Question 1 of 7: Proeutectoid Ferrite–Pearlite Micrograph, Its Formation, and Cementite Fraction Above the Eutectoid Temperature

Question I: Proeutectoid Ferrite–Pearlite Micrograph, Its Formation, and Cementite Fraction Above the Eutectoid Temperature (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.

I.1 — (i) Phase identification at the two arrows

The sample is stated to have a carbon content below the eutectoid composition (hypoeutectoid steel), and the micrograph shows exactly the two constituents expected of such a steel cooled to room temperature under near-equilibrium conditions. Both arrows point to the white, essentially featureless patches that outline what were the prior-austenite grain boundaries — this is proeutectoid (grain-boundary) ferrite ($\alpha$), a nearly carbon-free BCC phase that is soft and etches light under the etchant used. The surrounding darker, finely striated (lamellar) texture filling the grain interiors is pearlite, the cooperative eutectoid product of alternating ferrite and cementite (Fe$_3$C) lamellae, which etches dark because of the closely spaced lamellar boundaries scattering light. The proeutectoid ferrite forms a thin, continuous network tracing the original austenite grain boundaries, which is the classic morphology (grain-boundary allotriomorphs) for this phase.

I.2 — (ii) How the microstructure formed

The steel was first fully austenitized: heated above its upper critical temperature ($A_3$) and held long enough to produce a single-phase, homogeneous austenite ($\gamma$) structure. It was then cooled slowly — a furnace anneal or a mild air cool, not a quench — so that the transformation could proceed close to thermodynamic equilibrium. As the temperature dropped below $A_3$ and entered the $\alpha+\gamma$ two-phase field, proeutectoid ferrite nucleated preferentially at the high-energy prior-austenite grain boundaries (the lowest-barrier nucleation sites) and grew inward, rejecting carbon into the shrinking pool of remaining austenite and progressively enriching it as cooling continued. This partitioning continued until, at 727 °C, the remaining untransformed austenite had been driven to exactly the eutectoid composition; at that point it transformed isothermally (eutectoidally) into pearlite — the cooperative, lamellar co-precipitation of ferrite and cementite. The micrograph is therefore a two-stage record of cooling: the white, grain-boundary-tracing ferrite formed FIRST, above 727 °C, and the finely lamellar pearlite formed LAST, filling in what had been the remaining austenite.

I.3 — (iii) Weight fraction of cementite just above the eutectoid temperature

Given. Bulk carbon concentration of the second sample, $C_0 = 0.85\ \text{wt\%\,C}$, held for a long time at a temperature slightly higher than the eutectoid temperature (727 °C). Standard Fe–Fe$_3$C tie-line data just above 727 °C: eutectoid composition $C_{eutectoid} = 0.77\ \text{wt\%\,C}$ (the $\gamma$-phase boundary of the $\gamma$/($\gamma$+Fe$_3$C) field essentially starts at this value immediately above 727 °C); cementite composition $C_{Fe_3C} = 6.70\ \text{wt\%\,C}$.

Find. The equilibrium weight fraction of cementite, $W_{Fe_3C}$, in the material at that holding temperature.

Approach. Since $C_0 = 0.85\ \text{wt\%\,C} > C_{eutectoid}$, this is a hypereutectoid composition; held at a temperature just ABOVE 727 °C (rather than below), the equilibrium microstructure is entirely within the two-phase $\gamma+\text{Fe}_3\text{C}$ field — austenite plus proeutectoid cementite, with no pearlite present yet (pearlite forms only on cooling through 727 °C). Apply the ordinary two-phase lever rule on that $\gamma$–Fe$_3$C tie line.

1. Write the lever rule for the cementite fraction on the $\gamma$–Fe$_3$C tie line. $$W_{Fe_3C} = \dfrac{C_0 - C_{eutectoid}}{C_{Fe_3C} - C_{eutectoid}}$$
2. Substitute the tie-line and given values. $C_{Fe_3C} - C_{eutectoid} = 6.70 - 0.77 = 5.93\ \text{wt\%\,C}$, and $C_0 - C_{eutectoid} = 0.85-0.77 = 0.08\ \text{wt\%\,C}$, so $$W_{Fe_3C} = \dfrac{0.08}{5.93} = \boxed{0.0135 = 1.35\%}$$
3. Round-trip check. Substituting $W_{Fe_3C}=0.0135$ back into the lever rule with $W_\gamma = 1-W_{Fe_3C}=0.9865$ gives a mass balance $0.9865(0.77)+0.0135(6.70) = 0.7597+0.0905 = 0.850\ \text{wt\%\,C}$, reproducing $C_0$ and confirming internal consistency.