Calculating i v from the Virial EOS

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CHEE 311 J.S. Parent 1 Calculating i v from the Virial EOS We have used the virial equation of state to calculate the fugacity and fugacity coefficient of pure, non-polar gases at moderate pressures. Under these conditions, it represents non-ideal PVT behaviour of pure gases quite accurately We will generalize the virial equation to allow the calculation of mixture fugacities. For mixtures, we use the truncated virial equation: (3.31) where B is a function of temperature and composition according to: (10.65) B ij characterizes binary interactions between i and j; B ij =B ji RT BP 1 Z i n 1 n 1 j i B y y B

description

Calculating  i v from the Virial EOS. We have used the virial equation of state to calculate the fugacity and fugacity coefficient of pure, non-polar gases at moderate pressures. Under these conditions, it represents non-ideal PVT behaviour of pure gases quite accurately - PowerPoint PPT Presentation

Transcript of Calculating i v from the Virial EOS

Page 1: Calculating   i v  from the Virial EOS

CHEE 311 J.S. Parent 1

Calculating iv from the Virial EOS

We have used the virial equation of state to calculate the fugacity and fugacity coefficient of pure, non-polar gases at moderate pressures.

Under these conditions, it represents non-ideal PVT behaviour of pure gases quite accurately

We will generalize the virial equation to allow the calculation of mixture fugacities.

For mixtures, we use the truncated virial equation:

(3.31)

where B is a function of temperature and composition according to:

(10.65)Bij characterizes binary interactions between i and j; Bij=Bji

RTBP

1Z

ij

n

1

n

1ji ByyB

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CHEE 311 J.S. Parent 2

Calculating iv from the Virial EOS

Pure component coefficients (B11, B22, etc) are calculated as previously (Equations 3.48, 3.50, 3.51), and cross coefficients are found from:

(10.70)

where,

and(10.71-75)

Bo and B1 for the binary pairs are calculated using the standard equations 3.50 and 3.51at Tr=T/Tcij.

)BB(P

RTB 1

ijo

cij

cijij

33/1cj

3/1ci

cijcjci

cij

cij

cijcijcijijcjcicij

jiij

2

VVV

2

ZZZ

V

RTZP)k1(TTT

2

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CHEE 311 J.S. Parent 3

Calculating iv from the Virial EOS

We now have an equation of state that represents non-ideal PVT behaviour of mixtures:

or

We are equipped to calculate mixture fugacity coefficients from equation 10.60

RTBP

n)nnZ(

RTBP

1Z

nj,Ti

P

0 nj,Ti

P

0 nj,P,Ti

vi

n)nB(

RTP

dPn

)nB(RT1

PdP

n)nnZ(ˆln

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CHEE 311 J.S. Parent 4

Calculating iv from the Virial EOS

The result of differentiation is:

(10.69)

with the auxilliary functions defined as:

In the binary case, we have

(10.67)

(10.68)

kkiiijij

kkiiikik

ni

1iij

nj

1jikjikk

vk

BBB2

BBB2

)2(yy21

BRTPˆln

22111212

122122

v2

122211

v1

BBB2

yBRTPˆln

yBRTPˆln

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CHEE 311 J.S. Parent 5

6. Calculating iv from the Virial EOS

Method for calculating mixture fugacity coefficients:1. For each component in the mixture, look up:

Tc, Pc, Vc, Zc,

2. For each component, calculate the virial coefficient, B

3. For each pair of components, calculate:Tcij, Pcij, Vcij, Zcij, ij

andusing Tcij, Pcij for Bo,B1

4. Calculate ik, ij and the fugacity coefficients from:

)BB(P

RTB 1o

c

c

)BB(P

RTB 1o

cij

cijij

ni

1iij

nj

1jikjikk

vk )2(yy

21

BRTPˆln

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CHEE 311 J.S. Parent 6

6. Ideal Liquid Solutions

We have already developed a model for the chemical potential of ideal solutions.

Acknowledges the fact that molecules have finite volume and strong interactions, but assumes that these interactions are the same for all components of the mixture.

This are the same assumptions used in Section 7.2 for ideal mixtures of real gases.

The chemical potential of species i in an ideal solution is given by:(10.26)

where Gil (T,P) represents the pure liquid Gibbs energy at T,P.

This reference state can be shifted to (T,unit pressure) using:(10.37)

ilii

idi xlnRT)P,T(G)x,P,T(

lii

li flnRT)T()P,T(G

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CHEE 311 J.S. Parent 7

Ideal Liquid Solutions

Substituting for Gil (T,P) yields:

To estimate the chemical potential of component i in an ideal liquid solution, all we require is the composition (xi) and the pure liquid fugacity (fi

l ).

The fugacity of a pure liquid can be calculated using:

(10.41)

For those cases in which the ideal solution model applies, we require only pure component data to estimate the chemical potentials and total Gibbs energy of the liquid phase.

liiii

idi fxlnRT)T()x,P,T(

RT

)PP(VexpPf

satilsat

isati

li

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Non-Ideal Liquid Solutions

Relatively few liquid systems meet the criteria required by ideal solution theory. In most cases of practical interest, molecular interactions are not uniform between components, resulting in mixture behaviour that deviates significantly from the ideal case.

The approach for handling non-ideal liquid solutions is exactly the same as that adopted for non-ideal gas mixtures. We define a solution fugacity, fi

l as:

(10.42)

To use this approach, we require experimental data or correlations pertaining to the specific mixture of interest

li

lii

f̂lnRT)T()P,T(

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Lewis-Randall Rule

The ideal solution model developed in Sections 7.2 and 7.4 is known (in a slightly different form) as the Lewis-Randall equation:

(10.84)

The solution fugacity of component i in an ideal solution (gas or liquid) can be represented by the product of the pure component fugacity and the mole fraction.

Whenever you apply an ideal solution model, you are using the Lewis-Randall rule.

This is an approximation that yields reasonable results for similar compounds (benzene/toluene, ethanol/propanol)

However, it is important that you appreciate the limitations of this rule. When you cannot find mixture data, you may need to use it (but I suggest you look harder).

iiidi fxf̂

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Liquid Phase Activity Coefficients

Based on our definition of solution fugacity:(10.42)

we could define a liquid phase solution fugacity coefficient:

that reflects deviations of the solution fugacity from a perfect gas mixture.

A more logical approach is to measure the deviations of the solution fugacity from ideal solution behaviour. For this purpose, we define the activity coefficient:

(10.89)

this convenient parameter is used to correlate non-ideal liquid solution data, just as i is used for gas mixtures

li

lii

f̂lnRT)T()P,T(

Pxf̂ˆi

lil

i

lii

li

i fx

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Excess Properties of Non-Ideal Liquid Solutions

Most of the information needed to describe non-ideal liquid solutions is published in the form of the excess Gibbs energy, GE.

Excess properties are defined as the difference between the actual property value of a solution and the ideal solution value at the same T, P, and composition.

ME(T,P, xn) = M(T,P, xn) - Mid(T,P, xn) (10.86)

In defining excess properties, we use ideal solution behaviour as our reference. Pure components cannot have excess properties.

Partial excess properties can also be defined:Mi

E(T,P, xn) = Mi(T,P, xn) - Miid(T,P, xn) (10.87)

where

nj,P,Ti

EEi

n)nM(

M

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Excess Properties of Non-Ideal Liquid Solutions

The partial excess Gibbs energy is of primary interest:

where the actual partial molar Gibbs energy is provided by equation 10.42:

and the ideal solution chemical potential is:

Leaving us with the partial excess Gibbs energy:

(10.90)

idii

Ei GGG

li

li ii

f̂lnRT)T(G

liii

idi

idi fxlnRT)T(G

i

lii

li

lii

li

Ei

lnRT

fx

f̂lnRT

fxlnRTf̂lnRTG

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Excess Properties of Non-Ideal Liquid Solutions

Why do we define excess properties for liquid solutions? They are more easily applied to experimental data Activity coefficients can be treated as partial molar properties

with respect to excess properties. Three important results follow:

(10.94)

The Gibbs-Duhem equation:(10.98)

The summability relation, providing GE from lni data:

(10.97)

nj,P,Ti

E

i n)RT/nG(

ln

0lndxi

ii

i

ii

E

lnxRTG

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CHEE 311 J.S. Parent 14

Review of Thermodynamic Principles

Although thermodynamics applies to a great many problems of engineering importance, CHEE 311 focuses on phase equilibrium of multi-component systems.

In general, we are trying to describe systems at equilibrium

How many stable phases exist, what are their compositions?

What property of the system determines its state?

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The Fundamental Equation

By combining the 1st and 2nd Laws, we derived the fundamental equation, which relates changes in the internal energy of a system to variations in volume, entropy and composition.

In our calculations we are most interested not in changes of V,S and composition, but P,T and composition

We therefore defined the Gibbs energy, G

The fundamental equation in terms of Gibbs energy is:

This equation is useful because it relates changes of the pressure (dP), temperature (dT) and composition (dni) of a system to changes in Gibbs energy.

iidnnSdTnVdPdnG

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Chemical Equilibrium in terms of Gibbs EnergyOur first definition of chemical equilibrium is based on the total Gibbs energy (Section 4 notes, 14.1 text)

We considered a closed system (a vessel) in thermal and mechanical equilibrium with its surroundings. We charged components to the system, and watched it move towards a state of chemical equilibrium.

What is the change in Gibbs energy as it does so?

Since these changes take place spontaneously, the second law tells us that entropy must be created:

dSt + dSsurr 0where dSt is the entropy change of the system and dSsurr of the surroundings.

dSsurr is calculated from the heat transferred to maintain thermal and mechanical equilibrium:

dSsurr = dQsurr / T = - dQ / T

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Chemical Equilibrium in terms of Gibbs EnergySubstituting for dSsurr gives:

dSt -dQ / Tor

TdSt dQ (A)which describes the total entropy change of the system as it transfers heat to/from its surroundings in an effort to reach equilibrium.

How much heat is governed by the first law:dQ = dUt + P dVt (B)

Substituting (A) into (B) yields,TdSt dUt + P dVt

or,dUt + P dVt - TdSt 0 (14.2)

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Chemical Equilibrium in terms of Gibbs EnergyEquation 14.2 is a general equation which relates how the thermodynamic properties of the system change as it moves towards chemical equilibrium:

dUt + P dVt - TdSt 0 (14.2)

The Gibbs energy is defined as:Gt = Ut + PVt - TSt

from which changes to the Gibbs energy dG are:dGt = dUt + P dVt + VtdP - TdSt - StdT

= (dUt + P dVt - TdSt)+ VtdP - StdT

If the system approaches equilibrium at constant T,P dGt = dUt + P dVt - TdSt

which from equation 14.2 tells us:

dGtT,P 0 (14.3)

Spontaneous changes in the composition of the system that occur at constant T,P must decrease the Gibbs energy.

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CHEE 311 J.S. Parent 19

Equilibrium in Terms of Chemical Potential

If the approach towards equilibrium decreases the Gibbs energy, then the equilibrium state must reside at the minimum Gt. This occurs when:

dGtT,P = 0 (14.4)

If the equilibrium state is comprised of two phases we know any change of the total Gibbs energy is:

d(nG) = d(nG)+ d(n G)

Each phase is an open system:

Therefore, changes of the total Gibbs energy at a given T,P are:

or,

or,

i

i

dndTnSdPnVdnG

dndTnSdPnVdnG

i

i

0dndndG iit

ii

0dn)( iii

ii

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Equilibrium in Terms of Mixture Fugacity

We have developed equations to represent the chemical potential of vapour and liquid phases.For vapour mixtures:

For liquid mixtures:

Chemical equilibrium requires the equivalence of chemical potential for all species in both phases. Therefore,

or,

or,

Our working definition for equilibrium is now based on fugacity.

vii

vi f̂lnRT)T()P,T(

lii

li f̂lnRT)T()P,T(

li

vi

lii

vii

li

vi

f̂f̂

f̂lnRT)T(f̂lnRT)T(

)P,T()P,T(

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CHEE 311 J.S. Parent 21

Raoult’s Law

We can apply our new definition of chemical equilibrium to derive Raoult’s Law. In developing this simplified model, we assumed:

The vapour behaves as a perfect gas mixture The liquid behaves as an ideal solution

The mixture fugacity for a perfect gas mixture is:

because iv= 1 for this simplified case.

The solution fugacity for an ideal liquid solution is:

given that i=1, isat= 1, and we ignore the pressure dependence of

the Gibbs energy (Poynting factor = 1)

PyPyˆf̂ iivi

vi

satii

sati

lsati

satiii

li Px

RT

)PP(VexpPxf̂

i

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Raoult’s Law

Given our definition of chemical equilibrium in terms of fugacity:

we can apply our simplified expressions to derive Raoult’s Law:

or

for systems where a perfect gas mixture is in equilibrium with an ideal liquid solution.

When this is not the case, we must use the full expression:

li

vi f̂f̂

RT

)PP(VexpPxPˆy

sati

lsati

satiii

vii

i

PP

xy sat

i

i

i

RT

)PP(VexpP

xy

vi

sati

lsati

satii

i

i

i

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CHEE 311 J.S. Parent 23

Thermodynamic Calculations - Raoult’s Law

Bubble Point Pressure:Given a liquid composition at a specified temperature, find the composition of the vapour in equilibrium and the pressure

Given T, x1, x2, ... xn find P, y1, y2,... yn

Bubble Point Temperature:Given a vapour composition at a specified pressure, find the composition of the liquid in equilibrium and the temperature

Given P, x1, x2, ... xn find T, y1, y2,... yn

Governing Equation - Bubble Line Equation:

to find P or T

thento find yi,…yn

i

satiiPxP

P

Pxy

satii

i

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CHEE 311 J.S. Parent 24

Thermodynamic Calculations - Raoult’s Law

Dew Point Pressure:Given a vapour composition at a specified temperature, find the composition of the liquid in equilibrium and the pressure

Given T, y1, y2,... yn find P, x1, x2, ... xn,

Dew Point Temperature:Given a vapour composition at a specified pressure, find the composition of the liquid in equilibrium and the temperature

Given P, y1, y2,... yn find T, x1, x2, ... xn

Governing Equation - Dew Line Equation:

to find P or T

thento find xi,…xn

isati

i

Py1

P

sati

ii P

Pyx

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CHEE 311 J.S. Parent 25

Thermodynamic Calculations

“Classic” P-T Flash Calculation:Given an overall composition at a specified temperature and pressure, find the composition of the liquid and vapour phases as equilibrium and the relative amounts of vapour and liquid

Given T,P, z1, z2,... zn find x1... xn , y1,... yn, V, L

Governing Equation - Flash equation:

to find V0

where Ki = Pisat/P, V=vapour phase fraction

Then

to find xi,…xn

andto find yi,…yn

i i

i 01)1K(V1

z

)1K(V1

zx

i

ii

P

Pxy

satii

i