13 replies to this topic

[leyva_luis]

Hi,
I want to design a heat exchanger to cool GLP(it's a mixed mainly propane and propene) from 75.2°F to
44.6°F using propane at 15°F. The unit is to have propane in the tubes. Propane outlet is vapour saturated. This heat exchanger belongs to refrigeration system.

Thanks in advance

[Zauberberg]

Luis,

With such a poor description of your design task, you will hardly receive any feedback from the forum members. Your post is lacking of basic process information.

Please put some efforts from your side and provide us with the minimum required data.

Best regards,

[leyva_luis]

Ok Zauberberg, thanks for the advice.

I want to design a heat exchanger to cool LPG(Composition is in file attached).

inlet LPG Conditions: T = 75.2°F; P = 117.6 psig
outlet LPG Conditions: T = 44.6°F; P = 115.6 (I assumed pressure drop in shell side is 2 psi)
LPG is in the shell. mass Flow = 25 tonne/h

inlet propane Conditions: T = 23°F; P = 44.08 psig; vapor fraction = 0.3627(Mixed liquid-vapor)
outlet propane Conditions: T = 12.68°F; P = 34.08 psig; (saturated Vapor) (I assumed a pressure drop in the tube side as 10 psi).
Propane is in the tubes side. Propane mass Flow = 4.458 tonne/h

Basic assumptions:
Tubes: 3/4" OD x 14 BWG x 32 ft long
Pass Tubes: 2
Pass Shell: 1

How can I start the design?
How can I determined boiling coeficient?

Thanks in advance

Attached Files

•  GLP Composition.xlsx   9.21KB   442 downloads

[Art Montemayor]

Lucho:

I’m sure Zauber will help you out in your problem. He won’t solve it for you, but he’ll more than likely indicate and point you in the right direction with some helpful hints.

In reading your second posting (which is a BIG improvement in communications as compared to your first one) I have the following comments to further aid you get to the right answer:

• I edited your post to change the term GLP (Gas Licuado de Petroleo) to LPG (Liquefied Petroleum Gas). This is an important item because it can confuse your English readers on the Forum. Since it is LPG, they can immediately identify the state it is in: LIQUID (correct?). You don’t identify the phase in your basic data, and you probably are thinking in Spanish that your readers should know that – but in this English Forum they probably don’t.
• Since the tube side has LPG in it, I have to also assume that the inlet LPG is SATURATED (as LPG normally is). If that is the case, then I also have to assume that this saturated liquid will be cooled by the colder propane vapor in the shell side and, as a result, what you will have at the end of the tube side pass is SATURATED LPG also – but at a corresponding lower pressure (related to its exit temperature). This means that as the LPG is getting colder through its trajectory in the tube passes, its vapor pressure will be DECREASING due to equilibrium – and not due to frictional resistance through the tube passes. You will wind up with a pressure drop in the LPG, but it will be due to the lower exit temperature causing a lower vapor pressure in order to maintain the LPG in a state of saturation. I want to make sure that I make this point well so that you can fully understand what is confronting you. You don’t have a simple heat transfer application with a subcooled fluid. You are dealing with a saturated fluid that is subject to its vapor pressure. Now do you understand why it is so important to recognize that the fluid is a Liquefied Petroleum Gas? You cannot “assume the pressure drop in the shell side as 2 psi”; it is fixed for you by the corresponding exit vapor pressure. Many design engineers fail to recognize the importance of phase equilibria and wind up on the wrong page of a design fiasco in the field.
• Another important practical point to bear in mind when trying to design a TEMA type heat exchanger: obtain and read a good heat transfer text book or manual such as Don Kern’s classic, “Process Heat Transfer”, the GPSA Engineering Databook, Ernie Ludwig’s “Applied Process Design for Chemical & Petrochemical Plants”, and others. There you will discover that heat exchanger tubes are fabricated in standard lengths of 8, 10, 16, and 20 feet. There is no such length as 32 feet for exchanger tubes. I don’t know where you got that figure. Your resultant exchanger would wind up similar to a “weenie” or a piece of spaghetti. I recommend a length of either 10 or 20 feet – all depending on the length to diameter ratio that gives a reasonable stable structure for installation, fabrication, and maintenance.
• And now for the “kicker”: you are COOLING the LPG from 75.2 °F to 44.6 °F; to do that, you are taking Propane vapor and ALSO COOLING it from 23 °F to 12.68 °F. I DON’T BELIEVE YOU. If you can do that, all thermodynamic textbooks will have to be re-written.

Please check your post and your basic data.

[Propacket]

leyva_luis ,

• You have not described why are you cooling LPG? I presume that you want to subcool LPG to get its vapour pressure sufficiently below the saturation pressure.

• You said that LPG is on shell side and prpane is on tube side. Can you explain the philosophy of fluid allocation? Because If you want to vaporize a fluid by lantent heat transfer ,the best suited practice is to keep the vaporizing fluid (prpane in your case) in shell side and employ a kettle type shell. Reason of using kettle type shell herein the refrigeration loop is to completely eliminate the entrainment which will be destructive for your downstream equipment (refrigeration compressor). Another reason due to which I will not allocate boiling fluid to tube side is the fact that right from the inlet nozzle to outlet nozzle you don’t have any means to oust the vapors as soon as they are formed. These vapors will reduce the heat transfer area availabe for latent heat transfer and consequently, you will need to provide extra surface area during the design of the exchanger.This is not the case when you use a kettle type unit because vapors are removed from the boiling fluid as soon as they are formed providing efficient latent heat transfer.

• I agree with Art that 20 ft tubes are the most common. If i lave to purchase an exchanger, I shall prefer 20 ft long tubes for ease of transpotation and for ease of maitenance. But if am a heat exchanger manufacturer, I shall always prefer large L/D ratio because it is the least expensive option for me. For your information,I have seen as long as 40 ft tubes.

• Pressure drop through the exchanger is not just based on assumption. You will have to analyze how much pressure drop your system can bear. I shall disagree with 10 psi pressure drop you allocated to the tube side. This indicates that you are using a very small ID tube. I already described above that ,with your recommended design, you need to oversize the exchnager for good latent heat transfer. 10 psi is not possible with your design. However, if you employ kettle type unit ( and you should), pressure drop will be less than 1 psi.

Once you review your basic data and re-furnish the reviewd data, we will further explain how to size the exchanger.

Thanks

Edited by P.Engr, 11 August 2010 - 04:25 AM.

[Jason W.]

I've referenced "Heat Exchanger Design Handbook" by T. Kuppan quite a bit when I have had HE design questions. Might be worth taking a look at.

Jason

[leyva_luis]

I apologise everyone,

Art Montemayor/P.Engr

• You are right about the correct term LPG(Liquefied Petroleum Gas), but LPG is subcooled liquid at inlet of heat exchanger(because LPG is transfering from trucks tank with the aid of pump or compressor)
• I was wrong, I can't asummed drop pressure. Again, LPG is subcooled liquid at inlet of heat exchanger. While LPG is transfering (after heat exchanger, LPG is delivered to storage sphere. Where LPG will be liquid saturated)
• I selected kettle reboiler, TEMA type exchanger AKT. Lenght of tubes 20feet.
• I cooling LGP 75.2°F to 44.6°F. I am taking propane at 23°F(mixed liquid-vapor) and obtain propane as saturated vapour at 23°F.
• I want to cool LPG to get its vapour pressure sufficiently below design pressure of the storage sphere.
• I selected kettle reboiler, shell side is propane and tubes side is LPG. After reading your topics and Ludwig. I can understand why I have to select this type of exchanger(kettle reboiler).
I start again.

I want to design a heat exchanger to cool LPG (Liquefied Petroleum Gas)(Composition is in file attached).

inlet LPG Conditions: T = 75.2°F; P = 117.6 psig (subcooled liquid)
outlet LPG Conditions: T = 44.6°F; P = 117.6 psig (subcooled liquid)
LPG is in the tubes. Mass Flow = 25 tonne/h

inlet propane Conditions: T = 23°F; P = 44.08 psig; vapor fraction = 0.3627(Mixed liquid-vapor saturated)
outlet propane Conditions: T = 12.68°F; P = 44.08 psig; vapor fraction = 1 (saturated Vapor)
Propane is in the shell side. Mass Flow = 4.331 tonne/h

Basic assumptions:
Tubes: 3/4" OD x 14 BWG x 20 ft long
Pass Tubes: 2
Pass Shell: 1
Type Tema: AKT (Kettle Reboiler)

How can I start the design?
How can I determined boiling coeficient?

Thanks for your help.

Attached Files

•  GLP Composition.xlsx   9.21KB   364 downloads

[Art Montemayor]

Lucho:

I think we are all getting very confused. Your supply of data and information does not make sense and you do not furnish complete definitions or explanations of your proposal.

For example:

• You cannot have subcooled (supercooled) LPG in a tanker truck – unless you have a very special tanker truck (and a very expensive one, also). I believe you are receiving saturated LPG – and you have a supercooled LPG storage (or a low temperature, saturated LPG storage) that you want to put the LPG into without suffering a lot of expansion vapors.
• You say you selected an AKT exchanger. I don’t think you want to go down that road. This is a very special and troublesome type of tube bundle for this application. Why do you select a floating type of tube bundle in a Kettle shell? Do you seriously expect to have to rod out and inspect your LPG tubes? If you do, why would you be sending the LPG to storage at a low temperature? I would never advise you to use this type of bundle in this application. You should be using a normal, standard U-tube hairpin (“horquilla) type of bundle. This is much cheaper, simpler, and does not leak.
• Your description of the propane refrigerant is very confusing and lacks detail. It is impossible to have an exit propane stream at 12.68 °F and saturated at 44.08 psig as you describe. If you are going to have the exit propane vapor at 12.68 °F and it is saturated, then it MUST be at 34.2 psig (48.9 psia) – at least that is what the NIST database states. Please refer to a revised copy of your Workbook for a look at what I perceive as what you propose to do. Correct me if I have the wrong idea by indicating in the same sketch or another one exactly what you are proposing.

You can start your design by first firmly defining a credible and practical cooling method for your LPG. To do that you must find a reasonable and available refrigerant (coolant). The Propane is OK; except that you must define how exactly does it exist.

The so-called “boiling coefficient” becomes more understandable if you are trying to do what I have sketched out. I believe you mean to say that you want to know how to design the size of the tube bundle (surface area). Am I correct?

Await your reply.

Attached Files

•  LPG Subcooling Rev1.xlsx   78.82KB   537 downloads

[Zauberberg]

1. The refrigerant (Propane) does not meet the outlet conditions you specified. In reality - assuming that you are dealing with 100% pure Propane - the true conditions will be as indicated by Art; similar results are obtained by Aspen software (see attached sketch). What is the actual composition of refrigerant? It is very impractical to consider 100% pure Propane in industrial applications, unless you are in very specific position to have such a luxury.

2. Secondly, it's practically impossible to design S&T exchanger with no pressure drop in the tube-side. For a two-pass U-tube design (no reason not to employ AKU or BKU design instead of having floating head - I have almost made that mistake once), assuming 0.3 - 0.7 bar pressure drop is reasonable. You will calculate the actual pressure drop once when you confirm the tube layout resulting in the overall heat transfer coefficient you have selected for your final design. On the other hand, having virtually zero pressure drop across the shell is fairly OK for evaporators.

3. The outlet temperature approach seems to be unreasonably high (~18 degC), you may go for lower values and see what differences (in the overall heat transfer area requirements) you get. In reality, this temperature will be driven by the suction pressure of Propane refrigerant compressor, which should be an independent variable in the process, in order to allow for duty trimming and proper utilization of Propane condensers/Compressor power saving.

4. As for the boiling/overall heat transfer coefficient, you can refer to Ludwig's "Applied Process Design" and Kern's "Process Heat Transfer", as they offer complete procedure for evaporator/reboiler design. If you have any of the software tools such as Hetran/HTFS or HTRI in your company, you may run it a few times and seek for optimum design - but first make sure that you can do it by hand calculations in order to be 100% confident in what you are getting from the software.

Attached Thumbnails

• 

[Zauberberg]

A very important point I forgot to mention is that - there cannot be such a huge temperature drop across the shell-side in boiling applications, as shown on the sketch I have uploaded and which was based on your initial data. Both Propane inlet and outlet temperatures will be almost identical, since it is a latent heat transfer application.

Attached Thumbnails

• 

[leyva_luis]

I was wrong about outlet temperature propane. I corrected it

Inlet LPG Conditions: T = 75.2°F; P = 117.6 psig (subcooled liquid)
Outlet LPG Conditions: T = 44.6°F; P = 117.6 psig (subcooled liquid)
LPG is in the tube side. Mass Flow = 25 tonne/h

inlet propane Conditions: T = 23°F; P = 44.08 psig; vapor fraction = 0.3627(Mixed liquid-vapor saturated)
outlet propane Conditions: T = 23°F; P = 44.08 psig; vapor fraction = 1 (saturated Vapor)
Propane is in the shell side. Mass Flow = 4.331 tonne/h

Basic assumptions:
Tubes: 3/4" OD x 14 BWG x 20 ft long
Pass Tubes: 2
Pass Shell: 1
Type Heat Exchanger: floating type of tube bundle in a Kettle shell

Art Montemayor, I attached a LPG Subcooling Rev2. There is a diagram with description in Spanish

Thanks in advance

Attached Files

•  LPG Subcooling Rev2.xlsx   296.85KB   430 downloads
•  GLP Composition.xlsx   9.21KB   357 downloads

[Propacket]

luis,

• Now that you have corrected the basic data as per recommendations of Art and Zauberberg , you must have learnt how to analyze your basic data before you start designing any equipment. You must consult some heat transfer books for design of the kettle type vaporizer. Kern and Ludwig books are good to start with. However, a general guideline for design of this type of exchanger is as follows.

• First of all, study the basics of pool boiling. You will find different regions of pool boiling as convective boiling, nucleate boiling, transition boiling and film boiling. You must design the vaporizer to operate in nucleate boiling region as heat flux increases sharply with temperature difference in this region.

• Keep the heat flux below the critical heat flux. Beyond the critical heat flux, an insulating layer of bubbles is formed which decreases the heat flux.
• Estimate critical heat flux with Zuber equation. Kern recommends 12000 Btu/hr.ft2. However, it will depend upon your service. I have seen as much heat flux as 16,000 Btu/hr.ft2 recommended by heat exchanger manufacturers.

• Assume overall heat transfer coefficient as you do in design of all types of heat exchangers. For kettle type exchangers, overall heat transfer coefficient is around 100 Btu/hr.ft2.°F.
• Calculate the required surface area. Based on 3/4” ID, 20 ft length of tube, calculates number of tubes.
• Based on tube layout, find out the bundle diameter. Square layout is the most common for kettle type design as it gives minimum pressure drop.

• Using Mostinsiki’s equation, find out the boiling coefficient.

• Assuming reasonable values for tube side coefficient and fouling coefficient, calculate overall heat transfer coefficient. If it is in agreement with the assumed value, your design is OK. Otherwise, perform iterative calculation as you do in design of all types of heat exchangers.

• Now, you need to design kettle portion of the exchanger. Diameter of the kettle will be governed by maximum allowable vapour velocity. Velocity through the kettle is limited to 75 % of maximum allowable vapor velocity to avoid entrainment. For rough estimate of the maximum allowable vapour velocity, you can use famous Souder Brown equation with K value of 0.2 ft/s. This will complete thermal design of the kettle type vaporizer.

I have found the HTRI design manual the most detailed and most accurate for the design of exchangers. If you can find a copy of the manual, it will be really helpful for you.


Thanks

Edited by P.Engr, 12 August 2010 - 02:58 AM.

[Art Montemayor]

All:

Guys, before we spend any more time and effort on this topic, let’s get the basics straight – once and for all:

1) This thread has originated from one that had little or no basic data;
2) Then it went to an extended explanation of the scope of work;
3) The basic data furnished was erroneous;
4) The basic data and scope now start to appear as what I had suspected all along – this is NOT a heat exchange problem. This is an attempt to unload saturated LPG from a road tanker truck into a low-temperature LPG storage tank by using a Propane refrigeration system to lower the LPG vapor pressure.

The REAL story now appears as follows:

1) The LPG being unloaded is saturated at ambient temperature (as I suspected) and 100 psig. The low temperature Storage Sphere exists at 44 ^oF and 57 psig.
2) What is being proposed is to use a vapor displacement type of unloading system to transfer the tanker truck LPG through a Propane Vaporizer coil (the proposed “kettle exchanger”) and into the Sphere as sub-cooled LPG.
3) The Propane Vaporizer is a unit unto itself as it is part of a closed Propane refrigeration unit.
4) As has clearly been pointed out, this is not a conventional heat exchanger problem. This is a refrigeration vaporizer application – straight and simple.
5) What is getting very complex and mixed-up is the proposal to employ mechanical energy in the form of a vapor compressor to forcefully displace the vapors in the Sphere with higher pressure liquid that exists in the tanker truck. This, in my opinion, is totally illogical and wasteful. A pressure driving force already exists inherently in the proposed unloading system. There is no need for an unloading compressor if we are to believe the vapor pressure data.
6) All that is simply required is to employ a Propane vaporizer (a standard BKU design is fine and OK; a floating tube bundle type – such as AKT – is ludicrous because there is no justification for this type that is meant for dirty tube-side fluids and is prone to leakage through the internal gasketed joint, especially in a service that has a temperature gradient) and connect the road tanker directly to the vaporizer-subcooler with an expansion/control valve before the vaporizer to contol the pressure downstream. The vaporizer will subcool and condense the subsequent products of the LPG expansion.
7) Another option is to simply expand the 100 psig LPG DIRECTLY into the Sphere and use the Propane vaporizer as a condenser for the subsequent expansion vapors. I have favored this method in the past because of its simplicity and economy of capital and operating costs – especially when one already has a refrigeration system installed to control the Sphere vapor pressure – which would be a normal design criterion. However, this entails having an elevated condenser.

I have commented and marked up the Rev3 workbook that I am attaching. I have had to do this in Spanish out of courtesy and convenience for the originator.

The original thread has been completely turned around and mixed up. This, as usual, is due to a reluctance or un-willingness on the part of OP to share and contribute ALL of the basic data and scope of work at the very outset of the thread – as well pointed out by Zauberberg. I recommend we conserve our energy and comments until we get confirmation and agreement on the part of the OP regarding what I have stated here. This is a very practical and interesting problem to resolve – but we need FULL cooperation and input from the OP. It is too bad that we couldn’t get this at the very start.

Attached Files

•  LPG Subcooling Rev3.xlsx   298.9KB   372 downloads

[leyva_luis]

Thanks P.Engr/Art Montemayor/Zauberberg,

Kettle type vaporizer
Tubes: 3/4" OD x 14 BWG x 24 ft long
Pass Tubes: 2
Pass Shell: 1
Kettle-type-U-tube-evaporator

After iterations the values calculated are shown:

Heat Duty = 1,004,052 BTU/h
ho = 300 BTU/h.ft².°F
hi = 307 BTU/h.ft².°F
GLP side fouling = 0.002 h.ft2.°F/BTU
propane side fouling = 0.001 h.ft².°F/BTU
Ai(internal surface) = 0.1529 ft²/ft length
Ao(external surface) = 0.1963 ft²/ft length

U = 95.2 BTU/h.ft².°F
N° of tubes = 72 tubes
Area = 325 ft²
Heat Flux = 3,088 BTU/h.ft²


Art Montemayor, I attached Rev4 for your check.

Thanks for your support.

Attached Files

•  LPG Subcooling Rev4.xlsx   305.28KB   536 downloads