Local metal loss assessment of piping elbows 1

[etsen]

For a local metal loss assessment, the remaining life of piping elbows can be computed by using the method based upon computation of a minimum required thickness for the intended service conditions? For example, r=(t1-t2)/(2c)

where r=remaining life,years
t1 = the minimum remaining wall thickness determined at the time of inspection,mm
t2 = minimum required thickness required by code,mm
c=anticipated corrosion rate,mm/year.To be conservative,the anticipated corrosion rate is multiplied by 2.

 

[ione]

You should be able to find the answer you're looking for at this link

http://books.google.it/books?id=G1t...&resnum=1&ved=0CBUQ6AEwAA#v=onepage&q&f=false

 

[rneill]

the problem with elbows is that often they are thinned out on the backside during manufacture so unless you have baseline survey data of the thickness when new, you estimate of t1 may be off significantly. Of course, your calculation will be on the conservative side.

I have conducted many baseline assessments on new piping systems and have on occasion found brand new elbows that fail to meet the minimum specified wall thickness (they are below even the minimum tolerance value).

 

[racookpe1978]

I disagree with multiplying the assumed "corrosion rate" by 2.

Yes, it is a conservative method, but then it (incorrectly) cuts your lifetime by more than 1/2.

If metal is corroding at 1/8 inch per year, then at one you've lost 1/8", two years = 1/4 inch, three years = 3/8 inch, four years = 1/2 inch, 10 years = 1.25" inch.

If you double that rate, then you have predicted (with the formula) that your metal has lost 1/4 inch, 1/2 inch, 3/4 inch, 1 inch, and 2.50 inch at the same time intervals - and clearly you will calculate your pipe is too thin much too early.

If you know your corrosion rate, use that rate, but start with a single (conservative) amount of metal held as a safety factor.

That is: start with a 1/4 inch margin, then each year assume you will lose an additional 1/8 inch.
For example, after 1 year you predict a (1/4 + 1/8) loss, after 2 years you predict a (1/4 + 1/4) loss, and after ten years you have a (1/4 + 1.25) inch loss.

 

[SJones]

"anticipated corrosion rate" versus "measured corrosion rate"? How are you doing the 'anticipating'?

Steve Jones
Materials & Corrosion Engineer

http://www.oilandgaspeople.com/cv/11499664

http://www.linkedin.com/pub/8/83b/b04

 

[racookpe1978]

That's the problem. You can't measure something that's going to happen in the future.

My point was trying to show that arbitrarily doubling an assumed rate is too conservative, and not prudent and economical and practical, when you use that formula.

You have to find a way to estimate the rate.

 

[rneill]

It is very common practice to use an estimate of the "anticipated" corrosion rate multiplied by 2 in order to determine the re-inspection interval. For example, I might determine based on the actual anticipated corrosion rate that the remaining life is 10 years so I would re-inspect at 5.

In my opinion, it would not be wise to base your determination of the next inspection date on the actual corrosion rate as actual corrosion rates can vary significantly over time.

 

[racookpe1978]

OK - I can definitely see that. Inspecting "twice as early" makes sense if if you have no real or reliable data about what is really going on.

Designing for a "twice the expected corrosion rate" is not prudent.

 

[SJones]

Again, how will you 'estimate' the "anticipated corrosion rate"? There has to be some corrosion rate input from somewhere and its reliability is then accounted for in the analysis. What will happen if we also throw erosion into the ring?

Plus, we are assuming that this is for an RBI determination of next inspection - is it actually the case? Trending a remnant life on one thickness measurement and an 'estimated anticipated corrosion rate' doesn't appear to be a sound approach.

Steve Jones
Materials & Corrosion Engineer

http://www.oilandgaspeople.com/cv/11499664

http://www.linkedin.com/pub/8/83b/b04

 

[etsen]

The corrosion rate is determined as follows.
For example,the nominal thickness of a elbow is 6.3,and the minimum remaining wall thickness is 4.2 determined by inspection data from the elbow.
The inspection data :5.8,4.8,4.2,5.8,5.7.
The time:3 years
So,the corrosion rate = (6.3-4.2)/3 = 0.7 mm/year
To be conservative,the anticipated corrosion rate is multiplied by 2.
So,the anticipated corrosion rate = the corrosion rate * 2 = 0.7 * 2 = 1.4 mm/year

 

[jte]

... have on occasion found brand new elbows that fail to meet the minimum specified wall thickness (they are below even the minimum tolerance value).

Anyone care to tell me what the minimum specified wall thickness of an ASME B16.9 LR ell is? Pick a size - say 8" Sch 80 for starters. How 'bout a tee?

jt

 

[SJones]

Etsen,

Sure you can do it that way if you wish if you are happy that uniform thinning is the only damage mechanism that could be operative. It looks pretty clear that you would want to be running a more detailed examination some time soon. You might want to try an automated UT scan, for example.

Steve Jones
Materials & Corrosion Engineer

http://www.oilandgaspeople.com/cv/11499664

http://www.linkedin.com/pub/8/83b/b04

 

[rneill]

I have interpreted the following paragraph from ASME B16.9 to mean that B16.9 fittings must have the same "size" and "material" as the connecting pipe which if a correct interpretation would mean that these fittings would have to meet the minimum specified for the pipe. Perhaps this interpretation is not correct but I've yet to have a supplier or manufacturer ever take exception to this interpretation?

"Fitting pressure rating is associated with the connecting wall thickness of pipe of equivalent size and material"

I admit that it could also be interpreted that this means the fitting pressure rating is associated with connection material of equivalent (nominal) size.

 

[jte]

rneill-

Paragraph 1.4 as quoted above deals with Standard Units - metric and US Customary.

Take a look at 2.2, Design of Fittings and note the design method options. These include mathematical analysis (most likely FEA), and proof testing. It is further noted that the thickness is likely to vary within a fitting.

Moving on to paragraph 9, Design Proof Test and you'll find that the fitting must merely be sufficiently stout to sustain what would be the nominal burst pressure of its corresponding pipe.

B16.9 places no limitations, min or max, on wall thickness of fittings.

As for your question regarding suppliers taking exception... Well, didn't you state in your post of 1 June that you'd measured values which were less than those for the corresponding pipe? I'd suspect that if you ask your fabricators whether they meet B16.9, they'll state unequivocally "yes." They do. But thicknesses will vary as allowed by the standard.

jt

 

[rneill]

jte,

I have cut out and rejected 90 degree elbows that did not have a minimum wall thickness of 87.5% of the nominal thickness of equivalent pipe. I have done this on a couple of occasions and have not yet had a supplier or manufacturer raise an objection of any sort about replacing such fittings at their cost?

Upon review, I would concur with you that B16.9 does not require the fitting minimum wall thickness to meet or exceed the pipe minimum wall thickness. However, I would have thought that the manufacturers of the fittings I rejected would have raised this point if their design requirement had been something less than 87.5% of the equivalent pipe nominal value since I clearly stated this was why I was rejecting the fittings ?

Interestingly, CSA Z245.11 "Steel Fittings" para 10.3 "Wall thickness tolerances" says ...

"For sizes NPS 18 and smaller, the minimum acceptable wall thickness shall be 87.5% of the design wall thickness of the applicable portions of the fitting ..."

Para 4.3.3 indicates that fittings shall have a "design wall thickness" but leaves the responsibility for determining this (via mathematical analysis or proof testing) to the manufacturer of the fitting.

Myself, I will likely continue to raise objection to new elbows that do not meet 87.5% of the nominal wall thickness of equivalent straight pipe unless the manufacturer can substantiate for me that the fitting has the "design wall thickness" they require plus the corrosion allowance I have specified. Without the manufacturers design data, I think this is a reasonable starting point.

This does raise a good question ... if the minimum required design wall thickness is specific to each manufacturers "design" then how are all the integrity management folks out there assessing the remaining life of corroded/eroded elbows since it is extremely unlikely that they have that data. I'm willing to bet that they are assuming the throw away thickness is the minimum acceptable for straight pipe but I could be wrong on this ?

 

[MrMyers]

You may want to consider referring to API-579/ASME FFS-1, which offers many levels of procedures for computing the remaining life of thinning components.

The procedures allow walls mildly thinner than design code 'minimum thickness', depending on the situation and level of knowledge of the thinned patch.

 

[jte]

rneill-

I don't disagree with your approach. Just making the point that the minimum is not spelled out. As you noted, this makes inspection programs problematic - unless they have initial baselines.

But let me ask you this: Would you consider a 8 x 8 x 4 Sch 40 tee to be adequate if the run and branch were UT'd and were at 100% of Sch 40 thicknesses? Would you consider that sufficient? Would the shape of the branch to run intersection play a role? How 'bout if the run part is somewhat spherical (you may have seen some like this) rather than cylindrical? Would you run an area reinforcement calc? Etc, etc...

I don't have the answers. All I know is that when someone comes to me with a FFS on a B16.9 component, I'm pretty much straight into Level 3 (i.e. FEA) with as good an approximation of the actual geometry of that particular fitting as I can get.

jt