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CPE553 CHEMICAL ENGINEERING THERMODYNAMICS 3/5/2013

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VAPOR LIQUID EQUILIBRIUM

Vapor liquid equilibrium is a state where vapor and liquid phase are in

equilibrium with each other.

Many processes in chemical engineering do not only involve a single phase

but a stream containing both gas and liquid. It is very important to

recognize and be able to calculate the temperature, pressure and

composition of each phase at equilibrium.

VLE information is useful in separation processes, e.g. distillation, evaporation, liquid-liquid extraction, adsorption, etc.

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THE NATURE OF EQUILIBRIUM

THE PHASE RULE: DUHEM’S THEOREM

VLE: QUALITATIVE BEHAVIOR

SIMPLE MODELS FOR VAPOR LIQUID EQUILIBRUM: RAOULT’S LAW

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Equilibrium is a condition where no changes occur in the properties of a

system with time.

The temperature, pressure and phase compositions reach final values

which thereafter remain fixed.

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Mass or mole fraction is defined as the ratio of mass or number of

moles of a particular chemical species in a mixture or solution to the total

mass or number of moles of mixture or solution.

Molar concentration is defined as the ratio of mole fraction of a

particular chemical species in a mixture or solution to molar volume of

the mixture or solution.

or multiplying and dividing by molar flow rate,

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n

i ii

m mx

m m i i

i

n nx

n n

ii

xC

V

i ii

x n nC

V n q

The molar mass of a mixture or solution is the mole fraction-weighted

sum of the molar masses of all species.

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i ii

M x M

Phase rule:

F = no. of variables that may be independently fixed in a system at

equilibrium (degree of freedom).

= no. of phases

N = no. of chemical species

Duhem’s theorem:

The two independent variables subject to specification may in general be

either intensive or extensive. However, the number of independent intensive

variables is given by the phase rule.

Thus, when F = 1, at least one of the two variables must be extensive, and

when F = 0, both must be extensive.

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2F N

For any closed system, the equilibrium state is completely

determined when any two independent variables are fixed.

For a two phases (=2) system of a single component (N=1):

F = 2- + N

F = 2- 2 + 1 = 1

Therefore, one intensive variable must be specified to fix the state of the

system at equilibrium.

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VLE for Pure Components

0

200

400

600

800

270 320 370 420Temperature: K

Pre

ss

ure

: k

Pa

Acetonitrile Nitromethane


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For a two phases (=2), binary system (N=2):

F = 2- 2 + 2 = 2

Therefore, two intensive variables must be specified to fix the state of the

system at equilibrium.

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Figure 10.1 shows the three-dimensional

P-T-composition surfaces which contain

the equilibrium states of saturated vapor

and saturated liquid for species 1 and 2

of a binary system.

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Further explanation on

page 341- 342.

• Line AEDBLA in Fig. 10.1 can be represented by Fig 10.2(a) P-x1-y1 diagram at Ta.

• The horizontal lines are tie lines connecting the compositions of phase in equilibrium.

• The temperatures Tb and Td lie between the two pure species critical temperature identified by C1 and C2 in Fig. 10.1.

• Line KJIHLK in Fig. 10.1 can be represented by Fig. 10.2(b) T-x1-y1 diagram at Pa.

• Pressure Pb lies between the critical pressures of the two pure species at points C1 and C2 in Fig. 10.1.

• Pressure Pd is above the critical pressures of both pure species, therefore the T-x1-y1 diagram appears as an island.

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• Vertical plane and perpendicular to the composition axis, passes through points

SLMN and Q is shown as Fig. 10.3 PT diagram.

• Each interior loop represents the P-T behavior of saturated liquid and of saturated

vapor for a mixture of fixed composition. Different loops are for different

compositions.

Further explanation on page 343.


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• Fig. 10.4 shows the enlarged nose section of a single P-T loop.

Further explanation on page 344.

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Further explanation on page 344-345.

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Further explanation on page 345-346.

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Figure 10.8 shows P-x-y

diagrams at constant T for

four systems (much lower

temperature and pressure).

Further explanation on

page 345-347.

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Figure 10.9 shows t-x-y

behavior for four systems at

low pressure – 1 atm).

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Figure 10.10: The y1-x1 diagrams at constant P for four systems.

The point at which a curve crosses the diagonal line of the diagram

represents an azeotrope, for such a point x1 = y1 (for (b) and (d)).

VLE calculation provides information on temperatures, pressures, and compositions of phases in equilibrium.

Two simplest models, Raoult’s law and Henry’s law are used to predict the behavior of systems in vapor liquid equilibrium.

Raoult’s Law

Assumption ◦ Vapor phase is an ideal gas ◦ Liquid phase is an ideal solution

Mathematical expression reflecting the two assumptions above is expressed quantitatively in Raoult’s law as:

where xi = mole fraction of liquid phase

yi = mole fraction of vapor phase

Pisat = vapor pressure of pure species i at the temperature of the system

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yiP = xiPisat (i = 1,2, ..., N) (10.1)


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Bubble point is the point at which the first drop of a liquid mixture

begins to evaporate (the first bubble of vapor appears).

Dew point is the point at which the first drop of a gas mixture begins to

condense (the last drops of liquid disappear).

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BUBL P : Calculate {yi} and P, given {xi} and T

BUBL T : Calculate {yi} and T, given {xi} and P

DEW P : Calculate {xi} and P, given {yi} and T

DEW T : Calculate {xi} and T, given {yi} and P

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Application of Raoult’s law:

Because iyi = 1, eq. (10.1) may be summed over all species to yield

This equation applied in bubblepoint calculations, where the vapor phase

compositions are unknown.

For a binary system with x2 = 1-x1,

A plot of P vs. x1 at constant temperature is a straight line connecting P2sat at

x1 = 0 with P1sat at x1 = 1.

Equation (10.1) may also be solved for xi and summed over all species. With

ixi = 1, this yields

This equation applied in dewpoint calculations, where the liquid phase

compositions are unknown.

sat

i ii

P x P

1sat

i ii

Py P

2 1 2 1

sat sat satP P P P x

(10.2)

(10.3)

Antoine’s equation is used to calculate Pisat :

*Refer Table B.2 for values of Antoine parameters (A, B and C)

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iln P sat i

i

i

BkPa A

T K C

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Binary system acetonitrile(1)/nitromethane(2) conforms closely to Raoult’s

law. Vapor pressures for the pure species are given by the following Antoine

equations:

(a) Prepare a graph showing P vs x1 and P vs y1 for a temperature of

348.15K

(b) Prepare a graph showing T vs x1 and T vs y1 for a pressure of 70kPa

1

2

2945.47ln P 14.2724

49.15

2972.64ln P 14.2043

64.15

sat

sat

T

T


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Solution:

(a) Prepare P-x1-y1 diagram ( T is given 75oC (384.15K), x1 and y1 in the range 01)

BUBL P calculation

Calculate P1sat and P2

sat from Antoine equations

At 384.15K,

P1sat = 83.21 kPa and P2

sat = 41.98 kPa

Calculate P from equation (10.2):

E.g. At x1 = 0.6, P = 66.72 kPa

Calculate y1 from equation (10.1):

*At 75oC (384.15 K) a liquid mixture of 60 mole % acetonitrile and 40 mole % nitromethane is in equilibrium with a vapor containing 74.83 mole % acetonitrile at a pressure of 66.72 kPa.

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1 11 0.7483

satx Py

P

2 1 2 1

sat sat satP P P P x

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Similar calculation of P and y1 at various x1 (01).

The results of calculations:

x1 y1 P

0 0 41.98

0.1 0.1805 46.10

0.2 0.3313 50.23

0.3 0.4593 54.35

0.4 0.5692 58.47

0.5 0.6647 62.60

0.6 0.7483 66.72

0.7 0.8222 70.84

0.8 0.8880 74.96

0.9 0.9469 79.09

1 1 83.21

At 75oC and x1 = 0.45,

P = ?, y1 = ? BUBL P

At 75oC and y1 = 0.6,

P = ?, x1 = ? DEW P Either can be read from the graph or calculated.

P-xy diagram for acetonitrile(1)/nitromethane(2) at 75oC

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30

40

50

60

70

80

90

100

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x1, y1

P/k

Pa

x1 y1

P2sat=41.98

P1sat=83.21

Bubble point

Dew point

Superheated vapor

Subcooled liquid

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DEW P calculation

Calculate P from equation (10.3)

For y1 = 0.6 and T = 75oC (348.15K),

Calculate x1 by equation (10.1),

1 1 2 2

1satsat

Py P y P

159.74

0.6 83.21 0.4 41.98P kPa

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1

0.6 59.740.4308

83.21sat

y Px

P

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(b) Prepare T-x1-y1 diagram ( P is given (70 kPa), x1 and y1 in the range 01)

Calculate T1sat and T2

sat at the given pressure by using Antoine equation

For P = 70kPa,

T1sat = 342.99 K (69.84oC) and T2

sat = 362.73 K (89.58oC)

Select T1sat < T < T2

sat to calculate P1sat and P2

sat for these temperature by

using Antoine equation, and evaluate x1 by equation:

For example take T = 78oC (351.15K),

P1sat = 91.76 kPa P2

sat = 46.84 kPa x1 = 0.5156

Calculate y1 by equation (10.1),

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1 2

sat

sat sat

P Px

P P

1 11

0.5156 91.760.6759

70

satx Py

P

ln

sat ii i

i

BT C

A P

From eqn. (10.2)


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Similar calculation of x1 and y1 for T1sat < T < T2sat.

The results of calculations:

x1 y1 T (K)

1.0000 1.0000 342.99

0.8596 0.9247 345.15

0.7378 0.8484 347.15

0.6233 0.7656 349.15

0.5156 0.6759 351.15

0.4142 0.5789 353.15

0.3184 0.4742 355.15

0.2280 0.3614 357.15

0.1424 0.2401 359.15

0.0613 0.1098 361.15

0.0000 0.0000 362.73

At x1 = 0.6 and P = 70 kPa,

T = ?, y1 = ? BUBL T

At y1 = 0.6 and P = 70 kPa,

T = ?, x1 = ? DEW T

340

345

350

355

360

365

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

T/K

x1, y1

T-xy diagram for acetonitrile(1)/nitromethane(2) at 70 kPa

y1 x1

T2sat=362.73K

T1sat=342.99K

Dew point

Bubble point

Superheated vapor

Subcooled liquid

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BUBL T calculation

For x1 = 0.6 and P = 70kPa, T is determine by iteration.

Iterate as follow:

1. Calculate initial T using mole fraction-weighted average of T1sat and T2sat:

T = x1T1sat + x2T2

sat

or select T1sat < T < T2

sat as initial T.

2. Calculate P1sat and P2sat at initial T.

3. Calculate .

4. With the current value of , calculate P2sat .

5. Calculate T from Antoine equation for species 2:

6. Find a new value of (step 3).

(a) subtract ln P2sat from ln P1

sat as given by Antoine equations.

or

(b) calculate P1sat and P2sat at T, then find new value of .

7. Repeat step 4-5 and iterate to convergence for a final value of T.

2

1 2

sat PP

x x

2945.47 2972.64ln 0.0681

49.15 64.15T T

2

2972.6464.15

14.2043 ln satT

P

1 2

sat satP P

From eqn. (10.2)

Use Antoine eqn.

Use T

from

step 5

OR

*The result is T = 349.57K (76.42oC). From Antoine equation, P1sat = 87.17 kPa and by eq (10.1), y1 = 0.7472

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Iteration P2sat

(kPa) T

(K)

1 1.961 44.40 349.68

2 1.970 44.24 349.58

3 1.971 44.23 349.57

4 1.971 44.23 349.57

Iteration P1sat

(kPa)

P2sat

(kPa)

P2sat

(kPa)

T

(K)

1 90.99 46.40 1.961 44.40 349.68

2 87.47 44.40 1.970 44.24 349.58

3 87.20 44.24 1.971 44.23 349.57

4 87.18 44.23 1.971 44.23 349.57

Calculate new

by using step 6(a)

Calculate new

by using step 6(b)

Initial T = 350.89 K

P1sat = 90.99 kPa, P2sat = 46.40 kPa

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DEW T calculation

For y1 = 0.6 and P = 70kPa, T is determine by iteration .

Iterate as follow:

1. Calculate initial T using mole fraction-weighted average of T1sat and T2sat:

T = y1T1sat + y2T2

sat

or select T1sat < T < T2

sat as initial T.

2. Calculate P1sat and P2sat at initial T.

3. Calculate .

4. With the current value of , calculate P1sat .

5. Calculate T from Antoine equation for species 1:

6. Find a new value of (step 3).

(a) subtract ln P2sat from ln P1

sat as given by Antoine equations.

or

(b) calculate P1sat and P2sat at T, then find new value of .

7. Repeat step 4-5 and iterate to convergence for a final value of T.

sat

1 1 2P = P y +y

2945.47 2972.64ln 0.0681

49.15 64.15T T

1 2

sat satP P

From eqn. (10.3)

Use Antoine eqn.

1

2945.4749.15

14.2724 ln satT

P

Use T

from

step 5


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*The result is T = 352.73K (79.58oC) and P1sat = 96.53 kPa. From eq (10.1), x1 =

0.4351

Iteration P1sat (kPa) T (K)

1 1.961 96.91 352.85

2 1.947 96.51 352.72

3 1.948 96.53 352.73

4 1.948 96.53 352.73

Iteration P1sat

(kPa)

P2sat

(kPa)

P1sat

(kPa)

T

(K)

1 90.99 46.40 1.961 96.91 352.85

2 96.91 49.78 1.947 96.51 352.72

3 96.51 49.55 1.948 96.53 352.73

4 96.53 49.57 1.948 96.53 352.73

Calculate new

by using step 6(b)

Calculate new

by using step 6(a)

Initial T = 350.89 K

P1sat = 90.99 kPa, P2sat = 46.40 kPa

Smith, J.M., Van Ness, H.C., and Abbott, M.M. 2005. Introduction to

Chemical Engineering Thermodynamics. Seventh Edition. Mc

Graw-Hill.

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PREPARED BY:

MDM. NORASMAH MOHAMMED MANSHOR

FACULTY OF CHEMICAL ENGINEERING,

UiTM SHAH ALAM.

[email protected]

03-55436333/019-2368303

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