ASME B31G - Level 1 Evaluation of Remaining Pipeline Strength

[p6g2p6]

Hi,
I am using ASME B31G to estimate the remaining strength in a corroded area of a pipeline. I have all the info I need (corrosion depth, length, etc.). When calculating the hoop stress of the pipe at the operating pressure, should you use the nominal pipe wall thickness, or the remaining thickness? Obviously this can have a big impact on the results.

I have also used DNV-RP-F101 recommended practice for corroded pipelines, which is supposed to be more conservative for rectangular defects, however I am finding that ASME B31G, which assumes a river-bottom defect is actually much more conservative. Does anyone have any experience using the DNV method? If so, how do you find it compares to ASME B31G?

Thanks.

 

[1503-44]

When calculating the remaining wall strength, yes of course you must use the remaining wall thickness. However the formula does not use that value. They use the fraction corroded. Kind of similar, but different. So actually the pressure formula 4.2 itself does use nominal wall thickness, t, to set up that fraction. The remaining uncorroded wall thickness is expressed as (1-d/t), being the a fraction of the remaining wall and is in the numerator, so it is that which is directly related to remaining strength.

No DNV experience.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[p6g2p6]

1503-44, my version of the code (2012)shows how to calculate the failure stress, which uses similar equation to yours above. It says to convert it to the failure pressure, Pf, the formula is Pf = 2*Sf*t/D, which uses the nominal thickness. Does this seem right?

 

[1503-44]

There was some talk about the old one being too conservative.
I think this new one is probably less conservative than the old formula.
Can you post the page with the variable definitions.
I want to be sure that they didn't change those, because the M term is completely different.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[p6g2p6]

1503-44, here you go:

 

[1503-44]

Yes, t = nominal wall thickness

In the criteria for passing
1) P[sub]f[/sub] >= SF x P[sub]o[/sub]
Or
2) S[sub]f[/sub] >= SF x S[sub]o[/sub]
Those two are equivalent evaluation criteria.
#1 is written in terms of pressure, #2 in terms of hoop stress.

Operating Pressure will have a Hoop Stress, S[sub]h[/sub] = P[sub]o[/sub] x D /2 /t
In that equation t is absolutely known to be t, the nominal wall thickness.
So we can also say for certain that at failure
S[sub]f[/sub] = P[sub]o[/sub] x D /2 /t /SF
And t is also the nominal wall thickness there too.






--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[1503-44]

Found this comment on the relationship to DNV

https://link.springer.com/article/10.1007/s11668-011-9490-8

2011
The applications of ASME B31G-2009 and RSTRENG 0.85 dL effectively improve the pipe’s conveying efficiency and optimize the cost of managing the piping system. However, they both are applicable only for evaluating the medium- and low-strength pipe steels. In contrast, DNV RP-F101 is applicable to the medium- and high-strength pipe steels, but its results are often not safe for application to the lower-strength pipe steels.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."