Question

In a two-stage process, acid is extracted from water into hexanol in a liquid-liquid extraction unit (i.e. extractor) and the extract is sent to a distillation column. Assume that water is completely insoluble in hexanol (think about what this means- hint: what goes into the distillation column and what does not?). A mixture of18 wt% acid and the balance water is fed to the extractor. Pure hexanol is fed to the extractor to extract the acid. The water-rich stream leaving the extractor is 99.5 wt% water and the balanced acid. The hexanol-rich extract is the part (hint: that goes to the distillation column). The composition of the distillate is 96 wt% acid and the balance hexanol. The bottoms stream contains 97.2 wt% hexanol and recovers 95% of the hexanol fed to the extractor. 1. Draw the process and label every single info. Calculate the percentage of acid in the process feed that is recovered in the distillate stream. You must assume a basis. Hint: please assume a basis for the pure hexanol stream.

Answer 1

Given:

Feed to extractor, F, containing Water and Acid

Wt% composition of F:

Acid = 18% = 0.18

Therefore, water = 1 - 0.18 = 0.82

(Summation of wt fraction of the components present in a system/stream is equal to 1)

Pure solvent for extraction, S, containing Hexanol, therefore,

wt% of Hexanol is = 100% = 1

water-rich stream is the raffinate, R, containing water and acid

wt% composition of R:

water = 99.5% = 0.995

therefore, acid = 1 – 0.995 = 5 x 10-3

hexanol-rich stream is the extract, E, containing hexanol and acid

E goes for distillation and gives two streams, Distillate, D and Bottoms, B.

Wt% composition of D:

Acid = 96% = 0.96

Therefore, hexanol = 1 – 0.96 = 0.04

Wt% composition of B:

Hexanol = 97.2% = 0.972

Therefore, acid = 1 – 0.972 = 0.028

Also, hexanol present in B stream = 0.95 x S

To calculate the values of the unknowns, we will use the mass balances, i.e., the mass entering a system is equal to the mass leaving the system.

Basis: 100 kg of solvent, S

(Here we take our basis in mass because we have been provided with the wt% (or wt fraction) and thus it will be easier to calculate the mass flows of the components)