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Heat Exchanger Design


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#1 Guest_RN_*

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Posted 22 February 2006 - 10:56 PM

Hello,
I'm in the middle of my 4th year plant design course and I am left the challenge of designing a heat exchanger to heat toluene at T = 25.76 C & P = 950 kPa to T = 167 C & P = 900 kPa. I figured out the pressure drop using heuristics from my class. I was thinking of using steam for the hot stream into the exchanger to heat up the toluene, but if you have another idea, let me know. Again using heuristics, I know to bring the steam in at 400 psig and 448 F, but I dont know what to expect for my outlet temperature. I also need to figure out a flow. Also, I am in need of advice for the type of exchanger and material to be used. Any help in this matter would be greatly appreciated.

Thanks

#2 aliadnan

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Posted 23 February 2006 - 09:28 AM

Hi

First should consult "Process Heat Transfer" By Kern because it seems that you don't have much idea about heat exchanger design. Look you have to calculate the actual pressure drop and not to use a heuristics to predict it. As long as the use of Steam as a heating medium is considered then you should have the this steam in sufficent amount produced somewhere else in the process. You cannot use a boiler to produce steam at these conditions just to heat toluene because it is highly uneconomical. Since it is the final year plant design project so you should have idea of what other streams in your process can be used as a heating medium.
If your going to use Saturated Steam then its oulet temperature will be same since I think you just want to use the latent heat of it and don't want to subcool it.

Now if your talking about how to calculate flow of steam then it is a simple job.
Since you have the inlet and outlet temperatures of Toluene and you should also have its flowrate. Use the equation

Q = m*Cp*dt

Q= heat gained toluene
m= mass flowrate of toluene
Cp = heat capacity of toluene

then put this Q = m*lemda
lemda = latent heat of steam

since you know both lemda and Q, from the above equation you can calculate the flowrate of steam.

Consult a Heat Transfer Equipment Design book to know about different kinds of exchangers type and what you can use for your purpose.

Please correct me if I am wrong in my statements.
I hope this help.

Regards
Ali

#3 djack77494

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Posted 23 February 2006 - 02:13 PM

Hello,
Definitely put the steam on the shell side of the exchanger. I'd use something like a TEMA type A - E - S or U exchanger; in other words, use an AES or AEU type exchanger. You get very little energy from cooling superheated steam to saturation and you get very little energy from subcooling water below saturation. For this reason, on the steam side of the exchanger forget anything except for condensation. Steam should be a very effective heating medium for your application. Calculate its flowrate using the fact that the energy gained by the toluene must equal the energy supplied by the steam (like Ali has said). Keep things simple. Since your steam is saturated, set the steam inlet temperature and pressure = condensate outlet temperature and pressure. This is very realistic if you won't overheat your toluene, and it simplifies any heat transfer calculations you might do. A carbon steel exchanger should work fine.
Good luck,
Doug

#4 Guest_RN_*

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Posted 23 February 2006 - 09:06 PM

So you're saying that because my steam is saturated, I can assume that the temperature & pressure of what is coming out stays the same? Correct me if I'm wrong here.

#5 aliadnan

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Posted 24 February 2006 - 09:08 AM

Hi RN

The reason that your steam temperature at outlet remains the same as inlet is that your just using the LATENT heat of steam and condensing it. Your not subcooling the Steam i.e. your not lowering the temperature of steam below its saturation temperature at that pressure. It is not assumption infact it is reality that when only LATENT HEAT is removed there is not change in temperature, since Latent means HIDDEN and removing latent heat means only change in phase not change in temperature. The temperature will change if you remove sensible heat from the steam.

Doug don't you think that the steam should be on the TUBE side. I am saying this because the steam which RN wants to use is 400psig and it is a high pressure steam and it is much more economical to use high pressure steam in the Tube side rather than shell side.


Hope it helps.

Ali

#6 djack77494

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Posted 24 February 2006 - 10:07 AM

RN,
I think Ali has explained things quite well. My inclination would be to run the steam side of the exchanger dry. By that I mean that any water condensing from the steam would immediatly drain and be removed from the exchanger. There is an alternative approach where you would maintain a controlled level of water in the heater. You would do so to reduce the effective heat transfer area, since the part of the heater that was flooded would not be very effective.

I would concede to Ali that it is probably better to put the steam on the tube side. Let me blame the mix of units for not quickly realizing the much higher pressure on the steam side. That fact, of course, would point to putting steam on the tube side. Thought there are a lot of other considerations involved in this decision, given the system definition provided, I think Ali is correct. Steam should be put on the tube side.
Doug

#7 Guest_RN_*

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Posted 24 February 2006 - 10:22 AM

Thanks for all your help guys, but I have run into a problem with my calculations. I calculated a Q = 855 kW, so I did what Ali said and used the latent heat equation to find a flow for my steam. It was 0.47 kg/s or 1703.7 kg/h. I then used Q = mCdt to calculate the outlet temperature of my steam. My C came from the assumption that there should be no change in P or T between the inlet and outlet (like you guys said) and I ended up with a negative T. Any ideas? Or am I approaching it wrong, since I already assumed a P & T, does it matter about calculating the outlet based on the assumed C? Let me know guys, you've been great help. Thank you

#8 aliadnan

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Posted 24 February 2006 - 10:51 AM

RN

I don't understand why you want to calculate the outlet temperature of stem. As long as what I have understand, you want to heat toluene by using the latent heat steam. There is no involvement of sensible heat so there will be no change in temperature of outlet steam (which has condensed to water now).

Look when your only extracting the Latent heat then there will be no change in temperature. The temperature will change when there will be sensible heat transfer and in this case there is not sensible heat tansfer. Your just using the latent heat carried by the saturated steam and condensing it into water at the same temperature at which steam entered. THERE WILL BE NO CHANGE IN THE TEMPERATURE OF OUTLET OF STEAM SIDE.

#9 Guest_Guest_*

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Posted 24 February 2006 - 11:10 AM

After I wrote the post, I realized my error, thank you Ali. So now if I do a zone analysis, I should have two zones correct? One for the vapour phase and one for the condensed phase (with respect to saturated steam). If I am incorrect let me know. Also, the physical properties will change regardless of the temperature and pressure being the same because there is a phase change, correct?

#10 aliadnan

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Posted 24 February 2006 - 11:19 AM

Its good to hear the you understand now what I was trying to explain.
Yes your right that there will be 2 zones and the properties will change since steam and water at the same temperature have different properties.
"Process Heat Transfer" by Kern will help you in designing and also book by Ludwig has a nice literature on the exchanger design.
Well to be frank I have'nt design this kind of exchanger before during the study period. I hope other peoples on this Forum will help you in your design if you furnish them the exact problem and data you have.
Good luck for the design.

Regards
Ali

#11 Guest_Myra_*

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Posted 24 February 2006 - 03:51 PM

its really nice to larn all the stuff from your discussion, but one question arose in my mind is that.
what are the conditions when we assume steam must be assume in tubes? yes you people are saying absolutely right that when ever you removed latent heat than there is no change in temperature only phase change. and the properties of gas and liquid is different. so you have to calculate both these calculations, if you are using more than one pass exchanger Ft should be calculated in this case
best of luck

#12 Guest_RN_*

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Posted 25 February 2006 - 12:52 PM

I've run across a problem, HYSYS forces my steam stream outlet to be all vapour, when we know there will be a phase change and it will be condensed coming out at the same temperature and pressure. It does not allow me to force the stream to be liquid coming out, which means I can't find the physical properties. Any ideas where I could find the properties for water at 231.1C and 2853.6 kPa?

RN

#13 Guest_RN_*

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Posted 25 February 2006 - 02:51 PM

Ok, so I solved the last problem, now I'm deciding on material of construction and exchanger type. I'd like to have a few alternatives and reasons why I chose a specific one, so if you have any ideas, let me know. I'm operating with two non-corrosive components and they are at high pressure. I am thinking Carbon Steel, but I keep seeing articles saying replace the carbon steel tubes with an alloy for higher performance, any ideas?

RN

#14 Art Montemayor

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Posted 25 February 2006 - 11:30 PM

RN:

I would not fabricate the Toluene heater as has been described. This is not meant as criticism of what has been expounded, but I would do it differently and I would like to explain my reasons. Please refer to the attached Excel workbook and you will hopefully understand my reasoning.

1. The very first thing you should be doing is analyzing the process conditions you are imposing on the proposed exchanger. To do this you must identify the thermal conditions of the required steam. In order to do this, you must establish the conditions of the required supply steam. You must give the supply steam some reasonable driving force in order to heat the Toluene to the desired end temperature. Note that I am giving the steam an edge of approximately 4 oC. You may need more if the steam or the Toluene are “dirty” (cause surface fouling or have inerts).

2. Once you have identified the minimum level of saturated steam pressure and temperature that you need, you can then select the practical and most obvious TEMA type of exchanger. The choice of whether you employ the steam in the tube side or the shell side is decided by 2 factors: (a) the need to avoid an expansion joint on the shell due to excessive thermal stress between the shell and the tubes; and (cool.gif the need to have positive and quick draining of all formed condensate in order to ensure a constant, swept clean heat transfer area. The latter is very difficult to achieve in the tube side unless you employ sophisticated sloped draining tubes and a minimum of 2 passes. The shell side is a natural choice for quick and efficient condensate formation and evacuation.

3. There is no need to fear using the shell side for saturated steam at a pressure of 117 psig to 200 psig. I don’t know where you get the idea that you require 400 psig steam. That is an excessive level of steam pressure for this application. Depending on the size of the shell diameter (I suspect it would be between 18” to 36”) the shell would be fabricated of seamless pipe material (probably A516 Grade 70) which can easily handle such relatively low pressures. If you are told that you only have 400 psig steam available, then simply use your head and install a steam regulator that will regulate the steam supply at around 125 psig. It’s that simple.

4. I would not employ an external floating head (TEMA type AES) on this heater because it is excessively expensive and it also is prone to internal leaks. The leak is steam into the Toluene and this causes a lot of trouble. The higher expense comes from the need to make the shell bigger in diameter due to having to extract the tube bundle. Additionally, there is a lot of inefficiency you have to tolerate with this design due to the excessive clearance you have to give the tubes and the internal shell. Also, if you are using steam on the tube side, there is no need to have an expensive “A” bonnet cover in order to access the tubes. If the steam is so dirty that you require an “A” type head, then you have more troubles than you realize. You have to assume that you can supply clean steam.

5. All the material of construction should be mild carbon steel. This is an easy application that doesn’t require too many brain cells in that department. However, it does require engineering ingenuity in establishing positive, constant condensate flow and an efficient removal of any potential non-condensables in the shell side. You must provide a non-condensable vent mechanism at the end of the steam’s trajectory and where you drain the condensate. This will ensure that you maintain a 100% steam-condensate shell side environment, just as the calculations are based on. Again, use common sense and introduce the liquid Toluene at the bottom nozzle on the tube side. This technique ensures that the tube side is 100% liquid and devoid of any gases or non-condensables and it also concurs with what has been established in the calculations. Do not forget to incorporate “V” notches at the bottom of the vertical segmental baffles to allow for efficient passage and draining of all condensate formed in each shell partition.

6. The use of a U-tube bundle design not only ensures that there is 100% thermal differential expansion compensation (& no undue material stress), but it also is the least expensive of the tube bundle designs. Additionally it allows for tube bundle removal and replacement in the future – if need be.

7. The use of a U-tube bundle doesn’t mean that you are limited to only 2 tube side passes; you can use more if your design requires it or if you need to increase the tube side velocity in order to increase the inside heat transfer coefficient.

I hope these comments help to explain the application as I see it and as I would build it if it were mine. I have only focused on a TEMA type heat exchanger because that’s what you have implied. A spiral plate or spiral tube type of exchanger could be applied – perhaps even more efficiently. We can’t tell because you haven’t furnished all the necessary basic data to make that estimate. I’ll leave you with the thought in order to stimulate your ingenuity.

Art Montemayor

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#15 Guest_RN_*

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Posted 26 February 2006 - 11:39 AM

Thanks a lot for your input Art. It is more than appreciated. I chose the high temperature & pressure because in our course notes we were given certain heuristics in choosing utilities for the design. I arbitrarily chose this combination of T = 231.1 C & P = 2853.6 kPa (High Pressure Exchanger) because we were given a choice of 4 values. Mind you, if I find a better combination that I can justify, I can use it as well. I just played it safe with the heuristics is all. I thought of Carbon Steel as the Material of Construction because it was inexpensive and neither of my streams are corrosive, but since I am operating at a high pressure, I don't know that it would do the job. Right now, I am calculating based on a Single Pass, Two tube exchanger as a starting point. Let me know your thoughts or if you need anymore info to clear up questions.

RN

#16 Art Montemayor

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Posted 26 February 2006 - 03:21 PM

RN:

Your basic data is confusing to read. You explain why you chose your steam supply pressure level, but you fail to say if you have to employ that pressure on your exchanger as it exists or if it is supplied as all normal and conventional utilities are supplied in an industrial plant: they are subject to regulation and modification as it suits the specific application. In other words, while 22,000 volts electrical supply is furnished to an industrial plant it doesn’t have to be consumed at that level of voltage (nor is it done). Electrical transformers reduce the voltage according to the specific needs. The same applies to steam supply. Just because you selected 2,853.6 kPa as your supply level doesn’t necessarily mean that you have to employ it at that high a pressure. At your level of engineering studies I seriously doubt that your professors want to impress upon you that you must use that level of steam. On the contrary; I assume that your university is a noteworthy one and that your professors are knowledgeable and aware of what you are going to confront out in the real world when you finish this very important plant design course. In their eagerness to prepare you, I am sure they want you to analyze your needs as compared to what is being supplied to you for use in solving the problem. I am certain that they expect you to use your common sense and employ some means to condition your steam supply such that it fulfills your objective while keeping your equipment and costs down as much as possible. For this, you would employ a steam regulator that reduces the 2,853.6 kPa (414 psig) steam supply down to 812 kPa (118 psig). Of course, you would have to study how the regulator works and what happens to the regulated steam thermodynamically. (I think you know that I’m hinting about the isenthalpic expansion of the steam as it goes through the regulator and assumes different thermodynamic properties besides the enthalpy - which remains constant). Your profs may be forcing you to look at that typical and normal plant operation and take it into consideration.

[As an aside, let me mention the following important advice to you before proceeding further: Don’t ever fail to specifically identify what type of pressure you are talking about. Note that I have assumed that you are citing GAGE kPas in the above units’ conversions. But I really don’t know if that is true because you fail to state whether your values are gage or absolute. This is flawed engineering and I don’t mind telling you that if you continue to fail to identify your units when communicating, it could cause you and your career a lot of trouble in the future. Besides being flawed, this practice is erroneous and simply bad and confusing communication. The same advice applies to temperature values. It is sad, but I find that those engineers insisting on the use of SI units for the sake of unity in the field are the principal ones guilty of failing to specify which pressure value they mean – gage or absolute.]

There is no logical reason for employing the 2,853.6 kPa(g) steam in your application – unless you are specifically ordered to do so. If you do, it will seriously affect the size, the weight, the cost, and the operation of the exchanger. Logic and common sense rules that you regulate the steam down to a lower and more cost-effective pressure.

Metals used as materials of construction are not necessarily less strong or tougher due to their quality nature. Steel is not weaker at the higher pressures. What is done is that its inherent working stress value is used to calculate the correct and safe wall thickness that will contain and safely support the required internal pressure. A steel shell can be designed to safely withstand the 812 kPa(g) pressure as easily as it can the 2,853.6 kPa(g). It’s just a matter of shell wall thickness. Steel is the material of choice in your application.

Now for some more advice: always state the precise and exact type of exchanger that you are referring to, using conventional and recognized descriptions. I (& I assure you everyone on this forum) don’t know what is meant by “Single Pass, Two tube exchanger”. Single pass where? – shell or tube side? Do you mean that you are using only two tubes in your exchanger? Levity forces me to exclaim that while it surely is going to be quite small in diameter, it will extend in length from one county to the other! When you are describing a shell & tube exchanger always give your listener or reader an accurate description, using the TEMA type (which tells of the configuration) and specify the shell and tube passes with emphasis on the orientation (vertical or horizontal) as well as on the tube size, baffle type, and tube length. This tells it all and paints an accurate and precise picture to the listener/reader. That is the reason why I went to the lengths of preparing the Excel Workbook with the detailed TEMA descriptions. All you have to do in the future is point to the specific one you mean and furnish the rest of the details.

If your professors are worth their salt as Chemical Engineers (and I bet they are) they will be looking to grade you on your ability to use your common sense in selecting the best, safest, cost-effective TEMA type exchanger, complete with your specific reasons for doing so. The more detailed you are in describing your selection and your answer, the better the grade – mainly because this proves to them that you know just what the hell you are doing and where you are going with your proposed solution. There are many other factors that we haven’t covered in this discussion – and that will probably go “under the rug” for the time being. However, the major ones have been covered and I’ll leave you to contemplate and identify the remaining ones as an exercise in practical plant design --- after all, that is what it’s all about isn’t it. Plant design is not just about grinding out a bunch of pipe sizes and heat transfer coefficients using a programmed computer simulator. It’s really about designing a complex and expensive group of equipments with the scope that the end product should function exactly as proposed and under the same conditions as if you were operating the plant yourself. The plant (& its equipment) should be capable of being started up, operating, shutting down, and undergoing emergency procedures and safety reliefs with dependable, safe results. That’s the way you would want it if you were to operate it yourself and that’s what the end user (your client) should receive. That should be the objective of every plant design course --- and I’m sure that’s the one that you are laboring under.

I have gone to this length in responding to your post because I believe you recognize the value of experience and have demonstrated you are appreciative of our contributions.

Good Luck,
Art Montemayor

#17 Guest_RN_*

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Posted 26 February 2006 - 09:09 PM

Your expertise is quite appreciated in this matter Art. My apologies, but the pressures presented are all based on a conversion from psig to kPa (gauge). Now, I thought I explained how I chose the pressure for this exchanger properly, but I guess not. We were given a list of heuristics on steam utilities. One of which was T & P relationships for steam. There were 5 values listed for us to consider for use and 3 were above the T necessary for the toluene to reach. I chose the 231.1 C and 2853.6 kPa (gauge) because the other two values seemed too low and too high. The other two values were T = 366 F & P = 150 psig and T = 488 F & P = 600 psig. I know I'm really wet behind the ears when it comes to this stuff, so believe me, the criticism and comments are much appreciated.

RN




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