Solution Thermodynamic: Vapor/Liquid Equilibrium (VLE) PTT 201/4 THERMODYNAMICS SEM 1 (2013/2014)

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Solution Thermodynamic: Vapor/Liquid Equilibrium (VLE) PTT 201/4 THERMODYNAMICS SEM 1 (2013/2014)

Transcript of Solution Thermodynamic: Vapor/Liquid Equilibrium (VLE) PTT 201/4 THERMODYNAMICS SEM 1 (2013/2014)

Page 1: Solution Thermodynamic: Vapor/Liquid Equilibrium (VLE) PTT 201/4 THERMODYNAMICS SEM 1 (2013/2014)

Solution Thermodynamic:

Vapor/Liquid Equilibrium (VLE)

PTT 201/4 THERMODYNAMICSSEM 1 (2013/2014)

Page 2: Solution Thermodynamic: Vapor/Liquid Equilibrium (VLE) PTT 201/4 THERMODYNAMICS SEM 1 (2013/2014)

Nature of Equilibrium– Definition– Measures of composition

VLE : Qualitative behaviorSimple Models for VLE - Raoult’s Law - Dewpoint & Bubblepoint Calculations with Raoult’s Law - Henry’s LawVLE by modified Raoult’s lawVLE from K-value correlations

Chapter Outline (Smith)

Page 3: Solution Thermodynamic: Vapor/Liquid Equilibrium (VLE) PTT 201/4 THERMODYNAMICS SEM 1 (2013/2014)

THE NATURE OF EQUILIBRIUM

Equilibrium : A static condition in which no changes occur in the macroscopic properties of a system with time.

The T, P, composition reaches final value which will remain fixed: equilibrium

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m

m

m

mx iii

V

xC ii

iiiMxM

Measures of composition

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VLE: State of coexistence of L & V phases A condition where a liquid phase and vapor phase

are in equilibrium with each other At this condition: rate of evaporation (liquid → vapor) = rate of condensation (vapor → liquid)

VLE: QUALITATIVE BEHAVIOR

Binary mixture: Mixture that contains two constituents e.g: mixture of liquid and vapor at an equilibrium level takes place when liquid and vapor are allowed to contact to each other in a closed location

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• Under surface- sat. V states (P-T-y1)

• Upper surface- sat. L states (P-T-x1)

• Liquid at F, reduces pressure at constant T & composition along FG, the first bubble appear at L – bubble point: a point when a liquid forms the first bubble of vapor and begins to evaporate

• As pressure reduces, more & more L vaporizes until completed at W; point where last drop of L (dew) disappear – dew point: a point when a vapor forms the first droplet of liquid and begins to condense

Fig. 10.1 – Shows the P-T-composition surfaces of equilibrium states of

saturated V & saturated L of a binary system

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SIMPLE MODELS FOR VLE

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Raoult’s Law

• V phase is an ideal gas– Applicable for low to moderate

pressure• L phase is an ideal solution

– Valid only if the species are chemically similar (size, same chemical nature e.g. isomers such as ortho-, meta- & para-xylene)

Assumptions;

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NiPxPy satiii ,...,2,1

Where;

pressure Total :

species pure of pressureVapor :

fraction mole phase:

fraction mole phase:

P

iP

Vy

Lx

sati

i

i

1

Page 12: Solution Thermodynamic: Vapor/Liquid Equilibrium (VLE) PTT 201/4 THERMODYNAMICS SEM 1 (2013/2014)

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

Dewpoint & Bubblepoint Calculations with Raoult’s Law

FIND GIVEN

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For binary systems to solve for bubblepoint calculation (T is given);

1i iy

i

satiiPxP 1212 xPPPP satsatsat

PPx

ysat

11

1

2

3

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i

satii Py

P1

Raoult’s law equation can be solved for xi to solve for dewpoint calculation (T is given) 1i i

x

satsat PyPyP

2211//

1

satPPy

x1

1

1

4

5

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Example 1

Binary system acetonitrile(1)/nitromethane(2) conforms closely to Raoult’s law. Vapor pressure for the pure species are given by the following Antoine equations:

00.209

64.972,22043.14ln

00.224

47.945,22724.14ln

02

01

CtkPaP

CtkPaP

sat

sat

a)Prepare a graph showing P vs. x1 and P vs. y1 at temperature 750C

b)Prepare a graph showing t vs. x1 and t vs. y1 for a pressure of 70 kPa

i

ii

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a) BUBL P calculations are required. Since this is a binary system, Eq. 2 may be used.

)(1212 AxPPPP satsatsat

At 750C, the saturated pressure is given by Antoine equation;

98.4121.83 21 satsat PP

Substitute both values in (A) to find P;

kPaP

P

72.66

6.098.4121.8398.41

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The corresponding value of y1 is found from Eq. 1, sat

iii PxPy

x1 y1 P/kPa

0.0 0.0000 41.98

0.2 0.3313 50.23

0.4 0.5692 58.47

x1 y1 P/kPa

0.6 0.7483 66.72

0.8 0.8880 74.96

1.0 1.0000 83.21

7483.0

72.66

21.836.0111

P

Pxy

sat

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At point c, the vapor composition is y1=0.6, but the composition of liquid at c’ and the pressure must read from graph or calculated. Thus DEW P calculations are required. By using Eq. 3;

satsat PyPyP

2211

1

For y1=0.6 and t=750C

kPaP 74.5998.414.021.836.0

1

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And by Eq. 1,

4308.0

21.83

74.596.0

1

11

satP

Pyx

This is the liquid-phase composition at point c’

b) When P is fixed, the T varies along T1sat and

T2sat, with x1 & y1. T1sat & T2sat are calculated

from Antoine equation;

ii

isati C

PA

Bt

ln

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For P=70kPa, T1sat=69.840C, T2sat=89.580C. Select T between these two temperatures and calculate P1sat &

P2sat for the two temperatures.

Evaluate x1 by Eq. (A). For example;

satsat

sat

PP

PPx

21

21

5156.0

84.4676.91

84.46701

x

Get y1 from Eq. 1

6759.0

70

76.915156.0111

P

Pxy

sat

e.g; select T= 78˚C

Substituting T= 78˚C into (i) and (ii)

P1sat = 91.76 kPa

P2sat = 46.84 kPa

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Summary;

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)

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1. For pressure low It is so low that it can be assume as ideal gas

2. For species present as a very dilute solution in liquid phase

Assumptions;

Henry’s Law

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NiHxPy iii ,...,2,1

Where;

pressure Total :

constant sHenry' :

fraction mole phase:

fraction mole phase:

P

H

Vy

Lx

i

i

i

Henry’s Law

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Example 2

Assuming that carbonated water contains only CO2(1) and H2O(2), determine the compositions of the V & L phases in a sealed can of ‘soda’ & the P exerted on the can at 100C. Henry’s constant for CO2 in water at 100C is about 990 bar and x1=0.01.

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Henry’s law for species 1 & Raoult’s law for species 2 are written;

111 HxPy satPxPy 222

With H1=990 bar & P2sat = 0.01227 bar (from steam tables at 100C)

barP

P

912.9

01227.099.099001.0

satPxHxP 2211

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Then by Raoult’s law, Eq. 1 written for species 2;

0012.0

912.9

01227.099.0222

P

Pxy

sat

Whence y1=1-y2=0.9988, and the vapor phase is nearly pure CO2, as expected.

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The 2nd assumption of Raoult’s Law is abandoned, taking into account the deviation from solution

ideality in L phase.

Thus, activity coefficient is introduced in Raoult’s Law

NiPxPy satiiii ,...,2,1

VLE BY MODIFIED RAOULT’S LAW

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Activity coefficients are function of T & liquid phase composition, x

1i iy

i

satiii PxP

i

satiii Py

P

1

For bubble point

For dew point

Since;

1i ix

8

9

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AZEOTROPE

A mixture that has a constant composition of liquid and vapor phase

When x1=y1, the dew point and bubble point curves are tangent to the same horizontal line

A boiling L of this composition produce a vapor exactly the same composition; L does not change in composition as it evaporates

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VLE FROM K-VALUE CORRELATTIONS

The partition between liquid and vapor phases of a chemical species is equilibrium ratio, Ki.

i

ii x

yK

This quantity is called K-value.

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satiii PxPy K-value for Raoult’s Law

P

PK

sati

i

K-value for modified Raoult’s Lawsatiiii PxPy

P

PK

satii

i

11

12

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Hence,

For binary systems to solve for bubble point calculation;

1i iy

1 ii ixK

For binary systems to solve for dew point calculation;

1i ix

Hence, 1ii

i

K

y

13

14

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K-value from DePriester chart-Low T range

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K-value from DePriester chart-High T range

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When given a mixture of composition at certain T or P;

Bubble point

- System is almost vaporized

- The given mole fraction is yi

- Need to satisfy equation 13

- Composition of dew is xi=yi/Ki

Dew point

- System is almost condensed

- The given mole fraction is xi

- Need to satisfy equation 14

- Composition of bubble is yi=Kixi

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The End