VLE VAPOR LIQUID EQUILIBRIUM - Introduction with examples

1. VAPOR/LIQUID EQUILIBRIUM-
Introduction
ERT 206: Thermodynamics
Miss Anis Atikah Ahmad
Email: anis atikah@unimap.edu.my

2. OUTLINE
1. The Nature of Equilibrium
2. Duhem’s Theorem
3. Simple Models for VLE
4. VLE by Modified Raoult’s Law
5. VLE from K-value Correlations

3. 1. The Nature of Equilibrium
• Equilibrium is a static condition in which no
changes occur in the macroscopic properties
of a system with time.
– Eg: An isolated system consisting of liquid & vapor
phase reaches a final state wherein no tendency
exists for change to occur within the system. The
temperature, pressure and phase compositions
reach final values which thereafter remain fixed.

4. • At microscopic level, conditions are not static.
– Molecules with high velocities near the interface
overcome surface forces and pass into the other
phase.
– But the average rate of passage of molecules is
the same in both directions & no net interphase
transfer of material occurs.

5. Measures of Composition
1. Mass fraction: the ratio of the mass of a particular chemical
species in a mixture or solution to the total mass of mixture or
solution.
2. Mole fraction: the ratio of the number of moles of a
particular chemical species in a mixture or solution to the number
of moles of mixture or solution.
m
m
m
m
x ii
i



n
n
n
n
x ii
i




6. Measures of Composition
3. Molar concentration: the ratio of the mole fraction of a
particular chemical species in a mixture or solution to the molar
volume of mixture or solution.
4. Molar mass of mixture/solution: mole-fraction-
weighted sum of the molar masses of all species present.
q
n
V
x
C ii
i


i
i
i MxM 
in
q
Molar flow rate
Volumetric flow rate

7. 2. Duhem’s Theorem
• Duhem’s Theorem: for any closed system formed initially
from given masses of prescribed chemical species, the equilibrium
state is completely determined when any two independent
variables are fixed.
– Applies to closed systems at equilibrium
– The extensive state and intensive state of system are fixed
22  NNF 
Similar to phase
rule, but it
considers
extensive state.
No of
equations
No of
variables

8. 3. SIMPLE MODELS FOR
VAPOR/LIQUID EQUILIBRIUM
• Vapor/liquid equilibrium (VLE): the state of coexistence of
liquid and vapor phase.
• VLE Model: to calculate temperatures, pressures and
compositions of phases in equilibrium.
• The two simplest models are:
– Raoult’s law
– Henry’s law

9. 3. SIMPLE MODELS FOR
VAPOR/LIQUID EQUILIBRIUM
3.1 Raoult’s Law
• Assumptions:
– The vapor phase is an ideal gas (low to moderate pressure)
– The liquid phase is an ideal solution (the system are chemically similar)
*Chemically similar: the molecular species are not too different in size
and are of the same chemical nature.
eg: n-hexane/n-heptane, ethanol/propanol, benzene/toluene
 NiPxPy sat
iii ...,2,1
ix
iy
Liquid phase mole fraction
Vapor phase mole fraction
sat
iP Vapor pressure of pure species i at
system temperature

10. Pxy Diagram

12. 3.2 Dewpoint & Bubblepoint
Calculations with Raoult’s Law
4 Calculations
• BUBL P : Calculate {yi} and P, given {xi} and T
• DEW P : Calculate {xi} and P, given {yi} and T
• BUBL T : Calculate {yi} and T, given {xi} and P
• DEW T : Calculate {xi} and T, given {yi} and P
If the vapor-phase composition is unknown, may be assumed; thus

i
sat
ii PxP
 i iy 1
sat
iii PxPy 
For bubble point calculation

13. 3.2 Dewpoint & Bubblepoint
Calculations with Raoult’s Law
If the liquid-phase composition is unknown, may be assumed; thus


i
sat
ii Py
P
/
1
 i ix 1
sat
iii PxPy 
For dew point calculation

14. 3.2.1 BUBL P CALCULATION
(Calculate {yi} and P, given {xi} and T)
Find P1
sat &
P2
sat using
Antoine
equation
Find P Calculate yi

i
sat
ii PxP
satsat
PxPxP 2211 
  satsat
PxPxP 2111 1
  1212 xPPPP satsatsat

sat
iii PxPy 
sat
PxPy 111 
P
Px
y
sat
11
1 

15. Example 1
Binary system acetronitrile (1)/ nitromethane (2) conforms
closely to Raoult’s law. Vapor pressure for the pure species are
given by the following Antoine equations:
Prepare a graph showing P vs. X1 and P vs. Y1 for a temperature
of 75°C.
00.224/
47.2945
2724.14/ln 1


Ct
kPaPsat
00.209/
64.2972
2043.14/ln 2


Ct
kPaPsat

16. 3.2.1 BUBL P CALCULATION
(Calculate {yi} and P, given {xi} and T)
At 75°C, by Antoine Equations,
00.22475
47.2945
2724.14/ln 1


C
kPaPsat
00.20975
64.2972
2043.14/ln 2


C
kPaPsat
kPaPsat
21.831 
kPaPsat
98.412 
Find P1
sat &
P2
sat using
Antoine
equation
Find P Calculate yi

17. 3.2.1 BUBL P CALCULATION
(Calculate {yi} and P, given {xi} and T)
  1212 xPPPP satsatsat

  198.4121.8398.41 xP 
Taking at any value of x1, say x1=0.6,
  6.098.4121.8398.41 P
kPa72.66
Find P1
sat &
P2
sat using
Antoine
equation
Find P Calculate yi

18. 3.2.1 BUBL P CALCULATION
(Calculate {yi} and P, given {xi} and T)
Find P1
sat &
P2
sat using
Antoine
equation
Find P Calculate yi
P
Px
y
sat
11
1 
  7483.0
72.66
21.836.0

At 75°C, a liquid mixture of 60 mol-% acetonitrile and 40 mol-%
nitromethane is in equilibrium with a vapor containing 74.83 mol-%
acetonitrile at a pressure of 66.72 kPa

19. To draw P-x-y graph, repeat the calculation with different values of x;
x1 y1 P/kPa
0.0 0.0000 41.98
0.2 0.3313 50.23
0.4 0.5692 58.47
0.6 0.7483 66.72
0.8 0.8880 74.96
1.0 1.0000 83.21
P-x-y Diagram

20. P x y diagram for acetonitrile/nitromethane at 75°C as given by
Raoult’s law
0
20
40
60
80
100
0 0.2 0.4 0.6 0.8 1
P/kPa
x1, y1
P2
sat = 41.98
P1
sat = 83.21T= 75°C
Subcooled liquid
Superheated vapor

21. P x y diagram for
acetonitrile/nitromethane at 75°C as
given by Raoult’s law
0
20
40
60
80
100
0 0.2 0.4 0.6 0.8 1
P/kPa
x1, y1
P2
sat = 41.98
P1
sat = 83.21T= 75°C
Subcooled liquid
Superheated vapor
a
b
b'
c
d
c’
Point a is a subcooled
liquid mixture of 60 mol-
% acetonitrile and 40
mol-% of nitromethane
at 75°C.
Point b is saturated
liquid.
Points lying between b
and c are in two phase
region, where saturated
liquid and saturated
vapor coexist in
equilibrium.
Saturated liquid and
saturated vapor of the
pure species coexist at
vapor pressure P1
sat and
P2
sat

22. P x y diagram for
acetonitrile/nitromethane at 75°C as
given by Raoult’s law
0
20
40
60
80
100
0 0.2 0.4 0.6 0.8 1
P/kPa
x1, y1
P2
sat = 41.98
P1
sat = 83.21T= 75°C
Subcooled liquid
Superheated vapor
a
b
b'
c
d
c’
Point b: bubblepoint
P-x1 is the locus of
bubblepoints
As point c is approached,
the liquid phase has
almost disappeared, with
only droplets (dew)
remaining.
Point c: dewpoint
P-y1 is the locus of
dewpoints.

23. P x y diagram for
acetonitrile/nitromethane at 75°C as
given by Raoult’s law
0
20
40
60
80
100
0 0.2 0.4 0.6 0.8 1
P/kPa
x1, y1
P2
sat = 41.98
P1
sat = 83.21T= 75°C
Subcooled liquid
Superheated vapor
a
b
b'
c
d
c’
Once the dew has
evaporated, only
saturated vapor at point
c remains.
Further pressure
reduction leads to
superheated vapor at
point d

24. 0
20
40
60
80
100
0 0.2 0.4 0.6 0.8 1
P/kPa
x1, y1
P2
sat = 41.98
P1
sat = 83.21T= 75°C
Subcooled liquid
Superheated vapor
a
b
b'
c
d
c’
3.2.2 DEW P CALCULATION
(DEW P : Calculate {xi} and P, given {yi} and T)
What is x1 & P
at point c’?
Step 1: Calculate P
Step 2: Calculate x1
satsat
PyPy
P
2211 //
1


kPa74.59
98.41/4.021.83/6.0
1


sat
P
Py
x
1
1
1 
 
21.83
74.596.0

4308.0

25. 3.2.2 DEW P CALCULATION
(DEW P : Calculate {xi} and P, given {yi} and T)
Find P from Raoult’s
Law assuming Calculate xi
sat
iii PxPy 
sat
PxPy 111 
sat
P
Py
x
1
1
1 
 i ix 1


i
sat
ii Py
P
/
1
satsat
PyPy
P
2211 //
1



26. T-x-y Diagram
Find T1
sat
& T2
sat
using
Antoine
equation
Find P1
sat
& P2
sat
using T
btween
T1
sat &
T2
sat
Calculate
xi
Calculate
yi
i
i
isat
i C
PA
B
T 


ln
sat
iii PxPy 
sat
PxPy 111 
P
Px
y
sat
11
1 

i
sat
ii PxP
satsat
PxPxP 2211 
  satsat
PxPxP 2111 1
  1212 xPPPP satsatsat

 satsat
sat
PP
PP
x
21
2
1




27. Example 2
Binary system acetronitrile (1)/ nitromethane (2) conforms
closely to Raoult’s law. Vapor pressure for the pure species are
given by the following Antoine equations:
Prepare a graph showing T vs. X1 and T vs. Y1 for a pressure of
of 70kPa.
00.224/
47.2945
2724.14/ln 1


Ct
kPaPsat
00.209/
64.2972
2043.14/ln 2


Ct
kPaPsat

28. i
i
isat
i C
PA
B
T 


ln
CT sat


 84.69224
70ln2724.14
47.2945
1
CT sat


 58.89209
70ln2043.14
64.2972
2
Find T1
sat
& T2
sat
using
Antoine
equation
Find P1
sat
& P2
sat
using T
btween
T1
sat &
T2
sat
Calculate
xi
Calculate
yi
T-x-y Diagram

29. Find T1
sat
& T2
sat
using
Antoine
equation
Find P1
sat
& P2
sat
using T
btween
T1
sat &
T2
sat
Calculate
xi
Calculate
yi
T1
sat = 69.84°C, T2
sat = 89.58°C
Let T=78°C,
kPaP
C
kPaP
sat
sat
76.91
00.22478
47.2945
2724.14/ln
1
1



kPaP
C
kPaP
sat
sat
84.46
00.20978
64.2972
2043.14/ln
2
2



T-x-y Diagram

30. Find T1
sat
& T2
sat
using
Antoine
equation
Find P1
sat
& P2
sat
using T
btween
T1
sat &
T2
sat
Calculate
xi
Calculate
yi
P1
sat = 91.76kPa, P2
sat = 46.84kPa
 satsat
sat
PP
PP
x
21
2
1



5156.0
84.4676.91
84.4670




T-x-y Diagram

31. Find T1
sat
& T2
sat
using
Antoine
equation
Find P1
sat
& P2
sat
using T
btween
T1
sat &
T2
sat
Calculate
xi
Calculate
yi
P1
sat = 91.76kPa, x = 0.5156
P
Px
y
sat
11
1 
  6759.0
70
76.915156.0

T-x-y Diagram

32. To draw T-x-y graph, repeat the calculation with different values of T;
x1 y1 T/°C
0.0000 0.0000 89.58 (T2
sat)
0.1424 0.2401 86
0.3184 0.4742 82
0.5156 0.6759 78
0.7378 0.8484 74
1.0000 1.0000 69.84 (T1
sat)
T-x-y Diagram

33. 65
70
75
80
85
90
0 0.2 0.4 0.6 0.8 1
T/°C
x1, y1
Subcooled liquid
Superheated vapor
T2
sat = 89.58°C
T1
sat = 69.84°c
T x y diagram for acetonitrile/nitromethane at 70 kPa as
given by Raoult’s law

34. What is y1 and T
at point b’
(with x1=0.6 and
P= 70 kPa)?
65
70
75
80
85
90
0 0.2 0.4 0.6 0.8 1
T/°C
x1, y1
Subcooled liquid
Superheated vapor
T2
sat = 89.58°C
T1
sat = 69.84°c
3.2.3 BUBL T CALCULATION
(Calculate {yi} and T, given {xi} and P)
c’
c
b b'

35. 3.2.3 BUBL T CALCULATION
(Calculate {yi} and T, given {xi} and P)
21
2
xx
P
Psat



sat
sat
P
P
2
1

C
PA
B
T sat



2ln
00.209
64.2972
00.224
47.2945
0681.0ln




tt

The substraction of ln P1
sat & P2
sat from
Antoine Equation
00.209
ln2043.14
64.2972
2


 sat
P
Start with
α=1, find
P2
sat
Find T using
Antoine eq
&
substitute
P2
sat
obtained in
step 1
Find new α
by
substituting
T
Repeat step
1 by using
new α until
similar
value of α
is obtained
Find P1
sat &
find y1
using
Raoult’s
law
satsat
PxPxP 2211 
2
2
11
2
x
P
Px
P
P
sat
sat
sat


36. 3.2.3 BUBL T CALCULATION
(Calculate {yi} and T, given {xi} and P)
1
kPaPsat
702 
CT  58.89
88.1
88.1
kPaPsat
81.452 
CT  38.77
96.1
Iteration 1
Iteration 2
96.1
kPaPsat
41.442 
CT  53.76
97.1
Iteration 3
97.1
CT  43.76
97.1
Iteration 4
kPaPsat
24.442 
Start with
α=1, find
P2
sat
Find T using
Antoine eq
&
substitute
P2
sat
obtained in
step 1
Find new α
by
substituting
T
Repeat step
1 by using
new α until
similar
value of α
is obtained
Find P1
sat &
find y1
using
Raoult’s
law
satsat
PP 21 
 24.4497.1
kPa17.87
P
Px
y
sat
11
1 
 
70
17.876.0

7472.0

37. What is x1 and T
at point c’
(with y1=0.6 and
P= 70 kPa)?
65
70
75
80
85
90
0 0.2 0.4 0.6 0.8 1
T/°C
x1, y1
Subcooled liquid
Superheated vapor
T2
sat = 89.58°C
T1
sat = 69.84°c
3.2.4 DEW T CALCULATION
(Calculate {xi} and T, given {yi} and P)
c’
c
b b'

38. 3.2.4 DEW T CALCULATION
(Calculate {xi} and T, given {yi} and P)
Start with
α=1, find
P1
sat
Find T using
Antoine eq
&
substitute
P1
sat
obtained in
step 1
Find new α
by
substituting
T
Repeat step
1 by using
new α until
similar
value of α
is obtained
Find x1
 211 yyPPsat

sat
sat
P
P
2
1

C
PA
B
T sat



1ln
00.209
64.2972
00.224
47.2945
0681.0ln




tt

00.224
ln2724.14
47.2945
1


 sat
P
satsat
PyPy
P
2211
1


  2211
1
yPPy
P
P satsat
sat



39. 3.3 Henry’s Law
• Used for a species whose critical temperature
is less than the temperature of application, in
which Raoult’s Law could not be applied (since
Raoult’s Law requires a value of Pi
sat).
iii xPy 
Where Hi is Henry’s constant and obtained from experiment.

40. 4. VLE by Modified Raoult’s Law
• Used when the liquid phase is not an ideal
solution.
sat
iiii PxPy 
Where ɣi is an activity coefficient
(deviation from solution ideality in liquid phase).

41. 4. VLE by Modified Raoult’s Law
• For bubblepoint calculation, (assuming )
• For dewpoint calculation, (assuming )

i
sat
iii PxP 
 i iy 1
 i ix 1


i
sat
iii Py
P

1

42. 5. VLE from K-value Correlations
• Equilibrium ratio, Ki
• When Ki > 1, species exhibits a higher
concentration of vapor phase
• When Ki < 1, species exhibits a higher
concentration of liquid phase (is considered as heavy
constituent.)
i
i
i
x
y
K 

43. 5. VLE from K-value Correlations
• K value for Raoult’s Law
• K value for modified Raoult’s Law
P
P
x
y
K
sat
i
i
i
i  sat
iii PxPy since
P
P
K
sat
ii
i

 sat
iiii PxPy since

44. 5. VLE from K-value Correlations
• For bubblepoint calculations,
• For dewpoint calculations
i
i
i
x
y
K 
 i iy 1
1i
ii xK
 i ix 1
i
i
i
x
y
K  1i i
i
K
y

45. Example
For a mixture of 10 mol-% methane, 20 mol-%
ethane, and 70 mol-% propane at 50°F, determine:
(a) The dewpoint pressure
(b)The bubblepoint pressure

46. Example
(a) The dewpoint pressure
When the system at its dewpoint, only an insignificant amount
of liquid is present.
Thus 10 mol-% methane, 20 mol-% ethane, and 70 mol-%
propane are the values of yi.
assuming, thus, 1i i
i
K
y
 i ix 1
For a mixture of 10 mol-% methane, 20 mol-%
ethane, and 70 mol-% propane at 50°F,
determine:
By trial, find the value of pressure that satisfy 1i i
i
K
y

48. Species yi P=100psia P=150psia P=126psia
Ki yi/Ki Ki yi/Ki Ki yi/Ki
Methane 0.10 20.0 0.005 13.2 0.008 16.0 0.006
Ethane 0.20 3.25 0.062 2.25 0.089 2.65 0.075
Propane 0.70 0.92 0.761 0.65 1.077 0.762 0.919
828.0i
ii Ky 174.1i
ii Ky 000.1i
ii Ky
Thus, the dewpoint pressure is 126 psia.
Example
(a) The dewpoint pressure
For a mixture of 10 mol-% methane, 20 mol-%
ethane, and 70 mol-% propane at 50°F,
determine:

49. Example
(b)The bubblepoint pressure
assuming , thus
1i
ii xK
 i iy 1
For a mixture of 10 mol-% methane, 20 mol-%
ethane, and 70 mol-% propane at 50°F,
determine:
By trial, find the value of pressure that satisfy 1i
ii xK

51. Species xi P=380psia P=400psia P=385psia
Ki Kixi Ki Kixi Ki Kixi
Methane 0.10 5.60 0.560 5.25 0.525 5.49 0.549
Ethane 0.20 1.11 0.222 1.07 0.214 1.10 0.220
Propane 0.70 0.335 0.235 0.32 0.224 0.33 0.231
017.1i
ii xK 963.0i
ii xK 000.1i
ii xK
Thus, the bubblepoint pressure is 385 psia.
Example
(b) The bubble point pressure
For a mixture of 10 mol-% methane, 20 mol-%
ethane, and 70 mol-% propane at 50°F,
determine: