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Duplication Of Psv And Flow Rate For Exchanger Tube Rupture Scenario

tube rupture

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#1 Platonicus

Platonicus

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Posted 11 March 2024 - 03:10 AM

Hello!

 

In API 521 4.4.14.2.2 indicated:

 

In practice, an internal failure can vary from a pinhole leak to a complete tube rupture. For the purpose of
determining the required relieving flow rate for the steady-state approach, the following basis should be used.
a) The tube failure is a sharp break in one tube.
b ) The tube failure is assumed to occur at the back side of the tubesheet.
c) The high-pressure fluid is assumed to flow both through the tube stub remaining in the tubesheet and through
the other longer section of tube.
 
Some of our colleagues understand that the statement in point c means that PSVs must be installed both at the inlet and outlet of the heat exchanger (low pressure side).
In this case, the flow rate value, which is obtained according to the calculation of Wing Y. Wong, should be multiplied by 2, referring to the fact that both of these PSVs will operate simultaneously. This solution oversizes the discharge system from these PSVs.
 
Usually, in our practice we have never done this.
 
What do you think about this?

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Edited by Platonicus, 11 March 2024 - 03:28 AM.


#2 breizh

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Posted 11 March 2024 - 03:52 AM

Hi,

You may want to review the paper(s) from Wing .Y. Yong.

Breizh 

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#3 Platonicus

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Posted 11 March 2024 - 06:46 AM

Hi,

You may want to review the paper(s) from Wing .Y. Yong.

Breizh 

In this papers it is not indicated whether the resulting flow rate should be multiplied by two. Or I didn't read carefully.



#4 Pilesar

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Posted 11 March 2024 - 08:35 AM

When an exchanger tube is assumed broken, the calculated area is twice the cross-section area of one tube because there will be two open orifices - one on each side of the break. This is not the same thing as 'multiplying the flow rate by two.' Work through the calcs for each orifice to determine the flow rate for each orifice! There is no requirement to take short cuts.






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