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

 

[Latexman]

I don't understand a 2X safety factor on flow, but then I don't know the chemistry and kinetics and data quality of your scenario. Don't forget though, pressure vessel Codes have healthy safety factors already built-in. The vessel should not burst until 2-3X the design pressure.

1) Do you have choked flow? If so, where?

Good luck,
Latexman

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

 

[don1980]

1) For disk installations the pressure is transient - it starts high and then decreases. So I assume that you're asking, what is the maximum vessel pressure during relief. If we assume the disk is large enough, then the max pressure is no higher than 1.1MAWP or 1.21MAWP. The actual max pressure value is determined by performing a fluid flow calculation. Using the required relief flowrate, calculate the pressure drop in the relief line (including the burst disk). Once the disk bursts, the vessel pressure will continue to rise if the pressure drop is high (exceeds the initial burst pressure of the disk). The vessel pressure will immediately start to decrease if the pressure drop in the relief line is less than the initial burst pressure. If this is indeed 2-phase flow, and the liquid portion is flashing in the relief line, then you'll need to use a calculation tool such as HEM (homogeneous equilibrium model).

2) Of course, the maximum pressure in the relief line is at the location where the line is connected to the vessel. Effectively, this pressure is the same as the pressure calculated in #1.

 

[chemks2012]

Please ignore my original post as it’s very clear. I was looking for an 'EDIT' option but I could not find.


Latexman,

Thank you for your input. I have allowed safety factor of 2 as it’s required by the process uncertainties [client requirement].


don1980

To rephrase my question please.

Pipe stress engineers are asking ‘For what maximum pressure they should design the bursting disc [BD] discharge line for?’ Whether it is 2.75barg or 6.6barg or 5.3barg [6.6-1.3 = 5.3barg]? I am not sure and I believe it is 5.3barg or 6.6barg. Please help.

BD size is 10”
BD set pressure is 2.75barg
Vessel design pressure is 6barg
Vessel accumulation is 10% [6.6barg]
Overpressure allowed for bursting disc design is 140% to equate to 6.6barg


With above data, I can prove that existing above stated bursting disc is adequate for the required release rate of 300kg/sec.

With the required release rate of 300kg/sec, pressure drop at the discharge line is 1.3bar which also includes pressure drop through bursting disc.

Thanks,
KS

 

[Latexman]

What does this 1.3 bar pressure drop mean?

What is the pressure in the RX during this 300 kg/sec flow?

What is the pressure at the end of the RD line during this 300 kg/sec?

For example, is the RX pressure 1.3 barg and the pressure at the end of the RD line near atmospheric pressure? Or, is the RX pressure at 6.6 barg and the pressure at the end of the RD line is 5.3 barg (and at sonic velocity/choked flow). There is a wide range of solutions between my two examples.

Good luck,
Latexman

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

 

[chemks2012]

Hi Latexman,

Sorry what does it mean by RX? Did you mean reactor?

Thanks

 

[chemks2012]

I believe, you meant ‘RX = reactor’. If so….

What does this 1.3 bar pressure drop mean? This is the pressure drop in the bursting disc discharge line [built up back pressure] at 300kg/sec when reactor pressure is set to 5.3barg.

What is the pressure in the RX during this 300 kg/sec flow? Please see above

What is the pressure at the end of the RD line during this 300 kg/sec? Pressure at the end of bursting disc is atmospheric pressure i.e. 0barg

 

[Latexman]

Yes, RX = reactor.

If the RX P = 5.3 barg, and [Δ]P = 1.3 bar, how is pressure at end of the line 0 barg? 5.3 - 1.3 = 4.0

Good luck,
Latexman

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

 

[chemks2012]

Hi Latexman,

Thanks for that.
I meant, discharge line ends at the bursting disc vessel which is at atmospheric pressure i.e. there is no any superimposed back pressure.

So, in this case, Do you think the bursting disc discharge line will see pressure of 4barg?

Regards,
KS

 

[Latexman]

I'm confused. A sketch may help.

Good luck,
Latexman

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

 

[Latexman]

On second reading, I think you are talking about what I call a "catch tank" at the end of the relief line. Basically an open top, or mostly open top, vessel to catch the liquid and solids.

"So, in this case, Do you think the bursting disc discharge line will see pressure of 4barg?" It really depends on what is going on in the RX and line. If you have choked flow (usually at the end of the relief line where it goes into the catch tank), just inside the end of the relief line in/at the catch tank could be significantly above 0 barg.

You are going to have to communicate much more pertinent information on this to determine what you have. If you'd attach your calcs and a detailed sketch or communicate all pertinent facts in them, that would be a good start.

Good luck,
Latexman

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

 

[chemks2012]

Hi Latexman,
Agree with you on sketch but I can’t, until I understand the situation…

To elaborate more..

As I stated before, minimum required release rate is 150kg/sec and with safety factor of 2, it is 300kg/sec [client requirement].

I have checked BD relieving capacity by two methods. Agree, HEM is recommended but I wish to check by both methods..

1) By HEM Method

CASE 1 (HEM) Here, I have considered, upstream stagnation pressure of 7.6bara [taking advantage of 140% overpressure]. With these conditions, for a given pipe layout, the

​

bursting disc capacity is 195kg/sec which is greater than the minimum required release rate of 150kg/sec, therefore it is OK.

My confusion: though we are allowed to consider 140% overpressure [as per code], I am sure this is not logical to prove that the bursting disc is adequate as bursting disc will burst at 2.75barg anyways.


CASE 2 (HEM) However, if I consider, upsteram stagnation pressure of 3.75bara [no overpressure allowed assuming burst pressure of 2.75barg]. With conditions, for a given pipe layout, the bursting disc capacity is 95kg/sec which is less than the minimum required release rate of 150kg/sec, therefore it is NOT OK.


2) By Two Phase Flow Pressure Drop Method

CASE 3 (Pressure drop) Again, to take advantage of 140%, for the release rate of 300kg/sec, I have increased the reactor pressure step wise between 2.75barg and 6.6barg such that sum of that pressure and total back pressure should not increase more than 6.6barg.
Therefore, when I considered the physical properties at about 5.3barg, for 300kg/sec, the pressure drop is 1.3bar through bursting disc line.

If my understanding is correct, here bursting disc will see the pressure at the beginning of line [i.e. near reactor nozzle] as 5.3barg and at the end it will see 4barg.

OR

From Latexman’s reacent comment, I guess, I should consider following for this case ….[does this mean CASE#3 is incorrect?]

CASE 4 (Pressure Drop) I should consider all the physical properties at 6.6barg, for 300kg/sec and simply check pressure drop through the bursting disc discharge line. If I consider this, pressure drop is 1.2bar.

If my understanding is correct, here bursting disc will see the pressure at the beginning of line [i.e. near reactor nozzle] as 6.6barg and at the end it will see 5.4barg.

Thanks

 

[Duwe6]

Nope - if your R/D is at the end of the line, away from the Rx, the line sees the R/D burst pressure. Bear in mind that R/D's are imprecise, so add in the range given from the mfr: i.e. 250# +4%, -7% would be line pressure of up to 260 at burst. Further, the operating temperature of a disk changes the burst pressure a lot. Don't use the Rx temperature, use the disk holder temperature [end of line] when the Rx is being 'vigorously' operated.

 

[Latexman]

CASE 1 would be acceptable for my company. On "My confusion", just because the RD bursts at the right pressure does not prove adequacy; the relief line must also flow an adequate flow within the accumulation limits of the vessel. Because of the dynamics of some scenarios (runaway reaction), some companies develop a stradegy of bursting the disk well below the MAWP, so some of the vessel contents can be removed before the pressure goes higher than allowed. This is similar to removing some of the fuel from a fire. The result is a smaller fire.

CASE 3 the RD will be burst well before the pressure rises to 5.3 barg. It bursts at 2.75 barg. The scenario keeps building pressure as the contents flow out the relief line. The beginning of the relief line sees 5.3 barg and the end of the relief line sees 4.0 barg. CASE 4 is similar but the relief line sees 6.6 barg and the end of the relief line sees 5.4 barg.

You have choked flow at the end of the relief line.

Good luck,
Latexman

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

 

[chemks2012]

Hi Latexman,

Thanks for your reply, much appreciated.

Yes agree, for HEM method, it seems Case#1 should be selected. Thanks.

With the same analogy of Case#1, to choose between Case#3 and Case#4, I believe, Case#3 should be favoured and the reason is…

One of the requirement for bursting disc is to make sure that vessel pressure during relief [ 5.3barg here] PLUS back pressure [1.3barg here] should be <= allowable accumulated pressure [6.6barg]. Is this not correct? If this is correct, I will consider physical properties of relieving fluid at 5.3barg and not at 6.6barg unlike Case#4

Thanks,
KS

 

[don1980]

It seems to me that you're making this harder than it has to be.

Calculate the pressure drop in the relief line, using a flow of 150 kg/sec and 6.6 barg at the beginning of the pipe. If the resulting pressure drop is less than 6.6 bar then you're finished - the installation is OK.

The actual burst pressure of the disk is irrelevent, so long as it is less than or equal to the MAWP (6 barg). That is, there's no need to perform any calculation at 2.75 barg (the pressure you've chosen for bursting pressure).

Unless there's a specific reason for setting the disk this low (2.75 barg), I wouldn't do that. Disks are an inherent liability (as compared to safety valves) from the perspective of plant reliability, and that liability is magnified by setting the burst pressure lower than necessary.

 

[Latexman]

That is not correct. In my pressure relief device sizing world (API-520 and 521), back pressure is defined as the pressure at a pressure relief valve exit. Back pressure is not defined for a RD. I understand there is a general definition of back pressure, but it's best to not confuse the two issues. 1.3 barg is your CASE 3 relief line pressure drop. 5.3 barg is your CASE 3 relieving pressure. I can rationalize the CASE 3 pressure just inside the end of the relief line is 4.0 barg (choked flow). The requirement is - the relieving pressure must be <= the allowed accumulation. This requirement is met in CASE 1, CASE 3, and CASE 4 depending on your assumptions and 2-phase pressure drop method. API-520 recommends a few 2-phase pressure drop methods (ANNEX C). It does not mandate which method to use. It mentions there are other methods available. It mentions new methods may be developed in the future. Choosing the method is left to engineering judgement.

Good luck,
Latexman

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

 

[don1980]

Latexman - for a disk, all that matters is the flowrate through the open pipe. The acceptasnce criteria is simple: will the pipe relieve the necessary flow without excessively overpressuring the vessel.

In the case we're discussing, the pipe (and disk) are OK if 150 kg/sec doesn't cause the pressure to exceed it's max value (6.6 barg)

 

[Latexman]

Don, my reply was directed to chemks2012's "One of the requirement for bursting disc is to make sure that vessel pressure during relief [ 5.3barg here] PLUS back pressure [1.3barg here] should be <= allowable accumulated pressure [6.6barg]. Is this not correct? If this is correct, I will consider physical properties of relieving fluid at 5.3barg and not at 6.6barg unlike Case#4". You posted before I did and I did not even see your reply.

Good luck,
Latexman

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

 

[chemks2012]

don1980 & Latexman,

Thanks for your reply.
Yes, it make sense that I should simply consider pressure of 6.6barg and calculate pressure drop at 300kg/sec [PS: or let say 150kg/sec without any safety factor]. However, I was referring to following link and hence I thought if I have to select either CASE3 or CASE4, I believed, I should consider CASE 3 and not CASE 4.
Have I misunderstood below link?
Thanks

http://www.cheresources.com/content/articles/safety/rupture-disks-for-process-engineers-part-4?pg=2

A rupture disk is actually a differential pressure device where the specified burst pressure is equal to the difference between the desired upstream pressure (vessel) at the time of rupture disk burst and the downstream pressure (backpressure):

Pburst = Pvessel - Pbackpressure

Or alternately the desired upstream pressure (vessel) at the time of rupture disk burst is equal to the sum of the specified burst pressure and the downstream pressure (backpressure):

Pvessel = Pburst + Pbackpressure

Either way, it is apparent that the vessel pressure at the time the rupture disk bursts (commonly called the relief pressure) is directly dependent on backpressure.
When discussing relief systems, three types of backpressure are considered, these being constant, built-up and superimposed.