18 replies to this topic

[harshad katre]

Dear All,
I am studying LPG vaporizer system. I want to check the heat transfer through LPG vaporizer to calculate thermal efficiency. I am giving the parameters to calculate the efficiency.

LPG In at 30⁰C, 6 kg/cm²
Out at 70⁰C, 4 kg/cm²
Flow=200kg/hr

Water In at 70⁰C
Out at 58⁰C
Flow= 15 m³/hr, pressure=2.5kg/cm²

Please give me the calculation steps.

Thanking You

[Art Montemayor]

Harshad:

What you are proposing cannot be carried out as a heat transfer operation. It is impossible. Therefore, the efficiency is not applicable.

You are dealing with what you call “LPG”. This, in the USA, is a mixture of liquid butane and propane – usually 50% weight by each. Its formal name is Liquefied Petroleum Gas and it exists in the saturated, liquid state, usually stored in what we call “bullet” storage vessels. These vessels are normally designed for 300 – 350 psig and ambient temperature service. Pay careful attention to what I just wrote: the mixture of hydrocarbon gases exists as SATURATED LIQUID. I have written about this very subject on many past threads, stressing the fact that this is a saturated liquid and has to be dealt with as such.

Therefore, you cannot vaporize the saturated liquid at any other pressure except at its own saturated temperature – whatever that happens to be. We usually install a vaporizer besides a storage tank where the vaporizer is gravity fed by the tank and the vaporized portion is piped up to the vapor space of the storage tank. The product vapor is withdrawn from the tank vapor space and the vaporizing fluid (or other heat source) is regulated by the pressure in the storage tank. This is a very simple and trouble-free operation. HOWEVER, your heat source must be at a higher temperature level than the saturated temperature of the liquid LPG in order to have heat transfer. You need a DRIVING FORCE. At present, you have no driving force (you are trying to create LPG vapor at 70 ^oC by using water at 70 ^oC. That will never work.

Additionally, as I stated, you cannot create a vaporized LPG at a different pressure than that of its saturation pressure – whatever that turns out to be according to its temperature. It is only AFTER you vaporize the LPG at its saturated pressure that you regulate that pressure down to the desired level you need. That is the way LPG is vaporized.

By the way, how did you arrive at the flow rates for BOTH the LPG and the hot water? Please show us how you came up with those figures.

[harshad katre]

Sir,
I am attaching one file containing the general system diagram which i am studying. Please help me to find out how can i show heat released= heat absorbed.

Regards,
Harshad

Attached Files

•  Book1.xlsx   13.98KB   326 downloads

[Art Montemayor]

Harshad:

You told us you are studying an LPG vaporizer system and you want to check the heat transfer through the LPG vaporizer to calculate thermal efficiency. You gave us the parameters you intend to use to calculate the efficiency and then you submitted your block diagram, asking how you can show that the heat released equals the heat absorbed.

I made several comments and raised some questions in my prior post. You have not responded to these comments and questions, and have probably ignored them. This is not a good way to obtain help and recommendations from engineers who are trying to help you and know the answer to your query. Please address the detailed PFD and comments I have added to your worksheet. Note how my graphic depicts exactly what I wrote in my prior post. It is simple, direct, and it works. It is a system working around the world for a long time. This is the way it is done in industry.

Your query on the heat released is not logical. It is elementary observation that leads anyone to arrive at the conclusion that the heat added by the hot water to the LPG is equal to the heat required to vaporize the saturated LPG – which is the latent heat of vaporization for the LPG mixture at the pressure it exists at. There is a definite methodology used to arrive at the latent heat of vaporization for the LPG mixture but since you haven’t asked for that, I’ll omit detailing it out.

I hope you come to realize what important learnings you have deprived yourself of by not responding to our comments and questions.

Good Luck.

Attached Files

•  LPG Vaporization.xlsx   22.36KB   473 downloads

[harshad katre]

Sir:
Sorry for late reply. First of all, I want to clear misunderstanding about the answers to your questions in previous posts.

Regarding the flow, I am arriving to the flow rates with the help of flow meter and mass flow meter installed at the pump delivery and at the inlet of LPG vaporizer respectively which I have shown in the block diagram. I regret you about the temperature specifications given by me in the first comment because there is no DRIVING FORCE as you said.

The PFD you depicted is right but with the small change that the LPG mixture (60% propane & 40% butane) is supplied directly to the usage point where I am studying the system. Also there are no pressure controllers. Instead pressure regulators are used to serve the purpose.

So, please tell me the methodology to arrive at the latent heat of vaporization of LPG mixture for the following specification:

LPG In: Pressure- 6kg/cm2; Temperature- ambient
Out: Pressure- 4 kg/cm2; Temperature- 50 to 60 Deg C
Water In: Temperature- 70 to 80 Deg C
Out: Temperature- 55 to 65 Deg C

Thanking You.

Regards,
Harshad

Edited by harshad katre, 23 June 2012 - 02:37 AM.

[Art Montemayor]

Harshad:

I believe you know the fundamental concept of reciprocal respect when receiving help (especially for free). I could also bide my time and render a late reply. I won’t. My point here is that you come asking for free help – but you want help under YOUR conditions. That’s OK; there is nothing wrong in having great expectations. However, that’s not the way life is normally carried out.

You have a different scheme on how to vaporize LPG (which I don’t approve of because it doesn’t respond well to vaporizer demand) and you insist on doing it that way. I would advise you to change it because it is a cheap (low initial capital cost) way to vaporize LPG and it will not work well – if at all. Regulators are nothing more than cheap control valves without a pressure controller. If that is what you call a sound engineered installation, so be it. You haven’t even taken the time to draw a schematic diagram like I did. In fact, you haven’t even made the effort to use my drawing as your basis to make modifications or point out where your scheme is different.

Now you tell me you need for me to tell you the methodology to arrive at the latent heat of vaporization of LPG mixture. OK; I am outlining the method and equation (with a documented reference) in Rev 1 of the workbook. I hope this helps you out. I even furnish you with a table of recommended Overall “U”s.

Attached Files

•  LPG Vaporization Rev1.xlsx   30.19KB   346 downloads

[harshad katre]

Sir:
Thank You for the methodology to calculate the latent heat of vaporization.
But I could not find the figure nos 2, 6 and 7 required to obtain latent heat of vaporization for mixture and individual constituent & weight per gallon of individual at desired temperature.
So will you please tell me where will I get the book (soft copy) you mentioned (Handbook Butane-Propane Gases)?

Thanking You.

Regards,
Harshad

Attached Files

•  LPG vaporizer cals.xlsx   30.19KB   227 downloads

Edited by harshad katre, 09 July 2012 - 10:33 PM.

[harshad katre]

Dear Sir,
Thank You for the methodology to use LPG vaporizer in the way you suggested.
I think it serves two purposes.
1) We can get benefit of natural vaporization (evaporation).
2) We can avoid the flow of liquid to the end user if vaporizer doesn't work for small time.

Please tell me am I right?

Also there are questions,

1) Whether there is any energy savings in the use of heaters for vaporizer?
2) In your last post (I am attaching with this message), you have given me methodology to calculate latent heat of vaporization in which you missed some diagrams (mentioned as red text). Will you please give them to me?

Thank You once again for valuable information.

Regards,
Harshad Katre

Attached Files

•  LPG Vaporizer.xlsx   22.86KB   135 downloads

Edited by harshad katre, 27 July 2012 - 02:14 AM.

[breizh]

http://infosys.korea.../kdb/index.html

Using this data base you should be able to find the properties of the chemicals ( Hydocarbon >>>> pure properties)

Hope this helps
Breizh

[harshad katre]

Thank You..
But also I want to have the properties for mixture of LPG(composition same as that of above). In this site for binary mixtures it is not showing the properties.

Regards,
Harshad

[breizh]

Apply the mixing rules as per Art's suggestion or use google "mixing rules" for the property you want to cover .

Breizh

[Art Montemayor]

Harshad:

This is the Student Forum. By Student, we mean a person who is in the process of learning. We on the Forum don’t teach; that is not our goal or our challenge. Rather, we guide, advise, recommend, and share personal and professional experience with students in order to assist them in their learning process. Our goal is for students to learn Chemical Engineering and become good at this profession. But we do not offer our free and voluntary services on conditional terms put forth by students. Our help is offered as we see fit.

By referring to the attached Rev1 workbook, you will note that not only have I kept watch on the progress of your threads on this simple vaporizer application, but I have also shown how to resolve the problem. To date, you have meandered through:

• Trying to vaporize LPG on-line;
• Vaporizing LPG with Solar panels;

This has gone on for 1-1/2 months. This is far too much time for such a simple solution. The lesson to be learned here is to carefully read and heed the advice given. Note that you do not require any further information other than the usual thermodynamic properties of saturated Butane and Propane – as Breizh has so kindly given you a hint and a lead on how to procure this information. You can also use my favorite source for this type of valuable data – as clearly indicated in my example calculations for the latent heat of vaporization. You should proceed immediately to make the required calculations yourself, using a convenient conversion program – such as UCONEER, which you can obtain on the Internet. UCONEER is a product of one of our outstanding Forum members, Mr. Harvey Wilson and is, in my opinion, the best conversion program on the market.

To comment on your thoughts, I believe you leave out the most valuable engineering inputs in the design I have advocated:

• It is simple, and less complex;
• It uses less and smaller equipment;
• It maintains a constant vapor pressure in the storage vessel.

To answer your specific questions:

1) Whether there is any energy savings in the use of heaters for vaporizer?

I don’t understand the meaning of this question. A vaporizer needs heat to function as such; a vaporizer is a generic heater – although it only adds heat without a temperature increase.

2) In your last post (I am attaching with this message), you have given me methodology to calculate latent heat of vaporization in which you missed some diagrams (mentioned as red text). Will you please give them to me?

You err in stating I “missed” the diagrams. You do not need diagrams; you need the thermodynamic properties which you can obtain through conventional means. As one of our contributing members, Shivshankar, has so aptly stated on previous occasions: “(If you are to become a successful graduate Chemical Engineer) you’ve got to make an effort, man!” If you read the definitions of the terms that I’ve highlighted in BOLD, you will see that all you have to do is convert the thermo data you can easily obtain – for free.

I hope this information resolves your vaporizer design and gives you a Lesson Learned to apply on future similar problems.

Attached Files

•  LPG Vaporization Rev1.xlsx   453.8KB   270 downloads

[harshad katre]

Dear All,
Thank You Sir for the valuable input. I have learned lot many things from all of your post.

1) Whether there is any energy savings in the use of heaters for vaporizer?

This question does mean that the vaporizer will work only when the pressure in the storage tank falls below the set OR it will control the flow of hot water through the vaporizer. This means that heaters used for heating water will opearate according to pressure controlled inside the storage tank. Because of this will this lead to any energy savings?

Thanking You all.

Regards,
Harshad Katre

[Zahid Mauladad]

Hi, I was on the verge of having a LPG vaporiser fabricated for my company but after reading the discussion here on this forum in particular the input from Art I delayed my decision as I need to know more about the components and features of a LPG vaporiser..My company has boiler which is running on natural gas but during winter their is a shortage of natural gas therefore we are seriously thinking of having LPG as a standby fuel..But to have proper combustion we need to vaporise the LPG ..We have LPG available in 60kg cylinder pressurised at 100psi..
Please advise/guide me..
Zahid

[Neelakantan]

i was reading this just out of curiosity; i find fresh engineering students dont even have a basic idea of what is heat transfer, what is the driving force for heat exchange, simple stoichiometric and heat balance calculations! i am surprised ART explained so patiently to a non-receptive guy.

coming to the issue @ zahid!
understand LPG is liquid when pressurised; when released of the pressure at the PCV before burners, it is a mixture of butane and propane; having said that, we need to know if your burner is suitable for a rich gas as LPG; what is your ambient in winter?

then confirm if you really need to vaporise LPG and what pressure and temperature you need at the fuel gas header;
then we can inform you about the way of doing it; (i wont do it myself!)
regards
neelakantan

[kkala]

1. Having recently read the whole thread, I have had the opinion that the query aimed at checking consistency of adopted data (including a heat balance) rather than getting help to start basic design. First queries asked "how to show that heat released = heat absorbed" in an "LPG vaporizer system", apparently containing means of throttling. Some of the given data is result of measurements "with the help of flow meter and mass flow meter".
Harshad katre might not be prepared for design matters yet; but touching them in previous posts was admittedly beneficial. Let us check adopted data now, by using input and output enthalpies for the steams of LPG (60/40 w/w C3H8/C4H10) and of water.
2a. Heat gained by LPG, see attached "LPGef.xls" for enthalpies.
Enthalpy of LPG in (30 oC, 7.3 kgf/cm2 g, all liquid) : 505.4 kJ/kg
Input pressure has undergone "reconciliation" to 7.3 kgf/cm2 g, so that all ingoing LPG is liquid.
Enthapy of LPG out (60 oC, 4.0 kgf/cm2 g, all vapor) : 907.5 kJ/kg
Heat gained by LPG : 907.5 - 505.4 = : 402.1 kJ/kg
Heat gained by total mass of LPG stream : 200*402.1 kJ/h = 80420 kJ/h (22.34 kW).
2β. Heat lost from water (steam tables can be used)
Water density at 70 oC 977.7 kg/m3, mass of water in 15.0*977.7 =14666 kg/h
Enthalpy of water in (70 oC) : 293 kJ/kg
Enthalpy of water out (58 oC) : 242.7 kJ/kg
Heat lost from water : 293-242.7 = : 50.3 kJ/kg
Heat lost from total mass of water stream : 14666*50.3 = 737700 kJ/h (205 kW)
2γ. According to the above, there is error in adopted data to be radically revised. I would first check actual water flow rate to see if it is much lower.
When this procedure is repeated, figures should be fixed (avoid ranges). However "the methodology" ("the calculation steps") has been presently requested, not the balance itself. Used figures are a bit arbitrary (but conclusion rather clear); fixed figures may be ready later (by Harshal katre) for new heat balance check.
3. Comments on this post would be welcomed.

Attached Files

•  LPGef.xls   26.5KB   206 downloads

Edited by kkala, 26 January 2013 - 06:24 PM.

[Art Montemayor]

Kkala:

There seems to be a discrepancy between your Thermo data on Propane (and presumably also on Butane) and that of NIST Thermo data. This puts a question on the accuracy of your calculations based on saturated conditions within the vaporizer.

Please refer to the attached Rev of your workbook. Perhaps my version of the data is in error, but I would appreciate your thoughts and comments on this.

Thanks

 LPGef-Rev1.xls   54KB   177 downloads

[kkala]

1. After the comments on "LPGef.xls" by Art Montemayor, post no 17, attached "LPGef-Rev2.xls" was made to contain following worksheets.
(α) "LPGef", same as the one attached to post no 16 by kkala (for reference).
(β) "Thermo Data", containing the comments of post no 17.
(γ) "NIST" to calculate enthalpies of n-propane / n-boutane in and out, according to NIST, <http://webbook.nist.gov/chemistry>. It also contains some comparison notes.
(δ) "Perry" to represent "LPGef" after consideration of "Thermo Data". Changes are shown by pink letters.
2. Enthalpies in "Perry" have remained same as in "LPGef". The enthalpies agree to "NIST" quite well, thus there is no discrepancy between the two. Conclusion (2γ) of post no 16 remains as is.
It is clarified that incoming LPG is assumed 100% liquid, though not explicitly stated in the posts. In case this is not, the gap between heat lost from water and heat gained by LPG gets somehow greater.
As indicated in para 4 below, almost exact enthalpy agreement between "NIST" and "Perry" is accidental. There are some uncertainties detected in "Perry", still agreement is quite satisfactory.
3. Notes relevant to the comments of post no 17.
3.1 Liquid enthalpies depend practically only on temperature for limited pressure variations. For a given temperature it is sufficient to use available liquid enthalpy of C3H8 or C4H10 at their saturation pressure.
E.g. enthalpy of liquid C3H8 at 300 oK and saturation pressure (~10 bara) is 593.11 kJ/kg (Perry, 7th ed, Table 2-297). Liquid specific volume = 2.044E-3 m3/kg. Increasing pressure by 5 bar = 5E5 Nt/m2, enthalpy gets higher roughly by 5E5*2.044E-3 Ntm/kg = 1.02 kJ/kg, not a big difference. Difference per NIST is much lower, seeing that liquid C3H8 enthalpy at 300 oK is presented  on the enthalpy versus pressure (10-20 bara) NIST diagram by an essentially straight line parallel to x axis (enthalpy variation 270.39 - 270.41 kJ/kg).
3.2 Assume liquid mixture (60/40) of 60% C3H8 w/w + 40% C4H10 w/w  (ideal mixture and vapor) stored in a drum at 30 oC constant temperature. Mixture vapor pressure is 8.34 kgf/cm2 a (see "LPGef").
3.2.1 For exerted pressure higher than 11.1 kgf/cm2 a (C3H8 vapor pressure), there will be only liquid LPG without vapor phase over it.
3.2.2 In case of "free" space in the drum, total pressure over LPG liquid (60/40) is 8.34 kgf/cm2 a. Liquid flow to vaporizer (diagram "Book1.xlsx", no return to drum) will develop no flashing, as long as frictional pressure drop remains lower than liquid static pressure (between storage level and vaporizer inlet). This is the case assumed, where liquid LPG (60/40) is under pressure of 8.34 kgf/cm2 a in the drum.
Note: As the drum is emptied, vapor phase volume will be more and enriched in C3H8; yet change in liquid LPG composition will be limited, since mass in vapor phase remains quite small. 
3.2.3  Asking C3H8 enthalpy from NIST at 30 oC and 8.34 kg/cm2 g, it will supply enthalpy of C3H8 vapor, not liquid. Thus a pressure of 11.1 kgf/cm2 a is considered for enthalpies of liquid C3H8 and C4H10 in "Perry" and "NIST". This will not practically affect accuracy, according to para 3.1. 
3.3.1 Differences between "Perry" and "NIST" in enthalpy values are due to different enthalpy base (standard state convention, per NIST), that is the conditions where enthalpies are taken as 0. Comparison between "Perry" and "NIST" indicates this. Enthalpy difference between two states counts, difference in "absolute" enthalpy values does not.  Even different base between C3H8 and C4H10 enthalpies (as in "Perry") does not  render wrong results in the present case.
3.3.2 Additional examples of difference in enthalpies.
-Saturated propane, 300 oK : 593.11 kJ/kg (L), 926.41 kJ/kg (g), difference 333.3 kJ/kg  (Perry 7th ed).
 Saturated propane, 300 oK:  270.4 kJ/kg (L), 603 kJ/kg (g), difference 332.6 kJ/kg (NIST, see Propane_Butane_Thermo_Properties.xls in <http://www.cheresour...and-butane-mix/ >).

-See also example in para 3.1 above.
4.  "Perry" and "LPGef" use propane enthalpies from Perry, 7th and 4th edition, to be checked for consistency.
Saturated C3H8, 10 oC : 543.8 kJ/kg (L), 906.9 kJ/kg (g), difference 363.1 kJ/kg (Perry 4th ed, Table 3-254)
Saturated C3H8, 10 oC : ........................., 906.7 kJ/kg (g), (Perry 4th ed, Table 3-255, interpolation 41.69 - 55.62 oF)
Saturated C3H8, 10 oC: 548.1 kJ/kg (L),  909.3 kJ/kg (g), difference 361.2 kJ/kg (Perry 7th ed, Table 2-297)
Assuming same base, there is difference of about 5 kJ/kg in liquid and 3 kJ/kg in vapor enthalpies.  Uncertainty would be less in case only Perry 4th ed were used.
5. Hopefully above gives a clear view, comments by anybody are welcomed (especially on whole para 3.2)

Attached Files

•  LPGef-Rev2.xls   78KB   231 downloads

Edited by kkala, 06 February 2013 - 03:05 AM.

[hsaranha]

Dear Art & All,

 

Your experience is reflected by the very informative comments. Need guidance on this design problem.

 

I am designing an Propane Tank with Vaporizer. The ambient temperature is min -2.5 to max 38 C. The pressure required is 15 barg, for which the liquid needs to be heated to 47.5 C by external boiling (vaporizer).

 

The vendor has suggested a system where the liquid inlet line has a Flow control Valve that is controlled by the water bath temperature to be maintained between 50 to 55C. If temp goes below 50 C, that means that the flow exceeds the heater duty and the valve closes. So flow is regulated to maintain the bath temperature between 50 to 55 C.

 

After the required pressure is achieved if no gas is drawn, the pressure of the system which includes the vaporizer will increase based on liquid temperature in the tank.

 

The vendor suggests a pump upstream of the vaporizer during start up but is not too sure and we are not convinced. We think that as long as the liquid level in the tank when at its lowest position is higher than the vaporizer plus the liquid line pressure loss for suction and in the vaporizer which should be about 1ft head we may not need the pump, 4.5 ft head is provided for.

 

Please advice if a Propane liquid pump (vane, regenerative turbine, gear type) is required to ensure liquid flow through the vaporizer for this system.

 

Second question is that is the liquid level and vaporizer tank height assumption correct to ensure liquid flow.

 

Thanks