Recently, I'm working on a project about ASTM D86 distillation. I've theoretically calculated distillation curves for some binary mixtures (like 50/50 decane & dodecane) through equations. However, I found that the calculation is really complicated, especially for multi-component systems.
Can I use DWSIM to simulate the distillation curve? If yes, it will help a lot.
Thanks!
Qiwei
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ASTM D86 (see attachment) basically is standard test where heat is applied to the liquid mixture in the flask. When the mixture temperature reach the boiling point (total vapor pressure reach 1atm), we measurethe temperature & volume of liquid vaporized.
Another question, what is this vapor fraction? Say we have a mixture of decane & tetradecane, is it (decane vapor moles)/(decane vapor moles+tetradecane vapor moles) or (decane vapor moles+tetradecane vapor moles)/(decane vapor moles+tetradecane vapor moles+air moles)? In your method, I think it's the second one.
Thanks,
Qiwei
Last edit: Qiwei Xu 2017-02-03
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I fowllowed you method, but it seems that this method only simulates the process before boiling. However, the distillation only starts after boiling when the temperature reach the boiling point of the mixture.
Hi
i modified your sample. I think this is what you want to calculate.
Just run the sensitivity study. You may chart any result variable against every other variable.
Gregor
I learned from some ttorials that the molar vapor praction means how much mixture is in the vapor pahse (1 means the entire stream is a vapor). In your sensitivity study, I can't understand that as outlet molar vapor fraction rises to 1, why 'Molar Fraction (Overall Liquid Phase) / N-decane' is not going to be 0, and temperature not going to be 253oC (the boiling point of N-tetradecane). Because I think in a complete distillation process, the molar fraction of the more volatile component in the liquid phase will eventually drop to 0 and temperature rises to the highest boiling point.
the answer is "thats thermodynamics". Please have a look to the VLE diagram.
The global composition of your mixture is 50/50. After total evaporation you are at the dew point, which is the upper curve in the diagram. The vapour composition is 50/50 therefore. Then you go the left to find the composition at boiling point. This composition is representing the last drop of liquid to be evaporated. This is definitely not a pure componenten as you will find.
You only would get a single component in the liquid if the vapour is a pure component as well.
BTW, the pahse digram should be like this. The mile fraction of n-dodecane, more volatile component, should be lager in the vapor phase than that in the liquid phase.
Hi,
the correct answer is: "it depends"
My diagram depicts the end of evaporation process where the last drop of liquid is evaporating.
Your diagrams shows the start of evaporation with all liquid beeing at 50/50. The first buble at boiling point is at the point you marked at higher concentrations.
Gregor
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Your model sorks percetly great with both banary and multi-component systems. But the problem is that it can't reach the highest teperature. For a complete distillation process, the temperature of the mixture, in the end, will always reach the highest. For example, for decane-tetradecane mixture distillation, the temperature of the mixture will go up to 250oC, but in your model, it only goes to 231oC.
So, how to fix this problem to make the temperature of the mixture go up to the highest?
Hi Qiwei,
why should temperature go up to 250°C? This is the dew/boiling point of the pure component! You are operating with a mixture which has a dew point below that temperature.
If you have a look into the VLE diagram from my post above, you will find that dew temperature of 231°C as well. The result is correct.
Sure you may heat the vapour after complete evaporation even more. This will result in an overheated vapour at any desirable temperature (e.g. 250°C).
Gregor
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That's because I want to simulate a distillation process, in which the vapor is condensed in the condenser and received by a graduated cylinder.
In the VLE diagram, we can see that 50/50 decane-tetradecane mixture can produce 88/12 decane-tetradecane vapor, which leaves the flask then is condensed and collected. Then, the composition in the liquid mixture will change, say, to 45/55 decane-tetradecane (the mole fraction of more volatile component, decane inthis case,will decrease), and the 45/55 decane-tetradecane liquid mixture will produce, say, 80/20 decane-tetradecane.... When mixture becomes 14/86 decane-tetradcane (temperature is 231oC), and produce 50/50 decane-tetradecane vapor.
At this time, your model stops, but in the distillation, the 14/86 decane-tetradecane mixture will keep boiling, and the molar fraction of decane in the liquid mixture will keep going down (to 0 finally), the temperature going up (to 250oC finally). So your model can only simulate a part of the complete distillation.
And I want to simulate a complete distillation. I think I may use a distillation column.
Best
Last edit: Qiwei Xu 2017-02-08
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I've found a kind of dumb way to fix this problem. I just add a seperator after your model to seperate the liquid and the vapor. We will get a liquid which is richer in n-decane than the orifinal one (original is 50/50 n-decane/n-tetradecane). Then add another heater, and repeat your sensitivity model (change the vapor fraction in the second heater from 0 to 1), and in this way temperature will rise higher than before.
I'm sorry I couldn't help you earlier. For this task you'll need tro simulate a real distillation column with a very low reflux ratio, specifying the amount of distillate flow indirectly through the bottoms flow. Heat is applied over the resulting liquid, not over the overall mixture, that's why the sample from Gregor didn't work for you, neither your "fix".
For a 50/50 mixture, the initial boiling point is about 193 C. When the distillate flow rate goes above 50 mol/s (for a 100 mol/s 50/50 feed), there's almost only N-tetradecane left on the liquid, which is why the temperature remains near 253 C.
Check the attached simulation and run the sensitivity analysis case.
When I try to change the mixture to three-component mixture, (Ethanol 0.651000, N-heptane 0.136000 , 2,2,4-trimethylpentane 0.213000), it cannot calcualte the root under Peng-Robinson basis.
So I try to change the basis to UNIQUAC, and this time it shows no errors, but the calculation results show that there are just no mixture flow in the distillation column. Could you help me identify where's the problem.
Hi,
I am a new user of DWSIM.
Recently, I'm working on a project about ASTM D86 distillation. I've theoretically calculated distillation curves for some binary mixtures (like 50/50 decane & dodecane) through equations. However, I found that the calculation is really complicated, especially for multi-component systems.
Can I use DWSIM to simulate the distillation curve? If yes, it will help a lot.
Thanks!
Qiwei
The attachment is one of the distillation curves that I've calculated. (50/50 decane & tetradecane)
Thanks!
Qiwei
I don't know the inners of the ASTM D86 standard, but I believe that you can do it in DWSIM normally just like any other distillation modelling.
Thank you!
ASTM D86 (see attachment) basically is standard test where heat is applied to the liquid mixture in the flask. When the mixture temperature reach the boiling point (total vapor pressure reach 1atm), we measurethe temperature & volume of liquid vaporized.
But I don't know how to simulate this in DWSIM.
Qiwei
I don't know the exacte definition of this standard but it looks like a boiling curve. This may be calculated in the following way:
Gregor
Hi, Gregor
Another question, what is this vapor fraction? Say we have a mixture of decane & tetradecane, is it (decane vapor moles)/(decane vapor moles+tetradecane vapor moles) or (decane vapor moles+tetradecane vapor moles)/(decane vapor moles+tetradecane vapor moles+air moles)? In your method, I think it's the second one.
Thanks,
Qiwei
Last edit: Qiwei Xu 2017-02-03
Hi,
Thank you!
I fowllowed you method, but it seems that this method only simulates the process before boiling. However, the distillation only starts after boiling when the temperature reach the boiling point of the mixture.
Thanks
Qiwei
Last edit: Qiwei Xu 2017-02-03
Hi
i modified your sample. I think this is what you want to calculate.
Just run the sensitivity study. You may chart any result variable against every other variable.
Gregor
Hi, Gregor
Thank you for your reply!
I learned from some ttorials that the molar vapor praction means how much mixture is in the vapor pahse (1 means the entire stream is a vapor). In your sensitivity study, I can't understand that as outlet molar vapor fraction rises to 1, why 'Molar Fraction (Overall Liquid Phase) / N-decane' is not going to be 0, and temperature not going to be 253oC (the boiling point of N-tetradecane). Because I think in a complete distillation process, the molar fraction of the more volatile component in the liquid phase will eventually drop to 0 and temperature rises to the highest boiling point.
Qiwei
Last edit: Qiwei Xu 2017-02-04
Hi Qiwei,
the answer is "thats thermodynamics". Please have a look to the VLE diagram.
The global composition of your mixture is 50/50. After total evaporation you are at the dew point, which is the upper curve in the diagram. The vapour composition is 50/50 therefore. Then you go the left to find the composition at boiling point. This composition is representing the last drop of liquid to be evaporated. This is definitely not a pure componenten as you will find.
You only would get a single component in the liquid if the vapour is a pure component as well.
All your questions are answered now?
Gregor
Hi, Gregor
Yes, and great thanks to all your kind help!
Best
Qiwei
BTW, the pahse digram should be like this. The mile fraction of n-dodecane, more volatile component, should be lager in the vapor phase than that in the liquid phase.
Best
Hi,
the correct answer is: "it depends"
My diagram depicts the end of evaporation process where the last drop of liquid is evaporating.
Your diagrams shows the start of evaporation with all liquid beeing at 50/50. The first buble at boiling point is at the point you marked at higher concentrations.
Gregor
Yes. You're right.
Qiwei
Hi, Gregor
Sorry to bother you again.
Your model sorks percetly great with both banary and multi-component systems. But the problem is that it can't reach the highest teperature. For a complete distillation process, the temperature of the mixture, in the end, will always reach the highest. For example, for decane-tetradecane mixture distillation, the temperature of the mixture will go up to 250oC, but in your model, it only goes to 231oC.
So, how to fix this problem to make the temperature of the mixture go up to the highest?
Thank you!
Hi Qiwei,
why should temperature go up to 250°C? This is the dew/boiling point of the pure component! You are operating with a mixture which has a dew point below that temperature.
If you have a look into the VLE diagram from my post above, you will find that dew temperature of 231°C as well. The result is correct.
Sure you may heat the vapour after complete evaporation even more. This will result in an overheated vapour at any desirable temperature (e.g. 250°C).
Gregor
Hi, Gregor,
Thank you.
That's because I want to simulate a distillation process, in which the vapor is condensed in the condenser and received by a graduated cylinder.
In the VLE diagram, we can see that 50/50 decane-tetradecane mixture can produce 88/12 decane-tetradecane vapor, which leaves the flask then is condensed and collected. Then, the composition in the liquid mixture will change, say, to 45/55 decane-tetradecane (the mole fraction of more volatile component, decane inthis case,will decrease), and the 45/55 decane-tetradecane liquid mixture will produce, say, 80/20 decane-tetradecane.... When mixture becomes 14/86 decane-tetradcane (temperature is 231oC), and produce 50/50 decane-tetradecane vapor.
At this time, your model stops, but in the distillation, the 14/86 decane-tetradecane mixture will keep boiling, and the molar fraction of decane in the liquid mixture will keep going down (to 0 finally), the temperature going up (to 250oC finally). So your model can only simulate a part of the complete distillation.
And I want to simulate a complete distillation. I think I may use a distillation column.
Best
Last edit: Qiwei Xu 2017-02-08
Hi, Gregor
I've found a kind of dumb way to fix this problem. I just add a seperator after your model to seperate the liquid and the vapor. We will get a liquid which is richer in n-decane than the orifinal one (original is 50/50 n-decane/n-tetradecane). Then add another heater, and repeat your sensitivity model (change the vapor fraction in the second heater from 0 to 1), and in this way temperature will rise higher than before.
Hope you have a better method.
Thank you!
Qiwei
I'm sorry I couldn't help you earlier. For this task you'll need tro simulate a real distillation column with a very low reflux ratio, specifying the amount of distillate flow indirectly through the bottoms flow. Heat is applied over the resulting liquid, not over the overall mixture, that's why the sample from Gregor didn't work for you, neither your "fix".
For a 50/50 mixture, the initial boiling point is about 193 C. When the distillate flow rate goes above 50 mol/s (for a 100 mol/s 50/50 feed), there's almost only N-tetradecane left on the liquid, which is why the temperature remains near 253 C.
Check the attached simulation and run the sensitivity analysis case.
Regards
Daniel
Last edit: Daniel Medeiros 2017-02-09
Hi, Daniel
Thank you Daniel!
But your attached simulation cannot plot the attached curve.
Best
Qiwei
Hi, Daniel
I don't know where's wrong, but I reconnect the bottoms, and now the simualation works.
Thank you very much!
Best
Qiwei
Make sure to be running the latest DWSIM version (4.2 Update 1): http://dwsim.inforside.com.br/wiki/index.php?title=Downloads
Yeah, sure.
Anyway, thank again for your great help.
Best
Qiwei
Hi, Daniel
When I try to change the mixture to three-component mixture, (Ethanol 0.651000, N-heptane 0.136000 , 2,2,4-trimethylpentane 0.213000), it cannot calcualte the root under Peng-Robinson basis.
So I try to change the basis to UNIQUAC, and this time it shows no errors, but the calculation results show that there are just no mixture flow in the distillation column. Could you help me identify where's the problem.
Best
Qiwei
Last edit: Qiwei Xu 2017-02-10