4 replies to this topic

[Root]

Hi all, My derated pressure answer is coming in negative value according to B31 G derated equation according to following conditions, Line size:20" Design Pressure :730 psig Corrosion Depth%: 45 Wall thickness: 0.22" Depth to wall thickness:2.05 Function of calculation: 2.21 Defect Length: 1.18" According to B31 G and API 579, if corrosion depth is more than 20% this should be repaired or derate if this not possible. Now my question is with this much defect and corrosion my answer is negative then how can i further derate pressure. Any thought on this problem will be highly appreciated. thanks Toor

[kkala]

Not being a specialist, thoughts as below could be only introductory. I remember a steam elbow of similar pressure (60 barg) was ejected some 60 m far in a local electricity station (1960s)(*). Consultation from a responsible (piping) engineer should be taken.
A1. Pipe is understood to be of Sch10 with wall thickness tolerance 12.5%. Thus minimum thickness of non corroded pipe s = 0.250×(1-0.125) = 0.22 in, mentioned. Corrosion depth (maximum) is understood as 0.22×45% = 0.10 in.
A2. Depth to wall thickness may be 0.22/0.1=2.2 (versus 2.05). This and "function of calculation" are not clear (to me), but may not affect what is written below.
A3. Referring to <http://www.cheresour...e-and-pressure/>, post no 4 (by kkala), presented example of thickness calculation would give wall thickness of 20×730/2/(20000+730×0.4) = 0.36 in, plus corrosion allowance, for the present pipe. Perry's table in Process Plant Piping does not present steels of allowable stress much higher than 20000 psi. I may miss something, but can you also check given pipe data? Is 0.22 in thickness adequate for 730 psig?
Β. At any case new design pressure could be estimated according to the applicable formula in B31 G, more or less similarly to the example of above mentioned post no 4. The new thickness is understood to be 0.22-0.10=0.12 in, corrosion allowance (say 0.04 in) has to be deducted from it and find what max pressure (at design temperature) corresponds  to this thickness (say 0.08 in). Then a hydraulic test  has to be made to certify the new design pressure.
Although above seems reasonable, an experienced piping engineer should verify (or even modify) it, who should also check loads not due to pressure (e.g. compressive loads due to thermal expansion).

 

(*) The steam line with the elbow was in the 2nd or 3rd floor, elbow was ejected out of the building.

Edited by kkala, 18 February 2013 - 02:58 AM.

[Root]

Hi KKala, May be you did not understand my question very well,I mentioned my answer is in negative vlaue according to B31G equation for derating the corroded pipe. I would like to seek some guide line or alternative calculation with such a thin line so that we can operate it until we repaire it. Toor

[kkala]

Hi Toor, I had previously thought that new thickness is less than what required by 730 psig. Now I see  "negative answer" means negative value resulting from an equation of B31 G. Does this negative value concerns pressure? But the latter would be a result of negative thickness, after the deduction of actual corrosion depth and corrosion allowance. This is hard to believe, probably something else is not clear (to me).

Using previous nomenclature and some simplifications, new thickness s = PD/(2SE) + CA (corrosion allowance), so P=2(s-CA)SE/D (B31 G will be somehow different), which means s<CA. Is it so?

Can you give more clarifications, step by step, along with your results?  It may be useful to anybody intending to give a response.

[Root]

HI KKala, I have calculated the correct pressure, in my calculation was little mistake that i corrected and now my answer is ok. Thanks Toor