91Ó°ÊÓ

Briefly explain the difference between self diffusion and inter diffusion.

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

Briefly explain the concept of steady state as it applies to diffusion.

(a) Briefly explain the concept of a driving force. (b) What is the driving force for steady-state diffusion?

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 \(0.25 \mathrm{m}^{2}\) at \(600^{\circ} \mathrm{C}\). Assume a diffusion coefficient of \(1.7 \times 10^{-8} \mathrm{m}^{2} / \mathrm{s},\) that the concentrations at the high- and low-pressure sides of the plate are 2.0 and \(0.4 \mathrm{kg}\) of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.

A sheet of steel 2.5 \(\mathrm{mm}\) thick has nitrogen atmospheres on both sides at \(900^{\circ} \mathrm{C}\) and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is \(1.2 \times 10^{-10} \mathrm{m}^{2} / \mathrm{s},\) and the diffusion flux is found to be \(1.0 \times 10^{-7} \mathrm{kg} / \mathrm{m}^{2}\) -s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is \(2 \mathrm{kg} / \mathrm{m}^{3} .\) How far into the sheet from this highpressure side will the concentration be 0.5 \(\mathrm{kg} / \mathrm{m}^{3} ?\) Assume a linear concentration profile.

When \(\alpha\) -iron is subjected to an atmosphere of nitrogen gas, the concentration of nitrogen in the iron, \(C_{\mathrm{N}}\) (in weight percent), is a function of hydrogen pressure, \(p_{\mathrm{N}_{2}}(\text { in } \mathrm{MPa}),\) and absolute temperature \((T)\) according to $$C_{\mathrm{N}}=4.90 \times 10^{-3} \sqrt{p_{\mathrm{N}_{2}}} \exp \left(-\frac{37.6 \mathrm{kJ} / \mathrm{mol}}{R T}\right)$$.Furthermore, the values of \(D_{0}\) and \(Q_{d}\) for this diffusion system are \(3.0 \times 10^{-7} \mathrm{m}^{2} / \mathrm{s}\) and \(76,150 \mathrm{J} / \mathrm{mol}\), respectively. Consider a thin iron membrane \(1.5 \mathrm{mm}\) thick that is at \(300^{\circ} \mathrm{C}\) Compute the diffusion flux through this membrane if the nitrogen pressure on one side of the membrane is \(0.10 \mathrm{MPa}(0.99 \mathrm{atm}),\) and on the other side \(5.0 \mathrm{MPa}(49.3 \mathrm{atm})\).

An FCC iron-carbon alloy initially containing 0.55 wt \(\%\) C is exposed to an oxygen-rich and virtually carbon-free atmosphere at \(1325 \mathrm{K}\) \(\left(1052^{\circ} \mathrm{C}\right) .\) Under these circumstances the carbon diffuses from the alloy and reacts at the surface with the oxygen in the atmosphere that is, the carbon concentration at the surface position is maintained essentially at 0 wt \(\%\) C. (This process of carbon depletion is termed decarburization.) At what position will the carbon concentration be 0.25 wt\% after a 10-h treatment? The value of \(D\) at \(1325 \mathrm{K}\) is \(4.3 \times 10^{-11} \mathrm{m}^{2} / \mathrm{s}\).

Nitrogen from a gaseous phase is to be diffused into pure iron at \(675^{\circ} \mathrm{C}\). If the surface concentration is maintained at \(0.2 \mathrm{wt} \% \mathrm{N}\) what will be the concentration \(2 \mathrm{mm}\) from the surface after 25 h? The diffusion coefficient for nitrogen in iron at \(675^{\circ} \mathrm{C}\) is \(1.9 \times 10^{-11} \mathrm{m}^{2} / \mathrm{s}\).

Consider a diffusion couple composed of two semi-infinite solids of the same metal, and that each side of the diffusion couple has a different concentration of the same elemental impurity; furthermore, assume each impurity level is constant throughout its side of the diffusion couple. For this situation, the solution to Fick's second law (assuming that the diffusion coefficient for the impurity is independent of concentration), is as follows: $$C_{x}=\left(\frac{C_{1}+C_{2}}{2}\right)-\left(\frac{C_{1}-C_{2}}{2}\right) \operatorname{erf}\left(\frac{x}{2 \sqrt{D t}}\right).$$In this expression, when the \(x=0\) position is taken as the initial diffusion couple interface, then \(C_{1}\) is the impurity concentration for \(x<0\) likewise, \(C_{2}\) is the impurity content for \(x>0\).A diffusion couple composed of two platinum-gold alloys is formed; these alloys have compositions of \(99.0 \mathrm{wt} \% \mathrm{Pt}-1.0 \mathrm{wt} \%\) Au and 96.0 wt \(\%\) Pt- 4.0 wt\% Au. Determine the time this diffusion couple must be heated at \(1000^{\circ} \mathrm{C}(1273 \mathrm{K})\) in order for the composition to be 2.8 wt \(\%\) Au at the \(10 \mu \mathrm{m}\) position into the \(4.0 \mathrm{wt} \%\) Au side of the diffusion couple. Preexponential and activation energy values for Au diffusion in \(\mathrm{Pt}\) are \(1.3 \times 10^{-5}\) \(\mathrm{m}^{2} / \mathrm{s}\) and \(252,000 \mathrm{J} / \mathrm{mol},\) respectively.