23 replies to this topic

[Kogara]

Hello, I'm a third year chemical engineering student and i have to design a shell-and-tube heat exchanger.
My two fluids are:
- Tube side: Acetic acid at 8.90kg/s (from 303 to 343K).
- Shell side: Water at 2.5kg/s (from 350 to 310K).

This is my problem: When i want to find the correlation factor F, i do the calculation of R and S and i found R=1 and S=0.85.
When i have a look to the chart , i can see that this value of S is after the "R=1" curve.
Following the lecture of my teacher who said: "If F>0.75 inachievable, use single tube-side pass; then F become 1", i use this F=1 and Np=1.
I found an area A=125.208m².

The problem is when i do the calculation of the tube side velocity, i found (in the best case) u=0.234m/s and the suggested range is between 1 and 2 m/s...

This is because i use Np=1 because when i use Np=6 or 8, my heat exchanger is working...

So i really don't know what can i do for finding my tube side velocity between 1 and 2 if i have to use a single tube side pass...

I hope someone is going to help me about that and if i'm not clear, i can add all the informations you want..

[breizh]

HI ,

Let you try this resource ,it should help you : F calculation versus type of exchanger and simulation of heat exchanger .

http://www.chemsof.com/exch/exch.htm

Hope this helps
Breizh

[Kogara]

Breizh:

Thanks for your answer, I tried the software, but it's saying "temperature cross is occurring, recheck the data". And I don't know how I can avoid this problem...

[Zauberberg]

If temperature cross is occurring, there is a heat balance issue and you need to revise your input data.

Temperature cross means that the cold fluid outlet temperature is higher than the hot fluid inlet temperature, and if you look at your process data there is imbalance between hot- and cold-side duties. For given process information, calculated duty on the hot side (acetic acid) is 775 kW, while the cold side duty is 419 kW.

In simple words, you need to increase the hot fluid (water) flow if you want to heat the Acetic acid stream up to 343 degK.

As for the Ft correction factor, you have to take into account what type/configuration of heat exchanger you are considering for this application. If you are designing for 7 degC temperature approach, E-shell will not do the work.

Attached Files

•  Heat Balance.xls   20.5KB   663 downloads

[Art Montemayor]

Kogara:

Besides not communicating well, you are failing to state what – specifically – you are trying to do. In other words, you fail to state and identify your problem in a concise and accurate manner. That, more than anything else is probably leading you astray and confusing you in attacking your problem in a logical, engineering manner. I am being frank and candid with you because you have been kind enough to identify yourself as a 3rd year student and I am concerned that you should already be organizing and communicating your work problems in a correct manner. You may basically know and understand what you are trying to do, but your inability to describe it thoroughly is also a weakness that confuses you and makes you do unnecesary and incorrect decisions. Do not misinterpret my comments as derrogatory of your work up to now. My intention is to clearly show you how you are making wrong engineering statements and analyses – and going down the wrong road to a solution. Let’s go back to pure basics to get a correct orientation and definition of what you are trying to do:

Refer to Donald Q. Kern’s famous classic textbook on the subject: Process Heat Transfer, McGraw-Hill Publishers, 1950; page 139, “the true temperature difference of a 1-2 heat exchanger”. After derriving in detail the Correction Factor parameters, “R” and “S”, Kern goes on to carefully - and in detail – explain: “Accordingly it is not advisable or practical to use a 1-2 exchanger whenever the correction factor F_T is computed to be less than 0.75. Instead, some other arrangement is required which more closely resembles counterflow.”

You have failed to tell us the basic data straight up-front: you are trying to apply a 1-pass shell with a 2-pass tube side heat exchanger. You have also failed to state which TEMA configuration you are applying. (you obviously have not used our SEARCH engine and found my heat transfer workbook where I list all the TEMA heat exchanger configurations and types. You should do this ASAP)

What you are obviously trying to avoid is A TEMPERATURE CROSS. Yet, you fail to identify this in your post. I don’t know whether you are aware of this important fact or not, but I highly stress that you should be well aware of a temperature cross effect in a heat exchanger.

Of course a tubeside fluid velocity of 0.234 m/s is much too low and you should be applying the suggested velocity of 1 to 2 m/s. You do this just as Kern suggests: USE A DIFFERENT TEMA CONFIGURATION. I would try a BFM type. I personally do not usually recommend an F type of shell in real applications, but this is an academic exercise. You could also design the system to have two heat exchangers in series, one mounted on top of the other. If you used an Excel Workbook to do your sketches and calculations, I could easily sketch you the actual configuration – which would save us both more than a thousand words of trying to describe what I mean

When you discover that you have to increase your tubeside velocity, you are forced to reduce the number of tubes you use per pass. but you can also LENGTHEN THE SHELL (and consequently the length of the tubes). This effect gives you more area per pass – which means you can – guess what? – REDUCE THE NUMBER OF TUBES! So, as you can see, you have other tools you can use to solve the problem. There are more ways than one to skin a cat.

But it is basically a need to master the knowledge and description of what you are trying to apply. And in doing that, you have to develop your communicating skills in describing what you are trying to do. I urge you to obtain a copy of Don Kern’s classic book and study it well – particularly the pages I cited to you. You do not have a difficult problem. You have a difficulty in describing your problem – and that is a clue that you don’t understand the basic problem involved – yet. But you can quickly catch up in the required knowledge by reading Kern.

Good Luck.

[Kogara]

Thanks for your answers, zauberberg and art montemayor, now i think i've got two problems, one is communicating about what i want and explain it well and my other one is maybe my initials data.
First, english is not my native language (i'm french) and i'm really sorry for the mistakes and secondly, this my "first year" in chemical engineering, i was studying materials engineering in France, it's for that i'm not aware of some specifications like "temperature cross".

Now about the heat exchanger: My aim is to build a shell-and-tube heat exchanger with two fluids: Water (350 - 310K and 2.5kg/s)
Acetic acid (303 - 343K and flowrate unknown).
For finding the properties i need, i use different sources: Perry’schemical engineers’ handbook 7th ed, Thermodynamicsand transport properties of fluids (Fifth edition) and Coulson& Richardson’s Chemical Engineering Vol. 6.


The result on my researches is this one:


Water (shell side) from 350K to 310K:- k = 0.650W/mK. - Pr = 3.126.

- μ= 0.000485Pa.s. - Rf =0.00020 m²K/W.

- Cp= 4183.07 J/kg.K. - m’ = 2.5 kg/s.

- ρ= 984.45 kg/m³.

Duty is Q’ = 2.5 × 4183.07 × (350–310) = 418 307 Watts -> Aim to transfer 460 138W.

Acetic acid (tube side) from 303K to343K:

- k = 0.167W/mK. - Pr = 5.59.

- μ= 0.000795Pa.s. - Rf = 0.00018 m²K/W.

- Cp= 1174.61 J/kg.K.

- ρ= 1018 kg/m³.

For finding the flowrate of acetic acid: m' = Q'/Cp(Tout-Tin) = 418307/1174.61(343-303) = 8.90kg/s.

I can found a problem here now because in your excel file Zauberberg, you insert a Cp value of 2180 J/kgK ... I use a formulae with 4 differents coefficients in Coulson & Richardson’s Chemical Engineering Vol. 6. for finding this Cp value (and the Pr value) but maybe it's not right..so if i start with wrong value, it's sure that i can't achieve it...

Afterthat, i found a DeltaTlm = 7, a R = 1 and a S = 0.85, F is inachievable so it becomes 1 but i have to use a single tube-side pass (Np = 1) :"This is directly coming of the lecture of my teacher".

I found after that i have to use a BEM exchanger (a fixed tubesheet design).

I do the calculation for A = Q'/Usugg.F.DeltaT = 460138/530x1x7 = 124.03m². I found before that Usuggested = 530W/m²K.

•We started the design with L = 4.88 m, do = 20 mm:

•Area of one tube = π ×4.88 × 0.020 = 0.3066m²

•Number of tubes needed= 124.03 ÷ 0.3066 = 405

Uestimate = Q'/Nt.At.F.DeltaT = 460138/405x0.3066x1x7 = 529.37W/m²K.

Now, it's the time for the tubeside velocity: ut = (4.m't.Np)/(rhot.pi.di².Nt) = (4x8.90x1)/(1018xpix0.016²x405) = 0.107 m/s...

Suggested ranges: 1 to 2 m/s...

I built an excel spreadsheet with all those points plus Re plus Nu until hi but with this number of passes = 1 the maximum tubeside velocity that i can found is 0.229 m/s...

I maybe have errors on my original date but i don't think its going to change my velocity to 1..

I hope this description is more clear for you and you're understanding where is my problem. Thanks you for you help again

Attached Files

•  heat exchanger design.xlsx   15.46KB   434 downloads

[Art Montemayor]

Kogara:

Your problem is, as I mentioned before, communications.

Note how I have had to practically re-write your last post in order to make some sense out of it. If we are to communicate in the English language, we must abide by its rules:

Stop using the lower case (i) for the proper personal pronoun (I). It causes a lot of confusion in trying to read through a paragraph that continues to use the lower case instead of I. We are NOT “texting” on our Forums. If that is to be the case, then I personally will stop contributing my comments, quit the Forums, and retire from this work load. It is too taxing and wasteful to try to communicate in an immature and unorganized manner. You are not going to change the English language, so conform to its rules. The same rules hold in the French language as well, so the fact that you are not a native speaker doesn’t give you an excuse to stop using the proper rules and logical organization. The French language is well-known for its logic and its correct structure and grammar as well.

Do not use the written word to try to explain a mathematical calculation when you have the full use and resources of a spreadsheet. I fail to understand what you wrote down in the post explaining all the calculations you have done. Those, quite frankly are not needed to resolve your basic problem of a temperature cross. Your solution is much, much simpler and all it takes is common sense and practicality. There are no Nusselt, Prandlt, or Reynolds numbers involved – just common sense.

We place a space between each sentence and complete thought. We use paragraphs to unite a common thought or subject. We separate our units from our numbers. We do this because it doesn’t make any sense otherwise. You enter the numbers in a spreadsheet separately because the units don’t compute. They are not part of the number.

You need to make a correct and accurate heat balance (I have not done this for you) in order to compute and check the end temperatures of the exchanger. Zauberberg has already started to help you in this direction. This is a pre-requisite to checking to see if you have a potential temperature cross.

You also have to start thinking as a Chemical Engineer by VISUALIZING WHAT IS GOING ON INSIDE THE HEAT EXCHANGER. This type of mental exercise immediately alerts you to the fact that you can, in multi-tube pass apparatus, create a temperaure cross. You can also diagnose what is a logical way to defeat that effect in trying to apply a specified type heat exchanger. That is why I have stressed that you download the TEMA workbook that I uploaded in our forums. It will help you to better understand the differences and trade offs in each of the TEMA types of shell and tube arrangements.

You are now stating in your last post that the Acetic Acid flow rate is “unknown”. This comes after telling us in your first post that there are 8.9 kg/sec of Acetic Acid flowing inside the tubes. This is what causes a lot of confusion and wasted work effort in trying to help you out. We still don’t really know all the details of what you have been assigned and are trying to do.

I know – from practical experience – that you have stumbled into a specific problem area when trying to apply a multi-pass heat exchanger. This happens to almost ALL chemical engineering students when, for the first time, they have to confront the physical attributes and short comings of a mechanical device such as a “simple” heat exchanger. You have found out the bitter truth: heat exchangers are not that “simple”. It takes time, study, effort, research, and calculations to understand precisely how these devices work – and why.

I hope the work I put into the attached workbook is of help to you in explaining how to analyze and put together the essential parts of a heat exchanger. Do not fail to question or challenge any thing that you do not understand or that you have serious doubts about. We can help you help yourself, but we need your accurate and concise information communicated to us so that we can go directly to that which is troubling you. It’s tough, I know. But you must start communicating well because that is one part of engineering that is the basis for you getting jobs done when you graduate. As a professional engineer, you will not be allowed to do the basic work; rather, you are expected to explain to your workers and assistants exactly what needs to be done, when, where, why, and sometimes explained in detail. That is engineering in the real world. This need applies just as much in France as well. I know because I've been there and worked with excellent engineers who know the importance of communicating well to have a job well done.

Your edited prior post reads as follows (compare the differences):

"Thanks for your answers, zauberberg and art montemayor, now I think I've got two problems, one is communicating about what i want and explain it well and my other one is maybe my initials data.

First, english is not my native language (I'm French) and I'm really sorry for the mistakes and secondly, this my “first year” in chemical engineering, I was studying materials engineering in France, it's for that I'm not aware of some specifications like "temperature cross".

Now about the heat exchanger:
My aim is to build a shell-and-tube heat exchanger with two fluids:
Water (350 – 310 K and 2.5 kg/s)
Acetic acid (303 – 343 K and flowrate unknown).

For finding the properties I need, I use different sources:

• Perry’schemical engineers’ handbook 7th ed,
• Thermodynamicsand transport properties of fluids (Fifth edition) and
• Coulson& Richardson’s Chemical Engineering Vol. 6.

The result on my researches is this one:

Water (shell side) from 350K to 310K:
- k = 0.650 W/mK. - Pr = 3.126.
- μ = 0.000485 Pa.s. - Rf = 0.00020 m²K/W.
- Cp = 4183.07 J/kg.K. - m’ = 2.5 kg/s.
- ρ = 984.45 kg/m3.

Duty is Q’ = 2.5 × 4183.07 × (350–310) = 418 307 Watts -> Aim to transfer 460 138W.

Acetic acid (tube side) from 303K to343K:
- k = 0.167 W/mK. - Pr = 5.59.
- μ= 0.000795Pa.s. - Rf = 0.00018 m²K/W.
- Cp= 1174.61 J/kg.K.
- ρ= 1018 kg/m3.

For finding the flowrate of acetic acid:
m' = Q'/Cp (Tout-Tin) = 418307/1174.61(343-303) = 8.90 kg/s.

I can found a problem here now because in your excel file Zauberberg, you insert a Cp value of 2180 J/kgK. I use a formulae with 4 differents coefficients in Coulson & Richardson’s Chemical Engineering Vol. 6. for finding this Cp value (and the Pr value) but maybe it's not right. So if I start with wrong value, it's sure that I can't achieve it.

After that, I found a DeltaTlm = 7, a R = 1 and a S = 0.85, F is inachievable so it becomes 1 but I have to use a single tube-side pass (Np = 1) :"This is directly coming of the lecture of my teacher".

I found after that I have to use a BEM exchanger (a fixed tubesheet design).

I do the calculation for A = Q'/Usugg.F.DeltaT = 460138/530 x 1 x 7 = 124.03 m². I found before that Usuggested = 530 W/m²K.

We started the design with L = 4.88 m, do = 20 mm:
Area of one tube = π ×4.88 × 0.020 = 0.3066 m²
Number of tubes needed = 124.03 ÷ 0.3066 = 405
Uestimate = Q'/Nt.At.F DeltaT = 460138/405x0.3066 x 1 x 7 = 529.37W/m²K.

Now, it's the time for the tubeside velocity:
ut = (4.m't.Np)/(rhot.pi.di².Nt) = (4 x 8.90 x 1)/(1018 x pix 0.016²x405) = 0.107 m/s.
Suggested ranges: 1 to 2 m/s. I built an excel spreadsheet with all those points plus Re plus Nu until hi but with this number of passes = 1 the maximum tubeside velocity that i can found is 0.229 m/s. I maybe have errors on my original date but i don't think its going to change my velocity to 1. I hope this description is more clear for you and you're understanding where is my problem. Thanks you for you help again."

Attached Files

•  Designing a BEM Heat ExchangerRev1.xls   293KB   509 downloads

[Kogara]

I'm sorry but I'm feeling a little bit lost between the lecture about designing a heat exchanger (on the attaching file) and your design of the BFM heat exchanger.
I can't put a link between your work and this lecture, and now I'm feeling like I didn't understand at all the basis of the design of a heat exchanger.
If I follow the lecture, I'm stuck at the F coefficient factor but when you're doing the BFM you're directly using the velocity that you want... I'm sorry but I don't know which way I have to follow now.

Attached Files

•  TP - heat exchanger design basis.ppt   1.55MB   411 downloads

[Art Montemayor]

Kogara:

You may feel lost, but you are not lost. You are perplexed and confused because of one or more things, but you are doing something POSITIVE and CONSTRUCTIVE towards resolving your problem BY SEEKING HELP. This is a good and positive sign of a professional engineer. You cannot be expected to know everything; you ARE EXPECTED to know or find out where to obtain the correct answers. That is a large part of being a successful engineer. Now let me try to use the excellent communication information you sent in a timely manner and attempt to explain how you can resolve this problem:

First and foremost, concentrate on fulfilling what your instructor has assigned you – regardless if he/she knows what is happening or not. You must comply with the course’s requirements in order to graduate. When I was in your shoes, we called this requirement “cooperate and graduate”. It doesn’t mean that you have to surrender to an error and mistaken ideas that the instructor has (and instructors have a lot of those); it means that you should confront the apparent mistakes (or misconceptions) in an assignment, verify and prove to yourself that there are concerns in the assignment, and bring them to the instructor’s attention with some suggestions or recommendations of your own. THAT is what any reasonable instructor would relish and expect in a student.

I can only work with and comment on what you have communicated. And from the “lecture” (which I have to believe is the PowerPoint presentation you submitted), I can tell you that there are several factors that are not engineering “truths” in the material. But that is OK; university lecturers cannot be expected to know exactly what is happening in the industrial, real world. They teach theory. We engineers practice the REAL truths. And the real truths are vested in pure logical common sense. For example, every experienced engineer would never employ a bellows expansion joint in a heat exchanger shell. This was a practice that was stopped approximately 40 years ago because of the hazards of the corrosion taking place in the bellows. They just don’t work and are a hazard. Another more important thing that has not happened in this lecture is that certain, real-world facts are not stated:

In order to apply the basic heat transfer equations that involve the inevitable film heat transfer coefficients – which yield the Overall Heat Transfer Coefficient, “U” – you MUST FIRST FIX THE PHYSICAL DIMENSIONS OF THE PROPOSED HEAT TRANSFER DEVICE. This is simply common horse sense. You don’t have to be a Donald Kern or a Coulson & Richardson to know that if you are to calculate the convective heat transfer you have to know or fix the fluid velocity. In order to fix the fluid velocity you must also fix the size of the proposed tubes and their quantity per pass. This has to be obvious to any student – and especially to an instructor in heat transfer. You are expected to know that the effectiveness of fluid heat transfer depends to a large extent on the velocity it has as it goes through the tubes. Turbulent flow is desirable for a good “U”.

You have been given the “fixed” velocity, but you haven’t been told the condenser tube dimensions – which are fixed as common standard sizes by industry (something that students don’t know yet, but instructors should advise students on). This is clearly stated in my sketch and calculations and is something you should be instructed on. Otherwise, you can’t commence with the physical design.

Another item that fixes the physical size – and, as such, the feasibility of applying a certain type of design (such as the BEM) – is the LENGTH of the tubes (and the shell). The longer the tubes, the more economical the fabrication and the more area you can give a given shell diameter. This is common engineering sense that is always applied in real life.

It is common sense that tells us that you MUST COMPLY with the given fluid velocities (we have to assume that these velocities conform to turbulent flow and efficient convective film coefficients). If you employ velocities that yield a laminar flow (lower velocities, lower Reynolds Number) you obtain inefficient film coefficients (and a much lower “U”). Therefore, by common engineering sense, you have to FIRST MAKE A PRELIMINARY CALCULATION to find out what is the quantity of tubes that you will require to give you the pre-set fluid velocity per each tube pass. You don’t need a Logic Diagram to figure this out. This MUST be done FIRST and if you agree, you should bring this topic up with your instructor and point out that when you do this, you come up with a feasible BEM design that involves more than one exchanger – but you can “stack” these exchangers together to form one unit.

I am assuming that you have done the recommended, correct heat balance and have identified and confirmed the exit temperatures on the heat exchange. If you have done that, you can then confirm if you have an F_T that makes sense. If you don’t have an F_T that is above 0.75, than you can’t use one BEM unit with a 1-pass shell and 2-pass tubeside. It won’t work because of a temperature cross. There is, however, a way to avoid the temperature cross: use a “pure” counter-flow heat transfer approach – such as I depict in my sketches. The BFM TEMA type is designed to avoid such a temperature cross with a counter-current type of flow. Note that the internal, horizontal shell baffle makes the 2-pass shell flow go directly against the tubeside flow in a COUNTER-CURRENT direction. This is what makes the BFM work without a temperature cross and makes it more efficient. However, it has practical trade offs. The horizontal baffle is a pain in the backside if it isn’t seal-welded into the shell. And this is often the case in small diameter shells because of the inability to get a welder inside the shell to seal-weld it. However, in your case, you have a relatively large shell because of the large area required (125 m²) and therefore, your exchanger will have a rather large diameter that facilitates the welding of an internal horizontal baffle. That is why I mention the BFM as an option.

If the information you gave is correct, then your estimated area is correct and the calculations I followed up are also essentially correct (please check them out yourself). This means then that you CANNOT MAKE ONE BEM (1 SHELL PASS, 2-TUBE PASS) EXCHANGER WORK with your application. You should, after checking my calculations and agreeing with this, consult your instructor on what you have found upon doing these basic calculations. Also advise him/her that you CAN design a BEM type, but it would have to be composed of several shells stacked and operating as 1- shell pass and 1-tube pass.

The important point to make here is that you can’t practically build a 1-shell pass, 1-tube pass BEM unit that contains 125 m² because the tube side velocity would be too low and not give the required film coefficients. The ONLY way to abide by the constraints given by the velocity requirement would be to build BEM shells that contain 57 – ¾” tubes and stack them in series flow. This yields the required 1.5 m/sec fluid tube velocity. If you maximize the tube lengths at 20 feet, this yields six shells required. This is basically what my calculations reveal and what I try to explain to you of how to arrive at a feasible, practical design.

I hope I have succeeded in explaining what I feel you ought to know and apply to this problem.

[Zauberberg]

Art,

I have been following this thread carefully, and I have to say it is a true gem. Your continuous and crystal clear walk-through approach in the design procedure really emphasizes all the key aspects one needs to consider when dealing with heat transfer. It would be good to mark this tread in some way, and keep it available to everyone who faces with exchanger design.

Again, the way you explain and go through things is simply of outstanding quality. You should write your own process engineering textbook for sure.

Best regards,

[Art Montemayor]

Zauber:

Thank you for the kind words. I am only following up on the basic ground work you started with your heat balance spreadsheet. We have to put the sock on first, before putting on the shoe. As you well know, without the correct heat balance we don't have the correct end temperatures to identify if we have a cursed temperature cross in a multi-pass heat exchanger. And once a temperature cross is identified, the solution starts to evolve.

I did most of the posts during the Christmas - New Year rest period and I used a good bottle of Spanish Gran Reserva Rioja to fuel my writing. It works real good in stimulating what remains of my memory cells. I hope you had some restful Holidays.

Best Regards and A Prosperous New Year.

[Kogara]

I've got a doubt about this story of heat balance (we didn't do that on the lecture) because in this assignment the two temperatures of my fluids are:


Water 350 -310K
Acetic acid: 303 343K

And I have to "pick an appropriate shell-and-tube heat exchanger type and design it from standard shell and tube sizes".

We just saw the "E" shell on the lecture so I think he wants a BEM more than a BFM and for me and my knowledge it's mpossible at the moment to design a BFM, I don't know at all how can I do the calculations for the baffle spacing and others...

For Zauber:

On your first comment, you put on the spreadsheet a Cp of acetic acid = 2.18 KJ/kgK...
I found with Coulson and Richardson that, for a Taverage = 323K, Cp of acetic acid = 1174.61 J/kgK...

I'm not saying I'm right at all but if it's the case, you (or I) don't have a problem with the heat balance (confirm by your spreadsheet).

[breizh]

Attached the physical properties of acetic acid .
BTW you shoud have verified your data !

Breizh

[Kogara]

Thank you very much breizh for this confirmation of my datas: Cp for acetic acid is 2160.39 j/kgK.

My flowrate of acetic acid is now 4.84 kg/s instead of 8.90 kg/s.

The heat balance is ok now but the problem is : the biggest tubeside velocity that i can found now for the tubeside is : 0.125 m/s (with length of a pipe is 7.32m and the diameter is 16mm).

How can I resolve that if I have to use a sigle tube-side pass in just one unit and not creating a "heat-exchanger plant" ...

[CristianG]

Art,

I have been following this thread carefully, and I have to say it is a true gem. Your continuous and crystal clear walk-through approach in the design procedure really emphasizes all the key aspects one needs to consider when dealing with heat transfer. It would be good to mark this tread in some way, and keep it available to everyone who faces with exchanger design.

Again, the way you explain and go through things is simply of outstanding quality. You should write your own process engineering textbook for sure.

Best regards,

I would be the first in buying that book!

[Padmakar Katre]

If temperature cross is occurring, there is a heat balance issue and you need to revise your input data.

Temperature cross means that the cold fluid outlet temperature is higher than the hot fluid inlet temperature, and if you look at your process data there is imbalance between hot- and cold-side duties. For given process information, calculated duty on the hot side (acetic acid) is 775 kW, while the cold side duty is 419 kW.

Dear Zauberberg,
It's indeed pleasure to follow this thread and obviously the replies by you and Art. I am just confused about the temperature cross understanding. I think temperature cross is when we have cold fluid outlet temperature higher than the hot fluid outlet temperature in a single shell but it's totally possible with two shells in series or a single "F" type shell. The thermodynamics itself will not support this hypothetical of having cold outlet temperature more than the hot inlet temperature. Please understand me, I am doing some mistake in my understanding.

Attached Files

•  Heat Exchanger Network.pdf   98.51KB   308 downloads

Edited by Padmakar S Katre, 10 January 2011 - 08:11 PM.

[Zauberberg]

It's easier than that: if we look at given flowrates and inlet temperatures of both fluids, we can see - by calculating heat duty on both sides - that outlet temperatures of both fluids cannot be achieved in the way they were given. In other words, either the hot fluid inlet temperature would need to be higher (if we maintain the cold fluid outlet temperature), or the cold fluid outlet temperature will be actually much lower than the specified value. The first case is an obvious temperature cross(over), which is thermodynamically impossible.

[kkala]

Few notes (subject to comments) on this interesting thread, educating for a lot of us.
1. I remember the interpretation & check of an exchanger actual data in my forth year of Chemical Engineering (Greece); there was more confusion in our minds, compared to what Kogara patiently tries to clarify now. "Water cannot be clarified, unless it gets turbid", according to a Turkish proverb. There may be a reward in Kogara's career, if exchanger sizing is needed: parameters of the sizing software will have been understood.
2. Concerning the acetic acid heat capacity, G. E. Davis (HB of Chemical Engineering, 1904) reports 0.522 kcal/kg/oC at 96 oC, which agrees to Zauberberg & Breizh. So flow rate & temperatures of ingoing and outgoing streams have been already fixed. Besides no phase change occurs.
3. I think that temperature cross-over can be checked through heat balance in case of pure co-current or counter-current flow. But in multiple pass exchangers, there may be some temperature cross (i.e. local temperature of hot fluid lower than this of cold fluid, reducing heat transfer efficiency), even though heat balances are correct. This can be avoided (or at least reduced) by replacing the single exchanger with multiple ones. Advice would be welcomed on the subject. Having assumed pure counter-current flow, however, Kogara can neglect this matter just to simplify the case.
4. TP-heat exchanger design basis.ppm is guide for preliminary design as well as for "final" design by repeating the procedure with "corrected" values. In the "final" one, number of tubes per pass is not by calculation (e.g. 55 or 57), but dictated by selected TEMA exchanger. Tube length can be according to the needed transfer of heat, usually standardized to 8, 12, 16, 20 ft.
5. Step No 19 of design basis concerns selection of side for water and acetic acid. Probably the case of water in tubes / acetic acid in shell could be examined, to see whether higher velocities could be realized. Acetic acid can be corrosive, but acids generally do not create scales (used for descaling). It is noted that shell side should also have high velocity to avoid scales, though this is no so much pointed out in literature (not apparent in the design basis).
A recirculating pump could be a radical way out, e.g. for water: It recirculates enough water for a tube velocity ~ 1 m/s, 2.5 kg/s is introduced in pump suction and 2.5 kg/s (at 77 oC) is taken out from pump discharge. This would make the system complicated though.
Another way out is installation of standby exchanger, so that one of them could be under cleaning while the other is operating. But I have seen it only for the exchangers heating fuel oil for boiler burners (for water/acetic acid case it may be an exaggeration).
6. Threads http://www.cheresour...h__1#entry37241 and http://www.cheresour...1#entry37862 may be helpful concerning velocities, and http://www.cheresour...1#entry38892 might be of some help, giving an example of exchanger preliminary sizing (fluids are different).
7. And now what? I would specify an exchanger completely (from process viewpoint) and give it to the instructor, even if it did not comply to all design basis, clearly explaining assumptions and compromises.
Full investigation, apparently very hard without the appropriate software, seems beyond the intent of the exercise, especially if it concerns a task of some days (as supposed).

Edited by kkala, 09 January 2011 - 02:36 PM.

[sheiko]

Temperature cross means that the cold fluid outlet temperature is higher than the hot fluid inlet temperature,

I think temperature cross is when we have cold fluid outlet temperature less than the hot fluid outlet temperature in a single shell

I know another condition for temperature cross: when OUTLET temperature of the COLD side EXCEEDS the OUTLET temperature of the HOT side.

Edited by sheiko, 10 January 2011 - 04:48 PM.

[Zauberberg]

I know another condition for temperature cross: when OUTLET temperature of the COLD side EXCEEDS the OUTLET temperature of the HOT side.

That is thermodynamically possible if flow pattern is close to the true counter-current flow. I bet you can find a lot of examples of similar T-profiles when Plate/Frame or Brazed Aluminum Fin exchangers are used. What is important to observe is that, along the exchanger, heat is always transferred from warmer fluid to colder fluid. So it wouldn't be a cross(over). Consider, for example, Rich/Lean amine exchangers, heating rich amine from 60 degC to 105 degC while cooling lean amine from 115 degC to 70 degC: here, cold fluid outlet temperature (105 degC) is higher than hot fluid outlet temperature (70 degC) but it is thermodynamically possible case.

The temperature cross in this post would happen (for given flowrates) if we try to maintain cold fluid outlet temperature as specified. This would lead to hot fluid outlet temperature being lower than the cold fluid inlet temperature, and that is impossible: heat cannot be transferred from colder to warmer fluid, at least not in direct heat exchange applications.

[Art Montemayor]

The infamous "Temperature Cross" is created due to the physical arrangement of a multiple-pass shell and tube heat exchanger. It occurs due to a recurring fluid heating and cooling as it passes through the unit. This is plainly illustrated in "F" chart sketches that are issued with each exchanger configuration.

It is as simple as that. For example:

• A temperature cross is not possible in a spiral plate heat exchanger;
• A temperature cross is not possible in a double-pipe heat exchanger;
• A temperature cross is not possible in a single-pass tube, single-pass shell heat exchanger;
• A temperature cross is not possible in a two-pass tube, 2-pass shell (BFM) heat exchanger;

You will note that when you have perfect counter-current or parallel flow, you cannot have a temperature cross. It is when you vacilate between counter-current and parallel flow that you get into potential trouble. You should always strive to maintain a positive temperature driving force IN THE SAME DIRECTION as the fluid traverses the heat exchanger. Do that, and you will not have a temperature cross problem. Sometimes this can not be totally avoided due to physical constraints. The value of the "F" factor indicates to what level.

[Padmakar Katre]

Dear Art,

As usual very specific, precise and simply great reply on the topic. Thanks.

[Padmakar Katre]

sheiko:

Dear,
This is applicable with a single pass arrangement but if you see industrial configuration where based on DTmin approach, when more than 1 shell is augmented in the existing shell, it's quite possible. This is the basis i.e. overall approach or DTmin when we study the HEN pinch analysis. But anyways, a right argument.

[Padmakar Katre]

Dear Zauberberg,

A nice reply. It is always a pleasure to follow you and Mr. Art, since both reply so specific and precisely with examples, which makes everyone here to understand. Sometimes I think I was late to join this community, it's almost four years now but had I been since my engineering days today I might have seen myself a little more command on ChE.