Question

In the iron-carbon equilibrium phase diagram, the eutectoid reaction occurs at 723°C with the eutectoid composition of 0.83 weight % carbon. Ferrite and cementite phases are considered to contain 0.022 weight % carbon and 6.67 weight % carbon, respectively. If a steel specimen with 0.7 weight % carbon is cooled from 950°C to below 723°C, the fraction of eutectoid ferrite is .............

Hint:

The lever rule is a useful method for calculating the phase fractions in a two-phase region. The fraction of each phase is inversely proportional to the distance from the phase boundary.

Answer 1

To determine the fraction of eutectoid ferrite, we can use the lever rule from phase diagrams. The lever rule states that the fraction of a phase in a two-phase region is given by: \[ f_{\alpha} = \frac{C_{\beta} - C_0}{C_{\beta} - C_{\alpha}}, \] where: - \( f_{\alpha} \) is the fraction of eutectoid ferrite,
- \( C_0 \) is the composition of the specimen, which is 0.7 weight % carbon, 
- \( C_{\alpha} \) is the composition of ferrite, which is 0.022 weight % carbon, 
- \( C_{\beta} \) is the composition of cementite, which is 6.67 weight % carbon.
Now, substituting the values into the equation: \[ f_{\alpha} = \frac{6.67 - 0.7}{6.67 - 0.022} = 0.74. \] Thus, the fraction of eutectoid ferrite lies between 0.72 and 0.76. This is the portion of the steel that is in the eutectoid ferrite phase after cooling below the eutectoid temperature.