Mass Diffusion and Heat Conduction: A Steel Carburization Example

[Dustinsfl]

One can show that mass diffusion without chemical reactions obeys the same basic equation as heat conduction.
The one dimensional equation in dimensionless variables is given by
$$
D_{AB}\frac{\partial^2 C_A}{\partial x^2} = \frac{\partial C_A}{\partial t}
$$
where C_A is the concentration of the species A diffusing into a medium B and D_{AB} is the mass diffusivity.
You should compare this to the analogous heat conduction problem.
As an example problem, consider of steel carburization.
In this high-temperature process carbon is diffused into the steel to achieve certain desirable material properties (e.g., increase tensile strength).
When the carburization is performed at a temperature 1200^{\circ}C the value of the diffusion coefficient is approximately D_{AB}\approx 5.6\times 10^{-10} (\text{m}^2/\text{sec}).
Suppose that a 1cm thick slab initially has a uniform carbon concentration of C_A = 0.2%.
How much time is required to raise the carbon concentration at a depth of 1-mm below the slab surface to a value of 1%?
[Reminder: The transient solution requires homogeneous boundary conditions; the easiest way to achieve this is by rescaling/redefining the concentration variable.]

How do I find the time for when the value 1%

The boundary conditions are C_A(0,t) = C_A(1,t) = 0 and the initial condition is C_A(x,t) = .002.
We are looking for solutions of the form C_A(x,t) = \varphi(x)\psi(t).
We have that
\begin{alignat}{3}
\varphi(x) & = & A\cos x\lambda + B\sin x\lambda
\end{alignat}
and
$$
\psi(t) = C\exp\left[-D_{AB}\lambda^2 t\right].
$$
The family of solutions for \varphi(x) can be obtained by
\begin{alignat*}{3}
\varphi_1(0) = 1 & \quad & \varphi_2(0) = 0\\
\varphi_1'(0) = 0 & \quad & \varphi_2'(0) = 1
\end{alignat*}
This leads us to
\begin{alignat}{3}
\varphi(x) & = & A\cos x\lambda + B\frac{\sin x\lambda}{\lambda}.
\end{alignat}
Now, we can use the boundary conditions.
We have that
$$
\varphi_n(x) = \sin x\lambda = 0,\quad \lambda = \pi n, \quad n\in\mathbb{Z}.
$$
Our general solution is of the form
$$
C_A(x,t) = \frac{1}{\pi}\sum_{n = 1}^{\infty}A_n\frac{\sin xn\pi}{n}\exp\left[-D_{AB}\lambda^2 t\right].
$$
Finally, we can use the initial condition to solve for the Fourier coefficients.
$$
A_n = \frac{.004}{n\pi}\int_0^1\sin xn\pi dx = \begin{cases}
0, & \text{if n is even}\\
.008, & \text{if n is odd}
\end{cases}
$$
Therefore, the solution is
$$
C_A(x,t) = \frac{.008}{\pi}\sum_{n = 1}^{\infty}\frac{\sin x(2n - 1)\pi}{2n - 1}\exp\left[-D_{AB}(2n - 1)^2\pi^2 t\right].
$$

 

[Dustinsfl]

As an example problem, consider of steel carburization.
In this high-temperature process carbon is diffused into the steel to achieve certain desirable material properties (e.g., increase tensile strength).
When the carburization is performed at a temperature 1200^{\circ}C the value of the diffusion coefficient is approximately D_{AB}\approx 5.6\times 10^{-10} (\text{m}^2/\text{sec}).
Suppose that a 1cm thick slab initially has a uniform carbon concentration of C_A = 0.2\%.
Each side of the slab is exposed to a carbon-rich environment with
a constant concentration of C_A = 1.5\%. (Forgot this originally)
How much time is required to raise the carbon concentration at a depth of 1-mm below the slab surface to a value of 1\%?
[Reminder: The transient solution requires homogeneous boundary conditions; the easiest way to achieve this is by rescaling/redefining the concentration variable.]

1. Is this correct?
2. How do I solve for t when x = 0.1 and C_A(x,t) = .01?

The boundary conditions are C_A(0,t) = C_A(1,t) = 0.015 and the initial condition is C_A(x,t) = .002.
The solution to
$$
C_{A_{ss}} = 0.015.
$$
We are looking for solutions of the form C_{A_{\text{trans}}}(x,t) = \varphi(x)\psi(t).
We have that
\begin{alignat}{3}
\varphi(x) & = & A\cos x\lambda + B\sin x\lambda
\end{alignat}
and
$$
\psi(t) = C\exp\left[-d_{AB}\lambda^2 t\right]
$$
where d_{AB} = 5.6\times 10^{-6} (\text{cm}^2/\text{sec}).
The family of solutions for \varphi(x) can be obtained by
\begin{alignat*}{3}
\varphi_1(0) = 1 & \quad & \varphi_2(0) = 0\\
\varphi_1'(0) = 0 & \quad & \varphi_2'(0) = 1
\end{alignat*}
This leads us to
\begin{alignat}{3}
\varphi(x) & = & A\cos x\lambda + B\frac{\sin x\lambda}{\lambda}.
\end{alignat}
To see why this is correct, take \lambda = 0 for equation (1).
We would have \varphi = A which is not the general solution of \varphi'' = 0.
The general solution is
$$
\varphi = A + Bx.
$$
If we take equation (2), we get
$$
\varphi = A + B\lim_{\lambda\to 0}\frac{x\sin x\lambda}{x\lambda} = A + Bx
$$
as needed.
Now, we can use the boundary conditions.
We have that
$$
\varphi_n(x) = \sin x\lambda = 0,\quad \lambda = \pi n, \quad n\in\mathbb{Z}.
$$
Our general solution is of the form
$$
C_{A_{\text{trans}}}(x,t) = \frac{1}{\pi}\sum_{n = 1}^{\infty}A_n\frac{\sin xn\pi}{n}\exp\left[-d_{AB}\lambda^2 t\right].
$$
Using -0.013 as our initial condition, we can solve
$$
A_n = \frac{-0.026}{n\pi}\int_0^1\sin xn\pi dx = \begin{cases}
0, & \text{if n is even}\\
0.052, & \text{if n is odd}
\end{cases}
$$
Therefore, the solution is
$$
C_A(x,t) = C_{A_{ss}} + C_{A_{\text{trans}}} = 0.015 + \frac{0.052}{\pi}\sum_{n = 1}^{\infty}\frac{\sin x(2n - 1)\pi}{2n - 1}\exp\left[-d_{AB}(2n - 1)^2\pi^2 t\right].
$$

 

[Chestermiller]

One can show that mass diffusion without chemical reactions obeys the same basic equation as heat conduction.
The one dimensional equation in dimensionless variables is given by
$$
D_{AB}\frac{\partial^2 C_A}{\partial x^2} = \frac{\partial C_A}{\partial t}
$$
where C_A is the concentration of the species A diffusing into a medium B and D_{AB} is the mass diffusivity.
You should compare this to the analogous heat conduction problem.
As an example problem, consider of steel carburization.
In this high-temperature process carbon is diffused into the steel to achieve certain desirable material properties (e.g., increase tensile strength).
When the carburization is performed at a temperature 1200^{\circ}C the value of the diffusion coefficient is approximately D_{AB}\approx 5.6\times 10^{-10} (\text{m}^2/\text{sec}).
Suppose that a 1cm thick slab initially has a uniform carbon concentration of C_A = 0.2%.
How much time is required to raise the carbon concentration at a depth of 1-mm below the slab surface to a value of 1%?
[Reminder: The transient solution requires homogeneous boundary conditions; the easiest way to achieve this is by rescaling/redefining the concentration variable.]

How do I find the time for when the value 1%

The boundary conditions are C_A(0,t) = C_A(1,t) = 0 and the initial condition is C_A(x,t) = .002.
We are looking for solutions of the form C_A(x,t) = \varphi(x)\psi(t).
We have that
\begin{alignat}{3}
\varphi(x) & = & A\cos x\lambda + B\sin x\lambda
\end{alignat}
and
$$
\psi(t) = C\exp\left[-D_{AB}\lambda^2 t\right].
$$
The family of solutions for \varphi(x) can be obtained by
\begin{alignat*}{3}
\varphi_1(0) = 1 & \quad & \varphi_2(0) = 0\\
\varphi_1'(0) = 0 & \quad & \varphi_2'(0) = 1
\end{alignat*}
This leads us to
\begin{alignat}{3}
\varphi(x) & = & A\cos x\lambda + B\frac{\sin x\lambda}{\lambda}.
\end{alignat}
Now, we can use the boundary conditions.
We have that
$$
\varphi_n(x) = \sin x\lambda = 0,\quad \lambda = \pi n, \quad n\in\mathbb{Z}.
$$
Our general solution is of the form
$$
C_A(x,t) = \frac{1}{\pi}\sum_{n = 1}^{\infty}A_n\frac{\sin xn\pi}{n}\exp\left[-D_{AB}\lambda^2 t\right].
$$
Finally, we can use the initial condition to solve for the Fourier coefficients.
$$
A_n = \frac{.004}{n\pi}\int_0^1\sin xn\pi dx = \begin{cases}
0, & \text{if n is even}\\
.008, & \text{if n is odd}
\end{cases}
$$
Therefore, the solution is
$$
C_A(x,t) = \frac{.008}{\pi}\sum_{n = 1}^{\infty}\frac{\sin x(2n - 1)\pi}{2n - 1}\exp\left[-D_{AB}(2n - 1)^2\pi^2 t\right].
$$

Set x = 1 mm and make a graph of C vs t.

For this low value of the diffusion coefficient, you might consider treating the sample as a semi-infinite slab, and using the semi-infinite slab solution. In the semi-infinite slab solution, there is not enough time for the concentration profile to essentially penetrate to the centerline of the sample. This solution allows you to solve explicitly for the time. But, if the concentration does essentially penetrate to the centerline, the semi-infinite slab approximation becomes inaccurate.