Chapter 5: Problem 1
Briefly explain the difference between self diffusion and inter diffusion.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chapter 5: Problem 1
Briefly explain the difference between self diffusion and inter diffusion.
These are the key concepts you need to understand to accurately answer the question.
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Get started for freeA sheet of \(\mathrm{BCC}\) iron \(2 \mathrm{mm}\) thick was exposed to a carburizing gas atmosphere on one side and a decarburizing atmosphere on the other side at \(675^{\circ} \mathrm{C}\). After having reached steady state, the iron was quickly cooled to room temperature. The carbon concentrations at the two surfaces of the sheet were determined to be 0.015 and 0.0068 wt \(\% .\) Compute the diffusion coefficient if the diffusion flux is \(7.36 \times 10^{-9}\) \(\mathrm{kg} / \mathrm{m}^{2}\) -s. Hint: Use Equation 4.9 to convert the concentrations from weight percent to kilograms of carbon per cubic meter of iron.
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.
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 \(4.3 \times 10^{-11} \mathrm{m}^{2} / \mathrm{s}\).
The steady-state diffusion flux through a metal plate is \(7.8 \times 10^{-8} \mathrm{kg} / \mathrm{m}^{2}-\mathrm{s}\) at a tem- perature of \(1200^{\circ} \mathrm{C}(1473 \mathrm{K})\) and when the concentration gradient is \(-500 \mathrm{kg} / \mathrm{m}^{4}\). Calculate the diffusion flux at \(1000^{\circ} \mathrm{C}(1273 \mathrm{K})\) for the same concentration gradient and assuming an activation energy for diffusion of \(145,000 \mathrm{J} / \mathrm{mol}\).
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