Designing A Shell-And-Tube Heat Cheresources.com Community -> Student -> Problem On Designi Kogara Posted 30 December 2010 - 08:09 PM Hello, I'm a third year chemical engineering student an exchanger. My two fluids are: Tube side: Acetic acid at 8.90kg/s (from 303 to Shell side: Water at 2.5kg/s (from 350 to 310K This is my problem: When i want to find the correlation found R = 1 and S = 0.85. When i have a look to the chart , i can see that this valu lecture of my teacher who said: “If F>0.75 inachievabl use this F = 1 and Np = 1. I found an area A = 125.208 m². The problem is when i do the calculation of the tube sid 0.234m/s and the suggested range is between 1 and 2 m This is because i use Np = 1 because when i use Np = 6 So I really don't know what can i do for finding my tub single tube side pass. I hope someone is going to help me about that and if i'm want. breizh Posted 31 December 2010 - 01:05 AM HI , Let you try this resource ,it should help you : F calculat heat exchanger. http://www.chemsof.com/exch/exch.htm Hope this helps
Transcript
Designing A Shell-And-Tube Heat ExchangerCheresources.com Community -> Student -> Problem On Designing A Sheel-And-Tube Heat Exchanger
KogaraPosted 30 December 2010 - 08:09 PM 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 1use this F = 1 and Np = 1.
I found an area A = 125.208 m².
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.
breizhPosted 31 December 2010 - 01:05 AM 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
KogaraPosted 31 December 2010 - 12:59 PM Breizh:
Thanks for your answer, I try the software but its saying "temperature cross is occurring, recheck the data". And I dont know how can i avoid this problem.
ZauberbergPosted 31 December 2010 - 01:55 PM 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.
Art MontemayorPosted 31 December 2010 - 05:03 PM 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 andmakes 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 whever the correction factor FT 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 inyour 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 twoheat 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 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
Art MontemayorPosted 31 December 2010 - 05:03 PM 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 andmakes 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 whever the correction factor FT 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 inyour 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 twoheat 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 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
KogaraPosted 01 January 2011 - 03:11 PM 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 thatI'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) andCoulson& 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.
KogaraPosted 01 January 2011 - 03:11 PM 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 thatI'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) andCoulson& 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.
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 1to 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 m²Number of tubes needed = 124.03 ÷ 0.3066 = 405Uestimate = 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 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.
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 1to 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 m²Number of tubes needed = 124.03 ÷ 0.3066 = 405Uestimate = 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 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.
Designing A Shell-And-Tube Heat ExchangerCheresources.com Community -> Student -> Problem On Designing A Sheel-And-Tube Heat Exchanger
KogaraPosted 30 December 2010 - 08:09 PM 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 1use this F = 1 and Np = 1.
I found an area A = 125.208 m².
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.
breizhPosted 31 December 2010 - 01:05 AM 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
KogaraPosted 31 December 2010 - 12:59 PM Breizh:
Thanks for your answer, I try the software but its saying "temperature cross is occurring, recheck the data". And I dont know how can i avoid this problem.
ZauberbergPosted 31 December 2010 - 01:55 PM 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.
Art MontemayorPosted 31 December 2010 - 05:03 PM 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 andmakes 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 whever the correction factor FT 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 inyour 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 twoheat 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 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
Art MontemayorPosted 31 December 2010 - 05:03 PM 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 andmakes 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 whever the correction factor FT 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 inyour 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 twoheat 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 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
KogaraPosted 01 January 2011 - 03:11 PM 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 thatI'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) andCoulson& 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.
KogaraPosted 01 January 2011 - 03:11 PM 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 thatI'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) andCoulson& 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.
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 1to 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 m²Number of tubes needed = 124.03 ÷ 0.3066 = 405Uestimate = 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 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.
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 1to 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 m²Number of tubes needed = 124.03 ÷ 0.3066 = 405Uestimate = 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 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.
Area of a tube 0.016 0.02 0.025 0.031.83 0.0920 0.1150 0.1437 0.17252.44 0.1226 0.1533 0.1916 0.23003.66 0.1840 0.2300 0.2875 0.34494.88 0.2453 0.3066 0.3833 0.4599
Heat transfer area in 10-ft long shell = 103,151 10.3
Heat transfer area in 20-ft long shell = 206,302 20.6
From these preliminary calculations, it can be established that a BEM TEMA type of heat exchanger is not the appropriate type because of the Temperature Cross that takes place and the multiple exchangers quantity
Each shell diamter is estimated as 12". Allowing 12" for the nozzles, this makes the height of each shell 24".This means that if six BEMs are stacked as shown, the total height of the assembly would be 6 x2 = 12 feet = 3.7 meters.
This type of configuration is frequently found in MEA heat exchangers in order to avoid a temperature cross.In that application, normally two BEMs are stacked together and suffice to avoid the temperature cross.However, in this application, the quantity of units is 3 times that and the height might be a deterrent to use the BEM type of exchanger in this application.
cm2 =
cm2 =
required to make the total area, stacked as shown in the above sketch is SIX units.
m/sec
(per 1.0 foot of tube length)
Condenser tubes come in standard lengths of:10 feet
m2
cm2
cm2
1-pass Shell, 1-pass Tubes BEM Heat Exchanger
1-pass Shell, 1-pass Tubes BEM Heat Exchanger
1-pass Shell, 1-pass Tubes BEM Heat Exchanger
8.9 kg/s Acetic Acid, 69.85 oC
2.5 kg/s Water, 76.85 oC
2.5 kg/s Water, 36.85 oC
16 feet
20 feet
From these preliminary calculations, it can be established that a BEM TEMA type of heat exchanger is not the appropriate type because of the Temperature Cross that takes place and the multiple exchangers quantity
Each shell diamter is estimated as 12". Allowing 12" for the nozzles, this makes the height of each shell 24".This means that if six BEMs are stacked as shown, the total height of the assembly would be 6 x2 = 12 feet = 3.7 meters.
This type of configuration is frequently found in MEA heat exchangers in order to avoid a temperature cross.In that application, normally two BEMs are stacked together and suffice to avoid the temperature cross.However, in this application, the quantity of units is 3 times that and the height might be a deterrent to use the
m2
m2
Comments, Notes, & Calculations:
The BFM TEMA type of heat exchanger allows for a single shell to have multiple shell passes by introducing a horizontal baffle(or multiple horizontal baffles). This type of construction allows the designer to incorporate the pre-set number of tubes that yield the desired tube fluid velocity in each tube pass.
If the tube side fluid velocity is set at 1.5 m/sec (4.92126 ft/sec) and the tubes are also set as 3/4"OD, the size of the heat exchanger can be estimated, based on the total heat transfer area calculated.
Kogara calculates the required total heat transfer area, A = 125.2Quantity of 3/4" tubes for 1.5 m/sec = 57 tubes
Heat transfer area in 20-ft long tubes = 206,302 20.6
Quantity of tube passes required in one heat exchanger shell =
Note that in order to maintain the specified fluid velocity in the tubes, the quantity of tube passes is fixed at 6.It is possible to fabricate a BFM type of exchanger with 6 tube passes, but in a practical sense this seems to be too much welding within one shell. Two duplicate BFMs stacked together would be a much more practical and workable solution.
cm2 =
8.9 kg/s Acetic Acid, 29.85 oC
8.9 kg/s Acetic Acid, 69.85 oC
2.5 kg/s Water, 76.85 oC
2.5 kg/s Water, 36.85 oC
2-pass Shell, 2-pass Tubes BFM Heat Exchanger
The BFM TEMA type of heat exchanger allows for a single shell to have multiple shell passes by introducing a horizontal baffle(or multiple horizontal baffles). This type of construction allows the designer to incorporate the pre-set number of tubes that
If the tube side fluid velocity is set at 1.5 m/sec (4.92126 ft/sec) and the tubes are also set as 3/4"OD, the size of the
6
Note that in order to maintain the specified fluid velocity in the tubes, the quantity of tube passes is fixed at 6.It is possible to fabricate a BFM type of exchanger with 6 tube passes, but in a practical sense this seems to be too much welding within one shell. Two duplicate BFMs stacked together would be a much more practical
