- 2 days ago
In this session you will understand
1. Scenarios of Overpressure & how to calculate relief load for that
2. Relief load for Control valve Failure case
3. Relief load for blocked outlet
4. Reflux failure
5. Tube rupture
6. Cooling water failure
7. Fin fan cooler failure
1. Scenarios of Overpressure & how to calculate relief load for that
2. Relief load for Control valve Failure case
3. Relief load for blocked outlet
4. Reflux failure
5. Tube rupture
6. Cooling water failure
7. Fin fan cooler failure
Category
📚
LearningTranscript
00:01Hi friends welcome to the channel we are in module 1 for pressure relief falls and this
00:06is the session 3 again for the relief load for non-fire contingencies and this is a part 2 we
00:13are completed part 1 with non-fire contingencies and the first case is the fire case contingencies
00:20so let's start disclaimer for our channel and the content you shown in this videos
00:28the next scenario is the tube rupture or heat exchanger failure so in this case before
00:36knowing the formulas for calculating the tube rupture or the flow through the tube rupture or
00:42heat exchanger failure there are few exclusions in which we do need to no need to consider the tube
00:49rupture or a exchanger failure cases for relief load calculation as per epi the first rule is that
00:56if the exchanger is designed by 10 by 13th rule then you no need to calculate the relieving
01:02rate for the tube rupture case so what is 10 by 13th rule rule if the low pressure side is
01:09equal
01:10to or greater than equal to or a greater than 10 13th of the design pressure of the high pressure
01:16side
01:17then it considered 10 by 13th rule and no need to calculate the relieving rate for two rupture case
01:24also there are two more criteria which are excluded if it is for the tubular reactor or a waste heat
01:35boiler
01:35with a tube of 1.5 or 38 mm and larger in diameter which the tubes have wall thickness equal
01:44to the
01:45pipe process piping thickness and which are also welded to the tube sheet if this this type of tubular
01:52reactor is there then for that tube rupture case not to be considered also double pipe pipe exchanger
01:59tube rupture case not to be considered however if it is a multi tube double pipe multi tube exchanger is
02:06there then it is not excluded the tube rupture case can be considered in that
02:14similarly there are exclusions for shell and tube also however shell and tube exchanger is excluded
02:20only if the following all eight criteria criteria are meet so these are the listed criteria if you
02:28don't know whether these criteria are fulfilling or not then don't exclude your shell and tube exchanger
02:35from the calculation do the calculation if any one of this criteria if you don't know
02:41so first criteria is tube vibration is not likely not likely based on the rigorous tube vibration
02:47analysis tube wall thickness is at least one standard gauge thicker than the minimum required for the
02:54specific material the equipment strategy must be specially recognized in application of 6 mm corrosion hole
03:02concept the tubes are not subjected to erosion the tubes will be operated at temperature warmer than
03:08minus 101 degree centigrade the tubes are not subjected to fatigue or grip the process fluid will not cause
03:18aggressive corrosion degradation of tubes or tube sheets
03:24an appropriate tube inspection program will be developed for the exchanger bundle in consultation with
03:31the material engineering specialist if all these eight criteria are met then only you can exclude the shell and tube
03:37exchanger from the tube rupture or air exchanger filler case or other otherwise if it is a design for 10
03:44by 13th rule
03:48So, how to calculate so when the leakage whenever the leakage considered in any shell and tube exchanger
03:58it is a 6 mm hole it is a 6 mm hole or 0.25 inches to be considered and
04:03when you want to
04:05design a safety wall for the tube rupture case in that case two open ends of the tubes to be
04:12considered for
04:14the as a leakage as a leakage and that also from the means it is a single tube from break
04:23from the or cut from the tube sheet
04:27so basically it will be like a single tube which is getting cut from the tube sheet so it is
04:33a two
04:34opening two opening of stoop or two open cuts of the single tube to be considered and if it is
04:41leakage it
04:42is only a 6 mm of hole to be considered so calculating maximum flow rate the normal process flow on
04:51the low
04:52pressure side has to be stopped it has to be assumed that when we are calculating the relief load normal
04:57process flow on the lower pressure side has stopped and the pressure difference is
05:04between that the lower pressure side and the maximum operating pressure of the higher pressure side
05:11and design or a set pressure of the lower pressure side so
05:18the difference has to be calculated like p1 is the you can say pl is a lower pressure side and
05:24p2 will be
05:25difference is sorry p2 will be your out the higher pressure side pressure and the lower side pressure will
05:33be the set pressure of your psv so that is the difference between the pressures the maximum pressure will be
05:40the p2
05:41at the high pressure side and set pressure will be the
05:47the p2 pressure or a lower side pressure on the lower side pressure
05:54and the flow capacity for both side rupture tube is defined based on the single orifice equation
06:00with a discharge coefficient of 0.7
06:06so to calculate the liquid relief or you can say liquid flow from the two or one tube failure with
06:15the two openings it is calculated for the liquid flow rate it is calculated as relief load is equal to
06:220.7 into
06:23the low area square root of 2 into p1 minus p2 into density of p1
06:33similarly for the critical flow rate it is calculated already we know the critical flow rate
06:39critical flow rate is p2 is less than 0.5 of p1
06:43so p1 is the absolute upstream pressure based on the maximum operating pressure that is a high pressure side
06:49and p2 is your set pressure of the psv and density is the density of the fluid which is leaking
06:58in the lower pressure side
07:01and if it is a flashing liquid or a vapor which coming out of the tube leakage and if it
07:08is a critical vapor flow
07:09we have to see the pressures so p2 will be always the set point and p1 is always the maximum
07:15pressure on the
07:17higher pressure side so that formula to be used is w is equal to 0.7a square root of p1
07:24into rho 1 into k
07:26and k is basically a uh it's a ratio of specific heats so when when you consider this ratio of
07:35specific
07:35heat as 1.4 sorry 1.4 then that formula will be get simplified to 0.7 into 0.685
07:42into
07:43area this area this area is 2 cross sectionals of the your liquid tubes so always this area will be
07:54your cross sectional of 2 with 2 tubes
08:00k is the ratio of specific heats so w is relieving rate a is the cross sectional area across so
08:07everything is
08:08mentioned also here what if i miss something to talk it is already mentioned the units are also
08:13mentioned so unit has to be followed specifically which are the given units only
08:20and for non-critical vapors when the p2 is not less than half of the p1 then
08:28uh vapor flow rate will be calculated as 0.7 into y into a into square root of 2 into
08:37p1 p1 is again the
08:38highest operating pressure of the higher pressure site and p2 is the set pressure and p1 is the density
08:45of high pressure of high pressure side fluid and to calculate y this is also uh based on the your
08:53cp by cv so y is calculated based uh based on the this formula and this formula is based on
09:02the api only
09:04so r has to r is the ratio of uh set pressure to the highest inlet pressure or the high
09:11pressure side
09:12uh pressure so w is your relieving pressure a is your cross sectional area so whatever cross sectional
09:19area comes into two that will be the further as i told for the two tubes we have to consider
09:24one tube
09:25fail which will give you two openings then p1 is absolute upstream pressure based on the maximum
09:31operating pressure of higher pressure site and p2 is your psv set pressure r is your p2 by p1 which
09:38is
09:38the ratio and k is your specific heat ratio and rho is your density so based on that you can
09:44calculate
09:44and then put this in this formula you will get your relieving rate for the tube leakage case again unit
09:51has to be maintained you have to see the what are the units and accordingly you have to calculate
09:57you can make these formulas in excel and then it will be easy to calculate
10:03so similar to the shell and tube heat exchanger plate type and frame heat exchanger also can leak
10:09however the pin um the leakage is only because the plate type heat exchanger does not have any tube so
10:15it it's it has to be considered as a pin hole of 6 mm hole and 6 mm hole the
10:22the same formulas which are
10:24described in the shell and tube exchanger to be considered in place of area only you have to consider
10:30the area comes or based on the 6 mm hole gasket failure also to be considered if the gasket failure
10:39is
10:40considered then the orifice size has to be calculated in the same way as the shell and tube exchanger only
10:48the area has to be considered based on the gasket also the orifice size should be calculated for hydraulic
10:54equivalent a rectangle opening of 1 by 16 inch or 0.0625 meter square to be considered with length equal
11:05to the diameter of relevant inlet or outlet or header so when gasket failure comes the opening has to be
11:15considered 1 by 16 inch wide and the length is to be considered as an inlet of the either inlet
11:23or outlet
11:23diameter of the that particular exchanger based on that size and all the formula which is used in
11:31shell and tube exchanger will be used in plate and frame exchanger only the area has to be changed based
11:37on the 6 mm hole or based on the gasket opening or this rectangular opening a square will be changed
11:45power failure stream or electrical failure the load the load calculation for this case is very typical
11:52uh generally individual PSVs are not depending on power failure case the power failure case when it
11:59decided it it will have a impact on multiple PSVs multiple equipments so there are few guidelines how
12:07it has to be minimized and basically power failure case or a partial power power failure case is considered
12:16only while designing the unit so while that designing that unit you can change some some substations or
12:24something so that will if that one substation get stopped or tripped or grounded then it will have a minimum
12:33impact on your overall process and relief load should not be very high so what are this like individual failure
12:40of
12:40power supplies to any one item consuming equipment such as motor driver or pump or fan compressor so if it
12:47is a local power failure considered like compressor dish compressor trips because of the power failure then
12:53it can be individual PSV can be you can consider the vapor flow rate if you are condensing the vapor
12:59if
12:59it's a boil off compressor and it's get tripped so your boil off will be increased but generally boil off
13:04is a
13:05atmospheric tank if it is it will not be from the pressurized tank so there is also a PSV requirement
13:11is not required the total failure of power to all consuming equipments in a process unit supplied by
13:20the unit substation if a complete substation get tripped the whole unit get tripped then there is an impact
13:26on the multiple PSVs on the maybe multiple places you have to relieve the relief and accordingly you have to
13:33calculate that general power failure to equipment supplied for any one bus bar in a substation
13:40service serving one or more process unit note that some substation designed to include hereditary
13:47for the bus bar so like as i said when it's a power filler case we have to study the
13:53your substation
13:54as well well from where we are getting the power for the equipments
13:59so then we have to see that each each unit should get a power from a two or three substation
14:07which then if you are taking power from the two or three substations then it is likely it is unlikely
14:14that all the three substation trips are having a problem at the same time so your relief load will be
14:20always minimized in case of bus bar contingency the basic assumption for the contingency is
14:27ground fault in bus bar does the impact of it will have a equipment with a effect by a design
14:34or a subtraction of protective equipment provided so that will be calculated based on which equipment
14:40is getting filled and then for that individual PSVs to be considered some stations some substations are
14:49designed with a normally closed circuit breaker isolating adjacent bus bar when these are fed from
14:55the same electrical feeder when a ground fault occurs in a bus bar this circuit breakers open thus
15:02isolating the fault and preventing the ground fault from the exiting to the bus bars and perhaps causing
15:09the complete substation fail so plant-wide failure comes then following general power failure on plant-wide
15:22scale must be considered failure of purchased power supply from the plant so if you are purchasing power
15:29from the outside or the grid that that can be failed failure of in the internally generated the CPP
15:36captive power plant is there if the power stops from that that also to be considered the total power
15:42failure for any major substation there are multiple substation in any process unit so any major substation
15:48considered to be considered for plant-wide power failure and based on that you can design your relief loads
15:56but that will be like for the multiple cases
16:00the total electrical electrical power failure may be result in loss of sea water cooling water steam
16:06instrument air and utilities or electrical driven equipments
16:14in case parcel power failure equipment that are not affected by the failure concern will be considered to be
16:19remain operational and control will be assumed to be operated as design so if any equipments which are not
16:29run on the power like steam turbine driven pumps or steam turbine compressors they will not have any impact
16:36because of the power failure so while calculating the relief load for this power failure case or partial power
16:44failure case this we have to check
16:51so to minimize this low relief load in case of power failure you have to check that as I said
16:58the which
16:59substation we are getting the power we are getting the power and if better to give the power from the
17:04multiple sources or you should have a standby pump or standby machine or standby rotary equipment with the
17:12turbine driven so if power fails the turbine driven can work still work and you will not have a higher
17:19relief
17:20loads or relief conditions
17:26instrument air failure generally instrument air failure there is an inventory always there in
17:33instrument air which which is generally may be 15 to 20 minutes and that will allow your safe shutdown of
17:40the
17:40plant and as you know all the control walls
17:44our fail safe positions has been given so if they that wall get fail safe position then that will
17:51not increase your relief load so that's why instrument instrument air failure
17:56total instrument air failure will always will have a 20 minutes or 10 minutes or 15 minutes backup
18:01which will allow your control walls to be go to the safe positions
18:10the failure position of the control wall upon the instrument air fail shall be specified
18:14such that the potential hazards including over pressure are minimized
18:19it shall be assumed that upon partial or a total loss of instrument air
18:22all the control walls affected by the failure will assume their specific failure position
18:28so control wall that specified to initial fail stationary shall be either assumed to be dirty or either
18:37specified ultimate failure position or assumed to be remain at the last controlling position
18:43whichever condition is more restrictive for over pressure protection
18:46so in basically instrument air failure case
18:50there will not be increasing in your relief load but however this case has to be considered
18:56for the impact then cooling water failure
19:03in cooling water failure normally individual or a process event basis
19:07for the pressure relief to be consideration because cooling water is also fail if it is a local fail
19:13then it's not in problem
19:16the following design contingency shall be considered as the basis of violating overpressure that can result
19:21from cooling water failure if it is as i said if it is individual failure
19:25water supply to any one of the cooler or condenser will get stopped and if it gets stopped that
19:31individual load on that cooler the vapor loads coming because of the non-availability of that
19:38condenser that has to be considered for the relief load calculation
19:42if the total failure of any lateral
19:45supplying the process unit that can be isolated from the upside main
19:50or consider then consideration for the plant weight failure to be considered if
19:54if the complete cooling tower fails then there will be a
20:01there is no cooling available for the condensers so that will have a lot of impact and then based on
20:10that it has to be tripped or each individual vapor load for that condenser to be calculated which we
20:17already seen for the parcel condensing and all in the earlier slides so in that way so total
20:23failure also when we checked it for that also generally for a very critical exchangers we are
20:30giving the supply from the different different cooling towers so always provide a supply of cooling
20:36water not from the single cooling tower it should be from the multiple cooling towers or there should
20:41be a jump force between the two cooling towers so in case of emergency cooling water from one cooling
20:47tower can be transferred to the other cooling tower which is having more critical services
20:55the following general cooling water failure shall be considered failure of any section of the
20:59upside cooling water main and loss of cooling water pump that would result in any design
21:04contingency in utility system supplied by controlling the pump drivers even sometimes the cooling water
21:12goes to the seal system of the mini pumps because of whether maybe that unit may not have any condenser
21:22but
21:22if that pumps get trip that also will increase may increase your relief load so that way also to be
21:30considered the cooling water failure if it is going for any sealing system of the pump or compressor
21:38relief load calculation can be done based on the following total condenser the total normal
21:43incoming vapor and partial condenser normal condensing rate if it is for the condenser but i said it
21:49it can be going into some seal plants so if the pump drips then the next is the abnormal heat
21:58input
22:00the required capacity of the maximum rate of a vapor generation at relieving conditions
22:07less than the rate of normal condensation of the vapor so basically abnormal input is
22:13if you are you have a reboiler and the column and a reboiler and reboiler getting heat from the steam
22:25and steam having a control valve and it is getting it controlling the steam flow rate either from the
22:31condensate side or from the steam side if this valve fails open it will give additional inputs to your
22:37exchanger and this will be the abnormal heat input for that re-boiler so every case of potential behavior
22:45has to be studied and how to optimize the abnormal heat input the few few points are there like design
22:55value should be used for the for an atom for such walls like if my required cv for this re
23:04-boiler is
23:05for this control valve or steam valve coming to the re-boiler is suppose 50 i should not buy any
23:12valve which is
23:13giving like 100 or higher size like 150 which what it will happen once it fails open it will give
23:21a very
23:21high abnormal heat input and which will gives a very high vapor load so while such typical walls we can
23:33we can select when we want to select we have to select it as a optimum and your control valve
23:39when we are
23:41designing the relief load you have to consider not only the vapors coming from the tower you have to
23:48consider this abnormal input also for your psv sizing on the vapor side of the column
23:56also build up capacity to be used for burners and heaters like generally what happens we will take
24:03like burner for 110 percent capacity or 120 percent capacity and we use it at 100 percent so instead
24:11of that we should we should use that capacity means always it should not have an additional capacity
24:19available in that so when the additional capacity available it can be a source of additional abnormal
24:24heating
24:28where the limit stops are installed and in some cases if if suppose like in the same case of re
24:35-boiler
24:35if you have calculated cvs 50 your control valve is normally operated at 50 cv and if you have selected
24:41the cv of 150 then what you can do to avoid a failure case you can put a mechanical lock
24:48to this
24:50control valve so and and that control the limit may be around the 70 cv or some percentage like 60
24:5870
24:58percentage so in that case if you provide a mechanical lock this will not allow the additional abnormal
25:04input going to the re-boiler so that study has to be done during initial stage of the project only
25:13or if you find that this is not done so you can propose that also because mechanical stop can be
25:19added in the control valve so that will reduce your relief load also the shell and tube exchanger
25:27like heat input should be calculated basis on the clean rather than the foul condition generally when
25:32we design a heat exchanger we will calculate on the fouling condition we will give the fouling factor
25:38and then we will calculate the heat required or the heating material required based on the fouled
25:45condition not on the clean condition but these are the practical issues because if you
25:50calculate it on clean services and if the exchanger fouls then you will have a
25:57lower throughput of the plant so that's why you have to consider the this may not be feasible however
26:07some mechanical lock can be provided to avoid abnormal heat input from the control valves
26:14and if you put a mechanical lock it has it has been considered in api
26:19you no need to then calculate your calculate you no need if any
26:29mechanical stop is provided on the control valve then the control valve
26:33the flow rate from the control valve to be calculated on that mechanical lock position only not on the full
26:39cv
26:43check wall generally check wall not effective for preventing over pressure by reverse flow or a high pressure source
26:53experience indicates that substantial leakage through the check wall is happens
26:57so that's why check walls are not the not stopping your reverse flow or over pressurization
27:03the following guidelines are applied for the check walls if the relief device is not required to protect the piping
27:12against potential over pressure caused by reverse flow
27:16if the pressure of the high pressure source does not exceed the short term allowable piping pressure
27:21the short term allowable piping pressure generally be the 133 percent so you should not require a safety wall if
27:30your
27:34the high pressure the high pressure side is not more than your lower lower pressure side is designed as 133
27:41percent so again the 10 by 13 rule it is coming
27:46so the short term allowable over pressure for piping is a generally 133 percent of the maximum continuous pressure
27:56pressure and also if the pressure relief valve does not required to protect the any pressure vessel against
28:01the potential over pressure caused by reverse flow if the pressure of high if the pressure of the high
28:09pressure source does not exceed the maximum level working pressure so you no need to put considering the
28:16NRV failure and reverse flow and the vessel will be pressurized if it is not if the source pressure is
28:25lower than your MAWP similarly for the pipeline if your source pressure or the high pressure side is
28:32lower than 133 percent of your piping design then you no need to put in a safety wall
28:44so check wall malfunctioning for piping pressure vessel or a pressure relief device may be required to
28:52protect against the potential over pressure caused by over pressure for which case it is required if a
28:57partial failure of a check wall if you are having only a single check wall then assume that the failed
29:05check wall behaves as a restricted orifice generally any check wall if it is a fails it will act as
29:11a
29:11restricted orifice and diameter equal to one third of the nominal diameter of the check wall
29:18that should be the your orifice size and based on that orifice size you can calculate the
29:24flow rate going through the or reverse flow going through the NRV
29:29so the same formula for the tube tube rupture the orifice size the area will be one third of the
29:36nominal diameter of check wall
29:40and the partial failure of one check wall or two or more check walls like if the first check wall
29:46if the single check wall is available then you have to consider one third of nominal diameter of the
29:50check wall if the two check walls are provided in the series and that is also if it is
29:58dissimilar type then no need to consider but if it is a similar type then from the first wall it
30:05is
30:05one by third of the diameter and from the second wall is one by tenth so if two walls in
30:10series
30:11you have to just consider one by tenth size of your orifice restricted orifice or a area and then you
30:19can
30:19calculate the relief load also but you have to see that your higher pressure side
30:27which is which is protect higher pressure side is protected by NRV is higher than
30:32your 133 percent of your pipeline pressure or your vessel is the source pressure is higher than your
30:41MAWP of that vessel
30:45if it is higher then you have to consider these two cases for the orifice calculation and from the orifice
30:52calculation you can calculate the relieving load based on the same formula
30:58so loss of heat in series and fractionation system
31:03so generally in like in a gas plant the fractionations are in series
31:08so if heat input lost to the fractionator means suppose it is a de-methanizer and it is having a
31:18reboiler so what this reboiler does and then this is going to the second column as a feed
31:25so if it is a de-methanizer this will be de-ethanizer this will be de-methanizer so if the
31:33steam to this
31:35reboiler stops now this time it stops so when this steam stops to this reboiler the lighters from
31:44from the column will get dissolved in liquid and this liquid will be sent to the second fractionator
31:52and when it goes to the second fractionator the lighter loads will be increased in the second column
31:59and the condenser of the second column or a fractionator is designed for certain vapor load
32:08however because you have a you have a failure of your steam in the first fractionator the lighter
32:15goes into the second fractionator and it is increasing your load so that is the
32:22uh heat increase or loss of heat due to fractionation so sub so you have to understand that
32:29not always heat input is a problem if it stops also it may create a problem in refinery or in
32:36a gas unit
32:40it is possible that the loss of heat input to the column
32:44to overpressure is the following columns loss of heat result in a in a sum of the light ends remains
32:51in the bottom and begins to transfer to the next column as i said then under these circumstances overhead
32:58load of the second column will consist of the normal vapor load plus the light ends from the first column
33:05if the second column does not have the have the condensing capacity of the additional vapor load
33:11then excessive pressure could occur so if this condenser is not having the capacity then second
33:16column will be pressurized
33:21and then you then this psv on the second column has to be considered this vapor
33:30liquid overfill generally two scenarios must be considered the first is liquid overflow stops
33:37while liquid inflow continues so outflow is stops but inflow is continuous so that will
33:42uh the the case of liquid overfill or liquid inflow increases above the design flow rate
33:49for example due to control wall failure again or while liquid overflow continues to be a normal turn
33:55down rate and inlet is higher
34:00to determine the required relief capacity of a pressure relief device credit may be taken for a
34:05flow through normal open process channel that are not likely to be become a partially or totally blocked
34:12as consequences of overfill so while considering the overfill it can be happen in a tank or in a
34:19surge vessel if the inlet control wall fails open this can overfill your vessel
34:30so overfilling may require relief relief rate for your psv
34:39such as for example if steam drum is balanced directly on the steam collection header without any
34:45inverting uh control walls a failure of level control wall for a full open position will eventually
34:53cause the drum to overfill but credit may be taken up for the capacity of the steam piping for handling
35:00the
35:01combined flow of incoming water plus design the stream so basically liquid overfill happens in
35:10either it can be happen because of your outlet like if your outlet get stopped this wall get closed then
35:17this overfilling can happen so for that overfilling you have to calculate the relief flow that is also
35:23based on the judgment then last is thermal expansion or thermal relief wall
35:34thermal expansion is increase in liquid volume caused by increasing in temperature
35:39the most common cause is piping or a vessel are blocked while they are filled with a cold liquid and
35:48subsequently heated by a heat tracing or coil or ambient heat gate so if suppose any pipe is there
35:55that pipe is steam jacketed or it can be electrically heated so suppose it this line is get blocked
36:06from both end and your electrical tracing has not stopped by because of some fault then there is a chance
36:15of
36:16thermal expansion happen in this line and to relieve it you need a thermal expansion wall
36:23also an exchanger block on the cold side like if you have a cooler
36:32and at the cooler outlet suppose on the this is inlet this is outlet of for the cooler and then
36:37this is
36:38the process inlet process outlet whatever so if suppose your cooling water get isolated inlet get isolated
36:46and however your the process fluid is still passing through the exchanger then this will hit your
36:55water available in the exchanger and that need a thermal expansion so you have generally seen wherever
37:01the cooling water are supplied to any exchanger there will be a thermal relief wall on the outlet line of
37:07the
37:08cooling water that is before isolation
37:13it can it required for the piping or a vessel are blocked in in while
37:18they are filled with liquid and nearly at ambient temperature and heated by direct solar radiation
37:27thermal relief wall or a thermal trv is considered three by four into one
37:34it is commonly used only for the two cases when you have to consider the relief wall requirement instead
37:42of trv that is the long pipeline of large diameter and un in an insulated above ground installation and large
37:52vessel or exchanger operating at liquid full in that two cases you may required a psv instead of thermal relief
38:00wall
38:02if the block in liquid has vapor pressure higher than the relief design pressure then the pressure relief
38:09device should be capable see if the block liquid has a higher vapor pressure than the
38:15then the relieving design pressure then it is required a psv or pressure relief system
38:22if the line length is less than 100 feet or 30 meter then thermal relief may not required
38:31for a liquid full system expansion rate for the sizing of relief device
38:36generally we use three by four into one as a standard for thermal relief design however there is a
38:42top there is a formula to calculate that relief flow from that trv as well so this is the formula
38:51the
38:52v is the relieving rate at meter cube per second p is your cubical expansion for that liquid which is
38:58depending on your api gravity with that we will see h is the heat transfer rate for a heat exchanger
39:09that
39:10this can be taken from maximum exchanger duty so this particular it is for when you have a
39:15exchanger as i said on the cooling water side you have a trv here so that trv to calculate the
39:24relief
39:24load from this trv or a psv the the total heat transfer rate from this exchanger to be considered
39:34g is the specific gravity refer to the water f is the compressibility factor of liquid usually ignored
39:42because and then c is the p2 by p1 your p2 is your
39:55set pressure and p1 is your the abnormal operating pressure k is your specific heat
40:05trapped in the fluid
40:10so how to get that cubical expansion factor typically cubical expansion
40:14expansion coefficient for hydrocarbon liquid and water is
40:18at 60 degree heat it is depending on the gravity of api so based when you know the what is
40:25your fluid
40:26api gravity you can take the cubical expansion factor based on this
40:33so conclusion for this session key principles to be taken forward the majority complex scenario
40:41strictly formula fall shorts therefore the sound engineering judgment is crucial
40:45to accurate and define the required relief load so as you seen in this complete session only for the
40:54tube leakage there is some formulas otherwise all most of the places you have to do the engineering
40:59judgment for to calculate the relief load the total control flow is the sum of the selected cv
41:05capacity and unavoidable bypass flow the flow for a blocked outlet relief flow is defined
41:12as the system maximum available operating flow the relief load for a cooling water failure is based
41:18on the maximum heat input that can be no longer be removed the relief device on the low pressure side
41:25must be sized to prevent the over pressure during the tube rupture scenario and all the relief load
41:32has to be calculated based on the relieving load condition not on the operating conditions
41:38since the relief load calculation is the difference between the safe design and catastrophic failure
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41:53so thank you very much write your question and comment i will be happy to answer it
42:00you can reach us on conceptengineering2025 at gmail.com links and links for the other sessions are given in
42:10this description
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