Chapter 5.7, Problem 34SEP

To determine

The flux of carbon atoms between the surface and plane $25\text{μm}$ deep.

Answer to Problem 34SEP

The flux of carbon atoms between the surface and plane $25\text{μm}$ deep is $3.76\times {10}^{21}{\text{atoms/m}}^2\cdot \text{s}$.

Explanation of Solution

It is given that the 1018 steel having $0.8\text{wt\%}$ at $1000°\text{C}$. It means that $0.8\text{g}$ of carbon and $99.2\text{g}$ iron in a $100\text{g}$ steel.

The number of carbon $\left(\text{C}\right)$ atoms present in $0.8\text{g}$ of carbon is expressed as follows.

  $\begin{array}{c}{N}_{\text{C}}=\frac{0.8\text{g}}{\text{molecular weight of carbon}}\times \text{Avogadro's number}\\ =\frac{0.8\text{g}}{12.01\text{g}}\times \left(6.02\times {10}^{23}\right)\\ =4\times {10}^{22}\text{atoms}\end{array}$

The number of iron $\left(\text{Fe}\right)$ atoms present in $99.2\text{g}$ of iron is expressed as follows.

  $\begin{array}{c}{N}_{\text{Fe}}=\frac{99.2\text{g}}{\text{molecular weight of iron}}\times \text{Avogadro's number}\\ =\frac{99.2\text{g}}{55.85\text{g}}\times \left(6.02\times {10}^{23}\right)\\ =1.07\times {10}^{24}\text{atoms}\end{array}$

The atom % of carbon is expressed as follows.

  $\begin{array}{c}\text{atom}\text{\%}\text{carbon}=\frac{N_{\text{C}}}{N_{\text{C}}+N_{\text{Fe}}}\times 100\\ =\frac{4\times {10}^{22}\text{atoms}}{\left(4\times {10}^{22}\text{atoms}\right)+\left(1.07\times {10}^{24}\text{atoms}\right)}\times 100\\ =3.6\%\end{array}$

Thus, the concentration of carbon surface is $3.6a\%$.

Write the formula for lattice constant $\left(a\right)$ of unit of FCC iron.

  $a=\frac{4R}{\sqrt{2}}$        (I)

Here, the radius of the iron atom is $R$.

Write the formula for volume $\left(V\right)$ of the unit cell.

  $V=a^3$        (II)

Write the formula for Fick’s law of diffusion.

  $J=-D\frac{dC}{dx}$        (III)

Here, flux or flow of atoms is $J$ in $\frac{\text{atoms}}{{\text{m}}^2\cdot \text{s}}$, constant of diffusivity is $D$ in $\frac{{\text{m}}^2}{\text{s}}$ , and concentration gradient is $\frac{dC}{dx}$ in $\frac{\text{atoms}}{{\text{m}}^4}$.

Conclusion:

Refer Table 3.2, “Selected metals that have the BCC crystal structure at room temperature (20°C) and their lattice constants and atomic radii”.

The atomic radius of iron is $0.124\text{nm}$.

Substitute $0.124\text{}\text{nm}$ for $R$ in Equation (I).

  $\begin{array}{c}a=\frac{4\left(0.124\text{}\text{nm}\right)}{\sqrt{2}}\\ =0.351\text{nm}\end{array}$

Substitute $0.351\text{nm}$ for $a$ in Equation (II).

  $\begin{array}{c}V={\left(0.351\text{nm}\right)}^3\\ ={\left(0.351\text{nm}\times \frac{{\text{10}}^{-9}\text{m}}{1\text{nm}}\right)}^3\\ =4.32\times {10}^{-24}{\text{m}}^3\end{array}$

The number of atoms per unit volume is expressed as follows.

  $\begin{array}{c}\text{atoms per unit volume}=\frac{\text{no}\text{. of atoms in FCC crystal}}{\text{volume}\text{of unit cell}\left(V\right)}\\ =\frac{4\text{atoms}}{4.32\times {10}^{-24}{\text{m}}^3}\\ =9.25\times {10}^{28}{\text{atoms/m}}^3\end{array}$

The concentration of atoms at the surface $C_s$ is $3.6a\%$.

  $\begin{array}{c}{C}_s=3.6a\%\\ =\frac{3.6}{100}\left(9.25\times {10}^{28}{\text{atoms/m}}^3\right)\\ =3.33\times {10}^{27}{\text{atoms/m}}^3\end{array}$

The concentration of atom at the distance of $25\text{μm}$ below the surface $C_{25\text{μm}}$ is $0.18a\%$.

  $\begin{array}{c}{C}_{25\text{μm}}=0.18a\%\\ =\frac{0.18}{100}\left(9.25\times {10}^{28}{\text{atoms/m}}^3\right)\\ =1.66\times {10}^{26}{\text{atoms/m}}^3\end{array}$

Refer Table 5.2, “Diffusivities at 500°C and 1000°C for selected solute-solvent diffusion systems”.

The diffusivity of carbon in FCC iron at 1000°C is $3\times {10}^{-11}{\text{m}}^2\text{/s}$.

Here,

  $\begin{array}{c}dC=C_{25\text{μm}}-C_s\\ =\left(1.66\times {10}^{26}{\text{atoms/m}}^3\right)-\left(3.33\times {10}^{27}{\text{atoms/m}}^3\right)\\ =-3.134\times {10}^{27}{\text{atoms/m}}^3\end{array}$

Substitute $3\times {10}^{-11}{\text{m}}^2\text{/s}$ for $D$, $-3.134\times {10}^{27}{\text{atoms/m}}^3$ for $dC$, and $25\text{μm}$ for $dx$ in Equation (III).

  $\begin{array}{c}J=-\left(3\times {10}^{-11}{\text{m}}^2\text{/s}\right)\frac{-3.134\times {10}^{27}{\text{atoms/m}}^3}{25\text{μm}}\\ =-\left(3\times {10}^{-11}{\text{m}}^2\text{/s}\right)\frac{-4.575\times {10}^{26}{\text{atoms/m}}^3}{25\text{μm}\times \frac{{10}^{-6}\text{m}}{\text{1}\text{μm}}}\\ =3.76\times {10}^{21}{\text{atoms/m}}^2\cdot \text{s}\end{array}$

Thus, the flux of carbon atoms between the surface and plane $25\text{μm}$ deep is $3.76\times {10}^{21}{\text{atoms/m}}^2\cdot \text{s}$.