Carburizing a steel roller — semi-infinite diffusion: 23-Chem-A3 Heat and Mass Transfer, May 2015, Q6

Question 6: Carburizing a steel roller — semi-infinite diffusion (Part C — equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Carbon diffuses from a carbon-rich surface into steel. Over the short depth of interest (0.4 mm $\ll$ 5-mm radius) the roller behaves as a semi-infinite solid with a fixed surface concentration; the error-function solution applies.

Quantity	Value
Depth of interest $x$	0.4 mm $=4\times10^{-4}$ m
Surface carbon $c_s$	$\rho_\text{gas}=100$ kg/m³
Initial carbon $c_0$	0.2%$\times7850=15.7$ kg/m³
Target carbon $c_x$	0.5%$\times7850=39.25$ kg/m³
Diffusivity $D$	$2.0\times10^{-11}$ m²/s

Find. The carburizing time to reach 0.5% C at 0.4-mm depth.

Figure 6 — Carbon penetrates the steel from a carbon-rich surface. Because the target depth (0.4 mm) is small compared with the 5-mm radius, the roller is treated as a semi-infinite medium and the concentration follows the complementary-error-function profile.

Approach. Apply the semi-infinite constant-surface-concentration solution $\dfrac{c_s-c_x}{c_s-c_0}=\mathrm{erf}\!\left(\dfrac{x}{2\sqrt{Dt}}\right)$, invert the error function for the similarity variable $\eta$, then solve for $t$.

Concentrations on a consistent basis. Working in kg/m³ (the ratio is identical in mass-%): surface $c_s=100$, initial $c_0=0.002(7850)=15.7$, target $c_x=0.005(7850)=39.25$.
Error-function argument. $$\frac{c_s-c_x}{c_s-c_0}=\frac{100-39.25}{100-15.7}=0.721=\mathrm{erf}(\eta),\qquad \eta=\frac{x}{2\sqrt{Dt}}.$$
Invert the error function. $\mathrm{erf}(\eta)=0.721\Rightarrow\eta=0.765$ (by bisection / erf tables).
Solve for the time. $$t=\frac{1}{D}\left(\frac{x}{2\eta}\right)^2=\frac{1}{2.0\times10^{-11}}\left(\frac{4\times10^{-4}}{2(0.765)}\right)^2=3.42\times10^{3}\ \text{s}=\boxed{0.95\ \text{h}}.$$

Quantity	Result
erf argument (concentration ratio)	0.721
Similarity variable $\eta$	0.765
Carburizing time $t$	$3.4\times10^{3}$ s ≈ 0.95 h

Check

The two densities are not distractors: the surface carbon concentration is set by the carbon gas ($c_s=100$ kg/m³) while the initial and target carbon contents are mass fractions of the steel density. Because the ratio $(c_s-c_x)/(c_s-c_0)$ is dimensionless, it is unchanged whether concentrations are in kg/m³ or in mass-%, confirming the setup is self-consistent.

Concept & Theory

Topic. Transient diffusion into a semi-infinite solid with constant surface concentration (carburizing).

Key relations. $\dfrac{c_s-c_x}{c_s-c_0}=\mathrm{erf}\!\left(\dfrac{x}{2\sqrt{Dt}}\right)$; similarity variable $\eta=x/2\sqrt{Dt}$; penetration depth $\propto\sqrt{Dt}$.

Why this works. When the diffusion penetration ($\sqrt{Dt}\sim0.26$ mm here) is small compared with the body dimension (5-mm radius), the far boundary is never felt and the 1-D semi-infinite similarity solution applies exactly. The error function arises because the governing PDE reduces to an ODE in the single variable $\eta$.

Common pitfall. Mixing up which concentration is the surface value (the carbon-gas density, not a mass-% of steel), or using the wrong side of the erf definition. Also check the semi-infinite assumption by confirming $\sqrt{Dt}\ll$ radius before trusting the answer.

Source. Geankoplis, §7.1 (unsteady-state diffusion, semi-infinite medium); Welty et al., §27 (semi-infinite solid, error-function solution); Crank, §2.

Question 6 of 7: Carburizing a steel roller — semi-infinite diffusion

Question 6: Carburizing a steel roller — semi-infinite diffusion (Part C — equal value)