Examples of LLE and stability analyses Chapter 14-Part V Examples of LLE and stability analyses
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
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)
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)
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)
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
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?
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?