Case Hardening of a Steel Rod: 23-Chem-A3 Heat and Mass Transfer, May 2013, Q7

Question 7: Case Hardening of a Steel Rod (Part C — 30%)

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. Semi-infinite steel rod, initial carbon $C_0=0.2$ wt% (uniform), surface held at the equilibrium $C_i=1.5$ wt% from $t=0$. Diffusivity $D=5.6\times10^{-10}\ \text{m}^2/\text{s}$. Target $C=0.8$ wt% at depth $x=1.0\ \text{mm}=10^{-3}$ m.

Find. The time $t$ to reach 0.8 wt% carbon at 1.0 mm depth.

Figure 7 — Carbon penetrates from the fixed-composition surface into the rod. The dimensionless concentration (C−C₀)/(Cᵢ−C₀) is the erfc of the similarity variable x/2√(Dt) — the "∞" curve of the attached chart (surface concentration held constant).

Approach. The rod is semi-infinite with a fixed surface concentration, so use the complementary-error-function solution; form the dimensionless concentration, read (or invert) the similarity variable, and solve for $t$.

Dimensionless concentration. $$\frac{C-C_0}{C_i-C_0}=\frac{0.8-0.2}{1.5-0.2}=\boxed{0.4615}=\operatorname{erfc}(\eta),\qquad \eta=\frac{x}{2\sqrt{Dt}}.$$ This is the "$\infty$" (constant-surface-concentration) curve of the attached chart.
Invert for the similarity variable. $\operatorname{erfc}(\eta)=0.4615\Rightarrow\operatorname{erf}(\eta)=0.5385$, which gives $$\eta=0.521.$$
Solve for the time. From $\eta=x/2\sqrt{Dt}$, $$t=\frac{x^{2}}{4D\eta^{2}}=\frac{(10^{-3})^{2}}{4(5.6\times10^{-10})(0.521)^{2}}=\boxed{1.65\times10^{3}\ \text{s}\approx27.4\ \text{min}}.$$

Quantity	Result
Dimensionless concentration $(C-C_0)/(C_i-C_0)$	0.4615
Similarity variable $\eta$	0.521
Time to 0.8 wt% at 1.0 mm	$\approx1.65\times10^{3}$ s ($\approx27$ min)

Concept & Theory

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

Key relations. $\dfrac{C-C_0}{C_i-C_0}=\operatorname{erfc}\!\big(x/2\sqrt{Dt}\big)$; equivalently $\dfrac{C_i-C}{C_i-C_0}=\operatorname{erf}(\eta)$. Solve $\eta$ then $t=x^{2}/(4D\eta^{2})$.

Why this works. A "large-diameter" rod means diffusion is one-dimensional (into a half-space) over the short penetration depth, and the surface composition is pinned at the phase-equilibrium value — exactly the constant-surface-concentration semi-infinite solution. The attached chart is just a graphical erfc; the $\infty$ curve corresponds to negligible surface (convective) resistance.

Common pitfall. Choosing a finite-surface-coefficient curve instead of the $\infty$ curve, or mixing the erf and erfc forms; both give the same $\eta$ only if the ratio is set up consistently.

Source. Geankoplis §7.1 (unsteady-state diffusion, semi-infinite solid); Welty et al. §27; the attached dimensionless chart (Welty Fig. for a semi-infinite medium).

Question 7 of 7: Case Hardening of a Steel Rod

Question 7: Case Hardening of a Steel Rod (Part C — 30%)