Thin WallThick in Large Diameter

[Dj364]

Hello every one,I am a new member
As per B31.3 304.1.3, D/t > 100, I am following the procedure outlined in the BPVC-VIII-1, UG28,UG29,UG30 I am trying to determine the minimum wall thickness required under external pressure, and the required stiffening ring. "A" factor that I found falls to the left material/temperature line chart.
the design pressure in my case is 0.7 bar (internal pressure). does it mean that external pressure is just atmospherique pressure?
Anyone who has performed or familiar with such calculation, his/her help would be very appreciated.
NB:
- I made a search in this forum, I didnot find answers to my questions.
Thanks in advance for your help..
Dj

 

[BigInch]

External pressure is atmospheric pressure, unless its something else.... as if you place your pipe inside a heat exchanger where the shell side has a pressure higher than atmospheric, or the pipe is going down to rest on the sea floor at the bottom of 1000 feet of water, etc.

http://virtualpipeline.spaces.msn.com

 

[Dj364]

Thanks for your reply,
I am checking a 66" and 70" pipe under vaccum condition ( design pressure -0.21 barg), internal design pressure is 0.7 barg. to prevent from buckling and ovalization, i am calculating the required minimum thickness under external pressure and the required stiffening rings, using BPVC section VIII div 1, UG28,UG29 and UG30.
My concern is the external pressure, i assume is 0.79 barg, am I correct?
Thanks in advance
Dj

 

[BigInch]

If the pipe is to be placed in atmospheric air, yes, its atmospheric pressure.

http://virtualpipeline.spaces.msn.com

 

[Dj364]

Hi Biginch
the design pressure conditions include -0.21 barg then the external design pressure will be 0.21 barg. Am I correct?
Regards
Yani

 

[prex]

Normally a design pressure is referenced to the atmospheric pressure, and barg indeed means overpressure (or underpressure) with respect to atmospheric pressure. So your -0.21 barg would indicate an internal pressure 0.21 bar lower than the atmospheric pressure and you should use 0.21 bar as the calculation pressure for vacuum, if the outside of your vessel is exposed to atmosphere.


prex

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[BigInch]

As Prex notes, there is some confusion of the pressure reference. In your first post you didn't say either Gauge or Absolute.

The best thing to do is to Reference all pressures to Gauge or Absolute. If pressures involve - gage pressures its even better to convert the references to use ONLY absolute pressures.

Atmospheric is 100 kPaA or 1 BarA (or 0 BarG)
-0.21 BarG = 1.00 BarA -0.21 BarG = 0.79 BarA or 79 kPaA

Using an example of,

Internal Pressure -0.21 kPaG
Absolute = 1 -0.21 = 0.79 BarA = 79 kPaA

External Pressure of 200 kPaG
Absolute External Pressure = 100+200 = 300 kPaA


The Design Pressure should be the maximum pressure differentials between internal and external, thus you could have a maximum case for critical compressive hoop stress and a different maximum critical condition for tensile hoop stress.

Net Design pressure (a differential pressure) =
Internal Pressure - External Pressure
DP = 79 - 300
therefore, Net Design Pressure = -221 kPa

where,

A - sign indicates a net external to internal direction.
(creates compressive hoop stress in the pipe)
A + sign indicates a net internal to external direction
(creates tensile hoop stress in the pipe)

http://virtualpipeline.spaces.msn.com

 

[chicopee]

Dj364, from your second description the pipe is under a net internal pressure of .7barg-(-.21 barg)=.91barg which is the value required for your calculation.

 

[Dj364]

Thanks every one.I got the response. The external pressure I will consider for my calculation will be .21 barg or .79 bara
Kind Regards.
Dj.