Don't get in panic. I believe the problem is not that difficult to solve. Let's see:

1. Generally speaking, the chemical composition of E 308 is better than that of E 16-8-2.
Carbon: 0.08 % for 308 and 0.10 % for 16-8-2. Good.
Chromium: 19.5 to 22 % for 308 and 14.5 to 16.5 % for 16-8-2. Good.
Nickel: 9.0 to 11.0 % for 308 and 7.5 to 9.5 % for 16-8-2. Good.
Manganese. 1.0 to 2.5 % for both. Good.
Molybdenum. 0.5 % for 308 and 1.0 to 2.0 % for 16-8-2. Bad.

2. My advice is to organize a meeting between the client and the fabricator (you) and discuss frankly what real (not imaginary) problem may arise from the lower molybdenum content if the welds are left untouched. At the meeting, you should point out to the client that the mistake will involve not only a higher cost (it's your problem, the client will say) but also a longer delivery, a problem for both parties, as you will say to the client. Try by all means that the client's representative at the meeting is an open minded person, i.e., one who really wishes to solve problems, not to create new ones; one who takes into account not only what the specification says (specifications are sometimes unnecessarily stringent), but also his company interests, among which the supplier bankruptcy is not surely included, but is surely included a shorter delivery of the catalytic cracker.

3. At the meeting, discuss frankly the situation from the following points of view:
a) metallurgical. How will the slight molidbenum lower content really influence the performance of the equipment?
b) delivery. How long will it take to remove the mistaken welds to make new ones? Is the client willing to support it?
c) cost. Are you ready to make a discount on price to compensate for the unexpected mistake?

If all aspects of the problem are frankly and carefully reviewed, the best solution will be surely found. Chances are that it is to leave the welds as they are.

Giovanni S. Crisi
Sao Paulo - Brazil

I am not fammiliar with the E16-8-2 electrode, so I had to do a bit of research to find the chemical composition etc.

From the information I got, the major difference is that E16-8-2 contains more Mo and also Cu (3.25 - 4.0%) The other major chemical constituents are in general a little lower than for E308H.

As the material of construction is an H grade of 304, I believe that the application will be in a high temperature environment. Should the temperature lie in certain range, sigma phase embrittlement becomes a problem. Sigma phase is formed when a ferrite containing austenitic stainless steel is kept at the elevated temperature for a long time. The ferrite would then tend to transform to a sigma phase which is rather brittle, and can compromise the integrity of the vessel.

It is my guess that your customer is afraid of sigma phase imbrittlement. From my calculations, with a typical composition taken for the 304H and E308H and E16-8-2, you will have around 10% ferrite in the weld when using E308H, but only around 2% when using E16-8-2. (I assumed 40% dilution.) Welding with E16-8-2 also gives one a possibility of attaining a small amount of martensite. (Under very low dilution conditions.) This may also have the effect of giving a slightly stronger, but less ductile weld.

I suggest that you find out from your customer whether the sigma phase embrittlement is in fact their concern. If it is, then your only real "defence" would be to show that the welds have no more ferrite than the parent metal. This will vary from one cast of material to the next. If you are lucky, the plate has a high ferrite content, and your E308H filler was of such a composition that it resulted in a lower ferrite content. (This is however unlikely.) Should this be the case, you can argue that the parent metal will become embrittled at the same time, or before your weld metal. If the weld metal has a higher ferrite content, then you have a problem.

If I understand your question correctly, you are also looking for a way to tell the two deposits appart. I believe that the big differentiators will be the ferrite content and the Cu content. This means that you can use a standard ferrite measuring tool, (Generally opperate on a magnetic principle.) to establish which welds have let's say above 5% ferrite and which have less. The other option would be to use a portable "spark emmission spectometer" to determine which wleds have a lot of Cu in. (Possibly above 3%)

If you want to do the ferrite calcs yourself, to include in a report, complete with graph, you can download a spreadsheet that I designed for this purpose from:

www.proknownet.com

From the home page follow the "software" link. On the search page, place a * in the keyword field. Then download the spreadsheet.

Let me know if you hear from your customer what his real concern is.

Regards
Niekie Jooste
South Africa