CHAPTER 5

DIFFUSION

PROBLEM SOLUTIONS

5.3 (a) Compare interstitial and vacancy atomic mechanisms for diffusion.(b) Cite two reasons why interstitial diffusion is normally more rapid than vacancy diffusion.

Answer(a) With vacancy diffusion, atomic motion is from one lattice site to an adjacent vacancy. Self-diffusion and the diffusion of substitutional impurities proceed via this mechanism. On the other hand, atomic motion is from interstitial site to adjacent interstitial site for the interstitial diffusion mechanism.(b) Interstitial diffusion is normally more rapid than vacancy diffusion because: (1) interstitial atoms, being smaller, are more mobile; and (2) the probability of an empty adjacent interstitial site is greater than for a vacancy adjacent to a host (or substitutional impurity) atom.

5.8 The purification of hydrogen gas by diffusion through a palladium sheet was discussed in Section 5.3. Compute the number of kilograms of hydrogen that pass per hour through a 6-mm thick sheet of palladium having an area of 0.25 m2 at 600C. Assume a diffusion coefficient of 1.7 108 m2/s, that the respective concentrations at the high- and low-pressure sides of the plate are 2.0 and 0.4 kg of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.

SolutionThis problem calls for the mass of hydrogen, per hour, that diffuses through a Pd sheet. It first becomes necessary to employ both Equations 5.1 and 5.2. Combining these expressions and solving for the mass yields

5.15 Nitrogen from a gaseous phase is to be diffused into pure iron at 675C. If the surface concentration is maintained at 0.2 wt% N, what will be the concentration 2 mm from the surface after 25 h? The diffusion coefficient for nitrogen in iron at 675C is 2.8 1011 m2/s.

SolutionThis problem asks us to compute the nitrogen concentration (Cx) at the 2 mm position after a 25 h diffusion time, when diffusion is nonsteady-state. From Equation 5.5

= 1 erf (0.630)

Using data in Table 5.1 and linear interpolation

zerf (z)0.6000.60390.630y0.6500.6420

from whichy = erf (0.630) = 0.6268

Thus,

And solving for Cx gives

Cx = 0.075 wt% N

5.20 Cite the values of the diffusion coefficients for the interdiffusion of carbon in both -iron (BCC) and -iron (FCC) at 900C. Which is larger? Explain why this is the case.

Solution

We are asked to compute the diffusion coefficients of C in both and iron at 900C. Using the data in Table 5.2,

The D for diffusion of C in BCC iron is larger, the reason being that the atomic packing factor is smaller than for FCC iron (0.68 versus 0.74Section 3.4); this means that there is slightly more interstitial void space in the BCC Fe, and, therefore, the motion of the interstitial carbon atoms occurs more easily.

5.21 Using the data in Table 5.2, compute the value of D for the diffusion of magnesium in aluminum at 400C.

Solution

This problem asks us to compute the magnitude of D for the diffusion of Mg in Al at 400C (673 K). Incorporating the appropriate data from Table 5.2i.e.,D0 = 1.2 104 m2/sQd = 130,000 J/mol into Equation 5.8 leads to

5.25 The preexponential and activation energy for the diffusion of chromium in nickel are 1.1 104 m2/s and 272,000 J/mol, respectively. At what temperature will the diffusion coefficient have a value of 1.2 1014 m2/s?

Solution

We are asked to calculate the temperature at which the diffusion coefficient for the diffusion of Cr in Ni has a value of 1.2 1014 m2/s. Solving for T from Equation 5.9a

and using the data the problem statement (i.e., D0 = 1.1 104 m2/s and Qd = 272,000 J/mol) , we get

= 1427 K = 1154C