Piping backpressure included in sizing pressure for relief, API520 10

[Luuk21]

Hi,

I'm a bit confused about the impact of pressure drop (inlet & outlet) on the Relieving Pressure (P1 as defined in API520, chapter 5.4).

Chapter 5.4 API states that P1, the pressure that is used in the formula to calculate A relief, is equal to the Set Pressure + the allowable overpressure. If we look closely, this is (at first glance) ALWAYS equal to the MAWP of the equipment + allowable overpressure (either 10% or 21% for fire case).

For calculation purposes, P1 is always the same, no matter what set pressure you choose. However, pressure drop over piping is not included (or deducted) from P1, see f.e. Figure 15 (API 520-1, 2014).

When I calculate the required area, according to input from my fire case, I get the required area which corresponds with the flow at MAWP + 21%. However, with that corresponding flow comes a certain pressure drop over the inlet/outlet piping, which ofcourse brings the pressure in the vessel during relief above MAWP + 21%.

One could easily say: OK, problem solved, just deduct the pressure drop from the Allowed overpressure, use the new P1 and calculate again. This leads to the problem that the flow itself depends on P1 - so I have to iterate again and again.

I also know about the rule that the inlet pressure drop is not allowed to be higher than 3% of the SET pressure (not MAWP!), and the outlet not higher than 10% (in conventional PSV's). These numbers are strange if one considers that the maximum allowed overpressure is 10% in normal cases. If these rules are followed, piping itself would excess the allowed 10% (3+10=13), let alone that you need to calculate (again) with a new P1.

Question 1: Why is P1 always the design pressure + Overdesign. This makes no sense. I have f.e. a vessel that has a design pressure of 30 barg, but I use it as atmospheric storage (still rated at 30 barg), the relief calculations are based on a relief pressure of 33 barg - where my set pressure is 100mbar.

This is an extreme case, but same applies for a SV that has a set pressure of 25 barg, the set pressure and used pressure to calculate the reliefload differs.

Question 2: How to deal with pressure drop in inlet/outlet piping. Do you really need to iterate the calculation untill there is a balance between the piping pressure drop experienced and the maximum allowed pressure in the vessel - or am I interpreting the API-520 wrong and is it allowed to have:
- MAWP (design pressure)
- 21% / 10% overpressure
- 3%+10% backpressure (of set pressure, not design pressure, which is also confusing)
Leading to a total allowed pressure of +34% / +23% in the vessel (which is what my results are at the moment).

Thanks for your insights!

 

[Latexman]

Those results are suspicious. Are you 100% sure you are using the tool correctly? If not sure, I'd definitely probe that in your one-on-one. After you are certain of the tools operations, and you may be now, idk, I'd also compare your tool's results to 1-3 other methods, which some of us may be able to help with. We would need to know the basic input data for the problem, whatever was input to your tool. Like physical characterization of the inlet and outlet pipe and fittings, the PSV (make/model), the fluid, and the settings.

Have you gone through the FAQs in this forum? Like faq1203-1293?

Good Luck,
Latexman

 

[TiCl4]

Luuk21,

The 21%/10% are maximum allowable overpressures (of MAWP, not set pressure) in the vessel. You CANNOT exceed these values, and must account for discharge line losses in the design. If doing the calculations by hand, the normal process I like to follow is below. Assume for the moment set pressure = MAWP.

[ol 1]
[li]Use required relief flow rate to calculate PRV inlet and discharge line pressure losses. Set P1 = MAWP*1.1 (or 1.21) - inlet line losses and P2 = discharge pressure (atmos or header pressure) + estimated discharge line losses. If inlet losses are greater than 3% of set pressure for compressible fluids, you need to think about reworking the inlet.[/li]

[li]Use API 520 (assuming choked flow equation because of your MAWP and atmos discharge), P1, and estimated P2 to obtain a generic API area to get the approximate size of the valve. Find the next largest API orifice size. Use this info to find a specific relief valve model (see NB-18) to find a specific rated orifice size and discharge coefficient.[/li]

[li]Plug the rated orifice size and discharge coefficient back into API 520 equation to get a "rated" flow (still preliminary).[/li]

[li]Use this rated flow to calculate inlet line losses and discharge piping losses. You should end up with a different P2 (which should always be higher than your original P2 value because rated capacity will exceed your required relief capacity). If less than 3% inlet losses you can ignore the changes in P1.[/li]

[li]Use the new P2 and re-calculate outlet losses to get a new P2. Continue iterating until a stable value is achieved. Check to ensure this rated flow calculation exceeds your required flow. If rated flow is below required flow, select the next valve size up and redo the calculations. Check pressure ratio of P1 and P2 to ensure you are still choked through the PRV[/li]

[li]If rated flow is now > required flow, you now have your final design.[/li]
[/ol]

You will notice that P1 is always MAWP*1.1 (or 1.16 for multiple valves or 1.21 for fire case). It is NEVER allowed to exceed this. You seem confused on how the discharge pressure affects the sizing. It does not change the allowed overpressure. What it does is increase P2, effectively reduce the dP available for your PRV, requiring a larger orifice.

Your attached picture is confusing. Even assuming MAWP+overpressure = inlet pressure, you should have PRV relief pressure (P2) as well as relief line outlet pressure (P2 - line losses). There are only two pressures shown here. If you are going from 4.7 bara to 1 bara through the PRV, there is no question that you are choked. Most fluids critical pressure ratios fall between 0.4 -0.6. You are well beyond that pressure drop.

 

[Luuk21]

The 21%/10% are maximum allowable overpressures (of MAWP, not set pressure) in the vessel. You CANNOT exceed these values, and must account for discharge line losses in the design.

Thanks for confirming, this is what I couldn't find in API-520/521. So basically P1 should account for discharge losses, if f.e. inletpiping is above 3%, deduct the inlet piping pressure drop. The definition of P1 changes from 100% MAWP * 1.21 (fire case) to (100% - % discharge pressure drop) * 1.21 - correct?

If inlet losses are greater than 3% of set pressure for compressible fluids, you need to think about reworking the inlet.

That is not practical in my case. There are 150 SV's all existing. Old factory. Probably all inlets are (a little bit) too small. In the 4-7% range, and more important, only in the fire case (the leading case by far for most). So chattering or other mechanical issues during relief are not important, after a fire the SV's need to be checked (and likely replaced) anyway.

The API does not mention that Pdrop inlet piping above 3% is not allowed, it just states that you can not ignore it.

Furthermore - there are also overfill cases. The height of the SV is (sometimes) 1m above the vessel, which leads to a static height pressure of 0.1 bar (assuming density 1000 kg/m3). This pressure also needs to be accounted for i.m.o.

Sizing a new SV with API-520 is straightforward, checking an existing one is a bit more complicated it seems.

you should have PRV relief pressure (P2)

I think I understand what you mean, but to be sure (since P2 is not mentioned in API-520); its the pressure after the reliefvalve. With direct output to the atmosphere, P2 = P atmosphere. If there is piping involved, you need to calculate the pressure drop over the outlet piping, and add this up to P atmosphere. (I think that this is what the tool is doing).

Your attached picture is confusing. Even assuming MAWP+overpressure = inlet pressure, you should have PRV relief pressure (P2) as well as relief line outlet pressure (P2 - line losses). There are only two pressures shown here. If you are going from 4.7 bara to 1 bara through the PRV, there is no question that you are choked. Most fluids critical pressure ratios fall between 0.4 -0.6. You are well beyond that pressure drop.

I agree. But I think I understand what is happening. What the tool does is the following; it takes the pressure where it is relieved to (almost always atmospheric in my case) and using that pressure it calculates the pressure drop over the outlet piping. This is 0.063 bar (outlet is a DN100 in this case) so P2 becomes 1.076.

However, then it does something that is not completely clear; it calculates if the flow is choked at the very end (at atmospheric pressure) which it ofcourse never is. It does the same for the relief pressure P1 - and ofcourse that flow is also subsonic.

It is the flow that goes through the nozzle, from P1 to P2 (4.706 to 1.076) that IS sonic. So sizing calculations for chocked flow need to be used. Which the tool doesnt do since further in the tool it indicates the flow is subsonic.

See attachment for full data - unfortunately I cannot share the tool here, but if we contact one-on-one I can show the full thing, I will post all relevant data below.

 

[Latexman]

What is the fluid? Air? Steam? Other?

Is it just a rupture disk, a combination rupture disk/safety valve, or a safety valve? Inlet section “c” has a K = 1.5 for rupture disk.

Outlet pipe section “d” and “e” are “no”, but Pi’s are shown. And the Pi’s are < P2,f, which is double confusing.

If you want, take a look at the FAQ I linked above. Send a question, comment, or whatever to the FAQ author (me), and I’ll get it AND your registered email address. Just be sure your registered email address is valid; many are not. Then we can share more freely.

Good Luck,
Latexman

 

[Snickster]

Thanks for all the replies. Things are getting more clearer now.

Unfortunately things are just getting weirder - the tool I'm required to use (build by our senior PE) makes a bit of a mess regarding the piping pressure drop calculations. It does make a distinction between sonic/subsonic flows, however, it doesn't compare the relief pressure (pressure at nozzle SV, assumed to be the "point of most resistance") to the eventual pressure its going to (atmosphere).

Looking at the calculation output this is my understanding of the results:

At the outlet of the discharge pipe the flow is subsonic. The pressure indicated as "Critical" Pcrit = 0.137 bara is the pressure at the outlet of the discharge pipe needed to get the flow to critical sonic velocity. Since actual atmospheric pressure is 1.013 bara then the flow is subsonic. So the actual flow is subsonic with an outlet pressure of atmospheric at 1.013 bara. Next the pressure drop of 0.063 of the discharge pipe is added to get a pressure at the relief valve outlet of 1.076 bara.

On the inlet side the pressure in the vessel is 4.723 bara with 0.018 friction loss in the inlet pipe to get a pressure of 4.706 bara at the valve inlet. The flow is subsonic in the inlet pipe since the pressure required to reach sonic at upstream condions is Pcrit = 0.319 bara as indicated, and the actual pressure is much higher.

Since pressure downstream of the relief valve is less than the critical flow pressure Pc (not same as critical pressure indicated above which is critical flow pressure in the piping but Pc per API 520 which is critical sonic flow pressure in relief valve orifice) as defined per API 520 then flow is sonic in valve orifice and API sonic flow sizing equations should be used.

So most of my flows are calculated as subsonic, where I personally have big doubts if this is correct.

All flow are subsonic in the piping but it is sonic in the relief valve orifice where the API 520 relief valve sizing equations are performed at.

I will take some time to understand more, but if possible I'd like to discuss this one-on-one.

For now, see Attachment - our tool calculates the relief pressure (I agree with this calculation, inlet pressure drop + MAWP +20%) but then the outlet is regarded as subsonic, where the pressure drops from 4.7 bara to 1 bara.

The pressure drop across the relief valve of 4.706 bara inlet pressure and 1.076 outlet pressure is not due to friction. When flowing through the relief valve orifice the 4.706 pressure is reduced to critical flow pressure Pc, in the throat of the orifice, as defined by API 520 via an almost isentropic expansion (with just a little friction). Therefore the pressure in the orifice of the relief valve is Pc as defined by API 520. Also the temperature drops in accordance with the isentropic relationships, in addition the flow reaches sonic with value of SQRT(gkRT). Therefore at the orifice of the relief valve under critical flow all conditions are at sonic flow P,T,V, basically by an isentropic expansion.

The isentropically expanded pressure is at Pc which is about 0.5 to 0.6 times the upstream stagnation pressure absolute as can be seen in th API tables. So in your case at the orifice will be about 2.35 bara at sonic flow. Now as the flow exits the orifice the pressure immediately drops to the pressure in the discharge pipe via irreversible expansion and shock waves causing the pressure to reduce - not by friction - to the pressure just downstream of the relief valve of 1.076 bara which is the pressure calculated by the sum of the pressure at the outlet pipe discharge plus friction loss back to the relief valve. Note that if the actual calculated pressure (based on pipe outlet pressure plus friction loss) at the outlet of the relief valve were to be above Pc (approximately 2.35 bara) then flow through the relief valve orifice would be subsonic and therefore you would need to use the API sizing equations for subsonic flow.

I don't think you have a good grasp on compressible flow if you don't understand what I just wrote. Is there anyone where you work that you can discuss with. I learned mostly on my own but did have someone I worked with to ask questions.

 

[Snickster]

One more thing. The critical pressure/flow in the piping is same as the critical pressure/flow in the orifice of the relief valve - which is when the flow reaches sonic velocity. However there is one difference. In the orifice of the relief valve the flow reaches sonic via a basically adiabatic isentropic reversible expansion and that is basically how the API 520 sizing equations are based and derived on. However in the piping if the flow reaches sonic, it would be due to an irreversible expansion due to friction. This could be adiabatic or non adiabatic, or even isothermal. So different equations may be used to calculate the pressure drop based on assumptions of which type of friction flow is present - such as isothermal flow equations - friction flow but it is assumed heat is added during flow to keep temperature of flow constant even though energy is consumed in velocity increase as pressure decreases), Fanno flow equations - adiabatic flow with friction - no heat added or removed during flow, etc.

 

[TiCl4]

Luuke,

The definition of P1 changes from 100% MAWP * 1.21 (fire case) to (100% - % discharge pressure drop) * 1.21 - correct?

Almost. P1 is pressure at the PSV inlet, and is thus 1.21*MAWP - Inlet line pressure drop.

The API does not mention that Pdrop inlet piping above 3% is not allowed, it just states that you can not ignore it.

No, but ASME has this in Non-Mandatory Appendix M and is considered RAGAGEP. Despite being non-mandatory, OSHA has issued fines for companies that fail to comply with this RAGAGEP. (

https://www.osha.gov/enforcement/cwsa/bp-products-north-america-inc-08122010).

Proceed at your own risk if inlet pressure drop is >3%.

However, then it does something that is not completely clear; it calculates if the flow is choked at the very end (at atmospheric pressure) which it ofcourse never is. It does the same for the relief pressure P1 - and ofcourse that flow is also subsonic.

There is something wrong with your tool. The iterative process I described is mandatory (there is no explicit solution to the system of equations that I know of) when you have a tailpipe on the PSV, especially when using adiabatic expansion calculations. Are you clicking some kind of macro button to perform an iteration? As you describe it, you are dropping from ~4 to ~1 bara over the valve, so choked flow equations must be used. Check the equation used in the flow calculation to see if it using the choked or subsonic flow equation from API520. It could be something as simple as the wording "sub sonic" is wrong indicated, but the tool actually uses the sonic flow equation.

 

[Snickster]

TiCl4 There are a couple of things I want to point out in your original post that may be misleading:

1. Use required relief flow rate to calculate PRV inlet and discharge line pressure losses. Set P1 = MAWP*1.1 (or 1.21) - inlet line losses and P2 = discharge pressure (atmos or header pressure) + estimated discharge line losses. If inlet losses are greater than 3% of set pressure for compressible fluids, you need to think about reworking the inlet.

The pipe discharge pressure may be above atmospheric or header pressure if the flow at the discharge is sonic flow. This is why the OP program output sheet checks if the flow is sonic at the pipe exit. Depending on the flowrate the actual presure just inside the pipe tip can be much higher, and any value so that sonic flow will still exist in the relief valve orifice if exit pressure plus back pressure due to friction is still less than Pc the critical pressure at the relief valve flow orifice as defined by API 520. If pressure at exit plus back pressure is greater than Pc then subsonic flow will exist in relief valve. For instance if the OP spreadsheet indicated that Pcrit at pipe exit was 2.0 bara (then this would be the actual pressure at the exit not 1.013 bara) plus 0.063 friction loss then pressure at valve dischare would be 2.063 bara which is still less than Pc per API 520.

You will notice that P1 is always MAWP*1.1 (or 1.16 for multiple valves or 1.21 for fire case). It is NEVER allowed to exceed this. You seem confused on how the discharge pressure affects the sizing. It does not change the allowed overpressure. What it does is increase P2, effectively reduce the dP available for your PRV, requiring a larger orifice.

As long as Pc at the reliefe valve discharge (downstream of nozzle orifice) is less than the maximum allowable Pc per PAI 520 for sonic flow, there is no reduction in flow through the relief valve with changing P2 at the relief valve discharge. This is because as long as the P2 is less than Pc then sonic flow in the orifice will exist at same sonic conditions of Pc, Tc, and Vc.

 

[Snickster]

Quote (Luuke21)
However, then it does something that is not completely clear; it calculates if the flow is choked at the very end (at atmospheric pressure) which it ofcourse never is. It does the same for the relief pressure P1 - and ofcourse that flow is also subsonic.

The flow can be choked in the pipe at the exit as stated in previous post. If it is then the pressure at the exit will be equal to or above atmospheric or header exit pressure. This is why the program is checking for sonic flow at the exit because if it is sonic then your starting pressure at the exit will not be the exit pressure but the sonic flow critical pressure at the pipe exit.

The program also checks for sonic flow in the pipe inlet but this will never happen with design for 3% set pressure. I think they just put this in there as informational just to see how close to sonic the flow is at the inlet if you compare the Pcrit pressure shown in the spreadsheet to the actual pressure at the inlet.

 

[TiCl4]

Snickster,

Gave you a star - thanks for pointing both things out. I have, so far, managed to stay away from sonic flow in discharge piping.

The API formula for choked flow sizing doesn't even have P2, so yes, increasing P2 doesn't technically decrease choked flow capacity until your PSV outlet pressure rises above the critical pressure and turns the PSV into a sub-critical flow regime. Calculating P2 is still necessary to ensure the proper PSV selection of conventional (<10% backpressure) or bellows. It's also necessary to calculate the flow capacity of the discharge pipe from relief valve outlet pressure of P2 to discharge pressure (P3) to ensure it matches or exceeds the PSV flow capacity. In this case, P3 will be the greater of either P2 * critical pressure ratio or atmos/header pressure, correct?

 

[Snickster]

The API formula for choked flow sizing doesn't even have P2, so yes, increasing P2 doesn't technically decrease choked flow capacity until your PSV outlet pressure rises above the critical pressure and turns the PSV into a sub-critical flow regime.

Yes correct - there is sonic flow in relief valve orifice until P2 = Pc and same P1 is used in the sizing calcs. But as you pointed out below there is a limit of 10% back pressure for conventional relief valves and a reduction in capacity due to the K correction factor for balanced relief valves even though critical flow still exists in the relief valve orifice. This is not didrectly due to the backpressure P2 but indirectly because the backpressure causes the relief valve disc to close a little causing more restriction to flow.

Calculating P2 is still necessary to ensure the proper PSV selection of conventional (<10% backpressure) or bellows. It's also necessary to calculate the flow capacity of the discharge pipe from relief valve outlet pressure of P2 to discharge pressure (P3) to ensure it matches or exceeds the PSV flow capacity.

Yes you need to calculate P2 of course as P2 needs to below Pc for sonic flow equations to apply and for input into subsonic flow equations. Also need to be below 10% for conventional valve and required for calculating K correction factor for balanced bellows valve. 

In this case, P3 will be the greater of either P2 * critical pressure ratio or atmos/header pressure, correct?

If there is sonic flow at the pipe exit then the pressure at the pipe exit will be equal to or greater than the atmos/header pressure but not P2 * critical pressure ratio (at relief valve Pc). If sonic velocity exists at the exit of the pipe it and the actual sonic pressure P3 can be deternined as follows using the ideal gas equation:

For a given mass flow rate, temperature. gas MW, pipe exit area, etc. using the ideal gas laws it can be determined the actual pressure at the exit required to flow such mass of gas at these conditions. If the pressure calculated is higher than the atmos/header pressure then that is the actual pressure. If the pressure calculated is equal to or less than the atmos/header pressure than the actual pressure is the actual atmos/header pressure.

I am positng this now but will complete later so I wont lose what I have already typed which has happened a few time before by pressing wrong key.

 

[Snickster]

This is continuation from above with overlap of last paragraph:

For a given mass flow rate, temperature. gas MW, pipe exit area, etc. using the ideal gas laws it can be determined the actual pressure at the exit required to flow such mass of gas at these conditions. If the pressure calculated is higher than the atmos/header pressure then that is the actual pressure. If the pressure calculated is equal to or less than the atmos/header pressure than the actual pressure is the actual atmos/header pressure. Note that if it is equal to the atmos/header pressure then this means the flow just reaches sonic but any pressure less means the flow is subsonic.

Ideal Gas Equation for Flowing Fluid: in Ft.Lbs/sec, Ft/sec, Ft[sup]2[/sup], etc units

P*(144)*(V)*(A) = m*R*T

P is pressure in PSI - 144 converts to PSF.
V is velocity at pipe exit FT/sec - assume that it is sonic for purposes of checking what the pressure must then be for sonic to exists = SQRT(gkRT).
A is the flow area of the pipe exit in FT[sup]2[/sup]
m is the mass flow rate in pounds per second (pounds force/mass not slugs)
R is the universal gas constant = 1545/MW Where MW is the molecular weight of the gas, units of gas constant = 1545 FT-pounds/pound mole
T is the temperature at the exit in degrees Rankine = (2/(k+1))To - again assume sonic conditions exist at the exit so actual T is at sonic velocity.
To is stagnation temperature upstream of the relief valve - relieving temperature - degrees Rankine.
g is gravity constant = 32.2 ft/sec[sup]2[/sup]
k = ratio of specific heats

Therefore the ideal gas equation takes the following form:

P*(144)*SQRT(gkR(2/(k+1))To)*(A) = m*(1545/MW)*(2/(k+1))To

(Note that this is same equation, but in different form and metric units, in OP's spreadsheet to the right of Pcrit .319 in free area of spreadsheet)

Since everything above is known except for P then you can solve for P which is the actual pressure at the pipe exit if sonic flow exists. If P is above the atmos/header pressure then flow is sonic. This is because the fastest you can go is sonic in a pipe and it will always be at the end of the line, therefore if you get a higher pressure than atmos/header it means that even going as fast as you can, in order to pass that mass, you still have to have a higher pressure to do so. Therefore the pressure is higher inside the pipe tip because the upstream energy availability is able to keep pushing the flow through while creating a higher pressure just inside the pipe tip. It just a flow phenomenon.

Likewise if the pressure caluclated is equal to atmos/header pressure then the flow is just at sonic and any pressure calculated below atmos/header pressure is subsonic because this says that for the mass of gas flowing to actually be able to flow at the conditions indicated the pressure would need to be less so the fluid would be less dense because at atmos/header the fluid density is such that sonic velocity to pass it is not necessary. However the fluid cannot be at a pressure less than atmos/header pressure at exit so it must be at atmos/header pressure.

 

[Luuk21]

The pipe discharge pressure may be above atmospheric or header pressure if the flow at the discharge is sonic flow. This is why the OP program output sheet checks if the flow is sonic at the pipe exit. Depending on the flowrate the actual presure just inside the pipe tip can be much higher, and any value so that sonic flow will still exist in the relief valve orifice if exit pressure plus back pressure due to friction is still less than Pc the critical pressure at the relief valve flow orifice as defined by API 520. If pressure at exit plus back pressure is greater than Pc then subsonic flow will exist in relief valve. For instance if the OP spreadsheet indicated that Pcrit at pipe exit was 2.0 bara (then this would be the actual pressure at the exit not 1.013 bara) plus 0.063 friction loss then pressure at valve dischare would be 2.063 bara which is still less than Pc per API 520.

Exactly! And I'm quite sure the program does this the incorrect way. Let me elaborate, as seen in the attachment (picture) P2c (the pressure at the inlet of the relief valve in the excel sheet) is calculated by adding up the relief pressure + the pressure drop over the outlet.

But this is, as you stated, wrong. There is no pressure drop over the outlet. Well, yes, there is ofcourse, but that is less than Pcrit so there is no accumulation of pressure. The correct relieving pressure here, is exactly the set pressure.

It then calculates the vessel pressure by adding the inlet pressure drop to P2c, which is correct, but the end result is wrong. Since there is chocked flow - the outlet pressure can not be added up to determine the pressure at the inlet of the Safety Valve.

Next mistake is that the excel sheet assumes subsonic flow in the calculations of the PSV area - I can show you the next sheet if you want. This is also wrong. There is sonic flow. So the formula for critical relief should be used.

I hope my previous confusion becomes a bit more clear now - I think this is a major mistake in the tool.

 

[Snickster]

Exactly! And I'm quite sure the program does this the incorrect way. Let me elaborate, as seen in the attachment (picture) P2c (the pressure at the inlet of the relief valve in the excel sheet) is calculated by adding up the relief pressure + the pressure drop over the outlet.

The pressure at the inlet of the relief valve is 4.706 bara which is found by taking the relief pressure 4.72 bara minus the pressure drop across the inlet pipe of 0.018 bar. Why do you say relief pressure + pressure drop over the outlet? It don't work like that where you take the relief valve downstream pipe pressure and add the pressure drop across the valve to get the pressure at the relief valve inlet. This is because the presure drop from relief valve inlet to outlet is not due to friction but to:

1) A almost isentropic expansion in the nozzle of the relief valve from P upsteam to Pcf which is the critical flow pressure at nozzle orifice of the relief valve as defined by API 520. Ok say you have air at k = 1.4 then Pcf per API 520 is determined by equation (1) 0f 5.6.2.4 to be (.53)* T inlet = (.53)* (4.706) = 2.353 bara. This is the sonic flow pressure in the orifice of the relief valve. Sonic flow will exist in the orifice of the relief valve as long at the pressure in the downstream piping at the relief valve exit is below Pcf = 2.353 bara. If sum of back pressure at the relief valve exit flange is greater than this then flow in relief valve will be subsonic - then you need to use the subsonci flow sizing equations 5.6.4. As long as the flow is sonic in the relief valve then you use the sonic flow equations 5.6.3. During the isentropic expansion in the relief valve to sonic velocity the pressure at the orifice of the relief valve is Pcf, temperature Tcf is corresponding to sonic flow = 2/(k+1)* T inlet, and velocity is sonic = SQRT(gkRT)=SQRT(g*k*(1545/MW)*(2/(k+1))*(T inlet)). So at the orifice nozzle is sonic flow conditions for all parameters. Note that you can actually derive the API sizing equations for critical flow by plugging the above relations into the ideal gas equation and then simplifying believe it or not.

2) Irreversible constant stagnation enthalpy expansion from the orifice of the nozzle to the downstream piping pressure (not due to piping friction per se but due to an irreversible expansion wher there is alot of irreversible energy consumption of the fluid when it suddenly drops to the lower downstream pressure). The piping pressure at the outlet of the relief valve is produced by the sum of the of the backpressure in the downstream piping including pressure at exit of pipe plus friction loss in piping back to the relief valve. So the downstream pressure at the relief valve exit has nothing to do with the pressure drop in the relief valve - it is solely due to the backpressure developed in the piping. If the back pressure is equal to or less than Pcf then the flow will be sonic with same Pcf,Tcf,Vcf in the orifice as indicated above. If the pressure is greater than Pcf then the flow in the orifice will be subsonic. So the only time the pressure is continuous across the relief valve orifice is when the pressure at the outlet is equal to or greater than Pcf. If the pressure downstream of the orifice is less than Pcf then flow through the orifice is sonic and there is a discontinuity of pressure from Pcf to downstream piping pressure due to an irreversible increase in entropy expansion. For instance in your case the Pcf pressure at the orifice is 2.353 bara and the downstream pressure is 1.08 bara. So via an irreversible expansion of constant stagnation enthaply the pressure drops to the downstream piping pressure of 1.08 bara. And as long as the pressure downstream is less than Pcf = 2.353 bara then the actual flow in the orifice is sonic at Pcf = 2.353 bara.

The important point to remember here is that the pressure at the relief valve exit is determined by the pressure at the pipe exit, be it atmospheric, terminal header pressure, or choked critical flow pressure (above atmos/header pressure) if sonic flow exists at the pipe exit, plus the frictional back pressure in the piping back to the relief valve. On the inlet side the pressure at the relief valve inlet is the relieving pressure in the vessel minus the pressure drop in the inlet pipe. The pressure on the outlet side is not felt by the inlet side until Pcf is equaled or exceeded. Then pressure continuity across the orifice exists and the subsonic flow equations apply when Pcf is exceeded.


But this is, as you stated, wrong. There is no pressure drop over the outlet. Well, yes, there is ofcourse, but that is less than Pcrit so there is no accumulation of pressure. The correct relieving pressure here, is exactly the set pressure.

I am not totally sure what you mean here. Just rmember that in the isentropic expansion to Pcf in the relief valve orifice it is considered basically a reversible expansion to a the lower pressure Pcf and sonic velocity. Since reversible there really is no loss per se because no energy is lost (entropy remains constant) and you can reverse the process back to the original stagnation conditions in the vessel. Note that the API 520 coefficient of discharge for relief valve sizing accounts for the small irreversibilities in the expansion in the relief valve flow nozzle. The actual relieving pressure is the vessel pressure minus the inlet loss to the relief valve inlet. I believe that if you stay below 3% friciton loss in inlet per API 520 you can just consider the pressure at inlet of the relief valve to be the pressure in the vessel.

It then calculates the vessel pressure by adding the inlet pressure drop to P2c, which is correct, but the end result is wrong. Since there is chocked flow - the outlet pressure can not be added up to determine the pressure at the inlet of the Safety Valve.

True. As stated above when pressure at relief valve outlet is less than Pcf there is a pressure discontinuity across the relief valve. Pcf exists in the nozzle orifice and drops to downstream piping pressure due to an irreversible expansion. This irreversible expansion is not much different than if you have two vessels of equal size with one vessel pressurized and other at atmospheric with piping and isolation valve inbetween and you suddenly open the valve to equalize the pressure with no work being done but pressure drops irreversibly.

Next mistake is that the excel sheet assumes subsonic flow in the calculations of the PSV area - I can show you the next sheet if you want. This is also wrong. There is sonic flow. So the formula for critical relief should be used.

No the cell where it is indicating "subsonic flow" noted as "Pcrit" is in the piping at the inlet of the relief valve P2c (.319) and at the exit of the piping P2e (.137). This is in the piping not in the relief valve orifice. This calculation does not look at flow in relief valve orifice. You have to calculate the flow rate using API 520 critical flow sizing equations and plug into this spreadsheet which is only looking at what is going on in the piping. What Pcrit indicates is that on the inlet in order for sonic velocity to exist in the piping then the pressure must drop to 0.319 bara, and on the outlet pipe at the exit the pressure must drop to 0.137 bara for sonic flow to exist. Since in both cases the actual pressure is higher at those points then the flow is subsonic - in the piping.

So if you want you can calculate the Pcrit shown in the piping yourself or 0.319 on inlet and 0.137 on outlet using the ideal equations I indicated above. I am used to english units but you can convert to metric if you want and solve. Or if you give me all of the unknown parameters that go into the equation (such as inlet temperature, MW, mass flow rate, k, etc.) then I will solve and I bet I get the same values indicated for Pcrit in your spreadsheet.

Ideal Gas Equation for Flowing Fluid: in Ft.Lbs/sec, Ft/sec, Ft2, etc units

P*(144)*(V)*(A) = m*R*T

P is pressure in PSI - 144 converts to PSF.
V is velocity at pipe exit FT/sec - assume that it is sonic for purposes of checking what the pressure must then be for sonic to exists = SQRT(gkRT).
A is the flow area of the pipe exit in FT2
m is the mass flow rate in pounds per second (pounds force/mass not slugs)
R is the universal gas constant = 1545/MW Where MW is the molecular weight of the gas, units of gas constant = 1545 FT-pounds/pound mole
T is the temperature at the exit in degrees Rankine = (2/(k+1))To - again assume sonic conditions exist at the exit so actual T is at sonic velocity.
To is stagnation temperature upstream of the relief valve - relieving temperature - degrees Rankine.
g is gravity constant = 32.2 ft/sec2
k = ratio of specific heats

Therefore the ideal gas equation takes the following form:

P*(144)*SQRT(gk(1545/MW)*(2/(k+1))To)*(A) = m*(1545/MW)*(2/(k+1))To

 

[pierreick]

Hi,
Probably good to read this document with examples.
Pierre

 

[Luuk21]

Hi,
Probably good to read this document with examples.
Pierre

Thanks!

The pressure at the inlet of the relief valve is 4.706 bara which is found by taking the relief pressure 4.72 bara minus the pressure drop across the inlet pipe of 0.018 bar. Why do you say relief pressure + pressure drop over the outlet? It don't work like that where you take the relief valve downstream pipe pressure and add the pressure drop across the valve to get the pressure at the relief valve inlet.

No - the relief pressure at the inlet of the relief valve is not 4.706 bara, but 4.643 bara. See left upper corner - this is the Set Pressure (3 barg) + accumulation (21% = 0.63 bar) + P_atm (1.013 bar) + superimposed backpressure (0 bar) = 3 + 0.63 + 1.013 + 0 = 4.643 bar(a). This is exactly 21% above the MAWP. 4.72 would violate this.

4.723 is btw not the relief pressure but the pressure inside the vessel during relief. The tool makes the mistake by adding up the calculated pressure drop over the outlet, which is 0.063 bar, so 4.643 (actual relief pressure) + 0.018 (pdrop inlet) + 0.063 (pdrop outlet) = 4.723 leading to the conclusion that the relief pressure must be 4.706 (4.723 - 0.018), which is wrong.

I am not totally sure what you mean here. Just rmember that in the isentropic expansion to Pcf in the relief valve orifice it is considered basically a reversible expansion to a the lower pressure Pcf and sonic velocity. Since reversible there really is no loss per se because no energy is lost (entropy remains constant) and you can reverse the process back to the original stagnation conditions in the vessel.

See above, but indeed, it needs to be treated as an isentropic expansion. Calculate Pcf, check if Poutlet>Pcf, if not, critical flow.

The pressure on the outlet side is not felt by the inlet side until Pcf is equaled or exceeded.

Exactly! This is not what the tool does.

The actual relieving pressure is the vessel pressure minus the inlet loss to the relief valve inlet. I believe that if you stay below 3% friciton loss in inlet per API 520 you can just consider the pressure at inlet of the relief valve to be the pressure in the vessel.

Yes and no, as I understand it the actual relieving pressure is the set pressure + acc of the safety valve. The pressure in the vessel is just the inlet pdrop + relief pressure. At critical flow over the safety valve that is.

If flow through the relief valve is subsonic, pressure drop of outlet piping needs to be added. Then the relief pressure (pressure at inlet safety valve) becomes higher (and thus the pressure in the vessel).

No the cell where it is indicating "subsonic flow" noted as "Pcrit" is in the piping at the inlet of the relief valve P2c (.319) and at the exit of the piping P2e (.137). This is in the piping not in the relief valve orifice. This calculation does not look at flow in relief valve orifice.

Exactly! It doesnt look at the flow in the orifice itself. But on the next sheet, sizing equations are used for subsonic flow, based on the outcome of these pressure drop calculations (it checks the cell that indiciates if . This is just completely wrong - the conditions at the piping inlet/outlet do not determine if there is critical or non-critical flow over the relief device.

The Pdrop and check if this is below Pcf is never checked in the tool. It bases the design equations on the wrong input. These design equations are on the next sheet.

Or if you give me all of the unknown parameters that go into the equation (such as inlet temperature, MW, mass flow rate, k, etc.) then I will solve and I bet I get the same values indicated for Pcrit in your spreadsheet.

I'm quite sure that all equations used in the spreadsheet are correct. But the starting points are wrong.

I have included a file in which I have exaggerated the point I'm trying to make. I hope you have powerpoint, otherwise I'll convert it into PDF.

Thanks for your detailed explenation, I have a much better grip on the material now.

 

[Snickster]

I am not totally sure what the progrm is doing based on your further comments. I need to look at it and see if I can figure it out.

 

[Snickster]

Quote (Snickster)
The pressure at the inlet of the relief valve is 4.706 bara which is found by taking the relief pressure 4.72 bara minus the pressure drop across the inlet pipe of 0.018 bar. Why do you say relief pressure + pressure drop over the outlet? It don't work like that where you take the relief valve downstream pipe pressure and add the pressure drop across the valve to get the pressure at the relief valve inlet.

No - the relief pressure at the inlet of the relief valve is not 4.706 bara, but 4.643 bara. See left upper corner - this is the Set Pressure (3 barg) + accumulation (21% = 0.63 bar) + P_atm (1.013 bar) + superimposed backpressure (0 bar) = 3 + 0.63 + 1.013 + 0 = 4.643 bar(a). This is exactly 21% above the MAWP. 4.72 would violate this.

I see that 4.643 bara is equal to the Set Pressure (3 barg) + accumulation (21% = 0.63 bar) + P_atm (1.013 bar) + superimposed backpressure (0 bar) as you indicate. But what is this pressure? It is the pressure at the inlet of the relief valve if it was sized based on P1 being this pressure - this is because if the required relief flow is really what you designed for at P1 then you will get that flow through the valve with a resulting pressure P1 upstream and given orifice size. But I don't think you would include the superimposed backpressure term to get this pressure as downstream pressure is not transmitted through the valve unless above Pcf as discussed. However superimposed backpressure will make the conventional relief valve pop at an equal ammount to the superimposed back pressure. So the valve will pop higher due to the superimposed back pressure but slowly the pressure will drop to P1 = 4.643 bara when equillibrium of flow and P1 occurs, so maybe the program is looking at when the valve pops for a conventional relief valve.

4.723 is btw not the relief pressure but the pressure inside the vessel during relief. The tool makes the mistake by adding up the calculated pressure drop over the outlet, which is 0.063 bar, so 4.643 (actual relief pressure) + 0.018 (pdrop inlet) + 0.063 (pdrop outlet) = 4.723 leading to the conclusion that the relief pressure must be 4.706 (4.723 - 0.018), which is wrong.

Yes I see now that the program takes the 4.643 bara value and adds the pressure drop in the outlet which does not make sense to me. I also see that the program calculates the pressure drop using what appears to be the isothermal gas equation. This can be seen in the next section of the calculations noted as "Components of zero". Attached is a copy of the isothermal equation which is basically seen the the next section of the spreadsheet, so the spreadsheet appears to calculate the pressure drop by starting at the pipe outlet at atmospheric (known value) then adding the pressure drop to get 3.46 bara at relief valve discharge. In other words by knowing P2 the isothermal equation is solved for P1.

Quote (Snickster)
The actual relieving pressure is the vessel pressure minus the inlet loss to the relief valve inlet. I believe that if you stay below 3% friciton loss in inlet per API 520 you can just consider the pressure at inlet of the relief valve to be the pressure in the vessel.

Yes and no, as I understand it the actual relieving pressure is the set pressure + acc of the safety valve. The pressure in the vessel is just the inlet pdrop + relief pressure. At critical flow over the safety valve that is.

The relief valve should pop at set presure. Since you sized the valve for set pressure plus accumulation at a given flowrate then this will be the pressure required at the valve inlet to get that flow (assuming the orifice is not oversized but you can get exact area orifice required which you normally can't). Add the pressure drop in the inlet to get the pressure in the vessel.

If flow through the relief valve is subsonic, pressure drop of outlet piping needs to be added. Then the relief pressure (pressure at inlet safety valve) becomes higher (and thus the pressure in the vessel).

Correct because even the subsonic relief valve sizing equations of API 520 includes the actual P2 pressure to deterine the flow.

Quote (Snickster)
No the cell where it is indicating "subsonic flow" noted as "Pcrit" is in the piping at the inlet of the relief valve P2c (.319) and at the exit of the piping P2e (.137). This is in the piping not in the relief valve orifice. This calculation does not look at flow in relief valve orifice.

Exactly! It doesnt look at the flow in the orifice itself. But on the next sheet, sizing equations are used for subsonic flow, based on the outcome of these pressure drop calculations (it checks the cell that indiciates if . This is just completely wrong - the conditions at the piping inlet/outlet do not determine if there is critical or non-critical flow over the relief device.

Correct but the conditions at the piping inlet and outlet do determine if there is critical flow. If outlet pressure is less than or equal to Pcf based on the given inlet pressure then there is sonic flow in the relief valve.

The Pdrop and check if this is below Pcf is never checked in the tool. It bases the design equations on the wrong input. These design equations are on the next sheet.

Yes the program should divide the outlet prsssure calculated by the inlet pressure and determine if the critical pressure is such that flow is sonic - then if so then size relief valve flow based on critical flow. It would be an iteration since you would first assume critical flow - then calculate pressure drop - then resize relief valve and determine flow based on the backpressure considering any "K" correction factors.

Quote (Snickster)
Or if you give me all of the unknown parameters that go into the equation (such as inlet temperature, MW, mass flow rate, k, etc.) then I will solve and I bet I get the same values indicated for Pcrit in your spreadsheet.

I'm quite sure that all equations used in the spreadsheet are correct. But the starting points are wrong.

It appears the sheet of the program you posted only 1) Checks for critical flow at pipe exit and if not then considers pressure at exit atmos/header pressure, and if so considers pressure at exit Pcrit. 2) Calculates pressure drop using the isothermal flow equations to determine the frictional losses in the piping. This step appears correct. What don't appear correct is that if the flow is sonic then the downstream pressure drop should not be added to the 4.643 bara value to get the upstream pressure, as the bakpressure should not be felt by the upstream. The upstream pressure should be determined by the pressure at the inlet of the relief valve plus the pressure drop in the inlet line. The pressure at the inlet should be the P1 that was plugged into the relief valve sizing equation at the given flowrate.

I have included a file in which I have exaggerated the point I'm trying to make. I hope you have powerpoint, otherwise I'll convert it into PDF.

 

[Sebastien Pelletier P. Eng.]

Be careful with your design approach, I already see few misconceptions within the provided answer from previous contributors regarding of:
(1) 3% pressure drop rule at PSV inlet (PSV set pressure adjustment is not a proper cure by any mean for this topic);
(2) Some misconception about system's MAWP, design pressure and normal operating pressure have also been observed within previous post.

Once again, a face to face discussion will be beneficial.

Regards,
Sebastien Pelletier, P. Eng.