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Crystal Growth General Formalism    Phase growing into  with velocity v : f ( “site factor” ) : fraction of sites where a new atom can be incorporated.

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Presentation on theme: "Crystal Growth General Formalism    Phase growing into  with velocity v : f ( “site factor” ) : fraction of sites where a new atom can be incorporated."— Presentation transcript:

1 Crystal Growth General Formalism    Phase growing into  with velocity v : f ( “site factor” ) : fraction of sites where a new atom can be incorporated 0 < f < 1 : interatomic distance k : jump frequency   

2 *at equilibrium temperature  and  ( T  ) the net k must be zero.   kk k +  G* G  = G 

3 below T  :  G = G  - G    kk k +k +  G* GG GG

4 net jump rate = for small undercooling Same as for nucleation Collision limited :  s / diff limited : 6D / 2

5 For metal solidification: (Richards’ rule) Small undercooling

6 Crystal growth velocity normal to area A is v. Consider growth to occur by nucleation of monolayer patches. Growth governed by two – dimensional nucleation vlvl v area A area concentration of critical nuclei. # at. in cluster Nuc.rate : h

7 growth of patches occurs by lateral spreading at a velocity v l (ledge velocity) fraction of sites along ledge that can incorporate new atoms. : net jump rate to ledge vapor  solid; k is collision rate, surface migration The time for 1 layer of area A to form : for condensed phase

8 Number of nuclei formed in that amount of time Average area of each patch grown in that amount of time - In that time the whole area must be covered: # of nuclei so, or or

9 Large undercooling required 1 0 f TT v TT f(  G) Since nucleation of ledges is difficult at low undercoolings in many systems crystal growth is governed by intrinsic ledge structure. Plug in eqns. for I s and v l Probability that an atom is in a critical nucleus - very T dependent

10 Growth on Surface Defects - Observation : many crystals grow at small undercooling supersaturations F. C. Frank : Screw dislocation mechanism Spiral growth around screw dislocation

11 t0t0 t1t1 t2t2 t3t3 t4t4 ~ r* After some time t n a spiral forms. RR Top View time evolution Ledge Size ~ critical nucleus size K ~ 4  ; Archimedes spiral

12 Area fraction of growth sites : (Assume attachment at all step sites; f l = 1 ) : is inter-atomic spacing. and 1 0 f TT TT parabolic v

13 Fraction transformed in isothermal process – Avrami analysis Consider    transformation  How do you deal with the overlap? Mathematical device : extended volume fraction X ex  volume fraction transformed disregarding overlap.

14 The actual volume fraction grows in a relative amount to the unconsumed fraction, at the same rate the extended volume fraction does.: Unconsumed fraction Integrate Avrami equation Expand : dilute overlap of two overlap of three or

15 Application to nucleation & growth : ( Johnson - Mehl) Case (1) constant number of heterogeneous nuclei present from the beginning. concentration: N growth rate of crystals : v x t Plot of ln t vs ln[-ln(1-x)] should have slope of 3.

16 Case (2) Assume a constant nucleation rate I, # of nuclei formed between t’ and t’ + dt’ ; concentration, N = I dt’ and at some later time ( t > t’ ) the “radius” of transformed phase is v (t – t’) so Plot of ln t vs ln[-ln(1-x)]  slope of 4 These plots are called Johnson- Mehl –Arami plots (JMA plots)

17 Calorimetry results power DSC isothermals Time (min) 20406080100 329K 328K 327K 326K 325K 324K Time 1/2 1 X 329K 328K327K 326K 325K324K 0 Fraction transformed Case study : Devitrification of Au 65 Cu 12 Si 9 Ge 14 glass C. Thompson et. al., Acta Met., 31, 1883 (1983)

18 ln (1-  ) ln (t) ln [-ln(1-x)] JMA plot (327K) must be introduced N = I ss (t -  ) Slope = 4.0 ln [-ln(1-x)] slope = 4

19 Transient time ln [  (s)] 10 3 / T (K -1 ) activation energy : 2.0 eV

20 Interface Stability during Growth (i) Consider solidification in a one component system: The solidification process is controlled by the rate at which the latent heat of solidification can be conducted away from the solid / liquid interface. Solid growth into a liquid @ T > T m T SolidLiquid TmTm heat flux v - interface velocity (a)

21 The heat flux away from the interface through the solid must balance that from the liquid plus the latent heat generated at the interface.: Here K is the thermal conductivity, and L f the latent heat of fusion per unit volume.

22 heat Solid Liquid The dotted lines are isotherms. Consider the stability of a perturbation that may develop at the interface: Since the perturbation is an the high temperature side of the interface, the thermal gradient in the protuberance is less than it is in the planar portion of the interface and the protuberance will disappear. Rate limiting step in the growing phase.

23 T SolidLiquid heat flux v x (b) Solidification in a super-cooled liquid: The perturbation in the super-cooled liquid sees a higher temperature gradient than the planar portion of the interface  perturbation is stable. Rate limiting step in the shrinking phase. Solid Liquid heat flux

24 Sieradzki’s rule of interface stability: If the rate limiting step is in the phase that is growing a planar interface will be stable to a geometrical perturbation. If the rate limiting step is in the phase that is shrinking the planar interface will be unstable.

25 Alloy Solidification Def. Partition coefficient k x s and x L are the mole fraction of solute in the solid and liquid respectively. T3T3 T1T1 T2T2 kx 0 x0x0 x 0 / k

26 We will consider 3 limiting cases of the solidification process. : (a) Infinitely slow (equilibrium) solidification (b) Solidification with no diffusion in the solid and perfect mixing in the liquid. (c) Solidification with no diffusion in the solid and only diffusional mixing in the liquid.

27 (a) Equilibrium solidification. @ composition x 0 will be the composition of the 1 st amount of solid to solidify. * Note that k is constant for straight liquidus and solidus. As the temperature is lowered more solid forms. For slow enough cooling mixing in liquid and solid is perfect and x s and x L will follow the solidus and liquidus lines respectively At T 3 the last liquid to freeze out has a composition x o


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