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8/10/2019 Thermo PPT Utah

1/13

Chemical Engineering

ThermodynamicsA Brief Review

SHR Chapter 2

1 Thermo.key - February 7, 2014


2/13

Rate vs. StateThermodynamics tells us how things end up (state) but not how (or howfast) they get there (rate)

Limitations on getting to equilibrium:

Transport (mixing): diffusion, convection

!example: why doesnt more of the ocean evaporate to make higher humidity in the deserts? (heat transfer

alters dew point locally, atmospheric mixing limitations, etc.)

Kinetics (reaction): the rate at which reactions occur are too slow to get to equilibrium!example: chemical equilibrium says trees (and our bodies) should become CO2and H2O (react with air)

!example: chemical equilibrium says diamonds should be graphite (or CO2if in air)

Sometimes, mixing & reaction are sufficiently fast that equilibriumassumptions are good enough.

if we assume that we can use thermodynamics to obtain the state of the system, we avoid a lotof complexity.

!do it if it is good enough!

Most unit operations (e.g. distillation) assume phase equilibrium, ignoring transport limitations.! good enough if things are mixed well and have reasonable residence/contact times.



3/13

Some Terminology...

M =

CX

i=1

yiMi

Gibbs-DuhemEquation:

M

P

T,y

dP +

M

T

P,y

CXi=1

yid Mi = 0

Partial Molar property -

arbitrary thermodynamicproperty (per mole)

Ni -moles of species i

yi -mole fraction of species i(liquid or gas)

Note at constant Tand P,X

i=1

yid Mi = 0

Excess property ME

MMideal

how muchdeviates from

the ideal solution behavior.

Residual property

M

ideal

=

C

Xi=1

yiMi

how muchdeviates from

the ideal gas (ig) behavior.

Mig=

CX

i=1

yiMigiM

RMM

ig

change indue to adding a differential amount

of species i. For ideal mixtures, Mi =Mi

Mi

NM

Ni

T,P,Nj



4/13

Phase Equilibrium

G= G(T,P,N1,N2, . . . ,N C)

Gibbs Energy(thermodynamic property)

Phase Equilibriumis achieved whenGis minimized.

dG =

S dT+ V dP+

CXi=1

GNi

P,T,Nj

dNi

i

G

Ni

P,T,Nj

Chemical Potential

(thermodynamic property,partial molar Gibbs energy)

In phase equilibrium,

T(1)= T(2)== T(N)

P(1)= P(2)=

= P(N)

For multiplephases in a

closed system,

Gsystem =

NX

p=1

G(p)

dGsystem =

NX

p=1

dG(p)

For a nonreactingsystem, molesare conserved.

dGsystem =

NXp=2

" CXi=1

(p)i

(1)i

dN

(p)i

#T,P

dGsystem =

N

Xp=1" C

Xi=1

(p)i dN

(p)i #

T,P

To minimize G, dG=0. But eachterm in the summation for dGis

independent, so each must be zero.(p)i =

(1)i =

(2)i = =

(N)i

For phase equilibrium, thechemical potential of any

species is equal in all phases.

SHR 2.2

fi =Cexp iRT

Solution: Partial Fugacity

Cis a temperature-dependent constant.

f

(p)

i =

f

(1)

i =

f

(2)

i = =

f

(N)

i

For phase equilibrium, the fugacityof any species is equal in all phases.

Problem:i

P 0as

NX

p=1

dN(p)i = 0 dN

(1)i =

NX

p=2

dN(p)i



5/13

Chemical Potential, Fugacity, ActivitySHR 2.2.1 & Table 2.2

Ps

i vapor pressure of species i.

yi

xi

vapor mole fraction of species i.

liquid mole fraction of species i.

For phase equilibrium, the chemical potential, fugacity,activity(but not activity coefficient or fugacitycoefficient) of any species is equal in all phases.

Thermodynamic Quantity Definition Description Limiting case of ideal

gas and ideal solution

Chemical potential i GNi

T,P,Nj

Partial molar free energy, gi i =g i

Partial fugacity fi =C exp (i/(RT)) Thermodynamic pressure fiV=yiP

Fugacity coefficient of a

pure species

i fi/P Deviation of fugacity due to

pressure

iV=1.0

iL = Psi/P

Fugacity coefficient of a

species in a mixture

iV fiV/(yiP)

iL fiL/(xiP)

Deviations to fugacity due

to pressure and composition

iV=1.0

iL = Psi/P

Activity ai fi/foi Relative thermodynamic

pressure

aiV=yiaiL =xi

Activity coefficient iV aiV/yi

iL aiL/xi

Deviation of fugacity due to

composition.

iV=1.0

iL =1.0

fiL = iLxiP

=iLxifo

iL

fiV = iVyiP

=iVyifo

iV

fiL = fiV yi

xi=

iL

iV

we will use this onthe next slide...

fi =fi for a pure species



6/13

Phase Equilibrium - K-ValuesPhase equilibrium ratio:

How much species iisenriched in the vapor

Relativevolatility:

ij KiKj

=yi/xiyj/xj

"ij = 0 "ij = ""ij = 1

easy easyimpossible

Solve foryi/xifromthe previous slide:

Note: liquid-liquidcase equivalent is thepartition coefficient

KDi x(1)i /x(2)

i

ij Di

KDj

Note: liquid-liquid case equivalentis the relative selectivity#ij.

Ki yi

xi

Mix

&matchfordifferen

t

sp

eciesinamixture

Equation Application

Rigorous forms

Equation of state Ki = iL/iV Benedict-Webb-Rubin-Starling, SRK, PR

equations of state, Hydrocarbon and light gas

mixtures from cryogenic to critical

Activity coefficient Ki = iLiL/iV All mixtures from ambient to near critical.

Approximate forms

Raoults law (ideal) Ki = Psi/P Ideal solutions at low pressures

Modified Raoults law Ki = iLPsi/P Nonideal liquid solutions near ambient pressure

Poynting correction Ki = iLsiV

Psi

P

exp

1RT

RP

Psi

viLdP Nonideal liquid solutions at moderate pressures

and below Tc

Henrys law Ki = HiP Low to moderate pressures for species aboveTc

SHR 2.2.2 & Table 2.3

Concept: determine Kifromthermodynamic models.

This gives usyi/xi.

SHRTable2.3

Ki

yi

xi=

iL

iV

=iLf

o

iL

iVP =

iLiL

iV



7/13

Example: Raoults Law (Ideal Mixtures)

Need: Pis- saturation pressure for each species

Antoine equation:

Ki =

s

i

PRaoults law:

i = xiPs

i

partial pressure is theweighted saturation pressure

(ideal liquid mixture)

Derivation of Raoults Law:

i = yiPDaltons law

(ideal mixture of gases)

combineto find

yi

xi=

s

i

P=Ki

SHR Figure 2.3: Pisfor a few species

Atmospheric Pressure

For Raoults law, Kiis independent of composition,but varies exponentiallywith temperature!

log10 Ps

i = A

B

C+ T

Ki = iP

s

i

PModified Raoults law:



8/13

Liquid-Phase Activity Coefficients

ln k = 1 ln

CX

j=1

xjkj

CXi=1

xiikPCj=1 xjij

!Wilson 2-parameter model for mixtures:

For a binary system:ln 1= ln (x1+ x212) + x2

12

x1+ x212

21

x2+ x121

ln 2= ln (x2+ x121) x1

12

x1+ x212+

21

x2+ x121

#ijare functions of temperature

but not composition

$ij= $ij= 1implies idealsolution (%i = %j = 1)

ij = ji

ii = jj

ij 6= ji

ii = 1

ij = vjLviL

exp(ij

ii)

RT

ln

1 = 1 ln12 21

ln

2 = 1 ln21 12

Alternatively, if we can obtain %at infinite dilution,

2 equations,2 unknowns

Less accurate than regression.

Parameters #ijor $ijare typically determinedfrom experimental data via regression.



9/13

Example: Graphical Methods1. Locate appropriate species /mixture

2. Locate appropriate Tand P.

3. Draw a straight line between the two

points to determine Ki.

SHR Figure 2.4

Example: at T=60 Fand P=1atm,

Kpropane= 7Kisopentane= 0.64

NOTE: this can also be used to estimatethe bubble point for a pure fluid. (how?)



10/13

DePriester Chart

Empirical Correlation:

M. L. McWilliams, Chemical Engineering, 80(25), 1973 p. 138.

lnK= a1

T2+

a2

T +a3 + b1 lnp+

b2

p2+

b3

p

Chemical Engineering Progress(AIChE), Vol. 74(4), pp. 85-86

NOTE: Tin Randpin psia!



11/13

Thermo from Equations of State

iV = exp

( 1

RT

Z

V

"P

Ni

T,V,Nj

RT

V

#dV lnZ

)

SHR 2.5.2

For Redlich-Kwong (and S-R-K) EOS:

= exp

Z 1 ln (ZB)

A

Bln

1 +

B

Z

Note that these can be used in vapor or liquid phases, by

using appropriate values forZ, and using appropriate phase

mole fractions (yifor vapor,xifor liquid)

(because cubic EOS can approximate two-phase behavior.)

i= exp

"(Z 1)

Bi

B ln (ZB)

A

B

2

rAi

A

Bi

B

!ln

1 +

B

Z

#volume

vapor

V =

v

C

Xi=1

Ni

fugacitycoefficient

partial fugacitycoefficient

h=

C

Xi=1

(yiho

i ) + RTZ 13A

2B

ln 1 +B

Z

Z Z +AB B

ZAB = 0

B =bP

RTA =

aP

R2T2

a =

CXi=1

CX

j=1

yiyjaiaj b =

CX

i=1

yibimixing rules:

Ki yi

xi=

iL

iV

V = exp" 1RT

Z P0

v

RT

PdP

#= exp

1

RT

Z

v

P

RT

v

dv lnZ+ Z 1

R-K: S-R-K(much better):

P =RT

v b

a

v2 + bv

$- acentric factor

bi = 0.08664RTi,c

Pi,c

ai = 0.42748R2T2.5i,c

Pi,cT0.5

bi = 0.08664RTi,c

Pi,c

ai = 0.42748R2T2i,c

1 +f

1 T0.5i,r

2

Pi,c

f

= 0.48 + 1.574

0.1762



12/13

Example: K-Values from SRKGiven a mixture of propane(x=0.0166,y=0.3656) and

benzene (x=0.9834,y=0.6344)

at 300 K & 1 atm, find the K-value using SRK.

i = exp

"(Z 1)

Bi

B ln (ZB)

A

B

2

rAi

A

Bi

B

!ln

1 +

B

Z

#Z3 Z2 +AB

B2

ZAB = 0

1. Calculate ai, biand thenAi,Bi(tedious, but not difficult).

2. CalculateAandBusing mixing rules.

3. CalculateZin the vapor and liquid phases. This requiresus to solve the cubic equation and pick the appropriateroots.

4. Calculate partial fugacity coefficient in vapor and liquid.

5. Calculate Ki.

Ki yi

xi=

iL

iV

Requiresxi,yi.This is a problem since

we typically need Kitodetermine this ratio!

This is typically used withinanother solver to determine

xi,yi(more soon).



13/13

Heuristics for Property Model Selection

AqueousSolutions

NRTL Special

electrolytes and nopolar compounds?

NoYes

SHR 2.8

Polar Compounds

PSRK(see SHR

2.7.1)

UNIFACWilsonNRTL

NRTLUNIQUAC

with light gases?

liquid phaseseparation?

Yes No

No Yes

Hydrocarbonswith or without

light gases

CorrespondingStates

PR(all T, P)

SRK(non cryogenic

T, all P)

BWRS(all T, sub-critical P)

Wide boilingrange?

Yes No

See also http://www.che.utah.edu/~sutherland/PropertySelection.pdf for a

much more extensive property selection flowchart from Professor Ring.


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