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I have loads of books on the iron carbide phase diagram but none tell me what happens in the .1 to .5 part of the process can you please help meGlenn
- Middlesbrough, UK
I am presuming you are wanting to know what phase changes are taking place when slowly cooling or heating a 0.10% to 0.50% carbon steel. A 0.10% carbon steel upon slow (approaching equilibrium)cooling to just above the A3 line will be all austenite with 0.10 wt% carbon atoms combined interstitually in the face centered cubic (austenite) lattice structure. As this steel cools to the A3 line, ferrite (body centered cubic) crystals start to form. These ferrite crystals can accommodate up to 0.025 wt% carbon at 723 deg. C. and only .008 wt% C at room temp. As this steel slowly cools from the A3 line more and more ferrite grains form. The carbon atoms that cannot fit into the ferrite (body centered cubic) lattice structure enrich the remaining austenite (face centered cubic) lattice structure. As we cool to 723 deg. C. the remaining austenite which is rich in carbon atoms transforms to a ferrite/cementite laminar structure.
Cementite is a orthorombic lattice structure and is very hard. (Fe3C)Three atoms of iron unite with one atom of carbon. At room temperature this steel is composed mainly of ferrite grains with 0.008 wt% C. and the laminated structure consisting of ferrite and cementite. The laminated structure ferrite and cementite is called pearlite because it looks like mother of pearl when etched and viewed under the light microscope. A 0.5% C. steel will cool the same way except we will have less ferrite and more pearlite at room temperature. A 0.80% C. steel will be austenitic until it reaches 723 deg. C. At 723 deg. C. and below it will transform to 100% pearlite.Dave McGowan
- Port Coquitlam, B.C. Canada