The steel type 1.4571 / (X6CrNiMoTi17-12-2, ASTM: more or less 316Ti) is commonly used in wastewater applications due to it's good corrosion resistance. An alternative is 1.4404 / (X2CrNiMo17-12-2, ASTM: 316L), which is often said to be inferior. Several steel plants published information materials to the following effect (don't have an english source, so I summarize the key points):

  • Heat treatment of carbon steels would lead to the formation of Chrome-Carbides, these are attack angles for corrosion. Ni and Ti are added to form 1.4571 because these elements form carbides and so prevent the formation of Chrome-Carbides.

  • Current (since a few decades ...) manufacturing technology allows lower carbon content in steel production, so that addition of Ni & Ti are no longer neccessary to have corrosion resistance as good or better as 1.4571 in many cases.

  • 1.4571 is clearly superior in applications with T > 350°C

All information about the comparative corrosion resistance of these steels I found so far comes from steel plants and traders who have a vested interest in pushing 1.4404: 1.4571 is commonly used in Germany, Austria and some eastern European countries, apparantly the rest of the world moved on to 1.4404 et. al. This means german steel traders need to warehouse two types of steel.

Is, according to an independent assessment, the corrosion resistance of 1.4404 same or better than 1.4571 in most conditions?


That is some serious hair splitting. They have essentially the same corrosion resistance. One would need testing in a particular environment to POSSIBLY find some difference. The 316 L has low carbon. The Ti addition in 4571 combines with the carbon to make titanium carbide also resulting in very low C in solution. When welded ,the carbon in solution forms chromium carbide reducing chrome content in the grain boundaries ,thereby reducing corrosion resistance in the grain boundaries : So the result is very similar corrosion resistance. Both alloys would be poor choices for elevated service. ASME did not have allowable high temperature stresses for "L" grades , when I was working. In addition , Mo has a "volatile " oxide the can promote oxidation at high temperatures in some situations. American mills tend to make L grades while European mills added Ti or Nb (Cb), at least in the past. Also , Ni is not a factor in this consideration . Comment ; American mills add smaller amounts of Ti to deoxidize the melt, so metallography will find the orange Ti carbonitides in both alloys

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  • $\begingroup$ There is some "smoke and mirrors" in these material spec, The key is the 0.03 C max in 4404 and the 0,08 C max in 4571. BUT there is no minimum C in 4571 so you may get less than 0.03 C ; So it has NO guaranteed high temperature strength advantage over 4404. For high temp strength you need "H" grade ; carbon content of 0.08 MINIMUM. And , repeating , neither alloy is a good choice for high temperature. $\endgroup$ – blacksmith37 Mar 12 '18 at 19:23
  • $\begingroup$ OOPs; the minimum C for "H" grades is 0.04 % minimum, this changes nothing in the answer. $\endgroup$ – blacksmith37 Mar 13 '18 at 12:13

Background related to the first answer: Roughly 50 years ago Amoco Chemicals used 317 L , essentially the same as 316 L but with more Mo ( 3.5 %). Laboratory testing in the specific environment ( concentrated acetic plus halides) showed a small advantage for European 317 with the traditional Ti / Nb additions. The advantage of the European grades was found not to be the Ti / Nb . The advantage was that they had Mo content in the high end of range while American mills had Mo in the low end of range. And European mills also used a higher solution anneal temperature than the American mills. I doubt this was published , but if it was, the author was Dr. M R Berry , Amoco Chemicals. Amoco changed to Titanium equipment as time passed and no longer used 317 SS.

So the answer is the same ; very similar corrosion resistance. But testing in some specific environments might show differentiation.

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    $\begingroup$ Can you put all of your comments and answers together into one answer? Just edit your original answer. $\endgroup$ – hazzey Mar 13 '18 at 14:31

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