Case Study: 1 Use of WRC-1992 diagram for estimation of weld metal composition & Ferrite Number (FN)

Suppose we want to predict the FN of welded UNS S32205 Duplex Steel with Commercially available “matching” flux shielded electrode for duplex stainless – AWS A5.4/A5.4M E2209-XX (AWS A5.4/A5.4M, Specification for Stainless Steel Electrodes for Shielded Metal Arc Welding).

Let us assume 30% dilution (the 2205 base metal contributes total 30% of the union (15 % + 15 %) and the E2209-XX electrode the other 70%). The composition of the resultant weld metal can be predicted as below.

Step: 1 We should have a complete chemical composition of base metal (Mills / material suppliers Certificate / PMI Spectroscopy report)

Chemical Composition of UNS S32205 as per PMI Spectroscopy

C	Si	Mn	P	S	Cr	Ni	Mo	Nb	Cu	Co	N
0.023	0.37	1.50	0.018	0.001	22.37	5.72	3.21	0.11	0.14	0.08	0.177

Step 2

We should have filler/electrode chemical composition (As per ASME SEC IIC, Weld Pad Method/ Electrode manufacturer’s suppliers test report)

The Chemical compositions of the undiluted weld metal as per ASME SEC IIC weld pad test method

Elements Nominal Values in (Wt%)	C	Mn	Si	S	P	Cr	Ni	Mo	Cu	N
Commercial E2209-16	0.031	1.08	0.59	0.007	0.025	22.38	9.15	3.35	0.096	0.18

Step 3

Calculate the Chromium equivalent and Nickel equivalent for base metal and filler/electrode composition:

Cr eq = Cr + Mo +0.7Cb

Ni eq = Ni + 35C+20N+0.25Cu

Step 4

Plotting the values of Cr eq & Ni eq on the WRC-1992 diagram

The UNS S32205 composition is represented by point B (Cr equivalent 25.6%; Ni equivalent 10.1%) and the E2209-XX electrode composition by point A (Cr equivalent 25.7%; Ni equivalent 13.8%). As shown in Fig 1. At the same time Table 1. Lists the estimated composition of weld metal/synthetic base metal. Table 1 Calculates Weld metal composition / Synthetic base metal composition from given base metal and Filler/electrode composition, considering 30:70 dilution (Base metal: Filler Metal)

By Plotting the values of Creq & Nieq, it can be observed that any resultant weld metal from this mixture of A and B will be on the line that joins them. As we have assumed 30 per cent dilution, point C will give the resultant predicted FN = 42.7 approximately as shown in Fig 1. (See the calculated Weld metal composition point C is near thick orange line (FN=40) & between thin blue line (FN=46)