Mass Balance Approach for a 2-Stage Distillation Problem

[Rogue]

Homework Statement

180kmol/hr of a three component mixture made up of 60 mol % A, 25 mol % B and 15 mol %.
Component A is the most volatile ,component C is the least volatile.

The mixture is to be passed through a 2 stage distillation process.
80% of component C leaves the bottom of the first column in a 90% mixture with component B.

The top product leaving the first column is further separated in the second column to give a top product 95% A and a bottom product 85% B.

Determine the flows out each column and the mol % A in the bottom of the second column.

Assume none of component A leaves at the bottom of column 1 and none of component C leaves at the top of column 2.

Homework Equations

Mass Balance Diagram

The Attempt at a Solution

My approach is to calculate the kmol % of each component in the feed.

Component A = 108kmol /h
B = 45
C = 27

80% of C leaves at the bottom of column 1.
27 x 0.8 = 21.6

This makes a solution 90% C and 10% B,
So 21.6/0.9 = 24 (100% of product leaving bottom of column 1)
So B = 2.4kmol/hIs this the correct starting approach please?

 

[haruspex]

Is this the correct starting approach please?

Looks good so far.

 

[Rogue]

Are you sure?

As I follow up ,I can't seem to get it to balance.

For leaving the bottom of column 1 I have:

21.6kmol Component C
2.4kmol Component BFor entering column 2 I have :
5.4 kmol component C
42.6kmol component B
108 kmol component ALeaving the bottom of column 2:

5.4 kmol component C
36.21kmol component B
0.99 kmol component A

Leaving the top of column 2:

6.39kmol component B

But, 121.41 kmol component A.

Hence it doesn't balance so I must be doing something wrong? ??

 

[Chestermiller]

Let's see how you solved column 2. Please show your equations.

 

[Rogue]

Thanks gents.

Realised my mistake, sorted now.