Chapter 5: Problem 4
Briefly explain the concept of steady state as it applies to diffusion.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chapter 5: Problem 4
Briefly explain the concept of steady state as it applies to diffusion.
These are the key concepts you need to understand to accurately answer the question.
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Get started for freeWhen \(\alpha\) -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 \(5.0 \mathrm{MPa}(49.3 \mathrm{atm})\).
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.
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 \(1.9 \times 10^{-11} \mathrm{m}^{2} / \mathrm{s}\).
For a steel alloy it has been determined that a carburizing heat treatment of 15 h duration will raise the carbon concentration to 0.35 wt \(\%\) at a point \(2.0 \mathrm{mm}\) from the surface. Estimate the time necessary to achieve the same concentration at a 6.0 -mm position for an identical steel and at the same carburizing temperature.
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 \(4.3 \times 10^{-11} \mathrm{m}^{2} / \mathrm{s}\).
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