Chapter 5, Problem 5.38P

Interpretation Introduction

Interpretation:

The minimum thickness of iron needs to be determined.

Concept introduction:

Calculate the concentration $\left(c_1\right)$ of hydrogen in ${\text{atoms/cm}}^3$ at the surface of

  $0.05\text{}\text{H}\text{atom/unit}\text{cell}$

  $c_1=\frac{0.05\text{H}\text{atom unit cell}}{{\left(\text{Lattice parameter}\right)}^3\text{/unit cell}}\mathrm{.............}\left(1\right)$

Answer to Problem 5.38P

The minimum thickness $\Delta x$ of iron is $0.179\text{cm}$.

Explanation of Solution

Calculate the concentration $\left(c_1\right)$ of hydrogen in ${\text{atoms/cm}}^3$ at the surface of $0.05H\text{atom/unit cell}$.

  ${\text{c}}_1=\frac{0.05H\text{atom/unit cell}}{{\left(\text{Lattice parameter}\right)}^3/\text{unit}\text{cell}}\mathrm{.........}\left(1\right)$

Form the selected physical properties of the metals table, select the lattice parameter of BCC iron as $2.866\dot{A}$.

Convert the unit of lattice parameter of $Fe\text{ from }\dot{A}\text{ to cm}$.

  $\begin{array}{c}\text{Lattice parameter}=2.866\dot{A}\left(\frac{{10}^{-8}cm}{1\dot{A}}\right)\\ =2.866\times {10}^{-8}cm\end{array}$

Substitute $2.866\times {10}^{-8}cm$ for the lattice parameter in equation $\left(1\right)$.

  $\begin{array}{c}{c}_1=\frac{0.05\text{ H atom/unit cell}}{{\left(2.866\times {10}^{-8}cm\right)}^3/\text{unit}}\\ =212.4\times {10}^{19}{\text{ atoms/cm}}^3\end{array}$

Calculate the concentration $\left(c_2\right)$ of hydrogen in ${\text{atoms/cm}}^3$ at the other surface of

  $0.001Hatom/unitcell$.

  $c_2=\frac{\text{0}\text{.001}\text{H}\text{atom/unit}\text{cell}}{{\left(\text{Lattice parameter}\right)}^3/\text{unit cell}}$

Substitute $2.866\times {10}^{-8}cm$ for the lattice parameter.

  $\begin{array}{c}{c}_2=\frac{0.001\text{H}\text{atom/unit cell}}{{\left(2.866\times {10}^{-8}cm\right)}^3/\text{Unit cell}}\\ =4.25\times {10}^{19}{\text{ atoms/cm}}^3\end{array}$

Calculate the concentration gradient $\left(\Delta c/\Delta x\right)$ of hydrogen in ${\text{atoms/cm}}^3.cm$.

  $\frac{\Delta c}{\Delta x}=\frac{c_2-c_1}{\Delta x}$

Here, the difference in concentration is $\Delta c$ and distance is $\Delta x$.

Substitute $212.4\times {10}^{19}{\text{ atoms/cm}}^3\text{ for }{c}_1\text{ and 4}\text{.25}\times {10}^{19}{\text{ atoms/cm}}^3\text{ for }{c}_2$.

  $\begin{array}{c}\frac{\Delta c}{\Delta x}=\frac{\text{4}\text{.25}\times {10}^{19}{\text{ atoms/cm}}^3-212.4\times {10}^{19}{\text{ atoms/cm}}^3}{\Delta x}\\ =\frac{-2.081\times {10}^{21}{\text{ atoms/cm}}^3.\text{cm}}{\Delta x}\end{array}$

Calculate the flux $\left(J\right)$ of hydrogen.

  $J=\frac{\rho N_A}{{\text{u}}_H}$

Here, density of hydrogen is $\rho$, Avogadro number is $N_A$, and atomic mass of hydrogen in $g/mol{\text{ is u}}_H$.

From the physical properties of the elements table, select the atomic mass of hydrogen as

  $1.0079g/mol$.

Consider that loss of hydrogen will not be more than $50$ gram per year in each $c{m}^2$ of iron.

Substitute $50gram/c{m}^3/year\text{ for }\rho .6.02\times {10}^{23}\text{atoms/mol for }{N}_A\text{ and 1}\text{.0079}\text{g/mol}$ for ${\text{u}}_H$.

  $\begin{array}{c}J=\frac{\left(50gram/c{m}^3/year\right)\left(6.02\times {10}^{23}atoms/mol\right)}{\left(1.0079g/mol\right)}\left(\frac{1year}{31.536{\times }^{6}s}\right)\\ =9.47\times {10}^{17}{\text{ atoms/cm}}^3.s\end{array}$

Calculate the diffusion coefficient using the Arrhenius type equation relation.

  $D=D_0\exp \left(\frac{-Q}{RT}\right)\mathrm{.........}\left(2\right)$

Here, diffusion coefficient is $D$, constant is $D_0$, activation energy is $Q$, gas constant is $R$ and absolute temperature is $T$.

Convert the unit temperature form degree Celsius to degree Kelvin.

  $\begin{array}{c}T=\left(400+273\right)K\\ =673K\end{array}$

From the diffusion data of the selected materials table, select the diffusion constant $D_0$ and activation energy for $H$ in BCC iron as $0.0012{\text{ cm}}^2/s\text{ and 3600}\text{cal/mol}$.

Substitute $0.0012{\text{ cm}}^2/s$ for $D_0$, $\text{3600}\text{cal/mol}$ for $Q,1.987cal/mol.K\text{ for }R$, and $673K\text{ for }T$ in equation $\left(2\right)$.

  $\begin{array}{c}D=\left(0.0012c{m}^2/s\right)\exp \left(\frac{-3600cal/mol}{\left(1.987cal/mol.K\right)\left(673K\right)}\right)\\ =\left(0.0012\right)\left(0.0677\right)\\ =8.128\times {10}^{-5}c{m}^2/s\end{array}$

Express the flux $\left(J\right)$ of hydrogen related to concentration gradient.

  $J=-D\frac{\Delta c}{\Delta x}$

Here, diffusion coefficient is $D$ and concentration gradient is $\Delta c/\Delta x$.

Calculate the minimum thickness $\Delta c$ of iron.

Substitute $9.47\times {10}^{17}{\text{ atoms/cm}}^3.s\text{ for }J,8.128\times {10}^{-5}c{m}^2/s\text{ for }D$, and

  $\frac{-2.081\times {10}^{-21}{\text{atoms/cm}}^3.cm}{\Delta x}\text{ for }\Delta c/\Delta x$

  $-\left(8.128\times {10}^{-5}c{m}^2/s\right)\left(\frac{-2.081\times {10}^{21}atoms/c{m}^3.cm}{\Delta x}\right)=9.47\times {10}^{17}{\text{ atoms/cm}}^3.s$

  $\begin{array}{l}\Delta x=\frac{2.081\times {10}^{21}}{1.165\times {10}^{22}}\\ \Delta x=0.179cm\end{array}$

Therefore, the minimum thickness $\Delta x$ of iron is $0.179cm$.

Conclusion

The minimum thickness $\Delta x$ of iron is $0.179cm$.