Solution to IA1 Acetic Acid Acetic Anhydride problem

(Statement of problem: A mixture of Acetic Acid and Acetic Anydride containing 40 mol % Acetic Acid is to be separated by distillation. The top product is to be 90 mol % Acetic Acid and the bottom product 10 mol % Acetic Acid. The feed flowrate is 100 kmol/hr and enters the column as liquid at it’s boiling point.

The feed is heated to its boiling point. The vapour is condensed but not cooled and some is returned at a reflux ratio of 3 kmol/kmol product.

Determine the operating lines for the rectifying and stripping sections and draw them on the equilibrium curve.)

How do we do this?

Start with the rectifying line – it is the simplest to determine because we know the reflux ratio. The definition of this line is:

In this case R = 3 and the mole fraction of a (acetic acid) in the distillate (x_d) is 0.9 so the equation becomes

Half way there. All we need now is the equation of the stripping line. To write this we need vapour and liquid flowrates in the stripping section and the bottoms flowrate. To get these we must do a mass balance. We can write a mass balance on the whole column for each component:

Again, a is acetic acid, b is acetic anhydride (MVC first). We know the mole fractions, they are:

	Feed	Bottoms	Distillate
Acetic acid	0.4	0.1	0.9
Acetic anhydride	0.6	0.9	0.1

The flowrate of the feed, F, is 100 kmol/hr. Equations above become:

Two equations for two unknowns can be solved to give D = 37.5 kmol/hr and B = 62.5 kmol/hr. Draw a schematic of the column, add these numbers to it and see what flowrates are left.

We must next determine the internal flowrates in the column of vapour and liquid.

The reflux ratio is defined as R = L/D. In this case R = 3 and D = 37.5 kmol/hr. Therefore, L = 112.5 kmol hr.

The vapour flow of the top of the column is condensed to give the L and D flowrates so V = L + D = 150 kmol hr.

The feed is all liquid at boiling point, i.e. no vapour, only liquid but this liquid does not have to be heated to boiling point, it is already at it. This means that the feed only contributes to the liquid flowrate down the column and not to the vapour flowrate up the column. Therefore the vapour flowrate is the same in both the stripping and rectifying sections, i.e. V’ = V = 150 kmol/hr.

The liquid flowrate in the stripping is the sum of the feed and the rectifying flow, i.e. L’ = L + F = 212.5 kmol/hr.

Add these to the schematic:

Now, we can write the equation for the stripping line:

All that is left is to plot these on the VLE curve.