External pressure question UG-99 3

[ME231]

Could anybody please help me calculate the required hydrostatic test pressure for external pressure on a piece of tubing?
UG-99 seems to suggest that the hydrostatic test pressure due to external pressure is the same as that which would be required had the pressure been internal. Is this true?
Thanks!

 

[desertfox]

Hi ME321

How much external pressure will the vessel see?

desertfox

 

[ME231]

about 1000 psi external pressure

 

[desertfox]

Hi ME231

Sorry I have no references here to the one you refer but I suggest that what you have read is correct as I would expect a margin of safety.
Unless you can expand more on the UG-99 so I can interprete its definition I can't help you further.
I thought I had a copy of BS5500 here but I must of chucked it a while back.

desertfox

 

[desertfox]

Hi ME231

I found a pressure design reference for ASME here it says:-

Ps= 1.3Pm or 1.3Pw*[Sa/Sdt] where Ps is the shop test pressure.

hope this helps

desertfox

 

[Simdro]

Hi ME231,

as I wrote on my thread UG-99(f) and as it seemed, I have the same problem or something like that.
As I understood you have a differential external design pressure greater than the internal design pressure in respect to the normal atmospheric pressure.

Therefore the Hydrostatic test pressure should seem to be calculated as the same rule for the internal design pressure.

I'm not sure of the above sentence but it is the only help that I can give you.

I hope that somebody more experienceed than me can help us on the matter.

Bye bye

 

[CodeRef]

Yes, it is true.
See question 2 of interpretation VIII-80-44. issued on May 23, 1980 found here:

http://cstools.asme.org/csconnect/pdf/CommitteeFiles/864.doc

Question (2)How are hydrostatic test rules to be applied for a vessel which is designed for internal and external pressure?

Reply (2):A vessel which is designed for internal pressure and for external pressure must be examined as a combination unit according to UG-99(c) unless some design feature shows that the combination of internal and external pressure cannot occur independently. If each can occur independently from the other, the vessel must be hydrostatically tested for each independent design condition.

 

[CodeRef]

Also see...
Interpretation:VIII-1-92-115
Subject:Section VIII, Division 1 (1992 Edition); UG-99(f)
Date Issuedecember 11, 1992

Question:Are the requirements of UG-99(f) of Section VIII, Division 1 applicable for a single-wall vessel designed and stamped for internal pressure of 8.7 psi and full vacuum?

Reply:The hydrostatic test in the vessel shall be the larger of that obtained by the rules of UG-99(f) for external pressure, or UG-99(b) or (c) for internal pressure.

 

[ME231]

Thank you all for your responses. From the above I gather that regarless of whether or not the net pressure becomes internal or external on a particular chamber, the hydrostatic test is to be performed always using internal pressure. I guess for the case of smaller pressure vessels (or tubing) one could afford to ask a question as to whether or not using external pressure to hydro test a chamber would be a more appropriate approach/valid route. However, if I were to have a big pressure vessel designed for vacuum (and therefore with with net external presure), I could not possibly imagine making an even bigger chamber to test it with external pressure.
This explaination satisfies me and I think that I can actually proceed with both internal and external pressure for my case where I am testing just tubing. Please feel free to correct me if you feel I am wrong!
Thanks much again.

 

[Simdro]

Hi again CodeJackal,

QUOTE
If each can occur independently from the other, the vessel must be hydrostatically tested for each independent design condition.
UNQUOTE

This is my case, considering that the Vessel is designed for 7.25psi and full vacuum. Another important thing is that the Vessel during normal operating conditions will work under an external pressure close to the full vacuum condition (we are talking about a big crystallizr for chemical plant).

Therefore the design condition for internal pressure is just to assure that the vessel can withstand some internal pressure loads due to a mulfunctioning of vacuum system.
In fact the above sentence is not properly correct considering that the vacuum design condition governs the design.
In summary the Vessel is subjected on both independently design condition where the vacuum condition governs the design than,

how can I calculate the Hydrostatic test pressure?

Any kind of helps would be greatly appreciated.

 

[CodeRef]

Simdro,
For single chamber pressure vessels designed for full or partial vacuum and internal pressure, and where each operate independently from each other, please see UG-99(f) as you referenced earlier and of course UG-99(b)temperature correction for calculating the pressure test.

Except as covered in UG-99(f), there are no specific rules for external pressure testing in ASME Section VIII-1

If external pressure greater than vacuum, see U-2(g)per the following interpretation:

Interpretation: VIII-82-57
Subject Section VIII, Division 1, UG-99, External Pressure Testing
Date Issued:June 3, 1982
File:BC81-718

 

[Simdro]

Hi CodeJackal,

thank you again for your reply.

I have already sent to my client what I learned during our discussion.

I hope to solve this issue as soon as possible.
Bye Bye