20 replies to this topic

[panagiotis]

Hello everyone, let's assume that I have saturated steam at known T,p and I have calculated the relief load as well. How can I estimate the temperature at the outlet of the PSV? I had assumed as worst scenario to be equal to the tempeeature at the inlet. For sure the temperature will be lower during the expansion.

I am looking forward for your reply! ????

[Bobby Strain]

You can calculate the discharge temperature using adiabatic flash.

 

Bobby

[panagiotis]

But, also the outlet pressure is a kind of unknown.

[breizh]

Hi,

To get the pressure you can perform back calculation, from exit of the pipe (my guess atmospheric pressure) to the discharge of the PSV.

 

P exit +piping head losses = P discharge PSV.

 

My 2 cents

Breizh  

[panagiotis]

Hi,

What I want to say is that both T,p are unknown and to perform this pressure back calculation you need to know the temperature at the discharge of the PSV.
To perform the adiabatic flash calculation you need to specify Q=0 and pressure=unkonwn.

[breizh]

Hi,

It might be an iterative calculation using relief load to get at the Delta P from the outlet of the PSV to the exit.

You may find pointers in the document attached.

Good luck anyway.

Breizh 

[Bobby Strain]

Engineering is about "close enough".

 

Bobby

[panagiotis]

Actually, this was my first way the iterative way and the 2nd way the most conservative was just assuming same temp at the inlet and the discharge of PSV. Provided that the gas does not have negative joule thompson. Thanks for the document Breizh. I will read it even though when I looked through it at first it was full of complicated formulas. (Which means more questions ????????????????)

Bobby, yeah I agree!

[shvet1]

https://www.eng-tips....cfm?qid=501944

other reputable manufacturers have similar software, e.g. Emerson

 

Pressure at one point is not able to be different while mass flow is constant all the way, right? This is an important point, but not obvious one.

 

Pressure at the end of discharge pipe equals to 0 barg (assume it is ambient air). This means you know dP from inlet to discharge.

This means you know P, T and properties at inlet and dP. This data is enough to calculate T and properties at discharge. 

Then divide outlet pipe to 10 segments and assume that T and properties are constant at each one.

Then calculate dP of 10th segment knowing P, T and properties at discharge conditions. You know P at the end of 9th segment.

This means you are able to calculate T and properties at break point of 9th and 10th segments.

and so on up to outlet of valve. The less dP of segments the less inaccuracy.

 

Easy in theory, challenging in practice. This is the reason why many engineering companies purchase dedicated software for such calculations, e.g. see SimSci Inpant, Flare Analyzer, Flaretot. Expensive one I should say.

Edited by shvet1, 03 July 2023 - 12:14 AM.

[panagiotis]

Thank you for your useful answer!!! ????

[Mothilal]

What is allowable downcomer flooding and jet flooding velocity?

[breizh]

Hi Mothilal,

Please don't jump in an existing post. You need to start a new one.

Breizh 

[snickster]

For a compressible gas:

 

Relief valves are sized for sonic flow at the orifice, there the pressure and temperature will correspond to sonic flow conditions.  This is because relief valves are sized for sonic flow at their orifices and sonic flow is the highest velocity achievable in a converging nozzle, hence the highest flowrate possible.  Since this is an adiabatic almost isentropic expansion in the converging nozzle of the relief valve, then the critical flow pressure at the orifice of the relief valve is approximately:

 

P_cr=P_o (2/k+1)^k/k-1

 

P_cr   is the critical pressure at the orifice at sonic flow conditions

P_o  is the upstream static pressure conditions upstream of relief valve ( at zero velocity = set pressure plus overpressure)

k  is the ratio of specific heats

 

And the associated temperature at sonic flow conditions at the orifice of the relief valve is:

 

T_cr=T_o(2/k+1)

 

T_cr is temperature at orifice at sonic flow conditions also referred to as critical temperature

T_o is static temperature upstream of relief valve (at zero velocity)

k is ratio of specific heats

 

Once the flow enters the discharge piping the temperature will change from the orifice temperature based on how much backpressure develops and the resulting flowing velocity along the different sections of the discharge piping.  Typically the analysis begins at the pipe line exit.

 

If the piping is small diameter and relatively high pressure drop occurs then the flow will likely reach sonic flow again at the discharge of the line if it discharges to atmosphere, or sonic even if discharges to a higher pressure vessel.  In this case the temperature at sonic flow will again be as calculated above as this is the minimum possible temperature of flow for an adiabatic frictional flow process.  This is because the temperature at sonic flow will always be the same regardless of the pressure in accordance with the below energy equation.

 

The larger the discharge pipe diameter, the lower the flow velocity at the exit and therefore the higher the temperature per the below energy equation.  If the pipe is large enough diameter the velocity is subsonic at the exit and the temperature will be higher than the sonic flow temperature  T_cr depending on the velocity.  At a very large diameter pipe the velocity theoretically approaches zero at the outlet and temperature approaches T_o, the static pressure upstream of the relief valve. 

 

A back pressure calculation is performed from the end of the pipe to the exit of the relief valve starting with a pressure of:

 

Case 1: for sonic flow at exit discharging to atmosphere start with either atmospheric pressure or sonic flow pressure at pipe exit.  Since velocity cannot increase above sonic in a pipe, the pressure will increase inside the tip of the pipe for higher flowrates above that which sonic flow just begins.  For instance, pressure inside the tip of the pipe may reach hundreds of PSI while just outside the tip is atmospheric, depending on relief set pressure..

 

Case 2: for subsonic flow at pipe exit use the pressure at the exit of the pipe whether it be atmospheric or a higher pressure such as a pipe that discharges into a higher pressure vessel.

 

A more precise back pressure calculation is performed with the Fanno flow equations for adiabatic flow with friction (although an approximate frictional loss calculation can be performed using the isothermal flow equations using the temperature upstream of the relief valve T_o  or T_cr depending on engineering judgement as to what will be the closest approximation considering factors as describe herein).

 

Using the Fanno flow equations as the pressure increases, the velocity will decrease and the temperature will increase in accordance with the energy equation:

 

C_p(T2-T1) = V²/2

 

In other words, any velocity decrease is converted into a increase in the enthalpy and therefore temperature of the gas.

 

So the following will be true:

 

For smaller outlet pipe diameter the flow will be either close to sonic or sonic at the exit, the pressure will be either the pressure of what the pipe discharges to, or a higher sonic flow pressure inside the tip.  The temperature at the pipe exit will be very close to or equal to T_cr as calculated above and will increase as you go upstream and will get higher with absolute limit at  T_o  as the pressure increases due to high back pressure in the small diameter pipe.  However, a good designed pipe will be of a large enough diameter as to not produce a back pressure greater than about 10% of set pressure of relief valve (unless balanced relief valves are used) and an exit velocity of no greater than 0.5 Mach (half speed of sound) so in reality the temperature will be more closer to T_cr at the relief valve exit rather than T_o for a smaller diameter discharge pipe.

 

 

For a larger diameter discharge pipe the flow will likely be subsonic at exit and the pressure at pipe exit will equal the pressure of what it discharges into (be it atmosphere or higher pressure vessel).  The temperature at the tip will be in accordance with the above equation except T1 is replaced by T_o,   So the higher the exit velocity the lower temperature T2 with lower limit being at sonic flow temperature of T_cr.  If the discharge pipe diameter is sufficiently large and the velocity pressure drop is very low, the temperature will be closer to T_o at exit and increase upstream as the pressure increases with limit being T_o

 

 

In summary the limits of temperature at the exit of the relief valve will be between T_o and T_cr but much closer to T_o for typically large diameter low pressure drop piping and closer to T_cr for high pressure drop small diameter piping.

 

Note for air with k = 1.4 at 925 deg. Rankine upstream of relief valve (To) then the minimum critical flow temperature is 770.8 deg. Rankine (Tcr).

 

Note that sonic velocity is (gkRT_cr)^1/2  in feet per second.

 

You can determine if sonic velocity exist at the pipe exist using the equation

 

P(144)VA = mRT

 

P is pressure in psi

V is sonic velocity in ft per second

A is area of Pipe in sq. ft

m is mass flowrate in pounds per second

R is universal gas constant = 1545/MW  where MW is the molecular weight of gas

T is sonic flow temperature T_cr

 

Since everything is known except for pressure (if you first assume sonic velocity exists at pipe exit) then you can solve for pressure.  If pressure thus determined is greater than the pressure that the pipe relieves to then you must be at sonic flow and the pressure inside the pipe tip at the exit must be the pressure you calculated which is above the downstream pressure you are dumping into.  If the calculated pressure is lower than the downstream pressure then you are at subsonic flow and the pressure inside the pipe at the pipe exit is the downstream pressure but the actual flow velocity is below sonic - so plug in the downstream pressure into the equation then solve for the actual velocity. 

Edited by snickster, 09 July 2023 - 04:30 PM.

[snickster]

Note that if the flow is sonic at the exit then the temperature is Tcr as indicated above.  If the flow is subsonic at the exist then the temperature at the exit can be found by the energy equation.   In standard english units, which I am more familiar with:

 

For example say the upstream relieving temperature was 200 F, and pressure is 100 psig, and air at k=1.4, Cp=0.24, Cv=0.17, MW=29

 

For expansion in the nozzle of relief valve at sonic flow conditions in the orifice of the relief valve the critical sonic flow temperature is:

 

Tcr = To (2/k+1) = 660 ^oR (2/1.4+1) = 660 x .833 = 550 ^oR

 

Using the enegy equation to check this value as follows:

 

sonic velocity = (gkRT_cr)^1/2   = ((32.2)(1.4)(1545/29)(2/1.4+1)(660))^1/2   = 1149 ft/sec

 

Now find temperature predicted by the energy equation:

 

Cp ( To-Tcr) = V²/2g(778)    where Cp is specific heat in Btu/lb-deg R   and V is velocity in feet per second  g is gravitational constant 32.2                                                 and 778 is conversion factor for ft-lbs to BTU

 

Solve for T to check if it matches that calculated above:

 

0.24 (660-T) = (1149)²/2(32.2)(778)

 

T = 550 ^oR  Checks with above.

 

More to follow

Edited by snickster, 10 July 2023 - 05:05 PM.

[panagiotis]

Snickster,

I am impressed from your answer. I want to spend time and to go deeper into it.

Edited by panagiotis, 10 July 2023 - 05:12 PM.

[snickster]

Next I am going to give an example of what happens when the flow enters the discharge pipe. I need to set up an example and then I will post.

Edited by snickster, 10 July 2023 - 05:14 PM.

[snickster]

For the same values above we will look at what happens in the discharge piping:

 

Upstream relieving temperature 200 F, and pressure is 100 psig, and air at k=1.4, Cp=0.24, Cv=0.17, MW=29

 

First need to find the relieving rate through the relief valve.  To do this we need use the ideal gas equation at critical sonic flow condtions:

Also assume orifice of relief valve diameter is 1".

 

Ideal gas equation:

 

PQ=mRT for a 1" diameter orifice at sonic flow conditions =

 

Pcr(144)(Vcr)(A)=m(1545/29)Tcr      Note R is universal gas constant = 1545/MW in ft.lb units 

 

Pcr, Vcr, Tcr are sonic flow conditions at the relief valve orifice, m is the mass flow rate in lbs/sec (weight), A is area of the RV orifice

 

Since Vcr and Tcr have already been determined in previous post then need to calculate Pcr:

 

Pcr = Po (2/k+1) ^k/k-1 =                 Note that Po is static (stagnate) pressure upstream of RV:

 

Pcr = 114.7 (.5283) = 60.6 psia

 

Next calculate the mass flow rate through the RV using the ideal gas equation:

 

Pcr(144)(Vcr)(A)=m(1545/29)Tcr

 

60.6(144)(1149)(Pi/4)(1/12)²=m(1545/29)(550)

 

m=1.866 lbs/sec.

 

Note that this is the same mass flow rate predicted by API 520 relief valve sizing equations with the coefficient of discharge and compressibility factor = 1.0.

 

In other words the relief valve sizing equations are derived from the ideal gas equation.  If you substitute the values for Tcr, Pcr and Vcr  indicated already in previous post and repeated below, into the ideal gas equation shown directly above, and rearrange you will get the same API 520 equations

 

P_cr=P_o (2/k+1)^k/k-1

 

_Tcr=T_o(2/k+1)

 

Vcr = (gkRT_cr)^1/2

 

In next post I will go over what happens when the flow of 1.866 lbs/sec enters the discharge piping.

Edited by snickster, 10 July 2023 - 06:18 PM.

[snickster]

Now we will take a look at the discharge piping at the same flowing conditions above for two cases - one an undersized pipe where the flow reaches sonic at the end and one where the piping is sufficiently sized where the flow is subsonic at the end.  In both cases the pipe discharges into the atmosphere at 14.7 psia.

 

First an undersized pipe 1.5" Sch 40 with ID = 1.61"

 

Using the ideal gas equation pQ=mRT  assume that the flow conditions are sonic at Tcr and Vcr and solve for pressure:

 

P(144)(Vcr)(A)=m(1545/29)Tcr

 

P(144)(1149)(Pi/4)(1.61/12)²=1.866(1545/29)(550)

 

P=23.37 psia  Therefore since the pressure is above the atmospheric discharge pressure the flow is sonic at Pcr = 23.37 psia this is the actual pressure inside the tip of the pipe at the exit which immediately reduces to 14.7 psia exiting the pipe tip via irreversible expansion.

 

The reasoning is that if the flow is at Vcr sonic velocity (which is the maximum possible in a pipe) and temperature at Tcr which is the lowest temperature possible.  Then the only way the flow can adjust, considering that the relieve valve flow is fixed by the capacity of the orifice, is to increase in pressure at the exit tip of the line.  This is what actually happens

 

 

Next a sufficiently sized pipe 3" Sch 40 with ID = 3.068"

 

Pcr(144)(Vcr)(A)=m(1545/29)Tcr

 

In this case  first assume that the flow is sonic and solve for pressure P

 

P(144)(1149)(Pi/4)(3.068/12)²=1.866(1545/29)(550)

 

P = 6.437  psia

 

Since the calculated pressure is below the atmospheric discharge pressure, which is not possible - it must be either equal to discharge pressure for subsonic flow or greater than or equal to the discharge pressure for sonic flow - the actual pressure at the exit is 14.7 psia the actual atmospheric discharge pressure.

 

Now using the same ideal gas equation with know pressure of 14.7 psia and subsonic flow, solve for T.  This can be done by using the value of V as determined by the equation:

 

Cp ( To-T) = V²/2g(778)

 

0.24 (660-T) = (V)²/2(32.2)(778)

 

Solving for V:

 

V = ((0.24(660-T))(2)(32.2)(778))^1/2

 

Substitute V into the ideal gas equation:

 

14.7(144)(V)(Pi/4)(3.068/12)²=1.866(1545/29)(T)

 

Where V is the right side of the equation above - so now you can solve for T since that is the only unknown now

 

I will leave the math to you but I get that T = 632.5 ^oR  approx.

 

From this point you can do a pressure drop calculation back to the RV discharge either assuming isothermal flow at 632.5 ^oR or do a more precise adiabatic flow with friction using the Fanno flow equations, with a initial pressure of 14.7 psia at discharge.  The book Compressible Fluid Flow by Michel A. Saad gives a good discussion on compressible flow which is what I learned from.

 

Note that the smaller 1.5 inch pipe is a little undersized since I get a pressure drop of about 249 psi per 100 ft. for isothermal flow, so unless you have just a very short discharge pipe your pressure at RV discharge is going to be too high and reduce the flow.  However if you did have a shortdischarge pipe, of say 10 feet total equivalent lenght, than the pressure at the RV outlet considering isothermal flow is:

 

Pcr + dP friction = 23.37 + 24.9 = 48.3 psia

 

This is high for a conventional relief valve which limit is 10% of set pressure, however if you use a balanced diaphram pressure relief valve the back pressure allowed is higher but may be a correction factor if pressure is too great.

 

But usually you would not design for sonic flow at exit - check API 520 for maximum velocity recommendations.   Note that for normal gas lines the limit is 60 ft/sec based on noise generation, but since this is a short term very seldomly occuring event you may be able go get by but still it may just be too loud for safety reasons, as on blast could damage someones ears is close by.

 

This is basically concludes how to size a gas relief valve and discharge piping in a nutshell.

Edited by snickster, 10 July 2023 - 09:09 PM.

[breizh]

Hi snickster,

Thanks for the effort.

Breizh 

[snickster]

OK no problem - your welcome!

 

One final thing:

 

To find the conditions at the end of the pipe an adiabatic flow process was considered in the dischartge pipe such that any increase in velocity would have to come from the internal energy/enthalpy if no other heat was input to the discharge pipe from the outside..  This is true for an adiabatic process with or without friction. That is why the critical temperature is the same whether in the RV orifice for adiabatic isentropic flow or in the discharge pipe exit for adiabatic flow with friction.  Adiabatic flow is likely the case for short pipes in the atmosphere.  In reality some heat is picked up from the atmosphere from the discharge pipe so the actual process is somewhere between adiabatic and isothermal (polytropic) but closer to adiabatic I believe.  It is acceptable I understand and a close approximation to consider the flow in the discharge piping an isothermal process with much easier equations to solve than Fanno flow equations for adiabatic flow with friction, but I would start by calculating the end of pipe temperature as above then do an isothermal equation calculation back to the relief valve.  The closer you get to sonic conditions at the end of the line the more the temperature drops and can be much lower than the relieving temperature To, so more error is introduced.

Edited by snickster, 11 July 2023 - 12:13 AM.

[panagiotis]

Wow snickster, impressive! Thanks!!