Bursting disc line - maximum pressure to be seen - (this info is required by pipe stress engineers) 2

[chemks2012]

Hello all,

A reactor has an independent 10” bursting disc [BD] set at 2.75barg [with tolerance +0% to -10%]. Vessel design pressure is 6barg with allowable accumulation of 10%.

If I take advantage of overpressure allowable, I can allow 140% so that vessel pressure should be limited to 6.6barg during two phase relief i.e. 2.75 barg + 140%(2.75 barg) = 6.6barg

BD is designed for two phase flow and the required two phase flow release rate is 150kg/sec. If I wish to allow safety factor of 2 i.e. if I wish to design BD for the capacity of 300kg/sec.

I have checked pressured drop at the BD discharge line at 300kg/sec and it is about 1.3bar.

My queries:

Now pipe stress analyst need some info on maximum pressure that could be seen by the BD discharge line and I am stuck and need your help on following please.

1) What would be the vessel pressure during relief? I believe it is 5.5barg [6.6 – 1.3 = 5.3barg]. Please comment. However, I believe the pressure of 5.3barg is all THEORETICAL and vessel will never see pressure of 5.3barg as the set pressure of bursting disc is 2.75barg and it will burst at 2.75barg

2) What maximum pressure would be seen by BD discharge line? I believe it is 6.6barg [5.3 + 1.3 = 6.6barg]. Please comment. However, I believe bursting disc discharge line pressure of 6.6barg is all THEORETICAL and the discharge line will never see pressure of 6.6barg as the set pressure of bursting disc is 2.75barg and it will burst at 2.75barg and the line will see the pressure of 2.75barg + 0.5bar [assumed pressure drop] = 3.25barg


Your help would be greatly appreciated.
KS

 

[chemks2012]

don1980,

Thanks a ton for your help and agree what you stated earlier but then I followed the link posted above to keep vessel pressure of 6.6barg.

Pvessel = Pburst + Pbackpressure

6.6barg = 5.3barg + 1.3barg

To comply above, I have considered vapour phase [i.e. vapour phase part of two phase flow] properties at 5.3barg and not at 6.6barg.

Thanks once again.

 

[Latexman]

1.3 barg was the built-up backpressure in CASE 3, i.e. during flow.

The equation, Pvessel = Pburst + Pbackpressure, applies at and before the disk bursts, not after. After the disk bursts, it reduces to, Pvessel = Pbackpressure, flow in a pipe.



Good luck,
Latexman

Technically, the glass is always full - 1/2 air and 1/2 water.

 

[chemks2012]

Thanks Latexman,

If I understood correct then, the Pbackpressure in that equation [Pvessel = Pburst + Pbackpressure] is the superimposed back pressure. However, after bursting disc ruptures Pbuiltupbarkpressure)could be as high as Pvessel. I am correct?

Also, please let me know if I am correct on this : If I simply consider two phase flow properties at 6.6barg and simply calculate the builtup back pressure it is 1.2barg at 300kg/sec, then can i say 'PROTECTED EQUIPMENT MAXIMUM PRESSURE DURING RELIEF IS 6.6barg' while pressure at the end of relief line is 6.6 - 1.2 = 5.4barg?

Sorry, I was confused as one of my colleagues argued that the 'PROTECTED EQUIPMENT MAXIMUM PRESSURE DURING RELIEF IS 5.4barg and NOT 6.6barg which does not seems right to me.

 

[Latexman]

Yes, that is the intent of the Code in this case.

Good luck,
Latexman

Technically, the glass is always full - 1/2 air and 1/2 water.

 

[chemks2012]

Thanks a lot Latexman for all your help.

You are a star!!