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4

# Calculating Evaporation Rate For A Specific Liquid To Use For Condense

8 replies to this topic
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### #1 rosneft1

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Posted 04 June 2018 - 07:40 AM

Hi,

I will be boiling off 30 L of liquid x in a 100L jacketed vessel under vacuum connected to a condenser and a reflux drum. I was wondering how can I calculate the evaporation rate of the liquid which I would like to use to calculate the size of the condenser required. The physical properties of liquid x are as follows:

• At 0.2 bar boiling point Tb = 160 C - I have derived an antoine equation for the vapour pressure up until atmospheric pressure.
• Cp = 1.9 KJ/Kg.K
• Thermal Conductivity, k = 0.170 KJ/s.m.c
• Latent heat of vaporisation = 464 KJ/Kg

If my thinking is wrong, please feel free to let me know. Any guidance would be highly appreciated.

### #2 Art Montemayor

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Posted 04 June 2018 - 12:50 PM

### #3 rosneft1

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Posted 04 June 2018 - 02:04 PM

Thanks for the answer, however I'm not sure what you mean.

### #4 Art Montemayor

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Posted 04 June 2018 - 03:58 PM

What I meant is that you are not being candid - as a member - in giving us your profile information.

This information helps our members familiarize themselves as much as possible in order to give you as much information and comments as you can use and understand.  We don't want to pry into your personal affairs, but our members can help you more if they are more familiar with your background, training, and experience.

I believe it is our business - if we are to try to help you.  We share our knowledge and experience with you; why can't you share yours with us?

### #5 rosneft1

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Posted 05 June 2018 - 03:55 AM

Thank you, that is understandable. I can't really update my last name, I usually prefer not to share that information unless its with a specific person when on a forum. I have further updated the information on my profile. I have been away from real chemical engineering for a few years with limited post graduate experience, hence the rudimentary nature of the question.

As for the question, unfortunately the liquid's application is patented and covered by a confidentiality agreement due to the R&D nature of the project. Disclosing the process requirement together with the chemical name (single component) will infringe that agreement resulting in legal preceding. Its my intention to be as candid as possible.
I used to use this forum as a student, however I do not recall my account old account details. If the name is still an issue I will resign up and provide full details.

Thank you again for your time.

### #6 Pilesar

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Posted 05 June 2018 - 07:34 AM

Evaporation rate will be limited by your heat input rate. At the boiling point, you can calculate the amount boiled by taking the heat input divided by the latent heat. The only variable in that is the heat input -- to increase the rate of vaporization you turn up the heat. So size the condenser to handle the maximum heat input from your jacket. If you want maximum vaporization, you should design for low enough overhead pressure drop to not overload your vacuum system. Your condenser cost may not be very sensitive to size, so a 'too large' condenser may be better in the long run.

### #7 rosneft1

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Posted 06 June 2018 - 09:25 AM

Evaporation rate will be limited by your heat input rate. At the boiling point, you can calculate the amount boiled by taking the heat input divided by the latent heat. The only variable in that is the heat input -- to increase the rate of vaporization you turn up the heat. So size the condenser to handle the maximum heat input from your jacket. If you want maximum vaporization, you should design for low enough overhead pressure drop to not overload your vacuum system. Your condenser cost may not be very sensitive to size, so a 'too large' condenser may be better in the long run.

Thanks Pilesar,

Just to check in an ideal system:

If the boiling point of the liquid at atmospheric P is 235 C

The energy required to boil the liquid QB = 464 x 30 = 13920 KJ

In a Jacketed vessel with a 7 KW oil bath the time required to boil the liquid is 13920/7 = 33 min

Hence the flow rate in the condenser 0.015 Kg/s

Therefore the condenser requires to remove (0.015 x 1.9 X 10) + 7 = 7.285 KW (condense by decreasing temperature by 10 C).

Is this correct?

### #8 Pilesar

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Posted 06 June 2018 - 12:55 PM

Your solution indicates that the condenser removes slightly more heat than is used to boil the liquid? That sounds okay since you are subcooling the overhead.

### #9 breizh

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Posted 07 June 2018 - 04:13 AM

hi ,

I don't know how you are doing your calculation !

Probably the results are OK but units are not  consistant in SI system . No indication about the density of your material .

Breizh