Question

An equimolar mixture of A and B is flashed. What is the maximum possible liquid concentration of B in the product? Assume that A is more volatile and the relative volatility is 4.38.

The answer should be within 0.02 of the correct answer.

Answer 1

Given;

An equimolar mixture of A and B is flashed.

A is more volatile than B.

Relative volatility ; α = 4.38

As A is more volatile; all the calculations will be based on concentration of A.

Concentration of A in feed ; z = 0.5

Concentration of A in liquid product = x

Concentration of A in vapor product = y

As liquid and vapor product stream are in equilibrium; x and y are related as :

y = α x / (1 + (α - 1) x)

y = 4.38 x / (1 + 3.38 x)

Let the flowrate of feed , liquid and vapor streams be F, L and V respectively.

Overall material balance :

F = L + V

Therefore;

V = F - L

Component A balance:

z F = x L + y V

0.5 F = x L + ( 4.38 x / (1 + 3.38 x) ) (F - L)

Therefore;

(0.5 - 4.38 x / (1 + 3.38 x) ) F = (x - 4.38 x / (1 + 3.38 x)) L

which on simplification gives;

(0.5 + 1.69 x - 4.38 x) F = (x + 3.38 x^2 - 4.38 x) L

(0.5 - 2.69 x) F = 3.38 x (x - 1) L

L / F = (0.5 - 2.69 x) / 3.38 x (x - 1)

The concentration of A in liquid phase = x

Moles of A in liquid phase = L x

Moles of A in liquid phase per mole of feed = L x / F

Therefore;

Multiplying both sides in the equation by x :

L x / F = x (0.5 - 2.69 x) / 3.38 x (x - 1)

L x / F = (0.5 - 2.69 x) / 3.38 (x - 1)

To get the minimum amount of B in liquid phase; we need to find the minima point of the RHS.

Let f (x) = L x / F

Therefore;

f (x) =  (0.5 - 2.69 x) / 3.38 (x - 1)

To get the minima of this function; we plot this function and check for any point of minima.

The plot of f(x) vs x is given below :

From the graph we can see that the non-negative minimum value of f(x) is 0. This occurs at x = 0.2

Therefore;

Minimum possible concentration of A in liquid product = 0.2

Maximum possible concentration of B in liquid product = 0.8