Nucleation Science

4. Size and Density of Precipitates

  The Basics
We have a supersaturation of point defects, we have some heterogeneous nucleation - finally we can make our precipitate. Two (related) questions should now occur to you:
  1. How large will those precipitates be (on average, of course) and what determines their size?
  2. How many will I find in a given volume or in other words: what kind of precipitate density will I get?
Looking at energy balances, as in the preceding modules, will not do the trick here. If you just minimize the energy, you will of course end up with exactly one spherical precipitate, with a size that is just right to accommodate all the point defects that were supersaturated in your crystal. You can't have a better surface to volume ratio that that.
Equally of course, that is not what you will find in iron or steel at room temperature (or in most other crystals) if you wait less than a sizeable fraction of the age of the known universe (about 14 billion years, by the way).
Remember: if nothing moves, nothing happens.
The typical situation we have is:
  1. At high temperatures our point defects - vacancies, impurity atoms, whatever - can move vigorously and thus can go places. On the other hand, the supersaturation is low; definitely lower than if you have the same concentration at lower temperatures.
  2. The mobility of point defects is described by their diffusion coefficient D(T) = D0exp-(ED/kT); the average distance they covered after a time t is given by the diffusion length L = (Dt)½.
  3. At low temperatures, the supersaturation is high but movement is sluggish.
  4. Nothing moves at room temperature.
So how large can a precipitate be? There is an easy general answer: it can at best be so large as to contain all the point defects that had a chance to reach it. In other words:
All the point defects that were contained in a sphere with radius "total diffusion length" Lto determine the maximal size of a precipitate.
Note that the answer to the second question from above is also clear now. We have one precipitate in a volume of about (L3to) and thus a precipitate density of 1/(L3to).
All that remains to do is to calculate Lto. For that we need to know the temperature profile T(t), the way the specimen cools down with time.
Then we calculate the diffusion coefficient at some temperature, see how far a point defect could go in some small interval around that temperature for the time the specimen stayed in that interval, repeat the procedure for other temperatures, and add everything up.
In other words; we integrate!
It is easiest to do that for L2to; we can always take the root when we are done.
Quite generally, the (square) of the average distance L or total diffusion length that a diffusing object covers during cooling down in three dimensions is given by
 
L2to  =  2D · t  =     t' = ¥
   ó
    õ
t' = t0
D(t') · dt'
     
The diffusion coefficient D(T) is always given by D(T) = D0 · exp–(ED/kT) with ED = migration energy with typical values of (0.5 - 2.5) eV.
The most simple way of cooling is to put the specimen into some cold environment. The temperature profile then follows proper beer-froth dynamics (get a beer and look it up!) or T(t) = T0 · exp–(lt). T0 is the starting temperature, and l the "decay parameter".
While the "decay parameter" l is most convenient for the equations, the temperature gradient [dT/dt]t = 0 = –lT0 is easier to grasp. To give an example: If we cool down from 1800 K with a gradient of –100 K/s, we have a decay parameter l = 1/18 s–1 = 0.055 s–1.
Another convenient way for looking at cooling-down dynamics is to define a "cooling half-time" thalf = 1/l; i.e. the time it takes to cool to about half the starting temperature. 1/l actually takes you to 0.37 T0 - but what the heck. If you are a stickler for correctness or just anal-retentive, you must go with thalf = (1/l) · ln(0,5) » 0,69/l.
Combining the two, we can express the cooling half-time in terms of the temperature gradient as thalf = T0/[dT/dt]
Here is the illustration for all that:
 
Exponential decay
Exponential decay of temperature and various parameters to describe it
 
Writing out the integral by inserting the basic equation for D(T) and the exponential decay function of the temperature from above, we get a little beauty:
 
L2   = 2D 0  ¥
ó
õ
t0
exp – æ
ç
è
ED
kT0 exp–(lt)
ö
÷
ø
dt 
 
The integral is a little math beauty because it looks neat, approachable and easy to conquer. It's like female human beauties. Behind the comely surface often hides a strong mind, and conquering is not all that easy either. Strangely enough, those humans who have no problem with the integral (few and mostly male) typically have problems conquering that female beauty.
Our little math beauty is not easy to do but needs to be wooed by suitable substitutions and approximations.
What one gets for the total diffusion length Lto after a rather lengthy exercise in calculus is
 
Lto  =  æ
ç
è
2D0 · kT0
l · ED
ö
÷
ø
½  · exp –  ED
2kT0
     
That is an easy function to plot for some bandwidth of parameters.
Shown below are curves for diffusion energies from 0.5 eV to 2 eV. Cooling rates from 1 K/s to 50 K/s and a D0 = 10–5 cm2s–1 are assumed.
     
Total diffusion length with cooling down
Total diffusion length and cooling down
 
Of course, diffusion lengths below 10–4 µm = 1 Å are meaningless for atoms and just shown for completeness.
What we see quite generally is:
  • The cooling rate matters less than the migration energy.
  • Most everything happens at high temperatures.
  • 1 µm is already a large distance under many circumstances.
We knew that already to some extent. But now we have numbers for all kinds of circumstances.
Let's look at a few numbers for possible precipitate sizes. For simplicities sake we assume that the volume is L3to, that all the impurity atoms contained in that volume end up in a "cube" precipitate, and that the size of one "particle" is 0.3 nm3. That's what we get for the size of the precipitate-cube in nanometer:
     
Lto 10 nm
(0.01 µm)
100 nm
(0.1 µm)
1.000 nm
(1 µm)
10.000 nm
(10 µm)
Conc. impurity Size of precipitate [nm]
10–6 (1 ppm) 0,1 1 10 100
10–5 (10 ppm) 0,215 2,15 21.5 215
10–4 (100 ppm) 0,464 4,64 46,4 464
10–3 (0.1 %) 1 10 100 1.000
10–2 (1 %) 2,15 21,5 215 2.150
   
Only in the two red cases the size of the precipitates surpass 1 µm, making them just about visible in a light microscope.
Looking at carbon steel or "low-alloy" steel, containing fast-diffusing carbon in the 1 at% range, and slower diffusing stuff in the 0.01 at% to 1 at% range, we can expect to find carbon precipitates (Fe3C or cementite) in the µm range and thus visible in a light microscope. Everything else should be far smaller, however, and thus had to escape the attention of early steel scientists, who could not yet command electron microscopes.
   
  The Great Simplification
The total diffusion length Lto is quite important for many things related to iron and steel, and that's why I want to dwell on it a bit longer. I will get more specific for the atoms diffusing around in iron and steel, and more simplistic in how to look at the situation.
For that it is beneficial to rewrite the equation for Lto by introducing parameters that are more easy to relate to. My choice is:
  • The "cooling half-time" thalf = 1/l or l = 1/thalf.
  • The diffusion length L0 for a time t1 = 1s at the starting temperature T0. That is just the square root of the diffusion coefficient and directly obtainable from diffusion data. We then have L0 = (D(T0) · t1)½ = (D0 · t1)½ · exp–(ED/2kT0). This gives us (D0 )½ exp–(ED/2kT0) = Lto/(t1)½
  Inserting this in the equation above yields
     
   
Lto  =  L0 æ
ç
è
2thalf
t1
ö
÷
ø
½  ·    æ
ç
è
kT0
ED
ö
÷
ø
½
                       
   =  1,41 · L0 æ
ç
è
|thalf| ö
÷
ø
½  ·    æ
ç
è
kT0
ED
ö
÷
ø
½
                       
   =  L0  ·  g(thalf, T0, ED)
     
While this may still look complicated, it makes estimations of Lto very easy for all the atom where you have some Arrhenius diagram of their diffusion behavior. All we need to do is to make a little table for typical values of the number g(T0, ED, thalf) by using values typical for iron and steel for some parameters. Don't forget: if we have an Arrhenius plot, we also have the migration energy ED. So let's look at the parameters.
Looking at typical migrations energies ED for atoms moving around in iron, we find
  • Values close to 1 eV for the interstitials carbon (C), nitrogen (N) and oxygen (O).
  • Values roughly around 1.5 eV for most substitutional atoms.
Taking ED = 1 eV and 2 eV thus covers most everything.
Looking at typical T0 temperatures for iron, we can take the melting point of pure iron at around 1800 K as an upper limit, and the "working " temperature off a smith at around 1000 oC (1832 oF) as a lower limit - in the form of 1300 K, of course. This gives us kT0 values of 0.16 eV or 0,11 eV, respectively.
For cooling half-times we might pick 36.000 s (10 hr) 3600 s (1 hr), 360 s (6 min) and 36 s, again covering most practical situations in between the two extreme values.
Now we can come up with two tables for the values of the factor g(thalf, T0, ED) that (with interpolations) cover pretty much every cooling-down event.
     
g(T0= 1.800 K, ED, thalf)
ED          thalf 36.000 s 3600 s 360 s 36 s
1 eV 106 34 11 3
2 eV 75 24 7 2

g(T0= 1.300 K, ED, thalf)
ED          thalf 36.000 s 3600 s 360 s 36 s
1 eV 90 28 9 3
2 eV 64 20 6 2
 
What the numbers mean is clear. An atom that moves a certain distance Lto within 1 second at one of the two temperatures given, will go for a distance that is larger by the number in the table for the conditions indicated.
Let's look at an example. From the Arrhenius plots in the "diffusion in iron" module you can see that within 1 s carbon in iron covers roughly 50 µm at 1800 K, and 10 µm at 1300 K. It's migration energy is close to 1 eV.
Looking at the table you realize that carbon covers 53.300 µm = 53 mm for sluggish cooling from the melting temperature, and still 150 µm for extremely fast cooling. If the starting temperature is 1300 K, it moves 900 µm or 30 µm for the extremes in cooling, respectively.
The same exercise for manganese, starting with 70 nm or 2 nm, respectively, and a migration energy of 1 eV, gives 5.25 µm or 140 nm for the extremes and 1800 K, and 128 nm or 4 nm for 1300 K cooling.
Yes, I know it is not quite that easy. Cooling relatively pure iron from the melting point at 1800 K to room temperature involves a bcc- fcc phase change at high temperatures and a fcc- bcc at around 1.000 K, with the carbon moving much more sluggishly in the fcc austenite phase; this is clearly visible in the Arrhenius plot. So we overestimate the total diffusion length Lto somewhat - but who cares? We only get orders of magnitude anyway. The consequence of this little exercise are obvious and dramatic, no matter how you look at details:
     

The carbon concentration in iron / steel will equilibrate over macroscopic distances at least in the order of 0.1 mm if you heat it just once to 1000 0C

     
It doesn't matter how you cool down, and in most cases equilibration occurs over much larger distances. Provided, of course, the carbon is atomically dissolved and can move freely in all directions
In other words: banging together thin sheets of low carbon / high carbon steel for damascene pattern welding won't do you much good. You don't get a compound material but a rather uniform material with an medium carbon content. I have stated that before but now we are one step more quantitative. I'll come back to this topic later again.
     
Back to
  Nucleation Science Overview
  1. Global and local equilibrium for point defects
2. Homogeneous nucleation
3. Heterogeneous nucleation
On to
  5. Precipitation and structures
     

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