperature. Then with continuous heating, ablation takes place. The ablated 58

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59 surface moves inward with time and the imposed heat flux follows this boundary motion. As shown in Figure 4.1, it is assumed that the medium is homogeneous and isotropic. The thermophysical properties are constant. For the ablation problem, the ablated region is immediately removed upon formation. Radiation is taken to be a surface phenomenon. It is also assumed that the medium changes phase at a distinct temperature; that is, no mushy zone in the medium. Moreover, there is no volumetric heat generation in the medium. Under these assumptions, the ablation problem can be formulated as follows: Pre-ablation Stage Governing equation-V 2 T ,. t x_i«W) * l 0 ( r »t) — a f G (A US) t 0 > t > 0 (4.1) Initial condition-T o (r,0) = T,(r) (4-2) Boundary condition-3T 0 (f 0 ,t) dn t ’ (4.3)

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Figure 4.1 System analyzed

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61

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62 Ablation Stage Solid Region: Governing equation-V 2 T 5 (r,t) _ 1 ST s (r,t) Q dt r € S t > t 0 ( 4 . 4 ) Initial condition-T s ( r ,t 0 ) — T 0 (r,to) ( 4 . 5 ) Boundary conditions-T s (r s ,t) = t„ ( 4 . 6 ) G(t) = k + pLvSl ( 4 . 7 ) where all notations have their usual meaning; see nomenclature. Here r
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63 The ablation problem above is solved by using a source-andsink method. In this method, the domain for investigation is extended to cover AUS; the ablation front becomes an interior phasechange interface. Solution is thus rendered in a fixed domain, in which the ablated region is fictitious in the sense that it is physically nonexistent; however, the unablated (solid) region is real. The conditions imposed at the moving boundary in the original ablation problem are taken to be those imposed on the interior phase-change interface. Thus, for the new problem in the fixed domain, the interface position and the fictitious condition on the fixed boundary are the unknowns to be determined. Once they are found, they can be used for input in the determination of the temperature in the solid region. The problem is thus solved totally. An equivalent problem is formulated with a source-and-s ink method as follows: Equivalent Problem Governing equation-V 2 T(f,t) pL — r/ — — \ 1 v.n«5(r r 5 ) = s dT(r,t) dt FG(AuS) (4.8) t > t 0 Initial condition-T(r, t 0 ) = T 0 (r,t 0 ) (4.9) Interface conditions--

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64 T(? 5 ,t) = T„ (4.10) dT(fg,t) <9n + pLv.n (4.11) where 8(r—r s ) denotes a Dirac delta function. As has been shown in Chapter 3, equation (4.8) reduces to (4.4). Also integrating (4.8) across the interface at r^ enables the G(t) to represent the heat flux imposed on the interface at the fictitious side. Equation (4.8) can be solved by using (3.18) in which the domains for integration for the first and second terms on the right are changed to AUS. Also minus sign is taken for the Dirac delta function term in which fy is changed to r s as T(r,t)= G(f,t | r',t 0 )T .(f')dV -/\ A U 5 t G(r,t | r',r)v.n<5(r' — r^)dV'dr + | t Q Au5 1 (4.12) This temperature equation is forced to satisfy the interface temperature equation (4.10) by setting r to the ablated surface position, r 5 , and T(r 5 ,t) to the phase-change temperature, T^, as t V G(r 5 ,t | r / ,t 0 )T-(r / )dV / — ^ G(r 5 ,t | r',r)v.7t6(r' r 5 )dV'dr + ^( j (4.13) A U 5 t Q ,4 U 5 Equation (4.12) is also differentiated to satisfy (4.11) as

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65 G(t)= k< ™^T j(f Oav' t L c 'A u S % 3G(r 5 ,t| •/ % dn -i -v.n6( v' — r^)dV / dr t 0 AuS + }_ _ } i r=r s ) + pLv.n (4.14) where differentiation is to be effected at the solid side of the interface. Notice that the boundary condition is embodied in the summation term, which, according to Table 3.1, should be expressed in the form »= a r( t ,-/ r ^T(r 0 ,t) G(r,t I r 0) ^) Qn arpor (4.15) o r o where the following condition is taken at the fixed boundary: dT(rp,t) _ G(fQ,t) dn : k for t < t 0 and — ^ for t > t Q (4.16) where g(rg,t) represents the imposed fictitious heat flux. Clearly, there are two unknowns in (4.13) and (4.14): g(Fg,t) and v; there are two equations to solve them. In this effort, t 0 and T 0 (r,* 0 ) are to be found from the solution of a pre-ablation problem as described previously . 4.2 Illustrative Examples Use is now made of the solution methodology given in the preceding section to solve examples in semi-infinite medium. Four examples will be provided and they include medium with or without subcooling with the boundary imposed with constant or variable heat flux conditions.

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66 For a subcooled medium, the pre-ablation stage solution has been given in Chapter 3. The pre-ablation temperature can be taken from (3.22) as T 0 (x,t) = T, + i N a 7 r t 0 G(s) (t-s) 1/2 exp[4a(t — s) ]ch (4.17) Thus the time when ablation starts (t 0 ) can be found by solving the following equation implicitly T = T-h1 1 ; 1 * ' V kN a 7 r (4.18) Also at the moment when ablation starts, the temperature in the solid region is given by the relation To( x ?lo) ” o r a 7 r 0 G(s) 1/2 exp[X 4a(< 0 -s) ]ds (4.19) which serves as the initial condition for the ablation stage. For the problem at hand, (4.12) can be used to derive temperature as J 0 T(x,t) = T,+ i a * 1 T — 0 t exp[ —77 v ]dr -f r (t -r ) 1/2 4a(t — r) k t a g(' r ) kNir 172 ex P[ X T — t (t -r ) 1/2 4a(t — r) ]d0 L c T—t d#i( r ) ^J47ra(t — r) 1 |exp( ft — t) ^ (x-^i(r)) : 4o(t — r) ) + exp( (x+ii 1 (r)) 4a(t — r) 0 (4.20) Also (4.13) is used to derive

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67 t o 1 T = T + A v ^ v a G(r) ex p[R i(t) kN7r J (t-r) 1/2 4a(t — r) r=o v ’ ]dr + r a 1 k N7T g( r ) r u_ exp[ — T777 Z\ J dr T— t 0 t L c r= t dT ^47ra(t — r) L_ {exp( ft — t) ^ (^(t)-^)): 4a(t — r) •) + exp( 4a(t — r) (^(tj+^r)) 4a(t — r) 0 (4.21) Equation (4.14) follows as t o G(t) = G(r) r R\{ t) nj , R,( t) ex P[ ~ T7Tl ~ l dr + 2^f!r fil(t) 2>f7ra J ( t _ r \3/2 4a(t — r) r=o v ' t g( r ) r u_ ex P[ — J dr T — t (t-r) 3/2 4a(t — t) 0 Lp 4>j7ra d ^i ( r ) dr r=t * 3/2 {(fli(t) ^ 1 (r))exp( (t-r) 4a(t — r) ) 0 + (fl 1 (t)+ii 1 (r))exp((R,(t)+R,(r)) 4a(t — r) + * dt (4.22) Solution of (4.21) and (4.22) subject to the initial condition, f?i(t 0 )=0, yields the results for two unknown functions i?j(t) and SO)4.3 Numerical Solution of Ablated Front Equations (4.21) and (4.22) are coupled nonlinear integrodif f erential equations, which will be solved numerically. In this effort, a local linearization scheme is employed as described in Section 3.4. For the present problem, the entire time range is dR,(t) and g(t) are divided into small increments in which both dt treated as constants. They can thus be taken out of their respective integrals, and the convolution integrals in these equations are changed to summations as

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68 t o 1 T V = T t + fc a G(r) exp[k ri 0 (tjV -^ /2 4 a (^-r) ]d r t _ N -f £«Jf E g(t„) K n=l J T—t 1 r ^1 ( t 7 v) :j ' xp[ 53(i^) ]dr n1 _L gdiii(t„) n /* C ^ n— l ai . ^ 47 ra(t N -r) r=t n-l (^l(t n) ~ ^i( r )) 4 a(t n -t) ) + exp( (fi^t N )+R 1 (r))\ \ A _ Aa(t N -r) >!* (4.23) and G(t j\f) — ^l(tyv) { J 0 GM „ p[ _agg) |dT 2 ^ U (t w -r ) 3/2 4 o (t N~ t ) T— 0 V N 7 t N + E g(t n ) n=i r=t n 1 exp[ R i(t n) ( t N -rf /2 n-1 t Lp n 4 MWa n ^ dt J (tjv _ r) 1 f/p /, \ n / \\ / (•^l( t 7v) _ -^l( r ))\ 73372 4a(tjv _ r ) — ) n-l , ,r> u \ . n / \\ t (^l( t Af) + -^l( T )) 2 \\j_ , + (Ri(tN)+ R i( T )) e M 4a ( t N -7j — ) / tlr + pL dt (4.24) Notice that the removal of diJ 1 (t) dt and g(t) will cause a slight error in the numerical solution of the ablated front position. However, R 1 is a gradual function of t, and so is g(t) during the early stage of ablation. The error associated with the linearization is expected to be small. In practice, such an error can always be reduced by taking small time increments.

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69 It is also noted that the . ablated position iZj(t) in the GreenÂ’s function can be related to by means of the interface velocity, Â’ as s ^ own figure 3.2. The position R a at any time t should thus be discounted as unknown. Numerically, the R 1 ( t) and g(t) in these equations will be solved incrementally starting from t 0 until the final time tjy is reached. This corresponds to a time marching scheme in the numerical solution. Computer programs developed for the solution of the ablation problems given in this chapter are compiled in Appendix F. 4.4 Results and Discussion Four examples are tested and they include semi inf ini te medium with and without subcooling with the moving boundary imposed with constant and time-variant conditions (see Table 4.1 for summary). Notice that, in this table, under the column of problem description, the ablated region is always counted as one phase. Thus, for a medium initially at phase-change temperature imposed with a constant heat-flux condition, such as Example 1, there is one ablated phase. It is thus called one -phase problem in the table. Example 1 deals with an ablation problem that can be solved exactly. As shown in Carslaw and Jaeger [7], for a semi inf ini te medium ablated with a constant velocity U, the heat flux imposed on the boundary is given by the relation G = [L + c(T ( ,-T i )]pU (4-25) where T t is constant and the velocity is related to the ablation position as

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70 Table 4.1 Conditions tested in four examples Problem Description Material Heat Flux Condition Imposed G(t) (W/m 2 );! (S) Remarks 1. One-phase T=932 K=T m =T v Aluminum G(t)=5xl0 5 Exact solution is available for this case. 2. Two-phase T—882 K
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71 U _ (4.26) It is expected that, for a medium initially at the phase-change temperature, G is equal to /?LU product. It is also noted that equation (4.25) is strictly valid for the medium in a quasi-steady state. At that time, the temperature in the solid is stabilized given as T(x,t) = T v e ~ ~ Ut ) (4.27) which occurs for the medium having ablated for a long period of time. The first two examples are thus chosen to substantiate these points . For the first example, the ablation is due to melting. The medium is initially at the melting temperature so that the surface ablates as soon as the heat is applied. The position of the ablated surface is shown in Figure 4.2, where the curve appears to be linear. Using the exact solution (4.25) for comparison yields the error curves shown in Figure 4.3. Clearly, the results are stable and converge satisfactorily, errors being less than one tenth of one percent for all the time increments tested. Also the errors are slightly larger at small time, a result of the singularity associated with the heat flux that is abruptly appl ied on the surface at time zero . In what follows in the numerical computation of all the examples to be presented in this chapter, the time increment is taken as 0.5 sec, unless otherwise noted. Example 1 provides an excellent test for the algorithms

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0) P c 3 0 P aj P o3 Sh * rH 0) ~C U a c 0 E 0 P a) u p c X 0 0 3 • pH bO f-H P G P • pH o3 i to J3 P 0 0 o3 a l 0) > o rH P i-H o3 o3 o3 pH • iH X5 P P C$ • Hi »pH G 2 P • pH 0 TJ E 0) ~0 3 co C • pH O 0) tj a u 0) E H E • r “ l CN V U s bO • pH Uh

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73 o GO a CO A CM o # w o w o w o o o (ujo) ‘uoi^isod ooDj-ins p9}D|qy Time, t (s)

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U U P P p U cfi O O In 0)

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At=1.00 S 75 % J0JJ3 Time, t (s)

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76 developed for the solution of the ablation problems in this dissertation. In fact, the good results obtained in the first example are somewhat expected because of the linearization used in the solution of the integrodif f erential equations for which the ablation rate itself is linear. Example 2 is different because it deals with a medium which is initially subcooled (see Table 4.1). This time, there is no exact solution for checking the results, which have also been tested for convergence in Figure 4.4. Of interest in this figure is the slightly smaller ablation rate at small time, which can be attributed to the large slope of the temperature curve at the solid side of the moving boundary (not shown in dissertation). At large time, this slope diminishes, and finally it is stabilized as evidenced by the linearity of the position curve at large time. Using the numerical data over the last time step in Figure 4.4 yields an ablation rate of 0.00161 m/s, which is in good agreement with that computed using equation (4.25), error being less than 2.5 7c. This gives a good indication of the ablation approaching the quasi -steady state. An effort is also made to check the results of Example 2 with the analytical solutions reported in the literature. As shown in Figure 4.5, both the heat integral method (HIM) of Vallerani [40] and the moment method (MOM) of Zien [41] yield results that are in close agreement with the present source-and-s i nk method. LandauÂ’s results also fall right on the curve and have thus been omitted in the plot. Example 2 has thus been tested successfully. Examples 3 and 4 deal with a subcooled medium imposed with

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~0 0) SS ^ CD 0 X CD 0 0 r— H 0) „Q JS P 1 4^ CO ^P d P-< (3 CD 0 p 0 ^P CO 4h a 0) d

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78 o to o to o in o • • • • • • • m (N CN ^ o o (uuo) ‘uo^isod aoD^jns pepiqv Time, t (s)

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81 time-variant heat-flux conditions. A linear heat-flux is imposed in Example 3, while a quadratic heat-flux is imposed in Example 4. Like Example 2, the medium is subcooled initially; there are two phases in the medium for both examples. Comparison of the present solution with the moment method for the linear heat-flux condition is given in Figure 4.6; convergence test based on the change of time increments for this condition in given in Figure 4.7. As shown in both figures, the results for the source-and-s ink method are in close agreement with the moment method, and the results have converged with a time step as large as 0.5 sec. Similar results are seen in Figures 4.8 and 4.9 for the case of quadratic heat-flux. The tests for ablation problems with one moving boundary have thus been completed satisfactorily.

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83 (uio) ‘uoi^isod ©ODpns papiqv Time, t (a)

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87 (ujo) ‘uoi^isod ooD^jns pe}D|qv Time, t (s)

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89 (ujo) *uoi^sod ooD^jns po}D|qy Time, t (a)

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CHAPTER V SOLUTION OF ABLATION PROBLEMS WITH TWO MOVING BOUNDARIES BY A SOURCE-AND-SINK METHOD In this chapter, the source-and-s ink method is used to solve more general ablation problems in which ablation occurs at the surface, while, at the same time, the solid melts and the melting front moves inward with time. In these problems, there will be three phases occurring simultaneously: the ablation layer at the surface, the liquid layer near the surface, and the solid region deep inside the body. There are two moving boundaries: one is the moving ablated surface and the other is the moving solid-liquid interface. Their positions are unknown a priori and must be determined as a part of the solution. These problems can be considered as the combination problems since they incorporate all the features of the ablation and Stefan problems studied earlier. Using the same concept employed in solving the ablation problems with one moving boundary in Chapter IV, the ablated region is again considered as a fictitious region. The heat flux applied on the ablated boundary is then used to match the imposed condition at the moving boundary. This condition together with the vaporization and melting temperatures at the ablated and melting fronts, respectively, give sufficient information for the solution of the problem. 90

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91 5 . 1 General Analysis Considering a subcooled medium, the present ablation problem consists of three stages: pre-melt stage, melting stage, and ablation stage. In the pre-melt stage, heat is added to the surface of the medium to raise its temperature to the phase-change temperature. Then as heat continues, melting stage commences with the melting front starts at the surface and moves inward with time. With further addition of heat at a sufficiently high rate, vaporization may take place. It is assumed that the vapor is removed instantaneously upon formation (by being blown away, for instance) while the melting material stays in place. This is taken as the ablation stage. Use is made of the system shown in Figure 5.1 for analysis. It is assumed that the thermophysical properties of the liquid and solid states are constant and of equal value. The medium is homogeneous and isotropic; phase changes take place at distinct temperatures. Then, in the absence of radiation, convection, and volumetric heat generation, the formulation of the problem can be given as follows. Pre-melt Stage Equations for this stage are the same as those for the same stage given earlier in Chapter IV; see equations (4. 1-4.3). Note that, for the problem at hand, the domain in equation (4.1) should be changed to r E (AULUS).

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Figure 5.1 System analyzed

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£6

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94 Melting Stage Liquid Region: Governing equation-V *r L (r,t) = £ 1 <9T L (r,t) <9t Boundary conditions-f GL 1q < ^ ( 5 . 1 ) G(t) = _ k ™ ( 5 . 2 ) Solid Region: Governing equation-V *r 5 (r,t) = i 1 5T s (r,t) <9t Initial Condi t ionsr G S ^0 < 1 ^ ( 5 . 3 ) T$(r,t 0 ) — T 0 (r,t 0 ) ( 5 . 4 )

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95 Interface conditions-T L (f„t) = T s (f„t) = T ( 5 . 5 ) OT 5 (f„t) dT L (r„t) _ />L, " — “ 17 “ v /‘ n / dn l dn l ( 5 . 6 ) ( 5 . 7 ) Ablation Stage Liquid Region: Governing equation-v 2 T iL (r,t) 1 dT LL (r,t) _ « dt f €L ( 5 . 8 ) t < t V Boundary Condition-^'TLL( r 5,t) dn s + n s r ^LL( r 5> t ) _ ^ ( 5 . 10 ) Initial condition-T LL (r,t v ) — T l ( r y t v ) ( 5 . 11 )

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96 r 5 (t v) = r o Solid Region: Governing equation-( 5 . 12 ) dt r G S t < t V ( 5 . 13 ) Initial condition-( 5 . 14 ) Interface conditions-Tss( r >* v) v) TliX 1 /’*) — ?SsM ~ ( 5 . 15 ) ^ T Sg( r /» t ) _ ^ T Lt( r f’ t ) _ ^7 v jj drii dni k 1 1 ( 5 . 16 ) ( 5 . 17 ) where r 0 denotes the original surface position, and r t and r 5 represent the solid-liquid interface and ablated surface positions, respectively; Vj and v^ are the velocities of these two interfaces, which are unknown for the ablation problem. The time when melting starts, t 0 , and the temperature distribution in the medium at the

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97 onset of melting, T 0 (r,t 0 ), are to be found from the solution of the pre-melt stage. On the other hand, t v is the time when ablation begins; T L (r ,t v ) and T 5 (r,t v ) are the temperature distributions in the liquid and solid regions, respectively, at the beginning of the ablation stage. Like before, t v , T^r,^), and T^-(r ,t v ) will be found from the melting stage solution. The source-and-sink method is used to derive the temperature for the ablation problem. In this method, the solution will be sought in a fixed domain which is extended to cover AULUS; the ablation front is taken to be the second interface which separates the gas and liquid regions. The first interface appearing in the me 1 ting stage is the interface that separates the liquid and solid regions As is previously the case, the ablated region is a fictitious region, while the solid and liquid regions are real. The interface conditions imposed at the melt front in the original ablation problem are still those in the new problem. However, the conditions imposed at the surface are now taken to be the conditions at the interface between the liquid and gas regions. The equivalent problem is thus worked in a fixed domain so that the surface of this domain is exposed to an imaginary heat flux whose magnitude is adjusted to meet the interface conditions specified. In this method, the interface locations and the fictitious heat flux on the fixed boundary are treated as unknowns. Once they are found, they can be used in the temperature equation to complete the solution. Following this approach, the ablation problem is recast into an equivalent problem as follows.

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98 Equ i va 1 ent problem Governing equation-f G (A ULUS) V 2 T(f,t) — v M(rr j) pK k _ ^ _ v _ 1 dT(r,t) v s .n s o 2 (T t s ) — a , (5.18) t>t„ Initial condition-T(r,t J = T x (r,tJ (5.19) Interface conditions-G(t) = -k dT(r 5 ,t) dn c ( 5 . 20 ) T(rj,t) — T m ( 5 . 21 ) T(r s ,t) = T„ ( 5 . 22 ) Here equation (5.18) can be reduced to equations (5.8) and (5.13) by using the definition of the Dirac delta function. Furthermore, by integrating (5.18) across r j and r^, one obtains (5.9) and (5.16). Other equivalences for this stage are apparent. In (5.19), T 1 (r,t () ) is the temperature distribution in the medium at the onset of

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99 ablation, which is found by solving the melting stage. Using equation (3.18), the melting stage solution can be expressed as T x (r,t)= G(T,t | r , ,t 0 )T t (r , )dV / G(r,t | r , ,r)v i .n,<5(r' f z )dV'd r + E { } LuS t 0 Lu5 (5.23) Note that the minus sign has been taken for the Dirac-del ta-f unction term; r^ and L in that equation have been changed to r; and Ly, respectively. This equation is used to derive the solid-liquid interface position by setting r to r^ and T 1 (r^,t) to T rn as T = m L t G(r„t | r , ,t 0 )T i (f , )dV / 7 G(f„t | r',r)v,.n,6(r' r,)dV'd r + E { } LuS t 0 LuS (5.24) Also substituting t v for t in (5.23) gives the temperature at the onset of ablation as t V Tl( r ^v)“ 7 G(f,t„ | f / ,t 0 )T,(r')dV' G(r,t„ LuS t 0 LuS r',r) v pU;<5(r / r^dV'dr + E{ j (5.25) Changing r to r 0 and T(r 0 ,<„) to T,, gives T = V ^ V G(f 0 ,t„ | f',t 0 )T,(r')dV' 4 \ f G ( f o I r , ,r)v / .rV(f'-f,)dV'dr + E{ } LuS t 0 LuS (5.26) which will be solved implicitly for the time when ablation starts (*„)•

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100 In the ablation stage, the equivalent problem can be solved by again using equation (3.18) in which the domain of integration in the first and second terms on the right are changed to AULUS. Also the Dirac delta function term is modified to include both solidliquid and ablated interfaces; again minus signs are taken for the modified term, giving the result T(r,t)= G(r,t | f',tJT .(f')dV' Au LuS t L c t A U L U 5 v G(r,t | r',r)| Vj.n^r' r,) + v s .n s 6(i' r s )jdV'dT + £){ j ( 5 . 27 ) Then by setting r to r j and T(r^,t) to T m in this equation, there is derived /» G(f, ,t | ?',t 0 )T,.(r')dV' A u L U S G(r, ,t | r',r)| v^.«^(r / r ; ) + v s .h s 6(t' r 5 )|dV / dr + E{ } t A U L U 5 v ( 5 . 28 ) In a similar fashion, setting r to r^ and T(r^,t) to T v in (5.27) gives the following equation for T v as

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101 T = V A u L U S G (?5 ,t | f',tJT t (r')dV' t L c t A U L U S v G(r s ,t | r',r){ v,.n,S( r' r,) + v s .n s 6(r' r s )|dV , dr + } (5.29) Furthermore , equation (5.27) is differentiated and forced to satisfy (5.20) as G(t)=-k< dG(r s ,t | r ,tj diiQ T l (r , )dV / -^ L c 9G(r 5 ,t | r',r) r _ 8n | Vj.nj6(r / — T[) A u L U S t A u L u S v + v s .7i 5 5(r / -r 5 )}d v/ dr+ _ i + />I u v s .n, i S T — Tc ) (5.30) In the equations given above, the boundary condition is embodied in the summation term, which is to be taken from equations (4.15) and (4.16) in which t 0 and are changed to t v and n^, respectively. Clearly, equations (5.28) through (5.30) contain three unknowns (vj, v 5 , and g(rQ,t)); there are three equations to solve them. The to solve examples constant The 5.2 Examples general analysis given in the previous section is now used example problems in a semi inf ini te domain. Here three are given, and they are for a subcooled medium imposed with and time-variant heat-flux conditions, pre-melt stage can be solved by using (4.18) and (4.19). The melting stage can then be solved by using (5.23) as

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102 1 T 1 (x,t)=T i + A a G(r) exp[ — k N,r J (t-r) 1/2 4a(t — r) T— 0 V ’ 1 A j47ra(t — r) (x-R^r)) 2 4a(t — r) ) + exp( (x+R 1 (r)) 2 4a(t — r) r (5.31) Correspondingly, “the 'temperature distribution at t v can be derived by using (5.25) as 1 V /> T 1 (x,t v ) = T , + fa G(r) exp[ — X k N,r J ( t _ r )i/2 4 a(t v -r) T— O'' " ’ ]dr t V L / f dRj(r) dr hi (“ p( " T — t J47ra(t„ r) (x-R^r)) 1 4ar(t v -r) •) + exp( (x + R x (r)) 4a(t„ t) 0 (5.32) which is used as the initial condition tor the ablation stage. Also y using equation (5.26) gives t t, = T,+ 1 1 kN’r V r t G(r) t; 1/2 dr — 7 dRx(r) exp( R 2 j(r) r=t dr A J 7 ra(t t; -r) 4a(t„ t) ;)d' 0 (5.33) which is solved implicitly for t v , the time when ablation starts. The ablation-stage solution is derived by using (5.27), (5.28), (5.29) as

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103 t t; T(x,t) = T,+ i «' + W j (t^i/2 CXP[ " 4a(t-r) ]dr + J (t /2 M* ~ r ) t=O k ’ r— t„ g(r) ]dt T — t dR a (r) N j47ra(t — r) \ ( , (x-R^t))^ , , (x + R 1 (r))\' lj h7) W “ 4.(t-r) 1 + ' XP( " 4„(tV-)) d 0 t t; T = t dR 2 (r) dr A j47ra(t — r) 1 f , (x — R 2 (t)) 2 . (X+R 2 (r )) 2 -1 bri W + “ p( V (5.34) V 1 T mT i + V a G(r) exp[ — R 2 (t) t W* J (t _ r) l/2 4a(t — r) r=o v ' ]dr + p a 1 k^ 7 r r=t g( r ) (t-r) 1/2 exp[R?(t) 4a(t — r) ]d' t> r=t dR 1 ( r ) d?" ^j47Tor(t — r) ^ i exp( ft _ ^ I V (R^-R^ 4a(t — r) -) + exp( (R a (t) 4R 1 ( r )) o v -l_(exp( dr ^ 4 xa(t r) \ P( T — t 4a(t — r) (Ri(t) + R 2( r )) 2 A dr 4a(t — r) J V (5.35) V T = T + r a G(r) exp[ R 2 2 (t) t ’ kNrJ ( t _ r )l/2 KL 4a(t — r) T— 0 V ' ]d r + r cv 1 kNr g( r ) T — t (>-r) 1/2 exp[ — R&t) 4a(t — r) ]dr V L / f dRj(r) r=t dT — r) = . = iexp( — ft tH (R 2 (t) — Ri( r )) : 0 v T—t dR 2 (r) _ dr ^ 47 ra(t r) l i exp( — ft — r) * 4a(t — r) (R 2 (t) R 2 (r)) : 4a(t — r) -) + exp( (R 2 (t) + Rj(r)) ) + exp( 4a(t r) (R 2 (t) + R 2 (r)) 4a(t — r) V (5.36) Finally, using (5.30) gives the flux condition as

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104 t G « = £#{ t; G(r) r ex P[ r2(1) t 2>flra IJ ( t _ r \3/2 4a(t — r) r=0 v ' ]dr + T — t g( r ) (t rfl exp[R|(t) 4o(t — r) v t P L f f dR a (r) 1 4 r — t dr (t _ r) 3/2 |(R 2 (t) R 1 (r))exp( (R^Q-R^r )) 2 4a(t — r) o + (R 2 (t) + R 1 (r))exp( (R 2 (t) + R 1 (r)) 4a(t — r) ^L } y t pK A^Wa r — t dR 2 ( r ) 1 dr (t _ r )3/2 |(R 2 (t) R 2 (r))exp( (R 2 (t)-R 2 (r)) : 4a(t — r) •) t; + (R 2 (t) + R 2 (r))exp( (R 2 (t) + R 2 ( r )) 2 N \,,_ , „ T dR 2 (t) 4a(t _ r) )jdT + pL„ dt (5.37) Equations (5.35), (5.36), and (5.37) contain three unknowns: Rj , R 2 , and g(t). They must be solved simultaneously by a numerical method as shown in the following section. 5 . 3 Numerical Solution of the Ablation Problem A local linearization is again used for time (see Figure (5.2)), and the melting stage is solved first. In this effort, equation (5.24) is recast in a summation form as T m ~ T i + P a 1 kNTT G(r) R i( t N 1 ) ^ ’ exp[ 777 ; ^7 ]dr “ r)1/2 4«(t w -r) Nt L / -i dR t (t n ) C n=l dt n + exp( (Ri(t Ni )+Ri(r)) 1 x { ex P( • at — r) L 4™( t ;v 1 -r) (Rj(tyy^) R;(r)) 4a(ttf -r) ) 4a(t n ~ t ) (5.38)

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Figure 5.2 Linearization of solid-liquid interface and ablated surface position curves for the numerical solution

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106

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107 where R 1 (t 0 )=0; N^ = l ,2, • . . 9 N 1 ; This equation is to be used to solve implicitly for t v , the time when the ablation takes place. Following the similar approach, (5.35), (5.36), (5.37) are linearized to be t T = i m 1 oj n V G(r) r R l^N 2 ) expf-7777 =-^ ]dr kN7r| -J (t„ — r) 1 / 2 4< *0w T ) T= 0 V 1 N. t n + E g(t„) i n-Afj + 1 • (t N r) 1 / 2 T l n-l 2 exp[^-1 4a ( t 7V 2 _ T ) L, i? dR,(t n ) t n — r E C ^ n=l Ql 1 .4»a(t« -r) T= ^n-l ’ C (Ri^AT ,) — Ri( r )) 2 (R,(t N ) + R x (r)) . H 4aftiv r) > + “ P( 4 r) ^ 4tt ( t W 2 T ) k £ dR 2 (t n ) + 1 t n dt l 4va(t N -t) (R^J-R 2 (r)) 2 (R x (t N )+R 2 (r)) 2 — aoL — } + exp( 4a ^N 2 ~ T ) t T = 1 qJ n “ kNX U (t„ -r) 1 / 2 r=0 v ; G(r) r R 2( 4 JV 2 ) v ; exp[ Y777 =-^ ]dr 4a ( t N 2 r ) N. t n + E g^n) 1 n-N 1 + l | (t N -r) 1 / 2 r_ l n-l exp[ R 2 (
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108 R 2( t JV 2 )rf G(r) R 2 ^N 2 ) 1 , 2>f•a {]_ ( tjv . _ r )3/2 CXp ^ 4o(t Wj -r) N. t n + ]£ S^n) 1 n=^i + 1 * j ( fc JV 2 T ) 3/2 r ~ l n-l 2 exp[ R 2( t AT„) 4a ( t N 2 “ r ) pL f ga dR+J dr _* (^N~ r ) n y 4^J7fa n •e' 1 r=t n-l' J ' 2 {(^(‘nj) R i( r )) ex p( (R 2 (t^ 2 ) R 1 ( r )) 2 4a (tjv 2 r ) + ( R 2( t N 2 ) + R l( T )) eX P(“ (R 2 (< w ) + R^r)) 2 . 4«(t n 2 r) ) J dT PK 4 y[Wa ri _ N t E ? dR 2 (t n ) n dr n=N l+ i “* J (t N -r) l n-l 2 1 3/2 {( R 2( t N 2 ) “ R 2( T )) (R 2 (W ) R 2 (r)) 2 ( R 2 (tN 2 ) + R 2 M) 2 , exp( ~ ^77 ^ ) + ( R 2 (tN J + R 2 ( r )) ex P( 4tt( ^r) 4a ( t N 2 ~ T ) dR 2( t JV 9 ) + pL v dt (5.41) where R 2 (^n ) = 0> N 2 =N 1 +1 , N 1 +2,...,N 2 ; t v < t N . l ^ A time marching scheme has been developed for the solution of these equations. In this effort, #i(t), i^ 2 (t), and g(t) are solved simultaneously using (5.39), (5.40), and (5.41). Starting from t n they are solved incrementally step-by-step until the desired time is reached. A computer program (SSM) useful to solve the ablation problem with two moving boundaries is provided in Appendix G.

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109 5.4 Numerical Solution of Temperature and Energy Storage Once the interface positions and the condition (i^ 1 (t), g(t)) are found numerically, they can be used in (5.34) to determine the temperature in the medium. In the absence of solutions for the combination problems in the literature, this temperature distribution is vital for checking the accuracy of the solution in the present investigation. Use is thus made of thermodynamic principles to derive the total heat consumed in heating, melting and vaporizing the medium as R 2 (t) AV i W oo Qact = p{ [ [K + L/ + C(T„ T,)]dx + f (l, + c[T(x,t) Tj) dx + [ c[T(x,t) T,]dx} V^ ^ Ri(t) o R 2 (t) Ri(t) (5.42) The true heat input can be evaluated by using the imposed condition as t true G(t)dt 0 An overall error can then be defined as (5.43) E = (5-44) ^ true Which embodies the errors not only in the temperature distribution but in the interface and boundary positions as well. This error can thus be safely taken to be the upper bound for the error in the solution .

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110 5 . 5 Numerical Examples The analysis devoted in the previous section was employed to obtain the numerical results for three examples imposed with constant, linear, and quadratic flux condition at the boundary of a subcooled medium; see Table 5.1. Again, aluminum has been used for tests; its properties are given in Table 3.2. The algorithm developed for solution of these examples. The pre-melt stage is solved first to yield tf 0 , the time when melting starts, and T 0 (r,£o), which is the temperature in the medium at the onset of melting. This temperature is, in turn, used as the initial condition for the melting stage to solve for t) , the solid-liquid interface position, t v , the time when ablation starts, and T(r,< v ), which is the temperature in the medium at the onset of the ablation. Finally, this last temperature is used as the initial condition for the ablation stage to solve for i£j(t), R 2 {^) • In this effort, the heat flux imposed at the boundary at x equal to zero is changed to g(t) over the time interval ($ V ,J) to account for the fact that, during that interval, the imposed heat flux at the fixed boundary is hypothetical in the sense that it is nonexistent physically. However, this flux is unknown; equations, (5.39), (5.40), and (5.41), must therefore be used to solve them (/^(t), R 2 (t) , «(*>*!,))• For a subcooled medium exposed to a large heat flux till vaporization, three phases appear in the medium. Examples 1, 2, and 3 address three situations, in which Example 1 is for a medium exposed to a constant heat-flux, Example 2 is for a linear heatflux, and Example 3 is for a quadratic heat-flux (see Table 5.1).

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Ill Table 5.1 Conditions tested in three examples Problem Description Material Heat Flux condition Imposed G(t) (W/m 2 ); t (s) 1. Two-phase Ti=300 K
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112 Expectedly, in all these examples, the melt front appears first in the medium, and the ablation front follows sometime later. To show how these fronts move, Figures 5.3, 5.4, 5.5 are prepared for Examples 1, 2, and 3, respectively. It is interesting to note that, in all these figures, the melting front positions have very large curvature bended downward (see top curves in these figures), whereas for the ablation front positions all curves upward (see bottom curves) in duplication of what was seen earlier in Examples 3 and 4 in Chapter IV. A temperature plot is also provided for the constant heat flux case as shown in Figure 5.6. Here three temperature curves are drawn for three different times covering the moment when vaporization starts and two instants of time when vaporization is in progress To facilitate viewing the positions of the moving boundary as well as the solid-liquid interface, horizontal lines are drawn at the melting and vaporization temperatures. Thus the xposition of the intersection of the curves with the vaporization line locates the positions of the moving boundary, while that of the curves with the melting line locates the positions of the solidliquid interface. Not given in this figure are temperature curves in the hypothetical regions for obvious reasons. The numerical method developed for the solution of the combination problems in this chapter has also been tested for convergence and stability as shown in Figures, 5.7, 5.8, 5.9. They are good as shown in the figures. It should be noted that, for the numerical results presented in Figures 5.3 through 5.6, the time increment chosen is chosen to be 0.5 sec.

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T3 0) E 'P PDA r “ H ctf hh jD 4-5 -O O CO Ct P 3 a o "C U 3 c O C$ (D P X U Cj 4-5 b 3 c P"< »»H ^ P S
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fe Solid— Liquid Interface, Ri (cm) o o o o a o A o A o • o • o • o w o o o o o ro 04 CJ1 o> Ablated Surface, R 2 (cm) f’TI

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TJ 43 U 03 4J G »*H Hh • pH ^ p X 0) E T> p> O 0) g O CO •pH 0 a T3 E •PH P "H S3 O ctsh e g • pH 33 O p— H CO «PH • PH 1 g tj 'p TJ o a; • pH .pH £ ^ P—H 'P G 0 • pH -O O CO (0 0) u O r-H a o x 0 O G (D O *"H CO U pO Sh T5 G 1 G Sh m p 0) P o3 P G Hh 0) H (0 0 pG IO 0) P bO • iH Uh

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s o • pH p c$ S • pH P ,0 E O -3 E U P s •i— i cb ^ T3
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120 o ro o CO in CM CM o o o> CO K) O) CO CM o r; £ cm ro h* !£2 £2 H o> o ro m CM o CM in o in (>0 (V x )l 'oJnpjodoiej. Distance From Fixed Surface, x (cm)

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c d c$ 25 0 CD • rH X CD P p • rH D c X • rH p X! £ p 0 Cfi 0 0 0 a p > • rH P u p T5 • rH c 0 0) -o 0) • rH E fi to P 0 o3 TJ u r—H X r-H X c c$ 0 0 0 r—H u P o P 0 X i ”0 p P fl P a> «$ 0 d> • rH p X >» P 0 p 3 p • rH r—H E c r—H 0 0) c$ • rH w r—H P X X W o3 0) 0 c P X u 0 CD p a u LO 0) Sh 3 bO • rH Uh

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122 (uuo) ‘©oD^jns popiqv (uuo) * 90 Dj,J 8 ^u| pinbi“|-P!|os Time, t (s)

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oro 124 (iuo)*y ‘©ODjjns papiqy (wo) l y * 90 Dj,J 9 )U| pjnbn-P!|os Time, t(s)

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126 (lud) z y ‘soD^jns ps}D|qy (iuo) l y *ood^j®^u| pjnbn-pjios Time, t (s)

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127 To the knowledge of the author, the combination problems solved in this chapter have rarely been attempted in the literature. Use is thus made of (5.44) to test the overall accuracy in the solution. In this regard, an error plot is prepared for example as shown in Figure 5.10. Here the overall error is shown to reach one percent asymptotically. However, it should be noted that (5.44) encompasses errors not only in the temperature distribution but in the boundary positions as well. Its being one percent in the figure provides renewed assurance of the overall accuracy in the work. It can also be taken as the upper bound for the error in the method.

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129 % J0JJ3 IIDJ8AQ

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CHAPTER VI CONCLUSIONS AND RECOMMENDATIONS In the present study of the solution of phase change problems, it has been assumed that the thermophysical properties of the liquid and solid are constant and of equal value and convection and radiation effects are neglected. Based on the results obtained from the analysis, the following conclusions can be drawn: 1. The source and sink method has been developed for the solution of three phase-change problems, encompassing inverse Stefan problems, regular ablation problems, and the combination of ablation and Stefan problems. In each case, general solution methodologies are developed first, and they are applied to the solution of specific examples which are in semi inf ini te space and imposed with either constant or time-variant temperature and flux conditions. The medium may be subcooled or superheated, but the properties are constant and of equal value in the liquid and solid phases. 2. Source-and-s ink method has been used in the solution of all problems. In this method, a melting interface is taken to be a moving heat-sink front and a freezing interface is taken to be a moving heat-source front. Thus one set of equations is used for the solution of the temperature in all phase regions. In this effort, GreenÂ’s functions are used and one temperature equation is derived. 130

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131 Whether it is in the solid, liquid, or gas region depends on the position that is assigned in the equation. 3. For all the examples solved in this work, the interface positions are evaluated numerically. In this effort, a local linearization scheme is employed which divides the entire time range into small increments in which the interface velocities and the boundary condition (if unknown) are treated as constants. They can thus be taken out of their integrals, and the original convolution integrals are expressed in summations. This method is motivated by the fact that the interface position curves are usually a gradual function of time. They can thus be approximated as linear over small time increments. According to the present study, this local linearization only causes a slight error in the numerical solution of the interface position, which can still be reduced systematically by taking small time increments. 4. In the use of the source-and-sink method in the solution of the inverse Stefan problems, two approaches have been developed and they include the series expansion method and the time incremental method. In the series expansion method, the boundary condition is expanded in a power series, and the interface motion data at equal time intervals are used for input to solve for the coefficients in the series expansion. This power series is, in turn, used to generate the conditions at other times. On the other hand, in the time incremental approach, the interface motion data at consecutive times are used for input to determine the conditions imposed at the boundary incrementally. In either case, since the type of the condition imposed on the boundary is unknown a priori, a

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132 method has been developed for determining the boundary temperature via the evaluated heat flux. Test results with eight examples indicate that the methods converge and are stable. Of the two methods developed, the time incremental approach is particularly attractive. Not only is the method more accurate, but also for the fact that the Stefan problems can be solved with this method by simple algorithms. Test results also suggest that the two-step so lution of the flux then temperature may accumulate large error In fact, the time saved in such a two-step operation is insignificant as compared with the separate, one-step evaluation of the temperature and heat flux. Solution of the inverse Stefan problems is successful. 5. The source-and-sink method has also been used in the so lution of ablation problems and the combination of ablation and Stefan problems. In this application, the problems are solved in a fixed domain, and the condition that is originally imposed on the moving boundary is taken to be the condition imposed on the interior moving interface. Then by solving the motion of this interface together with the flux condition that is imposed on the fixed boundary, the temperature in the medium can be determined numerically. Seven examples have been provided that include one-, two-, and three-phase ablation for medium imposed with constant, linear, and quadratic flux conditions. The moving boundary positions have been compared with those evaluated with the methods documented in the literature. In all cases, the results are good. Errors are less than one percent, which include those in the

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133 boundary position as well as in the integrated temperature in the medium. Solution of the ablation problems are also successful. Following recommendation are made on the basis of the work done : 1. The analysis developed for the solution of inverse Stefan problems can be extended to the solution of inverse problems with multiple phases. For such problems, times for re-melt and re-freeze of the medium must be considered, and the analysis described in references 4 and 65 can be easily adapted for the solution of such problems. Also the problems solved in this work are in onedimensional semi inf ini te domain. Problems in finite domains can also be solved with the present method. For these problems, there will be two boundaries where two boundary conditions are imposed. Two unknowns must therefore be found simultaneously, and this requires the input of one additional condition at any interior point that is close to the boundary where no phase change takes place. 2. The thermal properties of the liquid and solid regions have been treated as constants and of equal value in this study. In practice, these properties may differ and may be a function of temperature. In these instances, double source and sink may be employed as suggested by Kolodner [17], and Kirchhoff transformation may be used for cases when the conductivity is a function of temperature. The property variations will be accounted for in future studies. 3. It will be of practical interest to solve the inverse Stefan problems and Stefan problems in two or three dimensions. It is expected that, because of the presence of the second (or third)

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134 dimensions in these problems, the solution will be difficult. A hybrid analytic-finite difference scheme may be needed to solve such a problem.

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APPENDIX A EXTENSION OF THE SSM For a Stefan problem in two-dimensional Cartesian system, the interface position can be represented as: y / = R(x / ,t) Then, according to References 69 and 72: v n (t)«(F-ry) = S( y-yy) Three-dimensional cases and problems in other coordinate systems can be formulated accordingly. 135

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o o o APPENDIX B SSM FORTRAN PROGRAM FOR INVERSE STEFAN PROBLEM C********************************************************************^ C SOLUTION OF ID INVERSE STEFAN PROBLEMS IN A SEMI-INFINITE C C MEDIUM WITH OR WITHOUT SUBCOOLING BY THE SOURCE-AND-SINK C C METHOD USING SERIES APPROACH C C C C INPUT FILE UNIT NUMBER IS SET AT 12 C C OUTPUT FILE UNIT IS SET AT 8 AND 16 C C C C NOTATIONS C C c C TO =TIME WHEN PHASE CHANGE STARTS C C DT =TIME-STEP SIZE C C N = NUMBER OF ENTRIES IN INPUT FILE C C ALPHA =THERMAL DIFFUSIVITY C C CK =THERMAL CONDUCTIVITY C C C =SPECIFIC HEAT C C CL =LATENT HEAT OF FUSION C C RHO =DENSITY C C TI =INITIAL TEMPERATURE C C TM =MELTING TEMPERATURE C C***************** **************************************** ***********^ MAIN PROGRAM IMPLICIT DOUBLE PRECISION(A-H,0-Z) PARAMETER (NP=250) DATA PI/3 . 14159265359D0/ DATA RO/O.ODO/ DIMENSION R(NP) ,UL(NP) ,CO(NP,NP) ,A(NP,NP) ,B1(NP) DIMENSION B3(NP) ,B23(NP) ,B2(NP,NP) ,X(NP) ,AA(NP,NP) DIMENSION GG(NP) ,GLL(NP) ,A23(NP) ,TEM(NP) ,B(NP) CHARACTER *40 FNAME CHARACTER* 12 MQ COMMON/MINA/CC COMMON/MIN/FF COMMON/AL/ALPHA COMMON/ADT/DT COMMON/AAD/TO COMMON/MON/AAA COMMON/MONA/BB COMMON/EE/E COMMON/FF/F COMMON/ICC/IC COMMON/SAR/TT 136

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ooo ooo o o o 137 COMMON/SARA/DD C0MM0N/M0N1/AA1 INTEGER RDSTAT,OPSTAT EXTERNAL FUN, FUN 1, DG , DG1 C 10 CONTINUE PRINT* , Â’ENTER FILE NAME OR MQ TO QUIT:Â’ READ Â’(A)Â’ , FNAME 0PEN(UNIT=1 , FILE=FNAME,STATUS=Â’OLDÂ’ , IOSTAT=OPSTAT) IF( .NOT. (OPSTAT .EQ. 0 .OR. FNAME .EQ. Â’MQÂ’)) GO TO 10 IF (FNAME .NE. Â’MQÂ’) THEN C PRINT*, Â’DT, N, TO ,TM ,TI : Â’ READ* , DT , N , TO , TM , TI READ(12,*)C,CL,CK,RH0 ALPHA=CK/RHO/C COEF=C/CL/CK*DSQRT( ALPHA/PI ) READ DATA FILE DO 1=1, N READ(UNIT=1 , FMT=* , I OSTAT=RDSTAT ) UL( I ) ,R( I ) ENDDO FINDING VECTOR B1 DO 1=1, N B1 ( I )=0 . ODO ENDDO DO 1=1, N B1 ( I )=C*TM/CL ENDDO PRINT*, Â’BlÂ’ ,B1(1) ,B1(2) FINDING MATRIX B2 DO K=1,N DO 1=1, N B2( I ,K)=0 . ODO ENDDO ENDDO CC=RO TT=TO CS=0 . ODO DO 1=1, N AAA=UL(I)-TO FF=R( I ) DO K=I,N BB=R(K) IF(K .EQ. 1) THEN CALL ROMBERG (DG,CS, A AA,ANTG) ELSE AA1=UL(K)-T0

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138 CALL ROMBERG (DG1,CS, AAA, ANTG) END IF B2(K , I )=ANTG ENDDO CS=UL(I)-TO CC=R( I ) TT=UL( I )-TO ENDDO PRINT* , Â’ B2 Â’ , ( (B2(K , I ) , 1=1 ,3) ,K=1 ,3) C C FINDING VECTOR B3 C DO K=1 ,N B3(K)=0 . ODO ENDDO DO K=1,N B3(K)=(R(K)-R(K-1 ) )/DT ENDDO PRINT *, Â’B3Â’ ,B3(1) ,B3(2) ,B3(3) C C FINDING MATRIX CO C DO 1=1, N DO J=1,N CO (I , J)=O.ODO ENDDO ENDDO C T0=0 . ODO DO J=1,N F=R( J) E=UL( J) DO 1=1, J IC=I CALL ROMBERG (FUN, TO, E, CO ANT) CO( J , I )=COANT ENDDO ENDDO PRINT*, Â’COÂ’ ,((A(I,J), J=1 ,3) , 1=1 ,3) C C MULTIPLY MATRIX CO BY THE COEFF. C DO 1=1, N DO J=1,N A( J , I )=0 . ODO ENDDO ENDDO DO 1=1, N DO J=1,N A( J , I )=CO( J , I )*COEF ENDDO ENDDO PRINT*, Â’AÂ’ ,((A(I,J) ,J=1,3),I=1,3) C

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o o o o o 139 MULTIPLY MATRIX B2 BY VECTOR B3 DO 1=1, N B23(I)=0.0D0 ENDDO DO 1=1, N DO J=1 ,N B23(I)=B23(I)+B2(I,J)*B3(J) ENDDO ENDDO PRINT* , Â’ B23 Â’ , B23( 1 ) , B23(2) FINDING VECTOR B DO 1=1, N B( I )=0 . ODO ENDDO DO 1=1, N B( I )=B1 ( I )+B23( I ) ENDDO DO 1=1, N X( I )=B( I ) DO J=1,N AA(I , J)=A(I , J) ENDDO ENDDO CALL LUDCMP(AA,N,NP, INDX,D) CALL LUBKSB(AA,N,NP, INDX,X) IDUM=-13 DO 1=1, N X( I )=X( I )*( 1 . 0+0 . 2*RAN3( IDUM) ) ENDDO CALL MPROVE( A , AA , N , NP , INDX , B , X ) ENDIF WRITE(16,*) Â’ ANÂ’ DO 1=1, N WRITE(16,*) X(I) ENDDO DO 1=1,20 PP=X(N) DO J=N-1 ,1,-1 PP=PP*I+X(J) ENDDO GG( I )=PP PRINT* , GG ( I ) ENDDO C C FINDING TEMPERATURE C DT=1 DO 1=1, NN READ(8 , *)UL( I ) ,R( I ) PRINT*, R( I) ENDDO

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140 DO 1=1, NN DO J=1 ,NN A( I , J)=0 . ODO ENDDO ENDDO IC=1 .ODO CSS=0 . ODO DO 1=1, NN AAAA=UL( I ) DO K=I ,NN F=0 . ODO E=UL(K) CALL ROMBERG(FUN, CSS, AAAA, COANT) A(K, I )=COANT ENDDO CSS=UL( I ) ENDDO DO 1=1, NN DO J=1 ,NN B2( I , J)=0 . ODO ENDDO ENDDO CC=RO TT=TO CS=O.ODO DO 1=1, NN AAA=UL(I)-TO FF=R( I ) DO K=I ,NN BB=0 . ODO IF(K.EQ.1)THEN CALL ROMBERG ( DG,CS, AAA, ANTG) ELSE AA1=UL(K)-T0 CALL ROMBERG ( DG 1 ,CS, AAA, ANTG) ENDIF B2(K, I )=ANTG ENDDO CS=UL(I)-TO CC=R(I) TT=UL( I )-T0 ENDDO DO K=1 ,NN B3(K)=0 . ODO ENDDO DO K=1,NN B3(K)=(R(K)-R(K-1))/DT ENDDO DO 1=1, NN B23(I)=0.0D0

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141 ENDDO DO 1=1, NN DO J=1 ,NN B23 ( I ) =B23 ( I ) +B2 ( I , J ) *B3 ( J ) ENDDO ENDDO C DO 1=1, NN A23(I)=O.ODO ENDDO C DO 1=1, NN DO J=1 ,NN A23( I )=A23( I )+A( I , J)*GG(J) ENDDO ENDDO C DO 1=1, NN TEM( I )=0 . ODO ENDDO DO 1=1, NN TEM( I )=1 /CK*DSQRT(ALPHA/PI )*A23( I )-CL/C*B23( I ) PRINT* ,TEM( I )+TI ENDDO C END C SUBPROGRAM FUNCTION C C**************************************************************^ FUNCTION FUN(T) IMPLICIT DOUBLE PRECISION(A-H,0-Z) COMMON/AL/ALPHA COMMON/ICC/IC COMMON/EE/E COMMON/FF/F IF(E.EQ.T)THEN FUN=0 . ODO ELSE FUN=T**( IC-1 )/DSQRT(E-T)*DEXP( (F**2)/4 . DO/ ALPHA/ (E-T ) ) END IF END C C**************************************************************^ C SUBPROGRAM FUNCTION C FUNCTION DG(T) IMPLICIT DOUBLE PRECISION(A-H,0-Z) COMMON/ADT/DT COMMON/AL/ALPHA COMMON/MON/AAA COMMON/MIN/FF COMMON/MONA/BB COMMON/AAD/TO DATA RO/O.DO/

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142 DATA PI/3 . 14159265359D0/ IF(AAA.EQ.T) THEN DG=O.ODO ELSE DG=1 . D0/DSQRT(4 . DO*PI*ALPHA*(AAA-T) )* k (DEXP(-(BB-FF/DT*T)**2/(4 . DO*ALPHA*(AAA-T) ) )+ k DEXP(-(BB+FF/DT*T)**2/ (4 . DO*ALPHA*(AAA-T) ) ) ) ENDIF END C C SUBPROGRAM FUNCTION C FUNCTION DG1(T) IMPLICIT DOUBLE PRECISION( A-H , O-Z) DATA PI/3.14159265359D0/ COMMON/ADT/DT COMMON/AL/ALPHA COMMON/MON/AAA COMMON/MONA/BB COMMON/AAD/TO COMMON/SARA/DD COMMON/MINA/CC COMMON/MIN/FF COMMON/SAR/TT COMMON /MON 1 / AA1 IF(AA1.EQ.T)TIIEN DG1=0 . ODO ELSE DG1=1 . D0/DSQRT(4*PI*ALPHA*(AA1-T) )* k (DEXP(-((BB-(CC+((FF-CC)/DT)*(T-TT)))**2)/(4.D0*ALPHA*(AA1-T)))+ k DEXP(-( (BB+(CC+( (FF-CC)/DT)*(T-TT) ) )**2)/ (4. D0*ALPHA*(AA1-T) ) ) ) ENDIF END C C SUBPROGRAM C C ROMBERG INTEGRATION C SUBROUTINE ROMBERG ( FUNC , A , B , RESULT) IMPLICIT DOUBLE PRECISION(A-H,0-Z) EXTERNAL FUNC PARAMETER(MAX=40 ,EPS=0 .OOIDO) DIMENSION T( MAX, MAX) C T( 1 , 1 )=(B-A)*(FUNC(A)+FUNC(B) )/2 . ODO T(1,2)=T(1,1) /2 .ODO+(B-A)*FUNC( ( A+B)/2 . 0D0)/2 . ODO T(2,1)=(4.0D0*T(1 ,2)-T(l ,1))/3.0D0 J=3 C C SUCCESSIVE APPLICATION OF TRAPEZOIDAL RULE C 50 DELX=(B-A)/2.0D0**(J-1) X=A-DELX

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o o n o o o 143 N=2**( J-2) SUM=0 . ODO DO 100 1=1, N X=X+2 . ODO*DELX SUM=SUM+FUNC(X) 100 CONTINUE T(l,J)=T(l,J-l)/2. ODO+DELX*SUM EXTRAPOLATION DO 200 L=2, J K=J+1-L T(L,K)=(4.0D0**(L-1)*T(L-1,K+1)-T(L-1,K))/ $ (4.0D0**(L-1)-1.0D0) 200 CONTINUE CHECK ACCURACY CRITERION IF(T( J , 1 ) .EQ. 0 . ODO) THEN RESULT=T( J , 1 ) GO TO 111 END IF C IF(DABS( (T( J , 1)-T(J-1,1))/T(J,1)) .GE. EPS) THEN J=J+1 IF( J.GT.MAX) THEN PAUSE Â’TOO MANY STEPSÂ’ GO TO 111 ELSE GO TO 50 END IF ELSE RESULT=T( J , 1 ) END IF 111 RETURN END C C SUBROUTINE MPROVE C c**************************************************************c SUBROUTINE MPROVE (A , ALUD , N , NP , INDX , B , X ) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) PARAMETER (NMAX=100) DIMENSION A(NP,NP) ,ALUD(NP,NP) , INDX(N) ,B(N),X(N) ,R(NMAX) REAL*8 SDP DO 12 1=1, N SDP=-B( I ) DO 11 J=1 ,N SDP=SDP+DBLE(A( I , J ) )*DBLE(X( J ) ) 11 CONTINUE R(I)=SDP 12 CONTINUE CALL LUBKSB(ALUD,N ,NP, INDX ,R) DO 13 1=1, N

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144 X ( I )=X( I ) — R( I ) 13 CONTINUE RETURN END C C *** ***** * *** *** * * * * ** *** ***************** >1= * ******************* C C SUBROUTINE LU DECOMPOSITION C SUBROUTINE LUDCMP(A,N,NP, INDX,D) IMPLICIT DOUBLE PRECISION(A-H,0-Z) PARAMETER (NMAX=100 ,TINY=1 . 0E-20) DIMENSION A(NP,NP) ,INDX(N) ,VV(NMAX) D=1.0D0 DO 12 1=1, N AAMAX=0 . DO DO 11 J=1,N IF (ABS(A(I , J)) .GT.AAMAX) AAMAX=ABS( A( I , J) ) 11 CONTINUE IF (AAMAX.EQ.O. ) PAUSE ’SINGULAR MATRIX.’ VV(I)=1 ./AAMAX 12 CONTINUE DO 19 J=1 ,N IF (J.GT.l) THEN DO 14 1=1, J-l SUM=A(I , J) IF (I.GT.l)THEN DO 13 K=1 ,1-1 SUM=SUM-A( I ,K)*A(K , J) 13 CONTINUE A ( I , J)=SUM END IF 14 CONTINUE ENDIF AAMAX=0. DO 16 I=J,N SUM=A(I , J) IF (J.GT.l)THEN DO 15 K=1 , J-l SUM=SUM-A(I,K)*A(K,J) 15 CONTINUE A(I,J)=SUM ENDIF DUM=VV( I )*ABS(SUM) IF (DUM.GE. AAMAX) THEN IMAX=I AAMAX=DUM ENDIF 16 CONTINUE IF (J.NE.IMAX)THEN DO 17 K=1,N DUM=A(IMAX,K) A( IMAX ,K)=A( J ,K) A( J ,K)=DUM 17 CONTINUE

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145 D=-D VV(IMAX)=VV(J) ENDIF INDX( J)=IMAX IF(J.NE.N)THEN IF(A(J, J).EQ.O. )A(J, J)=TINY DUM=1./A(J,J) DO 18 I=J+1 ,N A(I,J)=A(I,J)*DUM 18 CONTINUE ENDIF 19 CONTINUE IF(A(N,N).EQ.O.)A(N,N)=TINY RETURN END C C SUBROUTINE BACKSUBSTITUTION 54c 9(c s|c s|e s4c 5|c 3+c 3|e ajc 3#c af: s|c 3^c 3^: 3|c =+: 3+c sfc a+c a|c af= 3|c sf: s^c 3|e 3^c a|c a(*: s(c 3t= =+c a|c sf: a(c a+c 3#c 3^: a|= =4c s|c a^c 3|c 5+: a4c 3^c =|c 34c s|c 54c s4c Cy SUBROUTINE LUBKSB(A ,N ,NP , INDX,B) IMPLICIT DOUBLE PRECISION(A-H,0-Z) DIMENSION A(NP,NP) , INDX(N) ,B(N) 11=0 DO 12 1=1, N LL=INDX( I ) SUM=B(LL) B(LL)=B( I ) IF (II.NE.O)THEN DO 11 J=II ,1-1 SUM=SUM-A(I ,J)*B(J) 11 CONTINUE ELSE IF (SUM.NE.O.) THEN II=I ENDIF B( I )=SUM 12 CONTINUE DO 14 I=N, 1 , -1 SUM=B( I ) IF(I.LT.N)THEN DO 13 J=I+1 ,N SUM=SUM-A(I , J)*B( J) 13 CONTINUE ENDIF B( I )=SUM/A( I , I ) 14 CONTINUE RETURN END C c**************************************************************c C SUBROUTIN RANDOM C FUNCTION RAN3( IDUM) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) PARAMETER (MBIG=1000000000,MSEED=161803398,MZ=0,FAC=1 .E-9) O O

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146 DIMENSION MA(55) DATA IFF /O/ IF(IDUM.LT.O.OR. IFF.EQ.O)THEN IFF=1 MJ=MSEED-IABS( IDUM) MJ=MOD(MJ ,MBIG) MA(55)=MJ MK=1 DO 11 1=1,54 II=MOD(21*I ,55) MA( II )=MK MK=MJ-MK IF(MK . LT.MZ)MK=MK+MBIG MJ=MA( II) 11 CONTINUE DO 13 K=1 ,4 DO 12 1=1,55 MA( I )=MA( I )-MA( l+MOD( 1+30 ,55) ) IF(MA(I) .LT.MZ)MA(I)=MA(I)+MBIG 12 CONTINUE 13 CONTINUE INEXT=0 INEXTP=31 IDUM=1 END IF INEXT=INEXT+1 IF(INEXT.EQ .56)INEXT=1 INEXTP=INEXTP+1 IF( INEXTP . EQ . 56 ) INEXTP=1 M J=MA ( INEXT ) -MA ( INEXTP ) IF(MJ.LT.MZ)MJ=MJ+MBIG MA( INEXT)=MJ RAN3=MJ*FAC RETURN END

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o o o APPENDIX C SSM FORTRAN PROGRAM FOR INVERSE STEFAN PROBLEM C********************************************************************^ c c c c c c c c c c c c c c c c c c c SOLUTION OF ID INVERSE STEFAN PROBLEMS IN A SEMI-INFINITE MEDIUM WITH OR WITHOUT SUBCOOLING BY THE SOURCE-AND-SINK METHOD USING INCREMENTAL APPROACH INPUT FILE UNIT NUMBER IS SET AT 12 OUTPUT FILE UNIT IS SET AT 8 AND 16 TO DT N ALPHA CK C CL RHO TI TM NOTATIONS :TIME WHEN PHASE CHANGE STARTS :TIME-STEP SIZE ; NUMBER OF ENTRIES IN INPUT FILE : THERMAL DIFFUSIVITY : THERMAL CONDUCTIVITY ^SPECIFIC HEAT : LATENT HEAT OF FUSION ^DENSITY INITIAL TEMPERATURE :MELTING TEMPERATURE C C C C C C C C C C C C C c c c c c c c**** ****************** ******************* ********************** ****=t=C MAIN PROGRAM IMPLICIT DOUBLE PRECISION(A-H,0-Z) PARAMETER (NP=250) DATA PI/3.14159265359D0/ DATA RO/O.ODO/ DIMENSION R(NP) ,UL(NP) ,CO(NP,NP) ,A(NP,NP) DIMENSION B3(NP) ,B23(NP) ,B2(NP,NP) ,X(NP) DIMENSION GG(NP) ,GLL(NP) ,A23(NP) ,TEM(NP) CHARACTER *40 FNAME CHARACTER* 12 MQ COMMON/MINA/CC COMMON/MIN/FF COMMON/AL/ALPHA COMMON/ADT/DT COMMON/AAD/TO COMMON/MON/AAA COMMON/MONA/BB COMMON/EE/E COMMON/FF/F COMMON/ICC/IC COMMON/SAR/TT COMMON/SARA/DD 147

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o o o o o o 148 C0MM0N/M0N1/AA1 INTEGER RDSTAT,OPSTAT EXTERNAL FUN, FUN 1, DG , DG1 C 10 CONTINUE PRINT* , Â’ENTER FILE NAME OR MQ TO QUIT:Â’ READ Â’(A)Â’ , FNAME 0PEN(UNIT=1 , FILE=FNAME,STATUS=Â’OLDÂ’ , IOSTAT=OPSTAT) IF( .NOT. (OPSTAT .EQ. 0 .OR. FNAME .EQ. Â’MQÂ’)) GO TO 10 IF (FNAME .NE. Â’MQÂ’) THEN C PRINT*, Â’DT, N, TO ,TM ,TI : Â’ READ* , DT , N , TO , TM , TI READ( 12 , * )C , CL , CK , RHO ALPHA=CK/RHO/C COEF=CL*CK/C*DSQRT(PI /ALPHA) C0EF1=CK*TM*DSQRT(PI/ALPHA) READ DATA FILE DO 1=1, N READ(UNIT=1 ,FMT=*,IOSTAT=RDSTAT) UL(I) ,R(I) ENDDO FINDING MATRIX B2 DO K=1,N DO 1=1, N B2( I ,K)=0 . ODO ENDDO ENDDO CC=RO TT=TO CS=0 . ODO DO 1=1, N AAA=UL( I )-T0 FF=R( I ) DO K=I,N BB=R(K) IF(K .EQ. 1) THEN CALL ROMBERG (DG,CS, A AA,ANTG) ELSE AA1=UL(K)-T0 CALL ROMBERG ( DG 1,CS, AAA, ANTG) ENDIF B2(K , I )=ANTG ENDDO CS=UL(I)-TO CC=R( I ) TT=UL(I)-TO ENDDO PRINT*, Â’B2Â’ , ( (B2(K , I ) , 1=1 ,3) ,K=1 ,3)

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o o o o o o o c c FINDING VECTOR B3 149 DO K=1,N B3(K)=0 . ODO ENDDO DO K=1 ,N B3(K)=(R(K)-R(K-1))/DT ENDDO PRINT *,Â’B3>,B3(1),B3(2),B3(3) FINDING MATRIX A DO 1=1, N DO J=1 ,N A(I , J)=O.ODO ENDDO ENDDO CSS=0 . ODO DO 1=1, N AAAA=UL( I ) DO K=I,N F=R(K) E=UL(K) CALL ROMBERG ( FUN , CSS , A A A A , CO ANT ) A(K , I )=COANT ENDDO CSS=UL( I ) ENDDO PRINT*, Â’AÂ’ ,((A(I,J), J=1 ,3) ,1=1 ,3) MULTIPLY MATRIX B2 BY VECTOR B3 DO 1=1, N B23( I )=0 . ODO ENDDO DO 1=1, N DO J=1,N B23 ( I ) =B23 ( I ) +B2 ( I , J ) *B3 ( J ) ENDDO ENDDO PRINT* , Â’ B23Â’ , B23( 1 ) , B23(2) GG(l)=(COEFl+COEF*B23(l))/A(l,l) XX=0 . ODO DO 1=2, N DO K=1 ,1-1 GLL(I)=GG(K)*A(I,K)+XX XX=GLL( I ) ENDDO GG ( I ) = ( COEF1 +COEF*B23( I ) -GLL( I ) ) / A ( I , I ) XX=0 . ODO ENDDO WRITE(10,*) Â’ G( I ) Â’

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150 DO 1=1, N VRITE(10,*) GG( I ) ENDDO DT=1 DO 1=1, N DO J=1,N A(I , J)=O.ODO ENDDO ENDDO CSS=0 . ODO DO 1=1, N AAAA=UL( I ) DO K=I,N F=0 . ODO E=UL(K) CALL ROMBERG (FUN, CSS, AAAA,COANT) A(K, I )=COANT ENDDO CSS=UL( I ) ENDDO DO 1=1, N DO J=1,N B2(I ,J)=O.ODO ENDDO ENDDO CC=RO TT=TO CS=0 . ODO DO 1=1, N AAA=UL( I )-T0 FF=R( I ) DO K=I ,N BB=0 . ODO IF(K.Eq.l)THEN CALL ROMBERG (DG,CS, AAA, ANTG) ELSE AA1=UL(K)-T0 CALL ROMBERG (DG1,CS, AAA, ANTG) ENDIF B2(K , I )=ANTG ENDDO CS=UL( I )-T0 CC=R( I ) TT=UL(I )-T0 ENDDO DO K=1,N B3(K)=0 . ODO ENDDO DO K=1,N B3(K)=(R(K)-R(K-1))/DT

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151 ENDDO C DO 1=1, N B23( I )=0 . ODO ENDDO DO 1=1, N DO J=1 ,N B23( I )=B23( I )+B2( I,J)*B3(J) ENDDO ENDDO C DO 1=1, N A23( I )=0 . ODO ENDDO C DO 1=1, N DO J=1,N A23(I)=A23(I)+A(I,J)*GG(J) ENDDO ENDDO C DO 1=1, N TEM( I )=0 . ODO ENDDO DO 1=1, N TEM( I )=l/CK*DSqRT(ALPHA/PI )*A23( I )-CL/C*B23( I ) PRINT* ,TEM( I )+TI ENDDO END IF END C Es(c*************************************************************C C SUBPROGRAM FUNCTION C E**************************************************************^ FUNCTION FUN(T) IMPLICIT DOUBLE PRECISION(A-H , O-Z) COMMON/AL/ALPHA COMMON/EE/E COMMON/FF/F COMMON/ICC/IC IF(E.EQ.T)THEN FUN=0 . ODO ELSE FUN=1 .0D0/DSQRT(E-T)*DEXP(-(F**2)/4. DO/ALPHA/ (E-T) ) ENDIF END C C E**************************************************************U C SUBPROGRAM FUNCTION C c**************************************************************u FUNCTION DG(T) IMPLICIT DOUBLE PRECISION(A-H,0-Z) COMMON/ ADT/DT

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152 COMMON/AL/ALPHA COMMON/MON/AAA COMMON/MIN/FF COMMON/MONA/BB COMMON/AAD/TO DATA RO/O.DO/ DATA PI/3.14159265359D0/ IF(AAA.EQ.T) THEN DG=0 . ODO ELSE DG=1 . D0/DSQRT(4.D0*PI*ALPHA*(AAA-T) )* b (DEXP(-(BB-FF/DT*T)**2/(4 . DO*ALPHA*(AAA-T) ) )+ b DEXP( -(BB+FF/DT*T)**2/ (4. DO* ALPHA* (AAA -T) ) ) ) END IF END C C CsMoie***********************************************************^ C SUBPROGRAM FUNCTION C C**************************************************************^-' FUNCTION DG1(T) IMPLICIT DOUBLE PRECISION(A-H , O-Z) DATA PI/3.14159265359D0/ COMMON/ADT/DT COMMON/AL/ALPHA COMMON/MON/AAA COMMON/MONA/BB COMMON/AAD/TO COMMON/SARA/DD COMMON/MINA/CC COMMON/MIN/FF COMMON/SAR/TT COMMON/MON1/AA1 IF(AA1.EQ.T)THEN DG 1=0. ODO ELSE DG1=1 . D0/DSQRT(4*PI*ALPHA*(AA1-T) )* & (DEXP(-( (BB-(CC+( (FF-CC)/DT)*(T-TT) ) )**2)/ (4.D0*ALPHA*(AA1-T) ) )+ b DEXP(-( (BB+(CC+( (FF-CC)/DT)*(T-TT) ))**2)/(4. D0*ALPHA*(AA1-T) ) ) ) ENDIF END C C C SUBROUTINE C C ROMBERG INTEGRATION C C**************************************************************C SUBROUTINE ROMBERG ( FUNC ,A,B, RESULT ) IMPLICIT DOUBLE PRECISION( A-H , O-Z) EXTERNAL FUNC PARAMETER(MAX=50 , EPS=0 . 001 DO) DIMENSION T( MAX, MAX) C T(1,1)=(B-A)*(FUNC(A)+FUNC(B))/2.0D0

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153 T(1 ,2)=T(1 , 1 )/2 . ODO+(B-A)*FUNC( ( A+B)/2 . 0D0)/2 . ODO T(2,1)=(4.0D0*T(1 ,2)-T(l , 1))/3.0D0 J=3 C C SUCCESSIVE APPLICATION OF TRAPEZOIDAL RULE C 50 DELX=(B-A)/2 . 0D0**( J-l ) X=A-DELX N=2**( J-2) SUM=0 . ODO DO 100 1=1, N X=X+2 . ODO*DELX SUM=SUM+FUNC(X) 100 CONTINUE T(l,J)=T(l,J-l)/2. ODO+DELX*SUM C C EXTRAPOLATION C DO 200 L=2 , J K=J+1-L T(L,K)=(4. 0D0**(L-1 )*T(L-1 ,K+1 )-T(L-l ,K))/ $ (4.0D0**(L-1)-1 .ODO) 200 CONTINUE C C CHECK ACCURACY CRITERION C IF(T( J,l) .EQ. 0 . ODO) THEN RESULT=T( J,l) GO TO 111 END IF C IF(DABS((T(J,1)-T(J-1,1))/T(J,1)) .GE. EPS) THEN J=J+1 IF(J.GT.MAX) THEN PAUSE Â’TOO MANY STEPSÂ’ GO TO 111 ELSE GO TO 50 END IF ELSE RESULT=T( J , 1 ) END IF 1 1 1 RETURN END

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o o o APPENDIX D SSM FORTRAN PROGRAM FOR INVERSE STEFAN PROBLEM C******************************************************************** C c c c c c c c c c c c c c c c c c c c c SOLUTION OF ID INVERSE STEFAN PROBLEMS IN A SEMI -INFINITE MEDIUM WITH OR WITHOUT SUBCOOLING BY THE SOURCE-AND-SINK METHOD USING SERIES APPROACH. TEMPERATURE BOUNDARY IS RETRIEVED VIA THE ASSUMED FLUX CONDITION. INPUT FILE UNIT NUMBER IS SET AT 12 OUTPUT FILE UNIT IS SET AT 8 AND 16 TO DT N ALPHA CK C CL RHO TI TM NOTATIONS :TIME WHEN PHASE CHANGE STARTS :TIME-STEP SIZE rNUMBER OF ENTRIES IN INPUT FILE THERMAL DIFFUSIVITY THERMAL CONDUCTIVITY SPECIFIC HEAT :LATENT HEAT OF FUSION = DENSITY ^INITIAL TEMPERATURE ^MELTING TEMPERATURE C C C C C C C C C C C C C C C C C C C C c******************************************************************** ( : MAIN PROGRAM IMPLICIT DOUBLE PRECISION(A-H,0-Z) PARAMETER (NP=250) DATA PI/3.14159265359D0/ DATA RO/O.ODO/ DIMENSION R(NP) ,UL(NP) ,CO(NP,NP) ,A(NP,NP) ,B1(NP) DIMENSION B3(NP) ,B23(NP) ,B2(NP,NP) ,X(NP) ,AA(NP,NP) DIMENSION GG(NP) ,GLL(NP) ,A23(NP) ,TEM(NP ) ,B(NP) CHARACTER *40 FNAME CHARACTER* 12 MQ COMMON/MINA/CC COMMON/MIN/FF COMMON/AL/ALPHA COMMON/ ADT/DT COMMON/AAD/TO COMMON/MON/AAA COMMON/MONA/BB COMMON/EE/E COMMON/FF/F COMMON/ICC/IC COMMON/SAR/TT 154

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o o o ooo o o o 155 COMMON/SARA/DD C0MM0N/M0N1/AA1 INTEGER RDSTAT,OPSTAT EXTERNAL FUN, FUN 1, DG , DG1 C 10 CONTINUE PRINT* , Â’ENTER FILE NAME OR MQ TO QUIT:Â’ READ Â’(A)Â’ , FNAME 0PEN(UNIT=1 , FILE=FNAME,STATUS=Â’OLDÂ’ , IOSTAT=OPSTAT) IF( .NOT. (OPSTAT .EQ. 0 .OR. FNAME .EQ. Â’MQÂ’)) GO TO 10 IF (FNAME .NE. Â’MQÂ’) THEN C PRINT*, Â’DT, N, TO ,TM ,TI : Â’ READ* , DT , N , TO , TM , TI READ(12,*)C,CL,CK,RH0 ALPHA=CK/RHO/C COEF=C/CL/CK*DSQRT( ALPHA/PI ) READ DATA FILE DO 1=1, N READ(UNIT=1 , FMT=* , I OSTAT=RDSTAT) UL( I ) ,R( I ) ENDDO FINDING VECTOR B1 DO 1=1, N B1 ( I )=0 . ODO ENDDO DO 1=1, N B1 ( I )=C*TM/CL ENDDO PRINT*, Â’BlÂ’ ,B1(1) ,B1(2) FINDING MATRIX B2 DO K=1,N DO 1=1, N B2(I ,K)=O.ODO ENDDO ENDDO CC=RO TT=TO CS=0 . ODO DO 1=1, N AAA=UL( I )-T0 FF=R( I ) DO K=I,N BB=R(K) IF(K .EQ. 1) THEN CALL ROMBERG ( DG,CS, AAA, ANTG) ELSE AA1=UL(K)-T0

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156 CALL ROMBERG ( DG 1 , CS , A A A , ANTG ) ENDIF B2(K , I )=ANTG ENDDO CS=UL( I )-TO CC=R( I ) TT=UL( I )-TO ENDDO PRINT* , Â’ B2 Â’ , ( (B2(K , I ) , 1=1 ,3) ,K=1 ,3) C C FINDING VECTOR B3 C DO K=1 ,N B3(K)=0 . ODO ENDDO DO K=1,N B3(K)=(R(K)-R(K-1))/DT ENDDO PRINT *, Â’B3Â’ ,B3(1) ,B3(2) ,B3(3) C C FINDING MATRIX CO C DO 1=1, N DO J=1,N CO(I , J)=O.ODO ENDDO ENDDO C T0=0 . ODO DO J=1 ,N F=R( J) E=UL( J) DO 1=1, J IC=I CALL ROMBERG (FUN, TO, E, CO ANT) CO( J , I )=COANT ENDDO ENDDO PRINT*, Â’COÂ’ ,((A(I,J), J=1 ,3) , 1=1 ,3) C C MULTIPLY MATRIX CO BY THE COEFFICIENTS C DO 1=1, N DO J=1,N A( J , I )=0 . ODO ENDDO ENDDO DO 1=1, N DO J=1,N A(J,I)=CO(J,I)*COEF ENDDO ENDDO PRINT*, Â’AÂ’ ,((A(I,J),J=1,3),I=1,3) C MULTIPLY MATRIX B2 BY VECTOR B3

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o o o 157 DO 1=1, N B23( I )=0 .ODO ENDDO DO 1=1, N DO J=1 ,N B23( I )=B23( I )+B2( I , J)*B3( J) ENDDO ENDDO PRINT*, ’ B23’ , B23(l ) , B23(2) FINDING VECTOR B DO 1=1, N B( I )=0 . ODO ENDDO DO 1=1, N B( I )=B1 ( I )+B23( I ) ENDDO DO 1=1, N X( I ) — B ( I ) DO J=1,N AA(I,J)=A(I,J) ENDDO ENDDO CALL LUDCMP ( AA , N , NP , INDX , D) CALL LUBKSB(AA,N,NP, INDX,X) IDUM=-13 DO 1=1, N X(I)=X(I)*(1.0+0.2*RAN3(IDUM)) ENDDO CALL MPRO VE (A,AA,N,NP,INDX,B,X) ENDIF WRITE( 16,*) ’ AN’ DO 1=1, N WRITE(16,*) X(I) ENDDO DO 1=1,20 PP=X(N) DO J=N-1 ,1,-1 PP=PP*I+X(J) ENDDO GG( I )=PP PRINT* ,GG( I ) ENDDO DT=1 DO 1=1, NN READ(8 , *)UL( I ) ,R( I ) PRINT*, R( I) ENDDO DO 1=1, NN DO J=1 ,NN A(I , J)=O.ODO ENDDO

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158 ENDDO IC=1 . ODO CSS=0 . ODO DO 1=1, NN AAAA=UL( I ) DO K=I ,NN F=0 . ODO E=UL(K) CALL ROMBERG ( FUN , CSS , AAAA , COANT) A(K, I )=COANT ENDDO CSS=UL( I ) ENDDO DO 1=1, NN DO J=1,NN B2(I , J)=O.ODO ENDDO ENDDO CC=RO TT=TO CS=0 . ODO DO 1=1, NN AAA=UL(I)-TO FF=R( I ) DO K=I ,NN BB=0 . ODO IF(K.EQ.1)THEN CALL ROMBERG ( DG,CS, AAA, ANTG) ELSE AA1=UL(K) -TO CALL ROMBERG ( DG 1 , CS , A A A , ANTG ) ENDIF B2(K , I )=ANTG ENDDO CS=UL(I)-TO CC=R(I) TT=UL( I )-TO ENDDO DO K=1,NN B3(K)=0 . ODO ENDDO DO K=1,NN B3(K)=(R(K)-R(K-1))/DT ENDDO DO 1=1, NN B23( I )=0 . ODO ENDDO DO 1=1, NN DO J=1 ,NN B23(I)=B23(I)+B2(I,J)*B3(J)

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159 ENDDO ENDDO C DO 1=1, NN A23( I )=0 . ODO ENDDO C DO 1=1, NN DO J=1 ,NN A23( I )=A23( I )+A( I , J)*GG( J) ENDDO ENDDO C DO 1=1, NN TEM( I )=0 . ODO ENDDO DO 1=1, NN TEM( I )=1 /CK*DSQRT(ALPHA/PI )*A23( I )-CL/C*B23( I ) PRINT*,TEM(I)+TI ENDDO C END C SUBPROGRAM FUNCTION c C * 5|= **** ** * * **** ************ ****** ****************************** C FUNCTION FUN(T) IMPLICIT DOUBLE PRECISION(A-H,0-Z) COMMON/AL/ALPHA COMMON/ ICC/ I C COMMON/EE/E COMMON/FF/F IF(E.EQ.T)THEN FUN=0 . ODO ELSE FUN=T**( IC-1 ) /DSQRT(E-T)*DEXP( -(F**2)/4 . DO/ ALPHA/ (E-T) ) END IF END C C**************************************************************^ C SUBPROGRAM FUNCTION c FUNCTION DG(T) IMPLICIT DOUBLE PRECISION( A-H , O-Z) COMMON/ADT/DT COMMON/AL/ALPHA COMMON/MON/AAA COMMON/MIN/FF COMMON/MONA/BB COMMON/AAD/TO DATA RO/O.DO/ DATA PI/3.14159265359D0/ IF(AAA.EQ.T) THEN DG=0 . ODO ELSE

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160 DG=1.D0/DSQRT(4.D0*PI*ALPHA*( AAA-T))* k ( DEXP( ( BB-FF/DT*T)**2/(4 . D0*ALPHA*( AAA-T) ) )+ k DEXP(-(BB+FF/DT*T)**2/(4.D0*ALPHA*( AAA-T)))) ENDIF END C c**************************** ********************* *************c C SUBPROGRAM FUNCTION c C**************************************************************^ FUNCTION DG1(T) IMPLICIT DOUBLE PRECISION(A-H , 0-Z) DATA PI/3.14159265359D0/ COMMON/ADT/DT COMMON/AL/ALPHA COMMON/MON/AAA COMMON/MONA/BB COMMON/AAD/TO COMMON/SARA/DD COMMON/MINA/CC COMMON/MIN/FF COMMON/SAR/TT C0MM0N/M0N1/AA1 IF(AA1.EQ.T)THEN DG1=0 . ODO ELSE DG1=1 . D0/DSQRT(4*PI*ALPHA*(AA1-T) )* k (DEXP( ( (BB-(CC+( (FF-CC)/DT)*(T-TT) ))**2)/(4. DO*ALPHA*( AA1-T) ) )+ k DEXP( -( (BB+(CC+( (FF-CC)/DT)*(T-TT )))**2)/(4. DO*ALPHA*( AA1-T) ) ) ) ENDIF END C C**************************************************************C C SUBPROGRAM c C ROMBERG INTEGRATION c C**************************************************************^' SUBROUTINE ROMBERG ( FUNC , A , B , RESULT ) IMPLICIT DOUBLE PRECISION(A-H , 0-Z) EXTERNAL FUNC PARAMETER(MAX=40 , EPS=0 . 0001 DO ) DIMENSION T(MAX,MAX) C T( 1 , 1 )=(B-A)*(FUNC(A)+FUNC(B) )/2 . ODO T(l,2)=T(l,l)/2. ODO+(B-A)*FUNC( ( A+B)/2 . 0D0)/2 . ODO T(2,1)=(4.0D0*T(1 ,2)-T(l ,1))/3.0D0 J=3 C C SUCCESSIVE APPLICATION OF TRAPEZOIDAL RULE C 50 DELX=(B-A)/2.0D0**( J-l) X=A-DELX N=2** ( J-2 ) SUM=0 . ODO DO 100 1=1, N X=X+2 . ODO*DELX

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non o o o 161 SUM=SUM+FUNC(X) 100 CONTINUE T(l,J)=T(l,J-l)/2. 0D0+DELX*SUM EXTRAPOLATION DO 200 L=2 , J K=J+1-L T(L ,K)=(4. 0D0**(L-1 )*T(L-1 ,K+1 )-T(L-l ,K) )/ $ (4.0D0**(L-1)-1 .0D0) 200 CONTINUE CHECK ACCURACY CRITERION IF(T( J , 1 ) .EQ. O.ODO) THEN RESULT=T( J , 1 ) GO TO 111 END IF C IF(DABS( (T( J ,1)-T(J-1,1))/T(J,1)) .GE. EPS) THEN J=J+1 IF( J.GT.MAX) THEN PAUSE Â’TOO MANY STEPSÂ’ GO TO 111 ELSE GO TO 50 END IF ELSE RESULT=T( J , 1 ) END IF 111 RETURN END C SUBROUTINE MPROVE SUBROUTINE MPROVE( A , ALUD , N , NP , I NDX , B , X ) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) PARAMETER (NMAX=100) DIMENSION A(NP,NP) ,ALUD(NP,NP) , INDX(N) ,B(N) ,X(N) ,R(NMAX) REAL*8 SDP DO 12 1=1, N SDP=-B( I ) DO 11 J=1 ,N SDP=SDP+DBLE( A( I , J))*DBLE(X( J)) 11 CONTINUE R( I )=SDP 12 CONTINUE CALL LUBKSB(ALUD,N ,NP, INDX,R) DO 13 1=1, N X( I )=X( I )-R( I ) 13 CONTINUE RETURN END O O

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162 C SUBROUTINE LU DECOMPOSITION C C**************************************************************^ SUBROUTINE LUDCMP(A ,N ,NP , INDX , D) IMPLICIT DOUBLE PRECIS I ON (A-H , O-Z) PARAMETER (NMAX=100,TINY=1 .0E-20) DIMENSION A(NP,NP),INDX(N) , VV(NMAX) D=1 . ODO DO 12 1 = 1, N AAMAX=0 . ODO DO 11 J=1,N IF (ABS(A(I , J)) .GT.AAMAX) AAMAX=ABS( A( I , J) ) 11 CONTINUE IF (AAMAX.EQ.O. ) PAUSE Â’SINGULAR MATRIX.Â’ VV(I)=1 .ODO/AAMAX 12 CONTINUE DO 19 J=1,N IF (J.GT.l) THEN DO 14 1=1, J-l SUM=A( I , J) IF (I.GT.l)THEN DO 13 K=1 ,1-1 SUM=SUM-A(I ,K)*A(K, J) 13 CONTINUE A ( I , J)=SUM ENDIF 14 CONTINUE ENDIF AAMAX=0 . DO DO 16 I=J,N SUM=A( I , J ) IF (J.GT.l )THEN DO 15 K=1 , J-l SUM=SUM-A(I ,K)*A(K,J) 15 CONTINUE A ( I , J)=SUM ENDIF DUM=VV ( I ) * ABS ( SUM ) IF (DUM.GE.AAMAX) THEN IMAX=I AAMAX=DUM ENDIF 16 CONTINUE IF (J.NE.IMAX)THEN DO 17 K=1,N DUM=A( IMAX ,K) A(IMAX,K)=A(J,K) A( J ,K)=DUM 17 CONTINUE D=-D VV(IMAX)=VV( J) ENDIF INDX( J)=IMAX IF(J.NE.N)THEN

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163 IF(A(J,J).EQ.O.)A(J,J)=TINY DUM=1./A(J,J) DO 18 I=J+1,N A(I,J)=A(I,J)*DUM 18 CONTINUE ENDIF 19 CONTINUE IF(A(N,N).EQ.O.)A(N,N)=TINY RETURN END C SUBROUTINE BACKSUBSTITUTION C C >i= * ***** * ********* ** ******************************************* c SUBROUTINE LUBKSB(A,N,NP, INDX,B) IMPLICIT DOUBLE PRECISION(A-H,0-Z) DIMENSION A(NP,NP) , INDX(N) ,B(N) 11=0 DO 12 1=1, N LL=INDX( I ) SUM=B(LL) B(LL)=B( I ) IF (II .NE.O)THEN DO 11 J=II ,1-1 SUM=SUM-A(I,J)*B(J) 11 CONTINUE ELSE IF (SUM.NE.O.) THEN II = I ENDIF B( I )=SUM 12 CONTINUE DO 14 I=N , 1 , -1 SUM=B( I ) IF( I . LT. N)THEN DO 13 J=I+1,N SUM=SUM-A(I , J)*B( J) 13 CONTINUE ENDIF B( I )=SUM/A( I , I ) 14 CONTINUE RETURN END c**************************************************************c C SUBROUTINE RANDOM C FUNCTION RAN3( IDUM) IMPLICIT DOUBLE PRECISION( A-H , 0-Z) PARAMETER ( MBIG=1000000000 , MSEED=161803398 , MZ=0 , FAC=1 . E-9 ) DIMENSION MA(55) DATA IFF /O/ IF(IDUM.LT.O.OR. IFF.EQ.O)THEN IFF=1 MJ=MSEED-IABS( IDUM) MJ=MOD(MJ,MBIG)

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164 MA(55)=MJ MK=1 DO 11 1=1,54 II=MOD(21*I ,55) MA( II )=MK MK=MJ-MK IF(MK . LT.MZ)MK=MK+MBIG MJ=MA(II) 11 CONTINUE DO 13 K=1 ,4 DO 12 1=1,55 MA( I )=MA( I )-MA( l+MOD( 1+30 , 55) ) IF(MA( I ) . LT.MZ)MA( I )=MA( I )+MBIG 12 CONTINUE 13 CONTINUE INEXT=0 INEXTP=31 I DUM= 1 ENDIF INEXT=INEXT+1 IF(INEXT.EQ.56)INEXT=1 I NEXTP= I NEXTP+ 1 IF( INEXTP . EQ . 56 ) INEXTP=1 MJ=MA( INEXT)-MA( INEXTP) IF(MJ . LT. MZ)MJ=MJ+MBIG MA( INEXT)=MJ RAN3=MJ*FAC RETURN END

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o o o APPENDIX E SSM FORTRAN PROGRAM FOR INVERSE STEFAN PROBLEM U********************************************************************^ c c c c c c c c c c c c c c c c c c c SOLUTION OF ID INVERSE STEFAN PROBLEMS IN A SEMI -INFINITE MEDIUM WITH OR WITHOUT SUBCOOLING BY THE SOURCE-AND-SINK METHOD USING INCREMENTAL APPROACH. BOUNDARY CONDITION FOUND VIA ASSUMED FLUX CONDITION INPUT FILE UNIT NUMBER IS SET AT 12 OUTPUT FILE UNIT NUMBER IS SET AT 16 NOTATIONS TO =TIME WHEN PHASE CHANGE STARTS DT =TIME-STEP SIZE N =NUMBER OF ENTRIES IN INPUT FILE ALPHA =THERMAL DIFFUSIVITY CK =THERMAL CONDUCTIVITY C ^SPECIFIC HEAT CL =LATENT HEAT OF FUSION RHO = DENSITY TI =INITIAL TEMPERATURE TM =MELTING TEMPERATURE C C C C C C C C C C C C C C c c c c c c********************************************************************c MAIN PROGRAM IMPLICIT DOUBLE PRECISION(A-H ,0-Z) PARAMETER (NP=250) DATA PI/3. 14159265359D0/ DATA RO/O.ODO/ DIMENSION R(NP) ,UL(NP) ,CO(NP,NP) ,A(NP,NP) DIMENSION B3(NP) ,B23(NP) ,B2(NP,NP) ,X(NP) DIMENSION GG(NP) ,GLL(NP) ,A23(NP) ,TEM(NP) CHARACTER *40 FNAME CHARACTER* 12 MQ COMMON/MINA/CC COMMON/MIN/FF COMMON/ AL/ALPHA COMMON/ADT/DT COMMON/AAD/TO COMMON/MON/AAA COMMON/MONA/BB COMMON/EE/E COMMON/FF/F COMMON/ICC/IC COMMON/SAR/TT COMMON/SARA/DD 165

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o o C0MM0N/M0N1/AA1 INTEGER RDSTAT , OPSTAT EXTERNAL FUN, FUN 1, DG , DG1 166 10 CONTINUE PRINT* , Â’ENTER FILE NAME OR MQ TO QUIT:Â’ READ Â’(A)Â’ , FNAME 0PEN(UNIT=1 , FILE=FNAME,STATUS=Â’ OLDÂ’ , IOSTAT=OPSTAT) IF( .NOT. (OPSTAT .Eq. 0 .OR. FNAME .EQ. Â’MQÂ’)) GO TO 10 IF (FNAME .NE. Â’MQÂ’) THEN C PRINT*, Â’DT, N, TO ,TM ,TI : Â’ READ* , DT , N , TO , TM , TI READ(12,*)C,CL,CK,RH0 ALPHA=CK/RHO/C COEF=CL*CK/C*DSQRT( PI /ALPHA) C0EF1=CK*TM*DSQRT( PI /ALPHA) READ DATA FILE DO 1 = 1, N READ(UNIT=1 ,FMT=*,IOSTAT=RDSTAT) UL(I) ,R(I) C PRINT* ,UL( I ) ,R( I ) ENDDO C c FINDING MATRIX B2 c DO K=1,N DO 1=1, N B2( I ,K)=0 . ODO ENDDO ENDDO c CC=RO TT=TO CS=0 . ODO DO 1=1, N AAA=UL(I)-TO FF=R( I ) DO K=I,N BB=R(K) IF(K .EQ. 1) THEN CALL ROMBERG (DG,CS, AAA, ANTG) ELSE AA1=UL(K)-T0 CALL ROMBERG (DG1,CS, AAA, ANTG) ENDIF B2(K , I )=ANTG ENDDO CS=UL( I ) -TO CC=R( I ) TT=UL( I )-T0 ENDDO PRINT*, Â’B2Â’ , ( (B2(K , I ) , 1=1 ,3) ,K=1 ,3)

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o o 167 C c FINDING VECTOR B3 c DO K=1,N B3(K)=0 . ODO ENDDO DO K=1 ,N B3(K)=(R(K)-R(K-1 ) )/DT ENDDO PRINT *, >B3Â’ ,B3(1) ,B3(2) ,B3(3) FINDING MATRIX A c DO 1=1, N DO J=1 ,N A(I , J)=O.ODO ENDDO ENDDO C CSS=0 . ODO DO 1=1, N AAAA=UL( I ) DO K=I,N F=R(K) E=UL(K) CALL ROMBERG ( FUN , CSS , A AAA , CO ANT ) A(K , I )=COANT ENDDO CSS=UL( I ) ENDDO PRINT*, Â’AÂ’ ,((A(I,J), J=1 ,3) , 1=1 ,3) C c MULTIPLY MATRIX B2 BY VECTOR B3 c DO 1=1, N B23( I )=0 . ODO ENDDO DO 1=1, N DO J=1,N B23( I )=B23( I )+B2( I , J)*B3( J) ENDDO ENDDO PRINT* , Â’ B23 Â’ , B23( 1 ) , B23(2) C GG( 1 ) = (C0EF1+C0EF*B23( 1 ) )/A( 1 , 1 ) XX=0 . ODO DO 1=2, N DO K=1 ,1-1 GLL(I)=GG(K)*A(I,K)+XX XX=GLL( I ) ENDDO GG(I) = (COEFl+COEF*B23(I)-GLL(I ) )/A(I , I) XX=0 . ODO ENDDO

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o o o 168 WRITE(11 ,*) Â’ G( I ) Â’ DO 1=1, N VRITE(11,*) GG( I ) ENDDO FINDING TEMPERATURE DT=1 DO 1=1, N DO J=1,N A( I , J)=0 . ODO ENDDO ENDDO CSS=0 . ODO DO 1=1, N AAAA=UL( I ) DO K=I,N F=0 . ODO E=UL(K) CALL ROMBERG (FUN, CSS, AAAA, COANT) A(K , I )=COANT ENDDO CSS=UL( I ) ENDDO DO 1=1, N DO J=1,N B2( I , J)=0 . ODO ENDDO ENDDO CC=RO TT=TO CS=0 . ODO DO 1=1, N AAA=UL( I )-TO FF=R( I ) DO K=I,N BB=0 . ODO IF(K.EQ.1)TIIEN CALL ROMBERG ( DG , CS , A A A , ANTG ) ELSE AAl=UL(K)-TO CALL ROMBERG (DG1,CS, AAA, ANTG) END IF B2(K , I )=ANTG ENDDO CS=UL( I )-TO CC=R( I ) TT=UL( I )-TO ENDDO DO K=1,N B3(K)=0 . ODO ENDDO

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169 DO K=1,N B3(K)=(R(K)-R(K-1))/DT ENDDO C DO 1=1, N B23( I )=0 . ODO ENDDO DO 1=1, N DO J=1,N B23( I )=B23( I )+B2( I , J)*B3(J) ENDDO ENDDO C DO 1=1, N A23( I )=0 . ODO ENDDO DO 1=1, N DO J=1,N A23 ( I ) =A23 (I)+A(I,J) *.GG ( J ) ENDDO ENDDO C DO 1=1, N TEM( I )=0 . ODO ENDDO DO 1=1, N TEM( I )=1 /CK*DSQRT( ALPHA/PI )*A23( I )-CL/C*B23( I ) PRINT( 16 , *)TEM( I )+TI ENDDO ENDIF END C C SUBPROGRAM FUNCTION C c**************************************************************c FUNCTION FUN(T) IMPLICIT DOUBLE PRECISION(A-H,0-Z) COMMON/AL/ALPHA COMMON/EE/E COMMON/FF/F COMMON/ICC/IC IF(E.EQ.T)THEN FUN=0 . ODO ELSE FUN=1 . ODO/DSQRT(E-T) *DEXP( ( F**2 ) /4 . DO/ALPHA/ (E-T) ) ENDIF END C c**%***********************************************************u C SUBPROGRAM FUNCTION C FUNCTION DG(T) IMPLICIT DOUBLE PRECISION(A-H,0-Z) COMMON/ADT/DT

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170 COMMON/AL/ALPHA COMMON/MON/AAA COMMON/MIN/FF COMMON/MONA/BB COMMON/AAD/TO DATA RO/O.DO/ DATA PI/3.14159265359D0/ IF(AAA.EQ.T) THEN DG=0 . ODO ELSE DG=1 . DO/DSQRT( 4. DO*PI* ALPHA* (AAA-T) )* k (DEXP( -(BB-FF/DT*T)**2/ (4. DO*ALPHA*( AAA-T) ) ) + k DEXP(-(BB+FF/DT*T)**2/ (4. DO*ALPHA* ( AAA-T ) ) ) ) END IF END C C**************************************************************^ C SUBPROGRAM FUNCTION C C**************************************************************^ FUNCTION DG1(T) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) DATA PI/3 . 14159265359D0/ COMMON/ADT/DT COMMON/AL/ALPHA COMMON/MON/ AAA COMMON/MONA/BB COMMON/AAD/TO COMMON/SARA/DD COMMON/M I NA/CC COMMON/MIN/FF COMMON/SAR/TT COMMON /M0N1 /AA1 IF(AA1.EQ.T)THEN DG1=0 . ODO ELSE DG1=1.D0/DSQRT(4*PI*ALPHA*(AA1-T))* k (DEXP( ( (BB-(CC+( (FF-CC)/DT)*(T-TT) ) )**2)/ (4 . DO*ALPHA*( AA1-T) ) ) k +DEXP(-( (BB+(CC+( (FF-CC)/DT)*(T-TT)))**2)/(4. DO*ALPHA*( AA1-T) ) ) ) END IF END C C**************************************************^****^******^ C SUBROUTINE C C ROMBERG INTEGRATION C C************************************************** s * c ^ c ^ e5 * :! ^^ c ^ c3 ^* : ^ : ^ c ^ c ^-' SUBROUTINE ROMBERG ( FUNC , A , B , RESULT ) IMPLICIT DOUBLE PRECISION(A-H , 0-Z) EXTERNAL FUNC PARAMETER( MAX=50 , EPS=0 . 0001 DO ) DIMENSION T(MAX ,MAX) C T( 1 , 1 )=(B-A)*(FUNC(A)+FUNC(B) )/2 . ODO T(1,2)=T(1,1)/2.0D0+(B-A)*FUNC((A+B)/2.0D0)/2.0D0 T(2,l)=(4. ODO*T( 1 ,2)-T(l,l))/3. ODO

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ooo ooo ^ o o o 171 J=3 SUCCESSIVE APPLICATION OF TRAPEZOIDAL RULE 50 DELX=(B-A)/2 . 0D0**( J-l ) X=A-DELX N=2**( J-2) SUM=0 . 0D0 DO 100 1=1, N X=X+2 . 0D0*DELX SUM=SUM+FUNC(X) 100 CONTINUE T(l,J)=T(l,J-l)/2. 0D0+DELX*SUM EXTRAPOLATION DO 200 L=2 , J K=J+1-L T(L,K)=(4. 0D0**(L-1 )*T(L-1,K+1)-T(L-1 ,K))/ $ (4.0D0**(L-1 )-l .0D0) 200 CONTINUE CHECK ACCURACY CRITERION IF(T( J , 1 ) .EQ. O.ODO) THEN RESULT=T( J , 1 ) GO TO 111 END IF IF(DABS( (T(J,1)-T(J-1,1))/T(J,1)) .GE. EPS) THEN J=J+1 IF(J.GT.MAX) THEN PAUSE Â’TOO MANY STEPSÂ’ GO TO 111 ELSE GO TO 50 END IF ELSE RESULT=T( J , 1 ) END IF 111 RETURN END

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APPENDIX F SSM FORTRAN PROGRAM FOR ABLATION PROBLEM C SOLUTION OF ID ABLATION PROBLEM WITH ONE MOVING BOUNDARY C C IN A SEMI-INFINITE MEDIUM WITH OR WITHOUT SUBCOOLING C C BY THE SOURCE-AND-SINK METHOD C C C C INPUT FILE UNIT NUMBER IS SET AT 13 C C OUTPUT FILE UNIT IS SET AT 8 AND 14 C C c C NOTATIONS C C C C TMI =TIME WHEN PHASE CHANGE STARTS C C DELTM =TIME-STEP SIZE C C ALPHA =THERMAL DIFFUSIVITY C C CK =THERMAL CONDUCTIVITY C C CP =SPECIFIC HEAT C C CL =LATENT HEAT OF FUSION C C R =ABLATED SURFACE POSITION C C RV =SURFACE VELOCITY C C RHO =DENSITY C C TM =PHASE CHANGE TEMPERATURE C c c MAIN PROGRAM C IMPLICIT DOUBLE PRECISION(A-H,0-Z) PARAMETER (MAXF=2000 ,NP=15) DATA PI/3 . 14159265359D0/ COMMON/C/ALPHA DIMENSION XX(NP) COMMON /T/NF , I NF , TM I , TMF , DELTM COMMON /Cl /CK , RHO , CP , CL , TM , U COMMON /R1 /R( MAXF ) ,RV(MAXF) COMMON/XXF/XX(NP) COMMON/GH/NK COMMON /GG /G ( MAXF ) COMMON/SARA/NNN COMMON/EPSILON/EE COMMON/Ql/Q(MAXF) COMMON/GAS/SUMA , SUMB , SUMD , SUME , SAA , SDD , Q INTT , Q INT EXTERNAL TFUNC1 ,TFUNC2 C READ(13,*)CK,RHO,CP,CL,TM NTRIAL=1000 WRITE(*,*) Â’TMI ,DELTM=Â’ READ(* , *) TMI, DELTM 172

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ooo oo O o o 173 WRITE( * , * ) ’ TOTAL NO. OF ITERATION TIME STEP MF=?’ READ(* , *)MF WRITE (14 j*) ’ s**************************:********************* WRITE( 14 , * ) ’ INPUT DATA’ WRITE(14,*)’ ’ WRITE( 14,*) ’MELTING TEMPERATURE= ’ ,TM ,’C’ WRITE( 14,*) ’THERMAL CONDUCTIVITY^ ,CK ,’KJ/S.CM.C’ WRITE( 14 , *) ’ DENSITY= ’ ,RH0 ,’G/CUBIC CM’ WRITE(14,*) ’SPECIFIC HEAT CAPACITY= ’ ,CP ,’KJ/G.C’ WRITE( 14,*) ’LATENT HEAT= ’ ,CL ,’KJ/G’ WRITE(14,*)’ ’ WRITE! 14,*)’*********************************************** WRITE(14,200) 200 F0RMAT(8X, ’TIME’ ,17X, ’Q(T) ’ ,17X, ’R(T) ’ ,14X, ’DR/DT’ ) WRITE(14,*)’ ’ INITIALIZE R , RV , F DO 1=1, MAXF R( I )=0 . ODO RV( I )=0 . ODO Q( I )=0 . ODO ENDDO ALPHA=CK/RHO/CP DO NF=1 , MF XNF=NF TFF=TMI+XNF*DELTM ASSUME INTERFACE POSITION INTERVAL R(NF ) T0L=1 . D-20 IF(NF . EQ . 1 ) THEN WRITE(*,*)TFF, ’ ENTER INITIAL R1,R2’ READ(* , *)R1 ,R2 ELSE R1=R(NF-1 ) R2=5 . D0*R1 ENDIF EE=0 .ODO R(NF)=ZBRENT(TFUNC1 ,R1 ,R2,T0L) Q(NF)=(TM-SUMA+CL/CP*SUMB-QINT)/SAA WRITE(14,1000)TFF,Q(NF) ,R(NF) ,R(NF)/TFF 1000 FORMAT ( 3X , F9 . 4 , 5X , F23 . 9 , 5X , E14 . 7 , 5X , E14 . 7 ) ENDDO END C**************************************************************C C SUBPROGRAM FUNCTION C £**************************************************************C FUNCTION TFUNC1 (X) IMPLICIT DOUBLE PRECISION( A-H , 0-Z) PARAMETER (MAXF=2000 , PI=3 . 141592653589793D0 , NP=2) COMMON/C/ALPHA

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174 COMMON/EPS I LON/EE COMMON /T/NF , INF , TMI , TMF , DELTM COMMON/C1 /CK , RHO , CP , CL , TM , U COMMON/R1 /R(MAXF) ,RV(MAXF) COMMON/Ql/Q(MAXF) COMMON/GH/NK COMMON/GG/G(MAXF) COMMON/GAS /SUMA , SUMB , SUMD , SUME , SAA , SDD , q INTT , Q INT EXTERNAL FUNC1 , FUNC2A , FUNC2B , FUNC3 , FUNC33 , FUNC22A EXTERNAL FUNC11 ,FUNC22B EXTERNAL FUNC4 , FUNC5A , FUNC5B , FUNC6 , FUNC44 , FUNC66 EXTERNAL FUNC55A1 ,FUNC55B,FUNC55A2,FUNC5A1 ,FUNC5A2 R(NF)=X EE=0 . ODO C TFF2=TM I +NF*DELTM TFF=TFF2 TFF1=TMI+(NF-1)*DELTM IF(NF . EQ . 1 ) THEN RV(NF)=R(NF)/ (TFF2-TFF1 ) ELSE RV(NF)=(R(NF)-R(NF-1) )/ (TFF2-TFF1 ) END IF C C COMPUTE PART A C SUMA=0 . ODO DO INF=1 ,NF TF2=TM I + INF*DELTM TFl=TMI+( INF-1 )*DELTM IF( INF.EQ .NF)THEN AA=DSQRT(TFF2-TF1 ) BB=DSQRT(TFF2-TF2) CALL ROMBERG (FUNGI , AA ,BB, SAA) ELSE AA=DSQRT(TFF2-TF1 ) BB=DSQRT(TFF2-TF2) CALL ROMBERG ( FUNC1 , AA , BB , A ) A=Q( INF) *A*DSQRT( ALPHA/PI )/CK SUMA=SUMA+A ENDIF ENDDO SAA=DSQRT( ALPHA/PI )/CK*SAA C C COMPUTE PART B C SUMB1=0 . ODO SUMB2=0 . ODO DO INF=1 ,NF TF2=TMI+INF*DELTM TFl=TMI+( INF-1 ) *DELTM AA=DSQRT(TFF2-TF1 ) BB=DSQRT( TFF2-TF2 ) CALL ROMBERG (FUNC2A , AA ,BB, BB1 )

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175 CALL ROMBERG (FUNC2B, AA,BB,BB2) BB1=RV( INF)*BB1 BB2=RV ( INF)*BB2 SUMB1=SUMB1+BB1 SUMB2=SUMB2+BB2 ENDDO SUMB=SUMB1+SUMB2 C C COMPUTE PART C C IF(TMI .EQ. O.ODO) THEN QINT=0 . ODO ELSE CC1=DSQRT(TFF) CC2=DSQRT(TFF-TMI ) CALL ROMBERG (FUNC3,CC1 ,CC2,QINT) QINT=QINT*DSQRT(ALPHA/PI )/CK ENDIF C C COMPUTE PART D C SUMD=0 . ODO DO INF=1 ,NF-1 TF2=TMI+INF*DELTM TF1 =TM I + ( I NF1 )*DELTM Al=4 .ODO* DSQRT( ALPHA )/(R(NF)+EE) A2=(R(NF)+EE) /DSQRT(4 . 0D0*ALPHA*(TFF2-TF1 ) ) A3=(R(NF)+EE)/DSQRT(4 .0D0*ALPHA*(TFF2-TF2) ) CALL ROMBERG (FUNC5A,A2, A3, D) D=A1*D D=Q(INF)*D*R(NF)/2.0D0/DSQRT(ALPHA*PI) SUMD=SUMD+D ENDDO A2=(R(NF)+EE)/DSQRT(4 . 0D0*ALPHA*(TFF2-TFF1 ) ) SDD=ERFC(A2) C C COMPUTE PART E C SUME1=0 . ODO SUME2=0 . ODO DO INF=1 ,NF TF2=TM I + 1 NF* DELTM TF1 =TM I + ( I NF1 ) * DELTM IF( INF .EQ. 1 )THEN AM=RV( INF) BN=-R( INF)*TF1 / (TF2-TF1 ) ELSE AM=RV( INF) BN=(R( INF-1 )*TF2-R( INF)*TF1 )/(TF2-TFl ) ENDIF IF( INF .EQ. NF)THEN C1=(R(NF)+EE) -2 . 0D0*AM*TFF2+AM*TF1 -BN C2=DSQRT(4 . ODO* ALPHA* (TFF2-TF1 ) ) CC=C1/C2

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o o o o o o 176 C01=l.ODO/2.ODO*DEXP(-AM*((R(NF)+EE)-AM*TFF2-BN)/ALPHA) E1=C01*ERFC(CC) Fl=( (R(NF)+EE)+2 . 0D0*AM*TFF2-AM*TF1+BN) FF=F1/C2 C02=l . ODO /2 . ODO*DEXP( AM* ( ( R( NF )+EE ) +AM*TFF2+BN ) / ALPHA ) E2=C02*ERFC(FF) ELSE CO 1=1 .ODO/DSQRT(PI )*DEXP( -AM*( (R(NF) +EE)-AM*TFF2BN) /ALPHA) C02= 1 . ODO/DSQRT( PI ) *DEXP ( AM* ( ( R( NF ) +EE ) +AM*TFF2+BN ) / ALPHA ) Cl= ( R( NF) +EE) -2 . 0D0*AM*TFF2+AM*TF1 -BN C2=DSQRT(4.0D0*ALPHA*(TFF2-TF1 ) ) CC=C1/C2 C3=(R(NF)+EE) -2 . 0D0*AM*TFF2+AM*TF2-BN C4=DSQRT(4 . 0D0*ALPHA*(TFF2-TF2) ) CC1=C3/C4 CALL ROMBERG (FUNC5B,CC,CC1, El) E1=C01*E1 Fl=( (R(NF)+EE)+2.0D0*AM*TFF2-AM*TF1+BN) FF=F1/C2 F2=( (R(NF)+EE)+2 . 0D0*AM*TFF2-AM*TF2+BN) FF1=F2/C4 CALL ROMBERG (FUNC5B,FF,FF1,E2) E2=C02*E2 ENDIF E1=RV(INF)*E1 E2=RV( INF)*E2 SUME1=SUME1+E1 SUME2=SUME2+E2 ENDDO SUME=SUME1 +SUME2 COMPUTE TO TERM IF(TMI .EQ. O.ODO) THEN QINTT=O.ODO ELSE CC1=1.0D0/DSQRT(TFF) CC2=1 . ODO/DSQRT(TFF-TMI ) CALL ROMBERG ( FUNC6 , CC1 , CC2 , Q INTT ) QINTT=QINTT*R(NF)/2.0D0/DSQRT(ALPHA*PI) ENDIF q2=GX(TFF2)-qiNTT+RHO*CL*SUME-RHO*CL*RV(NF)-SUMD q3=TM-qiNT+CL/CP*SUMB-SUMA TFUNCl=SDD*q3-SAA*q2 RETURN END SUBPROGRAM FUNCTION FOR TFUNC1 FUNCTION FUNCl(X) IMPLICIT DOUBLE PRECISION( A-H , O-Z) PARAMETER (MAXF=2000) O O O O

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o o o 177 COMMON/C/ALPHA DATA PI/3.14159265359D0/ COMMON/C1 /CK , RHO , CP , CL , TM , U COMMON/T/NF, INF,TMI ,TMF, DELTM COMMON/R1 /R(MAXF) ,RV(MAXF) C TFF=TMI+NF*DELTM IF(X .EQ. O.ODO) THEN FUNC1=0 . ODO ELSE ARG=-R(NF)**2/4 . ODO/ALPHA/X/X FUNC1=-2.0D0*DEXP(ARG) END IF RETURN END **>|o(<>|o(o|o(o|c*****************************************************C SUBPROGRAM FUNCTION C FUNCTION FUNC2A(X) IMPLICIT DOUBLE PRECISION( A-H , O-Z) PARAMETER (MAXF=2000 , PI=3 . 141592653589793D0) COMMON/C/ALPHA COMMON /T/NF , I NF , TMI , TMF , DELTM COMMON/R1 /R(MAXF) ,RV(MAXF) C TFF=TMI+NF*DELTM TF2=TMI+INF*DELTM TFl=TMI+( INF-1)* DELTM IF( INF .EQ. 1) THEN A=RV(INF) Z=-R(NF)*TF1 / (TF2-TF1 ) ELSE A=RV( INF) Z=(R( INF-1 )*TF2-R( INF)*TF1 ) / (TF2-TF1 ) END IF RX=A*(TFF-X**2)+Z C IF(X .EQ. O.ODO) THEN FUNC2A=-2.0D0/DSQRT(4*PI*ALPHA) ELSE FUNC2A=-2 . ODO/DSQRT(4*PI*ALPHA) $ *DEXP(-(R(NF)-RX)**2/4. ODO/ALPHA/X/X) END IF RETURN END C C**************************************************************^ C SUBPROGRAM FUNCTION C c**************************************************************c FUNCTION FUNC2B(X) IMPLICIT DOUBLE PIlECISI0N(A-H,0-Z) PARAMETER (MAXF=2000 , PI=3 . 141592653589793D0) COMMON/C/ALPHA

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o o o o 178 COMMON/T/NF, INF,TMI ,TMF,DELTM COMMON /R1 /R( MAXF ) ,RV(MAXF) I TFF=TM I +NF*DELTM TF2=TM I + INF*DELTM TFl=TMI+( INF-1 )*DELTM IF( INF .EQ. 1) THEN A=RV( INF) Z=-R(NF)*TF1 / (TF2-TF1 ) ELSE A=RV( INF) Z=(R( INF-1 )*TF2-R( INF)*TF1 ) / (TF2-TF1 ) END IF RX=A*(TFF-X**2)+Z IF(X .EQ. O.ODO) THEN FUNC2B=0 . ODO ELSE FUNC2B=-2.0D0/DSQRT(4*PI*ALPHA) $ *DEXP( -(R(NF)+RX)**2/4 . ODO/ALPHA/X/X) END IF RETURN END **************************************************************C SUBPROGRAM FUNCTION C **************************************************************C FUNCTION FUNC3(X) IMPLICIT DOUBLE PRECISION(A-H,0-Z) PARAMETER (MAXF=2000) COMMON/C/ALPHA DATA PI/3.14159265359D0/ COMMON /Cl /CK , RHO , CP , CL , TM , U COMMON /T/NF , INF , TM I , TMF , DELTM COMMON/R1 /R(MAXF) ,RV(MAXF) C TFF=TMI+NF*DELTM FUNC3=-2 . 0D0*GX(TFF-X**2)*DEXP( -R(NF)**2/4 . ODO/ALPHA/X/X) RETURN END C C C C=t==l<*=(!=l=*^**=l=********^*!(<=|5******^=K***;**:*>|t*******=*'>t:*>l'******=t I: * !: * :> t < ****C C SUBPROGRAM FUNCTION C C******+;**^*!i<**>l==t!****^********>t : **=* t ***************** : +* :(c ******* !, '*C FUNCTION FUNC4(X) IMPLICIT DOUBLE PRECISION(A-H , O-Z) PARAMETER (MAXF=2000) COMMON/C/ALPHA COMMON /T /NF , INF , TMI , TMF , DELTM COMMON/Rl/R(MAXF) ,RV(MAXF) COMMON/EPSILON/EE C

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o o o 179 TFF=TM I +NF*DELTM FUNC4=2 . ODO*DEXP( -(R(NF)+EE)**2/4 . ODO/ALPHA*X*X) RETURN END SUBPROGRAM FUNCTION C FUNCTION FUNC5A(X) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) PARAMETER (MAXF=2000) DATA PI/3 . 14159265359D0/ COMMON/C/ALPHA COMMON/T/NF, INF,TMI ,TMF,DELTM COMMON /R1 /R( MAXF ) ,RV(MAXF) COMMON/EPSILON/EE FUNC5A=DEXP( -X*X) RETURN END C C SUBPROGRAM FUNCTION C C>t e3 t !5 4 e5 ) e + a 4 e3 f :5 l c3 t e3 t c3 ^^ : 4 c5 4 e;, K 5 4 cs ( e3 4 e, t ei 4 <5 ^ , t cs ^ :, ( e3 ^ i:a ^ i ^ 3 t ( Â’t e5 ^^* 3 * e * s ^^ c ^ c5 ^ <3 ^ i ^ :s * es * c, * eJ * c3 * 1 **^ c ^ ea * c5 * :: * : *^
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180 C**************************************************************^ C SUBPROGRAM FUNCTION c C**************************************************************^ FUNCTION GX(X) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) GX=0 . 01D0 END C* ************************************************************ * C SUBPROGRAM C ROMBERG INTEGRATION C ******* *********** * ****************************************** * SUBROUTINE ROMBERG ( FUNC , A , B , RESULT) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) EXTERNAL FUNC PARAMETER ( MAX=50 , EPS=0 . 0001 DO ) DIMENSION T(MAX ,MAX) C T(1,1)=(B-A)*(FUNC(A)+FUNC(B))/2.0D0 T(l,2)=T(l,l)/2. ODO+(B-A)*FUNC( ( A+B)/2 . 0D0)/2 . ODO T(2, 1 )=(4. ODO*T( 1 ,2)-T(l,l))/3. ODO J=3 C C SUCCESSIVE APPLICATION OF TRAPEZOIDAL RULE C 50 DELX=(B-A)/2 . 0D0**( J-l ) X=A-DELX N=2**( J-2) SUM=0 . ODO DO 100 1=1, N X=X+2 . ODO*DELX SUM=SUM+FUNC ( X ) 100 CONTINUE T(l,J)=T(l,J-l)/2. ODO+DELX*SUM C C EXTRAPOLATION C DO 200 L=2 , J K=J+1-L T(L,K)=(4. 0D0**(L-1 )*T(L-1,K+1)-T(L-1,K))/ $ (4.0D0**(L-1)-1 .ODO) 200 CONTINUE C C CHECK ACCURACY CRITERION C IF(T( J , 1 ) .EQ. 0 . ODO) THEN RESULT=T( J , 1 ) GO TO 111 END IF C IF(DABS( (T( J , 1 )-T( J-l ,1))/T(J,1)) .GE. EPS) THEN J=J+1 IF(J.GT.MAX) THEN PAUSE Â’TOO MANY STEPSÂ’ GO TO 111 O O O O

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o o 181 ELSE GO TO 50 END IF ELSE RESULT=T( J , 1 ) END IF 111 RETURN 222 END C C************************************************************** SUBPROGRAM ERROR FUNCTION C************************************************************** FUNCTION ERF(X) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) IF(X . LT. 0 . ODO)THEN ERF=-GAMMP( . 5D0,X**2) ELSE ERF=GAMMP( . 5D0,X**2) ENDIF RETURN END C FUNCTION ERFC(X) IMPLICIT DOUBLE PRECISION( A-H , 0-Z) IF(X . LT . 0 . ODO)THEN ERFC=1 .ODO+GAMMP( . 5D0,X**2) ELSE ERFC=GAMMQ( .5D0,X**2) ENDIF RETURN END c**************************************************************c FUNCTION GAMMP(A,X) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) IF(X . LT . 0 . ODO .OR. A.LE.O.ODO)PAUSE IF(X.LT. A+l .ODO)THEN CALL GSER(GAMSER,A,X,GLN) GAMMP=GAMSER ELSE CALL GCF(GAMMCF, A ,X ,GLN) GAMMP=1 .ODO-GAMMCF ENDIF RETURN END C**************************************************************^ FUNCTION GAMMQ(A,X) IMPLICIT DOUBLE PRECISION( A-H , 0-Z) IF(X.LT.O.ODO .OR. A.LE.O.ODO)PAUSE IF(X . LT. A+l . ODO)THEN CALL GSER(GAMSER, A,X,GLN) GAMMQ=1 .ODO-GAMSER ELSE O O O O

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182 CALL GCF( GAMMCF , A , X , GLN ) GAMMQ=GAMMCF END IF RETURN END SUBROUTINE GCF( GAMMCF, A, X, GLN) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) PARAMETER ( ITMAX=200 ,EPS=3 . D-12) GLN=GAMMLN( A) G0LD=0 . ODO AO=l . ODO A1=X B0=0 . DO Bl=l . DO FAC=1 . DO DO 11 N=1 , ITMAX AN=DFLOAT(N) ANA=AN-A AO=( Al+AO*ANA)*FAC BO=(Bl+BO*ANA)*FAC ANF=AN*FAC Al=X*AO+ANF*Al Bl=X*BO+ANF*Bl IF(A1 .NE.O.ODO)THEN FAC=1 . ODO/A1 G=B1*FAC IF(DABS( (G-GOLD)/G) . LT.EPS)GO TO 1 G0LD=G ENDIF 11 CONTINUE PAUSE Â’A TOO LARGE, ITMAX TOO SMALLÂ’ I GAMMCF=DEXP( -X+A*DLOG ( X ) -GLN ) *G RETURN END SUBROUTINE GSER( G AMSER , A , X , GLN ) IMPLICIT DOUBLE PRECIS I ON (A-H , O-Z) PARAMETER ( ITMAX=200 , EPS=3 .D-12) GLN=GAMMLN( A) IF(X . LE. 0 . ODO)THEN IF(X.LT.O. ODO) PAUSE Â’X LT O.ODOÂ’ GAMSER=0 . ODO RETURN ENDIF AP=A SUM=1 . ODO/ A DEL=SUM DO 11 N=l, ITMAX AP=AP+1 . ODO DEL=DEL*X/AP SUM=SUM+DEL I F( DABS (DEL) . LT. DABS(SUM)*EPS)GO TO 1 II CONTINUE

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o o 183 PAUSE Â’A TOO LARGE, UMAX TOO SMALLÂ’ I GAMSER=SUM*DEXP(-X+A*DLOG(X)-GLN) RETURN END FUNCTION GAMMLN(XX) IMPLICIT DOUBLE PRECISION(A-H,0-Z) DIMENSION COF(6) DATA COF , STP/76 . 180091729406D0 , -86 . 505320327 1 12D0 , $ 24 . 014098222230D0 , -1 . 231739516140D0 , . 120858003D-2 , $ . 536382D-5 , 2 . 50662827465D0/ DATA HALF , ONE , FPF/O . 5D0 , 1 . ODO , 5 . 5D0/ X=XX-ONE TMP=X+FPF TMP=(X+HALF)*DLOG(TMP)-TMP SER=ONE DO 11 J=1 ,6 X=X+ONE SER=SER+COF(J)/X II CONTINUE GAMMLN=TMP+DLOG(STP*SER) RETURN END USING BRENTÂ’S METHOD FOR FINDING THE ROOT OF FUNCTION TFUNC FUNCTION ZBRENT(TFUNC,X1 ,X2,T0L) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) EXTERNAL TFUNC1 PARAMETER ( ITMAX=100 ,EPS=3 . D-16) A=X1 B=X2 FA=TFUNC( A) FB=TFUNC(B) IF(FB*FA . GT. 0 . DO) PAUSE Â’ROOT MUST BE BRACKETED FOR ZBRENT. FC=FB DO 11 ITER=1 , ITMAX IF(FB*FC . GT . 0 . DO) THEN C=A FC=FA D=B-A E=D END IF IF(DABS(FC) . LT. DABS(FB) ) THEN A=B B=C C=A FA=FB FB=FC FC=FA ENDIF T0L1=2 . D0*EPS*DABS(B)+0 . 5DO*TOL XM= . 5D0*(C-B) o o o o

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184 IF(DABS(XM) .LE.T0L1 .OR. FB.EQ.O.DO)THEN ZBRENT=B RETURN END IF IF(DABS(E) .GE.TOL1 .AND. DABS (FA) . GT . DABS(FB) ) THEN S=FB/FA IF(A.EQ.C) THEN P=2.D0*XM*S Q=1 .DO-S ELSE Q=FA/FC R=FB/FC P=S*(2.D0*XM*Q*(Q-R)-(B-A)*(R-1 .DO)) Q=( Q-l . DO)*(R-l . DO)*(S-l . DO) END IF IF(P.GT.O.DO) Q=-Q P=DABS(P) IF(2.D0*P .LT. DMIN1 (3. *XM*Q-DABS(TOLl*Q) ,DABS(E*Q) ) ) $ THEN E=D D=P/Q ELSE D=XM E=D ENDIF ELSE D=XM E=D ENDIF A=B FA=FB IF(DABS(D) .GT. T0L1) THEN B=B+D ELSE B=B+DSIGN(TOLl ,XM) ENDIF FB=TFUNC(B) 11 CONTINUE PAUSE Â’ZBRENT EXCEEDING MAXIMUM ITERATIONS.Â’ ZBRENT=B RETURN END C SUBROUTINE ZBRAK(FX,X1 ,X2,N,XB1 ,XB2,NB) IMPLICIT DOUBLE PRECISION(A-H,0-Z) DIMENSION XB1(1) ,XB2(1) NBB=NB NB=0 X=X1 DX=(X2-X1 )/N FP=FX(X) DO 11 1=1, N X=X+DX

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FC=FX(X) IF(FC*FP.LT.O. ) THEN NB=NB+1 XB1 (NB)=X-DX XB2(NB)=X ENDIF FP=FC IF(NBB.EQ.NB)RETURN CONTINUE RETURN END

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APPENDIX G SSM FORTRAN PROGRAM FOR COMBINATION PROBLEM C SOLUTION OF ID ABLATION PROBLEM WITH TWO MOVING BOUNDARIES C IN A SEMI -INFINITE MEDIUM WITH SUBCOOLING BY THE SOURCEC AND-SINK METHOD C C INPUT FILE UNIT NUMBER IS SET AT 15 C OUTPUT FILE UNIT IS SET AT 16 C c c c TMI NOTATIONS =TIME WHEN PHASE CHANGE STARTS c DELTM =T I MESTEP SIZE c ALPHA =THERMAL DIFFUSIVITY c CK =THERMAL CONDUCTIVITY c CP =SPECIFIC HEAT c CLF =LATENT HEAT OF FUSION c CLV =LATENT HEAT OF VAPORIZATION c TIN =TEMPERATURE AT INTERIOR POINTS c R1 =SOLID-LIQUID INTERFACE POSITION c R2 =ABLATED SURFACE POSITION c R1V =SOLID-LIQUID INTERFACE VELOCITY c R2V =SURFACE VELOCITY c RHO = DENSITY c TM =MELTING TEMPERATURE c TM ^VAPORIZATION TEMPERATURE c c MAIN PROGRAM C C c c c c c c c c c c c c c c c c c c c c c c c c c IMPLICIT DOUBLE PRECISION(A-H,0-Z) PARAMETER(MAXF=2500,NP=3, PI=3 . 141592653589793D0) COMMON /T/NF , INF , TMI , TMF , DELTM , TRTM PARAMETER (TOLX=l . OD-12) PARAMETER (N=3,TOLF=l .OD-12) COMMON /C/ALPHA DIMENSION XX(NP) COMMON /Cl /CK , RHO , CP , CLF , CLV , TM , TV COMMON /R1/R1(MAXF) ,RV1(MAXF) ,Q(MAXF) COMMON /R2/R2(MAXF) ,RV2(MAXF) ,MF,MT COMMON /RINT/RIX COMMON /DIF/CKA,CKB,DIFF COMMON /HEAT1/RX(300) ,TIN(300) COMMON/HEAT/ERRX , ERRF EXTERNAL TFUNC 1 , TFUNC2 , FUNC 1 2 , FUNC22 A , TFUNC3 , FUNC 1 EXTERNAL FUNC3B , FUNC3A , FUNC2A , FUNC2B , FUNCl 1 , TFUNCA 186

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o o o 187 c READ(15,*) CK , RIIO , CP , CLF , CLV , TM , TV VRITE(* , * ) Â’ TMI , EPSILON , DELTM= Â’ READ(* , *) TMI , EPSILON, DELTM WRITE( * , * ) Â’ TOTAL NO. OF ITERATION TIME STEP MF=? Â’ READ(* , *)MF MT=10000 WRITE(*,*) Â’NO. OF INTERNAL COMPUTATION STEP NPS=?Â’ READ(*,*)NPS VRITE(16,*) Â’ WRITE(16,*) Â’ INPUT DATAÂ’ WRITE( 16 , 111 )TM WRITE( 16 , 122)TV WRITE( 16 , 133)CK VRITE( 16 , 144)RHO WRITE( 16 , 155) CP WRITE( 16 , 166)CLF WRITE( 16 , 177) CLV WRITE (16, 188) DELTM WRITE( 16 , *) Â’ WRITE( 16 , * ) Â’ TIME INTERFACEÂ’ WRITE( 16,*) Â’ Â’ VRITE(16,*)Â’ R1 $ GÂ’ WRITE(77,*) Â’ TIME X TEMPÂ’ INITIALIZE R1 ,R2 ,RV1 ,RV2 ,RX ,TIN R2 WRITE(66,*) Â’ Â’ WRITE(66 , *) Â’ CKA=RHO*LV*DR2/DT Â’ WRITE(66,*) Â’ CKB=-K*DT(R2,T)/DTÂ’ WRITE(66 , * ) Â’ WRITE(66 , *) Â’ Â’ WRITE(66,*) Â’ TIMEÂ’,Â’ G(T) Â’,Â’ CKAÂ’,Â’ $ CKBÂ’,Â’ G(T)-(CKA+CKB) Â’ VRITE(66,*)Â’ Â’ DO 1=1 ,MAXF R1 ( I )=0 . ODO RV1 ( I )=0 . ODO R2( I )=0 . ODO RV2( I )=0 . ODO Q(NF)=0 . ODO ENDDO DO 1=1,300 RX( I )=0 . ODO TIN( I )=0 . ODO ENDDO C ALPHA=CK/RHO / CP C DO 100 NF=1 ,MF XNF=NF TFF=TM I +XNF* DELTM C

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o o o 188 T0L=1 . D-15 IF(NF .LT. MT )THEN ASSUME INTERFACE POSITION INTERVAL Rl(NF) IF(NF .Eq. 1) THEN WRITE(*,*)TFF, Â’ ENTER INITIAL RA ,RB=? Â’ READ(* , *) RA,RB ELSE RA=R1 (NF-1 ) /2 . ODO RB=5 . ODO*Rl (NF-1 ) END IF C 88 R1(NF)=ZBRENT(TFUNCA,RA,RB,T0L) AA=DSQRT(TFF) CALL ROMBERG ( FUNC12 , AA , 0 . ODO , qUI CK ) qUICK=qUICK*DSqRT(ALPHA/PI )/CK SUMP=0 . ODO DO INF=1 ,NF TF2=TM I + I NF*DELTM TFl=TMI+( INF-1 )*DELTM AA=DSqRT(TFF-TFl ) BB=DSqRT(TFF-TF2) CALL ROMBERG (FUNC22A,AA,BB,P) P=P*RV1 ( INF) SUMP=SUMP+P ENDDO SUMP=SUMP*CLF/2 . ODO/CP/DSqRT( PI* ALPHA ) SP=+SUMP-qUICK+TV WRITE(*,*)SP IF(SP .LE. EPSILON)THEN MT=NF WRITE(*,*) NF,R1(NF) WRITE( 16 , 1000)TFF , R1 (NF) C UPDATE INTERFACE POSITION VELOCITY C IF(NF .Eq. 1) THEN RV1 (NF)=R1 (NF)/DELTM ELSE RV1 (NF)=(R1 (NF)-Rl (NF-1 ) )/DELTM END IF C C INTERNAL POINT TEMP. COMPUTATION C IF(MOD(NF,NPS) .NE. 0) GO TO 100 RIX=0 . ODO RX( 1 )=RIX CALL INTERNAL(TINTL) TIN( 1 )=TINTL DO 1=1,10 RIX=R1 (NF)/10 . ODO* I RX( 1+1 )=RIX CALL INTERNAL(TINTL) TIN(I+1)=TINTL

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189 ENDDO DO 1=1,42 RIX=R1 (NF)+R1 (NF)/2 . ODO*I RX( 1+1 1 )=RIX CALL INTERNAL(TINTL) TIN( 1+11 )=TINTL ENDDO C C COMPUTATION OF STORED HEAT C CALL HEATS GO TO 100 ELSE C UPDATE INTERFACE POSITION VELOCITY C IF(NF .EQ. 1) THEN RV1 (NF)=R1 (NF) /DELTM ELSE RV1 (NF)=(R1 (NF)-Rl (NF-1 )) /DELTM END IF C c INTERNAL POINT TEMP. COMPUTATION C IF(MOD(NF,NPS) .NE. 0) GO TO 100 RIX=0 . ODO RX( 1 )=RIX CALL INTERNAL(TINTL) TIN( 1 )=TINTL DO 1=1,10 RIX=R1(NF)/10.0D0*I RX( 1+1 )=RIX CALL INTERNAL(TINTL) TIN( 1+1 )=TINTL ENDDO DO 1=1,42 RIX=R1 (NF)+R1 (NF)/2 . ODO* I RX( 1+1 1 )=RIX CALL INTERNAL(TINTL) TIN( 1+1 1 )=TINTL ENDDO C C COMPUTATION OF STORED HEAT C CALL HEATS GO TO 100 END IF ELSE TRTM=TM I +MT*DELTM IF(NF .EQ. MT+1 ) THEN C C COMPUTE TWO INTERFACE POSITIONS R1,R2 C INITIAL GUESS OF INTERFACE POSITION R2(NF) C WRITE(* ,*)NF, Â’ ENTER ASSUMED INITIAL Rl=? , R2=? AND QÂ’

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ooo o o o o o o o o o 190 READ(* ,*)R1(NF),R2(NF),Q(NF) ELSE R2 ( NF ) = 1 . 002D0*R2 ( NF1 ) R1 (NF) = 1 . 002D0*R1 (NF-1 ) q(NF)=1.001D0*Q(NF-l) ENDIF CREATE VECTOR XX XX(1)=R1(NF) XX(2)=R2(NF) XX(3)=Q(NF) COMPUTE B(NF ) AND Q(NF) NTRIAL=1000 CALL MNEWT( NTRI AL , XX , N , TOLX , TOLF ) WRITE(16,1200)TFF,R1(NF) ,R2(NF) ,Q(NF) UPDATE INTERFACE POSITION VELOCITY RV1(NF)=(R1(NF)-R1(NF-1)) /DELTM IF(NF .EQ. MT+1 ) THEN RV2(NF)=R2(NF)/DELTM ELSE RV2(NF)=(R2(NF)-R2(NF-1 ) ) /DELTM END IF INTERNAL POINTS TEMP. COMPUTATION IF(MOD(NF,NPS) .NE. 0) GO TO 100 RIX=0 . ODO RX( 1 )=RIX CALL INTERNAL1 (TINTL) TIN( 1 )=TINTL DO 1=1,10 RIX=R2(NF)/10 . ODO* I RX( 1+1 )=RIX CALL INTERNAL1 (TINTL) TIN(I+1)=TINTL ENDDO DO 1=1,50 RIX=(R1(NF)-I12(NF))/50.0D0*I+R2(NF) RX( 1+11 )=RIX CALL INTERNAL1 (TINTL) TIN( 1+1 1 )=TINTL ENDDO DO 1=1,30 RIX=R1(NF)+R1(NF)/4.0D0*I RX( 1+61 )=RIX CALL INTERNAL1 (TINTL) TIN ( 1+61 )=TINTL ENDDO CALL HEATS

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191 END IF 100 CONTINUE WRITE( 16 , 199)TRTM C C FORMATS C 111 FORMAT(// Â’MELTING TEMPERATURES ,F7.2) 122 FORMAT( Â’VAPORIZATION TEMPERATURE= Â’ , F7 . 2) 133 FORM AT ( Â’THERMAL CONDUCTIVITY^ Â’ , F10 . 6) 144 FORMAT ( Â’DENSITY=Â’ , FI 0.6) 155 FORMAT (Â’SPECIFIC HEAT CAPACITYS ,F10.6) 166 FORM AT ( Â’FUSION LATENT HEAT=Â’ ,F10.7) 177 FORMAT( Â’VAPORIZATION LATENT HEATS ,F10.7) 188 FORMAT(/ Â’TIME STEP SIZES ,F10.7) 199 FORMAT( Â’TRANSIENT TIMES ,F10.7/ /) 1000 FORMAT( 3X , F13 . 7 , 5X , E14 . 7 ) 1200 FORM AT (2X ,F13.7,5X,E14.7,5X,E14.7,5X,E14.7) 1300 FORMAT ( 2X , FI 0 . 7 ,2X ,F14.7 ,2X,F14.7 ,2X ,F14. 7 ,5X ,E14 . 7) END C C SUBPROGRAM MNEVTON C C****************************** ***************** ***************C SUBROUTINE MNEWT(NTRIAL , XX ,N ,TOLX ,TOLF) IMPLICIT DOUBLE PRECISION( A-H , 0-Z) PARAMETER (NP=3) PARAMETER (MAXF=2500 , ADS . IDO, J=2) COMMON /HEAT/ERRX ,ERRF DIMENSION ALPIIAl (NP,NP) ,BETA(NP) , INDX(NP) ,XX(NP) C DO 13 I<=1 ,NTRIAL CALL USRFUN(XX, ALPIIAl , BETA) ERRF=0 . ODO DO 11 1=1, N ERRF=ERRF+DABS ( BETA ( I ) ) 11 CONTINUE IF(ERRF .LE. TO LF) RETURN CALL LUDCMP( ALPIIAl ,N , NP , INDX , D) CALL LUBKSB( ALPHA1 , N , NP , INDX , BETA ) ERRX=0 . ODO DO 12 1=1, N ERRX=ERRX+DABS ( BETA ( I ) ) XX ( I )=XX( I )+BETA( I ) 12 CONTINUE I F ( ERItX . LE.TOLX) RETURN 13 CONTINUE RETURN END C SUBPROGRAM USERFUNCTION C SUBROUTINE USRFUN (XX, ALPIIAl , BETA) IMPLICIT DOUBLE PRECISION( A-H , 0-Z) PARAMETER(MAXF=2500 , PI =3 . 141 592653589793D0 , NP=3 )

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192 DIMENSION ALPHA1(NP,NP) ,BETA(NP) ,XX(NP) COMMON /T/NF , INF ,TMI ,TMF , DELTM , TRTM COMMON /R1/R1(MAXF),RV1(MAXF),Q(MAXF) COMMON /R2/R2(MAXF) ,RV2(MAXF) ,MF,MT C CH=1 . D-10 C R1 (NF)=XX( 1 ) R2(NF)=XX(2) Q(NF)=XX(3) C Fl=TFUNCl (R1 (NF) ) R1(NF)=R1 (NF)+CH ALPHA1(1,1)=(TFUNC1(R1(NF))-F1)/CH R1(NF)=R1(NF)-CH R2(NF)=R2(NF)+CH ALPHA 1 ( 1 , 2)=(TFUNCl (R1 (NF) )-Fl )/CH R2(NF)=R2(NF)-CH Q(NF)=Q(NF)+CH ALPHA1 (1 , 3)=(TFUNC1(R1 (NF) )-Fl )/CH Q(NF)=Q(NF)-CH C F2=TFUNC2(R2(NF) ) R2 ( NF ) =R2 ( NF ) +CH ALPHA1 (2 ,2)=(TFUNC2(R2(NF) )-F2)/CH R2(NF)=R2(NF ) -CH R1(NF)=R1(NF)+CH ALPHA! (2 , 1 )=(TFUNC2(R2(NF) ) -F2)/CH R1(NF)=R1 (NF)-CH Q(NF)=Q(NF)+CH ALPHA1 (2,3)=(TFUNC2(R2(NF) )-F2)/CH Q(NF)=Q(NF)-CH C F3=TFUNC3(Q(NF) ) Q(NF)=Q(NF)+CH ALPHA1 (3 , 3)=(TFUNC3(Q(NF) )-F3)/CH Q(NF)=Q(NF)-CH R1 (NF)=R1 (NF)+CH ALPHA1 (3 , 1 )=(TFUNC3(Q(NF) )-F3)/CH R1 (NF)=R1 (NF)-CH R2(NF)=R2(NF)+CH ALPHA1 (3 ,2)=(TFUNC3(Q(NF) )-F3)/CH R2(NF)=R2(NF)-CH C BETA ( 1 )=-Fl BETA(2)=-F2 BETA(3)=-F3 C END C*>|e*>(o(=>K>l<*****=(<*>t:**>K*****=l==t:=t=>t ! *****X“(!*********=t! > t t ************** ,, ‘**C C SUBPROGRAM TFUNCA C FUNCTION TFUNCA (X) IMPLICIT DOUBLE PRECISION(A-H,0-Z)

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193 PARAMETER^ MAXF=2500 , PI=3 . 141592653589793D0 ) COMMON /T/NF, INF,TMI , TMF , DELTM , TRTM COMMON /C/ALPHA COMMON /Cl /CK , RHO , CP , CLF , CLV , TM , TV COMMON /R1 /R1 (MAXF) ,RV 1 (MAXF) ,q(MAXF) COMMON /R2/R2(MAXF) , RV2 ( M AXF ) , MF , MT COMMON /REST/Q I NTT , SUM A , SUMB , SUMC , SUMAA EXTERNAL FUNCl , FUNC2A , FUNC2B R1(NF)=X TFF2=TM I +NF* DELTM TFF1 =TM I + ( NF1 )* DELTM IF(NF .EQ. 1) THEN RV1(NF)=R1(NF)/ (TFF2-TFF1 ) ELSE RV1 (NF)=(R1 (NF)-Rl (NF-1 ) )/ (TFF2-TFF1 ) END IF IF(TMI .EQ. 0 . ODO)THEN QINT=0 . ODO ELSE AA=DSQRT(TFF2) BB=DSQRT(TFF2-TMI ) CALL ROMBERG (FUNCl ,AA,BB,QINT) END IF qiNTl=O.ODO DO INF=1 ,NF TF2=TMI+INF*DELTM TFl=TMI+( INF-1 )*DELTM AA=DSQRT(TFF2-TF1 ) BB=DSQRT(TFF2-TF2) CALL ROMBERG ( FUNCl ,AA,BB, A) qiNTl=QINTl+A ENDDO SUMB1=0 . ODO SUMB2=0 . ODO DO INF=1 ,NF TF2=TM I + I NF*DELTM TFl=TMI+( INF-1 )*DELTM AA=DSQRT(TFF2-TF1 ) BB=DSQRT(TFF2-TF2) CALL ROMBERG (FUNC2A , AA ,BB , B1 ) CALL ROMBERG (FUNC2B,AA,BB,B2) B1=RV1 ( INF)*B1 B2=RV1 ( INF)*B2 SUMBUSUMB1+B1 SUMB2=SUMB2+B2 ENDDO SUMB=SUMB1+SUMB2 TFUNCA=(2 . ODO*ALPHA/CK*(QINT+QINTl )-CLF/CP*SUMB) $ /2 . ODO/DSQRT(PI*ALPHA)-TM

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RETURN END 194 C SUBPROGRAM TFUNCl C************************************************************** FUNCTION TFUNCl (X) IMPLICIT DOUBLE PRECISION( A-H , O-Z) PARAMETER( MAXF=2500 , PI =3 . 141 592653589793D0 ) COMMON /T/NF, INF,TMI , TMF , DELTM , TRTM COMMON /C/ALPHA COMMON /C 1 /CK , RHO , CP , CLF , CLV , TM , TV COMMON /R1/R1(MAXF) ,RV1 (MAXF) , Q(MAXF) COMMON /R2/R2(MAXF) ,RV2(MAXF) ,MF,MT EXTERNAL FUNC1 , FUNC4 , FUNC5A , FUNC5B , FUNC6A , FUNC6B $ , FUNC2A , FUNC2B , FUNC3A , FUNC3B , FUNC44 , FUNC1 1 R1(NF)=X C ' TFF2=TM I +NF* DELTM TFF1=TMI+(NF-1 )*DELTM IF(NF .EQ. MT+1 ) THEN RV2(NF)=R2(NF)/ (TFF2-TFF1 ) ELSE RV2(NF)=(R2(NF)-R2(NF-1 ) )/ (TFF2-TFF1 ) END IF C RV1 (NF)=(R1 (NF) -R1 (NF-1))/ (TFF2-TFF1 ) C AA=DSQRT(TFF2) BB= DS QRT ( TFF2 TRTM ) CALL ROMBERG (FUNC1 ,AA,BB,QINTT) C SUMA=0 . ODO DO INF=MT+1 ,NF TF2=TM I + INF*DELTM TFl=TMI+( INF-1 )*DELTM AA=DSQRT(TFF2-TF1) BB=DSQRT(TFF2-TF2) CALL ROMBERG (FUNCll ,AA,BB,A) A=A*Q( INF) SUMA=SUMA+A ENDDO C SUMB1=0 . ODO SUMB2=0 . ODO DO INF=1 ,NF TF2=TM I + I NF*DELTM TFl=TMI+( INF-1 )*DELTM AA=DSQRT(TFF2-TF1 ) BB=DSQRT(TFF2-TF2) CALL ROMBERG (FUNC2A , AA , BB ,B1 ) CALL ROMBERG ( FUNC2B , AA , BB , B2 ) B1=RV1(INF)*B1 B2=RV1( INF)=kB2 o o

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195 SUMB1 =SUMB1 +B1 SUMB2=SUMB2+B2 ENDDO SUMB=SUM B1 +SUM B2 C SUMC1=0 . ODO SUMC2=0 . ODO DO INF=MT+1 ,NF TF2=TM I + INF*DELTM TFl=TMI+( INF-1 )*DELTM AA=DSQRT(TFF2-TF1 ) BB=DSQRT(TFF2-TF2) CALL ROMBERG ( FUNC3A , A A , BB , Cl ) CALL ROMBERG (FUNC3B,AA,BB,C2) C1=RV2( INF)*C1 C2=RV2( INF)*C2 SUMC1=SUMC1+C1 SUMC2=SUMC2+C2 ENDDO SUMC=SUMCl+SUMC2 C TFUNC1 =TM-DSQRT( ALPHA/PI ) /CK* ( QINTT+SUMA )+l . 0D0/2 . ODO/CP/ $ DSQRT( PI* ALPHA ) * ( CLF*SUMB+CLV*SIJMC ) RETURN END C**************************************************************^ C SUBPROGRAM TFUNCT2 C C**************************************************************^-' FUNCTION TFUNC2(X) IMPLICIT DOUBLE PRECIS I ON (A-H , O-Z) PARAMETER(MAXF=2500 , PI =3 . 141592653589793D0 ) COMMON /T/NF, INF,TMI , TMF , DELTM , TRTM COMMON /C/ALPHA COMMON /Cl /CK , RHO , CP , CLF , CLV , TM , TV COMMON /R1/R1(MAXF) ,RV1 (MAXF) , Q(MAXF) COMMON /R2/R2(MAXF) ,RV2(MAXF) ,MF,MT EXTERNAL FUNC4 , FUNC44 , FUNC5A , FUNC5B , FUNC6A , FUNC6B R2(NF)=X C TFF2=TM I +NF* DELTM TFF 1 =TM I + ( N F1 ) *DELTM IF(NF .EQ. MT+1 ) THEN RV2(NF)=R2(NF)/ (TFF2-TFF1 ) ELSE RV2(NF)=(R2(NF) -R2(NF-1 ) ) / (TFF2-TFF1 ) END IF C RV1(NF)=(R1(NF)-R1(NF-1))/ (TFF2-TFF1 ) C AA=DSQRT(TFF2) BB=DSQRT(TFF2-TRTM) CALL ROMBERG ( FUNC4 , A A , BB , QINTTT) C SUMD=0 . ODO

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196 DO INF=MT+1,NF TF2=TMI+INF*DELTM TFl=TMI+( INF-1 )*DELTM AA=DSQRT(TFF2-TF1 ) BB=DSQRT ( TFF2-TF2 ) CALL ROMBERG (FUNC44,AA,BB,D) D=D*Q( INF) SUMD=SUMD+D ENDDO 1 SUME1=0 . ODO SUME2=0 . ODO DO INF=1 ,NF TF2=TM I + I NF* DELTM TFl=TMI+( INF-1 )*DELTM AA=DSQRT(TFF2-TF1) BB=DSQRT(TFF2-TF2) CALL ROMBERG ( FUNC5A , A A , BB , El ) CALL ROMBERG ( FUNC5B , A A , BB , E2 ) E1=RV1(INF)*E1 E2=RV1 ( INF)*E2 SUME1=SUME1+E1 SUME2=SUME2+E2 ENDDO SUME=SUME1+SUME2 SUMFl=O.ODO SUMF2=0 . ODO DO INF=MT+1 ,NF TF2=TM I + 1 NF* DELTM TF 1 =TM I+( INF1 )* DELTM AA=DSQRT(TFF2-TF1 ) BB=DSQRT(TFF2-TF2) CALL ROMBERG ( FUNC6A , AA , BB , FI ) CALL ROMBERG (FUNC6B,AA,BB,F2) F1=RV2(INF)*F1 F2=RV2(INF)*F2 SUMF1=SUMF1+F1 SUMF2=SUMF2+F2 ENDDO SUMF=SUMF1 +SUMF2 C TFUNC2=TV-DSQRT( ALPHA/PI )/CK*(QINTTT+SUMD)+l . 0D0/2 . ODO/CP/ $ DSQRT( P I * ALPHA ) * ( CLF*SUME+CLV *SUMF ) RETURN END C**************************************************************^ C SUBPROGRAM TFUNC3 C C**************************************************************^ FUNCTION TFUNC3(X) IMPLICIT DOUBLE PRECISION(A-H,0-Z) PARAMETER( MAXF=2500 , PI =3 . 141 592653589793D0 ) COMMON /T/NF , I NF , TM I , TMF , DELTM , TRTM COMMON /C/ ALPHA

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197 COMMON /DIF/CKA ,CKB,DIFF COMMON /Cl /CK , RIIO , CP , CLF , CLV , TM , TV COMMON /R1/R1(MAXF) ,RV1(MAXF) ,Q(MAXF) COMMON /R2/R2(MAXF) ,RV2(MAXF) ,MF ,MT EXTERNAL FUNC7,FUNC8 Q(NF)=X TFF2 =TM I +N F* DELTM TFF1 =TM I + ( NF1 ) * DELTM IF(NF .EQ. MT+1) THEN RV2(NF)=R2(NF)/ (TFF2-TFF1 ) ELSE RV2(NF)=(R2(NF)-R2(NF-1 ) ) / (TFF2-TFF1 ) END IF RV1 (NF)=(R1 (NF)-Rl (NF-1 ) ) / (TFF2-TFF1 ) CALL ROMBERG ( FUNC7 , 0 . ODO , TRTM , Q INTI V ) SUMG=0 . ODO DO INF=MT+1 ,NF TF2=TMI+INF*DELTM TF 1 =TM I + ( I NF1 ) * DELTM IF( INF .EQ. NF)THEN Al=2 . ODO*DSQRT( ALPHA*PI )/R2(NF) A2=R2(NF) /DSQRT(4 . ODO* ALPHA* (TFF2-TFF1 ) ) G=A1*ERFC( A2) ELSE Al=4 . ODO*DSQRT( ALPHA )/R2(NF ) A2=R2(NF)/DSQRT(4 . 0D0*ALPHA*(TFF2-TF1 ) ) A3=R2(NF) /DSQRT(4 . 0D0*ALPHA*(TFF2-TF2) ) CALL ROMBERG (FUNC8,A2, A3, G) G=A1*G ENDIF G=Q( INF)*G SUMG=SUMG+G ENDDO SUMH1=0 . ODO SUMH2=0 . ODO DO INF=1 ,NF TF2=TM I + I NF*DELTM TF 1 =TM I + ( I N F1 ) * DELTM IF( INF .EQ. 1 )THEN AM=RV1 (INF) BN=-R1 ( INF)*TF1/(TF2-TF1 ) ELSE AM=RV1 ( INF) BN=(R1 ( INF-1 )*TF2-R1 ( INF)*TF1 )/(TF2-TFl ) ENDIF IF( INF .EQ. NF )THEN CC= ( R2 ( NF ) -2 . 0D0*AM*TFF2+AM*TF1 -BN ) / $ DSQRT(4.0D0*ALPHA*(TFF2-TF1 )) C01=DEXP( AM* ( R2 (NF) -AM*TFF2-BN) /ALPHA )/2 . ODO HI =C0 1 * ( 1 . ODO-ERF(CC) )

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198 FF=(R2(NF)+2.0D0*AM*TFF2-AM*TF1+BN)/ $ DSQRT(4.0D0*ALPHA*(TFF2-TF1 ) ) C02=DEXP ( AM* ( R2(NF)+AM*TFF2+BN)/ ALPHA )/2.0D0 I F ( (R2(NF)+( AM*TFF2+BN ) ) . LT . 0 . ODO ) THEN H2=C02*(-1 . ODO-ERF(FF) ) ELSE H2=C02*(1 . ODO-ERF(FF) ) ENDIF ELSE C01=DEXP(-AM*(R2(NF)-AM*TFF2-BN)/ALPHA)/DSQRT(PI) CC= (R2(NF)-2 . ODO* AM*TFF2+AM*TF1 -BN ) / $ DSQRT(4.0D0*ALPHA*(TFF2-TF1)) CC1=(R2(NF)-2.0D0*AM*TFF2+AM*TF2-BN)/ $ DSQRT(4 . 0D0*ALPHA*(TFF2-TF2) ) CALL ROMBERG (FUNC8,CC,CC1, HI) H1=C01*H1 C02=DEXP( AM*(R2(NF)+AM*TFF2+BN)/ALPHA)/DSQRT(PI ) FF= ( R2 ( NF) +2 . ODO* AM*TFF2AM*TF 1 +BN ) / $ DSQRT(4. 0D0*ALPHA*(TFF2-TF1 ) ) FF1 = ( R2 ( NF ) +2 . 0D0*AM*TFF2AM*TF2+BN ) / $ DSQRT(4 . 0D0*ALPHA*(TFF2-TF2) ) CALL ROMBERG ( FUNC8 , FF , FF1 , H2 ) H2=C02*H2 ENDIF H1=RV1(INF)*H1 H2=RV1 ( INF)*H2 SUMH1=SUMH1+H1 SUMH2=SUMH2+H2 ENDDO SUMH=SUMH 1 +SUMH2 SUM I 1=0. ODO SUMI2=0 . ODO DO INF=MT+1 , NF TF2=TM I + I NF*DELTM TFl=TMI+( INF-1 )*DELTM IF(NF .EQ. MT+1 ) THEN AM=RV2( INF) BN=-R2( INF)*TF1/(TF2-TF1 ) ELSE AM=RV2( INF) BN=(R2( INF-1 )*TF2-R2( INF)*TF1 )/ (TF2-TF1 ) ENDIF IF( INF .EQ. NF)TIIEN Cl 1=-ERF( -AM*(TFF2-TFF1 ) /DSQRT $ (4 . 0D0*ALPHA*(TFF2-TFF1 ) ) )/2 . ODO FF=(R2(NF)+2 . 0D0*AM*TFF2-AM*TF1+BN)/ $ DSQRT(4 . 0D0*ALPHA*(TFF2-TF1 ) ) C02=DEXP( AM*(R2(NF)+R2(NF) ) /ALPHA )/2 . ODO CI2=C02*( 1 . ODO-ERF(FF) ) ELSE CO 1=DEXP( -AM*(R2(NF)-AM*TFF2-BN)/ALPHA)/DSQRT(PI )

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199 CC= (R2 (NF ) -2 . 0D0*AM*TFF2+AM*TF1 -BN ) / $ DSQRT(4.0D0*ALPHA*(TFF2-TF1)) CCl=(R2(NF)-2. 0D0*AM*TFF2+AM*TF2-BN)/ $ DSQRT(4.0D0*ALPHA*(TFF2-TF2) ) CALL ROMBERG ( FUNC8, CC, CC1 , Cl 1) CIl=COl*CIl C02=DEXP( AM* ( R2(NF)+AM*TFF2+BN ) / ALPHA ) /DSQRT( P I ) FF=(R2(NF)+2 . 0D0*AM*TFF2-AM*TF1+BN)/ $ DSQRT(4 . 0D0*ALPHA*(TFF2-TF1 ) ) FF1=(R2(NF)+2.0D0*AM*TFF2-AM*TF2+BN)/ $ DSQRT(4 . 0D0*ALPHA*(TFF2-TF2) ) CALL ROMBERG (FUNC8,FF,FF1 ,CI2) CI2=C02*CI2 ENDIF CI1=RV2(INF)*CI1 CI2=RV2( INF)*CI2 SUMI1=SUMI1+CH SUM I 2=SUM I 2+C I 2 ENDDO SUM I =SUM 1 1 +SUM 1 2 TFUNC3=GX(TFF2) -R2(NF)/2 . ODO/DSQRT( PI* ALPHA)* $ ( Q I NT I V+SUMG ) +RHO* ( CLF*SUMH+CLV*SUM I -CLV*RV2 ( NF ) ) RETURN END C**************************************************************^ FUNCTION FUNCl(X) IMPLICIT DOUBLE PIlECISION( A-H , O-Z) PARAMETER(MAXF=2500) COMMON /T/NF , INF , TMI , TMF , DELTM , TRTM COMMON /C/ALPHA COMMON /R1/R1(MAXF) ,RV1 (MAXF) , Q(MAXF) TFF=TMI+NF* DELTM IF(X .EQ. O.ODO) THEN FUNC 1=0.0 DO ELSE FUNCl =-2 . 0D0*GX(TFF-X**2)*DEXP( -R1 (NF ) **2/ 4 . ODO/ ALPHA/X/X) END IF RETURN END C**************************************************************^ FUNCTION FUNC4(X) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) PARAMETER( M AXF=2500 ) COMMON /T/NF , I NF , TM I , TMF , DELTM , TRTM COMMON /C/ALPHA COMMON /Rl/Rl (MAXF) ,RV1 (MAXF) ,Q(MAXF) COMMON /R2/R2(MAXF) ,RV2(MAXF) ,MF ,MT TFF=TM I +N F* DELTM IF(X .EQ. O.ODO) THEN FUNC4=0 . ODO ELSE FUNC4=-2 . 0D0*GX(TFF-X**2)*DEXP( -R2(NF)**2/4 . ODO/ ALPHA/X/X) END IF

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o o RETURN END 200 FUNCTION FUNC7(X) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) PARAMETER(MAXF=2500 ) COMMON /T/NF , INF ,TMI ,TMF , DELTM , TRTM COMMON /C/ALPHA COMMON /R1 /R1 (MAXF ) ,RV1 (MAXF) , Q(MAXF) COMMON /R2/R2(MAXF) ,RV2(MAXF) ,MF ,MT TFF=TMI+NF*DELTM FUNC7=GX(X)*DEXP(-R2(NF)**2/4. ODO/ ALPHA/ (TFF-X) ) $ /DSQRT(TFF-X)**3 RETURN END C FUNCTION FUNC12(X) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) PARAMETER( M AXF=2500 ) COMMON /T/NF , I N F , TM I , TM F , DELTM , TRTM COMMON /C/ALPHA COMMON /Rl/Ill(MAXF) ,RV1(MAXF) ,Q(MAXF) TFF=TM I +NF* DELTM FUNC12=-2.0D0*GX(TFF-X**2) RETURN END C FUNCTION FUNC8(X) IMPLICIT DOUBLE PRECISI0N(A-II,0-Z) PARAMETER (MAXF=2500) DATA PI/3. 14159265359D0/ COMMON/C/ ALPHA COMMON /T/NF , I NF , TM I , TMF , DELTM , TRTM COMMON /Rl/Rl (MAXF) ,RV1 (MAXF) ,Q(MAXF) FUNC8=DEXP( -X*X) RETURN END FUNCTION FUNCll(X) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) PARAMETER( M AXF=2500 ) COMMON /T/NF , I NF , TM I , TMF , DELTM , TRTM COMMON /C/ALPHA COMMON /R1 /R1 (MAXF ) ,RV1(MAXF) ,Q(MAXF) TFF=TM I +NF*DELTM IF(X .EQ. O.ODO) THEN FUNC11=0.0D0 ELSE FUNC1 1=-2 . ODO*DEXP( -III (NF)**2/4 . ODO/ALPHA/X/X) END IF RETURN

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o o 201 END FUNCTION FUNC44(X) IMPLICIT DOUBLE PRECISION( A-H , 0-Z) PARAMETER(MAXF=2500 ) COMMON /T/NF, INF,TMI ,TMF,DELTM,TRTM COMMON /C/ALP1IA COMMON /R1 /R1 ( MAXF ) , RV1 ( M AXF ) , Q ( MAXF ) COMMON /R2/R2(MAXF) ,RV2(MAXF) ,MF,MT TFF=TMI+NF*DELTM IF(X .Eq. O.ODO) THEN FUNC44=0 . ODO ELSE FUNC44=-2 . ODO*DEXP( -R2(NF)**2/4.0D0/ ALPHA/X/X) END IF RETURN END C FUNCTION FUNC2A(X) IMPLICIT DOUBLE PRECISI0N(A-II,0-Z) PARAMETER(MAXF=2500) COMMON /T/NF, INF, TMI , TMF , DELTM , TRTM COMMON /C/ALPHA COMMON /R1 /R1 (MAXF) ,RV1 (MAXF) , Q(MAXF) C TFF=TM I +N F* DELTM TF2=TM I + I NF* DELTM TF1 =TM I + ( I NF1 )* DELTM IF( INF .EQ. 1) THEN A=RV1 ( INF) B=-R1 ( INF)*TF1/ (TF2-TF1 ) ELSE A=RV1 ( INF ) B=(R1 ( INF1 )*TF2-R1 ( INF)*TF1 )/(TF2-TFl ) END IF RX=A*(TFF-X**2)+B C IF(X .Eq. O.ODO) THEN FUNC2A=0 . ODO ELSE FUNC2A=-2 . 0D0*DEXP(-(Rl(NF)+RX)**2/4 . ODO/ALPHA/X/X) END IF RETURN END FUNCTION FUNC5A(X) IMPLICIT DOUBLE PRECISION( A-H , 0-Z) PARAMETER(MAXF=2500) COMMON /T/NF , I NF , TM I , TMF , DELTM , TRTM COMMON /C/ALPHA COMMON /R1/R1(MAXF) ,RV1(MAXF) ,Q(MAXF) COMMON /R2/R2(MAXF) ,RV2(MAXF) ,MF,MT

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202 TFF=TMI+NF*DELTM TF2=TM I + I NF*DELTM TFl=TMI+( INF-1 )*DELTM IF( INF .EQ. 1) THEN A=RV1 ( INF) B=-R1 ( INF)*TF1/ (TF2-TF1 ) ELSE A=RV1 ( INF) B=(R1 ( INF1 )*TF2-R1 ( INF)*TF1 )/ (TF2-TF1 ) END IF RX=A*(TFF-X**2)+B IF(X .EQ. 0.0D0) THEN FUNC5A=0 . ODO ELSE FUNC5A=-2 . ODO*DEXP( (R2(NF)+RX)**2/4 . ODO/ ALPHA/X/X) END IF RETURN END FUNCTION FUNC6A(X) IMPLICIT DOUBLE PRECISION( A-H , 0-Z) PARAMETER ( M AXF=2500 ) COMMON /T/NF, INF,TMI , TMF , DELTM , TRTM COMMON /C/ALPHA COMMON /R1/R1(MAXF) ,RV1 (MAXF) , Q(MAXF) COMMON /R2/I12(MAXF) ,RV2(MAXF) ,MF,MT C TFF=TM I +NF*DELTM TF2=TMI+INF*DELTM TFl=TMI+( INF-1 )*DELTM IF( INF .EQ. MT+1 ) THEN A=RV2( INF ) B=-R2( INF)*TF1/(TF2-TF1 ) ELSE A=RV2(INF) B=(R2( INF-1 )*TF2-R2( INF)*TF1 )/ (TF2-TF1 ) END IF RX=A*(TFF-X**2)+B IF(X .EQ. 0 . ODO) THEN FUNC6A=0 . ODO ELSE FUNC6A=-2 . ODO*DEXP( -(R2(NF)+RX)**2/4 . ODO/ ALPHA/X/X) END IF RETURN END FUNCTION FUNC6B(X) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) PARAMETER(MAXF=2500)

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203 COMMON /T/NF , INF ,TMI ,TMF, DELTM ,TRTM COMMON /C/ALPHA COMMON /R1 /R1 (MAXF ) ,RV1 (MAXF) , Q(MAXF) COMMON /R2/R2(MAXF) ,RV2(MAXF) ,MF,MT C TFF=TM I +NF* DELTM TF2=TM I + 1 N F* DELTM TFl=TMI+( INF1 )* DELTM IF( INF .EQ. MT+1 ) THEN A=RV2( INF) B=-R2( INF)*TF1/(TF2-TF1 ) ELSE A=RV2( INF ) B=(R2( INFI )*TF2-R2( INF)*TF1 )/(TF2-TFl ) END IF RX=A*(TFF-X**2)+B C IF(X .EQ. O.ODO) THEN FUNC6B=-2 . ODO ELSE FUNC6B=-2 . ODO*DEXP( (R2 ( NF) -RX ) **2/ 4 . ODO/ ALPHA/X/X ) END IF RETURN END FUNCTION FUNC22A(X) IMPLICIT DOUBLE PRECISION(A-H , 0-Z) PARAMETER(MAXF=2500 ) COMMON /T/NF , I NF , TM I , TMF , DELTM , TRTM COMMON /C/ ALPHA COMMON /R1/R1(MAXF) ,RVI(MAXF) ,Q(MAXF) C TFF=TMI+NF*DELTM TF2=TM I + I NF*DELTM TFl=TMI+( INF-1 )*DELTM IF( INF .EQ. 1) THEN A=RV1 (INF ) B=-R1 ( INF)*TF1/(TF2-TF1 ) ELSE A=RV1 ( INF) B=(R1( INF-1 )*TF2-R1(INF)*TF1)/(TF2-TF1) END IF RX=A*(TFF-X**2)+B C IF(X .EQ. O.ODO) THEN FUNC22A=0 . ODO ELSE FUNC22A=-4 . ODO*DEXP( -RX**2/4 . ODO/ALPHA/X/X ) END IF RETURN END C

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204 C**************************************************************^ FUNCTION FUNC2B(X) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) PARAMETER(MAXF=2500) COMMON /T/NF , INF ,TMI ,TMF , DELTM , TRTM COMMON /C/ALPHA COMMON /R1/R1(MAXF) ,RV1(MAXF) ,Q(MAXF) C TFF=TM I +NF* DELTM TF2=TM I + I NF*DELTM TFl=TMI+( INF-1 )*DELTM IF( INF .EQ. 1) THEN A=RV1 ( INF) B=-R1 ( INF)*TF1/ (TF2-TF1 ) C B=0 . ODO ELSE A=RV1 ( INF ) B=(R1 (INF-1 ) *TF2-Ri ( I NF ) *TF 1 ) / ( TF2-TF1 ) END IF RX=A*(TFF-X**2)+B ' C IF(X .EQ. 0 . ODO) THEN FUNC2B=-2 . ODO ELSE FUNC2B=-2 . 0D0*DEXP(-(R1 (NF)-RX)**2/4.0D0/ ALPHA/X/X) END IF RETURN END C c**************************************************************c FUNCTION FUNC5B(X ) IMPLICIT DOUBLE PIiECISI0N(A-H,0-Z) PARAMETER/ MAXF=2500 ) COMMON /T/NF, INF,TMI ,TMF, DELTM, TRTM COMMON /C/ALPHA COMMON /R1/R1(MAXF) ,RV1(MAXF) ,Q(MAXF) COMMON /R2/R2(MAXF) ,RV2(MAXF) ,MF ,MT C TFF=TMI+NF*DELTM TF2=TMI+INF*DELTM TF 1 =TM I + ( I N F1 ) * DELTM IF( INF .EQ. 1) THEN A=RV1 ( INF) B=-R1 ( INF)*TF1/(TF2-TF1 ) ELSE A=RV1 ( INF ) B=(R1 ( INF1 )*TF2-Ili ( INF)*TF1 )/ (TF2-TF1 ) END IF RX=A*(TFF-X**2)+B C IF(X .EQ. 0 . ODO) THEN IF(R1 (NF) .EQ. R2(NF) ) THEN FUNC5B=-2 . ODO ELSE

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205 FUNC5B=0 . 0D0 ENDIF ELSE FUNC5B=-2 . 0D0*DEXP(-(R2(NF)-RX)**2/4 . ODO/ALPHA/X/X) END IF RETURN END C FUNCTION FUNC3A(X) IMPLICIT DOUBLE PRECISION( A-H , 0-Z) PARAMETER(MAXF=2500 ) COMMON /T/NF, INF,TMI ,TMF,DELTM ,TRTM COMMON /C/ ALPHA COMMON /R1/R1(MAXF) ,RV1(MAXF) ,Q(MAXF) COMMON /R2/R2(MAXF) , RV2 ( M AXF ) , MF , MT C TFF=TM I +NF* DELTM TF2=TM I + I NF*DELTM TFl=TMI+( INF-1 )*DELTM IF( INF .Eq. MT+1 ) THEN A=RV2( INF ) B=-R2( INF)*TF1 / (TF2-TF1 ) ELSE A=RV2( INF ) B=(R2( INF-1 )*TF2-R2( INF)*TF1 )/ (TF2-TF1 ) END IF RX=A*(TFF-X**2)+B C IF(X .Eq. O.ODO) THEN FUNC3A=0 . ODO ELSE FUNC3A=-2 . ODO*DEXP( ( R1 ( NF)+RX )**2/ 4 . ODO/ ALPHA/X/X ) END IF RETURN END C FUNCTION FUNC3B(X) IMPLICIT DOUBLE PRECISION(A-H , 0-Z) PARAMETER(MAXF=2500) COMMON /T/NF, INF,TMI ,TMF , DELTM ,TRTM COMMON /C/ALPHA COMMON /R1 /R1 (MAXF) , RV1 (MAXF) , q(MAXF) COMMON /R2/R2(MAXF ) , RV2(MAXF) ,MF ,MT C TFF=TM I +NF+DELTM TF2=TMI+INF* DELTM TFl=TMI+( INF-1 )*DELTM IF( INF .Eq. MT+1) THEN A=RV2( INF) B=-R2( INF)*TF1 / (TF2-TF1 ) ELSE A=RV2( INF)

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o o o o 206 B=(R2( INF1 )*TF2-R2( INF)*TF1 )/(TF2-TFl ) END IF RX=A*(TFF-X**2)+B IF(X .EQ. 0.0D0) THEN IF(R1 (NF ) .EQ. R2(NF) ) THEN FUNC3B=-2 . ODO ELSE FUNC3B=0 . ODO END IF ELSE FUNC3B=-2.0D0*DEXP(-(R1(NF)-RX)**2/4.0D0/ALPHA/X/X) END IF RETURN END SUBPROGRAM BRENTÂ’S TO FIND THE ROOT OF FUNCTION TFUNCA FUNCTION ZBRENT(TFUNC,X1,X2,T0L) IMPLICIT DOUBLE PRECISI0N(A-II,0-Z) EXTERNAL TFUNC PARAMETER ( ITMAX=100 ,EPS=3 . D-16) A=X1 B=X2 FA=TFUNC(A) FB=TFUNC( B) IF(FB*FA . GT. 0 . DO) PAUSE Â’ROOT MUST BE BRACKETED FOR ZBRENT. FC=FB DO II ITER=1 , ITMAX IF(FB*FC . GT . 0 . DO) THEN C=A FC=FA D=B-A E=D ENDIF IF(DABS(FC) . LT . DABS(FB) ) THEN A=B B=C C=A FA=FB FB=FC FC=FA ENDIF TO LI =2 . DO*EPS*DABS( B)+0 . 5D0*T0L XM=.5D0*(C-B) IF(DABS(XM) . LE.T0L1 .OR. FB . EQ . 0 . DO)THEN ZBRENT=B RETURN ENDIF IF(DABS(E) .GE.T0L1 .AND. DABS(FA) .GT.DABS(FB)) THEN S=FB/FA IF(A.EQ.C) THEN O O

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ooo o o o o o 207 P=2 . D0*XM*S Q=1 .DO-S ELSE Q=FA/FC R=FB/FC P=S*(2.D0*XM*Q*(Q-R)-(B-A)*(R-1 .DO)) Q=(Q-1 .D0)*(R-1 .D0)*(S-1.D0) ENDIF IF(P.GT.O.DO) Q=-q P=DABS(P) IF(2 . D0*P .LT. DMIN1 (3. *XM*Q-DABS(T0L1*Q) , DABS(E*Q) ) ) $ THEN E=D D=P/Q ELSE D=XM E=D ENDIF ELSE D=XM E=D ENDIF A=B FA=FB IF(DABS(D) .GT. T0L1) THEN B=B+D ELSE B=B+DSIGN(T0L1 ,XM) ENDIF FB=TFUNC(B) 11 CONTINUE PAUSE Â’ZBRENT EXCEEDING MAXIMUM ITERATIONS.Â’ Z BRENT =B RETURN END C************************************************************** ROMBERG INTEGRATION PROGRAM REFERRED TO HORNBECK Â’ BOOK PP.154 INPUT--A , B , EPS (INTERVAL AND CRITERION) ************************************************************ SUBROUTINE ROMBERG ( FUNC , A , B , RESULT) IMPLICIT DOUBLE PRECISI0N(A-1I,0-Z) EXTERNAL FUNC PARAMETER(MAX=40 , EPS=0 . 000 IDO) DIMENSION T( MAX, MAX) T(1,1)=(B-A)*(FUNC(A)+FUNC(B))/2.0D0 T(1,2)=T(1 ,1)/2.0D0+(B-A)*FUNC((A+B)/2.0D0)/2.0D0 T(2,l)=(4. ODO*T( 1 ,2)-T(l,l))/3. ODO J=3 SUCCESSIVE APPLICATION OF TRAPEZOIDAL RULE 50 DELX=(B-A)/2.0D0**(J-1) O O O O O

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o o o o o o 208 X=A-DELX N=2**( J-2) SUM=0 . 0D0 DO 100 1=1, N X=X+2 . ODO*DELX SUM=SUM+FUNC ( X ) 100 CONTINUE T(1 , J)=T(1 , J-1)/2.0D0+DELX*SUM EXTRAPOLATION DO 200 L=2 , J K=J+1 -L T(L,K)=(4. 0D0**(L-1 )*T(L-1 ,K+1 )-T(L-l ,K) )/ $ (4.0D0**(L-1)-1 .ODO) 200 CONTINUE CHECK ACCURACY CRITERION IF(T( J , 1 ) .EQ. 0 . ODO) THEN RESULT=T( J , 1 ) GO TO 111 END IF IF(DABS( (T( J , 1 )-T( J-l , 1 ) )/T( J , 1) ) .GE. EPS) THEN J=J+1 IF( J.GT.MAX) THEN PAUSE Â’TOO MANY STEPSÂ’ GO TO 111 ELSE GO TO 50 END IF ELSE RESULT=T( J , 1 ) END IF 111 RETURN END C**************************************************************^ FUNCTION GX(X) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) GX=0 . 5D0 END FUNCTION ERF(X) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) IF(X . LT . 0 . ODO)TIIEN ERF=-GAMMP( .5D0,X**2) ELSE ERF=GAMMP( . 5D0,X**2) END IF RETURN END C

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209 C****************** *********************** ****** ***************0 FUNCTION ERFC(X) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) IF(X . LT . 0 . ODO)THEN ERFC=1 .ODO+GAMMP( . 5D0,X**2) ELSE ERFC=GAMMQ( . 5D0 ,X**2) END IF RETURN END C * * ******* ******************************************** ****** * ** C FUNCTION GAMMP(A,X) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) IF(X.LT.O.ODO .OR. A.LE.O.ODO)PAUSE IF(X.LT.A+1 .ODO)THEN CALL GSER( GAMSER , A , X , GLN ) GAMMP=GAMSER ELSE CALL GCF( G AMMCF , A , X , GLN ) GAMMP=1 .ODO-GAMMCF ENDIF RETURN END U**************************************************************^ FUNCTION GAMMq(A ,X) IMPLICIT DOUBLE PRECISION(A-H , 0-Z) IF(X . LT . 0 . ODO .OR. A.LE.O.ODO)PAUSE IF(X . LT. A+l . ODO)THEN CALL GSER( GAMSER, A, X, GLN) GAMMQ=1 .ODO-GAMSER ELSE CALL GCF(GAMMCF, A ,X ,GLN) GAMMQ=GAMMCF ENDIF RETURN END C**************************************************************^' SUBROUTINE GCF( G AMMCF , A , X , GLN ) IMPLICIT DOUBLE PRECISION( A-Il , 0-Z) PARAMETER ( ITMAX=200 , EPS=3 . D12 ) GLN=GAMMLN( A) G0LD=O . ODO A0=1 .ODO A1=X B0=0 . DO Bl=l .DO FAC=1 .DO DO 11 N=1 , ITMAX AN=DFLOAT(N) ANA=AN-A A0=( A1+A0*ANA)*FAC B0=(B1+B0*ANA)*FAC ANF=AN*FAC A 1 =X*A0+ANF*A1

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210 B1=X*B0+ANF*B1 IF(A1.NE.0.0D0)THEN FAC=1 . 0D0/A1 G=B1*FAC IF(DABS( (G-GOLD)/G) . LT. EPS)GO TO 1 GOLD=G ENDIF 11 CONTINUE PAUSE Â’A TOO LARGE, ITMAX TOO SMALLÂ’ 1 GAMMCF=DEXP( -X+A*DLOG(X) -GLN)*G RETURN END SUBROUTINE GSER( G AMSER , A , X , GLN ) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) PARAMETER ( ITMAX=200 , EPS=3 . D12 ) GLN=GAMMLN(A) IF(X . LE . 0 . ODO)THEN IF(X . LT. 0 . ODO) PAUSE Â’X LT O.ODOÂ’ GAMSER=0 . ODO RETURN ENDIF AP=A SUM=1 .ODO/A DEL=SUM DO 11 N=l, ITMAX AP=AP+1 .ODO DEL=DEL*X/AP SUM=SUM+DEL I F( DABS (DEL) . LT . DABS(SUM)*EPS)GO TO 1 11 CONTINUE PAUSE Â’A TOO LARGE, ITMAX TOO SMALLÂ’ I GAMSER=SUM*DEXP( -X+A*DLOG ( X ) -GLN ) RETURN END FUNCTION GAMMLN(XX) IMPLICIT DOUBLE PRECISION(A-H ,0-Z) DIMENSION C0F(6) DATA COF , STP/76 . 1 80091 729406D0 , -86 . 505320327 1 1 2D0 , $ 24 . 014098222230D0 , -1 . 231739516140D0 , . 120858003D-2 , $ . 536382D-5 , 2 . 50662827465D0/ DATA HALF , ONE , FPF/O . 5D0 , 1 . ODO , 5 . 5D0/ X=XX-ONE TMP=X+FPF TMP= ( X+HALF ) *DLOG (TMP ) -TMP SER=ONE DO 11 J=1 ,6 X=X+ONE SER=SER+COF( J)/X II CONTINUE GAMMLN=TMP+DLOG ( STP*SER) RETURN END

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211 C * * ** * ** * * * * ** * * * ***** * ******** * * ********** * ****************** C C SUBPROGRAM LU DECOMPOSITION SUBROUT I NE LUDCMP ( A , N , NP , I NDX , D ) IMPLICIT DOUBLE PRECISION( A-H , O-Z) PARAMETER (NMAX=100 ,TINY=1 . 0D-20) DIMENSION A(NP,NP) , INDX(N) ,VV(NMAX) D=1 .ODO DO 12 1=1, N AAMAX=0 . ODO DO 11 J=1,N IF (DABS(A( I , J) ) . GT. AAMAX) A AMAX=DABS ( A ( I , J ) ) 11 CONTINUE IF (AAMAX. EQ.O. ODO) PAUSE Â’SINGULAR MATRIX.Â’ VV(I)=1 .ODO/ AAMAX 12 CONTINUE DO 19 J=1 ,N IF (J.GT.l) THEN DO 14 1=1, J-l SUM=A( I , J) IF (I.GT.l)THEN DO 13 K=1 , 1-1 SUM=SUM-A(I ,K)*A(K, J) 13 CONTINUE A( I , J)=SUM END IF 14 CONTINUE ENDIF AAMAX=0 . ODO DO 16 I=J ,N SUM=A(I , J) IF (J.GT.l )THEN DO 15 K=1 , J-l SUM=SUM-A(I ,K)*A(K, J) 15 CONTINUE A ( I , J ) =SUM ENDIF DUM=VV(I)*DABS(SUM) IF (DUM.GE. AAMAX) THEN IMAX=I AAMAX=DUM ENDIF 16 CONTINUE IF ( J .NE. I MAX) THEN DO 17 K=1 ,N DUM=A( I MAX , K) A( IMAX ,K)=A( J , K ) A( J ,K)=DUM 17 CONTINUE D=-D VV( IMAX)=VV( J ) ENDIF INDX( J)=IMAX IF( J . NE . N )THEN

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212 IF(A(J,J).EQ.O.ODO)A(J,J)=TINY DUM=1.0D0/A(J, J) DO 18 I=J+1,N A(I ,J)=A(I,J)*DUM 18 CONTINUE ENDIF 19 CONTINUE IF(A(N,N).EQ.O. ODO)A(N ,N)=TINY RETURN END C SUBPROGRAM BACKSUBSTITUTION C C**************************************************************C SUBROUTINE LUBKSB ( A , N , NP , I NDX , B ) IMPLICIT DOUBLE PRECISION(A-H,0-Z) DIMENSION A(NP,NP),INDX(N),B(N) 11=0. ODO DO 12 1 = 1, N LL=INDX( I ) SUM=B(LL) B(LL)=B( I ) IF ( 1 1 . NE . 0 . ODO)THEN DO 11 .1 = 11,1-1 SUM=SUM-A(I , J)*B(J) 11 CONTINUE ELSE IF (SUM . NE . 0 . ODO) THEN II = I ENDIF B( I )=SUM 12 CONTINUE DO 14 I =N , 1 , 1 SUM=B( I ) IF( I . LT. N)TIIEN DO 13 J = I + 1 , N SUM=SUM-A( I , J)*B( J) 13 CONTINUE ENDIF B(I)=SUM/A(I ,1) 14 CONTINUE RETURN END C**************************************************************^ FUNCTION RAN3( IDUM) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) C PARAMETER (MBIG=4000000 . ,MSEED=1618033. ,MZ=0. ,FAC=2.5E-7) PARAMETER ( MB I G= 1 000000000 , MSEED= 1 6 1 803398 , MZ=0 , FAC= 1 . E-9 ) DIMENSION MA(55) DATA IFF /O/ IF( IDUM . LT . 0 . OR. I FF . EQ . 0)THEN IFF=1 MJ=MSEEDI ABS( IDUM) MJ=MOD(MJ ,MBIG) MA(55)=MJ

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o o 213 MK=1 DO 11 1=1,54 II=MOD(21*I ,55) MA( I I )=MK MK=MJ-MK IF(MK.LT.MZ)MK=MK+MBIG MJ=MA( II) 11 CONTINUE DO 13 K=1 ,4 DO 12 1=1,55 MA( I )=MA( I )-MA( l+MOD( 1+30 ,55) ) IF(MA( I ) . LT. MZ)MA( I )=MA( I )+MBIG 12 CONTINUE 13 CONTINUE INEXT=0 INEXTP=31 IDUM=1 END IF INEXT=INEXT+1 I F ( INEXT. EQ . 56) INEXT=1 I NEXTP= I NEXTP+ 1 IF( INEXTP . EQ . 56) INEXTP= 1 MJ=MA( INEXT)-MA( INEXTP) IF(MJ.LT.MZ)MJ=MJ+MBIG MA( INEXT)=MJ RAN3=MJ*FAC RETURN END SUBROUTINE M PROVE (A , ALUD , N , NP , I NDX , B , X ) IMPLICIT DOUBLE PRECISI0N(A-11 , 0-Z) PARAMETER (NMAX=100) DIMENSION A(NP,NP) , ALUD(NP,NP) , INDX(N) ,B(N),X(N) ,R(NMAX) REALMS SDP DO 12 1 = 1, N SDP=-B( I ) DO 11 .J = 1 , N SDP=SDP+DBLE( A( I , J) )*DBLE(X( J ) ) 11 CONTINUE R( I )=SDP 12 CONTINUE CALL LUBKSB(ALUD,N,NP, INDX,R) DO 13 1 = 1, N X ( I )=X( I )-R( I ) 13 CONTINUE RETURN END SUBPROGRAM INTERNAL POINTS TEMPERATURE COMPUTATION SUBROUTINE INTERNAL(TINTL) IMPLICIT DOUBLE PRECISION(A-H , 0-Z) PARAMETER(MAXF=2500 , PI=3 . 141592653589793D0) o o o o

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214 COMMON /T/NF, INF.TMI , TMF , DELTM , TRTM COMMON /C/ALPHA COMMON /Cl /CK , RHO , CP , CLF , CLV , TM , TV COMMON /Rl/Rl (MAXF) ,RV1 (MAXF) , tj(MAXF) COMMON /R2/R2(MAXF) ,RV2(MAXF) ,MF ,MT COMMON /RINT/RIX EXTERNAL FINTL1 , FINL2A , FINL2B TFF=TMI+NF*DELTM IF(TMI .EQ. 0 . ODO)THEN QINT=0 . ODO ELSE AA=DSQRT(TFF) BB=DSQRT(TFF-TMI ) CALL ROMBERG (FINTL1 ,AA,BB,QINT) END IF QINT1=0 . ODO DO INF=1 , NF TF2=TMI+INF*DELTM TF1 =TM 1+ ( INF1 ) * DELTM AA=DSQRT(TFF-TF1 ) BB=DSQRT(TFF-TF2) CALL ROMBERG (FINTL1 ,AA,BB,A) QINT1=QINT1+A ENDDO SUMB1=0 . ODO SUMB2=0 . ODO DO INF=1,NF TF2=TM I + 1 NF* DELTM TFl=TMI+( INF-1 )*DELTM AA=DSQRT(TFF-TF1 ) BB=DSQRT(TFF-TF2) CALL ROMBERG (FINL2A , AA , BB , B1 ) CALL ROMBERG ( FINL2B, AA , BB , B2) B1=RV1(INF)*BI B2=RVI ( INF)*B2 SUM B 1 = SUM B 1 + B 1 SUMB2=SUMB2+B2 ENDDO SUMB=SUMB1+SUMB2 TINTL=(2.0D0*ALP11A/CK*(QINT+QINT1 )-CLF/CP*SUMB) $ /2.0D0/DSQRT(PI*ALPHA) TIME=TM I +DELTM*NF VRITE(* , *)TIME,RIX ,TINTL WRITE( 77 , 555 )TI ME , RI X , TI NTL WRITE(71,*)RIX WRITE (90 , *)TIME WRITE (72 , *)TINTL 555 FORM AT (2X , F10 . 7 ,5X ,E14 . 7 ,5X ,E14 . 7) RETURN

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o o 215 END C C**************************************************************^ FUNCTION FINTL1 (X) IMPLICIT DOUBLE PRECISION(A-H,0-Z) PARAMETER(MAXF=2500 ) COMMON /T/N F , I N F , TM I , TMF , DELTM , TRTM COMMON /C/ALPHA COMMON /R1/R1(MAXF) ,RV1(MAXF) ,Q(MAXF) COMMON /RINT/RIX C TFF=TMI+NF*DELTM IF(X .EQ. O.ODO) THEN FINTLUO.ODO ELSE FINTLl=-2 . 0D0*GX(TFF-X**2)*DEXP( -RIX**2/4 . ODO/ALPHA/X/X) END IF RETURN END ************************************************************** ^-' FUNCTION FINL2A(X) IMPLICIT DOUBLE PRECISION( A-H , O-Z) PARAMETER(MAXF=2500 ) COMMON /T/NF , I NF , TM I , TMF , DELTM , TRTM COMMON /C/ALPHA COMMON /Ri/Rl (MAXF) ,RV1 (MAXF) ,Q(MAXF) COMMON /RINT/RIX TFF=TM I +NF* DELTM TF2=TM I + 1 NF* DELTM TFI =TM I+( INF1 )* DELTM IF( INF .EQ. 1) THEN A=RV1 (INF) B=-R1 ( INF)*TF1 / (TF2-TF1 ) ELSE A=RV1 ( INF) B=(RI (INF-1 )*TF2-RI ( INF)*TF1 )/ (TF2-TF1 ) END IF RX=A*(TFF-X**2)+B IF(X .EQ. O.ODO) THEN FINL2A=0 . ODO ELSE FINL2A=-2 . ODO*DEXP( ( RIX+RX) **2/4 . ODO/ALPHA/X/X) END IF RETURN END C**************************************************************^ FUNCTION FINL2B(X) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) PARAMETER(MAXF=2500 ) COMMON /T/NF , INF ,TMI , TMF , DELTM , TRTM COMMON /C/ALPHA

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216 COMMON /R1/R1(MAXF) ,RV1(MAXF) ,Q(MAXF) COMMON /RINT/RIX C TFF=TMI+NF*DELTM TF2=TM I + 1 NF * DELTM TFl=TMI+( INF-1 )*DELTM IF( INF .EQ. 1) THEN A=RV1 ( INF ) B=-R1 ( INF)*TF1 / (TF2-TF1 ) ELSE A=RV1( INF) B=(R1 ( INF-1 )*TF2-R1(INF)*TF1)/ (TF2-TF1 ) END IF RX=A*(TFF-X**2)+B C IF(X .Eq. O.ODO) THEN IF(RIX .EQ. Rl(NF)) THEN FINL2B=-2 . ODO ELSE FINL2B=0 . ODO ENDIF ELSE FINL2B=-2 . ODO*DEXP( (RIX-RX)**2/4 . ODO/ALPHA/X/X ) END IF RETURN END C SUBPROGRAM C STORED HEAT COMPUTATION C************************************************************** SUBROUTINE HEATS IMPLICIT DOUBLE PRECISION( A-H , 0-Z) PARAMETER/ MAXF=2500 , PI=3 . 141592653589793D0) COMMON /T/NF, INF,TMI , TMF , DELTM , TRTM COMMON /Cl /CK , RHO , CP , CLF , CLV , TM , TV COMMON /R1 /111 (MAXF) , RVl(MAXF) ,q(MAXF) COMMON /R2/R2(MAXF) ,RV2(MAXF) ,MF ,MT COMMON /HEAT1 /RX(300) ,TIN(300) C TFF=TM I +NF* DELTM C C COMPUTE THE STORED LATENT HEAT IN THE MATERIAL C SLH=RHO*(Rl (NF)-R2(NF) )*CLF+RHO*R2(NF)*(CLV+CLF+CP*TV) C C COMPUTE THE STORED SENSIBLE HEAT IN THE MATERIAL C (SIMPSONÂ’S 1/3 RULE) C IF(NF .LE. MT) THEN SSH1 = (RX(2)-RX( l))/3.0D0*(TIN(l)+4. 0D0*TIN(2) + $ 2 . 0D0>t'TIN(3)+4 . 0D0*TIN(4)+2 . 0D0=t=TIN(5)+4 . 0D0*TIN(6) + $ 2.0D0*TIN(7)+4.0D0*TIN(8)+2.0D0*TIN(9)+4.0D0*TIN(10) $ TIN(ll)) SSH2=(RX( 12)-RX( 1 1 ) )/3 . ODO*(TIN( 11 )+4. ODO*TIN( 12)+ o o o o

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217 $ 2.0D0*TIN(13)+4.0D0*TIN(‘14)+2.0D0*TIN(15)+4.0D0*TIN(16) + $ 2.0D0*TIN(17)+4.0D0*TIN(18)+2.0D0*TIN(19)+4.0D0*TIN(20)+ $ 2.0D0*TIN(21 )+4 . 0D0*TIN(22)+2 . 0D0*TIN(23)+4 . 0D0*TIN(24)+ $ 2.ODO*TIN(25)+4.0DO*TIN(26)+2.ODO*TIN(27)+4.0DO*TIN(28)+ $ 2 . 0D0*TIN(29)+4 . 0D0*TIN(30)+2 . 0D0*TIN(31 )+4 . 0D0*TIN(32)+ $ 2.0D0*TIN(33)+4.0D0*TIN(34)+2.0D0*TIN(35)+4.0D0*TIN(36)+ $ 2.0D0*TIN(37)+4.0D0*TIN(38)+2.0D0*TIN(39)+4.0D0*TIN(40)+ $ 2 . 0D0*TIN(41 )+4 . 0D0*TIN(42)+2 .0D0*TIN(43)+4 . 0D0*TIN(44)+ $ 2.0D0*TIN(45)+4.0D0*TIN(46)+2.0D0*TIN(47)+4.0D0*TIN(48)+ $ 2.0D0*TIN(49)+4.0D0*TIN(50)+2.0D0*TIN(51)+4.0D0*TIN(52)+ $ TIN(53) ) SSH=SSH1+SSH2 SSH=RHO*CP*SSH ELSE SSH 1=0.0 DO SSH2=(RX( 12)-RX('l 1 ) )/3 . ODO*(TIN( 1 1 )+4 . ODO*TIN( 12)+ $ 2 . ODO*TIN( 13)+4 . ODO*TIN( 14)+2 . ODO*TIN( 15)+4 . ODO*TIN( 16)+ $ 2.0D0*TIN(17)+4.0D0*TIN(18)+2.0D0*TIN(19)+4.0D0*TIN(20)+ $ 2.0D0*TIN(21)+4.0D0*TIN(22)+2.0D0*TIN(23)+4.0D0*TIN(24)+ $ 2.0D0*TIN(25)+4.0D0*TIN(26)+2.0D0*TIN(27)+4.0D0*TIN(28)+ $ 2 . 0D0*TIN(29)+4 . 0D0*TIN(30)+2 . 0D0*TIN(31 )+4 . 0D0*TIN(32)+ $ 2 . 0D0*TIN(33)+4 . 0D0*TIN(34)+2 . 0D0*TIN(35)+4 . 0D0*TIN(36)+ $ 2.0D0*TIN(37)+4.0D0*TIN(38)+2.0D0*TIN(39)+4.0D0*TIN(40)+ $ 2.0D0*TIN(41)+4.0D0*TIN(42)+2.0D0*TIN(43)+4.0D0*TIN(44)+ $ 2 .0D0*TIN(45)+4 . 0D0*TIN(46)+2.0D0*TIN(47)+4.0D0*TIN(48)+ $ 2.0D0*TIN(49)+4.0D0*TIN(50)+2.0D0*TIN(51 )+4. 0D0*TIN(52)+ $ 2 . 0D0*TIN(53)+4 . 0D0*TIN(54)+2 . 0D0*TIN(55)+4 . 0D0*TIN(56)+ $ 2 . 0D0*TIN(57 )+4 . 0D0*TIN(58)+2 . 0D0*TIN(59)+4 . 0D0*TIN(60)+ $ TIN(61 ) ) SSII3=(RX(62)-RX(61 ) )/3.0D0*(TIN(61 )+4.0D0*TIN(62)+ $ 2 . 0D0*TIN(63)+4 . 0D0*TIN(64)+2 . 0D0*TIN(65)+4 . 0D0*TIN(66)+ $ 2 . 0D0*TIN(67)+4 . 0D0*TIN(68)+2 . 0D0*TIN(69)+4 . 0D0*TIN(70)+ $ 2.0D0*TIN(71)+4.0D0*TIN(72)+2.0D0*TIN(73)+4.0D0*TIN(74)+ $ 2.0D0*TIN(75)+4.0D0*TIN(76)+2.0D0*TIN(77)+4.0D0*TIN(78)+ $ 2 . 0D0*TIN(79)+4 . 0D0*TIN(80)+2 . 0D0*TIN(81 )+4 . 0D0*TIN(82)+ $ 2 . 0D0*TIN(83)+4 . 0D0*TIN(84)+2 . 0D0*TIN(85)+4 . 0D0*TIN(86)+ $ 2.0D0*TIN(87)+4.0D0*TIN(88)+2.0D0*TIN(89)+4.0D0*TIN(90)+ $ TIN(91 ) ) SSH=SSH 1 +SSH2+SSH3 SSH=RHO*CP*SSH END IF TOTAL=SLH+SSH WRITE( 77 , 200 )TOTAL , SLR , SSH 200 FORMAT(/ IX, ’STORED HEAT=’ ,E'14.7,2X, ’LATENT=’ ,E14.7 ,2X $ , ’ SENSIBLE= ’ , E'14.7) RETURN END C SUBPROGRAM C C INTERNAL POINTS TEMPERATURE COMPUTATION C SUBROUTINE INTERNALl (TINTL) IMPLICIT DOUBLE PRECISI 0N( A-H , 0-Z) PARAMETER( M AXF=2500 , P I =3 . 14 1 592653589793D0 )

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218 COMMON /T/NF, INF,TMI , TMF , DELTM , TRTM COMMON /C/ALPHA COMMON /Cl /CK , RHO , CP , CLF , CLV , TM ,TV COMMON /R1 /Rl (MAXF ) , RV1 (MAXF) , Q(MAXF ) COMMON /R2/R2(MAXF) ,RV2(MAXF) ,MF,MT COMMON /RINT/RIX EXTERNAL FINTL1 , FINL2A , FINL2B , FINTL3 , FINTL4 ,FINTL5A , $ FI NTL5B , F I NTL6B , FI NTL6A TFF=TM I +NF* DELTM AA=DSQRT(TFF) BB=DSQRT(TFF-TRTM) CALL ROMBERG ( FI NTL3 , A A , BB , Q I NTT ) SUMA=0 . ODO DO INF=MT+1 ,NF TF2=TMI+INF*DELTM TFl=TMl+( INF-1 )* DELTM AA=DSQRT(TFF-TF1) BB=DSQRT(TFF-TF2) CALL ROMBERG(FINTL4,AA,BB,A) A=A*Q( INF) SUMA=SUMA+A ENDDO SUMB1=0 . ODO SUMB2=0 . ODO DO INF=1 ,NF TF2=TM I + I NF* DELTM TFI=TMI+( INF-1 )*DELTM AA=DSQRT(TFF-TF 1 ) BB=DSQRT ( TFF-TF2 ) CALL ROMBERG ( FI NTL5A , AA , BB , B1 ) CALL ROMBERG ( FI NTL5B , A A , BB , B2 ) B1=RV1 ( INF)*B1 B2=RV1(INF)*B2 SUM B 1 = SUM B 1 + B 1 SUMB2=SUMB2+B2 ENDDO SUMB=SUMB 1 +SUMB2 SUMC1=0 .ODO SUMC2=0 . ODO DO INF=MT+1 ,NF TF2=TMI+INF*DELTM TF 1 =TM I + ( I NF1 ) *DELTM AA=DSQRT ( TFF -TF 1 ) BB=DSQRT(TFF-TF2) CALL ROMBERG (FINTL6A , AA , BB , Cl ) CALL ROMBERG ( F I NTL6B , AA , BB , C2 ) C1=RV2( INF)*C1 C2=RV2 (INF) *C2 SUMC 1 =SUMC 1 +C 1

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219 SUMC2=SUMC2+C2 ENDDO SUMC=SUMC1+SUMC2 C TINTL=DSQRT (ALPHA/PI ) /CK*(QI NTT+SUMA ) 1 . ODO/2 . ODO/CP/ $ DSQRT( PI *ALPIIA ) * ( CLF*SUMB+CLV*SUMC) TIME=TMI+DELTM*NF WRITE ( * , * )TIME,RIX ,TINTL WRITE (77 ,556)TIME,RIX ,TINTL WRITE(90,*)TIME WRITE(71 , *)RIX WRITE( 72 , * )TINTL+300 . DO 556 F0RMAT(2X ,F10 . 7 ,5X ,E14 . 7 ,5X ,E14 .7) RETURN END c**************************************************************c FUNCTION FINTL3(X) IMPLICIT DOUBLE PRECISION(A-H , 0-Z) PARAMETER(MAXF=2500) COMMON /T/NF , I NF , TM I , TMF , DELTM , TRTM COMMON /C/ ALPHA COMMON /R1 /III (MAXF) , RV1 (MAXF) , Q(MAXF ) COMMON /RINT/RIX C TFF=TM I +NF* DELTM IF(X .EQ. O.ODO) THEN FINTL3=0 . ODO ELSE FINTL3=-2 . 0D0*GX(TFF-X**2)*DEXP( -RIX**2/4 . ODO/ ALPHA/X/X) END IF RETURN END C c**************************************************************c FUNCTION FINTL4(X) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z) PARAMETER (MAXF=2500) COMMON /T/NF, INF, TM I , TMF, DELTM, TRTM COMMON /C/ALPHA COMMON /Rl/Rl (MAXF) ,RV1 (MAXF) ,Q(MAXF) COMMON /RINT/RIX C TFF=TM I +N F* DELTM IF(X .EQ. O.ODO) THEN FINTL4=0 . ODO ELSE FINTL4=-2 . ODO=*DEXP( -RIX**2/4 . ODO/ALPHA/X/X) END IF RETURN END C FUNCTION FINTL5A(X) IMPLICIT DOUBLE PRECISI0N(A-H,0-Z)

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220 PARAMETER MAXF=2500) COMMON /T/NF, INF,TMI , TMF , DELTM , TRTM COMMON /C/ALPHA COMMON /R1/R1(MAXF) ,RV1(MAXF) ,Q(MAXF) COMMON /RINT/RIX C TFF=TM I +NF*DELTM TF2=TMI+INF* DELTM TFl=TMI + ( INF-1 )* DELTM IF( INF .EQ. 1) THEN A=RV1 (INF) B=-R1 ( INF)*TF1 / (TF2-TF1 ) ELSE A=RV1 ( INF ) B=(R1(INF-1)*TF2-R1(INF)*TF1)/(TF2-TF1) END IF RX=A*(TFF-X**2)+B C IF(X .EQ. O.ODO) THEN FINTL5A=0 . ODO ELSE FINTL5A=-2 . ODO*DEXP( (RIX+RX)**2/4 . ODO/ ALPHA/X/X) END IF RETURN END C C* * * * * * * * ********* ********** *************************** ******** G FUNCTION FINTL5B(X) IMPLICIT DOUBLE PIlECISI0N(A-H,0-Z) PARAMETEIl(MAXF=2500) COMMON /T/NF, I NF , TM I , TMF , DELTM , TRTM COMMON /C/ALPHA COMMON /R1/R1(MAXF) ,RV1(MAXF) ,Q(MAXF) COMMON /I12/R2(MAXF) ,RV2(MAXF) ,MF ,MT COMMON /RINT/RIX C TFF=TMI+NF*DELTM TF2=TM I + 1 NF* DELTM TFl=TMI+( INF-1 )*DELTM IF( INF .EQ. 1) THEN A=RV1 (INF) B=-R1 ( INF)*TF1/(TF2-TF1 ) ELSE A=RV1 (INF) B=(R1 (INF1 )*TF2-R1 ( INF)*TF1 )/ (TF2-TF1 ) END IF RX=A*(TFF-X**2)+B C IF(X .EQ. O.ODO) THEN IF(RIX .EQ. Rl(NF)) THEN FINTL5B=-2 . ODO ELSE FINTL5B=0 . ODO ENDIF

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o o ELSE FINTL5B=-2 . ODO*DEXP( (RIX-RX)**2/4 . ODO/ ALPHA/X/X) END IF RETURN END C FUNCTION FINTL6A(X) IMPLICIT DOUBLE PRECISION( A-H , O-Z) PARAMETER(MAXF=2500 ) COMMON /T/NF , I NF , TM I , TMF , DELTM , TRTM COMMON /C/ALPHA COMMON /R1 /R1 (MAXF) ,RV1 (MAXF) , Q(MAXF) COMMON /R2/R2(MAXF) , RV2(MAXF) ,MF,MT COMMON /RINT/RIX TFF=TMI+NF*DELTM TF2=TM I + I NF*DELTM TF1 =TM I + ( I NF1 ) * DELTM IF( INF .EQ. MT+1) THEN A=RV2( INF) B=-R2( INF)*TF1 / (TF2-TF1 ) ELSE A=RV2( INF) B=(R2( INF1 )*TF2-I12( INF)*TF1 )/ (TF2-TF1 ) END IF RX=A*(TFF-X**2)+B IF(X .EQ. 0 . ODO) THEN FINTL6A=0 . ODO ELSE FI NTL6A=-2 . ODO*DEXP( (RIX+RX )**2/4 . ODO/ ALPHA/X/X) END IF RETURN END FUNCTION FINTL6B(X) IMPLICIT DOUBLE PRECISION( A-H ,0-Z) PARAMETER( M AX F =2500 ) COMMON /T/NF, INF, TMI ,TMF, DELTM , TRTM COMMON /C/ALPHA COMMON /Rl/Rl (MAXF) ,RV1 (MAXF) ,Q(MAXF) COMMON /R2/R2(MAXF) ,RV2(MAXF) ,MF,MT COMMON /RINT/RIX I TFF=TM I +N F* DELTM TF2=TM I + 1 N F* DELTM TF1 =TM I + ( I N F 1 ) * DELTM IF( INF .EQ. MT+1) THEN A=RV2(INF) B=-R2( INF)*TF1/ (TF2-TF1 ) ELSE A=RV2( INF)

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222 B=(R2( INF-1 )*TF2-R2( INF)*TF1 )/ (TF2-TF1 ) END IF RX=A*(TFF-X**2)+B IF(X .EQ. O.ODO) THEN IF(RIX .EQ. R2(NF) ) THEN FINTL6B=-2 . ODO ELSE FINTL6B=0 . ODO END IF ELSE FINTL6B=-2.0D0*DEXP(-(RIX-RX)**2/4.0D0/ALPHA/X/X) END IF RETURN END

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BIOGRAPHICAL SKETCH Mehdi Akbari , was born on April 23, 1954, in Hamedan , Iran. He is the first son of six children in the Akbari family. Mehdi attended Hamedan public schools and was graduated from Dr. Shariati High School in May 1973. He entered the Sharif University of Technology, Tehran, Iran, in September 1973 where he received a degree of Bachelor of Science in mechanical engineering in August 1979. In May 1986, after six years working in the Technical Institutions and National Iranian Oil company in Iran, he entered the Graduate School at the University of Florida and earned an Master of Science with a thesis on the conversion of solar energy to electricity. He was admitted to the candidacy for Ph.D in 1991. He is married to Vahideh Lamian and has two daughters with the names of Sara and Mona. 229

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I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. /) Chung K. Hsieh, Chairman Professor of Mechanical F,n or i n ppr i np" I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. Professor of Materials Science and Engineering I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. Gerard G. Emch Professor of Mathematics I certify that I conforms to acceptable adequate, in scope and Doctor of Philosophy. have read this study and that in my opinion it standards of scholarly presentation and is fully quality, as a dissertation for the degree of Roger A./Gater Associates Professor of Mechanical Engineering I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. Associate Professor of Mechanical Engineering

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This dissertation was submitted to the Graduate Faculty of the College of Engineering and to the Graduate School and was accepted as partial fulfillment of the requirements for the degree of Doctor of Phi losophy . May 1993 £ Winfred M. Phillips Dean, College of Engineering Madelyn M. Lockhart Dean, Graduate School