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#1 icasensio

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Posted 13 March 2012 - 07:14 AM

Dear all,

I need to know how calculate the volume flow that is formed in return line when LPG ( 70% C3 and 30% C4) is been loading in a truck, (300gpm).

I appreciate your help, comments and answers.

Thanks in advance.

#2 ankur2061

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Posted 13 March 2012 - 12:17 PM

icasensio,

This is a simple isenthalpic flash problem. Since you are pumping the LPG into the tanker you are expanding or flashing the LPG liquid from the pressure at the inlet of the tanker to the pressure inside the tanker which is at much lower pressure than the LPG inlet pressure. This expanding or flashing will generate LPG vapors in the tanker vapor space which will be reflected as vapor flow in the return line.

You can use a simulator such as HYSYS or any other to calculate the vapor flow in the return line by simulating it as a simple isenthalpic flash using a separator vessel. For your case if you flash the LPG from 260 psig to 30 psig the vapor flow rate will be approximately 9 cfm in the vapor line. Note that if the outlet pressure is decreased from 30 psig to 15 psig the vapor flow rate increases from 9 to 11.5 cfm. So higher the pressure drop or pressure differential higher will be the vapor flow rate in the return line.

Hope this helps.

Regards,
Ankur.

Edited by ankur2061, 13 March 2012 - 12:23 PM.


#3 S.AHMAD

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Posted 13 March 2012 - 08:16 PM

Dear icasensio
1. As mentioned by Ankur, you need to do flash calculation to determine the vapor composition, pressure and equilibrium temperature
2. The tanker pressure will built up under its own vapor pressure and the vapor is contained inside the tanker. Once the vapor pressure is higher than the storage pressure (to overcome friction), only then the vapor starts to flow and return to the storage tank.
3. The volumetric flowrate of the vapor return is determined from the rate of pumping.
4. For example, if the pumping/loading rate is 100 m3/h, then the volumetric flowrate of vapor is also 100m3/h at the flashing conditions (displacement volume - the liquid inside the tanker will displace the vapor space above the liquid level at the same volumetric rate).
5. By knowing the composition, pressure and temperature you can determine the flowrate in kg/h or nM3/h, and other properties that you may require for line sizing calculation.
6. Take note that the volumetric flowrate is the same for loading and vapor return but their mass rate is not the same.

Edited by S.AHMAD, 13 March 2012 - 08:23 PM.


#4 icasensio

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Posted 14 March 2012 - 05:22 AM

Ankur2061, thank you so much for your help.

I have calculated volumen flow rate with the following parameters and the results from Hysys are after them:

- Min. Pressure at line : 90 psig / Min. Pressure at tanker : 30 psig --> 9.6 cfm/s
- Max. Pressure at line : 195 psig / Max. Pressure at tanker : 120 psig --> No flashing

Instalation is existing and we have to define normal flow in order to specify flow meters for these lines (1 1/2" nominal diameter)

First result seems quite large for a 1 1/2" pipe line. May be I am not considering well max and min pressure.

What do you think about it? Could you help me with it?

#5 fallah

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Posted 14 March 2012 - 06:25 AM

3. The volumetric flowrate of the vapor return is determined from the rate of pumping.
4. For example, if the pumping/loading rate is 100 m3/h, then the volumetric flowrate of vapor is also 100m3/h at the flashing conditions (displacement volume - the liquid inside the tanker will displace the vapor space above the liquid level at the same volumetric rate).


S.AHMAD,

IMO, the above mentioned balance in volumetric flow rates of liquid LPG and its returned vapor would be stablished if the returned line to be sized such that at the flashing conditions the tanker due to, let say, small size of returned line not to be pressurized more and more i.e. a steady state condition being existed in returned vapor flow rate.

Fallah

Edited by fallah, 14 March 2012 - 06:38 AM.


#6 icasensio

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Posted 14 March 2012 - 06:25 AM

Dear icasensio
1. As mentioned by Ankur, you need to do flash calculation to determine the vapor composition, pressure and equilibrium temperature
2. The tanker pressure will built up under its own vapor pressure and the vapor is contained inside the tanker. Once the vapor pressure is higher than the storage pressure (to overcome friction), only then the vapor starts to flow and return to the storage tank.
3. The volumetric flowrate of the vapor return is determined from the rate of pumping.
4. For example, if the pumping/loading rate is 100 m3/h, then the volumetric flowrate of vapor is also 100m3/h at the flashing conditions (displacement volume - the liquid inside the tanker will displace the vapor space above the liquid level at the same volumetric rate).
5. By knowing the composition, pressure and temperature you can determine the flowrate in kg/h or nM3/h, and other properties that you may require for line sizing calculation.
6. Take note that the volumetric flowrate is the same for loading and vapor return but their mass rate is not the same.


S. Ahmad, many thanks for your reply. Actually, my first consideration was the same as yours: "The volumetric flowrate of the vapor return is determined from the rate of pumping"

However, I can´t find the relationship between this and flashing (except vapor composition) in that trouble. Could you explain it for me?

Regards.

#7 ankur2061

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Posted 14 March 2012 - 08:43 AM

icasensio,

There is no such unit as cfm/s. "cfm" means "cubic feet per minute".

1 cfm = 1.7 m3/h if you prefer it in Metric units. So 9.6 cfm = 16.3 m3/h.

This is not a very large vapor flow rate for a 1-1/2" pipe.

Regards,
Ankur.

#8 icasensio

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Posted 14 March 2012 - 12:25 PM

Ankur, I apologize for my error, but I am still quite confused.

On one hand, Hysys does not operate correctly or I am doing somenthing wrong because vapour flow rate is higher than your values. (See picture attached)

On the other hand, Don´t you think that vapour will be confinated in the tanker during the loading operation?. At the beginning, flashing will occurre and vapour flow rate will increase the presure inside the tanker. This will occurr untill pressure inside exceeds pressure in the storage tank. Then, vapour flow rate (volumen) should be the same flow rate of pump, under equilibrium condition (pressure and temperature).

Actually, I don´t know if I am wrong, but right sense doesn´t let me think in your way.

Thank you again.

icasensio.

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#9 ankur2061

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Posted 14 March 2012 - 02:35 PM

icasensio,

Your calculation in HYSYS is absolutely right. But your conversion is not correct. Also note that vapor flow rate is never expressed in gpm. In SI units it is expressed as m3/h and in English units as cfm. For your 90 psig -30 psig case the vapor flow rate is 952.9 m3/h or 561 cfm.

Also as the tanker pressure starts to rise due to LPG entering the tanker and flashing inside it the differential pressure or pressure difference between the inlet and outlet connections start reducing. The reduction in this pressure difference causes the vapor flow rate in the vapor flow line to reduce as a time dependent variable.

When there is no pressure difference or in other words if the pressure equalizes between the tanker vapor space and the vapor space above the sphere or bullet from which you are filling the tanker the vapor flow rate in the return line drops sharply until it becomes negligible. I am using the term negligible because there will always be some vapor flow from the tanker vapor space through the return line to the sphere or bullet as long as LPG liquid can enter the tanker due to displacement of vapor by the LPG liquid.

In your case when I reduce the pressure drop in the separator from 60 psi to 50 psi the volume flow rate reduces from 952.9 m3/h (561 cfm) to 633.4 m3/h (373 cfm) which clearly illustrates the point I have made in the previous paragraph regarding the drop in the vapor flow.

Hope this explanation helps.

Regards,
Ankur.

Edited by ankur2061, 14 March 2012 - 02:37 PM.


#10 S.AHMAD

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Posted 14 March 2012 - 07:06 PM

1. Flashing will only occur when the tanker pressure is below the vapor pressure of LPG which is depending on temperature after pump discharge.
2. This means that flashing will occur initially, but once the pressure of tanker equalizes with that of storage (actual pressure is higher to overcome friction), flashing will not occur and hence no more vapor flow due to flashing (unless the LPG temperature at pump discharge is higher than that of storage)
3. The only vapor is then the vapor from liquid displacement.
4. If no return line provided, the pressure in the tanker will rise due to compression of vapor inside the tanker. That is one of the reason to have LPG return line so that the compressed vapor can be released back to the storage and prevent tanker overpressure
5. Flash calculation is for determining the vapor composition that is in equilibrium with the liquid. This is the vapor that will be "pushed" by the rising liquid.
6. Please do a flash calculation ans set the flash drum pressure slightly higher that the vapor pressure, and see how much vapor generated.
7. During initial loading when the tanker pressure is lower than the vapor pressure, vapor will be flashed off. If the return line has a check valve, the vapor does not go anywhere, it stays inside the tanker. Until the tanker pressure is greater than the storage pressure, only then the vapor starts to flow. At this pressure, vapor is no longer formed unles the LPG temperature is higher due to pumping. Then you may need to consider the flashed vapor due to this higher temperature.
8. Effect of pressure on liquid enthalpy is negligible but not of temperature.

Edited by S.AHMAD, 14 March 2012 - 07:41 PM.


#11 S.AHMAD

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Posted 14 March 2012 - 08:43 PM

Dear Icasensio
1. I have checked using PRO-II for 70%C3 and 30%C4 by liquid volume, the vapor pressure is 108 psia at 77F. I believe this is the storage pressure and temperature.
2. Let say for the vapor to return to storage we have 3 psi pressure drop. This means that the tanker pressure shall be 111 psia.
3. Do a flash calculation at P=111 psia and T=77F. See what vapor flowrate do you get? My PRO-II simulation results show none. What is yours? (I set feed at P=120 psia and T=77F)

Edited by S.AHMAD, 14 March 2012 - 09:13 PM.


#12 icasensio

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Posted 16 March 2012 - 02:01 AM

icasensio,

Your calculation in HYSYS is absolutely right. But your conversion is not correct. Also note that vapor flow rate is never expressed in gpm. In SI units it is expressed as m3/h and in English units as cfm. For your 90 psig -30 psig case the vapor flow rate is 952.9 m3/h or 561 cfm.

Also as the tanker pressure starts to rise due to LPG entering the tanker and flashing inside it the differential pressure or pressure difference between the inlet and outlet connections start reducing. The reduction in this pressure difference causes the vapor flow rate in the vapor flow line to reduce as a time dependent variable.

When there is no pressure difference or in other words if the pressure equalizes between the tanker vapor space and the vapor space above the sphere or bullet from which you are filling the tanker the vapor flow rate in the return line drops sharply until it becomes negligible. I am using the term negligible because there will always be some vapor flow from the tanker vapor space through the return line to the sphere or bullet as long as LPG liquid can enter the tanker due to displacement of vapor by the LPG liquid.

In your case when I reduce the pressure drop in the separator from 60 psi to 50 psi the volume flow rate reduces from 952.9 m3/h (561 cfm) to 633.4 m3/h (373 cfm) which clearly illustrates the point I have made in the previous paragraph regarding the drop in the vapor flow.

Hope this explanation helps.

Regards,
Ankur.


Ankur,

Your explanations have been helpfully. Let say, the maximum flow rate that is generated (from 90 psig to 30 psig) is 952.9 m3/h, (obviously, in case of return line is under 30 psig or less).

However, the sphere pressure is higher and the return line has a check valve to avoid return vapor to the tanker. In this situation, vapor will not flow until pressure inside the tanker is higher than pressure inside the sphere plus pressure drop.

Thank you very much for your help.

Regards.

#13 icasensio

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Posted 16 March 2012 - 02:15 AM

Dear Icasensio
1. I have checked using PRO-II for 70%C3 and 30%C4 by liquid volume, the vapor pressure is 108 psia at 77F. I believe this is the storage pressure and temperature.
2. Let say for the vapor to return to storage we have 3 psi pressure drop. This means that the tanker pressure shall be 111 psia.
3. Do a flash calculation at P=111 psia and T=77F. See what vapor flowrate do you get? My PRO-II simulation results show none. What is yours? (I set feed at P=120 psia and T=77F)


S. Ahmad

As you have wrote in the first point of your post, flashing will only occur when the tanker pressure is below the vapor pressure of LPG, so result of my simulation is same as yours: no flashing so no vapor.

Now I have to know the composition of the vapor return line. Hysys can calculate vapor composition after flashing, and I deduce it is the same I am looking for, but I don’t know how to get the density. ¿Could it be the density at pressure that vapor starts flow through return line? (In your example, 111 psia)

I appreciate so much your answer.

Many thanks and regards.

#14 S.AHMAD

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Posted 16 March 2012 - 03:19 AM

Icasensio
1. If you are using HYSIS, I suggest that you determine the vapor composition by flashing at a slightly higher temperature than the saturated temperature of LPG.
2. Alternatively, you can calculate manually using Roult's and Dalton's Law
3. Or assume a worst case of 100% C3 or 100% C4 whichever is the worst case. I think 100% C3 is the worst case for line sizing.
4, Why you need to know the density? for gas what we need is the molecular weight since density is changing with pressure and temperature. MW can be calculated from composition.

Edited by S.AHMAD, 16 March 2012 - 03:36 AM.


#15 kkala

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Posted 20 March 2012 - 06:28 AM

Following notes can be useful on the topic, according to my understanding.
1. http://www.quantity...lculations.pdf , section "LPG tank measurements - stock control", indicates that "empty" trucks still contain a small stock of liquid LPG. Gas pressures in both truck and sphere are the saturated pressures of their LPG liquids, not far from each other. Consequently flashing (if any) in the truck just at the start of loading is insignificant.
2. Return line intends to "equalize" gas pressures (well, neglecting ΔP along it) in truck and sphere. Check valve on the line has not been heard; but it exists in the specific case, to prevent gas flow from sphere to truck. This will be an obstacle in case of truck unloading into the sphere through compressed LPG vapor; see http://www.tankstora...tions/Satam.pdf '> http://www.tankstora...tions/Satam.pdf , section "Tank unloading operation (1)".
2. Very soon after liquid LPG pumping, steady state will be approached: gas pressure in the truck will be slightly higher than in the sphere (hence temperatures too, if detectable), the difference being ΔP along the return line. See post by S.AHMAD explaining it. Concerning the truck, no flashing takes place in it; volume of liquid "LPG in" equals volume of gaseous "LPG out", being 300*3.785*60/1000=68.1 actual m3/h. Design flow of gas return line may include some margin.
3. A 1.5" (Sch40) return line will result in 14.4 m/s gas velocity (acceptable) for 68.1 m3/h; assuming 40 oC max operating temperature and pure butane vapor, estimated frictional ΔP can be about 2.7 psi/100 ft (0.6 bar/100 m), which is judged high (according to line sizing criteria). Detailed calculation of return line ΔP has to be elaborated, based on actual specific conditions (max operating temperature, estimated gas composition based on LPG liquid per post No 1, total length of return line, line sizing criteria of your company) in order to assess suitability of existing 1.5" return line for the required gas flow rate.

#16 icasensio

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Posted 28 March 2012 - 04:12 AM

Icasensio 1. If you are using HYSIS, I suggest that you determine the vapor composition by flashing at a slightly higher temperature than the saturated temperature of LPG. 2. Alternatively, you can calculate manually using Roult's and Dalton's Law 3. Or assume a worst case of 100% C3 or 100% C4 whichever is the worst case. I think 100% C3 is the worst case for line sizing. 4, Why you need to know the density? for gas what we need is the molecular weight since density is changing with pressure and temperature. MW can be calculated from composition.


S. Ahmad, many thanks for your answer, but I have some questions for you related with your explanation:

1.- HYSYS gives the composition of liquid and vapor at determined composition.

If I define Pressure 107,6 psia, and Temperature 77 F, phase is liquid and composition will be as follow (Mole Fractions).

Liquid Phase
Propane 0,7276
n-Butane 0,2723

If I define a slightly higher temperature (80 F) and the same Pressure 107,6 psia, vapor will appear (0,2256 vapor fraction) and composition will be as follow (Mole Fractions).

Vapour Phase Liquid Phase
Propane 0,8723 0,6855
n-Butane 0,1276 0,3144

Then, is this the composition of vapor phase I have to considerate?

2.- I need vapor density to get mass flow rate. I will multiply vapor density and volume flow rate to get mass flow rate.

Vapor will start to flow when pressure (at ambient temperature) will be slightly higher than sphere pressure. However, if tank pressure is equilibrium pressure at ambient temperature, if I simulate with a higher pressure vapor will no appear. Could you explain me why I should use a higher temperature as you told me before?

Thanks in advance

#17 icasensio

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Posted 28 March 2012 - 04:34 AM

Following notes can be useful on the topic, according to my understanding. 1. http://www.quantity...lculations.pdf , section "LPG tank measurements - stock control", indicates that "empty" trucks still contain a small stock of liquid LPG. Gas pressures in both truck and sphere are the saturated pressures of their LPG liquids, not far from each other. Consequently flashing (if any) in the truck just at the start of loading is insignificant. 2. Return line intends to "equalize" gas pressures (well, neglecting ΔP along it) in truck and sphere. Check valve on the line has not been heard; but it exists in the specific case, to prevent gas flow from sphere to truck. This will be an obstacle in case of truck unloading into the sphere through compressed LPG vapor; see http://www.tankstora...tions/Satam.pdf '> http://www.tankstora...tions/Satam.pdf , section "Tank unloading operation (1)". 2. Very soon after liquid LPG pumping, steady state will be approached: gas pressure in the truck will be slightly higher than in the sphere (hence temperatures too, if detectable), the difference being ΔP along the return line. See post by S.AHMAD explaining it. Concerning the truck, no flashing takes place in it; volume of liquid "LPG in" equals volume of gaseous "LPG out", being 300*3.785*60/1000=68.1 actual m3/h. Design flow of gas return line may include some margin. 3. A 1.5" (Sch40) return line will result in 14.4 m/s gas velocity (acceptable) for 68.1 m3/h; assuming 40 oC max operating temperature and pure butane vapor, estimated frictional ΔP can be about 2.7 psi/100 ft (0.6 bar/100 m), which is judged high (according to line sizing criteria). Detailed calculation of return line ΔP has to be elaborated, based on actual specific conditions (max operating temperature, estimated gas composition based on LPG liquid per post No 1, total length of return line, line sizing criteria of your company) in order to assess suitability of existing 1.5" return line for the required gas flow rate.



Kkala, thank you for your explanation, it has been clear and very usefull.

I only want underline two points:

1.- This instalation will be used only as a loding tanker area, then (as you said) a valve check will be necesary to avoid return vapor from spheres to tanker .

2.- When I define the composition of vapor, I will check pressure drop in the return line in HYSYS.

Regards.

#18 S.AHMAD

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Posted 28 March 2012 - 08:54 PM

Dear Icansensio
1.The idea of flashing at slightly higher temperature is to get the vapor composition (not the flowrate) that is in equilibrium with the liquid. If you do flash at bubble point, you will not get the vapor composition.
2. What I suggest is that you do flash at 77.01F, 77.02F etc until you get some vapor fraction probably 0.0000001 mole fraction. Use this vapor composition as the design basis for you line size.
3. Once we know the vapor composition, we know the MW and hence density (HYSIS will give you the composition and properties).
4. To obtain the mass rate, multiply volumetric rate with density at conditions.
5. The actual composition will varies from start of run to completion of loading.
6. Initially, vapor downstream of check valve will be that in equilibrium with the liquid in the storage. Vapor upstream check valve will be that of vapor in equilibrium with liquid in the tanker which is depending on pressure. If the initial pressure in the tanker is atmospheric (as an example), the vapor will be 100% LPG composition that is all LPG vaporizes. As the pressure build-up, the C3 fraction in the vapor will be higher than that of C4. The higher the pressure the more C3 fraction in the vapor. It is for this reason, I suggest that you use the vapor composition as obtained in item 2 above, as the design basis. The worst case scenario will be that of 100% C3 (if the tanker pressure keeps rising, more and more C4 will condense).
7. The line size must be selected on the basis that the tanker pressure (storage pressure + pressure drop) is well below the design limit of the tanker design pressure.
8. If you still confuse, do not hesitate to submit your detailed calculation and relevant information for further comments.

Edited by S.AHMAD, 28 March 2012 - 08:58 PM.


#19 icasensio

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Posted 29 March 2012 - 04:25 AM

Dear Icansensio 1.The idea of flashing at slightly higher temperature is to get the vapor composition (not the flowrate) that is in equilibrium with the liquid. If you do flash at bubble point, you will not get the vapor composition. 2. What I suggest is that you do flash at 77.01F, 77.02F etc until you get some vapor fraction probably 0.0000001 mole fraction. Use this vapor composition as the design basis for you line size. 3. Once we know the vapor composition, we know the MW and hence density (HYSIS will give you the composition and properties). 4. To obtain the mass rate, multiply volumetric rate with density at conditions. 5. The actual composition will varies from start of run to completion of loading. 6. Initially, vapor downstream of check valve will be that in equilibrium with the liquid in the storage. Vapor upstream check valve will be that of vapor in equilibrium with liquid in the tanker which is depending on pressure. If the initial pressure in the tanker is atmospheric (as an example), the vapor will be 100% LPG composition that is all LPG vaporizes. As the pressure build-up, the C3 fraction in the vapor will be higher than that of C4. The higher the pressure the more C3 fraction in the vapor. It is for this reason, I suggest that you use the vapor composition as obtained in item 2 above, as the design basis. The worst case scenario will be that of 100% C3 (if the tanker pressure keeps rising, more and more C4 will condense). 7. The line size must be selected on the basis that the tanker pressure (storage pressure + pressure drop) is well below the design limit of the tanker design pressure. 8. If you still confuse, do not hesitate to submit your detailed calculation and relevant information for further comments.


S. Ahmad,

First of all, I appreciate your answers, help, and your favor. It is very kind of you.

Secondly, I have attached a sketch with considerations and results of the calulation with HYSYS. I would appreciate so much if you could tell me what do you think about them, or if I am wrong and where.

Many thanks and regards.

Attached Files



#20 kkala

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Posted 29 March 2012 - 04:42 PM

In my understanding vapor pressure of the specific LPG grade in the sphere has been estimated at 100.9 psia at 77 oF (25 oF), temperature that can be assumed constant.
Gas flow in the return line (RL) will start as soon as gas pressure in the truck gets higher than 100.9 psia, but will increase till the value of 68.1 actual m3 /h coming out of the truck. Gas pressure in truck will increase to 100.9 psia+ΔP, which means increase of truck LPG temperature accordingly. So problem at steady state can be solved through a trial and error procedure (e.g. starting from an assumed LPG truck temperature). Gas LPG composition can be estimated depending on truck temperature (ΔP may not be very sensitive to it).
It seems that first trial assumed for truck LPG: temperature=89.4 oF, gas presure=120 psia; then hydraulic calculation of ΔP in RL has to be elaborated for estimated LPG composition out of the truck (68.1 m3/h at 89.4 oF and 121 psia), to see whether this is close to 121 - 100.9 = 20.1 psi, and assumptions be revised accordingly.
However line sizing criteria are applicable for the RL. We would allow frictional ΔP in the line not higher than 1.5 psi/100 ft (pressure 50-150 psig, pipe up to 300 ft); Norsok standard P-001 (para 6.3.3) would recommend lower than 0.5 psi /100 ft. So if actually ΔP=20.1 psi, this might be high (also depending on line length), meaning small line diameter.
It is noted that check as above should be elaborated for max LPG temperature in sphere (77 oF or higher?).

Edited by kkala, 29 March 2012 - 04:45 PM.


#21 S.AHMAD

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Posted 29 March 2012 - 09:00 PM

icasensio
1. Congratulation for the good effort. Your understanding on the system and process is improving but not well enough.
2. Kkala comments are valid. The actual calculation is “trial and error”. However, before you can do this iteration, you need to use the right method.
3. The moment LPG liquid from storage is pumped into the tanker, if the tanker pressure is lower than the VP, the liquid will flash and vapor will be formed. Temperature will be lower due to auto-refrigeration. However, to simplify our calculation, assume that the initial tanker pressure is very close or equal to VP such that auto-refrigeration does not exist and the process can be assumed to be isothermal.
4. The BIG question now is to determine the vapor composition.
There are two methods that you can use. Either using pressure slightly lower than the VP at the storage temperature or using slightly higher temperature at the storage pressure (I.e. the VP).
5. The reason I suggested using the second approach is due to the fact that temperature is normally higher after pumping. Therefore, increase the flash temperature by small increment (e.g. 0.01 etc) until you has some small fraction of vapor. Use this vapor composition as the BASIS of DESIGN.
6. What happens next is a process of COMPRESSION. You can visualize a reciprocating compressor in a single stroke where the liquid LPG is the “piston” of the compressor.
7. As the vapor is compressed, pressure increases such that the system is in “hydraulic equilibrium”, meaning the pressure (let say P2) of the vapor is sufficient to overcome the piping friction. Again, to simplify Calculation, we can assume an “isothermal process”. Of course, if you want more challenging calculation, you can assume adiabatic or polytrophic. However, let us work smart in this case.
9. Once the tanker pressure attain P2, pressure will be maintained at the storage temperature (We assume isothermal process, right?).
10. I stop here and let you figure out on the method to determine the vapor composition at P2.
11. GOOD LUCK. Let us know your results.

#22 kkala

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Posted 30 March 2012 - 02:22 AM

Post No 21 by S.AHMAD clarifies a lot of points of the matter. As explained in point 7, P2=gas pressure in the truck (tanker) is higher than P1=gas pressure in the sphere during steady state. I think this results in temperature T2 a bit higher than T1 (steady state), since gas pressures in both truck and sphere are the equilibrium pressures and liquid phase composition remains practically same in both. This can be considered in calculating equilibrium pressures, but hydraulic calculation of ΔP in the return line had better assume isothermal flow for simplicity.
ΔP in the return line will most probably be lower than assumed 20.1 psi in the attached "esquema1.pdf", so difference between T2, T1 will be lower than 89.4-77=12.4 oF, understood from the attachment (post No 20 also tries to comment esquema1.pdf).

#23 icasensio

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Posted 30 March 2012 - 04:42 AM

In my understanding vapor pressure of the specific LPG grade in the sphere has been estimated at 100.9 psia at 77 oF (25 oF), temperature that can be assumed constant. Gas flow in the return line (RL) will start as soon as gas pressure in the truck gets higher than 100.9 psia, but will increase till the value of 68.1 actual m3 /h coming out of the truck. Gas pressure in truck will increase to 100.9 psia+ΔP, which means increase of truck LPG temperature accordingly. So problem at steady state can be solved through a trial and error procedure (e.g. starting from an assumed LPG truck temperature). Gas LPG composition can be estimated depending on truck temperature (ΔP may not be very sensitive to it). It seems that first trial assumed for truck LPG: temperature=89.4 oF, gas presure=120 psia; then hydraulic calculation of ΔP in RL has to be elaborated for estimated LPG composition out of the truck (68.1 m3/h at 89.4 oF and 121 psia), to see whether this is close to 121 - 100.9 = 20.1 psi, and assumptions be revised accordingly. However line sizing criteria are applicable for the RL. We would allow frictional ΔP in the line not higher than 1.5 psi/100 ft (pressure 50-150 psig, pipe up to 300 ft); Norsok standard P-001 (para 6.3.3) would recommend lower than 0.5 psi /100 ft. So if actually ΔP=20.1 psi, this might be high (also depending on line length), meaning small line diameter. It is noted that check as above should be elaborated for max LPG temperature in sphere (77 oF or higher?).


Hello Kkala,

1.- I realized I should considerate an iteration calculation, because I have to know AP in return line (for which I need composition, pressure and temperature) in order to calculate composition of vapor line… Because of that and to understand well the problem, firstly I have considered an approximated AP. When I have the clue to resolve the calculations I will analyze the pressure drop in the return line.

2.- I would like to know why you define temperature 89.4ºF, and pressure 120 psia inside the tanker at the beginning. In my understanding, and as you have post before, there will be vapor and some liquid in equilibrium inside the tanker. Therefore, if we assume surrounding temperature is the same as tanker temperature, 77ºF, pressure inside will be 100.9 psia approximately.

3.- I have read somewhere (among thousand docs about LPG storage) LPG is stored in spheres at boiling point conditions, so if surrounding temperature is 77 ºF and it is assumed is the same as the sphere, pressure will be the pressure at bubble point (100.9 psia). Am I in the right way?

Thanks in advance.

#24 icasensio

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Posted 30 March 2012 - 07:33 AM

icasensio 1. Congratulation for the good effort. Your understanding on the system and process is improving but not well enough. 2. Kkala comments are valid. The actual calculation is “trial and error”. However, before you can do this iteration, you need to use the right method. 3. The moment LPG liquid from storage is pumped into the tanker, if the tanker pressure is lower than the VP, the liquid will flash and vapor will be formed. Temperature will be lower due to auto-refrigeration. However, to simplify our calculation, assume that the initial tanker pressure is very close or equal to VP such that auto-refrigeration does not exist and the process can be assumed to be isothermal. 4. The BIG question now is to determine the vapor composition. There are two methods that you can use. Either using pressure slightly lower than the VP at the storage temperature or using slightly higher temperature at the storage pressure (I.e. the VP). 5. The reason I suggested using the second approach is due to the fact that temperature is normally higher after pumping. Therefore, increase the flash temperature by small increment (e.g. 0.01 etc) until you has some small fraction of vapor. Use this vapor composition as the BASIS of DESIGN. 6. What happens next is a process of COMPRESSION. You can visualize a reciprocating compressor in a single stroke where the liquid LPG is the “piston” of the compressor. 7. As the vapor is compressed, pressure increases such that the system is in “hydraulic equilibrium”, meaning the pressure (let say P2) of the vapor is sufficient to overcome the piping friction. Again, to simplify Calculation, we can assume an “isothermal process”. Of course, if you want more challenging calculation, you can assume adiabatic or polytrophic. However, let us work smart in this case. 9. Once the tanker pressure attain P2, pressure will be maintained at the storage temperature (We assume isothermal process, right?). 10. I stop here and let you figure out on the method to determine the vapor composition at P2. 11. GOOD LUCK. Let us know your results.


Hello S.Ahmad,

Following your instructions, I have redone the calculations and I think I am in the right direction. (See attached file) Let’s see:

1.- I have assumed that pressure inside the tanker is 100.9 psia and temperature 77.001ºF (so LPG will flash inside).

Considering pressure inside the sphere 100.9 psia and temperature 77 ºF (saturated liquid), pressure at suction line of pump is 122.4 psia, and 235.2 at discharge line.
Then, I have modeled in Hysys a flash tank (from 77 ºF and 235.2 psia to 100.9 psia) and composition of vapor has resulted C3 82.7% and C4 17.3%. (Basis Design composition)

2.- In order to consider the compression of vapor in the tanker during its filling, I have modeled an isentropic compressor and a cooler after it in Hysys.
Compressor raises the pressure from 100.9 psia to 121 psia (pressure from sphere plus friction loss AP=20.1 psia) and temperature at the end is the same. 77.006.

With that composition, I will calculate the friction pressure loss in order to redo the calculations.

What do you think? Am I right?
Thank you for your patient, aid and time.

Attached Files



#25 kkala

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Posted 30 March 2012 - 11:41 AM

Icasensio, below is response to queries of post No 23, according to my understanding.
2. Concerning steady state, gas pressure in truck (tanker) was understood to be assumed 120 psia and corresponding saturation pressure of LPG 89.4 oF, from attached "esquema1.pdf". Truck LPG gas temperature corresponds to saturation pressure of 100.9 psia+ΔP, so it will be a bit higher than the temperature in sphere (assumed as saturation temperature of 100.9 psia).
3. We can assume ambient temperature for LPG in the sphere, but LPG in the truck will have somehow higher temperature (steady state), so that flow in return line will be caused.
It is noted that liquid LPG is assumed of same composition in truck and sphere; mass of gas is much less than mass of liquid (so flashing, if any, does not change liquid composition).




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