Winter 2009

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Solid State Diffusion

1) Figure 2.3 illustrates the location of the octahedral interstices in FCC and BCC crystals. For a crystal of Al, which has an FCC crystal structure at room temperature, use the hard sphere model to answer the following questions:

a) Given that the lattice parameter of Al is a = 404.959 [pm], determine the atomic size of an Al atom.

b) What is the size of the largest interstitial atom that could fit in an octahedral interstice in Al? (Look for the radius of a hard sphere that would just fit in the interstitial space. IMPORTANT: Be sure to use the smallest spacing between lattice atoms to determine the interstitial size.)

c) How does this interstitial size compare to the size of a small atom like C?

d) When an interstitial atom moves to a neighboring octahedral interstice it doesn't move directly. Look at figure 2.3a. Assume the interstitial at the center of the unit cell is going to move to the "destination" interstitial space to the right front (let the origin be at the rear bottom left of the unit cell; the center interstice is at position 1/2, 1/2, 1/2, and the destination is at position 1, 1, 1/2). What is the position of the interstitial atom when it is in the intermediate "activated" state? (HINT: the easiest path involves moving through a hole created by the three atoms at 1, 1/2, 1/2; 1/2, 1, 1/2; and 1, 1, 1.)

e) Determine the size of the largest interstitial atom that could fit through the hole in part (d) (the activated state) without straining the lattice atoms. The ratio of your answer for part (b) to that for part (e) gives an estimate of the lattice strain needed for the diffusion to take place.

2) A diffusion couple has been fabricated to measure the diffusivity in a binary alloy of A and B. The left side of the diffusion couple has an initial uniform composition of 70wt%A, and the right side of the couple has an initial uniform composition of 72wt%A.

a) After a 200 [hour] diffusion anneal at 1300 [K], the composition at a distance 200 [µm] to the left of the original interface is measured to be 70.6wt%A. Use the Grube method to find the diffusivity at 1300 [K].

b) Make a plot of the composition at the position X

_{1}= 200 [µm] to the left of the original interface as a function of time from 0 - 200 [hours]. What is the significance of the composition that is approached at long times? NOTE: All plots must be produced using a computer graphing software package, unless otherwise instructed.c) On the same plot as part (b) SKETCH by hand (do not calculate) the composition at a position X

_{2}= 400 [µm] to the left of the original interface.

3) Consider a compositionally uniform bar made of a binary alloy of A with interstitial B atoms that has a TEMPERATURE gradient along its length.

a) Would you expect to see a net flux of B atoms moving along the bar? If so, in which direction, relative to the temperature gradient? EXPLAIN in a few sentences.

b) Using a similar derivation to that used in class (and in section 2.2.1 of Porter and Easterling), derive Ficks First Law for this case and determine an expression for the thermal diffusion coefficient, D

_{T}, due to the temperature gradient. Is D_{T}independent of position along the bar? Why?

4) We want to examine the de-carburization of a tool steel
bar (diffusing carbon out from the surface) at a temperature of 773 K for
5 hours. The initial concentration of carbon in the iron bar is 0.86wt%, and
the surface concentration due to gas atmosphere is constant at
0.12wt%. The diffusivity of C in Fe is given by D_{o} =
2 X 10^{-5} m^{2}/s, and Q =142,000 J/mole.

a) Determine the depth below the surface at which the carbon concentration is equal to 0.4wt%,

b) What temperature would be required to get the same concentration profile in the iron in one-fifth the time?

c) Using reasonable values for vibration frequency and jump distance, what is the average residence time of an interstitial C atom in the Fe at this temperature?

1) Figure 2.3 illustrates the location of the octahedral interstices in FCC and BCC crystals. For a crystal of Al, which has an FCC crystal structure at room temperature, use the hard sphere model to answer the following questions:

1a) *Given that the lattice parameter of Al is a = 404.959 [pm], determine
the atomic size of an Al atom.*

From the hard sphere model, we know the relationship between the lattice parameter, a, and the atomic radius of an Al atom, R, for the FCC crystal. The face diagonal of the unit cell is 4 atomic radii long, and is equal to the square root of 2 times the lattice parameter:

1b) *What is the size of the largest interstitial atom that
could fit in an octahedral interstice in Al?*

From Figure 2.3, and looking at the octahedral interstice at the body center, we can see that the interstitial atom is surrounded by three pairs of lattice atoms (above and below, left and right, and front and back), each of which are spaced by one lattice parameter. This means that the size of the largest interstitial atom that will fit in this space is the one that fills the hole between any pair of atoms (see figure to the right.) |

1c) *How does this interstitial size compare to the size of
a small atom like C?*

C has an atomic size of 77 [pm], according to my reference source, so C is 30% bigger than the interstitial size in Al.

1d) *What is the position of the interstitial
atom when it is in the intermediate "activated" state?*

This is actually a multi-step process, and is more complicated than Porter and Easterling suggest in their figure 2.4 (though the starting and ending points are correct in their figure.) The figure to the right shows a portion of the FCC cell, with the interstitial start and end positions marked by the BLACK circles and the numbers "1" and "3," respectively. I have indicated the FCC lattice atoms on the top center ("T"), right center ("R"), front center ("F"), and the front upper corner ("C"). The interstitial atom at the body center of the FCC lattice begins in position 1, and must first squeeze though the small hole formed by the three atoms forming the equilateral triangle T,F,R (step 1). This is the first activated complex state, and is the smallest spacing the atom has to navigate. In step 2, the interstitial atom moves from the activated complex into the hole created by the four atoms T,F, R and C. This space is the TETRAHEDRAL interstice in FCC. Since it is bigger than the hole in step 1 (and smaller than the octahedral interstice), this is an intermediate (lower energy state) activated complex. Finally, moving from the tetrahedral interstice through the opening formed by C,F, and R allows the interstitial atom to finish up in the end position, 3, another octahedral interstice. The activated complex in step 3 is the same as in step 1. |

The question is not simply answered...the position of the atom in the middle step (step 2) is in the center of the tetrahedral interstice, and is at a location of 3/4, 3/4, 3/4 in the unit cell. The position of the activated complexes is given by the center of the hole between the three atoms in steps 1 and 3, and that's more complicated to figure out.

1e) *Determine the size of the largest interstitial atom that
could fit...*

The size of the hole between the atoms T, F, and R is found to be (see figure): The ratio mentioned in the problem is 59.3 [pm]/22.15 [pm] = 2.68, which is a strain of 168%! That's WAY bigger than any elastic strain (and most plastic strains too.) |

2a) *Use the Grube method to find the diffusivity
at 1300 [K].*

The Grube solution for Fick's second law is given as (for an infinite diffusion couple):

Looking up (or calculating) the argument that solves the error function equation gives

2b) *Make a plot of the composition at the position X _{1} =
200 [µm] to the left of the original interface...*

I used Mathematica to calculate and plot the answer from the Grube method.

Plotting again to long times shows the composition approached is 71%, which is the average composition of the diffusion couple:

2c) *On the same plot as part (b) SKETCH by hand...*

Because the solutions scale as , we will get the same composition at 400 [um] as we did at 200 [um] when the time is increased by a factor of FOUR:

I used this fact to find the composition at a time of 50 [hours] and X = -200[um], and shifted this composition to a time of 200 [hours] for the X = -400[um] line, as shown below:

3a) *Would you expect to see a net flux of B atoms moving
along the bar?*

YES, there should be a net flux of B atoms. This will happen because, at the hot end of the bar the jump rate for the B atoms will be higher since there is more thermal energy to allow the atoms to get over the diffusive energy barrier. At the cold end of the bar, the jump rate is smaller, so that B atoms will tend to jump in a direction down the temperature gradient (from hot to cold.)

3b) *Using a similar derivation to that used in class...*

Look at two neighboring planes in the crystal, plane 1 and plane 2. We can write the flux of B atoms that jump from plane 1 to 2 as:

We can write the

This gives us Fick's First Law for the thermal gradient case.

Given the expression for the jump frequency as a function of temperature, we can evaluate the partial derivative in terms of the temperature gradient:

Is D_{T} independent of position along the bar? NO.
The reason is that the temperature varies along the bar, and so the jump frequency
will vary along the bar as well. The temperature dependence is clearly seen
in the value determined above.

4a) *Determine the depth below the surface at which the carbon
concentration is equal to 0.4wt%,*

Because we have a semi-infinite bar of tool steel, the correct solution to FFL is given by the Grube equation:

We want to solve for x, so we need to know what is the value of erf(z) to find the value of the argument, z:

4b) *What temperature would be required to get the same concentration
profile in the iron in one-fifth the time? *

Because the concentration is the same, the value of the error function is the same, and therefore the argument of the error function is the same. Since the value of z is the same, and x is the same, the quantity "Dt" is the same. A decrease in the time of a factor of five must be offset by an increase in the diffusivity by a factor of five. This now lets us determine the temperature from the equation for the diffusivity:

An increase of about 60 degrees decreases the diffusion time by a factor of five. Exponentials are powerful (but not as powerful as Chuck Norris, I'm told...)

4c) *Using reasonable values for vibration frequency and jump
distance...*

We have the relationship between the diffusivity and the jump frequency, GAMMA as:

For Fe, a reasonable value for the jump distance of a C interstitial is a/2 = 286.65/2 [pm] = 143.33 [pm] (see Figure 2.3, P&E). Thus the residence time, TAU, is:

End of File.