A question concerning the avidity of carbide formers.

Walt Welch · Apr 18, 2000

A recent thread about alloys containing varying amounts of carbide formers created a question in my mind.

There is in medicine a difference in which certain molecules will compete for a binding site; some will more avidly bind than others. For example, ethyl alcohol is used for methanol (wood alcohol) ingestions, as the ethyl alcohol will compete for the binding sites on alcohol dehydrogenase, and therefore prevent the methanol from binding, and thereby prevent the formation of the toxic compounds that can cause blindness.

My question is this: Is there an analagous situation with carbide formers? I know that C, Cr, and V will form carbides. However, is there a way to determine, from the chemical formula of an alloy, what the quantity of the various carbides will be? For example, if you have so much C, and so much Cr, and so much V, what must the proportions be in order to have enough free Cr to provide corrosion resistance?

The specific alloy which brought about my question is CPM420V, which has the following composition: C 2.2%, Cr 13%, V 9%, Mo 1%. Would this alloy be corrosion resistant?

I realize that there is enough Fe to make carbides of all the carbide formers, but this does not happen. Stainless steels depend on free Cr for corrosion resistance, hence not all Cr is bound up in carbides. On the other hand, there is one whopping lot of Fe which could be bound in carbides with all that C and V. Would this, then, make less Fe available for the Cr?

Any assistance in alleviating my confusion would be welcomed. Walt

[This message has been edited by Walt Welch (edited 04-18-2000).]

RJ Martin · Apr 18, 2000

Walt: Without digging out the Metallurgy textbooks, the answer is "YES"-you can predict percentages of elements tied up as carbides, although, I'm not sure how well the calculation translates into real-life.
And, "YES" different elements have different affinities for Carbon. The heat treatment also affects carbide type/size/quantity and thus, corrosion resistance.
In 420V, the high Vanadium content actually improves the corrosion resistance by scking up Carbon that would otherwise form Chromium Carbides. This leaves the Cr in the "free state", as you alluded. In Crucible's datasheets, they reference that fact as contributing to the excellent stain-resistance in 420V.

RJ Martin

Apr 18, 2000

In a general way (there are exceptions) steel contains enough Carbon to form carbides with all carbide-formers, that's the idea behind the "carbide-formers".
When austenitising and "holding", most of the carbides will dissolve, [some easier as Fe, Cr, Mo, some less easy as Ti, V, W] in the austenite.
Upon quenching the carbon will mostly be trapped in Martensite and the carbide-formers as "substitutes" in the iron lattice.
When "low" tempering (below 300°C) carbon will form Fe3C "cementite", while, when tempering "high" (above 400°C) secondary "hardening" will occur, where the Iron is exchanged in the carbides by the exotic "carbide-formers". The chemical composition tends to be VERY complicated, depending on many factors.
Stainless steel ist "most stainless" when hardened and "low tempered" and much "less" stainless when "fully soft" or secondary hardened, because the chromium will then (mostly) be caught in the carbides.
CPM 420 is QUITE "stainless" but not "really corrosion resistant", no knife-steel is.

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D.T. UTZINGER

Osbourn · Apr 18, 2000

Very cool question.
RJ, what metallurgy books would you be checking? I've been looking for some books that would address subjects such as this.

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Oz

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[This message has been edited by Osbourn (edited 04-18-2000).]

Walt Welch · Apr 18, 2000

ULP!! I goofed. I meant to say that C (not Fe) combines with the carbide formers. Thank you all for overlooking my error, and answering my question by reading between the lines. I shall not edit my post, as this would, I think, confuse people re: Oz's post.

I am sure he meant, as did I, that carbide formers combine with C to form carbides. It is the avidity of the carbide formers to carbon about which my question pertains.

Thanks for the rapid and erudite answers.
Walt