Question 4 of 8: Reading the Fe–Fe 3 C Phase Diagram — Phases at 1200 °C, a Full Cooling Sequence, and a Critical Hypereutectoid Composition

Question 4: Reading the Fe–Fe₃C Phase Diagram — Phases at 1200 °C, a Full Cooling Sequence, and a Critical Hypereutectoid Composition (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.

Given. The attached partial Fe–Fe₃C diagram is drawn with straight-line boundaries between its own printed invariant points: peritectic 1495 °C ($\delta$ 0.09, L 0.53, $\gamma$ 0.17 wt%C), eutectic 1148 °C ($\gamma$ 2.11, L 4.30 wt%C), eutectoid 727 °C ($\alpha$ 0.022, $\gamma$ 0.77 wt%C), and cementite at 6.70 wt%C. Reading compositions by linear interpolation between these printed points (exactly as a candidate reads a straight-edged schematic diagram with a ruler) is therefore exact, not an approximation.

Find. (a) the phase(s) and their compositions at 1200 °C for Fe–3.0wt%C; (b) the complete sequence of phase transformations and resulting microstructures for Fe–1.5wt%C cooled from 1600 to 400 °C; (c) the critical hypereutectoid %C giving exactly 10 wt% total cementite at room temperature.

Approach. Read every boundary composition/temperature by linear interpolation along the diagram's own printed straight-line segments between invariant points, then apply the lever rule on the appropriate tie line for each part.

1. (a) Locate 1200 °C on the $\gamma$+L field and read its two boundary compositions. Both boundaries are straight lines between the peritectic (1495 °C) and eutectic (1148 °C) points. Fractional distance down from 1495 to 1148: $f=(1495-1200)/(1495-1148)=295/347=0.850$. $$C_\gamma(1200)=0.17+0.850\,(2.11-0.17)=1.82\ \text{wt\%C}\qquad C_L(1200)=0.53+0.850\,(4.30-0.53)=3.74\ \text{wt\%C}$$ Since $C_\gamma=1.82 < C_0=3.0 < C_L=3.74$, the alloy is genuinely inside the two-phase $\gamma$+L field at 1200 °C, confirming the reading.
2. (a) Lever rule for the phase amounts. With $C_0=3.0$ wt%C on the tie line $C_\gamma=1.82$–$C_L=3.74$: $$W_L=\frac{C_0-C_\gamma}{C_L-C_\gamma}=\frac{3.0-1.82}{3.74-1.82}=\frac{1.18}{1.92}=0.615\qquad W_\gamma=1-W_L=0.385$$ $$\boxed{\text{At 1200°C: }\gamma\text{ (1.82 wt\%C, 38.5\% by mass)}+L\text{ (3.74 wt\%C, 61.5\% by mass)}}$$
3. (b) Liquidus and solidus temperatures for Fe–1.5wt%C. Interpolating along the SAME two boundary segments, now solving for $T$ at $C_0=1.5$: $$T_{\text{liquidus}}=1495-\frac{1.5-0.53}{4.30-0.53}(1495-1148)=1495-0.257(347)=1406\ ^{\circ}\text{C}$$ $$T_{\text{solidus}}=1495-\frac{1.5-0.17}{2.11-0.17}(1495-1148)=1495-0.686(347)=1257\ ^{\circ}\text{C}$$
4. (b) Onset of proeutectoid cementite (the Acm line). The $\gamma$/($\gamma$+Fe₃C) boundary runs from (2.11, 1148 °C) to (0.77, 727 °C): $$T_{Acm}(1.5)=1148-\frac{2.11-1.5}{2.11-0.77}(1148-727)=1148-0.455(421)=956\ ^{\circ}\text{C}$$ Below 956 °C, proeutectoid cementite begins precipitating at prior-austenite grain boundaries, and the remaining $\gamma$ composition slides down the Acm line toward 0.77 wt%C as $T\to727\,{}^{\circ}$C.
5. (b) Assemble the full transformation sequence, 1600→400 °C. $$1600\text{--}1406\,{}^{\circ}\text{C: 100\% L}\ \to\ 1406\text{--}1257\,{}^{\circ}\text{C: L}+\gamma\ (\text{dendritic solidification})\ \to\ 1257\text{--}956\,{}^{\circ}\text{C: 100\%}\ \gamma\ (\text{single-phase austenite})$$ $$\to\ 956\text{--}727\,{}^{\circ}\text{C: }\gamma+\text{Fe}_3\text{C}\ (\text{proeutectoid cementite network at grain boundaries, }\gamma\to0.77\text{ wt\%C})$$ $$\to\ \text{at }727\,{}^{\circ}\text{C: eutectoid reaction}\ \gamma(0.77)\to\alpha(0.022)+\text{Fe}_3\text{C}(6.70)\ (\text{pearlite})$$ $$\to\ 727\text{--}400\,{}^{\circ}\text{C: }\boxed{\alpha+\text{Fe}_3\text{C}\ (\text{proeutectoid Fe}_3\text{C network}+\text{pearlite, unchanged in amount below 727°C})}$$ Lever rule on the tie line 0.77 ($\gamma\to$pearlite)–6.70 (Fe₃C), just below 727 °C: $W_{\text{pro-Fe}_3\text{C}}=(1.5-0.77)/(6.70-0.77)=0.123$ (12.3%), $W_{\text{pearlite}}=0.877$ (87.7%).
6. (c) Set up the total-cementite lever rule at room temperature. Below 727 °C the two phases in equilibrium are $\alpha$ (0.022 wt%C) and Fe₃C (6.70 wt%C) for ANY hypo- or hyper-eutectoid composition, so the TOTAL cementite fraction for an alloy of composition $C_0$ is $$W_{\text{Fe}_3\text{C,total}}=\frac{C_0-0.022}{6.70-0.022}$$ Setting this to the 10% cap and solving for $C_0$: $$C_0=0.022+0.10\,(6.70-0.022)=0.022+0.668=\boxed{0.69\ \text{wt\%C}}$$
7. (c) Check the result against the eutectoid point — the teaching point of this part. The critical composition, 0.69 wt%C, is LESS than the eutectoid composition (0.77 wt%C) — i.e. it is HYPOEUTECTOID, not hypereutectoid. Evaluating the same formula exactly AT the eutectoid composition (pure pearlite, no proeutectoid phase at all): $$W_{\text{Fe}_3\text{C}}(0.77)=\frac{0.77-0.022}{6.678}=0.112\ (11.2\%)$$ Since total cementite fraction increases monotonically with $C_0$ across the whole $\alpha$+Fe₃C field, and pure pearlite ALONE already carries 11.2% total cementite — above the 10% cap — NO hypereutectoid alloy ($C_0>0.77$) can ever satisfy the stated <10% criterion; every hypereutectoid alloy carries even more total cementite than pearlite does. The honest answer is that the criterion as posed cannot be met by any hypereutectoid steel.