Question

Question 2:

A train of two distillation colum

ns is being used to separate a mixture of hexane, heptane and octane.

The feed to the first column contains 40 mole% hexane, 30 mole% heptane and 30 mole% octane and

enters at a rate of 2500 kmol/h. In the first column 98% of the octane fed is recovered

in the bottom

product. Essentially no hexane goes with the bottom product and the mole fraction of octaine in the

bottom is 99.5%. The overhead product flows to a second column. The overhead product from the

second column contains 98 mole% hexane and no

octane. In addition, 98% of the hexane fed to this

column is recovered in the overhead.

a) Draw a diagram of the process, labeling all of the streams.

b) Calculate the molar flow rates and mole fractions for each stream in the process

Answer 1

First column

Feed F = 2500 kmol/h

Composition

Hexane = 0.40

Heptane = 0.30

Octane = 0.30

Molar flow rate of hexane = 2500*0.40 = 1000 kmol/h

Molar flow rate of heptane = 2500*0.30 = 750 kmol/h

Molar flow rate of octane = 2500*0.30 = 750 kmol/h

Bottom product

Composition

Octane = 0.995

Hexane = 0

Heptane = 1 - 0.995 = 0.005

Molar flow rate

Octane = 0.98*750 = 735 kmol/h

Hexane = 0

Bottom flow rate B = 735/0.995 = 738.69 kmol/h

heptane = 738.69*(1 - 0.995) = 3.69 kmol/h

Overhead product

Molar flow rate

Octane = octane in feed - octane in bottom

= 750 - 735 = 15 kmol/h

Hexane = 1000 - 0 = 1000 kmol/h

Heptane = 750 - 3.69 = 746.31 kmol/h

Distillate D = 15 + 1000 + 746.31 = 1761.31 kmol/hr

Composition

Octane = molar flow of octane / D = 15/1761.31 = 0.0085

Hexane = 1000/1761.31 = 0.5680

Heptane = 746.31/1761.31 = 0.4235

Second column

Overhead product

Composition

Hexane = 0.98

Octane = 0

Heptane = 1 - 0.98 = 0.02

Molar flow rate

Hexane = 0.98*1000 = 980 kmol/h

Distillate = 980/0.98 = 1000 kmol/h

Heptane = 0.02*1000 = 20 kmol/h

Bottom product

Molar flow rate

Hexane = hexane in overhead from first column - hexane in overhead of second column

= 1000 - 980 = 20 kmol/h

Octane = 15 - 0 = 15 kmol/h

Heptane = 746.31 - 20 = 726.31 kmol/h

Total bottom = 20 + 15 + 726.31 = 761.31 kmol/h

Composition

Hexane = 20/761.31 = 0.026

Heptane = 726.31/761.31 = 0.954

Octane = 15/761.31 = 0.02