Jump to content



Featured Articles

Check out the latest featured articles.

File Library

Check out the latest downloads available in the File Library.

New Article

Product Viscosity vs. Shear

Featured File

Vertical Tank Selection

New Blog Entry

Low Flow in Pipes- posted in Ankur's blog

Condenser Design For Distillation

distillation condenser

This topic has been archived. This means that you cannot reply to this topic.
8 replies to this topic
Share this topic:
| More

#1 Sureesh91

Sureesh91

    Brand New Member

  • Members
  • 5 posts

Posted 05 May 2017 - 02:52 AM

Hi All,

 

I am looking to design a condenser for a distillation column. I am condensing methanol, using water. Can anyone suggest a method of approach? Or how I can go about the design. I am finding it difficult to find the dew point also.



#2 Bobby Strain

Bobby Strain

    Gold Member

  • Members
  • 3,526 posts

Posted 05 May 2017 - 07:06 PM

As one member says, "Good for you." You haven't told us anything. Give some details.

 

Bobby



#3 Sureesh91

Sureesh91

    Brand New Member

  • Members
  • 5 posts

Posted 08 May 2017 - 04:00 AM

I am carrying out a continuous distillation process.  From a Mass balance I have calculated the following:

 

Feed 0.591362609 kmol/h Xf (Methanol Mole Fraction Feed) 0.306161746   Xd (Methanol Mole Fraction Distillate) 0.78   Xb (Methanol Mole Fraction Bottoms) 0.1   B (Bottoms Flow) 0.412073862 kmol/h D (Tops Flow) 0.179288747 kmol/h

 

Using McCabe Thiele I have calculated 8 stages, and a reboiler and want a total condenser. The column will be operated under atmospheric pressure. Is this information okay?



#4 Art Montemayor

Art Montemayor

    Gold Member

  • Admin
  • 5,779 posts

Posted 08 May 2017 - 10:10 AM

Sureesh91:

 

Your post is perplexing.  You seem to have already taken Unit Operation courses, enabling you to undertake a distillation process design.  However, you seem to lack any knowledge of process heat transfer.  Please explain what is your academic situation.  Have you taken heat transfer courses?  If so, did you understand the material presented?

 

The majority of the basic data you present is simply useless and doesn't apply to any necessary input in process designing an overhead methanol total condenser.  It shows that you don't know the basics of what a condenser is supposed to do and how it goes about doing it if designed appropriately.  Please explain your lack of knowledge in heat transfer.

 

Our members can help you with useful advice but they won't do your work for you.  You must have studied the basics in order to discuss and understand what they are writing and explaining.  If you lack background study, obtain a copy of Kern's classic book, "Process Heat Transfer" and study it carefully.  In it you will find the method used to process design an overheads condenser.

 

For a typical explanation of how condensation is treated in process heat transfer design, read and study the attached document.  This may help orient you.  I hope this helps.

 

 

 

Attached Files



#5 Sureesh91

Sureesh91

    Brand New Member

  • Members
  • 5 posts

Posted 09 May 2017 - 02:47 AM

Thanks for the document! I am a final year student, yes I have taken a heat transfer course but seems like I'll have to revise the material. I took the course in my first year and seems like I have forgotten some things. I thought the flows and mass fractions maybe key to allow the energy required to condense the stream. But reading it seems like I need the enthalpy of vaporization instead? Thank you so much for the document again, I will obtain the heat transfer book.  



#6 Sureesh91

Sureesh91

    Brand New Member

  • Members
  • 5 posts

Posted 15 May 2017 - 07:23 AM

So if my understanding is correct these are my calculations:

 

Latent heat of feed: 1041 kJ/kg

Feed Flowrate: 29.8 kg/hr

 

Q = 29.8 x 1041 = 31020.26 kJ/hr

 

31020.26/3600 = 8.62 kW duty

 

Water inlet temp: 30 degrees C

Assumed water outlet: 35 degrees C

 

Vapour inlet temp: 78 degrees C

Condensate temp: 74 degrees C

 

LMTD: 43.5

U= 500W/m2 C

 

Area = ((8.62 x 1000)/(43.5 x 500))= 0.4 m2

 

Is this methodology used correct?



#7 Art Montemayor

Art Montemayor

    Gold Member

  • Admin
  • 5,779 posts

Posted 15 May 2017 - 08:42 AM

Sureesh91:

 

Please be correct about your terms.  You have not furnished any calculations.  What you have furnished our members is your (or a computer's) calculation RESULTS.  There is a big difference between both terms.

 

What our members need are detailed, explained, accurate calculations that contain the logic and explained steps executed to arrive at the results.  They need this type of information in order to give you intelligent, detailed comments and recommendations on your work.

 

They can't make any intelligent and meaningful comments on what you present if they don't know how you went about arriving at the results.  If you want intelligent and meaningful advice and comments on your work, you should present your calculations in an organized, legible, and detailed manner in an Excel (or equal) spreadsheet that will allow our members to quickly determine if the correct, applicable logic has been applied and the expected accuracy is obtained.

 

Anything less than that means our members have to do the logic application and all the math.  In other words: our members have to do your work for you to determine if your results are valid.  That, I don't think, is going to happen.

 

Await your detailed and explained calculations.



#8 Sureesh91

Sureesh91

    Brand New Member

  • Members
  • 5 posts

Posted 15 May 2017 - 10:07 AM

Okay , I have worked out the latent heat of the feed by multiplying the latent heat of methanol by its mole fraction at 78 degrees C. Then the same with water. Eg.

 

(Xm * Lm) + (Xw + Lw) where Xm is methanol mole fraction, Lm is methanol latent heat, Xw is water mole fraction, Lw is water latent heat.

 

I have then found the total latent heat of the feed. As condensing and vaporization require the same amount of energy.

 

I then converted this value into kW , to work out the duty of the condenser required.

 

Then assuming an inlet water temperature of 30 degrees and outlet temperature of 35 degrees and a requirement of the vapour to go from 78 to 74 degrees I have calculated the log mean temperature difference.

 

Using the equation Q=U*A*LMTD 

Assuming 500 for U from C&R Vol 6, I have calculated the area of condenser required.

 

Is this description okay? Does my methodology make sense? 

 

I apologise if it is still not clear to follow


Edited by Sureesh91, 15 May 2017 - 10:08 AM.


#9 Art Montemayor

Art Montemayor

    Gold Member

  • Admin
  • 5,779 posts

Posted 15 May 2017 - 10:36 AM

Sureesh91:

 

I am going to take a further step in trying to impress upon you - and all other potential or future engineers presently studying Chemical Engineering on what I mean by typical, detailed, and legible engineering calculations.  Some years ago I took the time to transcribe a typical condenser calculation found in Donald Kern's famous book into step-wise, detailed calculations on a spread sheet in order to show the detailed calculations and how they are justified and carried out for the benefit of Chemical Engineering students.  I am attaching this spread sheet for your (and all other interested students') benefit to see exactly what I mean by engineering calculations.

 

The example worked out is based on a pure propanol total condenser.  Note that your assignment may not involve a pure methanol vapor.  Because you continue to be deficient in describing your specific problem, we don't know the actual composition of your condensed vapor.  You will have to furnish that.  Kern spends two chapters in discussing and calculating condensers - one for pure vapors, the other for mixed vapors.  As I recommended before (and you said you would follow up), you should obtain Kern's book and study it diligently if you want to fully understand what is going on in a condenser and how the problem is attacked - and why.  Engineering is NOT about simply using equations to solve problems.  You have to first fully understand the applicable equation and WHY it can be applied.  You also have to fully understand and know how and why a unit operation works the way it is supposed to operate.  If you don't know the workings and details of a unit operation, you will always fail in trying to calculate any operation taking place within it.  I hope you take my words seriously and recognize how much you have to study and master before you can seriously take on an engineering application.

 

Engineering is hard, dedicated, and serious work.  That is why many may be called to the profession, but few are chosen.  I hope you take on this challenge and respond with positive steps in taking on this unit operations problem with positive results.

 

 

Attached File  Kerns Propanol Condensers.xlsx   67.96KB   70 downloads






Similar Topics