At approximately what temperature would a specimen of \(\gamma\)-iron have to be carburized for \(2 \mathrm{~h}\) to produce the same diffusion result as at \(900^{\circ} \mathrm{C}\) for \(15 \mathrm{~h}\) ?

Short Answer

Expert verified
Answer: Approximately \(1161^{\circ}\mathrm{C}\).

Step by step solution

01

Understand the given information

We are given that diffusion at \(900^{\circ}\mathrm{C}\) for 15 hours yields a certain result. We need to find the temperature at which diffusion will take 2 hours to achieve the same result.
02

Write down the Arrhenius equation for diffusion coefficient (D)

The Arrhenius equation for the diffusion coefficient (D) is given by: \(D = D_0 \cdot \exp{(-\frac{Q}{RT})}\), where \(D_0\) is the pre-exponential factor, \(Q\) is the activation energy, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.
03

Determine the relationship between the two scenarios

The relationship between the two scenarios is given by: \(D_1t_1 = D_2t_2\), where \(D_1\) and \(D_2\) are the diffusion coefficients at temperatures \(T_1\) and \(T_2\), and \(t_1\) and \(t_2\) are the respective times.
04

Rewrite the relationship in terms of the Arrhenius equation

Using the Arrhenius equation for diffusion coefficient, we can rewrite the relationship as: \((D_0 \cdot \exp{(-\frac{Q}{R{T_1}}})t_1) = (D_0 \cdot \exp{(-\frac{Q}{R{T_2}}})t_2)\)
05

Cancel out common factors and solve for the unknown temperature \(T_2\)

Cancelling out the common factor \(D_0\) and substituting the given temperatures and times, we get: \(\exp{(-\frac{Q}{R\cdot(900+273.15)})\cdot(15 \cdot 3600)} = \exp{(-\frac{Q}{R \cdot T_2})\cdot(2 \cdot 3600)}\) Now, we need to isolate \(T_2\). Divide both sides by \(2 \cdot 3600\) and take the natural logarithm of both sides: \(-\frac{Q}{R\cdot(900+273.15)}\cdot 15\cdot 3600 = -\frac{Q}{R \cdot T_2} \cdot 2 \cdot 3600\) Next, divide both sides by \(-\frac{Q}{R} \cdot 3600\): \(\frac{1}{(900+273.15) \cdot 15} = \frac{1}{T_2 \cdot 2}\) Now, find \(T_2\): \(T_2 = \frac{1}{\frac{1}{(900+273.15) \cdot 15} \cdot 2}\)
06

Calculate the temperature in Celsius and conclude the solution

Calculating \(T_2\), we get \(T_2 \approx 1434.09 \mathrm{K}\), which is equal to \(T_2 \approx 1161^{\circ}\mathrm{C}\) in Celsius. So, a temperature of approximately \(1161^{\circ}\mathrm{C}\) would have to be applied to the specimen of \(\gamma\)-iron for 2 hours to produce the same diffusion result as at \(900^{\circ}\mathrm{C}\) for 15 hours.

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Most popular questions from this chapter

Aluminum atoms are to be diffused into a silicon wafer using both predeposition and drive-in heat treatments; the background concentration of \(\mathrm{Al}\) in this silicon material is known to be \(3 \times 10^{19}\) atoms/m \(^{3}\). The drive-in diffusion treatment is to be carried out at \(1050^{\circ} \mathrm{C}\) for a period of \(4.0 \mathrm{~h}\), which gives a junction depth \(x_{j}\) of \(3.0 \mu \mathrm{m}\). Compute the predeposition diffusion time at \(950^{\circ} \mathrm{C}\) if the surface concentration is maintained at a constant level of \(2 \times 10^{25}\) atoms \(/ \mathrm{m}^{3}\). For the diffusion of \(\mathrm{Al}\) in \(\mathrm{Si}\), values of \(Q_{d}\) and \(D_{0}\) are \(3.41\). \(\mathrm{eV} /\) atom and \(1.38 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\), respectively.

The outer surface of a steel gear is to be hardened by increasing its carbon content. The carbon is to be supplied from an external carbon-rich atmosphere, which is maintained at an elevated temperature. A diffusion heat treatment at \(850^{\circ} \mathrm{C}(1123 \mathrm{~K})\) for \(10 \mathrm{~min}\) increases the carbon concentration to \(0.90 \mathrm{wt} \%\) at a position \(1.0 \mathrm{~mm}\) below the surface. Estimate the diffusion time required at \(650^{\circ} \mathrm{C}(923 \mathrm{~K})\) to achieve this same concentration also at a) 1.0-mm position. Assume that the surface carbon content is the same for both heat treatments, which is maintained constant. Use the diffusion data in Table \(5.2\) for C diffusion in \(\alpha\)-Fe.

Nitrogen from a gaseous phase is to be dif? fused into pure iron at \(700^{\circ} \mathrm{C}\). If the surface concentration is maintained at \(0.1 \mathrm{wt} \% \mathrm{~N}\) what will be the concentration \(1 \mathrm{~mm}\) from the surface after \(10 \mathrm{~h}\) ? The diffusion coefficient for nitrogen in iron at \(700^{\circ} \mathrm{C}\) is \(2.5 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\).

The steady-state diffusion flux through a metal plate is \(5.4 \times 10^{-10} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\) at a temperature of \(727^{\circ} \mathrm{C}(1000 \mathrm{~K})\) and when the concentration gradient is \(-350 \mathrm{~kg} / \mathrm{m}^{4} .\) Calculate the diffusion flux at \(1027^{\circ} \mathrm{C}(1300 \mathrm{~K})\) for the same concentration gradient and assuming an activation energy for diffusion of \(125,000 \mathrm{~J} / \mathrm{mol}\).

Briefly explain the difference between selfdiffusion and interdiffusion.

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