Solvent Swap Distillation

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Consider a binary mixture containing a solvent (60 mole% methanol) and a solute (40 mole% 1-hexanol). We carry out a solvent swap using batch rectification. A second solvent (1-propanol) is fed into the still to replace the methanol. This is possible since 1-propanol has a higher boiling point than the original solvent. The flow rate of the fresh 1-propanol stream is such that the volume of the still is kept constant during the entire duration of the solvent swap. For simplicity, assume that the mixture obeys Raoult's law (all three species are aliphatic alcohols). You can set the reflux ratio, the number of stages of the column, and the holdup of the trays and condenser.


The first snapshot shows the compositions in the still of methanol, 1-propanol, and 1-hexanol in red, blue, and green, respectively. It is clear that methanol in the still is being depleted. By contrast, the 1-propanol composition increases. At the end of the operation, only 1-hexanol and 1-propanol are present.

The second snapshot shows the composition in the distillate. The lighter component (methanol) distills first (see red curve). The solute is the highest-boiling chemical species and remains in the apparatus, as indicated by the green curve.

If you decrease the reflux ratio (see the third snapshot), the distillation operation is conducted faster and the solvent swap is more rapidly achieved. You can also vary the number of stages of the column and observe that the lower the number of stages, the sloppier the separation. Finally, it is possible to see how the still's volume and the molar flow rate of the stream of 1-propanol vary with time. As an extension of the present calculation, one can find: (1) the required time to replace 99.9% of methanol with 1-propanol; and (2) the volume of 1-propanol needed to swap 99.9% of the original solvent.


Contributed by: Housam Binous, Mamdouh Al-Harthi, and Ahmed Bellagi (October 2015)
Open content licensed under CC BY-NC-SA




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