Breaking the Azeotrope between Methanol and Acetone with an Entrainer

The binary mixture of acetone (boiling point
56.10°C
) and methanol (boiling point
64.45°C
) is difficult to separate by distillation because of the presence of a minimum boiling azeotrope (
79.07
mole % acetone at
55.35°C
). This Demonstration shows that MEK (boiling point
79.74°C
) cannot be used as an entrainer to break this azeotrope at 1 atm. Indeed, MEK forms with methanol a minimum boiling azeotrope (
22.70
mole % MEK at
63.1°C
), and a distillation boundary (shown in cyan) divides the composition space into two distillation regions. Methanol and acetone are separated by this boundary. The temperature versus warped time clearly shows the nature of the pure components and azeotropes.
A
2
, indicated by a green line, is an unstable node. Acetone and
A
1
, indicated by a blue line, are saddle points. MEK and methanol, indicated by a red line, are stable points. Two snapshots show what happens if you choose a simple binary mixture. Of course, for a ternary mixture, you have to select a locator position inside the triangular diagram.

References

[1] V. Julka, M. Chiplunkar, and L. O'Young, "Selecting Entrainers for Azeotropic Distillation," Chemical Engineering Progress, 105(3), 2009 pp. 47–53.

Permanent Citation

Housam Binous
​
​"Breaking the Azeotrope between Methanol and Acetone with an Entrainer"​
​http://demonstrations.wolfram.com/BreakingTheAzeotropeBetweenMethanolAndAcetoneWithAnEntrainer/​
​Wolfram Demonstrations Project​
​Published: November 29, 2010