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Relief Vs Bernoulli

power failure bernoulli psv

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#1 JanPau

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Posted 16 December 2024 - 04:55 PM

Hello everyone,
 
I am studying flarenet on my own these days and would like to ask for support for a question. I apologize if it may be trivial.
 
There is a total power failure scenario, and I have seven different sources of relief, each with a single PSV. Out of pure curiosity I did a simulation 1 and simulation 2 , in the second one I increased the mass flow rate  of only one PSV ( let's call it PSV #1). As a consequence, there was naturally an increase in velocity in the tailpipe relative to PSV#1 and the main header ( there's no subheader in my network ). I would have expected to have along the tailpipe #1 a lower pressure ( Bernoulli Theom ) than what happened during simulation 1 for the same tailpipe, but this did not occur. Why?  As I understand it, flarenet starts counting in reverse, i.e., it creates the pressure profile starting to count from the fixed point of the tip ( 0 barg ) and it is therefore natural that in simulation 2 having more mass flow ( and therefore volumetric ) and therefore more pressure drop, the stream pressure at the valve outlet will be higher. However, I do not understand why, apparently, Bernoulli seems to me not respected ( High P and High Velocity...)
 
Problem of settings? Or concepts relative to backpressure I'm not taking in consideration?
 
Thanks a lot.
 
JP
 
 


#2 Pilesar

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Posted 16 December 2024 - 07:25 PM

A diagram might help your description. When a PSV relieves, the pressure is increased everywhere in the header. The header at the junction of another PSV outlet will therefore have a higher pressure. At that node, the pressure in the header and the pressure in the PSV outlet line will be the same. Therefore the PSV outlet line will have higher pressure when the first PSV relieves.



#3 shvet1

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Posted 16 December 2024 - 11:19 PM

 

in simulation 2 having more mass flow ( and therefore volumetric ) and therefore more pressure drop

 

 

Did you check volumetric flow profile along the header in sim 1 vs sim 2? Do you have a choked flow somwhere in the header, especially at flare tip? Do PSVs relieves have equal specs? Is PSV #1 located upstream or downstream of PSVs #2,3..?

 

Pressure drop is irreleant to mass flowrate. The higher pressure drop = the higher pressure upstream of #1 calc point = the higher gas density = the lower volumetric flow = the lower pressure drop = the lower pressure drop upstream of #2 calc point. And so on.

 

If you want to push through a pipe more gas all you need just increase the pressure at source. Not decrease.


Edited by shvet1, 16 December 2024 - 11:38 PM.


#4 shvet1

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Posted 16 December 2024 - 11:29 PM

 

As I understand it, flarenet starts counting in reverse, i.e., it creates the pressure profile starting to count from the fixed point of the tip ( 0 barg )  

 

 

Correct. Gas pipe profile simulation is started from a point where pressure is fixed/irrelevant to calculation results. In your case this point is atmosphere. Whatever flow/pipe would be the pressure downstream of the tip is 0 barg.



#5 snickster

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Posted 17 December 2024 - 04:01 PM

The pressure at the discharge of the relief valve is not set but is based on the built-up frictional pressure drop in the piping due to flow. The higher the flow the greater the frictional pressure drop in the piping and the greater the built-up back pressure at the discharge of the relief valve. 

 

In the Simulation 1 there was a pressure drop in the piping created by the lower flowrate of Simulation 1 relative to Simulation 2 so there was a lower built up pressure at the discharge of the relief valve in Simulation 1.  In both cases the exit of the line to the atmosphere is at 0 psig or at the operating pressure of the vessel if the relief system discharges into a vessel.  The pressure drop calculation starts at the point of known pressure which is the termination pressure. So if you are discharging into the atmosphere at 0 psig and add frictional pressure drop back to the discharge of the relief valve you will get a higher pressure at a higher flowrate at the discharge of the relief valve.

 

The pressure of the discharge of the relief valve is not set but is the unknown, so it is not set at the value obtained in Simulation 1.  Starting at the termination point of the flare header you work backwards performing a pressure drop calculation to get the pressure at the exit of the relief valve.  The pressure at the exit of the relief valve can build up to any value below the relieving pressure upstream of the relief valve depending on the flow.  The worst case if if the pressure at the discharge of the relief valve would be equal to the upstream relief pressure but this cannot be possible as then there would be no flow possible as there would no longer be any differential pressure for flow to exists.


Edited by snickster, 17 December 2024 - 04:04 PM.


#6 JanPau

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Posted 18 December 2024 - 04:28 PM

Thanks you all for the kind replies.

 

Is the built-up pressure mentioned the dynamic pressure?



#7 snickster

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Posted 18 December 2024 - 06:18 PM

No - dynamic pressure refers to KE kinetic energy contained in the fluid velocity which KE = V2/2g in ft-lbs/lb (feet) units. 

 

Built-up back pressure is pressure that only exist by pressure drop caused by flowing fluid.  It is called built-up because it don't exist on the discharge of the relief valve unless there is flow and hence frictional pressure drop.  However frictional pressure drop is related to dynamic pressure in that it is a multiple of dynamic velocity pressue per the following equation:

 

Head Loss Friction (in Feet) = fL/D(V2/2g)

 

Also in the relief valve discharge there is what is called superimposed back pressure which exist even if flow does not exist.  This may be when the termination point of the relief header is not into atmosphere but into a vessel or system that is at a constant elevated pressure.



#8 JanPau

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Posted 22 December 2024 - 04:45 AM

Thank you for the answers, unfortunately I can't write here in time.
 
From what I understand: simulation 2 has at the valve outlet and along the tailpipe a higher built up pressure because since there is a higher volumetric flow rate, there is a higher velocity and therefore the kinetic energy term ( v^2 / 2g ) is higher.
 
This means that there will be a higher pressure drop ( in general regarding friction ) . Of course proceeding from 0 barg, the delta P will be larger than in simulation 1 ( because indeed high is the loss observed from the discharge to the flare ) .
 
Another directly related question: 
 
I imagine that I have a vessel of liquid at a fixed pressure and I have to take this liquid along a pipe and reach the inlet of fully closed control valve. I want to compute the pressure i will have there. I also do two simulations here, the second one with changing exclusively mass flow rate ( higher ). 
I know that in the second simulation at the outlet of the vessel and at the inlet of the pipe ( in respect of first simulation ) I would have a higher velocity, so
a lower pressure and in fact at the point of arrival I will record a lower pressure  ( P vs velocity for Bernoulli Theorem )
But this seems contradictory, because I would be told that in the second simulation I have a higher Bernoulli velocity energy component exiting the vessel ( as in the case of flarenet ) and I expect at pipe inlet an higher pressure and so a higher pressure at the inlet of the valve.
 
Why?
 
Thanks again.


#9 snickster

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Posted 22 December 2024 - 04:04 PM

For one thing you would not be able to get flow if you have a long pipe coming from a vessel and at the end were a fully closed control valve.  That is like trying to flow in a pipe that has a cap at the end.  Initially if there were air in the pipe the liquid will not flow until the air is pushed out of the inlet into the vessel very slowly, then flow would stop when it hits the location of the closed valve.

 

Basically though you want to know what the flow conditions will be along the pipe at a given distance from the vessel for two conditions.  The conditions in the vessel is a fixed pressure but in simulation 2 you increase the flow relative to simulation 1. Under steady flow conditions once the flow has stabilized, at a selected point along the outlet pipe the pressure will be reduced due to frictional energy losses and loss of energy due to kinetic energy KE of the velocity increase of the fluid from zero velocity in the tank.  However under steady flow conditions this lower pressure will be a constant value at any given point downstream in the pipe, and also a constant velocity of flow, at a given flowrate from the tank. 

 

Starting the tank nozzle connection/inlet of pipe there is a head loss due to friction due to the exit opening of the tank (= fL/D(V2/2g) in feet of head where f is friction factor, L is the equivalent length pipe of tank opening, D is diameter of pipe, V is velocity, and g is gravitational constant 32.2 ft/sec2) and a pressure drop due to velocity increase from zero velocity in the tank to the velocity in the pipe (=V2/2g also in feet of head per the Bernoulli equation).  Then from there, there is an additional pressure drop to the point of interest downstream based on the length of pipe using the same frictional equation above.   This is the only head loss (energy loss) in the system from the energy in the tank at tank pressure and zero velocity in the tank.

 

From the Bernoulli equation (not considering potential energy of elevation difference), the energy (= Head in ft-lb/lb foot pounds per pound weight) in the tank at zero velocity is just due to pressure and is equal to:

 

Head (energy ft-lb/lb) = P(144)/Density   Where P is pressure in PSI and density is density of fluid which for water is 62.4 lbs/ft3 (this is in pounds weight not mass).

 

To find the resulting pressure at your point of interest downstream of the outlet subtract the energy due to friction and velocity increase indicated above (from the energy in the tank due to pressure head) to get the remaining head (energy) at the point of interest.  You don't have to subtract the velocity head again just the additional pressure drop in the pipe from the outlet to the point of interest.  Find the pressure at the downstream point by converting the remaining head(energy) to pressure using the same equation above.

 

If you have higher flow you will get higher frictional losses and higher velocity head so you will have less remaining head for the pressure term in the Bernoulli equation and therefore less pressure at the downstream point.


Edited by snickster, 22 December 2024 - 04:35 PM.


#10 JanPau

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Posted 23 December 2024 - 03:27 PM

Thanks, Snickster.

 

I don't know if what I am about to ask has already been expressed in your answer, but my point is:
 
“If the flow rate is higher, the friction losses are higher and the velocity head is higher, so the residual head for the pressure term in Bernoulli's equation is lower and therefore the pressure at the downstream point is lower.”
 
In my excel calculations for the vessel, where P1 is the fixed pressure and the target is P2 .
Sym 1: Q=15000 kg/h P1=5 bar P2 = 2.5 bar  
Sym 2: Q=40000 kg/h P1=5 bar P2= 1.5 bar
 
In flarenet, where P inside the tank is 5 bar, P1 is relief valve outlet and taipipe beginning and P2 is tailpipe end and I have gas only outlet:
 
Sym1 Q= 15000 kg/h P1= 3 bar P2 = 1.5 bar
Sym2 Q = 40000 kg/h P1= 4 bar P2= 2.5 bar
 
I can't really understand the discrepancy in P2 between the two cases, sorry. Is it because I have another fixed point in flarenet simulation?
 
Thanks again.


#11 snickster

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Posted 23 December 2024 - 06:52 PM

“If the flow rate is higher, the friction losses are higher and the velocity head is higher, so the residual head for the pressure term in Bernoulli's equation is lower and therefore the pressure at the downstream point is lower.”
 
Correct.
 
In my excel calculations for the vessel, where P1 is the fixed pressure and the target is P2 .
Sym 1: Q=15000 kg/h P1=5 bar P2 = 2.5 bar  
Sym 2: Q=40000 kg/h P1=5 bar P2= 1.5 bar
 
Is this a liquid or a gas?  I thought you were talking about flow of a liquid?  Something seems wrong with your calculations.  Pressure drop/head loss is proportional to velocity squared which is in turn proportional to flowrate squared.  So for Sym 1 pressure drop is 2.5 bar, Therefore for Sym 2 pressure drop should be approximately (4/1.5)2*2.5 = 17.7 bar.
 
In flarenet, where P inside the tank is 5 bar, P1 is relief valve outlet and taipipe beginning and P2 is tailpipe end and I have gas only outlet:
 
Sym1 Q= 15000 kg/h P1= 3 bar P2 = 1.5 bar
Sym2 Q = 40000 kg/h P1= 4 bar P2= 2.5 bar
 
I can't really understand the discrepancy in P2 between the two cases, sorry. Is it because I have another fixed point in flarenet simulation?
 
Is this liquid or gas flow?  If gas flow you can have a higher pressure at the exit (which is greater than 0 bar atmospheric) if your flow reaches sonic velocity at the exit.  Gas pressure drop at sonic flow gets a little complicated
 
Something again is wrong with your calculations.  In both cases your pressure drop is equal to 1.5 bar.  Again if you increase the flow from 15000 to 40000 you new pressure drop should be approximately (4/1.5)2*1.5 = 10.7 bar.
 
Where is the termination point of the tail pipe? To atmosphere at 0 barg or into a flare header that also has a pressure drop.  If the termination point is into a flare header that also has a pressure drop then it makes sense that at the connection point of the tail pipe to the header there would be a higher pressure at a higher flow due to higher pressure drop in the header.  But at higher flow you would also get a higher pressure drop of about 10.7 bar between header connection point and discharge of relief valve/beginning of tail pipe not same 1.5 bar drop for both high and low flow.
 
Looks like you are modelling something incorrectly.

Edited by snickster, 23 December 2024 - 07:57 PM.


#12 JanPau

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Posted 30 December 2024 - 11:19 AM

Hi Snickster,
 
thank you for your response.
 
You are correct, the results on the P's are incorrect. The numbers I put in my post are purely symbolic, but they reflect the discrepancy I had.
 
Basically assuming the statement you wrote: 
 
“If the flow rate is greater, the friction losses are greater and the velocity head is greater, so the residual head for the pressure term in Bernoulli's equation is less and therefore the pressure at the downstream point is less.”
 
Again, orders of magnitude are symbolic:
what I cannot understand is: why in the simulations of higher flow rates, and analyzing P1 just after the efflux source, I have a pressure in flarenet which is higher ( e.g. P inside tank 5 bar, P1 = 4 bar, deltaP =1bar ), than lower flow rate ( e.g. P inside tank 5 bar, P1=2 bar, deltaP=3 bar) while in the vessel case RIGHTLY is lower ( e.g. P inside tank 5 bar, P1 = 2.5 bar, deltaP =2.5 bar) than lower flow rate ( e.g. P inside tank 5bar, P1 =3.5 bar, delta P =1.5 bar ) ? 
 
I do understand that in flarenet coming from 0 bar g and moving backwards I should have i higher pressure out of relief in order to justify the bigger deltaP.  But I'm trying to picture moving from vessel as like i'm a molecule.
 
"From what I understand: simulation 2 has at the valve outlet and along the tailpipe a higher built up pressure because since there is a higher volumetric flow rate, there is a higher velocity and therefore the kinetic energy term ( v^2 / 2g ) is higher. " or contradicts your statement?
 
 
 
Thanks.

Edited by JanPau, 30 December 2024 - 11:58 AM.


#13 snickster

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Posted 30 December 2024 - 05:06 PM   Best Answer

I see what you are getting stuck on now as follows. 

 

The basic idea you don't understand is that there is no pressure drop across the relief valve regardless of flowrate, therefore there is no head losses in the bernoulli equation from upstream of the relief valve to the tail pipe.  The relief valve can be thought of as a perfect nozzle covered by a disc (there is a very very small pressure drop cross the nozzle which is accounted for in the coefficient of discharge of the relief valve sizing equations).  When the pressure reaches the set point of the relief valve the disc is removed which exposes a perfect nozzle.  The flow then proceeds through the nozzle converting all of the upstream pressure into velocity/kinetic energy of the jet exiting the nozzle without any bernoulli equation energy/head loss - only total conversion of pressure head to velocity head without any loss of energy.

 

If the relief valve discharged directly into the atmosphere the jet would just disperse into the atmosphere and onto the ground giving up its kinetic energy along the way and no pressure would build up at the relief valve discharge. However if this jet is discharged into a tail pipe/header the kinetic energy is recaptured and converted into frictional energy and kinetic energy of flow in the discharge pipe.  The smaller the diameter of the discharge pipe the more friction pressure is built up and more kinetic energy of velocity is recaptured from the kinetic energy of the jet exiting the relief valve nozzle.  The maximum built up back pressure possible is the pressure upstream of the nozzle or relief valve set pressure, since if this were to happen there would be no differential pressure for flow, and also the maximum pressure recaptured from the jet kinetic energy can only be the pressure existing upstream of the relief valve that created the jet kinetic energy to begin with.

 

So when you perform your flarenet calculations, flarenet calculates the built up back pressure due to friction at P1 for both cases starting at the end of the discharge piping and working back.  The larger the flow the more of the jet velocity kinetic energy exiting the nozzle is converted to frictional losses due to flow plus the kinetic energy of velocity in the pipe (which is much much lower than the kinetic energy of velocity in the jet).  The lower the flow or the larger the discharge pipe the less of the jet velocity kinetic energy is converted to frictional losses (built-up back pressure) and the more jet velocity kinetic energy there is wasted and not converted back to anything when it exits the tail pipe/header as lost energy.

 

In the "vessel" case it is not the same as with a relief valve in the system as there is a frictional and velocity head loss from the vessel to the inlet of the pipe at P1, so there is an energy loss in the bernoulli equation.  However in the relief valve case there is no pressure loss (no bernoulli equation head loss) through the relief valve as flow enters the pipe at P1 but only a conversion of pressure energy into velocity/kiinetic energy.


Edited by snickster, 30 December 2024 - 05:50 PM.


#14 JanPau

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Posted 05 January 2025 - 02:32 PM

Thanks very much to everyone, in particular to Snickster.






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