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Calculating solvent losses in a batch vacuum distillation.

Calculating solvent losses in a batch vacuum distillation.

Calculating solvent losses in a batch vacuum distillation.

(OP)
In a batch process kettle with overhead condensors connected to single stage steam ejector for vacuum generation I would like to calculate how much xylene
(the only low boiler) would be lost from the reactor
at kettle vacuum of 150 mbar. & 110 deg. C. (ejector
side vacuum 0 mbar).  
Any ideas how this can be done ?  The vacuum in the
keetle is controlled by means of throttling of a control valve at suction to the ejector.

RE: Calculating solvent losses in a batch vacuum distillation.

My suggestion is to start by treating it as a one stage flash separator and assume that you reach equilibrium, with the liquid phase staying behind in the kettle, and the vapour being removed.  This can be modelled with a process simulator, if you have acce$$ to one, or you can buy simulator time over the web (eg: Hysys for flash calculations), or you can "do it yourself".  "Flash Calculations" are well described in chemical engineering textbooks, handbooks (eg:  Perry's handbook), on the web, etc.

The hard work will be in determining the thermodynamic properties of your mixture and its components, which are not specified.  Regarding the xylene itself, is it ortho, meta or para xylene or a mixture.  

Perhaps for your needs, treating the mixture as ideal is sufficient.  Calculate the xylene boiling point at your pressure (eg:  with Clausius Clapeyron equation and basic properties data) to see if it would boil.  If not, you may have a more complicated diffusion calculation, and low vapour flow rates, and not a flash calculation.

Perry's handbook shows ortho-xylene vapour pressure of about 300 mmHg at 110C (didn't need to estimate it as it was given in table of vapour pressure vs temperature).  So, for an ideal mixture, with negligible other vapours, the xylene should be boiling at your 0.15 Bar (114 mmHg) and 110C.

Rate of boiling is another question, and would depend upon heat transfer rates and the vapour's heat of vaporization.

Hope this may help.  Good luck!

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