I have a safety valve exhaust system. The piping at the safety valve exhaust(8") is connected(horizontally) through an 8" pipe(5-6m in length with few bends)to another bigger exhaust pipe(20")which extends 25-30m to the atmosphere and has a silencer mounted(welded to it) at the top. There is no drip pan in the system all the connections are welded.

First Question: Is it an Open Discharge system or a closed discharge system as per B31.1 (Non mandatory appendix)

second Question: For calculating the safety valve reaction forces in system, can i use the same formula as given for calculating reaction force at at the discharge elbow in B31.1 (for open discharge system). Shall i assume this force to be acting vertically on the exhaust pipe (20") and no force acting on 8" pipe? (using, Dynamic load factor=2)

Third question: Will there be any reaction force due to silencer, which needs to be added to the reaction force calculated above.

Fourth question: Many times horizontal reaction is given as acting on safety valve along with the vertical reaction, but there is no such force given in 31.1, how can we calculate horizontal reaction forces.

Think i am asking a lot many questions, but will be very thankful for your replies

it would be very educating to know about basis for calculation of horizontal reaction forces.

To the extent I know the origin of the horizontal force equation is D Laval rocket eqn.

Dear Anindya.....
Can you explore more regarding your answer.

Hi Adi
Regarding other points.
1) You may consider it as close system.
2) Open discharge calculation is quite conservative one. You may use it to calculate force unless force value supplied by PSV vendor. I think you have two PSV with 8" & 20", in that case you have to prepare OCC load case for both of them. You may consider both case non-concurrent in case both are connected to same source.
3) No need to consider additional force for silencer.

Appreciate comment on above statement from Anindya or other expert.

Regards

Habib

Thanks anindya!
but it would help if you could explain it a bit.for i believe if there is a jet from elbow(curved)the reaction force would be vertical as is covered in the code and not horizontal.Please forgive my ignorance..i may be quite wrong.
I also came across one formula for horizontal reaction force, but it had no mention of the source.

Dear habib!

please clarify one doubt, for the closed system, shall he consider the reaction force at the first elbow(8")after exit from safety valve or at the last/silencer end(20")(where it meets the atmosphere).

Manvender [:)]

Hi Manu,

For the safety valve forces you have to consider them at the first elbow after the safety valve. As what I see Mr. Adi was asking of some formula mention in ASME B31.1. You can use the same formula for determining the magnitude of force for the close discharge system. It is mention in the code. If I remember correctly then it is nonmendatory appendices under II-2.3.2

Regards,
RK

Dear RK

u see the formula is ok!, but where should one apply the force?(in a closed system)..

at the exit to atmosphere(i feel so!) or at the first elbow after the exhayst valve(assuming there are a few bends before it reaches atmosphere)..i read in one of the posts..mr brown had mentioned that it should be applied at all the elbows ..but what magnitude..i couldn't get it.

Manu,

I think you can consider the first elbow after the PSV. After the pressure thrust hits the first elbow, the reaction force would not be significant at the exit to atm.

Thanks,

Thanks Piper!

i searched through the forrum and found a very similar problem. It was very informative. But now i am more confused than i was before

http://www.coade.com/ubbthreads/ubbthreads.php?ubb=showflat&Number=18466&page=4

Dear Manu,
You can add the force at the first elbow. There is no point in adding the force at the exit to ATM.

Regards,
RK

Well in the Caesar technical reference manual it is written than if a manifold piping is attached and the discharge of psv fills the exit and manifold piping than the force should act on the last elbow of discharge to the atmosphere whereas in case of the manifold not being filled completely than the reaction force should act on the first elbow on the exit piping. In another post in the thread mentioned below Mr.Loren has suggested that in an open system this force should act on each elbow.

http://www.coade.com/ubbthreads/ubbthreads.php?ubb=showflat&Number=11770&page=1#Post11770

Please clarify

Regards,
Umair

I would like to say please refer ASME 31.1 nonmendatory app. for more clarification on this issue.

Thanks umair

It's more clear now.
I think the first time i read that thread i was very much sleeping.

Thanks RK,
I had read nonmendatory app.ASME 31.1, it doesn't say much about closed systems, does it? though it was very informative, for sure.

..also there is a doubt I couldn't clarify
why is:
p3=p1(A1/A3) ......(II-2.2.1)
v3=v1 ......(II-2.2.1)

It’s a fluid mechanics model.
If you are outside B31.1 scope, you can follow your own model using software like AFT Arrow. Inside B31.1 scope it’s better to follow the appendix.

Just few words.
You may consider a famous work on this subject “75-WA/FE-23 Steam Flow through Safety Valve Vent Pipes by H. E. Brandmaier and M. E. Knebel”. Please see

http://www.coade.com/ubbthreads/ubbthreads.php?ubb=showflat&Board=1&Number=21607

I second Mr. G.S. Liao criticism developed in "Discussion" section of the above article. Arrow software solves the problem just as Mr. Liao hoped.

I’ve tried to make some comments on this subject at:

http://www.coade.com/ubbthreads/ubbthreads.php?ubb=showflat&Number=21000
It would be "informative" or not, it’s not more than my opinion.

regards

Grazie Mariog!

I have downloaded it, i will read it and will come back to u for help.

(..was in italy(genoa) last year..this time..and i love that country)

ciao

Dear Adi,

The origin of th horizontal reaction force is somewhat like this ( this derivation was told to me by a person who I consider as my guru in stress analysis):


The second half of the equation – AP -- is obvious. The first part -- 129W√[kt/((k+1)M)] – represents the momentum part of the equation (i.e., F = Aρv2, where A = cross-sectional area, ρ = density, v = velocity). Writing this another way:



F = Aρv*v = W*v (where W = Aρv, mass flow in kg/sec).



So the question is, what is the maximum possible velocity when a vent is opened, releasing a hot gas into the atmosphere (or an open system)? To estimate that, we can turn to de Laval’s nozzle equation (from rocketry):







where:


Ve
= Exhaust velocity at nozzle exit, m/s

T
= absolute temperature of inlet gas, K

R
= Universal gas law constant = 8314.5 J/(kmol·K)

M
= the gas molecular mass, kg/kmol (also known as the molecular weight)

k
= cp / cv = isentropic expansion factor

cp
= specific heat of the gas at constant pressure

cv
= specific heat of the gas at constant volume

Pe
= absolute pressure of exhaust gas at nozzle exit, Pa

P
= absolute pressure of inlet gas, Pa




If we assume that the valve vents into the atmosphere, then Pe approaches 0, so this equation approaches:



V = √(TR/M)(2k)/(k-1)



So:



F = W*√(TR/M)(2k)/(k-1)



Since R = 8314.5, √2R = 129, reducing the equation to:



F = 129W*√(kT)/[(k-1)M]



This is almost exactly the API-520 equation (we have “k-1” in the denominator inside the radical, while API-520 has “k+1”. I don’t know if that is an error on their part, or if they are using a completely different basis for their estimate than I am, or if the Pe/P part of the equation that I dropped out makes a difference…


Regards

Dear Anindya,

A "k+1" in denominator formula is naturally coming-up from Laval nozzle theory. See, for example, Roberson, Crowe‘s "Engineering Fluid Mechanics", paragraph "mass-flow rate through a Laval nozzle".

API formula is correct against the assumptions made: a perfect gas evolving isentropic from stagnation state (M=0) to critical state (M=1). You can correct this formula for a specific real gas. But staying under API calculation it means to follow the formula.

Now few ideas not related to your comment but to Laval theory.

Although isentropic flow is an idealization, it often is a good approximation for the actual behavior of nozzles. Since a nozzle is a device that accelerates a flow, the internal pressure gradient is favorable. This trends to keep the wall boundary layers thin and to minimize the effects of friction, Fox- McDonald "Introduction to fluid mechanics"

In my opinion that’s why we cannot extrapolate directly the Laval theory as valid for a PSV model (I refer here only to the PSV). A lot of friction is generated in the PSV's body (and we want to have that friction!) and it’s hard to think that after the PSV throat the fluid accelerates to supersonic flow.

It is one reason I don’t agree with Brandmaier and Knebel’s article…They said "The flow leaving the valve orifice can be characterized as an underexpanded jet exhausting into a larger diameter cylindrical pipe. The flow expands across a series of expansion waves to supersonic velocities until its static pressure equals the pressure surrounding the jet…."
IMO, there are few IFs here: IF the wall boundary layer is thin; IF the friction effects are kept to minimum, IF the internal pressure gradient is favorable… Nevertheless, with such turbulence and friction developed in PSV body, for me it’s hard to believe such behavior.

I reattach this article; I remember you have posted it some time ago. Thank you for your generosity!

My kind regards

Thanks anindya!
Thanks mariog!

seems like i have a lot of homework to do

Mariog,

Thanks a lot.

I will go through all the references cited by you.I also agree with you on your view on the particular aspect of Brandmaier and Knebel's article.

Regards

Mariog,

I have seen some companies use an expression in lieu of the (Pexit-Patm)* Exit Area ( for the Reaction force at the open end of an PSV tail pipe) :

Force= mass flow rate times SQRT( GAMMA *g*R1*T2)/g

where GAMMA is the ratio Cp/Cv, R1 is gas constant, g=acceleration due to gravity , T2= Max. Exit temp.

I am not sure how is this expression derived.

Kind help from Mariog/any member of this forum will be highly appreciated.

Regards

In continuation of the post above, my understanding is, the expression is nothing but mass flow rate times sonic velocity at exit. But , my question is, why this expression should be a substitute for (Pexit-Patm)* Exit area?

Regards

Any thoughts from members of this forum?

Regards

anindya,

The different terms for the derivation of the load might be from the equation for conservation of momentum, or impulse = chamge in momentum,
F*t = mv2-mv1 .

Dear sir,

refer API RP520 for PSV Reaction force calculation..

Richard,

I agree with you that it is the momentum eqn.My question is , the expression mentioned in my post, in what way it can be a substitute for Pressure differential at exit times exit area and how is it derived in that form?

Dear Stress Engineer,

My question is not about where to get the PSV reaction force from . I also know that it is there in API 520. My question is in the previous paragraph.If you have answer to this specific question , only then respond.

Regards

Dear anindya,

You can consider my understanding of this reaction force derivation in
http://www.coade.com/ubbthreads/ubbthreads.php?ubb=showflat&Main=4654&Number=21003#Post21003

The reactive force has two components, derived from Fluid mechanics model.
Reactive_Force= [mass flow-rate]*[jet_velocity]+ [p_jet]*[area_jet]

The first term has that expression which appears as an "enigma" for a regular engineer. However, jet_velocity is just the critical speed when the jet gas flow has Mach=1; the expression is not exactly the sound speed in resting fluid but is closed to that value- anyway this is just a detail.
The second term, even familiar, cannot be evaluated without a fluid mechanics model. The "best" is to have a result from a flow simulation, but depending on the application you can see some theoretical results in some articles.

So in my understanding both terms are to be considered.

Sorry for the answer delay!
Best regards.

Mariog,

Thanks for the input.

Reaction force should be equal to

= Mass flow rate times velocity at Relief valve outlet + ( Exit pressue-atmospheric pressure) x exit area. This is precisely what API 520 expression says.

The flow process to bring the fluid from the vessel stagnation condition to the nozzle throat can be assumed to be isentropic i.e adiabatic and reversible, the flow from nozzle throat to the relief valve can be considered to be adiabatic only but irreversible and finally the flow through the outlet nozzle can be considred to be isentropic.

The first part of API 520 expression is = mass flow rate times exit velocity.There can be three different scenarios.

1) flow through the relief valve is choked at the relief valve nozzle and relief valve exit plane.

2) Flow through the relief valve nozzle is still choked but the flow at the relief valve exit plane remains unchoked.

3) flow throughout the relief valve is unchoked.

It can be shown that the API 520 equation fits case 1 and the velocity at exit is the sonic velocity. API expresses it in terms of stagnation temp. in the vessel which can be coverted to temp. at exit and this conversion shows the well known expression for sonic velocity ( SQRT GAMMA X R X T/M)in terms of exit temperature.

My doubt is: the expression mentioend in my original post is sonic velocity. So if this one used along with the first part i.e. momentum part of API 520 eqn. is not not a doubling of momentum expression and secondly how do we justify this expression as a substitute for the second term of API expression?

Regards

Dear anindya,

I’m not sure I can follow your doubts.
I’m not blaming you for this… it’s just my incapacity to resolve your doubts vs. my "knowledge".

In API 520 "2.4.1 DETERMINING REACTION FORCES IN AN OPEN DISCHARGE SYSTEM" first paragraph is saying "The following formula is based on a condition of critical steady-state flow of a compressible fluid that discharges to the atmosphere through an elbow and a vertical discharge pipe. The reaction force ( F ) includes the effects of both momentum and static pressure; thus, for any gas, vapor, steam,….etc"

So it’s clear that the intention is to consider a steady-state flow counted as critical where the jet exists in the atmosphere. In brackets…the formula is not looking to what is happening in PSV. Why? In my opinion because is nothing serious to give a reaction force there, at least in steady state flow. It’s funny sometimes Vendor is giving a reaction force in PSV but saying it’s counted as PSV is free discharging at outlet flange. In my opinion, in this case the real effect is "free jet" at outlet flange, not specifically the PSV presence…

In Fluid Mechanics the terminology "critical" is referring to Mach=1 in that section and the critical velocity is counted as sqrt(kRT*) with R=Ru/M.
To obtain T* , the theory is counting an isentropic evolution between stagnation conditions in vessel (T0 @ Mach=0) and critical condition (T* @ Mach=1) in exit section. The Fluid Mechanics rule for that is
T*=2*T0/(k+1).
Now you can get the API result, and you can calculate c*= the critical velocity sqrt(kRT*) as
sqrt(2Ru)* sqrt[kT0/(k+1)/M].
In SI, Ru=8314.47 J/kgmol/K so numerically sqrt(2Ru)=129.
So the first term of API's SI formula is
Wxc*=129*W*sqrt[kT0/(k+1)/M] with W "flow of any gas or vapor, in kilograms per second" as API says.

API is not entering in details for the second part of formula- effect of static. They are just saying is A = area of the outlet at the point of discharge, multiply by P = static pressure within the outlet at the point of discharge.

The formula has two parts and must include the effects of both momentum and static pressure.
Nothing is double counted- in my opinion. Both expressions must be evaluated- API says.

Now I’m not sure what’s your doubt.
Could be the assumption that the gas/vapor expands isentropically from protected vessel/ pipeline to the exit? True, it seems that the PSV existence is not specifically counted… And true again, that pipe will choke at more than one location along its length.
But this assumption (isentropic evolution on all path) just provides a convenient mean to determine the maximum capacity / reaction force, in my opinion.
And- again in my opinion- "choke" is not synonym with "free jet"...the real troubles are where we have a free jet giving a reaction force, so exactly where API is looking for...

You said "But , my question is, why this expression should be a substitute for (Pexit-Patm)* Exit area?"
No, shouldn't replace the second term. Just the fact the second term is not so easy to be counted and is moderate as value...

Mariog,

Thanks for your input.

I will try to put my question in more clear terms.

I am totally in agreement with the first part of the API expression i.e. the momentum part.

My question is about the second term i.e. Area at exit x Exit Pressure.

API does not provide any expression for exit pressure.

Considering choked flow and adiabatic conditions, we can write an expression for exit pressure as :

P exit= (W/Ae)x SQRT( 2RTo/(k(k+1)M)

Where is the mass flow rate (kg/sec)
Ae= exit area
R=universal gas constant
k=Cp/Cv
M=molecular weight
To= vessel stagnation temp

Using this expression and also the ratio of To/Te where Te is the exit temp ( which is =(k+1)/2),I am getting the following expression for Pexit x A exit

=WXSQRT(R'Te/k)...(1)

where R' is the characteristic gas constant=R/M

This expression is close to but not same as what is there in my original post, which I am rewritting for ready reference.This time I am writing the full expression with the units.

WX1.88 SQRT(k.g.R'.Te)/g....(2)This is in Newton

where g stands for gravitaional acceleration in m/sec^2. Unit of R' in (2) is in J/kg K , Te is in K

I am not sure how to prove that (1)=(2)

Kindly throw some light.

Regards

I understand you are looking to the second term of API formula.
I think your problem hasn’t a simple answer- and for this fact API doesn’t enter in details.

Let’s simplify the discussion looking to a "dummy "fluid mechanics model.
One pipe is connected to a large vessel and the other end of a pipe exits into the atmosphere.
The vessel is large enough so even in steady-state flow the vessel pressure ("large enough") remains constant.
So the pipe is connected to an "infinite source" with density0, p0 ,T0 and discharge into atmosphere with p_atm. T_atm. The pipe is adiabatic insulated.

There is a steady-state flow trough the pipe. At the endpoint, the gas pressure cannot drop to match the atmospheric pressure without the gas accelerating to critical conditions. A shock wave forms at the end of the pipe, resulting in a pressure discontinuity. In my opinion that discontinuity results in energy dissipated in the atmospheric turbulence so the pressure that would be taken into consideration in API reactive force formula is the "upstream shock" static pressure. This approach would be questionable but anyway is conservative.

It is true also that, the choked flow at the endpoint can be count as Laval. However it is difficult to apply these equations to choked conditions, because the local conditions upstream are not known at the point of choking.
Anyway, to perform this calculation, one must be able to calculate the stagnation pressure and temperature at the end of the pipe, upstream of the shock wave. How?

The real behavior of gas flow in adiabatic pipe is the gas accelerates along the length of the pipe. As the pressure drops, the gas density will also drop and the dropping density must be balanced by an increase in velocity to maintain mass balance. So to make a calculation means to evaluate pressure, temperature and gas density as "local conditions" upstream of the shock wave. In this case, important is to simulate the flow trough the pipe.
Even so, the fluid mechanics model is a little bit strange since you know one boundary condition at the end of the pipe (Mach=1).
BTW, if you take a look in the section "Discussions" of the famous article "Steam Flow Through Safety valve Vent Pipes" by Brandmayer and Knebel, Mr. G.S.Liao said "In any case, the conditions at the vent pipe inlet can not be determined without knowing the conditions at the vent pipe outlet. Therefore, the calculations are always backward". True, that’s why I think that article is just a fluid mechanics model that makes assumptions hard to be confirmed.

For adiabatic conditions, the calculation for this model can be performed by hand. Some guidance you can found in the article "Gas-Flow calculations: Don’t Choke". I’ve downloaded the article from AFT site. Mr. Trey Walters is the President of AFT (Applied Flow Technology).

Anyway, the real case of PSV is a much more complicated simulation, so I would prefer to have the results from AFT package- Arrow is the module for Gas-Flow calculations.

I don’t know if what I've written clarifies your doubts. Probably neither of your formulas is what you need.
In my understanding it's important to simulate what is happening through the piping system rather to know what is happening in final shock wave.

Mariog,

Thanks for your input. I would like to share two papers with you on this subject. Due to copyright issues, I cannot post it on the forum.

It will be great if you can share your email with me.

Regards

I’ll be honored to share ideas about the subject.
I’ve tried to send you a message via My stuff/ Messages.
I get a nice "Private messages are disabled".
It is a setting of my profile or Coade service is disabled?

Dear Mariog,

I also could not send you a private message as against your name it is disabled.

My email id is: abhattacharya_uk@hotmail.com

You can send a test mail at this address and I will Send you the papers.

Regards

Mariog,

I will send you the papers today. Regarding my post, this is how I could get 1=2 well at least nearly.

Since k= 1.4 ,the expression for PX A that I got earlier can be written as :

= W x 0.71 X SQRT( k x R' X Texit)

The expression in my post is :

(W x 1.88/g) X SQRT( k x g XR' X Texit)

This is equal to :

((W X 1.88 )/ SQRT(g)) X SQRT( kxg XR'XTexit)

Since g=9.81m/sec^2 and inverse of SQRT g= 0.319, the above expression comes to :

WX1.88X0.319 X SQRT( kx R'XTexit) = W x 0.6 X SQRT( kx R'XTexit)

Hope my algebra above was correct.

Regards

Dear all,

Anindya has sent me two articles about the subject. They are valuable, no doubt. However, the author has developed a quite complex system – eight equations, all based on the assumptions of perfect fluid in steady-state, adiabatic, frictionless and isentropic flow.

If the goal is to evaluate the choked sonic speed and choked fluid pressure, a serious simplification can be applied. There is a p_choked formula that supplements API formula being developed in the same way as that leading to the choked sonic speed expression. I have no idea why API does not detail the reaction force formula with p_choked expression.

I attach my interpretation. The result is identical with the articles’ conclusion and is- in fact- exactly what Anindya has posted previously.
So my contribution was just to simplify everything to "one page" presentation. In addition it is not subject to copyright.

This discussion is not about where the choked flow and reaction force may appear. The formulas are evaluating the sonic speed and choked pressure in isentropic choked flow section- they are not showing where is that point.
About this aspect, I think the reaction force must be count just where there is "the free jet" effect- at the end of open discharge piping and in that point there is a "choked" flow in many cases.
It may be a long discussion about this aspect- anyway is what I am thinking.

The formulas are not counting friction effects.
Are interesting also the remarks I found in the articles "As mentioned before, the relief valve body has a very complex geometry. This complex geometry causes the relieving fluid to have a complex flow pattern. The fluid must change direction several times and there are certainly some frictional effects within the relief valve body. However, by making these assumptions, analytical expressions can be developed and these expressions yield interesting information and valuable insight about a rather complex problem. More complicated methods can certainly be utilized to analyze the body of the relief valve. However, these added complexities will certainly require a more detailed description of the geometry within the relief valve body and will almost certainly require numerical methods for computations."

I second this interpretation, the formulas are just what we can do analytically and this fluid mechanics model is just one possible. The fact API has accepted this way is encouraging.

Thank you, Anindya for your help.

Best regards.

Piping and Pipe Support Design and Engineering by Pau R. Smith had pointed out this question and gave the equation to calculation reaction fore

... or B31.1, NONMANDATORY APPENDIX II, RULES FOR THE DESIGN OF SAFETY VALVE INSTALLATIONS, II-2.2.1 Design Pressure and Velocity for Open Discharge Installation Discharge Elbows and Vent Pipes....

sasklimin,

Because you’ve mentioned the formulas given by Pipe Support Design and Engineering by Smith & Van Laan/ B31.1 "PSV steam case", activating this old thread, I have some remarks addressed to B31.1/ P1 and V1 expressions.

First, it is clear that B31.1/ Appendix II considers that not all stagnation enthalpy (h0) can be transformed to a state based on chocked parameters, so there is a correction as "h0-a" instead of "h0". Here I can speculate that this difference is because steam is not a "perfect gas", so to the domain of lower temperatures/ pressures we may have steam/water mixture or water and the idea is PSV can "consume" a quantity of enthalpy up to such fluid state; maybe this approach is not conservative, but I don’t need to comment it too much; just consider "h0-a" as the enthalpy that is transformed.

It is interesting that, excepting the above mentioned deviation, the B31.1 model is still based on the "perfect gas" model and there is equivalence between the "b" factor of B31.1 and "k"- the specific heat ratio (or "adiabatic index").
The corresponding formula is k=b/(b-1); I guess this info was lost from the original work (unknown for me), but this is evident checking the values of "k" versus "b" factors given in B31.1. i.e. k=11/(11-1)=1.10 for "wet steam" and k=4.33/(4.33-1)=1.30 for both saturated steam "> 90% quality" and "Superheated steam".

With these remarks, one can compare directly the results (between the paper with "perfect gas" model and B31.1 model) because (k-1)/(k+1)=1/(2b-1) appears in B31.1/"V1" expression and 1/k=(b-1)/b appears in "P1" expression.

More specific, following the paper I attached some time ago, in SI units we obtain h0-a=(1/2)*(k+1)/(k-1)*V1^2=(1/2)*(2b-1)*V1^2 that would lead to a SI correspondent of B31.1/(A.2) expression; also P1=(W/A)*(1/k)*V1 =(W/A)*((b-1)/b)*V1 i.e. you obtain the SI correspondent of (A.1) expression.

You may note that "gc" and "J" appears there because B31.1 is based on US Customary Units, while the "SI formulas" coefficients in Smith &Van Laan’s book must be "2" (from theory) instead 2.0085 and 1.995 (from units conversions). But this is a secondary remark and anyway the results are quite accurate vs. the assumptions made; more interesting is that "the perfect gas model" of "P1" pressure gets results in-line with B31.1.

What I try to say is that P=(W/A)*(1/k)*V (where P and V are characteristic to isentropic choked status) is confirmed indirectly by B31.1 and obviously it is a more general formula. In fact, this long discussion was just intended to clarify which pressure "p" must be considered as the design pressures in the end of PSV discharge piping and vent pipe, for open discharge installation; the goal is to check the reaction force there.

My best regards.