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Chapter 5: Problem 14

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 \(3.3 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\).

Short Answer

Expert verified
Answer: The position from the surface where the carbon concentration will be 0.25 wt% C after a 10-hour treatment is approximately 1.31 mm.

Step by step solution

01

Identify the Parameters

We are given: - Initial carbon concentration \(C_0 = 0.55\) wt% C, - Surface carbon concentration \(C_s = 0\) wt% C, - Treatment time \(t = 10 \ \mathrm{h}\) (which we'll convert into seconds), - Diffusion coefficient \(D = 3.3 \times 10^{-11} \ \mathrm{m}^2/\mathrm{s}\), - Desired carbon concentration after treatment \(C(x, t) = 0.25\) wt% C.
02

Convert Time to Seconds

We need to use the same units for time as in the diffusion coefficient, which is in seconds. Thus, we convert the treatment time from hours to seconds: \(t = 10 \ \mathrm{h} \times \frac{3600 \ \mathrm{s}}{1 \ \mathrm{h}} = 36000 \ \mathrm{s}\)
03

Rearrange Fick's Second Law

We rearrange Fick's second law to isolate x: \(x = 2\sqrt{Dt} \times erf^{-1}\left(\frac{C_0 - C(x, t)}{C_0 - C_s}\right)\)
04

Plug in the Values and Calculate x

Plug in the values and calculate x: \(x = 2\sqrt{(3.3 \times 10^{-11} \ \mathrm{m}^2/\mathrm{s})(36000 \ \mathrm{s})} \times erf^{-1}\left(\frac{0.55 - 0.25}{0.55 - 0}\right)\) \(x = 2\sqrt{1.188 \times 10^{-6} \ \mathrm{m}^2} \times erf^{-1}(0.545)\) \(x \approx 2(0.00109 \ \mathrm{m}) \times 0.601\) \(x \approx 0.00131 \ \mathrm{m}\) So, after a 10-hour treatment, the position where the carbon concentration will be 0.25 wt% C is approximately 1.31 mm from the surface.

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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.

Antimony atoms are to be diffused into a silicon wafer using both predeposition and drive-in heat treatments; the background concentration of \(\mathrm{Sb}\). in this silicon material is known to be \(2 \times 10^{20}\) atoms/m^3. The predeposition treatment is to be conducted at \(900^{\circ} \mathrm{C}\) for \(1 \mathrm{~h}\); the surface concentration of \(\mathrm{Sb}\) is to be maintained at a constant level of \(8.0 \times 10^{25}\) atoms \(/ \mathrm{m}^{3}\). Drive-in diffusion will be carried out at \(1200^{\circ} \mathrm{C}\) for a period of \(1.75 \mathrm{~h}\). For the diffusion of \(\mathrm{Sb}\) in Si, values of \(Q_{d}\) and \(D_{0}\) are \(3.65 \mathrm{eV} /\) atom and \(2.14 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), respectively. (a) Calculate the value of \(Q_{0}\) - (b) Determine the value of \(x_{j}\) for the drive-in diffusion treatment. (c) Also, for the drive-in treatment, compute the position \(x\) at which the concentration of \(S b\) atoms is \(5 \times 10^{23}\) atoms/m \(^{3}\)

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 \(6-\mathrm{mm}\) 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 respective 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 \(1.85 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{s}\), and the diffusion flux is found to be \(1.0 \times 10^{-7} \mathrm{~kg} / \mathrm{m}^{2} \cdot\) 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 high-pressure side will the concentration be \(0.5 \mathrm{~kg} / \mathrm{m}^{3}\) ? Assume a linear concentration profile.

The diffusion coefficients for nickel in iron are given at two temperatures, as follows: \begin{tabular}{cc} \hline \(\boldsymbol{T}(\boldsymbol{K})\) & \(\boldsymbol{D}\left(\mathrm{m}^{2} / \mathrm{s}\right)\) \\ \hline 1473 & \(2.2 \times 10^{-15}\) \\ \hline 1673 & \(4.8 \times 10^{-14}\) \\ \hline \end{tabular} (a) Determine the values of \(D_{0}\) and the activation energy \(Q_{d^{\prime}}\) (b) What is the magnitude of \(D\) at \(1300^{\circ} \mathrm{C}(1573 \mathrm{~K}) ?\)
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