Rated working pressure for flange according to ASME B16.5-2003

[Berto57]

The following is explain in Annex B of ASME B16.5-2003

Flange characteristics:
12’’ 600# in A182 Gr. F53 (Group n°2 material)


Conditions:
Internal Design pressure = 84,8 bar (not relevant for the calculation)

Design Temperature = 204°C (approx. 400°F)

Test temperature = 20°C (approx. 68°F)


By applying the formula (3) from ASME B16.5-2003 Annex B:
Pc = 97,2 barg (1410 psig)

Pr = 600 psi

S1 =minimum of the following (Group 2 material)

· 70% of Yield Strength at 100°F (40°C) = 0,7 x 80,06 = 56 ksi

· 70% of Yield Strength at T (204°C) = 0,7 x 60,82 = 42,6 ksi

· 1,25 times the allowable stress at T (204°C) = 1,25 x 30,17 = 37,7 ksi

Therefore, S1 = 37,7 ksi


Then PT = 600 x 37700 / 8750 = 2585 psig


Our problem is then: PT is not below Pc.


The table 2-2.8 gives the value we should find: we should find 1230 psig @ 400°F. Where is our mistake?

The most disturbing is that by doing the same method with a more standard material (A105), we manage to find the same results than in the Table 2-1.1 specific to A105.



So does anybody know why we do not find 1230 psig even though we follow the method? Does it come from another consideration as explained in B1.1?

Thanks for your help.
Bertrand

 

[doct9960]

ASME B16.5-2003 was issued October 29, 2004. At that time, SA-182 Gr. F53 was not listed in ASME Section II Part D (2001 Edition & later Addenda). Data for material properties (allowable stress, UTS, YS) were most probably given directly by the ASME BPVC Subcommittee on Materials per ASME B16.5 Annex B1.4. So whatever those data were, that was the basis for the pressure rating. That's my guess.

 

[Berto57]

Thanks for this answer doct9960.

If what you say is correct, then the material properties used for calculating a flange are not the same than the one available in the table from B31.3 or BPV Section II-D

I looked at ASME B16.5-1996. The same methode is given in Annex D and the material A182 Gr. 53 is listed in the table 2-2.8.

So if someone knows how the value 1230 psig is found and which values are used... I will be very interested

Bertrand