UNIFAC Help needed urgently

[7604]

Can anyone please explain to me how to calculate the groups quantities or the group mole fractions in UNIFAC? I have tried "Wallas" but it is not explicity at all.
There is an exemple here that says: for a equimolar mixture of benzene(1) and n-propanol(2)binary mixture. bemzene has six ACH groups N#3.Thus v3(benzene)=6 R3=0.5313 and Q3=0.400
n-propanolhas one CH3 group (1A), one CH2 group(1B) and one COH group (5A). V(1A)n-propanol=1, V(1B)n-propanol=1 and V(5A)n-propanol=1
Because it is an equimolar mixture X1=X2=1/2
Until here it is clear for me. But how do they came up with this quantification such as:
X(1A)=(1/2)/(6/2+1/2+1/2+1/2)=1/9
and
X(3)=(6/2)/(6/2+1/2+1/2+1/2)=3/2
Can anyone explain that to me?

Cheers

 

[rbcoulter]

I am currently writing a spreadsheet for UNIFAC.
I think I can answer your question:

X(1A) and X(3) are the group mole fractions for groups 1A and 3 respectively. The "total moles" of all groups in the mixture is (6/2 + 1/2 + 1/2 + 1/2). The "6/2" is the moles of
ACHs (6 * 1/2). Each of the other "1/2"s represent the moles of each of the three groups in the propanol.

This appears to be the beginning of the calculation for
the residual part of the activity coefficient.

My reference is "Properties of Gases and Liquids" by Poling, Prausnitz , and O'Connell.

The UNIFAC calculation is complicated but apparently very useful.

 

[AndreChE]

rbcoulter is correct!

My UNIFAC Excel spreadsheet is already done.

AndreChE