Bubble Point of Liqued Propane

1

Vapour/Liquid Equilibrium

(VLE): Introduction

Equilibrium = a static condition in

which no changes occur in a

macroscopic properties of a system

with time

2

The Phase Rule. Duhem’s Theorem

Recall from Chapter 1

The phase rule:

by J. Willard Gibbs

2F Nπ= − + (1.1)

F = degree of freedom (dof)

π = # of phase(s)

N = # of chemical species

Example Determine a dof of a

system of a single species of 2 phases

(e.g., liquid and vapour)

3

From the given information

π =

N =

Thus, from the phase rule

This means that we need only one

Thermodynamic property in order to

identify the other properties of the

system

4

Consider a P-T diagram

On the line separating vapour and

liquid zones (on which vapour and liquid

coexist in equilibrium), when we specify

the value of T, the value of P will

automatically be fixed, or vice versa

P

T

S

L V

5

Example Determine a dof of a

system of a single species and a

single phase (e.g., either liquid or

vapour)

From the given information

π =

N =

Thus, from the phase rule

6

When considering a P-T diagram,

in order to be able to specify the

exact point on the liquid (i.e. LHS of

the dome) or vapour (i.e. RHS of the

dome) zone, we need both P and T;

only P or T does not suffice

7

Example Determine a dof of a “triple

point” of a pure species

At the triple point, all 3 phases (S,

L, & V) coexist in equilibrium)

Thus,

π =

Since it is a pure species,

N = 1

Hence, from the phase rule

8

This implies that, for any pure species

(a single species), when we know the

triple point, we will automatically know

the name of the substance, or vice

versa

We do NOT need to know any other

Thermodynamic properties

P

T

S

L

V

Triple point

9

Duhem’s Theorem

Applied to “closed” systems

“For any closed system formed

initially from given masses (or

moles) of prescribed chemical

species, the equilibrium state is

“completely determined” when

any “two” independent variables

are fixed”

10

For a 2 species system (i.e. a

binary mixture) with 2 phases (e.g.,

V+L), from the phase rule, we obtain

22 2 22

F Nπ= − += − +=

In addition to T or P, we need to know

a chemical composition of a mixture, in

the from of either mole or mass fraction

Since it is a binary mixture, we need

to know mole or mass fraction of only

one substance; mole or mass fraction of

another substance will automatically be

fixed

Hence, dof = 1+1 = 2

11

For an n species system (i.e. N = n),

from the phase rule, we get

22 2

F Nn

n

π= − += − +=

This means that we need to know

either P or T, and mole (or mass)

fraction of 1n− components

Hence, dof = ( )1 1n n+ − =

12

VLE: Qualitative Behaviour

Consider VLE of a binary mixture

Let

mole fraction of A in a liquid

phase = Ax

and

mole fraction of A in a gaseous

phase = Ay

A+B

Liquid phase

Gaseous phase

A+B

13

At any given P, we vary T and

measure the values of Ax and Ay , and

then plot a graph between T and Ax or

Ay , or a Txy diagram as illustrated below

T

xA or yA

T1

Bubble-point curve

Dew-point curve

xA1 yA1

A B

C T2

T3 D

Vapour zone

Liquid zone

14

From a Txy diagram,

at any given T (e.g., T1)

Point A = saturated liquid state

of a liquid mixture at which the mole

fraction of species A = xA1

Point B = saturated vapour state

of a gaseous mixture where the mole

fraction of species A = yA1

At T = T1, for the mole fraction of

species A in the range of xA1—yA1, the

mixture is in the form of a vapour-

liquid mixture

15

Additionally, at T = T1

• a mixture with a mole fraction

of A lower than xA1 is in the form

of compressed or sub-cooled

liquid

• a mixture with a mole fraction

of A higher than yA1 is in the

form of superheated vapour

16

At a given xA (e.g., xA1)

when T is increased from the

state of compressed liquid (e.g., at T

= T3) to the point where T = T1, the

mixture starts to boil at the point A

If T is kept increasing, the

mixture is in the form of the vapour-

liquid mixture until T = T2 (point C),

where the mixture changes to the

form of a saturated vapour

and if T is still increased, the

mixture is in the form of superheated

vapour

17

On the other hand, if T is

decreased from the vapour zone,

where the mole fraction of species A

= yA1, to the point at which T = T1

(point B), the mixture turns to be

saturated vapour (there is a first drop

of liquid at this point)

If T is kept decreasing, the mixture

is in the form of liquid-vapour mixture

until T = T3 (point D), where the

mixture turns to be liquid completely

(or becomes saturated liquid)

18

When we draw Txy diagrams at

various P, we get

T

xA or yA

CA

CA

CB

P3

P2

P1

19

Note that

CA & CB = critical P of A & B,

respectively

that

P1 < P2 < P3

and that

P1 < Pc of A & B

P2 is in between Pc of A & B

P3 > Pc of A & B

20

At any given T, a Pxy diagram can

be drawn as follows

P

xA or yA

Bubble-point curve

Dew-point curve

Liquid zone

Vapour zone

L+V

21

Pxy diagrams at various T are as

illustrated below

P

xA or yA

CA

CA CB T3

T2

T1

22

Note that

CA & CB = critical T of A & B,

respectively

that

T1 < T2 < T3

and that

T1 < Tc of A & B

T2 is in between Tc of A & B

T3 > Tc of A & B

23

In the case where either a Txy or a

Pxy diagram is as follows

Point Q is called an azeotropic

point

T

xA or yA

Q

xAQ

24

It is the point where the mixture

behaves as it is a pure substance

Thus, at this point (or at xA = xAQ),

we cannot separate the mixture with

by distillation (WHY?)

25

At any mole fraction of A, we can

draw a P-T diagram as follows

Note that a critical point needs

not be the highest T or P

P

T

Bubble-point curve

Dew-point curve

Critical point

Liquid Vapour L+V

26

When we combine a P-T diagram at

different mole fractions of A, we obtain

Note that CA & CB = critical points of

species A & B, respectively

P

T

CA

CB

Critical locus

27

When we draw a graph of xA and

yA at a given T & P, we get

yA

xA

yA = xA

28

At the point where xA = yA, i.e. the

point where the composition of a

mixture in the gaseous phase is the

same as that in the liquid phase, we

cannot separate the mixture by

distillation

29

Simple Models for VLE Calculations

Two SIMPLEST models:

• Raoult’s law

• Henry’s law

Raoult’s law

Assumptions for a Raoult’s law

• The VAPOUR PHASE is an IDEAL

GAS

• The LIQUID PHASE is an IDEAL

SOLUTION

30

Hence,

• Raoult’s law is applicable for a

system where P is not too high

(in order that a vapour phase

behaves as an ideal gas)

• The mixture should be in similar

chemical nature and not too

different in size – in order that it

behaves as an ideal solution

31

Raoult’s law

sati i iy P x P= (4.1)

where

iy = mole fraction of species i

in a gaseous phase

ix = mole fraction of species i

in a liquid phase

P = total pressure of a system

satiP = vapour pressure of

species i

32

Since

i iy P P= (4.2)

where

iP = partial pressure of species i

in a gaseous phase

sati i iP x P= (4.3)

33

How can we obtain the value of satiP ?

From a Thermodynamic table

Example

Saturated propane: T-table

Specific Volume T (oC) P (bar)

vf vg

0

..

..

20

..

..

40

4.743

…

…

8.362

…

…

13.69

…

…

…

…

…

…

…

….

….

….

….

….

….

….

34

From Charts (e.g., a Cox chart)

Note: Try using the Cox chart above

to determine vapour pressures of

propane at 0, 20, and 40 oC (32, 68,

and 104 oF), and compare the read

values with those from the Table on

Page 33

35

Using Empirical Equations

Example

521 3 4exp ln Csat CP C C T C T

T⎛ ⎞= + + +⎜ ⎟⎝ ⎠

(4.4)

where

satP is in the unit of Pa

T is in the unit of K

1 5C C− are properties of a

substance

Note that C1 – C5 can be obtained

from, e.g., Table 2.6, pp. 2-50 – 2-54 of

Perry’s Chemical Engineers’ Handbook (7th

ed.)

36

Example The values of 1 5C C− of

propane are as follows:

1 59.078C = ; 2 3492.6C = − ;

3 6.0669C = − ; 54 1.0919 10C −= × ;

5 2C =

From Eq. 4.4, at T = 20 oC, we obtain

Note that the value of Psat from the

Table is 8.362 bar

37

Another Empirical Equation

log sat BP AT C

= −+

(4.5)

where

satP is in the unit of bar

T is in the unit of oC

A C− are properties of a substance Note that A – C are available in, e.g.,

Section D of Appendix A, pp. A.48 – A.60 of

The Properties of Gases and Liquids (5th ed.)

Eq. 4.5 = “Antoine equation”

38

Example The values of A C− for

propane are as follows:

3.92828A =

803.9970B =

247.040C =

From Eq. 4.5, at T = 20 oC, we get

39

Applying a summation for all

species to a Raoult’s law (Eq. 4.1)


yields

sati i iy P P x P= =∑ ∑ (4.6)

For a binary mixture, we obtain

1 1 2 2sat satP x P x P= + (4.7)

and

2 11x x= − (4.8)

40

Combining Eqs. 4.7 & 4.8 gives

( )1 1 1 2

1 1 2 1 2

1sat sat

sat sat sat

P x P x P

x P P x P

= + −

= + −

( )1 2 1 2sat sat satP P P x P= − + (4.9)

When plotting a graph of P against

1x , we obtain a straight line with

• a slope of 1 2sat satP P−

• a Y-intercept of

2satP at 1 0x =

1satP at 1 1x =

Examples of a plot between P and x1

are illustrated in the next page

41

Note that dashed lines are the

1P x− lines from a Raoult’s law

42

Since liquid boils when 1iy =∑ , Eq. 4.6

is for bubble-point calculations

On the other hand, vapour starts

to condense when 1ix =∑ or when

( )1/ sat

i i

Py P

=∑

(4.10)

Hence, Eq. 4.10 is for dew-point

calculations

43

Example Draw a Pxy diagram of

propane for a propane + ethane

mixture at T = 75 oC using a Raoult’s

law and an Antoine equation

The values for an Antoine

equation for propane and ethane

are as follows

Propane:

3.92828A =

803.9970B =

247.040C =

Ethane:

3.95405A =

663.720B =

256.681C =

44

Employing an Antoine equation to

calculate satP of propane and ethane

at 75 oC yields

45

Applying Eq. 4.9

( )1 1 2 2sat sat satP x P P P= − + (4.9)

to this Question gives

( )sat sat satpropane propane ethane ethaneP x P P P= − +

Substituting the numerical values

of satpropaneP and sat

ethaneP into the above

equation results in

46

Using Eq. 4.11 to calculate the

values of P at various propanex results in

the following Table

propanex P [bar]

0

0.2

0.4

0.6

0.8

1.0

89.7

77.2

64.6

52.1

39.5

27.0

47

When the values of P at different

propanex are obtained, the values of propaney

of various P can be calculated using

Eq. 4.1 (i.e. a Raoult’s law), as follows

satpropane propane propaney P x P=

and ........ propane

propane

xy

P=

The results are as follows

propanex P [bar] propaney 0

0.2 0.4 0.6 0.8 1.0

89.7 77.2 64.6 52.1 39.5 27.0

0 0.07 0.17 0.31 0.55 1.00

48

From the Table in the previous

page, we can plot a Pxy diagram as

follows

0

20

40

60

80

100

0 0.2 0.4 0.6 0.8 1

xpropane or ypropane

P [b

ar]

P-x

P-y

Bubble-point curve

Dew-point curve

49

Example Draw Txy diagram of

propane for a propane + ethane

mixture at P = 10 bar using a Raoult’s

law and an Antoine equation

We can rearrange an Antoine

equation as follows

log

log

1log

sat

sat

sat

BP AT C

BA PT C

T CB A P

= −+

− =+

+=

−

log satBT C

A P= −

−

50

Thus, satT at a given P can be

calculated from the following

equation

log

sat BT CA P

= −−

(4.12)

At P = 10 bar

51

We, then, pick T between -32.0 –

27.5 oC, for example,

-32.0, -22.0, -12.0, -2.0, 8.0, 18.0, 27.5 oC

Calculating satpropaneP and sat

ethaneP at these

T’s, for instance, at T = -22.0 oC, yields

52

Rearranging Eq. 4.9 results in

21

1 2

sat

sat satP Px

P P−

=−

Hence,

sat

ethanepropane sat sat

propane ethane

P PxP P

−=

− (4.13)

Substituting the resulting values of sat

propaneP and satethaneP , & the value of P

(= 10 bar in the case) into Eq. 4.13

yields propanex =

53

The value of propaney can then be

calculated as follows (using Eq. 4.1)

satpropane propane

propane

x Py

P=

==

Carrying out the same calculations

for other T’s results in

54

T

[oC]

satpropaneP

[bar]

satethaneP

[bar] propanex propaney

-32.0

-22

-12

-2

8

18

27.5

1.547

2.268

3.218

4.438

5.968

7.849

10.000

10.00

13.42

17.44

22.28

27.95

34.49

41.54

0

0.307

0.523

0.688

0.817

0.919

1.000

0

0.070

0.168

0.306

0.487

0.722

1.000

55

A Txy diagram can then be drawn

as follows

-40

-20

0

20

40

0.0 0.2 0.4 0.6 0.8 1.0

xpropane or ypropane

T [o C

]

Bubble-point curve

Dew-point curve

T-x

T-y

56

Bubble-Point Calculations

When we know mole fraction of a

substance in liquid phase ( ix ) and T, we

can calculate a “bubble-point P”

Vice versa, when we know mole

fraction of a substance in liquid phase

( ix ) and P, we can calculate a “bubble-

point T”

Example Calculate a bubble-point T

of a propane + ethane mixture at

P = 10 bar and propanex = 0.3

57

At a bubble point, 1iy =∑


species to a modified Raoult’s law

sati i iy P x P=

sat

i ii

x PyP

= (4.1a)

11

sati i

i

sati i

x PyP

x PP

=

=

∑ ∑

∑

yields

sati iP x P= ∑ (4.6)

Since this is a binary mixture,

sat satpropane propane ethane ethaneP x P x P= + (4.14)

58

Given

10 barP = 0.3propanex = 0.7ethanex =

We need T to calculate the values

of satpropaneP and sat

ethaneP

Hence, we have to start our

calculations by guessing a value of T,

e.g., T = -20 oC, and then calculate

the values of satpropaneP and sat

ethaneP at the

guessed T, as follows

59

Substituting the values of resulting sat

propaneP and satethaneP into Eq. 4.14 yields

60

Since the resulting P is higher than

10.0 bar, it indicates that the guessed

value of T (-20 oC) are too high we

need a NEW GUESS

We have to perform a Trial & Error

iteration until we obtain T that makes sat

i ix P∑ = 10 bar

TRY DOING IT YOURSELF

61

Dew-Point Calculations

When we know mole fraction of a

substance in gaseous phase ( iy ) and

T, we can calculate a “dew-point P”

Vice versa, when we know mole

fraction of a substance in gaseous

phase ( iy ) and P we can calculate a

“dew-point T”

62

Example Calculate a dew-point P of

a propane + ethane mixture at T =

0 oC and propaney = 0.6

At a dew point, 1ix =∑


species to a modified Raoult’s law

sati i iy P x P=

ii sat

i

y PxP

=

1

ii sat

i

isat

i

y PxP

yPP

=

=

∑ ∑

∑

63

results in

1i

sati

Py

P

=⎛ ⎞⎜ ⎟⎝ ⎠

∑ (4.10)

Since it is a binary mixture,

1propane ethanesat sat

propane ethane

P y yP P

=+

(4.15)

Given

o0 CT = 0.6propaney = 0.4ethaney =

64

Thus,

Substituting corresponding values

into Eq. 4.15 gives

65

Precautions for the use of Raoult’s law • Can be used at not-too-high P

• Can be used at T < Tc

• Can be used with species whose

mole fractions approach 1 (if mole

fractions are too low, e.g., 5% or

0.05, Raoult’s law is NOT suitable)

• Can be used with a mixture whose

components are in chemical similar

nature and shape

66

Henry’s Law

From a Raoult’s law


in the case where a concentration

( ix ) of a solute is low (e.g., CO2 or air

dissolves in water) and the case

where T > Tc, a Raoult’s law is NOT

suitable and/or applicable for VLE

calculations

67

For example, a Raoult’s law is NOT

applicable when we want to perform

VLE calculations for the CO2 (Tc =

304.12 K) + water system at 373.15 K,

since, by using a Raoult’s law, we

shall NOT have vapour pressure data

of CO2 at the temperatures above its

critical point

If we, however, desperately need

to do the VLE calculations at those

conditions, what should we do?

68

Henry’s law can help you

Henry’s law

“The partial pressure of the

species in the vapour phase is

directly proportional to its

liquid-phase mole fraction”

i i iy P x= Η (4.16)

where

iΗ = Henry’s constant

“Henry’s constant ( iΗ )” replaces

vapour pressure ( satP ) in a Raoult’s

law equation (Eq. 4.1)

69

Thus, a unit of Henry’s constant is

the same as that of PRESSURE

Henry’s constant ( iΗ ) depends on T

as same as vapour pressure ( satP )

For instances,

iΗ of CO2 at 10 oC = 900 bar

iΗ of CO2 at 25 oC = 1670 bar

Henry’s constant ( iΗ ) also depends

on type of substance and is obtained

experimentally

70

Example Determine why we should

keep pop or soda at a low

temperature, using both Henry’s &

Raoult’s laws. Henry’s constants ( )iΗ

of CO2 at 25 and 10 oC are 1,670 and

900 bar, respectively. Vapour

pressures of water at 10 oC and 25 oC

are 0.01228 and 0.03169 bar,

respectively

71

Since CO2 can scarcely be dissolved

in water, its mole fraction of CO2 in

water is low

Hence, we should use a Henry’s law

for VLE calculations of CO2

On the other hand, we can use a

Raoult’s law for VLE calculations of

water, as the mole fraction of water in

the CO2-water solution approaches 1

72

Hence,

for CO2:

2 2 2 2CO co co coP y P x= = Η

for water:

water water water watersatP y P x P= =

Accordingly, the total pressure of the

gaseous phase can be calculated from

the following equation

2

2 2

CO water

co co water watersat

P P P

x x P

= +

= Η +

73

Assume that 5% (by mol) of CO2 is

dissolved in water (2CO 0.05x =

water 0.95x = )

Thus, at 10 oC

However, if T is increased to 25 oC,

while the total pressure is kept

constant (at 45.01 bar), mole fraction

of CO2 of the same solution changes

to

74

(Note that, since it is the same bottle, the

total pressure of gaseous phase is still the

same)

75

Approximately, the amount of CO2

dissolved in water reduces by half

(5% to 2.7%) when T is increased from

10 to 25 oC

If a refrigerator is not available

(hence, we cannot keep the

temperature at 10 oC), but we have

a compressor,

“can we use a compressor to

keep CO2 in the solution?”

76

Assume that we use a compressor

to increase the total pressure of the

system from ~45 bar to 80 bar

Thus, the amount of CO2 dissolved

in the solution, at 25 oC, can be

calculated as follows

Evidently, in order to keep CO2 in

the solution, we can do 2 things:

77

• Keep T of a solution (soda/pop)

at low temperatures (and a

bottle should also be capped)

• Keep P of the system at high

pressures, using, e.g., a

compressor

78

VLE Calculations by a MODIFIED

Raoult’s Law

From the assumptions of a Raoult’s

law, in which

• VAPOUR Phase = IDEAL GAS

• LIQUID Phase = IDEAL SOLUTION

it is found that, in the real situations,

assuming that vapour phase = ideal

gas causes only a small error in VLE

calculations, as, for most interested

systems:

• T are usually high

• P are normally low

79

hence, it is justified to assume that

VAPOUR Phase = IDEAL GAS

The main cause of deviations,

however, comes from the assumption

that liquid phase = ideal solution

To enable a Raoult’s law to be

applicable for real solutions (by

having an acceptable error) the

“original” Raoult’s law has to be

modified, as follows

sati i i iy P x Pγ= (4.17)

where iγ = activity coefficient

80

From Lecture 3, we knew that

lnEi

igRT

γ = (3.135)

where E idi i ig g g= − , which indicates the

deviation from the ideal solution

Additionally, we learned that

ii

i i

fx f

γ =

which also indicates the deviation

from the ideal solution

Hence, an activity coefficient ( )iγ

serves as a correction factor for the

non-idealness of real solutions

81

We can use Eq. 4.17 to calculate a

bubble point and a dew point in the

same way we use Eqs. 4.6 and 4.10,

by keeping in mind that

• at a bubble point: 1iy =∑

• at a dew point: 1ix =∑

82


Thus,

sati i iP x Pγ= ∑ (4.18)


Hence,

1isat

i i

PyPγ

=⎛ ⎞⎜ ⎟⎝ ⎠

∑ (4.19)

83

How can we calculate activity

coefficients ( iγ )???

Activity coefficient depends on T

and composition of each species in

the solution

Activity Coefficient Calculations

Activity coefficient ( iγ ) at given T

and composition (in case of binary

mixtures) can be calculated using

the following equations:

84

21 2ln Axγ =

22 1ln Axγ =

where

A P QT= +

( & constantP Q = and depend

on type of a binary system)

85

Example For a methanol (1) + methyl

acetate (2) binary mixture, activity

coefficients of methanol and methyl

acetate can be calculated using the

following equations: 21 2ln Axγ = and

22 1ln Axγ = , where 2.771 0.00523A T= − (T in

K)

The values of A, B, & C for Antoine

equation are as follows

A = 5.20277

B = 1580.080

C = 239.500

A = 4.18621

B = 1156.430

C = 219.690

86

Determine

a) bubble-point P at T = 300 K and

1 0.2x =

At T = 300 K (27 oC)

A ==

Thus,

and

87

Vapour pressure of each

species can be calculated using an

Antoine equation, as follows

88

Hence, a bubble-point P can

be calculated as follows

We can then calculate

composition of gaseous phase at the

bubble point, using Eq. 4.17

89

and

90

b) dew point T at P = 1 bar and 1 0.3y =

Since we do not have the data

of ix , we cannot calculate the values

of iγ

We, however, can start with the

“original” Raoult’s law by assuming

that 1iγ =

In addition, since we do not

know the value of T, we cannot

calculate satiP

91

Thus, we need to employ a

TRIAL & ERROR

technique

Modifying an Antoine equation

gives (Eq. 4.12)

log

sat BT CA P

= −−

(4.12)

Thus,

92

Accordingly, the first guess of T

should be between 56.6 and 64.2 oC

(WHY??)

Let the first guess be that

T = 60 oC (333 K)

Hence, from Eq. 4.5 (Antoine

equation), we get

93


Thus,

1isat

i i

PyPγ

=⎛ ⎞⎜ ⎟⎝ ⎠

∑ (4.20)

Substituting the corresponding

values (at T = 60 oC) into Eq. 4.20

yields

94

However, the total pressure has

to be 1 bar

Hence, we need another trial &

error, but we can use P obtained

recently to calculate the values of ix

as follows

95

and

We can use these 1x and 2x to

calculate the values of iγ at 60 oC, as

follows

A ==

96

Since T is still the same (= 60 oC), sat

iP is also still the same

97

Substituting new corresponding

values into Eq. 4.20 results in

Close to 1 bar, but not 1 bar yet

another Trial & Error

We can use P obtained recently

(= 1.07 bar) to calculate 1x and 2x as

follows

98

and

99

Trial & Error Procedure

• Calculate 1γ and 2γ from x1

and x2

• Guess new T

• Calculate satiP at new guessed T

• Calculate P and determine if it

is close to 1 bar (in an

acceptable level), if NOT,

another Trial & Error

100

c) if an azeotropic point exists at T = 300

K. If so, determine the composition of

each species at the azeotropic point

We have known that, at the

azeotropic point

1 1x y= 2 2x y=

Rearranging Eq. 4.17

sati i i iy P x Pγ= (4.17)

results in

sat

i i i

i

y Px P

γ=

101

Thus,

1 1 1

1

saty Px P

γ= (4.21)

and 2 2 2

2

saty Px P

γ= (4.22)

(4.21)/(4.22) yields

1 1 1

1

2 2 2

2

sat

sat

y Px Py Px P

γ

γ=

1 1 1 112

2 2 2 2

//

sat

saty x Py x P

γαγ

= = (4.23)

102

From the relationships

21 2ln Axγ =

22 1ln Axγ =

when 1 0x = 2 1x = (binary solutions),

we obtain

22 1ln 0Axγ = = 2 1γ =

and 1ln Aγ = ( )1 exp Aγ =

and when 1 1x = 2 0x = , we get

21 1ln 0Axγ = = 1 1γ =

and 2ln Aγ = ( )2 exp Aγ =

103

Hence, in the case where 1 0x =

( )112

2

expsat

sat

P AP

α = (4.24)

and when 1 1x =

( )

112

2 exp

sat

satP

P Aα = (4.25)

At an azeotropic point, 1 1x y= &

2 2x y= 12 1α =

From Question a, at T = 300 K

(27 oC), 1 0.188 barsatP = , 2 0.315 barsatP = ,

and 1.202A =

104

Substituting corresponding

values into Eqs. 4.24 & 4.25 yields

105

From the values of ( )12 1@ 0xα =

and ( )12 1@ 1xα = , it was found that

( )12 1@ 0xα = > 1 while ( )12 1@ 1xα = < 1

indicating that there is a point

between 1 0x = and 1 1x = where 12 1α =

This, in turn, means that there

exists an azeotropic point between

1 0x = and 1 1x =

106

At the azeotropic point, 12 1α =

Hence,

1 112

2 2

1sat

satPP

γ αγ

= =

1 2

12

sat

satPP

γγ

=

Substituting corresponding

values (from Question a)) gives

1

2

sat

satPP

=

107

21 2ln Axγ = and 2

2 1ln Axγ =

Hence,

( )

( )( )

( )( )

2 222 1 1 2

1

21 2 1 2

1

1 2

1

ln ln ln

ln

2 1

A x x

A x x x x

A x xA x

γγ γγ

γγ

− = = −

= − +

= −

= −

Thus,

1

108

The composition at the

azeotropic point of species 1 is

………………………………………………

109

VLE from K-value Correlations

Although a modified Raoult’s law

yields the results close to

experimental data, VLE calculations

using modified Raoult’s law are

relatively complicated

Is there any other (easier) method?

110

DePriester, C.L. (1953) constructed

the relationship between iy and ix in the

form of iK , as follows

ii

i

yKx

= (4.26)

Rearranging Eq. 4.26 results in

ii

i

yxK

= (4.27)

i i iy x K= (4.28)

111


Hence, from Eq. 4.28,

1i i iy x K= =∑ ∑ (4.29)


Thus, from Eq. 4.27,

1ii

i

yxK

= =∑ ∑ (4.30)

112

Example Determine a bubble-point T

at 150 psia of a mixture comprising

10 mol% methane, 20% ethane, and

70% propane

At bubble point, 1i i iy x K= =∑ ∑

Given

4

2 6

3 8

CH

C H

C H

0.10

0.20

0.70

x

x

x

=

=

=

113

Try T = 50 oF as the first guess

Reading the iK values from the

chart at T = 50 oF and P = 150 psia

results in

4

2 6

3 8

CH

C H

C H

K

K

K

=

=

=

Thus,

Guessed T was too high

114

Guess a new T T = 0 oF

Reading the iK values from the

chart at T = 0 oF & P = 150 psia gives

4

2 6

3 8

CH

C H

C H

K

K

K

=

=

=

Hence,

Still to high guess new T, until we

obtain T that makes 1iy =∑

115

The resulting T is bubble-point T at

P = 150 psia

Try doing it yourself

(the answer is -50 oF)

Although we have to use a trial &

error technique, the calculations are not

too complicated, but rather

straightforward

Additionally, this technique is good

for the mixtures with more than 2

components

Accordingly, this technique is widely

used for VLE calculations, especially for

hydrocarbon mixtures

118

Flash Calculations

When a liquid feed is introduced

into a low-P tank (called a “flash

tank”), the feed divides into 2

phases: liquid & vapour

In other words, the feed, especially

for components whose vapour

pressures are high, partially

evaporates

119

Suppose F mol of feed with a

composition of each species of iz is

introduced into a flask tank

Let

V is the number of moles of feed

that evaporates

and

L is the number of moles of liquid

(or the number of moles of feed that

does not evaporate)

Mole fraction of each species in the

vapour and liquid phases are denoted

as iy and ix , respectively

120

Material Balance

Overall balance

F V L= + (4.31)

Species balance

i i i

i i i

z F y V x Lf v l= +

= + (4.32)

Substituting Eq. 4.28

i i iy K x= (4.28)

into Eq. 4.32 yields

( )

1

i i i i

i i

i i

z F K x V x Lx K V L

Vx K LL

= +

= +

⎛ ⎞= +⎜ ⎟⎝ ⎠

121

Rearranging gives

1

ii i

i

z Fl LxVKL

= =⎛ ⎞+⎜ ⎟⎝ ⎠


Thus,

1i

i i i

i

z Fl Lx L x LVKL

= = = =⎛ ⎞+⎜ ⎟⎝ ⎠

∑ ∑ ∑ ∑

Rearranging the above equation

results in

1

ii

i

z FL lVKL

= =⎛ ⎞ +⎜ ⎟⎝ ⎠

∑ ∑ (4.33)

122

Substituting Eq. 4.26

ii

i

yxK

= (4.26)

into Eq. 4.32 yields

( )

ii i

i

i i i

i

i i i i i

i i

i i

yz F y V LK

y K V y LK

z K F y K V y Ly K V L

Ly K VV

= +

+=

= +

= +

⎛ ⎞= +⎜ ⎟⎝ ⎠

Rearranging gives

i ii i

i

z K Fv y VLKV

= =⎛ ⎞+⎜ ⎟⎝ ⎠

123


Hence,

i ii i i

i

z K Fv y V V y VLKV

= = = =⎛ ⎞+⎜ ⎟⎝ ⎠

∑ ∑ ∑ ∑

Rearranging the above equation

gives

i ii

i

z K FV v LKV

= =+

∑ ∑ (4.34)

124

Example Perform a flash calculation

for the following mixture:

methane 10% mol

ethane 20% mol

propane 70% mol

fed into a tank with P = 200 psia and

T = 50 oF

Basis: 100 kmol of mixture (F)

Thus,

100 V L= +

125

Since there are 2 unknowns (V &

L), but we have only ONE equation,

we need to use a trial & error

technique

We start by guessing a value of V

(or L). Let’s start with 30 kmolV =

70 kmolL =

Component if

[kmol] zi Ki

i ii

i

z K Fv LKV

=+

Methane

Ethane

Propane

10

20

70

0.10

0.20

0.70

9.8

1.75

0.52

( )( )( )0.10 9.8 1008.11709.8

30

=+

8.57

12.76

100 1.00 Σ = 29.44

126

The resulting value of iv∑ or V is

close to 30 (the guessed value), but it

is still not close enough

Try another guess, another guess,

and another guess, if the resulting

and guessed values are still not close

to each other

It was found that, when the

guessed value of V is 28.7, we get

127

Component if

[kmol] zi Ki

i ii

i

z K Fv LKV

=+

Methane

Ethane

Propane

10

20

70

0.10

0.20

0.70

9.8

1.75

0.52

( )( )( )0.10 9.8 1008.0471.39.8

28.7

=+

8.35

12.30

100 1.00 Σ = 28.69

CLOSE ENOUGH!!

128

The results of the flash calculation

are summarised in the following Table

Component if

[kmol] iv

[kmol]

il

[kmol]

Methane

Ethane

Propane

10

20

70

8.04

8.35

12.30

10 – 8.04 = 1.96

11.65

57.70

100 28.69 71.31

Bubble Point of Liqued Propane

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