Question

A 1.10 % C hypereutectoid plain-carbon steel is slowly cooled from 900°C to a temperature just slightly below 727°C.

a. Calculate the weight percent proeutectoid cementite present in the steel.
b. Calculate the weight percent eutectoid cementite and the weight percent eutectoid ferrite present in the steel.

Answer 1

We need the Carbon Iron diagram to be able to solve this exercise.

Basically we make a line that is at the temperature of 727 ° C

If we make the line in such a way that it crosses the entire plane we can notice that there is no formation of Proeutectoid ferrite and only there is eutectoid ferrite.

PART A) From that line we can find the weight percent of proeutectid cementite, at which is:

`\eta_(pc) = (1.10-0.8)/(6.67-0.8)*100\%`

`\eta_(pc) = 5.11\%`

PART B) At the same way the total cementite would be

`\eta_(c) = (1.10-0.02)/(6.67-0.02)*100\%`

`\eta_(c) = 16.24\%`

The the weight percent of total eutectoid cementite would be

`\eta_(ec) = \eta{c}-\eta{pc}`

`\eta_(ec) = 11.13\%`

At the end the percent of eutectoid ferrite is

`\eta{ef} = (6.67-1.10)/(6.66-0.02)*100\%`

`\eta{ef} = 83.76\%`