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1. From the Last Class

2. Vapor Pressure Equations
 Cox chart & Duhring plots: log p* vs. T
 Antoine equation: (Table B.4) ☺☺
 Wagner equation
 “Properties of Gases and Liquids”
C
T
B
A
*
P
log10



r
r
6
3
5
.
1
T
1
T
/
)
D
C
B
A
(
*
P
ln











Note : the vapor pressure equations should be used
within the specified temperature range !

3. 6.2 Gibbs Phase Rule ☺☺☺
 Types of Process Variables
 Extensive Variables
 depend on the size of the system (mass and volume)
 Intensive Variables
 do not depend on the size of the system (T,P, density, and mass fraction)
 Gibbs Phase Rule
 Degree of freedom (F) is the number of intensive variables that can
be specified independently for a system at equilibrium.



 m
F 2
The number of phases
The number of chemical species
The number of degree of freedom
No reaction

4. Raoult’s Law for GLE:
Single Condensable Component ☺☺☺
 Water evaporation into dry air
 Saturation (GLE)
 Gibbs Phase Rule
 Raoult’s Law for GLE ☺☺☺
2
2
2
2
2 






 m
F
두 개의 변수가 결정되면 다른 값들은 결정될 수 있다.
)
(
*
T
p
P
y
p i
i
i 
 P & T  y
P & y  P
T & y  P
Partial pressure of vapor in the gas
= Pure-component vapor pressure at the system temperature
GLE
A+B
B

5. Example 6.3.-2 Material Balance Around a Condenser
 A stream of air at 100oC contains 5260 mmHg contains
10% water by volume.
(a) Calculate the dew point and degrees of superheat of the
air.
)
(
*
2
2
2 dp
O
H
O
H
O
H T
p
P
y
p 

mmHg
526
5260
1
.
0
2 


O
H
p
증기압이 526mmHg인 온도를 찾음.
Antoine equation
C
T
B
A
p



*
10
log
C
C
p
A
B
T o
dp 90
228
526
log
96681
.
7
21
.
1668
log 10
*
10







degree of superheat = 100 -90 = 10 oC
Table B-4
☺☺☺

6. Example 6.3.-2: A stream of air at 100oC contains 5260
mmHg contains 10% water by volume.
(b) Calculate the percent of vapor that condenses and the
final composition of the gas phase if the gas is cooled to
80oC at constant pressure.
Basis : 100 mol feed gas
100 mol feed
0.1 H2O
0.9 BDA
Q1 mol H2O
Q2 mol
y H2O
1-y BDA
Number of component = 2
Number of unknown = (Q1, Q2, y) = 3 D.O.F. = 1: cannot be solved
Should we stop?
BDA: Bone-Dry Air
☺☺☺

7. Basis : 100 mol feed gas
100 mol feed
0.1 H2O
0.9 BDA
Q1 mol H2O
Q2 mol
y H2O
1-y BDA
A gas in equilibrium with liquid must be saturated with the liquid .
T = 80oC, P = 5260 mmHg
Use thermodynamic information:
thus, y is a saturated condition at 80 oC, 5260 mmHg
0675
.
0
5260
/
355
/
)
80
(
*
2
2 

 P
C
p
y o
O
H
O
H
9325
.
0
1 2 

 O
H
BDA y
y
Material Balance
BDA Balance 100×0.9 = Q2×0.9325 Q2=96.5 mol
Total Balance 100=Q1+Q2
Q1=3.5 mol
% Condensed 3.5/(100×0.1) ×(100) = 35 %
T = 100oC, P = 5260 mmHg
)
(
*
T
p
P
y
p i
i
i 

Number of component = 2
Number of unknown = (Q1, Q2) = 2
D.O.F. = 0

8. (c) Calculate the percentage condensation and the final
gas-phase composition if the gas is compressed
isothermally to 8500 mmHg.
Basis : 100 mol feed gas
100 mol feed
0.1 H2O
0.9 BDA
Q1 mol H2O
Q2 mol
y H2O
1-y BDA
Number of component = 2
Number of unknown = (Q1, Q2, y) = 3 Cannot be solved !
BDA: Bone Dry Air
Example 6.3.-2: A stream of air at 100oC contains 5260
mmHg contains 10% water by volume. ☺☺☺

9. Basis : 100 mol feed gas
100 mol feed
0.1 H2O
0.9 BDA
Q1 mol H2O
Q2 mol
y H2O
1-y BDA
A gas in equilibrium with liquid must be saturated with the liquid .
T = 100 oC, P = 8500 mmHg
Use thermodynamic information :
thus, y is a saturated condition at 100oC, 8500 mmHg
0894
.
0
8500
/
760
/
)
100
(
*
2
2 

 P
C
p
y o
O
H
O
H
9106
.
0
1 2 
 O
H
y
Material Balance
BDA Balance 100×0.9=Q2×0.9106 Q2=98.8 mol
Total Balance 100=Q1+Q2
Q1=1.2 mol
% Condensed 1.2/(100×0.1) ×(100) = 12 %
)
(
*
T
p
P
y
p i
i
i 

Number of component = 2
Number of unknown = (Q1, Q2) = 2
D.O.F. = 0
T = 100oC, P = 5260 mmHg

10. Let’s Start!

11. 6.4 Multicomponent Vapor-Liquid Equilibria
 Gas-Liquid Processes
 Chemical reactions
 Distillation (증류)
 Gas  Liquid : Absorption (흡수)
 Liquid Gas : Stripping (탈기)
 VLE information
 From literature, databases
 Raoult’s Law & Henry’s Law ☺☺☺
 Rigorous calculation using model equations
 Distribution of components between vapor and liquid
phases
 Phase-Equilibrium Thermodynamics

12. Raoult’s Law and Henry’s Law ☺☺☺
 Raoult’s Law
 Valid for almost pure liquid (xi 1)
 Valid for mixture of similar substances
(over entire range of compositions: 0 < xi < 1)
 Henry’s Law
 Valid for dilute solution (xi0)
)
(
*
T
p
x
P
y
p i
i
i
i 

)
(T
H
x
P
y i
i
i 
Vapor
(P and yi)
Liquid
(xi)
 Ideal gas
 Ideal Solution
Hi = Henry’s law constant
pi* = vapor pressure

13. Example 6.4-2 ☺☺
Use either Raoult’s law or Henry’s law to solve the following problems.
1. A gas containing 1 mole% ethane is in contact with water at 20oC
and 20 atm. Estimate the mole fraction of dissolved ethane.
N2, O2, CO2, ….
CH4, C2H6, ….
Dilute solution Apply Henry’s Law
From Perry’s Handbook
fraction
atm/mole
10
63
.
2
)
20
( 4
6
2 

C
H o
H
C
)
(T
H
x
P
y i
i
i 
6
4
10
6
.
7
10
63
.
2
20
01
.
0
)
(
/ 





 T
H
P
y
x i
i
i

14. Example 6.4-2 ☺☺
Use either Raoult’s law or Henry’s law to solve the following problems.
2. An equimolar liquid mixture of benzene (B) and toluene (T) is in
equilibrium with its vapor at 30oC. What is the system pressure and
the composition of the vapor?
Benzene + Toluene Similar Substances Apply Raoult’s Law




i
i
i
i
i
p
P
T
p
x
P
y
p )
(
*
mmHg
7
.
36
)
30
(
mmHg
119
)
30
(
*
*


C
p
C
p
o
T
o
B
Table B.4
Antoine Eq’n
mmHg
35
.
18
7
.
36
5
.
0
mmHg
5
.
59
119
5
.
0
*
*








T
T
T
B
B
B
p
x
p
p
x
p mmHg
77.9
mmHg
35
.
18
mmHg
5
.
59 


P
236
.
0
9
.
77
/
35
.
18
/
764
.
0
9
.
77
/
5
.
59
/
*
*






P
p
x
y
P
p
x
y
T
T
T
B
B
B

15. Phase diagrams for binary VLE ☺☺
 Txy diagram
(at a fixed P)
T P
x or y x or y
vapor
liquid
vapor
liquid
V+L
V+L
Dew T
Bubble T
Dew P
Bubble P
x1 y1 x1 y1
 Pxy diagram
(at a fixed T)

16. Bubble and Dew Points
 Bubble Point Temperature : Constant P, T
 Bubble Point Pressure : Constant T, P
 Dew Point Temperature : Constant P, T
 Dew Point Pressure : Constant T, P
Bubble point Dew point

17. Bubble P
Liquid
Vapor
yi xi
P
Given T,x  Calculate P,y
Pressure
Composition

18. Dew P
Liquid
Vapor
xi
yi
P
Given T,y  Calculate P,x
Pressure
Composition

19. Bubble T
Vapor
Liquid
yi
xi
T
Given P,x  Calculate T,y
Temperature
Composition

20. Dew T
Vapor
Liquid
xi yi
T
Given T,x  Calculate P,y
Temperature
Pressure

21. VLE Calculations
 Bubble Point Temperature Calculation
 Given P, x  Calculate T,y
 Dew Point Temperature Calculation
 Given P,y  Calculate T,x
...
)
(
)
(
)
(
*
*
*




bp
b
b
bp
a
a
bp
i
i
i
T
p
x
T
p
x
P
P
T
p
x
y
  


i i dp
i
i
i
dp
i
i
i
T
p
P
y
x
T
p
P
y
x
1
)
(
)
(
*
*

22. VLE Calculation
 Iterative calculation required
 Not explicit form
 Iterative calculation techniques
 Trial and error method
 Newton-Raphson Method
 Secant Method
0
)
( 
X
OBJ find X that satisfies given relation

23. Algorithm for Bubble/Dew Point Calculations




i
i
i
i
y
x
1
1
Objective Function

 


i
i
i
i y
x
OBJ 0
)
(
*
T
p
x
P
y i
i
i 
Phase Equilibrium
P
T
p
x
y
K i
i
i
i /
)
(
/ *


Start
Given P,x
Assume T
Calculate OBJ
Calculate new T,y
|DT| <e
End
P
T
p
x
y i
i
i /
)
(
*


 

i
i
i
i y
x
OBJ
Newton Raphson
Secant iteration
…
Example : Bubble T Calculation

24. VLE Calculations for Nonideal Systems
 Phase equilibrium relations
 Ideal Gas + Ideal Solution
 Nonideal Gas + Nonideal Solution
 Fugacity coefficient : gas phase nonideality
 Activity coefficient : liquid phase nonideality
)
(
*
T
p
x
P
y i
i
i 
)
(
*
T
p
x
P
y i
i
i
i
i 
 
from equation of state models : SRK, PR ,….
from activity models : WILSON, NRTL, UNIQUAC, ,….

25. Example 6.4-4
 Bubble- and Dew- point calculation using Txy diagrams
1. Using the Txy diagram, estimate the bubble-point temperature and
the equilibrium vapor composition associated with a 40 mol %
benzene-60 mol % toluene liquid mixture at 1 atm. If the mixture is
steadily vaporized until the remaining liquid contains 25% benzene,
what is the final temperature?
95oC
100oC

26. 102oC
x=0.2
96oC
Example 6.4-4
 Bubble- and Dew- point calculation using Txy diagrams
1. Using the Txy diagram, estimate the dew-point temperature and the
equilibrium liquid composition associated with a vapor mixture of
benzene abd toluene containing 40 mol % benzene at 1 atm. If
condensation proceeds until the remaining vapor contains 60 %
benzene, what is the final temperature?

27. 6.5 Solutions of Solids in Liquid Solution
 Solubility (용해도)
 Limits on the amount of solid that can be dissolved
 Solubility strongly depends on T
 Example)
222 g AgNO3 / 100 g H2O at 20 o C
952 g AgNO3 / 100 g H2O at 100 oC
 Crystallization
 Separation of solids and liquids
 Driving force = solubility differences
 A solute in equilibrium with a crystal must be
saturated

28. Example 6.5-1
 150 kg of a saturated aqueous solution of AgNO3 at 100oC is cooled
to 20oC, thereby forming AgNO3 crystals, which are filtered from the
remaining solution.
 The wet filter cake, which contains 80% solid crystals and 20%
saturated solution by mass, passes to a dryer in which the remaining
water is vaporized.
 Calculate the fraction of the AgNO3 in the feed stream eventually
recovered as dry crystals and the amount of water that must be
removed in the drying stage.
Cooler
Crystallizer
Fiter
Evaporator
150 kg Saturated Solution
Water
Filter Cake
Filter Cake + Solution

29. Cooler
Crystallizer
Fiter
Evaporator
150 kg Q1 kg
Q2 kg filter cake Q3 kg solution
Q4 kg H2O
Q5 kg filter cake
952 g AgNO3 / 100 g H2O at 100 o C
x2 = 952 / (100+952) = 0.905
1-x2 = 0.095
222 g AgNO3 / 100 g H2O at 20 o C
x1 = 222/(100+222) = 0.689
1-x1 = 0.311
0.905 AgNO3
0.095 H2O
0.689 AgNO3
0.311 H2O
0.689 AgNO3
0.311 H2O
Additional Information
Q2 = 0.8 (Q2 + Q3)
 Q2 = 4Q3
Basis: 150 kg Feed

30. Cooler
Crystallizer
Fiter
Evaporator
150 kg Q1 kg
4Q3 kg filter cake Q3 kg solution
Q4 kg H2O
Q5 kg filter cake
0.905 AgNO3
0.095 H2O
0.689 AgNO3
0.311 H2O
0.689 AgNO3
0.311 H2O
Unknown = 3 Unknown = 3
Unknown = 2

31. Cooler
Crystallizer
Fiter
Evaporator
150 kg Q1 kg
4Q3 kg filter cake Q3 kg solution
Q4 kg H2O
Q5 kg filter cake
0.905 AgNO3
0.095 H2O
0.689 AgNO3
0.311 H2O
0.689 AgNO3
0.311 H2O
Water balance
150 * 0.095 = Q1 * 0.689 + Q3 *0.311
Total Balance
150 = Q1 + 4Q3 + Q3
Q1 = 20 kg
Q2 = 104 kg
Q3 = 26 kg

32. Cooler
Crystallizer
Fiter
Evaporator
150 kg 20 kg
104 kg filter cake 26 kg solution
Q4 kg H2O
Q5 kg filter cake
0.905 AgNO3
0.095 H2O
0.689 AgNO3
0.311 H2O
0.689 AgNO3
0.311 H2O
Water balance
26 * 0.311 = Q4
Total Balance
104 + 26 = Q4 + Q5
Q4 = 8 kg
Q5 = 122 kg
Recovery % = 122 / (150*0.905) * 100 % = 89.9 %

33. Hydrated Salts
 Several structures can be produced for water-
salt systems.
 Example ) Solid magnesium sulfate
MgSO4 anhydrous magnesium sulfate
MgSO4·H2O magnesium sulfate monohydrate
MgSO4·6H2O magnesium sulfate hexahydrate
MgSO4·7H2O magnesium sulfate heptahydrate
MgSO4·12H2O magnesium sulfate dodecahydrate

34. Colligative Solution Properties
 Colligative solution properties (용액의 총괄성)
 Property change of a solution
 Vapor pressure lowering
 Boiling point elevation
 Melting point depression
 Depends only on molar concentration
 Not depends on solute and solution

35. Colligative Solution Properties
 Vapor pressure lowering
 Boiling point elevation
 Melting point depression
*
*
*
*
*
*
*
)
(
)
(
)
1
(
)
(
)
(
)
1
(
)
(
s
e
s
s
x
s
s
e
s
s
s
xp
p
p
p
T
p
x
p
p
T
p
x
T
p



D





x
H
RT
T
v
b
b
ˆ
2
0
D

D
x
H
RT
T
m
m
m
ˆ
2
0
D

D

36. 6.6 Immiscible and Partially Miscible Liquids
 Terminology
 Immiscibility (불혼화성)
 Partial miscibility (부분혼화성)
 Liquid extraction (액체 추출)
 Distribution coefficient
WATER
MIBK
x
x
K
)
(
)
(

Water Phase
Chloroform Phase
Water-Rich Phase
MIBK-Rich Phase
immiscible partially miscible
Acetone
Distribution
of
Acetone

37. Phase Diagram for Partially Miscible
Ternary Systems
 Equilateral ternary LLE phase diagram

38. Phase Diagram for LLE Systems
Miscibility limit for water-furfural
Tie Lines
Two liquids have identical
compositions
Single Phase
Region
Two liquid
Phase Region

39. Phase Diagram for LLE Systems
Feed
EG : 30 %
Furfural : 30 %
Water : 40 %
Extract
EG : 44 %
Furfural : 46 %
Water : 10 %
Raffinate
EG : 12 %
Furfural : 8 %
Water : 80 %