1. Examples of LLE and stability analyses
Chapter 14-Part V
Examples of LLE and stability analyses

2. Example LLE
A binary liquid exhibits LLE at 25oC. Determine from each of the following sets of miscibility data estimates for parameters A12 and A21 in the Margules equation at 25oC:
(a) x1a =0.1, x1b = 0.90
(b) x1a =0.20; x1b = 0.90
(c) x1a=0.10; x1b =0.80

3. The activity coefficients are functions of A12 and A21
a) Given:
x1a =0.1, x2a =0.9
x1b = 0.90, x2b =0.1
LLE equations:
The activity coefficients are functions of A12 and A21
Solution: A12 = A21 = 2.747
Why are the two coefficients equal?
Plot the T-x diagram for this case (schematic)

4. Why are the two coefficients different?
b) Given:
x1a =0.2, x2a =0.8
x1b = 0.90, x2b =0.1
LLE equations:
Solution: A12 = 2.148; A21 = 2.781
Why are the two coefficients different?
Plot the T-x diagram for this case (schematic)

5. Why are the two coefficients the same as in the previous example?
c) Given:
x1a =0.1, x2a =0.9
x1b = 0.80, x2b =0.2
LLE equations:
Solution: A12 = 2.781; A21 = 2.148
Why are the two coefficients the same as in the previous example?
Plot the T-x diagram for this case (schematic)

6. 2nd example
It is shown in example 14.7 that the Wilson equation is not able of representing LLE. Show that a simple modification of the Wilson equation:
can represent LLE, C is a constant.
Solution:
First, compare this equation with the original Wilson equation

7. Modified equation = C (Original equation)
(GE/RT)m = C (GE/RT)
How are the activity coefficients calculated from a GE equation?
ln (g1)m = C ln g1
How do we prove that the modified model is able to predict LLE?

8. Lets add and subtract ln x1 to the lhs, and C ln x1
to the rhs of the equation:
ln x1 + ln (g1)m -ln x1 = C ln g1 + C ln x1 – C ln x1
ln (x1g1)m – ln x1 = C(ln g1x1) – C ln x1
ln (x1g1)m = C(ln g1x1) – (C-1) ln x1
Lets take the derivatives wrt x1:
Using the stability criterium, what can we conclude with respect to the ability of the modified equation to predict LLE?