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Open discharge system: Pressure value at point of discharge??Helpful Member!(2) 

tr6 (Mechanical)
28 Jan 11 16:44
API 520, 4.4.1.2: (Reaction force in an open discharge system) the value of "P" is the "static pressure within the outlet at the point of discharge".

How is that pressure determined?


4Pipes (Mechanical)
30 Jan 11 8:49
Fanno line calc.
You might be able to use ASME B31.1 App 2 if appropriate.
jamesbanda (Chemical)
31 Jan 11 7:56
surely it is atmospheric pressure as you are discharging to atmosphere ?  

tr6 (Mechanical)
31 Jan 11 13:38
Jamesbanda,
That is what I was thinking. I just needed confirmation.

Thanks for the reply.
CJKruger (Petroleum)
31 Jan 11 21:35
jamesbanda, yes it is atmospheric pressure except if the pipe chokes at the outlet.
jamesbanda (Chemical)
2 Feb 11 8:01
CJKruger,

Surely, it is still atomspheric pressure the chokes just creates back pressure upstream.  

For any calculation basis it is atm pressure  

Petroprocess (Chemical)
3 Feb 11 17:25
There was a similar discussion in our company. The situation was that there was a scenario where a single valve was relieving gas to an atmospheric vent. A couple guys argued that the pressure "available" at the tip of vent is basically pressure at the inlet of relieving valve (not PSV, consider ESD butterfly valve) minus the frictional/static losses in the vent headers till tip. Hence they come up with larger pressure at the vent tip. But actually this calculation has to be done in the reverse, i.e. take atmospheric pressure at the tip and work backwards adding the losses untill you reach the outlet of the relieving source. This is similar to what Flarenet does. The pressure at the vent tip will always be atmospheric. If we want to know pressure at some point closer to vent tip, it will be much lesser that your source outlet.
 
tr6 (Mechanical)
5 Feb 11 18:07
If the pressure at the discharge is 0, what is the purpose of the last term in the API 520, 4.4.1.1 formula for the reaction force(A*P) (Area at point of the outlet x static pressure at the point of discharge)?

As a point of reference, Coade performs dynamic analysis and produces an output with conditions along the vent pipe (P, Vel. T)at the valve orifice, vent pipe exit gas conditions. For a current analysis (steam) it reports an pressure at the exit from the vent as 7.8 psig.

I discovered this after my original post.
DSB123 (Mechanical)
17 Feb 11 1:35
The pressure "just inside" the RV tailpipe must be greater than atmospheric and it is this pressure which is to be used in the API 520, 4.4.1.1 equation. Most people beleive that this pressure is the atmospheric pressure and hence forget about the pressure term of this equation. The number of Excel spreadsheets I have come across neglecting the pressure term to calculate the reaction force is ever increasing at large companies. Without this term the reaction forces could be underestimated by a significant margin. Also people beleive that the forces in a closed system are negligible - they are under steady state - but the transient situation is significantly different. Again the number of companies that ignore the transient condition I find is ever increasing.
tr6 (Mechanical)
19 Feb 11 18:35
ASME B31.1 II2.1.1 provides a formula to determine the pressure at the exit of a discharge elbow. This formula does not take into account any pressure losses along the length of the vent pipe, so it should give a conservative result to be used in determining the vent pipe thrust.
 
Helpful Member!(2)  zdas04 (Mechanical)
21 Feb 11 9:39
The reason that choked flow is choked is a standing pressure wave that is persistent at the point that pressure goes below critical pressure.  Like any compressible flow problem, the arithmetic on this is pretty ugly.  At the PSV there will be a standing wave which maintains the pressure at the critical pressure based on PSV set point (i.e., P(crit)=P(setpoint)(2/(k+1))^(k/(k-1)).  But if that critical pressure is high enough to create another standing wave then you will have a second choked stream within the pipe.  For a high pressure PSV and a discharge pipe with any length, you can get several standing waves within the pipe.

Since sonic velocity is not a function of pressure, you might think that these standing waves don't matter, but they do.  We calculate PSV capacity based on pressure upstream of the PSV (velocity doesn't change, but density does change and that changes mass flow rate for a given velocity), but if the pipe is very long we can be significantly overstating capacity.

The way I approach this kind of long-pipe problem is to assume that at the pipe outlet P(crit)=P(atm), so

P(upstream) = P(atm)*(2/(k+1))^(k/(1-k))

And that pressure is the important one controlling mass flow rate out of the PSV.  It is also the pressure used to calculate the reactive force on the pipe.  

If the pipe is short enough to fail to support a secondary standing wave, then the reactive force is critical pressure based on PSV set point.

David

 
tr6 (Mechanical)
21 Feb 11 15:53
David,
Thanks for your response. I agree that there are two points for choked flow: through the valve and in the vent pipe. And that if the vent pipe is too long, or too small in diameter, that capacity will be affected.

If I did my calculation correctly for your second formula, using k=1.31, then P(upstream) = 0.85 psia. Less than atmospheric. (I am assuming the last term is supposed to be (k-1)??). Physically, where is point?

Have you seen the paper by Brandmaier, "Steam Flow Through Safety Vent Pipes"? I am attaching for your reference if not. His model shows super-sonic conditions in the vent pipe. Which, as you have stated, will cause the second point of choked flow as a result of the shock wave.

Even after trying to get through the derivations in that paper, I am still having difficulty not accepting that the exit pressure from the vent pipe is atmospheric.

I would appreciate your comments.
Regards,
Bill
zdas04 (Mechanical)
21 Feb 11 17:30
The last term is 1-k, because a lot of people find it confusing to see

P(choke)=P(up)(2/(k+1)^(k/(k-1))
P(up)=P(choke)/[(2/(k+1)^(k/(k-1))]
P(up)=P(choke)[(2/(k+1)^-(k/(k-1))]

So I took the minus sign into the expression (^-k/(k-1)=^+k/(1-k)

When I do the calc with air (k=1.4) at sea level (P(atm)=14.73 psia) I get P(up)=27.88 psia = 13.1 psig.  Different gases and different elevations give you different numbers.

We all accept the term "choked flow" without asking "Why?".  I did it for 20 years until a @$%@$% student asked me "Why is that?" and I had to think about it.  Thinking about it for the last year has made my head ache, but I think I have some insight.  

For approximately compressible flow (think velocity less than 0.6 Mach), gas sort of acts like an avalanche--the upstream gas "falls" into the downstream lower pressure.  As the velocity approaches Mach 1.0, that stops happening as simply.  Something limits the velocity to the speed of sound (unless you do something special that I'll get to in a minute).  That "something" is a standing wave.  The standing wave is a largely impermeable barrier to flow.  Gas at Mach 1.0 can escape, but anything slower is trapped and builds up pressure.  In that model, pressure upstream of the standing wave is at P(choked) and pressure downstream is at some lower pressure.  So if you have a 10,000 psig PSV discharging into 50 ft of 2-inch pipe,
  • then critical pressure (right at the PSV) is 5276 psia.
  • That is a bunch of pressure, so it will set up a second standing wave in the pipe with a critical pressure at 2780 psig.
  • Still too much so there would be a third wave at 1462 psig.
  • If I'm not at the end of the pipe yet, then I'll have a 4th wave maintaining 765 psig.
  • Next would be 397 psig,
  • followed by 203 psig,
  • followed by 100 psig,
  • then 46 psig,
  • then 17 psig which would probably tend to move back up the pipe a bit and actually discharge to atmosphere.
The velocity at every point in the pipe is equal to every other point because they're all at sonic velocity.  The theoretical mass flow rate is significantly lower at each standing wave, and you'll eventually reach a pseudo-steady-state point where the mass flow rate (which actually must be the same at every point in the pipe) will reach an equilibrium somewhere around v(sonic)*A(pipe)*rho(last wave).

We all learned that in incompressible flow, pressure is directly related to the square of velocity.  Sonic flow is not incompressible, and as cross sectional area increases the standing wave is spread thinner and velocity is able to increase above Mach 1.0 (velocity increases to the point that the standing wave for that cross section is adequate to hold back the new velocity > 1.0 Mach).  That is the only way I know of to get greater than the speed of sound in a pipe.  Contentions that the traffic jam of standing waves I described above result in super sonic flow aren't supported in the data.

David Simpson, PE
MuleShoe Engineering
www.muleshoe-eng.com
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Latexman (Chemical)
22 Feb 11 8:58
David,

This is an excellent description of a complex system.  In certain services, you can see evidence of the upstream mechanisms you mentioned at the end of the tailpipe – many individual, expanding, billowing "clouds" – definitely not a single, homogenous, steady-state stream.  The avalanche comes in waves.  I see it every year at my voluntary fire extinguisher training.

Good luck,
Latexman

tr6 (Mechanical)
22 Feb 11 9:44
Thanks David,
A couple more questions:
Your first post said that if the discharge pipe is short enough where it fails to support the secondary standing wave then the reactive force is the critical pressure based on the PSV set point. That makes sense.

So, if the pipe is only long enough to support a certain number of waves, then the reactive force would be whatever pressure is there at the end of the pipe as described in your previous post. ???

How would you model the distance between the standing waves to know how many waves the discharge pipe could support?
zdas04 (Mechanical)
22 Feb 11 16:00
Compressible flow calcs are UGLY.  I read a rule of thumb once that said that the minimum distance between stable standing waves is around 5 pipe diameters.  I don't remember where I saw that reference or if I made any effort to verify it.  It has been too long ago and I can't find where I wrote down where I heard it.

Other than some lame rule of thumb, your other option is building a CFD model.  I don't think you could do the itterations required to approach "steady" flow in a compressible stream by hand in this life time.

Yes, if I could get to a reasonable confidence level that I had properly estimated the number of waves, then I would use the critical pressure of the last wave for reactive forces.  Otherwise, I would calculate it twice--once with P(crit) calculated from the PSV setpoint and once with it calculated from exhaust conditions.  I would then build for the higher reactive force.

David
CJKruger (Petroleum)
22 Feb 11 21:30
zdas, this is a very strange explanation of choked flow. Some proof is needed before this explanation can be regarded as correct.

I think what really happens is much simpler. Taking your example of 10,000 psig PSV discharging into 50 ft of 2-inch pipe:

(a) If the PSV nozzle area is very small, then the flow rate is small. The PSV nozzle is choked, but the velocity everywhere in the pipe is small and far below sonic.

(b) If you increase the PSV size, the flow through the 2 inch pipe will increase untill it chokes at the pipe outlet also.

(c) Increasing the PSV size further, still increases the flow and the pressure before the outlet choke point increases.

(d) Increasing the PSV size even further, will increase the flow till the PSV outlet pressure is high enough so that the PSV nozzle is no longer choked. Only choked point is now the pipe exit.

  
zdas04 (Mechanical)
22 Feb 11 23:47
Mr. Kruger,
I have a hard and fast rule--I provide documentation to paying clients, not on free sites.  Putting together "proof" is boring, I don't do it for free.  Like anything else on the Internet, you can accept it or not.  That is up to you.

I'm not going to argue with you about your simplistic "definition", if it works for you then that is what you should use for an explanation of the concepts.

Have a nice evening.

David
zdas04 (Mechanical)
23 Feb 11 16:33
OK, I've been thinking about how to describe this in a way that people can accept.

Let's start with the 10,000 psig application.  If the PSV has an "H" orifice then the standard PSV calcs say that it will move 219,196 MCF/day at 10% overpressure.  Now, look at the end of the pipe.  Volume flow rate at standard conditions into the pipe has to be the same as volume flow rate out of the pipe at standard conditions, the pipe diameter is 1.996 inches, and pressure outside the pipe is 0 psig.  That is enough information to calculate a velocity out the end of the pipe.  120,344 ft/sec.  That is 112 Mach.  Really fast.  Not really possible.

Pressure drop is nearly 100% so you can't do incompressible flow calcs.  What is happening?  It has to be compressible flow.  That means Mach numbers and shock waves.

David
mariog123 (Mechanical)
28 Jul 11 8:39
Dear All,

I think I have an answer to your question and I attach a paper with mathematical derivation of it. It is just one page so you can evaluate it.

I posted it in 2010 on Coade forum, i.e.
http://65.57.255.42/ubbthreads/ubbthreads.php?ubb=showflat&Number=33428&page=3

It would be an answer to your original question, valid only if you know the "free jet" exists with Mach=1.

My best regards.


 

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