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!

 

[The Obturator]

To perform a Pressure-relief Valve sizing calculation to API-520, it is necessary to know if there is an inlet pressure drop greater than 3% of set pressure. API-520 does not calculate pressure drop.

API-520 States that if the pressure drop is no greater than 3% then the effect of the pressure drop can be ignored in sizing.

Set Pressure is normally the MAWP (MAWP is not the design pressure).

If you have a pressure drop greater than 3%, the approach is to subtract the pressure drop value from the set pressure then add the overpressure - some manufacturers sizing calculations allow you to input this.

Depending on how high the pressure drop is relative to the set pressure, your valve selection may need attention. Typically applications with inlet pressure drops greater than 3% are fixed by the use of a pilot operated Pressure-relief Valve with remote sensing. That is having the pilot sense the vessel pressure directly to the protected vessel, rather than the inlet of the valve. In all such cases consult the manufacturer for further confirmation/advice.

Inlet pressure drop is discussed further in API-520 Part II.


*** Per ISO-4126, the generic term
'Safety Valve' is used regardless of application or design ***

*** 'Pressure-relief Valve' is the equivalent ASME/API term ***

 

[mk3223]

IMO, for Q-1, P1 is to be the design process pressure with a "proper overdesign margin" as defined by the Engineer. It's okay to use a pressure vessel as a storage vessel as the oration pressure is not higher than the vessel design/MAWP pressure.

For the Q-2, yes, the relief analysis may need to iterate the calculation or using CFD to have a better accuracy.

 

[Latexman]

You have a misconception.

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).

Negative. The maximum set pressure = MAWP. Set pressure can be < MAWP.

Good Luck,
Latexman

 

[Luuk21]

API-520 States that if the pressure drop is no greater than 3% then the effect of the pressure drop can be ignored in sizing.

Hi, thanks for all the replies. I understand now that MAWP <> P design

This leads to multiple questions:

Q1: Which chapter/page of API 520 states that piping pressure drop can be ignored?
Q2: If I understand correctly, pressure drop can be ignored as long as it is under the 3% set pressure treshold for inlet piping. Set pressure, not MAWP or P design?
Q3: Is this threshold 10% for outlet piping?
Q4: Can the allowable overpressure over the MAWP also be ignored, compared to P design?

So, for example, I have a vessel with P design 3 barg and a MAWP of 3 barg. What is my maximum allowable set pressure and overpressure?
Scenario 1:
Set pressure 3 barg - overpressure .63 barg (21% MAWP, fire case) - piping pressure drop .39 barg (13% set pressure)
Total pressure in vessel during relief: 4.02 barg (34% over design pressure).
In this case the overpressure & piping pressure drop is ignored

Scenario 2:
Set pressure 2.71 barg - overpressure .57 barg (21% MAWP, fire case) - piping pressure drop .35 barg (13% set pressure)
Total pressure in vessel during relief: 3.63 barg (21% over design pressure).
In this case the piping pressure drop is accounted for
(Note that MAWP changes to set pressure, if it would stay at 3 barg, MAWP is above set point PSV)

Scenario 3:
Set pressure 2.24 barg - overpressure .47 barg (21% MAWP, fire case) - piping pressure drop .23 barg (13 % set pressure)
Total pressure in vessel during relief: 3.00 barg (equal to design pressure).
In this case all over pressure is accounted for.

Which scenario is, according to API 520, applicable?

Thanks again for your input!

 

[The Obturator]

To answer your last questions above;-
A1. Refer to API-520 Part I, 10th Edition (2020) Paragraph 5.4.1.1.
A2. See above referenced paragraph (...3% of set pressure).
A3. ? Please clarify question. I think you are confusing with back pressure limits. Please refer to section 5.3 in above API-520 edition and familiarise yourself with the various scenarios.
A4. Overpressure cannot be ignored - it is required for the PRV to open properly (capacity certification is also done at a prescribed overpressure value).

Any inlet pressure drop greater than 3% should be subtracted from the set pressure before applying overpressure. This is in order to calculate that the valve can pass the minimum required capacity with the pressure drop. This is also all before the impact of any back pressure.

Note also that API-520 is only for preliminary sizing calculations using assumed coefficient of discharge and nozzle areas. Once a valve design/style/manufacturer is determined, calculations should be redone using actual coefficient and nozzle area.


*** Per ISO-4126, the generic term
'Safety Valve' is used regardless of application or design ***

*** 'Pressure-relief Valve' is the equivalent ASME/API term ***

 

[The Obturator]

With 3.0 barg set pressure, 21% Overpressure (fire) and a predetermined inlet pressure drop of 13% (0.39 barg).
P1 - 'Upstream Relieving Pressure' as defined by API -520 and used in compressible fluid sizing calculations is;-
(3.0 - 0.39) x 1.21 + 1.01 = 4.17 bara (note absolute pressures are used for P1)

Per ASME VIII (2023) UG-135 (2) "...the pressure relief device(s) shall be capable of preventing the pressure from rising more than more than 21% above the maximum allowable working pressure (MAWP)"




*** Per ISO-4126, the generic term
'Safety Valve' is used regardless of application or design ***

*** 'Pressure-relief Valve' is the equivalent ASME/API term ***

 

[Luuk21]

Hi,

Thanks for your reply. Things are getting more clear, but I want to confirm that I have things straight:

Definitions:
Pdesign barg - design pressure (not mentioned or used in API-520)
MAWP barg - maximum allowed working pressure
P1 bara - reliefpressure used for calculation purposes
Pset barg - set pressure
Pvessel barg - pressure in vessel during relief
Pinlet bar - inlet pressure drop
Pbackp bar - superimposed backpressure + outlet pressure drop (mentioned as build-up backpressure in API 520)
Prelief barg - pressure at inlet PSV
Pmax barg - maximum allowed pressure at inlet PSV - MAWP +21%

Scenario: Vessel with design pressure of 3 barg, MAWP of 3 barg needs a SV for the fire case. A set pressure of 3 barg is assumed. First, a pressure drop calculation is performed:
Pdesign | 3 barg / 4.01 bara (3.63 barg / 4.64 bara incl allowed overpressure fire case)
MAWP(1) | 3 barg (initial)
Pset(1) | 3 barg (initial)
P1(1) | 3.00 + 0.63(21%) + 1.01 = 4.64 bara (initial)
Pinlet(1) | 0.089 bar (calculated at Pset (1), 3barg. This is below 3% of the set pressure. Ignore (for calculation purposes or totally?)
Pbackp(1) | 0.63 bar (21% of Pset(1), conventional PSV is allowed but pressure cannot be ignored).
Pvessel(1) | 4.26 barg (3 + 0.63 + 0.63)
Prelief (1) | 3.63 barg
Pmax(1) | 3.63 barg

P1(1) is too high since pressure in vessel will exceed Pdesign/MAWP +21% (4.64 + 0.63 = 5.27 bara) & - Pback needs to be deducted
--------------------
Pdesign (2) = 3 barg
MAWP(2) | 3 - 0.63 = 2.37 barg (This is leading to conflict, see below)
Pset(2) | 3 - 0.63 = 2.37 barg
P1(2) | 2.37 + 0.5(21%) + 1.01 = 3.88 bara
Pinlet(2) | 0.1 bar (calculated at Pset (2), 2.37 barg. This is ABOVE 3% of set pressure. Do not ignore. Pdrop increases!)
Pbackp(2) | 0.63 bar (27% of Pset(2), use balanced PSV)
Pvessel(2) | 3.60 barg (2.37 + 0.5 + 0.63 + 0.1)
Prelief (2) | 2.97 barg
Pmax(2) | 2.87 barg

P1(2) is too high since pressure in vessel will exceed MAWP +21%) (3.88 + 0.63 + 0.1 = 4.61 bara)

Note that Pinlet(2) increases due to lower pressure at inlet reliefvalve (which leads to more volume in compressed gas).
Note that Pbackp(2) stays the same due to the same pressure at the end of the outlet (atmospheric).

This leads to an endless loop - since P1(2) will ALWAYS exceed the MAWP + 21% limit, pressure drop cannot be ignored. The solution is to NOT change the MAWP. However, the set pressure HAS to be altered - if that stays at 3 barg, the actual vessel pressure will be Pvessel(1)

Redo calculation with MAWP(2) | 3 barg and account for inlet pressure
Pdesign (3) = 3 barg
MAWP(3) | 3 barg (I think API is confusing at this point)
Pset(3) | 3 - 0.63 - 0.1 = 2.27 barg
P1(3) | 2.27 + 0.48(21%) + 1.01 = 3.76 bara
Pinlet(3) | 0.105 bar (calculated at Pset (2), 2.27 barg. This is ABOVE 3% of set pressure. Do not ignore.)
Pbackp(3) | 0.63 bar (28% of Pset(3), use balanced PSV)
Pvessel(3) | 3.485 barg (2.27 + 0.48 + 0.63 + 0.105)
Prelief (3) | 2.855 barg (2.27 + 0.48 + 0.105)
Pmax(3) | 3.63 barg (confusing, Pset is lower than MAWP).

Now the calculation is correct, Pdesign is not violated, and Prelief(3) is well below Pmax(3)

Is this the correct way to go? Basically I think MAWP is confusing, since you will always violate it if Pset is not (significantly) lower. In my case there is always a piping loss.

How to deal with this Pset<MAWP<Pset<MAWP loop?

 

[Luuk21]

With 3.0 barg set pressure, 21% Overpressure (fire) and a predetermined inlet pressure drop of 13% (0.39 barg).
P1 - 'Upstream Relieving Pressure' as defined by API -520 and used in compressible fluid sizing calculations is;-
(3.0 - 0.39) x 1.21 + 1.01 = 4.17 bara (note absolute pressures are used for P1)

Per ASME VIII (2023) UG-135 (2) "...the pressure relief device(s) shall be capable of preventing the pressure from rising more than more than 21% above the maximum allowable working pressure (MAWP)"

This confuses me. The MAWP is 3 bar, as is Pdesign as is Pset. Pdesign and the MAWP + 21% is clearly violated in the example above. We need to lower the set pressure. However, this leads to set pressure below MAWP. This looks strange to me.

If I lower the MAWP as well, I will still violate the MAWP + 21% rule of ASME. Can MAWP be higher than Pset?

 

[Latexman]

P1(1) is too high since pressure in vessel will exceed Pdesign/MAWP +21% (4.64 + 0.63 = 5.27 bara)

Negative. You added dP[sub]outlet[/sub] twice.

Here's how I do it:

Scenario: Vessel with design pressure of 3 barg, MAWP of 3 barg needs a SV for the fire case. A set pressure of 3 barg is assumed.
Pdesign | 3 barg / 4.01 bara (3.63 barg / 4.64 bara incl allowed overpressure fire case)
MAWP(1) | 3 barg
Pset(1) | 3 barg
First, size the safety valve.
Then, pick the smallest commercially available orifice that is > A[sub]calculated[/sub].
Then, calculate the rated flow of the selected PSV. Use the PSV's full orifice area (not A[sub]calculated[/sub]) and the rated flow coefficient.
P1(1) | 3.00 + 0.63(21%) + 1.01 = 4.64 bara. Yes.
Pinlet(1) | Design the inlet pipe and fittings such that dP[sub]inlet[/sub] < 0.09 bar
Pbackp(1) | Design the outlet pipe and fittings such that dP[sub]outlet[/sub] < 0.63 bar
You are done.

Btw, your company should be training you to do this, or you should be learning from a Senior Engineer. Not SGOTI. My $0.02.

Good Luck,
Latexman

 

[Luuk21]

Negative. You added dPoutlet twice.

No - I've taken the 21% of the MAWP (according to API 520-1 figure 15 (2014)) - the pressure drop is by pure chance also equal to 21%. The only reason I took 21% is that this allowes to use a conventional PSV, above 21% a balanced SV is adviced (with fire case). Piping is existing so there is no new engineering possible.

I understand that I need to do a Rated calculation as soon as I would design a new SV, but the SV is already installed, I only need to check if it's area is sufficient.

Let me clearify my example by taking a different piping pressure drop.

Pdesign | 3 barg / 4.01 bara (3.63 barg / 4.64 bara incl allowed overpressure fire case)
MAWP(1) | 3 barg
Pset(1) | 3 barg
P1(1) | 3.00 + 0.63(21%) + 1.01 = 4.64 bara (according to API-520)
Pinlet(1) | 0.1 bar (calculated at Pset (1), 3barg. This is above 3% of the set pressure.)
Pbackp(1) | 0.45 bar (15% of Pset(1), normal SV allowed).
Pvessel(1) | 4.08 barg (3 + 0.63 + 0.45 + 0.1)
Prelief (1) | 3.73 barg (3 + 0.63 + 0.10) (pressure at inlet SV)
Pmax(1) | 3.63 barg (MAWP + 21%, AMSE Code)

According to API 520, P1 is MAWP + 21%, set pressure is irrelevant. If the set pressure is lower than MAWP, P1 is still the same. See attachement or Table 2 under 5.4.2.1.3 API-520-1.

If I then perform the calculation as you indicated in the post above, my Area required is based on a relief pressure of 4.64 bara - so for all intents and purposes, the SV opens at a pressure of 4.64 bara (it needs the 21% overpressure to open properly). Then material starts flowing and the pressure in the vessel increases due to the pressure drop in the piping.

The pressure drop accumulates to 0.55 barg, so the pressure in the vessel rises to 4.18 barg. This is 15% over the max allowed pressure of 3.63 barg.

How to solve this? Lowering the set pressure and "accept" that the MAWP is higher than the set pressure? I don't see any resolution on this issue in API-520/521.

Regarding training: I've asked the Senior Engineer - he doesn't know (by head) and refers to API-520.

 

[Latexman]

Well, if dP[sub]inlet[/sub] < 3% SP and dP[sub]outlet[/sub] < 21% SP (fire), then you can ignore the pressure drops, right? If so, you added dP[sub]outlet[/sub] (or 21% overpressure) twice. That's how I see it.

Now, I didn't realize this was an existing PSV. I thought is was a new design. May I suggest we stop using fictitious examples and see the real numbers, please. As you can see, I am not The Amazing Kreskin. If your mother tongue is not English, and/or my humor does not translate, I apologize. A lot of times humor doesn't translate so well.

And, yes, you are right, the relieving/sizing pressure does not change if set pressure changes. The allowable accumulation (based on vessel) remains the same, and the % over pressure (based on PSV) changes.

Good Luck,
Latexman

 

[LittleInch]

I've got confused here and I think you're over complicating this.

API 520 is there to size relief valves.

The requirements are set by the code of the thig you're protecting.

For ASME vessels this is set pressure cannot exceed MAWP.
The max pressure in the vessel in a fire case with one valve is MAWP + 21%

Your system needs to work within those two limits.

If your back pressure is > 21% of the MAWP in a fire case flow rate then you have a problem.
You will either need to increase the capacity of the relief system or reduce the set pressure which could impact on the process in the vessel you're protecting.

table 4 seems to lay put some examples.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Luuk21]

If your back pressure is > 21% of the MAWP in a fire case flow rate then you have a problem.
You will either need to increase the capacity of the relief system or reduce the set pressure which could impact on the process in the vessel you're protecting

I don't think this is correct. Back pressure can be 50% - see API-520-1 (2014) chapter 5.3.3.2. The amount of backpressure is only relevant for the type of PSV you install.

With a back pressure above 21% of the MAWP in a fire case it is adviced to install a balanced SV.

For ASME vessels this is set pressure cannot exceed MAWP.
The max pressure in the vessel in a fire case with one valve is MAWP + 21%

Exactly! And this is so confusing, since P1 (Relief pressure in API-520) is used to calculate required Area, and P1 is MAWP + 21% in a fire case. When I follow the API-520 I achieve a pressure in my vessel that is way above the design pressure - certainly above MAWP +21%. It looks like the norm contradicts itself in this matter.

You will either need to increase the capacity of the relief system or reduce the set pressure which could impact on the process in the vessel you're protecting.

Lowering the set pressure does not solve anything, see attachment of my previous post - API-520 doesn't care about set pressure, only about MAWP.

table 4 seems to lay put some examples.

Table 4 makes it more confusing - as you can see the set pressure has no impact on the calculation, P1 is the same for whatever set pressure you use, so the relief pressure stays the same.

 

[Latexman]

Exactly! And this is so confusing, since P1 (Relief pressure in API-520) is used to calculate required Area, and P1 is MAWP + 21% in a fire case. When I follow the API-520 I achieve a pressure in my vessel that is way above the design pressure - certainly above MAWP +21%. It looks like the norm contradicts itself in this matter.

Not if dP[sub]inlet[/sub] < 3% SP and dP[sub]outlet[/sub] < 21% SP (fire), and you can ignore the pressure drops, right? You may need to change the PSV to balanced bellows or pilot operated, or change the pipe and fittings on the inlet and/or outlet.

Options to Correct High Inlet Pressure Drop
The following actions will either reduce inlet pressure drop or allow the system to adequately
function with the calculated pressure drop:
• Relocate the safety valve to a larger vessel nozzle, and increase inlet pipe diameter.
• Shorten the pipe length.
• Remove pipe fittings (e.g., use a 90° elbow rather than a tee with branch flow).
• Remove a rupture disk in series with the safety valve.
• Install a non-modulating pilot-operated safety valve with a remote pressure tap (safety valve
must still be able to handle the required capacity with a lower static pressure at the safety
valve inlet nozzle).
• Mitigating the worst credible scenario may result in a smaller safety valve and a lower inlet
pressure drop.

Options to Correct High Discharge Piping Pressure Drop
The following actions will either reduce outlet pressure drop or allow the system to
adequately function with the calculated pressure drop:
• Mitigating the worst credible scenario may result in a smaller safety valve and a
lower inlet pressure drop.
• Increase pipe diameter. Expand pipe diameter (expander) at the safety valve outlet
nozzle.
• Shorten the pipe length.
• Remove pipe fittings.
• Remove a conventional safety valve and replace with a pilot-operated or bellows
type.
• Install a pilot-operated safety valve in place of a bellows safety valve.

Also, can you change to a smaller nozzle size on the existing PSV, or install a restricted lift limiting device on the existing PSV? These reduce the rated capacity and reduces the dP's.

Good Luck,
Latexman

 

[Luuk21]

Well, if dPinlet < 3% SP and dPoutlet < 21% SP (fire), then you can ignore the pressure drops, right? If so, you added dPoutlet (or 21% overpressure) twice. That's how I see it.

This is the big question. I think this is incorrect. API only states that if dPinlet < 3% (API-520-1 5.4.1.1) it can be ignored. It does not mention anything about ignoring outlet pressure - or I have not found it yet.

In my example I use 2 times 21% because:
P1 is MAWP + 21% (see API-520)
pressure drop is by pure chance 21% of Pset, Pset = MAWP. You can change this to 1%, 10% or whatever positive number.

So if I would ignore the outlet pressure, the pressure in my vessel becomes MAWP + 42%, which is way above design pressure.

If your mother tongue is not English, and/or my humor does not translate, I apologize. A lot of times humor doesn't translate so well.

Don't worry, just happy that you respond! I'm trying to learn here. API-520 is a well established code made by the top experts in the field - let's be real, the chance that I'm making a mistake somewhere is way more obvious. I'm trying to find the mistake

There are about 150 SV's to check, the issue is that the tool we use, doesn't consider backpressure as something to account for. P1 = MAWP + 21%, and just ignore backpressure up to 21%. This doesn't make sense for 2 reasons:
1. Pressure in vessel comes (way) above MAWP + 21%
2. API states no limit to backpressure - they even give a correction factor Kb in figure 30 for backpressures over 30% so it is defenitely accounted for.

If you want real numbers, I have a case here that is basically similar:

Pdesign | 3 barg
MAWP | 3 barg
Pset | 3 barg
Pinlet | 0.241 bar (DN65)
Poutlet | 0.470 bar (DN100)
d orifice | 57mm (2552mm2)
Relief 5418 kg/h

What should I choose for P1, fire case?

 

[Latexman]

API only states that if dPinlet < 3% (API-520-1 5.4.1.1) it can be ignored. It does not mention anything about ignoring outlet pressure - or I have not found it yet.

You are correct!

Here's the model I use. Usually, the exit of the PSV's flow nozzle (not the PSV's exit flange) is choked (Mach 1) at roughly 0.5 P1 when the valve disc has lifted and you have full flow. Right at/after the PSV's flow nozzle exit, there will be a shock wave (pressure discontinuity) and the pressure at the PSV's exit flange will be the pressure needed to push the fluid out of the discharge pipe and fittings to atmosphere. Some people say, calculate the discharge pipe and fitting dP backwards since we know it is atmosperic pressure at the end/tailpipe exit "back to" the unknown backpressure at the PSV's exit flange. It is trial and error calculation, so it is iterated until it converges.

Is that clear?

Also, You cannot transmit the backpressure (discharge pipe and fitting dP) upstream to the vessel, because it is Mach 1 at the PSV's flow nozzle exit, and pressure waves cannot exceed Mach 1 in this situation. That is why you should not add the backpressure to try to determine the pressure inside the vessel.

Good Luck,
Latexman

 

[Snickster]

Here is my take on your questions the way I understand.

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).

Overpressure refers to the relief valve pressure above the set pressure during relief. Accumulation refers to allowed pressure in the vessel during relief per ASME Code - this is 10% above MAWP for single relief valve installations non fire case. If set pressure equals MAWP then both over pressure and accumulation are 10%. If set pressure is below MAWP then the pressure can build up at the relief valve inlet to whatever as long as the accumulation in the vessel does not exceed 10%. So the overpressure used to calculate relieving pressure P1 may be 30% or more of set presssure depending on how low the set pressure of the relief valve is relative to the MAWP of the vessel.

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).

P1 is always set pressure + allowed overpressure + atm. Pressure drop in piping is accounted for in different ways.

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%.

Correct - Area required is based on P1 which is set pressure plus allowed overpressure + atm.

Discharge pressure drop should never be added to the vessel pressure during relief as the vessel does not see this pressure. With a conventional relief valve the back pressure is handled as follows:

For superimposed back pressure that is constant (like relief into a pressurized header at constant pressure) this back pressure will cause the valve to open at a higher pressure of the backpressure. So if you have a PSV set at 300 psig and there is 50 psig always in the relief header, the relief valve will open at 350 psig due to the superimposed back pressure acting on the backside of the disc. To compensate so the relieve valve still opens at 300 psig then the valve spring force must be set lower so that the valve is set at 250 psig and the 50 psig backpressure causes it to pop at 300 psig. In this case the relieving pressure P1 will be 300 + 30 + 14.7 assuming 10% over pressure. This because the flow under critical flow conditions in the orifice of the relief valve will be at sonic flow coditions as long as the downstream pressure remains below the critical flow pressure Pc - this downstream pressure now is the sum of the superimposed plus built up pressure when calculation critical pressure. So even with 50 psig TOTAL back pressure or even 100 psig TOTAL back pressure the flow will still be same and only dependent on the upstream pressure. The flow is only reduced when the TOTAL back pressure exceeds the critical flow pressure Pc (see API 520 5.6.2) because that is when the flow in the orifice becomes subsonic - as long as the flow is sonic the downstream pressure does not affect the flowrate of a gas. However note that the critical flow pressure increases with decreasing values of the ratio of specific heats k (see Table 7)

For built up backpressure as long as the backpressure due to friction is less than 10% and the sum of built up back pressure plus superimposed pressure does not exceed the critical flow pressure (as noted above) then no corretion factor is required. This is because all this built-up back pressure does is tend to oppose the opening force of the pressure on the upstream side of the disc. When the built up back pressure is equal to the over pressure then there is just about equilibrium between forces keeping the disc open and those trying to close it so it starts to flutter and chatter which can cause damage to the valve and reduce capacity.

In any case the back pressure is not seen by the piping or vessel upstream of the valve

CONTINIUED NEXT POST

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).

 

[Snickster]

CONTINUED FROM PREVIOUS POST

A quick note on balanced relief valves - balance relief valves have disc pressure balanced so that the back pressure does not have an effect on the opening of the valve so no need to compensate for superimposed back pressure. However the total backpressure does have an effect on the capacity of the valve as shown in API 520 Figure 30, for different values of overpressure as the valves are not totally balanced so there is some effect of backpressure.
 
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.

The discharge pressure was discussed in the previous post. For the inlet pressure drop with new installations you would definitely design for the 3% pressure drop rule and therefore there would be no compensations required to set pressure or P1 relieving pressure. However if you have an existing system and you exceed 3% rule you may have an issue because then valve will chatter and lose capacity and maybe be damaged, and possibly slightly overpressure the vessel. This is because the valve pops at the set pressure. Then as the flow rapidly increases to full flow (as determnined by upstream pressure and orifice area) and the pressure drop exceeds 3% the valve will want to shut. Fortunately there is blowdown which is the diffrence between pressure where the valve pops and resets which I believe is about 5% to 8%. So if the pressure drop is maintained in inlet of 3% then valve should not reseat but it might want to start chattering if gets too close to blowdown point.

I never had to design for this issue. I think what API 520 is stating that if the pressure drop does exceed 3% you would need to resize your relief valve based on subtracting the actual pressure drop when calculating P1. In this case the pressure in the vessel will still be no higher but you would just have more pressure drop in the line to the relief valve so the vessel won't care. But the relief valve flow will be derated by the pressure drop of the inlet. Still you have to consider that there may be some chatter if you get too close to the blowdown point. I think you can get blowdown rings for relief valves that increase the blowdown. I don't know if you can fit them to an existing valve though. You might just need to replace the valve with a higher blowdown valve or use a pilot operated valve.

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.

I believe you can have a set pressure of 100mbar and an overpressure such that the vessel allowed accumulation is not exceeded which is 33 bar normal operation and 36.3 bar fire for sizing the relief valve P1. In this case the 3% of set pressure inlet rule will still govern which will be 3mbar. But you need to check through the codes to see of there is any restrictions. It has been a long time since I read through the applicable API and ASME Codes.

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).

One final comment on discharge piping pressure drop. If the flow in the discharge pipe is subsonic then you would do a pressure drop calculation from the pipe exit back to the relief valve exit by starting at the end of the pipe and working backwards. If flow is subsonic then the pressure at the exit of the pipe is atmospheric or whatever the pressure is that you are dumping into, so this will be the start pressure and then you would sum all pressure losses back to the outlet of the relief valve to get the built up backpressure.

However if you reach sonic flow at the pipe exit you will will not be starting at atmospheric pressure at the pipe exit but at the critical sonic flow pressure which could be much higher than atmospheric inside the tip of the pipe just at the exit. This is a phenomenon that happens with sonic flow in pipes. So you would start with this higher than atmospheric pressure then add to this the sum of the frictional losses to get the backpressure at the valve exit. I just wanted to mention this because I don't know how familiar you are with compressible flow and you might overlook this. If you are not familiar with this then you need to study up on compressible flow in pipes.

 

[Luuk21]

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).

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

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.

I have another question regarding liquid reliefs (also part of the check) - in this case pressure drop inlet/outlet can not be ignored (non-flashing liquid) - but for now I will just adjust the MAWP and pretend that it is allowed to go over 21% of the MAWP.