microstructure simulation

Microstructure Simulations

B. Nestler, F. Wendler, A. Choudhury

December 17, 2011

Contents

1 30.10.10 191.1 Lecture 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2 1.11.10 442.1 lecture 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3 1.11.10 553.1 Lecture 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4 1.11.10 664.1 lecture 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

5 1.11.10 745.1 lecture 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

6 1.11.10 796.1 lecture 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 796.2 Tasks to perform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

Introduction to solidification

Microsstructure simulations

Content

Introduction to solidificationTechnical importance of solidificationGeneral terms and basics in solidificationThe role of heat extractionDirectional solidificationSpecial solidification techniquesAim of computational materials scienceDendritic solidificationEutectic solidificationPeritectic solidificationMonotectic solidificationCourse outline

B. Nestler, A. Choudhury, F. Wendler WS 2010-2011 2/33



Literature recommendations

A large part of this course is treated in [Kurz and Fisher, 1998] and[Porter et al., 2009]. A good german textbook is [Gottstein, 2007].

Gottstein, G. (2007).

Physikalische Grundlagen der Materialkunde.

Springer Verlag Berlin Heidelberg.

Kurz, W. and Fisher, D. (1998).

Fundamentals of solidification.

Trans Tech Publications ltd, Switzerland Germany UK USA.

Porter, D. A., Easterling, K. E., and Sherif, M. Y. (2009).

Phase transformations in metals and alloys (third edition).

CRC Press, Taylor & Francis Group, Boca Raton, London, New York.




General problem definition

During processing of materials an inhomogeneous microstructure evolves,originating from the kinetics of the processes. This may lead to potentially weakpoints and eventually failure.In this lecture, there is a focus on metal solidification. Here crystal nucleation,a necessary prerequisite to solidification, is a ubiquitous event. Glass, ametastable phase, is formed only at extremely high cooling rates.

Understand microstructure evolution depending on different solidificationconditions. → optimize processing parameters.

Modeling goal:




Effect of microstructure - example

Microstructure formed after casting may have severe effects on mechanicalbehaviour.Example: Ingot casting of Al-7%Si alloy with subsequent ECAP processing(severe plastic deformation)

J.M. Garcia-Infanta et al., Scripta Mater. 58 (2008) 138.




Technical importance of solidification

Solidification is one of the most importantforming processes in materials science,involved in the production of almost alltechnical products. It is a very economicmetal forming process (espec. for complexshaped pieces)Solidification as a phase transition occursduring

metal casting (ingot casting,continuous casting)

metal welding, soldering

polymer injection moulding

growth of silicon single or poly- crystals




Technical solidification processes (incomplete)

Casting continuous castingingot castinginvestment castingprecision castingdie casting

Welding arc-weldingresistance weldinglaser/electron beam welding

Soldering/BrazingRapid solidification melt-spinning

atomisationDirectional solidification Bridgman

Czochralski




Some general terms used in the courseOverview, most of the terms will be filled with ’live’ during the following lectures.

Phase: A phase is a homogeneous part in a possibly heterogeneoussystem (Gibbs1). Here we have solid (s ) and liquid (l ). Can be related tothe local atomic arrangement→ construction of order parameters.

Equilibrium: State of minimal free energy (G or F ). Phases in contact dohave enough time (locally) to reach this state.

Transition: may arise by changing a control parameter (temperature,pressure ...), during which latent heat is released/absorbed.

1Josiah Willard Gibbs, 1839 - 1903, american physicist, chemist and mathematicianB. Nestler, A. Choudhury, F. Wendler WS 2010-2011 8/33



Solidification basics - casting

Casting of a melt of composition C0 andtemperature T into the mold.

Cooling: A heat flux q depending on heatdiffusivity k gives rise to undercooling ∆Tof melt.

Then, due to the difference of chemicalpotentials µ between s - l crystallizationstarts, the speed of which is given by T .

For the (macro) flow of melt the viscosity ηand liquid-vapour surface tension σ maybe important.

Kurz, Fisher, Fig. 0B. Nestler, A. Choudhury, F. Wendler WS 2010-2011 9/33



Solidified ingots - morphology

Melt is usually cooled from thecrucible wall

This results in growth, beingfastest parallel to the temperaturegradient G.

Mushy zone: the solid/liquidcontact region, in which all growthdetermining kinetic processes takeplace→ evolution of complicateinterface structure.

Density changes givr rise toshrinkage and porosity.




Solidified ingots - morphology IIMicrostructure zones in cast ingot,schematicKurz and Fisher, Fig. 1.6

Aluminum ingot (5 cm)H. K. D. H. Bhadeshia, University of Cambridge




The role of heat extraction

After casting the liquid metal, heat of different physicalorigin must be extracted:

Heat related to the specific heat capacity of onephase: enthalphy decrease ∆H =

∫ T2T1

CdT

Latent heat of fusion, released in the phasetransition: ∆Hf

Later, we will use the specific latent heat ∆hf = ∆Hf/Vm (Vm: molar volume)and specific heat capacity c = C/Vm.




Solidification stages

After casting, three different steps in thesolidification can be observed in the coolingcurve:

1. Cooling from pouring temperature,nucleation starts after falling belowmelting temperature Tm

2. Reheating to Tm (recalescence) due torelease of latent heat

3. After complete transformation to solid,continued cooling sets on.

Figure: Quasi 1-D experiment,P. Thevoz et al., Metal Transact. 20 A (1989) 311




Directional solidification - experimentsExperimental setup: Controlled solidification using optical microscopywith transparent binary analog substances.

2D:

3D:Figure: Eutectic growth in SCN - DC,[Akamatsu et al., Journal of Crystal Growth 299 (2007) 418]




Directional solidification - principlesSteady heat extraction at pulling velocities v small enough to reach steady state.Growth rate (= v ) and temperature gradient G can be separately controlled. Ifheat flux qext and temperature gradient are parallel:

Ts+z =∂T∂z ′· ∂z ′

∂ t

∣∣s+z = G · v

∣∣s+z

But: due to different heat conductivity in solid/liquid and release of latent heat:

Gsolid 6= Gliquid

Experimental setup foralloys:




Directional solidification - technical applicationsProduction setup (schematic): Adjust(and move) temperature field bycontrolling current in electric coils.

Figure: Al alloy - polycrystal [Tom Quested,Microstructural kinetics group - Univ. of Cambridge.]

Figure: Solar grade silicon, directionallysolidified using the Bridgman method [B. Ryningen].




Special solidification methods - metal matrix composites

Gas pressure infiltration of melt into fibre matrix to produce compositematerials:

Figure: Steps of gas pressure infiltration: a)porous preform preheating; b) immersion ofporous preform into the melt and inlet ofpressurized gas; c) infiltrated sample inmolten matrix and d) withdrawal ofinfiltrated sample from melt.[Institute of materials & machine mechanics, Slovakian academy ofsciences]

Figure: Pure Al matrix reinforcedby unidirectional carbon fibres (65vol %)[G. Requena et al., Composites: Part A, 40 (2009) 152.]




Special solidification methods - ThixocastingProcessing of metals (mostly Al, Cu, Mg) resulting in granular semi-solidswith liquid vol. fractions of 30% - 60%.

Sample in solid state ascast (a) and semi-solid (b) a

material with rheology likebutter.[H. V. Atkinson, Prog. Mat. Sci. 50 (2005) 341.]




What is the aim of computational materials science?

The macroscopic properties are aa function of the microstructure

Dependence between themicrostructure and the mechanicalproperties are generallydetermined experimentally

The dependence of the processingparameters and the microstructure→determined eitherexperimentally or throughmathematical modeling




Why is study of solidification important to physicists andengineers?

Gain an understanding ofthe physics behind thepattern and microstructureformation

This would enable controlof processing parameters

Hence, the microstructureand the correspondingmechanical properties

[W. Losert et al., Phys. Rev. E 58(6) (1998) 7492.]




Dendritic Solidification

Rather a ubiquitous type of front instability than a solidification mode

Co-Cr alloy[Labo des Micro-Analyses des Surfaces, Besancon]

MOVIESuccinonitrile-Aceton[H. Esaka, J. Stramke and W. Kurz, Department of

Materials, Swiss Federal Institute of Technology

Lausanne, Switzerland]




Dendritic Solidification - phase-field simulations

Solidification of pure Ni into undercooled melt (∆T = 130K ,∆T/ρcp = 0.3)

.with P. Galenko, D. Herlach, (DLR Koln), E. Brener, H. Muller-Krumbhaar (FZJulich)




Eutectic Solidifications - Types and Instabilities

What is eutecticgrowth/reaction;L−→ S1 + S2?

Below the eutectictemperature the liquid isunstable to decompositionto two solids

What is the volumefraction of the solid phasesgrowing together?




Eutectic Solidification - Length Scales

What are the differentlength scales for diffusion?

Is there a physical basisfor the choice of the lengthscale?

Mechanical properties −→function of the length scale

Is it possible to determinethis length scale as afunction of the processingconditions?




Eutectic Solidification - Instabilities

Is it possible to get coupledgrowth over a whole rangeof length scales?

What makes coupledgrowth unstable?

What are the instabilitiesfor large scalemicrostructure?

What are the instabilitiesfor small scalemicrostructure?




Peritectic Solidification

Reaction and Interesting Questions

What is the peritecticreaction,L + S1 −→ S2?

Why is peritectic reactionimportant to study?

- Steels (Fe−C)- Superalloys in NiAl system(Al3Ni)- Nb- Perovskite (YBaCuO)




Peritectic Solidification - Microstructures IIsland structures, Banded structures and coupled growth:




Peritectic Solidification - Microstructures II

[R. Siquieri et al., J. Phys. Cond. Mat. 21 (2009) 464112.]




Peritectic Solidification - Interesting questions?

Can two solid peritectic phases grow together?

How is it different from eutectic growth?

What are the patterns and instabilities in such microstructure?

What are the microstructures and the corresponding instabilities in3D?




Monotectic Solidification - Challenges, microstructures

What is the monotecticreaction, L1 −→ L2 + S?

Why is it interesting tolearn about this reaction?

- Lubrication phenomena(oil-oil mixtures)

- Cu-Pb (Tmono = 935), Pb-Fe




Monotectic Solidification - Microstructures

What are theinterestingmicrostructures?

What are thephysics behind theformation of each ofthesemicrostructures?




Monotectic Solidification

What are the open questions?

Is nucleation important?

How is the phenomena offluid flow important?

No validated theoryexisting until now

What are the instabilities?(Rayleigh)




Outline of the course

(Detailed sequence in handout)

1. Introduction to the process of solidification

2. Thermodynamics and Transport Phenomena, necessary tounderstand the process

3. Introduction to phase-field modeling

4. Numerical Techniques

5. Theoretical Analysis, of Dendritic, Eutectic and PeritecticSolidification

6. Verification of theoretical concepts through modeling

7. Outlook to the open questions and challenges


1 30.10.10

1.1 Lecture 2

Classification of materials (mat. sci.)

Materials can be classified into the following basic classes,

• metals

• ceramics

• polymers

• composites.

The aim of studying the materials science is in general to understand the functioningof the materials, their properties, the pathway to process materials and optimize per-formance, through an understanding of the correlation between the structure and theperformance. In order to perform this task, it is essential to gain insights into the ther-modynamics of materials, which enables us to qualify and predict the equilibrium statesof a material given, a set of conditions.

Introduction to theromodynamics

Figure 1.1: What is a system? The diagram illustrates the structure of the system in relationto the surroundings and the means of interaction.

1 30.10.10 20

microscopic state ↔ macroscopic state

Figure 1.2: The state of the system can be completely defined at two levels. One is at themicroscopic level called the microscopic state, which contains the information aboutthe mass, velocity and position of the particles. The other is at the contnuum scale,called the macroscopic state of the system, defined by variable, pressure, volumeand temperature, where two variables are independent, and related to the thirdthrough a state function

An example of a ’simple system’ consisting of 3 variables (2 indep.) can be for examplewhere the volume, V=V(P,T), which is a state function of the pressure and temperatureof the system. A characteristic of a state function is that for any reversible process, thechange of state describe by the state function depends only on the end states. This canbe seen as follows: Process from 1→ 2: volume change

∆V = V2 − V1

= (Va − V1)︸︷︷︸P=const.=P1

+ (V2 − Va)︸︷︷︸T=const.=T2

1→ a→ 2

∆V = V2 − V1 =

∫ T2

T1

(∂V

∂T

)

P1

dT +

∫ P2

P1

(∂V

∂P

)

T2

dP (1)

1→ b→ 2

∆V =

∫ P2

P1

(∂V

∂P

)

T1

dP +

∫ T2

T1

(∂V

∂T

)

P2

dT (2)

1 30.10.10 21

Figure 1.3: An exemplary process showing two reversible process from the start state (1) toend state (2), through the routes 1→ a→ 2 or 1→ b→ 2.

This means that V=V(P,T) is a state function, as the integral does not depend on theintegration path. The same result is found integrating the complete/total differential ofV,

dV =

(∂V

∂P

)

T

dP +

(∂V

∂P

)

P

dT,

between the limits P2, T2 and P1, T1.

Equation of state for an ideal gas

• Robert Boyle (1660) P ∼ 1V (T=const)

• Alexandre-Ciser Charles (1787) V ∼ T (P=const)

• Joseph-Linis Gray-Lussac (1802)

• Coefficient of thermal expansion

α =1

V o

(∂V

∂T

)

P

≈ 1

273at 0C

1 30.10.10 22

good for gases with low boiling points

model: ideal gas α = 1273,15

V(T=0 K) = 0 absolute Zero of Temp. Scale Combination of (Boyle) and (Charles)gives,

P0V0(T1P0) = PV (T1, P ) and

V0︷︸︸︷V0(P0, T0)

T0=

︷︸︸︷V (P0, T )

T

⇒ PV

T=P0V0

T0= const. (∗)

(a) (b)

Figure 1.4: (a)Variation of the state function along given isotherms, (b)Variation of state func-tion relation V and T at given isobars.

Extensive / intensive state variables

Intensive state variables are independent of the system size (P,T), extensive variablesdepend on size → can be transformed into intensive variables by division through mass(e.g. specific volume) or mole number (e.g. molar volume V = V ′

n ).

Phase diagrams, terms + examples A phase-diagram, shows the regions of stability ofthe different phases in a given systems. Comlexity of the phase diagram depends on

1 30.10.10 23

no. of components (chemical compositions): chemical elements (Ni,Cu), stochiometriccompounnds (H2O), and general categorization can be given as follows:

• one-component (pure) system

• binary system

• ternary system

• quaternary system

One-comp. system (20 repr.)

A one component system is characterized by two independent variables, T, P. An exem-plary phase diagram showing the stability regions of the phases solid, liquid and vaporis shown in figure

Ice-Water-VaporIce-Water-Vapor1.5.

within,AOB−→ liquidCOA−→ sCOB−→ v

one-phase eq.(system:homogeneous)

on the line AO :two phases (s/l) coexist

→ two - phase eq. (system: heterogeneous)

→ pressure- temp. relation for eq. of sol/liq H2O

CO: saturation vapor pressure of solid OB: saturation vapor pressure of liquid

Figure 1.5: Phase diagram of the ice-water system.Ice-Water-Vapor

1 30.10.10 24

Two-component system → additional degree of freedom, composition

The complete diagam is three dimensional consisting of the variables, C,T,P,V. At const.pressure the diagram at (1 atm ∼ 1 bar) can be drawn as follows,

Figure 1.6: The phase diagram of a two component system.

example: system Al2O3-Cr2O3 (1 atm)

at T < Tm (Al2O3)= 2050C and

T > Tm (Cr2O3)= 2265C completely

miscible in all proportions ((Al2O3,Cr2O3 have similar crystalstructure and Al3+, Cr3+

similar size)

• single phase region, l,s

• two-phase region, l+s equil.

At any temperature and given composition, the relative fractions of two phases in equi-librium are derived through the well known lever rule. This can be written as follows.Let us say, that at the temperature T1, the equilibrium compositions of the solid andliquid are Cs and Cl respectively. Then, the fractions of solid and liquid must satisfy,the following mass balance relation:

C = Csfs + Clfl,

1 30.10.10 25

and since, there are only two phases in equilibrium, implies that fs + fl = 1. Using thisthe following relation for the solid and the liquid fractions can be derived,

fs =(C − Cl)Cs − Cl

.

The first law of thermodynamics

The first law of thermodynamics is motivated from the basic concept of conservation ofenergy. The sum total change of the internal energy U which is the combination of thevibration, rotational and kinetic energies of the atoms, when a system goes undergoes achange of state equals the difference between the total amount of heat transferred to thebody q and the work done by the body w. An expansive work is treated as work doneby the body. Conversely, a compressive work is treated as work done on the body w. Adifferential change in the internal energy U can thus be written as,

dU = δq − δw.

Please note: the differential sign on the internal energy U, which the operator δ on thevariables q and w. This signifies, a fundamental difference between a state function,whose change is independent of the path of integration, while, the quantities, q and wdepend on the path of integration. This can be immediately be appreciated from figure,WorkWork1.7. For the three paths, a, b, c the change of state, U, is the same (U2−U1). However,the work done, which is area under the curve is not.

Figure 1.7: Three process paths taken by a fixed quantitiy of gas in moving from the state1 tothe state2.Work

1 30.10.10 26

For a cyclic process: 1 → 2 →1

∆U =

∫ 2

1dU +

∫ 1

2dU

= (U2 − U1) + (U1 − U2) = 0 =

∮dU

For a simple one-component system, the internal energy is competely defined by twoindependent variables. If temperature and volume are chosen, then, the complete differ-ential can be written as follows,

→ two independent state︸︷︷︸ variables U=U(V,T)

dU =

(∂U

∂V

)

T

dV +

(∂U

∂T

)

V

dT.

Examples of different processes:

• T=const. - isothermal

• P=const. - isobaric

• V=const. - isochoric

• (∆ Q = 0 - adiabatic)

∗ Constant - volume processes

∆W =

∫PdV = 0 dU = δQV , ⇒ ∆U = ∆QV =

∫ 2

1CV dT

chage of U = absorbed/rejected heat

∗ Constant - pressure processes

∆W =

∫ 2

1PdV = P

∫ 2

1dV = P (V2 − V1)

U2 − U1 = ∆QP − P (V2 − V1)

1 30.10.10 27

( U2︸︷︷︸+PV2︸︷︷︸)− ( U1︸︷︷︸+PV1︸︷︷︸) = ∆QP

all are state functions

In the above, all are state functions, hence, the resulting function is also a state function.We define new state function H=U+PV : which is called the enthalpy of the system.

At P= const. H2-H1= ∆ H =∆ U + P∆ V = ∆ QP =∫ 2

1 CPdT

Heat capacity

The heat capacity, C, of a system is the ration of the heat added to or withdrawn fromthe system to the resultant change in the temperature of the system. Thus,

C =∆Q

∆T; small changes ∆T : C =

δQ

dT

This concept is however useful when there is no phase change. When a phase changeoccurs, the heat goes in the change of state while the tempersture stay constant, whichwould result, in an infinite heat capacity by the above formula. Since, the state ofa simple system is defined by two independent variables, the measurement of the heatcapacity is completely defined, only when the variation of the another variables (pressureor volume) is defined. This leads to the following definitiions of the heat capacities,

CV =( δQdT )V = (dUdT )V ⇔ dU= CV dT (const. volume)

CP=( δQdT )P = (dHdT )P ⇔ dH= CPdT (const. pressure).

C is extensive quantity, depending on system size (mass, volume, mole no.). One canmake it intensive by dividing through respective number of moles. This gives us molarheat capacities,

cP=CPn ; cV =CV

n .

For, cP > cV , as additional heat is required to provide work during expansion of thesystem,

1 30.10.10 28

cp =

(∂H

∂T

)

P

=

(∂U

∂T

)

P

+ P

(∂V

∂T

)

P

(1mole)

cv =

(∂U

∂T

)

V

cp − cv =

(∂U

∂T

)

P

+ P

(∂V

∂T

)

P

−(∂U

∂T

)

V

with dU =

(∂U

∂V

)

T

dV +

(∂U

∂T

)

V

dT

⇒(∂U

∂T

)

P

=

(∂U

∂V

)

T

(∂V

∂T

)

P

+

(∂U

∂T

)

V

⇒ cp − cv =

(∂V

∂T

)

P︸︷︷︸1

P +

(∂U

∂V

)

T︸︷︷︸2

To determine, the term ∂U∂V T

, Joule performed the experiment, as shown in figureJoule’s experimentJoule’s experiment1.8,

which involved filling a copper vessel with a gas at some pressure and connecting thisvessel via a stopcock to a similar but evacuated vessel. The two vessel system wasimmersed in a quantity of adiabatically contained water and the stopcock was opened,thus allowing free expansion of the gas into the evacuated vessel. After this expansion,Joule could not detect any change in temperature of the system, As the system, wasadiabatically contained, and no work was performed and hence, from first law of ther-modynamics, it follows that the total change in internal energy ∆U = 0. From the total

differential of U , one can then work out that,

(∂U

∂V

)

T

= 0.

ideal gas:

This gives the following relation for gases:

(∂U

∂V

)

T

= 0⇒ CP − CV = P

(∂V

∂T

)

P

.

In addition for 1 mole of ideal gas,: PV=RT and hence ⇒ cP − cV =R

1 30.10.10 29

Figure 1.8: Joule’s experimentJoule’s experiment

The term P

(∂V

∂T P

)represents work done in expanding gas against an constant external

pressure P. The term(∂U∂V

)T

(∂V∂T

)P

represents the work done against internal cohesiveforces.

(∂U

∂V

)

T

≈ 0 fo gases, but

(∂U

∂V

)

T

P for liquids, solids

Reversible adiabatic process

∆W =

∫ 2

1PdV only for reversible process

δQ = 0⇒ dU = −δW(1 mole) = CV dT δW = PdV

CV =

(dU

dT

)

V

CV dT = −PdV = −RTVdV

CV lnT1

T2= R ln

V1

V2

1 30.10.10 30

(T2

T1

)CV=

(V1

V2

)R⇔ T2

T1=(V1V2

) RCV

[CPCV

= γ; CP − CV = R;cPcV− 1 =

R

CV= γ − 1

]

T2

T1=

(V1

V2

)γ−1

=P2V2

P1V1⇒ P2

P1=(V1V2

)γ

⇔ P·V γ = const (adiabatic)

Reversible isothermal process, ∆P, ∆V (ideal gas)

dU = δQ− δW

isothermal dT = 0⇒ dU = 0⇒ δW = δQ = PdV =RT

VdV

1→ 2 ∆W = ∆Q = RTln

(V2

V1

)= RTln

(P1

P2

)

∆P fixed1→ 2, 1→ 3

∆W isothermal > ∆Wadiab

↓= ∆Q isothermal(absorbed), U=const ∆U= -∆W

Second Law of Thermodynamics

In previous chapter, only extreme processes ∆W=0 or ∆Q=0 were studied. For ∆W6=0and ∆Q 6=0, is there a definite amount of work, which the system can do?

Here we need to,

→ identify two classes of processes - reversible and irreversible

→ introduce a state function, entropy S (gives the degree of irreversibility of a process)

1 30.10.10 31

Figure 1.9: Decription of a isothermal process 1→ 2 at T=1000K, while process 1→ 3 repre-sents an adiabatic process.

Definition of Spontaneity

Spontaneous (=natural) processes Left to itself a sytem has two possibilities a) it staysin its initial state (=if it is in equilibrium; = state of rest) or

b) moves towards its equilibrium state, if it is away from equilibrium.

Hence, a process which involves, the movement of a system away from equilibrium to-wards its equilibrium state, is called a spontaneous process. Since, this process cannot bereversed without the application of an external agency, it is called an irreversible process,as it would require an irreversible change in the properties of the external agency.

Examples of two spontaneous processes,

1. Mixing of gases

2. Flow of heat down a temperature/conc. gradient

In both cases, the state of equilibrium is in (a) where the composition is uniform in thegas mixture, and in (b) the temperature of the system becomes uniform. A knowledgeof the equilibrium allows us to decide, the direction of the spntaneous reaction given theinitial state of the the system. During a spontaneous process, work is performed and asequilibrium is reached the ability to perform work is reduced and finally at equilibrium,the system is at rest. As woek is performed, there is degradation of energy to heat andthe ability to do work is lost.

Entropy and the quantitfication of irreversibility

1 30.10.10 32

There exist two distinct types of spontaneous processes,

• Conversion of work to heat (i.e degradation of mechanical energy to thermal en-ergy)

• Flow of heat down a temperature gradient

If it is considered that during an irreversible process, the ability to do work is de-graded,then it is possible to quantify the extent of degradation or the extent of irre-versibility of the system. Consider the weight-heat reservoir system as shown in figurereservoirsystemreservoirsystem1.10. The system consists of a weight-pulley arangement which is coupled to a constant-temperature heat reservoir , and the system is at equilibrium when an upwrd forceacting on the weight exacty balances the downward force, W, of the weight. If the up-ward force is removed then the equilibrium is upse and the wieght spontaneously fallsthus performing work which is converted by means of a suitable system of paddle wheelsto heat which enters the constant temperature heat reservoir. Equilibrium is reattainedwhen the upward force acting on the weight is replaced and the net effect of this processis that mechanical energy has been converted to thermal energy. Now three differentexperiments can be considered with this setup.

1. work performed by weight ∆W → transformed to heat ∆Q entering reservoir atT2

2. same amount of heat flows to a different reservoir at temperature T1 where T1 <T2 which is in thermal contact

3. same as 1. , but now the main reservoir is at T1 and heat flows into heat reservoir.

Figure 1.10: Weight-heat reservoir systemreservoirsystem

It can be percieved that essence process 3. is the sum of 1. and 2. And hence, process(3) should be more irreversible. Examining the processes closely, shows, that the amount

1 30.10.10 33

of heat Q and the temperature between which the heat flows, is imporatant in finding aquantitative measure of irreversiblity. Hence, we derive a term of the form ∆Q

T , definedby the amount of heat ∆Q and (absol.)temeratures T1, T2. With this definitiion wedirectly see that process (3) is more irreversible than (1) for T1 < T2. This degree ofirreversibility we call as the increase of entropy S and define,

Entropy produced: ∆S = ∆QT .

If irrversibility is quantified, how do we visualize a reversible process?

A reversible process, is one where the degree of irreversibility is minimized. The ultimateminimization would occur when the concept of spontaneity is no longer applicable, whichapparently implies that the system is only infinitesimaly away from equilibrium such thatthe extend of degradation is as small as we please. This a reversible process, wherein, asystem is taken from state A to state B, through string of states all at equlibrium. Anillustration of a reversible process can be seen as follows.

Consider a system as shown inReversibleprocessReversibleprocess1.11. it consists of water in equilibrium with its vapor

at given saturated pressure PH20 (T ), which is function of temperature T. Now thesystem is at rest where the Pext exactly balances the saturated vapor pressure. Nowconsider, we reduce the external pressure by ∆P , which would then cause the gas toexpand owing to mismatch of pressure, and the water would start to evaporate to restoreequilibrium. Since, evaporation is an endothermic process, heat would be drawin infrom the reservoir to maintain constant temperature T, by providing the required latentheat for the process. If after one mole of evaporation, the pressure is agian increasedto its original value, then the evaporation stops, and the work done by the system is(Pext−∆P )V, where V is the molar volume of the gas. Conversely, when the pressure ofthe gas is now increased by an amount ∆P , this would cause the piston to fall rapidly, andthe water-vapor would start to condense to restore equilibrium between the water- andwater vapor. This is an exothermic process, which would need a transfer of heat to thereseervoir to maintain the same temperature T. After one mole of gas has condensed, theexternal pressure is again increased to the saturation vapor pressure at the temperatureT. This restores equilibrium and the total work done is (Pext + ∆P ) V. In the totalcyclic process, the total work done by the external agency is 2δPV . Now a reversibleprocess would be such, where this term is minimized. Or in other words, a system wherethe change in the external pressure is an infinitesimal δP , and the process is carried outslowly, such that the change in saturation is reduced and the heat flow keeps up wth theequilibrium.

It is thus seen that reversibility is approached when the evaporation or condensationprocesses are carried out in such a manner that the pressure exerted by the vapor isnever more than infinitesimaly different from its value at the saturation temperature T.It is also seen that as complete reversibility approached, the process becomes infinitelyslow.

1 30.10.10 34

The whole process is summarized below:

example equilibrium: Pext = PH2O(T) and T=TH2O= Tvaporn

1. Pressure drop Pext −∆P , work done

∆W = (Pext −∆P )V (evaporation)

2. Pressure increase Pext + ∆P

3. ∆W = (Pext + ∆P )V (condensation)

4. ∆W12 = 2∆P ·V

Figure 1.11: A hypothetical system of water in equilibrium with its vapor at saturated va-por pressure PH20 (T ), which is a function of temperature T. The system is keptin contact with a heat reservoir, and an external system to adjust the externalpressure Pext.Reversibleprocess

1 30.10.10 35

Entropy and reversible heat Consider just the process of evaporation in the previoustext. For a reversible process, the work done by the system is PextV . For any irreversibleprocess, the work done by the system is (Pext −∆P )V . It is clearly seen that maximaumwork Wmax occurs during a reversible process.

1. Reversible process (∆P → δP ):

∆Wmax = PextV ; ∆Qrev = ∆U + ∆Wmax

∆U − indep. of process

∆W = (Pext −∆P )V ; ∆Q = ∆U + ∆W

difference∆Wmax −∆W = ∆Qrev −∆Q

is degraded into heat within the irrev. process

Rev.: entropy change of heat reservoir: ∆Sheat res. = −∆QrevT

of H2O/ vapour: ∆SH2O+vap. = ∆QrevT

∆Stotal = 0

Irrev.: ∆Q < ∆Qrev; ∆Sheat res = −∆QT

∆SH2O+vap. = ∆QT + ∆Qrev−∆Q

T

∆Stotal = ∆Qrev−∆QT = ∆Sirrev > 0

for each possible process ∆SH2O+vap. = ∆QT + ∆Sirrev

2. Condensation, reversible:

∆SH2O+vap. = −∆QrevT

∆Sheat+res. = ∆QrevT

∆Stotal = 0

irreversible

∆SH2O+vap. = −∆Q

T︸︷︷︸leaving

+∆Q−∆Qrev

T︸︷︷︸heat created

∆SH2O+vap. = −∆QrevT

∆Stotal = ∆Q−∆Qrev

T ∆Q > ∆Qrev, > 0

for each poss.: process ∆SH2O+vap. = −∆QT + ∆Sirrev

1 30.10.10 36

important ∆S = SB − SA = ∆QT + ∆Sirr = ∆Qrev

T , indep. of the path/process.

S is state function, only measurable for reversible process

Reversible isothermal compression, ideal gas (1 mole)

eq. path (VA,T) → (VB,T)

isotermal comp ∆U = 0 ; ∆Wmax = ∆Qrev

∆W =∫ VBVA

PdV =∫ VBVA

RTV dV = RTlnVBVA < 0

entropy change of gas ∆Sgas = ∆QrevT = RlnVBVA < 0

∆Sheat res = −∆Sgas = RlnVAVB

Figure 1.12

Reversible adiabatic expansion, ideal gas

eq. path (PA, TA) → (PB, TB), ∆Q=0 if rev. proc. (quasistatic) and isolated (adia-batic)

∆S = 0, const. entropy PV γ = const γ = CPCV

= 52

∆Wmax = ∆U

1 30.10.10 37

Sudden increase of PA → PB from (PA, TA): System leaves the P-V-T surface, entropyis produced, TB > TB,rev

∆W = ∆U < ∆Wmax

Summary

1. Entropy S increases for irrev. process

2. ∆Srev = 0 (entropy is transfered from one part to another of the system)

3. S is a state function, S = ∆QrevT

Heat engines

Figure 1.13

efficiency

η = work obtainedheat input

= ∆W∆Q2

Studied by Sadi Carnot (1824)

1. A→B isoth. expansion at T2

2. B→C adiab exp. (rev.) T2 → T1

3. C→D isoth. compression at T1

4. D→A adiab. comp. T1 → T2

One cycle: ∆W = ∆W1 + ∆W2 −∆W3 −∆W4

∆Q = ∆Q2 −∆Q1

∆U = 0⇒ ∆Q = ∆W = ∆Q2 −∆Q1

η = ∆W∆Q2

= ∆Q2−∆Q1

∆Q2

two heat engines 1 and 2, η2 > η1 : reverse direction=heat pump

1 30.10.10 38

Figure 1.14

2: ∆W ′ = ∆Q2 −∆Q′11: −∆W ′ = −∆Q2 + ∆Q1

∆W ′ −∆W = ∆Q1 −∆Q′1 heat transferred directly to

work other changes (Perp. mobile of 2nd kind)

2: ∆W = ∆Q′2 −∆Q′11:−∆W ′ = −∆Q2 + ∆Q1

∆Q2 − ∆Q′2= ∆Q1 −∆Q′1︸︷︷︸

−(−∆Q1+∆Q′1)

= ∆Q heat flows from low

temp.T1 to high temp.T2 without other changes

2nd Law of Thermodynamics

There is no cyclic process to convert heat from reservoir directly into work without trans-ferring heat to a cold reservoir

- To transfer heat from cold to hot reservoir, some amount of work is necessary.

(discuss.) All rev. Carnot cycles (T2, T1) have the same efficiency (the max.) and depend

1 30.10.10 39

Figure 1.15

Figure 1.16

1 30.10.10 40

only on temperature

η = 1− ∆Q1

∆Q2:= 1− f(T1, T2)

∆Q1

∆Q2= f(T1, T2)

∆Q2

∆Q3= f(T2, T3)

∆Q1

∆Q3= f(T1, T3)

⇒ ∆Q1

∆Q3· ∆Q3

∆Q2=

∆Q1

∆Q2=f(T1, T3)

f(T2, T3)= f(T1, T2)indep. of

T3 ⇒ f(T1, T3) = F (T1)/F (T3)

⇒ ∆Q1

∆Q2=F (T1)

F (T2)

Kelvin:=

T1

T2

efficiency

η =∆Q2 −∆Q1

∆Q2=T2 − T1

T2

defines an absolute temperature scale (indep. of warley substance)(Kelvin scale)

to ideal gas (eq. of state)

A→ Bisoth. expansion ∆U = 0; ∆Q2 = ∆W1 = RT2lnVBVA

B → Cadiab exp. ∆Q = 0; ∆W2 = −∆U = −T1∫

T2

CV dT

C → Disoth. compression ∆U = 0; ∆Q1 = ∆W3 = RT1lnVDVC

D → Aadiab. comp. ∆Q = 0; ∆W4 = −T2∫

T1

CV dT

total work done by gas:

∆W = ∆W1 + ∆W2 + ∆W3 + ∆W4 = RT2lnVBVA

+RT1lnVDVC

total heat absorbed from hot reservoir

∆Q2 = RT2lnVBVA

within VBVA

= VCVD

(T · V γ−1 = const)⇒ ∆W = R(T2 − T1)lnVBVA

1 30.10.10 41

this η = ∆W∆Q2

= T2−T1T2

The second law of thermodynamics

Carnot process ∆Q2−∆Q1

∆Q2= T2−T1

T2⇔ ∆Q1

T1− ∆Q2

T2= 0

Any cyclic process can be broken down in a number of Carnot cyces for Zig-Zag path:∑ ∆QT = 0

∆Q > 0 heat entering, ∆Q < 0 heat leaving system

can be replaced in the limit by

∮δQ

T︸︷︷︸total diff. of state function

= 0

1. the entropy S, defined by dS = δQrevT is a function of state

(eg loop ABA∮dS = 0 =

B∫A

dS +A∫B

dS = (SB − SA) + (SA − SB) = 0)

2. For an adiabatically enclosed system, the entropy can never decrease

irreversible p. : ∆S > 0

reversible p. : ∆S = 0

Maximum work A→ B UB − UA = ∆W −∆Q

2nd Law limits the maximum amount of work extractable

dS =δQ

T+ dSirr, δQ = dU + δW

dS =dU + δW

T+ dSirr

δW = TdS − dU − TdSirrδW 6 TdS − dU

example: isothermal process, ∆Wmax = T (SB − SA)− (UB − UA) if proc. is revers.

S is state function→ ∆S it the sane if expansion is used to create work or free expanded

Equilibrium= reached. when entropy reaches a maximum value

in an isolated system of constant int. energy U and volume V, eq. is attained at maxi-mum entropy

1 30.10.10 42

example: reaction

A+B C+D

Figure 1.17

Figure 1.18

for closed system

dU = δQ− δW ; δW = PdV

dS =δQ

T⇒ dU = TdS − PdV (∗)

U = U(S, V )with dU = (∂U

∂S)V dS + (

∂U

∂V)SdV

1 30.10.10 43

compare to (∗)

T = (∂U

∂S)V temperature def.

P = −(∂U

∂V)S pressure def.

S = S(U, V )

dS =

(∂S

∂U

)

V

dU +

(∂S

∂V

)

U

dV(∗)=dU

T+PdV

T

(∂S

∂V)U =

P

T

Equilibrium occurs:

- in a system with fixed U and V, if entropy is maximized

- in a system with fixed S and V, if internal energy is minimized

Further development: S and V are inconvenient variables to be controlled in real world.

2 1.11.10

2.1 lecture 3

Entropy from statistical mechanics

Motivation: What is the physical significance of entropy? The second law of thermo-dynamics is only valid because of the failure of human ingenuity to invent a perpetualmachine. Thus, within the scope of classical thermodynamics, the second law is a ”Law”only because it has not yet been disproved. Physical interpretation of entropy came fromthe development of quantum theory and statistical mechanics. So, in the present classwe search an interpretation of entropy in the framework of statical mechanics.

Definition from Gibbs:Entropy is the ”degree of disorder” or the ”degree of mixedness” Correlation with macro-scopic picture:→ Melting of a substance −→ absorption of quantity of heat (Latent)

→ ∆q ⇒ ∆s = ∆qTm

=∆HfTm

∆sm↑ ⇒ ∆sm↑ atTm

Only true however for reversible process. For irreversible process the ”surroundings +system” should be consideredConcept of Microstate:1. Quantification of the idea of ”degree of disorder or “mixedness” In statistical me-chanics, ⇒ equilibrium state of the system in the ”most probable”of all its possible states.2. Energy is quantized. Particle in a box, has discrete values of energy. Spacing betweenthe discrete values of energy decreases and the energy space becomes continuous whenthe space available for the particle increases and the restriction placed on the positionof the particle reduces.3. Description of Microstates:. Example: consider a crystal with 3 non-identical particleslocated on different lattice sites. Suppose, for simplicity, that the quantization is suchthat the energy levels are equally spaced as below Fig.

EnergyLevelsEnergyLevels2.1.

Total energy U = 3u.Three possibilities exist.

2 1.11.10 45

Figure 2.1: Energy levels of the different possible lattice sites.EnergyLevels

Figure 2.2: The different possibilities for the arrangement of the particles in the different energylevels with the total energy 3u.Configurations

Possibility (a):-1 Probability of (a):-1

10

Possibility (b):-3 Probability of (b):-3

10

Possibility (c):-6 Probability of (c):-6

10

Each of the individual possibilities is a ”microstate” also called a ”complexion”, whichare shown in Fig.

ConfigurationsConfigurations2.2

4. Relation between ”microstate and macrostate”. The number of a arrangements withina given distribution Ω, that have ’n’ particles among energy levels n0 in ε0, n1 in ε1, n2

in ε2 and so on can be written as,

Ω =n!

n0!n1!n2!...nr!=

n!

Πri=0ni!

.

ln Ω =n lnn− n−i=r∑

i=0

(ni lnni − ni).

U =n0ε0 + n1ε1 + n2ε2....nrεr ← Macrostate property fixed

2 1.11.10 46

=i=r∑

i=0

niεi(Energy constant)

n =constant = n0 + n1 + n2...+ nr.

=i=r∑

i=0

ni(Particles Constant).← Macrostate property fixed

Any interchange of particles, among the energy levels must confirm with the condition

δU =∑

i

εiδni = 0 and δn =∑

i

δni = 0.

If Ω is maximum

δ ln Ω =−∑

(δni lnni +niδnini− δni).

=−∑

(δnilnni) = 0

1, 2, 3 need to be simultaneously satisfied for the most probable distribution. ”Methodof undetermined Coefficients”

∑βεiδni = 0;

∑αδni = 0.

1,2,3 together gives:-

i=r∑

i=0

(lnni + α+ βεi)δni = 0.

This needs to be true for all ”δni”; hence.

2 1.11.10 47

lnni + α+ βεi = 0.

ni = e−αe−βεi .

Summing over all ”r” energy levels.

n =i=r∑

i=0

ni = e−α

i=r∑

i=0

e−βεi

.

∑i=ri=0 e

−βεi is called the partition function P.

e−α =n

p.

ni =n

pe−βεi ; β is the Boltzmann constant =

1

KT

The distribution of the particles in the ”r” energy levels reduces exponentially with theenergy value of the ”r” energy level.

Variation with Temperature The relation the occupancy of the different energy levelsat different temperatures is shown in Fig.

nivsepsilonnivsepsilon2.3. At higher temperatures the occupancy

shifts towards higher energy levels.

Relation between macroscopic variables and microscopic property Ω.For large no of particles, the maximum possible distribution Ωmax Ω min and Ωmax∼

2 1.11.10 48

Figure 2.3: Plot showing the variation of the particles in the various energy levels at different

temperatures, through the choice of different values of β =1

KT.

nivsepsilon

Ω total.

ln Ωtotal = ln Ωmax = n lnn−∑

ni lnni.

β =1/KT.

ln Ωtotal =n lnn−∑ n

pe−εi/KT ln

(n

pe−εi/KT

).

=n lnn− n

p(lnn− ln p)

∑e−εi/KT +

n

pKT

∑εie

−εiKT .

But

U =∑

niεi =∑ n

pεie

−εiKT =

n

p

∑εie−εi/KT .

∑εie−εi/KT =

UP

n.

ln Ω = n ln p+U

KT.

Thermal. Equilibrium and Boltzmam Equation.

2 1.11.10 49

Consider equilibrium between reservoir and particles.U = U particles + U neat bathV = V particles + V heat bathAt Thermodynamic Equilibrium;

δ ln Ω =δU

KT

At constant n, P(εi, T )

at constant volume;

δU = δQ.

δ ln Ω =δQ

KT.

For reversible process; δs = δQKT = Kδ ln Ω.

Since S and Ω are state functions, the above expression can be written as a differen-tial equation, integration of S gives, S = K ln Ω. This is also known as configurationalentropy. An exemplary calculation of configurational entropy can be seen in the follow-ing:

The distribution of particles at the available sites can be visualized in the followingbinary alloy consisting of two types of atoms A and B as shown in the following:

Figure 2.4: Configurations of atoms before mixing.

During the mixing process: A+B (divided) → A/B mixed, different possibilities exist.For example, when there is no mixing, there is just one possible state written as Ω4 :0 = 1. With one atom transfer, the number of possibilities of arrangement on each sideof XY plane is 4 and hence, the total number of possibilities of such an arrangementis 16, given by Ω3 : 1 = 16. With an exchange of two atoms, we have six possibilities,which gives a total of 36 possibilities Ω2 : 2 = 36. Finally with 3 atoms exchanged,

2 1.11.10 50

we have a total of 16 possibilities Ω1 : 3 = 16. Among the 70 possibilities, the onewith highest probability is Ω2 : 2 = 36 as this maximizes the entropy and hence is theequilibrium state of the system. From the above discussion, we derive that the totalnumber of possibilities for a given configuration of na:A atoms and nb:B atoms can be

given by(na + nb)!

na!nb!. The change in the configurational entropy going from a non-mixed

state can be calculated as:

∆Sconf = Sconf2 − Sconf1 =K ln Ωconf2 −K ln Ωconf1

=K lnΩconf2

Ωconf1

.

=K ln(na + nb)!

na!nb!

Other potentials:

dU = TdS − PdV +∑

i

(∂U

∂ni

)

s,v,nj

dni. U

dH = TdS − V dP +∑

i

(∂H

∂ni

)

s,p,nj

dni. H = U + PV

dF = −SdT − PdV +∑

i

(∂F

∂ni

)

T,V,nj

dni. F = U − TS(dF |T,V,nj = −

(∂F

∂ni

)

T,V,nj

dni = dU |T,V,nj − TdS|T,V,nj =

(∂U

∂ni

)

S,V,nj

).

compare, with (*)

µi =(∂G∂ni

)T,P,nj

=(∂U∂ni

)S,V,nj

=(∂H∂ni

)S,P,nj

=(∂F∂ni

)T,V,nj

Complete set of the potentials:

2 1.11.10 51

dU = TdS − PdV +∑

i

µidni

dH = TdS + V dP +∑

µidni

dF = −SdT − PdV +∑

µidni

dG = −SdT + V dP +∑

µidni

fundamental relation

U is characteristic function of the independent variables S, V and nj . First law:dU = δQ− δW with δQ = TdS(reversible)⇒ δW = PdV −∑µidni

Thermodynamic relations (from (1-4))

T =

(∂U

∂S

)

v,nj

=

(∂U

∂S

)

P,nj

.

P = −(∂U

∂V

)

s,nj

= −(∂F

∂V

)

T,nj

.

V =

(∂H

∂P

)

s,nj

=

(∂G

∂P

)

T,nj

.

S = −(∂F

∂T

)

v,nj

= −(∂G

∂T

)

P,nj

.

Maxwell relations

Z is a state function of independent variables X and Y :Z=Z(X,Y)

dZ =

(∂Z

∂X

)

Y

dX +

(∂Z

∂Y

)

X

dY

=L(X,Y )dX +M(X,Y )dY

2 1.11.10 52

order of differentiation doesn’t make a change:

[∂

∂Y

(∂Z

∂X

)

Y

]

X

=

[∂

∂X

(∂Z

∂Y

)

X

]

Y

=∂2Z

∂X∂Y

⇔(∂L

∂U

)

X

=

(∂M

∂X

)

Y

applied to (1)-(4): Maxwell relations:

(∂T

∂V

)

S

= −(∂P

∂S

)

V

.

(∂T

∂P

)

S

=

(∂V

∂S

)

P

.

(∂S

∂V

)

T

=

(∂P

∂T

)

V

.

(∂S

∂P

)

T

= −(∂V

∂T

)

P

.

Example 1 Entropy of ideal gas S = S(T, V )

dS =

(∂S

∂T

)

v

dT +

(∂S

∂V

)

T

dV

heat capacity at const.V : TdS = ∂Qv = dU = nCvdT

⇒(∂S

∂T

)

v

=nCvT

;

(∂S

∂V

)

T

=

(∂P

∂T

)

v

=nR

V

dS =nCvT

dT +nR

VdV

S2 − S1 = nCv lnT2

T1+ nR ln

V2

V1

is-entropic processT2

T1=

(V1

V2

) RCv

2 1.11.10 53

Example 2 closed system / fixed comp.

dU = TdS − PdV(∂U

∂V

)

T

= T

(∂S

∂V

)

T

− P = T

(∂P

∂T

)

V

− P

equation of state, relation of U to measurable quantities P, V, T

Upstairs - downstairs-inside- out formula

three state variables X,Y,Z with X=X(Y,Z)

dX =

(∂X

∂Y

)

Z

dY +

(∂X

∂Z

)

Y

dZ

forX = const :

(∂X

∂Y

)

Z

dY = −(∂X

∂Z

)

Y

dZ

(∂X

∂Y

)

Z

(∂Y

∂Z

)

X

= −(∂X

∂Z

)

Y(∂X

∂Y

)

Z

(∂Y

∂Z

)

X

(∂Z

∂X

)

Y

= −1

Phase Equilibrium in a one component system

two phase in contact: open to was exchangeintensive Thermodynamic properties of a system are ”temperature” ”pressure” and”chemical potential”. There properties are measures of potentials, spatial gradientsof whichTemperature⇒ Potential of heat→ decides direction of heat fluxPressure⇒Potential for expansions or contractionchemical potential⇒Tendency of species i to leave the system. Measure of the ”chemicalpressure” exerted by ’i’ in the phase → direction of diffusive mass flux

2 1.11.10 54

Variation of Gibbs free energy as Temperature

dG = V dP − SdT(∂G

∂T

)

P

= −S;

(∂2G

∂T 2

)

P

= −(∂S

∂T

)

P

= −CPT

Figure 2.5

Phase transition dG<0fixed -S:= Slope of the line G(T).First order transition: There is a change in the slope of the line as there is a phasetransition Sl 6= Ss

Second order Transition: Slopes are continuous; but a discontinuity in the second deriva-

tive(∂2G∂T 2

). C l1P 6= C l2P Example:-Glass-Transition

3 1.11.10

3.1 Lecture 4

Gibbs Free energy as a function of temperature and pressure.

Consider equilibrium between solid and liquid. Gs = Gl

For infinitesimal changes: dGs = dGl

dGs = −S(s)dT + V(s)dP

dGl = −SldT + VldP

−S(s)dT + V(s)dP = −SldT + VldP

V(s) − V(l)dP = (S(s) − S(l))dT(dP

dT

)

eq.

=∆S

∆V=

(Ss − SlVs − Vl

)

at eq.: ∆S =∆Hf

T

Clausius Clapeyron equation:

(∂P

∂T

)

eq.

=∆Hf

T∆V

This gives the dependence of pressure on temperature (or the relation between the two)to maintain equilibrium. Alternatively this defines the properities of the equilibrium linedenoting the co-existence between the solid and liquid phase.

Behavior of Gases, solutions, Phase-Diagrams:

P:-Pressure of Gas.

V:-Molar Volume.

R:-Gas constant.

3 1.11.10 56

T:-Absolute temperature

(pV

RT→ 1

)

P→0

Real gases

Mixture of Ideal Gases - (a model for solubility and mixing)

How do we treat a mixture of gases? For a pure gas:- what is the relation of Gibbs freeenergy and pressure in a closed isothermal system given by dT=0, (Isothermal changefrom P1 → P2)

dG = V dP.

For 1 mole of “ideal” gas,

dG =RT

PdP = RTdlnP.

G(P2, T )−G(P1, T ) = RTlnP2

P1− (I)

G = G0 +RTlnP

P0

G0 → (Free energy of 1 mole, P0= 1 atm “Standard state”)

Mixture of gases:

To describe a mixture of gases we need to derive the concept of “mole fraction”, “partialpressure” and “partial molar quantities”.

Mole fraction:

XA = nAnA+nB+nC

XB = nBnA+nB+nC

; XC = nCnA+nB+nC

XA +XB +XC = 1 (representing composition)

Partial pressure: PA:-Pressure exerted by component A, if it were alone present in acontainer,

Total pressure P = pA + pB + pC

3 1.11.10 57

pA =nART

V ′

pB =nBRT

V ′

P = pA + pB =(nA + nB)RT

V ′

Dalton’s law of partial pressures pA = XAP

Partial molar quantities

Partial molar value of on extensive quantity Q, in a mixture i, j, k denoting (compo-nents),

Qi =

(∂Q′

∂ni

)

T,P,nj ,nk

= rate of change of Q′ with ni at T, P, nj , nk = const.

Q′:- Value of Q for an arbitrary quantity system size of mixture for Q ≡ G.

Gi =

(∂G′

∂ni

)

T,P,nj ,nk

= µi(chemical potential of the component in mixture)

“Maxwell relations are also applicable for partial molar quantities”from fundamentalrel.

(∂G′

∂P

)

T,comp

= V ′

(∂

∂ni

(∂G′

∂P

)

T,comp

)

T,P,nj ,nk

= V

=

(∂

∂P

(∂G′

∂ni

)

T,P,nj

)

T,comp

=

(∂Gi∂P

)

T,

Solutions

γe(A):- equilibrium rate of evaporation (A).

3 1.11.10 58

Figure 3.1: Example of mixture of gas molecules A and B.

In pure A, we have the activation energy as EA the rate of evaporation is proportionalto ∼ e(−EA

KT ).

(1) γe(A) = kp0A := p0

A (saturated vap. pressure)

Similarly in pure B we have the rate of evaporation (B) given by - γe(B),

γe(B) = k′p0B,

where k, k′ are the rate constants related to the process of evaporation.

In a mixture, the rate of evaporation of A is reduced by XA (the mole fraction). (withthe assumption of identical bond energies A-A, A-B, B-B and composition of suface likein bulk) while,

(2)γe(A)XA = kpA

γe(B)XB = k′pB

giving us the Raoult’s Law pA = XAp0A pB = XBp

0B .

Gibbs duhem Equation

How are partial molar quantities of components in a system related?

3 1.11.10 59

Let Q’ be an extensive property

Q′ = Q′(T, P, ni, nj , nk...)

at const. T,P

dQ′ =

(∂Q′

∂ni

)

T,nj ,nk

dni +

(∂Q′

∂ni

)

T,nj ,nk

dnj +

(∂Q′

∂nk

)

T,ni,nj

dnk...

dQ′ = Qidni +Qjdnj +Qkdnk...(1)

also from definition:

Q′ = Qini +Qjnj ...

dQ′ = (Q1dni +Qjdnj + ...) + dQini + dQjnj + ...)(2)

(1) and (2) gives.

∑

i

dQini = 0 ; or dividing by no. of moles of system⇒∑

i

dQixj = 0

Binary solution(A,B) if Q′ ≡ G′; T, P =const:

G′ = nAGA + nBGB G′ : −Gibbs energy

(1)G ≡ XAGA +XBGB G = molar Gibbs energy

Gi = Excess enthalpy

Gibbs - Duhem:

XAdGA +XBdGB = 0

dG = GAdXA +GBdXB

dG

dXA= GA −GB

(2) XBdG

dXA= XBGA −XBGB

(1) + (2) G+XBdG

dXA= GA; GB = G+XA

dG

dXB

Properties of “Raoultion” “Ideal” Solutions

3 1.11.10 60

Figure 3.2: Definition of partial molar Gibbs-free energies given the free energy at a givencomposition

Some terminology

ai(activity of component i) =fif0i

When the vapor in equilibrium is “ideal”

ai =pip0i

....(II)

In addition if the component i exhibits Raoultion bahavior ai = Xi.

Change in Gibbs-free energy due to the formation of a solution: (A,B)

Deriving the free energy change in a mixture of gases (I)

∆Gi = Gisolution −Gi(pure) = RTlnpip0i

= RTlna0i

= Gi −G0i = RTlnai = RTlnXi

Free energy before mixing:-(na : A,nB : B)

nAG0A + nBG

0B

3 1.11.10 61

Free energy after mixing:-nAGA + nBGB

Free energy change accompanying mixing:

∆G′ = nA(GA −G0A) + nB(GB −G0

B).

= nA∆GA + nB∆GB.

= nARTlnaA + nBRTlnaB.

= RT (nAlnaA + nBlnaB).

For Ideal Solution

∆G′ = RT (nAlnXA + nBlnXB).

for a mole

∆G = RT (XAlnXA +XBlnXB).

Figure 3.3: Concept of the change in free energies

Change in volume of an ideal Solution.

3 1.11.10 62

Using:-

(∂G

∂P

)

T,comp

= V

(∂Gi∂P

)

T,C

= Vi;

(∂G0

i

∂P

)

T,C

= V 0i

(∂(Gi −G0

i )

∂P

)

T,C

= Vi − V 0i

(∂∆Gi∂P

)

T,C

= Vi − V 0i

∆Gi = RTlnXi ∴

(∂∆Gi∂P

)

T,C

= 0 or V i − V 0i = 0

⇒ ∆Vi = 0

From this it follows:-

change of volume is zero in ideal mixing

⇒ ∆V ′M = nA∆V A + nB∆V B

∆V A = 0 and ∆V B = 0

Heat of formation of an ideal solution.

[∂Gi/T

∂T

]

P,comp

= −H i

T 2;

[∂G

0i /T

∂T

]

P,comp

= −H0i

T 2

−(H i −H0i )

T 2= −∆H i

T 2=

∂(Gi−G0

i )T

∂T

=

[∂∆G

i/T

∂T

]

⇒ ∂(RlnXi)

∂T=

∆Hi

T 2= 0

∆H ′ = nA∆HA + nB∆HB = 0 ⇒ ∆HA → 0 and ∆HB → 0.

Entropy of formation

3 1.11.10 63

(∂G

∂T

)

P,comp

= −S.(∂∆GM

∂T

)

P,comp

= −∆SM = −R(XAlnXA +XBlnXB) · · · III

From the definition of configurational entropy.

∆S′ = k ln(NA +NB)!

NA!NB!

= k[ln(NA +NB)!− lnNA!− lnNB!]

Apply stirlings’s approximation

∆S′ = k[(NA +NB)ln(NA +NB)−NA −NB.

−NAlnNA +NA −NBlnNB +NB]

= −k[NAln

(NA)

(NA +NB)+NBln

NB

(NA +NB)

]

NA

NA +NB=

nAnA + nB

= XA.

NA :NA

N0moles = nA; NB :

NB

N0moles = nB.

↓Avogadro no

∆S′ = −kN0(nAlnXA + nBlnXB)

1 mole:- ∆S = −R(XAlnXA +XBlnXB) · · · IV

III ≡ IV !!!

The thermodynamic activity of a comp in solution

at temperature T: activity is the ratio of fugacity of a substance

ai =fif0i

at T

for solution↔ vapor fi = Pi ⇒ ai =PiP 0i

Raoultian behaviour ai = Xi

3 1.11.10 64

Henry behaviour ai = kiXi

The Gibbs-Duhem equation

[∂(Gi/T )

∂T

]

P,comp

= −Hi

T 2

G = H − TS

= H + T

(∂G

∂T

)

P

G

T=H

T+

(∂G

∂T

)

∂(GT )

∂T= −H

T 2+

1

T

(∂H

∂T

)

P

+∂2G

∂T 2

∂2G

∂T= −∂S

∂T

S =dH

T∂S

∂T= − 1

T

∂H

∂T

∂(GT )

∂T= −H

T 2

Gibbs-Helmoltz relation.

Gibbs-Phase Rule.

* N phases, K componentsNK→ degrees of freedom(N-1)K→ Equations of equilibriumNK-(N-1)K=KN→ Equations of sum constraintF=K-NFor Non, constant T,P F=K-N+2

CAGαA + (1− CA)GαB +RT (CAlnCA + CBlnGB)

G = CA(GαA +RTlnCA) + (1− CA)(GαB +RTln(1− CA)).

3 1.11.10 65

≡ CAGAα + (1− CA)GBα

GαA = GαA +RTlnCA

GAB = GαB +RTlnCB

4 1.11.10

4.1 lecture 5

Figure 4.1: A solution is different from a mixture of pure A and pure B. Therefore for a com-postion of ‘XA‘ in binary system A, B,

G0AXA +G0

B(1−XA) 6= G′(XA).

G′(XA) = GAXA +GBXB.

∆G(XA) = (GA −G0A)XA + (GB −G0

B)XB.

= ∆GAXA + ∆GBXB.

Where ∆GA is the partial molar Gibbs free energy change of the system for the compo-nent A.

For Ideal Solution (“Raoultian“)

∆GA = RTlnXA.; ∆GBRTlnXB.

∆GMIX(XA) = XARTlnXA +XBRTlnXB

4 1.11.10 67

Figure 4.2

Free energy change of entropy of missing ∆S(X)

∆SMIX = −R(XAlnXA +XBlnXB)

Maximum randomners for ”equal“ mole fractions of components A, B. This holds also

Figure 4.3

for more than 2 components (A,B,C...)⇒ (XA = XB = XC ...)

4 1.11.10 68

To get the free energy of the system,

G = G0AXA +G0

BXB + ∆GAXA + ∆GBXB.

Nobody can (calculate)measure ”absolute“ ’G’, but only change. So one would need a ”reference“

state. For the discussion, we fix the reference state as T0 = 298K, P=1 atm. becausemost measurements are done at 1 atm.

Then at const P=1 atm.

∆G = XA

∫ T

T0=298dG0

A +XB

∫ T

T0=298dG0

B + ∆GAXA + ∆GBXB.

G = H − TS.∆GS = ∆HS − T∆SS (Free energy change)

= XA

∫ T

T0=298CpSAdT +XB

∫ T

T0=298CpSBdT − T

∫ T

T0=298

CpSAT

dT −∫ T

T0=298

CpSBT

dT.

∆Gl = ∆H l − T∆Sl.

= XA

∫ Tm

T0=298CpSAdT +XB

∫ Tm

T0=298CpSBdT − T

∫ Tm

T0=298

CpSAT

dT − T∫ Tm

T0=298

CpSBT

dT.

+(∆HAf −

T∆HAf

Tf)XA + (∆HB

f −T∆HB

f

TBf)XB.

∆Gl = XA

∫ T

T0=298CplAdT +XB

∫ T

T0=298CplBdT − TXA

∫ T

T0=298

CplAT

dT − TXB∫ T

T0=298

CplBT

dT

+(∆HAf −

T∆HAf

Tf)XA + (∆HB

f −T∆HB

f

TBf)XB

= X

∫ T

TmC lV dT − T

∫ T

Tm

C lVT

dT

d+ ∆Hf − T

∆Hf

Tf.

∆Gl = XA

∫ T

TmCplAdT +XB

∫ T

TmCplBdT −XAT

∫ T

TmCplA

dT

T−XBT

∫ T

TmCplB

dT

T

+(∆HAf −

T∆HAf

Tf)XA + (∆HB

f −T∆HB

f

Tf)XB.

∆GS = XA

∫ T

TmCpSAdT +XB

∫ T

TmCpSBdT −XAT

∫ T

TmCpSA

dT

T−XBT

∫ T

TmCpSB

dT

T.

∆Gl −∆GS .

XA

∫ l

Tm∆CpAdt−XAT

∫∆CplA

dT

T⇒ 0 for ∆CplsA = 0 for ∆CplsB = 0.

4 1.11.10 69

XB

∫ l

Tm∆CpBdT −XBT

∫∆CpTB

dT

T⇒ 0.

∆Gl→s = ∆HAf

(Tf − TTf

)XA + ∆HB

f

(Tf − T )

TfXB.

Taking liquid as reference.

∆GS : ∆HAf

(T − Tf )

TfXA + ∆HB

f

(T − Tf )

TfXB.

∆Gl = 0.

∆GS = XA∆HAf

(T − TAf )

TAf+XB∆HB

f

(T − TBf )

TBf+ ∆G

SAXA + ∆G

SBXB.

∆Gl = ∆GlAXA + ∆G

lBXB.

Points to note:

∆HAf : - Latent heat of fusion. for S → L transition. In principle, the latent heat of

transformation will then depend on the solid phase and the liquid phase.

∴ We name ∆Hf ≡ LαA ≡ α→ L transformation for pure A.

What is the latent heat of transformation for an alloy of comprosition ’XA‘?We can analogously derive as for the case of pure components at equilibrium;

∆GS−L = 0.

∆HS−L = T (∆SS −∆SL).

∆SS = −∂∆GS

∂T.

∆Sl = −∂∆Gl

∂T.

LS−L = − 1

T

(∂∆GS

∂T− ∂∆Gl

∂T

)

XA

.

How do we ”geometrically“ see the equilibrium between tow phases in a binary system?

∆GSA = ∆G

lA

∆GSB = ∆G

lB

4 1.11.10 70

Figure 4.4

Thus, for a binary system (A,B) phase equilibrium between two phases (α, l) implies thepartial molar Gibbs free energy of each component in each phase is equal.

≡ Equivalently, the compositions of the phases ’α’ and ’l’ at which these conditions holdcorrespond to the common tangent constructions.

For a system Ni-Cu:

Can we derive the phase-diagram?

∆Gα = XαAL

α (T − TαA)

TαA+Xα

BLαB

(T − TαB)

TαB+RT (Xα

AlnXαA +Xα

BlnXαB)

∆Gl = RT (X lAlnX

lA +X l

BlnXlB)

At a Temperature ’T’

∆GαA = ∆G

lA.

LαA(T − TαA)

TαA+RTlnXα

A = RTlnX lA...(1)

LαB(T − TαB)

TαB+RTlnXα

B = RTlnX lB...(2)

For Ni-Cu:

4 1.11.10 71

LαNi = 2310 J/cm3 LαCu = 1728 J/cm3.R = 8.314 J/Kmol v = 7.42 cm3.

Figure 4.5

For phase-field purposes we always take free energy density so if we divide the free energyby the molar volume we get the free energy pure unit volume. The equations for thefree energies per-unit volume become,

LαA(T − TαA)

TαA+RT

VmlnXα

A =RT

VmlnX l

A

LαB(T − TαB)

TαB+RT

VmlnXα

B =RT

VmlnX l

B.

We can now non-dimensionalize our system by dividing throughout withRTαAVm

and choos-ing TαA as reference temperature

LA(T − TαA)

TαA+ T lnXα

A = T lnX lA

LB(T − TαB)

TαB+ T lnXα

B = T lnX lB

lnXαA = lnXα

n − LA(T − TαA)

T TαA

4 1.11.10 72

= lnXαA − LA

(T − TαA)

T TαA.

Similarly;

ln(1−XαA) = lnXα

B − LB(T − TαB)

T TαB

XαA

(1−XαA)

=exp

(lnX l

A −LA(T−TαA)

T TαA

)

exp(lnX l

B −LB(T−TαB)

T TαB

)

XαA =

exp(lnX l

A −LA(T−TαA)

T TαA

)

1 + exp(lnX l

B −LB(T−TαB)

T TαB

)

Equivalently the liquidus can be written as:

X lA =

exp(lnXα

A +LA(T−TαA)

T TαA

)

[1 + exp

(lnXα

B +LB(T−TαB)

T TαB

)]

Slopes of solidus and liquidus.

I− LαA(T − TαA)

TαA+ T lnXα

A = T lnX lA

II− LαB(T − TαB)

TαB+ T lnXα

B = T lnX lB

XαA +Xα

B = 1

X lA +X l

B = 1

Variable T ,XαA, X

αB, X

lA, X

lB

4 1.11.10 73

There are four equations V Degree of freedom 5-4=1.

We choose X lB a8 our independent variable.

Differentiate I

LαAT 2

(∂T

∂X la

)+

1

XαA

∂XαA

∂X lB

=−1

X lA

III XαA

LαAT 2·(∂T

∂X lB

)+∂Xα

A

∂X lB

= −XαA

X lA

Similarly for II

IV XαB

LαBT 2·(∂T

∂X lB

)+∂Xα

B

∂X lB

=XαB

X lB

Adding III and IV;

∂XαA

∂X lB

+∂Xα

B

∂X lB

= 0

1

T 2

∂T

∂X lA

(XαAL

αA +Xα

BLαB) =

XαB

X lB

− XαA

X lA

∂T

∂X lB

=

XαB

XlB

− XαA

XlA

)

1T 2

(XαAL

αA +Xα

BLαB

)

Similarly for solidus

∂T

∂XαB

=−(

XlB

XαB− Xl

AXαA

)

1T 2

X lAL

αA +X l

BLαB

5 1.11.10

5.1 lecture 6

Phase-field Evolutions:

1. Van-der Waals: - 1893 (Density variations as a functions of ‘h‘

h→ distance along gas-liquid interface.

Introduction of square gradient energy term).

2. Ginzburg-Landau: - 1920 (Free energy of a non-uniform system).

Cahn-Hilliard: - 1958 (Free energy of a non-uniform system).

3. Allen, Langer: - First phase-field formulations of dynamics

in phase transitions (solidification,Langer).

Free energy of a non-uniform system:

Postulate: - The free energy of a system dependent on an intensive property of thesystem depends on the local value and on its variation a cross a system (gradients).

f0: - Equilibrium energy of a uniform system at ‘c‘.

f(c,∇c,∇2c · · · )Expanding ‘f ‘ around ‘f0‘ as a Taylor series would allow us to write ‘f ‘ as a function ofcompositions and its gradients. Since ‘f ‘ is a scalar, its value should be independent ofthe directions of the gradient: - hence only even powers of the gradient can occur in theexpansion.

f(c,∇c,∇2c) = f0(c) + k1∇2c+ k2(∇c)2 · · ·

Integrating over a volume,

F = NV

∫

VfdV = NV

∫

V(f0(c) + k1∇2c+ k2(∇c)2 · · · )dV.

5 1.11.10 75

NV : - Number of molecules per unit volume.

∫

Vk1∇2cdV = −

∫

V

∂k1

∂c(∇c)2dV +

∫

Sk1(∇c · n)dS

One can always choose a surface such that the second integral goes to zero.

F = NV

∫

Vfo(c) +

(k2 −

∂k1

∂c

)(∇c)2dV ;

(k2 −

∂k1

∂c

)⇒ k

F = Nv

∫

v(fo(c) + k(∇c)2)dV

For convenience, and in cases where partial molar volumes of components are equal, onecan write ‘f0‘ also as a energy E/vol. such that,

F =

∫

V(f ′o(c) + k′(∇c)2)dV

‘f ′0‘ and k′(∇c)2 are quantities per unit vol. of the system.

What is the interfacial energy with this definition of energies?

Interface energy of the system by definition is the difference per unit area of the actualfree energy of the system with the interface and what the system would have, if it werecontinuous.

The ”continuous” term is uniquely defined at equilibrium between the phases α and β.

∴ ‘σ‘ can be written as: -

F = A

∫

x[f ′o(c) + k′(∇c)2 − (cµB(e) + (1− c)µA(e))]dx.

σ =

∫

x[f ′o(c) + k′(∇c)2 − (cµB(e) + (1− c)µA(e))]dx.

f ′o(c) = cµA + (1− c)µB.

σ =

∫

xµB − µB(e)) + (1− c)(µA − µA(e)) + k′(∇c)2dx.

∆f(c) = c(µB − µB(e)) + (1− c)(µA − µA(e)).

∆f(c) > 0.

5 1.11.10 76

σ =

∫

x∆f(c) + k′

(∂c

∂x

)2

dx.

Figure 5.1

If one reduces k′( ∂c∂x)2 by making the interface broader if will be at the expense of incor-porating more material at the interface, such that ∆f(c) ↑(goes up). On the other hand,reduction of material at the interface increases the gradient. A compromise between twoterms gives an equilibrium interface.

Concept of variational derivative/(Derivation of the Euler -Lagrange equation).

F =

∫f(c,∇c)dV

F (c+ δc) =

∫ (f +

∂f

∂c· δc+

∂2f

∂∇c · δ∇c)dV

F =

∫ (∂f

∂c· δc+

∂f

∂∇c · δ∇c)dV

Weak approximation δ∇c = ∇δc

δF =

∫ [∂f

∂c· δc+

∂f

∂∇c · ∇δc]dV

=

∫

V

∂f

∂c· δcdV +

∫

S

(∂f

∂∇c · n)dSδc−

∫∇ · ∂f

∂∇cδcdV

δF =

∫

V

∂f

∂c· δcdV −

∫

S∇ · ∂f

∂∇cδcdV

5 1.11.10 77

The above is obtained by choosing a surface over which the surface integral goes to zero,this is always possible.

‘δc‘ is a delta function around ‘c‘ where we compute the variation. Using the propertyof ‘δ‘ function

∫ ∞

−∞f(x)δ(x− x0)dx = f(x0)δx

δF =

(∂f

∂c

)

c0

−(∇ · ∂f

∂∇c

)δc.

δF

δc=∂f

∂c−∇ · ∂f

∂∇c

δF

δc=

[∂

∂c−∇ · ∂

∂∇c

]f

we name this operator : -δ

δc=

∂

∂c−∇ · ∂

∂∇c

.

Using this operator on,

σ =

∫

x

[∆f(c) + k′

(∂c

∂x

)2]dx.

∂∆f

∂c= 2k′

∂2c

∂x2.

∫∂∆f

∂c· dcdx

=

∫2k′

∂2c

∂x2· dcdx.

∆f = k′(∂c

∂x

)2

+ const.

At x = ±∞(∂c

∂x

)2

= 0 and ∆f = 0.

Therefore: const = 0

5 1.11.10 78

σV ∆f = k′(∂c

∂x

)2

: −Equipartition of energy

∴ σ =

∫

x∆f(c) + k′

(∂c

∂x

)2

dx.

= 2

∫

xk′(∂c

∂x

)2

dx.

= 2

∫

xk′(∂c

∂x

)· ∂c∂xdx ≡ 2

∫

xk′√

∆f/k′dc.

σ = 2

∫

x

√k′∆fdc.

Also∆f

k′=

(dc

dx

)2

√∆f/k′ =

dc

dx

dx =dc√

∆f/k′∫ λ

ox =

∫ cβ

cα

dc ·√k′√

∆f

λ =

∫ cβ

cα

√k′ · dc√∆f

.

λ is the interface width.We see - ‘λ‘ is inversely proportional to

∫ cβ

cα

dc√∆f(c)

6 1.11.10

6.1 lecture 7

In the Cahn-Hilliard approach, the surface energy ‘σ‘ is fixed by the form of the freeenergy. While this is true for certain systems, in general the surface excess is a propertyindependent of the free energy parameters. Hence one would like to set this propertyindependent of the bulk system.The way to do this, is to choose two fields. 1) denotingthe volume fraction of a phase and the second 2) the mole fraction of the components.Let us say the volume fraction corresponding to a phase is φ. Then in a two-phasesystem, the second phase has volume fractions (1-φ). This system can have minimum oftwo components. We denote A,B, with mole fractions cA and cB such that cA + cB = 1in each phase and we can denote the ‘concentration‘ field in the system as ‘c‘.

Following the same principles to write the energy of the system as the Cahn-Hilliardequation, we can write,

F =

∫ [f(c, φ) + kφ|Oφ|2

]dx.

Points to note:

1) We have chosen the energy functional to be independent of the gradient in the concen-tration. This is true for small concentration gradients and corresponding small gradientenergy coefficients for the concentration field.

2) kφ, is the gradient energy coefficient for the gradients in the order parameter ‘φ‘. Thiscan be tuned to set the right surface energies.

3) f(c,φ) can be broken down into: -

fb(c, φ) + w(φ).

6 1.11.10 80

where w(φ) denotes an equivalent term to ∆f(c) in the Cahn.Hilliard equation.

Figure 6.1

To define our surface energy we would need to define our energy of a uniform homoge-neous system; consisting of a mixture of two phases. This can be done similar to whatwe have encountered in the case of the Cahn-Hilliard equation.

Figure 6.2

∴ ∆f(c, φ) = f(c, φ)− (fαeq(c) + (1− c)f leq) + w(φ).

Or alternatively in terms of the Grand Chemical potential or the partial molar Gibbsfree energy of the system, we can write;

∆f(c, φ) = f(c, φ)− (µeqA c+ µeqB (1− c)) + w(φ).

6 1.11.10 81

Then, the surface energy reduces to:

σ =

∫ ∞

−∞

[f(c, φ)− (µeqA c+ µeqB (1− c))] + w(φ) + kφ|Oφ|2

dx.

Notice, we have not lost the dependence of ‘σ‘ on the free energy density of the phases.However, there exists now another tunable function w(φ) to get the surface energies thatis representative of an interface.The extremizer of the above function would be the Euler-Lagrange equations, we hadfor the Cahn-Hilliard equation.

δσ

δφ=∂f(c, φ)

∂φ+ w′(φ)− 2kφ

d2φ

dx2= 0 · · · (1)

δσ

δc=∂f

∂c− (µB − µA) = 0.

⇒ ∂f

∂c= (µB − µA) = µeq ⇒ chemical potential of the system.

∂f(φ, c)

∂c= (µB − µA) · · · (2).

To integrate (φ) we need to write (∂f(c,φ)∂φ ) in terms of the total derivative of ‘φ‘.

∂f

∂φ=df

dφ− df

dc· dcdφ

From (2) the minimizing condition for∂f

∂c= (µB − µA) = µeq.

∴∂f

∂φ=df

dφ− µeq ·

dc

dφ⇒ d

dφ(f − µc).

Equation (1) becomes

d

dφ(f − µc) + w′(φ)− 2kφ

d2φ

dx2= 0.

d

dφ(f − µc) + w′(φ) = 2kφ

d2φ

dx2.

Multiplying

(dφ

dx

)on both sides.

6 1.11.10 82

2kφd2φ

dx2

dφ

dx=

d

dφ(f − µc) · dφ

dx+ w′(φ) · dφ

dx.

Integrate both sides.

kφ

(dφ

dx

)2

= (f − µc)− (f(0)− µc(0)) + w(φ).

We denote F (c, φ) = (f − µc).

kφ

(dφ

dx

)2

= F (c, φ)− F (c(0), 0) + w(φ).

(dφ

dx

)=

1√kφ

√F (c, φ)− F (c(0), 0) + w(φ).

The surface energy ‘σ‘ can then be derived using,

F (c, φ)− F (c(0), 0) = f(c, φ)− (f(0) + µepc− µc(0))

= f(c, φ)− (µeqA c+ µeqB (1− c)).

∴ σ = 2

∫ ∞

−∞kφ

(dφ

dx

)2

dx ≡ 2

∫ 1

0kφ

(dφ

dx

)dφ

= 2

∫ 1

0kφ ·

1√kφ

√F (c, φ)− F (c(0), 0) + w(φ) · dφ.

= 2

∫ 1

0

√kφ√F (c, φ)− F (c(0), 0) + w(φ) · dφ · · · (3)

The interface width can also be derived as;

dφ

dx=

1√kφ

√F (c, φ)− F (c(0), 0) + w(φ).

∫ 1

0

√kφdφ√

F (c, φ)− F (c(0), 0) + w(φ)= λ · · · (4)

Notice we can also use (2) to derive the equilibrium concentration profile.

∂f

∂c(φ, c) = (µB − µA). Hence c can be written as a function from‘φ‘.

In systems, where the free energy of the (phases) are closely related the term

F (c(φ))− F (c(0), 0)

is generally small, which implies equations (3) and (4) can be simplified to,

6 1.11.10 83

Figure 6.3

σ = 2

∫ 1

0

√kφ√w(φ) · dφ · · ·(5) σα

√k · √w

λ =

∫ 1

0

√kφ√w(φ)

· dφ · · ·(6) λα

√k√w

σλαkφ · · ·(7) kασλ

σ/λαw · · ·(8) wασ/λ

Therefore one can propose to write down F as;

F =

∫

Ω[fb(c, φ) + σλ|Oφ|2 +

σ

λw(φ)]dΩ.

where w(φ) is dimensionless term.

We will use this formulation of the model throughout in our discussion.

Points to discuss:1)what is the form of w(φ)?If we see (5) and (6) and use (7) and (8) we get;

σ =2σ

∫ 1

0

√w(φ)dφ.

∴ we want 2

∫ 1

0

√w(φ)dφ = 1.

6 1.11.10 84

Posible constructions are;

w(φ) =16

π2φ(1− φ) (Obstacle) and 9φ2(1− φ)2 (Double-well)

Integrate and verify the ansatz

DynamicsWe have two fields; c, φ in our system, while ‘c‘ is a conserved quantity, ‘φ‘ is non-conserved as we have transition from one phase to another.The field ‘c‘ evolves through ”gradient dynamics”. This implies the driving force for fluxis the gradient of a potential.Conservation law for ‘c‘ can be writing down as

∂c

∂t= −O · Je

From ”irreversible thermodynamics” one can write down.Je; as a result of a linear combination of driving forces in the system. the driving foresare the gradients of the intensive properties of the system.

Jeα− L0O(T )L1O

(∂f

∂c

),

∂f

∂c⇒ chemical potential

Jeα− L0O(T ).

∴∂c

∂t= O ·

(L0O(T ) + L1O

(∂f

∂c

))= O ·

(L0OT + L1O

(∂f

∂c

))

‘L0‘ represents the coupling between the temperature and concentration field. L0, L1

are mobility coefficients which denote velocity/(unit driviforce).

The system is uniform and stationary when there are no gradients in the system:

O(T ) = 0;

O

(∂f

∂c

)= 0.

6 1.11.10 85

In general for a multi-component system;

∂ci∂t

= O · (L00O(T )) +∑

j

LijO

(∂f

∂cj

).

The field ‘φ‘ has a dynamics based on a phenomenological minimization of energy,

∂φ

∂t= −M δF

δφ.

6 1.11.10 86

6.2 Tasks to perform

1. Make an infile, with the free energies of the form:

fα (c, T ) = cαNiLαNi

T − TαNiTαNi

+ cαCuLαCu

T − TαCuTαCu

+RT

vm(cαNi log cαNi + cαCu log cαCu) ,

where α stands for both the solid and the liquid phases. The parameters for eachof the phases are listed below:

LαNi LαCuα=Solid 235e7 172.8e7

α=Liquid 0.0 0.0

TαNi TαCuα=Solid 1728K 1358K

α=Liquid 1.0 1.0

Table 6.1: Table containing the values for the latent heats and the temperatures for the Ni-Cualloy.

2. Plot the phase diagram, get the concentrations of the liquidus and the solidus atthe temperature 1500K. You can input the temperature in the program as 1500Kand plot the full phase diagram. Use the tool “binphasediag ”.

3. Using the concentrations at the solidus and the liquidus at the temperature 1500Kcalculate the slope of the liquidus using the formula:

mLiquidusNi =

RT

vm

csolidusNi

cliquidusNi

−(1− csolidusNi

)(

1− cliquidusNi

)

[1

T

((1− csolidusNi )LsolidusCu + csolidusNi LsolidusNi

)]

4. Compute the Latent heat of transformation at 1500K. Use the equation:

L = T(Sliquidus − Ssolidus

),

where, Sliquidus and the solidus Ssolidus denote the entropy and can be written as

−∂fα

∂T, α standing for both solid and the liquid.

5. Compute the parameter τ in the following fashion from the Sharp Interface limit.Propose that the Grand Chemical Potential as the difference as the following:

(f − µc) |10= L(∆T )

|mliquidusNi ∆cNi|

6 1.11.10 87

where, (∆T ) is the equivalent undercooling that would give rise to the same drivingforce than that given by the grand chemical potential difference of the phases. Thenuse the following relation:

− τV = L(∆T )

|mliquidusNi ∆cNi|

∆T =−τVL|mliquidus

Ni ∆cNi| = −βV,

where β is the kinetic coefficient of the phases. and ∆cNi is the difference inthe concentrations of the solidus and the liquidus at the given temperature. Thisdenotes the freezing range of the alloy. The above expression gives an expressionfor the paramete τ ,

τ =βL

T |mliquidusNi ∆cNi|

,

where β is given as (0.33e-2 m/K/s) for the present system. We derive this fromthe fact that for the pure materials Ni the measured β is (0.29e-2m/K/s) and forCu it is (0.37e-2m/K/s), and so for the present case we approximate an alloy of thefollowing to be an average of the two. YOU CAN THINK OF BETTER ONES.

6. Next we input the surface energies and the diffusivities at the particular temper-ature given as below. Note, here too we use an average value for the case of thesurface energy σ = 0.335 J/m2. In the solver we input surface entropy densitiesinstead of the surface energy. Hence to retrieve the same surface energy, one wouldneed to divide the surface energy by the temperature at which one is simulating,which would help retrieve the same surface energy one needs to input in the sim-ulation γ = σ/T .

In a binary alloy there is only one independent inter-diffusivity in a given phase.We use the following values for the diffusivities.

Solid Liquid

DNi = DCu 1e-13 m2/s 1e-9 m2/s

7. Start a simulation with this infile, for the case of a one dimensional simulation withthe solid and the liquid phases at the equilibrium concentrations, and verfiy that

6 1.11.10 88

indeed the calculations of the phase diagram and the equilibrium concentrationsare correct.

8. Next change the concentration of the liquid towards that of the liquid and redothe simulation. One should observe that now the interface profile moves. Notedown the profile of the concentration that you see. This is called giving a super-saturation to the liquid, that is the liquid has more Cu, than the solid can dissolveat equilibrium which gives a driving force for the system. Compare the soluteprofiles with those that you observe for the case of equilbrium.

9. Change the simulation domain to a two dimensional one and perform the simulationby placing the nuclei at the center. In a two dimensional case, the driving force forthe transformation needs to exceed a critical driving force required to overcomethe capillary tension. This gives the size of the critical radius at a given supersaturation. For the given setting of the grid-sizes the nucleus is far beyond thecritical size, and hence you see it growing. Note, it is growing circular.

10. Change Function A in the infile to 1, and put the strength of the anisotropy in theinfile to 0.04 and put the nucleus at the corner of the domain. What shape do yousee? The change in shape is due to surface energy anisotropy in the system. Thiswill eventually give rise to a dendrite if run long enough.

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