Question

A heated piece of steel, with a uniform initial carbon concentration of \(0.20\) percent by mass, was exposed to a carburizing atmosphere for an hour. Throughout the entire process, the carbon concentration on the surface was \(0.70\) percent. If the mass diffusivity of carbon in steel in this process was uniform at \(1 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\), determine the percentage of mass concentration of carbon at \(0.2 \mathrm{~mm}\) and $0.4 \mathrm{~mm}\( below the surface after the process. Answers: \)0.428$ percent, \(0.268\) percent

Short answer

Based on the given information and calculated results, write a short answer conclusion: After the carburizing process, the mass concentrations of carbon at depths of \(0.2\mathrm{~mm}\) and \(0.4\mathrm{~mm}\) are found to be \(0.428\%\) and \(0.268\%\) respectively, which were calculated using Fick's second law of diffusion. This indicates that the carbon diffused into the steel and its concentration varies with depth.

Step-by-step solution

Understand the problem and parameters

The given parameters are: - Initial carbon concentration: \(0.20\%\) - Surface carbon concentration: \(0.70\%\) - Mass diffusivity of carbon in steel: \(1 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\) - Depths below the surface: \(0.2 \mathrm{~mm}\) and \(0.4 \mathrm{~mm}\) - Duration of the process: \(1\) hour. We have to find the mass concentration of carbon at the depths of \(0.2 \mathrm{~mm}\) and \(0.4 \mathrm{~mm}\) after the carburizing process.

Apply Fick's second law of diffusion

Fick's second law relates the concentration of a diffusing species with time, position, and the diffusivity of the species. It can be written in the form: \(C(x,t) = C_s + (C_0 - C_s) \mathrm{erf} \left( \dfrac{x}{2\sqrt{Dt}} \right)\) Where: - \(C(x,t)\) is the concentration at depth \(x\) and time \(t\) - \(C_s\) is the surface concentration - \(C_0\) is the initial concentration - \(D\) is the mass diffusivity of the species - erf is the error function. In our case, we need to find \(C(0.2\mathrm{mm},1\mathrm{hour})\) and \(C(0.4\mathrm{mm},1\mathrm{hour})\).

Convert units

To use the given values and keep units consistent, we need to convert the time and depth into seconds and meters, respectively. \(0.2 \mathrm{~mm} = 2 \times 10^{-4} \mathrm{~m}\) \(0.4 \mathrm{~mm} = 4 \times 10^{-4} \mathrm{~m}\) \(1 \mathrm{~hour} = 3600 \mathrm{~s}\)

Calculate the concentrations using Fick's second law

We have now all the required information to use Fick's second law. We can find the concentration at each depth by substituting the given parameters and converted units into the equation: \(C(2 \times 10^{-4}\mathrm{m}, 3600\mathrm{s}) = 0.70 + (0.20 - 0.70) \mathrm{erf} \left( \dfrac{2 \times 10^{-4}\mathrm{m}}{2\sqrt{1 \times 10^{-11}\mathrm{m^2/s} \times 3600\mathrm{s}} } \right)\) \(C(2 \times 10^{-4}\mathrm{m}, 3600\mathrm{s}) = 0.428 \%\) \(C(4 \times 10^{-4}\mathrm{m}, 3600\mathrm{s}) = 0.70 + (0.20 - 0.70) \mathrm{erf} \left( \dfrac{4 \times 10^{-4}\mathrm{m}}{2\sqrt{1 \times 10^{-11}\mathrm{m^2/s} \times 3600\mathrm{s}} } \right)\) \(C(4 \times 10^{-4}\mathrm{m}, 3600\mathrm{s}) = 0.268 \%\) Thus, the percentage of mass concentrations of carbon at a depth of \(0.2\mathrm{~mm}\) and \(0.4\mathrm{~mm}\) after the carburizing process are \(0.428\%\) and \(0.268\%\), respectively.