Qualification of tube/tubesheet with groove weld 1

[JuF]

Hello everyone,

My field of experince is situated in EN ISO standards. In my early career however I did work with ASME but it is quiete a long time ago. So I can use a little help with the following because I'm not sure if I interpreted correct: a procedure qualification of a WPS constits of a tube (D:88.9mm (3.5")/ T: 2.8mm (0.11")) into (set-in) a tubesheet (T:24mm (app. 1")acc. to ASME VIII, both TP316L (P8). There is only one such tube in the tubesheet.

ASME VIII, UW-20 doesn't mention some specific limitations regarding the procedure qualification so according to QW-202.6, ASME IX gives me 3 options to qualify this WPS. Option 3 is not relevant since I need a welding thickness of 3 mm which is not achievable with a fillet weld on a tube with T:2.8 mm. This leaves option 1 and 2 where option 2 (10 mock-up) is less intresting. So option 1 (groove weld) has my preference. However, to qualify the WPS from above, 2 dissimilar thicknesses should be joined with brings me to QW-202.4. When I read this para, I interpret that if I weld a coupon of 6 mm (1/4") to 6mm (1/4"), I fulfill the requirements of QW-202.4a and QW-202.4b and subsequently the WPS mentioned above is covered? Is this correct?

Thank you!

 

[RandySavage]

JuF,

I like your breakdown of how you went about looking into the problem, it's very helpful to see someone's process and it adds a lot of clarity to the situation.
Also very much appreciate that you listed your materials and also converted the dimensions.

As for your WPS, looking into 202.4 we know that the material thickness falls within ranges of QW-451.1 for both members and that (a) is satisfied.
Since we know we are working with a P8 material then as per (b)(1) using a thicker coupon of 1/4" would satisfy the requirements and give you the ability to weld unlimited thickness.

My only confusion here lies in using a 1/4" to a 1/4" coupon as you are now not qualifying base metals of dissimilar thickness. To satisfy the conditions of this you may be better off using a thinner coupon of 1/8" (~3mm) and a thicker coupon of 1/4" (~6mm) for your PQR.

With that your WPS would qualify you for a minimum base metal thickness of 1/16" (1.5875 mm) to unlimited (Per section QW-451.1 and 202.4 respectively).
Also qualifying a groove weld will allow "All fillet sizes on all base metal thicknesses and all diameters" (QW 451.3)

Hope this helps a bit, as the more I read these books the more I realize how ambiguous and open to interpretation they can be.

 

[david339933]

For QW-202.4 only the thinner member has to be in the range of QW-451.1 (for P8) and you don't need to weld dissimilar thicknesses for qualification. 1/4" to 1/4" will suffice.

 

[RandySavage]

David,

Probably a dumb question but, is it possible to fall outside the ranges of QW-451.1 since it covers from Less than 1/16" to over 6"? Seems like the table would allow for any thickness imaginable.

Also thank you for the clarification on the dissimilar thicknesses, is it a situation where it doesn't specify that they shall be of different thicknesses so they don't need to be?

 

[david339933]

An example may be better:
Say you had a WPS qualified on a groove weld test of 1/4" to 1/4" thick P8 material. This would be qualified for 1/16" to 1/2" (2T) per QW-451.1.

Now if you wanted to use this WPS to weld dissimilar thicknesses, the thinner member must be within the 1/16" to 1/2" limit, per QW-202.4(a). Since it is P8 material we don't need to be concerned with the maximum thickness per QW-202.4(b)(1).

Now if the material was P1 rather than P8, both thicknesses would have to be within the 1/16" to 1/2" limit in order to use this WPS (assuming P1 instead of P8 qualification material).

 

[RandySavage]

That was a perfect example, thank you.