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Psv Fire Load Calculation


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#1 ANUJ M AGARWAL

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Posted 09 June 2008 - 06:06 AM

I was determining the PSV load for fire case around a surge drum which is getting feed of a blend of three different gas oils. Though I have the ASTM D86 assays of these three streams but not able to find the right way to calculate latent heat of the blend.
1) The surge drum is at 10 ft elevation and it's height & Normal liquid height are 24 ft & 12 ft respectively. I calculated the wetted area for 12 ft normal height only, is that O.K or I should take 25ft-10ft=15ft?
2) How to find the blend Latent heat? I think to finalise 50 btu/lb is not justified. I just take the initial enthalpy H1 of the lightest gasoil at drum temperature then h1 (find out the sensible heat) required to bring it upto boiling point at the pressure upto PSV set pressure +21% . Then adjust the vapor fraction to vaporise 1 lb of gasoil and noted enthalpy H2. Now the latent heat is (H2-H1-h1). I don't know is it o.k or not?

Regards,
Anuj

#2 djack77494

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Posted 09 June 2008 - 12:52 PM

Anuj,
You should use 12 ft as the height of liquid in your horizontal vessel. By coincidence, it is also 1/2 the vessel's diameter, so I think you could use 12 ft even if your liquid level had been higher.

Regarding calculation of the latent heat, first do NOT include the sensible heat in going from normal operating temperature to the boiling point (@121% of setpoint). That is not latent heat, and that heat will be absorbed by the vessel's contents prior to the PSV popping. I would calculate the latent heat separately for all three blend components and then use the minimum value. I suspect it will be the heaviest, not the lightest, of the three. 50 btu/lb is not a good assumption in your case, since it sounds as if you can get a realistic value.
Good luck,
Doug

#3 JoeWong

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Posted 09 June 2008 - 06:05 PM

Anuj,

For your first question, i have different opinion than Doug.

As per API, pool fire would lead to flame height of 25 ft. Flame with high momentum passing the vessel body would "bend" into the vessel body and lead to high heat flux to vessel body above maximum vessel cross sectional area. Secondly by principle, you have normal level and you also have maximum expected operating level (before trip function take place), thus fire could happen when the liquid level is reaching (just before) high level tripping point.

For second question, I wish you have a simulator with you. You may consider to do rigorous method to determine the latent heat. Bring pressure and temperature upto the maximum accumulated pressure (121% of set pressure if vessel designed to ASME 8 Div II, other may have checked accordingly) by considering constant volume and mass. Then marginally increase the temperature by 2 degC, partially release vapor from vessel until the volume is same as initial volume, check out quantity of heat required to release the vapor mass. Latent heat can be estimated by dividing the heat with vapor mass.

Good luck.

#4 ANUJ M AGARWAL

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Posted 10 June 2008 - 01:04 AM

Dear Doug,

Thanks for your prompt reply.
1) Actually the drum is vertical with dia of 11 ft, TL-TL=24 ft, Normal level=12 ft & HHLL = 22.8 ft. In this case what should be the hight input for wetted area calculation, considering the elevation of 10 ft from ground. Operating temperature & pressure are 135 degF, 2 bara.
2) Actually the calculated absorbed heat by 21000F(A)0.82 is the total heat and part of it will be utilised as sensible heat so why we should not subtract the sensible heat from this total absobed heat to calculate the vapor Lb/hr mass.
3) I calculated the latent heat of all three separately and found following.
**Oil-1 of Mwt=155: At Vapour fraction=0, temperature is 378 degF and enthalpy=-781 btu/lb. At VF=1, temp. & enthalpy are 538 degF & -577 btu/lb. So it's latent heat is -577-(-781)=204 btu/lb. All these calculations are done at PSV set pressure + 21%(around 5 bara).
**Oil-2 of Mwt=324: At VF=0, temp & enthalpy are 308 degF & -814 btu/lb. At VF=1, temp & enthalpy are 917 degF & -313 btu/lb.So it's latent heat is -313-(-814)=501 btu/lb. All these calculations are done at PSV set pressure + 21%(around 5 bara).
**OIL-3 of Mwt=237:At VF=0, temp & enthalpy are 307 degF & -817btu/lb. At VF=1, temp & enthalpy are 761 degF & -428 btu/lb.So it's latent heat is -428-(-817)=389 btu/lb. All these calculations are done at PSV set pressure + 21%(around 5 bara).

So as you stated, 204 btu/lb the lowest one for Oil-1 can be taken as overall latent heat for fire case calculation. But as per mol wt, this latent heat is of lightest component but as per IBP it is of heaviest but as per FBP it is again of lightest component. I think with the rate of absorbed heat, some amount of all the three oils will vaporise so in that case what should be the relieving temperature and Mol.wt at 5 bara?

#5 ANUJ M AGARWAL

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Posted 10 June 2008 - 02:19 AM

Dear JoeWong,

Thanks for your reply.
I have written in detail about drum and feed oils in reply to Mr. Dong. Please take reference from there.
1) As per your opinion, I should consider 25 ft height but I have to subtract 10 ft height of elevation and in that case, 25 ft-10ft=15 ft height should be considered for wetted area calculation which is more than normal liquid level but less than HHLL. Is that o.k OR irrespective to drum elevation, should I consider 24 ft TL-TL height for wetted area calculation? OR Should I consider 15ft height with bottom piping of around 10 ft length to calculate wetted area?
2) Another view is that generally double jeopardy is not being considered. I mean HHLL and external fire both happen together should be considered as double jeopardy or not.
3) I am using HYSYS. Dear Joe, things are not clear but I followed you as below.
3.1) I feed the pressure upto Set point +21% in the stream of blend composition of all three oils. As operating temperature was sub cool so I put VF=0 and corresponding temp came 306 degF. At this time heat flow was 819 btu/hr.
3.2) Increased temperature to 308 degF, VF came .0004 & Heat flow = 817 btu/hr. Not much change and no flow in vapor phase.
3.3) Actually what do you mean by "partially release vapor from vessel until the volume is same as initial volume". I tried to bring VF=0.01, at this VF, temp came out 434 degF and heat flow is 739 btu/hr and mass flow in vapor is 0.2 lb/hr. Now If I subtract 817-739=78 and latent heat = 78/0.2=390 btu/lb, looks very high.
3.4) I increased VF to 0.1 then temp=604 degF, Vapor mass flow=6.2 lb/hr, heat flow=615 btu/hr. In this way, Latent heat = (817-615)/6.2=32 btu/lb which looks low. At VF=0.2, LH= (817-585)/14.3=16 btu/lb.
3.5) At VF=.027, vapour flow=1 lb, temp.=527, heat flow=675 btu/hr so LH=(817-675)/1=142 btu/lb.
I don't know what should be the right approach?
4) How to decide relieving temperature & mol wt. in this case?
5) How can we reach upto Set pr.+21% pressure in Hysys by adjusting VF & temperature as we don't know temperature & VF at relieving condition and to get this pressure, number of combinations of temp. & VF are possible.

Regards,
Anuj

#6 djack77494

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Posted 10 June 2008 - 08:34 AM

QUOTE (ANUJ M AGARWAL @ Jun 9 2008, 10:04 PM) <{POST_SNAPBACK}>
1) Actually the drum is vertical with dia of 11 ft, TL-TL=24 ft, Normal level=12 ft & HHLL = 22.8 ft. In this case what should be the hight input for wetted area calculation, considering the elevation of 10 ft from ground. Operating temperature & pressure are 135 degF, 2 bara.


I'm sorry, Anuj, for making the unjustified assumption that we were discussing a horizontal drum. I assume your liquid levels are all relative to the bottom tangent line, as is common practice. If the bottom of your head is located 10ft above grade (which is not the more common useage but is interesting to discuss) then the calculated effective area is as follows:

Allowing 1/2 * diameter for the "height" of the head, the NLL is located 10 + (1/2)(11) + 12 = 27.5ft above grade. You need to consider the wetted area UP TO 25ft in your fire imposed heat gain. See API RP521 Section 3.15.1 (1997 ed.). Since even the NLL is above the 25ft "limit", you need only consider from grade to grade+25ft.

If the bottom TANGENT LINE of your drum is located 10ft above grade (which IS the more common useage) then the calculated effective area is as follows:

The NLL is located 10 + 12 = 22ft above grade, but the HLL is located 10 + 22.8 = 32.8ft above grade. I would always consider that a fire may occur with the level anywhere within a normal operating range of the vessel. So, I would need you to justify why the liquid contents would never be between NLL and HLL at the same time as a fire occurs (which is always at the least opportune moment). Assuming that liquid can indeed be at any point up to the HLL, then you must consider the first 25ft (measured from grade) to be exposed to fire. Just to make matters interesting, let's say that you content that liquid can never be above the 12ft mark. Then the WETTED surface area that is effectively exposed to fire would be the area of the bottom head + the lowest 12ft of circumferencial area (which we note is less than 25ft above grade). The 3ft of VAPOR SPACE that is located above the liquid level but still less than 25ft above grade is not effective in heat absorption and may be neglected.

I'm sorry that this seems to have gotten more confusing than the subject really is. There are a number of cases to keep in mind, and it is difficult to express some of these concepts in words. Also, due to time constraints, I cannot address some of the other issues at this time. I believe Joe Wong's description of latent heat is quite complete and accurate. Sensible heat must be neglected basically because when the heat being absorbed is utilized to raise the liquid temperature, then it is NOT being used to generate vapors. Vapor generation governs the PSV sizing. Joe, I invite your further comments and thank you for clarifying the latent heat calc.

Regards,
Doug

#7 ANUJ M AGARWAL

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Posted 12 June 2008 - 01:30 AM

Dear Mr. Doug,

Thanks for your detailed analysis. I will consider 25 ft from grade not just NLL so the height for wetted area will be 25ft - 10 ft elevation TL-TL = 15 ft.

I am waiting for Mr. Joe reply regarding latent heat. Actually as he said to increase 2 degC temperature( I assume it is above IBP) and find latent heat but I found that with every 1 deg change in temperature, latent heat is changing so just to consider at 2 degF rise in temperature is not looking me justifiable. As you said, we can choose the lowest LH out of individual LH but More over during absorbing entire calculated heat of fire, all the three component will vaporise above there IBP so in that case how to judge the relieving temperature and Mol wt. Moreover with the blend, how to know that at what VF of blend, pressure will reach at PSV set point? After all these excercise, I am still not reach on any conclusion regarding LH.

Regards,
Anuj

#8 JoeWong

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Posted 12 June 2008 - 06:59 AM

QUOTE
I have written in detail about drum and feed oils in reply to Mr. Dong. Please take reference from there.
1) As per your opinion, I should consider 25 ft height but I have to subtract 10 ft height of elevation and in that case, 25 ft-10ft=15 ft height should be considered for wetted area calculation which is more than normal liquid level but less than HHLL. Is that o.k OR irrespective to drum elevation, should I consider 24 ft TL-TL height for wetted area calculation? OR Should I consider 15ft height with bottom piping of around 10 ft length to calculate wetted area?


I am getting confuse with the provided information at different posts.
Simple sketch tell all stories. Can you please kindly roughly draw and upload for evaluation ?

QUOTE
2) Another view is that generally double jeopardy is not being considered. I mean HHLL and external fire both happen together should be considered as double jeopardy or not.


System will not trip prior to HHLL. HHLL (just before touching) can be reached before a fire is started. Two scenario are exclusive and sequential between each and other. So, they are not double jeopardy.


QUOTE
3) I am using HYSYS. Dear Joe, things are not clear but I followed you as below.
3.1) I feed the pressure upto Set point +21% in the stream of blend composition of all three oils. As operating temperature was sub cool so I put VF=0 and corresponding temp came 306 degF. At this time heat flow was 819 btu/hr.

3.2) Increased temperature to 308 degF, VF came .0004 & Heat flow = 817 btu/hr. Not much change and no flow in vapor phase.

3.3) Actually what do you mean by "partially release vapor from vessel until the volume is same as initial volume". I tried to bring VF=0.01, at this VF, temp came out 434 degF and heat flow is 739 btu/hr and mass flow in vapor is 0.2 lb/hr. Now If I subtract 817-739=78 and latent heat = 78/0.2=390 btu/lb, looks very high.

3.4) I increased VF to 0.1 then temp=604 degF, Vapor mass flow=6.2 lb/hr, heat flow=615 btu/hr. In this way, Latent heat = (817-615)/6.2=32 btu/lb which looks low. At VF=0.2, LH= (817-585)/14.3=16 btu/lb.

3.5) At VF=.027, vapour flow=1 lb, temp.=527, heat flow=675 btu/hr so LH=(817-675)/1=142 btu/lb.
I don't know what should be the right approach?


4) How to decide relieving temperature & mol wt. in this case?
5) How can we reach upto Set pr.+21% pressure in Hysys by adjusting VF & temperature as we don't know temperature & VF at relieving condition and to get this pressure, number of combinations of temp. & VF are possible.


In my opinion, the initial system definition at 3.1 is not representing the real system.

(i-1) Define fluid in the vessel is at maximum Operating pressure (P0) & operating temperature (T0) and maximum liquid level (L0) prior to fire start. The system volume (V0) will be defined by the physical vessel dimension and piping volume. The system volume (V0) can be vapor volume (Vv0) and liquid volume (Vl0)

(i-2) Adjust the inventory (vapor mass and liquid mass) in the vessel until you achieve the vapor volume (Vv0) and liquid volume (Vl0).

(ii) Input some heat into the system to bring the the system upto relieving pressure (Pr0=121% of Pset). By maintaining system volume at V0. At this point, your system is at relieving condition. The temperature at this point is Tr0.

(iii) Next step is further input heat (Q1) to achieve 1-2 deg C above Tr1 but maintain pressure at Pr0. Normally the total volume at this point potentially higher than V0. If yes, proceed to (iv). if not, further increase the temperature (Tr1) until the liquid start to flash but maintian pressure at Pr0.

(iv) Remove part of the vapor (m1) from the system to bring the system volume back to V0.

(v) the latent heat is approximate with Q1/m1. If you found the LH change significantly, change the 1-2 degC according.

(vi) Calculate the time (t1) required base on Q1 and fire load calc from API equation.

(vii) Calculate the PSV size base on m1, t1 and other properties.

(vi) Repeat (iii) - (vii) until you get the maximum PSV area.

#9 ANUJ M AGARWAL

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Posted 17 June 2008 - 03:32 AM

Dear Mr. Joe,

Sorry to write you late as I was on emmergency vacation.

I tried to do what you said but could not understand how to start with. As per (i-1), I defined the blend composition as feed to a two phase separator as surge drum( in hysys there is no separate surge drum) with feed temperature of 135 degF and drum operating pressure of 15 psig. I put dia & height & HHLL corresponding liquid fill %.

As per (i-2), how to adjust the inventory?

As per (i-3), I don't know how to increase pressure by itself by providing heat in it. Pressure is not changing.

I just stuck-up. I have attached an excel sheet. Please go through it and reply. Without LH, I am just stuck up.

Regards,
Anuj Mohan Agarwal

Attached Files

  • Attached File  psv.xls   63KB   282 downloads


#10 ANUJ M AGARWAL

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Posted 19 June 2008 - 11:19 PM

Dear Mr. JoeWong,

I am waiting for your reply.

There is one correction in the attached sheet. The flow of oils in ft3/hr are to be considered half of that. Actully it will be divided into two different headers before entering into surge drum.

One more point I would like to add.

After the surge drum, feed will be boosted up by a pump and goes into another vessel V2. This vessel PSV set point is 230 PSIG. But here around 200 ft/hr water is being added before going into the vessel V2. So water composition is more in vessel V2. I found through Hysys that changing the water concentration or taking dry oil, there is a big change in enthalpy and thereby there should be big change in Latent Heat.

Please reply on attached PSV sheet and about LH in V2 at above condition to calculate the same for fire case in V2.

Regards,
Anuj

#11 JoeWong

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Posted 22 June 2008 - 07:23 PM

Anuj,

Check this post. "Determine Latent Heat for Multi-Component and Relieving Area Using Rigorous Method in HYSYS"

Good luck.

#12 Sevail

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Posted 30 June 2008 - 11:24 AM

QUOTE (JoeWong @ Jun 22 2008, 07:23 PM) <{POST_SNAPBACK}>



Dear Joe

With your mentioned guidlines, we can calculate PSV load for Critical condition.
if we have Multi-Component ,and our relifing condition below the critical condition , for calculation of Latent heat only it is necassry to increased feed stream pressure to vessel up to Relifing pressure then at relifing pressure & bubble point we extract MAssheat of vaporization from HYSYS. it give us LAtent heat.

if my method is wrong please help me.

#13 JoeWong

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Posted 30 June 2008 - 02:18 PM

QUOTE (Sevail @ Jun 30 2008, 12:24 PM) <{POST_SNAPBACK}>
With your mentioned guidlines, we can calculate PSV load for Critical condition.
if we have Multi-Component ,and our relifing condition below the critical condition , for calculation of Latent heat only it is necassry to increased feed stream pressure to vessel up to Relifing pressure then at relifing pressure & bubble point we extract MAssheat of vaporization from HYSYS. it give us LAtent heat.

if my method is wrong please help me.


Theoretically the guidelines and method can be applied for Full liquid, liquid boiling, full vapor (as well gas in critical condition). HOWEVER, the limitation could come from the thermodynamic model used, the calculation error near/at critical point may be significant Where many thermodynamic model may be capable of handling it, thus, a more conservative, empirical and experience based approach should be taken for critical and supercritical fluid.

For latent heat, it is pretty much same as what you have mentioned.

#14 Sevail

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Posted 01 July 2008 - 11:29 AM

Dear Joe

thank very much..

may you let me about reference of ""Determine Latent Heat for Multi-Component and Relieving Area Using Rigorous Method in HYSYS" ?

#15 JoeWong

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Posted 08 July 2008 - 05:07 AM

QUOTE (Sevail @ Jul 1 2008, 11:29 AM) <{POST_SNAPBACK}>
Dear Joe

thank very much..

may you let me about reference of ""Determine Latent Heat for Multi-Component and Relieving Area Using Rigorous Method in HYSYS" ?


Sevail,

It was the output of personal inhouse studies.
You may refer to "Rigorously Size Relief Valves for Supercritical Fluids" by Ryan Ouderkirk.




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