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#10854 - 04/20/07 05:09 AM BS 806 Unreinforced Tee
Magus Offline
Member

Registered: 08/11/06
Posts: 2
Loc: RSA
Good Day,


Please advise what is the correct way to model an unreinforced tee in Caesar for BS 806. Doing it the conventional way shows that there are pressure failures (sustained case) around the tee. Checking this manually shows that in fact this is incorrect and there are no failures.

The piping system has also been installed and operating for 15 years and has had no failures to date.

The parameters are :-


Pressure 125 bar

Temperature 514°C

Main branch : 273.05 OD and wall thickness 39.69 mm

Branch : 193.68 OD and wall thickness 42.86 mm

The material is BS 3604, HF 622.


Modeling is as follows:

Node 10 to node 20 (main pipe). Indicate a unreinforced Tee at node 20. The main pipe continues from Node 20 to Node 30. Branch is then Node 20 to Node 40. Lengths used do not matter as failures occur no matter what one puts in.


Please advise if possible
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DO NOT STRESS ABOUT IT

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#10856 - 04/20/07 07:24 AM Re: BS 806 Unreinforced Tee [Re: Magus]
Richard Ay Offline
Member

Registered: 12/13/99
Posts: 6226
Loc: Houston, Texas, USA
That is all you have to do to model the tee.

I presume you considered the "pressure stress multiplier (m)" that comes into play at branches?
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Regards,
Richard Ay - Consultant

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