Metallurgy Question:
A plain carbon steel contains 45 wt% of proeutectoid ferrite. What is its average carbon content in weight percent? I know that you need to use a tie line in the phase diagram, but my book does not really give a good example to go by
Answer:
The calculation is so easy if you have the iron-carbon diagram in your mind. Proeutectoid ferrite is ferrite formed before eutectoid transformation. At 0.8 wt% carbon, we got 100% austenite before the transformation and at 0.02wt% carbon, we got 100% ferrite, and between these two values of carbon content, we have different amounts of proeutectoid ferrite.
Considering that, we have x wt% carbon we calculate proeutectoid ferrite using the tie line.
Proeutectoid ferrite amount = (0.8-x)/ (0.8-0.02)*100=45
==> x=0.45 wt%
You can check it with eyes. At the middle of the tie line, we must have 50% austenite, 50% ferrite; and it is at (0.8-0.02)/2=0.38%C.
We have 45% ferrite, which is less than 50% so we are closer to eutectoid point (0.8%C); so the carbon content must be more than 0.38%.
Considering that, we have x wt% carbon we calculate proeutectoid ferrite using the tie line.
Proeutectoid ferrite amount = (0.8-x)/ (0.8-0.02)*100=45
==> x=0.45 wt%
You can check it with eyes. At the middle of the tie line, we must have 50% austenite, 50% ferrite; and it is at (0.8-0.02)/2=0.38%C.
We have 45% ferrite, which is less than 50% so we are closer to eutectoid point (0.8%C); so the carbon content must be more than 0.38%.
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