Phase Diagram related

[kakalee1]

Hi,

Imagine if I had a plate of stainless steel 316L 1"W x 5"L x.25" thick. On top of the stainless steel plate was a small graphite sheet with the same size and only .04" thick.

I put them inside a furnace and applied pressure (8 psi) of both side to keep them together. I ran the furnace up to 1200 C. Then I cool it down to room temp. When I take the piece down, the two pieces stick together, and there was an alloy formed between them.

Now, I am trying to use the phase diagram of Carbon-Iron to explain the process of how the alloy were formed. I would like to see what the expert think.

Here is my two cents: The alloy is Pearlite (alpha + Fe3C). It was created because the temperature exceed the eutectic temperature (1147 C). Also, it was formed because of the Iron from the stainless steel react with the Carbon from the graphite. What I don't know is what the composition for this alloy is. Is there anyway to find out?

 

[CoryPad]

You need a different phase diagram. Type 316 stainless steel is austenitic, so ferrite (alpha phase) is not likely to form. You are correct that iron carbide would form. Your composition question does not have a lot of physical sense if only a thin reaction layer is present. Bulk composition can be different than local composition.

 

[EdStainless]

And remember that C is a austenite stabilizer.
If you removed it from the furnace and water quenched it then there would only be carbides very near the interface, where the C was very high.
As you progressed into the 316 you would just have higher C 316.

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P.E. Metallurgy, Plymouth Tube

 

[MagBen]

most likely Cr carbides, less likely Fe3C, unlikely ferrite.