F r o m t h e L a s t C la las s
Vapor Pressure Equations
Cox chart & Duhring Duhring plots: log p* vs. T
Antoine
equation: (Table equation: (Table B.4) ☺☺ log log10 P* A
B TC
Wagner equation
“Properties of Gases and Liquids” ln P* (A B1.5 C3 D6 ) / Tr 1 Tr
Note : the vapor pressure equations should be used within the specified temperature range !
Vapor Pressure Equations
Cox chart & Duhring Duhring plots: log p* vs. T
Antoine
equation: (Table equation: (Table B.4) ☺☺ log log10 P* A
B TC
Wagner equation
“Properties of Gases and Liquids” ln P* (A B1.5 C3 D6 ) / Tr 1 Tr
Note : the vapor pressure equations should be used within the specified temperature range !
6.2 Gibbs Phase Rule
Types of Process Variables
Extensive Variables
Intensive Variables
depend on the size of the system (mass and volume) 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. No reaction
F 2 m The number of phases The number of chemical species The number of degree of freedom
Raoult’s Law for GLE: Single Condensable Component
Water evaporation into dry air
A+B
Saturation (GLE) B
Gibbs Phase Rule
GLE
F 2 m 2 2 2 2 .
Raoult’s Law for GLE ☺☺☺ Partial pressure of vapor in the gas = Pure-component vapor pressure at the system temperature
pi yi P pi* (T )
P&Ty P&yP T&yP
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. * p H 2O y H 2O P p H 2O (T dp )
p H 2O 0.1 5260 526 mmHg 증기압이 526mmHg인 온도를 찾음.
Antoine equation
T dp
log10 p* A
B A log10 p*
C
B
Table B-4
T C
1668.21 7.96681 log10 526
degree of superheat = 100 -90 = 10 oC
o
228 90 C
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
Q2 mol y H2O 1-y BDA
0.1 H2O 0.9 BDA
Q1 mol H2O Number of component = 2 Number of unknown = (Q1, Q2, y) = 3
BDA : Bone-Dry Air
D.O.F. = 1: cannot be solved
Basis : 100 mol feed gas
100 mol feed
Q2 mol y H2O 1-y BDA
0.1 H2O 0.9 BDA T = 100oC, P = 5260 mmHg
T = 80oC, P = 5260 mmHg Q1 mol H2O
Use thermodynamic information: A gas in equilibrium with liquid must be saturated with the liquid . thus, y is a saturated condition at
80 oC,
5260 mmHg
* o y H 2O p H 2O (80 C ) / P 355 / 5260 0.0675 y BDA 1 y H 2O 0.9325
Number of component = 2 Number of unknown = (Q1, Q2) = 2
D.O.F. = 0
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 %
pi yi P pi* (T )
Example 6.3.-2: A stream of air at 100oC contains 5260 mmHg contains 10% water by volume. ☺☺☺ (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
Q2 mol y H2O 1-y BDA
0.1 H2O 0.9 BDA
Q1 mol H2O Number of component = 2 Number of unknown = (Q1, Q2, y) = 3
BDA : Bone Dr y Air
Cannot be solved !
Basis : 100 mol feed gas
100 mol feed
Q2 mol y H2O 1-y BDA
0.1 H2O 0.9 BDA T = 100oC, P = 5260 mmHg
T = 100 oC, P = 8500 mmHg Q1 mol H2O
Use thermodynamic information : A gas in equilibrium with liquid must be saturated with the liquid . thus, y is a saturated condition at * H 2 O
y H 2O p
100oC,
8500 mmHg
o
(100 C ) / P 760 / 8500 0.0894
1 y H 2O 0.9106 Number of component = 2 Number of unknown = (Q1, Q2) = 2
D.O.F. = 0
Material Balance Q2=98.8 mol Q1=1.2 mol
BDA Balance
100×0.9=Q2×0.9106
Total Balance
100=Q1+Q2
% Condensed
1.2/(100×0.1) ×(100) = 12 %
pi yi P pi* (T )
Let’s Start!
6.4 Multicomponent Vapor-Liquid Equilibria
Gas-Liquid Processes
VLE information
Chemical reactions Distillation (증류) Gas Liquid : Absorption ( 흡수) Liquid Gas : Stripping ( 탈기)
From literature, databases Raoult’s Law & Henry’s Law ☺☺☺ Rigorous calculation using model equations
Distribution of components between vapor and liquid phases P h a s e -E q u i l i b r i u m Th e r m o d y n a m i c s
Raoult’s Law and Henry’s Law
Raoult’s Law pi yi P xi pi* (T )
Vapor (P and yi)
Ideal gas
Liquid (xi)
Ideal Solution
pi* = vapor pressure
Valid for almost pure liquid (xi 1)
Valid for mixture of similar substances (over entire range of compositions: 0 < xi < 1)
Henry’s Law yi P xi H i (T )
Hi = Henry’s law constant
Valid for dilute solution (x i0)
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 20 oC and 20 atm. Estimate the mole fraction of dissolved ethane.
N2, O2, CO2, …. CH4, C2H6, ….
From Perry’s Handbook
Dilute solution
Apply Henry’s Law
H C 2 H 6 (20o C ) 2.63104 atm/mole fraction yi P xi H i (T ) xi yi P / H i (T )
0.01 20 2.63 104
7.6 10
6
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 30 oC. What is the system pressure and the composition of the vapor?
Benzene + Toluene
Similar Substances
Apply Raoult’s Law
pi yi P xi pi* (T )
Table B.4
p B* (30o C ) 119 mmHg
P pi
Antoine Eq’n
o * pT (30 C ) 36.7 mmHg
p B x B p B* 0.5 119 59.5 mmHg * pT xT pT 0.5 36.7 18.35 mmHg
y B x B p B* / P 59.5 / 77.9 0.764 * yT xT pT / P 18.35 / 77.9 0.236
P 59.5 mmHg 18.35 mmHg 77.9 mmHg
Phase diagrams for binary VLE
Txy diagram
Pxy diagram
(at a fixed P)
T
(at a fixed T)
P
vapor
liquid
Bubble P
Dew T
V+L
Bubble T
Dew P
V+L
liquid vapor
x1
x or y
y1
x1
y1
x or y
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
Bubble P Pressure
Liquid
P
Vapor
y i
x i
Given T,x Calculate P,y
Composition
Dew P Pressure
Liquid
P
Vapor
y i
x i
Given T,y
Calculate P,x
Composition
Bubble T Temperature
Vapor
T
Liquid
x i
y i
Given P,x
Calculate T,y
Composition
Dew T Temperature Vapor
T
Liquid
x i
y i
Given T,x
Calculate P,y
Pressure
VLE Calculations
Bubble Point Temperature Calculation
Given P, x Calculate T,y yi
xi pi* (T bp ) P
* * P xa pa (T bp ) xb pb (T bp ) ...
Dew Point Temperature Calculation
Given P,y
Calculate T,x xi
yi P *
pi (T dp )
xi i
i
yi P pi* (T dp )
1
VLE Calculation
Iterative calculation required
Not explicit form
Iterative calculation techniques
Trial and error method
Newton-Raphson Method
Secant Method OBJ ( X ) 0
find X that satisfies given relation
Algorithm for Bubble/Dew Point Calculations Example : Bubble T Calculation Objective Function Start
x
i
1
y
1
Given P,x
i i
i
OBJ xi yi 0 i
Assume T
yi xi pi* (T ) / P
i
Calculate OBJ
i
Phase Equilibrium Calculate new T,y
yi P xi pi* (T )
K i yi / xi pi* (T ) / P
OBJ xi yi
|DT|
End
i
Newton Raphson Secant iteration …
VLE Calculations for Nonideal Systems
Phase equilibrium relations
Ideal Gas + Ideal Solution yi P xi pi* (T )
Nonideal Gas + Nonideal Solution
Fugacity coefficient : gas phase nonideality
Activity
coefficient : liquid phase nonideality i yi P i xi pi* (T )
from activity models : WILSON, NRTL, UNIQUAC, ,…. from equation of state models : SRK, PR ,….
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?
100oC 95oC
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?
102oC 96oC
x=0.2
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) o
222
g AgNO 3 / 100 g H2O at 20
952
g AgNO 3 / 100 g H2O at 100
C oC
Crystallization
Separation of solids and liquids
Driving force = solubility differences
A
solute in equilibrium with a crystal must be saturated
Example 6.5-1
150 kg of a saturated aqueous solution of AgNO 3 at 100oC is cooled to 20oC, thereby forming AgNO 3 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 AgNO 3 in the feed stream eventually recovered as dry crystals and the amount of water that must be removed in the drying stage. 150 kg
Cooler Crystallizer
Saturated Solution
Fiter
Filter Cake + Solution
Evaporator
Filter Cake
Water
Basis: 150 kg Feed
150 kg 0.905 AgNO3 0.095 H2O
Q1 kg
Cooler Crystallizer
Fiter
Q2 kg filter cake
Q3 kg solution
0.689 AgNO3 0.311 H2O
0.689 AgNO3 0.311 H2O Q4 kg H2O Evaporator
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
Additional Information Q2 = 0.8 (Q2 + Q3)
Q2 = 4Q3
Unknown = 2 150 kg
Q1 kg
Cooler Crystallizer
0.905 AgNO3 0.095 H2O
Fiter
4Q3 kg filter cake
Q3 kg solution
0.689 AgNO3 0.311 H2O
0.689 AgNO3 0.311 H2O Q4 kg H2O Evaporator
Q5 kg filter cake Unknown = 3
Unknown = 3
150 kg 0.905 AgNO3 0.095 H2O
Q1 kg
Cooler Crystallizer
Fiter
4Q3 kg filter cake
Q3 kg solution
0.689 AgNO3 0.311 H2O
0.689 AgNO3 0.311 H2O Q4 kg H2O Evaporator
Q5 kg filter cake 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
150 kg
Cooler Crystallizer
0.905 AgNO3 0.095 H2O
20 kg Fiter
104 kg filter cake
26 kg solution
0.689 AgNO3 0.311 H2O
0.689 AgNO3 0.311 H2O Q4 kg H2O Evaporator
Q5 kg filter cake Water balance 26 * 0.311 = Q4 Total Balance 104 + 26 = Q4 + Q5
Recovery % = 122 / (150*0.905) * 100 % = 89.9 %
Q4 = 8 kg Q5 = 122 kg
Hydrated Salts
Several structures can be produced for watersalt 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
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
Colligative Solution Properties
Vapor pressure lowering p s (T ) (1 x) p s* (T )
( p s* ) e p s (1 x) p s* (T ) *
*
*
*
D p x p s ( p s ) e xp s
Boiling point elevation 2
DT b
RT b0 D H v
x
ˆ
Melting point depression 2
DT m
RT m0 D H m ˆ
x
6.6 Immiscible and Partially Miscible Liquids
Terminology
Immiscibility ( 불혼화성)
Partial miscibility ( 부분혼화성)
Liquid extraction ( 액체 추출)
Distribution coefficient
Water Phase
Water-Rich Phase
Chloroform Phase
MIBK-Rich Phase
immiscible
partially miscible
K
( x) MIBK ( x)WATER
Distribution of Acetone
Phase Diagram for Partially Miscible Ternary Systems
Equilateral ternary LLE phase diagram
Phase Diagram for LLE Systems Single Phase Region Two liquids have identical compositions
Two liquid Phase Region
Tie Lines
Miscibility limit for water-furfural
