Slow cooling of 0.4% C steel

Cooling path

1. At 950 °C the structure is 100% austenite ($\gamma$) of 0.4% C.
2. On cooling to the A$_3$ line (about 790 °C), pro-eutectoid (primary) ferrite nucleates at austenite grain boundaries. The remaining austenite is enriched in carbon.
3. Between A$_3$ and 727 °C more ferrite forms; austenite composition follows the GOS line toward 0.76% C.
4. Just above 727 °C the austenite has 0.76% C. At 727 °C it transforms by the eutectoid reaction $\gamma \rightarrow \alpha + \text{Fe}_3\text{C}$ into pearlite.
5. Below 727 °C: pro-eutectoid ferrite + pearlite, little change on cooling to room temperature.

 950C: austenite     ->  790C: ferrite at grain boundaries
 +-----------+            +-----------+
 | gamma     |            |f gamma  f |
 +-----------+            +-----------+
 727C and below: ferrite (white) + pearlite (dark)

Phase fractions just below 727 °C

Pro-eutectoid ferrite (using tie line from 0.4 to 0.76 and 0.022):

$$W_{\alpha'} = \frac{0.76 - 0.40}{0.76 - 0.022} = \frac{0.36}{0.738} = 0.488$$

Pearlite (formed from the remaining austenite):

$$W_{P} = \frac{0.40 - 0.022}{0.76 - 0.022} = \frac{0.378}{0.738} = 0.512$$

Total ferrite and cementite (phases, including those inside pearlite):

$$W_{\alpha} = \frac{6.67 - 0.40}{6.67 - 0.022} = \frac{6.27}{6.648} = 0.943$$ $$W_{Fe_3C} = \frac{0.40 - 0.022}{6.67 - 0.022} = \frac{0.378}{6.648} = 0.057$$

Answer: 48.8% pro-eutectoid ferrite and 51.2% pearlite; total phases 94.3% ferrite and 5.7% cementite.