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Chapter 5: Problem 15

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 \(25 \mathrm{~h}\) ? The diffusion coefficient for nitrogen in iron at \(675^{\circ} \mathrm{C}\) is \(2.8 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\).

Short Answer

Expert verified
Answer: Approximately 0.0798 wt% N.

Step by step solution

01

Write down the given information

- Surface concentration, \(C_{s} = 0.2 \mathrm{wt} \% \mathrm{~N}\) - Distance from the surface, \(x = 2 \mathrm{~mm}\) - Diffusion time, \(t = 25 \mathrm{~h}\) - Diffusion coefficient, \(D = 2.8 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\) - Temperature, \(T = 675^{\circ} \mathrm{C}\) (given, but not required for this problem)
02

Convert the units for distance and time

Convert distance to meters and time to seconds for compatibility. - Distance: \(x = 2 \mathrm{~mm} \times \dfrac{1 \mathrm{~m}}{1000 \mathrm{~mm}} = 2 \times 10^{-3} \mathrm{~m}\) - Time: \(t = 25 \mathrm{~h} \times \dfrac{3600 \mathrm{s}}{1 \mathrm{~h}} = 90,000 \mathrm{s}\)
03

Use Fick's Second Law of Diffusion

We will use the equation for non-steady-state diffusion and approximate it with an error function since the surface concentration is maintained constant: \(C(x,t) = C_{s} \cdot erf\left(\dfrac{x}{2\sqrt{Dt}}\right)\)
04

Calculate the concentration at the given distance and time

Plug in the values for \(C_{s}\), \(D\), \(x\), and \(t\) and calculate the concentration \(C(x, t)\): \(C(x,t) = 0.2 \times erf\left(\dfrac{2 \times 10^{-3} \mathrm{~m}}{2\sqrt{2.8 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s} \times 90,000 \mathrm{s}}}\right)\) \(C(x,t) \approx 0.2 \times erf(0.0357)\) \(C(x,t) \approx 0.2 \times 0.399\) \(C(x,t) \approx 0.0798 \mathrm{wt} \% \mathrm{~N}\)
05

Interpret the result

The concentration of nitrogen 2 mm from the surface of the iron after 25 hours will be approximately 0.0798 wt% N.

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Most popular questions from this chapter

Indium atoms are to be diffused into a silicon wafer using both predeposition and drive-in heat treatments; the background concentration of In in this silicon material is known to be \(2 \times 10^{20}\) atoms \(/ \mathrm{m}^{3}\). The drive-in diffusion treatment is to be carried out at \(1175^{\circ} \mathrm{C}\) for a period of \(2.0 \mathrm{~h}\), whichgives a junction depth \(x_{j}\) of \(2.35 \mu \mathrm{m}\). Compute the predeposition diffusion time at \(925^{\circ} \mathrm{C}\) if the surface concentration is maintained at a constant level of \(2.5 \times 10^{26}\) atoms \(/ \mathrm{m}^{3}\). For the diffusion of In in Si. values of \(Q_{d}\) and \(D_{0}\) are \(3.63 \mathrm{eV} /\) atom and \(7.85 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), respectively.

An FCC iron-carbon alloy initially containing \(0.55 \mathrm{wt} \% \mathrm{C}\) 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 \(0 \mathrm{wt} \%\) C. (This process of carbon depletion is termed decarburization.)At what position will the carbon concentration be \(0.25 \mathrm{wt} \%\) after a \(10-\mathrm{h}\) treatment? The value of \(D\) at \(1325 \mathrm{~K}\) is \(3.3 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\).

Self-diffusion involves the motion of atoms that are all of the same type; therefore, it is not subject to observation by compositional changes, as with interdiffusion. Suggest one way in which selfdiffusion may be monitored.

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, \(p_{\mathrm{N}_{2}}\) (in \(\left.\mathrm{MPa}\right)\), and absolute temperature \((T)\) according to $$ C_{\mathrm{N}}=4.90 \times 10^{-3} \sqrt{p_{\mathrm{N}_{2}}} \exp \left(-\frac{37,600 \mathrm{~J} / \mathrm{mol}}{R T}\right) $$ Furthermore, the values of \(D_{0}\) and \(Q_{d}\) for this diffusion system are \(5.0 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) and \(77,000 \mathrm{~J} / \mathrm{mol}\), respectively. Consider a thin iron membrane 1.5-mm thick 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

The steady-state diffusion flux through a metal plate is \(7.8 \times 10^{-8} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\) at a temperature 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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