Homogeneous nucleation

Homogeneous nucleation of a solid from its melt

Course Name: transformations and treatment

Authors M P Gururajan

Learning Objectives

After interacting with this Learning Object, the learner will be able to: • Derive the for the formation of a homogeneous nucleus from the melt • Show how the undercooling values influence the critical radius and the critical free change

Definitions of the components/Keywords:

Interfacial energy: excess free energy associated with the 1 formation of interfaces Equilibrium (Tm): the at which the 2 solid and its melt have the same free energy

Driving force (∆G): the reduction in free energy due to phase transformation

3 Undercooling (∆T): the process of cooling a melt below its point without the formation of the solid

Nucleation: the process of the formation of a solid in its melt 4 which grows

Clusters/embryos: solids that are formed in the melt which may or may not grow into a nucleus 5 Definitions of the components/Keywords:

Critical radius: the radius of (a spherical) embryo/cluster which 1 can become a nucleus with the addition of an atom

2 : the release of energy associated with the formation of a solid from its melt

3 F r e ∆G e

e n e r g 4 y Solid ∆T

Temperature T 5 m Derivation

Consider the undercooled liquid of VL and free energy per unit volume GL. Hence, its free energy G1 is given by

G1 =V LG L Now, let a volume VS of the melt transform into a solid of free energy per unit volume GS. Also, the formation of solid involves the formation of a solid-melt interface of area A and interfacial free nergy per unit area of γ. Then, the energy of this system of melt with a small solid is given by

G2= V L−V S  G L+V S G S +A S γ

Thus, the net free energy change is given by G2-G1:

ΔG=G 2−G1=−V S  G L−G S  +A S γ=−V S ΔGv +A S γ Assuming spherical solid of radius r, −4 ΔG= πr 3 ΔG 4 πr 2 γ 3 v Derivation

Hence, for the change in free energy to be minimized

dΔG 2 =0=−4 πr ΔGv8 πrγ [ dr ]r=r¿ Hence, 2γ r ¿= ΔGv Since LΔT ΔG v= T m where L is the latent heat, Tm is the melting temperature and ∆T is the undercooling, ∗ 2T m 1 r =  L T The free energy change at r = r* is called ∆G* and is obtained using 3 2 16 πγ T m 1 ΔG¿=  3L2   ΔT 2 Master Layout 1 2

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5 1 Analogy / Scenario / Action

Show the formation of solid from an undercooled melt 2 Show the free energy changes associated with the process 3

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5 T1: Undercooled melt 1 Step 1:

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5 Description of the action

Draw a rectangle on the top left hand side of the animation area. Mark the inside of the rectangle as “Liquid below its Tm”

Write G1 below the rectangle

Audio narration

Consider an undercooled melt – that is a liquid which is kept below its freezing point without it being allowed to turn into a solid. Let the free energy of this liquid be G1.

T1: Nucleation 1 Step 2:

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5 Description of the action

Copy an identical rectangle on the right hand side at the top animation area. Mark the inside of the rectangle as “Undercooled liquid”. Draw a circle inside the rectangle (near about the centre) and mark it as solid

Write G2 below this rectangle

Audio narration

Let a spherical solid of radius r be formed from the undercooled liquid. This formation is driven by the fact that turning the liquid into the solid will reduce the free energy; this driving force is proportional to the volume of the sphere. On the other hand, the formation of the solid also leads to the formation of the solid-melt interface. This interfacial energy increases the overall free energy and is propotional to the surface area of the sphere. Let us plot these two contributions and see how the curves look like.

T1: The energetics 1 Step 3:

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5 Description of the action

Below the rectangles, draw two axes. The y-axes should run from negative to positive and should be labeled ∆G. The x-axis should be marked r. Then, draw a curve given by the equation r^2 for the values of r ranging from 0 to 1. Mark this curve as Interfacial energy (Area). Now, draw another curve given by -r^3 for the same range of r. Mark this curve as bulk free energy (Volume). Add these two curves and plot the third curve and mark it as net free energy change. On this curve, at r = 2/3, mark r* and ∆G* as shown in the figure.Note that neither the r nor the ∆G numerical values should be marked on the axes; it is enough to show the qualitative trends.

Audio narration

The free energy change as a function of the radius of the spherical solid that is formed is shown in the figure. The interfacial energy is positive and raises as r-squared since it is proportional to the surface area. On the other hand, the bulk free energy change associated with the transformation is negative and decreases as the cube or r. The net change, thus, for small sizes of the solid are positive (since the square term dominates) but is negative for relatively larger sizes of the solid (wherein the cube term takes over). Hence, the net energy change is as shown. The critical sized particle is that for which the overall energy change is minimised. Any small addition of atoms to the critical nucleus leads to decrease of volume free energy which can more than compensate for the gain in interfacial energy. Hence, the nucleus will grow if atoms are added. To see the effect of undercooling on r* and ∆G* in various metals, choose a metal and give the undercooling!

Expressions to show on the animation area

∗ 2T 1 r = m  L  T

3 2 16 πγ T m 1 ΔG∗=  3L2   ΔT 2

Credits

What will you learn Radio buttons (if any)/Drop down (if any) Play/pause Restart

Lets Learn! Interactivity options Definitions

Concepts Input Boxes(IO2) /Drop down(IO3) Assumptions (if any) Animation Area

Formula with derivation (if any) Graphs/Diagram (for reference) Test your understanding (questionnaire)

Lets Sum up (summary)

Want to know more… (Further Reading)

Output result of interactivity (if any)

Instructions/ Working area

1 Interactivity and Boundary limits •In this section, I want to add the ‘Interactivity’ options to the animation. The interactivity involves a drop-down menu to 2 choose any particular element and an input box to choose a given undercooling.

• The undercooling value should be between 1 and 300. • Results: For the given material and undercooling the critical 3 radius as well as critical free energy change should be reported in metres and Joules respectively

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In metres In Joules 5 Tm in K, γ in J/m^2 and L in J/m^3 Interactivity option 1: 1 Step No: 1

2 Drop down menu of a list of elements

Interactivity Instruction Boundary type to the Instructions for the animator Results and Output limits 3 (IO1/IO2..) learner

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5 Interactivity type: Drop down menu

Instruction to the learner: Choose a material from this list.

Boundary limits: Al (Aluminium), Ag (Silver), Au (Gold), Co (Cobalt), Cu (Copper), Fe (Iron), Mn (Manganese), Ni (Nickel), Pb (Lead), Pt (Platinum), Sn (Tin) are the elements that should show up in the drop down menu Instructions to the animator: Depending on the choice of the user, the values to be used in the next step are determined. These values are melting temperature Tm in Kelvin, the interfacial energy γ in Joules per square metre, and the latent heat in Joules per cubic metre. They have to be shown to the user on the animation screen at the bottom. In the next page, these values are listed in this order.

Choose the following values of Tm, γ, and L (given in that order below) for the given elements and show it on the screen : Al: 931.7 K, 0.093 J/m2, 1.075x109 J/m3 Ag: 1233.7 K, 0.126 J/m2, 1.1644x109 J/m3 Au: 1336 K, 0.132 J/m2, 1.2172x109 J/m3 Co: 1763 K, 0.234 J/m2, 0.2213x109 J/m3 Cu: 1356 K, 0.177 J/m2, 1.8368x109 J/m3 Fe: 1803 K, 0.204 J/m2, 2.141x109 J/m3 Mn: 1493 K, 0.206 J/m2, 1.991x109 J/m3 Ni: 1725 K, 0.255 J/m2, 2.643x109 J/m3 Pb: 600.7 K, 0.0333 J/m2, 0.261x109 J/m3 Pt: 2043 K, 0.240 J/m2, 2.145x109 J/m3 Sn: 505.7 K, 0.0545 J/m2, 0.431x109 J/m3

Interactivity option 1: 1 Step No: 2

2 Box for uncercooling

Interactivity Instruction Boundary type to the Instructions for the animator Results and Output limits 3 (IO1/IO2..) learner

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5 Interactivity type: Input box

Instruction to the learner: Choose an undercooling. Typical undercoolings that can be achieved are in the range of 0 to 200 or so. However, here you may choose any value above zero and less than 300.

Boundary limits: If the user chooses zero, tell that the calculations will blow-up. Ask to choose a number above zero, however small. Do not allow the range go beyond 300.

Instructions to the animator: Depending on the value of ∆T chosen by the user, and the material chosen in the earlier step, use the following formulae to calculate r* and ∆G* and show them on the screen. 3 2 ∗ 2T m 1 16 πγ T m 1 r = ΔG∗=  L   T  3L2   ΔT 2 INSTRUCTIONS SLIDE Self- Assessment Questionnaire for Learners • At the critical size of the nucleus, the overall change in free energy is extemised. True or False? • The volume free energy gain is directly proportional to the undercooling. True or false? • The critical nucleus size decreases with increasing undercooling. True or false? • The critical free energy increases with increasing interfacial energy. True or false? • The critical free energy increases with increasing undercooling. True or false?

Questionnaire 1 1. True. 2. True; it is directly propotional to L and undercooling and inversely propotional to Tm. 2 3. True 4. True 5. False 3

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5 Links for further reading

Reference websites: The latent heat of fusion values are taken from http://www.engineeringtoolbox.com/fusion-heat-metals-d_1266.html Books: V Raghavan, Materials Science and Engineering: A first course, Third edition, Prentice-Hall of India Pvt. Ltd., 1995. Phase tranasformations in metals and alloys, D A Porter, K E Easterling and M Y Sherif, Third edition, CRC Press, 2009 (First Indian edition).

Research papers: INSTRUCTIONS SLIDE

Summary

• The critical radius and critical free energy for homogeneous nucelation are determined by the latent heat, undercooling, interfacial energy and melting point. • With increasing undercooling both the critical nucleus size and the the critical free energy decrease. • The critical radius is determined by extremising the negative (Volume) contribution from the formation of the solid and the positive (Area) contribution from the formation of the interface associated with the formation of the solid.