API 653 Shell thickness determination 1

[EmilianoS87]

It's pretty clear the procedure to find t2 in an inspection plane for a localized corroded area. My doubt about the procedure, once I get t2 and calculate the value of L with the 4.3.2.1.b equation. When I calculate t1, I have to get all the measurements 0.5 L on top of t2 location and 0.5 L on the bottom of t2 location? My question is, does it have to be centered? If not how do you determine which span of L length that includes the point of t2 measurement over the inspection plane to choose, with which criteria?

Another doubt I have is when you calculate the tmin value from 4.3.3.1 it is clear that you have to compare it with t1 and t2 with what is stated in 4.3.2.1. That seems pretty straightforward for localized corroded areas, but what about if I want to evaluate the entire shell course? How do I get t1 and t2 for the entire shell course?

Thanks for your time and help in advance.

 

[HTURKAK]

- No..The inspector shall decide the location of L at corroded area. Pls look Figure 4.1—Inspection of Corrosion Areas,please notice that , the t2 is located along profile (c) and please also look 4.3.2.1( c) The authorized inspector shall visually or otherwise decide which vertical plane(s) in the area is likely to be the most affected by corrosion. Profile measurements shall be taken along each vertical plane for a distance, L. In the
plane(s), determine the lowest average thickness, t1, averaged over a length of L, using at least five equally
spaced measurements over length L.

- The procedure described Figure 4.1

1) Determine t2.
2) Calculate L = 3.7 Dt2, but not more than 40 in.
2a) Locate inspection planes a – e ,
2b) Measure the thk. at least 5 pts with equal spacing over a length L,
2c) Determine the lowest average thickness, for each plane ( ax, bx, ...)
3) Choose the min . of five lowest average thickness and get minimum tavg, which is t1.
....

Pls look item 4.3.3.1 again,
- a) When determining the minimum acceptable thickness for an entire shell
course, tmin is calculated as follows ;tmin = 2.6 (H – 1)DG/SE
- b )b) When determining the minimum acceptable thickness for any other portions of a shell course (such as a locally
thinned area or any other location of interest), tmin is calculated as follows: tmin = 2.6 (H)DG/SE
...
I hope my respond answers your question.





He is like a man building a house, who dug deep and laid the foundation on the rock. And when the flood arose, the stream beat vehemently against that house, and could not shake it, for it was founded on the rock..

Luke 6:48

 

[Andres Romero]

Hello, my English is not very good, I hope you understand what I mean.
In my opinion it does not necessarily have to be centered. (example) If t2 is displaced above the corrosion area you can make one measurement above t2 and three below t2.
If the corrosion area is greater than 1000 mm then it should be considered generalized corrosion and evaluated according to 4.3.3.1.



If you want to evaluate the entire ring according to 4.3.3.1, you do not have to average the thicknesses as 4.3.2.1 says. This is only for localized areas.

According to 4.3.3.1 if current thickness is less than tmin then it is not acceptable

 

[HTURKAK]

True ..
I am not sure i understand this . Pls look again 4.3.2 Actual Thickness Determination and Figure 4.2—Pit Measurement.


Again i am not sure that i got your point.
IMO, the code is clear and the minimum acceptable thickness calc. can be summarized ;

-t1 ≥ tmin
-tmin =2.6(H-1)DG/SE + CA, shell course
-tmin =2.6(H)DG/SE + CA , local .
....



He is like a man building a house, who dug deep and laid the foundation on the rock. And when the flood arose, the stream beat vehemently against that house, and could not shake it, for it was founded on the rock..

Luke 6:48