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INTERFACIAL INSTABILITIES DURING THE SOLIDIFICATION OF A PURE MATERIAL FROM ITS MELT By SAURABH AGARWAL 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 2007 O 2007 Saurabh Agarwal To my Mom and my Dad ACKNOWLEDGMENTS First of all, I thank Professor Ranga Narayanan and Professor Lewis Johns for their constant support and advice throughout my graduate studies. They have been my mentors as well as friends. Their enthusiasm and tenacity for work has inspired me to accept every challenge in research. In particular, their philosophy has taught me to constantly "learn to learn." I have truly enjoyed working with them. I am grateful to the members of my PhD committee: Prof. Dinesh Shah, Prof. Jason Butler and Prof. Sergei Pilyugin. I have really enjoyed being a teaching assistant for Prof. Dickinson and Prof. Crisalle. I also thank Prof. Chauhan for his helpful discussions. Many thanks go to my friends and colleagues Dr. Ojzgiir Ozen, Dr. Kerem Uguz and Dr. Weidong Guo for numerous helpful discussions. I have the highest appreciation for my mother, father and brother for their love and support throughout my educational career. Despite the large geographical distances, they have always stood by me. They have encouraged me at every step of my career and given me the courage to overcome every obstacle. I feel very exceptional to have a partner as special as Smita. I thank her for her constant support during the long working hours put in the final writing of this dissertation. I express my gratitude to my cousin brother and sisterinlaw, Shekhar and Vaishali, who took time to care for me in every possible way. I also thank my research group and other fellow graduate students. I owe much to my friends who have always been there when I needed help and friendship. My final acknowledgements go to the University of Florida for an Alumni Fellowship. TABLE OF CONTENTS page ACKNOWLEDGMENTS .............. ...............4..... LIST OF TABLES ............ ...... .__ ...............8.... LIST OF FIGURES .............. ...............9..... AB S TRAC T ............._. .......... ..............._ 1 1.. CHAPTER 1 INTRODUCTION ................. ...............13.......... ...... 2 PRECIPITATION: LINEAR THEORY............... ...............23. Introducti on ................ ........._ ...............23... The Cause of the Instability ............ ..... ._ ...............23.. The Nonlinear Equations .............. ...............26.... Scaled Nonlinear Equations ............... ............_..... .....___. ...........2 Theoretical Approach and the Linearization of Model Equations ................. .........._.._. ...29 The Solutions to the Base and the Perturbation Problems ........._.._. ......___ ................30 The o vs. k2 CUTVCS .............. ...... ...............3 The Neutral Curves: U vs. k 2 at o = 0 .............. ...............37.... Concluding Remarks .............. ...............38.... 3 SOLIDIFICATION: LINEAR THEORY............... ...............41. Introducti on ................ ........._ ...............41... The Cause of the Instability ............ ..... ._ ...............43.. The Nonlinear Equations .............. ...............47.... Scaled Nonlinear Equations.............___.........__ ............___.........4 The Solutions to the Base and the Perturbation Problems ......____ ........_ ........._.....50 The o vs. k 2 CUTVCS .............. ...............53.... The Neutral Curves: U vs. k 2 at o = 0 .............. ...............65.... Concluding Remarks .............. ...............69.... 4 ON THE IMPORTANCE OF BULK TRANSPORT ON INTERFACIAL INSTABI ITY .............. ...............71.... Introducti on .................. ........ ..... ........ ...... ..... .............7 Case 1: Equilibrium Precipitation Problem in Rectangular Geometry ................. ................71 Case 2: Nonequilibrium Precipitation Problem in a Rectangular Geometry ........................78 Concluding Rem arks .................... ........... ....... .. .. ............8 Endnote 1: Equilibrium Precipitation Problem in Cylindrical Geometry .............. ................82 Endnote 2: Equilibrium Solidification Problem in Cylindrical Geometry .............................86 5 WEAK NONLINEAR ANALYSIS INT PRECIPITATION: GEOMETRIC EFFECTS ON ROUGHNE SS............... ...............9 Introducti on .................. ...............92................. Rectangular Cross Section ............... ... .... ........... ...............94...... Circular Cross Section with Axisymmetric Disturbances .............. ..... ............... 10 A Cross Section with an Arbitrary Shape ................ ........ ...............109... Cross Sections on which the Integral Does Not Vanish ................. ......... ................1 14 Cross Sections on which the Integral Vanishes ................. ........... ....... .......... .....11 An Equilateral Triangular and a Regular Hexagonal Cross Section ................. ................11 9 Concluding Remarks ................... ...... ....... .. .............. ... .. .........1 Endnote 1: A Circular Cross Section with the Possibility of Nonsymmetric Di sturb ance s ................ ...............119............... 6 SOLIDIFICATION: WEAK NONLINEAR ANALYSIS .............. .....................2 Introducti on ............... ..... ..... .. .... ........_ ..... ... .. .......... .......12 Case 1: Latent Heat Rej ected Only to the Subcooled Liquid, Both Phases of Finite E xtent ........._..... .. ... ...... .._ ..... .. ..... ...._._ ... .. ..... .. .. ... ............12 Case 2: Latent Heat Rej ected Only to the Subcooled Liquid, Solid Phase of Infinite E xtent .........._.. .... ... ..._._. ... ..... .. ... ...... .. .... ........ .. ......... 5 Case 3: Latent Heat Rej ected to the Frozen Solid as Well as to the Subcooled Liquid, Both Phases of Finite Extent............._ ... .. .......... ... .... ........ ._.. ............6 Case 4: Solid Phase Ignored, Latent Heat Rejected Only through the Subcooled Liquid....182 Case 5: Latent Heat Rej ected to Frozen Solid as Well as to the Subcooled Liquid, Both Phases of Finite Extent, U Dropped from Domain Equations............... ..... .............9 Case 6: Latent Heat Rej ected to the Subcooled Liquid Only, Both Phases of Finite Extent, U Dropped From Domain Equations .............. ...............210.... Concluding Remarks .............. ...............219.... 7 CONCLUSIONS AND RECOMMENDATIONS .............. ...............220.... APPENDIX A SURFACE VARIABLE S ........._.___..... .___ ...............223.... A. 1 The Unit Normal Vector ........._.___......___ ...............223.. A.2 The Surface Speed .............. ...............224.... A.3 The M ean Curvature .............. ...............225.... B THE PERTURBATION EQUATIONS AND THE MAPPINGS ................. ................ ..227 B.1 The Expansion of a Domain Variable and its Derivatives along the Mapping .........227 B.2 The Higher Order Mappings ............_ .......__ ...............231. C EIGENFUNCTIONS OF AN EQUILATERAL TRIANGLE AND A REGULAR HEXAGON SATISFYINTG NEUMANN BOUNDARY CONDITIONS ................... .........232 C.1 Eigenfunctions of an Equilateral Triangle .............. ...............232.... C.2 Eigenfunctions of a Regular Hexagon .............. ...............233.... REFERENCES .............. ...............234.... BIOGRAPHICAL SKETCH ............. ..............237..... LIST OF TABLES Table page 2.1 Thermophysical properties of CuSO4.5H20.......___ ............ ......_.._..........3 3.1 Thermophysical properties of succinonitrile used in calculations............... ..............5 4.1 Thermophysical properties of CuSO4.5H20 ................. ...............74............... 5.1 Some integrals concerning Bessel's functions............... ...............10 6.1 Case 1 Values of Uo and A2 for inputs L = 1, 2 and k2 ............ .....................15 6.2 Case 3 Some derivatives of base temperatures in two phases in solidification ............164 6.3 Case 6 Values of Uo and A2 for inputs L = 1, 2 and for large values of k2 ...............213 6.4 Case 6 Values of Uo and A2 for inputs L = 1, 2 and for small values of k2 .............214 6.5 Case 6 Values of Uo and A2 for inputs L = 1, 2 and k2 ............ .....................21 LIST OF FIGURES Figure page 1.1 A picture argument for the effect of longitudinal diffusion in solidiaication ....................16 1.2 A picture argument for the effect of surface curvature in solidification...........................1 1.3 A picture argument for the effect of transverse diffusion in solidiaication. ................... ....17 2.1 A precipitation front growing into a supersaturated solution .............. ....................2 2.2 A picture argument for the effect of longitudinal diffusion in precipitation .....................24 2.3 A picture argument for the effect of surface curvature in precipitation ............................25 2.4 A picture argument for the effect of transverse diffusion in precipitation ................... .....26 2.5 A cartoon illustrating the nonuniform concentration gradient in the supersaturated solution in precipitation .............. ...............34.... 2.6 A picture argument illustrating the destabilizing effect of surface tension in precipitation .............. ...............34.... 2.7 Typical growth curve in precipitation............... .............3 2.8 Typical neutral curve in precipitation ...........__.......___..... ...........3 3.1 A moving solidiaication front. ..........._ ..... ..__ ...............43.. 3.2 A picture argument for the effect of longitudinal diffusion in solidiaication ....................44 3.3 A picture argument for the effect of surface curvature in solidification...........................4 3.4 A picture argument for the effect of transverse diffusion in solidification. ................... ....46 3.5 a v. k fr te lw Umoel ithS =Ic = cm T = .2 .......................1 3.5 o vs. k 2 for the low U model with S = Icm L = cm Ts = 0.2 ............. .................56 3. avs k frth cmpet odl it S=Ic ,L ~m 02 .................25 3.6 o vs. k 2 for the low Uet model with S = Icm L = 2cm Ts = 0.2 .............. ...............57 3.9 An unusual neutral curve in solidification ....._.._.. .... ... .__. ......._...........6 3.10 Typical neutral curve in solidification ................ ...............66............... 4.1 A precipitation front at rest ................. ...............72............... 4.2 a vs. k" for the eqluilibriumn precipitation problem with Lee, = 4.67x10 meters ..........77 4.3 a vs. k" for the equilibrium precipitation problem with Leen = 10 meters .............. .....78 4.4 a vs. k' for the nonequilibrium precipitation problem with Lee = 4.67x10 meters ...81 5.1 A precipitation front growing into a supersaturated solution .............. ....................9 6.1 A moving solidification front ................. ...............123....._... ... 6.2 Typical neutral curve for solidification when the latent heat is rej ected only through the subcooled liquid ................. ...............130._._.. ...... 6.3 An unusual neutral curve in solidification when the latent heat is rej ected through both phases ................. ...............168............... 6.4 Typical neutral curve in solidification when the latent heat is rej ected through both phases ................ ...............168............... 6.5 Typical neutral curve for solidification when the latent heat is rej ected through both phases; U is dropped from the domain equations............... ...............19 6.6 Typical neutral curve for solidification when the latent heat is rej ected through the subcooled liquid only; U is dropped from the domain equations. .............. ... .............211 6.7 Transition diagram of L versus k2 showing regions corresponding to forward and backward pitchfork .............. ...............215.... 6.8 A cartoon illustrating a forward/supercritical branching ................. ................ ...._216 6.9 A cartoon illustrating a backward/suberitical branching .............. ....................21 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 INTERFACIAL INSTABILITIES DURING THE SOLIDIFICATION OF A PURE MATERIAL FROM ITS MELT By Saurabh Agarwal August 2007 Chair: Ranganathan Narayanan Cochair: Lewis E. Johns Major: Chemical Engineering Solidification of materials is important in the semiconductor and electronics industry. It is ordinarily required that crystals grown industrially have uniform mechanical and electrical properties to ensure better quality and reproducibility of the Einal devices that are produced from them. Thus there is a considerable economic incentive in producing uniform crystals. However, the occurrence of morphological defects at the growing solidiaication interface arising from growth instabilities leads to undesirable electromechanical properties of such crystals. It is the goal of this work to thoroughly understand the physics and causes of morphological instability in solidiaication. Inasmuch as the problem of morphological instability of a solidliquid interface has been considered in the past, many questions remain unanswered. Of central interest is the pattern that evolves when an instability occurs. To begin to answer this question one begins to address the earlier question of what the magnitude of the wavelength of the initial instability might be. My research focuses on this earlier issue. And a set of questions begin to present themselves. They are: What is the reason for the instability at a solidliquid interface? When does the instability set in? How can one explain the dependence of roughness growth rate on its wave number? Why is a maximum growth rate seen in problems like precipitation and solidiaication? What happens after the instability sets in? Does the front speed up or slow down compared to its predicted base value? Why are cellular patterns observed in solidiaication whereas precipitation is incapable of giving anything as interesting? Can one determine the nature of the branching to the nonplanar steady state in the postonset regime? Many important Eindings have been made in the course of answering these questions. The primary discoveries of this work are: First, the shapes of the disturbance growth curves in solidification require the effect of transverse heat diffusion for their explanation. The neutral curves in solidiaication are found to give rise to the possibility of a cellular pattern at the onset of instability whereas for other solidliquid front growth problems independent of thermal transport, such as precipitation from a supersaturated solution or electrodeposition of a metal from an ionic electrolyte, the onset of an instability from a planar state leads to a single crest and trough. Second and more important, a solidliquid front that grows on account of temperature gradients is the only one amongst all solidliquid front growth problems where a critical wavelength, independent of surface energy, can occur. This critical wavelength is of the order of the size of the container in which the growth takes place and as such is the only critical point with which fluid convection can interact. Third, a solidification front always demonstrates a "speedup" upon becoming unstable, which is to say that it always moves ahead of its predicted base value. This is in contrast with other solidliquid growth instability problems like precipitation. Finally the nature of the bifurcation in a solidliquid growth problem where growth occurs on account of thermal gradients may be a forward pitchfork, leading to cellular growth, or a backward pitchfork, leading to dendritic growth. This stands in contrast to other growth problems where the post onset region is ordinarily characterized by dendritic growth. CHAPTER 1 INTRODUCTION This dissertation focuses on investigating the instability of the solidliquid interfaces that occurs during precipitation from a supersaturated solution or during solidiaication from an undercooled melt. In both ways of growing a solid, an erstwhile planar interface can suddenly lose its stability and change its shape as a control parameter is advanced beyond its critical value. An instability occurring in such a fashion is often called a morphological instability [1]. Now the occurrence of morphological instability during solidification has implications in crystal growth as it leads to undesirable electromechanical properties of crystals that are often needed in the semiconductor industry [2]. To tackle this problem requires an understanding of the physics of the instability. Now solidiaication is a typical 'moving front' problem where the liquid phase turns into solid phase by rej ecting the latent heat from the interface [3]. The speed of the solidiaication front depends on how fast the latent heat released there can be conducted through one or both phases. The process of precipitation, on the other hand, occurs when a solid crystal is in contact with its supersaturated solution. In solidification as well as precipitation, the speed and the direction of the front determine its stability, and hence the state beyond which an erstwhile planar front loses its stability. Hence, in both, the solid surface can acquire roughness and understanding the shape of the growth rate curve, where growth rate is plotted against the disturbance wave number, is an important step in understanding the source of the instabilities that arise during the growth. Finding how roughness growth rate, cr, varies with the wave number of the roughness, k has been the obj ect of many experimental studies [4,5]. One of the main goals of the present work is to explain the correspondence between these growth curves and the factors contributing to the stability or instability of the surface, such as diffusion normal to the surface (longitudinal diffusion), diffusion parallel to the surface (transverse or lateral diffusion), surface tension etc. In fact, a greatest growth rate occurs at a Einite wavelength in some, but not all moving front problems, and the appearance of such a maximum growth rate in solidiaication is an example of an observation that needs to be explained. As discussed by Coriell and coworkers [6], the wavelength having the largest initial growth rate might be expected to be predominant during the evolution of the instability, and it is therefore important from the point of view of pattern selection. Therefore, while explaining the shapes of the growth curves in precipitation and solidification, we emphasize the reason for the existence of a maximum growth rate. We will see later that this has to do with transverse diffusion of heat in the liquid at the interface. Now the cr vs. k2 CUTVCS at various values of a control variable tell us what the critical wavelengths are, these being the wavelengths at zero growth rate. Thus they predict the shape of the neutral curve, a curve where the control variable is given as a function of k2 This curve is useful in understanding what might be seen in a series of experiments as the control variable slowly increases to, and just beyond its critical value. Of central interest is the pattern that evolves during the early stages of instability. We will find that the neutral curve in precipitation is monotonically increasing, hence only one crest and trough will be seen in precipitation. In solidiaication, when heat is withdrawn only through the liquid, the neutral curves are as simple as those in precipitation. Upon withdrawing heat through the solid as well, neutral curves like those seen in the RayleighBenard convection problem [7] can be found. And by adjusting the depths of the phases, we can go beyond this and get something new, viz., critical points that are due to transverse diffusion in the solid phase, and not due to surface tension. Hence, the central result of our linear theory is this: In the solidification of a pure material it is possible to obtain zero, one, two, or three critical points, and a critical wavelength independent of surface energy can occur. This critical wavelength is of the order of the size of the container in which the growth takes place. The reader is encouraged to refer to Cross and Hohenberg [8] who review pattern formation at great length, never finding neutral curves like the new ones discovered by us. In fact, it turns out that getting three critical points is a feature typical of pure solidification when the latent heat is rej ected through both phases, when the control variable is made explicit, and when the depths of the two phases are taken to be finite. The Mullins and Sekerka [9] model for the solidification of a dilute binary alloy and the Wollkind and Notestine [10] model for pure solidification fail to give anything as interesting; they can give at most two critical points. Now to understand why one might expect three critical points in solidification, let us see why a solidification front might be unstable and how factors such as surface curvature and transverse diffusion affect the stability. The usual explanation for the instability of a growing interface from an undercooled melt (cf., Langer [1 l]) is that a displacement of the surface into the melt in the form of a crest increases the temperature gradient at its tip, which reinforces the growth of the crest. Figure 1.1 depicts a growing front in an undercooled melt and one sees that longitudinal diffusion in the liquid phase is strengthened at a crest thereby destabilizing the surface. It is also clear from Figure 1.1 that longitudinal diffusion in the solid phase acts to stabilize the interface. The effect of longitudinal diffusion does not depend on the wavelength of the disturbance, i.e., it is a k independent effect. Now the melt temperature at a solidliquid interface depends on its morphology. In other words, on account of the curvature of the nonplanar interface itself, the melt temperature at a curved interface is different from the normal melt temperature at a planar interface. This is explained by Delves [12]. Figure 1.2 shows this effect where surface curvature lowers the melt temperature at a crest while increasing it at a trough. Hence, surface tension weakens the instability caused by longitudinal diffusion in the liquid, while strengthening its stabilizing effect in the solid. The stabilizing effect of surface tension is k dependent, the stabilization being stronger for shorter wavelengths. Solid Liquid Temp. gradient ~~j( stregthened Temp. gradient weakened Moving Front Temp. gradient weakened Temp. gradient strengthened Figure 1.1 A picture argument for the effect of longitudinal diffusion in solidification Solid Liquid Melting point inraeMelting points.=` Moving Front Figure 1.2 A picture argument for the effect of surface curvature in solidification Finally, let us look at the effect of transverse diffusion in the two phases. Setting surface tension aside, the effect of the displacement is as shown in Figure 1.3. On the melt side, the displacement of the surface strengthens the temperature gradient at a crest, weakening it at a trough; this induces transverse temperature gradients, carrying heat away from crests toward troughs. This tends to build crests up. Transverse diffusion is therefore destabilizing on the melt side of the surface. A similar argument made in the solid phase tells us that transverse diffusion is stabilizing in the solid phase. In both phases, the effect of transverse diffusion, whether stabilizing or destabilizing, is k dependent. Solid Liquid higher Wavelength  higher Figure 1.3 A picture argument for the effect of transverse diffusion in solidification This is enough to speculate the occurrence of three critical points in solidification. Here is what is required: The longitudinal diffusion in the liquid should lead to positive values of o at k2 = 0, then the transverse diffusion in the solid should drive 0 negative, providing the lowk2 critical point. Thereafter the transverse diffusion in the liquid should drive the o positive for further increase in k resulting in a second critical point. Finally, the stabilizing effect of surface tension should drive o down to zero for large enough k providing a third critical point. From a mathematical point of view, we first write the governing domain equations along with appropriate interface conditions and farfield conditions. These model equations are non linear as the position of the interface depends upon the temperature gradient there, and the gradient, in turn, depends upon the position and morphology of the interface. This nonlinearity is the reason for the interface to depart from its flat base state, trying to become unstable. We write the model equations in terms of an observer moving at the speed of the front. Such an observer can find a steady planar solution to our nonlinear equations. To obtain the cr vs. k2 CUTVCS in solidification, we introduce a perturbation of a steady base solution, as did Mullins and Sekerka [9], whose work grew out of the earlier work of Rutter et al [13] and Tiller et al [14]. More detail can be found in the book by Davis [15] who presents the theory of front instabilities in the case of binary alloys. Many of the models [91 1,16,17] proposed for pure solidification are based on local equilibrium at the interface and isotropic interfacial properties. Convection induced by density differences is neglected and steady thermal and diffusion fields are assumed [9, 11] in the interior of the solid and liquid phases. These assumptions are made by us as well. To maintain our focus on the effects that longitudinal and transverse diffusion have on the growth rate of surface roughness, most other effects, however important they may be, are omitted. These effects have been dealt elsewhere. For example, the effect of convective flow on morphological instability was studied by Coriell and Sekerka [18]. The effect of solutal convection has been studied by Hurle et al [19], while Favier and Rouzaud [20] have given a detailed account of the effect of a deformable boundary layer. The effect of Soret diffusion has been explained by Hurle [21]. The effect of departure from local equilibrium has been considered by Coriell and Sekerka [22]. In a different paper, Coriell and Sekerka [23] have considered the effect of anisotropic surface tension and interface kinetics. Wollkind and coworkers [10, 24] have considered surface tension as a function of temperature. Now we may have neglected the effect of convection and the anisotropy of interfacial properties. But we do take into account the finite thickness of the solid and liquid phases. They are not taken to be of infinite extent as they are by Wollkind et al [10]. This gives us a way of altering the relative strengths of transverse diffusion in the two phases and hence of altering the shapes of the growth curves, and this, in turn, gives us a way to obtain three critical points in pure solidification. Up till now our emphasis has been on the linear stability analysis in investigating the interfacial instability during precipitation and solidification. It turns out that we discover two more important results in the post onset regime after the instability sets in. These results emanate from a weakly nonlinear calculation which is done by going to second and third orders in the amount by which the control parameter is increased. The first of these results has to do with the correction to the front speed on account of morphological instability. We find that once the instability sets in, a solidification front always speeds up compared to what its base value would have been, i.e., it moves ahead of its predicted base value. This is in contrast with a precipitation front which demonstrates an unconditional "slowdown" upon becoming unstable. The second important result of the nonlinear calculations has to do with the nature of the branching to the new nonplanar solutions once the instability sets in. The book by Strogatz [25] gives an excellent description of bifurcations in several applied problems. Bifurcations are important scientifically as they provide essential information about the kind of transition from one steady state to another as some control parameter is varied. We discover that in solidification, the nature of the bifurcation depends on the dimensions of the container in which the growth takes place. The branching may be a forward pitchfork, leading to cellular growth, or a backward pitchfork, leading to a dendritic growth. This is in contrast to the precipitation problem, where the onset of instability is ordinarily characterized by a dendritic growth on account of a backward pitchfork. However, precipitation is also capable of demonstrating a transcritical bifurcation. In fact, we discover that the nature of bifurcation in precipitation depends on the cross section of the growth container. We deduce a condition to decide whether a container of a given cross sectional shape will demonstrate a backward pitchfork or a transcritical bifurcation. Finally, before moving on to explain the organization of this dissertation, we would like to emphasize that a thorough understanding of theoretical results mentioned above is extremely helpful in designing good experiments. As explained in the book by Montgomery [26], a successful design of experiments requires a good theoretical knowledge of all important factors. And it is our belief that the proposed theory should lead to a more rational design and interpretation of solidification experiments. Organization of the Research: The remainder of this dissertation is dedicated to theoretical investigation of interfacial instability in precipitation and solidification via mathematical models and to explain the results using simple physical arguments. As mentioned, our primary goal is to understand the underlying causes and implications of such instabilities. Chapter 2 delineates the physics of instability in precipitation. Being a onephase problem, precipitation offers a simple way of understanding the concept of longitudinal diffusion, transverse diffusion, and surface tension etc. We also learn how the various factors compete with each other to result in an instability. The governing equations along with the relevant interface and boundary conditions are presented. This chapter also explains the theoretical concept of a linear stability analysis which is used to predict whether or not a planar interface can remain planar in the face of small disturbances. This chapter forms the framework for understanding the physics of instability in the more complicated problem of solidiaication. Chapter 3 focuses on the interfacial instability during solidiaication. The first half of this chapter is devoted to understanding the physics of the instability and in deriving the formula for the growth rate of a disturbance. This is followed by an analysis that predicts the wavelength of the cellular pattern observed at the onset of instability. We show how one can obtain interesting growth curves having up to three critical points. Chapter 4 addresses the question whether the domain transport dynamics is important compared to the transport dynamics at a solidliquid interface. A simple precipitation model is studied to investigate whether the time derivatives in the domain equations are important relative to the time derivative of the interface position introduced via interfacial mass balance. Chapter 5 presents a weak nonlinear calculation in the problem of precipitation, where the goal is to examine what happens beyond the onset of the instability. We first work out the problem in a rectangular geometry and then in a circular geometry. In each case, we first Eind the critical value of the control parameter using the linear stability analysis. This is followed by a weak nonlinear calculation where we look for steady solutions just beyond this critical point by going to second and third orders in the amount by which the control parameter is increased. Our main goal is to Eind out the nature of the bifurcation to the new nonplanar solution. Through our calculations, we also wish to discover whether the front speeds up or slow down compared to its base value once the instability sets in. Toward the end of the chapter, we deduce a condition that determines whether a cross section of an arbitrary shape will lead to a pitchfork or a transcritical bifurcation. Chapter 6 deals with the weak nonlinear calculation in the problem of solidification. The goal is to understand the early stages of the growth of surface roughness in solidification. Again, as in Chapter 5, we try to answer two key questions: Does a solidification front speed up or slow down compared to its predicted base value once the instability sets in? Is the branching to the new nonplanar solution a forward or a backward pitchfork? Chapter 7 presents a general conclusion along with a scope for future studies. CHAPTER 2 PRECIPITATION: LINEAR THEORY Introduction In this chapter, our goal is to investigate the causes of instability during precipitation from a supersaturated solution and to quantify the growth and neutral curves by using a linear perturbation theory. First, picture arguments are given to establish the role of longitudinal diffusion, transverse diffusion, and surface tension with regard to the observed instability. To quantify our arguments, we then give the nonlinear model equations which are written in a moving frame. These equations are then scaled and linearized around a base state. This base state solution is slightly perturbed to obtain the formula for o,the growth constant. The playoff between the various terms in this formula is then explained in terms of the effect of diffusion (in two directions), surface tension, and surface speed. Based on this, the shapes of the growth curve and the neutral curve are then explained. The chapter is concluded by predicting the number of crests observed in a precipitation experiment when the supersaturation in the solution is slowly increased to its critical value. It is found that at most one crest and one trough will be seen at the onset of instability. We now move on to discuss the physics of the instability observed when precipitation occurs from a supersaturated solution. The Cause of the Instability Consider the schematic shown in Figure 2.1 which shows a moving precipitation front. Let a solid of density cs lie to the left of a supersaturated solution. The two phases are divided by a plane interface that advances to the right as the reservoir at z = L supplies the solute at a fixed concentration c,, which is higher than c, the saturation concentration at a planar interface. reservoir maintained at Solid /I Solution a fixed concentration Moving front cSAT: Solute concentration in z c cEQZxcs at a plane surface z = Z (~t) n cEQ: Solute concentration in the liquid in equilibrium with cs at a curved surface c =cSAT z=L z = Z =0O u = nI. u surface velocity 2H = "32,Surface curvature Figure 2.1 A precipitation front growing into a supersaturated solution Soli Liqid Concentration source Concentration gradient iCrest strengthened Trough wI Moving Front Figure 2.2 A picture argument for the effect of longitudinal diffusion in precipitation Let c, be the control variable. Increasing it increases the speed of the front. At some value of c, the plane front loses its stability and becomes nonplanar. The explanation for the instability can be understood from Figure 2.2. Consider a perturbation of the planar front to the right, a crest where the local speed runs above the speed of the planar front. To sustain this increased speed, the crest must be supplied more solute from the reservoir, which is reinforced by the increased concentration gradient due to the perturbation. Similarly, a displacement in the form of a slower moving trough is reinforced by a decreased concentration gradient at its tip. Longitudinal diffusion in the liquid is therefore destabilizing; it acts more strongly the faster the front is moving and the steeper the solute gradient feeding the growth. But it does not depend on the wavelength of the disturbance, i.e., it is k independent. Operating against this is the effect of surface curvature which plays a stabilizing role. This is shown in Figure 2.3. At a crest, cEQ the equilibrium concentration, is higher than c,, [3]. By reducing the local concentration gradient at a crest, surface tension weakens the instability caused by longitudinal diffusion. For short wavelengths, the stabilization is strong; for large wavelengths, it is weak; hence surface tension is a k dependent stabilizing effect. Soli Liqid ,Concentration source /oi Eiquiliru oc Moving Front Figure 2.3 A picture argument for the effect of surface curvature in precipitation These two effects would then predict instability at low values of k, and stability for high values of k But by themselves they cannot explain an increase in instability as k increases from zero. There is, however, another effect, and it has to do with transverse diffusion in the liquid. Setting surface tension aside, the displacement of the surface strengthens the solute gradient at a crest, weakening it at a trough (cf., Figure 2.4). This induces transverse solute gradients, carrying solute away from troughs toward crests. This is destabilizing. It tends to build crests up and the effect is stronger the higher the k, i.e., the closer together the crests and troughs. Also, this effect is stronger the faster the front is moving. The effect of surface tension on transverse diffusion is again stabilizing. Its effect on the surface concentration works against transverse diffusion just as it works against longitudinal diffusion. By reducing the local concentration gradient at a crest while increasing it at a trough (cf., Figure 2.3), surface tension weakens the solute gradients in the transverse direction, hence weakening the instability caused by transverse diffusion. Soli Liqid Concentration source concentration gradient J c~,strengthened Moving Front Figure 2.4 A picture argument for the effect of transverse diffusion in precipitation Now in precipitation the solid phase plays no role. Consequently, the effects that will concern us are three in number: the destabilizing effect of longitudinal and transverse diffusion in the liquid and the stabilizing effect of surface tension which offsets each of these. To make these qualitative ideas quantitative, we turn to a model. The Nonlinear Equations The sketch given in Figure 2.1 fixes the basic ideas and presents some formulas. The solid phase is passive and the solute concentration in the solid is cs everywhere and for all time. Of interest are the equation of solute diffusion in the liquid phase, and the interface equations for the local motion of the front. The liquid phase depth is maintained constant and the problem is taken to be twodimensional, x and z being parallel and perpendicular, respectively, to the base surface. The equations will be written in a frame moving with a constant velocity U = Uk, where U is the base state speed of the precipitation front as seen by a laboratory observer. Then, in the liquid phase, the equation for the solute concentration is given by DVe= Oc + U c (2.1) dt 8z for Z < z < L and 0 < x < W, where L is the depth of the liquid phase and W is the width of the precipitation cell; D denotes the diffusivity of the solute in the liquid phase. The solute concentration must meet two requirements along the solidsolution interface, viz., along z = Z(x, t) First, phase equilibrium requires, after accounting for the correction due to surface curvature c= cm _at[2H] (2.2) I c where ~at denotes "A and where y denotes the surface tension. Second, the solute RT cs surplus or deficit induced by the motion of the interface must be made good by solute diffusion into or out of the solution, i.e., [csc U.n+ u = Dn.Vc (2.3) where u denotes the local surface speed of the perturbed front. The far field condition is given by c(z = L) = cL (2.4) Finally we impose side wall conditions such that both Sc/8x and 8Z/8x vanish at x = 0 and at x = W The formulas for the surface normal, n the surface speed, u, and the mean curvature 2H are given in Appendix A. Scaled Nonlinear Equations The nonlinear equations will now be presented in dimensionless form. All lengths are scaled by L and all speeds are scaled by D/ L Time is scaled by L2 / D Concentration is measured from a reference point corresponding to cSAT, and is scaled by cs cSAT, i.e., scaled SAT All variables used hereon are scaled variables. cs cSAT Then, in the liquid phase, the equation for the scaled solute concentration is given by icV'c + U~ (2.5) dt 8z for Z < z <1 and 0 < x < W The equations along the interface, viz., along z = Z(x, t) now become c = l[2H](2.6) and n.Vc= [1c] Ukrnnuu (2.7) where 11, a dimensionless parameter introduced by our scaling, denotes 7 SAT L RT cs cs cSAT The far field condition is given by c(z = 1) =cL (2.8) dc 8Z Side wall conditions are Neumnann, i.e., (x = 0, W) = (x = 0, W) = 0 . Dx Dx Notice that these model equations are nonlinear as the position of the interface depends upon the concentration gradient there, and this gradient, in turn, depends upon the position and morphology of the interface. This is clearly seen in Eq. (2.6) which couples the interface concentration c to the interface position Z(x,t) via the surface curvature 2H This coupling leads to nonlinearity, which forms the heart of the problem. It is the reason for the interface to depart from its flat base state, trying to become unstable. Now to investigate the instability during precipitation, we will study the effect of small disturbances on a planar front which is moving with a constant speed U. To do so, we now turn to the process of linearization of the equations. Theoretical Approach and the Linearization of Model Equations A system in steady state is said to be stable if all disturbances imposed on it decay with time; it is called unstable if any disturbance grows. Such an instability arises when a steady state system is driven away from its steady state upon imposing small perturbations. Very often, an instability arises in growth problems when a control variable exceeds its critical value. For instance, in the problem of precipitation, the supersaturation in the liquid phase may be taken to be the control variable, which, when exceeded beyond a critical value, will lead to instability. To study instability in precipitation, our plan is to evaluate the time dependence of the amplitude of a sinusoidal perturbation of infinitesimal initial amplitude introduced on the planar interface. For this purpose, it is sufficient to analyze a linearized model where the linearization is done over a base state. The idea behind linearizing the equations is this: Once all equations are linearized, the evolution of a sinusoidal perturbation is simply a superposition of the evolution of its Fourier components. The interface is unstable if any Fourier component of the sinusoidal wave grows; it is stable if all decay with time. Now a base state can often be guessed for most problems; for precipitation, it is a planar front which is moving with a constant speed. Denoting the base state concentration by co, the domain concentration c can be expanded as c = co +e c,+zc (2.9) where e denotes the amplitude of the perturbation; z, represents the first order mapping from the current domain to the reference domain. The first order mapping at the interface is denoted Z,, and it needs to be determined as a part of the solution. Because the current domain is unknown (it is a part of the solution), it becomes imperative that we solve the problem on a known reference domain, i.e., one that has all planar boundaries. The meaning of the mappings z, and Z,, and of the correction c, is explained in Appendix B. 1. The first order variables can be further expanded using a normal mode expansion [9] as c1 = c1 cos(kcx)eet and Z, = Z, cos(kx)e"t (2.10) where k is the wave number of the disturbance and o denotes the growth (or decay) constant, also known as the inverse time constant. It is an output variable, and a positive value of o for a given k means that the disturbance grows, leading to instability. The Solutions to the Base and the Perturbation Problems The base state is one in which the precipitation front is a plane at rest in the moving frame. The base state variables are denoted by the subscript 0 and they satisfy oC + U oe = 0 (2.11) & 2 dz in the liquid phase, O < z <1 The equations at the planar interface, z = 0 are given by co = 0 (2.12) and dc a= [1cjU (2.13) dz The far field condition is co (Z = 1)= cL (2.14) Then a solution to equations (2. 112. 14) is given by c,(z)=1e (2.15) where U is given by Ui= In 1cL] (2. 16) Besides the physical properties, the input variables in the base state are L and cL, and the output is the front speed, U To determine if this base solution is stable, we impose a small displacement on the surface at fixed values of cL and U, and determine its growth rate, o, as a function of its wave number k, whence k is an input to the perturbation problem. We solve the perturbation problem on the base domain by expanding the surface shape and the solute concentration at the surface in the form Z = Z, + EZ, = Z, + E Z, cos(kCx)e"t (2.17) and c = co +E c, +Zd, = co +E c,+Z, coeke (2.18) dz dz where the side wall conditions determine the admissible values of k. The subscript 1 denotes a perturbation variable. Then using the expansions given above, the perturbation equations for c, and Z, are given by Oc k2 c,]+U de (2.19) dz2 dz in the solution, O < z <1 The equations that need to be satisfied across the interface, z = 0, are c, +~ dZ = Ak~; Z, (2.20) and 0Z, = + Uc, (2.21) The farfield equation is c,(z = 1) = 0 (2.22) A solution to the domain equation satisfying Eqs. (2.20) and (2.22) is given by dc Ck z 0 1 = em'" where 2m, = U+ U +4 ki +oj .Finally, using the mass balance across the interface, Eq. (2.21), there obtains 0 A +A U (2.24) dz dz2 I I III IV where N= ,,m ) Eq. (2.24) gives o implicitly in terms of k2 and the derivatives of the base concentration at the reference surface, viz., at z = 0. A graph of o vs. k2 can be drawn from this equation, but the equation is not explicit in o Now o appears in the problem in two ways. It appears in the domain equation where it has to do with solute depletion in the liquid as the front advances. It also appears in the surface solute balance where it has to do with the local speed of a crest or a trough. We will drop o on the domain (cf., Chapter 4) on the grounds that the solute depleted by the growth of a slowly moving front is easily supplied by diffusion. Then the dominant growth rate is given by Eq. (2.24), where N remains as defined, but m, are now given by 2m = U+~ U 4 and where m is always positive and m is always negative. Consequently N is always negative, and terms I and IV in Eq. (2.24) are negative, hence stabilizing, while terms II and III are positive, hence destabilizing. Other than the wave number of the displacement, the growth rate depends only on the base dc, d2c solute gradient, ''(z = 0) = Ui, and its correction (z = 0) = U The physical factors that dz dz2 each term takes into account are these: Term II, the essential destabilizing term, combines the two effects of the longitudinal solute gradient. The primary effect is that a displacement strengthens the gradient at a crest, and the secondary effect is that a strengthened gradient at a crest and a weakened gradient at a trough induce transverse solute diffusion from a trough to a crest; both destabilizing. Term IV which is stabilizing, is in fact a correction to term II, weakening its destabilizing effect. It corrects for the fact that the solute gradient is not uniform, being stronger at the base surface than in the nearby solution (cf., Figure 2.5). Hence a crest proj ected into the supersaturated solution is fed by a less steep gradient than dc'(z = 0) . Term I, the essential stabilizing term, combines two of the three effects of surface tension, cutting into the destabilizing effect of both longitudinal diffusion and transverse diffusion. Term III, the remaining effect of surface tension is destabilizing. It accounts for the fact that at a crest, the amount of solute that must be supplied to move the crest is lessened as the solute concentration rises at the surface (cf., Figure 2.6). Such a term will not be seen in solidification (cf., Chapter 3) unless the latent heat is perturbed. Solid Liquid SCrest being fed by a less steep gradient Concentration source I Crest Moving Front Figure 2.5 A cartoon illustrating the nonuniform concentration gradient in the supersaturated solution in precipitation Soli Liqid ~ $ suplieOld difference to be New difference Sto be supplied cs Moving Front Figure 2.6 A picture argument illustrating the destabilizing effect of surface tension in precipitation A significant algebraic simplification can be obtained by observing that cs, the density of the solid, is usually very high (cf., Table 2.1) compared to [c, csa i, the supersaturation. The scaled value of c, therefore is ordinarily small ( c, = 0.1i), and consequently U must also be small (U 0.1i). The base state concentration profile in the liquid phase must then be nearly linear and a low U result may be obtained by dropping U from the domain equation. The base speed and the o formula are then given by U L n 0 A U k + ~ak U (2.25) and at the small values of ~a ordinarily encountered, viz., a =109, the following formulas are useful in sketching a graph of o vs. k2 0(k" = 0) = U1 (2.26) k )=U (2.27) dk2I 3o= 0 k" large but not too large) = kU (2.28) and 0 k' very large) = ak3 (2.29)) Table 2.1 Thermophysical properties of CuSO4.5H20 Saturation concentration, c, 243 (kg/m3) Density of solid, cs 2284 (kg/m3) Diffusivity, D 109 (m2/S) Gas constant, R 8.314 (J/mol K) Surface Tension, y 8.94 x 103 (N/m) The o vs. k2 Curves A typical growth curve is shown in Figure 2.7. It starts positive. Then o increases as k2 increases and the curve passes through a maximum before decreasing and passing through zero. The value of o at k2 = 0 is positive, and this is due entirely to longitudinal diffusion (cf., Figure 2.2). In fact, one might have predicted intuitively that the value of o at k2 = 0 could never have been negative for the precipitation model presented here. Now at k2 = 0, only longitudinal diffusion is important. And since precipitation has only one active phase, and longitudinal diffusion is destabilizing there, we do not have any reason to expect stability at k2 = Notice that the analog will not hold in the solidification problem (cf., Chapter 3) where the latent heat can be rej ected through the solid as well as via the subcooled liquid. The increase in o as k2 increases is due to the fact that the longitudinal solute gradient induces transverse diffusion (cf., Figure 2.4) from a trough to the adj acent crest. This is k  dependent, and o increases linearly in k. Eventually surface tension comes into play, driving o downward as k' The separation of the effects of transverse diffusion and surface tension allows us to see first an increase in o followed by a decrease. This is due to the small value of ~a, which delays the effect of surface tension. Hence a range of k2 exists upward from zero where o increases due to transverse diffusion. And this explains the existence of a greatest growth rate, max,, at a value of k denoted by kisx 0 k2 Figure 2.7 Typical growth curve in precipitation Now the effect of surface tension cannot become important until k2 becomes large. Consequently, the value of the term 1U +4k coming via m will be large at the point of maximum growth rate, and N may then be approximated as m .With this simplification it is easy to obtain SU 2U1/ k = and cr (2.30) whence for most cases of interest, k ex is onethird the critical value of k denoted k ,,lt which is the value of k2 at o = 0 . Finally observe that in the model of precipitation presented here, we can have one and only one critical point. Again, this is because of the fact that the growth curve must start positive at zero wave number (due to longitudinal diffusion in the solution), rise initially as the wave number increases (due to transverse diffusion in the solution), and eventually fall down (due to surface tension), giving a critical point. Again notice that since the solidification problem (cf., Chapter 3) has two active phases through which the latent heat can be rej ected, we have a possibility of seeing more than one critical point. But before going on to solidification, let us look at the neutral curves in precipitation. The Neutral Curves: U vs. k2 at o = 0 Turning our attention to the neutral curve, viz., the U vs. k2 curve at o= 0, we have, from Eq. (2.24), [ak2 U[N+U]= 0 (2.31) But N is always less than U for all positive values of k hence the root N = U can be ruled out. There obtains U = ak2 and the neutral curve is a straight line as shown in Figure 2.8. For a fixed k the system is initially stable for small values of U, becoming unstable as U increases. 0 k2 Figure 2.8 Typical neutral curve in precipitation The width of the precipitation cell plays an important role in determining the stability of a system. For a precipitation cell of scaled width W the maximum admissible wavelength of a disturbance is 2W and the admissible values of the wave number are restricted to k = ns/rW , n = 1, 2..; the case n = 0 is ruled out on the grounds that the volume of the supersaturated solution is held constant. One can then find the lowest possible value of U for which o = 0 It occurs for the lowest allowable value of k viz., for k2 = t / W2 Therefore, in an experiment, we expect to see at most one crest when the front just becomes unstable. The critical U is U,,w= 71 x /Wl (2.32) and the corresponding critical value of the control parameter cL is given by cL = 1 ea[?u 2 (2.33) Concluding Remarks What we have found in precipitation is this: the shape of the growth curve requires the effect of transverse diffusion for its explanation. The corresponding neutral curve is monotonically increasing, implying only one crest at critical. In other words, since the neutral curve in precipitation does not have a "dip", it is not possible to observe a "cellular" pattern or multiple cells in precipitation when the instability is approached by raising a control parameter and crossing its critical value. Next let us try to get an idea as to when exactly does the instability set in when one creeps up on the control parameter. Using the thermophysical properties given in Table 2.1, assuming the temperature to be T = 300K and taking the depth of the liquid solution to be L = Icn the scales for concentration difference, velocity and time are: cs cs,, 2041 kg / na D/ L 107 m/s and L2 /D 105 s. The value of the surface tension group is ~a 4.67 x 109. Let us take the width of the precipitation cell to be W =1cnz Then Eq. (2.32) predicts the scaled value of Uc,, = ~a tr / W' to be around 4.6x104. This corresponds to a scaled value of cLen =1 ep[~:~' to be 4.6x104, i~e., c,,te, = ""SA = 4.6x104 This means that unsealed cs csA cent is cs,, + 4.6x10" x (2284243) kgm" csw .. Because cent is Very close to c,,, ,it is extremely hard to run reasonable experiments to study precipitation as the front will become unstable as soon as we tweak up cL CVen Slightly above cs, .Notice that this is due to the small value of the surface tension group, ~a Because of reasons to be discussed later, the physical numbers for parameters like critical undercooling etc. in solidification turn out to be much more reasonable and interesting. Nevertheless precipitation was interesting in itself and presented us with the opportunity to understand concepts such as longitudinal diffusion, transverse diffusion and surface tension in a relatively easier onephase set up. Finally notice that because the precipitation model is "onesided", there are no domain length scales to adjust. One can only have a scaled capillary length. The reader' s attention is drawn to the fact that the results will change substantially for solidification (cf., Chapter 3) where the latent heat can be rej ected through two phases. In solidification, we will have the ratio of the liquid to solid phase depth as an important control, as also a capillary length, in addition to the control variables arising from the solid and liquid farfield temperatures. We will revisit precipitation when we discuss solidification and recall some of the results discussed here. CHAPTER 3 SOLIDIFICATION: LINEAR THEORY Introduction The nature of instability during solidification of a pure substance is investigated in this chapter. The latent heat is assumed to be rej ected through the subcooled liquid as well as through the solid phase. Consequently, unlike precipitation, solidification is a twophase problem where the solid phase is not passive any more. We find that transverse diffusion, as well as longitudinal diffusion, have opposite effects in the two phases, and hence, compared to precipitation, we find a greater variety of growth curves in solidification. As in Chapter 2, we start by giving picture arguments, where we explain the role of longitudinal diffusion, transverse diffusion, and surface tension with regard to the observed instability during solidification. An important difference from precipitation is that we now need to understand the effects of diffusion and surface tension in two phases. We give a simple model for pure solidification where we assume local equilibrium at the interface and isotropic interfacial properties; convection induced by density differences is neglected. Then we derive the o formula by using linear stability methods, and explain the various terms in the formula. Next we draw several o vs. k2 CUTVCS, and our emphasis is on the importance of transverse diffusion in explaining the shapes of these curves. The growth curves imply the shape of the neutral stability curve. This curve is useful in understanding what might be seen in a series of experiments as the control variable increases to, and just beyond its critical value. When heat is withdrawn only through the liquid, the neutral curves in solidification are as simple as those in precipitation. Upon withdrawing heat through the solid as well, neutral curves like those seen in the problem of heating a fluid from below [7] can be found as shown by Wollkind et al. [10]. And by adjusting the depths of the phases, we can go beyond this and get something new, viz., critical points that are not due to surface tension. This means that interesting experiments can be run by controlling the transverse extent of the solid, thereby making the allowable values of the wavelength an input variable. We draw some conclusions along these lines toward the end of the chapter. The central result is this: In the solidification of a pure material it is possible to obtain zero, one, two, or three critical points, at least one having nothing to do with the surface curvature. In the solidification of a dilute binary alloy [9] only zero or two can be found. Before moving on to discuss the causes of instability in solidification, we would like to make a comment on the reason for considering the finite depths of the solid and the liquid phase. We have already mentioned that we do not account for convection. Now the wavelengths characteristic of convection are much longer than those of interest in solidification, in the absence of convection, dominated as they are by surface curvature, cf., [15]. Our plan is to show that in solidification of a pure material, there are critical wave numbers in the convection range and that these can be found in ordinary experiments if one knows where to look. But if convection is ultimately to be of interest, bounded experiments are essential, hence we investigate solidification, by itself, in the case where the depths of the liquid and the solid phases are finite. There is an advantage in doing this. The theory of solidification is ordinarily presented for phases of infinite depth in terms of the gradients at the solidmelt interface, cf., [9]. These gradients can be controlled in two ways in experiments at finite depths, by depth of the phase and by the temperature at its edge. With this additional flexibility of control, critical points can be found at much longer wave lengths than those determined by surface curvature. It is these critical points that ought to couple strongly to convection. Coriell et al. [27] study phases of finite depth in the solidification of a dilute binary alloy. Davis [15] explains how such experiments can be set up. We now move on to discuss the various causes of instability during solidification from a subcooled melt. Whenever possible, we will try to point out similarities and differences from precipitation. The Cause of the Instability solid sink Solid Liquid liquid sink T=T, IT=TL moving front II : = Z (, t) Oz= Z i I.U= ufck vlct u Z = n J'u = staface ~uvelocit Figure 3.1 A moving solidification front Let a solid lie to the left of its subcooled melt. This is shown in Figure 3.1 where the two phases are divided by a planar interface that advances to the right as the liquid turns into solid, releasing its latent heat to both phases. Taking TL to be the control variable, decreasing TL increases the speed of the front, and for some value of TL, the planar front becomes unstable, acquiring a nonplanar shape. Solid Liquid Crest Temp. gradient ~~i~ strengthened Trough / Temp. gradient ~ A weakened Temp. gradient ,ep rain strengthened weakened Moving Front Figure 3.2 A picture argument for the effect of longitudinal diffusion in solidification The effect of longitudinal diffusion is shown in Figure 3.2. A displacement of the surface in the form of a crest increases the liquid side temperature gradient at its tip, which reinforces the growth of the crest; hence longitudinal diffusion in the liquid acts to destabilize the surface, just as it did in precipitation. However, the solid side temperature gradient is weakened at the crest, and this tends to flatten out the crest; hence longitudinal diffusion is stabilizing in the solid phase. A similar effect was obviously not seen in precipitation as the solid phase was passive there. Again, the strength of longitudinal diffusion does not depend on the wavelength of the disturbance; hence it is a k independent effect. Setting aside the effect due to nonuniform temperature gradients, longitudinal diffusion by itself acts more strongly the faster the front moves. Next let us consider the effect of surface tension on the stability of the front. It is stabilizing. This can be understood from the fact (cf., Figure 3.3) that the melt temperature depends on the shape of the solidliquid interface; the surface curvature lowers the melt temperature at a crest while increasing it at a trough. Hence, surface tension weakens the instability caused by longitudinal diffusion in the liquid, while strengthening its stabilizing effect in the solid. Again, the stabilizing effect of surface tension is k dependent, the stabilization being stronger for shorter wavelengths. Solid Liquid p:~tnMeltingn point Moving Front Figure 3.3 A picture argument for the effect of surface curvature in solidification Finally, let us look at the effect of transverse diffusion in the two phases. Setting surface tension aside, the effect of the displacement is as shown in Figure 3.4. On the melt side, the displacement of the surface strengthens the temperature gradient at a crest, weakening it at a trough; this induces transverse temperature gradients, carrying heat away from crests toward troughs. This tends to build crests up. Transverse diffusion is therefore destabilizing on the melt side of the surface. A similar argument can be made in the solid phase, where the displacement of the surface weakens the temperature gradient at a crest, strengthening it at a trough. This tends to flatten out the displacement by carrying heat away from troughs toward crests. Hence transverse diffusion is stabilizing in the solid phase. In both phases, the effect of transverse diffusion, whether it is stabilizing or destabilizing, is stronger the higher the k, i.e., the closer together the crests and troughs. Having learned the effect of transverse diffusion in the two phases in the absence of surface tension, here is how surface tension modifies the argument. By reducing the melt temperature at a crest, surface tension weakens the instability caused by transverse diffusion in the liquid while strengthening its stabilizing effect in the solid. Hence surface tension is seen to be stabilizing yet again. Solid Liquid higher Wavelength  higher Figure 3.4 A picture argument for the effect of transverse diffusion in solidification What we have learned so far is this: On the liquid side, we have both longitudinal and transverse diffusion, and they are destabilizing just as they were in precipitation. And surface tension weakens both effects. On the solid side, we also have both longitudinal and transverse diffusion. It turns out that these are stabilizing, and surface tension strengthens their stabilizing effect. Thus in solidification, diffusion (in both directions) has opposite effects in the two phases, and we might therefore expect growth curves that are significantly different from those found in precipitation. What is more interesting is that the introduction of the solid phase gives us an extra parameter in the form of the ratio of the depths of the two phases, and this can be seen as an additional degree of control. It would appear, then, by fixing Ts to establish a stable framework, that we can arrange to find diffusive critical states, independent of capillary effects by controlling the strength of liquid and solid transverse diffusion, which, in turn, are controlled by the depths of the two phases. To quantify this, we turn to a model. The Nonlinear Equations The sketch given in Figure 3.1 Eixes the principal ideas and gives some useful formulas. The liquid and solid phase depths are maintained constant and all equations are written in a frame moving with a constant velocity U = U k, where UI is the base state speed of the solidiaication front. All variables introduced are movingframe variables; the superscript * denotes a liquid phase variable. Then, the temperature in the solid phase must satisfy aV 2 T + U? (3.1) dt 8z for S < z < Z and 0 < x < W, where S is the depth of the solid phase and W is the width of the solidiaication cell and a is the thermal diffusivity of the solid phase. In the liquid phase aT=L'V2 +UI (3.2) dt 8z must hold for Z < z < L and 0 < x < W where a" is the thermal diffusivity of the liquid phase. Two demands must be met along the solidliquid interface, viz., along z = Z(x, t) First, phase equilibrium, taking into account the curvature due to the interface, requires T = T, + 7T 2H = T* (3.3) LH where LH is latent heat of solidiaication per unit volume of the solid, and where y denotes the surface tension; T, is the melt temperature at a flat surface. Second, the rate at which the latent heat is being released due to the motion of the solidiaication front must be the same as the rate at which heat is being conducted away from the interface, viz., naln.VT= ~rl 'nV'=L .n+u (3.4) where ii and A'* denote the thermal conductivity of the solid and liquid phase respectively and u denotes the local surface speed of the perturbed front. Farfield conditions are given by T(z = S) = Ts (3.5) and T*(z = L) = T, (3.6) Finally we impose side wall conditions such that 8T/8x, dT'/8x and 8Z/8x all vanish atx = 0 and at x = W . Eq. (3.4) in solidification corresponds to Eq. (2.3) in precipitation (cf., Chapter 2); however, in Eq. (2.3), the term [cs c] was perturbed and this introduced a term in the perturbation equations that will not be seen in solidification unless the latent heat is perturbed (cf., [3]). Scaled Nonlinear Equations To scale the modeling equations we introduce the length scale via S, the depth of the solid phase. All speeds are scaled by a /S while time is scaled by S2 / a Temperature is measured T T below a reference point corresponding to T,, and it is scaled by aLH / ii, i.e., TscaledM aLH i Then, assuming both the thermal conductivity and the thermal diffusivity to be the same in the two phases, an approximation, that is sometimes reasonable, sometimes not, the scaled temperature in the solid phase must satisfy irV2T+UI (3.7) dt 8z for 1 < z < Z and 0 < x < W, while in the liquid phase i=V2 + U? (3.8) dt 8z must hold for Z < z < L and 0 < x < W The equations along the interface, viz., along z = Z(x, t) now become T= _l[2H]= T (3.9) and n.VTn.VT' = Uk.n+u (3.10) where ~a, a dimensionless parameter introduced by our scaling, denotes T .~ Farfied conditions are given by T(z = 1)= Ts (3.11) and T*(z = L) = TL (3.12) Again, the equations of the model are nonlinear as the interface temperature gradients and the position of the interface are coupled to each other, which is clear from Eq. (3.9). Hence the problem is capable of admitting a nontrivial solution besides the trivial base solution. This is to say that the problem is capable of departing from its steady base state solution (a planar front at rest in the moving frame), leading to an unstable nonplanar front. We take the scaled solid far field temperature, the scaled liquid depth and ~a to be input variables while U and the temperature fields as well as amplitudes of the temperature and interface deflection to be outputs that depend on these inputs. To investigate this instability during solidification, we now turn to study the effect of small disturbances on the base state. (3.13) (3.14) The Solutions to the Base and the Perturbation Problems The base state variables satisfy d T, dT S+ U = 0 dz2 dz in the solid phase, 1< z <0O and 0 < x < W, while in the liquid phase d T' dT' S+ U = 0 dz2 dz must hold for 0 < z < L and 0 < x < W The equations at the planar interface, z 0 are given by T, = 0 = T,' (3.15) dT, dT~ dz dz (3.16) The farfield conditions are given by TO z = 1 = Ts and ,' (z = L) = TL (3.17) (3.18) A solution to equations (3.133.18) is given by , (z) = 1 eUT ] l; and T,,()=[ 1 e where U is obtained from (3.19) (3.20) TT 7 s + = 1 (3.21) eU 1 1e " This corresponds to a planar solidification front at rest in the moving frame. To determine the growth rate of a small disturbance of wave number k imposed on the steady base solution, we expand the surface variables Z, T and T' as Z = Zo + EZ, = Zo + E Z, cos(kCx)e"t (3.22) P=t~it~ Tdz ]=;+ dz (.3 and T* = To' + E (' +, d~do ]= To' + E (+Z dTo coskxe (3.24) dz dz where the side wall conditions determine the admissible values of k. Then the perturbation equations for T, 7;' and Z, are given by 07d2 I d T ag = k2 (U '(3.25) in the solid phase, 1< z <0O and 0 < x < W, while in the liquid phase 07d2 ^ d T'~/d ag= 2*+ (3.26) must hold for 0 < z < L and 0 < x < W The equations at the planar interface, z = 0 are given by ~+; fJd + Ak2 1 = 0 (3.27) '+Z +Ak21 z= 0 (3.28) and crZ (3.29) d dz dz dz2 The farfield equations are 7(z = 1)= 0 (3.30) 7* (z = L) = 0 (3.3 1) A solution to the domain equations satisfying Eqs. (3.27, 3.28) and (3.30, 3.31) is given by 7;(z) = a 1 e"'"''l"' ,; (3.32) * (z = 0) g n dT* ak? (z ( 1 e['] [ngn;_]L n Z, (3.33) where 2m, = U U+ k +oj.. Finally, using the energy balance across the interface, Eq. (3.29), there obtains dz ~dz2 ak2 + M i dz I II (3.34) e~m"] dT, (z =o 0)e e"' et[ , dz III IV where M~= m~['] and N= [I 1 e[nin 1 Eq. (3.34) gives o in terms of k2 and the derivatives of the base temperatures at the reference surface, viz., at z = 0. Again o's on the domain are not important (cf., Chapter 4), and dropping them, we Eind that the dominant growth rate is given explicitly by Eq. (3.34), but m, are now given by 2m1 = UI + U +4k2 Then mn is always positive and m is always negative. Consequently, M2 is always positive and N is always negative, and the signs of the terms I, II, III and V are negative while term IV is positive. Term IV in Eq. (3.34) accounts for the effect of longitudinal and transverse diffusion in the liquid, both destabilizing. Term II accounts for the stabilizing effect of longitudinal and transverse diffusion in the solid; a term like this did not appear in precipitation. Term V is stabilizing and it is like term IV in precipitation. It takes account of the fact that the base temperature gradient gets weakened in the liquid on moving away from the interface, while it gets strengthened on the solid side. The stabilizing effect of surface tension appears in terms I and III. Term III accounts for the fact that surface tension cuts into the destabilizing effect of longitudinal and transverse diffusion in the liquid. Term I accounts for the fact that surface tension enhances the stabilizing effect of longitudinal and transverse diffusion in the solid. Next, we move on to draw and explain the o vs. k2 curves obtained from Eq. (3.34). The o vs. k2 Curves Our aim is to explain how the various terms in Eq. (3.34) get together to produce a curve of o vs. k The values of Ts, TL ~a and L define a given curve. First we set the value of Ts, which for Ts > 0, imposes a basic stabilization on the curves. Then we set the value of TL > 0, and hence of U > 0 The higher the value of TL the stronger the destabilization. The remaining two inputs are ~a and L The first determines the value of k2 where surface tension begins to exert its effect; the second, the ratio of the depths of the phases, establishes the relative importance of diffusion in the two phases. Now most of the difficulty in seeing what Eq. (3.34) is trying to tell us stems from the effect of U on the domain in both the base and the perturbation problems. As a first step therefore, we look at Eq. (3.34) in the low U limit. A low U result obtains by dropping the term in U from Eqs. (3.1) and (3.2), whence the base speed and the growth rate are given by U =Ts L (3.35) and k TL kL1 1 tanh k Ltanh kLtanh k tanh kL (.6 Notice that if k2 is large enough that tanh k and tanh kL are nearly 1, we obtain o= k Ts + 2ak and we see that surface tension becomes important only at values of k2 above 1~ T TL Now the value of ~a is typically about 106. Hence, unless Ts TL is close to zero, there is a range of k2 up to around 10000 on which the third term is not important, and we can deduce the shape of the o vs. k2 curve from the first two terms in Eq. (3.36), which have to do only with diffusion in the two phases. To illustrate the predictions of Eqs. (3.34) and (3.36) we base our calculations on the physical properties of succinonitrile (SCN) [28] which are listed in Table 3.1. Taking the depth of the solid to be S =1 cm the scales for temperature difference, velocity and time are aLH /Al = 24 K a / S 10' m/s and S2 / a 900 s. The value of the surface tension group is ~a 0.26 x 106 The o vs. k2 curves are presented in Figures 3.5, 3.6, 3.7 and 3.8 as qualitatively correct sketches. Figures 3.5 and 3.6 illustrate the low U formula, whether or not U is low; Figures 3.7 and 3.8 include the effect of U on the domain. The values of Ts, TL L and k2 given on the sketches were used in Eqs. (3.34) and (3.36) to determine the trends, and these input values are experimentally realizable. To indicate the importance of the depths of the two phases, curves are drawn for two values of L In each figure, a value of Ts is set, which gives stabilization at TL = 0 Then a value of L is set to strengthen or weaken the diffusion effects in the two phases. Curves are then drawn for increasing values of TL, increasing the speed of the front. To understand the curves in Figures 3.5 and 3.6, it is helpful to note the following formulas which hold for low U, and which can be obtained from Eq. (3.36):  k low) =[ Ts +; TL Ts*T (3.37) 0 kintemedite) k s +(3.38) and  (k? high)= 2Lak3 (3.39) Table 3.1 Thermophysical properties of succinonitrile used in calculations Melting Point, T,, 331 (K) Density of solid, p 1016 (kg/m3) Density of solid, p* 988 (kg/m3> Thermal Conductivity of solid, Ai 0.225 (W/m K) Thermal Conductivity of liquid, A2* 0.223 (W/m K) Thermal Diffusivity of solid, a 1.16 x 10' (m2/S) Thermal Diffusivity of liquid, a* 1.12 x10 (m2/S) Surface Tension (solidliquid), y 8.94 x 103 (N/m) Latent Heat of Fusion, LH 0.48 x 10 (J/m3) 1 rsig due to L = lateral diflusionl a T.= 0 2 in the liquid ...4 = 0.25x10 ,~ k'180lkCill falling due tor the effect of curviature on the nickingp point T, = 0.2 L T, L = 0.12 fa the~ due to lateral diffursion in th~e schid Ti=0 . T, T, = 0.08: IT, = 0.05 T, LL 1 cm Ts = 0.2 2 Figure 3.5 o vs. k2 for the low U model with S = Icm L \ T,= 0., = Ts9 T, = 0.32 T~r = 0~ 2,L T L'L T,=0 Figure 3.6 o vs. k2 for the low U model with S = Icm L = 2cm Ts = 0.2 T, = 0. 1,0. Figue 37 o s. forthecomlet moel ith = cm L m Ts = 0.2 8 2 ,=0.7 SDDDD r, = 0.3 T,=0.2475 r, = o.z ~~=D Figure 3.8 a vs. k2 for the complete model with S = Icm L = 2cm T, = 0.2 d0 These formulas show that for sufficiently low values of TL, O is negative at k2 = 0, dk2 is negative at k2 = 0, and o remains negative in the intermediate range of k2 The effect of surface tension is always to make o more and more negative at larger and larger values of k2 d0 As TL increases, and before surface tension comes in, o at k2 = 0, at k2 = 0, and o at dk2 intermediate values of k all pass through zero and become positive, and they do this for physically interesting values of TL The order in which these variables turn positive depends on the value of L Notice that in the low U formulas for low and intermediate ranges of k2, we see Ts, the solid side base temperature gradient, combined with L the liquid side base LL and 1 in the three terms, whence the ratio of the depths of the two phases plays a very important role as far as the growth curves are concerned. For L <1, cf., Figure 3.5, o at k2 = 0 turns positive first, then o for intermediate values of k and finally at k2 = 0 For L > 1, cf., dk2 Figure 3.6, the order gets reversed. These facts alone permit the o vs. k2 curve to be sketched correctly . The intermediate formula for o,cf.., Eq (3.38), holds once the value of k2 is large enough that the wavelength of a displacement is smaller than the depths of the two phases, but not so small that surface curvature begins to exert its influence by affecting the melt temperature. In this range, the value of o is determined entirely by diffusion parallel to the surface, viz., by transverse diffusion in the two phases. The strength of the base gradients combines directly with the surface displacement to determine the strength of the troughtocrest diffusion in the solid phase, and the cresttotrough diffusion in the liquid phase. This is illustrated in Figure 3.4. We also note that the value of o for intermediate values of k2, iS a multiple of k or Hence o is positive and rising or negative and falling for these wavelength of displacement intermediate values of k2, according as T +~ is positive or negative. As k continues to increase and the melt temperature begins to respond to the curvature, the rising curves turn downward causing a greatest growth rate to appear in the intermediate range of k2 The falling curves simply fall faster once the effect of surface curvature takes over at larger values of k2 The low k2 Variation of o is also due to transverse diffusion, but the wavelength of the surface displacement is now greater than the depths of the phases. The nearsurface region is therefore influenced by the sinks, and an explanation based on base gradients is no longer easy to present in picture form. Nonetheless, at low U, the second term in Eq. (3.37), and Eq. (3.38) indicate that stabilizing diffusion in the solid can dominate destabilizing diffusion in the liquid at low values of k 2, before their contributions are reversed at intermediate values of k 2, and this depends on L i.e., on whether the depth of the liquid is greater or lesser than the depth of the solid. The curves show a least o at low k2 if L <1 (Figure 3.5) and a greatest o at low k2 i L > 1 (Figure 3.6). To see the effect of the ratio of the phase depths, Figures 3.5 and 3.6 are T, T, drawn at values of T, chosen so that and therefore Ts + is the same in both figures. L L Hence the intermediate and large k2 regions of the curves in the two figures at the same values of Ts + i are the same. They differ only at small values of k2 Say up to k2 = 100 The curves that fall for low values of k2 in Figure 3.5 do not fall as strongly and may even rise in Figure 3.6. Again, this may not be easily explained from a picture argument, but the second term in Eq. (3.37) shows that at small values of k larger L 's weigh diffusion in the liquid more strongly than diffusion in the solid. Next we take U into account on the domain, and the corresponding curves are shown in Figures 3.7 and 3.8. Again, we draw two sets of curves: Figure 3.7 is drawn for L = _, whereas Figure 3.8 shows the results for L = 2 .All the values of Ts L in Figures 3.5 and 3.6 are in Figures 3.7 and 3.8, but a few more are included. Given Ts, L and TL the effect of including U on the domain, if it is large enough, is to weaken the base gradient at the planar surface in the solid phase and to strengthen it in the liquid. This in turn weakens the effect of transverse diffusion in the solid, strengthening it in the liquid. This is enough to explain the maj ority of changes in going from Figures 3.5 and 3.6 to Figures 3.7 and 3.8. Hence, wherever o is falling in Figures 3.5 and 3.6, in Figures 3.7 and 3.8 it is not falling as fast and may even rise, and wherever o is rising in Figures 3.5 and 3.6, it is now rising more strongly in Figures 3.7 and 3.8. As a result, the intermediate region grows into both the small k2 and high k2 regions and this delays the effect of surface tension, pushing the values of k2 at the greatest growth rate to the right. In the intermediate region, o is given by 1 U U +4k 1 2Ts 1 U whence o rises or falls depending on whether TL is greater or lower than 1 12T L Or wha isth same, whether U is greater or lower than In(1+2Ts). For small values of L cf., Figure 3.7, we see that for small values of TL O is negative for all values of k2 Upon increasing TL, Say for 0.05 < TL < 0.078, o turns positive for small values of k2 and a critical point appears in the range 0 < k2 < 25. It is due to transverse diffusion in the solid phase and it is independent of surface tension. It is seen in Figure 3.5 as well, whence it is not due to U on the domain. However, if U on the domain is not accounted for (cf., Figure 3.5), no further critical points can be seen, the only one found, at first due to solid transverse diffusion, eventually is due to melting point lowering, the transition being due to the growing destabilizing effect of transverse diffusion in the liquid phase as T, increases. The effect of U on the domain is to turn base states stabilized by solid phase transverse diffusion at low k2 into unstable states at higher value of k2 an effect that cannot be seen if U on the domain is not taken into account. Hence, with U on the domain, we find that further increasing T, leads to three critical points as liquid transverse diffusion is strengthened. In Figure 3.7, these occur for 0.078 < T, < 0.085, and the three critical points are due to solid transverse diffusion, liquid transverse diffusion and surface tension. This is not seen in Figure 3.5 due to the fact that the liquid transverse diffusion comes in too late, i.e., after o at k2 = 0 has already become strongly positive. For yet higher values of T,, the two critical points due to solid and liquid transverse diffusion disappear, giving only one critical point due to surface tension. What we have found is this: Upon increasing T, at small L viz., L = _, the number of critical points is first zero, then one, then three, and finally, again one. It turns out that getting three critical points is a feature typical of pure solidification when the depths of the two phases are taken to be finite, and when the control variable is defined explicitly. The Mullins and Sekerka [9] model for the solidification of a dilute binary alloy fails to give anything as interesting as has been found above; this is mainly because the depths of the two phases are taken to be of infinite extent in their analysis. Also there is no thermal undercooling in the Mullins and Sekerka model. Similarly, the Wollkind and Notestine [10] model for pure solidification gives at most two critical points. The fact that we get three critical points has important implications from the point of view of an experiment where the liquid side undercooling is slowly increased up to its critical value, and this will be explained in detail toward the end of the chapter where we discuss the neutral curves. Turning our attention to Figure 3.8 to see the effect of increasing L, many changes from Figure 3.7 are like those seen in going from Figure 3.5 to Figure 3.6. Most interesting, however, are the changes at low values of k2 Increasing L weakens transverse diffusion in the liquid which might be expected to enlarge the region of solid transverse induced critical points seen in Figure 3.7. But increasing L also weakens longitudinal diffusion in the liquid causing the values of o at low values of k2 to be negative, thereby eliminating the possibility of solid transverse diffusion induced critical points. Hence at large L ,we find no critical points for small values of T,, then upon increasing T, we find two critical points, the first of these due to liquid transverse diffusion, followed by a second critical point due to melting point lowering. So at L = 2 there are no critical points at low values of T,, two at higher values of T,. An important effect of increasing L is this: at low L ,there are critical points at small values of T,, while on increasing L the value of T, at which critical points are first seen increases. If Ts were lower, the curves at L = 2 might be more like those at L = _, but then increasing L above 2 ought to restore the curves at L = 2 at the higher values of Ts  Next, we move on to understand the neutral curves in solidification, which are derived from growth curves, though they are much more convenient while trying to understand what goes on in an experiment where the undercooling in the melt is slowly crept up to its critical value. The Neutral Curves: U vs. k2 at o= 0 In neutral curves, the speed of the front U is plotted against k2 at o= 0 Each set of o vs. k2 CUTVCS at a given Ts and L leads to a U vs. k2 CUTVe. For example, Figure 3.7 leads to Figure 3.9, and Figure 3.8 leads to Figure 3.10. Again, no scale is given on the ordinate and the reader' s attention is drawn only to the shapes of the curves. As long as Ts is not zero, all possible neutral curves look like Figure 3.9 or 3.10. The first thing to notice is this: no matter the value of Ts > 0 and L, once k2 is large enough, U is a decreasing then increasing function of k2, i.e., the curves all have a dip. This is neither true in precipitation nor in solidification at Ts = 0, where the neutral curve is a straight line as shown in Figure 2.8 of Chapter 2. The curve in Figure 3.10, at a high value of L, shows only the dip, and its shape is like the neutral curve obtained in the problem of heating a fluid from below [7]. But here, the rising side has a very gradual slope due to the small value of ~a, and this leads to a very flat minimum. The dip is not new; it was seen by Wollkind et al. [10] but it was not explained. Two things are required for a dip: First, o should be negative at small values of k2 In Figure 3.8, o is negative at k2 = 0, because large values of L weaken longitudinal diffusion in the liquid; in Figure 3.7, solid transverse diffusion can cause positive o's at k2 = 0 to become negative for small values of k2 Second, transverse diffusion in the liquid should cause o to increase and become positive, before surface tension drives it back to zero. Hence, the falling branch of the dip is due to liquid transverse diffusion while the rising branch is due to surface tension. U /I\falling due to lateral on the melting point diffusion mn the liquid o= 0 k2 = 1500 k2 Figure 3.9 An unusual neutral curve in solidification S = kmr L =2cm~ T = 0.2 rising due to the Y effect of curvature U falling due to lateral on the mnelting point diffusion in the liquid 0 = k = 1500 k2 Figure 3.10 Typical neutral curve in solidification Figure 3.9, at small L, shows something new; it appears at small values of k2 and it is due to transverse diffusion in the solid. The curve shows a sharply rising branch feeding into the falling branch of the dip. At small L longitudinal diffusion in the liquid leads to positive values of o at k2 = 0 Solid transverse diffusion then drives o negative, providing a basis for the dip. In the course of doing this, another critical point is produced. This is seen in Figure 3.9 as the initially rising branch. The reader' s attention is drawn to the fact that a neutral curve like the one shown in Figure 3.9 cannot be obtained if U on the domain is not accounted for. In fact, one can prove that if U is dropped from the domain equations, then one can get zero, one or at most two critical points, but never three. Figure 6.5 in Chapter 6 shows this. Notice that a neutral curve like the one seen in Figure 3.9 is special to solidiaication. Nothing like it is observed in the closely related diffusioncontrolled problems: * precipitation, a one phaseone surface problem * electrodeposition, a one phasetwo surface problem * viscous fingering, a two phaseone surface problem The setup in viscous Eingering is most like that in solidiaication, but solidiaication presents us with an extra degree of freedom. In solidification, unlike these other problems, we have three k dependent effects determining the o vs. k2 curves. Also, solidification takes place at a thermodynamically specified temperature, giving us the option of setting the control temperature above or below the phase change value. The reader is referred to Cross and Hohenberg [8] who review pattern formation at a great length in a variety of problems, never finding neutral curves like the new one given in Figure 3.9. It is clear by now that the value of L is important. The transition between the two types of curves shown in Figures 3.9 and 3.10 might seem to be at L = 1. And that would be true if U on the domain is not accounted for. But with U on the domain, it is only roughly true. It depends on Ts and a, and for Ts = 0.2, it is around L = 0.85 for succinonitrile. The neutral curves can be used to answer the question: What does one expect to see in an experiment? For example, how many crests are seen when a plane front loses its stability? In a sequence of experiments, let the values of ~a, S, L and T, be set. Then W must be set and its role is to determine the admissible values of the wave number as k = nzr/W, n = 1, 2..Hence not all points on the neutral curve are of interest. The value of TL and therefore the value of U, is then increased from one experiment to another until a critical point is reached, i.e., a point where a line of constant U first crosses the neutral curve at an admissible value of k Our aim is to predict the effect of W on the number of crests appearing at a critical point. Now Figures 3.9 and 3.10 both show a dip and this dip implies that the critical value of U need not correspond to the least allowable value of k In fact, n might be any of the numbers: 1, 2, 3.., and hence multiple crests might be seen. This is not true of precipitation, where we found that one can get at most one crest at critical. A neutral curve like the one shown in Figure 3.10 can be obtained for S =1 Icn L = 2cn? , Ts = 326.2K (4.8 K solid cooling). The value of k2 at the dip would then be around 1.5 x 107 na The critical wave number will depend on the value of W. For W = 0.08cn the critical value of k" occurs near k~ viz., k 2~,= 1.54 x 107 n It corresponds to n = 1, whence only one crest will be seen at critical, and this is true for all smaller values of W The corresponding critical interface speed is 3.8 pum / s and the critical undercooling is 5.9 K. For a larger width, more crests will be seen at critical. For example, for W = 0.5cm, the critical value of k2 corresponds to n = 6 As W increases further, we would expect to see even more crests. A neutral curve like the one in Figure 3.9 can be obtained for input values: S =1 Icn , L = 0.5 cn Ts = 326.2K (4.8 K solid cooling). Let us again start with a small value of W and see what happens as W is increased. For W = 0.08cna, the critical value of k2 corresponds to n = 1 giving one crest at critical. For a higher value of W viz., for W = 0.5 cm the critical k2 corresponds to multiple crests, viz., n = 6 Here too, increasing the width further will give more crests until W is chosen large enough that the first allowable value of k viz., corresponds W2 to a point on the sharply increasing part of the neutral curve which is below the dip. For example, for W = 2cn2, the first allowable value of k2 is given by 2.5 x 104 n2 ,and it corresponds to an onset of instability giving only one crest. And the critical value of k2 is now independent of the value of ~a, and hence is not like those seen above. The critical interface speed is now 3.5 pn2 / s and the critical undercooling is 1.6 K. Hence for small enough values of W the expectation is one crest at critical, as in precipitation. For larger values of W, the expectation is multiple crests at critical, until, at small values of L for high enough values of W, the expectation returns to one crest at critical at a value of k2 now independent of surface tension. Concluding Remarks Our view is that problems of phase change often lead to problems in pattern formation. Hence one role of theoretical work is to decide what experiments might be interesting. We suggest that experiments ought to be run by increasing the control variable, here T, TL Of U, to its critical value and then just beyond. Looked at in this way there is only one critical point in any given experiment, notwithstanding the fact that our a versus k2 curves show three critical points at small values of ,two at low values of k and two critical points at large values of S' ,one at a low value of k In both cases, the low k2 critical points are determined by heat S' conduction in its various manifestations, independent of surface tension. It is the neutral curve, not the a curves, that bears on the design and interpretation of experiments. As U is increased, at the wave numbers consistent with the width of the cell, the lowest intersections of lines of constant U and the neutral curve predict the critical point that ought to be observed. Hence at small values of ,there may be a critical point at low values of S' k2, independent of surface tension, at large values of ~there cannot be such a low k2 critical pomnt. We present a simple model that includes only the most important factors: heat conduction on the domain and surface tension at the moving front not unlike the models used by Wollkind and Notestine [10] and Mullins and Sekerka [9]. But instead of introducing control via the gradients at the front, we propose farfield conditions more akin to how an experiment might be run. This adds the depths of the two phases as interesting input variables and these variables influence the strength of the effect of cresttotrough heat conduction in both phases, an effect that is stabilizing in the solid, destabilizing in the melt. It is by doing this that we discover the existence of a critical point at small wave numbers independent of the effect of surface tension. The width of the solidification cell strongly affects the patterns that can be obtained in an experiment and we explain what ought to be seen as a function of the three lengths that can be controlled: the two depths and the width. CHAPTER 4 ON THE IMPORTANCE OF BULK TRANSPORT ON INTERFACIAL INSTABILITY Introduction Our goal is to understand whether or not domain transport dynamics is important compared to the dynamics at the interface. In other words we wish to investigate if it is reasonable to drop a s from the domain equations. To study this we will consider two problems using precipitation physics as a guide. Both problems are done in rectangular geometry. First we consider an equilibrium problem where the planar front is at rest. Second, we consider a nonequilibrium problem where the front is moving with a nonzero base speed. In each case we derive a condition that dictates whether or not a s can be dropped from the domain equations. In both cases we illustrate that the transport dynamics at the interface is ordinarily much more important compared to the domain transport dynamics, whence dropping a s from the domain equations is a good approximation. In precipitation, this result is due to the extremely small value of capillary length. The endnote discusses two problems. First we consider a solid cylinder in equilibrium with its solution at a uniform concentration. Second we consider a solid cylinder in equilibrium with its subcooled melt at a uniform temperature. In each case we derive a formula for the growth rate a and hence a condition that must be met if a s are to be dropped from the domain equations. Case 1: Equilibrium Precipitation Problem in Rectangular Geometry Consider the schematic shown in Figure 4. 1 which shows a solid of density cs in equilibrium with its solution which is held at a uniform concentration cs, . Then, in the liquid phase, the equation for the solute concentration is given by = DV c (4.1) for Z < : < L and 0 < x < W where L is the depth of the liquid phase and W is the width of the precipitation cell; D denotes the diffusivity of the solute in the liquid phase. The equations along the interface, viz., along : = Z(x, t) are given by c = s, 7 c' 2H(4.2) RT cs and [cs c]u= Dn.Vc (4.3) where 7 denotes the surface tension, and where u denotes the local surface speed of the perturbed front. The far field condition is given by c(Z = L) = c,, (4.4) Finally we impose side wall conditions so that Sc/8x and 8Z/8x vanish at x = 0 and at x = W . Solid Solution c = c,4 c = cs front at rest z; EQ Z, i +k z= Z(x,t) n=iI c = car z = Figure 4.1 A precipitation front at rest The Base State and the Perturbation Eigenvalue Problem Since U = 0, the concentration gradient in the base state is zero. The concentration in the liquid is constant everywhere, viz., c, (z) = cs, . To determine if this base solution is stable, we impose a small displacement on the surface at fixed values of cL and U, and determine its growth rate, a as a function of its wave number k We expand Z = Z, + EZ, = Z, + E Z, cos(kCx)e"t (4.5) and c = co+E[4 c,+ ,d a =Co+e c+Z, cosnkxxe" (4.6) whence the perturbation equations for c, and Z, are given by ks2 t + c, = 0 (4.7) in the solution, O < : <1, O < x < W The equations that need to be satisfied across the interface, z =0, are [cs c,4, i a Z, = Ddc (4.8) dz and c, =7c"kZ, (4.9) RT cs The farfield equation is c, (z = 1) = 0 (4. 10) These equations can be solved in the usual manner to obtain a formula for the growth rate: aC = D 7c,, k N (4.11) wher N= a 1 [mv; L and na = + .The formula for the growth rate becomes much simpler if we let L 4 00 whereupon the factor N reduces to n_, whence = Lci k' (4.12) where Lc, represents the capillary length, defined as Los r 7T c,4, 1 ex etu RT csc c,4, e l u define os as the value of a due to the surface only. Hence os represents the value of the growth rate if a is dropped from the domain equation. Hence we have s =L Lk2 (4.13) Dividing Eq. (4. 12) by Eq. (4.13), there obtains = 1+ s (4.14) as k as Dk2 Hence the sufficient condition for a = os iirsh givn b <<1 or kL <<1. Hence the quantity kLc, should be much less than unity if one is to justify dropping a from the domain equation. Therefore k <~ is a sufficient condition for a = os or equivalently for dropping a cap from the domain equation. Table 4. 1 Thermophysical properties of CuSO4.5H20 Saturation concentration, c, 243 (kg/m3) Density of solid, cs 2284 (kg/m3) Diffusivity, D 109 (m2/S) Gas constant, R 8.314 (J/mol K) Surface Tension, 7 8.94 x 103 (N/m) As an example, let us consider the precipitation of CuSO4.5H,O Using the thermophysical properties given in Table 4.1, and assuming T = 300K, the capillary length is found to be Leaop = 4.67x0 I"meters which is extremely small. Hence the a vs. k: curve should be the same as the os vs. k2 curve so long as k2 << 102 ". Consequently, for most practical purposes we can easily drop o from the domain equations and this is due to the ordinarily small values of Lcos In fact it is easy to get an explicit formula for a vs. k From Eq. (4.12), we get a = DLcapk? (4.15) Squaring both sides we get (X L2 apk4L" pk6 =0 (4.16) which is a quadratic m Keeping in mind that the growth rate must be negative, we get D al 4 k4L 1p 1+ 4 P(4.17) D 2 'k L2 Next let us try to approximate the formula given by Eq. (4.17) for large and small values of kL cap For small values of kLcap We can rearrange Eq. (4.17) to get a 1 2 k2L 12 k k4Lca 1~c 1+ o' ~ k4Lp 1LC" 1+ k2Lr (4.18) D 2 o kL4 2 k2 4 Hence c~~~ 1 01 1  k4 2 kL > + kL ) kL > k4L2 1 (419 D 2 cyP cap cap 4 o 2 kL cap hr gher order term]_ or  kL 1 kL 1 kL (4.20) D "[ 2 ] D This means that a is less negative compared to s In other words, neglecting a on the domain overpredicts the stability. However it is clear from Eq. (4.20) that the correction is negligible. For large values of kLeap We can rearrange Eq. (4. 17) to get _1k4L2 1 1+ 4 k4L2 1 +12(.1 D 2 o'k2L2ca 2 a ,I 2 k2L,2~n 2(. or k2 ap S3 (4.22) D Lcepk D Lceak Again a is less negative compared to as and neglecting a on the domain overpredicts the stability. And it is clear from Eq. (4.22) that the correction can be significant if the quantity kLcg is large enough. And we again conclude that dropping a from the domain is a bad approximation if k > or equivalently if the wavelength of the disturbance is much less than cap the capillary length. The graph given in Figure 4.2 is drawn for Leap = 4.67x10"1 meters and it illustrates the a vs. k2 CUTVe Superposed over the os vs. k2 CUTVe. a is drawn from Eq. (4.17) and os is drawn from Eq. (4.13). Clearly we cannot see any difference as a and os are in excellent agreement. Figure 4.3 is drawn for a much larger value of Leap Viz., Lcp = 103meters, to emphasize that the magnitude of capillary length plays a key role in deciding the importance of bulk transport compared to the interface transport. We see that the difference between a and os is very subtle for a high value of Lc,  Next we move on to the usual precipitation case where a planar front is moving at a constant speed in the base state. Sigma vs. k_square 1 E1 2E1 3E1 4E1 Sigmas on the domain Sigmas off the domarin Figure 4.2 a vs. k2 for the equilibrium precipitation problem with L~c,p = 4.67x10"l meters 0 200000 400000 600000 800000 1000000 Sigma vs. k_square 0.001 "A 0.002 Oa Oa 0.0031 o 0.004 , 0 5E5 1 E6 1.5EB 2EB 2.5EB k2 Sigmas on the domain a Sigmas off the domain Figure 4.3 c vs. k2 for the equilibrium precipitation problem with Leap = 103 meters Case 2: Nonequilibrium Precipitation Problem in a Rectangular Geometry This is identical to the previous case except that the precipitation front is now moving with a speed U. Then, in the liquid phase, the equation for the solute concentration is given by ICDV'c + UI (4.23) dt 8z for Z < z < L and 0 < x < W where L is the depth of the liquid phase and W is the width of the precipitation cell. The equations along the interface, viz., along z = Z(x, t) are given by c = ,4, 7 c" 2H(4.24) RT cs and [cs c] Uk.n+u =D .c(4.25) The far field condition is given by c(Z = L) = cL (4.26) The side wall conditions are Neumann, whence dc/8x and 8Z/8x vanish at x = 0 and at x = W . The Base State and the Perturbation Eigenvalue Problem A base state solution is given by cl(z) = cs + [cs,, cs e[Ulo D]" (4.27) where U, is given by D c, cs U. = I 1 L s4T (4.28) Expanding Z and c as in Eqs. (4.5, 4.6), and solving the perturbation eigenvalue problem, there obtains the following formula for the growth, a : o= D Lgk N (4.29) whreL~p R7c c ,, 1 n nz e[ntng]L1 ndm,~ I r2 whee = N= an a + "+ 4k" c" ~~ ~ [ RTc sc, ,tn]L D D2 D U,L The formula for the growth rate becomes much simpler if we let 4 00 whereupon the factor N reduces to na ,whence S= D Look? [ m_+ (4.30) or a=~~~ Uo2 Lek1 14 (4.31) Next define os as the value of a due to the surface only. Then s~ = Uoz Lok1 + (4.32) 2=o[ D2 Un 14k Dividing Eq. (4.31) by Eq. (4.32), there obtains 4433 1 1+4 1+ s "?, UUZ "~ +4 dropig frm heDoa euto ti odin sife. Hetance the sufficient condition for & Us is given by D, <<1 k and ohn e ane justify Los = 4.67x0 medters s shwhich Fisexremel y small. o Taing cL =h 2cs= 486g/ ,L = I.iscm w enc 4 "s is of the order of 109 2 But the interesting k2 S are of the order of 10122 Hence D 4 Gs 2 D is ordinarily much less than unity. o + 4k2 D2 0.16. oo 0.1 o oo sO U 1E12 2E12 3E12 4E12 5E12 k2 Figure 4.4 o vs. k2 for the nonequilibrium precipitation problem with Leap= 4.67x10 meters Therefore dropping 0 from the domain equation is again found to be a good approximation for precipitation under nonequilibrium conditions. In fact, for a given set of input parameters, after evaluating os using Eq. (4.32), one can use Eq. (4.33) to solve for the ratio and hence to find . It is found that for unstable wave numbers, is slightly greater than unity, hence putting 0 s on the domain predicts more positive growth rates compared to their value had a s been dropped from the domain equation. For stable wave numbers, is slightly less than unity, whence in the stable regime, a is less negative compared to as  Concluding Remarks We have demonstrated that the transport dynamics in the bulk of the domain is ordinarily insignificant compared to the interfacial transport dynamics, whence dropping a s from the domain equations is a good approximation for most cases of interest. Endnote 1: Equilibrium Precipitation Problem in Cylindrical Geometry Consider a vertical solid cylinder of radius R, and of density cs in equilibrium with its solution which is held at a uniform concentration. Then, in the liquid phase, the equation for the solute concentration is given by = DV c (4.34) for R, < r < co and 0 < : < W where W is the height of the cylinder. The equations along the interface, viz., along r = R (z, t) are given by c s 7 c 2H (4.35) RT cs and [cs c]u = Dn .Vc (4.36) where 7 denotes the surface tension, and where u denotes the local surface speed of the perturbed front. 2H denotes the curvature of the surface. The far field condition is given by c(r a co)= finite (4.37) Assuming axisymmetry, the surface normal and surface speed are respectively given by n = ii and u = The side wall conditions require that both Sc/S: and dR/8z vanish at : = 0 and at : = W. The Base State and the Perturbation Eigenvalue Problem Since U = 0, the concentration gradient in the base state is zero. The concentration in the liquid is cnonsantevryrwhere,vz, c,(r i= c4 + 1_ 7 c" Notice that the base state concentration is not cs,, as it used to be for the planar case, cf., case 1. The appropriate correction is present due to the fact that the base surface is itself curved. To determine whether or not this base state is stable, let a small disturbance of amplitude F be imposed on the steady base solution. Then expanding R and c at r = R (z) as R = 4, + ER, (4 38) c = c, +E[ c, +R, (4.39) where the term involving the surface displacement, R,, has been introduced so that the perturbation problem can be solved on the reference domain. To find the critical base state, we must solve the perturbation eigenvalue problem. It is given by c = DV c, (4.40) [Lcs csX, = D (4.41) and c, = S 72 c ~2a R, (4.42) The farfield equation is c, (r > c)= 0 (4.43) Expanding R, = R;, cos(kz)e"', c, = c(r)cos(kz)e"' (4.44) the perturbation equations for c, and Z, are given by d2 id a + k c, = 0 (4.45) drr dr D in the solution, R, < r < co. The equations that need to be satisfied across the interface, r = R , are [ cs,]oR, = D (4.46) dr and cA = 7 ,__ k'R (4.47) RT cs The farfield equation is c, (r > c)= 0 (4.48) Defining A1 = k2 +, a solution to the domain equation, Eq. (4.45) is given as c, (r) = AI, (Ar) +BK, (Ar) (4.49) where I, and K, are modified Bessel's functions of zeroth order. Since we require the solution to be bounded at ra 00c, we have A = 0, whence c,(r= BK Ar(4.50) and using Eqs. (4.46) and (4.47), there obtains [ cs, CoR, = DBA1Kn AR, (4.5 1) and BKA (A, = 1 kR (4.52) RT cs respectively, where K, (x)= [K. (x) Dividing Eq. (4.51) by(4.52), we get L rKo (AR,) (. a = Dl o' [1 k' R (453 RX K, AR,) where Lc, represents the capillary length, defined as: L = c7_ c, 1. Next let us define os as the value Lea RT cs cs c, RTc s cs csA RT cs As of a due to the surface only. Hence os represents the value of the growth rate if a is dropped from the domain equation. Hence we have L K, (klG) (.4 as = Dk p [1 k Rz (454 Dividing Eq. (4.53) by Eq. (4.54), there obtains a TK(/R ) 1 K(k) ) (4.55) as k LK, A4, )IK klg, Substituting 1 .r+ there obtains k DkZ K, 1+ s ~~s V s Dk2 K, o (k 1+s Dk~x, Lok~> Hence the sufficint conition~ for *us is givenby i <<1. Therefore 1L 1 k K,(ky,) <<1 is a sufficient condition to justify dropping a from the k4, LK, (kE,) domain equation. Endnote 2: Equilibrium Solidification Problem in Cylindrical Geometry Consider a solid cylinder of radius R, and at a uniform temperature in equilibrium with its subcooled melt which is held at a uniform temperature. The cylindrical front is at rest in the base state. We will scale the temperature so that it is measured from a reference point corresponding to T,, and it is scaled by aLH / ii Where a is the thermal diffusivity of the solid phase Ai is the thermal conductivity of the solid phase, and LH denotes the latent heat of solidification per T T unit volume of the solid, viz., Tscoale We assume the thermal conductivities and the aLH i thermal diffusivities of the two phases to be the same. Then, in the solid phase, the equation for the scaled temperature is given by dT = aV T (4.57) for 0 < r < R, and 0 < : < W where W is the height of the cylinder. The equation for the scaled temperature in the liquid phase is given by dT* = aV2 (4.58) The equations along the interface, viz., along r = R (z, t) are given by T = T* = Lc 12H1 (4.59) and n .VT n.VT =[ u1; (4.60) where L,,, represents the capillary length, defined asLcg = entste ufc tension and u denotes the local surface speed of the perturbed front. The farfield condition is given by T(r = 0) = finite (4.61) and T'(r a oo) = L (4.62) The side wall conditions require that dT/8z, dT'/8z and 8R/8z vanish at z = 0 and at z = W. The Base State and the Perturbation Eigenvalue Problem Since U = 0, the temperature gradient in the base state is zero in both phases. The temperature is constant everywhere in the solid and the liquid phase. It is given by To(r) = To'(r) = LcosTt Notice that an appropriate correction is present in the base itself because the base surface is curved. To determine whether or not this base state is stable, let a small disturbance of amplitude F be imposed on the steady base solution. Then expanding R T and T' at r = R (z) as R = 4, + ER, (4.63) T = + e + RT,5 (4.64) and T = 4 ,' e + R d y ,' (4.65) where the term involving the surface displacement, R,, has been introduced so that the perturbation problem can be solved on the reference domain. To find the critical base state, we must solve the perturbation eigenvalue problem. It is given by = aV (4.66) '= aV ', (4.67) "R, R, = ( =Le% (4.68) and  a (4.69)9 The farfield equations are 7(r = 0) = 0 (4.70) and 7*(ra oo) = (4.71) Expanding R, = R cos(kr)e"', I= j(r) cos(kr)e"', I; = I' (r) cos(A)e"' (4.72) the perturbation equations for 7 7; and Z, are given by + k+ T=0 d r dr a 7= in the solid phase, 0 < r < % and + k+ T=0 d r dr in the solution, R, < r < c The equations across the interface, r (4.73) (4.74) R,, are (4.75) (4.76) d T d T' dz dz "i The farfield equations are 7(r = 0) = 0 and (4.77) (4.78) kZ +, a solution to the domain equations, Eqs. (4.73) and (4.74) is given as Defining A 2 *;(r) = BC~ + A"K, (Alr) (4.79) (4.80) and ~.82) L~ r~J k:l where I, and K, are modified Bessel's functions of zeroth order. Since we require 7; to be bounded at r 4 0, we have B = 0 Similarly since 7;* must be bounded at ra 00c we have B' = 0 whence (r) =AI, Ar) 4 and *(r) A'K,(Ar)(4 ~.81) where A and A* are given by Eq. (4.75) as A = ` 'R, and A* = ` In AR,> K, AR,) respectively. Substituting in Eq. (4.76) there obtains Leo~ ~L" I (ARl) Kd (AR,) (. where Ir,(x) = [I, (x)1 and KI (x) = [K, (x)1 Denoting as as the value of the growth rate if a is dropped from the domain equation there obtains Os =a k~R 1 kR (4.8 R ~ I, kR, K, kR) Dividing Eq. (4.83) by Eq. (4.84), there obtains I nl(AR,) K, (R )l (4.8 s, k I, (kR,) K, (kR )1 I, kR,) K, kR~) Substituting 1 .I+ there obtains k ak2 ;3) ;4) ;5) Leap r Rz1 k'l R, sDks Dk2 lo kho 1+ s oko1 a ass Dk s, Dk2 0= 1II S s 2 s, s Dk 2I,(:) ko Llo(kRo) K (kFo)I (4.86) Hence the sufficient condition for a = os isgiven by os <<1 or equivalently by K (kyo)l I Leap ) (kb2~Rl~ko) % ~~1k 4 ~L~k2 CHAPTER 5 WEAK NONLNEAR ANALYSIS INT PRECIPITATION: GEO1VETRIC EFFECTS ON ROUGHNESS Introduction In Chapter 2, linear stability methods were used to analyze the precipitation of a solid from a supersaturated solution. The physics of the instability in precipitation was explained in terms of diffusion and surface tension. We concluded that a steady planar interface becomes unstable and transforms into a nonplanar shape as the solute concentration in the supersaturated solution is increased beyond its critical value. In this chapter, we will be interested in examining what happens beyond the onset of instability. In doing so, our goal is to understand the early stages of the growth of surface roughness in precipitation. The notation used in this chapter is the same as that used in Chapter 2, and the schematic is again shown in Figure 5.1. reservoir maintained at Solid /I Solution c = cs Ia fixed concentration Moving front cSAT: Solute concentration in i, ;the liqulid in equilibrium wlith z c= EQ Zx csa paesufc z =Z (~t)n =7I cEQ Solute concetration in the liquid in equilibrium with cs at a curved surface c =cSAT z=L z=Z, =0 Z u = nI.u surface velocity 2H = xx ,surface curvature [Z, +1]3/2 Figure 5.1 A precipitation front growing into a supersaturated solution Again, c,, the solute concentration, is taken to be the control variable. Now to observe surface roughness, one might imagine running a series of experiments in which c, is increased to, and just beyond its critical value. In studying precipitation, we would like to know: Can the critical point be detected in an experiment? What is the nature of the branching to the new non planar solution? Is it a pitchfork? Is it forward or backward? Or is it a transcritical bifurcation? What is the effect on the mean front speed as c, just crosses its critical value? Is it higher or lower than the values predicted by the base formulas? Our plan is to begin by presenting again the nonlinear equations in scaled form, and working out the base solution and the solution to the perturbation eigenvalue problem to find the critical value of c,. To do this we follow the same steps as in Chapter 2, but nonetheless, we will rewrite the perturbation equations, now for o = 0, so as to formulate an eigenvalue problem. Having found the critical value of c,, we look for steady solutions just beyond this critical point by going to second and third orders in the amount by which c, is increased. We discover that the crosssectional shape of the precipitation cell is important in deciding the nature of the bifurcation. Ordinarily, a rectangular cross section leads to a backward pitchfork; a circular cross section with symmetric disturbances gives a transcritical bifurcation. Toward the end of the chapter, we deduce a symmetry condition that determines whether a cross section of an arbitrary shape will lead to a pitchfork or a transcritical bifurcation. Then some other specific cross sectional shapes are studied and it is found that a transcritical bifurcation is more likely to occur. Finally, regardless of the nature of the branching, it is concluded that the surface speed always decreases compared to what its base value would have been. Rectangular Cross Section The Steady State Nonlinear Equations The dimensionless nonlinear model equations are the same as given in Chapter 2. However, we set all terms to zero as our goal is to investigate the nature of the steady state solutions to the problem upon crossing the threshold. Again, all variables used here are scaled. In the liquid phase, the equation for the scaled solute concentration is given by V2 c+U = 0 (5.1) for Z < z <1 and 0 < x < W The equations along the interface, viz., along z = Z(x) are c = _l[2H] (5.2) and n .Vc = [1 c]Uk. n (_5.3) where 11 denotes 7 1c,, The solute concentration at the reservoir is given by L RT cs cs c,4, c(z = 1)= cL (5.4) Sc 8Z Side wall conditions are Neumnann, i.e., (x = 0, W) = (x = 0, W) = 0 Finally, the Dx Dx volume of the supersaturated solution must be maintained fixed, whence IZ(x) dx= 0 (5.5) Once a material is chosen, the physical properties are fixed. Once a cell is built, other variables such as L and W remain fixed. In running a sequence of experiments, c, is increased from one steady experiment to another. In each experiment, the mean interface speed, U, and the interface shape, Z (x), are the outputs. The Base Solution The base state variables are denoted by the subscript 0. A base state solution is given by cO(z)=1eo"" (5.6) where U, is given by U, = In(1 cL) (5.7) Besides the physical properties, the input variables in the base state are L and cL, and the output is LL,. The Eigenvalue Problem at Neutral Conditions On increasing cL, We expect the planar interface to lose its stability. To determine the critical value of cL let a small disturbance be imposed on the steady base solution at fixed values of cL and Ui,,where cL and UI, satisfy Eq. (5.7). Then expanding z = Z(x) as Z = Z, + EZ, Zo = 0 (5.8) the solute concentration at the surface is given by c = c, + +[ Ze," (5.9) where the term involving the surface displacement, Z,, has been introduced so that the perturbation problem can be solved on the base domain (cf., Appendix B. 1). To find the critical base state, the perturbation eigenvalue problem must be solved at zero growth rate, whence V cr, +U '= (5.10) on the domain, O < z <1, O < x < W, where, at z = 0, d2Z, dc c1 = 1 Z o (5.11) 1 ~21 dz and + +Uoc = 0 (5.12) must hold, and where, at the reservoir, z = 1, c, = 0 (5.13) must hold. The sidewall conditions require Sc, /Dx = 0 and dZ, /dxe = 0 at x = 0 and at x = W . Thus the problem is homogeneous in c, and Z,, and it has solutions of the form c, = c, (z) cos(kxr) and Z, = A cos(loc) (5.14) which satisfy the side wall conditions so long as k = nzr/W, n = 1, 2..., where k is the wave number of the disturbance and n = 0 is ruled out by the condition IZ,dx = (5.15) which holds the volume of the solution fixed. Equations (5.10), (5.12) and (5.13) imply that c, must be zero, and therefore, at z = 0, Z, must satisfy dc, d2Z, Z o = )1 (5.16) 1dz & ~2 where do(z = 0) = Uo Then Eq. (5.16) is satisfied for all values of A so long as Uo = ~ak2 (5.17) The corresponding neutral curve is a straight line as shown in Figure 2.7 (b) (cf., Chapter 2), where the ordinate must be replaced by I now. Given W, and hence a value of k Eq. (5.17) determines the critical value of the interface speed, U, The most dangerous value of k2 corresponds to n = 1 or to k = yir/W, hence only one crest is seen at critical. As the control variable cL increases upward from zero, the output variable 1 increases from zero, until there comes a point when Eq. (5.17) is satisfied. The corresponding value of U, will be referred as its critical value. Henceforth the subscript zero will refer to this critical state. At values of U higher than LL,, the base solution loses its stability. The critical value of cL denoted c~o COrresponds to the critical value of U via Eq. (5.7). Ordinarily then, one might increase the control parameter cL to its critical value, and then slightly beyond, in order to discover the new steady states that ensue. Assuming with some evidence that the bifurcation would be pitchfork in nature, one can only hope to get a locally forward pitchfork by advancing the control variable beyond its critical value. However it turns out that our hope of finding a forward pitchfork is not realized upon proceeding in this way. The branching to nonplanar steady solution is, in fact, a backward pitchfork. Hence we introduce 1 E as an input giving the amount by which cL is decreased from its critical value (note that e will no longer be used as a measure of the amplitude of a disturbance as it was in the eigenvalue problem). Now the new subcritical solution will not have the speed of the corresponding base solution, hence we first find the change in the base speed corresponding to a given change in cL ' Eq. (5.7) indicates that for a small decrease in cL fTOm c~ the corresponding change in U from LL,, per unit change in cL is given by e " The First and Higher Order Problems To Eind the steady backward branches, we write 1 where E2 is an mnput. We then expand U and Z as 1 1 U = U, + EU, +EU2 +E U3 + ... (5.19) 2 6 and 1 1 Z = Z, +EZ, + E Z + _EZ,+ ... (5.20) 2 6 Using the formula for the surface normal vector given in Figure 5.1, and observing that the factor JZ1 in the denominator cancels from both sides of Eq. (5.3), only the numerator of n.U is needed, where U = Uk Hence the expansion of n.U is given by Eq. (5.19). This and the other expansions (cf., Appendix B.2) lead to the respective problems at first and higher orders in e . The First Order Problem The first order problem differs from the eigenvalue problem in that the system now must respond to a change in cL and this may lead to a change in U, whence, at first order, we must determine cy, Z, and also U The first order problem is given by Sc, dc, V c, + UO = Uz (5.21) 8: dz on the domain, O < z <1, O < x < W, d Z, dc, cl = 1 Z, (5.22) dx~C dz and + +Ep, = U, (5.23) at z =0, c, = 0 (5.24) at z =1, and IZ,dx = (5.25) The sidewall conditions require dc, /Dx and dZ, /dx to vanish at x = 0 and at x = W Now Eqs. (5.21), (5.23) and (5.24) have a particular solution depending only on z ,while the corresponding homogeneous problem is just the eigenvalue problem whose solution is c, = 0 . Hence we have c, = Uz [z 1]e"o= (_5.26) and we can turn to Eq. (5.22), which is a differential equation for Z, Using c, (z = 0) = U,, along with Eq. (5.25) and dZ,/dxL~ = 0 at x = 0, W and integrating Eq. (5.22) over 0 < x < W, we find that U, must be zero. Hence c, must also be zero, whereupon Eq. (5.22) must have the solution Z, = A cos(kx) (5.27) What we have so far is this: c, = 0, U, = 0 Z, = A cos(kx) And we go on to the second order problem to find the value of A The speed of the surface does not change at order e upon 1 decreasing cL fTOm cL. by 2E The Second Order Problem At second order, the new unknowns to be determined are c2, Z2 and U2 and the unknown A which carries over from first order. The second order problem is V2 2 +U = U o (5.28) 0 Z 2 dz in the liquid solution, O < z <1, O < x < W, 2d2c, d2Z, dc, c2 + Z,2 0 Z2 (5.29) dz 2 LC2 dz and dec 2+ Uoc2 = U2 (5.30) at z =0, c2 = 1 (5.31) at z =1, and IZ~dx= 0 (5.32) The sidewall conditions require Sc2le dZ2 /dx~ = 0 at x = 0 and at x = W Solving Eqs. (5.28), (5.30) and (5.31), we get c2 = [U2 Z 1Uo Uoz (5.33) And it remains to solve Eqs. (5.29) and (5.32) for Z2 To find Z2, a solvability condition must be satisfied. It is c2Z +,2 Z =0)cos (kx) dx= 0 (5.34) 02 and it is satisfied, no matter the value of A, due to cos (kx) dr cos kx) d = (5.35) Hence A cannot be found at second order, but we can find U, and Z, To get U,, integrate Eq. (5.29) over 0 < x < W, and use Eq. (5.32), along with dZ,/dx~ = 0 at x = 0, W The result is U, = ec. _UA2 (5.36) where the first term is what U, would be if the base solution obtains at the new value of cL This term was derived earlier. The second term is the correction corresponding to a curved front. It is negative. If a curved front could exist, its speed would be slower than that of the corresponding planar front. In other words, the mean speed of the interface runs behind its base value. Before moving on to solve the third order problem, we would like to emphasize that an analogous weak nonlinear calculation will be performed in solidification (cf., Chapter 6) where we will demonstrate that once the instability sets in, the mean speed of the solidification front always runs ahead of its base value. In an attempt to delineate the key disparity between the two problems, differences will be traced to difference in the interfacial mass balance in one versus the energy balance in the other. Now to go to third order, we need to solve Eq. (5.29) for Z, Substituting for c, (z = 0) and U,, the constant term disappears, and we see that Z, can be written as Z, = Z, cos(kx)i + cos(2kx) (5.37) where, while Z, cannot be determined at this order, Z, is given by Zz = 2" (5.38) Hence, Z, and A carry on to third order. The Third Order Problem The equations at the third order are given by dec dc V c, + UO = U3 (5.39) on the domain, O < z <1, O < x < W, dc, d2c d3c dZ, d Z,dZ d cz +3Z, + +3ZZ2 +Z: 9 : = ~aZ, (.40 dz dz2 dz~C dx x dx dz and dc, + Uoc3 U3 (5.4 1) at z =0, c, = 0 (5.42) at z =1, and Zmdx= 0 (5.43) The sidewall conditions require dc, /8x = dZ, /dx~ = 0 at x = 0 and at x = W Again, c ,, like c, and c, is only a function of z, and it is given by c, Us z 1 emo= (_5.44) which satisfies Eqs. (5.39), (5.41) and (5.42). This leaves Eqs. (5.40) and (5.43) to determine U, and Z,. Now Eq. (5.40) is a differential equation for Z,. And again, a solvability condition must be satisfied if Z, is to be solved for. This condition is dc, dc dcd, d Z\17 c, (z= 0)+ 3Z, ((z=0)+ 3Z,Z, "(z =0)+ lTL~ Z3 "( )9 ~ Lcosk2d wTOLdz d: dz' (= sa dx, \ dxz ,(,,= (5.45) and it is an equation for A Substituting our results at orders zero, one and two, we find A2 to be given by 6eU" A2 (5.46) U ,r [,3 1 U Ordinarily, ~a is of the order of 109, and Un is of the order of 0. 1. Hence A2 is positive and a backward pitchfork is confirmed. The value of U, can be found in terms of A2 by integrating Eq. (5.40) over 0 < x < W, and using Eq. (5.43), along with the side wall condition, dZ, dx~ = 0 at x = 0, W . We started the nonlinear calculation by guessing the expansion to be cL Lo .An the reason we find a pitchfork branching in the case of a onedimensional rectangle is because at second order, solvability does not determine A This, in turn, is due to the fact that the eigenfunction cos (kx) satisfies n cos3 (kx) dx = 0 Had the solvability condition determined A at second order, it would have determined A = 0, and the hypothesis of a pitchfork bifurcation would have failed. If the cross section were a twodimensional rectangle, a pitchfork would also have been found due to cos' (la)cos (ky) dxdy = 0 Indeed a twodimensional rectangle, which is thin no enough such that there is no y variation, reduces to the onedimensional case presented above. But there is no reason to believe that an eigenfunction f of the Laplacian operator, V on an arbitrary cross section, satisfying Neumann conditions at the edges, should satisfy the condition Sf 'dA = 0 Hence there is reason to imagine that for some cross section, we can determine the Area value of A to be zero at second order, whence we will have to choose a different expansion, and the branching will not be a pitchfork. To see this, we work out the case of a circular cross section with axisymmetric disturbances. Circular Cross Section with Axisymmetric Disturbances Now let the precipitation cell have a circular cross section of radius R To demonstrate the existence of a transcritical bifurcation, we will focus our attention on axisymmetric displacements, which may not be the most dangerous; a more general case allowing for the non axisymmetric disturbances is also very interesting and it is discussed in the endnote to this chapter. Since the base state variables depend only on z ,the base state remains the same as it was in the case of a rectangular cross section. The perturbation eigenvalue problem is nearly as it was before, only the curvature is a little different, hence the equilibrium condition at z = 0 is now c, =)1 r Z, "c (5.47) r dr dr dz where the B dependence of the variable Z, has been ignored. The fixed volume condition is now given by Z7,rdr= 0 (5.48) and the sidewall conditions require that Sc/8r and dZ/dr vanish at r = R, whence Sc, dZ, 0 and = 0 at r =R (5.49) dr dr The remaining equations lead, as before, to c, = 0, whence Eq. (5.47) has the solution Z, = l, (kr) (5.50) where k can be any positive root of Jz (kR) = 0, and where the neutral curve is again given by U, = ak2 Hence, given R ,the critical value of U, is determined by the smallest positive root of J (kR) = 0, k = 0 being ruled out by Eq. (5.48). Again, we start with the guess that the steady branches leave the critical point as 1 cL L" 2 +E Then using the expansions given in Eqs. (5.19) and (5.20) we find at first order that c, = 0, U, = 0, and Z, = AJ,(kr) At second order, the solvability condition is not satisfied unless A3 is zero, and therefore Z, is zero. And going to higher orders, Z,, Z, all turn out to be zero, indicating that our original guess for expansion of cL WaS incorrect, and also indicating that the expansion cL = L, + E (5.51) will work out. Introducing F as cL Lo and retaining Eqs. (5.19) and (5.20), the eigenvalue problem remains as before. It is written holding U = U,, and it gives us the critical value of U, in terms of k again U, = ak2 where the subscript zero henceforth denotes critical values. The surface noral ectr i nw gvenby =Z, i +i Then observing that the denominator J~i cancels from both sides ofEql. (5.3), only the numnerator of n.U is needed, and its expansion is given by Eqs. (5.19). This leads to the problems at first and higher orders in e, and we solve them as c, increases by e beyond its critical value. The First Order Problem The first order problem is dec dc, V2C 1 +U = U o (5.52) a z dz on the domain, O < z <1, c, = 11 r ac (5.53) r dr dr dz and dec S+ Uoc1 = U, (5.54) at z =0, c = 1 (5.55) at z =1, and Z,rdr= 0 (5.56) The sidewall conditions require that dc, /Br and dZ, /dr vanish at r = R Again, c, can depend at most on z and it is given by c, = U, [z ]+eL'o] Lvo (5.57) which satisfies Eqs. (5.52), (5.54) and (5.55). Eqs. (5.53) and (5.56) then determine both U, and Z, .Multiplying Eq. (5.53) by r ,integrating the result over 0 < r < R, and using Eq. (5.56) along with the side wall condition, dZ,/dr = 0 at r = R, there obtains jc, (z=0)r dr =0 (5.58) whence Uz = ev (5.59) whereupon U, is no longer zero and the expression for c, can be simplified to c1 = zeUogvoz (5.60) The fact that c, and U, are not zero is another sign that a circular cross section differs from a rectangular cross section. However, c, is zero at z = 0, and the solution to Eq. (5.53) is Z, = Mo,(kr) (5.61) but A cannot be determined at first order, and we turn to the second order problem. The Second Order Problem At second order, we must solve dec d, dec V2C 2 +U2 = U2 o 2U (5.62) 0 Z 2 dz 8z on the domain, O < z <1, d d2 0 1~ Id dZz dc, c+ +2 Z2 2 (5.63) 2 1dz dz 2 dr dr 2dz and dec 2+ Uoc2 = U2 (5.64) at z =0, c2 = 0 (5.65) at z =1, and Z~~rdr = 0 (5.66) The sidewall conditions require Sc, /Sr = dZ,/ldr = 0 at r = R. Solving Eqs. (5.62), (5.64) and (5.65), we get and it remains to solve Eq. (5.63) for Z,, while satisfying Eq. (5.66). It is our hope that this will give us the value of A, as well as the value of U,. Now Eq. (5.63) can be solved for Z, if and only if a solvability condition is satisfied. It is jc,(z=0)+2Z, (z==0)+Z (z =0 J(rdr0(.8 and this is an equation for A Substituting our results at zeroth and first orders, we get A (5.69) whereupon using the formulas given in Table 5.1, we get A = eU11. (5.70) Table 5.1 Some integrals concerning Bessel's functions rf~\~ (kr dr R /I (kR) where Jz, (kR) = rF kr) dr~\ n =~ R ~ 0.02) whre J (kR) = Hence the branching to the new steady state is not a pitchfork; it is transcritical. Multiplying Eq. (5.63) by r ,integrating the result over 0 <; r < R and using the constant volume conditions, Eqs. (5.48) and (5.66), and side wall condition, dZ,/dr = 0 at r = R, we get R2R c2 (z = 0) 2= U rZ drUI (_5.71) whence U.? = e "' U J (kR) A2 (5.72) The first term on the right hand side, eU"', is what U, would be if the base solution obtains at the new value of cL The second term is the correction corresponding to a curved front. It is again negative, as it was for a rectangular cross section, and we conclude again that the mean speed of the interface runs behind its base value. What we have found so far is this: the nature of the branching to nonplanar steady states is ordinarily a pitchfork for a rectangular cross section, while it is transcritical in the case of a circular cell with axisymmetric disturbances. The cross section dependence raises the question: How can we predict the nature of the branching for an arbitrary cross section? It is to this question that we now turn. A Cross Section with an Arbitrary Shape Let the cell have an arbitrary cross section, but we will work in Cartesian coordinates. The nonlinear equations, Eqs. (14) remain the same, but the Eixed volume condition is now given by ~Z (x,y)dx y = 0 (5.73) where z = Z (x, y) defines the shape of the solidliqluid interface, and where the integral is carried out over the cross section S The mean curvature of the surface z = Z (x, y) is given by rl1+ZnZ 2ZxZZ + 1+Z Z,"1 2H = 32(5.74) 1+Z: + Zz1 The sidewall conditions are now p.Vc = 0 and p.VZ = 0 on aS (5.75) where aS denotes the boundary of S, and p denotes the outward normal to aS. The solution to base state equations, which depends only on z ,is given by Eqs. (5.6) and (5.7). The Eigenvalue Problem at Neutral Conditions The eigenvalue problem at neutral condition does not change much. It is given by V2 1 + Uo I= (5.76) on the domain, O < z <1, where, at z = 0, c, + Z, a= ~aV2Z, = a + 5.7 dedz 2 1x 2 27 and dec S+ Uoc, = 0 (5.78) must hold, and where, at the reservoir, z = 1, c, = 0 (5.79) must hold. This problem has solutions of the form c, = c, (z) f (x, y) and Z, = Af(x, y) (5.80) whereI f (xY y)is anl eigefjluncionLV of V2 On S subject to Neumann sidewall conditions. It satisfies p.Vf = 0 on dS V f (x, y)= k f (x, y) on S; (5.81) Hence, denoting the eigenvalue by k the solution f = const., k2 = 0 is ruled out by the condition ~Z,(x, y) dx dy= (5.82) Eqs. (5.76), (5.78) and (5.79) imply that c, must be zero, and therefore Z, must satisfy dc dz, where dc'(z = 0) = Ui,. And we Eind that the neutral curve is again given by U, = ak' It holds dz regardless of the shape of the cross section. But the allowable values of k2 do depend on the geometry. Again, the lowest positive value of k2 determines the critical speed of the front. 1 Let us begin by assuming the branching to be a pitchfork, and expanding cL as c~o 2 2 The expansions for U and Z are given by Eqs. (5.19) and (5.20) and we move on to solve the problem at the first and higher orders to Eind the steady subcritical solution. The First Order Problem The first order problem is given by V c, + UO dc = e U, ''(5.84) 8: dz on the domain, O < z <1, Sdc, c, = ,aV Z, ''Z, (5.85) dz + +Up, = U, (5.86) and ~Z,(x, y) dx dy= (5.87) at z =0, and c, = 0 (5.88) at z =1. Eqs. (5.84), (5.86) and (5.88) have the solution c, = Uz [z]ewo= (_5.89) Using this in Eq. (5.85), we find from Eq. (5.87) that U, must be zero, and hence c, must be zero, whereupon Z, is given by Z, =Af (x, y) (5.90) where f (x, y) is an eigenfunction of V', on S, satisfying Neumann conditions on dS and it corresponds to the critical, or lowest, value of k2, with k2 = 0 ruled out. We go to second order to find the value of A but we already see that the speed of the interface does not change at order 1 e upon decreasing cL fTOm cL. by E. The Second Order Problem At second order, we must determine c., Z, and U,, and the unknown A which carries over from first order. The second order problem is VBc3 dcu Vcz + Uo = U (5.91) 8: dz on the domain, O < z <1, ,d2c dco c + Z, =~aV,Z "Z, (5.92) dz2 dz dec 2+ Uoc2 = U2 (5.93) and ~Z2(x, y) dxdy= (5.94) at z =0, and c2 = 1 (5.95) at z = 1. Solving Eqs. (5.91), (5.93) and (5.95), we get c2 = U122 Uo Uoz :0 (5.96) and it remains to solve Eqs. (5.92) and (5.94) for Z2 To find Z2, a solvability condition must be satisfied. It is c2I Z ) 2C Z )(,y)dd (5.97) whence we have A23 (x, y)dtxdyi= (5.98) At this point, there are two possibilities. One possibility is that the integral vanishes, whereupon solvability is satisfied no matter the value of A And as in the case of a rectangular cross section, we go on to third order to find A This is the pitchfork case. The second possibility is that the integral does not vanish and we must conclude that A is zero, whence the branching is not a pitchfork and we must propose a new expansion in F, cf., [29]. Cross Sections on which the Integral Does Not Vanish If the integral Jf3 (x, y)dxdy is not zero, as in the case of a circular cross section, we assume the branching is transcritical and propose to move beyond critical via the expansion cL = cLo + E (5.99) The base problem and the eigenvalue problem remain unchanged. But the first order problem is now dec dc, V2C 1 +U = U o (5.100) d z dz on the domain, O < z <1, Sdc, c, = aV2,z, aZ, (5.101) dedz S+ Uoc1 = U, (5.102) and Z, (x, y) dx y = (5.103) at z =0, and c = 1 (5.104) at z = 1. Eqs. (5.100), (5.102) and (5.104) have the solution c, = Uz[z 1]+ evo ecoz (5. 105) whereupon integrating Eq. (5.101) over S and using Eq. (5.103), we have Uo = ~ak2 (5.106) and c, simplifies to U, U c1 = ze oc (5.107) whence c, must be zero at z = 0, and Z, is then Z, = Af (x, y) To find A we go to second order, where we must solve (5.108) (5.109) V2C 2 +UO 23 = U o ~ 0 Z 2 dz on the domain, O < z <1, d, ,d2 , c2 + 2Z, + ,2 0 dz dz ac, 2U ' 8z Sdco 2a,Z2 0Z2 dz (5.110) (5.111) 2+ Uoc2 = U2 and ~Z2 (x,y) dxdy = 0 at z =0, and c2 = 0 at z = 1. Eqs. (5.109), (5.111) and (5.113) are satisfied by (5.112) (5.113) (5.114) and it remains to solve Eq. (5.110) for Z2, While satisfying Eq. (5.112). The solvability condition for Eq. (5.110) is 0)+~ Z2 Z 0) + 2Z, (z (5.115) This is an equation for A in which U2 does not appear and we find c2=Z Z82euUo +eU2 2Uo U2 U c2 (Z 0) f(x, y) dxdy= 0 A = r e"IS~'c,>~s (5.116) UZ Sff (x,y) dx dy: whence a transcritical bifurcation is verified. The value of U, obtains upon integrating Eq. (5.110) over S to get Uj = e24 UA2 f1 S'(x, y)dxdy (5.117) where the first term on the right hand side gives the value U, if the base solution persists at the new value of cL The second term is a correction due to the displacement of the front. It is always negative, predicting a slow down. Thus if the integral lf' (xw, y:)dxdy is not zero, the branching is transcritical and the front slows down. Cross Sections on which the Integral Vanishes If we have a cross section where the integral ~f (x, y)dxdy: is zero, then A cannot be determined at second order, but the solvability condition for Eq. (5.92) is satisfied no matter the value of A . Our j ob then is to solve Eq. (5.92) for Z, and then go to third order hoping to determine A To do this we first obtain U? by integrating Eq. (5.92) over S, using Z, = A f(x, y), c? (z = 0) = U. e"", Neumann side wall conditions and Eq. (5.9)4). The result is U, = e"o U f, (x Axy 518 whence we have cz (z =0) = UA' ~Jf(x, y) dx dy Sg and Eq. (5.92) is then  xydxy f(, ) SaJJ\,l~y ~y (5.119) [V + k? Z, (5.120) whereupon Z, can be obtained as a series in the eigenfunctions of V~ on S subject to Neumann boundary conditions, viz., Z2 = Bf + B, Rf, (5.121) where the coefficients B,, depend on the expansion of the left hand side of Eq. (5.121) and where the eigenfunction 1 does not appear. The equations to be solved at third order are dc, U 0 B'c, II, c V c3+U (5.122) on the domain, O < z <1, dc, d2c c, + 3Z, + 3Z,Z" + dz dz ,d3c Z, "+~aN(Z,) dz dz (5.123) (5.124) Sc, and ~Z, (x, y)dxdy = 0 at z =0, and c, = 0 (5.125) (5.126) at z = 1 where c, = 0 = U, has been used and where N(Z,) denotes the nonlinear part of 2H3.I depends on Z, and introduces the factor A3 Eqs. (5.122), (5.124) and (5.126) can be solved for c3, ViZ., c3 U3 Uoz"" (5.127) leaving Eqs. (5.123) and (5.125) to be solved for U3 and Z3 Where Z3 can be obtained if and only ofa solvability condition is satisfied. It is wrc3 Z=0)+ 3Zd, (z =0)+ 3Z,2, Z0 )+Z,\1 3 dZ=0+NZ)fxydd (5.128) and it is an equation for A2 .If A2 is not zero, the branching is a pitchfork, as assumed, and if A2 is positive the pitchfork is backward, as assumed. The equation for A2 can be simplified somewhat without making S definite. Using d2c d3C dc,=[+U]U ~U ehv Z, =Af (x, y) and, at z =0, ar = Ui a U ad 2=[+U 2+ ov ehv 3A [1+UojU2Ti +Uoie""o 2fhdxd 3UfAnT f.2 Z2,auardxdy + A3Uo'T f4 Ldy +A N(Z, fddy =0 ss s(5.129) where Z2,parhcular iS a multiple of A2, Where N(Z, ) is a multiple of A3 and where [1+UjU2+Uoco s gvenby evo [+Uo~fA 2r ,(x, y)dxdy Hence the term in A S  does not vanish and a pitchfork is confirmed. Thus if the integral Slf3 (x, y:)dxdy is zero, the branching is pitchfork and the front slows down. An Equilateral Triangular and a Regular Hexagonal Cross Section We explore two other cross sections: an equilateral triangle and a regular hexagon. Appendices C.1 and C.2 give the solution to the eigenvalue problem, Eq. (5.81), for these geometries. In the case of a regular hexagon, we find f3~~ (, dyhc~i= 4 L24 f 0, whence the branching is transcritical. The branching is found to be transcritical on an equilateral triangle as well. Concluding Remarks We have given a condition that determines the nature of branching of a steady planar crystalsolution interface to a nonplanar shape for an arbitrary cross section of a precipitation cell. Given any cross section of the cell, if one can find the solution f (x, y) to the eigenvalue problem, Eq. (5.81), then the nature of the branching to the steady nonplanar solution can be predicted by finding whether or not the integral If3 (x, y;)dxdy3 is zero. The branching is usually a pitchfork if this integral is zero; if not, we expect a transcritical bifurcation. We find that for a rectangular cross section, the branching is ordinarily a backward pitchfork; it is transcritical for all other shapes that we have studied: circular (symmetric disturbances), triangular (equilateral), and hexagonal (regular). And it is our belief that transcritical bifurcation is what will be found in the case of a more general cross section. We also find that the surface speed always decreases compared to what its base value would have been. Endnote 1: A Circular Cross Section with the Possibility of Nonsymmetric Disturbances Let us consider the possibility of nonsymmetric displacements. If nonsymmetric disturbances are admitted, the critical value of k2 is given by the smallest positive root of Jz (kR = 0 (5.130) and this eigenvalue corresponds to the eigenfunction f = J(kr)cos8 (5.131) where R 2x2 r Fz3 (kr) cos30d~rd6 = 0 (5.132) Hence the branching must be a pitchfork. Because nonsymmetric disturbances are more dangerous, the branching in the case of a circular cross section must be a pitchfork. CHAPTER 6 SOLIDIFICATION: WEAK NONLINEAR ANALYSIS Introduction We concluded in Chapter 3 that a steady planar interface becomes unstable and changes to a nonplanar shape as the undercooling in the subcooled liquid is increased beyond a critical value. Linear stability methods were used to analyze the shapes of the various disturbance growth curves. In addition, the shape of the neutral curves showed that as many as three neutral points could be obtained, each depending upon the longitudinal and transverse and sizes of the solid and liquid phases. The occurrence of three possible neutral points was explained in terms of longitudinal diffusion, transverse diffusion and surface tension. In this chapter, we are interested in examining the postonset regime in solidification of a pure material. We shall use the techniques learned from our analysis of the precipitation problem in the vicinity of the instability. Again, our goal is to understand the early stages of the growth of surface roughness. The notation used in this chapter is the same as that used in Chapter 3. The liquid phase sink temperature, T,, is taken to be the control variable. The idea is to imagine running a series of experiments in which T, is decreased to, and just below its critical value. We try to answer questions like: What is the effect on the mean front speed as T, just crosses its critical value? In other words, does the front speed up or slow down compared to its base value? Is the branching to the new nonplanar solution a forward or a backward pitchfork? We will consider three maj or cases. First, we consider a case where the latent heat is rej ected only though the subcooled liquid while the solid phase is held at the melt temperature; both phases are taken to be of finite extent. Second, we consider a simplification of the first case where the solid phase is taken to be of infinite extent. This may be useful when the solid sink is far away from the interface. Third, we consider the most general case where the latent heat is rej ected though the frozen solid as well as the subcooled liquid and both phases are taken to be of Einite extent. In all of these cases, we begin by presenting the nonlinear equations in scaled form. We work with a rectangular cross section and all derivations are done in Cartesian coordinates. We first work out the base solution and the solution to the perturbation eigenvalue problem to Eind the critical value of TL Having found the critical value of TL We l00k for steady solutions near this critical point by going to second and third orders in the amount by which TL is decreased from its critical value. All derivations are done in a manner similar to Chapter 5. Therefore the details of the derivations will not be presented in their entirety as they are straightforward enough that a reader can easily reproduce them. Our emphasis will be to point out important differences from the precipitation problem. In all of these cases we will find that the mean speed of the surface is higher than its base speed. This is to say that once the instability sets in, the surface speeds up compared to its base value. This result is striking in its contrast to what was obtained in Chapter 5 for precipitation, where we found that the surface unconditionally slows down. The obvious question is: Can we explain this basic difference of a "speedup" in solidification versus a "slowdown" in precipitation? In our quest to answer this question, we will go to a yet simpler model in a fourth case where the solid phase is completely ignored. It is as if the solid phase is there only for the sake of having an interface, but it is ignored for all other purposes such as transport etc. One might also look at it as if the solid is being chopped off as soon as it forms. We do not learn anything new, and again the front speeds up compared to its base value. But the model (although an artificial one) is now as close as one can get to the precipitation model, and the key disparity between the two problems becomes clear. We learn that the difference is posed by the heat balance (in solidification) versus mass balance (in precipitation) at the interface. And this is what leads to the essential dissimilarity between the two problems regarding the correction to the front speed. Now in order to learn the nature of the branching to the new nonplanar solution, one must go to third order in the amount by which the undercooling in the subcooled liquid is increased. In general, we will find that one can obtain a forward as well as a backward pitchfork depending on the input parameters. We will identify regions in the input parameter space where the branching is forward and regions where it is backward. To do all of this, let us start by performing a weak nonlinear analysis for a case where the frozen solid is maintained at its melt temperature and the latent heat is rej ected only to the subcooled liquid, both phases being of Einite extent. Case 1: Latent Heat Rejected Only to the Subcooled Liquid, Both Phases of Finite Extent solid sink Solid Liquid liquid sink T=T,  IT=TL moving front II z =S z = Zo=0 z=L U = nI. U = staface velocity JZ~ 2H = staccrate [Z +1]J s~fc uvli Figure 6.1 A moving solidiaication front The sketch given in Figure 6.1 Eixes the principal ideas. The liquid and solid phase depths are maintained constant and all equations are written in a frame moving with a constant velocity, U = U k where U is the base state speed of the solidification front. All variables introduced are movingframe variables; the superscript denotes a liquid phase variable. All lengths are scaled by S, the depth of the solid phase. All speeds are scaled by a / S where a is the thermal diffusivity of the solid phase. Temperature is measured from a reference point corresponding to T,, and it is scaled by aLH / ii, Where ii is the thermal conductivity of the solid phase, and LH T T denotes the latent heat of solidification per unit volume of the solid, viz., Tscaled M. Then, atLH the scaled temperature in the solid phase must satisfy dT V2T +U = 0 (6.1) for 1< z < Z, while in the liquid phase dT* V2 +U = 0 (6.2) must hold for Z < z < L Two demands must be met along the solidliquid interface, viz., along z = Z(x) First, phase equilibrium, taking into account the curvature due to the interface, requires T= Jl2H]= T* (6.3) where 11 denotes r ts1T,, a dimensionless parameter introduced by our scaling, and where y denotes the surface tension. Second, the rate at which the latent heat is being released due to the motion of the solidification front must be the same as the rate at which heat is being conducted away from the interface, viz., norz.VT .V' =nU (6.4) FarHield conditions are given by T(z = 1) = 0 (6.5) and T(z = L) = T, (6.6) dT dT' 8Z Side wall conditions are Neumann, i.e., (x = 0, W) = (x = 0, W) = (x = 0, W) = 0 . Dx 8x Dx Finally, the volume of the subcooled liquid must be maintained Eixed, whence Sz (x) dx = (6.7) Notice that these model equations are nonlinear as the position of the interface depends upon the temperature gradients there, and these gradients, in turn, depend upon the position and morphology of the interface. This nonlinearity forms the heart of the problem without which instability would not be possible. The Base Solution The base state is one in which the solidiaication front remains planar and at rest in the moving frame. The base state variables are denoted by the subscript 0, and they satisfy d2T, dT o+ U, =0 (6.8) dz 2 0dz in the solid phase, 1< z <0, < x d2T' dT O +U, O = 0 (6.9) dz 2 0dz in the liquid phase, O < z < L, O < x < W The equations at the planar interface, z = 0 are given by To = To' = 0 (6 10) and dT, dT' = U (6.1 dz dz The farfield conditions are given by T ( z = 1) = 0 (6 1 and To'(z =L) =T, (6.1 Then, a solution to Eqs. (6.86.13) is given by To(z) = (6.1 To*(z) =1 ecoZ (6.1 where Uo is given by Uo = In(1 T,) (6.1 dT' d2T' d3T' d4 * whence o (z = 0) = U,, a (z = 0) = Uf, a (z = 0) = U~ a (z = 0) = Uf. Beside dz dz2 dz3 dz4 the physical properties, the input variables in the base state are S, L and T,, and the output is 1) 2) 3) 4) 5) 6) es Uo  The Perturbation Eigenvalue Problem On increasing T,, we expect the planar interface to lose its stability. To determine the critical value of T,, let a small disturbance of amplitude E be imposed on the steady base solution at fixed values of T, and Uo, which satisfy Eq. (6.16). Then expanding Z, T, and T*, at z =Z (x) as Z = Zo +EZ, Zo = 0 (6.17) T= Toeii (+Z,d (6.18) and T'=To+e Z,, (6.19) T*=7 +E 7 dz1 where the term involving the surface displacement, Z,, has been introduced so that the perturbation problem can be solved on the reference domain (cf., Appendix B.1). To find the critical base state, we must solve the perturbation eigenvalue problem. It is given by V2 ;+ U,"o =0(6.20) on the domain, 1< z <0, < x V 2 + Uo I = (6.21) on the domain, O < z < L, O < x < W, where, at the base surface z = 0, = A~l~ Z (6.22) j Z = (6.23) and +Ii Z2 o = (6.24) must hold, and where, 7(z = 1)= 0 (6.25) and 7* (z = L) = 0 (6.26) The stricken terms are shown to emphasize that for this case the solid temperature is taken to be uniform in the base state. The sidewall conditions require 84/8x~ = 0 84'/8x~ = 0 and dZ, /dx~ = 0 at x = 0 and at x = W Thus the problem is homogeneous in 7;, 7;* and Z,, and it has solutions of the form ;= 7(z) cos(kx), 7;' = 7;' (z) cos(kx) and Z, = A cos(kx) (6.27) which satisfy the side wall conditions so long as k = nzr/W, n = 1, 2..., where k is the wave number of the disturbance. n = 0 is ruled out by the condition SZ~dx = (6.28) The xpasion prsentd aovegive the equations for~ I (' Z, They are k2 d7+U = 0 (6.29) in the solid phase, 1< z <0, < x d2 ^ dT k +o I= 0 (6.30) in the liquid phase, O < z < L, O < x < W, (6.31) 7;= Ik2 1 (6.32) S7 7 + ,[ dT5= (6.33) and (6.34) 7(z and 1)= 0 (6.35) 7*(z = L) = 0 Defining m (6.36) 2 ,a solution to Eqs. (6.296.33, 6.35, 6.36) is given by 1 e() em, z [m.]mz Z (6.37) ,~ e 1m,m]e z e~,m ]Le= z i (6.38) We can then turn to Eq. (6.34), which gives us  ak2 M ak2 2] U = 0 (6.39) m_ m emm m_ m e[m m,]L where M~ = > 0 and N = <0O Figure 6.2 shows the corresponding 1em.m 1e[m m]L neutral curve. The graph of Uo versus k2 is nearly a straight line. In fact it is easy to prove the following asymptotic limits for small values of k2 : For small k2, Eq. (6.39) may be simplified to dT; dT' d2~ d2 )*; dz dz 2dz 2 at z =0, ;(z) = A, em~z e[ mm]e m dT ~ak21 0 X (Z = 0) 1[ ~[m m]L e * (z) = A. [em get UI, = ak' [1+ L] at neutral while for large values of k we get 1 = 2Lak This gives us some idea about the shape of the curve without having to actually plot it. Figure 6.2 Typical neutral curve for solidification when the latent heat is rej ected only through the subcooled liquid Now given W and hence k Eq. (6.39) determines the critical value of the interface speed, U, The most dangerous value of n is one, which corresponds to k = yr/W, giving only one crest and one trough at critical. As the control variable TL increases upward from zero, the output variable U, increases from zero, until there comes a point when Eq. (6.39) is satisfied. The corresponding value of L will be referred as its critical value. Henceforth the subscript zero will refer to this critical state. At values of U higher than LL,, the base solution loses its stability. Now on increasing TL to its critical value T~ and then slightly beyond, it might be thought that the branching to the new steady solution would be a forward pitchfork. Hence we now introduce E2 as an input giving the amount by which TL is increased from its critical value. Observe that E takes on a different meaning from earlier. Now the new supercritical solution will not have the speed of the corresponding base solution, hence we first find the change in U corresponding to a given change in T,. Eq. (6.16) indicates that for a small increase in T, from Tro, the corresponding change in U from Uo, per unit change in T, is given by eUoL Before moving on to higher order problems, observe that neither i; nor 1' is zero. It is worth noticing here that in the analogous precipitation problem in Chapter 5, the variable c, was found to be zero. This, in turn, was due to the fact that there were several cancellations of mappings in the interfacial mass balance equation, viz., n .Vc = [1 c] Uk. n. The corresponding cancellations do not occur in solidification in the interfacial energy balance equation, viz., nV.V n.VT' = nl/.Uk.Not only does this make the analysis more cumbersome in solidification, but it also leads to a very fundamental difference between the two problems: Once the instability sets in, the front slows down in precipitation, while it speeds up in solidification. And to prove the fact of a "speedup" in solidification, we move on to solve the higher order problems in E . The First and Higher Order Problems We begin by making a guess as to how the steady states emanate from the planar morphology, i.e., how do the steady branches leave the critical point. Thus, to find the steady forward branches, we write 1 T, = Tq + 62 (6.40) where E2 L TU (6.41) is an input. We then expand U and Z as 1 1 U = U, + EU, +EU2 +E U3 + ... (6.42) 2 6 and 1 1 Z = Z,+ EZ, + EZ + _EZ,+ ... (6.43) 2 6 This guess is known to be correct in some other phase change problems (cf., Q. Buali, L.E. Johns and R. Narayanan) [30] and we begin by seeing if it works here. Using the formula for the surface normal vector given in Figure 6.1, and observing that the factor J in the denominator cancels from both sides of Eq. (6.4), only the numerator of n;.U is needed, where U = Uk The expansion of n.U is given by Eq. (6.42). This and the other expansions (cf., Appendix B.2) lead to the respective problems at first and higher orders in e. The First Order Problem The first order problem differs from the eigenvalue problem in that the system now responds to a change in TL and this may require a change in U at first order, whence, at first order, we must determine 7;, (', Z, and also U The first order problem is given by V (; + U, = UI (6.44) on the domain, 1< z< 0, < x 8: dz on the domain, O < : < L, O < x < W, d2Z, 7;= 11a (6.46) dxC2 d 7 7; + Z = 0 (6.47) and + Z U (6.48) 8z 8z 2 dz 21 at z =0, 7(z = 1)= 0 (6.49) and 7* (z = L,)= 0 (6.50) The volume condition is given by SZ,dx = (6.5 1) The sidewall conditions are Neumann. Now Eqs. (6.44, 6.45, 6.4751) have a particular solution depending only on z and the corresponding homogeneous problem is just the eigenvalue problem. Hence we have solutions of the form 7;(x, z)= 7;a(z)+ 7, (z)cos(kx) (6.52) 7* (x z)= 7;0 (z)+ 7'F(z)cos(kx) (6.53) Z, (x) = A cos(kr7)= Z, cos(kx?) = Z,, cos(kx) (6.54) where the notation for the subscript indices needs some clarification. The first index of a subscript will be used to represent the order of the problem and the second index will denote the harmonic of which the variable in question is a coefficient. For example, 7;, represents a first order variable, and it is the coefficient of the first harmonic, viz., cos(kx) Next, the expansions IPpresente aboveT areP usedT to obtain the { 10\ } ndl {ll\ }hprobes The {10} problem The { 10 } problem is given by d2T;, dT 10+U (6.55) dz 2 0dz ~'dz in the solid phase, 1< z < 0, O < x < W, and d2T'~ dT' dT' 10 + U = U, o (6.56) dz 2 0dz dz in the liquid phase, O < z < L, O < x < W, 7;0 7;0 = 0 (6.57) and dT; dT' 10 t = U~ (6.58) dz dz at z =0, o;(z = 1)= 0 (6.59) and 7"(z = L) = 0 (6.60) The general solution to Eqs. (6.55, 6.56) is given by I0 (z) = Aoecoz +ecoz e"'o Biod{ (6.61) S(z) = Azel'eo + ecoz e"o B'o U, 1 evocEr (6.62) dT; dT' and their derivatives are given by to(z = 0) = Bio UoAo, and 'o (z = 0) = B' UoA' , dz dz whereupon using Eqs. (6.576.59) to eliminate Bo, Az' and B there obtains 40U~ (z) =e~u Aio e _1 evozI (6.63) 4(z =Ai ecz 1 ecozI + U,zeoz (6.64) UL where Ao is found by using Eq. (6.60) as Ao 1 + eU e v oL 1;;; The {11} problem The { ll }problem is similar to the eigenvalue problem. It is given by k 1 +Uo l= 0(6.65) in the solid phase, 1< z < 0, O < x < W, and k 2 + U o l = ( 6 .6 6 ) in the liquid phase, O < z < L, O < x < W, ~TXd 1n+Z (6.67) and [+ Z_ o = (6.68) at z =0, 7 (z = 1) = 0 (6.69) and 7;(z = L) = 0 (6.70) Finally we turn to Eq. (6.46), which is a differential equation for Z, .Using Eq. (6.51) and dZ,/dxL~ = 0 at x = 0, W, and integrating Eq. (6.46) over 0 < x < W, we find that U, must be zero, and therefore 7;0 and 7;,must both be identically zero. Then the first order problem is exactly like the eigenvalue problem, and the condition at neutrality is given by Eq. (6.3 9), viz.,  ak2 M _a k Akh~2 + UO 1N U2L = 0 What we have so far is this: U, = 0, I(x, z) =,(3 ) cos (kx =j ( z) cos(kx), I;' (x\, z) = I (z) cos (kx)= ( z) cos(kx), and Z, (x) = A cos(kx) = Z, cos(kx) And we go on to the second order problem to find the value of A The speed of the surface does not change at order E upon increasing T,; from T,_ by 2 2 The Second Order Problem At second order, the new unknowns to be determined are T2 T2', Z2 and U2 and the unknown A comes over from first order. The second order problem is V 2T2 + Uo = U~ (6.7 1) on the domain, 1< z <0, < x 8T2' dT' V2T2' + Uo = U2 0 (6.72) 8z dz on the domain, < z T2 + 2Z, + Z.2 = _aZ (6.73) T2 T2 + 2Zc + 2 2X +l Z2 = 0 (6.74) and z + 2 Y + Z 2 IY + Z 2d (6.75) at z =0, T2Z "= 1)= 0 (6.76) and T2 (Z = L) = 1 (6.77) The volume condition is given by SZdx = 0 (6.78) Observing the nature of the above equations, we assume the following form for the solution T2 X, Z) = T20 Z) + T21 (Z) COs(kx) + T22 Z) COs(2kx) (6.79) T2 (X, Z) = ~Z) + T Z) COs(kx)+ +7 (z) cos(2kx) (6.80) Z2 (x) = Z21 COs(kx) + Z22 COs(2kx) (6.81) And again, we use these expansions to deduce the {20} and {21} and {22} problems. At z = 0, the following formulae may be useful: 2ZZ = UOL k 2~ + Uj 1 + cos(2kx)~ 2Zix =k2oZiL 1cos(2kx The {20} problem The {20} problem is~ given by d2T,, dT dz 2 0dz ~"dz in the solid phase, 1< z <0, < x (6.82) (6.83) d2T* dT' dz 2 0dz dT* U, o in the liquid phase, O < z < L, O < x < W, T20 T' = ~Uf,2 2~ and 20 UI U 2 Sddz ddz2 at z =0, T20 (Z = 1) = 0 and (6.84) (6.85) (6.86) T' (z = L)= 1 (I A general solution to Eqs. (6.82, 6.83) is given by T20 (Z) = A20 Uo Uoz Ue~oFB20d4 (( 7'(z) =A oe "oz+eczeo 2_od dTz dT'~/_\, and their derivatives are given by 20Z = 0) = B20 UoA20 and 20 Z~ i = ) O UA' , dz dz 6.87) 6.88) 6.89) whereupon using Eqs. (6.846.86) to eliminate B,0, A1 and B~,\we get T20 (Z) = A20 ec U, (6.90) 1 , U,,e 2 Uo= Z,2 + U,zec, (6.91) where A,, is found by using Eq. (6.87) as UL + U, Z1 + eUoL 1+ eL The (21) problem The (21) problem is similar to the p eignaolue problem. It is given by dz2 i 1] ii dT, Oz in the solid phase, 1< z< 0, < x k 7 + UO 2 = 0 in the liquid phase, O < : < L, O < x < W, 7' Z = 0 and (6.92) (6.93) (6.94) (6.95) d2 o  2 d2T dz2 (6.96) dT dr' at := 0, T,, (z = 1) = 0 (6.97) 1 e 1 e o= _ and 7' (z = L) = 0 (6.98) U, + U +4~k2 Defining m+21 + m, a solution to Eqs. (6.936.98) is given by T21 (Z) =A21ell? em+21 21 +21 21 (6.99) 7'l (z) =A21' e[m+21' +21" 21z2 (6.100) where _ak2 A2 =Z21 (6.101) [1e"lm21 +2 ~ak2 U Az" oZ, (6. 102) 1= 1 e[m+2 21 21 The {22} problem The {22} problem is given by 2 4k2 d2 +UO22= 0= (6.103) in the solid phase, 1< z <0, < x 2s 4k2 T* + UO 22= 0 (6.104) in the liquid phase, O < z < L, O < x < W, T22 T' = Uf ,2 2+ UoZ22 (6.105) and Sddz ddz 261(6 at := 0, (6.107) T,, (z = 1) = 0 and T (z = L) (6.108) U, + JU +16k 2 Defining na,, ,a solution to Eqs. (6.1036.108) is given by T2 (Z= A22[ni2 e"'" " "'e" "' T~ (z) =A e"'" ; e n L n e, where 1le[" N_,L]~ Mzz (6.109) (6.110) (6.111) (6.112) n? na 7e "zz' "'"L and N = 1 e "' m,, L na, e"' "'  e" "' na 1 and where Ms Using the curvature equation Finally we turn to Eq. (6.73), which is a differential equation for Z, .Using Eq. (6.78) and dZ,/dx~ = 0 at x = T,0 ,(=0)+Z, Ul(z 0, W, and integrating Eq. (6.73) over 0 < x < W, we find =0)+ =0 7 d?\ (6.113) dT; Using (z dz 0) = ~ak M Z, and T,, (z ~ak22 M Z Comparing with 0) = A~,, we get Az, Eq. (6.92), there obtains U = 1evo 1 8 AkM euo ]eUoL1 Z,2 (6.114) L L[' 2 + a2\ Ievo 1 Z2planar UZcorrection>0 where the first term is what U2 WOuld be if the base solution obtains at the new value of T,, and this term was derived earlier. The second term is the correction corresponding to a curved front. It is always positive. Hence the mean speed of a curved front would be faster than that of the corresponding planar front. In other words, the mean speed of the interface runs ahead of its base value. To go to third order, we need to solve Eq. (6.73) for Z2 To find Z2,, a solvability condition must be satisfied. It is T2%Z = 0) 2 + 2 aZ, (z = 0) Zdx (z 0)Z6x = (6.115) 0L 0 It is easy to verify that this condition is automatically satisfied, whence we cannot learn the value of A at this order. Finally, substituting the expansions from Eqs. (6.79, 6.81) into Eq. (6.73), the constant term disappears due to the result obtained earlier using integrability, viz., due to Eq. (6. 113). The coefficient of the first harmonic vanishes due to solvability, and there obtains by equating the coefficients of the second harmonic dT; T22 (Z= 0)+ '(z =0)Z, 4~ak2Z22 =0 (6.116) dz Substituting for~ T2 Z=0) = A22l [m22+2 and using id(z = 0)= ~ak2M~ Z, there obtains [dk2M]Z Z+ 4dk2 Z22 A, (6.117) 22 22 +"221 Comparing with Eq. (6.111l), and simplifying, we get (6.118) (6.119) Uf~' N22 +2Uok2' +UJ+ ak2M [N22 M22, Q 4ak2 [N22 M22] UjNr22 + UJ] Hence, Z21 and A carry on to third order. The Third Order Problem The equations at the third order are given by (6.120) on the domain, 8T3d' V273 +U, 0 Z 1 a7T' dTX; 3U U o 2 Z 3 dz (6.121) on the domain, < z 9d,: d2Z, T3 + 3Z, d + 2.3 dz3 + 32 O 2 dZ2 +3Z2 +3Z2 (6.122) dx 2 ZZ Z '0 dz (6.123) +3~Z Z2 + Z,3 + Z3 o  d27'* o dz2 Z22 = Z,2 where 8"T T2 8 T2* 2 2 T2 T31: r3'? +3Z~"L~T +3ZZ 2'~Y +3Z2? ~ 1Z 1'~: = 0 ]; 3cr ri, +3sZ 2 +3Z2 1 +3Z (311 BZ a Z at Z2 dZ2 1 Z L B Z3 L Z2 dZ2 +3Z,2 + Z3 +Z3 = U3 (6. 124) 3Z2x2, ~ d7I I +2Z, Zx 1 at z =0, T3 (z = i1 = (6. 125) and T3'(z = L)= 0 (6.126) The volume condition is given by ij zdx =0 (6.127) We assume the following form for the solution T3 X, Z) = T30 Z) + T31 (Z) COs(kx) + T32 Z) COs(2kx) + T33 Z) COs(3kx) (6.128) T3 (X, Z) = 730 Z) + ,3 (Z) COs(kx) + T32 Z) COs(2kx) + T3 (Z) COs(3kx) (6.129) Z3 (x) = Z31 COs(kx) + Z32 COs(2kx) + Z33 COs(3krX) (6.130) Now this problem can be sonrplve comnpletly, but; it sffces o snolvep the {31} problem to learn the nature of bifurcation and the value of A2, and it is to this that we turn our attention next. The {31} problem The {31} problem is~ given by d2 *] r dT d k2+ 3+U 3+3U '= (6.131) dz2 0dz 2dz in the solid phase, 1< z <0, < x (6.132) d ,'* dz3 d 3 d3 dz2 4 3 (6.133) ?~ Z,, dz 3d3 + 0 2 3 d3* 3 d4 ~ dz3 4 4 at z =0, T3(Z = 1)= 0 and (6.135) T' (z = L) = 0 A general solution to Eqs. (6.131, 6.132) is given by 3(Z) =[ A ze"'+z +B, e" ]+[ Cze' + +DZe"' T' (z) = A'*mze+ +B*,e"'1 + [C'ze"'?+ +D*ze"'l i (6.136) (6.137) (6.138) k 7' + U, ~t i + U '= in the liquid phase, O < : < L, O < x < W, dTz dr', ] 1 dT,, dT' 9 d d dz dz 2dz dz 4dz2 dz2 3 d T d T* 3d2 + Z z+ a 2 dz dz 2 2 +T Z, = 0 and dT, d ,1+Id Tz d T', 1d d T,, dz dz dz2 dz 2 dz2 dz 4dzd 3 d T +  2 dz d4 * dz 4 (6.134) + Z3;], 3 k 7, T 3,,Z+k Z 3 k2 7* Z, whee =3172 A m,D= 3U2 A, m(m m,), C* 3 A/ an m_m m_m m_ m D* = 3172 A; mem ]. Eliminating B31 and BF using Eqls. (6.135, 6.136), we get mm T (z)= A31 [em z[ a][(tmii m)zin_n+D e"'z Czem" z +Dze"'z (6. 139) 1' (z) = A', [e"' ze[" ]e"z L [C'e[" ""' +D'" enez + [C'ze"z +D')ze"' (6.140) Now Eqs. (6.133, 6.134) can be written as A31[I e""] m_m 1I m",mL = R, (6.141) A31 m + 7"'"'1 n2_n? m e '" "' = R2 (6.142) where R = Ce[" +D +L C'elm,m "] +Dj dr d'1 T2 T91 d2 7;d2 7* 3: d( d(.43 dz dz 2dz dz 4dz2 dz2 2d dz 2 + 223 and d3 0 d3' 3d40 dz3 4 4 d23, where all terms in R, and R, are evaluated at z = 0. Now R, and R, can be simplified using results from previous orders. At : dT , dz dz dT * a dz dT " dz 1 U, U~ Z 2 1 = 2UnU, +U,4 Z, 2 UZ,[+U1 = Un 2k + U Z~ dT; dT; dz dz 2 7 dz dz dT ; dz2 dz2 dT, dT"I dz dz LA Z, d2T dz d2T " dz Hence R,= m [Ce(""'"l+D +[C+D1 Lm_ [Ce[ni,n;]L+D']+[ C+D'] dl' ZZ dz3  d"Tz d2T, +1 d2T, d2T "I dz2 dz2 2 dz2 dz (6.144) 3 d ( + 2 dz2 +  dz4 *  3k L, Tj Z + k Z 3k Z, 0, the following formulae may be helpful: [4k l +1 ZI ] +1 [l +4k ] Z: R, = Ce[m m.1]+D +L Cie[m.m "]' +D'] 3U2; 1 Uo [k2 + U02 i:3 U~ Z 22 UoZ31 R = m_ Ce[m m] +D +[C+D] Lm_ [C'e[m. +] +Dj'+ C'+D +6UoU2ri4 Z+ U k2 +U Z3 Uo Uf +k LZ1 Z22 +Uf Z31 24u1 1 2'0'O L1 (6.145) (6.146) Then solving Eqs. (6.141, 6. 142) for A31 and A3, ,there obtains [R,N4Z] A3= (6.147) 1 [e[m m.] [MN] [RM ] A3 (6.148) jl 1e[m,m'] [MN] Using the curvature equation Finally we turn to Eq. (6. 122), which is a differential equation for Z3 To find Z3, a solvability condition must be satisfied. It is o~odZ aT, 82 = 0)+3Z2, (Z= 0)+ 3Z2 1 Z 1z BZ 2 = 0)Z~dx = 0 S) Z ,dx = ~W Z T3, 1Z = 0 ) 0)+ 3Z (z 2 Z (6.149) Using W 1(= 3Z, 0 ^i dT ,d 8z 1 dT, 0) + 22Z 2 dz 1 2 82 a7 3 2, 1 Z , 1 ^9 ^~ d2 ( 2 4 dz 2 2d2Z, LC2 0) dx " = iZ' lW2 Z, Z7 (z 22 dz ak Z;1 S(z 0)Z,dx = IW Z, 1WZ 2 9 A 4 there obtains 1 dT, S(z 2 dz 0):3 dl T~ 2 dz 0)Z22 (6.150) + ak4 Z + [ak ]Z;, = 0 which can be written as (6.151) where 1 d T, 0)+ "(z 2 dz 0)Z3 (6.152) 3 d T +2 (z whence u z = 0) Z + 4 kZ + Ak 23;, sing 0) = U1 [ bak M Z dT , "(z d dT, (z dz dl(z dz Ak M Z3 +4Ak Z ~ak" MZ, 3~ jl a;Z (z = 0) dx S9A dZ, 7 d Z zdx dx dx rr(z=0)+3 "T(z=0)+ 9~l d T A 1[ e "' "' ] d T R Ce"'"' +D + "(z "=[~'''] ~ dz 9 d Ti( 0) =rl M d2 7 (z = 0) ~ak2 [k2 Z, there obtains R3LI mm]Dli + 3 ak2M ~ak2 22,,+ Ak kUMZ + 3k2 M + 2M22, 1 Z22+ Ak2Z3 Comparing the value of A31 fTOm Eq. (6.147) and Eq. (6.151), we get [R,N R2]i+ R3 [M~ N]= 0 (6.153) (6.154) [UoZ31 R2 = R2 UfZ3, and R3 = R3 ak2 Z31 Next, we define: R, = R Substituting in Eq. (6.154), and using the relation at neutrality, viz., Eq. (6.39), there obtains RNR2 +R3[MN]=0 (6.155) where now R, = C[ml i m.] +D + )] LCe[m.mei ]L +D' 13U2 1 Uo, k 2 +Uj2'{ Uf1Z; 22 R2[ ce[m m ]+D +[C+D]] Lml [C'e[mm]' +Dj]+ C'+ D'] 3'i~~ 3'iL' ^ 3Z Z (6.156) (6.157) k2 22 + Ak2 2k2 (6.158) R3 =[m_ +D + 3 jl~~]ak2M UoM Z, +[ 3AkL2 M + 2M~ 122]l 1 Z2 Now,, Eq. (6155 isE an equatio~n for ,2, Whence, upon combining terms, we get ;+ T,L/2 Z,+ TZ+ +T4Z, Z22 (6.159) where 3 172 Z 3N 6Uo (6.162) 3 Lak2~\L, 229 ak2 2k2G 2 _i4cl LI T4 =NU0 (6.163) Hence T 17 l2 Z +7, 172 +E 32, + T4Z22 (6.165) 7;= LN C'e[mm']L+D" ] m [Ce["' +D+[C'+D]] + Lm [ C'e"''"'] +D ~"+ C' +D'+M [Ce[m m.+]+D (6.160) 1ak e a e UL[_+ _  [m m + Uo M m_+ 1e "L ~l~~1~: (6.161) 3 3 3 T3= N1V io k 2+ U~ k2 _ U4 4 24 [M N] 170 Ak2M 1 Uo Uk2 +l~i UQ;]~ + llx[ I [MN 3k 2M22 l (6.164) 7 12planar 1 T 2orecon4 Z 2 u2, U2Zu, zZ ~ ~ 172 (6.166) Hence, using Eq. (6. 114) to substitute for 172planar and L/2corechonand Eq. (6. 119) for , we finally obtain ke" "Lak m U +e[ mi e n, Il UoL[ N m_]+ 1;e ~ak2 e m,[ +e..lu[mm] G[Mm _]+ ke" "Lak m U +e[ m ] _OLNm i.l i ,[ o +mr 1e" m +eII U.[Mm +m] U +A 2 + L (6.167) + 7, U 2 planar U2 Z 3 1 r m m 1[3N +6Uo ] eoL m [3N +6Uo + Mu, N]k Z1Ak2M k222 + Lak2 [2k2 OMil 41 Uo 2r 4v~ UN22 + 2Uok2 +U +A2 N22 2vl ~a22~[, 2 4ak2 N22 M22i O~N:22+U Table 6. 1 Case 1 Values of Uo and A2 for inputs L = 1, 2 and k2 L=1 L=2 Uo A 2 Uo A 2 10 5 x 10 5.99 x 100 7.5 x 10 3.998 x 100 103 5 x 1010 5.99 x 109 7.5 x 1010 3.987 x 109 10 5 x 10 5.94 x 109 7.45 x 10 3.849 x 10 101 5 x 10s 5.41 x 109 7.07 x 10s 3.342 x 107 1 5 x 10 2.215 x 106 5.66 x 10 3.579 x 106 10 5 x 106 2.662 x 104 5 x 106 1.487 x 104 102 5 x 10 2.667 x 10 5 x 10 1.379 x 102 103 5 x104 2.667 5 x104 1.348 104 5 x 10 2.68 x 10 5 x 10 1.35 x 10 105 5 x 102 2.802 x 104 5 x 102 1.474 x 104 106 0.5 4.392 x 106 0.5 3.621 x 106 The sign of Z,2 is of interest. If it is positive, the pitchfork is forward, as assumed, otherwise it is backward. In the latter case, T, must be expanded as T, = T~o 2 2 Clearly it is a formidable task to determine the sign of Z,2 and one can possibly not determine its sign by inspection. We will therefore have to do some calculations in order to learn the nature of the bifurcation. Table 6.1 gives the sign of Z,2 foT Various input values of k2, L Clearly Z,2 can have either sign, whence the pitchfork can be backward as well as forward depending on the input parameters. However, the formula given in Eq. (6. 167) is not very useful if one wants to learn how, the;,, sinofZ2 depends upon the inputs. It is hard to figure out the point of crossover from a backward to a forward pitchfork. Hence we will consider a case later where we will look at a special case of case I where the model equations will not have U on the domain. We will obtain a formula for Z,2 Which would be much simpler compared to Eq. (6.167). This case is deferred until the~ endC of the~ chapterL, where~ we will fIndI out the~ anlalytilcal dependencec~ of Z,2 OH inputs k2 and L . For now, we continue in the spirit of hoping to find the reason for a "speedup" in pure solidification. So we consider a simplification of case I where the solid phase is of infinite extent. This case might indeed be more realistic in that solids of extent much greater than the capillary length ought to be considered infinite. Case 2: Latent Heat Rejected Only to the Subcooled Liquid, Solid Phase of Infinite Extent The notation remains the same as in case 1. Since the solid phase depth is of infinite extent, all lengths are scaled by L, the depth of the liquid phase. All speeds are scaled by a /L . T T The scaled temperature is given by Tscaled M. Then, the scaled temperature in the solid phase must satisfy dT V2T +U = 0 (6.168) for oo < z < Z and 0 < x < W, while in the liquid phase dT* V2T* +U = 0 (6. 169) must hold for Z < z <1 and 0 < x < W The equations along the interface, viz., along z = Z(x) are (6.170) and n.VTn.V'= .Uk(6.171) where 1a denotes T~,L Farfield conditions are given by T(z = o) = 0 (6. 172) and T(z = 1) = T, (6.173) Side wall conditions are Neumann. The volume of the subcooled liquid must be maintained fixed, whence Sz (x) dx= 0 (6.174) The Base Solution A base state solution is given by To(z)= 0 (6.175) To*(z) =1 eoz (6. 176) where Uo is given by Uo = In(1 T,) (6. 177) The Perturbation Eigenvalue Problem Expanding Z, T',and T', at z = Z(x) as in Eqs. (6.176.19), the perturbation eigenvalue problem is given by V24 + Uo I = 0 (6.178) T = _l[2H] = T' on the domain, co < z < 0, O < x < W, and dT' V 2 + Uo I = (6. 179) on the domain, O < z <1, O < x < W, where, at the base surface z = 0, d2Z, 7;= 11a (6.180) LC21 ' ;+Z = 0 (6.181) and 87 T 8* d 2~ d 2TO* +Z o = 0(6.182) 8z 8z2 dz 2 must hold, and where, 7(z = o) = 0 (6.183) and 7* (z = 1)= 0 (6.184) also hold. The sidewall conditions are Neumann. Thus the problem is homogeneous in 7;, 7* and Z and it has solutions of the form 7;= (z) cos(kx), 7;* = 7' (z) cos(kx) and Z, = A cos(kx) (6.185) which satisfy the side wall conditions so long as k = nzr/W, n = 1, 2..., where k is the wave number of the disturbance. n = 0 is ruled out by the condition SZ,dx= 0 (6.186) The expansions presented above give the equations for ;, 7; and Z They can be solved just as in case 1 to get ~(z) =A, e"''"= Ak?[e"' Z (6.187) and (z) = z' e"I= e 'e"=k" e"'+" ~~_ e "~~~~' "'e"'Z, (6.188) where nz = .We can then turn to Eq. (6. 182), which gives us  ak] nLak2 +UIJDNUi = 0 (6. 189) na n e mm, where N = Notice that the neutrality relation given by Eq. (6.189) can be 1e[n; n;] obtained from Eq. (6.39) of case 1 simply by observing that the asymptotic limit of M~ as the solid depth approaches infinity is given by na The neutral curve is again as shown in Figure 6.2 (for case 1) and only one crest is seen at critical. Now on increasing TL to its critical value T~o , and then slightly beyond, the branching to the new steady solution is expected be a forward pitchfork. So as before we write 1 TL To 2 (6. 190) and expand U and Z as in Eqs. (6.42, 6.43) to get the first and higher order problems. Eq. (6.177) indicates that for a small increase in TL fTOm T~ the corresponding change in U from U,, per unit change in TL is given by eU. The First Order Problem The first order problem is given by V 27 + Uo (7 = ~ (6.191) on the domain, aT' V2 ;* + Uo co < z < 0, O < x < W, and dT* U, o (6.192) on the domain, O < z <1, O < x < W,  Z 7;= 11 (6.193) (6.194) + Z [/~ = 0 ]= and 8z' 8z dz 2 (6.195) at z =0, 7(z = m) = 0 and 7* (z = 1)= 0 The volume condition is given by IZ,dx = 0 The sidewall conditions are Neumann. Hence we have solutions of the form (6.196) (6.197) (6.198) 7;(x, z)= 7;a(z)+ 7, (z)cos(kx) (6.199) 7;'(x, z)= 7;0 (z)+ 7T(z)cos(kx) (6.200) Z, (x) = A cos(kx) = Z, cos(kx) (6.201) The expansions presented above are used to obtain the { 10: }, and { ll } pobemsAai, as,; in case 1, U, turns out to be zero, and therefore 7;0 and 7;" must both be zero. Then the first order problem is exactly like the eigenvalue problem, and the condition at neutrality is given by Eq. (6. 189), viz., ak2 m ak2 + Uog N U2 = 0. And1 we~ goV on1 to theC scondII order problem to find the value of A . The Second Order Problem The second order problem is V22 U = (6.202) on the domain oo < z < 0, O < x < W, and 8T2 dT' V2T2' + Uo = U2 0 (6.203) 8z dz on the domain 0 < z <1, O < x < W, T2 ; +2Z + A Z (.24 T2 2'+ 2 Z Z + Z2 LI = 0_ (6.205) and z +2Z z +Z2 Z +Z2 (6.206) 2Z, x U Dx Dx 7 a =1 at z =0, T2 Z = o) = 0 (6.207) and T2 (Z = 1) =1 (6.208) The volume condition is given by IZ~dx= (6.209) Again, assuming the solution has the form T2 X, Z) = T20 Z) + T21 (Z) COs(kx) + T22 Z) COs(2kx) (6.210) T2 (X,Z) = T~Z)+ T Z) COs(kx)+ T'(z) cos(2kx) (6.211) Z2 (x) = Z21 COs(kx) + Z22 COs(2kx) (6.2 12) we deduce the solutions to the {20}\ andl {21}\ andl {22} nhprobes At z = 0,he following~ formulae may be useful: 2Zixc? 1 = k2Uo Z~ 1 cos (2kx)] The {20} problem The {20} problem is~ given by d2T, dT MX 20 + (6.213) dz 2 0dz ~"dz in the solid phase, co < z < 0, O < x < W, and d2T'* dT' 20 + UO 20 dz 2 0dz dT'* U, o (6.214) in the liquid phase, O < z <1, O < x < W, 1 T20 7 =~ Uf ,2 21 and (6.215) 1 2 (6.216) (6.217) dTd dT' at z =0, T20 (Z = 0) = 0 and r' (z = 1) = 1 (6.218) A solution to Eqs. (6.213, 6.214), subj ect to the far field condition at the solid sink is given by T20 Z) = A20 , and (6.219) 7'; (z) = A oeoz + eCoz (6.220) dT' 20 = 0) dz Bo UoA' and using Eqs. where A20 iS Some negative constant. Noticing that (6.215, 6.216) to eliminate Ao and B0, we get T20 (Z) = A20 T7* (z) = A,,~20 fo,+28Uz where A20 iS found by using Eq. (6.218) as (6.221) (6.222) evo B'o, U2 _e Uo ] 0 AzO = evo U, + e (6.223) Now the (21) and {22) problems can be sonrplve entirely, but itis sufficient to solve the (20) problem to learn whether the front speeds up or slows down. To do this, we integrate Eq. (6.204) over < x T,, (z = 0) + Z,' ; (z = )+ Z d = 0 (6.224) Using (z = 0) = ak m, Z and T,, (z = 0)= Az,,, there obtains Az~, = Jrk m ZZ dz Comparing with Eq. (6.223), there obtains U rm r =rV e vo + + A k e Z 3 (6 .2 2 5 ) 72correchon The correction term is always positive, whence the mean speed of the interface runs ahead of its base value, and our hypothesis of a "speedup" in solidification is again confirmed. But one might ask: Is this result limited to cases where the latent heat is rej ected only to the subcooled liquid? It is to this question that we turn next. Hence we consider a third case where the latent heat is rejected to the subcooled liquid as well as to the frozen solid, and both phases are taken to be finite. Indeed, one might intuitively expect that as rej ecting the latent heat to the solid phase is stabilizing in contrast to rej ecting heat to the liquid, it might also have an opposite effect in regard to the front speed correction, i.e., a large enough Ts might be expected to slow down the front compared to its base value. However, it will be shown in the next section that this is not so, and that once the instability sets in, the front speeds up, no matter the value of Ts  Case 3: Latent Heat Rejected to the Frozen Solid as Well as to the Subcooled Liquid, Both Phases of Finite Extent The notation, including all scales and references, remains the same as in case 1. Then in the solid phase, the equation for the scaled temperature is given by dT V2T+U = 0 (6.226) for 1< z < Z and 0 < x < W, while in the liquid phase dT* V2T* +U = 0 (6.227) must hold for Z < z < L and 0 < x < W The equations along the interface, viz., along z = Z(x) are T= ~a[2H]= T (6.228) and n .VT n.VT* = U k. n (6.229) where ~a denotes TL,~, Frfeld ioinnditinsar given by T(z = 1)= Ts (6.230) and T'(z = L) = T, (6.231) Side wall conditions are Neumann. The volume of the subcooled liquid must be maintained fixed, whence Iz(x)dx= 0 (6.232) The Base Solution A base state solution is given by (6.233) (6.234) where Uo is given by T, s " TL = 1 1 evoL (6.235) Some useful derivatives and their differences are given in Table 6.2. Table 6.2 Case 3 Some derivatives of base temperatures in two phases in solidification dT o(z d2T (z d3T (z d4T (z dT' O Z= a z d2 '* S(z d3 * (: (z d4 * (z 0)= UoAo 0) U A O = U > 0 0) U A O = U > 0 dT Z= dz d2T o(z dz 2 o2 (z dT' 0) a (z = dz d2 * = 0) z dz 2 d3 '* dz3(Z 0) =  0)> = U,4 d4T, d4 *; o(z = 0) (z dz4 dz4 The Perturbation Eigenvalue Problem Z (x) as in Eqs. (6. 176. 19), the perturbation eigenvalue Expanding Z, T, and T', at z problem is given by 7;/r)=,~~~~~~ [e.l= U,] U" To'(z)= A;I 1e oz 1 Go 1 eo eOT . eUT 1 U 0 S) = Uf A, = Uf 1 eT e, ]>or =r 0)= A = U 1 eT UoL > = 0) = Uf A =O U 1 e o V2 ;+ Uo = 0 (6.236) on the domain, 1< z <0, < x dT'* V 2 + Uo = 0 (6.237) on the domain, O < z < L, O < x < W, where, at the base surface z = 0, d2Z, dT, 7;= 11 Z (6.238) dx C21 dz O ]+, a = (6.239) [". IdTdz ddz and + Z, = 0(6.240) 8z 8z dz 2 dz 2 must hold, and where, 7(z = i1) 0 (6.241) and 7* (z = L,)= 0 (6.242) The sidewall conditions are Neumann. Thus the problem is homogeneous in 7;, 7;* and Z,, and it has solutions of the form ;= 7(z) cos(kx), 7;' = 7;' (z) cos(kx) and Z, = A cos(kx) (6.243) which satisfy the side wall conditions so long as k = nzr/W, n = 1, 2..., where k is the wave number of the disturbance. n = 0 is ruled out by the condition SZ,dx= 0 (6.244) The expansions presented above give the equations for 7, 7; and Z They are (6.245) (6.246) (6.247) (6.248) (6.249) (6.250) 7*(z = L) = 0 Defining m (6.251) Uo, E U2 + 4k2 ,a solution to Eqs. (6.2456.248, 6.250, 6.251) is given by 2 in the solid phase, 1< z < 0, O < x < W, and in the liquid phase, O < z < L, O < x < W, = Ik2 To Z, andz dz dz2 dz ddz* 2 dz2T;= at z =0, 7(z = 1) = 0 and dT ~ak2d,[ o,(z = 0) 71(z)=~[ A ,z e"'"e"' "' e' =, 1 e e'e" 622 dT' *~~~~~~~~~~ (z =~ A "+e" nL ;c;] t_[m,ni_]L n";_ Z (6.253) We can then turn to Eq. (6.249), which gives us A +d M ak2 + dTX; N= U (6.254) or [ ? ek + 1 M+ i Ak + "U N= L2 (6.255) where M~= adN le["" 1e["] Figures 6.3 and 6.4 show the corresponding neutral curves. As the control variable TL increases upward from zero, the output variable U, increases from zero, until there comes a point when a line of constant 1 first crosses the neutral curve at an admissible value of k The corresponding value of U, will be referred as its critical value. Henceforth the subscript zero will refer to this critical state. At values of U higher than [ ,, the base solution loses its stability. The "dip" is present in both kinds of neutral curves and this implies that the critical value of U need not correspond to the least allowable value of k In fact, n might be any of the numbers: 1, 2, 3.., and hence multiple crests might be seen. Figure 6.3   An unusual neutral curve in solidification when the latent heat is rejected through both phases Figure 6.4 Typical neutral curve in solidification when the latent heat is rej ected through both phases Now on increasing T, to its critical value Tro, and then slightly beyond, the branching to the new steady solution is expected be a forward pitchfork. Hence we write 1 T, = T, + 2 (6.256) and expand U and Z as in Eqs. (6.42, 6.43) to get the first and higher order problems. Eq. (6.235) indicates that for a small increase in T, from Tro, the corresponding change in U from Uo, per unit change in T, is given by LeUoL + 1 2 8z on the domain, aT' V2 ;* + Uo 8z dT, dz (6.257) 1 dT' U, o (6.258) on the domain. O < z < L, O < x < W, where, at the base surface z = 0 d2Z, dT + Z o dz (6.259) (6.260) (6.261) (6.262) (6.263) T UoL Lo i" e[oL  T Sel 2T 1 eUo The First Order Problem The first order problem is given by ii. i IdT~d dT~d 1o and I~ I must hold, and where 7(z = 1)= 0 and 7* (z = L) = 0 The volume condition is given by IZ,dx = (6.264) The sidewall conditions are Neumann, and Eqs. (6.257, 6.258, 6.2606.264) can be solved by writing 7;(x, z) = 7;0 (z) + 7;, (z) cos(kx) (6.265) 7* (x, z) = 7;0 (z)+ 7' (z) cos(kxc) (6.266) Z, (x) = A cos(kxc) = Z, cos(kx) (6.267) The epansions~n IPpresente aboveT areP usedT to deducelr the {10\ } ndl {ll\ }hprobes The {10} problem The { 10 } problem is given by d2T; dT; dT, 10+ U = U o (6.268) dz 2 0dz dz in the solid phase, 1< z <0, < x d2T'~ dT' dT' +u0 +U = U, o (6.269) dz 2 0dz dz in the liquid phase, O < z < L, O < x < W, 7;0 7;0 = 0 (6.270) and dT; dT* 10 t = U1 (6.271) dz dz at z =0, 70 (z = 1) = 0 (6.272) and o;(z = L)= 0 (6.273) The general solution to Eqs. (6.268, 6.269) is given by: 0; (z) =Aloe 'oz + e "oz e"o Bio and their derivatives are given by (6.274) (6.275) dz dT' 0) = B UoAo,, and "o (z dz 0) = B'xo UoA', whereupon using Eqs. (6.2706.272) to eliminate Bo, A' and B' we get (6.276) (6.277) +U(:zevo (6.278) (6.279) U/, Ao 1 eo ]u5 UzAf1 levot ] /10 (z)= A loeco + Uzo i 1e where Ao is found by using Eq. (6.273) as Aio uoj+UzAOzevoz Uoji+ U,Azeo u I r e U lo 1 ; 1 The {11} problem The { ll }problem is similar to the eigenvalue problem. It is given by dz x~ 27; =dz in the solid phase, 1< z < 0, O < x < W, and in the liquid phase, O < z < L, O < x < W, An +Z = 0(6.280) dz dz and dT; d7; ] d T, d2X + Z,, = 0(6.281) dz dz dz2 dz2 at := 0, 7;(z= 1)= 0 (6.282) and 7;(z = L) = 0 (6.283) Finally using Eq. (6.264) and dZ, /dxe = 0 at x = 0, W, and integrating Eq. (6.259) over 0 < x < W, we find that the constant A,, must be zero. Hence U, must be zero, and therefore 7;, and 7;, must both be identically zero. Then the first order problem is exactly like the eigenvalue problem, and the condition at neutrality is given by Eq. (6.255), viz., [; eAk + 1 M + U N = U, What we have so far is this: U, = 0, I(x,z) = (l) cos~(k) = j (z) cos(kxi), I' (x, z) = I; (z) cos (kx) = (' (z) cos (k), and Z, (x) = A cos(kx) = Z, cos(kx) And we go on to the second order problem to find the value of A. The Second Order Problem The second order problem is 8To dT, V .+, = U (6.284) 8: dz on the domain, 1< z< 0, < x V2T2 + Uo 8z 0 (6.285) dz on the domain, O < z < L, O < x < W, ,d2T 1dz 2 d2Z LC2 aT T2 + 2Z ' 1 Z dT, 2 , (6.286) (6.287) d3TO' d dz3 ]I dz22T (6.288) at z =0, T2 Z = 1) = 0 and (6.289) T2 (Z = L) = 1 The volume condition is given by Assuming the solution to be of the form T2 (X, Z) = ~Z) + T Z) COs(kx)+ +7 (z) cos(2kx) Z2 (x) = Z21 COs(kx) + Z22 COs(2kx) we deduce the {20} and {21} and {22} nhprobes At z (6.290) (6.291) (6.292) (6.293) (6.294) 0, the following formulae may be 8z1Y 8zZ dz 2dz 2d dz [T "8T2* 82 2 d3T ~ ;21 +Y 2Z,2 + Z2 BZ~ ~ ~ ?ZB2B2d3 useful : 27 ";Z k2Uo Z2 [1 COs (2kxc) d2T , dz 2 dT U, 20 0dz dT, 2dz (6.295) in the solid phase, 1< z <0, < x dz 2 dTz' U 20 0dz dT' U, o 2dz (6.296) in the liquid phase, O < z < L, O < x < W, ,,1 T20 7' = Uf,2 21 and (6.297) 2 (6.298) (6.299) dTd dTr' at z =0, T20 (Z = 1) = 0 and T' (z = L) = 1 A general solution to Eqs. (6.295, 6.296) is given by (6.300) T20 Z) = A20"' Uoz' Uoz~ UoB20 0 (6.301) a7*= Uo k2+U 2Z ,2 [1+COs (2kx)] 2Zix~  The {20} problem The {20} problem is given by U2,A [1 evot u' 7' (z) =A oe "oz+ eo i, enlo B~ U2Ao[ 1 evot (6.302) and their derivatives are given by 20Z=0 2 o2 n 0Z = 0) = Bo1 i UA o,, dz dz whereupon using Eqs. (6.2976.299) to eliminate B2 0 and B 0,we get T2 Z=A2eo +[U _oz +UAzoz (6.303) T,, (z)= A20eU oz +U [U Ai]1 +U2A~zev ,, eY U~z(6.304) 1 where A20 iS found by using Eq. (6.300) as 1 + I U e v oL 1 1 + r e v o L 1 The {21} problem The {21} problem is similar to the eigenvalue problem. It is given by d2 dT 2 k 21+UO2 = 0 (6.306) dz2 dz in the solid phase, 1< z <0, < x d2 dT'* k2T*T +~ UO2 = 0 (6.307) dz 2 dz in the liquid phase, O < z < L, O < x < W, T21Zj+Z2 0Td d~,=0 (6.308) where dk2 0 Z = 0) A'*1 Z [ 1 e[m+21 21'" 21 The {22} problem The {22} problem is~ given by and (6.309) (6.310) (6.311) U + U2 +4k2 2 Defining m+21 + a solution to Eqs. (6.3066.311) is given by (6.312) (6.313) [ Z21 dz [1e"m21 +2 (6.314) (6.315) d2 2 dz Jkl7;dTi:i (6.316) in the solid phase, 1< z <0, < x SdTd d7r' d2Toz d2TO~z at z =0, T21 (Z = 1) = 0 and T'(z = L) = 0 d2 2 dz dT' (6.317) (6.318) in the liquid phase, O < z < L, O < x < W, T2~:2 1_] 21 = Uf 2+UoZ22 and dT22 dz dTdz ]iri SU03 2~2+ U0 Z22 2 (6.319) at z =0, T22 (Z = 1) = 0 (6.320) T22 Z = L) = 0 Defining m+22 (6.321) Uo + JU +16k2 ,a solution to Eqs. (6.3166.321) is given by 2 T22 (Z) = A22 [m+22 2 2 222 (6.322) (6.323) 7' (z) =A22 m+2 +22 22e22' i where (6.324) Uf N22 +2Uok2 + U ,2+ UoN22 +Uj ]Z22 [1 1? e 022 +2 22 M2 (6.325) +22 [mm2 +2 _[m22 +22z +2m+22 22 z m+22 22 l? 22 22 and N, 22 and where M22 , Using the curvature equation Finally, using Eq. (6.291) and dZ2 /dx~ = 0 at x = 0, W, and integrating Eq. (6.286) over 0 < x < W, we find 1 ^d2T 0) + 2Z,2 Z = 0) = 0 T20 (Z = 0) + Z,(z (6.326) Using (z dz dT, 0 Z dz 1 = 0) 2 , and T20 (Z 0) = A20 there obtains A20= d2 T dz2 dT o(z dz ak2M~+M (6.327) Comparing with Eq. (6.305), and simplifying, there obtains Le':or +s UZplanar Uo 2 (6.328) [c'"~ U M+ UO 1+ eo 1 Lo LeUOr s uo oL 1 2 Uo" 2 UZcorrection>0) UfIIM22 + 2Uk2 +: U .Z2+ UoM22 + UJ Z22 A2' 2le" m+22 L] 22 22M22 0) = ~ak2 0) M Z, UoL eUoL 1 U2 Lo Uo 2 U1 + Ak2 + where the correction term is always positive, whence the mean speed of the interface runs ahead of its base value. We have learned that in pure solidification, once the front becomes unstable, its mean speed is higher than its base value. Hence the result of a "speed up" in pure solidification is unconditional, viz., the front always speeds up regardless of whether or not the latent heat is being rejected to the solid phase. Also, the reader is encouraged to verify that as Ts goes to zero, the limiting value of U2 is given by 11 1 eUo lim U, = evL +2+Al2 +evr 2 639 [, 2r planar2 UZcorrection> which is the same expression as what was obtained in Eq. (6. 114) for case I where the latent heat was rej ected only through the subcooled liquid. For this case then, the only other information that one might be interested in seeking is the nature of the branching to the new steady solution, i.e., whether the bifurcation is a forward or a backward pitchfork. To that end, we need to go to third order, but before that, we need to solve Eq. (6.286) for Z2 .Now to find Z2,, a solvability condition must be satisfied. It is ST2Z =0) + Z, d7 (z =, 0) + ,2d Z =0) zd 4(z = )Z~d = 0 (6.330) 0L 0 It is easy to verify that this condition is automatically satisfied. Hence, Eq. (6.286) can be solved for Z2 Substituting the expansions from Eqs. (6.2926.294), the constant term disappears due to the result obtained earlier using integrability, viz., Eq. (6.326). The coefficient of the first harmonic vanishes due to solvability, and there obtains by equating the coefficients of the second harmonic 0)Z,+ a dZ;(z 2 dz dT T22 (Z = 0)+ (z dz dT, 0),2+ 0Z = 0)Z22 dz 0) =ak2 , (6.331) (Z Substituting T22 (Z = 0) there obtains 1 d2T 2 d'z' 2 0)M o; (z dz (6.332) 1l em 22+22 Comparing with Eq. (6.324), and simplifying, we learn that Z22 can be found in terms of Z,2 as Z22= Z,2 (6.333) where dT, 0)M o (z = dz  UoN22 +UQJ (6.334) Hence, Z21 and A carry on to third order. The Third Order Problem The equations at the third order are given by aT V2T3 + U 3 0 Z on the domain 8T3' V2T3 + U 0 Z a7T dT1 3U 'Uo 2 Z 3 dz 1 a7T' dT'; I3U U o 2 Z 3 dz (6.335) (6.336) on the domain 0 < z < L, O < x < W, 0) M , A22[ 22 +22 ,1 and using (z Z,2 ~ak2M~+ 0) 221[ 0Z = 0)4ak2'1 0) [N22 M22 dZ( )2 d2Z t7 i "T2 SLC2 1 Z 82 a7 3Z21 1 Z a7T d2T 2 Z 2 dz 2 (6.337) (6.338) d2Z3 A ~C dT, Zo 3dz T3~~c7 a3' +Z~ 3 Z, +7 3 a22 + Z d2T d2T' d3T d3T' dT d~ Uo U3 (6.339) (6.340) i3 iZ +IY i3 Z 2 + 3 2 1 +LIiZ_ 2 3i Z2I C1~ d3T[ d3TO* d4T d4TO' d2T d2TO; dz3 ~~ dz3 dz4 dxz4 z z at z =0, T3 (z = i1 = 0 and T3'(z = L) = 0 The volume condition is given by We assume the following form for the solution T3 X, Z) = T30 Z) + T31 (Z) COs(kx) + T32 Z) COs(2kx) + T33 Z) COs(3kx) (6.342) (6.343) Z3 1 3Tdz3 (6.34 1) T, (x, z) = T,1(z)+ T, (z) cos(kx)+ T, (z) cos(2kx)+ TL(z) cos(3kx) (6.344) Z, (x) = Z,, cos(kx) + Z,, cos(2kx) + Z,, cos(3kCx) (6.345) This problem can be solved completely, but it is sufficient to solve the (31) problem to learn the nature of bifurcation and the value of A This can be done exactly as in case 1. However, the algebra is cumbersome and we will stop here as far as case 3 is concerned. Up till now we have worked out three cases of pure solidification, and we have learned that the solidification front speeds up, no matter what. By contrast we learned in Chapter 5 that a precipitation front always slows down once the instability sets in. Now these two problems are closely related and their physics are very similar. Then one might ask: What is the reason for this fundamental difference? In other words, can we explain this basic difference of a "speedup" in solidification versus a "slowdown" in precipitation? In our quest to answer this question, we go to a yet simpler model, where the solid phase is completely ignored. Nothing new will be learned, and we will again see that the front speeds up compared to its base value. But the model is now as close as one can get to the precipitation model, and we can therefore hope to learn the fundamental difference between the two problems. Case 4: Solid Phase Ignored, Latent Heat Rejected Only through the Subcooled Liquid The notation remains the same as in case 1. Since the solid phase is altogether ignored, all lengths are scaled by L the depth of the liquid phase. All speeds are scaled by a / L The scaled T T temperature is given by Tscaled AI. Then the equation for the scaled liquid side aLH i temperature is given by dT* V T + U = 0 (6.346) for Z < : <1 and 0 < x < W The equations along the interface, viz., along : = Z(x) are T' = [2H](6.347) and n .VT*= Uk. n (6.348) where ~a denotes aLLT, The temperature at the liquid side sink is given by T*(z =1) =T, (6.349) Side wall conditions are Neumann. The volume of the subcooled liquid must be maintained fixed, whence Iz(x)dx= 0 (6.350) The Base Solution A base state solution is given by To*(z) =1 evoZ (6.351) where Uo is given by Uo = In(1 T,) (6.352) dT'* d27'* d37'* d4 '* whence o(z = 0) = Uo, a (z = 0)\ = U z=0)=U, a( =0 U dz dz2 \ V dz3 \~V Odz4 \ I The Perturbation Eigenvalue Problem Expanding Z, T, and T*, at z = Z(x) as in Eqls. (6.17, 6.19), the perturbation eigenvalue problem is given by dT' V27;* + Uo I = (6.353) 8z on the domain, O < z <1, O < x < W, where, at z = 0, 7 d2Z, dT (=A 'Z, o (6.354) dx C21 dz and a7T* d2 * S+ Z, O = 0 (6.355) 8z dz 2 must hold, and where, 7* (z = 1) = (6.356) must hold. The sidewall conditions are Neumann. Then the problem has solutions of the form 7* = ('(z) cos(kx) and Z, =A cos(kx) (6.357) which satisfy the side wall conditions so long as k = nzr/W, n = 1, 2..., where k is the wave number of the disturbance and n = 0 is ruled out by the condition SZ,dx = (6.35 8) The expansions presented above give the equations for 7;' and Z, They are k2 ,*+U = 0 (6.359) on the domain, O < z <1, O < x < W, ;* =k2dTod Z, (6.360) and d T d2 '* S+ Z, a = 0 (6.361) dz dz 2 at z =0, and 7;(z = 1) = (6.362) U, + U+4k' Defining m, = ", a solution to Eqs. (6.359, 6.361, 6.362) is given by (z) = k e "'mm_ e"' Z,~1] (6,.363) We can then turn to Eq. (6.361), which gives us Ik 1 N= X (6.364) where N=. 1e[non] The neutral curve is as shown in Figure 6.2, whence only one crest is seen at critical. Now on increasing TL to its critical value T~ and then slightly beyond, the branching to the new steady solution is expected be a forward pitchfork. Hence we write TL TL + ( 6. 3 65) and expand U and Z as in Eqs. (6.42, 6.43) to get the first and higher order problems. Eq. (6.352) indicates that for a small increase in TL fTOm T~ the corresponding change in U from LL,, per unit change in TL is given by eU. Before moving on to solve the higher order problems, observe that even for this simple case where the solid phase has been completely ignored, 7;' turns out to be nonzero. Now in Chapter 5, the variable c, was found to be zero due to cancellations of mappings in the mass balance equation at the interface, viz., n2.Vc = [1 c]Uk.nr The corresponding cancellations are not seen in the interfacial energy balance equation in solidification. And mathematically, this is what leads to the key difference between the two problems. The First Order Problem The first order problem is given by dT' dT* V 2 + Uo = Uz O (6.3 66) 8z dz on the domain, O < z <1, O < x < W, 7 d2Z, dT ('=1 'Z, a (6.367) dx C2 dz and BT' d2 * S+Z1 O = U, (6.368) 8z dz 2 at z =0, 7* (z = 1) = (6.369) and IZ,dx = (6.370) The sidewall conditions are Neumann, and Eqs. (6.366, 6.3686.370) can be solved by writing 7;'(x, z)= 7;0 (z)+ 7'T(z)cos(kx) (6.371) Z, (x) = A cos(kx) = Z, cos(kx) (6.3 72) The {10} problem The { 10 } problem is given by d27'* dT'* dT'* i0 + U = U, O (6.373) dz2 0dz dz on the domain, O < z <1, O < x < W, dT* 10 = U1 (6.3 74) dz at z =0, and 7;(z = 1) = (6.375) A general solution to Eq. (6.373) is given by I,(z) = Az'eo "+e e~i~'[oz evo B' U, ev (6.376) dT'* whence 'o (z = 0) = B'~ UA' Using Eqs. (6.374, 6.375) to eliminate Az' and B we get dz oa (z) = U,[~ zeoz_ (6.377) The {11} problem The { ll } problem is si;mlarto+ the eigenvallue problem. It+ isgivn by~ d 2 '''i dT' k2 U l= 0 (6.378) dz 2 dz on the domain, O < z <1, O < x < W, dT' d2 * 11 + Z,, a = 0 (6.379) dz dz 2 at z =0, and 7;(z = 1) = (6.380) Finally using Eq. (6.370) and dZ, /dxe = 0 at x = 0, W, and integrating Eq. (6.367) over 0 < x < W, we find that U, must be zero, and therefore 7;0 must be identically zero. Then the first order problem is exactly like the eigenvalue problem, and the condition at neutrality is given by Eq. (6.364), viz., [lk2 Uo N= U2 What we have so far is this: UI, = 0, 7* (x, z) = 7;' (z) cos (kx) = 7;' (z) cos (x) and Z, (x) = A cos(hx) = Z, cos(x) And we go on to the second order problem to find the value of A . The Second Order Problem The second order problem is 8T2' dT* V2T2' + Uo = U2 0 (6.381) 8z dz on the domain, O < z <1, O < x < W, 7T* ,d2T*; d2 T T2+Z Z __ 2Z o (6.382) d z dz 2 2 2C dz and aT' 82T* ,d3T' d27 7 2 +2Z +Z2 0 + Z2 0 2Z, = U2 (6.383) 8z 8z2 1dz3 dz2 1x at z =0, and T2 (Z = 1) = 1 (6.3 84) and IZ~dx= (6.385) Then, we assume the following form for the solution: T2 (X, Z) = T(Z) + T~(Z) COs(kx)+ +7 (z) cos(2kx) (6.386) Z2 (x) = Z21 COs(kx) + Z22 COs(2kx) (6.387) Andl again, wer use these ex~pansins to deducel the {20} and {21} and {22} nhprobes The {20} problem The {20} equations~ ~ a;re ivn byr d2T* dT' dz 2 0dz dT* U, o (6.388) on the domain, O < z <1, O < x < W, dT' 1 20 = U2 + Z,2 z dz 2 at z =0, and 7' (z = 1) = 1 A general solution to Eq. (6.388) is given by (6.389) (6.390) I' (z) = A' eo+eoeo Bo U2 ']Uo (6.391) dT'* whence 20 (Z = 0) = dz 7 (z) = 1+ Uf co B20 UoA20 Using Eqs. (6.389, 6.390) to eliminate Ao and B0, we get Z8Uz (6.392) whence r 1 7,, (z = 0) = Uf~ Z,2+ Uo " 2 (6.393) (6.394) " C'ozZ]2 U2[ Uo 1ii; i: The {21} problem The {21} problem is~ given by d2 dT* dz 2 dz on the domain, O < z <1, O < x < W, dT 21 dz d2 '* + Zz O 21dz2 (6.395) at z =0, and T'(z = 1)= 0 The {22} problem The {22} problem is given by (6.396) d2 2 dz 4k2 T* + U=O22=0 (6.397) on the domain, O < z <1, O < x < W, _ak2 +U Z,2];;+U Z2 22 (6.398) (6.399) at z =0, and 1' (z = 1) = Using the curvature equation Finally using Eq. (6.385) and dZ2/dx e 0 at x = 0, W, and integrating Eq. (6.367) over 0 < x < W, we find '* (z=0+ '(z 1 ^d2 *; )+ , 2 0 (Z 2 dz 2 (6.400) d T' Using (z dz 0) = UfZ, and a j(z Uf, there obtainsJ T;: (z = 0) = UfZ, 2 (6.401) d'22=2k Uo dz =2?( Comparing with Eq. (6.393), we get U = evo +U,2 Z12 (6.402) 2 U2planar U~correckon> The correction term is always positive, whence the mean speed of the interface runs ahead of its base value. Now the model given in case 4 is the closest that one can get to the model of precipitation. Yet, we find a "speedup" in case 4 while the precipitation problem exhibits an unconditional "slowdown" once the instability sets in. The only difference between the two models is that the energy balance equation in the present solidification model is different from the mass balance equation of the precipitation model. Hence we have learned that the key disparity between the two problems is that mass is fundamentally different from heat. In the most basic terms, there is no such thing as a hidden mass or latent mass. Now notice that the result of a "speedup" in solidification is obtained at second order, and this result holds regardless of the values of the input parameters k2 and L But recall from the third order derivations for case 1, that the sign of A2 depends strongly on the inputs k2 and L and hence the nature of the bifurcation is a function of the input variables. Although this was clear from Table 6.1, we had to cater to numerical calculations and it was hard to tell the functional dependence of A2 On the inputs simply by inspection. In our quest to delineate this dependence, we move on to the next case where the latent heat is rej ected to the frozen solid as well as to the subcooled liquid, as in case 3, but now we drop U from the domain equations. It will be seen that this simplifies the model equations and also the formula for A2 Case 5: Latent Heat Rejected to Frozen Solid as Well as to the Subcooled Liquid, Both Phases of Finite Extent, U Dropped from Domain Equations In this section we work out a special case of case 3 where U has been dropped from the domain regardless of whether or not it is small. For small values of growth speed U, this is a reasonable approximation. This is done solely with the purpose of making the algebra more tractable. Once again, we will learn that the front speeds up at the onset of instability. What we also wish to learn is whether A2 is found to be positive or negative at third order. In other words, we wish to find whether the pitchfork bifurcation is forward or backward. We will find that A2 can have either sign, whence the bifurcation can be forward as well as backward depending on the inputs k2 and the ratio of the depth of the two phases. To see this, we start by rewriting the nonlinear model. All scales remain the same as in case 3. Then, the scaled temperature in the solid phase must satisfy V2T = 0 (6.403) for 1< z < Z and 0 < x < W, while in the liquid phase V2 = 0 (6.404) must hold for Z < z < L and 0 < x < W The equations along the interface, viz., along z = Z(x) are T= JI[2H]= T (6.405) and n.VTn.VT = .Uk(6.406) where 1a denotes r ts1T, Farfield conditions are given by T(z = 1) = Ts (6.407) and T(z = L) = T, (6.408) Side wall conditions are Neumann. The volume of the subcooled liquid must be maintained fixed, whence Sz(x)dx= 0 (6.409) The Base Solution A base state solution is given by To(z)= Tsz (6.410) and T*(z) = z (6.411) where Uo is given by Uo = Ts+ (6.412) Notice that both phases have linear base temperature gradients, and this leads to a lot of simplification at higher orders as all second and higher derivatives of To and To' vanish. The Perturbation Eigenvalue Problem Expanding Z, T',and T*, at z = Z(x) as in Eqs. (6.176.19), the perturbation eigenvalue problem is given by V27 = 0 (6.413) on the domain, 1< z <0, < x V2;* = 0 (6.414) on the domain, O < z < L, O < x < W, where, at the base surface z = 0, d2Z, dT, 7;= 11 Z o (6.415) dx C21 dz ".] j+Z,d a = (6.416) and = 0 (6.417) 8z 8z must hold, and where, 7(z = 1)= 0 (6.418) and 7* (z = L) = (6.419) The sidewall conditions are Neumann. Thus the problem is homogeneous in 7;, 7;" and Z,, and it has solutions of the form 7;= (z) cos(kx), 7;' = 7;' (z) cos(kx) and Z, = A cos(kx) (6.420) which satisfy the side wall conditions so long as k = nzr/W, n = 1, 2..., where k is the wave number of the disturbance. n = 0 is ruled out by the condition IZ,dx = (6.421) The expansions presented above give the equations for 7 7;' and Z, They are d2 k2l ( =0 (6.422) in the solid phase, 1< z < 0, O < x < W, and d2 k2*=0(6.423) in the liquid phase, O < z < L, O < x < W, (6.424) (6.425) (6.426) (6.427) 7*(z = L) = 0 A solution to Eqs. (6.4226.425, 6.427, 6.428) is given by (6.428) 7;(z) = A, [e z e2k k _ 0) I[ dT"( d z 1 ek 2k e Z, (6.429) and _ak2 7* (z) = A; [e e"zue = 0) e dT'* (z dzza 1 e~ Pe e Z , (6.430) We can then turn to Eq. (6.426) which gives us c~=[a~;r,][ l~tanh kL tn The following formulae might be helpful while drawing a curve of Un versus k2 (6.431) andz U'7;_d Td 7;*= ddz dz at := 0, 7(z = 1)= 0 and ek tanh k = ek k k1 k' 3 1 lim tanh k kZ small lim tanh k kZ large dU ~~" 2 [ak2 + Ts] for all values of L . Figure 6.5 Typical neutral curve for solidification when the latent heat is rej ected through both phases; U is dropped from the domain equations kZ~malim [ tanh kL1 k smll tanh k3 lim U, =Z lim lak +Ts 1+tanhkLl~ =~ .k +Ts [1+L]+k 1L[1L][1Lr k sml ksmall tanh k 3" ,,,, '1 lImU, = Ts[1+ L] 0> = 3L~[1L[1+ L] Ts+_a[+ L] lim li T~k2~[1tanh kL kZ large kZ>1arge tanh k L>1 UI r=i Figure 6.5 shows the corresponding neutral curve. Given W, and hence k2, Eq. (6.431) determines the critical value of the interface speed, For L <; the curve is monotonically increasing, hence the most dangerous value of n is one, which corresponds to k = yr/W, giving only one crest at critical. For L >1, the curve must show a dip whence multiple crests are possible. As the control variable TL increases upward from zero, the output variable U, increases from zero, until there comes a point when Eq. (6.431) is satisfied. The corresponding value of LL, will be referred as its critical value. Henceforth the subscript zero will refer to this critical state. At values of U higher than LL,, the base solution loses its stability. Now on increasing TL to its critical value T~ and then slightly beyond, the branching to the new steady solution is expected be a forward pitchfork. As before, we write TL TL +r (6.432) and expand U and Z as in Eqs. (6.42, 6.43) to get the first and higher order problems. Eq. (6.412) indicates that for a small increase in TL fTOm T~ the corresponding change in U from EL,, per unit change in TL is given by The First Order Problem The first order problem is given by V (; = 0 (6.433) on the domain, 1< z< 0, < x V (' = 0 (6.434) on the domain, O < : < L, O < x < W, d2Z 7;= 11 ' dxC2 dT, 1dz (6.435) (6.436) (6.437) (6.438) at z =0, and 7(z = 1)= 0 and 7* (z = L) = 0 (6.439) The volume condition is given by IZ,dx = 0 The sidewall conditions are Neumann. Hence we have solutions of the form 7(x, z)= 7;0 (z)+ 7, (z)cos(kx) 7* (x, z)= 7;0 (z)+ 7'T(z)cos(kx) Z, (x) = A cos(kx) = Z, cos(kx) The expansions presented above are used to deduce the { 10: }, and { l }problems The {10} problem The { 10: }prolem is~ given by~ d27; dz 2 (6.440) (6.441) (6.442) (6.443) (6.444) dT do [7;~ ddz ddz, in solid phase, 1 = 0 dz2 in liquid phase, O < : 7;o 7;, = 0 and (6.445) (6.446) (6.447) d;dz d7dz at = 0, 7;(z = 1)= 0 and (6.448) 7;(z = L) = 0 A solution to Eqs. (6.4446.449) is given by '; (z)= Uz [z 1] 1 The (11) problem The (11) problem is si;mlarto+ the eigenva~lue problem. It+ is givn by~ dlU~I JJIIIL VLI ljll LU lU~I.I JflV~IU k2 =l~ 0 (6.449) (6.450) (6.451) (6.452) in the solid phase, ds li.= 1<:<0, O (6.453) in the liquid phase, O < z < L, O < x < W, dT dTo dz dz and dT; dT' "= (6.455) dz dz at z =0, 7;(z= 1) =0 (6.456) and 7;(z = L) = 0 (6.457) Finally we can turn to Eq. (6.435), which is a differential equation for Z, Using Eq. (6.440) and dZ,/dxL~ = 0 at x = 0, W, and integrating Eq. (6.435) over 0 < x < W, we find that U, must be zero, and therefore 7;0 and 7;0 must both be identically zero. Then the first order problem is exactly like the eigenvalue problem, and the condition at neutrality is given by Eq. tanh kL (6.41), iz.,Uo Alk + T 1+ What we have so far is this: tanh k ;= 7; (z) cos (kx) = 7 (z)cos (kx) 7;' = 7;' (z) cos (kx) = 7 (z) cos (kx), U, = 0 and Z, (x) = A cos(kx) = Z, cos(kx) And we go on to the second order problem to find the value of A. The Second Order Problem The second order problem is V2T2 = 0 (6.458) 200 on the domain, 1< z< 0, < x V 7' = 0 on the domain, O < : < L, O < x < W, (6.459) dx , aT +2, + 22 8: dT, Z 0 dz (6.460) (6.461) T, Ii7'd~d + Z,+Z = and z +2 2Z h' d (6.462) at the base surface : 0 The farfield conditions are given by r, Z = 1 = 0 and T' (z = L = 1 The volume condition is given by SZ,dx = 0 (6.463) (6.464) (6.465) The stricken terms are zero because of results from previous orders. Assuming that the above equations have a solution of the form T,(x, z) = T,, (z)+ T,,(z) cos(kx)+ T,,(z) cos(2kx) T' (x, z) = 7',(z)+ T'(z) cos(kx)+ ,',(z) cos(2kx) Z, ( = Z,, cos(kCx)+ Z,, cos(2kx) we deduce the (20) and (21) and (22) problems. At : (6.466) (6.467) (6.468) 0, the following formulae may be useful : 82 a7;Z k2UZ,2 Z 1+COs (2kx)  k2Uo, Z2 [1 COs (2kx)l (6.469) (6.470) T20 T' = 0 and dT, dT' dz dz (6.471) (6.472) at z =0, T20 Z, = 1) = 0 and (6.473) T2' (z = L,) (6.474) A solution to Eqs. (6.4696.474) is given by 202 8 T 8 (' The {20} problem The {20} problem is given by d2T, dz 2 in the solid phase, 1< z <0, < x dz 2 in the liquid phase, O < z < L, O < x < W, T20 Z) = (6.475) (6.476) (6.477) (6.478) (6.479) dT' 21 =0 dz dT 21 dz (6.480) at z =0, T21 (Z = 1) = 0 (6.481) m7 i (z=L)=0 (6.482) A solution to Eqs. (6.4776.482) is given by T21 Z) = 1z [ekz ekkz_ 0) kz dT, SZ dz2k 1 ek 2k kzZ21 (6.483) 203 U2L+1 [z +1] L+1 T20 Z)= [zL+1 LL+11 The {21} problem The {21} problem is similar to the p eignaolue problem. It is given by [d2 ; = in the solid phase, 1< z <0, < x sd k2 T* = 0 in the liquid phase, O < z < L, O < x < W, ,,dT, dTo; dz dz and * (z)= Az* [e" ee k"= ak The (22) problem The (22) problem is~ given by~ dlU~I JflV~IU 4kT= 0 1]ek dT'* (z 1 _za 2&e eZ (6.484) (6.485) in the solid phase, 1 (6.486) (6 .4 87 ) dT, dT~ dz dz 2Unk2 Z (6.488) at = 0, T, (z = 1)= 0 and (6.489) T,,(z = L) = 0 A solution to Eqs. (6.4856.490) is given by T,? =') A^ [e e4k ~ T,, (= A., [e c e4kL lb (6.490) (6.491) (6.492) 204 in the liquid phase, O < : < L, O < x < W, T,, T,, = UnZ , and where 1 tanh 2k 643 1, e 4 tanh 2k + tanh 2kL]r, a 2 Z 1 tanh 2kL A' =[ 4kL Uo Z22 k+ ah k + k tanh 2k 2,2 (6.494) Using the curvature equation Finally we can turn to Eq. (6.460), which is a differential equation for Z2 .Using Eq. (6.465) and dZ2 /dx~ = 0 at x = 0, W and integrating Eq. (6.460) over 0 < x < W, we find T20 Z = 0) + ~, '( = 0) = 0 (6.495) dz Usig (z =0) = tn k]ak2~; ZS and T20Z = 0) =[, ,1+1 there obtains dz n tanhl;[ l k1 L +19 planarr U2correction>0 where the correction term is again seen to be always positive, whence the mean speed of the interface runs ahead of its base value. To go to third order, we need to solve Eq. (6.460) for Z2 To find Z2 a solvability condition must be satisfied. It is T2(Z =0) +zz 27;Z, (z = 0 Zd 4 z= )Zd = 0 (6.497) 0L 0 Again this condition is automatically satisfied, whence Eq. (6.460) can be solved for Z2 ' Substituting the expansions from Eqs. (6.4666.468) into Eq. (6.460), the constant term disappears due to the result obtained earlier using integrability, viz., Eq. (6.495). The coefficient 205 of the first harmonic vanishes due to solvability, and there obtains by equating the coefficients of the second harmonic T22 Z=0)+ '(z= dz (6.498) Substituting T22 Z = 0) ,we get (6.499) (6.500) (6.501) (6.502) (6.503) on the domain, < z dZ 2 d2Z BT 9 + T3 +3Z 2 2 1 BZ +3Z2 1 d2Z LC2 dTo Z 3dz (6.504) 206 dTIj~oJa].= O=tankhk[J kI] A22 _l 4k and using (I z 1e4 e [~r; la tanh k r 12 Comparing with Eq. (6.493) and simplifying, there obtains where k tanh 2k P tan~h k]l il]l, o Itanh 2k + tanh2kL] e [4aPI~C~tr[ tanh 2k + tanh 2kLtah2 Hence, Z21 and A carry on to third order. The Third Order Problem The equations at the third order are given by V2T3 = 0 on the domain, 1< z <0, < x V2T3' = 0 84 3 T3Z l3ly +3Z ? +3Z2~ +3Z (6.505) dT[ dTo; and 371: +3Z1~ 7 + C3Z2 1 +3Z2* 1 8z z 8z 2~Y i3Z BZ 3 B 2 B 3Zi 2Z 3Z~x =U3(6.506) at z =0, T3 (z = 1) =0 (6.507) and T3'(z = L,)= 0 (6.508) The volume condition is given by j zdx= 0 (6.509) The stricken terms vanish at z = 0 on account of results from previous orders. We assume the following form for the solution T3 X, Z) = T30 Z) + T31 Z) COs(kx) + T32 Z) COs(2kx) + T33 Z) COs(3kx) (6.510) T3 (X,Z) = 730 Z)+ 3 Z) COs(kx)+ T3 Z) COs(2kx)+ T3 Z) COs(3kx) (6.511) Z3 (x) = Z31 COs(kCx)+ Z32 COs(2kCx)+ Z33 COs(3krX) (6.512) This problem can be solved completely but it is sufficient to solve the {31} problem to learn the nature of bifurcation and the value of A2 207 The (31) problem is~ given by dlU~I JflV~IU 2 k T = 0 (6.513) in the solid phase, 1<:<0, O (6.5 14) (6.515)  k U, Z, Z 2 (6.516) dT dr 1 = dz dz at z =0, T 1(z = 1)= 0 and (6.517) T~ (z = L)=0 O A solution to Eqs. (6.5136.518) is given by T, (z) = A21 ek:_62 k e"] and T3 (z) Az, [ek: e2kL k= where (6.518) (6.519) (6.520) 208 The (31) problem Sdzd x T in the liquid phase, O < : < L, O < x < W, ,, 3 T,, T,, = U 23, + 3U, Z +k U, Z, and (6.521) (6.522) Using the curvature equation Finally we turn to Eq. (6.504), which is a differential equation for Z3 To find Z3, a solvability condition must be satisfied. It is 8z 2 7 aT 0)+3Z, 2 Z 8z 0) +3Z2, 1Z dZ (6.523) ;(z = 0)Z~dx = whence 0) 7,+ 9) d' 2 T (z 4 dz dT 7 22Z dz 0)13 dl T~ 2 dz 0) Z22 (6.524) (6.525) + ~ak4 Z,3 [ak2 + T, Z31 = 0 which can be written as A3, 1 _e2k = R3 where 9 d2 T +4 1' (z 1 dT , 0) + 22Z 2 dz 0)~3 dl Tz 2 dz 0) Z22 (6.526) + ak4 Z,3 ak2 + T Z17 which can be simplified to get 1 tanh k 3 3 1 e[2kI tanh k + tanh kL 42,Z 2 i tnlL,1 1 tanh kL 3 3 1 e tn k+tahk 42 S ~( oi. dZ 2 d2Z, 0)9A dx 1?,(30,3~~dT j R= dT ,d UL1 k k R3=3 1r UL +1 tanh 2kyr tanh k]a: lf1 (6.527) +9 [_k2 +k TS] Z,3 3 k Ak2 S 1 Z22 9 _ak4 Z,3 [ak2 TS] Z31 Comparing the value of A31 fTOm Eq. (6.521) and (6.525), and using Eq. (6.527), we get after some simplification P2k tanh kL ah2 1 k+~ Q~~ ~~~ 2 ah2 A2T tanh k Ak S 1+A (6.528) Uo tanh kk ]r]L 1 Uo k k k 4s tanh k tanh kI tanh 2k 1Iak2 TS where T is given by Eq. (6.501). Again, the sign of A2 is of interest. If it is positive, the pitchfork is forward. On doing calculations it will be learned that A2 can be positive as well as negative depending upon the input parameters. But in order to make our analysis yet simpler, let us put Ts = 0 in our results for this case. Case 6: Latent Heat Rejected to the Subcooled Liquid Only, Both Phases of Finite Extent, U Dropped From Domain Equations Putting Ts = 0 in the results of case 5, there obtains at the base To(z)= 0 (6.529) and To*(z = z(6.530) where U, is given by _T U, =L(6.531) The condition at neutrality is given by tanh kL U, = A' 1+ (6.532) tanh k Figure 6.6 shows the corresponding neutral curve. The curve is monotonically increasing for all values of L Hence the most dangerous value of n is one, which corresponds to k = yr/W, giving only one crest at critical. L>1 L=1 L<1 Figure 6.6 Typical neutral curve for solidification when the latent heat is rej ected through the subcooled liquid only; U is dropped from the domain equations At second order, we will find a "speedup" and U, is given by the formula 1 k1 Ut = + k 1Z (6.533) L tanh k],:, L Z2planar U~correchon Z,, is found as Z,,P = Z where k k tanh 2kL 1+ tanh 2kL.tanh 2k P tanh k 1+ tanh 2kL.tanh 2k (6.534) Q! 1+ tanh 2kL.tanh 2k 1 + tanh 2kL.tanh 2 And at third order we find A2 as P ktnhk 4 k 1r k Q] 2 ahk tanh 2k 2 tanh kI 1 k1 1 Utah k 1,~ +i A (6.535) k k k 7 tanh k tanh kI tanh 2k 4 k Our j ob then is to find the sign of A2 To alccomplish this, we will consider two subcases: one where k2 is large, and another where k2 is small. Analysis for Large k : Since tanh k approaches 1 for large values of k we have lim U, = 2ak2 and kZ large P 1ec 1sn 1q 363) 1e 1e 1~ lim =0hecusnEq(655,wge k k kA >~large Q! 2 2 L 4 LU, L 2~ak2 whence neglecting the lower order terms in k2 there obtains 2111 A2 < 0 (6.536) 3a L k4 We learn that the bifurcation is always a backward pitchfork for large values of inputs k2 no matter the value of L Table 6.3 gives the value of the critical speed U, and A2 for ~a = 0.26 x106 and for two different values of L The value of U, is calculated using Eq. (6.532). The values of A2 were obtained from Eq. (6.535) but they can be calculated directly from Eq.(6.536). Notice that since k2 is large the value of the critical speed Uo does not depend on L . Table 6.3 Case 6 Values of Uo and A2 for inputs L = 1, 2 and for large values of k2 L=1 L=2 k2 Uo A 2 Uo A 2 103 5 x104 2.667 5 x104 1.333 104 5 x 10 2.667 x 10 5 x 10 1.333 x 10 105 5 x 102 2.667 x 104 5 x 102 1.333 x 104 106 0.5 2.667 x 106 0.5 1.333 x 106 10 5 2.667 x 10 5 1.333 x 10 10s 50 2.667 x 1010 50 1.333 x 1010 Analysis for Small k2: Since tanhk approaches k for small values of k2 We have lim Uo, = [1+L]Lak2 and k2 small P l lim ,hence using Eq. (6.535), we get k241arge Q! 3) 3 L1 1 1 1 1 A2 = 3 a (6.537) Clearly the nature of the bifurcation depends on the ratio of the depths of the two phases. For 3 3 L < we have a backward pitchfork whereas the pitchfork is forward for L > . Table 6.4 2 2 gives the value of A2 for ~a = 0.26 x106 and for two different values of L The values of A2 were obtained from Eq. (6.535) but they can be calculated directly from Eq. (6.537). 213 Table 6.4 Case 6 Values of Uo and A2 for inputs L = 1, 2 and for small values of k2 L=1 L=2 Uo A 2 Uo A 2 10 5 x 10 6 x 101 7.5 x 10 4 x 101 107 5 x 1014 6 x 1013 7.5 x 1014 4 x 1013 106 5 x 101 6 x 101 7.5 x 101 4 x 101 105 5 x 1012 6 x 1011 7.5 x 1012 4 x 1011 10 5 x 10" 6 x 100 7.5 x 10" 4 x 100 103 5 x 1010 6 x 109 7.5 x 1010 4 x 109 L=1 L=2 k2 Uo A 2 Uo A 2 104 5 x 10ii 6 x 1010 7.499 x 10"1 4 x 1010 10 5 x 101 5.99 x 109 7.495 x 101 4 x 109 102 5 x 109 5.94 x 10s 7.451 x 109 3.85 x 10s 10 5 x 10 5.41 x 10 7.07 x 10 3.34 x 10 1 5 x 107 2.215 x 106 5.664 x 107 3.59 x 106 10 5 x 106 2.66 x 104 5.01 x 106 1.49 x 104 102 5 x 105 2.66 x 102 5 x 105 1.37 x 102 103 5 x104 2.66 5 x104 1.3547 104 5 x 10 2.67 x 10 5 x 10 1.34 x 10 105 5 x 102 2.67 x 104 5 x 102 1.33 x 104 106 0.5 2.66 x 106 0.5 1.33 x 106 1, 2 and k2 Table 6.5 Case 6 Values of Uo and A2 for inputs L = 214 Then defining Lcrsoe we have the following result: For L < Lcrossove the pitchfork is backward for small as well as for large values of k2 For L > Lrossov,, the pitchfork is forward for small values of k2 While it is backward for large values of k2 This is shown in Table 6.5 which shows the values of A2 for 11 = 0.26 x106 and for L = 1, 2 over a wide range of k2 Observe that for L > Lrossov,, A2 flipS sign for some intermediate value of k2 and the bifurcation type changes from a forward to a backward pitchfork. 21 A2 O A2 OC+ 2k4 Branching flips from L Forward forward to backward Backward Pitchfork >Pitchfork 1 1 A2 OC  A2  k2 k4 Backward Backward Pitchfork Pitchfork Small k2 k2 Large k2 Figure 6.7 Transition diagram of L versus k2 Showing regions corresponding to forward and backward pitchfork The reader might also be interested in comparing the results of Table 6.5 with those calculated in Table 6.1 where U was not dropped from the domain equations. The two tables list the outputs Uo and A2 for the same values of inputs k2 and L Observe that the results are in excellent agreement unless k2 is large enough that Uo is not small anymore and dropping U from the domain equations is not a good approximation. 215 Finally we show a transition diagram, cf., Figure 6.7 which shows windows in the parameter space (L, k2) Where we have forward and backward pitchfork. Discussion We have identified regions where the bifurcation might be backward and regions where it might be a forward pitchfork. We obtain a backward pitchfork for most cases, unless L is greater than Lorossove and the input values of k2 are small. Hence a forward pitchfork is easy to get for experimentally realizable values of L and k2 if One just knows where to look. A forward pitchfork is also called a supercritical bifurcation (pitchfork opening to the right) while a backward pitchfork is often termed as a subcritical bifurcation (pitchfork opening to the left). Figures 6.8 and 6.9 show schematics representing the two types of bifurcations. In both these cartoons we plot the control parameter ii on the horizontal axis and the square of the amplitude of the nonplanar solution A2 On the vertical axis. For the solidification problem our control parameter is the undercooling in the sub cooled liquid. i.e., Ai = T, One might study the stability of the new branches (new nonplanar solutions) to prove that subcritical solutions are unstable while supercritical solutions are stable [6]. Hence the solid lines in Figures 6.8 and 6.9 signify a stable solution while the dotted lines are indicative of unstable solutions. stable A2 stable unstable stable Figure 6.8 A cartoon illustrating a forward/supercritical branching 216 stable A2 stbe ntal nsal unstable stable Figure 6.9 A cartoon illustrating a backward/suberitical branching In the case of a supercritical bifurcation, as we increase the control parameter, the instability will set in only after one crosses the critical value of the control parameter Ac Hence the supercritical case is in good agreement [3 1] with the predictions of the linear theory and gives us a solution when the planar solution is no longer stable to infinitesimal perturbations. And this corresponds to a smooth transition from a planar to a nonplanar solution. In the case of a subcritical bifurcation however, the solution can jump discontinuously to the new nonplanar steady state even before the critical point predicted by the linear theory [6]. However some other researchers [32] have emphasized that the prediction of a subcritical bifurcation should not be understood as evidence that finite amplitude solutions do not exist. Ordinarily then, the implication of a subcritical case is that the planar solution might be unstable to finite amplitude perturbations even though it is stable to infinitesimal perturbations according to the linear stability theory [10]. 217 Now Wollkind and Segel [33] have argued that the occurrence of a forward pitchfork is related to the formation of stable bands and a periodic structure. Hence the region of a forward pitchfork is also referred as a cellular regime while the region of a subcritical bifurcation is ordinarily considered to physically correspond to dendritic growth. Next, going back to the analysis of case 6, let us try to advance some arguments to relate the predictions of the linear and the nonlinear theory. The predictions of the linear theory are seen from the neutral curve, cf., Figure 6.6. It is clear that for small values of k2 the curve is very sensitive to the value of L A possible explanation is that for low values of k2 the effect of surface tension cannot be significant. What is important is the diffusion in the two phases and hence the ratio of the depths of the two phases plays a key role. For large values of k2 we see from Figure 6.6 that the neutral curve becomes asymptotic to U, = 2)1k2 for all values of L, hence the ratio of the depths of the two phases is not of much importance. This is reasonable as the effect of surface tension becomes dominating for large enough values of k . Now the prediction of the nonlinear theory for large values of k2 is that the nature of the branching is independent of L and it is always a backward pitchfork. And this is consistent with the predictions of the linear theory in that the ratio of the depths of the two phases cannot possibly have a significant effect on the critical front speed if one is running an experiment in the large k2 regime. However, for small values of k diffusion in the two phases is important compared to surface tension and hence the nature of branching depends on L What is more interesting is 3 3 this: We learned that for L < we have a backward pitchfork while for L > the branching is 2 2 forward. Now the linear stability theory predicted (cf., Figure 6.6) that a larger value of L ought 218 to give us more stability compared to a smaller L Then the fact that the weak nonlinear analysis predicts~~ ~~~ a agrvleo as opposed to a smaller value L ) to give a forward bending pitchfork is interesting because a forward pitchfork is ordinarily associated with the formation of cellular patterns (as opposed to dendrites) which are in turn associated with a lesser degree of disorder. A backward pitchfork is, on the other hand, related to the formation of dendrites which certainly have a higher degree of disorder, and that is what we get for smaller value of L < for small values of k2 Concluding Remarks We have performed a weak nonlinear analysis for the problem of solidification of a pure material. We have found that once the instability sets in, the front speed always runs ahead of its base value regardless of the values of input parameters. We have also learned that the nature of the branching of the planar solution to the nonplanar solution depends on the inputs k2 and L . We have identified regions in the (L,k2) plane where the pitchfork is forward and regions where it is backward. It is possible to obtain both in physically realizable experiments. 219 CHAPTER 7 CONCLUSIONS AND RECO1V1VENDATIONS In this chapter, we summarize the main results of this dissertation and propose relevant future work. This dissertation has focused on understanding the underlying physics and causes of morphological instability in two phasechange problems, namely precipitation and solidiaication. The theoretical methods involve linear stability calculations and weak nonlinear calculations. The maj or Eindings of the linear stability analysis are these: First, the solidiaication instability can give rise to the possibility of a multiplecell pattern at the onset of instability whereas for other growth problems involving solidliquid interfaces, such as precipitation from a supersaturated solution, the onset of an instability from a planar state can result in at most a single crest and trough. Second, we have found that a solidliquid front that grows on account of temperature gradients is the only one amongst all solidliquid front growth problems where a critical wavelength, independent of surface energy, can occur. This critical wavelength which is of the order of the size of the container can be obtained for reasonable and measurable values of growth speed for most systems of practical interest. This fact alone sets solidification apart from any other growth problem involving a solidliquid interface where the critical wavelength must obtain for unreasonably low values of growth speeds. These conclusions demand that careful experiments be designed in order to validate the Endings. The experiment must be run so that one is able to creep up on the instability by slowly increasing the value of the control parameter to its critical value. It would be valuable to demonstrate in an experiment that the pattern evolving at the onset of instability is indeed consistent with the theoretical predictions. If the prediction of the nonlinear results is a forward pitchfork, then one might want to confirm that the critical value of the control parameter seen in the experiment is consistent with the theoretical value predicted by the linear stability analysis. 220 Also, one might want to check in the case of a backward pitchfork if the critical value of the control parameter seen in an experiment is very different from the prediction of the linear analysis. The main findings of the nonlinear calculations are these: First, a solidification front always moves ahead of its predicted base value upon becoming unstable. This is in contrast with other solidliquid growth instability problems such as precipitation where an unstable front always slows down compared to its predicted base value. Second, the nature of the bifurcation in solidification may be a forward pitchfork, leading to cellular growth, or a backward pitchfork, leading to dendritic growth. In contrast, other growth problems such as precipitation and electrodeposition are ordinarily characterized by dendritic growth in the post onset region. These findings raise several interesting questions that ought to be answered in future studies. One of the most important questions is this: Although the two phenomenon are very similar, why is the interfacial instability in solidification so different from that in precipitation? For instance, can we find out a physical reason as to why a solidification front always speeds up upon becoming unstable whereas a precipitation front slows down compared to its predicted base value? Why do we only see a backward pitchfork in a precipitation instability whereas a solidification instability gives rise to the possibility of a backward as well as a forward pitchfork? Then, one might investigate the question of crosssectional dependence on the nature of bifurcation beyond the onset of instability. In other words, can we ascribe a physical reason as to why the nature of the branching should depend on the shape of the cross section of the container in which the growth takes place? One might also want to validate such a cross sectional dependence in an experiment. Another important question that one might ask is whether we can specifically point out the algebraic terms in the nonlinear calculations that tend to favor a backward bifurcation and terms that want to make the branching forward. Finally, one might want to trace out the stable steady state solution in the post onset regime in the case of a backward pitchfork, and one might wonder why the morphology of the interface changes in a discontinuous and abrupt fashion instead of changing slowly and continuously. 222 APPENDIX A SURFACE VARIABLES In this appendix, we give the formulas for three surface variables, namely the unit normal vector, the surface speed and the mean curvature. This is sufficient as far as the theoretical work in this dissertation is concerned. The reader is referred to the book by Johns and Narayanan [3] for further details. A.1 The Unit Normal Vector L~et us define a free surface by z = Z (x, t) in Cartesian coordinates or z = Z (r, 0, t) in cylindrical coordinates. Also we define a functional that vanishes on the surface as .f = zZ (x,tf)=0O in Cartesian coordinates, and f=z zZ (r,0, t)= 0 in cylindrical coordinates. Then the normal vector pointing into the region where f is positive is given by Here, 7/ = fix+ 1 Ox 8z in Cartesian coordinates and V f = i, + i + 7 dr r 80 8 z in cylindrical coordinates. Then the equation for the normal is given by 223 8Z . dr in cylindrical coordinates, if I independence is assumed. A.2 The Surface Speed Let a surface be denoted by f (F, t) = Let the surface move a small distance As along its normal in time At Then, f (rf + s n, t + At) is given by af (r, t) nVf(r,t)+At +... at f(r+~s n, t + t)= f(r, t)+ s whence f (rE As n, t + At) = 0 = f (r, t) requires af (r, t) +As nVf(r,t) The normal speed of the surface, u, is then given by af (r, t) t 7f (r, t) Hence, using the definition of the unit normal given earlier we get 224 8Z . lx +1I in Cartesian coordinates, and at u= For the problems concerned in this dissertation, the definition of u becomes 8Z u= in Cartesian coordinates, and 8Z u= in cylindrical coordinates, if I independence is assumed. A.3 The Mean Curvature The book by Johns and Narayanan [3] gives a detailed derivation of the curvature for a general surface. In this appendix, we simply provide the formulas used in this dissertation. For a Cartesian surface defined as the mean curvature is given by 2H = Z2 3/2; where the subscripts denote the derivative of Z with respect to that variable. For a Cartesian surface defined as .f = zZ (x, y,t)= 0 the mean curvature is given by 2H =[1+2 ~ 2Z;~ 1+Z ~ 1l+ Z + Z2 3/2 APPENDIX B THE PERTURBATION EQUATIONS AND THE MAPPINGS In this appendix, we explain the perturbation equations and the mappings used in this dissertation. The reader is referred to the book by Johns and Narayanan [3] for further details. Higher order mappings are given in Appendix B.2. B.1 The Expansion of a Domain Variable and its Derivatives along the Mapping Let u denote the solution of a problem on a domain D Now the domain D might itself be inconvenient. It may not be specified and must be determined as a part of the solution. This becomes clear when we study a moving front problem such as precipitation or solidification where the shape of the interface is not known and it is a part of the solution. It is however possible to obtain the solution u as well as to solve for the domain shape D by solving a new and easier problem defined on a regular and specified reference domain Do Do may be determined by the original problem or it may be chosen so as to simplify our work. The key idea is to imagine a family of domains D, growing out of the reference domain Do and to imagine that u must be determined on each of these, one being the domain of interest, D Now imagine that the domain D being in the vicinity of the reference domain Do, can be expressed in terms of the reference domain Do via a small parameter e Therefore the solution u and the domain D are solved simultaneously in a series of companion problems. What needs to be done is to discover how to determine u in terms of the solutions to problems defined on Do .Let the points on Do be denoted by the coordinate yo and those on D by the coordinate y . The x coordinate is assumed to remain unchanged. Let u be a function of the spatial coordinate y Then u must be a function of E directly because it lies on D and also because it is a function of y The point y of the domain D, is then determined in terms of the point yo of the reference domain Do by the mapping y = f (y, ,) The function f can be expanded in powers of E as 8~f (yo~E = 0) 1 2 C2 0,e~~= 0) f~(y~E)= ~f(yo, E= 0)+e +e 2 " dE 2 dE2 where f (yo, e = 0) = yo and the derivatives of f are evaluated holding yo fixed. Then in terms of the notation ~f (yo,e = 0) y:, (yo F)= 82f 0y, E = 0) y2 0Yo 2) etc., the mapping can be written as y = f (yo, E) = Y0+ lyl(yo0,=0O)+e2y (n 2 ,= 0)+... The boundary of the reference domain must be carried into the boundary of the present domain by the same mapping. At the boundary of the new domain, the function y is replaced by Y to point out the difference. Its expansion in powers of E can be written similarly as Y = Yo + Y(Yo, = 0)+ eF2Y 2Y, ,= 0)+... Notice that it is the Y 's that need to be determined to specify the domain D in terms of the domain Do  Finally, the variable u (y, e) can be expanded in powers of e along the mapping as du*(y = yo~F = 0) 1 2 d'2* 0o,F= 0) u (y, e)= ut(y = yo, e= 0)+ E +E 2 " dE 2 de2 du where denotes the derivative of the function u depending on y and E taken along the dE du* ( y = yo, E = 0) mapping. To obtain a formula for differentiate u along the mapping taking dE y to depend on e, holding yo fixed. Using the chain rule, this gives du*(y,E) Cu(y,e) Bu;1(y,e) f (yo/~) dE dE By e Evaluating the above equation at e = 0, we get du*(y =yo,e=0) rBu(yo,,E= ) cu(y,,e= 0) + y, ( yo, E dE dE B Then, introducing the definition of u, via bu(yo,e = 0) and observing that bu(yo,eF= 0) uyo we get du*(y = yo,e = 0) (Iu,,(yo) All higher order derivatives of u can be determined the same way. If a domain variable needs to be specified at the boundary it is written similarly as du (y = Yo,e = 0) uYo de 1yo, When additional derivatives are obtained and substituted into the expansion of u, it becomes Co, 1 2 S+ u, 2 d2u, c u~~e=u~euzy, +62 2+y +y1 +y2 1 ". ao 2 dyo The reader might wonder that the mapping y, does not appear in the domain equations given in this dissertation. In fact, the mappings on the domain can not be determined and they are never required. This can be proved rigorously as done in the book by Johns and Narayanan [3], but in the interest of continuity, let us work out an example to show that one need not worry about the mappings on the domain. Once we have gone through this example, we can use this as a rule of thumb. Let u satisfy c~=0 in the inconvenient domain D. Using chain rule du du dyo where, holding E fixed, ais given by Dy By 2 Dy Thus, up to the first order in e, the domain equation becomes du cFu,, r Bu, 8Y dU2 0 .. Therefore, the domain equation at the zeroth order in E is "n,=0 d2U, whereas the domain equation at first order in E, using a= 0, becomes 0y Notice that the mapping is lost from domain equations. If surface variables were considered the mapping would not be lost, as is evident from all the problems studied in this dissertation. B.2 The Higher Order Mappings Up to the fourth order, the expansion of the domain variable u along the mapping is Fa, 1 2 S+ y u, 82 du / u = uo+e u y, + e2 2+2y ,+ +yz2 + 2 02 0o a 0 1 3 1 12 u 1u d2u d2 2 1 3 d3u 0 ( ,, +3 3 +3 3y2 + 3y y2 32 +3y 2y 1y 3 6 8 o if i a31 ijf a?I ifa?~j ;3: u4 +10Yy, y2 +6y2 2 +6y12 2 0 +4y 3+4y2 4z3 +yy +E 24 d2U 2 0 4 d4U 0 II~ +3y2 2 +2y3 +1 ,,4 +4 and the expansion of its derivative is given as du cFu,, ru, 8Y 2U 0 2 u, dU 2 1 3U 0 2 +e +y +E2 +2y, ~ y ~ + y2~~ S0 1 y, 2 dyo 2y 1y 3 2 S3d3 d22 d21 d30O 2 31 3 d40 d20O +3 +3y, + 3y2 + 3y,y2 y +3y,3 +y y, ~ + y3 6~ 8~ yoo Vif Vif I i i 83U 11~23U d4U d23 d22 l34 1 d3U u4 +10~y, y2 +62 +6~y2 2 0 +4yl 2~ +4y'2 ,~2 +4y3 ~4 +4y)1? y3 0 +E 24 ~2 d30 d21 4 50O d20 +3y2 3 +2y3 1, y ~ +4 2 APPENDIX C EIGENFUNCTIONS OF AN EQUILATERAL TRIANGLE AND A REGULAR HEXAGON SATISFYING NEUMANN BOUNDARY CONDITIONS In this appendix, we give the function f (x, y) which is an eigenfunction of V2, On an arbitrary cross section S, subject to Neumann sidewall conditions. It satisfies V2 f(x, y)= k2f (X, y) On S; p.Vf = 0 on dS where dS denotes the boundary of S and p denotes the outward normal to dS. Appendix C.1 gives the eigenfunctions and eigenvalues corresponding to an equilateral triangle. Appendix C.2 gives the eigenfunctions and eigenvalues of a regular hexagon. C.1 Eigenfunctions of an Equilateral Triangle Consider an equilateral triangle of side h whose vertices are given by (0, 0) (h, 0) and (h/2, Ih/2 The eigenfunction f (x, y) that satisfies V2 f(x, y)=k2/ xf) On S;, p.Vf = 0 on dS is given by [34] as f """ (x, y) = f """ (x, y) + fdn" (x, y) where f,"""(x,y)=cos [[3ry] cos [xh/2 ] 3r 9r 3r 9r and fd'" (x,y)=cosI' [3ry ( Js~in x h/2] r + O3r 9 I r 5i Y] I ;l SC i'Y3r 9 I sr I ;l where l= [nz+n] and r = 2~ The corresponding eigenvalue is given by C.2 Eigenfunctions of a Regular Hexagon Consider a regular hexagon of side L and centered at the origin (0, 0) The eigenfunction f (x, y) that satisfies V f (x, y)= k f (x, y) on S; p.Vf = 0 on c;S is given by [7] as The corresponding eigenvalue is given by 4i\3L REFERENCES [1] D.T.J. Hurle, Handbook of Crystal Gi ,1ral t, Vol. 1, Elsevier, Amsterdam (1 993). [ 2] A. Leal, Enhanced Morphological Stability in Sbdoped Ge Single Crystals, Thesis, University of Florida (2004). [3] L.E. Johns and R. Narayanan, InterfaciallInstability, Springer (2001). [4] S.C. Hardy and S.R. Coriell, "Morphological Stability and the IceWater Interfacial Free Energy," Journal of Crystal Gi ,~ r l t, 42, 24 (1 977). [5] S.C. Hardy and S.R. Coriell, "Morphological Stability of a Growing Cylindrical Crystal of Ice," Journal ofApplied Physics, 39, 3505 (1968). [6] S.R. Coriell, G.B. McFadden, and R.F. Sekerka, "Cellular Growth during Directional Solidification," Annual Reviews of2aterial Science, 15, 119 (1985). [7] S. Chandrasekhar, Hydrodynamnic and Hydromagnetic Stability, Dover (198 1). [8] M.C. Cross and P.C. Hohenberg, "Pattern Formation Out of Equilibrium," Reviews of Modern Physics, 65, 851 (1993). [9] W.W. Mullins and R.F. Sekerka, "Stability of a Planar Interface during Solidification of a Dilute Binary Alloy," Journal ofAppliedPhysics, 35, 444 (1964). [10] D.J. Wollkind and R.D. Notestine, "A Nonlinear Stability Analysis of the Solidification of a Pure Substance," IM4r Journal ofApplied Ma'~thematics, 27, 85 (1981). [1l] J.S. Langer, "Instabilities and Pattern Formation in Crystal Growth," Reviews of2~odern Physics, 52, 1 (1980). [12] R.T. Delves, Theory oflnterface Stability, In Crystal Growth (Pamplin R., Ed.), pp. 40 103, Oxford: Pergamon Press (1975). [13] J.W. Rutter and B. Chalmers, "A Prismatic Sub structure Formed During Solidification of Metals," Canadian Journal ofPhysics, 31, 15 (1953). [14] W.A. Tiller, K.A. Jackson, J.W. Rutter, and B. Chalmers, "The Redistribution of Solute Atoms during the Solidification of Metals," ActaMetallurgica, 1, 428 (1953) [15] S.H. Davis, Theory ofSolidification, Cambridge University Press (2001). [16] D.J. Wollkind and R.N. Maurer, "The Influence on a Stability Analysis Involving a Prototype Solidification Problem of Including an Interfacial Surface Entropy Effect in the Heat Balance Relationship at the Interface," Journal of Crystal Gi ,~ ral t, 42, 24 (1 977). [17] D.B. Oulton, D.J. Wollkind, and R.N Maurer, "A Stability Analysis of a Prototype Moving Boundary Problem in Heat Flow and Diffusion," The American Mathematical Monthly, 86, 175 (1979). [18] S.R. Coriell and R.F. Sekerka, "Effect of Convective Flow on Morphological Stability," Physicochens. Hydrodyn., 2, 28 1(198 1). [19] D.T.J. Hurle, E. Jakeman, and A.A. Wheeler, "Effect of Solutal Convection on the Morphological Stability of a Binary Alloy," Journal ofCrystal Gi ,1ral t, 58, 163 (1982). [20] J.J. Favier and A. Rouzaud, "Morphological Stability of the Solidification Interface under Convective Conditions," Journal ofCrystal Gi ,~ ral 64, 367 (1983). [21] D.T.J. Hurle, "The Effect of Soret Diffusion on the Morphological Stability of a Binary Alloy," Journal ofCrystal Gi ,1ral t, 61, 463 (1983). [22] S.R. Coriell and R.F. Sekerka, "Oscillatory Morphological Instabilities due to NonEquilibrium Segregation," Journal ofCrystal Gi ,~ ral 61, 499 (1983). [23] S.R. Coriell and R.F. Sekerka, "Effect of the Anisotropy of the Surface Tension and Interface Kinetics on Morphological Stability," Journal ofCrystal Gi ,~ r l t, 34, 157 (1976). [24] R. Sriranganathan, D.J. Wollkind, and D.B. Oulton, "A Theoretical Investigation of the Development of Interfacial Cells during the Solidification of a Dilute Binary Alloy: Comparisons with the Experiments of Morris and Winegard," Journal of Crystal Gi 1,n thr, 62, 265 (1983). [ 25] S. Strogatz, Nonlinear Dynamics and' Chaos: With Applications to Physics, Biology, Chentistly, and Engineering, Perseus Books Group (2001). [26] D.C. Montgomery, Design and Analysis ofExperintents, John Wiley and Sons Inc. (2000). [27] S.R. Coriell, R.F. Boisvert, G.B. McFadden, L.N. Brush, and J.J. Favier, "Morphological Stability of a Binary Alloy during Directional Solidification: Initial Transient," Journal of Crystal Gi ,1ral t, 140, 139 (1994). [28] R. Tonhardt and G. Amberg, "Simulation of Natural Convection Effects on Succinonitrile Crystals," Physical Review E, 62, 828 (2000). [ 29] P. Grindrod, Pattern and Waves: The Theory and Application of ReactionDiffusion Equations, Oxford University Press, (1996). [30] Q. Buali, L.E. Johns, and R. Narayanan, "Geometric Effects on Surface Roughness in Electrodeposition," IM4r Journal ofApplied2\a~thentatics, 71, 692 (2006). [31] S. de Cheveigne, C. Guthmann, P. Kurowski, E. Vicente, and H. Biloni, "Directional Solidification of Metallic Alloys: The Nature of the Bifurcation from Planar to Cellular Interface," Journal of Crystal Gi 1,n thr, 92, 616 (1988). [32] D. Wollkind, R. Sriranganathan, and D. Oulton, "Interfacial Patterns During Plane Front Alloy Solidification," Physica D, 12, 215 (1984). [33] D. Wollkind and L. Segel, "A Nonlinear Stability Analysis of the Freezing of a Dilute Binary Alloy," Philosophical Transactions of the Royal Society ofLond'on. Series A, Mathematical and' Physical Sciences, 268, 3 5 1 (1970). [34] B.J. McCartin, "Eigenstructure of the Equilateral Triangle, Part II: The Neumann Problem," Mathematical Problems in Engineering, 8, 517 (2002). BIOGRAPHICAL SKETCH Saurabh Agarwal was born in 1980 in New Delhi, India. In 2002, he received a B.S. in chemical engineering from the Indian Institute of Technology, Kanpur. He joined the University of Florida as a graduate student in 2002. During his graduate studies, he worked under the supervision of Prof. Ranga Narayanan and Prof. Lewis E. Johns. In 2007, he graduated from the University of Florida with a Ph.D in chemical engineering. PAGE 1 1 INTERFACIAL INSTABILITIES DURING THE SOLIDIFICATION OF A PURE MATERIAL FROM ITS MELT By SAURABH AGARWAL A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLOR IDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA 2007 PAGE 2 2 2007 Saurabh Agarwal PAGE 3 3 To my Mom and my Dad PAGE 4 4 ACKNOWLEDGMENTS First of all, I thank Professor Ranga Nara yanan and Professor Lewis Johns for their constant support and advice throughout my graduate studies. They have been my mentors as well as friends. Their enthusiasm and tenacity for work has inspired me to accept every challenge in research. In particular, their philo sophy has taught me to constantly learn to learn. I have truly enjoyed working with them. I am grateful to the members of my PhD co mmittee: Prof. Dinesh Shah, Prof. Jason Butler and Prof. Sergei Pilyugin. I have really enjoye d being a teaching assistant for Prof. Dickinson and Prof. Crisalle. I also thank Prof Chauhan for his helpful discussions. Many thanks go to my friends and co lleagues Dr. zgr Ozen, Dr. Kerem U uz and Dr. Weidong Guo for numerous helpful discussions. I have the highest appreciation for my mother, father and brother for their love and support throughout my educational career. Despite the la rge geographical distances they have always stood by me. They have encouraged me at every st ep of my career and given me the courage to overcome every obstacle. I feel very exceptional to have a partner as special as Smita. I thank her for her constant support during the long working hours put in the final writing of this dissertation. I express my gratitude to my cousin brother and sisterinlaw, Shekhar and Vaishali, who took time to care for me in every possible way. I also thank my research group and other fe llow graduate students. I owe much to my friends who have always been there when I needed help and friendship. My final acknowledgements go to the University of Florida for an Alumni Fellowship. PAGE 5 5 TABLE OF CONTENTS page ACKNOWLEDGMENTS...............................................................................................................4 LIST OF TABLES................................................................................................................. ..........8 LIST OF FIGURES................................................................................................................ .........9 ABSTRACT....................................................................................................................... ............11 CHAPTER 1 INTRODUCTION..................................................................................................................13 2 PRECIPITATION: LINEAR THEORY.................................................................................23 Introduction................................................................................................................... ..........23 The Cause of the Instability................................................................................................... .23 The Nonlinear Equations........................................................................................................26 Scaled Nonlinear Equations....................................................................................................28 Theoretical Approach and the Li nearization of Model Equations..........................................29 The Solutions to the Base and the Perturbation Problems......................................................30 The vs. 2k Curves.............................................................................................................35 The Neutral Curves: U vs. 2k at 0 ................................................................................37 Concluding Remarks............................................................................................................. .38 3 SOLIDIFICATION: LINEAR THEORY...............................................................................41 Introduction................................................................................................................... ..........41 The Cause of the Instability................................................................................................... .43 The Nonlinear Equations........................................................................................................47 Scaled Nonlinear Equations....................................................................................................48 The Solutions to the Base and the Perturbation Problems......................................................50 The vs. 2k Curves.............................................................................................................53 The Neutral Curves: U vs. 2k at 0 ................................................................................65 Concluding Remarks............................................................................................................. .69 4 ON THE IMPORTANCE OF BULK TRANSPORT ON INTERFACIAL INSTABILITY.................................................................................................................... ...71 Introduction................................................................................................................... ..........71 Case 1: Equilibrium Precipitation Problem in Rectangular Geometry...................................71 Case 2: Nonequilibrium Precipitation Problem in a Rectangular Geometry........................78 Concluding Remarks............................................................................................................. .82 Endnote 1: Equilibrium Precipitation Problem in Cylindrical Geometry..............................82 PAGE 6 6 Endnote 2: Equilibrium Solidification Problem in Cylindrical Geometry.............................86 5 WEAK NONLINEAR ANALYSIS IN PRECIPITATION: GEOMETRIC EFFECTS ON ROUGHNESS..................................................................................................................92 Introduction................................................................................................................... ..........92 Rectangular Cross Section......................................................................................................94 Circular Cross Section with Axisymmetric Disturbances....................................................104 A Cross Section with an Arbitrary Shape.............................................................................109 Cross Sections on which the Integral Does Not Vanish.......................................................114 Cross Sections on which the Integral Vanishes....................................................................116 An Equilateral Triangular and a Re gular Hexagonal Cross Section....................................119 Concluding Remarks............................................................................................................119 Endnote 1: A Circular Cro ss Section with the Possibi lity of Nonsymmetric Disturbances................................................................................................................... ...119 6 SOLIDIFICATION: WEAK NONLINEAR ANALYSIS...................................................121 Introduction................................................................................................................... ........121 Case 1: Latent Heat Rejected Only to th e Subcooled Liquid, Both Phases of Finite Extent......................................................................................................................... .......123 Case 2: Latent Heat Rejected Only to th e Subcooled Liquid, Solid Phase of Infinite Extent......................................................................................................................... .......154 Case 3: Latent Heat Rejected to the Frozen Solid as Well as to the Subcooled Liquid, Both Phases of Finite Extent.............................................................................................163 Case 4: Solid Phase Ignored, Latent Heat Rejected Only through the Subcooled Liquid....182 Case 5: Latent Heat Rejected to Frozen So lid as Well as to the Subcooled Liquid, Both Phases of Finite Extent, U Dropped from Domain Equations.........................................191 Case 6: Latent Heat Rejected to the Subc ooled Liquid Only, Both Phases of Finite Extent, U Dropped From Domain Equations..................................................................210 Concluding Remarks............................................................................................................219 7 CONCLUSIONS AND RECOMMENDATIONS...............................................................220 APPENDIX A SURFACE VARIABLES.....................................................................................................223 A.1 The Unit Normal Vector............................................................................................223 A.2 The Surface Speed.....................................................................................................224 A.3 The Mean Curvature..................................................................................................225 B THE PERTURBATION EQUA TIONS AND THE MAPPINGS........................................227 B.1 The Expansion of a Domain Variable a nd its Derivatives along the Mapping.........227 B.2 The Higher Order Mappings......................................................................................231 PAGE 7 7 C EIGENFUNCTIONS OF AN EQUILATER AL TRIANGLE AND A REGULAR HEXAGON SATISFYING NEUM ANN BOUNDARY CONDITIONS............................232 C.1 Eigenfunctions of an Equilateral Triangle.................................................................232 C.2 Eigenfunctions of a Regular Hexagon.......................................................................233 REFERENCES..................................................................................................................... .......234 BIOGRAPHICAL SKETCH.......................................................................................................237 PAGE 8 8 LIST OF TABLES Table page 2.1 Thermophysical properties of CuSO4.5H2O.....................................................................35 3.1 Thermophysical properties of su ccinonitrile used in calculations....................................55 4.1 Thermophysical properties of CuSO4.5H2O.....................................................................74 5.1 Some integrals concer ning Bessels functions.................................................................108 6.1 Case 1 Values of 0U and 2A for inputs 1,2 L and 2k.............................................153 6.2 Case 3 Some derivatives of base temp eratures in two phases in solidification............164 6.3 Case 6 Values of 0U and 2A for inputs 1,2L and for large values of 2k...............213 6.4 Case 6 Values of 0U and 2A for inputs 1,2L and for small values of 2k..............214 6.5 Case 6 Values of 0U and 2A for inputs 1,2 L and 2k.............................................214 PAGE 9 9 LIST OF FIGURES Figure page 1.1 A picture argument for the effect of longitudinal diffusion in solidification....................16 1.2 A picture argument for the effect of surface curvature in solidification............................16 1.3 A picture argument for the effect of transverse diffusion in solidification........................17 2.1 A precipitation front growing in to a supersaturated solution............................................24 2.2 A picture argument for the effect of longitudinal diffusion in precipitation.....................24 2.3 A picture argument for the effect of surface curvature in precipitation............................25 2.4 A picture argument for the effect of transverse diffusion in precipitation........................26 2.5 A cartoon illustrating the nonuniform concentration gradient in the supersaturated solution in precipitation.....................................................................................................34 2.6 A picture argument illustrating the de stabilizing effect of surface tension in precipitation.................................................................................................................. .....34 2.7 Typical growth curve in precipitation................................................................................36 2.8 Typical neutral curve in precipitation................................................................................38 3.1 A moving solidification front.............................................................................................43 3.2 A picture argument for the effect of longitudinal diffusion in solidification....................44 3.3 A picture argument for the effect of surface curvature in solidification............................45 3.4 A picture argument for the effect of transverse diffusion in solidification........................46 3.5 vs. 2k for the low U model with 1 Scm 1 2 L cm 0.2ST ................................56 3.6 vs. 2k for the low U model with 1 Scm 2 L cm 0.2ST .................................57 3.7 vs. 2k for the complete model with 1 Scm 1 2 L cm 0.2ST .............................58 3.8 vs. 2k for the complete model with 1 Scm 2 L cm 0.2ST ..............................59 3.9 An unusual neutral curve in solidification.........................................................................66 PAGE 10 10 3.10 Typical neutral curve in solidification...............................................................................66 4.1 A precipitation front at rest.............................................................................................. ..72 4.2 vs. 2k for the equilibrium precipitation problem with 114.67x10 meterscapL ..........77 4.3 vs. 2k for the equilibrium precipitation problem with 310 meterscapL ....................78 4.4 vs. 2k for the nonequilibrium pr ecipitation problem with 114.67x10 meterscapL ...81 5.1 A precipitation front growing in to a supersaturated solution............................................92 6.1 A moving solidification front...........................................................................................123 6.2 Typical neutral curve for so lidification when the latent h eat is rejected only through the subcooled liquid.........................................................................................................130 6.3 An unusual neutral curve in solidification when the latent heat is rejected through both phases.................................................................................................................... ...168 6.4 Typical neutral curve in so lidification when the latent heat is rejected through both phases......................................................................................................................... ......168 6.5 Typical neutral curve for so lidification when the latent heat is rejected through both phases; U is dropped from the domain equations...........................................................196 6.6 Typical neutral curve for so lidification when the latent heat is rejected through the subcooled liquid only; U is dropped from the domain equations...................................211 6.7 Transition diagram of L versus 2k showing regions corresponding to forward and backward pitchfork..........................................................................................................215 6.8 A cartoon illustrating a forward/supercritical branching.................................................216 6.9 A cartoon illustrating a backward/subcritical branching.................................................217 PAGE 11 11 Abstract of Dissertation Pres ented to the Graduate School of the University of Florida in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy INTERFACIAL INSTABILITIES DUR ING THE SOLIDIFICATION OF A PURE MATERIAL FROM ITS MELT By Saurabh Agarwal August 2007 Chair: Ranganathan Narayanan Cochair: Lewis E. Johns Major: Chemical Engineering Solidification of materi als is important in the semiconducto r and electronics industry. It is ordinarily required that crysta ls grown industrially have unifo rm mechanical and electrical properties to ensure bette r quality and reproducibility of the fi nal devices that are produced from them. Thus there is a considerable economic ince ntive in producing uniform crystals. However, the occurrence of morphological defects at the growing solidification interface arising from growth instabilities leads to undesi rable electromechanical properties of such crystals. It is the goal of this work to thoroughly understand the physics and causes of morphological instability in solidification. Inasmuch as the problem of morphological inst ability of a solidliq uid interface has been considered in the past, many questions remain unans wered. Of central interest is the pattern that evolves when an instability occurs. To begin to answer this question one begins to address the earlier question of what the magnitude of the wa velength of the initial instability might be. My research focuses on this earlier issue. And a set of questions begin to present themselves. They are: What is the reason for the instability at a solidliquid interface? When does the instability set in? How can one explain the dependence of roughne ss growth rate on its wave number? Why is a PAGE 12 12 maximum growth rate seen in problems like precip itation and solidification? What happens after the instability sets in? Does the front speed up or slow down compared to its predicted base value? Why are cellular patterns observed in solidification whereas precipitation is incapable of giving anything as interesting? Can one determine the nature of the branch ing to the nonplanar steady state in the postonset regime? Many importan t findings have been made in the course of answering these questions. The primary discoveries of this work are: Fi rst, the shapes of the disturbance growth curves in solidification require th e effect of transverse heat di ffusion for their explanation. The neutral curves in solidification ar e found to give rise to the possibi lity of a cellular pattern at the onset of instability whereas for other solidliqui d front growth problems independent of thermal transport, such as precipitation from a supersat urated solution or electr odeposition of a metal from an ionic electrolyte, the onset of an instabil ity from a planar state leads to a single crest and trough. Second and more important, a solidliqui d front that grows on account of temperature gradients is the only one amongst all solidliq uid front growth probl ems where a critical wavelength, independent of surface energy, can occu r. This critical wavelengt h is of the order of the size of the container in which the growth take s place and as such is the only critical point with which fluid convection can interact. Third, a solidification front always demonstrates a speedup upon becoming unstable, which is to sa y that it always moves ahead of its predicted base value. This is in contrast with othe r solidliquid growth instability problems like precipitation. Finally the nature of the bifurcati on in a solidliquid growth problem where growth occurs on account of thermal gradients may be a fo rward pitchfork, leading to cellular growth, or a backward pitchfork, leading to dendritic grow th. This stands in contrast to other growth problems where the post onset region is ordi narily characterized by dendritic growth. PAGE 13 13 CHAPTER 1 INTRODUCTION This dissertation focuses on inve stigating the instability of th e solidliquid interfaces that occurs during precipitati on from a supersaturated solution or during solidifi cation from an undercooled melt. In both ways of growing a soli d, an erstwhile planar interface can suddenly lose its stability and change its shape as a contro l parameter is advanced beyond its critical value. An instability occurring in such a fashion is of ten called a morphological in stability [1]. Now the occurrence of morphological instab ility during solidification has im plications in crystal growth as it leads to undesirable electrom echanical properties of crystals that are often needed in the semiconductor industry [2]. To tackle this prob lem requires an understanding of the physics of the instability. Now solidification is a typical moving front problem where the liquid phase turns into solid phase by rejecting the latent heat from th e interface [3]. The speed of the solidification front depends on how fast the latent heat rele ased there can be conduc ted through one or both phases. The process of precipitati on, on the other hand, occurs when a solid crystal is in contact with its supersaturated solution. In solidifica tion as well as precipitation, the speed and the direction of the front determin e its stability, and hence the state beyond which an erstwhile planar front loses its stability. Hence, in both, the solid surface can acquire roughness and understanding the shape of the growth rate curve, where growth rate is plo tted against the disturbance wave number, is an important step in understanding th e source of the inst abilities that aris e during the growth. Finding how roughness growth rate, varies with the wave number of the roughness, k, has been the object of many experimental studies [4,5]. One of the main goals of the present work is to explain the correspondence between these grow th curves and the fact ors contributing to the PAGE 14 14 stability or instability of the surface, such as diffusion normal to the surface (longitudinal diffusion), diffusion parallel to the surface (transvers e or lateral diffusion), surface tension etc. In fact, a greatest growth rate occurs at a fin ite wavelength in some, but not all moving front problems, and the appearance of such a maximum gr owth rate in solidification is an example of an observation that needs to be explained. As discussed by Coriell and coworkers [6], the wavelength having the largest initia l growth rate might be expected to be predominant during the evolution of the instability, and it is therefore important from the point of view of pattern selection. Therefore, while explaining the shap es of the growth curves in precipitation and solidification, we emphasize the reason for the ex istence of a maximum growth rate. We will see later that this has to do with transverse diffu sion of heat in the liquid at the interface. Now the vs. 2k curves at various values of a contro l variable tell us what the critical wavelengths are, these being the wa velengths at zero growth rate. Thus they predict the shape of the neutral curve, a curve where the cont rol variable is given as a function of 2k. This curve is useful in understanding what might be seen in a series of experiments as the control variable slowly increases to, and just be yond its critical value. Of central interest is the pattern that evolves during the early stages of instability. We will find that the neutral curve in precipitation is monotonically increasing, hence only one cres t and trough will be seen in precipitation. In solidification, when heat is wit hdrawn only through the liquid, the ne utral curves are as simple as those in precipitation. Upon wit hdrawing heat through the solid as well, neutral curves like those seen in the RayleighBnard c onvection problem [7] can be found. And by adjusting the depths of the phases, we can go beyond this and get someth ing new, viz., critical points that are due to transverse diffusion in the solid phase, and not due to surface tension. Hence, the central result of our linear theory is this: In the solidification of a pure material it is possible to obtain zero, one, PAGE 15 15 two, or three critical points, and a critical wavelength independ ent of surface energy can occur. This critical wavelength is of th e order of the size of the cont ainer in which the growth takes place. The reader is encouraged to refer to Cr oss and Hohenberg [8] who review pattern formation at great length, never finding neutral curves like the new ones discovered by us. In fact, it turns out that getting thr ee critical points is a feature ty pical of pure solidification when the latent heat is rej ected through both phases, when the control variable is made explicit, and when the depths of the two phases are taken to be finite. The Mullins and Sekerka [9] model for the solidification of a dilute binary alloy a nd the Wollkind and Notes tine [10] m odel for pure solidification fail to give anything as interesting; they can give at most two critical points. Now to understand why one might expect three cr itical points in solidif ication, let us see why a solidification front might be unstable a nd how factors such as surface curvature and transverse diffusion affect the stability. The usua l explanation for the instability of a growing interface from an undercooled melt (cf., Langer [11]) is that a di splacement of the surface into the melt in the form of a crest increases the temp erature gradient at its tip, which reinforces the growth of the crest. Figure 1.1 depicts a growing front in an u ndercooled melt and one sees that longitudinal diffusion in the liquid phase is stre ngthened at a crest thereby destabilizing the surface. It is also clear from Figure 1.1 that longitudinal diffusion in the solid phase acts to stabilize the interface. The effect of longitudinal diffusion does not depend on the wavelength of the disturbance, i.e., it is a kindependent effect. Now the melt temperature at a solidliquid interface depends on its morphology. In other words, on account of the curvature of the nonpl anar interface itself, the melt temperature at a curved interface is different from the normal me lt temperature at a planar interface. This is PAGE 16 16 explained by Delves [12]. Figure 1.2 shows this effect where surface curvature lowers the melt temperature at a crest while increasing it at a trough. Hence, surface tension weakens the instability caused by longitudinal diffusion in the liquid, while strengthening its stabilizing effect in the solid. The stabilizing effect of surface tension is kdependent, the stabilization being stronger for shorter wavelengths. Figure 1.1 A picture argument for the effect of longitudinal diffusion in solidification Figure 1.2 A picture argument for the effect of surface curvature in solidification Liquid Solid Crest Trough Moving Front Temp. gradient strengthened Temp. gradient weakened Temp. gradient strengthened Temp. gradient weakened Liquid Solid Moving Front Melting point decreases Melting point increases PAGE 17 17 Finally, let us look at the effect of transver se diffusion in the two phases. Setting surface tension aside, the effect of th e displacement is as shown in Figure 1.3. On the melt side, the displacement of the surface strengt hens the temperature gradient at a crest, weakening it at a trough; this induces transverse te mperature gradients, carrying heat away from crests toward troughs. This tends to build crests up. Transverse diffusion is therefore de stabilizing on the melt side of the surface. A similar argument made in th e solid phase tells us that transverse diffusion is stabilizing in the solid phase. In both phase s, the effect of transverse diffusion, whether stabilizing or destabilizing, is kdependent. Figure 1.3 A picture argument for the effect of transverse diffusion in solidification This is enough to speculate the occurrence of three critical points in solidification. Here is what is required: The longitudinal diffusion in the liquid should lead to positive values of at 20k, then the transverse diffusion in the solid should drive negative, providing the low2k critical point. Thereafter the transverse diffusion in the liquid should drive the positive for further increase in 2k, resulting in a second critical point. Fi nally, the stabilizing effect of surface tension should drive down to zero for large enough 2k, providing a third critical point. Liquid Solid higher lower lower lower higher higher 2 wavelength= k PAGE 18 18 From a mathematical point of view, we fi rst write the governing domain equations along with appropriate interf ace conditions and farfield conditi ons. These model equations are nonlinear as the position of the in terface depends upon the temperature gradient there, and the gradient, in turn, depends upon the position and morphology of the interface. This nonlinearity is the reason for the interface to depart from its flat base state, trying to become unstable. We write the model equations in terms of an observer m oving at the speed of the front. Such an observer can find a steady planar solution to our nonlinear equations. To obtain the vs. 2k curves in solidification, we introduce a pe rturbation of a steady base soluti on, as did Mullins and Sekerka [9], whose work grew out of the earlier work of Ru tter et al [13] and Tiller et al [14]. More detail can be found in the book by Davis [15] who presents the theory of front inst abilities in the case of binary alloys. Many of the models [911,16,17] proposed for pure solidification are based on local equilibrium at the interface and isotropic interf acial properties. Convection induced by density differences is neglected and stea dy thermal and diffusion fields are assumed [9,11] in the interior of the solid and liquid phases. These assumptions are made by us as well. To maintain our focus on the effects that longitudinal and transverse diffusion have on the growth rate of surface roughness, most other effects, however important they may be, are omitted. These effects have been dealt elsewhere. For example, the effect of convective flow on morphological instability was studied by Coriell and Sekerk a [18]. The effect of solutal convection has been studied by Hurle et al [19], while Favier a nd Rouzaud [20] have given a detail ed account of the effect of a deformable boundary layer. The effect of Soret diffusion has been explained by Hurle [21]. The effect of departure from local equilibrium has been considered by Coriell and Sekerka [22]. In a different paper, Coriell and Sekerka [23] have considered the effect of anisotropic surface PAGE 19 19 tension and interface kinetics. Wollkind and cowork ers [10, 24] have considered surface tension as a function of temperature. Now we may have neglected the effect of convection and the anis otropy of interfacial properties. But we do take into account the finite th ickness of the solid and liquid phases. They are not taken to be of infinite ex tent as they are by Wollkind et al [10]. This gives us a way of altering the relative strengths of transverse di ffusion in the two phases and hence of altering the shapes of the growth curves, and this, in turn, gi ves us a way to obtain th ree critical points in pure solidification. Up till now our emphasis has been on the linear stability analysis in investigating the interfacial instability during preci pitation and solidification. It tu rns out that we discover two more important results in the post onset regime af ter the instability sets in. These results emanate from a weakly nonlinear calculation which is done by going to second and third orders in the amount by which the control parameter is increased. The first of these resu lts has to do with the correction to the front speed on account of mor phological instability. We find that once the instability sets in, a solidification front always speeds up compared to what its base value would have been, i.e., it moves ahead of its predicted base value. This is in contrast with a precipitation front which demonstrates an unconditi onal slowdown upon becoming unstable. The second important result of th e nonlinear calc ulations has to do with the nature of the branching to the new nonplanar solutions once th e instability sets in. The book by Strogatz [25] gives an excellent description of bifurcations in several applied problems. Bifurcations are important scientifically as they provide essential information a bout the kind of transition from one steady state to another as some control parameter is varied. We discover that in solidification, the nature of the bifurcation depe nds on the dimensions of the container in which PAGE 20 20 the growth takes place. The branching may be a fo rward pitchfork, leading to cellular growth, or a backward pitchfork, leading to a dendritic grow th. This is in contrast to the precipitation problem, where the onset of instability is ordi narily characterized by a dendritic growth on account of a backward pitchfork. However, prec ipitation is also capable of demonstrating a transcritical bifurcation. In f act, we discover that the nature of bifurcation in precipitation depends on the cross section of the growth contai ner. We deduce a condition to decide whether a container of a given cross sectional shape will demonstrate a backward pitchfork or a transcritical bifurcation. Finally, before moving on to explain the organiza tion of this dissertation, we would like to emphasize that a thorough understanding of theore tical results mentioned above is extremely helpful in designing good experiments. As explained in the book by Montgomery [26], a successful design of experiments requires a good theoretical knowledge of all important factors. And it is our belief that the proposed theory should lead to a more rational design and interpretation of solidification experiments. Organization of the Research: The remainder of this di ssertation is dedicated to theoretical investigation of interfacial inst ability in precipitation and solidification via mathematical models and to explain the results using simple physical arguments. As mentioned, our primary goal is to understand the underlying causes and implicat ions of such instabilities. Chapter 2 delineates the physics of instability in precipitati on. Being a onephase problem, precipitation offers a simple way of understa nding the concept of longitudinal diffusion, transverse diffusion, and surface tension etc. We also learn how the various factors compete with each other to result in an instability. The gove rning equations along with the relevant interface and boundary conditions are presented. This chapte r also explains the th eoretical concept of a PAGE 21 21 linear stability analysis which is used to pred ict whether or not a planar interface can remain planar in the face of small disturbances. This chapter forms the framework for understanding the physics of instability in the more co mplicated problem of solidification. Chapter 3 focuses on the interfacial instability during solidification. The first half of this chapter is devoted to understanding the physics of the instability and in deriving the formula for the growth rate of a disturbance. This is followe d by an analysis that predicts the wavelength of the cellular pattern observed at the onset of inst ability. We show how on e can obtain interesting growth curves having up to three critical points. Chapter 4 addresses the question whether th e domain transport dynamics is important compared to the transport dynamics at a solidliquid interface. A simple precipitation model is studied to investigate whether the time derivatives in the domain equations are important relative to the time derivative of the interface position introduced via interfacial mass balance. Chapter 5 presents a weak nonlinear calculation in the problem of precipitation, where the goal is to examine what happens beyond the onset of the instability. We first work out the problem in a rectangular geometry and then in a circular geometry. In each case, we first find the critical value of the control pa rameter using the linear stability analysis. This is followed by a weak nonlinear calculation where we look for stea dy solutions just beyond this critical point by going to second and third orders in the amount by which the control para meter is increased. Our main goal is to find out the nature of the bifu rcation to the new nonplanar solution. Through our calculations, we also wish to discover whether th e front speeds up or slow down compared to its base value once the instability sets in. Toward th e end of the chapter, we deduce a condition that determines whether a cross section of an arbitrary shape will lead to a pitc hfork or a transcritical bifurcation. PAGE 22 22 Chapter 6 deals with the weak nonlinear calculation in the problem of solidification. The goal is to understand the early stages of the gr owth of surface roughness in solidification. Again, as in Chapter 5, we try to answer two key ques tions: Does a solidificatio n front speed up or slow down compared to its predicted base value once th e instability sets in? Is the branching to the new nonplanar solution a forwar d or a backward pitchfork? Chapter 7 presents a general conclusion along with a scope for future studies. PAGE 23 23 CHAPTER 2 PRECIPITATION: LINEAR THEORY Introduction In this chapter, our goal is to investigate th e causes of instability during precipitation from a supersaturated solution and to quantify the growth and neutra l curves by using a linear perturbation theory. First, pict ure arguments are given to esta blish the role of longitudinal diffusion, transverse diffusion, and surface tensi on with regard to the observed instability. To quantify our arguments, we then give the nonlin ear model equations which are written in a moving frame. These equations are then scaled and linearized around a base state. This base state solution is slightly perturbed to obtain the formula for the growth constant. The playoff between the various terms in this formula is then explained in terms of the effect of diffusion (in two directions), surface tension, and surface speed. Based on this, th e shapes of the growth curve and the neutral curve are then explained. The ch apter is concluded by predicting the number of crests observed in a precipitation experiment when the supersaturation in the solution is slowly increased to its critical value. It is found that at most one crest and one trough will be seen at the onset of instability. We now move on to discuss the physics of th e instability observed when precipitation occurs from a supersaturated solution. The Cause of the Instability Consider the schematic shown in Figure 2.1 which shows a moving pr ecipitation front. Let a solid of density Sc lie to the left of a supersaturated solution. The two phases are divided by a plane interface that advances to the right as the reservoir at zL supplies the solute at a fixed concentration Lc, which is higher than SATc, the saturation concentrati on at a planar interface. PAGE 24 24 Figure 2.1 A precipitati on front growing into a supersaturated solution Figure 2.2 A picture argument for the effect of longitudi nal diffusion in precipitation Let Lc be the control variable. Increasing it increa ses the speed of the front. At some value of Lc the plane front loses its stability and becomes nonplanar. The explanation for the instability can be understood from Figure 2.2. Cons ider a perturbation of the planar front to the right, a crest where the local spee d runs above the speed of the planar front. To sustain this increased speed, the crest must be supplied more solute from the reservoir, which is reinforced Solution Solid movin g front Scc zZxt 21x x Z ik n Z 2., 1t xZ unusurfacevelocity Z 23/22, [1]xx xZ Hsurfacecurvature Z z x E Qcc zL reservoir maintained at a fixed concentrationSAT Sc: solute concetration in the liquid in equilibrium with c at a plane surfaceEQ Sc: solute concetration in the liquid in equilibrium with c at a curved surface L cc SATcc 00 zZ Liquid Solid Moving Front Crest Trough Concentration source Concentration gradient strengthened PAGE 25 25 by the increased concentration gradient due to th e perturbation. Similarly, a displacement in the form of a slower moving trough is reinforced by a decreased concentration gradient at its tip. Longitudinal diffusion in the liquid is therefore de stabilizing; it acts more strongly the faster the front is moving and the steeper the solute grad ient feeding the growth. But it does not depend on the wavelength of the disturbance, i.e., it is kindependent. Operating against this is the effect of surf ace curvature which plays a stabilizing role. This is shown in Figure 2.3. At a crest, EQc, the equilibrium concentration, is higher than SATc [3]. By reducing the local concentration gradient at a crest, surface tension weakens the instability caused by longitudinal diffusion. For short wavele ngths, the stabilization is strong; for large wavelengths, it is weak; he nce surface tension is a kdependent stabilizing effect. Figure 2.3 A picture argume nt for the effect of surface curvature in precipitation These two effects would then predict instability at low values of k, and stability for high values of k. But by themselves they cannot expl ain an increase in instability as k increases from zero. There is, however, another e ffect, and it has to do with transverse diffusion in the liquid. Setting surface tension aside, the displacement of the surface strengthens the solute gradient at a crest, weakening it at a trough (cf ., Figure 2.4). This induces transv erse solute gradients, carrying Liquid Solid Moving Front Concentration source Equilibrium conc. increases PAGE 26 26 solute away from troughs toward crests. This is de stabilizing. It tends to build crests up and the effect is stronger the higher the k, i.e., the closer together the cr ests and troughs. Also, this effect is stronger the faster the front is moving. The effect of surface te nsion on transverse diffusion is again stabilizing. Its effect on the surface concentr ation works against transverse diffusion just as it works against longitudinal diffusion. By reduci ng the local concentration gradient at a crest while increasing it at a trough (cf ., Figure 2.3), surface tension weakens the solute gradients in the transverse direction, hence weakening th e instability caused by transverse diffusion. Figure 2.4 A picture argument for the effect of transverse diffusion in precipitation Now in precipitation the solid phase plays no role. Consequen tly, the effects that will concern us are three in number: the destabilizing effect of longitudinal and transverse diffusion in the liquid and the stabilizing effect of surface tension which o ffsets each of these. To make these qualitative ideas quantita tive, we turn to a model. The Nonlinear Equations The sketch given in Figure 2.1 fixes the basi c ideas and presents some formulas. The solid phase is passive and the solute concentration in the solid is Sc everywhere and for all time. Of interest are the equation of solute diffusion in the liquid phase, and the interface equations for the Liquid Solid Moving Front Concentration source Concentration gradient strengthened SATc SATc Solute Flow PAGE 27 27 local motion of the front. The liqui d phase depth is maintained cons tant and the problem is taken to be twodimensional, x and z being parallel and perpendicula r, respectively, to the base surface. The equations will be written in a frame moving with a constant velocity k U U, where U is the base state speed of the precipitatio n front as seen by a laboratory observer. Then, in the liquid phase, the equation for the solute concentration is given by 2cc DcU tz (2.1) for Z zL and 0 x W where L is the depth of the liquid phase and W is the width of the precipitation cell; D denotes the diffusivity of the solu te in the liquid phase. The solute concentration must meet two requirements along the solidsolu tion interface, viz., along ) (t x Z z. First, phase equilibrium requires, afte r accounting for the correction due to surface curvature 2SATccHA (2.2) where A denotes 1SAT Sc R Tc, and where denotes the surface tension. Second, the solute surplus or deficit induced by the motion of the interface must be made good by solute diffusion into or out of the solution, i.e., ..SccUknuDnc (2.3) where u denotes the local surface spee d of the perturbed front. The far field condition is given by ()LczLc (2.4) Finally we impose side wall conditions such that both cx and Z x vanish at0 x and at PAGE 28 28 x W The formulas for the surface normal, n, the surface speed, u, and the mean curvature 2 H are given in Appendix A. Scaled Nonlinear Equations The nonlinear equations will now be presente d in dimensionless form. All lengths are scaled by L and all speeds are scaled by / D L. Time is scaled by 2/ L D. Concentration is measured from a reference point corresponding to SATc, and is scaled by SSATcc i.e., SAT scaled SSATcc c cc All variables used hereon are scaled variables. Then, in the liquid phase, the equation for th e scaled solute concentration is given by 2cc cU tz (2.5) for 1 Z z and 0 x W The equations along the interface, viz., along ) (t x Z z now become 2 cH A (2.6) and .1. nccUknu (2.7) where A, a dimensionless parameter introduced by our scaling, denotes 11SAT SSSATc L RTccc The far field condition is given by (1)Lczc (2.8) Side wall conditions are Neumann, i.e., 0,0,0 cZ xWxW x x PAGE 29 29 Notice that these model equations are nonlin ear as the position of the interface depends upon the concentration gradient there, and this gradient, in turn, depends upon the position and morphology of the interface. This is clearly seen in Eq. (2.6) which couples the interface concentration c to the interface position Z xt via the surface curvature 2 H This coupling leads to nonlinearity, which forms the heart of the problem. It is the reason for the interface to depart from its flat base stat e, trying to become unstable. Now to investigate the instability during precipitation, we will study the effect of small disturbances on a planar front which is moving with a constant speed U. To do so, we now turn to the process of linearization of the equations. Theoretical Approach and the Line arization of Model Equations A system in steady state is said to be stable if all disturba nces imposed on it decay with time; it is called unstable if any disturbance grows. Such an instability arises when a steady state system is driven away from its steady state upon imposing small perturbations. Very often, an instability arises in growth problems when a c ontrol variable exceeds its critical value. For instance, in the problem of precipitation, the supe rsaturation in the liquid phase may be taken to be the control variable, which, when exceeded beyond a critical value, wi ll lead to instability. To study instability in precipitation, our plan is to evaluate the time dependence of the amplitude of a sinusoidal perturbation of infin itesimal initial amplitude introduced on the planar interface. For this purpose, it is sufficient to analyze a linearized model where the linearization is done over a base state. The idea behind linearizing the equations is this : Once all equations are linearized, the evolution of a sinusoidal perturbati on is simply a superpositi on of the evolution of its Fourier components. The interface is unstable if any Fourier component of the sinusoidal wave grows; it is stable if all decay with time. PAGE 30 30 Now a base state can often be guessed for most problems; for precipitation, it is a planar front which is moving with a constant speed Denoting the base state concentration by 0c, the domain concentration c can be expanded as 0 011.. c cccz z (2.9) where denotes the amplitude of the perturbation; 1z represents the first order mapping from the current domain to the reference domain. Th e first order mapping at the interface is denoted 1 Z and it needs to be determined as a part of the solution. Because the current domain is unknown (it is a part of the solution), it become s imperative that we solve the problem on a known reference domain, i.e., one that has all planar boundaries. The meaning of the mappings 1z and 1 Z and of the correction 1c is explained in Appendix B.1. The first order variables can be further expanded using a normal mode expansion [9] as 11cos()tcckxe and 11cos()t Z Zkxe (2.10) where k is the wave number of the disturbance and denotes the growth (or decay) constant, also known as the inverse time constant. It is an output variable, and a positive value of for a given k means that the disturbance gr ows, leading to instability. The Solutions to the Base and the Perturbation Problems The base state is one in which the precipitation front is a plane at re st in the moving frame. The base state variables are denoted by the subscript 0 and they satisfy 2 00 20dcdc U dzdz (2.11) in the liquid phase, 01 z The equations at the planar interface, 0 z are given by 00c (2.12) PAGE 31 31 and 0 01 dc cU dz (2.13) The far field condition is 0(1)Lczc (2.14) Then a solution to equations (2.112.14) is given by 0()1Uzcze (2.15) where U is given by ln1LUc (2.16) Besides the physical properties, the i nput variables in the base state are L and Lc, and the output is the front speed, U. To determine if this base solution is stable, we impose a small displacement on the surface at fixed values of Lc and U, and determine its growth rate, as a function of its wave number k, whence k is an input to the perturbation problem. We solve the perturbation problem on the base domain by expanding the surface shape and the solute concentration at the surface in the form 0101cos()t Z ZZZZkxe (2.17) and 00 011011cos()tdcdc cccZccZkxe dzdz (2.18) where the side wall conditions determine the admissible values of k. The subscript 1 denotes a perturbation variable. Then us ing the expansions gi ven above, the perturbation equations for 1cand 1 Z are given by PAGE 32 32 2 2 1 11 2dc d ckcU dzdz (2.19) in the solution, 01 z The equations that need to be satisfied across the interface, 0 z are 2 0 111dc cZkZ dz A (2.20) and 1 11dc Z Uc dz (2.21) The farfield equation is 1(1)0cz (2.22) A solution to the domain equation satisfying Eqs. (2.20) and (2.22) is given by 2 0 110 1mzmmmz mmdc kz dz ceeeZ e A (2.23) where 2224 mUUk Finally, using the mass bala nce across the interface, Eq. (2.21), there obtains 2 22 00 2dcdc kNkU dzdz AA (2.24) I II III IV where () ()1mm mmmme N e Eq. (2.24) gives implicitly in terms of 2k and the derivatives of the base concentration at the reference surface, viz., at 0 z A graph of vs. 2k can be drawn from this equation, PAGE 33 33 but the equation is not explicit in Now appears in the problem in two ways. It appears in the domain equation where it has to do with solute depletion in th e liquid as the front advances. It also appears in the surface solute balance where it has to do with the loca l speed of a crest or a trough. We will drop on the domain (cf., Chapter 4) on the grounds that the solute depleted by the growth of a slowly moving front is easily supplied by diffusion. Then the dominant growth rate is given by Eq. (2.24), where N remains as defined, but m are now given by 2224mUUk, and where m is always positive and m is always negative. Consequently N is always negative, an d terms I and IV in Eq. (2.24) are negative, hence stabilizing, while terms II and III are positive, hence destabilizing. Other than the wave number of the displacemen t, the growth rate depends only on the base solute gradient, 00dc zU dz and its correction 2 2 0 20 dc zU dz The physical factors that each term takes into account are th ese: Term II, the essential de stabilizing term, combines the two effects of the longitudinal solute gradient The primary effect is that a displacement strengthens the gradient at a crest, and the second ary effect is that a stre ngthened gradient at a crest and a weakened gradient at a trough induce transverse solu te diffusion from a trough to a crest; both destabilizing. Term IV which is stabilizing, is in fact a correction to term II, weakening its destabilizing effect. It corrects for the fact that the solute gradient is not uniform, being stronger at the base surface than in th e nearby solution (cf., Figure 2.5). Hence a crest projected into the supersat urated solution is fed by a less steep gradient than 00dc z dz Term I, the essential stabilizing term, combines two of the three eff ects of surface tension, cutting into the destabil izing effect of both longitudinal diffu sion and transverse diffusion. Term III, the remaining effect of surface tension is destabilizing. It account s for the fact that at a crest, PAGE 34 34 the amount of solute that must be supplied to move the crest is lessened as the solute concentration rises at the surface (cf., Figure 2.6). Such a term will not be seen in solidification (cf., Chapter 3) unless the la tent heat is perturbed. Figure 2.5 A cartoon illu strating the nonuniform concentration gradient in the supersaturated solution in precipitation Figure 2.6 A picture argument illustrating the destabilizing e ffect of surface tension in precipitation A significant algebraic simplificatio n can be obtained by observing that Sc the density of the solid, is usually very high (cf., Table 2.1) compared to LSATcc the supersaturation. The scaled value of Lc therefore is ordinarily small (0.1Lc ), and consequently U must also be Crest being fed by a less steep gradient Liquid Solid Moving Front Concentration source Crest S cLiquid Solid Moving Front Lc Old difference to be supplied New difference to be supplied PAGE 35 35 small (0.1 U ). The base state concentration profile in the liquid phase must then be nearly linear and a low U result may be obtained by dropping U from the domain equation. The base speed and the formula are then given by LUc and 22tanh k kUkU k AA (2.25) and at the small values of A ordinarily encountered, viz., 910 A, the following formulas are useful in sketching a graph of vs. 2k: 20kU (2.26) 2 21 0 3d kU dk (2.27) 2large but not too largekkU (2.28) and 23very largekkA (2.29) Table 2.1 Thermophysi cal properties of CuSO4.5H2O The vs. 2k Curves A typical growth curve is shown in Fi gure 2.7. It starts positive. Then increases as 2k increases and the curve passes through a maximum before decr easing and passing through zero. The value of at 02k is positive, and this is due enti rely to longitudinal diffusion (cf., Figure 2.2). In fact, one might have pr edicted intuitively that the value of at 02k could Saturation concentration, SATc 243 (kg/m3) Density of solid, Sc 2284 (kg/m3) Diffusivity, D 109 (m2/s) Gas constant, R 8.314 (J/mol K) Surface Tension, 8.94 x 103 (N/m) PAGE 36 36 never have been negative for the prec ipitation model presented here. Now at 02k, only longitudinal diffusion is importa nt. And since precipitation ha s only one active phase, and longitudinal diffusion is destabilizing there, we do not have any reason to expect stability at 02k. Notice that the analog will not hold in the solidification problem (cf., Chapter 3) where the latent heat can be reject ed through the solid as well as via the subcooled liquid. The increase in as 2k increases is due to the fact that the longitudinal solute gradient induces transverse diffusion (cf ., Figure 2.4) from a trough to the adjacent crest. This is kdependent, and increases linearly in k. Eventually surface tension comes into play, driving downward as 3k. The separation of the effects of transv erse diffusion and surface tension allows us to see first an increase in followed by a decrease. This is due to the small value of A, which delays the effect of su rface tension. Hence a range of 2k exists upward from zero where increases due to transverse diffusion. And this explains the existence of a greatest growth rate, max at a value of 2k, denoted by 2 maxk. Figure 2.7 Typical growth curve in precipitation Now the effect of surface tension cannot become important until 2k becomes large. Consequently, the value of the term 224 Uk coming via m will be large at the point of 2k 0 PAGE 37 37 maximum growth rate, and N may then be approximated as m With this simplification it is easy to obtain 2 max3U k A and 3/2 max2 33UA (2.30) whence for most cases of interest, 2 maxk is onethird the critical value of 2k, denoted 2critk, which is the value of 2k at 0 Finally observe that in the model of precipita tion presented here, we can have one and only one critical point. Again, this is because of the fact that the growth curve must start positive at zero wave number (due to longitudinal diffusion in the solution), rise initially as the wave number increases (due to transverse diffusion in the solution), and eventually fall down (due to surface tension), giving a critical point. Again notice that since the solidification problem (cf., Chapter 3) has two active phases through which the latent heat can be rejected, we have a possibility of seeing more than one critical poi nt. But before going on to solidification, let us look at the neutral curves in precipitation. The Neutral Curves: U vs. 2k at 0 Turning our attention to th e neutral curve, viz., the U vs. 2k curve at 0 we have, from Eq. (2.24), 20kUNU A (2.31) But N is always less than U for all positive values of 2k, hence the root NU can be ruled out. There obtains 2UkA and the neutral curve is a straight line as shown in Figure 2.8. For a fixed 2k, the system is initially stable for small values of U becoming unstable as U increases. PAGE 38 38 Figure 2.8 Typical neutra l curve in precipitation The width of the precipitation cell plays an impor tant role in determin ing the stability of a system. For a precipitation cell of scaled width W, the maximum admissible wavelength of a disturbance is 2 W, and the admissible values of the wave number are restricted to knW 1,2..n; the case 0 n is ruled out on the grounds that the volume of the supersaturated solution is held constant. One can then find the lowest possible value of U for which 0 It occurs for the lowest allowable value of 2k, viz., for 222/kW. Therefore, in an experiment, we expect to see at most one crest when th e front just becomes uns table. The critical U is 22/critUW A (2.32) and the corresponding critical va lue of the control parameter Lc is given by 22/1critW Lce A (2.33) Concluding Remarks What we have found in precipitation is this: the shape of the grow th curve requires the effect of transverse diffusion for its e xplanation. The corresp onding neutral curve is monotonically increasing, implying only one crest at critical. In other words, since the neutral curve in precipitation does not have a dip, it is not possible to observe a cellular pattern or U0 2k 0 PAGE 39 39 multiple cells in precipitation when the instabilit y is approached by raising a control parameter and crossing its critical value. Next let us try to get an idea as to when exact ly does the instability set in when one creeps up on the control parameter. Using the thermophys ical properties given in Table 2.1, assuming the temperature to be 300 TK and taking the depth of the liquid solution to be 1 L cm, the scales for concentration differe nce, velocity and time are: 32041/SSATcckgm, / D L 107 m/s and 2/ L D 105 s. The value of the surface tension group is A4.67 x 109. Let us take the width of the precipitation cell to be 1Wcm Then Eq. (2.32) predicts the scaled value of 22/critUW A to be around 4.6x104. This corresponds to a scaled value of 22/1critW Lce A to be 4.6x104, i.e., 44.6x10SAT scaled SSATcc c cc This means that unscaled critLc is 43+ 4.6x10 x (2284243) kgmSATSATcc. Because critLc is very close to SATc, it is extremely hard to run reasonabl e experiments to study precipitation as the front will become unstable as soon as we tweak up Lc even slightly above SATc Notice that this is due to the small value of the surface tension group, A. Because of reasons to be discussed later, the physical numbers for parameters like critical undercooling et c. in solidification turn out to be much more reasonable and interesting. Nevert heless precipitation was interest ing in itself and presented us with the opportunity to understand concepts such as longitudinal diffusi on, transverse diffusion and surface tension in a relatively easier onephase set up. Finally notice that because the precipitation model is onesided, there are no domain length scales to adjust. One can only have a scal ed capillary length. The readers attention is drawn to the fact that the resu lts will change substa ntially for solidification (cf., Chapter 3) where the latent heat can be rejected through two phases. In solidification, we will have the ratio PAGE 40 40 of the liquid to solid phase depth as an important c ontrol, as also a capillary length, in addition to the control variables arising from the solid a nd liquid farfield temperatures. We will revisit precipitation when we discuss solidification a nd recall some of the results discussed here. PAGE 41 41 CHAPTER 3 SOLIDIFICATION: LINEAR THEORY Introduction The nature of instability during solidification of a pure substance is investigated in this chapter. The latent heat is assumed to be rej ected through the subcooled liquid as well as through the solid phase. Consequently, un like precipitation, solidification is a twophase problem where the solid phase is not passive any more. We find th at transverse diffusion, as well as longitudinal diffusion, have opposite effects in the two phases, and hence, compared to precipitation, we find a greater variety of growth curves in solidification. As in Chapter 2, we start by giving pictur e arguments, where we explain the role of longitudinal diffusion, transverse diffusion, a nd surface tension with regard to the observed instability during solidification. An important di fference from precipitation is that we now need to understand the effects of diffusion and surface tension in two phases. We give a simple model for pure solidification where we assume local equilibrium at the interface and isotropic interfacial properties; convection induced by density differences is neglected. Then we derive the formula by using linear stabilit y methods, and explain the various terms in the formula. Next we draw several vs. 2k curves, and our emphasis is on the importance of transverse diffusion in explaining the shapes of these curves. The growth curves imply the shape of the neutra l stability curve. This curve is useful in understanding what might be seen in a series of experiments as th e control variable increases to, and just beyond its critical value. When heat is withdrawn only through the liquid, the neutral curves in solidification are as simple as those in precipitati on. Upon withdrawing heat through the solid as well, neutral curves like those seen in the problem of heating a fluid from below [7] can be found as shown by Wollkind et al. [10]. And by adjusting the depths of the phases, we can PAGE 42 42 go beyond this and get something new, viz., criti cal points that are not due to surface tension. This means that interesting experiments can be run by controlling the tran sverse extent of the solid, thereby making the allowabl e values of the wavelength an input variable. We draw some conclusions along these lines towa rd the end of the chapter. The central result is this: In th e solidification of a pure material it is possible to obtain zero, one, two, or three critic al points, at least one having nothing to do with the surface curvature. In the solidification of a dilute binary al loy [9] only zero or two can be found. Before moving on to discuss the causes of inst ability in solidification, we would like to make a comment on the reason for considering the fi nite depths of the solid and the liquid phase. We have already mentioned that we do not account for convection. Now the wavelengths characteristic of convection are much longer than those of interest in solidification, in the absence of convection, dominated as they are by su rface curvature, cf., [15]. Our plan is to show that in solidification of a pure material, there are critical wave numbers in the convection range and that these can be found in ordinary experiments if one knows where to look. But if convection is ultimately to be of interest, bounded expe riments are essential, hence we investigate solidification, by itself, in the ca se where the depths of the liquid and the solid phases are finite. There is an advantage in doing th is. The theory of solidification is ordinarily presented for phases of infinite depth in terms of the gradients at the solidm elt interface, cf., [9]. These gradients can be controlled in two ways in experiments at finite depths, by depth of the phase and by the temperature at its edge. With this additional flexibility of control, critical points can be found at much longer wave lengths than t hose determined by surface cu rvature. It is these critical points that ought to c ouple strongly to convection. PAGE 43 43 Coriell et al. [27] study phases of finite depth in the solidification of a dilute binary alloy. Davis [15] explains how such experiments can be set up. We now move on to discuss the various causes of instability during solidification from a subcooled melt. Whenever possible, we will try to point out similarities and differences from precipitation. The Cause of the Instability Figure 3.1 A moving solidification front Let a solid lie to the left of its subcooled me lt. This is shown in Figure 3.1 where the two phases are divided by a planar inte rface that advances to the right as the liquid turns into solid, releasing its latent heat to both phases. Taking LT to be the control variable, decreasing LT increases the speed of the front, and for some value of LT the planar front becomes unstable, acquiring a nonplanar shape. Liquid Solid z Zxt 00 zZ z S z L solidsinkSTT L TT liquidsinkmoving front 21x x Z ik n Z 2., 1t xZ unusurfacevelocity Z 23/22, [1]xx xZ Hsurfacecurvature Z PAGE 44 44 Figure 3.2 A picture ar gument for the effect of longit udinal diffusion in solidification The effect of longitudinal diffusion is shown in Figure 3.2. A displa cement of the surface in the form of a crest increases the liquid side te mperature gradient at its tip, which reinforces the growth of the crest; hence longitudinal diffusion in the liquid acts to dest abilize the surface, just as it did in precipitation. However, the solid side temperature gradient is weakened at the crest, and this tends to flatten out the crest; hence l ongitudinal diffusion is st abilizing in the solid phase. A similar effect was obviously not seen in precipitation as the solid phase was passive there. Again, the strength of longitudinal di ffusion does not depend on the wavelength of the disturbance; hence it is a kindependent effect. Setting asid e the effect due to nonuniform temperature gradients, longitudi nal diffusion by itself acts more st rongly the faster the front moves. Next let us consider the eff ect of surface tension on the stab ility of the front. It is stabilizing. This can be understood from the f act (cf., Figure 3.3) that the melt temperature depends on the shape of the solidliquid inte rface; the surface curvat ure lowers the melt temperature at a crest while increasing it at a trough. Hence, surface tension weakens the instability caused by longitudinal diffusion in the liquid, while strengthening its stabilizing effect Liquid Solid Crest Trough Moving Front Temp. gradient strengthened Temp. gradient weakened Temp. gradient strengthened Temp. gradient weakened PAGE 45 45 in the solid. Again, the stabilizing effect of surface tension is kdependent, the stabilization being stronger for shorter wavelengths. Figure 3.3 A picture argument for the effect of su rface curvature in solidification Finally, let us look at the eff ect of transverse diffusion in the two phases. Setting surface tension aside, the effect of th e displacement is as shown in Figure 3.4. On the melt side, the displacement of the surface strengt hens the temperature gradient at a crest, weakening it at a trough; this induces transverse te mperature gradients, carrying h eat away from crests toward troughs. This tends to build crests up. Transverse diffusion is therefore de stabilizing on the melt side of the surface. A similar argument can be made in the solid phase, where the displacement of the surface weakens the temperat ure gradient at a cres t, strengthening it at a trough. This tends to flatten out the displacement by carrying h eat away from troughs toward crests. Hence transverse diffusion is stabilizing in the solid ph ase. In both phases, the effect of transverse diffusion, whether it is stabilizing or destabilizing, is st ronger the higher the k, i.e., the closer together the crests and troughs. Having learned the effect of transverse diffusion in the two phases in the absence of surface tension, here is how surface tension modifies the argument. By reducing the melt temperature at a crest, surf ace tension weakens the instability caused by Liquid Solid Moving Front Melting point decreases Melting point increases PAGE 46 46 transverse diffusion in the liquid while strengthen ing its stabilizing effect in the solid. Hence surface tension is seen to be stabilizing yet again. Figure 3.4 A picture ar gument for the effect of transverse diffusion in solidification What we have learned so far is this: On the liquid side, we have both longitudinal and transverse diffusion, and they are destabilizing just as they were in precipitation. And surface tension weakens both effects. On the solid side, we also have both long itudinal and transverse diffusion. It turns out that thes e are stabilizing, and surface tens ion strengthens their stabilizing effect. Thus in solidification, diffusion (in both directions) has opposite eff ects in the two phases, and we might therefore expect gr owth curves that are significan tly different from those found in precipitation. What is more interesting is that th e introduction of the solid phase gives us an extra parameter in the form of the ratio of the depths of the two phases, and th is can be seen as an additional degree of control. It would appear, then, by fixing ST to establish a stable framework, that we can arrange to find diffusive critical states, independent of capillary effects by controlling the strength of liquid and solid transverse diffusion, which, in turn, are controlled by the depths of the two phases. To quantify this, we turn to a model. Liquid Solid higher lower lower lower higher higher 2 wavelength=k PAGE 47 47 The Nonlinear Equations The sketch given in Figure 3.1 fixes the prin cipal ideas and gives some useful formulas. The liquid and solid phase depths are maintained constant and all equations are written in a frame moving with a constant velocity k U U, where U is the base state speed of the solidification front. All variables introduced ar e movingframe variables; the superscript denotes a liquid phase variable. Then, the temperature in th e solid phase must satisfy 2TT TU tz (3.1) for SzZ and 0 x W where S is the depth of the solid phase and W is the width of the solidification cell and is the thermal diffusivity of th e solid phase. In the liquid phase 2TT TU tz (3.2) must hold for Z zL and 0 x W where is the thermal diffusivity of the liquid phase. Two demands must be met along the solidliquid interface, viz., along ) (t x Z z First, phase equilibrium, taking into account the cu rvature due to the interface, requires 2M M HT TTHT L (3.3) where H L is latent heat of solidification per unit volume of the solid, and where denotes the surface tension; M T is the melt temperature at a flat surface. Second, the rate at which the latent heat is being released due to the motion of the soli dification front must be the same as the rate at which heat is being conducted away from the interface, viz., ...HnTnTLUknu (3.4) PAGE 48 48 where and denote the thermal conductivity of the solid and liquid phase respectively and u denotes the local surface speed of the perturbed front. Farf ield conditions are given by ()STzST (3.5) and () L TzLT (3.6) Finally we impose side wa ll conditions such that Tx Tx and Z x all vanish at0 x and at x W Eq. (3.4) in solidification corresponds to Eq. (2.3) in precipitation (cf., Chapter 2); however, in Eq. (2.3), the term Scc was perturbed and this introduced a term in the perturbation equations that will not be seen in solidification unless the latent heat is perturbed (cf., [3]). Scaled Nonlinear Equations To scale the modeling equations we introduce the length scale via S, the depth of the solid phase. All speeds are scaled by / S while time is scaled by 2/S Temperature is measured below a reference po int corresponding to M T and it is scaled by /HL i.e., /M scaled HTT T L Then, assuming both the thermal conductivity and the thermal diffu sivity to be the same in the two phases, an approximation, that is sometimes reasonable, sometimes not, the scaled temperature in the solid phase must satisfy 2TT TU tz (3.7) for 1zZ and 0 x W while in the liquid phase 2TT TU tz (3.8) PAGE 49 49 must hold for Z zL and 0 x W The equations along the interface, viz., along ) (t x Z z now become 2 THTA (3.9) and ...nTnTUknu (3.10) where A, a dimensionless parameter introduced by our scaling, denotes 2 M HT LS Farfield conditions are given by (1)STzT (3.11) and () L TzLT (3.12) Again, the equations of the model are nonlin ear as the interface temp erature gradients and the position of the interface are coupled to each other, which is clear from Eq. (3.9). Hence the problem is capable of admitting a nontrivial solution besides the trivial base solution. This is to say that the problem is capable of departing from its steady base st ate solution (a planar front at rest in the moving frame), leadi ng to an unstable nonpl anar front. We take the scaled solid farfield temperature, the scaled liquid depth and A to be input variables while U and the temperature fields as well as amplitudes of th e temperature and interface deflection to be outputs that depend on these inputs. To investigate this instability during solidification, we now turn to study the effect of small disturbances on the base state. PAGE 50 50 The Solutions to the Base and the Perturbation Problems The base state variables satisfy 2 00 20 dTdT U dzdz (3.13) in the solid phase, 10 z and 0 x W while in the liquid phase 2 00 20 dTdT U dzdz (3.14) must hold for 0 z L and 0 x W The equations at the planar interface, 0 z are given by 000 TT (3.15) and 00dTdT U dzdz (3.16) The farfield conditions are given by 01STzT (3.17) and 0 LTzLT (3.18) A solution to equations (3.133.18) is given by 0()1 1Uz S UT Tze e (3.19) and 0()1 1Uz L ULT Tze e (3.20) where U is obtained from PAGE 51 51 1 11S L UULT T ee (3.21) This corresponds to a planar solidificati on front at rest in the moving frame. To determine the growth rate of a small disturbance of wave number k imposed on the steady base solution, we expand the surface variables Z T and T as 0101cos()t Z ZZZZkxe (3.22) 00 011011cos()tdTdT TTTZTTZkxe dzdz (3.23) and 00 011011cos()tdTdT TTTZTTZkxe dzdz (3.24) where the side wall conditions determine the admissible values of k. Then the perturbation equations for 1T, 1T and 1 Z are given by 2 2 1 11 2dT d TkTU dzdz (3.25) in the solid phase, 10 z and 0 x W while in the liquid phase 2 2 1 11 2dT d TkTU dzdz (3.26) must hold for 0 z L and 0 x W The equations at the planar interface, 0 z are given by 2 0 1110dT TZkZ dz A (3.27) 2 0 1110dT TZkZ dz A (3.28) and PAGE 52 52 22 00 11 11 22dTdT dTdT Z Z dzdzdzdz (3.29) The farfield equations are 1(1)0Tz (3.30) and 1()0 TzL (3.31) A solution to the domain equations satisfying Eqs. (3.27, 3.28) and (3.30, 3.31) is given by 2 0 1 10 1mm mzmz mmdT kz dz TzeeeZ e A (3.32) and 2 0 1 10 1mmL mzmz mmLdT kz dz TzeeeZ e A (3.33) where 2224 mUUk Finally, using the energy bala nce across the interface, Eq. (3.29), there obtains 22 22 0000 22dTdTdTdT kMkN dzdzdzdz AA (3.34) I II III IV V where 1mm mmmme M e and 1mmL mmLmme N e PAGE 53 53 Eq. (3.34) gives in terms of 2k and the derivatives of the base temperatures at the reference surface, viz., at 0 z Again s on the domain are not important (cf., Chapter 4), and dropping them, we find that the dominant grow th rate is given explicitly by Eq. (3.34), but m are now given by 2224 mUUk. Then m is always positive and m is always negative. Consequently, M is always positive and N is always negative, and the signs of the terms I, II, III and V are negative while term IV is positive. Term IV in Eq. (3.34) accounts for the effect of long itudinal and transverse diffusion in the liquid, both destabili zing. Term II accounts for the stabili zing effect of longitudinal and transverse diffusion in the solid; a term like this did not appear in precipitation. Term V is stabilizing and it is like term IV in precipitation. It takes account of the fact that the base temperature gradient gets weakened in the li quid on moving away from the interface, while it gets strengthened on the solid side. The stabilizi ng effect of surface tension appears in terms I and III. Term III accounts for the fact that surface tension cuts into the destabilizing effect of longitudinal and transverse diffusion in the li quid. Term I accounts for the fact that surface tension enhances the stabilizing e ffect of longitudinal and transver se diffusion in the solid. Next, we move on to draw and explain the vs. 2k curves obtained from Eq. (3.34). The vs. 2k Curves Our aim is to explain how the various terms in Eq. (3.34) get together to produce a curve of vs. 2k. The values of ST LT A and L define a given curve. Fi rst we set the value of ST which for 0ST imposes a basic stabilization on the curves. Then we set the value of 0LT and hence of 0 U The higher the value of LT the stronger the destabilization. The remaining two inputs are A and L The first determines the value of 2k where surface tension begins to PAGE 54 54 exert its effect; the second, th e ratio of the depths of the phases, establishes the relative importance of diffusion in the two phases. Now most of the difficulty in seeing what Eq. (3.34) is trying to tell us stems from the effect of U on the domain in both the base and the pe rturbation problems. As a first step therefore, we look at Eq. (3.34) in the low U limit. A low U result obtains by dropping the term in U from Eqs. (3.1) and (3.2), whence the base speed and the growth rate are given by L ST UT L (3.35) and 3 211 tanhtanhtanhtanhL ST kkL Tk kLkLkkL A (3.36) Notice that if 2k is large enough that tanhk and tanhkL are nearly 1, we obtain 3 22L ST kTk L A and we see that surface tension becomes important only at values of 2k above 1 2 L ST T L A. Now the value of A is typically about 610 Hence, unless L ST T L is close to zero, there is a range of 2k up to around 10000 on which the third term is not important, and we can de duce the shape of the vs. 2k curve from the first two terms in Eq. (3.36), which have to do only w ith diffusion in the two phases. To illustrate the predictions of Eqs. (3.34) and (3.36) we base our calculations on the physical properties of succinonitr ile (SCN) [28] which are listed in Table 3.1. Taking the depth of the solid to be 1Scm, the scales for temperature difference, velocity and time are 24HLK / S 105 m/s and 2/S 900 s. The value of the surface tension group is A0.26 x 106. PAGE 55 55 The vs. 2k curves are presented in Figures 3.5, 3.6, 3.7 and 3.8 as qualitatively correct sketches. Figures 3.5 and 3.6 illustrate the low U formula, whether or not U is low; Figures 3.7 and 3.8 include the effect of U on the domain. The values of ST LT L and 2k given on the sketches were used in Eqs. (3.34) and (3.36) to determine the trends and these input values are experimentally realizable. To indicate the importa nce of the depths of th e two phases, curves are drawn for two values of L In each figure, a value of ST is set, which gives stabilization at 0LT Then a value of L is set to strengthen or weaken th e diffusion effects in the two phases. Curves are then drawn for increasing values of LT increasing the speed of the front. To understand the curves in Figures 3.5 a nd 3.6, it is helpful to note the following formulas which hold for low U, and which can be obtained from Eq. (3.36): 22low11 3LL SSTT kTkTL LLL (3.37) 2intermediate L ST kkT L (3.38) and 23high2 kkA (3.39) Table 3.1 Thermophysical properties of succinonitrile used in calculations Melting Point, MT 331 (K) Density of solid, 1016 (kg/m3) Density of solid, 988 (kg/m3) Thermal Conductivity of solid, 0.225 (W/m K) Thermal Conductivity of liquid, 0.223 (W/m K) Thermal Diffusivity of solid, 1.16 x 107 (m2/s) Thermal Diffusivity of liquid, 1.12 x 107 (m2/s) Surface Tension (solidliquid), 8.94 x 103 (N/m) Latent Heat of Fusion, H L 0.48 x 108 (J/m3) PAGE 56 56 Figure 3.5 vs. 2k for the low U model with 1Scm 1 2 L cm 0.2ST PAGE 57 57 Figure 3.6 vs. 2k for the low U model with 1Scm 2 L cm 0.2ST PAGE 58 58 Figure 3.7 vs. 2k for the complete model with 1Scm 1 2 L cm 0.2ST PAGE 59 59 Figure 3.8 vs. 2k for the complete model with 1Scm 2 L cm 0.2ST PAGE 60 60 These formulas show that for sufficiently low values of LT is negative at 20 k 2d dk is negative at 20 k and remains negative in the intermediate range of 2k. The effect of surface tension is always to make more and more negative at larger and larger values of 2k. As LT increases, and before surface tension comes in, at 20 k 2d dk at 20 k and at intermediate values of 2k, all pass through zero and become positive, and they do this for physically interesting values of LT The order in which these variables turn positive depends on the value of L Notice that in the lowU formulas for low and intermediate ranges of 2k, we see ST the solid side base temperature gradient, combined with L T L the liquid side base temperature gradient in three ways. The liqui d side temperature gradient is multiplied by 1 L L and 1 in the three terms, whence the ratio of the depths of the two phases plays a very important role as far as the growth curves are concerned. For 1 L cf., Figure 3.5, at 20 k turns positive first, then for intermediate values of 2k, and finally 2d dk at 20 k For 1 L cf., Figure 3.6, the order gets revers ed. These facts alone permit the vs. 2k curve to be sketched correctly. The intermediate formula for cf.., Eq (3.38), holds once the value of 2k is large enough that the wavelength of a displacement is smaller than the depths of the two phases, but not so small that surface curvature begins to exert its influence by affecti ng the melt temperature. In this range, the value of is determined entirely by diffusion parallel to the surface, viz., by transverse diffusion in the two phases. The strength of the base gradients combines directly with the surface displacement to determ ine the strength of the troughto crest diffusion in the solid PAGE 61 61 phase, and the cresttotrough diffusi on in the liquid phase. This is il lustrated in Fi gure 3.4. We also note that the value of for intermediate values of 2k, is a multiple of k or 1 wavelength of displacement. Hence is positive and rising or negative and falling for these intermediate values of 2k, according as L ST T L is positive or negative. As k continues to increase and the melt temperature begins to re spond to the curvature, the rising curves turn downward causing a greatest gr owth rate to appear in the intermediate range of 2k. The falling curves simply fall faster once the effect of su rface curvature takes over at larger values of 2k. The low 2k variation of is also due to transverse diffu sion, but the wavelength of the surface displacement is now greater than the dept hs of the phases. The nearsurface region is therefore influenced by the sinks, and an explana tion based on base gradie nts is no longer easy to present in picture form. Nonetheless, at low U, the second term in Eq. (3.37), and Eq. (3.38) indicate that stabilizing diffusion in the solid can dominate destabilizing di ffusion in the liquid at low values of 2k, before their contributions are re versed at intermediate values of 2k, and this depends on L i.e., on whether the depth of the liquid is greater or lesser th an the depth of the solid. The curves show a least at low 2k if 1 L (Figure 3.5) and a greatest at low 2k if 1 L (Figure 3.6). To see the effect of the rati o of the phase depths, Figures 3.5 and 3.6 are drawn at values of LT chosen so that L T L and therefore L ST T L is the same in both figures. Hence the intermediate and large 2k regions of the curves in the two figures at the same values of L ST T L are the same. They differ only at small values of 2k, say up to 2100 k The curves that fall for low values of 2k in Figure 3.5 do not fall as strong ly and may even rise in Figure PAGE 62 62 3.6. Again, this may not be easil y explained from a picture argument, but the second term in Eq. (3.37) shows that at small values of 2k, larger L s weigh diffusion in the liquid more strongly than diffusion in the solid. Next we take U into account on the domain, and th e corresponding curves are shown in Figures 3.7 and 3.8. Again, we draw two sets of curves: Figure 3.7 is drawn for 1 2L whereas Figure 3.8 shows the results for 2 L All the values of L ST T L in Figures 3.5 and 3.6 are in Figures 3.7 and 3.8, but a fe w more are included. Given ST L and LT the effect of including U on the domain, if it is large enough, is to weaken the base gradient at the planar surface in the solid phase and to strengthen it in the liquid. This in turn weakens th e effect of transverse diffusion in the solid, strengthening it in the liq uid. This is enough to explain the majority of changes in going from Figures 3.5 and 3.6 to Figures 3.7 and 3.8. Hence, wherever is falling in Figures 3.5 and 3.6, in Figures 3.7 and 3.8 it is not falling as fast and may even rise, and wherever is rising in Figures 3.5 and 3.6, it is now rising more str ongly in Figures 3.7 and 3.8. As a result, the intermediate region grows into both the small 2k and high 2k regions and this delays the effect of surface te nsion, pushing the values of 2k at the greatest growth rate to the right. In the intermediate region, is given by 2222 11 41 212S UT UUkU e whence rises or falls depending on whether LT is greater or lower than 1 112 2 L ST or what is the same, whether U is greater or lower than ln12ST For small values of L cf., Figure 3.7, we see that for small values of LT is negative for all values of 2k. Upon increasing LT say for 0.050.078LT turns positive for small PAGE 63 63 values of 2k and a critical point appears in the range 2025 k It is due to transverse diffusion in the solid phase and it is indepe ndent of surface tension. It is seen in Figure 3.5 as well, whence it is not due to U on the domain. However, if U on the domain is not accounted for (cf., Figure 3.5), no further critical points can be seen, the only one found, at first due to solid transverse diffusion, eventually is due to melting point lowering, the transition being due to the growing destabilizing effect of transverse diffusion in the liquid phase as LT increases. The effect of U on the domain is to turn base states stabili zed by solid phase transverse diffusion at low 2k into unstable states at higher value of 2k, an effect that cannot be seen if U on the domain is not taken into account. Hence, with U on the domain, we find that further increasing LT leads to three critical points as liquid tr ansverse diffusion is strengthened. In Figure 3.7, these occur for 0.0780.085LT and the three critical points are due to solid transverse diffusion, liquid transverse diffusion and surface tension. This is no t seen in Figure 3.5 due to the fact that the liquid transverse diffusion comes in too late, i.e., after at 20 k has already become strongly positive. For yet higher values of LT the two critical points due to solid and liquid transverse diffusion disappear, giving onl y one critical point due to surfa ce tension. What we have found is this: Upon increasing LT at small L viz., 1 2L the number of critical poi nts is first zero, then one, then three, and finally, again one. It turns out that getting three critical points is a feature typical of pur e solidification when the depths of the two phases are taken to be fi nite, and when the control variable is defined explicitly. The Mullins and Sekerka [9] model for th e solidification of a dilute binary alloy fails to give anything as interesting as has been found above; this is mainly because the depths of the two phases are taken to be of infinite extent in their analysis. Also there is no thermal PAGE 64 64 undercooling in the Mullins and Sekerka model. Similarly, the Wollki nd and Notestine [10] model for pure solidification gives at most two critical points. The fact that we get three critical points has important implications from the point of view of an e xperiment where the liquid side undercooling is slowly increased up to its critical value, and this will be explained in detail toward the end of the chapter wher e we discuss the neutral curves. Turning our attention to Figure 3.8 to see the effect of increasing L many changes from Figure 3.7 are like those seen in going from Figure 3.5 to Figure 3.6. Most interesting, however, are the changes at low values of 2k. Increasing L weakens transverse diffusion in the liquid which might be expected to enlarge the region of solid transverse induced critical points seen in Figure 3.7. But increasing L also weakens longitudinal diffusi on in the liquid causing the values of at low values of 2k to be negative, thereby eliminating the possibility of solid transverse diffusion induced critical points. Hence at large L we find no critical points for small values of LT then upon increasing LT we find two critical points, the first of these due to liquid transverse diffusion, followed by a second critical point due to melting point lowering. So at 2 L there are no critical points at low values of LT two at higher values of LT An important effect of increasing L is this: at low L there are critical points at small values of LT while on increasing L the value of LT at which critical points are first seen increases. If ST were lower, the curves at 2 L might be more like those at 1 2L but then increasing L above 2 ought to restore the curves at 2 L at the higher values of ST Next, we move on to understa nd the neutral curves in soli dification, which are derived from growth curves, though they are much more convenient wh ile trying to understand what PAGE 65 65 goes on in an experiment where th e undercooling in the melt is sl owly crept up to its critical value. The Neutral Curves: U vs. 2k at 0 In neutral curves, the speed of the front U is plotted against 2k at 0 Each set of vs. 2k curves at a given ST and L leads to a U vs. 2k curve. For example, Figure 3.7 leads to Figure 3.9, and Figure 3.8 leads to Figure 3.10. Again, no scale is given on the ordinate and the readers attention is drawn only to th e shapes of the curves. As long as ST is not zero, all possible neutral curves look like Figure 3.9 or 3.10. The first thing to noti ce is this: no matte r the value of 0ST and L once 2k is large enough, U is a decreasing then increasing function of 2k, i.e., the curves all have a dip. This is neither tr ue in precipitation nor in solidification at 0ST where the neutral curve is a straight line as shown in Figure 2.8 of Chapter 2. The curve in Figure 3.10, at a high value of L shows only the dip, and its shape is like the neutral curve obtained in the prob lem of heating a fluid from belo w [7]. But here, the rising side has a very gradual slope due to the small value of A, and this leads to a very flat minimum. The dip is not new; it was seen by Wollkind et al. [10] but it was not explained. Two things are required for a dip: First, should be negative at small values of 2k. In Figure 3.8, is negative at 20 k because large values of L weaken longitudinal diffu sion in the liquid; in Figure 3.7, solid transverse diffusion can cause positive s at 20 k to become negative for small values of 2k. Second, transverse diffusion in the liquid should cause to increase and become positive, before surface tens ion drives it back to zero. He nce, the falling branch of the dip is due to liquid transverse diffusion while the rising branch is due to surface tension. PAGE 66 66 Figure 3.9 An unusual neutra l curve in solidification Figure 3.10 Typical neutra l curve in solidification Figure 3.9, at small L shows something new; it ap pears at small values of 2k and it is due to transverse diffusion in the solid. The curve shows a sharply rising branch feeding into the falling branch of the dip. At small L longitudinal diffusion in the liquid leads to positive values of at 20 k Solid transverse diffusion then drives negative, providing a basis for the dip. PAGE 67 67 In the course of doing this, another critical point is produced. This is se en in Figure 3.9 as the initially rising branch. The readers attention is drawn to the fact that a neutral curve like the one shown in Figure 3.9 cannot be obtained if U on the domain is not accoun ted for. In fact, one can prove that if U is dropped from the domain equations, then one can get zero, one or at most two critical points, but never three. Figure 6.5 in Chapter 6 shows this. Notice that a neutral curve like the one seen in Figure 3.9 is special to solidification. Nothing like it is observed in the closel y related diffusioncontrolled problems: precipitation, a one phaseone surface problem electrodeposition, a one phasetwo surface problem viscous fingering, a two phaseone surface problem The setup in viscous fingering is most like that in solidificatio n, but solidification presents us with an extra degree of freedom. In solidificat ion, unlike these other problems, we have three kdependent effects determining the vs. 2k curves. Also, solidification takes place at a thermodynamically specified temperature, giving us the option of setting the control temperature above or below the phase change value. The reader is referred to Cross and Hohenberg [8] who review pattern formation at a great length in a variety of problems, never finding neutral curves like the new one given in Figure 3.9. It is clear by now that the value of L is important. The transition between the two types of curves shown in Figures 3.9 and 3.10 might seem to be at 1 L And that would be true if U on the domain is not accounted for. But with U on the domain, it is only roughly true. It depends on ST and A, and for 0.2ST it is around 0.85 L for succinonitrile. The neutral curves can be used to answer the question: What does one expect to see in an experiment? For example, how many crests are s een when a plane front loses its stability? In a sequence of experiments, let the values of A, S, L and ST be set. Then W must be set and its PAGE 68 68 role is to determine the admissible values of the wave number as knW 1,2.. n Hence not all points on the neutral curve are of interest. The value of LT and therefore the value of U, is then increased from one experiment to another un til a critical point is reached, i.e., a point where a line of constant U first crosses the neutral curv e at an admissible value of k. Our aim is to predict the effect of W on the number of crests appe aring at a critical point. Now Figures 3.9 and 3.10 both show a dip and th is dip implies that the critical value of U need not correspond to the least allowable value of 2k. In fact, n might be any of the numbers: 1,2,3.., and hence multiple crests might be seen. Th is is not true of precipitation, where we found that one can get at mo st one crest at critical. A neutral curve like the one shown in Figure 3.10 can be obtained for 1 Scm 2 L cm 326.2STK (4.8 K solid cooling). The value of 2k at the dip would then be around 1.5 x 107 2m. The critical wave number will depend on the value of W. For 0.08 Wcm the critical value of 2k occurs near 2 dipk, viz., 2 critk= 1.54 x 107 2m It corresponds to 1 n whence only one crest will be seen at critical, and this is true for all smaller values of W. The corresponding critical interface speed is 3.8 / ms and the critical undercooling is 5.9 K. For a larger width, more crests will be seen at critical. For example, for 0.5 Wcm the critical value of 2k corresponds to 6 n As W increases further, we would e xpect to see even more crests. A neutral curve like the one in Figure 3.9 can be obtained for input values: 1 Scm 0.5 L cm 326.2STK (4.8 K solid cooling). Let us ag ain start with a small value of W and see what happens as W is increased. For 0.08 Wcm the critical value of 2k corresponds to 1 n giving one crest at critical. For a higher value of W, viz., for 0.5 Wcm the critical 2k corresponds to multiple crests, viz., 6 n Here too, increasing the width further will give more PAGE 69 69 crests until W is chosen large enough that the first allowable value of 2k, viz., 2 2W corresponds to a point on the sharply increasi ng part of the neutral curve which is below the dip. For example, for 2 Wcm the first allowable value of 2k is given by 2.5 x 104 2m and it corresponds to an onset of instability giving only one crest. And the critical value of 2k is now independent of the value of A, and hence is not like those seen above. The critical interface speed is now 3.5 / ms and the critical und ercooling is 1.6 K. Hence for small enough values of W the expectation is one cr est at critical, as in precipitation. For larger values of W, the expectation is multiple cres ts at critical, until, at small values of L for high enough values of W, the expectation returns to one crest at critical at a value of 2k now independent of surface tension. Concluding Remarks Our view is that problems of phase change of ten lead to problems in pattern formation. Hence one role of theoretical work is to decide what experiments might be interesting. We suggest that experiments ought to be run by increasing th e control variable, here M LTT or U, to its critical value and th en just beyond. Looked at in this way there is only one critical point in any given experiment, notwithst anding the fact that our versus 2k curves show three critical points at small values of L S, two at low values of 2k, and two critical point s at large values of L S, one at a low value of 2k. In both cases, the low2k critical points are determined by heat conduction in its various manifestati ons, independent of surface tension. It is the neutral curve, not the curves, that bears on the design and interpretation of experiments. As U is increased, at the wave numbers consistent with the width of the cell, the PAGE 70 70 lowest intersections of lines of constant U and the neutral curve predict the critical point that ought to be observed. Hence at small values of L S, there may be a critical point at low values of 2k, independent of surface tension, at large values of L S there cannot be such a low2k critical point. We present a simple model that includes only the most important f actors: heat conduction on the domain and surface tension at the moving fr ont not unlike the models used by Wollkind and Notestine [10] and Mullins and Sekerka [9]. But instead of introducing control via the gradients at the front, we propose farfield conditions more akin to how an experiment might be run. This adds the depths of the two phases as interesting input variables and these variables influence the strength of the eff ect of cresttotrough h eat conduction in both phases, an effect that is stabilizing in the solid, de stabilizing in the melt. It is by doing this that we discover the existence of a critical point at small wave number s independent of the effect of surface tension. The width of the solidification cell strongly affects the patterns that can be obtained in an experiment and we explain what ou ght to be seen as a function of the three lengths that can be controlled: the two depths and the width. PAGE 71 71 CHAPTER 4 ON THE IMPORTANCE OF BULK TRANSPO RT ON INTERFACIAL INSTABILITY Introduction Our goal is to understand whether or not domai n transport dynamics is important compared to the dynamics at the interface. In other words we wish to inves tigate if it is reasonable to drop s from the domain equations. To study this we will consider two problems using precipitation physics as a guide. Both problem s are done in rectangular geometry. First we consider an equilibrium problem where the pl anar front is at rest. Second, we consider a nonequilibrium problem where the front is moving with a nonze ro base speed. In each case we derive a condition that dictat es whether or not s can be dropped from th e domain equations. In both cases we illustrate that the transport dynamics at the interface is ordinar ily much more important compared to the domain transport dynamics, whence dropping s from the domain equations is a good approximation. In precipitation, this result is due to the ex tremely small value of capillary length. The endnote discusses two problems. First we c onsider a solid cylinder in equilibrium with its solution at a uniform concentr ation. Second we consider a solid cylinder in equilibrium with its subcooled melt at a uniform temperature. In e ach case we derive a formula for the growth rate and hence a condition that must be met if s are to be dropped from the domain equations. Case 1: Equilibrium Precipitation Problem in Rectangular Geometry Consider the schematic shown in Figur e 4.1 which shows a solid of density Sc in equilibrium with its solution which is held at a uniform concentration SATc. Then, in the liquid phase, the equation for the solute concentration is given by 2c Dc t (4.1) PAGE 72 72 for Z zL and 0 x W, where L is the depth of the liquid phase and W is the width of the precipitation cell; D denotes the diffusivity of the solute in the liquid phase. The equations along the interface, viz., along ) (t x Z z are given by 2SAT SAT Sc ccH RTc (4.2) and .SccuDnc (4.3) where denotes the surface tension, and where u denotes the local surface speed of the perturbed front. The far fi eld condition is given by ()SATczLc (4.4) Finally we impose side wall conditions so that cx and Z x vanish at0x and at x W Figure 4.1 A precipita tion front at rest The Base State and the Pertur bation Eigenvalue Problem Since 0U, the concentration gradient in the base state is zero. The concentration in the liquid is constant everywhere, viz., 0()SATczc Scc Solution Solid front at rest ,zZxt 21x x Z ik n Z z x E Qcc zL SATcc SATcc 00 zZ PAGE 73 73 To determine if this base so lution is stable, we impose a small displacement on the surface at fixed values of Lc and U, and determine its growth rate, as a function of its wave number k. We expand 0101cos()t Z ZZZZkxe (4.5) and 00 011011cos()tdcdc cccZccZkxe dzdz (4.6) whence the perturbation equations for 1c and 1 Z are given by 2 2 1 20 d kc dzD (4.7) in the solution, 01 z 0 x W The equations that need to be satisfied across the interface, 0z, are 1 1SSATdc ccZD dz (4.8) and 2 11SAT Sc ckZ RTc (4.9) The farfield equation is 1(1)0 cz (4.10) These equations can be solved in the usual ma nner to obtain a formula for the growth rate: 21SAT SSSATc DkN RTccc (4.11) PAGE 74 74 where 1mmL mmLmme N e and 2mk D The formula for the growth rate becomes much simpler if we let L, whereupon the factor N reduces to m whence 2capLk mD (4.12) where capL represents the capillar y length, defined as 1SAT cap SSSATc L RTccc Next let us define S as the value of due to the surface only. Hence S represents the value of the growth rate if is dropped from the domain equation. Hence we have 2 S capLk kD (4.13) Dividing Eq. (4.12) by Eq. (4.13), there obtains 21S SSm kDk (4.14) Hence the sufficient condition for S is given by 21SDk or 2 capkLDk 2Dk 1 Hence the quantity capkL should be much less than unity if one is to justify dropping from the domain equation. Therefore 1capk L is a sufficient condition for S or equivalently for dropping from the domain equation. Table 4.1 Thermophysical properties of CuSO4.5H2O Saturation concentration, SATc243 (kg/m3) Density of solid, Sc 2284 (kg/m3) Diffusivity, D 109 (m2/s) Gas constant, R 8.314 (J/mol K) Surface Tension, 8.94 x 103 (N/m) PAGE 75 75 As an example, let us co nsider the precipitation of 42CuSO.5HO. Using the thermophysical properties given in Table 4.1, and assuming 300TK the capillary length is found to be 114.67x10meterscapL which is extremely small. Hence the vs. 2k curve should be the same as the S vs. 2k curve so long as 22010k Consequently, for most practical purposes we can easily drop from the domain equations and this is due to the ordinarily small values of capL. In fact it is easy to get an explicit formula for vs. 2k. From Eq. (4.12), we get 22capDLkk D (4.15) Squaring both sides we get 2 24260capcapLkLk DD (4.16) which is a quadratic in D Keeping in mind that the growth rate must be negative, we get 42 2214 11 2cap capkL DkL (4.17) Next let us try to approximate the formula given by Eq. (4.17) for large and small values of capkL For small values of capkL we can rearrange Eq. (4.17) to get 2222 424212121 1111 24224capcap capcap capcapkLkL kLkL DkLkL (4.18) Hence PAGE 76 76 101 42421112 21 242capcapcapcapcap cap leadingterm higherordertermkLkLkLkLkL DkL (4.19) or 311 11 22S capcapcapkLkLkL DD (4.20) This means that is less negative compared to S In other words, neglecting on the domain overpredicts the stability. However it is clear from Eq. (4.20) that the correction is negligible. For large values of capkL we can rearrange Eq. (4.17) to get 42422 222214114 1111 222capcap capcapkLkLk DkLkL (4.21) or 3 21cap S capcapLk k DLkDLk (4.22) Again is less negative compared to S and neglecting on the domain overpredicts the stability. And it is clear from Eq. (4.22) that the correction can be significant if the quantity capkL is large enough. And we ag ain conclude that dropping from the domain is a bad approximation if 1capk L or equivalently if the wavelength of the disturbance is much less than the capillary length. The graph given in Figure 4.2 is drawn for 114.67x10 meterscapL and it illustrates the vs. 2k curve superposed over the S vs. 2k curve. is drawn from Eq. (4.17) and S is drawn from Eq. (4.13). Clearly we cannot see any difference as and S are in excellent PAGE 77 77 agreement. Figure 4.3 is drawn for a much larger value of capL viz., 310meterscapL to emphasize that the magnitude of capillary length plays a key role in de ciding the importance of bulk transport compared to th e interface transport. We see that the difference between and S is very subtle for a high value of cap L Next we move on to the usual precipitation case where a planar front is moving at a constant speed in the base state. Figure 4.2 vs. 2k for the equilibrium preci pitation problem with 114.67x10 meterscapL 2k PAGE 78 78 Figure 4.3 vs. 2k for the equilibrium preci pitation problem with 310 meterscapL Case 2: Nonequilibrium Precipitation Prob lem in a Rectangular Geometry This is identical to the previous case except th at the precipitation front is now moving with a speed U. Then, in the liquid phase, the equation fo r the solute concentration is given by 2cc DcU tz (4.23) for Z zL and 0 x W where L is the depth of the liquid phase and W is the width of the precipitation cell. The equations al ong the interfa ce, viz., along ) ( t x Z z are given by 2k PAGE 79 79 2SAT SAT Sc ccH RTc (4.24) and ..SccUknuDnc (4.25) The far field condition is given by ()LczLc (4.26) The side wall conditions are Neumann, whence cx and Z x vanish at0x and at x W The Base State and the Pertur bation Eigenvalue Problem A base state solution is given by 00()UDz SSATSczccce (4.27) where 0U is given by 0ln1LSAT SSATcc D U Lcc (4.28) Expanding Z and c as in Eqs. (4.5, 4.6), and solving th e perturbation eigenvalue problem, there obtains the following formula for the growth, : 2 00 capUU DLkN DD (4.29) where 1SAT cap SSSATc L RTccc 1mmL mmLmme N e and 2 2 00 244 UU mk DDD. The formula for the growth rate becomes much simpler if we let 0UL D, whereupon the factor N reduces to m, whence PAGE 80 80 2 00 capUU DLkm DD (4.30) or 22 2 0 0 2 2 2 0 0 24 1 1141 2 4capU kD D ULk U DU k D (4.31) Next define S as the value of due to the surface only. Then 22 2 0 0 2 01 114 2ScapU kD ULk DU (4.32) Dividing Eq. (4.31) by Eq. (4.32), there obtains 22 2 2 2 0 0 2 22 2 04 1141 4 114S S SD kD U U k D kD U (4.33) Hence the sufficient condition for S is given by 2 2 0 24 1 4SD U k D and one can justify dropping from the domain equation if th is condition is satisfied. Again considering the precipitation of 42CuSO.5HO at 300TK we have 114.67x10 meterscapL which is extremely small. Taking 32= 486kg/m LSATcc 1Lcm we obtain from Eq. (4.28), 7 02.12x10m/secU A graph of S vs. 2k can then be plotted using Eq. (4.32) and this is shown in Figure 4.4. Typically S is of the order of S0.1/sec Hence PAGE 81 81 4SD is of the order of 9210m. But the interesting 2ks are of the order of 12210m. Hence 2 2 0 24 4SD U k D is ordinarily much less than unity. Figure 4.4 vs. 2k for the nonequilibrium pr ecipitation problem with 114.67x10 meterscapL Therefore dropping from the domain equation is again found to be a good approximation for precipitation under nonequilibrium conditions. In fact, for a given set of input parameters, after evaluating S using Eq. (4.32), one can use Eq. (4.33) to solve for the ratio S and hence to find It is found that for unstable wave numbers, S is slightly greater than unity, hence putting s on the domain predicts more positiv e growth rates compared to their S 2k PAGE 82 82 value had s been dropped from the domain equation. For stable wave numbers, S is slightly less than unity, whence in the stable regime, is less negative compared to S Concluding Remarks We have demonstrated that the transport dynamics in the bulk of the domain is ordinarily insignificant compared to the interfac ial transport dynamics, whence dropping s from the domain equations is a good approxima tion for most cases of interest. Endnote 1: Equilibrium Precipitation Problem in Cylindrical Geometry Consider a vertical so lid cylinder of radius 0 R and of density Sc in equilibrium with its solution which is held at a uniform concentration. Then, in the liquid phase, the equation for the solute concentration is given by 2c Dc t (4.34) for 0Rr and 0 zW where W is the height of the cylinder. The equations along the interface, viz., along rRzt are given by 2SAT SAT Sc ccH RTc (4.35) and .SccuDnc (4.36) where denotes the surface tension, and where u denotes the local surface speed of the perturbed front. 2H denotes the curvature of the surface. The far field condition is given by ()finitecr (4.37) PAGE 83 83 Assuming axisymmetry, the surface normal a nd surface speed are respectively given by 21rzz ziRi n R and 21t zR u R The side wall conditions require that both cz and R z vanish at 0 z and at zW The Base State and the Pertur bation Eigenvalue Problem Since 0U, the concentration gradient in the base state is zero. The concentration in the liquid is constant everywhere, viz., 0 01 ()SAT SAT Sc crc R RTc Notice that the base state concentration is not SATc as it used to be for the planar case, cf., case 1. The appropriate correction is present due to the fact that the base surface is itself curved. To determine whether or not this base state is stable, let a small disturbance of amplitude be imposed on the steady base solution. Then expanding R and c at rRz as 01 R RR (4.38) 0 011dc cccR dr (4.39) where the term involving th e surface displacement, 1 R has been introduced so that the perturbation problem can be solved on the referen ce domain. To find the critical base state, we must solve the perturbation eige nvalue problem. It is given by 2 1 1c Dc t (4.40) 11 SSAT R c ccD tr (4.41) PAGE 84 84 and 2 11 1 22 0 SAT Sc R R c R TczR (4.42) The farfield equation is 1()0cr (4.43) Expanding 11cos()t R Rkze 11cos()tccrkze (4.44) the perturbation equations for 1cand 1 Z are given by 2 2 1 21 0 dd kc drrdrD (4.45) in the solution, 0Rr. The equations that need to be satisfied across the interface, 0rR are 1 01 Sdc ccRD dr (4.46) and 2 11 2 01SAT Sc ckR RTcR (4.47) The farfield equation is 1()0 cr (4.48) Defining 22k D a solution to the domain equation, Eq. (4.45) is given as 100() crAIrBKr (4.49) PAGE 85 85 where 0 I and 0K are modified Bessels functions of zero th order. Since we require the solution to be bounded at r we have 0A whence 10() crBKr (4.50) and using Eqs. (4.46) and (4.47), there obtains 0100SccRDBKR (4.51) and 2 001 2 01SAT Sc B KRkR RTcR (4.52) respectively, where 00d KxKx dx Dividing Eq. (4.51) by (4.52), we get 00 22 0 2 0001capL KR DkR RKR (4.53) where capL represents the capillar y length, defined as: 0 011 1SATSAT cap SAT SSS SSAT Scc L c RTcccRTc cc R TcR Next let us define S as the value of due to the surface only. Hence S represents the value of the growth rate if is dropped from the domain equation. Hence we have 00 22 0 2 0001cap SL KkR DkkR RKkR (4.54) Dividing Eq. (4.53) by Eq. (4.54), there obtains 0000 0000SKRKkR kKRKkR (4.55) PAGE 86 86 Substituting 21 kDk there obtains 00 2 00 2' 00 00 21 1 1S S S SS S SKkR Dk KkR DkKkR KkR Dk (4.56) Hence the sufficient condition for S is given by 21SDk Therefore 00 22 0 2 0001 11capL KkR kR kRKkR is a sufficient condition to justify dropping from the domain equation. Endnote 2: Equilibrium Solidification Problem in Cylindrical Geometry Consider a solid cylinder of radius 0 R and at a uniform temperature in equilibrium with its subcooled melt which is held at a uniform temperatur e. The cylindrical front is at rest in the base state. We will scale the temperat ure so that it is measured fr om a reference point corresponding to M T and it is scaled by /HL where is the thermal diffusivity of the solid phase is the thermal conductivity of the solid phase, and H L denotes the latent heat of solidification per unit volume of the solid, viz., /M scaled HTT T L We assume the thermal conductivities and the thermal diffusivities of the two phases to be the same. Then, in the solid phase, the equation fo r the scaled temperature is given by 2T T t (4.57) for 00 rR and 0zW where W is the height of the cylinder. The equation for the scaled temperature in the li quid phase is given by PAGE 87 87 2T T t (4.58) The equations along the interface, viz., along rRzt are given by 2capTTLH (4.59) and 1 .. nTnTu (4.60) where capL represents the capill ary length, defined as 2capM HLT L denotes the surface tension and u denotes the local surface speed of the perturbed front The farfield condition is given by (0)finite Tr (4.61) and 01 ()capTrL R (4.62) The side wall condi tions require that Tz Tz and R z vanish at 0z and at zW The Base State and the Pertur bation Eigenvalue Problem Since 0U the temperature gradient in the base state is zero in both phases. The temperature is constant everywhere in th e solid and the liquid phase. It is given by 00 01 ()()capTrTrL R Notice that an appropriate correction is present in the base itself because the base surface is curved. To determine whether or not this base state is stable, let a small disturbance of amplitude be imposed on the steady base solution. Then expanding R T and T at rRz as PAGE 88 88 01 R RR (4.63) 0 011dT TTTR dr (4.64) and 0 011dT TTTR dr (4.65) where the term involving th e surface displacement, 1 R has been introduced so that the perturbation problem can be solv ed on the reference domain. To fi nd the critical base state, we must solve the perturbation eige nvalue problem. It is given by 2 1 1T T t (4.66) 2 1 1T T t (4.67) 2 11 11 22 0cap R R TTL zR (4.68) and 1111 TTR rrt (4.69) The farfield equations are 1(0)0 Tr (4.70) and 1()0 Tr (4.71) Expanding 11cos()t R Rkze 11cos()tTTrkze 11cos()tTTrkze (4.72) PAGE 89 89 the perturbation equations for 1T, 1T and 1 Z are given by 2 2 1 21 0 dd kT drrdr (4.73) in the solid phase, 00 rR and 2 2 1 21 0 dd kT drrdr (4.74) in the solution, 0Rr The equations across the interface, 0rR are 2 111 2 01capTTLkR R (4.75) and 11 1dTdT R dzdz (4.76) The farfield equations are 1(0)0 Tr (4.77) and 1()0 Tr (4.78) Defining 22k a solution to the domain equations, Eqs. (4.73) and (4.74) is given as 100() TrAIrBKr (4.79) 10() TrBIr 0AKr (4.80) PAGE 90 90 where 0 I and 0K are modified Bessels functions of zeroth order. Since we require 1T to be bounded at 0 r we have 0 B Similarly since 1T must be bounded at r we have 0 B whence 10() TrAIr (4.81) and 10() TrAKr (4.82) where A and A are given by Eq. (4.75) as 2 2 0 1 001capLk R AR IR and 2 2 0 1 001capLk R AR KR respectively. Substituting in Eq. (4.76) there obtains '' 0000 22 00 3 000001capL IRKR RkR RIRKR (4.83) where 00d I xIx dx and 00d KxKx dx Denoting S as the value of the growth rate if is dropped from the domain equation there obtains '' 0000 22 00 3 000001cap SL IkRKkR kRkR RIkRKkR (4.84) Dividing Eq. (4.83) by Eq. (4.84), there obtains '' 0000 0000 '' 0000 0000 SIRKR IRKR k IkRKkR IkRKkR (4.85) Substituting 21 kk there obtains PAGE 91 91 '' 0000 22 0000 22 2 '' 0000 000011 11 1SS SS SS SS S SSIkRKkR DkDk IkRKkR DkDk Dk IkRKkR IkRKkR (4.86) Hence the sufficient condition for S is given by 21Sk or equivalently by '' 0000 22 0 0000001 11capL IkRKkR kR RkRIkRKkR PAGE 92 92 CHAPTER 5 WEAK NONLINEAR ANALYSIS IN PRECIPITATION: GEOMETRIC EFFECTS ON ROUGHNESS Introduction In Chapter 2, linear stability methods were used to analyze the pr ecipitation of a solid from a supersaturated solution. The physics of the instability in precipitation was explained in terms of diffusion and surface tension. We conc luded that a steady pl anar interface becomes unstable and transforms into a nonplanar shape as the solute concentration in the supersaturated solution is increased beyond its critical value. In this chapter, we will be interested in examining what happens beyond the onset of instability. In doing so, our goal is to understand the early stages of the growth of surface roughness in precipitation. The notati on used in this chapter is the same as that used in Chapter 2, and the schematic is again shown in Figure 5.1. Figure 5.1 A precipitati on front growing into a supersaturated solution Again, Lc, the solute concentration, is taken to be the control variab le. Now to observe surface roughness, one might imagine runni ng a series of experiments in which Lc is increased Solution Solid movin g front Scc ,zZxt 21x x Z ik n Z 2., 1t xZ unusurfacevelocity Z 23/22, [1]xx xZ Hsurfacecurvature Z z x E Qcc zL reservoir maintained at a fixed concentrationSAT Sc: solute concetration in the liquid in equilibrium with c at a plane surfaceEQ Sc: solute concetration in the liquid in equilibrium with c at a curved surface L cc SATcc 00zZ PAGE 93 93 to, and just beyond its critical value. In studying preci pitation, we would like to know: Can the critical point be detected in an experiment? What is the nature of the branching to the new nonplanar solution? Is it a pitchfork? Is it forward or backward? Or is it a transcritical bifurcation? What is the effect on the mean front speed as Lc just crosses its critical value? Is it higher or lower than the values predicted by the base formulas? Our plan is to begin by presenting again th e nonlinear equations in scaled form, and working out the base solution and the solution to the perturbation eigenvalue problem to find the critical value of Lc. To do this we follow the same steps as in Chapter 2, but nonetheless, we will rewrite the perturbation equations, now for 0 so as to formulate an eigenvalue problem. Having found the critical value of Lc, we look for steady solutions ju st beyond this critical point by going to second and third orders in the amount by which Lc is increased. We discover that the crosssectional shape of the pr ecipitation cell is important in deciding the nature of the bifurcation. Ordinarily, a rectangul ar cross section leads to a backwa rd pitchfork; a circular cross section with symmetric disturban ces gives a transcritical bifur cation. Toward the end of the chapter, we deduce a symmetry condition that determ ines whether a cross section of an arbitrary shape will lead to a pitchfork or a transcritical bifurcation. Then some other specific cross sectional shapes are studied and it is found that a transcritical bifurcation is more likely to occur. Finally, regardless of the nature of the branchi ng, it is concluded that the surface speed always decreases compared to what its base value would have been. PAGE 94 94 Rectangular Cross Section The Steady State Nonlinear Equations The dimensionless nonlinear model equations are the same as given in Chapter 2. However, we set all t terms to zero as our goal is to inve stigate the nature of the steady state solutions to the problem upon crossi ng the threshold. Again, all va riables used here are scaled. In the liquid phase, the equation for the s caled solute concentr ation is given by 20c cU z (5.1) for 1 Z z and 0 x W The equations along the interface, viz., along () zZx are 2 cH A (5.2) and .1. nccUkn (5.3) where A denotes 11SAT SSSATc L RTccc The solute concentration at the reservoir is given by (1)Lczc (5.4) Side wall conditions are Neumann, i.e., 0,0,0cZ xWxW x x Finally, the volume of the supersaturated solution must be maintained fixed, whence 00WZxdx (5.5) Once a material is chosen, the physical properties are fixed. Once a cell is built, other variables such as L and W remain fixed. In running a sequence of experiments, Lc is increased from one PAGE 95 95 steady experiment to another. In each experiment, the mean interface speed, U, and the interface shape, Z x, are the outputs. The Base Solution The base state variables are denoted by the subs cript 0. A base state solution is given by 00()1Uzcze (5.6) where 0U is given by 0ln(1)LUc (5.7) Besides the physical properties, the i nput variables in the base state are L and Lc, and the output is 0U. The Eigenvalue Problem at Neutral Conditions On increasing Lc, we expect the planar interface to lose its stability. To determine the critical value of Lc, let a small disturbance be imposed on th e steady base solution at fixed values of Lc and 0U, where Lc and 0U satisfy Eq. (5.7). Then expanding z Zx as 01 Z ZZ 00Z (5.8) the solute concentration at the surface is given by 0 011dc cccZ dz (5.9) where the term involving th e surface displacement, 1 Z has been introduced so that the perturbation problem can be solved on the base domain (cf., Appendix B.1). To find the critical base state, the perturbation ei genvalue problem must be solved at zero growth rate, whence 2 1 100 c cU z (5.10) PAGE 96 96 on the domain, 01 z 0 x W where, at 0 z 2 0 1 11 2dc dZ cZ dxdz A (5.11) and 1 010 c Uc z (5.12) must hold, and where, at the reservoir, 1z 10c (5.13) must hold. The sidewa ll conditions require 10cx and 10dZdx at 0 x and at x W Thus the problem is homogeneous in 1c and 1 Z and it has solutions of the form 11cos() cczkx and 1cos() Z kx A (5.14) which satisfy the side wall conditions so long as ,1,2... knWn where k is the wave number of the disturbance and 0 n is ruled out by the condition 1 00WZdx (5.15) which holds the volume of the solution fixed. Equations (5.10), (5.12) and (5.13) imply that 1c must be zero, and therefore, at 0 z 1 Z must satisfy 2 0 1 1 2dc dZ Z dzdx A (5.16) where 0 00 dc zU dz Then Eq. (5.16) is satisfied for all values of so long as 2 0Uk A (5.17) PAGE 97 97 The corresponding neutral curve is a straight line as shown in Figure 2.7 (b) (cf., Chapter 2), where the ordinate must be replaced by 0U now. Given W, and hence a value of 2k, Eq. (5.17) determines the critical value of the interface speed, 0U. The most dangerous value of 2k corresponds to 1 n or to kW hence only one crest is seen at critical. As the control variable Lc increases upward from zero, the output variable 0U increases from zero, until there comes a point when Eq. (5.17) is satisfied. The corresponding value of 0U will be referred as its critical value. Henceforth the subscript zero will refer to this critical state. At values of U higher than 0U, the base solution loses its stability. The critical value of Lc, denoted 0 L c, corresponds to the critical value of U via Eq. (5.7). Ordinarily then, one might increase the control parameter Lc to its critical value, and then slightly beyond, in order to discover the new steady states that ensue. Assuming with some evidence that the bifurcation would be pitchfork in nature, one can only hope to get a locally forward pitchfork by advancing th e control variable beyond its crit ical value. However it turns out that our hope of find ing a forward pitchfork is not realiz ed upon proceeding in this way. The branching to nonplanar steady so lution is, in fact, a backward pitchfork. Hence we introduce 21 2 as an input giving the amount by which Lc is decreased from its critical value (note that will no longer be used as a measure of the amplitude of a disturbance as it was in the eigenvalue problem). Now the new subcritical solution will not have the speed of the corresponding base solution, hence we first find the change in the base speed corresponding to a given change in Lc. Eq. (5.7) indicates that for a small decrease in Lc from 0 L c, the corresponding change in U from 0U, per unit change in Lc is given by 0Ue PAGE 98 98 The First and Higher Order Problems To find the steady backward branches, we write 021 2LLcc (5.18) where 21 2 is an input. We then expand U and Z as 23 012311 ... 26 UUUUU (5.19) and 23 012311 ... 26 ZZZZZ (5.20) Using the formula for the surface normal vector given in Figure 5.1, and observing that the factor 21xZ in the denominator cancels from both sides of Eq. (5.3), only the numerator of nU is needed, where UUk. Hence the expansion of nU is given by Eq. (5.19). This and the other expansions (cf., Appendix B.2) lead to the respective problem s at first and higher orders in The First Order Problem The first order problem differs from the eige nvalue problem in that the system now must respond to a change in Lc, and this may lead to a change in U, whence, at first order, we must determine 11,cZ and also 1U. The first order problem is given by 2 0 1 101dc c cUU zdz (5.21) on the domain, 01 z 0 x W 2 0 1 11 2dc dZ cZ dxdz A (5.22) PAGE 99 99 and 1 011c UcU z (5.23) at 0 z 10c (5.24) at 1z, and 1 00WZdx (5.25) The sidewall conditions require 1cx and 1dZdx to vanish at 0 x and at x W Now Eqs. (5.21), (5.23) and (5.24) have a particular solution depending only on z, while the corresponding homogeneous problem is just the eigenvalue problem whose solution is 10c Hence we have 0111UzcUze (5.26) and we can turn to Eq. (5.22), which is a differential equation for 1 Z Using 110 czU along with Eq. (5.25) and 10dZdx at 0, x W and integrating Eq. (5.22) over 0 x W we find that 1U must be zero. Hence 1c must also be zero, whereupon Eq. (5.22) must have the solution 1cos() Z kx A (5.27) What we have so far is this: 10c 10U 1cos() Z kx A. And we go on to the second order problem to find the value of A. The speed of the surface does not change at order upon decreasing Lc from 0 L c by 21 2 PAGE 100 100 The Second Order Problem At second order, the new unknowns to be determined are 22,cZand 2U, and the unknown A which carries over from first orde r. The second order problem is 2 0 2 202dc c cUU zdz (5.28) in the liquid solution, 01 z 0 x W 2 2 2 00 2 212 22dcdc dZ cZZ dzdxdz A (5.29) and 2 022c UcU z (5.30) at 0 z 21c (5.31) at 1z, and 2 00WZdx (5.32) The sidewall conditions require 220cxdZdx at 0 x and at x W Solving Eqs. (5.28), (5.30) and (5.31), we get 00221UUzcUzee (5.33) And it remains to solve Eqs. (5.29) and (5.32) for 2 Z To find 2 Z a solvability condition must be satisfied. It is 2 2 0 21 2 000cos0Wdc czZzkxdx dz (5.34) PAGE 101 101 and it is satisfied, no matter the value of A, due to 3 00coscos0WWkxdxkxdx (5.35) Hence A cannot be found at second order, but we can find 2U and 2 Z To get 2U, integrate Eq. (5.29) over 0 x W and use Eq. (5.32), along with 20dZdx at 0, x W The result is 022 201 2UUeU A (5.36) where the first term is what 2U would be if the base solution obtains at the new value of Lc. This term was derived earlier. The second term is the correction corresponding to a curved front. It is negative. If a curved front coul d exist, its speed would be slow er than that of the corresponding planar front. In other words, the mean speed of the interface runs behind its base value. Before moving on to solve the third order pr oblem, we would like to emphasize that an analogous weak nonlinear calculation will be performed in solidific ation (cf., Chapter 6) where we will demonstrate that once the instability sets in, the mean sp eed of the solidification front always runs ahead of its base value. In an atte mpt to delineate the key disparity between the two problems, differences will be traced to differen ce in the interfacial mass balance in one versus the energy balance in the other. Now to go to third order, we need to solve Eq. (5.29) for 2 Z Substituting for 20 cz and 2U, the constant term disappears, and we see that 2 Z can be written as 222cos()cos(2) Z ZkxZkx (5.37) where, while 2 Z cannot be determined at this order, 2 Z is given by PAGE 102 102 2 0 2 2 2 01 2 4U Z Uk AA (5.38) Hence, 2 Z and A carry on to third order. The Third Order Problem The equations at the third order are given by 2 30 303cdc cUU zdz (5.39) on the domain, 01 z 0 x W 2 232 2 3 0030 211 311213 2322339 dcdcdZdc dcdZdZ cZZZZZ dzdzdzdxdxdxdz AA (5.40) and 3 033c UcU z (5.41) at 0 z 30c (5.42) at 1z, and 3 00WZdx (5.43) The sidewall conditions require 330cxdZdx at 0 x and at x W Again, 3c, like 1c and 2c is only a function of z, and it is given by 0331UzcUze (5.44) which satisfies Eqs. (5.39), (5.41) and (5.42). This leaves Eqs. (5.40) and (5.43) to determine 3U PAGE 103 103 and 3 Z Now Eq. (5.40) is a differential equation for 3 Z And again, a solvability condition must be satisfied if 3 Z is to be solved for. This condition is 2 23 2 3 00 2 11 31121 232 00303009cos0Wdcdc dcdZdZ czZzZZzZzkxdx dzdzdzdxdx A(5.45) and it is an equation for 2A. Substituting our results at orders zero, one and two, we find 2A to be given by 02 2 006 3 31 2Ue UU AA (5.46) Ordinarily, A is of the order of 910 and 0U is of the order of 0.1. Hence 2A is positive and a backward pitchfork is confirmed. The value of 3U can be found in terms of 2A by integrating Eq. (5.40) over 0 x W and using Eq. (5.43), along with the side wall condition, 30dZdx at 0, x W We started the nonlinear calculati on by guessing the expansion to be 021 2LLcc And the reason we find a pitchfork branching in the ca se of a onedimensional rectangle is because at second order, solvabil ity does not determine A. This, in turn, is due to the fact that the eigenfunction cos kx satisfies 3 0cos0Wkxdx Had the solvability condition determined A at second order, it would have determined 0 A, and the hypothesis of a pitchfork bifurcation would have failed. If the cross section were a tw odimensional rectangle, a pitc hfork would also have been found due to 33 00coscos0y xW Wkxkydxdy Indeed a twodimensional rectangle, which is thin PAGE 104 104 enough such that there is no yvariation, reduces to the oned imensional case presented above. But there is no reason to belie ve that an eigenfunction f of the Laplacian operator, 2, on an arbitrary cross section, satisfying Neumann cond itions at the edges, s hould satisfy the condition 30AreafdA Hence there is reason to imagine that for some cross section, we can determine the value of A to be zero at second order, whence we will have to choose a different expansion, and the branching will not be a pitchfork. To see this, we work out the case of a circular cross section with axisymmetric disturbances. Circular Cross Section with Axisymmetric Disturbances Now let the precipitation cell have a circular cross section of radius R To demonstrate the existence of a transcritical bifurcation, we will focus our attention on axisymmetric displacements, which may not be the most dange rous; a more general case allowing for the nonaxisymmetric disturbances is also very intere sting and it is discussed in the endnote to this chapter. Since the base state variables depend only on z, the base state remains the same as it was in the case of a rectangular cross section. The pe rturbation eigenvalue problem is nearly as it was before, only the curvature is a little diffe rent, hence the equilibrium condition at 0 z is now 0 1 111 dc dZ d crZ rdrdrdz A (5.47) where the dependence of the variable 1 Z has been ignored. The fixed volume condition is now given by 1 00RZrdr (5.48) and the sidewall conditions require that cr and dZdr vanish at rR whence PAGE 105 105 10 c r and 10 dZ dr at rR (5.49) The remaining equations lead, as before, to 10c whence Eq. (5.47) has the solution 10() Z Jkr A (5.50) where k can be any positive root of 1()0JkR and where the neutral curve is again given by 2 0Uk A. Hence, given R the critical value of 0U is determined by the smallest positive root of 1()0JkR 0 k being ruled out by Eq. (5.48). Again, we start with the guess that the st eady branches leave the critical point as 021 2LLcc Then using the expansions given in Eqs. (5.19) and (5.20) we find at first order that 10c 10U and 10() Z Jkr A. At second order, the solvabil ity condition is not satisfied unless 3A is zero, and therefore 1 Z is zero. And going to higher orders, 2 Z 3 Z all turn out to be zero, indicating that our orig inal guess for expansion of Lc was incorrect, and also indicating that the expansion 0LLcc (5.51) will work out. Introducing as 0 L Lcc and retaining Eqs. (5.19) and (5.20), the eigenvalue problem remains as before. It is written holding 0UU and it gives us the critical value of 0U in terms of 2k, again 2 0Uk A where the subscript zero henceforth denotes critical values. The surface normal vector is now given by 21rrz r Z ii n Z Then observing that the denominator 21rZ cancels from both sides of Eq. (5.3), only the numerator of nU is needed, and its expansion is PAGE 106 106 given by Eqs. (5.19). This leads to the problem s at first and higher orders in and we solve them as Lc increases by beyond its critical value. The First Order Problem The first order problem is 2 0 1 101dc c cUU zdz (5.52) on the domain, 01 z 0 1 111 dc dZ d crZ rdrdrdz A (5.53) and 1 011c UcU z (5.54) at 0 z 11c (5.55) at 1z, and 1 00RZrdr (5.56) The sidewall conditions require that 1cr and 1dZdr vanish at rR Again, 1c can depend at most on zand it is given by 00111UUzcUzee (5.57) which satisfies Eqs. (5.52), (5.54) and (5.55). Eqs. (5.53) and (5.56) then determine both 1U and 1 Z Multiplying Eq. (5.53) by r, integrating the result over 0 rR and using Eq. (5.56) along with the side wall condition, 10dZdr at rR there obtains PAGE 107 107 1 000Rczrdr (5.58) whence 01UUe (5.59) whereupon 1U is no longer zero and the expression for 1c can be simplified to 001UUzczee (5.60) The fact that 1c and 1U are not zero is another sign that a circular cross section differs from a rectangular cross section. However, 1c is zero at 0 z and the solution to Eq. (5.53) is 10() Z Jkr A (5.61) but A cannot be determined at first order, and we turn to the second order problem. The Second Order Problem At second order, we must solve 2 0 21 20212 dc cc cUUU zdzz (5.62) on the domain, 01 z 2 2 00 12 2112 21 2dcdc dcdZ d cZZrZ dzdzrdrdrdz A (5.63) and 2 022c UcU z (5.64) at 0 z 20c (5.65) at 1z, and PAGE 108 108 2 00RZrdr (5.66) The sidewall conditions require 220crdZdr at rR Solving Eqs. (5.62), (5.64) and (5.65), we get 00022 222 UUUzczzeUeUe (5.67) and it remains to solve Eq. (5.63) for 2 Z while satisfying Eq. (5.66). It is our hope that this will give us the value of as well as the value of 2U. Now Eq. (5.63) can be solved for 2 Z if and only if a solvability condition is satisfied. It is 2 2 0 1 2110 2 002000Rdc dc czZzZzrJkrdr dzdz (5.68) and this is an equation for A. Substituting our results at ze roth and first orders, we get 2 0 0 1 2 3 0 0 02R RrJkrdr U U rJkrdr (5.69) whereupon using the formulas gi ven in Table 5.1, we get 02 0 2 00.02UJkR e U (5.70) Table 5.1 Some integrals concerning Bessels functions 222 001 01 0 2RrJkrdrRJkRwhereJkR 32 01 00.020RrJkrdrRwhereJkR Hence the branching to the new steady state is not a pitchfork; it is transcritical. PAGE 109 109 Multiplying Eq. (5.63) by r, integrating the result over 0 rR and using the constant volume conditions, Eqs. (5.48) and (5.66), and side wall condition, 20dZdr at rR we get 2 22 201 00 2RR czUrZdr (5.71) whence 02 222 200 UUeUJkR A (5.72) The first term on the right hand side, 02Ue, is what 2U would be if the base solution obtains at the new value of Lc. The second term is the co rrection corresponding to a cu rved front. It is again negative, as it was for a rectangula r cross section, and we conclude again that the mean speed of the interface runs behind its base value. What we have found so far is this: the nature of the branching to nonpl anar steady states is ordinarily a pitchfork for a recta ngular cross section, while it is transcritical in the case of a circular cell with axisymmetric disturbances. Th e cross section dependenc e raises the question: How can we predict the nature of the branching fo r an arbitrary cross section? It is to this question that we now turn. A Cross Section with an Arbitrary Shape Let the cell have an arbitrary cross section, but we will work in Cartesian coordinates. The nonlinear equations, Eqs. (14) remain the same, but the fixed volume condition is now given by ,0SZxydxdy (5.73) where z Zxy defines the shape of the solidliquid interface, and where the integral is carried out over the cross section S. The mean curvature of the surface z Zxy is given by PAGE 110 110 22 32 22121 2 1yxxxyxyxyy xy Z ZZZZZZ H ZZ (5.74) The sidewall conditions are now .0 pc and .0 pZ on S (5.75) where S denotes the boundary of S, and p denotes the outward normal to S The solution to base state equations, which depends only on z, is given by Eqs. (5.6) and (5.7). The Eigenvalue Problem at Neutral Conditions The eigenvalue problem at ne utral condition does not change much. It is given by 2 1 100 c cU z (5.76) on the domain, 01 z where, at 0 z 22 2 0 1111 22Hdc cZZZ dzxy AA (5.77) and 1 010 c Uc z (5.78) must hold, and where, at the reservoir, 1z 10c (5.79) must hold. This problem has solutions of the form 11, cczfxy and 1(,) Z fxy A (5.80) where f xy is an eigenfunction of 2 H on S subject to Neumann sidewall conditions. It satisfies PAGE 111 111 22,, f xykfxy on S; .0 pf on S (5.81) Hence, denoting the eigenvalue by 2k the solution f const 20 k is ruled out by the condition 1,0SZxydxdy (5.82) Eqs. (5.76), (5.78) and (5.79) imply that 1c must be zero, and therefore 1 Z must satisfy 2 0 1100Hdc Z zZ dz A (5.83) where 0 00 dc zU dz And we find that the neut ral curve is again given by 2 0Uk A. It holds regardless of the shape of the cross section. But the allowable values of 2k do depend on the geometry. Again, the lo west positive value of 2k determines the critical speed of the front. Let us begin by assuming the branching to be a pitchfork, and expanding Lc as 021 2Lc The expansions for U and Z are given by Eqs. (5.19) and (5.20) and we move on to solve the problem at the first and higher orders to find the steady subcritical solution. The First Order Problem The first order problem is given by 2 0 1 101dc c cUU zdz (5.84) on the domain, 01 z 2 0 111Hdc cZZ dz A (5.85) 1 011c UcU z (5.86) and PAGE 112 112 1,0SZxydxdy (5.87) at 0 z and 10c (5.88) at 1 z Eqs. (5.84), (5.86) and (5.88) have the solution 0111UzcUze (5.89) Using this in Eq. (5.85), we find from Eq. (5.87) that 1U must be zero, and hence 1c must be zero, whereupon 1 Z is given by 1, Z fxy A (5.90) where f xy is an eigenfunction of 2 H on S, satisfying Neumann conditions on S and it corresponds to the critical or lowest, value of 2k, with 20 k ruled out. We go to second order to find the value of A but we already see that the speed of the interface does not change at order upon decreasing Lc from 0 L c by 21 2 The Second Order Problem At second order, we must determine 22,cZand 2U, and the unknown A which carries over from first order. The second order problem is 2 0 2 202dc c cUU zdz (5.91) on the domain, 01 z 2 22 00 2122 2 Hdcdc cZZZ dzdz A (5.92) PAGE 113 113 2 022c UcU z (5.93) and 2,0SZxydxdy (5.94) at 0 z and 21c (5.95) at 1z. Solving Eqs. (5.91), (5.93) and (5.95), we get 00221UUzcUzee (5.96) and it remains to solve Eqs. (5.92) and (5.94) for 2 Z To find 2 Z a solvability condition must be satisfied. It is 2 2 0 21 200,0Sdc czZzfxydxdy dz (5.97) whence we have 23,0Sfxydxdy A (5.98) At this point, there are two possibilities. On e possibility is that the integral vanishes, whereupon solvability is satis fied no matter the value of A. And as in the case of a rectangular cross section, we go on to third order to find A. This is the pitchfork case. The second possibility is that the integral does no t vanish and we must conclude that A is zero, whence the branching is not a pitchfork and we must propose a new expansion in cf., [29]. PAGE 114 114 Cross Sections on which the Integral Does Not Vanish If the integral 3,S f xydxdy is not zero, as in the case of a circular cross section, we assume the branching is transc ritical and propose to move be yond critical via the expansion 0LLcc (5.99) The base problem and the eigenvalue problem remain unchanged. But the first order problem is now 2 0 1 101dc c cUU zdz (5.100) on the domain, 01 z 2 0 111Hdc cZZ dz A (5.101) 1 011c UcU z (5.102) and 1,0SZxydxdy (5.103) at 0 z and 11c (5.104) at 1z. Eqs. (5.100), (5.102) and (5.104) have the solution 00111UUzcUzee (5.105) whereupon integrating Eq. (5.101) over S and using Eq. (5.103), we have 2 0Uk A (5.106) and 1c simplifies to PAGE 115 115 001UUzczee (5.107) whence 1c must be zero at 0 z and 1 Z is then 1, Z fxy A (5.108) To find A we go to second order, where we must solve 2 0 21 20212 dc cc cUUU zdzz (5.109) on the domain, 01 z 2 22 00 1 21122 22Hdcdc dc cZZZZ dzdzdz A (5.110) 2 022c UcU z (5.111) and 2,0SZxydxdy (5.112) at 0 z and 20c (5.113) at 1z. Eqs. (5.109), (5.111) and (5.113) are satisfied by 00022 222 UUUzczzeUeUe (5.114) and it remains to solve Eq. (5.110) for 2 Z while satisfying Eq. (5.112). The solvability condition for Eq. (5.110) is 2 2 0 1 211 20200,0Sdc dc czZzZzfxydxdy dzdz (5.115) This is an equation for A in which 2U does not appear and we find PAGE 116 116 02 2 3 0, 2 ,U S S f xydxdy e U f xydxdy A (5.116) whence a transcritical bifurcation is verified. The value of 2U obtains upon integrating Eq. (5.110) over S to get 02 222 201 ,U SUeUfxydxdy S A (5.117) where the first term on the ri ght hand side gives the value 2U if the base solution persists at the new value of Lc. The second term is a correction due to the displacement of the front. It is always negative, predicting a slow down. Thus if the integral 3,S f xydxdy is not zero, the branching is transcritical and the front slows down. Cross Sections on which the Integral Vanishes If we have a cross section where the integral 3,S f xydxdy is zero, then A cannot be determined at second order, but the solvability condition for Eq. (5.92) is satisfied no matter the value of A. Our job then is to solve Eq. (5.92) for 2 Z and then go to third order hoping to determine A. To do this we first obtain 2U by integrating Eq. (5.92) over S, using 1, Z fxy A, 0220UczUe Neumann side wall conditions and Eq. (5.94). The result is 0222 201 ,U SUeUfxydxdy S A (5.118) whence we have PAGE 117 117 222 201 0,SczUfxydxdy S A (5.119) and Eq. (5.92) is then 22 2222 0 21 ,,H SU f xydxdyfxykZ S AA (5.120) whereupon 2 Z can be obtained as a series in the eigenfunctions of 2H on S subject to Neumann boundary conditions, viz., 2 2nn n Z BfBf (5.121) where the coefficients n B depend on the expansion of the left hand side of Eq. (5.121) and where the eigenfunction 1 does not appear. The equations to be solved at third order are 2 30 303cdc cUU zdz (5.122) on the domain, 01 z 23 32 000 2 31121133 2333Hdcdcdc dc cZZZZNZZZ dzdzdzdz AA (5.123) 3 033c UcU z (5.124) and 3,0SZxydxdy (5.125) at 0 z and 30c (5.126) PAGE 118 118 at 1 z where 110cU has been used and where 1NZ denotes the nonlinear part of 32H. It depends on 1 Z and introduces the factor 3A. Eqs. (5.122), (5.124) and (5.126) can be solved for 3c, viz., 0331UzcUze (5.127) leaving Eqs. (5.123) and (5.125) to be solved for 3U and 3 Z where 3 Z can be obtained if and only of a solvability condition is satisfied. It is 23 3 00 2 311211 23 0030300,0Wdcdc dc czZzZZzZzNZfxydxdy dzdzdz A(5.128) and it is an equation for 2A. If 2A is not zero, the branching is a pitchfork, as assumed, and if 2A is positive the pitchfork is backward, as assumed. The equation for 2A can be simplified somewhat without making S definite. Using 1, Z fxy A and, at 0 z 2 2 0 0 2dc U dz 3 3 0 0 3dc U dz and 02 0201Udc UUUe dz we have 0222334 02002,0 1313 0U particular SSS SUUUefdxdyUfZdxdyUfdxdy NZfdxdy AAAA (5.129) where 2,particularZ is a multiple of 2A, where 1NZ is a multiple of 3A and where 00201UUUUe is given by 0222 001 1,U SeUUfxydxdy S A. Hence the term in A does not vanish and a pitchfork is confirmed. Thus if the integral 3,S f xydxdy is zero, the branching is pitchfork and the front slows down. PAGE 119 119 An Equilateral Triangular and a Regular Hexagonal Cross Section We explore two other cross sections: an e quilateral triangle and a regular hexagon. Appendices C.1 and C.2 give the solu tion to the eigenvalue problem, Eq. (5.81), for these geometries. In the case of a regular hexagon, we find 3293 ,0 4SfxydxdyL whence the branching is transcritical The branching is found to be transc ritical on an equila teral triangle as well. Concluding Remarks We have given a condition that determines th e nature of branching of a steady planar crystalsolution interface to a nonplanar shape fo r an arbitrary cross se ction of a precipitation cell. Given any cross section of th e cell, if one can find the solution f xy to the eigenvalue problem, Eq. (5.81), then the nature of the branchi ng to the steady nonplanar solution can be predicted by finding whether or not the integral 3,S f xydxdy is zero. The branching is usually a pitchfork if this integral is zero; if not, we expect a transcritic al bifurcation. We find that for a rectangular cross section, the branch ing is ordinarily a backward pitchfork; it is transcritical for all other shapes that we have studied: circular (sym metric disturbances), triangular (equilateral), and hexagona l (regular). And it is our belief that transcritical bifurcation is what will be found in the case of a more gene ral cross section. We al so find that the surface speed always decreases compared to what its base value would have been. Endnote 1: A Circular Cross Section with the Possibility of Nonsy mmetric Disturbances Let us consider the possibility of nonsymmetric displacements. If nonsymmetric disturbances are admitted, the critical value of 2k is given by the smallest positive root of 10 JkR (5.130) PAGE 120 120 and this eigenvalue corres ponds to the eigenfunction 1cos fJkr (5.131) where 2 33 1 00cos0RrJkrdrd (5.132) Hence the branching must be a pitchfork. Because nonsymmetric disturbances are more dangerous, the branching in the case of a ci rcular cross section mu st be a pitchfork. PAGE 121 121 CHAPTER 6 SOLIDIFICATION: WEAK NONLINEAR ANALYSIS Introduction We concluded in Chapter 3 that a steady planar interface becomes unstable and changes to a nonplanar shape as the underc ooling in the subcooled liquid is increased beyond a critical value. Linear stability methods were used to analyze the shapes of the various disturbance growth curves. In addition, the sh ape of the neutral curves showed that as many as three neutral points could be obtained, each depending upon the longitudinal and transverse and sizes of the solid and liquid phases. The occurr ence of three possible neutral poi nts was explained in terms of longitudinal diffusion, transverse diffusion and surface tension. In this chapter, we are interested in exam ining the postonset regime in solidification of a pure material. We shall use the techniques learned from our analysis of the precipitation problem in the vicinity of the instability. Again, our goal is to understand the early st ages of the growth of surface roughness. The notation used in this chapter is the same as that used in Chapter 3. The liquid phase sink temperature, LT, is taken to be the control variable. The idea is to imagine running a series of experiments in which LT is decreased to, and just below its critical value. We try to answer questions like: What is the effect on the mean front speed as LT just crosses its critical value? In other words, does the front speed up or slow dow n compared to its base value? Is the branching to the new nonplanar solu tion a forward or a backward pitchfork? We will consider three major cases. First, we consider a case where the latent heat is rejected only though the subcooled liquid while the solid phase is held at the melt temperature; both phases are taken to be of fin ite extent. Second, we consider a simplification of the first case where the solid phase is taken to be of infinite extent. This may be useful when the solid sink is far away from the interface. Third, we consider the most general case where the latent heat is PAGE 122 122 rejected though the frozen solid as well as the subcooled liquid a nd both phases are taken to be of finite extent. In all of these cases, we begin by presenting the nonlinear equations in scaled form. We work with a rectangular cross section and al l derivations are done in Cartesian coordinates. We first work out the base solution and the so lution to the perturbation eigenvalue problem to find the critical value of LT. Having found the cr itical value of LT, we look for steady solutions near this critical point by going to sec ond and third orders in the amount by which LT is decreased from its critical value. All derivations are done in a manner similar to Chapter 5. Therefore the details of the derivations will not be presented in their entirety as they are straightforward enough that a reader can easily re produce them. Our emphasis will be to point out important differences fr om the precipitation problem. In all of these cases we will find that the m ean speed of the surface is higher than its base speed. This is to say that once th e instability sets in, the surfac e speeds up compared to its base value. This result is striking in its contrast to what was obtained in Chapter 5 for precipitation, where we found that the surface unconditionally slows down. The obvious question is: Can we explain this basic difference of a speedup in solidification versus a slowdown in precipitation? In our quest to answer this questio n, we will go to a yet simpler model in a fourth case where the solid phase is completely ignored. It is as if the solid phas e is there only for the sake of having an interface, but it is ignored fo r all other purposes such as transport etc. One might also look at it as if the solid is being chopped off as soon as it forms. We do not learn anything new, and again the front speeds up compar ed to its base value. But the model (although an artificial one) is now as clos e as one can get to the precipita tion model, and the key disparity between the two problems becomes clear. We lear n that the difference is posed by the heat balance (in solidification) versus mass balance (in precipitation) at the inte rface. And this is what PAGE 123 123 leads to the essential dissimilar ity between the two problems regard ing the correction to the front speed. Now in order to learn the nature of the bran ching to the new nonplan ar solution, one must go to third order in the amount by which the underc ooling in the subcooled liquid is increased. In general, we will find that one can obtain a forw ard as well as a backward pitchfork depending on the input parameters. We will identify regions in the input parameter space where the branching is forward and regions where it is backward. To do all of this, let us start by performing a weak nonlinea r analysis for a case where the frozen solid is maintained at its melt temperatur e and the latent heat is rejected only to the subcooled liquid, both phases being of finite extent. Case 1: Latent Heat Rejected Only to the Subcooled Liquid, Both Pha ses of Finite Extent Figure 6.1 A moving solidification front The sketch given in Figure 6.1 fixes the prin cipal ideas. The liquid a nd solid phase depths are maintained constant and all equations are writ ten in a frame moving with a constant velocity, Liquid Solid z Zx 00zZ z S z L solidsink M TT L TT liquidsinkmoving front 21x x Z ik n Z 2., 1t xZ unusurfacevelocity Z 23/22, [1]xx xZ Hsurfacecurvature Z PAGE 124 124 k U U where U is the base state speed of the solidif ication front. All vari ables introduced are movingframe variables; the superscript denotes a liquid phase variable. All lengths are scaled by S, the depth of the solid pha se. All speeds are scaled by / S where is the thermal diffusivity of the solid phase. Temperature is measured from a reference point corresponding to M T, and it is scaled by /HL where is the thermal conductivity of the solid phase, and H L denotes the latent heat of solidificat ion per unit volume of the solid, viz., /M scaled HTT T L Then, the scaled temperature in the solid phase must satisfy 20 T TU z (6.1) for 1 zZ while in the liquid phase 20T TU z (6.2) must hold for Z zL Two demands must be met along the solidliquid interface, viz., along () zZx First, phase equilibrium, taking into account the curvature du e to the interface, requires 2 THTA (6.3) where A denotes 2 M HT LS a dimensionless parameter intro duced by our scaling, and where denotes the surface tension. Second, the rate at which the latent heat is being released due to the motion of the solidification front must be th e same as the rate at which heat is being conducted away from the interface, viz., ... nTnTnUk (6.4) PAGE 125 125 Farfield conditions are given by (1)0 Tz (6.5) and ()LTzLT (6.6) Side wall conditions are Neumann, i.e., 0,0,0,0TTZ xWxWxW xxx Finally, the volume of the subcooled liqui d must be maintained fixed, whence 00WZxdx (6.7) Notice that these model equations are nonlin ear as the position of the interface depends upon the temperature gradients there, and these gr adients, in turn, depend upon the position and morphology of the interface. This nonlinearity fo rms the heart of the problem without which instability would not be possible. The Base Solution The base state is one in which the solidifica tion front remains planar and at rest in the moving frame. The base state variables are de noted by the subscript 0, and they satisfy 2 00 0 20 dTdT U dzdz (6.8) in the solid phase, 10 z 0 x W and 2 00 0 20 dTdT U dzdz (6.9) in the liquid phase, 0 zL 0 x W The equations at the planar interface, 0 z are given by 000 TT (6.10) PAGE 126 126 and 00 0dTdT U dzdz (6.11) The farfield conditions are given by 0(1)0 Tz (6.12) and 0() L TzLT (6.13) Then, a solution to Eqs. (6.86.13) is given by 0()0 Tz (6.14) 00()1UzTze (6.15) where 0U is given by 0ln(1)LUT (6.16) whence 0 00 dT zU dz 2 2 0 0 20 dT zU dz 3 3 0 0 30 dT zU dz 4 4 0 0 40 dT zU dz Besides the physical properties, the input variables in the base state are S, L and LT and the output is 0U The Perturbation Eigenvalue Problem On increasing LT we expect the planar interface to lose its stability. To determine the critical value of LT let a small disturbance of amplitude be imposed on the steady base solution at fixed values of LT and 0U which satisfy Eq. (6.16). Then expanding Z T, and T at z Zx as 01 Z ZZ 00 Z (6.17) PAGE 127 127 0 011dT TTTZ dz (6.18) and 0 011dT TTTZ dz (6.19) where the term involving th e surface displacement, 1 Z has been introduced so that the perturbation problem can be solved on the re ference domain (cf., Appendix B.1). To find the critical base state, we must solve the pe rturbation eigenvalue pr oblem. It is given by 2 1 100 T TU z (6.20) on the domain, 10 z 0 x W and 2 1 100 T TU z (6.21) on the domain, 0 zL 0 x W where, at the base surface 0 z 2 0 1 11 2dT dZ TZ dxdz A (6.22) 0 111dT TTZ dz 00 dT dz (6.23) and 2 0 11 1 2dT TT Z zzdz 2 0 20 dT dz (6.24) must hold, and where, 110 Tz (6.25) and PAGE 128 128 10 TzL (6.26) The stricken terms are shown to emphasize that fo r this case the solid temperature is taken to be uniform in the base state. The sidewall conditions require 10 Tx 10 Tx and 10 dZdx at 0 x and at x W Thus the problem is homogeneous in 1T 1T and 1 Z and it has solutions of the form 11cos() TTzkx, 11cos() TTzkx and 1cos() Z kx A (6.27) which satisfy the side wall conditions so long as ,1,2...knWn where k is the wave number of the disturbance. 0 n is ruled out by the condition 1 00WZdx (6.28) The expansions presented a bove give the equations for1T 1T 1 Z They are 2 2 1 10 20 dT d kTU dzdz (6.29) in the solid phase, 10 z 0 x W and 2 2 1 10 20dT d kTU dzdz (6.30) in the liquid phase, 0 zL 0 x W (6.31) 2 0 1dT Tk dzA 1 Z (6.32) 0 111dT TTZ dz 00dT dz (6.33) and PAGE 129 129 2 0 11 1 2dT dTdT Z dzdzdz 2 0 20 dT dz (6.34) at 0z 1(1)0 Tz (6.35) and 1()0TzL (6.36) Defining 22 004 2 UUk m a solution to Eqs. (6.296.33, 6.35, 6.36) is given by 2 0 110mm mzmzdT kz dz TzAeee A 11mm mzmz mmeeeZ e (6.37) and 2 0 11 10 1mmLmmL mzmzmzmz mmLdT kz dz TzAeeeeeeZ e A (6.38) We can then turn to Eq. (6.34), which gives us 222 000 kMkUNU AA (6.39) where 0 1mm mmmme M e and 0 1mmL mmLmme N e Figure 6.2 shows the corresponding neutral curve. The graph of 0U versus 2k is nearly a straight line. In fact it is easy to prove the following asymptotic limits for small values of 2k: For small 2k, Eq. (6.39) may be simplified to PAGE 130 130 get 2 01UkL A at neutral while for large values of 2k we get 2 02 Uk A. This gives us some idea about the shape of the curve without having to actually plot it. Figure 6.2 Typical neutral curve fo r solidification when the latent heat is rejected only through the subcooled liquid Now given W, and hence 2k, Eq. (6.39) determines the critic al value of the interface speed, 0U The most dangerous value of n is one, which corresponds to kW giving only one crest and one trough at critical. As the control variable LT increases upward from zero, the output variable 0U increases from zero, until there comes a point when Eq. (6.39) is satisfied. The corresponding value of 0U will be referred as its critical value. Henceforth the subscript zero will refer to this criti cal state. At values of U higher than 0U the base solution loses its stability. Now on increasing LT to its critical value 0LT, and then slightly beyond, it might be thought that the branchin g to the new steady solution would be a forward pitchfork. Hence we now introduce 21 2 as an input giving the amount by which LT is increased from its critical value. Observe that takes on a different meaning from earlier. U0 2k PAGE 131 131 Now the new supercritical solution will not have the speed of the corresponding base solution, hence we first find the change in U corresponding to a given change in LT Eq. (6.16) indicates that for a small increase in LT from 0LT, the corresponding change in U from 0U per unit change in LT is given by 01ULe L Before moving on to higher orde r problems, observe that neither 1T nor 1T is zero. It is worth noticing here that in the analogous preci pitation problem in Chapter 5, the variable 1c was found to be zero. This, in turn, was due to the fact that there were several cancellations of mappings in the interfacial mass balance equation, viz., .1.nccUkn. The corresponding cancellations do not occur in solidification in the interfacial energy balance equation, viz., ... nTnTnUk Not only does this make the analysis more cumbersome in solidification, but it also leads to a very fundamental difference between the two problems: Once the instability sets in, the front slows down in pr ecipitation, while it speeds up in solidification. And to prove the fact of a speedup in solidi fication, we move on to solve the higher order problems in The First and Higher Order Problems We begin by making a guess as to how the steady states emanate from the planar morphology, i.e., how do the steady branches leav e the critical point. Thus, to find the steady forward branches, we write 021 2LLTT (6.40) where PAGE 132 132 021 2 L LTT (6.41) is an input. We then expand U and Z as 23 012311 ... 26 UUUUU (6.42) and 23 012311 ... 26 ZZZZZ (6.43) This guess is known to be correct in some ot her phase change problems (cf., Q. Buali, L.E. Johns and R. Narayanan) [30] and we begin by seei ng if it works here. Using the formula for the surface normal vector given in Figure 6.1, and observing that the factor 21xZ in the denominator cancels from both sides of Eq. (6.4), only the numerator of nU is needed, where UUk. The expansion of nU is given by Eq. (6.42). This and the other expansions (cf., Appendix B.2) lead to the respective pr oblems at first and higher orders in The First Order Problem The first order problem differs from the ei genvalue problem in that the system now responds to a change in LT and this may require a change in U at first order, whence, at first order, we must determine 111,, TTZ and also 1U The first order problem is given by 2 0 1 101dT T TUU zdz (6.44) on the domain, 10z 0 x W and 2 0 1 101dT T TUU zdz (6.45) on the domain, 0 zL 0 x W PAGE 133 133 2 0 1 11 2dT dZ TZ dxdz A (6.46) 0 111dT TTZ dz 00dT dz (6.47) and 2 0 11 1 2dT TT Z zzdz 2 0 1 2dT U dz (6.48) at 0z 110Tz (6.49) and 10TzL (6.50) The volume condition is given by 1 00WZdx (6.51) The sidewall conditions are Neum ann. Now Eqs. (6.44, 6.45, 6.4751) have a particular solution depending only on z, and the corresponding homogeneous problem is just the eigenvalue problem. Hence we have solutions of the form 11011,cos()TxzTzTzkx (6.52) 11011,cos()TxzTzTzkx (6.53)1111cos()cos()cos() Z xkxZkxZkxA (6.54) where the notation for the subscript indices n eeds some clarification. The first index of a subscript will be used to represent the order of the problem and the second index will denote the PAGE 134 134 harmonic of which the variable in question is a coefficient. For example, 11T represents a first order variable, and it is the coefficient of the first harmonic, viz., cos() kx Next, the expansions presented above are used to obtain the {10} and {11} problems. The {10} problem The {10} problem is given by 2 10100 01 2dTdTdT UU dzdzdz (6.55) in the solid phase, 10z 0 x W and 2 10100 01 2dTdTdT UU dzdzdz (6.56) in the liquid phase, 0 zL 0 x W 10100 TT (6.57) and 1010 1dTdT U dzdz (6.58) at 0z 10(1)0 Tz (6.59) and 10()0 TzL (6.60) The general solution to Eqs. (6.55, 6.56) is given by 000101010 0z UzUzUTzAeeeBd (6.61) 00001010101 01z UzUzUUTzAeeeBUed (6.62) PAGE 135 135 and their derivatives are given by 10 100100dT zBUA dz and 10 100100dT zBUA dz whereupon using Eqs. (6.576.59) to eliminate 10 B 10A and 10 B there obtains 0 00 010101 1U UzUz Ue TzAee e (6.63) 0 000 0101011 1U UzUzUz Ue TzAeeUze e (6.64) where 10A is found by using Eq. (6.60) as 0 0 01 1011 1U UL UUL A e e e The {11} problem The {11} problem is similar to th e eigenvalue problem. It is given by 2 2 11 110 20dT d kTU dzdz (6.65) in the solid phase, 10 z 0 x W and 2 2 11 110 20dT d kTU dzdz (6.66) in the liquid phase, 0 zL 0 x W 0 111111dT TTZ dz 00dT dz (6.67) and 2 0 1111 11 2dT dTdT Z dzdzdz 2 0 20dT dz (6.68) at 0 z 11(1)0Tz (6.69) PAGE 136 136 and 11()0TzL (6.70) Finally we turn to Eq. (6.46), which is a differential equation for 1 Z Using Eq. (6.51) and 10dZdx at 0, x W and integrating Eq. (6.46) over 0 x W we find that 1U must be zero, and therefore 10T and 10T must both be identically zero. Then the first order problem is exactly like the eigenvalue problem, and the condition at neutrality is given by Eq. (6.39), viz., 222 000 kMkUNU AA. What we have so far is this: 10 U 1111,coscosTxzTzkxTzkx, 1111,coscos TxzTzkxTzkx and 11cos()cos() Z xkxZkxA. And we go on to the second order problem to find the value of A. The speed of the surface does not change at order upon increasing LT from 0LT by 21 2 The Second Order Problem At second order, the new unknowns to be determined are 2T 2T 2 Z and 2U and the unknown A comes over from first order. The second order problem is 2 0 2 202dT T TUU zdz (6.71) on the domain, 10 z 0 x W and 2 0 2 202dT T TUU zdz (6.72) on the domain, 0 zL 0 x W 2 2 0 1 211 22 dT T TZZ zdz 2 0 2 2 2dT dZ Z dxdz A (6.73) PAGE 137 137 2 2 0 11 2211 22 dT TT TTZZ zzdz 2 00 2 2dTdT Z dzdz 00 dT dz (6.74) and 3 22 2 0 2211 11 2232 dT TTTT ZZ zzzzdz 32 00 2 32dTdT Z dzdz 2 0 2 11 122xdT dz TT ZU xx (6.75) at 0 z 210 Tz (6.76) and 21 TzL (6.77) The volume condition is given by 2 00WZdx (6.78) Observing the nature of the above equations, we assume the following form for the solution 2202122(,)()()cos()()cos(2) TxzTzTzkxTzkx (6.79)2202122(,)()()cos()()cos(2) TxzTzTzkxTzkx (6.80)22122cos()cos(2) Z xZkxZkx (6.81) And again, we use these expansions to dedu ce the {20} and {21} and {22} problems. At 0 z the following formulae may be useful: 22 222 11 1001 2221cos2 TT Z UkUZkx zz 22 11 10121cos2xTT Z kUZkx xx PAGE 138 138 The {20} problem The {20} problem is given by 2 20200 02 2dTdTdT UU dzdzdz (6.82) in the solid phase, 10 z 0 x W and 2 20200 02 2dTdTdT UU dzdzdz (6.83) in the liquid phase, 0 zL 0 x W 22 2020011 2 TTUZ (6.84) and 32 2020 2011 2 dTdT UUZ dzdz (6.85) at 0 z 2010 Tz (6.86) and 201 TzL (6.87) A general solution to Eqs. (6.82, 6.83) is given by 000202020 0z UzUzUTzAeeeBd (6.88) 00002020202 01z UzUzUUTzAeeeBUed (6.89) and their derivatives are given by 20 200200 dT zBUA dz and 20 200200 dT zBUA dz PAGE 139 139 whereupon using Eqs. (6.846.86) to eliminate 20 B 20A and 20 B we get 0 00 020201 1U UzUz Ue TzAee e (6.90) 0 0000 022 20200121 1 12U UzUzUzUz Ue TzAeeUeZUze e (6.91) where 20A is found by using Eq. (6.87) as 0 0 0 022 201 201 2 11 1UL U UL UULUZe A e e e (6.92) The {21} problem The {21} problem is similar to the eigenvalue problem. It is given by 2 2 21 210 20 dT d kTU dzdz (6.93) in the solid phase, 10 z 0 x W and 2 2 21 210 20 dT d kTU dzdz (6.94) in the liquid phase, 0 zL 0 x W 0 212121dT TTZ dz 00 dT dz (6.95) and 2 0 2121 21 2dT dTdT Z dzdzdz 2 0 20 dT dz (6.96) at 0 z 21(1)0 Tz (6.97) PAGE 140 140 and 21()0 TzL (6.98) Defining 22 00 214 2 UUk mm a solution to Eqs. (6.936.98) is given by 2121 21212121 mm mzmzTzAeee (6.99)2121 21212121 mmL mzmzTzAeee (6.100) where 21212 21211mmk A Z e A (6.101) 21212 0 21211mmLkU A Z e A (6.102) The {22} problem The {22} problem is given by 2 2 22 220 240 dT d kTU dzdz (6.103) in the solid phase, 10 z 0 x W and 2 2 22 220 240 dT d kTU dzdz (6.104) in the liquid phase, 0 zL 0 x W 22 2222010221 2 TTUZUZ (6.105) and 22322 2222 01010221 20 2 dTdT UkZUZUZ dzdz (6.106) PAGE 141 141 at 0 z 22(1)0 Tz (6.107) and 22()0 TzL (6.108) Defining 22 00 2216 2 UUk m a solution to Eqs. (6.1036.108) is given by 222222222222 mzmmmzTzAeee (6.109) 2222 22222222 mmL mzmzTzAeee (6.110) where 222222322 022001022022 22 222211 2 22 1mmUNUkUZUNUZ A eNM (6.111) 222222322 022001022022 22 222211 2 22 1mmLUMUkUZUMUZ A eNM (6.112) and where 2222 22222222 221mm mmmme M e and 2222 22222222 221mmL mmLmme N e Using the curvature equation Finally we turn to Eq. (6.73), which is a differential equation for 2 Z Using Eq. (6.78) and 20 dZdx at 0, x W and integrating Eq. (6.73) over 0 x W we find 2 2 0 1 2011 21 000 2 dT dT TzZzZz dzdz 0 (6.113) Using 2 1 10 dT zkMZ dz A and 20200 TzA we get 22 201 A kMZA. Comparing with PAGE 142 142 Eq. (6.92), there obtains 0 0 0 0 2 2222 201 0111 11 21planar correctionU ULUL U U Ue UeUkMeZ LLe A (6.114) where the first term is what 2U would be if the base solution obtains at the new value of LT and this term was derived earlier. The second term is the correction correspon ding to a curved front. It is always positive. Hence the mean speed of a curved front would be faster than that of the corresponding planar front. In other words, the mean speed of the interface runs ahead of its base value. To go to third order, we need to solve Eq. (6.73) for 2 Z To find 2 Z a solvability condition must be satisfied. It is 1 21112 0002000WWT TzZzZdxTzZdx z (6.115) It is easy to verify that this condition is au tomatically satisfied, wh ence we cannot learn the value of A at this order. Finally, substituting the expansions from Eqs. (6.79, 6.81) into Eq. (6.73), the constant term disappear s due to the result obt ained earlier using integrability, viz., due to Eq. (6.113). The coefficient of the first harmonic va nishes due to solvabili ty, and there obtains by equating the coefficients of the second harmonic 2 1 221220040 dT TzzZkZ dz A (6.116) Substituting for 2222222201mmTzAe and using 2 1 10 dT zkMZ dz A, there obtains 2222222 122 224 1mmkMZkZ A e AA (6.117) PAGE 143 143 Comparing with Eq. (6.111), and simplifying, we get 2 221P Z Z Q (6.118) where 2232 022002222 22 2222022011 2 22 4 UNUkUkMNM P Q kNMUNU A A (6.119) Hence, 21 Z and A carry on to third order. The Third Order Problem The equations at the third order are given by 2 30 1 30233 TdT T TUUU zzdz (6.120) on the domain, 10 z 0 x W and 2 30 1 30233 TdT T TUUU zzdz (6.121) on the domain, 0 zL 0 x W 2 2 22 2 0 11211 311212 2 293333 dT dZdZTTT TZZZZZ dxdxzzzdz A 3 3 0 1 3dT Z dz 2 30 3 2dZdT Z dxdz A (6.122) 22 2 222211 33112 2 0 12 2333 3 TTTTTT TTZZZ zzzzzz dT ZZ dz 23 3 00 1 23dTdT Z dzdz 3 00 3 3dTdT Z dzdz 00 dT dz (6.123) and PAGE 144 144 223322 2 33 221111 112 223322 3 0 12 3333 3 TT TTTTTT ZZZ zzzzzzzz dT ZZ dz 34 3 00 1 34dTdT Z dzdz 42 00 3 42dTdT Z dzdz 2 0 3 2 22 221111 112323xxdT U dz TTTTTT ZZZ xxxzxzxx (6.124) at 0 z 310Tz (6.125) and 30TzL (6.126) The volume condition is given by 3 00WZdx (6.127) We assume the following form for the solution 330313233(,)()()cos()()cos(2)()cos(3) TxzTzTzkxTzkxTzkx (6.128)330313233(,)()()cos()()cos(2)()cos(3) TxzTzTzkxTzkxTzkx (6.129)3313233cos()cos(2)cos(3) Z xZkxZkxZkx (6.130) Now this problem can be solved completely, but it suffices to solve the {31} problem to learn the nature of bifurcation and the value of 2A, and it is to this that we turn our attention next. The {31} problem The {31} problem is given by 2 2 31 1 3102 230 dT dT d kTUU dzdzdz (6.131) in the solid phase, 10 z 0 x W and PAGE 145 145 2 2 31 1 3102 230 dT dT d kTUU dzdzdz (6.132) in the liquid phase, 0 zL 0 x W 22 2 2020 222211 313111 22 2 0 11 22 219 3 24 33 22 dTdT dTdTdTdT TTZZ dzdzdzdzdzdz dT dTdT Z dzdzdz 23 00 122 233 4 dTdT ZZ dzdz 3 3 0 1 3 0dT Z dz dT dz 0 310 dT Z dz (6.133) and 22 2233 2 31312020 222211 11 222233 3 22 0 11 22 22319 3 24 33 22 dTdTdTdT dTdTdTdT ZZ dzdzdzdzdzdzdzdz dT dTdT Z dzdzdz 34 00 122 343 4 dTdT ZZ dzdz 4 3 0 1 4 2 0 2dT Z dz dT dz 22 22 00 3122221 223 3 2 dTdT ZkTTZk dzdz 2 32 0 11122 230 dT ZkTTZ dz (6.134) at 0 z 31(1)0 Tz (6.135) and 31()0 TzL (6.136) A general solution to Eqs. (6.131, 6.132) is given by 313131mzmzmzmzTzAeBeCzeDze (6.137) 313131mzmzmzmzTzAeBeCzeDze (6.138) PAGE 146 146 where 213 UA Cm mm 213mmUA Dme mm 213 UA Cm mm and 213mmLUA Dme mm Eliminating 31 B and 31 B using Eqs. (6.135, 6.136), we get 3131 mmmm mzmzmzmzmzTzAeeeCeDeCzeDze (6.139)3131 mmLmmL mzmzmzmzmzTzAeeeLCeDeCzeDze (6.140) Now Eqs. (6.133, 6.134) can be written as 3131111mmmmL A eAeR (6.141)31312mmmmLAmmeAmmeR (6.142) where 1 22 2 2020 22221111 1122 22 2 0 2193 3 242 3 2mmmmLRCeDLCeD dTdT dTdTdTdTdTdT Z ZZ dzdzdzdzdzdzdzdz dT dz 23 00 122 233 4 dTdT ZZ dzdz 3 3 00 1 3dTdT Z dzdz 0 31dT Z dz (6.143) and PAGE 147 147 2 22 2233 2 2020 222211 11 222233 3 22 0 11 22 22319 3 24 33 22mmL mmRmCeDCDLmCeDCD dTdT dTdTdTdT ZZ dzdzdzdzdzdz dT dTdT Z dzdzdz 34 00 122 343 4 dTdT ZZ dzdz 4 3 0 1 4 2 0 2dT Z dz dT dz 22 22 00 3122221 223 3 2 dTdT ZkTTZk dzdz 2 32 0 11122 23 dT Z kTTZ dz (6.144) where all terms in 1 R and 2 R are evaluated at 0 z Now 1 R and 2 R can be simplified using results from previous orders. At 0 z the following formulae may be helpful: 32 2020 2011 2 dTdT UUZ dzdz 22 42 2020 02011 2 2 dTdT UUUZ dzdz 2 11 01dTdT UZ dzdz 22 22 11 001 22dTdT UkUZ dzdz 33 222 11 001 222dTdT UkUZ dzdz 2322 2222 0010221 2 2dTdT UkUZUZ dzdz 22 224222 2222 00100221 44 2dTdT kUUZUUkZ dzdz Hence PAGE 148 148 2232 1 210010122 03133 3 42mmmmL R CeDLCeDUZUkUZUZZ UZ (6.145) 2 2243222 02100100122031333 6 242mmmmLRmCeDCDLmCeDCD UUZUkUZUUkZZUZ (6.146) Then solving Eqs. (6.141, 6.142) for 31A and 31 A there obtains 12 311mmRNR A eMN (6.147) 12 311mmLRMR A eMN (6.148) Using the curvature equation Finally we turn to Eq. (6.122), which is a differential equation for 3 Z To find 3 Z a solvability condition must be satisfied. It is 2 22 2 21111 31121 22 0 13 003030309 00W WTTTdZdZ TzZzZzZzZdx zzzdxdx TzZdx A (6.149) Using 31131 01 00 2WTzZdxWZTz 20 2 22 1111 011 30300 22WdT TdT ZzZdxWZZzz zdzdz 22 22 11 1111 22 019 300 24WTdT ZzZdxWZZz zdz PAGE 149 149 11 21122 013 300 22WTdT ZzZdxWZZz zdz 2 13131 01 0 2WTzZdxWZkZ A 2 2 43 11 111 2 019 9 24WdZdZ AZdxWZkZ dxdx A there obtains 2 2 20 2211 311122 2 432 131193 030000 242 9 0 4 dT dTdTdT TzzzZzZzZ dzdzdzdz kZkZ AA (6.150) which can be written as 3131mmAeR (6.151) where 2 2 20 221 3 11 2 432 1 2213119 3000 24 39 0 24mmdT dTdT R CeDzzZzZ dzdzdz dT zZkZkZ dz AA (6.152) whence using 022 20 011 0 1UdT zUkMZ dze A 222 22 2212204dT zMkMZkZ dz AA 2 1 10 dT zkMZ dz A PAGE 150 150 2 22 1 01 20 dT zkkUMZ dz A there obtains 22223 0 3 2201 0 22 221223139 32 124 1 32 2mmU R CeDkMkMMkkUMZ U kMMZZkZ AAA AA (6.153) Comparing the value of 31 A from Eq. (6.147) and Eq. (6.151), we get 1230 RNRRMN (6.154) Next, we define: __ 11031 R RUZ __ 2 22031 R RUZ and __ 2 3331 R RkZ A. Substituting in Eq. (6.154), and using the relation at neutrality, viz., Eq. (6.39), there obtains ______ 1230RNRRMN (6.155) where now __ 2232 1 21001012233 3 42mmmmL R CeDLCeDUZUkUZUZZ (6.156) __ 2 224322 02100100122333 6 242mmmmLRmCeDCDLmCeDCD UUZUkUZUUkZZ (6.157) __ 22223 0 3 2201 0 2 2212239 32 124 1 32 2mmU R CeDkMkMMkkUMZ U kMMZZ AAA A (6.158) Now Eq. (6.155) is an equation for 2 1 Z whence, upon combining terms, we get PAGE 151 151 3 12213141220 TTUZTZTZZ (6.159) where 1 2 0 0 2 01 1mmLmm mmLmm mmL mmL mm mmTLNCeDmCeDCD LmCeDCDMCeD kU meULNmm e k meUMmm e A A 213 UZ mm (6.160) 2036 TNU (6.161) 22224 30000 2222 0 220 0333 424 39 32 124 TNUkUkUU U MNkMkMMkkUM U AAA (6.162) 2222 400022331 32 222 TNUUkUMNkMM A (6.163) Hence 2 1 2231422 210 T TUTZTZ UZ (6.164) or 2 2 11 222341 2 2121 10correction planarU TTP TUTTTZ Q UZUZ Z (6.165) PAGE 152 152 or 1 22 21 2 1 2 1 234 2 21 1 planar correctionT TU UZ Z U TP TTT Q UZ Z (6.166) Hence, using Eq. (6.114) to substitute for 2 planarU and 2 correctionU and Eq. (6.119) for P Q we finally obtain 02 0 0 0 2 0 2 1 2 0 01 31 36 1 1mmL mmL UL mm mm mmL mmLkU meULNmm e NUe mmL k meUMmm e Z kU meULNmm e A A A 0 0 00 2 0 22 0 22224 0000 0 03 36 1 11 .11 21 333 424 3 1mm mm U UL UNU mm k meUMmm e e UkMe Le NUkUkUU U MN U A A 2222 220 2222 00022 2232 022002222 22 2222022039 2 24 331 32 222 11 2 22 4 kMkMMkkUM NUUkUMNkMM UNUkUkMNM kNMUNU AAA A A A (6.167) PAGE 153 153 Table 6.1 Case 1 Values of 0U and 2A for inputs 1,2 L and 2k 1L 2L 2k 0U 2A 0U 2A 104 5 x 1011 5.99 x 1010 7.5 x 1011 3.998 x 1010 103 5 x 1010 5.99 x 109 7.5 x 1010 3.987 x 109 102 5 x 109 5.94 x 109 7.45 x 1093.849 x 108 101 5 x 108 5.41 x 109 7.07 x 1083.342 x 107 1 5 x 107 2.215 x 106 5.66 x 1073.579 x 106 10 5 x 106 2.662 x 104 5 x 106 1.487 x 104 102 5 x 105 2.667 x 102 5 x 105 1.379 x 102 103 5 x 104 2.667 5 x 104 1.348 104 5 x 103 2.68 x 102 5 x 103 1.35 x 102 105 5 x 102 2.802 x 1045 x 102 1.474 x 104 106 0.5 4.392 x 1060.5 3.621 x 106 The sign of 2 1 Z is of interest. If it is positive, the pitchfork is forward, as assumed, otherwise it is backward. In the latter case, L T must be expanded as 021 2LLTT Clearly it is a formidable task to determine the sign of 2 1 Z and one can possibly not determine its sign by inspection. We will therefore have to do some calc ulations in order to learn the nature of the bifurcation. Table 6. 1 gives the sign of 2 1 Z for various input values of 2k, L Clearly 2 1 Z can have either sign, whence the pitchfork can be backward as well as forward depending on the input parameters. However, the formula given in Eq. (6.167) is not very useful if one wants to PAGE 154 154 learn how the sign of 2 1 Z depends upon the inputs. It is hard to figure out the point of crossover from a backward to a forward pitchfork. Hence we will consider a case later where we will look at a special case of case 1 where the model equations will not have U on the domain. We will obtain a formula for 2 1 Z which would be much simpler compared to Eq. (6.167). This case is deferred until the end of the chapter, where we will find out the analytical dependence of 2 1 Z on inputs 2k and L For now, we continue in the spirit of hopi ng to find the reason for a speedup in pure solidification. So we consider a simplification of case 1 where the solid phase is of infinite extent. This case might indeed be more realistic in that solids of extent much greater than the capillary length ought to be considered infinite. Case 2: Latent Heat Rejected Only to the Subc ooled Liquid, Solid Phase of Infinite Extent The notation remains the same as in case 1. Since the solid phase depth is of infinite extent, all lengths are scaled by L, the depth of the liquid phas e. All speeds are scaled by /L The scaled temperature is given by /M scaled HTT T L Then, the scaled temperature in the solid phase must satisfy 20 T TU z (6.168) for zZ and 0 x W while in the liquid phase 20 T TU z (6.169) must hold for 1 Z z and 0 x W The equations along the interface, viz., along () zZx are PAGE 155 155 2 THTA (6.170) and ... nTnTnUk (6.171) where A denotes 2 M HT LL Farfield conditions are given by ()0 Tz (6.172) and (1)LTzT (6.173) Side wall conditions are Neumann. The volume of the subcooled liquid must be maintained fixed, whence 00WZxdx (6.174) The Base Solution A base state solution is given by 0()0 Tz (6.175) 00()1UzTze (6.176) where 0U is given by 0ln(1)LUT (6.177) The Perturbation Eigenvalue Problem Expanding Z T, and T, at z Zx as in Eqs. (6.176.19), the perturbation eigenvalue problem is given by 2 1 100 T TU z (6.178) PAGE 156 156 on the domain, 0 z 0 x W and 2 1 100 T TU z (6.179) on the domain, 01 z 0 x W where, at the base surface 0 z 2 0 1 11 2dT dZ TZ dxdz A (6.180) 0 111dT TTZ dz 00 dT dz (6.181) and 2 0 11 1 2dT TT Z zzdz 2 0 20 dT dz (6.182) must hold, and where, 10 Tz (6.183) and 110 Tz (6.184) also hold. The sidewall condi tions are Neumann. Thus the problem is homogeneous in 1T 1T and 1 Z and it has solutions of the form 11cos() TTzkx, 11cos() TTzkx and 1cos() Z kx A (6.185) which satisfy the side wall conditions so long as ,1,2... knWn where k is the wave number of the disturbance. 0 n is ruled out by the condition 1 00WZdx (6.186) PAGE 157 157 The expansions presented a bove give the equations for1T 1T and 1 Z They can be solved just as in case 1 to get 2 111 mzmzTzAekeZ A (6.187) and 2 0 11 11mmmm mzmzmzmz mmkU TzAeeeeeeZ e A (6.188) where 22 004 2 UUk m We can then turn to Eq. (6.182), which gives us 222 000 kmkUNU AA (6.189) where 1mm mmmme N e Notice that the neutrality relation given by Eq. (6.189) can be obtained from Eq. (6.39) of case 1 simply by observi ng that the asymptotic limit of M as the solid depth approaches infinity is given by m The neutral curve is agai n as shown in Figure 6.2 (for case 1) and only one crest is seen at critical. Now on increasing LT to its critical value 0LT, and then slightly beyond, the br anching to the new steady soluti on is expected be a forward pitchfork. So as before we write 021 2LLTT (6.190) and expand U and Z as in Eqs. (6.42, 6.43) to get th e first and higher order problems. Eq. (6.177) indicates that for a small increase in LT from 0LT, the corresponding change in U from 0U per unit change in LT is given by 0Ue. PAGE 158 158 The First Order Problem The first order problem is given by 2 0 1 101dT T TUU zdz (6.191) on the domain, 0 z 0 x W and 2 0 1 101dT T TUU zdz (6.192) on the domain, 01z, 0 x W, 2 0 1 11 2dT dZ TZ dxdz A (6.193) 0 111dT TTZ dz 00 dT dz (6.194) and 2 0 11 1 2dT TT Z zzdz 2 0 1 2dT U dz (6.195) at 0 z 10 Tz (6.196) and 110 Tz (6.197) The volume condition is given by 1 00WZdx (6.198) The sidewall conditions ar e Neumann. Hence we have solutions of the form 11011,cos() TxzTzTzkx (6.199) PAGE 159 159 11011,cos() TxzTzTzkx (6.200)11cos()cos() Z xkxZkxA (6.201) The expansions presented above ar e used to obtain the {10} and {11} problems. Again, as in case 1, 1U turns out to be zero, and therefore 10T and 10T must both be zero. Then the first order problem is exactly like the eigenvalue problem, a nd the condition at neutrality is given by Eq. (6.189), viz., 222 000 kmkUNU AA. And we go on to the second order problem to find the value of A. The Second Order Problem The second order problem is 2 0 2 202dT T TUU zdz (6.202) on the domain 0 z 0 x W and 2 0 2 202dT T TUU zdz (6.203) on the domain 01z, 0 x W, 2 2 0 1 211 22 dT T TZZ zdz 2 0 2 2 2dT dZ Z dxdz A (6.204) 2 2 0 11 2211 22 dT TT TTZZ zzdz 2 00 2 2dTdT Z dzdz 00 dT dz (6.205) and 3 22 2 0 2211 11 2232 dT TTTT ZZ zzzzdz 32 00 2 32dTdT Z dzdz 2 0 2 11 122xdT dz TT ZU xx (6.206) PAGE 160 160 at 0z, 20 Tz (6.207) and 211 Tz (6.208) The volume condition is given by 2 00WZdx (6.209) Again, assuming the solution has the form 2202122(,)()()cos()()cos(2) TxzTzTzkxTzkx (6.210)2202122(,)()()cos()()cos(2) TxzTzTzkxTzkx (6.211)22122cos()cos(2) Z xZkxZkx (6.212) we deduce the solutions to the {20} and {21} and {22} problems. At 0z the following formulae may be useful: 22 222 11 1001 2221cos2 TT Z UkUZkx zz 22 11 10121cos2xTT Z kUZkx xx The {20} problem The {20} problem is given by 2 20200 02 2dTdTdT UU dzdzdz (6.213) in the solid phase, 0 z 0 x W and PAGE 161 161 2 20200 02 2dTdTdT UU dzdzdz (6.214) in the liquid phase, 01z, 0 x W, 22 2020011 2 TTUZ (6.215) and 32 2020 2011 2 dTdT UUZ dzdz (6.216) at 0 z 200 Tz (6.217) and 2011 Tz (6.218) A solution to Eqs. (6.213, 6.214), subject to the far field condition at the solid sink is given by 2020TzA (6.219) and 00002020202 01z UzUzUUTzAeeeBUed (6.220) where 20A is some negative constant. Noticing that 20 200200 dT zBUA dz and using Eqs. (6.215, 6.216) to eliminate 20A and 20 B we get 2020TzA (6.221) 0022 20200121 2UzUzTzAUeZUze (6.222) where 20A is found by using Eq. (6.218) as PAGE 162 162 0022 202011 2UUAeUUZe (6.223) Now the {21} and {22} problems can be solved en tirely, but it is sufficient to solve the {20} problem to learn whether the front speeds up or slows down. To do this, we integrate Eq. (6.204) over 0 x W, whence 2 2 0 1 2011 21 000 2 dT dT TzZzZz dzdz 0 (6.224) Using 2 1 10 dT zkmZ dz A, and 20200 TzA there obtains 22 201 A kmZ A. Comparing with Eq. (6.223), there obtains 00 2 2222 201 01 2planar correctionUU U UUeUkmeZ A (6.225) The correction term is always positive, whence the mean speed of the interface runs ahead of its base value, and our hypothesis of a speedup in solidification is again confirmed. But one might ask: Is this result limited to cases where the latent heat is rejected only to the subcooled liquid? It is to this question that we turn next Hence we consider a third case where the latent heat is rejected to the subcooled liquid as well as to the frozen solid, and both phases are taken to be finite. Indeed, one might intuitiv ely expect that as rejecting the latent heat to the solid phase is stabilizing in contrast to rejec ting heat to the liquid, it might also have an opposite effect in regard to the front speed co rrection, i.e ., a large enough ST might be expected to slow down the front compared to its base value. However, it will be shown in the next sec tion that this is not so, and that once the instability sets in, the front speeds up, no matter the value of ST PAGE 163 163 Case 3: Latent Heat Rejected to the Frozen Solid as Well as to the Subcooled Liquid, Both Phases of Finite Extent The notation, including all scales and referenc es, remains the same as in case 1. Then in the solid phase, the equation for the scaled temperature is given by 20 T TU z (6.226) for 1 zZ and 0 x W while in the liquid phase 20 T TU z (6.227) must hold for Z zL and 0 x W The equations along the interface, viz., along ()zZx are 2 THTA (6.228) and ... nTnTUkn (6.229) where A denotes 2 M HT LS Farfield conditions are given by (1)STzT (6.230) and ()LTzLT (6.231) Side wall conditions are Neumann. The volume of the subcooled liquid must be maintained fixed, whence 00WZxdx (6.232) PAGE 164 164 The Base Solution A base state solution is given by 00 000()11 1UzUz S UT TzAee e (6.233) and 00 000()11 1UzUz L ULT TzAee e (6.234) where 0U is given by 001 11S L UULT T ee (6.235) Some useful derivatives and their di fferences are given in Table 6.2. Table 6.2 Case 3 Some derivatives of base temperatures in two phases in solidification 00 00000 1S UdTT zUAU dze 02 22 0 000 200 1S UdTT zUAU dze 03 33 0 000 300 1S UdTT zUAU dze 04 44 0 000 400 1S UdTT zUAU dze 00 00000 1L ULdT T zUAU dze 02 22 0 000 200 1L ULdT T zUAU dze 03 33 0 000 300 1L ULdT T zUAU dze 04 44 0 000 400 1L ULdT T zUAU dze 00 000 dTdT zzU dzdz 22 2 00 0 2200 dTdT zzU dzdz 33 3 00 0 3300 dTdT zzU dzdz 44 4 00 0 4400 dTdT zzU dzdz The Perturbation Eigenvalue Problem Expanding Z T, and T, at z Zx as in Eqs. (6.176.19), the perturbation eigenvalue problem is given by PAGE 165 165 2 1 100 T TU z (6.236) on the domain, 10 z 0 x W and 2 1 100 T TU z (6.237) on the domain, 0 zL 0 x W where, at the base surface 0 z 2 0 1 11 2dT dZ TZ dxdz A (6.238) 00 1110 dTdT TTZ dzdz (6.239) and 22 00 11 1 220 dTdT TT Z zzdzdz (6.240) must hold, and where, 110 Tz (6.241) and 10 TzL (6.242) The sidewall conditions ar e Neumann. Thus the problem is homogeneous in 1T 1T and 1 Z and it has solutions of the form 11cos() TTzkx, 11cos() TTzkx and 1cos() Z kx A (6.243) which satisfy the side wall conditions so long as ,1,2... knWn where k is the wave number of the disturbance. 0 n is ruled out by the condition 1 00WZdx (6.244) PAGE 166 166 The expansions presented a bove give the equations for 1T 1T and 1 Z They are 2 2 1 10 20 dT d kTU dzdz (6.245) in the solid phase, 10 z 0 x W and 2 2 1 10 20 dT d kTU dzdz (6.246) in the liquid phase, 0 zL 0 x W 2 0 11dT TkZ dz A (6.247) 00 1110 dTdT TTZ dzdz (6.248) and 22 00 11 1 220 dTdT dTdT Z dzdzdzdz (6.249) at 0 z 1(1)0 Tz (6.250) and 1()0 TzL (6.251) Defining 22 004 2 UUk m a solution to Eqs. (6.2456.248, 6.250, 6.251) is given by PAGE 167 167 2 0 11 10 1mmmm mzmzmzmz mmdT kz dz TzAeeeeeeZ e A (6.252) 2 0 11 10 1mmLmmL mzmzmzmz mmLdT kz dz TzAeeeeeeZ e A (6.253) We can then turn to Eq. (6.249), which gives us 222 00 0dTdT kMkNU dzdz AA (6.254) or 00222 00 0011SS UUUTUT kMkUNU ee AA (6.255) where 1mm mmmme M e and 1mmL mmLmme N e Figures 6.3 and 6.4 show the corresponding ne utral curves. As the control variable LT increases upward from zero, the output variable 0U increases from zero, until there comes a point when a line of constant 0U first crosses the neutral curv e at an admissible value of k. The corresponding value of 0U will be referred as its critical value. Henceforth the subscript zero will refer to this critical state. At values of U higher than 0U the base solution loses its stability. The dip is present in both kinds of neutral curves and this implies that the critical value of U need not correspond to the least allowable value of 2k. In fact, n might be any of the numbers: 1,2,3.., and hence multiple crests might be seen. PAGE 168 168 Figure 6.3 An unusual neutral curve in solidifica tion when the latent heat is rejected through both phases Figure 6.4 Typical neutral curve in solidification when the latent heat is rejected through both phases Now on increasing LT to its critical value 0LT, and then slightly beyond, the branching to the new steady solution is expected be a forward pitchfork. Hence we write 021 2LLTT (6.256) and expand U and Z as in Eqs. (6.42, 6.43) to get th e first and higher order problems. Eq. PAGE 169 169 (6.235) indicates that for a small increase in LT from 0LT, the corresponding change in U from 0U per unit change in LT is given by 0 0 000 00221 11UL L ULULU S ULUe T T eLee ee The First Order Problem The first order problem is given by 2 0 1 101dT T TUU zdz (6.257) on the domain, 10 z 0 x W and 2 0 1 101dT T TUU zdz (6.258) on the domain. 0 zL 0 x W where, at the base surface 0 z 2 0 1 11 2dT dZ TZ dzdx A (6.259) 00 1110 dTdT TTZ dzdz (6.260) and 22 00 11 11 22dTdT TT Z U zzdzdz (6.261) must hold, and where 110 Tz (6.262) and 10 TzL (6.263) The volume condition is given by PAGE 170 170 1 00WZdx (6.264) The sidewall conditions are Ne umann, and Eqs. (6.257, 6.258, 6.2606.264) can be solved by writing 11011,cos() TxzTzTzkx (6.265)11011,cos() TxzTzTzkx (6.266)11cos()cos() Z xkxZkxA (6.267) The expansions presented above are used to deduce the {10} and {11} problems. The {10} problem The {10} problem is given by 2 10100 01 2dTdTdT UU dzdzdz (6.268) in the solid phase, 10 z 0 x W and 2 10100 01 2dTdTdT UU dzdzdz (6.269) in the liquid phase, 0 zL 0 x W 10100 TT (6.270) and 1010 1dTdT U dzdz (6.271) at 0 z 10(1)0 Tz (6.272) and PAGE 171 171 10()0 TzL (6.273) The general solution to Eqs. (6.268, 6.269) is given by: 000010101010 01z UzUzUUTzAeeeBUAed (6.274) 000010101010 01z UzUzUUTzAeeeBUAed (6.275) and their derivatives are given by 10 100100 dT zBUA dz and 10 100100 dT zBUA dz whereupon using Eqs. (6.2706.272) to eliminate 10 B 10A and 10 B we get 0 000 010101010101 1U UzUzUz Ue TzAeUAAeUAze e (6.276) 0 0000 0101010101011 1U UzUzUzUz Ue TzAeUAAeUAzeUze e (6.277) where 10A is found by using Eq. (6.273) as 0 0 0 0 0 000 1011 1 11 1U UL U U UL Ue AeALL e AU e e e The {11} problem The {11} problem is similar to th e eigenvalue problem. It is given by 2 2 11 110 20 dT d kTU dzdz (6.278) in the solid phase, 10z, 0 x W, and 2 2 11 110 20 dT d kTU dzdz (6.279) in the liquid phase, 0zL, 0 x W, PAGE 172 172 00 1111110 dTdT TTZ dzdz (6.280) and 22 00 1111 11 220 dTdT dTdT Z dzdzdzdz (6.281) at 0 z 11(1)0 Tz (6.282) and 11()0 TzL (6.283) Finally using Eq. (6.264) and 10 dZdx at 0, x W and integrating Eq. (6.259) over 0 x W we find that the constant 10A must be zero. Hence 1U must be zero, and therefore 10T and 10T must both be identically zero. Then the fi rst order problem is exactly like the eigenvalue problem, and the condition at neutrality is given by Eq. (6.255), viz., 00222 00 0011SS UUUTUT kMkUNU ee AA. What we have so far is this: 10 U 1111,coscos TxzTzkxTzkx, 1111,coscos TxzTzkxTzkx and 11cos()cos() Z xkxZkxA. And we go on to the second order problem to find the value of A. The Second Order Problem The second order problem is 2 0 2 202dT T TUU zdz (6.284) on the domain, 10 z 0 x W and PAGE 173 173 2 0 2 202dT T TUU zdz (6.285) on the domain, 0zL, 0 x W, 2 2 2 00 12 2112 222 dTdT TdZ TZZZ zdzdxdz A (6.286) 22 2 0000 11 22112 2220 dTdTdTdT TT TTZZZ zzdzdzdzdz (6.287) and 3322 22 2 0000 2211 112 223322 11 122 2xdTdTdTdT TTTT ZZZ zzzzdzdzdzdz TT ZU xx (6.288) at 0 z 210 Tz (6.289) and 21 TzL (6.290) The volume condition is given by 2 00WZdx (6.291) Assuming the solution to be of the form 2202122(,)()()cos()()cos(2)TxzTzTzkxTzkx (6.292)2202122(,)()()cos()()cos(2)TxzTzTzkxTzkx (6.293)22122cos()cos(2) Z xZkxZkx (6.294) we deduce the {20} and {21} and {22} problems. At 0 z the following formulae may be PAGE 174 174 useful: 22 222 11 1001 2221cos2 TT Z UkUZkx zz 22 11 10121cos2xTT Z kUZkx xx The {20} problem The {20} problem is given by 2 20200 02 2dTdTdT UU dzdzdz (6.295) in the solid phase, 10 z 0 x W and 2 20200 02 2dTdTdT UU dzdzdz (6.296) in the liquid phase, 0 zL 0 x W 22 2020011 2 TTUZ (6.297) and 32 2020 2011 2 dTdT UUZ dzdz (6.298) at 0z, 2010 Tz (6.299) and 201 TzL (6.300) A general solution to Eqs. (6.295, 6.296) is given by 000020202020 01z UzUzUUTzAeeeBUAed (6.301) PAGE 175 175 000020202020 01z UzUzUUTzAeeeBUAed (6.302) and their derivatives are given by 20 200200dT zBUA dz and 20 200200dT zBUA dz whereupon using Eqs. (6.2976.299) to eliminate 20 B 20 A and 20 B we get 0 000 020202020201 1U UzUzUz Ue TzAeUAAeUAze e (6.303) 0 000 0 002020201020 22 0121 1 1 2U UzUzUz U UzUze TzAeUAAeUAze e UZeUze (6.304) where 20A is found by using Eq. (6.300) as 0 0 0 0 00 00 0022 00 01 2021 1 1 2 1111 11U UL UL U UU ULUL UUe AeALL UZe e AU ee ee ee (6.305) The {21} problem The {21} problem is similar to the eigenvalue problem. It is given by 2 2 21 210 20 dT d kTU dzdz (6.306) in the solid phase, 10z, 0 x W, and 2 2 21 210 20 dT d kTU dzdz (6.307) in the liquid phase, 0zL, 0 x W, 00 2121210 dTdT TTZ dzdz (6.308) PAGE 176 176 and 22 00 2121 21 220 dTdT dTdT Z dzdzdzdz (6.309) at 0 z 21(1)0Tz (6.310) and 21()0TzL (6.311) Defining 22 00 214 2UUk mm a solution to Eqs. (6.3066.311) is given by 2121 21212121 mm mzmzTzAeee (6.312) 2121 21212121 mmL mzmzTzAeee (6.313) where 21212 0 21210 1mmdT kz dz A Z e A (6.314) 21212 0 21210 1mmLdT kz dz A Z e A (6.315) The {22} problem The {22} problem is given by 2 2 22 220 240 dT d kTU dzdz (6.316) in the solid phase, 10 z 0 x W and PAGE 177 177 2 2 22 220 240 dT d kTU dzdz (6.317) in the liquid phase, 0 zL 0 x W 22 2222010221 2 TTUZUZ (6.318) and 22322 2222 01010221 20 2 dTdT UkZUZUZ dzdz (6.319) at 0z, 22(1)0Tz (6.320) and 22()0TzL (6.321) Defining 22 00 2216 2UUk m a solution to Eqs. (6.3166.321) is given by 2222 22222222 mm mzmzTzAeee (6.322) 2222 22222222 mmL mzmzTzAeee (6.323) where 222222322 022001022022 22 222211 2 22 1mmUNUkUZUNUZ A eNM (6.324) PAGE 178 178 222222322 022001022022 22 222211 2 22 1mmLUMUkUZUMUZ A eNM (6.325) and where 2222 22222222 221mm mmmme M e and 2222 22222222 221mmL mmLmme N e Using the curvature equation Finally, using Eq. (6.291) and 20dZdx at 0, x W and integrating Eq. (6.286) over 0 x W, we find 2 2 0 1 2011 21 0000 2 dT dT TzZzZz dzdz (6.326) Using 2 0 1 100dT dT zkzMZ dzdz A, and 20200 TzA there obtains 2 22 00 20 1 21 00 2 dTdT AkMMzzZ dzdz A (6.327) Comparing with Eq. (6.305), and simplifying, there obtains 0 0 0 00 00 2 0 0 0000 0 02 22 222 0 001 11 111 11 121211 1planarUL UL L ULU S ULU U U UL SS ULUUU L ULe e U T T Lee ee UTT e UkMUe eeee T e A 00 0 22 1 22 01correctionULU S U U Z T Lee e (6.328) PAGE 179 179 where the correction term is always positive, wh ence the mean speed of the interface runs ahead of its base value. We have learned that in pure solidificati on, once the front becomes unstable, its mean speed is higher than its base value. Hence the result of a speed up in pure solidification is unconditional, viz., the front always speeds up regardless of whethe r or not the latent heat is being rejected to the solid phase. Also, the reader is encouraged to verify that as ST goes to zero, the limiting value of 2U is given by 0 0 0 0 2 2222 201 0 0111 lim11 21S planar correctionU ULUL U T U Ue UeUkMeZ LLe A (6.329) which is the same expression as what was obtained in Eq. (6.114) for case 1 where the latent heat was rejected only throug h the subcooled liquid. For this case then, the only other information th at one might be interest ed in seeking is the nature of the branching to the new steady solution, i.e., whether th e bifurcation is a forward or a backward pitchfork. To that end, we need to go to third order, but before that, we need to solve Eq. (6.286) for 2 Z Now to find 2 Z a solvability condition mu st be satisfied. It is 2 2 0 1 211112 2 0 0020000W WdT T TzZzZzZdxTzZdx zdz (6.330) It is easy to verify that this conditio n is automatically satisfied. Hence, Eq. (6.286) can be solved for 2 Z Substituting the expansions from Eqs. (6.2926.294), the constant term disappears due to the result obtained earlier using integrability, viz., Eq. (6.326). The coefficient of the first harmonic vanishes due to solvability, and there obtains by equating the coefficients of the second harmonic PAGE 180 180 2 22 00 1 22112222 21 000040 2 dTdT dT TzzZzZzZkZ dzdzdz A (6.331) Substituting 2222222201mmTzAe and using 2 0 1 100 dT dT zkzMZ dzdz A, there obtains 22222 22 2 000 1 22 2 221 0004 2 1mmdTdTdT Z kMzMzZzk dzdzdz A e AA (6.332) Comparing with Eq. (6.324), and simplifying, we learn that 22 Z can be found in terms of 2 1 Z as 2 221P Z Z Q (6.333) where 2 2232 00 02200 2222 2 22 0 22220220111 200 222 40 dTdT UNUkUkMzMzNM dzdz P dT Q kzNMUNU dz A A (6.334) Hence, 21 Z and A carry on to third order. The Third Order Problem The equations at the third order are given by 2 30 1 30233TdT T TUUU zzdz (6.335) on the domain 10 z 0 x W and 2 30 1 30233 TdT T TUUU zzdz (6.336) on the domain 0zL, 0 x W, PAGE 181 181 2 23 22 23 00 11211 3112121 2 23 2 30 3 293333 dTdT dZdZTTT TZZZZZZ dxdxzzzdzdz dZdT Z dxdz A A (6.337) 22 2 222211 33112 2233 3 000000 1213 2233333 30 TTTTTT TTZZZ zzzzzz dTdTdTdTdTdT ZZZZ dzdzdzdzdzdz (6.338) and 223322 2 33 221111 112 223322 334422 3 000000 1213 334422 22 1333 3 3xTT TTTTTT ZZZ zzzzzzzz dTdTdTdTdTdT ZZZZ dzdzdzdzdzdz TT Z xx 3 22 1111 1223xU TTTT ZZ xzxzxx (6.339) at 0 z 310 Tz (6.340) and 30 TzL (6.341) The volume condition is given by 3 00WZdx (6.342) We assume the following form for the solution 330313233(,)()()cos()()cos(2)()cos(3) TxzTzTzkxTzkxTzkx (6.343) PAGE 182 182 330313233(,)()()cos()()cos(2)()cos(3) TxzTzTzkxTzkxTzkx (6.344)3313233cos()cos(2)cos(3) Z xZkxZkxZkx (6.345) This problem can be solved completely, but it is sufficient to solve the {31} problem to learn the nature of bifu rcation and the value of 2A. This can be done ex actly as in case 1. However, the algebra is cumbersome and we will stop here as far as case 3 is concerned. Up till now we have worked out three cases of pure solidification, and we have learned that the solidification front speeds up, no matter what. By contrast we learned in Chapter 5 that a precipitation front always slows down once the instability sets in. Now these two problems are closely related and their physics are very similar. Then one might as k: What is the reason for this fundamental difference? In other words, can we explain this basi c difference of a speedup in solidification versus a slowdown in precipitation? In our quest to answer this question, we go to a yet simpler model, where the solid phase is completely ignored. Nothing new will be learned, and we will again see that the front spee ds up compared to its base value. But the model is now as close as one can get to the precipitation model, and we can therefore hope to learn the fundamental difference between the two problems. Case 4: Solid Phase Ignored, Latent Heat Rejected Only through the Subcooled Liquid The notation remains the same as in case 1. Si nce the solid phase is altogether ignored, all lengths are scaled by L, the depth of the liquid pha se. All speeds are scaled by /L The scaled temperature is given by /M scaled HTT T L Then the equation for the scaled liquid side temperature is given by 20 T TU z (6.346) for 1 Z z and 0 x W The equations along the interface, viz., along () zZx are PAGE 183 183 2 THA (6.347) and .. nTUkn (6.348) where A denotes 2 M HT LL The temperature at the li quid side sink is given by (1)LTzT (6.349) Side wall conditions are Neumann. The volume of the subcooled liquid must be maintained fixed, whence 00WZxdx (6.350) The Base Solution A base state solution is given by 00()1UzTze (6.351) where 0U is given by 0ln(1)LUT (6.352) whence 0 00 dT zU dz 2 2 0 0 20 dT zU dz 3 3 0 0 30 dT zU dz 4 4 0 0 40 dT zU dz The Perturbation Eigenvalue Problem Expanding Z T, and T, at z Zx as in Eqs. (6.17, 6.19), the perturbation eigenvalue problem is given by 2 1 100 T TU z (6.353) on the domain, 01z, 0 x W, where, at 0z PAGE 184 184 2 0 1 11 2dT dZ TZ dxdz A (6.354) and 2 0 1 1 20 dT T Z zdz (6.355) must hold, and where, 110 Tz (6.356) must hold. The sidewall conditions are Neumann. Then the problem has solutions of the form 11cos() TTzkx and 1cos() Z kx A (6.357) which satisfy the side wall conditions so long as ,1,2... knWn where k is the wave number of the disturbance and 0n is ruled out by the condition 1 00WZdx (6.358) The expansions presented a bove give the equations for 1T and 1 Z They are 2 2 1 10 20 dT d kTU dzdz (6.359) on the domain, 01z, 0 x W, 2 0 11dT TkZ dz A (6.360) and 2 0 1 1 20 dT dT Z dzdz (6.361) at 0z, and PAGE 185 185 1(1)0 Tz (6.362) Defining 22 004 2 UUk m a solution to Eqs. (6.359, 6.361, 6.362) is given by 2 0 111mm mzmz mmkU TzeeeZ e A (6.363) We can then turn to Eq. (6.361), which gives us 22 00kUNU A (6.364) where 1mm mmmme N e The neutral curve is as shown in Figure 6.2, whence only one cr est is seen at critical. Now on increasing LT to its critical value 0LT, and then slightly beyond, the branching to the new steady solution is expected be a fo rward pitchfork. Hence we write 021 2LLTT (6.365) and expand U and Z as in Eqs. (6.42, 6.43) to get th e first and higher order problems. Eq. (6.352) indicates that for a small increase in LT from 0LT, the corresponding change in U from 0U per unit change in LT is given by 0Ue Before moving on to solve the higher order pr oblems, observe that even for this simple case where the solid phase ha s been completely ignored, 1T turns out to be nonzero. Now in Chapter 5, the variable 1c was found to be zero due to cancell ations of mappings in the mass balance equation at the interface, viz., .1. nccUkn. The corresponding cancellations are PAGE 186 186 not seen in the interfacial ener gy balance equation in solidification. And mathematically, this is what leads to the key difference between the two problems. The First Order Problem The first order problem is given by 2 0 1 101dT T TUU zdz (6.366) on the domain, 01z, 0 x W, 2 0 1 11 2dT dZ TZ dxdz A (6.367) and 2 0 1 11 2dT T Z U zdz (6.368) at 0z, 110 Tz (6.369) and 1 00WZdx (6.370) The sidewall conditions are Neum ann, and Eqs. (6.366, 6.3686.370) can be solved by writing 11011,cos() TxzTzTzkx (6.371)11cos()cos() Z xkxZkxA (6.372) The {10} problem The {10} problem is given by 2 10100 01 2dTdTdT UU dzdzdz (6.373) PAGE 187 187 on the domain, 01z, 0 x W, 10 1dT U dz (6.374) at 0 z and 10(1)0 Tz (6.375) A general solution to Eq. (6.373) is given by 00001010101 01z UzUzUUTzAeeeBUed (6.376) whence 10 100100 dT zBUA dz Using Eqs. (6.374, 6.375) to eliminate10A and 10 B we get 00101UzUTzUzee (6.377) The {11} problem The {11} problem is similar to the eigenvalue problem. It is given by 2 2 11 110 20 dT d kTU dzdz (6.378) on the domain, 01 z 0 x W 2 0 11 11 20 dT dT Z dzdz (6.379) at 0 z and 11(1)0 Tz (6.380) Finally using Eq. (6.370) and 10 dZdx at 0, x W and integrating Eq. (6.367) over 0 x W we find that 1U must be zero, and therefore 10T must be identically zero. Then the first order problem is exactly lik e the eigenvalue problem, and the c ondition at neutrality is given PAGE 188 188 by Eq. (6.364), viz., 22 00kUNU A. What we have so far is this: 10 U 1111,coscos TxzTzkxTzkx and 11cos()cos() Z xkxZkxA. And we go on to the second order problem to find the value of A. The Second Order Problem The second order problem is 2 0 2 202dT T TUU zdz (6.381) on the domain, 01 z 0 x W 2 2 2 00 12 2112 222 dTdT TdZ TZZZ zdzdxdz A (6.382) and 32 2 2 00 211 11212 23222xdTdT TTT Z ZZZU zzdzdzx (6.383) at 0 z and 211 Tz (6.384) and 2 00WZdx (6.385) Then, we assume the following form for the solution: 2202122(,)()()cos()()cos(2) TxzTzTzkxTzkx (6.386)22122cos()cos(2) Z xZkxZkx (6.387) And again, we use these expansions to de duce the {20} and {21} and {22} problems. PAGE 189 189 The {20} problem The {20} equations are given by 2 20200 02 2dTdTdT UU dzdzdz (6.388) on the domain, 01z, 0 x W, 32 20 2011 2 dT UUZ dz (6.389) at 0 z and 2011 Tz (6.390) A general solution to Eq. (6.388) is given by 00002020202 01z UzUzUUTzAeeeBUed (6.391) whence 20 200200 dT zBUA dz Using Eqs. (6.389, 6.390) to eliminate 20A and 20 B we get 000022 200121 1 2UUzUUzTzUeeZUeze (6.392) whence 02222 200101211 01 22UTzUZeUZU (6.393) The {21} problem The {21} problem is given by 2 2 21 210 20 dT d kTU dzdz (6.394) on the domain, 01 z 0 x W PAGE 190 190 2 0 21 21 20 dT dT Z dzdz (6.395) at 0z, and 21(1)0 Tz (6.396) The {22} problem The {22} problem is given by 2 2 22 220 240 dT d kTU dzdz (6.397) on the domain, 01 z 0 x W 22322 22 0010221 2 2 dT kUkUZUZ dz A (6.398) at 0 z and 22(1)0 Tz (6.399) Using the curvature equation Finally using Eq. (6.385) and 20 dZdx at 0, x W and integrating Eq. (6.367) over 0 x W, we find 2 2 0 1 2011 21 0000 2 dT dT TzZzZz dzdz (6.400) Using 2 1 010 dT zUZ dz and 2 2 0 0 20 dT zU dz there obtains 22 20011 0 2 TzUZ (6.401) PAGE 191 191 Comparing with Eq. (6.393), we get 0 2 222 201 01 2planar correctionU U UUeUZ (6.402) The correction term is always positive, whence the mean speed of the interface runs ahead of its base value. Now the model given in case 4 is the closest that one can get to the model of precipitation. Yet, we find a speedup in case 4 while the precipitation problem exhibits an unconditional slowdown once th e instability sets in. The onl y difference between the two models is that the energy balance equation in th e present solidification m odel is different from the mass balance equation of the precipitation model. Hence we have learned that the key disparity between the two problems is that mass is fundamentally different from heat. In the most basic terms, there is no such thi ng as a hidden mass or latent mass. Now notice that the result of a speedup in solidification is obtained at second order, and this result holds regardless of th e values of the input parameters 2k and L But recall from the third order derivations for case 1, that the sign of 2A depends strongly on the inputs 2k and L and hence the nature of the bifurcation is a f unction of the input vari ables. Although this was clear from Table 6.1, we had to cater to numer ical calculations and it was hard to tell the functional dependence of 2A on the inputs simply by inspection. In our quest to delineate this dependence, we move on to the next case where the latent heat is rejected to the frozen solid as well as to the subcooled liquid, as in case 3, but now we drop U from the domain equations. It will be seen that this simplifies the m odel equations and also the formula for2A. Case 5: Latent Heat Rejected to Frozen So lid as Well as to the Subcooled Liquid, Both Phases of Finite Extent, U Dropped from Domain Equations In this section we work out a special case of case 3 where U has been dropped from the domain regardless of whether or not it is small. For small values of growth speed U, this is a PAGE 192 192 reasonable approximation. This is done solely with the purpose of making the algebra more tractable. Once again, we will learn that the fron t speeds up at the onset of instability. What we also wish to learn is whether 2A is found to be positive or negative at third order. In other words, we wish to find whether the pitchfork bifurcat ion is forward or backward. We will find that 2A can have either sign, whence the bifurcation can be forward as well as backward depending on the inputs 2k and the ratio of the depth of the two phases. To see this, we start by rewriting the nonlinear model. All scales remain the same as in case 3. Then, the scaled temperature in the solid phase must satisfy 20T (6.403) for 1 zZ and 0 x W, while in the liquid phase 20T (6.404) must hold for Z zL and 0 x W. The equations along the interface, viz., along ()zZx are 2 THTA (6.405) and ...nTnTnUk (6.406) where A denotes 2 M HT LS Farfield conditions are given by (1)STzT (6.407) and ()LTzLT (6.408) Side wall conditions are Neumann. The volume of the subcooled liquid must be maintained PAGE 193 193 fixed, whence 00WZxdx (6.409) The Base Solution A base state solution is given by 0()STzTz (6.410) and 0()LT Tzz L (6.411) where 0U is given by 0 L ST UT L (6.412) Notice that both phases have linear base temperature gradients, and this leads to a lot of simplification at higher orders as all second and higher derivatives of 0T and 0T vanish. The Perturbation Eigenvalue Problem Expanding Z T, and T, at z Zx as in Eqs. (6.176.19), the perturbation eigenvalue problem is given by 2 10 T (6.413) on the domain, 10z, 0 x W, and 2 10 T (6.414) on the domain, 0 zL 0 x W where, at the base surface 0 z 2 0 1 11 2dT dZ TZ dxdz A (6.415) PAGE 194 194 00 1110 dTdT TTZ dzdz (6.416) and 110 TT zz (6.417) must hold, and where, 110 Tz (6.418) and 10 TzL (6.419) The sidewall conditions ar e Neumann. Thus the problem is homogeneous in 1T 1T and 1 Z and it has solutions of the form 11cos() TTzkx, 11cos() TTzkx and 1cos() Z kx A (6.420) which satisfy the side wall conditions so long as ,1,2... knWn where k is the wave number of the disturbance. 0n is ruled out by the condition 1 00WZdx (6.421) The expansions presented a bove give the equations for 1T 1T and 1 Z They are 2 2 1 20 d kT dz (6.422) in the solid phase, 10z, 0 x W, and 2 2 1 20 d kT dz (6.423) in the liquid phase, 0zL, 0 x W, PAGE 195 195 2 0 11dT TkZ dz A (6.424) 00 1110 dTdT TTZ dzdz (6.425) and 110 dTdT dzdz (6.426) at 0 z 1(1)0 Tz (6.427) and 1()0 TzL (6.428) A solution to Eqs. (6.4226.425, 6.427, 6.428) is given by 2 0 22 11 1 20 1kzkkzkzkkz kdT kz dz TzAeeeeeeZ e A (6.429) and 2 0 22 11 1 20 1kzkLkzkzkLkz kLdT kz dz TzAeeeeeeZ e A (6.430) We can then turn to Eq. (6.426) which gives us 2 0tanh 1 tanhSkL UkT k A (6.431) The following formulae might be helpful while drawing a curve of 0U versus 2k: PAGE 196 196 tanhkk kkee k ee 23 small1 limtanh 3kkkk 2largelimtanh1kk 22tanh1 lim1111 tanh3ksmallkL LkLLL k 22222 0 smallsmalltanh1 limlim1111 tanh3SS kkkL UkTkTLkLLL k AA 20 0lim1S kUTL 2 0 21 0111 3SdU kLLLTL dk A 2222 0 largelargetanh limlim12 tanhSS kkkL UkTkT k AA for all values of L. Figure 6.5 Typical neutral curve for solidification when the latent heat is reje cted through both phases; U is dropped from the domain equations L > 1 L < 1 L = 1 U k2 PAGE 197 197 Figure 6.5 shows the corres ponding neutral curve. Given W, and hence 2k, Eq. (6.431) determines the critical value of the interface speed, 0U For 1 L the curve is monotonically increasing, hence the most dangerous value of n is one, which corresponds to kW giving only one crest at critical. For 1 L the curve must show a dip whence multiple crests are possible. As the control variable LT increases upward from zer o, the output variable 0U increases from zero, until there comes a point when Eq. (6.431) is satisfied. The corresponding value of 0U will be referred as its critical value. Hen ceforth the subscript zero will refer to this critical state. At values of U higher than 0U the base solution loses its stability. Now on increasing LT to its critical value 0LT, and then slightly beyond, the branching to the new steady solution is expected be a fo rward pitchfork. As before, we write 021 2LLTT (6.432) and expand U and Z as in Eqs. (6.42, 6.43) to get th e first and higher order problems. Eq. (6.412) indicates that for a small increase in LT from 0LT, the corresponding change in U from 0U per unit change in LT is given by 1 L The First Order Problem The first order problem is given by 2 10 T (6.433) on the domain, 10 z 0 x W and 2 10 T (6.434) on the domain, 0 zL, 0 x W, PAGE 198 198 2 0 1 11 2dT dZ TZ dxdz A (6.435) 00 1110 dTdT TTZ dzdz (6.436) and 11 1TT U zz (6.437) at 0 z and 110 Tz (6.438) and 10 TzL (6.439) The volume condition is given by 1 00WZdx (6.440) The sidewall conditions ar e Neumann. Hence we have solutions of the form 11011,cos() TxzTzTzkx (6.441)11011,cos() TxzTzTzkx (6.442)11cos()cos() Z xkxZkxA (6.443) The expansions presented above are used to deduce the {10} and {11} problems. The {10} problem The {10} problem is given by 2 10 20 dT dz (6.444) PAGE 199 199 in solid phase, 10 z, 0 x W, and 2 10 20 dT dz (6.445) in liquid phase, 0 zL 0 x W 10100 TT (6.446) and 1010 1dTdT U dzdz (6.447) at 0 z 10(1)0 Tz (6.448) and 10()0 TzL (6.449) A solution to Eqs. (6.4446.449) is given by 1011 1 L TzUz L (6.450) 1011 1 TzUzL L (6.451) The {11} problem The {11} problem is similar to th e eigenvalue problem. It is given by 2 2 11 20 d kT dz (6.452) in the solid phase, 10 z 0 x W and 2 2 11 20 d kT dz (6.453) PAGE 200 200 in the liquid phase, 0 zL, 0 x W, 00 1111110 dTdT TTZ dzdz (6.454) and 11110 dTdT dzdz (6.455) at 0 z, 11(1)0 Tz (6.456) and 11()0 TzL (6.457) Finally we can turn to Eq. (6.435), which is a differential equation for 1 Z Using Eq. (6.440) and 10 dZdx at 0, x W and integrating Eq. (6.435) over 0 x W we find that 1U must be zero, and therefore 10T and 10T must both be identically zero. Then the first order problem is exactly like the eige nvalue problem, and the condition at neutrality is given by Eq. (6.431), viz., 2 0tanh 1 tanhSkL UkT k A. What we have so far is this: 1111coscos TTzkxTzkx, 1111coscos TTzkxTzkx 10 U and 11cos()cos() Z xkxZkxA. And we go on to the second order problem to find the value of A. The Second Order Problem The second order problem is 2 20 T (6.458) PAGE 201 201 on the domain, 10 z, 0 x W, and 2 20 T (6.459) on the domain, 0 zL 0 x W 2 0 12 212 22 dT TdZ TZZ zdxdz A (6.460) 11 2212 TT TTZ zz 00 20 dTdT Z dzdz (6.461) and 22 221111 112 2222xTTTTTT Z ZU zzzzxx (6.462) at the base surface 0 z The farfield conditions are given by 210 Tz (6.463) and 21 TzL (6.464) The volume condition is given by 2 00WZdx (6.465) The stricken terms are zero because of results from previous orders. Assuming that the above equations have a solution of the form 2202122(,)()()cos()()cos(2) TxzTzTzkxTzkx (6.466)2202122(,)()()cos()()cos(2) TxzTzTzkxTzkx (6.467)22122cos()cos(2) Z xZkxZkx (6.468) we deduce the {20} and {21} and {22} problems. At 0 z the following formulae may be PAGE 202 202 useful: 22 22 11 101 2221cos2 TT Z kUZkx zz 22 11 10121cos2xTT Z kUZkx xx The {20} problem The {20} problem is given by 2 20 20 dT dz (6.469) in the solid phase, 10 z 0 x W and 2 20 20 dT dz (6.470) in the liquid phase, 0 zL 0 x W 20200 TT (6.471) and 2020 2dTdT U dzdz (6.472) at 0 z, 2010 Tz (6.473) and 201 TzL (6.474) A solution to Eqs. (6.4696.474) is given by PAGE 203 203 2 201 1 1 UL Tzz L (6.475) 2 201 1 1 U TzzL L (6.476) The {21} problem The {21} problem is similar to th e eigenvalue problem. It is given by 2 2 21 20 d kT dz (6.477) in the solid phase, 10 z 0 x W and 2 2 21 20 d kT dz (6.478) in the liquid phase, 0 zL 0 x W 00 2121210 dTdT TTZ dzdz (6.479) and 21210 dTdT dzdz (6.480) at 0 z 21(1)0 Tz (6.481) and 21()0 TzL (6.482) A solution to Eqs. (6.4776.482) is given by 2 0 22 2121 21 20 1kzkkzkzkkz kdT kz dz TzAeeeeeeZ e A (6.483) PAGE 204 204 and 2 0 22 2121 21 20 1kzkLkzkzkLkz kLdT kz dz TzAeeeeeeZ e A (6.484) The {22} problem The {22} problem is given by 2 2 22 240 d kT dz (6.485) in the solid phase, 10 z, 0 x W, and 2 2 22 240 d kT dz (6.486) in the liquid phase, 0 zL, 0 x W, 2222022TTUZ (6.487) and 22 2222 012 dTdT UkZ dzdz (6.488) at 0 z 22(1)0 Tz (6.489) and 22()0 TzL (6.490) A solution to Eqs. (6.4856.490) is given by 242 2222 kzkkzTzAeee (6.491) 242 2222kzkLkzTzAeee (6.492) PAGE 205 205 where 2 220221 41tanh2 tanh2 1tanh2tanh2kk A UZkkLZ ekkL (6.493) 2 220221 41tanh2 tanh2 1tanh2tanh2kLkL A UZkkZ ekkL (6.494) Using the curvature equation Finally we can turn to Eq. (6.460), which is a differential equation for 2 Z Using Eq. (6.465) and 20 dZdx at 0, x W, and integrating Eq. (6.460) over 0 x W we find 1 201000 dT TzZz dz (6.495) Using 2 1 10 tanhSdTk zkTZ dzk A, and 2 201 0 1 UL Tz L there obtains 2 222 21 011 1 tanhplanar correctionS U Uk UkTZ LkL A (6.496) where the correction term is again seen to be always positive, whence the mean speed of the interface runs ahead of its base value. To go to third order, we need to solve Eq. (6.460) for 2 Z To find 2 Z a solvability condition must be satisfied. It is 1 21112 0002000WWT TzZzZdxTzZdx z (6.497) Again this condition is automatically satisfied, whence Eq. (6.460) can be solved for 2 Z Substituting the expansions from Eqs. (6.4666.468) into Eq. (6.460), the constant term disappears due to the result obtained ear lier using integrability, viz., Eq. (6.495). The coefficient PAGE 206 206 of the first harmonic vanishes due to solvability, and there obtain s by equating the coefficients of the second harmonic 2 0 1 2212200040 dT dT TzzZzkZ dzdz A (6.498) Substituting 4 222201kTzAe and using 2 1 10 tanhSdT k zkTZ dzk A, we get 222 22221 41 4 1tanhSS kk A kTZkTZ ek AA (6.499) Comparing with Eq. (6.493) and simplifying, there obtains 2 221P Z Z Q (6.500) where 2 0 2 0tanh2 tanh2 tanhtanh2tanh2 tanh2 4 tanh2tanh2S Skk kTkUkL P kkkL k Q kTU kkL A A (6.501) Hence, 21 Z and A carry on to third order. The Third Order Problem The equations at the third order are given by 2 30 T (6.502) on the domain, 10 z, 0 x W, and 2 30 T (6.503) on the domain, 0 zL 0 x W 2 2 22 2 30 11211 31123 2229333 dZdT dZdZTTT TZZZZ dxdxzzzdxdz AA (6.504) PAGE 207 207 22 2 222211 33112 22333 TTTTTT TTZZZ zzzzzz 00 30 dTdT Z dzdz (6.505) and 2233 2 33 2211 11 223333 TT TTTT ZZ zzzzzz 22 11 2 22 22 2211 113 32xTT Z zz TTTT ZZ xxxzxz 11 233xTT ZU xx (6.506) at 0 z 310 Tz (6.507) and 30 TzL (6.508) The volume condition is given by 3 00WZdx (6.509) The stricken terms vanish at 0 z on account of results from previous orders. We assume the following form for the solution 330313233(,)()()cos()()cos(2)()cos(3) TxzTzTzkxTzkxTzkx (6.510) 330313233(,)()()cos()()cos(2)()cos(3) TxzTzTzkxTzkxTzkx (6.511)3313233cos()cos(2)cos(3) Z xZkxZkxZkx (6.512) This problem can be solved completely but it is su fficient to solve the {31} problem to learn the nature of bifurcation and the value of 2A. PAGE 208 208 The {31} problem The {31} problem is given by 2 2 31 20 d kT dz (6.513) in the solid phase, 10 z 0 x W and 2 2 31 20d kT dz (6.514) in the liquid phase, 0 zL 0 x W 23 313103121013 3 4 TTUZUZkUZ (6.515) and 2 3131 01223 2 dTdT kUZZ dzdz (6.516) at 0 z 31(1)0 Tz (6.517) and 31()0 TzL (6.518) A solution to Eqs. (6.5136.518) is given by 2 3121kzkkzTzAeee (6.519) and 2 3131 kzkLkzTzAeee (6.520) where PAGE 209 209 23 3103121010122 21tanh33 3tanh 1tanhtanh42kk A UZUZkUZkUkLZZ ekkL (6.521) and 23 3103121010122 21tanh33 3tanh 1tanhtanh42kLkL A UZUZkUZkUkZZ ekkL (6.522) Using the curvature equation Finally we turn to Eq. (6.504), which is a differential equation for 3 Z To find 3 Z a solvability condition must be satisfied. It is 2 22 2 21111 31121 22 0 13 003030309 00W WTTTdZdZ TzZzZzZzZdx zzzdxdx TzZdx A (6.523) whence 2 2 20 2211 311122 2 432 131193 030000 242 9 0 4SdT dTdTdT TzzzZzZzZ dzdzdzdz kZkTZ AA (6.524) which can be written as 2 3131kAeR (6.525) where 2 2 20 2211 31122 2 432 131193 30000 242 9 4SdT dTdTdT R zzZzZzZ dzdzdzdz kZkTZ AA (6.526) which can be simplified to get PAGE 210 210 222 2 31221 2232432 11221311 34 1tanh2tanh 939 42tanh4SS SSSUL kk RZkTZkTZ Lkk k kkTZkTZZkZkTZ k AA AAAA (6.527) Comparing the value of 31 A from Eq. (6.521) and (6.525), and using Eq. (6.527), we get after some simplification 2 2 2 0 2 2 24 11 tanh 2tanh22tanh 11 1 tanh 7 4 tanhtanhtanh2S S S S SkT Pkk kkL QkkTk k kT UkL kT kkk k kkkkT A A A A A 2 01 LU A (6.528) where P Q is given by Eq. (6.501). Again, the sign of 2A is of interest. If it is positive, the pitchfork is forward. On doing calcu lations it will be learned that 2A can be positive as well as negative depending upon the i nput parameters. But in order to ma ke our analysis yet simpler, let us put 0ST in our results for this case. Case 6: Latent Heat Rejected to the Subcooled Liquid Only, Both Phases of Finite Extent, U Dropped From Domain Equations Putting 0ST in the results of case 5, there obtains at the base 0()0 Tz (6.529) and 0()LT Tzz L (6.530) PAGE 211 211 where 0U is given by 0LT U L (6.531) The condition at neutrality is given by 2 0tanh 1 tanh kL Uk k A (6.532) Figure 6.6 shows the corresponding neutral curve. The curve is monotonically increasing for all values of L. Hence the most dangerous value of n is one, which corresponds to kW giving only one cres t at critical. Figure 6.6 Typical neutral curve for solidification when the latent heat is reje cted through the subcooled liquid only; U is dropped from the domain equations At second order, we will find a speedup and 2U is given by the formula 2 222 21 011 1 tanhplanar correctionU Uk UkZ LkL A (6.533)22 Z is found as 2 221P Z Z Q where L > 1 L < 1 L = 1 U k2 PAGE 212 212 1tanhtanh tanh2 tanh1tanh2tanh2 1tanhtanh 4 1tanh2tanh2 kkLk kkL kkLk P Q kLk kLk (6.534) And at third order we find 2A as 22 0 0 211 tanh4 2tanh22tanh 111 1 tanh 7 tanhtanhtanh24 Pkk kkL Qkk k k UkLLU kkk k kkk AA (6.535) Our job then is to find the sign of 2A To accomplish this, we will consider two subcases: one where 2k is large, and another where 2k is small. Analysis for Large 2k: Since tanh k approaches 1 for large values of 2k we have 22 0 largelim2kUkA and 2largelim0kP Q hence using Eq. (6.535), we get 22 2 01113111 2242 kkk LLULk AA, whence neglecting the lo wer order terms in 2k there obtains 2 42111 0 3 Lk AA (6.536) We learn that the bifurcation is always a b ackward pitchfork for large values of inputs 2k no matter the value of L. Table 6.3 gives the value of the critical speed 0U and 2A for 60.26x10 A and for two different values of L The value of 0U is calculated using Eq. (6.532). The values of 2A were obtained from Eq. (6.535) but they can be calculated directly PAGE 213 213 from Eq.(6.536). Notice that since 2k is large the value of the critical speed 0U does not depend on L. Table 6.3 Case 6 Values of 0U and 2A for inputs 1,2 L and for large values of 2k 1 L 2 L 2k 0U 2A 0U 2A 103 5 x 104 2.667 5 x 1041.333 104 5 x 103 2.667 x 102 5 x 1031.333 x 102 105 5 x 102 2.667 x 104 5 x 1021.333 x 104 106 0.5 2.667 x 106 0.5 1.333 x 106 107 5 2.667 x 108 5 1.333 x 108 108 50 2.667 x 101050 1.333 x 1010 Analysis for Small 2k: Since tanh k approaches k for small values of 2k we have 22 0 smalllim1kULkA and 2large1 lim 3kP Q hence using Eq. (6.535), we get 2 231111 2132 LLk AA (6.537) Clearly the nature of the bifur cation depends on the ratio of the depths of the two phases. For 3 2 L we have a backward pitchfork whereas the pitchfork is forward for 3 2 L Table 6.4 gives the value of 2A for 60.26x10 A and for two different values of L The values of 2A were obtained from Eq. (6.535) but they can be calculated directly from Eq. (6.537). PAGE 214 214 Table 6.4 Case 6 Values of 0U and 2A for inputs 1,2L and for small values of 2k 1L 2L 2k 0U 2A 0U 2A 108 5 x 10156 x 10147.5 x 10154 x 1014 107 5 x 10146 x 10137.5 x 10144 x 1013 106 5 x 10136 x 10127.5 x 10134 x 1012 105 5 x 10126 x 10117.5 x 10124 x 1011 104 5 x 10116 x 10107.5 x 10114 x 1010 103 5 x 10106 x 109 7.5 x 10104 x 109 Table 6.5 Case 6 Values of 0U and 2A for inputs 1,2 L and 2k 1 L 2 L 2k 0U 2A 0U 2A 104 5 x 1011 6 x 1010 7.499 x 10114 x 1010 103 5 x 1010 5.99 x 109 7.495 x 10104 x 109 102 5 x 109 5.94 x 108 7.451 x 109 3.85 x 108 101 5 x 108 5.41 x 107 7.07 x 108 3.34 x 107 1 5 x 107 2.215 x 1065.664 x 107 3.59 x 106 10 5 x 106 2.66 x 104 5.01 x 106 1.49 x 104 102 5 x 105 2.66 x 102 5 x 105 1.37 x 102 103 5 x 104 2.66 5 x 104 1.3547 104 5 x 103 2.67 x 102 5 x 103 1.34 x 102 105 5 x 102 2.67 x 104 5 x 102 1.33 x 104 106 0.5 2.66 x 106 0.5 1.33 x 106 PAGE 215 215 Then defining 3 2crossoverL we have the following result: For crossoverLL the pitchfork is backward for small as well as for large values of 2k. For crossoverLL the pitchfork is forward for small values of 2k while it is backward for large values of 2k. This is shown in Table 6.5 which shows the values of 2A for 60.26x10 A and for 1,2L over a wide range of 2k Observe that for crossoverLL 2A flips sign for some intermediate value of 2k and the bifurcation type changes from a forward to a backward pitchfork. Figure 6.7 Transition diagram of L versus 2k showing regions corresponding to forward and backward pitchfork The reader might also be interested in co mparing the results of Table 6.5 with those calculated in Table 6.1 where U was not dropped from the domain equations. The two tables list the outputs 0U and 2A for the same values of inputs 2k and L. Observe that the results are in excellent agreement unless 2k is large enough that 0U is not small anymore and dropping U from the domain equations is not a good approximation. Forward Pitchfor k Backward Pitchfork Backward Pitchfor k Backward Pitchfork Branching flips from forward to backward 21 k 2A 41 k 2A 41 k 2A 21 k 2A L k2 crossover3 2 L Small k2 Large k2 PAGE 216 216 Finally we show a transition diagram, cf., Figure 6.7 which shows windows in the parameter space 2, Lk where we have forward and backward pitchfork. Discussion We have identified regions where the bifurcat ion might be backward and regions where it might be a forward pitchfork. We obtain a backward pitchfork for most cases, unless L is greater than crossoverL and the input values of 2k are small. Hence a forward pitchfork is easy to get for experimentally realizable values of L and 2k if one just knows where to look. A forward pitchfork is also called a supercri tical bifurcation (pitc hfork opening to the right) while a backward pitchfork is often termed as a subcritical bifurcation (pitchfork opening to the left). Figures 6.8 and 6.9 show schematic s representing the two types of bifurcations. In both these cartoons we pl ot the control parameter on the horizontal axis and the square of the amplitude of the nonplanar solution 2A on the vertical axis. For the solidification problem our control parameter is the unde rcooling in the subcooled liquid. i.e., LT One might study the stability of the new branches (new nonplanar solutions) to prove that subcritical solutions are unstable while supercritical solutions are stable [6]. Hence the solid lines in Figures 6.8 and 6.9 signify a stable solution while the dotted lines are indicative of unstable solutions. Figure 6.8 A cartoon illustrating a forward/supercritical branching stable stable stable unstable c 2 A PAGE 217 217 Figure 6.9 A cartoon illustrating a backward/subcritical branching In the case of a supercritical bifurcation, as we increase the control parameter, the instability will set in only after one crosses the critical value of the control parameter c Hence the supercritical case is in good agreement [31] with the predictions of the linear theory and gives us a solution when the plan ar solution is no longer stable to infinitesimal perturbations. And this corresponds to a smooth transition from a planar to a nonplanar so lution. In the case of a subcritical bifurcation however the solution can jump disconti nuously to the new nonplanar steady state even before the critic al point predicted by the linear theory [6]. However some other researchers [32] have emphasized that the predic tion of a subcritical bifu rcation should not be understood as evidence that finite amplitude so lutions do not exist. Ordinarily then, the implication of a subcritical case is that the planar solution might be unstable to finite amplitude perturbations even though it is stable to infinitesimal pertur bations accordi ng to the linear stability theory [10]. stable unstable unstable unstable stable stable 2 A c PAGE 218 218 Now Wollkind and Segel [33] have argued that the occurrence of a forward pitchfork is related to the formation of stable bands and a periodic structure. Hence the region of a forward pitchfork is also referred as a cellular regime while the region of a subcritical bifurcation is ordinarily considered to physica lly correspond to dendritic growth. Next, going back to the analysis of case 6, le t us try to advance some arguments to relate the predictions of the linear and the nonlinear theo ry. The predictions of the linear theory are seen from the neutral curve, cf., Figure 6.6. It is clear that for small values of 2k the curve is very sensitive to the value of L A possible explanation is that for low values of 2k the effect of surface tension cannot be significant. What is im portant is the diffusion in the two phases and hence the ratio of the depths of the two pha ses plays a key role. For large values of 2k we see from Figure 6.6 that the neutra l curve becomes asymptotic to 2 02 Uk A for all values of L, hence the ratio of the depths of the two phases is not of much importance. This is reasonable as the effect of surface tension becomes dominating for large enough values of 2k Now the prediction of the nonlinear theory for large values of 2k is that the nature of the branching is independent of L and it is always a backward pitchf ork. And this is consistent with the predictions of the linear theory in that th e ratio of the depths of the two phases cannot possibly have a significant effect on the critical fr ont speed if one is running an experiment in the large 2k regime. However, for small values of 2k diffusion in the two phases is important compared to surface tension and hence the na ture of branching depends on L What is more interesting is this: We learned that for 3 2 L we have a backward pitchfork while for 3 2 L the branching is forward. Now the linear stability theory predic ted (cf., Figure 6.6) th at a larger value of L ought PAGE 219 219 to give us more stability compared to a smaller L Then the fact that the weak nonlinear analysis predicts a larger value of L 3 2 L (as opposed to a smaller value L ) to give a forward bending pitchfork is interesting b ecause a forward pitchfork is or dinarily associated with the formation of cellular patterns (as opp osed to dendrites) which are in turn associated with a lesser degree of disorder. A backward pitchfork is, on the other hand, related to the formation of dendrites which certainly have a higher degree of disorder, and that is what we get for smaller values of L 3 2 L for small values of 2k. Concluding Remarks We have performed a weak nonlinear analysis for the problem of so lidification of a pure material. We have found that once the instability se ts in, the front speed al ways runs ahead of its base value regardless of the values of input parameters. We have al so learned that the nature of the branching of the planar solution to the nonplanar solu tion depends on the inputs 2k and L We have identified regions in the 2, Lk plane where the pitchfork is forward and regions where it is backward. It is possible to obtain both in physically realizable experiments. PAGE 220 220 CHAPTER 7 CONCLUSIONS AND RECOMMENDATIONS In this chapter, we summarize the main resu lts of this dissertati on and propose relevant future work. This dissertation has focused on understanding the underlying physics and causes of morphological instability in two phasechange problems, namely pr ecipitation and solidification. The theoretical methods involve linear stability calculations a nd weak nonlinear calculations. The major findings of the linear stability anal ysis are these: First, the solidification instability can give rise to the possibility of a multiplecell pattern at the onset of instability whereas for other growth problems involving solidliquid interfaces, such as precipitation from a supersaturated solution, the onset of an instabilit y from a planar state can result in at most a single crest and trough. Second, we have found that a solidliquid front that grows on account of temperature gradients is the only one amongst a ll solidliquid front growth problems where a critical wavelength, independent of surface energy, can occur. This critical wavelength which is of the order of the size of the container can be obtained for reasonable and measurable values of growth speed for most systems of practical interest. This fact alone sets solidification apart from any other growth problem involving a solidliqui d interface where the cr itical wavelength must obtain for unreasonably low values of growth speeds. These conclusions demand that careful experime nts be designed in order to validate the findings. The experiment must be run so that one is able to creep up on the instability by slowly increasing the value of the control parameter to its critical value. It would be valuable to demonstrate in an experiment that the pattern evolving at the onset of instability is indeed consistent with the theoretical predictions. If th e prediction of the nonlinear results is a forward pitchfork, then one might want to confirm that the critical value of the control parameter seen in the experiment is consistent with the theoretical value predicted by the li near stability analysis. PAGE 221 221 Also, one might want to check in the case of a backward pitchfork if the cr itical value of the control parameter seen in an experiment is ve ry different from the prediction of the linear analysis. The main findings of the nonlinear calculations are these: First, a solidification front always moves ahead of its predicted base value u pon becoming unstable. This is in contrast with other solidliquid growth instab ility problems such as precipit ation where an unstable front always slows down compared to its predicted base value. Second, th e nature of the bifurcation in solidification may be a fo rward pitchfork, leading to cellular growth, or a backward pitchfork, leading to dendritic growth. In contrast, ot her growth problems such as precipitation and electrodeposition are ordinarily characterized by dendritic grow th in the post onset region. These findings raise several in teresting questions that ought to be answered in future studies. One of the most important questions is this: Although the two phenomenon are very similar, why is the interfacial instability in solid ification so different from that in precipitation? For instance, can we find out a physical reason as to why a solid ification front always speeds up upon becoming unstable whereas a precipitation front slows down compared to its predicted base value? Why do we only see a backward pitchf ork in a precipitation instability whereas a solidification instability gives rise to the possibility of a b ackward as well as a forward pitchfork? Then, one might inves tigate the question of crosssect ional dependence on the nature of bifurcation beyond the onset of instability. In other words, can we ascribe a physical reason as to why the nature of the branching should depe nd on the shape of the cross section of the container in which the growth takes place? On e might also want to validate such a crosssectional dependence in an experiment. Anothe r important question that one might ask is whether we can specifically point out the algebraic terms in the nonlinear calculations that tend PAGE 222 222 to favor a backward bifurcation and terms that wa nt to make the branching forward. Finally, one might want to trace out the stable steady state so lution in the post onset re gime in the case of a backward pitchfork, and one might wonder w hy the morphology of the interface changes in a discontinuous and abrupt fashion instead of changing slowly and continuously. PAGE 223 223 APPENDIX A SURFACE VARIABLES In this appendix, we give the formulas for three surface variables, namely the unit normal vector, the surface speed and the mean curvature. Th is is sufficient as far as the theoretical work in this dissertation is concerned. The reader is referred to the book by Johns and Narayanan [3] for further details. A.1 The Unit Normal Vector Let us define a free surface by z Zxt in Cartesian coordinates or ,, z Zrt in cylindrical coordinates. Also we define a functional that vanishes on the surface as ,0 fzZxt in Cartesian coordinates, and ,,0 fzZrt in cylindrical coordinates. Then the norma l vector pointing into the region where f is positive is given by f n f Here, x z f f f ii x z in Cartesian coordinates and 1rz f ff f iii rrz in cylindrical coordinates. Then the equation for the normal is given by PAGE 224 224 1 2 21xzZ ii x n Z x in Cartesian coordinates, and 1 2 21rzZ ii r n Z r in cylindrical coordinates, if independence is assumed. A.2 The Surface Speed Let a surface be denoted by (,)0 frt Let the surface move a small distance s along its normal in time t Then, (,) f rsntt is given by (,) (,)(,) (,)... t frt frsnttfrtsnfrtt whence ( ,)0(,) f rsnttfrt requires (,) (,) t f rt snfrtt The normal speed of the surface, u, is then given by (,) (,) f rt s t u t nfrt Hence, using the definition of th e unit normal given earlier we get PAGE 225 225 f t u f For the problems concerned in this dissertation, the definition of u becomes 1 2 21 Z t u Z x in Cartesian coordinates, and 1 2 21 Z t u Z r in cylindrical coordinates, if independence is assumed. A.3 The Mean Curvature The book by Johns and Narayanan [3] gives a de tailed derivation of the curvature for a general surface. In this appendix, we simply provi de the formulas used in this dissertation. For a Cartesian surface defined as ,0 fzZxt the mean curvature is given by 32 22 1xx xZ H Z where the subscripts denote the derivative of Z with respect to that variable. For a Cartesian surface defined as ,,0 fzZxyt PAGE 226 226 the mean curvature is given by 22 32 22121 2 1yxxxyxyxyy xy Z ZZZZZZ H ZZ PAGE 227 227 APPENDIX B THE PERTURBATION EQUATIONS AND THE MAPPINGS In this appendix, we explain the perturbati on equations and the mappings used in this dissertation. The reader is re ferred to the book by Johns and Nara yanan [3] for further details. Higher order mappings are given in Appendix B.2. B.1 The Expansion of a Domain Variable and its Derivatives along the Mapping Let u denote the solution of a problem on a domain D. Now the domain D might itself be inconvenient. It may not be specified and must be determined as a part of the solution. This becomes clear when we study a moving front prob lem such as precipitation or solidification where the shape of the interface is not known and it is a part of the solution. It is however possible to obtain the solution u as well as to solve for the domain shape D by solving a new and easier problem defined on a regular and specified reference domain 0D .0D may be determined by the original problem or it may be chosen so as to simplify our work. The key idea is to imagine a family of domains D growing out of the reference domain 0D and to imagine that u must be determined on each of these, one being the domain of interest, D Now imagine that the domain D being in the vicinity of the reference domain 0D can be expressed in terms of the reference domain 0D via a small parameter Therefore the solution u and the domain D are solved simultaneously in a series of companion problems. What needs to be done is to discover how to determine u in terms of the solutions to problems defined on 0D Let the points on 0D be denoted by the coordinate 0 y and those on D by the coordinate y. The x coordinate is assumed to remain unchanged. Let u be a function of the spatial coordinate y. Then umust be a function of directly because it lies on D and also because it is a function of y The point y of the domain D is PAGE 228 228 then determined in terms of the point 0 y of the reference domain 0D by the mapping 0, yfy The function f can be expanded in powers of as 2 00 2 0 2,0,0 1 ,,0... 2 fyfy fyfy where 00,0 f yy and the derivatives of f are evaluated holding 0 y fixed. Then in terms of the notation 0 10,0 fy yy 2 0 20 2,0 fy yy etc., the mapping can be written as 2 0010201 ,,0,0... 2 yfyyyyyy The boundary of the reference domain must be carried into the bounda ry of the present domain by the same mapping. At the bounda ry of the new domain, the function y is replaced by Y to point out the difference. Its expansion in powers of can be written similarly as 2 010201 ,0,0... 2 YYYYYY Notice that it is the iY s that need to be determined to specify the domain D in terms of the domain 0D Finally, the variable uy can be expanded in powers of along the mapping as 2 00 2 0 2,0,0 1 ,,0... 2 duyyduyy uyuyy dd PAGE 229 229 where du d denotes the derivative of the function u depending on y and taken along the mapping. To obtain a formula for 0,0 duyy d differentiate u along the mapping taking y to depend on holding 0 y fixed. Using the chain rule, this gives 0,,,, duyuyuyfy dy Evaluating the above equation at 0 we get 000 10,0,0,0 duyyuyuy yy dy Then, introducing the definition of 1u via 0 10,0 uy uy and observing that 000 0,0 uyuy yy we get 000 1010 0,0 duyyuy uyyy dy All higher order derivatives of u can be determined the same way. If a domain variable needs to be specified at the boundary it is written similarly as 000 1010 0,0 duyYuY uYYY dy PAGE 230 230 When additional derivatives are obtained and substituted into the expansion of u, it becomes 2 22 000 1 0112112 2 00001 ,2... 2 uuu u uyuuyuyyy yyyy The reader might wond er that the mapping 1 y does not appear in the domain equations given in this dissertation. In f act, the mappings on the domain can not be determined and they are never required. This can be proved rigorously as done in the book by Johns and Narayanan [3], but in the interest of continuity, let us work out an example to show that one need not worry about the mappings on the domain. Once we have gone through this example, we can use this as a rule of thumb. Let u satisfy 0 u y in the inconvenient domain D. Using chain rule 0 0 y uu y yy where, holding fixed, 0 y y is given by 2 0 121 1... 2 y yy yyy Thus, up to the first order in the domain equation becomes 2 00 1 1 2 000... uu u u y yyyy Therefore, the domain equation at the zeroth order in is 0 00 u y PAGE 231 231 whereas the domain equa tion at first order in using 2 0 2 00 u y becomes 1 00 u y Notice that the mapping is lost from doma in equations. If surface variables were considered the mapping would not be lost, as is evident from all the problems studied in this dissertation. B.2 The Higher Order Mappings Up to the fourth order, the e xpansion of the domain variable u along the mapping is 2 22 000 1 0112112 2 0000 23 2 323 000 211 31212113 223 000000 3 22 22 0 12 412112 223 000 41 2 2 1 3333 6 1066 1 24 uuu u uuuyuyyy yyyy uuu uuu uyyyyyyy yyyyyy u uu uyyyyy yyy 2 3 3 30 21 12113 32 0000 24 24 000 1 2314 24 00004444 32 uu uu yyyyy y yyy uuu u yyyy yyyy and the expansion of its derivative is given as 232 2 22 0000 121 1112 2232 0000000 342 223 323 3000 211 1212113 223342 0000000 3 1 412 41 2 2 1 3333 6 10 1 24 uuuu uuu u yyyy yyyyyyyy uuuu uuu yyyyyyy yyyyyyy u uyy y 423 324 223 030 221 11212113 3342243 0000000 352 2 24 000 1 2314 3252 0000664444 32 uuu uuu yyyyyyyy y yyyyy uuu u yyyy yyyy PAGE 232 232 APPENDIX C EIGENFUNCTIONS OF AN EQUILATERAL TRIANGLE AND A REGULAR HEXAGON SATISFYING NEUMANN BOUNDARY CONDITIONS In this appendix, we give the function f xy which is an eigenfunction of 2 H on an arbitrary cross section S, subject to Neumann sidewa ll conditions. It satisfies 22,, f xykfxy on S; .0 pf on S where S denotes the boundary of S and p denotes the outward normal to S Appendix C.1 gives the eigenfunctions and eige nvalues corresponding to an equ ilateral triangle. Appendix C.2 gives the eigenfunctions and eige nvalues of a regular hexagon. C.1 Eigenfunctions of an Equilateral Triangle Consider an equilateral triangle of side h, whose vertices are given by 0,0, ,0 h and /2,3/2 hh The eigenfunction f xy that satisfies 22,, f xykfxy on S; .0 pf on S is given by [34] as ,,,,,,mnmnmn sa f xyfxyfxy where ,3 ,cos3cos/2 39 3 cos3cos/2 39 3 cos3cos/2 39mn smn l fxyryxh rr nl m ryxh rr lm n ryxh rr and PAGE 233 233 ,3 ,cos3sin/2 39 3 cos3sin/2 39 3 cos3sin/2 39mn amn l fxyryxh rr nl m ryxh rr lm n ryxh rr where lmn and 23 h r The corresponding eigenvalue is given by 2 2224 27 kmmnn r C.2 Eigenfunctions of a Regular Hexagon Consider a regular hexagon of side L and centered at the origin 0,0. 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Oulton, Interfacial Patterns During Plane Front Alloy Solidification, Physica D 12, 215 (1984). [33] D. Wollkind and L. Segel, A Nonlinear St ability Analysis of the Freezing of a Dilute Binary Alloy, Philosophical Transactions of the Ro yal Society of London. Series A, Mathematical and Physical Sciences 268, 351 (1970). [34] B.J. McCartin, Eigenstructure of the Equilateral Triangle, Part II: The Neumann Problem, Mathematical Problem s in Engineering 8, 517 (2002). PAGE 237 237 BIOGRAPHICAL SKETCH Saurabh Agarwal was born in 1980 in New Delhi, India. In 2002, he received a B.S. in chemical engineering from the I ndian Institute of Technology, Ka npur. He joined the University of Florida as a graduate stude nt in 2002. During his graduate studies, he worked under the supervision of Prof. Ranga Nara yanan and Prof. Lewis E. Johns. In 2007, he graduated from the University of Florida with a Ph.D in chemical engineering. 