21-Mat-A5 Phase Transformations and Thermal Treatment · December 2018

Question 2 of 8: Solubility Product of Carbides and Nitrides in Austenite

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, December 2018 — 12-Mtl-A5, Phase Transformations of Metals, Glasses and Ceramics. Three hours, closed book, approved Casio or Sharp calculator only. Eight questions of 20 marks each; the rubric states that five questions constitute a complete paper and only the first five appearing in the answer book are marked. All eight are answered here, because this set is a study resource rather than an exam script. Several sub-parts explicitly call for an essay-format answer, and the rubric rewards clarity and organisation, so those answers are written as structured prose with supporting sketches rather than as note form.

Note on the exam code

This December 2018 sitting's printed header reads 12-Mtl-A5, Phase Transformations of Metals, Glasses and Ceramics, covering the Fe-C phase diagram and microstructural design, precipitate solubility, precipitation hardening and spinodal decomposition, interfaces and precipitate-free zones, grain growth and Zener pinning, nucleation mechanisms, classical nucleation theory and constitutional supercooling, and glass and glass-ceramic processing.

Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:

D. A. Porter, K. E. Easterling and M. Y. Sherif, Phase Transformations in Metals and Alloys, 3rd ed. — nucleation and growth theory (Ch. 4), precipitation hardening and spinodal decomposition (Ch. 5), solidification (Ch. 4), recrystallization and grain growth (Ch. 3).
W. D. Callister and D. G. Rethwisch, Materials Science and Engineering: An Introduction, 10th ed. — the Fe-Fe₃C phase diagram and microstructural development (Ch. 9–10), glasses and glass-ceramics (Ch. 12–13).
R. E. Reed-Hill and R. Abbaschian, Physical Metallurgy Principles, 4th ed. — binary phase diagrams and the lever rule, ordering and superlattices, nucleation kinetics.
F. C. Campbell (ed.), Phase Diagrams: Understanding the Basics, ASM International — precipitation sequences, coherent/semi-coherent/incoherent interfaces.
ASM Handbook, Vol. 4 (Heat Treating) and Vol. 9 (Metallography and Microstructures) — microconstituent identification and heat-treatment practice.
J. E. Shelby, Introduction to Glass Science and Technology, 2nd ed. — glass transition, glass-ceramic nucleation and crystallization schedules.

Question 2: Solubility Product of Carbides and Nitrides in Austenite (20 marks)

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.

2.1 — (a) Deriving Eq. (1) from the reaction free energy, and its assumptions

Precipitation of a carbide or nitride $MX$ from dissolved $M$ and $X$ in austenite is the reaction $[M]_\gamma+[X]_\gamma=[MX]_\gamma$ (or, more precisely, $[M]_\gamma+[X]_\gamma\rightarrow MX_{(s)}$, since $MX$ leaves solid solution as a distinct precipitate phase). The standard free-energy change of this reaction is

$$\Delta G^{\circ}=-RT\ln K=-RT\ln\dfrac{a_{MX}}{a_M a_X}$$

where $a_M$, $a_X$ are the activities of dissolved $M$ and $X$ in austenite and $a_{MX}$ is the activity of the precipitate. Substituting the equilibrium condition and converting to base-10 logarithms gives $\log_{10}K=-\Delta G^{\circ}/(2.303\,RT)$, and since $\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}$ (both taken as approximately temperature-independent over the range of interest):

$$\log_{10}K=\dfrac{\Delta S^{\circ}}{2.303\,R}-\dfrac{\Delta H^{\circ}}{2.303\,R}\cdot\dfrac{1}{T}$$

With $a_{MX}=1$, $K=1/(a_Ma_X)$, so the solubility product is $\log_{10}(a_Ma_X)=-\log_{10}K$:

$$\log_{10}(a_Ma_X)=\dfrac{-\Delta S^{\circ}}{2.303\,R}-\dfrac{-\Delta H^{\circ}}{2.303\,R}\cdot\dfrac{1}{T}$$

which has exactly the form of Eq. (1), with $A=-\Delta S^{\circ}/2.303R$ and $B=-\Delta H^{\circ}/2.303R$ for the precipitation reaction (equivalently, the entropy and enthalpy of the reverse, dissolution, reaction divided by $2.303R$). Precipitation is exothermic ($\Delta H^{\circ}<0$), which is why $B$ is positive and the solubility falls as $T$ falls; for AlN, $B=7060$ K corresponds to $\Delta H^{\circ}\approx-135$ kJ/mol. $A$ and $B$ are therefore not free-fitting constants. Reaching Eq. (1) from this general relation requires four specific assumptions:

The precipitate is a pure, stoichiometric compound of unit activity. Setting $a_{MX}=1$ (rather than carrying a separate activity term for the precipitate) requires $MX$ to be present as its own separate phase, not dissolved back into austenite or into a mixed carbonitride of variable composition — then $K=1/(a_Ma_X)$ and $\log_{10}(a_M a_X)=-\log_{10}K$, giving the product form on the left of Eq. (1).
Dilute (Henrian) solution behaviour. Replacing activities $a_M,a_X$ with weight-percent concentrations $[\text{wt.\%}M],[\text{wt.\%}X]$ requires the activity coefficients $\gamma_M,\gamma_X$ to be constant (Henry's-law region) over the composition range considered, so that $a_i=\gamma_i[\text{wt.\%}i]\approx\text{const}\times[\text{wt.\%}i]$ and the constant activity coefficients simply fold into $A$. This is reasonable for the ppm-to-low-percent solute levels typical of microalloyed steels.
$\Delta H^{\circ}$ and $\Delta S^{\circ}$ are independent of temperature over the range fitted (i.e. no significant $\Delta C_p$ for the reaction), so a single pair of constants $A,B$ describes the solubility over the whole austenitizing range.
No interaction (cross) terms between $M$, $X$ and the other alloying elements present (e.g. Mn, Si do not measurably shift $\gamma_M$ or $\gamma_X$), so the two-species product law holds without additional correction terms.

Check: assumes a single, stoichiometric MX precipitate (e.g. AlN, not a mixed (Al,Nb)(C,N) carbonitride) and Henrian solute behaviour at the ppm-level Al and N contents typical of hot-rolled steel — both standard simplifications for this class of solubility-product equation (Irvine/Hansen-type relations used across the HSLA-steel literature).

2.2 — (b) Maximum tolerable Al at 900°C, 40 ppm N

Given. $A=1.55$, $B=7060$ (K), $T=900{}^{\circ}\text{C}$, N content $=40$ ppm.

Find. The maximum wt% Al that keeps the steel undersaturated in AlN at 900°C (i.e. that avoids AlN precipitation).

Approach. Convert $T$ to kelvin and N to wt%, evaluate Eq. (1) for the solubility product $[\text{wt.\%Al}][\text{wt.\%N}]$, then divide by the fixed N content to isolate the maximum Al.

Convert units. $T=900+273.15=1173.15\ \text{K}$. $40\ \text{ppm}=40\times10^{-4}\ \text{wt\%}=0.0040\ \text{wt\%N}$.
Evaluate the solubility product. $$\log_{10}[\text{Al}][\text{N}]=1.55-\dfrac{7060}{1173.15}=1.55-6.018=-4.468$$ $$[\text{wt.\%Al}][\text{wt.\%N}]=10^{-4.468}=3.40\times10^{-5}$$
Solve for the maximum Al content. At the solubility limit, precipitation is just avoided when the product does not exceed $3.40\times10^{-5}$: $$[\text{wt.\%Al}]_{\max}=\dfrac{3.40\times10^{-5}}{0.0040}=\boxed{0.0085\ \text{wt\%}\ (\approx 85\ \text{ppm})}$$

Final results — maximum Al before AlN precipitation, 900°C, 40 ppm N
Quantity	Value
$T$	1173.15 K (900°C)
Solubility product $[\text{Al}][\text{N}]$	$3.40\times10^{-5}$
Maximum Al content	0.0085 wt% (≈ 85 ppm)

Topic. Precipitate solubility-product design limits in microalloyed steel processing.

Key relations. $\log_{10}[\text{wt.\%}M][\text{wt.\%}X]=A-B/T$; at fixed $[\text{wt.\%}X]$, $[\text{wt.\%}M]_{\max}=10^{A-B/T}/[\text{wt.\%}X]$.

Why this works. The product $[\text{Al}][\text{N}]$ is the equilibrium solubility limit of AlN in austenite at temperature $T$; any composition combination whose product exceeds this value is supersaturated and will drive AlN precipitation (undesirable during hot rolling, since AlN pins grain boundaries and can promote surface/edge cracking or unwanted fine-grain hardening at the wrong stage). Because $B>0$, the solubility product falls sharply as $T$ decreases — hence AlN control is most critical at LOWER finishing/reheating temperatures, not higher ones.

Common pitfall. Forgetting to convert ppm to wt% (a factor of $10^{-4}$, not $10^{-6}$, because wt% is itself parts per hundred) is the single most common numeric slip on this class of question; a second common error is using $T$ in °C directly in the $B/T$ term instead of converting to kelvin.

Source. Porter, Easterling & Sherif, Phase Transformations in Metals and Alloys, 3rd ed., §5 (precipitation in microalloyed steels); Callister & Rethwisch §9 (solubility and the Gibbs phase rule).

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