Thermodynamics 2

Generalized Phase Equilibria Models, Colombia, Summer 2000 -

Author: Dr. Maria Barrufet - Summer, 2000 Page 1/77

Instructional Objectives • Understand the Principle of Corresponding States.

• Calculate the compressibility factor using different correlations and models.

• Understand phase equilibrium.

• Determine the number of variables required to define a system in equilibrium (Phase

Rule).

• Evaluate energy relationships using the First and Second Law of thermodynamics.

• Evaluate dew and bubble points given pressure or temperature as independent

variables.

Generalized Phase Equilibria Models The Principle of Corresponding States. Correlations and Models. Extension of Corresponding States to Mixtures. Phase equilibrium. Phase rule. Thermodynamic Properties of Homogeneous and Heterogeneous Systems. Phase Equilibrium: Vapor-Liquid-Equilibrium (VLE), Liquid-Liquid Equilibrium (LLE), Solid-Liquid-Equilibrium (SLE). Phase Equilibrium Models: Single Components. Reduced Equations of State (EOS). Multicomponents. Mixing Rules. Types of VLE Computations: Dew Point and Bubble Point Calculations. Multiphase Flash. Low Pressure Phase Equilibria Computations (Surface Separators). Ideal Systems. K-value correlations. Empirical methods to determine equilibrium ratios (K-values). Suggested reading: EL, WM, MAB



• Evaluate flash separation processes.

The Principle of Corresponding States The compressibility factor, or Z factor, of all pure species (C1, C2, N2, CO2 etc) can be

read from charts which are presented as a function of reduced properties Tr and Pr.

Any correlation, or model, which expresses the Z factor as function of Tr and Pr is said

to be generalized. Modern equation of state (EOS) can be put into this form, thus

providing a generalized correlation for the compressibility factor.

One needs only the critical temperature and the critical pressure of the fluid. This is the

basis for the two-parameter theorem of corresponding states.

“All fluids when compared at the same reduced temperature and reduced pressure, have approximately the same compressibility factor, and all deviate from ideal gas behavior to about the same degree”

The Principle of Corresponding states (POC) originated with single component fluids.

We, engineers, stretched it to multicomponent systems.

Generalized Corresponding States The principle of corresponding states says that all material properties when expressed

in terms of reduced parameters such as:



Reduced Temperature:

( )cr TTT /=

Reduced Pressure:

( )cr P/PP = ,

Reduced Molar Volume :

( )cr V~/V~V~ =

obey a similar or corresponding behavior.

Note that below Tc and Pc the reduced properties are lower than one and at these

conditions, for a single component fluid, there are two coexisting phases. The Z-factor

charts DO NOT provide information for saturation properties.

Nomenclature:

=V~ molar volume [=] cm3/gmol, ft3/lbmol (an intensive property.)

The Van der Waals (VdW) and the Redlich-Kwong equations of state are two-parameter

corresponding state equations, these two parameters are the critical temperature and

the critical pressure of the component in question. The critical volume is determined

once these two parameters )P,T( cc are fixed.

Further improvement, in terms of describing the fluid behavior for a broad spectrum of

pressures and temperatures is achieved by adding a third parameter. These models are

called three-parameter corresponding state equations. This third parameter is called the

acentric factor and was introduced by Pitzer and coworkers. It takes into account the

non-spherical nature of molecules. The Peng Robinson and the Soave Redlich Kwong

equations of state (EOS) are examples of three parameter corresponding states

models.



The acentric factor ω is tabulated and is defined as

( ) 701 .Tsat

r rPlog =−−=ω (1)

For ideal gases such as He, Ar, and Kr, the acentric factor is zero. For methane, which

is a nearly spherical molecule the acentric factor is nearly zero (0.0104).

1/Tr

log(PrSat)

-1

1.431.0

Slope = -2.3 ( Ar, Kr, Xe)

Figure 1 - Acentric factor definition.

Compressibility Factor Charts Following the POC only one compressibility factor chart can be used to determine

volumetric properties of any pure fluid using its reduced properties. The shape of this

chart is in general.



T r

P r

Z

1

Figure 2 - Compressibility factor as a function of reduced properties..

Corresponding States Correlations & Models

Equations of State for Gases: Virial Equations The objective is then to find a model (models) to predict the Z factor. The ideal gas

behavior is described from the ideal gas Equation of State (EOS) with a compressibility

factor of 1.

1=RT

V~P (2)

For real gases the analogous expression is



ZRT

V~P = (3)

where Z is the compressibility factor. The compressibility factor can also be defined as

the ratio of the �real molar volume� over the �ideal molar volume� of a substance

measured at the same pressure and temperature.

ZZV~P

RTRT

V~P

id

==

1

(4)

Now the deviation of the compressibility from ideal behavior (Z = 1) can be expressed in

terms of an infinite series (in practice only two or three terms are used). Two equations

were devised for this purpose

The pressure virial equation is a polynomial expansion in pressure

(1) Pressure Virial Equation

...P'DP'CP'BRT

V~P ++++= 321 (5)

with B�, C�, D� called �pressure virial coefficients�. These are determined from

experimental data and are temperature dependent. This equation is used for moderate

pressures (P < 15 bar at subcritical temperatures). Only two pressure-virial coefficients

are enough.

For higher pressures we would require more terms in the series, but these would be

difficult to determine experimentally, thus other models are used instead.



(2) Density Virial Equation

...V~D

V~C

V~B

RTV~P ++++= 321 (6)

This equation is used for higher pressures (P between 15 and 50 bar, with three

density-virial coefficients being enough for this pressure range).

The coefficients from the two expansions (pressure and density virials) are related.

The higher the pressure the higher the deviation from ideality and the more terms are

required (in either) expansion to describe the compressibility factor.

These virial coefficients have been obtained experimentally for various substances and

are a function of temperature. Additionally, several correlations exist for them. (see EL).

Cubic Equations of State Cubic polynomials in molar density (or molar volume) are the most popular

EOS for many industrial processes (petroleum production, reservoir compositional

simulation, refining, distillation, separation processes, petroleum recovery, etc.) These

equations describe reasonably well the pressure-volume-temperature (PVT) behavior of

fluids in both the gas and the liquid region. They can also be used for mixtures,

provided certain mixing rules are applied (these will be seen later).

The most well-known and older EOS is the Van der Waals equation, which is:

2V~a

bV~RTP −−

= (7)



The parameter a corrects for attraction forces between molecules, while the parameter b

corrects for repulsion forces by taking into account the volume of the molecules. Recall

that the ideal gas assumption was that gas molecules had zero volume.

Probable the most widely used EOS in the gas and petroleum industry is the Peng-

Robinson EOS

V~)bV~()bV~(V~a

bV~RTP

−++−

−= (8)

The two parameters (a and b) in these EOS can be determined from least squares

regression (fitting) of data at a broad range of Pressure and Temperature for the

substance in question. However most of the time, that is not affordable.

Critical properties, however, are known for a variety of substances and these have been used to determine the constants a and b using theoretical constraints.

The critical point observed in a )V~P( diagram or a (PT) diagram exhibits a maximum in

pressure and an inflection point on the critical isotherm (Tc). Figure 3 shows these

conditions for a pure substance in a )V~P( diagram.



01200

300

400

500

600

700

12

14

Pres

sur e

Molar Volume

Tc

T2

T1

P1v

L

2 - Phases

CP

V

L

V

01200

300

400

500

600

700

12

14

Pres

sur e

Molar Volume

Tc

T2T2

T1T1

P1v

L

2 - Phases

CP

V

L

V

Figure 3 - Pressure-volume behavior indicating isotherms for a pure component system.

These two conditions are expressed mathematically in two equations that are used to

solve the constants a and b in terms of the critical properties.

For example let�s use the VdW EOS.

2V~a

bV~RTP −−

= (9)



The first and second derivatives of pressure with respect to volume at constant

temperature are:

32

2V~

a)bV~(

RTV~P

T

+−

−=

∂∂ (10)

432

2 62V~

a)bV~(

RTV~

P

T

−−

=

∂∂ (11)

These two derivatives must vanish at the critical point

32

20cc

c

TT V~a

)bV~(RT

V~P

c

+−

−==

∂∂ (12)

432

2 620

cc

c

T V~a

)bV~(RT

V~P −

−==

∂∂ (13)

Now we have two equations and two unknowns.

332

23

2

2

32

33

32c

cc

c

c

c

cc

c

c

cc

c

V~b V~

)bV~(RT

)bV~(RT

V~

aV~)bV~(

RTV~

a )bV~(

RT

=→−

=−

→

=−

=−

(14)

Once b has been found, a can be obtained from either Eq. (12) or (13).



( ) 89

89

322 2

3

2

33

2cc

c

cc

cc

ccc

c

c V~RTV~

V~RT

/V~V~V~RTV~

)bV~(RTa ==

−=

−= (15)

The critical compressibility factor is

cc

c ZRT

V~P = (16)

Replacing the values found for a and b and using the EOS at the critical point we obtain

c

c

c

c

c

c

c

cc

cc

cc V~

RTV~RT

V~RT

V~V~RT

V~V~RT

P83

89

23

89

3 2 =−=−−

= (17)

And the critical compressibility calculated from VdW EOS is

375083 .Z

RTV~P

cc

c === (18)

Using this we can express the a and b constants as

( ) ( )c

c

c

cc

c

cccc

PRT

PRTZ

PRTRTV~RTa

6427

83

89

89

89 22

==

== (19)

and



c

c

c

cc

c

cc

PRT

PRTZ

PRTV~b

883

333=

=== (20)

The Z factor is then evaluated as

RTV~a

bV~V~

RTV~PZ −

−== (21)

Note that this EOS is a cubic polynomial in volume, therefore three possible real roots

could be obtained from the equation. The root selection will be discussed in future

lectures. To solve cubic equations there are analytical techniques. The following web

site will provide you the computer codes to solve the roots of polynomials up to quintic

degree.

http://www.uni-koeln.de/math-nat-fak/phchem/deiters/quartic/quartic.html

Since web sites change addresses quite frequently I recommend to copy and test the

course codes as soon as possible.

http://www.uni-koeln.de/math-nat-fak/phchem/deiters/quartic/quartic.html



Correlations for the Compressibility Factor The simplest correlation for the compressibility factor is expressed in terms of the

second virial coefficient.

r

r

c

c

TP

RTBP

RTBPZ

+=+= 11 (22)

And the term (BPc/RTc) is determined using Pitzer�s Correlation as follows,

10 BBRTBP

c

c ω+=

(23)

with

6.10 422.0083.0

rTB −= (24)

2.41 0172139.0

rTB −= (25)

Therefore the compressibility factor is expressed as:

r

r

r

r

TPB

TPBZ 101 ω++= (26)



The following example illustrates the use of the compressibility factor (Z) in a design

problem

Example Problem Example Problem Example Problem Example Problem

Mr. Jones wants to use some 30 liter cans to ship ethane from College Station to Conroe. He

would like to fill each of these cylinders with 10 kg of ethane, but he does not know the

pressure at which he needs to fill these tanks or if the walls of the tanks will be able to

withstand that kind of pressure. The shipping should be done at an average temperature of 25 oC.

Use three different methods:

(a)(a)(a)(a) Ideal gas EOS

(b)(b)(b)(b) Z factor compressibility correlations given in class

(c)(c)(c)(c) Z factor charts using chart given in class notes (you will do this one)

(d)(d)(d)(d) Z factor from a cubic EOS (you will do this)

(e)(e)(e)(e) Z factor using properties evaluated from NIST website seen in Module 1 (you will do

this)

The Critical properties and acentric factor for ethane are:

Mw = 30 g / mol

Tc = 305.5 ºK

Pc = 48.8 bar

ω = 0.098

(a) Ideal Gas EOS(a) Ideal Gas EOS(a) Ideal Gas EOS(a) Ideal Gas EOS



( )( ) ) K.(

mol Kbar cm.

g/mol g,

MwmRTnRTPV 25152731483

3000010 3

+

===

bar.cm

barcm,

.MwVmRTP 424275

0003010262738

3

36

=

×== (about 4,000 psia)

(b) Z (b) Z (b) Z (b) Z----factor Correlationsfactor Correlationsfactor Correlationsfactor Correlations

To use Pitzer correlation we must calculate the reduced temperature which is

9755.05.305

15.27325 =

+==

cr T

TT

The correlation is

r

r

c

c

TP

RTBPZ

+=1 (A)

we also know that

mRTVMwPP

nRTPVZ cr==

rr PPZ 177181.0

)2515.273)(14.83)(000,10()30)(000,30)(8.48( =+

= (B)



with

10 BBRTBP

c

c ω+=

The acentric factor for ethane can be obtained from the tables provided with properties for pure

components.

( ) 3561.09755.0

422.0083.0422.0083.0 6.16.10 −=−=−=

rTB

( ) 0519.09755.00172139.00172139.0 2.42.4

1 −=−=−=rT

B

( ) 3612.00519.0098.03561.010 −=−×+−=ω+=

BB

RTBP

c

c

and

−=

+=

9755.03612.011 r

r

r

c

c PTP

RTBPZ

Combining the 2 equations A and B, we have

rPZ 37027.01−=

and

rPZ 177181.0=



Equating these two equations (A) & (B) we solve for the reduced pressure, and then for the

pressure P.

bar...PPP.P

cr

r

12898488263182631

=×===

There is a substantial difference from the ideal gas model!

(c)(c)(c)(c), (d) (d) (d) (d) and (f)(f)(f)(f) are part of your homework assignment # 2homework assignment # 2homework assignment # 2homework assignment # 2.

Extension of Corresponding States to Mixtures We, engineers, would love to stretch the corresponding states principle to mixtures and

we do.

Z factor charts (all built from EOS) are also used for multicomponent systems in this

case the coordinates used are �pseudo-reduced properties�. You can use the same

charts for a pure component.

For mixtures the same type of charts apply but using �pseudoreduced properties� which

are defined similarly as the ratio of pressure (or temperature) with �pseudoreduced

critical pressure" (or temperature). These pseudocritical properties are an average of

the critical properties of the components in the mixture. Charts for mixtures can also be

used for single component fluids.

A typical chart using an EOS is shown in Figure 4.



Figure 4- Compressibility factor Z as a function or pseudoreduced pressure.

The same models are used to determine the gas compressibility factor for mixtures. The

extension is through some �mixing rules�

The accuracy will depend largely from model used and information input to the model.



Pseudocritical Properties of Natural Gases

Pseudoreduced Pressure

pcpr P

PP = (27)

Pseudoreduced Temperature

pcpr T

TT = (28)

If only the specific gravity of the gases is known then charts are available to estimate

these pseudocritical properties (undergraduate material, review McCain).

Naturally the degree of accuracy is reduced substantially. We well see methods when

compositional information is available, in this case:

( )cii

N

ipc PyP

c

Σ=

=1

(29)

( )cii

N

ipc TyT

c

Σ=

=1

(30)

Once Z is evaluated you can find the gas density as

( )3/ ftlbmVM

g =ρ (31)

gnMwM =

(32)



(Mwg is the average molecular weight of the gas evaluated as�)

( )∑=

=Nc

iiig MwyMw

1 (33)

and the total volume (extensive property is)

PZnRTV =

(34)

Therefore,

ZRTPMw

VM g

g ==ρ

(35)

So far we were just determining properties either for a �gas� or a highly compressed

fluid (liquid like density) in the SINGLE PHASE REGION.

Notice that Z- factor charts DO NOT HELP AT ALL IN DETERMINING

PROPERTIES OF GAS AND LIQUID COEXISTING PHASES

Figure 5 shows a chart of the compressibility factor for low reduced pressures.



Figure 5- Compressibility factor chart for low reduced pressures.

There is an undefined region that corresponds to the two-phase region.



Phase Equilibrium and the Phase Rule Equilibrium indicates static conditions, the absence of change. In thermodynamics is

taken no mean not only the absence of change, but the absence of any tendency to

change. Therefore a system that is in equilibrium is one in which under such conditions

that there is no tendency for a change to state to occur.

Tendencies toward a change are caused by a driving force of any kind, the absence of

such a tendency indicates also the absence of any driving force, or that all forces are in

exact balance.

Typical driving forces include mechanical forces such as pressure on a piston tend to

cause energy transfer as work; temperature differences tend to cause the flow of heat;

chemical potentials tend to cause mass transfer from one phase to another or cause

substances to react chemically.

In reservoir engineering applications we assume that reservoir fluids are at equilibrium,

we do not say how long the equilibrium will last. Therefore as a reservoir block

changes pressure due to production (injection) we assume that equilibrium is reached

instantly. Fluid properties in reservoir cells are evaluated using a sequence of

connected equilibrium stages.

The Phase Rule As mentioned earlier, the state of a system is determined when all intensive properties

are defined. The intensive properties are related, for example for a single component in the single phase region providing P and T is enough to define the state of the system,

since the molar volume and the compressibility can be evaluated as a function of these

two variables.



For multicomponent systems, we need to find out what is the minimum number of

properties (variables) is required to define the state of the system. The phase rule

provides the answer. Let�s begin by describing different cases.

Single component – Single Phase To define the state of the system need to provide two coordinates. The number of

independent variables is two, and these are usually pressure and temperature. The

molar volume is another variable, but that is not independent, because a relationship

exists between P, T, and V.

We have two degrees of freedom. Once a pressure and a temperature are selected, the

state of our single component system is defined (i.e. all intensive properties).

Single component – Two Phases Assume the system exhibits VLE (vapor-liquid-equilibrium). Here we just need to define

either the saturation pressure or the saturation temperature. Only one variable is

needed to specify the state of the system. We have one degree of freedom. These two

are related through the vapor pressure equation.

Single component – Three Phases Here we have a unique point in space named the triple point. The system is an

invariant. We cannot specify any variable. We have zero degrees of freedom.

Two components – Single Phase To define the state of the system need to provide three coordinates usually (P ,T & z1).

We have three degrees of freedom. The state of our binary system is defined (i.e. all

intensive properties).



Two components – Two Phase Assume the system exhibits VLE (vapor-liquid-equilibrium). Here we just need to define two variables to define the state of the system The choices could be (P,y1) or (P,x1),or,

(T,x1), or (T,y1). Note that the overall composition is not phase rule variables when more

than a phase are present. We have three degrees of freedom. Following this reasoning,

Table 1 shows the degrees of freedom, or number of independent variables required to

define the system, of different non-reacting components and number of phases

Number of components Number of phases Degrees of Freedom

1 1 2

1 2 1

1 3 0

2 1 3

2 2 2

2 3 1

� � �

Nc Np (Nc-Np)+2

Table 1 - Generalization of the phase rule for Nc non reacting components.

Thus for non-reacting systems

2+−= pc NNF (36)

This is the so called phase rule presented by an American mathematical physicist, J.

Willard Gibbs (1839-1903).



The number of independent variables that must be arbitrarily fixed to establish the intensive state of any system is called the degrees of freedom F of the system. Another

way to derive this equation (phase rule) in a more general way is:

F = # of variables � # of Independent equations relating these variables

These independent equations are: an EOS (equation of state), material constraints

(sum of mole fractions = 1), chemical reactions.

Note that the phase rule does not tell which variables to chose, it only says how many

and these must be independent. The choice of independent variables depends upon the

type of model available and the simplicity, or complexity of the calculations.

Thermodynamic Properties of Homogeneous and Heterogeneous Systems The objective of this section is to present the most widely used thermodynamic

properties of homogeneous and heterogeneous systems. These properties are

functions of primary variables such as pressure, temperature, molar volume, and

compositions (the last for multicomponent systems). These properties are based upon

the first and the second law of thermodynamics and are used to evaluate energy

requirements for a variety of processes and to derive models to evaluate phase

equilibrium.

First Law and Fundamental Thermodynamic Relationships Closed Systems

The system does not exchange matter with the surroundings, but it can exchange

energy.



The first law is a generalization of the conservation of energy and can be defined by the

following equation

dWdQdUt −= (37)

where dUt is the change in internal energy as a result of dQ, heat absorbed (or

released) by the system, and dW, work done (or provided) by the system on the

surroundings. By convention work done on the system is negative, and heat released by

the system is positive. Figure 6 indicates this convention, a gas contained in a vessel

with a movable piston. Compressing the gas (-) will cause the system to increase its

temperature and heat will be released by the system (+). The opposite process involving

gas expansion has the opposite signs for work and heat.

- dW

+ dQ

Compression

+ dW

- dQ

Expansion

- dW

+ dQ

Compression

+ dW

- dQ

Expansion



Figure 6 - Compression and expansion work in a gas container indicating the

convention used for heat and work.

For a reversible process, dQ = TdSt thus,

dWTdSdU tt −= (38)

If the work of expansion or compression is the only kind of work allowed then:

tPdVdW = (39)

Replacing Eq. (39) into Eq. (38)

ttt PdVTdSdU −= (40)

Thus

( )tttt VSUU ,= (41)

Equation (40) applies to any process in a closed PVT system that results in a

differential change from one equilibrium state to another.

Equation (40) could also be written as:



( ) ( ) ( )nVPdnSTdnUd −= (42)

again this equation applies to ANY CLOSED single or multicomponent system.

Equations (40) or (42) are exact differentials therefore one can identify,

( )( ) TnSnU

nnV

=

∂∂

,

and ( )( ) PnVnU

nnS

−=

∂∂

,

(43)

The primary thermodynamic properties are internal energy, volume and entropy (Ut, Vt,

St , respectively). For convenience a set alternate thermodynamic properties are

defined. These are the Entalphy, the Gibbs's energy and the Helmholtz energy (Ht, Gt,

At, respectively). The enthalpy is used for flow processes while the Gibbs�s energy is

used for equilibrium computations. The Helmholtz energy does not have a lot of use in

Petroleum engineering type calculations.

By definition:

Mt, = nM with M = U, H, A, G, S (A also known as F in European notation). The

relationship among these properties is:

ttt PVUH += (44)

tttttt TSUTSPVHF −=−−= (45)

and

ttt TSHG −= (46)



The same relationship holds for the intensive properties (M = Mt /n)

Expressions similar to Eq. (40) can be derived for equations (44) to (46).

( )PSHHdPVTdSdH tttttt , i.e. =→+= (47)

( )tttttt VTFFPdVdTSdF , i.e. =→−−= (48)

(((( ))))TPGGdP VdTSdG ttttt , i.e. ====→→→→++++−−−−==== (49)

The (Ut ,Ht ,Ft ,Gt ,St ) are STATE properties which means independent of path.

Open Systems For an open system, the basic thermodynamic functions Ut, Ht, Ft, and Gt in addition to

the two independent variables outlined above, will also depend on the concentration of

each of the components.

The number of moles of each specie may change due to:

• Chemical reaction within system

• Interchange of matter with surroundings

• Interchange and chemical reaction.

In this course we will not consider chemical reactions. However; the treatment for these

is similar.



The functional form of Ut, Ht, Ft, and G t for open systems are,

( )cNtttt nnnVSUU ...,,,, 21= (50)

( )cNttt nnnPSHH ...,,,, 21= (51)

( )cNttt nnnVTFF ...,,,, 21= (52)

( )cNtt nnnPTGG ...,,,, 21= (53)

The differential form of the above equations are,

iinVS

N

i i

tttt dn

nUPdVTdSdU

jtt

c

≠=∑

∂∂+−=

,,1

(54)

iinPS

N

i i

tttt dn

nHdPVTdSdH

jt

c

≠=∑

∂∂++=

,,1

(55)

iinVT

N

i i

tttt dn

nFPdVdTSdF

jt

c

≠=∑

∂∂+−−=

,,1

(56)

iinPT

N

i i

tttt dn

nGdPVdTSdG

j

c

≠=∑

∂∂++−=

,,1

(57)

let the chemical potential of component " i " , be defined by

inPTi

t

inVTi

t

inPSi

t

inVSi

ti

jjtjtjttnG

nF

nH

nU

≠≠≠≠

∂∂=

∂∂=

∂∂=

∂∂=

,,,,,,,,

µ̂ (58)



The chemical potential was introduced first by Gibbs.

From all the alternative expressions in Eq. (58) for the chemical potential the last is the

most useful for phase equilibrium computations. The reason is that the independent variables (P, and T) are readily measured.

Second Law and the Equilibrium Criteria One form to state the second law is that for an isolated system, all real processes occur

with a zero or positive entropy change. Figure 7. shows the entropy evolution with time

and Eq, (59) puts the above statement in mathematical form.

0≥tdS (59)

In Eq. (59) dSt is positive for an irreversible process and zero for a reversible process.



Equilibrium

dS = 0

Time

Entro

py, S

Figure 7 - Entropy versus time for any physical process.

Figure 7 shows the trend to equilibrium. It then follows that for an isolated system to be

at equilibrium, the entropy must have reached the maximum value. Therefore, at

equilibrium,

0=tdS (60)

subject to the constraints

0=tdU (61)



0=tdV (62)

cj ,...N,idn 21 0 == (63)

Note that if the system is composed of several phases (and it is a

heterogeneous system) St ,Ut ,Vt , and ni in Eqs. (54) to (57) are the

summation over the values in all parts or phases.

The criteria of equilibrium of a system can also be stated in terms of Ut, Ht, Ft, and Gt as follows • The internal energy, Ut , must be a minimum at constant St, Vt, and ni.

• The enthalpy, Ht, must be a minimum at constant St, P, and ni.

• The Helmholtz free energy, Ft , must be a minimum at constant T, Vt, and ni.

• The Gibbs free energy, Gt, must be a minimum at constant T, P, and ni.

Therefore the equilibrium problem is evaluated by minimizing either one of these

thermodynamic functions. We require a thermodynamic model to evaluate these

functions and EQUATIONS OF STATE are these models.

The choice of the function will depend upon the selection for dependent/independent

variables. The most popular one is the Gibbs�s energy because of its natural dependent

variables. It�s the easier from a computational view point.

Chemical and Phase Equilibria Criteria for an Open System Using Intensive Properties Consider a closed PVT system consisting of two phases in equilibrium. Each phase

may be considered a single-phase OPEN system.



VaporPv

Tv

niv

LiquidPl

Tl

nil

VaporPv

Tv

niv

LiquidPl

Tl

nil

Figure 8 - Two phases in equilibrium.

The differential equation for the internal energy applied to each phase is,

( ) ( ) ( ) vi

N

i

vi

vvv dnnVPdnSTdnUdc

∑=

µ+−=1

ˆ (64)

( ) ( ) ( ) li

N

i

li

lll dnnVPdnSTdnUdc

∑=

µ+−=1

ˆ (65)

And the total energy is the sum of Eqs. (25) and (26).

( ) ( ) ( ) li

N

i

li

vi

N

i

vi dndnnVPdnSTdnUd

cc

∑∑==

µ+µ+−=11

ˆˆ (66)



Comparing Eq. (67) with Eq. (42) for a closed system we must have that at equilibrium;

0ˆˆ11

=+∑∑==

li

N

i

li

vi

N

i

vi dndn

cc

µµ (67)

but from mass conservation;

li

vi dndn −= (68)

Thus

0ˆˆ11

=−∑∑==

vi

N

i

li

vi

N

i

vi dndn

cc

µµ (69)

replacing Eq. (58) into Eq. (59),

( ) 0ˆˆ1

=µ−µ∑=

vi

li

N

i

vi dn

c

(70)

For more than two phases in equilibrium, successive applications of Eqs. (64) and (65)

lead to.

ciiii Ni ,...2,1 ˆ...ˆˆˆ =µ==µ=µ=µ πδβα

(71)



Where the phases could be two liquids, solid-liquid, vapor-solid-liquid etc.

Additional primary thermodynamic functions are the mole fractions defined as,

and

11

l

li

N

i

li

li

iv

vi

N

i

vi

vi

i nn

n

nxnn

n

nycc

====

∑∑==

(72)

conventionally "vapor" and "liquid" molar compositions of component "i" are denoted as

"xi" and "yi" respectively.

Now rewrite Eq. (65) using (72). (We could have chosen a gas phase as well).

( ) ( ) ( ) li

N

i

li

lll ndxnVPdnSTdnUdc

∑=

µ+−=1

ˆ (73)

by using the chain rule for differentiation, expanding and collecting terms, we obtain:

0ˆˆ11

=

µ−+−+

µ−+− ∑∑

==

li

N

i

li

lllli

N

i

li

lll dnxPVTSUndxPdVTdSdUcc

(74)

Since nl and dnl are arbitrary, the terms inside the brackets must both be zero, and this

provides the following identities in terms of intensive properties instead of total

properties.

i

N

i

li

lll dxPdVTdSdUc

∑=

µ+−=1

ˆ (75)



and

i

N

i

li

lll xPVTSUc

∑=

µ+−=1

ˆ (76)

Similar expressions are obtained with the gas phase.

Recall that for a constant composition fluid

( )ttttttt VSUUPdVTdSdU , i.e. =→−= (77)

( )PSHHdPVTdSdH tttttt , i.e. =→+= (78)

( )tttttt VTFFPdVdTSdF , i.e. =→−−= (79)

( )TPGGdP VdTSdG ttttt , i.e. =→−−= (80)

The (Ut ,Ht ,Ft ,Gt ,St ) are STATE properties which means independent of path. These

sets of equations are exact differential expressions.

For example let�s take the last set

and tT

tt

P

t VPGS

TG =

∂∂−=

∂∂ (81)

From math relations we can see that cross derivatives are the same regardless of the

order of differentiation.



12

2

21

2

xxy

xxy

∂∂∂=

∂∂∂ (82)

therefore

TPG

PTG tt

∂∂∂=

∂∂∂ 22

(83)

That means

P

t

T

t

TV

PS

∂∂=

∂∂− (84)

These identities are very handy and are used frequently in deriving suitable expressions

for a variety of processes (heat transfer, evaluation of energy requirements in flow

processes, phase equilibria, etc).

These identities are called the �The Maxwell Equations�.

Some Mathematical Relations of Thermodynamics The next part of this handout will give you a flavor of some of the mathematical

manipulations that will be used in many of the mathematical derivations that will follow

in the course.

Appendix A from Van Ness and Abbot (Classical Thermodynamics of Non Electrolyte

Solutions, Mc Graw Hill is also a helper for some of these math rules



In the derivation of a thermodynamic identity, you start with one or more equations (i.e.,

a fundamental eq. or definition) and apply the following mathematical relations:

Maxwell Equations.

P

t

T

t

TV

PS

∂∂=

∂∂− (85)

tVTt

t

TP

VS

∂∂=

∂∂ (86)

tt VtSt SP

VT

∂∂−=

∂∂ (87)

and

Pt

t

S SV

PT

t

∂∂=

∂∂ (88)

The Maxwell identities can be written also for constant composition fluids, in this case

we replace Mt by M, with M any of the functions in Eq.(85) to (88).

Minus One Rule If P (T, V) then:

(∂P/∂T)v (∂T/∂V)p (∂V/∂P)T = - 1

If S(T, V) then

(∂S/∂T)v (∂T/∂V)s (∂V/∂S)T = -1

or

(CvT) (∂T/∂V)s = - (∂S/∂V)T = - (∂P/∂T)v



Straight Chain Rule

(∂P/∂T)v (∂T/∂S)v = (∂P/∂S)v

(the partial differentials may be cancelled but always the same property must be held

constant)

Change of Variable Rule if P (U, V)

dP = (∂P/∂U)v dU + (∂P/∂V)U dV

or

(∂P/∂V)T = (∂P/∂U)V (∂U/∂V)T + (∂P/∂V)U

You can use this rule to switch from (∂P/∂V)U to (∂P/∂V)T.

Inversion Rules First derivatives can be inverted:

[1/(∂P/∂T)V] = (∂T/∂P)V

Second derivatives cannot be directly inverted:

[1/(∂ 2P/∂T2)V] ≠ (∂ 2T/∂P2)V

rather

(∂ 2P/∂T2)V (∂T/∂P)2V = - (∂ 2T/∂P2)V

Integration by Parts.

( )[ ] ∫∫ −=2

1

2

1

21 VdPPVPdV



Measurable Thermodynamics Quantities & Their Definitions

1. Ci = (∂Q/∂T)i

Isobaric Heat Capacity

Cp = (∂Q/∂T)p = T (∂S/∂T)p = (∂H/∂T)p

Isochoric Heat Capacity

Cv = (∂Q/∂T)v = T(∂S/∂T)v= (∂U/∂T)v

Saturated Heat Capacity

Cσ = (∂Q/∂T)σ = T(∂S/∂T)σ = (dH/dT)σ - Vσ (dPσ/dT)

2. Compressibilities

Isobaric Compressibility

β = V-1 (∂V/∂T)p = - ρ-1 (∂ρ-/∂T)p

Isothermal Compressibility

KT = V-1 (∂V/∂P)T = ρ-1 (∂ρ/∂P)T

Isentropic Compressibility

Ks = V-1 (∂V/∂P)s = ρ-1 (∂ρ/∂P)s = (ρU2s)-1, where Us = sonic velocity



3. Joule Thomson and Isothermal Throtling coefficients

(∂H/∂T)p (∂T/∂P)H (∂P/∂H)T = - 1,

where

Cp = (∂H/∂T)p,

µJT = (∂T/∂P)H the Joule Thomson coefficient

ΘT = (∂H/∂P)T, the isothermal throttling coefficient.

4. More Relations

Cp - Cv = TVβ 2/KT = TVγ 2v/KT,

γV = (∂P/∂T)V , the isochoric slope.

(∂Cp/∂P)T = - T(∂ 2V/∂T2)p

and (∂Cv/∂V) = Tγ 2v.Ks - KT = - VTβ 2/Cp.

For a perfect gas (P.G)

U2s = RTγo, γo = (Cpo/Cvo)



Phase Equilibria Models There are many models to describe fluid phase equilibria these could be classified

according to the type of fluids (hydrocarbons, alcohols, electrolytes, water and other

non-hydrocarbon species), and to the pressure and temperature ranges of interest. In

this session we will describe models that are used in petroleum engineering

applications. These models are for low-pressure ranges, such as those of separator and

surface conditions and models for high pressures which apply to the reservoir. The type

of reservoir fluid, whether a black oil or a volatile oil, also determines the type of model

that can be used.

We will start with the simpler models first, the ones for lower pressures.

Residual Properties To derive the phase equilibria models we define the residual properties for

mathematical convenience as the difference between the actual (real) property minus

the same property, evaluated at the same pressure, temperature, and composition, but

evaluated using the ideal gas equation. That is

MR = M-Mig M=U, H, G, S, F (F is A in American Notation)

M: Real Property @ (T, P) of the system

MR: Residual Property

Mig: Property @ (T, P) of the system evaluated as if the fluid were an ideal gas

Note: there is no TR or PR



Recall for a constant composition closed system

SdTVdPdG −= (89)

dTSdPVdG igigig −=− (90)

dTSdPVdG RRR −= (91)

Note that the properties used in these equations are intensive properties, that is the

volume is the molar volume, G and S are expressed in BTU/lb-mol and BTU/lb-mol-R,

respectively, (or in cal/g-mol, cal/g-mol K in the SI system of units).

At constant temperature,

dPVdG RR = (92)

we divide by RT

∫∫ =→=P RG RRR

dPRTV

RTdGdP

RTV

RTdG

R

00

(93)

From previous lectures we had:

1 , ==RT

PVzRTPV ig

(94)

Thus,



Pz

RTV R )1( −= (95)

and,

( )∫=PR

PdPz-

RTG

0

1 (96)

Phase Equilibrium of a Single Component We will start deriving the general expressions for phase equilibrium of a pure

component.

Recall

lvlll

vvv GGdTdP-S V dGdTdP-S V dG

=⇒==

(97)

We also know that,

vlvl

lll

vvv

S-TH

-TS H- G -TS HG

∆∆=

==

0

(98)

At saturation, the pressure and temperature of the liquid and gas phases are the same



ldGdTSdPVdTSdPVdG llvvv =−=−= σσσσ (99)

Thus,

vl

vl

lv

lv

VTH

VVSS

dTdP

∆∆=

−−=σ

σ

(100)

This is nothing else but the Clapeyron Equation seen without derivation previously.

Reduced Equations of State and Maxwell Equal Area Rule If states 1 and 5 in Figure 9 are in equilibrium, then

51 GG = (101)



-100

0

100

200

300

400

500

600

700

2 4 6 8 10

12

14

A1

A2

Pres

sure

Molar Volume

Tc

T2

T1P1

v

L

2 - Phases

CP

V

LV

1

2

34

7 6

5

0>

∂∂

TV~P

-100

0

100

200

300

400

500

600

700

2 4 6 8 10

12

14

A1

A2

Pres

sure

-100

0

100

200

300

400

500

600

700

2 4 6 8 10

12

14

A1

A2

Pres

sure

Molar Volume

Tc

T2

T1P1

v

L

2 - Phases

CP

V

LV

1

2

34

7 6

5

0>

∂∂

TV~P

Figure 9 - Predicted isotherms from a cubic EOS.

Or

∫ =5

1

0dG (102)

For constant temperature,

0)( =−== PdVPVdVdPdG (103)



∫ ∫ =−−=5

1

5

11155 0 dVPVPVPdG (104)

At equilibrium P1=P5=Pσ

∫ =−−σ5

115 0 )( dVPVVP (105)

By inspection,

)176531()( 15 −−−−−=−σ AreaVVP (106)

And also,

)17654321(5

1

−−−−−−−=∫ AreaPdV (107)

)3543()1231()176531()17654321(

−−−+−−−−−−−−−=−−−−−−−

AreaAreaAreaArea

(108)

Combining Equations (106), (107) and (108), we get

)3543()1231( −−−=−−− AreaArea (109)



The liquid and vapor saturated volumes lVVV == 17 and gVVV == 65 are at a specified

isotherm and at pσ.

VLE in Dimensionless or Reduced Form

Write the EOS in dimensionless form using Tr=T/Tc, Pr=P/Pc, Vr=V/Vc, and the values for

a and b found from the critical constraints

0 ,0 2

2

=

∂∂=

∂∂

cTcT VP

VP (110)

For Van der Waals EOS

2Va

bVRTP −−

= (111)

with

ccVRTa89= (112)

c

cc

PRTVb83

== (113)

c

ccc RT

VPz ==83 (114)



c

cc P

RTV83= (115)

2289

3rc

cc

ccr

crcr VV

VRTVVV

TRTPP −−

= (116)

After rearranging and simplifying

23

138

rr

rr VV

TP −−

= (117)

Equation (117) applies to the gas and liquid. For the liquid, we will use a reduced

volume Vrl=Vl/Vc, and for the gas Vrg=Vg/Vc.

Apply equal area rule

0=dG (118)

At T constant

0 )( =−== dVPPVdVdPdG (119)

or

( ) ∫=−σrg

rl

V

Vrrrlrgr dVPVVP (120)



Replace Equation (117) in the integrand of Equation (120) and integrate

( )

−+

−−

=−σ

rlrgrl

rgrrlrgr VVV

VTVVP 331313

ln3

8 (121)

Since P is constant, 0=dP

Thus, ∫ = 0 dPV

0 r

=

∂∂= ∫∫

rg

rl r

rg

rl

V

Vr

T

rr

V

Vrr dV

VPVdPV

∫

+=

rg

rl

V

Vr

rrr

dVV)-V(

Tr-V 613

24 32 (122)

From integral tables,

+++=

+∫ bxaabxa

bbxaxdx )ln(1

)( 22 (123)

etc.,

01149

131

131

1313

ln =

−−

−

+

−

−

−−

rgrlrrlrgrg

rg

VVTVVVV

(124)



So, we have three equations to work with,

EOS (117)

Maxwell Equal Area (121) ! unknowns Prσ, Vrl, Vrg.

and ∫ = 0 dPV (124)

In density form

=ρ

rr V

1

( )( ) 033

1833 =

ρ−ρ−

−ρρ−σ

rlrgrrlrgr T

p (125)

combining ∫ = 0 dPV and Maxwell equal area

( )( ) 0)(33

8 =ρ+ρ−ρ−ρ− rgrl

rlrg

rT (126)

EOS for liquid and gas.

And Eq. (124) in density form

033

)(49

33ln =

ρ−

ρ−

ρ−

ρ+ρ−ρ+

ρρ

ρ−ρ−

rl

rl

rg

rgrgrl

rrl

rg

rl

rl

T (127)



Change of Variables

rlrgs ρ+ρ=

rgrlw ρ−ρ=

2ws

rl+=ρ

2ws

rg−=ρ

Write Equations (125), (126) and (127) in terms of s and w.

Advantage 1: Good starting point! At the critical point s = 2 and w = 0. VLE calculation

starts from the critical point and the algorithm steps down in temperature.

Tr =1 ! solution - s = 2 and w = 0

0.99

0.98

0.97

�.

0.4

Keep lowering the Tr until it gets close to the triple point, which in reduced coordinates is

pretty close to 0.4. For each Tr, you will obtain s and w, which will give provide the

densities for the gas and the liquid.

Advantage 2: Reduced form makes it universal. VLE calculated only once.



Disadvantage: Algebraic manipulations.

AssignmentAssignmentAssignmentAssignment

Read paper “Generalized Saturation Properties of Pure Fluids via Cubic Equations of State”

by Barrufet & Eubank – Chemical Engineering Education – Summer 1989. You can access

this paper through the link in your Calendar of Events in the WebCT site.

• One derivation will be assigned following this paper plus these notes.

Systems of Variable Composition - Mixtures We will start with the simplified assumption that the gas phase behaves as an Ideal Gas

and the liquid phase exhibits Ideal Solution Behavior.

We have seen in previous lectures that the equilibrium criteria between 2 phases α and

β was,

cii Ni

TTPP

,....2,1,ˆˆ =µ=µ

=

=

βα

βα

βα

(128)

where α and β could be: vapor, liquid 1, liquid 2, solid 1, solid 2, etc.

We will mainly deal with vapor and liquid equilibria: VLE. Thus, at constant T and P,



∑

∑

=

=

µ=

µ=

c

c

n

i

li

li

l

n

i

vi

vi

v

dnnGd

dnnGd

1

1

ˆ)(

ˆ)( (129)

The simplest model is to assume that the gas phase behaves as an ideal gas

(IG), and that the liquid phase behaves as an ideal solution (IS). The assumptions imply

that

IG: molecular interactions are zero, molecules have no volume.

IS: forces of attraction/repulsion between molecules are the same regardless of

molecular species. Volumes are additive (Amagat�s Law).

A A B B A B

Figure 10 - Forces between molecular species.

ABBBAA FFF ==

Ideal Gas Mixture



P pk

T1 T1

n1,n2, nk

…,nk

Figure 11 - Comparison of the pressure in a constant volume vessel between a mixture

of gases and a pure component.

The pressure in a vessel containing an ideal gas mixture (n) or a single gas component

(nk) is

t

kk

t

VRTnP

VnRTP

=

= (130)

kkk y

nn

PP == (131)

Pk is the partial pressure of component k, and by definition, for ideal behavior

∑=

=cN

ik PP

1

(132)

Same vessel volume



Equation (132) can be generalized to any thermodynamic property for an ideal gas

mixture.

∑=

=cn

kk

igkk

ig PTMnPTnM1

),(),( (133)

or,

�A total thermodynamic property (nU, nG, nS, nH, nF) of an ideal gas

mixture is the Σ of the total properties of the individual species each

evaluated at the T of the mixture and at its own partial pressure.�

To derive the equilibrium relations we are interested in using equation (133) for Hig and

Sig because,

igigig TSHG −= (134)

For an ideal gas, the enthalpy is independent of pressure, thus,

),(),( kig

kig

k PTHPTH = (135)

For the entropy, we must express ),( PTSS igig =

dTTSdP

pSdS

PT

∂∂+

∂∂= (136)



Recall Maxwell Rules

dTTc

dPTVdS p

P

+

∂∂−= (137)

For ideal gas,

dTT

cdP

PRdS

igpig

k +−= (138)

at constant temperature,

∫∫ −=

−=−=

P

P

P

P

igk

igk

kk

PdRdS

PRddPPRdS

ln

ln (139)

kk

kig

kig

k yRPPRPTSPTS lnln),(),( −=

−=− (140)

We also know from equation (133) applied to entropy,

∑=

=cN

kk

igkk

ig PTSnPTnS1

),(),( (141)

or



∑=

=cN

kk

igkk

ig PTSyPTnS1

),(),( (142)

Substitute ),( kig

k PTS from equation (140) into equation (142) after Σ over the

components,

∑∑==

−=cc N

kkk

N

kk

igkk

ig yyRPTSyPTS11

ln),(),( (143)

which means that the entropy change of mixing the ideal gases is not zero, and it is

greater than zero,

∑∑==

>=−cc N

k kk

N

k

igkk

ig

yyRSyS

11

01ln (144)

Now, we can build the expression for the Gibbs energy using equations (135) and (144).

∑∑∑===

+−=ccc N

kkk

N

k

igkk

N

k

igkk

ig yyRTPTSyTPTHyPTG111

ln),(),(),( (145)

Remember that expression (145) is for an ideal gas.

Now, recall the expression for the chemical potential,



ijnPTii n

nG

≠

∂∂=µ

,,

ˆ (146)

Using equation (145) expressed in terms of n (yk=nk/n),

∑∑==

+=cc N

k

kkN

k

igk

kig

nn

nnRTG

nnPTG

11

ln),( (147)

−+= ∑ ∑∑

= ==

c cc N

i

n

ikkk

N

i

igkk

ig nnnnRTGnPTnG1 11

ln)(ln),( (148)

Recall,

∑=

=cN

iinn

1

(149)

kjnn

kjnnnn

j

k

j

k

k

==∂∂

≠=∂∂

=∂∂

,1

,0

1

(150)

Therefore,

+−−+=

∂

∂=µ ∑≠

nnn

nnnRTG

nnG k

i

ii

igi

npTi

igig

i

ij

lnlnˆ,,

(151)



iig

iig

i yRTG lnˆ +=µ (152)

Keep in mind equation(152) since it will be used later in the development of Raoult�s

Law for phase equilibria of IG+IS.

So far, we have worked with an ideal mixture of gases. Now, we will work with an ideal

solution of liquids.

Ideal Solution We have seen that Amagat�s Law is followed.

∑= iiid VxV @ same T and P of mixture (153)

Following the same reasoning as for gases, we have that,

∑∑ −= iiiid xRVxS (154)

iiiiid xxRTGxG ln∑∑ += (155)

iiid

i RTxG +=µ̂ (156)

Here, Si and Gi are the properties of the pure species in the liquid state at the T and P of

the mixture.



Raoult’s Law It is a combination of IG + IS models. If we consider VLE for a mixture made up of Nc

components

cidl

iigv

il

iv

i Ni ,...,1;)ˆ()ˆ(ˆˆ =µ=µ→µ=µ (157)

Thus, at T and P,

il

iiig

i xRTGyRTG lnln +=+ (158)

),(),(ln PTGPTGxyRT ig

il

ii

i −= (159)

The right hand side of Eq. (159) indicates pure species properties evaluated at the

equilibrium T and P of the mixture

�More simplifications

Assume negligible effect of pressure on Gil, (nearly incompressible fluid far away from

the critical point).

),(),( σ≅ il

il

i PTGPTG (160)

where Piσ is the pure species saturation pressure at T.

For an ideal gas, we have,



∫ ∫∫σ σσ

== i ii P

P

P

P

igi

P

P

igi dP

PRTdPVdG @ T constant (161)

PPRTPTGPTG iig

iiig

i

σσ =− ln),(),( (162)

Now, combine Eqs. (159) to (162),

PPRTPTGPTG

xyRT i

iig

iil

ii

iσ

σσ +−= ln),(),(ln (163)

As we seen before for a pure component,

0),(),( ====−−−− σσσσσσσσi

igii

li PTGPTG (164)

So, Eq. (163) leads to Eq (165), which is known as Raoult’s Law.

σ= iii PxPy (165)

This equation will be used (as a warm-up) to evaluate all the phase equilibrium

computations (dew point, bubble point, and flashes).

Equation (165)must hold for all species if equilibrium exists. Its validity is for low

pressures (P < 100 psia) and high temperatures (T > 70ºF).



Types of Calculations Given Variables

(independent)

Unknown Variables

(dependent) Problem Type

Example Application

P, zi = xi T, yi Bubble Point

T, zi = xi P,yi Bubble Point Gas injection,

production

P, zi = yi T,xi Dew Point

T, zi = yi P,yi Dew Point

Gas

Condensates,

Production

P, T, zi xi, yi, fv Flash Production

Separation

Bubble Point Evaluation given T, zi and σiP

We need a model for each Piσ as a function of temperature.

ii xz = (166)

Find BP pressure and equilibrium gas compositions

σ

σ

=

=

iii

iii

PzPy

PxPyor (167)



The bubble point pressure at a given T is

∑∑ σ= iibpi PzPy (168)

∑ σ= iibp PzP (169)

Under Raoult�s law, the bubble point has a linear dependence with the vapor pressures

of the pure components.

Once the bubble point pressure is found, the equilibrium vapor compositions are found

from Eq. (167).

The dew point curve (lower black curve) in Figure 12 is always curved regardless

whether the mixture is ideal or not.

P2σ

P1σT

x1,y1

Figure 12 - Deviations from Raoult's law.

The red curves in Figure 12 indicate deviations from Raoult's law. When the bubble

point curve is above the straight line, we will have positive deviations from Raoult's Law.



When the bubble point curve is below the straight line, we will have negative deviations

from Raoult's Law. This happens for non-ideal mixtures and may lead to azeotropy.

Dew Point Calculation given T, zi and σiP

At the dew point the overall fluid composition coincides with the gas composition. That

is.

ii yz = (170)

Find DP pressure and equilibrium liquid compositions

σ

σ

=

=

iii

iii

PxPz

PxPy (171)

∑∑ =σ Px

Pz i

i

i (172)

1

1

−

=σ

= ∑

cN

i i

idp P

zP (173)

Once Pdp is found, the equilibrium liquid compositions are found from Eq. (171).

Bubble Point Temperature Given P, zi and σiP as a function of T

Here we need to find the corresponding bubble point temperature which enters into the

equation indirectly and we must follow an iterative procedure.

ii xz = (174)



Find TB pressure and equilibrium gas compositions

σ

σ

=

=

iii

iii

PzPy

PxPyor (175)

The problem is that we do not know yet at what temperature to evaluate the pure

component vapor pressures. See the following diagram

∑ σ= )( bpiibp TPzP (176)

For well-behaved systems (no azeotropes), the searched temperature will be bounded

by the highest and lowest saturation temperature of the components in the mixture at

the selected system pressure, as shown in Figure 13.

T2σ

T1σ

P

x1,y1

P

Figure 13 - Binding of the searched temperature in a well-behaved system.

For demonstration purposes, let�s select a binary and a model such as Antoine�s for

vapor pressures.



Procedure

1. Evaluate σ1T and σ

2T at the given pressure P, which is a saturation pressure.

ii

iii cT

baP+

−= σσln " i

ii

ii c

PabT +

−= σ

σ

ln (177)

2. Choose your first guess bubble point temperature as

σ

=∑= i

n

ii

obp TzT

c

1

)( (178)

using the overall composition of the mixture and the saturation temperatures evaluated

in (177).

3. Define a relative volatility using a reference substance such that all relative volatilities

are either > 0 or < 0 (i.e., monotonically increasing or decreasing).

σ

σ

=αj

iij P

P (179)

with the saturation pressures evaluated at the guess temperature evaluated in (178).

4. Expand the volatility as

2

2

1

1212112 lnlnln

cTb

cTbaaPP

++

+−−=−=α σσ (180)

+

++

−−=α2

2

1

12112 exp

cTb

cTbaa (181)

with T from Eq. (177).



Then write the bubble point equation in terms of the volatilities and the reference vapor

pressure for the selected component.

For a binary, you would have only one volatility and this expression would be,

[ ]2121222

1122211 zzPz

PPzPPzPzP +α=

+=+= σ

σ

σσσσ (182)

Thus,

[ ]21212 zz

PP+α

=σ (183)

this is your first guessed saturation pressure for the reference component (here �2�) at

the first guessed temperature evaluated in (177).

From the saturation pressure evaluated in (182) use the Antoine equation to find a new

temperature (Eq. (177)).

This new T " new 12α " new σ2P " iterate until two successive temperatures do not

change by a specified tolerance.

The Excel file provided in our WEB site illustrates this procedure for a ternary mixture.

You can modify it and extend it to multicomponents.



Dew Point Temperature given P, zi and σiP as a function of T

You can follow a very similar reasoning as the one developed for the bubble point and

devise the algorithm required to solve this problem using relative volatilities.

Flash Calculations In this type of calculations, the work-horse of reservoir simulation packages,the

objective is to find fraction of vapor vaporized as well an equilibrium gas and liquid compositions given the overall mixture composition, P and T.

Start with the same equation as usual

σ= iii PxPy (184)

Material balance

( ) vivivilii fyfxfyfxz +−=+= 1 (185)

Now replace either liquid or gas compositions using equation (184) in equation (185)

( ) vivi

ii fyfPPyz +−= σ 1 (186)

( ) vvi

ii

ffPP

zy+−

=σ 1

(187)



( )∑∑

+−=

σ vvi

ii

ffPP

zy1

(188)

zi(T1,P1)

xi(T1,P2)

yi(T1,P2)

P1 > P2T1,P2

Figure 14 - Separation process.

Objective function (flash function) is

( )01

1)( =−

+−= ∑

σ vvi

iv

ffPP

zfF (189)

There are several equivalent expressions for the flash function

(a) 01 =−∑ iy (190)

(b) 01 =−∑ ix (191)



(c) 0=−∑∑ ii xy " the best well behaved for numerical solution (Rachford- Rice

function) (192)

Once fv is found the equilibrium gas and liquid compositions are evaluated from Eqs.

(189) and (186), respectively.

Systems of Variable Composition - Non-Ideal Behavior Here, we will lay the foundation for a general treatment of VLE, LLE, SLE, etc. by

introducing an auxiliary thermodynamic function related to the Gibbs�s energy, this is

called the fugacity coefficient.

We have seen that

inpTi

i GnnG

ij

≡

∂∂=µ

≠,,

ˆ Partial Molar Gibbs Energy

In general

ijnpTii n

nMM≠

∂

∂=,,

M=U, H, A, G, S, V

Definition of Fugacity and Fugacity Coefficient For a constant composition fluid at constant T

SdTVdPdG −= (193)



∫∫ =P

P

P

P

VdPdG**

(194)

So, ∞→→ VP ,0*

∫=−P

P

VdPPGPG*

)()( * (195)

Need a function that behaves better and is equivalent.

Recall for an IG @ T=constant

PRTdP

dPRTdPVdG igig ln=== (196)

In a similar fashion, we define,

fRTddG ln= (197)

Then,

φ==− lnln RTdPfRTddGdG ig with

Pf=φ (198)

φ= lnRTddGR (199)



Integration provides,

)(ln TCRTddGR +φ= (200)

Let�s complete the definition of the fugacity by setting the limit of the fugacity of an ideal

gas equal to its pressure,

0)(C and 01 ==⇒=φ⇒= TGPf Rig (201)

General Expressions

φ= lnRTGR

Applies to a gas mixture and �φ� is for mixture. (202)

Similarly,

Pf

RTG i

ii lnln =φ= pure component (203)

We also saw that,

∫ −=PR

PdPz

RTG

0

)1( (204)

∫ −=φP

PdPz

0

)1(ln (205)



Similarly, for a pure component,

∫ −=φP

ii PdPz

0

)1(ln zi = compressiblity factor of component I (206)

Redefining equilibrium for a pure species @ σ= iPP

∫∫ =v

i

li

f

fi

v

li fRTddG ln (207)

li

vil

ivi f

fRTGG ln=− (208)

using σ=φi

ii P

f (209)

li

vil

ivi RTGG

φφ=− ln (210)

So, at equilibrium,

li

vi φ=φ needs models to evaluate this expression (211)



Fugacity for species “i” in a Mixture Analogous to previous equations, we can derive a fugacity for mixtures from,

iigi

igi

iigi

igi

fRTddGd

PyRTddGdˆln)ˆ(

ln)ˆ(

=µ=

=µ= (212)

Integration provides,

ii fRT ˆlnˆ =µ (213)

For VLE,

li

vi ff ˆˆ = i = 1, 2,…,Nc (214)

where,

Pyf ivi

vi φ= ˆˆ (215)

Pxf ili

li φ= ˆˆ (216)

Thus, the General VLE problem becomes

ilii

vi xPyP φ=φ ˆˆ (217)



what results in,

ilii

vi xy φ=φ ˆˆ (218)

To find the fugacity coefficients there are several models available. For hydrocarbon

fluids we shall use EOS.

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Thermodynamics 2