13 replies to this topic

[colt16]

Dear friends, 

 

Our plant is currently down and there is thus no refinery fuel gas. 

 

However we still have loading to be performed via our jetty. We need to supply fuel gas to our Vapour Destruction Unit (located at the jetty) to destroy the returned vapour from the marine vessel to comply with environmental laws. 

 

One such idea is to depressure our LPG sphere (C3) to supply fuel to the Vapour Destruction Unit. 

 

I have modeled the pressure drop and temperature drop in PRO\II using the Depressuring module. The temperature drop and pressure drop seems okay for the hours we intend to do this. 

 

I would just like to know however if there are concerns that may have escaped my attention.

[Art Montemayor]

Colt:

 

Please explain your proposal more clearly.

 

Do you mean that you have an existing LPG storage sphere that is currently NOT CONTAINING LPG (Liquefied Petroleum Gas - not just propane) but is, instead, pressurized with hydrocarbon vapors?  In other words, THERE IS NO HYDROCARBON LIQUIDS in the sphere - just hydrocarbon vapor.

 

If the above is the true case, then what quantity of hydrocarbon gas (presumably all propane) is in the sphere and at what pressure and temperature?  Is there sufficient available sphere gas to keep your vapor destruction unit operating while your refinery fuel gas is unavailable?  What fuel gas supply pressure is required by your burner at the vapor destruction unit?

[Bobby Strain]

Impromptu, temporary provisions sometimes lead to disasters. So, proceed with deliberation.

 

Bobby

[colt16]

 

Sorry.  There is still C3 Propane liquid remaining in the sphere.  The sphere is 21.0 m in ID and the liquid level is about 4.4 m.  This liquid level indicates about 275 ton of liquid C3.  Although I call it an LPG sphere, it is essentially pure C3 (98%) because we have a unit that fractionates LPG to C3 and C4.

 

Pressure of C3 sphere = 9.75 kg/cm²G

Temperature of C3 sphere = 29.4 degC

 

Using an estimated depressurizing, I figured that depressurizing this at a small flow rate of about 350 kg/h (roughly the VDU requirement for this LHV) causes the pressure to be 9.55 kg/cm²G and 28.8 degC after 12 h. (assuming there is no heat input, but actually there could be some as the ambient temp = 35 degC).

 

Fuel Gas pressure required at the VDU is quite low.  Just 1.5 kg/cm²G is required.  We have a PRV at the fuel line which controls the pressure of the fuel to the VDU at 1.5 kg/cm²G and we intend to tie-in just before this PRV.  We intend to install an RO at the outlet of the C3 sphere.  The whole line to the VDU can handle original sphere pressure (300#).

Edited by Art Montemayor, 05 November 2015 - 11:09 AM.

[Art Montemayor]

Colt:

 

Once again, get your facts and story straight.  As Bobby expertly points out:  proceed with deliberation - and with accurate, concise information.

 

You are not "depressurizing" the saturated propane contents in your sphere.  In order to understand what I am stating as fact, refer to your propane thermodynamic properties or to a propane T-S diagram.  As you extract propane vapors from a tank holding saturated liquid propane at what appears to be ambient temperature, the liquid contents will start to evaporate - much as they do in a refrigerant evaporator, dropping down in pressure AND TEMPERATURE.  If your storage sphere is constructed of carbon steel (like A-516, Grade 70) - as your storage pressure and temperature seem to indicate, then you will reach a minimum temperature within the tank through the evaporation where your carbon steel material cannot go any lower without exceeding the safe stress limits of the material.  THAT IS PROBABLY WHAT BOBBY IS ALERTING YOU TO.

 

If you don't know or understand the thermodynamics of the "depressurizing" process you are contemplating, you should not go any further with the idea.  There is another way to skin this cat and that is to rent a portable steam generator that is either electric powered or utility natural gas powered and generate the steam to vaporize the liquid propane contents at the saturated pressure in the sphere.

[colt16]

 

 

Yes, I understand that there will be a temperature drop.  I believe I have already given that temperature drop value. 

 

Anyway, the link below is the image of what it will look like for 96 hours. From the chart you see the temperature does not drop so severely, if the RO is sized correctly. It drops to 23.4 degC which is still acceptable.

 

http://postimg.org/image/pprr1qmbh/

[Art Montemayor]

Colt:

 

Your submitted graph supposedly showing the temperature drop in the storage tank doesn’t mean anything without reviewing the detailed calculations or source of the data.  Without the ability to confirm that the presented data is valid and accurate, our Forum members can’t do anything else except acknowledge that it is your opinion or results and nothing more.  Your reported partial tonnage of 275 tonnes of liquid Propane in the tank appears to be in error.  My calculations, as shown in the attached workbook, show it to be 266.9 tonnes.  Out of this liquid quantity you propose to consume 33.6 tonnes over a span of 4 days - approximately 12.6 %.

 

This may be, as you state, a very small amount of the stored propane that is extracted but it must be shown clearly that the remaining stored liquid propane temperature resulting from the internal vaporization does, in fact, not cause a low temperature problem.  I have to explain all this because you don’t supply a flow diagram.  And that is why I suggested an external evaporator that would vaporize the required 33.6 tonnes of liquid without lowering the tank’s temperature.  My reason for suggesting this is that by vaporizing only the liquid required for gaseous fuel and sending this to the VDU unit, you keep the propane tank under normal conditions.  Since you only require 350 kg/hr of propane, vaporizing this quantity should not be a problem.  You could use an electric steam generator or even a hot water heated evaporator coil.  I mention the hot water coil since you infer that your storage tank’s walls are enough to heat the tank and maintain a tolerable vapor pressure inside.

 

I doubt that you could expect to have the tank’s vapor contents heat up and maintain a tolerable vapor pressure using only the heat transfer from the ambient air.  This is one of the most difficult heat transfer operations in industry because it involves surface heat transfer film coefficients on both sides of the metal wall that rely solely on gas films.  Even if your outside air is strong enough to cause turbulence, you still have to contend with the fact that the vapor inside the tank is essentially static and has one of the poorest heat transfer coefficients to be found.  All this means very poor and inefficient heat transfer over 60% of the sphere’s surface area.  The remaining 40% of the sphere’s surface area is wetted with liquid propane and, although better than the vapor portion, is still a bad heat transfer condition because the contained liquid is static.

 

Your problem is one of unsteady state heat transfer.  The tank’s contents are differentially cooling down due to the liquid content’s evaporation rate that is caused by the extraction of its saturated vapor content.  This is the same effect that takes place in any mechanical refrigeration evaporation coil - except that your liquid refrigerant (the evaporating propane) is not being replenished but is, instead, a batch content of liquid that will differentially continue to cool down while its saturated vapor continues to be extracted.  This is not a simple calculation and is going to involve some careful modeling of the operation in order to predict the resultant end temperature.  That is why the detailed calculations are required for checking and review.  A related flow diagram would be of value in following the assumptions and proposal.

 

Why are you proposing to employ an “RO”?  I presume you mean a Resistance Orifice.  I don’t understand the need for this because the orifice is not going to accomplish any control of the vapor extracted.  That is the role of the proposed PCV (pressure control valve) - NOT A PRV - that is controlling the vapor pressure fed to the burner at the required 1.5 kg/cm².  The RO cannot do anything to control the drop in tank temperature.  The rate at which the vapors are extracted is what affects the temperature of the liquid in the tank - and that rate is controlled by the PCV.  Are you confusing a PCV with a Pressure Relief Valve?  These are clear examples of why all engineers should always identify all acronyms employed.  Again, if you supply a clear and detailed flow diagram of your proposed operation, these questions can be answered very quickly and accurately.

 

One safety instrument that I would install into a system that you are proposing is a pressure control on the required PCV feeding the VDU burner.  The PCV should be fail closed and set to open only when the vapor pressure in the tank is above a set, safe value with respect to the safe saturated temperature.  This would be a safeguard against drawing down the tank pressure lower than that pressure corresponding to the lowest acceptable saturated temperature.

If you have any follow up queries, comments, or statements on this topic then I highly recommend that you take the time and make the effort to submit detailed calculations and diagrams of the proposed operation.  As Bobby Strain stated: “Proceed with deliberation”.

 

 Mensuration Calculations.xlsx   29.17KB   41 downloads

[Bobby Strain]

You don't have to search around to find the charasterictics of a sphere or cylinder. You can use the calculator at my site. Here is an illustration using your dimensional data. So, now you have at least two sources, both with the same result.

 

Bobby

 

 Bobby Strain Group.pdf   50.48KB   48 downloads

Edited by Bobby Strain, 05 November 2015 - 04:55 PM.

[colt16]

Hi Art, 

 

Thank you for taking the time to reply to this. 

I will try to supply the necessary flow scheme and diagrams but meanwhile let me reply to some of the things you brought up. 

 

Colt:

 

Your submitted graph supposedly showing the temperature drop in the storage tank doesn’t mean anything without reviewing the detailed calculations or source of the data.  Without the ability to confirm that the presented data is valid and accurate, our Forum members can’t do anything else except acknowledge that it is your opinion or results and nothing more.  Your reported partial tonnage of 275 tonnes of liquid Propane in the tank appears to be in error.  My calculations, as shown in the attached workbook, show it to be 266.9 tonnes.  Out of this liquid quantity you propose to consume 33.6 tonnes over a span of 4 days - approximately 12.6 %.

You are right and actually I calculated the same. When I typed what I typed I was recalling the problem statement and calculations from memory.

 

This may be, as you state, a very small amount of the stored propane that is extracted but it must be shown clearly that the remaining stored liquid propane temperature resulting from the internal vaporization does, in fact, not cause a low temperature problem.  I have to explain all this because you don’t supply a flow diagram.  And that is why I suggested an external evaporator that would vaporize the required 33.6 tonnes of liquid without lowering the tank’s temperature.  My reason for suggesting this is that by vaporizing only the liquid required for gaseous fuel and sending this to the VDU unit, you keep the propane tank under normal conditions.  Since you only require 350 kg/hr of propane, vaporizing this quantity should not be a problem.  You could use an electric steam generator or even a hot water heated evaporator coil.  I mention the hot water coil since you infer that your storage tank’s walls are enough to heat the tank and maintain a tolerable vapor pressure inside.

I don't think this is necessary because while there is 270 ton of liquid in the C3, the remaining vapor space when calculated is about 98.5 ton of C3. Now when the valves are opened for the removal of C3 vapors, some of the C3 will be from the vapour space and some of the C3 will be from evaporation of liquid C3 due to the drop in pressure. The reality is that it will be somewhere in between these two extremes. In either case however, we do not need to provide external heating source to obtain the vapour given that the initial state of the tank has 98.5 ton of C3 vapour in the sphere. While you are right 33 ton will be the calculated figure based on the depressuring, we expect to use far less, maybe around 15 ton. The model I provided was simply for the extreme case to give reasonable assurance that even in this misoperation, the temperature drop is not so severe.

 

I doubt that you could expect to have the tank’s vapor contents heat up and maintain a tolerable vapor pressure using only the heat transfer from the ambient air.  This is one of the most difficult heat transfer operations in industry because it involves surface heat transfer film coefficients on both sides of the metal wall that rely solely on gas films.  Even if your outside air is strong enough to cause turbulence, you still have to contend with the fact that the vapor inside the tank is essentially static and has one of the poorest heat transfer coefficients to be found.  All this means very poor and inefficient heat transfer over 60% of the sphere’s surface area.  The remaining 40% of the sphere’s surface area is wetted with liquid propane and, although better than the vapor portion, is still a bad heat transfer condition because the contained liquid is static.

Again see my reply above.

 

Your problem is one of unsteady state heat transfer.  The tank’s contents are differentially cooling down due to the liquid content’s evaporation rate that is caused by the extraction of its saturated vapor content.  This is the same effect that takes place in any mechanical refrigeration evaporation coil - except that your liquid refrigerant (the evaporating propane) is not being replenished but is, instead, a batch content of liquid that will differentially continue to cool down while its saturated vapor continues to be extracted.  This is not a simple calculation and is going to involve some careful modeling of the operation in order to predict the resultant end temperature.  That is why the detailed calculations are required for checking and review.  A related flow diagram would be of value in following the assumptions and proposal. Yes, that is why PRO/II's depressuring module has been very useful here. It models the rate of pressure drop and temperature drop with a fixed depressuring valve size or orifice size and backpressure and proceeds in time steps specified. By altering the valve/orifice size we can see what will happen with a faster rate and slower rate.

 

Why are you proposing to employ an “RO”?  I presume you mean a Resistance Orifice.  I don’t understand the need for this because the orifice is not going to accomplish any control of the vapor extracted.  That is the role of the proposed PCV (pressure control valve) - NOT A PRV - that is controlling the vapor pressure fed to the burner at the required 1.5 kg/cm².  The RO cannot do anything to control the drop in tank temperature.  The rate at which the vapors are extracted is what affects the temperature of the liquid in the tank - and that rate is controlled by the PCV.  Are you confusing a PCV with a Pressure Relief Valve?  These are clear examples of why all engineers should always identify all acronyms employed.  Again, if you supply a clear and detailed flow diagram of your proposed operation, these questions can be answered very quickly and accurately.  I proposed the use of the Restriction Orifice because if there is some failure of instruments (fail open or what not) or operational error, then the depressuring will occur at a very fast rate causing the temperature to cool down too fast. The line out of the sphere is 6" and that is going to give you way above 350 kg/h without a restriction orifice. I "stole" the idea of using an Restriction Orifice because I thought through this concept in terms of sizing an emergency blowdown valve (minus the heat input that happens with a fire) where the typical configuration involves an on-off valve with a restriction orifice downstream of the on-off valve to control the depressuring rate to reach 7 kg/cm2G after 15 minutes. Similarly, the same principles are at play here in this situation without the heat input to the sphere. 

 

 

[Neelakantan]

While reading my reply, keep in mind that I have not looked at the volume and depressurization numbers that you have provided;

 

When I joined the oil/gas industry way back, I worked in the operations of an LPG plant  and during maintenance one of our senior engineers decided to depressurize the insulated propane and LPG spheres (after a blowdown to 15% volume) to the flare before using nitrogen purging for some work.  The next day, in the morning, when our guys saw that the sphere pressure was at the flare pressure, they thought the spheres had been depressurized and only had vapours.  Being a new engineer, I went around the sphere and saw that the temperature needle was struck at -12 deg C (that was the lowest temperature indicatable in these field thermometers); and when I tried to check the discharge pump suction, I found it was full of liquid!   The fire protection insulation of the supports was swollen with ice and the feed lines of the pumps were all ice laden.

 

It was then that I remembered the principles of thermodynamics to understand what had happened.

 

Now, coming to your situation, the easiest  and probably the most sensible way of supplying fuel is to pump LPG and flash the discharge LPG as vapours at the point of use.  If required, you may need to have a vaporizer heater.

 

regards

neekantan

[Bobby Strain]

And, there has been no mention whether the burner can even use propane fuel. If it is designed for lighter gas, there may be a problem when using propane. So, look at both ends. With deliberation.

 

Bobby

[Art Montemayor]

Again, experience proves its worth.  What Bobby Strain rightfully states about the need to ensure that the burner can handle the heavier propane fuel safely is just another "detail" that must be addressed.  I suspect that there are other details - some missing, and others overlooked in this proposal.  This is main reason why I have spent any time at all on this thread and the concern I have made for the lack of detailed calculations and a decent flow diagram.

 

A detailed, marked up P&ID together with all applicable Data Sheets and calculations is what I always required from plant process engineers who worked in engineering departments I managed and some I organized.  All this information was done and subjected to a meticulous checking, review, discussion, and HAZOP prior to submittal to plant management in a detailed Management of Change (MOC) package.  This, in my opinion, should be the minimum requirements for any changes proposed within a processing plant environment.  This concern, I believe is also the reason for Bobby's comments.   I think that, between he and I, we must have close to a sum total of close to 100 years of process engineering experience behind us.  I have seen and heard of far too many engineering mistakes, errors, omissions, and wrong decisions in my time and I have learned that no process - no matter how simple it may appear - is really simple and can be taken lightly.  Use caution, common sense, detailed scrutiny, experienced advice, and peer checks and support.

 

Since I don't think there will be any of the requested P&ID, flow diagram, or detailed calculations for this thread being submitted, I'll close my comments on this topic and hope that all turns out safe, well, and successful.

[colt16]

Hi Art, 

 

Again to reply to you, this is a VERY initial phase of the idea. We are definitely subjecting all this to a detailed MOC process. 

The organization that I am from is not the best (my personal view), but we have some standards 

 

Please do not go away with the idea that I am doing this tomorrow. No, no. 

 

Having questions like "Did we consider whether this can take propane". Yes this is what I like about a forum - bringing up details that we may have missed. To answer, yes. The original operation of this burner like I said is for refinery fuel gas.

 

I am not sure why the interrogation that is happening at the moment thinks that we will carry out all of this stuff tomorrow. I have ~2 months (estimated) before we actually load this stuff. 

 

Diagrams and all of that stuff is great, but I am sure you can understand that I can't upload confidential stuff online. (P & IDs, etc.).

 

Anyway I will supply the stuff and would love to have more discussion. If however you feel that it is not worth discussing then I see no point in uploading it either. I do wonder though, what is the point of a forum if discussions die so easily?

Edited by colt16, 07 November 2015 - 12:53 AM.

[colt16]

I just like to share that we completed this operation successfully. Propane was delivered from the C3 sphere to the Furnace. The flowrates achieved was from 0.2 - 0.7 ton/h. However we couldn't reach the target temperature in our stack but we believe that is due to the excess air coming from the blower. 

 

Thank you all for allowing me to bounce off ideas with you.