Depressuring Tutorial

8/10/2019 Depressuring Tutorial

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Depressuring TutorialIn this posting, we will learn the depressuring simulation study using HYSYS simulator. The

simulation is mainly divided into two steps, fre case and adiabatic case. Commonly, the

fre case simulation result in the peak owrate to are, whereas the adiabatic case

simulation is used as basis in determining !inimum "esign !etal Temperature

#!"!T$. The %ollowing picture shows the summary o% the simulation procedure using

HYSYS simulator

See the picture below. Hope it help

&irst, simulate the fre case. 'e(t, simulate the adiabatic case, #the Cv result %rom fre case

shall be used in adiabatic case$. )nd fnally, the conclusion can be made based on thesimulation result.

!y %riend, I have made a tutorial %or depressurring simulation. The tutorial comprises step

by step both %or fre and adiabatic case. “You can download the tutorial here”. The

tutorial is dedicated to you, my %riends, hope it use%ul %or you.

http://code.google.com/p/tutorial-depressuring2/downloads/detail?name=TUTORIAL%20DEPRESSURING%205.pdf&can=2&q=

http://code.google.com/p/tutorial-depressuring2/downloads/detail?name=TUTORIAL%20DEPRESSURING%205.pdf&can=2&q=



Relieving Condition Exceed Design Temperature

*+uipment design temperature is usually determined based on the ma(imum temperature

plus a certain margin. The margin could be dierent %or each pro-ect, say /&, 0 & or 1/&.

Conducting rigorous simulation %or fre case, the temperature in each stage simulation is

set higher than the previous stage by certain interval. 2et say, using 3/& in interval

temperature will result in temperature about 1/& higher than the operating temperature at

third stage simulation. That is why4 relieving temperature %or fre case is very o%ten higher

than the design temperature. 5hether or not increasing rating pipe is re+uired6.

7S8 "oesn9t 7rovide )de+uate 7rotection o% &ire Case

See the picture below

http://process-eng.blogspot.com/2012/04/rigorous-method-for-fire-case.html

http://process-eng.blogspot.com/2012/04/rigorous-method-for-fire-case.html



) Separator has operating temperature o% 3/ & with design temperature o% 3:0&. 2ooking

at simulation above, in only two stages o% temperature increase, the temperature will

reach 30/&.

)%ter fre, the temperature will increase commonly above the design temperature. )nd

even in some cases, the vessel could rupture be%ore increasing pressure reaches 7S8;s

setting pressure. In other word, the vessel will have ruptured be%ore the 7S8 open. Thus,

the 7S8 doesn;t provide ade+uate protection %or the vessel in case o% fre.

&or e(ample, the %ollowing picture show that the metal plate temperature reaches 3// &

in only 30 minutes a%ter fre.



That is, actually, the 7S8 doesn;t provide ade+uate protection %or the vessel in case o% fre.

There%ore, some other options %or vessel protection %rom fre case beside the 7S8 should be

considered, such as<

3. "epressuring

. *(ternal cooling

1. *(ternal insulation

:. 7rovide proper drainage

!y %riend, %or now, we already know the %act that 7S8 will not give ade+uate protection %or vessel in

fre case. =ut, why do we always provide 7S8 as protection devise o% the vessel6 5hy do we

consider %or fre case too6 That could be a -oke, right6 In my opinion, that is all to meet the C>"*

re+uirement

!y %riend, a note shall be made, although the relieving temperature is higher than the design

temperature, and the 7S8 is not e?cient to protect the vessel against fre, the 7S8 remain to be

designed based on the relieving temperature, since the 7S8 is installed as a sa%e guard.

Example Case

I still remember, one o% our %riends has ever asked me4 at that time, he did calculation o% fre case

%or instrument air receiver, the relieving temperature was e(ceeding the ma(imum allowable %or

30/@ rating pipe. The +uestion was, whether the 30/@ pipe rating o% inlet 7S8 would be needed to



be increased to 1//@ or not6

>% course '>T, the pipe rating class o% 30/@ does not need to be increased to 1//@, since the

determination o% pipe rating shall based on ma(imum condition without consider fre case.

=esides that, it is not common to design 7S8 %or instrument air system with fre case. 7ersonally, Ipre%er calculate the 7S8 load o% instrument air receiver based on blocked outlet case #i% it is

possible$. &urthermore, In some cases, it might not be applicable to siAe 7S8 o% instrument air

receiver based on fre case. )ctually, it shall be based on &*B) #&ire and *(plosion Bisk )ssessment$

-ustifcation whether the instrument air receiver included in fre Aone or not.

The block outlet case has more possibility occurred than the fre case. =ased on my e(perience, the

load o% block outlet is also smaller than unwetted fre case. There%ore, it is better to design 7S8 o%

instrument air based on block outlet case than fre case. !oreover, i% you still design 7S8 based on

fre case to consider worst case load, that 7S8 will not provide ade+uate protection %or the vessel<

".

"o you agree with me, don9t you6

Talk back in comment section below and let me know your opinion D

Basic Depressuring - Wh !" minutes #

In previous posting, we have discussed that 7S8 wouldn;t provide ade+uate protection %or

vessel o% fre case. There%ore, depressurring can be applied %or another sa%ety layer.

Commonly the plant area is divided into the *S" Eone. *ach *S" Aone may contain one or



more e+uipments. *S"8 or S"8 valves are provided in each *S" Eone to isolate the

system Aone. In case o% fre, a system will be isolated by those S"8 valves. Then the

inventory uid #commonly gas phase only$ in the system will be released to are through

="8 valve. Commonly one ="8 is provided %or one system Aone, but in some cases, it is

possible to provide more than one ="8s in one system Aone.

See the picture.

*"7 #*mergency "epressuring$ is generally initiated by manual push button. In case o%

fre, the operator will push the *"7 push button in the control room. That will initiate S"8

valves closing and ="8 valves opening. The hydrocarbon uid will be released to are so

that the pressure o% the system will be depressuriAed to lower pressure at certain time

#recommendation %rom )7I ST" 03, decrease the pressure to 3// psig or 0/ F o% the

system design pressure within 30 minutes$

&or e(ample, I% a pool fre e(poses the unwetted carbon steel vessel, it will take about 30

minutes to heat the vessel wall to around 3 // G& #very close to material;s allowable

stress condition$. I% the vessel is depressuriAed within the 30 min to 0/ F o% the initial

pressure, then the time to rupture would increase to about 1 hour



Hope this picture will give better e(planation

&or thickness o% vessel less than 3 inch, the system is depressuriAed to 3// psig, and %or

more than 3 inch thickness can be depressuriAed to 0/F o% design pressure. The

depressuring time can be longer and less than 30 minutes. The depressuring time o% 30

minutes is only an e(ample in )7I ST" 03 which is applicable %or carbon steel vessel withhas thickness greater than 3 inch.

Consideration o% limiting are capacity, the depressuring time longer than 30 minutes may



be applied. It will result in lower depressuring load. Considering o% the ma(imum reduction

o% the vessel stress, vessel with thickness less than 3 inch, generally re+uires %aster

depressuring rate. The %aster the depressuring time, the higher the depressuring load. )nd

%or the vessel with stainless steel material, the depressuring rate may be longer than 30

minutes %or 3 inch thickness or more.

=ased on my e(perience, many companies have their own manual %or conducting

depressuring study. enerally, the ma(imum depressuring time o% 30 minutes is applied.

=ut each company has dierence consideration o% thickness vessel and depressuriAed

pressure. Some companies apply that %or thickness greater than :.0 mm, the pressure is

depressuriAed to 0/F o% design pressure, but other companies apply that %or thickness

greater than J/ mm the pressure can be depressuriAed to 0/F o% design pressure.

I have ever read discussion in Cheresources # check here and here$ about the depressuring

time, one o% the participants says that his company applies the depressuring time which

depends on the vessel thickness. &or thickness greater than 0.: mm, 30 minutes

depressuring time is applied. The depressuring time will be decreased 1 minutes %or each

0 mm decrease in thickness.

Yesterday, I checked to )7I ST" 03, #&i%th *dition, Kan //L$, fgure 3 #section .30.3..$,

the graphic show M 7late Temperature vs Time )%ter &ire M %or carbon steel 1. mm, 3.L

mm and 0.: mm thickness. It is very logic that the re+uired depressuring time %or those

vessel are dierent each other. In my opinion, it is better to state in manual Nthe

depressuring time will be decreased, say or 1 or (( minutes, %or each 0mm decrease in

thicknessO.

>h,,I miss something important. *ven though the depressuring time o% 30 minutes is used,

the depressuriAation will not stop a%ter 30 minutes and that the pressure will continue to

decline.

Ha ha ha,,I guess you already know that <"

2et;s imagine, a 72)'T is shutdown %or annual maintenance purpose, the fre does not

e(ist, then the system is to be depressuriAed to atmosphere condition. In this case, the

system is depressuriAed in adiabatic condition, which means no heat input to the system.

"uring depressuriAation, the pressure decreases, and the temperature decreases as well.

The fnal temperature o% adiabatic depressuring could be very low. )s 7rocess *ngineer,

we have responsibility to determine the !inimum !etal "esign Temperature #!"!T$ %or

each system Aone based on the adiabatic depressuring case

!y %riend, that;s all I can share today. Hope%ully it is use%ul.

http://www.cheresources.com/invision/topic/14895-api-521-automatic-blowdown-time/

http://www.cheresources.com/invision/topic/12954-blowdown-in-fire-case/

http://materias.fi.uba.ar/7699/API_521_07.pdf

http://www.cheresources.com/invision/topic/14895-api-521-automatic-blowdown-time/

http://www.cheresources.com/invision/topic/12954-blowdown-in-fire-case/

http://materias.fi.uba.ar/7699/API_521_07.pdf



Depressuring - $preadsheet %ethod

In Hysys "epressuring utility, the model is only specifed as one horiAontal or vertical

vessel. "imension o% that single vessel, is back calculated based on total li+uid and vapour

volume. 5etted area calculation is di?cult %or horiAontal vessel. It is di?cult to match

wetted area and li+uid volume o% vessel with wetted area and volume o% a system.

)lso, In Hysys "epressuring Ptility, the heat input model, especially o% &IB* )7I03 method

only provide e+uation %or heat u( o% 3./// =TPQ%tR3.J:Qhr. HYSYS doesn;t provide -et

fre case

How to solve those problems6 To overcome those problems, we can use spreadsheetmethod as e(plained in my tutorial. “You can download the tutorial here”. )nd %or

better report o% depressuring simulation study, I make a simple tutorial here.

That;s all,,

http://code.google.com/p/depressuringtutorialkedua/downloads/detail?name=TUTORIAL%20DEPRESSURING%20second%20%5BCompatibility%20Mode%5D.pdf&can=2&q=

http://code.google.com/p/depressuringtutorialketiga/downloads/detail?name=TUTORIAL%20DEPRESSURING%20thirth%20%5BCompatibility%20Mode%5D.pdf&can=2&q=



http://code.google.com/p/depressuringtutorialkedua/downloads/detail?name=TUTORIAL%20DEPRESSURING%20second%20%5BCompatibility%20Mode%5D.pdf&can=2&q=




&ire Case - 'eat (nput Rate

&ire can cause overpressure o% storage or vessel. *ither li+uid vaporiAation o% wetted

vessel or vapor e(pansion o% unwetted vessel due to heat input will increase the pressure.

Heat input rate o% fre e(posure is not calculable %rom standard o% heat trans%er

%undamental. &ortunately, >SH) regulations speci%y standards which are to be %ollowed %or

particular material in storagevessel and )7I also provides %ormulas %or calculating heat

input rate which are to be %ollowed %or particular condition o% process vessel.

The %ollowing picture presents >SH) 8enting Be+uirement o% &ire *(posure to Storage

8essel.

&or ammable and combustible li+uid, heat input re%ers to '&7) 1/ as %ollow.



'ote <

3. In%ormation in table applies to li+uifed components 4 see original standard %ornonli+uifed gases

. C) Compressed as )ssociation

'&7) 'ational &ire 7rotection )ssociation

)'SI )merican 'ational Standard )ssociation

1.Heat Input e(pressed in =TPQhr, with area in s+uare %eet.

:.See source documents %or details o% area calculations

!y %riend, i% you want to check the correctness o% value in above table, please re%ers to

original re%erence, uidelines %or 7ressure Belie% and *uent Handling System, )!*BIC)'I'STITPT* >& CH*!IC)2 *'I'**BS #)ICH*$, Table 1.13 #page 313 o% 01U$. You can fnd

that book in our library.

)7I ST" 03 provides %ormulas %or calculating heat input rate to process vessel containing

li+uid.

V 3./// &)R/.U

5here ade+uate drainage and frefghting e+uipment do not e(ist, e+uation below should

be used,

V 1:,0// &)R/.U

Begarding a drainage, as long as the amount o% li+uid which is spilled out around the

vessel is possibly decrease, use the frst %ormula # heat u( 3///$.

& is environment %actor. The value & 3 %or bare vessel. Be%er to )7I ST" 03 Table 0

#page 3L$ %or other than bare vessel.

) is wetted area o% the vessel.



The value o% ) R/.U reect that not 3//F vessel area e(posed by fre. It is true %or large

vessel, %or small vessel, we can use assumption that 3// F vessel area is totally e(posed

by fre.

&or e(ample, %or small bare vessel, the e+uation to be V 3./// ).

!y %riend, please note, based on the %ormula, when UF %raction area is applied the heat

u( is 3./// =TPQ)t*!+,Qhr or 1:0// =TPQ)t*!+,Qhr. )nd when 3//F %raction area is

applied, the unit will be =TPQ)t*.Qhr. This understanding will use%ul in per%orm

depressuring simulation using HYSYS.

The above %ormula is %or pool fre case which has heat u( o% 3./// =TPQ%t3.J:Qhr.

5hereas, based on )7I ST" 03, the heat u( %or -et fre is W0.0// =TPQ%tQhr #average$.

&rom the unit o% heat u( %or -et fre, its show that 3//F %raction area is used #%or -et fre is

localiAed heat u($.

Hope this drawing helps you get better understanding o% pool fre and -et fre

)nd below is -et fre



!y %riend, in this posting we discuss about heat input rate o% fre e(posure %or process vessel. The

%ollowing are the important point,

3. &or ade+uate drainage and fre fghting e(ist, use %ormula V 3./// &)R/.U

. &or small vessel, use 3//F %raction area is e(posed by fre, there%ore V 3./// &). In my

understanding, vessel with wetted area less than // %t can be considered small vessel #see '&7)

1/$

1. &or wetted vessel, overpressure caused by li+uid evaporation, the relieving load capacity can be

calculated as heat input divided by latent heat o% vaporiAation. #5 VQHv$.

:. Heat u( o% -et fre is very high #W0.0// =TPQ%tQhr average$, and in localiAed area.

0. I suggest you to siAing 7S8 on fre case %or pool fre case only.

J. The relieving temperature can be higher than the vessel collapse temperature. 8essel will

collapse be%ore increasing pressure reach 7S8;s set pressure. There)ore/ actuall/ 0$1 doesn2t



provide su3cient protection+ Depressuring is one o) the additional protections against

4re. #5e will discuss about fre case depressuring both %or pool and -et fre in other posting$

!y %riend, thank you %or reading, please correct me i% I9am 5rong.

Rigorous %ethod )or &ire Case



In previous article, I have e(plained that relieving load %or fre case can be calculated as heat input

divided by latent heat o% vaporiAation. #5 VQHv$. The heat input have been discussed be%ore,

see fre case Xheat input rate. The latent heat o% vaporiAation is rather di?cult to be determined.

"uring fre, li+uid in the vessel will be vaporiAed, but the amount o% vapor %ormed is not f(ed,

because the li+uid composition is change overtime. In this article, I will e(plain you step by step the

rigorous method %or fre case

2et imagine the system consist o% a three phase separator below, the fre case calculationprocedure is as %ollow4

3. "etermine the inventory volume o% isolated system, both %or condensate, water, and gas based

on actual siAe o% the vessel and pipe.

. Calculate the wetted area.

http://www.process-eng.blogspot.com/2012/04/fire-heat-input-rate.html





1. 7er%orm simulation<

a. "efne stream %or 5)T*B, C>'"*'S)T*, and )S. Input composition, pressure, and temperature

as operating condition o% the three phase separator.

b. )d-ust mass ow %or those streams, to achieve actual volume %or each stream in accordance with

inventory volume calculation #3$. !i( those streams to be STB*)! >7*B)TI'

c. #a$ and #b$ is model o% the separator at normal operating condition based on actual condition.

d. Install three phase separator, 8/. The STB*)! >7*B)TI' is as inlet stream to the separator.

)d-ust V3// #heat input to vessel$ to achieve pressure B*2I*8I' C>'"ITI>'



#e$. The temperature and pressure will increase to initial B*2I*8I' C>'"ITI>' due to fre. #This

simulation assumes the 7S8 instantly reach relieving condition a%ter opening$. The outlet stream

%rom separator is mi(ed.

e. =alance the mi(ing stream #d$ to stream B*2I*8I' C>'"ITI>'. Input the pressure o% relieving

pressure value #3.3 ( set pressure$.

%. #d$ and #e$ is model o% the separator at relieving condition as initial stage which is 7S8 will start

open

g. Install a three phase separator #83$. The stream B*2I*8I' C>'"ITI>' #e$ is as inlet stream to

the separator.

h. Heat input due to fre case is calculated by increasing temperature with interval 3/ &. Thus, set

the temperature outlet separator #83$ o% 3/ & higher than the B*2I*8I' C>'"ITI>'.

i. The outlet stream is C>'"*'S)T* , 5)T*B and 8)7>B. The 8)7>B will be divided into two

streams, 2>)" ST)* 3 and 8)7>B

-. !i( C>'"*'S)T* , 5)T*B and 8)7>B to be I'2*T ST)* .

k. )d-ust mass ow rate o% 8)7>B to achieve actual volume ow o% I'2*T ST)* is e+ual with

initial actual volume STB*)! >7*B)TI'.

l. The stream 2>)" ST)* 3 is relieving load fre case



m. The heat input to 83 in HYSYS simulation is V3 =TPQhr, whereas the actual heat input is V )7I 3./// & ) R/.U =TPQhr

n. Calculate the re+uired time o% increasing temperature at stage 3, that Is #V3$ Q #3./// &.)

R/.U$

o. Install a three phase separator 8. The stream I'2*T ST)* is as inlet stream to the separator.

p. Bepeat the procedure %rom #g$ to #n$, %or stage , stage 1, stage : ....with temperature is

increased by 3/& in every stage. 5e can stop the procedure i% the cumulative time is one hour #it;s

assumed that the fre has been suppressed during 3 hr, you can use more than 3 hour i% you want$

+. Calculate the re+uired area %or each stage, #with the relieving load are 2>)" S)T* 3, 2>)"

ST)* , 2>)" ST)* 1, $

r. Pse the stage re+uiring the ma(imum orifce area as basis datasheet o% the 7S8.

fnish <"

ood morning my %riend, That9s all I can share today, I;m very happy today because I fnd a new

spirit %or better li%e. "on9t regret what was loss4 it didn9t disappear actually, -ust in use by more

appropriate user to make it more use%ul thing. Sometimes we lose something, but at the same time,

we get the other thing that better, at least a lesson.

Hope%ully you never bored with my article.

Thank %or reading,



0$1 Calculation - 5d/5c/56 )actor

Hi %riends, how are you today6 Hope%ully, you are not already bored with my articles. 'ow,

we will learn the concept o% relie% valve calculation. The spreadsheets not tell you

anything, but we should know what really it does. )%ter read this %ully article, I hope you

get a better understanding o% the relie% valve calculation concept, and then know well with

your spreadsheet.

7ressure is e+ual to the %orce divided by the area #7Z&Q)$. That is a very basic concept. I

am sure you already %amiliar with this %ormula in the very beginning o% -unior high school.

In talking 7S8, the %orce is generated by mass relieving rate. I am sure, all o% us know well

that relieving rate is depends on cases that to be considered #e.g. fre, blocked discharge,

gas blow by etc.$

5e need some corrections since the uid ows through a relie% valve noAAle orifce rather

than an ideal noAAle. &or the same area, at certain condition, the ow capacity o% relie%

valve orifce must be less compared with the ideal one X Its mean that more area is

re+uired to handle the same mass relieving rate . That;s why, there are some correction

%actors are re+uired such as4 [c, [d, [b, [v, [p, [h, [n etc.

2et;s %ocus to [c, [d and [b. I will make other posting %or the e(planation o% the other

%actors.

E7ective Discharge 8 5d

[d is eective discharge coe?cient used %or the mass u( capacity correction %or the real

noAAle. The higher the [d value, the closer the mechanical to an ideal design ideal

noAAle, [d 3. It is very obvious that the 7S8 has [d value lower than 3. &or instance, )7I

relie% valve has [d /.WL0 and /.J0 %or gas and li+uid respectively.



[d is depends on the mechanical design. In other word, every vendor has a specifc value

o% [d. It is important to be realiAed, so you are not always input /.WL0 in the spreadsheet

whereas you use 7S8 other than )7I.

5e can use )7I data as preliminary calculation only when we don;t have any re%erence o%[d value. 5e usually do this at proposal stage.

Some vendors have a better design 7S8 than standard )7I. In other word, they have [d

value higher than /.WL0. So, I suggest you to ask your vendor the e(act value o% [d,

especially at pro-ect stage. There is potential that the smaller orifce area is re+uired.

Com6ination with Rupture Dis9 8 5c

[c is correction %actor when rupture disk to be installed upstream o% the 7S8.

Bupture disk is re+uired to be installed at upstream o% the 7S8 %or system contain solid

that may plug the 7S8 over time. )t H)EI"QH)E>7, %or to(ic service, potential leaking o%relie% valve shall be considered, and then rupture disk at upstream 7S8 can be used as

positive seal %or the sa%eguard. That is why, based on my e(perience, the combination

rupture disk and 7S8 is very seldom to be applied in the gas processing.

)ctually, the [c value is comple(. =ut I am sure, we seldom use it. So, I am not too interest

makes longer e(planation. I understand many engineers hate with the long article with

something not practical, me too. I% you want to know more detail about [c, please read )7I

0/ part I by yoursel%.

(mpact o) Bac9 0ressure 8 56

=ack pressure is defned as a pressure e(isting at 7S8;s outlet. It impact to opening

pressure, reduction capacity, instability or may combination o% all. [b is re+uired %or

correction o% reducing capacity.

&or the low back pressure system that the impact is not signifcant, conventional type can

be used. )nd %or the e(cessive back pressure service, pilot type 7S8 is re+uired to

overcome it. Then, imagine that low back pressure %or the conventional type and no

impact back pressure on pilot type due to mechanical design. That;s all, I think, very clear

why [b is re+uired %or balance below type only.

'ote that the back pressure correction %actor [b %or conventional and pilot type is not

re+uired, and then use [b3. )sk your vendor o% [b value Xas graphic %or using below

7S8 type or re%er to )7I 0/ part 3 %or preliminary design.

)ctually, %or conventional type, when the condition is noncritical due to superimposed

back pressure, [b is re+uired. =ut %or now, rather than it will make you con%use, %orget it

since there are not much system likes that.





In this e(planation I use a general are system which the disposal is discharged to atmosphere via

are stack. 2et imagine, when relie% valve open, relieving uid ow thru Tail 7ipe XSub Header X

!ain Header&lare "rumSeal 5ater "rum X &lare Stack X &lare Tip and fnally discharged to

atmosphere at certain height # %or sa%e disposal purpose$

The =uilt up back pressure can be calculated by count up total pressure drop at the pipe, drum,

stack and tip. Start calculation %rom the pressure drop at the tip, then at stack until tail pipe o% the

relie% valve. The built up back pressure is pressure that e(ist at outlet relie% valve # that is same as

with total pressure drop %rom the tail pipe to the are tip$

&or more clear, see this picture below

To calculate the back pressure o% the 7S8, determine the pressure drop at tip, stack, 5ater Seal,

[>" and pressure drop along the pipe %rom point to point L. Since the pressure at outlet tip is

atmospheric pressure condition, we can calculate the pressure at outlet 7S8, which is a total

pressure drop along the system %rom point 3 to point L.

The basis back pressure calculation is rated ow. Bated ow is ow rate based on actual orifce area

o% 7S8. !y %riend, once the back pressure is calculated, then we can determine the 7S8 type. &or

built up back pressure less than 3/F o% 7S8;s set pressure, a conventional type can be used.

=alance below type can be used %or back pressure up to 1/F, and pilot type shall be used %orhigher back pressure. 'ote shall be made, even though the selection o% 7S8 type is based on the

back pressure consideration, it is not the only consideration in 7S8 selection

)%ter the back pressure calculated, we can also check the uid condition due to decreasing pressure

%rom inlet 7S8 #relieving pressure$ to outlet 7S8 #back pressure$ using HYSYS simulator. Is hydrate

%ormed there6 is the icing occurred6 )nd %or li+uid phase, is the li+uid will be ashed6



The back pressure calculation is so simple, isn;t6 How about i% there is any 7S8 open

simultaneously6 The dierence is only the combined owrate %rom others 7S8 shall be considered

in pressure drop calculation. Hope%ully this picture help you more understand with the concept o%

back pressure calculation.

See the picture above. I will e(plain you why the calculation o% built up back pressure is very

important not only %or 7S8 design but also %or are system design. 2et imagine4

7S8 /3 is set at 0/ psig with relieving ow o% // lbQhr.

7S8 / is set at 3// psig with relieving ow o% 3/// lbQhr.

In case the two 7S8 open simultaneously, what do you think what happen with the 7S8 /3 i% the

back pressure o% 7S8 / is more than 0/ psig6 >% course the 7S8 /3 cannot open properly

5hat is the solution6 Hmm,, look at this picture, it will show you a better confguration o% are

header system to avoid that condition



The other options are increase the are header siAe or provide are header %or high pressure and

and low pressure separately.

The above sample is a very simple case. Sometimes, %or instance in power %ailure case, there is

many 7S8 will open at the same time. The back pressure %or each 7S8 shall be checked to ensure

each 7S8 can properly operated. The back pressure profle o% each 7S8 can be used to determine

the confguration o% the are system

!y %riend, that;s all, I can share to you. Hope%ully, this topic reminds us that the back pressure issomething very important to be considered both in 7S8 and are system design

Thank you %or your attention.

=ack 7ressure *ect

In the previous article, two kinds o% back pressure have been discussed. 5e have already

understood that back pressure has several impacts to 7S8 per%ormance. In this article, we will learn

the impact o% back pressure on the relie% valve opening, operation and ow capacity.



!y %riend, I% you don;t have a good understanding on the defnition o% terminology such as4

overpressure, accumulation, set pressure, re seat pressure etc, I guest you will be con%used with my

e(planation. In this article, I won;t to e(plain each o% them, so please re%er to )7I 0/ part I by your

sel%. It;s very important to understand the meaning o% its defnition.

7B*SSPB* B*2I*& 8)28* >7*'I'

Superimposed back pressure has impact to opening o% conventional relie% valve type. This back

pressure will give additional spring %orce onto valve disk in closed position. There%ore, the actual

spring setting can be reduced by an amount e+ual to the amount o% superimposed back pressure.

Hope%ully, this picture will help us.

&or balance and pilot type, the compensation o% the constant superimposed back pressure is not

re+uired

7B*SSPB* B*2I*& 8)28* >7*B)TI>'

*(cessive o% built up back pressure has impact the conventional valve operates in unstable

condition. Its may be chatter or utter. Chatter is rapid motion o% closing and opening valve where

the disc contacts with the relie% valve seat during cycling, whereas utter is not contact with the

seat. Chatter cause damage to the valve.

7B*SSPB* B*2I*& 8)28* C)7)CITY

=uilt up back pressure has impact reducing the valve capacity. High back pressure reduces the

li%ting o% disc result in reduction o% ow capacity. &or conventional type, built up back pressure shall

not e(ceed 3/F o% set pressure at 3/F allowable overpressure. &or application that allowable

overpressure is higher than 3/F, say 3JF o% multiple valve application, then the built up back

pressure up to 3JF o% set pressure is allowed %or conventional type.

See the picture below o% capacity correction %actor due to back pressure %or conventional type

where the spring setting compensation %or superimposed back pressure is re+uired.



The fgure below shows that the capacity o% balance below type will be reduced signifcantly due to

back pressure. 2ook at this picture, hope%ully you get the answer why we choose to use balance

below type %or back pressure up to 1/F at 3/F allowable overpressure.

&inally, I can share this topic %or you. Hope%ully we get better understanding on the importance o%

the back pressure. The back pressure is one o% consideration %or optimiAation are system. &or new

design system, which one do you pre%er, use larger siAe o% tail pipe to reduce back pressure or use

balance below type to overcome it6

&riends, Thank you %or reading. 7lease correct me i% I am wrong.

Critical - $u6critical

Hi my %riend, how are you today6 I am sorry, It;s been a while my blog has not been updated. I was

very busy this week4 there%ore I had no time to write a new article. >k, 2et9s continue the discussion

about back pressure. 'ow, by this article I will e(plain you that the critical and subcritical condition

depend on the back pressure.



=e%ore I make %urther e(planation, please note, e(planation in this article will be limited %or gas

vapor phase application only.

)7I 0/ part 3 provides calculation procedure %or siAing o% eective area. The procedure is divided

to critical and subcritical condition that depends on the back pressure that e(ists at outlet 7S8.

Critical condition is considered when the back pressure is lower than the critical ow pressure.

Critical &low Rate

How critical condition occurred6 I% a compressible gas is e(panded across a noAAle or orifce, at

constant upstream condition, its velocity increases with the decreasing downstream pressure. The

increasing velocity means that the mass ow rate increases. The ow rate will increase until a

limiting value.

)t sonic velocity, the ow rate could never increase anymore even though the downstream

pressure is much lower. This ma(imum ow rate at sonic velocity is known as critical ow rate.

see this picture.

Critical &low 0ressure

To avoid con%usion, terminology o% critical ow pressure is used instead o% critical pressure. Critical

ow pressure is defned as an absolute pressure at noAAle e(it at critical ow rate. The actual

pressure at noAAle e(it cannot %all below the critical ow pressure even though the downstream

pressure is much lower



Critical ow pressure can be calculated using the ideal gas relationship.

See above fgure. &or 7S8, pressure at outlet is known as back pressure. =ased on the back

pressure value, we can determine whether the condition is critical or subcritical.

Critical condition is considered i% the back pressure is less than the critical ow pressure. In other

word, i% the back pressure is greater than the critical ow pressure, it will be considered as

subcritical condition.

The orifce calculation procedures both %or critical and sub critical condition shall be used in eachappropriate condition.

!y %riend, that;s all, I can share to you. Critical and subcritical condition depend on the back

pressure. Hope%ully, this topic reminds us that the back pressure is something very important to be

considered when we conduct 7S8 siAing.




Built :p Bac9 0ressure Calculation

!y %riend, do you still remember6 >ne o% consideration in selection o% relie% valve type is built up

back pressure. In this article I will e(plain you how to calculate the built up back pressure. >h, do

you still remember what is built up back pressure6 Is it a constant or variable6

=uilt up back pressure is pressure at outlet 7S8 in open mode, a variable back pressure since it is

depend on the relieving ow. Bead my previous post about back pressure %or more detail in built up

and superimposed back pressure

In this e(planation I use a general are system which the disposal is discharged to atmosphere via

are stack. 2et imagine, when relie% valve open, relieving uid ow thru Tail 7ipe XSub Header X

!ain Header&lare "rumSeal 5ater "rum X &lare Stack X &lare Tip and fnally discharged toatmosphere at certain height # %or sa%e disposal purpose$

The =uilt up back pressure can be calculated by count up total pressure drop at the pipe, drum,

stack and tip. Start calculation %rom the pressure drop at the tip, then at stack until tail pipe o% the

relie% valve. The built up back pressure is pressure that e(ist at outlet relie% valve # that is same as

with total pressure drop %rom the tail pipe to the are tip$

&or more clear, see this picture below

http://process-eng.blogspot.com/2011/02/built-up-and-superimposed-back-pressure_2304.html






To calculate the back pressure o% the 7S8, determine the pressure drop at tip, stack, 5ater Seal,

[>" and pressure drop along the pipe %rom point to point L. Since the pressure at outlet tip is

atmospheric pressure condition, we can calculate the pressure at outlet 7S8, which is a total

pressure drop along the system %rom point 3 to point L.

The basis back pressure calculation is rated ow. Bated ow is ow rate based on actual orifce area

o% 7S8. !y %riend, once the back pressure is calculated, then we can determine the 7S8 type. &or

built up back pressure less than 3/F o% 7S8;s set pressure, a conventional type can be used.

=alance below type can be used %or back pressure up to 1/F, and pilot type shall be used %or

higher back pressure. 'ote shall be made, even though the selection o% 7S8 type is based on the

back pressure consideration, it is not the only consideration in 7S8 selection

)%ter the back pressure calculated, we can also check the uid condition due to decreasing pressure

%rom inlet 7S8 #relieving pressure$ to outlet 7S8 #back pressure$ using HYSYS simulator. Is hydrate

%ormed there6 is the icing occurred6 )nd %or li+uid phase, is the li+uid will be ashed6

The back pressure calculation is so simple, isn;t6 How about i% there is any 7S8 open

simultaneously6 The dierence is only the combined owrate %rom others 7S8 shall be considered

in pressure drop calculation. Hope%ully this picture help you more understand with the concept o%

back pressure calculation.





The other options are increase the are header siAe or provide are header %or high pressure and

and low pressure separately.

The above sample is a very simple case. Sometimes, %or instance in power %ailure case, there is

many 7S8 will open at the same time. The back pressure %or each 7S8 shall be checked to ensure

each 7S8 can properly operated. The back pressure profle o% each 7S8 can be used to determine

the confguration o% the are system

!y %riend, that;s all, I can share to you. Hope%ully, this topic reminds us that the back pressure issomething very important to be considered both in 7S8 and are system design


0$1 (nstallation-;uide



!y %riend, 2et me share a simple material about 7S8 installation. However it will only %ocus on

designing the system. This is become my frst posting in this year. Hope%ully, this material is use%ul

%or you in developing the system around 7S8.

!y %riends, in developing or reviewing the 7\I", especially in 7S8 system, please consider the

%ollowing 4

!+ 0$1 normall 6e installed close to protected e<uipment+

The sa%ety valve is installed %or protecting the e+uipment, so that the closer to the e+uipment is

better. &or e(ample, %rom sa%ety point o% view, to install the 7S8 directly at vessel is better than at

line outlet vapor.

. 7S8 %or vapor application at vessel shall be connected to the vessel in the vapor space higher

than HH22.

1. () an demister/ 0$1 shall 6e connected to the vessel at 6elow the demister since there

is any potential blockage o% demister, and i% the 7S8 be installed at downstream demister, it willnot protect the vessel in that case.

:. 7S8 may be installed at outlet vapor line. &or this case, take care with the 1F pressure drop

limitation.



0. &or inlet line, please consider the total pressure drop o% inlet line less than 1F o% set pressure o%

the 7S8.

This re+uirement #1F$ is to avoid chattering on 7S8.

J. (nlet 0$1 shall =)ree draining= to avoid li<uid accumulation+

7lease consider put '>T* on the 7\I", and as process engineer, you should check the fnal design

by 7iping engineer.

L. Bloc9 valve at inlet line > i) an? shall 6e @A and @C )or spare 0$1+

Bloc9 valve at outlet line >i) an ? shall 6e @A



U. i% discharged to )T! # please indicate on 7\I" )S2 #at sa%e location$ instead o% )T!$. The block

valve is not re+uired. The weep hole to 6e provided at lowest point o) discharge pipe to

drain the li<uid+

W. >utlet line to be designed with !ach 'o ] /.L

Check the momentum # Bho v $ i% the value o% #Bho v$ ^ //./// 7a..please take care since it

may be vibration. Psually 7iping engineer will also check the re+uirement o% piping support.

=ack pressure to be considered based on 7S8 type.

'oise %or emergency case typically more than U0 d= is still accepted.

Check the two phase ow pattern, i% there is any slug ow please take care, you must state in the

7\I" so that 7iping *ngineer will consider the strengthening o% support.



3/. >utlet line shall onl have downward elevation change to the are subheader.

33. Consider minimum distance )rom 6 pass 6loc9 valve to discharge line )or heav

hdrocar6on service+

3. @ateral connection to 6e designed with " deg or deg to the header+

31 7lease consider, >ther system may have specifc re+uirement. Think about it by yoursel%.



I think that all, that I can share to you by today.

!y %riend, how are you today6 It;s been long time I don;t update my blog, really sorry guys. I was

very busy doing my pro-ect X urea plant in 7alembang. Thank you very much %or many o% you

who9ve contacted me via email and give me %eedback and muchmuch spirit %or me to keep

updating this blog Nagain;

Bloc9 Autlet Case

!y %riend, regarding the overpressure protection system design, one o% important responsibility %or

process engineer is determining the cause o% overpressure. In this posting, we will discuss whether

a kind o% case is possible occurred.

There are many causes o% overpressure. *very possible overpressure condition shall be reviewed to

ensure overpressure protection device capacity is ade+uate to protect the system. The %ollowing

picture presents the causes o% overpressure.



!y %riend, let discuss how block outlet is applicable to be considered.

#other cases in ne(t posting.$

B@AC5ED A:T@ET

&or review whether this case applicable or not, check < is there any valve that possible closed6 >r

any control valves which is N%ail to close; position6 is there any ob-ect that possible block the uid

ow 6 I% Y*S, it may be blocked and lead to pressure increase.

The re+uired load capacity o% 7S8 can be determined based on the mass balance in the system

a%ter block outlet occurred.

Here is -ust simple case.



&or system which comprises o% many vessel Q separator with same design pressure, block outlet

case is only considered %or 7S8 in the most upstream location. 5hen block outlet occurred at

downstream the system pressure increase lead to open 7S8 at the upstream.

I think, it is better to e(plain by this picture

There%ore, %or system with same design pressure, the block outlet only applicable %or one 7S8 only.

!y %riend, that;s all, I can share to you today. Hope%ully, this simple article use%ul %or you.



Beu( &ailure Case

Beu( %ailure case is the ma-or case o% all tower or column. Beu( %ailure can be caused by the

%ollowing<

7ower %ailure lead to reu( pump o

7ump %ailure Control valve %ailure # %ail closed or stuck closed$

>perator error etc

5hen reu( pump %ail, there no reu( ow inlet to column. =asically the reu( ow is own back to

column #at recti%ying section$ to increase separation e?ciency. >ther that, the reu( ow is use%ul

also %or cooling the vapor ow to the top o% the column. Beu( pump %ailure cause N2oss o% Cooling;,

as a conse+uence, the vapor temperature will not be cooled down and overpressure will occur.

Control valve %ailure has same eect as a pump %ailure.

7ower %ailure case cause trip o% the air cooler, cooling %ailure and overpressure will occur.

=ased on my e(perience, the simplifcation o% the calculation load capacity %or reu( %ailure has

been made. ndtray vapor ow is assumed as basis load capacity. In HYSYS, the column model



never convergent i% the reu( ow /, there%ore in HYSYS simulation, input reu( ow with very

small number. )%ter column model convergent, we can use nd tray vapor ow as load capacity o%

the 7S8.

)ctually, the determining load capacity o% reu( %ailure is very comple(. 2et imagine, once the

pressure increase in the column, the boiling point o% uid in the reboiler increase, the number o%

vapor will decrease. It has many eects to separation and e+uilibrium condition.

I have %ound much discussion regarding reu( %ailure case, especially the method %or determining

load capacity. Pn%ortunately, until now, there is no agreement reached, no method f(ed, there%ore I

can;t make a -ustifcation which one is the best approach %or calculating load capacity %or reu(

%ailure.

7lease %ollow the %ollowing link %or more discussion about reu( %ailure

3."iscussion 3 %rom Cheresources

."iscussion %rom Cheresources

1.) sample dynamic method %or determining load capacity reu( %ailure

!y %riend, please share to me i% you have other method %or calculating load capacity %or reu(%ailure case

Thank %or reading, hope%ully this posting use%ul %or you

Tube Bupture Case

Tube rupture may be occurred %or shell and tube heat e(changer type. =ased on my e(perience,

some design philosophy using Q1 rule, and some other using 3/Q31 rule %or the criteria o%re+uirement 7S8 %or tube rupture. 5hich one is correct6 Should we apply Q1 or 3/Q31 rule6

Tube rupture is possible cause overpressure in shell and tube heat e(changer type i% Test 7ressure

o% 2ow 7ressure Side 2*SS than design pressure o% High 7ressure Side. That is where Q1 and 3/Q31

rule coming %rom. &or e+uipment with test pressure 3, 0 ( design pressure, the Q1 #3/Q30$ rule

is applied, whereas %or test pressure 3, 1 ( design pressure, the 3/Q31 rule is applied

http://www.cheresources.com/invision/topic/2937-reflux-failure-case/


http://www.cheresources.com/invision/topic/8817-psv-sizing-for-distillation-column/


http://iom.invensys.com/EN/pdfLibrary/AppSolution_SimSci_DYNSIMColumnReliefAnalysis_01-10.pdf









Blowdown 0$1

I can9t sleep again, till this midnight , almost //.// 7! #or )! 6 $, I -ust con%use what to do. I think it

will better %or me to do something use%ul. =ut what 6 I don9t get any idea. Pntil fnally, I remember

this blog.

!y blog save my brain <".

Psually, when I don9t know what to do, the most o%ten I decide is check my %acebook, -ust want to

know my %riend9s status.



>h !y od,,why I share my %eeling to you.,I should share my knowledge instead o% my %eeling, right

6 I am so sorry.

!y %riend,there are many terminology related to 7S8. I give you a summary as %ollowing picture.

7lease comment to this posting %or only item which you don9t understand <",

I will answer your +uestion later.

ho,, ho,, sorry %or this disappointed posting,,hope%ully you understand me, I am a%raid will make

many mistakes e(planation in this midnight.

>kay,,I give one item only. I will e(plain you about =2>5">5'.

=lowdown is dierential pressure #can be in percentage$ between set pressure and re-seat

pressure. 0$1 start to open at set pressure and will ! 6ac9 closed at re-seat

pressure. 5hat do you think, why the re seat pressure lower than set pressure 6 and 5hat impact

blowdown value to 7S89s operation 6

See the picture below

)nd this picture below will show why the re seat pressure is lower than set pressure.





>ne o% the eects o% gas blow by can be overpressure in a downstream component. >% course, the

low level is the detectable condition that indicates gas blowby may be occurred.

This picture below show gas blowby case

&or instance in a separator system, to determine whether this case applicable or not, check design

pressure o% the e+uipment in downstream control valve. I% the design pressure is e+ual, the gas

blowby will not cause overpressure in the downstream e+uipment. Conversely, I% the design

pressure o% the downstream e+uipment is lower than the upstream e+uipment, when control valve

%ail, the gas will ow to the downstream e+uipment and cause overpressure.

Please note, gas blowby case is applicable if only the design pressure of downstream equipment

lower than the upsteam pressure.

The re+uired load capacity o% 7S8 can be determined based on the mass balance in the system

a%ter control valve %ail is occurred.

Here is -ust simple case.



The most o%ten +uestions which is raised by new process engineer in calculation load capacity gas

blowby is the %ormula %or determining the ma(imum ow through the control valve

The ma(imum ow through control valve is depends on the control valve Cv and the pressure drop

across the control valve. The %ollowing picture show an e(ample %ormula %rom one o% control valve

8endor4 please note, each 8endor has their own %ormula that might dierent in each other. I give

this %ormula to show you what the variable that has aect to owrate through the control valve

!y %riend, %or detail calculation in pro-ect stage, I suggest you to use I'STBPC)2 so%tware %or

determining the ma(imum ow through the control valve.

)%ter the ma(imum ow through control valve is calculated, the re+uired capacity o% 7S8 at

downstream e+uipment can be calculated #%or simple, 2oad !a(imum ow stream 1 X 'ormal

ow stream :$



Thank you %or reading my %riend, that;s all I can share to you today. Correct me i% I am wrong

Selamat liburan pan-ang <" , semoga ada pencerahan hidup yg berman%aat yang kita dapatkan... #

don9t know how to e(press this in english <"$

0$1 $iing



SiAing o% pressure relie% valve is clearly e(plained in )7I ST" 0/ part 3. Some %ormula is provided

%or calculation the orifce siAe either %or vapor or li+uid application, critical or non critical condition.

!y %riend, there is no di?cult thing in siAing 7S8 as long as all data re+uired is already provided.

The all we need is -ust inputting the data in the %ormula or in the spreadsheet <" , then select the

appropriate standard orifce siAe # re%er to )7I ST" 0J$

The %ollowing pictures, show the %ormula in )7I ST" 0/





&or non critical condition, the outlet pressure has impact in siAing orifce area but %or critical

condition is not. In critical condition, the outlet pressure is lower than the critical ow pressure

there%ore there will be no impact to the siAing.

'ote this, in critical condition, the ow through the noAAle will be inuenced by the inlet pressure

#relieving pressure$. In non critical condition, the ow will depend on the dierential pressure

between inlet and outlet. #Believing pressure X back pressure$.

Pse correct %ormula D

&or you who looking %or the siAing e(ample, click the %ollowing link.

3. asQ8apor Critical and an e(ample %rom 2eser

. as Subcritical and an e(ample %rom 2eser

1. Steam and an e(ample %rom 2eser

:. 2i+uid, an e(ample %rom 2eser


http<QQprocesseng.blogspot.comQ/3:Q3/Qsthebcstream.html

http://www.enggcyclopedia.com/2011/11/pressure-relief-valve-sizing-calculations-critical-gas-flow/



http://frame.leser.com/engineering/file/EHB_en_file_7.4.3-Gases-and-Vapors-Critical-Flow.pdf

http://www.enggcyclopedia.com/2011/11/pressure-relief-valve-sizing-calculations-subcritical-gas-flow/


http://frame.leser.com/engineering/file/EHB_en_file_7.4.4-Gases-and-Vapors-Subcritical-Flow.pdf

http://www.enggcyclopedia.com/2011/11/pressure-relief-valve-sizing-calculations-steam-service/

http://frame.leser.com/engineering/file/EHB_en_file_7.4.5-Steam.pdf

http://frame.leser.com/engineering/file/EHB_en_file_7.4.6-Liquids.pdf

http://process-eng.blogspot.com/2014/10/sthe-b-c-stream.html


http://frame.leser.com/engineering/file/EHB_en_file_7.4.3-Gases-and-Vapors-Critical-Flow.pdf


http://frame.leser.com/engineering/file/EHB_en_file_7.4.4-Gases-and-Vapors-Subcritical-Flow.pdf

http://www.enggcyclopedia.com/2011/11/pressure-relief-valve-sizing-calculations-steam-service/

http://frame.leser.com/engineering/file/EHB_en_file_7.4.5-Steam.pdf

http://frame.leser.com/engineering/file/EHB_en_file_7.4.6-Liquids.pdf

http://process-eng.blogspot.com/2014/10/sthe-b-c-stream.html

Documents

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