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Chapter 14: Phase Equilibria Applications Part II

Chapter 14: Phase Equilibria Applications Part II

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Page 1: Chapter 14: Phase Equilibria Applications Part II

Chapter 14: Phase Equilibria Applications

Part II

Page 2: Chapter 14: Phase Equilibria Applications Part II

If the component is supercritical, then the vapor pressure is not defined

111

101

ˆlim

1f

x

fH x

P

Pxy

P

HxyPyHx

Hxf

sat

l

2

2222

1

11

11

11111

11

11

111

ˆˆ

ˆ

2

Page 3: Chapter 14: Phase Equilibria Applications Part II

example• A binary methane (1) and a light oil (2) at 200K and 30 bar

consists of a vapor phase containing 95% methane and a liquid phase containing oil and dissolved methane. The fugacity of methane is given by Henry’s law, and at 200 K, H1 = 200 bar. Estimate the equilibrium mole fraction of methane in the liquid phase. The second virial coefficient of methane at 200K is -105 cm3/mol

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Page 4: Chapter 14: Phase Equilibria Applications Part II

solution

:

ˆˆ

ˆ

111

111

sAssumption

Pyf

Hxfv

l

4

Page 5: Chapter 14: Phase Equilibria Applications Part II

Solution (cont)

PyPyf

Hxfv

l

11111

111

ˆˆ

ˆ

Need equation for the fugacity coefficient vapor phase

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Page 6: Chapter 14: Phase Equilibria Applications Part II

Solution (cont)

827.0)/exp( 111 RTPB

How do we solve for the mole fraction of the solutein the liquid phase?

x1 =0.118

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Page 7: Chapter 14: Phase Equilibria Applications Part II

Another example

• For chloroform(1)/ethanol(2) at 55oC, the excess Gibbs energy is

2121 )59.042.1( xxxxRT

G E

• The vapor pressures of chloroform and ethanol at 55oC are P1

sat = 82.37 kPa, and P2sat = 37.31 kPa

• Make BUBLP calculations, knowing that B11=-963 cm3/mol, B22 =-1,523 cm3/mol, B12 =52 cm3/mol

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Page 8: Chapter 14: Phase Equilibria Applications Part II

)1)((2()(exp

)(2()1(exp

12112212

12

11221122

11

xAAAx

xAAAx

satsat2

22

1

11

ˆ;

ˆ

Need equations for the fugacity coefficients

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Page 9: Chapter 14: Phase Equilibria Applications Part II

2

satsat

and calculate

BBB

RT

PyPB

RT

PB

1

22111212

122211

1

1111

2

expˆ

exp

But we don’t know P

Guess P (avg sat. pressures)

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Page 10: Chapter 14: Phase Equilibria Applications Part II

12

21222122

11112111

1

)(),,,(

)(),,,(

yy

PxxPyyTPy

PxxPyyTPysat

sat

For example, at x1 =0.25, solve for y1, y2, P

y1 = 0.558y2 = 0.442P = 63.757 kPa

In our web site, there are model spreadsheets that you can download 10

Page 11: Chapter 14: Phase Equilibria Applications Part II

VLE from cubic EOS

PxPy

NDESCRIPTIO EALTERNATIV

...N 2, 1, i ff

lii

vii

li

vi

ˆˆ

,ˆˆ

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Page 12: Chapter 14: Phase Equilibria Applications Part II

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Page 13: Chapter 14: Phase Equilibria Applications Part II

Vapor pressure pure species

Liquid branch

Vapor branch

LV transition

li

vi

Using equation (1), the cubic EOS yields Pisat= f(T)

(1)

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Page 14: Chapter 14: Phase Equilibria Applications Part II

14

Page 15: Chapter 14: Phase Equilibria Applications Part II

Compressibility factors

ii

lii

ilii

lii

li q

ZZZZ

1

))((1

ci

cirii

i

ii

ci

cii

ii

P

TRTTa

RTb

Taq

P

RTb

RT

Pb

22)()(

)(

For the vapor phase there is anotherexpression, (14.36)

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Page 16: Chapter 14: Phase Equilibria Applications Part II

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Page 17: Chapter 14: Phase Equilibria Applications Part II

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Page 18: Chapter 14: Phase Equilibria Applications Part II

How to calculate vapor pressure from a cubic EOS

ili

ilil

i

iiiiii

Z

ZI

where

IqZZ

ln1

)ln(1ln

We solve for the saturation pressure at a given T such that the fugacity coefficients are equal in the two phases: 2 compressibility eqns., two fugacity coefficient eqns., equality of fugacity coefficients, 5 unknowns: Psat, Zl, Zv, l, v

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Page 19: Chapter 14: Phase Equilibria Applications Part II

Mixture VLE from a cubic EOS

• Equations for Zl and Zv have the same form as for pure components

• However the parameters a & b are functions of composition

• The two phases have different compositions, therefore we could think in terms of two P-V isotherms, one for each composition

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Page 20: Chapter 14: Phase Equilibria Applications Part II

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Page 21: Chapter 14: Phase Equilibria Applications Part II

Mixing rules for parameters

2/1jijiij

iji j

ji

ii

i

aaaa

axxa

bxb

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Page 22: Chapter 14: Phase Equilibria Applications Part II

Also we need “partial” parameters

j

j

j

nTii

nTii

nTii

n

nqq

n

nbb

n

naa

,

,

,

)(

)(

)(

22

Page 23: Chapter 14: Phase Equilibria Applications Part II

The partial parameters are used for the calculation of fugacity

coefficients• Because i is related to the partial molar

property of GR (residual G)

b

b

a

aqq

I

IqZZb

b

iii

ii

i

1

different) are (equations or Z Zis Z

17) (slide before as defined

)ln()1(ˆln

vl

23

Page 24: Chapter 14: Phase Equilibria Applications Part II

example

• Vapor mixture of N2(1) and CH4 (2) at 200K and 30 bar contains 40 mol% N2. Calculate the fugacity coefficients of nitrogen and methane using the RK equation of state.

vvvv

vvvvvv

ZZ

ZqZ

1

For RK, =0 and =1

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Page 25: Chapter 14: Phase Equilibria Applications Part II

molcmP

Tb

molcm barP

TTa

b

b

a

aayayqq

aaayaya

ayaayyaya

ci

cii

6

ci

cirii

/14.83

08664.0

/)14.83(

42748.0

22

22

2

3

2222/1

1212111

212111

22221211

21

25

Page 26: Chapter 14: Phase Equilibria Applications Part II

Calculate P-x-y diagram at 100 oF for methane(1)-n butane (2) using SRK and mixing rules (14.42) to

(14.44)

Compare with published experimental data (P, x, y)

Initial values for P and yi can be taken from given experimental data

First read critical constants, , from Table B.1 and from Table 3.1

22/12 )1)(176.0574.148.0(1 rSRK T

Calculate b1, b2, a1, a2

In this case T > Tc1 26

Page 27: Chapter 14: Phase Equilibria Applications Part II

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Page 28: Chapter 14: Phase Equilibria Applications Part II

K value given by

vi

li

iKˆ

ˆ

Step 1

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Page 29: Chapter 14: Phase Equilibria Applications Part II

The equations for are valid only up to the critical temperature; however is OK toextend the correlation slightly above the critical temperature

Lets calculate the mixture parameters (for step 1). When applied to the liquid phase we use the xi mole fractions

RTb

aq

b

b

a

aaxaxqq

b

b

a

aaxaxqq

bxbxb

axaaxxaxa

l

ll

llll

llll

l

l

2211222

1212111

2211

22221211

21

22

22

2

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Page 30: Chapter 14: Phase Equilibria Applications Part II

Follow diagram Fig. 14.9

Assume P and yi

Calculate Zl and Zv, and the mixture fugacity coefficients

Calculate K1 and K2 and the Kixi

Calculate normalized yi=Kixi/ Kixi

Reevaluate fugacity coefficients vapor phase, etc

If Kixi > 1, P is too low so increase P; if Kixi < 1, then reduce P

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Page 31: Chapter 14: Phase Equilibria Applications Part II

Results:Rms % difference between calculated and exp. P is3.9%

Rms deviation between calculated and exp. y1 is0.013

Note that the system consistsof two similar molecules

Where are the largestdiscrepancies with the experimental data?

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