Chemical Engineering Thermodynamics: Vapor/liquid Equilibrium

Chemical Engineering Thermodynamics Chapter 8

VAPOR/LIQUID EQUILIBRIUM Sumber: Http://portal.unimap.edu.my/.../CHAPTER%204%20VAPORLIQU..

• So far we have only dealt with pure substances and constant composition mixtures. • We will move a step further where the desired outcome is the composition change. • In system such as distillation & absorption, if the system is not in equilibrium, the mass transfer between system will alter their composition.

Chapter Outline (Smith) 10.1 Nature of Equilibrium – Definition – Measures of composition

10.2 The Phase Rule – Duhem’s Theorem

10.3 VLE : Qualitative behavior 10.4 Simple Models for VLE

- Raoult’s Law -Dewpoint & Bubblepoint Calculations with Raoult’s Law - Henry’s Law

10.1 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

Measures of composition Measures of composition

Mass or mole fraction

Molar concentration

Molar mass for a mixture or solution

10.2 PHASE RULE & DUHEM’S THEORY Equilibrium states are determined by; –Phase Rule –Duhem’s Theory

The Phase Rule Number of variables that may be independently fixed in a system at equilibrium = Difference between total number of variables that characterize the intensive state of the system and number of independent equation

F = 2-π+N Where : F – degrees of freedom π – No of phase N – No of species

Duhem’s Theory For any closed system formed initially from given masses of prescribed chemical species, the equilibrium state is completely determined when any two (2) independent variables are fixed

10.3 VLE: QUALITATIVE BEHAVIOR VLE: State of coexistence of L & V phases

Fig. 10.1 – Shows the P-Tcomposition surfaces of equilibrium states of saturated V & saturated L of a binary system

• 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 • As pressure reduces, more & more L vaporizes until completed at W; point where last drop of L (dew) disappear – dew point

10.4 SIMPLE MODELS FOR VLE Simple Models For VLE :

Find T, P, composition

Raoult’s Law

Henry’s Law

Raoult’s Law Assumptions; • 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- & paraxylene)

Where;

Dewpoint & Bubblepoint Calculations with Raoult’s Law FIND

GIVEN

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

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

y i

P   xi Pi i

sat

PP

sat 2

i

1



 P

x1 P1 sat y1  P

sat 1

P

sat 2

x

1

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

i xi

P

1

sat y P  i i i

1

1 P sat sat y1 / P1  y2 / P2

y1 P x1  sat P1

Example 10.1 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: sat 1

ln P

sat 2

ln P

2,945.47 kPa  14.2724  0 t C  244.00 2,972.64 kPa  14.2043  0 t C  209.00

• Prepare a graph showing P vs. x1 and P vs. y1 at temperature 750C • Prepare a graph showing t vs. x1 and t vs. y1 for a pressure of 70 kPa

a) BUBL P calculations are required. Since this is a binary system, Eq. 10.2 may be used.

PP

sat 2



 P

sat 1

P

sat 2

x

( A)

1

At 750C, the saturated pressure is given by Antoine equation; sat 1

P

 83.21 P

sat 2

 41.98

Substitute both values in (A) to find P;

P  41.98  83.21  41.980.6  P  66.72kPa

The corresponding value of y1 is found from Eq. 10.1.

yi P  xi Pi

sat

x1 P1sat 0.6 83.21 y1    0.7483 66.72 P x1

y1

P/kPa

x1

y1

P/kPa

0.0

0.0000

41.98

0.6

0.7483

66.72

0.2

0.3313

50.23

0.8

0.8880

74.96

0.4

0.5692

58.47

1.0

1.0000

83.21

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. This is DEW P, by Eq. 10.3;

P

sat 1

y1 P

1 sat  y2 P2

For y1=0.6 and t=750C

1 P  59.74kPa 0.6 83.21  0.4 41.98

And by Eq. 10.1,

y1 P 0.6 59.74  x1  sat   0.4308 P1 83.21

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;

t

sat i

Bi   Ci Ai  ln P

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;

P  P2sat x1  sat sat P1  P2

70  46.84 x1   0.5156 91.76  46.84

Get y1 from Eq. 10.1

x1 P1sat 0.5156 91.76  y1    0.6759 P 70

Summary; x1

y1

P/kPa

0.0000

0.0000

0.1424

0.2401

89.58 (t2sat) 86

0.3184

0.4742

82

0.5156

0.6759

78

0.7378

0.8484

74

1.0000

1.0000

69.84 (t1sat)

For x1=0.6 & P=70kPa, T is determined by BUBL T calculation, which requires iteration. Eq. 10.2 is rewritten; sat 2

P

P  x1  x2

sat P ( B) Where;   1sat P2

Subtracting lnP2sat from lnP1sat as given by Antoine equations yields;

2,945.47 2,972.64 ln   0.0681   t  224.00 t  209.00

(C )

Initial value for α is from arbitrary intermediate t

With α, calculate P2sat by Eq. (B)

Calculate T from Antoine eq. for species 2

Find new α by Eq. (C)

Return to initial step and iterate until converge for final value of T

The result is t=76.420C. From Antoine eq., P1sat=87.17kPa and by (10.1), the composition at b’ is; sat 1 1

xP y1  P

 0.6 87.17    0.7472 70

Vapor composition at point c is y=0.6. P is known (p=70kPa), a DEW T calculation is possible. The steps are the same as BUBL T, but it is based on P1sat, rather than P2sat. The result is t=79.580C. From Antoine eq., P1sat=96.53kPa and by (10.1), the composition at c’ is;

y1 P 0.6 70  x1  sat   0.4351 P1 96.53 This shows that the temperature rises from 76.420C to 79.580C during vaporization step from point b to c. Continued heating simply superheats the vapor to point d.

Henry’s Law Assumptions; • For pressure low

It is so low that it can be assume as ideal gas

• For species present as a very dilute solution in liquid phase

Henry’s Law

Where;

Example 10.2 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.

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

y1 P  x1 H1

y2 P  x P

sat 2 2

P  x1 H1  x P

sat 2 2

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

P  0.01990   0.99 0.01227  P  9.912 bar

Then by Raoult’s law, Eq. 10.1 written for species 2;

x2 P2sat 0.99 0.01227  y2    0.0012 P 9.912 Whence y1=1-y2=0.9988, and the vapor phase is nearly pure CO2, as expected.

Review • What is bubble point? • What is dew point? • We have previously went through the 2 simplest models for solving VLE problems – Raoult’s Law – Henry’s Law

Chapter Outline 10.5 VLE by modified Raoult’s law 10.6 VLE from K-value correlations - Flash calculation

VLE

Raoult’s Law

Henry’s Law

Modified Raoult’s Law

K-Values

10.5 VLE BY MODIFIED RAOULT’S LAW 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

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

P   xi i Pi

y i

sat

i

1 For bubble point

i

i xi P

y

1 i

 i Pi

1 For dew point

sat

i

(See Example 10.3)

BUBL P

BUBL P CALCULATION

Find Pi sat Find  i

Find P from equation 10.6 P  x1 1P1sat  x2 2 P2sat

Find yi from equation 10.5 yi  xi i Pi sat P

DEW P CALCULATION

DEW P

Find Pi sat Find  1 &  2 . Initial guess its=1 Find P using equation 10.7 1 P sat y1  1 P1  y2  2 P2sat

Find xi using equation 10.5 xi  yi P  i Pi sat

Evaluate from the given equation

NO

Converge?

YES

It is the P dew. Find liquid phase mole fraction yP x1  1 sat & x2  1  x1  1 P1

BUBL T CALCULATION

BUBL T

Find Ti sat

Find initial T from mole-fraction weighted average T  x1T1sat  x2T2sat

For current T, find A,  1 ,  2 ,

P1 sat , P2sat

  P1sat P2sat

Find new value for P1sat from equation 10.6; P P1sat  x1 1  x2 2  

Find new T from Antoine equation for species 1 B1 T  C1 A1  ln P1sat

NO

Converge?

YES

It is the T bubble. Find Pi sat , A and  1 &  2

Find vapor phase mole fraction y1  x1 1 P1sat P & y 2  1  y1

DEW T CALCULATION

DEW T

Find Ti sat

Find initial T from mole-fraction weighted average T  y1T1sat  y 2T2sat

For current T, find A, P1sat , P2sat   P1sat P2sat

Find x1  y1 P  1 P1sat & x2  1  x1

Calculate  1 &  2 from given correlation

Find new value for P1sat from equation 10.7;

y y  P1sat  P 1  2    1  2 

Find new T from Antoine equation for species 1 B1 T  C1 A1  ln P1sat

NO

Converge? It is the T bubble. Find Pi sat , A and  1 &  2 YES Find vapor phase mole fraction y1  x1 1 P1sat P & y 2  1  y1

AZEOTROPE 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

Relative volatility;

y1 x1 12  y 2 x2

12 x 0  1

sat 1

P

exp A sat P2 sat 1

P 12 x1 1  sat P2 exp A

(10.8)

If one limit is <1 & the other limit is <1; azeotrope exists.

10.6 VLE FROM K-VALUE CORRELATTIONS The partition between liquid and vapor phases of a chemical species is equilibrium ratio, Ki.

yi Ki  xi

This quantity is called K-value.

K-value for Raoult’s Law

yi P  xi Pi

sat

sat

Pi Ki  P

K-value for modified Raoult’s Law

Ki 

 i Pi

sat

P

yi P  xi i Pi

sat

For binary systems to solve for bubble point calculation;

y i

Hence,

i

Kx

i i

i

1 1

For binary systems to solve for dew point calculation;

x i

Hence,

i

1

yi i K  1 i

K-value from DePriest er chart -Low T range

K-value from DePriest er chart -High T range

When given a mixture of composition at certain T or P; Bubble point - Insignificant L • The given mole fraction is yi • Need to satisfy equation 10.14 - Composition of dew is xi=yi/Ki

Dew point • System is almost condensed • The given mole fraction is xi

• Need to satisfy equation 10.13 - Composition of buble is yi=Kixi

Flash Calculation The most important application of VLE. Originates from a fact that a liquid at a pressure equal to or greater that its bubble point pressure ‘flashes’ or evaporates when the pressure is reduced, producing a two-phase system of vapor and liquid in equilibrium.

FLASH CALCULATION Vapor, V

Feed, F

Liquid at P < Pbubble partially evaporates when P is reduced, producing 2-phase system of V & L in equilibrium

V L Liquid, L

Find; T, P, z

In a system with one mole chemical species with an overall composition by set of mole fraction, zi. Li would be the moles of liquid with mol fraction xi and V be the moles of vapor with the mol fraction of yi: z

L V  1 zi  xi L  yiV

i  1,2,.....N 

From

zi  xi L  yiV

Eliminate for L gives:

zi  xi 1  V   yiV From K-value

yi Ki  xi

i  1,2,.....N  yi xi  Ki

Hence solving for yi,

zi K i yi  1  V K i  1

i  1,2,.....N 

Because Hence,

y i

i

1

zi K i  yi  1  V K  1  1 i

(See Example 10.5 and 10.6)

Flowchart for flash pressure

FLASH CALCULATION Find BUBL P with zi  xi ; Pbubble Find DEW P with zi  yi ; Pdew

Is the given P between Pbubble & Pdew ?

NO

No need for flash calculation YES

Using equation 10.11, find K i

Substitute K i in equation 10.17. By trial & error, solve for V. Then L=1-V

Solve equation 10.16 for each component - yi

Solve equation 10.10 for each component - xi

The End