Transcript VAPOR/LIQUID EQUILIBRIUM

VAPOR/LIQUID EQUILIBRIUMIntroduction
ERT 206: Thermodynamics
Miss Anis Atikah Ahmad
Email: anis [email protected]
OUTLINE
1.
2.
3.
4.
5.
The Nature of Equilibrium
Duhem’s Theorem
Simple Models for VLE
VLE by Modified Raoult’s Law
VLE from K-value Correlations
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.
• 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.
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.
mi m i
xi 

m m
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.
ni ni
xi  
n n
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.
xi ni
Ci  
V q
ni 
Molar flow rate
q
Volumetric flow rate
4. Molar mass of mixture/solution: mole-fractionweighted sum of the molar masses of all species present.
M   xi M i
i
2. Duhem’s Theorem
Similar to phase
rule, but it
considers
extensive state.
• 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
F  2  N  N  2
No of
variables
No of
equations
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
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
yi P  xi Pi sat
xi
Liquid phase mole fraction
yiVapor phase mole fraction
i  1,2..., N 
Pi sat Vapor pressure of pure species i at
system temperature
Pxy Diagram
3.2 Dewpoint & Bubblepoint
Calculations with Raoult’s Law
yi P  xi Pi sat
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,
P   xi Pi sat
i
y
i
i
1
may be assumed; thus
For bubble point calculation
3.2 Dewpoint & Bubblepoint
Calculations with Raoult’s Law
yi P  xi Pi sat
If the liquid-phase composition is unknown,
P

i
1
yi / Pi sat
x
i
i
1
may be assumed; thus
For dew point calculation
3.2.1 BUBL P CALCULATION
(Calculate {yi} and P, given {xi} and T)
Find P1sat &
P2sat using
Antoine
equation
Find P
P   xi Pi sat
yi P  xi Pi sat
i
y1P  x1 P1sat
PxP
 x2 P
PxP
 1 x1 P
sat
1 1
sat
1 1
sat
2
sat
2

Calculate yi

P  P2sat  P1sat  P2sat x1
x1 P1sat
y1 
P
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:
sat
1
2945 .47
/ kPa  14.2724 
t / C  224 .00
sat
2
2972 .64
/ kPa  14.2043 
t / C  209 .00
ln P
ln P
Prepare a graph showing P vs. X1 and P vs. Y1 for a temperature
of 75°C.
3.2.1 BUBL P CALCULATION
(Calculate {yi} and P, given {xi} and T)
Find P1sat &
P2sat using
Antoine
equation
Find P
At 75°C, by Antoine Equations,
sat
1
ln P
2945 .47
/ kPa  14.2724 
75C  224 .00
P1sat  83.21kPa
sat
2
ln P
2972 .64
/ kPa  14.2043 
75 C  209 .00
P2sat  41.98kPa
Calculate yi
3.2.1 BUBL P CALCULATION
(Calculate {yi} and P, given {xi} and T)
Find P1sat &
P2sat using
Antoine
equation
Find P


P  P2sat  P1sat  P2sat x1
P  41.98  83.21 41.98x1
Taking at any value of x1, say x1=0.6,
P  41.98  83.21 41.980.6
 66.72 kPa
Calculate yi
3.2.1 BUBL P CALCULATION
(Calculate {yi} and P, given {xi} and T)
Find P1sat &
P2sat using
Antoine
equation
Find P
Calculate yi
x1 P1sat
y1 
P
0.683 .21

 0.7483
66 .72
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
P-x-y Diagram
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 for acetonitrile/nitromethane at 75°C as given by
Raoult’s law
100
T= 75°C
P1sat = 83.21
80
Subcooled liquid
P/kPa
60
40
P2sat = 41.98
20
Superheated vapor
0
0
0.2
0.4
0.6
x1, y1
0.8
1
P x y diagram for
acetonitrile/nitromethane at 75°C as
given by Raoult’s law
100
T= 75°C
Point a is a subcooled
liquid mixture of 60 mol% acetonitrile and 40
mol-% of nitromethane
at 75°C.
P1sat = 83.21
a
80
Subcooled liquid
Point b is saturated
liquid.
b
c’
Points lying between b
and c are in two phase
region, where saturated
liquid and saturated
vapor coexist in
equilibrium.
c
P/kPa
60
b'
d
40
P2sat = 41.98
20
Superheated vapor
0
0
0.2
0.4
0.6
x1, y1
0.8
1
Saturated liquid and
saturated vapor of the
pure species coexist at
vapor pressure P1sat and
P2sat
P x y diagram for
acetonitrile/nitromethane at 75°C as
given by Raoult’s law
Point b: bubblepoint
100
T= 75°C
P1sat = 83.21
P-x1 is the locus of
bubblepoints
a
80
Subcooled liquid
b
c’
b'
c
P/kPa
60
As point c is approached,
the liquid phase has
almost disappeared, with
only droplets (dew)
remaining.
d
40
Point c: dewpoint
P2sat = 41.98
P-y1 is the locus of
dewpoints.
20
Superheated vapor
0
0
0.2
0.4
0.6
x1, y1
0.8
1
P x y diagram for
acetonitrile/nitromethane at 75°C as
given by Raoult’s law
100
T= 75°C
P1sat = 83.21
a
80
Subcooled liquid
b
c’
b'
c
P/kPa
60
Once the dew has
evaporated, only
saturated vapor at point
c remains.
d
Further pressure
reduction leads to
superheated vapor at
point d
40
P2sat = 41.98
20
Superheated vapor
0
0
0.2
0.4
0.6
x1, y1
0.8
1
3.2.2 DEW P CALCULATION
(DEW P : Calculate {xi} and P, given {yi} and T)
What is x1 & P
T= 75°C
P1sat = 83.21 at point c’?
100
a
80
Subcooled liquid
b
c’
b'
c
P/kPa
60
Step 1: Calculate P
1
P
y1 / P1sat  y2 / P2sat
1

0.6 / 83.21  0.4 / 41.98
d
40
 59.74 kPa
P2sat = 41.98
Step 2: Calculate x1
y1 P
x1 
P1sat
20
Superheated vapor
0
0
0.2
0.4
0.6
x1, y1
0.8
1
0.659.74 
83.21
 0.4308

3.2.2 DEW P CALCULATION
(DEW P : Calculate {xi} and P, given {yi} and T)
Find P from Raoult’s
Law assuming

xi  1

1
yi / Pi sat
i
P
i
P
y1 / P1sat
1
 y2 / P2sat
Calculate xi
yi P  xi Pi sat
y1P  x1 P1sat
x1 
y1 P
P1sat
T-x-y Diagram
Find T1sat
& T2sat
using
Antoine
equation
Ti
sat
Bi

 Ci
Ai  ln P
Find P1sat
& P2sat
using T
btween
T1sat &
T2sat
Calculate
xi
P   xi Pi sat
i
P  x1P1sat  x2 P2sat
P  x1P1sat  1 x1 P2sat


P  P2sat  P1sat  P2sat x1
P  P2sat
x1  sat
P1  P2sat


Calculate
yi
yi P  xi Pi sat
y1P  x1 P1sat
x1 P1sat
y1 
P
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:
sat
1
2945 .47
/ kPa  14.2724 
t / C  224 .00
sat
2
2972 .64
/ kPa  14.2043 
t / C  209 .00
ln P
ln P
Prepare a graph showing T vs. X1 and T vs. Y1 for a pressure of
of 70kPa.
T-x-y Diagram
Find T1sat
& T2sat
using
Antoine
equation
Ti sat 
sat
1
T
Find P1sat
& P2sat
using T
btween
T1sat &
T2sat
Calculate
xi
Bi
 Ci
Ai  ln P
2945 .47

 224  69 .84 C
14 .2724  ln 70
T2sat 
2972 .64
 209  89 .58 C
14 .2043  ln 70
Calculate
yi
T-x-y Diagram
Find T1sat
& T2sat
using
Antoine
equation
T1sat = 69.84°C,
Let T=78°C,
Find P1sat
& P2sat
using T
btween
T1sat &
T2sat
Calculate
yi
T2sat = 89.58°C
2945.47
ln P / kPa  14.2724
78C  224.00
P1sat  91.76kPa
sat
1
Calculate
xi
2972.64
ln P / kPa  14.2043
78C  209.00
P2sat  46.84kPa
sat
2
T-x-y Diagram
Find P1sat
& P2sat
using T
btween
T1sat &
T2sat
Find T1sat
& T2sat
using
Antoine
equation
P1sat = 91.76kPa,
P2sat = 46.84kPa
P  P2sat
x1  sat
P1  P2sat


70  46.84

 0.5156
91.76  46.84
Calculate
xi
Calculate
yi
T-x-y Diagram
Find T1sat
& T2sat
using
Antoine
equation
P1sat = 91.76kPa,
Find P1sat
& P2sat
using T
btween
T1sat &
T2sat
Calculate
xi
x = 0.5156
x1 P1sat
y1 
P

0.5156 91 .76 
 0.6759
70
Calculate
yi
T-x-y Diagram
To draw T-x-y graph, repeat the calculation with different values of T;
x1
y1
T/°C
0.0000
0.0000
89.58 (T2sat)
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 (T1sat)
T x y diagram for acetonitrile/nitromethane at 70 kPa as
given by Raoult’s law
90
85
Superheated vapor
T1sat = 69.84°c
80
T/°C
T2sat = 89.58°C
Subcooled liquid
75
70
65
0
0.2
0.4
0.6
x1, y1
0.8
1
3.2.3 BUBL T CALCULATION
(Calculate {yi} and T, given {xi} and P)
90
Superheated vapor
85
T1sat = 69.84°c
T2sat = 89.58°C
T/°C
80
c
c’
b
75
b'
Subcooled liquid
70
65
0
0.2
0.4
0.6
x1, y1
0.8
1
What is y1 and T
at point b’
(with x1=0.6 and
P= 70 kPa)?
3.2.3 BUBL T CALCULATION
(Calculate {yi} and T, given {xi} and P)
Start with
α=1, find
P2sat
P  x1P1sat  x2 P2sat
P
x1P1sat
 sat  x2
sat
P2
P2
P1sat
  sat
P2
P
sat
P2 
x1  x2
Find T using
Antoine eq
&
substitute
P2sat
obtained in
step 1
Repeat step
1 by using
new α until
similar
value of α
is obtained
Find new α
by
substituting
T
Find P1sat &
find y1
using
Raoult’s
law
B
C
sat
A  ln P2
2972.64

 209.00
sat
14.2043 ln P2
T
2945 .47
2972 .64
ln   0.0681 

t  224 .00 t  209 .00
The substraction of ln P1sat & P2sat from
Antoine Equation
3.2.3 BUBL T CALCULATION
(Calculate {yi} and T, given {xi} and P)
Start with
α=1, find
P2sat
Find T using
Antoine eq
&
substitute
P2sat
obtained in
step 1
Iteration 1
 1
P2sat  70kPa
T  89 .58 C
  1.88
Iteration 2
  1.88
P2sat  45.81kPa
T  77 .38 C
  1.96
Find new α
by
substituting
T
Iteration 3
  1.96
P2sat  44.41kPa
T  76 .53 C
  1.97
Iteration 4
  1.97
P2sat  44.24kPa
T  76 .43 C
  1.97
Repeat step
1 by using
new α until
similar
value of α
is obtained
Find P1sat &
find y1
using
Raoult’s
law
P1sat  P2sat
 1.9744.24
 87.17 kPa
x1 P1sat
y1 
P
0.687.17 

70
 0.7472
3.2.4 DEW T CALCULATION
(Calculate {xi} and T, given {yi} and P)
90
Superheated vapor
85
T1sat = 69.84°c
T2sat = 89.58°C
80
c
T/°C
c’
b
75
b'
Subcooled liquid
70
65
0
0.2
0.4
0.6
x1, y1
0.8
1
What is x1 and T
at point c’
(with y1=0.6 and
P= 70 kPa)?
3.2.4 DEW T CALCULATION
(Calculate {xi} and T, given {yi} and P)
Find T using
Antoine eq
&
substitute
P1sat
obtained in
step 1
Start with
α=1, find
P1sat
P
P
y1 P1sat
1
 y2 P2sat
sat
1
sat
sat
1
2
P
y1  P P

P1sat
  sat
P2
y
T

Repeat step
1 by using
new α until
similar
value of α
is obtained
Find new α
by
substituting
T
Find x1
B
C
sat
A  ln P1
2945.47
 224.00
sat
14.2724 ln P1
2
P1sat  P y1  y2 
ln   0.0681 
2945 .47
2972 .64

t  224 .00 t  209 .00
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 Pisat).
yi P  xi  i
Where Hi is Henry’s constant and obtained from experiment.
4. VLE by Modified Raoult’s Law
• Used when the liquid phase is not an ideal
solution.
yi P  xi i Pi
sat
Where ɣi is an activity coefficient
(deviation from solution ideality in liquid phase).
4. VLE by Modified Raoult’s Law
• For bubblepoint calculation, (assuming
y
i
i
1
P   xi i Pi sat
i
• For dewpoint calculation, (assuming i xi  1 )
P
y
i
i
1
 i Pi
sat
)
5. VLE from K-value Correlations
• Equilibrium ratio, Ki
yi
Ki 
xi
• 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.)
5. VLE from K-value Correlations
• K value for Raoult’s Law
yi Pi sat
Ki 

xi
P
since yi P  xi Pi sat
• K value for modified Raoult’s Law
Ki 
 i Pi sat
P
since
yi P  xi i Pi sat
5. VLE from K-value Correlations
• For bubblepoint calculations, i yi  1
yi
Ki 
Ki xi  1

xi
i
• For dewpoint calculations
yi
Ki 
xi
x
i
i
1
yi
i K  1
i
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
Example
For a mixture of 10 mol-% methane, 20 mol-%
ethane, and 70 mol-% propane at 50°F,
determine:
(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,
i xi  1 thus,
yi
1

i Ki
yi
By trial, find the value of pressure that satisfy  K  1
i
i
Example
Species
For a mixture of 10 mol-% methane, 20 mol-%
ethane, and 70 mol-% propane at 50°F,
determine:
(a) The dewpoint pressure
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
y
i
i
Ki  0.828
y
i
Ki  1.174
i
Thus, the dewpoint pressure is 126 psia.
y
i
i
Ki  1.000
Example
For a mixture of 10 mol-% methane, 20 mol-%
ethane, and 70 mol-% propane at 50°F,
determine:
(b)The bubblepoint pressure
assuming
K x
i
i
y
i
i
 1 , thus
1
i
By trial, find the value of pressure that satisfy
K x
i
i
i
1
Example For a mixture of 10 mol-% methane, 20 mol-%
ethane, and 70 mol-% propane at 50°F,
(b) The bubble point pressure
determine:
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
 0.963
K x
K x
i
i
i
 1.017
K x
i
i
i
Thus, the bubblepoint pressure is 385 psia.
i
i
i
 1.000