Chapter 3. Volumetric Properties of Pure Fluids

Chapter 3. Volumetric Properties of

Pure Fluids

Introduction

Thermodynamic properties (U, H and thus Q, W) are

calculated from PVT data

PVT data are important for sizing vessels and pipelines

Subjects

PVT behavior of pure fluids

Ideal gas behavior

Real gas behavior

Generalized correlation when experimental data are lacking

3.1 PVT Behavior of Pure Substances

P-T Phase Diagram

P

T

Solid

Phase (subcooled)

Liquid

Phase

(superheated)

Vapor

Phase

Pc

Tc

critical point

triple

point

Gas

Super-critical (fluid) phase

Vapor vs. Gas?

- Gas : noncondensable (cannot become liquid)

- Vapor : condensable


P-T Phase Diagram

P

T

Solid

Phase

Liquid

Phase

Vapor

Phase

T1

pvap

vapor pressure curve

Vaporization curve

• Component =1

• Phase = 2 (V and L)

• F = 1

• Vapor pressure at T1


P

T

Solid

PhaseLiquid

Phase

Vapor

Phase

Pc

Tc

triple

point

Gas

sublimation pressure

sublimation

point

freezing pressure

freezing

point

melting curve

fusion curve

sublimation curve

P-T Phase Diagram


P

T

Solid

Phase

Liquid

Phase

Vapor

Phase

Pc

Tc

triple

point

Gas


P

T

Solid

PhaseLiquid

Phase

Vapor

Phase

Pc

Tc

triple

point

Gas


PV Phase Diagram

Compare with PT diagram

- triple point?

- vapor pressure line?


PV diagram vs. PT diagram

heating heating

Single Phase Region

Equation relating P, V, T Equation of State

Volume expansivity

Isothermal compressibility

0),,( TVPf

dPP

VdT

T

VdV

TP

PT

V

V

1

TP

V

V

1

dPdTV

dV

3.2 Virial Equation of State

PVT behavior is complex cannot be represented by simple

equation

Gas phase alone can be correlated with simple equation

PV term is expanded as a function of P by a power series

...2 cPbPaPV

...)'''1( 32 PDPCPBaPV

At low pressure, truncation after two terms usually provide satisfactory result

Ideal-Gas Temperature : Universal Gas Constant

B’, C’, D’, …. depends on component, temperature

a same for all species, function of temperature only

...)'''1( 32 PDPCPBaPV

aPVP ,0

)()( * TfaPV

Useful for defining the

temperature scale

(PV)*, the limiting value of PV as P 0, is

independent of the gas


Ideal Gas Temperature Scale

Make (PV)* proportional to T, (R: proportionality constant)

Assign the value 273.16 K to triple point of water

RTaPV *)(

KRPV t 16.273)( *

K

KT

PV

PV

t 16.273

/

)(

)(*

*

*

*

)(

)(16.273/

tPV

PVKT

Kelvin scale temperature


P 0

Molecular volume becomes smaller

Molecules are separated infinite distances

Intermolecular forces approaches zero

Universal Gas Constant

K.

)PV(R

*

t

16273

mol.R.psia/lbft 10.73R

mol.R.atm/lbft 0.7302R

mol.RBtu/lb 1.987R

cal/mol.K 1.987R

.atm/mol.Kcm 82.06R

kPa/mol.Kcm 8314R

bar/mol.Kcm 83.14R

Pa/mol.Km 8.314R

J/mol.K 8.314R

3

3

3

3

3

3

(PV)t* = 22,711.8 m3bar/molK

Two forms of the Virial Equation

Compressibility

B, B’ : the second virial coefficient

C, C’ : the third virial coefficient

RT

PVZ

...'''1 32 PDPCPBZ

...132

V

D

V

C

V

BZ

Virial expression

RT

BB '

2

2

)('

RT

BCC

3

323

)RT(

BBCD'D

Importance of the Virial Equation

Many form of equation of state have been proposed

Virial Equation has a firm basis in theory

Statistical mechanics

B/V : interaction between two molecules (two-body interaction)

C/V2 : three-body interaction

...132

V

D

V

C

V

BZ

3.3 The Ideal Gas

No interaction between molecules (B/V, C/V2 ,…=0)

Internal energy of gases

Real gas : function of T and P

Ideal gas : function of T only

- Pressure dependency resulting from intermolecular force

- No intermolecular force in ideal gas no dependency in P

- The Equation of State

- Internal Energy

RTPVZ or 1 Ideal gas

RTPVZ or 1

)(TUU

Implied Property Relations for an Ideal Gas

Heat capacity at const. volume is a function of temperature only

Enthalpy is also a function of temperature only

Heat capacity at const. pressure is a function of temperature only

Heat Capacity Relationship

)(TCdT

dU

dT

dUC v

V

v

)(THRTUPVUH

)(TCdT

dH

dT

dHC P

P

P

RCdT

RTUd

dT

PVUd

dT

dHC VP

)()(Caution:

Cv and Cp are not constant,

they vary with T while

keeping Cp = Cv + R

Equations for Process Calculation for Ideal Gases

From the first law of thermodynamics

Combine ideal gas law

PdVdQdWdQdTCdU V

VRTP / RCC VP

V

dVRTdTCdQ V

V

dVRTdW

P

dPRTdTCdQ P

P

dPRTRdTdW

T, V

T, P

PdVR

CVdP

R

CdQ PV PdVdW P,V


- (1) Isothermal Process

PdVdQdWdQdTCdU V

sP dWdQdTCdH

0U

0H

dWdQ

1

2

1

2 lnlnP

PRT

V

VRTQ

1

2

1

2 lnlnP

PRT

V

VRTW

DO NOT memorize the equation !

Memorize the sequences of derivation !

T= const.

V

dVRTdTCdQ V

P

dPRTdTCdQ P

V

dVRTdW

P

dPRTRdTdW


- (2) Isobaric Process

PdVdQdWdQdTCdU V dTCU V

dTCH P

QH

dTCHQ P

)( 12 TTRW

P= const.

sP dWdQdTCdH

V

dVRTdW

P

dPRTRdTdW

PdVdW


- (3) Isochoric Process


dTCH P

QU

dTCUQ V

0W

V= const.

sP dWdQdTCdH

V

dVRTdW

P

dPRTRdTdW

PdVdW


- (4) Adiabatic Process


dTCH P

0Q dQ=0

sP dWdQdTCdH

V

dV

C

R

T

dT

V

integration

VCR

V

V

T

T/

2

1

1

2

0

V

dVRTdTCdQ V

0P

dPRTdTCdQ P

0 PdVR

CVdP

R

CdQ PV

P

dP

C

R

T

dT

P

integration

PCR

P

P

T

T/

1

2

1

2

V

dV

C

C

P

dP

V

P integrationVP CC

V

V

P

P/

2

1

1

2



VCR

V

V

T

T/

2

1

1

2

PCR

P

P

T

T/

1

2

1

2

VP CC

V

V

P

P/

2

1

1

2

VVP CCC

V

V

T

T/)(

2

1

1

2

PVP CCC

P

P

T

T/)(

1

2

1

2

.1 constTV

./)1( constTP

.constPV

VP CC /

0Q

TCdTCW VV

These equations are restricted

to const. heat capacities and

reversible, adiabatic processes



Other expression for WORK calculation

TCdTCW VV

1

RCV

1

TRTCW V

11

111

)(/)1(

1

21

/)1(

1

21112

P

PRT

P

PVPTTRTCW V

Use P instead of T1 or T2



Values of Cp/Cv

Monatomic gases : 1.67 (He, Ne, Kr,…)

Diatomic gases : 1.4 (H2, N2, O2, …)

Simple polyatomic gases : 1.3 (CO2, SO2, NH3, CH4,…)


- (5) Polytropic Process

Polytropic : “Turning many ways”

.constPV

0 constP

1 constRTPV

constPV

constV

isobaric

isothermal

adiabatic

isochoric

Can be used to represent

a in-between processes


- (5) Polytropic Process

Using ideal gas equation,

.constPV

VRTP /

.1 constTV

.constTP /)1(

1P

P

1

RTPdVW

/)1(

1

21

1P

P

)1)(1(

RT)(Q

/)1(

1

21

constant heat capacity

Irreversible Processes

Equations in (1)-(5)

Only valid for mechanically reversible, closed system for ideal

gases

Properties changes (U, H) are also same regardless of the

process State Properties

Heat and Work amount depends on the nature of process

(reversible/irreversible, closed/open) Path function

For irreversible processes, the following procedures are

commonly employed

W is determined

For reversible process

Efficiency (h)

is multiplied

Example 3.2

Air is compressed from an initial state of 1 bar and 25 oC to a final state of

5 bar and 25 oC by three different mechanically reversible processes in a

closed system.

(a) Heating at constant volume followed by cooling at constant pressure.

(b) Isothermal compression.

(c) Adiabatic compression followed by cooling

at constant volume.

Assume air to be an ideal gas with the constant

heat capacities, Cv = 5/2R and Cp = 7/2R.

Calculate W, Q, U, and H.

Example 3.2 – solution

Choose the system as 1 mol of air (basis)

Cv = 20.785 Cp = 29.099 J/molK

Since T is same, U = H = 0

Also, U = Q + W = 0 Q = -W

(although the absolute value of Q & W can be different –

path function)

P1 = 1 bar

T1 = 298.15 K

V1 = 0.02479 m3

P2 = 5 bar

T2 = 298.15 K

V2 = 0.04958 m3

Example 3.3

An ideal gas undergoes the following sequence of mechanically reversible

processes in a closed system.

(a) From an initial state of 70 oC and 1 bar, it is compressed adiabatically to

150 oC.

(b) It is then cooled from 150 to 70 oC at constant P.

(c) Finally, it is expanded isothermally to

its original state.

Calculate W, Q, U, and H for each of

the three processes and for the entire cycle.

Assume air to be an ideal gas with the constant

heat capacities, Cv = 3/2R and Cp = 5/2R.

Example 3.3 – solution

Choose the system as 1 mol of gas (basis), R = 8.314 J/molK

Cv = 12.471 Cp = 20.785 J/molK = Cp/Cv = 5/3

(a) Adiabatic compression, Q = 0

U = W = Cv T = 12.471(150 - 70) = 998 J

H = Cp T = 20.785(150 - 70) = 1663 J

P2 from PT/(-1) = const P2 = 1.689 bar

(b) Const P process

Q = H = Cp T = 20.785(70 - 150) = -1663 J

U = Cv T = 12.471(70 - 150) = -998 J

W = U – Q = -998 – (-1663) = 665 J or W = -P V

(c) Isothermal process

U = H = 0 const T

Q = -W = RTln(P3/P1) = RTln(P2/P1) = (8.314)(343.15)ln(1.689/1) = 1495 J

P1 = 1 bar

P2

3.4 Application of the Virial Equations

Virial Equation

Infinite Series

Only useful for engineering purpose when convergence is rapid

- Two or three terms

Derivatives of compressibility

Compressibility-factor graph for methane

...'''1 32 PDPCPBZ

...'3'2' 2

PDPCB

P

Z

T

'0;

BP

Z

PT

RT

BP

RT

PVZ 1

Truncation equation

Virial Equation for Engineering Purpose

Truncated two terms

Truncated three terms

Extended Virial equation

(Benedict-Webb-Rubin Equation)

V

BZ 1PBZ '1

21

V

C

V

B

RT

PVZ 2''1 PCPBZ

similar accuracy

but convenient

more accurate

32

2

000 /

V

abRT

V

TCARTB

V

RTP

22236exp1

VVTV

c

V

a

eight parameters

,,,,,,, 000 cbaCBA

Example 3.8

Reported values for the virial coefficients of isopropanol vapor at

200 oC are:

B = -388 cm3/mol C = -26,000 cm6/mol2

Calculate V and Z for isopropanol vapor at 200 oC and 10 bar by:

(a) the ideal-gas equation

(b) Second virial coefficient, B

(c) Second virial coefficient, B, and third virial coefficient, C

Example 3.8 - solution

T = 473.15 K and R = 83.14 cm3bar/mol K

(a) For ideal gas, Z = 1

(b) Using B,

(c) with B and C, use the iteration or solver to solve for V

mol/cm,).)(.(

P

RTV 39343

10

154731483

mol/cm,,BP

RTV 354633889343

21

V

C

V

B

RT

PVZ

3.5 Cubic Equations of State

PVT behavior of liquid & vapor over wide range

Equation must not be so complex

Polynomial equation at least cubic equation

Cubic EOS are simplest equation

P

V

P

Vreal behavior cubic EOS representation

The van der Waals Equation of State

J. D. Van der Waals

First proposed in 1873

Won Novel Prize in 1910

2V

a

bV

RTP

2V

a

bV

RTP

V

RTP

. .

. .

.

.

Ideal gas : no volume, no interactionVan der Walls gas

volume + interaction


T > Tc ; monotonically

decreasing function

T = Tc ; Critical Region

Saturated liquid and vapor

T < Tc ; Liquid region

T < Tc ; Unrealistic Behavior

T < Tc ; Vapor region

T < Tc ; Real behavior : two phase region


2V

a

bV

RTP

b

Molar volume cannot be smaller than b

b often called “hard sphere volume”

At T = Tc and T > Tc, only single root

exists

At T<Tc, three roots exist

Smaller : liquid-like volume

Middle : no significance

Larger : vapor-like volume

Inaccurate critical point prediction

Inaccurate vapor pressure prediction

a parameters are not optimized to fit vapor pressure

Only historical interest

Limitation of Van der Waals EOS

375.0c

ccc

RT

VPZ For all fluids

Zc = 0.24 to 0.29 for real fluid

(mainly hydrocarbons)

Cubic Equation of State - History

Improvement over Van der Waals EOS

Redlich and Kwong (1949)

- Slightly different volume dependence in attractive term

- Improved critical compressibility (0.3333) and Second Virial Coefficient

- Vapor pressure and liquid density are still inaccurate

Wilson (1964)

- First introduced T dependency of a parameter

)]11

)(62.157.1(1[ r

rT

T )70at ( 1ln c

c

T.TP

P

Acentric factor

2V

a

bV

RTP

)bV(VT

a

bV

RTP

/

21

)bV(V

)T(a

)bV(

RTP c

a(T) = ac(T)


Soave (1972)

More refined temperature dependency of a parameter

Improved vapor-pressure calculation

Also called Soave-Redlich-Kwong (SRK) EOS

Peng-Ronbinson (1976)

)bV(V

a)T(

bV

RTP c

)bV(VT

a

bV

RTP

/

21

2

2

156130551711485090

11

...m

)T/T(m c

)bV(b)bV(V

)T(a

)bV(

RTP c


Equation of State with more parameters

Scmidt and Wenzel (1980)

Hamens and Knapp (1980)

Patel and Teja (1982)

22 331 bbv)(v

)T(a

)bv(

RTP

22 1 b)c(cbvv

)T(a

)bv(

RTP

bcv)cb(v

)T(a

)bv(

RTP

2

Results can be

improved …

But meaning of the

third parameter is not

clear…

It is not easy to write a

meaningful mixing

rule for the third

parameter

Note:

a(T) = ac(T)


a(T) law improvement

Stryjek-Vera (1986)

Soave (1984)

Carrier (1988)

Mathias and Copeman (1983)

Numerous functional forms have been proposed …

)1/1()1(1 21 rr TCTC

25.0)1(1 rT

)1()1(1 21 rr TCTC

23

3

2

211 ccc rT1

)7.0)(1(5.0

10 rr TT

32

0

0196554.017131848.0

4897153.1378893.0


Ref) J. O. Valderrama, Ind. Eng. Chem. Res., 42,1603 (2003)

- Reference in textbook, page 92

A Generic Cubic Equation of State

Several hundred cubic equation of state have been proposed

Most EOS can be reduced to a generic form;

))((

)(

)( 2

h

VVbV

V

bV

RTP

))((

)(

)( bVbV

Ta

bV

RTP

and : pure number (same for all substance)

a(T) and b : substance dependent (a(T) = f(Tr, Tc, Pc) and b = f(Tc, Pc))

e.g., for van der Waals EOS,

a(T) = a, = = 0

A Generic Cubic Equation of State

Determination of Equation-of-State Parameters

How to determine a and b parameters ?

An EOS should represent PVT behavior of pure fluid

Two conditions at critical point

At critical point ; T = Tc, P = Pc, V = Vc

0;

crTV

P0

;

2

2

crTV

P

c

c

P

TRa

64

27 22

c

c

P

RTb

8

93-94 page derivation (Van der Waals EOS)

375.0c

ccc

RT

VPZ


For other EOS (RK, SRK, PR), similar equation can be obtained

c

c

P

TRa

64

27 22

c

c

P

RTb

8

c

c

P

TRa

22

c

c

P

RTb


For better representation of vapor pressure, a parameters are

assumed to be temperature dependent

c

c

P

TRa

22

)()(

)(22

rc

c

cr TaP

TRTTa

Theorem of Corresponding State: Acentric Factor

a(T) law can be fitted using vapor pressure data

Is there any method not using vapor pressure ?

Corresponding state theorem acentric factor ( )

)()(

)(22

rc

c

cr TaP

TRTTa

)1/1()1(1 21 rr TCTC


Reduced Properties : How far from critical point?

Theorem of Corresponding State (Simplest Form)

All fluids, when compared at the same reduced

temperature and pressure, have approximately the

same compressibility factor and all deviate from ideal

gas behavior to about the same degree.

Only valid for simple fluid (argon, krypton and xenon)

Systematic deviations are observed

cr TTT / cr PPP /


Acentric Factor

Introduced by K. S. Pitzer and coworkers

Plot of log (Prsat) vs. (1/Tr)

STd

Pd

r

sat

r )/1(

log

Slope varies depending on

the chemical species


Acentric Factor

Basis : = 0 for Ar, Kr, Xe

This parameter represent the degree how far from simple gases

Example: CH4 : 0.012, C2H5: 0.100, C3H8: 0.152, C4H10: 0.200, …., C10H22: 0.492

See Table B.1 for , Tc, Pc, Vc, and Zc for many chemical species

7.0)log(0.1 rT

sat

rP

(1/Tr)

log(1/Prsat)

S=-2.3 (for Ar, Kr, Xe)

Slope varies depending on

the chemical species

(1/0.7)

-1


Three-Parameter Theorem of Corresponding State

All fluids having the same value of , when compared at the same

Tr and Pr, have about the same value of Z, and all deviate from

ideal gas behavior to about the same degree.

Use of acentric factor in Cubic EOS

How to Solve Cubic EOS ?

Solution Technique

Root formula for Cubic equation

Iterative calculation

- Newton-Raphson iteration or Secant iteration

- Successive Substitution

Use the solver function

- Engineering calculator

- Computer program: Mathematica, Matlab, etc

Initial guess

- For gas phase : Ideal Gas Root (V = RT/P)

- For liquid phase : Hard Sphere Volume (V = b)

Solution Method (3) Successive Substitution

Vapor or Vapor-Like Roots

Start with Ideal Gas Root :V=RT/P

Liquid and Liquid-Like Roots

Start with : V=b

))((

)(

bVbV

bV

P

Tab

P

RTV

)())((

Ta

VPbPRTbVbVbV

Example 3.9

Given that the vapor pressure of n-butane at 350 K is 9.4573 bar, find

the molar volume of (a) saturated-vapor and (b) saturated-liquid n-

butane using the Redlich/Kwong equation


Tc and Pc from Table B.1, Tc = 425.1 K and Pc = 37.96 bar

From Table 3.1,

(a) For vapor-like root, start with V = RT/P = 83.14×350/9.4573 = 3157.5 cm3/mol

Using solver with initial value of V = 3157.5 cm3/mol, then V = 2555 cm3/mol

(b) For liquid-like root, start with b = 80.667 cm3/mol

Using solver with initial value of V = 80.667 cm3/mol, then V = 133.3 cm3/mol

823301425

350.

.T

TT

c

r 249109637

45739.

.

.

P

PP

c

r

26722212221

105519637

1425148382330427480 mol/barcm.

.

....

P

TRTa

/

c

c

/

r

mol/cm667.8096.37

1.42514.8308664.0

P

RTb 3

c

c

).V(V

.V

.

..

.

.

)bV)(bV(

bV

P

)T(ab

P

RTV

66780

66780

45739

1055166780

45739

3501483 7

3.6 Generalized Correlations for Gases

Pitzer correlations for the Compressibility factor

Z0 and Z1 are given as a generalized function of Tr and Pr

- They are tabulated

- See Appendix E (Table E.1 – E.4)

Lee-Kesler Method (1975)

- Given by chart (Z0 and Z1) : Table E.1 – E.4

- Basis : Benedict/Webb/Rubin Equation

For quantum gases (hydrogen, helium, neon)

- Small molecules

- Do not conform corresponding behavior

- Use effective critical parameters (3.58-3.60)

Errors

- 2-3 percent for nonpolar or slightly polar small molecules (mainly hydrocarbons)

- Larger errors for polar/associating components, large molecules

10 ZZZ

3.6 Generalized Correlations for Gases

Pitzer correlations for the Second Virial Coefficient

Not recommended

Not accurate for highly polar or associating molecules

r

r

T

PB

RT

BP

RT

PVZ ˆ11

10ˆ BBB

6.1

0 422.0083.0

rTB 2.4

1 172.0139.0

rTB

10 ZZZ

3.7 Generalized Correlation for Liquids

Cubic EOS

Not accurate for liquid phase molar volume (5-10 % error)

Lee-Kesler correlation

Accurate liquid volume but not accurate for polar species

Saturated molar volume

Rackett equation (1978)

7/2)1( rT

cc

sat ZVV

])1(1[ 7/2

rT

c

r

rsat ZT

PZ

Example 3.10

Determine the molar volume of n-butane at 510 K and 25 bar by

each of the following:

(a) The ideal-gas equation

(b) The generalized compressibility-factor correlation


(a) For ideal gas,

(b) Tc and Pc from Table B.1, Tc = 425.1 K and Pc = 37.96 bar

from Table E.1 and E.2 for Z0 and Z1,

Z0 = 0.865, Z1 = 0.038, and = 0.200 (Table B.1)

Z = Z0 + Z1 = 0.865 + (0.200)(0.038) = 0.873

mol/cm.,))(.(

P

RTV 316961

25

5101483

20011425

510.

.T

TT

c

r 65909637

25.

.P

PP

c

r

mol/cm.,))(.)(.(

P

ZRTV 374801

25

51014838730

Homework

Homework

3.8, 3.10, 3.14, 3.20, 3.32, 3.44, 3.68

Due:

Other Recommend Problems

3.2, 3.9, 3.12, 3.15, 3.16, 3.17, 3.18, 3.21, 3.22, 3.23, 3.24, 3.27, 3.36

Chapter 3. Volumetric Properties of Pure Fluids

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Transcript of Chapter 3. Volumetric Properties of Pure Fluids