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Microstructure-Properties: II The KJMA Equation

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Microstructure-Properties: II The KJMA Equation

27-302

Lecture 5

Fall, 2002

Prof. A. D. Rollett

Processing

Performance

Properties

Microstructure

- The objective of this lecture is to introduce the concept of phase transformation kinetics as described by the Kolmogorov-Johnson-Mehl-Avrami equation.
- Part of the motivation for this lecture is to prepare the class for a Lab on the crystallization of glass-ceramics.

- Phase transformations in metals and alloys, D.A. Porter, & K.E. Easterling,Chapman & Hall, 0-412-45030-5, 669.94 P84P2: page 289.
- Kolmogorov, A. (1937). “A statistical theory for the recrystallization of metals.” Akad. nauk SSSR, Izv., Ser. Matem.1: 355.
- Johnson, W. and R. Mehl (1939). “Reaction kinetics in processes of nucleation and growth.” Trans AIME135: 416.
- Avrami, M. (1939). “Kinetics of Phase Change. I: General Theory.” J. Chem. Phys.7: 1103.
- Avrami, M. (1940). “Kinetics of Phase Change. II: Transformation-Time relations for random distribution of nuclei.” J. Chem. Phys.8: 212.
- Avrami, M. (1941). “Kinetics of Phase Change. III: Granulation, Phase Change an Microstructures.” J. Chem. Phys.9: 177.
- Anderson, W. and R. Mehl (1945). “Recrystallization of Al in terms of the rate of nucleation and growth.” Trans. AIME161: 140.

- The kinetics of transformation are typically described by a standard equation known as the Kolmogorov-Johnson-Mehl-Avrami equation, named after the individuals who derived it.
- The characteristic of the kinetics is that of the “S-curve”, i.e. slow at first, then accelerating, then decelerating.

- The kinetics of transformation are universal.
- Consider this example of the kinetics of cell growth.
- High-Throughput Assay System for the Discovery Of Anti-Bacterial Drugs1J. Bruce Pitner, 2Mark R. Timmins, 2Maurice Kashdan, 3Mandar Nagar, and 3David T. Stitt, 1BD Technologies 21 Davis Drive, Research Triangle Park, NC 27709, 2BD Biosciences, Two Oak Park, Bedford, MA 01730, 3BD Biosciences, 250 Schilling Circle, Cockeysville, MD 21030; Presented at AAPS - New Orleans, LA November, 1999

Note the “S-curve” kinetics

- First, a remark on derivations.
- The objective of a derivation is to build a (mathematical) bridge between basic concepts (axioms in math) and a result (generally, an equation, with parameters or variables corresponding to physical quantities).

Continuous nucleation:nuclei added during transformation.Site Saturated:all nuclei presentat t=0.Cellular:recrystallization,for example. Kinetics same as forsite saturated case.

- The central idea in the derivation of the KJMA equation is to focus on the increment in the (volume) fraction transformed and to relate it to the current value of the fraction transformed.
- Notation:f fraction transformedttimet50%time required for 50% transformationrradiusVvolumevgrowth rate (speed)tincubation/delay timeNrate of nucleation, or, density of nuclei per unit volume

- The relationship between volume and fraction transformed is simple.Fraction transformed = volume / total volume, or,f = V / Vtotal
- Similarly for area (2D), line (1D), etc.

- To understand the concept of an extended fraction transformed, imagine that each patch of new phase can overlap with another one as they grow (ignore the effect of impingement):

V2

fext = (V1+V2)/Vtotal

V1

f = (V1V2)/Vtotal

The true fraction transformed counts only the actual volume transformed: the extended fraction counts all volume as if no impingement occurs.

“union of”

- There is one key assumption in the derivation of the KJMA equation:the nuclei are distributed randomly in space.
- This assumption allows us to make a quantitative relation between the true increment in fraction transformed, a fictitious or extended fraction transformed and the current fraction:df = dfext(1-f)
- Why does this work? The reason is that the volume that each patch can grow into is decreased from the total in proportion to the fraction that has already transformed.

- The KJMA derivation is therefore a bridge between the differential equation just stated and the final equation that we use.

df = dfext(1-f)

- Step 1: define the differential equation (above).
- Step 2: describe the growth rate of an individual patch/region.
- Example: 3D, isotropic growth, site saturated nucleation:V = 4π/3 r3(t) = 4π/3 (vt)3
- Step 3: multiply the individual volume by the number density of nuclei, N:fext = SVi /Vtotal = 4π/3 N (vt)3
- Step 4: obtain the extended fraction increment:dfext = V/Vtotal = 4π Nv3 t2 dt

- Step 5: insert the extended fraction increment into the differential equation:df = dfext(1-f)df = 4π Nv3 t2 dt (1-f)df/ (1-f) = (4π N v3) t2dt
- Step 6: collect the nucleation and growth terms into a constant (which varies depending on the conditions of nucleation and growth):k = 4π/3 N v3

- Step 7: solve the differential equation:recognize that df = -1 * d(1-f), and that the fraction transformed is zero at t=0, so that we are dealing with a logarithmic solution.- ln(1 - f ) = k t3
- Re-arrange to obtain the final result:

- general result

- site saturated, 3D growth

- In general, the k value contains all the temperature dependent terms because thermal activation affects the growth strongly through boundary/interface mobility, and because the nucleation density depends very strongly on driving force.See P&E p269:v = v(T) = v0 exp-(Q/RT)
- In general, the exponent n in the equation is related to the geometry of the transformation.

- Site saturated:1D growth12D growth23D growth3
- Continuous nucleation, constant nucleation rate:1D growth22D growth33D growth4
- CAUTION: you cannot always deduce the geometry of transformation from the value of the exponent.

- Kolmogorov was a Russian mathematician who work is much referenced in statistics. He worked out this relation for the case of continuous nucleation.
- Johnson was a graduate student at Carnegie Tech under R.F. Mehl as his adviser. He studied recrystallization in aluminum.
- Avrami was a chemist and worked out the most general approach: his work is known in the chemical engineering world.
- Porter & Easterling describe the equation but do not explain it in detail. Other texts provide more detail.

- A very useful way to analyze the kinetics of transformation (e.g. recrystallization) is to plot the quantity -ln(1-f) versus time on a double-logarithmic plot. The slope of the line is then the exponent, n.

[Humphreys]

n = slope = 2

- The fraction transformed can be measured in almost any conceivable way.
- From micrographs
- Hardness
- Electrical resistivity
- Optical properties
- Calorimetry

- Step 1: define the differential equation (above).
- Step 2: describe the growth rate of an individual patch/region.
- Example: 3D, isotropic diffusion controlled growth, site saturated nucleation:V = 4π/3 r3(t) = 4π/3 {∆X0/(Xppt-Xe)}3 (√{Dt})3= 4π/3 {∆X0/(Xppt-Xe)}3 (Dt)1.5
- Step 3: multiply the individual volume by the number density of nuclei, N:fext = SVi /Vtotal = N 4π/3 {∆X0/(Xppt-Xe)}3 (Dt)1.5
- Step 4: obtain the extended fraction increment:dfext = V/Vtotal = N 2π {∆X0/(Xppt-Xe)}3 D1.5 √t dt

- Step 5: insert the extended fraction increment into the differential equation:df = dfext(1-f)df = 2π N {∆X0/(Xppt-Xe)}3 D1.5 √t dt (1-f)df/ (1-f) = 2π N {∆X0/(Xppt-Xe)}3 D1.5 √t dt
- Step 6: collect the nucleation and growth terms into a constant (which varies depending on the conditions of nucleation and growth):k = 4π/3 ND1.5 {∆X0/(Xppt-Xe)}3

- Step 7: solve the differential equation:recognize that df = -1 * d(1-f), and that the fraction transformed is zero at t=0, so that we are dealing with a logarithmic solution.- ln(1-f) = 4π/3 N {∆X0/(Xppt-Xe)}3 (Dt)1.5
- Re-arrange to obtain the final result:

- general result

- site saturated, 3D diffusioncontrolled growth

- If we need to find the time for a fixed fraction transformed, this is easily accomplished by manipulation of the basic equation, e.g. for 10% and 90% transformed.

- What is the connection between KJMA kinetics and TTT diagrams?
- Answer: once you have defined the relevant quantities in the KJMA equation, i.e. the nucleation density and the growth rate (and exponent) as a function of temperature, then you can calculate the time required to achieve a certain fraction transformed (previous slide).
- Armed with a set of times for a fixed fraction transformed, draw the locus of points that is a curve on the TTT diagram. Repeat for each volume fraction of interest.