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Help With 3 Phase Separator

Started by Guest_workingbugs_*, Feb 16 2011 03:08 PM

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#1 Guest_workingbugs_*

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Posted 16 February 2011 - 03:08 PM

Ok, So I have been poking around the sight reading as many threads as I can about 2 and 3 phase separator design. I also have a paper written by Monnery and Svrcek about specifying three phase separators and I'm still having trouble with this. What I am trying to do is seperate a mixture of butanol and water after butanol/water vapors have gone through a condenser. They will be fed into a decanter or in this case a 3 phase seperator before sending them to two seperate distillation columns. The problem I am having is with all these guidelines it says the size is based on the vapor flow of the inlet gas but I don't think that I have any gas/vapor flow into my seperator. I should only have condensed liquid, I think or at least I hope so. How do I go about designing something that can separate the two phases (butanol and water)? My inlet stream is a total of 669 L/day or 617 kg. It is 64.7% water by weight, 29.4% butanol, 3.9% ethanol and 2% acetone. After a materials balance, the top phase should be a total volume of 210 L with 78.6% butanol by weight,15.6% water, 3.9 % ethanol and 1.9% acetone. The bottom phase should be 458 L total volume being 84.2% water by weight, 9.8% butanol 3.9% ethanol and 2.1% acetone. What I would like to do is size this so both the top phase and bottom phase are pumped into their respective columns and eventually overhead vapors are returned to the separator after being condensed so they can phase out again. Can anyone help me with this. I have never sized anything like a separator and am looking for all the help I can get. Attached is a picture of what I am envisioning. Thanks for any help

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#2 Zauberberg

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Posted 16 February 2011 - 03:17 PM

Have a look at the referenced document, page 18 (Paragraph 6.3) and Appendix F:

http://217.174.18.60...pr/e-pr-850.pdf

#3 Art Montemayor

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Posted 16 February 2011 - 04:19 PM


Bugsy:

Why do you waste your time reading and trying to apply a Unit Operation (3-phase separation) to what is clearly and directly nothing more than a Decantation of two liquid phases? The two unit operations are different.

What I would design is a classical Decanter – a vessel that has enough residence time and volume to ensure that the two phase separate and are held there. This decanter would feed (by level controls) two feed vessels with sufficient residence time to ensure good instrumentation contol feed to the related distillation columns.

Stop calling this “a separator”. It is a DECANTER. Do some research on this unit operation if you have not received any lecture or instructions on decanting of liquids. The residence time necessary to carry out a good decantation depends on the difference in specific gravities between the two liquid layers. You can do research on your liquids or do an experiment in a glass container to see how long it takes for the two liquid phases to separate out. If I had to take a guess, I would use 45 minutes to an hour as a residence time.

You are trying to do a separation of both liquid phases – WHY DO YOU JOIN BOTH LIQUIDS THAT GO INTO THE DECANTATION? That is not logical or smart engineering. Introduce the liquids separately and with laminar flow design (very s-l-o-w-ly) to aid in the separation. You don’t want mixing or turbulence.

#4 Guest_workingbugs_*

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Posted 18 February 2011 - 07:54 AM

OK, I went and tried to find some information on decanter unit operations per the advice of Art and I found a book on Google books, Chemical Engineering Design; Principles, Practice and Economics of Plant and Process Design, and they have a decent section in there on decanters. It gives a basic design with an example but I do not understand something I guess. It first says Uc = Lc / Ai < Ud meaning velocity of continuous phase in m/s is equal to the continuous phase volumetric flow rate in m3/s divided by the area of the interface if it is less than the settling velocity of the dispersed phase droplets in m/s. The value of Ud is than calculated by stokes law of Ud = dd^2*g*(rhod-rhoc) / 18*Muc which I understand as well. The problem I don't understand is in the example problem they calculate Ud and then Lc then out of nowhere say that Uc is not greater than Ud and then go on to calculating the area using Lc / Ud or at least the value they obtained from calculating Ud. Can someone please explain this to me? Or give me another reference I can look at to try and figure it out. Thanks
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