Mass Diffusion and Heat Conduction: A Steel Carburization Example

In summary, the equation for mass diffusion without chemical reactions is similar to that of heat conduction. For an example problem of steel carburization, the concentration of carbon can be increased by diffusing it into the steel at a temperature of 1200^{\circ}C with a diffusion coefficient of D_{AB}\approx 5.6\times 10^{-10} (\text{m}^2/\text{sec}). To find the time required to reach a concentration of 1% at a depth of 1-mm below the surface, we use the boundary conditions and initial condition to solve for the coefficients in the general solution, which is given by C_A(x,t) = \frac{.008}{\pi
  • #1
Dustinsfl
2,281
5
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 [itex]C_A[/itex] is the concentration of the species [itex]A[/itex] diffusing into a medium [itex]B[/itex] and [itex]D_{AB}[/itex] 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 [itex]1200^{\circ}[/itex]C the value of the diffusion coefficient is approximately [itex]D_{AB}\approx 5.6\times 10^{-10}[/itex] [itex](\text{m}^2/\text{sec})[/itex].
Suppose that a 1cm thick slab initially has a uniform carbon concentration of [itex]C_A = 0.2[/itex]%.
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 [itex]C_A(0,t) = C_A(1,t) = 0[/itex] and the initial condition is [itex]C_A(x,t) = .002[/itex].
We are looking for solutions of the form [itex]C_A(x,t) = \varphi(x)\psi(t)[/itex].
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 [itex]\varphi(x)[/itex] 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
$$
C_A(x,t) = \frac{.008}{\pi}\sum_{n = 1}^{\infty}\frac{\sin x(2n - 1)\pi}{2n - 1}\exp\left[-D_{AB}(2n - 1)^2\pi^2 t\right].
$$
 
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  • #2
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 [itex]1200^{\circ}[/itex]C the value of the diffusion coefficient is approximately [itex]D_{AB}\approx 5.6\times 10^{-10}[/itex] [itex](\text{m}^2/\text{sec})[/itex].
Suppose that a 1cm thick slab initially has a uniform carbon concentration of [itex]C_A = 0.2\%[/itex].
Each side of the slab is exposed to a carbon-rich environment with
a constant concentration of [itex]C_A = 1.5\%[/itex]. (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[itex]\%[/itex]?
[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 [itex]C_A(0,t) = C_A(1,t) = 0.015[/itex] and the initial condition is [itex]C_A(x,t) = .002[/itex].
The solution to
$$
C_{A_{ss}} = 0.015.
$$
We are looking for solutions of the form [itex]C_{A_{\text{trans}}}(x,t) = \varphi(x)\psi(t)[/itex].
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 [itex]d_{AB} = 5.6\times 10^{-6}[/itex] [itex](\text{cm}^2/\text{sec})[/itex].
The family of solutions for [itex]\varphi(x)[/itex] 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 [itex]\lambda = 0[/itex] for equation (1).
We would have [itex]\varphi = A[/itex] which is not the general solution of [itex]\varphi'' = 0[/itex].
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
$$
C_A(x,t) = C_{A_{ss}} + C_{A_{\text{trans}}} = 0.015 + \frac{0.052}{\pi}\sum_{n = 1}^{\infty}\frac{\sin x(2n - 1)\pi}{2n - 1}\exp\left[-d_{AB}(2n - 1)^2\pi^2 t\right].
$$
 
  • #3
Dustinsfl said:
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 [itex]C_A[/itex] is the concentration of the species [itex]A[/itex] diffusing into a medium [itex]B[/itex] and [itex]D_{AB}[/itex] 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 [itex]1200^{\circ}[/itex]C the value of the diffusion coefficient is approximately [itex]D_{AB}\approx 5.6\times 10^{-10}[/itex] [itex](\text{m}^2/\text{sec})[/itex].
Suppose that a 1cm thick slab initially has a uniform carbon concentration of [itex]C_A = 0.2[/itex]%.
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 [itex]C_A(0,t) = C_A(1,t) = 0[/itex] and the initial condition is [itex]C_A(x,t) = .002[/itex].
We are looking for solutions of the form [itex]C_A(x,t) = \varphi(x)\psi(t)[/itex].
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 [itex]\varphi(x)[/itex] 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.
 

1. What is mass diffusion and heat conduction?

Mass diffusion is the movement of particles from an area of high concentration to an area of low concentration, resulting in the mixing of substances. Heat conduction is the transfer of thermal energy from a region of higher temperature to a region of lower temperature through a medium.

2. How does mass diffusion and heat conduction apply to steel carburization?

In the steel carburization process, carbon is diffused into the surface of the steel to increase its hardness. This is achieved through the use of heat, which allows for the carbon molecules to diffuse into the steel. The rate of diffusion and the resulting hardness of the steel are dependent on the temperature and time of the carburization process.

3. What factors affect the rate of mass diffusion and heat conduction in steel carburization?

The rate of mass diffusion and heat conduction in steel carburization is affected by the temperature and time of the process, as well as the carbon potential of the carburizing atmosphere. Other factors such as the composition and microstructure of the steel also play a role in the diffusion process.

4. How is mass diffusion and heat conduction measured in steel carburization?

The rate of mass diffusion and heat conduction in steel carburization can be measured through techniques such as thermogravimetry, which measures the weight change of the steel due to the diffusion of carbon, and dilatometry, which measures the change in dimensions of the steel due to thermal expansion and contraction.

5. Are there any other industrial applications of mass diffusion and heat conduction?

Yes, mass diffusion and heat conduction are important processes in a variety of industrial applications such as surface hardening, welding, and sintering. They are also essential in fields such as materials science, chemical engineering, and metallurgy.

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