PETE 310
Lectures # 25 -26
Chapter 12
Gas-Liquid Equilibrium
Gas-Liquid Equilibrium
Ideal Behavior
Applications to low pressures
Simplifications
the gas phase behaves as an Ideal
Gas
the liquid phase exhibits Ideal
Solution Behavior
Ideal Behavior
The equilibrium criteria between 2
phases a and b is,
a
P
T
a
ˆ i
a
 P
b
T
b
b
 ˆ i , i  1 , 2 ,.... N c
Equilibrium Conditions
The last criteria implies “tendency
of a component to be in phase a or
b is balanced” – “net mass transfer
across phases is zero”
ˆ i
a
b
 ˆ i , i  1, 2 ,.... N c
Ideal Behavior Model
Gas phase behaves as an ideal gas (IG),
and liquid phase behaves as an ideal
solution (IS).
These assumptions imply that
IG: molecular interactions are zero,
molecules have no volume.
IS: forces of attraction/repulsion
between molecules are the same
regardless of molecular species. Volumes
are additive (Amagat’s Law).
Forces between molecular
species
A
A
B
B
A
FAA  FBB  FAB
B
Statement of Equilibrium
y i P  x i Pi

IG/IS Raoult’s law
1
P1

2
P
T
3
Types of VLE Calculations
CP1
Ta
Liquid
P1v
Pressure
P1v
P2
Flash
CP2
P2v
v
Ta
Temperature
0
Vapor
x1, y1
1
Recall Molar Compositions
By convention liquid compositions (mole
fractions) are indicated with an x and gas
compositions with a y.

n1
x 1  
 n1  n 2


 liquid

n1
y 1  
 n1  n2


 gas
Mathematical Relationships
z 1  x 1f l  y 1fv
with
fv 
z1  x
1
y1  x1
In general
z 1  x 1 (1  fv )  y 1fv
fv 
( n 1  n 2 )v
n 1  n 2 v
fv 
  n 1  n 2 l
zi  x i
yi  xi
Depletion Path
Isothermal Reservoir Depletion Process for a
Reservoir Oil with 2 Components
z1 = fixed
T = Ta
CP M
Pressure
PB
A
B
C
PD
Ta
Temperature
z1=overall mole fraction of [1],
0
x1
y1=vapor mole fraction of [1],
z1
y1 1
x1=liquid mole fraction of [1]
Quantitative Phase Equilibrium
Exercise
P -xy D ia g ra m
2000
P r essu r e (p sia)
1600
T=160F
1200
800
400
0
0 .0
0 .1
0 .2
0 .3
0 .4
0 .5
C o m p o sitio n (% C 1)
0 .6
0 .7
0 .8
Bubble Point Evaluation
(Ideal Behavior Model)
The bubble point pressure at a
given T is

yi P
Pbp 
bp


z
i
Pi
z i Pi


Bubble Point from Raoult's law
T
P1

zi=xi

P2
x1,y1
Bubble Point Evaluation
Under Raoult’s law, the bubble point has
a linear dependence with the vapor
pressures of the pure components.
Once the bubble point pressure is found,
the equilibrium vapor compositions are
found from Raoult’s law.
Dew Point Calculation
At the dew point the overall fluid
composition coincides with the gas
composition. That is.
zi  yi
Dew Point Calculation
(Ideal Behavior Model)
Find DP pressure and equilibrium
liquid compositions
y i P  x i Pi
z i P  x i Pi


Pdp

zi
Pi
 Nc zi 
   
 i  1 Pi 

1


xi
P
Dew Point from Raoult's law
T
P1

zi=yi

P2
x1,y1
Flash Calculations
In this type of calculations the
objective is to:
find fraction of vapor vaporized (fv)
and equilibrium gas and liquid
compositions
given the overall mixture composition,
P and T.
Flash Calculations
(Ideal Behavior Calculations)
Start with the equilibrium equation
y i P  x i Pi

Material balance
z i  x i f l  y i f v  x i 1  f v   y i f v
Flash Calculations
Now replace either liquid or gas
compositions using equilibrium
equation
zi  yi
P
Pi

 yi P




x
i 
 P
 i

1 
fv   yi fv
Here replaced xi
Flash Calculations
 Rearrange and sum over all yi
yi 
zi
P
Pi

yi 


1 
fv   fv




zi


 P 1  f   f 
v
v
 Pi 



Separation process
yi(T1,P2)
zi(T1,P1)
T1,P2
P1 > P 2
xi(T1,P2)
Flash Calculations
Objective function (flash function)
is




zi
 1  0
F ( fv )   
 P 1  f   f 
v
v
 Pi 



This is 1/ki – ideal equilibrium ratio
Flash Calculations
There are several equivalent expressions
for the flash function
(a)

yi  1  0
(b)

xi  1  0
(c)

yi 

xi  0
Flash Calculations
Once fv is found the equilibrium gas
and liquid compositions are
evaluated from
yi 
zi
P
Pi

1 
fv   fv
and
 yi P




x
i 
 P
 i

Vapor Pressure Models
(Antoine Equation)
1. Constants depend upon the
component – Different Units
bi

ln Pi  a i 

Ti  c i
Example in our web site excel file VLE_310
F la s h F u n c tio n s a n d R a c h fo rd R ic e
F u n c tio n


zi
  1
 
i 1  f v  k i  1   1 
Nc
S um X i
6 .0 0
S um Yi
R a chfo rd R ice
4 .0 0


zi ki
  1
 
i 1  f v  k i  1   1 
Nc
F (fv )
2 .0 0
0 .0 0
-2 .0 0
Nc

-4 .0 0
i 1
-6 .0 0
0.00
0.20
0.40
0.60
0.80
M o la r F ra c tio n o f v a p o r (fv )
1.00
z i ( k i  1)
f v  k i  1  1