21-Mat-A6 Materials Selection and Design for Materials Processing · May 2013
Question 8 of 8: Carbide/Nitride Solubility in Austenite; AlN in Hot-Rolled Steel
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
10-Met-A6 — Phase Transformation & Thermal Treatment of Metals & Alloys — National Exams, May 2013 — 3 hours — 8 questions printed, first 5 as answered are marked (all 8 answered below as a complete study resource).
Reference texts: Callister & Rethwisch, Materials Science and Engineering: An Introduction, 10th ed.; Porter, Easterling & Sherif, Phase Transformations in Metals and Alloys, 3rd ed.; ASM Handbook Vol. 4, Heat Treating; Krauss, Steels: Processing, Structure, and Performance.
Check: this paper's printed exam code is 10-Met-A6 and its printed title is “Phase Transformation & Thermal Treatment of Metals & Alloys.” Every question below is genuinely phase-transformation/heat-treatment content (Fe-C diagram heat treatments, TTT curves, precipitation/spinodal decomposition, nucleation mechanisms, Cu-alloy heat treatments, furnace design, carbide/nitride solubility).
Question 8: Carbide/Nitride Solubility in Austenite; AlN in Hot-Rolled Steel (20 marks)
For the dissolution/precipitation equilibrium $[MX]_\gamma \rightleftharpoons [M]_\gamma + [X]_\gamma$, the standard free energy of reaction relates to the equilibrium constant by
$$\Delta G^{\circ} = -RT\ln K, \qquad K = \frac{a_M\,a_X}{a_{MX}}$$
where $a_M$, $a_X$ are the activities of dissolved M and X in austenite and $a_{MX}$ is the activity of the (solid) compound MX. Taking $\log_{10}$ and substituting $\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}$:
$$\log_{10}\!\left(\frac{a_{MX}}{a_M a_X}\right) = \frac{\Delta S^{\circ}}{2.303\,R} - \frac{\Delta H^{\circ}}{2.303\,R\,T}$$
Matching this to the printed form $\log_{10}[\text{wt.\%M}][\text{wt.\%X}] = A - B/T$ requires three simplifying assumptions:
MX is a stoichiometric, essentially pure solid compound. Taking $a_{MX}=1$ removes it from the equilibrium expression entirely, leaving $K^{-1}=a_M a_X$, so the left side of Eq. (1) is $-\log_{10}(a_M a_X)$ rather than a genuine equilibrium constant — valid only if MX carries negligible solubility for other species and departs negligibly from ideal stoichiometry.
M and X obey Henry's law (dilute-solution behaviour) in austenite. At the low concentrations typical of microalloying/interstitial additions, the activities are taken proportional to concentration, $a_M=\gamma_M[\text{wt.\%M}]$ with $\gamma_M$ treated as a constant independent of composition; this lets the printed equation use raw weight-percent values in place of thermodynamic activities, with the (assumed T-independent, or only present through $A$) activity coefficients absorbed into the fitted constant $A$.
$\Delta H^{\circ}$ and $\Delta S^{\circ}$ of the reaction are independent of temperature over the fitted range (equivalent to neglecting any heat-capacity difference between products and reactants, $\Delta C_p\approx0$), which is what allows $B=\Delta H^{\circ}/2.303R$ and $A=\Delta S^{\circ}/2.303R$ to be reported as single constants rather than functions of $T$.
Under these three assumptions, $A=\Delta S^{\circ}/2.303R$ and $B=\Delta H^{\circ}/2.303R$, i.e. Eq. (1) is exactly the van't-Hoff form of the solubility product, re-expressed in base-10 log and weight-percent units.
8.2 — (b) Maximum tolerable Al at 900 °C for 40 ppm N, without AlN precipitation
Given. $\log_{10}[\text{wt.\%Al}][\text{wt.\%N}] = A - B/T$ with $A=1.55$, $B=7060$; hot-rolling temperature $900\,{}^{\circ}\text{C} = 1173.15\ \text{K}$; nitrogen content $40\ \text{ppm} = 0.0040\ \text{wt.\%N}$.
Find. The maximum wt.%Al that keeps the steel just below the AlN solubility limit (i.e. the solubility PRODUCT, not exceeded) at 900 °C.
Approach. Evaluate the solubility product from Eq. (1) at $T=1173.15\ \text{K}$, then divide by the known wt.%N to isolate the maximum wt.%Al.
Evaluate the right-hand side of Eq. (1) at 900 °C.
$$\log_{10}[\text{Al}][\text{N}] = 1.55 - \frac{7060}{1173.15} = 1.55 - 6.018 = -4.468$$
Convert to the solubility product.
$$[\text{wt.\%Al}][\text{wt.\%N}] = 10^{-4.468} = 3.404\times10^{-5}$$
Solve for the maximum wt.%Al at the given nitrogen level. With $[\text{wt.\%N}]=0.0040$,
$$[\text{wt.\%Al}]_{\max} = \frac{3.404\times10^{-5}}{0.0040} = 8.511\times10^{-3}\ \text{wt.\%}$$
$$\boxed{[\text{wt.\%Al}]_{\max} = 0.00851\ \text{wt.\%} \approx 85.1\ \text{ppm}}$$
Provided the steel's total aluminum content stays at or below about 85 ppm (0.0085 wt.%) at a 40 ppm nitrogen level, the product $[\text{Al}][\text{N}]$ remains below the Eq. (1) solubility limit at 900 °C and AlN will not precipitate during hot rolling; any aluminum content above this — common in fully aluminum-killed steels, which typically carry well over 200–300 ppm total Al — will drive AlN precipitation at this temperature, which is in fact the intended microalloying effect used for austenite grain-pinning in those grades, at the cost of the free (uncombined) nitrogen available for other reactions.
