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API 570 exam, ASME B 16.5 question

API 570 exam, ASME B 16.5 question

(OP)
Dear All

I'm having difficulties to solve the problem below (which i get from API 570 exam sample), can someone (API 570 Inspector) explain how to solve these problem, it is refer to formula areas of subminimum thickness with formula at paragraph 6.1.2 Fitting Local Areas at ASME b 16.5 Standard. But until now i can't solve these problem


28. A local area has been thinned on the wall of a flanged fitting. The fitting is NPS 12 Class 900, and the local
area has been thinned to 0.945”. What is the minimum acceptable thickness for this thinned area per
ASME B16.5?
a. 0.9375”
b. 1.250”
c. 1.750”
d. Cannot be calculated from information given.

29. From the information in Question #28, what is the maximum circular area of subminimum thickness allowed
in square inches?
a. 2.75”
b. 1.33”
c. 1.85”
d. 0.431”

30. Using the information in questions #28 and #29, if two areas of subminimum thickness are observed on the
fitting, what is the minimum distance between the edges of these circles?
a. 3.0”
b. 2.70”
c. 6.52”
d. 4.70”


Thanks for help and your answer
Regards,

RE: API 570 exam, ASME B 16.5 question

Please recheck which edition of B16.5 this exam sample refer to.
Paragraphs 6.1.1 and 6.1.2 regarded only to Table 9 (Class 150) and 12 (Class 300) flanged fittings.
Flanged fittings of class 400 and higher are listed only with Nonmandatory annexes since the edition of 2003

RE: API 570 exam, ASME B 16.5 question

I checked my 1996 edition of ASME B16.5, and it has the dimensions for Class 900 fittings in Table 22. Local area thinning criteria and separation distances between areas of subminimum thickness are in Para 6.1.1 (and applicable to more tables than current edition, including Table 22). The formulae are the same though, so even though the test questions seemingly are based on an earlier edition of ASME B16.5 than present, answers can be determined.
Using the given formulae, I get the following answers-
28. b. 1.250"
29. b. 1.33in2
30. c. 6.52"
Cheers,
John

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