An FCC iron-carbon alloy initially containing 0.10 wt \(\% \mathrm{C}\) is carburized at an elevated temperature and in an atmosphere wherein the surface carbon concentration is maintained at 1.10 wt\%. If after 48 h the concentration of carbon is \(0.30 \mathrm{wt} \%\) at a position \(3.5 \mathrm{mm}\) below the surface, determine the temperature at which the treatment was carried out.

Short Answer

Expert verified
Answer: The carburization temperature for the given FCC iron-carbon alloy is approximately 1226 Kelvin.

Step by step solution

01

Identify relevant equations and constants

To solve this problem, we need Fick's second law of diffusion, as well as the diffusion coefficient equation for carbon in iron, which depends on temperature. Here are the equations: 1. Fick's second law of diffusion: \(C(x,t) = C_{\infty} - (C_{\infty} - C_{0}) \cdot \operatorname{erf}\Big(\frac{x}{2\sqrt{Dt}}\Big)\) 2. Diffusion coefficient equation for carbon in FCC iron: \(D = D_{0} \cdot \exp\Big(-\frac{Q_{\text{d}}}{RT}\Big)\) For carbon in FCC iron, the constants are as follows: - \(D_{0} = 2.3 × 10^{-5} \, \mathrm{m^2/s}\) - \(Q_{\text{d}} = 1.4 \times 10^{5} \, \mathrm{J/mol}\)
02

Calculate the diffusion coefficient

We can use Fick's second law of diffusion to find the diffusion coefficient \(D\), which is needed to find the temperature. First, rearrange the equation to solve for D: \(D = \frac{x^2}{4t}\Big(\operatorname{erf}^{-1}\Big(\frac{C_{\infty} - C(x,t)}{C_{\infty} - C_{0}}\Big)\Big)^2\) Substitute the given values: - \(C_{\infty} = 1.10 \,\mathrm{wt\%}\) - \(C_{0} = 0.10 \,\mathrm{wt\%}\) - \(C(x,t) = 0.30 \,\mathrm{wt\%}\) - \(x = 3.5 \,\mathrm{mm} = 3.5 \times 10^{-3} \,\mathrm{m}\) - \(t = 48 \,\mathrm{h} = 48 × 3600 \,\mathrm{s}\) - \(\operatorname{erf}^{-1}()\) is the inverse error function \(D = \frac{(3.5 \times 10^{-3}\,\mathrm{m})^2}{4(48 × 3600 \,\mathrm{s})(\operatorname{erf}^{-1}(\frac{1.10 - 0.30}{1.10 - 0.10}))^2} ≈ 5.74 × 10^{-12} \, \mathrm{m^2/s}\)
03

Calculate the carburization temperature

With the diffusion coefficient calculated, we can now use the diffusion coefficient equation for carbon in FCC iron to find the temperature of the treatment. Rearrange the diffusion coefficient equation to solve for the temperature T: \(T = \frac{Q_{\text{d}}}{R\cdot \ln(D_{0}/D)}\) Use the known values: - \(D = 5.74 × 10^{-12} \, \mathrm{m^2/s}\) - \(D_{0} = 2.3 × 10^{-5} \, \mathrm{m^2/s}\) - \(Q_{\text{d}} = 1.4 \times 10^{5} \, \mathrm{J/mol}\) - \(R = 8.314 \, \mathrm{J/(mol\cdot K)}\) \(T = \frac{1.4 × 10^5\,\mathrm{J/mol}}{8.314\,\mathrm{J/(mol\cdot K)}\cdot \ln\Big(\frac{2.3 × 10^{-5}\,\mathrm{m^2/s}}{5.74 × 10^{-12}\,\mathrm{m^2/s}}\Big)} ≈ 1226 \, \mathrm{K}\) The carburization temperature for the given FCC iron-carbon alloy is approximately 1226 Kelvin.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Briefly explain the concept of a driving force. (b) What is the driving force for steady-state diffusion?

Briefly explain the difference between self diffusion and inter diffusion.

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

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 \(5.0 \mathrm{MPa}(49.3 \mathrm{atm})\).

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 \(1.9 \times 10^{-11} \mathrm{m}^{2} / \mathrm{s}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free