For example take P1 material as ASTM 106 Gr. B and P8 as ASTM A 312 TP 304. For SMAW process an approximate 30% dilution is expected. Therefore the resultant weld metal would be 70% electrode chemistry with 15% ASTM A 312 TP 304 and
15 % ASTM A 106 Gr B pipe chemistry. Chemical constituents are relevant to Schaeffler diagram. From the chemical composition of both metal (Austenitic S.S and CS) and electrodes (E 308 L and E309L) we can find the Nickel and Chromium equivalents.
Cr Eqv = %Cr+%Mo+1.5 X %Si+0.5 X %Cb
Ni Eqv = %Ni+30X%C+0.5 X %Mn
When you are calculating consider the percentage of different chemistries ,ie, 70 % weld metal and 15 % each dilution and add all three values. After getting Ni&Cr Eqv values of each product, plot it on schaffler diagram. When plotting you can find that the weld deposit structure of E 308 L is austenite, martensite and 2% ferrite. The weld deposit structure of E 309 L is austenite and 7% delta ferrite.
In welding Type 304 to mild steel for instance, the choice of filler metal is a key point. If you use Type E308 filler metal, the diluted weld metal with the formation of martensite (a brittle structure) may contain cracking, because the filler metal cannot tolerate dilution by both base metals. A proper filler metal, in this case, is Type E309 (typically contains 24% Cr and 13% Ni.), for applications below 315 deg C in general. The procedure of estimating the micro structure of the E309-type diluted weld metal can be done by using a Schaeffler diagram ; the diluted weld metal will contain about 4% of ferrite and no martensite in the austenitic matrix, which is resistible to cracking
On Wed, Jun 30, 2010 at 6:57 PM, limesh M <limesh78@gmail.com> wrote:--
Dear All
What would be the problem if we use E 308L electrode instead of E 309L
to join P1 to P8 dissimilar material joining in oilfield sour service?
On what basis we are selecting E 309 L over E 308 L in P1 to P8
dissimilar metal joining?
Thanks and Regards
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