- #1

- 699

- 5

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].

$$