Introductory ChemicalEngineering Thermodynamics

By J.R. Elliott and C.T. Lira

Chapter 6 - Engineering Equations of State

Chapter 6 - Engineering Equations of State Slide 1

II. Generalized Fluid PropertiesThe principle of two-parameter corresponding states

-10

0

10

20

30

40

50

0 0.1 0.2 0.3 0.4

Density (g/cc)

T=133K

T=191K

T=286K ρL

ρV-10

0

10

20

30

40

50

0 0.1 0.2 0.3 0.4 0.5

De nsity (g/cc)

T=329KT=470

T=705K

VdW Pressure (bars) in Methane VdW Pressure (bars) in PentaneCritical Definitions:Tc - critical temperature - the temperature above which no liquid can exist.Pc - critical pressure - the pressure above which no vapor can exist.ω - acentric factor - a third parameter which helps to specify the vapor pressure curvewhich, in turn, affects the rest of the thermodynamic variables.

Note: at the critical point, cc

TT

PandTPP

at 0 and 02

2

=

=

∂ρ∂

∂ρ∂

Chapter 6 - Engineering Equations of State Slide 2

II. Generalized Fluid PropertiesThe van der Waals (1873) Equation Of State (vdW-EOS)

Based on some semi-empirical reasoning about the ways that temperature and densityaffect the pressure, van der Waals (1873) developed the equation below, which heconsidered to be fairly crude. We will discuss the reasoning at the end of the chapter, butit is useful to see what the equations are and how we use them before deriving the details.The vdW-EOS is:

( ) ( )Zb

b

a

RT b

a

RT= +

−− =

−−1

1

1

1

ρρ

ρρ

ρ

van der Waals’ trick for characterizing the difference between subcritical andsupercritical fluids was to recognize that, at the critical point,

cc

TT

, PTPP

at 0 and 02

2

=

=

∂ρ∂

∂ρ∂

Since there are only two “undetermined parameters” in his EOS (a and b), he has reducedthe problem to one of two equations and two unknowns. Running the calculus gives:a = 0.475 R2Tc

2/Pc; b = 0.125 RTc/Pc

The capability of this simple approach to represent all of the properties and processesthat we will discuss below is a tribute to the genius of van der Waals.

Chapter 6 - Engineering Equations of State Slide 3

II. Generalized Fluid PropertiesThe principle of three-parameter corresponding statesReduced vapor pressure behavior:

1 / T r

1 1 .2 1 .4 1 .6 1 .8 2 2 .2 2 .4

To improve our accuracy over the VdW EOS, we can generate a different set of PVTcurves for each family of compounds. We specify the family of compounds via the"third parameter" i.e. ω. Note: The specification of Tc , pc , and ω provides two pointson the vapor pressure curve. The key to accurate characterization of the vapor liquidbehavior of mixtures of fluids is the accurate characterization of the vapor pressure ofpure fluids. VLE was central to the development of distillation technology for thepetrochemical industry and provided the basis for most of today’s process simulationtechnology.

Chapter 6 - Engineering Equations of State Slide 4

II. Generalized Fluid PropertiesThe Peng-Robinson (1976) Equation of State

( ) ( ) 2222

2

211

1= Zor

211 ρρρ

ρρρ

ρρ

bb

b

bRT

a

-bbb

a

b

RTp

−+⋅−

−+−

−=

where ρ = molar density = n/VBy fitting the critical point, where ∂P/∂V = 0 and ∂2P/∂V2 = 0,a = 0.457235528 αR2Tc

2/Pc; b = 0.0777960739 RTc/Pc

α = [1+ κ (1-√Tr)]2 ; κ = 0.37464 + 1.54226ω - 0.26992ω2

ω ≡ -1 - log10(Psat/Pc)Tr =0.7 ≡ “acentric factor”

Tc , Pc , and ω are reducing constants according to the principle of corresponding states.By applying Maxwell's relations, we can calculate the rest of the thermodynamicproperties (H,U,S) based on this single equation.

Chapter 6 - Engineering Equations of State Slide 5

II. Generalized Fluid PropertiesSolving the Equation of State for Z

( )Z =1

1- / ZB

A

B

B Z

B Z B Z− ⋅

+ −/

/ ( / )1 2 2

bρ ≡ B/Zwhere B ≡ bP/RT Z ≡ P/ρRT

A ≡ aP/R2T

2

Rearranging yields a cubic function in Z

Z3 -(1-B)Z2 + (A-3B2-2B)Z - (AB-B2-B3) = 0

Naming this function F(Z), we can plot F(Z) vs. Z to gain some understanding about itsroots

Chapter 6 - Engineering Equations of State Slide 6

II. Generalized Fluid PropertiesSolving the Equation of State for Z

-0.04

-0.02

0

0.02

0.04

0.06

0.08

0.1

0 0.5 1

P= PvapT< Tc

F

Z -0.04

-0.02

0

0.02

0.04

0.06

0.08

0.1

0 0.5 1

P> > PvapT< Tc

F

Z

-0.04

-0.02

0

0.02

0.04

0.06

0.08

0.1

0 0.5 1

P< < PvapT< Tc

F

Z-0.06

-0.04

-0.02

0

0.02

0.04

0.06

0.08

0.1

0 0.5 1

T> TcF

Z

Chapter 6 - Engineering Equations of State Slide 7

II. Generalized Fluid PropertiesSolving the Equation of State for Z (cont.)1. Guess Zold=1 or Zold=0 and compute Fold(Zold).2. Compute Fnow(Z) at Z=1.0001 or Z=0.00013. "Interpolate" between these guesses to estimate where F(Z)=0.

∆Z = (0 - Fnow)*(Znow-Zold)/(Fnow - Fold).4. Set Fold=Fnow, Zold=Znow, Znow=Zold+∆Z5. If |∆Z/Znow| < 1.E-5, print the value of Znow and stop.6. Compute Fnow(Znow) and return to step 3 until step 5 terminates.

Chapter 6 - Engineering Equations of State Slide 8

II. Generalized Fluid PropertiesAn Introduction to the Radial Distribution Function

Nc g ro

R o

= ∫ρ π( ) 4 r dr;2

where g(r) is our "weighting factor" henceforth referred to as the radial distributionfunction (rdf).g

r / s

1

1 2

The body centered cubicunit cell

The radial distribution functionfor the bcc hard sphere fluid

Chapter 6 - Engineering Equations of State Slide 9

An Introduction to the Radial Distribution Function

drrrgNcoR

o

24 )( πρ ∫=

g

r / s

1

0

1

2

3

4

0 1 2 3 4r/σ

g

The low density hard-sphere fluid The high density hard sphere fluid

g

r / s

1

0

1

2

3

4

0 1 2 3 4

r/σ

gη=0 βε =1.0

η=0.4

The low density square-well fluid The high density square-well fluid

Chapter 6 - Engineering Equations of State Slide 10

II. Generalized Fluid PropertiesThe connection from the molecular scale to the macroscopic scale

The Energy Equation

U U

nRT

N

RTN u g 4

idA

A

− =∞

∫ρ

π2 0

r dr2

The Pressure Equation

P

RT

N

RTrN

du

dr

AA

ρρ π= −

∞

∫16 0

g 4 r dr2

P RTN

RTrN

du

dr

NrN

du

drg 4

AA

AA/ ρ ρ π ρ π

σ

σ

= −

∫ ∫

∞

16 0

g 4 r dr - 6RT

r dr2 2

Chapter 6 - Engineering Equations of State Slide 11

II. Generalized Fluid PropertiesThe Van der Waals Equation of State

( )−

≈∫N

RTrN

du

dr

A

O

Aρ π ρ

ρ

σ

6 g 4 r dr

b

1-b2

where b~close-packed volume

N

RTrN

du

drg 4

N N

RTx

du

dxg 4

AA

A Aρ π ρ σ ε π σσ6 6 1

∞ ∞

∫ ∫

≡ r dr = x dx where x r /23

2

aN N

xdu

dxg 4

A A≡

∞

∫σ ε ε π

σ

3

6

/ x dx2

The resulting equation of state is:

( ) ( )Zb

b

a

RT b

a

RT= +

−− =

−−1

1

1

1

ρρ

ρρ

ρ

By fitting the critical point, where ∂P/∂V = 0 and ∂2P/∂V2 = 0,a = 0.475 R2Tc

2/Pc; b = 0.125 RTc/Pc

Chapter 6 - Engineering Equations of State Slide 12

II. Generalized Fluid PropertiesExample. From the molecular scale to the continuumSuppose that the radial distribution function can be reasonably represented by:

gb

kT x≈ +1

14

ρεwhere x ≡ r/σ and b ≡ π/6 NAσ3

Derive expressions for the compressibility factors of fluids that can be accurately representedby the square-well potential. Evaluate this expression at bρ = 0.2 and ε/kT = 1.Solution: First read Appendix C then note that, for the square-well potential,exp(-u/kT) = exp(ε/kT) H(r-σ) + [exp(ε/kT)-1] [1-H(r-1.5σ)]Taking the derivative of the Heaviside function gives the Dirac delta in two places:

( ) ( )[ ][ ]∫ −−−−+= drrkTryrkTryrN

Z A 341/exp)()5.1(/exp)()(6

1 πεσδεσδρ

( ) ( )[ ]{ }1/exp)5.1(5.1/exp)(6

41 3

3

−−+= kTykTyN

Z A εσεσρσπ

Noting that y(r) ≡ g(r) exp(u/kT) and that exp(u/kT) is best evaluated inside the well:Z = 1+ 4bρ { g(σ+) - 1.53[1-exp(-ε/kT)] g(1.5σ-)} This is true for SW with any g(r).For the above expression: g(σ+) = 1 + bρ ε/kT and g(1.5σ-) = 1 + 0.198 bρ ε/kTZ = 1+ 4bρ {1 + bρ ε/kT - 1.53[1-exp(-ε/kT)]( 1 + 0.198 bρ ε/kT)}Z = 1+4(0.2){1+0.2-2.1333*1.0396} = 0.1858